Solucionario_Felder.pdf

2-1
CHAPTER TWO
2.1 (a)
3 24 3600
1
18144 109
wk 7 d h s 1000 ms
1 wk 1 d h 1 s
ms
= ×
.
(b)
38 3600
2598 26 0
.
. .
1 ft / s 0.0006214 mi s
3.2808 ft 1 h
mi / h mi / h
= ⇒
(c)
554 1 1
1000 g
3
m 1 d h kg 10 cm
d kg 24 h 60 min 1 m
85 10 cm g
4 8 4
4
4 4
⋅
= × ⋅
. / min
2.2 (a)
760 mi
3600
340
1 m 1 h
h 0.0006214 mi s
m / s
=
(b)
921 kg
35.3145 ft
57 5
2.20462 lb 1 m
m 1 kg
lb / ft
m
3
3 3 m
3
= .
(c)
537 10 1000 J 1
1
119 93 120
3
.
.
× ×
= ⇒
kJ 1 min .34 10 hp
min 60 s 1 kJ J / s
hp hp
-3
2.3 Assume that a golf ball occupies the space equivalent to a 2 2 2
in in in
× × cube. For a
classroom with dimensions 40 40 15
ft ft ft
× × :
nballs
3 3
6
ft (12) in 1 ball
ft in
10 5 million balls
=
× ×
= × ≈
40 40 15
2
518
3
3 3 3 .
The estimate could vary by an order of magnitude or more, depending on the assumptions made.
2.4 4 3 24 3600 s
1 0 0006214
.
.
light yr 365 d h 1.86 10 mi 3.2808 ft 1 step
1 yr 1 d h 1 s mi 2 ft
7 10 steps
5 16
× = ×
2.5 Distance from the earth to the moon = 238857 miles
238857 mi 1
4 1011
1 m report
0.0006214 mi 0.001 m
reports
= ×
2.6
19 0 0006214 1000
26417
44 7
500
25 1
14 500 0 04464
700
25 1
21 700 0 02796
km 1000 m mi L
1 L 1 km 1 m gal
mi / gal
Calculate the total cost to travel miles.
Total Cost
gal (mi)
gal 28 mi
Total Cost
gal (mi)
gal 44.7 mi
Equate the two costs 4.3 10 miles
American
European
5
.
.
.
$14,
$1.
, .
$21,
$1.
, .
=
= + = +
= + = +
⇒ = ×
x
x
x
x
x
x

2-2
2.7
6 3
3
5
5320 imp. gal 14 h 365 d 10 cm 0.965 g 1 kg 1 tonne
plane h 1 d 1 yr 220.83 imp. gal 1 cm 1000 g 1000 kg
tonne kerosene
1.188 10
plane yr
⋅
= ×
⋅
9
5
4.02 10 tonne crude oil 1 tonne kerosene plane yr
yr 7 tonne crude oil 1.188 10 tonne kerosene
4834 planes 5000 planes
× ⋅
×
= ⇒
2.8 (a)
250
250
.
.
lb 32.1714 ft / s 1 lb
32.1714 lb ft / s
lb
m
2
f
m
2 f
⋅
=
(b)
2
2
25 N 1 1 kg m/s
2.5493 kg 2.5 kg
9.8066 m/s 1 N
⋅
= ⇒
(c)
10 1000 g 980.66 cm 1
9 109
ton 1 lb / s dyne
5 10 ton 2.20462 lb 1 g cm / s
dynes
m
2
-4
m
2
× ⋅
= ×
2.9
50 15 2 853 32174
1
4 5 106
× ×
⋅
= ×
m 35.3145 ft lb ft 1 lb
1 m 1 ft s 32.174 lb ft s
lb
3 3
m f
3 3 2
m
2 f
. .
/
.
2.10
500 lb
5 10
1
2
1
10
25
2
m
3
m
3
1 kg 1 m
2.20462 lb 11.5 kg
m
≈ ×
F
HG I
KJF
HG I
KJ ≈
2.11 (a)
m m V V h r H r
h
H
f f c c f c
c
f
displaced fluid cylinder
3
3
cm cm g / cm
30 cm
g / cm
= ⇒ = ⇒ =
= =
−
=
ρ ρ ρ π ρ π
ρ
ρ
2 2
30 141 100
053
( . )( . )
.
(b) ρ
ρ
f
c H
h
= = =
( )( . )
.
30 053
171
cm g / cm
(30 cm - 20.7 cm)
g / cm
3
3
H
h
ρf
ρc
2.12
V
R H
V
R H r h R
H
r
h
r
R
H
h
V
R H h Rh
H
R
H
h
H
V V
R
H
h
H
R H
H
H
h
H
H
H h h
H
s f
f
f f s s f s
f s s s
= = − = ⇒ =
⇒ = −
F
HG I
KJ = −
F
HG
I
KJ
= ⇒ −
F
HG
I
KJ =
⇒ =
−
=
−
=
−
F
HG I
KJ
π π π
π π π
ρ ρ ρ
π
ρ
π
ρ ρ ρ ρ
2 2 2
2 2 2 3
2
2 3
2
2
3
2
3
3 3 3
3 3 3
3 3 3
3 3
1
1
; ;
ρf
ρs
R
r
h
H

2-3
2.13 Say h m
( ) = depth of liquid
A( )
m 2
h
1 m
⇒
y
x
y= 1
y= 1
– h
x = 1
– y 2
dA
( )
( ) ( ) ( ) ( )
2
2
1 1
2 2 2
1
1
1
2
2 2 1 1
1
Table of integrals or trigonometric substitution
2 1 2 1
m 1 sin 1 1 1 sin 1
2
π
− − +
−
− −
−
− −
−
= ⋅ = − ⇒ = −
⎤
= − + = − − − + − +
⎥
⎦
⇓
∫ ∫
y h
y
h
dA dy dx y dy A m y dy
A y y y h h h
W N
A
A
A
g g
b g=
×
= ×
E
4 0879 1
1 10 345 10
2
3 4
0
m m g 10 cm kg 9.81 N
cm m g kg
Substitute for
2 6
3 3
( ) .
.
N
W h h h
N
b g b g b g b g
= × − − − + − +
L
NM O
QP
−
345 10 1 1 1 1
2
4 2 1
. sin
π
2.14 1 1 32174 1
1
1
32174
lb slug ft / s lb ft / s slug = 32.174 lb
poundal =1 lb ft / s lb
f
2
m
2
m
m
2
f
= ⋅ = ⋅ ⇒
⋅ =
.
.
(a) (i) On the earth:
M
W
= =
=
⋅
= ×
175 lb 1
544
175 1
1
m
m
m
2
m
2
3
slug
32.174 lb
slugs
lb 32.174 ft poundal
s lb ft / s
5.63 10 poundals
.
(ii) On the moon
M
W
= =
=
⋅
=
175 lb 1
544
175 1
1
m
m
m
2
m
2
slug
32.174 lb
slugs
lb 32.174 ft poundal
6 s lb ft / s
938 poundals
.
( ) /
b F ma a F m
= ⇒ = =
⋅
=
355 pound 1 1
als lb ft / s 1 slug m
25.0 slugs 1 poundal 32.174 lb 3.2808 ft
0.135 m / s
m
2
m
2
y= –1
y= –1+h
x
dA

2-4
2.15 (a) F ma
= ⇒
F
HG I
KJ = ⋅
⇒
⋅
1
1
6
53623
1
fern = (1 bung)(32.174 ft / s bung ft / s
fern
5.3623 bung ft / s
2 2
2
) .
(b) On the moon:
3 bung 32.174 ft 1 fern
6 s 5.3623 bung ft / s
fern
On the earth: =18 fern
2 2
W
W
=
⋅
=
=
3
3 32174 53623
( )( . ) / .
2.16 (a) ≈ =
=
( )( )
( . )( . )
3 9 27
2 7 8 632 23
(b)
4
5
4 6
4.0 10
1 10
40
(3.600 10 ) / 45 8.0 10
−
−
− −
×
≈ ≈ ×
× = ×
(c) ≈ + =
+ =
2 125 127
2 365 1252 127 5
. . .
(d) ≈ × − × ≈ × ≈ ×
× − × = ×
50 10 1 10 49 10 5 10
4 753 10 9 10 5 10
3 3 3 4
4 2 4
.
2.17
1 5 4
2 3
6
3
(7 10 )(3 10 )(6)(5 10 )
42 10 4 10
(3)(5 10 )
3812.5 3810 3.81 10
exact
R
R
−
× × ×
≈ ≈ × ≈ ×
×
= ⇒ ⇒ ×
(Any digit in range 2-6 is acceptable)
2.18 (a)
A: C
C
C
o
o
o
R
X
s
= − =
=
+ + + +
=
=
− + − + − + − + −
−
=
731 72 4 0 7
72 4 731 72 6 72 8 730
5
72 8
72 4 72 8 731 72 8 72 6 72 8 72 8 72 8 730 72 8
5 1
0 3
2 2 2 2 2
. . .
. . . . .
.
( . . ) ( . . ) ( . . ) ( . . ) ( . . )
.
B: C
C
C
o
o
o
R
X
s
= − =
=
+ + + +
=
=
− + − + − + − + −
−
=
1031 973 58
973 1014 987 1031 1004
5
1002
973 1002 1014 1002 987 1002 1031 1002 1004 1002
5 1
23
2 2 2 2 2
. . .
. . . . .
.
( . . ) ( . . ) ( . . ) ( . . ) ( . . )
.
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.

2-5
2.19 (a)
X
X
s
X
X s
X s
i
i i
= = =
−
−
=
= − = − =
= + = + =
= =
∑ ∑
1
12
2
1
12
12
735
735
12 1
12
2 735 2 12 711
2 735 2 12 759
.
( . )
.
. ( . ) .
. ( . ) .
C
C
min=
max=
(b) Joanne is more likely to be the statistician, because she wants to make the control limits
stricter.
(c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor
temperature (failure of reactor control system), problems with the color measurement
system, operator carelessness
2.20 (a), (b)
(c) Beginning with Run 11, the process has been near or well over the upper quality assurance
limit. An overhaul would have been reasonable after Run 12.
2.21 (a)
4 2 2 2
2
2.36 10 kg m 2.20462 lb 3.2808 ft 1 h
'
h kg m 3600 s
Q
−
× ⋅
=
(b)
4
( 4 3) 6 2
approximate 3
6 2 2
exact
(2 10 )(2)(9)
' 12 10 1.2 10 lb ft /s
3 10
' =1.56 10 lb ft /s 0.00000156 lb ft /s
Q
Q
−
− − −
−
×
≈ ≈ × ≈ × ⋅
×
× ⋅ = ⋅
(a) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
X 134 131 129 133 135 131 134 130 131 136 129 130 133 130 133
Mean(X) 131.9
Stdev(X) 2.2
Min 127.5
Max 136.4
(b) Run X Min Mean Max
1 128 127.5 131.9 136.4
2 131 127.5 131.9 136.4
3 133 127.5 131.9 136.4
4 130 127.5 131.9 136.4
5 133 127.5 131.9 136.4
6 129 127.5 131.9 136.4
7 133 127.5 131.9 136.4
8 135 127.5 131.9 136.4
9 137 127.5 131.9 136.4
10 133 127.5 131.9 136.4
11 136 127.5 131.9 136.4
12 138 127.5 131.9 136.4
13 135 127.5 131.9 136.4
14 139 127.5 131.9 136.4
126
128
130
132
134
136
138
140
0 5 10 15

2-6
2.22 N
C
k
C
C
N
p
o
o
Pr
Pr
. .
.
( )( )( )
( )( )( )
. . .
= =
⋅
⋅ ⋅
≈
× × ×
× ×
≈
×
≈ × ×
−
−
μ 0583 1936 32808
0 286
6 10 2 10 3 10
3 10 4 10 2
3 10
2
15 10 163 10
1 3 3
1 3
3
3 3
J / g lb 1 h ft 1000 g
W / m ft h 3600 s m 2.20462 lb
The calculator solution is
m
m
2.23
Re
. . .
.
Re
( )( )( )( )
( )( )( )( )
(
= =
× ⋅
≈
× ×
× ×
≈
×
≈ × ⇒
−
− −
−
− −
Duρ
μ
0 48 2 067 0805
0 43 10
5 10 2 8 10 10
3 4 10 10 4 10
5 10
3
2 10
3
1 1 6
3 4
1 3)
4
ft 1 m in 1 m g 1 kg 10 cm
s 3.2808 ft kg / m s 39.37 in cm 1000 g 1 m
the flow is turbulent
6 3
3 3
2.24
1/ 2
1/3
1/3 1/ 2
5 2 3
3 5 2 5 2
(a) 2.00 0.600
1.00 10 N s/m (0.00500 m)(10.0 m/s)(1.00 kg/m )
2.00 0.600
(1.00 kg/m )(1.00 10 m /s) (1.00 10 N s/m )
(0.00500
44.426
ρ
μ
ρ μ
−
− −
⎛ ⎞
⎛ ⎞
= + ⎜ ⎟
⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎡ ⎤ ⎡ ⎤
× ⋅
= + ⎢ ⎥ ⎢ ⎥
× × ⋅
⎣ ⎦ ⎣ ⎦
= ⇒
g p p
g
k d y d u
D D
k
5 2
m)(0.100)
44.426 0.888 m/s
1.00 10 m /s
−
= ⇒ =
×
g
k
(b) The diameter of the particles is not uniform, the conditions of the system used to model the
equation may differ significantly from the conditions in the reactor (out of the range of
empirical data), all of the other variables are subject to measurement or estimation error.
(c)
dp (m) y D (m2
/s) μ (N-s/m2
) ρ (kg/m3
) u (m/s) kg
0.005 0.1 1.00E-05 1.00E-05 1 10 0.889
0.010 0.1 1.00E-05 1.00E-05 1 10 0.620
0.005 0.1 2.00E-05 1.00E-05 1 10 1.427
0.005 0.1 1.00E-05 2.00E-05 1 10 0.796
0.005 0.1 1.00E-05 1.00E-05 1 20 1.240
2.25 (a) 200 crystals / min mm; 10 crystals / min mm2
⋅ ⋅
(b) r =
⋅
−
⋅
= ⇒ =
200 10
4 0
crystals 0.050 in 25.4 mm
min mm in
crystals 0.050 in (25.4) mm
min mm in
238 crystals / min
238 crystals 1 min
60 s
crystals / s
2 2 2 2
2 2
min
.
(c) D
D
D
mm
in mm
1 in
b g b g
=
′
= ′
254
254
.
. ; r r r
crystals
min
crystals 60 s
s 1 min
F
HG I
KJ = ′ = ′
60
⇒ ′ = ′ − ′ ⇒ ′ = ′ − ′
60 200 254 10 254 84 7 108
2 2
r D D r D D
. . .
b g b g b g

2-7
2.26 (a) 705
. / ;
lb ft 8.27 10 in / lb
m
3 -7 2
f
×
(b)
7 2 6 2
f
3
m 2 5 2
f
3 3
m
3 3 6 3
m
8.27 10 in 9 10 N 14.696 lb /in
(70.5 lb /ft )exp
lb m 1.01325 10 N/m
70.57 lb 35.3145 ft 1 m 1000 g
1.13 g
ft m 10 cm 2.20462 lb
ρ
−
⎡ ⎤
× ×
⎢ ⎥
=
×
⎢ ⎥
⎣ ⎦
= = 3
/cm
(c) ρ ρ ρ
lb
ft
g lb cm
cm g 1 ft
m
3
m
3
3 3
F
HG I
KJ = ′ = ′
1 28 317
453593
62 43
,
.
.
P P P
lb
in
N .2248 lb m
m N 39.37 in
f
2
f
2
2 2 2
F
HG I
KJ = = × −
' . '
0 1
1
145 10
2
4
⇒ ′ = × × ⇒ ′ = ×
− − −
62 43 705 8 27 10 145 10 113 120 10
7 4 10
. . exp . . ' . exp . '
ρ ρ
d id i d i
P P
P' . ' . exp[( . )( . )] .
= × ⇒ = × × =
−
9 00 10 113 120 10 9 00 10 113
6 10 6
N / m g / cm
2 3
ρ
2.27 (a) V
V
V
cm
in 28,317 cm
in
3
3 3
3
d i d i
= =
'
. '
1728
16 39 ; t t
s hr
b g b g
= ′
3600
⇒ = ′ ⇒ = ′
16 39 3600 0 06102 3600
. ' exp ' . exp
V t V t
b g b g
(b) The t in the exponent has a coefficient of s-1
.
2.28 (a) 300
. mol / L, 2.00 min-1
(b) t C
C
= ⇒ =
⇒ =
0 300
300
.
.
exp[(-2.00)(0)] = 3.00 mol / L
t =1 exp[(-2.00)(1)] = 0.406 mol / L
For t=0.6 min: C
C
int
. .
( . ) . .
.
=
−
−
− + =
=
0 406 300
1 0
0 6 0 300 14
300
mol / L
exp[(-2.00)(0.6)] = 0.9 mol / L
exact
For C=0.10 mol/L: t
t
int
exact
min
= -
1
2.00
ln
C
3.00
= -
1
2
ln
0.10
3.00
=1.70 min
=
−
−
− + =
1 0
0 406 3
010 300 0 112
.
( . . ) .
(c)
0
0.5
1
1.5
2
2.5
3
3.5
0 1 2
t (min)
C
(mol/L)
(t=0.6, C=1.4)
(t=1.12, C=0.10)
Cexact vs. t

2-8
2.29 (a) p*
. .
( . )
=
−
−
− + =
60 20
199 8 166 2
185 166 2 20 42 mm Hg
(b) c MAIN PROGRAM FOR PROBLEM 2.29
IMPLICIT REAL*4(A–H, 0–Z)
DIMENSION TD(6), PD(6)
DO 1 I = 1, 6
READ (5, *) TD(I), PD(I)
1 CONTINUE
WRITE (5, 902)
902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X,
* ‘ (C) (MM HG)’/)
DO 2 I = 0, 115, 5
T = 100 + I
CALL VAP (T, P, TD, PD)
WRITE (6, 903) T, P
903 FORMAT (10X, F5.1, 10X, F5.1)
2 CONTINUE
END
SUBROUTINE VAP (T, P, TD, PD)
DIMENSION TD(6), PD(6)
I = 1
1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2
I = I + 1
IF (I.EQ.6) STOP
GO TO 1
2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I))
RETURN
END
DATA OUTPUT
98.5 1.0 TEMPERATURE VAPOR PRESSURE
131.8 5.0 (C) (MM HG)
# # 100.0 1.2
215.5 100.0 105.0 1.8
# #
215.0 98.7
2.30 (b) ln ln
(ln ln ) / ( ) (ln ln ) / ( ) .
ln ln ln . ( ) . .
y a bx y ae
b y y x x
a y bx a y e
bx
x
= + ⇒ =
= − − = − − = −
= − = + ⇒ = ⇒ = −
2 1 2 1
0.693
2 1 1 2 0 693
2 0 63 1 4 00 4 00
(c) ln ln ln
(ln ln ) / (ln ln ) (ln ln ) / (ln ln )
ln ln ln ln ( ) ln( ) /
y a b x y ax
b y y x x
a y b x a y x
b
= + ⇒ =
= − − = − − = −
= − = − − ⇒ = ⇒ =
2 1 2 1 2 1 1 2 1
2 1 1 2 2
(d) / /
2 1 2 1
3 /
ln( ) ln ( / ) ( / ) [can't get ( )]
[ln( ) ln( ) ]/[( / ) ( / ) ] (ln807.0 ln 40.2)/(2.0 1.0) 3
ln ln( ) ( / ) ln807.0 3ln(2.0) 2 2
[can't solve explicitly for
by x by x
y x
xy a b y x xy ae y a x e y f x
b xy xy y x y x
a xy b y x a xy e
= + ⇒ = ⇒ = =
= − − = − − =
= − = − ⇒ = ⇒ =
( )]
y x

2-9
2.30 (cont’d)
(e) ln( / ) ln ln( ) / ( ) [ ( ) ]
[ln( / ) ln( / ) ] / [ln( ) ln( ) ]
(ln . ln . ) / (ln . ln . ) .
ln ln( / ) ( ) ln . . ln( . ) .
/ . ( ) . ( )
/
.33 / .
y x a b x y x a x y ax x
b y x y x x x
a y x b x a
y x x y x x
b b
2 2 1 2
2
2
2
1 2 1
2
2 4 1 2 2 165
2 2 2
2 2
807 0 40 2 2 0 10 4 33
2 807 0 4 33 2 0 40 2
40 2 2 6 34 2
= + − ⇒ = − ⇒ = −
= − − − −
= − − =
= − − = − ⇒ =
⇒ = − ⇒ = −
2.31 (b) Plot vs. on rectangular axes. Slope Intcpt
2 3
y x m n
= = −
,
(c)
1 1 1 a 1
Plot vs. [rect. axes], slope = , intercept =
ln( 3) ln( 3) b b
a
x x
y b b y
= + ⇒
− −
(d)
1
1
3
1
1
3
2
3
2
3
( )
( )
( )
( ) , ,
y
a x
y
x a
+
= − ⇒
+
−
Plot vs. [rect. axes] slope = intercept = 0
OR
2 1 3 3
1 3
2
ln( ) ln ln( )
ln( ) ln( )
ln
y a x
y x
a
+ = − − −
+ −
⇒ − −
Plot vs. [rect.] or (y +1) vs. (x - 3) [log]
slope =
3
2
, intercept =
(e) ln
ln
y a x b
y x y x
= +
Plot vs. [rect.] or vs. [semilog ], slope = a, intercept = b
(f)
Plot vs. [rect.] slope = a, intercept = b
log ( ) ( )
log ( ) ( )
10
2 2
10
2 2
xy a x y b
xy x y
= + +
+ ⇒
(g) Plot vs. [rect.] slope = , intercept =
OR
b
Plot
1
vs.
1
[rect.] , slope = intercept =
1
1 1
2 2
2 2
y
ax
b
x
x
y
ax b
x
y
x a b
y
ax
b
x xy
a
x xy x
b a
= + ⇒ = + ⇒
= + ⇒ = + ⇒
,
,

2-10
2.32 (a) A plot of y vs. R is a line through ( R = 5, y = 0 011
. ) and ( R = 80 , y = 0169
. ).
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0 20 40 60 80 100
R
y
y a R b a
b
y R
= + =
−
−
= ×
= − × = ×
U
V
|
W
|
⇒ = × + ×
−
− −
− −
0169 0 011
80 5
211 10
0 011 211 10 5 4 50 10
211 10 4 50 10
3
3 4
3 4
. .
.
. . .
. .
d ib g
(b) R y
= ⇒ = × + × =
− −
43 211 10 43 4 50 10 0 092
3 4
. . .
d ib g kg H O kg
2
1200 0 092 110 kg
kg h kg H O kg H O h
2 2
b gb g
. =
2.33 (a) ln ln ln
(ln ln ) / (ln ln ) (ln ln ) / (ln ln ) .
ln ln ln ln ( . )ln( ) . . .
T a b T a
b T T
a T b a T
b
= + ⇒ =
= − − = − − = −
= − = − − ⇒ = ⇒ = −
φ φ
φ φ
φ φ
2 1 2 1
119
120 210 40 25 119
210 119 25 9677 6 9677 6
(b) T T
T C
T C
T C
= ⇒ =
= ⇒ = =
= ⇒ = =
= ⇒ = =
−
9677 6 9677 6
85 9677 6 85 535
175 9677 6 175 291
290 9677 6 290 19 0
119 0.8403
0.8403
0.8403
0.8403
. . /
. / .
. / .
. / .
.
φ φ
φ
φ
φ
b g
b g
b g
b g
o
o
o
L / s
L / s
L / s
(c) The estimate for T=175°C is probably closest to the real value, because the value of
temperature is in the range of the data originally taken to fit the line. The value of T=290°C
is probably the least likely to be correct, because it is farthest away from the date range.

2-11
2.34 (a) Yes, because when ln[( ) / ( )]
C C C C
A Ae A Ae
− −
0 is plotted vs. t in rectangular coordinates,
the plot is a straight line.
-2
-1.5
-1
-0.5
0
0 50 100 150 200
t (min)
ln
((C
A
-C
Ae
)/(C
A0
-C
Ae
))
Slope = -0.0093 k = 9.3 10 min
-3
⇒ × −1
(b)
3
0 0
(9.3 10 )(120) -2
-2
ln[( )/( )] ( )
(0.1823 0.0495) 0.0495 9.300 10 g/L
9.300 10 g 30.5 gal 28.317 L
= / = 10.7 g
L 7.4805 gal
−
−
− ×
− − = − ⇒ = − +
= − + = ×
×
⇒ = =
kt
A Ae A Ae A A Ae Ae
A
C C C C kt C C C e C
C e
C m V m CV
2.35 (a) ft and h , respectively
3 -2
(b) ln(V) vs. t2
in rectangular coordinates, slope=2 and intercept= ln( . )
353 10 2
× −
; or
V(logarithmic axis) vs. t2
in semilog coordinates, slope=2, intercept= 353 10 2
. × −
(c) V ( ) . exp( . )
m t
3 2
= × ×
− −
100 10 15 10
3 7
2.36 PV C P C V P C k V
k k
= ⇒ = ⇒ = −
/ ln ln ln
lnP = -1.573(lnV ) + 12.736
6
6.5
7
7.5
8
8.5
2.5 3 3.5 4
lnV
lnP
slope (dimensionless)
Intercept = ln mm Hg cm4.719
k
C C e
= − = − − =
= ⇒ = = × ⋅
( . ) .
. .
.736
1573 1573
12 736 340 10
12 5
2.37 (a)
G G
G G K C
G G
G G
K C
G G
G G
K m C
L
L
m
L
L
m
L
L
−
−
= ⇒
−
−
= ⇒
−
−
= +
0
0 0
1
ln ln ln
ln(G0-G)/(G-GL)= 2.4835lnC - 10.045
-1
0
1
2
3
3.5 4 4.5 5 5.5
lnC
ln(G
0
-G)/(G-G
L
)

2-12
2.37 (cont’d)
m
K K
L L
= =
= − ⇒ = × −
slope (dimensionless)
Intercept = ln ppm-2.483
2 483
10 045 4 340 10 5
.
. .
(b) C
G
G
G
= ⇒
− ×
× −
= × ⇒ = ×
−
−
− −
475
180 10
300 10
4 340 10 475 1806 10
3
3
5 2 3
.
.
. ( ) .
.483
C=475 ppm is well beyond the range of the data.
2.38 (a) For runs 2, 3 and 4:
Z aV p Z a b V c p
a b c
a b c
a b c
b c
= ⇒ = + +
= + +
= + +
= + +
ln ln ln ln
ln( . ) ln ln( . ) ln( . )
ln( . ) ln ln( . ) ln( . )
ln( . ) ln ln( . ) ln( . )
35 102 91
2 58 102 112
372 175 112
b
c
=
⇒ = −
⋅
0 68
146
.
.
a = 86.7 volts kPa / (L / s)
1.46 0.678
(b) When P is constant (runs 1 to 4), plot ln
Z vs. lnV . Slope=b, Intercept= ln ln
a c p
+
lnZ = 0.5199lnV + 1.0035
0
0.5
1
1.5
2
-1 -0.5 0 0.5 1 1.5
lnV
lnZ
b
a c P
= =
+ =
slope
Intercept = ln
052
10035
.
ln .
When
V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept=ln ln
a c V
+
lnZ = -0.9972lnP + 3.4551
0
0.5
1
1.5
2
1.5 1.7 1.9 2.1 2.3
lnP
lnZ
c slope
a b V
= = − ⇒
+ =
0 997 10
34551
. .
ln .
Intercept = ln
Plot Z vs
V P
b c
. Slope=a (no intercept)
Z = 31.096Vb
Pc
1
2
3
4
5
6
7
0.05 0.1 0.15 0.2
Vb
Pc
Z
a slope
= = ⋅
311
. volt kPa / (L / s).52
The results in part (b) are more reliable, because more data were used to obtain them.

2-13
2.39 (a)
s
n
x y
s
n
x
s
n
x s
n
y
a
s s s
s s
xy i i
i
n
xx i
i
n
x i
i
n
y i
i
n
xy x y
xx x
= = + + =
= = + + =
= = + + = = = + + =
=
−
−
=
−
=
=
= =
∑
∑
∑ ∑
1
0 4 0 3 21 19 31 32 3 4 677
1
0 3 19 32 3 4 647
1
0 3 19 32 3 18
1
0 4 21 31 3 1867
4 677 18 1
1
2
1
2 2 2
1 1
2
[( . )( . ) ( . )( . ) ( . )( . )] / .
( . . . ) / .
( . . . ) / . ; ( . . . ) / .
. ( . )( .
b g
867
4 647 18
0 936
4 647 1867 4 677 18
4 647 18
0182
0 936 0182
2
2 2
)
. ( . )
.
( . )( . ) ( . )( . )
. ( . )
.
. .
−
=
=
−
−
=
−
−
=
= +
b
s s s s
s s
y x
xx y xy x
xx x
b g
(b) a
s
s
y x
xy
xx
= = = ⇒ =
4 677
4 647
10065 10065
.
.
. .
y = 1.0065x
y = 0.936x + 0.182
0
1
2
3
4
0 1 2 3 4
x
y
2.40 (a) 1/C vs. t. Slope= b, intercept=a
(b) b a
= ⋅ =
slope = 0.477 L / g h Intercept = 0.082 L / g
;
1/C = 0.4771t + 0.0823
0
0.5
1
1.5
2
2.5
3
0 1 2 3 4 5 6
t
1/C
0
0.5
1
1.5
2
1 2 3 4 5
t
C
C C-fitted
(c) C a bt
t C a b
= + ⇒ + =
= − = − =
1 1 0 082 0 477 0 12 2
1 1 0 01 0 082 0 477 209 5
/ ( ) /[ . . ( )] .
( / ) / ( / . . ) / . .
g / L
h
(d) t=0 and C=0.01 are out of the range of the experimental data.
(e) The concentration of the hazardous substance could be enough to cause damage to the
biotic resources in the river; the treatment requires an extremely large period of time; some
of the hazardous substances might remain in the tank instead of being converted; the
decomposition products might not be harmless.

2-14
2.41 (a) and (c)
1
10
0.1 1 10 100
x
y
(b) y ax y a b x a
b
= ⇒ = +
ln ln ln ; Slope = b, Intercept = ln
ln y = 0.1684ln x + 1.1258
0
0.5
1
1.5
2
-1 0 1 2 3 4 5
ln x
ln
y
b
a a
= =
= ⇒ =
slope
Intercept = ln
0168
11258 308
.
. .
2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0
(b)
Lab 1
ln(1-Cp/Cao) = -0.0062t
-4
-3
-2
-1
0
0 200 400 600 800
t
ln(1-Cp/Cao)
Lab 2
ln(1-Cp/Cao) = -0.0111t
-6
-4
-2
0
0 100 200 300 400 500 600
t
ln(1-Cp/Cao)
k = 0 0062
. s-1
k = 0 0111
. s-1
Lab 3
ln(1-Cp/Cao) = -0.0063t
-6
-4
-2
0
0 200 400 600 800
t
ln(1-Cp/Cao)
Lab 4
ln(1-Cp/Cao)= -0.0064t
-6
-4
-2
0
0 200 400 600 800
t
ln(1-Cp/Cao)
k = 0 0063
. s-1
k = 0 0064
. s-1
(c) Disregarding the value of k that is very different from the other three, k is estimated with
the average of the calculated k’s. k = 0 0063
. s-1
(d) Errors in measurement of concentration, poor temperature control, errors in time
measurements, delays in taking the samples, impure reactants, impurities acting as
catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty
reactor.

2-15
2.43 y ax a d y ax
d
da
y ax x y x a x
a y x x
i i i
i
n
i i
i
n
i i
i
n
i i i
i
n
i
i
n
i i
i
n
i
i
n
= ⇒ = = − ⇒ = = − ⇒ − =
⇒ =
= = = = =
= =
∑ ∑ ∑ ∑ ∑
∑ ∑
φ
φ
( )
/
2
1
2
1 1 1
2
1
1
2
1
0 2 0
b g b g
2.44 DIMENSION X(100), Y(100)
READ (5, 1) N
C N = NUMBER OF DATA POINTS
1FORMAT (I10)
READ (5, 2) (X(J), Y(J), J = 1, N
2FORMAT (8F 10.2)
SX = 0.0
SY = 0.0
SXX = 0.0
SXY = 0.0
DO 100J = 1, N
SX = SX + X(J)
SY = SY + Y(J)
SXX = SXX + X(J) ** 2
100SXY = SXY + X(J) * Y(J)
AN = N
SX = SX/AN
SY = SY/AN
SXX = SXX/AN
SXY = SXY/AN
CALCULATE SLOPE AND INTERCEPT
A = (SXY - SX * SY)/(SXX - SX ** 2)
B = SY - A * SX
WRITE (6, 3)
3FORMAT (1H1, 20X 'PROBLEM 2-39'/)
WRITE (6, 4) A, B
4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/)
C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF
RESIDUALS
SSQ = 0.0
DO 200J = 1, N
YC = A * X(J) + B
RES = Y(J) - YC
WRITE (6, 5) X(J), Y(J), YC, RES
5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X
* 'RESIDUALb =', F6.3)
200SSQ = SSQ + RES ** 2
WRITE (6, 6) SSQ
6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3)
STOP
END
$DATA
5
1.0 2.35 1.5 5.53 2.0 8.92 2.5 12.15
3.0 15.38
SOLUTION: a b
= = −
6536 4 206
. , .

2-16
2.45 (a) E(cal/mol), D0 (cm2
/s)
(b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0.
(c) Intercept = ln = -3.0151 = 0.05 cm / s
2
D D
0 0
⇒ .
Slope = = -3666 K = (3666 K)(1.987 cal / mol K) = 7284 cal / mol
− ⇒ ⋅
E R E
/
ln D = -3666(1/T) - 3.0151
-14.0
-13.0
-12.0
-11.0
-10.0
2.0E-03
2.1E-03
2.2E-03
2.3E-03
2.4E-03
2.5E-03
2.6E-03
2.7E-03
2.8E-03
2.9E-03
3.0E-03
1/T
ln
D
(d) Spreadsheet
T D 1/T lnD (1/T)*(lnD) (1/T)**2
347 1.34E-06 2.88E-03 -13.5 -0.03897 8.31E-06
374.2 2.50E-06 2.67E-03 -12.9 -0.03447 7.14E-06
396.2 4.55E-06 2.52E-03 -12.3 -0.03105 6.37E-06
420.7 8.52E-06 2.38E-03 -11.7 -0.02775 5.65E-06
447.7 1.41E-05 2.23E-03 -11.2 -0.02495 4.99E-06
471.2 2.00E-05 2.12E-03 -10.8 -0.02296 4.50E-06
Sx 2.47E-03
Sy -12.1
Syx -3.00E-02
Sxx 6.16E-06
-E/R -3666
ln D0 -3.0151
D0 7284
E 0.05

3-1
CHAPTER THREE
3.1 (a) m =
× ×
≈ × ≈ ×
16 6 2 1000
2 10 5 2 10 2 10
3 5
m kg
m
kg
3
3 b gb gb gd i
(b)
m = ≈
×
×
≈ ×
8 10
2 32
4 10
3 10 10
1 10
6 6
3
2
oz 1 qt cm 1 g
s oz 1056.68 qt cm
g / s
3
3
b gd i
(c) Weight of a boxer 220 lbm
≈
Wmax ≥
×
≈
12 220
220 stones
lb 1 stone
14 lb
m
m
(d) dictionary
V
D L
= =
≈
× × × × × × ×
× ×
≈ ×
π 2 2
2 3
7
4
314 4 5
4
3 4 5 8 10 5 10 7
4 4 10
1 10
. . ft 800 miles 5880 ft 7.4805 gal 1 barrel
1 mile 1 ft 42 gal
barrels
2
3
d i d i
(e) (i)V ≈
× ×
≈ × × ≈ ×
6
3 3 10 1 10
4 5
ft 1 ft 0.5 ft 28,317 cm
1 ft
cm
3
3
3
(ii)V ≈ ≈
× ×
≈ ×
150 28 317 150 3 10
60
1 10
4
5
lb 1 ft cm
62.4 lb 1 ft
cm
m
3 3
m
3
3
,
(f) SG ≈ 105
.
3.2 (a) (i)
995 1 0 028317
0 45359 1
6212
kg lb m
m kg ft
lb / ft
m
3
3 3 m
3
.
.
.
=
(ii)
995 62 43
1000
6212
kg / m lb / ft
kg / m
lb / ft
3
m
3
3 m
3
.
.
=
(b) ρ ρ
= × = × =
H O SG
2
62 43 57 360
. .
lb / ft lb / ft
m
3
m
3
3.3 (a)
50
10
35
3
L 0.70 10 kg 1 m
m L
kg
3 3
3
×
=
(b)
1150
1 60
27
kg m 1000 L 1 min
0.7 1000 kg m s
L s
3
3
min ×
=
(c)
10 1 0 70 62 43
2 7 481 1
29
gal ft lb
min gal ft
lb / min
3
m
3 m
. .
.
×
≅

3-2
3.3 (cont’d)
(d) Assuming that 1 cm3
kerosene was mixed with Vg (cm3
) gasoline
V V
g g
cm gasoline g gasoline
3
d i d i
⇒ 0 70
.
1 082
cm kerosene g kerosene
3
d i d i
⇒ .
SG
V
V
V
g
g
g
=
+
+
= ⇒ =
−
−
=
0 70 082
1
0 78
082 0 78
0 78 0 70
05
. .
.
. .
. .
.
d id i
d i
g blend
cm blend
3
0 cm
3
Volumetric ratio
cm
cm
cm gasoline / cm kerosene
gasoline
kerosene
3
3
3 3
= = =
V
V
050
1
050
.
.
3.4 In France:
50 0 5
0 7 10 1 522
42
. $1
. . .
$68.
kg L Fr
kg L Fr
×
=
In U.S.:
50 0 1 20
0 70 10 37854 1
64
. $1.
. . .
$22.
kg L gal
kg L gal
×
=
3.5
(a) .
V =
×
=
700
1319
lb ft
h 0.850 62.43 lb
ft / h
m
3
m
3

.
m
V
B
B
=
×
=
3
m
3 B
ft 0.879 62.43 lb
h ft
V kg / h
d i
b g b g
54 88
. . .
m V V
H H H
= × =
d hb g b g
0 659 62 43 4114 kg / h
. /
V V
B H
+ = 1319 ft h
3
. .
m m V V
B H B H
+ = + =
54 88 4114 700 lbm
⇒ . / /
V m
B B
= ⇒ =
114 ft h 628 lb h benzene
3
m
. / . /
V m
H H
= ⇒ =
174 716
ft h lb h hexane
3
m
(b) – No buildup of mass in unit.
– ρB and ρH at inlet stream conditions are equal to their tabulated values (which are
strictly valid at 20
o
C and 1 atm.)
– Volumes of benzene and hexane are additive.
– Densitometer gives correct reading.
( ), ( )
V m
H H
ft / h lb / h
3
m
( ), ( )
V m
B B
ft / h lb / h
3
m
700 lb / h
m
( ), .
V SG
ft / h
3
= 0850

3-3
3.6 (a) V =
×
=
1955 1
0 35 12563 1000
445
.
. . .
kg H SO kg solution L
kg H SO kg
L
2 4
2 4
(b)
Videal
2
2 4 2
2 4
kg H SO L
kg
kg H SO kg H O L
kg H SO kg
L
=
×
+ =
1955
18255 100
1955 0 65
0 35 1000
470
4
.
. .
. .
. .
% .
error =
−
× =
470 445
445
100% 5 6%
3.7 Buoyant force up Weight of block down
b g b g
=
E
Mass of oil displaced + Mass of water displaced = Mass of block
ρ ρ ρ
oil H O c
2
0542 1 0542
. .
b g b g
V V V
+ − =
From Table B.1: g / cm , g / cm g / cm
3 3
o
3
ρ ρ ρ
c w il
= = ⇒ =
2 26 100 3325
. . .
m V
oil oil
3 3
g / cm cm g
= × = × =
ρ 3325 353 117 4
. . .
moil + flask g g g
= + =
117 4 124 8 242
. .
3.8 Buoyant force up = Weight of block down
b g b g
⇒ = ⇒ =
W W Vg Vg
displaced liquid block disp. Liq block
( ) ( )
ρ ρ
Expt. 1: ρ ρ ρ ρ
w B B w
A g A g
15 2
15
2
.
.
b g b g
= ⇒ = ×
ρ
ρ
w
B B
SG
=
= ⇒ =
1
0 75 0 75
.00
. .
g/cm
3
3
g / cm b g
Expt. 2: ρ ρ ρ ρ
soln soln
3
soln
g / cm
A g A g SG
B B
b g b g b g
= ⇒ = = ⇒ =
2 2 15 15
. .
3.9
W + W hs
A B
hb
hρ1
Before object is jettisoned
1
1
Let ρw = density of water. Note: ρ ρ
A w
(object sinks)
Volume displaced: V A h A h h
d b si b p b
1 1 1
= = −
d i (1)
Archimedes ⇒ = +
ρw d A B
V g W W
1
weight of displaced water

Subst. (1) for Vd1 , solve for h h
p b
1 1
−
d i
h h
W W
p gA
p b
A B
w b
1 1
− =
+
(2)
Volume of pond water: V A h V V A h A h h
w p p d
i
w p p b p b
= − ⇒ = − −
1 1 1 1 1
b g
d i
for
subst. 2
b h
w p p
A B
w
p
w
p
A B
w p
p b
V A h
W W
p g
h
V
A
W W
p gA
1 1
1 1
−
= −
+
⇒ = +
+
b g
(3)
2 solve for
subst. 3 for in
b g
b g b g
, h
h
b
w
p
A B
w p b
b
p
h
V
A
W W
p g A A
1
1
1
1 1
= +
+
−
L
N
MM
O
Q
PP
(4)

3-4
3.9 (cont’d)
W
hs
B
hb
hρ2
After object is jettisoned
WA
2
2
Let VA = volume of jettisoned object =
W
g
A
A
ρ
(5)
Volume displaced by boat:V A h h
d b p b
2 2 2
= −
d i (6)
Archimedes ⇒ =
E
ρW d B
V g W
2
Subst. forVd2 , solve for h h
p b
2 2
−
d i
h h
W
p gA
p b
B
w b
2 2
− = (7)
Volume of pond water: V A h V V V A h
W
p g
W
p g
w p p d A w p p
B
w
A
A
= − − = − −
2 2
5 6 7
2
b g b g b g
,
2
solve for
2
p
w B A
p
h
p w p A p
V W W
h
A p gA p gA
⇒ = + + (8)
⇒ = + + −
for in 7 solve for
subst. 8
h h
b
w
p
B
w p
A
A p
B
w b
p b
h
V
A
W
p gA
W
p gA
W
p gA
2 2
2
b g
b g
,
(9)
(a) Change in pond level
( ) ( ) ( )
8 3
2 1
1 1
0
W A
A W A
A
p p
p A W A W p
W p p
W
h h
A g p p p p gA
ρ ρ
−

−
⎡ ⎤
− = − = ⎯⎯⎯⎯
→
⎢ ⎥
⎣ ⎦
⇒ the pond level falls
(b) Change in boat level
h h
W
A g p A p A p A
V
A
p
p
A
A
p p
A
p A p W p W b
A
p
A
W
p
b
2 1
9 4 5
0 0
1 1 1
1 1 0
− = − +
L
N
MM
O
Q
PP
=
F
HG
I
KJ + −
F
HG I
KJ
F
HG
I
KJ
L
N
MMMM
O
Q
PPPP

−

b g b g b g

⇒ the boat rises
3.10 (a) ρbulk
3 3
3
2.93 kg CaCO 0.70 L CaCO
L CaCO L total
kg / L
= = 2 05
.
(b) W Vg
bag bulk
= =
⋅
= ×
ρ
2 05 1
100 103
.
.
kg 50 L 9.807 m / s N
L 1 kg m / s
N
2
2
Neglected the weight of the bag itself and of the air in the filled bag.
(c) The limestone would fall short of filling three bags, because
– the powder would pack tighter than the original particles.
– you could never recover 100% of what you fed to the mill.

3-5
3.11 (a) W m g
b b
= =
⋅
=
122 5
1202
. kg 9.807 m / s 1 N
1 kg m / s
N
2
2
V
W W
g
b
b I
w
=
−
=
⋅
×
=
ρ
(
.
1202
0 996 1
119
N - 44.0 N) 1 kg m / s
kg / L 9.807 m / s N
L
2
2
ρb
b
b
m
V
= = =
122 5
103
.
.
kg
119 L
kg / L
(b) m m m
f nf b
+ = (1)
x
m
m
m m x
f
f
b
f b f
= ⇒ = (2)
( ),( )
1 2 1
⇒ = −
m m x
nf b f
d i (3)
V V V
m m m
f nf b
f
f
nf
nf
b
b
+ = ⇒ + =
ρ ρ ρ
⇒ +
−
F
HG
I
KJ = ⇒ −
F
HG
I
KJ = −
2 3 1 1 1 1 1
b gb g
,
m
x x m
x
b
f
f
f
nf
b
b
f
f nf b nf
ρ ρ ρ ρ ρ ρ ρ
⇒ =
−
−
xf
b nf
f nf
1 1
1 1
/ /
/ /
ρ ρ
ρ ρ
(c) xf
b nf
f nf
=
−
−
=
−
−
=
1 1
1 1
1 103
0 31
/ /
/ /
/ .
.
ρ ρ
ρ ρ
1/1.1
1/ 0.9 1/1.1
(d) V V V V V
f nf lungs other b
+ + + =
m m
V V
m
m
x x
V V m
f
f
nf
nf
lungs other
b
b
mf mbx f
mnf mb x f
b
f
f
f
nf
lungs other b
b nf
ρ ρ ρ
ρ ρ ρ ρ
+ + + =
−
−
F
HG
I
KJ+ + = −
F
HG
I
KJ
=
= −
( )
( )
1
1 1 1
⇒ −
F
HG
I
KJ = − −
+
x
V V
m
f
f nf b nf
lungs other
b
1 1 1 1
ρ ρ ρ ρ
⇒ =
−
F
HG
I
KJ−
+
F
HG I
KJ
−
F
HG
I
KJ
=
−
F
HG I
KJ−
+
F
HG I
KJ
−
F
HG I
KJ
=
x
V V
m
f
b nf
lungs other
b
f nf
1 1
1 1
1
103
1
11
12 01
122 5
1
0 9
1
11
0 25
ρ ρ
ρ ρ
. .
. .
.
. .
.

3-6
3.12 (a)
From the plot above, r = −
5455 539 03
. .
ρ
(b) For = g / cm , 3.197 g Ile /100g H O
3
2
ρ 0 9940
. r =

.
.
mIle = =
150 0 994
4 6
L g 1000 cm 3.197 g Ile 1 kg
h cm L 103.197 g sol 1000 g
kg Ile / h
3
3
(c) The measured solution density is 0.9940 g ILE/cm3
solution at 50o
C. For the calculation
of Part (b) to be correct, the density would have to be changed to its equivalent at 47o
C.
Presuming that the dependence of solution density on T is the same as that of pure water,
the solution density at 47o
C would be higher than 0.9940 g ILE/cm3
. The ILE mass flow
rate calculated in Part (b) is therefore too low.
3.13 (a)
y = 0.0743x + 0.1523
R2
= 0.9989
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0.0 2.0 4.0 6.0 8.0 10.0 12.0
Rotameter Reading
Mass
Flow
Rate
(kg/min)
y = 545.5x - 539.03
R2
= 0.9992
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0.987 0.989 0.991 0.993 0.995 0.997
Density (g/cm3)
Conc.
(g
Ile/100
g
H2O)

3-7
3.13 (cont’d)
From the plot, = . / min
R m
53 0 0743 53 01523 055 kg
⇒ = + =
. . . .
b g
(b)
Rotameter
Reading
Collection
Time
(min)
Collected
Volume
(cm3)
Mass Flow
Rate
(kg/min)
Difference
Duplicate
(Di)
Mean Di
2 1 297 0.297
2 1 301 0.301 0.004
4 1 454 0.454
4 1 448 0.448 0.006
6 0.5 300 0.600
6 0.5 298 0.596 0.004 0.0104
8 0.5 371 0.742
8 0.5 377 0.754 0.012
10 0.5 440 0.880
10 0.5 453 0.906 0.026
Di = + + + + =
1
5
0 004 0 006 0 004 0 012 0 026 0 0104
. . . . . .
b g kg / min
95% confidence limits: ( . . . .
0 610 174 0 610 0 018
± = ±
Di ) kg / min kg / min
There is roughly a 95% probability that the true flow rate is between 0.592 kg / min
and 0.628 kg / min.
3.14 (a)
150
117 103
.
.
kmol C H 78.114 kg C H
kmol C H
kg C H
6 6 6 6
6 6
6 6
= ×
(b)
150 1000
15 104
.
.
kmol C H mol
kmol
mol C H
6 6
6 6
= ×
(c)
15 000
3307
,
.
mol C H lb - mole
453.6 mol
lb - mole C H
6 6
6 6
=
(d)
15 000 6
1
90 000
,
,
mol C H mol C
mol C H
mol C
6 6
6 6
=
(e)
15 000 6
1
90 000
,
,
mol C H mol H
mol C H
mol H
6 6
6 6
=
(f)
90 000
108 106
,
.
mol C 12.011 g C
mol C
g C
= ×
(g)
90 000
9 07 104
,
.
mol H 1.008 g H
mol H
g H
= ×
(h)
15 000
9 03 1027
,
.
mol C H 6.022 10
mol
molecules of C H
6 6
23
6 6
×
= ×

3-8
3.15 (a)
m = =
175
2526
m 1000 L 0.866 kg 1 h
h m L 60 min
kg / min
3
3
(b)
n = =
2526
457
kg 1000 mol 1 min
min 92.13 kg 60 s
mol / s
(c) Assumed density (SG) at T, P of stream is the same as the density at 20o
C and 1 atm
3.16 (a)
200 0 0150
936
. .
kg mix kg CH OH kmol CH OH 1000 mol
kg mix 32.04 kg CH OH 1 kmol
mol CH OH
3 3
3
3
=
(b)
mmix = =
100.0 lb - mole MA 74.08 lb MA 1 lb mix
h 1 lb - mole MA 0.850 lb MA
/ h
m m
m
m
8715 lb
3.17 M = + =
0 25 28 02 0 75 2 02
852
. . . .
.
mol N g N
mol N
mol H g H
mol H
g mol
2 2
2
2 2
2

. .
mN2
3000 0 25 28 02
2470
= =
kg kmol kmol N kg N
h 8.52 kg kmol feed kmol N
kg N h
2 2
2
2
3.18 Msuspension g g g
= − =
565 65 500 , MCaCO3
g g g
= − =
215 65 150
(a)
V = 455 mL min ,
m = 500 g min
(b) ρ = = =
/ .
m V 500 110
g / 455 mL g mL
(c) 150 500 0 300
g CaCO g suspension g CaCO g suspension
3 3
/ .
=
3.19 Assume 100 mol mix.
mC H OH
2 5 2 5
2 5
2 5
2 5
10.0 mol C H OH 46.07 g C H OH
mol C H OH
g C H OH
= = 461
mC H O
4 8 2 4 8 2
4 8 2
4 8 2
4 8 2
75.0 mol C H O 88.1 g C H O
mol C H O
g C H O
= = 6608
mCH COOH
3 3
3
3
3
15.0 mol CH COOH 60.05 g CH COOH
mol CH COOH
g CH COOH
= = 901
xC H OH 2 5
2 5
461 g
g + 6608 g + 901 g
g C H OH / g mix
= =
461
0 0578
.
xC H O 4 8 2
4 8 2
6608 g
g + 6608 g + 901 g
g C H O / g mix
= =
461
08291
.
xCH COOH 3
3
901 g
g + 6608 g + 901 g
g CH COOH / g mix
= =
461
0113
.
MW = =
461
79 7
g + 6608 g + 901 g
100 mol
g / mol
.
m = =
25
75
2660
kmol EA 100 kmol mix 79.7 kg mix
kmol EA 1 kmol mix
kg mix

3-9
3.20 (a)
Unit Function
Crystallizer Form solid gypsum particles from a solution
Filter Separate particles from solution
Dryer Remove water from filter cake
(b) mgypsum
4 2
4 2
L slurry kg CaSO H O
L slurry
kg CaSO H O
=
⋅
= ⋅
1 0 35 2
0 35 2
.
.
Vgypsum
4 2 4 2
4 2
4 2
kg CaSO H O L CaSO H O
2.32 kg CaSO H O
L CaSO H O
=
⋅ ⋅
⋅
= ⋅
035 2 2
2
0151 2
.
.
CaSO in gypsum:
kg ypsum 136.15 kg CaSO
172.18 kg ypsum
kg CaSO
4
4
4
m
g
g
= =
0 35
0 277
.
.
CaSO in soln.:
L sol 1.05 kg kg CaSO
L 100.209 kg sol
kg CaSO
4
4
4
m =
−
=
1 0151 0209
000186
. .
.
b g
(c) m = = ×
0 35 0 209
0 95
384
. .
.
.
kg gypsum 0.05 kg sol g CaSO
kg gypsum 100.209 g sol
10 kg CaSO
4 -5
4
% .
recovery =
0.277 g + 3.84 10 g
0.277 g + 0.00186 g
-5
×
× =
100% 99 3%
3.21
CSA:
45.8 L 0.90 kg kmol
min L 75 kg
kmol
min
FB:
55.2 L 0.75 kg kmol
min L 90 kg
kmol
min
mol CSA
mol FB
=
=
U
V
|
|
W
|
|
⇒ =
0 5496
0 4600
0 5496
0 4600
12
.
.
.
.
.
She was wrong.
The mixer would come to a grinding halt and the motor would overheat.
3.22 (a)
150
6910
mol EtOH 46.07 g EtOH
mol EtOH
g EtOH
=
6910 g EtO
10365
H 0.600 g H O
0.400 g EtOH
g H O
2
2
=
V = + = ⇒
6910 g EtO 10365
19123 191
H L
789 g EtOH
g H O L
1000 g H O
L L
2
2
. .
SG = =
(6910+10365) g L
L 1000 g
191
0903
.
.
(b) ′ =
+
= ⇒
V
( )
.
6910 10365
18 472
g mix L
935.18 g
L 18.5 L
%
( . . )
.
.
error
L
L
=
−
× =
19 123 18 472
18 472
100% 35%

3-10
3.23 M = + =
0 09 0 91
27 83
. .
.
mol CH 16.04 g
mol
mol Air 29.0 g Air
mol
g mol
4
700 kg kmol 0.090 kmol CH
h 27.83 kg 1.00 kmol mix
2.264 kmol CH h
4
4
=
2 264
. kmol CH 0.91 kmol air
h 0.09 kmol CH
22.89 kmol air h
4
4
=
5% CH
2.264 kmol CH 0.95 kmol air
h 0.05 kmol CH
43.01 kmol air h
4
4
4
⇒ =
Dilution air required:
43.01- 22.89 kmol air
h 1 kmol
mol air h
b g 1000 mol
20200
=
Product gas: 700 20.20 kmol 29
1286
kg
h
Air kg Air
h kmol Air
kg h
+ =
43.01 kmol Air 0.21 kmol O 32.00 kg O h
h 1.00 kmol Air 1 kmol O 1286 kg total
0.225
kg O
kg
2 2
2
2
=
3.24 x
m
M
m
V
i
i i
i
= =
, , =
M
V
i
ρ ρ
A
m
M
m
V M
m
V
i
i i
i
i
i
: x Not helpful.
i ρ ρ
∑ ∑ ∑
= = ≠
1 2
B
x m
M
V
m M
V
V
M
i
i
i i
i
i
: Correct.
ρ ρ
∑ ∑ ∑
= = = =
1 1
1 0 60
0 791
0 25
1 049
0 15
1595
1 091 0 917
ρ ρ
ρ
= = + + = ⇒ =
∑
xi
i
.
.
.
.
.
.
. . g / cm3
3.25 (a) Basis 100 mol N 20 mol CH
mol CO
mol CO
2 4
2
: ⇒ ⇒
× =
× =
R
S
|
T
|
20
80
25
64
20
40
25
32
Ntotal = + + + =
100 20 64 32 216 mol
x x
CO CO 2
mol CO / mol , mol CO mol
2
= = = =
32
216
0 15
64
216
0 30
. . /
x x
CH 4 N 2
4 2
mol CH mol , mol N mol
= = = =
20
216
0 09
100
216
0 46
. / . /
(b) M y M
i i
= = × + × + × + × =
∑ 015 28 0 30 44 0 09 16 0 46 28 32
. . . . g / mol

3-11
3.26 (a)
Samples Species MW k Peak Mole Mass moles mass
Area Fraction Fraction
1 CH4 16.04 0.150 3.6 0.156 0.062 0.540 8.662
C2H6 30.07 0.287 2.8 0.233 0.173 0.804 24.164
C3H8 44.09 0.467 2.4 0.324 0.353 1.121 49.416
C4H10 58.12 0.583 1.7 0.287 0.412 0.991 57.603
2 CH4 16.04 0.150 7.8 0.249 0.111 1.170 18.767
C2H6 30.07 0.287 2.4 0.146 0.123 0.689 20.712
C3H8 44.09 0.467 5.6 0.556 0.685 2.615 115.304
C4H10 58.12 0.583 0.4 0.050 0.081 0.233 13.554
3 CH4 16.04 0.150 3.4 0.146 0.064 0.510 8.180
C2H6 30.07 0.287 4.5 0.371 0.304 1.292 38.835
C3H8 44.09 0.467 2.6 0.349 0.419 1.214 53.534
C4H10 58.12 0.583 0.8 0.134 0.212 0.466 27.107
4 CH4 16.04 0.150 4.8 0.333 0.173 0.720 11.549
C2H6 30.07 0.287 2.5 0.332 0.324 0.718 21.575
C3H8 44.09 0.467 1.3 0.281 0.401 0.607 26.767
C4H10 58.12 0.583 0.2 0.054 0.102 0.117 6.777
5 CH4 16.04 0.150 6.4 0.141 0.059 0.960 15.398
C2H6 30.07 0.287 7.9 0.333 0.262 2.267 68.178
C3H8 44.09 0.467 4.8 0.329 0.380 2.242 98.832
C4H10 58.12 0.583 2.3 0.197 0.299 1.341 77.933
(b) REAL A(10), MW(10), K(10), MOL(10), MASS(10), MOLT, MASST
INTEGER N, ND, ID, J
READ (5, *) N
CN-NUMBER OF SPECIES
READ (5, *) (MW(J), K(J), J = 1, N)
READ (5, *) ND
DO 20 ID = 1, ND
READ (5, *)(A(J), J = 1, N)
MOLT = 0 0
.
MASST = 0 0
.
DO 10 J = 1, N
MOL(J) =
MASS(J) = MOL(J) * MW(J)
MOLT = MOLT + MOL(J)
MASST = MASST + MASS(J)
10 CONTINUE
DO 15 J = 1, N
MOL(J) = MOL(J)/MOLT
MASS(J) = MASS(J)/MASST
15 CONTINUE
WRITE (6, 1) ID, (J, MOL(J), MASS (J), J = 1, N)
20 CONTINUE
1 FORMAT (' SAMPLE: `, I3, /,
∗ ' SPECIES MOLE FR. MASS FR.', /,

3-12
3.26 (cont’d)
∗ 10(3X, I3, 2(5X, F5.3), /), /)
END
$DATA
∗
4
16 04 0 150
30 07 0 287
44 09 0 467
58 12 0 583
5
. .
. .
. .
. .
3 6 2 8 2 4 1 7
7 8 2 4 5 6 0 4
3 4 4 5 2 6 0 8
4 8 2 5 1 3 0 2
6 4 7 9 4 8 2 3
. . . .
. . . .
. . . .
. . . .
. . . .
[OUTPUT]
SAMPLE: 1
SPECIES MOLE FR MASS FR
1 0.156 0.062
2 0.233 0.173
3 0.324 0.353
4 0.287 0.412
SAMPLE: 2
(ETC.)
3.27 (a)
(8.
. .
7 10
12
128 10 2 9 10
6
7 5
× ×
= × ⇒ ×
0.40) kg C 44 kg CO
kg C
kg CO kmol CO
2
2 2
( .
. .
11 10
6 67 10 2 38 10
6
5 4
× ×
= × ⇒ ×
0.26) kg C 28 kg CO
12 kg C
kg CO kmol CO
( .
. .
3 8 10
5 07 10 317 10
5
4 3
× ×
= × ⇒ ×
0.10) kg C 16 kg CH
12 kg C
kg CH kmol CH
4
4 4
m =
× + × + ×
=
( . . . )
,
128 10 6 67 10 5 07 10
13 500
7 5 4
kg 1 metric ton
1000 kg
metric tons
yr
M y M
i i
= = × + × + × =
∑ 0 915 44 0 075 28 0 01 16 42 5
. . . . g / mol
3.28 (a) Basis: 1 liter of solution
1000
0 525 0 525
mL 1.03 g 5 g H SO mol H SO
mL 100 g 98.08 g H SO
mol / L molar solution
2 4 2 4
2 4
= ⇒
. .

3-13
3.28 (cont’d)
(b) t
V
V
= = =
min
55 60
144
gal 3.7854 L min s
gal 87 L
s
55
23 6
gal 3.7854 L 10 mL 1.03 g 0.0500 g H SO 1 lbm
gal 1 L mL g 453.59 g
lb H SO
3
2 4
m 2 4
= .
(c) u
V
A
= =
×
=

( / )
.
87
4
0513
L m 1 min
min 1000 L 60 s 0.06 m
m / s
3
2 2
π
t
L
u
= = =
45
88
m
0.513 m / s
s
3.29 (a)

.
.
n3
150
1147
= =
L 0.659 kg 1000 mol
min L 86.17 kg
mol / min
Hexanebalance: 0 (mol C H / min)
Nitrogenbalance: 0.820 (mol N
6 14
2
. . .
. / min)
180 0050 1147
0950
1 2
1 2
n n
n n
= +
=
U
V
W ⇒
=
R
S
|
T
|
solve mol / min
= 72.3 mol / min
.

n
n
1
2
838
(b) Hexane recovery= × = × =

.
. .
n
n
3
1
100%
1147
0180 838
100% 76%
b g
3.30
30mL 1L 0.030mol 172g
10 mL l L 1mol
gNauseum
3 =0155
.
0.180 mol C6H14/mol
0.820 mol N2/mol
1.50 L C6H14(l)/min

n3 (mol C6H14(l)/min)

n2 (mol/min)
0.050 mol C6H14/mol
0.950 mol N2/mol

n1 (mol/min)

3-14
3.31 (a) kt k
is dimensionless (min-1
⇒ )
(b) A semilog plot of vs. t is a straight line
CA ⇒ ln ln
C C kt
A AO
= −
k = −
0 414 1
. min
ln . .
C C
AO AO
3
lb-moles ft
= ⇒ =
02512 1286
(c) C C C
A A A
1b - moles
ft
mol liter 2.26462 lb - moles
liter 1 ft mol
3 3
F
HG I
KJ = ′ = ′
28 317
1000
0 06243
.
.
t
t s
t
C C kt
A A
min
exp
b g b g
=
′
= ′
( )
= −
1
60
60
0
min
s
0 06243 1334 0 419 60 214 0 00693
. . exp . . exp .
′ = − ′ ⇒ = −
C t C t
A A
b g b g b g
drop primes
mol / L
t CA
= ⇒ =
200 530
s mol / L
.
3.32 (a)
2600
50 3
mm Hg 14.696 psi
760 mm Hg
psi
= .
(b)
275 ft H O 101.325 kPa
33.9 ft H O
kPa
2
2
= 822 0
.
(c)
3.00 atm N m m
1 atm cm
N cm
2 2
2
2
101325 10 1
100
30 4
5 2
2
.
.
×
=
(d)
280 cm Hg 10 mm dynes cm cm
1 cm m
dynes
m
2 2
2 2
101325 10 100
760 mm Hg 1
3733 10
6 2
2
10
.
.
×
= ×
(e) 1
20 1
0 737
atm
cm Hg 10 mm atm
1 cm 760 mm Hg
atm
− = .
y = -0.4137x + 0.2512
R2
= 0.9996
-5
-4
-3
-2
-1
0
1
0.0 5.0 10.0
t (min)
ln(CA)

3-15
3.32 (cont’d)
(f)
25.0 psig 760 mm Hg gauge
14.696 psig
1293 mm Hg gauge
b g b g
=
(g)
25.0 psi 760 mm Hg
14.696 psi
2053 mm Hg abs
+
=
14 696
.
b g b g
(h) 325 435
mm Hg 760 mm Hg mm Hg gauge
− = − b g
(i)
2 3 2
f m
2 2 2
m f
4
35.0 lb 144 in ft s 32.174 lb ft 100 cm
Eq. (3.4-2)
in 1 ft 1.595x62.43 lb 32.174 ft s lb 3.2808 ft
1540 cm CCl
P
h
g
ρ
⋅
⇒ = =
⋅
=
3.33 (a) P gh
h
g = =
×
⋅
ρ
0 92 1000
. kg 9.81 m / s (m) 1 N 1 kPa
m 1 kg m / s 10 N / m
2
3 2 3 2
⇒ =
h Pg
(m) (kPa)
0111
.
P h
g = ⇒ = × =
68 0111 68 7 55
kPa m
. .
m V
oil = = ×
F
HG I
KJ× × ×
F
HG I
KJ = ×
ρ π
0 92 1000 7 55
16
4
14 10
2
6
. . .
kg
m
m kg
3
3
(b) P P P gh
g atm top
+ = + ρ
68 101 115 0 92 1000 9 81 103
+ = + × ×
. . /
b g b g h ⇒ =
h 598
. m
3.34 (a) Weight of block = Sum of weights of displaced liquids
( )
h h A g h A g h A g
h h
h h
b b
1 2 1 1 2 2
1 1 2 2
1 2
+ = + ⇒ =
+
+
ρ ρ ρ ρ
ρ ρ
(b)
, ,
,
top atm bottom atm b
down atm up atm
down up block liquiddisplaced
P P gh P P g h h gh W h h A
F P gh A h h A F P g h h gh A
F F h h A gh A gh A W W
b
b
b
= + = + + + = +
⇒ = + + + = + + +
= ⇒ + = + ⇒ =
ρ ρ ρ ρ
ρ ρ ρ ρ
ρ ρ ρ
1 0 1 0 1 2 2 1 2
1 0 1 2 1 0 1 2 2
1 2 1 1 2 2
( ) ( )
( ) ( ) [ ( ) ]
( )
h
Pg

3-16
3.35 ΔP P gh P
= + −
atm inside
ρ
b g
= −
1 atm 1 atm +
⋅
105 1000
.
b g kg 9.8066 m 150 m 1 m 1 N
m s 100 cm 1 kg m / s
2 2
3 2 2 2 2
F = = × ×
F
HG I
KJ=
154
100 10
022481
1
2250
4
N 65 cm
cm
N
lb
N
lb
2
2
f
f
.
.
3.36 m V
= =
× ×
= ×
ρ
14 62 43
2 69 107
. .
.
lb 1 ft 2.3 10 gal
ft 7.481 gal
lb
m
3 6
3 m
P P gh
= +
0 ρ
= +
×
⋅
14 7
14 62 43 12
.
. .
lb
in
lb 32.174 ft 30 ft 1 lb ft
ft s 32.174 lb ft / s 12 in
f
2
m f
2
3 2
m
2 2 2
= 32 9
. psi
— Structural flaw in the tank.
— Tank strength inadequate for that much force.
— Molasses corroded tank wall
3.37 (a) mhead
3 3
m
3 3 3 m
in 1 ft 8.0 62.43 lb
12 in ft
lb
=
× × ×
=
π 24 3
4
392
2
W m g
s
= =
⋅
=
head
m f
m
2 f
lb 32.174 ft lb
32.174 lb ft / s
lb
392 1
392
2
/
( ) 2 2
f
net gas atm 2
2 2
f 3
2 f f
30 14.7 lb 20 in
in 4
14.7 lb 24 in
392 lb 7.00 10 lb
in 4
F F F W
π
π
+ ×
⎡ ⎤
⎣ ⎦
= − − =
×
− − = ×
The head would blow off.
Initial acceleration:
3 2
f m 2
net
m f
head
7.000 10 lb 32.174 lb ft/s
576 ft/s
392 lb 1 lb
F
a
m
× ⋅
= = =
(b) Vent the reactor through a valve to the outside or a hood before removing the head.

3-17
3.38 (a)
P gh P P P
a atm b atm
= + =
ρ ,
If the inside pressure on the door equaled Pa , the force on
the door would be F A P P ghA
door a b door
= − =
( ) ρ
Since the pressure at every point on the door is greater than
Pa , Since the pressure at every point on the door is greater
than Pa , F ρghAdoor
(b) Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to
fill.
.
. . . / min
V
V
t
V
tub = ≈
× ×
= ⇒ = × =
5 25 2
25 5 25 125
ft
10 min
ft / min ft
3
3 3
(i) For a full room, 7 m
h =
⇒
2
5
3 2 2
1000 kg 9.81m 1 N 7 m 2 m
1.4 10 N
m s 1 kg m/s
F F
⇒ ×
⋅
The door will break before the room fills
(ii) If the door holds, it will take
t
V
V
fill
room
3 3
3 3
m 35.3145 ft 1 h
12.5 ft 1 m min
h
= =
× ×
=
/ min
5 15 10
60
31
b g
He will not have enough time.
3.39 (a) Pg tap
d i = =
25
10 33
245
m H O 101.3 kPa
m H O
kPa
2
2
.
Pg junction
d i b g
=
+
=
25 5
294
m H O 101.3 kPa
10.33 m H O
kPa
2
2
(b) Air in the line. (lowers average density of the water.)
(c) The line could be clogged, or there could be a leak between the junction and the tap.
3.40 Pabs = 800 mm Hg
Pgauge = 25 mm Hg
Patm = − =
800 25 775 mm Hg
a b
2 m
1 m

3-18
3.41 (a) P g h h P gh gh
A B C
1 1 2 2 1 2
+ + = + +
ρ ρ ρ
b g
⇒ − = − + −
P P gh gh
B A C A
1 2 1 2
ρ ρ ρ ρ
b g b g
(b) P1 121
10 0 792 137 0 792
=
−
L
NM +
− O
QP
kPa +
g 981 cm 30.0 cm
cm s
g 981 cm 24.0 cm
cm s
3 2 3 2
. . . .
b g b g
×
⋅
F
HG
I
KJ ×
F
HG
I
KJ
dyne
1 g cm / s
kPa
1.01325 10 dynes / cm
2 6 2
1 101325
.
= 1230
. kPa
3.42 (a) Say ρt (g/cm3
) = density of toluene, ρm (g/cm3
) = density of manometer fluid
(i) Hg: cm cm
(ii) H O: cm
2
ρ ρ
ρ
ρ
ρ ρ
ρ ρ
t m
m
t
t m
t m
g h R gR R
h
h R
h R
( )
. , . , .
. , . ,
500
500
1
0866 136 150 238
0866 100 150 2260 cm
− + = ⇒ =
−
−
= = = ⇒ =
= = = ⇒ =
Use mercury, because the water manometer would have to be too tall.
(b) If the manometer were simply filled with toluene, the level in the glass tube would be at
the level in the tank.
Advantages of using mercury: smaller manometer; less evaporation.
(c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen,
minimizing the risk of combustion.
3.43 P g
P
f f
atm
atm
m
7.23 g
= ⇒ =
ρ ρ
7 23
.
b g
P P g
P
g
a b f w w
− = − = −
F
HG I
KJ
ρ ρ ρ
d i b g b g
26 26
cm
7.23 m
cm
atm
= −
⋅ ×
F
HG
I
KJ
756mmHg 1m
7.23m 100cm
kg 9.81m/s N 760mmHg 1m
m 1kg m/s 1.01325 10 N m cm
cm
2
3 2 5 2
1000 1
100
26
b g
⇒ − =
P P
a b 81
. mm Hg
3.44 (a) Δh h h
= − = = ⇒ −
900
75
388
l l
psi 760 mmHg
14.696 psi
mmHg =900 388=512 mm
.
(b) Δh Pg
= − × = ⇒ =
388 25 2 338 654
mm =
338 mm Hg 14.696 psi
760 mm Hg
psig
.

3-19
3.45 (a) h = L sin θ
(b) h = ° = =
8 7 15 2 3 23
. sin .
cm cm H O mm H O
2 2
b g b g
3.46 (a) P P P P
atm oil Hg
= − −
= − −
⋅ ×
765 365
920 1
kg 9.81 m/ s 0.10 m N 760 mm Hg
m 1 kg m/ s 1.01325 10 N / m
2
3 2 5 2
= 393 mm Hg
(b) — Nonreactive with the vapor in the apparatus.
— Lighter than and immiscible with mercury.
— Low rate of evaporation (low volatility).
3.47 (a) Let ρ f = manometer fluid density 110
. g cm3
c h, ρac = acetone density
0 791
. g cm3
c h
Differential manometer formula: ΔP gh
f ac
= −
ρ ρ
d i
ΔP mmHg
g 981cm h(mm) 1cm dyne 760mmHg
cm s 10mm 1g cm/s 1.01325 10 dyne/cm
3 2 2 6 2
b g b g
=
−
⋅ ×
110 0791 1
. .
= 0 02274
. h mm
b g

. . . . .
V
h
P
mL s 62 87 107 123 138 151
mm 5 10 15 20 25 30
mm Hg 0.114
b g
b g
b g
Δ 0 227 0 341 0 455 0568 0 682
(b) ln ln ln
V n P K
= +
Δ
b g
From the plot above, ln . ln .
V P
= +
04979 52068
Δ
b g
⇒ ≈
n = ,
04979 05
. . ln . .
K K
= ⇒ =
52068 183 0 5
ml s
mm Hg
b g
y = 0.4979x + 5.2068
4
4.5
5
5.5
6
-2.5 -2 -1.5 -1 -0.5 0
ln( P)
ln(V)

3-20
3.47 (cont’d)
(c) h P V
= ⇒ = = ⇒ = =
23 002274 23 0523 183 0523 132
0 5
Δ . . .
.
b gb g b g
mm Hg mL s
132
104 180
mL 0.791 g
s mL
g s
104 g 1 mol
s 58.08 g
mol s
= = .
3.48 (a) T= ° + = ° = − = °
85 4597 18 273 30
F 544 R 303K C
. / .
(b) T =− ° + = × = ° − = °
10 273 18 460 14
C 263K 474 R F
.
(c) ΔT =
° °
°
= °
° °
°
= °
° °
°
= °
85 10
10
85
85 18
1
153
85
10
C K
C
K
C F
C
F
C 1.8 R
C
153 R
.
.
;
.
;
.
(d)
150 R 1 F
1 R
F;
150 R 1.0 K
1.8 R
K;
150 R 1.0 C
1.8 R
83.3 C
° °
°
= °
°
°
= °
° °
°
= °
150 833
D
.
3.49 (a) T = × + = ⇒ ×
0 0940 1000 4 00 98 0
. . .
D D D
FB C T = 98.0 1.8 + 32 = 208 F
(b) Δ Δ Δ
T T T
C) FB) C (K) K
( . ( . .
D D D
= = ⇒ =
0 0940 0 94 0 94
Δ Δ
T T
F)
C 1.8 F
1.0 C
F ( R) R
(
.
. .
D
D D
D
D D D
= = ⇒ =
0 94
169 169
(c) T1 15
= ⇒
D D
C 100 L ; T2 43
= ⇒
D D
C 1000 L
T aT b
C) L)
( (
D D
= +
a =
−
=
F
HG
I
KJ
43 15
0 0311
b g
b g
D
D
D
D
C
1000 -100 L
C
L
. ; b = − × =
15 0 0311 100 119
. . DC
⇒
o o
( C) 0.0311 ( L) 11.9 and
1
( L) 0.0940 ( FB)+4.00-11.9 3.023 ( FB)-254
0.0311
T T
T T T
= +
⎡ ⎤
= =
⎣ ⎦
D D
D
(d) Tbp = − ⇒ ⇒ ⇒
88 6
. D D D
C 184.6 K 332.3 R -127.4 F ⇒ − ⇒ −
9851 3232
. D D
FB L
(e) ΔT = ⇒ ⇒ ⇒ ⇒ ⇒
50 0 16 6 156 2 8 2 8
. . . . .
D D D D D
L 1.56 C FB K F R

3-21
3.50 T T
b m
b g b g
H O AgCl
2
C C
= ° = °
100 455
(a) V mV C
b g b g
= ° +
aT b
527 100
24 88 455
0 05524
0 2539
.
.
.
.
= +
= +
⇒
= °
= −
a b
a b
a
b
mV C
mV
V T
mV C
b g b g
= ° −
0 05524 0 2539
. .
⇓
T V
° = +
C mV
b g b g
1810 4 596
. .
(b) 100 136 1856 2508
. . . .
mV mV C C
→ ⇒ ° → ° ⇒ =
− °
= °
C
s
C/ s
dT
dt
2508 1856
20
326
. .
.
b g
3.51 (a) ln ln ln
T K n R T KRn
= + =
n = =
ln . .
ln . .
.
250 0 110 0
40 0 20 0
1184
b g
b g
ln ln . . ln . ) . . . .
K K T R
= − = ⇒ = ⇒ =
(
1100 1184 200 1154 3169 3169 1184
(b) R =
F
HG I
KJ =
320
3169
49 3
1 1184
.
.
/ .
(c) Extrapolation error, thermocouple reading wrong.
3.52 (a) PV nT
= 0 08206
.
P
P
V
atm
psig
, V L ft
ft
L
3
3
b g b g b g d i
=
′ +
= ′ ×
14696
14696
28317
.
.
.
n n
T
mol lb - moles
453.59 mol
lb moles
, T( K)
F)
1.8
b g b g
= ′ ×
−
=
′ −
+
D
D
(
.
32
27315
⇒
′ +
× ′ × = × ′ × ×
′ −
+
L
NM O
QP
453.59
1.8
P
V n
T
14696
14696
28317 008206
1
32
27315
.
.
. .
( )
.
b g
⇒ ′ + × ′ =
× ×
×
× ′ × ′ +
P V n T
14 696
0 08206 14 696
28 317 18
459 7
.
. .
. .
.
b g b g
453.59
⇒ ′ + ′ = ′ ′ +
P V n T
14696 1073 4597
. . .
b g b g

3-22
3.52 (cont’d)
(b) ′ =
+ ×
× +
=
ntot
500 14 696 35
10 73 85 459 7
0 308
. .
. .
.
b g
b g lb - mole
mCO = =
0308
26
.
.
lb-mole 0.30 lb- mole CO
lb-mole
28 lb CO
lb- mole CO
lb CO
m
m
(c) ′ =
+ ×
×
− =
T
3000 14 696 35
10 73 0 308
459 7 2733
. .
. .
.
b g DF
3.53 (a) T ° = × +
C ohms
b g b g
a r b
0 23624
100 33028
10634
25122
10634 25122
= +
= +
U
V
W
⇒
=
=−
⇒ ° = −
.
.
.
.
. .
a b
a b
a
b
C ohms
T r
b g b g
(b)

min

n
n n
kmol
s
(kmol) min
60 s
F
HG I
KJ =
′
=
′
1
60
P
P P
T T
atm
mmHg atm
760 mmHg
, K C
b g b g b g b g
=
′
=
′
= ′ ° +
1
760
27316
.

V V
V
m
s
m min
min 60 s
3 3
F
HG
I
KJ= ′ =
′
1
60
.
.

.
.
′
=
′ ′
′ +
⇒ ′ =
′ ′
′ ° +
n P V
T
n
P V
T
60
12186
760 27316 60
0016034
27316
mmHg m min
C
3
b g d i
b g
(c) T r
= −
10 634 25122
. .
⇒
r T
r T
r T
1 1
2 2
3 3
26159 26 95
26157 26 93
44 789 2251
= ⇒ = °
= ⇒ = °
= ⇒ = °
. .
. .
. .
C
C
C
P h P h h
(mm Hg) in Hg)
760 mm Hg
29.92 in Hg
atm
= + = +
F
HG I
KJ = +
( . .
29 76 7559
⇒
h P
h P
h P
1 1
2 2
3 3
232 9879
156 9119
74 8299
= ⇒ =
= ⇒ =
= ⇒ =
mm mmHg
mm mmHg
mm mmHg
.
.
.

3-23
3.53 (cont’d)
(d)
. .
. .
. min
n1
0 016034 987 9 947 60
26 95 27316
08331
=
+
=
b gb gb g kmol CH4

. .
. .
. min
n2
0 016034 9119 195
26 93 27316
9 501
=
+
=
b gb gb g kmol air
. min
n n n
3 1 2 10 33
= + = kmol
(e) V
n T
P
3
3 2
3
27316
0 016034
10 33 2251 27316
0 016034 829 9
387
=
+
=
+
=
.
.
. . .
. .
min
b g b gb g
b gb g m3
(f)
08331 16 04
1336
. .
.
kmol CH kg CH
min kmol
kg CH
min
4 4 4
=
2 2 2 2
2 2
0.21 9.501 kmol O 32.0 kg O 0.79 9.501 kmol N 28.0 kg N kg air
274
min kmol O min kmol N min
× ×
+ =
xCH4
1336
1336 274
0 0465
=
+
=
. min
( . )
.
kg CH
kg / min
kg CH kg
4
4
3.54 REAL, MW, T, SLOPE, INTCPT, KO, E
REAL TIME (100), CA (100), TK (100), X (100), Y(100)
INTEGER IT, N, NT, J
READ 5,∗
b g MW, NT
DO 10 IT=1, NT
READ 5,∗
b g TC, N
TK(IT) = TC + 273.15
READ 5,∗
b g (TIME (J), CA (J), J = 1, N)
DO 1 J=1, N
CA J CA J / MW
b g b g
=
X J TIME J
b g b g
=
Y J 1./CA J
b g b g
=
1 CONTINUE
CALL LS (X, Y, N, SLOPE, INTCPT)
K IT SLOPE
b g=
WRITE (E, 2) TK (IT), (TIME (J), CA (J), J = 1, N)
WRITE (6, 3) K (IT)
10 CONTINUE
DO 4 J=1, NT
X J 1./TK J
b g b g
=
Y J LOG K J
b g b g
c h
=

3-24
3.54 (cont’d)
4 CONTINUE
CALL LS (X, Y, NT, SLOPE, INTCPT)
KO EXP INTCPT
= b g
E 8.314 SLOPE
= − =
WRITE (6, 5) KO, E
2 FORMAT (' TEMPERATURE (K): ', F6.2, /
* ' TIME CA', /,
* ' (MIN) (MOLES)', /
* 100 (IX, F5.2, 3X, F7.4, /))
3 FORMAT (' K (L/MOL – MIN): ', F5.3, //)
5 FORMAT (/, ' KO (L/MOL – MIN) : ', E 12.4, /, ' E (J/MOL): ', E 12.4)
END
SUBROUTINE LS (X, Y, N, SLOPE, INTCPT)
REAL X(100), Y(100), SLOPE, INTCPT, SX, SY, SXX, SXY, AN
INTEGER N, J
SX=0
SY=0
SXX=0
SXY=0
DO 10 J=1,N
SX = SX + X(J)
SY = SY + Y(J)
SXX = SXX + X(J)**2
SXY = SXY + X(J)*Y(J)
10 CONTINUE
AN = N
SX = SX/AN
SY = SY/AN
SXX = SXX/AN
SXY = SXY/AN
SLOPE = (SXY – SX*SY)/(SXX – SX**2)
INTCPT = SY – SLOPE*SX
RETURN
END
$ DATA [OUTPUT]
65.0 4 TEMPERATURE (K): 367.15
94.0 6 TIME CA
10.0 8.1 (MIN) (MOLS/L)
20.0 4.3 10.00 0.1246
30.0 3.0 20.00 0.0662
40.0 2.2 30.00 0.0462
50.0 1.8 40.00 0.0338

3-25
3.54 (cont’d)
60.0 1.5 50.00 0.0277
60.00 0.0231
K L/ MOL MIN : 0.707 at 94 C
⋅ °
b g b g
110. 6
10.0 3.5
20.0 1.8 TEMPERATURE (K): 383.15
30.0 1.2 #
40.0 0.92 K L/ MOL MIN : 1.758
⋅
b g
50.0 0.73
60.0 0.61 #
127. 6
# K0 L/ MOL MIN : 0.2329E 10
− +
b g
# ETC E J / MOL 0.6690E
b g: + 05

4-1
CHAPTER FOUR
4.1 a. Continuous, Transient
b. Input – Output = Accumulation
No reactions ⇒ Generation = 0, Consumption = 0
6 00 300 300
. . .
kg
s
kg
s
kg
s
− = ⇒ =
dn
dt
dn
dt
c.
t = =
100
1
1
300
333 s
.
.
m 1000kg
m
s
kg
3
3
4.2 a. Continuous, Steady State
b. k k
= ⇒ = = ∞ ⇒ =
0 0
C C C
A A0 A
c. Input – Output – Consumption = 0
Steady state ⇒ Accumulation = 0
A is a reactant ⇒ Generation = 0

V C V C kVC C
C
kV
V
A A A A
A
m
s
mol
m
m
s
mol
m
mol
s
3
3
3
3
F
HG I
KJ F
HG I
KJ =
F
HG I
KJ F
HG I
KJ+
F
HG I
KJ ⇒ =
+
0
0
1
4.3 a.
100kg / h
0.550kg B / kg
0.450kgT / kg

mv kg / h
0.850kg B / kg
0.150kgT / kg
b g

ml kg / h
0.106kg B / kg
0.894kgT / kg
b g
Input – Output = 0
Steady state ⇒ Accumulation = 0
No reaction ⇒ Generation = 0, Consumption = 0
(1) Total Mass Balance: 100 0
.
kg / h = +
m m
v l
(2) Benzene Balance: 0550 100 0 0850 0106
. . . .
× = +
kgB / h m m
v l
Solve (1) (2) simultaneously ⇒ . .
m m
v l
= =
59 7 40 3
kg h, kg h
b. The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by
masses (kg). The balance equations are also identical (initial input = final output).
c. Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state,
the feed composition is incorrect, the flow rates are not what they are supposed to be, other
species are in the feed stream, measurement errors.

4-2
4.4 b. n(mol)
mol N mol
mol CH mol
2
4
0 500
0 500
.
.
0 500 28 1
0 014
.
.
n
n
mol N g N
mol N
kg
1000g
kg N
2 2
2
2
b g b g
=
c. 100 0
. g / s
g C H g
g C H g
g C H g
2 6
3 8
4 10
x
x
x
E
P
B
b g
b g
b g

.
.
n
x
x
E
E
E
=
=
100 1
453593 30
3600
26 45
g C H
s
lb
g
lb - mole C H
lb C H
s
h
lb- mole C H / h
2 6 m 2 6
m 2 6
2 6
b g
b g
d. lb - moleH O s
lb - moleDA s
lb - moleO lb - moleDA
lb - moleN lb - moleDA
2
2
2

.
.
n
n
1
2
021
079
b g
b g
R
S
|
T
|
U
V
|
W
|
( )
2
2
2
O 2 2
1 2
H O
1 2
2 2
O
1 2
0.21 lb-mole O /s
lb-mole H O
lb-mole
0.21 lb-mole O
lb-mole
n n
n
x
n n
n
x
n n
=
⎛ ⎞
= ⎜ ⎟
+ ⎝ ⎠
⎛ ⎞
= ⎜ ⎟
+ ⎝ ⎠

e. ( )
( )
( )
2
2
NO 2
NO 2 4
mol
0.400mol NO mol
mol NO mol
0.600 mol N O mol
n
y
y
−
( )
2 4 2
N O NO 2 4
0.600 mol N O
n n y
⎡ ⎤
= −
⎣ ⎦
4.5 a.
1000 lb C H / h
m 3 8
Still

.
.
n7 m
m 3 8 m
m 3 6 m
lb / h
lb C H / lb
lb C H / lb
b g
097
0 03

.
.
n6 m
m 3 8 m
m 3 6 m
lb / h
lb C H / lb
lb C H / lb
b g
0 02
098

n
n
1 m 3 8
2 m 3 6
lb C H / h
lb C H / h
b g
b g

n
n
n
n
1 m 3 8
2 m 3 6
3 m 4
4 m 2
lb C H / h
lb C H / h
lb CH / h
lb H / h
b g
b g
b g
b g

n
n
3 m 4
4 m 2
lb CH / h
lb H / h
b g
b g
n5 m
lb / h
b g

n
n
n
1 m 3 8
2 m 3 6
5 m
lb C H / h
lb C H / h
lb oil / h
b g
b g
b g
Reactor
Absorber
Stripper
Compressor
Basis: 1000 lbm C3H8 / h fresh feed
(Could also take 1 h operation as basis -
flow chart would be as below except
that all / h would be deleted.)
Note: the compressor and the off gas from
the absorber are not mentioned explicitly
in the process description, but their presence
should be inferred.

4-3
4.5 (cont’d)
b. Overall objective: To produce C3H6 from C3H8.
Preheater function: Raise temperature of the reactants to raise the reaction rate.
Reactor function: Convert C3H8 to C3H6.
Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other
components.
Stripping tower function: Recover the C3H8 and C3H6 from the solvent.
Distillation column function: Separate the C3H5 from the C3H8.
4.6 a. 3 independent balances (one for each species)
b. 7 unknowns ( , , , , , ,
m m m x y y z
1 3 5 2 2 4 4 )
– 3 balances
– 2 mole fraction summations
2 unknowns must be specified
c. y x
2 2
1
= −
A Balance: 5300 1200 0 70
2 3
x m
kg A
h
kg A
h
F
HG I
KJ = +
F
HG I
KJ
.
b gb g
Overall Balance:
m m m
1 3 5
5300 1200
+
F
HG I
KJ = + +
F
HG I
KJ
kg
h
kg
h
B Balance: 0 03 5300 1200 0 60
1 2 4 5
. .
m x y m
+
F
HG I
KJ = +
F
HG I
KJ
kg B
h
kg B
h
z y
4 4
1 0 70
= − −
.
4.7 a. 3 independent balances (one for each species)
b.
Water Balance:
400 0885 0995
356
g m g
m
R
R
min
.
min
.

g H O
g
g H O
g
g min
2 2
= ⇒ =
b g
b g
Acetic Acid Balance: 400 0115 0 005 0 096
b gb g
. . .
gCH OOH
min
gCH OOH
min
3 3
F
HG I
KJ = +
F
HG I
KJ
m m
R E
⇒ =

mE 461g min
Overall Balance:
m m m m
C R E C
+
F
HG I
KJ = +
F
HG I
KJ⇒ =
400 417
g
min
g
min
g min
c.
0115 400 0 005 356 0 096 461 44 44
. . .
b gb g b gb g b gb g
−
F
HG I
KJ =
F
HG I
KJ ⇒
g
min
g
min
g min = g min

4-4
4.7 (cont’d)
d.
Extractor
CH COOH
H O
3
2
H O
someCH COOH
2
3
C H OH
CH COOH
4 9
3
C H OH
4 9
Distillation
Column
C H OH
4 9
CH COOH
3
4.8 a.
120 eggs/min
0.30 broken egg/egg
0.70 unbroken egg/egg
X-large: 25 broken eggs/min
35 unbroken eggs/min
Large: broken eggs/min
unbroken eggs/min
n1
n2
b. ( )
( )( )
1 2 1 2 1
2
1
120 25 45 eggs min 50 11
39
0.30 120 25
n n n n n
n
n
⎫
= + + + ⇒ + = =
⎪
⇒
⎬
=
= + ⎪
⎭
c. n n
1 2 50
+ = large eggs min
n1 large eggs broken/50 large eggs = =
11 50 0 22
b g .
d. 22% of the large eggs (right hand) and 25 70 36%
b g⇒ of the extra-large eggs (left hand)
are broken. Since it does not require much strength to break an egg, the left hand is probably
poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed.
4.9 a. m1
015
0 85
lb strawberries
lb S / lb
lb W / lb
m
m m
m m
b g
.
.
m2 lb S sugar
m
c h
m3 lb W evaporated
m
b g
100
0 667
0 333
.
.
.
lb jam
lb S / lb
lb W / lb
m
m m
m m
b. 3 unknowns (m m m
1 2 3
, , )
– 2 balances
– 1 feed ratio
0 DF
c. 1 2
1 2
1 m 2 m
Feed ratio: / 45 / 55 (1)
S balance: 0.15 0.667 (2)
Solve simultaneously 0.49 lb strawberries, 0.59 lb sugar
m m
m m
m m
=
+ =
⇒ = =
X-large: 25 broken eggs/min
45 unbroken eggs/min

4-5
4.10 a.
300
0 750
0 250
1
gal
lb
lb C H OH / lb
lb H O / lb
m
m 2 5 m
m 2 m
m b g
.
.
V
m
40
2
0 400
0 600
gal
lb
lb C H OH / lb
lb H O / lb
m
m 2 5 m
m 2 m
b g
b g
.
.
m3
0 600
0 400
lb
lb C H OH / lb
lb H O / lb
m
m 2 5 m
m 2 m
b g
.
.
4 unknowns (m m V m
1 2 40 3
, , , )
– 2 balances
– 2 specific gravities
0 DF
b.
m1
300 1
7 4805
0877 62 4
2195
=
×
=
gal ft
gal
lb
ft
lb
3
m
3 m
.
. .
Overall balance: m m m
1 2 3
+ = (1)
C2H5OH balance: 0 750 0 400 0 600
1 2 3
. . .
m m m
+ = (2)
Solve (1) (2) simultaneously ⇒ = =
m m
2 3
1646 3841
lb lb
m, m
,
V40
1646 7 4805
1
207
=
×
=
lb ft
0.952 62.4lb
gal
ft
gal
m
3
m
3
.
4.11 a.

.
.
n1
0 0403
09597
mol/ s
mol C H / mol
mol air / mol
3 8
b g

.
.
n2
0 21
0 79
molair / s
molO / mol
mol N / mol
2
2
b g

.
.
n3
0 0205
09795
mol/ s
molC H / mol
molair / mol
3 8
b g
3 unknowns ( , ,
n n n
1 2 3 )
– 2 balances
1 DF
b. Propane feed rate: 0 0403 150 3722
1 1
.
n n
= ⇒ = mol / s
b g
Propane balance: 0 0403 0 0205 7317
1 3 3
. .
n n n
= ⇒ = mol / s
b g
Overall balance: 3722 7317 3600
2 2
+ = ⇒ =

n n mol / s
b g
c. . The dilution rate should be greater than the value calculated to ensure that ignition is not
possible even if the fuel feed rate increases slightly.

4-6
4.12 a.
1000
0500
0500
kg / h
kgCH OH / kg
kgH O / kg
3
2
.
.

.
.
m kg / h
kgCH OH / kg
kgH O / kg
3
2
b g
0960
0 040
673
1
kg / h
kgCH OH / kg
kgH O / kg
3
2
x
x
b g
b g
−
2 unknowns ( ,
m x )
– 2 balances
0 DF
b. Overall balance: 1000 673 327
= + ⇒ =

m m kg / h
Methanol balance: 0500 1000 0960 327 673 0 276
. . .
b g b g b g
= + ⇒ =
x x kgCH OH / kg
3
Molar flow rates of methanol and water:
673 0 276 1000
32 0
580 10
673 0 724 1000
18
2 71 10
3
4
kg
h
kgCH OH
kg
g
kg
molCH OH
gCH OH
molCH OH / h
kg
h
kgH O
kg
g
kg
molH O
gH O
molH O / h
3 3
3
3
2 2
2
2
.
.
.
.
.
= ×
= ×
Mole fraction of Methanol:
580 10
580 10 2 71 10
0176
3
3 4
.
. .
.
×
× + ×
= molCH OH / mol
3
c. Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the
system is not at steady state.
4.13 a.
Feed Reactor effluent
Product
Waste
2253 kg 2253 kg
R = 388
1239 kg
R = 583
mw kg
R = 140
b g
Reactor Purifier
Analyzer Calibration Data
xp = 0.000145R
1.364546
0.01
0.1
1
100 1000
R
xp

4-7
4.13 (cont’d)
b. Effluent: xp = =
0 000145 388 0 494
1 3645
. .
.
b g kgP / kg
Product: xp = =
0 000145 583 0861
1 3645
. .
.
b g kgP / kg
Waste: xp = =
0 000145 140 0123
1 3645
. .
.
b g kgP / kg
Efficiency = × =
0861 1239
0 494 2253
100% 958%
.
.
.
b g
b g
c. Mass balance on purifier: 2253 1239 1014
= + ⇒ =
m m
w w kg
P balance on purifier:
Input: 0 494 2253 1113
. kgP / kg kg kgP
b gb g=
Output: 0861 1239 0123 1014 1192
. .
kgP / kg kg kgP / kg kg kgP
b gb g b gb g
+ =
The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation
beyond analyzer calibration data is risky -- recalibrate; get data for R 583; not at steady
state; additional reaction occurs in purifier; normal data scatter.
4.14 a.

.
.
n1
00100
09900
lb-mole/ h
lb-moleH O/lb-mole
lb-moleDA/ lb-mole
2
b g

n
v
2
2
lb-mole HO/ h
ft / h
2
3
b g
d i

.
.
n3
0100
0900
lb-mole/ h
lb-moleH O/lb-mole
lb-moleDA/ lb-mole
2
b g
4 unknowns ( , , ,
n n n v
1 2 3 ) – 2 balances – 1 density – 1 meter reading = 0 DF
Assume linear relationship:
v aR b
= +
Slope: a
v v
R R
=
−
−
=
−
−
=
. .
.
2 1
2 1
96 9 40 0
50 15
1626
Intercept: b v aR
a
= − = − =
. . .
1 40 0 1626 15 15 61
b g
. .
v2 1626 95 15 61 170
= + =
b g c h
ft / h
3

.
n2
170 62 4
589
= =
ft
h
lb
ft
lb - mol
18.0 lb
lb - moles H O / h
3
m
3
m
2
b g
DA balance: 0 9900 0 900
1 3
. .
n n
= (1)
Overall balance:
n n n
1 2 3
+ = (2)
,
n n
1 3
5890 6480
lb - moles / h lb - moles / h
b. Bad calibration data, not at steady state, leaks, 7% value is wrong,
v − R relationship is not
linear, extrapolation of analyzer correlation leads to error.

4-8
4.15 a.
100
0 600
0 050
0 350
kg / s
kg E / kg
kgS / kg
kg H O / kg
2
.
.
.

.
.
m kg / s
kg E / kg
kg H O / kg
2
b g
0 900
0100

m
x
x
x x
E
S
E S
kg / s
kg E / kg
kgS / kg
kg H O / kg
2
b g
b g
b g
b g
1− −
3 unknowns ( , ,
m x x
E S )
– 3 balances
0 DF
b. Overall balance: 100 2 50 0
= ⇒ =
.
m m kg / s
b g
S balance: 0 050 100 50 0100
. .
b g b g b g
= ⇒ =
x x
S S kgS / kg
E balance: 0 600 100 0900 50 50 0300
. . .
b g b g b g
= + ⇒ =
x x
E E kgE / kg
kgEin bottomstream
kgEinfeed
kgEin bottomstream
kgEinfeed
= =
0300 50
0 600 100
0 25
.
.
.
b g
b g
c. x aR x a b R
b
x x
R R
a x b R a
x R
R
x
a
b
b
= ⇒ = +
= = =
= − = − = − ⇒ = ×
= ×
=
F
HG I
KJ =
×
F
HG I
KJ =
−
−
−
ln ln ln
ln /
ln /
ln . / .
ln /
.
ln ln ln ln . . ln . .
.
.
.
.
.
.
b g b g b g
b g
b g
b g
b g
b g b g b g b g b g
2 1
2 1
1 1
3
3 1 491
1
3
1
1 491
0400 0100
38 15
1491
0100 1491 15 6340 1764 10
1764 10
0900
1764 10
655
d. Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass
fractions – measure against known standard. Impurities in the stream – analyze a sample.
Mixture is not all liquid – check sample. Calibration data are temperature dependent – check
calibration at various temperatures. System is not at steady state – take more measurements.
Scatter in data – take more measurements.

4-9
4.16 a. 4 00 0 098
1213
0323
. .
.
.
molH SO
L of solution
kgH SO
molH SO
L of solution
kgsolution
kgH SO / kgsolution
2 4 2 4
2 4
2 4
= b g
b.
v1
100
0 200
0 800
1139
L
kg
kg H SO / kg
kg H O / kg
SG
2 4
2
b g
.
.
.
=
v
m
2
2
0 600
0 400
1498
L
kg
kg H SO / kg
kg H O / kg
SG
2 4
2
b g
b g
.
.
.
=
v
m
3
3
0 323
0 677
1213
L
kg
kg H SO / kg
kg H O / kg
SG
2 4
2
b g
b g
.
.
.
=
5 unknowns (v v v m m
1 2 3 2 3
, , , , )
– 2 balances
0 DF
Overall mass balance:
Water balance
kg
kg
100
0800 100 0 400 0 677
44 4
144
2 3
2 3
2
3
+ =
+ =
U
V
W
⇒
=
=
m m
m m
m
m
: . . .
.
b g
v1
100
1139
8780
= =
kg L
kg
L20%solution
.
.
v2
44 4
1498
29 64 60%
= =
.
.
.
kg L
kg
L solution
v
v
1
2
8780
29 64
296
60%
= =
.
.
.
L 20%solution
L solution
c. 1250 44 4
144 1498
257
kgP
h
kg60%solution
kgP
L
kgsolution
L / h
.
.
=
4.17 m1
025
075
kg @$18/ kg
kgP / kg
kgH O / kg
2
b g
.
.
m2
012
088
kg @$10/ kg
kgP / kg
kgH O / kg
2
b g
.
.
100
017
083
.
.
.
kg
kg P/ kg
kgH O / kg
2
Overall balance: m m
1 2 100
+ = . (1)
Pigment balance: 0 25 012 017 100
1 2
. . . .
m m
+ = b g (2)
Solve (1) and (2) simultaneously ⇒ = =
m m
1 2
0385 0 615
. , .
kg25%paint kg12%paint
Cost of blend: 0 385 00 0 615 00 08
. $18. . $10. $13.
b g b g
+ = per kg
Selling price:110 08 39
. $13. $14.
b g= per kg

4-10
4.18 a.
100
0800
0 200
kg
kgS / kg
kgH O / kg
2
.
.
m
m
2
3
kgS
kg H O
2
b g
b g
m1 kgH O 85%of entering water
2
b gb g
85% drying: m1 0850 0 200 100 17 0
= =
. . .
b gb g kgH O
2
Sugar balance: m2 0800 100 80 0
= =
. .
b g kgS
Overall balance: 100 17 80 3
3 3
= + + ⇒ =
m m kgH O
2
xw =
+
=
3
3 80
0 0361
kgH O
kg
kgH O / kg
2
2
b g .
m
m m
1
2 3
17
80 3
0 205
+
=
+
=
kgH O
kg
kgH O / kgwetsugar
2
2
b g .
b. 1000 3
100
30
tonswet sugar
day
tonsH O
tonswet sugar
tons H O / day
2
2
=
1000 0800 2000 15 365
8 107
tonsWS
day
tonsDS
ton WS
lb
ton lb
days
year
per year
m
m
. $0.
$8.
= ×
c.
x x x x
x x x x
w w w w
w w w w
= + + + =
= − + + − =
= ±
= =
1
10
0 0504
1
9
0 00181
0 0504 3 0 00181
0 0450 0 0558
1 2 10
1
2
10
2
... .
... .
. .
. , .
b g
b g b g
b g
kg H O / kg
SD kg H O / kg
Endpoints
Lower limit Upper limit
2
2
d. The evaporator is probably not working according to design specifications since
xw =
0 0361 0 0450
. . .
4.19 a. v
m
SG
1
1
1 00
m
kg H O
3
2
c h
b g
= .
v
SG
2
400
7 44
m
kg galena
3
d i
= .
v
m
SG
3
3
1 48
m
kg suspension
3
d i
b g
= .
5 unknowns (v v v m m
1 2 3 1 3
, , , , )
– 1 mass balance
– 1 volume balance
0 DF
Total mass balance: m m
1 3
400
+ = (1)

4-11
4.19 (cont’d)
Assume volume additivity:
m m
1 3
1000
400
7440 1480
kg m
kg
kg m
kg
kg m
kg
3 3 3
b g b g
+ = (2)
m m
1 3
668 1068
kgH O kgsuspension
2 ,
v1
668
1000
0 668
= =
kg m
kg
m waterfed totank
3
3
.
b. Specific gravity of coal 1.48 Specific gravity of slate
c. The suspension begins to settle. Stir the suspension. 1.00 Specific gravity of coal 1.48
4.20 a.

.
.
n1
0 040
0960
mol / h
molH O / mol
molDA / mol
2
b g
n
x
x
2
1
mol / h
molH O / mol
molDA / mol
2
b g
b g
b g
−

n3
97%
molH Oadsorbed / h
of H Oin feed
2
2
b g
Adsorption rate:
. .
.
.
n3
354 340
0 0180
1556
=
−
=
b gkg
5 h
molH O
kgH O
molH O / h
2
2
2
97% adsorbed: 156 097 0 04 401
1 1
. . . .
= ⇒ =
n n
b g mol / h
Total mole balance: . . .
n n n n
1 2 3 2 401 1556 3854
= + ⇒ = − = mol / h
Water balance: ( ) ( ) ( )
3
2
0.040 40.1 1.556 38.54 1.2 10 molH O/mol
x x −
= + ⇒ = ×
b. The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction
will reach that of the inlet stream, i.e. 4%.
4.21 a. 300
055
0 45
lb / h
lb H SO / lb
lb H O / lb
m
m 2 4 m
m 2 m
.
.

.
.
mB lb / h
lb H SO / lb
lb H O / lb
m
m 2 4 m
m 2 m
b g
090
010

.
.
mC lb / h
lb H SO / lb
lb H O / lb
m
m 2 4 m
m 2 m
b g
0 75
0 25
Overall balance: 300+ =

m m
B C (1)
H2SO4 balance: 055 300 0 90 0 75
. . .
b g+ =
m m
B C (2)
,
m m
B C
400 700
lb / h lb / h
m m

4-12
4.21 (cont’d)
b.
. .
m R m R
A A A A
− =
−
−
− ⇒ = −
150
500 150
70 25
25 7 78 44 4
b g
.
m R m R
B B B B
− =
−
−
− ⇒ = −
200
800 200
60 20
20 150 100
b g
ln ln
ln ln
ln . . .
x R x R x e
x x
Rx
− =
−
−
− ⇒ = + ⇒ =
20
100 20
10 4
4 0 2682 1923 6841 0.2682
b g
m R m R
x R
A A B B
x
= ⇒ =
+
= = ⇒ =
+
=
= ⇒ =
F
HG I
KJ =
300
300 44 4
7 78
443 400
400 100
150
333
55%
1
0 268
55
6841
7 78
.
.
. ,
.
. ,
.
ln
.
.
c. Overall balance:
m m m
A B C
+ =
H2SO4 balance: 0 01 0 90 0 75 0 75
0 75 0 01
015
. . . .
. .
.
xm m m m m m
x m
A B C A B B
A
+ = = + ⇒ =
−
b g b g
⇒ − =
− −
⇒ = − + −
150 100
0 75 0 01 6841 7 78 44 4
015
259 0 236 135 813
0.2682
0.2682 0.2682
.
. . . . .
.
. . . .
R
e R
R e R e
B
R
A
B
R
A
R
x
x x
d ib g
d i
Check: R R R e e
A x B
= = ⇒ = − + − =
443 7 78 259 0 236 443 135 813 333
0.2682 7.78 0.2682 7.78
. , . . . . . . .
b g b g
e j
4.22 a.

.
.
nA kmol / h
kmolH / kmol
kmol N / kmol
2
2
b g
010
090

.
.
nB kmol / h
kmolH / kmol
kmol N / kmol
2
2
b g
050
050
100
0 20
080
kg / h
kmol / h
kmolH / kmol
kmol N / kmol
2
2

.
.
nP b g
MW kg / kmol
= + =
0 20 2 016 080 28 012 22813
. . . . .
b g b g
⇒ = =

.
.
nP
100
22813
438
kg
h
kmol
kg
kmol / h
Overall balance: .
n n
A B
+ = 438 (1)
H2 balance: 010 050 0 20 438
. . . .
n n
A B
+ = b g (2)
. , .
n n
A B
329 110
kmol / h kmol / h

4-13
4.22 (cont’d)
b.

.
n
m
P
P
=
22 813
Overall balance:

.
n n
m
A B
P
+ =
22 813
H2 balance: x n x n
x m
A A B B
P P

.
+ =
22 813
⇒ =
−
−
=
−
−

.

.
n
m x x
x x
n
m x x
x x
A
P B P
B A
B
P P A
B A
22813 22813
b g
b g
b g
b g
c. Trial XA XB XP mP nA nB
1 0.10 0.50 0.10 100 4.38 0.00
2 0.10 0.50 0.20 100 3.29 1.10
3 0.10 0.50 0.30 100 2.19 2.19
4 0.10 0.50 0.40 100 1.10 3.29
5 0.10 0.50 0.50 100 0.00 4.38
6 0.10 0.50 0.60 100 -1.10 5.48
7 0.10 0.50 0.10 250 10.96 0.00
8 0.10 0.50 0.20 250 8.22 2.74
9 0.10 0.50 0.30 250 5.48 5.48
10 0.10 0.50 0.40 250 2.74 8.22
11 0.10 0.50 0.50 250 0.00 10.96
12 0.10 0.50 0.60 250 -2.74 13.70
The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot
blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture.
d. Results are the same as in part c.
4.23
Arterial blood
ml / min
mg urea / ml
200 0
190
.
.
Dialyzing fluid
ml / min
1500
Venous blood
ml / min
mg urea /ml
195 0
175
.
.
Dialysate
ml / min
mg urea / ml

v
c
b g
b g
a. Water removal rate: 2000 1950 50
. . .
− = ml / min
Urea removal rate: 190 200 0 175 1950 388
. . . . .
b g b g
− = mg urea / min
b. . / min
v = + =
1500 50 1505ml
38.8mg urea/min
0.0258mg urea/ml
1505ml/min
c = =
c. 2 7 11
206
. .
min
−
=
b gmg removed 1 min 10 ml 5.0 L
ml 38.8 mg removed 1 L
(3.4 h)
3

4-14
4.24 a.
.
n1
20 0
kmol / min
kgCO / min
2
b g

.
n2
0 015
kmol / min
kmolCO / kmol
2
b g

.
n3
0 023
kmol / min
kmolCO / kmol
2
b g

.
min .
.
n1
20 0
44 0
0 455
= =
kgCO kmol
kgCO
kmolCO / min
2
2
2
.
+ =
n n (1)
CO2 balance: 0 455 0 015 0 023
2 3
. . .
+ =
n n (2)
. , .
n n
2 3
556 561
kmol / min kmol / min
b.
u = =
150
18
833
m
s
m / s
.
A D D
= = ⇒ =
1
4
561
0123
1
60 833
108
2
π
.
min .
min
.
.
kmol m
kmol s
s
m
m
3
4.25 Spectrophotometer calibration: C kA C A
A
C
= ==== =
=
=
0.9
3
3333
g / L
μ
b g .
Dye concentration: A C
= ⇒ = =
018 3333 018 0 600
. . . .
b gb g g / L
μ
Dye injected = =
0.60 cm L 5.0 mg 10 g
10 cm L 1 mg
g
3 3
3 3
1
1
30
μ
μ
.
⇒ = ⇒ =
30 0 600 50
. . .
g L g / L L
μ μ
b g b g
V V
4.26 a.

V
n
y
y
1
1
1
1
1
m / min
kmol / min
kmol SO / kmol
kmol A / kmol
3
2
d i
b g
b g
b g
−
1000
2
LB / min
kgB / min

m b g

n
y
y
3
3
3
1
kmol / min
kmol SO / kmol
kmol A / kmol
2
b g
b g
b g
−

m
x
x
4
4
4
1
kg / min
kg SO / kg
kg B / kg
2
b g
b g
b g
−

4-15
4.26 (cont’d)
8 unknowns ( , , , , , , ,
n n v m m x y y
1 3 1 2 4 4 1 3 )
– 3 material balances
– 2 analyzer readings
– 1 meter reading
– 1 gas density formula
– 1 specific gravity
0 DF
b. Orifice meter calibration:
A log plot of vs. is a line through the points and
, , .
V h h V h V
1 1 2 2
100 142 400 290
= = = =
d i d i
ln ln ln
ln
ln
ln
ln
.
ln ln ln ln . ln . . .
.
V b h a V ah
b
V V
h h
a V b h a e V h
b
= + ⇒ =
= = =
= − = − = ⇒ = = ⇒ =
2 1
2 1
1 1
2 58 0.515
290 142
400 100
0515
142 0515 100 2 58 132 132
d h
b g
b g
b g
b g
Analyzer calibration:
ln ln
y bR a y aebR
= + ⇒ =
b
y y
R R
a y bR
a
y e R
=
−
=
−
=
= − = − = −
E
= ×
U
V
|
|
|
W
|
|
|
⇒ = ×
−
−
ln ln . .
.
ln ln ln . . .
.
.
2 1
2 1
1 1
4
4 0.0600
01107 0 00166
90 20
0 0600
0 00166 0 0600 20 7 60
500 10
500 10
b g b g
b g b g
c. h V
1 1
0.515
210 132 210 207 3
= ⇒ = =
mm m min
3
. .
b g
ρfeed gas
3
3
3
atm
K
mol / L = 0.460 kmol / m
m
min
kmol
m
kmol min
=
+
+
=
E
= =
12 2 150 14 7 14 7
75 460 18
0 460
2073 0 460
9534
1
. . .
.
.

. .
.
b gb g b g
b g b g
n
R y
R y
m
1 1
4
3 3
4
2
824 500 10 00600 824 00702
116 500 10 00600 116 000100
1000 130
1300
= ⇒ = × × =
= ⇒ = × × =
= =
−
−
. . exp . . .
. . exp . . .

.
b g
b g
kmol SO kmol
kmol SO kmol
L B
min
kg
L B
kg / min
2
2

4-16
4.26 (cont’d)
A balance: 1 0 0702 9534 1 0 00100 88 7
3 3
− = − ⇒ =
. . . .
b gb g b gn n kmol min
SO balance: kg / kmol) (1)
B balance: 1300 = (2)
Solve (1) and (2) simultaneously =1723 kg / min, = 0.245 kg SO absorbed / kg
SO removed = kg SO / min
2
2
2 2
00702 9534 640 000100 887 64
1
422
4 4
4 4
4 4
4 4
. . ( . . . ( )
( )

b gb g b gb g
= +
−
⇒
=
m x
m x
m x
m x
d. Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a
higher rate of transfer of SO2 from the gas to the liquid phase.
4.27 a.

, , ,
V
n
y
y
P T R h
1
1
1
1
1 1 1 1
1
m / min
kmol / min
kmolSO / kmol
kmol A / kmol
3
2
d i
b g
b g
b g
−

V
m
2
2
m / min
kg B / min
3
d i
b g

n
y
y
R
3
3
3
3
1
kmol / min
kmolSO / kmol
kmol A / kmol
2
b g
b g
b g
−

m
x
x
4
4
4
1
kg / min
kgSO kg
kg B / kg
2
b g
b g
b g
−
b. 14 unknowns ( , , , , , , , , , , , , ,
n V y P T R h V m n y R m x
1 1 1 1 1 1 1 2 2 3 3 3 4 4 )
– 3 material balances
– 3 analyzer and orifice meter readings
– 1 gas density formula (relates
n V
1 1
and )
– 1 specific gravity (relates
m V
2 2
and )
6 DF
A balance: 1 1
1 1 3 3
− = −
y n y n
b g b g
(1)
SO2 balance: y n y n
x m
1 1 3 3
4 4
64

= +
kgSO / kmol
2
(2)
B balance:
m x m
2 4 4
1
= −
b g (3)
Calibration formulas: y e R
1
4 0.060
500 10 1
= × −
. (4)
y e R
3
4 0.060
500 10 3
= × −
. (5)
.
V h
1 1
0.515
132
= (6)
Gas density formula:
. . / .
/ .

n
P
T
V
1
1
1
1
12 2 14 7 14 7
460 18
=
+
+
b g
b g (7)
Liquid specific gravity: SG V
m
= ⇒ =
130
1300
2
2
. kg
h
m
kg
3
b g (8)

4-17
4.27 (cont’d)
c. T1 75 °F y1 0.07 kmol SO2/kmol
P1 150 psig V1 207 m3/h
h1 210 torr n1 95.26 kmol/h
R1 82.4
Trial x4 (kg SO2/kg) y3 (kmol SO2/kmol) V2 (m3/h) n3 (kmol/h) m4 (kg/h) m2 (kg/h)
1 0.10 0.050 0.89 93.25 1283.45 1155.11
2 0.10 0.025 1.95 90.86 2813.72 2532.35
3 0.10 0.010 2.56 89.48 3694.78 3325.31
4 0.10 0.005 2.76 89.03 3982.57 3584.31
5 0.10 0.001 2.92 88.68 4210.72 3789.65
6 0.20 0.050 0.39 93.25 641.73 513.38
7 0.20 0.025 0.87 90.86 1406.86 1125.49
8 0.20 0.010 1.14 89.48 1847.39 1477.91
9 0.20 0.005 1.23 89.03 1991.28 1593.03
10 0.20 0.001 1.30 88.68 2105.36 1684.29
V2 vs. y3
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
0.000 0.020 0.040 0.060
y3 (kmol SO 2/kmol)
V
2
(m
3
/h)
x4 = 0.10 x4 = 0.20
For a given SO2 feed rate removing more SO2 (lower y3) requires a higher solvent feed
rate (
V2 ).
For a given SO2 removal rate (y3), a higher solvent feed rate (
V2 ) tends to a more dilute
SO2 solution at the outlet (lower x4).
d. Answers are the same as in part c.
4.28 Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3
Overall mass balance ⇒
m3
Mass balance - Unit 1 ⇒
m1
A balance - Unit 1 ⇒ x1
Mass balance - mixing point ⇒
m2
A balance - mixing point ⇒ x2
C balance - mixing point ⇒ y2

4-18
4.29 a. 100
0300
0 250
0 450
mol / h
molB / mol
molT / mol
molX / mol
.
.
.

n
x
x
x x
B
T
B T
2
2
2
2 2
1
mol / h
molB / mol
molT / mol
molX / mol
b g
b g
b g
b g
− −

.
.
n4
0940
0 060
mol / h
molB / mol
molT / mol
b g

.
.
n3
0 020
0980
mol / h
molT / mol
molX / mol
b g
n
x
x
x x
B
T
B T
5
5
5
5 5
1
mol / h
molB / mol
molT / mol
molX / mol
b g
b g
b g
b g
− −
Column 1 Column 2
Column 1 Column 2:
4 unknowns ( , , ,
n n x x
B T
2 3 2 2 ) 4 unknowns ( , , ,
n n n yx
3 4 5 )
–3 balances – 3 balances
– 1 recovery of X in bot. (96%) – 1 recovery of B in top (97%)
0 DF 0 DF
Column 1
96% X recovery: 096 0 450 100 098 3
. . .
b gb g= n (1)
Total mole balance: 100 2 3
= +

n n (2)
B balance: 0300 100 2 2
.
b g= x n
B (3)
T balance: 0 250 100 0 020
2 2 3
. .
b g= +
x n n
T (4)
Column 2
97% B recovery: 097 0940
2 2 4
. .
x n n
B = (5)
Total mole balance:
n n n
2 4 5
= + (6)
B balance: x n n x n
B B
2 2 4 5 5
0940
.
= + (7)
T balance: x n n x n
T T
2 2 4 5 5
0 060
.
= + (8)
b. ( ) . ( ) .
( ) . ( ) .
( ) . ( ) .
( ) . ( ) .
1 441 2 559
3 0536 4 0 431
5 30 95 6 24 96
7 0 036 8 0892
3 2
2 2
4 5
5 5
⇒ = ⇒ =
⇒ = ⇒ =
⇒ = ⇒ =
⇒ = ⇒ =
n n
x x
n n
x x
B T
B T
mol / h mol / h
mol B / mol molT / mol
mol / h mol / h
mol B / mol molT / mol
Overall benzene recovery:
0940 3095
0300 100
100% 97%
. .
.
b g
b g × =
Overall toluene recovery:
0892 2496
0 250 100
100 89%
. .
.
b g
b g × =

4-19
4.30 a.
100
0 035
0965
kg / h
kg S / kg
kg H O / kg
2
.
.

m
x
x
3
3
3
1
kg / h
kg S / kg
kg H O / kg
2
b g
b g
b g
−

m
x
x
4
4
4
1
kg / h
kg S / kg
kg H O / kg
2
b g
b g
b g
−

.
.
m10
0 050
0950
kg / h
kg S / kg
kg H O / kg
2
b g
0100
.
mw kg H O / h
2
b g 0100
.
mw kg H O / h
2
b g 0100
.
mw kg H O / h
2
b g

mw kg H O / h
2
b g
1 4 10
b. Overall process
100 kg/h (
m10 kg / h)
0.035 kg S/kg 0.050 kg S/kg
0.965 kg H2O/kg 0.950 kg H2O/kg
( )
mw kg H O / h
2
Salt balance: 0 035 100 0 050 10
. .
b g= m
Overall balance: 100 10
= +

m m
w
H2O yield: Y
m
w
w
=

.
kgH Orecovered
kgH Oinfreshfeed
2
2
b g
b g
965
First 4 evaporators
100
0035
0965
kg/ h
kg S/ kg
kg H O / kg
2
.
.
4 0100
× .

mw kg H O / h
2
b g

m
x
x
4
4
4
1
kg/ h
kg S/ kg
kg H O / kg
2
b g
b g
b g
−
= +
.
b gm m
w
Salt balance: 0 035 100 4 4
.
b g= x m
c. Y
x
w =
=
0 31
0 0398
4
.
.

4-20
4.31 a.
100
050
050
mol
molB / mol
molT / mol
.
.
2
097
0 03
1

.
.
n mol
molB / mol
molT / mol
b g

.
.
n1
097
0 03
mol
molB / mol
molT / mol
b g ( )
.
.
n1
097
0 03
mol 89.2%of Binfeed
molB / mol
molT / mol
b g

n
y
y
B
B
4
1
mol 45%of feed toreboiler
molB / mol
molT / mol
b gb g
b g
b g
−

n
z
z
B
B
2
1
mol
molB / mol
molT / mol
b g
b g
b g
−

n
x
x
B
B
3
1
mol
molB / mol
molT / mol
b g
b g
b g
−
Still
Condenser
Reboiler
Overall process: 3 unknowns ( , ,
n n xB
1 3 ) Still: 5 unknowns ( , , , ,
n n n y z
B B
1 2 4 )
– 2 balances – 2 balances
– 1 relationship (89.2% recovery) 3 DF
0 DF
Condenser: 1 unknown (
n1 ) Reboiler: 6 unknowns ( , , , , ,
n n n x y z
B B B
2 3 4 )
1 DF – 2 relationships (2.25 ratio 45% vapor)
3 DF
Begin with overall process.
b. Overall process
89.2% recovery: 0 892 0 50 100 0 97 1
. . .
b gb g= n
Overall balance: 100 1 3
= +

n n
B balance: 050 100 0 97 1 3
. .
b g= +
n x n
B
Reboiler
Composition relationship:
y y
x x
B B
B B
/
/
.
1
1
2 25
−
−
=
e j
b g
Percent vaporized: .
n n
4 2
0 45
= (1)
Mole balance:
n n n
2 3 4
= + (2)
(Solve (1) and (2) simultaneously.)
B balance: z n x n y n
B B B

2 3 4
= +

4-21
4.31 (cont’d)
c. B fraction in bottoms: xB = 0100
. molB / mol
Moles of overhead: .
n1 46 0
= mol Moles of bottoms: .
n3 54 0
= mol
Recovery of toluene:
1
050 100
100%
1 010 54 02
050 100
100% 97%
3
−
× =
−
× =
x n
B
b g
b g
b gb g
b g

.
. .
.
4.32 a.
100
012
088
kg
kg S / kg
kg H O / kg
2
.
.
m1
012
088
kg
kg S/ kg
kg H O / kg
2
b g
.
.
m4
058
0 42
kg
kg S/ kg
kg H O / kg
2
b g
.
.
m2
012
088
kg
kg S / kg
kg H O / kg
2
b g
.
.
m5
0 42
058
kg
kg S / kg
kg H O / kg
2
b g
.
.
m3 kg H O
2
b g
Mixing point
Bypass
Evaporator
Overall process: 2 unknowns (m m
3 5
, ) Bypass: 2 unknowns (m m
1 2
, )
– 2 balances – 1 independent balance
0 DF 1 DF
Evaporator: 3 unknowns (m m m
1 3 4
, , ) Mixing point: 3 unknowns (m m m
2 4 5
, , )
1 DF 1 DF
Overall S balance: 012 100 0 42 5
. .
b g= m
Overall mass balance: 100 3 5
= +
m m
Mixing point mass balance: m m m
4 2 5
+ = (1)
Mixing point S balance: 058 012 0 42
4 2 5
. . .
m m m
+ = (2)
Solve (1) and (2) simultaneously
Bypass mass balance: 100 1 2
= +
m m
b. m m m m m
1 2 3 4 5
90 05 9 95 714 18 65 28 6
= = = = =
. , . , . , . , .
kg kg kg kg kg product
Bypass fraction:
m2
100
0 095
= .
c. Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a
stream consisting of 90% solids could be hard to transport.
Basis: 100 kg

4-22
4.33 a.

.
.
m1
0 0515
09485
kg / h
kgCr / kg
kgW / kg
b g
.
.
m2
0 0515
09485
kg / h
kgCr / kg
kgW / kg
b g

.
.
m3
0 0515
09485
kg / h
kgCr / kg
kgW / kg
b g

m4 kgCr / h
b g

m
x
x
5
5
5
1
kg / h
kgCr / kg
kgW / kg
b g
b g
b g
−

m
x
x
6
6
6
1
kg / h
kgCr / kg
kgW / kg
b g
b g
b g
−
Treatment
Unit
b.
m m
1 2
6000 4500
= ⇒ =
kg / h kg / h maximumallowed value
b g
Bypass point mass balance:
m3 6000 4500 1500
= − = kg / h
95% Cr removal: . . .
m4 095 0 0515 4500 220 2
= =
b gb g kg Cr / h
Mass balance on treatment unit: . .
m5 4500 220 2 42798
= − = kg / h
Cr balance on treatment unit: x5
0 0515 4500 220 2
47798
0 002707
=
−
=
. .
.
.
b g kgCr / kg
Mixing point mass balance: . .
m6 1500 42798 57798
= + = kg / h
Mixing point Cr balance: x6
0 0515 1500 0 0002707 42798
57798
0 0154
=
+
=
. . .
.
.
b g b g kgCr / kg
c. m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h) x 5 m 6 (kg/h) x 6
1000 1000 0 48.9 951 0.00271 951 0.00271
2000 2000 0 97.9 1902 0.00271 1902 0.00271
3000 3000 0 147 2853 0.00271 2853 0.00271
4000 4000 0 196 3804 0.00271 3804 0.00271
5000 4500 500 220 4280 0.00271 4780 0.00781
6000 4500 1500 220 4280 0.00271 5780 0.0154
7000 4500 2500 220 4280 0.00271 6780 0.0207
8000 4500 3500 220 4280 0.00271 7780 0.0247
9000 4500 4500 220 4280 0.00271 8780 0.0277
10000 4500 5500 220 4280 0.00271 9780 0.0301

4-23
4.33 (cont’d)
m1 vs. x6
0.00000
0.00500
0.01000
0.01500
0.02000
0.02500
0.03000
0.03500
0 2000 4000 6000 8000 10000 12000
m1 (kg/h)
x
6
(kg
Cr/kg)
d. Cost of additional capacity – installation and maintenance, revenue from additional
recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon,
regulatory limits on Cr emissions.
4.34 a.
Evaporator

.
.
m1
0196
0804
kg / s
kg K SO / kg
kg H O / kg
2 4
2
b g

m
m
4
5
kgK SO / s
kgH O / s
2 4
2
b g
b g
Filtrate
kg / s
kg K SO / kg
kg H O / kg
2 4
2

.
.
m3
0 400
0 600
b g
175 kg H O / s 45% of water fed to evaporator
2 b g
Filtercake
kgK SO / s
kgsoln / s
kg K SO / kg
kg H O / kg
2 4
2 4
2
10
0 400
0 600
2
2

.
.
m
m
b g
b g
R
S
T
U
V
W

m
m
6
7
kgK SO / s
kgH O / s
2 4
2
b g
b g Crystallizer
Filter
Let K = K2SO4, W = H2 Basis: 175 kg W evaporated/s
Overall process: 2 unknowns ( ,
m m
1 2 ) Mixing point: 4 unknowns ( , , ,
m m m m
1 3 4 5 )
- 2 balances - 2 balances
0 DF 2 DF
Evaporator: 4 unknowns ( , , ,
m m m m
4 5 6 7 ) Crystallizer: 4 unknowns ( , , ,
m m m m
2 3 6 7 )
– 1 percent evaporation 2 DF
1 DF
Strategy: Overall balances
% evaporation
Balances around mixing point
Balances around evaporator
verify that each
chosen subsystem involves
no more than two
unknown variables
⇒
⇒
⇒
⇒
U
V
|
|
W
|
|
,

,
,
m m
m
m m
m m
1 2
5
3 4
6 7

4-24
4.34 (cont’d)
Overall K balance:

. .
m m m
m m m
1 2 2
1 2 2
175 10
0196 10 0 400
= + +
= +
U
V
|
W
|
Production rate of crystals = 10 2

m
45% evaporation: 175 0 450 5
kg evaporated min = .
m
W balance around mixing point: 0804 0 600
1 3 5
. .
m m m
+ =
Mass balance around mixing point:
m m m m
1 3 4 5
+ = +
K balance around evaporator:
m m
6 4
=
W balance around evaporator:
m m
5 7
175
= +
Mole fraction of K in stream entering evaporator =

m
m m
4
4 5
+
b. Fresh feed rate:
m1 221
= kg / s
Production rate of crystals kg K s s
= =
10 416
2
.
m b g
Recycle ratio:

.
.
.
m
m
3
1
3523
2208
160
kg recycle s
kg fresh feed s
kg recycle
kg fresh feed
b g
b g = =
c. Scale to 75% of capacity.
Flow rate of stream entering evaporator = . ( kg / s) = kg / s
. .
0 75 398 299
463% K, 537% W
d. Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and
the cooling cost for the crystallizer.

4-25
4.35 a. Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge
CO2 to the atmosphere.
Absorber function: Separates CO2 from CH4.
Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused.
b. The top streams are liquids while the bottom streams are gases. The liquids are heavier than
the gases so the liquids fall through the columns and the gases rise.
c.
100
0300
0 700
mol / h
molCO / mol
molCH / mol
2
4
.
.

.
.
n1
0 010
0990
mol / h
molCO / mol
molCH / mol
2
4
b g

.
.
n2
0 005
0995
mol / h
molCO / mol
molCH OH / mol
2
3
b g

n
n
3
4
molCO / h
molCH OH / h
2
3
b g
b g
/

n
n
5
6
mol N h
molCO / h
2
2
b g
b g
/
n5 mol N h
2
b g
Absorber Stripper
Overall: 3 unknowns ( , ,
n n n
1 5 6 ) Absorber: 4 unknowns ( , , ,
n n n n
1 2 3 4 )
1 DF 1 DF
Stripper: 4 unknowns ( , , ,
n n n n
2 3 4 5 )
– 2 balances
– 1 percent removal (90%)
1 DF
Overall CH4 balance: 0 700 100 0990 1
. .
b gb gb g
molCH / h
4 = n
Overall mole balance: 100 1 6
mol / h
b g= +

n n
Percent CO2 stripped: 090 3 6
.
n n
=
Stripper CO2 balance: .
n n n
3 6 2
0 005
= +
Stripper CH3OH balance: .
n n
4 2
0995
=
d. . , . , . , . ,
.
n n n n
n
1 2 3 4
6
70 71 6510 3255 647 7
29 29
= = = =
=
mol / h mol / h mol CO / h mol CH OH / h
mol CO / h
2 3
2
Fractional CO2 absorption: f
n
CO 2
2
. molCO absorbed / mol fed
=
−
=
30 0 0 010
30 0
0 976
1
. .
.

4-26
4.35 (cont’d)
Total molar flow rate of liquid feed to stripper and mole fraction of CO2:
,

.
n n x
n
n n
3 4 3
3
3 4
680 0 0478
+ = =
+
=
mol / h molCO / mol
2
e. Scale up to 1000 kg/h (=106
g/h) of product gas:
MW g CO / mol g CH / mol g / mol
g / h g / mol mol / h
mol / h mol / h) mol / h) mol / h
2 4
1
feed
1
6 4
4 4
001 44 099 16 1628
10 10 1628 6142 10
100 6142 10 7071 869 10
= + =
= × = ×
= × = ×
. . .
. . .
( . / ( . .
b g b g
b g d ib g
b g b g
n
n
new
new
f. T T
a s
The higher temperature in the stripper will help drive off the gas.
P P
a s
The higher pressure in the absorber will help dissolve the gas in the liquid.
g. The methanol must have a high solubility for CO2, a low solubility for CH4, and a low
volatility at the stripper temperature.
4.36 a. Basis: 100 kg beans fed
m
1
kg C
6
H
14
e j 300 kg C H
6 14
130
87 0
.
.
kg oil
kg S
m
x
y
x y
2
2
2
2 2
1
kg
kg S / kg
kg oil / kg
kg C H / kg
6 14
b g
b g
b g
b g
− −
m
5
kgC
6
H
14
e j
m
y
y
3
3
3
0 75
0 25
kg
kg S / kg
kg oil / kg
kg C H / kg
6 14
b g
b g
b g
.
. −
m
y
y
4
4
4
1
kg
kg oil / kg
kg C H / kg
6 14
b g
b g
b g
−
m6 kg oil
b g
Ex F Ev
Condenser
Overall: 4 unknowns (m m m y
1 3 6 3
, , , ) Extractor: 3 unknowns (m x y
2 2 2
, , )
1 DF 0 DF
Mixing Pt: 2 unknowns (m m
1 5
, ) Evaporator: 4 unknowns (m m m y
4 5 6 4
, , , )
– 1 balance – 2 balances
1 DF 2 DF
Filter: 7 unknowns (m m m x y y y
2 3 4 2 2 3 4
, , , , , , )
– 3 balances
– 1 oil/hexane ratio
3 DF
Start with extractor (0 degrees of freedom)
Extractor mass balance: 300 87 0 130 2
+ + =
. . kg m

4-27
4.36 (cont’d)
Extractor S balance: 87 0 2 2
. kg S = x m
Extractor oil balance: 130 2 2
. kg oil = y m
Filter S balance: 87 0 0 75 3
. .
kg S = m
Filter mass balance: m m m
2 3 4
kg
b g= + Oil / hexane ratio in filter cake:
y
y
y
x y
3
3
2
2 2
0 25 1
. −
=
− −
Filter oil balance: 130 3 3 4 4
. kg oil = +
y m y m
Evaporator hexane balance: 1 4 4 5
− =
y m m
b g
Mixing pt. Hexane balance: m m
1 5 300
+ = kg C H
6 14
Evaporator oil balance: y m m
4 4 6
=
b.
Yield
kg oil
kg beans fed
kg oil / kg beans fed
= = =
m6
100
118
100
0118
.
. b g
Fresh hexanefeed
kg C H
kg beans fed
kg C H kg beans fed
6 14
6 14
= = =
m1
100
28
100
0 28
. /
b g
Recycleratio
kg C H recycled
kg C H fed
kg C H recycled / kg C H fed
6 14
6 14
6 14 6 14
= = =
m
m
5
1
272
28
9 71
. b g
c. Lower heating cost for the evaporator and lower cooling cost for the condenser.
4.37
100
2
98
lbm
lb dirt
lb dry shirts
m
m
m
2
lb Whizzo
m
b g m
3
0 03
0 97
lb
lb dirt / lb
lb Whizzo / lb
m
m m
m m
b g
.
.
m
4
013
0 87
lb
lb dirt / lb
lb Whizzo / lb
m
m m
m m
b g
.
.
m
5
0 92
0 08
lb
lb dirt / lb
lb Whizzo / lb
m
m m
m m
b g
.
.
m
x
x
6
1
lb
lb dirt / lb
lb Whizzo/ lb
m
m m
m m
b g
b g
b g
−
m
1
98
3
lb dirt
lb dry shirts
lb Whizzo
m
m
m
b g
Tub
Filter
Strategy
95% dirt removal ⇒ m1 (= 5% of the dirt entering)
Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling
the chart) ⇒ m m
2 5
, (solves Part (a))

4-28
4.37 (cont’d)
Balances around the mixing point involve 3 unknowns m m x
3 6
, ,
b g, as do balances
around the filter m m x
4 6
, ,
b g, but the tub only involves 2 m m
3 4
,
b g and 2 balances are
allowed for each subsystem. Balances around tub ⇒ m m
3 4
,
Balances around mixing point ⇒ m x
6, (solves Part (b))
a. 95% dirt removal: m1 0 05 2 0 010
= =
. . .
b gb g lb dirt
m
Overall dirt balance: 2 0 010 092 2 065
5 5
. . . .
= + ⇒ =
b gm m lb dirt
m
Overall Whizzo balance: m2 3 0 08 2 065 317
= + =
. . .
b gb g b g
lb Whizzo lb Whizzo
m m
b. Tub dirt balance: 2 0 03 010 013
3 4
+ = +
. . .
m m (1)
Tub Whizzo balance: 097 3 087
3 4
. .
m m
= + (2)
Solve (1) (2) simultaneously ⇒ m m
3 4
20 4 193
= =
. , .
lb lb
m m
Mixing pt. mass balance: 317 20 4 173
6 6
. . .
+ = ⇒ =
m m
lb lb
m m
Mixing pt. Whizzo balance:
( ) ( )( ) m m
3.17 17.3 0.97 20.4 0.961 lb Whizzo/lb 96% Whizzo, 4% dirt
x x
+ = ⇒ = ⇒
4.38 a.
C
mixer 3
Filter 3
Discarded
3 kg L
L
C3 kg S
S
2720 kg S
F3 kg L
L
F3 kg S
S
C2 kg L
L
C2 kg S
S
mixer 2
Filter 2
mixer 1
Filter 1
C1 kg L
L
C1 kg S
S
3300 kg S
F2 kg L
L
F2 kg S
S
F1 kg L
L
F1 kg S
S
To holding tank
620 kg L
mixer filter 1 kg L
balance: kg L
mixer filter 2
balance:
mixer filter 3
kg L
kg L
kg L
balance: 613.7 = 6.1+ C kg L
3L
: . .
. .
:
:
. .
.
.
.
.
.
.
0 01 620 6 2
620 6 2 6138
0 01 6138
6138
0 01
6 2
6137
61
607 6
1 1
1 1
3 2
3 2 3
2 3
2
2
3
3
b g
b g
= ⇒ =
= + ⇒ =
+ =
+ = +
=
U
V
|
W
|
⇒
=
=
=
⇒ =
F F
C C
F F
F F C
C F
F
C
F
C
L L
L L
L L
L L L
L L
L
L
L
L

4-29
4.38 (cont’d)
Solvent
m f 1 kg S
balance: kg S
m f 2
balance:
m f 3
balance: 2720+ C
kg S
kg S
kg S
kg S
2S
: .
:
:
.
.
.
.
.
.
015 3300 495
3300 495 2805
015 495
495
015 2720
482 6
2734 6
480 4
2722 2
1 1
1 1
3 2
3 2 2
2 3
3 3
2
2
3
3
b g
b g
b g
= ⇒ =
= + ⇒ =
+ =
+ = +
+ =
= +
U
V
|
|
W
|
|
⇒
=
=
=
=
C C
F F
F C
F C F
C C
F C
C
F
C
F
S S
S S
S S
S S S
S S
S S
S
S
S
S
Holding Tank Contents
6 2 6 2 12 4
2805 2734 6 5540
. . .
.
+ =
+ =
kg leaf
kg solvent
b.
Extraction
Unit
Steam
Stripper
5540
0165
0835
kgS
kgE / kg
kgW / kg
.
.
Q
Q
D
F
kgD
kgF
b g
b g Q
Q
Q
E
D
F
kgE
kgD
kgF
b g
b g
b g
QR kg
kgE / kg
0.15kgF / kg
0.855kgW / kg
b g
013
.
Q0
0 200
kg
kgE / kg
0.026kgF / kg
0.774kgW / kg
b g
.
QB kg
kgE / kg
0.987kgW / kg
b g
0 013
.
Q3 kg steam
b g
Mass of D in Product:
1 kg D 620 kg leaf
1000 kg leaf
kg D
= =
0 62
. QD
Water balance around extraction unit: 0835 5540 0855 5410
. .
b g= ⇒ =
Q Q
R R kg
Ethanol balance around extraction unit:
0165 5540 013 5410 211
. .
b g b g b g
= + ⇒ =
Q Q
E E kg ethanol in extract
c. F balance around stripper
0 015 5410 0 026 3121
0 0
. .
b g b g
= ⇒ =
Q Q kg mass of stripper overhead product
E balance around stripper
013 5410 0 200 3121 0 013 6085
. . .
b g b g b g
= + ⇒ =
Q Q
B B kg mass of stripper bottom product
W balance around stripper
0855 5410 0 774 3121 0 987 6085 3796
. . .
b g b g b g
+ = + ⇒ =
Q Q
S S kg steam fed to stripper
4.39 a. C H 2 H C H
mol H react / mol C H react
kmol C H formed / kmol H react
2 2 2 2 6
2 2 2
2 6 2
+ →
2
05
.

4-30
4.39 (cont’d)
b. n
n
H
C H
2
2 2 2 2 2
2 2
2
2 2
H islimitingreactant
molH fed molC H fed molC H required (theoretical)
excess C H
mol fed mol required
mol required
= ⇒
⇒ ⇒
=
−
× =
15 2 0
15 10 0 75
10 0 75
0 75
100% 333%
. .
. . .
%
. .
.
.
c. 4 10
300 24 3600
1000 1
30 0
2
1
2 00
1
20 6
6
×
=
tonnes C H
yr
1 yr
days
1 day
h
1 h
s
kg
tonne
kmolC H
kgC H
kmolH
kmolC H
kgH
kmolH
kg H / s
2 6 2 6
2 6
2
2 6
2
2
2
.
.
.
d. The extra cost will be involved in separating the product from the excess reactant.
4.40 a. 4 5 4 6
4
125
NH O NO H O
5 lb - mole O react
lb - mole NO formed
lb - mole O react / lb - mole NO formed
3 2 2
2
2
+ → +
= .
b.
n
n
O theoretical
3 2
3
2
2 O fed 2 2
2
2
kmol NH
h
kmol O
kmol NH
kmol O
excess O kmol O kmol O
d i
d i b g
= =
⇒ = =
100 5
4
125
40% 140 125 175
.
c. 50 0 17 294
100 0 32 3125
3125
294
106
5
4
125
. / .
. / .
.
.
. .
kg NH 1 kmol NH kg NH kmol NH
kgO 1 kmolO kgO kmolO
O is the limiting reactant
3 3 3 3
2 2 2 2
O
NH fed
O
NH stoich
2
2
3
2
3
b gb g
b gb g
=
=
F
HG
I
KJ = =
F
HG
I
KJ = =
⇒
n
n
n
n
Required NH3:
3125 4
5
250
.
.
kmolO kmol NH
kmolO
kmol NH
2 3
2
3
=
%excess NH excess NH
3 3
=
−
× =
294 250
250
100% 17 6%
. .
.
.
Extent of reaction: n n v
O O O
2 2 2
kmol mol
= − ⇒ = − − ⇒ = =
d i b g
0
0 3125 5 0 625 625
ξ ξ ξ
. .
Mass of NO:
3125 4
5
30 0
1
750
. .
.
kmol O kmol NO
kmol O
kg NO
kmol NO
kg NO
2
2
=
4.41 a. By adding the feeds in stoichometric proportion, all of the H2S and SO2 would be consumed.
Automation provides for faster and more accurate response to fluctuations in the feed stream,
reducing the risk of release of H2S and SO2. It also may reduce labor costs.

4-31
4.41 (cont’d)
b.

. .
.
nc =
×
=
300 10 085 1
127 5
2
kmol
h
kmol H S
kmol
kmol SO
2 kmol H S
kmolSO / h
2 2
2
2
c.
C alibration C urve
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0.0 20.0 40.0 60.0 80.0 100.0
R a (m V )
X
(mol
H
2
S/mol)
X Ra
= −
0 0199 0 0605
. .
d.

n
x
f kmol / h
kmolH S / kmol
2
b g
b g

nc kmol SO / h
2
b g
Blender
Flowmeter calibration:

,

n aR
n R
n R
f f
f f
f f
=
= =
U
V
W
=
100 15
20
3
kmol / h mV
Control valve calibration:
. .
. , .

n
n R
n R
c
c c
c c
= =
= =
U
V
W
= +
250 10 0
60 0 250
7
3
5
3
kmol / h,R mV
kmol / h mV
c
Stoichiometric feed: . .
n n x R R R
c f c f a
= ⇒ + =
F
HG I
KJ −
1
2
7
3
5
3
1
2
20
3
0 0119 0 0605
b g
⇒ = − −
R R R
c f a
10
7
0 0119 0 0605
5
7
. .
b g
.
n R n
f f f
= × ⇒ = =
300 10
3
20
45
2
kmol / h mV

4-32
4.41 (cont’d)
R mV
kmol / h
c = − − =
⇒ = + =
10
7
45 0 0119 765 0 0605
5
7
539
7
3
539
5
3
127 4
b gb gb g
b g
. . . .
. .
nc
e. Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond
range of calibration data, system had not reached steady state yet.
4.42
165
1
mol / s
mol C H / mol
mol HBr / mol
2 4
x
x
b g
b g
−

.
.
n mol / s
mol C H / mol
mol HBr / mol
0.517 mol C H Br / mol
2 4
2 5
b g
0310
0173
C H HBr C H Br
2 4 2 5
+ →
C balance:
165 2
0310 2 0517 2
mol
s
molC H
mol
molC
molC H
2 4
2 4
x
n n
b g b gb g b gb g
= +
. . (1)
Br balance: ( )( ) ( )( ) ( )( )
165 1 1 0.173 1 0.517 1
x n n
− = +
(2)
(Note: An atomic H balance can be obtained as 2*(Eq. 2) + (Eq. 1) and so is not
independent)
. .
n x
108 77 0545
mol / s, molC H / mol
2 4
⇒ − =
1 0 455
x
b g . molHBr / mol
Since the C2H4/HBr feed ratio (0.545/0.455) is greater than the stoichiometric ration (=1),
HBr is the limiting reactant .
. .
nHBr fed
mol / s molHBr / mol molHBr
b g b gb g
= =
165 0 455 7508
( )( )
( )
( ) ( )( )
2 4
2 4
C H 2 4
stoich
C H 2 4 2 4
fed
2 4
75.08 0.173 108.8
Fractional conversion of HBr 0.749 molHBr react/molfed
75.08
75.08molC H
165mol/s 0.545molC H /mol 89.93molC H
89.93 75.08
% excess of C H 19.8%
75.08
Extentof reacti
n
n
−
= =
=
= =
−
= =

( ) ( )( ) ( )
2 5 2 5 2 5
C H Br C H Br C H Br
0
on: 108.8 0.517 0 1 56.2mol/s
n n v ξ ξ ξ
= + ⇒ = + ⇒ =

4-33
4.43 a.
2HCl
1
2
O Cl H O
2 2 2
+ → + Basis: 100 mol HCl fed to reactor
100 mol HCl
n1
0 21
0 79
mol air
mol O / mol
mol N / mol
35% excess
2
2
b g
.
.
n
n
n
n
n
2
3
4
5
6
molHCl
molO
mol N
mol Cl
molH O
2
2
2
2
b g
b g
b g
b g
b g
O stoic
100 mol HCl 0.5 mol O
2 mol HCl
mol O
2
2
2
b g = = 25
35% excess air: 0 21 135 25 160 7
1 1
. . .
n n
mol O fed mol air fed
2
b g= × ⇒ =
85% conversion ⇒ ⇒ =
85 15
2
mol HCl react mol HCl
n
n5 425
= =
85 mol HCl react 1 mol Cl
2 mol HCl
mol Cl
2
2
.
n6 85 1 2 425
= =
b gb g . mol H O
2
N balance:
2 160 7 0 79 127
4 4
. .
b gb g= ⇒ =
n n mol N2
O balance:
160.7 mol O 2 mol O
1 mol O
42.5 mol H O 1 mol O
1 mol H O
mol O
2
2
2
2
2
b gb g
0 21
2 125
3 3
.
.
= + ⇒ =
n n
Total moles:
nj
j=
∑ = ⇒ =
2
5
239 5 0 063 0 052 0530
0177 0177
. . , . , . ,
. , .
mol
15 mol HCl
239.5 mol
mol HCl
mol
molO
mol
mol N
mol
molCl
mol
mol H O
mol
2 2
2 2
b. As before, n n
1 2
160 7 15
= =
. ,
mol air fed molHCl
2HCl
1
2
O Cl H O
2 2 2
+ → +
n n v
i i i
= +
E
= − ⇒ =
b g0
15 100 2 42 5
ξ
ξ ξ
HCl mol
: .

4-34
4.43 (cont’d)
O mol O
N mol N
Cl mol Cl
H O mol H O
2 2
2 2
2 2
2 2
: . . .
: . .
: .
: .
n
n
n
n
3
4
5
6
0 21 160 7
1
2
12 5
0 79 160 7 127
42 5
42 5
= − =
= =
= =
= =
b g
b g
ξ
ξ
ξ
These molar quantities are the same as in part (a), so the mole fractions would also be the
same.
c. Use of pure O2 would eliminate the need for an extra process to remove the N2 from the
product gas, but O2 costs much more than air. The cheaper process will be the process of
choice.
4.44 FeTiO 2H SO TiO SO FeSO 2H O
Fe O 3H SO Fe SO H O
TiO SO 2H O H TiO s H SO
H TiO s TiO s H O
3 2 4 4 4 2
2 3 2 4 2 4 3 2
4 2 2 3 2 4
2 3 2 2
+ → + +
+ → +
+ → +
→ +
b g
b g
b g b g
b g b g
3
Basis: 1000 kg TiO2 produced
1000 kg TiO kmol TiO 1 kmol FeTiO
79.90 kg TiO 1 kmol TiO
kmol FeTiO decomposes
2 2 3
2 2
3
= 12 52
.
12.52 kmol FeTiO dec. 1 kmol FeTiO feed
0.89 kmol FeTiO dec.
kmol FeTiO fed
3 3
3
3
=14 06
.
14.06 kmol FeTiO 1 kmol Ti 47.90 kg Ti
1 kmol FeTiO kmol Ti
kg Ti fed
3
3
= 6735
.
673 5 0 243 2772
. .
kg Ti / kg ore kg ore fed
M M
b g= ⇒ =
Ore is made up entirely of 14.06 kmol FeTiO3 + n kmol Fe O
2 3
b g (Assumption!)
n = − =
2772 6381
kg ore
14.06 kmol FeTiO 151.74 kg FeTiO
kmol FeTiO
kg Fe O
3 3
3
2 3
.
638.1 kg Fe O kmol Fe O
159.69 kg Fe O
kmol Fe O
2 3 2 3
2 3
2 3
= 4 00
.
14.06 kmol FeTiO 2 kmol H SO
1 kmol FeTiO
4.00 kmol FeTiO 3 kmol H SO
1 kmol Fe O
kmol H SO
3 2 4
3
3 2 4
2 3
2 4
+ = 4012
.
50% excess: 15 4012 6018
. . .
kmol H SO kmol H SO fed
2 4 2 4
b g=
Mass of 80% solution:
60.18 kmol H SO 98.08 kg H SO
1 kmol H SO
kg H SO
2 4 2 4
2 4
2 4
= 5902 4
.
5902 4 080 7380
. / .
kg H SO kg soln kg 80% H SO feed
2 4 2 4
M
a
M
a
b g= ⇒ =

4-35
4.45 a. Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line through
R C
1 1
10 0 30
= =
, . g m3
d i and R C
2 2
48 2 67
= =
FH IK
, . g m3
ln ln
ln . .
. ln ln . . . .
.78
C bR a C ae
b a a e
br
= + ⇔ =
=
−
= = − = − ⇒ = =
−
2 67 0 30
48 10
0 0575 2 67 0 0575 48 178 0169
1
b g b g b g
,
⇒ =
=
′
= ′
E
= ⇒ ′ = × −
g m
lb g 35.31 ft
ft 1 lb m
' lb SO ft
3 m
3
3
m
3
m 2
3
C e
C
C
C
C e C e
R
R R
0169
4536
1
16 020
16 020 0169 1055 10
0.0575
0.0575 5 0.0575
.
( ) .
,
, . .
d i
d i
b. 2867 60
1250
138
ft s s min
lb min
ft lb coal
3
m
3
m
d ib g =
R C e
= ⇒ ′ = × − = ×
−
37 1055 10 5 886 10
0.0575 37 5
lb SO ft lb SO ft
m 2
3
m 2
3
d i b gb g
. .
886 10
1
0 012 0 018
5
.
. .
×
=
−
lb SO 138 ft
ft lb coal
lb SO
lb coal
compliance achieved
m 2
3
3
m
m 2
m
c. S O SO
2 2
+ →
1250 lb
1
1249
m m m 2
m m
m 2
coal 0.05 lb S 64.06 lb SO
min lb coal 32.06 lb S
lb SO generated min
= .
2867 ft s lb SO
s min ft
lb SO in scrubbed gas
3
m 2
3 m 2
60 886 10
1
152
5
.
. min
×
=
−
furnace
ash
air
1250 lb coal/min
m
62.5 lb S/min
m
stack gas
124.9 lb SO /min
m 2
scrubber
liquid effluent
scrubbing fluid
(124.9 – 15.2)lb SO (absorbed)/min
m 2
scrubbed gas
15.2 lb SO /min
m 2
%
. .
.
removal
lb SO scrubbed min
lb SO fed to scrubber min
m 2
m 2
=
−
× =
1249 152
1249
100% 88%
b g
d. The regulation was avoided by diluting the stack gas with fresh air before it exited from the
stack. The new regulation prevents this since the mass of SO2 emitted per mass of coal
burned is independent of the flow rate of air in the stack.

4-36
4.46 a. A B ===== C + D
Total
+
= −
= − = −
= + = −
= + =
= = +
= ∑
+
n n
n n y n n
n n y n n
n n y n n
n n y n n
n n
A A
B B A A T
C C B B T
D D C C T
I I D D T
T i
0
0 0
0 0
0 0
0 0
ξ
ξ ξ
ξ ξ
ξ ξ
ξ
e j
e j
e j
e j
At equilibrium:
y y
y y
n n
n n
C D
A B
C c D c
A c B c
=
+ +
− −
=
0 0
0 0
487
ξ ξ
ξ ξ
b gb g
b gb g . (nT ’s cancel)
387 487 487 0
0
2
0 0 0 0 0 0 0 0
2
. . .
[ ]
ξ ξ
ξ ξ
c C D A B c C D A B
c c
n n n n n n n n
a b c
− + + + − − =
+ + =
b g
c h b g
∴ = − ± −
=
= − + + +
= − −
ξc C D A B
C D A B
a
b b ac
a
b n n n n
c n n n n
1
2
4
387
487
487
2
0 0 0 0
0 0 0 0
e j b g
where
.
.
.
b. Basis: 1 mol A feed nA0 1
= nB0 1
= 0 0 0 0
C D I
n n n
= = =
Constants: 3.87 9.74 4.87
a b c
= = − =
( )
( ) ( )( )
( )
( )
2
1
2
1
9.74 9.74 4 3.87 4.87 0.688
2 3.87
1.83 is also a solution but leads to a negative conversion
e e
e
ξ ξ
ξ
= ± − ⇒ =
=
Fractional conversion: ( ) 0 1
0 0
0.688
A A e
A B
A A
n n
X X
n n
ξ
−
= = = =
c. 0 0 0 0
80, 0
B C D J
n n n n
= = = =
n n
n
n n n
n n
n n
n n
y y
y y
n n
n n n
n
C C c
C
c
A A c A
B B c
C C c
D D c
C D
A B
C D
A B A
A
= = + =======
=
=
= − = −
= − = − =
= + =
= + =
= = ⇒
−
= ⇒ =
70
0
70
70
80 70 10
70
70
487
70 70
70 10
487 1706
0
0
0 0
0
0
0
0
0
ξ ξ
ξ
ξ
ξ
ξ
mol
mol
mol
mol
mol
mol methanol fed
. . .
b gb g
b gb g

4-37
4.46 (cont’d)
Product gas mol
mol
mol
mol
mol CH OH mol
mol CH COOH mol
mol CH COOCH mol
mol H O mol
mol
3
3
3 3
2
n
n
n
n
y
y
y
y
n
A
B
C
D
A
B
C
D
total
= − =
=
=
=
U
V
|
|
W
|
|
⇒
=
=
=
=
=
170 6 70 100 6
10
70
70
0 401
0 040
0 279
0 279
250 6
. . .
.
.
.
.
d. Cost of reactants, selling price for product, market for product, rate of reaction, need for
heating or cooling, and many other items.
4.47 a. CO
(A)
H O
(B)
CO
(C)
H
(D)
2 2 2
+ ← →
⎯ +
100
0 20
010
0 40
030
.
.
.
.
.
mol
mol CO / mol
mol CO / mol
mol H O / mol
mol I / mol
2
2
n
n
n
n
n
A
B
C
D
I
mol CO
mol H O
mol CO
mol H
mol I
2
2
2
b g
b g
b g
b g
b g
Degree of freedom analysis: 6 unknowns ( n n n n n
A B C D I
, , , , ,ξ )
– 4 expressions for ni ξ
b g
– 1 balance on I
– 1 equilibrium relationship
0 DF
b. Since two moles are prodcued for every two moles that react,
n n
total out total in
mol
b g b g b g
= =100
.
nA = −
0 20
. ξ (1)
nB = −
0 40
. ξ (2)
nC = +
010
. ξ (3)
nD = ξ (4)
nI = 030
. (5)
ntot =100
. mol
At equilibrium:
y y
y y
n n
n n
C D
A B
C D
A B
= =
+
− −
=
F
HG I
KJ ⇒ =
010
0 20 0 40
0 0247
4020
1123
0110
.
. .
. exp .
ξ ξ
ξ ξ
ξ
b gb g
b gb g mol
y n
D D
= = =
ξ 0110
. molH / mol
2
b g
c. The reaction has not reached equilibrium yet.

4-38
4.47 (cont’d)
d. T (K) x (CO) x (H2O) x (CO2) Keq Keq (Goal Seek) Extent of Reaction y (H2)
1223 0.5 0.5 0 0.6610 0.6610 0.2242 0.224
1123 0.5 0.5 0 0.8858 0.8856 0.2424 0.242
1023 0.5 0.5 0 1.2569 1.2569 0.2643 0.264
923 0.5 0.5 0 1.9240 1.9242 0.2905 0.291
823 0.5 0.5 0 3.2662 3.2661 0.3219 0.322
723 0.5 0.5 0 6.4187 6.4188 0.3585 0.358
623 0.5 0.5 0 15.6692 15.6692 0.3992 0.399
673 0.5 0.5 0 9.7017 9.7011 0.3785 0.378
698 0.5 0.5 0 7.8331 7.8331 0.3684 0.368
688 0.5 0.5 0 8.5171 8.5177 0.3724 0.372
1123 0.2 0.4 0.1 0.8858 0.8863 0.1101 0.110
1123 0.4 0.2 0.1 0.8858 0.8857 0.1100 0.110
1123 0.3 0.3 0 0.8858 0.8856 0.1454 0.145
1123 0.5 0.4 0 0.8858 0.8867 0.2156 0.216
The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of
carbon monoxide and water also maximizes the extent of reaction.
4.48 a. A 2B C
+ →
ln ln
K A E T K
e = +
0 b g
E
K K
T T
e e
=
−
=
×
−
=
−
ln / ln . / .
1 2
1 2
4
1 1
105 2316 10
1 373 1 573
11458
b g d i
ln ln ln . . .
A K T A
e
0 1 1 0
13
11458 105 11458 373 2837 4 79 10
= − = − = − ⇒ = × −
K T K K K
e e
= × ⇒ =
− − −
4 79 10 11458 450 00548
13 2 1
. exp ( ) .
b g
c hatm atm
b. n n
n n
n n
n n
y n n
y n n
y n n
n n n n
A A
B B
C C
T T
A A T
B B T
C C T
T A B C
= −
= −
= +
= −
U
V
|
|
W
|
|
⇒
= − −
= − −
= + −
= + +
0
0
0
0
0 0
0 0
0 0
0 0 0 0
2
2
2
2 2
2
ξ
ξ
ξ
ξ
ξ ξ
ξ ξ
ξ ξ
b g b g
b g b g
b g b g
b g
At equilibrium,
y
y y P
n n
n n P
K T
C
A B
C e T e
A e B e
e
2 2
0 0
2
0 0
2 2
1 2
2
1
=
+ −
− −
=
ξ ξ
ξ ξ
b gb g
b gb g b g (substitute for K
e
T
b g from Part a.)
c. Basis: 1 mol A (CO)
n n n n
A B C T
0 0 0 0
1 1 0 2
= = = ⇒ = , P = 2 atm , T = 423K
ξ ξ
ξ ξ
e e
e e
e
K
2 2
1 1 2
1
4
423 0 278
2
2
−
− −
= =
b g
b gb g b g
atm
atm
2
-2
. ⇒ ξ ξ
e e
2
01317 0
− + =
.

4-39
4.48 (cont’d)
(For this particular set of initial conditions, we get a quadratic equation. In general, the
equation will be cubic.)
ξe = 0156
. , 0.844 Reject the second solution, since it leads to a negative nB.
y y
y y
y y
A A
B B
C C
= − − ⇒ =
= − − ⇒ =
= + − ⇒ =
1 0156 2 2 0156 0500
1 2 0156 2 2 0156 0 408
0 0156 2 2 0156 0 092
. . .
. . .
. . .
b g b g
c h
b g
c h b g
c h
b g b g
c h
Fractional Conversion of CO A
n n
n n
A A
A A
A A
b g=
−
= =
0
0 0
0156
ξ
. mol reacted / mol feed
d. Use the equations from part b.
i) Fractional conversion decreases with increasing fraction of CO.
ii) Fractional conversion decreases with increasing fraction of CH3OH.
iii) Fractional conversion decreases with increasing temperature.
iv) Fractional conversion increases with increasing pressure.
*
REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI,
FN, FDN, NT, CON, YA, YB, YC
INTEGER NIT, INMAX
TAU = 0.0001
INMAX = 10
A = 4.79E–13
E = 11458.
READ (5, *) YA0, YB0, YC0, T, P
KE = A * EXP(E/T)
P2KE = P*P*KE
C0 = YC0 – P2KE * YA0 * YB0 * YB0
C1 = 1. – 4. * YC0 + P2KE * YB0 * (YB0 + 4. * YA0)
C2 = 4. * (YC0 –1. – P2KE * (YA0 + YB0))
C3 = 4. * (1. + P2KE)
EK = 0.0 (Assume an initial value ξe = 0 0
. )
NIT = 0
1 FN = C0 + EK * (C1 + EK * (C2 + EK * C3)) FDN = C1 + EK * (2. * C2 +
EK * 3. * C3) EKPI = EK - FN/FDN NIT = NIT + 1 IF (NIT.EQ.INMAX)
GOTO 4 IF (ABS((EKPI – EK)/EKPI).LT.TAU) GOTO 2 EK =
EKPI GOTO 1
2 NT = 1. – 2. * EKPI
YA = (YA0 – EKPI)/NT
YB = (YB0 – 2. + EKPI)/NT
YC = (YC0 + EKPI)/NT

4-40
4.48 (cont’d)
CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP
4
3
WRITE (6, 5) INMAX, EKPI
FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X)) FORMAT ('DID NOT
CONVERGE IN', I3, 'ITERATIONS',/,
* 'CURRENT VALUE = ', F6.3) END
$ DATA 0.5 0.5 0.0 423. 2.
RESULTS: YA = 0.500, YB = 0.408, YC = 0.092, CON = 0.156
Note: This will only find one root — there are two others that can only be found by
choosing different initial values of ξa
4.49 a.
CH O HCHO H O
4 2 2
+ ⎯ →
⎯ + (1)
CH 2O CO 2H O
4 2 2 2
+ ⎯ →
⎯ + (2)
100
0 50
0 50
mol / s
mol CH / mol
mol O / mol
4
2
.
.

n
n
n
n
n
1
2
3
4
5
mol CH / s
mol O / s
mol HCHO / s
mol H O / s
mol CO
4
2
2
2
b g
b g
b g
b g
b g
7 unknowns ( , , , , , ,
n n n n n
1 2 3 4 5 1 2
ξ ξ )
–5 1 2
equations for ,
ni ξ ξ
e j
2 DF
b.
n1 1 2
50
= − −
ξ ξ (1)

n2 1 2
50 2
= − −
ξ ξ (2)

n3 1
= ξ (3)

n4 1 2
2
= +
ξ ξ (4)

n5 2
= ξ (5)
c.
Fractional conversion:
50
50
0 900 5 00
1
1
−
= ⇒ =

. .
n
n
b g mol CH
4
/ s
Fractional yield:
. .
n
n
3
3
50
0 855 42 75
= ⇒ = mol HCHO / s
Equation 3
Equation 1
Equation 2
Equation 4
Equation 5
molCH / mol
molO / mol
mol HCHO / mol
mol H O / mol
molCO / mol
CH4 4
O2 2
HCHO
H2O 2
CO2 2
⇒ =
⇒ =
⇒ =
⇒ =
⇒ =
U
V
|
|
|
W
|
|
|
⇒
=
=
=
=
=
ξ
ξ
1
2
2
4
5
42 75
2 25
2 75
47 25
2 25
0 0500
0 0275
0 4275
0 4725
0 0225
.
.
.
.
.
.
.
.
.
.
n
n
n
y
y
y
y
y
Selectivity: 2 2
[(42.75molHCHO/s)/(2.25molCO /s) 19.0molHCHO/molCO
=
5 2
(mol CO / s)
n

4-41
4.50 a. Design for low conversion and feed ethane in excess. Low conversion and excess ethane
make the second reaction unlikely.
b. C2H6 + Cl2 → C2H5Cl + HCl, C2H5Cl + Cl2 → C2H4Cl2 + HCl
Basis: 100 mol C2H5Cl produced
n1 (mol C2H6) 100 mol C2H5Cl 5 unknowns
n2 (mol Cl2) n3 (mol C2H6) –3 atomic balances
n4 (mol HCl) 2 D.F.
n5 (mol C2H5Cl2)
c. Selectivity: 100 14 5
mol C H Cl (mol C H Cl
2 5 2 4 2
= n ) ⇒ =
n5 7143
. mol C H Cl
2 4 2
15% conversion:
C balance:
1 015
2 2 100 2 2 7143
1 3
1 3
− =
= + +
U
V
|
W
|
.
.
b g
b g b g
n n
n n
⇒
=
=
n
n
1
3
714 3
114 3
.
.
mol C H in
mol C H out
2 6
2 6
H balance: 6 714 3 5 100 6 114 3 4 7143
4
. . .
b g b g b g b g
= + + +
n ⇒ =
n4 6071
. mol HCl
Cl balance: 2 100 6071 2 7143
2
n = + +
. .
b g ⇒ =
n2 114 3
. mol Cl2
Feed Ratio: 1143 016
6 6
. / . /
mol Cl 714.3 mol C H mol Cl mol C H
2 2 2 2
=
Maximum possible amount of C2H5Cl:
nmax = =
114.3 mol Cl 1 mol C H Cl
1 mol Cl
114.3 mol C H Cl
2 2 5
2
2 5
Fractional yield of C2H5Cl:
n
n
C H Cl
2 5 100
114 3
0875
max .
.
= =
mol
mol
d. Some of the C2H4Cl2 is further chlorinated in an undesired side reaction:
C2H5Cl2 + Cl2 → C2H4Cl3 + HCl
4.51 a. C2H4 + H2O → C2H5OH, 2 C2H5OH → (C2H5)2O + H2O
Basis: 100 mol effluent gas
n
n
1
2
(mol C2H4 )
[mol H2O (v)]
n3 (mol I)
100 mol
0.433 mol C2H4 / mol
0.025 mol C2H5OH / mol
0.0014 mol (C
2
H
5
)
2
O / mol
0.093 mol I / mol
0.4476 mol H2O (v) / mol
3 unknowns
-2 independent atomic balances
-1 I balance
0 D.F.
(1) C balance: 2 100 2 0 433 2 0 025 4 0 0014
1
n = ∗ + ∗ + ∗
. . .
b g
(2) H balance: 4 2 100 4 0 433 6 0 025 10 0 0014 2 0 4476
1 2
n n
+ = ∗ + ∗ + ∗ + ∗
. . . .
b g
(3) O balance: n2 100 0 025 0 0014 0 4476
= + +
. . .
b g
Note; Eq. ( ) ( ) ( )
Eq. Eq.
1 2 3 2 2
∗ + ∗ = ⇒2 independent atomic balances
(4) I balance: n3 = 9.3

4-42
4.51 (cont'd)
b.
(1) 46.08 mol C H
(3) 47.4 mol H O
(4) 9.3 mol I
Reactor feed contains 44.8% C H , 46.1% H O, 9.1% I
1 2 6
2 2
3
2 6 2
⇒ =
⇒ =
⇒ =
U
V
|
W
|
⇒
n
n
n
% conversion of C2H4:
46 08 433
46 08
100% 6 0%
. .
.
.
−
× =
If all C2H4 were converted and the second reaction did not occur, nC H OH
2 5
46 08
d imax
.
= mol
⇒ Fractional Yield of C2H5OH: n n
C H OH C H OH
2 5 2 5
2 5 46 08 0 054
/ . / . .
max
d i b g
= =
Selectivity of C2H5OH to (C2H5)2O:
2.5 mol C H OH
0.14 mol (C H ) O
17.9 mol C H OH / mol (C H ) O
2 5
2 5 2
2 5 2 5 2
=
c. Keep conversion low to prevent C2H5OH from being in reactor long enough to form
significant amounts of (C2H5)2O. Separate and recycle unreacted C2H4.
4.52 CaF s H SO l CaSO s 2HF g
2 2 4 4
b g b g b g b g
+ → +
1 metric ton acid 1000 kg acid 0.60 kg HF
1 metric ton acid 1 kg acid
kg HF
= 600
Basis: 100 kg Ore dissolved (not fed)
(kg H O)
n1
(kg HF)
n2
(kg H SO )
2
n4
(kg CaSO )
n5
4
n3 (kg H SiF )
4 6
100 kg Ore dissolved
0.96 kg CaF /kg
2
0.04 kg SiO /kg
2
n
0.93 H SO kg/kg
4
0.07 H O kg/kg
2
A (kg 93% H SO )
2 4
2
2 4
Atomic balance - Si:
( ) 3 2 6
2
3 2 6
2 6
2
(kg H SiF ) 28.1 kg Si
0.04 100 kg SiO 28.1 kg Si
9.59 kg H SiF
144.1 kg H SiF
60.1 kg SiO
n
n
= ⇒ =
Atomic balance - F:
( ) 2
2
2
2 6
2
2 6
(kg HF) 19.0 kg F
0.96 100 kg CaF 38.0 kg F
20.0 kg HF
78.1 kg CaF
9.59 kg H SiF 114.0 kg F
41.2 kg HF
144.1 kg H SiF
n
n
=
+ ⇒ =
600
1533 kg or
kg HF 100 kg ore diss. 1 kg ore feed
41.2 kg HF 0.95 kg ore diss.
e
=
n1 (kg CaSO4)
n2 (kg HF)
n3 (kg H2SiF6)
n4 (kg H2SO4)
n5 (kg H2O)

4-43
4.53 a. C H Cl C H Cl HCl
C H Cl Cl C H Cl HCl
C H Cl Cl C H Cl HCl
6 6 2 6 5
6 5 2 6 4 2
6 4 2 2 6 3 3
+ → +
+ → +
+ → +
Convert output wt% to mol%: Basis 100 g output
species g Mol. Wt. mol mol %
C H
6 6
65.0 78.11 0.832 73.2
C H Cl
6 5
32.0 112.56 0.284 25.0
C H Cl
6 4 2
2.5 147.01 0.017 1.5
C H Cl
6 3 3
0.5 181.46 0.003 0.3
total 1.136
Basis: 100 mol output
n1
(mol HCl(g))
n4
(mol C H )
6 n3 (mol I)
6
n2 (mol Cl )
2
n3 (mol I)
32.0
65.0 mol C H
6 6
mol C H Cl
6 5
0.5
2.5 mol C H Cl
6 4
mol C H Cl
6 3
2
3
4 unknowns
-3 atomic balances
-1 wt% Cl2 in feed
0 D.F.
b. C balance: ( )
1
6 6 73.2 25.0 1.5 0.3
n = + + + ⇒ =
n1 100 mol C H
6 6
H balance: ( ) ( ) ( ) ( ) ( ) 4
6 100 6 73.2 5 25.0 4 1.5 3 0.3 n
= + + + + 4 28.9 mol HCl
n
⇒ =
Cl balance: ( ) ( )
2
2 28.9 25.0 2 1.5 3 0.3
n = + + + 2 2
28.9 mol Cl
n
⇒ =
Theoretical C H
6 6 ( )
2 6 6 2 6 6
28.9 mol Cl 1 mol C H 1 mol Cl 28.9 mol C H
= =
Excess C H
6 6: ( ) 6 6
100 28.9 28.9 100% 246% excess C H
− × =
Fractional Conversion: ( ) 6 6
100 73.2 100 0.268 mol C H react/mol fed
− =
Yield: 6 5 6 5
(25.0 mol C H Cl) (28.9 mol C H Cl maximum)=0.865
( )
2 2
2 2
6 6
6 6
6 6
28.9 mol Cl 70.91 g Cl 1 g gas
Gas feed: 2091 g gas
mole Cl 0.98 g Cl g gas
0.268
g liquid
78.11 g C H
Liquid feed: 100 mol C H 7811 g liquid
mol C H
⎫
= ⎪
⎪
⇒
⎬
⎛ ⎞ ⎪
=
⎜ ⎟ ⎪
⎝ ⎠ ⎭
c. Low conversion ⇒ low residence time in reactor ⇒ lower chance of 2nd and 3rd reactions
occurring. Large excess of C H Cl
6 6 2
⇒ much more likely to encounter C H
6 6
than substituted C H
6 6 ⇒ higher selectivity.
d. Dissolve in water to produce hydrochloric acid.
e. Reagent grade costs much more. Use only if impurities in technical grade mixture affect the
reaction rate or desired product yield.
73.2 mol C6H6
25.0 mol C6H5Cl
1.5 mol C6H4Cl2
0.3 mol C6H3Cl3

4-44
4.54 a. 2CO 2CO O 2A 2B C
O N 2NO C D 2E
2 2
2 2
⇔ + ⇔ +
+ ⇔ + ⇔
n n
n n
n n
n n
n n
y n n
y n n
y n n
y n n
y n n
A A e
B B e
C C e e
D D e
E E e
A A e T e
B B e T e
C C e e T e
D D e T e
E E e T e
= −
= +
= + −
= −
= +
= − +
= + +
⇒ = + − +
= − +
= + +
0 1
0 2
0 1 2
0 2
0 2
0 1 0 1
0 1 0 1
0 1 2 0 1
0 2 0 1
0 2 0 1
2
2
2
2
2
1
2
ξ
ξ
ξ ξ
ξ
ξ
ξ ξ
ξ ξ
ξ ξ ξ
ξ ξ
ξ ξ
b g b g
b g b g
b g b g
b g b g
b g b g
ntotal = n n n n n n n
T e T A B C D E
0 1 0 0 0 0 0 0
+ = + + + +
ξ b g
Equilibrium at 3000K and 1 atm
y y
y
n n
n n
B C
A
B e C e e
A e T e
2
2
0 1
2
0 1 2
0 1
2
0 1
2
2
01071
=
+ + −
− +
=
ξ ξ ξ
ξ ξ
b g b g
b g b g
.
( )
( )( )
2
2
0 2
0 1 2 0 2
2
0.01493
E e
E
C D A e e D e
n
y
y y n n
ξ
ξ ξ ξ
+
= =
+ − −
E
= − + − + + − =
= + − − − + =
U
V
|
W
|
f n n n n
f n n n
f
f
A e T e B e C e e
C e e D e E e
1 0 1
2
0 1 0 1
2
0 1 2
2 0 1 2 0 2 0 2
2 1 1 2
2 1 2
01071 2 2 0
0 01493 2 0
.
.
,
,
ξ ξ ξ ξ ξ
ξ ξ ξ ξ
ξ ξ
ξ ξ
b g b g b g b g
b gb g b g b g
b g
Defines functions
and
b. Given all nio’s, solve above equations for ξe1 and ξe2 ⇒ nA, nB, nC, nD, nE ⇒ yA, yB, yC, yD, yE
c. nA0 = nC0 = nD0 = 0.333, nB0 = nE0 = 0 ⇒ ξe1 =0.0593, ξe2 = 0.0208
⇒ yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393
d. a d a d f a d a d f
d
a f a f
a a a a
d
a f a f
a a a a
d d
e e e e
11 1 12 2 1 21 1 22 2 2
1
12 2 22 1
11 22 12 21
2
21 1 11 2
11 22 12 21
1 1 1 2 1 2
+ = − + = −
=
−
−
=
−
−
= + = +
ξ ξ ξ ξ
b g b g
new new
(Solution given following program listing.)
. IMPLICIT REAL * 4(N)
WRITE (6, 1)
1 FORMAT('1', 30X, 'SOLUTION TO PROBLEM 4.57'///)
30 READ (5, *) NA0, NB0, NC0, ND0, NE0
IF (NA0.LT.0.0)STOP
WRITE (6, 2) NA0, NB0, NC0, ND0, NE0

4-45
2 FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/)
NTO = NA0 + NB0 + NC0 + ND0 + NE0
NMAX = 10
X1 = 0.1
X2 = 0.1
DO 100 J = 1, NMAX
NA = NA0 – X1 – X1
NB = NB0 + X1 + X1
NC = NC0 + X1 – X2
ND = ND0 – X2
NE = NE0 + X2 + X2
NAS = NA ** 2
NBS = NB ** 2
NES = NE ** 2
NT = NT0 + X1
F1 = 0.1071 * NAS * NT – NBS * NC
F2 = 0.01493 * NC * ND – NES
A11 = –0.4284 * NA * NT * 0.1071 * NAS – 4.0 * NB * NC – NBS
A12 = NBS
A21 = 0.01493 * ND
A22 = –0.01493 * (NC + ND) – 4.0 * NE
DEN = A11 * A22 – A12 * A21
D1 = (A12 * F2 – A22 * F1)/DEN
D2 = (A21 * F1 – A11 * F2)/DEN
X1C = X1 + D1
X2C = X2 + D2
WRITE (6, 3) J, X1, X2, X1C, X2C
3 FORMAT(20X, 'ITER *', I3, 3X, 'X1A, X2A =', 2F10.5, 6X, 'X1C, X2C =', * 2F10.5)
IF (ABS(D1/X1C).LT.1.0E–5.AND.ABS(D2/X2C).LT.1.0E–5) GOTO 120
X1 = X1C
X2 = X2C
100 CONTINUE
WRITE (6, 4) NMAX
4
120
5
FORMAT('0', 10X, 'PROGRAM DID NOT CONVERGE IN', I2, 'ITERATIONS'/)
STOP
YA = NA/NT
YB = NB/NT
YC = NC/NT
YD = ND/NT
YE = NE/NT
WRITE (6, 5) YA, YB, YC, YD, YE
FORMAT ('0', 15X, 'YA, YB, YC, YD, YE =', 1P5E14.4///)
GOTO 30
END
$DATA
0.3333 0.00 0.3333 0.3333 0.0
0.50 0.0 0.0 0.50 0.0
0.20 0.20 0.20 0.20 0.20
SOLUTION TO PROBLEM 4.54
NA0, NB0, NC0, ND0, NE0 = 0.33 0.00 0.33 0.33 0.00
ITER = 1 X1A, X2A = 0.10000 0.10000 X1C, X2C = 0.06418 0.05181
ITER = 2 X1A, X2A = 0.06418 0.05181 X1C, X2C = 0.05969 0.02986
ITER = 3 X1A, X2A = 0.05969 0.02486 X1C, X2C = 0.05937 0.02213
4.54 (cont’d)

4-46
ITER = 4 X1A, X2A = 0.05437 0.02213 X1C, X2C = 0.05931 0.02086
ITER = 5 X1A, X2A = 0.05931 0.02086 X1C, X2C = 0.05930 0.02083
ITER = 6 X1A, X2A = 0.05930 0.02083 X1C, X2C = 0.05930 0.02083
YA, YB, YC, YD, YE = 2.0270E 1 E 3 E
2 E 3 E
− − −
− −
01 1197 01 5100 01
9501 01 9319 02
. .
. .
NA0, NB0, NC0, ND0, NE0 = 0.20 0.20 0.20 0.20 0.20
ITER = 1 X1A, X2A = 0.10000 0.10000 X1C, X2C = 0.00012 0.00037
↓
ITER = 7 X1A, X2A = –0.02244 –0.08339 X1C, X2C = –0.02244 –0.08339
YA, YB, YC, YD, YE= 2.5051E 01 1.5868E 01 2.6693E 01
2.8989E 01 3.3991E 02
− − −
− −
4.55 a.
0 (kg/h)
0
(B)
(1 )
R
x
f m
=
−
0
(A)
(kg/h)
(kg R/kg)
RA
m
x
0
(kg R/kg)
(kg/h)
(1 )
RA
x
f m
− 1
1
(kg/h)
(kg R/kg)
R
m
x
(P)
(kg/h)
0.0075 kg R/kg
P
m
0 (kg/h)
(kg R/kg)
RA
fm
x
Mass balance on reactor: 0 1
2(1 ) (1)
f m m
− =
99% conversion of R: 1 1 0
0.01(1 ) (2)
R RA
m x f m x
= −
Mass balance on mixing point: 1 0 (3)
P
m fm m
+ =
R balance on mixing point: 1 1 0 0.0075 (4)
R RA P
m x fm x m
+ =
The system has 6 unknowns 0 1 1
( , , , , , )
RA R P
m x f m x m and four independent equations relating
them, so there must be two degrees of freedom.
b. 0 1
2(1 )
f m m
− =
1 1 0
0.01(1 )
R RA
m x f m x
= −
1 0 P
m fm m
+ =
1 1 0 0.0075
R RA P
m x fm x m
+ =
4850
0.0500
P
RA
m
x
=
=
4.54 (cont’d)
Reactor: 99%
conv. of R
E-Z Solve
⎯⎯⎯⎯
→
0 2780 kg/h
0.254 kg bypassed/kg fresh feed
m
f
=
=

4-47
4.55 (cont’d)
c. mP xRA mA0 mB0 f
4850 0.02 3327 1523 0.54
4850 0.03 3022 1828 0.40
4850 0.04 2870 1980 0.31
4850 0.05 2778 2072 0.25
4850 0.06 2717 2133 0.21
4850 0.07 2674 2176 0.19
4850 0.08 2641 2209 0.16
4850 0.09 2616 2234 0.15
4850 0.10 2596 2254 0.13
mP xRA mA0 mB0 f
2450 0.02 1663 762 0.54
2450 0.03 1511 914 0.40
2450 0.04 1435 990 0.31
2450 0.05 1389 1036 0.25
2450 0.06 1359 1066 0.22
2450 0.07 1337 1088 0.19
2450 0.08 1321 1104 0.16
2450 0.09 1308 1117 0.15
2450 0.10 1298 1127 0.13
f vs. xRA
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.00 0.02 0.04 0.06 0.08 0.10 0.12
x R A (kg R/kg A)
f
(kg
bypass/kg
fresh
feed)

4-48
4.56 a. 900 1
30 03
30 0
kg HCHO kmol HCHO
kg HCHO
kmol HCHO / h
h .
.
=
n
1
(kmol CH
3
OH / h)
30.0 kmol HCHO / h
(kmol H2 / h)
(kmol CH3OH / h)
2
3
n
n
% conversion:
30 0
0 60 50 0
1
1
.
. .
n
n
= ⇒ = kmol CH OH / h
3
b.
n
1
(kmol CH
3
OH / h) 30.0 kmol HCHO / h
2 (kmol H2 / h)
3 (kmol CH3OH /h)
n
n
30.0 kmol HCHO / h
2 (kmol H2 / h)
n
n
3
(kmol CH
3
OH / h)
Overall C balance: n1 (1) = 30.0 (1) ⇒ n1 = 30.0 kmol CH3OH/h (fresh feed)
Single pass conversion:
30 0
1 3
0 60 3 20 0
.
. .
n n
n
+
= ⇒ = kmol CH3OH / h
n1 + n3 = 50.0 kmol CH3OH fed to reactor/h
c. Increased xsp will (1) require a larger reactor and so will increase the cost of the reactor and
(2) lower the quantities of unreacted methanol and so will decrease the cost of the
separation. The plot would resemble a concave upward parabola with a minimum
around xsp = 60%.
4.57 a. Convert effluent composition to molar basis. Basis: 100 g effluent:
2 2
2
2
3 3
3
3
10.6 g H 1 mol H
5.25 mol H
2.01 g H
64.0 g CO 1 mol CO
2.28 mol CO
28.01 g CO
25.4 g CH OH 1 mol CH OH
32.04 g CH OH
0.793 mol CH OH
=
=
=
⇒
H
2
: 0.631 mol H
2
/ mol
CO: 0.274 mol CO / mol
CH
3
OH: 0.0953 mol CH
3
OH / mol

4-49
4.57 (cont’d)
4
0.004 mol CH OH(v)/mol
3
(mol CO/mol)
(0.896 - ) (mol H /mol)
2
(mol/min)
x
x
n
1
2
(mol CO/min)
(mol H /min)
2
n
n
CO H2 CH3OH
+ →
350 mol/min
0.631 mol CH3OH(v)/mol
0.274 mol CO/mol
0.0953 mol H2 mol
/
3 (mol CH OH(l)/min)
3
n
Condenser Overall process
3 unknowns 3 4
( , , )
n n x 2 unknowns 1 2
( , )
n n
–3 balances –2 independent atomic balances
0 degrees of freedom 0 degrees of freedom
Balances around condenser
3 3
4
4 4
3 4
32.1 mol CH OH(l)/min
CO: 350 0.274
H : 350 0.631 (0.996 ) 318.7 mol recycle/min
2
.301 molCO/mol
CH OH : 350 0.0953 0.004
3
n
n x
n x n
x
n n
⎫ =
∗ = ∗ ⎪
⎪
∗ = ∗ − ⇒ =
⎬
⎪ =
∗ = + ∗ ⎪
⎭
Overall balances
1
2
1 3
2 3 2
32.08 mol/min CO in feed
64.16 mol/min H in feed
C: =
H : 2 =4
n
n
n n
n n
=
=
⎫
⇒
⎬
⎭
Single pass conversion of CO:
( )
( )
%
07
.
25
%
100
3009
.
0
72
.
318
08
.
32
274
.
0
350
3009
.
0
72
.
318
08
.
32
=
×
∗
+
∗
−
∗
+
Overall conversion of CO: %
100
%
100
08
.
32
0
08
.
32
=
×
−
b. – Reactor conditions or feed rates drifting. (Recalibrate measurement instruments.)
– Impurities in feed. (Re-analyze feed.)
– Leak in methanol outlet pipe before flowmeter. (Check for it.)
Reactor
Cond.

4-50
4.58 a. Basis: 100 kmol reactor feed/hr
Reactor Cond. Absorb
Still
n1
(kmol CH4
/h)
n2 (kmol Cl2 /h)
100 kmol /h
80 kmol CH4
/h
20 kmol Cl2
/h
n3
(kmol CH4
/h)
n4
(kmol HCl /h)
5n5 (kmol CH3Cl /h)
n5
(kmol CH2
Cl 2
/h)
n3
(kmol CH4
/h)
n4
(kmol HCl/h)
Solvent
n4
(kmol HCl/h)
n3
(kmol CH4
/h)
5n5 (kmol CH3Cl /h)
n5
(kmol CH2
Cl 2
/h)
n5
(kmol CH2
Cl 2
/h)
5n5 (kmol CH3Cl /h)
Overall process: 4 unknowns (n1, n2, n4, n5) -3 balances = 1 D.F.
Mixing Point: 3 unknowns (n1, n2, n3) -2 balances = 1 D.F.
Reactor: 3 unknowns (n3, n4, n5) -3 balances = 0 D.F.
Condenser: 3 unknowns (n3, n4, n5) -0 balances = 3 D.F.
Absorption column: 2 unknowns (n3, n4) -0 balances = 2 D.F.
Distillation Column: 2 unknowns (n4, n5) -0 balances = 2 D.F.
Atomic balances around reactor:
5
n
,
4
n
,
3
n
for
Solve
5
2n
5
5n
4
n
40
:
balance
Cl
)
3
5
2n
5
15n
4
n
3
4n
320
:
balance
H
)
2
5
n
5
5n
3
n
80
:
balance
C
1)
⇒
⎪
⎭
⎪
⎬
⎫
+
+
=
+
+
+
=
+
+
=
CH4 balance around mixing point: n1 = (80 – n3) Solve for n1
Cl2 balance: n2 = 20
b. For a basis of 100 kmol/h into reactor
n1 = 17.1 kmol CH4/h
n2 = 20.0 kmol Cl2/h
n3 = 62.9 kmol CH4/h
n4 = 20.0 kmol HCl/h
5n5 = 14.5 kmol CH3Cl/h
c. (1000 kg CH3Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH3Cl/h
Scale factor = 366
.
1
Cl/h
3
CH
kmol
5
.
14
Cl/h
3
CH
kmol
81
.
19
=
Fresh feed:
2
Cl
mole%
54.0
,
4
CH
mol%
0
.
46
kmol/h
6
.
50
tot
n
/h
2
Cl
kmol
3
.
27
)
366
.
1
)(
0
.
20
(
2
n
/h
4
CH
kmol
3
.
23
)
366
.
1
)(
1
.
17
(
1
n =
⇒
⎭
⎬
⎫
=
=
=
=
Recycle: n3 = (62.9)(1.366) = 85.9 kmol CH4 recycled/h

4-51
4.59 a. Basis: 100 mol fed to reactor/h ⇒ 25 mol O2/h, 75 mol C2H4/h
reactor Seperator
n1
(mol C2
H 4
//h)
n2
(mol O2
/h)
n1
(mol C 2
H 4
//h)
n2
(mol O2
/h)
n3 (mol C2H 4O /h)
n4
(mol CO2
/h)
n5
(mol H2
O /h)
n4
(mol CO2
/h)
n5 (mol H2O /h)
n3 (mol C2H 4O /h)
75 mol C2
H 4
//h
25 mol O2
/h
nC2H4 ( mol C2H 4 /h)
nO2 (mol O2
/h)
Reactor
5 unknowns (n1 - n5)
-3 atomic balances
-1 - % yield
-1 - % conversion
0 D.F.
Strategy: 1. Solve balances around reactor to find n1- n5
2. Solve balances around mixing point to find nO2, nC2H4
(1) % Conversion ⇒ n1 = .800 * 75
(2) % yield: )
O
H
C
of
rate
n
(productio
n
H
C
mol
100
O
H
C
mol
90
H
C
mol
)
75
)(
200
(. 4
2
3
4
2
4
2
4
2 =
×
(3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4
(4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5
(5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5
(6) O2 balance (mix pt): nO2 = 25 – n2
(7) C2H4 balance (mix pt): nC2H4 = 75 – n1
Overall conversion of C2H4: 100%
b. n1 = 60.0 mol C2H4/h
n2 = 13.75 mol O2 /h
n3 = 13.5 mol C2H4O/h
n4 = 3.00 mol CO2/h
n5 = 3.00 mol H2O/h
nO2 = 11.25 mol O2/h
nC2H4 = 15.0 mol C2H4/h
100% conversion of C2H4
c.
Scale factor =
h
/
mol
h
/
mol
lb
363
.
3
O
H
C
mol
5
.
13
h
O
H
C
lbm
05
.
44
O
H
C
mole
-
lb
1
h
O
H
C
lbm
2000
4
2
4
2
4
2
4
2 −
=
nC2H4 = (3.363)(15.0) = 50.4 lb-mol C2H4/h
nO2 = (3.363)(11.25) = 37.8 lb-mol O2/h
separator

4-52
4.60 a. Basis: 100 mol feed/h. Put dots above all n’s in flow chart.
reactor cond.
100 mol/h
32 mol CO/h
64 mol H2 / h
4 mol N2
/ h
500 mol / h
x1 (mol N2 /mol)
x2
(mol CO / mol)
1-x1-x2 (mol H2 / h)
n1
(mol /h)
.13 mol N2
/mol
n3
(mol / h)
x1 (mol N2 /mol)
x2
(mol CO / mol)
1-x1
-x2
(mol H2
/ h)
n3
(mol CH3
OH / h)
Purge
Mixing point balances:
total: (100) + 500 = 1
n ⇒ 1
n = 600 mol/h
N2: 4 + x1 * 500 = .13 * 600 ⇒ x1 = 0.148 mol N2/mol
Overall system balances:
N2: 4 = .148 * 3
n ⇒ 3
n = 27 mol/h
Atomic C: 32 = 2
n + x2*27 2
n = 24.3 mol CH3OH/h
Atomic H: 2 * 64 = 4*24.3 + 2*(1-0.148-x2)*27 x2 = 0.284 mol CO/mol
Overall CO conversion: 100*[32-0.284(27)]/32 = 76%
Single pass CO conversion: 24.3/ (32+.284*500) = 14%
b. Recycle: To recover unconsumed CO and H2 and get a better overall conversion.
Purge: to prevent buildup of N2.
4.61 a.
Reactor Condenser
1 mol
(1-XI0
)/4 (mol N2
/ mol)
3/4 (1-XI0
) (mol H2
/ mol)
XI0
(mol I / mol)
nr (mol)
n1 (mol N2)
3n1 (mol H2)
n2 (mol I)
nr (mol)
(1-fsp) n1 (mol N2)
(1-fsp) 3n1 (mol H2)
n2 (mol I)
2 fsp n1 (mol NH3)
np (mol)
2 fsp n1 (mol NH3)
yp
(1-fsp
) n1
(mol N2
)
yp
(1-fsp
) 3n1
(mol H2
)
yp
n2
(mol I)
(1-fsp) n1 (mol N2)
(1-fsp) 3n1 (mol H2)
n2 (mol I)
(1-yp
) (1-fsp
) n1
(mol N2
)
(1-yp
) (1-fsp
) 3n1
(mol H2
)
(1-yp
) n2
(mol I)
2N2
+ 3H2
- NH3
n2 (mol CH3OH/h)
⇒
N2 + 3H2 Æ 2NH3

4-53
4.61 (cont’d)
At mixing point:
N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1
I: XI0 + (1-yp) n2 = n2
Total moles fed to reactor: nr = 4n1 + n2
Moles of NH3 produced: np = 2fspn1
Overall N2 conversion: %
100
4
/
)
X
1
(
n
)
f
1
(
y
4
/
)
X
1
(
0
I
1
sp
p
0
I
×
−
−
−
−
b. XI0 = 0.01 fsp = 0.20 yp = 0.10
n1 = 0.884 mol N2
n2 = 0.1 mol I
nr = 3.636 mol fed
np = 0.3536 mol NH3 produced
N2 conversion = 71.4%
c. Recycle: recover and reuse unconsumed reactants.
Purge: avoid accumulation of I in the system.
d. Increasing XI0 results in increasing nr, decreasing np, and has no effect on fov. Increasing fsp
results in decreasing nr, increasing np, and increasing fov.
Increasing yp results in decreasing nr, decreasing np, and decreasing fov.
Optimal values would result in a low value of nr and fsp, and a high value of np, this would
give the highest profit.
XI0 fsp yp nr np fov
0.01 0.20 0.10 3.636 0.354 71.4%
0.05 0.20 0.10 3.893 0.339 71.4%
0.10 0.20 0.10 4.214 0.321 71.4%
0.01 0.30 0.10 2.776 0.401 81.1%
0.01 0.40 0.10 2.252 0.430 87.0%
0.01 0.50 0.10 1.900 0.450 90.9%
0.10 0.20 0.20 3.000 0.250 55.6%
0.10 0.20 0.30 2.379 0.205 45.5%
0.10 0.20 0.40 1.981 0.173 38.5%

4-54
4.62 a. i - C H C H C H
4 10 4 8 8 18
+ = Basis: 1-hour operation
reactor (n-C H )
4
n5 10
(i-C H )
4
n6 10
(C H )
8
n7 18
(91% H SO )
2
m8 4
(n-C H )
4
n2 10
(i-C H )
4
n3 10
(C H )
8
n1 18
(91% H SO )
2
m4 4
decanter still
(C H )
8
n1 18
(n-C H )
4
n2 10
(i-C H )
4
n3 10
(C H )
8
n1 18
(n-C H )
4
n2 10
F
P
D
E
C
B
A
(kg 91% H SO )
2
m4 4
(i-C H )
4
n3 10
40000 kg
kmol
n0
0.25
0.50
0.25
i-C H
4 10
n-C H
4 10
C H
4 8
Units of n: kmol
Units of m: kg
Calculate moles of feed
M M M M
= + + = +
=
− −
0 25 050 0 25 0 75 5812 0 25 5610
57 6
. . . . . . .
.
L C H n C H C H
4 10 4 10 4 8
kg kmol
b gb g b gb g
n0 40000 1 694
= =
kg kmol 57.6 kg kmol
b gb g
Overall n - C H balance:
4 10 n2 050 694 347
= =
.
b gb g kmol n - C H in product
4 10
C H balance:
8 18
n1
694
1735
= =
0.25 kmol C H react 1 mol C H
1 mol C H
kmol C H in product
4 8 8 18
4 8
8 8
b gb g .
At (A), 5 mol - C H 1 mole C H n mol - C H kmol
4 10 4 8 4 10 A
moles C H at
A=173.5
-C H at
A and B
4 8
4 10
i i
i
⇒ = =
b g b gb gb g
b g b g
5 0 25 694 867 5
. .
Note: n mol C H 173.5
4 8
b g= at (A), (B) and (C) and in feed
i n
n i
- C H balance around first mixing point
kmol - C H recycled from still
4 10
4 10
⇒ + =
⇒ =
0 25 694 867 5
694
3
3
. .
b gb g
At C, 200 mol -C H mol C H
n mol - C H kmol - C H
4 10 4 8
4 10 C 4 10
i
i i
⇒ = =
b g b gb g
200 1735 34 700
. ,

4-55
4.62 (cont’d)
i n
n
- C H balance around second mixing point
l C H in recycle E
4 10
4 10
⇒ + =
⇒ =
867 5 34 700
33,800 kmo
6
6
. ,
Recycle E: Since Streams (D) and (E) have the same composition,
n
n
n i
n i
n
5
2
6
3
5
moles n - C H
moles n - C H
moles - C H
moles - C H
16,900 kmol n - C H
4 10 E
4 10 D
4 10 E
4 10 D
4 10
b g
b g
b g
b g
= ⇒ =
n
n
n
n
n
7
1
6
3
7
moles C H
moles C H
8460 kmol C H
8 18 E
8 18 D
4 18
b g
b g = ⇒ =
Hydrocarbons entering reactor:
347 16900 5812
+
F
HG I
KJ
b gb g
kmol n - C H
kg
kmol
4 10 .
+ +
F
HG I
KJ +
F
HG I
KJ
867 5 33800 5812 1735 5610
. . . .
b gb g
kmol - C H
kg
kmol
kmol C H
kg
kmol
4 10 4 8
i
+
F
HG I
KJ = ×
8460 kmol 114.22 4 00 106
C H
kg
kmol
kg
8 18 . .
H SO solution entering reactor
and leaving reactor
4.00 10 kg HC 2 kg H SO aq
1 kg HC
kg H SO aq
2 4
6
2 4
2 4
b g
b g
b g
=
×
= ×
8 00 106
.
m n
n n
m
8
6
5
2 5
8
6
8 00 10
7 84 10
H SO in recycle
H SO leaving reactor
n - C H in recycle
n - C H leaving reactor
kg H SO aq in recycle E
2 4
2 4
4 10
4 10
2 4
b g
b g
b g
b g
b g
.
.
×
=
+
⇒ = ×
m4
5
16 10
= −
= ×
H SO entering reactor H SO in E
kg H SO aq recycled from decanter
2 4 2 4
2 4
. b g
⇒ 16 10 0 91 1480
5
. .
× =
d ib g b g
kg H SO 1 kmol 98.08 kg kmol H SO in recycle
2 4 2 4
16 10 0 09 799
5
. .
× =
d ib g b g
kg H O 1 kmol 18.02 kg kmol H O from decanter
2 2
Summary: (Change amounts to flow rates)
Product: 173.5 kmol C H h 347 kmol - C H h
Recycle from still: 694 kmol - C H h
Acid recycle: 1480 kmol H SO h 799 kmol H O h
Recycle E: 16,900 kmol n - C H h 33,800 kmol L - C H h, 8460 kmol C H h,
kg h 91% H SO 72,740 kmol H SO h, 39,150 kmol H O h
8 18 4 10
4 10
2 4 2
4 10 4 10 8 18
2 4 2 4 2
,
,
,
.
n
i
7 84 106
× ⇒

4-56
4.63 a. A balance on ith tank (input = output + consumption)
,
,
,
v C vC kC C V
C C k C C
A i Ai Ai Bi
A i Ai Ai Bi
v V v
L min mol L mol liter min L
note /
b g b g b g b g
−
−
= + ⋅
E
= +
÷ =
1
1
τ
τ
B balance. By analogy, C C k C C
B i Bi Ai Bi
, − = +
1 τ
Subtract equations ⇒ − = − = − = = −
− −
A
−
− −
C C C C C C C C
Bi Ai B i A i
i
B i A i B A
, , , ,
1 1
1
2 2 0 0
from balances on
tank
st
b g
…
b. C C C C C C C C
Bi Ai B A Bi Ai B A
− = − ⇒ = + −
0 0 0 0 . Substitute in A balance from part (a).
C C k C C C C
A i Ai Ai Ai B A
, − = + + −
1 0 0
τ b g . Collect terms in CAi
2
, CAi
1
, CAi
0
.
C k C k C C C
C C k k C C C
Ai AL B A A i
AL AL B A A i
2
0 0 1
2
0 0 1
1 0
0 1
τ τ
α β γ α τ β τ γ
+ + − − =
⇒ + + = = = + − = −
−
−
b g
b g
,
,
, ,
where
Solution: CAi =
− + −
β β αγ
α
2
4
2
(Only + rather than ±: since αγ is negative and the
negative solution would yield a negative concentration.)
(xmin = 0.50, N = 1), (xmin = 0.80, N = 3), (xmin = 0.90, N = 4), (xmin = 0.95, N = 6),
(xmin = 0.99, N = 9), (xmin = 0.999, N = 13).
As xmin → 1, the required number of tanks and hence the process cost becomes infinite.
d. (i) k increases ⇒ N decreases (faster reaction ⇒ fewer tanks)
( )
ii increases increases (faster throughput less time spent in reactor
lower conversion per reactor)
v N
⇒ ⇒
⇒
(iii) V increases ⇒ N decreases (larger reactor ⇒ more time spent in reactor
⇒ higher conversion per reactor)
c.
k = 36.2 N gamma CA(N) xA(N)
v = 5000 1 -5.670E-02 2.791E-02 0.5077
V = 2000 2 -2.791E-02 1.512E-02 0.7333
CA0 = 0.0567 3 -1.512E-02 8.631E-03 0.8478
CB0 = 0.1000 4 -8.631E-03 5.076E-03 0.9105
alpha = 14.48 5 -5.076E-03 3.038E-03 0.9464
beta = 1.6270 6 -3.038E-03 1.837E-03 0.9676
7 -1.837E-03 1.118E-03 0.9803
8 -1.118E-03 6.830E-04 0.9880
9 -6.830E-04 4.182E-04 0.9926
10 -4.182E-04 2.565E-04 0.9955
11 -2.565E-04 1.574E-04 0.9972
12 -1.574E-04 9.667E-05 0.9983
13 -9.667E-05 5.939E-05 0.9990
14 -5.939E-05 3.649E-05 0.9994

4-57
4.64 a. Basis: 1000 g gas
Species m (g) MW n (mol) mole % (wet) mole % (dry)
C3H8 800 44.09 18.145 77.2% 87.5%
C4H10 150 58.12 2.581 11.0% 12.5%
H2O 50 18.02 2.775 11.8%
Total 1000 23.501 100% 100%
Total moles = 23.50 mol, Total moles (dry) = 20.74 mol
Ratio: 2.775 / 20.726 = 0.134 mol H O / mol dry gas
2
b. C3H8 + 5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
Theoretical O2:
C H
100 kg gas
h
80 kg C H
100 kg gas
1 kmol C H
44.09 kg C H
5 kmol O
1 kmol C H
9.07 kmol O / h
3 8
3 8 3 8
3 8
2
3 8
2
: =
C H
100 kg gas
h
15 kg C H
100 kg gas
1 kmol C H
58.12 kg C H
6.5 kmol O
1 kmol C H
1.68 kmol O / h
4 10
4 10 4 10
4 10
2
4 10
2
: =
Total: (9.07 + 1.68) kmol O2/h = 10.75 kmol O2/h
Air feed rate:
10.75 kmol O
h
1 kmol Air
.21 kmol O
1.3 kmol air fed
1 kmol air required
66.5 kmol air / h
2
2
=
The answer does not change for incomplete combustion
4.65 5 L C H 0.659 kg C H
L C H
1000 mol C H
86 kg C H
38.3 mol C H
6 14 6 14
6 14
6 14
6 14
6 14
=
4 L C H 0.684 kg C H
L C H
1000 mol C H
100 kg C H
27.36 mol C H
7 16 7 16
7 16
7 16
7 16
7 16
=
C6H14 +19/2 O2 → 6 CO2 + 7 H2O C6H14 +13/2 O2 → 6 CO + 7 H2O
C7H16 + 11 O2 → 7 CO2 + 8 H2O C7H16 + 15/2 O2 → 7 CO + 8 H2O
Theoretical oxygen:
38.3 mol C H 9.5 mol O
mol C H
27.36 mol C H 11 mol O
mol C H
mol O required
6 14 2
6 14
7 16 2
7 16
2
+ = 665
O2 fed: (4000 mol air )(.21 mol O2 / mol air) = 840 mol O2 fed
Percent excess air:
840 665
665
100% 26 3%
−
× = . excess air

4-58
4.66
CO
1
2
O CO H
1
2
O H O
2 2 2 2 2
+ → + →
175 kmol/h
0.500 kmol N2/kmol
x (kmol CO/mol)
(0.500–x) (kmol H2/kmol)
20% excess air
Note: Since CO and H2 each require 0. /
5 mol O mol fuel
2 for complete combustion, we can
calculate the air feed rate without determining xCO . We include its calculation for illustrative
purposes.
A plot of x vs. R on log paper is a straight line through the points R x
1 1
10 0 0 05
= =
. , .
b g and
R x
2 2
99 7 10
= =
. , .
b g.
ln ln ln ln . . ln . . .
ln ln . . ln . . . .
exp . .
. .
x b R a
x a Rb
b
a x R
a
R x
= +
=
= =
= − = − ⇒ = × −
= − = × −
= ⇒ =
@
10 0 05 99 7 10 0 1303
10 1303 99 7 6 00 2 49 10 3 1303
6 00 2 49 10 3
38 3 0 288
b g b g
b g b g
b g
moles CO
mol
Theoretical O
2
175 kmol kmol CO 0.5 kmol O
2
h kmol kmol CO
175 kmol kmol H
2
0.5 kmol O
2
h kmol kmol H
2
kmol O
2
h
: .
.
.
0 288
0 212
4375
+ =
Air fed:
43.75 kmol O
2
required 1 kmol air 1.2 kmol air fed
h 0.21 kmol O
2
1 kmol air required
kmol air
h
= 250
4.67 a. CH 2O CO 2H O
C H
7
2
O 2CO 3H O
C H 5O 3CO 4H O
C H
13
2
O 4CO 5H O
4 2 2 2
2 6 2 2 2
3 8 2 2 2
4 10 2 2 2
+ → +
+ → +
+ → +
+ → +
100 kmol/h
0.944 CH4
0.0340 C2H6
0.0060 C3H8
0.0050 C4H10
17% excess air
na (kmol air/h)
0.21 O2
0.79 N2
Theoretical O
0.944 100 kmol CH kmol O
h 1 kmol CH
0.0340 100 kmol C H .5 kmol O
h 1 kmol C H
0.0060 100 kmol C H kmol O
h 1 kmol C H
0.0050 100 kmol C H .5 kmol O
h 1 kmol C H
kmol O h
2
4 2
4
2 6 2
2 6
3 8 2
3 3
4 10 2
4 10
2
:
.
b g b g
b g b g
2 3
5 6
207 0
+
+ +
=

4-59
4.67 (cont’d)
Air feed rate: nf = =
207.0 kmol O 1 kmol air 1.17 kmol air fed
h 0.21 kmol O kmol air req.
kmol air h
2
2
1153
b. n n x x x x P
a f xs
= + + + +
2 35 5 65 1 100 1 0 21
1 2 3 4
. . .
b gb gb g
c. , ( . ) .
, ( ) .
n aR n R n R
n bR n R n R
f f f f f f
a a a a a a
= = = ⇒ =
= = = ⇒ =
kmol / h,
kmol / h,
750 60 125
550 25 22 0
x kA x k A k
A
x
A
A
i
i i i i
i
i
i
i
= ⇒ = = ⇒ =
⇒ =
∑ ∑
∑
∑
, = CH C H C H C H
i i
i
i
4 2 4 3 8 4 10
1
1
, , ,
d. Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the
flowmeter calibration formulas might not be linear, or the stack gas analysis could be
incorrect.
4.68 a. C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
Basis: 100 mol C4H10 nCO2 (mol CO2)
nH2O (mol H2O)
nC4H10 (mol C4H10)
Pxs (% excess air) nO2 (mol O2)
nair (mol air) nN2 (mol N2)
0.21 O2
0.79 N2
D.F. analysis
6 unknowns (n, n1, n2, n3, n4, n5)
-3 atomic balances (C, H, O)
-1 N2 balance
-1 % excess air
-1 % conversion
0 D.F.
Run Pxs Rf A1 A2 A3 A4
1 15% 62 248.7 19.74 6.35 1.48
2 15% 83 305.3 14.57 2.56 0.70
3 15% 108 294.2 16.61 4.78 2.11
Run nf x1 x2 x3 x4 na Ra
1 77.5 0.900 0.0715 0.0230 0.0054 934 42.4
2 103.8 0.945 0.0451 0.0079 0.0022 1194 54.3
3 135.0 0.926 0.0523 0.0150 0.0066 1592 72.4

4-60
4.68 (cont’d)
b. i) Theoretical oxygen = (100 mol C4H10)(6.5 mol O2/mol C4H10) = 650 mol O2
n mol O mol air / 0.21 mol O mol air
air 2 2
= =
( )( )
650 1 3095
100% conversion ⇒ nC4H10 = 0, nO2
= 0
n
n
n
N2
CO2 4 10 2 4 10 2
H2O 4 10 2 4 10 2
2
2
2
mol
100 mol C H react 4 mol CO mol C H mol CO
100 mol C H react 5 mol H O mol C H mol H O
73.1% N
12.0% CO
14.9% H O
= =
= =
= =
U
V
|
W
|
0 79 3095 mol 2445
400
500
.
b gb g
b gb g
b gb g
ii) 100% conversion ⇒ nC4H10 = 0
20% excess ⇒ nair = 1.2(3095) = 3714 mol (780 mol O2, 2934 mol N2)
Exit gas:
400 mol CO2
500 mol H2O
130 mol O2
2934 mol N2
10.1% CO2
12.6% H2O
3.3% O2
74 0%
. N2
iii) 90% conversion ⇒ nC4H10 = 10 mol C4H10 (90 mol C4H10 react, 585 mol O2 consumed)
20% excess: nair = 1.2(3095) = 3714 mol (780 mol O2, 2483 mol N2)
Exit gas:
10 mol C4H10
360 mol CO2
450 mol H2O (v)
195 mol O2
2934 mol N2
0.3% C4H10
9.1% CO2
11.4% H2O
4.9% O2
74 3%
. N2
4.69 a. C3H8 + 5 O2 → 3 CO2 + 4 H2O H2 +1/2 O2 → H2O
C3H8 + 7/2 O2 → 3 CO + 4 H2O
Basis: 100 mol feed gas
100 mol
0.75 mol C3H8 n1 (mol C3H8)
0.25 mol H2 n2 (mol H2)
n3 (mol CO2)
n4 (mol CO)
n0 (mol air) n5 (mol H2O)
0.21 mol O2/mol n6 (mol O2)
0.79 mol N2/mol n7 (mol N2)
Theoretical oxygen
75 mol C H mol O
mol C H
mol H mol O
mol H
mol O
3 8 2
3 8
2 2
2
2
:
.
.
5 25 050
387 5
+ =

4-61
4.69 (cont’d)
Air feed rate
mol O
h
kmol air
0.21 kmol O
kmol air fed
1 kmol air req'd.
mol air
2
2
:
. .
.
n0
387 5 1 125
23065
= =
90% 0100 75
85% 0150 25
95%
0 95 67 5 3
192 4
5%
0 05 67 5 3
101
1
2
3
3
propane conversion mol C H ) = 7.5 mol C H
(67.5 mol C H reacts)
hydrogen conversion mol C H ) = 3.75 mol H
CO selectivity
mol C H react) mol CO generated
mol C H react
mol CO
CO selectivity
mol C H react) mol CO generated
mol C H react
mol CO
3 8 3 8
3 8
3 8 2
2
3 8 2
3 8
2
3 8
3 8
⇒ =
⇒ =
⇒ =
=
⇒ = =
n
n
n
n
. (
. (
. ( .
.
. ( .
.
H balance (75 mol C H
mol H
mol C H
mol H
mol C H mol H mol H O)(2) mol H O
3 8
3 8
2
3 8 2 2 2
: ) ( )( )
( . )( ) ( . )( ) ( .
8 25 2
7 5 8 375 2 2912
5 5
F
HG I
KJ +
= + + ⇒ =
n n
O balance ( .21 2306.5 mol O )(2
mol O
mol O
) 4 mol CO
mol CO mol H O)(1) + 2 mol O mol O
2
2
2
2 2 2
: ( . )( )
( . )( ) ( . ( ) .
0 192 2
101 1 2912 1413
6 6
× =
+ + ⇒ =
n n
N balance: mol N mol N
2 2 2
n7 0 79 23065 1822
= =
. ( . )
Total moles of exit gas = (7.5 + 3.75 + 192.4 + 10.1 + 291.2 + 141.3 + 1822) mol
= 2468 mol
CO concentration in exit gas =
101
10 4090
6
. mol CO
2468 mol
ppm
× =
b. If more air is fed to the furnace,
(i) more gas must be compressed (pumped), leading to a higher cost (possibly a larger
pump, and greater utility costs)
(ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the
product gas temperature decreases and less steam is produced.

4-62
4.70 a. C5H12 + 8 O2 → 5 CO2 + 6 H2O
Basis: 100 moles dry product gas
n1 (mol C5H12) 100 mol dry product gas (DPG)
0.0027 mol C5H12/mol DPG
Excess air 0.053 mol O2/mol DPG
n2 (mol O2) 0.091 mol CO2/mol DPG
3.76n2 (mol N2) 0.853 mol N2/mol DPG
n3 (mol H2O)
3 unknowns (n1, n2, n3)
-3 atomic balances (O, C, H)
-1 N2 balance
-1 D.F. ⇒ Problem is overspecified
b. N2 balance: 3.76 n2 = 0.8533 (100) ⇒ n2 = 22.69 mol O2
C balance: 5 n1 = 5(0.0027)(100) + (0.091)(100) ⇒ n1 = 2.09 mol C5H12
H balance: 12 n1 = 12(0.0027)(100) + 2n3 ⇒ n3 = 10.92 mol H2O
O balance: 2n2 = 100[(0.053)(2) + (0.091)(2)] + n3 ⇒ 45.38 mol O = 39.72 mol O
Since the 4th
balance does not close, the given data cannot be correct.
c.
n1 (mol C5H12) 100 mol dry product gas (DPG)
0.00304 mol C5H12/mol DPG
Excess air 0.059 mol O2/mol DPG
n2 (mol O2) 0.102 mol CO2/mol DPG
3.76n2 (mol N2) 0.836 mol N2/mol DPG
n3 (mol H2O)
N2 balance: 3.76 n2 = 0.836 (100) ⇒ n2 = 22.2 mol O2
C balance: 5 n1 = 100 (5*0.00304 + 0.102) ⇒ n1 = 2.34 mol C5H12
H balance: 12 n1 = 12(0.00304)(100) + 2n3 ⇒ n3 = 12.2 mol H2O
O balance: 2n2 = 100[(0.0590)(2) + (0.102)(2)] + n3 ⇒ 44.4 mol O = 44.4 mol O √
Fractional conversion of C5H12: fed
react/mol
mol
870
.
0
344
.
2
00304
.
0
100
344
.
2
=
×
−
Theoretical O2 required: 2.344 mol C5H12 (8 mol O2/mol C5H12) = 18.75 mol O2
% excess air: air
excess
%
6
.
18
%
100
required
O
mol
18.75
required
O
mol
18.75
-
fed
O
mol
23
.
22
2
2
2
=
×

4-63
4.71 a.
h
/
OH
CH
mol
6
.
296
g
04
.
32
mol
ml
g
792
.
0
L
ml
1000
h
OH
CH
L
12
3
3
=
CH3OH + 3/2 O2 → CO2 +2 H2O, CH3OH + O2 → CO +2 H2O
296.6 mol CH3OH(l)/h (
n2 mol dry gas / h)
0.0045 mol CH3OH(v)/mol DG
0.0903 mol CO2/mol DG
/
n1 (mol O h)
2 0.0181 mol CO/mol DG
376 1
. /
n (mol N h)
2 x (mol N2/mol DG)
(0.8871–x) (mol O2/mol DG)
(
n3 mol H O(v) / h)
2
4 unknowns ( , , , )
n n n x
1 2 3 – 4 balances (C, H, O, N2) = 0 D.F.
b. Theoretical O2: 296.6 (1.5) = 444.9 mol O2 / h
C balance: 296.6 =
n 2 (0.0045 + 0.0903 + 0.0181) ⇒
n 2 = 2627 mol/h
H balance: 4 (296.6) =
n 2 (4*0.0045) + 2
n 3 ⇒
n 3 = 569.6 mol H2O / h
69.6
5
x)]
-
2(0.8871
.0181
0
2(0.0903)
627[0.0045
2
1
n
2
296.6
:
balance
O +
+
+
+
=
+
N2 balance: 3.76
n 1 = x ( 2627)
Solving simultaneously ⇒ . / /
n x
1 574 3
= mol O h, = 0.822 mol N mol DG
2 2
Fractional conversion: fed
react/mol
OH
CH
mol
960
.
0
6
.
296
)
0045
.
0
(
2627
6
.
296
3
=
−
% excess air: %
1
.
29
%
100
9
.
444
9
.
444
3
.
574
=
×
−
Mole fraction of water: /mol
O
H
mol
178
.
0
mol
)
6
.
569
2627
(
O
H
mol
6
.
569
2
2
=
+
c. Fire, CO toxicity. Vent gas to outside, install CO or hydrocarbon detector in room, trigger
alarm if concentrations are too high
4.72 a. G.C. Say ns mols fuel gas constitute the sample injected into the G.C. If xCH4
and xC H
2 6
are
the mole fractions of methane and ethane in the fuel, then
n x
n x
x
x
s
s
mol mol C H mol 2 mol C 1 mol C H
mol mol CH mol 1 mol C 1 mol CH
mol C H mol fuel
mol CH mol fuel
mole C H mole CH in fuel gas
C H 2 2 2 6
CH 4 4
C H 2 6
CH 4
2 6 4
2 6
4
2 6
4
b g b gb g
b g b gb g
b g
b g
=
E
=
20
85
01176
.

4-64
4.72 (cont’d)
Condensation measurement:
1.134 g H O 1 mol 18.02 g
mol product gas
mole H O
mole product gas
2 2
b gb g
050
0126
.
.
=
Basis: 100 mol product gas. Since we have the most information about the product stream
composition, we choose this basis now, and would subsequently scale to the given
fuel and air flow rates if it were necessary (which it is not).
CH 2O CO 2H O
C H
7
2
O 2CO 3H O
4 2 2 2
2 6 2 2 2
+ → +
+ → +
n1 (mol CH4 )
0.1176 n1 (mol C2H6)
n2 (mol CO2)
n3 (mol O2 / h)
376 n3 (mol N2 / h)
100 mol dry gas / h
0.126 mol H
2O / mol
0..874 mol dry gas / mol
0.119 mol CO
2 / mol D.G.
x (mol N
2 / mol)
(0.881-x) (mol O
2 / mol D.G.)
Strategy: H balance n
⇒
1
; C balance ⇒ n2 ;
N balance
O balance
n
2
3
U
V
W
⇒ , x
H balance: mol CH in fuel
4
4 6 01176 100 0126 2 5356
1 1 1
n n n
+ = ⇒ =
b gb g b gb gb g
. . .
⇒ 0.1176(5.356) = 0.630 mol C2H6 in fuel
C balance: 5356 2 0 630 100 0874 0119 3784
2 2
. . . . .
+ + = ⇒ =
b gb g b gb gb g
n n mol CO in fuel
2
Composition of fuel: 5356
. mol CH4 , 0 630
. mol C H
2 6 , 3784
. mols CO2
⇒ 0548
. CH , 0.064 C H , 0.388 CO
4 2 6 2
N balance:
2 376 100 0874
3
. .
n x
= b gb g
O balance: 2 3784 2 100 0126 100 0874 2 0119 0881
3
b gb g b gb g b gb gb g b g
. . . . .
+ = + + −
n x
Solve simultaneously: n3 1886
= . mols O fed
2 , x = 0813
.
Theoretical O
5.356 mol CH mol O
1 mol CH
0.630 mol C H .5 mol O
1 mol CH
mol O required
2
4 2
4
2 6 2
4
2
:
.
2 3
12 92
+
=
Desired O2 fed:
air
mol
O
mol
0.21
fuel
mol
1
air
mol
7
fuel
mol
3.784)
0.630
(5.356 2
+
+
= 14.36 mol O2
Desired % excess air: %
11
%
100
92
.
12
92
.
12
36
.
14
=
×
−
b.
Actual % excess air: %
46
%
100
92
.
12
92
.
12
86
.
18
=
×
−
Actual molar feed ratio of air to fuel: 1
:
9
feed
mol
77
.
9
air
mol
)
21
.
0
/
86
.
18
(
=

4-65
4.73 a. C3H8 +5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
Basis 100: mol product gas
n1 (mol C3H8) 100 mol
n2 (mol C4H10) 0.474 mol H2O/mol
x (mol CO2/mol)
n3 (mol O2) (0.526–x) (mol O2/mol)
Dry product gas contains 69.4% CO2 ⇒ /mol
CO
mol
365
.
0
x
6
.
30
4
.
69
x
526
.
0
x
2
=
⇒
=
−
3 unknowns (n1, n2, n3) – 3 balances (C, H, O) = 0 D.F.
O balance: 2 n3 = 152.6 ⇒ n3 = 76.3 mol O2
10
H
4
C
34.9%
,
8
H
3
C
%
1
.
65
10
H
4
C
mol
3.8
2
n
8
H
3
C
mol
7.1
1
n
94.8
2
n
10
1
n
8
:
balance
H
36.5
2
n
4
1
n
3
:
balance
C
⇒
=
=
⇒
⎭
⎬
⎫
=
+
=
+
b. nc=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C
nh = 100 mol (0.474 mol H2O/mol)(2mol H/mol H2O)=94.8 mol H
⇒ 27.8%C, 72.2% H
From a:
C
%
8
.
27
%
100
H
C
mol
H)
(C
mol
14
H
C
mol
3.80
H
C
mol
H)
(C
mol
11
H
C
mol
7.10
H
C
mol
C
mol
4
H
C
mol
3.80
H
C
mol
C
mol
3
H
C
mol
7.10
10
4
10
4
8
3
8
3
10
4
10
4
8
3
8
3
=
×
+
+
+
+
4.74 Basis: 100 kg fuel oil
Moles of C in fuel: C
kmol
08
.
7
C
kg
12.01
C
kmol
1
kg
C
kg
0.85
kg
100
=
Moles of H in fuel: H
kmol
0
.
12
H
kg
1
H
kmol
1
kg
H
kg
0.12
kg
100
=
Moles of S in fuel: S
kmol
053
.
0
S
kg
32.064
S
kmol
1
kg
S
kg
0.017
kg
100
=
1.3 kg non-combustible materials (NC)

4-66
4.74 (cont’d)
100 kg fuel oil
7.08 kmol C n2 (kmol N2)
12.0 kmol H n3 (kmol O2)
0.053 kmol S C + O2 → CO2 n4 (kmol CO2)
1.3 kg NC (s) C + 1/2 O2 → CO (8/92) n4 (kmol CO)
2H + 1/2 O2 → H2O n5 (kmol SO2)
20% excess air S + O2 → SO2 n6 (kmol H2O)
n1 (kmol O2)
3.76 n1 (kmol N2)
Theoretical O2:
2
2
2
2
O
kmol
133
.
10
S
kmol
1
O
kmol
1
S
kmol
.053
0
H
kmol
2
O
kmol
5
.
H
kmol
2
1
C
kmol
1
O
kmol
1
C
kmol
08
.
7
=
+
+
20 % excess air: n1 = 1.2(10.133) = 12.16 kmol O2 fed
O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n3 ⇒ n3 = 2.3102 kmol O2
C balance: 7.08 = n4+8n4/92 ⇒ n4 = 6.514 mol CO2
⇒ 8 (6.514)/92 = 0.566 mol CO
S balance: n5 = 0.53 kmol SO2
H balance: 12 = 2n6 ⇒ n6 = 6.00 kmol H2O
N2 balance: n2 = 3.76(12.16) = 45.72 kmol N2
Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol
= 61.16 kmol
⇒ 10.7% CO, 0.92% CO, 0.087% SO 9.8% H O, 3.8% O , 74.8% N
2 2 2 2
,
4.75 a. Basis: 5000 kg coal/h; 50 kmol air min kmol air h
= 3000
C + 02 -- CO2
2H + 1/2 O2 --H2O
S + O2 -- SO2
C + 1/2 O2 -- CO
5000 kg coal / h
0.75 kg C / kg
0.17 kg H / kg
0.02 kg S / kg
0.06 kg ash / kg
3000 kmol air / h
0.21 kmol O
2
/ kmol
0.79 kmol N
2 / kmol
n1 (kmol O2 / h)
n2 (kmol N2 / h)
n3 (kmol CO
2 / h)
0.1 n3 (kmol CO / h)
n4
(kmol SO
2
/ h)
n5
(kmol H2
O / h)
mo kg slag / h
Theoretical O2:
C:
0.75 5000 kg C 1 kmol C 1 kmol O
h kg C 1 kmol C
2
b g
12 01
.
= 312 2
. kmol O h
2

4-67
4.75 (cont’d)
H:
0.17 5000 kg H 1 kmol H 1 kmol H O 1 kmol O
h kg H 2 kmol H 2 kmol H O
2 2
2
b g
101
.
= 210 4
. kmol O h
2
S:
0.02 5000 kg S 1 kmol S 1 kmol O
h 1 kmol S
2
b g
32.06 kg S
= 3.1 kmol O2/h
Total = (312.2+210.4 + 3.1) kmol O2/h = 525.7 kmol O h
2
O fed
2 = 0 21 3000 630
. b g= kmol O h
2
Excess air:
630 5257
5257
100% 19 8%
−
× =
.
.
. excess air
b. Balances:
C:
0 94 0 75
12 01
. .
.

b gb gb g
5000 kg C react 1 kmol C
h kg C
0.1
3 3
= +
n n
⇒ =
.
n3 2
kmol CO h
2668 , 01 26 7
. .
n3 kmol CO h
=
H:
017
101
.
.
b gb g
5000 kg H 1 kmol H 1 kmol H O
h kg H 2 kmol H
2
5
= n ⇒ =
n5 2
kmol H O h
4208
.
S: (from part a)
3.1 kmol O for SO 1 kmol SO
h kmol O
2 2 2
2
4
b g
1
=
n ⇒ =
.
n4 2
kmol SO h
31
N2: 0 79 3000
.
b gb gkmol N h n
2 2
= ⇒ =

n2 2
kmol N h
2370
O: 0 21 3000 2 2 2 2668 1 26 68 2 31 1 4208
. ( ) . . . .
b g b g b g b g b g b gb g
= + + + +
n1
⇒ . /
n1 136 4
= kmol O h
2
Stack gas total = 3223 kmol h
Mole fractions:
xCO mol CO mol
= = × −
26 7 3224 8 3 10 3
. .
xSO 2
2
mol SO mol
= = × −
31 3224 9 6 10 4
. .
c. SO O SO
SO H O H SO
2 2 3
3 2 2 4
+ →
+ →
1
2
3.1 kmol SO 1 kmol SO 1 kmol H SO 98.08 kg H SO
h kmol SO 1 kmol SO kmol H SO
304 kg H SO h
2 3 2 4 2 4
2 3 2 4
2 4
1
=

4-68
4.76 a. Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile
matter
100 g c.a.r. 1.147 g a.d.c.
1.207 g c.a.r.
g air - dried coal; 4.97 g H O lost by air drying
2
= 9503
.
95.03 g a.d.c 1.234 g H O
1.234 g a.d.c.
g H O lost in second drying step
2
2
−
=
1204
2 31
.
.
b g
Total H O g g g moisture
2 = + =
4 97 2 31 7 28
. . .
95.03 g a.d.c g v.m. H O
1.347 g a.d.c.
g H O g volatile matter
2
2
1347 0811
2 31 3550
. .
. .
− +
− =
b g b g
95.03 g a.d.c .111 g ash
1.175 g a.d.c.
.98 g ash
0
8
=
Fixed carbon = − − − =
100 7 28 3550 8 98 48 24
. . . .
b gg g fixed carbon
7.28 g moisture
48.24 g fixed carbon
35.50 g volatile matter
8.98 g ash
100 g coal as received
7.3% moisture
48.2% fixed carbon
35.5% volatile matter
9.0% ash
⇒
b. Assume volatile matter is all carbon and hydrogen.
C CO CO
2 2
+ → :
1 mol O 1 mol C 10 g 1 mol air
1 mol C 12.01 g C 1 kg 0.21 mol O
mol air kg C
2
3
2
= 3965
.
2H O H O
2
+ →
1
2
2 :
0.5 mol O 1 mol H 10 g 1 mol air
2 mol H 1.01 g H 1 kg 0.21 mol O
mol air kg H
2
3
2
= 1179
Air required:
1000 kg coal 0.482 kg C 396.5 mol air
kg coal kg C
+
1000 kg 0.355 kg v.m. 6 kg C 396.5 mol air
kg 7 kg v.m. kg C
+ = ×
1000 kg 0.355 kg v.m. 1 kg H 1179 mol air
kg 7 kg v.m. kg H
mol air
372 105
.

4-69
4.77 a. Basis 100 mol dry fuel gas. Assume no solid or liquid products!
C + 02
-- CO2
C + 1/2 O2 -- CO
2H + 1/2 O2 --H2O
S + O2 -- SO2
n1 (mol C)
n2 (mol H)
n3
(mol S)
n4 (mol O2)
(20% excess)
100 mol dry gas
0.720 mol CO
2 / mol
0.0257 mol CO / mol
0.000592 mol SO
2 / mol
0.254 mol O
2
/ mol
n5 (mol H2O (v))
⎪
⎭
⎪
⎬
⎫
=
+
+
+
+
+
+
=
=
n
]
n
0.25
0.0592
(74.57
(1.20)
:
O
excess
%
20
n
(0.254)]
2
(0.000592)
2
0.0257
2(0.720)
[
100
n
2
:
balance
O
n
2
n
:
balance
H
4
2
2
5
4
5
2
⇒ n2 = 183.6 mol H, n4 = 144.6 mol O2, n5 = 91.8 mol H2O
Total moles in feed: 258.4 mol (C+H+S) ⇒ 28.9% C, 71.1% H, 0.023% S
4.78 Basis: 100 g oil
0.87 g C/g
mol O
n1 2
(25% excess)
100 g oil
0.10 g H/g
0.03 g S/g
mol N
n1 2
3.76
furnace
mol N
n2 2
mol O
n3 2
mol CO
n4 2
mol SO
n5 2
mol H O
n6 2
(1 – )
x mol SO
n5 2
(N , O , CO , H O)
2
2
2
2
scrubber
(1 – )
x mol SO
n5 2
0.90
Alkaline solution
x mol SO
n3 2
(N , O , CO , H O)
2
2
2
2
(1 – )
x mol SO
n5 2
0.10
(N , O , CO , H O)
2
2
2
2
Stack
N , O , CO , H O
2
2
2
2
SO ,
2
(612.5 ppm SO )
2
CO2:
0.87 100 g C 1 mol C 1 mol CO
g C 1 mol C
mol CO
mol O
consumed
2
4 2
2
b g
12 01
7 244
7 244
.
.
.
⇒ =
F
HG I
KJ
n
H O
2 :
0.10 100 g H 1 mol H 1 mol H O
g H 2 mol H
mol H O
.475 mol O
consumed
2
6 2
2
b g
101
4 95
2
.
.
⇒ =
F
HG I
KJ
n

4-70
4.78 (cont’d)
SO2:
0.03 100 g S 1 mol S 1 mol SO
g S 1 mol S
n mol SO
.0956 mol O
consumed
2
5 2
2
b g
32 06
0 0936
0
.
.
⇒ =
F
HG I
KJ
25% excess O2 : n1 2
mol O
= + + ⇒
125 7 244 2 475 0 0936 12 27
. . . . .
b g
O balance: 12.27 mol O fed mol O consumed
mol O
2 3 2 2
2
n = − + +
=
7 244 2 475 0 0936
2 46
. . .
.
b g
N balance:
2 n mol mol N
2 2
= =
376 12 27 4614
. . .
b g
SO in stack SO balance around mixing point
2 2
b g:
x x x
n
0 0936 010 1 0 0936 0 00936 0 0842
5
. . . . .
F
H
I
K+ − = +
b gb g b g
mol SO2
Total dry gas in stack (Assume no CO2, O2 , or N2 is absorbed in the scrubber)
7 244 2 46 4614 0 00936 0 0842 5585 0 0842
. . . . . . .
CO O N
SO
2 2 2
2
mol dry gas
b g b g b g b g
b g b g
+ + + + = +
x x
612 5
. ppm SO dry basis in stack gas
2 b g
0 00936 0 0842
5585 0 0842
612 5
10 10
0 295 30%
6
. .
. .
.
.
.
+
+
=
×
⇒ = ⇒
x
x
x bypassed
4.79 Basis: 100 mol stack gas
C + O2 CO2
→
2H + O2
1
2
H O
2
→
S + O2 SO2
→
(mol C)
n1
(mol H)
n2
(mol S)
n3
(mol O )
n4 2
(mol O )
n4 2
3.76
100 mol
0.7566 N2
0.1024 CO2
0.0827 H O
2
0.0575 O2
0.000825 SO2
a.
b.
C balance: mol C
H balance: mol H
mol C
mol H
mol C
mol H
1
2
n
n
= =
= =
⇒ =
100 01024 10 24
100 0 0827 2 1654
10 24
1654
0 62
b gb g
b gb gb g
. .
. .
.
.
.
The C/H mole ratio of CH4 is 0.25, and that of C H
2 6 is 0.333; no mixture of the two could
have a C/H ratio of 0.62, so the fuel could not be the natural gas.
S balance: n mol S
3 = =
100 0 000825 0 0825
b gb g
. .
10 24 122 88
1654 16 71
0 2 65
122 88
16 71
7 35
2 65
142 24
100% 1
. .
. .
.
.
.
.
.
.
mol C 12.0 g 1 mol g C
mol H 1.01 g 1 mol g H
.0825 mol S 32.07 g 1 mol g S
g C g H
.9% S
No. 4 fuel oil
b gb g
b gb g
b gb g
=
=
=
U
V
|
W
|
⇒
=
× =
⇒

4-71
4.80 a. Basis: 1 mol CpHqOr
C + 02 -- CO2
2H + 1/2 O2 --H2O
S + O2 -- SO2
1 mol CpHqOr
no (mol S)
Xs (kg s/ kg fuel)
P (% excess air)
n1 (mol O2)
3.76 n1 (mol N2)
n2 (mol CO2)
n3 (mol SO2)
n4 (mol O2)
3.76 n1 (mol N2)
n5 (mol H2O (v))
Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C)
q (mol H) (1 g / mol) = q (g H) ⇒ (12 p + q + 16 r) g fuel
r (mol O) (16 g / mol) = 16 r (g O)
S in feed:
no= S)
(mol
)
X
-
(1
07
.
32
r)
16
q
p
(12
X
S
g
32.07
S
mol
1
fuel)
(g
)
X
-
(1
S)
(g
X
fuel
g
16r)
q
p
(12
s
s
s
s +
+
=
+
+
(1)
Theoretical O2:
O
mol
2
O
mol
1
O)
mol
r
(
H
mol
2
O
mol
0.5
H)
(mol
q
C
mol
1
O
mol
1
C)
(mol
p 2
2
2
−
+
required
O
mol
r)
/2
1
q
1/4
(p 2
−
+
=
% excess ⇒ n1 = (1 + P/100) (p +1/4 q – ½ r) mol O2 fed (2)
C balance: n2 = p (3)
H balance: n5 = q/2 (4)
S balance: n3 = n0 (5)
O balance: r + 2n1 = 2n2 + 2n3 + 2n4 + n5 ⇒ n4 = ½ (r+2n1-2n2-2n3-n5) (6)
Given: p = 0.71, q= 1.1, r = 0.003, Xs = 0.02 P = 18% excess air
(1) ⇒ n0 = 0.00616 mol S
(2) ⇒ n1 = 1.16 mol O2 fed
(3) ⇒ n2 = 0.71 mol CO2
(5) ⇒ n3 = 0.00616 mol SO2
(6) ⇒ n4 = 0.170 mol O2
(4) ⇒ n5 = 0.55 mol H2O
(3.76*1.16) mol N2 = 4.36 mol N2
Total moles of dry product gas = n2 + n3 + n4 + 3.76 n1=5.246 mol dry product gas
Dry basis composition
yCO2 = (0.710 mol CO2/ 5.246 mol dry gas) * 100% = 13.5% CO2
yO2 = (0.170 / 5.246) * 100% = 3.2% O2
yN2 = (4.36 / 5.246) * 100% = 83.1% N2
ySO2 = (0.00616 / 5.246) * 106
= 1174 ppm SO2

5-1
CHAPTER FIVE
5.1 Assume volume additivity
Av. density (Eq. 5.1-1):
1 0 400
0
0 600
0
0 719
ρ
ρ
= + ⇒ =
. .
.
.703 kg L .730 kg L
kg L
A
ρO
A
ρD
a. m mt m m =
kg
kg min m mass flow rate of liquid
mass of tank
at time
0
mass of
empty tank
A A
= + ⇒
−
−
= =
t

min
.
250 150
10 3
14 28
b g
b g b g
⇒ ⇒ = =

(
.
V(L / min) =
m(kg / min)
kg / L)
V
kg 1 L
min 0.719 kg
L min
ρ
14 28
19.9
b. m m(t) - mt kg
0 150 14 28 3 107
= = − =
. b g
5.2 void volume of bed: 100 2335 184 505
3 3 3
cm cm cm
− − =
. .
b g
porosity: 505 184 0 274
3 3
. .
cm void cm total cm void cm total
3 3
=
bulk density: 600 g 184 326
3
cm g cm3
= .
absolute density: 600 g 184 505 4
3
− =
.
b gcm .49 g cm3
5.3
C H
6 6 ( )
l

m (kg / min)
V = 20.0 L / min
B
B

m (kg / min)
V (L / min)

m (kg / min)
V (L / min)
T
T
( . .
. / min
V =
V
t
D h
t
m)
4
m
60 min
m
2 2
3
Δ
Δ
Δ
Δ
= = =
π π
4
55 015
0 0594
Assume additive volumes
. .
V V - V L / min = 39.4 L / min
T B
= = −
59 4 20 0
b g
. . . .
.
m V V
kg
L
L
min
kg
L
L
min
kg / min
B B T T
= ⋅ + ⋅ = + =
ρ ρ
0879 20 0 0866 39 4
517
x
m
m
kg / L)(20.0 L / min)
kg / min)
kg B / kg
B
B
= = =

( .
( .
.
0879
517
0 34
C H
7 8 ( )
l

5-2
5.4 a.
P P gh
P P gh
h = h h
P = P P g h m
1 N 1 Pa
gh
1 0 sl 1
2 0 sl 2
1 2
1 2 sl
kg
m
m
s kg m
s
N
m
sl
3 2
2 2
= +
= +
−
U
V
|
W
|
⇒ − =
F
HG
I
KJ
F
HG
I
KJ =
⋅
ρ
ρ ρ ρ
Δ e j e j b g 1 1
b.
1 1
ρ ρ ρ
sl
c
c
c
l
x x
check units!
= +
−
⇒
b g
1
kg slurry / L slurry
kg crystals / kg slurry
kg crystals / L crystals
kg liquid / kg slurry
kg liquid / L liquid
= +
L slurry
kg slurry
L crystals
kg slurry
L liquid
kg slurry
L slurry
kg slurry
= + =
c. i.) ρsl
3
P
gh
kg / m
= = =
Δ 2775
9 8066 0 200
1415
. .
b gb g
ii.)
1 1 1 1 1
ρ ρ ρ ρ ρ ρ ρ
sl
c
c
c
l
c
c l sl l
x 1- x
x
= + ⇒ −
F
HG I
KJ = −
F
HG I
KJ
b g
x
kg / m kg / m
kg / m kg / m
kg crystals / kg slurry
c
3 3
3 3
=
−
F
H
GG
I
K
JJ
−
F
H
GG
I
K
JJ
=
1
1415
1
12 1000
1
2 3 1000
1
12 1000
0 316
.
. .
.
d i
d i d i
iii.) V
m kg
1415 kg / m
L
m
L
sl
sl
sl
3 3
= = =
ρ
175 1000
1238
.
iv.) m x m kg crystals / kg slurry kg slurry kg crystals
c c sl
= = =
0 316 175 553
. .
b gb g
v.) m
kg CuSO H O kmol
249 kg
kmol CuSO
kmol CuSO H O
kg
kmol
kg CuSO
CuSO
4 2 4
4 2
4
4
=
⋅
⋅
=
553 5 1 1
1 5
159 6
1
354
. .
.
vi.) m x m kg liquid / kg slurry kg slurry kg liquid solution
l c sl
= − = =
1 0 684 175 120
b g b gb g
.
vii.) V
m kg
1.2 kg / m
L
m
L
l
l
l
3 3
= = =
ρ
120
1000
1000
100
b gd i
d.
h(m) 0.2
ρl(kg/m^3) 1200
ρc(kg/m^3) 2300
ΔP(Pa) 2353.58 2411.24 2471.80 2602.52 2747.84 2772.61 2910.35 3093.28
xc 0 0.05 0.1 0.2 0.3 0.316 0.4 0.5
ρsl(kg/m^3) 1200.00 1229.40 1260.27 1326.92 1401.02 1413.64 1483.87 1577.14
Effect of Slurry Density on Pressure Measurement
0
0.1
0.2
0.3
0.4
0.5
0.6
2300.00 2500.00 2700.00 2900.00 3100.00
Pressure Difference (Pa)
Solids
Fraction
ΔP = 2775, = 0.316
ρ

5-3
5.4 (cont’d)
e. Basis 1 kg slurry x kg crystals V m crystals
x kg crystals
kg / m
c c
3 c
c
3
: ,
⇒ =
b g d i b g
d i
ρ
1- x kg liquid V m liquid
1- x kg liquid
kg / m
kg
V V m x x
c l
3 c
l
3
sl
c l
3
c
c
c
l
b gb g d i b gb g
d i
b gd i b g
, =
=
+
=
+
−
ρ
ρ
ρ ρ
1 1
1
5.5 Assume P atm
atm = 1
PV RT V =
0.08206 m atm
kmol K
K
4.0 atm
kmol
10 mol
m mol
3
3
3
.
.
= ⇒
⋅
⋅
=
3132 1
0 0064
ρ = =
1
0 0064 10
4 5
3
mol 29.0 g 1 kg
m air mol g
kg m
3
3
.
.
5.6 a. V =
nRT
P
mol L atm
mol K
373.2 K
10 atm
L
=
⋅
⋅
=
100 0 08206
306
. .
.
b. %
.
.
error =
3.06L - 2.8L
L
b g
2 8
100% 9 3%
× =
5.7 Assume P bar
atm = 1013
.
a.
PV nRT n
bar m kmol K
25+ 273.2 K .08314 m bar
kg N
kmol
kg N
3
2
2
= ⇒ =
+ ⋅
⋅
=
10 1013 20 0
0
28 02
249
3
. . .
b g
b g
b.
PV
P V
nRT
n RT
n V
T
T
P
P
n
V
s s s s
s
s
s
s
= ⇒ = ⋅ ⋅ ⋅
n
m 273K bar 1 kmol
298.2K 1.013 bar 22.415 m STP
kg N
kmol
3
2
=
+
=
20 0 10 1013 28 02
249 kg N
3 2
. . .
b g
b g
5.8 a. R =
P V
n T
atm
1 kmol
m
273 K
atm m
kmol K
s s
s s
3 3
= = ×
⋅
⋅
−
1 22 415
8 21 10 2
.
.
b. R =
P V
n T
atm
1 lb - mole
torr
1 atm
ft
492 R
torr ft
lb - mole R
s s
s s
3 3
= =
⋅
⋅
1 760 359 05
555
.
D D

5-4
5.9 P =1 atm +
10 cm H O m
10 cm
atm
10.333 m H O
atm
2
2
2
1 1
101
= .
T = 25 C = 298.2 K , V =
2.0 m
5 min
m min = 400 L min
3
3
D .
= 0 40

m = n mol / min MW g / mol
b g b g
⋅
a.
.
m =
PV
RT
MW =
1.01 atm
0.08206 298.2 K
g min
L atm
mol K
L
min
g
mol
⋅ =
⋅
⋅
400 28 02
458
b.
.
min
m =
400 K
298.2 K
mol
22.4 L STP
g
L
min
g
mol
273 1 28 02
458
b g =
5.10 Assume ideal gas behavior: u
m
s
V m s
A m
nRT P
D 4
u
u
nR
nR
T
T
P
P
D
D
3
2 2
2
1
2
1
1
2
1
2
2
2
F
HG I
KJ = = ⇒ = ⋅ ⋅ ⋅

d i
d i π
( ) ( )
( ) ( )
2
2
2 1 1
2 1 2
2
1 2 2
60.0 m 333.2K 1.80 1.013 bar 7.50 cm
T P D
u u 165 m sec
T P D sec 300.2K 1.53 1.013 bar 5.00 cm
+
= = =
+
5.11 Assume ideal gas behavior: n
PV
RT
atm 5 L
300 K
mol
L atm
mol K
= =
+
=
⋅
⋅
100 100
0 08206
0 406
. .
.
.
b g
MW g 0.406 mol g mol Oxygen
= = ⇒
130 32 0
. .
5.12 Assume ideal gas behavior: Say mt = mass of tank, n mol
g = of gas in tank
N : 37.289 g m n g mol
CO : 37.440 g m n mol
n mol
m g
2 t g
2 t g
g
t
= +
= +
U
V
|
W
|⇒
=
=
28 02
44.1 g
0 009391
37 0256
. .
.
b g
b g
unknown: MW
g
mol
g mol Helium
=
−
= ⇒
37 062 37 0256
0 009391
39
. .
.
.
b g
5.13 a. .
V cm STP min
V liters 273K mm Hg cm
t min 296.2K mm Hg 1 L
V
t
std
3
3 3
763 10
760
9253
b g = =
Δ
Δ
Δ
Δ
φ
φ

.
.
.
. .
V cm STP min
straight line plot
V
std
3
std
b g
50 139
9 0 268
12 0 370
0 031 0 93
U
V
|
|
W
|
| = +
E
b.
.
min
.
/
/ . .
V
mol N liters STP
mole
cm
L
cm min
= 0.031 224 cm min
std
2 3
3
= =
+ =
0 010 22 4
1
10
1
224
0 93 7 9
3 3
b g
d i
φ

5-5
5.14 Assume ideal gas behavior ρ kg L
n kmol M(kg / kmol)
V L
PM
RT
n
V
P
RT
b g b g
b g
=
=
====
V cm s V cm s V P M T P M T
2
3
1
3
1 1 1 2 2 2 1
d i d i
= ⋅
F
HG I
KJ =
ρ
ρ
1
2
1 2
1 2
a. V
cm
s
mm Hg 28.02 g mol 323.2K
2.02 g mol 298.2K
cm s
H
3
3
2
=
L
NM O
QP =
350
758
1800 mm Hg
881
1 2
b. M 0.25M 0.75M g mol
CH C H
4 3
= + = + =
8
0 25 16 05 0 75 4411 3710
. . . . .
b gb g b gb g
V
cm
s 1800
cm s
g
3
=
L
NM O
QP =
350
758 28 02 3232
3710 298 2
205
3
1 2
b gb gb g
b gb gb g
. .
. .
5.15 a.
Δh
b.

. .
. /
n
PV
RT
V =
R h
t
m m
7.4 s
s
min
m min
CO
2 2
3
2
= ⇒ = = × −
π π
Δ
Δ 4
0 012 12 60
11 10
2
3
d i
3
2
-3 3
CO m atm
kmol K
755 mm Hg 1 atm 1.1 10 m / min 1000 mol
n 0.044 mol/min
760 mm Hg 300 K 1 kmol
0.08206 ⋅
⋅
×
= =

5.16
.

m kg / h
air
n (kmol / h)
air
= 10 0

/
n (kmol / h)
y (kmol CO kmol)
CO 2
2
/

V = 20.0 m h
CO
3
2
n (kmol / h)
150 C, 1.5 bar
CO
o
2
Assume ideal gas behavior

.
.
n
kg
h
kmol
29.0 kg air
kmol air / h
air = =
10 0 1
0 345

. . /
. /
n
PV
RT
bar
8.314
kPa
1 bar
m h
423.2 K
kmol CO h
CO m kPa
kmol K
3
2
2 3
= = =
⋅
⋅
15 100 20 0
0853
y
kmol CO h
kmol CO h + kmol air h
CO
2
2
2
× = × =
100%
0853
0853 0 345
100% 712%
. /
. / . /
.
b g
Reactor
soap

5-6
5.17 Basis: Given flow rates of outlet gas. Assume ideal gas behavior

m (kg / min)
0.70 kg H O / kg
0.30 kg S / kg
1
2
311 m 83 C, 1 atm
n (kmol / min)
0.12 kmol H O / kmol
0.88 kmol dry air / kmol
3 o
3
2
/ min,

/ min)
n (kmol air / min)
V (m
C, - 40 cm H O gauge
2
2
3
o
2
167

m (kg S / min)
4
a.
3
3
3
1 atm 311 m kmol K
n 10.64 kmol min
356.2K min 0.08206 m atm
⋅
= =
⋅

2
2 1
1
0.12 kmol H O 18.02 kg
10.64 kmol
H O balance : 0.70 m
kmol kmol
min
m 32.9 kg min milk
=
⇒ =

S olids balance 0.30 32.2 kg min m m kg S min
b g b g
: .
= ⇒ =
4 4 9 6
( )
2 2
Dry air balance : n 0.88 10.64 kmol min n 9.36 kmol min air
= ⇒ =

( )
3
2
2
2
3
9.36 kmol 0.08206 m atm 440K 1033 cm H O
V
min kmol K 1033 40 cm H O 1 atm
352 m air min
⋅
=
⋅ −
=

3 3
air
air 2 2
4
V (m /s) 352 m 1 min
u (m/min)= 0.21 m/s
A (m ) min 60 s (6 m)
π
= =
⋅

b. If the velocity of the air is too high, the powdered milk would be blown out of the reactor
by the air instead of falling to the conveyor belt.
5.18 SG
M
M
kg / kmol
kg / kmol
CO
CO
air
PM
RT
PM
RT
CO
air
2
2
CO2
air
2
= = = = =
ρ
ρ
44
29
152
.
5.19 a. x x
CO air
2
= = − =
0 75 1 0 75 0 25
. . .
Since air is 21% O , x mole% O
2 O 2
2
= = =
( . )( . ) . .
0 25 0 21 0 0525 525
b. m = n x M
atm
0.08206
1.5 3 m
K
kmol CO
kmol
kg CO
kmol CO
kg
CO CO CO m atm
kmol K
3
2 2
2
2 2 2 3
⋅ ⋅ =
× ×
=
⋅
⋅
1 2
298 2
0 75 44 01
12
b g
.
. .
More needs to escape from the cylinder since the room is not sealed.

5-7
5.19 (cont’d)
c. With the room closed off all weekend and the valve to the liquid cylinder leaking, if a
person entered the room and closed the door, over a period of time the person could die
of asphyxiation. Measures that would reduce hazards are:
1. Change the lock so the door can always be opened from the inside without a key.
2. Provide ventilation that keeps air flowing through the room.
3. Install a gas monitor that sets off an alarm once the mole% reaches a certain amount.
4. Install safety valves on the cylinder in case of leaks.
5.20 n
kg 1 kmol
44.01 kg
kmol CO
CO 2
2
= =
157
0 357
.
.
Assume ideal gas behavior, negligible temperature change T C K
= ° =
19 292 2
.
b g
a.
P V
P V
n RT
n 0.357 RT
n
n 0.357
P
P
102kPa
3.27 10 kPa
n kmol air in tank
1
2
1
1
1
1
1
2
3
=
+
⇒
+
= =
×
⇒ =
b g
1 0 0115
.
b. V
n RT
P
0.0115 kmol 292.2 K 8.314 m kPa
102kPa kmol K
L
m
L
tank
1
1
3
3
= =
⋅
⋅
=
10
274
3
ρf
2
g CO +11.5 mol air (29.0 g air / mol)
274 L
g / L
=
⋅
=
15700
585
.
c. CO2 sublimates ⇒ large volume change due to phase change ⇒ rapid pressure rise.
Sublimation causes temperature drop; afterwards, T gradually rises back to room
temperature, increase in T at constant V ⇒ slow pressure rise.
5.21 At point of entry, P ft H O in. Hg 33.9 ft H O in. Hg in. Hg
2 2
1 10 29 9 28 3 371
= + =
b gb g
. . . .
At surface, P in. Hg, V bubble volume at entry
2 2
28 3
= =
.
Mean Slurry Density:
1 x x
g / cm g / cm
sl
solid
solid
solution
solution
3 3
ρ ρ ρ
= + = +
0 20
12 100
080
100
.
( . )( . )
.
( . )
cm
g
g
cm
2.20 lb
g
ton
1 lb
cm
264.17 gal
ton / gal
3
sl 3
3
= ⇒ =
×
= ×
−
−
0 967
103
1000
5 10 10
4 3 10
4 6
3
.
.
.
ρ
a.
300
4 3 10
40 0 534 7
492
29 9
371
2440
3
ton
hr
gal
ton
ft (STP)
1000 gal
R
R
in Hg
in Hg
ft hr
3 o
o
3
.
. . .
.
/
×
=
−
b.
P V
P V
nRT
nRT
V
V
P
P
D 1.31D D mm
2 2
1 1
2
1
1
2
4
3
D
2
3
4
3
D
2
3 2
3
1
3
D 2 mm
2
2
1
1
= ⇒ = ⇒ = ⇒ = =
=
==
π
π
e j
e j
371
28 3
2 2
.
.
.
% change =
2.2 - 2.0 mm
mm
b g
2 0
100 10%
.
× =

5-8
5.22 Let B = benzene
n n n moles in the container when the sample is collected, after
the helium is added, and after the gas is fed to the GC.
1 2 3
, , =
n moles of gas injected
inj =
n n n moles of benzene and air in the container and moles of helium added
B air He
, , =
n m moles, g of benzene in the GC
BGC BGC
, =
y mole fraction of benzene in room air
B =
a. P V n RT (1 condition when sample was taken): P = 99 kPa, T K
1 1 1 1 1 1
= ≡ = 306
n
kPa
101.3
L
306 K
mol K
.08206 L atm
mol = n n
1 kPa
atm
air B
=
⋅
⋅
= +
99 2
0 078
.
P V n RT (2 condition when charged with He): P = 500 kPa, T K
2 2 2 2 2 2
= ≡ = 306
n
kPa
101.3
L
306 K
mol K
.08206 L atm
mol = n + n n
2 kPa
atm
air B He
=
⋅
⋅
= +
500 2
0 393
.
P V n RT (3 final condition in lab): P = 400 kPa, T K
3 3 3 3 3 3
= ≡ = 296
n
kPa
101.3
L
296 K
mol K
.08206 L atm
mol = (n n n n
3 kPa
atm
air B He inj
=
⋅
⋅
= + + −
400 2
0 325
. )
n = n n mol
inj 2 3
− = 0 068
.
n n
n
n
mol
0.068 mol
m g B) mol
78.0 g
m
B BGC
2
inj
BGC
BGC
= × = = ⋅
0 393 1
0 0741
. (
.
y (ppm) =
n
n
m
0.078
m
B
B
1
BGC
BGC
× =
⋅
× = × ⋅
10
0 0741
10 0 950 10
6 6 6
.
.
9 0 950 10 0 656 10 0 623
1 0 950 10 0 788 10 0 749
0 950 10 0 910 10 0864
6 6
6 6
6 6
am: y ppm
pm: y ppm
5 pm: y ppm
The avg. is below the PEL
B
B
B
= × × =
= × × =
= × × =
U
V
|
|
W
|
|
−
−
−
( . )( . ) .
( . )( . ) .
( . )( . ) .
b. Helium is used as a carrier gas for the gas chromatograph, and to pressurize the container
so gas will flow into the GC sample chamber. Waiting a day allows the gases to mix
sufficiently and to reach thermal equilibrium.
c. (i) It is very difficult to have a completely evacuated sample cylinder; the sample may
be dilute to begin with. (ii) The sample was taken on Monday after 2 days of inactivity
at the plant. A reading should be taken on Friday. (iii) Helium used for the carrier gas is
less dense than the benzene and air; therefore, the sample injected in the GC may be He-
rich depending on where the sample was taken from the cylinder. (iv) The benzene may
not be uniformly distributed in the laboratory. In some areas the benzene concentration
could be well above the PEL.

5-9
5.23 Volume of balloon m m3
= =
4
3
10 4189
3
πb g
Moles of gas in balloon
n kmol
m 492 R atm 1 kmol
535 R 1 atm m STP
kmol
3
3
b g b g
=
°
°
=
4189 3
22 4
5159
.
.
a. He in balloon:
m kmol kg kmol kg He
= ⋅ =
5159 4 003 2065
. .
b g b g
m
kg m
s
N
1 kg m / s
N
g 2 2
=
⋅
=
2065 9 807 1
20 250
.
,
b.
P V n RT
P V n RT
n
P
P
n
atm
3 atm
kmol kmol
gas in balloon gas
air displaced air
air
air
gas
gas
d i
d i
=
=
⇒ = ⋅ = ⋅ =
1
5159 172 0
. .
Fbuoyant
Fcable
Wtotal
F W
kmol 29.0 kg 9.807 m
1 kmol s
N
1
N
buoyant air displaced 2 kg m
s2
= = =
⋅
172 0 1
48 920
.
,
Since balloon is stationary, F1 0
=
∑
F F W N
kg 9.807 m
s
N
1
cable buoyant total 2 kg m
s2
= − = −
+
=
⋅
48920
2065 150 1
27 20
b g ,
c. When cable is released, F = 27200 N M a
net tot
A =
d i
⇒ =
⋅
=
a
N 1 kg m / s
2065+150 kg N
m s
2
2
27200
12 3
b g .
d. When mass of displaced air equals mass of balloon + helium the balloon stops rising.
Need to know how density of air varies with altitude.
e. The balloon expands, displacing more air ⇒ buoyant force increases ⇒ balloon rises
until decrease in air density at higher altitudes compensates for added volume.
5.24 Assume ideal gas behavior, P atm
atm = 1
a.
3
3
N N
N N c c c
c
5.7 atm 400 m / h
P V
P V P V V 240 m h
9.5 atm
P
= ⇒ = = =
b. Mass flow rate before diversion:
( )
3
3 6
3
400 m 273 K 5.7 atm 1 kmol 44.09 kg kg C H
4043
h
h 303 K 1 atm 22.4 m STP kmol
=

5-10
5.24 (cont’d)
Monthly revenue:
( )( )( )( )
4043 kg h 24 h day 30 days month $0.60 kg $1,747,000 month
=
c. Mass flow rate at Noxious plant after diversion:
3
3
400 m 273 K 2.8 atm 1 kmol 44.09 kg
1986 kg hr
hr 303 K 1 atm 22.4 m kmol
=
( )
Propane diverted 4043 1986 kg h 2057 kg h
= − =
5.25 a. P y P = 0.35 (2.00 atm) = 0.70 atm
He He
= ⋅ ⋅
P y P = 0.20 (2.00 atm) = 0.40 atm
CH CH
4 4
= ⋅ ⋅
P y P = 0.45 (2.00 atm) = 0.90 atm
N N
2 2
= ⋅ ⋅
b. Assume 1.00 mole gas
0 35 140
0 20 321
0 45 12 61
17 22
321
0186
. .
. .
. .
.
.
.
mol He
4.004 g
mol
g He
mol CH
16.05 g
mol
g CH
mol N
28.02 g
mol
g N
g mass fraction CH
g
17.22 g
4 4
2 2
4
F
HG I
KJ =
F
HG I
KJ =
F
HG I
KJ =
U
V
|
|
|
W
|
|
|
⇒ = =
c. MW
g of gas
mol
g / mol
= = 17 2
.
d. ρgas
m atm
kmol K
3
m
V
n MW
V
P MW
RT
atm kg / kmol
0.08206 K
kg / m
3
= = = = =
⋅
⋅
d i d i b gb g
e jb g
2 00 17 2
3632
115
. .
.
.
5.26 a. It is safer to release a mixture that is too lean to ignite.
If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the
C3H8 mole fraction can drop below the UFL, thereby producing a fire hazard.
b.
fuel-air mixture
(
. /

n mol / s)
y mol C H mol
n mol C H / s
1
C H 3 8
C H 3 8
3
3
8
8
0 0403
150
=
=
(
. /
n mol / s)
mol C H mol
3
3 8
0 0205
diluting air
(
n mol / s)
2

n
mol C H
s
mol
0.0403 mol C H
mol / s
1
3 8
3 8
= =
150
3722
Propane balance 150 = 0.0205 n n mol / s
3 3
:
⋅ ⇒ = 7317

5-11
5.26 (cont’d)
Total mole balance n n n n mol air / s
1 2 3 2
:
+ = ⇒ = − =
7317 3722 3595
c. .
n n mol / s
2 2 min
= =
13 4674
b g
. .
/
. .
. /

.
V
4674 mol / s m Pa
mol K
K
131,000 Pa
m s
V
3722 mol
s
m Pa
mol K
K
110000 Pa
m s
V
V
m diluting air
m fuel gas
2
3
3
1
3
3
2
1
3
3
=
⋅
⋅
=
=
⋅
⋅
=
U
V
|
|
W
|
|
=
8 314 398 2
118
8 314 298 2
839
141
y
mol / s
n n
mol / s
mol / s + 4674 mol / s
2
1 2
=
+
= × =
150 150
3722
100% 18%

.
b g
d. The incoming propane mixture could be higher than 4.03%.
If ,
min
n n
2 2
= b g fluctuations in the air flow rate would lead to temporary explosive
conditions.
5.27 ( )( )
Basis: 12 breaths min 500 mL air inhaled breath 6000 mL inhaled min
=
24o
C, 1 atm
6000 mL / min 37o
C, 1 atm

.
.
n (mol / min)
0.206 O
N
H O
in
2
2
2
0 774
0 020
blood

.
.
.
n (mol / min)
0.151 O
CO
N
H O
out
2
2
2
2
0 037
0 750
0 062
a. .
n
6000 mL 1 L 273K 1 mol
min 10 mL 297K 22.4 L STP
mol min
in 3
= =
b g 0 246
N balance: 0.774 n n mol exhaled min
2 out out
b gb g
0 246 0 750 0 254
. . .
= ⇒ =
O transferred to blood: 0.246 mol O 32.0 g mol
g O
2 2
2
b gb g b gb gb g
0 206 0 254 0151
0 394
. . . min
. min
−
=
CO transferred from blood: mol CO 44.01 g mol
0.414 g CO
2 2
2
0 254 0 037
. . min
min
b gb gb g
=
H O transferred from blood:
0.246 mol H O 18.02 g mol
g H O
2
2
2
0 254 0 062 0 020
0195
. . . min
. min
b gb g b gb gb g
−
=
lungs

5-12
5.27 (cont’d)
PV
PV
n RT
n RT
V
V
n
n
T
T
0.254 mol min
0.246 mol min
310K
297K
mL exhaled ml inhaled
in
out
in in
out out
out
in
out
in
out
in
=
⇒ =
F
HG I
KJF
HG I
KJ =
F
HG I
KJF
HG I
KJ = 1078
.
b. 0 414 0195 0 394 0 215
. . . .
g CO lost min g H O lost min g O gained min g min
2 2 2
b g b g b g
+ − =
5.28
2
Ts (K)
Ms (g/mol)
Ps (Pa)
STACK
Ta (K)
Ma (g/mol)
Pc (Pa)
L( )
M
Ideal gas:
PM
RT
ρ =
a. D gL gL
P M
RT
gL
P M
RT
gL
P gL
R
M
T
M
T
combust. stack
a a
a
a s
s
a a
a
a
s
= − = − = −
L
NM O
QP
ρ ρ
b g b g
b. M g mol
s = + + =
018 441 0 02 32 0 080 28 0 310
. . . . . . .
b gb g b gb g b gb g , T 655K
s = ,
P mm Hg
a = 755
M g mol
a = 29 0
. , T K
a = 294 , L 53 m
=
D
mm Hg 1 atm 53.0 m 9.807 m kmol - K
mm Hg s 0.08206 m atm
kg kmol
294K
kg kmol
655K
N
1 kg m / s
N
m
cm H O
1.013 10 N m
cm H O
2 3
2 2
2
5 2
2
=
−
× −
L
NM O
QP×
⋅
F
HG I
KJ =
×
=
755
760
29 0 310 1 323 1033
33
. .
.
5.29 a. ρ
ρ
ρ
= = =
=
=======
P MW
RT
MW /mol
air
CCl2O
CCl2O
b g 98.91 g
98 91
29 0
341
.
.
.
Phosgene, which is 3.41 times more dense than air, will displace air near the ground.
b. V
D L
4
cm- 2 0.0559 cm cm cm
tube
in 3
= = =
π π
b g b g b g
2
2
4
0 635 150 322
. . .
m V
cm L
10 cm
atm
0.08206
g / mol
K
0.0131 g
CCl2O CCl2O
tube
3
3 3 L atm
mol K
= ⋅ = =
⋅
⋅
ρ
322 1 1 98 91
296 2
. .
.
c. n
cm g
cm
mol
mol CCl O
CCl O(l)
3
3 2
2
=
×
=
322 137 1000
98 91 g
0 0446
. . .
.
.
L(m)

5-13
5.29 (cont’d)
n
PV
RT
atm ft
296.2K
L
ft
mol K
L atm
mol air
air
3
3
= =
⋅
⋅
=
1 2200 28 317
08206
2563
.
.
n
n
ppm
CCl O
air
2
= = × =
−
0 0446
2563
17 4 10 17 4
6
.
. .
The level of phosgene in the room exceeded the safe level by a factor of more than 100.
Even if the phosgene were below the safe level, there would be an unsafe level near the
floor since phosgene is denser than air, and the concentration would be much higher in
the vicinity of the leak.
d. Pete’s biggest mistake was working with a highly toxic substance with no supervision or
guidance from an experienced safety officer. He also should have been working under a
hood and should have worn a gas mask.
5.30 CH O CO H O
2 2
4 2
2 2
+ → +
C H O CO H O
2 2 2
6 2
7
2
2 3
+ → +
C H O CO H O
3 2 2
8 2
5 3 4
+ → +
1450 m / h @ 15 C, 150 kPa
n (kmol / h)
3 o
1

086
. , ,
CH 0.08 C H 0.06 C H
4 2 6 3 8

n (kmol air / h)
2
8% excess, 0.21 O 0.79 N
2 2
,

.
.
n
m 273.2K kPa 1 kmol
h 288.2K 101.3 kPa m STP
kmol h
1
3
3
=
+
=
1450 1013 150
22 4
152
b g
b g
Theoretical O2:
152
086 0 08 0 06 349 6
kmol
h
2 kmol O
kmol CH
3.5 kmol O
kmol C H
5 kmol O
kmol C H
kmol h O
2
4
2
2 6
2
3 8
2
. . . .
F
HG I
KJ+
F
HG I
KJ+
F
HG I
KJ
L
NMM
O
QPP=
Air flow: V
kmol O 1 kmol Air m STP
h kmol O kmol
.0 10 m STP h
air
2
3
2
4 3
. . .
.
= = ×
108 349 6 22 4
0 21
4
b g b g b g

5-14
5.31 Calibration formulas
T 25.0; R 14
T
= =
b g, T 35.0, R 27 T C 0.77R 14.2
T T
= = ⇒ ° = +
b g b g
P 0; R 0
g p
= =
d i, P 20.0, R 6 P kPa 3.33R
g r gauge p
= = ⇒ =
d i b g

V 0; R 0
F p
= =
d i,
V 2.0 10 , R 10 V m h 200R
F
3
F F
3
F
= × = ⇒ =
d i d i

V 0; R 0
A A
= =
d h,
V 1.0 10 , R 25 V m h 4000R
A
5
A A
3
A
= × = ⇒ =
d i d i
/
V (m h), T, P
F
3
g

/
/
/
/
/
n (kmol / h)
x (mol CH mol)
x (mol C H mol)
x (mol C H mol)
x (mol n - C H mol)
x (mol i - C H mol)
F
A 4
B 2 6
C 3 8
D 4 10
E 4 10
CH O CO H O
C H O CO H O
C H O CO H O
C H O CO H O
4 2 2 2
6 2 2 2
8 2 2 2
10 2 2 2
+ → +
+ → +
+ → +
+ → +
2 2
7
2
2 3
5 3 4
13
2
4 5
2
3
4
( /
V m h) (STP)
A
3

n
V m h 273.2K P 101.3 kPa 1 kmol
T 273.2 K 101.3 kPa 22.4 m STP
0.12031V P 101.3
T + 273
kmol
h
F
F
3
g
3
F g
=
+
+
=
+ F
HG I
KJ
d i d i
b g b g
d i
b g
Theoretical O2:

n n 2x 3.5x 5x 6.5 x x kmol O req. h
o Th F A B C D E 2
2
d i b g
c h
= + + + +
Air feed: n
n kmol O req. 1 kmol air 1 P 100 kmol feed
h 0.21 kmol O 1 kmol req.
4.762 1
P
100
n
A
o Th 2 x
2
x
o Th
2
2

=
+
= +
F
HG I
KJ
d i b g
d i

V n kmol air h 22.4 m STP kmol 22.4n m STP h
A a
3
A
3
= =
b g b g
d i b g
R
T T(C
) R
p P
g(kP
a) R
f xa xb xc xd xe P
X(%
) nF nO
2,th nA Vf(m
3/h) Va(m
3/h) R
a
23.1 32.0 7.5 25.0 7.25 0.81 0.08 0.05 0.04 0.02 15 72.2 183.47 1004.74 1450 22506.2 5.63
7.5 20.0 19.3 64.3 5.8 0.58 0.31 0.06 0.05 0.00 23 78.9 226.4 1325.8 1160 29697.8 7.42
46.5 50.0 15.8 52.6 2.45 0.00 0.00 0.65 0.25 0.10 33 28.1 155.2 983.1 490 22022.3 5.51
21 30.4 3 10.0 6 0.02 0.4 0.35 0.1 0.13 15 53.0 248.1 1358.9 1200 30439.2 7.6
23 31.9 4 13.3 7 0.45 0.12 0.23 0.16 0.04 15 63.3 238.7 1307.3 1400 29283.4 7.3
25 33.5 5 16.7 9 0.5 0.3 0.1 0.04 0.06 15 83.4 266.7 1460.8 1800 32721.2 8.2
27 35.0 6 20.0 10 0.5 0.3 0.1 0.04 0.06 15 94.8 303.2 1660.6 2000 37196.7 9.3

/
/
n (kmol / h)
0.21 mol O mol
0.79 mol N mol
A
2
2

5-15
5.32 NO O NO
+ ⇔
1
2 2 2
1 mol
0 20
380
. mol NO / mol
0.80 mol air / mol
0.21 O
0.79 N
P kPa
2
2
0
R
S
|
T
|
U
V
|
W
|
=
n (mol NO)
n (mol O
n (mol N
n (mol NO
P (kPa)
1
2 2
3 2
4 2
f
)
)
)
a. Basis: 1.0 mol feed
90% 010 0 20 0 020
NO conversion: n mol NO NO reacted = 0.18 mol
1 = = ⇒
. ( . ) .
O balance: n
mol NO mol O
mol NO
mol O
2 2
2
2
= − =
080 0 21
018 05
0 0780
. ( . )
. .
.
N balance: n mol N
2 3 2
= =
080 0 79 0 632
. ( . ) .
n
mol NO mol NO
1 mol NO
mol NO n n n n n mol
4
2
2 f 1 2 3 4
= = ⇒ = + + + =
018 1
018 0 91
.
. .
y
mol NO
0.91 mol
mol NO
mol
y
mol O
mol
y
mol N
mol
y
mol NO
mol
NO
O
2
N
2
NO
2
2 2 2
= =
= = =
0 020
0 022
0 086 0 695 0198
.
.
. . .
P V
P V
=
n RT
n RT
P P
n
n
kPa
0.91 mol
mol
f
0
f
0
f 0
f
0
⇒ = =
F
HG I
KJ =
380
1
346 kPa
b. n = n
P
P
mol)
360 kPa
mol
f 0
f
0
= =
( .
1
380 kPa
0 95
n n
n (mol NO)
n mol O
n (mol N )
n mol NO
n
n mol NO n mol O n mol N
n mol NO y , y y y
i i0 i
1
2 2
3 2
4 2
f
1 2 2 3 2
4 2 NO O N NO
2 2 2
= +
E
= −
= −
=
=
= − = ⇒ =
⇒ = = =
= ⇒ = = = =
υ ξ
ξ
ξ
ξ
ξ ξ
0 20
0 21 080 05
0 79 080
1 05 0 95 010
010 0118 0 632
010 0105 0124 0 665 0105
.
( ) ( . )( . ) .
( . )( . )
( )
. . .
. , . , . ,
. . . , . , .
NO conversion =
0.20- n1
b g
0 20
100% 50%
.
× =
P (atm) =
360 kPa
101.3
atm
kPa
atm
= 355
.
K
(y P
y P y P
(y
y y P
105
0.124 3.55
atm
p
NO
NO O
0.5
NO
NO O
0.5 0.5
2
2
2
2
1
2
= = = =
)
( )( )
)
( )( )
.
( . )
.
0
0105
151
0.5 0.5
b g b g

5-16
5.33
Liquid composition
100 kg liquid
49.2 kg M kmol
112.6 kg
kmol M kmol M / kmol
29.6 kg D kmol
147.0 kg
kmol D 0.221 kmol D / kmol
21.2 kg B kmol
78.12 kg
kmol B 0.298 kmol B / kmol
0.909 kmol
:
. .
.
.
⇒ =
= ⇒
=
1
0 437 0 481
1
0 201
1
0 271
a. Basis 1 kmol C H fed
6 6
:
V (m @ 40 C, 120 kPa
n (kmol)
1
3 o
1
)
0 920
. HCl
0.080 Cl2
1 7812
kmol C H kg)
6 6 ( .
n (kmol Cl
0 2 ) n (kmol)
2
0 298
0 481
0 221
.
.
.
C H
C H Cl
C H Cl
6 6
6 5
6 4 2
C H Cl C H Cl + HCl C H Cl Cl C H Cl + HCl
6 6 2 6 5 6 5 2 6 5 2
+ → + →
C balance
1 kmol C H kmol C
1 kmol C H
n
n kmol
6 6
6 6
2
2
: . . .
.
6
0 298 6 0 481 6 0 221 6
100
= × + × + ×
⇒ =
H balance
1 kmol C H kmol H
1 kmol C H
n
n n kmol
6 6
6 6
1
2 1
: . ( )
. . . .
6
0 920 1
0 298 6 0 481 5 0 221 4 100
=
+ × + × + × ⇒ =
b g
V
n RT
P
kmol
kPa
kPa
atm
m atm
kmol K
K
m
V
m
m
kg B
m kg B
1
1
3
3
1
B
3
3
= =
⋅
⋅
=
⇒ = =
100
120
1013
1
0 08206 3132
217
217
7812
0 278
. . . .
.
.
.
. /
b. ( / )

V m s) u(m / s) A(m u
d
d =
4 V
u
gas
3 2
2
2 gas
= ⋅ = ⋅ ⇒
⋅
⋅
π
π
4
d =
4m kg B) 0.278 m
kg B
s
m
min
60 s
cm
m
m (cm
d(cm) 2.43 m
2 B0
3 2
2 B0
2
B0
(
min ( )
. )

π 10
1 10
590
4
1
2
=
⇒ = ⋅b g
c. Decreased use of chlorinated products, especially solvents.

5-17
5.34
374
. SCMM V m @900 C, 604 mtorr
b
3
( / min) D
60% DCS conversion
n (mol DCS / min)
n (mol N O / min)
n (mol N / min)
n (mol HCl(g) / min)
n (mol / min)
1
2 2
3 2
4
b

U
V
|
W
|
SiH Cl N O SiO N HCl
2 2(g) 2 (g) 2(s) 2(g) (g)
+ → + +
2 2 2
a.
.
n
m (STP)
min
mol
22.4 m (STP)
mol / min
a
3
3
= =
374 10
167
3
60% conversion: n = 1- 0.60
mol DCS
mol
/ min mol DCS / min
1

.
.
b g b g
0 220
167 mol 14 7
F
HG I
KJ =
DCS reacted: 0.60
mol DCS
mol DCS reacted / min
b gb gb g
0 220 167 22 04
.
min
.
=
N O balance: n
mol N O
min
DCS mol N O
mol DCS
mol N O / min
2 2
2
2
2
.
min
.
=
− =
0 780 167
22.04 mol 2
8618
b g
N balance: n
DCS mol N
mol DCS
mol N
2 3
2
2

min
. / min
= =
22.04 mol 2
44 08
HCl balance: n
DCS mol HCl
mol DCS
mol HCl / min
4

min
.
= =
22.04 mol 2
44 08

.
/ min
n n n n n mol / min
V
n RT
P min
torr
mol K
m
L
K
0.604 torr
.29 10 m
B 1 2 3 4
B
B
3
4 3
= + + + =
⇒ = =
⋅
⋅
= ×
189
189 mol 62.36 L 0 001 1173
2
b. 1
DCS DCS
B
n 14.7 mol DCS/min
p x P= P= 604 mtorr=47.0 mtorr
n 189 mol/min
= ⋅ ⋅

2 2
2 2
N O N O
B
n 86.2 mol N O/min
p x P= P= 604 mtorr=275.5 mtorr
n 189 mol/min
= ⋅ ⋅

( )( )
2
0.65
-8 0.65 -8 5 2
DCS N O 2
mol SiO
r=3.16 10 p p 3.16 10 47.0 275.5 5.7 10
m s
−
× ⋅ ⋅ = × = ×
⋅
2
5 10
2
2 6 3
SiO
(Table B.1)
5
5.7 10 mol SiO
MW 60 s 120 min 60.09 g/mol 10 A
c h(A)=r t
min 1 m
m s 2.25 10 g/m
=1.1 10 A
ρ
−
×
⋅ ⋅ =
⋅ ×
×
.

The films will be thicker closer to the entrance where the lower conversion yields higher
pDCS and pN O
2
values, which in turn yields a higher deposition rate.
(
n mol / min)
0.220 DCS
0.780 N O
a
2

5-18
5.35
Basis: 100 kmol dry product gas
n (kmol
m (kg C H )
1
1 x y
C H )
x y
a. N balance: 0.79n n =106.6 kmol air
2 2 2
= ⇒
0842 100
. ( )
O balance: 2 0.21n n n kmol H O
2 3 3 2
b g b g b g
= + + ⇒ =
100 2 0105 2 0 053 1317
. . .
C balance:
n kmol C H x kmol C
kmol C H
n x =10.5 1
1 x y
x y
1
d i b g
d i b g b g
= ⇒
100 0105
.
H balance: n y = 2n n y 2
1 3
n
1
3
==== =
=13 17
26 34
.
. b g
Divide 2 by 1
y
x
mol H / mol C
b g b g⇒ = =
26 34
105
2 51
.
.
.
O fed: 0.21 106.6 kmol air kmol
2 b g= 22 4
.
O in excess = 5.3 kmol Theoretical O = 22.4 -5.3 kmol =17.1 kmol
% excess =
5.3 kmol O
17.1 kmol O
excess air
2 2
2
2
⇒
× =
b g
100% 31%
b. V
N m (STP)
kmol
kPa
kPa
K
273 K
m
2
2
3
3
= =
106.6 kmol 22 4 1013
98
303
2740
. .
m =
n x kmol C 12.0 kg
kmol
n y kmol H kg
kmol
m kg
1
1 1
n y=26.34
n x=10.5
1
1
1
b g b g
+ ===== =
101
152 6
.
.
V
m
=
2740 m air
kg fuel
m air
kg fuel
2
1
3 3
152 6
18 0
.
.
=
5.36 3 6 1 2 4 4
N H xH x)N x)NH
2 4 2 2 3
→ + + + −
( (
a. 0 ≤ ≤
x 1
b. n
L kg
L
1 kmol
32.06 kg
kmol
N H
2 4
= =
50 082
128
.
.
n kmol N H
x kmol H
3 kmol N H
x kmol N
3 kmol N H
x kmol NH
3 kmol N H
6x 1 2x 4 4x x + 2.13 kmol
product 2 4
2
2 4
2
2 4
3
2 4
= +
+
+
−
L
NM O
QP
= + + + − =
128
6 1 2 4 4
128
3
1707
.
.
.
b g b g
b g
)
V (m
2
3
n (kmol air)
0.21 O
N
2
2
2
o
30 C, 98 kPa
0 79
.
100
0105
kmol dry gas
n (kmol H O)
CO
0.053 O
0.842 N
2
2
2
3 2
.
R
S
|
T
|
U
V
|
W
|

5-19
5.36 (cont’d)
x nproduct Vp (L)
0 2.13 15447.92
0.1 2.30 16685.93
0.2 2.47 17923.94
0.3 2.64 19161.95
0.4 2.81 20399.96
0.5 2.98 21637.97
0.6 3.15 22875.98
0.7 3.32 24113.99
0.8 3.50 25352.00
0.9 3.67 26590.01
1 3.84 27828.02
Volume of Product Gas
0.00
5000.00
10000.00
15000.00
20000.00
25000.00
30000.00
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
V
(L)
c. Hydrazine is a good propellant because as it decomposes generates a large number of
moles and hence a large volume of gas.
5.37
(
m g A / h)
A

V m / h
air
3
c h
a. (i) Cap left off container of liquid A and it evaporates into room, (ii) valve leak in
cylinder with A in it, (iii) pill of liquid A which evaporates into room, (iv) waste
containing A poured into sink, A used as cleaning solvent.
b.
m
kg A
h
m
kg A
h
V
m
h
C
kg A
m
A
in
A
out
air
3
A 3
F
HG I
KJ =
F
HG I
KJ =
F
HG
I
KJ F
HG I
KJ
c. y
mol A
mol air
C V
M n
A
A
g A
m
A
g A
mol air
3
= =
⋅
⋅
e j
e j
===================
=
⋅
=
C
m
k V
n
PV
RT
A
A
air
air

;
y
m
k V
RT
M P
A
A
air A
=
⋅

d. y m g/ h
A A
= × =
−
50 10 90
6
.
. .
/
min
V
m
ky
RT
M P
g / h
0.5 101.3 10 Pa
K
104.14 g / mol
m h
air
A
A A
m Pa
mol K
3
3
3
d h d i
= =
× ×
=
−
⋅
⋅
9 0
50 10
8314 293
83
6
Concentration of styrene could be higher in some areas due to incomplete mixing (high
concentrations of A near source); 9.0 g/h may be an underestimate; some individuals
might be sensitive to concentrations PEL.
e. Increase in the room temperature could increase the volatility of A and hence the rate of
evaporation from the tank. T in the numerator of expression for
Vair : At higher T, need
a greater air volume throughput for y to be PEL.
C (g A / m )
A
3

5-20
5.38 Basis: 2 mol feed gas C H H C H
3 6 3
+ ⇔
2 8
1
1
25
6
2
mol C H
mol H
C, 32 atm
3
D
n (mol C H
1- n )(mol C H
1- n )(mol H
C, P
p 3 8
p 3 6
p 2
2
n n 2(1 n ) 2 n
2 p p p
)
( )
( )
U
V
|
W
|
= + − = −
235D
a. At completion, n 1 mol
p = , n 2 1 1 mol
2 = − =
P V
P V
n RT
n RT
P
n
n
T
T
P
1 mol 508K atm
2 mol 298K
atm
2
1
2 2
1 1
2
2
1
2
1
1
= ⇒ = = =
32 0
27 3
.
.
b. P 35.1 atm
2 =
n
P
P
T
T
n
35.1 atm 298K 2 mol
32.0 atm 508K
1.29 mol
2
2
1
1
2
1
= = =
1.29 2 n n 0.71 mol C H produced
1- 0.71 mol C H unreacted 71% conversion of propylene
p p 3 8
3 6
= − ⇒ =
⇒ = ⇒
b g 0 29
.
c.
P2 (atm) n2 C3H8 prod. %conv.
27.5 1.009 0.99075 99.075
28.0 1.028 0.9724 97.24
29.5 1.083 0.91735 91.735
30.0 1.101 0.899 89.9
31.5 1.156 0.84395 84.395
32.0 1.174 0.8256 82.56
33.0 1.211 0.7889 78.89
33.5 1.229 0.77055 77.055
34.0 1.248 0.7522 75.22
34.5 1.266 0.73385 73.385
35.0 1.285 0.7155 71.55
37.0 1.358 0.6421 64.21
39.0 1.431 0.5687 56.87
40.0 1.468 0.532 53.2
Pressure vs Fraction Conversion
0
20
40
60
80
100
120
25.0 27.0 29.0 31.0 33.0 35.0 37.0 39.0 41.0
Pressure (atm)
%
conversion
%conv.

5-21
5.39 Convert fuel composition to molar basis
Basis: 100 g
g CH 1 mol 16.04 g mol CH
g C H 1 mol 30.07 g mol C H
mol % CH
2.8 mol % C H
4 4
2 6 2 6
4
2 6
⇒
=
=
U
V
W
⇒
95 592
5 017
97 2
b g
b g
.
.
.
500 m h
3
n (mol / h)
0.972 CH
1
4
/

n (kmol CO / h)
n (kmol H O / h)
2 2
3 2

V (SCMH)
25% excess air
air

.
.
.
n
P V
RT
bar 500 m kmol K
313K h m bar
kmol h
1
1 1
1
3
3
= =
⋅
⋅
=
11
0 08314
211
CH 2O CO H O C H O CO H O
2 2 6 2
4 2 2 2 2
2
7
2
2 3
+ → + + → +
Theoretical O =
21.1 kmol
h
kmol CH 2 kmol O
kmol 1 kmol CH
kmol C H 3.5 kmol O
kmol 1 kmol C H
kmol O
h
2
4 2
4
2 6 2
2 6
2
0 972
0 028
431
.
.
.
L
NM
+
O
QP =
Air Feed:
kmol O 1 kmol Air m STP
h 0.21 kmol O 1 kmol
SCMH
2
3
2
125 431 22 4
5700
. . .
b g b g=
5.40 Basis: 1 m gas fed @ 205 C, 1.1 bars Ac acetone
3
° =
1 m @205 C, 1.1 bar
3 D
n (kmol), 10 C, 40 bar
3
D
n (kmol)
y (kmol Ac / kmol)
(1- y )(kmol air / kmol)
p bar
1
1
1
AC = 0100
.
n (kmol Ac(l))
2
a. n
m 273K bars 1 kmol
478K bars 22.4 m STP
kmol
1
3
3
= =
100 110
10132
0 0277
. .
.
.
b g
y
bar
1.1 bars
kmol Ac kmol
1 = =
0100
0 0909
.
. , y
.379 bar
40.0 bars
kmol Ac kmol
3 = = × −
0
9 47 10 3
.
Air balance: )n n 0 kmol
3 3
0 0277 0 910 1 9 47 10 0254
3
. . ( . .
b gb g= − × ⇒ =
−
Mole balance: n n kmol Ac condensed
2 2
0 0277 0 0254 0 0023
. . .
= + ⇒ =
Acetone condensed
kmol Ac 58.08 kg Ac
1 kmol Ac
kg acetone condensed
= =
0 0023
0133
.
.
condenser y (kmol Ac / kmol)
(1- y )(kmol air / kmol)
p bar
3
3
AC = 0 379
.
0.028 C H
C, 1.1 bar
2 6
40
D

n (kmol O / h)
n (kmol N / h)
4 2
5 2

5-22
5.40 (cont’d)
Product gas volume
.0254 kmol 22.4 m STP 283K bars
273K
m
3
3
= =
0 10132
40.0 bars
0 0149
b g .
.
b.
20 0 0 0277 0 0909 58 08
0
196 kg
. . . .
m effluent kmol feed kmol Ac kg Ac
h .0149 m effluent kmol feed kmol Ac
Ac h
3
3
=
5.41 Basis: 1.00 10 gal. wastewater day. Neglect evaporation of water
6
× .
100 106
. × gal / day Effluent gas: 68 F, 21.3 psia(assume)
D
1 2
1 3
(lb-moles H O/day)
0.03 (lb-moles NH /day)
n
n

2
3 3
(lb-moles air/day)
(lb-moles NH /day)
n
n

300 106
× ft air / day
3
Effluent liquid
2
68 F, 21.3 psia
(lb-moles air/day)
n
D

1 2
4 3
(lb-moles H O/day)
(lb-moles NH /day)
n
n

a. Density of wastewater: Assume lb ft
m
3
ρ = 62 4
.
3
1 2 m 1 m 3 m
3
m
6
lb-moles H O 18.02 lb 0.03 lb NH 17.03 lb 1 ft 7.4805 gal
day 1 lb-mole day 1 lb-mole 62.4 lb 1 ft
gal
1.00 10
day
n n
⎡ ⎤
+ ×
⎢ ⎥
⎢ ⎥
⎣ ⎦
= ×

⇒ 5
1 2
4.50 10 lb-moles H O fed day
n = ×
, 4
1 3
0.03 1.35 10 lb-moles NH fed day
n = ×

( )
6 3
6
2 3
300 10 ft 492 R 21.3 psi 1 lb-mole
1.13 10 lb-moles air day
day 527.7 R 14.7 psi 359 ft STP
n
×
= = ×
D
D

3 3 3
93% stripping: 0.93 13500 lb-moles NH fed day 12555 lb-moles NH day
n = × =

Volumetric flow rate of effluent gas
( )
6 3 6
out out out
out in 6
in in
in
6 3
300 10 ft 1.13 10 12555 lb-moles day
day 1.13 10 lb-moles day
303 10 ft day
PV n RT n
V V
n RT n
PV
× × +
= ⇒ = =
×
= ×

Partial pressure of NH y P
- moles NH day
1.129 10 lb - moles day
psi
psi
3 NH
3
6
3
= =
× +
×
=
12555 lb
12555
213
0 234
d i
.
.

5-23
5.42 Basis: 2 liters fed / min
Cl ads.=
2.0 L soln
min
g
L
g NaOH
g soln
mol
40.0 g
NaOH ads.
mol NaOH
mol Cl
2 mol NaOH
mol
min
2
60
1130 012 1 0 23 1
0 013
. .
.
=
2 L / min @ 23 C, 510 mm H O
2
D
n (mol air / mol)
2

/
n (mol / min)
y (mol Cl mol)
(1- y)(mol air / mol)
1
2
0013
. /
mol Cl min
2
Assume P 10.33 m H O P 10.33 510 m H O 10 m H O
atm 2 abs in 2 2
= ⇒ = + =
b g b g
0 84
. .
.
n
L 273K 10.84 m H O 1 mol
min 296K 10.33 m H O 22.4 L STP
mol min
1
2
2
= =
2
0 0864
b g
Cl balance: 0.0864y y
mol Cl
mol
, specification is wrong
2
= ⇒ = ∴
0 013 0150
. .
5.43
a. Hygrometer Calibration ln y bR ln a y aebR
= + =
d i
b
ln y y
R R
1 2
2 1
=
−
=
−
=
−
b g d i
ln .
.
0 2 10
90 5
0 08942
4
ln a ln y bR ln 10 0.08942 5 a 6.395 10 y 6.395 10 e
1 1
4 5 5 0.08942R
= − = − ⇒ = × ⇒ = ×
− − −
b g
b.
n
125 L 273K 105 kPa 1 mol
min 298K 101 kPa 22.4 L STP
5.315 mol min wet gas
1 = =
b g

n
355 L 273K 115 kPa 1 mol
min 348K 101 kPa 22.4 L STP
14.156 mol min wet air
2 = =
b g
R 86.0 y 0.140
1 1
= → = , R 12.8 y 2.00 10 mol H O mol
2 2
4
2
= → = × −
125 L / min @ 25 C, 105 kPa
o
n mol / min)
y (mol H O / mol)
1- y mol dry gas / mol)
0.235 mol C H mol DG
0.765 mol C H / mol DG
1
1 2
1
2 6
2 4
(
(
/
b g
R
S
|
T
|
U
V
|
W
|

V (L / min) @ 65 C, 1 atm
3
o
/
/
/
/
n (mol C H min)
n (mol C H min)
n (mol air min)
n (mol H O min)
C H 2 6
C H 2 4
air
H O 2
2 6
2 4
2
355 L / min air @ 75 C, 115 kPa
n (mol / min)
y mol H O / mol)
(1- y mol dry air / mol)
o
2
2 2
2

(
)(

5-24
5.43 (cont’d)
C H balance: n 5.315 mol min 1 0.140
mol DG
mol
0.235
mol C H
mol DG
1.07 mol C H min
2 6 C H
2 6
2 6
2 6
= −
F
HG I
KJF
HG I
KJ
=
b g b g
C H balance: n 5.315 0.860 0.765 3.50 mol C H min
2 4 C H 2 4
2 4
= =
b gb gb g
Dry air balance: n 14.156 1 2.00 10 14.15 mol DA min
air
4
= − × =
−
b gd i
Water balance: n 5.315 0.140 14.156 1.00 10 0.746 mol H O min
H O
4
2
2
= + × =
−
b gb g b gd i

n 1.07 3.50 14.15 mol min 18.72 mol min
dry product gas = + + =
b g ,

n 18.72 0.746 19.47 mol min
total = + =
b g

V
19.47 mol min 22.4 L STP 338K
mol 273K
540 liters min
3 = =
b g
Dry basis composition:
1.07
18.72
100% 5.7% C H , 18.7% C H , 75% dry air
2 6 2 4
F
HG I
KJ× =
c. p y P
0.746 mol H O
19.47 mol
1 atm 0.03832 atm
H O H O
2
2 2 l
= ⋅ = × =
y 0.03832 R
1
0.08942
ln
0.03832
6.395 10
71.5
H O 5
2
= ⇒ =
×
F
HG I
KJ =
−
5.44 CaCO CaO CO
3 2
→ +
.
n
1350 m 273K 1 kmol
h 1273K 22.4 m STP
kmol CO h
CO
3
3 2
2
= =
b g 12 92
12.92 kmol CO kmol CaCO 100.09 kg CaCO 1 kg limestone
h kmol CO 1 kmol CaCO 0.95 kg CaCO
limestone h
2 3 3
2 3 3
1
1
1362 kg
=
1362 kg limestone 0.17 kg clay
h 0.83 kg limestone
279 kg clay h
=
Weight % Fe O
2 3
279 0 07
1362 279 12 92 441
100% 18%
.
. .
.
b g
b g
kg Fe O kg clay
kg limestone kg clay
kg CO evolved
2 3
2 3
2
Fe O
+ −
× =

5-25
5.45
Basis: 1 kg Oil
864.7 g C 1 mol 12.01 g mol C
116.5 g H 1 mol 1.01 g mol H
13.5 g S 1 mol 32.06 g mol S
5.3 g I
⇒
=
=
=
R
S
|
|
T
|
|
b g
b g
b g
72 0
1153
0 4211
.
.
.
72 0
. mol C
115.3 mol H
0.4211 mol S
5.3 g I
53
.
)
)
)
)
g I
n (mol CO
n (mol CO)
n (mol H O)
n (mol SO
n (mol O
n (mol N
1 2
2
3 2
4 2
5 2
6 2
C O CO
C +
1
2
O CO
S O SO
2H
1
2
O H O
2 2
2
2 2
2 2
+ →
→
+ →
+ →
n (mol), 0.21 O , 0.79 N
excess air
175 C, 180 mm Hg (gauge)
a 2 2
15%
D
a. Theoretical O2:
72.0 mol C 1 mol O 115.3 mol H 0.25 mol O
2 2
1 mol C 1 mol H
0.4211 mol S 1 mol O
2 101.2 mol O
2
1 mol S
+
+ =
( )
1.15 101.2 mol O 1 mol Air
2
Air Fed: 554 mol Air n
a
0.21 mol O
2
= =
( ) 3
3
3
554 mol Air 22.4 liter STP 1 m 448K 760 mm Hg
16.5 m air kg oil
1 kg oil mol 10 liter 273K 940 mm Hg
=
b. S balance: n 0.4211 mol SO
4 2
=
H balance: 115.3= 2n n 57.6 mol H O
3 3 2
⇒ =
C balance: 0.95 72.0 = n n 68.4 mol CO
1 1 2
b g ⇒ = 2
0.05(72.0) n 3.6 mol CO
⇒ = =
( )
2 6 6 2
N balance: 0.79 554 =n n 437.7 mol N
⇒ =
( ) ( ) 5 5 2
O balance: 0.21 554 2=57.6+3.6+2(68.4)+2 0.4211 +2n n 16.9 mol O
⇒ =
( )
Total moles excluding inerts wet: 585 mols dry: 527 mols
dry basis: 3 4
2 2
0.4211 mol SO mol SO
3.6 mol CO mol CO
6.8 10 , 7.2 10
527 mol mol 527 mol mol
− −
= × = ×
wet basis: 6 6
2
2
0.4211 mol SO
3.6 mol CO
10 6150 ppm CO , 10 720 ppm SO
585 mol 585 mol
× = × =

5-26
5.46 Basis: 50.4 liters C H
5 12 l
b g min
1 2
n , n

C H CO H O
5 12 2 2
+ → +
80 5 6
2
a.
n
50.4 L 0.630 kg 1 kmol
min L 72.15 kg
0.440 kmol min
1 = =

.
.
n
kg
min
kmol
72.15 kg
kmol / min
3 = =
3175 1
0 044
frac. convert =
0.440- 0.044 kmol
0.440
C H converted
5 12
× =
100 90%
( )
5 12 2
2
5 12 2
0.440 kmol C H 1.15 8 kmol O 1 mol air
n 19.28 kmol air min
min kmol C H 0.21 mol O
= =

V
22.4 L STP 336K 101 kPa
min mol 273K 309.6 kPa
mol
kmol
173000 L min
air = =
19.28 kmol 1000
b g
2
4 2
kmol O
n [(0.21)(19.28) (0.90)(0.440)(8)] 0.882 kmol O / min
min
= − =

2
5 2
0.79 kmol N
19.28 kmol air
n 15.23 kmol N / min
min kmol air
= =

. ( . )
min
.
n
kmol C H kmol CO
kmol C H
kmol CO / min
6
5 12 2
5 12
2
= =
0 90 0 440 5
198
5
gas
0.882+15.23+1.98 kmol 22.4 L(STP) 275 K 1000 mol
V 4.08 10 L/min
min mol 273 K kmol
= = ×

heater Combustion
chamber
Condenser
5 12
1 5 12
n
50.4 L C H ( ) / min
(kmol C H / min)
l

15%
0 21
336
excess air, V (L / min)
n kmol air
O
0.79 N
K, 208.6 kPa (gauge)
air
2
2
2

. / min)
/ min)
/ min)
/ min)
/ min)
n (kmol C H
n (kmol O
n (kmol N
n (kmol CO
n (kmol H O
3 5 12
4 2
5 2
6 2
7 2
5 12
3 12
7 2
liq
5
V (L/min)
m=3.175 kg C O / min
n (kmol C O / min)
n (kmol H O( ) / min)
l

/ min)
/ min)
/ min)
n (kmol O
n (kmol N
n (kmol CO
4 2
5 2
6 2
gas
V (L/min), 275 K, 1 atm

5-27
5.46 (cont’d)

. ( . )
min
.
n
kmol C H kmol H O
kmol C H
kmol H O( ) / min
7
5 12 2
5 12
2
= =
0 9 0 440 6
2 38 l
Condensate:
.
.
V
kmol 72.15 kg L
min kmol 0.630 kg
L min
C H
5 12
= =
0 044
504
.
V
.38 kmol 18.02 kg L
min kmol 1 kg
L min
H O
2
= =
2
42 89
Assume volume additivity (liquids are immiscible)
. . .
V L min
liq = + =
504 42 89 47 9
b.
5.47

.
n (kmol / min), 25 C, 1 atm
0.21 O
N
air
2
2
D
0 79

n (kmol / min)
0
n (kmol H S / min)
1 2
0 20
080
.
.
kmol H S / mol
kmol CO / mol
2
2
10 0
. / min

m @ 380 C, 205 kPa
n (kmol / min)
n (kmol N / min)
n (kmol H O / min)
n (kmol CO / min)
n (kmol S / min)
3
exit
3 2
4 2
5 2
6
D

. / min
.
n
PV
RT
kPa
8.314
m
K
kmol / min
exit m kPa
kmol K
3
3
= = =
⋅
⋅
205 10 0
653
0 377
.
n n / 3= 0.0667n ; n n = 0.133n
1 0 0 2 1 0
= =
0 20 2
b g
Furnace
H S + O SO H O
2
3
2 2 2 2
→ +
Reactor
2H S + SO S(g) H O
2 2 2
→ +
3 2
C H ( )
5 12 l
H O
2 l
b g
H O
2 l
b g
C H
5 12 l
b g

n (kmol H S / min)
2 2

5-28
5.47 (cont’d)
Air feed to furnace: n
n (kmol H S fed)
(min)
kmol O
kmol H S
kmol air
0.21 kmol O
n kmol air / min
air
0 2 2
2 2
0

. .
.
=
=
0 0667 15
1
1
0 4764
Overall N balance: n
n (kmol air)
(min)
kmol N
min
n kmol N
2 3
0 2
0 2

. .
. ( / min)
= =
0 4764 0 79
0 3764
Overall S balance: n
0.200n (kmol H S)
(min)
kmol S
1 kmol H S
n (kmol S
6
0 2
2

. / min)
= =
1
0 200 0
Overall CO balance n n (kmol CO min)
2 5 0 2
: . /
= 0800
Overall H balance:
0.200n (kmol H S)
(min)
kmol H
1 kmol H S
n kmol H O
min
kmol H
1 kmol H O
n = 0.200n (kmol H O / min)
0 2
2
4 2
2
4 0 2

2 2
=
⇒
. . . .
n n + + = 0.377 kmol / min n = 0.24 kmol / min
exit 0
= + ⇒
0 376 0 200 0 200 0800 0
b g
.
n = 0.4764(0.24 kmol air / min) kmol air / min
air = 0114
5.48 Basis: 100 kg ore fed 82.0 kg FeS s), 18.0 kg I.
2
⇒ (
n fed = 82.0 kg FeS kmol /120.0 kg kmol FeS
FeS 2 2
2
b gb g
1 0 6833
= .
n (kmol SO )
n (kmol SO )
n (kmol O )
n (kmol N )
2 2
3 3
4 2
5 2
m (kg FeS
m (kg Fe O
kg I
6 2
7 2 3
)
)
18
2 4
2 4
11
2
15
2
FeS O Fe O SO
FeS O Fe O SO
2(s) 2(g) 2 3(s) 2(g)
2(s) 2(g) 2 3(s) 3(g)
+ → +
+ → +
a. n
kmol FeS .5 kmol O
2 kmol FeS
1 kmol air req'd
0.21 kmol O
kmol air fed
kmol air req'd
kmol air
1
2 2
2 2
= =
0 6833 7 140
17 08
. .
.
V kmol SCM / kmol SCM /100 kg ore
1 = =
17 08 22 4 382
. .
b gb g
n
kmol FeS kmol SO
2 kmol FeS
kmol SO
2
2 2
2
2
= =
( . )( . ) .
.
085 0 40 0 6833 4
0 4646
100 kgore
06833
18
. kmol FeS
kg I
2
40% excess air
n (kmol)
0.21 O
0.79 N
V m (STP)
1
2
2
1
3
V m (STP)
out
3

5-29
5.48 (cont’d)
n
kmol FeS kmol SO
2 kmol FeS
kmol SO
3
2 2
2
3
= =
( . )( . ) .
.
085 0 60 0 6833 4
0 6970
( )
.4646 kmol SO 5.5 kmol O
2 2
n 0.21 17.08 kmol O fed
4 2 4 kmol SO
2
.697 kmol SO 7.5 kmol O
3 2 1.641 kmol O
2
4 kmol SO
3
= × −
− =
n kmol N kmol N
5 2 2
= × =
0 79 17 08 1349
. . .
b g
( ) [ ]
out
V = 0.4646+0.6970+1.641+13.49 kmol 22.4 SCM (STP)/kmol
365 SCM/100 kg ore fed
⎡ ⎤
⎣ ⎦
=
2 3 2 2
2
SO SO O N
0.4646 kmol SO
y 100% 2.9%; y 4.3%; y 10.1%; y 82.8%
16.285 kmol
= × = = = =
b.
Let (kmol) = extent of reaction
ξ
2
2 3
3
2
2 2 2
SO
SO SO
1 1
SO
2 2
1
O 2 1
2
N O N
1 1
1 2 2
2
n 0.4646 0.4646 0.697
y , y
n 0.697 16.29- 16.29-
n 1.641
1.641 13.49
n 13.49 y , y
16.29- 16.29-
n=16.29-
ξ ξ ξ
ξ ξ ξ
ξ
ξ
ξ ξ
ξ
⎫
= − − +
⎪ = =
= +
⎪
= − ⇒
⎬
−
⎪
= = =
⎪
⎭
( )
( )
1
2
1
3 2
1 1
2 2
2 2
1
SO -
2
p p
1
SO O 2
P y (0.697 ) 16.29
K (T)= P K (T)
P y (P y ) (0.4646 ) 1.641
ξ ξ
ξ ξ
⋅ + −
⇒ ⋅ =
⋅ ⋅ − −
( )
1
2
2 2
-
o
p
2
SO SO
2
P=1 atm, T=600 C, K 9.53 atm 0.1707 kmol
0.4646 0.2939 kmol SO reacted
n 0.2939 kmol f 0.367
0.4646 kmol SO fed
ξ
= ⇒ =
−
⇒ = ⇒ = =
1
2
2 2
-
o
p
SO SO
P=1 atm, T=400 C, K 397 atm 0.4548 kmol
n 0.0098 kmol f 0.979
ξ
= ⇒ =
⇒ = ⇒ =
The gases are initially heated in order to get the reaction going at a reasonable rate. Once
the reaction approaches equilibrium the gases are cooled to produce a higher equilibrium
conversion of SO2.
0 4646
0 697
1633
1349
.
.
.
.
kmol SO
kmol SO
kmol O
kmol N
2
3
2
2
n (kmol)
n (kmol)
n (kmol)
n (kmol)
SO
SO
O
N
2
3
2
2
Product gas, T C
o
e j
Converter

5-30
5.48 (cont’d)
c. SO leaving converter: (0.6970 + 0.4687) kmol =1.156 kmol
3
1.156 kmol SO
min
kmol H SO
kmol SO
kg H SO
kmol
kg H SO
3 2 4
3
2 4
2 4
⇒ =
1
1
98
1133
.
Sulfur in ore:
0.683 kmol FeS kmol S
kmol FeS
kg S
kmol
kg S
2
2
2 321
438
.
.
=
1133
2 59
.
.
kg H SO
43.8 kg S
kg H SO
kg S
2 4 2 4
=
100% conv.of S:
0.683 kmol FeS kmol S
kmol FeS
kmol H SO
1 kmol S
kg
kmol
kg H SO
kg H SO
43.8 kg S
kg H SO
kg S
2
2
2 4
2 4
2 4 2 4
2 1 98
1339
1339
306
=
⇒ =
.
.
.
The sulfur is not completely converted to H2SO4 because of (i) incomplete oxidation of
FeS2 in the roasting furnace, (ii) incomplete conversion of SO2 to SO3 in the converter.
5.49 N O NO
2 4 2
⇔ 2
a. n
P 1.00 V
RT
2.00 atm 2.00 L
473K 0.08206 L atm mol - K
0.103 mol NO
0
gauge
0
2
=
+
=
⋅
=
d i b gb g
b gb g
b. n mol NO
1 2
= , n mol N O
2 2 4
=
p y P
n
n n
P
NO NO
1
1 2
2 2
= =
+
F
HG I
KJ , p
n
n n
P K
n
n n n
P
N O
2
1 2
p
1
2
2 1 2
2 4
=
+
F
HG I
KJ ⇒ =
+
b g
Ideal gas equation of state ⇒ = + ⇒ + =
PV n n RT n n PV / RT 1
1 2 1 2
b g b g
Stoichiometric equation ⇒ each mole of N O
2 4 present at equilibrium represents a loss
of two moles of NO2 from that initially present ⇒ + =
n 2n 0.103 2
1 2 b g
Solve (1) and (2) ⇒ n 2(PV / RT) 0.103 3
1 = − b g, n 0.103 (PV / RT 4
2 = − ) b g
Substitute (3) and (4) in the expression for Kp , and replace P with P 1
gauge +
K
2n 0.103
n 0.103 n
P 1
p
t
2
t t
gauge
=
−
−
+
b g
b gd i where n
P 1 V
RT
n
24.37 P 1
T
t
gauge
V 2 L
t
g
=
+
⇒ =
+
=
d i d i
T(K) Pgauge(atm) nt Kp(atm) (1/T) ln(Kp)
350 0.272 0.088568 5.46915 0.002857 1.699123
335 0.111 0.080821 2.131425 0.002985 0.756791
315 -0.097 0.069861 0.525954 0.003175 -0.64254
300 -0.224 0.063037 0.164006 0.003333 -1.80785
Variation of Kp with Temperature
y = -7367x + 22.747
R2
= 1
-2
-1
0
1
2
0.0028 0.003 0.0032 0.0034
1/T
ln
Kp

5-31
5.49 (cont’d)
c. A semilog plot of Kp vs.
1
T
is a straight line. Fitting the line to the exponential law
yields
ln K
7367
T
22.747 K 7.567 10 exp
7367
T
a 7.567 10 atm
b = 7367K
p p
9
9
= − + ⇒ = ×
−
F
HG I
KJ⇒
= ×
5.50
A + H2 S
Overall A balance
kmol S 1 kmol A react
h 1 kmol S form
kmol A / h
Overall H balance
kmol S 1 kmol H react
h 1 kmol S form
kmol H / h
2
2
2
:
.
.
:
.
.
n
n
1
2
500
500
500
500
= =
= =
Extent of reaction equations n n
A + H S
i i0 i
2
:
= +
↔
ν ξ
A: n n
H n n
S: 5.00 = n n
n n
n = 5.00
n n .00
p y P =
n
n
P
n - 5.00
n
p y P =
n
n
P
n - 5.00
n
4 3
2 5 3
4 3
5 3
S
tot 3
A A
4
tot
3
3
H H
5
tot
3
3
2 2

:
.
.

.
.

.
.
= −
= −
===== = −
= −
= −
U
V
|
|
W
|
|
⇒ = =
−
= =
−
3
3 500
500
4 5
3
4 500
10 0
4 500
10 0
ξ
ξ
ξ
p y P =
5.00
n
S S
3
=
−
4 500
10 0
.
.
K
p
p p
5.00 4n
10.0 3n n
n kmol H h
p
S
A H
3
3 3
3 2
2
= =
−
− −
= ⇒ =
.
. .
. . /
500
500 500
0100 1194
b g
b gb g
( . .
. . . /
. ( ) /
n ) - 5.00 kmol A / h
n kmol H h
V + 6.94 kmol / h m STP kmol
4
5 2
rcy
3
= =
= − =
= =
3 1194 3082
1194 500 6 94
30.82 22 4 846 SCMH
b g d i

n (kmol A / h)
n (kmol H / h)
(SCMH)
4
5 2
Vrcy
10 00
500
.
.
atm
kmol S / h
3n (kmol A / h)
n (kmol H / h)
3
3 2

500
. kmol S / h

/
n (kmol A / h)
n (kmol H h)
1
2 2

n (kmol A / h)
n (kmol H / h)
4
5 2

5-32
5.51

n (kmol CO / h)
n (kmol H / h)
4
5 2
a.
5.0% XS H2:
2 2 2
3
2
100 kmol CO fed 2 kmol H reqd 1.05 kmol H fed kmol H
210
h 1 kmol CO fed 1 kmol H reqd h
n = =

( )
4 6 4 6
100 kmol CO 1 kmol C
C balance: (1) (1) 100 1
h 1 kmol CO
n n n n
= + ⇒ = +

5 6 5 6
H balance: 210(2) (2) (4) 210 2 (2)
n n n n
= + ⇒ = +

4 6
(O balance: 100 identical to C balance not independent)
n n
= + ⇒ ⇒

4 6 5 6
4 5 6 6 6 6 6
(1) 100 , (2) 210 2
(100 ) (210 2 ) 310 2
tot
n n n n
n n n n n n n n
⇒ = − ⇒ = −
= + + = − + − + = −

( )
( )
( ) ( )
4 7 -2
p
2
-3 -8
9143.6
21.225+ 7.492ln 500K
500 K
K T=500K 1.390 10 exp 9.11 10 kPa
+4.076 10 500K -1.161 10 500K
− −
⎛ ⎞
−
⎜ ⎟
= × = ×
⎜ ⎟
⎜ ⎟
× ×
⎝ ⎠
K
y P
y P y P
K P
y
y y
n
n
n
n
n
n
K P kPa kPa
n n
n n
p
M
CO H
p
2 M
CO H
6
6
6
6
6
6
p
2 -2 6 6
6 6
2 2
= ⇒ = ====
−
−
−
−
−
= × = =
−
− −
−
−
d i d i
b g
b g
b g
b g
b g
b g b g
b gb g
2 2
1 3)
2
2
7 2
2
2
310 2
100
310 2
210 2
310 2
911 10 5000 22 775
310 2
100 210 2
( ) (

. .

6 6 3 4 6
5 6 2
1 6 1
2
Solving for 75.7 kmol CH OH/h , 100 24.3 kmol CO/h
210 2 58.6 kmol H / h
Overall C balance: (1) (1) 75.7 kmol CO/h
Overall H balance: (2)
n n n n
n n
n n n
n
⇒ = = − =
= − =
= ⇒ =

( )
6 2 2
3
rec 4 5
(4) n 151 kmol H /h
22.4 m (STP)
1860 SCMH
kmol
n
V n n
= ⇒ =
= + =

Reactor Separator

n (kmol CO / h)
n (kmol H / h)
1
2 2
6 3
n (kmol CH OH/h)

100 kmol CO / h
H (% H excess)
n (kmol H / h)
T (K), P (kPa)
3 2
xs 2
4
5 2
6 3
n (kmol CO/h)
n (kmol H /h)
n (kmol CH OH/h)
T, P

Balances on reactor ⇒ 4 equations in 3 4 5 6
, , , and .
n n n n

5-33
5.51 (cont’d)
b.
`
P(k
P
a
) T(K
) H
xs(%
) K
p
(T
)E
8 K
p
P
^2
n
3(k
m
o
l
H
2
/h
)
n
4(k
m
o
l
C
O
/h
)
n
5(k
m
o
l
H
2
/h
)
1
0
0
0 5
0
0 5 9
.1
E
+
0
1 0
.9
1 2
1
0 7
4
.4
5 1
5
8
.9
0
5
0
0
0 5
0
0 5 9
.1
E
+
0
1 2
2
.7
8 2
1
0 9
1
.0
0 1
9
2
.0
0
1
0
0
0
0 5
0
0 5 9
.1
E
+
0
1 9
1
.1
1 2
1
0 1
3
.2
8 3
6
.5
6
5
0
0
0 4
0
0 5 3
.1
E
+
0
4 7
8
4
9
.7
7 2
1
0 1
.0
7 1
2
.1
5
5
0
0
0 5
0
0 5 9
.1
E
+
0
1 2
2
.7
8 2
1
0 2
4
.3
2 5
8
.6
4
5
0
0
0 6
0
0 5 1
.6
E
+
0
0 0
.4
1 2
1
0 8
5
.4
2 1
8
0
.8
4
5
0
0
0 5
0
0 0 9
.1
E
+
0
1 2
2
.7
8 2
0
0 2
6
.6
5 5
3
.3
0
5
0
0
0 5
0
0 5 9
.1
E
+
0
1 2
2
.7
8 2
1
0 2
4
.3
2 5
8
.6
4
5
0
0
0 5
0
0 1
0 9
.1
E
+
0
1 2
2
.7
8 2
2
0 2
2
.2
3 6
4
.4
5
n
6(k
m
o
l
M
/h
)
n
to
t
(k
m
o
l/h
) K
p
c
E
8
K
p
P
^2
-
K
p
c
P
^2
n
1(k
m
o
l
C
O
/h
)
n
2(k
m
o
l
H
2
/h
)
V
re
c
(S
C
M
H
)
2
5
.5
5 2
5
8
.9
0 9
.1
E
-0
1 1
.3
E
-0
5 2
5
.5
5 5
1
.1
0 5
2
2
7
9
.0
0 2
9
2
.0
0 2
.3
E
-0
1 2
.3
E
+
0
1 9
.0
0 1
8
.0
0 6
3
3
9
8
6
.7
2 1
3
6
.5
6 9
.1
E
+
0
1 4
.9
E
-0
3 8
6
.7
2 1
7
3
.4
4 1
1
1
6
9
8
.9
3 1
1
2
.1
5 7
.8
E
+
0
3 3
.2
E
-0
8 9
8
.9
3 1
9
7
.8
5 2
9
6
7
5
.6
8 1
5
8
.6
4 2
.3
E
+
0
1 3
.4
E
-0
3 7
5
.6
8 1
5
1
.3
6 1
8
5
8
1
4
.5
8 2
8
0
.8
4 4
.1
E
-0
1 -2
.9
E
-0
4 1
4
.5
8 2
9
.1
6 5
9
6
4
7
3
.3
5 1
5
3
.3
0 2
.3
E
+
0
1 9
.8
E
-0
3 7
3
.3
5 1
4
6
.7
0 1
7
9
1
7
5
.6
8 1
5
8
.6
4 2
.3
E
+
0
1 3
.4
E
-0
3 7
5
.6
8 1
5
1
.3
6 1
8
5
8
7
7
.7
7 1
6
4
.4
5 2
.3
E
+
0
1 -3
.1
E
-0
3 7
7
.7
7 1
5
5
.5
5 1
9
4
2
c. Increase yield by raising pressure, lowering temperature, increasing Hxs. Increasing the
pressure raises costs because more compression is needed.
d. If the temperature is too low, a low reaction rate may keep the reaction from reaching
equilibrium in a reasonable time period.
e. Assumed that reaction reached equilibrium, ideal gas behavior, complete condensation of
methanol, not steady-state measurement errors.
5.52
10
10
10
.
.
.
mol CO
mol O
mol N
T = 3000 K, P = 5.0 atm
2
2
2
A B
1
2
C
⇔ + A CO2
− , B CO
− , C O2
− , D N2
− , E NO
− ξ1 - extent of rxn 1
1
2
C
1
2
D E
+ = n n n 1
A0 C0 D0
= = = , n n 0
B0 E0
= = ξ2 -extent of rxn 2
CO CO + O
K
p p
p
atm
O N NO
K
p
p p
2
1
2 2
1
CO O
CO
1/2
1
2 2
1
2 2
2
NO
N O
2
2
2 2
⇔
= =
+ ⇔
= =
1/2
1/2
0 3272
01222
d i
d i
.
.

5-34
5.52 (cont’d)
n
n
n
n
n
n
y n n
y
y
y
y
p y P
A
B
C
D
E
tot
A A tot
B
C
D
E
i i
= −
=
= + −
= −
=
= + =
+
U
V
|
|
|
|
|
W
|
|
|
|
|
= = − +
= +
= + − +
= − +
= +
=
1
1
1
2
1
2
1
1
2
3
1
2
6
2
2 1 6
2 6
2 6
2 6
2 6
1
1
1 2
2
2
1
1
1 1
1 1
1 2 1
2 1
2 1
ξ
ξ
ξ ξ
ξ
ξ
ξ
ξ
ξ ξ
ξ ξ
ξ ξ ξ
ξ ξ
ξ ξ
b g b g
b g
b g b g
b g b g
b g
K
p p
p
y y
y
p
(1)
1
CO O
1 2
CO
B C
1 2
A
1 1
2
2
= = =
+ −
− +
=
⇒ − + = + −
+ −
1
2
2 2
2 1 6
5 0 3272
0 3272 1 6 2 236 2
1 1 2
1 2
1 1
1 2
1 2
1 1
1 2
1 1 2
1 2
b g b g
b gb g b g
b gb g b g
ξ ξ ξ
ξ ξ
ξ ξ ξ ξ ξ
.
. .
K
p
p p
y
y y
p
(2)
2
NO
O N
1 2
E
C
1 2
D
1 2
1 1 2 1 2
2 2
= = =
+ − −
=
⇒ + − − =
− −
d i b g b g
b g b g
2
2 2
01222
01222 2 2 2
2
1 2
1 2
2
1 2
1 2
1 2
2
1 2
2
ξ
ξ ξ ξ
ξ ξ ξ ξ
.
.
Solve (1) and (2) simultaneously with E-Z Solve⇒ = =
ξ ξ
1 2
0 20167 012081
. , . ,
y mol CO mol y mol N mol
y mol CO mol y mol NO mol
y mol O mol
A 2 D 2
B E
C 2
= − + = =
= =
=
2 1 6 0 2574 0 3030
0 0650 0 0390
0 3355
1 1
ξ ξ
b g b g . .
. .
.
5.53 a. 8 10 8 6 4
PX=C H , TPA=C H O , S=Solvent

V (m / h) @105 C, 5.5 atm
n (kmol O / h)
n (kmol N / h)
n (kmol H O(v) / h)
3
3O
3N
3 o
2
2
3W 2

.
V (m / h) at 25 C, 6.0 atm
n (kmol / h)
0.21 O
N
2
3 o
2
2
2
0 79

.
.
n (kmol / h)
4
O
N
2
2
0 04
0 96

n (kmol H O(v) / h)
V (m / h)
3W
3W
2
3
condenser
100 mol TPA / s
( )

n kmolPX / h
100 kmol TPA / h
3p
s
m (kg S / h)
reactor
( )

n n kmol PX / h
m (kg S / h)
3 kg S / kg PX
1 3p
s
+

n (kmol PX / h)
1 separator

n (kmol PX / h)
m (kg S / h)
3p
s

5-35
5.53 (cont’d)
b. Overall C balance:

n
kmol PX
h
kmol C
kmol PX
kmol TPA
h
kmol C
kmol TPA
n kmol PX / h
1 1
F
HG I
KJ = ⇒ =
8 100 8
100
c. 2
2 2
3 kmol O
100 kmol TPA
O consumed = 300 kmol O /h
h 1 kmol TPA
=
2
2 2 4 2
4
2 2 4
kmol O
Overall O balance: 0.21n 300 +0.04n n 1694 kmol air/h
h
n 1394 kmol/h
Overall N balance: 0.79n 0.96n
⎫
= =
⎪
⇒
⎬
=
⎪
= ⎭

Overall H O balance: n
kmol TPA
h
kmol H O
1 kmol TPA
kmol H O / h
2 3W
2
2
= =
100 2
200
3
3 3
2
2
n RT 1694 kmol 0.08206 m atm 298 K
V 6.90 10 m air/h
P h kmol K 6.0 atm
⋅
= = = ×
⋅

( ) ( ) 3
3W 4 3
3
n n RT 200+1394 kmol 0.08206 m atm 378 K
V 8990 m /h
P h kmol K 5.5 atm
+ ⋅
= = =
⋅

.
.
V
kmol H O (l)
h
kg
kmol
1 m
1000 kg
m H O(l) / h leave condenser
3W
2
3
3
2
= =
200 18 0
360
d. 90% single pass conversion n = 0.10 n n ====n kmol PX / h
3p 1 3p
n
3p
1
⇒ + =
=
.

d i
100
111
recycle 3
4
(100 11.1) kg PX 106 kg PX 3 kg S 11.1 kmol PX 106 kg PX
h 1 kmol PX kg PX h 1 kmol PX
= 3.65 10 kg/h
S P
m m m
+
= + = +
×

e. O2 is used to react with the PX. N2 does not react with anything but enters with O2 in the air.
The catalyst is used to accelerate the reaction and the solvent is used to disperse the PX.
f. The stream can be allowed to settle and separate into water and PX layers, which may then
be separated.
5.54
Separator
Separator
0 90
2
.
/
n
kmol N h
2
2
/
n (kmol CO / h), n (kmol H / h), 0.10n (kmol H h)
1 3 2 2 2

n (kmol CO / h)
n (kmol H / h)
n (kmol CO / h)
2 kmol N / h
6
7 2
8 2
2
0.300 kmol CO / kmol
0.630 kmol H / kmol
0.020 kmol N / kmol
2
2
2
Reactor

n (kmol CO / h)
n (kmol H / h)
n (kmol CO / h)
n (kmol M / h)
n (kmol H O / h)
2 kmol N / h
1
2 2
3 2
4
5 2
2

n (kmol M / h)
n (kmol H O / h)
4
5 2
, ,
/
n n n
kmol N h
1 2 3
2
2

5-36
5.54 (cont’d)
CO + 2H CH OH(M)
CO H CH OH + H O
2 3
2 2 3 2
⇔
+ ⇔
3
a. Let kmol / h) extent of rxn 1, kmol / h) extent of rxn 2
1 2
ξ ξ
( (
= =
CO: n = 30-
H : n = 63- 2
CO : n = 5-
M: n =
H O: n =
N : n = 2
n 100- 2
K
P y
P y P y
K
P y P y
P y P y
1 1
2 2 1
2 3 2
4 1
2 5
2 N
tot 1
p
M
CO H
p
M H O
CO H
2
2
2
2 2

,
ξ
ξ ξ
ξ
ξ ξ
ξ
ξ ξ
−
+
= −
U
V
|
|
|
W
|
|
|
⇒ =
⋅
⋅ ⋅
=
⋅ ⋅
⋅ ⋅
3
2
2
2
2
2
1 2 2 3
d i
d i
d i b gd i
d id i
K P =
n
n
n
n
n
n
(1)
p
2
4
tot
1
tot
2
tot
1
d i b gb g
b gb g
1 2
2 1 2
2
1 1 2
2
100 2 2
30 63 2 3
84 65
⋅
F
HG I
KJ
=
+ − −
− − −
=

.
ξ ξ ξ ξ
ξ ξ ξ
K P =
n
n
n
n
n
n
n
n
(2)
p
2
4
tot
5
tot
3
tot
2
tot
1
d i b gb g
b gb g
2 3
2 2 1 2
2
2 1 2
2
100 2 2
5 63 2 3
1259
⋅
F
HG I
KJF
HG I
KJ
F
HG I
KJF
HG I
KJ
=
+ − −
− − −
=

.
ξ ξ ξ ξ ξ
ξ ξ ξ
Solve (1) and (2) for = 25.27 kmol / h = 0.0157 kmol / h
1 1 2
ξ ξ ξ ξ
, 2 ⇒
⇒
. . .
. ( . ) ( . )
. .
. .
.
.
n kmol CO / h 9.98% CO
n H / h 26.2% H
n .98 kmol CO / h 10.5% CO
n M / h 53.4% M
n .0157 kmol H O / h 0.03% H O
n kmol / h
1
2 2 2
3 2 2
4
5 2 2
total
= − =
= − − =
= − =
= + = ⇒
= =
=
30 0 2527 4 73
630 2 2527 3 0 0157 12.4 kmol
50 0 0157 4
2527 0 0157 25.3 kmol
0 0157 0
49 4
C balance: n kmol / h
O balance: n n n n mol / s
n kmol CO / h
n = 0.02 kmol CO h
4
6 8 4 5
6
8 2
.
.
.
/
=
+ = + =
U
V
W
⇒
=
253
2 2544
254
H balance: 2n n n n n mol H s
7 2 4 5 7 2
( . ) . . /
= + + = ⇒ =
2 0 9 4 2 1237 618
b. (n kmol M / h
4 process
) = 237
⇒ Scale Factor =
237 kmol M / h
kmol / h
253
.

5-37
5.54 (cont’d)
Process feed: 0 .0
kmol / h
25.3 kmol / h
m (STP)
kmol
SCMH
3
254 618 02 2
237 22 4
18 700
. . .
.
,
+ + +
F
HG I
KJF
HG
I
KJ =
b g
Reactor effluent flow rate: / h
kmol / h
25.3 kmol / s
kmol / h
444
kmol
h
m (STP)
kmol
m STP
h
K
273.2 K
kPa
4925 kPa
m h
std
3
actual
3
3
49.4 kmol
237
444
22 4
9946 SCMH
9950 4732 1013
354
b gF
HG I
KJ =
⇒
F
HG I
KJF
HG
I
KJ =
⇒ = =
.
( ) . .
/
V
V
c.

/
.
V =
V
n
h
444 kmol / h
L
m
kmol
1000 mol
L / mol
3
3
= =
354 m 1000 1
08

V 20 L / mol====ideal gas approximation is poor
(5.2-36)
Most obviously, the calculation of
V from
n using the ideal gas equation of state is likely
to lead to error. In addition, the reaction equilibrium expressions are probably strictly
valid only for ideal gases, so that every calculated quantity is likely to be in error.
5.55 a.
PV
RT
B
V
B =
RT
P
B B
c
c
o 1

= + ⇒ +
1 ω
b g
From Table B.1 for ethane: T K, P atm
From Table 5.3-1 = 0.098
B
T K
K
B
T K
305.4K
c c
o
r
1
r
= =
= − = − = −
= − = − = −
3054 48 2
0 083
0 422
0 083
0 422
308 2
3054
0 333
0139
0172
0139
0172
308 2
0 0270
1 1
4 4
. .
.
.
.
.
.
.
.
.
.
.
.
.
.
.6 .6
.2 .2
ω
e j
e j
B(T) =
RT
P
B B
L atm
mol K
K
48.2 atm
L / mol
c
c
o 1
+ =
⋅
⋅
− −
= −
ω
b g b g
0 08206 3054
0 333 0 098 0 0270
01745
. .
. . .
.
PV
RT
V - B =
10.0 atm
308.2K
mol K
L atm
V V + 0.1745= 0
2
2

.

−
⋅
⋅
F
HG I
KJ −
0 08206
⇒
±
=

.
.
V =
1 1- 4 0.395 mol / L L / mol
2 0.395 mol / L
L / mol, 0.188 L / mol
b gb g
b g
01745
2 343
/ . . / . . ,
V RT P
ideal so the second solution is
likely to be a mathematical artifact.
= = × =
0 08206 308 2 10 0 2 53
b. z =
PV
RT
atm
0.08206
.343 L / mol
308.2K
L atm
mol K
.
.
= =
⋅
⋅
10 0 2
0 926
c.

.
m =
V
V
MW =
1000 L
h
mol
.343 L
g
mol
kg
1000 g
/ h
2
30 0 1
12.8 kg
=

5-38
5.56
PV
RT
B
V
B =
RT
P
B B
c
c
o 1

= + ⇒ +
1 ω
b g
From Table B.1 T CH OH K, P atm
T C H K, P atm
From Table 5.3-1 CH OH = 0.559, C H = 0.152
B CH OH)
T
B C H )
T
B CH OH)
T
c 3 c
c 3 8 c
3 3 8
o 3
r
o 3 8
r
1 3
r
b g
b g
b g b g
e j
e j
= =
= =
= − = − = −
= − = − = −
= − = −
5132 7850
369 9 42 0
0 083
0 422
0 083
0 422
373.2K
513.2K
0 619
0 083
0 422
0 083
0 422
373.2K
369.9K
0 333
0139
0172
0139
0172
373.2K
1 1
1 1
4
. .
. .
( .
.
.
.
.
( .
.
.
.
.
( .
.
.
.
.6 .6
.6 .6
.2
ω ω
513.2K
B C H )
T
369.9K
1 3 8
r
e j
e j
4
4 4
0516
0139
0172
0139
0172
373.2K
0 0270
.2
.2 .2
.
( .
.
.
.
.
= −
= − = − = −
B(CH OH) =
RT
P
B B
L atm
mol K 78.5 atm
B(C H ) =
RT
P
B B
L atm
mol K 42.0 atm
3
c
c
o 1
L
mol
3 8
c
c
o 1
L
mol
+
=
⋅
⋅
− − = −
+
=
⋅
⋅
− − = −
ω
ω
b g
b g
c h
b g
b g
c h
0 08206 513.2K
0 619 0559 0516 0 4868
0 08206 369.9 K
0 333 0152 0 0270 0 2436
.
. . . .
.
. . . .
B y y B B B B
B L / mol = -0.3652 L / mol
B
L / mol
mix i j ij
j
i
ij ii jj
ij
mix
= ⇒ = +
= − −
= − + − + −
= −
∑
∑ 05
05 0 4868 0 2436
0 30 0 30 0 4868 2 0 30 0 70 0 3652 0 70 0 70 0 2436
0 3166
.
. . .
. . . . . . . . .
.
d i
b g
b gb gb g b gb gb g b gb gb g
PV
RT
V - B =
10.0 atm
373.2K
mol K
L atm
V V + 0.3166 = 0
2
mix
2

.

−
⋅
⋅
F
HG I
KJ −
0 08206
Solve for V:V =
1 1- 4 0.326 mol / L .3166 L / mol
2 0.326 mol / L
.70 L / mol, 0.359 L / mol

±
=
b gb g
b g
0
2
. .
. .
V
RT
P
L atm
mol K
K
10.0 atm
L / mol V L / mol
ideal virial
= =
⋅
⋅
= ⇒ =
0 08206 3732
306 2 70

.
.
V = Vn
L / mol 15.0 kmol CH OH / h
kmol CH OH / kmol
1000 mol
kmol
m
1000 L
/ h
3
3
3
3
= =
2 70
0 30 1
1
135 m

5-39
5.57 a. van der Waals equation: P =
RT
V - b
a
V
2
2

d i
−
Multiply both sides by V V - b PV PV b = RTV aV + ab
PV + -Pb - RT V aV - = 0
c P = 50.0 atm
c -Pb - RT atm L / mol K L atm / mol
c a = atm L mol
c ab = - atm L mol L / mol
atm L
mol
2 3 2 2
3 2
3
2
L atm
mol K
1
2 2
0
2 2
3
3

ab
. . . .
. /
. / .
.
d i
b g
b g b gb g c hb g
d ib g
⇒ − −
+
=
= = − − = − ⋅
= − ⋅
= − ⋅
= −
⋅
⋅
⋅
50 0 0 0366 0 08206 223 201
133
133 0 0366
0 0487
b. .
.
V
RT
P
L atm
mol K
K
50.0 atm
L / mol
ideal = =
⋅
⋅
=
0 08206 223
0 366
c.
T(K) P(atm) c3 c2 c1 c0 V(ideal) V f(V) %error
(L/mol) (L/mol)
223 1.0 1.0 -18.336 1.33 -0.0487 18.2994 18.2633 0.0000 0.2
223 10.0 10.0 -18.6654 1.33 -0.0487 1.8299 1.7939 0.0000 2.0
223 50.0 50.0 -20.1294 1.33 -0.0487 0.3660 0.3313 0.0008 10.5
223 100.0 100.0 -21.9594 1.33 -0.0487 0.1830 0.1532 -0.0007 19.4
223 200.0 200.0 -25.6194 1.33 -0.0487 0.0915 0.0835 0.0002 9.6
d. 1 eq. in 1 unknown - use Newton-Raphson.
1 0 133
b g d i b g b g
⇒ = +
g V 50 V + -20.1294 V V-.0487 = 0
3 2
. .
Eq. (A.2-13)⇒ =
∂
∂
= −
a
g
V
V V +1.33
2

.
150 40 259
Eq. (A.2-14)⇒ = − ⇒ =
−
ad g d
g
a
solve
Then
V V d
(k+1) (k)
= + Guess .
V V L / mol
(1)
ideal
= = 0 3660 .

V(k)
V(k+1)
1 0.3660 0.33714
2 0.33714 0.33137
3 0.33137 0.33114
4 0.33114 0.33114 converged
b

5-40
5.58 C H T K P atm 4.26 10 Pa
3 8 C C
6
: . . .
= = × =
369 9 42 0 0152
d i ω
Specific Volume
5.0 m 44.09 kg 1 kmol
75 kg 1 kmol 10 mol
m mol
3
3
3
= × −
2 93 10 3
.
Calculate constants
a
0.42747 8.314 m Pa mol K K
Pa
m Pa mol
b
0.08664 8.314 m Pa mol K K
Pa
m mol
m
3
6 2
3
3
=
⋅ ⋅
×
= ⋅
=
⋅ ⋅
×
= ×
= + − =
= + − =
−
d i b g
d i b g
b g b g
e j
2 2
6
6
5
2
2
369 9
4 26 10
0 949
369 9
4 26 10
6 25 10
0 48508 155171 0152 015613 0152 0 717
1 0 717 1 298 2 369 9 115
.
.
.
.
.
.
. . . . . .
. . . .
α
SRK Equation:
P
m Pa mol K K
m mol
.949 m Pa mol
m mol m mol
P Pa 7.30 atm
3
3
6 2
3 3
=
⋅ ⋅
× − ×
−
⋅
× × + ×
⇒ = × ⇒
− − − − −
8 314 298 2
2 93 10 6 25 10
115 0
2 93 10 2 93 10 6 25 10
7 40 10
3 5 3 3 5
6
. .
. .
.
. . .
.
d ib g
d i
d i
d i
Ideal: P
RT
V
m Pa mol K K
m mol
Pa 8 atm
3
3
= =
⋅ ⋅
×
= × ⇒
−

. .
.
. .
8 314 298 2
2 93 10
8 46 10 35
3
6
d ib g
Percent Error:
( . . )
.
.
8 35 7 30
7
100% 14 4%
−
× =
atm
30 atm
5.59 CO : T 304.2 K P 72.9 atm
2 C C
= = =
ω 0 225
.
Ar: T 151.2 K P 48.0 atm
C C
= = = −
ω 0 004
.
P atm
= 510
. , . .
V L / 50.0 mol L mol
= =
350 0 70
Calculate constants (use R L atm mol K
= ⋅ ⋅
0 08206
. )
CO : a
L atm
mol
m , b
L
mol
T
Ar: a
L atm
mol
m , b
L
mol
T
2
2
2
2
2
=
⋅
= = = + −
=
⋅
= = = + −
365 0826 0 0297 1 0826 1 304 2
137 0 479 0 0224 1 0 479 1 1512
2
2
. , . . , . .
. , . . , . .
α
α
e j
e j
f T
RT
V b
a
V V b
1 m 1 T T P = 0
C
2
b g d i e j
=
−
−
+
+ − −

Use E-Z Solve. Initial value (ideal gas):
T atm
L
mol
L atm
mol K
K
ideal =
F
HG I
KJ ⋅
⋅
F
HG I
KJ =
510 0 70 0 08206 4350
. . . .
b g
E - Z Solve T K , T 431.2 K
max CO max Ar
⇒ = =
b g b g
2
4554
.

5-41
5.60 O :
2 T K
C = 154 4
. ; P atm
C = 49 7
. ; ω = 0 021
. ; T K 65 C
= °
208 2
. b g; P atm
= 8 3
. ;

m kg h
= 250 ; R L atm mol K
= ⋅ ⋅
0 08206
.
SRK constants: a L atm mol
2 2
= ⋅
138
. ; b L mol
= 0 0221
. ; m = 0517
. ; α = 0840
.
SRK equation: f V
RT
V b
a
V V b
P = 0=====V = 2.01 L / mol
E-Z Solve

d i d i d i
=
−
−
+
−
α
⇒ = =

V
250 kg kmol 10 mol 2.01 L
h 32.00 kg 1 kmol mol
L h
3
15,700
5.61
F P A - W = 0 where W = mg = 5500 kg 9.81 N
y CO
m
s
2 2
∑ = ⋅ =
e j 53900
a. P
W
A
N
m
atm
1.013 10 N / m
atm
CO
piston 4
5 2
2
= =
×
=
53900
015
1
301
2
π
.
.
b g
b. SRK equation of state: P =
RT
V - b
a
V V + b

d i d i
−
α
For CO : T , P atm
2 c c
= =
304 2 72 9
. . ,ω = 0.225
a = m atm / kmol , b = m / kmol, m C)
6 2 3 o
3654 0 02967 08263 25 1016
. . . , ( .
⋅ = =
α
301
298 2 1016 3654
3
.
.

. .

atm =
0.08206 K
V - 0.02967 V V + 0.02967
===== V = 0.675 m / kmol
m atm
kmol K
m
kmol
m atm
kmol
m
kmol
E-Z Solve
3
3
6
2
6
2
⋅
⋅
⋅
−
e jb g
e j
b ge j
d i
V before expansion m
V after expansion m m m m
3
3 3
b g
b g b g b g
=
= + =
0 030
0 030 015 15 0 0565
4
2
.
. . . .
π
m
V
V
MW =
0.0565 m
m / kmol
kg
kmol
kg
CO
3
3
2
= =
.
.
.
0 675
44 01
368
m initially) =
PV
RT
MW =
1 atm m
298.2 K
kg
kmol
kg
CO m atm
kmol K
3
2 3
(
.
. .
.
0 08206
0 030 44 01
0 0540
⋅
⋅
=
m added) = 3.68- 0.0540 kg = 3.63 kg
CO2
(
W
P A
CO2
⋅

5-42
5.61 (cont’d)
c.
Given T, V h, find d
Initial: n
V
RT
P
o
o
o
o
,
= = 1
b g
Final: V = V
d h
4
n = n
(kg)
44 (kg / kmol)
V
RT
o
2
o
o
+ + = +
π
,
.
.
363
0 0825

.

,
/
V =
V
n
V
d h
4
V
RT
P =
W
A
RT
V - b
a
V V + b d
RT
V - b
a
V V + b
1
o
2
o
piston
2
=
+
+
= − ⇒ = −
π
α
π
α
0 0825
53 900
4
d i d i
b g
Substitute expression for V in 1 one equation in one unknown. Solve for d
b g⇒ .
5.62 a. Using ideal gas assumption:
P
nRT
V
P
lb O lb - mole
32.0 lb
ft psia
lb - mole R
R
2.5 ft
psia = 2400 psig
g atm
m 2
m
3
o
o
3
= − =
⋅
⋅
−
353 1 10 73 509 7
14 7
. . .
.
b. SRK Equation of state: P =
RT
V - b
a
V V + b

d i d i
−
α
3 3
m
ideal
m
32.0 lb /lb-mole
2.5 ft ft
V̂ = 2.27
35.3 lb lb-mole
(Use as a first estimate when solving the SRK equation)
=
For O : T R, P psi, = 0.021
2 c
o
c
= =
277 9 730 4
. . ω
a =
ft psi
lb - mole
, b
ft
lb - mole
m = F
6
2
3
o
52038 0 3537 0518 50 0 667
. . , . , .
⋅
= =
αd i
2400 +14.7 psi =
10 R
V - 0.3537 V V + 0.3537
ft psi
lb-mole R
o
ft
lb-mole
ft psi
lb-mole
ft
lb-mole
3
o
3
6
6
b g
e jd i
d i
b ge j
d i
. .

. .

73 509 7 0 667 52038 2
2
⋅
⋅
⋅
−
E - Z Solve V = 2.139 ft lb - mole
3
⇒ /
n (kmol)
V (m
atm, 25 C
o
o
3
o
)
1
ho
d(m)
========
add 3.63 kg CO2
d(m)
ho
h
n (kmol)
P (atm), 25 C
o
V
W = 53,900 N

5-43
5.62 (cont’d)
m
V
V
MW =
2.5 ft
2.139 ft lb - mole
lb
lb - mole
lb
O
3
3
m
m
2
= =
/
.
.
32 0
37 4
Ideal gas gives a conservative estimate. It calls for charging less O2 than the tank can
safely hold.
c. 1. Pressure gauge is faulty
2. The room temperature is higher than 50°F
3. Crack or weakness in the tank
4. Tank was not completely evacuated before charging and O2 reacted with something in
the tank
5. Faulty scale used to measure O2
6. The tank was mislabeled and did not contain pure oxygen.
5.63 a. SRK Equation of State: P =
RT
V - b
a
V V + b

d i d i
−
α
⇒
− + =
= − +
multiply both sides of the equation by V V - b V + b
f V = PV V - b V + b RTV V + b a V - b
f V PV RTV a - b P - bRT V - ab = 0
3 2 2
:

d id i
d i d id i d i d i
d i d i
α
α α
0
b.
Problem5.63-SRKEquationSpreadsheet
Species CO2
Tc(K) 304.2 R=0.08206m^3atm/kmol K
Pc(atm) 72.9
ω 0.225
a 3.653924 m^6atm/kmol^2
b 0.029668 m^3/kmol
m 0.826312
f(V)=B14*E14^3-0.08206*A14*E14^2+($B$7*C14-$B$8^2*B14-$B$8*0.08206*A14)*E14-C14*$B$7*$B$8
T(K) P(atm) alpha V(ideal) V(SRK) f(V)
200 6.8 1.3370 2.4135 2.1125 0.0003
250 12.3 1.1604 1.6679 1.4727 0.0001
300 6.8 1.0115 3.6203 3.4972 0.0001
300 21.5 1.0115 1.1450 1.0149 0.0000
300 50.0 1.0115 0.4924 0.3392 0.0001
c. E-Z Solve solves the equation f(V)=0 in one step. Answers identical to VSRK values in part b.
d. REAL T, P, TC, PC, W, R, A, B, M, ALP, Y, VP, F, FP
INTEGER I
CHARACTER A20 GAS
DATA R 10.08206/
READ (5, *) GAS
WRITE (6, *) GAS
10 READ (5, *) TC, PC, W

5-44
5.63 (cont’d)
READ (5, *) T, P
IF (T.LT.Q.) STOP
R = 0 42747
. *R*R/PC*TC*TC
B = 0 08664
. *R*TC/PC
M W W
= + = − ∗
0 48508 155171 015613
. . .
b g
ALP M T / TC
= + ∗ − ∗∗ ∗∗
1 1 05 2
. .
b g
c h
d i .
VP R T / P
= ∗
DO 20 I = 7 15
,
V = VP
F = R * T/(V – B) – ALP * A/V/(V + B) – P
FP = ALP * A * (2. * V + B)/V/V/(V + B) ** 2 – R * T/(V – B) ** 2.
VP = V – F/FP
IF (ABS(VP – V)/VP.LT.0.0001) GOTO 30
20 CONTINUE
WRITE (6, 2)
2 FORMAT ('DID NOT CONVERGE')
STOP
30 WRITE (6, 3) T, P, VP
3 FORMAT (F6.1, 'K', 3X, F5.1, 'ATM', 3X, F5.2, 'LITER/MOL')
GOTO 10
END
$ DATA
CARBON DIOXIDE
304.2 72.9 0.225
200.0 6.8
250.0 12.3
300.0 21.5
–1 0.
RESULTS
CARBON DIOXIDE
200.0 K 6.8 ATM 2.11 LITER/MOL
250.0 K 12.3 ATM 1.47 LITER/MOL
300.0 K 6.8 ATM 3.50 LITER/MOL
300.0 K 21.5 ATM 1.01 LITER/MOL
300.0 K 50.0 ATM 0.34 LITER/MOL
5.64 a.
N : T 126.2 K
P 33.5 atm
T
P
MPa 10 atm
33.5 atm 1.013 MPa
z
2 C
C
r
r
Fig. 5.4-4
=
=
⇒
= + =
= =
U
V
|
W
|
⇒ =
40 2732 126 2 2 48
40
1178
12
. . .
.
.
b g
b. He: T K
P atm
T
P
z
C
C
r
r
Fig. 5.4-4
=
=
⇒
= − + + =
= + =
U
V
|
W
| ⇒ =
526
2 26
200 2732 526 8 552
350 2 26 8 3411
16
.
.
. . .
. .
.
b g b g
b g
↑
Newton’s correction

5-45
5.65 a. ρ kg / m
m (kg)
V (m )
(MW)P
RT
3
3
d i= =
= =
⋅
⋅
30 kg kmol
465 K
9.0 MPa
0.08206
10 atm
1.013 MPa
kg m
m atm
kmol K
3
3 69 8
.
b.
T
P
r
r
Fig. 5.4-3
= =
= =
U
V
W
⇒ =
465 310 15
9 0 4 5 2 0
084
.
. . .
.
z
ρ = = =
(MW)P
zRT
kg m
kg m
3
3
69 8
084
831
.
.
.
5.66 Moles of CO :
2
100 lb CO lb - mole CO
lb CO
lb - moles
m 2 2
m 2
1
44 01
2 27
.
.
=
( )
C
r C
C
3
C
r 3
C
1600 14.7 psi 1 atm
T 304.2 K
P P P 1.507
72.9 atm 14.7 psi
P 72.9 atm
ˆ 10.0 ft 72.9 atm lb-mole R 1 K
VP
V̂ 0.80
RT 2.27 lb-moles 304.2 K 0.7302 ft atm 1.8 R
+
= ⎫
⇒ = = =
⎬
= ⎭
⋅°
= = =
⋅ °
Fig. 5.4-3: Pr = 1507
. , V z
r = ⇒ =
080 085
. .
T
PV
znR
1614.7 psi 10.0 ft lb - mole R 1 atm
0.85 2.27 lb - moles 0.7302 ft atm 14.7 psi
R F
3
3
= =
⋅°
⋅
= ° = °
779 320
5.67 O : T 154.4 K
P 49.7 atm
T 298 154.4 1.93
P 1 49.7 0.02
z 1.00 (Fig. 5.4 - 2)
T 358 154.4 2.23
P 1000 49.7 20.12
z 1.61 Fig. 5.4 - 4
2 C
C
r
r
1
r
r
2
1
1
2
2
=
=
= =
= =
U
V
|
W
| =
= =
= =
U
V
|
W
| = b g
V V
z
z
T
T
P
P
V
127 m 1.61 358 K 1 atm
h 1.00 298 K 1000 atm
m h
2 1
2
1
2
1
1
2
2
3
3
=
= = 0 246
.
5.68 O : T 154.4 K
P 49.7 atm
T 27 273.2 154.4 1.94
P 175 49.7 3.52 z 0.95
P 1.1 49.7 0.02 z 1.00
2 C
C
r
r 1
r 2
1
2
=
=
= + =
= = ⇒ =
= = ⇒ =
b g
(Fig. 5.3-2)
n n
V
RT
P
z
P
z
10.0 L mol K
300.2 K 0.08206 L atm
atm
0.95
atm
1.00
1 2
1
1
2
2
2
− = −
F
HG I
KJ =
⋅
⋅
−
F
HG I
KJ =
175 11
74.3 mol O
.

5-46
5.69 a. .
.
.
.
V =
V
n
mL
g
g
mol
mL / mol
= =
50 0
500
44 01
4401
P =
RT
V
82.06 mL atm
mol K 440.1 mL / mol
atm

=
⋅
⋅
=
1000 K
186
b. For CO : T K, P atm
2 c c
= =
304 2 72 9
. .
T
T
T
K
304.2 K
V
VP
RT
mL
mol
atm
304.2 K
mol K
82.06 mL atm
r
c
r
ideal c
c
= = =
= =
⋅
⋅
=
1000
32873
4401 72 9
128
.
. .
.
ideal
r r
Figure 5.4-3: V 1.28 and T 3.29 z=1.02
zRT 1.02 82.06 mL atm mol 1000 K
P= 190 atm
ˆ mol K 440.1 mL
V
= = ⇒
⋅
= =
⋅
c. a = mL atm / mol , b = mL / mol, m
2 2
3654 10 29 67 08263 1000 01077
6
. . . , ( ) .
× ⋅ = =
α K
P =
82.06
440.1- 29.67 440 440.1+ 29.67
atm
mL atm
mol K
mL
mol
mL atm
mol
mL
mol
2
2
2
2
⋅
⋅
⋅
−
×
=
c hb g
b g
b ge j
b g
1000 K 01077 3654 10
1
198
6
. .
.
5.70 a. The tank is being purged in case it is later filled with a gas that could ignite in the
presence of O2.
b. Enough N2 needs to be added to make xO2
= × −
10 10 6
. Since the O2 is so dilute at this
condition, the properties of the gas will be that of N2.
T K, P atm, T
c c r
= = =
126 2 335 2 36
. . .
n n
PV
RT
atm
0.08206
L
298.2 K
mol
n mol air
0.21 mol O
mol air
mol O
n
n
n mol
initial 1 L atm
mol K
O
2
2
O
2
2
2
2
= = = =
=
F
HG I
KJ =
= × ⇒ = ×
⋅
⋅
− −
1 5000
204 3
204 3 42 9
10 10 4 29 10
6 6
.
. .
.
.

. .
.

. .
.
.
.
V =
5000 L
4.29 10 mol
10 L / mol
V
VP
RT
L
mol
mol K
0.08206 L atm
atm
126.2 K
not found on compressibility charts
Ideal gas: P =
RT
V
L atm
mol K
K
L / mol
atm
The pressure required will be higher than atm if z 1, which from
Fig. 5.3- 3 is very likely.
6
-3
r
ideal c
c
×
= ×
= =
× ⋅
⋅
= ×
⇒
=
⋅
⋅ ×
= ×
× ≥
−
−
−
116
116 10 335
38 10
0 08206 298 2
116 10
21 10
21 10
3
3
3
4
4
n mol N kg N mol kg N
added 2 2 2
= × − ≅ × = ×
4 29 10 204 3 4 29 10 0 028 120 10
6 6 5
. . . . / .
d ib g

5-47
5.70 (cont’d)
c.
N at 700 kPa gauge = 7.91 atm abs. P T ======= z = 0.99
2 r r
Fig 5.4-2
⇒ = =
0 236 2 36
. , .
n
P V
zRT
atm
0.99
L
0.08206 .2 K
kmol
y
y n
1.634
y
y n
1.634
y
n
y y
n
n =
ln
y
y
ln
n
1.634
Need at least 5 stages
Total N kmol N kg / kmol
2
2
L atm
mol K
1
init init
2
1 init
init
init
n init
init
n
n
init
init
2 2 2
= = =
= = =
= =
F
HG I
KJ =
=
F
HG I
KJ ⇒
F
HG I
KJ
F
HG I
KJ
= ⇒
= =
⋅
⋅
7 91 5000
298
1633
0 21 0 204
1634
0 026
1634
0 0033
1634
4 8
5 143 28 0 200 kg N
2
.
.
. .
.
.
.
.
.
.
. .
b g
b gb g
d. Multiple cycles use less N2 and require lower operating pressures. The disadvantage is
that it takes longer.
5.71 a.

. /
.

m = MW
PV
RT
Cost ($ / h) = mS = MW
SPV
RT
lb lb - mol
0.7302
SPV
T
SPV
T
m
ft atm
lb-mol R
3
o
⇒ =
F
HG
I
KJ =
⋅
⋅
44 09
60 4
b. T K = 665.8 R T
P atm P
z = 0.91
c
o
r
c r
Fig. 5.4-2
= ⇒ =
= ⇒ =
U
V
W
⇒
369 9 085
42 0 016
. .
. .

.
m = 60.4
PV
zT
m
z
m
Delivering 10% more than they are charging for (undercharging their customer)
ideal
ideal
= =
⇒
110
n kmol
y kmol O kmol
initial
O2 2
=
=
0 204
0 21
.
. /
143
. kmol N
y
2
1
143
. kmol N
y
2
2
143
. kmol N2
143
. kmol N2

5-48
5.72 a. For N T K R, P atm
2 c
o
c
: . . .
= = =
126 20 22716 335
After heater T
R
R
P
psia
atm
atm
14.7 psia
z 1.02
r
o
o
r
:
.
.
.
.
.
= =
= =
U
V
|
|
W
|
|
⇒ =
609 7
22716
2 68
600
335
1
12
.
n =
150 SCFM
359 SCF / lb - mole
lb - mole / min
= 0 418
. . .
/ min
V =
zRTn
P
lb - mole
min
psia
lb - mole R
R
600 psia
.
3
o
o
3
=
⋅
⋅
=
102 0 418 10.73 ft 609 7
4 65 ft
b. tank =
0.418 lb - mole
min
lb lb - mole
0.81 lb ft
min
h
h
day
days
week
weeks
m
m
3
28
62 4
60 24 7 2
/
. /
b g
= =
4668 34 900
ft gal
3
,
5.73 a. For CO T K, P atm
c c
: . .
= =
1330 34 5
Initially T
K
133.0 K
P
2514.7 psia
34.5 atm
atm
14.7 psia
z =1.02
r1
r1
Fig. 5.4-3
: .
.
= =
= =
U
V
|
|
W
|
|
⇒
300
2 26
1
50
n
psia
1.02
L
300 K
atm
14.7 psia
mol K
0.08206 L atm
mol
1 =
⋅
⋅
=
2514 7 150 1
1022
.
After 60h T
K
133.0 K
P
2258.7 psia
34.5 atm
atm
14.7 psia
z =1.02
r1
r1
Fig. 5.4-3
: .
.
= =
= =
U
V
|
|
W
|
|
⇒
300
2 26
1
4 5
n
psia
1.02
L
300 K
atm
14.7 psia
mol K
0.08206 L atm
mol
2 =
⋅
⋅
=
2259 7 150 1
918
.
.
n =
n n
60 h
mol / h
leak
1 2
−
= 173
b. n y n y
PV
RT
mol CO
mol air
atm
0.08206
m
K
L
m
mol
2 2 air 2 L atm
mol K
3
3
= = =
×
=
−
⋅
⋅
200 10 1 30 7
300
1000
0 25
6
.
.
t
n
n
mol
1.73 mol / h
h
t would be greater because the room is not perfectly sealed
min
2
leak
min
= = =
⇒

.
.
0 25
014
c. (i) CO may not be evenly dispersed in the room air; (ii) you could walk into a high
concentration area; (iii) there may be residual CO left from another tank; (iv) the tank
temperature could be higher than the room temperature, and the estimate of gas escaping
could be low.

5-49
5.74 CH4 : T K
c = 190 7
. , P atm
c = 458
.
C H
2 6 : T K
c = 3054
. , P atm
c = 48 2
.
C H
2 4 : T K
c = 2831
. , P atm
c = 505
.
Pseudocritical temperature: ′ = + + =
T K
c 0 20 190 7 0 30 3054 050 2831 2713
. . . . . . .
Pseudocritical pressure: ′ = + + =
P atm
c 0 20 458 0 30 48 2 050 505 48 9
. . . . . . .
Reduced temperature:
Reduced pressure:
T
K
K
P
200 bars 1 atm
48.9 atm 1.01325 bars
r
r
Figure 5.4-3
=
+
=
= =
U
V
|
|
W
|
|
⇒ =
90 2732
2713
134
4 04
0 71
.
.
.
.
.
b g
z
Mean molecular weight of mixture:
M 0.20 M 0.30 M 0.50 M
0.20 16.04 0.30 30.07 0.50 28.05
26.25 kg kmol
CH C H C H
4 2 6 2 4
= + +
= + +
=
b g b g b g
V
znRT
P
0.71 10 kg 1 kmol 0.08314 m bar 90 + 273 K
26.25 kg kmol K 200 bars
m L)
3
= =
⋅
⋅
=
b g 0 041 41
3
. (
5.75 N : T 126.2 K, P 33.5 atm
N O: T 309.5 K, P 71.7 atm
T 0.10 309.5 0.90 126.2 144.5 K
P 0.10 71.7 0.90 33.5 37.3 atm
2 c C
2 c C
c
c
= =
= =
U
V
W
′ = + =
′ = + =
b g b g
b g b g
M 0.10 44.02 0.90 28.02 29.62
n 5.0 kg 1 kmol 29.62 kg 0.169 kmol 169 mol
= + =
= = =
b g b g
b g
a. T 24 273.2 144.5 2.06
V
30 L 37.3 atm mol K
169 mol 144.5 K 0.08206 L atm
0.56
z 0.97 Fig. 5.4 - 3
r
r
= + =
=
⋅
⋅
=
U
V
|
W
|
⇒ =
b g
b g

P
0.97 169 mol 297.2 K 0.08206 L atm
30 L mol K
atm 132 atm gauge
=
⋅
⋅
= ⇒
133
b. P 273 37.3 7.32
V 0.56 from a.
z 1.14 Fig. 5.4 - 3
r
r
= =
=
U
V
|
W
|⇒ =
b g b g
T
atm
1.14 mol
mol K
0.08206 L atm
518 K 245 C
=
⋅
⋅
= ⇒ °
273 30 L
169

5-50
5.76 CO: T 133.0 K, P 34.5 atm
H : T 33 K, P 12 atm
T 0.60 133.0 0.40 33 8 96.2 K
P 0.60 34.5 0.40 12.8 8 29.0 atm
c c
2 c c
c
c
= =
= =
U
V
W
′ = + + =
′ = + + =
.8
b g b g
b g b g
Turbine inlet: T 150 273.2 96.2 4.4
P
2000 psi 1 atm
29.0 atm 14.7 psi
4.69
z 1.01
Turbine exit: T 373.2 96.2 3.88
P 1 29.0 0.03
r
r
Fig. 5.4-1
r
r
= + =
= =
U
V
|
W
|
⎯ →
⎯⎯
⎯ ≈
= =
= =
b g
P V
P V
z nRT
z n RT
V V
P
P
z
z
T
T
15,000
ft
min
14.7 psia
2000 psia
1.01
1.00
423.2K
373.2
ft
in in
out out
in in
out out
in out
out
in
in
out
in
out
3
3

/ min
= ⇒ = × =
= 126
If the ideal gas equation of state were used, the factor 1.01 would instead be 1.00
⇒ −1% error
5.77 CO: T 133.0 K, P 34.5 atm
CO : T 304.2 K, P 72.9 atm
T 0.97 133.0 0.03 304.2 138.1 K
P 0.97 34.5 0.03 72.9 35.7 atm 524.8 psi
c c
2 c c
c
c
= =
= =
U
V
W
′ = + =
′ = + = =
b g b g
b g b g
Initial: T 303.2 138.1 2.2
P 2014.7 524.8 3.8
z 0.97
r
r
Fig. 5.4-3
1
= =
= =
U
V
W
⎯ →
⎯⎯
⎯ =
Final: P z
r 1
= = ⇒ =
1889 7 524 8 36 0 97
. . . .
Total moles leaked:
n n
P
z
P
z
V
RT
2000 1875 psi 30.0 L 1 atm mol K
0.97 303 K 14.7 psi 0.08206 L atm
10 mol leaked
1 2
1
1
2
2
− = −
F
HG I
KJ =
− ⋅
⋅
=
b g
.6
Moles CO leaked: mol CO
0 97 10 6 10 3
. . .
b g=
Total moles in room:
24 2 10
303
9734
3
.
.
m L 273 K 1 mol
1 m K 22.4 L STP
mol
3
3
b g=
Mole% CO in room =
10 3
9734
100% 10%
.
.
.
mol CO
mol
CO
× =
⇒ z=1.0

5-51
5.78 Basis: 54.5 kmol CH OH h
3 CO 2H CH OH
2 3
+ →
a.
n
54.5 kmol CH OH 1 kmol CO react 1 kmol CO fed
h 1 kmol CH OH 0.25 kmol CO react
kmol h CO
1
3
3
= = 218
2n kmol H h
1 2
= =
2 218 436
b g ⇒ 218 436 654
+ =
b g kmol h (total feed)
CO: T K
c = 1330
. P atm
c = 34 5
.
H2 : T K
c = 33 P atm
c = 12 8
.
⇓ Newton’s corrections
′ = + + =
T K
c
1
3
1330
2
3
33 8 717
. .
b g b g
′ = + + =
P atm
c
1
3
34 5
2
3
12 8 8 254
. . .
b g b g
T 644 71.7 8.98
P
34.5 MPa 10 atm
24.5 atm 1.013 MPa
13.45
z 1.18
r
r
Fig. 5.4-4
1
= =
= =
U
V
|
W
|
⎯ →
⎯⎯
⎯ =

V
1.18 654 kmol 644 K 0.08206 m atm 1.013 MPa
h 34.5 MPa kmol K 10 atm
m h
feed
3
3
=
⋅
⋅
= 120
V
m h 1 m cat
25,000 m / h
m catalyst (4.8 L)
cat
3 3
3
3
= =
120
0 0048
.
b.
Overall C balance ⇒ =
.
n mol CO h
4 54 5
Fresh feed: 54.5 kmol CO h
109.0 kmol H h
163.5 kmol feed gas h
2

V
1.18 163.5 kmol 644 K 0.08206 m atm 1.013 MPa
h 34.5 MPa kmol K 10 atm
m h
feed
3
3
=
⋅
⋅
= 29.9

n (kmol CO / h)
1
2n (kmol H h)
644 K
34.5 MPa
1 2
/
CO, H2
54 5
. ( ) /
kmol CH OH h
3 l
Catalyst
Bed Condenser

n kmol CO / h
4
2n kmol H h
4 2
/
54 5
. kmol CH OH ( ) / h
3 l
CO, H2

5-52
5.79 H T K = 41.3 K 1- butene T K
P atm = 20.8 atm P atm
2 c c
c c
: ( . ) : .
( . ) .
= + =
= + =
333 8 419 6
12 8 8 39 7
T K) + 0.85(419.6 K) = 362.8 K T
P atm) + 0.85(39.7 atm) = 36.9 atm P
z = 0.86
c r
c r
Fig. 5.4-2
' . ( . ' .
' . ( . ' .
= =
= =
U
V
W
⇒
015 413 089
015 208 0 27
. .
. / min
V =
znRT
P
kmol
h
m atm
kmol K
K
10 atm
h
60 min
m
3
3
=
⋅
⋅
=
086 35 0 08206 323 1
133

. / min
.
V
m
min
= u
m
min
A m = u
d
4
d =
4V
u
m
m / min
cm
m
cm
3
2
2 3
F
HG
I
KJ F
HG I
KJ × ⇒ =
F
HG I
KJ =
d i d i
b g
π
π π
4 133
150
100
10 6
5.80 4 c c
2 4 c c
2 6 c c
CH : T 190.7 K P 45.8 atm
C H : T 283.1 K P 50.5 atm
C H : T 305.4 K P 48.2 atm
= =
= =
= =
T K) + 0.60(283.1 K) + 0.25(305.4 K) = 274.8 K==== T
P atm) + 0.60(50.5 atm) + 0.25(48.2 atm) = 49.2 atm===== P
z = 0.67
c
T=90 C
r
c
P=175 bar
r
Fig. 5.4-3
o
' . ( . ' .
' . ( . ' .
= =
= =
U
V
|
W
|
⎯ →
⎯⎯
⎯
015 190 7 132
015 458 35
. .
V
m
= u
m
s
A m = 10
m
s
s
min 4
m
m
min
3
2
3
s
F
HG
I
KJ F
HG I
KJ F
HG I
KJF
HG I
KJ =
d i b g
60 0 02 0188
2
π

. / min
.
n =
PV
zRT
bar
0.67
atm
1.013 bar
kmol K
.08206 m atm
m
K
kmol / min
3
3
=
⋅
⋅
=
175 1 0188
363
163
5.81 N T K = 227.16 R P atm
acetonitrile T K = 986.4 R P atm
2 c
o
c
c
o
c
: . .
: .
= =
= =
126 2 335
548 47 7
Tank 1 (acetonitrile) T = 550 F, P = 4500 psia T P z = 0.80
n =
P V
z RT
306 atm
0.80
0.200 ft
R
lb - mole R
0.7302 ft atm
= 0.104 lb - mole
1
o
1 r1 r1
Fig. 5.4-3
1
1
1 1
1 1
3
o
o
3
: . .

.
⇒ = = ⇒
⇒ =
⋅
⋅
102 6 4
1009 7
Tank 2 (N ) T = 550 F, P = 10 atm T , P z =1.00
n =
P V
z RT
10.0 atm
1.00
2.00 ft
R
lb - mole R
.7302 ft atm
= 0.027 lb - mole
2 2
o
2 r2 r2
Fig. 5.4-3
2
2
2 2
2 2
3
o
o
3
: . .

.
⇒ = = ⇒
⇒ =
⋅
⋅
4 4 6 4
1009 7

5-53
5.81 (cont’d)
Final T R + R = 830 R T
P + = 44.8 atm
c
o o o T=550 F
r
c
o
: '
.
.
.
.
.
. ' .
'
.
.
.
.
=
F
HG I
KJ F
HG I
KJ ⎯ →
⎯⎯
⎯ =
=
F
HG I
KJ F
HG I
KJ
0104
0131
986 4
0 027
0131
22716 122
0104
0131
47.7 atm
0 027
0131
33.5 atm

'
'
. . . .
.
V =
VP
RT
=
2.2 ft
0.131 lb - mole
44.8 atm
830 R
lb - mole R
0.7302 ft atm
=1.24 z = 0.85
P =
znRT
V
lb - mole ft atm
lb - mole R
R
2.2 ft
atm
r
ideal
c
c
3
o
o
3
Fig. 5.4-2
3
o
o
3
d i ⋅
⋅
⇒
=
⋅
⋅
=
085 0131 7302 1009 7
37 3
5.82
a. Volume of sample: g 1 cm g cm
3 3
342 159 215
. . .
d i=
O in Charge:
2
n
1.000 L 2.15 cm 10 L km 499.9 kPa 1 atm
0.08206
L atm
mol K
K 101.3 kPa
0.200 mol O
O
3 3 3
2
2
=
−
⋅
⋅
=
−
d i
300
Product
n
1.000 L 1950 kPa 1 atm
0.08206
L atm
mol K
756.6 K 101.3 kPa
0.310 mol product
p = ⋅
⋅
=
Balances:
O: 2 0.200 n 0.310 2 0.387 2 0.258 0.355 n 0.110 mol O in sample
O O
b g b g b g
+ = + + ⇒ =
C: n 0.387 0.310 0.120 mol C in sample
C = =
b g
H: n 2 0.355 0.310 0.220 mol H in sample
H = =
b gb g
Assume c 1 a 0.120 0.110 1.1 b 0.220 0.110 2
= ⇒ = = = =
Since a, b, and c must be integers, possible solutions are (a,b,c) = (11,20,10), (22,40,20),
etc.
b. MW 12.01a 1.01b 16.0c 12.01 1.1c 1.01 2c 16.0c 31.23c
= + + = + + =
b g b g
300 MW 350
⇒ c 10 C H O
11 20 10
= ⇒
348 268
. , .
g C H O C, 499.9 kPa
n (mol C), n (mol H), n (mol O)
a b c
o
c H O
n (mol O
C, 499.9 kPa
O 2
o
2
)
.
268
1 L @483.4 C, 1950 kPa
n (mol)
0.387 mol CO mol
0.258 mol O mol
mol H O / mol
o
p
2
2
2
/
/
.
0 355

5-54
5.83 Basis: 10 mL C H charged to reactor
5 10 l
b g C H
15
2
O 5CO 5H O
5 10 2 2 2
+ → +
a. n
10.0 mL C H l 0.745 g 1 mol
mL 70.13 g
0.1062 mol C H
1
5 10
5 10
= =
b g
Stoichiometric air: n
0.1062 mol C H 7.5 mol O 1 mol air
1 mol C H 0.21 mol O
3.79 mol air
2
5 10 2
2 10 2
= =
P
nRT
V
3.79 mol 0.08314 L bar 300K
11.2 L mol K
8.44 bars
o = =
⋅
⋅
=
(We neglect the C H
5 10 that may be present in the gas phase due to evaporation)
Initial gauge pressure bar 1 bar bar
= − =
8 44 7 44
. .
b.
n
0.1062 mol C H 5 mol CO
1 mol C H
0.531 mol CO
n
0.531 mol CO 1 mol H O
1 mol CO
0.531 mol H O
n 0.79 3.79 2.99 mol N
4.052 mol product gas
3
5 10 2
5 10
2
4
2 2
2
2
5 2
= =
= =
= =
U
V
|
|
|
W
|
|
|
⇒
b g
CO y = 0.531/ 4.052 = 0.131 mol CO / mol, T = 304.2 K P = 72.9 atm
H O y = 0.531/ 4.052 = 0.131 mol H O / mol, T = 647.4 K P = 218.3 atm
N y = 2.99 / 4.052 = 0.738 mol N / mol, T =126.2 K P = 33.5 atm
2 3 2 c c
2 4 2 c c
2 5 2 c c
:
:
:
T K) + 0.131(647.4 K) + 0.738(126.2 K) = 217.8 K
P atm) + 0.131(218.3 atm) + 0.738(33.5 atm) = 62.9 atm P
c
c r
' . ( .
' . ( . ' .
=
= ⇒ =
0131 304 2
0131 72 9 121

'
'
.
.
.
. .
V
VP
RT
11.2 L
4.052 mol
atm
K
mol K
.08206 L atm
z 1.04 (Fig. 5.4 - 3)
T =
PV
znR
bars
1.04
L
4.052 mol
mol K
0.08314 L bar
K - 273= 2166 C
r
ideal c
c
o
= =
⋅
⋅
= ⇒ ≈
=
+ ⋅
⋅
=
62 9
217 8
9 7
753 1 112
2439
b g
10 mL C H
n (mol C H
5 10
1 5 10
l
b g
)
n (mol air)
0.21 O
N
C, 11.2 L, P (bar)
2
2
2
o
o
0 79
27
.
n (mol CO
n mol H O(v)
n (mol N )
75.3 bar (gauge), T C
3 2
4 2
5 2
ad
o
)
b g
d i

6-1
CHAPTER SIX
6.1 a. AB: Heat liquid - - constant
V ≈
BC: Evaporate liquid - - increases, system remains at point on vapor - liquid equilibrium
curve as long as some liquid is present. C.
o
V
T = 100
CD: Heat vapor - -T increases, V increases .
b. Point B: Neglect the variation of the density of liquid water with temperature, so ρ = 1.00 g/mL
B
and 10 mL
V =
Point C: H2O (v, 100°C)
10 mL 1.00 g 1 mol
0.555 mol
mL 18.02 g
n = =
0.555 mol 0.08206 L atm 373 K
17 L
1 atm mol K
C
C C C C
C
nRT
P V nRT V
P
⋅
= ⇒ = = =
⋅
6.2 a. mm Hg
final
P = 243 . Since liquid is still present, the pressure and temperature must lie on the
vapor-liquid equilibrium curve, where by definition the pressure is the vapor pressure of the
species at the system temperature.
b. Assuming ideal gas behavior for the vapor,
m(vapor)
(3.000 - 0.010) L mol K 243 mm Hg 1 atm 119.39 g
(30 + 273.2) K 0.08206 L atm 760 mm Hg mol
g
=
⋅
⋅
= 4 59
.
m(liquid)
10 mL 1.489 g
mL
g
= = 14 89
.
m m(vapor) + m(liquid) =19.5 g
total =
x =
4.59
g vapor / g total
vapor
19 48
0 235
.
.
=
6.3 a. log .
.
. .
* .370
10
2
7 09808
1238 71
45 217
2 370 10 234 5
p p
∗
= −
+
= ⇒ = = mm Hg
b. ln
ln / ln / .
* *
.0 .
p
H
R T
B
H
R
p p
v v
T T
∗
+ +
= − + ⇒ − =
−
=
−
= −
Δ Δ
1 760 118 3
4151
2 1
5
2 1
d i b g
b g b g
1 1 1
77 273.2 K
1
29 273.2 K
K
B p
H R
T
v
= + = +
+
=
ln( )
/
ln . .
*
1
1
118 3 18 49
Δ
b g b g
4151 K
29.5 273.2 K

6-2
6.3 (cont’d)
ln ( . .
. .
.
.
p p
∗ ∗
= −
+
+ ⇒ =
−
× = −
45 18 49 2310
2310 234 5
234 5
100% 15%
o
C)
4151
45 273.2
mm Hg
error
b g
c. p∗
=
−
−
F
HG I
KJ − + =
118 3 760
29 5 77
45 29 5 118 3 327.7 mm H
.
.
. .
b g g
327 7 234 5
234 5
100% 39 7%
. .
.
.
−
× = error
6.4 Plot p∗
+
log
.
scale vs
1
T
b g 2732
(rect. scale) on semilog paper
⇒ straight line: slope = − =
7076 2167
K , intercept .
ln
.
. exp
.
.
p
T
p
T
∗ ∗
=
−
+
+ ⇒ =
−
+
+
L
NM O
QP
mm Hg
7076
( C)
mm Hg
7076
( C)
o o
b g b g
2732
2167
2732
2167
ΔH
R
v
= 7076K ⇒ Δ
Hv =
⋅
=
7076 K 8.314 J 1 kJ
mol K 10 J
58.8 kJ mol
3
6.5 ln p* = A/T(K) + B
T(
o
C) p*(mm Hg) 1/T(K) ln(p*) p*(fitted) T(o
C) p*(fitted)
79.7 5 0.002834 1.609 5.03 50 0.80
105.8 20 0.002639 2.996 20.01 80 5.12
120.0 40 0.002543 3.689 39.26 110 24.55
141.8 100 0.002410 4.605 101.05 198 760.00
178.5 400 0.002214 5.991 403.81 230 2000.00 Least confidence
197.3 760 0.002125 6.633 755.13 (Extrapolated)
y = -7075.9x + 21.666
0
1
2
3
4
5
6
7
0.002
0.0022
0.0024
0.0026
0.0028
0.003
1/T
ln(p*)

6-3
6.6 a.
T(°C) 1/T(K) p*
(mm Hg)
=758.9 + hright -hleft
42.7 3.17×10-3
34.9
58.9 3.01×10-3
78.9
68.3 2.93×10-3
122.9
77.9 2.85×10-3
184.9
88.6 2.76×10-3
282.9
98.3 2.69×10-3
404.9
105.8 2.64×10-3
524.9
b.Plot is linear, ln

ln
.
.
p
H
RT
B p
K
T
v
∗= − + ⇒ ∗=
−
+
Δ 51438
19 855
At the normal boiling point, p Tb
∗= ⇒ = °
760 116
mmHg C
Δ
Hv =
⋅
=
8.314 J 5143.8 K 1 kJ
mol K 10 J
42.8 kJ mol
3
c. Yes — linearity of the ln /
p T
∗ vs 1 plot over the full range implies validity.
6.7 a. ln . ln ; .
p a T b y ax b y p x T
∗= + + ⇒ = + = ∗ = +
2732 1 2732
b g b g
Perry's Handbook, Table 3- 8:
T1 39 5
= °
. C , p x
1 1
3
400 31980 10
∗= ⇒ = × −
mm Hg . , y1 599146
= .
T2 565
= °
. C , p x
2 2
3
760 30331 10
∗= ⇒ = × −
mm Hg . , y2 6 63332
= .
T x
= ° ⇒ = × −
50 30941 10 3
C .
y y
x x
x x
y y p e
= +
−
−
F
HG I
KJ − = ⇒ ∗ ° = =
1
1
2 1
2 1
6.39588
6 39588 50 599
b g b g
. C mm Hg
b. 50 122
° = °
C F Cox chart
12 psi 760 mm Hg
625 mm Hg
14.6 psi
p
⎯⎯⎯⎯
→ ∗ = =
c. log .
.
. .7872
p p
∗= −
+
= ⇒ ∗= =
7 02447
11610
50 224
2 7872 10 613
2
mm Hg
6.8 Estimate C Assume
p p
a
T K
b
∗ ° ∗= +
35
b g b g
: ln , interpolate given data.
a
p p
b p
a
T
p
p e
T T
=
∗ ∗
−
=
−
= −
= ∗− = +
+
=
U
V
|
|
W
|
|
⇒
∗ ° = −
+
+ =
∗ ° = =
+ +
ln ln
.
ln ln
.
.
.
ln
.
.
. .
.
.2 .2
.630
2 1
1 1 1
45 273
1
25 273
1
1
4
2 1
200 50
65771
50
65771
25 2732
2597
35
65771
35 2732
2597 4 630
35 102 5
b g b g
b g
b g
b g
C
C mm Hg
Moles in gas phase:
150 mL 273 K 102.5 mm Hg 1 L 1 mol
35 + 273.2 K 760 mm Hg 10 m 22.4 L STP
mol
3
n
L
=
= × −
b g b g
8 0 10 4
.

6-4
6.9 a. m F
= = ⇒ = + − =
2 2 2 2 2 2
π . Two intensive variable values (e.g., T P) must be
specified to determine the state of the system.
b. log .
.
.
. .
p p
MEK MEK
∗ = −
+
= ⇒ ∗ = =
6 97421
1209 6
55 216
2 5107 10 324
2 5107
mm Hg
Since vapor liquid are in equilibrium p p
MEK MEK
= ∗ = 324 mm Hg
⇒ = = =
y p P
MEK MEK / . .
324 1200 0 27 0115 The vessel does not constitute an explosion
hazard.
6.10 a. The solvent with the lower flash point is easier to ignite and more dangerous. The solvent with
a flash point of 15°C should always be prevented from contacting air at room temperature. The
other one should be kept from any heating sources when contacted with air.
b. At the LFL, y p p
M M M
= ⇒ = = ×
0 06 0 06 760
. .
*
mm Hg = 45.60 mm Hg
Antoine = 7.87863-
1473.11
T + 230
C
⇒ ⇒ = °
log . .
10 4560 685
T
c. The flame may heat up the methanol-air mixture, raising its temperature above the flash point.
6.11 a. At the dew point,
p p T
∗
× ⇒ °
( (
H O) = H O) = 500 0.1= 50 mm Hg = 38.1 C from Table B.3.
2 2
b. VH O
2
3
3
2
30.0 L 273 K 500mmHg 1 mol 0.100mol H O
(50+273) K 760 mmHg 22.4 L(STP) mol
18.02 g
mol
1 cm
g
cm
= =134
.
c. (iv) (the gauge pressure)

6-5
6.12 a. T1 58 3
= °
. C , p1 755 747 52 60
∗= − − =
mm Hg mm Hg mm Hg
b g
T2 110
= °C , p2 755 577 222 400
∗= − − =
mm Hg mm Hg mm Hg
b g
ln p
a
T K
b
∗= +
b g
a
p p
T T
=
∗ ∗
−
=
−
= −
+ +
ln ln
.
.2 .2
2 1
1 1 1
110 273
1
58.3 273
2 1
400 60
46614
b g b g
b p
a
T
= ∗− = +
+
=
ln ln
.
. .
.
1
1
60
46614
58 3 2732
18156
b g
ln
.
.
p
T
∗=
−
+
46614
18156
ln . .
p p
∗ ° = ⇒ ∗ ° = =
130 6595 130 7314
C C e mm Hg
6.595
b g b g
b. Basis: 100 mol feed gas CB denotes chlorobenzene.
Saturation condition at inlet: C
mm Hg
760 mm Hg
mol CB mol
CB
y P p y
o o
= ∗ ° ⇒ = =
130
731
0 962
b g .
Saturation condition at outlet: C
mm Hg
760 mm Hg
mol CB mol
CB
y P p y
1 1
58 3
60
0 0789
= ∗ ° ⇒ = =
. .
b g
Air balance: 100 mol
1 1 100 1 0 962 1 0 0789 4126
1 1 1
− = − ⇒ = − − =
y n y n
o
b g b g b gb g b g
. . .
Total mole balance: 100 mol CB
= + ⇒ = − =
n n n l
1 2 2 100 4126 9587
. . b g
% condensation:
95.87 mol CB condensed
0.962 100 mol CB feed
×
× =
b g 100% 99 7%
.
c. Assumptions: (1) Raoult’s law holds at initial and final conditions;
(2) CB is the only condensable species (no water condenses);
(3) Clausius-Clapeyron estimate is accurate at 130°C.
6.13 T = ° °
78 F = 25.56 C , Pbar = 29 9
. in Hg = 759.5 mm Hg , hr = 87%
y p
H O
2
P 0.87 25.56 C
= ∗ °
b g ( )
2
H O 2
0.87 24.559 mm Hg
0.0281 mol H O mol air
759.5 mm Hg
y = =
( ) ( )
Dew Point: 0.0281 759.5 21.34 mm Hg
dp
p T yp
∗ = = = 23.2 C
dp
T = °
n1 mol @ 58.3°C, 1atm
100 mol @ 130°C, 1atm
y0 (mol CB(v)/mol) (sat’d)
(1-y0) (mol air/mol)
y1 (mol CB(v)/mol) (sat’d)
(1-y1) (mol air/mol)
n2 mol CB (l)
T=130o
C=403.2 K
Table B.3
Table B.3

6-6
6.13 (cont’d)
2
0.0281
0.0289 mol H O mol dry air
1 0.0281
m
h = =
−
2 2
2
2
0.0289 mol H O 18.02 g H O mol dry air
0.0180 g H O g dry air
mol dry air mol H O 29.0 g dry air
a
h = =
( ) ( ) [ ]
0.0289
100% 100% 86.5%
24.559 759.5 24.559
25.56 C 25.56 C
m
p
h
h
p P p
= × = × =
−
⎡ ⎤
∗ ° − ∗ °
⎣ ⎦
6.14 Basis I : 1 mol humid air @ 70 F (21.1 C), 1 atm,
° ° =
hr 50%
h y P p
r H O H O
50% 0.50 21.1 C
2 2
= ⇒ = ∗ °
b g
yH O
2
2
mm Hg
760.0 mm Hg
mol H O
mol
=
×
=
050 18 765
0 012
. .
.
Mass of air:
mol H O 18.02 g
1 mol
mol dry air 29.0 g
1 mol
g
2
0 012 0 988
2887
. .
.
+ =
Volume of air:
1 mol 22.4 L STP 273.2 21.1 K
1 mol 273.2K
L
b g b g
+
= 2413
.
Density of air
g
24.13 L
g L
= =
2887
1196
.
.
Basis II 1 mol humid air @ 70 F (21.1 C), 1 atm,
: ° ° =
hr 80%
h y P p
r H O H O
80% 0.80 21.1 C
2 2
= ⇒ = ∗ °
b g
yH O
2
2
mm Hg
760.0 mm Hg
mol H O
mol
=
×
=
080 18 765
0 020
. .
.
Mass of air:
mol H O 18.02 g
1 mol
mol dry air 29.0 g
1 mol
g
2
0 020 0 980
28 78
. .
.
+ =
Volume of air:
1 mol 22.4 L STP 273.2 21.1 K
1 mol 273.2K
L
b g b g
+
= 2413
.
Density of air
g
24.13 L
g L
= =
28 78
1193
.
.
Basis III: 1 mol humid air @ 90 F (32.2 C), 1 atm,
° ° =
hr 80%
h y P p
r H O H O
80% 0.80 32.2 C
2 2
= ⇒ = ∗ °
b g
yH O
2
2
mm Hg
760.0 mm Hg
mol H O
mol
=
×
=
080 36 068
0 038
. .
.
Table B.3
Table B.3
Table B.3

6-7
6.14 (cont’d)
Mass of air:
mol H O 18.02 g
1 mol
mol dry air 29.0 g
1 mol
g
2
0 038 0 962
2858
. .
.
+ =
Volume of air:
1 mol 22.4 L STP 273.2 32.2 K
1 mol 273.2K
L
b g b g
+
= 2504
.
Density
g
25.04 L
g L
= =
2858
1141
.
.
Increase in increase in decrease in density
Increase in more water (MW =18), less dry air (MW = 29)
decrease in m decrease in density
Since the density in hot, humid air is lower than in cooler, dryer air, the buoyancy force
on the ball must also be lower. Therefore, the statement is wrong.
T V
hr
⇒ ⇒
⇒
⇒ ⇒
6.15 a. h y P p
r H O H O
50% 0.50 90 C
2 2
= ⇒ = ∗ °
b g
yH O 2
2
mm Hg
760.0 mm Hg
mol H O / mol
=
×
=
050 52576
0 346
. .
.
Dew Point: y p p T 0.346 760 mm Hg
H O dp
2
= ∗ = =
d i b g 262 9
. T 72.7 C
dp = °
Degrees of Superheat = − = °
90 72 7 17 3
. . C of superheat
b. Basis:
1 m feed gas 10 L 273K mol
m 363K 22.4 L STP
mol
3 3
3
b g= 336
.
Saturation Condition:
C
mol H O mol
H O
2
2
y
p
P
1
25 23756
760
0 0313
=
°
= =
*
.
.
b g
( ) ( )
1 1
Dry air balance: 0.654 33.6 1 0.0313 22.7 mol
n n
= − ⇒ =
3
2 2 2
Total mol balance: 33.6=22.7+ 10.9 mol H O condense/m
n n
⇒ =
c. y P p P
p
y
H O
H O
2
2
C
C) mmHg
mm Hg = 2.00 atm
= ∗ ° ⇒ =
°
= =
90
90 52576
0 346
1520
b g
*
( .
.
Table B.3
Table B.3

6-8
6.16 T = ° °
90 F = 32.2 C, p = 29 7
. in Hg = 754.4 mm Hg , hr = 95%
Basis: 10 gal water condensed/min
3
2 m
3
m
10 gal H O 1 ft 62.43 lb 1 lb-mol
4.631 lb-mole/min
min 7.4805 gal ft 18.02 lb
condensed
n = =

95% at inlet: 0.95 32.2 C
H O
2
h y P p
r = ∗ °
b g
yH O 2
2
mm Hg
754.4 mm Hg
lb - mol H O lb - mol
= =
0 95 36 068
0 045
. .
.
b g
( )
*
2 2 2
6.274
Raoult's law: 4.4 C 0.00817 lb-mol H O lb-mol
754.4
y P p y
= ° = =
1 2 1
1 2 2
Mole balance: 4.631 124.7 lb-moles/min
Water balance: 0.045 0.00817 4.631 120.1 lb-moles/min
n n n
n n n
= + =
⎫ ⎧
⇒
⎬ ⎨
= + =
⎭ ⎩

3 o
o
4 3
124.7 lb-moles 359 ft (STP) (460+90) R 760 mm Hg
Volume in: =
min lb-moles 492 R 754 mm Hg
5.04 10 ft / min
V
= ×

6.17 a. Assume no water condenses and that the vapor at 15°C can be treated as an ideal gas.
p p
p p
final 2
2
760 mm Hg K
(200 + 273) K
mm Hg ( mm Hg
C) =12.79 mm Hg Impossible condensation occurs.
H O final
*
H O
=
+
= ⇒ = × =
° ⇒
( )
. ) . . .
( .
15 273
462 7 0 20 462 7 92 6
15
( ) ( ) ( . ) .
p p
T
T
air final air initial
final
initial
mm Hg
288 K
473 K
mm Hg
= = × × =
080 760 370 2
P p p
= + = + =
H O air
2
mm Hg
370 2 12 79 383
. .
b. Basis:
L 273 K mol
473 K 22.4 L (STP)
mol
1
0 0258
= .

n2 (lb - moles / min)
y2 (lb-mol H2O (v)/lb-mol) (sat’d)
(1-y2) (lb-mol DA/lb-mol)
40o
F (4.4o
C), 754 mm Hg
y1 (lb-mol H2O (v)/lb-mol)
(1-y1) (lb-mol DA/lb-mol)
hr=95%, 90o
F (32.2o
C),
29.7 in Hg (754 mm Hg)
/ min)

V
n
1
1
(ft
(lb - moles / min)
3
4.631 lb-moles H2O (l)/min
Table B.3
Table B.3

6-9
6.17 (cont’d)
Saturation Condition:
C mm Hg
mm Hg
mol H O mol
H O
2
2
y
p
P
1
15 12 79
3831
0 03339
=
°
= =
*
.
.
.
b g
c. Dry air balance: 0.800 0.0258 mol
b g b g
= − ⇒ =
n n
1 1
1 0 03339 0 02135
. .
Total mole balance: 0.0258 = 0.02135 + mol
n n
2 2 0 00445
⇒ = .
Mass of water condensed =
0.00445 mol 18.02 g
mol
g
= 0 0802
.
6.18 Basis: 1 mol feed
(mol), 15.6°C, 3 atm
1 mol, 90°C, 1 atm
0.10 mol H O (v)/mol
0.90 mol dry air/mol
n2
y2 (mol H O (v)/mol)(sat'd)
(1 – )
y2 (mol DA/mol)
(mol) H O( ), 15.6°C, 3 atm
n3 l
2
2
2
heat
100°C, 3 atm
V (m )
1
(mol)
n2
3
V (m )
3
2
Saturation:
C mm Hg atm
3 atm 760 mm Hg
H O
2
y
p
P
y
2 2
156 1329
0 00583
=
°
= =
*
. .
.
b g
Dry air balance: mol
0 90 1 1 0 00583 0 9053
2 2
. . .
b g b g
= − ⇒ =
n n
H O mol balance: 0.10 mol
2 1 0 00583 0 9053 0 0947
3 3
b g b g
= + ⇒ =
. . .
n n
Fraction H O condensed:
mol condensed
mol fed
mol condense mol fed
2
0 0947
0100
0 947
.
.
.
=
h
y P
p
r =
×
∗ °
= × =
2 100%
100
0 00583 3
100% 175%
C
atm
1 atm
b g
b g
.
.
V2
3 3
9 24 10
= = × °
−
0.9053 mol 22.4 L STP 373K 1 atm 1 m
mol 273K 3 atm 10 L
m outlet air @ 100 C
3
3
b g .
V1
2 3
2 98 10
= = × °
−
1 mol 22.4 L STP 363K 1 m
mol 273K 10 L
m feed air @ 90 C
3
3
b g .
V
V
2
1
3
2
9 24 10
2 98 10
0 310
=
×
×
=
−
−
.
.
.
m outlet air
m feed air
m outlet air m feed air
3
3
3 3
Table B.3
n1 mol @ 15°C,
383.1 mm Hg
0.0258 mols @ 200°C,
760 mm Hg
0.200 H2O mol /mol
0.800 mol air/mol
y1 (mol H2O (v)/mol) (sat’d)
(1-y1) (mol dry air/mol)
n2 mol H2O (l)

6-10
6.19 Liquid H O initially present:
L kg 1 kmol
L kg
kmol H O l
2 2
25 100
18 02
1387
.
.
.
= b g
Saturation at outlet:
C mm Hg
mol H O mol air
H O
H O
2
2
2
y
p
P
=
°
=
×
=
*
.
.
.
25 2376
15 760 mm Hg
0 0208
b g
⇒
0 0208
1 0 0208
0 0212
.
.
.
−
= mol H O mol dry air
2
Flow rate of dry air:
15 L STP 1 mol
min 22.4 L STP
mol dry air min
b g
b g= 0 670
.
Evaporation Rate:
mol dry air mol H O
min mol dry air
mol H O min
2
2
0 670 0 0212
0 0142
. .
.
=
Complete Evaporation:
1.387 kmol mol 1 h
kmol mol 60 min
h days
3
10
0 0142
1628 67 8
min
.
.
= b g
6.20 a. Daily rate of octane use =
4
30 (18 8)
7.069 10 ft 7.481 gal
day ft
5.288 10 gal / day
2
3 3
3
4
π
⋅ ⋅ − =
×
= ×
( ) .
SG C H
8 18
= 0 703 ⇒
5288 10
3
4
.
/
× ×
= ×
gal 1 ft 0.703 62.43 lb
day 7.481 gal ft
.10 10 lb C H day
3
m
3
5
m 8 18
b. m
2
2
m f
lb ft
3 2 2 2
f
s
0.703 62.43 lb 32.174 ft 1 lb (18-8) ft 1 ft 29.921 in Hg
6.21 in Hg
ft s 32.174 144 in 14.696 lb /in
p ⋅
×
Δ = =
c. Table B.4: p p y P
C H
o
f
2
octane octane
8 18
F)
20.74 mm Hg 14.696 psi
760 mm Hg
lb in
*
( . /
90 0 40
= = = =
Octane lost to environment = octane vapor contained in the vapor space displaced by liquid
during refilling.
Volume:
5.288 10 gal 1 ft
7.481 gal
ft
4 3
3
×
= 7069
Total moles
(16.0 +14.7) psi 7069 ft
ft psi / (lb - mole R) (90 + 460) R
lb - moles
3
3 o o
:
.
.
n
pV
RT
= =
⋅ ⋅
=
10 73
36 77
Mole fraction of C H =
psi
(16.0 +14.7) psi
lb - mole C H / lb - mole
8 18
C H
8 18
8 18
:
.
.
y
p
P
= =
0 40
0 0130
Octane lost lb - mole lb - mole lb kg)
m
= = = =
0 0130 36 77 0 479 55 25
. ( . ) . (
d. A mixture of octane and air could ignite.

6-11
6.21 a. Antoine equation⇒ p p p
tol tol tol
* *
( ( .
85 29 44
o o
F) = C) = 35.63 mmHg =
Mole fraction of toluene in gas: y
p
P
tol
= = =
3563
0 0469
.
.
mmHg
760 mmHg
lb - mole toluene / lb - mole
Toluene displaced = =
yn
yPV
RT
total
=
+
=
⋅
⋅
0.0469 lb - mole tol 1 atm gal 1 ft 92.13 lb tol
lb - mole 0.7302 R 7.481 gal lb - mole
.31 lb toluene displaced
3
m
o
m
ft atm
lb - mole R
3
o
900
85 460
1
( )
b.
90% condensation⇒ = =
n l
L 0 90 0 0469 1 0 0422
. ( . )( ) . ( )
mol C H mol C H
7 8 7 8
Mole balance: 1 0 0422 0 9578
= + ⇒ =
n n
V V
. . mol
Toluene balance: 0 0469 1 0 9578 0 0422 0 004907
. ( ) ( . ) . . /
= + ⇒ =
y y mol C H mol
7 8
Raoult’s law: p yP p T
tol tol
= = × =
( . )( ) . ( )
*
0 004907 5 760 18 65 mmHg =
Antoine equation:
*
o o
10 10
*
10
10
( log ) 1346.773 219.693(6.95805 log 18.65)
17.11 C=62.8 F
6.95805 log 18.65
log
B C A p
T
A p
− − − −
= = =
−
−
6.22 a. Molar flow rate: =
m kmol K 2 atm
h 82.06 10 m atm (100 + 273) K
6.53 kmol / h
3
-3 3

n
VP
RT
=
⋅
× ⋅
=
100
b. Antoine Equation:
*
10
*
1175.817
log (100 C)=6.88555- 3.26601
100+224.867
1845 mm Hg
Hex
p
p
° =
⇒ =
p y P p
Hex Hex Hex
= ⋅ = =
0.150(2.00) atm 760 mm Hg
atm
mm Hg
228 *
⇒ not saturated
*
10
1175.817
( ) 228 mm Hg log 228=6.88555- 2.35793 34.8 C
T+224.867
Hex
p T T
= ⇒ = ⇒ = °
nV (mol)
y (mol C7H8(v)/mol)
(1-y) (mol G/mol)
T(o
F), 5 atm
nL [mol C7H8 (l)]
90% of C7H8 in feed
Basis: 1mol
0.0469 mol C7H8(v)/mol
0.9531 mol G/mol
Assume G is
noncondensable

6-12
6.22 (cont’d)
c.
80% condensation: n l
L = 080 015 653
. ( . )( . ( ) /
kmol / h) = 0.7836 kmol C H h
6 14
Mole balance: 653 0 7836 5746
. . .
= + ⇒ =
n n
V V kmol / h
Hexane balance: 015 653 5746 0 7836 0 03409
. ( . ) ( . ) . .
= + ⇒ =
y y kmol C H / kmol
6 14
Raoult’s law: p yP p T
Hex Hex
= = ×
( . )( ( )
*
0 03409 2 760 mmHg) = 51.82 mmHg =
Antoine equation: o
10
1175.817
log 51.82 6.88555 2.52 C
224.867
T
T
= − ⇒ =
+
6.23 Let H=n-hexane
a.
50% relative saturation at inlet: y P p
o H
= 0500 80
. (
* o
C)
yo =
( . )(
.
0500 1068
0 703
mmHg)
760 mmHg
= kmolH / kmol
Saturation at outlet: 0 05 0 05 760
1 1
. ( ) ( ) . (
* *
P p T p T
H H
= ⇒ = mmHg) = 38 mmHg
Antoine equation: o
10 1
1
1175.817
log 38 6.88555 3.26 C
224.867
T
T
= − ⇒ = −
+
Mole balance:
N balance:
kmol / min
kmol / min
2
.
( . ) .
.
.
n n
n n
n
n
0 1
0 1
0
1
150
1 0 703 0 95
218
0 682
= +
− =
U
V
W
⇒
=
=
R
S
T
N2 volume: ( )
.
VN2
14 5
= =
(0.95)0.682 kmol 22.4 m STP
min kmol
SCMM
3
Table B.4
nL (kmol C6H14 (l)/h)
80% of C6H14 in feed
nV (kmol/h)
y (kmol C6H14 (v)/kmol), sat’d
(1-y) (kmol N2/kmol)
T (o
C), 2 atm
6.53 kmol/h
0.15 C6H14 (v)
0.85 N2
1.50 kmol H(l)/min
( )
n1 kmol / min
0.05 kmol H(v)/kmol, sat’d
0.95 kmol N2/kmol
T (o
C), 1 atm
( )
n0 kmol / min
y0 (kmol H(v)/kmol
(1-y0) (kmol N2/kmol)
80o
C, 1 atm, 50% rel. sat’n
Condenser

6-13
6.23 (cont’d)
b. Assume no condensation occurs during the compression
50% relative saturation at condenser inlet:
0500 0 703 7600 1068 10
0 0
4
. ( ) . ( ( ) .
* *
p T p T
H H
= ⇒ = ×
mmHg) mmHg T0 187
= o
C
Saturation at outlet: 0 050 7600 1
. ( ( )
*
mmHg) = 380 mmHg = p T
H T1 48 2
= °
. C
Volume ratio:
3
1 1 1 1 1
3
0 0 0 0 0
/ ( 273.2) 0.682 kmol/min 321 K m out
0.22
/ ( 273.2) 2.18 kmol/min 460 K m in
V n RT P n T
V n RT P n T
+
= = = × =
+

c. The cost of cooling to o
3.26 C
− (installed cost of condenser + utilities and other operating
costs) vs. the cost of compressing to 10 atm and cooling at 10 atm.
6.24 a. Maximum mole fraction of nonane achieved if all the liquid evaporates and none escapes.
(SG)nonane
n
l
max
9 20
9 20
9 20
L C H 0.718 1.00 kg kmol
L C H 128.25 kg
0 kmol C H
=
×
=
15
084
( )
.
Assume T = 25o
C, P =1 atm
ngas =
×
×
=
2 10
818
4
L 273 K 1 kmol
298 K 22.4 10 L(STP)
0 kmol
3
.
y
n
ngas
max
max .
. /
= = =
0 084
010
kmol C H
0.818 kmol
kmol C H kmol (10 mole%)
9 20
9 20
As the nonane evaporates, the mole fraction will pass through the explosive range (0.8% to
2.9%). The answer is therefore yes .
The nonane will not spread uniformly—it will be high near the sump as long as liquid is present
(and low far from the sump). There will always be a region where the mixture is explosive at
some time during the evaporation.
b. ln . .
* *
p
A
T
B T p
= − + = =
C = 299 K, mmHg
o
1 1
258 500
T p
2 2
66 0 40 0
= =
. .
*
o
C = 339 K, mmHg
Antoine
Antoine

V1 (m / min)
3
0.682 kmol/min
0.05 H(v), sat’d
0.95 N2
T1 (o
C), 10 atm
1.5 kmol H(l)/min
( )
V0 m / min
3
2.18 kmol/min
0.703 H(v)
0.297 N2
T0 (o
C), 10 atm, 50% R.S.
2.18 kmol/min
0.703 H(v)
0.297 N2
80o
C, 1 atm
Compressor Condenser

6-14
6.24 (cont’d)
− =
−
⇒ = = ⇒ = −
A A B p
T
ln( . / . )
, ln . exp( .
( )
)
*
40 0 500
1
339
1
299
5269 500
5269
299
19 23 19 23
5269
= ( . ) +
K
At lower explosion limit, y p T yP
= ⇒ = =
0 008 0 008 760
. / ( ) ( . )(
*
kmol C H kmol mm Hg)
9 20
= 6.08 mm Hg T = 302 K = 29 C
o
c. The purpose of purge is to evaporate and carry out the liquid nonane. Using steam rather
than air is to make sure an explosive mixture of nonane and oxygen is never present in the
tank. Before anyone goes into the tank, a sample of the contents should be drawn and
analyzed for nonane.
6.25 Basis: 24 hours of breathing
23°C, 1 atm
y1 (mol H O/mol)
+ O2 , CO2
n1 (mol) @ hr = 10%
0.79 mol N /mol
2
2
Lungs
O2 CO2
37°C, 1 atm
y2 (mol H O/mol)
+ O2 , CO2
n2 (mol), saturated
0.75 mol N /mol
2
2
n0 (mol H O)
2
Air inhaled:
12 breaths 500 ml 1 liter 273K 1 mol 60 min 24 hr
min breath 10 ml 23 273 K 22.4 liter STP 1 hr 1 day
mol inhaled day
1 3
n =
+
=
b g b g
356
Inhaled air - -10% r.h.:
C mm Hg
760 mm Hg
mol H O
mol
H O 2
2
y
p
P
1
3
010 23 010 2107
2 77 10
=
∗ °
= = × −
. . .
.
b g b g
Inhaled air - -50% r.h.:
C mm Hg
760 mm Hg
mol H O
mol
H O 2
2
y
p
P
1
2
050 23 050 2107
139 10
=
∗ °
= = × −
. . .
.
b g b g
H O balance:
mol
day
mol H O
mol
g
1 mol
g / day
2 rh rh
2
n n y n y n n n y n y
0 2 2 1 1 0 10% 0 50% 1 1 50% 1 1 10%
356 0 0139 0 00277
18 0
71
= − ⇒ − = −
=
F
HG I
KJ −
L
NM O
QPF
HG I
KJ =
( ) ( ) ( ) ( )
( . . )
.
Although the problem does not call for it, we could also calculate that n2 = 375 mol exhaled/day,
y2 = 0.0619, and the rate of weight loss by breathing at 23o
C and 50% relative humidity is
n0 (18) = (n2y2 - n1y1)18 = 329 g/day.
Formula for p*

6-15
6.26 a. To increase profits and reduce pollution.
b.
Assume condensation occurs. A=acetone
1 mol @ 90o
C, 1 atm
0.20 mol A(v)/mol
0.80 mol N2/mol
n1 mol @ To
C, 1 atm
y1 mol A(v)/mol (sat’d)
(1-y1) mol N2/mol
n2 mol A(l)
For cooling water at 20o
C
( ) ( )
* o * o
10
1210.595
log 20 C 7.11714 2.26824 20 C 184.6 mmHg
20 229.664
A A
p p
= − = ⇒ =
+
Saturation: y P p y
A
1 1
20
184 6
760
0 243 0 2
⋅ = ⇒ = =
* .
. .
o
C
d i , so no saturation occurs.
For refrigerant at –35o
C
( ) ( )
* o * o
10
1210.595
log 35 C 7.11714 0.89824 35 C 7.61 mmHg
35 229.664
A A
p p
− = − = ⇒ − =
− +
(Note: -35o
C is outside the range of validity of the Antoine equation coefficients in Table
B.4. An alternative is to look up the vapor pressure of acetone at that temperature in a
handbook. The final result is almost identical.)
Saturation: y P p y
A
1 1
35
7 61
760
0 0100
⋅ = − ⇒ = =
* .
.
o
C
d i
N2 mole balance: 1 08 1 0 01 0808
1 1
. . .
b g b g
= − ⇒ =
n n mol
Total mole balance: 1 0808 0192
2 2
= + ⇒ =
. .
n n mol
Percentage acetone recovery:
0.192
100% 96%
2
× =
c.
Costs of acetone, nitrogen, cooling tower, cooling water and refrigerant
d.
The condenser temperature could never be as low as the initial cooling fluid temperature
because heat is transferred between the condenser and the surrounding environment. It
will lower the percentage acetone recovery.

6-16
6.27
Basis:
12500 1 273 K 103000
5285
L
h
mol
22.4 L(STP) 293 K
Pa
101325 Pa
mol / h
= .
no (mol/h) @ 35o
C, 103 KPa
yo mol H2O(v)/mol
(1-yo) mol DA/mol
hr=90%
528.5 (mol/h) @ 20o
C, 103 KPa
y1 mol H2O(v)/mol (sat’d)
(1-y1) mol DA/mol
n2 mol H2O(l)/h
Inlet:
( )
2
* o
r H O
o 2
35 C 0.90 42.175 mmHg 101325 Pa
0.04913 mol H O/mol
103000 Pa 760 mmHg
h p
y
P
⋅ ×
= = =
Outlet: y
p
P
1
20 17 535 101325
0 02270
= = =
H O
o
2
2
C mmHg
103000 Pa
Pa
760 mmHg
mol H O / mol
*
.
.
d i
Dry air balance: 1 0 04913 1 0 02270 5285 5432
− = − ⇒ =
. . . .
b g b gb g
n n
o o mol / h
Inlet air:
5432 22 4 308 101325
13500
. .
mol
h
L(STP)
mol
K
273 K
Pa
103000 Pa
L / h
=
Total balance: 5432 5285 14 7
2 2
. . .
= + ⇒ =
n n mol / h
Condensation rate:
14 7 18 02 1
0 265
. .
.
mol
h
g H O
1 mol H O
kg
1000 g
kg / h
2
2
=
6.28
Basis:
10000 1 492 29 8
24 82
ft lb - mol
359
R
550 R
in Hg
29.92 in Hg
lb - mol / min
3 o
o
ft (STP)
3
min
.
.
=
n1 lb-mole/min n1 lb-mole/min
24.82 lb-mole/min 40o
F, 29.8 in.Hg 65o
F, 29.8 in.Hg
90o
F, 29.8 in.Hg y1 [lb-mole H2O(v)/lb-mole] y1 [lb-mole H2O(v)/lb-mole]
y0 [lb-mole H2O(v)/mol 1- y1 (lb-mole DA/mol) 1- y1 (lb-mole DA/lb-mole)
1- y0 (lb-mole DA/mol)
hr = 88%
n2 [lb-mole H2O(l)/min]
Inlet: y
h p
P
r
o
H O
* o
2
2
F mmHg
in Hg
1 in Hg
25.4 mmHg
lb - mol H O / lb - mol
=
⋅
= =
90 088 36 07
29 8
0 0419
d i b g
. .
.
.
Outlet: y
p
P
1
H O
* o
2
2
F mmHg
in Hg
1 in Hg
25.4 mmHg
lb - mol H O / lb - mol
= = =
40 6 274
29 8
0 00829
d i .
.
.
Dry air balance: 24 82 1 0 0419 1 0 00829 2398
1 1
. . . .
− = − ⇒ =
b g b g
n n lb - mol / min
2 2
. . .
= + ⇒ =
n n lb - mole / min
y0 [mol H2O(v)/mol]
1– y0 (mol DA/mol)
hr=90%
y1 [mol H2O(v)/mol]
1– y1 (mol DA/mol)
n2[mol H2O(l)/h]

6-17
6.28 (cont’d)
Condensation rate:
084 18 02 7 48
181
. . .
.
lb - mol
min
lb
lb mol
1 ft
62.4 lb
gal
1 ft
gal / min
m
3
m
3
−
=
Air delivered @ 65o
F:
23.98 lb - mol
min
359 ft (STP)
1 lb mol
525 R
R
in Hg
29.8 in Hg
ft / min
3 o
o
3
−
=
492
29 92
9223
.
6.29 Basis: 100 mol product gas
no mol, 32o
C, 1atm
yo mol H2O(v)/mol
(1-yo) mol DA/mol
hr=70%
100 mol, T1, 1atm
y1 mol H2O(v)/mol, (sat’d)
(1-y1) mol DA/mol
n2 lb-mol H2O(l)/min
100 mol, 25o
C,1 atm
y1 mol H2O(v)/mol,
(1-y1) mol DA/mol
hr=55%
Outlet: y
h p
P
r
1
25 055 23756
760
0 0172
=
⋅
= =
H O
o
2
2
C
mol H O / mol
*
. .
.
d i b g
Saturation at 15.3 C
H O
o
2
T p T T
1 1 1
0 0172 760 1307
: . . *
b g b g
= = ⇒ =
Inlet: y
h p
P
r
o
H O
o
2
2
C
mol H O / mol
=
⋅
= =
*
. .
.
32 0 70 35663
760
0 0328
d i b g
Dry air balance: n n
o o mol
1 0 0328 100 1 0 0172 1016
− = − ⇒ =
. . .
b g b g
2 2
. . .
+ = ⇒ = −
n n mol (i.e. removed)
kg H O removed
2 :
16 18 02 1
0 0288
. .
.
mol g
1 mol
kg
1000 g
kg H O
2
=
kg dry air:
100 1 0 0172 29 0 1
2 85
−
=
. .
.
b gmol g
1 mol
kg
1000 g
kg dry air
Ratio:
0 0288
2 85
0 0101
.
.
.
= kg H O removed / kg dry air
2
n2(mol H2O(l))

6-18
6.30 a. Room air C
− = °
T 22 , P = 1 atm , hr = 40% :
( )
( )
2
1 H O 1 2
0.40 19.827 mm Hg
0.40 22 C 0.01044 mol H O mol
760 mm Hg
y P p y
= ∗ ° ⇒ = =
Second sample C
− = °
T 50 , P = 839 mm Hg , saturated:
( )
2
2 H O 2 2
92.51 mm Hg
50 C 0.1103 mol H O mol
839 mm Hg
y P p y
= ∗ ° ⇒ = =
ln ln
y bH a y aebH
= + ⇔ = , y H
1 1
0 01044 5
= =
. , , y H
2 2
01103 48
= =
. ,
b
y y
H H
=
−
=
−
=
ln ln . .
.
2 1
2 1
01103 0 01044
48 5
0 054827
b g b g
ln ln ln . . . exp . .
a y bH a
= − = − = − ⇒ = − = × −
1 1
3
001044 0054827 5 48362 48362 7 937 10
b g b gb g b g
⇒ y H
= × −
7 937 10 0 054827
3
. exp .
b g
b.
Basis:
1 m delivered air 273K 1 k mol 10 mol
22 273 K 22.4m STP 1 kmol
mol air delivered
3 3
3
+
=
b g b g 4131
.
Saturation condition prior to reheat stage:
( ) ( )( )
2 2
*
H O H O 0.01044 760 mm Hg 7.93 mm Hg
7.8 C (from Table B.3)
y P p T
T
= ⇒ =
⇒ = °
Humidity of outside air: mol H O mol
Part a
2
H y
= ⇒ =
30 0 0411
0
b g
.
( ) ( )
( )( )
( )
0 0 0
41.31 0.9896
Overall dry air balance: 1 41.31 0.9896 42.63 mol
1 0.0411
n y n
− = ⇒ = =
−
( )( ) ( )( ) ( )( )
0 0 1 1
2
Overall water balance: 41.31 0.0104 42.63 0.0411 41.31 0.0104
1.32 mol H O condensed
n y n n
= + ⇒ = −
=
Mass of condensed water
mol H O g H O 1 kg
1 mol H O 10 g
kg H O condensed m air delivered
2 2
2
3
2
3
=
=
132 18 02
0 024
. .
.
no mol, 35o
C, 1 atm
yo mol H2O(v)/mol
(1-yo) mol DA/mol
H=30
41.31 mol, T, 1 atm
0.0104 mol H2O(v)/mol, (sat’d)
0.09896 mol DA/mol
n1 mol H2O(l)
41.4 mol, 22o
C,1 atm
0.0104 mol H2O(v)/mol
0.09896 mol DA/mol
41.31 mol, 22o
C, 1 atm
0.0104 mol H2O(v)/mol,
sat’d
0.9896 mol DA/mol
0.9896 mol DA/mol

6-19
6.31 a. Basis: mol feed gas
0

n . S solvent
= , G solvent - free gas
=
Inlet dew point = 0
T y P p T y
p T
P
o o do o
do
o
⇒ = ∗ ⇒ =
∗
b g b g (1)
Saturation condition at outlet: y P p T y
p T
P
f f
f
f
1 1
= ∗ ⇒ =
∗
d i d i (2)
Fractional condensation of S = 0
f n n y f
⇒ =
2 0
( )
1
⎯ →
⎯ n n fp T P
2 0 0 0
= ∗b g (3)
Total mole balance:
Eq. 3 for

n n n n n n n n
n fp T
P
n
do
o
0 1 2 1 0 2 1 0
0
1
= + ⇒ = − ⇒ = −
∗
b g b g (4)
S balance: 0
(1) - (4)
n y n y n
n p T
P
n
n fp T
P
p T
P
n fp T
p
do
o
do
o
f
f
do
o
b gb g
b g b g d i b g
0 1 1 2
0
0
0 0
= +
∗
= −
∗
L
NMM
O
QPP
∗
F
H
GG
I
K
JJ +
∗

⇒
− ∗
= −
∗
L
NMM
O
QPP
∗
1
1
f p T
P
fp T
P
p T
P
do
o
do
o
f
f
b g b g b g d i ⇒ =
∗ −
∗
L
N
MMM
O
Q
PPP
−
∗
P
p T
fp T
do
P
o
f
p T
do
P
o
f
f
d i e j
b g e j
1
1
y1 [mol S(v)/mol] (sat’d)
(1–y1) (mol G/mol)
n0 (mol) @ T0 (°C), P0 (mm Hg)
y0 (mol S/mol)
(1-y0) (mol G/mol)
Td0 (°C) (dew point)
n2 (mol S (l))
n1 (mol) @ Tf (°C), P4 (mm Hg)
C
ondensationof ethylbenzenefromnitrogen
Antoineconstantsforethylbenzene
A= 6.9565
B= 1423.5
C
= 213.09
R
un T0 P0 Td0 f Tf p*(Td0) p*(Tf) Pf C
refr C
com
p C
tot
1 50 765 40 0.95 45 21.472 27.60 19139 2675 107027109702
2 50 765 40 0.95 40 21.472 21.47 14892 4700 83329 88029
3 50 765 40 0.95 35 21.472 16.54 11471 8075 64239 72314
4 50 765 40 0.95 20 21.472 7.07 4902 26300 27582 53882
b.

6-20
6.31 (cont’d)
c.
d.
When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both to
increase the fractional condensation. When you decrease Tf, less compression is required to
achieve a specified fractional condensation.
A lower Tf requires more refrigeration and therefore a greater refrigeration cost (Crefr).
However, since less compression is required at the lower temperature, Ccomp is lower at the
lower temperature. Similarly, running at a higher Tf lowers the refrigeration cost but raises
the compression cost. The sum of the two costs is a minimum at an intermediate temperature.
6.32 a. Basis : 120m min feed @ 1000 C(1273K), 35 atm
3 o
. Use Kay’s rule.
Cmpd. atm Apply Newton's corrections for H
H
CO 133.0 34.5
CO
CH
2
2
2
4
T K P T P
c c c corr c corr
b g b g b g b g b g
.
. . . .
. .
. .
332 12 8 413 208
304 2 72 9
190 7 458
− −
− −
− −
′ = = + + + =
∑
T y T K
c i ci 0 40 413 0 35 1330 0 20 304 2 0 05 190 7 1334
. . . . . . . . .
b g b g b g b g
′ = = + + + =
∑
P y P
c i ci 0 40 208 0 35 34 5 0 20 72 9 0 05 458 37 3
. . . . . . . . .
b g b g b g b g atm
Feed gas to cooler
1273 133.4 9.54 Generalized compressibility charts (Fig. 5.4-3)
35.0 atm 37.3 atm 0.94 1.02
r
r
T K K
P z
= = ⎫
⎬
= = ⇒ =
⎭
. .
.
V =
⋅
⋅
= × −
1 02 8 314 1273
101325
3 04 10 3
N m K 1 atm
35 atm mol K N m
m mol
3
3
120
3 04 10 10
39 5
3 3
m mol 1 kmol
m mol
kmol min
3
3
min .
.
×
=
−
Feed gas to absorber
283 133.4 2.12 Generalized compressibility charts (Fig. 5.4-3)
35.0 atm 37.3 atm 0.94 0.98
r
r
T K K
P z
= =
= = ⇒ =
⎫
⎬
⎭
4 3
3
0.98 8.314 N m 283 K 1 atm
ˆ 6.50 10 m mol
35 atm mol K 101325 N m
V −
⋅
= = ×
⋅
3 -4 3 3
39.5 kmol 10 mol 6.50 10 m m
25.7
min
min 1 kmol mol
V
×
= =

6-21
6.32 (cont’d)
Saturation at Outlet:
mm Hg
35 atm 760 mm Hg atm
mol MeOH mol
McOH
MeOH
y
p K
P
=
∗
=
= ×
− − +
−
261 10
4 97 10
7 87863 1473 11 12 2300
4
b g
b g
b g
. .
.
y
n
n n n n
n
n
n
CO
McOH
MeOH
MeOH
input
H
of input
CH
input
MeOH
MeOH
MeOH
+
kmol min MeOH in gas
2 4
=
+ +
=
=
+ + +
A
=
A
=
A
=
E
( )
( )
0.02
0 0148
39 5 0 40 0 02 0 05 0 35
.
. . . . .
b. The gas may be used as a fuel. CO2 has no fuel value, so that the cost of the added energy
required to pump it would be wasted.
1
n
(kmol/min), 261 K, 35 atm
1.2(39.5) kmol/min
39.5 kmol/min, 283K, 35 atm
0.40 mol H2/mol
0.35 mol CO/mol
0.20 mol CO2/mol
0.05 mol CH4/mol
yMeOH sat’d
yH2
yCH4 (2% of feed)
yCO
2
n
(kmol/min), liquid
xMeOH
xCO2
xCH4 (98% of feed)

6-22
6.33
Dry pulp balance: 1500
1
1 0 75
1 0 0015 858
1 1
×
+
= − ⇒ =
.
( . )
m m kg / min
50% rel. sat’n at inlet: 2
* o
1 H O 1
2
0.50 (28 C) 0.50(28.349 mm Hg)/(760 mm Hg)
= 0.0187 mol H O/mol
y P p y
= ⇒ =
40o
C dew point at outlet: y P p y
2 2
40 324 770
= ⇒ =
H O
* o
2
2
C) 55 mm Hg) / ( mm Hg)
= 0.0718 mol H O / mol
( ( .
Mass balance on dry air:
( . ) ( . ) ( )
n n
0 1
1 0 0187 1 0 0718 1
− = −
Mass balance on water:
( . )( . ) ( . / . ) ( . )( ) ( . ) ( )
n n
0 1
0 0187 18 0 1500 0 75 1 75 0 0718 18 858 0 0015 2
kg / kmol + = +
Solve (1) and (2) ⇒ . .
n n
0 1
622 8 658 4
= =
kmol / min, kmol / min
Mass of water removed from pulp: [1500(0.75/1.75)–858(.0015)]kg H2O = 642 kg / min
Air feed rate: . ( )
min
/ min
V0
622 8
= = ×
kmol 22.4 m STP (273 + 28) K
kmol 273 K
1.538 10 m
3
4 3

n1 (kmol/min wet air) @ 80°C, 770
0.0015 kg H2O/kg
0.9985 kg dry pulp/kg

n0 (kmol/min wet air) @ 28°C, 760 mmHg
y1 (mol H2O/mol)
50% rel. sat.
1500 kg/min wet pulp
0.75 /(1 + 0.75) kg H2O/kg
1/1.75 kg dry pulp/kg

m1 (kg/min wet pulp)
y2 (mol H2O/mol)
Tdew point = 40.0o
C

6-23
6.34 Basis: 500 lb hr dried leather
m (L)
lb- moles/ h)@130 F, 1 atm
lb- moles dryair / h)@140 F, 1 atm lb- moles H O/ lb- mole)
(1- lb- moles dryair / lb- mole)
lb h) 500 lb h
0.61 lb H O(l) / lb 0.06 lb H O(l) / lb
0.39 lb L/ lb 0.94 lb
o
o
2
m m
m 2 m m 2 m
m m
(
( (
)(
( / /
n
n y
y
m
1
0 1
1
0
m L/ lb
Dry leather balance: lb wet leather hr
m
0 39 0 94 500 1205
0 0
. .
m m
= ⇒ =
b gb g
Humidity of outlet air: F
mm Hg)
760 mmHg
mol H O
mol
H O
2
2
y P p y
1 1
050 130
050 115
0 0756
= ∗ ° ⇒ = =
.
. (
.
b g
H O balance: 0.61 lb hr lb
m
hr
lb - moles H O 18.02 lb
hr 1 lb - mole
lb - moles hr
2 m
2 m
b gb g b g
a f
1205 0 06 500
0 0756
517 5
1
1
= +
E
=
( )
.
.
.
n
n
Dry air balance: lb - moles hr lb - moles hr
n0 1 0 0756 517 5 478 4
= − =
. ( . ) .
b g
Vinlet
3
5 3
lb - moles 359 ft STP R
hr 1 lb - mole 492 R
.09 10 ft hr
=
+ °
°
= ×
478 4 140 460
2
. b g b g
6.35 a. Basis: 1 kg dry solids
1.00 kg solids
n1
dryer
(kmol)N , 85°C
2
0.78 kg Hex
n2 (kmol) 80°C, 1 atm
y2 (mol Hex/mol)
(1 – )
y (mols N /mol)
2 2
70% rel. sat.
0.05 kg Hex
1.00 kg solids
condenser
n3 (kmol) 28°C, 5.0 atm
y3 (mol Hex/mol) sat'd
(1 – )
y (mols N /mol)
3 2
n4 (kmol) Hex(l)
Mol Hex in gas at 80°C:
0.78 kg kmol
86.17 kg
kmol Hex
−
= × −
0 05
8 47 10 3
.
.
b g
( ) ( ) ( )
Antoine eq.
6.88555 1175.817 80 224.867
2
0.70 80 C 0.70 10
70% rel. sat.: 0.984 mol Hex mol
760
hex
p
y
P
↓
− +
∗ °
= = =

6-24
6.35 (cont’d)
n2
3
8 47 10
0 0086
=
×
=
−
.
.
kmol Hex 1 kmol
0.984 kmol Hex
kmol
N balance on dryer: kmol
2 n1
4
1 0 984 0 0086 1376 10
= − = × −
. . .
b g
( ) ( )
( )
Antoine Eq.
6.88555 1175.817 28 224.867
3
28 C 10
Saturation at outlet: 0.0452 mol Hex mol
P 5 760
hex
p
y
↓
− +
∗ °
= = =
Overall N balance: .376 10 kmol
2
-4
1 1 0 0452 144 10
3 3
4
× = − ⇒ = × −
n n
. .
b g
Mole balance on condenser: kmol
0 0086 144 10 0 0085
4
4 4
. . .
= × + ⇒ =
−
n n
Fractional hexane recovery:
kmol cond. 86.17 kg
0.78 kg feed kmol
kg cond. kg feed
0 0085
0 939
.
.
=
b. Basis: 1 kg dry solids
1.00 kgsolids
n1
dryer
(kmol)N
85°C
2
0.78 kgHex
n2
(kmol) 80°C, 1 atm
y2
(mol Hex/mol)
(1 – )
y (mols N /mol)
2 2
70% rel. sat.
0.05 kgHex
1.00 kgsolids
condenser
n3 (kmol) 0.1n
y3
(1 – )
y3
3
n4
(kmol) Hex(l)
heater 0.9n (kmol) @ 28°C, 5.0 atm
3
(mol Hex/mol) sat'd
3
y
(mol N /mol)
3
(1 – )
y
y 3
2
y
(1 – )
3
3
0.9n
Mol Hex in gas at 80°C: 8.47x10-3
+ 0.9n3(0.0452) = n2(0.984) (1)
N2 balance on dryer: n n n
1 3 2
0 9 1 0 0452 1 0 984 2
+ − = −
. ( . ) ( . ) ( )
Overall N2 balance: 1 3
0.1 (1 0.0452) (3)
n n
= −
Equations (1) to (3) ⇒
5
1
2
4
3
1.38 10 kmol
0.00861 kmol
1.44 10 kmol
n
n
n
−
−
⎧ = ×
⎪
=
⎨
⎪
= ×
⎩
-4 5
4
1.376 10 1.38 10
Saved fraction of nitrogen= 100% 90%
1.376 10
−
−
× − ×
× =
×
Introducing the recycle leads to added costs for pumping (compression) and heating.

6-25
6.36 b.
Strategy: Overall balance⇒
m n
1 2
;
Relative saturation⇒y1;, Gas and liquid equilibrium⇒y3
Balance over the condenser⇒
n n
1 3

Toluene Balance:
Dry Solids Balance:
lb / h
lb - mole / h
m
300 0167 00196 9213
300 0833 09804
255
0488
1 2
1
1
2
× = × + ×
× = ×
U
V
W
⇒
=
=
R
S
T
. . .
. .

.
m n
m
m
n
70% relative saturation of dryer outlet gas:
7 8
1346.773
(6.95805 )
* O O 65.56 219.693
(150 F=65.56 C)=10 172.47 mmHg
C H
p
−
+ =
y P pC H
1 0 70 150
7 8
= . (
* O
F) ⇒ y
p
P
C H
1
070 070 172 47
12 760
01324
7 8
= =
×
=
. ( . )( . )
.
.
*
lb - mole T(v) / lb - mole
Saturation at condenser outlet:
7 8
7 8
1346.773
(6.95805 )
* O O 65.56 219.693
*
3
(90 F=32.22 C)=10 40.90 mmHg
40.90
0.0538 mol T(v)/mol
760
C H
C H
p
p
y
P
−
+ =
= = =
Condenser Toluene Balance:
Condenser N Balance:
lb - mole / h
lb - mole / h
2
. . .
( . ) ( . )
.
.
n n
n n
n
n
1 3
1 3
1
3
01324 0 488 0 0538
1 01324 1 0 0538
5875
5387
× = + ×
× − = × −
U
V
W
⇒
=
=
R
S
T

n1 (lb-mole/h)

n2 ( lb-mole T(l)/h
y3 (lb-mole T/lb-mole)
(1-y3)( lb-mole N2/lb-mole)

n3 (lb-mole/h) @ 200O
F,
002 102 00196
09804
. / ( . ) .
.
= lb T(l) /
lb D / lb
m
m m

m1 (lbm/h)
y1 (lb-mole T(v)/lb-mole)
(1–y1) (lb-mole N2/lb-
mole)
70% r.s.,150o
F, 1.2 atm
T=toluene
D=dry solids
0 2
1 0 2
0167
.
.
.
+
= lb T(l) / lb
0.833 lb D / lb
m m
m m
300 lbm/h wet product
Dryer
Heater
y3 (lb-mole T(v)/lb-mole)
(1-y3) (lb-mole N2/lb-mole)

n
3 (lb-mole/h)
Condenser Eq.@ 90O
F,
1atm

6-26
6.36 (cont’d)
Circulation rate of dry nitrogen = 5.875 (1- 0.1324) =
5.097 lb - mole lb - mole
h 28.02 lb
lb / h
m
m
×
= 0182
.
Vinlet
3
3
.387 lb - moles 359 ft STP R
hr 1 lb - mole 492 R
ft h
=
+ °
°
=
5 200 460
2590
b g ( )
6.37
Basis: 100 mol C H
6 14 C H
19
2
O 6CO 7H O
6 14 2 2 2
+ → +
n0
100 mol C H
6 14
(mol) air
0.21 mol O /mol
2
0.79 mol N /mol
2
n1 (mol) dry gas, 1 atm
0.821 mol N /mol D.G.
2
0.069 mol CO /mol D.G.
2
0.021 mol CO/mol D.G.
0.086 mol O /mol D.G.
2
0.00265 mol C H /mol D.G.
6 14
n2 (mol H O)
2
C balance: mol dry gas
CO CO
C H
2
6 14
6 100 0 069 0 021 6 0 00265 5666
1 1
b g b g
b g b g b g
= + +
L
N
MM
O
Q
PP
⇒ =
n n
. . .
Conversion:
100 0.00265 5666 mol reacted
100 mol fed
−
× =
b g 100% 850%
.
H balance: 14 100 mol H O
2
b g b gb g
= + ⇒ =
2 5666 14 0 00265 595
2 2
n n
.
Dew point:
595
595 + 5666 mm Hg
mm Hg C
H O
Table B.3
2
y
p T
p T T
dp
dp dp
= =
∗
⇒ ∗ = ⇒ = °
d i d i
760
72 2 451
. .
2 0
0
N balance: 0.79 5666(0.907 )
O balance: 0.21( )(2) 5666[(0.069)(2) 0.021 2 ) 595
n x
n x
= −
= + + +
Solve simultaneously to obtain n0 = 5888 mol air, x = 0.086 mol O2/mol
Theoretical air:
mol C H 19 mol O 1 mol air
2 mol C H 0.21 mol O
mol air
2 14 2
2 14 2
100
4524
=
Excess air: excess air
5888 4524
4524
100% 30 2%
−
× = .
0.069 mol CO2/mol D.G.
0.00265 mol C6H14/mol
x (mol O2/mol)
(0.907–x) (mol N2/mol)
n2 (mol H2O)

6-27
6.38 Basis: 1 mol outlet gas/min
(
( /
( /
/
/ min) /
/ min) ) /
n
y
y
y
n y
n y y
0
0
0
1
1 2
1 1 2
mol / min)
mol CH mol)
(1 mol C H mol) 1 mol / min @ 573K, 105 kPa
(mol CO mol)
(mol O (mol H O mol)
3.76 (mol N (1 mol N mol
4
2 6
2
2 2
2 2
−
− −
CH 2O CO 2H O C H
7
2
O 2CO 3H O
4 2 2 2 2 6 2 2 2
+ → + + → +
pCO2
mmHg
= 80 ⇒ y1
80 101325
01016
= =
mmHg
105000 Pa
Pa
760 mmHg
mol CO / mol
2
.
100%O conversion
2 : 2
7
2
1 1
n y n y n
o o o o
+ − =
b g (1)
C balance: n y n y
o o o o
+ − =
2 1 01016
b g . (2)
N2 balance: 376 1
1 1 2
. n y y
= − − (3)
H balance: 4 6 1 2 2
n y n y y
o o o o
+ − =
b g (4)
Solve equations 1 to 4 ⇒
=
=
=
=
R
S
|
|
T
|
|
n
y
n
o
o
0 0770
0 6924
01912
01793
1
.
.
.
.
mol
mol CH / mol
mol O
y mol H O / mol
4
2
2 2
Dew point:
p T
H O dp
2
01793 760
1412 588
* .
. .
d i b g b g
= = ⇒ =
105000 Pa mmHg
101325 Pa
mmHg T C Table B.3
dp
o
6.39 Basis: 100 mol dry stack gas
nP
(mol C H )
3 8
Stack gas:
0.21 O2
nB
(mol C H )
4 10
n out (mol)
0.79 N2
T = 46.5°C
dp
P = 780 mm Hg
100 mol dry gas
0.000527 mol C H /mol
3 8
0.000527 mol C H /mol
4 10
0.0148 mol CO/mol
0.0712 mol CO /mol
2
+ O , N
2 2
nw (mol H2O)
C H 5O 3CO 4H O C H
13
2
O 4CO 5H O
3 8 2 2 2 4 10 2 2 2
+ → + + → +

6-28
Dew point C C
mm Hg
mm Hg
mol H O
mol
2
= ° ⇒ = ∗ ° ⇒ = =
465 465
77 6
780
0 0995
. .
.
.
y P p y
w w w
b g
But 2
0.0995 11.05 mol H O
100
w
w w
w
n
y n
n
= = ⇒ =
+
(Rounding off strongly affects the result)
C balance: 3 4 100 0 000527 3 0 000527 4 0 0148 0 0712
n n
p B
+ = + + +
b gb gb g b gb g
. . . .
⇒ 3 4 8 969 1
n n
p B
+ = . b g
( ) ( )( ) ( )( ) ( )( )
H balance: 8 10 100 0.000527 8 0.000527 10 11.05 2
p B
n n
+ = ⎡ + ⎤ +
⎣ ⎦
⇒ ( )
8 10 23.047 2
p B
n n
+ =
( ) ( )
( )
3 8
3 8
4 10
4 10
49% C H
1.25 mol C H
Solve 1 2 simultaneously:
51% C H
1.30 mol C H
Answers may vary 8% due to loss of precision
p
B
n
n
⎧
=
⎧ ⎫
⎪ ⎪ ⎪
⇒ ⇒
⎨ ⎬ ⎨
=
⎪ ⎪
⎩ ⎭ ⎪
⎩
±
6.40 a.
/ /

L L
x
x
G G
y
y
1 2
2
2
1 2
1
1
1
1
(lb - mole C H h) (lb - mole h)
(lb - mole C H / lb - mole)
(lb - mole C H / lb - mole)
(lb - mole / h) = 1 lb - mole / h
(lb - mole C H / lb - mole) 0.07 (lb - mole C H / lb - mole)
(lb - mole N / lb - mole) 0.93 (lb - mole N / lb - mole)
10 22
3 8
10 22
3 8 3 8
2 2
−
−
Basis: lb - mole h feed gas

G2 1
=
N balance: 1
2 b gb g b g b g b g
0 93 1 1 0 93 1
1 1 1 1
. .
= − ⇒ − =
G y G y
98.5% propane absorption ⇒ = − ⇒ = × −
. . .
G y G y
1 1 1 1
3
1 0 985 1 0 07 105 10 2
b gb gb g b g
1 2 lb - mol h , mol C H mol
1 3 8
b g b g⇒ = = × −
. .
G y
0 93105 1128 10
1
3
Assume streams are in equilibrium

G L
2 2
−
From Cox Chart (Figure 6.1-4), p F
C H
o
* ( ) .
3 8
80 160 1089
= =
lb / in atm
2
Raoult's law: F
atm
10.89 atm
mol H O
mol
C H
2
3 8
x p p x
2 2
80 0 07
0 07 10
0 006428
∗ ° = ⇒ = =
b g b gb g
.
. .
.
Propane balance:
lb - mole h
0 07 1
0 07 0 93105 1128 10
0 006428
10 726
1 1 2 2 2
3
.
. . .
.
.
b gb g b gd i
= + ⇒ =
− ×
=
−
G y L x L
Decane balance: lb - mole h
1
. . .
L x L
= − = − =
1 1 0 006428 10 726 10 66
2 2
b gd h b gb g
⇒ / .
L G
1 2 10 7
d hmin
mol liquid feed / mol gas feed
=
6.39 (cont’d)

6-29
6.40 (cont’d)
b. The flow rate of propane in the exiting liquid must be the same as in Part (a) [same feed
rate and fractional absorption], or
.
nC H
3 3
3 8
3 8
- mole 0.006428 lb - mole C H
h lb - mole
lb - mol C H h
= =
10.726 lb
0 06895
The decane flow rate is 1.2 x 10.66 = 12.8 lb-moles C10H22/h
⇒ x2
0 06895
0 00536
= =
.
. /
lb - mole C H h
0.06895 +12.8 lb - moles h
lb - mole C H lb - mole
3 8
3 8
b g
c. Increasing the liquid/gas feed ratio from the minimum value decreases the size (and
hence the cost) of the column, but increases the raw material (decane) and pumping costs.
All three costs would have to be determined as a function of the feed ratio.
6.41 a. Basis: 100 mol/s liquid feed stream Let B n - butane
= , HC = other hydrocarbons
pB
*
(30 41 2120
o 2
C) lb / in mm Hg
≅ = (from Figure 6.1-4)
Raoult's law: y C) y
C)
P
4
o
4
o
P x p
x p
B B
B B
= ⇒ = =
×
=
*
*
(
( .
.
30
30 0125 2120
760
0 3487
95% n-butane stripped: . . . .
n n
4 4
0 3487 12 5 0 95 34 06
⋅ = ⇒ =
b g b gb g mol / s
Total mole balance: 3 3
100 34.06 88.125 22.18 mol/s
n n
+ = + ⇒ =

⇒
mol gas fed 22.18 mol/s
0.222 mol gas fed/mol liquid fed
mol liquid fed 100 mol/s
= =
b. If y4 = × =
08 0 3487 0 2790
. . . , following the same steps as in Part (a),
95% n-butane is stripped: . . . .
n n
4 4
0 2790 12 5 0 95 42 56
⋅ = ⇒ =
b g b gb g mol / s
Total mole balance: 100 42 56 88125 30 68
3 3
+ = + ⇒ =
. . .
n n mol / s
⇒
mol gas fed 30.68 mol/s
0.307 mol gas fed/mol liquid fed
mol liquid fed 100 mol/s
= =
c. When the N2 feed rate is at the minimum value calculated in (a), the required column length
is infinite and hence so is the column cost. As the N2 feed rate increases for a given liquid
feed rate, the column size and cost decrease but the cost of purchasing and compressing
(pumping) the N2 increases. To determine the optimum gas/liquid feed ratio, you would
need to know how the column size and cost and the N2 purchase and compression costs
depend on the N2 feed rate and find the rate at which the cost is a minimum.
y4 (mol B/mol)
(1-y4) (mol N2/mol)

n4 (mol/s) @ 30°C, 1 atm
100 mol/s @ 30o
C, 1 atm
xB =12.5 mol B/s
87.5 mol other hydrocarbon/s
0.625 mol B/s (5% of B fed)
87.5 mol HC/s

n3 (mol N2/s)
88.125 mol/s

6-30
6.42 Basis: 100 mol NH3
( )
aq
100 mol NH 3
780 kPa sat'd
converter
Preheated
air
n1 (mol) O2
n1 (mol) N2
3.76
n2 (mol) H O
2
1 atm, 30°C
hr
= 0.5
n3
(mol NO)
n4 (mol N )
n5 (mol O )
n6 (mol H O)
2
2
2
absorber
n7 (mol H O)
2
n8 (mol HNO )
3
n9 (mol H O)
2
55 wt% HNO 3
N2
O2
a. i) NH feed: kPa mm Hg atm
3 P P Tsat
= ∗ = = =
b g 820 6150 8 09
.
Antoine:
log . . . .
10 6150 7 55466 1002 711 247 885 18 4 291
b g b g
= − + ⇒ = ° =
T T
sat sat C .6 K
Table B.1
atm
⇒
= ⇒ = =
= ⇒ = =
U
V
W
⇒ =
P P
T K T
z
c r
c r
1113 8 09 1113 0 073
4055 2916 4055 0 72
0 92
. . / . .
. . / . .
. (Fig. 5.3-1)
VNH
3
3
3
mol Pa K
mol - K Pa
m NH
=
×
=
0 92 100 8 314 2916
820 10
0 272
3
. . .
.
b g
Air feed: NH O HNO H O
3 3 2
+ → +
2 2
n1
100
200
= =
mol NH 2 mol O
mol NH
mol O
3 2
3
2
Water in Air:
C
p 760
0.02094 mol H O
H O
mol air mol O
2
2
2
y
h p
n
n
n
r
=
⋅ °
=
×
=
⇒ =
+
⇒ =
A
( )
*
. .
.
.
.
( .76 )
30 0500 31824
0 02094
4 76 200
20 36
2
2
4
2
b g
Vair
3
3
3
4.76 200 20.36 mol 22.4 L STP 303K 1 m
1 mol 273K 10 L
m air
=
+
=
b g b g 24 2
.
ii) Reactions: NH O NO H O NH O N H O
2 2 2
4 5 4 6 4 3 2 6
3 2 3 2
+ → + + → +
,
Balances on converter
NO:
mol NH 4 mol NO
4 mol NH
mol NO
3
3
n3
97
97
= =

6-31
6.42 (cont’d)
N : 3 mol +
mol NH 2 mol N
4 mol NH
mol N
O : 200 mol
mol NH 5 mol O
4 mol NH
mol NH 3 mol O
4 mol NH
mol O
H O: mol +
6 mol H O
4 mol NH
mol H O
mol
= mol converter effluent
68.7% N 7.0% O 15.5% H O
2
3 2
3
2
2
3 2
3
3 2
3
2
2
3 2
3
2
total
2 2 2
n
n
n
n
4
5
6
76 2 00
3
7535
97
3
765
20 36
100 mol NH
170 4
97 7535 765 170 4
1097
88% NO,
= =
= −
− =
= =
⇒ = + + +
. . .
.
. .
( . . . )
. , ,
b g
iii) Reaction: NO O H O HNO
2
4 3 2 4
2 3
+ + →
HNO bal. in absorber:
mol NO react 4 mol HNO
4 mol NO
mol HNO
3
3
3
n8
97
97
= =
H O in product:
mol HNO 63.02 g HNO g H O 1 mol H O
mol 55 g HNO 18.02 g H O
mol H O
2
3 3 2 2
3 2
2
n9
97 45
277 56
=
= .
H balance on absorber: mol H
mol H O added
2
170 4 2 2 97 277 6 2
1557
7
7
. .
.
b gb g b gb gb g
+ = +
⇒ =
n
n
V l
H O
2 2
3 3
6 3
3
2
2
mol H O 18.02 g H O 1 cm 1 m
1 mol 1 g 10 cm
m H O
= = × −
1557
2 81 10 3
.
. b g
b. 3 3 2 2
acid
97 mol HNO 63.02 g HNO 277.6 mol H O 18.02 g H O
in old basis
mol mol
11115 g 11.115 kg
M = +
= =
Scale factor
metric tons kg metric ton
11.115 kg
= = ×
1000 1000
8 997 104
b gb g .
V
V
V l
NH
3
3
3
3
air
3 3
H O
3
2
3
2
3
2
m NH m NH
m air m air
2.81 10 m H O 253 m H O
= × = ×
= × = ×
= × × =
−
8 997 10 0 272 2 45 10
8 997 10 24 2 218 10
8 997 10
4 4
4 6
4 3
. . .
. . .
.
d id i
d id i
d id i b g

6-32
6.43 a. Basis: 100 mol feed gas
100 mol
0.10 mol NH /mol
3
0.90 mol G/mol
n1 (mol H O( ))
2 l
G = NH -free gas
3
n2 (mol)
yA (mol NH /mol)
3
yW (mol H O/mol)
2
yG (mol G/mol)
n3 (mol)
xA (mol NH /mol)
3
xA (mol H O/mol)
2
(1 – )
in equilibrium
at 10°C(50°F)
and 1 atm
Absorber
Composition of liquid effluent . Basis: 100 g solution
Perry, Table 2.32, p. 2-99: T = 10o
C (50o
F), ρ = 0.9534 g/mL ⇒ 0.120 g NH3/g solution
⇒
12 0
17 0
88 0
18 0
.
( .
,
.
( .
g NH
g / 1 mol)
= 0.706 mol NH
g H O
g / 1 mol)
= 4.89 mol NH
3
3
2
3
⇒ mole% NH aq), 87.4 mole% H O(l)
3 2
12 6
. (
Composition of gas effluent
p
T x p
p
y
y
y y y
A
Perry
A
W
G A W
NH
o
H O
total
3
2
3
2
psia Table 2 - 23
F, psia Table 2 - 21
psia
mol NH mol
mol H O mol
mol G mol
=
= = ⎯ →
⎯
⎯ =
=
U
V
|
W
|
⇒
= =
= =
= − − =
121
50 0126 0155
14 7
121 14 7 0 0823
0155 14 7 0 0105
1 0 907
.
. .
.
. / . .
. / . .
.
b g
b g
G balance: 100 mol
b gb g b gb g b g
0 90 100 0 90 0 907 99 2
2 2
. . . .
= ⇒ = =
n y n
G
NH absorbed mol NH
3 in out 3
= − =
100 010 99 2 0 0823 184
b gb g b gb g
. . . .
% absorption
.84 mol absorbed
100 mol fed
= × =
1
010
100% 18 4%
b gb g
.
.
b. If the slip stream or densitometer temperature were higher than the temperature in the
contactor, dissolved ammonia would come out of solution and the calculated solution
composition would be in error.
6.44 a.
15% oleum: Basis -100kg
kg SO
kg H SO kmol H SO kmol SO O
SO kmol H SO kmol SO
kg
84.4% SO
3
2 4 2 4 3 3
2 4 2 4 3
3
15
85 1 1 80.07 kg S
98.08 kg H 1 1
84 4
+ =
⇒
.

6-33
6.44 (cont’d)
b. Basis 1 kg liquid feed
no (mol), 40o
C, 1.2 atm
0.90 mol SO3/mol
0.10 mol G/mol
1 kg 98% H2SO4
0.98 kg SO3
0.02 kg H2O
n1 (mol), 40o
C, 1.2 atm
y1 mol SO3/mol
(1-y1) mol G/mol
m1 (kg) 15% oleum
0.15 kg SO3/kg
0.85 kg H2SO4/kg
Equilibrium @ 40o
C
i) y
p
P
1
3
3
40 115
760
151 10
=
°
= = × −
SO
3
C, 84.4%
mol SO mol
b g .
.
ii)
H balance:
kg H SO 2.02 kg H
98.08 kg H SO
kg H O 2.02 kg H
18.02 kg H O
m H SO 2.02 kg H
98.08 kg H SO
kg
2
2 4
2
2
1 2
2 4
0 98 0 02
085
128
4
4
1
. .
.
.
+
= ⇒ =
m
But since the feed solution has a mass of 1 kg,
SO absorbed kg
0.28 kg SO 10 g 1 mol
kg 80.07 g
mol
3.5 mol
balance: 0.10
mol
mol
3
3
3
= − = =
⇒ = −
= − ×
=
=
−
E
128 10 350
1 151 10
389
0 39
0 1
0
3
1
0
1
. . .
.
.
.
b g
d i
n n
G n n
n
n

V =
= ×
3.89 mol 22.4 L STP 313K 1 atm 1 m
1 kg liquid feed mol 273K 1.2 atm 10 L
.33 10 m kg liquid feed
3
3
-2 3
b g
8
6.45 a. Raoult’s law can be used for water and Henry’s law for nitrogen.
b. Raoult’s law can be used for each component of the mixture, but Henry’s law is not valid
here.
c. Raoult’s law can be used for water, and Henry’s law can be used for CO2.
6.46 ( ) ( )
( )
100 C 10 6.89272 1203.531 100 219.888 1350.1 mm Hg
B
p∗ ∗∗
° = − + =
( ) ( )
( )
100 C 10 6.95805 1346.773 100 219.693 556.3 mm Hg
T
p∗ ∗∗
° = − + =
( )
( )
( )
( )
2
N 2
0.40 1350.1
Raoult's Law: 0.0711 mol Benzene mol
10 760
0.60 556.3
0.0439 mol Toluene mol
10 760
1 0.0711 0.0439 0.885 mol N mol
B B B B
T
y P x p y
y
y
∗
= ⇒ = =
= =
= − − =

6-34
6.47 N - Henry's law: Perry's Chemical Engineers' Handbook, Page. 2 -127, Table 2 -138
H C atm mole fraction
2
N2
⇒ ° = ×
80 12 6 104
b g .
⇒ p x
N N N
2 2 2
H atm
= = × =
0 003 12 6 10 378
4
. .
b gd i
H O - Raoult's law: C
mm Hg 1 atm
760 mm Hg
atm
2 H O
2
p∗
° = =
80
3551
0 467
b g .
.
⇒ p x p
H O H O H O
2 2 2
atm
= = =
∗
d id i b gb g
0 997 0 467 0 466
. . .
Total pressure: atm
N H O
2 2
P p p
= + = + =
378 0 466 3785
. .
2 2
2 2
3
H O H O 2
N H O 2
Mole fractions: 0.466/378.5 1.23 10 mol H O mol gas
1 0.999 mol N mol gas
y p P
y y
−
= = = ×
= − =
6.48 H O - Raoult's law: C
mm Hg 1 atm
760 mm Hg
atm
2 H O
2
p∗
° = =
70
2337
0 3075
b g .
.
⇒ p x p xm
H O H O H O
2 2 2
= = −
∗
1 0 3075
b gb g
.
Methane Henry's law: m
− = ⋅
p x H
m m
Total pressure:
mol CH mol
m H O
4
2
P p p x x
x
m m
m
= + = ⋅ × + − =
⇒ = × −
6 66 10 1 0 3075 10
146 10
4
4
. ( )( . )
. /
6.49 a.
Moles of water
cm 1 g mol
cm 18.02 g
55.49 mol
H O
3
3
2
: n = =
1000
Moles of nitrogen
1- 0.334) 14.1 cm STP 1 mol 1 L
22.4 L (STP) 1000 cm
4 mol
N
3
3
2
:
( ( )
.
n =
×
= × −
192 10 4
Moles of oxygen
n
0.334) 14.1 cm STP mol L
22.4 L (STP) 1000 cm
mol
O
3
3
2
:
( ( )
.
=
⋅
= × −
2102 10 4
Mole fractions of dissolved gases:
2
2
2 2 2
2
2
2 2 2
4
N 6 2
4 4
H O N O
4
O 6
2
4 4
H O N O
mol N
4.192 10
7.554 10
mol
55.49 4.192 10 2.102 10
2.102 10
3.788 10 mol O / mol
55.49 4.192 10 2.102 10
N
O
n
x
n n n
n
x
n n n
−
−
− −
−
−
− −
×
= = = ×
+ + + × + ×
×
= = = ×
+ + + × + ×

6-35
6.49 (cont’d)
Henry's law
Nitrogen atm / mole fraction
Oxygen atm / mole fraction
N
O
2
2
:
.
.
.
:
.
.
.
H
p
x
H
p
x
N
N
O
O
= =
⋅
×
= ×
= =
⋅
×
= ×
−
−
2
2
2
2
0 79 1
7 554 10
1046 10
0 21 1
3788 10
5544 10
6
5
6
4
b. Mass of oxygen dissolved in 1 liter of blood:
m
2.102 10 mol 32.0 g
mol
g
O
-4
2
=
×
= × −
6 726 10 3
.
Mass flow rate of blood:
0.4 g O 1 L blood
min 6.72 10 g O
59 L blood / min
blood
2
-3
2

m =
×
=
c. Assumptions:
(1) The solubility of oxygen in blood is the same as it is in pure water (in fact, it is much greater)
(2) The temperature of blood is 36.9°C.
6.50 a. Basis 1 cm H O
1 g H O 1 mol
18.0 g
mol H O
0.0901 cm STP CO 1 mol
22,400 cm STP
mol CO
3
2
(SG) 2
2
SC) 0.0901
3
2
3 2
H2O
CO2
:
.
.
.0
(
l
b g
b g
b g
=
= −
⎯ →
⎯⎯⎯
⎯ =
⎯ →
⎯⎯⎯⎯
⎯ = ×
1
6
0 0555
4 022 10
p x
CO CO
2
2
2 2
atm
mol CO
0.0555 + 4.022 10 mol
mol CO mol
= ⇒ =
×
×
= ×
−
−
−
1
4 022 10
7 246 10
6
6
5
.
.
d i
d i
p x H H
CO CO CO CO
2 2 2 2
C
atm
7.246 10
atm mole fraction
= ⇒ ° =
×
=
−
20
1
13800
5
b g
b. For simplicity, assume mol
total H O
2
n n
≈ b g
x p H
CO CO 2
2 2
atm atm mole fraction mol CO mol
= = = × −
35 13800 2 536 10 4
. .
b g b g
nCO
3
2 2 2 2
2 2 2
2
2
12 oz 1 L 10 g H O 1 mol H O 2.536 mol CO 44.0 g CO
33.8 oz 1 L 18.0 g H O 1 mol H O 1 mol CO
g CO
=
×
=
−
10
0 220
4
.
c. V =
+
= =
0.220 g CO 1 mol CO 22.4 L STP 273 37 K
44.0 g CO 1 mol 273K
L cm
2 2
2
3
b g b g 0127 127
.

6-36
6.51 a. – SO2 is hazardous and should not be released directly into the atmosphere, especially if
the analyzer is inside.
– From Henry’s law, the partial pressure of SO2 increases with the mole fraction of SO2 in
the liquid, which increases with time. If the water were never replaced, the gas leaving
the bubbler would contain 1000 ppm SO2 (nothing would be absorbed), and the mole
fraction of SO2 in the liquid would have the value corresponding to 1000 ppm SO2 in the
gas phase.
b. Calculate mol SO mol in terms of g SO g H O
2 2 2
x r
b g b g
100
Basis: 100 g H O 1 mol 18.02 g mol H O
(g SO 1 mol 64.07 g (mol SO
mol SO
mol
2 2
2 2
SO
2
2
b g
b g
=
=
⇒ =
+
F
HG I
KJ
555
0 01561
0 01561
555 0 01561
.
) . )
.
. .
r r
x
r
r
From this relation and the given data, p x
SO SO 2
2 2
mmHg mol SO mol
= ⇔ =
0 0
42 1.4 x 10–3
85 2.8 x 10–3
129 4.2 x 10–3
176 5.6 x 10–3
A plot of pSO2
vs. xSO2
is a straight line. Fitting the line using the method of least squares
(Appendix A.1) yields p H x
SO SO SO
2 2 2
=
d i , H
mm Hg
mole fraction
SO2
= ×
3136 104
.
c.
( )( )
2
2 2
4
2 2
2 SO 6
4
SO SO
100 mol SO mol SO
100 ppm SO 1.00 10
mol
10 mols gas
1.0 10 760 mm Hg 0.0760 mm Hg
y
p y P
−
−
⇒ = = ×
⇒ = = × =
Henry's law
H
mm Hg
3.136 10 mm Hg mole fraction
mol SO mol
SO
SO
SO
4
2
2
2
2
⇒ = =
×
= × −
x
p 0 0760
2 40 10 6
.
.
Since xSO2
is so low, we may assume for simplicity that V V
final initial L
≈ = 140 , and
n n
l
final initial
3
2
140 L 10 g H O 1 mol
1 L 18 g
moles
≈ = = ×
b g 7 78 103
.
⇒ =
× ×
=
−
nSO
2
2
2
mol solution mol SO
1 mol solution
mol SO dissolved
7 78 10 2 40 10
0 0187
3 6
. .
.
0 0187
134 10 4
.
.
mol SO dissolved
140 L
mol SO L
2
2
= × −
Raoult’s law for water:
2 2
2
2 2
* o
H O H O 2
H O
SO H O
(30 C) mol H O(v)
(1)(31.824 mm Hg)
0.419
P 760 mm Hg mol
mol dry air
1 0.958
mol
air
x p
y
y y y
= = =
= − − =
d. Agitate/recirculate the scrubbing solution, change it more frequently. Add a base to the
solution to react with the absorbed SO2.

6-37
6.52 Raoult’s law + Antoine equation (S = styrene, T = toluene):
( )
7.06623 1507.434 214.985
0.650(150 mm Hg)
10
S S S S T
y P x p x
∗
− +
= ⇒ =
( )
6.95334 1343.943 219.377
0.350(150 mm Hg)
10
T T T T T
y P x p x
∗
− +
= ⇒ =
( ) ( )
7.06623 1507.434 214.985 6.95334 1343.943 219.377
0.65(150) 0.35(150)
1
10 10
86.0 C (Determine using E-Z Solve or a spreadsheeet)
S T T T
x x
T
− + − +
= + = +
⇒ = °
( )
7.06623 1507.434 86.0 214.985
0.65(150)
0.853 mol styrene/mol
10
S
x − +
= =
1 1 0.853 0.147 mol toluene/mol
T S
x x
= − = − =
6.53 ( ) ( )
6.89272 1205.531 85 219.888
85 C 10 881.6 mm Hg
B
p
− +
∗
° = =
( ) ( )
6.95805 1346.773 85 219.693
85 C 10 345.1 mm Hg
T
p
− +
∗
° = =
( )
( )
( )
( )
2
N 2
0.35 881.6
Raoult's Law: 0.0406 mol Benzene mol
10 760
0.65 345.1
0.0295 mol Toluene mol
10 760
1 0.0406 0.0295 0.930 mol N mol
B B B B
T
y P x p y
y
y
∗
= ⇒ = =
= =
= − − =
6.54 a. From the Cox chart, at 77 F, p psig p psig, p psig
P
*
nB
*
iB
*
° = = =
140 35 51
,
* * *
p p nB nB iB iB
Total pressure P=x p +x p +x p
0.50(140) 0.30(35) 0.20(51) 91 psia 76 psig
⋅ ⋅ ⋅
= + + = ⇒
P 200 psig, so the container is technically safe.

b. From the Cox chart, at 140 F psig psig, psig
P
*
nB
*
iB
*
° = = =
, ,
p p p
300 90 120
Total pressure P = psig
050 300 0 30 90 0 20 120 200
. ( ) . ( ) . ( )
+ + ≅
The temperature in a room will never reach 140o
F unless a fire breaks out, so the container
is adequate.
6.55 a. Antoine: ( ) ( )
6.84471 1060.793 120 231.541
120 C 10 6717 mm Hg
np
p
− +
∗
° = =
( ) ( )
6.73457 992.019 120 231.541
120 C 10 7883 mm Hg
ip
p
− +
∗
° = =
(Note: We are using the Antoine equation at 120o
C, well above the validity ranges in Table B.4
for n-pentane and isopentane, so that all calculated vapor pressures must be considered rough
estimates. To get more accuracy, we would need to find a vapor pressure correlation valid at
higher temperatures.)
When the first bubble of vapor forms,

6-38
6.55 (cont’d)
x n
np = 0500
. mol - C H (l) / mol
5 12 5 12
0.500 mol -C H (l)/mol
ip
x i
=
* *
Total pressure: = + 0.50(6717) 0.50(7883) 7300 mm Hg
np np ip ip
P x p x p
⋅ ⋅ = + =
*
5 12
0.500(6717)
0.46 mol -C H (v)/mol
7300
np np
np
x p
y n
P
⋅
= = =
5 12
1 1 0.46 0.54 mol -C H (v)/mol
ip np
y y i
= − = − =
When the last drop of liquid evaporates,
y n
np = 0500
. mol - C H (v) / mol
5 12 5 12
ip
y i
=
x x
y P
p
y P
p
P P
P
np ip
np
np
ip
ip
+ = + = + = ⇒ =
* *
( (
. .
120 120
0500
6725
0500
7960
1 7291
o o
C) C)
mm Hg
5 12
0.5*7250 mm Hg
0.54 mol n-C H (l)/mol
6717 mm Hg
np
x = =
5 12
1 1 0.54 0.46 mol -C H (l)/mol
ip np
x x i
= − = − =
b. When the first drop of liquid forms,
ynp = 0500
. mol n - C H (v) / mol
5 12 y i
ip = 0500
. mol - C H (v) / mol
12 12
P = (1200 + 760) = 1960 mm Hg
* * 6.84471 1060.793/( 231.541) 6.73457 992.019/( 231.541)
o
0.500 0.500 980 980
1
( ) ( ) 10 10
63.1 C
dp dp
np ip T T
np dp ip dp
dp
P P
x x
p T p T
T
− + − +
+ = + = + =
⇒ =
( )
6.84471 1060.793 63.1 231.541
10 1758 mm Hg
np
p
− +
∗
= =
( )
6.73457 992.019 63.1 231.541
10 2215 mm Hg
ip
p
− +
∗
= =
5 12
* o
0.5*1960 mm Hg
0.56 mol -C H /mol
(63.1 C)
np
np
x n
p
= =
5 12
1 1 0.56 0.44 mol -C H /mol
ip np
x x i
= − = − =
When the last bubble of vapor condenses,
xnp = 0500
. mol n - C H (l) / mol
5 12 x i
ip = 0500
. mol - C H (l) / mol
5 12
( )
* *
6.73457 992.019/( 231.541)
6.84471 1060.793 231.541
Total pressure: = +
1960 (0.5)10 (0.5)10 62.6 C
bp
np np ip ip
T
T
P x p x p
T
− +
− +
⋅ ⋅
⇒ = + ⇒ = °
* o
5 12
(62.6 C) 0.5(1734)
1960
np np
np
x p
y n
P
⋅
= = =
5 12
1 1 0.44 0.56 mol -C H (v)/mol
ip np
y y i
= − = − =

6-39
6.56 B = benzene, T = toluene
( / min)
nv mol at 80 C, 3 atm
o
yN2 (mol N2/mol)
10 L(STP)/min xB [mol B(l)/mol] yB [mol B(v)/mol]
(
nN mol / min)
2
xT [mol T(l)/mol] yT [mol T(v)/mol]

.
nN 2
2
L(STP) / min
22.4 L(STP) / mol
= 0.4464 mol N / min
=
10 0
Antoine: ( ) ( )
6.89272 1203.531 80 219.888
80 C 10 757.6 mm Hg
B
p
− +
∗
° = =
( ) ( )
6.95805 1346.773 80 219.693
80 C 10 291.2 mm Hg
T
p
− +
∗
° = =
a. Initially, xB = 0.500, xT = 0.500.
N balance mol N / min = mol / min
mol
min
mol B
mol
mol B(v)
min
mol
min
mol B
mol
mol T(v)
min
2 2 v v
B
T
: . ( . . ) .
. . .
. . .
0 4464 1 0166 0 0639 05797
05797 0166 0 0962
05797 0 0639 0 0370
0
0
n n
n
n
− − ⇒ =
⇒ =
F
HG I
KJF
HG I
KJ =
=
F
HG I
KJF
HG I
KJ =
b. Since benzene is evaporating more rapidly than toluene, xB decreases with time and xT (=
1–xB) increases.
c. Since xB decreases, yB (= xBpB*/P) also decreases. Since xT increases, yT (= xTpT*/P) also
increases.
6.57 a. P x p T x p T y
x p T
P
hex hex bp hep hep bp i
i i bp
= + =
∗ ∗
∗
d i d i d i
, , Antoine equation for i
p∗
bp bp
6.88555 1175.817/( 224.867) 6.90253 1267.828/( 216.823)
760 mm Hg 0.500 10 0.500 10
T T
− + − +
⎡ ⎤ ⎡ ⎤
= +
⎣ ⎦ ⎣ ⎦
E-Z Solve or Goal Seek ⇒ T y y
bp hex hep
= ° ⇒ = =
805 0 713 0 287
. . , .
C
b. x
y P
p T
x P
y
p T
i
i
i dp
i
i
i
i dp
i
= ⇒ = =
∗ ∗
∑ ∑
d i d i
1
dp dp
6.88555 1175.817/( 224.867) 6.90253 1267.828/( 216.823)
0.30 0.30
760 mmHg 1
10 10
T T
− + − +
⎡ ⎤
+ =
⎢ ⎥
⎣ ⎦
E-Z Solve or Goal Seek ⇒ T x x
dp hex hep
= ° ⇒ = =
711 0 279 0 721
. . .
C ,

6-40
6.58 a. f T P x p T T p T
i i
i
N
i
A
B
T C
i
i
i
( ) ( ) , ( )
* *
= − = ⇒ =
=
−
+
F
HG I
KJ
∑
1
0 10
where
y i N
x p T
P
i
i i
( , , , )
( )
*
= =
1 2
b.
Calculation of Bubble Points
A B C
Benzene 6.89272 1203.531 219.888
Ethylbenzene 6.95650 1423.543 213.091
Toluene 6.95805 1346.773 219.693
P(mmHg)= 760
xB xEB xT Tbp(o
C) pB pEB pT f(T)
0.226 0.443 0.331 108.09 378.0 148.2 233.9 -0.086
0.443 0.226 0.331 96.47 543.1 51.6 165.2 0.11
0.226 0.226 0.548 104.48 344.0 67.3 348.6 0.07
When pure benzene C
When pure ethylbenzene C
When pure toluene C
C H
o
C H
o
C H
o
6
8 10
7
x T T
x T T
x T T
B bp bp
EB bp bp
T bp bp
= = =
= = =
= = =
1 801
1 136 2
1 110 6
6
8
b g d i
b g d i
b g d i
, .
, .
, .
⇒ T T T
bp EB bp T bp B
, , ,

Mixture 1 contains more ethylbenzene (higher boiling point) and less benzene (lower bp)
than Mixture 2, and so (Tbp)1 (Tbp)2 . Mixture 3 contains more toluene (lower bp) and
less ethylbenzene (higher bp) than Mixture 1, and so (Tbp)3 (Tbp)1. Mixture 3 contains
more toluene (higher bp) and less benzene (lower bp) than Mixture 2, and so (Tbp)3
(Tbp)2

6-41
6.59 a. Basis: 150.0 L/s vapor mixture
Gibbs phase rule: F=2+c- =2+2-2=2
π
Since the composition of the vapor and the pressure are given, the information is enough.
Equations needed: Mole balances on butane and hexane, Antoine equation and Raoult’s law
for butane and hexane
b. 0
150.0 L 273 K mol
Molar flow rate of feed: n 4.652 mol/s
s 393 K 22.4 L (STP)
= =

6.82485 943.453/( 239.711)
2
Raoult's law for butane: 0.600(1100)=x 10 (1)
T
− +
⋅
6.88555 1175.817/( 224.867)
2
Raoult's law for hexane: 0.400(1100)=(1-x ) 10 (2)
T
− +
⋅
1 2 2
Mole balance on butane: 4.652(0.5)=n 0.6 n x (3)
⋅ + ⋅

1 2 2
Mole balance on hexane: 4.652(0.5)=n 0.4 n (1 x ) (4)
⋅ + ⋅ −

c.
1100(0.6) 1100(0.4)
From (1) and (2), 1=
943.453 1175.817
10**(6.82485 ) 10**(6.88555 )
239.711 224.867
T T
+
− −
+ +
⇒ °
T = 57.0 C
2 6.82485 943.453/(57.0 239.711)
1100(0.6)
0.149 mol butane /mol
10
x − +
= =
Solving (3) and (4) simultaneously ⇒ 1 4 10 2 6 14
3.62 mol C H /s; 1.03 mol C H /s
n n
= =

d. Assumptions: (1) Antoine equation is accurate for the calculation of vapor pressure;
(2) Raoult’s law is accurate;
(3) Ideal gas law is valid.
6.60 P = n-pentane, H = n-hexane
x2 [mol B(l)/mol]
(1- x2) [mol H(l)/mol]

n1 (mol/s) @ T(o
C), 1100 mm Hg

n0 (mol/s)@120°C, 1 atm
0.500 mol B(v)/mol
0.500 mol H(v)/mol
0.600 mol B(v)/mol
0.400 mol H(v)/mol

n2 (mol/s)
170.0 kmol/h, T1a (o
C), 1 atm
85.0 kmol/h, T1b (o
C), 1

n0 (kmol/h)
0.45 kmol P(l)/kmol
0.55 kmol H(l)/kmol
0.98 mol P(l)/mol
0.02 mol H(l)/mol
x2 (kmol P(l)/kmol)
(1- x2) (kmol H(l)/kmol)

n2 (kmol/h) (l),

6-42
6.60 (cont’d)
a. Molar flow rate of feed: n n 195 kmol / h
0 0
( . )( . ) ( . )
0 45 0 95 85 0 98
= ⇒ =
Total mole balance : n n 110 kmol / h
2 2
195 850
= + ⇒ =
.
Pentane balance: 195 x x 0.0405 mol P / mol
2
( . ) . ( . )
0 45 850 0 98 110 2
= + ⋅ ⇒ =
b. Dew point of column overhead vapor effluent:
1a 1a
o
1
6.84471 1060.793/( 231.541) 6.88555 1175.817/( 224.687)
Eq. 6.4-7, Antoine equation
0.98(760) 0.02(760)
1 37.3 C
10 10
a
T T
T
− + − +
⇒ + = ⇒ =
Flow rate of column overhead vapor effluent. Assuming ideal gas behavior,

Vvapor
3
3
kmol 0.08206 m atm (273.2 + 37.3) K
h kmol K 1 atm
/ h
=
⋅
⋅
=
170
4330 m
Flow rate of liquid distillate product.
Table B.1 ⇒ ρ ρ
P H
= 0.621 g / mL, = 0.659 g / mL
. ( ) .
. ( )
.
Vdistillate
kmol P
h
kg P
kmol P
L
0.621 kg P
+
kmol H
h kmol H
L
0.659 kg H
L / h
=
= ×
0 98 85 7215
0 02 85 86.17 kg H
9 9 103
c. Reboiler temperature.
2
2 6.88555 1175.817/( 224.867)
2
6.84471 1060.793/( 231.541)
0.04 10 0.96 10 760 T =66.6 C
T
T − +
− +
⋅ + ⋅ = ⇒ °
Boilup composition.
* o )
2 P
2
6.84471 1060.793/(66.6 231.541
(66.6 C) 0.04 10
0.102 mol P(v)/mol
760
x p
y
P
− +
⋅
= = =
⇒ (1- y 0.898 mol H(v) / mol
2 ) =
d. Minimum pipe diameter
( )
/
.
V u
D
D
V
u
m
s
m
s 4
m
m h
m / s
h
3600 s
m (39 cm)
3
max
min
2
2
min
vapor
max
3
F
HG
I
KJ =
F
HG I
KJ ×
⇒ =
⋅
= =
π
π π
4 4 4330
10
1
0 39
Assumptions: Ideal gas behavior, validity of Raoult’s law and the Antoine equation,
constant temperature and pressure in the pipe connecting the column and the condenser,
column operates at steady state.

6-43
6.61 a.
(mol)
F
(mol butane/mol)
x0
T
P
Condenser
(mol)
V
0.96 mol butane/mol
(mol)
R
(mol butane/mol)
x1
Partial condenser: C is the dew point of a 96% C H 4% C H vapor mixture at
4 10 5 12
40° −
=
P Pmin
Total condenser: C is the bubble point of a 96% C H - 4% C H liquid mixture at
4 10 5 12
40°
=
P Pmin
Dew Point:
C C
min
(Raoult's Law)
1
40
1
40
= =
°
⇒ =
°
∑ ∑
∑
∗ ∗
x
y P
p
P
y p
i
i
i i i
b g b g
Antoine Eq. for p H
i
∗
−
+
F
HG I
KJ
= =
C mmHg
4 10
6.82485
943
40 239
10 2830 70
b g
.453
.711
.
Antoine Eq. for ( )
1060.793
6.84471
40 231.541
5 12
C 10 867.22 mmHg
i
p H
⎛ ⎞
−
⎜ ⎟
∗ +
⎝ ⎠
= =
⇒ =
+
=
Pmin Hg partial condenser
1
0 96 2830 70 0 04 867 22
2595.63 mm
. . . .
b g
Bubble Point: C
P y P x p
i i i
= = °
∑ ∑ ∗
40
b g
P = + =
0 96 2830 70 0 04 867 22 275216
. . . . .
b g b g b g
mm Hg total condenser
b.
V = 75 kmol / h , . .
R V R
= ⇒ = × =
15 75 15 112.5 kmol
kmol / h / h
Feed and product stream compositions are identical: 0.96 kmol butane kmol
y =
Total balance: / h
.
F = + =
75 112 5 187.5 kmol
c. Total balance as in b. kmol / h kmol / h
. .
R F
= =
112 5 187 5
Equilibrium: .
Raoult's law
mm Hg
mol butane mol
Butane balance: mol butane mol reflux
0 96 2830 70
0 04 1 867 22
2596
08803
187 5 112 5 08803 0 96 75 0 9122
1
1 1
0 0
P x
P x
P
x
x x
=
= −
U
V
W
=
=
= + ⇒ =
.
. . .
. . . . .
b g
b g b gb g
b g b g
6.62 a. Raoult's law:
y
x
p
P
y x
y x
p P
p P
p
p
i
i
i
AB
A A
B B
A
B
A
B
AB
= ⇒ = = = =
∗ ∗
∗
∗
∗
α α
( )
1507.434
7.06623
o 85 214.985
1423.543
6.95650
o 85 213.091
1203.531
6.89272
o 85 219.888
*
*
*
85 C 10 109.95 mm Hg
(85 C) 10 151.69 mm Hg
(85 C) 10 881.59 mm Hg
S
EB
B
p
p
p
⎛ ⎞
−
⎜ ⎟
+
⎝ ⎠
⎛ ⎞
−
⎜ ⎟
+
⎝ ⎠
⎛ ⎞
−
⎜ ⎟
+
⎝ ⎠
= =
= =
= =
b.

6-44
6.62 (cont’d)
S,EB B,EB
* *
* *
109.95 881.59
0.725 , 5.812
151.69 151.69
S B
EB EB
p p
p p
α α
= = = = = =
Styrene ethylbenzene is the more difficult pair to separate by distillation
because is closer to 1 than is
S,EB B,EB
−
α α .
c. α α
α
α
ij
i i
j j
y y
x x
ij
i i
i i
i
ij i
ij i
y x
y x
y x
y x
y
x
x
j i
j i
= =
− −
⇒ =
+ −
= −
= −
⇒
1
1
1 1 1 1
( ) b g d i
d. α B EB
, .
= 5810⇒ y
x
x
x
x
P x p x p
B
B B EB
B EB B
B
B
B B B EB
=
+ −
=
+
= + −
α
α
,
,
( )
.
.
( )
* *
1 1
581
1 4 81
1
,
x
y
P
B l
B v
B
B
0 0 0 2 0 4 0 6 08 10
0 0 0592 0 795 0897 0 959 10
152 298 444 5900 736 882
. . . . . .
. . . . . .
mol mol
mol mol
mmHg
b g
b g
6.63 a. Since benzene is more volatile, the fraction of benzene will increase moving up the
column. For ideal stages, the temperature of each stage corresponds to the bubble point
temperature of the liquid. Since the fraction of benzene (the more volatile species)
increases moving up the column, the temperature will decrease moving up the column.
b. Stage 1: mol / h, mol / h
l v

n n
= =
150 200 ; x1 055 0 45
= ⇒
. .
mol B mol mol S mol ;
y0 0 65 0 35
= ⇒
. .
mol B mol mol S mol
Bubble point :
T P x p T
i i
= ∑ ∗
b g
( ) ( ) 7.06623 1507.434/( 214.985)
1
E-Z Solve o
1
6.89272 1203.531/( 219.888)
(0.400 760) mmHg 0.55 10 0.45 10
67.6 C
T
T
P
T
− +
− +
= × = +
⎯⎯⎯⎯
→ =
⇒ y
x p T
P
B
1
1 055 508
0 400 760
0 920 0 080
= =
×
= ⇒
∗
b g b g
.
.
. .
mol B mol mol S mol
B y n x n y n x n x
v l v l
balance: mol B mol mol S mol
0 2 1 1 2 0 910 0 090
. .
+ = + ⇒ = ⇒
Stage 2:
mmHg C
E-Z Solve o
( . ) . ( ) . ( ) .
* *
0 400 760 0 910 0 090 553
2 2 2
× = + ⎯ →
⎯⎯
⎯ =
p T p T T
B S
y2
0 910 3310
760 0 400
0 991 0 009
=
×
= ⇒
. .
.
. .
b g mol B mol mol S mol
B y n x n y n x n x
v l v l
balance: 1 mol B mol mol S mol
1 3 2 2 3 0

+ = + ⇒ ≈ ⇒ ≈
c. In this process, the styrene content is less than 5% in two stages. In general, the
calculation of part b would be repeated until (1–yn) is less than the specified fraction.

6-45
6.64 Basis: 100 mol/s gas feed. H=hexane.
a. ( ) ( )
( )( )
( )
2
4
1 1
4
95.025 mol s
N balance: 0.95 100 1
99.5% absorption: 0.05 100 0.005 2.63 10 mol H(v) mol
Mole Balance: 100 200 95.025 205 mol s
Hexane Balance: 0.05 100 2.63 10 95.02
GN
N GN
N GN N
L L
n
y n
y n y
n n
−
−
=
⎫
= − ⎪
⇒
⎬
= = ×
⎪
⎭
+ = + ⇒ =
= ×

( ) ( )
( ) ( )
1 1
5 204.99 0.0243 mol H(l) mol
1 1
200 205 202.48 mol s , 100 95.025 97.52 mol s
2 2
L L G G
x x
n n n n
+ ⇒ =
= + ⇒ = = + ⇒ =

b. y x p P
H
1 1 50 0 0243 40373 760 0 0129
= ° = =
B
∗
Antoine
C mol H(v) mol
b g b g
/ . . / .
H balance on 1 Stage: mol H(l) mol
st
y n x n y n x n x
v l v l
0 2 1 1 2 0 00643
.
+ = + ⇒ =
c. The given formulas follow from Raoult’s law and a hexane balance on Stage i.
d.
Hexane Absorption
P= 760 PR= 1
y0= 0.05 x1= 0.0243 yN= 2.63E-04
nGN= 95.025 nL1= 204.98 nG= 97.52 nL= 202.48
A= 6.88555 B= 1175.817 C= 224.867
T p*(T) T p*(T) T p*(T)
30 187.1 50 405.3059 70 790.5546
i x(i) y(i) i x(i) y(i) i x(i) y(i)
0 5.00E-02 0 5.00E-02 0 5.00E-02
1 2.43E-02 5.98E-03 1 2.43E-02 1.30E-02 1 2.43E-02 2.53E-02
2 3.10E-03 7.63E-04 2 6.46E-03 3.45E-03 2 1.24E-02 1.29E-02
3 5.86E-04 1.44E-04 3 1.88E-03 1.00E-03 3 6.43E-03 6.69E-03
4 7.01E-04 3.74E-04 4 3.44E-03 3.58E-03
5 3.99E-04 2.13E-04 5 1.94E-03 2.02E-03
... ... ...
21 4.38E-04 4.56E-04
nGN (mol/s)
yN (mol H/mol)
(1–yN) (mol N2/mol)
nL (mol/s)
xi (mol H/mol)
nL (mol/s)
xi–1 (mol H/mol)
nG (mol/s)
yi+1 (mol H/mol)
nL1 (mol/s)
x1 (mol H/mol) (99.5% of H in feed)
(1–x1) (mol oil/mol)
nG (mol/s)
yi (mol H/mol)
200 mol oil/s
100 mol/s
0.05 mol H/mol
0.95 mol N2/mol
Stage i

6-46
6.64 (cont’d)
e. If the column is long enough, the liquid flowing down eventually approaches equilibrium
with the entering gas. At 70o
C, the mole fraction of hexane in the exiting liquid in
equilibrium with the mole fraction in the entering gas is 4.56x10–4
mol H/mol, which is
insufficient to bring the total hexane absorption to the desired level. To reach that level at
70o
C, either the liquid feed rate must be increased or the pressure must be raised to a
value for which the final mole fraction of hexane in the vapor is 2.63x10–4
or less. The
solution is min 1037 mm Hg.
P =
6.65 a. Intersection of vapor curve with yB = 0 30
. at T = ° ⇒
104 13% B(l),
C 87%T(l)
b. ( ) ( )
100 C 0.24 mol B mol liquid , 0.46 mol B mol vapor
B B
T x y
= ° ⇒ = =
Basis: 1 mol
0.30 mol B(v)/mol
nV
(mol vapor)
0.46 mol B(v)/mol
nL
(mol liquid)
0.24 mol B(l)/mol
Balances
Total moles:
B: . .
mol
mol
mol vapor
mol liquid
1
0 30 0 46 0 24
0 727
0 273
0 375
= +
= +
U
V
W
⇒
=
=
⇒ =
n n
n n
n
n
n
n
V L
V L
L
V
V
L
.
.
.
.
c. Intersection of liquid curve with xB = 0 3
. at T = ° ⇒
98 50% B(v),
C 50%T(v)
6.66 a. P yB
= =
798 050
mm Hg, mol B(v) mol
.
b. 690 mm Hg, 0.15 mol B(l) mol
B
P x
= =
c. B B
750 mm Hg, 0.43 mol B(v) mol, 0.24 mol B(l) mol
P y x
= = =
3 mol B
7 mol T
nV (mol)
0.43 mol B/mol
nL (mol)
0.24 mol B/mol
Mole bal.:
B bal.: .43
mol
mol
mol vapor
mol liquid
10
3 0 0 24
316
684
0 46
= +
= +
U
V
W
⇒
=
=
⇒ =
n n
n n
n
n
n
n
V L
V L
V
L
v
l
.
.
.
.
Answers may vary due to difficulty of reading chart.
d. i) P = ⇒
1000 mm Hg all liquid . Assume volume additivity of mixture components.
V = + =
− −
3 mol B 78.11 g B 10 L
mol B 0.879 g B
7 mol T 92.13 g T 10 L
mol T 0.866 g T
L
3 3
10
.
ii) 750 mmHg. Assume liquid volume negligible

6-47
6.66 (cont’d)
V =
⋅
⋅
− =
3.16 mol vapor 0.08206 L atm 373 K 760 mm Hg
mol K 750 mm Hg 1 atm
L L
0 6 97 4
. .
(Liquid volume is about 0.6 L)
iii) 600 mm Hg
v =
⋅
⋅
=
10 mol vapor 0.08206 L atm 373K 760 mm Hg
mol K 600 mm Hg 1 atm
L
388
6.67 a. M = methanol
nV (mol)
n f (mol)
xF (mol M(l)/mol)
y (mol M(v)mol)
nL (mol)
x (mol M(l)/mol)
Mole balance:
MeOH balance:
n n n
x n yn xn
x n x n yn xn f
n
n
x x
y x
f V L
F f V L
F V F L V L
V
L
F
= +
= +
U
V
W
⇒ + = + ⇒ = =
−
−
x x y f
F = = = ⇒ =
−
−
=
0 4 0 23 0 62
0 4 0 23
0 62 0 23
0 436
. , . , .
. .
. .
.
b. T f
min ,
= =
75 0
o
C , T f
max ,
= =
87 1
o
C
6.68 a.
b. x y
A A
= =
0 47 0 66
. ; .
Txy diagram
50
55
60
65
70
75
80
0 0.2 0.4 0.6 0.8 1
Mole fraction of Acetone
T(
o
C)
Vapor
liquid
(P=1 atm)

6-48
6.68 (cont’d)
c. (i) x y
A A
= =
0 34 055
. ; .
(ii) Mole bal.: 1
0.762 mol vapor, 0.238 mol liquid
A bal.: 0.50 0.55 0.34
76.2 mole% vapor
V L
V L
V L
n n
n n
n n
= + ⎫
⎪
⇒ = = ∂
⎬
= + ⎪
⎭
⇒
(iii) 3 3 3
( ) E(l)
0.791 g/cm , 0.789 g/cm 0.790 g/cm
A l l
ρ ρ ρ
= = ⇒ ≈
(To be more precise, we could convert the given mole fractions to mass fractions and
calculate the weighted average density of the mixture, but since the pure component
densities are almost identical there is little point in doing all that.)
( )( ) ( )( )
A E
58.08 g/mol, M 46.07 g/mol
M 0.34 58.08 1 0.34 46.07 50.15 g/mol
l
M = =
⇒ = + − =
Basis: 1 mol liquid (0.762 mol vapor / 0.238 mol liquid) = 3.2 mol vapor
Liquid volume:
mol g / mol
g / cm
cm
Vapor volume
=
mol 22400 cm (STP) (65 + 273)K
mol 273K
cm
Volume percent of vapor volume vapor
3
3
3
3
⇒
= =
=
=
+
× =
V
V
l
v
( )( . )
( . )
.
:
.
,
,
.
. %
1 5015
0 790
6348
32
88 747
88 747
88747 6348
100% 99 9
d. For a basis of 1 mol fed, guess T, calculate nV as above; if nV ≠ 0.20, pick new T.
T xA yA fV
65 °C 0.34 0.55 0.333
64.5 °C 0.36 0.56 0.200
e. Raoult's law: = *
E E
y P x p P x p x p
i i i A A
⇒ = +
* *
o
*
7.11714 1210.595/( 229.664) 8.11220 1592.864/( 226.184)
7.11714 1210.595/(66.25 229.664)
760 0.5 10 0.5 10 66.16 C
0.5 10
0.696 mol acetone/mol
760
bp bp
bp
A
T T
T
xp
y
P
− + − +
− +
= × + × ⇒ =
×
= = =
o
A
A A
A
66.25 61.8
The actual 61.8 C 100% 7.20% error in
(real) 61.8
0.696 0.674
0.674 100% 3.3% error in
(real) 0.674
bp
bp bp
bp
T
T T
T
y
y y
y
Δ −
= ⇒ = × =
Δ −
= ⇒ = × =
Acetone and ethanol are not structurally similar compounds (as are, for example, pentane
and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s
law to be valid for acetone mole fractions that are not very close to 1.

6-49
6.69 a. B = benzene, C = chloroform. At 1 atm, (Tbp)B = 80.1o
C, (Tbp)C = 61.0o
C
The Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1o
C when
xC = 0 and at 61.0o
C when xC = 1. (See solution to part c.)
b.
Txy Diagram for an Ideal Binary Solution
A B C
Chloroform 6.90328 1163.03 227.4
Benzene 6.89272 1203.531 219.888
P(mmHg)= 760
x T y p1 p2 p1+p2
0 80.10 0 0 760 760
0.05 78.92 0.084 63.90 696.13 760.03
0.1 77.77 0.163 123.65 636.28 759.93
0.15 76.66 0.236 179.63 580.34 759.97
0.2 75.58 0.305 232.10 527.86 759.96
0.25 74.53 0.370 281.34 478.59 759.93
0.3 73.51 0.431 327.61 432.30 759.91
0.35 72.52 0.488 371.15 388.79 759.94
0.4 71.56 0.542 412.18 347.85 760.03
0.45 70.62 0.593 450.78 309.20 759.99
0.5 69.71 0.641 487.27 272.79 760.07
0.55 68.82 0.686 521.68 238.38 760.06
0.6 67.95 0.729 554.15 205.83 759.98
0.65 67.11 0.770 585.00 175.10 760.10
0.7 66.28 0.808 614.02 145.94 759.96
0.75 65.48 0.844 641.70 118.36 760.06
0.8 64.69 0.879 667.76 92.17 759.93
0.85 63.93 0.911 692.72 67.35 760.07
0.9 63.18 0.942 716.27 43.75 760.03
0.95 62.45 0.972 738.72 21.33 760.05
1 61.73 1 760 0 760
Txy diagram
60
65
70
75
80
85
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Mole fraction of chloroform
T(
o
C)
Vapor
Liquid
(P=1 atm)

6-50
6.69 (cont’d)
d.
Raoult’s law: T y
bp
o
= C, = .
71 058 ⇒
Δ
Δ
T
T
T
y
y
y
actual
bp
actual
error in
error in
=
−
× = −
=
−
× = −
71 753
753
100% 57%
058 0 60
0 60
100% 333%
.
.
.
. .
.
.
Benzene and chloroform are not structurally similar compounds (as are, for example,
pentane and hexane or benzene and toluene). There is consequently no reason to expect
Raoult’s law to be valid for chloroform mole fractions that are not very close to 1.
6.70 P x p T x p T
m m bp m P
≈ = = + −
1 760 1
atm mm Hg bp
* *
d i b g d i
7.87863 1473.11/( 230) 7.74416 1437.686/( 198.463) E-Z Solve o
760 0.40 10 0.60 10 79.9 C
bp bp
T T
T
− + − +
= × + × ⎯⎯⎯⎯
→ =
We assume (1) the validity of Antoine’s equation and Raoult’s law, (ii) that pressure head and
surface tension effects on the boiling point are negligible.
The liquid temperature will rise until it reaches 79.9 °C, where boiling will commence. The
escaping vapor will be richer in methanol and thus the liquid composition will become richer
in propanol. The increasing fraction of the less volatile component in the residual liquid will
cause the boiling temperature to rise.
Txy diagram
60
65
70
75
80
85
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Mole fraction of choloroform
T(
o
C)
(P=1 atm)
yc xc
x y

6-51
6.71 Basis: 1000 kg/h product
nA1 (mol A/h)
nE1 (mol E/h)
280°C
reactor
nA2 (mol A/h)
nE2 (mol E/h)
condenser
nH2 (mol H /h)
2
nC (mol/h)
still
0.550 A
0.450 E
liquid, –40°C
n0 (mol E/h)
Fresh feed
n3 (mol/h)
vapor, –40°C
yA3 (mol A/mol), sat'd
yE3
(mol E/mol), sat'd
yH3 (mol H /h)
2
scrubber
nH4 (mol H /h)
2
Product
1000 kg/h
np (mol/h)
0.97 A
0.03 E
Scrubbed
Hydrocarbons
nA4 (mol A/h)
nE4 (mol E/h)
E = C H OH (
2 5 M = 46.05)
A = CH CHO (
3 M = 44.05)
P = 760 mm Hg
nr (mol/h)
0.05 A
0.95 E
Strategy
• Calculate molar flow rate of product
np
d i from mass flow rate and composition
• Calculate yA3 and yE3 from Raoult’s law: y y y
H3 A3 E3
= − −
1 . Balances about
the still involve fewest unknowns ( )
n n
c r
and
•
Total mole balance about still
A balance about still
U
V
W
⇒ ,
n n
c r
• A, E and H2 balances about scrubber ⇒ ,
n n
A4 E4 , and
nH4 in terms of
n3 .
Overall atomic balances on C, H, and O now involve only 2 unknowns ( ,
n n
0 3 )
•
Overall C balance
Overall H balance
U
V
W
⇒ ,
n n
0 3
• A balance about fresh feed-recycle mixing point ⇒
nA1
• E balance about fresh feed-recycle mixing point ⇒
nE1
• A, E, H2 balances about condenser , ,
n n n
A2 E2 H2
• All desired quantities may now be calculated from known molar flow rates.
a. Molar flow rate of product
M M M
= + = + =
0 97 0 03 0 97 44 05 0 03 46 05 4411
. . . . . . .
A E g mol
b gb g b gb g
.
np = =
1000 1
22 67
kg
h
kmol
44.11 kg
kmol h
Table B.4 (Antoine) ⇒ ( )
*
A 40 C 44.8 mm Hg
p − ° =
( )
*
E 40 C 0.360 mm Hg
p − ° =
Note: The calculations that follow can at best be considered rough estimates, since we are
using the Antoine correlations of Table B.4 far outside their temperature ranges of validity.
Raoult’s law ⇒
( )
*
A
A3
0.550 40 C 0.550(44.8)
0.03242 kmol A/kmol
760
p
y
P
− °
= = =

6-52
6.71 (cont’d)
( )
*
E 4
E3
0.450 40 C 0.450(0.360)
2.13 10 kmol E kmol
760
p
y
P
−
− °
= = = ×
H3 A3 E3 2
1 0.9674 kmol H kmol
y y y
= − − =
Mole balance about still:
A balance about still:
kmol / h recycle
kmol / h
.
. . ( . ) .
.
.
n n n n n
n n
n
n
c p r c r
c r
r
c
= + ⇒ = +
= +
U
V
W
⇒
=
=
22 67
0550 0 97 22 67 0 05
29 5
521
A balance about scrubber: A4 3 A3 3
0.03242
n n y n
= =
(1)
E balance about scrubber: 4
E4 3 E3 3
2.13 10
n n y n
−
= = ×
(2)
H balance about scrubber:
2 H4 3 H3 3
0.9764
n n y n
= =
(3)
Overall C balance:

. .
n
n n n n
p p
0
2 2 0 97 2 0 03 2
(mol E) 2 mol C
h 1 mol E A4 E4
= + + +
b gb g b gb g d ib g d ib g
⇒ .
n n n
0 4 4 22 67
= + +
A E (4)
Overall H balance:
6 2 4 6 0 97 4 0 03 6
0
. .
n n n n np
= + + + +
H4 A4 E4 b gb g b gb g (5)
Solve (1)–(5) simultaneously (E-Z Solve):
0 H4 2
23.4 kmol E/h (fresh feed), 22.7 kmol H /h (in off-gas)
n n
= =

3 A4 E4
= 23.3 kmol/h, = 0.755 kmol A/h, = 0.00496 kmol E/h
n n n

A balance about feed mixing point: A1 0.05 1.475 kmol A h
r
n n
= =

E balance about feed mixing point: E1 0 0.95 51.5 kmol E h
r
n n n
= + =

E balance about condenser: E2 3 E3 0.450 23.5 kmol E h
c
n n y n
= + =

( ) ( ) ( )
3
3 3
reactor feed
Ideal gas equation of state:
1.47 51.5 kmol 22.4 m STP 273+280 K
2.40 10 m h
h 1 kmol 273K
V
+
= = ×
b. Overall conversion
( )( )
0
0
0.03 23.4 0.03 22.67
100% 100% 97%
23.4
p
n n
n
− −
= × = × =

Single-pass conversion E1 E2
E1
51.5 23.5
100% 100% 54%
51.5
n n
n
− −
= × = × =

Feed rate of A to scrubber: A4 =0.76 kmol A/h
n

Feed rate of E to scrubber: E4 0.0050 kmolE h
n =

6-53
6.72 a. G = dry natural gas, W = water

n
n
3
4
m
6
o
(lb - mole G / d)
(lb - mole W / d)
10 lb W /10 SCF gas
90 F, 500 psia
Absorber
(
n7 lb - mole W / d)
4 0 106
1
2
.
/
(
[
×
×
SCF / d
4 80 = 320 lb W d
lb - mole G / d)
lb - mole W(v) / d]
m
n
n
Overall system D.F. analysis 5 unknowns (
feed specifications (total flow rate, flow rate of water)
water content of dried gas
balances (W, G)
0 D.F.
: , , , , )
n n n n n
1 2 3 4 7
2
1
2
−
−
−
Water feed rate: .
n2 17 78
= =
320 lb W 1 lb - mole
d 18.0 lb
lb - moles W / d
m
m
Dry gas feed rate:

.
. .
n1
6
4
40 10
1778 1112 10
=
×
− = ×
SCF 1 lb- mole
d 359 SCF
lb- moles W
d
lb- moles G / d
Overall G balance: .
n n n
1 3 3
4
1112 10
= ⇒ = × lb - moles G / d
Flow rate of water in dried gas:

( )

.
n
n n
n
n
4
3 4
1112 10
4
3
4
=
+
⎯ →
⎯⎯⎯
⎯ =
= ×
lb - moles 359 SCF gas 10 lb W 1 lb - mole W
d lb - mole 10 SCF 18.0 lb
2.218 lb - mole W(l) / d
m
6
m
Overall W balance:

( . . )
n7
1778 2218
280
=
−
= ×
F
HG
I
KJ =
lb - moles W 18.0 lb
d 1 lb - mole
lb W
d
1 ft
62.4 lb
4.5
ft W
d
m m
3
m
3

n
n
5
6
lb - mole TEG
d
lb - mole W
d
F
HG I
KJ
F
HG I
KJ

n
n
5
8
lb-mole TEG
d
lb-mole W
d
F
HG I
KJ
F
HG I
KJ
Distillation
Column

6-54
6.72 (cont’d)
b. Mole fraction of water in dried gas =
y
n
n n
w 4
lb - moles W / d
(2.218 + 1.112 10 lb - moles / d
lb - moles W(v)
lb - mole
=
+
=
×
= × −

.
)
.
4
3 4
4
2 218
199 10
Henry’s law: ywP = Hwxw ⇒
( )
xw max =
( . )(
.
199 10 500
0 0170
4
×
=
−
psia)(1 atm / 14.7 psia)
0.398 atm / mole fraction
lb - mole dissolved W
lb - mole solution
c. Solvent/solute mole ratio

.
. ( . . ) .
n
n n
n
5
2 4
5
37 1
4434
4434 1778 222 690
−
= =
⇒ = − =
lb TEG lb- mole TEG 18.0 lb W
lb W 150.2 lb TEG 1 lb W
lb- mole TEG
lb- mole Wabsorbed
lb- moles TEG / d
m m
m m m
(xw)in = 0.80(0.0170) 5 69.0
8
8
5 8
lb-mole W
=0.0136 = 0.951lb-mole W/d
lb-mole
n
n
n
n n
=
⎯⎯⎯⎯
→ =
+

Solvent stream entering absorber

/
m= +
×
0 69.0 lb
.951 lb- moles W 18.0 lb
d lb- mole
- moles TEG 150.2 lb
d lb- mole
= 1.04 10 lb d
m m
4
m
W balance on absorber
6 (17.78 0.95 2.22) lb-moles W/d=16.51lb-moles W/d
n = + −

⇒ out
16.51 lb-moles W/d
( ) 0.19 lb-mole W/lb-mole
(16.51 + 69.9) lb-moles/d
w
x = =
d. The distillation column recovers the solvent for subsequent re-use in the absorber.
6.73 Basis: Given feed rates
absorber
n3
(mol/h)
x3 (mol H S/mol)
2
(1 – )
x3
(mol solvent/mol)
0°C
0°C
n4
(mol/h)
0.002 H S
2
0.998 solvent
stripper
40°C
n3
(mol/h)
x3 (mol H S/mol)
2
(1 – )
x3
(mol solvent/mol)
40°C
heater
100 mol/h
0.96 H2
0.04 H S, sat'd
2
1.8 atm
0.999 H 2
0.001 H S
2
n1
(mol/h) 200 mol air/h
0.
n2 mol H S/mol
2
200 mol air/h
G2
G1 G3 G4
L2 L1
40°C, 1 at m

6-55
6.73 (cont’d)
Equilibrium condition: At G1, pH S
2
atm atm
= =
0 04 18 0 072
. . .
b gb g
⇒ = = = × −
x
p
H
3
3
0 072
2 67 10
H S
H S
2
2
2
atm
27 atm mol fraction
mole H S mole
.
.
Strategy: Overall H2 and H S
2 balances ⇒ ,
n n
1 2

n2 + air flow rate ⇒ volumetric flow rate at G4
H S
2 and solvent balances around absorber ⇒ ,
n n
3 4
0 998 4
.
n = solvent flow rate
Overall H balance:
2 100 0 96 0 999 961
1 1
b gb g
. . .
= ⇒ =
n n mol h
Overall H S balance:
2 100 0 04 0 001 390
1 2
96.1
2
1
b gb g
. . .

= + ⇒ =
=
n n n
n
mol H S h
2
Volumetric flow rate at stripper outlet

VG4
200 + 3.90 mol 22.4 liters STP K
h 1 mol 273 K
L hr
=
+
=
b g b g b g
273 40
5240
H S
2 and solvent balances around absorber:
100 0 04 0 002 0 001 1335 1952
0 998 1 2 67 10
5830
4 1 3 3 4 3
4 3
3 3 4
b gb g
d i
. . . .
. .

+ = + ⇒ = −
= − ×
U
V
|
W
|⇒ ≈ =
−
n n n x n n
n n
n n mol h
Solvent flow rate = =
0 998 4
.
n 5820 mol solvent h
6.74 Basis: 100 g H O
2
Sat'd solution @ 60°C
100 g H O
2
16.4 g NaHCO3
Sat'd solution @ 30°C
100 g H O
2
11.1 g NaHCO3
ms (g NaHCO ( ))
3 s
NaHCO balance g NaHCO s
3 3
⇒ = + ⇒ =
16 4 111 53
. . .
m m
s s b g
% crystallization = × =
5
100% 32 3%
.3 g crystallized
16.4 g fed
.
6.75 Basis: 875 kg/h feed solution
1.03
(1 – )
(kg KOH/kg)
(kg H O/kg)
Sat'd solution 10°C
m3 (kg KOH-2H O( )/h)
x0
(kg H O( )/h)
2 v
m2 (kg H O(1)/h)
2
m2 (kg KOH/h)
875 kg/h
x0 2
m1
2 s
60% of KOH in feed

6-56
6.75 (cont’d)
Analysis of feed: 2KOH H SO K SO 2H O
2 4 2 4 2
+ → +
x0
0 427
=
=
22.4 mL H SO l 1 L 0.85 mol H SO 2 mol KOH 56.11 g KOH
5 g feed soln 10 mL L 1 mol H SO 1 mol KOH
g KOH g feed
2 4 2 4
3
2 4
b g
.
60% recovery: ( )( )
875 0.427 0.60 224.2 kg KOH h
=
( )
2
3 2 2
224.2 kg KOH 92.15 kg KOH 2H O
368.2 kg KOH 2H O h 143.8 kg H O h
h 56.11 kg KOH
m
⋅
= = ⋅
KOH balance: ( ) 2 2
0.427 875 224.2 1.03 145.1 kg h
m m
= + ⇒ =
Total mass balance: ( ) 1 1 2
875 368.2 2.03 145.1 212kg H O h evaporated
m m
= + + ⇒ =
6.76 a.
C
R
C
C R C R
A
A
A A
g A dissolved
mL solution
Plot vs. =
=
⇒
0 30 45
0 0 200 0 300
150
. .
/
b. Mass of solution:
500 mol 1.10 g
ml
g
= 550 (160 g A, 390 g S)
The initial solution is saturated at 10.2 °C.
Solubility @ 10.2 °C = = = °
160
0 410 410
g A
390 g S
g A g S g A 100 g S @ 10.2 C
. .
At 0°C, R = 17 5
. ⇒ CA
g A 1 mL soln
mL soln g soln
g A g soln
= =
17 5 150
110
0106
.
.
.
Thus 1 g of solution saturated at 0°C contains 0.106 g A 0.894 g S.
Solubility @ 0°C
0106
0118 118
.
. .
g A
0.894 g S
g A g S g A 100 g S @ 0 C
= = °
Mass of solid A: 160 114
g A
390 g S 11.8 g A
100 g S
g A s
− = b g
c. 160 114
0 5 390
23 0
− −
×
=
b g b g
g A
g S 11.8 g A
100 g S
g A s
g A initial
g A remaining in soln

.
.
6.77 a. Table 6.5-1 shows that at 50o
F (10.0o
F), the salt that crystallizes is MgSO H O
4 2
⋅ 7 , which
contains 48.8 wt% MgSO4.
b. Basis: 1000 kg crystals/h.

m0 (g/h) sat’d solution @ 130o
F
m1 (g/h) sat’d solution @ 50o
F
0.35 g MgSO4/g 0.23 g MgSO4/g
0.65 g H2O/g 0.77 g H2O/g
1000 kg MgSO4·7H2O(s)/h

6-57
6.77 (cont’d)
Mass balance kg / h
MgSO balance: 0.35 kg MgSO h
kg feed / h
kg soln / h
The crystals would yield 0.488 1000 kg / h = 488
kg anhydrous MgSO
h
4 4
4
:
. . ( ) /

m m
m m
m
m
0 1
0 1
0
1
1000
0 23 0 488 1000
2150
1150
= +
= + ⇒
=
=
×
6.78 Basis: 1 lbm feed solution.
Figure 6.5-1 ⇒ a saturated KNO3 solution at 25o
C contains 40 g KNO3/100 g H2O
⇒ xKNO
3
3 m 3 m
3
g KNO
(40 +100) g solution
g KNO g = 0.286 lb KNO lb
= =
40
0 286
. / / x
1 lbm solution @ 80o
C
0.50 lbm KNO3/lbm m1(lbm) sat’d solution @ 25o
C
0.50 lbm H2O/lbm 0.286 lbm KNO3/lbm soln
0.714 lbm H2O/lbm soln
m2 [lbm KNO3(s)]
Mass balance 1 lb
KNO balance 0.50 lb KNO
= 0.700 lb solution / lb feed
0.300 lb crystals/ lb feed
Solid / liquid mass ratio =
0.300 lb crystals / lb feed
0.700 lb solution / lb feed
= 0.429 lb crystals/ lb solution
m
3 m 3
m m
m m
m m
m m
m m
:
: .
= +
= +
⇒
=
m m
m m
m
m
1 2
1 2
1
2
0286
6.79 a. Basis: 1000 kg NaCl(s)/h.
Figure 6.5-1 ⇒ a saturated NaCl solution at 80o
C contains 39 g NaCl/100 g H2O
⇒ x g
NaCl
(39 +100) g solution
g NaCl g = 0.281 kg NaCl k
= =
39 g NaCl
0 281
. / /

m2 [kg H O(v) / h]
2

m0 (kg/h) solution
m1 (kg/h) sat’d solution @ 80o
C
0.100 kg NaCl/kg 0.281 kg NaCl/kg soln
0.900 kg H2O/kg 0.719 kg H2O/kg soln
1000 kg NaCl(s)/h
Mass balance
NaCl balance 0.100 kg NaCl
=0.700 lb solution / lb feed
Solid / liquid mass ratio =
0.700 lb solution / lb feed
=0.429 lb crystals/ lb solution
m m
m m
m m
m m
m m
:
: .

m m m
m m
m
m
0 1 2
1 2
1
2
0281
= +
= +
⇒
=
The minimum feed rate would be that for which all of the water in the feed evaporates to
produce solid NaCl at the specified rate. In this case

6-58
0100 1000 10 000
9000
0
0 0
2
1
. ( ) ( ) ,
:
:
min min
m m
m
m
= ⇒ =
=
=
kg NaCl / h kg / min
Evaporation rate kg H O / h
Exit solution flow rate
2
b.
m2 [kg H O(v) / h]
2

m0 (kg/h) solution
m1 (kg/h) sat’d solution @ 80o
C
0.100 kg NaCl/kg 0.281 kg NaCl/kg soln
0.900 kg H2O/kg 0.719 kg H2O/kg soln
1000 kg NaCl(s)/h
40% solids content in slurry ⇒ 1000 2500
1 1
kg NaCl
h
= 0.400( ) ( )
kg
h
max max

m m
⇒ =
NaCl balance 0.100 7025 kg / h
Mass balance kg H O evaporate / h
2
: . ( )
:
m m
m m m
0 0
0 2 2
0281 2500
2500 4525
= ⇒ =
= + ⇒ =
6.80 Basis: 1000 kg K Cr O s h
2 2 7 ( ) . Let K = K Cr O
2 2 7 , A = dry air, S = solution, W = water.
Composition of saturated solution:
020 020
01667
. .
.
kg K
kg W
kg K
1+ 0.20 kg soln
kg K kg soln
⇒ =
b g

me [kg W(v) / h)
(
)(
.
n
y
y
T
2
2
2
392
mol / h)
(molW(v) / mol)
(1 molA/ mol)
90 C, 1atm, C
o
dp
o
−
=
kg/ h) (kg/ h)
0.210kgK/ kg
0.790kgW(l) / kg
(
m m m
f f r
+
(

m
na
1 kg / h)
0.90 kgK(s) / kg 1000 kg K(s) / h
0.10 kgsoln / kg
0.1667kgK/ kg
0.8333kgW/ kg
(mol A / h)

mr (kg recycle / h)
0.1667 kg K / kg
0.8333 kg W / kg
Dryer outlet gas: y P p y
2 2
39 2
5301
00698
= ° ⇒ = =
W
*
C
mm Hg
760 mm Hg
mol W mol
.
.
.
b g
Overall K balance: 0210 1000 4760
.
m m
f f
= ⇒ =
kg K h kg h feed solution
CRYSTALLIZER-
CENTRIFUGE
DRYER
6.79 (cont’d)

6-59
6.80 (cont’d)
K balance on dryer: 090 01667 010 1000 1090
1 1 1
. . .
m m m
+ = ⇒ =
b gb g kg h kg h
Mass balance around crystallizer-centrifuge

m m m m m m
f r e r e
+ = + + ⇒ = − =
1 4760 1090 3670 kg h water evaporated
95% solution recycled
0.10 1090 kg h not recycled kg recycled
5 kg not recycled
kg h recycled
r
⇒ =
×
=

m
b g 95
2070
Water balance on dryer
08333 010 1090
1801 10
00698 7 225 10
3 2 2
4
. .
.
. .
b gb gb g kg W h
kg mol
mol h
×
= ⇒ = ×
−
n n
Dry air balance on dryer
na =
− ×
= ×
1 00698 7
151 106
.
.
b g b g b g
.225 10 mol 22.4 L STP
h 1 mol
L STP h
4

6-60
6.81. Basis : 100 kg liquid feed. Assume Patm=1 atm
Degree of freedom analysis:
Reactor Filter
6 unknowns (n1, n2, y2w, y2c, m3, m4) 2 unknowns
–4 atomic species balances (Na, C, O, H) –2 balances
–1 air balance 0 DF
–1 (Raoult's law for water)
0 DF
Na balance on reactor
100 kg 0.07 kg Na CO 46 kg Na
kg 106 kg Na CO
( 0.024 ) kg NaHCO 23kg Na
84 kg NaHCO
3.038 0.2738( 0.024 ) (1)
2 3
2 3
3 4 3
3
3 4
=
+
⇒ = +
m m
m m
Air balance: 0 300 2
1 2
. ( )
n n a
=
C balance on reactor :
n
n m m n n m m
c c
1 2
2
2 3
2 3
(kmol) 0.700 kmol CO 12 kg C
kmol 1kmol CO
100 kg 0.07 kg Na CO 12 kg C
kg 106 kg Na CO
+
= + + ⇒ + = + +
( )( ) ( . )( ) . . . ( . ) ( )
2 3 2 1 2 3 4
12 0024
12
84
840 07924 12 01429 0024 3
H balance :
( )( . )( ) ( )( ) ( . )( ) . ( )
100 0 93
2
18
2 0 024
1
84
0 976
2
18
2 3 4 4
= + + +
n m m m
w
⇒ = + + +
10 33 2 0 01190 0 024 01084 4
2 3 4 4
. . ( . ) . ( )
n m m m
w
n (kmol H O )(sat'd)
n (kmol CO )
n (kmol Air)
70 C, 3 atm(absolute)
2w 2
2c 2
2a
o
Filter
Filtrate
m (kg)
0.024 kg NaHCO / kg
0.976 kg H O / kg
5
3
2
Reactor
e
100 kg Feed
0.07 kg Na CO / kg
0.93 kg H O / kg
2 3
2
n (kmol)
0.70 kmol CO / kmol
0.30 kmol Air / kmol
1
2
Filter cake
m (kg)
0.86 kg NaHCO
0.14 kg solution
0.024 kg NaHCO / kg
0.976 kg H O / kg
6
3
3
2
( ) /
s kg
R
S
|
T
|
U
V
|
W
|
m kg NaHCO s
3 3
( ( ))
m (kg solution)
0.024 kg NaHCO / kg
0.976 kg H O / kg
4
3
2
R
S
|
T
|
U
V
|
W
|
Reactor

6-61
6.81(cont'd)
O balance (not counting O in the air):
n1 0 700 932 100 0 07
48
106
100 0 93
16
18
( . )( ) ( . )( ) ( . )( )
+ +
= + + + +
( )( ) ( ) ( . )( ) . ( )
n n m m m
w c
2 2 3 4 4
16 32 0 024
48
84
0 976
16
18
⇒ + = + + + +
22 4 8584 16 32 05714 0 024 08676 5
1 2 2 3 4 4
. . . ( . ) . ( )
n n n m m m
w c
Raoult's Law :
y P p C
n
n n n
n n n n
w w
o w
w c a
w w c a
= ⇒
+ +
=
⇒ = + +
*
( )
. ( ) ( )
70
01025 6
2
2 2 2
2 2 2 2
233.7 mm Hg
(3*760) mm Hg
Solve (1)-(6) simultaneously with E-Z solve (need a good set of starting values to
converge).
n n n
n m m
1 08086
= = =
= = =
. kmol, 0.2426 kmol air, 0.500 kmol CO ,
0.0848 kmol H O(v), 8.874 kg NaHCO (s), 92.50 kg solution
2a 2c 2
2w 2 3 3 4
NaHCO3 balance on filter:
m m m m
m
m
3 4 5 6
92 50
8.874
0 024 0 024 086 014 0 024
4
3
+ = + +
= +
=
=
. . [ . ( . )( . )]
)
.
11.09 0.024m 0.8634m (7
5 6
Mass Balance on filter: 8874 92 50 1014 8
5 6
. . . ( )
+ = = +
m m
Solve (7) (8) ⇒
=
=
⇒ =
m
m
5
6
3
91.09 kg filtrate
10.31 kg filter cake
(0.86)(10.31) 8.867 kg NaHCO (s)
Scale factor = = −
500 kg / h
8.867 kg
56.39 h 1
(a) Gas stream leaving reactor

n
n
n
2w 2
2c 2
2a
2
2
(0.0848)(56.39) 4.78 kmol H O(v) / h
(0.500)(56.39) 28.2 kmol O / h
(0.2426)(56.39) 13.7 kmol air / h
46.7kmol / h
0.102 kmol H O(v) / kmol
0.293 kmol Air / kmol
= =
= =
= =
U
V
|
W
|
⇒
R
S
|
|
T
|
|

V
n RT
P
2
2
= = ⋅ =
(46.7 kmol / h)(0.08206
m atm
kmol K
)(343 K)
3atm
438 m / h
3
3
(b) Gas feed rate:
V1 =
×
=
56.39 0.8086 kmol 22.4 m (STP) 1h
h kmol 60 min
17.0 SCMM
3

6-62
6.81(cont'd)
(c) Liquid feed: ( )( . )
100 56 39 5640
= kg / h
To calculate
V , we would need to know the density of a 7 wt% aqueous Na2CO3 solution.
(d) If T dropped in the filter, more solid NaHCO3 would be recovered and the residual
solution would contain less than 2.4% NaHCO3.
(e)
Benefit: Higher pressure greater higher concentration of CO in solution
higher rate of reaction smaller reactor needed to get the same conversion lower cost
Penalty Higher pressure greater cost of compressing the gas (purchase cost of compressor,
power consumption)
CO
Henry's law
2
2
⇒
⇒ ⇒ ⇒
⇒
p
:
6.82
a. Heating the solution dissolves all MgSO4; filtering removes I, and cooling recrystallizes
MgSO4 enabling subsequent recovery.
(b) Strategy: Do D.F analysis.
Dissolution
Tank
Dissolution
Tank

m
O
F
2 (lb soln / h)
0.32 kg MgSO / kg
0.68 kg H / kg
6000 lb I / h
110
m
4
2
m
o
R
S
|
T
|
U
V
|
W
|
( / )
( )
. /
. /
m lb MgSO H O h
m lb
lb lb
lb O lb
m
m
m m
m m
4 4 2
5
4
2
7
0 23
0 77
⋅
R
S
|
|
T
|
|
U
V
|
|
W
|
|
soln
MgSO
H
600 lb / h
0.90 MgSO4 7H
m
2
⋅ O
I
010
.
Filter I 6000
300
0 32
0 68
4
2
lb I h
lb h
MgSO
H O
m
m
/
/
.
.
soln
R
S
|
T
|
U
V
|
W
|
Crystallizer
( ln/ )
.
.
m lb so h
MgSO
H O
m
3
4
2
0 32
0 68
Filter II
( )
. ( )
.
m lb MgSO H O
m lb
O
m
m
4 4 2
4
7
0 05
0 77
⋅
R
S
|
T
|
U
V
|
W
|
soln
0.23 lb MgSO / lb
lb H / lb
m 4 m
m 2 m
( / )
.
m lb h
m
6
0 23 lb MgSO / lb
0.77 lb H O / lb
m 4 m
m 2 m
( / )
m lb H O h
m
1 2
Dissolution
Tank

6-63
6.82(cont'd)
Overall mass balance
Overall MgSO balance
4
U
V
W
⇒ ,
m m
1 4
Diss. tank overall mass balance
Diss. tank MgSO balance
4
U
V
W
⇒ ,
m m
2 6
( ) ( . . . ) . , ( ) ( . * . ) .
MW MW
MgSO MgSO 7H O
4 4 2
= + + = = + =
⋅
2431 3206 6400 12037 12037 7 1801 24644
Overall MgSO4 balance:
60,000 lb 0.90 lb MgSO 7H O 120.37 lb MgSO
h lb 246.44 lb MgSO 7H O
(300 lb / h)(0.32 lb MgSO / lb )
m m 4 2 m 4
m m 4 2
m m 4 m
⋅
⋅
= + +
( . / . ) . ( . )
m m
4 4
120 37 246 44 0 05 0 23
⇒ =
.
m x
4
4
5257 10 lb crystals / h
m
Overall mass balance: 60 000 6300 105 1494
1 4
5 10
1
4
4
, .
.257 /
+ = + =
=
m m m
m x h
lb
m 2
m
lb H O / h
c.
Diss. tank overall mass balance:
Diss. tank MgSO balance
lb / h
lb / h recycle
4
m
m
60 000 6000
54 000 120 37 246 44 0 23 0 32
1512 10
9 575 10
1 6 2
6 2
2
5
6
4
,
: , ( . / . ) . .
.
.
+ + = +
+ =
U
V
W
⇒
=
=
m m m
m m
m x
m x
Recycle/fresh feed ratio = =
9.575x10 lb / h
1494 lb / h
64 lb recycle / lb fresh feed
4
m
m
m m
6.83 a.
Solution composition:
X
X
X
a 4
a 2
a 3 2
(kg CaSO / kg)
500 (kg H O / kg)
(1 501 )(kg Ca(NO ) / kg)
−
R
S
|
T
|
U
V
|
W
|

n (kmol CO / h)
1 2

m (kg CaCO / h)
0 3 2m (kg solution / h)
0

m (kg CaSO / h)
m (kg Ca(NO ) / h)
m (kg H O/ h)
2 4
3 3 2
4 2
Filter cake
m (kg / h)
0.96 kg CaSO (s) / kg
0.04 kg soln / kg
5
4

1000 kg H SO / h (10 wt%)
1000 kg HNO / h
m (kg H O / h)
2 4
3
w 2

Cryst
Filter

m (kg CaCO / h)
2m (kg solution / h)
0 3
0

m (kg soln / h)
8

6-64
6.83 (cont’d)
b. Acid is corrosive to pipes and other equipment in waste water treatment plant.
c. Acid feed:
1000 kg H SO / h
(2000 ) kg / h
0.10 8000 kg H O / h
2 4
w
w 2
+
= ⇒ =

m
m
Overall S balance:
1000 kg H SO 32 kgS
h 98 kg H SO
(kg / h) (0.96 0.04 ) (kg CaSO ) 32 kgS
kg 136 kg CaSO
(kg / h) (kg CaSO ) 32 kgS
kg 136 kg CaSO
2 4
2 4
5 a 4
4
8 a 4
4
=
+
+
⇒ = + +

. . ( . . ) . ( )
m X
m X
m X m X
a a
3265 02353 096 004 02353 1
5 8
Overall N balance:
1000 kg HNO 14 kg N
h 63kg HNO
0.04 (kg / h) (1 501 ) (kg Ca(NO ) ) 28 kg N
kg 164 kg Ca(NO )
(kg / h) (1 501 ) (kg Ca(NO ) ) 28 kg N
kg 164 kg Ca(NO )
3
3
5 a 3 2
3 2
8 a 3 2
3 2
=
−
+
−
⇒ = − + −

. . ( ) . ( ) ( )
m X
m X
m X m X
a a
222 2 000683 1 501 0171 1 501 2
5 8
Overall Ca balance:

. . ( . . ) . ( )
. .
m (kg / h) 40 kg Ca
100 kg CaCO
m (kg / h) (0.96 0.04X ) (kg CaSO ) 40 kg Ca
kg 136 kg CaSO
(1 501X ) (kg Ca(NO ) ) 0.04m (kg / h) 40 kg Ca
kg 164 kg Ca(NO )
m (kg / h) X (kg CaSO ) 40 kg Ca
kg 136 kg CaSO
m (kg / h) (1 501X ) (kg Ca(NO ) ) 40 kg Ca
kg 164 kg Ca(NO )
0
3
5 a 4
4
a 3 2 5
3 2
8 a 4
4
8 a 3 2
3 2
=
+
+
−
+
+
−
⇒ = + + −
+ +
0 40 0 294 0 96 0 04 0 00976 1 501
0 294 0 244
0 5 5
8 8
m m X m X
m X m
a a
a ( ) ( )
1 501 3
− Xa
Overall C balance :

. ( )
m (kg / h) 12 kg C
100 kg CaCO
n (kmol CO / h) 1kmol C 12 kg C
1kmol CO 1kmol C
0
3
1 2
2
=
⇒ =
0 01 4
0 1
m n

6-65
6.83 (cont’d)
Overall H balance :
1000(kg H SO ) 2 kg H
h 98 kg H SO
1000 kg HNO 1kg H
h 63kg HNO
(kg / h) 2 kg H
18 kg H O
0.04 (kg / h) 500 (kg H O) 2 kg H
kg 18 kg H O
(kg / h) 500 (kg H O) 2 kg H
kg 18 kg H O
2 4
2 4
3
3
w
2
5 a 2
2
8 a 2
2
+ +
= +
⇒ = +

. . . ( )
m
m X m X
m X m X
a a
92517 222 5556 5
5 8
Solve eqns. (1)-(5) simultaneously, using E-Z Solve.

, .
m m m
n Xa
0 3 5 8
1 2 4
1812.5 kg CaCO (s) / h, 1428.1kg / h, 9584.9 kg soln / h,
18.1kmol CO / h(v) kg CaSO / kg
= = =
= = 0 00173
Recycle stream = =
2 3625
0
*
m kg soln / h
0.00173(kg CaSO / kg)
500*0.00173(kg H O / kg)
(1 501*0.00173)(kg Ca(NO ) / kg)
CaSO
H O
Ca(NO
4
2
3 2
4
2
3
−
R
S
|
T
|
U
V
|
W
|
⇒
R
S
|
|
T
|
|
U
V
|
|
W
|
|
0173%
865%
133% 2
.
.
. )
d. From Table B.1, for CO2:
T K P atm
T
T
T
K
P
atm
atm
c c
r
c
r
= =
⇒ = =
+
= = =
304 2 72 9
40 2732
304 2
103
30
72 9
0 411
. , .
( . )
.
. ,
.
.
From generalized compressibility chart (Fig. 5.4-2):
z V
zRT
P
= ⇒ = =
⋅
⋅
=
086
. 0.86 0.08206 L atm 313.2 K
mol K 30 atm
0.737
L
mol CO2
Volumetric flow rate of CO2:
*
V n V
= = =
1
18.1 kmol CO 0.737 L 1000 mol
h mol CO 1 kmol
1.33x10 L / h
2
2
4
e. Solution saturated with Ca(NO3)2:
⇒
−
= ⇒ =
1 501X (kg Ca(NO ) / kg)
500Xa (kg H O / kg)
1.526 X 0.00079 kg CaSO / kg
a 3 2
2
a 4
Let
m1 (kg HNO3/h) = feed rate of nitric acid corresponding to saturation without
crystallization.

6-66
6.83 (cont’d)
Overall S balance:
1000kg H SO 32 kgS
h 98kg H SO
(kg / h) (0.96 (0.04)(0.00079)) (kgCaSO ) 32 kgS
kg 136kgCaSO
(kg / h) 0.00079(kgCaSO ) 32 kgS
kg 136kgCaSO
326.5 0.226 0.000186 (1')
2 4
2 4
5 4
4
8 4
4
5 8
=
+
+
⇒ = +

m
m
m m
Overall N balance:

. . . ( ')
m m
m
m m m
1 3
3
5 3 2
3 2
8 3 2
3 2
(kgHNO ) 14kgN
h 63kgHNO
0.04 (kg/ h) (1 (501)(0.00079)) (kgCa(NO ) ) 28kgN
kg 164kgCa(NO )
(kg/ h) (1 (501)(0.00079))(kgCa(NO ) ) 28kgN
kg 164kgCa(NO )
=
−
+
−
⇒ = +
0222 000413 0103 2
1 5 8
Overall H balance:
1000(kg H SO ) 2 kg H
h 98 kg H SO
kg HNO 1kg H
h 63kg HNO
8000(kg / h) 2 kg H
18 kg H O
0.04 (kg / h) 500(0.00079) (kg H O) 2 kg H
kg 18 kg H O
(kg / h) 500(0.00079) (kg H O) 2 kg H
kg 18 kg H O
2 4
2 4
1 3
3
2
5 2
2
8 2
2
+
+ =
+
⇒ + = +

. . . . ( ')
m
m
m
m m m
909 30 00159 000175 00439 3
1 5 8
Solve eqns (1')-(3') simultaneously using E-Z solve:
. ; . .
m x x x
1
4 3 4
1155 10 1424 10 2 484 10
= = =
kg / h m kg / h; m kg / h
5 8
Maximum ratio of nitric acid to sulfuric acid in the feed
= =
1155 10
1000
115
4
.
. /
x kg / h
kg / h
kg HNO kg H SO
3 2 4

6-67
6.84
Moles of diphenyl (DP):
Moles of benzene (B):
g
154.2 g mol
mol
550.0 ml 0.879 g 1 mol
ml 78.11 g
mol
mol DP mol
DP
56 0
0 363
619
0 363
619 0 363
0 0544
.
.
.
.
. .
.
=
=
U
V
|
|
W
|
|
⇒ =
+
=
x
p T x p T
DP
B
*
B
*
mm Hg mm Hg
b g b g b g
= − = =
( ) . . .
1 0 945 120 67 114 0
Δ
Δ
T
RT
H
x C T
m
m0
m
DP
o
m
K = 3.6 C
= =
+
= ⇒ = − = °
2 2
8 314 2732 55
9837
0 0554 36 55 36 19

. . .
. . . . .
b g b g
Δ
Δ
T
RT
H
x
T
bp
b0
v
DP
o
b
K =1.85 C
C
= =
+
=
⇒ = + = °
2 2
8 314 2732 801
30 765
0 0554 185
801 185 82.0

. . .
,
. .
. .
b g b g
6.85
o o
0
Eq. 6.5-5
Table B.1 2 2
0
2 2
0
0.0 C, 4.6 C=4.6K
ˆ (4.6K)(600.95 J/mol)
0.0445 mol urea/mol
( ) (8.314 J/mol K)(273.2K)
(8.314)(373.2)
Eq. (6.5-4) 0.0445 1.3K
ˆ 40,656
m m
m m
u
m
b
b u
v
T T
T H
x
R T
RT
T x
H
= Δ =
Δ Δ
= = =
⋅
⇒ Δ = = = =
Δ
⎯⎯⎯
→
o
1.3 C
1000 grams of this solution contains mu (g urea) and (1000 – mu) (g water)
1 1
1 1
1
1 1 1
1 1
(g) (1000 )(g)
(mol urea) (mol water)
60.06 g/mol 18.02 g/mol
(mol urea)
60.06
0.0445 134 g urea, 866 g water
(1000 )
(mol solution)
60.06 18.02
u u
u w
u
u u w
u u
m m
n n
m
x m m
m m
−
= =
= = ⇒ = =
−
+
⎡ ⎤
⎢ ⎥
⎣ ⎦
o
2 2 2
0
2
2 2
2
ˆ (3.0K)(40,656 J/mol)
3.0 C 3.0K 0.105 mol urea/mol
( ) (8.314 J/mol K)(373.2K)
(mol urea)
60.06
0.105 339 g urea
866
(mol solution)
60.06 18.02
Add (339-134) g
b v
b u
b
u
u u
u
T H
T x
R T
m
x m
m
Δ Δ
Δ = = ⇒ = = =
⋅
= = ⇒ =
+
⇒
⎡ ⎤
⎢ ⎥
⎣ ⎦
urea = 205 g urea

6-68
6.86 xa
I
0.5150 g g mol
0.5150 g g mol 100.0 g g mol
mol solute mol
=
+
=
b g b g
b g b g b g b g
1101
1101 9410
0 00438
.
. .
.
Δ
Δ
Δ
Δ
Δ
Δ
T
RT
H
x
T
T
x
x
x x
T
T
m
m
m
s
m
m
s
s
s s
m
m
= ⇒ = ⇒ = =
°
°
=
0
2
0 00438
0 49
0 00523

.
.
.
I
II
I
II
II I
II
I
C
0.41 C
mol solute
mol solution
⇒
1 mol solvent 94.10 g solvent 0.4460 g solute
0.00523 mol solute 1 mol solvent 95.60 g solvent
g solute mol
−
=
0 00523
8350
.
.
b g
Δ
Δ
. . .
.
. .
H
RT
T
x
m
m
m
s
= =
−
= =
0
2 2
8 314 2732 500
0 49
0 00523 6380 6 38
b g b g J mol kJ / mol
6.87 a. ln *
p T
H
RT
B
s b
v
b
0
0
b g= − +
Δ I
, ln *
p T
H
RT
B
s bs
v
bs
b g= − +
Δ II
Assume Δ Δ
H H
v v
I II
≅ ; T T T
s
0 0
2
≅
⇒ ln ln
* *
P T P T
H
R T T
H
R
T T
T
s b bs
v
b bs
v bs b
b
0 0
0
0
0
2
1 1
b g b g
− = − −
F
HG I
KJ ≅
−
Δ Δ
b. Raoult’s Law: p T x p T
s b bs
* *
0 0
1
b g b g b g
= − ⇒ ln 1
0
2
0
2
− ≈ − = − ⇒ =
x x
H T
RT
T
RT
H
x
v b
b
b
b
v
b g Δ Δ
Δ
Δ
6.88
90 g ethylbenzene
m1 (g styrene)
100 g EG
30 g styrene
90 g ethylbenezene
m2 (g styrene)
100 g EG
Styrene balance: m m
1 2
+ = 30 g styrene
Equilibrium relation:
m
m
m
m
2
2
1
1
100
019
90
+
=
+
F
HG I
KJ
.
solve simultaneously
m
m
1
2
256
4 4
=
=
.
.
g styrene in ethylbenzene phase
g styrene in ethylene glycol phase

6-69
6.89 Basis: 100 kg/h. A=oleic acid; C=condensed oil; P=propane
a. 90% extraction: ( . )( . )(
m3 0 09 0 05 100
= kg / h) = 4.5 kg A / h
Balance on oleic acid: ( . )( ) . .
0 05 100 4 5 05
2 2
= + ⇒ =
m m
kg A / h kg A / h
Equilibrium condition: 015
05 05
4 5 4 5 95
732
1
1
.
. / ( . )
. / ( . )
.
=
+
+
⇒ =
n
n kg P / h
b. Operating pressure must be above the vapor pressure of propane at T=85o
C=185o
F
Figure 6.1-4 ⇒ ppropane
*
psi 34 atm
= =
500
c. Other less volatile hydrocarbons cost more and/or impose greater health or environmental
hazards.
6.90 a. Benzene is the solvent of choice. It holds a greater amount of acetic acid for a given mass
fraction of acetic acid in water.
Basis: 100 kg feed. A=Acetic acid, W=H2O, H=Hexane, B=Benzene
Balance on W: 100 0 70 0 90 77 8
1 1
* . * . .
= ⇒ =
m m kg
Balance on A: 100 0 30 77 8 010 22 2
2 2
* . . * . .
= + ⇒ =
m m kg
Equilibrium for H:
K
m m m
x
m
m x
H
H
A
H
H
=
+
=
+
= ⇒ =
2 2 4
22 2 22 2
010
0 017 130 10
/ ( ) . / ( . )
.
. . kg H
Equilibrium for B:
K
m m m
x
m
m x
B
B
A
B
B
=
+
=
+
= ⇒ =
2 2 3
22 2 22 2
010
0 098 2 20 10
/ ( ) . / ( . )
.
. . kg B
(b) Other factors in picking solvent include cost, solvent volatility, and health, safety, and
environmental considerations.
95.0 kg C / h
m kg A / h
2

100 kg / h
0.05 kg A / kg
0.95 kg C / kg
/
m kg P h
1

m kg A / h
m kg P / h
3
1
m (kg A)
m (kg H) or m (kg B)
2
H B
100 (kg)
0.30 kg A / kg
0.70 kg W / kg
m (kg H)
or m (kg B)
H
B
m (kg)
0.10 kg A / kg
0.90 kg W / kg
1

6-70
6.91 a. Basis: 100 g feed 40 g acetone, 60 g H O.
2
⇒ A = acetone, H = n - C H
6 14 , W = water
40 g A
60 g W
100 g H
25°C
100 g H
r1 (g A)
60 g W
75 g H
25°C
75 g H
r2 (g A)
e1 (g A)
60 g W
e2 (g A)
x x
A in H phase A in W phase
/ .
= 0343 x mass fraction
=
b g
Balance on A stage 1:
Equilibrium condition stage 1:
g acetone
g acetone
− = +
−
+
+
=
U
V
|
W
|
⇒
=
=
40
100
60
0 343
27 8
12 2
1 1
1 1
1 1
1
1
e r
r r
e e
e
r
b g
b g .
.
.
g acetone
g acetone
− = +
−
+
+
=
U
V
|
W
|
⇒
=
=
27 8
75
60
0 343
7 2
20 6
2 2
2 2
2 2
2
2
.
.
.
.
e r
r r
e e
r
e
b g
b g
% acetone not extracted = × =
20 6
100% 515%
.
.
g A remaining
40 g A fed
b.
g acetone
g acetone
− = +
−
+
+
=
U
V
|
W
|
⇒
=
=
40 0
175
60
0 343
17 8
22 2
1 1
1 1
1 1
1
1
.
.
.
.
e r
r r
e e
r
e
b g
b g
% acetone not extracted = × =
22 2
100% 555%
.
.
g A remaining
40 g A fed
Equilibrium condition:
20 6 20 6
19 4 60 19 4
0 343 225
. / ( . )
. / ( . )
.
m
m
+
+
= ⇒ = g hexane
d. Define a function F=(value of recovered acetone over process lifetime)-(cost of hexane
over process lifetime) – (cost of an equilibrium stage x number of stages). The most cost-
effective process is the one for which F is the highest.
40 g A
60 g W
m (g H)
19.4 g A
60 g W
20.6 g A
m (g H)
40 g A
60 g W
175 g H
e1 g A
60 g W
r1 g A
175 g H
c.

6-71
6.92 a. P--penicillin; Ac--acid solution; BA--butyl acetate; Alk--alkaline solution
b. In Unit I, 90% transfer ⇒ m P
3 0 90 15 135
= =
. ( . ) . kg P
P balance: 15 135 015
2 2
. . .
= + ⇒ =
m m
P P kg P
pH=2.1⇒ = =
+
+
⇒ =
K
m
m
250
135 135
015 015 985
3416
1
1
.
. / ( . )
. / ( . . )
. kg BA
In Unit II, 90% transfer: m m kg P
P P
5 3
0 90 1215
= =
. ( ) .
P balance: m m m kg P
P P P
3 6 6
1215 0135
= + ⇒ =
. .
pH=5.8⇒ = =
+
+
⇒ =
K
m m
m
m
P P
010
3416
1215 1215
29 65
6 6
4
4
.
/ ( . )
. / ( . )
. kg Alk
m
m
1
4
100
100
= =
= =
34.16 kg BA
100 kg broth
0.3416 kg butyl acetate / kg acidified broth
29.65 kg Alk
100 kg broth
0.2965 kg alkaline solution / kg acidified broth
Mass fraction of P in the product solution:
x
m
m m
P
P
P
=
+
= =
5
4 5
1215
0 394
.
.
P
(29.65 +1.215) kg
kg P / kg
c. (i). The first transfer (low pH) separates most of the P from the other broth constituents,
which are not soluble in butyl acetate. The second transfer (high pH) moves the
penicillin back into an aqueous phase without the broth impurities.
(ii). Low pH favors transfer to the organic phase, and high pH favors transfer back to the
aqueous phase.
(iii).The penicillin always moves from the raffinate solvent to the extract solvent.
100 kg
0.015 P
0.985 Ac
m4 (kg Alk)
m5P (kg P)
m4 (kg Alk)
pH=5.8
m1 (kg BA)
m3P (kg P)
98.5 (kg Ac)
pH=2.1
m6P (kg P)
m1 (kg BA)
Mixing tank
Broth
Acid
Extraction Unit II (consider m1, m3p)
3 unknowns
–1 balance (P)
–1 distribution coefficient
–1 (90% transfer)
0 DF
D.F. analysis:
Extraction Unit I
3 unknown (m1, m2p, m3p)
–1 balance (P)
–1 distribution coefficient
–1 (90% transfer)
0 DF
Extraction Unit I
Extraction
II

6-72
6.93 W = water, A = acetone, M = methyl isobutyl ketone
x
x
x
x x x
x x x
Figure
W
A
M
6.6-1
W A M
W A M
Phase 1:
Phase 2:
=
=
=
U
V
|
W
|
= = =
= = =
⇒
0 20
0 33
0 47
0 07 0 35 058
0 71 0 25 0 04
.
.
.
. , . , .
. , . , .
Basis: 1.2 kg of original mixture, m1=total mass in phase 1, m2=total mass in phase 2.
H O Balance:
Acetone balance:
kg in MIBK - rich phase
kg in water - rich phase
2 12 0 20 0 07 0 71
12 0 33 0 35 0 25
0 95
0 24
1 2
1 2
1
2
. * . . .
. * . . .
.
.
= +
= +
⇒
=
=
R
S
|
T
|
m m
m m
m
m
6.94 Basis: Given feeds: A = acetone, W = H2O, M=MIBK
Overall system composition:
5000 g 30 wt% A, 70 wt% W 1500 g A, 3500 g W
3500 g 20 wt% A, 80 wt% M 700 g A, 2800 g M
2200 g A
3500 g W
2800 g M
25.9% A, 41.2% W, 32.9% M
Phase 1: 31% A, 63% M, 6% W
Phase 2: 21% A, 3% M, 76% W
Fig. 6.6-1
b g
b g
⇒
⇒
U
V
|
W
|
⇒
U
V
|
W
|
⇒
Let m1=total mass in phase 1, m2=total mass in phase 2.
H O Balance:
Acetone balance:
g in MIBK - rich phase
g in water - rich phase
2 3500 0 06 0 76
2200 0 31 0 21
4200
4270
1 2
1 2
1
2
= +
= +
⇒
=
=
R
S
|
T
|
. .
. .
m m
m m
m
m
6.95 A=acetone, W = H2O, M=MIBK
Figure 6.6-1⇒ Phase 1: x x x
M w A
= ⇒ = =
0 700 0 05 0 25
1 1
. . ; .
, , ;
Phase 2: x x x
w A M
, , ,
. ; . ; .
2 2 2
081 081 0 03
= = =
Overall mass balance: lb / h lb h
MIBK balance
lb MIBK / h
19.1lb h
m m
m
m
32 0 410
410 0 7 0 03
281
1 2
1 2
1
2
. .
: . * . * .
.

+ = +
= +
U
V
W
⇒
=
=
m m
m m
m
m

m h
x
m
A W M
2
2 2 2
lb /
, x , x
, , ,

m1 (lb M / h)
m
32 lb / h
x (lb A / lb
x (lb W / lb
m
AF m m
WF m m
)
)
41.0 lb / h
x , x 0.70
m
A,1 W,1,

6-73
6.96 a. Basis: 100 kg; A=acetone, W=water, M=MIBK
System 1: = 0.375 mol A, = 0.550 mol M, x = 0.075 mol W
= 0.275 mol A, = 0.050 mol M, = 0.675 mol W
a,org m,org w,org
a,aq m,aq w,aq
x x
x x x
Mass balance:
Acetone balance
kg
kg
aq,1
m m
m m
m
m
aq org
aq org org
, ,
, , ,
: * . * . .
.
.
1 1
1 1 1
100
0275 0375 3333
417
583
+ =
+ =
U
V
W
⇒
=
=
System 2: = 0.100 mol A, = 0.870 mol M, x = 0.030 mol W
= 0.055 mol A, = 0.020 mol M, = 0.925 mol W
a,org m,org w,org
a,aq m,aq w,aq
x x
x x x
Mass balance
Acetone balance
m kg
kg
aq,2
org,2
:
: * . * .
.
.
, ,
, ,
m m
m m m
aq org
aq org
2 2
2 2
100
0 055 0100 9
22 2
77 8
+ =
+ =
U
V
W
⇒
=
=
b. K
x
x
K
x
x
a
a org
a aq
a
a org
a aq
,
, ,
, ,
,
, ,
, ,
.
.
. ;
.
.
.
1
1
1
2
2
2
0 375
0 275
136
0100
0 055
182
= = = = = =
High Ka to extract acetone from water into MIBK; low Ka to extract acetone from MIBK
into water.
c. β β
aw
x x
x x aw
a org w org
a aq w aq
,
/
/
. / .
. / .
. ;
,
. / .
. / .
.
, ,
, ,
1
0 375 0 075
0 275 0 675
12 3
2
0100 0 040
0 055 0 920
418
= = = = =
If water and MIBK were immiscible, x
aw
w org
, = ⇒ → ∞
0 β
d. Organic phase= extract phase; aqueous phase= raffinate phase
βa w
a w org
a w aq
a org a aq
w org w aq
a
w
x x
x x
x x
x x
K
K
,
( / )
( / )
( ) / ( )
( ) / ( )
= = =
When it is critically important for the raffinate to be as pure (acetone-free) as possible.
6.97 Basis: Given feed rates: A = acetone, W = water, M=MIBK

r2 (kg / h)
y (kg A / kg)
y (kg W / kg)
y (kg M / kg)
2A
2W
2M

r1 (kg / h)
y (kg A / kg)
y (kg W / kg)
y (kg M / kg)
1A
1W
1M
Stage
IIS
300 kg W / h

e1 (kg / h)
x (kg A / kg)
x (kg W / kg)
x (kg M / kg)
1A
1W
1M

e2 (kg / h)
x (kg A / kg)
x (kg W / kg)
x (kg M / kg)
2A
2W
2M
200 kg / h
0.30 kg A / kg
0.70 kg M / kg
300 kg W / h
Stage I Stage II

6-74
6.97(cont'd)
Overall composition of feed to Stage 1:
200 0 30 60
200 60 140
300
12%
b gb g
. =
− =
U
V
|
W
|
⇒
kg A h
kg M h
kg W h
500 kg h
A, 28% M, 60% W
Figure 6.6-1 ⇒
= = =
= = =
Extract:
Raffinate:
A W
A W
x x x
y y y
1 1 1M
1 1 1M
0 095 0880 0 025
015 0 035 0815
. , . , .
. , . , .
Mass balance
Acetone balance:
kg / h
kg / h
500
60 0 095 015
273
227
1 1
1 1
1
1
= +
= +
⇒
=
=
R
S
|
T
|

. .

e r
e r
e
r
Overall composition of feed to Stage 2:
227 015 34
227 0815 185
227 0 035 300 308
65%
b gb g
b gb g
b gb g
.
.
.
.
=
=
+ =
U
V
|
W
|
⇒
kg A h
kg M h
kg W h
527 kg h
A, 35.1% MIBK, 58.4% W
Figure 6.6-1 ⇒
= = =
= = =
Extract:
Raffinate:
A W M
A W M
x x x
y y y
2 2 2
2 2 2
0 04 0 94 0 02
0 085 0 025 089
. , . , .
. , . , .
Mass balance:
Acetone balance:
kg / h
kg / h
527
34 0 04 0 085
240
287
2 2
2 2
2
2
= +
= +
⇒
=
=
R
S
|
T
|

. .

e r
e r
e
r
Acetone removed:
[ ( . )( )]
.
60 0 085 287
059
−
=
kg A removed / h
60 kg A / h in feed
kg acetone removed / kg fed
Combined extract:
Overall flow rate = kg / h
Acetone:
kg A
kg A / kg
Water
kg W
kg W / kg
MIBK
kg M
kg
kg M / kg

( ) . * . *
.
:
( )

. * . *
.
:
( )
( )
. * . *
.
e e
x e x e
x e x e
e e
x e x e
e e
A A
w w
M M
1 2
1 1 2 2
1 1 2 2
1 2
1 1 2 2
1 2
273 240 513
0 095 273 0 04 240
513
0 069
088 273 0 94 240
513
0 908
0 025 273 0 02 240
513
0 023
+ = + =
+
=
+
=
+
+
=
+
=
+
+
=
+
=

6-75
6.98. a.

n
PV
RT
0 = =
⋅ ⋅
=
(1atm)(1.50 L / min)
(0.08206 L atm / mol K)(298 K)
0.06134 mol / min
r.h.=25%⇒
p
p
H O
H O
* o
2
2
C)
(
.
25
025
=
Silica gel saturation condition: X
p
p
*
*
. . * . .
= = =
12 5 12 5 0 25 3125
H O
H O
2
2
2
g H O ads
100 g silica gel
Water feed rate: y
p C
p
o
0
0 25 25 0 25 23756
760
0 00781
= = =
. ( ) . ( . )
.
*
H O 2
2
mm Hg
mm Hg
mol H O
mol
⇒
mH2O
0.06134 mol 0.00781mol H O 18.01g H O
min mol mol H O
0.00863 g H O / min
2 2
2
2
= =
Adsorption in 2 hours = =
( .
0 00863 g H O / min)(120min) 1.035 g H O
2 2
Saturation condition:
1.035g H O
(g silica gel)
3.125g H O
100 g silica gel
33.1g silica gel
2 2
M
M
= ⇒ =
Assume that all entering water vapor is adsorbed throughout the 2 hours and that P and
T are constant.
b. Humid air is dehumidified by being passed through a column of silica gel, which absorbs a
significant fraction of the water in the entering air and relatively little oxygen and nitrogen.
The capacity of the gel to absorb water, while large, is not infinite, and eventually the gel
reaches its capacity. If air were still fed to the column past this point, no further
dehumidification would take place. To keep this situation from occurring, the gel is
replaced at or (preferably) before the time when it becomes saturated.
6.99 a. Let c = CCl4
Relative saturation
C
(169 mm Hg) 50.7 mm Hg
c
c
* o
= ⇒ ⇒ = =
0 30
34
0 30
.
( )
. *
p
p
pc
b. Initial moles of gas in tank:
n
P V
RT
0
0 0
0
1
1985
= =
⋅ ⋅
=
atm 50.0 L
0.08206 L atm / mol K 307 K
mol
.
Initial moles of CCl4 in tank:
n y n
p
P
n
c c
c
0 0 0
0
0
0
50 7
1985 01324
= = = × =
.
. .
mm Hg
760 mm Hg
mol mol CCl4
1.50 L/ min
25 C, 1atm, rh=25%
(mol / min)
(mol H O/ mol)
(1- ) (mol dryair / mol)
o
0
0 2
0

n
y
y
M (g gel)
Ma (g H2O)

6-76
6.99 (cont’d)
50% CCl4 adsorbed ⇒ n nc
c 4
mol CCl
= =
0500 0662
0
. (= nads)
Total moles in tank: n n n
tot ads mol =1.919 mol
= − = −
0 1985 0 0662
( . . )
Pressure in tank. Assume T = T0 and V = V0.
P
n RT
V
= =
F
HG I
KJ F
HG I
KJ =
tot
atm
760 mm Hg
atm
735 mm Hg
0
0
1919 0 08206 307
50 0
( . )( . )( )
.
y
n
n
p
C
c
= = =
⇒ =
c
tot
4 4
mol CCl
mol
mol CCl
mol
mm Hg) = 26.2 mm Hg
0 0662
1919
0 0345
0 0345 760
.
.
.
. (
c. Moles of air in tank mol air = 1.853 mol air
: ( . . )
n n n
a c
= − = −
0 0 1985 01324
y
n
n
n
n n n
p y P
n RT
V
c
c
c
c
c
c c
tot
=
+
= ⇒ = ×
⇒ = + =
= =
L
NM O
QP × ⋅
⋅
=
−
−
1853
0 001 1854 10
1854
0 001
1854 10 0 08206
50 0 1
3
0
0
3
.
. .
.
.
. .
.
mol CCl
mol
mol CCl
mol
=
mol L atm 307 K 760 mm
L mol K atm
0.710 mm Hg
4
4
tot air
X
p
p
X
c
* *
.
.
. ( . )
. ( . )
.
g CCl
g carbon
g CCl adsorbed
g carbon
4 c 4
F
HG I
KJ =
+
⇒ =
+
=
0 0762
1 0 096
0 0762 0 710
1 0 096 0 710
0 0506
Mass of CCl4 adsorbed
m n n MW
c c c
ads
4
4
4
mol CCl 153.85 g
1 mol CCl
20.3 mol CCl adsorbed
= − =
−
=
( )( )
( . . )
0
01324 0 001854
Mass of carbon required: mc = =
20.3g CCl ads
0.0506
gCCl ads
gcarbon
400 gcarbon
4
4
6.100 a. X K p X K p
F NO F NO
* *
ln ln ln
= ⇒ = +
2 2
β β
β
y = 1.406x - 1.965
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 1 2 3
ln(X*)
ln(PNO2)

6-77
6.100 (cont’d)
ln . ln . .
* * .965 .406 .406
X p X e p p
NO NO NO
= − ⇒ = =
−
1406 1965 0140
2 2 2
1 1 1
KF = =
−
0.140 (kg NO / 100kg gel)(mmHg)
2
1.406
; .
β 1406
b. Mass of silica gel : mg = =
π *(0.05m) (1m) 10 L 0.75kggel
1m L
5.89 kggel
2 3
3
Maximum NO2 adsorbed :
p
mads
NO
1.406
2
2
2
in feed 0.010(760 mmHg) 7.60 mmHg
0.140(7.60) kg NO 5.89 kggel
100 kggel
0.143 kg NO
= =
= =
Average molecular weight of feed :
MW MW MW
NO air
= + = + =
0 01 0 99 0 01 46 01 0 99 29 0 2917
2
. ( ) . ( ) ( . )( . ) ( . )( . ) .
kg
kmol
Mass feed rate of NO2:

m = =
8.00 kg 1kmol 0.01kmol NO 46.01kg NO
h 29.17 kg kmol kmol NO
0.126
kg NO
h
2 2
2
2
Breakthrough time: tb = = =
0.143 kg NO
0.126 kg NO / h
1.13 h 68 min
2
2
c. The first column would start at time 0 and finish at 1.13 h, and would not be available for
another run until (1.13+1.50) = 2.63 h. The second column could start at 1.13 h and finish
at 2.26 h. Since the first column would still be in the regeneration stage, a third column
would be needed to start at 2.26 h. It would run until 3.39 h, at which time the first
column would be available for another run. The first few cycles are shown below on a
Gantt chart.
Run Regenerate
Column 1
0 1.13 2.63 3.39 4.52 6.02
Column 2
1.13 2.26 3.76 4.52 5.65
Column 3
2.26 3.39 4.89 5.65 6.78

6-78
6.101 Let S=sucrose, I=trace impurities, A=activated carbon
Assume no sucrose is adsorbed
solution volume (V) is not affected by addition of the carbon
•
•
a. R(color units/kg S) = kCi (kg I / L) = k
m
V
I
(1)
⇒ − = − =
= −
= (
Δ Δ
R k C C
k
V
m m R
km
V
i i I I
mIA mI mI
IA
0 0
0
) ( ) (2)
% removal of color = = =
ΔR
R
x
km V
km V
x
m
m
IA
I
IA
I
0 0 0
100% 100 100
/
/
(3)
Equilibrium adsorption ratio: X
m
m
i
IA
A
*
= (4)
Normalized percentage color removal:
υ = =
%
/
/
/
(
removal
=
m
m m
m
m m
m
m
m
m
A S
IA I
A S
IA
A
S
I
3)
0
0
100
100
⇒ ⇒ =
=100Xi
*
υ υ
m
m
X
m
m
S
I
i
I
S
0
0
100
*
(5)
Freundlich isotherm X K C
m
m
K
R
k
i F i
I
S
F
*
( ),(
( )
= =
β β
υ
1 5)
0
100
⇒ =
=
100
υ β
β β
m K
m k
R K R
S F
I
F
0
'
A plot of lnυ vs. ln R should be linear: slope ; intercept = lnKF
'
= β
Add mA (kg A)
Come to equilibrium
m
m
R
V
S
I0
0
(kg S)
(kg I)
(color units / kg S)
(L)
m (kg S)
m (kg I)
R (color units / kg S)
V (L)
S
I
mA (kg A)
m (kg I adsorbed)
IA
y = 0.4504x + 8.0718
8.000
8.500
9.000
9.500
0.000 1.000 2.000 3.000
ln R
ln
v

6-79
ln . ln .
υ υ
= + ⇒ = =
0 4504 8 0718 3203
2
8.0718 0.4504 0.4504
p e R R
NO
⇒ KF
'
, .
= =
3203 0 4504
β
b. 100 kg 48% sucrose solution ⇒ =
m kg
S 480
95% reduction in color ⇒ R = 0.025(20.0) = 0.50 color units / kg sucrose
υ β
= = =
⇒ = ⇒ =
K R
m m m
F
A S A
'
( . )
/
.
/
.
3203 050 2344
97 5
480
20 0
0.4504
2344 =
% color reduction
m kg carbon
A
6.101 (cont’d)

7-1
CHAPTER SEVEN
7.1
080 35 10 0 30
2 33 2 3
4
. .
. .
L kJ . kJ work 1 h 1 kW
h L 1 kJ heat 3600 s 1 k J s
kW kW
×
= ⇒
2
1
312 3
3
.33 kW 10 W 1.341 10 hp
kW 1 W
hp .1 hp
3
×
= ⇒
−
.
7.2 All kinetic energy dissipated by friction
(a) E
mu
k =
=
×
⋅ ⋅
=
−
2
2 2 2 2 4
2 2
2
5500 lb 5280 1 9 486 10
715
m
2 2
f
2 2 2
m
2
f
55 miles ft 1 h lb Btu
2 h 1 mile 3600 s 32.174 lb ft / s 0.7376 ft lb
Btu
.
(b)
8
4 6
3 10 brakings 715 Btu 1 day 1 h 1 W 1 MW
2617 MW
day braking 24 h 3600 s 9.486 10 Btu/s 10 W
3000 MW
−
×
=
×
⇒
7.3 (a) Emissions:
Paper
1000 sacks oz 1 lb
sack 16 oz
lb
m
m
⇒
+
=
( . . )
.
0 0510 0 0516
6 41
Plastic
2000 sacks oz 1 lb
sack 16 oz
lb
m
m
⇒
+
=
( . . )
.
0 0045 0 0146
2 39
Energy:
Paper
1000 sacks Btu
sack
Btu
⇒
+
= ×
( )
.
724 905
163 106
Plastic
2000 sacks Btu
sack
Btu
⇒
+
= ×
( )
.
185 464
130 106
(b) For paper (double for plastic)
Raw Materials
Acquisition and
Production
Sack
Production and
Use
Disposal
Materials
for 400 sacks
1000 sacks 400 sacks

7-2
7.3 (cont’d)
Emissions:
Paper
400 sacks .0510 oz 1 lb
sack 16 oz
1000 sacks .0516 oz 1 lb
sack 16 oz
lb
reduction
m m
m
⇒ + =
⇒
0 0
4 5
30%
.
Plastic
800 sacks .0045 oz 1 lb
sack 16 oz
2000 sacks .0146 oz 1 lb
sack 16 oz
lb
reduction
m m
m
⇒ + =
⇒
0 0
2 05
14%
.
Energy:
Paper
400 sacks Btu
sack
1000 sacks Btu
sack
Btu; 27% reduction
⇒ + = ×
724 905
119 106
.
Plastic
800 sacks Btu
sack
2000 sacks Btu
sack
Btu; 17% reduction
⇒ + = ×
185 464
108 106
.
(c) .
3 10 persons 1 sack 1 day 1 h 649 Btu 1 J 1 MW
person - day h 3600 s 1 sack 9.486 10 Btu J / s
MW
8
-4
×
×
=
24 10
2 375
6
,
Savings for recycling: 017 2 375
. ( , MW) = 404 MW
(d) Cost, toxicity, biodegradability, depletion of nonrenewable resources.
7.4 (a) Mass flow rate:
gal 1 ft (0.792)(62.43) lb 1 min
min 7.4805 gal 1 ft 60 s
lb s
3
m
3 m

.
.
m = =
300
0 330
Stream velocity:
gal 1728 in 1 1 ft min
min 7.4805 gal 0.5 in 12 in 60 s
ft s
3
2
u = =
300 1
1225
2
.
.
Πb g
Kinetic energy:
.330 lb ft 1 1 lb
s s 32.174 lb ft / s
ft lb
s
ft lb s
hp
ft lb s
hp
m
2
f
2
m
2
f
f
f
E
mu
k = =
⋅
= ×
⋅
= × ⋅
×
⋅
F
HG
I
KJ= ×
−
−
−
−
2 2
3
3
3
5
2
0 1225
2
770 10
770 10
1341 10
07376
140 10
.
.
. /
.
. /
.
b g
d i
(b) Heat losses in electrical circuits, friction in pump bearings.

7-3
7.5 (a) Mass flow rate:
.
m = =
42.0 m 0.07 m 10 L 273 K 130 kPa 1 mol 29 g
s 4 1 m K 101.3 kPa 22.4 L STP mol
g s
3
3
πb g
b g
2
573
127 9
.
E
mu
k = =
⋅ ⋅
=
2 2
2
1 42 0
113
127.9 g kg m 1 N 1 J
2 s 1000 g s 1 kg m / s N m
J s
2
2 2
(b)
( ) 3
3 2 2
127.9 g 1 mol 673 K 101.3 kPa 22.4 L STP 1 m 4
49.32 m s
s 29 g 273 K 130 kPa 1 mol 10 L (0.07) m
π
=
.
.
) )
E
mu
E E E
k
k k k
= =
⋅ ⋅
=
⇒
2 2
2
1 49 32
1558
127.9 g kg m 1 N 1 J
2 s 1000 g s 1 kg m / s N m
J / s
= (400 C - (300 C = (155.8 -113) J / s = 42.8 J / s 43 J / s
2
2 2
Δ D D
(c) Some of the heat added goes to raise T (and hence U) of the air
7.6 (a)
Δ Δ
E mg z
p = =
−
⋅
= − ⋅
1 gal 1 ft 62.43 lb ft ft lb
7.4805 gal 1 ft s 32.174 lb ft / s
ft lb
3
m f
3 2 m 2 f
32174 10 1
834
.
.
(b) E E
mu
mg z u g z
k p
= − ⇒ = − ⇒ = − =
F
HG I
KJ
L
NM O
QP =
Δ Δ Δ
2
1 2
1 2
2
2 2 32174 10 254
b g b g b g
. .
ft
s
ft
ft
s
2
(c) False
7.7 (a) Δ
E positive
k ⇒ When the pressure decreases, the volumetric flow rate increases, and
hence the velocity increases.
Δ
E negative
p ⇒ The gas exits at a level below the entrance level.
(b)
.
.
m =
=
5 m 1.5 cm 1 m 273 K 10 bars 1 kmol 16.0 kg CH
s 10 cm 303 K bars 22.4 m STP 1 kmol
kg s
2 3
4
4 2 3
πb g
b g
2
101325
0 0225
( )
2
out out out out
in in
2
in in in out in out
in
out in
out
(m/s) A(m )
(m/s) A(m )
10 bar
5 m s 5.555 m s
9 bar
P V V u
P P
nRT
P V nRT V P u P
P
u u
P
⋅
= ⇒ = ⇒ =
⋅
⇒ = = =

2 2 2
2 2
2 2
1
2
0.5(0.0225) kg (5.555 5.000 )m 1 N 1 W
( )
s s 1 kg m/s 1 N m/s
0.0659 W
0.0225 kg 9.8066 m -200 m 1 N 1 W
( )
s
k out in
p out in
E m u u
E mg z z
−
Δ = − =
⋅ ⋅
=
Δ = − =

2
s kg m/s 1 N m/s
44.1 W
⋅ ⋅
= −

7-4
7.8
Δ Δ

.
.
E mg z
p = =
− × ⋅
⋅ ⋅
=− × ⋅
−
10 m 10 L kgH O m m N 1J 2.778 10 kW h
h 1 m L s 1kg m/ s 1N m 1J
kW h h
5 3 3
2
3 2 2
1 981 75 1
1
204 10
7
4
The maximum energy to be gained equals the potential energy lost by the water, or
2.04 10 kW h 24 h 7 days
h 1 day 1 week
kW h week (more than sufficient)
4
× ⋅
= × ⋅
343 106
.
7.9 (b) Q W U E E
k p
− = + +
Δ Δ Δ
Δ
Δ
E
E
k
p
=
=
0
0
system is stationary
no height change
b g
b g
Q W U Q W
− =
Δ , ,
0 0
(c) Q W U E E
k p
− = + +
Δ Δ Δ
Q W
E
E
k
p
= =
=
=
0 0
0
0
adiabatic , no moving parts or generated currents
no height change
b g b g
b g
b g
Δ
Δ
ΔU = 0
(d). Q W U E E
k p
− = + +
Δ Δ Δ
W
E
E
k
p
=
=
=
0
0
0
no moving parts or generated currents
no height change
b g
b g
b g
Δ
Δ
Q U Q
=
Δ , 0 Even though the system is isothermal, the occurrence of a chemical
reaction assures that ΔU ≠ 0 in a non-adiabatic reactor. If the
temperature went up in the adiabatic reactor, heat must be transferred
from the system to keep T constant, hence Q 0 .
7.10 4.00 L, 30 °C, 5.00 bar ⇒ V (L), T (°C), 8.00 bar
(a). Closed system: Δ Δ Δ
U E E Q W
k p
+ + = −
initial / final states stationary
by assumption
Δ
Δ
Δ
E
E
U Q W
k
p
=
=
R
S
T
|
= −
0
0
b g
b g
(b)
Constant T ⇒ = ⇒ = =
− ⋅
⋅
= −
ΔU Q W
0
765
765
. L bar 8.314 J
0.08314 L bar
J
transferred from
gas to
surroundings
(c) Adiabatic ⇒ = ⇒ = − = ⋅ °
Q U W T
0 7 65 30
Δ . L bar 0, C
final

7-5
7.11 A = = ×
2
−
π 3 cm m
cm
m
2 2
2
2
b g 1
10
2 83 10
4
3
.
(a) Downward force on piston:
F P A m g
d = +
=
× ×
+
⋅
=
−
atm piston+weight
5 2 2
2 2
1 atm 1.01325 10 N / m m
atm
24.50 kg 9.81 m 1 N
s kg m / s
N
283 10
1
527
3
.
Upward force on piston: F AP P
u g
= = × −
gas
2 2
m N m
2 83 10 3
.
d i d i
Equilibrium condition:
F F P P
u d
= ⇒ × = ⇒ = × = ×
−
⋅
2 83 10 527 186 10 186 10
3
0 0
5 5
. . .
m N m Pa
2 2
V
nRT
P
0
0
1
0677
= =
× ⋅
× ⋅
=
1.40 g N mol N 303 K 1.01325 10 Pa 0.08206 L atm
28.02 g 1.86 10 Pa 1 atm mol K
L
2 2
5
5
.
(b) For any step, Δ Δ Δ Δ
Δ
Δ
U E E Q W U Q W
k p
E
E
k
p
+ + = − ⇒ = −
=
=
0
0
Step 1: Q U W
≈ ⇒ = −
0 Δ
Step 2: ΔU Q W
= − As the gas temperature changes, the pressure remains constant, so
that V nRT Pg
= must vary. This implies that the piston moves, so that W is not zero.
Overall: T T U Q W
initial final
= ⇒ = ⇒ − =
Δ 0 0
In step 1, the gas expands ⇒ ⇒ ⇒
W U T
0 0
Δ decreases
(c) Downward force Fd = × × + =
−
100 101325 10 2 83 10 4 50 9 81 1 331
5 3
. . . . .
b gd id i b gb gb g N (units
as in Part (a))
Final gas pressure P
F
A
f = =
×
= ×
−
331
10
116 10
3
5
N
2.83 m
N m
2
2
.
Since T Tf
0 30
= = °C , P V P V V V
P
P
f f f
f
= ⇒ = =
×
×
=
0 0 0
0
5
5
0 677
186 10
116 10
108
.
.
.
.
L
Pa
Pa
L
b g
Distance traversed by piston = =
−
×
=
−
ΔV
A
1.08 L 1 m
L 2.83 10 m
m
3
2
0677
10
0142
3 3
.
.
b g
⇒ W Fd
= = = ⋅ =
331 0142 47 47
N m N m J
b gb g
.
Since work is done by the gas on its surroundings, W Q
Q W
= + ⇒ = +
− =
47 47
0
J J
(heat transferred to gas)
7.12 .
V = =
32.00 g 4.684 cm 10 L
mol g 10 cm
L mol
3 3
6 3
01499

H U PV
= + = +
⋅
⋅ ⋅
=
1706 2338
J mol
41.64 atm 0.1499 L 8.314 J / (mol K)
mol 0.08206 L atm / (mol K)
J mol

7-6
0
7.13 (a) Ref state
U = ⇒
0
d i liquid Bromine @ 300 K, 0.310 bar
(b) Δ . . .
U U U
= − = − = −
final initial kJ mol
0 000 28 24 28 24
Δ Δ Δ Δ Δ

H U PV U P V
= + = +
d i (Pressure Constant)
( )
3
0.310 bar 0.0516 79.94 L 8.314 J 1 kJ
ˆ 28.24 kJ mol 30.7 kJ mol
mol 0.08314 L bar 10 J
H
−
Δ = − + = −
⋅
Δ Δ
H n H
= = − = − ⇒ −
. . .
500 30 7 15358 154
mol kJ / mol kJ kJ
b gb g
(c)
U independent of P U U
⇒ = =
, . , . .
300 0 205 300 0 310 28 24
K bar K bar kJ mol
b g b g
, , .
U P U
f
340 340 1 29 62
K K .33 bar kJ mol
d i b g
= =
Δ
Δ

. . .
U U U
U
= −
= − =
E final initial
kJ mol
29 62 28 24 1380

.
.
V changes with pressure. At constant temperature PV = P' V' V'= PV / P'
V'(T = 300K, P = 0.205 bar) =
(0.310 bar)(79.94 L / mol)
bar
L / mol
⇒ ⇒
=
0 205
12088
n = =
500
0 0414
.
.
L 1 mol
120.88 L
mol
Δ Δ
U n U
= = =
. .
0 0414 138 0
mol kJ / mol .0571 kJ
b gb g
Δ Δ Δ
U E E Q W
k p
+ + = − ⇒ =
Q 0 0571
. kJ
(d) Some heat is lost to the surroundings; the energy needed to heat the wall of the container is
being neglected; internal energy is not completely independent of pressure.
7.14 (a) By definition
H U PV
= + ; ideal gas PV RT H U RT

= ⇒ = +
, ,
U T P U T H T P U T RT H T
b g b g b g b g b g
= ⇒ = + = independent of P
(b) Δ Δ Δ
.
H U R T
= + = +
⋅
=
3500
1987
3599
cal
mol
cal 50 K
mol K
cal mol
Δ Δ
H n H
= = = ⇒ ×
.
2 5 3599 8998
mol cal / mol cal 9.0 10 cal
3
b gb g
7.15 Δ Δ Δ
U E E Q Ws
k p
+ + = −
Δ
Δ
Δ
E m u
E
W P V
k
p
s
=
=
=
0
0
no change in and
no elevation change
since energy is transferred from the system to the surroundings
b g
b g
b g
Δ Δ Δ Δ Δ Δ Δ
U Q W U Q P V Q U P V U PV H
= − ⇒ = − ⇒ = + = + =
( )
0
0

7-7
7.16. (a) Δ
Δ
Δ
Δ
E u u
E
P
W P V
k
p
s
= = =
=
=
=
0 0
0
0
1 2
no elevation change
(the pressure is constant since restraining force is constant, and area is constrant)
the only work done is expansion work
b g
b g
b g
.
.
.
.

H T
T
= +
=
×
⋅
=
⇒ =
⋅
34980 355
125 10 1
8 314
0 0295
480
3
2
(J / mol), V = 785 cm , T = 400 K, P =125 kPa, Q = 83.8 J
n =
PV
RT
Pa 785 cm m
m Pa / mol K 400 K 10 cm
mol
Q = H = n(H - H ) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol)
83.8 J = 0.0295 35.5T - 35.5(400) K
1
3
1
3 3
3 6 3
2 1 2
2
Δ
i V
nRT
P
ii W P V
iii Q U P V U Q PV
)
. .
) .
) . . .
mol m Pa cm K
Pa mol K m
cm
N (941- 785)cm m
m cm
J
J J J
3 3
3
3
3 3
2 3
= =
⋅
× ⋅
=
= =
×
=
= + ⇒ = − = − =
0 0295 8 314 10 480
125 10 1
941
125 10 1
10
19 5
838 19 5 64 3
6
5
5
6
Δ
Δ Δ Δ Δ
(b) ΔEp = 0
7.17 (a) The gas temperature remains constant while the circuit is open. (If heat losses could
occur, the temperature would drop during these periods.)
(b) Δ Δ Δ Δ Δ
U E E Q t W t
p R
+ + = −

Δ Δ
E E W U t
Q
p k
= = = = =
=
×
=
0 0 0 0 0
0 90 1
126
, , , ( )

.
.
1.4 W J s
1 W
J s
U t
( ) .
J = 126
Moles in tank:
1 atm 2.10 L 1 mol K
K L atm
mol
n PV RT
= =
⋅
+ ⋅
=
25 273 0 08206
0 0859
b g .
.
.
.
U
U
n
t
t
= = =
126
14 67
(J)
0.0859 mol
Thermocouple calibration: C mV
T aE b T E
T E
T E
= + ° = +
= =−
= =
0 0.249
100 5
181 4 51
,
, .27
. .
b g b g
.
. .
U t
T E
=
= +
14 67 0 440 880 1320
181 4 51 25 45 65 85
(c) To keep the temperature uniform throughout the chamber.
(d) Power losses in electrical lines, heat absorbed by chamber walls.
(e) In a closed container, the pressure will increase with increasing temperature. However, at
the low pressures of the experiment, the gas is probably close to ideal ⇒ =

U f T
b g only.
Ideality could be tested by repeating experiment at several initial pressures ⇒ same
results.

7-8
7.18 (b) Δ Δ Δ

H E E Q W
k p s
+ + = − (The system is the liquid stream.)
Δ
Δ

Ek m u
Ep
Ws
=
=
=
0
0
0
no change in and
no elevation change
no moving parts or generated currents
c h
c h
c h
Δ ,
H Q Q
= 0
(c) Δ Δ Δ

H E E Q W
k p s
+ + = − (The system is the water)
Δ
Δ
Δ
~

H T a P
Ek m u
Q T
=
=
=
0
0
0
nd constant
no change in and
no between system and surroundings
c h
c h
c h
Δ ,
E W W
p s s
= − for water system
0 b g
(d) Δ Δ Δ

H E E Q W
k p s
+ + = − (The system is the oil)
Δ
Ek =0 no velocity change
c h
Δ Δ

H E Q W
p s
+ = −
Q 0 (friction loss);
Ws 0 (pump work).
(e) Δ Δ Δ

H E E Q W
k p s
+ + = − (The system is the reaction mixture)
Δ Δ
Δ

Ek Ep
Ws
= =
=
0
0
given
no moving parts or generated current
c h
c h
Δ ,
H Q Q
= pos. or neg. depends on reaction
7.19 (a) molar flow:
m 273 K kPa 1 mol L
423 K 101.3 kPa 22.4 L STP m
mol min
3
125 122 10
1
434
3
3
.
min
.
b g =
Δ Δ Δ

H E E Q W
k p s
+ + = −
Δ Δ

Ek Ep
Ws
= =
=
0
0
given
no moving parts
c h
c h
.
.
Q H n H
= = = =
Δ Δ
4337
2 63
mol 1 min 3640 J kW
min 60s mol 10 J / s
kW
3
(b) More information would be needed. The change in kinetic energy would depend on the
cross-sectional area of the inlet and outlet pipes, hence the internal diameter of the inlet
and outlet pipes would be needed to answer this question.

7-9
7.20 (a) .
H T H
= ° −
104 25
C in kJ kg
b g
. .
Hout 9.36 kJ kg
= − =
104 34 0 25
. . .
Hin 5 kJ kg
= − =
104 30 0 25 20
Δ . . .
H = − =
9 36 520 16
4 kJ kg
Δ Δ Δ

H E E Q W
k p s
+ + = −
assumed
no moving parts
Δ Δ

Ek Ep
Ws
= =
=
0
0
c h
c h

Q H n H
= =
Δ Δ
1.25 kW kg 1 kJ / s 10 g 1 mol
4.16 kJ kW 1 kg 28.02 g
mol s
3
⇒ = = =

.
n
Q
H
Δ
10 7
=
10.7 mol 22.4 L STP 303 K kPa
s mol 273 K 110 kPa
L / s L s
⇒ = ⇒
.
.
V
b g 1013
2455 246
(b) Some heat is lost to the surroundings, some heat is needed to heat the coil, enthalpy is
assumed to depend linearly on temperature and to be independent of pressure, errors in
measured temperature and in wattmeter reading.
7.21 (a)
. .
.
. . .
. .
H aT b a
H H
T T
b H aT
H T
= + =
−
−
=
−
−
=
= − = − = −
U
V
|
W
|
⇒ = ° −
2 1
2 1
1 1
129 8 258
50 30
52
258 52 30 130 2
52 130 2
b gb g
b g b g
kJ kg C
.
.
H T
= ⇒ = = °
0
130 2
52
25
ref C
Table B.1 ⇒ = ⇒ = = × −
S G V
. . . .
b g b g
C H l
3
3
6 14
m
659 kg
m kg
0 659
1
152 10 3
. .
U H PV T
kJ kg kJ / kg
1 atm 1.0132 10 N / m 1.52 10 m J kJ
1 atm 1 kg 1 N m 10 J
5 2 3
3
b g b g
= − = −
−
× ×
⋅
−
52 130 2
1 1
3
⇒ = −
. .
U T
kJ kg
b g 52 130 4
(b)
Δ Δ
Δ
E E W
Q U
k p
, ,
Energy balance:
20 kg [(5.2 20 -130.4) - (5.2 80 -130.4)] kJ
1 kg
kJ
=
= =
× ×
= −
0
6240
Average rate of heat removal
kJ min
5 min 60 s
kW
= =
6240 1
208
.

n
(m
ol/
s)
N2
30
o
C
34
o
C

Q
=1
.25
k
W
P=
11
0
kP
a

7-10
7.22
0

m
(kg/s)
260°C, 7 bars
H = 2974 kJ/kg

u = 0
(kg/s)
200°C, 4 bars
H = 2860 kJ/kg

u (m/s)

m
Δ Δ Δ
Δ Δ
Δ

( )
.

H E E Q W
E H
mu
m H H
u H H u
k p s
k
E p Q Ws
+ + = −
=− ⇒ =− −
= − =
− ⋅ ⋅
= × ⇒ =
= = = 0
2
2 5
2
2
2 2974 2860 1
228 10 477
kJ 10 N m kg m/ s
kg 1 kJ 1 N
m
s
m/ s
out in
in out
3 2 2
2
d i
d i b g
7.23 (a)
in
5 L/min
0 mm Hg (gauge)
Q
100 mm Hg (gauge)
out
Q
5 L/min
Since there is only one inlet stream and one outlet stream, and
m m m
in out
= ≡ ,
Eq. (7.4-12) may be written

m U m PV
m
u mg z Q Ws
U
m PV mV P P V P
u
z
W
Q Q Q
s
Δ Δ Δ Δ
Δ
Δ Δ
Δ
Δ
+ + + = −
( )
( )
( )
=
= − =
=
=
=
= −
d i d i
a f
2
2
0
0
0
0
2
given
assume for incompressible fluid
all energy other than flow work included in heat terms
out in
in out

V P Q Q
Δ = −
in out
(b) Flow work:
5 L mm Hg 1 atm 8.314 J
min 760 mm Hg 0.08206 liter atm
J min
.
V P
Δ =
−
⋅
=
100 0
667
b g
2
in
2
5 ml O 20.2 J
Heat input: 101 J min
min 1 ml O
Q = =

in
66.7 J min
Efficiency: 100% 66%
101 J min
V P
Q
Δ
= × =

7-11
7.24 (a) Δ Δ Δ

H E E Q W
k p s
+ + = − ; Δ
Ek , Δ
Ep ,
W H Q
s = ⇒ =
0 Δ

H 400 3278
° =
C, 1 atm kJ kg
b g (Table B.7)

H 100 2676
° ⇒ =
C, sat'd 1 atm kJ kg
b g (Table B.5)
2 2
o o
100 kg H O(v)/s 100 kg H O(v)/s
100 C, saturated 400 C, 1 atm
(kW)
Q

.
.
Q =
−
= ×
100 kg kJ 10 J
s kg 1 kJ
J s
3
3278 2676 0
6 02 107
b g
(b) Δ Δ Δ
U E E Q W
k p
+ + = − ; ΔEk , ΔEp , W U Q
= ⇒ =
0 Δ
( ) ( ) ( )
3
final
kJ m
ˆ ˆ ˆ
Table B.5 100 C, 1 atm 2507 , 100 C, 1 atm 1.673 400 C,
kg kg
U V V P
⇒ ° = ° = = °
Interpolate in Table B.7 to find P at which ˆ
V =1.673 at 400o
C, and then interpolate again
to find Û at 400o
C and that pressure:
3 o
final
3.11 1.673
ˆ ˆ
1.673 m /g 1.0 4.0 3.3 bar , (400 C, 3.3 bar) = 2966 kJ/kg
3.11 0.617
V P U
−
= ⇒ = + =
−
⎛ ⎞
⎜ ⎟
⎝ ⎠
( )
[ ]( )
3 7
ˆ 100 kg 2966 2507 kJ kg 10 J kJ 4.59 10 J
Q U m U
⇒ = Δ = Δ = − = ×
The difference is the net energy needed to move the fluid through the system (flow work).
(The energy change associated with the pressure change in Part (b) is insignificant.)
7.25 , .
H l
H O C kJ kg
2 b g
c h
20 83 9
° = (Table B.5)
.
H steam , sat' d kJ kg
20 bars, 2797 2
b g = (Table B.6)
2 2
o
[kg H O(l)/h] [kg H O(v)/h]
20 C 20 bar (sat'd)
=0.65(813 kW) 528 kW
m m
Q =

(a) Δ Δ Δ

H E E Q W
k p s
+ + = − ; Δ
Ek , Δ
Ep ,
W H Q
s = ⇒ =
0 Δ
Δ Δ

H m H
=

. .
m
Q
H
= =
−
=
Δ
528
2797 2 83 9
701
kW kg 1 kJ / s 3600 s
kJ 1 kW 1 h
kg h
b g
(b) . .
V = =
A
701 0 0995 69 7
kg h m kg m h sat'd steam @ 20 bar
Table B.6
3 3
b gd i
(c)
.
V
nRT
P
= =
⋅
⋅
=
701
785
kg / h 10 g / kg 485.4 K 0.08314 L bar 1 m
18.02 g / mol 20 bar mol K 10 L
m / h
3 3
3
3
The calculation in (b) is more accurate because the steam tables account for the effect of
pressure on specific enthalpy (nonideal gas behavior).
(d) Most energy released goes to raise the temperature of the combustion products, some is
transferred to the boiler tubes and walls, and some is lost to the surroundings.

7-12
7.26 , .
H l
H O C, 10 bar kJ kg
2 b g
c h
24 100 6
° = (Table B.5 for saturated liquid at 24o
C; assume
H
independent of P).
.
H 10 bar, 2776 2
sat'd steam kJ kg
b g= (Table B.6) ⇒ Δ . . .
H = − =
2776 2 100 6 26756 kJ kg
2 2
o 3
[kg H O(l)/h] [kg H O(v)/h]
24 C, 10 bar 15,000 m /h @10 bar (sat'd)
(kW)
m m
Q

. .
m = = ×
A
15000
01943 7 72 104
m kg
h m kg h
3
3
Table 8.6
b g
Energy balance Δ Δ Δ
, :
E W H E Q
p s k
= + =
0
d i
Δ Δ

E E E E E
k k k k k
final initial final
Ekinitial
= − =
≈0
Δ

.
.
E
mu
A D
k
f
= =
×
=
⋅
= ×
A
2
2 3
2 2 3
5
2
1 1 1
4
0 15 4 2 1
5 96 10
2
7.72 10 kg 15,000 m h h J
h m 3600 s kg m / s
J / s
4 3
2 3 2 2
d i
π
π
. .
Q m H Ek
= + =
×
+
×
= = ×
Δ Δ
7 72 10 596 10
57973 5
4 5
kg 2675.6 kJ 1 h
h kg 3600 s
J 1 kJ
s 10 J
kJ s .80 10 kW
3
4
7.27 (a) 228 g/min 228 g/min
25o
C T(o
C)
(
Q kW)
Δ Δ
, ,
Ex Ep Ws
Energy balance:
=0

Q H Q
= ⇒ =
Δ W
b g 228 g min J
min 60 s g
1 ( )
H H
out in
−
⇒ =
.
H Q W
out J g
b g b g
0 263
T
H Q W
°
=
C
J g
b g
b g b g
25 26 4 27 8 29 0 32 4
0 263 0 4 47 9 28 13 4 24 8
. . . .
. . . . .
(b) .
H b T b H T T
i
i i i
i
= − = − − =
∑ ∑
25 25 25 334
2
b g b g b g
Fit to data by least squares (App. A.1)
⇒ .
H T
J g C
b g b g
= ° −
334 25
(c)
Q H
= =
− ⋅
=
Δ
350 kg 10 g 1 min 3.34 40 J kW s
min kg 60 s g 10 J
kW
3
3
20
390
b g heat input to liquid
(d) Heat is absorbed by the pipe, lost through the insulation, lost in the electrical leads.
=0
= 5.96x105
J/s

7-13
7.28
[ [
[ [
m m
m m
w 2 w 2
o
e 2 6 e 2 6
o o
kg H O(v) / min] kg H O(l) / min]
3 bar, sat'd 27 C
kg C H / min] kg C H / min]
16 C, 2.5 bar 93 C, 2.5 bar
(a) C H mass flow:
m L 2.50 bar 1 K - mol 30.01 g kg
min m 289 K 0.08314 L - bar mol 1000 g
kg min
2 6
3
3

.
me =
= ×
795 10 1
2 487 10
3
3
,
H H
ei ef
= =
941 1073
kJ kg kJ kg
Energy Balance on C H :
kg
min
kJ
kg
kJ min
min 60 s
kW
2 6
3
Δ Δ Δ
, ,
.
.
.
E W E Q H
Q
p s k
= ≅ ⇒ =
= × −
L
NM O
QP=
×
= ×
0 0
2 487 10 1073 941
2 487 10 1
547 10
3
3
b g
(b) .
Hs1
3 2724 7
.00 bar, sat'd vapor kJ kg
b g= (Table B.6)
.
Hs2
1131
liquid, 27 C kJ kg
° =
b g (Table B.5)
Assume that heat losses to the surroundings are negligible, so that the heat given up by the
condensing steam equals the heat transferred to the ethane 547 103
. × kW
d i
Energy balance on H O:
2

Q H m H H
s s
= = −
Δ 2 1
d i
⇒ =
−
=
− ×
−
=

.
. .
.
m
Q
H H
s s
2 1
547 10
1131 2724 7
2 09
3
kJ kg
s kJ
kg s steam
b g
⇒ = =
A
. .
Vs 2 09 0 606 1
kg / s m kg .27 m s
Table B.6
3 3
b gd i
Too low. Extra flow would make up for the heat losses to surroundings.
(c) Countercurrent flow Cocurrent (as depicted on the flowchart) would not work, since it
would require heat flow from the ethane to the steam over some portion of the exchanger.
(Observe the two outlet temperatures)
( )
Q kW

7-14
7.29
250 kg H O( )/min
v
2
40 bar, 500°C
H1
(kJ/kg)
Turbine
W s =1500 kW
250 kg/min
5 bar, T2 (°C), H2
(kJ/kg)
Heat
exchanger
Q(kW)
250 kg/min
5 bar, 500°C
H3
(kJ/kg)
H O 40 bar, 500 C kJ kg
2 v H
, :
° =
b g 1 3445 (Table B.7)
H O 5 bar, 500 C kJ kg
2 v H
, :
° =
b g 3 3484 (Table B.7)
(a) Energy balance on turbine: Δ Δ
, ,
E Q E
p k
= = ≅
0 0 0
Δ
H W m H H W H H W m
s s s
= − ⇒ − = − ⇒ = −
= − =
2 1 2 1
3445 1500
3085
d i
kJ
kg
kJ min 60 s
s 250 kg 1 min
kJ kg

H P
= = ⇒ °
3085 5
kJ kg and bars T = 310 C (Table B.7)
(b) Energy balance on heat exchanger: s
Δ Δ
, ,
E W E
p k
= = ≅
0 0 0

Q H m H H
= = − =
−
=
Δ 3 2
250 3484 3085
1663
d i b g
kg kJ 1 min 1 kW
min kg 60 s 1 kJ / s
kW
(c) Overall energy balance: Δ Δ
,
E E
p k
= ≅
0 0
Δ
H Q W m H H Q W
s s s
= − ⇒ − = −
3 1
d i

Q H Ws
= + =
−
+
= √
Δ Δ
250 3484 3445 1500
1663
kg kJ 1 min 1 kW
min kg 60 s 1 kJ / s
kJ 1 kW
s 1 kJ / s
kW
b g
(d) H O 40 bar, 500 C m kg
2
3
v V
, : .
° =
b g 1 0 0864 (Table B.7)
H O 5 bar, 310 C m kg
2
3
v V
, : .
° =
b g 2 05318 (Table B.7)
u1
4
183
= =
250 kg 1 min 0.0864 m 1
min 60 s kg 0.5 m
m s
3
2 2
π
.
u2
4
113
= =
250 kg min 0.5318 m 1
min 60 s kg 0.5 m
m s
3
2 2
π
.
Δ . .
.
E
m
u u
k = − =
− ⋅
⋅ ⋅
=
2
250 113 183
0 26
2
2
1
2
2 2 2
kg 1 1 min m 1 N 1 kW s
min 2 60 s s 1 kg m / s 10 N m
kW 1500 kW
2 2 3
b g b g

7-15
7.30 (a) Δ
Ep , Δ
Ek , .
W Q H hA T T h T T
s s o s o
= ⇒ = ⇒ − − = − ⇒ − =
0 300 18 300
Δ b g b g
kJ h
kJ
h
(b) Clothed: 8
= .
C
s
h
T
To
= ⇒ = °
34 2
134
.
Nude, immersed: 64
= .
C
s
h
T
To
= ⇒ = °
34 2
316
. (Assuming Ts remains 34.2°C)
(c) The wind raises the effective heat transfer coefficient. (Stagnant air acts as a thermal
insulator —i.e., in the absence of wind, h is low.) For a given To , the skin temperature
must drop to satisfy the energy balance equation: when Ts drops, you feel cold.
7.31 Basis: 1 kg of 30°C stream
1 kg H2O(l)@30o
C
3 kg H2O(l)@Tf(o
C)
2 kg H2O(l)@90o
C
(a) Tf = + =
1
3
30
2
3
90 70
o
C C C
b g b g
D D
(b) Internal Energy of feeds: C, liq. kJ kg
C, liq. kJ kg
.
.
U
U
30 1257
90 376 9
° =
° =
U
V
|
W
|
b g
b g
(Table B.5 - neglecting effect of P on
H )
Energy Balance: - = + +
= = =
Q W U E E U
p k
Q W E p Ek
Δ Δ Δ Δ
Δ Δ =
=
0
0
⇒ − − =
3 1 1257 2 376 9 0
( . ( .
U f kg) kJ / kg kg) kJ / kg
b g b g
⇒ = ⇒ = °
. .
U T
f f
2932 70 05
kJ kg C (Table B.5)
Diff.=
−
× =
70 05 70 00
70 05
100% 0 07%
. .
.
. (Any answer of this magnitude is acceptable).
7.32
.
.
.
.
Q (kW)
52.5 m3
H2O(v)/h
m(kg/h)
5 bar, T(o
C)
m(kg/h)
0.85 kg H O( )/kg
v
2
0.15 kg H O( )/kg
l
2
5 bar, saturated, T(o
C)
(a) Table B.6 C
bars
P
T
=
= °
5
1518
. , . .
H H
L V
kJ kg, kJ kg
= =
6401 2747 5

.
V(5 bar, sat'd) = 0.375 m / kg m
m 1 kg
h 0.375 m
kg h
3
3
3
⇒ = =
52 5
140
(b) H O evaporated kg h kg h
2 = =
015 140 21
.
b gb g
Energy balance: . .
Q H
= =
−
=
Δ
21 kg kJ 1 h 1 kW
h kg 3600 s 1 kJ s
kW
2747 5 6401
12
b g

7-16
7.33 (a) P T o
= =
5 bar C
Table B.6
saturation 1518
. . At 75°C the discharge is all liquid
(b) Inlet: T=350°C, P=40 bar Table B.7
in = 3095 kJ / kg

H ,
Vin
3
= 0.0665 m / kg
Outlet: T=75°C, P=5 bar Table B.7
out = 314.3 kJ / kg

H ,
Vout
-3 3
=1.03 10 m / kg
×
u
V
A
u
V
A
in
in
in
3
2 2
out
out
out
3
2 2
kg 1 min m / kg
min 60 s 0.075) / 4 m
m / s
kg 1 min 0.00103 m / kg
min 60 s 0.05) / 4 m
m / s
= = =
= = =
.
(
.

(
.
200 0 0665
5018
200
175
π
π
Energy balance: ( )

( )
Q W H E m H H
m
u u
s k
− ≈ + = − + −
Δ Δ 2 1 2
2
1
2
2
2 2 2
2
200 kg 1 min (314-3095) kJ 200 kg 1 min (1.75 -50.18 ) m
min 60 s kg 2 min 60 s s
13,460 kW ( 13,460 kW transferred from the turbi
s
Q W
− = +
= − ⇒

ne)
7.34 (a) Assume all heat from stream transferred to oil
1.00 10 kJ 1 min
min 60 s
kJ s
4

Q = × = 167
25 bars, sat'd
100 kg oil/min

m(kg H2O(v)/s)
25 bars, sat'd
135°C
100 kg oil/min
185°C

m(kg H2O(l)/s)
Energy balance on H O:
2 out in

, ,
Q H m H H
E E W
p k s
= = −
=
Δ
Δ Δ
d i
0
, .
H l 25 bar, sat' d kJ kg
( ) = 962 0 , , .
H v 25 bar, sat' d kJ kg
( ) = 2800 9 (Table B.6)

. .
.
m
Q
H H
=
−
=
−
−
=
out in
kJ kg
s kJ
kg s
167
962 0 2800 9
0 091
b g
Time between discharges:
g 1 s 1 kg
discharge 0.091 kg 10 g
s discharge
3
1200
13
=
(b) Unit Cost of Steam:
$1 kJ 0.9486 Btu
10 Btu kg kJ
/ kg
6
2800 9 839
6 10 3
. .
$2.
−
= × −
b g
Yearly cost:
1000 traps 0.091 kg stream 0.10 kg last 2.6 10 $ 3600 s 24 h 360 day
trap s kg stream kg lost h day year
year
×
⋅
= ×
−3
5
4 10
$7. /

7-17
7.35 Basis: Given feed rate
200 kg H2O(v)/h
10 bar, sat’d, .
H = 2776 2 kJ / kg [
n3 kg H O(v) / h]
2
10 bar, 250o
C,
H = 2943 kJ / kg
[

n
H
2
3052
kg H O(v) / h]
10 bar, 300 C, kJ / kg
2
o
=

Q(kJ / h)

H from Table B.6 (saturated steam) or Table B.7 (superheated steam)
Mass balance: 200 2 3
+ =

n n (1)
Δ Δ
Δ
, ,
. ,
E E W
K p
Q H n n Q
Energy balance: in kJ h
=
= = − −
0
3 2
2943 200 2776 2 3052
b g b g b g (2)
(a)
n3 300
= kg h
( )

1
100
2
n = kg h
( ) .
2
2 25 104
Q = × kJ h
(b)
Q = 0
( ),( )

1 2
306
2
n = kg h ,
n3 506
= kg h
7.36 (a) Tsaturation @ 1.0 bar = 99.6 °C⇒ =
Tf 99 6
. D
C
H O (1.0 bar, sat'd) kJ / kg, kJ / kg
H O (60 bar, 250 C) kJ / kg
2
2
⇒ = =
=
. .
.
H H
l v
417 5 26754
10858
D
Mass balance: kg
Energy balance:
kg)(1085.8 kJ / kg) = 0 (2)
,
m m
H
m H m H m H m H m H
v l
E Q E W
v v l l v v l l
K p
+ =
=
⇒ + − = + −
=
100 1
0
100
0
1 1
( )

(
, ,
Δ Δ
Δ
( , )
. .
1 2
70 4 29 6
m m
l v
= =
kg, kg ⇒ yv = =
29 6
0 296
.
.
kg vapor
100 kg
kg vapor
kg
(b) is unchanged.
T The temperature will still be the saturation temperature at the given final
pressure. The system undergoes expansion, so assuming the same pipe diameter, 0.
k
E
Δ

would be less (less water evaporates) because some of the energy that would have
vaporized water instead is converted to kinetic energy.
v
y
(c) Pf = 39 8
. bar (pressure at which the water is still liquid, but has the same enthalpy as the feed)
(d) Since enthalpy does not change, then when Pf ≥ 39 8
. bar the temperature cannot increase,
because a higher temperature would increase the enthalpy. Also, when Pf ≥ 39 8
. bar , the product
is only liquid ⇒ no evaporation occurs.

7-18
7.36 (cont’d)
0
0.1
0.2
0.3
0.4
0 20 40 60 80
Pf (bar)
y
0
50
100
150
200
250
300
1 5 10 15 20 25 30 36 39.8 60
Pf (bar)
Tf
(C)
7.37 10 m3
, n moles of steam(v), 275°C, 15 bar⇒10 m3
, n moles of water (v+l), 1.2 bar
Q
1.2 bar, saturated
10.0 m3 H2O (v)
min (kg)
275oC, 1.5 bar
10.0 m3
mv [kg H2O (v)]
ml [kg H2O (l)]
(a) P=1.2 bar, saturated, C
Table B.6
T2 104 8
= . D
(b) Total mass of water: min
3
3
=
10 m kg
0.1818 m
kg
1
55
=
Mass Balance:
Volume additivity: m m kg) m kg)
kg, kg condensed
3 3 3
m m
V V m m
m m
v l
v l v l
v l
+ =
+ = = +
⇒ = =
550
10 0 1428 0 001048
7 0 48 0
.
. ( . / ( . /
. .
(c) Table B.7 = 2739.2 kJ / kg; = 0.1818 m / kg
Table B.6 = 439.2 kJ / kg; = 0.001048 m / kg
= 2512.1 kJ / kg; =1.428 m / kg
in in
3
3
3
⇒
⇒
R
S
|
T
|

U V
U V
U V
l l
v v
Δ Δ
Δ
E E W
v v l l
p k
Q U m U m U m U
, ,

[( . ) ( . )( .
in in
5
Energy balance: = =
(2512.1 kJ / kg) + ) - kg (2739.2)] kJ
= 1.12 10 kJ
=
+ −
=
− ×
0
7 0 48 0 439 2 55
7.38 (a) Assume both liquid and vapor are present in the valve effluent.
1 kg H O( ) / s
15 bar, T C
2
sat
o
v
+150 [
[
m l
m v
l
v
2
2
kg H O( ) / s]
kg H O( ) / s]
1.0 bar, saturated
(b) Table B.6 T bar) =198.3 C T C
Table B.7 C, 15 bar) 3149 kJ / kg
Table B.6 1.0 bar, sat'd) = 417.5 kJ / kg; 1.0 bar, sat'd) = 2675.4 kJ / kg
sat'n
o
in
o
in
⇒ ⇒ =
⇒ = ≈
⇒
( .
( .
( (
15 348 3
348 3
H H
H H
l v
D
, 15 bar

7-19
7.38 (cont’d)
Δ Δ
Δ
, , ,

( . ) ( )( . )

E E Q W
l l v v
l l v v l l
p k s
H m H m H m H
m H m H m H m m
m m
v l
in in
in in
Energy balance:
kJ / kg
=
= ⇒ + − =
⇒ = + = + −
+
0
0 0
3149 417 5 1 26754
There is no value of
ml between 0 and 1 that would satisfy this equation. (For any value
in this range, the right-hand side would be between 417.5 and 2675.4). The two-phase
assumption is therefore incorrect; the effluent must be pure vapor.
(c) Energy balance ⇒ =
= =
≈
( )
m H m H
m m
H
T
out out in in
in out
out
out
3149 kJ / kg = bar, T
337 C
Table B.7
1
1
D
(This answer is only approximate, since Δ
Ek is not zero in this process).
7.39 Basis: 40 lb min circulation
m
(a) Expansion valve
R = Refrigerant 12
40 lbm R(l)/min
H
= 27.8 Btu/lbm
93.3 psig, 86°F
40
1
77 8 9 6
lb min
lb R lb
lb R( / lb
Btu / lb Btu / lb
m
m m
m m
m m
/
( ) /
( ) )
. , .
x v
x l
H H
v
v
v l
−
= =
Energy balance: neglect
out in
Δ Δ Δ
, , ,
E W Q E H n H n H
p s k i i i i
= ⇒ = − =
∑ ∑
0 0
40 77 8 40 1 9 40 27.8 Btu
0
X R v X R l
v v
lb Btu
min lb
lb .6 Btu
min lb
lb
min lb
m
m
m
m
m
m
b g b g b g
.
+
−
− =
Xv =
E
0 267
. 26.7% evaporates
b g
(b) Evaporator coil
40 lb /min
m
H
v = 77.8 Btu/lb ,
11.8 psig, 5°F
m
40
H
= 77.8 Btu/lb
11.8 psig, 5°F
m
0.267 R( )
v
0.733 R( )
l
H
l = 9.6 Btu/lbm
lb R( )/min
m v
Δ Δ Δ
, ,
E W E Q H
p s k
= ⇒ =
0
. .
Q
R v R l
= − −
=
40 77 8 40 0 267 77.8 Btu 9
2000
lb Btu
min lb
lb
min lb
40 0.733 lb .6 Btu
min lb
Btu min
m
m
m
m
m
m
b gb g b g b gb g b g

7-20
7.39 (cont’d)
(c) We may analyze the overall process in several ways, each of which leads to the same
result. Let us first note that the net rate of heat input to the system is

Q Q Q
= − = − = −
evaporator condenser Btu min
2000 2500 500
and the compressor work Wc represents the total work done on the system. The system is
closed (no mass flow in or out). Consider a time interval Δt min
b g. Since the system is at
steady state, the changes ΔU , ΔEk and ΔEp over this time interval all equal zero. The
total heat input is
Q t
Δ , the work input is
W t
c Δ , and (Eq. 8.3-4) yields
.
.
.
Q t W t W Q
c c
Δ Δ
− = ⇒ = =
− ×
×
=
−
−
0
500 1341 10
9 486 10
118
3
4
Btu 1 min hp
min 60 s Btu s
hp
7.40 Basis: Given feed rates

.
n1
08
0
(mol / h)
0.2 C H
C H
C, 1.1 atm
3 8
4 10
o

n
n
C H 3 8
C H 4 10
o
3 8
4 10
(mol C H / h)
(mol C H / h)
227 C

Q (kJ / h)

n2
3 8
4 10
o
(mol / h)
0.40 C H
.60 C H
25 C, 1.1 atm
0
Molar flow rates of feed streams:
300 L 1.1 atm 1 mol
hr 1 atm 22.4 L STP
mol h
.
n1 14 7
= =
b g
.
n2 9 00
= =
200 L 273 K 1.1 atm 1 mol
hr 298 K 1 atm 22.4 L STP
mol h
b g
Propane balance
14.7 mol 0.20 mol C H
h mol
9.00 mol 0.40 mol C H
h mol
mol C H h
C H
3 8 3 8
3 8
3 8
⇒ = +
=

.
n
654
Total mole balance: mol mol
C H h C H h
C H 4 20 4 20
4 10
( . . . ) .
n = + − =
14 7 9 00 654 1716
Δ Δ Δ
, ,
E W E Q H
p s k
= ⇒ =
0

Q H N H N H
i i i i
= = − = +
−
×
−
×
=
∑ ∑
Δ
out in
3 8 4 10
3 8 4 10
6.54 mol C H 20.685 kJ
h mol
17.16 mol C H 27.442 kJ
h mol
0.40 9.00 mol C H 1.772 kJ
h mol
0.60 9.00 mol C H 2.394 kJ
h mol
kJ h
b g b g 587
(
Hi = 0 for components of 1st feed stream)

7-21
7.41 Basis:
510 m 273 K L 1 mol 1 kmol
min 291 K m 22.4 L STP 10 mol
kmol min
3
3 3
10
214
3
b g = .
(a)
38°C,
(mol H O/mol)
2
18°C, sat'd
y
(mol dry air/mol)
x
(1 –y0)
21.4 kmol/min
(mol dry air)
n0 (kmol/min)
hr = 97%
0
(kmol H O( )/mol)
2
n2 l
18°C
y
(1 – )
1
1
y
(mol H O/mol)
2
.
.
( ) ( )
2
H O
2
38 C 0.97 49.692 mm Hg
Inlet condition: 0.0634 mol H O mol
760 mm Hg
r
o
h P
y
P
∗
°
= = =
( )
2
H O
1 2
18 C 15.477 mm Hg
Outlet condition: 0.0204 mol H O mol
760 mm Hg
P
y
P
∗
°
= = =
Dry air balance: kmol min
1 0 0634 1 0 0204 214 22 4
− = − ⇒ =
. . . .
b g b g
n n
o o
Water balance: 0.98 kmol min
kmol kg
kmol
18 kg / min H O condenses
2
0 0634 22 4 0 0204 214
0 98 18 02
2 2
. . . .
. .
min
b g b g
= + ⇒ =
=
n n
(b). Enthaphies: C kJ mol
air
. .
H 38 0 0291 38 25 0 3783
° = − =
b g b g
. .
Hair C kJ mol
18 0 0291 18 25 0 204
° = − = −
b g b g
,
.
.
,
.
.
, .
H v
H v
H l
H O 3
H O 3
H O 3
2
2
2
38 C
kJ 1 kg 18.02 g
kg 10 g mol
kJ mol
18 C
kJ 1 kg 18.02 g
kg 10 g mol
kJ mol
18 C
1 kg 18.02 g
kg 10 g mol
kJ mol
Table B.5
° = =
° = =
° = =
U
V
|
|
|
W
|
|
|
b g
b g
b g
25708
46 33
2534 5
4567
75.5 kJ
136
Δ Δ
Δ

, ,
. . .
. . . . . . . .
. . . .
E W E
p s k
Q H n H n H Q
i i i i
Energy balance:
kJ min
out in
= ≅
= = − ⇒ = − × −
+ × + × − − ×
− × = − ×
∑ ∑
0 0
1 0 0204 214 10 0 204
0 0204 214 10 4567 0 98 10 136 1 0 0634 22 4 10 0 3783
0 0634 22 4 10 46 33 567 10
3
3 3 3
3 4
b gd ib g
b gd ib g d ib g b gd ib g
b gd ib g
⇒ × =
567 10 270 tons o
4
. kJ 60 min 0.9486 Btu 1 ton cooling
min h kJ 12000 Btu
f cooling
(kJ/min)
Q

7-22
7.42 Basis: 100 mol feed
n
0.65 A( )
l
2 (mol), 63.0°C
100 mol, 67.5°C
0.35 B( )
l
0.98 A( )
v
0.02 B( )
v
n2 (mol)
0.98 A( )
l
0.02 B( )
l
0.5
n5 (mol), 98.7°C
0.544 A( )
v
0.456 B( )
v
56.8°C
Qc (cal)
n2 (mol)
0.98 A( )
l
0.02 B( )
l
0.5
Qr (cal)
n5 (mol), 98.7°C
0.155 A( )
l
0.845 B( )
l
A - Acetone
B - Acetic Acid
(a) Overall balances:
Total moles: 100
0.65 100
mol
mol
= +
= +
U
V
W
=
=
05
0 98 05 0155
120
40
2 5
2 5
2
5
.
: . . .
n n
A n n
n
n
b g b g
Product flow rates: Overhead 0.5 120 mol
0.5 120 mol
b g
b g
0 98 588
0 02 12
. .
. .
=
=
A
B
Bottoms 0 40 mol
0 40 mol
. .
. .
155 6 2
845 338
b g
b g
=
=
A
B
Δ Δ
Δ
E W E
p x
Q H n H n H
i i i i
, ,

Overall energy balance:
out in
2
0 0
= ≅
= = −
∑ ∑
⇒ ( ) ( ) ( ) ( ) ( ) ( )
interpolate in table interpolate in table
4
58.8 0 1.2 0 6.2 1385 33.8 1312 65 354 35 335 1.82 10 cal
Q
↓ ↓
= + + + − − = ×
(b) Flow through condenser: mols
2 588 117 6
. .
b g= A
2 12 2 4
. .
b g= mols B
Δ Δ
Δ
E W E
p k
Q H
c
, ,
Energy balance on condenser:
3
0 0
= ≅
=
Qc = − + − = − ×
117 6 0 7322 2 4 0 6807 8 77 105
. . .
b g b g cal heat removed from condenser
Assume negligible heat transfer between system surroundings other than Qc Qr
( )
4 5 5
1.82 10 8.77 10 8.95 10 cal
r c
Q Q Q
= − = × − − × = × heat added to reboiler
7.43
Q= 0
1.96 kg, P1= 10.0 bar, T1
1.00 kg, P2= 7.0 bar, T2
2.96 kg, P3= 7.0 bar, T3=250o
C

7-23
7.43 (cont’d)
(a) T T P
2 7 0
= =
( . bar, sat'd steam) =165.0 C
o
( ( ),
( ( ), (
H v
H v
3
2
2954
2760
H O P = 7.0 bar, T = 250 C) kJ kg (Table B.7)
H O P = 7.0 bar, sat'd) kJ kg Table B.6)
2
o
2
=
=
Δ Δ
Δ
E Q W E
p s k
H H H H H
H
, ,
,
. . . . .
( . )
Energy balance
kg(2954 kJ / kg) -1.0 kg(2760 kJ / kg)
bar, T kJ / kg T C
1 1
≅
= = − − ⇒ =
⇒ = ⇒ ≅
0
0 2 96 196 10 196 2 96
10 0 3053 300
3 1 2 1
1
D
(b) The estimate is too low. If heat is being lost the entering steam temperature would have to
be higher for the exiting steam to be at the given temperature.
7.44 (a) T T P
V P
V P
l
v
1 30
30
30
= =
=
=
( .
( .
( .
bar, sat'd.) =133.5 C
bar, sat'd.) = 0.001074 m / kg
bar, sat'd.) = 0.606 m / kg
3
3
D
V
V
m
l
space
v
= =
=
= =
0 001074
177 2
200 0
22.8 L 1
0 606
0 0376
.
.
.
.
.
m 1000 L 165 kg
kg m
L
L -177.2 L = 22.8 L
m 1 kg
1000 L m
kg
3
3
3
3
m=165.0 kg
P=3 bar
V=200.0 L
Pmax=20 bar
Vapor
Liquid
(b) P P mtotal
= = = + =
max . . . .
20 0 1650 0 0376 16504
bar; kg
T T P
V P V P
l v
1 20 0
20 0 20 0
= =
= =
( .
( . ( .
bar, sat'd.) = 212.4 C
bar, sat'd.) = 0.001177 m / kg; bar, sat'd.) = 0.0995 m / kg
3 3
D
V m V m V m V m m V
m m
m m
total l l v v l l total l v
l l
l v
= + ⇒ + −
⇒ =
⇒ = =
( )
. ( . / ) ( . /
. .
L m
L
kg m kg) + (165.04 - kg m kg)
kg; kg
3
3 3
200 0 1
1000
0 001177 0 0995
164 98 0 06
V V
m
l space
evaporated
= = =
= =
0 001177
194 2 200 0
1000
20
.
. ; .
(
m 1000 L 164.98 kg
kg m
L L -194.2 L = 5.8 L
0.06 - 0.04) kg g
kg
g
3
3
(c)
Δ Δ
Δ
E W E
p s k
U U P U P
, ,
( . ( .
Energy balance Q = bar, sat'd) bar, sat'd)
≅
= = − =
0
20 0 30
( . ( .
( . ( .
U P U P
U P U P
l v
l v
= =
= =
20 0 20 0
30 30
bar, sat'd.) = 906.2 kJ / kg; bar, sat'd.) = 2598.2 kJ / kg
bar, sat'd.) = 561.1 kJ / kg; bar, sat'd.) = 2543 kJ / kg
Q =
− ×
0 06
. kg(2598.2 kJ / kg) +164.98 kg(906.2 kJ / kg) - 0.04 kg(2543 kJ / kg)
165.0 kg (561.1 kJ / kg) = 5.70 10 kJ
4
Heat lost to the surroundings, energy needed to heat the walls of the tank

7-24
7.44 (cont’d)
(d) (i) The specific volume of liquid increases with the temperature, hence the same mass of
liquid water will occupy more space; (ii) some liquid water vaporizes, and the lower
density of vapor leads to a pressure increase; (iii) the head space is smaller as a result of
the changes mentioned above.
(e) – Using an automatic control system that interrupts the heating at a set value of pressure
– A safety valve for pressure overload.
– Never leaving a tank under pressure unattended during operations that involve
temperature and pressure changes.
7.45 Basis: 1 kg wet steam
(a) H O
2
1 kg 20 bars
0.97 kg H O(v)
2
0.03 kg H O(l)
2
H
1 (kJ/kg)
H O,(v) 1 atm
2
1 kg
H
2 (kJ/kg)
H O
2
1 kg
Tamb, 1 atm
Enthalpies: bars, sat'd kJ kg
bars, sat'd kJ kg
Table B.7
, .
, .
H v
H l
20 2797 2
20 908 6
b g
b g b g
=
=
U
V
|
W
|
Δ Δ
Δ
E E Q W
p K
H H H
H T
, , ,
. . . .

=
Energy balance on condenser:
= 2740 kJ / kg C
Table B.7
o
3
0
0 0 97 2797 2 0 03 908 6
132
2 1
2
= ⇒ = = +
⇒ ≈
b g b g
(b) As the steam (which is transparent) moves away from the trap, it cools. When it reaches
its saturation temperature at 1 atm, it begins to condense, so that T = °
100 C . The white
plume is a mist formed by liquid droplets.
7.46 Basis:
oz H O 1 quart 1 m 1000 kg
32 oz 1057 quarts m
kg H O
2
3
3 2
8
0 2365
l
l
b g b g
= .
(For simplicity, we assume the beverage is water)
0.2365 kg H2O (l)
18°C
32°F (0°C)
4°C
m (kg H2O (s))
(m + 0.2365) (kg H2O (l))
Assume P = 1 atm
Enthalpies (from Table B.5):
( ) ( ) ( )
2 2 2
ˆ ˆ ˆ
H O( ), 18 C 75.5 kJ/kg; H O( ), 4 C 16.8 kJ/kg; H O( ), 0 C =-348 kJ/kg
H l H l H s
° = ° = °
( )
out in
, , , 0
ˆ ˆ
Energy balance closed isobaric system : 0
p k
i i i i
E E Q W
H n H n H
Δ Δ =
⇒ Δ = − =
∑ ∑
⇒ +
( . ) ( .
m m
0 2365 168
kg kJ / kg) = 0.2365 kg(75.5 kJ / kg) + kg (-348 kJ / kg)
⇒ =
m 0 038 38
. kg = g ice
Q=0 Q

7-25
7.47 (a) When T H
= = ⇒ =
0 0 0
o
ref
o
C, T C
,
(b) Energy Balance-Closed System: ΔU = 0
Δ Δ
E E Q W
k p
, , , = 0
(°C)
25 g Fe, 175°C
T
20°C
25 g Fe
f
1000 g H2O(l) 1000 g H2O
U T U T U U
f f
Fe H O Fe H O
2 2
C C, 1 atm
d i d i b g b g
+ − ° − ° =
175 20 0 or Δ Δ
U U
Fe H O
2
+ = 0
ΔU
T
T
f
f
Fe
g 4.13 cal 4.184 J
g cal
J
=
−
= −
250 175
432 175
. d i
Table B.5
1.0 L g J
1 L g
J
432
H O
H O
H O
H O
2
2
2
2
⇒ =
−
= −
⇒ + − × = =
ΔU
U T
U T
T U T f T
f
f
f f f
10 839
1000 839
1000 160 10 0
3
5
. .
.
d i
e j d i
e j
d i d i
⇒
T
f T
T
f
f
f
°
− × + × −
= °
C Interpolate
C
30 40 35 34
21 10 2 5 10 1670 2612
34 6
4 4
d i . .
.

7-26
7.48
H O( )
v
T0
I
2
760 mm Hg
100°C
H O( ), 100 °C
l
2
H O( )
v
Tf
II
2
(760 + 50.1) mm Hg
H O( ),
l
2
⇒
Tf
Tf
⇒
⇒
1.08 bar sat'd
Tf = 101.8°C (Table 8.5)
Energy balance - closed system:
Δ Δ
E E W Q
p K
, , , = 0
ΔU m U m U m U m U m U m U
v v l
I
l b b v v l l b b
v
l
b
= = + + − − −
0 II II I II II II I I I I I I
-vapor
-liquid
-block

I II
V
V
U
U
l
v
l
v
101 108
1044 1046
1673 1576
419 0 426 6
25065 2508 6
. .
. .

. .
. .
bar, 100 C bar, 101.8 C
L kg
L kg
L kg
L kg
° °
b g b g
b g
b g
b g
b g
Initial vapor volume: 20.0 L L
kg 1 L
8.92 kg
L H O
I
2
V v
v = − − =
50
50
14 4
. . b g
Initial vapor mass: = 14.4 L 1673 L kg kg H O
I
2
m v
v b g b g
= × −
8 61 10 3
.
Initial liquid mass: 5 L 1.044 L kg kg H O
I
2
m l
l = =
. .
0 4 79
b g b g
Final energy of bar: .36 101.8 kJ kg
II
.
Ub = =
0 36 6
b g
Assume negligible change in volume liquid ⇒ =
Vv
II
L
14 4
.
Final vapor mass: 14.4 L 1576 L kg kg H O
II
2
m v
v = = × −
b g b g
914 10 3
.
Initial energy of the bar:

.
. . . . . . . . . .
.
Ub
I
kg
kJ kg
= × + + − × −
=
− −
1
50
914 10 25086 4 79 4266 50 366 861 10 25065 4 79 419 0
441
3 3
b g b g b g b g b g
d i
(a) Oven Temperature:
kJ / kg
0.36 kJ / kg C
C
o
To =
⋅
= °
441
122 5
.
.
H O kg - kg = kg = 0.53 g
2 evaporated
II I
= − = × × ×
− − −
m m
v v 914 10 8 61 10 530 10
3 3 4
. . .
(b) . . . .
Ub
I
kJ kg
= + =
441 8 3 50 458
To = = °
458 0 36 127 2
. . . C
(c) Meshuggeneh forgot to turn the oven on (To °
100 C )

7-27
7.49 (a) Pressure in cylinder = +
weight of piston
area of piston
atmospheric pressure
P
Tsat
= + =
⇒ = °
30.0 kg 9.807 N cm bar
400.0 cm kg 1 m 10 N m
atm 1.013 bar
atm
bar
C
2 2 5 2
100 10 1
108
1018
2
2
b g
b g
.
.
.
Heat required to bring the water and block to the boiling point
Q U m U U m U T U
w wl wl Al Al sat Al
= = − ° + − °
=
−
+
−
=
Δ . ,
. . ( . )
108 20
839 0 94 1018 20
2630
bar, sat'd l 20 C C
7.0 kg 426.6 kJ
kg
3.0 kg [ ]kJ
kg
kJ
b g b g
d i b g b g
d i
b g
2630 kJ 3310 kJ Sufficient heat for vaporization
⇒
(b) T T
f sat
= = °
1018
. C . Table B.5 ⇒
= =
= =
. , .
, .
V U
V U
l l
v v
1046 426 6
1576 2508 6
L kg kJ kg
L kg kJ kg
T 101.8°C
P 1.08 bars
1576 L/kg, 2508.6 kJ/kg
1.046 L/kg, 426.6 kJ/kg
Q (kJ) W (kJ)
≡
≡
m v
v (kg H O( ))
2
m l
l (kg H O( ))
2
7.0 kg H O( )
2 l
.

H = 4266 kJ / kg
V = 1.046 L / kg
(Since the Al block stays at the same temperature in this stage of the process, we can
ignore it -i.e.,
U U
in out
= )
Water balance: 7 0
. = +
m m
l v (1)
Work done by the piston:
bar L
8.314 J / mol K 1 kJ
0.08314 liter - bar / mol K 10 J
kJ
piston atm
atm
3
W F z w P A z
w
A
P A z P V W m m
m m
v l
v l
= = +
= +
L
NM O
QP = ⇒ = + −
×
⋅
⋅
= + −
Δ Δ
Δ Δ
b g b g b gb g
b g
108 1576 1046 1046 7 0
170 2 0113 0 7908
. . . .
. . .
Energy balance: Δ
Δ
U Q W
m m m m
m m
v L
U Q
v L
W
v L
= −
⇒ + − = − − + −
⇒ + − =
2508 6 426 6 426 6 7 3310 2630 170 2 0113 0 7908
2679 426 7 3667 0
. . . ( ) ( . . . )
.
b g

(2)
Solving (1) and (2) simultaneously yields mv = 0 302
. kg , ml = 6 698
. kg
Liquid volume kg L kg L liquid
= =
6 698 1046 7 01
. . .
b gb g
Vapor volume .302 kg L kg 476 L vapor
= =
0 1576
b gb g
Piston displacement:
L 10 cm 1
1 L 400 cm
3 3
2
Δ
Δ
z
V
A
= =
+ −
=
7 01 476 7 0 1046
1190 cm
. . .
b gb g
(c) Tupper ⇒ All 3310 kJ go into the block before a measurable amount is transferred to the
water. Then ΔU Q T T
AL u u
= ⇒ − = ⇒ = °
30 094 20 3310 1194
. .
kg kJ kg C
b g b g if melting is
neglected. In fact, the bar would melt at 660o
C.

7-28
7.50 100
. ),
L H O( 25 C
m (kg)
2
o
v1
v
4 00
. ),
L H O( 25 C
m (kg)
2
o
L1
l
m [kg H O( ]
= m m
v2 2
v1 e
v)
+
m [kg H O( ]
= m m
L2 2
L1 e
l)
+
Q=2915 kJ
U
V
W =
Assume not all the liquid
is vaporized. Eq. at
kg H O vaporized.
2
T P m
f f e
, .
Initial conditions: Table B.5 kJ kg
⇒ =
.
U L1 104 8 , .
VL1 1003
= L kg P = 0 0317
. bar
T = °
25 C, sat'd ⇒ .
Uv1 2409 9
= kJ kg , ,
VL1 43 400
= L kg
mv1
5
43400 2 304 10
= = × −
1.00 l l kg kg
b g b g . , mLI = =
4 00 1 3988
. .
l .003 l kg kg
b g b g
Energy balance:
ΔU Q m U T m U T
e v f e L f
= ⇒ × + + − − ×
− −
2 304 10 3988 2 304 10 2409 9
5 5
. . . .
d i d i b g d i d ib g
− =
3988 104 8 2915
. ( . )
b g kJ
⇒ 2 304 10 3988 3333
5
. .
× + + − =
−
E
m U T m U T
e v f e v f
d i d i b g d i
⇒ =
− × −
−
−
m
U U
U U
e
v L
v L
3333 2 304 10 3988
5
. .

d i (1)
V V V m V T m V T
m
V V
V V
L v e L f e L f
e
v L
v L
+ = ⇒ × +
F
H
GG
I
K
JJ + − =
⇒ =
− × −
−
−
A A
−
tan . . .
. . .

k
kg liters kg
L
2 304 10 3988 500
500 2 304 10 3988
5
5
d i b g d i
d i b g
2
1 2
3333 2 304 10 3988
500 2 304 10 3988
0
5
5
b g b g d i d i d i d i
d i
− ⇒ =
− × −
−
−
− × −
−
=
−
−
f T
U T U T
U U
V V
V V
f
v f L f
v L
v L
v L
. .

. . .

Procedure: Assume
Table 8.5
T U U V V f T
f v L v L f
⇒ ⇒
, , , d i Find Tf such that f Tf
d i= 0
T U U V V f
T P
f v L v L
f f

. . . . . .
. . . . . .
. . . . . .
. . . . . . . .
2014 25938 856 7 1237 1159 512 10
198 3 2592 4 842 9 1317 1154 193 10
1950 25908 8285 140 7 1149 134 10
196 4 25915 834 6 136 9 1151 4 03 10 196 4 14 4
2
2
2
4
− ×
− ×
×
− × ⇒ ≅ ° =
−
−
−
−
C, bars
or Eq 2
Eq 1
kg 2.6 g evaporated
b g
b g
me = × ⇒
−
2 6 10 3
.

7-29
7.51. Basis: 1 mol feed
(1 – )
B = benzene
z
T = toluene
1 mol @ 130°C
B (mol B(l)/mol)
z B (mol T(l)/mol)
(1 – )
nV
y
(mol vapor)
B(mol B(v)/mol)
yB (mol T(v)/mol)
(1 – )
nL
x
(mol liquid)
B(mol B(l)/mol)
xB (mol T(l)/mol)
in equilibrium
at T(°C), P(mm Hg)
(a) 7 variables: ( , , , ,
n y n x Q T P
V B L B , , )
–2 equilibrium equations
–2 material balances
–1 energy balance
2 degrees of freedom. If T and P are fixed,we can calculate and .
n y n x Q
V B L B
, , , ,
(b) Mass balance: n n n n
V L V
+ = ⇒ = −
1 1 2 (1)
Benzene balance: z n y n x
B V B L B
= + (2)
C H :
6 6 l T H
b g d i
= =
0 0
, , T H H T
BL
= = ⇒ =
80 1085 01356
, . .
d i (3)
C H :
6 6 v T H
b g d i
= =
80 4161
, . , T H H T
BV
= = ⇒ = +
120 4579 01045 3325
, . . .
d i (4)
C H
7 8 l T H
b g d i
: ,
= =
0 0 , T H H T
TL
= = ⇒ =
111 1858 01674
, . .
d i (5)
C H
7 8 v T H
b g d i
: , .
= =
89 4918 , T H H T
TV
= = ⇒ = +
111 52 05 01304 37 57
, . . .
d i (6)
Δ Δ
E W E
p s k
, ,
= 0
Q H n y H n y H n x H n x H z H T
z H T
V B BV V B TV L B BL L B TL B BL F
B TL F
= = + − + + − −
− −
Δ

1 1 1
1 1
b g b g b g b g
b gb g b g (7)
Raoult's Law:
(1-
y P x p
y P x p
B B B
B B T
=
= −
*
*
) ( )
1
(8)
(9)
Antoine Equation. For T= 90°C and P=652 mmHg:
* [6.89272 1203.531/(90 219.888)]
* [6.95805 1346.773/(90 219.693)]
(90 C) 10 1021 mmHg
(90 C) 10 406.7 mmHg
o
B
o
T
p
p
− +
− +
= =
= =
Adding equations (8) and (9) ⇒
P x p x p x
P p
p p
P p
p p
l
y
x p
P
v
B B B T B
T
B T
T
B T
B
B B
= − ⇒ =
−
−
=
−
−
=
−
=
= = =
* *
*
* *
*
* *
*
( )
.
.
. ( )
.
+
1021-406.7
mol B( ) / mol
mmHg
mmHg
mol B( ) / mol
1
652 4067
0399
0399 1021
652
0625
Solving (1) and (2) ⇒ =
−
−
=
−
−
=
= − = − =
mol vapor
mol liquid
n
z x
y x
n n
V
B B
B B
L V
05 0 399
0 625 0 399
0 446
1 1 0 446 0554
. .
. .
.
. .

7-30
7.51 (cont’d)
Substituting (3), (4), (5), and (6) in (7) ⇒
Q
Q
= + + − +
+ + − −
− ⇒ =
0 446 0 625 01045 90 3325 0 446 1 0 625 01304 90 37 57
0554 0 399 01356 90 0554 1 0 399 01674 90 05 01356 130
05 01674 130 814
. ( . )[ . ( ) . ] . ( . )[ . ( ) . ]
. ( . )[ . ( )] . ( . )[ . ( )] . [ . ( )]
. [ . ( )] . kJ / mol
(c). If PPmin, all the output is vapor. If PPmax, all the output is liquid.
(d) At P=652 mmHg it is necessary to add heat to achieve the equilibrium and at P=714
mmHg, it is necessary to release heat to achieve the equilibrium. The higher the pressure,
there is more liquid than vapor, and the liquid has a lower enthalpy than the equilibrium
vapor: enthalpy out enthalpy in.
zB T P pB pT xB yB nV nL Q
0.5 90 652 1021 406.7 0.399 0.625 0.446 0.554 8.14
0.5 90 714 1021 406.7 0.500 0.715 -0.001 1.001 -6.09
0.5 90 582 1021 406.7 0.285 0.500 0.998 0.002 26.20
0.5 90 590 1021 406.7 0.298 0.516 0.925 0.075 23.8
0.5 90 600 1021 406.7 0.315 0.535 0.840 0.160 21.0
0.5 90 610 1021 406.7 0.331 0.554 0.758 0.242 18.3
0.5 90 620 1021 406.7 0.347 0.572 0.680 0.320 15.8
0.5 90 630 1021 406.7 0.364 0.589 0.605 0.395 13.3
0.5 90 640 1021 406.7 0.380 0.606 0.532 0.468 10.9
0.5 90 650 1021 406.7 0.396 0.622 0.460 0.540 8.60
0.5 90 660 1021 406.7 0.412 0.638 0.389 0.611 6.31
0.5 90 670 1021 406.7 0.429 0.653 0.318 0.682 4.04
0.5 90 680 1021 406.7 0.445 0.668 0.247 0.753 1.78
0.5 90 690 1021 406.7 0.461 0.682 0.176 0.824 -0.50
0.5 90 700 1021 406.7 0.477 0.696 0.103 0.897 -2.80
0.5 90 710 1021 406.7 0.494 0.710 0.029 0.971 -5.14
(e). P = 714 mmHg, P = 582 mmHg
max min
nV vs. P
0
0.2
0.4
0.6
0.8
1
582 632 682 732
P (mm Hg)
n
V
n P
V = ≅
05 640
. @ mmHg

7-31
7.52 (a). Bernoulli equation:
Δ Δ
Δ
P u
g z
ρ
+ + =
2
2
0
ΔP
ρ
=
× − ×
×
= −
−
0 977 10 15 10
46 7
5 5 2
. .
.
d iPa 1 N / m m
Pa 1.12 10 kg
m
s
3
3
2
2
g z
Δ = =
(9.8066 m / s m m s
2 2 2
) . /
6 588
b g
Bernoulli m / s m / s
2 2 2 2
⇒ = − ⇒ = + −
Δu
u u
2
2
2
1
2
2
46 7 588 2 121
. . .
b g d i
= − = ⇒ =
500 2 121 0800 0894
2 2 2 2
2
. / ( )( . ) / . / .
b g m s m s m s m / s
2 2 2
u
(b). Since the fluid is incompressible,
V m s d u d u
3
1
2
1 2
2
2
4 4
d i= =
π π
⇒ d d
u
u
1 2
2
1
6
0894
500
2 54
= = =
cm
m s
m s
cm
b g .
.
.
7.53 (a).
V A u A u u u
A
A
u u
A A
m s m m s m m s
3 2 2
d i d i b g d i b g
= = ⇒ = =
=
1 1 2 2 2 1
1
2
4
2 1
1 2
4
(b). Bernoulli equation (Δz = 0)
Δ Δ
Δ
P u
P P P
u u
ρ
ρ
+ = ⇒ = − = −
−
2
2 1
2
2
1
2
2
0
2
d i
Multiply both sides by
Substitute
Multiply top and bottom of right - hand side by
−
=
1
16
2
2
1
2
1
2
u u
A
note
V A u
2
1
2
1
2
=
P P
V
A
1 2
2
1
2
15
2
− =
ρ
(c) P P gh
V
A
V
A gh
1 2
2
1
2
2 1
2
15
2
2
15
1
− = − = ⇒ = −
F
HG
I
KJ
ρ ρ
ρ ρ
ρ
Hg H O
H O Hg
H O
2
2
2
d i

.
.
V 2
2 2
4
3
2 75 1
1955 10
=
−
= × −
πb g b g
cm 1 m 9.8066 m 38 cm 1 m 13.6
15 10 cm s 10 cm
m
s
4
8 4 2 2
6
2
⇒ = =
.
V 0 044 44
m s L s
3

7-32
7.54 (a). Point 1- surface of fluid . P1 31
= . bar , z1 7
= + m , u1 0
= m s
b g
Point 2 - discharge pipe outlet . P2
1
1
=
=
atm
bar
.013
b g
, z2 0
= m
b g, u2 = ?
Δρ
ρ
=
−
⋅ ×
= −
1013 31
2635
. .
.
b gbar 10 N 1 m
m bar 0.792 10 kg
m s
5 3
2 3
2 2
g z
Δ =
−
= −
9.8066 m m
s
68.6 m s
2
2 2
7
Bernoulli equation m s m s
2 2 2 2
⇒ = − − = + =
Δ Δ
Δ
u P
g z
2
2
2635 68 6 3321
ρ
. . .
b g
Δu u
2
2
2 2
0
= −
u u
2
2
2
2 3321 664 2 258
= = ⇒ =
( . ) . .
m s m s m / s
2 2 2 2
( . )
V = =
π 100 2580
122
2
cm cm 1 L 60 s
4 1 s 10 cm 1 min
L / min
2
3 3
(b) The friction loss term of Eq. (7.7-2), which was dropped to derive the Bernoulli equation,
becomes increasingly significant as the valve is closed.
7.55 Point 1- surface of lake . P1 1
= atm , z1 0
= , u1 0
=
Point 2 - pipe outlet . P2 1
= atm , z z
2 = ft
b g
u
V
A
2 2 2
95
05 1049
353
= =
×
=

. .
.
gal 1 ft 1 144 in 1 min
min 7.4805 gal in 1 ft 60 s
ft s
3 2
2
πb g
Pressure drop: Δ P P P
ρ = =
0 1 2
b g
L
Z
z
F z z
=
°
=
F
HG I
KJ
= ⋅ = ⋅
sin
. . )
30
2
0 041 2 0 0822
Friction loss: ft lb lb (ft lb lb
f m f m
b g
Shaft work:
-8 hp 0.7376 ft lb s 7.4805 gal 1 ft 60 s
1.341 10 hp gal 1 ft 62.4 lb 1 min
ft lb lb
f
3
3
m
f m

/ min
W
m
s
=
⋅
×
= − ⋅
−
1
95
333
3
Kinetic energy:
ft 1 lb
2 s lb ft / s
ft lb lb
2
f
2
m
2
f m
Δ u2
2 2
2
353 0
32174
19 4
=
−
⋅
= ⋅
.
.
.
b g
Potential energy: g
ft ft lb
s lb ft / s
ft lb lb
f
2
m
2 f m
Δz
z
z
=
⋅
= ⋅
32174 1
32174
.
.
b g b g
Eq. 7.7 - 2 ft
b g⇒ + + + =
−
⇒ + + = ⇒ =
Δ Δ
Δ
P u
g z F
W
m
z z z
s
ρ
2
2
19 4 0 082 333 290

. .

7-33
7.56 Point 1- surface of reservoir . P1 1
= atm (assume), u1 0
= , z1 60
= m
Point 2 - discharge pipe outlet . P2 1
= atm (assume), u2 = ?, z2 0
=
ΔP ρ = 0
Δu u V A V
V
2
2
2
2 2 6 4
2 2
4
2
2 2 2
1
35 1
3376
= = =
⋅
= ⋅
/ )
.
d h
b g
b g
(m s 10 cm 1 N
(2) cm 1 m kg m / s
N m kg
2 8
4 2
π
g z
Δ =
−
⋅
= − ⋅
9.8066 m m N
s 1 kg m / s
637 N m kg
2 2
65 1

.

W
m V
V
s
=
× ⋅
= ⋅
080 10
800
6
W 1 N m / s s 1 m
W m 1000 kg
N m kg
3
3
d i b g
Mechanical energy balance: neglect Eq. 7.7 - 2
F b g
Δ Δ
Δ
P u
g z
W
m
V
V
V
s
T E
ρ
+ + =
−
⇒ − = − = =
⇒
+
2
2
2
3376 637
800 127
76 2

.

.
.
m 60 s
s 1 min
m min
3
3
Include friction (add F 0 to left side of equation) ⇒
V increases.
7.57 (a). Point 1: Surface at fluid in storage tank, P1 1
= atm , u1 0
= , z H
1 = m
b g
Point 2 (just within pipe): Entrance to washing machine. P2 1
= atm , z2 0
=
u2 2
4 0 4
7 96
= =
600 L 10 cm 1 min 1 m
cm 1 L 60 s 100 cm
m s
3 3
min .
.
πb g
ΔP
ρ
= 0 ;
Δu u
2
2
2 2
2 2
7 96 1
2
317
= =
⋅
=
.
.
m s J
1 kg m / s
J / kg
2 2
b g
g z
H
H
Δ =
−
⋅
= −
9.807 m m J
s 1 kg m / s
(J / kg)
2 2 2
0 1
9 807
b g
c h .
Bernoulli Equation: m
Δ Δ
Δ
P u
g z H
ρ
+ + = ⇒ =
2
2
0 323
.
(b). Point 1: Fluid in washing machine. P1 1
= atm , u1 0
≈ , z1 0
=
Point 2: Entrance to storage tank (within pipe). P2 1
= atm , u2 7 96
= . m s , z2 323
= . m
ΔP
ρ
= 0 ;
Δu2
2
317
= .
J
kg
; g z
Δ = − =
9 807 323 0 317
. . .
b g J
kg
; F = 72
J
kg
Mechanical energy balance:
W m
P u
g z F
s = − + + +
L
NM O
QP
Δ Δ
Δ
ρ
2
2
⇒ = − = −
.
min
.
Ws
600 L kg 1 min 31.7 + 31.7 + 72 J 1 kW
L 60 s kg J s
1 kW
0 96
10
30
3
b g
(work applied to the system)
Rated Power kW . 1.7 kW
= =
130 0 75
.

7-34
7.58 Basis: 1000 liters of 95% solution . Assume volume additivity.
Eq. 6.1-1
Density of 95% solution
l
kg
kg liter
b g
:
.
.
.
.
. .
1 0 95
126
0 05
100
0804 124
ρ ρ
ρ
= = + = ⇒ =
∑
xi
i
Density of 35% solution
l
kg
kg liter
:
.
.
.
.
. .
1 0 35
126
0 65
100
0 9278 108
ρ
ρ
= + = ⇒ =
Mass of 95% solution:
liters 1.24 kg
liter
kg
1000
1240
=
G = glycerol
m
W = water
1240 kg (1000 L)
1 (kg)
0.95 G
0.05 W
0.35 G
0.65 W
23 m
m2 (kg)
0.60 G
0.40 W
5 cm I.D.
Mass balance: 1240
Glycerol balance:
kg 35% solution
kg 60% solution
+ =
+ =
U
V
W
⇒
=
=
m m
m m
m
m
1 2
1 2
1
2
0 95 1240 0 35 0 60
1740
2980
. . .
Volume of 35% solution added
kg 1 L
1.08 kg
L
= =
1740
1610
⇒ = + =
Final solution volume L L
1000 1610 2610
b g
Point 1. Surface of fluid in 35% solution storage tank. P1 1
= atm , u1 0
= , z1 0
=
Point 2. Exit from discharge pipe. P2 1
= atm , z2 23
= m
u2 2
1
2 5
1051
= =
1610 L 1 m 1 min 10 cm
13 min 10 L 60 s cm 1 m
m s
3 4 2
3 2 2
π .
.
b g
Δ P ρ = 0 ,
Δ Δ
u u
2
2
2 2
2 2
1051
1
= =
⋅
= ⋅
. /
b g m s 1 N
(2) kg m / s
0.552 N m kg
2 2
2
g z
Δ =
⋅
= ⋅
9.8066 m 23 m 1 N
s 1 kg m / s
225.6 N m kg
2 2 , F = = ⋅
50 J kg 50 N m kg
Mass flow rate:
kg 1 min
13 min 60 s
kg s
.
m = =
1740
2 23
Mechanical energy balance Eq. 7.7 - 2
b g

.
.
W m
P u
g z F
s = − + + +
L
NM O
QP= −
⋅
⋅
= − ⇒
Δ Δ
Δ
ρ
2
2
2 23
0 62
kg 0.552 + 225.6 + 50 N m 1 J 1 kW
s kg 1 N m 10 J s
kW 0.62 kW delivered to fluid by pump.
3
b g

8-1
CHAPTER EIGHT
8.1 a. ( ) . .
U T T T
= +
2596 0 02134 2
J / mol
( ) ( )
U U Tref
0 0 100 2809 0
o o o o
C J / mol C J / mol C (since U(0 C) = 0)
= = =
b. We can never know the true internal energy. ( )
U 100o
C is just the change from ( )
U 0o
C to
( )
U 100o
C .
c. Q W U E E
k p
− = + +
Δ Δ Δ
Δ Δ
E E W
k p
= = =
0 0 0
, ,
Q U
= = − = ⇒
Δ ( . )[( )
30 2809 0 8428 8400
mol J / mol] J J
d. C
U
T
dU
dT
T
v
V
=
∂
∂
F
HG
I
KJ = = + ⋅

[ . . ]

2596 0 04268 J / (mol C)
o
Δ
Δ Δ
( ) ( . . ) . .
( . (
( . [ . ( ) . ( )]
U C T dT T dT T
T
U U
v
T
T
= = + = +
O
QPP
F
H
GG
I
K
JJ
= ⋅
= ⋅ − + − = ⇒
z z
1
2
2596 0 04268 2596 0 04268
2
30
30 2596 100 0 0 02134 100 0 8428 8400
0
100 2
0
100
2
J / mol
mol) J / mol)
mol) (J / mol) J J
8.2 a. C C R C T
v p v
= − ⇒ = + ⋅° − ⋅ °
353 0 0291 8 314
. . [ .
b g b gb g
J / (mol C)] [J / (mol K)] 1 K 1 C
⇒ = + ⋅°
C T
v 27 0 0 0291
. . [J / (mol C)]
b. ]
100
100 2
100
25
25 25
ˆ 35.3 0.0291 2784 J mol
2
p
T
H C dT T
⎤
Δ = = + =
⎥
⎦
∫
c. Δ Δ Δ
.
U C dT C dT RdT H R T
v p
= = − = − = − − =
z z z
25
100
25
100
25
100
2784 8 314 100 25 2160
b gb g J mol
d.
H is a state property
8.3 a. C T T
v [ ] . . .
kJ / (mol C)
o
⋅ = + × − ×
− −
0 0252 1547 10 3012 10
5 9 2
atm L
atm L / (mol K) K
mol
mol) kJ / mol kJ
) kJ
kJ
n
PV
RT
Q n U dT
Q n U T dT
Q n U T T dT
= =
⋅ ⋅
=
= = ⋅ =
= = ⋅ + × =
= = ⋅ + × − × =
z
z
z
−
− −
( . )( . )
( . [ ]( )
.
( . . ( ) .
( . [ . . ] .
( . ) [ . . . ] .
2 00 300
0 08206 298
0 245
0 245 0 0252 6 02
0 245 0 0252 1547 10 7 91
0 245 0 0252 1547 10 3012 10 7 67
1 1
25
1000
2 2
5
25
1000
3 3
5 9 2
25
1000
Δ
Δ
Δ
% .
error in =
6.02 -7.67
7.67
Q1 100% 215%
× = −
% .
error in =
7.91-7.67
7.67
Q2 100% 313%
× =

8-2
8.3 (cont’d)
b. C C R
p v
= +
C T T
T T
p [ ] ( . . . ) .
. . .
kJ / (mol C)
o
⋅ = + × − × +
= + × − ×
− −
− −
0 0252 1547 10 3012 10 0 008314
0 0335 1547 10 3012 10
5 9 2
5 9 2
Q H n C dT
T T dT
P
T
T
= =
= ⋅ + × − × ⋅ = ×
z
z − −
Δ
1
2
0 245 0 0335 1547 10 3012 10 9 65
5 9 2
25
1000
mol [kJ / (mol C)] 10 J
Piston moves upward (gas expands).
o 3
( . ) [ . . . ] .
c. The difference is the work done on the piston by the gas in the constant pressure process.
8.4 a. ( ) ( )
( ) ( )
6 6
o 5
C H
40 C 0.1265 23.4 10 40 0.1360 [kJ/(mol K)]
p l
C −
= + × = ⋅
b. Cp v
d i b g b g b g b g
b g
C H
o
6 6
C
[kJ / (mol C)
40 0 07406 32 95 10 40 2520 10 40 77 57 10 40
0 08684
5 8 2 12 3
° = + × − × + ×
= ⋅
− − −
. . . .
. ]
c. Cp s
d i b g b g b g
b g
C
313 K 0.009615 kJ / (mol K)]
= + × − × = ⋅
− −
0 01118 1095 10 313 4 891 10 313
5 2 2
. . . [
d. Δ .
. . .
.
H T T T T
v
C H
6 6
3
kJ mol
b g = +
×
−
×
+
× O
QPP =
− − −
007406
32 95 10
2
2520 10 7757 10
4
3171
5
2
8
3
12
4
40
300
e. Δ .
.
. .
H T T T
C s
b g = +
×
+ ×
O
QPP =
−
−
0 01118
1095 10
2
4 891 10 3459
5
2 2 1
313
573
kJ / mol
8.5 H O (v, 100 C, 1 atm) H O (v, 350 C, 100 bar)
2
o
2
o
→
a.
H = − =
2926 2676 250
kJ kg kJ kg kJ kg
b. . . . .
.
H T T T dT
= + × + × − ×
= ⇒
− − −
z0 03346 0 6886 10 0 7604 10 3593 10
8845
5 8 2 12 3
100
350
kJ mol 491.4 kJ kg
Difference results from assumption in (b) that
H is independent of P. The numerical difference
is Δ
H for H O v, 350 C, 1 atm H O v, 350 C, 100 bar
2 2
° → °
b g b g
8.6 b. Cp
d in C H (l)
o
6 14
kJ / (mol C)
−
= ⋅
0 2163
. ⇒ Δ [ . ] .
H dT
= =
z0 2163 1190
25
80
kJ / mol
The specific enthalpy of liquid n-hexane at 80o
C relative to liquid n-hexane at 25o
C is 11.90 kJ/mol
c. C T T T
p
d in C H (v)
o
6 14
kJ / (mol C)
−
− − −
⋅ = + × − × + ×
[ ] . . . .
013744 4085 10 2392 10 57 66 10
5 8 2 12 3
Δ [ . . . . ] .
H T T T dT
= + × − × + × −
− − −
z013744 4085 10 2392 10 57 66 10 110 7
5 8 2 12 3
500
0
= kJ / mol
The specific enthalpy of hexane vapor at 500o
C relative to hexane vapor at 0o
C is 110.7 kJ/mol. The
specific enthalpy of hexane vapor at 0o
C relative to hexane vapor at 500o
C is –110.7 kJ/mol.

8-3
8.7 T T T
° = ′ ° − = ′ ° −
C F F
b g b g b g
1
18
32 05556 17 78
.
. .
C T T
p cal mol C F F
⋅° = + ′ ° − = + ′ °
b g b g b g
6890 0 001436 05556 17 78 6864 0 0007978
. . . . . .
′ ⋅° =
°
⋅° °
=
E
C C C
p p p
Btu lb - mole F
cal 453.6 mol 1 Btu 1 C
mol C 1 lb - mole 252 cal 1.8 F
drop primes
b g b g
100
.
C T
p Btu lb - mole F F
⋅° = + °
b g b g
6864 0 0007978
. .
8.8 C T T T
p
d i b g b g
CH CH OH(l)
o
3 2 100
[kJ / (mol C)]
= +
−
= + ⋅
01031
01588 01031
01031 0 000557
.
. .
. .
Q H T T
= = +
F
HG O
QP
×
Δ
550 789
01031
0 000557
2
2
20
78 5
.
.
.
.
L
s
g
1 L
1 mol
46.07 g
= 941.9 7.636 kJ / s = 7193 kW
kJ mol

8.9 a.
, . . . .
,
Q H T T T dT
= = ⋅ + × − × + ×
=
− − −
z
Δ 5 000 0 03360 1367 10 1607 10 6 473 10
17 650
5 8 2 12 3
100
200
mol s
kW
kJ mol
b g

b. Q U H PV H nR T
= = − = − = − ⋅ ⋅ ⋅
=
Δ Δ Δ Δ Δ 17 650 50
13 490
, . ]
,
kJ kmol 8.314 [kJ / (kmol K) 100 K
kJ
b g b g b g
The difference is the flow work done on the gas in the continuous system.
c. Qadditional = heat needed to raise temperature of vessel wall + heat that escapes from wall to
surroundings.
8.10 a. C C
p p
is a constant, i.e. is independent of T.
b. Q mC T C
Q
m T
p p
= ⇒ =
Δ
Δ
C
Q
m T
C
p
p
= = ⋅
⇒ = ⋅ = ⋅
Δ
(16.73-6.14) kJ 1 L 86.17 g 10 J
(2.00 L)(3.10 K) 659 g 1 mol 1 kJ
= 0.223 kJ / (mol K)
Table B.2 kJ / (mol C) kJ / (mol K)
3
o
0 216 0 216
. .
8.11

H U PV H U RT
H
T
U
T
R C
U
T
R
PV RT T
p p
p
p
P
= + ===== = + =====
∂
∂
=
∂
∂
+ ⇒ =
∂
∂
+
= ∂ ∂
F
HG I
KJ F
HG I
KJ F
HG I
KJ
a f
But since
U depends only on T,
∂
∂
F
HG
I
KJ = =
∂
∂
F
HG
I
KJ ≡ ⇒ = +

U
T
dU
dT
U
T
C C C R
p V
v p v

8-4
8.12 a. Cp
d iH O(l)
o
2
kJ / (kmol C)
= ⋅
754
. =75.4 kJ/(kmol.o
C) V = 1230 L ,
n
V
M
= = =
ρ 1230 1 1
68 3
L kg
1 L
kmol
18 kg
kmol
.
. . ( )
.
`
Q
Q
t
n C dT
t
p
T
T
= =
⋅
=
⋅
−
=
zd iH O(l)
o
o
2
kmol kJ
kmol C
C
8 h
h
3600 s
kW
2
68 3 754 40 29 1
1967
b.
Q Q Q
total to the surroundings to water
= + , .
Qto the surroundings kW
= 1967
. .
.
( )
Q
Q
t
n C dT
t
P H O
to water
to water
o o
kmol
3 h
kJ / (kmol C)
3600 s/ h
C
kW
= =
⋅
=
⋅
=
z 2
29
40
68 3 754 11
5245
. .
Q E
total total
kW kW 3 h = 21.64 kW h
= ⇒ = × ⋅
7 212 7 212
c. Costheating up from 29 C to 40 C
o o 21.64 kW h $0 / (kW h) = $2.16
= ⋅ × ⋅
.10
keeping temperature constant for 13 h
total
1.967 kW 13 h $0.10/(kW h)=$2.56
$2.16 $2.56 $4.72
Cost
Cost
= × × ⋅
= + =
d. If the lid is removed, more heat will be transferred into the surroundings and lost, resulting in higher
cost.
8.13 a. Δ . .
H H H
N (25 C) N (700 C) N (700 C) N (25 C)
2
o
2
o
2
o
2
o kJ mol
→
= − = − =
2059 0 2059
b g
b. Δ
H H H
H (800 F) H (77 F) H (77 F) H (800 F)
2
o
2
o
2
o
2
o Btu / lb - mol
→
= − = − = −
0 5021 5021
b g
c. Δ . . .
H H H
CO (300 C) CO (1250 C) CO (1250 C) CO (300 C)
2
o
2
o
2
o
2
o kJ mol
→
= − = − =
6306 1158 5148
b g
d. Δ
H H H
O (970 F) O (0 F) O (0 F) O (970 F)
2
o
2
o
2
o
2
o Btu / lb - mol
→
= − = − − = −
539 6774 7313
b g
8.14 a.
m = 300 kg / min .
n = =
300 1 1000 1
1785
kg
min
min
60 s
g
1 kg
mol
28.01 g
mol / s

( . [ . . . . ]
( . ) .
Q n H n C dT
T T T dT
p
T
T
= ⋅ = ⋅
= ⋅ + × + × − ×
= − −
z
z − − −
Δ
1
2
1785 0 02895 0 411 10 0 3548 10 2 22 10
1785 12 076
5 8 2 12 3
450
50
mol / s) [kJ / mol]
mol / s [kJ / mol] = 2,156 kW
b g
b.
( (
Q n H n H H
= ⋅ = ⋅ −
Δ 50 450
o o
C) C)
mol / s)(0.73-12.815[kJ / mol]) = kW
= −
( . ,
1785 2 157
8.15 a.
n = 250 mol / h
i) ( ) .
Q n H
= =
−
= −
Δ
250 2676 3697 1 1 18 02
mol
h
kJ
1 kg
kg
1000 g
h
3600 s
g
1 mol
1.278 kW
ii)

[ . . . . ]
Q n H n C dT
T T T
p
T
T
= = ⋅
+ × + × − × = −
z
z − − −
Δ
1
2
250 1
003346 06880 10 07604 10 3593 10
5 8 2 12 3
600
100
=
mol
h
h
3600 s
1.274 kW

8-5
8.15 (cont’d)
iii) . . .
Q = ⋅ − = −
250
2 54 20 91 1276
mol
3600 s
[kJ / mol] kW
b g
b. Method (i) is most accurate since it takes into account the dependence of enthalpy on pressure and
(ii) and (iii) do not.
c. The enthalpy change for steam going from 10 bar to 1 atm at 600o
C.
8.16 Assume ideal gas behavior, so that pressure changes do not affect Δ
H .

.
.
n
R
R
= =
200 492 12 1
0 6125
ft
h 537
atm
1 atm
lb - mol
359 ft (STP)
lb - mole / h
3 o
o 3
( . ( )
Q n H
= = ⋅ − =
Δ 0 6125 2993 0 1833
lb - mole
h
) [Btu / lb - mole] Btu / h
b g
8.17 a.
50 kg 1.14 kJ 50 C
kg C
kJ
− °
⋅°
=
10
2280
b g
b.
( ) ( ) ( ) ( ) ( ) ( )
2 3
Na CO Na C O
2 3 2 0.026 0.0075 3 0.017 0.1105 kJ mol C
p p p p
C C C C
≈ + + = + + = ⋅°
50,000 g 0.1105 kJ 1 mol 50 C
mol C 105.99 g
2085 kJ
% error error
− °
⋅°
=
=
−
× = −
10
2085 2280
2280
100% 8 6%
b g
.
8.18 Cp
d i b g b g b g
C H O(l)
o
6 14
kJ / (mol C)
= + + = ⋅
6 0 012 14 0 018 1 0 025 0 349
. . . . (Kopp’s Rule)
Cp
d iCH COCH (l)
3 3
T kJ (mol C
= + × ⋅°
−
01230 18 6 10 5
. . )
Assume ΔHmix ≅ 0
↓ CH COCH
3 3 ↓ C H O
6 14
( ) ( )
5
7
20
7
45
0.30 0.1230+18.6 10 kJ 0.70 0.349 kJ
1 mol 1 mol
mol C 58.08 g mol C 102.17 g
[0.003026 9.607 10 T] kJ (g C)
ˆ [0.003026 9.607 10 T] 0.07643 kJ g
pm
T
C
H dT
−
−
−
×
= +
⋅° ⋅°
= + × ⋅°
Δ = + × = −
∫
8.19 Assume ideal gas behavior, ΔHmix ≅ 0
Mw = + =
1
3
16 04
2
3
32 00 26 68
. . .
b g b g g
mol
Δ Δ
. .
H C dT H C dT
p p
O O CH CH
2
2
4
4
kJ / mol, kJ / mol
= = = =
z z
d i d i
25
350
25
350
10 08 14 49
. .
H = +
L
NM O
QP
F
HG I
KJF
HG I
KJ =
1
3
14 49
2
3
10 08
1000 1
433
kJ / mol kJ / mol
g
1 kg
mol
26.68 g
kJ kg
b g b g

8-6
8.20 n = = =
1000 m 1 min 273 K 1 kmol
min 60 s 303 K 22.4 m STP
kmol s mol / s
3
3
b g 0 6704 670 4
. .
Energy balance on air:
Q H n H Q
H
= = = =
Δ Δ
Δ 670.4 mol 0.73 kJ 1 kW
s mol 1 kJ s
kW
Table B.8 for
.
489 4
Solar energy required = =
489.4 kW heating 1 kW solar energy
0.3 kW heating
kW
1631
Area required
2
2
1631 kW 1000 W 1 m
1813 m
1 kW 900 W
= =
8.21 C H 5O 3CO 4H O
3 8 2 2 2
+ → +

.

.
.
n
n
fuel 3
air
2
3 8 2
SCFH
h
lb - mol
359 ft
lb - mol
h
lb mol
h
lb - mol O
1b - mol C H
1 lb - mol air
0.211b - mol O
lb mol
h
=
×
=
=
−
= ×
−
135 10 1
376
376 5 115
103 10
5
4
2
1
302
4 5 8 2 12 3
0
4 -1
7
= =
lb mol
= 1.03 10 [0.02894 0.4147 10 0.3191 10 1.965 10 ]
h
1.03 10 lb-mol 8.954 kJ 453.593 mol 9.486 10 Btu
= =3.97 10 Btu/h
h mol lb-mol kJ
T
p
T
Q H n C dT
T T T dT
− − −
Δ ⋅
−
⎛ ⎞
× ⋅ + × + × − ×
⎜ ⎟
⎝ ⎠
× ×
×
∫
∫

8.22 a. Basis: 100 mol feed (95 mol CH4 and 5 mol C2H6)
CH O CO 2H O C H
7
2
O 2CO 3H O
4 2 2 2 2 6 2 2 2
+ → + + → +
2
nO
4 2
4
2 6 2
2 6
2
2
mol CH mol O
1 mol CH
mol C H mol O
1 mol C H
mol O
= ⋅ +
L
N
MM
O
Q
PP=
125
95 2 5 35
259 4
.
.
.
Product Gas:
2 2 2 2
2 2 2 2
CO : 95(1)+5(2)=105 mol CO H O: 95(2)+5(3)=205 mol H O
O : 259.4-95(2)-5(3.5)=51.9 mol O N : 3.76(259.4)=975 mol N
Energy balance (enthalpies from Table B.8)
Δ
Δ
Δ
Δ
Δ

. .
. .
. .
H H H 18.845 42.94 24.09 kJ / mol
H H H 18.20 kJ / mol
Q = H 105(-24.09) 205(-18.20) 51.9(-15.51) 975(-14.49)
Q 21,200 kJ /100 mol feed
CO (CO , 450 C) (CO , 900 C)
H O (H O, 450 C) (H O, 900 C)
O (O , 450 C) (O , 900 C)
N (N , 450 C) (N , 900 C)
2 2
o
2
o
2 2
o
2
o
2 2
o
2
o
2 2
o
2
o
= − = − = −
= − = − = −
= − = − = −
= − = − = −
= + + +
=
1512 3332
13375 2889
12 695 2719
b. From Table B.5:
H (40 C) 167.5 kJ / kg; H (50 bars) 2794.2 kJ / kg;
liq
o
vap
= =
Q = n H = n(2794.2 -167.5) = 21200 n = 8.07 kg /100 mol feed
⋅ ⇒
Δ

8-7
8.22 (cont’d)
c. From part (b), 8.07 kg steam is produced per 100 mol feed

.
.
n feed = = × −
1250 01 1
4 30 10 3
kg steam
h
kmol feed
8.07 kg steam
h
3600 s
kmol / s
. .
.
Vproduct gas
3
5
3
mol feed
s
mol product gas
100 mol feed
8.314 Pa m
mol K
723 K
1.01325 10 Pa
m / s
=
⋅
⋅ ×
=
4 30 1336 9
341
d. Steam produced from the waste heat boiler is used for heating, power generation, or
process application. Without the waste heat boiler, the steam required will have to be
produced with additional cost to the plant.
8.23 Assume Δ Δ Δ Δ
H H H H
mix C H O C H
≅ ⇒ = +
0 10 12 2 6 6
Kopp’s rule: Cp C H O
d i e j e j
10 12 2
10 12 12 18 2 25 386 2 35
= + + = ⋅ = ⋅
( ) ( ) ( ) .
J mol C J g C
o o
Δ
Δ
Δ
H
H
H
C H O
C H
10 12 2
6 6
20 0 1021 1 2 35 71 25
2207
150 879 1
0 06255 234 10 1166
2207 1166 3373
5
298
348
=
⋅
−
=
= ⋅ + ×
L
NM O
QP=
= + =
−
z
. . ( )
.
[ . .
L g
L
kJ
10 J
J
g C
C
kJ
L g
L
mol
78.11 g
T] dT kJ
kJ
3 o
o
b. References: H2O (l, 0.01 o
C), C3H8 (gas, 40 o
C)
C H kJ / mol kJ mol ( fromTableB.2)
3 8 in ou p
C3H8
: ; .
H H C dT C
t p
= = =
z
0 1936
40
240
2 in out
ˆ ˆ
H O: 3065 kJ/kg (Table B.7); 640.1 kJ/kg (Table B.6)
H H
= =
c.
3 8
C
ˆ ˆ
19.36 kJ/mol, (640.1 3065) kJ/kg 2425 kJ/kg
H w
H H
Δ = Δ = − = −
Q H H m H
= = + =
Δ Δ Δ
100 0

C H w w
3 8
0.798 kg
w
m
⇒ =
From Table B.7: . .
Vsteam
3
bar, 300 C m kg
50 0522
° =
b g
.
VC H
3 8
40 0 0104
° =
⋅ ⋅
=
C, 250 kPa
0.008314 m kPa (mol K) 313 K
250 kPa
m mol C H
3
3
3 8
b g
3
3 8 3 3
3 3 8
3 8 3 8
0.798 kg steam 0.522 m steam 1 mol C H
0.400 m steam m C H
100 mol C H 1 kg steam 0.0104 m C H
=
d. 3 8
3
w w 3
3 8 3 8 3 8
0.798 kg steam 2425 kJ 1 mol C H kJ
ˆ 1860
100 mol C H kg steam 0.0104 m C H m C H fed
Q m H
= Δ = =
e. A lower outlet temperature for propane and a higher outlet temperature for steam.
100 mol C3H8 @ 40 o
C, 250 kPa
VP1(m3
)
100 mol C3H8 @ 240 o
C, 250 kPa
VP2(m3
)
mw kg H2O(l, sat‘d) @ 5.0 bar
Vw2(m3
)
mw kg H2O(v) @ 300 o
C, 5.0 bar
Vw1(m3
)
8.24 a.

8-8
8.25 a.
2 3
5500 L(STP) 1 mol
245.5 mol CH OH(v)/min
min 22.4 L(STP)
n = =
An energy balance on the unit is then written, using Tables B.5 and B.6 for the specific enthalpies
of the outlet and inlet water, respectively, and Table B.2 for the heat capacity of methanol vapor.
The only unknown is the flow rate of water, which is calculated to be 2
1.13 kg H O/min.
b. kg kJ 1 min 1 kW
1.13 2373.9 44.7 kW
min kg 60 sec 1 kJ/s
Q
⎛ ⎞
⎛ ⎞ ⎛ ⎞⎛ ⎞
= =
⎜ ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
⎝ ⎠

8.26 a.
100 mol/s (30o
C)
0.100 mol CO/mol
0.800 mol CO2/mol
n2 mol/s (30o
C)
y2 mol CO/s
(0.980-y2) mol CO2/s
m4 kg humid air/s (30o
C)
y4 kg H2O(v)/kg humid air
(1-y4) kg dry air/kg humid air
m3 kg humid air/s (50o
C)
(0.002
/1.002) kg H2O(v)/kg humid air
(1.000
/1.002) kg dry air/kg humid air
H2O(v) only
Basis: 100 mol gas mixture/s
5 unknowns: n2, m3, m4, y2, y4
– 4 independent material balances, H2O(v), CO, CO2 , dry air
– 1 energy balance equation
0 degrees of freedom all unknowns may be determined)
(
b. (1) CO balance: (100)(0.100) =
(2) CO balance: (100)(0.800) =
mol / s, mol CO / mol
(3) Dry air balance:
(4) H O balance:
(100)(0.100)(18)
1000
2
2 2
2

( )
. .
.
.
( )

.
.
.
( . )( )

n y
n y
n x
m m y
m m y
2
2
2 2
3 4 4
3 4 4
1
9184 01089
1000
1002
1
0 002
1002
9184
0 020 18
1000
−
U
V
|
W
|⇒ = =
= −
+ = +
References: CO, CO2, H2O(v), air at 25o
C (
H values from Table B.8 )
substance ( )
nin mol / s
Hin (kJ / mol) ( )
nout mol / s
Hout (kJ / mol)
H2O(v) 10 0.169 91.84(0.020) 0.169
CO 10 0.146 10 0.146
CO2 80 0.193 80 0.193
H2O(v) m3(0.002
/1.002)(1000
/18) 0.847 m4y4(1000
/18) 0.779
dry air m3(1.000
/1.002) (1000
/29) 0.727 m4(1-y4) (1000
/29) 0.672
(mol CO/mol)
(mol CO2/mol)
(48o
C)
5500 L(STP)/min CH3 OH (v) 65o
C
n2 (mol/min)
n2 mol/min CH3OH (v) 260o
C
mw kg/min H2O(l, sat‘d) @ 90o
C
Vw2(m3
/min )
mw kg/min H2O(v, sat‘d) @ 300o
C
Vw1(m3
/min )

8-9
8.26 (cont’d)
(5) Energy balance:
=
10 0169
0 002
1002
1000
18
0847
1000
1002
1000
29
0 727
91.84 0 020 0169 0 779
1000
18
1 0 672
1000
29
3 3
4 4 4 4
( . )
.
.
( . )
.
.
( . )
( . )( . ) ( . ) ( )( . )
+
F
HG I
KJF
HG I
KJ +
F
HG I
KJF
HG I
KJ
+
F
HG I
KJ+ −
F
HG I
KJ
m m
m y m y
Solve Eqs. (3)–(5) simultaneously ⇒ m3 = 2.55 kg/s, m4 = 2.70 kg/s, y4 = 0.0564 kg H2O/kg
2
0 0255
.55 kg humid air / s
100 mol gas/ s
kg humid air
mol gas
= .
Mole fraction of water :
kg H O
(1-.0564) kg dryair
kg DA
kmol DA
kmol H O
kg H O
kmol H O
kmol DA
.0963 kmol H O
kmol humid air
kmol H O
kmol humid air
2 2
2
2
2 2
00564 29 1
18
0963
0
1 00963
00878
.
.
( . )
.
=
⇒
+
=
Relative humidity:
C
mm Hg
mm Hg
H O
H O
o
2
2
p
p*
( . )( )
.
.
48
0 0878 760
8371
100% 79 7%
e j
= × =
c. The membrane must be permeable to water, impermeable to CO, CO2, O2, and N2, and
both durable and leakproof at temperatures up to 50o
C.
8.27 a. y
p
P
H O 2
2
C mm Hg
760 mm Hg
mol H O mol
=
°
= =
*
.
.
57 129 82
0171
b g
↓
28.5 m STP 1 mol
h 0.0224 m STP
mol h mol H O h
3
3 2
b g
b g= ⇒
1270 217 2
. 391
. kg H O h
2
b g
1270 217 2 1053
− = =======
R
S
|
|
T
|
|
.
mol dry gas
h
89.5 mol CO h
110.5 mol CO h
5.3 mol O h
847.6 mol N h
percentages
given
2
2
2
m (kg H O( )/h), 20°C
1270 mol/h, 620°C
425°C
l
2
References for enthalpy calculations:
CO, CO2 , O2 , N2 at 25°C (Table B.8); H O 0.01 C
2
o
l,
e j (steam tables)
substance nin

Hin
nout

Hout
CO
CO2
O2
N2
89.5
110.6
5.3
847.6
18.22
27.60
19.10
18.03
89.5
110.6
5.3
847.6
12.03
17.60
12.54
11.92
U
V
|
W
|
n
H
in mol h
in kJ mol

H O
2 v
b g
H O
2 l
b g
3.91
m
3749
83.9
391
. + m
--
3330
--
U
V
W
n
H
in kg h
in kJ kg

8-10
ΔH n H n H m m
i i i i
= − = ⇒ − + = ⇒ =
∑ ∑
.
out in
kg h
0 8504 3246 0 2 62
b. When cold water contacts hot gas, heat is transferred from the hot gas to the cold water
lowering the temperature of the gas (the object of the process) and raising the temperature
of the water.
8.28 2°C, 15% rel. humidity ⇒ = =
pH O
2
mm Hg mm Hg
015 5294 0 7941
. . .
b gb g
yH O
inhaled
2
2
mol H O mol inhaled air
d i b g b g
0 7941 760 1045 10 3
. .
= × −
.
ninhaled 3
5500 ml 273 K 1 liter 1 mol
min 275 K 10 ml 22.4 liters STP
mol air inhaled min
= =
b g 0 2438
Saturation at 37 °C ⇒ =
°
= =
y
p
H O 2
2
C
mm Hg
mol H O mol exhaled dry gas
*
.
.
37
760
47 067
760
0 0619
b g
n1 mol H2O(l)/min 22o
C
n2 kmol/min 37o
C
0.0619 H2O
0.9381 dry gas
0.2438 mol/min 2o
C
1.045 x 10-3
H2O
0.999 dry gas
Mass of dry gas inhaled (and exhaled) = =
0 2438 0 999
7 063
. .
.
b gb gmol dry gas 29.0 g
min mol
g min
Dry gas balance: 0 999 0 2438 0 9381 0 2596
2 2
. . . .
b gb g= ⇒ = mols exhaled min
n n
H O balance:
2 0 2438 1045 10 0 2596 0 0619 0 0158
3
1 1
. . . . .
b ge j b gb g
× + = ⇒ =
−
n n mol H O min
2
References for enthalpy calculations: H O
2 l
b g at triple point, dry gas at 2 °C
substance
min

Hin

mout

Hout
Dry gas
H O
2 v
b g
H O
2 l
b g
7.063
0.00459
0.285
0
2505
92.2
7.063
0.290
—
36.75
2569
—

m
H
in g min
in J g
( )
2 2
2
H O H O
H O
dry gas
18.02
ˆ from Table 8.4
ˆ 1.05 2
m n
H
H T
=
= −

Q H m H m H
i i i i
= = − = = ×
∑ ∑
Δ
out in
6
966.8 J 60 min 24 hr
min 1 hr 1 day
.39 10 J day
1
8.27 (cont’d)

8-11
8.29 a. 75 liters C H OH 789 g 1 mol
liter 46.07 g
mol C H OH
2 5
2 3
l
l
b g b g
= 1284
( ) . .
C T
p CH OH
o
3
kJ / (mol C)
= + × ⋅
−
01031 0557 10 3
e j (fitting the two values in Table B.2)
55 L H O 1000 g 1 mol
liter 18.01 g
mol H O
2
2
l
l
b g b g
= 3054 ( ) .
Cp H O
2
kJ mol C
= ⋅°
0 0754 b g
1284 mol C2H5OH(l) (70.0o
C)
3054 mol H2O(l) (20.0o
C)
1284 mol C2H5OH (l) (To
C)
3054 mol H2O(l) (To
C)
( )
( )
( ) ( )
3
70 25
o
Integrate, solve quadratic equation
0 1284 0.1031 0.557 10 3054 0.0754
liquids
0 adiabatic
T=44.3 C
T T
T dT dT
Q U H
Q
−
= + × +
= Δ ≅ Δ ⎫
⎪
⇒ ⇓
⎬
= ⎪
⎭
∫ ∫
b. 1. Heat of mixing could affect the final temperature.
2. Heat loss to the outside (not adiabatic)
3. Heat absorbed by the flask wall thermometer
4. Evaporation of the liquids will affect the final temperature.
5. Heat capacity of ethanol may not be linear; heat capacity of water may not be
constant
6. Mistakes in measured volumes initial temperatures of feed liquids
7. Thermometer is wrong
8.30 a.
1515 L/s air
500o
C, 835 tor,
Tdp=30o
C
110 g/s H2O, T=25o
C
1515 L/s air , 1 atm
110 g/s H2O(v)
Let
n1 (mol / s) be the molar flow rate of dry air in the air stream, and
n2 (mol / s) be the
molar flow rate of H2O in the air stream.
.

* .
.
. .
n n
n
n n
y
p
P
n n
1 2
2
1 2
1 2
1515 835
26 2
31824
0 0381
252 10
+
L
s
mm Hg
773 K
mol K
62.36 L mm Hg
mol / s
+
= =
(30 C) mmHg
835 mmHg
mol H O / mol air
mol dry air / s; mol H O / s
o
total
2
2
=
⋅
⋅
=
= =
⇒ = =

8-12
References: H2O (l, 25o
C), Air (v, 25o
C)
substances
nin (mol / s)
Hin (kJ / mol)
nout (mol / s)
Hout (kJ / mol)
dry air 25.2 14.37 25.2
C dT
p air
T
d i
25
z
H2O(v) 1.0
C dT H
C dT
p H O l vap
p H O v
d i
d i
2
2
25
100
100
500
( )
( )

z
z
+
7.1
C dT H
C dT
p H O l vap
p H O v
T
d i
d i
2
2
25
100
100
( )
( )

z
z
+
H2O(l) 6.1 0 -- --
ΔH n H n H
C dT C dT H C dT
C dT H C dT
out out in in
p air
T
p H O l vap p H O v
T
p H O l vap p H O v
= = ⋅ − ⋅
F
HG I
KJ+ + +
F
HG I
KJ
− − + +
F
HG I
KJ =
z z z
z z
0
252 71
252 14 37 100 0
25 25
100
100
25
100
100
500
2 2
2 2

. .
. . .
( ) ( )
( ) ( )
b g d i b g d i d i
b gb g b g d i d i
Integrate, solve : T = 139o
C
b.
( ) ( ) ( ) ( ) 2
139 139
( )
500 500
25.2 1.00 290 kW
p p
air H O v
Q C dT C dT
= − − = −
∫ ∫

This heat goes to vaporize the entering liquid water and bring it to the final temperature
of 139o
C.
c. When cold water contacts hot air, heat is transferred from the air to the cold water mist,
lowering the temperature of the gas and raising the temperature of the cooling water.
8.30 (cont’d)

8-13
8.31
3
3
3
520 kg NH 10 g 1 mol 1 h
Basis: 8.48 mol NH s
h 1 kg 17.03 g 3600 s
=
8.48 mol NH /s
3
n1
25°C
(mol air/s)
T °C
n2 (mol/s)
0.100 NH
0.900 air
600°C
3
Q = –7 kW
NH balance: mol s
3 8 48 0100 84 8
2 2
. . .
= ⇒ =
n n
Air balance: mol air s
1
n = =
0 900 84 8 76 3
. . .
b gb g
References for enthalphy calculations: NH g
3b g, air at 25°C
NH
kJ mol
3 in
out NH Table B.2
from
out
3
.
.
H
H C dT H
p
Cp
=
= ⇒ =
z
0 0
2562
25
600
d i
Air: ( ) 5 8 2 12 3
J mol C 0.02894 0.4147 10 0.3191 10 1.965 10
p
C T T T
− − −
⋅° = + × + × − ×
( )( )
in 25
12 4 8 3 5 2
ˆ
0.4913 10 0.1064 10 0.20735 10 0.02894 0.7248 kJ mol
T
p
H C dT
T T T T
− − −
=
= − × + × + × + −
∫
600
out 25
ˆ 17.55 kJ mol
p
H C dT
= =
∫
Energy balance:
out in
ˆ ˆ
i i i i
Q H n H n H
= Δ = −
∑ ∑
⇓
( )( ) ( )( ) ( )( )
( )( )
3
12 4 8 3 5 2
7 kJ s 8.48 mols NH s 25.62 kJ mol 76.3 mols air s 17.55 kJ mol 8.48 0.0
76.3 0.4913 10 0.1064 10 0.20735 10 0.02894 0.7248
T T T T
− − −
− = + −
− − × + × + × + −
Solve for T by trial-and-error, E-Z Solve, or Excel/Goal Seek ⇒ o
691 C
T =
8.32 a. Basis: 100 mol/s of natural gas. Let M represent methane, and E for ethane
100 mol/s
0.95 mol M/mol
0.05 mol E/mol
Furnace
Heat
Exchanger
Stack gas (900o
C)
n3 mol CO2/s
n4 mol H2O/s
n5 mol O2/s
n6 mol N2/s
air (245o
C)
n1 mol O2/s
n2 mol N2/s
20 % excess air (20o
C)
n1 mol O2/s
n2 mol N2/s
Stack gas (To
C)
n3 mol CO2/s
n4 mol H2O/s
n5 mol O2/s
n6 mol N2/s
CH 2O CO 2H O
C H 7 / 2 O 2CO 3H O
4 2 2 2
2 6 2 2 2
+ → +
+ → +
b g

8-14
2 2
2 2
1 2 2 2
2
3
2 mol O 3.5 mol O
95 mol M 4.76 mol air 5 mol E 4.76 mol air
1.2
s 1 mol M mol O s 1 mol E mol O
1185 mol air/s
0.21 1185 249 mol O /s, 0.79 1185 936 mol N /s
1 mol CO
95 mol M
s 1 m
air
air
n
n
n n
n
⎡ ⎤
= +
⎢ ⎥
⎣ ⎦
=
= × = = × =
=

2
2
2 2
4 2
2 2
5 2
6 2
2 mol CO
5 mol E
105 mol CO /s
ol M s 1 mol E
2 mol H O 3 mol H O
95 mol M 5 mol E
205 mol H O/s
s 1 mol M s 1 mol E
2 mol O 3.5 mol O
95 mol M 5 mol E
249 41.5 mol O /s
s 1 mol M s 1 mol E
936 mol N
n
n
n n
+ =
= + =
= − + =
= =

2/s
245
20
mol air kJ kJ
( ) 1185 6.649 7879 ( 7879 kW)
s mol air s
air p air
Q n C dT
⎛ ⎞⎛ ⎞
= = = =
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
∫

Energy balance on stack gas:
( )
( )
( ) ( ) ( ) ( )
2 2 2 2
6
900
3
3 4 5 6
( )
900 900 900 900
7879
T
i p i
i
T T T T
p p p p
CO H O v O N
Q H n C dT
n C dT n C dT n C dT n C dT
=
= −Δ = −
− = + + +
∑ ∫
∫ ∫ ∫ ∫

Substitute for the heat capacities (Table B.2), integrate, solve for T using E-Z Solve⇒ o
T 732 C
=
b. 350 1000 1
4 34
0 0434
0 0434 7851 341
m (STP)
h
mol
22.4 L(STP)
L
m
h
3600 s
mol / s
Scale factor =
4.34 mol / s
100 mol / s
kW
3
3
=
=
′ = =
.
.
.
Q b g
8.33 a. Δ . . . . . . .
H C dT
p
= = + + + + + =
z0
600 100
3
335 4 351 38 4 42 0 2 36 7 40 2 439 23100
b g b g J mol
Q H n H
= = = =
Δ Δ 150 23100
3465
mol
s
J
mol
1 kW
1000 J / s
kW
b. The method of least squares (Equations A1-4 and A1-5) yields (for X T
= , y Cp
= )
C T
p = + × ° ⋅°
−
00334 1732 10 5
. . C kJ (mol C)
b g ⇒ Q T dT
= + × =
−
z
150 00334 1732 10 3474
5
0
600
. . kW
The estimates are exactly identical; in general, (a) would be more reliable, since a linear fit is
forced in (b).
8.34 a. ln ln exp
C bT a C a bT
p p
= + ⇒ =
1 2 1 2
e j, T1 71
= . , Cp1 0 329
= . , T2 17 3
= . , Cp2 0533
= .
b
C C
T T
a C b T a e
C T
p p
p
p
=
−
=
= − = − ⇒ = =
U
V
|
|
W
|
|
⇒ =
−
ln
.
ln ln . .
. exp .
.
2 1
2 1
1 1
1 4475
1 2
0 0473
14475 0 235
0 235 0 0473
e j
8.32 (cont’d)

8-15
8.34 (cont’d)
b. 0 235 0 0473
0 235 2
0 0473
0 473
1
0473
1730
1800
150 1 2 1 2 1 2
1800
150
. exp .
.
.
exp .
.
z = −
L
NM O
QP
R
S
T
U
V
W
= −
T dT T T
e j b gb g e j cal g
DIMENSIONS CP(101), NPTS(2)
WRITE (6, 1)
1 FORMAT (1H1, 20X'SOLUTION TO PROBLEM 8.37'/)
NPTS(1) = 51
NPTS(2) = 101
DO 200K = 1, 2
N = NPTS (K)
NM1 = N – 1
NM2 = N – 2
DT = (150.0 – 1800.0)/FLOAT (NM1)
T = 1800.0
DO 20 J = 1, N
CP (J) = 0.235*EXP(0.0473*SQRT(T))
20 T = T + DT
SUMI = 0.0
DO 30 J = 2, NM1, 2
30 SUMI = SUMI + CP(J)
SUM2 = 0.0
DO 40 J = 3, NM2, 2
40 SUM2 = SUM2 + CP (J)
DH = DT*(CP(1) + 4.0 = SUM1 + 2.0 = SUM2 + CP(N))/3.0
WRITE (6, 2) N, DH
2 FORMAT (1H0, 5XI3, 'bPOINT INTEGRATIONbbbDELTA(H)b= ', E11.4,'bCAL/G')
200 CONTINUE
STOP
END
Solution: N H
= ⇒ = −
11 1731
Δ cal g
N H
= ⇒ = −
101 1731
Δ cal g
Simpson's rule with N = 11 thus provides an excellent approximation
8.35 a.
. . .
.
.
m
M W
H
Q H
v
=
=
=
U
V
|
W
|
⇒ = = =
175
62 07
56 9
175 1000 1 56 9 1
2670
kg / min
g / mol
kJ / mol
kg
min
g
kg
mol
62.07 g
kJ
mol
min
60 s
kW
Δ
Δ
b. The product stream will be a mixture of vapor and liquid.
c. The product stream will be a supercooled liquid. The stream goes from state A to state B as shown
in the following phase diagram.
T
P
A
B

8-16
8.36 a. Table B.1 T 68.74 C, H (T ) 28.85 kJ / mol
Assume: n - hexane vapor is an ideal gas, i.e. H is not a function of pressure
C H C H
H H
C H C H
b
o
v b
6 14 l, 20 C
H
6 14 v, 200 C
1 2
6 14 l, 68.74 C
H T
6 14 v, 68.74 C
o
Total
o
o
v b
o
⇒ = =
⎯ →
⎯⎯
⎯
B A
⎯ →
⎯⎯
⎯
Δ
Δ
Δ Δ
Δ
Δ

b g b g
b g b g
b g
Δ
Δ
Δ
Δ Δ Δ Δ
. .
. . . .
.
. . . .
.
.
H dT
H T T T dT
H
H H H H T
Total v b
1 20
68 74
2
5 8 2 9 3
68 74
200
2
1 2
0 2163 1054
013744 4085 10 2392 10 57 66 10
24 66
1054 24 66 2885 64 05
= =
= + × − × + ×
=
= + + = + + =
z
z − − −
kJ / mol
kJ / mol
kJ / mol
b g
b. Δ .
H = −64 05 kJ / mol
c. ,
.
. . .
U C H PV
PV RT
U
o
200
393
64 05 393 6012
2 atm
Assume ideal gas behavior kJ / mol
kJ / mol
e j= −
⇒ = =
= − =
8.37 Tb = °
100 00
. C Δ .
H t
v b
b g= 40 656 kJ mol
H O l, 50 C H O v, 50 C
H H
H O l, 100 C H O v, 100 C
2
o H 50 C
2
o
1 2
2
o H 100 C
2
o
v
o
v
o
e j e j
e j e j
e j
e j
Δ
Δ
Δ Δ

⎯ →
⎯⎯⎯
⎯
B A
⎯ →
⎯⎯⎯
⎯
Δ .
H C dT
p l
1
25
100
377
= =
z H O
2
kJ mol
b g
Δ .
H C dT
p v
2
100
25
169
= = −
z H O
2
kJ mol
b g
Δ . . . .
Hv 50 377 40 656 169 42 7
° = + − =
B
C kJ mol
Table B.1
b g
Steam table:
( )
2547.3 104.8 kJ 18.01 g 1 kg
44.0 kJ mol
kg 1 mol 1000 g
−
=
The first value uses physical properties of water at 1 atm (Tables B.1, B.2, and B.8), while the heat of
vaporization at 50o
C in Table B.5 is for a pressure of 0.1234 bar (0.12 atm). The difference is ΔH for
liquid water going from 50o
C and 0.1234 bar to 50o
C and 1 atm plus ΔH for water vapor going from
50o
C and 1 atm to 50o
C and 0.1234 bar.
8.38
3
3
1.75 m 879 kg 1 kmol 1000 mol 1 min
164.1 mol/s
2.0 min m 78.11 kg 1 kmol 60 s
=
Tb = °
801
. C , Δ .
H T
v b
b g= 30 765 kJ mol

8-17
8.38 (cont’d)
C H v, 580 C C H l, 25 C
H H
C H v, 80.1 C C H l, 80.1 C
6 6
o
6 6
o
1 2
6 6
o
6 6
o
e j e j
e j e j
⎯ →
⎯
B A
⎯ →
⎯
Δ Δ

Δ .
.
H C dT
p v
1
580
80 1
77 23
= = −
z C H
6 6
kJ mol
b g
Δ .
.
H C dT
p l
2
3531
298
7 699
= = −
z C H
6 6
kJ mol
b g
Δ Δ Δ Δ
Δ Δ
. .
. . .
H H H H
Q H n H x
v
= − + = −
= = = − = − −
1 2
4
801 1157
1641 1157 190 10
o
C kJ / mol
mol / s kJ / mol kW
d i
b gb g
35 C
15% relative saturation
C
1 atm
mm Hg
760 mm Hg
mol CCl mol
CCl 4
4
Antoine
° U
V
W
⇒ =
°
= =
B
∗
y
P
V
015
25
015
176 0
0 0347
. .
.
.
b g
( ) . .
Δ Δ
H Q H
v CCl
Table B.1
4
4
4
kJ
mol
10 mol 0.0347 mol CCl 30.0 kJ
min mol mol CCl
kJ min
= ⇒ = = =
30 0 10 4
Time to Saturation
6 kg carbon 0.40 g CCl 1 mol CCl 1 mol gas 1 min
g carbon 153.84 g CCl mol CCl mol gas
min
4 4
4 4
0 0347 10
450
.
.
=
8.40 a.
CO g, 20 C CO s, 78.4 C C
2 2 CO g
2
° → − ° = − − °
−
z
b g b g d i b g
b g
: .
.
Δ Δ
H C dT H
p sub
20
78 4
78 4
In the absence of better heat capacity data; we use the formula given in Table B.2 (which is strictly
applicable only above 0°C ).
Δ . . . .
.
.
.
H T T T dT
≈ + × − × + ×
F
HG I
KJ
−
×
= −
− − −
−
−
z 03611 4 233 10 2 887 10 7 464 10
6030
4184 10
28 66
5 8 2 12 3
20
78 4
3
kJ
mol
cal kJ
mol 1 cal
kJ mol
Q H n H
= = = = ×
Δ Δ .
300 kg CO 10 g 1 mol 28.66 kJ removed
h 1 kg 44.01 g mol CO
k J h
2
3
2
195 105
(or 6 23 107
. × cal hr or 72.4 kW )
b. According to Figure 6.1-1b, Tfusion=-56o
C

.
.
Q H n H
H C dT H C dT
Q n C dT H C dT
p v p
p v p
= =
= + − +
= + − +
L
NM O
QP
−
−
−
−
−
−
z z
z z
Δ Δ
Δ Δ
Δ
where, C
C
CO (v)
o
CO (l)
CO (v)
o
CO (l)
2 2
2 2
d i e j d i
d i e j d i
20
56
56
78 4
20
56
56
78 4
56
56
8.39
lv
H
−Δ

8-18
8.41 a. C a bT
p = +
b
a
C T
p
=
−
−
=
= − =
U
V
|
|
W
|
|
⇒ ⋅ = +
5394 50 41
500 300
0 01765
5394 0 01765 500 4512
4512 0 01765
. .
.
. . .
. .
b gb g
b g b g
J mol K K
NaCl s s l
, , ,
300 1073 1073
K NaCl K NaCl K
b g b g b g
→ →
Δ Δ
. .
.
.
H C dT H T dT
ps m
= + = +
L
NM O
QP +
= ×
z z
300
1073
300
1073
4
4512 0 01765
30 21
7 44 10
1073 K
J
mol
kJ 10 J
mol 1 kJ
J mol
3
b g b g
b. Q U n C dT U
v m
v p
m m
C C
U H
= = +
z
≈
≅
Δ Δ
Δ Δ
300
1073
1073 K
b g
Q H n H
≈ = = = ×
Δ Δ .
200 kg 10 g 1 mol 74450 J
1 kg 58.44 g mol
J
3
2 55 108
c. t =
×
×
=
2.55 10 J s 1 kJ
0.85 3000 kJ 10 J
100 s
8
3
8.42 Δ .
Hv = 3598 kJ mol , Tb = ° =
136 2
. C 409.4 K , Pc = 37 0
. atm , Tc = 619.7 K (from Table B.1)
Trouton's rule: Δ . . . .
H T
v b
≈ = =
0 088 0 088 36 0 01%
b gb g b g
409.4 K kJ mol error
Chen's rule:
Δ
. . . log
.
H
T
T
T
P
T
T
v
b
b
c
c
b
c
≈
F
HG I
KJ− +
L
N
MM
O
Q
PP
−
F
HG I
KJ
0 0331 0 0327 0 0297
107
10
= 357
. kJ mol (–0.7% error)
Watson’s correlation : Δ .
. .
. .
.
.
Hv 100 3598
619 7 3732
619 7 409 4
38 2
0 38
° ≈
−
−
F
HG I
KJ =
C kJ mol
b g
8.43 C H N
7 2 : Kopp's Rule ⇒ ≈ + + = ⋅°
Cp 7 0 012 12 0 018 0 033 0 333
. . . .
b g b g k J (mol C)
Trouton's Rule ⇒ ° =
Δ
Hv C 0.088 200 + 273.2 = 41.6 kJ mol
200
b g b g
C H N , 25 C C H N , 200 C C H N , 200 C
7 12 7 12 7 12
l l v
° → ° → °
b g b g b g
( )
200
25
kJ kJ
ˆ ˆ 200 C 0.333(200 25) 41.6 100 kJ mol
mol mol
p v
H C dT H
Δ = + Δ ° ≈ − + =
∫

8-19
8.44 a. Antoine equation: Tb ° =
−
− = °
C C
b g b g
1211033
6 90565 100
220 790 261
.
. log
. .
Watson Correction: Δ . .
. .
. .
.
.
Hv 261 30 765
562 6 299 3
562 6 3531
336
0 38
° =
−
−
F
HG I
KJ =
C kJ mol
b g
b. Antoine equation: Tb 50 mm Hg C
b g= °
118
. ; Tb 150 mm Hg C
b g= °
352
.
Clausius-Clapeyron: ln

ln
p
H
RT
C H R
p p
T T
v
v
= − + ⇒ = −
−
Δ
Δ
2 1
2 1
1 1
b g
Δ .
ln
.
.
Hv = −
⋅ −
R
S
|
T
|
U
V
|
W
|=
0 008314
150 50
1 308 4 1
34 3
kJ
mol K K 285.0 K
kJ mol
b g
c. ( , 26.1°C)
C H l
6 6
( , 80.1°C)
C H l
6 6
ΔH1

( , 26.1°C)
C H v
6 6
( , 80.1°C)
C H v
6 6
ΔH2

ΔHv
(80.1°C)
Δ .
.
.
H C dT
p l
1
26 1
80 1
7 50
= =
zd i kJ mol
Δ .
.
.
H C dT
p v
2
80 1
26 1
4 90
= = −
zd i kJ mol
Δ . . . . .
Hv 261 7 50 30 765 4 90 334
° = + − =
C kJ mol
b g
8.45 a. Tout = 49.3o
C. The only temperature at which a pure species can exist as both vapor and liquid at 1
atm is the normal boiling point, which from Table B.1 is 49.3o
C for cyclopentane.
b. Let and denote the molar flow rates of the feed, vapor product, and liquid product
streams, respectively.
, ,
n n n
f v l
Ideal gas equation of state
.
n f = =
1550 273
44 66
L K 1 mol
s 423 K 22.4 L(STP)
mol C H (v) / s
5 10
55% condensation: . ( . ( ) /
nl = 0550 44 66 mol / s) = 24.56 mol C H l s
5 10
Cyclopentane balance ⇒ ( . . ) /
nv = −
44 66 24 56 mol C H s = 20.10 mol C H (v) / s
5 10 5 10
Reference: C5H10(l) at 49.3o
C
Substance
nin
(mol/s)

Hin
(kJ/mol)

nout
(mol/s)

Hout
(kJ/mol)
C5H10 (l) — — 24.56 0
C5H10 (v) 44.66
H f
20.10
Hv
H H C dT
i v p
Ti
= + z
Δ
. o
C
49 3

8-20
8.45 (cont’d)
Substituting for from Table B.1 and for from Table B.2
kJ / mol, kJ / mol
v
Δ
. .
H C
H H
p
f v
⇒ = =
38 36 27 30
Energy balance: . .
Q n H n H
= − = − × − ×
∑ ∑
out out in in kJ / s = kW
116 10 116 10
3 3
8.46 a. Basis: 100 mol humid air fed
100 mol
y1 (mol H2O/mol)
1-y1 (mol dry air/mol)
50o
C, 1 atm, 2o
superheat
n2 (mol), 20o
C, 1 atm
y2 (mol H2O/mol), sat’d
1-y2 (mol dry air/mol)
n3 (mol H2O(l))
There are five unknowns (n2, n3, y1, y2, Q) and five equations (two independent material
balances, 2o
C superheat, saturation at outlet, energy balance). The problem can be solved.
b.
2 C superheat
C
° ⇒ =
∗ °
y
p
p
1
48
b g
saturation at outlet ⇒ =
∗ °
y
p
p
2
20 C
b g
dry air balance: 100 1 1
1 2 2
b gb g b g
− = −
y n y
H O balance:
2 100 1 2 2 3
b gb g b gb g
y n y n
= +
c. References: Air 25°C
b g, H O C
2 l, 20°
b g
Substance
Air in mol
H O in kJ mol
H O
in in out out
2
2
n H n H
y H n y H n
v y H n y H H
l n

100 1 1
100
0
1 1 2 2 3
1 2 2 2 4
3
⋅ − ⋅ −
⋅ ⋅
− −
b g b g
b g
b g
. . . .

. .
. . . .
H C dT T T T dT
H C dT H C dT
dT
T T T dT
p
p v p
1 25
50 5 8 2 12 3
25
50
2 20
100
100
50
20
100
5 8 2 12 3
100
50
0 02894 0 4147 10 0 3191 10 1965 10
100
0 0754 40 656
0 03346 0 688 10 0 7604 10 3593 10
= = + × + × − ×
= + +
+ +
+ × + × − ×
z z
z z
z
z
− − −
− − −
d i
d i e j d i
air
H O(l)
o
H O(v)
2 2
C
=
Δ

H C dT
p
3 25
20
= z d iair

H C dT H C dT
p v p
2 20
100
100
20
100
= + +
z z
d i e j d i
H O(l)
o
H O(v)
2 2
C
Δ
Q(kJ)

8-21
8.46 (cont’d)
c.
Q H n H n H V
i i
out
i i
in
= = − =
⋅
⋅ ×
∑ ∑
Δ .
.
mol Pa m
mol K
K
Pa
air
3
100 8 314 323
101325 105
⇒
Q
V
n H n H
i i
out
i i
in
air
3
mol Pa m
mol K
K
Pa
=
−
⋅
⋅ ×
∑ ∑

.
.
100 8 314 323
101325 105
d.
2 C superheat
C mm Hg
760 mm Hg
mol H O mol
2
° ⇒ =
∗ °
= =
y
p
p
1
48 8371
0110
b g .
.
saturation at outlet ⇒ =
∗ °
= =
y
p
p
2
20 17 535
0 023
C mm Hg
760 mm Hg
mol H O mol
2
b g .
.
dry air balance: 100 1 0110 1 0 023 9110
2 2
b gb g b g
− = − ⇒ =
. . .
n n mol
H O balance:
mol H O 0.018 kg
1 mol
kg H O condensed
2
2
2
100 0110 9110 0 023
8 90
0160
3 3
b gb g b gb g
. . .
.
.
= + ⇒ =
=
n n
Q H n H n H
i i
out
i i
in
= = − = −
∑ ∑
Δ .
4805 kJ
Vair
3
3
2
3 2
3
3
3
mol Pa m
mol K
K
Pa
m
kg H O condensed
2.65 m air fed
kg H O condensed / m air fed
kJ
2.65 m air fed
kJ / m air fed
=
⋅
⋅ ×
=
⇒ =
⇒
−
= −
100 8 314 323
101325 10
2 65
0160
0 0604
4805
181
5
.
.
.
.
.
.
e. Solve equations with E-Z Solve.
f.
Q =
−
= −
181 250 1 1
12 6
kJ
m air fed
m air fed
h
h
3600 s
kW
1 kJ / s
kW
3
3
.

8-22
8.47 Basis:
226 m K 10 mol
min K 22 m STP
mol humid air min
3 3
3
273
309 415
8908
. b g
= . DA = Dry air
(
Q kJ / min)
8908 mol
0
0
/ min
mol H O(v) / mol]
(1- (mol DA / mol)
36 C, 1 atm, 98% rel. hum.
2
o
y
y
[
)
(
[
)
n
y
y
1
1
1
mol / min)
mol H O(v) / mol]
(1- (mol DA / mol)
10 C, 1 atm, saturated
2
o
[
n2 mol H O(l) / min], 10 C
2
o
a. Degree of freedom analysis: 5 unknowns – (1 relative humidity + 2 material balances + 1 saturation
condition at outlet + 1 energy balance) = 0 degrees of freedom.
b. Inlet air: y P p y
w
0 0
0 98 36
0 98 44 563
0 0575
= ° ⇒ = =
B
.
. ( .
.
*
C
mm Hg)
760 mm Hg
mol H O(v) mol
Table B.3
2
b g
Outlet air: y p P
1 10 9 760 mm Hg 0 0121
= = =
∗
( .
o
2
C) / .209 mm Hg mol H O(v) mol
b g b g
Air balance: 1 0 0575 1 0 0121 8499
1 1
− = − ⇒ =
. (8908 .
b g b g
mol / min) mol / min
n n
H O balance: 0.0575
mol
min
= 0.0121(8499
mol
min
) H O(l) min
2 2
8908 409 mol
2 2
F
HG I
KJ + ⇒ =

n n
References: H O triple point air 77 F
2 l, ,
b g b g
°
( )
( )
in in out out
2
2
ˆ ˆ
Substance
Air 8396 0.3198 8396 0.4352 in mol min
.
ˆ
H O 512 46.2 103 45.3 in kJ/mol
H O 409 0.741
n H n H
n
v H
l
−
− −

Air:
H from Table B.8
H O: kJ / kg) from Table B.5 (0.018 kg / mol)
2
(
H ×
Energy balance:
4 4
out in
2.50 10 kJ 60 min 9.486 10 Btu 1 ton
ˆ ˆ 119 tons
min 1 h 0.001 kJ 12000 Btu h
i i i i
Q H n H n H
−
− × ×
=Δ = − = =
−
∑ ∑

8-23
8.48
Basis:
746.7 m outlet gas / h atm 1 kmol
1 atm 22.4 m STP
kmol / h
3
3
3
100 0
b g
= .
Antoine:
( ) ( )
1175.817
log 6.88555 0 C 45.24 mm Hg, 75 C 920.44 mm Hg
224.867
v v v
p p p
T
∗ ∗ ∗
= − ° = ° =
+
y
p
P
v
out 6 14
C
kmol C H kmol
=
°
= =
∗
0 4524
3 760
0 0198
b g
b g
.
. ,
y
p
P
v
in
6 14
C kmol C H
kmol
=
°
= =
∗
0 90 75 0 90 920 44
3 760
0 363
. . .
.
b g b gb g
b g
N balance: kmol h
2 . . .
n n
1 1
1 0 363 100 1 0 0198 1539
− = − ⇒ =
b g b g
C H balance: kmol C H h
6 14 6 14
1539 0 363 100 0 0198 5389
2 2
. . . .
b gb g b gb g b g
= + ⇒ =
n n l
Percent Condensation: 5389 0 363 1539 100% 965%
. . . .
kmol h condense kmol h in feed
b g b gb g
× × =
References: N2(25o
C), n-C6H14(l, 0o
C)
Substance
N in mol h
-C H in kJ mol
-C H
in in out out
2
6 14
6 14
n H n H
n
n r H
n l

. .
. .
.
98000 146 98000 0 726
55800 44 75 2000 3333
53800 0 0
−
− −
b g
b g
N : C H (v):
2 6 14
, .
.
.
H C T n H C dT H C dT
p p v pv
T
= − − = + +
z z
25 68 7
0
68 7
68 7
b g b g
A Δ
Energy balance: Q H
= = − × ⇒ −
Δ ( . )(
2 64 10 1 733
6
kJ h h / 3600 s) kW
n H n H
i i i i
out in
∑ ∑
−

o
1
6 14
2
(kmol/h) @ 75 C, 3 atm
(kmol C H (v)/kmol), 90% sat'd
(1 )(kmol N /kmol)
in
in
n
y
y
−

o
6 14
2
100 kmol/h @ 0 C, 3 atm
(kmol C H (v)/kmol), sat'd
(1 )(kmol N /kmol)
out
out
y
y
−
o
2 6 14
(kmol C H (l)/h), 0 C
n

8-24
8.49 Let A denote acetone.
(
Q Ws
kW) 25.2 kW
= −
142 L / s @ C, 1.3 atm
mol / s)
[mol A(v) / mol], sat'd
(1 mol air / mol)
o
150
0
0
0
(
)(
n
y
y
−

)(
n
y
y
1
1
1
(mol / s) @ 18 C, 5 atm
[mol A(v) / mol], sat'd
(1 mol air / mol)
o
−
−
[
n2 18
mol A(l) / s]@ C, 5 atm
o
−
a. Degree of freedom analysis: 6 unknowns ( , , , , , )
n n n y y Q
0 1 2 0 1
–1 equation of state for feed gas
–1 sampling result for feed gas
–1 saturation condition at outlet
–1 energy balance
0 degrees of freedom
b. Ideal gas equation of state Raoult’s law
(1)

n
P V
RT
0
0 0
0
= (2) y
p
p
A
A
1
18
=
−
*
*
(
)
o
C)
5 atm
(Antoine equation for
Feed stream analysis
(3) y
P RT
0
0 0
4 017
300
mol A
mol
[(4.973 g A][1 mol A / 58.05 g]
L) mol feed gas
F
HG I
KJ =
− . )
[( . / ]
Air balance:
( )
( )
n
n y
y
1
0 0
1
1
1
=
−
−
(4) Acetone balance:
n n y n y
2 0 0 1 1
= − (5)
Reference states: A(l, –18
o
C), air(25
o
C)
out
in out
in
2
0 0 0 1 1 1
0 0 0 1 1 1
ˆ
ˆ
Substance
(mol/s) (mol/s)
(kJ/mol) (kJ/mol)
A(l) 0
ˆ ˆ
A(v)
ˆ ˆ
air (1 ) (1 )
A A
a a
n
n H
H
n
n y H n y H
n y H n y H
− −
− −

(6) + ( +
A(v) A(l)
C
C
A A(v)
C
o
o
o
Table B.2 Tab le B.1 Ta ble B.2
( ) ( ) ) ( )
H T C dT H C dT
p v p
T
=
−
z z
18
56
56
Δ
(7) ( )
H T
air from Table B.8
(8) .
Q W n H n H W
s s
= + − = −
∑ ∑
out out in in ( kJ / s)
252

8-25
8.49 (cont’d)
c.
3
0 1
0
1 2
0
(1) 5.32 mol feed gas/s (2) 6.58 10 mol A(v)/mol outlet gas
(3) 0.147 mol A(v)/mol feed gas
(4) 4.57 mol outlet gas/s (5) 0.75 mol A(l)/s
ˆ
(6) 48.1 kJ/
A
n y
y
n n
H
−
⇒ = ⇒ = ×
⇒ =
⇒ = ⇒ =
⇒ =

1
0 1
ˆ
mol, 34.0 kJ/mol
ˆ ˆ
(7) 3.666 kJ/mol, 1.245 kJ/mol
(8) 84.1 kW
A
a a
H
H H
Q
=
⇒ = = −
⇒ = −

8.50 a. Feed:
2 2 2 3
4 2 3
3 m (35) cm 1 m 273 K 850 torr 1 kmol 10 mol mol
50.3
s 10 cm (273+40)K 760 torr 22.4 m (STP) 1 kmol s
π
=
Assume outlet gas is at 850 mm Hg.
Degree-of-freedom analysis
6 unknowns 0 1 2 3
( , , , , , )
y n n n T Q

– 2 independent material balances
– 2 Raoult’s law (for feed and outlet gases)
– 1 60% recovery equation
– 1 energy balance
0 degrees of freedom ⇒ All unknowns can be calculated.
b. Let H = C6H14 Antoine equation, Table B.4
( ) ( )
*
dp 0
feed
25 C 151 mm Hg
25 C 0.178 mol H mol
850 mm Hg
H
p
T y
P
°
= ° ⇒ = = =
60% recovery
( )( )
( )
1
0.600 50.3 0.178 mols H feed
5.37 mol H s
s
n l
⇒ = =

Hexane balance: ( )
2 2
(0.178)(50.3) = 5.37 + 3.58 mol H s
n n v
⇒ =

o
o 6 14
o
o
50.3 mol/s @ 40 C, 850 mm Hg
(mol C H (v)/mol)
(1 )(mol air/mol)
25 C
dp
y
y
T
−
=
o
2 6 14
3
(mol C H (v)/s), sat'd at ( C) 850 torr
(mol air/s)
n T
n

o
1 6 14
(mol C H (l)/s), ( C)
60% of hexane in feed
n T

(kW)
Q

8-26
8.50 (cont’d)
Air balance: ( )( )
3 50.3 1 0.178 41.3 mol air s
n = − =

Mole fraction of hexane in outlet gas:
( )
( )
( )
2
2 3
3.58
67.8 mm Hg
3.58 41.3 850 mm Hg
H
H
p T
n
p T
n n
= = ⇒ =
+ +

Saturation at outlet: Table B.4
*
H ( ) ( ) 67.8 mm Hg 7.8 C
H
p T p T T
= = ⎯⎯⎯⎯
→ = °
Reference states: C H C
6 14 l, .
7 8°
b g, air (25°C)
Substance
nin

Hin

nout

Hout
( )
6 14
C H v
( )
6 14
C H l
Air
8.95
—
41.3
37.5
—
0.435
3.58
5.37
41.3
32.7
0
–0.499

n in mol/s

H in kJ/mol
( )
6 14
C H v : . ,

.
.
.
H C dT H C dT
C
H
pl v pv
T
p
v
= + ° +
z z
7 8
68 74
68 74
68 74
Δ
Δ
C
from Table B.2
from Table B.1
b g
Air:
H from Table B.8
Energy balance: Q H n H n H
i i i i
= = − =
−
−
=
∑ ∑
Δ
out in
kJ s kW cooling
kJ s
kW
257 1
1
257
c.
u A u A A
D
D D u u
⋅ = ⋅ =
⋅
=
U
V
|
W
|⇒ = ⋅ =
' '; ; ' ' . m / s
π 2
4
1
2
4 12 0

8-27
8.51
( (
[
)
n P
y
y
v mol / min) @ 65 C, atm)
mol P(v) / mol], sat'd
(1- (mol H(v) / mol)
o
0
100 mol 80
0
/ s @ C, 5.0 atm
.500 mol P(l) / mol
0.500 mol H(l) / mol
o
(
Q kJ / s)
( (
n P
l mol / min) @ 65 C, atm)
.41 mol P(l) / mol
0.59 mol H(l) / mol
o
0
0
a. Degree of freedom analysis
5 unknowns – 2 material balances – 2 equilibrium relations (Raoult’s law) at outlet – 1 energy balance
= 0 degrees of freedom
Antoine equation (Table B.4) ⇒ p p
P H
* *
( (
65 65
o o
C) = 1851 mm Hg, C) = 675 mm Hg
Raoult's law for pentane and hexane
0.410 C) =
0.590 C) = (1
mol P(v) / mol
mm Hg (1.52 atm)
o
o
p yP
p y P
y
P
P
H
*
*
(
( )
.
65
65
0 656
1157
0
0 0
−
⇒
=
=
Total mole balance 100 mol =
Pentane balance 50 mole P = 0.656 + 0.410
mol vapor / s
mol liquid / s
:
:
.
.
n n
n n
n
n
v l
v l
v
l
+
⇒
=
=
36 6
634
mol 0.08206 L atm (65+ 273)K
s mol K atm
L / s
:
.
.
V
n RT
P
v
v
= =
⋅
⋅
=
0
36 6
152
667
Fractional vaporization of propane:
(0.656 36.6) mol P(v)/s mol P vaporized
0.480
50.0 mol P(l) fed/s mol fed
f
×
= =
References: P(l), H(l) at 65 C
o
Substance
P(v) in mol s
P(l) in kJ / mol
H(v)
H(l)
in in out out

. .
. .
. .
. .
n H n H
n
H
− −
− −
24 0 24 33
50 2 806 26 0 0
12 6 29 05
50 3245 37 4 0
Vapor: ( ) ( )
H T C dT H T C dT
pl
T
v b pv
T
T
=
65 C
o
b
b
z z
+ +
Δ
Liquid: T) =
65 C
o
(
H C dT
pl
T
z
T H C
b v p
and from Table B.1, from Table B.2
Δ
Energy balance: out out in in
ˆ ˆ 647 kW
Q n H n H
= − =
∑ ∑

8-28
8.52 a. B=benzene; T=toluene
Q
1320 mol/s 25o
C
0.500 mol B/mol
0.500 mol T/mol
n2 mol/s 95o
C
0.735 mol B/mol
0.265 mol T/mol
n3 mol/s 95o
C
0.425 mol B/mol
0.575 mol T/mol
Total mole balance:
Benzene balance: 1320(0.500) =
mol / s
mol / s
1320
0 735 0 425
319
1001
2 3
2 3
2
3
= +
+
U
V
W⇒
=
=
R
S
T
n n
n n
n
n
( . ) ( . )
References: B(l, 25o
C), T(l, 25o
C)
Substance
nin (mol / s) (
Hin kJ / mol)
nout (mol / s) (
Hout kJ / mol)
B(l) 660 0 425 9.838
B(v) -- -- 234 39.91
T(l) 660 0 576 11.78
T(v) -- -- 85 46.06
4
ˆ ˆ 2.42 10 kW
i i i i
out in
Q n H n H
= − = ×
∑ ∑
b. Antoine equation (Table B.4) C torr , C torr
Raoult's law
Benzene: 0.425 torr
Toluene: 0.575 torr
Analyses are inconsistent.
o o
⇒ = =
= ⇒ =
= ⇒ =
U
V
|
W
|⇒ ≠
⇒
p p
P P
P P
P P
B T
* *
.
.
. . ' '
'
95 1176 95 476 9
1176 0 735 680
476 9 0 265 1035
e j e j
b gb g b g
b gb g b g
Possible reasons: The analyses are wrong; the evaporator had not reached steady state when the
samples were taken; the vapor and liquid product streams are not in equilibrium; Raoult’s law is
invalid at the system conditions (not likely).
8.53 Kopp’s rule (Table B.10): C H O
5 12 s
b g — Cp = + + =
5 7 5 12 9 6 17 170
b gb g b gb g
. . J mol
C H O
5 12 l
b g — Cp = + + =
5 12 12 18 25 301
b gb g b gb g J mol
Trouton’s rule — Eq. (8.4-3): ΔHv = + =
0109 113 273 421
. .
b gb g kJ mol
Eq. (8.4-5) ⇒ Δ . .
Hm = + =
0 050 52 273 16 25
b gb g k J mol
Basis:
235 m 273 K 1 kmol 10 mol 1 h
h 389 K 22.4 m STP 1 kmol 3600 s
mol s
3 3
3
b g
= 2 05
.
Neglect enthalpy change for the vapor transition from 116°C to 113°C.
C H O C C H O C C H O C
C H O s C C H O s C
5 12 5 12 5 12
5 12 5 12
v l v
, , ,
, ,
113 113 52
52 25
° → ° → °
→ ° → °
b g b g b g
b g b g

8-29
8.53 (cont’d)
Δ Δ Δ

. .
H H C H C
v pl m ps
= − + − − + −
= − − − + × = −
52 113 25 52
421 301 61 170 27
1
813
b g b g
b gb g b gb g
kJ
mol
16.2
kJ
mol
J
mol
kJ
10 J
kJ mol
3
Required heat transfer: Q H n H
= = =
−
= −
Δ Δ
.
2.05 mol kJ 1 kW
s mol 1 kJ s
kW
813
167
8.54
Basis: 100 kg wet film ⇒
95 kg dry film
5 kg acetone
0.5 kg acetone remain in film
4.5 kg acetone exit in gas phase
90% A evaporation
a.
= 35°C
n1
5 kg C H O( )
Tf
95 kg DF
6 l
3
mol air
, 1.01 atm
n1
0.5 kg C H O( )
95 kg DF
6 l
3
mol air
= 49°C, 1.0 atm
T
1
Ta1
Tf2
4.5 kg C H O( ) (40% sat'd)
6 v
3
a2
Antoine equation (Table B.4) mm Hg
C H O
3 6
⇒ =
p*
.
59118
4.5 kg C H O 1 kmol 10 mol
58.08 kg kmol
mol C H O in exit gas
3 6
3
3 6
= 77 5
. v
b g
⇒ y =
775
775
040 59118
760
405
1
1
.
.
. .
.
+
= ⇒ = =
n
n
mm Hg
mm Hg
171.6 mol 22.4 L STP
mol 95 kg DF
L STP
kg DF
b g b g b g
b. References: Air 25 C C H O 35 C DF 35 C
3 6
° ° °
b g b g b g
, , ,
l
Substance nin

Hin
nout

Hout
DF 95 0 95 1.33 Tf 2 − 35
d i n in kg

H in kJ/kg
C H O
6 14 l
b g
C H O
6 14 v
b g
Air
86.1
—
171.6
0
—
C dT
p
d iair
Ta1
25
z
8.6
77.5
171.6
0.129 Tf 2 − 35
d i
32.3
0.70
n in mol

H in kJ/mol

H C dT H C dT H C T
p l v p v p
A(v) DF
,
= + + = −
z z
d i d i b g
35
86
86
49
35
Δ
Energy balance
ΔH n H n H T T C dT
C dT
T
i i
out
i i
in
f f p
p
f
= − = − + − + − =
⇒ =
− +
∑ ∑ z
z
. . ( ) . .
. .
.
1264 35 111 35 26234 1716 0
1275 35 26234
1716
2 2
25
25
2
d i d i
d i d i
air
T
air
T
a1
a1
c. Ta1
120
= °C ⇒ C dT T
p f
d i d i
air
Ta1
kJ mol C C
25
2
2 78 35 168
z = ⇒ − ° = − °
. .

8-30
8.54 (cont’d)
d. T T
f a
2 34 506
1
= ° ⇒ = °
C C
TE
, T T
f a
2 1
36 552
= ° ⇒ = °
C C
TE
e. In an adiabatic system, when a liquid evaporates, the temperature of the remaining condensed
phase drops. In this problem, the heat transferred from the air goes to (1) vaporize 90% of the
acetone in the feed; (2) raise the temperature of the remaining wet film above what it would be if
the process were adiabatic. If the feed air temperature is above about 530 °C, enough heat is
transferred to keep the film above its inlet temperature of 35 °C; otherwise, the film temperature
drops.
8.55 T p
set psia F
= ≈ °
200 100
b g (Cox chart – Fig. 6.1-4)
a. Basis:
3.00 10 SCF 1 lb - mole
h 359 SCF
lb mole h C H
3
3 8
×
= ⋅
8 357
.
8.357 lb-mole C3H8(v)/h 8.357 lb-mole C3H8(l)/h
200 psia, 100o
F 200 psia, 100o
F

m(lb - mole H O(l) / h
2
m(lb - mole H O(l) / h
2
70o
F 85o
F
The outlet water temperature is 85o
F. It must be less than the outlet propane temperature;
otherwise, heat would be transferred from the water to the propane near the outlet, causing
vaporization rather than condensation of the propane.
b. Energy balance on propane:

Q H n Hv
= = −
Δ Δ = − −
⋅
= − ×
B
8.357 lb moles kJ 0.9486 Btu 453.593 mol
h mol kJ 1 lb mole
Btu
h
Table B.1
1877 675 104
. .
Energy balance on cooling water: Assume no heat loss to surroundings.

Q H mC T
p
= =
Δ Δ ⇒
4
m m
6.75 10 Btu lb F lb cooling water
4500
h 1.0 Btu 15 F h
m
× ⋅°
= =
°

8.56
o
2 2
[kg H O(v)/h]@100 C, 1 atm
m

1000 kg/h, 30o
C
0.200 kg solids/kg
0.800 kg H2O(l)/kg
o
3
2
(kg/h) @ 100 C
0.350 kg solids/kg
0.650 kg H O(l)/kg
m

[
m1 kg H O(v) / h], 1.6 bar, sat'd
2 [
m1 kg H O(l) / h], 1.6 bar, sat'd
2
a. Solids balance: 200 0 35 3
= . m ⇒ =
m3 5714
. kg h slurry
H O balance:
2 800 0 65 5714
2
= +
m . .
b g ⇒ =
m v
2 428 6
. kg h H O
2 b g

Q

8-31
8.56 (cont’d)
References: Solids (0.01°C), H O
2 (l, 0.01o
C)
Substance in
m

Hin out
m

Hout
Solids
H O
2 l
b g
H O
2 v
b g
200
800
—
62.85
125.7
—
200
571.4
428.6
209.6
419.1
2676
( )
kg h
m

H kJ kg
b g

HH O
2
from steam tables
H O
2 , 1.6 bar 1
m
2696.2 1
m
475.4
E.B. 6
1 1
out in
ˆ ˆ 0 1.315 10 2221 0 592 kg steam h
i i i i
Q H m H m H m m
= Δ = − = ⇒ × − = ⇒ =
∑ ∑

b. ( )
592.0 428.6 163 kg h additional steam
− =
c. The cost of compressing and reheating the steam vs. the cost of obtaining it externally.
8.57 Basis: 15,000 kg feed/h. A = acetone, B = acetic acid, C = acetic anhydride
(kg A( )/h)
15000 kg/h
l
0.46 A
0.27 B
0.27 C
348 K, 1 atm
still
n1
303 K
(kg A( )/h)
v
n1
329 K
2
condenser
(kg A( )/h)
l
n1
303 K
Q (kJ/h)
c
reboiler
Q (kJ/h)
r
1% of A in feed
(kg A( )/h)
l
n2
(kg B( )/h)
l
n3
(kg C( )/h)
l
n4
398 K
a. . . ,
n2 0 01 0 46 15 000 69
= =
b gb gb g
kg h kg A h
Acetic acid balance: . ,
n3 0 27 15 000 4050
= =
b gb g kg B h
Acetic anhydride balance: . ,
n4 0 27 15 000 4050
= =
b gb g kg h
Acetone balance: 0 46 15 000 69 6831
1 1
. ,
b gb g= + ⇒ =
n n kg h `
⇓
Distillate product: 6831 kg acetone h
Bottoms product: 69 4050 4050
8169
08%
+ + =
b g kg h
kg h
acetone
49.6% acetic acid
49.6% acetic anhydride
.
b. Energy balance on condenser

8-32
8.57 (cont’d)
C H O K C H O K C H O K
K kJ kg
kg kJ
h kg
kJ h
3 6 3 6 3 6
v l l
H H C dT
Q H n H
v pl
c
, , ,
. . .
.
.
329 329 303
329 520 6 2 3 26 580 4
2 6831 580 4
7 93 10
329
303
6
b g b g b g
b g b gb g
b g
→ →
= − + = − + − = −
= = =
× −
= − ×
z
Δ Δ
Δ Δ
c. Overall process energy balance
Reference states: A(l), B(l), C(l) at 348 K (All
Hm = 0 )
Substance
nin

Hin

nout

Hout
A l, 303 K
b g
A l, 398 K
b g
B l, 398 K
b g
C l, 398 K
b g
—
—
—
—
0
0
0
0
6831
69
4050
4050
–103.5
115.0
109.0
113

n in kg/h

H in kJ/kg
Acetic anhydride (l):
Cp ≈ × + × + ×
⋅°
= ⋅°
4 12 6 18 3 25
2 3
b g b g b g J 1 mol 10 g 1 kJ
mol C 102.1 g 1 kg 10 J
kJ kg C
3
3
.

H T C T
p
b g b g
= − 348 (all substances)
. .
.
Q H Q Q n H n H Q Q n H
c r i i i i r c i i
= ⇒ + = − ⇒ = − + = × + ×
A= = ×
∑ ∑ ∑
Δ
out in out
kJ h
kJ h
7 93 10 2 00 10
0 813 10
6 5
6
e j
(We have neglected heat losses from the still.)
d. H O
2 (saturated at ≈ 11 bars): Δ
Hv = 1999 kJ kg (Table 8.6)

.
Q n H n
r v
= ⇒ =
×
=
H O H O
2 2
kJ h
kJ kg
kg steam h
Δ
813 10
1999
4070
6
8.58 Basis: 5000 kg seawater/h
a. S = Salt
0.945 H O( )
0.965 H O( )
5000 kg/h @ 300 K
l
2
(kg H O( )/h @ 4 bars)
l
n3
0.035 S
113.1 kJ/kg
2
2738 kJ/kg
(kg H O( )/h @ 4 bars)
l
n5
2
605 kJ/kg
(kg/h @ 0.6 bars)
n1
0.055 S
360 kJ/kg
n2
2654 kJ/kg
(kg H O( )/h @ 0.6 bars)
v
2
(kg H O( )/h @ 0.6 bars)
n2
360 kJ/kg
l
2 l
2
(kg H O( )/hr)
l
2
(kg/h @ 0.2 bars)
n3
(kg S/kg)
252 kJ/kg
x
(1 – )
x
n4 kg H O( )/h @ 0.2 bars
v
2
2610 kJ/kg
b. S balance on 1st effect: 0 035 5000 0 055 3182
1 1
. .
b gb g= ⇒ =
n n kg h
Mass balance on 1st effect: 5000 3182 1818
2 2
= + ⇒ =

n n kg h

8-33
8.58 (cont’d)
Energy balance on 1st effect:
Δ .

H n n n
n v
n
n
= ⇒ + + − − =
=
=
=
0 2654 360 605 2738 5000 1131 0
2534
2 1 5
3182
1818
5
1
2
b gb g b gb g b gb g b gb g
b g
kg H O h
2
c. Mass balance on 2nd effect: 3182 3 4
= +

n n (1)
Energy balance on 2nd effect: ΔH = 0
b g
n n n n
n n
n n
4 3 2 1
1 2
6
3 4
2610 252 360 2654 360 0
3182 1818
5316 10 252 2610
+ + − − =
E = =
× = +
,
. (2)
Solve (1) and (2) simultaneously:

n3 1267
= kg h brine solution

n4 1915
= kg h H O
2 v
b g
Production rate of fresh water = + = + =

n n
2 4 1818 1915 3733
b g kg h fresh water
Overall S balance: 0 035 5000 1267 0138
. .
b gb g= ⇒ =
x x kg salt kg
d. The entering steam must be at a higher temperature (and hence a higher saturation pressure) than
that of the liquid to be vaporized for the required heat transfer to take place.
e.
0.965 H O( )
5000 kg/h
l
2
(kg H O( )/h)
v
n5
0.035 S
113.1 kJ/kg
2
2738 kJ/kg
(kg H O( )/h)
n5
605 kJ/kg
(kg brine/h @ 0.2 bar
n1
252 kJ/kg
2610 kJ/kg
3733 kg/h H O( ) @ 0.2 bar
v
2
l
2
Q3
Mass balance: 5000 3733 1267
1 1
= + ⇒ =

n n kg h
Energy balance: Δ
H = 0
d i
3733 2610 1267 252 605 2738 5000 1131 0
4452
5
5
b gb g b gb g b g b gb g
b g
+ + − − =
⇒ =
.

n
n v
kg H O h
2
Which costs more: the additional 1918 kg/hr fresh steam required for the single-stage process, or
the construction and maintenance of the second effect?

8-34
8.59 a. Salt balance: x n x n n
L L L L L
7 7 1 1 1
0 035 5000
0 30
583

.
.
= ⇒ = =
b gb g kg h
Fresh water produced: n n
L L
7 1 5000 583 4417 kg
− = − = fresh water h
b. Final result given in Part (d).
c. Salt balance on effect:
th
i

n x n x x
n x
n
i i L i L i Li
L i L i
Li
L L = ⇒ =
+ +
+ +
b g b g b g b g
1 1
1 1
θ
(1)
Energy balance on effect:
th
i
Δ

H n H n H n H n H n H
n
n H n H n H
H H
vi vi v L v
L
Li Li L L L
L
v L v
L
v L
vi vi Li Li L i L
i
v
i
L
L
= ⇒ + + − − =
⇒ =
+ −
−
− − + + − −
−
+ +
− −
0 0
1 1 1 1 1 1
1
1 1
1 1
b g e j b g e j b g e j
b g
b g e j
e j e j
(2)
Mass balance on effect:
th
i −1
b g

n n n n n n
Li v i L i L i Li v i
= + ⇒ = −
− − − −
b g b g b g b g
1 1 1 1
(3)
d.
P T nL xL nV HL HV
(bar) (K) (kg/h) (kg/h) (kJ/kg) (kJ/kg)
Fresh steam 2.0 393.4 --- --- 981 504.7 2706.3
Effect 1 0.9 369.9 584 0.2997 934 405.2 2670.9
Effect 2 0.7 363.2 1518 0.1153 889 376.8 2660.1
Effect 3 0.5 354.5 2407 0.0727 809 340.6 2646.0
Effect 4 0.3 342.3 3216 0.0544 734 289.3 2625.4
Effect 5 0.2 333.3 3950 0.0443 612 251.5 2609.9
Effect 6 0.1 319.0 4562 0.0384 438 191.8 2584.8
Effect (7) 1.0 300.0 5000 0.0350 --- 113.0 ---

8-35
8.60 a. C C
p v p l
d i d i
= = ⋅°
20 cal (mol C) ; C C R
v v p v
b g d i b g
≈ − ≈ −
⋅°
= ⋅°
10 2 8
cal
mol C
cal (mol C)
b.
n0 (mol N2) n0 (mol N2)
3.00 L@ 93o
C, 1 atm n2 [mol A(v)]
85o
C, P(atm)
n1 (mol A(l) n3 [mol A(l)]
0.70 mL, 93o
C 85o
C, P(atm)
n0
273 93
0100
=
+
=
3.00 L 273 K 1 mol
K 22.4 L STP
mol N2
b g b g .
n A l
1
0 90
15
= =
70.0 mL g 1 mol
mL 42 g
mol
.
. b g
Energy balance ⇒ = ⇒ − =
∑ ∑
ΔU n U n U
i i i i
0 0
out in

c. References: N g C, 1 atm
2 b g b gb g
, A l 85°
Substance
N in mol
in cal mol
in in out out
2
n U n U
n
A l n U
A v n

. . .
.
010 39 8 010 0
15 160 0
20050
3
2
b g
b g − −
C
A l, 93°
b g and N C
2 g U Cv
, :
93 93 85
° = −
b g b g
C cal mol
A v U A v
, : ,
( )
85 20 90 85 20 000 10 85 90 20050
° = − + + − =
b g b g b g
ΔU n n
v v
= ⇒ − − = ⇒ =
0 20050 010 39 8 15 160 0 0 012
1 1
b g b gb g b gb g
. . . . mol A evaporate
⇒
0.012 mol 42 g
mol
g evaporate
A A
A
= 051
.
d. Ideal gas equation of state
P
n n RT
V
=
+
=
+ ⋅
⋅
=
0 2 273 85
1097
b g b g
0.112 mol K 0.08206 L atm
3.00 liters mol K
atm
.
Raoult’s law
p y P
n
n n
P
A A
∗
° = =
+
= = =
85
1097
0 89 3
2
0 2
C
0.012 mol atm
0.112 mol
.117 atm mmHg
b g b g
.
.

8-36
8.61 (a) i) Expt 1 ⇒
F
HG I
KJ =
−
= ⇒ =
m
V
SG
liquid
liquid
kg
2.000 L
kg
L
4 4553 32551
0 600 0 600
. .
. .
b g b g
ii) Expt 2 ⇒ Mass of gas = − = =
32571 32551 0 0020 2 0
. . . .
b gkg kg g
Moles of gas =
−
=
2.000 L 273 K mm Hg 1 mol
363 K 760 mm Hg 22.4 liters STP
mol
763 500
0 0232
b g
b g .
Molecular weight = =
2 0 86
. g 0.0232 mol g mol
b g b g
iii) Expt. 1
2.000 liters 10 cm 0.600 g 1 mol
1 liter cm 86 g
mol
liquid
3 3
3
b g
⇒ = =
n 14
Energy balance: The data show that Cv is independent of temperature
Q U nC T
C
Q
n T
v
v
= =
⇒ = = = ⋅
= = ⋅
Δ Δ
Δ
J
14 mols 2.4 K
J mol K 284.2 K
J
14 mols 2.4 K
J mol K 331.2 K
liquid
b g b gb g
b gb g
800
24
800
24
@
@
⇒ ≡ ⋅
Cv
b gliquid
J mol K
24
Expt. 2 mol from ii
vapor
b g
b g
⇒ =
n 0 0232
.
C a bT Q a bT dT a T T
b
T T
a
b
v T
T
= + ⇒ = + = − + −
L
NM O
QP
−
L
NM O
QP
−
L
NM O
QP
U
V
|
|
W
|
|
⇒
= −
=
z
0 0232 0 0232
2
130 366 9 3630
130 492 7 490 0
4 069
0 05052
2 1 2
2
1
2
2 2
2 2
1
2
. ( ) . ( ) ( )
. ( . . )
. ( . . )
.
.
J = 0.0232 a(366.9 - 363.0) +
b
2
J = 0.0232 a(492.7 - 490.0) +
b
2
⇒ C T
v
b g b g
vapor
J / mol K) K
( . .
⋅ = − +
4 069 0 05052
iv) Liquid: C C
p v
≈ ≡ ⋅
24 J mol K
Vapor: Assuming ideal gas behavior, C C R C
p v v
= + = + ⋅
8 314
. J mol K
⇒ C T
p J mol K K
⋅ = +
b g b g
4 245 0 05052
. .
v) Expt. 3 mm Hg mm Hg
mm Hg
mm Hg
mm Hg
⇒ = = − =
= =
= =
= =
∗
∗
∗
∗
T K p
T K p
T K p
T K p
315 763 564 199
334 401
354 761
379 1521
,
,
,
,
b g

8-37
8.61 (cont’d)
Plot p∗
(log scale) vs. 1 T (linear scale); straight line fit yields
ln .
p
T
∗
=
−
+
3770
17 28
K
b g or p T
∗
= × −
3196 10 3770
7
. expb g
vi) p
T
T
b
b
∗
A
− −
= ⇒ =
−
= × ⇒ =
760
1 17 28 760
3770
2 824 10 3 1
mm Hg K 354 K
Part v
. ln
.
b g
vii)
Δ
Δ Δ

. ,
H
R
H H
v
v v
= ⇒ = ⋅ ⇒ =
A
Part v
K 3770 K J mol K J mol
3770 8 314 31 300
b g b gb g
(b) Basis:
3.5 L feed 273 K 1 mol
s 510 K 22.4 STP
mol s feed gas
lb g= 0 0836
.
Let A denote the drug
0.80 N
0.0836 mol/s @ 510 K
2
[mol A(v)/s]
n1
0.20 A [mol N /s]
n2 2
T(K), saturated with A
(mols A( )/s), 90% of A in feed
n3 l
T(K)
Q(kW) .
.
.
N balance: mol s mol N s
2 2
. . .
n2 0800 0 0836 0 0669
= =
b gb g
90% condensation: . . . .
n A l
3 0 900 0 200 0 0836 0 01505
= × =
b gb g b g
mol s
. . . .
n A v
1
3
0100 0 200 0 0836 167 10
= × = × −
b gb g b g
mol s
Partial pressure of A in outlet gas:
p
n
n n
P p T
A A
=
+
=
×
= =
−
∗

.
( .
1
1 2
3
167 10
760 185
b g b g
mol
0.0686 mol
mm Hg) mm Hg
EPart (a) - (v)
1 17 28 185
3770
381 10 3 1
T
=
−
= × − −
. ln .
.
b g K
⇓
T = 262 K
(c) Reference states: N at 262 K
2 , A l
b g
substance
N in mol s
in J mol
in in out out
2

. .
. .
.
n H n H
n
A v H
A l
0 0669 7286 0 0669 0
0 0167 37575 167 10 31686
0 01505 0
3
b g
b g
×
− −
−

8-38
8.61 (cont’d)
N 510 K K) - K) = C) - C)
[6.24 - (-1.05)] kJ / mol = 7.286 kJ / mol = 7286 J / mol
2
o o
Table B.8
b g: ( ( ( (
H H H H
N N N N
2 2 2 2
510 262 237 11
−
=
B
A(v, 262K):
H C T H K C dT
pl b v pv
Tb
= − + + z
262 359
262
b g b g
Δ
Part (a) results for Tb Cpl Cpv Hv
, , ,
Δ
. .
H
T
= − + + +
L
NM O
QP =
24 354 262 31300 4 245 0 05052
2
31686
2
354
262
b g J mol
A(v, 510K):
H C T H K C dT
pl b v pv
Tb
= − + + =
z
262 354 37575
510
b g b g
Δ J mol
Energy balance: .
Q H n H n H
i i i i
= = − =
−
−
=
∑ ∑
Δ
out in
J s 1 kW cooling
kJ s
kW
1060
10
106
3
8.62 a. Basis: 50 kg wet steaks/min
D.M. = dry meat
0.28 D.M.
50 kg/min @ –26°C
(kg H O( )/min)
m1
0.72 H O( )
60°C
Q(kW)
s
2
v
2 (96% of H O in feed)
2
(kg D.M./min)
m2
(kg H O( )/min)
m3 l
2
50°C
96% vaporization:
. . . min
m v
1 0 96 0 72 50 34 56
= × =
kg min kg H O
2
b g b g
. . . min
m l
3 0 04 0 72 50 144
= × =
kg min kg H O
2
b g b g
Dry meat balance:
. .
m2 0 28 50 14 0
= =
b gb g kg D.M. min
Reference states: Dry meat at − °
26 C , H O 0 C
2 l, °
b g
substance
dry meat in kg min
H O s, C in kJ kg
H O , C
H O , C
in in out out
2
2
2

. .
.
.
.
m H m H
m
H
l
v
14 0 0 14 0 105
26 36 0 390
50 144 209
60 34 56 2599
− ° − − −
° − −
° − −
b g
b g
b g
Dry meat:
H Cp
50 50 26 105
° = − − =
°
⋅ °
=
C
1.38 kJ 76 C
kg C
kJ kg
b g b g
H O s, C H O , C H O s, C H O s, 26 C
2 2 2 2
− ° ° → ° → − °
26 0 0
b g b g b g b g
: l

8-39
8.62 (cont’d)
Δ Δ
.
H H C dT
m p
= − ° + =
−
+
− °
⋅°
= −
−
A
z
0
217
390
0
26
C
6.01 kJ 1 mol 10 g
mol 18.02 g 1 kg
kJ 26 C
kg C
kJ kg
3
Table B.1
b g
H O C
2 l, :
50°
b g H O C H O C
2 2
l l
, ,
0 50
° → °
b g b g
Δ
H C dT
p
= =
− °
° =
z A
0
50 0
209
0.0754 kJ 50 C 1 mol 1000 g
mol C 18.02 g 1 kg kJ kg
Table B.2
b g
H O , C H O , C H O , 100 C H O , C H O , C
2 2 2 2 2
v l l v v
60 0 100 60
° ° → ° → ° → °
b g b g b g b g b g
:
Δ
Δ

.
.

H C dT
Hv
p
=
− °
⋅° + +
= =
A A A
z
0 0754 kJ 100 0
40 656
2599
100
60
b g
d i
d i
C
mol C
kJ
mol
46.830 kJ 1 mol 1000 g
mol 18.02 g 1 kg
kJ kg
Table B.2
Table B.1
Table B.2
2
H O(v)
Energy balance:
Q H m H m H
i i i i
= = − =
×
=
∑ ∑
Δ
out in
5
1.06 10 kJ 1 min 1 kW
min s 1 kJ s

60
1760 kW
8.63 Basis: 20,000 kg/h ice crystallized. S = solids in juice. W = water
0.88 H O( )(W)
(kg/h) juice
m1
l
2
0.12 solids(S)
20°C
(kg/h)
m2
preconcentrate
(kg S/kg)
x2
(kg W/kg)
x2
(1 – )
freezer
Qf
Slurry(10% ice), –7°C
20,000 kg W( )/h
s
kg residue/h
m4
0.45 kg S/kg
0.55 kg W( )/kg
l
filter
(kg/h) product
m5
0.45 kg S/kg
0.55 kg W/kg
20,000 kg W( )/h
s
(kg/h), 0.45 S, 0.55 W
m4
separator
20,000 kg W( )/h
s
(kg/h), 0°C
m3
0.45 kg S/kg
0.45 kg W( )/kg
l
. .
.
.
.
.
.
(a) 10% ice in slurry ⇒ = ⇒ =
20000 10
90
180000
4
4

m
m kg h concentrate leaving freezer
Overall S balance:
27273 kg h feed
7273 kg h concentrate product
012 0 45
20000
1 5
1 5
1
5
. .

m m
m m
m
m
=
= +
U
V
W
⇒
=
=
Mass balance on filter: 20000 20000 172730
4 5 6
180000
7273
6
4
5
+ + + + ⇒ =
=
=

m m m m
m
m
kg h recycle
Mass balance on mixing point:
27273 172730 2 000 10
2 2
5
+ = ⇒ = ×
.
m m kg h preconcentrate

8-40
8.63 (Cont’d)
S balance on mixing point:
012 27273 0 45 172730 2 000 10 100% 405%
5
2 2
. . . .
b gb g b gb g
+ = × ⇒ ⋅ =
X X S
(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs:
fresh feed and recycle streams; output; slurry leaving freezer)
Refs: S, H O
2 l
b g at − °
7 C
substance
12% soln kg h
45% soln kJ kg
H O s
in in out out
2

m H m H
m
H
27273 108
172730 28 180000 0
20000 337
− −
− − −
b g
b g
b g
Solutions: .
H T T
b g b g
= − −
4 00 7 kJ kg
Ice:
.
H H T H
m m
= − − ° ≈ − °
= − ⇒ −
D
Δ Δ
C C
kJ mol kJ kg
Table B.1
b g b g
0
6 0095 337
E.B. .
Q H m H m H
c i i i i
= = − =
− ×
= −
∑ ∑
Δ
out in
kJ 1 h 1 kW
h 3600 s 1 kJ s
kW
1452 10
4030
7
8.64 a. B=n-butane, I=iso-butane, hf=heating fluid. ( ) .
Cp hf = ⋅
2 62 kJ / kg C
o
d i
24.5 kmol/h @ 10o
C, P (bar) 24.5 kmol/h @ 180o
C
0.35 kmol B(l)/h 0.35 kmol B(l)/h
(
Q kW)

m (kg HF / h), T( C)
o

m (kg HF / h), 215 C
o
From the Cox chart (Figure 6.1-4)
p p
p p p x p x p
B
o
I
o
min B I B B I I
C psi, C psi
psi
1.01325 bar
14.696 psi
bar
* *
* *
. .
10 22 10 32
285 196
d i d i
= =
= + = + =
F
HG I
KJ =
b. B l, 10 C B v, 10 C B v, 180 C
I l, 10 C I v, 10 C I v, 180 C
o o o
o o o
d i d i d i
d i d i d i
Δ Δ
Δ Δ

H H
H H
v
v
⎯ →
⎯
⎯ ⎯ →
⎯
⎯
⎯ →
⎯
⎯ ⎯ →
⎯
⎯
1
2
Assume temperature remains constant during vaporization.
Assume mixture vaporizes at 10o
C i.e. won’t vaporize at respective boiling points
as a pure component.

8-41
8.64 (cont’d)
References: B(l, 10o
C), I(l, 10o
C)
substance
nin mol / h
b g
Hin kJ / mol
b g
nout mol / h
b g
Hout kJ / mol
b g
B (l) 8575 0 -- --
B (v) -- -- 8575 42.21
I (l) 15925 0 -- --
I (v) -- -- 15925 41.01
.
.
. .
H H C
H H C
H n H n H
out v p
out v p
i i
out
i i
in
d i d i d i
d i d i d i
b g b g
B B B
I I I
kJ / mol
kJ / mol
= + =
= + =
= − = −
z
z
∑ ∑
Δ
Δ
Δ
10
180
10
180
42 21
4101
8575 42 21 15825 4101
Δ .
H = ×
1015 106
kJ / h
c. Q m
= × ⋅ −
1015 10 2 62 215 45
6
. .
kJ / h = kJ / kg C C
hf
o o
d i b g

mhf kg / h
= 2280
d. 2540 2 62 215 45 1131 106
kg / h kJ / kg C C kJ / h
o o
b g d i b g
. .
⋅ − = ×
Heat transfer rate kJ / h
= × − × = ×
1131 10 1015 10 116 10
6 6 5
. . .
e. The heat loss leads to a pumping cost for the additional heating fluid and a greater
heating cost to raise the additional fluid back to 215o
C.
f. Adding the insulation reduces the costs given in part (e). The insulation is
probably preferable since it is a one-time cost and the other costs continue as long
as the process runs. The final decision would depend on how long it would take for
the savings to make up for the cost of buying and installing the insulation.
8.65 (a) Basis: 100 g of mixture, SGBenzene=0.879: SGToluene=0.866
ntotal
total 3 3
3
g
78.11 g / mol
g
92.13 g / mol
mol mol
V
g
0.879 g / cm
g
0.866 g / cm
cm
= + = + =
= + =
50 50
0 640 0542 1183
50 50
114 6
( . . ) .
.
xf
d iC H
6 6
6 6
6 6
mol C H
1.183 mol
mol C H mol
= =
0 640
0541
.
.
Actual feed:
32 5 1183 1
114.6 cm
9319
3
. .
.
m 10 cm mol mixture h
h 1 m mixture 3600 s
mol / s
3 6
3 3
=
T p
= ° ⇒ =
∗
90 1021
C mm Hg
C H
6 6
, pC H
7 8
mm Hg
∗
= 407 (from Table 6.1-1)
Raoult's law:
mmHg atm
760 mmHg
atm atm
C H C H C H C H
6 6 6 6 7 8 7 8
p x p x p
P
tot = + = +
= = ⇒
∗ ∗
0541 1021 0 459 407
739 2 1
0 973 0 973
0
. .
.
. .
b gb g b gb g

8-42
8.65 (cont’d)
(b) T p
= ° ⇒ =
∗
75 648
C mm Hg
C H
6 6
, pC H
7 8
mm Hg
∗
= 244 (from Table 6.1-1)
Raoult's law ⇒ = + = +
∗ ∗
p x p x p
tank C H C H C H C H
6 6 6 6 7 8 7 8
0 439 648 0561 244
. .
b gb g b gb g
= + ⇒ =
284 137
b gmm Hg = 421 mmHg P 0.554 atm
tank
y v
C H 6 6
6 6
284 mm Hg
421 mm Hg
mol C H mol
= = 0 675
. b g
0.541 C H ( )
nv
l
6
93.19 mol/s
6
0.459 C H ( )
l
7 8
90°C, P0 atm
(mol/s), 75°C
0.675 C H ( )
v
6 6
0.325 C H ( )
v
7 8
nL (mol/s), 75°C
0.439 C6 H6 (l )
0.541 C7H8 (l )
0.554 atm
Mole balance: =
C H balance: 0.541
40.27 mol vapor s
52.92 mol liquid s
6 6
9319
9319 0 675 0 439
.
. . .
n n
n n
n
n
v L
v L
v
L
+
= +
U
V
W
⇒
=
=
b gb g
(c) Reference states: C H , C H at 75 C
6 6 6 6
l l
b g b g °
Substance
C H in mol s
C H in kJ mol
C H
C H
in in out out
6 6
6 6
7 8
7 8

. .
. . .
. .
. . .
n H n H
v n
l H
v
l
b g
b g
b g
b g
− −
− −
2718 310
50 41 216 2323 0
1309 353
42 78 2 64 29 69 0
C H , 90 C kJ mol
6 6 l H
° = − =
b g b gb g
: . .
0144 90 75 216
C H , 90 C kJ mol
7 8 l H
° = − =
b g b gb g
: . .
0176 90 75 2 64
C H , 75 C
kJ mol
6 6
C
v H T dT
Hv
° = − + + + ×
=
A
−
°
z
b g b gb g
b g
: . . . . .
.
.
0144 801 75 30 77 0 074 0 330 10
310
80 1
3
80.1
75
Δ
C H , 75 C
kJ mol
7 8 v H T dT
° = − + + + ×
=
−
z
b g b gb g
: . . . . .
.
0176 110 6 75 3347 0 0942 0 380 10
353
3
110.6
75
Energy balance:
Q H n H n H
i i i i
= = − = =
∑ ∑
Δ
out in
1 kW
s 1 kJ s
1082 kW
1082 kJ
(d) The feed composition changed; the chromatographic analysis is wrong; the heating rate
changed; the system is not at steady state; Raoult’s law and/or the Antoine equation are
only approximations; the vapor and liquid streams are not in equilibrium.
(e) Heat is required to vaporize a liquid and heat is lost from any vessel for which TTambient.
If insufficient heat is provided to the vessel, the temperature drops. To run the experiment
isothermally, a greater heating rate is required.

8-43
8.66 a. Basis: 1 mol feed/s
1 mol/s @ TF
o
C
xF mol A/mol
(1-xF) mol B/mol
nV mol vapor/s @ T, P
y mol A/mol
(1-y) mol B/mol
nL mol vapor/s @ T, P
x mol A/mol
(1-x) mol B/mol
vapor and liquid streams
in equilibrium
Raoult's law ⇒ ⋅ + − ⋅ = ⇒ =
−
−
∗ ∗
∗
∗ ∗
x p T x p T P x
P p T
p T p T
A B
B
A B
b g b g b g b g
b g b g
1 (1)
p y P x p T y
x p T
P
A A
A
= ⋅ = ⋅ ⇒ =
⋅
∗
∗
b g b g (2)
Mole balance: 1 1
= + ⇒ = −

n n n n
L V V L (4)
A balance: x y n x n n
y x
y x
F V L
n
L
F
b gb g
1 = ⋅ + ⋅ ⎯ →
⎯⎯⎯⎯⎯⎯
⎯ =
−
−

Substitute for from (4)
v
(3)
Energy balance: Δ
H n H n H
i i i i
= − =
∑ ∑
out in
0 (5)
b.
Tref(deg.C) = 25
Compound A B C al av bv Tbp DHv
n-pentane 6.84471 1060.793 231.541 0.195 0.115 3.41E-04 36.07 25.77
n-hexane 6.88555 1175.817 224.867 0.216 0.137 4.09E-04 68.74 28.85
xF 0.5 0.5 0.5
Tf(deg.C) 110 110 150
P(mm Hg) 760 1000 1000
HAF(kJ/mol) 16.6 16.6 24.4
HBF(kJ/mol) 18.4 18.4 27.0
T(deg.C) 51.8 60.0 62.3
pA*(mm Hg) 1262 1609 1714
pB*(mm Hg) 432 573 617
x 0.395 0.412 0.349
y 0.656 0.663 0.598
nL(mol/s) 0.598 0.648 0.394
nV(mol/s) 0.402 0.352 0.606
HAL(kJ/mol) 5.2 6.8 7.3
HBL(kJ/mol) 5.8 7.6 8.0
HAV(kJ/mol) 31.4 32.5 32.8
HBV(kJ/mol) 42.4 43.7 44.1
DH(kJ/s) 0.00 0.00 0.00

8-44
8.66 (cont’d)
c.
C* PROGRAM FOR PROBLEM 8.66
IMPLICIT REAL (N)
READ (5, 1) A1, B1, C1, A2, B2, C2
C* ANTOINE EQUATION COEFFICIENTS FOR A AND B
1 FORMAT (8F10.4)
READ (5, 1) TRA, TRB
C* ARBITRARY REFERENCE TEMPERATURES (DEG.C.) FOR A AND B
READ (5, 1) CAL, TBPA, DHVA, CAV1, CAV2
READ (5, 1) CBL, TBPB, DHVB, CBV1, CBV2
C* CP(LIQ, KS/MBL-DEG.C.), NORMAL BOILING POINT (DEG.C), HEAT
OF
VAPORIZATION
C* (KJ/MOL), COEFFICIENTS OF CP(VAP., KJ/MOL-DEG.C) = CV1 +
CV2*T(DEG.C)
READ (5, 1) XF, TF, P
C* MOLE FRACTION OF A IN FEED, FEED TEMP.(DEG.C), EVAPORATOR
PRESSURE (MMHG)
WRITE (6, 2) TF, XF, P
2 FORMAT (1H0, 'FEEDbATb', F6.1, 'bDEG.CbCONTAINSb', F6.3,'
bMOLESbA/MOLEbT
*OTAL'//1X'EVAPORATORbPRESSUREb=', E11.4, 'bMMbHG'/)
ITER = 0
DT = 0.5
HAF = CAL*(TF – TRA)
HBF = CBL*(TF – TRB)
F1 = XF*HAF + (1.0 – XF)*HBF
F2 = CAL*(TBPA – TRA) + DHVA – CAV1*TBPA – 0.5*CAV2*TBPA**2
F3 = CBL*(TBPB – TRB) + DHVB – CBV1*TBPB – 0.5*CBV2*TBPB**2
T = TF
20 INTER = ITER + 1
IF(ITER – 200) 30, 30, 25
25 WRITE (6, 3)
3
30
FORMAT (1H0, 'NO CONVERGENCE')
STOP
PAV = 10.0** (A1 – B1/(T + C1))
PAV = 10.0** (A2 – B2/(T + C2))
XL = (P – PBV)/(PAV – PBV)
XV = XL*PAV/P
NL = (XV – XF)/(XV – XL)
NV = 1.0 – NL
IF (XL.LE.00.OR.XL.GE.1.0.OR.NL.LE.0.0.OR.NL.GE.1.0) GO TO 45
HAL = CAL*(T – TRA)
HBL = CBL*(T – TRB)
HAV = F2 + CAV1*T + 0.5*CAV2*T**2
HBV = F3 + CBV1*T + 0.5*CBV2*T**2

8-45
8.66(cont’d)
DELH = NL *(XL*HAL + (1.0 – XL)*HBL) + NV*(XV*HAV + (1.0 –
XV)*HBV) – F1
WRITE (6, 4) T, NL, NV, DELH
4 FORMAT (1Hb, 5X' Tb=', F6.1, 3X' NLb=', F7.4, 3X' NVb=', F7.4, 3X'DELHb
=',* E11.4)
WRITE (6, 5) PAV, PBV, XL, HAL, HBL, XV, HAV, HBV
5 FORMAT (1Hb, 5X' PAV, PBVb=', 2F8.1, 3X' XL, HAL, HBLb=', F7.4,
2E13.4,3X' XV, HAV, HBVb=', F7.4, 2E13.4/)
IF (DELH) 50, 50, 40
40 DHOLD = DELH
TOLD = T
45 T = T – DT
GO TO 20
50 T = (T*DHOLD – TOLD*DELH)/(DHOLD – DELH)
PAV = 10.0**(A1 – B1/(T + C1))
PBV = 10.0**(A2 – B2/(T + C2))
XL = (P – PBV)/(PAV – PBV)
XV = XL * PAV/P
NL = (XV – XF)/(XV – XL)
NV = 1.0 – NL
WRITE (6, 6) T, NL, XL, NV, XV
6 FORMAT (1H0, 'PROCEDUREbCONVERGED'//3X'EVAPORATORb
TEMPERATUREb=', F6.
*1//3X' LIQUIDbPRODUCTb--', F6.3, 'bMOLEbCONTAININGb', F6.3,
'bMOLEbA/
*MOLEbTOTAL'//3X' VAPORbPRODUCTb--', F6.3,
MOLEbCONTAININGb,' F6.3,
*'bMOLEbA/MOLEb TOTAL')
STOP
END
$DATA (Fields of 10 Columns)
Solution:
Tevaportor C
= °
52 2
.
n x
L = =
0552 0 383
. .
mol, mol C H mol liquid
C H liquid 5 12
5 12
d i
n x
v = =
0 448 0 644
. .
mol, mol C H mol liquid
C H vapor 5 12
5 12
d i

8-46
8.67 Basis:
2500 kmol product 1 kmol condensate
h .25 kmol product
kmol h fed to condenser
=10 000
,
1090 kmol/h C H ( )
v
3 8
7520 kmol/h -C H ( )
v
4 10
i
1390 kmol/h -C H ( )
v
4 10
n
saturated vapor at Tf, P
P (mm Hg)
m1 , (kg/h) at T1
1090 kmol/h C H ( )
l
3 8
7520 kmol/h -C H ( )
l
4 10
i
1390 kmol/h -C H ( )
l
4 10
n
Tout
m1 (kg/h) at T2
2
.
.
(a) Refrigerant: T o
out C
= 0 , T T o
1 2 6
= = − C .
Antoine constants A B C
C H
3 8
7.58163 1133.65 283.26
i − C H
4 10
6.78866 899.617 241.942
n − C H
4 10
6.82485 943.453 239.711
Calculate for out bubble pt.
P T T
=
( ) ( ) ( ) ( )
*
0 C 0.109 3797 mm Hg 0.752 1176 mm Hg 0.139 775 mm Hg
1406 mm Hg
i i
i
P x p
P
= ° = + +
⇒ =
∑
Dew pt. ( )
( )
dp *
1 0
i
f f
i i f
y
T T f T P
p T
= ⇒ = − =
∑ trial error to find o
5.00 C
f f
T T
⇒ =
Refs: C H
3 8 l
b g, C H
4 10 l
b g at 0 °C, Refrigerant @ –6°C
Assume: Δ
H T
v b
b g, Table B.1
substance
nin

Hin

nout

Hout
↓
C H
3 8
i − C H
4 10
n − C H
4 10
1090
7520
1390
19110
21740
22760
1090
7520
1390
0
0
0

n (kmol/h)

H (kJ/kmol)
U
V
|
W
|
= ° +
z

.95
H H
C dT
v
p
2
0
4
0
vapor C
Table B.2
b g b g
b g
Δ
Refrigerant
m1
0
m1
151
m (kg/h)

H (kJ/kmol)
U
V
W
=

H Hv
Δ
E.B.:
ΔH n H n H m m
i i i i
= − = ⇒ − × = ⇒ = ×
∑ ∑
. .
out in
kg h refrigerant
0 151 216 10 0 143 10
1
6
1
6

8-47
8.67 (cont’d)
(b) Cooling water: Tout C
= °
40 , T2 34
= °C , T1 25
= °C
( ) ( ) ( ) ( )
( )
( )
*
*
40 C 0.109 11,877 0.752 3961 0.139 2831 4667 mm Hg
1 0 45.7 C
i i
i
T E
i
f f
i i f
P x p
y
f T P T
p T
+
= ° = + + =
= − = ⇒ = °
∑
∑
Refs: C H
3 8 l
b g, C H
4 10 l
b g @ 40°C, H O
2 l
b g @ 25°C.
Δ . . .
H m m
= ⇒ − × = ⇒ = ×
0 37 7 217 10 0 574 10
1
8
1
6
kg H O / h
2
(c) Cost of refrigerant pumping and recompression, cost of cooling water pumping, cost of
maintaining system at the higher pressure of part (b).
8.68 Basis: 100 mol leaving conversion reactor
(mol H O( ))
0.37 g HCHO/g ( mol/min)
conversion
n3 (mol O )
2
n3 (mol N )
2
3.76
n4 (mol H O( ))
2 v
reactor
100 mol, 600°C, 1 atm
0.199 mol HCHO/mol
0.0834 mol CH OH/mol
3
0.303 mol N /mol
2
0.0083 mol O /mol
2
0.050 mol H /mol
2
0.356 mol H O( )/mol
2 v
n1 (mol CH OH( ))
3 l n2 (mol CH OH( ))
3 l
n8 (mol CH OH( ))
3 l
H O( )
3.1 bars, sat'd
2 v
mw1 (kg H O( ))
2 l
3.1 bars, sat'd
H O( )
45°C
2 l
mw2 (kg H O( ))
2 l
30°C
Q(kJ)
n8 (mol CH OH)
3
2.5
distillation
sat'd, 1 atm
CH OH( ), 1 atm, sat'd
3 l
Product solution
n7 (mol)
x1
0.01 g CH OH/g ( mol/min)
x2
3
0.82 g H O/g ( mol/min)
x3
3
(mol HCHO)
3
n6a
(mol CH OH( ))
n6b l
2
n6c l
88°C, 1 atm
absorption
Absorber off-gas
(mol N )
n5 2
(mol O )
n5 2
(mol H )
n5 2
(mol H O( )), sat'd
n5 2 v
(mol HCHO( )), 200 ppm
n5 v
27°C, 1 atm
mw3 (kg H O( ))
2 l
30°C
145°C 100°C
a
b
c
d
e
( )
l
a. Strategy
C balance on conversion reactor ⇒ n2 , N2 balance on conversion reactor ⇒ n3
H balance on conversion reactor ⇒ n4 , (O balance on conversion reactor to check
consistency)
N2 balance on absorber ⇒ n a
5 , O2 balance on absorber ⇒ n b
5
H2 balance on absorber ⇒ n e
5
H O saturation of absorber off - gas
200 ppm HCHO in absorber off - gas
2 U
V
W
⇒ n n
d b
5 5
,
20o
C

8-48
8.68 (cont’d)
HCHO balance on absorber ⇒ n a
6 , CH OH
3 balance on absorber ⇒ n b
6
Wt. fractions of product solution ⇒ x x x
1 2 3
, ,
HCHO balance on distillation column ⇒ n7
CH OH
3 balance on distillation column ⇒ n8
CH OH
3 balance on recycle mixing point ⇒ n1
Energy balance on waste heat boiler ⇒ mw1 , E.B. on cooler ⇒ mw2
Energy balance on reboiler ⇒ Q
C balance on conversion reactor:
n2 19 9 8 34 28 24
= + =
. . .
mol HCHO mol CH OH mol CH OH
3 3
N balance on conversion reactor:
2
376 30 3 8 06
3 3
. . .
n n
= ⇒ = mol O2 , 376 8 06 30 3
. . .
× = mol N feed
2
H balance on conversion reactor:
n n
4 4
2 28 24 4 19 9 2 8 34 4 5 2 356 2 20 7
b g b g b g b g b g b g
+ − + + + ⇒ =
. . . . . mol H O fed
2
O balance: 65.1 mol O in, 65.5 mol O out. Accept (precision error)
N balance on absorber:
2 30 3 30 3
5 5
. .
= ⇒ =
n n
a a mol N2
O balance on absorber:
2 083 083
5 5
. .
= ⇒ =
n n
b b mol O2
H balance on absorber:
2 500 500
5 5
. .
= ⇒ =
n n
c c mol H2
H O saturation of off - gas:
2
y
p
P
n
n n
w
w d
d e
=
°
= =
+ + + +
L
NM O
QP
*
.
. . .
27 26 739
30 3 083 500
5
5 5
C mm Hg
760 mm Hg
b g
⇒ = + +
⇒
+ +
=
U
V
|
|
|
W
|
|
|
⇒
=
= × −
n n n
n
n n
n
n
d d e
e
d e
d
e
5 5 5
5
5 5
6
5
5
3
0 03518 3613 1
3613
200
10
2
1318
7 49 10
. .
.
.
.
b g
200 ppm HCHO in off gas:
mol H O
mol HCHO
solve
2
Moles of absorber off-gas = + + + =
n n n n
a b c e
5 5 5 5 37 46
. mol off - gas
HCHO balance on absorber: 19 9 7 49 10 19 89
6
3
6
. . .
= + × ⇒ −
−
n n
a a mol HCHO
CH OH balance on absorber:
3 8 34 8 34
6 6
. .
= ⇒ =
n n
b b mol CH OH
3
Product solution
Basis -100 g 37.0 g HCHO mol HCHO
1.0 g CH OH 0.031 mol CH OH
62.0 g H O 3.441 mol H O
mol HCHO mol
mol CH OH mol
mol H O mol
%MW
3 3
2 2
3
2
⇒ ⇒
⇒
⇒
U
V
|
|
W
|
|
⇒
=
=
=
1232 0 262
0 006
0 732
1
2
3
. .
.
.
x
x
x

8-49
8.68 (cont’d)
HCHO balance on distillation column (include the condenser + reflux stream within the
system for this and the next balance):
19 89 0 262 759
7 7
. . .
= ⇒ =
n n mol product
CH OH balance on distillation column:
3
8 34 0 006 759 7 88
8 8
. . . .
= + ⇒ =
b g n n mol CH OH
3
CH OH balance on recycle mixing point:
3
n n n n
1 8 2 1 28 24 7 83 20 36
+ = ⇒ = − =
. . . mol CH OH fresh feed
3
Summary of requested material balance results:
n l
n
n l
n
1
2
3
4
20 4
75.9 mol p
7
37.5 mol a
=
=
=
=
. mol CH OH fresh feed
roduct solution
.88 mol CH OH recycle
bsorber off - gas
3
3
b g
b g
Waste heat boiler:
Refs: HCHO C
v,145°
b g, CH OH C
3 v,145°
b g; N2 , O2 , H2 , H O
2 v
b g at 25°C for product
gas, H O triple point
2 l,
b g for boiler water
substance nin

Hin
nout

Hout
HCHO
CH OH
3
N2
O2
H2
H O
2
19.9
8.34
30.3
0.83
5.0
35.6
22.55
32.02
17.39
18.41
16.81
20.91
19.9
8.34
30.3
0.83
5.0
35.6
0
0
3.51
3.60
3.47
4.09
n (mol)

H (kJ/mol)
U
V
W
= z

H C dT
p
T
145
U
V
|
|
W
|
|
= −

H C T T
p b g 25
H O
2
(boiler)
mw1
566.2 mw1
2726.32 m (kg)

H (kJ/kg)
U
V
W

H from steam tables
E.B. ΔH n H n H m m
i i i i w w
= − = ⇒ − + = ⇒ =
∑ ∑
.
out in
kg 3.1 bar steam
0 1814 2160 0 084
1 1

8-50
8.68 (cont’d)
Gas cooler: Same refs. as above for product gas, H O C
2 l, 30°
b g for cooling water
substance nin

Hin
nout

Hout
HCHO
CH OH
3
N2
O2
H2
H O
2
19.9
8.34
30.3
0.83
5.0
35.6
0
0
3.51
3.60
3.47
4.09
19.9
8.34
30.3
0.83
5.0
35.6
–1.78
–2.38
2.19
2.24
2.16
2.54
n (mol)

H (kJ/mol)
H O
2
(coolant)
mw2
0 mw2
62.76 m (kg)

H (kJ/kg)
.
H T
=
⋅°
− °
4184 30
kJ
kg C
C
b g
E.B. ΔH n H n H m m
i i i i w w
= − = ⇒ − + = ⇒ =
∑ ∑
. .
out in
.52 kg cooling water
0 1581 62 6 0 2
2 2
Condenser: CH OH
3 condensed = + = =
n n
8 8
2 5 35 7 88 27 58
. . . .
b gb g mol CH OH condensed
3
E.B.:
Q n Hv
= − = −
= −
Δ . .
1 27 58 3527
973
atm mol kJ mol
kJ (transferred from condenser)
b g b gb g
b.
3.6 10 tonne / y 10 g 1 yr 1 d
1 metric ton 350 d 24 h
g h product soln
4 6
×
= ×
4 286 106
.
⇒
× = × ⇒ ×
× = × ⇒
× = × ⇒ ×
U
V
|
|
W
|
|
⇒ × ⇒
×
= −
0 37 4 286 10 1586 10 5281 10
0 01 4 286 10 4 286 10 1338
0 62 4 286 10 2 657 10 1475 10
2 016 10
2 016 10
759
2657
6 6 4
6 6
6 6 5
5
5
1
. . . .
. . .
. . . .
.
.
.
b gd i
b gd i
b gd i
g HCHO h mol HCHO h
g CH OH h mol CH OH h
g H O h mol H O h
mol h Scale factor =
mol h
mol
h
3 3
2 2
8.69 (a) For 24°C and 50% relative humidity, from Figure 8.4-1,
Absolute humidity = 0.0093 kg water / kg DA, Humid volume 0.856 m kg DA
Specific enthalpy = (48 - 0.2) kJ / kg DA = 47.8 kJ / kg DA Dew point =13 C C
3
o o
≈
=
/
, , Twb 17
(b) 24o
C (Tdb )
(c) 13o
C (Dew point)
(d) Water evaporates, causing your skin temperature to drop. T Twb
skin
o
C (
≈ 13 ). At 98%
R.H. the rate of evaporation would be lower, T T
skin ambient
would be closer to , and you
would not feel as cold.

8-51
8.70 V
m
h
room
3
DA
3 o
3
m
o m
a
m 2
m
m 2 m
ft . DA = dry air.
=
ft lb - mol R
0.7302 ft atm
lb DA
lb - mol
atm
550 R
lb DA
lb H O
lb DA
lb H O / lb DA
=
⋅
⋅
=
= =
141
140 29 1
101
0 205
101
0 0203
.
.
.
.
From the psychrometric chart, F,
h T F V ft / lb DA
T F H Btu / lb
db
o
a
r wb
o 3
m
dew point
o
m
T h
= =
= = =
= = − ≅
90 0 0903
67% 805 14 3
77 3 44 0 011 439
.
. .
. . . .
8.71
T
T
hr
db
ab
C
C
He wins
= °
= °
⇒ =
35
27
55%
8.72 a. T T
h h
T
r a
wb
db dew point
Fig. 8.4-1 2
C, C
kg H O kg dry air
C
= ° = ° ⇒
= =
= °
40 20
33%, 0 0148
255
.
.
b. Mass of dry air: mda = = × −
2.00 L 1 m 1 kg dry air
10 L 0.92 m
kg dry air
3
3 3
2 2 10 3
.
↑ from Fig. 8.4-1
Mass of water:
2.2 10 kg dry air 0.0148 kg H O 10 g
1 kg dry air 1 kg
g H O
2
3
2
×
=
−3
0 033
.
c. . . .
H 40 78 0 0 65 77 4
° ≈ − =
C, 33% relative humidity kJ kg dry air kJ kg dry air
b g b g
.
H 20 57 5
° ≈
C, saturated kJ kg dry air
b g (both values from Fig. 8.4-1)
ΔH40 20
3
57 5 77 4
44
→
−
=
× −
= −
2.2 10 kg dry air kJ 10 J
kg dry air 1 kJ
J
3
. .
b g
d. Energy balance: closed system
n
Q U n U n H R T H nR T
=
×
+ =
= = = − = −
− −
− °
⋅ °
= −
−
2.2 10 kg dry air 10 g 1 mol
1 kg 29 g
0.033 g H O 1 mol
18 g
mol
= J
0.078 mol 8.314 J 20 C 1 K
mol K 1 C
J(23 J transferred from the air)
3
2
3
0 078
44
40
31
.

Δ Δ Δ Δ Δ Δ
d i
b g

8-52
8.73 (a)
400 2 44
97 56
10 0
kg kg water
kg air
kg water evaporates / min
.
min .
.
=
(b) ha = =
10
400
0 025
kg H O min
kg dry air min
kg H O kg dry air
2
2
. , Tdb C
= °
50
Fig. 8.4-1
dew point
kJ kg dry air C, C
. , .
H T h T
wb r
= − = = ° = = °
116 11 115 33 32%, 285
b g
(c) Tdb C
= °
10 , saturated ⇒ = =
h H
a 0 0077 29 5
. , .
kg H O kg dry air kJ kg dry air
2
(d)
400 kg dry air 0.0250 kg H O
min kg dry air
kg H O min condense
2
2
−
=
0 0077
6 92
.
.
b g
References: Dry air at 0 C, H O at 0 C
2
° °
l
b g
substance
min

Hin

mout

Hout
Air
H O
2 l
b g
400
—
115
—
400
6.92
29.5
42

mair in kg dry air/min,
mH O
2
in kg/min

Hair in kJ/kg dry air,
HH O
2
in kJ/kg
H O C H O C
2 2
l l
, ,
0 20
° → °
b g b g:
.
H =
− °
⋅°
=
754 10 0
42
J 1 mol C 1 kJ 10 g
mol C 18 g 10 J 1 kg
kJ kg
3
3
b g
out in
34027.8 kJ 1 min 1 kW
ˆ ˆ 565 kW
min 60 s 1 kJ/s
i i i i
Q H m H m H
−
= Δ = − = = −
∑ ∑

(e) T50°C, because the heat required to evaporate the water would be transferred from the
air, causing its temperature to drop. To calculate (Tair)in, you would need to know the
flow rate, heat capacity and temperature change of the solids.
8.74 a. Outside air: Tdb F
= °
87 , h h
r a
= ⇒ =
80% 0 0226
. lb H O lb D.A.
m 2 m ,
. . .
H = − =
455 0 01 455 Btu lb D.A.
m
Room air: Tdb F
= °
75 , h h
r a
= ⇒ =
40% 0 0075
. lb H O lb D.A.
m 2 m ,
. . .
H = − =
26 2 0 02 26 2 Btu lb D.A.
m
Delivered air: Tdb F
= °
55 , ha = 0 0075
. lb H O lb D.A.
m 2 m
⇒ . . .
H = − =
214 0 02 214 Btu lb D.A.
m , .
V = 1307 ft lb D.A.
3
m
Dry air delivered:
1,000 ft lb D.A.
min 13.07 ft
lb D.A. min
3
m
2 m
1
765
= .
H O condensed:
2
76.5 lb D.A. 0.0226 lb H O
min lb D.A.
lb H O mincondensed
m m 2
m
m 2
−
=
0 0075
12
.
.
b g

8-53
8.74 (cont’d)
The outside air is first cooled to a temperature at which the required amount of water is
condensed, and the cold air is then reheated to 55°F. Since ha remains constant in the
second step, the condition of the air following the cooling step must lie at the intersection
of the ha = 0 0075
. line and the saturation curve ⇒ = °
T 49 F
References: Same as Fig. 8.4-2 [including H O F
2 l, 32°
b g]
substance
min

Hin

mout

Hout
Air
H O F
2 l, 49°
b g
76.5
—
45.5
—
76.5
1.2
21.4
17.0

mair in lbm D.A./min

Hair in Btu/lbm D.A.

mH O
2
in lbm /min,
HH O
2
in Btu/lbm
Q H
= =
−
−
=
Δ
765 214 455
12 000
9
. . .
,
b g +1.2(17.0) (Btu) 60 min 1 ton cooling
min 1 h Btu h
.1 tons cooling
b.
6
m
7
m 2 m
o
m
(76.5 lb DA/min)
40%, 0.0075 lb H O/lb DA
75 F, 26.2 Btu/lb DA
r a
h h
= =
1
m
7
m 2 m
o
m
(76.5 lb DA/min)
80%, 0.0226 lb H O/lb DA
r a
h h
= =
m
m 2 m
o
m
76.5 lb DA/min
0.0075 lb H O/lb DA
a
h =
lab
Q

2
H O 2
(kg H O(l)/min) (tons)
m Q

Water balance on cooler-reheater (system shown as dashed box in flow chart)
( )( )
2
2
2
m H O
m 6
1
H O
7 7
m
H O 2
lb
lb DA
76.5 0.0226 76.5 0.0075 (76.5)(0.0075)
min lb DA
0.165 kg H O condensed/min
m
m
⎛ ⎞
⎛ ⎞
+ = +
⎜ ⎟
⎜ ⎟
⎝ ⎠⎝ ⎠
⇒ =

Cooler-
reheater
Lab

8-54
8.74 (cont’d)
Energy balance on cooler-reheater
References: Same as Fig. 8.4-2 [including H2O(l, 32o
F)]
Substance
in in
ˆ
m H
out out
ˆ
m H

Fresh air feed 10.93 45.5 — — DA m
in lb dry air/min
m

Recirculated air feed 65.57 26.2 — —
air m
ˆ in Btu/lb dry air
H
Delivered air — — 76.5 21.4 2
H O(l) m
in lb /min
m

Condensed water (49o
F) — — 0.165 17.0
2
H O(l) m
ˆ in Btu/lb
H
i i i i
out in
575.3 Btu 60 min 1 ton cooling
ˆ ˆ 2.9 tons
min 1 h -12,000 Btu/h
Q H m H m H
−
= Δ = − = =
∑ ∑

Percent saved by recirculating =
(9.1 tons 2.9 tons)
100% 68%
9.1 tons
−
× =
Once the system reaches steady state, most of the air passing through the conditioner is
cooler than the outside air, and (more importantly) much less water must be condensed
(only the water in the fresh feed).
c. Total recirculation could eventually lead to an unhealthy depletion of oxygen and buildup
of carbon dioxide in the laboratory.
8.75 Basis: 1 kg wet chips. DA = dry air, DC = dry chips
Outlet air: Tdb=38o
C, Twb=29o
C Inlet air: 11.6 m3
(STP), Tdb=100o
C
m2a (kg DA) m1a (kg DA)
m2w [kg H2O(v)]
1 kg wet chips, 19o
C m3c (kg dry chips)
0.40 kg H2O(l)/kg m3w [kg H2O(l)]
0.60 kg DC/kg T (o
C)
(a) Dry air: m1a =
11.6 m STP DA 1 kmol 29.0 kg
22.4 m STP 1 kmol
kg DA = m
3
3 2a
b g
b g = 1502
.
Outlet air:
( ) 2
Fig. 8.4-1 2
db wb 2
kg H O
kJ
ˆ
38 C, 29 C (95.3 0.48) 94.8 0.0223
kg DA kg DA
a
T T H h
= ° = ° ⎯⎯⎯⎯
→ = − = =
Water in outlet air: m h m
w a a
2 2
2
0 0223 1502 0 335
= = =
. . .
b g kg H O
2
(b) H O balance:
2 0 400 0 065
3 3
. .
kg = 0.335 kg + kg H O
2
m m
w w
⇒ =

8-55
8.75 (cont’d)
Moisture content of exiting chips:
0.065 kg water
100% 9.8% 15% meets design specification
0.600 kg dry chips + 0.065 kg water
× = ∴
(c) References: Dry air, H O
2 l
b g, dry chips @ 0°C.
substance min

Hin mout

Hout
Air
H O
2 l
b g
dry chips
15.02
0.400
0.600
100.2
79.5
39.9
15.02
0.065
0.6
94.8
4.184T
2.10T
mair in kg DA,
Hair in kJ/kg DA
m in kg DC,
Hin in kJ/kg DC
Energy Balance:
out out in in
ˆ ˆ 0 136.8 1.532 0 89.3 C
H m H m H T T
Δ = − = ⇒ − + = ⇒ = °
∑ ∑
8.76 a. T
h
T T h
r
as wb a
db
Fig. 8.4-1
2
C
C kg H O kg DA
= °
=
= = ° =
45
10%
210 0 0059
. .
b. T
h
T h
wb
r
db a
= °
=
= ° =
210
60%
268 0 0142
.
. .
C
C kg H O kg DA
Fig. 8.4-1
2
H O added:
2
15 kg air 1 kg D.A. kg H O
min 1.0059 kg air 1 kg D.A.
kg H O min
2
2
0 0142 0 0059
012
. .
.
−
=
b g
8.77 Inlet air: C
C
m kg D.A. , C
kg H O kg D.A.
db
dew pt.
Fig. 8.4-1
3
wb
2
T
T
V T
ha
= °
= °
= = °
=
50
4
0 92 22
0 0050
.
.
11.3 m 1 kg D.A.
min 0.92 m
D.A. min
3
3
= 12.3 kg
Outlet air: C
saturated
C kg H O kg D.A.
wb as
2
T T
T ha
= = °
⇒ = ° =
22
22 0 0165
.
Evaporation:
12.3 kg D.A. kg H O
min kg D.A.
kg H O min
2
2
0 0165 0 0050
014
. .
.
−
=
b g

8-56
8.78 a. T
T
h
T V
a
wb
db
dew point Fig. 8.4-1
in 2
3
C
C
kg H O kg D.A.
C, m kg D.A.
= °
= °
U
V
W
=
= ° =
45
4
0 0050
20 4 0 908
b g .
. .
T T h
wb as a
= = ° ⇒ =
20 4 0 0151
. .
C, saturated kg H O kg D.A.
out 2
b g
b. Basis: 1 kg entering sugar (S) solution
m1 (kg D.A.) m1 (kg D.A.)
0.0050 kg H2O/kg DA 0.0151 kg H2O(v)/kg
1 kg m2 (kg)
0.05 kg S/kg 0.20 kg S/kg
0.95 kg H2O/kg 0.80 kg H2O/kg
Sugar balance: 0 05 1 0 20 0 25
2 2
. . .
b gb g b g
= ⇒ =
m m kg
Water balance: m m
1 1
0 0050 1 0 95 0 0151 0 25 080
. . . . .
+ = +
⇒
m
V
1 74
67
=
= =
kg dry air
74 kg dry air 0.908 m
1 kg D.A.
m
3
3
8.79
a a 3
1 lb D.A.
m
h (lb H O)
m 2
Td = 20°F
hr = 70%
Coil
bank
1 lb D.A.
m
h (lb H O)
m 2
Td = 75°F
Spray
chamber
H O
2
1 lb D.A.
m
h (lb H O)
m 2
Coil
bank
1 lb D.A.
m
ha (lb H O)
m 2
Td = 70°F
hr = 35%
A B C D
3
a
2
1
Inlet air (A):
T
h
h
V
db
r
a
= °
=
U
V
W
≈
≈
20
70%
0 0017
12 2
1
F lb H O lb D.A.
ft lb D.A.
Fig. 8.4-2
m 2 m
3
m
.
.
Outlet air (D):
T
h
h
r
a
db
Fig. 8.4-2
m 2 m
F
0.0054 lb H O lb D.A.
= °
=
U
V
W
=
70
35% 3
a. Inlet of spray chamber (B):
h
T
T
a
wb
=
= °
U
V
W
⇒ = °
0 0017
75
49 5
.
.
lb H O lb D.A.
F
F
m 2 m
db
The state of the air at (C) must lie on the same adiabatic saturation curve as does the state
at (B), or Twb = °
49 5
. F . Thus,
Outlet of spray chamber (C):
h
T
h
a
wb
r
=
= °
U
V
W
⇒ =
0 0054
49 5
52%
.
.
lb H O lb D.A.
F
m 2 m
At point C, Tdb F
= °
585
.
b.
h h
V
a a
A
3 1 4
0 0054 0 0017
12 2
30 10
−
=
−
= × −
b g
d i
b g
lb H O evaporate lb DA
lb DA ft inlet air
lb H O
ft air
m 2 m
m
3
m 2
3

. .
.
.

8-57
8.79 (cont’d)
c. Q H H H
Q H H H
BA B A
DC D C
= = − ≅
= = − ≅
Δ
Δ
(
/
(
/
20
23
- 6.4) Btu / lb dry air
12.2 ft lb dry air
=1.1 Btu / ft
- 20) Btu / lb dry air
12.2 ft lb dry air
= 0.25 Btu / ft
m
3
m
3
m
3
m
3
d.
A B
C D
20 58.5 75
70
70%
52%
35%
8.80 Basis: 1 kg D.A.
a.
a
a
1 kg D.A.
h (kg H O/kg D.A.)
2
Tdb= 40°C, Tab= 18°C
mw kg H O
2
1 kg D.A.
h (kg H O/kg D.A.)
2
20°C,
1 2
Inlet air:
T
T
ha
db
wb
2
C
C
kg H O kg D.A.
= °
= °
⇒ =
40
18
0 0039
1 .
Outlet air:
T
T
ha
db
wb
2
C
C adiabatic humidification
kg H O kg D.A.
= °
= °
⇒ =
20
18
0 0122
2
b g .
Overall H O balance: kg H O kg D.A.
kg H O kg D.A.
2 2
2
m h h m
w a a n
+ = ⇒ = −
=
1 1 0 0122 0 0039
0 0083
1 2
b gb g b gb g b g
. .
.
Qc (Btu/h)
mc (lbm H2O/h)
liquid, 12°C
ma (lbm H2O/h)
T=15o
C, sat’d
1250 kg/h
T=37o
C, hr=50%
b.

8-58
8.80 (cont’d)
Inlet air:
T
h
h
H
r
a
db
Fig. 8.4-1
2
C 0.0198 kg H O kg DA
88.5- 0.5 kJ kg DA kJ kg DA
= °
=
⇒
=
= =
R
S
T
37
50% 88 0
1
1
.
b g
Moles dry air:
ma = =
1250 kg 1 kg DA
h 1.0198 kg
kg DA h
1226
Outlet air: T
h
H
a
db
Fig. 8.4-1 2
C, sat'd
kg H O kg DA
kJ kg DA
= ° ⇒
=
=
R
S
T
15
0 0106
421
2
.
.
Overall water balance ⇒ =
−

. .
mc
1226 kg DA kg H O
h kg DA
2
0 0198 0 0106
b g
= 113
. kg H O h withdrawn
2
Reference states for enthalpy calculations: H O
2 l
b g, dry air at 0o
C. (Cp)H2O(l) = 4.184
kJ
kg C
o
⋅
H O 12 C kJ / kg
2 l H C dT
p
, : .
° = =
z
b g 0
12
50 3
Overall system energy balance:

Q H m H m H
c i i i i
= = −
∑ ∑
Δ
out in
= +
−
L
NM O
QPF
HG I
KJF
HG I
KJ
= −
113 50 3 421 88 1 1
155
. . .
.
kg H O
h
kJ
kg H O
1226 kg DA
h
kJ
kg DA
h
3600 s
kW
1 kJ / s
kW
2
2
b g
8.81 ΔH =
−
= −
400 mol NH 78.2 kJ
mol NH
kJ
3
3
31 280
,
8.82 a. HCl , 25 C H O , 25 C HCl 25 C,
2
g l r
° ° → ° =
b g b g b g
, 5 .
Δ Δ Δ
.
H H r H
s
= ° = ⎯ →
⎯⎯
⎯ = −
25 5 64 05
C, kJ mol HCl
Table B.11
b g
b. HCl aq, = HCl H O
2
r r l
∞ → =
b g b g b g
5 ,
Δ Δ Δ

. . .
H H n H n
s s
= ° = − ° = ∞
= − + =
25 5 25
64 05 7514 1109
C, C,
kJ mol HCl kJ / mol HCl
b g b g
b g

8-59
8.83 Basis: 100 mol solution⇒ 20 mol NaOH, 80 mol H2O
⇒ r = =
80
4 00
mol H O
20 mol NaOH
mol H O mol NaOH
2
2
.
Refs: NaOH(s), H O @25 C
2 l
b g °
substance
NaOH in mol
H O in kJ mol
NaOH in mol NaOH
in in out out
2
n H n H
s n
l H
r n

. .
. .
. . .
b g
b g
b g
20 0 0 0
80 0 0 0
4 00 20 0 34 43
− −
− −
= − − − ←
.
H r
NaOH, 4.00 kJ mol
= = −
b g 34 43 NaOH (Table B.11)
ΔH n H n H
i i i i
= − = − =
− ×
= −
∑ ∑
−
−
out in
kJ 9.486 10 Btu
kJ
Btu
( )( . )
.
.
20 34 43
688 6
10
6532
4
3
Q =
−
+
= −
6532 10
20 0 40 00 80 0 18 01
132 3
3
.
. . . .
.
Btu g
g 2.20462 lb
Btu lb product solution
m
m
b g b g
8.84 Basis: 1 liter solution
nH SO 2 4 2 4
2 4
1 L 8 g - eq 1 mol
L 2 g - eq
4 mol H SO
kg
1 mol
kg H SO
= = ×
F
HG I
KJ =
0 09808
0 392
.
.
mtotal
L 1.230 kg
L
kg solution
= =
1
1230
.
nH O
2 2
2
2
2
kg H O 1000 mol H O
18.02 kg H O
mol H O
=
−
=
1230 0 392
465
. .
.
b g
⇒ = = =
r
n
n
H O
H SO
2
2 4
2
2 4
2
2 4
46.49 mol H O
4 mol H SO
11.6
mol H O
mol H SO
H SO aq, = C H SO aq, = C + H O C
(
kJ
mol H SO
2 4
o
2 4
o
2
o
Table B.11
2 4
r r l
H H r H r
s s
∞ →
= = − = ∞ = − + =
, ,
( . ) ( ) . . ) .
25 11.6, 25 25
116 67 6 9619 28 6
1
d i d i d i
Δ Δ Δ
, . ,
(
.
.
H r
n H m C dT
n
p
H SO C
kJ
mol H SO )
4 mol H SO
4 mol H SO
mol H SO
1.230 kg 00 kJ 60 C
kg C
kJ mol H SO
2 4
H SO
H SO 2 4
2 4
2 4
2 4
2 4
2 4
2 4
= ° =
+
L
NM O
QP
=
R
S
T
+
− °
⋅°
U
V
W
=
z
116 60
1 28.6 kJ 3 25
60 9
1
25
60
b g
b g
Δ

8-60
8.85 2 mol H SO mol H O
mol H O
mol H SO
2 4 H O H O 2
2
2 4
2 2
= + ⇒ = ⇒ = =
0 30 2 00 4 67
4 67
2
2 33
. . .
.
.
n n r
d i
a. For this closed constant pressure system,
Q H n H r
s
= = ° = =
−
= −
Δ Δ
H SO
2 4
2 4
2 4
C,
2 mol H SO kJ
mol H SO
kJ
.
.
.
25 2 33
44 28
88 6
b g
b. msolution
2 4 2 4 2 2
2 mol H SO 98.08 g H SO
mol
4.67 mol H O 18.0 g H O
mol
g
= + = 280 2
.
Δ Δ
H n H r m C dT
s p
T
= ⇒ ° = + =
z
0 25 2 33 0
25
H SO
2 4
C,
.
b g
−
+ − °
⋅°
= ⇒ = °
88 6
280 6 150 25
0 87
.
.
kJ +
g 3.3 J C 1 kJ
g C 1000 J
C
b g b g
T
T
8.86 a.
Basis:
1 L product solution 1.12 10 g
L
g solution
3
e j= 1120
1 L 8 mol HCl 36.47 g HCl
L mol HCl
g HCl
= 292
46.0 mol H2O(l, 25°C)
8.0 mol HCl(g , 20°C, 790 mm Hg) 1 L HCl (aq)
1120 g 292 g 828 g H O
2
− =
828 g H O mol
18.0 g
mol H O
2
2
= 46 0
.
n = =
46 0
575
.
.
mol H O
8.0 mol HCl
mol H O mol HCl
2
2
Assume all HCl is absorbed
Volume of gas:
8 mol 293 K 760 mm Hg 22.4 L STP
273 K 790 mm Hg mol
liter STP gas feed L HCl solution
b g b g
= 185
b. Ref: 25°C
substance
H O in mol
HCl in kJ mol
HCl
in in out out
2
n H n H
l n
g H
n

. .
. .
. . .
b g
b g
b g
46 0 0 0
8 0 015
575 8 0 59 07
− −
− − −
= − − −

8-61
. .
.
.
H n H n
n
mC dT
s p
HCl, C,
kJ mol
1120 g 0.66 cal C J kJ
8 mols g C cal 10 J
HCl
3
= = ° = +
= − +
− °
⋅°
z
575 25 575
1
64 87
40 25 4184
25
40
b g b g
b g
Δ
. . . .
H T T T dT
HCl, 20 C
= -0.15 kJ / mol
o
25
20
e j= − × + × − ×
− − −
z 0 02913 01341 10 0 9715 10 4 335 10
5 8 2 12 3
Q H
= = −
Δ 471 kJ L product
c. Q H H
H
T
T
= = = − −
− = = − +
⋅
−
=
0 8 8 015
015 64 87
1120 0 66 25 1
192
Δ .
. .
.
e j b g
b g
g
8 mol
cal
g C
C 4.184 J
cal
kJ
1000 J
C
o
o
o
8.87 Basis: Given solution feed rate
0.999 H O
(mol air/min)
2
na
200°C, 1.1 bars
150 mol/min solution
0.001 NaOH
25°C 0.95 H O
(mol air/min)
2
na
(mol H O( )/min)
0.05 NaOH
n1 2 v
saturated @ 50°C, 1 atm
(mol/min) @ 50°C
n2
. .
.
.
NaOH balance: 0 001 150 0 05 30
2 2
. . .
b gb g= ⇒ =
n n mol min
H O balance: mol H O min
2 2
0 999 150 0 95 30 147
1 1
. . .
b gb g b g
= + ⇒ =
n n
Raoult’s law: y P
n
n n
P p n
a
n
P
a
H O H O
Table B.4
2 2
C mm Hg
mol air
min
=
+
= ° = ⇒ =
∗
=
=

.

1
1
147
760
50 92 51 1061
1
b g
.
,
Vinlet air
1061 mol 22.4 L STP 473 K bars
min 1 mol 273 K 1.1 bars
L min
= =
b g 1013
37 900
References for enthalpy calculations: H O NaOH s air @ 25 C
2 l
b g b g
, , °
0.1% solution @ 25°C: r Hs
= ⇒ ° = −
999
25 42 47
mol H O
1 mol NaOH
C kJ mol NaOH
2
Table B.11
Δ .
b g
5% solution @ 50°C: r Hs
= = ⇒ ° = −
95 19
25 42 81
mol H O
5 mol NaOH
mol H O
mol NaOH
C
kJ
mol NaOH
2 2
Δ .
b g
Solution mass:
1 mol NaOH 40.0 g
1 mol
19 mol H O 18.0 g
1 mol
g solution
mol NaOH
2
m= + = 382

.
.
.
H H m C dT
s p
50 25
42 81
4184 50 25
2 85
25
50
° = ° +
= − +
− °
⋅°
= −
z
C C
kJ
mol NaOH
382 g J C 1 kJ
mol NaOH 1 g C 10 J
kJ
3
b g b g
b g
Δ
8.86 (cont’d)

8-62
8.87 (cont’d)
Air @ 200°C: Table B.8 ⇒ =
.
H 515 kJ mol
Air (dry) @ 50°C: Table B.8 ⇒ =
.
H 0 73 kJ mol
H O , 50 C
2 v °
b g: Table B.5 ⇒ =
−
=
.
H
2592 104.8 kJ 1 kg 18.0 g
kg 10 g 1 mol
kJ mol
3
b g 44 81
substance
NaOH in mol min
H O in kJ mol
Dry air
in in out out
2

. . . .
.
. .
n H n H
aq n
v H
b g
b g
015 42 47 015 2 85
147 44 81
1061 515 1061 0 73
− −
− −
neglect out in
Energy balance: 1900 kJ min transferred to unit
Δ
Δ
E
i i i i
n
Q H n H n H
b g

= = − =
∑ ∑
8.88 a. Basis: 1 L 4.00 molar H2SO4 solution (S.G. = 1.231)
1 L 1231 g
L
g
4.00 mol H SO
392.3 g H SO
1231 g H O
46 mol H O
mol H O / mol H SO kJ / mol H SO
2 4
2 4
2
2
2 2 4
Table B.11
s 2 4
= ⇒
=
⇒
− =
=
⇒ = ⎯ →
⎯⎯
⎯ = −
1231
392 3 838 7
57
1164 67 6
. .
.
. .
r H
Δ
Ref: H O , 25 C
2 l °
b g, H SO C
2 4 25°
b g
substance
H O in mol
H SO in kJ mol
H SO C,
in in out out
2
2 4
2 4
n H n H
l T n
l H
n

. .
.
. . .
b g b g
b g
b g
4657 0 0754 25
4 00 0
25 1164 4 00 67 6
− − −
− −
° = − − −
Q H T T
= = = − − − ⇒ = − °
Δ 0 4 00 67 6 4657 0 0754 25 52
. . . .
b g b gb g C
(The water would not be liquid at this temperature ⇒ impossible alternative!)
b. Ref: H O , 25 C
2 l °
b g, H SO C
2 4 25°
b g
substance
H O in mols
H O in kJ mol
H SO
H SO C,
in in out out
2
2
2 4
2 4
n H n H
l n n
s n H
l
n
l
s

.
. .
( ) .
. . .
b g b g
b g b g
b g
0 0754 0 25
6 01 0 0754 0 25
4 00 0
25 1164 4 00 67 61
− − −
− + − − −
− −
° = −
Δ .
Hm H O, 0 C kJ mol
2
Table B.1
° =
A
b g 6 01
n n
H n n
n
n
s
s
l l
l
s
A
A
+ =
= = − − − − − −
U
V
W
⇒
=
=
⇒ + °
4657
0 4 00 67 61 1885 4657 7 895
1618
30 39
2914 547 3 0
.
. . . . .
.
.
. . @
Δ b g b g b gb g
b g b g
mol liquid H O
mol ice
g H O g H O C
2
2 2

8-63
8.89 P O 3H O 2H PO
2 5 2 3 4
+ →
a. wt% P O wt% H PO
2 5 3 4
mol H3PO4 g H3PO4 mol
g total
= × = ×
B B
A
n
m
n
m
t c
14196
100%
2 98 00
100%
.
,
.
b g b g
P
where n mt
= =
mol P O and total mass
2 5 .
wt% H PO wt% P O wt% P O
3 4 2 5 2 5
= =
2 98 00
14196
1381
.
.
.
b g
b. Basis: 1 lbm feed solution 28 wt% P O wt% H PO
2 5 3 4
⇒ 38 67
.
(lb H O( )),
1 lb solution, 125°F
m
m1 2 v
0.3867 lb H PO
m 3 4
0.6133 lb H O
m 2
m T , 3.7 psia
(lb solution),
m2 m T
0.5800 lb H PO /lb
m 3 4
0.4200 lb H O/lb
m 2
m
m
H PO balance 0.3867
3 4 : . .
= ⇒
05800 0 667
2 2
m m lbm solution
Total balance: 1 0 3333
1 2 1
= + ⇒ =
m m m r
. lb H O
m 2 b g
Evaporation ratio: 0.3333 lb H O v lb feed solution
m 2 m
b g
c. Condensate:
P
T
l
=
⇒ = = =
37
654
000102 353145
2205
00163
.
.
. . /
. /
.
psia 0.255 bar
C=149 F, V
m ft m
kg lb kg
ft
lb H O( )
Table B.6
sat
o o
liq
3 3 3
m
3
m 2
b g
.
m =
×
=
100
1
46 3
tons feed 2000 lb 1 lb H O 1 day
day ton 3 lb (24 60) min
lb / min
m m 2
m
m
.
.
V = =
46 3
565
lb 0.0163 ft 7.4805 gal
min lb ft
gal condensate / min
m
3
m
3
Heat of condensation process:
46.3lbm H2O(v)/min
(149+37)°F, 3.7 psia
46.3lbm H2O(l)/min
149°F, 3.7psia
Q (Btu/min)
.

8-64
8.89 (cont’d)
Table B.6
F = 85.6 C) = (2652 kJ / kg) 0.4303
Btu
lb
kJ
kg
Btu / lb
F = 65.4 C) = (274 kJ / kg) 0.4303 Btu / lb
o o m
m
o o
m
⇒
F
H
GG
I
K
JJ =
=
R
S
|
|
|
T
|
|
|
(
(
( )
( )
H
H
H O v
H O l
2
2
186 1141
149 118
b g
( . ,
.
Q m H
= = −
L
NM O
QP= −
⇒ ×
Δ 46 3 47 360
4 74 104
lb
min
) (118 1141)
Btu
lb
Btu / min
Btu minavailable at 149 F
m
m
o
d. Refs: H PO , H O F
3 4 2
l l
b g b g@77°
substance
H PO in lb
H PO in Btu lb
H O
in in out out
3 4 m
3 4 m
2
m H m H
m
H
v

. .
. .
.
28% 100 1395
42% 0 667 3413
0 3333 1099
b g
b g
b g
− −
− −
− −

.
H H PO , 28%
Btu 1 lb - mole H PO 0.3867 lb H PO
lb - mole H PO 98.00 lb H PO 1.00 lb soln
0.705 Btu F
lb F
Btu lb soln
3 4
3 3 m 3 4
3 4 m 3 4 m
m
m
b g
b g
=
−
+
− °
⋅°
=
5040
125 77
1395

.
.
H H PO , 42%
Btu 1 lb - mole H PO 0.5800 lb H PO
lb - mole H PO 98.00 lb H PO 1.00 lb sol.
0.705 Btu F
lb F
Btu lb soln
3 4
3 4 m 3 4
3 4 m 3 4 m
m
m
b g
b g
=
−
+
− °
⋅°
=
5040
186 7 77
3413
. , .
H H H l
H O psia, 186 F 77 F kJ kg Btu lb
2 m
b g b g b g b g
= ° − ° = − ⇒
37 2652 104 7 1096
At 27.6 psia (=1.90 bar), Table B.6 ⇒ =
Δ
Hv 2206 kJ / kg = 949 Btu / lbm
Δ Δ
H n H n H H
i i i i v
= ∑ − ∑ = ⇒ = =
⇒
×
=
⇒
×
=
out in
steam steam
m
m
m m 3 4
m 3 4
m
m
m 2
m
m 2
Btu = m m
Btu
949 Btu / lb
lb steam
lb steam lb H PO 1 day
lb 28% H PO day 24 h
lb steam / h
lb steam
(46.3 60) lb H O evaporated / h
lb steam
lb H O evaporated
.
.
.
375
375
0395
0395 100 2000
3292
3292
118

8-65
8.90 Basis: 200 kg/h feed solution. A = NaC H O
2 3 2
(kmol A-3H O( )/h)
200 kg/h @ 60°C
(kmol/h)
n0
0.20 A
0.80 H O
2
(kmol H O( )/h)
n1 2 v
50°C, 16.9% of H O
2
in feed
Product slurry @ 50°C
n2 2 v
(kmol solution/h)
n3
0.154 A
0.896 H O
2
(kJ/hr)
Q
.
.
.
.
a. Average molecular weight of feed solution: M M M
A
= +
0 200 0800
. . H O
2
= + =
0 200 82 0 0800 18 0 308
. . . . .
b gb g b gb g kg k
Molar flow rate of feed: n0 6 49
= =
200 kg 1 kmol
h 30.8 kg
kmol h
.
b. 16.9% evaporation ⇒ = =
n v
1 0169 080 6 49 0877
. . . .
b gb gb g b g
kmol h kmol H O h
2
A balance: 0 20 6 49
3
0154
2
3
. . .
b gb g b g
kmol h
kmol H O 1 mole
h 1 mole 3 H O
2
2
=
⋅
⋅
+
E
n A A
A
n
⇒ + =
n n
2 3
0154 130
. . (1)
H O balance: kmol h
kmol H O 3 moles H O
h 1 mole 3 H O
2
2 2
2
080 6 49 0877
3
0846 3 0846 4 315
2
3 2 3
. . .
. . .
b gb g b g
b g
= +
⋅
⋅
+ ⇒ + =
n A
A
n n n 2
Solve 1 and simultaneously kmol 3H O s h
kmol solution h
2
b g b g b g
2 113
1095
2
3
⇒ = ⋅
=
n A
n
.
.
Mass flow rate of crystals
1.13 kmol 3H O 136 kg 3H O
h 1 kmol
154 kg NaC H O 3H O s
h
2 2 2 3 2 2
A A
⋅ ⋅
=
⋅ b g
Mass flow rate of product solution
200 154 kg cry 0877 18 0
30
kg feed
h
stals
h
kg H O
h
kg solution h
2
− − =
. .
b gb g b g
v
c. References for enthalpy calculations: NaC H O s H O C
2 3 2 2
b g b g
, @
l 25°
Feed solution: nH n H m C dT
A s p

= ° + z
Δ 25
25
60
C
b g (form solution at 25°C , heat to 60°C)
nH
A
A
. .
=
− ×
+
− °
⋅°
=
0.20 kmol kJ
h kmol
200 kg 3.5 kJ 60 C
hr kg C
kJ h
b g b g
649 171 10 25
2300
4

8-66
8.90 (cont’d)
Product solution: nH n H m C dT
A s p

= ° + z
Δ 25
25
50
C
b g
=
− ×
+
− °
⋅°
= −
0.154 .095 kmol kJ
h kmol
30 kg 3.5 kJ 50 C
h kg C
kJ h
b g b g
1 171 10 25
259
4
A
A
.
Crystals: nH n H m C dT
A p

= + z
Δ hydration
25
50
(hydrate at 25°C , heat to 50°C )
=
⋅ − ×
+
− °
⋅°
= −
1.13 kmol 3H O s kJ
h kmol
154 kg 1.2 kJ C
h kg C
kJ h
2
A b g b g
366 10 50 25
36700
4
.
H O , 50 C
2 v n H n H C dT
v p
° = +
L
NM O
QP
z
b g:
Δ Δ
25
50
(vaporize at 25°C , heat to 50°C )
=
× + −
=
0.877 kmol H O kJ
h
kJ h
2 4 39 10 32 4 50 25
39200
4
. .
b gb g
neglect out in
Energy balance: kJ h
60 kJ h (Transfer heat from unit)
Δ
Δ
E
i i i i
R
Q H n H n H
b g
b g b g
= = − = − − + −
= −
∑ ∑
259 36700 39200 2300
8.91 50 mL H SO g
mL
g H SO mol H SO
84.2 mLH O g
mL
g H O mol H O
mol H O mol H SO
2 4
2 4 2 4
2
2 2
2 2 4
1834
917 0935
100
842 4678
500
.
. .
.
. .
.
= ⇒
= ⇒
U
V
|
|
W
|
|
⇒ =
l
l l
r
bg bg bg
Ref: H O
2 , H SO
2 4 @ 25 °C
( ( ), [ .
H l
H O 15 C) kJ / (mol C)](15 25) C = 0.754 kJ / mol
2
o o o
= ⋅ − −
0 0754
, . .
( . . )( )
H r
T
T
H SO
kJ
mol
(91.7 + 84.2) g 2.43 J C 1 kJ
0.935 mol H SO g C 10 J
kJ / mol H SO
2 4
2 4
3
2 4
= = − +
− °
⋅°
= − +
500 58 03
25
69 46 0 457
b g b g
substance nin

Hin
nout

Hout
H O
2 l
b g
H SO
2 4
H SO
2 4 r = 4 00
.
b g
4.678
0.935
—
–0.754
0.0
—
—
—
0.935
—
—
− +
69 46 0 457
. . T
b g
n in mol

H in kJ/mol
n mol H SO
3 4
b g
Energy Balance: ΔH T T
= = − + − − ⇒ = °
0 0 935 69 46 0 457 4 678 0 754 144
. . . . .
b g b g C
Conditions: Adiabatic, negligible heat absorbed by the solution container.

8-67
8.92 a.
mA (g A) @ TA0 (o
C)
nA (mol A) nS (mol solution) @ Tmax (o
C)
mB (g B) @ TB0 (o
C)
nB (mol B)
Refs: A(l), B(l) @ 25 °C
substance nin

Hin nout

Hout
A nA

HA — — n in mol
B nB

HB — —
H in J / mol
S — — nA

HS (J mol A)
Moles of feed materials: n
m
M
n
m
M
A (mol A) =
(g A)
(g A / mol A)
,
A
A
B
B
B
=
Enthalpies of feeds and product
( (
/
/

(
( ( )
( )( ( )(
( ) ( ) ( )
max
max
H m C T H m C T
r n n
m M
m M
H
n
n H r
m m C T
H
n
n H r m m C T
A A pA A B B pB B
B A
B B
A A
S
A
A m
A B ps
S
A
A m A B ps
= − = −
=
F
HG I
KJ =
×
F
HG I
KJ
+ + ×
⋅
F
HG
I
KJ × −
L
N
MMMMM
O
Q
PPPPP
⇒ = + + −
0 0
25 25
1
25
1
25
o o
o
o
C), C)
(mol B mol A) =
J
mol A mol A)
mol A)
J
mol A
g soln)
J
g soln C
C)
Δ
Δ
Energy balance
Δ
Δ
Δ
H n H n H n H
m
M
H r m m C T m C T m C T
T
m C T m C T
m
M
H r
m m C
A S A A B B
A
A
m A B ps A pA A B pB B
A pA A B pB B
A
A
m
A B ps
= − − =
⇒ + + − − − − − =
⇒ = +
− + − −
+

( ) ( )

( )
max
max
0
25 25 25 0
25
25 25
0 0
0 0
b g b g b g
b g b g b g
Conditions for validity: Adiabatic mixing; negligible heat absorbed by the solution container,
negligible dependence of heat capacities on temperature between 25o
C and TA0 for A, 25o
C
and TB0 for B, and 25o
C and Tmax for the solution.
b.
m M T C
m M T C
r
A A A pA
B B B pB
= = = ° =
= = = ° = ⋅°
U
V
W
|⇒ =
100 0 40 00 25
2250 18 01 40 418
500
0
0
. . ?
. . .
.
g C irrelevant
g C J (g C)
mol H O
mol NaOH
2
b g
Cps = ⋅°
335
. )
J (g C Δ . ,
H n
m = = −
500 37 740
b g J mol A ⇒ Tmax C
= °
125

8-68
8.93 Refs: Sulfuric acid and water @ 25 °C
b. substance nin

Hin
nout

Hout
H2SO4
H2O
1
r
M C T
A pA 0 25
−
b g
M C T
w pw 0 25
−
b g
—
—
—
—
n in mol

H in J/mol
H SO aq
2 4 b g — — 1 Δ
H r M rM C T
m A w ps
b g b g b g
+ + −
s 25
(J/mol H2SO4)
Δ Δ
Δ
H H r M rM C T M C T rM C T
H r r C T C rC T
m A w ps A pa w pw
m ps pa pw
= = + + − − − − −
= + + − − + −
0 25 25 25
98 18 25 98 18 25
0 0
0

( )
b g b g b g b g b g
b g b g b g b g
s
s
⇒ = +
+
+ − −
T
r C
C rC T H r
ps
pa pw m
s 25
1
98 18
98 18 25
0
( )
( )
b g b g
Δ
c.
Cp
(J/mol-K)
Cp
(J/g-K)
H2O(l) 75.4 4.2
H2SO4 185.6 1.9
r Cps Δ ( )
H r
m Ts
0.5 1.58 -15,730 137.9
1 1.85 -28,070 174.0
1.5 1.89 -36,900 200.2
2 1.94 -41,920 205.7
3 2.1 -48,990 197.8
4 2.27 -54,060 184.0
5 2.43 -58,030 170.5
10 3.03 -67,030 121.3
25 3.56 -72,300 78.0
50 3.84 -73,340 59.6
100 4 -73,970 50.0
0
50
100
150
200
250
0.1 1 10 100
r
Ts
d. Some heat would be lost to the surroundings, leading to a lower final temperature.

8-69
8.94 a. Ideal gas equation of state n P V RT
A g
0 0 0
= / (1)
Total moles of B: n
V SG
M
B
l B
B
0
3
1 10
(
(
mol B)
L) kg / L g / kg
(g / mol B)
=
× ×
b gd i (2)
Total moles of A: n n n
Ao Av Al
= + (3)
Henry’s Law: r k p
n
n
c c T
n RT
V
s A
Al
B
Av
g
mol A(l)
mol B
F
HG I
KJ = ⇒ = +
0
0 1
b g (4)
Solve (3) and (4) for nAl and nAv.
n
n RT
V
c c T
n RT
V
c c T
Al
B
g
B
g
=
+
+ +
L
N
MM
O
Q
PP
0
0 1
0
0 1
1
b g
b g
(5)
n
n
n RT
V
c c T
Av
Ao
B
g
=
+ +
L
N
MM
O
Q
PP
1 0
0 1
b g
(6)
P
n RT
V
n RT
V n RT c c T
Av
g
A
g B
= =
+ +
( )
6
0
0 0 1
b g (7)
Refs: A g B l
b g b g
, @ 298 K
substance nin

Uin neq

Ueq
A g
b g nAo M C T
A vA 0 298
−
b g nAv M C T
A vA − 298
b g n in mol

U in kJ/mol
B l
b g nB0 M C T
B vB 0 298
−
b g — —
Solution — — nAl

U1 (kJ/mol A)

U U
n
n M n M C T
s
Al
Al A B B vs
1 0
1
298
= + + −
Δ b g b g
E.B.: ΔU n U n U
i i i i
= = −
∑ ∑
0
out in
0 298 298
298
298
0
0
= + + − + − + −
⇒ = +
− + + −
+ +
n C n M n M C T n U n C n C T
T
n U n C n C T
n C n M n M C
Av vA Al A B B vs Al s Ao vA B vB
Al s Ao vA B vB
Av vA Al A B B vs
b g
c hb g b gb g
d i b gb g
b g
Δ
Δ

8-70
8.94 (cont’d)
b.
Vt MA CvA MB CvB SGB c0 c1 Dus Cvs
20.0 47.0 0.831 26.0 3.85 1.76 0.00154 -1.60E-06 -174000 3.80
Vl T0 P0 Vg nB0 nA0 T nA(v) nA(l) P Tcalc
3.0 300 1.0 17.0 203.1 0.691 301.4 0.526 0.164 0.8 301.4
3.0 300 5.0 17.0 203.1 3.453 307.0 2.624 0.828 3.9 307.0
3.0 300 10.0 17.0 203.1 6.906 313.9 5.234 1.671 7.9 313.9
3.0 300 20.0 17.0 203.1 13.811 327.6 10.414 3.397 16.5 327.6
3.0 330 1.0 17.0 203.1 0.628 331.3 0.473 0.155 0.8 331.3
3.0 330 5.0 17.0 203.1 3.139 336.4 2.359 0.779 3.8 336.4
3.0 330 10.0 17.0 203.1 6.278 342.8 4.709 1.569 7.8 342.8
3.0 330 20.0 17.0 203.1 12.555 355.3 9.381 3.174 16.1 355.3
c.
C* REAL R, NB, T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS
REAL NA0, T, DEN, P, NAL, NAV, NUM, TN
INTEGER K
R = 0.08206
1 READ (5, *) NB
IF (NB.LT.0) STOP
READ (1, *) T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS
WRITE (6, 900)
NA0 = P0 * VG/R/T0
T = 1.1 * T0
K = 1
10 DEN = VG/R/T/NB + C + D * T
P = NA0/NB/DEN
NAL = (C + D * T) * NA0/DEN
NAV = VG/R/T/NB * NA0/DEN
NUM = NAL * (–DUS) + (NA0 * CVA + NB * CVB) * (TO – 298)
DEN = NAV * CVA + (NAL * MA + NB * MB) * CVS
TN = 298 + NUM/DEN
WRITE (6, 901) T, P, NAV, NAL, TN
IF (ABS(T – TN).LT.0.01) GOTO 20
K = K + 1
T = TN
IF (K.LT.15) GOTO 10
WRITE (6, 902)
STOP
20 WRITE (6, 903)
GOTO 1
900 FORMAT ('T(assumed) P Nav Nal T(calc.)'/
* ' (K) (atm) (mols) (mols) (K)')
901 FORMAT (F9.2, 2X, F6.3, 2X, F7.3, 2X, F7.3, 2X, F7.3, 2)
902 FORMAT (' *** DID NOT CONVERGE ***')
903 FORMAT ('CONVERGENCE'/)
END
$ DATA
300
291 10.0 15.0 1.54E–3 –2.6E–6 –74
35.0 18.0 0.0291 0.0754 4.2E–03

8-71
8.94 (cont’d)
300
291 50.0 15.0 1.54E–3 –2.6E–6 –74
35.0 18.0 0.0291 0.0754 4.2E–03
–1
Program Output
T (assumed) P Nav Nal T(calc.)
(K) (atm) (mols) (mols) (K)
321.10 8.019 4.579 1.703 296.542
296.54 7.415 4.571 1.711 296.568
296.57 7.416 4.571 1.711 296568
.
Convergence
T (assumed) P Nav Nal T(calc.)
(K) (atm) (mols) (mols) (K)
320.10 40.093 22.895 8.573 316.912
316.91 39.676 22.885 8.523 316.942
316.94 39.680 22.885 8.523 316 942
.
8.95
350 mL 85% H2SO4
ma(g), 60 oF, ρ=1.78
H2O, Vw(mL),
mw(g), 60 oF
30% H2SO4
ms(g), T(o
F)
Q=0
a.
Vw =
−
=
350 178 015 1
1140
mL feed g 0.85(70 / 30) g H O added mL water
1 mL feed g feed 1 g water
mL H O
2
2
. .
b. a m water m
ˆ ˆ
Fig. 8.5-1 103 Btu/lb ; Water: 27 Btu/lb
H H
⇒ ≈ − ≈
Mass Balance: mp=mf+mw=(350 mL)(1.78 g/mL)+(1142 mL)(1 g/mL)=623+1142=1765 g
Energy Balance: f f
product a a
product m
ˆ ˆ
ˆ ˆ ˆ ˆ
0
(623)( 103) (1140)(27)
ˆ 18.9 Btu/lb
1765
w w
p w w s
p
m H m H
H m H m H m H H
m
H
+
Δ = = − − ⇒ =
− +
⇒ = = −
c. o
m
ˆ
( 18.9 Btu/lb ,30%) 130 F
T H = − ≈
d. When acid is added slowly to water, the rate of temperature change is slow: few isotherms
are crossed on Fig. 8.5-1 when xacid increases by, say, 0.10. On the other hand, a change
from xacid=1 to xacid=0.9 can lead to a temperature increase of 200°F or more.

8-72
8.96 a. 2.30 lb 15.0 wt% H SO
@ 77 F H Btu / lb
m (lb ) 80.0 wt% H SO
@ 60 F H Btu / lb
m ( lb ) 60.0 wt% H SO @ T F, H
m 2 4
o
1 m
2 m 2 4
o
2 m
adiabatic mixing
3 m 2 4
o
3
⇒ = −
⇒ = −
U
V
|
|
|
W
|
|
|
⎯ →
⎯⎯⎯⎯
⎯

10
120
Total mass balance: 2.30 +
H SO mass balance: 2.30 0.150
lb (80%)
lb (60%)
2
2 4 2
m
m
m m
m m
m
m
=
+ =
U
V
|
W
|⇒
=
=
R
S
|
T
|
3
3
2
3
0800 0 600
517
7 47
b g b g
. ( . )
.
.
b. Adiabatic mixing =
Btu / lb
Figure 8.5-1
T =140 F
m
o
⇒ =
− − − − = ⇒ = −
E
Q H
H H
Δ 0
7 47 2 30 10 517 120 0 861
3 3
. . . .
b g b gb g b gb g
c.
. .
H
Q m H H
60 wt%, 77 F Btu / lb
60 wt%, 77 F Btu
o
m
o
d i
d i b gb g
= −
= − = − + = −
130
7 475 130 861 328
3 3
d. Add the concentrated solution to the dilute solution . The rate of temperature rise is
much lower (isotherms are crossed at a lower rate) when moving from left to right on
Figure 8.5-1.
8.97 a. x y T
NH
Fig. 8.5-2
NH m 3 m
3 3
lb NH lb vapor F
= = = °
0 30 0 96 80
. . ,
b. Basis: 1 lb system mass
m ⇒
=
0 90
0.30
. lb liquid
0.27 lb NH
0.63 lb H O
m
m 3
m 2
NH3
x
⇒
=
010
0.96
. lb vapor
0.096 lb NH
0.004 lb H O
m
m 3
m 2
NH3
x
Mass fractions: zNH
m 3
m
m 3 m
3
lb NH
1 lb
lb NH lb
=
+
=
0 27 0 096
0 37
. .
.
b g
1 0 37 0 63
− =
. . lb H O lb
m 2 m
Enthalpy:
H =
−
+ =
0.90 lb liquid Btu
1 lb lb liquid
0.10 lb vapor 670 Btu
1 lb lb vapor
Btu lb
m
m m
m
m m
m
25
1 1
44

8-73
8.98
T = °
140 F
Vapor: 80% NH 20% H O
Liquid: 14% NH 86% H O
Fig. 8.5-2
3 2
3 2
,
,
Basis: 250 g system mass
⇒ m m
v L
( (
g vapor), g liquid) .14 .60 .80 xNH3
B
A
C
Mass Balance: m m
v L
+ = 250
NH3 Balance: 080 014 0 60 250 175 75
. . ( . )( )
m m m m g
L v L
g g,
+ = ⇒ = =
Vapor: mNH 3 2
3
g g NH 35 g H O
= =
080 175 140
. ,
b gb g
Liquid: mNH 3 2
3
g 64.5 g H O Liquid
= =
014 75 10.5 g NH
. ,
b gb g
8.99 Basis: 200 lb feed h
m

mv (lb h)
m
xv(lbm NH3(g)/lbm)
200 lbm/h (
H v Btu lb )
m in equilibrium
0.70 lbm NH3(aq)/lbm at 80o
F
0.30 lbm H2O(l)/lbm
ml (lb h)
m

H f = −50 Btu lbm
xl[lbm NH3(aq)/lbm]
(
H l Btu lb )
m
(
Q Btu h)
Figure 8.5-2⇒ =
Mass fraction of NH in vapor: lb NH lb
3 m 3 m
xv 0 96
.
Mass fraction of NH in liquid: lb NH lb
3 m 3 m
xl = 0 30
.
Specific enthalpies:
Hv = 650 Btu lbm ,
Hl = −30 Btu lbm
Mass balance: 200
Ammonia balance:
120 lb h vapor
80 lb h liquid
m
m
= +
= +
U
V
W
⇒
=
=

. . .

m m
m m
m
m
v l
v l
v
l
0 70 200 0 96 0 30
b gb g
Energy balance: Neglect Δ
Ek .

,
Q H m H m H
i i f f
= = − = +
−
−
−
=
∑
Δ
out
m
m
m
m
m
m
120 lb Btu
h lb
80 lb Btu
h lb
200 lb Btu
h lb
Btu
h
650 30 50
86 000

9-1
CHAPTER NINE
9.1
9.2
a.
b.
c.
d.
e.
f.
a.
b.
c.
4 5
904 7
3
NH g O g) 4NO(g) + 6H O(g)
kJ / mol
2 2
r
o
( ) (
.
+ →
= −
ΔH
When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25°C and 1 atm react to form 4 g-moles of
NO(g) and 6 g-moles of water vapor at 25°C and 1 atm, the change in enthalpy is -904.7 kJ.
Exothermic at 25°C. The reactor must be cooled to keep the temperature constant. The temperature
would increase under adiabatic conditions. The energy required to break the molecular bonds of the
reactants is less than the energy released when the product bonds are formed.
2
5
2
3
NH g O g) 2NO(g) + 3H O(g)
2 2
( ) (
+ →
Reducing the stoichiometric coefficients of a reaction by half reduces the heat of reaction by half.
Δ .
.
Hr
o
kJ / mol
= − = −
904 7
2
452 4
NO(g) +
3
2
H O(g) NH g O g)
2 2
→ +
3
5
4
( ) (
Reversing the reaction reverses the sign of the heat of reaction. Also reducing the stoichiometric
coefficients to one-fourth reduces the heat of reaction to one-fourth.
Δ ( . )
.
Hr
o
kJ / mol
= −
−
= +
904 7
4
226 2
3
3
3
3
NH
NH
o
3
NH r 3
3
NH
m 340 g/s
340 g 1 mol
n 20.0 mol/s
s 17.03 g
ˆ 20.0 mol NH 904.7 kJ
n H
Q H= 4.52 10 kJ/s
s 4 mol NH
ν
=
= =
−
Δ
= Δ = = − ×

The reactor pressure is low enough to have a negligible effect on enthalpy.
Yes. Pure water can only exist as vapor at 1 atm above 100°C, but in a mixture of gases, it can
exist as vapor at lower temperatures.
C H l O g) 9CO (g) +10H O(l)
kJ / mol
9 2 2 2
r
o
20 14
6124
( ) (

+ →
= −
ΔH
When 1 g-mole of C9H20(l) and 14 g-moles of O2(g) at 25°C and 1 atm react to form 9 g-moles of
CO2(g) and 10 g-moles of water vapor at 25°C and 1 atm, the change in enthalpy is -6124 kJ.
Exothermic at 25°C. The reactor must be cooled to keep the temperature constant. The temperature
would increase under adiabatic conditions. The energy required to break the molecular bonds of the
reactants is less than the energy released when the product bonds are formed.

.
Q H =
n H 25.0 mol C H 6124 kJ 1 kW
s 1 mol C H 1 kJ / s
kW
C H r
0
C H
9 20
9 20
9 20
9 20
= =
−
= − ×
Δ
Δ
ν
153 105

9-2
d.
e.
Heat Output = 1.53×105
kW.
The reactor pressure is low enough to have a negligible effect on enthalpy.
C H g O g) 9CO (g) +10H O(l) (1)
kJ / mol
C H l O g) 9CO (g) +10H O(l) (2)
kJ / mol
(2) C H l C H g
C H C) kJ / mol kJ / mol) = 47 kJ / mol
9 2 2 2
r
o
9 2 2 2
r
o
9 9
v
o
9
20
20
20 20
20
14
6171
14
6124
1
25 6124 6171
( ) (

( ) (

( ) ( ) ( )
( , (
+ →
= −
+ →
= −
− ⇒ →
= − − −
Δ
Δ
Δ
H
H
H D
Yes. Pure n-nonane can only exist as vapor at 1 atm above 150.6°C, but in a mixture of gases, it
can exist as a vapor at lower temperatures.
9.3
9.4
a.
b.
c.
a.
b.
Exothermic. The reactor will have to be cooled to keep the temperature constant. The temperature
would increase under adiabatic conditions. The energy required to break the reactant bonds is less
than the energy released when the product bonds are formed.
C H g O g CO g H O g
C H l O g CO g H O l Btu lb - mole
6 14 2 r
o
6 14 2 r
o
b g b g b g b g b g
b g b g b g b g b g
+ → + =
+ → + = = − ×
19
2
6 7 1
19
2
6 7 2 1791 10
2 2
2 2 2
6
Δ
Δ Δ
?
.
H
H H
C H g C H l Btu lb - mole
6 14 6 14 v
C H
2 14
b g b g b g e j
→ = − = −
3 13 550
3
Δ Δ
,
H H
H O l H O g Btu lb - mole
2 2 v
H O
2
b g b g b g e j
→ = =
4 18 934
4
Δ Δ
,
H H
1 2 3 7 4 7 1672 10
1 2 3 4
6
b g b g b g b g
= + + × ⇒ = + + = − ×
Hess's law
Btu lb - mole
Δ Δ Δ Δ
.
H H H H
/ .
m n
= ⇒ =
120 375
lb lb
M =32.0
O2
m s - mole / s.
( )
2
2
6
o
O r 5
2
O
ˆ 3.75 lb-mole/s 1.672 10 Btu
6.60 10 Btu/s from reactor
9.5 1 lb-mole O
n H
Q H
v
− ×
Δ
= Δ = = = − ×

CaC s 5H O l CaO s 2CO g 5H g
2 2 2 2
b g b g b g b g b g
+ → + + , Δ .
Hr
o
kJ kmol
= 69 36
Endothermic. The reactor will have to be heated to keep the temperature constant. The temperature
would decrease under adiabatic conditions. The energy required to break the reactant bonds is more
than the energy released when the product bonds are formed.
Δ Δ
.
.
.
U H RT i i
r
o
r
o
gaseous
products
gaseous
reactants
3
kJ
mol
J 1 kJ 298 K 7 0
mol K 10 J
kJ mol
= − −
L
N
MMMM
O
Q
PPPP
= −
−
⋅
=
∑ ∑
ν ν 69 36
8 314
52 0
b g
9.2 (cont'd)

9-3
9.5
9.6
9.7
c.
a.
b.
a.
b.
a.
b.
c.
d.
Δ
Ur
o
2 2
2 2
is the change in internal energy when 1 g-mole of CaC (s) and 5 g-moles of H O(l) at 25 C and
1 atm react to form 1 g-mole of CaO(s), 2 g-moles of CO (g) and 5 g-moles of H (g) at 25 C
and 1 atm.
D
D
Q U
n U
v
= = = =
Δ
Δ
CaC r
o
CaC
2
2
2
2
150 g CaC 1 mol 52.0 kJ
64.10 g 1 mol CaC
121.7 kJ

Heat must be transferred to the reactor.
Given reaction = (1) – (2) ⇒ = − = −
Hess's law
r
o
r
o
r
o
Btu lb - mole
Δ Δ Δ
,
H H H
1 2 1226 18 935
b g
= −17 709
, Btu lb - mole
Given reaction = (1) – (2) ⇒ = − = − +
Hess's law
r
o
r
o
r
o
Btu lb - mole
Δ Δ Δ
, ,
H H H
1 2 121 740 104 040
b g
= −17 700
, Btu lb - mole
Reaction (3) ( )
Hess's law
o
r
kJ kJ kJ
ˆ
0.5 (1) 2 0.5 326.2 285.8 122.7
mol mol mol
H
⎛ ⎞ ⎛ ⎞
= × − ⇒ Δ = − − − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Reactions (1) and (2) are easy to carry out experimentally, but it would be very hard to decompose
methanol with only reaction (3) occurring.
N g O g 2NO g
2 2
b g b g b g
+ → , Δ Δ
. .
H H
r
o
f
o
NO(g)
kJ mol kJ mol
Table B.1
= =
F
H
GG
I
K
JJ =
B
2 2 90 37 180 74
e j
n − + → +
C H g
11
2
O g 5CO g 6H O l
5 12 2 2
b g b g b g b g
Δ Δ Δ Δ

. . . .
H H H H
n
r
o
f
o
CO(g)
f
o
H O l
f
o
C H g
2 5 12
kJ mol kJ mol
= + −
= − + − − − = −
−
5 6
5 11052 6 28584 146 4 21212
e j e j e j
b gb g b gb g b g
b g b g
C H l O g 6CO g 7H O g
6 14 2 2 2
b g b g b g b g
+ → +
19
2
( ) ( ) ( )
( ) ( )
( )( ) ( ) ( )
2 2 6 14
o o o o
r f f f
CO H O g C H l
ˆ ˆ ˆ ˆ
6 7
6 393.5 7 241.83 198.8 kJ mol 3855 kJ mol
H H H H
Δ = Δ + Δ − Δ
= ⎡ − + − − − ⎤ = −
⎣ ⎦
Na SO CO( Na S( CO (
2 4 2 2
( ) ) ) )
l g l g
+ → +
4 4
Δ Δ Δ Δ Δ

( . . ) . . . ( . .
( ) )
H H H H H
r
o
f
o
l
f
o
CO g
f
o
l
f
o
CO(g
2
kJ mol kJ mol
= + − −
= − + + − − − + − − = −
e j e j e j e j
b gb g b g
Na S( ) Na SO ( )
2 2 4
4 4
3732 6 7 4 3935 1384 5 24 3 4 11052 138 2
9.4 (cont’d)

9-4
9.8
9.9
a.
b.
c.
a.
b.
c.
d.
Δ Δ Δ Δ
. . .
( ) ( ) ( )
H H H H
r
o
f
o
C H Cl l
f
o
C H g
f
o
C H Cl l
2 2 4 2 4 2 2 4
kJ mol
1 38576 52 28 33348
= − ⇒ = − + = −
e j e j e j
Δ Δ Δ Δ
. . . .
)
H H H H
r
o
f
o
C HCl l
f
o
HCl g
f
o
C H Cl (l
2 3 2 2 4
kJ mol
2 276 2 92 31 33348 3503
= + − = − − + = −
e j e j e j
b g b g
Given reaction = + ⇒ − − = −
1 2 38576 3503 420 79
b g b g . . . kJ mol
.
.
Q H
= =
−
= − × = −
Δ
300 mol C HCl kJ
h mol
kJ h kW
2 3 420 79
126 10 35
5
b g
Heat is evolved.
C H O CO ( ) + H O( )
2 2 2 2 2
( ) ( ) .
g g g l kJ mol
c
o
+ → = −
5
2
2 1299 6
ΔH
The enthalpy change when 1 g-mole of C2H2(g) and 2.5 g-moles of O2(g) at 25°C and 1 atm react
to form 2 g-moles of CO2(g) and 1 g-mole of H2O(l) at 25°C and 1 atm is -1299.6 kJ.
Δ Δ Δ Δ

. . . .
( )
H H H H
c
o
f
o
CO g
f
o
H O l
f
o
C H g
2 2 2 2
Table B.1
kJ
mol
kJ
mol
= + −
= − + − − = −
B
2
2 3935 28584 226 75 1299 6
e j e j e j
b g b g b g
b g b g
(
. . .
(
. ( . ) . .
( )
( ) ( )
i)
kJ
mol
kJ
mol
ii)
kJ
mol
kJ
mol
r
o
f
o
C H g
f
o
C H g
r
o
c
o
C H g
c
o
g
c
o
C H g
2 6 2 2
Table B.1
2 2 2 6
Table B.1
Δ Δ Δ
Δ Δ Δ Δ
H H H
H H H H
= −
= − − = −
= + −
= − + − − − = −
B
B
d i d i
b g b g
d i d i d i
b g b g
b g
b g
84 67 226 75 3114
2
1299 6 2 28584 1559 9 3114
H2
C H O CO ( ) + H O( ) (1)
H O H O( ) (2)
C H O CO ( ) + 3H O( ) (3)
2 2 2 2 2
2 2 2
2 6 2 2 2
( ) ( ) .
( ) ( ) .
( ) ( ) .
g g g l kJ mol
g g l kJ mol
g g g l kJ mol
c
o
c
o
c
o
+ → = −
+ → = −
+ → = −
5
2
2 1299 6
1
2
28584
7
2
2 1559 9
1
2
3
Δ
Δ
Δ
H
H
H
The acetylene dehydrogenation reaction is (1) + 2 (2) (3)
kJ mol kJ / mol
Hess's law
r
o
c
o
c
o
c
o
× −
⇒ = + × −
= − + − − − = −
Δ Δ Δ Δ

. ( . ) ( . ) .
H H H H
1 2 3
2
1299 6 2 28584 1559 9 3114
b g

9-5
9.10
a.
b.
c.
C H l
25
2
O g) 8CO g 9H O g kJ / mol
8 18 2 2 2 r
o
b g b g b g
+ → + = −
(
ΔH 4850
When 1 g-mole of C8H18(l) and 12.5 g-moles of O2(g) at 25°C and 1 atm react to form 8 g-moles
of CO2(g) and 9 g-moles of H2O(g), the change in enthalpy equals -4850 kJ.
Energy balance on reaction system (not including heated water):
Δ Δ Δ Δ
E E W Q U n U
k p
, ,
mol C H consumed kJ mol
8 18 c
o
= ⇒ = =
0 b g b g
( .
Cp H O(l)
) from Table B.2 kJ / mol. C
2
= × −
754 10 3 D
− = =
× °
×
=
−
−
Q m C T
H O p H O(l)
2 2
1.00 kg 1 mol 75.4 10 kJ 21.34 C
18.0 10 kg mol. C
kJ
( ) .
Δ
3
3
89 4
D
Q U
U
U
= ⇒ − =
⇒ = −
Δ
Δ
Δ
89 4
5079
.

kJ
2.01 g C H consumed 1 mol C H (kJ)
114.2 g 1 mol C H
kJ mol
8 18 8 18 c
o
8 18
c
o
Δ Δ

.
H U RT i i
c
o
c
o
gaseous
products
gaseous
reactants
3
kJ mol
8.314 J 1 kJ 298 K
mol K 10 J
= + −
L
N
MMMM
O
Q
PPPP
=− +
+ −
⋅
∑ ∑
ν ν
5079
8 9 12 5
b g
o
c
ˆ 5068 kJ mol
H
⇒ Δ = −
( 5068) ( 4850)
% difference = 100 = 4.3 %
5068
− − −
× −
−
Δ Δ Δ Δ

H H H H
f
c
o
f
o
CO g
f
o
H O g
f
o
C H l
2 2 8 18
= + −
8 9
e j e j e j
b g b g b g
( ) ( )
( ) ( )
8 18
o
f
C H l
ˆ 8 393.5 9 241.83 5068 kJ/mol 256.5 kJ/mol
H
⇒ Δ = ⎡ − + − + ⎤ = −
⎣ ⎦
There is no practical way to react carbon and hydrogen such that 2,3,3-trimethylpentane is the only
product.

9-6
9.11 a.
b.
c.
d.
n i
− → −
C H g C H g
4 10 4 10
b g b g
Basis: 1 mol feed gas
0.930 mol n-C4H10 (nn- C4H10)out
0.050 mol i-C4H10 ( ni-C4H10)out
0.020 mol HCl 0.020 mol HCl
149°C Q(kJ/mol) 149°C
(n mol
(n mol
n-CH H out
i-CH H out
4 10
4 10
) . ( .400) .
) . . .400 .420
= − =
= + × =
0 930 1 0 0 560
0 050 0 930 0 0
ξ
ν
mol
H10 H10
10
H
=
−
=
−
=
( ) ( )
0.560 0.930
n-C4 n-C4
out in
n-C4
n n
1
0 370
.
Δ Δ Δ Δ
. . .
H H H H
i n
r
o
f
o
C H
f
o
C H
Table B.1
r
o
4 10 4 10
kJ mol kJ mol
= − = − − − = −
− − ⇒
e j e j b g
134 5 124 7 9 8
References: n i
− − °
C H g C H g at 25 C
4 10 4 10
b g b g
,
substance
(mol) kJ mol (mol) kJ mol
C H 1
C H
in in out out
4 10 1 1
4 10 2
n H n H
n H H
i H

.
.
b g b g
−
− − −
0 600
0 400
.
H C dT
p
1
Table B.2
kJ
mol
kJ mol
=
L
N
MM
O
Q
PP =
B
z25
149
14 29 .
H C dT
p
2
Table B.2
kJ
mol
kJ mol
=
L
N
MM
O
Q
PP =
B
z25
149
1414
( )( ) ( )( )
o
r
out in
ˆ ˆ ˆ
[ ] 0.370 9.8 1 14.142 1 14.287 kJ
3.68 kJ
i i i i
Q H H n H n H
ξ
= Δ = Δ + − = ⎡− + − ⎤
⎣ ⎦
= −
∑ ∑
For 325 mol/h fed,
9.8 kJ 325 mol feed 1 h 1 kW
= 0.90 kW
1 mol feed h 3600 s 1 kJ/s
Q
−
= −

( )
r
3.68 kJ
ˆ 149 C 9.95 kJ/mol
0.370 mol
H
−
Δ ° = = −

9-7
9.12 a.
b.
c.
1 m3
at 298K, 3.00 torr Products at 1375K, 3.00 torr
n0 (mol) n1 (mol O2)
0.111 mol SiH4/mol n2 (mol SiO2)
0.8889 mol O2/mol n3 (mol H2)
SiH g)+ O g) SiO s)+ 2H g)
4 2 2 2
( ( ( (
→
3
-3 3
o
i
4
2 1 2
2 2
1 m 273 K 3.00 torr 1 mol
Ideal Gas Equation of state : 0.1614 mol
298 K 760 torr 22.4 10 m
SiH : 0=0.1111(0.1614 mol) 0.0179 mol
O : 0.8889(0.1614 mol) 0.1256 mol O
SiO : 0.
io i
n
n n
n
n
ν ξ
ξ ξ
ξ
ξ
= =
×
= +
− ⇒ =
= − =
= = 2
2 3 2
0179 mol SiO
H : =2 =0.0358 mol H
n ξ
Δ Δ Δ
( ) ( )
( . )] .
( (
H H H
r
o
f
o
SiO s) f
o
SiH g)
2 4
= [ kJ mol kJ / mol
= −
− − − = −
851 619 7891
References : SiH g),O (g),SiO g),H g) at 298 K
4 2 2 2
( ( (
out
in out
in
4
2 1
2 2
2 3
ˆ
ˆ
Substance
(mol h) (mol h)
(kJ mol) (kJ mol)
SiH 0.0179 0
ˆ
O 0.1435 0 0.1256
ˆ
SiO 0.0179
ˆ
H 0.0358
n
n H
H
H
H
H
− −
− −
− −
O g,1375K): C) = kJ / mol
SiO s,1375K): kJ / mol
H g,1375K): C) = / mol
2 O
o
2 SiO s)
298
2 H
o
2
Table B.8
2
2
Table B.8
( ( .
( ( ) .
( (
(
H H
H C dT
H H
p
1
2
1375
3
1102 3614
7918
1102 32.35 kJ
=
= =
=
B
B
z
o 3
r
out in
3
3
ˆ ˆ ˆ 7.01 kJ/m feed
7.01 kJ 27.5 m 1 h 1 kW
Q= 0.0536 kW (transferred from reactor)
m h 3600 s 1 kJ/s
i i i i
Q H H n H n H
ξ
= Δ = Δ + − = −
−
= −
∑ ∑

9-8
9.13 a.
b.
c.
d.
Fe O s C s Fe s + CO g
2 3b g b g b g b g
+ →
3 2 3 , Δ ( .
Hr 77 2111 105
D
F) Btu lb - mole
= ×
Basis:
2000 lb Fe 1 lb - mole
55.85 lb
lb - moles
m
m
= 3581
. Fe produced
5372
. lb - moles CO produced
17.9 lb - moles Fe O fed
53.72 lb - moles C fed
2 3
17.9 lb-moles Fe2O3 (s) 35.81 lb-moles Fe (l)
77° F 2800° F
53.72 lb-moles C 53.72 lb-moles CO(g)
77° F 570° F
Q (Btu/ton Fe)
References: Fe O s C s Fe s , CO g at F
2 3b g b g b g b g
, , 77°
Substance
(lb - moles) (Btu lb - mole) (lb - moles) (Btu lb - mole)
Fe O s,77 F 17.91 0
C s,77 F
Fe l,2800 F
CO g,570 F
in in out out
2 3
n H n H
H
H

.
.
.
° − −
° − −
° − −
° − −
b g
b g
b g
b g
5372 0
3581
5372
1
2
Fe(l,2800 F): F Btu lb - mole
1 Fe s Fe l
D
H C dT H C dT
p m p
= + ° + =
z z
d i b g d i
b g b g
77
2794
2794
2800
2794 28400
Δ
CO(g,570 F): F) Btu lb - mole
2 CO
interpolating
from Table B.9
D D
(
H H
= =
A
FH IK
570 3486
Q H
n H
n H n H
r
i i i i
= = + −
∑ ∑
Δ
Δ
Fe
o
Fe out in

ν
Btu / ton Fe produced
=
×
+ + − = ×
3581 2111 10
2
3581 28400 5372 3486 0 4 98 10
5
6
. .
. . .
b ge j b gb g b gb g
Effect of any pressure changes on enthalpy are neglected.
Specific heat of Fe(s) is assumed to vary linearly with temperature from 77°F to 570°F.
Specific heat of Fe(l) is assumed to remain constant with temperature.
Reaction is complete.
No vaporization occurs.

9-9
9.14 a. C H g C H CH g) H g)
7 16 6 5 3 2
b g→ +
( (
4
Basis: 1 mol C7H16
b.
c.
d.
1 mol C7H16 1 mol C6H5CH3
400°C 4 mol H2
400° C
Q (kJ/mol)
References: C s ,H g at 25 C
2
b g b g °
substance
mol kJ mol mol kJ mol
C H 1
C H
H
in in out out
7 16 1
7 8 2
2 3
n H n H
H
H
H
b g b g b g b g

− −
− −
− −
1
4
( )
0.2427
7 16
400
7 16 1 f C H (g) 25
ˆ ˆ
C H g,400 C : ( )
( 187.8 +91.0) kJ/mol= 96.8 kJ/mol
p
H H C dT
↓
⎡ ⎤
⎢ ⎥
° = Δ +
⎢ ⎥
⎣ ⎦
= − −
∫
D
C H CH g, C :
(+50 + 60.2) kJ / mol = 110.2 kJ / mol
6 5 3 f C H CH g)
6 5 3
Table B.2
400 2
25
400
° = +
L
N
MM
O
Q
PP
=
B
z
b g ( ) (
H H C dT
p
Δ D
H g, C : C kJ mol
2 H2
Table B.8
400 400 1089
3
° = =
B
b g ( ) .
H H D
( )( ) ( )( ) ( )( )
out in
ˆ ˆ
1 110.2 4 10.89 1 96.8 kJ 251 kJ (transferred to reactor)
i i i i
Q H n H n H
= Δ = −
= ⎡ + − − ⎤ =
⎣ ⎦
∑ ∑
7 16
251 kJ
ˆ (400 C)= 251 kJ/mol
1 mol C H react
r
H
Δ =
D

9-10
9.15 a.
b.
c.
CH O g CH g H g CO g
3 2 4 2
b g b g b g b g b g
→ + +
Moles charged: (Assume ideal gas)
2.00 liters 273 K 350 mm Hg mol
873 K 760 mm Hg 22.4 liters STP
mol CH O
3
1
0 01286 2
b g b g
= .
Let x = fraction CH O
3
b g2
decomposed (Clearly x1 since P P
f 3 0 )
0.01286(1 – x ) mol
(CH ) O
3
0.01286 mol
2
600°C, 350 mm Hg
(CH3 )2O
0.01286 x mol CH4
0.01286 x mol H2
0.01286 x mol CO
600°C
875 mm Hg
Total moles in tank at t h x x x
= = − + = +
2 0 01286 1 3 0 01286 1 2
. .
b g b gmol
P V
P V
n RT
n RT
n
n
P
P
x
x
f f f f
0 0 0 0
0 01286 1 2
0 01286
875
350
0 75 75%
= ⇒ = ⇒
+
= ⇒ = ⇒
.
.
.
b g decomposed
References: ( ) ( ) ( )
2 2
C s , H g , O g at 25 C
D
substance
(mol) kJ / mol) mol) (kJ / mol)
CH O g 0.01286
CH g
H g
CO g
in in out out
3 2 1 1
4 2
2 3
4
n H n H
H H
H
H
H

( (

. .
. .
. .
. .
b g b g
b g
b g
b g
0 25 0 01286
0 75 0 01286
0 75 0 01286
0 75 0 01286
×
− − ×
− − ×
− − ×
CH O(g,600 C):
J
mol
kJ
10 J
62.40 kJ / mol
118 kJ mol
3 f
o
CH O 3
3
given
b g b g
2 1
298
873
2
1
18016
D ( ) ( . )
H H C dT
p
= +
L
N
MM
O
Q
PP × = − +
= −
B
z
Δ
Table B.2
4
600
o
4 2 f CH 25
ˆ ˆ
CH (g,600 C): ( ) 74.85 29.46 45.39 kJ mol
p
H H C dT
↓
⎡ ⎤
⎢ ⎥
= Δ + = − + = −
⎢ ⎥
⎣ ⎦
∫
D
H ( : ( .
2 3 600 1681
g,600 C) C) kJ mol
H2
Table B.8
D D
H H
= =
B
CO(g,600 C): C) kJ mol 92.95 kJ mol
f
o
CO CO
Table B.8
Table B.1
Table B.8
D D
( ) ( . .
H H H
4 600 11052 17 57
= + = − + = −
B B
Δ
For the reaction of parts (a) and (b), the enthalpy change and extent of reaction are :
[ ]
out out in in
ˆ ˆ 1.5515 ( 1.5175) kJ 0.0340 kJ
H n H n H
Δ = − = − − − = −
∑ ∑

9-11
d.
( ) ( )
4 4
4
CH out CH in
CH
r r
(n ) (n ) 0.75 0.01286
mol 0.009645 mol
1
0.0340 kJ
ˆ ˆ
600 C 600 C 3.53 kJ/mol
0.009645
H H H
ξ
ν
ξ
− ×
= = =
−
Δ = Δ ° ⇒ Δ ° = = −
Δ Δ
[
U H RT
r r i i
600 600
° = ° − −
∑ ∑
C C
gaseous
products
gaseous
reactants
]
b g b g ν ν
( )
3
8.314 J 1 kJ 873 K 1 1 1 1
3.53 kJ mol 18.0 kJ mol
mol K 10 J
+ + −
= − − = −
⋅
( )
ˆ 600 C (0.009645 mol)( 18.0 kJ/mol) 0.174 kJ (transferred from reactor)
r
Q U
ξ
= Δ ° = − = −
9.16 a.
SO g)
1
2
O g) SO (g)
2 2 3
( (
+ →
Basis :
l00 kg SO 10 mol SO
min 80.07 kg SO
mol SO min
3
3
3
3
3
= 1249
n1 mol O /min
2
n0
mol SO /min
2
450°C
100% excess
i
n1
mol O /min
2
3.76
450°C
1249 mol SO /min
3 s
n0
mol SO /min
2
n3
mol O /min
2
n1
mol O
/ i
2
3.76
550°C
mw
(kg H O( /h)
l)
2
25°C
mw
(kg H O(l) /h)
2
40°C
Assume low enough pressure for
H to be independent of P.
Generation output
3
2 2 3
2 2
3
2
SO balance :
(mol SO fed) 0.65 mol SO react 1 mol SO produced
min mol SO fed 1 mol SO react
mol SO
min
mol SO / min fed
=
=
⇒ =
b g

n
n
0
0
1
1249
1922
100% excess air:
n1
1922 mol S
1
1922
= =
O 0.5 mol O reqd 1+1 mol O fed
min mol SO 1 mol O reqd
mol O min fed
2 2 2
2 2
2
b g
N balance : mol / min in out
2 376 1922 7227
. b g b g
=
65% conversion : mol s mol SO min out
2
.
n2 1922 1 0 65 673
= − =
b g
O balance: 2 1922 2 1922 3 1249 2 673 2 1298
3 3
+ = + + ⇒ =

n n mol / min out
9.15 (cont’d)
o
0 2
o
1 2
1 2
(mol SO /min), 450 C
100% excess air, 450 C
(mol O /min)
3.76 (mol N /min)
n
n
n

3
2 2
3 2
1 2
o
1249 mol SO /min
(mol SO /min)
(mol O /min)
3.76 (mol N / min)
550 C
n
n
n

. .

9-12
b.
c.
Extent of reaction : ξ
ν
mol / min
SO2
.
. .
=
−
=
−
=
( ) ( )
673 1922
SO out SO in
2 2
n n
1
1249
Δ Δ Δ
( ( . ( . ) .
( (
H H H
r
o
f
o
SO g) f
o
SO g)
) ) kJ / mol
3 2
Table B.1
= − = − − − = −
B
39518 296 9 99 28
References : SO g O g N g SO g at C
2 3
2 2 25
b g b g b g b g
, , , D
Substance
mol / min) kJ / mol) mol / min) kJ / mol)
SO 1922
O
N
SO
in in out out
2
2
2
3

(

(

(

(

n H n H
H H
H H
H H
H
1 4
2 5
3 6
7
673
1922 1298
7227 7227
1249
− −
SO (g,450 C) : kJ / mol
2
Table B.2
D .
H C dT
p
1
25
450
19 62
= =
B
z
O g C C) kJ / mol
2 O2
Table B.8
( , ) ( .
450 450 1336
2
D D
= = =
B
H H
N g C C) kJ / mol
2 N2
Table B.8
( , ) ( .
450 450 12 69
3
D D
= = =
B
H H
Out :
Table B.2
550
2 4 25
ˆ
SO (g,550 C) : 24.79 kJ/mol
p
H C dT
↓
= =
∫
D
O g C C) kJ / mol
2 O2
Table B.8
( , ) ( .
550 550 16 71
5
D D
= = =
B
H H
N g C C) kJ / mol
2 N2
Table B.8
( , ) ( .
550 550 1581
6
D D
= = =
B
H H
SO (g,550 C) : kJ / mol
3
Table B.2
D .
H C dT
p
7
25
550
3534
= =
B
z
( )( ) ( )( ) ( )( ) ( )( ) ( )( ) (
( ) ( )( )
o
r
out in
4
ˆ ˆ ˆ
1249 98.28 673 24.796 179.8 16.711 7227 15.808 1249 35.336 192
1922 13.362 7227 12.691
8.111 10 kJ 1 min 1 kW
1350 kW
min 60 s 1 kJ/s
i i i i
Q H H n H n H
ξ
= Δ = Δ + −
= − + + + + −
− −
− ×
= = −
∑ ∑

Assume system is adiabatic, so that
Q Q
lost from reactor gained by cooling water
=
9.16 (cont’d)

9-13
d.

Q H m H H
w w w
= = −
L
N
MMM
O
Q
PPP
A A
Δ
Table B.5
Table B.5
l, 40 C l, 25 C
D D
e j e j
⇒ × =
F
HG I
KJ − ⇒ =
8111 10 167 5 104 8 1290
4
. . .
kJ
min
kg
min
kJ
kg
kg min cooling water
m m
w w
If elemental species were taken as references, the heats of formation of each molecular species would
have to be taken into account in the enthalpy calculations and the heat of reaction term would not have
been included in the calculation of Δ
H .
9.17
a.
CO(g) H O v H g CO (g)
2 2 2
+ → +
b g ( ) ,
Δ Δ Δ Δ
.
(
H H H H
r
o
f
o
CO g)
f
o
CO(g)
f
o
H O v
2 2
Table B.1
kJ
mol
= − − = −
B
e j e j e j b g 4115
Basis : 2 5 1116
. .
m STP product gas h 1000 mol 22.4 m STP mol h
3 3
b g b g =
n0 (mol CO/h)
25°C
150°C
111.6 mol/h
0.40 mol H /mol
2
500°C
Qr(kW)
n2 (mol H O( )/h)
v
2
reactor
0.40 mol CO /mol
2
0.20 mol H O( )/h
v
2
Qc(kW)
condenser
n3 (mol CO /h)
2
n4 (mol H /h)
2
n5 (mol H O(v)/h),
2 sat'd
15°C, 1 atm
15°C, 1 atm
n6 (mol H O( )/h)
l
2
C balance on reactor : . . .
n1 0 40 1116 44 64
= =
b gb g
mol h mol CO h
H balance on reactor : 2 1116 2 0 40 2 0 20 66 96
2 2
. . . .
n n
= + ⇒ =
b gb g b gb g b g
mol h mol H O v h
2
Steam theoretically required = =
h 1 mol CO
mol H O
2
2
44 64
.
% excess steam =
−
× =
66 96 44 64
100% 50%
. .
b g mol h
44.64 mol h
excess steam
CO balance on condenser : mol h mol CO h
2 2
. . .
n3 0 40 1116 44 64
= =
b gb g
H balance on condenser: mol h mol H h
2 2
. . .
n4 0 40 1116 44 64
= =
b gb g
Saturation of condenser outlet gas:
y
p n
n
n
w
H O
2
2
2
C
p
mol H O h
. + . + mol h
mm Hg
760 mm Hg
mol H O v h
=
°
⇒ = ⇒ =
∗
15
44 64 44 64
12 788
153
5
5
5
b g b g
b gb g b g

.
.
H O balance on condenser: 111.6 mol H O h
mol H O h condensed = 0.374 kg / h
2 2
2
b gb g
0 20 153
208
6
6
. .
.
= +
⇒ =
n
n
9.16 (cont’d)
n1 (mol CO/h)
25o
C

9-14
b.
c.
Energy balance on condenser
References : H g) CO (g) at C, H O
2 2
( , 2 25
D
at reference point of steam tables
Substance
mol / h kJ / mol mol / h kJ / mol
CO g 44.64
H g
H O v
H O l
in in out out
2 1 4
2 2 5
2 3 6
2 7

( ) .
( ) . .
. .
.
n H n H
H H
H H
H H
H
44 64
44 64 44 64
22 32 153
2080
b g
b g − −
Enthalpies for CO2 and H2 from Table B.8
CO g,500 C) : C) kJ / mol
2 CO2
( ( .
D D
H H
1 500 2134
= =
H g,500 C) : C) kJ / mol
2 H2
( ( .
D D
H H
2 500 1383
= =
H O(v,500 C) :
kJ
kg
kg
10 mol
kJ mol
2 3 3
D .
H = ×
F
HG I
KJ =
3488
18
62 86
2
2 4 CO
ˆ ˆ
CO (g,15 C) : (15 C) 0.552 kJ/mol
H H
= = −
D D
H g,15 C) : C) kJ / mol
2 H2
( ( .
D D
H H
5 15 0 432
= = −
H O(v,15 C) :
kJ
kg
kg
mol
kJ mol
2 6
D .
.
H = ×
F
HG I
KJ =
2529
18 0
10
4552
3
H O(l,15 C) :
kJ
kg
kg
mol
kJ mol
2 7
D .
.
.
H = ×
F
HG I
KJ =
62 9
18 0
10
113
3
.
.
Q H n H n H
i i i i
= = − =
−
= −
∑ ∑
Δ
out in
49.22 kJ 1 h 1 kW
h 3600 s 1 kJ s
kW
heat transferred from condenser
29718
0812
b g
b g
Energy balance on reactor :
References : H g) C(s), O (g) at C
2 2
( , 25°
Substance
mol / h) kJ / mol) mol / h) kJ / mol)
CO g 44.64
H O v
H g
CO g
in in out out
1
2 2 3
2 4
2 5

(

(

(

(
( )
( ) . .
.
.
n H n H
H
H H
H
H
− −
− −
− −
66 96 22 32
44 64
44 64
b g
b g
CO g,25 C) : kJ / mol
f CO
Table B.1
( ( ) .
D D
H H
1 11052
= = −
Δ
H O(v,150 C) : = C) kJ mol
2 2 f H O(v) H O
Tables B.1, B.8
2 2
D D D
( ) ( .
H H H
Δ + = −
150 237 56
9.17 (cont’d)

9-15
d.
H O(v,500 C) : = C) kJ mol
2 3 f H O(v) H O
Tables B.1, B.8
2 2
D D D
( ) ( .
H H H
Δ + = −
500 224 82
H g,500 C) : C) kJ / mol
2 H
Table B.8
2
( ( .
D D
H H
4 500 1383
= =
CO g,500 C) : C) 372.16 kJ / mol
2 CO CO
Tables B.1, B.8
2 2
( ( ) (
D D D
H H H
f
5 500
= + = −
Δ
Q H n H n H
i i i i
= = − =
− − −
= −
∑ ∑
Δ . ( . )
.
out in
kJ 1 h 1 kW
h 3600 s 1 kJ s
kW
heat transferred from reactor
2101383 20839 96
0 0483
b g
Benefits
Preheating CO ⇒ more heat transferred from reactor (possibly generate additional steam for plant)
Cooling CO ⇒ lower cooling cost in condenser.
9.17 (cont’d)

9-16
9.18 b.
References : FeO(s), CO(g), Fe(s), CO g) at 25 C
Substance
mol) kJ / mol) mol) kJ / mol)
FeO 1.00
CO
Fe
CO
2
o
in in out out
2
(
(

( (

(

n H n H
n H
n H n H
n H
n H
0 1 1
0 0 2 2
3 3
4 4
− −
− −
Q H n H n H
Q H n H n H n H n H n H
X
n
n X
= + −
⇒ = + + + + −
=
−
⇒ = −
∑ ∑
ξ
ξ
Fractional Conversion :
r
o
out out in in
r
o
Δ
Δ

( . )
.
1 1 2 2 3 3 4 4 0 0
1
1
100
100
1
CO consumed
mol CO (1 mol FeO consumed
mol FeO consumed
mol CO
Fe produced : =
mol Fe (1 mol FeO consumed
mol FeO consumed
mol Fe =
:
)
( )
( )
)
( )
1
1
1
1
1
1
1
1
1
2 0 1 0
3
1
1
−
= −
⇒ = − − = −
−
= −
n
n
n n n n X
n
n
n X
CO produced : =
mol CO (1 mol FeO consumed
mol FeO consumed
mol CO
2
2
2
n
n
n X
4
1
1
1
1
1
−
= − =
)
( )
Extent of reaction :
CO out CO in
CO
ξ
ν
=
−
=
−
=
( ) ( )
n n n n
X
2 0
1
, , , ,
H C dT i
i pi
T
= =
z25
0 1 2 3 4
for
. ( ) . ( )
( . . . )
H T T
H T T
0 0
6
0
2 2
0 0
6
0
2
0 02761 298 2 51 10 298
8 451 0 02761 2 51 10
= − + × −
⇒ = − + + ×
−
−
kJ / mol
. ( ) . ( ) . ( / / )
( . . . . / )
( . ( ) . ( )
. . . )
. ( ) . ( )

H T T T
H T T T
H T T
H T T
H T T
H
1
6 2 2 2
1
6 2 2
2
6 2 2
2
6 2
3
5 2 2
3
0 0528 298 31215 10 298 3188 10 1 1 298
17 0814 0 0528 31215 10 3188 10
0 02761 298 2 51 10 298
8 451 0 02761 2 51 10
0 01728 298 1335 10 298
= − + × − + × −
⇒ = − + + × + ×
= − + × −
⇒ = − + + ×
= − + × −
⇒ =
−
−
−
−
−
kJ / mol
kJ / mol
( . . . )
. ( ) . ( ) . ( / / )
( . . . . / )
− + + ×
= − + × − + × −
⇒ = − + + × + ×
−
−
−
6 335 0 01728 1335 10
0 04326 298 0573 10 298 818 10 1 1 298
16145 0 04326 0573 10 818 10
5 2
4
5 2 2 2
4
5 2 2
kJ / mol
kJ / mol
T T
H T T T
H T T T

9-17
c.
d.
0 0
1 2 3 4
0 1 2
3
2.0 mol CO, 350 K, 550 K, and 0.700 mol FeO reacted/mol FeO fed
1 0.7 0.3, 2 0.7 1.3, 0.7, 0.7, 0.7
ˆ ˆ ˆ
Summary: 1.520 kJ/mol, 13.48 kJ/mol, 7.494 kJ/mol,
ˆ 7.207
n T T X
n n n n
H H H
H
ξ
= = = =
⇒ = − = = − = = = =
= = =
= 4
ˆ
kJ/mol, 10.87 kJ/mol
H =
o
r
ˆ 16.48 kJ/mol
(0.7)( 16.48) (0.3)(13.48) (1.3)(7.494) (0.7)(7.207) (0.7)(10.87) (2)(1.520)
11.86 kJ
H
Q
Q
Δ = −
= − + + + + −
⇒ =
no To X T Xi n1 n2 n3 n4 H0 H1 H2 H3 H4 Q
1 400 1 298 1 0 0 1 1 2.995 0 0 0 0 -19.48
1 400 1 400 1 0 0 1 1 2.995 5.335 2.995 2.713 4.121 -12.64
1 400 1 500 1 0 0 1 1 2.995 10.737 5.982 5.643 8.553 -5.279
1 400 1 600 1 0 0 1 1 2.995 16.254 9.019 8.839 13.237 2.601
1 400 1 700 1 0 0 1 1 2.995 21.864 12.11 12.303 18.113 10.941
1 400 1 800 1 0 0 1 1 2.995 27.555 15.24 16.033 23.152 19.71
1 400 1 900 1 0 0 1 1 2.995 33.321 18.43 20.031 28.339 28.895
1 400 1 1000 1 0 0 1 1 2.995 39.159 21.67 24.295 33.663 38.483
1 298 1 700 1 0 0 1 1 0 21.864 12.11 12.303 18.113 13.936
1 400 1 700 1 0 0 1 1 2.995 21.864 12.11 12.303 18.113 10.941
1 500 1 700 1 0 0 1 1 5.982 21.864 12.11 12.303 18.113 7.954
1 600 1 700 1 0 0 1 1 9.019 21.864 12.11 12.303 18.113 4.917
1 700 1 700 1 0 0 1 1 12.11 21.864 12.11 12.303 18.113 1.83
1 800 1 700 1 0 0 1 1 15.24 21.864 12.11 12.303 18.113 -1.308
1 900 1 700 1 0 0 1 1 18.43 21.864 12.11 12.303 18.113 -4.495
1 1000 1 700 1 0 0 1 1 21.67 21.864 12.11 12.303 18.113 -7.733
1 400 0 500 0 1 1 0 0 2.995 10.737 5.55 5.643 8.533 13.72
1 400 0.1 500 0.1 0.9 0.9 0.1 0.1 2.995 10.737 5.55 5.643 8.533 11.82
1 400 0.2 500 0.2 0.8 0.8 0.2 0.2 2.995 10.737 5.55 5.643 8.533 9.92
1 400 0.3 500 0.3 0.7 0.7 0.3 0.3 2.995 10.737 5.55 5.643 8.533 8.02
1 400 0.4 500 0.4 0.6 0.6 0.4 0.4 2.995 10.737 5.55 5.643 8.533 6.12
1 400 0.5 500 0.5 0.5 0.5 0.5 0.5 2.995 10.737 5.55 5.643 8.533 4.22
1 400 0.6 500 0.6 0.4 0.4 0.6 0.6 2.995 10.737 5.55 5.643 8.533 2.32
1 400 0.7 500 0.7 0.3 0.3 0.7 0.7 2.995 10.737 5.55 5.643 8.533 0.42
1 400 0.8 500 0.8 0.2 0.2 0.8 0.8 2.995 10.737 5.55 5.643 8.533 –1.48
1 400 0.9 500 0.9 0.1 0.1 0.9 0.9 2.995 10.737 5.55 5.643 8.533 –3.38
1 400 1 500 1 0 0 1 1 2.995 10.737 5.55 5.643 8.533 –5.28
0.5 400 0.5 400 0.5 0.5 0.0 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
0.6 400 0.5 400 0.5 0.5 0.1 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
0.8 400 0.5 400 0.5 0.5 0.3 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
1.0 400 0.5 400 0.5 0.5 0.5 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
9.18 (cont'd)

9-18
1.2 400 0.5 400 0.5 0.5 0.7 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
1.4 400 0.5 400 0.5 0.5 0.9 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
1.6 400 0.5 400 0.5 0.5 1.1 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
1.8 400 0.5 400 0.5 0.5 1.3 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
2.0 400 0.5 400 0.5 0.5 1.5 0.5 0.5 2.995 5.335 2.995 2.713 4.121 -3.653
9.19 a. Fermentor capacity : 550,000 gal
Solution volume : (0.9 550,000) gal
Final reaction mixture :
lb C H OH / lb solution
lb (yeast, other species) / lb solution
lb H O / lb solution
m 2 5 m
m m
2 m
× =
R
S
|
T
|
495 000
0 071
0 069
086
,
.
.
.
Mass of tank contents :
gal 1 ft 65.52 lb
7.4805 gal 1 ft
lb
3
m
3 m
495 000
4335593
,
=
Mass of ethanol produced :
4.336 10 lb solution 0.071 lb C H OH
lb solution
3.078 10 lb C H
3.078 10 lb C H OH 1 lb - mole C H OH
46.1 lb C H OH
lb - mole C H OH
307827 lb C H OH 1 ft C H OH 7.4805 gal
49.67 lb C H OH 1 ft
gal C H OH
6
m m 2 5
m
5
m 2 5
5
m 2 5 2 5
m 2 5
2 5
m 2 5
3
2 5
m 2 5
3 2 5
×
= ×
⇒
×
=
⇒ =
6677
46 360
,
-30
-20
-10
0
10
20
30
40
50
0 500 1000 1500
T(K)
Q
-10
-5
0
5
10
15
20
0 500 1000 1500
To(K)
Q
-6
-5
-4
-3
-2
-1
0
0 0.2 0.4 0.6 0.8 1
X
Q
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
0 0.5 1 1.5 2 2.5
no
Q
9.18 (cont'd)

9-19
b.
c.
d.
Makeup water required : 495,000 gal
gal C H OH 25 gal mash
2.6 gal C H OH
gal
2 5
2 5
− = ×
46 360
4 9 104
,
.
Acres reqd.:
gal C H OH bu 1 acre 1 batch 24 h 330 days
batch 2.6 gal C H OH 101 bu h day 1 year
acres
year
2 5
2 5
46360 1
1 8 1
175 105
,
.
= ×
C H O s) O g) CO g H O(l) kJ / mol
CO H O C H O
C H O kJ / mol
C H O s) H O l) C H OH l 4CO (g)
C H OH 4 CO C
12 22 11 2 2 2 c
o
c
o
f
o
2 f
o
2 f
o
12 22 11
f
o
12 22 11
12 22 11 2 2 5 2
r
o
f
o
2 5 f
o
2 f
o
12
( ( ( ) .
( ) ( ) ( )
( ) .
( ( ( )
( ) ( ) (
+ → + = −
= + −
⇒ = −
+ → +
= + −
12 12 11 56491
12 11
221714
4
4
Δ
Δ Δ Δ Δ
Δ
Δ Δ Δ Δ
H
H H H H
H
H H H H H O H O = kJ / mol
kJ 453.6 mol 0.9486 Btu
1 mol 1 lb - mole kJ
Btu / lb - mole
Moles of maltose :
lb solution 0.071 lb C H OH 1 lb - mole C H OH lb - mole C H O
lb solution 46.1 lb C H OH 4 lb - mole C H OH
lb - moles C H O lb - moles
= F-85 F)
= (1669 lb - moles)(
22 11 f
o
2
r
o
m 2 5 2 5 12 22 11
m 2 5 2 5
12 22 11 C H O
10 22 11
) ( ) .
.
.
.
(
.
− −
⇒ =
−
= − ×
×
= ⇒ = =
+
−
Δ
Δ
Δ
H
H
n
Q H mC
r p
1845
1815
1
7811 10
4 336 10 1
1
1669 1669
95
7811
4
6
ξ
ξ D D
× + ×
− ×
10 4336 10 095 10
89 10
4 6
7
Btu
lb - mole
lb
Btu
lb - F
F)
= Btu heat transferred from reactor)
Brazil has a shortage of natural reserves of petroleum, unlike Venezuela.
m
) ( . )( . )(
. (
D
D
9.20 a. 4NH 5O 4NO 6H O,
2NH
3
2
O N 3H O
3 2 2
3 2 2 2
+ → +
+ → +
References: N g , H g , O (g), at 25 C
2 2 2
b g b g °
Substance
(mol min (kJ mol) (mol min) (kJ mol)
NH 100
Air
NO
H O
N
O
in in out out
3 1
2
3
2
2 5
2 6

)

n H n H
H
H
H
H
H
H
− −
− −
− −
− −
− −
− −
900
90
150
716
69
4

H H C dT
i i pi
T
= + z
Δ f
o
25
NH g, 25 C kJ mol
3 f
o
NH
Table B.1
3
° = = −
B
b g: ( ) .
H H
1 4619
Δ
9.19 (cont’d)

9-20
b.
c.
Air g, 150 C .67 kJ mol
Table B.8
° =
B
b g:
H2 3
NO g, 700 C kJ mol
Table B.1,Table B.2
° = + =
z B
b g: .
H C dT
p
3
25
700
90 37 111.97
H O g, 700 C 216.91 kJ mol
2
Table B.1,
Table B.8
° = −
B
b g:
H4
N g, 700 C kJ mol
2
Table B.8
° =
B
b g:
H5 20.59
O g, 700 C kJ mol
2
Table B.8
° =
B
b g:
H6 21.86
( .
Q H n H n H
i i i i
= = − = − × = −
∑ ∑
Δ
out in
kJ min min / 60s) kW
4890 1 815
(heat transferred from the reactor)
If molecular species had been chosen as references for enthalpy calculations, the extents of each
reaction would have to be calculated and Equation 9.5-1b used to determine Δ
H . The value of
Q
would remain unchanged.
9.21 a. Basis: 1 mol feed
1 mol at 310°C Products at 310°C
0.537 C2H4 (v) n1 (mol C2H4 (v))
0.367 H20 (v) n2 (mol H2O(v))
0.096 N2(g) 0.096 mol N2 (g)
n3 (mol C2H5OH (v))
n4 (mol (C2H5)2O) (v))
C H v) H O(v) C H OH(v)
2C H OH(v) C H O(v) H O(v)
2 4 2 2 5
2 5 2 5 2 2
( + ⇔
⇔ +
b g
5% ethylene conversion: 0537 0 05 0 02685
. . .
b gb g= mol C H consumed
2 4
⇒ = =
n1 4
0 95 0537 0510
. . .
b gb g mol C H
2
90% ethanol yield:
n3 0 02417
= =
0.02685 mol C H consumed 0.9 mol C H OH
1 mol C H
mol C H OH
2 4 2 5
2 4
2 5
.
C balance : 2 0537 2 0510 2 0 02417 4 1415 10
4 4
3
2
b gb g b gb g b gb g b g
. . . .
= + + ⇒ = × −
n n mol C H O
2 5
O balance : 0 367 0 02417 1415 10
2
3
2 2
. . .
= + + × ⇒ =
−
n n 0.3414 mol H O
9.20 (cont’d)

9-21
b.
( ) ( ) ( ) ( )
2 2 2
References: C s , H g , O g at 25 C, N g at 310 C
D D
( )
out
in out
in
2 4 1 1
2 2 2
2
2 5 3
3
2 5 4
2
ˆ
ˆ
substance
(mol) (mol)
(kJ/mol) (kJ/mol)
ˆ ˆ
C H 0.537 0.510
ˆ ˆ
H O 0.367 0.3414
N 0.096 0 0.096 0
ˆ
C H OH 0.02417
ˆ
C H O 1.415 10
n
n H
H
H H
H H
H
H
−
− −
− − ×
( ) ( )
o
2 4
f
p
310
o
2 4 1 f C H ˆ
25 Table B.1 for
Table B.2 for
ˆ ˆ
C H g, 310 C : ( ) 52.28 16.41 68.69 kJ mol
p
H
C
H H C dT
Δ
° = Δ + ⇒ + =
∫
H O g, 310 C C) kJ mol
2 f
o
H O(v) H O(v)
Table B.1
Table B.8
2 2
° = + ⇒ − + = −
b g b g
: ( ) ( . . .
H H H
2 310 24183 9 93 23190
Δ D
C H OH g, 310 C : kJ mol
2 5 f
o
C H OH(g)
Table B.1
Table B.2
2 5
° = + ⇒ − + = −
z
b g b g
( ) . . .
H H C dT
p
3
25
310
23531 2416 21115
Δ
C H O g, 310 C : C
kJ mol
2 5 2 f
o
(C H )O(l)
v
2 5
b g b g e j b g b g
° = + ° + = − + +
= −
z
. . .
.
H H H C dT
p
4
25
310
25 272 8 26 05 42 52
204 2
Δ Δ
i i i i
= = − =− ⇒
∑ ∑
Δ .
out in
kJ 1.3 kJ transferred from reactor mol feed
13
To suppress the undesired side reaction. Separation of unconsumed reactants from products and
recycle of ethylene.
9.22 C H CH O C H CHO H O
C H CH 9O 7CO 4H O
6 5 3 2 6 5 2
6 5 3 2 2 2
+ → +
+ → +
Basis: 100 lb-mole of C H CH
6 5 3 fed to reactor.
n0
100 lb-moles C H CH
(ft )
V0
6
reactor
5 3
(lb-moles O )
2
3.76n0 (lb-moles N )
2
350°F, 1 atm
3
jacket
Q(Btu)
mw(lb H O( )),
m 2 l mw
n2 (lb-moles O )
2
3.76n0 (lb-moles N )
2
(ft ) at 379°F, 1 atm
Vp
3
n1 (lb-moles C H CH )
6 5 3
n3 (lb-moles C H CHO)
6 5
n4 (lb-moles CO )
2
n5 (lb-moles H O)
2
80°F 105°F
(lb H O( )),
l
2
m
Strategy:
All material and energy balances will be performed for the assumed basis of 100 lb-mole
C H CH
6 5 3 . The calculated quantities will then be scaled to the known flow rate of water in
the product gas 29.3 lb 4 h
m
b g.
9.21 (cont'd)

9-22
Plan of attack: excess air Ideal gas equation of state
13% C H CHO formation Ideal gas equation of state
0.5% CO formation E.B. on reactor
C balance E.B. on jacket
H balance Scale , by actual / basis
O balance
6 5
2
%
, ,
⇒ ⇒
⇒ ⇒
⇒ ⇒
⇒ ⇒
⇒
⇒
n V
n V
n Q
n m
n V V Q m n n
n
p
w
p w
0 0
3
4
1
5 0 5 5
2
b g b g
100% excess air:
n0
1 1
200
=
+
=
100 lb - moles C H CH 1 mol O reqd mole O fed
1 mole C H CH 1 mol O reqd
lb - moles O
6 5 3 2 2
6 5 3 2
2
b g
N feed output lb - moles N lb - moles N
2 2 2
b g b g
= =
376 200 752
.
6 5 3 6 5 3 6 5
6 5 3
6 5 3 6 5 3
6 5
100 lb-moles C H CH 0.13 mole C H CH react 1mole C H CHO formed
13% C H CHO
1 mole C H CH fed 1mole C H CH react
=13 lb-moles C H CHO
n
→ ⇒ =
0.5% CO
100 0.005 lb- moles C H CH react 7 moles CO
1 mole C H CH
lb- moles CO
2
6 5 3 2
6 5 3
2
→ ⇒ = =
n4 35
b gb g .
C balance: 100 7 7 13 7 3 5 1 86 5
1 1
a fa f a fa f a fa f 6 5 3
mol C mole C7H8
lb - moles C lb - moles C H CH
B
= + + ⇒ =
n n
. .
H balance: 100 8 865 8 13 6 2 150
5 5
b gb g b gb g b gb g b g
lb - moles H lb - moles H O v
2
= + + ⇒ =
. .
n n
O balance: 200 2 2 13 1 35 2 15 1 182 5
2 2
lb - moles O lb - moles O2
= + + + ⇒ =
n n
. .
Ideal gas law − inlet:
V0
5
350 460
6 218 10
=
+
= ×
100 + 200 + 752 lb - moles 359 ft STP R
1 lb - moles 492 R
ft
3
3
b g b g b gD
D
.
Ideal gas law – outlet:
Vp =
+ + + + +
F
HG I
KJ +
= ×
865 182 5 13 35 15 752 379 460
6 443 105
. . .
.
C H O C H O CO H O N
3
3
7 8 2 7 8 2 2 2
lb - moles 359 ft R
1 lb - mole 492 R
ft
b gD
D
9.22 (cont'd)

9-23
Energy balance on reactor (excluding cooling jacket)
References : C s H g , O g , N g at C 77 F
2 2
b g b g b g b g e j
, 2 25D D
substance
lb - moles Btu lb - mole lb - moles Btu lb - mole
C H CH 100
O
N
C H CHO
CO
H O
in in out out
6 5 3 1 4
2 2 5
2 3 6
6 5 7
2 8
2 9
n H n H
H H
H H
H H
H
H
H
b g b g b g b g

.
.

.

865
200 182 5
752 752
13
35
15
− −
− −
− −
Enthalpies:
C H CH g,T): kJ mol
Btu lb - mole
1 kJ mol
Btu
1b - mole F
F
C H CH g,350 F): 10 Btu lb - mole
C H CH g,379 F): Btu lb - mole
6 5 3 f
o
6 5 3
4
6 5 3
Table B.1
( .
( .
( .
H T H T
H
H
b g b g b g
= × +
⋅°
−
L
N
MMM
O
Q
PPP
= ×
= ×
B
Δ
430 28
31 77
2 998
3088 10
1
4
4
D
D
D
C H CHO(g,T): F Btu lb - mole
10 Btu lb - mole
6 5
3

.
H T T
H
b g b g
= − + −
⇒ = − ×
17200 31 77
783
7
D
O g, F : F Btu / lb mole
N g, F : F Btu / lb mole
O g, F : F Btu / lb mole
2 O
2 N
2 O
2
Table B.9
2
Table B.9
2
Table B.9
350 350 1972 10
350 350 1911 10
379 379 2186 10
2
3
3
3
5
3
D D
D D
D D
e j
e j
e j
( ) .
( ) .
( ) .
H H
H H
H H
= = × −
= = × −
= = × −
B
B
B
N g, F : F Btu / lb mole
CO g, F : F Btu / lb mole
H O g, F : F Btu / lb mole
2 N
2 f CO g) CO
2 f H O g) H O
2
Table B.9
2 2
Table B.1 and B.9
2 2
Table B.1 and B.9
379 379 2116 10
379 379 1664 10
379 379 1016 10
6
3
8
5
9
5
D D
D D D
D D
e j
e j
b g
( ) .
( ) ( ) .
( ) ( ) .
(
(
H H
H H H
H H H
= = × −
= + = − × −
° = + = − × −
B
B
B
Δ
Δ
Energy Balance :
Q H n H n H
i i i i
= = − = − ×
∑ ∑
Δ .
out in
Btu
2 376 106
Energy balance on cooling jacket:
Q H m C dT
w p
= = z
Δ d i b g
H O l
2
80
105
Q = + ×
2 376 104
. Btu , C p = ⋅
1 0
. Btu (lb F)
m
D
2 376 10 10 105 80
6
. .
× = ×
⋅°
× −
Btu lb
Btu
lb F
F
m
m
mw b g b gD
⇒ mw = ×
9504 104
. lb H O l
m 2 bg
9.22 (cont'd)
⇓

9-24
a.
b.
Scale factor:

.
n
n
5
5 4
0 02711
b g
b g
actual
basis
m 2 2
m 2 2
1
29.3 lb H O 1b - mole H O 1
h 18.016 lb H O 15.0 lb - moles H O
h
= = −
V0
5 1 4
6 218 10 0 02711 169 10
= × = ×
−
. . .
ft h ft h feed
3 3
d id i b g
Vp = × = ×
−
6 443 10 0 175 10
5 1 4
. .
ft .02711 h ft h product
3 3
d id i b g
Q = − × = − ×
−
2 376 10 0 02711 6 10
6 1 4
. . .44
Btu h Btu / h
d id i
. . .
mw = × = =
−
9504 10 002711
2577 1
515
4 1
Btu h
lb 1 ft 7.4805 gal
h 62.4 lb 1 ft
h
60 min
gal H O min
m
3
m
3 2
d id i
9.23 a. CaCO s) CaO(s) +CO g)
3 2
( (
→
CaO(s)
900°C
CaCO3(s) CO2(g)
25°C 900°C
Q (kJ)
Basis : 1000 kg CaCO
kg mol
0.100 kg
kmol CaCO
10.0 kmol CaO(s) produced
10.0 kmol CO g) produced
10.0 kmol CaCO s) fed
3 3 2
3
= = ⇒
1000 1
10 0
. (
(
References: Ca(s), C(s), O2(g) at 25°C
out
in out
in
4
3 1
4
2
4
2 3
ˆ
ˆ
Substance (mol) (mol)
(kJ/mol) (kJ/mol)
ˆ
CaCO 10
ˆ
CaO 10
ˆ
CO 10
n
n H
H
H
H
H
− −
− −
− −
CaCO s, 25 C) : kJ / mol
CaO s, 900 C) : dT kJ / mol kJ / mol
CO g, 900 C) : C) kJ / mol kJ / mol
3
o
f
o
CaCO s)
Table B.1
o
f
o
CaO s)
Table B.1,
Table B.2
2
o
f
o
CO g) CO
o
Table B.1,
Table B.8
3
2 2
( ( ) .
( ( ) ( . . ) .
( ( ) ( ( . . ) .
(
(
(
H H
H H C
H H H
p
1
2
298
1173
3
1206 9
6356 4854 587 06
900 3935 42 94 35056
= = −
= + = − + = −
= + = − + = −
B
B
B
z
Δ
Δ
Δ
= = −
F
HG
I
KJ = ×
∑
∑
Δ i i i i
in
out
2.7 kJ
106
9.22(cont’d)

9-25
b.
c.
Basis : 1000 kg CaCO3 fed ⇒ 10.0 kmol CaCO3
CaCO s) CaO(s) + CO g)
2CO + O 2CO
3 2
2 2
( (
→
→
10 kmol CaCO3
25 o
C Product gas at 900o
C
200 kmol at 900o
C n2 (kmol CO2 )
n3 (kmol N2 )
0.75 N2 n4 (kmol CO)
0.020 O2
0.090 CO
0.14 CO2 n1 [kmol CaO(s)]
10 kmol CaCO react kmol CaO
3 ⇒ =
n1 10 0
.
n2 014 200
10 0 4
46
= + + =
( . )( )
. kmol CaCO react 1 kmol CO
1 kmol O
kmol O react 2 kmol CO
1 kmol O
kmol CO
3 2
2
2 2
2
2
n3 0 75 200 150
= =
( . )( ) kmol N2
C balance (10.0)(1) + (200)(0.09)(1) + (200)(0.14)(1) = 46(1) + kmol CO
: ( ) .
n n
4 4
1 10 0
⇒ =
References : Ca(s), C(s), O g), N g) at 25 C
2 2
( ( D
out
in out
in
3 1
2
2 2
2 3
2 4 4
ˆ
ˆ
Substance (mol) (mol)
(kJ/mol) (kJ/mol)
ˆ
CaCO 10.0
CaO 10 587.06
CO 28 350.56 46 350.56
ˆ ˆ
CO 18 10
ˆ
O 4.0
ˆ ˆ
N 150 150
n
n H
H
H
H H
H
H H
− −
− − −
− −
− −
3
2
Table B.1
o o
3 1 f CaCO (s)
Table B.1,
Table B.8
o o o
1 f CO(g) CO
Table
o o
2 2 O
ˆ ˆ
CaCO (s, 25 C) : ( ) 1206.9 kJ/mol
ˆ ˆ ˆ
CO(g, 900 C) : ( ) (900 C) ( 110.52 27.49) kJ/mol 83.03 kJ/mol
ˆ ˆ
O (g, 900 C) : (900 C)
H H
H H H
H H
↓
↓
↓
= Δ = −
= Δ + = − + = −
= =
2
B.8
Table B.8
o o
2 3 N
28.89 kJ/mol
ˆ ˆ
N (g, 900 C) : (900 C) 27.19 kJ/mol
H H
↓
= =
Q H n H n H
= = −
F
HG
I
KJ = ×
∑
∑
Δ i i i i
in
out
0.44 kJ
106
%
. .
.
.
reduction in heat requirement =
× − ×
×
× =
2 7 10 0 44 10
2 7 10
100 838%
6 6
6
The hot combustion gases raise the temperature of the limestone, so that less heat from the outside
is needed to do so. Additional thermal energy is provided by the combustion of CO.
9.23 (cont'd)

9-26
9.24 a. A + B C (1)
2C D + B (2)
→
→
Basis: 1 mol
x n
x n
x n
n
n
T
AO A
BO B
IO C
D
I
(mol A / mol) (mol A)
(mol B / mol) (mol B)
(mol I / mol) (mol C)
(mol D)
(mol I)
( C)
D
Fractional conversion: f
x n
x
n x f
A
AO A
AO
A AO A
= =
−
⇒ = −
mol A consumed
mol A feed
( )
1
C generated:
(mol A fed) (mol A consumed) (mol C generated)
mol A fed mol A consumed
n
x f Y
n x f Y
A A C
C AO A C
0
0
=
⇒ =
D generated: =0.5 mol C consumed =(1 2) (mol A consumed mol C out)
2
n
n x f n
D
D AO A C
× × −
⇒ = −
( )( )
1
Balance on B: mol B out = mol B in mol B consumed in (1) + mol B generated in (2)
= mol B in mol A consumed in (1) + mol D generated in (2)
−
−
⇒ = − +
n x x f n
B BO AO A D
Balance on I: mol I out = mol I in ⇒ =
n x
I IO
Species Formula DHf a b c d
A C2H4(v) 52.28 0.04075 1.15E-04 -6.89E-08 1.77E-11
B H2O(v) -241.83 0.03346 6.88E-06 7.60E-09 -3.59E-12
C C2H5OH(v) -235.31 0.06134 1.57E-04 -8.75E-08 1.98E-11
D C4H10)O(v -246.75 0.08945 4.03E-04 -2.24E-07 0
I N2(g) 0 0.02900 2.20E-05 5.72E-09 -2.87E-12
Tf Tp xA0 xB0 xI0 fA YC
310 310 0.537 0.367 0.096 0.05 0.90
n(in) H(in) n(out) H(out)
Species (mol) (kJ/mol) (mol) (kJ/mol)
A 0.537 68.7 0.510 68.7
B 0.367 -231.9 0.341 -231.9
C 0 -211.2 0.024 -211.2
D 0 -204.2 0.001 -204.2
I 0.096 9.4 0.096 9.4
Q(kJ) = -1.31
c. For = 125 C, = 7.90 kJ
o
T Q
f . Raising Tp, lowering fA, and raising YC all increase Q.
b.

9-27
9.25 a. CH g) O g) HCHO(g) + H O(g)
4 2 2
( (
+ →
10 L, 200 kPa n3 (mol HCHO)
n0 (mol feed gas) at 25°C n4 (mol H2O)
0.851 mol CH4/mol n5 (mol CH4)
0.15 mol O2 /mol T (°C), P(kPa), 10L
Q (kJ)
Basis
kPa 1000 Pa 10 L 10 m mol K
1 kPa 1 L 8.314 m Pa K
mol feed gas mixture
3
3
:
.
n0
3
200 1
298
08072
=
=
−
08072
. ,
mol feed gas mixture (0.85)(0.8072) = 0.6861 mol CH
(0.15)(0.8072) = 0.1211 mol O
4
2
⇒
⇒
CH consumed :
1 mol CH 0.1211 mol O fed
mol O fed
mol CH
mol CH mol CH
4
4 2
2
4
4 4
1
01211
0 6861 01211 05650
5
=
⇒ = − =
.
( . . ) .
n
HCHO produced :
mol HCHO mol CH consumed
1 mol CH consumed
mol HCHO
4
4
n3
1 01211
01211
= =
.
.
H O produced :
mol H O mol CH consumed
1 mol CH consumed
mol H O
Extent of reaction : mol
2
2 4
4
2
O out O in
O
2 2
2
n
n n
4
1 01211
01211
0 01211
1
01211
= =
=
−
=
−
=
.
.
( ) ( ) .
.
ξ
ν
References: CH g) O g), HCHO(g), H O(g), at 25 C
4 2 2
o
( , (
Substance
CH
O
HCHO
H O
in in out out
4
2
n U n U
U
U
U

. .
.
.
.
0 6861 0 05650
01211 0
01211
01211
1
2
2
3
− −
− −
− −
( ) ( ) , ,
( . . . . . )
( . . . . )
( . . .
U C dT C R dT i
C
U T T T T
U T T T
U T T
i v i p i
T
T
p i
= = − =
× ⋅
= + × + × − × −
= + × − × −
= + × + ×
z
z
−
− − −
− −
− −
Using ( ) from Table B.2 and R = 8.314 10 kJ / mol K:
kJ / mol
kJ / mol
25
1 2 3
0 02599 2 7345 10 01220 10 2 75 10 0 6670
0 02597 21340 10 21735 10 0 6623
0 02515 0 3440 10 0 2535 10
25
3
1
5 2 8 3 12 4
2
5 2 12 4
3
5 2 8
kJ / mol
T T
3 12 4
08983 10 0 6309
− × −
−
. . )

9-28
b.
c.
Q = =
100 1
85
J 85 s kJ
s 1000 J
kJ
.
Δ Δ Δ Δ
( ) ( ) ( ) ( . ) ( . ) ( . )
.
H H H H
r
o
f
o
HCHO f
o
H O f
o
CH
Table B.1
2 4
kJ / mol
kJ / mol
= + − = − + − − −
= −
B
11590 24183 7485
28288
b g
Δ Δ

.
. )
.
( )
U H RT
r
o
r
o
i i
gaseous
reactants
gaseous
products
3
kJ / mol
J 298 K (1 + 1 kJ
mol K 10 J
kJ / mol
= − −
= − −
− −
= −
∑
∑ ν ν
282 88
8 314 1 1 1
282 88
Energy Balance :
r
o
out out in in
Q U n U n U
i i i i
= + −
∑ ∑
ξΔ ( ( ) ( ( )
) )
(0.1211) kJ / mol) +0.5650
Substitute for through and
= − + +
( . . .

28288 01211 01211
1 2 3
1 3
U U U
U U Q
0 002088 1845 10 009963 10 1926 10 4329
1091 1364
08072 1364
10 10
915 10 915
5 2 8 3 12 4
3
3
= + × + × − × −
⇒ = =
⇒ = =
⋅
⋅
= × =
− − −
−
. . . . .
/
.
kJ / mol
Solve for using E - Z Solve C K
mol 8.314 m Pa K 1 L
mol K L m
Pa kPa
o
3
3
T T T T
T T
P nRT V
Add heat to raise the reactants to a temperature at which the reaction rate is significant.
Side reaction : CH O CO H O. would have been higher (more negative heat of
reaction for combustion of methane), volume and total moles would be the same, therefore
would be greater.
4 2 2 2
+ → +
=
2 2 T
P nRT V
/
9.25 (cont’d)

9-29
9.26 a.
b.
Basis: 2 mol C H fed to reactor
2 4
C H g
1
2
O g) C H O g
C H 3O 2CO 2H O
2 4 2 2 4
2 4 2 2 2
b g b g
+ →
+ → +
(
n1 (mol C H )
2
reactor
4
n2 (mol O )
2
25°C
2 mol C H
2 4
1 mol O2
450°C
heat
n3 (mol C H )
2 4
n4 (mol O )
2
450°C
n5 (mol C H O)
2 4
n6 (mol CO )
2
n7 (mol H O)
2
Qr (kJ)
separation
process
n3 (mol C H )
2 4
n4 (mol O )
2
n5 (mol C H O( ))
2 4 g
25°C
25°C
n6 (mol CO )
2
n7 (mol H O( ))
2 l
25% conversion ⇒ ⇒ =
0500 150
3 4
. .
mol C H consumed mol C H
2 4 2
n
70% yield ⇒ = =
n5 0 350
0.500 mol C H consumed 0.700 mol C H O
1 mol C H
mol C H O
2 4 2 4
2 4
2 4
.
C balance on reactor: 2 2 2 150 2 0 350 0 300
6 6
= + + ⇒ =
. . .
n n mol CO2
Water formed: n7 0 300
= =
1 mol CO
mol H O
2 2
2
2
.
O balance on reactor: 2 1 2 0 350 2 0 300 0 300 0 375
4 4
b gb g b gb g
= + + + ⇒ =
n n
. . . . mol O2
Overall C balance: 2 2 0 300 2 0 350 0500
1 6 5 1
n n n n
= + = + ⇒ =
. . .
b gb g mol C H
2 4
Overall O balance: 2 2 2 0 300 0 300 0 350 0 625
2 6 7 5 2
n n n n n
= + + = + + ⇒ =
b gb g b g b g
. . . . mol O2
Feed stream: 44.4% C H , O Reactor inlet: 66.7% C H O
Recycle stream C H , 20.0% O
Reactor outlet: C H , 13.3% O 12.4% C H O, 10.6% CO 10.6% H O
2 4 2 2 4 2
2 4 2
2 4 2 2 4 2 2
556% 333%
80 0%
53.1%
. , .
: .
, ,
Mass of ethylene oxide = =
0.350 mol C H O 44.05 g 1 kg
1 mol 10 g
kg
2 4
3 0 0154
.
References for enthalpy calculations :C s , H g , O g at 25 C
2 2
b g b g b g °

H T H C dT
i fi p
T
b g= + z
Δ o
2 4
for C H
25
= +
+
z
Δ
H C dT
f p
T
0
298
273
for C H O
2 4
= +
=
Δ
Δ
(

H H
H
fi i
f
o
o
2
table B.8)
for H O l
b g
for O , CO , H O g
2 2 2 b g

9-30
c.
Overall Process
Substance
mol kJ / mol) mol kJ / mol)
C H 0.500
O
C H O
CO
H O
in in out out
2 4
2
2 4
2
2
n H n H
l
( )

( ( )

(
.
.
. .
. .
. .
5228
0625 0
0350 5100
0300 3935
0300 28584
− −
− −
− − −
− − −
− − −
bg
Reactor
substance
mol) kJ / mol) mol) kJ / mol)
C H 2
O
C H O
CO
H O
in in out out
2 4
2
2 4
2
2
n H n H
g
(

( (

(
. . .
. . .
. .
. .
. .
7926 150 7926
1 1337 0375 1337
0350 1999
0300 37466
0300 22672
− − −
− − −
− − −
b g
Energy balance on process: Q H n H n H
i i i i
= = − = −
∑ ∑
Δ
out in
kJ
248
Energy balance on reactor: Q H n H n H
i i i i
= = − = −
∑ ∑
Δ
out in
kJ
236
Scale to kg C H O day
C H O production for initial basis mol)(
44.05 kg
10 mol
kg C H O
Scale factor
kg day
.01542 kg
day
2 4
2 4 3 2 4
1500
0 350 0 01542
1500
0
9 73 104 1
:
( . ) .
.
= =
⇒ = = × −
In initial basis, fresh feed contains
0.500 mol C H
0.625 mol O
g C H mol g O mol
= kg
2 4
2
2 4 2
U
V
|
W
|
= +
× −
M 0500 28 05 0 625 32 0
34 025 10 3
. . . .
.
b gb g b gb g
Fresh feed rate = × × =
− −
34 025 10 9 73 10 3310
3 4 1
. . )
kg day kg day (44.4% C H , 55.6% O
2 4 2
e je j
Qprocess
4
kJ .73 10 day 1 day 1 hr 1 kW
24 hr 3600 s 1 kJ s
kW
=
− ×
= −
−
248 9
279
1
b ge j
Qreactor
4
kJ 73 10 day 1 day 1 hr 1 kW
24 hr 3600 s 1 kJ s
kW
=
− ×
= −
−
236 9
265
1
b ge j
.
9.26 (cont’d)

9- 31
9.27 a.
b.
Basis:
1200 lb C H 1 lb - mole
h 120 lb
lb - moles cumene produced h
m 9 12
m
= 10 0
.
Overall process :
0.75 C H
n1 (lb-moles/h)
3 6
0.25 C H
4 10
n2 (lb-moles C H /h)
6 6
n3
10.0 lb-moles C H /h
9 12
(lb-moles C H /h)
3 6
n4 (lb-moles C H /h)
4 10
C H l C H l C H l , F Btu lb - mole
3 6 6 6 9 12 r
b g b g b g b g
+ → ° = −
Δ
H 77 39520
input consumption
9 12 6 6
9 12
6 6 m 6 6
m 6 6
Benzene balance:
10.0 lb - moles C H produced 1 mole C H consumed
h 1 mole C H produced
10.0 lb - moles C H lb C H
h 1 lb - mole
lb C H h
=
=
= =
b g

.
n2
781
781
input output consumption
9 12 3 6
9 12
Propylene balance: 0.75
10.0 lb - moles C H 1 mole C H
h 1 mole C H
= +
= +
b g

n n
1 3
C H unreacted
lb - moles h
lb - moles C H h
3 6 3 6
20%
0 75 10
0 20 0 75
16 67
2 50
1 3
3 1
1
3
⇒ = +
⇒ =
U
V
W
⇒
=
=
.
. .
.
.
n n
n n
n
n
b g
Mass flow rate of C H / C H feed
0.75 lb - moles C H 42.08 lb C H
h 1 lb - mole
0.25 lb - moles C H 58.12 lb C H
h 1 lb - mole
lb h
3 6 4 10
3 6 m 3 6
4 10 m 4 10
m
=
+ =
b gb g
b gb g
16 67
16 67
768
.
.
Reactor :
Benzene feed rate
10.0 lb - moles fresh feed moles fed to reactor
h 1 mole fresh feed
lb - moles C H h
6 6
=
+
=
3 1
40
a f
16.67 lb-moles/h @ 77o
F
0.75 C3H6 10.0 lb-moles C9H12/h 46.7 lb-moles/h
0.25 C4H10 2.50 lb-moles C3H6/h 21.4% C9H12
4.17 lb-moles C4H10/h 5.4% C3H6
40.0 lb-moles C6H6/h 30.0 lb-moles C6H6/h 8.9% C4H10
400o
F 64.3% C6H6
Overhead from T1 ⇒
U
V
W
⇒
2 50
4
6 67
.
.
lb - moles C H h
.17 lb - moles C H h
lb - moles h
37.5% C H
62.5% C H
3 6
4 10
3 6
4 10
Heat exchanger :
10.0 lb-moles C9H12 /h
Reactor effluent at 400°F 200°F
40.0 lb-moles C H /h
6 6
77°F
(°F)
T

9- 32
Energy balance: ΔH n H H n C T T
i i i i pi i
= ⇒ − = − =
∑ ∑
0 0

, out , in out in
e j b g
(Assume adiabatic)
10 lb - moles C H 120 lb 0.40 Btu
h 1 lb - mole 1b F
F F F
F F F F
F F
9 12 m
m
C3H6
C4H10
C6H6 in
effluent
C6H6 fed
to reactor
⋅
L
NM O
QP − + −
+ − + −
+ − = ⇒ = °
B
B
A
A
D
D D D
D D D D
D
200 400 2 50 42 08 057 200 400
417 5812 055 200 400 30 0 7811 0 45 200 400
40 0 7811 0 45 77 0 323
e j b gb gb ge
b gb gb ge j b gb gb ge j
b gb gb ge j
. . .
. . . . . .
. . . T T
(Refer to flow chart of Part b: T = °
323 F )
References : C H l , C H l , C H l , C H l at 77 F
3 6 4 10 6 6 9 12
b g b g b g b g °

H C M T
i pi i
Btu lb - mole Btu lb F lb lb - mole F
m m
b g b g b gb gb g
= ⋅° − °
77
Substance
(lb - mole / h) (Btu / lb - mole) (lb - mole / h) (Btu / lb - mole)
C H 12.0
C H
C H
C H
in in out out
3 6
4
6 6
9 12

.
. .
. .
.
n H n H
0 2 50 7750
417 0 417 10330
40 0 8650 30 0 11350
10 0 15530
10
− −
Energy balance on reactor :
Q H
n H
v
n H n H
i i i i
= = + −
∑ ∑
Δ
Δ

C H r
o
C H out in
9 12
9 12
=
−
+ + + +
− = −
10 0 39520
1
2 50 7750 417 10330 30 0 11350 10 0 15530
40 0 8650 183000
.
. . . .
.
b gb g
b g b gb g b gb g b gb g b gb g
b gb g b g
Btu h heat removal
9.28
a.
Basis :
100 kg C H 10 g 1 mol
h 1 kg 104.15 g
mol h
8 8
3
= 960 styrene produced
C H g) C H (g) H g)
8 10 8 8 2
( (
→ +
Overall system
960 mol C8H8 /h
n1 (mol C8H10/h)
Fresh feed
n2 (mol H2 /h)
C H balance
8 8 8 10
8 8
8 10
8 10
Fresh feed rate:
960 mol C H 1 mol C H
h 1 mol C H
mol C H h fresh feed
b g

n1 960
= =
H balance :
960 mol C H 1 mol H
h 1 mol C H
mol H h
2
8 10 2
8 10
2

n2 960
= =
9.27 (cont'd)

9- 33
b.
Reactor :
n3 (mol C8H10 /h)
n4 (mol H2O( v )/h)
600°C
.
n5 (mol C8H10 /h)
n4 (mol H2O(v)/s)
v
560°C
960 (mol C8H8 /s)
960 (mol H2 /s)
Qc (kJ/h)
35% 1-pass conversion ⇒ =
0 35
960
3
. n mol C H react 1 mol C H
h 1 mol C H
mol C H h
8 10 8 8
8 10
8 8
b g
⇒ =

n3 2740 mol C H h fed to reactor
8 10
⇒ Recycle rate mol C H h recycled
8 10
= − =
2740 960 1780
a f
Reactor feed mixing point
2740 mol C8H10(v)/h
500o
C
2740 mol C8H10(v)/h
n4 [mol H2O(v)/h]
n4 [mol H2O(v)/h] 600o
C
700o
C
Neglect ,
C H H O
Energy balance: kJ h
8 10 2
Q Ek
H H n H
Δ
Δ Δ Δ
b g
b g
= + =
2740 0
4

Δ . .
H T dT
Cp
C H 3
8 10
J
mol C
1 kJ
10 J
kJ mol
= +
L
N
MMM
O
Q
PPP ⋅
× =
z 118 0 30 28 3
500
600
b g
D
Δ .
H
P
H O
bar
Table B.8
2
kJ mol
⇒ = −
=1
39
2740 28 3 3 9 0 199 10
4 4
4
a fa f a f
. . .
+ − = ⇒ = ×
n n mol H O / h
2
Ethylbenzene preheater A
b g :
960 1780 mol r 2740 mol E
mol fresh feed
h
ecycled
h
B l
h
at 25 C
2740 mol EB v
h
at 500 C
+ = °
⇒ °
b g
b g
Δ Δ
. . . .
H C dT H C dT
pi pv
= + ° + = + + =
z z
25
136
136
500
136 20 2 36 0 77 7 133 9
v C kJ mol kJ mol
a f a f
.
Q H
A = = = ×
Δ
2740 mol C H 133.9 kJ
h mol C H
kJ h preheater
8 10
8 10
3 67 105
b g
Steam generator F
b g :
19400 19400
mol h H O l, 25 C mol h H O v, 700 C, 1 atm
2 2
° → °
b g b g
Table B.5 ⇒ ° =
.
H l, 25 C kJ kg
b g 104 8 ;
Table B.7 ⇒ ° ≈ =

H v, 700 C, 1 atm 1 bar kJ kg
b g 3928
9.28 (cont'd)

9- 34
c.
.
.
Q H
F = =
−
= ×
Δ
19400 mol H O 18.0 g 1 kg kJ
h 1 mol 10 g kg
kJ h steam generator
2
3
3928 104 8
134 106
a f
b g
Reactor C
b g :
References: C H v , C H v , H g , H O v at 600 C
8 8 8 10 2 2
b g b g b g b g °

H C dT
i pv i
560
600
560
D
C
e j d i
= z for C H , C H
8 10 8 8
≈ ( ,
H T) for H H O (interpolating from Table B.8)
2 2
Substance
(mol h (kJ mol (mol h (kJ mol
C H 2740
H O
C H
H
in in out out
8 10
2
8 8
2

)

)

)

)
.
.
.
.
n H n H
0 1780 1168
19900 0 19900 156
960 10 86
960 119
−
−
− − −
− − −
Energy balance :

.
Q H n H n H
c i i i i
= = + −
= ×
∑ ∑
Δ
960 mol C H produced 124.5 kJ
h 1 mol C H
kJ h reactor
8 8
8 8 out in
5 61 104
a f
This is a poorly designed process as shown. The reactor effluents are cooled to 25D
C , and
then all but the hydrogen are reheated after separation. Probably less cooling is needed, and
in any case provisions for heat exchange should be included in the design.
9.29
a.
b.
CH OH HCHO H , H
1
2
O H O
3 2 2 2 2
→ + + →
(mol CH OH/h)
0.42 mol CH OH/mol
nf (mol/h) at 145°C, 1 atm
3
reactor
ns mol H O( )/h
2
0.58 mol air/mol
0.21 mol O /mol air
2
0.79 mol N /mol air
2
v
saturated at 145°C
reactor
product gas, 600°C
n1 3
n2 (mol O /h)
2
n3 (mol N /h)
2
n4 (mol HCHO/h)
n5 (mol H /h)
2
n6 (mol H O/h)
2
waste
heat
boiler
mb (kg H O( )/h)
2 v
30°C
mb (kg H O( )/h)
2 v
sat'd at 3.1 bars
product gas
145°C separation
units
CH OH
3
O , N
2 2
H2
0.37 kg HCHO/h
0.63 kg H O/h
2
In the absence of data to the contrary, we assume that the separation of methanol from
formaldehyde is complete.
Methanol vaporizer:
The product stream, which contains 42 mole % CH OH v
3 b g, is saturated at Tm
D
C
e j and 1 atm.
9.28 (cont'd)
mb(kg H2O(l)/h)
30o
C

9- 35
c.
y P p T p T
m m m m m
= ⇒ =
∗ ∗
b g b gb g b g
0 42 760 319 2
. .
mmHg mmHg =
Antoine equation
mmHg 44.1 C
⎯ →
⎯⎯⎯⎯
⎯ = ⇒ =
∗
p T
m m
319 2
. D
Moles HCHO formed :
=
×
=
36 10
30 03
52 80
6
kg solution 0.37 kg HCHO 1 kmol 1 day
350 days 1 kg solution kg HCHO 24 h
kmol HCHO
h
.
.
but if all the HCHO is recovered, then this equals
n4 , or .
n4 52 80
= kmol HCHO h
70% conversion :
52.80 kmol HCHO 1 kmol CH OH react 1 kmol CH OH fed kmol feed gas
h 1 kmol HCHO formed 0.70 kmol CH OH react 0.42 kmol CH OH
3 3
3 3
1
=
nf
⇒ =
.
nf 179 59 kmol h
Methanol unreacted:

. . .
.
n1
0 42 179 59 1 0 70
22 63
=
−
=
b gb g b g
kmol CH OH fed kmol CH OH fed
h 1 kmol CH OH fed
kmol CH OH
h
3 3
3
3
N balance: kmol h kmol N h
2 2
. . . .
n3 179 6 058 0 79 82 29
= =
b gb gb g
Four reactor stream variables remain unknown — , ,
n n n
s 2 5 , and
n6 — and four relations are
available — H and O balances, the given H2 content of the product gas (5%), and the energy
balance. The solution is tedious but straightforward.
H balance: 179 6 0 42 4 2 22 63 4 52 8 2 2 2
5 6
. . . .
b gb gb g b gb g b gb g
+ = + + +
n n n
s
⇒ .
n n n
s = + −
5 6 52 80 (1)
O balance: 179 6 0 42 1 179 6 058 0 21 2 22 63 1 2 80 1
2 6
. . . ( . ) . ( . )( ) (52. )( )
b gb gb g b g b gb g
+ + = + + +
n n n
s
⇒ = + −
.
n n n
s 2 4375
2 6 (2)
H content:
2

. . .
. .
n
n n n
n n n
5
2 5 6
5 2 6
22 63 82 29 52 89
0 05 19 157 72
+ + + + +
= ⇒ − − = (3)
References : C s , H g , O g , N g at 25 C
2 2 2
b g b g b g b g °
H H C dT
p
T
= +
B
z
Δ
f
o
Table B.2
25
or Table B.8 for O , N and H
2 2 2
9.29 (cont'd)

9- 36
d.
substance
kmol / h kJ / kmol kmol / h kJ / kmol
CH OH 75.43
O
N
H O
HCHO
H
in in out out
3
2
2
2

.
.
. .
.
n H n H
n
n n
n
s
− −
− −
− − −
− −
195220 22 63 163200
2188 3620 18410
82 29 3510 82 29 17390
237740 220920
52 80 88800
16810
2 2
6
5
Energy Balance :
ΔH n H n H
i i i i
= − =
∑ ∑

out in
0 ⇒ + − + = − ×
18410 16810 220920 237704 7406 10
2 5 6
6
n n n ns . (4)
We now have four equations in four unknowns. Solve using E-Z Solve.

ns = =
58.8 kmol H O v 18.02 kg
h 1 kmol
kg steam fed h
2 b g 1060
.
n2 2 26
= kmol O h
2 , .
n5 1358
= kmol H h
2 , .
n6 98 00
= kmol H O h
2
Summarizing, the product gas component flow rates are 22.63 kmol CH3OH/h, 2.26 kmol O2/h,
82.29 kmol N2/h, 52.80 kmol HCHO/h, 13.58 kmol H2/h, and 98.02 kmol H2O/h
⇒
272 kmol h product gas
8% CH OH, 0.8% O 30% N 19% HCHO, 5% H 37% H O
3 2 2 2 2
, , ,
Energy balance on waste heat boiler. Since we have already calculated specific enthalpies of all
components of the product gas at the boiler inlet (at 600°C), and for all but two of them at the
boiler outlet (at 145°C), we will use the same reference states for the boiler calculation
Reference States: C s , H g , O g , N g at 25 C for reactor gas
2 2 2
b g b g b g b g °
H O l
2 b g at triple point for boiler water
out
in out
in
3
2
2
2
2
ˆ
ˆ
Substance
kmol/h mol
kJ/kmol kJ/mol
CH OH 22.63 163200 22.63 195220
O 2.26 18410 2.26 3620
N 82.29 17390 82.29 3510
H O 98.02 220920 98.02 237730
HCHO 52.80 88800 52.80 111350
H 13.58 16810 13.58
n
n H
H
− −
− −
− −

2
3550
H O 125.7 2726.1
(kg/h) (kJ/kg) (kg/h) (kJ/kg)
b b
m m
9.29 (cont'd)

9- 37
Energy Balance :
kg steam h
out in
ΔH n H n H
m
m
i i i i
b
b
= − =
⇒ − − × =
⇒ =
∑ ∑

. . .
0
27261 1257 4 92 10 0
1892
6
b g
9.30 a.
b.
C H HCl C H Cl
2 4 2 5
+ →
Basis:
1600 kg C H Cl l 10 g 1 mol
h 1 kg 64.52 g
mol h C H Cl
2 5
3
2 5
b g = 24800
(mol HCl( )/h)
reactor
n1 g
0°C
A
(mol/h) at 0°C
n2
0.93 C H
B
2 4
0.07 C H
2 6
(mol HCl( )/h)
n3 g
(mol C H ( )/h)
n4 g
2 4
(mol C H ( )/h)
n5 g
2 6
(mol C H Cl( )/h)
n6 g
2 5
50°C
condenser
n6
( – 24,800) (mol C H Cl( )/h)
l
2 5
0°C
(mol C H Cl( )/h)
n6 g
2 5
(mol HCl( )/h)
n3 g
(mol C H ( )/h)
n4 g
2 4
(mol C H ( )/h)
n5 g
2 6
C
0°C
24,800 mol C H Cl( )/h
l
2 5
D
Product composition data:
n n
3 1
0 015 1
= . b g
n n n
4 2 2
0 015 0 93 0 01395 2
= =
. . .
b g b g
n n
5 2
0 07 3
= . b g
Overall Cl balance :
n
n
1
3 1 24800 1 4
mol HCl h 1 mol Cl
1 mol HCl
b g b gb g b gb g b g
= +
Solve (4) simultaneously with (1) mol h kmol HCl fed / h
⇒ = =
n1 25180 2518
.
n3 378
= mol HCl g h
b g
Overall C balance :
n n n n
2 2 4 5
0 93 2 0 07 2 2 2 2 24800
. .
+ = + +
From Eqs. (2) and (3) ⇒ + − −
L
NM O
QP=
2 0 93 0 07 0 0139 0 07 2 24800
2
n . . . . b gb g
n2 27070 27 07
= =
mol fed h kmol h of Feed B
.
n
n
n
3
4
5
378
0 01395 27070
0 07 27070
=
=
=
U
V
|
W
|
mol HCl h
= 378 mol C H h
= 1895 mol C H h
2 4
2 6
.
.
b g
b g
2 65
14 3%
.
. ,
kmol / h of Product C
HCl, 14.3% C H 71.4% C H
2 4 2 6
9.29 (cont'd)
n6 (mol C2H5Cl(l)/h

9- 38
c.
d.
References : C H g , C H g , C H Cl g , HCl g at 0 C
2 4 2 6 2 5
b g b g b g b g D
C H g, 50 C kJ mol
2 4
Table B.2
D
e j: .
H C dT
p
= ⇒
z0
50
2181
( )
Table B.2
50
2 6 0
ˆ
C H g, 50 C : 2.512 kJ mol
p
H C dT
= ⇒
∫
D
HCl g, 50 C .456 kJ mol
Table B.2
D
e j:
H C dT
p
= ⇒
z0
50
1
C H Cl l, 0 C C kJ mol
2 5 v
D D
e j e j
: .
H H
= − = −
Δ 0 24 7
C H Cl g, 50 C kJ mol
2 5
D
e j: .
H C dT
pv
= =
z0
50
2 709
substance
mol kJ / mol mol kJ / mol
HCl 25180
C H
C H
C H Cl
in in out out
4
2 6
2 5
n H n H
n n

.
.
.
. .
0 378 1456
25175 0 378 2181
1895 0 1895 2 512
24800 24 7 2 709
2
6 6
− −
Energy balance:
Δ
Δ
H
n H
n H n H
i i i i
= ⇒ + − =
∑ ∑
0
0
0
A r
A out in
C

D
e j
ν
( )
( )( ) ( )( ) ( )( )
( )( )
6 6 6 2 5
25180 378 mol HCl react 64.5 kJ
378 1.456 378 2.181 1895 2.512
h 1 mol HCl
2.709 24800 24.7 0 80490 mol C H Cl h in reactor effluent
n n n
− −
⇒ + + +
+ − − − = ⇒ =
2 5
80490 mol condensed 24800 mol product mol
C H Cl recycled 55690
h h h
kmol recycled
55.7
h
= − =
=
Cp is a linear function of temperature.
Δ
Hv is independent of temperature.
100% condensation of ethylbenzene in the heat exchanger is assumed.
Heat of mixing and influence of pressure on enthalpy is neglected.
Reactor is adiabatic.
No C2H4 or C2H6 is absorbed in the ethyl chloride product.
9.31 a. 4NH3(g) + 5O2(g) Æ 4NO(g) +6H2O(g) Δ .
Hr
o
kJ / mol
= −904 7
Basis : 10 mol/s Feed gas
9.30 (cont'd)

9- 39
b.
c.
4
6
3
4
5
mol / s NH (mol O
mol / s O (mol NO)
= 200 C (mol H O)
3 2
2
o
2
)

n
n
T n
T
in
out
O consumed :
mol O mol NH fed
4 mol NH s
mol / s mol O / s mol O /
NO produced :
mol NO produced 4 mol NH fed
4 mol NH s
= 4 mol NO / s
2
2 3
3
2 2
3
3
5 4
5 6 1 1
4
3
4
= ⇒ = − =
=
( )

n
n
H O produced :
mol H O produced 4 mol NH fed
4 mol NH s
= 6 mol H O / s
Extent of reaction : =
( (
mol / s
2
2 3
3
2
NH out NH in
NH
3 3
3

) )
n
n n
5
6
0 4
4
1
=
−
=
−
=
ξ
ν
Well-insulated reactor, so no heat loss
No absorption of heat by container wall
Neglect kinetic and potential energy changes;
No shaft work
No side reactions.
References : NH g), O g), NO(g), H O(g) at 25 C, 1atm
Substance
mol / s) kJ / mol) mol / s) kJ / mol)
NH g)
O g
NO g)
H O(g)
3 2 2
o
in in out out
3 1
2 2 3
4
2 5
( (

(

(

(

(
( .
( ) . .
( .
.
n H n H
H
H H
H
H
4 00
6 00 100
4 00
6 00
− −
− −
− −
( ) . ( .
H C dT H H
p
1
25
200
2
6 74 200 531
= = = =
B B
z NH
Table B.2
O
o
Table B.8
3 2
kJ / mol, C) kJ / mol
5 2 8 3 12 4
3 out out out out
5 2 8 3 12 4
4 out out out out
Using ( ) from Table B.2 :
ˆ (0.0291 0.5790 10 0.2025 10 0.3278 10 0.7311) kJ/mol
ˆ (0.0295 0.4094 10 0.0975 10 0.0913 10 0.7400) kJ/mol
p i
C
H T T T T
H T T T T
− − −
− − −
= + × − × + × −
= + × − × + × −
5 2 8 3 12 4
5 out out out out
ˆ (0.03346 0.3440 10 0.2535 10 0.8983 10 0.8387) kJ/mol
H T T T T
− − −
= + × + × − × −
Energy Balance:
r
o
out out in in
Δ
Δ Δ

( ( ) ( ( )
) )
H
H H n H n H
i
i
i i
i
i
=
= + −
= =
∑ ∑
0
3
5
1
2
ξ
9.31 (cont'd)

9- 40
d.
e.
o
r 3 4 5 1 2
o
ˆ ˆ ˆ
Substitute for , , and through
r 1 6
5 2 8 3 12 4
out out out out
ˆ ˆ ˆ ˆ ˆ ˆ
(1.00) (4.00) (6.00) (4.00) (6.00)
(0.3479 4.28 10 0.9285 10 4.697 10 )
H H H
H H H H H H H
H T T T T
ξ
ξ
Δ
− − −
⇒ Δ = Δ + + + − −
Δ = + × + × − ×
⇓

o
out
972.24 kJ/mol = 0
E-Z Solve 2223 C
T
−
⇒ =
If only the first term from Table B.2 is used,
kJ / mol, kJ / mol, ,
,
out
out out
( ) ( )
. ( ) . . . ( )
. ( ) . ( )
H C dT C T
H H H T
H T H T
i pi pi
T
= = −
= − = = = −
= − = −
z 25
003515 200 25 615 531 00291 25
00295 25 003346 25
25
1 2 3
4 5
o
r 3 4 5 1 2
o
ˆ ˆ ˆ
Substitute for (=1 mol/s), ( 904.7 kJ/mol) and through
r 1 6
o
out out
ˆ ˆ ˆ ˆ ˆ ˆ
E.B. (1.00) (4.00) (6.00) (4.00) (6.00) 0
2788
0=0.3479 969.86 2788 C % error=
H H H
H H H H H H H
T T
ξ
ξ
Δ = −
Δ = Δ + + + − − =
− ⇒ = ⇒
⇓

o o
o
C 2223 C
100 25%
2223 C
−
× =
If the higher temperature were used as the basis, the reactor design would be safer (but more
expensive).
9.32
a.
Basis : 100 lbm coke fed
⇒ ⇒ ⇒
84 lb C 7.00 lb - moles C fed 7.00 lb - moles CO fed
m 2
400°F
7.00 lb-moles CO2 (lb-moles CO)
n1
77°F
7.00 lb-moles(84 lb )C/hr
m
16 lb ash/hr
m
(lb-moles CO )
n2 2
1830°F
lb-moles C( )/hr
n3 s
1830°F
16 lb ash/hr
m
585,900 Btu
C s CO g CO g
b g b g b g
+ →
2 2 ,
Δ Δ Δ

. .
,
H H H
r
o
C
c
o
CO g
c
o
CO g
F
kJ 0.9486 Btu 453.6 mols
mol 1 kJ 1 lb - mole
Btu lb - mole
2
77 2
39350 2 282 99
74 210
25
D
e j e j e j
b gb g
b g b g
= °
= −
=
− − −
=
Let x = fractional conversion of C and CO2 :
E
= =
n
x
x
1 14 0
7.00 lb - moles C reacted 2 lb - moles CO formed
1 lb - mole C reacted
lb - moles CO
b g .
n x
n x
2
3
7 00 1
7 00 1
= −
= −
.
.
b g
b g b g
lb - moles CO
lb - moles C s
2
References for enthalpy calculations: C s CO g , CO g ash at F
2
b g b g b g
, , 77D
9-31 (cont’d)

9- 41
b.
CO g F
2 400
, :
°
b g (
H H
= ⇒
CO
Table B.9
2
F) Btu lb - mole
400 3130
D
CO g, F
2 1830°
b g: (
H H
= ⇒
CO
Table B.9
2
F) Btu lb - mole
1830 20,880
D
CO g, F
1830°
b g: (
H H
= ⇒
CO
Table B.9
F) Btu lb - mole
1830 13,280
D
Solid 1830°F
b g: .
H =
− °
⋅
=
0 24 1830 77
420
Btu F
lb F
Btu lb
m
m
b g
D
Mass of solids (emerging)
=
−
+ = −
7 00 1
16 100 84
. x
x
b g b g
lb - moles C 12.0 lb
1 lb - mole
lb lb
m
m m
substance
(lb moles) (Btu lb - mole) (lb moles) (Btu lb - mole)
CO 7.00 3130
CO
solid
(lb (Btu lb (lb (Btu lb
in in out out
2
m m m m
n H n H
x
x
x
− −
−
− −
−

. ,
. ,
) ) ) )
7 00 1 20 890
14 0 13 280
100 0 100 84 420
b g
Extent of reaction: n n lb- moles) = 7.0
CO CO o CO
= + ⇒ = ⇒
( ) . (
ν ξ ξ ξ
14 0 2
x x
Energy balance:
Q H H n H n H
i i i i
= = + −
∑ ∑
Δ Δ
ξ r
o
out in

585 7 00 1 20 880
14 0 13 100 84 420 7 00 3130
0 801 801%
,900 . ,
. ,280 .
. .
Btu
7.0 (lb - moles) 74,210 Btu
lb - mole
conversion
= + −
+ + − −
E
= ⇒
x
x
x x
x
a fa f
a fa f a fa f a fa f
Advantages of CO. Gases are easier to store and transport than solids, and the product of the
combustion is CO2, which is a much lower environmental hazard than are the products of
coke combustion.
Disadvantages of CO. It is highly toxic and dangerous if it leaks or is not completely burned,
and it has a lower heating value than coke. Also, it costs something to produce it from coke.
9.32 (cont'd)

9- 42
9.33
Basis :
17.1 m L 273 K 5.00 atm 1 mol
h 1 m 298 K 1.00 atm 22.4 L STP
mol h feed
3
3
10
3497
3
a f=
CO g H g CH OH g
2 3
b g b g b g
+ →
2 ,
Δ Δ Δ
.
H H H
r
o
f
o
CH OH g
f
o
CO(g)
3
kJ mol
= − = −
e j e j
b g 90 68
127°C, 5 atm
25°C, 5 atm
2
(mol CH OH /h)
n
(mol CO/h)
n2
(mol H /h)
n3
3497 mol/h
0.333 mol CO/mol
0.667 mol H /mol
1 3
2
= –17.05 kW
Q
Let f = fractional conversion of CO (which also equals the fractional conversion of H2, since
CO and H2 are fed in stoichiometric proportion).
CO reacted : =
( )( ) ( )
= ( )
3497 mol CO feed mol react
mol feed
mol CO react
0 333
1166
. f
f
CH OH produced :
3
n
f
f
1 1166
= =
1166 mol CO react 1 mol CH OH
1 mol CO
mol CH OH h
3
3
CO remaining :
n f
2 1166 1
= −
a f mol CO h
H remaining : mol H fed
1166 mol CO react 2 mol H react
1 mol CO react
mol H h
2 2
2
2
.
n
f
f
3 3497 0 667
2332 1
= −
= −
b gb g
b g
Reference states : CO(g), H g
2 b g, CH OH g
3 b g at 25°C
Substance
mol h kJ mol mol h kJ mol
CO 1166 0
H
CH OH
in in out out
1
2 2
3 3

n H n H
f H
f H
f H
b g b g b g b g
a f
a f
1166 1
2332 0 2332 1
1166
−
−
− −
CO g,127 C : C kJ mol
H g, C : C 2.943 kJ mol
CH OH(g,127 C): kJ / mol
CO
2 H
3
Table B.8
2
Table B.8
Table B.2
D D
D D
D
e j
e j
( ) .
( )
.
H H
H H
H C dT
p
1
2
3
25
122
127 2 99
127 127
5009
= =
= =
= =
B
B
B
z
Energy balance :
Q H H n H n H
i i i i
= = + −
∑ ∑
Δ Δ
ξ r
o
out in
⇒
−
= − + −
+ − +
⇒ × = × ⇒ =
17 05
1166 90 68 1166 1 2 99
2332 1 2 993 1166 5009
1102 10 7173 10 0 651
5 4
.
( )( . ) .
. .
. . .
kJ 3600 s
s 1 h
kJ
h
kJ h
mol CO or H converted mol fed
2
f f
f f
f f
b gb g
b gb g b gb g
b g

9- 43
. .
. .
. .
.
n
n
n
n V
1
2
3
1166 0 651 7591
1166 1 0 651 406 9
2332 1 0 651 8139
1980 130
= =
= − =
= − =
E
= ⇒ = =
b g
b g
b g
b g
mol h
mol h
mol h
mol
h
1980 mol 22.4 L STP 400 K 1.00 atm 1 m
h 1 mol 273 K 5.00 atm 10 L
m h
tot out
3
3
3
9.34 a. CH g 4S g CS g H S g
2
4 2 2
b g b g b g b g
+ → + , Δ
Hr 700 274
°
( ) = −
C kJ mol
Basis : 1 mol of feed
1 mol at 700°C
4 (mol CS2)
n1
0.20 mol CH /mol
0.80 mol S/mol
Reactor
Product gas at 800°C
(mol H S)
n2 2
(mol CH )
n3 4
n4 (mol S (v))
= –41 kJ
Q
Let f = fractional conversion of CH4 (which also equals fractional conversion of S, since the
species are fed in stoichiometric proportion)
Moles CH reacted Extent of reaction = (mol) = 0.20
mol CH
mol S fed
0.20 mol CH react mol S react
1 mol CH react
mol S
0.20 mol CH react mol CS
1 mol CH
mol CS
0.20 mol CH react mol H S
1 mol CH
mol H S
4
4
4
4
4 2
4
2
4 2
4
2
=
= −
= − = −
= =
= =
0 20
0 20 1
080
4
080 1
1
0 20
2
0 40
3
4
1
2
. ,
.
. .
.
.
f f
n f
n
f
f
n
f
f
n
f
f
ξ
b g
b g b g
References: CH (g), S g , CS (g), H S(g)
4 2 2
b g at 700°C (temperature at which Δ
Hr is known)
substance
CH 0.20 0
S
CS
H S
in in out out
4 1
2
2 3
2 4
n H n H
f H
f H
f H
f H
b g b g b g b g
b g
b g

.
. .
.
.
0 20 1
080 0 080 1
0 20
0 40
−
−
− −
− −

H Cpi
out = −
( )
800 700 ⇒
CH g, C : 7.14 kJ / mol
S g, C : 3.64 kJ / mol
CS g, C : 3.18 kJ / mol
H S g, C : 4.48 kJ / mol
4
2
2
800
800
800
800
1
2
3
4
° =
° =
° =
° =
b g
b g
b g
b g

H
H
H
H
9.33 (cont’d)

9- 44
b.
Energy balance on reactor:

. .
. . . . . . . .
.
Q H H n H n H
f
f f f f
f
r i i i i
= = + − =
=
−
+ − + − + +
⇒ =
∑ ∑
Δ Δ
ξ
out in
kJ
s
41
0 20 274 0
1
0 20 1 7140 080 1 3640 0 20 3180 0 40 4 480
0800
b gb g
b g b gb g b gb g b g b g
preheater
0.32 mol H2S
150°C
0.20 mol CH4
0.80 mol S(l )
(kJ)
Q
800°C
0.04 mol CH4
0.16 mol S(g )
0.16 mol CS2
0.32 mol H S
200°C
0.04 mol CH4
0.16 mol S(l )
0.16 mol CS2
2
0.20 mol CH4
0.80 mol S( g)
T (°C) 700°C
0.20 mol CH4
0.80 mol S(l )
System: Heat exchanger-preheater combination. Assume the heat exchanger is adiabatic, so
that the only heat transferred to the system from its surroundings is Q for the preheater.
References : CH (g), S l , CS (g), H S(g)
4 2 2
b g at 200°C
Substance
CH 0.20
CH
S l
S g
CS
H S
in in out out
4 1 7
4 2
3
4 8
2 5
2 6
n H n H
H H
H
H
H H
H
H
b g b g b g b g
b g
b g
b g
b g

.
. .
. .
. .
. .
. .
,
,
150 700
800 200
0 20
0 04 0 04 0
080 016 0
016 080
016 016 0
0 32 0 32 0
° °
° °

. .
.
H C T
C T
C H T C T
i pi
p
p
T
v b p
b
= −
= −
= −
F
HG I
KJ+ + −
=
200
200
444 6 200 444 6
83 7
a f
d i a f a f
d i b g d i a f b g
a f
a f b g
for all substances but S
for S l
for S g
S l
S l
kJ mol
S g
Δ
9.34 (cont’d)

9- 45
c.
CH g, C : 3.57 kJ / mol CS g, C : 19.08 kJ / mol
CH g, C : 42.84 kJ / mol H S g, C : 26.88 kJ / mol
S l, C : 1.47 kJ / mol CH g, C : 35.7 kJ / mol
S g, C : 103.83 kJ / mol S g, C : 100.19 kJ / mol
4 2
4 2
4
150 800
800 800
150 700
800 700
1 5
2 6
3 7
4 8
° = − ° =
° = ° =
° = − ° =
° = ° =
b g b g
b g b g
b g b g
b g b g

H H
H H
H H
H H
Energy balance: Q n H n H
i i i i
kJ
out in
b g= −
∑ ∑
⇒ = ⇒
Q 59 2
. kJ 59.2 kJ mol feed
The energy economy might be improved by insulating the reactor better. The reactor effluent will
emerge at a higher temperature and transfer more heat to the fresh feed in the first preheater,
lowering (and possibly eliminating) the heat requirement in the second preheater.
9.34 (cont’d)

9-46
9.35
a.
b.
Basis : 1 mol C H fed to reactor
2 6
1273 K, P atm
(mol H )
1 mol C H
2
2
(mols) @
n
nH 2
6 T(K), P atm
(mol C H )
2
nC H
2 6 6
(mol C H )
2
nC H
2 4 4
C H C H H
2 6 2 4 2
⇔ + , K
x x
x
P T K
p = = × −
C H H
C H
2 4 2
2 6
7 28 10 17 000
6
. exp[ , / ( )] (1)
Fractional conversion = f mols C H react mol fed
2 6
b g
ξ(mol)
mol C H
mol C H
mol H
mols
mol C H
mol
mol C H
mol
mol H
mol
C H 2 6
C H 2 4
H 2
C H
2 6
C H
2 4
H
2
2 6
2 4
2
2 6
2 4
2
=
= −
=
=
= +
U
V
|
|
|
W
|
|
|
⇒
=
−
+
=
+
=
+
f
n f
n f
n f
n f
x
f
f
x
f
f
x
f
f
1
1
1
1
1
1
b gb g
b g
b g
b g
K
x x
x
K
f
f f
f
f
p p
f
f
f
f
= ⇒ = =
− +
=
−
+
−
+
C H H
C H
2 4 2
2 4
P
P
P
P
2
2
1
1
1
2 2
2
1 1 1
b g
b g
b g
b gb g
1 2
2 2
1 2
− = ⇒ =
+
F
HG
I
KJ
f K f f
K
K
p
p
p
e j b g
P
P
References : C H g C H g H g at 1273 K
2 6 2 4 2
b g b g b g
, ,
Energy balance:
Δ Δ
H H n H n H
r i i i i
= ⇒ + −
∑ ∑
0 ξ 1273 K
out in

b g

Hi
e j b g
in
inlet temperature reference temperature
= =
0

H C dT
i pi
T
e jout
= z1273
⇓ energy balance
f H f C dT f C dT f C dT
r p
T
p
T
p
T
1273 K kJ
C H C H H
2 6 2 4 2
Δ b g b g d i d i d i
+ − + + =
z z z
1 0
1273 1273 1273
rearrange, reverse limits and change signs of integrals
1
1273
3
1273 1273
1273
−
=
− −
z z
z
f
f
H K C dT C dT
C dT
r p
T
p
T
p
T
T
Δ b g d i d i
d i
b g
b g
C H H
C H
2 4 2
2 6
φ

1
1
1
1
4
−
= ⇒ − = ⇒ =
+
f
f
T f f T f
T
φ φ
φ
b g b g b g b g

9-47
c.
d.
φ T
T dT T dT
T dT
T T
T
b g
b g e j
b g
=
− + − + ×
+
z z
z
−
145600 9 419 01147 26 90 4167 10
1135 01392
1273
3
1273
1273
. . . .
. .
⇒ φ T
T T
T T
b g=
+ +
− −
3052 36 2 0 05943
127240 113 0 0696
2
2
. .
. .
K
K T
K
K T
T
p
p
p
p
1
1
1 1
1
1
0
1 2 1 2
+
F
HG
I
KJ =
+
⇒
+
F
HG
I
KJ −
+
= =
φ φ
ψ
b g b g b g
φ T
b g given by expression of Part b. K T
p b g given by Eq. (1)
P T f Kp Phi Psi
(atm) (K) (atm)
0.01 794 0.518 0.0037 0.93152 -0.0001115
0.05 847.4 0.47 0.0141 1.12964 -0.0002618
0.1 872.3 0.446 0.025 1.24028 0.00097743
0.5 932.8 0.388 0.0886 1.57826 3.41E-05
1 960.3 0.36 0.1492 1.77566 4.69E-05
5 1026 0.292 0.4646 2.42913 -2.57E-05
10 1055 0.261 0.7283 2.83692 -7.54E-05
Plot of T vs ln P
700
800
900
1000
1100
-3 -2 -1 0 1 2
ln P(atm)
T(K)
Plot of f vs. ln P
0
0.1
0.2
0.3
0.4
0.5
0.6
-3 -2 -1 0 1 2
ln P(atm)
f
e. C **PROGRAM FOR PROBLEM 9-35
WRITE (5, 1)
1 FORMAT ('1', 20X, 'SOLUTION TO PROBLEM 9-35'//)
T = 1200.0
TLAST = 0.0
PSIL = 0.0
9.35 (cont'd)

9-48
C **DECREMENT BY 50 DEG. AND LOOK FOR A SIGN IN PSI
DO 10I =1, 20
CALL PSICAL (T, PHI, PSI)
IF ((PSIL*PSI).LT.0.0) GO TO 40
TLAST = T
PSIL = PSI
T = T – 50.
10 CONTINUE
40 IF (T.GE.0.0) GO TO 45
WRITE (3, 2)
2 FORMAT (1X, 'T LESS THAN ZERO -- ERROR')
STOP
C **APPLY REGULA-FALSI
45 DO 50 I = 1, 20
IF (I.NE.1) T2L = T2
T2 = (T*PSIL-TLAST*PSI)/(PSIL-PSI)
IF (ABS(T2-T2L).LT.0.01) GO TO 99
CALL PSICAL (T2, PHIT, PSIT)
IF (PSIT.EQ.0) GO TO 99
IF ((PBIT*PBIL).GT.0.0) PSIL = PSIT
IF ((PSIT*PSIL).GT.0.0) TLAST = T2
IF ((PSIT*PSI).GT.0.0) PSI = PSIT
IF ((PSIT*PSI).GT.0.0) T = T2
50 CONTINUE
IF (I.EQ.20) WRITE (3, 3)
3 FORMAT ('0', 'REGULA-FALSI DID NOT CONVERGE IN 20 ITERATIONS')
93 STOP
END
1
*
SUBROUTINE PSICAL (T, PHI, PSI)
REAL KF
PHI = (3052 + 36.2*T + 36.2*T + 0.05943*T**2)/(127240. – 11.35*T
– 0.0636*T**2)
KP = 7.28E6*EXP(-17000./T)
FBI = SQRT((KP/(1. + KP)) – 1./12. + PHI)
WRITE (3, 1) T, PSI
FORMAT (6X, 'T =', F6.2, 4X, 'PSI =', E11,4)
RETURN
END
OUTPUT: SOLUTION TO PROBLEM 9-35
T PSI E
T PSI E
T PSI E
T PSI E
T PSI E
T PSI E
T PSI E
T PSI E
T PSI E
= = +
= = +
= = +
= = +
= = +
= = − −
= = − −
= = − −
= = − −
1200 00 08226 00
1150 00 0 7048 00
1100 00 05551 00
1050 00 0 3696 00
1000 00 01619 00
950 00 0 3950 01
959 80 01824 02
960 25 0 7671 04
960 27 0 3278 05
. .
. .
. .
. .
. .
. .
. .
. .
. .
Solution: T f
= =
960.3 K mol C H reacted mol fed
2 6
, .
0 360
9.35 (cont'd)

9-49
9.36
a.
b.
2 3
2
CH C H H
C H C(s) + H
4 2 2 2
2 2 2
→ +
→
4 1 4
o
2 2 2
Basis: 10.0 mol CH (g)/s (mol CH /s)
1500 C (mol C H /s)
n
n

3 2
4
(mol H (s)/s)
(mol C(s)/s)
n
n

o
1500 C
60% conversion ⇒ = − =
. .
n1 10 1 0 600 4 00
b g mol CH s
4
C balance: (1)
H balance: (2)
10 1 4 1 2 2 6
10 4 4 4 2 2 2 2 24
2 4 2 4
2 3 2 3
b g b g
b g b g
= + + ⇒ + =
= + + ⇒ + =

n n n n
n n n n
References for enthalpy calculations : C(s), H g
2 b g at 25°C
H H C
i
i
pi
= + −
Δ
f
o
e j b g
1500 25 , i = CH , C H , C, H
4 2 2 2
Substance
mol s kJ mol mol s (kJ mol
CH g 10 41.68
C H g
H g
C s
in in out out
4
2 2
2

( )

( )

( )

)
.
.
.
.
n H n H
n
n
n
b g
b g
b g
b g
4 4168
30345
4572
32 45
2
3
4
− −
− −
− −
Energy Balance: kJ / s (3)
mol C H s
Solve (1) -(3) simultaneously mol H s
mol C s
out in
2 2
2
Q H n H n H
n
n
n
i i i i
= ⇒ = −
=
⇒ =
=
∑ ∑
Δ 975
2 50
9 50
100
2
3
4

. /
. /
. /
Yield of acetylene = =
2 50
6
0 417
.
.
mol C H s
.00 mol CH consumed s
mol C H mol CH consumed
2 2
4
2 2 4
If no side reaction,
. ( . ) . /
. / , . /
n
n n n
1
3 2 4
10 0 1 0 600 4 00
0 300 9 00
= − =
= ⇒ = =
mol CH s
mol C H s mol H s
4
2 2 2
Yield of acetylene = =
300
6
0500
.
.
mol C H s
.00 mol CH consumed s
mol C H mol CH consumed
2 2
4
2 2 4
Reactor Efficiency = =
0 417
0500
0834
.
.
.
975 kW

9-50
9.37 C H g 3H O v 3CO g 7H g
3 8 2 2
b g b g b g b g
+ → +
CO g H O v CO g H g
2 2
b g b g b g b g
+ → +
2
Basis : 1 mol C H fed
3 8
1 mol C H (g )
4.94 m at 1400°C, 1 atm
3 (mol), 900°C
n g
Heating gas
3 8
6 mol H O( g )
2
125°C
a
Product gas, 800°C
(mol C H ) = 0
n 1 3 8
(mol H O)
n 2 2
(mol CO)
n 3
(mol CO )
n 4 2
(mol H )
n 5 2
(mol)
ng
ng = =
4.94 m L 273 K 1 mol
1 m 1673 K 22.4 L
mol heating gas
3
3
10
3599
3
.
Let ξ1 and ξ2 be the extents of the two reactions.
n
n
1 1
0
1
1 1
1
= − ⇒ =
=
ξ ξ mol n4 2
= ξ
n n
2 1 2
1
2 2
6 3 3
1
= − − ⇒ = −
=
ξ ξ ξ
ξ
n n
5 1 2
1
5 2
7 7
1
= + ⇒ = +
=
ξ ξ ξ
ξ
n n
3 1 2
1
3 2
3 3
1
= − ⇒ = −
=
ξ ξ ξ
ξ
References : C(s), H g O g
2 2
b g b g
, at 25°C, heating gas at 900°C

H H C dT
C dT C T
i pi
T
p
T
p
= +
=
= = −
z
z
Δ fi
o
3 8
2 2 2
for C H
Table B.8 for CO , H , H O, CO
for heating gas
25
900
900
b g
Substance
C H 1
H O
CO
CO
H
heating gas
in in out out
3 8
2
2
2
n H n H

.
. .
.
.
.
. . .
− −
− − −
− − − −
− − −
− − +
9539 0
6 238 43 3 212 78
3 86 39
35615
7 22 85
3599 200 00 3599 0
2
2
2
2
ξ
ξ
ξ
ξ
Energy Balance :
n H n H n n
n n
i i i i
.
, .
out in
2
2 2 2 2 2
mol mol H O, mol CO,
mol CO mol H mol % H O, 7.7% CO, 15.4% CO , 69.2% H
∑ ∑
− = ⇒ = ⇒ = =
= = ⇒
0 2 00 1 1
1 9 7 7
2 2 3
4 5
ξ

9-51
9.38 a.
b.
Any C consumed in reaction (2) is lost to reaction (1). Without the energy released by reaction
(2) to compensate for the energy consumed by reaction (1), the temperature in the adiabatic
reactor and hence the reaction rate would drop.
Basis : 1.00 kg coal fed (+0.500 kg H20)
0.500 kg H20
⇒
1.0 kg coal
0.105 kg H2O/kg coal
0.226 kg ash/kg coal
0 669
. kg combustible / kg coal
0.812 kg C / kg combustible
0.134 kg O / kg combustible
0.054 kg H / kg combustible
R
S
|
|
T
|
|
U
V
|
|
W
|
|
n
n
n
n
f
f
f
f
1
2
3
3
3
4
56
3577
= [ (1.00)(0.669)(0.812) kg C][1 mol C /12.01 10 kg] = 45.23 mol C
= (1.00)(0.669)(0.134) /16.0 10 mol O
= (1.00)(0.669)(0.054) /1.01 10 mol H
= [ (0.500 + 0.105) kg][1 mol H O /18.016 10 kg] = 33.58 mol H O
3
2
3
2
×
× =
× =
×
−
−
−
−
.
.
n0 (mol O2) 25°C Product gas at 2500°C
1 kg coal + H2O, 25°C
0
2
0 0
2
Reactive oxygen (O) available (2 5.60) mol O
35.77 mol H 1 mol O
Oxygen consumed by H ( 2H+O H O) : 17.88 mol O
2 mol H
Reactive O remaining =(2 5.60) 17.88 (2 12.28) mol O
CO formed ( C
n
n n
= +
→ =
⇒ + − = −
0 2
2 1 0 2
(2 12.28) mol O 1 mol CO
+2O CO ) : ( 6.14) mol CO
2 mol O
n
n n
−
→ = = −
1 0
1 0
2 0
4 0
6.14
1 2 2 0
6.14
0 1 2 4 4 0 2
51.37
0.06
3 4 3 0 2
C balance : 45.23= (51.37 ) mol CO
O balance : 2 5.60 33.58 2 ( 0.06) mol H O
H balance :35.77+ 2(33.58)=2 2 (51.37 ) mol H
n n
n n
n n
n n
n n n n
n n n n n n
n n n n
= −
= −
= −
= +
+ = −
+ + = + + = +
+ = −
⇒
⇒
⇒
nf1 (mol C)
nf2 (mol O)
nf3 (mol H)
nf4 (mol H2O)
0.226 kg ash
n1 (mol CO2)
n2 (mol CO)
n3 (mol H2)
n4 (mol H2O)
45.23 mol C
5.60 mol O
35.77 mol H
33.58 mol H2O
0.226 mol kg ash
0.226 kg slag
2500°C

9-52
c. 1 kg coal contains 45.23 mol C and 35.77 mol H
1 kg coal + nO CO mol H O (l)
kJ = 45.23( ( (
( kJ / kg
2 2 2
f
o
CO f
o
H O(l) f
o
coal
f
o
coal
2 2
⇒ → +
= − + −
⇒ = −
4523 3577 2
21 400 3577 2
1510
. ( . / )
, ) ( . / ) ) )
)
Δ Δ Δ Δ
Δ
H H H H
H
r
o
2 2
2 0 1
0 2
2 0 3
2 0 0 4
References :C(s), O (g), H (g), ash(s) at 25 C
ˆ ˆ
Substance
(mol) (mol)
(kJ/mol) (kJ/mol)
ˆ
CO 6.14
ˆ
CO 51.37
ˆ
H 51.37
ˆ ˆ
H O 33.58 0.06
Coal 1 kg 1510 kJ/kg
Ash
in out
in out
n n
H H
n H
n H
n H
H n H
− − −
− − −
− − −
+
− − −
5
ˆ
(slag) (in coal) 0 0.266 kg (kJ/kg)
H
o
1 2
2
3 2
4 2
5 m ash
ˆ ˆ (2500 25), 1,3
ˆ 393.5 0.0508(2475) 267.8 kJ/mol CO
ˆ 110.52 0.0332(2475) 28.35 kJ/mol CO
ˆ 0.0300(2475) 74.25 kJ/mol H
ˆ 241.83 0.0395(2475)= 144.07 kJ/mol H O
ˆ ˆ
( )
i fi pi
H H C i
H
H
H
H
H H
= Δ + − =
= − + = −
= − + = −
= =
= − + −
= Δ
0 2
1.4(2475) 710 1.4(2475) 4175 kJ/kg ash
Energy Balance
ˆ ˆ 0 35.4 mol O
out out in in
H n H n H n
+ = + =
Δ = − = ⇒ =
∑ ∑
9.38 (cont’d)

9-53
9.39
Mass of H SO
m 10 L 1 mol H SO
1 m L
mol H SO
98.02 g
1 mol
g H SO
Mass of solution
m 10 L 10 mL 1.064 g
1 m L 1 mL
g solution
2 4
3 3
2 4
3 2 4 2 4
3 3 3
3
= =
F
HG I
KJ = ×
= = ×
3
3000 2 941 10
3
3192 10
5
6
.
.
⇒ = × − × = ×
Moles of H O g H O(
1 mol
18.02 g
mol H O
2 2 2
( . . ) ) .
3192 10 2 941 10 161 10
6 5 5
n
mol H O
mol H SO
mol H O
mol H SO
mol H O mol H SO
2
2 4
2
2 4
2 2 4
F
HG I
KJ =
×
=
161 10
3000
536
5
.
.
Δ Δ Δ
. . .
, . ., .
H H H
f
H SO aq. r
f
o
H SO l
s
H SO aq r
2 2
Table B.1
2
Table B.11
kJ
mol
kJ mol
e j e j e j b g
b g b g b g
4 4 4
53 6 53 6
81132 7339 884 7
=
A
=
A
= + = − − = −
H = ×
(3000 mol H SO )(-884.7 kJ / mol H SO ) = -2.65 10 kJ
2 4 2 4
6
9.40
HCl (aq): Δ Δ Δ
. . .
H H H
f
o
f
o
HCl g
s
o
Tables B.1, B11
kJ mol
= + = − − = −
∞
e j e j
b g 92 31 7514 167 45
NaOH (aq): Δ Δ Δ
. . .
H H H
f
o
f
o
NaOH s
s
o
Tables B1, B.11
kJ mol
= + = − − = −
∞
B
e j e j
b g 426 6 42 89 469 49
NaCl (aq): Δ Δ Δ
. . .
H H H
f
o
f
o
NaCl s
s
o
Table B.1 Given
kJ mol
= + = − + = −
∞
B B
e j e j
b g 4110 4 87 4061
HCl aq NaOH aq NaCl aq H O l
2
b g b g b g b g
+ → +
Δ . . . . .
Hr
o
kJ mol
= − − − − − − = −
4061 28584 167 45 469 49 550
b g b g
HCl g NaOH s NaCl s H O l
2
b g b g b g b g
+ → +
Δ Δ Δ

. . . . .
H v H v H
i i
r
o
f
o
products
f
o
reactants
kJ mol kJ mol
= −
= − − − − − − = −
∑ ∑
4110 28584 92 31 426 6 177 9
b g b g
The difference between the two calculated values equals
Δ Δ Δ

H H H
s
NaCl
s
HCl
s
NaOH
e j e j e j
{ }
− − .
9.41 a. H SO aq 2NaOH aq Na SO aq 2H O l
2 4 2 2
b g b g b g b g
+ → +
4
Basis mol H SO soln
0.10 mol H SO g mol g H SO
0.90 mol H O g mol g H O
g soln 1 cm
g
cm
2 4
2 4 2 4
2 2
3
:
. .
. .
.
.
.
1
98 08 9 808
18 02 16 22
26 03
127
20 49 3
⇒
× =
× =
U
V
|
W
|
⇒ =
b g
b g
⇒
0.10 mol H SO 2 mol NaOH 1 liter caustic soln 10 cm
1 mol H SO 3 mol NaOH 1 L
cm NaOH aq
2 4
3 3
2 4
3
= 66 67
. b g

9-54
b.
Volume ratio
cm NaOH(aq)
20.49 cm H SO aq)
cm caustic solution / cm acid solution
3
3
2 4
3 3
= =
66 67
325
.
(
.
H SO aq
2 4 b g: r mol H O /1 mol H SO
2 2 4
= 9
Δ Δ Δ
. .
.,
H H H
f
o
soln
f
o
H SO l
f
o
H SO aq r
2 4
2 2
kJ
mol
kJ mol H SO
e j e j e j b g
b g b g
= + = − − = −
=
4 4 9
81132 6523 877
NaOH(aq) : The solution fed contains 66 67 113 7534
. . .
cm g cm g
3 3
e je j= , and
(0.2 mol NaOH) g mol g NaOH
75.34 g H O 67.39 g H O mol 18.02 g mol H O
mol H O 0.20 mol NaOH mol H O / mol NaOH
2 2 2
2 2
40 00 8 00
8 00 1 374
374 18 7
. .
. .
. .
b g
b g b gb g
=
⇒ − ⇒ =
⇒ = =
r
Δ Δ Δ
. . .
., .
H H H
f
o
soln
f
o
NaOH s
s
o
NaOH s aq r
kJ
mol
kJ mol NaOH
e j e j e j b g
b g b gb g
= + = − − = −
=18 7
426 6 42 8 469 4
Na SO aq
2 4 b g:
Δ Δ Δ
. . .
H H H
f
o
soln
f
o
Na SO s
f
o
Na SO aq
2 4
2 2
kJ
mol
kJ mol Na SO
e j e j e j b g
b g b g
= + = − − = −
4 4
1384 5 117 13857
Extent of reaction mol (1) mol
H SO final H SO fed H SO
2 2 4 2 4
: ( ) ( ) . .
n n
4
0 010 010
= + ⇒ = − ⇒ =
ν ξ ξ ξ
Energy Balance
= (0.10 mol) 1385.7 + 2(
kJ
mol
kJ
r
o
f
o
Na SO aq) f
o
H O l) f
o
H SO aq) f
o
NaOH aq)
2 4 2 2 4
:
( ) ( ) ( ) ( )
. ) ( . ) ( )( . ) .
( ( ( (
Q H H H H H H
= = = + − −
− − − − − − = −
Δ Δ Δ Δ Δ Δ
ξ ξ 2 2
28584 87655 2 469 4 14 2
9.42 a.
b.
NaCl(aq): Δ Δ Δ
. . .
H H H
f
o
f
o
NaCl s
s
o
Table B.1
given
,
kJ / mol kJ mol
= + = − + = −
∞
B
e j e j b g
b g 4110 4 87 4061
NaOH(aq):
Δ Δ Δ
. . .
H H H
f
o
f
o
NaOH s
s
o
Table B.1
kJ / mol kJ mol
= + = − − = −
∞
B
e j e j b g
b g 426 6 42 89 469 5
NaCl aq H O l
1
2
H g
1
2
Cl g NaOH aq
2 2 2
b g b g b g b g b g
+ → + +
Δ . . .
Hr
o
kJ mol kJ mol
= − − − − − =
469 5 4061 28584 222.44
b g b g
8500 ktonne Cl tonne 10 kg 10 g 1 mol Cl 222.44 kJ
yr 1 ktonne 1 tonne 1 kg 70.91 g Cl 0.5 mol Cl
2
3 3
2
2 2
103
10 J 2.778 10 kW h 1 MW h
1 kJ 1 J 10 kW h
MW h / yr
3
3
× ⋅ ⋅
⋅
= × ⋅
−7
7
148 10
.
9.41 (cont'd)

9-55
9.43 a.
b.
CaCl s O l CaCl aq r kJ mol
2 2 2 r
o
b g b g b g b g
+ → = = −
10H 10 1 64 85
1
, .
ΔH
CaCl H O s H O l CaCl aq r kJ mol
2 2 2 2 r
o
⋅ + → = = +
6 4 10 2 32 41
2
b g b g b g b g
, .
ΔH
1 2 6 6
b g b g b g b g b g
− ⇒ + → ⋅
CaCl s H O l CaCl H O s
2 2 2 2 (3)
⇒ = − = −
Δ Δ Δ
.
H H H
r
o
r
o
r
o
Hess's law kJ mol
3 1 2 97 26
b g
From (1), Δ Δ Δ

,
H H H
r
o
f
o
CaCl aq r
f
o
CaCl s
2 2
1
10
= −
=
e j e j
b g b g
⇒ = − − = −
=
Δ . . .
,
Hf
o
CaCl aq r
2
kJ mol kJ mol
e j b g
b g
10
64 85 794 96 859 81
9.44
a.
b.
c.
Basis: 1 mol NH SO produced
4 4
b g2
2 mol NH3 (g)
75°C 1 mol (NH4)2SO4 (aq)
1mol H2SO4 (aq)
25°C 25°C
2NH g H SO aq NH SO aq
3 2 4 4 2 4
b g b g b g b g
+ →
References : Elements at 25°C
NH g C
3 , :
75°
b g . . . (
H H C dT
p
= + = − +
F
HG I
KJ = −
z
Δ f
o
kJ / mol kJ mol Table B.1, B.2)
25
75
4619 183 44 36
H SO aq C
2 4 , :
25°
b g .
H H
= = −
Δ f
o
H SO aq
2 4
2 4
kJ mol H SO
e j b g 907 51 (Ta.ble B.1)
NH SO aq C
4 2 4
b g b g
, :
25° .
H H
= = −
Δ f
o
NH SO aq
4 2 4
4 2 4
kJ mol NH SO
e j b g
b g b g 11731 (Table B.1)
Energy balance:
Q H n H n H
i i i i
= = − = − − − − −
= − ⇒
∑ ∑
Δ . . .
out in
4 2 4
kJ
kJ 177
kJ withdrawn
mol NH SO produced
1 11731 2 44 36 1 907 51
177
b g
1 132
1914
177
1914 1 4184 25
471
mole % (NH ) SO solution
1 mol (NH ) SO 132 g
mol
g (NH ) SO
99 mol H O 18 g
mol
=
1782 g H O
g solution
The heat transferred from the reactor in part (a) now goes to heat the product solution from
25 C to kJ =
g kg kJ C
10 g kg C
C
4 2 4
4 2 4
4 2 4
2 2
final 3 final
⇒ =
⇒
−
⇒ =
D
D
D
D
T
T
T
. . ( )
.
In a real reactor, the final solution temperature will be less than the value calculated in part b, due
to heat loss to the surroundings. The final temperature will therefore be less than 47.1o
C.

9-56
9.45 a.
b.
H SO aq NaOH aq Na SO aq H O aq
2 4 2 4 2
b g b g b g b g
+ → +
2 2 Basis : 1 mol H SO fed
2 4
2 mol NaOH
25°C
40°C
1 mol H SO
2 4
25°C
1 mol Na SO
2 4
49 mol H O
2
38 mol H O
2
89 mol H O
2
Reference states : Na s H g S s O g
2
b g b g b g b g
, , , 2 at 25°C
H SO aq r C
2 4 , , :
= 49 25D
e j
n mol H SO r kJ mol
kJ
2 4 f
o
H SO l
s
o
f
o
H O l
f
o
H O l
2 4 2
2

. . .
H H H H
H
= + =
L
NM O
QP +
= − − = − +
1 49 49
1 8113 733 884 6 49
b g e j b g b g e j
b g e j
b g b g
b g
Δ Δ Δ
Δ
NaOH aq r C
, , :
= 19 25D
e j
n mol NaOH r kJ mol
kJ
f
o
NaOH s
s
o
f
o
H O l
f
o
H O l
2
2

. . .
H H H H
H
= + =
L
NM O
QP +
= − − = − +
2 19 38
2 426 6 42 8 9388 38
b g e j b g b g e j
b g e j
b g b g
b g
Δ Δ Δ
Δ
Na SO aq r C
2 4 , , :
= 89 40D
e j
1 kmol Na SO 142.0 kg
1 kmol
kg,
89 kmol H O 18.02 kg
1 kmol
kg kg
2 4 2
= = ⇒
0142 1604 1746
. . .
n mol Na SO
n kJ
2 f
o
Na SO
s
o
Na SO
f
o
H O l
f
o kJ mol Table B.1
s
o kJ mol
=1.746 kg,
H2O l
kJ (kg C)
f
o
H O l
2 2 2
2

.
.
.
H H H H mC
H H
p
H
H
m Cp Cp
= +
L
NM O
QP+ + −
= − +
=−
=−
≈FH IK = ⋅
1 89 40 25
1276 89
4
1384 5
1 2
4 814
4 4
b g e j e j e j b g
e j
b g
b g
b g
b g
Δ Δ Δ
Δ
Δ
Δ
D
Energy balance: Q H n H n H H
i i i i
= = − = + = −
∑ ∑
−
Δ Δ
. .
.
out in
f
o
H O l
kJ mol
2
kJ
547 4 2 24 3
285 84
e j b g
Mass of acid fed
1 mol H SO 98.08 g H SO
1 mol
49 mol H O 18.02 g H O
1 mol
g = 0.981 kg
kJ
0.981 kg acid
kJ / kg acid transferred from reactor contents
2 4 2 4 2 2
+ =
⇒ =
−
⇒
981
24 3
24 8
Q
Macid
.
.
If the reactor is adiabatic, the heat transferred from the reactor of Part(a) instead goes to heat
the product solution from 40°C to Tf
⇒ 24 3 10
1746 40
43
3
.
.
× =
−
⋅
⇒ =
J
kg 4.184 kJ C
kg C
C
T
T
f
f
d iD
D
D

9-57
9.46 a.
b.
H SO aq 2NaOH aq Na SO aq 2H O l
2 4 2 2
b g b g b g b g
+ → +
4
H SO solution:
2 4 :
75
4
1 10
0 30
ml of 4M H SO solution
mol H SO 1 L 75 mL
L acid soln mL
mol H SO
2 4
2 4
3 2 4
⇒ = .
75 123 92 25 98 08 29 42
29 42 1 349
349 1163
mL g mL g, (0.3 mol H SO ) g mol g H SO
mol H O 0.30 mol H SO mol H O / mol H SO
2 4 2 4
2 2 2
2 2 4 2 2 4
b gb g b g
b g b gb g
. . . .
. .
. .
= =
⇒ − ⇒ =
⇒ = =
r
Δ Δ Δ . .
.
., .
H H H
f
o
soln
f
o
H SO l
f
o
H SO aq r
Table B.1,
Table B.11
2 4
2 2
kJ
mol
kJ mol H SO
e j e j e j b g
b g b g
= + = − −
= −
=
B
4 4 11 63
81132 67 42
878 74
NaOH solution required:
0.30 mol H SO 2 mol NaOH 1 L NaOH(aq) 10 mL
1 mol H SO 12 mol NaOH 1 L
mL NaOH aq
2 4
3
2 4
= 50 00
. b g
50 00 137 685
. . .
mL g mL g
b gb g=
12
1 10
0 60 24 00
24 00 1 2 47
2 47 0 6
4
40
mol NaOH 1 L 50 mL
L NaOH(aq) mL
mol NaOH g NaOH
mol H O mol NaOH
.12 mol H O
mol NaOH
3
g/mol NaOH
2 2 2
2
2
=
⇒ − ⇒ =
⇒ = =
⇒
. .
. .
. .
b g b gb g
r
Δ Δ Δ . .
.
., .
H H H
f
o
soln
f
o
NaOH s
s
o
NaOH s aq r
kJ
mol
kJ mol NaOH
e j e j e j b g
b g b gb g
= + = − −
= −
=4 12
426 6 3510
46170
Na SO aq
2 4 b g:
Δ Δ Δ . . .
H H H
f
o
soln
f
o
Na SO s
f
o
Na SO aq
2 4
2 2
kJ
mol
kJ mol Na SO
e j e j e j b g
b g b g
= + = − − = −
4 4
1384 5 117 13857
mtotal = total mass of reactants or products = (92.25g H SO soln +68.5g NaOH) =160.75g = 0.161 kg
2 4
Extent of reaction mol (1) mol
H SO final H SO fed H SO
2 2 4 2 4
: ( ) ( ) . .
n n
4
0 0 30 0 30
= + ⇒ = − ⇒ =
ν ξ ξ ξ
Standard heat of reaction
r
o
f
o
Na SO aq
f
o
H O l
f
o
H SO aq
f
o
NaOH aq
2 2 2
Δ Δ Δ Δ Δ
H H H H H
= + − −
e j e j e j e j
b g b g b g b g
4 4
2 2
Energy Balance C
mol) kJ / mol) kg) 4.184 C = 0 C
r
o
kJ
kg C
: ( )
( . ( . ( . ( )
Q H H m C T
T T
total p
= = + −
= +
F
HG
I
KJ − ⇒ =
Δ Δ
ξ 25
030 1552 0161 25 94
Volumes are additive.
Heat transferred to and through the container wall is negligible.

9-58
9.47 Basis : 50,000 mol flue gas/h
0.100 (NH ) SO
2
50°C
50,000 mol/h
0.00300 SO
2
0.997 N
n1 (mol solution/h)
4 2 3
0.900 H O( )
2 l
25°C
n4 (mol SO /h)
2
n5 (mol N /h)
2
35°C
1.5
n2
n2
(mol NH HSO /h)
4 3
(mol (NH ) SO /h)
4 3
2
n3 (mol H O( )/h)
2
35°C
l
90% SO removal mol h mol SO h
2 2
: . . , .
n4 0100 0 00300 50 000 15 0
= =
a fb g
N balance:
2 . , ,
n5 0 997 50 000 49 850
= =
a fb g
mol h mol N h
2
NH balance:
S balance:
mol h
mol NH HSO h
4
+
4 3
2 0100 15 2 20
0100 000300 50 000 150 15
5400
270
1 2 2 1 2
1 2 2
1
2
b gb gb g b gb g
b gb g
. .
. . , . .
n n n n n
n n n
n
n
= + ⇒ =
+ = + +
U
V
|
W
|
⇒
=
=
H O balance
270 mol NH HSO produced 1 mol H O consumed
h 2 mol NH HSO produced
mol H O l h
2
4 3 2
4 3
2
: .
n3 0900 5400
4725
= −
=
b gb g
bg
Heat of reaction:
Δ Δ Δ Δ Δ
. . .
( )
H H H H H
r
o
f
o
NH HSO aq
f
o
NH SO aq
f
o
SO g
f
o
H O(l)
4 4 2 3
kJ mol kJ mol
= − − −
= − − − − − − − = −
2
2 760 890 296 90 28584 47 3
4 2 2
e j e j e j e j
b g b g b g b g
b g b g b g
References : N g SO g NH SO aq NH HSO aq H O l
2 4 2 3 4 2
2 3
, , , , at 25°C
SO g C
2 50
, :
e j .
H C dT
p
= =
zd iSO2
kJ mol
25
50
101 ( Cp from Table B.2)
SO g C
2 35
, :
e j .
H C dT
p
= =
zd iSO2
kJ mol
25
35
0 40
N g C
2 50
, :
e j .
H = 073 kJ mol (Table B.8)
N g C
2 35
, :
e j .
H = 0 292 kJ mol
Entering solution: H = 0
Effluent solution at 35°C
,
m g h
270 mol NH HSO g
h mol
1.5 270 mol NH SO
h mol
4725 mol H O
h 1 mol
g
h
4 3
4 2 3 2
b g
b g
=
+
×
+ =
99
116 g 18 g
159 000
nH mC T
p
= =
− °
⋅°
=
Δ
159,000 g 4 J 35 C 1 kJ
h g C 10 J
kJ / h
3
25
6360
a f
Extent of reaction:
( ) ( )
n n
out in
NH HSO NH HSO NH HSO
4 3 4 3 4 3
mol / h = 0 + 2 = 135 mol / h
= + ⇒ ⇒
ν ξ ξ ξ
270

9-59
Energy balance: Q H H n H n H
i i i i
= = + −
∑ ∑
Δ Δ
ξ r
o
out in
Q =
−
+ +
+ − − =
−
= −
135 mol kJ
h mol
kJ 1 h 1 kW
h 3600 s 1 kJ s
kW
SO out N out
effluent solution
2 2
47 3
15 0 49 850 0 292
6360 50 000 0 003 101 49 850 0 73
22 000
611
.
.40 , .
, . . , .
,
.
a fa f a fa f
a fa fa f a fa f
9.48 a.
b.
c.
CH g + 2O g CO g 2H O v
4 2 2 2
b g b g b g
→ +
( )
HHV LHV H H
= = − − = −
=
B B
°
890 36 2 890 36 2 44 01
802 34
. . .
.
kJ / mol, kJ mol
kJ mol CH
c
o
Table B.1 HHV
v
at 25 C
H O
4
2
Δ Δ
e j b g
C H g +
7
2
O g) CO g) 3H O(v)
2 4 2 2 2
b g ( (
→ +
2
HHV LHV
= = − =
1559 9 1559 9 3 44 01 1427 87
. . . .
kJ / mol, kJ mol kJ mol C H
2 6
a f
C H g +5O g) CO g) 4H O(v)
3 8 2 2 2
b g ( (
→ +
3
HHV LHV
= = − =
2220 0 2220 0 4 44 01 2043 96
. . . .
kJ / mol, kJ mol kJ mol C H
3 8
a f
HHV
b g b gb g b gb g b gb g
natural gas
0.875 890.36 kJ mol 1559.9 kJ mol kJ mol
kJ mol
= + +
=
0 070 0 020 2200 00
933
. . .
LHV
b g b gb g b gb g b gb g
natural gas
0.875 802.34 kJ mol 1427.87 kJ mol kJ mol
= kJ mol
= + +
0 070 0 020 204396
843
. . .
1 mol natural gas 0.875 mol CH
g
mol
mol C H
g
mol
mol C H
g
mol
mol N
g
mol
kg
10 g
kg
843 kJ 1 mol
mol 0.01800 kg
kJ kg
4 2 6
3 8 2 3
⇒
F
HG I
KJ+
F
HG I
KJ
+
F
HG I
KJ+
F
HG I
KJ × =
⇒ =
[ . . .
. . . . ] .
b g b g
b g b g
16 04 0 070 30 07
0 020 44 09 0 035 28 02
1
0 01800
46800
The enthalpy change when 1 kg of the natural gas at 25o
C is burned completely with oxygen at
25o
C and the products CO2(g) and H2O(v) are brought back to 25o
C.
9.49
C s + O g) CO g),
kJ 1 mol 10 g
mol 12.01 g 1 kg
kJ kg C
2 2 c
o
f
o
CO g)
3
Table B.1
2
b g e j
( (
.
,
(
→ = =
−
= −
B
Δ Δ
H H
3935
32 764
S s + O g) SO g kJ mol kJ / kg S
2 2 c
o
f
o
SO
Table B.1
2
MSO2
b g e j
( ( ), .
.
→ = = − ⇒ −
B B
=
Δ Δ
H H 296 90 9261
32 064
H g +
1
2
O g H O l kJ mol H kJ kg H
2 2 c
o
f
o
H O l
2
Table B.1
2
MH2
2 28584 141 790
1 008
b g b g e j b g
( ) , . ,
.
→ = = − ⇒ −
B B
=
Δ Δ
H H
9.47 (cont'd)

9-60
a.
b.
c.
H available for combustion = total H – H in H O
2 ; latter is
x0
16
(kg O) 2 kg H
kg coal kg O
in water
A
Eq. (9.6-3) ⇒ HHV = + −
F
HG I
KJ+
32 764 141 790
8
9261
, ,
C H
O
S
This formula does not take into account the heats of formation of the chemical constituents of
coal.
C = 0 758
. , H = 0 051
. , O = 0 082
. , S = 0 016
. ⇒ =
HHV
b gDulong
kJ kg coal
31 646
,
1 kg coal
0.016 kg S 64.07 kg SO formed
32.06 kg S burned
kg SO kg coal
2
2
⇒ = 0 0320
.
φ = = × −
0 0320
101 10 6
.
.
kg SO kg coal
31,646 kJ kg coal
kg SO kJ
2
2
Diluting the stack gas lowers the mole fraction of SO2, but does not reduce SO2 emission rates. The
dilution does not affect the kg SO2/kJ ratio, so there is nothing to be gained by it.
9.50 CH + 2O CO 2H O l
4 2 2 2
→ + b g, HHV H
= − =
Δ .
c
o
kJ mol Table B.1
890 36 b g
C H +
7
2
O CO 3H O l
2 6 2 2 2
→ +
2 b g, HHV = 1559 9
. kJ mol
CO +
1
2
O CO
2 2
→ , HHV = 282 99
. kJ mol
(Assume ideal gas)
Initial moles charged:
2.000 L 273.2K 2323 mm Hg 1 mol
25+ 273.2 K 760 mm Hg L STP
mol
a f a f
22
0 25
.4
.
=
Average mol. wt.: ( . ( .
4 929 0 25
g) mol) = 19.72 g / mol
Let x1 = mol CH mol gas
4 , x x x
2 1 2
1
= ⇒ − −
mol C H mol gas mol CO mol gas
2 6 b g b g
c h
MW x x x x
= ⇒ + + − − =
19 72 16 04 30 07 1 28 01 19 72 1
1 2 1 2
. . . . .
g mol CH4
b g b g b gb g b g
HHV x x x x
= ⇒ + + − − =
9637 890 36 1559 9 1 282 99 9637 2
1 2 1 2
. . . . .
kJ mol b g b g b gb g b g
Solving (1) (2) simultaneously yields
x x x x
1 2 1 2
0 725 0188 1 0 087
= = − − =
. . .
mol CH mol, mol C H mol, mol CO mol
4 2 6
9.51 a. Basis : 1mol/s fuel gas
CH (g) 2O (g) CO (g) 2H O(v), kJ / mol
C H (g)
7
2
O (g) 2CO (g) 3H O(v), kJ / mol
4 2 2 2 c
o
2 6 2 2 2 c
o
+ → + = −
+ → + = −
Δ
Δ
.
.
H
H
890 36
1559 9
n
n
n
2 2
3 2
4 2
, mol CO
, mol H O
, mol O
9.49 (cont'd)
1 mol/s fuel gas, 25°C
85% CH4
15% C2H6
Excess O2, 25°C
25°C

9-61
b.
1 mol / s fuel gas 0.85 mol CH / s 0.15 mol C H / s
4 2 6
⇒ ,
Theoretical oxygen
2 mol O 0.85 mol CH
1 mol CH s
3.5 mol O 0.15 mol C H
1 mol C H s
mol O / s
2 4
4
2 2 6
2 6
2
= + = 2 225
.
Assume 10% excess O O fed = 1.1 2.225 = 2.448 mol O / s
2 2 2
⇒ ×
C balance : mol CO / s
2
. . .
n n
2 2
085 1 015 2 115
= + ⇒ =
b gb g b gb g
H balance mol H O / s
2
: . . .
2 085 4 015 6 215
3 3
n n
= + ⇒ =
b gb g b gb g
10% 01 2 225 0 223
4
excess O mol O / s mol O / s
2 2 2
⇒ = =
. . .
n b gb g
Extents of reaction: . .
ξ ξ
1 2
085 015
= = = =
n n
CH C H
4 2 6
mol / s, mol / s
Reference states: CH g , C H g , N g , O g , H O l , CO (g) at 25 C
4 2 6 2 2 2 2
o
b g b g b g b g b g
(We will use the values of ΔHc
o
given in Table B.1, which are based on H O l
2 b g as a
combustion product, and so must choose the liquid as a reference state for water)
Substance
CH
C H
O
CO
H O v
in in out out
4
2 6
2
2
.
.
. .
.
.
n H n H
H
085 0
015 0
2 225 0 0 223 0
115 0
215
2
1
− −
− −
− −
− −
b g
.
H H
1 25 44 01
= =
Δ v
o
C kJ / mol
e j
Energy Balance :
Q n H n H n H n H
i i i i
= + + −
∑ ∑
CH c
o
CH
C H c
o
C H
out in
4
4
2 6
2 6
Δ Δ
e j e j
mol / s CH kJ mol mol / s C H kJ mol
mol / s H O kJ / mol kW
kW (transferred from reactor)
4 2 6
2
= − + −
+ = −
⇒ − =
085 890 36 015 1559 9
215 44 01 896
896
. . . .
. .
b gb g b gb g
b gb g
Q
Constant Volume Process. The flowchart and stoichiometry and material balance calculations are
the same as in part (a), except that amounts replace flow rates (mol instead of mol/s, etc.)
1 mol fuel gas 0.85 mol CH 0.15 mol C H
4 2 6
⇒ ,
Theoretical oxygen mol O2
= 2 225
.
Assume 10% excess O O fed = 1.1 2.225 = 2.448 mol O
2 2 2
⇒ ×
C balance : mol CO2
n n
2 2
085 1 015 2 115
= + ⇒ =
. . .
b gb g b gb g
H balance mol H O
2
: . . .
2 085 4 015 6 215
3 3
n n
= + ⇒ =
b gb g b gb g
10% 01 2 225 0 223
4
excess O mol O mol O
2 2 2
⇒ = =
n . . .
b gb g
9.51 (cont’d)

9-62
c.
Reference states: CH g , C H g , N g , O g , H O l , CO (g) at 25 C
4 2 6 2 2 2 2
o
b g b g b g b g b g
For a constant volume process the heat released or absorbed is determined by the internal
energy of reaction.
Substance
CH
C H
O
CO
H O v
in in out out
4
2 6
2
2
n U n U
U
.
.
. .
.
.
085 0
015 0
2 225 0 0 223 0
115 0
215
2
1
− −
− −
− −
− −
b g
.
.
.
U U H RT
1 25 25 44 01
8 314
1000
4153
= = − = − =
Δ Δ
v
o
v
o
C C kJ / mol
J 1 kJ 298 K
mol K J
kJ
mol
e j e j
Eq. (9.1-5) ⇒ Δ Δ ( )
U H RT
c
o
c
o
i
gaseous
products
i
gaseous
reactants
= − −
∑ ∑
ν ν
⇒ = − −
− −
= −
= − −
− −
= −
Δ
Δ
.
. )
.
.
. . )
.
U
U
c
o
CH 3
c
o
C H 3
4
2 6
kJ mol
J 298 K (1+ 2 kJ
mol K 10 J
kJ
mol
kJ mol
J 298 K (3+ 2 kJ
mol K 10 J
kJ
mol
e j b g
e j b g
890 36
8 314 1 2 1
890 36
1559 9
8 314 35 1 1
156114
Energy balance:
Q U n U n U n U n U
Q
i i i i
= = + + −
= − + −
+ = −
⇒ − =
∑ ∑
Δ Δ Δ
CH c
o
CH
C H c
o
C H
out in
4 2 6
2
4
4
2 6
2 6
mol / s CH kJ mol mol / s C H kJ mol
mol / s H O kJ / mol kJ
kJ (transferred from reactor)
. . . .
. .
e j e j
b gb g b gb g
b gb g
085 890 36 015 156114
215 4153 902
902
Since the O2 (and N2 if air were used) are at 25°C at both the inlet and outlet of this process, their
specific enthalpies or internal energies are zero and their amounts therefore have no effect on the
calculated values of Δ Δ .
H U
and
9.52 a.
b.
c.
( )
n H W Q
fuel s l
− = −
Δ c
o
(Rate of heat release due to combustion = shaft work + rate of heat loss)
.
.
V
V
(gal) L 0.700 kg 10 g 49 kJ
h 7.4805 gal L kg g
100 hp J / s 1 kJ 3600 s
1.341 10 hp 10 J h
15 10 kJ
298 h
gal / h
3
3
6
28 317
1
1
2 5
3
=
×
+
×
⇒ =
−
The work delivered would be less since more of the energy released by combustion would go into
heating the exhaust gas.
Heat loss increases as Ta decreases.
Lubricating oil becomes thicker, so more energy goes to overcoming friction.
9.51 (cont'd)

9-63
9.53 a.
b.
Energy balance: Δ
Δ
U
n U
mC T
v
= ⇒ + − ° =
0 77 0
lb fuel burned (Btu)
lb
F
m c
o
m
out
b g b g
⇒ + ⋅° ° − ° =
0 00215 4 62 0 900 87 06 77 00 0
. . . . .
a f b gb ga f
ΔUc
o
m m
lb Btu lb F F F
⇒ = −
ΔUc
o
m
Btu lb
19500
The reaction for which we determined ΔUc
o
is
1 lb oil + O g) CO g)+ H O(v) (1)
m 2 2 2
a b c
( (
→
The higher heating value is ΔHr for the reaction
1 lb oil + O g) CO g)+ H O(l) (2)
m 2 2 2
a b c
( (
→
Eq. (9.1-5) on p. 441 ⇒ Δ Δ ( )
H U RT b c a
c1
o
c1
o
= + + −
Eq. (9.6-1) on p. 462 ⇒ − = − +
Δ Δ Δ (
) )
H H c H
HHV LHV
c2
o
(
c1
o
(
v 2
H O, 77 F)
To calculate the higher heating value, we therefore need
a
b
c
=
=
=
lb - moles of O that react with 1 lb fuel oil
lb - moles of CO formed when 1 lb fuel oil is burned
lb - moles of H O formed when 1 lb fuel oil is burned
2 m
2 m
2 m
9.54 a.
CH OH v + O g CO g 2H O l
3 2 2 2
3
2
b g b g
( ) ( )
→ + Δ Δ .
H H
r
o
c
o
CH OH v
3
kJ
mol
= = −
e j b g 764 0
Basis : 1 mol CH OH fed and burned
3
1 mol CH OH( )
25°C, 1.1 atm
n0
3 l
(mol O )
2
vaporizer
1 mol CH OH( )
3 v
100°C
1 atm reactor
Q1(kJ)
3.76n0 (mol N )
2
100°C
Effluent at 300°C, 1 atm
np (mol dry gas)
2
2
2
nw (mol H O)
2
Overall C balance:
1 mol CH OH 1 mol C
1 mol CH OH
3
3
=np 0 048 1
.
b gb g ⇒ np = 2083
. mol dry gas
N balance: mol O
2 2
376 2083 0809 4 482
0 0
. . . .
n n
= ⇒ =
b gb g
Theoretical O : mol CH OH mol O mol CH OH mol O
2 3 2 3 2
1 15 15
b gb g
. .
=
% excess air =
−
× =
( . . )
.
4 482 15
15
100% 200%
mol O
mol O
excess air
2
2
H balance: 1 mol CH OH 4 mol H 1 mol CH OH mol H O
3 3 2
b gb g b g
= ⇒ =
n n
w w
2 2
(An atomic O balance⇒ =
mol O mol O
9 96 9 96
. . , so that the results are consistent.)
p
n
n n
P p T T
w
w
w p
w dp dp
∗ ∗
=
+
× =
+
× = = ⇒ = °
2 mol H O
mol
mm Hg mm Hg C
2
Table B.3
2 20 83
760 66 58 441
.
. .
a f d i
Q2(kJ)

9-64
b. Energy balance on vaporizer:
Q H n H C dT H C dT
pl v pv
1 1 40 33
= = = + +
L
N
MMM
O
Q
PPP
=
A A A
z z
Δ Δ Δ .
mol
kJ
mol
kJ
Table B.2
25
64.7
Table B.1 Table B.2
64.7
100
References : CH OH v , N (g), O (g), CO (g), H O l at 25 C
3 2 2 2 2
a f a f °
Substance
(mol) (kJ / mol) (mol) (kJ / mol)
CH OH 1.00
N
O
CO
H O
in in out out
3
2
2
2
n H n H
.
. . . .
. . . .
. .
. .
3603
1685 2187 1685 8118
4 482 2 235 2 98 8 470
100 11578
2 00 5358
2
− −
− −
− −
H T Hi
a f= for N , O , CO (Table B.8)
2 2 2
= +
ΔH H
v i
25 C for H O v (Eq. 9.6 -2a on p. 462, Table B.8)
2
d i a f
= z C dT
p
T
25
for CH OH v (Table B.2)
3 a f
(Note: H O l
2 b g was chosen as the reference state since the given value of ΔHc
o
presumes liquid
water as the product.)
Extent of reaction: ( ) ( )
n n
out in
CH OH CH OH CH OH
3 3 3
mol mol
= + ⇒ = − ⇒ =
ν ξ ξ ξ
0 1 1
Energy balance on reactor: Q H n H n H
i i i i
2 = + −
∑ ∑
ξΔ c
o
out in
= − + + −
= − ⇒
1 764 0 1685 8118 4 482 2 235
1
b g
. . . . .
Table B.
kJ
534 kJ 534 kJ transferred from reactor
…
9.55 a.
CH 2O CO 2H O CH O CO 2H O
4 2 2 2 4
3
2
2 2
+ → + + → +
Basis : 1000 mol CH h fed
4
1000 mol CH /s
25°C
n1
4
Q(kJ/s)
n0 (mol O /s)
2
3.76n0 (mol N /s)
2
100°C
(mol CH /s)
4
Stack gas, 400°C
n2 (mol O /s)
2
n3 (mol CO/s)
3.76n0 (mol N /s)
2
10 n3 (mol CO/s)
n4 (mol H O/s)
2
90% combustion ( )
1 4
0.10 1000 100 mol CH s
n
⇒ = =
Theoretical O2 required = 2000 mol/s
9.54 (cont'd)

9-65
b.
10% excess O2 2
O fed=1.1(2000 mol/s)=2200 mol/s
⇒
C balance:
( )( ) ( )( ) ( ) ( )
4 4 3 3 3
3 2
1000 mol CH s 1 mol C mol CH 100 1 1 10 1 81.8 mol CO s
10 818 mol CO s
n n n
n
= + + ⇒ =
⇒ =
H balance: ( )( ) ( )( ) 4 4 2
1000 4 100 4 2 1800 mol H O s
n n
= + ⇒ =
O balance: ( )( ) ( )( ) ( )( ) ( )( )
2 2 2
2200 2 2 81.8 1 818 2 1800 1 441 mol O s
n n
= + + + ⇒ =
References :C s , H g , O g , N g at 25 C
2 2 2
a f b g b g b g
( ) ( ) ( ) ( )
out
in out
in
4
2
2
2
2
ˆ
ˆ
Substance
mol s mol s
kJ mol kJ mol
CH 1000 74.85 100 57.62
O 2200 2.24 441 11.72
N 8272 2.19 8272 11.15
CO 81.8 99.27
CO 818 377.2
H O 1800 228.63
n
n H
H
− −
− − −
− − −
− − −
for CH
T) for others
Table B.1 Table B.2
Table B.8
f
o
4
f
o
i (
H H C dT
H H
p
T
= +
= +
B B
B
z
Δ
Δ
25
Energy balance: 5
out in
ˆ ˆ 5.85 10 kJ s (kW)
i i i i
Q H n H n H
= Δ = − = − ×
∑ ∑
(i) T Q
air (increases)
A ⇒ − A
(ii) %XS A⇒ − B
Q (more energy required to heat additional O2 and N2 to 400o
C, therefore
less energy transferred.)
(iii) S Q
CO CO
2
A⇒ − A(reaction to form CO2 has a greater heat of combustion and so releases
more thermal energy)
(iv) T Q
stack A⇒ − B (more energy required to heat combustion products)
9.55 (cont’d)

9-66
9.56
a.
b.
CH 2O CO 2H O, C H
7
2
O CO 3H O
4 2 2 2 2 6 2 2 2
+ → + + → +
2
Basis : 100 mol stack gas. Assume ideal gas behavior.
n1 (mol CH )
4
n3 (mol O )
2
n2 (mol C H )
2 6
Vf (m at 25°C, 1 atm)
3
3.76n3 (mol N )
2
200°C, 1 atm
100 mol at 800°C, 1 atm
0.0532 mol CO /mol
2
0.0160 mol CO/mol
0.0732 mol O /mol
2
0.1224 mol H O/mol
2
0.7352 mol N /mol
2
N balance: mol N mol O fed
2 2 2
376 100 0 7352 19 55
3 3
. . .
n n
= ⇒ =
b gb g
C balance:
H balance:
mol CH
mol C H
4
2 6
n n
n n
n
n
1 2
1 2
1
2
1 2 100 0 0532 1 100 0 0160 1
4 6 100 01224 2
372
160
b g b g b gb gb g b gb gb g
b g b g b gb gb g
+ = +
+ =
U
V
|
W
|
⇒
=
=
. .
.
.
.
Vf =
+
=
372 160 1
10
0130
3
. .
.
b g b g
mol fuel gas 22.4 L STP 298.2 K m
1 mol 273.2 K L
m
3
3
Theoretical O
3.72 mol CH 2 mol O
1 mol CH
1 mol C H
mol O
2
4 2
4
2 6 2
2 6
2
= + = 1304
.
Fuel composition:
372 69 9
301
. .
.
mol CH
1.60 mol C H
mole% CH
mole% C H
4
2 6
4
2 6
U
V
W
⇒
% Excess air:
19 55 1304
100% 50%
. .
−
× =
b gmol O in excess
13.04 mol O required
excess air
2
2
2 2 2
b g b g b g b g °
Substance
CH 3.72
C H
O
N
CO
CO
H O
in in out out
4
2 6
2
2
2
n H n H
.
. .
. . . .
. . . .
. .
. .
. .
− − −
− − −
− − −
− − −
− − −
74 85
160 84 67
19 55 531 7 32 2535
7352 513 7352 2386
160 86 39
532 3561
12 24 212 78
2

9-67
H H C dT
H H
p
T
= +
+
B B
zB
Table B.1
Table B.2, for
CH4 , C2H6
Table B.8
f
o
f
o
i 2 2 2 2
= (T) for O , N , CO, CO , H O v
Δ
Δ
25
b g
Energy balance:
Q H n H n H
i i i i
= = − =
−
= − ×
∑ ∑
Δ
out in
3
3
kJ
0.130 m fuel
kJ m fuel
.
2764
213 104
9.56 (cont’d)

9-68
9.57
a.
Basis : 50000 lb coal fed h
m ⇒
0.730 50000 lb C 1b - mole C
h 12.01 lb
1b - mole C h
m
m
b gb g = 3039
0 047 50000 101 2327
. .
b gb g = lb - moles H h (does not include H in water)
0 037 50000 32 07 57 7
. . .
b gb g = lb - moles S h
0 068 50000 18 02 189
. .
b gb g = lb - moles H O h
2
0118 50000 5900
.
b gb g= lb ash h
m
n1 (lb-moles air/h)
50,000 lb coal/h
m
3039 lb-moles C/h
2327 lb-moles H/h
57.7 lb-moles S/h
189 lb-moles H O/h
2
5900 lb ash/h
m
77°F, 1 atm (assume)
0.210 O 2
0.790 N 2
Stack gas at 600°F, 1 atm (assume)
n2 (lb-moles CO /h)
2
n3 (lb-moles H O/h)
2
n4 (lb-moles SO /h)
2
n5 (lb-moles O /h)
2
n6 (lb-moles N /h)
2
m7(lb fly ash/h)
m
m8(lb slag/h) at 600°F
m
0.287 lb C/lb
m m
0.016 lb S/lb
m m
0.697 lb ash/lb
m m
77°F, 1 atm (assume)
Feed rate of air :
O required to oxidize carbon C + O CO
lb - moles C 1 lb - mole O
h 1 lb - mole C
lb - moles O h
2 2 2
2
2
→ =
=
b g 3039
3039
Air fed:
n1 21710
=
×
=
1.5 3039 lb - moles O fed 1 mole air
h 0.210 mole O
lb - moles air h
2
2
30% ash in coal emerges in slag ⇒ = ⇒ =
0697 030 5900 2540
8 8
. .
m m
lb h lb slag / h
m m
b g
⇒ .
m7 0 700 5900 4130
= =
b g lb fly ash h
m
C balance: 3039 0 287 2540 12 01
2
lb - moles C h
b g b gb g
= +
. .
n
⇒ = ×
=
.
.
n2
44 01
5
2978 131 10
lb - moles CO h lb CO h
2
M
m 2
CO2
H balance: 2327 189 2 2 3
lb - moles H h
b g b gb g
+ =
n
⇒ = ×
=
. .
.
n3
18 02
4
1352 5 2 44 10
lb - moles H O h lb H O h
2
M
m 2
H2O
N balance: lb-moles h lb-moles N h lb N h
2 2
M
m 2
N2
. .
.
n6
28 02
5
0790 21710 17150 481 10
= = ×
=
b g
S balance: 57 7 1 0 016 2540 32 06
4
. . .
lb - moles S h
b g b g b g
= +
n
⇒ =
=
.
.
n4
64 2
56 4 3620
lb - moles SO h lb SO h
2
M
m 2
SO2
O balance:
coal air CO H O SO O
2 2 2 2
b g b g b g e j b g b g
b gb g b gb gb g b gb g b gb g b gb g
189 1 0 21 21710 2 2978 2 1352 5 1 56 4 2 2 5
+ = + + +
. . .
n
⇒ = ⇒
lb - moles O h lb O h
2 m 2

n5 943 30200

9-69
b.
c.
d.
Summary of component mass flow rates
Stack gas at 600 F, 1 atm
2978 lb - moles CO h 131000 lb CO h
1352.5 lb - moles H O h 24400 lb H O h
56.4 lb - moles SO h 3620 lb SO h
943 lb - moles O h 30200 lb O h
17150 lb - moles N h 48100 lb N h
lb fly ash h
2 m 2
2 m 2
2 m 2
2 m 2
2 m 2
m
°
⇒
⇒
⇒
⇒
⇒
4130
674,350 lbm stack gas/h
Check: 50000 21710 29 674350 2540
+ ⇔ +
b gb g in out
⇒ 679600 676900
b g b g
in out
⇔ (0.4% roundoff error)
Total molar flow rate = °
22480 lb - moles h at 600 F , 1 atm (excluding fly ash)
⇒ =
°
°
= ×
V
22480 lb - moles 359 ft STP R
h 1 lb - mole R
ft h
3
3
a f 1060
492
1 74 107
.
References: Coal components, air at 77°F ⇒ =
∑n H
i i
in
0
Stack gas: nH

.
.
=
− °
⋅°
= ×
674350 lb 7.063 Btu 1 lb - mole F
h lb - mole F lb
Btu h
m
m
600 77
28 02
8 90 107
b g
Slag: nH
.
=
− °
⋅°
= ×
2540 lb 0.22 Btu F
h lb F
Btu h
m
m
600 77
2 92 105
b g
Energy balance: Q H n H n H n H
i i i i
= = ° + −
∑ ∑
Δ Δ
coal burned c
o
out in
F

77
b g
=
× − ×
+ × + ×
= − ×
5 10 lb Btu
h lb
Btu h
Btu h
4
m
m
18 10
8 90 10 2 92 10
811 10
4
7 5
8
.
. .
.
e j
Power generated =
×
×
=
−
0 35 811 10
831
8
. .
.
b ge jBtu 1 hr 1 W 1 MW
h 3600 s 9.486 10 Btu s 10 W
MW
4 6
. .
Q = − × = − ×
811 10 5000 162 10
8 4
Btu h lb coal h Btu lb coal
m m
e j b g
⇒
−
=
×
×
=
.
.
.
Q
HHV
1 62 10
1 80 10
0 901
4
4
Btu lb
Btu lb
m
m
Some of the heat of combustion goes to vaporize water and heat the stack gas.
−
Q HHV would be closer to 1. Use heat exchange between the entering air and the stack gas.
9.57 (cont'd)

9-70
9.58 b.
c.
Basis : 1 mol fuel gas/s
(mol O )
3.76 (mol N ) Stack gas, ( C)
( C) (mol O s)
3.76 (mol N s)
1 mol / s @ 25 C (mol CO s)
(mol CH mol) (mol CO s)
(mol Ar mol) (mol H O s)
(1 ) (mol C H mol) (mol Ar s)
2
2
o
o
O 2
0 2
o
CO
4 CO 2
H O 2
2 Ar
2
2
6

/
/
/
/ /
/ /
/ /
n s
n s T
T n
n
n
x rn
x n
x x n
s
a
m
a
m a
0
0
− −
CH O CO H O
C H O CO H O
4 2 2 2
2 6 2 2 2
+ → +
+ → +
2 2
2 3
7
2
Percent excess air
C balance:
H balance: 4
O balance: 2
xs
CO CO
H O H O
O CO CO H O O CO
2 2
2 2 2
: ( ) . ( )
( ) ( )
( )
( )
( ) ( )
( ) /
n
P
x x x
x x x r n n
x x x
r
x x x n n x x x
n n n r n n n n n r
m m a
m m a
m m a
m m a m m a
0
0 0
1
100
2 35 1
2 1 1
2 1
1
6 1 2 2 3 1
2 2 1 2 2
= + + − −
+ − − = + ⇒ =
+ − −
+
+ − − = ⇒ = + − −
= + + + ⇒ = − + − nH O
2
/ 2
References : C(s), H2(g), O2(g), N2(g) at 25°C
Substance
CH
C H
A
O
N 3.76
CO
CO
H O
4
2 6
A
2 O
2
CO
2 CO
2 H O
2
2
n H n H
x
x x
x x H
n H n H
n H n H
n H
r n H
n H
in in out out
m
m A
A
o
o o

( )

.

0
1 0
0
376
3
1 4
2 5
6
7
8
− −
− − − −
− −
− −
− −
( ) ,
H H C dT
i i p i
T T
a s
= +
B
z
Δ f
Table B.2
or
25
Given : , C C
0.0955,
(kJ / mol) = 8.091, = 29.588, = 0.702, = 3.279,
= 166.72, = , = 345.35, = 433
o o
CO H O O
2 2
x x Px r T T
n n n n
H H H H
H H H H
m a s a s
o
= = = = = =
⇒ = = = =
− − −
085 0 05 5%, 10 0 150 700
2153 2 00 01500
8567 82
1 2 3 4
5 6 7 8
. . , . , ,
. , . , .

. .
Energy balance: kW

Q n H n H
out out in in
= − = −
∑ ∑ 655

9-71
d. Xa Pxs r Ta Ts Q
0.0 5 10 150 700 -
0.0
1
5 10 150 700 -
996
0.0
1
5 10 150 700 -
905
0.0
1
5 10 150 700 -
813
0.0
1
5 10 150 700 -
722
0.0
1
5 10 150 700 -
631
0.1 5 10 150 700 -
905
0.1 10 10 150 700 -
0.1 20 10 150 700 -
869
0.1 50 10 150 700 -
799
0.1 100 10 150 700 -
0.1 5 10 150 700 -
0.2 5 10 150 700 -
893
0.3 5 10 150 700 -
0.4 5 10 150 700 -
871
0.5 5 10 150 700 -
860
-1200
-1000
-800
-600
-400
-200
0
0 0.2 0.4 0.6 0.8 1 1.2
Xm
Q
-
1000
-
800
-
600
-
400
-
0
0 20 40 60 80 100 120
Pxs
Q
-910
-900
-890
-880
-870
-860
-850
0 0.1 0.2 0.3 0.4 0.5 0.6
x
Q
0.1 5 1 150 700 -722
0.1 5 2 150 700 -796
0.1 5 3 150 700 -834
0.1 5 4 150 700 -856
0.1 5 5 150 700 -871
0.1 5 10 150 700 -905
0.1 5 20 150 700 -924
0.1 5 50 150 700 -936
0.1 5 100 150 700 -941
0.1 5 10 25 700 -852
0.1 5 10 100 700 -883
0.1 5 10 150 700 -905
0.1 5 10 200 700 -926
0.1 5 10 250 700 -948
0.1 5 10 150 500 -1014
0.1 5 10 150 600 -960
0.1 5 10 150 700 -905
0.1 5 10 150 800 -848
0.1 5 10 150 900 -790
0.1 5 10 150 1000 -731
-1000
-800
-600
-400
-200
0
0 20 40 60 80 100 120
r
Q
-1000
-950
-900
-850
-800
0 50 100 150 200 250 300
Ta
Q
-1500
-1000
-500
0
0 200 400 600 800 1000 1200
Ts
Q
9.58 (cont'd)

9-72
9.59 a.
b.
Basis:
207.4 liters 273.2 K 1.1 atm 1 mol
s 278.2 K 1.0 atm liters STP
mols s
22 4
10 0
.
.
b g= fuel gas to furnace
H C H ; M CH
6 14 4
= =
(1 – )
n0
(mol/s)
y0
(mol C H /mol)
6 14
y0
(mol CH /mol)
4
(1 – )
60°C, 1.2 atm
T
dp
= 55°C
condenser
10.0 mol/s at 5°C, 1.1 atm
y2
(mol C H /mol)
6 14
y2
(mol CH /mol)
4
sat'd with C H
6 14
nb
(mol C H ( )/s)
l
6 14
Q (kW)
c
reactor
25°C
m
w (kg H O( )/s)
l
2
10 bars, sat'd
mw (kg H O( )/s)
v
2
Stack gas at 400°C, 1 atm
n3
(mol O /s)
2
n4
(mol N /s)
2
n5
(mol CO /s)
2
n6
(mol H O( )/s)
v
2
na
(mol air/s) @ 200°C
0.21 mol O /mol
2
0.79 mol N /mol
2
100% excess
T y P p
y
dp H
= ° ⇒ = ° =
⇒ =
×
= ⇒
B
55 55 4833
4833
0530 0470
0
0
C C mm Hg
mm Hg
1.2 760 mm Hg
mol C H mol mol CH mol
Antoine Eq.
6 14 4
α
b g .
.
. .
Saturation at condenser outlet:
y
pH
2
5 5889
1
0 0 93%
=
°
=
×
= =
∗
C
P
mm Hg
.1 760 mm Hg
.070 mol C H mol mol CH mol
6 14 4
b g .
.
Methane balance on condenser: . .
.
.
n y y n
y
y
0 0 2
0 070
0 530
0
1 10 0 1 19 78
2
0
− = − ⇒ =
=
=
b g b g mol s
Hexane balance on condenser: . .
.
.
.
n y n y n
n
y
y
0 0 2
19 78
0 530
0 070
10 0 9 78
0
0
2
= + ⇒ =
=
=
=
b b 6 14
mol C H s condensed
Volume of condensate =
A A
9.78 mol C H l 86.17 g cm 1L 3600 s
s mol 0.659 g 10 cm 1 h
6 14
3
3 3
Table B.1 Table B.1
b g
= 4600 L C H h
6 14 ( )
l
References : CH g, 5 C , C H l, 5 C
4 6 14
D D
e j e j
Substance
(mol / s) (kJ / mol) (mol / s) (kJ / mol)
CH 9.30
C H v
C H l
in in out out
4
6 14
6 14

. .
. . . .
.
n H n H
1985 9 30 0
10 48 41212 0 70 32 940
9 78 0
b g
b g − −
CH g
4
Table B.2
b g:
H C dT
p
T
=
B
z5
C H v
6 14
Table B.1 Table B.1
b g:
H C dT H C dT
pR
T
v pv
T
T
b
b
= + +
B B
z z
5
Δ
Condenser energy balance:
Q H n H n H
c i i i i
= = − = −
∑ ∑
Δ
out in
kW
427

9-73
CH 2O CO 2H O
4 2 2 2
+ → + , C H O 6CO 7H O
6 14 2 2 2
19
2
+ → +
Theoretical O
9.30 mol CH 2 mol O
s 1 mol CH
s 1 mol C H
mol O s
2
4 2
4
6 14 2
6 14
2
: .
+ = 253
100% excess ⇒ = × ⇒ = × ⇒ =
O 2 O mol air s
2 fed 2 theor.
b g b g 0 21 2 253 240 95
. . .
n n
a a
N balance: mol N s
2 2
0 79 240 95 190 35
4 4
. . .
b g= ⇒ =
n n
C balance:
9.30 mol CH 1 mol C
s 1 mol CH
0.70 mol C H 6 mol C
1 mol C H
mol CO 1 mol C
1 mol CO
mol CO s
4
4
6 14
6 14
2
2
2
+ =
⇒ =
n
n
5
5 135
b g
.
H balance:
9 30 4 0 70 14 2 235
6
. . .
mol CH s mol H mol CH mol H O
4 4 6 2
b gb g b gb g b g
+ = ⇒ =
n n
Since combustion is complete, O O O mol O s
2 remaining 2 excess 2 fed 2
b g b g b g
= = ⇒ =
1
2
253
3
.
n
2 2 2
b g b g b g b g ° for reactor side, H O l
2 b g at triple point for
steam side (reference state for steam tables)
Substance
mol / s kJ / mol mol / s kJ / mol
CH 9.30
C H v
O
N
CO
H O v
H O boiler water (kg / s) kg / s)
in in out out
4
6 14
2
2
2
2 w w

.
. .
. . . .
. . . .
. .
. .
. ( .
n H n H
m m
− − −
− − −
− − −
− − −
75553
0 70 170 07
50 6 531 253 1172
190 35 513 190 35 1115
135 37715
235 228 60
104 8 2776 2
2
b g
b g
b g

,
H T H C dT
H H T
p
T
i
b g
b g b g
= for CH , C H
= for O N , CO , H O v
Table B.1 and B.2
f
o
4 6 14
Table B.1 and B.8
f
o
2 2 2 2
B
B
+
+
z
Δ
Δ
25
Energy balance on reactor (assume adiabatic):
Δ . .
H n H n H m
i i i i
= − = ⇒ − + − =
∑ ∑
out in
w
0 8468 2776 2 104 8 0
b g ⇒ =
.
mw kg steam s
32
9.59 (cont'd)

9-74
9.60 a. Basis: 450 kmol CH fed h
4 CH 2O CO 2H O
4 2 2 2
+ → +
kmol air / h)@25 C
.21 kmol O kmol Stack gas@ C
.79 kmol N kmol (kmol CO h)
(kmol H O h)
450 kmol CH / h @ 25 C kJ / h) (kmol O h)
(kmol N h)
[kg H O(l) / h] [kg H O(v) / h]
25 C 17 bar, 250 C
a
o
4 2
2
2 2
o o
2
o
2 1 2
2 2
o
3
4
n
n
n
Q n
n
m m
w w
(
/
/ /
/
( /
/

0 300
0
Air fed
450 kmol CH 2 kmol O req'd 1.2 kmol O fed 1 kmol air
h 1 kmol CH 1 kmol O req'd 0.21 kmol O
kmol air h
4 2 2
4 2 2
:
na =
= 5143
450 450 900
1 2
kmol / h CH react kmol CO h kmol H O h
4 2 2
⇒ = =
,
n n
N balance mol h kmol N h
2 2
: . .
n4
6
0 79 5143 10 4060
= × =
b ge j
Molecular O balance
mol O fed
h
450 kmol CH react 2 mol O
h 1 mol CH
kmol O h
2
2 4 2
4
2
:
.
n3 0 21 5143 180
= − =
b gb g
450 kmol CO h
900 kmol H O h
4060 kmol N h
180 kmol O h
kmol / h
2
2
2
2
CO
H O
N
O
2
2
2
2
U
V
|
|
|
W
|
|
|
⇒
=
=
=
=
y
y
y
y
0 0805
0161
0 726
0 0322
5590
.
.
.
.
Mean heat capacity of stack gas
C y C
p i pi
= = + + +
= ⋅
∑ 0 0805 0 0423 0161 0 0343 0 726 0 0297 0 0322 0 0312
0 0315
. . . . . . . .
.
kJ mol C
D
Energy balance on furnace (combustion side only)
References: CH g ,CO g , O g , N g , H O l at 25 C
4 2 2 2 2
b g b g b g b g b g D
Substance (kmol / h) (kJ / kmol) (kJ / h)
CH 450
Air
Stack gas
in in out out
4

n H n H
H p
0
5143 0
−
−
− −
Extent of reaction:

ξ = =
nCH4
kmol / h
450

9-75
b.
( ) ( ) (
.
H n H n C T
p v p
= + −
+
⋅
×
2 H O(25 C) stack gas stack gas stack gas
o
2
3 3 o
o
7
2
o C)
= 180 kmol H O 10 mol kJ
h 1 kmol mol
kmol 10 mol 0.0315 kJ (300-25) C
h 1 kmol mol C
= 5.63 10 kJ / h
Δ 25
44 01 5590
( )
. . .
Q H H n H n H
i i i i
= = + −
F
HG I
KJF
HG I
KJ −
F
HG I
KJ+ × = − ×
∑ ∑
Δ Δ
ξ c
o
CH
out in
4
= 450
kmol
h
mol
kmol
kJ
mol
kJ
h
kJ
h
1000 890 36 563 10 344 10
7 8
Energy balance on steam boiler

.
Q m H m
m
w w w
w
= ⇒ × =
F
HG I
KJ
L
NM O
QP −
L
NM O
QP
⇒ = ×
Δ + 3.44 10
kJ
h
kg
h
kJ
kg
kg steam / h
8
Table B.7 Table B.6
2914 105
123 105
b g
45 kmol CH 4 /h
25°C
furnace
Liquid, 25°C
mw (kg H O/h)
2
vapor, 17 bars
mw
(kg H O/h)
2
250°C
n1
(mol O /h)
2
n2
(mol N /h)
2
n3
(mol CO /h)
2
n4
(mol H O/h)
2
na (mol air/h) at Ta (°C)
air
preheater
na (mol air/h) at 25°C
0.21 O 2
0.79 N 2
n1
(mol O /h)
2
n2
(mol N /h)
2
n3
(mol CO /h)
2
n4
(mol H O/h)
2
Stack gas
300°C 150°C
E.B. on overall process: The material balances and the energy balance are identical to those of part
(a), except that the stack gas exits at 150o
C instead of 300o
C.
References: CH g ,CO g , O g , N g , H O l at 25 C furnace side
4 2 2 2 2
D
H O l
2 a f at triple point (steam table reference) (steam tube side)
Substance
(kmol / h) (kJ / kmol) (kJ / h)
CH 450
Air
Stack gas
H O kg / h) 105 kJ / kg kg / h) 2914 kJ / kg
in in out out
4
2

( (
n H n H
H
m m
p
w w
0
5143 0
−
−
− −
( ) ( ) (
.
.
H n H n C T
p v p
= + −
+
⋅
×
2 H O(25 C) stack gas stack gas stack gas
o
2
3 3 o
o
7
2
o C)
= 180 kmol H O 10 mol kJ
h 1 kmol mol
kmol 10 mol 0.0315 kJ (150-25) C
h 1 kmol mol C
= 2 10 kJ / h
Δ 25
44 01 5590
99
9.60 (cont’d)

9-76
c.
Δ Δ
( )
. .

H H n H n H
m m
i i i i
w w
= + − =
⇒
F
HG I
KJF
HG I
KJ −
F
HG I
KJ+ ×
+
F
HG I
KJ
L
NM O
QP −
L
NM O
QP= ⇒ ×
∑ ∑
ξ c
o
CH
out in
5
4
450
kmol
h
mol
kmol
kJ
mol
kJ
h
kg
h
kJ
kg
= 1.32 10 kg steam / h
0
1000 890 36 2 99 10
2914 105 0
7
b g
Energy balance on preheater: Δ Δ Δ

H H H
= + =
d i d i
stack gas air
0
Δ Δ
H nC T
p
b g b g
stack gas
3
5590 kmol 10 mol 0.0315 kJ C
h 1 kmol mol C
kJ
h
= =
−
⋅
= − ×
150 300
2 64 107
D
D
.
− = = ⇒ = ×
= =
Δ Δ
H H n H T H T
H T
a a a
a
b g b g
stack gas air air air 3
air
Table B.8
o
kJ / h kmol
kmol / h 10 mol
= 5.133
kJ
mol
kJ / mol C
( ) ( ) .
.
2 64 10 1
5143
5133 199
7
The energy balance on the furnace includes the term −∑n H
in in
. If the air is preheated and the
stack gas temperature remains the same, this term and hence
Q become more negative, meaning
that more heat is transferred to the boiler water and more steam is produced. The stack gas is a
logical heating medium since it is available at a high temperature and costs nothing.
9.61
a.
Assume coal enters at 25 C
3
Basis: 40000 kg coal h
kg C 10 g 1 mol C
h 1 kg 12.01 g
mol C h
°
⇒
×
= ×
0 76 40000
2 531 106
.
.
b g
0 05 4000 10 101 198 10
3 6
. . .
× = ×
b g e j
kg H h mol H h
0 08 4000 10 16 0 2 00 10
3 5
. . .
× = ×
b g e j
kg O h mol O h
011 40000 4400
. × =
b g kg ash h
2.531 10 mol C/h
furnace
40,000 kg coal/h
6
×
1.98 10 mol H/h
6
×
2.00 10 mol O/h
5
×
4400 kg ash/h
25°C
4400 kg ash/h, 450°C
Flue gas at 260°C
(mol dry gas/h)
n3
2
2
2
(mol H O/h)
n4 2
preheater
Preheated air at Ta (°C)
Cooled flue gas at 150°C
0.078 CO
(mol dry gas/h)
n3
2
0.012 CO
0.114 O 2
0.796 N 2
(mol H O/h)
n4 2
air at 30°C, 1 atm, = 30%
hr
(mol O /h)
n1 2
3.76 (mol N /h)
n1 2
(mol H O/h)
n2 2
Q to steam
Overall system balances
C balance: 2 531 10 0 078 0 012 2 812 10
6
3 3 3
7
. . . .
× = + ⇒ = ×
n n n mol h dry flue gas
N balance: mol O h
mol N h
2 2
2
376 0 796 2 812 10 595 10 376 595 10
224 10
1
7
1
6 6
7
. . . . . .
n n
= × ⇒ = × ×
= ×
b ge j b ge j
9.60 (cont’d)

9-77
b.
30% relative humidity (inlet air):
y P p
n
n
H O H O
2 2
Table B.3
C mm Hg mm Hg
= ° ⇒
× + × +
=
∗
B
0 30 30
595 10 2 24 10
760 0 300 31824
2
6 7
2
.

. .
. .
b g b g b g
⇒ = ×
.
n2
5
361 10 mol H O h
2
Volumetric flow rate of inlet air:
. .
.
V =
× + × + ×
= ×
595 10 224 10 361 10
6 43 10
6 7 5
5
e j b g
mol 22.4 liters STP 1 m
h 1 mol 10 liters
SCMH
3
3
Air/fuel ratio:
6 43 10
161
5 3
.
.
×
=
m air h
40000 kg coal h
SCM air kg coal
H balance: 198 10 2 361 10 2 1351 10
6 5
4 4
6
. . .
× + × = ⇒ = ×
mol H h mol H h mol H O h
H in coal H in water vapor
2

e j n n
H O content of stack gas
mol H O h
mol h
H O
2
2
2
=
×
× + ×
× =
1357 10
1357 10 2 812 10
100% 4 6%
6
6 7
.
. .
.
e j
Energy balance on stack gas in preheater
References : CO , CO, O , N , H O v at 25 C
2 2 2 2 b g D
Substance
mol h kJ mol mol h kJ mol
CO 2.193 10 2.193 10
CO 0 10 0 10
O 3 10 3 10
N 22 10 72 10
H O 1 10 1 10
in in out out
2
6 6
6 6
6 6
2
6 6
2
6 6
n H n H

. .
. . .
. . .
. . .
. .
× ×
× ×
× ×
× ×
× ×
4 942 9 738
337 3669 337 6 961
706 3758 206 7193
38 3655 38 6 918
357 4266 351 8135
2
( ) ( )
H T H T C dT
i i p
from Table B.8 for inlet = for outlet
Table B.2
B
z
Q n H n H
i i i i
= − = − ×
∑ ∑
out in
kJ h Heat transferred from stack gas
.
101 108
b g
Air preheating
2.83 10 mol dry air/h
7
×
3.61 10 mol H O/h
5
× 2
30°C
1.01 10 kJ/hr
8
×
2.83 10 mol dry air/h
7
×
3.61 10 mol H O/h
5
× 2
Ta (°C)
(We assume preheater is adiabatic, so that Q Q
stack gas air
= − )
Q H n C dT n C dT n C dT
i p i
T
dry air p dry air
T
p
T
a a a
= ⇒ × = = +
z z
∑ z
B B
Δ 101 108
30 30 30
. ( ) ( ) ( )
kJ hr
Table B.2
H O H O
Table B.2
2 2
9.61 (cont'd)

9-78
c.
⇒ × = × − + − + − − × −
⇒ =
−
101 10 8 31 10 30 59 92 30 0 031 30 142 10 30
150
8 5 2 2 3 3 5 4 4
. . ( ) . ( ) . ( ) . ( )
T T T T
T
a a a a
a
D
C
5.95 10 mol O /h
40,000 kg coal/h
6
×
2.24 10 mol N /h
7
×
3.61 10 mol H O( )/h
5
×
150°C(= 149.8°C)
25°C
4400 kg ash/h at 450°C
2
2
2 v
Flue gas at 260°C
2193 10 mol CO /h
6
× 2
0.337 10 mol CO/h
6
×
3.206 10 mol O /h
6
× 2
22.38 10 mol N /h
6
× 2
1.351 10 mol H O( )/h
6
× 2 v
50°C
m (kg H O( )/h)
2 l
30 bars, sat'd
m (kg H O( )/h)
2 v
References for energy balance on furnace: CO , CO, O , N , H O l , coal at 25 C
2 2 2 2 b g °
(Must choose H O l
2 b g since we are given the higher heating value of the coal.)
substance
Coal kg h
Ash kJ kg
O 10 3 10
N 2 24 10 10 mol h
CO 10 kJ mol
CO 10
H O 10
in in out out
6 6
2
6 7
2
6
6
2
6
n H n H
n
H
n
H

.
. . . .
. . . .
. .
. .
. . . .
40000 0
4400 412 25
595 3758 206 7193
3655 2 24 6 918
2193 9 738
0 337 6 961
361 10 48 28 1351 5214
2
5
− −
− −
× ×
× ×
− − ×
− − ×
× ×
b g
b g
b g
b g
(Furnace only — exclude boiler water)
Heat transferred from furnace
Q n H n H n H
i i i i
H
= + −
= ×
F
HG I
KJ − ×
F
HG I
KJ+ × − ×
F
H
GG
I
K
JJ
= − ×
∑ ∑
A
coal i
o
out in
kg
h
kJ
kg
kJ
kg
kJ h
of preheated air
Δ
. . .
.

4 10 2 5 10 2 74 10 122 10
8 76 10
4 4 3 8
8
Heat transferred to boiler water: 0.60(8.76x10
8
kJ/h) = 5.25x10
8
kJ/h
Energy balance on boiler:
Q m H b H
kJ h
kg
h
H O l , 30 , sat'd H O l , C
2 2
b g b g
c h b g
e j
=
F
HG I
KJ − 50D
⇒ × = −
L
N
MM
O
Q
PP ⇒ = ×
A A
525 10 2802 3 209 3 2 02 10
8 5
. . . .
kJ h
kJ
kg
kg steam h
Table B.6 Table B.5
m m
9.61 (cont'd)

9-79
9.62
a.
b.
Basis : 1 mol CO burned. CO O CO
1
2
2 2
+ → , Δ .
Hc
o
kJ mol
= −282 99
( – 0.5)
n
1 mol CO
0 mol O2
3.76n0 mol N2
25°C
n
1 mol CO
0 mol O2
3.76n0 mol N2
1400°C
2
Oxygen in product gas: n n n
1 0 0 0 5
= − = −
mol O fed
1 mol CO react 0.5 mol O
1 mol CO
2
2
b g .
References: CO, CO , O , N at 25 C
2 2 2
D
Substance mol kJ mol mol kJ mol
CO
O
N 3
CO
in in out out
1
2 2
2 3
n H n H
n n H
n n H
H
b g b g b g b g

.
. .

1 0
0 05
76 0 376
1
2 0 0
0 0
− −
−
− −
O g,1400 C : C kJ mol
N g,1400 C : C kJ mol
CO g,1400 C : C kJ mol
2 O
2 N
2 CO
2
Table B.8
2
Table B.8
2
Table B.8
° = =
° = =
° = =
B
B
B
b g
b g
b g
( ) .
( ) .
( ) .
H H
H H
H H
1
2
3
1400 47 07
1400 44 51
1400 7189
D
D
D
E.B.:
Δ Δ
H n H n H n H n n
CO c i i i i
= + − = − + − + + =
∑ ∑
. . . . . .
o
out in
282 99 47 07 0 5 44 51 3 76 7189 0
0 0
b g b g
⇒ =
n0 1094
. mol O2
Theoretical O mol CO mol O mol CO mol O
2 2 2
= =
1 05 0500
b gb g
. .
Excess oxygen:
1094 0500
100% 119%
. .
mol fed mol reqd.
0.500 mol
excess oxygen
−
× =
Increase %XS air ⇒ Tad would decrease, since the heat liberated by combustion would go into
heating a larger quantity of gas (i.e., the additional N2 and unconsumed O2 ).
9.63 a. Basis : 100 mol natural gas ⇒ 82 mol CH 18 mol C H
4 2 6
,
CH (g) 2O (g) CO (g) 2H O(v), kJ / mol
C H (g)
7
2
O (g) 2CO (g) 3H O(v), kJ / mol
4 2 2 2 c
o
2 6 2 2 2 c
o
+ → + = −
+ → + = −
Δ
Δ
.
.
H
H
890 36
1559 9
82 mol CH4
18 mol C2H6
298 K Stack gas at T(°C)
n2 (mol CO2)
n0 (mol air) at 423 K n3 (mol H2O (v))
0.21 O2 (20% XS) n4 (mol O2)
0.79 N2 n5 (mol N2)

9-80
b.
Theoretical oxygen
2 mol O 82 mol CH
1 mol CH
3.5 mol O 18 mol C H
1 mol C H
mol O
2 4
4
2 2 6
2 6
2
= + = 227
Air fed : n1 129714
=
×
=
1.2 227 mol O 1 mol air
0.21 mol O
mol air
2
2
.
C balance : n n
2 2
82 00 1 18 00 2 118 00
= + ⇒ =
. . .
b gb g b gb g mol CO2
H balance : 2 82 00 4 18 00 6 218 00
3 3
n n
= + ⇒ =
. . .
b gb g b gb g mol H O
2
20% excess air, complete combustion ⇒ = =
n4 2
0 2 227 4540
. .
b gb gmol O mol O2
N balance : mol N
2 2
n5 0 79 129714 1024 63
= =
. . .
b gb g
Extents of reaction: ξ ξ
1 1
82 18
= = = =
n n
CH C H
4 2 6
mol, mol
Reference states: CH g , C H g , N g , O g , H O l at 298 K
4 2 6 2 2 2
b g b g b g b g b g
(We will use the values of Δ
Hc
o
2 b g as a combustion
product, and so must choose the liquid as a reference state for water.)

,
H T C T
H C T
i pi
p
b g b g
b g b g
= −
= + −
298
298 298
K for all species but water
K K for water
v,H O H O
2 2
Δ
Substance
CH
C H
O
N
CO
H O v
in in out out
4
2 6
2
2
2
n H n H
T
T
T
T

.
.
. . . .
. . . .
. .
. . .
82 00
18 00 0
272 40 414 4540 0 0331 298
1024 63 391 1024 63 0 0313 298
118 00 0 0500 298
218 00 44 013 0 0385 298
2
− −
− −
−
−
− − −
− − + −
b g
b g
b g
b g b g
Energy balance : ΔH = 0
ξ ξ
1 2 0
Δ Δ

H H n H n H
i i i i
c
o
CH
c
o
C H
out in
4 2 6
e j e j
+ + − =
∑ ∑
mol CH kJ mol mol C H kJ mol
4 2 6
⇒ − + −
+ + + + −
+ − − =
82 00 890 36 18 00 1559 90
4540 0 0331 1024 63 0 0313 118 00 0 0500 218 00 0 0385 298
218 00 44 01 272 40 414 1024 63 391 0
. . . .
. . . . . . . .
. . ( . )( . ) ( . )( . )
b gb g b gb g
b gb g b gb g b gb g b gb gb g
b gb g
T
Solving for using E - Z Solve K
T T
⇒ = 2317
Increase % excess air ⇒ T decreases.
out (Heat of combustion has more gas to heat)
% methane increases ⇒ Tout might decrease. (lower heat of combustion, but heat released goes
into heating fewer moles of gas.)
9.63 (cont’d)

9-81
9.64
a.
b.
C H O g 4O g 3CO g 3H O l , kJ mol
3 6 2 2 2
o
b g b g b g a f
+ → + = −
Δ .4
Hi 1821
Basis :
1410 m STP feed gas 10 mol 1 min
22.4 m STP 60 s
mol s feed gas
3 3
3
b g
b g
min
= 1049
Stochiometric proportion:
1 mol C H O 4 mol O 4 3.76 15.04 mol N mol
3 6 2 2
⇒ ⇒ × = ⇒ + + =
1 4 1504 20 04
. .
b g
yC H O
3 6 3 6
3 6
1 mol C H O
20.04 mol
mol C H O
mol
= = 0 0499
. , yO 2
2
4
20.04
mol O mol
= = 0 1996
.
Feed gas
1049 mol/s
0.0499 C H O
3 6
0.1496 O2
0.7505 N2
T (°C), 150 mm Hg
f
Rel. satn = 12.2%
Preheat
Q (kW)
1
Feed gas
562°C
Reaction
Product gas
n1 (mol CO /s)
2
n2 (mol H O/s)
2
n3 (mol N /s)
2
T (°C)
a
Cooling
Product gas
350°C
Q (kW)
2
Relative saturation
mm Hg
0.122
mm Hg C
C H O C H O
Table B.4
o
3 6 3 6
= ⇒ =
⇒ = = =
∗
∗
12 2% 0122
0 0499 1500
61352 50 0
. .
.
. .
y P p T
p T
f
f
d i
b gb g
Feed contains 1049 mol s 0.0499 C H O mol mol C H O s
3 6 3 6
b gb g= 52 34
.
1049 0.1996 mol O s
2
b gb g= 209 4
.
1049 0.7505 mol N s
2
b gb g= 787 3
.
⇒ Product contains
n
n
n
1
2
3
52 34 3 157 0
52 34 3 157 0
787 3
14 25
14 25% H
71.5% N
= =
= =
=
U
V
|
W
|
⇒
. .
. .
.
.
.
b gb g
b gb g
mol CO s
mol H O s
mol N s
mole% CO
O
2
2
2
2
2
2
References : C H O g , O , N , H O l , CO at 25 C
3 6 2 2 2 2
b g b g D
( )
( )
( )
out out
in in
3 6
2
2
2
2
ˆ
ˆ
Substance
(mols) (kJ/mol) (mols) (kJ/mol)
(562 C)
C H O 52.34 67.66
O 209.4 17.72
N 787.3 17.18 787.3 0.032 25
CO 157.0 0.052 25
H O 157.0 44.013 0.040 25
a
a
a
a
n H
n H
T
T
T
T
− −
− −
−
− − −
− − + −
D

Energy balance on reactor:
Δ Δ
H n H n H n H
c i i i i
= + − =
∑ ∑
C H O
o
out in
3 6
kJ s
0 b g
( ) ( ) ( ) 4
kJ
1821.1 39.638 25 157.0 44.013 2.078 10 0 2780 C
5234 mol s
mol
a a
T T
⎛ ⎞
− + − + − × = ⇒ = °
⇒ ⎜ ⎟
⎝ ⎠
0.1996 O2

9-82
c.
Preheating step: References: C H g , O , N at 25 C
3 6 2 2
b g °
Substance
mol / s) (kJ / mol)
(50 C)
(mol / s) kJ / mol)
(562 C)
C H O
O
N
in in out out
3 6
2

(

(
. . . .
. . . .
. . . .
n H n H
D D
52 34 315 52 34 67 66
209 4 0826 209 4 17 72
787 3 0 775 787 3 16 65
2
E.B. ⇒ .
Q n H n H
i i i i
1
4
194 10
= − = ×
∑ ∑
out in
kW
Cooling step. References: CO (g), H O v , N (g) at 25 C
2 2 2
a f D
Substance
mol) (kJ / mol)
C
(mol) (kJ / mol)
C
CO
H O
N
in in out out
2
2
n H n H
(

. . . .
. . . .
. . . .
2871 350
157 0 142 3 157 0 16 25
157 0 10815 157 0 12 35
787 3 88 23 787 3 10 08
2
D D
e j e j
E.B. ⇒ Q n H n H
i i i i
2 = − = − ×
∑ ∑

out in
4
9.64 10 kW
Exchange heat between the reactor feed and product gases.
9.65 a. Basis : 1 mol C5H12 (l)
C H (l) O (g) 5CO (g) 6H O(v), kJ / mol
5 12 2 2 2 c
o
+ → + = −
8 3509 5
Δ .
H
1 mol C5H12 (l) n2(mol CO2)
n3 (mol H2O (v))
n4 (mol O2)
n0 (mol O2) , 75°C Tad(o
C)
30% excess
Theoretical oxygen
1 mol C H 8 mol O
1 mol C H
mol O
5 12 2
5 12
2
= = 8
30% 13 8 10 4
0
excess mol O2
⇒ = × =
n . .
C balance: n n
2 2
1 5 5
= ⇒ =
b gb g mol CO2
H balance: 2 1 12 6
3 3
n n
= ⇒ =
b gb g mol H O
2
30% excess O2, complete combustion ⇒ = =
n4 0 3 8 2 4
. .
b gb gmol O mol O
2 2
Reference states: C H l , O g , H O l , CO (g) at 25 C
5 12 2 2 2
o
b g b g b g
Hc
0
2 b g as a
combustion product, and so must choose the liquid as a reference state for water)
9.64 (cont'd)

9-83
substance
C H
O
CO
H O
in in out out
5 12
2
n H n H
H H
H
H

.
. .
.
.
100 0
10 40 2 40
500
6 00
2 1 2
2 3
4
− −
− −
− −
( ) ,
( )
H C dT i
H C dT
i p i
T
p
T
= =
= +
z
z
C for H O(v)
v
o
H O(v) 2
2
2 3
25
25
25
Δ e j
( .
H H
1 75 148
=
B
O
o
Table B.8
2
C) = kJ / mol
Substituting ( ) from Table B.2 :
kJ
mol
kJ
mol
C
H T T T T
H T T T T
p i
ad ad ad ad
ad ad ad ad
( . . . . . )
( . . . . . )
2
5 2 8 3 12 4
3
5 2 8 3 12 4
0 0291 0579 10 0 2025 10 0 3278 10 0 7311
0 03611 21165 10 0 9623 10 1866 10 0 9158
= + × − × + × −
= + × − × + × −
− − −
− − −
. ( . . . . . )
H T T T T
ad ad ad ad
4
5 2 8 3 12 4
4401 003346 03440 10 02535 10 08983 10 0838
= + + × + × − × −
− − − kJ
mol
⇒ = + + × + × − ×
− − −
. ( . . . .
H T T T T
ad ad ad ad
4
5 2 8 3 12 4
4317 003346 03440 10 02535 10 08983 10 )
kJ
mol
Energy balance :ΔH = 0
n H n H n H
i i i i
C H c
o
C H l)
out in
5 12
5 12
Δ
(
e j + − =
∑ ∑ 0
( )( . ) ( . ) ( . ) ( . ) ( . )( )
1 3509 5 2 40 500 6 00 10 40 0
2 3 4 1
mol C H kJ / mol
5 12 − + + + − =
H H H H
Substitute for through
ad ad ad ad
ad ad ad ad ad
ad ad
o
kJ / mol = 0
Check :
Solving for using E-Z Solve C

( . . . . ) .
( ) . . . . .
.
.
.
H H
H T T T T
f T T T T T
T T
1 4
5 2 8 3 12 4
5 2 8 3 12 4
12
14
04512 14 036 10 3777 10 4 727 10 3272 20
3272 20 04512 14 036 10 3777 10 4 727 10 0
3272 20
4727 10
6922 10
4414
Δ = + × − × + × −
⇒ = − + + × − × + × =
−
×
= − ×
⇒ =
− − −
− − −
−
b. Terms Tad % Error
1 7252 64.3%
2 3481 –21.1%
3 3938 –10.8%
9.65 (cont'd)

9-84
c. T f(T) f'(T) Tnew
7252 6.05E+03 3.74 5634
5634 1.73E+03 1.82 4680
4680 3.10E+02 1.22 4426
4426 1.41E+01 1.11 4414
4414 3.11E-02 1.11 4414
d. The polynomial formulas are only applicable for T ≤ 1500°C
9.66 5.5 L/s at 25°C, 1.1 atm

n 1(mol CH4/s)
25% excess air
n 3 (mol O2/s)

n 2 (mol O2/s) 3.76
n 2 (mol N2/s)
3.76
n 2 (mol N2/s)
n 4 (mol CO2/s)
n 5 (mol H2O/s)
150°C, 1.1 atm T(°C), 1.05 atm
2 2 2
CH O CO H O
4 2 2 2
+ → +
Fuel feed rate :
L 273 K 1.1 atm mol
s 298 K 1.0 atm 22.4 L(STP)
mol CH s
4
= =
550
0 247
.
. /
Theoretical O mol O s
excess air mol O / s
mol N / s
Complete combustion = 0.247 mol / s, mol CO / s, mol H O / s
mol O fed / s 0.494 mol consumed / s
mol O s
2 2
2
2
4 2 5 2
3 2
2
= × =
⇒ = =
⇒ × =
⇒ = = =
= −
=
2 0 247 0 494
25% 125 0 494 0 6175
376 0 6175 2 32
0 247 0 494
0 6175
0124
2
1
. . /
. ( . ) . ,
. . .
. .
.
. /
n
n n n
n
ξ
Re , , , ,

(

(

(

(
.
. .
. .
.
.
ferences: CH O N CO H O at 25 C
Substance
mol / s) kJ / mol) mol / s) kJ / mol)
CH
O
N
CO
H O
4 2 2 2 2
o
in in out out
4
2
2
2
2
n H n H
H H
H H
H
H
0 247 0
0 6175 0124
2 32 2 32
0 247
0 497
1 3
2 4
5
6
− −
− −
− −
4
Table B.8
o
1 2,
Table B.8
o
2 2,
o
c CH
25
ˆ ˆ (O 150 C) 3.78 kJ/mol
ˆ ˆ (N 150 C) 3.66 kJ/mol
ˆ
( ) 890.36 kJ/mol
ˆ , 3 5
T
i pi
H H
H H
H
H C dT i
=
=
Δ = −
= = −
∫
9.65 (cont’d)
Adiabatic
Reactor
4 2
(mol CO /s)
n

9-85
a.
b.
( ) ( )
H H C dT
b p
T
= + z
Δ v H O(25 C) H O(v)
25
2 2
D
4
o
c CH out out in in
ˆ
Table B.2 for ( ), ( ) 44.01 kJ/mol
v H O
2
5 2 2 8 3 3
12 4
Energy Balance
ˆ ˆ ˆ
( ) 0
0.247( 890.36) 0.494(44.01) 0.0963( 25) 1.02 10 ( 25 ) 0.305 10 ( 25 )
1.61 10 ( 25
pi
C T H
H H n H n H
T T T
T
ξ
Δ =
− −
−
Δ = Δ + − =
− + + − + × − + × −
− × −
∑ ∑

4
) 0.6175(3.78) 2.32(3.66) 0
− − =
5 2 8 3 12 4 o
211.4 0.0963 1.02 10 0.305 10 1.61 10 0 1832 C
ad ad ad ad
T T T T T
− − −
⇒ − + + × + × − × = ⇒ =
2
2 2 2
o
H O 2
Table B.3
* *
H O H O H O
In product gas,
1832 C, 1.05 760 798 mmHg
0.494 mol/s
y 0.155 mol H O/mol
(0.124 2.32 0.247 0.494) mol/s
Raoult's law : y ( ) (0.155)(798) 124 mmHg 56 C
Degr. superheat
dp dp
T P
P p T p T
= = × =
= =
+ + +
= ⇒ = = =
⇒
D
= 1832 C 56 C = 1776 C
−
D D D
9.67 a. CH l O g CO (g) + 2H O(v)
4 2 2 2
( ) ( )
+ →
2
Basis 1 mol CH4
:
Theoretical oxygen
1 mol CH 2 mol O
1 mol CH
mol O
4 2
4
2
= = 2 00
.
30 130 2 00 2 60 376 2 60 9 78
excess air mol O mol N
2 2
% . ( . ) . , . . .
⇒ = ⇒ × =
1 mol CH4 n2 (mol CO2)
2.60 mol O2 n3 (mol H2O)
9.78 mol N2 n4 (mol O2 )
25° C, 4.00 atm
Complete combustion mol CO , mol H O
2.00 mol O consumed mol O mol O
2 2 3 2
2 4 2 2
⇒ = =
⇒ = − =
n n
n
100 2 00
2 60 2 00 0 60
. .
( . . ) .
Internal energy of reaction: Eq. (9.1-5) ⇒ Δ Δ

U H RT
c
o
c
o
i
gaseous
products
i
gaseous
reactants
= − −
F
H
GGG
I
K
JJJ
∑ ∑
ν ν
⇒ = −
F
HG I
KJ−
− −
= −
Δ .
. )
.
Uc
o
CH
3
4
kJ
mol
J 298 K (1+ 2 kJ
mol K 10 J
kJ
mol
e j 890 36
8 314 1 2 1
890 36
( ) ( )
( )(
U C dT C R dT
C T U C R T
v p
T
T
p p
= −
⇒ = − −
⇒ z
z Ideal Gas
g
25
g
o
If is independent of C)
25
25
9.66 (cont'd)

9-86
b.
c.
Reference states: CH g , N g , O g , H O l , CO (g) at 25 C
4 2 2 2 2
o
b g b g b g b g
Hc
0
2 b g as a
combustion product, and so must choose the liquid as a reference state for water.)
Substance
CH
O
N
CO
H O v
in in out out
4
2
2
n U n U
U
U
U
U

.
. .
. .
.
.
100 0
2 60 0 0 60
9 78 0 9 78
100
2 00
2 1
2 2
3
4
− −
− −
− −
b g
( )
( ) ( )
) .
( . . )( ) ( . . )
( . . )( ) ( .
U C R T
U C R T H R T C R T
C R
U T T
U T T
i p
p p
p i
= − −
= + − − = − + − −
= ×
= − × − = −
= − × − = −
B
−
−
−
Part a
g 2
v
o
g v
o
g ref g 2
g
)( for all species except H O(v)
C )( C )( for H O(v)
Substituting given values of ( and kJ / mol yields
10 kJ / mol kJ / mol
10 kJ / mol
25
25 25 25 25
8 314 10
0 033 8 314 25 0 02469 0 6172
0 032 8 314 25 0 02369 0
3
1
3
2
3
Δ Δ
e j e j
. )
5922 kJ / mol
( . . )( ) ( . . )
U T T
3
3
0 052 8 314 25 0 04369 10922
= − × − = −
−
10 kJ / mol kJ / mol
. . ( . . )( )
U T
4
3 3
44 01 8 314 0 040 8 314 25
= − ×
⋅
F
HG I
KJ
L
NM O
QP − × −
− −
kJ
mol
10
kJ
mol K
(298 K) + 10
kJ
mol
⇒ = − × − = −
−
. ( . . )( ) ( . . )
U T T
4
3
4153 0 052 8 314 25 0 03167 40 74
kJ
mol
+ 10
kJ
mol
kJ
mol
Energy Balance
kJ / mol
C
Ideal Gas Equation of State
K
K
atm atm
CH c
o
CH
out in
1 2 3 4
Substituting 1 through 4
o
f
i
f
i
f
4
4
Q n U n U n U
Q U U U U
T T
P
P
T
T
P
i i i i
U U
= + − =
⇒ = − + + + + =
− = ⇒ =
⇒ = ⇒ =
+
+
F
HG I
KJ× =
∑ ∑
Δ
( . ) . ( . ) ( . ) ( . ) ( . )
. .
( )
( )
. .

e j
b g
0
100 890 36 0 60 9 87 100 2 00 0
0 3557 81619 0 2295
2295 273
25 273
4 00 34 5
– Heat loss to and through reactor wall
– Tank would expand at high temperatures and pressures
9.67 (cont’d)

9-87
9.68
b. 1 mol natural gas
y n
y n
y n
n
CH 4 CO 2
C H 2 6 H O 2
C H 3 8 N 2
O 2
4 2
2 6 2
3 8 2
2
(mol CH / mol) (mol CO
(mol C H / mol) (mol H O)
(mol C H / mol) (mol N
mol O
)
)
)
Humid air
na (mol air)
ywo (mol H20(v)/mol)
(1-ywo) (mol dry air/mol)
0.21 mol O2/mol DA
0.79 mol N2/mol DA
Basis : 1 g-mole natural gas
CH (g) O (g) CO (g) H O(v)
C H (g) O (g) CO (g) H O(v)
C H (g) O (g) CO (g) H O(v)
4 2 2 2
2 6 2 2 2
3 8 2 2 2
+ → +
+ → +
+ → +
2
7
2
2 3
5 3 4
Theoretical oxygen
2 mol O (mol CH
1 mol CH
3.5 mol O (mol C H
1 mol C H
5 mol O (mol C H
1 mol C H
2 CH 4
4
2 C H 2 6
2 6
2 C H 3 8
3 8
4 2 6 3 8
:
) ) )
y y y
+ +
= +
( 2 +5
CH C H C H
4 2 6 3 8
y y y
35
. )
Excess oxygen:
100
( 2 +5 mol O
xs
CH C H C H 2
4 2 6 3 8
0 21 1 1 35
. ( ) . )
n y
P
y y y
a wo
− = +
F
HG I
KJ +
=
100
( 2 +5
1
0.21(1
mol air
Feed components
( ( (
N in product gas: = ( mol N
xs
CH C H C H
O in N in H O in
2 N N in 2
4 2 6 3 8
2 2 2
2 2
⇒ +
F
HG I
KJ +
−
= − = − =
n
P
y y y
y
n n y n n y n n y
n n
a
w
a wo a wo a wo
1 35
0 21 1 0 79 1
0
. )
)
) . ( ), ) . ( ), )
)
CO in product gas
1 mol CO (mol CH
1 mol CH
2 mol CO (mol C H
1 mol C H
3 mol CO (mol C H
1 mol C H
( mol CO
2
CO
2 CH 4
4
2 C H 2 6
2 6
2 C H 3 8
3 8
CH C H C H 2
2
4 2 6 3 8
4 2 6 3 8
:
) ) )
)
n
n n n
n n n
= + +
= + +
2 3
H Oin product gas:
1 mol H O (mol CH
1 mol CH
3 mol H O (mol C H
1 mol C H
4 mol O (mol C H
1 mol C H
[2 + (1- )] mol H O
2
H O
2 CH 4
4
2 C H 2 6
2 6
2 C H 3 8
3 8
CH C H C H a wo 2
2
4 2 6 3 8
4 2 6 3 8
n
n n n
n n n n y
= + +
= + +
) ) )
3 4
O in product gas :
P
( 2 +5 mol O
2 O
xs
CH C H C H 2
2 4 2 6 3 8
n n n n
= +
100
35
. )

9-88
c. References : C(s), H g) at 25 C
( (
2
o
CH f
o
CH CH
4 4 4
(
( ) )
H T) H C dT
p
T
= + z
Δ
25
Using ( from Table B.1 and ( from Table B.2
f
o
CH CH
4 4
Δ H Cp
) )
kJ / mol+ (0.03431+5.469 10 kJ / mol
CH
25
4
( ) . . . )
H T T T T dT
T
= − × + × − ×
F
H
GG
I
K
JJ
− − −
z
7485 03661 10 1100 10
5 8 2 12 3
⇒ = − × × + × − ×
− − − −
( ) [ . . . ]
H T T T T T
CH
2
4
+3.431 10 +2.734 10 kJ / mol
7572 0122 10 2 75 10
5 2 8 3 12 4
Substance
CH
C H
C H
O
N
CO
H O
in in out out
4 1
2 6 2
3 8 3
2 7
2 5 8
2 9
2 10
n H n H
n H
n H
n H
n H n H
n H n H
n n H
n H

1
2
3
4 4 7
5 8
6 9
10
− −
− −
− −
−
− −
Δ
Δ
H n H n H
H a b T c T d T e T
n H n H T n H T
H n a b T c T d T e T n H T n H T
i i
i
i i
i
i i i i i i
i i
i
i i
i
i i
i
i i i i i i
i
i i
i
i i
i
= −
= + + + +
= +
⇒ = + + + + − −
= =
= = =
= =
∑ ∑
∑ ∑ ∑
∑ ∑
( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
out out in in
in in in f in a
out out in f in a
4
7
1
6
2 3 4
1
6
1
3
4
6
2 3 4
4
7
1
3
=
= = = =
=
= =
= =
∑
∑ ∑ ∑ ∑
∑
∑ ∑
∑ ∑
⇒ = + + + +
− −
= + + + +
= −
4
6
1
7
2
1
7
3
1
7
4
1
7
4
7
1
3
4
6
0 1 2
2
3
3
4
4
0
1
7
1
3
ΔH n a n b T n c T n d T n e T
n H T n H T
T T T T
n a n H T
i i
i
i i i i
i
i i
i
i i
i
i
i i
i
i i
i
i i
i
i i
i
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
out out out out out
in f in a
out in f
where
α α α α α
α −
= =
= =
=
= =
= =
∑
∑ ∑
∑ ∑
( ) ( )
( ) ( )
( ) ( )
n H T
n b n c
n d n e
i i
i
i i
i
i i
i
i i
i
i i
i
in a
out out
out out
4
6
1
1
7
2
1
7
3
1
7
4
1
7
α α
α α
.
9.68 (cont’d)

9-89
Run 1 Run 2 Run 3 Run 4 Run 5 Run 6
yCH4 0.75 0.86 0.75 0.75 0.75 0.75
yC2H6 0.21 0.1 0.21 0.21 0.21 0.21
yC3H8 0.04 0.04 0.04 0.04 0.04 0.04
Tf 40 40 150 40 40 40
Ta 150 150 150 250 150 150
Pxs 25 25 25 25 100 25
ywo 0.0306 0.0306 0.0306 0.0306 0.0306 0.1
nO2i 3.04 2.84 3.04 3.04 4.87 3.04
nN2 11.44 10.67 11.44 11.44 18.31 11.44
nH2Oi 0.46 0.43 0.46 0.46 0.73 1.61
HCH4 -74.3 -74.3 -70 -74.3 -74.3 -74.3
HC2H6 -83.9 -83.9 -77 -83.9 -83.9 -83.9
HC3H8 -102.7 -102.7 -93 -102.7 -102.7 -102.7
HO2i 3.6 3.6 3.6 6.6 3.6 3.6
HN2i 3.8 3.8 3.8 6.9 3.8 3.8
HH2Oi -237.6 -237.6 -237.6 -234.1 -237.6 -237.6
nCO2 1.29 1.18 1.29 1.29 1.29 1.29
nH2O 2.75 2.61 2.75 2.75 3.02 3.9
nO2 0.61 0.57 0.61 0.61 2.44 0.61
nN2 11.44 10.67 11.44 11.44 18.31 11.44
Tad 1743.1 1737.7 1750.7 1812.1 1237.5 1633.6
alph0 -1052 -978.9 -1057 -1099 -1093 -1058
alph1 0.4892 0.4567 0.4892 0.4892 0.7512 0.5278
alph2 0.0001 1.00E-04 0.0001 0.0001 0.0001 0.0001
alph3 -3.00E-08 -3.00E-08 -3.00E-08 -3.00E-08 -4.00E-08 -2.00E-08
alph4 3.00E-12 3.00E-12 3.00E-12 3.00E-12 4.00E-12 2.00E-12
Delta H 3.00E-07 9.00E-06 -4.00E-07 -1.00E-04 -1.00E-05 6.00E-04
Species a b c d e
x 10^2 x 10^5 x 10^8 x 10^12
CH4 -75.72 3.431 2.734 0.122 -2.75
C2H6 -85.95 4.937 6.96 -1.939 1.82
C3H8 -105.6 6.803 11.3 -4.37 7.928
O2 -0.731 2.9 0.11 0.191 -0.718
N2 -0.728 2.91 0.579 -0.203 0.328
H20 -242.7 3.346 0.344 0.254 -0.898
CO2 -394.4 3.611 2.117 -0.962 1.866
d.
9.68 (cont’d)

9-90
9.69
(mol CH /h)
n14
25°C
4
(mol/h)
n15
25°C
0.96 mol O /mol
2
0.04 mol N /mol
2
(mol CH /h)
n14 4
0.96n15 (mol O /h)
2
0.04n15 (mol N /h)
2
converter
Feed gas, 650°C
Preheaters
Converter
product
Tad (°C)
quench
Converter
product
38°C
filter
(mol CH /h)
n6 4
(mol C H /h)
n7 2 2
(mol H /h)
n8 2
(mol CO/h)
n9
n10 (mol CO /h)
2
(mol H O/h)
n11 2
(mol N /h)
n12 2
(mol C( )/h)
n13 s
(mol C( )/h)
n13 s
(mol CH /h)
n6 4
(mol C H /h)
n7 2 2
(mol H /h)
n8 2
(mol CO/h)
n9
n10 (mol CO /h)
2
(mol H O/h)
n11 2
(mol N /h)
n12 2
absorber
0.917 (mol DMF/h)
n1
stripper
Lean solvent
Rich solvent
(mol/h)
n1
0.0155 mol C H /mol
2 2
0.0063 mol CO /mol
2
0.00055 mol CO/mol
0.00055 mol CH /mol
4
0.0596 mol H O/mol
2
0.917 mol DMF/mol
Absorber off-gas
(mol H /h)
n8 2
(mol N /h)
n12 2
0.988 (mol CO/h)
n9
0.950 (mol CH /h)
n6 4
0.006 (mol C H /h)
n7 2 2
0.991 mol C H ( )/mol
2 2
0.00059 mol H O/mol
2
0.00841 mol CO /mol
2
(mol/h)
n1
5000 kg/h Product gas
Basis:
(mol CO/h)
n2
(mol CH /h)
n3 4
(mol H O( )/h)
n4 v
2
n5 (mol CO /h)
2
Stripper off-gas
g
Average M.W. of product gas:
M = + + =
0 991 26 04 0 00059 18 016 0 00841 44 01 2619
. . . . . . .
b g b g b g g mol
Molar flow rate of product gas: n0 7955
= =
5000 kg 10 g 1 mol 1 day
day 1 kg 26.19 g 24 h
mol h
3
Material balances -- plan of attack (refer to flow chart):
Stripper balances: C H
2 2 ⇒ n1 , CO ⇒ n2 , CH4 ⇒ n3 , H O
2 ⇒ n4 , CO2 ⇒ n5
Absorber balances: CH4 ⇒ n6 , C H
2 2 ⇒ n7 , CO ⇒ n9 , CO2 ⇒ n10 , H O
2 ⇒ n11
5.67% soot formation
converter C balance
converter H balance
R
S
T
U
V
W
⇒ ⇒
n n n
13 14 8
, ,
Converter O balance converter N balance
2
⇒ ⇒
n n
15 12
,
Stripper balances:
C H : 0.0155 mol h mol h
2 2 n n
1 1
5
0 991 7955 5086 10
= ⇒ = ×
. .
b g
CO: 0.00055 mol CO h
b ge j
5086 10 79 7
5
2 2
. .
× = ⇒ =
n n
CH : 0.00055 mol CH h
4 4
b ge j
5086 10 79 7
5
3 3
. .
× = ⇒ =
n n
H O: 0.0596 = 30308 mol H O h
2 2
b ge j b gb g
5086 10 0 00059 7955
5
4 4
. .
× = + ⇒
n n
CO : 0.0068 = 3392 mol CO h
2 2
b ge j b gb g
5086 10 0 00841 7955
5
5 5
. .
× = + ⇒
n n
Absorber balances
CH : 0.00055 mol CH h
4 4
n n n
6 6
5
6
0 950 5086 10 5595
= + × = ⇒
. .
b ge j

9-91
a.
b.
C H : mol C H h
2 2 2 2
n n n
7
5
7 7
0 0155 5086 10 0 006 7931
= × + ⇒ =
. . .
b ge j
CO: 0.00055 = 23311 mol CO h
n n n
9 9
5
9
0 988 5086 10
= + × ⇒
. .
b ge j
CO : 3458 mol CO h
2 2
n10
5
0 0068 5086 10
= × =
. .
b ge j
H O: 0.0596 30313 mol H O h
2 2
n11
5
5086 10
= × =
b ge j
.
Soot formation:
n n
n n
13 14
13 14
0 0567
0 0567 1
=
⇒ =
. )
.
b g b g
(mol CH 1 mol C
h 1 mol CH
4
4
Converter C balance:
n n
n n
14 13
14 13
5595 1 7931 2 23311 1 3458 1
48226 2
= + + + +
⇒ = +
mol CH h mol C mol CH
4 4
b g
n n
13 14
2899 51120
mol C s h mol CH h
4
b g ,
Converter H balance:
51120 mol CH 4 mol H
h 1 mol Ch
4
4
CH C H H H O
4 2 2 2 2
= + + +
5595 4 7931 2 2 30313 2
8
n
⇒ n8 52816
= mol H h
2
Converter O balance: 0 96 2 3458 2 30313 1
15
. n
= + +
23311 mol CO 1 mol O
h 1 mol CO
CO H O
2 2
⇒ =
n15 31531 mol h
Converter N balance: 0.04 mol N h
2 2
b gb g
31531 1261
12 12
n n
⇒ =
Feed stream flow rates
VCH
4
3
4
4
mol CH 0.0244 m STP
h 1 mol
SCMH CH
= =
51120
1145
b g
VO
2
3
2
2
O N 0.0244 m STP
h 1 mol
N
=
+
= +
31531 mol
706 SCMH O
2
2
b g b g b g
Gas feed to absorber
5595 mol CH h
7931 mol C H h
23311 mol CO h
3458 mol CO h
30313 mol H O h
52816 mol H h
1261 mol N h
1.2469 10 mol h
kmol h ,
mole% CH , .4 % C H , 18.7% CO ,
.8% CO , O , , .0% N
4
2 2
2
2
2
2
5
4 2 2
2 2 2 2
×
U
V
|
|
|
|
W
|
|
|
|
⇒ 125
4 5 6
2 24.3% H 42 4% H 1
.
.
Absorber off-gas
52816 mol H h
1261 mol N h
23031 mol CO h
5315 mol CH h
41.6 mol C H h
8 10 mol h
kmol h
H .5% N
.4% CH .06% C H
2
2
4
2
4
2 2
4 2
2
2
2471
82.5
64.1 mole% 1 27.9% CO,
6 0
.
,
, ,
,
×
U
V
|
|
|
W
|
|
|
⇒
9.69 (cont'd)

9-92
c.
d.
e.
f.
Stripper off-gas
279.7 mol CO h
279.7 mol CH h
30308 mol H O h
3392 mol CO h
10 mol h
kmol h .82% CO, 0.82% CH O, 9.9% CO
4
2
4
4 2
34259
34.3 0 88.5% H2
.
, ,
×
U
V
|
|
W
|
|
⇒
DMF recirculation rate = ×
F
HG I
KJF
HG I
KJ
0917 5086 10
1
5
. .
mol
h
kmol
10 mol
3
= 466 kmol DMF h
Overall product yield = =
0 991 7955
0154
.
.
b gb gmol C H in product gas
51120 mol CH in feed h
mol C H
mol CH
2 2
4
2 2
4
The theoretical maximum yield would be obtained if only the reaction 2CH C H 3H
4 2 2 2
→ +
occurred, the reaction went to completion, and all the C H
2 2 formed were recovered in the
product gas. This yield is (1 mol C2H2/2 mol CH4) = 0.500 mol C2H2/2 mol CH4.
The ratio of the actual yield to the theoretical yield is 0.154/0.500 = 0.308.
Methane preheater

Q H n C dT
p
CH
Table B.2
CH
3
4
4
mol 32824 J 1 h 1 kJ
h mol 3601 s 10 J
kW
= = = =
B
z
Δ 14
25
650 51120
466
d i
Oxygen preheater
. ( , . ( ,
. . . .
Q H n H n H
O
Table B.8
2
Table B.8
2
2
O C) N C)
mol
h
kJ
mol C
h
3600 s
= = +
=
F
HG I
KJ × + ×
⋅
L
NM O
QPF
HG I
KJ =
B B
Δ 0 96 650 0 04 650
31531 0 96 20135 0 04 18 99
1
176 kW
15 15
D D
D
b g
2 2 2
b g b g b g b g °
Substance C
CH 51120
O
N
C H
H mol h
CO kJ mol
CO
H O
C s
in in out out out
4
2
2 2
2
2
2

. .
.
.
.
.
.
.
n H n H T
C dT
C dT
C dT
C dT n
C dT H
C dT
C dT
C dT
p
T
p
T
p
T
p
p
p
p
p
a
a
a
650
42 026 5595 74 85
30270 20125
1261 18 988 1261
7931 226 75
52816
23311 11052
3458 3935
30313 24183
2899
25
2
35
25
°
− − +
− −
− − +
− −
− − − +
− − − +
− − − +
− −
z
z
z
z
z
z
z
z
b g b g
b g
b g
b g
9.69 (cont'd)

9-93

H H C dT
i i pi
T
= +
⋅°
z
kJ mol
kJ mol C
Δ 0
25
.
n H
i i
in
kJ h
∑ = − ×
1575 106
.
n H C C C
C C C C dT
C dT
i i p p p
T
p p p p v
p s
Tad
out
CH N C H
H CO CO H O 3
C 3
kJ h
kJ
10 J
kJ
10 J
4 2 3 2
out
2 2 2
∑ z
z
= − × + + +
L
NM
+ + + +
O
QP
+ ×
+
9 888 10 5595 1261 7931
52816 23311 3458 3013
1
1
6
25
298
273
d i d i d i
d i d i d i d i
d i
b g
b g
We will apply the heat capacity formulas of Table B.2, recognizing that we will probably
push at least some of them above their upper temperature limits
. . . .
. .
.
n H T T dT
T
T
dT
i i
T
T
ad
ad
out
kJ h
∑ z
z
= − × + + − × − ×
+ + −
×
F
HG
I
KJ
− −
+
9 888 10 3902 12185 59885 10 10162 10
32 411 0 031744
14179 10
6 4 2 7 3
25
6
2
298
273
e j
. . . .
.
n H T T T T
T
i i a a a a
a
out
∑ = − × + + − × − × +
×
+
− −
1000 10 3943 0 6251 1996 10 2 5405 10
1418 10
273
7 2 4 3 8 4
6
Energy balance: Δ
H n H n H
i i i i
= − =
∑ ∑
out in
0
⇒ f T T T T T
T
c c c c c
c
b g= − × + + − × − × +
×
+
=
− −
8485 10 3943 06251 1996 10 25405 10
1418 10
273
0
6 2 4 3 8 4
6
. . . .
.
C
E-Z Solve
o
Tc = 2032 .
9.69 (cont’d)

9-94
9.70 a.
W = H2O 1
o
[kg W (v)/d]
100 C
m

F
24 000
, kg sludge / d, 22 C
0.35 solids, 0.65 W(l)
o
DRYER
o
2 (kg conc. sludge/d), 100 C
0.75 solids, 0.25 W(l)
m

INCINERATOR
Waste gas
3 [kg W(v)/d]
4B, sat'd
m

3
o
C
[kg W(l)/d]
20 C
m

(
Q
3 kJ/d) 3 [kg W(l)/d]
4B, sat'd
m

6
4
2 6
(kg gas/d)
kmol CH
0.90
kmol
kmol C H
0.10
kmol
m

110o
C
D
Q4 (kJ / d)
4
2 2
o
2
2 2
(kg oil/d) Stack gas
0.87 C CO , H O(v)
0.10 H 125 C SO
0.0084 S O , N
0.0216 ash ash
m

7
o
(kg air/d)
25 C
m

0
5
o
(kJ/d)
E (kg air/d)
25 C
Q
m

Q2
BOILER

Q1

9-95
Solids balance on dryer:
0 35 24 000 0 75 11200
2 2
. , .
× = ⇒ = ⇒
kg / d kg / d F 11.2 tonnes/ d (conc. sludge)
n n
Mass Balance on dryer: 24 000 11200 12 800
1 1
, ,
= + ⇒ =
n n kg / d
Energy balance on sludge side of dryer:
References : H O(l,22 C), Solids(22 C)
2
D D
out
in out
in
1
2 2
2 3
ˆ
ˆ
Substance
(kg d) (kg d)
(kJ kg) (kJ kg)
ˆ
Solids 8400 0 8400
ˆ
H O(l) 15600 0 2800
ˆ
H O(v) 12800
n
n H
H
H
H
H
− −

1
2
3
water
ˆ 2.5(100 22) 195.0 kJ/kg
ˆ (419.1 92.2) 326.9 kJ/kg
ˆ (2676 92.2) 2584 kJ/kg
ˆ
( from Table B.5)
H
H
H
H
= − =
= − =
= − =
.
.
.
. .
Q m H m H Q
Q Q
i i i i
2 2
7
7
7
3
7
356 10
356 10
055
6 47 10 2 91 10
= − ⇒ = ×
=
×
= × ⇒ = ×
∑ ∑
out in
steam
kJ day
kJ / d kJ / d
Energy balance on steam side of dryer:
6 47 10 2133
1
10
30 3
7
3 3 3
. .

×
F
HG I
KJ×
F
HG I
KJ
F
HG
I
KJ ⇒ =
B
kJ
d
=
kg
d
kJ
kg
tonne
kg
tonnes / d (boiler feedwater)
for
H O(sat'd, )
v
2
n n
H
Δ
Energy balance on steam side of boiler:
Q1
7
30300 2737 6 839 8 04 10
= − = ×
( )( . . ) .
kg
d
kJ
kg
kJ / d
62% efficiency ⇒ Fuel heating value needed =
8 04 10
0 62
13 10
7
8
.
.
.
×
= × kJ / d
⇒ =
×
×
= ⇒

.
.
n4
8
4
130 10
375 10
3458
kJ / d
kJ / kg
kg / d D = 3.5 tonnes / day (fuel oil)
Air feed to boiler furnace: C + O CO , 4H + O 2H O, S + O SO
2 2 2 2 2 2
→ → →
2
2
O theo
2
1 kmol O
kg kgC 1 kmol C 1 1 1 1
( ) 3458 (0.87 )( )( )+(0.10)( )( ) (0.0084)( )( )
d kg 12 kg 1 kmol C 1 4 32 1
338 kmol O /d
n
⎡ ⎤
= +
⎢ ⎥
⎣ ⎦
=
9.70 (cont'd)

9-96
b.
c.
d.
Air fed(25%excess) =1.25(4.76
kmol air
kmol O
kmol O
d
kmol air
d
2011kmol 29kg 1 tonne
d kmol 10 kg
E =58.3tonnes/ d air toboiler)
2
2
3
)( )
(
338 2011
=
⇒ ⇒
Energy balance on boiler air preheater:
3
o 6
0
2011 kmol 10 mol 2.93 kJ
kJ
ˆ
Table B.8 (125 C) 2.93 5.89 10 kJ/d
d 1 kmol mol
mol
air
H Q
⇒ = ⇒ = = ×

Supplementary fuel for incinerator:
.
n6 97 5
= =
11.2 tonne sludge 195 SCM 1 kmol
d tonne 22.4 SCM
kmol d
MW MW MW
gas = + = + =
0 90 010 0 90 16 010 30 17 4
. . ( . )( ) ( . )( ) .
CH C H
4 2 6
kg kmol
Mgas = ⇒
( . )( . )
97 5 17 4 G = 1.7 tonne / d (natural gas)
CH O CO + 2H O, C H O CO + 3H O
4 2 2 2 2 6 2 2 2
+ → + →
2
7
2
2
Air feed to incinerator:
3
4 3
,
11200 kg sludge 0.75 kg sol 19000 kJ 2.5 m (STP) air 1 kmol kmol air
(air) : 1781
d kg sludge 1 kg sol 10 kJ 22.4 m (STP) d
th sludge =
4 2
,
4 2
kmol CH 2 kmol O
kmol 4.76 kmol air kmol air
(air) : 97.5 0.90 (0.10)(3.5) 998
d kmol kmol CH 1 kmol O d
th gas
⎡ ⎤⎛ ⎞
× + =
⎜ ⎟
⎢ ⎥
⎣ ⎦⎝ ⎠
7
3
kmol air
100% excess air: 2(1781 998) 5558 kmol air/d
d
5558 kmol air 29.0 kg air 1 tonne
161 tonne air/d (incinerator air)
d 1 kmol air 10 kg
n = + =
⇒ =

Energy balance on air preheater :
3
o 7
4
5558 kmol 10 mol 2.486 kJ
kJ kJ
ˆ
Table B.8 (110 C) = 2.486 1.38 10
d 1 kmol mol
mol d
air
H Q
⇒ ⇒ = = ×

Cost of fuel oil, natural gas, fuel oil and air preheating, pumping and compression, piping, utilities,
operating personnel, instrumentation and control, environmental monitoring. Lowering environmental
hazard might justify lack of profit.
Put hot product gases from boiler and/or incinerator through heat exchangers to preheat both air streams.
Make use of steam from dryer.
Sulfur dioxide, possibly NO2, fly ash in boiler stack gas, volatile toxic and odorous compounds in gas
effluents from dryer and incinerator.
9.70 (cont’d)

10- 1
CHAPTER TEN
10.1 b. Assume no combustion
(mol C H /mol)
(mol gas),
4
(°C)
n1 T1
(mol CH /mol)
x1
2
x2 6
1 – – x2
x1 (mol C H /mol)
3 8
(mol air), (°C)
n2 T2
(mol C H /mol)
(mol), 200°C
4
n3
(mol CH /mol)
y1
2
y2 6
1 – – –
y2
y1 (mol air/mol)
(mol C H /mol)
3
y3 8
y3
(kJ)
Q
11
5
6
4
1 2 3 1 2 1 2 3 1 2
variables
relations
degrees of freedom
material balances and 1 energy balance
−
n n n x x y y y T T Q
, , , , , , , , , ,
b g
b g
A feasible set of design variables: n n x x T T
1 2 1 2 1 2
, , , , ,
l q
Calculate n3 from total mole balance, y y y
1 2 3
, , and from component balances,
Q from energy balance.
An infeasible set: n n n x x T
1 2 3 1 2 1
, , , , ,
l q
Specifying n n
1 2
and determines n3 (from a total mole balance)
c.
(mol C H /mol)
(mol gas),
n1 T , P
y 1 6 14
1 – y 1
(mol C H /mol)
(mol gas),
n2
y 2 6 14
1 – y 2
(kJ)
Q
(mol N /mol)
2
T , P
2
1 (mol N /mol)
2
n3 (mol C H ( )/mol),
6 14 T , P
2
l
9
4
5
1 2 3 1 2 1 2
2 2
variables
relations
degrees of freedom
2 material, 1 energy, and 1 equilibrium: C H
6 14
− =
n n n y y T T Q P
y P P T
, , , , , , , ,
*
b g
b g
d i
A feasible set: n y T P n
, , , ,
1 1 3
l q
Calculate n2 from total balance, y2 from C H
6 14 balance, T2 from Raoult’s law:
[ y P P T
2 2
= ∗
C H
6 4
b g], Q from energy balance
An infeasible set: n y n P T
2 2 3 2
, , , ,
l q
Once y P
2 and are specified, T2 is determined from Raoult’s law

10-2
10.2 10
2
2 1 1
6
1 2 3 4 1 2 3 4
3 4 3 4
variables
material balances
equilibrium relations: [
degrees of freedom
n n n n x x x x T P
x P x P T x P x P T
B C
, , , , , , , , ,
, ]
* *
b g
b g b g b g b g
−
− = − = −
a. A straightforward set: n n n x x T
1 3 4 1 4
, , , , ,
l q
Calculate n2 from total material balance, P from sum of Raoult's laws:
P x p T x P T
B c
= + −
∗ ∗
4 4
1
b g b g b g
x3 from Raoult's law, x2 from B balance
b. An iterative set: n n n x x x
1 2 3 1 2
, , , , , 3
l q
Calculate n4 from total mole balance, x4 from B balance.
Guess P, calculate T from Raoult's law for B, P from Raoult’s law for C, iterate until
pressure checks.
c. An impossible set: n n n n T P
1 2 3 4
, , , , ,
l q
Once n n n
1 2 3
, , and are specified, a total mole balance determines n4 .
10.3 2BaSO s 4C s 2BaS s 4CO g
4 2
b g b g b g b g
+ → +
a.
(kg BaSO /kg)
100 kg ore,
n0
T
xb 4
(kg CO )
(kg BaS)
n2
n3 2
(kJ)
Q
0 (K)
(kg coal), T0 (K)
(kg C/kg)
xc
(% excess coal)
Pex
(kg C)
n1
(kg other solids)
n4
Tf (K)
11
5
1
1
1
5
0 1 2 3 4 0
0
variables
material balances C, BaS, CO BaSO other solids
energy balance
reaction
relation defining in terms of and
degrees of freedom
b c ex
2 4
ex b c
n n n n n x x T T Q P
P n x x
f
, , , , , , , , , ,
, ,
, ,
d i
b g
−
−
+
−
b. Design set: x x T T P
f
b c ex
, , , ,
0
n s
Calculate n0 from x x P
b c ex
and
, , ; n n
1 4
through from material balances,
Q from energy balance

10- 3
10.3 (cont’d)
c. Design set: x x T n Q
B c
, , , ,
2
0
l q
Specifying xB determines n2 ⇒ impossible design set.
d. Design set: x x T P Q
B c ex
, , , ,
0
l q
Calculate n2 from xB , n3 from xB
n0 from x x
B c
, and Pex
n1 from C material balance, n4 from total material balance
Tf from energy balance (trial-and-error probably required)
10.4 2C H OH O 2CH CHO 2H O
2 5 2 3 2
+ → +
2CH COH O 2CH CHOOH
3 2 3
+ →
T
(kJ)
Q
0
(mol solution),
nf
x ef (mol EtOH/mol)
1 – x ef (mol H O/mol)
2
P ,
xs
(mol air),
nw T0
0.79 (mol N )
n 2
0.21 (mol O )
nair 2
T
(mol EtOH),
ne
nah (mol CH CHO)
3
nea (mol CH COOH)
3
nw (mol H O)
2
nax (mol O )
2
nn (mol N )
2
Pxs = % excess air)
(
air
a. 13
6
1
1
2
7
0 0
variables
material balances
energy balance
relation between and
reactions
degrees of freedom
air
n n n n n n n n x T T Q P
P n x n
f aw e eh ea w ex ef xs
xs f ef
, , , , , , , , , , , ,
, , ,
d i
−
−
−
+
b. Design set: n x P n n T T
f ef xs e ah
, , , , , ,
0
n s
Calculate nair from n x
f ef
, and Pxs ; nn from N2 balance;
naa and nw from n x n n
f ef e ah
, , , and material balances;
nex from O atomic balance; Q from energy balance
c. Design set: n x T n Q n n
f ef e w
, , , , , ,
0 air
n s
Calculate Pxs from n x
f ef
, and nair ; n’s from material balances; T from energy
balance (generally nonlinear in T)
d. Design set: n nn
air , , …
l q. Once nair is specified, an N2 balance fixes nn

10-4
10.5
a. (mol CO)
n1
(mol H )
n2 2
(mol C H )
n3 3 6
reactor
(mol C H )
n4 3 6
(mol CO)
n5
(mol H )
n6 2
(mol C H O)
n7 7 8
(mol C H OH)
n8 4 7
(kg catalyst)
n9
Flash
tank
(mol C H )
n11 3 6
(mol CO)
n12
(mol H )
n13 2
(mol C H O)
n14 7 8
(mol C H OH)
n15 4 7
Separation
(mol C H )
n16 3 6
(mol CO)
n17
(mol H )
n18 2
(mol C H O)
n19 7 8
(mol C H OH)
n20 4 7
Hydrogenator
(mol H )
n21 2 (mol H )
n22 2
(mol C H OH)
n20 4 7
Reactor: 10
6
2
6
1 16
variables
material balances
reactions
degrees of freedom
n n
−
−
+
b g
Flash Tank: 12
6
4 15
variables
material balances
n n
−
−
b g
Separation: 10
5
11 20
variables
material balances
n n
−
−
b g
Hydrogenator: 5
3
1
3
19 23
variables
material balances
reaction
degrees of freedom
n n
−
−
+
b g
Process: 20
14
Local degrees of freedom
ties
6 overall degrees of freedom
−
The last answer is what one gets by observing that 14 variables were counted two times
each in summing the local degrees of freedom. However, one relation also was counted
twice: the catalyst material balances on the reactor and flash tank are each n n
9 10
= . We
must therefore add one degree of freedom to compensate for having subtracted the same
relation twice, to finally obtain 7 overall degrees of freedom (A student who gets this one
has done very well indeed!)
b. The catalyst circulation rate is not included in any equations other than the catalyst balance
(n9 = n10). It may therefore not be determined unless either n9 or n10 is specified.
n10 (kg catalyst)

10- 5
10.6 n i n B i B
− → − − = −
C H C H
4 10 4 10 b g
(mol n-B)
n1
mixer
(mol n-B)
n2
(mol i-B)
n3
reactor
(mol n-B)
n4
(mol i-B)
n5
still
(mol)
n6
(mol n-B/mol)
x 6
(1 – )(mol i-B/mol)
x 6
(mol)
nr
(mol n-B/mol)
x r
(1 – )(mol i-B/mol)
x r
a. Mixer: 5
2
1 2 3
variables
material balances
n n n n x
r r
, , , ,
b g
−
Reactor: 4
2
1
3
2 3 3 5
variables
material balances
reaction
degrees of freedom
n n n n
, , ,
b g
−
+
Still: 6
2
4 5 6 6
variables
material balances
n n n x n x
r r
, , , , ,
b g
−
Process: 10
6
Local degrees of freedom
ties
4 overall degrees of freedom
−
b. n n
1 100
= −
mol C H
4 10 , x n
6 0115
= −
. mol C H mol
4 10 , x n
r = −
085
. mol C H mol
4 10
Overall C balance: 100 4 0115 4 0885 4 100
6 6
= + ⇒ =
n n
. . mol C mol overhead
Overall conversion =
−
× =
100 100 0115
100
100% 885%
mol - fed mol - unreacted
mol - fed
n B n B
n B
b gb g
.
.
Mixer n-B balance: 100 085 1
2
+ =
. n n
T b g
35% S.P. conversion: n n n nr
4 2
1
4
0 65 65 05525 2
= ⇒ = +
. .
b g
b g
Still n – B balance:
n n x n x n n n
r r r r r
4 6 6
2
65 05525 0115 100 085 179 83
= + ⇒ + = + ⇒ =
b g
b gb g
. . . . mol
Recycle ratio = =
179 83 100 179
. .
mol recycle mol fresh feed
mol recycle
mol fresh feed
b g b g

10-6
10.6 (cont’d)
c.
n
n n
n n
n n
n n n n
n n n n
n n n
n
n
r
r
r
r
r r
2
3
4 2
5 2 3 4
4 5 6
4 6
6
100 085
1 085
0 65
0115 085
= +
= −
=
= + −
+ = +
= +
U
V
W
⇒
=
=
.
.
.
. .
b g
k k k
= = =
1 2 3
100 0 132 3 1515
1850 212 5 2288
150 19 85 22 73
120 25 1381 148 7
79 75 94 21 102 8
67 69 80 76 8855
132 3 1515 163 0
. . .
. . .
. . .
. . .
. . .
. . .
. . .
Error:
179 83 1630
179 83
100 9 3%
. .
.
.
−
× = error
d. w =
−
−
=
1515 132 3
132 3 100 0
0595
. .
. .
.
q =
−
= −
0595
0595 1
1470
.
.
.
nr
3
1470 132 3 1 1470 1515 179 8
b g b g b g
c hb g
= − + − − =
. . . . .
Error:
179 8 179 8
179 8
100 01%
. .
.
.
−
× = error
e. Successive substitution, Iteration 32: nr = 179.8319 Æ nr = 179.8319
Wegstein, Iteration 3: nr = 179.8319 Æ nr = 179.8319
10.7
SF
Split
S1
S2
a.
A B C D
1 X1 = 0.6
2 Molar flow rates (mol/h)
3 SF S1 S2
4 nA 85.5 51.3 34.2
5 nB 52.5 31.5 21.0
6 nC 12.0 7.2 4.8
7 nD 0.0 0.0 0.0
8 T(deg.C) 315 315 315
Formula in C4: = $B$1*B4
Formula in D4: = B4-C4

10- 7
10.7 (cont’d)
b. C **CHAPTER 10 -- PROBLEM 7
DIMENSION SF(8), S1(8), S2(8)
FLOW = 150.
N = 3
SF(1) = 0.35*FLOW
SF(2) = 0.57*FLOW
SF(3) = 0.08*FLOW
SF(8) = 315.
X1 = 0.60
CALL SPLIT (SF, S1, S2, X1, N)
WRITE (6, 900)' STREAM 1', S1(1), S1(2), S1(3), S1(B)
WRITE (6, 900)' STREAM 2', S2(1), S2(2), S2(3), S2(B)
900
*
*
*
FORMAT (A10, F8.2,' mols/h n-octane', /,
10X, F8.2,' mols/h iso-octane', /,
10X, F8.2,' mols/h inerts', /,
10X, F8.2,' K')
END
C
C SUBROUTINE SPLIT
C
SUBROUTINE SPLIT (SF, S1, S2, X1, N)
DIMENSION SF(8), S1(8), S2(8)
D0 100 J = 1, N
S1(J) = X1*SF(J)
100 S2(J) = SF(J) – S1(J)
S1(8) = SF (8)
S2(8) = SF (8)
RETURN
END
Program Output: Stream 1 3150
. mols h n-octane
51.30 mols h iso-octane
7.20 mols h inerts
315.00 K
Stream 2 21.00 mols h n-octane
34.20 mols h iso-octane
4.80 mols h inerts
315.00 K

10-8
10.8
a. Let Bz = benzene, Tl = toluene
* 6.89272 1211.033/( 220.790)
* 6.95805 1346.773/( 219.693)
* * *
Antoine equations: 10 (=1350.491)
10 (=556.3212)
Raoult's law: ( )/( - ) (=0.307) ,
T
Bz
T
Tl
Bz Tl Bz Tl B
p
p
x P p p p y
− +
− +
=
=
= − *
/ ( 0.518)
Total mole balance: 100 =
Benzene balance: 40 =
40 100
(=44.13), 100 (=55.87)
Fractional
z Bz Bz
v l
Bz v Bz l
Bz
v l v
Bz Bz
x p P
n n
y n x n
x
n n n
y x
= =
+ ⎫
⎬
+ ⎭
−
⇒ = = −
−
benzene vaporization : /40 (=0.571)
Fractional toluene vaporization : (1 )/60 (=0.354)
B v Bz
T v Bz
f n y
f n y
=
= −
The specific enthalpies are calculated by integrating heat capacities and (for vapors)
adding the heat of vaporization.
Q n H n H
out out in in
= −
∑ ∑ (= 1097.9)
b. Once the spreadsheet has been prepared, the goalseek tool can be used to determine the
bubble-point temperature (find the temperature for which nv=0) and the dew-point
temperature (find the temperature for which nl =0). The solutions are
T T
bp dp
= =
96 9 1032
. .
o o
C, C
c. C **CHAPTER 10 PROBLEM B
DIMENSION SF(3), SL(3), SV(3)
DATA A1, B1, C1/6.90565, 1211.033, 220.790/
DATA A2, B2, C2/6.95334, 1343.943, 219.377/
DATA CP1, CP2, HV1, HV2/ 0.160, 0.190, 30.765, 33.47/
COMMON A1, B1, C1, A2, B2, C2, CP1, CP2, NV1, NV2
FLOW = 1.0
SF(1) = 0.30*FLOW
SF(2) = 0.70*FLOW
T = 363.0
P = 512.0
CALL FLASH2 (SF, SL, SV, T, P, Q)
WRITE (6, 900) 'Liquid Stream', SL(1), SL(2), SL(3)
WRITE (6, 900) 'Vapor Stream', SV(1), SV(2), SV(3)
900 FORMAT (A15, F7.4,' mol/s Benzene',/,
* 15X, F7.4, mol/s Toluene',/,
* 15X, F7.2, 'K')
WRITE (6, 901) Q

10- 9
10.8 (cont’d)
901 FORMAT ('Heat Required', F7.2,' kW')
END
C
SUBROUTINE FLASN2 (SF, SL, SV, T, P, Q)
REAL NF, NL, NV
DIMESION SF(3), SL(3), SV(3)
COMMON A1, B1, C1, C2, CP1, CP2, NV1, NV2
C Vapor Pressure
PV1 = 10.**(A1 – B1/(T – 273.15 + C1))
PV2 = 10.**(A2 – B2/(T – 273.15 + C2))
C Product fractions
XL1 = (P – PV2)/(PV1 – PVS)
XV1 = XL1*PM/P
C Feed Variables
NF = SF(1) + SF(2)
XF1 = SF(1)/NF
C Product flows
NL = NF*(XF1 – XV1)/(XL1 – XV1)
NV = NF – NL
SL(1) = XL1*NL
SL(2) = NL – SL(1)
SY(1) = XY1*NY
SY(2) = NV – SY(1)
SL(3) = T
SV(3) = T
C Energy Balance
Q = CP1*SF(1)*SF(1) + CP2*SF(2)
Q = Q*(T – SF(3)) + (NV1*XV1 + HV2*(1 – XV1))*NV
RETURN
END
10.9 a. Mass Balance: NF NL NV
= + 1
b g
XF I NF XL I NL XV I NV I n
b g b g b g b g
∗ = ∗ + ∗ = −
1 2 1 2
, …
Energy Balance: Q T TF CP I XL I NL XV I NV
I
N
= − ∗ ∗ ∗ + ∗
=
∑
b g b g b g b g
c h
1
+ ∗ ∗
=
∑
NV HV I XV
I
N
1
1 3
b g b g b g
where: XL N XL I XV N XV I
I
N
I
N
b g b g b g b g
= − = −
=
−
=
−
∑ ∑
1 1
1
1
1
1
Raoult’s law: P XL I PV I
I
N
= ∗
=
∑
1
4
b g b g b g
XV I P XL I PV I I N
b g b g b g b g
∗ = ∗ = −
1 2 1 5
, ,…

10-10
10.9 (cont’d)
where: PV I A I B I C I T I N
b g b g b g b g
c h
d i
= ∗∗ − + = −
10 1 2 1
, ,…
3 3 1 4
1
3
+ − + +
−
−
−
−
+
N N NF NL NV XF I XL I XV I PV I TF T P Q
N
N
N
N
b g b g
variables
mass balance
energy balances
equilibrium relations
Antoine equations
degrees of freedom
, , , ( ), ( ), ( ), ( ), , , ,
Design Set TF T P NF XF I
, , , , b g
m r
Eliminate NL form (2) using (1)
Eliminate XV(I) form (2) using (5)
Solve (2) for XL(I)
XL I XF I NF NF NV PV I P
b g b g b g
c h
d i b g
= ∗ + ∗ − 1 6
Sum (6) ove I to Eliminate XL(I)
f NV NF XF I NF NV PV I P
I
N
b g b g b g
c h
d i b g
= − + ∗ + ∗ − =
=
∑
1 1 0 7
1
Use Newton's Method to solve (7) for NV
Calulate NL from (1)
XL(I) from (2)
XV(I) from (5)
Q from (3)
b. C **CHAPTER 10 - - PROBLEM 9
DIMENSION A(7), B(7), C(7), CP(7), HV(7)
COMMON A, B, C, CP, NV
DATA A/6.85221, 6.87776, 6.402040, 0., 0., 0., 0./
DATA B/1064.63, 1171.530, 1268.115, 0., 0., 0., 0./
DATA C/232.00, 224.366, 216.900, 0., 0., 0., 0./
DATA CP/0.188, 0.216, 0.213, 0., 0., 0., 0./
DATA NV/25.77, 28.85, 31.69, 0., 0., 0., 0./
FLOW = 1.0
N*3
SF(1) = 0.348*FLOW
SF(2) = 0.300*FLOW
SF(3) = 0.352*FLOW
SF(4) = 363
SL(4) = 338
SV(4) = 338
P*611
CALL FLASHN (SF, SL, SV, N, P, Q)
WRITE (6, 900)' Liquid Stream', (SL(I), I = 1, N + 1)
WRITE (6, 900)' Vapor Stream', (SV(I), I = 1, N + 1)

10- 11
10.9 (cont’d)
900
*
*
*
FORMAT (A15, F7.4,' mols/s n-pentane', /,
15X, F7.4,' mols/s n-hexane', /,
15X, F7.4,' mols/s n-hephane', /,
15X, F7.2,' K')
WRITE (6, 901) Q
901 FORMAT ('Heat Required', F7.2, 'kW')
END
C SUBROUTINE FLASHIN (SF, SL, SV, N, P, Q)
REAL NF, NL, NV, NVP
DIMENSION XF(7), XL(7), XV(7), PV(7)
DIMENSION A(7), B(7), C(7), CP(7), HV(7)
COMMON A, B, C, CP, HV
TOL = 1,5 – 6
C Feed Variables
NF = 0.
DO 100 I = 1, N
100 NF = NF + SF(I)
DO 200 I = 1, N
200 XF(I) = SF(I)/NF
TF = SF (N + 1)
T = SL (N + 1)
TC = T – 273.15
C Vapor Pressures
DO 300 I = 1, N
300 PV(I) = 10.**(A(I) – B(I)/(TC + C(I)))
C Find NV -- Initial Guess = NF/2
NVP = NF/2
DO 400 ITER = 1, 10
NV = NVP
F = –1.
FP = 0.
DO 500 I = 1, N
PPM1 = PV(I)/P – 1.
F = F + NF*XF(I)/(NF + NV*PPM1)
500 FP = FP – PPM1*XF(I)/(NF + NV*PPM1)**2.
NVP = NV – F/FP
IF (ABS((NVP – NV)/NVP).LT.TOL) GOTO 600
400 CONTINUE
WRITE (6, 900)
900 FORMAT ('FLASHN did not converge on NV')
STOP
C Other Variables

10-12
10.9 (cont’d)
600 NL = NF – NVP
DO 700 I = 1, N
XL(I) = XF(I)*NF/(NF + NV**(PV(I)/P – 1))
SL(I) = XL(I)*NL
XV(I) = XL(I)*PV(I)/P
700 SV(I) = SF(I) – SL(I)
Q1 = 0.
Q2 = 0.
DO 800 I = 1, N
Q1 = Q1 + CP(I)*SF(I)
800 Q2 = Q2 + HV(I)*XV(I)
Q = Q1*(T – TF) + Q2*NVP
RETURN
END
Program Output: Liquid Stream 0.0563 mols s n-pentane
0.1000 mols s n-hexane
0.2011 mols s n-heptane
338.00 K
Vapor Stream 0.2944 mols s n-pentane
0.2000 mols s n-hexane
0.1509 mols s n-heptane
338.00 K
Heat Required 13.01 kW
10.10
a.
(kW) (mol / s)
mol A(v) / mol)
1 mol B(g) / mol)
(K), (mm Hg)
(mol / s)
(mol A(v) / mol)
1 (mol B(g) / mol)
(K), (mm Hg)
(mol A(l) / s)
(
(
Q n
x
x
T P
n
x
x
T P
n
v
v
v
F
F
F
F
l
−
−

10- 13
10.10 (cont’d)
10 variables ( , , , , , , , , , )
*,
n x T P n x T n p Q
F F F v v l A
–1 Antoine equation
–1 Raoult’s law
–1 energy balance
b.
References: A(l), B(g) at 25o
C
Substance nin Hin
nout Hout
A(l) — — nl H3
A(v) n x
F F H1
n x
v v H4
B(g) ( )
n x
F F
1− H2
( )
n x
v v
1− H5
Given and (or and (fractional condensation),
Fractional condensation
Mole balance
balance
Raoult's law
Antoine's equation
Enthalpies
), , ,
( ) /
log
: ( ), ( ), ( ),
( ),
*
*
n x n n T P y
n y n x
n n n
A x n x n n
p x P
T
B
A p
C
H H C T H C T H C T
H H C T H
F F AF BF F c
l c F F
v F l
v F F l v
A v
A
v pv F pg F pl
v pv
⇒ =
⇒ = −
⇒ = −
⇒ =
⇒ =
−
−
= + − = − = −
= + − =
10
1 2 3
4 5
25 25 25
25
Δ
Δ C T
Q n H n H
pg
out out in in
( )
:
−
= −
∑ ∑
25
Energy balance
c.
nAF nBF nF xF TF P yc nL
0.704 0.296 1.00 0.704 333 760 0.90 0.6336
nV xV A B C pA* T Cpl
0.3664 0.1921 7.87863 1473.11 230 146.0 300.8 0.078
Cpv Cpg H1 H2 H3 H4 H5 Q
0.050 0.030 37.02 1.05 0.2183 35.41 0.0839 –23.7
Greater fractional methanol condensation (yc) ⇒ lower temperature (T). (yc = 0.10 ⇒
T = 328o
C.)

10-14
10.10 (cont’d)
e. C **CHAPTER 10 -- PROBLEM 10
DIMENSION SF(3), SV(3), SL(2)
COMMON A, B, C, CPL, HV, CPV, CPG
DATA A, B, C / 7.87863, 1473.11, 230.0/
DATA CPL, HV, CPV, CPG,/ 0.078, 35.27, 0.050, 0.029/
FLOW = 1.0
SF(1) = 0.704*FLOW
SF(2) = FLOW – SF(1)
YC = 0.90
P = 1.
SF(3) = 333.
CALL CNDNS (SF, SV, SL, P, YC, Q)
WRITE (6, 900) SV(3)
WRITE (6, 401) 'Vapor Stream', SV(1), SV(2)
WRITE (6, 401) 'Liquid Stream', SL(1)
WRITE (6, 902)Q
900 FORMAT ('Condenser Temperature', F7.2,' K')
901
*
FORMAT (A15, F7.3,' 'mols/s Methyl Alcohol', /,
15X, F7.3, 'mols/s air')
902 FORMAT ('Heat Removal Rate', F7.2,' kW')
END
C SUBROUTINE CNDNS (SF, SV, SL, P, YC, Q)
REAL NF, NL, NV
DIMENSION SF(3), SV(3), SL(2)
COMMON A, B, C, CPL, HV, CPV, CPG
C Inlet Stream Variables
NF = SF(1) + SF(2)
TF = SF(3)
XF = SF(1)/NF
C Solve Equations
NL = YC * XF * NF
NV = NF - NL
XV = (XF*NF - NL)/NV
PV = P * XV * 760.
T = B/(A - LOG(N)/LOG (10.)) - C
T = T + 273.15
Q = ((CPV * XV + CPG * (1 - XY)) * NV + CPL * NL) * (T - TF) - NL * HV
C Output Variables
SL(1) = NL
S2(2) = T
SV(1) = XV*NV
SV(2) = NV - SV(1)
SV(3) = T
RETURN
END

10- 15
10.11 η η η η
1 1 2 2 3 3 0
A A A A
m m
+ + + =
…
a. Extent of reaction equations:
ξ = − ∗
[ ]
SF IX X NU IX
b g b g
SP I SF I NU I I N
b g b g b g
= + ∗ =
ξ 1 2
, ,…
Energy Balance: Reference states are molecular species at 298K.
TF SF N TP SP N
= + = +
1 1
b g b g
ΔH HF I NU I
r
I
N
= ∗
=
∑
1
b g b g
Q H TP SP I CP I TF SF I CP I
r
I
N
I
N
= ∗ + − ∗ ∗ − − ∗ ∗
= =
∑ ∑
ξ Δ ( )
298 298
1 1
b g b g b g b g b g
b. C H 5O 3CO 4H O
3 8 2 2 2
+ → +
Subscripts: 1 = C3H8, 2 = O2, 3 = N2, 4 = CO2, 5 = H2O
270 m 1 atm mol K 1000 liter h
h K 0.08206 liter atm m 3600 s
mol C H s
3
3 3 8
⋅
⋅
=
273
3348
. [=SF(1)]
3.348 mol C H 1.2 5 mol O
sec mol C H
mol O s [= SF(2)] mol N s [= SF(3)]
3 8 2
3 8
2 2
b g= ⇒
2009 7554
. .
X n
C H C H
3 8 3 8
0 90 010 3348 0 3348
= ⇒ = =
. . ( . ) . /
mol C H s in product gas [= SP(1)]
3 8
ξ = − ∗
[ ]
SF IX X NU IX
b g b g = –(3.348 mol/s)(0.90)/(–1) = 3.013 mol/s
For the given conditions, Q = −4006 kJ / s . As Tstack increases, more heat goes into the
stack gas so less is transferred out of the reactor: that is, Q becomes less negative.
1-C3H8 2-O2 3-N2 4-CO2 5-H2O(v)
Nu -1 -5 0 3 4
nin (SF) 3.348 20.09 75.54
X 0.90
Xi 3.01
nout (SP) 0.3348 5.024 75.54 9.0396 12.0528
Cp 0.1431 0.033 0.0308 0.0495 0.0375
Tin 423
Hin 17.9 4.1 3.9 6.2 4.7
Tout 1050
Hout 107.6 24.8 23.2 37.2 28.2
HF -103.8 0 0 -393.5 -241.83
DHr -2044
Q -4006

10-16
10.11 (cont’d)
C **CHAPTER 10 PROBLEM 11
DIMENSION SF(8), SP(8), CP(7), HF(7)
REAL NU(7)
DATA NU/–1., –5, 0., 3., 4., 0., 0./
DATA CP/0.1431, 0.0330, 0.0308, 0.0495, 0.0375, 0., 0./
DATA HF/–103.8, 0., 0., –393.5, –241.83, 0., 0./
COMMON CP, HF
SF(1) = 3.348
SF(2) = 20.09
SF(3) = 75.54
SF(4) = 0.
SF(5) = 0.
SF(6) = 423.
SP(6) = 1050.
IX = 1
X = 0.90
N = 5
CALL REACTS (SF, SP, NU, N, X, IX, Q)
WRITE (6, 900) (SP(I), I = 1, N + 1), Q
900
*
*
*
*
*
*
FORMAT ('Product Stream', F7.3, ' mols/s propane', /,
15X, F7.3,' mols/s oxygen', /,
15X, F7.3,' mols/s nitrogen', /,
15X, F7.3,' mols/s carbon dioxide', /,
15X, F7.3,' mols/s water', /,
15X, F7.2,'K', /,
Heat required', F8.2, 'kW')
END
C SUBROUTINE REACTS (SF, SP, NU, N, X, IX, Q)
DIMENSION SF(8), SP(8), CP(7), HF(7)
REAL NU(7)
COMMON CP, HF
C Extent of Reaction
EXT = –SF(IX)*X/NU(IX)
C Solve Material Balances
DO 100 I = 1, N
100 SP(I) = SF(I) + EXT = NU(I)
C Heat of Reaction
HR = 0
DO 200 I = 1, N
200 HR = HR + NF(I)*NU(I)
C Product Enthalpy (ref * inlet)
HP = 0.
DO 300 I = 1, N
300 HP = HP + SP(I)*CP(I)
HP = HP + (SP(N + 1) – SF (N + 1))
Q = EXT * HR + HP
RETURN
END

10- 17
10.12 a. Extent of reaction equations:
ξ = − ∗
SF IX X NU IX
b g b g
SP I SF I NU I I N
b g b g b g
= + ∗ =
ξ 1,
Energy Balance: Reference states are molecular species at feed stream temperature.
Q H H n H
r
= = + =
∑
Δ Δ
ξ out out 0 ⇒ = +
= =
∑ ∑ zfeed
0
1 1
ξ NU I HF I SP I CP I dT
i
N
I
N
T
T
b g b g b g b g
CP(I) = ACP(I) + BCP(I)*T + CCP(I)*T2
+ DCP(I)*T3
f T NU I HF I AP T T
BP
T T
CP
T T
DP
T T
I
N
b g b g b g
= ∗ + ∗ − + ∗ −
+ ∗ − + ∗ − =
=
∑
ξ * ( ) ( )
( ) ( )
1
2 2
3 3 4 4
2
3 4
0
feed feed
feed feed
where: AP SP I ACP I
I
N
= ∗
=
∑
1
b g b g, and similarly for BP, CP, DP
Use goalseek to solve f T
( ) = 0 for T [= SP(N+1)]
b. 2CO O 2CO
2 2
+ →
Temporary basis: 2 mol CO fed
2 mol CO 1.25 1 mol O
2 mol CO
mol O mol N
2
2 2
b g= ⇒
125 470
. .
⇒ Total moles fed = (2.00 + 1.25 + 4.70) mol = 7.95 mol
Scale to given basis:
( .
.
( ) .
( )
( )
230
08036
1 1607
2 1
3 3
kmol
h
)(
1 h
3600 s
)(
10 mol
1 kmol
)
7.95 mol
mol CO fed s
.004 mol O fed s
.777 mol N fed s
3
2
2
= ⇒
=
=
=
SF
SF
SF

10-18
10.12 (cont’d)
The adiabatic reaction temperature is 1560o
C .
As X increases, T increases. (The reaction is exothermic, so more reaction means
more heat released.)
d.
C **CHAPTER 10 -- PROBLEM 12
DIMENSION SF(8), SP(B), NU(7), ACP(7), BCP(7), CCP(7), DCP(7), HF(7)
COMMON ACP, BCP, CCP, DCP, NF
DATA NU / –2., –1., 0., 2., 0., 0., 0./
DATA ACP/ 28.95E-3, 29.10E-3, 29.00E-3, 36.11E-3, 0., 0., 0./
DATA BCP/ 0.4110E-5, 1.158E-5, 0.2199E-5, 4.233E-6, 0., 0., 0./
DATA CCP/ 0.3548E-B, –0.6076E-8, 0.5723E-8, –2.887E-8, 0., 0., 0./
DATA DCP/ –2.220 E-12, 1.311E-12, –2.871E-12, 7.464E-12, 0., 0., 0./
DATA HF / –110.52, 0., 0., –393.5, 0., 0., 0./
SF(1) = 1.607
SF(2) = 1.004
SF(3) = 3.777
SF(4) = 0.
SF(5) = 650.
IX = 1
X = 0.45
N = 4
CALL REACTAD (SF, SP, NU, N, X, IX)
WRITE (6, 900) (SP(I), I = 1, N + 1)
Solution to Problem 10.12
1-CO 2-O2 3-N2 4-CO2
Nu -2 -1 0 2
nin (SF) 1.607 1.004 3.777 0
X 0.45
Xi 0.36
nout (SP) 0.88385 0.642425 3.777 0.72315
ACP 0.02895 0.0291 0.029 0.03611
BCP 4.11E-06 1.16E-05 2.20E-06 4.23E-05
CCP 3.55E-09 -6.08E-09 5.72E-09 -2.89E-08
DCP -2.22E-12 1.31E-12 -2.87E-12 7.46E-12
AP 0.1799
BP 5.00E-05
CP -2.90E-11
DP -6.57E-12
Tfeed 650
DHF -110.52 0 0 -393.5
DHr -566
T 1560
f(T) -4.7E-08

10- 19
10.12 (cont’d)
900
*
*
*
*
FORMAT ('Product Stream', F7.3, ' mols/s carbon monoxide', /,
15X, F7.3, 'mols/s oxygen', /.
15X, F7.3, 'mols/s nitrogen', /.
15X, F7.3, 'mols/s carbon dioxide', /,
15X, F7.2, 'C')
END
C SUBROUTINE REACTAD (SF, SP, NU, N, X, IX)
DIMENSION SF(8), SP(8), NU(7), ACP(7), BCP(7), CCP(7), DCP(7), HF(7)
COMMON ACP, BCP, CCP, DCP, NF
TOL = 1.E-6
C Extent of Reaction
EXT = –SF(IX)*X/NU(IX)
C Solve Material Balances
DO 100 I = 1, N
100 SP(I) = SF(I) + EXT*NU(I)
C Heat of Reaction
HR = 0
DO 200 I = 1, N
200 HR = HR + HF(I) * NU(I)
HR = HR * EXT
C Product Heat Capacity
AP = 0.
BP = 0.
CP = 0.
DP = 0.
DO 300 I = 1, N
AP = AP + SP(I)*ACP(I)
BP = BP + BP(I)*BCP(I)
CP = CP + SP(I)*CCP(I)
300 DP = DP + SP(I)*DCP(I)
C Find T
TIN = SF (N + 1)
TP = TIN
D0 400 ITER = 1, 10
T = TP
F = HR
FP = 0.
F = F +T*(AP + T*(BP/2. + T*(CP/3. + T*DP/4.)))
*–TIN*(AP + TIN*(BP/2. + TIN*(CP/3. + TIN*DP/4.)))
FP = FP + AP + T *(BP + T*(CP + T*DP))
TP = T – F/FP
IF(ABS((TP – T)/T).LT.TOL) GOTO 500
400 CONTINUE
WRITE (6, 900)
900 FORMAT ('REACTED did not converge')
STOP

10-20
10.12 (cont’d)
500 SP(N + 1) = T
RETURN
END
Program Output:
0.884 mol/s carbon monoxide
0.642 mol/s oxygen
3.777 mol/s nitrogen
0.723 mol/s carbon dioxide
T = 1560.43 C

21
10.13
The second reaction consumes six times more oxygen per mole of ethylene consumed. The lower the single pass ethylene oxide
yield, the more oxygen is consumed in the second reaction. At a certain yield for a specified ethylene conversion, all the oxygen in
the feed is consumed. A yield lower than this value would be physically impossible.
37.5 mol C2H4O
a. Separator
50 mol C2H4 208.3333 mol C2H4 166.6667 mol C2H4 8.333333 mol C2H4
50 mol O2 50 mol O2 18.75 mol O2 18.75 mol O2
37.5 mol C2H4O 8.333333 mol CO2
8.333333 mol CO2 8.333333 mol H2O
Reactor 8.333333 mol H2O
Xsp = 0.2
Ysp = 0.9
158.3333 mol C2H4 (Ra) 158.3333 mol C2H4 (Rc)
Rc-Ra = 0
Procedure: Assume Ra, perform balances on mixing point, then reactor, then separator. Rc is recalculated recycle rate.
Use goalseek to find the value of Ra that drives (Rc-Ra) to zero.
b. Xsp Ysp Yo no
0.2 0.72 0.6 158.33
0.2 1 0.833 158.33
0.3 0.75333 0.674 99.25
0.3 1 0.896 99.25
10-21

10-22
10.14 C **CHAPTER 10 -- PROBLEM 14
DIMENSION XA(3), XC(3)
N = 2
EPS = 0.001
KMAX = 20
IPR = 1
XA(1) = 2.0
XA(2) = 2.0
CALL CONVG (XA, XC, N, KMAX, EPS, IPR)
END
C SUBROUTINE FUNCGEN(N, XA, XC)
XC(1) = 0.5*(3. – XA(2) + (XA(1) + XA(2))**0.5
XC(2) = 4. – 5./(XA(1) + XA(2))
RETURN
END
C SUBROUTINE CONVG (XA, XC, N, KMAX, EPS, IPR)
DIMENSION XA(3), XC(3), XAH(3), XCM(3)
K = 1
CALL FUNCGEN (N, XA, XC)
IF (IPR.EQ.1) CALL IPRNT (K, XA, XC, N)
DO 100 I = 1, N
XAM(I) = XA(I)
XA(I) = XC(I)
100 XCM(I) = XC(I)
110 K = K + 1
CALL FUNCGEN (N, XA, XC)
IF (IPR.EQ.1) CALL IPRNT (K, XA, XC, N)
D0 200 I = 1, N
IF (ABS ((XA(I) - XC(I))/XC(I)).GE.EPS) GOTO 300
200 CONTINUE
C Convergence
RETURN
300 IF(K.EQ.KMAX) GOTO 500
DO 400 I = 1, N
W = (XC(I) – XCM(I))/(XA(I) – XAM(I))
Q = W/(W – 1.)
IF (Q.GT.0.5) Q = 0.5
IF (Q.LT.–5) Q = –5.
XCM(I) = XC(I)
XAM(I) = XA(I)
400 XA(I) = Q = XAM(I) + (1. – Q)*XCM(I)
GOTO 110
500 WRITE (6, 900)
900 FORMAT (' CONVG did not converge')
STOP
END

10- 23
10.14 (cont’d)
C SUBROUTINE IPRNT (K, XA, XC, N)
IF (K.EQ.1) WRITE (6, 400)
IF (K.NE.1) WRITE (6, *)
DO 100 I = 1, N
100 WRITE (6, 901) K, I, XA(I), XC(I)
RETURN
900 FORMAT (' K Var Assumed Calculated')
901 FORMAT (I4, I4, 2E15.6)
END
Program Output: K Var Assumed Calculated
1 1 0.200000E + 01 0.150000E + 01
1 2 0.200000E + 01 0.275000E + 01
2 1 0.150000E + 01 0.115578E + 01
2 0.275000E + 01 0.282353E + 01
0.395135E + 00 0.482384E + 00
3 2 0.283152E + 01 0.245041E + 01
8 1 0.113575E + 01 0.113289E + 01
8 2 0.269023E + 01 0.269315E + 01
0.113199E + 01 0.113180E + 01
9 2 0.269186E + 01 0.269241E + 01
2
3 1
4 1

11-1
CHAPTER ELEVEN
11.1 a. The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:
x
M
M
p
p
=
Therefore, the leakage rate of hydrogen peroxide is /
m M M
p
1
b. Balance on mass: Accumulation = input – output
E
= −
= =
dM
dt
m m
t M M
,
0 1
0
0 (mass in tank when leakage begins)
Balance on H O
2 2: Accumulation = input – output – consumption
E
= −
F
HG I
KJ−
= =
dM
dt
m x m
M
M
kM
t M M
p
p
p
p
p p
,
0 0 1
0
0
11.2 a. Balance on H3PO4: Accumulation = input
Density of H3PO4: ρ =1834
. g / ml.
Molecular weight of H3PO4: M = 98 00
. g / mol .
Accumulation =
dn
dt
(kmol / min)
Input =
20.0 L 1000 ml 1.834 g mol 1 kmol
min L ml 98.00 g 1000 mol
kmol / min
dn
dt
n kmol
p
p
p0
=
E
=
= = × =
03743
03743
0 150 0 05 75
.
.
, . .
t
b. dn dt n t
p
n t
p
p
7 5 0
03743 75 03743
.
. . . )
z z
= ⇒ = + (kmol H PO in tank
3 4
x
n
n
n
n n n
t
t
p
p p
p p
= =
+ −
=
+
+
0 0
75 03743
150 03743
. .
.
kmol H PO
kmol
3 4
c. 015
75 03743
150 03743
471
.
. .
.
.
=
+
+
⇒ =
t
t
t min

11-2
11.3 a. m a bt
w = + t mw
= =
0 750
,
b g t m m t
w w
= = ⇒ = +
5 1000 750 50
,
b g b g b g
kg h h
Balance on methanol: Accumulation = Input – Output
M
dM
dt
m m t
dM
dt
t
t M
f w
=
= − = − +
E
= −
= =
kg CH OH in tank
kg h kg h
kg h
kg
3
,
1200 750 50
450 50
0 750
b g
b g
b. dM t dt
M t
750 0
450 50
z z
= −
b g
E
− = −
E
= + −
M t t
M t t
750 450 25
750 450 25
2
2
Check the solution in two ways:
( ) ,
1 0 750
450 50
t M
t
= = ⇒
= − ⇒
kg satisfies the initial condition;
(2)
dM
dt
reproduces the mass balance.
c.
dM
dt
t M
= ⇒ = = ⇒ = + − =
0 450 50 9 750 450 9 25 9 2775
2
h kg (maximum)
( ) ( )
M t t
= = + −
0 750 450 25 2
t =
− ± +
−
⇒
450 450 4 25 750
2 25
2
b g b gb g
b g t = –1.54 h, 19.54 h
d.
3.40 m 10 liter kg
1 m 1 liter
kg
3 3
3
0 792
2693
.
= (capacity of tank)
M t t
= = + −
2693 750 450 25 2
t =
− ± + −
−
⇒
450 450 4 25 750 2693
2 25
2
b g b gb g
b g t = 719 1081
. , .
h h
Expressions for M(t) are:
M(t) =
750+ 450t - 25t and (tank is filling or draining)
(tank is overflowing)
(tank is empty, draining
as fast as methanol is fed to it)
2
0 719 1081 1954
2693 719 1081
0 1954 2054
≤ ≤ ≤ ≤
≤ ≤
≤ ≤
R
S
|
T
|
|
t t
t
t
. . .
( . . )
( . . )
b g

11-3
11.3 (cont’d)
11.4 a. Air initially in tank: N0
492
0 0258
=
°
°
=
10.0 ft R 1 lb - mole
532 R 359 ft STP
lb - mole
3
3
b g .
Air in tank after 15 s:
P V
PV
N RT
N RT
N N
P
P
f f
f
f
0 0
0
0
0 0258
0 2013
= ⇒ = = =
.
.
lb- mole 114.7 psia
14.7 psia
lb- mole
Rate of addition:
. .
n =
−
=
0 2013 0 0258
0
b glb- mole air
15 s
.0117 lb- mole air s
b. Balance on air in tank: Accumulation = input
dN
dt
= 0 0117
. lb- moles s
b g; t N
= =
0 0 0258
, . lb- mole
c. Integrate balance: dN n dt N t
N t
0 0258 0
0 0258 0 0117
.
. .
z z
= ⇒ = + lb- mole air
b g
( ) = , = . lb- mole satisfies the initial condition
lb- moleair / s reproduces the mass balance
1 0 0 0258
2 0 0117
t N
dN
dt
⇒
= ⇒
( ) .
d. t N
= ⇒ = + =
120 0 0258 0 0117 120 143
s lb- moles air
. . .
b gb g
O in tank lb - mole O
2 2
= =
0 21 143 0 30
. . .
b g
0
500
1000
1500
2000
2500
3000
0 5 10 15 20
t(h)
M(kg)

11-4
11.5 a. Since the temperature and pressure of the gas are constant, a volume balance on the gas
is equivalent to a mole balance (conversion factors cancel).
Accumulation = Input – Output
dV
dt
t V t
dV dt V t dt t
w
V
w
t
w
t
= −
= = × =
= − ⇒ = × + −
×
z z z
540 1
0 300 10 0
9 00 300 10 9 00
3
3 00 10 0
3
0
3
m h
h 60 min
m min
m corresponds to 8:00 AM
m in minutes
3
3
3
3
, .
. . .
.
ν
ν ν
e j
b g
b g e j
b. Let νwi = tabulated value of νw at t i
= −
10 1
b g i = 1 2 25
, , ,
…
. . . .
. .
, , , ,
ν ν ν ν ν
w w w wi
i
wi
i
dt
V
0
240
1 25
2 4
24
3 5
24
3
10
3
4 2
10
3
114 98 4 124 6 2 1134
2488
300 10 9 00 240 2488 2672
z ∑ ∑
≅ + + +
L
N
MM
O
Q
PP= + + +
=
= × + − =
= =
… …
b g b g
b g
m
m
3
3
c. Measure the height of the float roof (proportional to volume).
The feed rate decreased, or the withdrawal rate increased between data points,
or the storage tank has a leak, or Simpson’s rule introduced an error.
d. REAL VW(25), T, V, V0, H
INTEGER I
DATA V0, H/3.0E3, 10./
READ (5, *) (VW(I), I = 1, 25)
V= V0
T=0.
WRITE (6, 1)
WRITE (6, 2) T, V
DO 10 I = 2, 25
T = H * (I – 1)
V = V + 9.00 * H – 0.5 * H * (VW(I – 1) + VW(I))
WRITE (6, 2) T, V
10 CONTINUE
1 FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)')
2 FORMAT (F8.2, 7X, F6.0)
END
$DATA
11.4 11.9 12.1 11.8 11.5 11.3
Results:
TIME (MIN) VOLUME (CUBIC METERS)
0.00 3000.
10.00 2974.
20.00 2944.
230.00 2683.
240.00 2674.
Vtrapezoid
3
m
= 2674 ; VSimpson
3
m
= 2672 ;
2674 2672
2672
100% 0 07%
−
× = .
Simpson’s rule is more accurate.

11-5
11.6 a. .
ν ν
ν
out
V
out
kV V
out
L min L
b g b g
= ⇒ =
=
=
300
60
0 200 .
νout s
V
= ⇒ =
20 0 100
L min L
b. Balance on water: Accumulation = input – output (L/min).
(Balance volume directly since density is constant)
dV
dt
V
t V
= −
= =
20 0 0 200
0 300
. .
,
c.
dV
dt
V V
s s
= = − ⇒ =
0 200 0 200 100
. L
The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is
20 0 0 200 300 40 0
. . ( ) . .
− = − As t increases, V decreases. ⇒ = −
dV dt V
/ . .
20 0 0 200
becomes less negative, approaches zero as t → ∞ . The curve is therefore concave up.
d.
dV
V
dt
V t
20 0 0 200
300 0
. .
−
=
z z
⇒ −
−
−
F
HG I
KJ =
⇒ − + = − ⇒ = + −
= = ⇒
= + − ⇒ =
−
=
1
0 200
20 0 0 200
40 0
05 0 005 0 200 100 0 200 0 0 200
101 100 101
101 100 200 0 200
1 200
0 200
265
.
ln
. .
.
. . exp . . . exp .
.
exp .
ln
.
.
V
t
V t V t
V
t t
b g b g
b g b g
b g b g
L 1% from steady state
min
t
V

11-6
11.7 a. A plot of D (log scale) vs. t (rectangular scale) yields a straight line through the points ( t = 1 week,
D = 2385 kg week ) and ( t = 6 weeks, D = 755 kg week ).
ln ln
ln ln
.
ln ln ln . . .
.
D bt a D ae
b
D D
t t
a D bt a e
D e
bt
t
= + ⇔ =
=
−
=
−
= −
= − = + = ⇒ = =
E
= −
2 1
2 1
1 1
8 007
0 230
755 2385
6 1
0 230
2385 0 230 1 8 007 3000
3000
b g
b g b gb g
b. Inventory balance: Accumulation = –output
dI
dt
e
t I
t
= −
= =
−
3000
0 18 000
0 230
.
, ,
kg week
kg
b g
dI e dt I e I e
I
t
t
t t t
18 000
0 230
0
0 230
0
0 230
3000 18 000
3000
0 230
4957 13 043
,
. . .
,
.
,
z z
= − ⇒ − = ⇒ = +
− − −
c. t I
= ∞ ⇒ = 4957 kg
11.8 a. Total moles in room: N = =
1100 m K 10 mol
295 K 22.4 m STP
mol
3 3
3
273
45 440
b g ,
Molar throughput rate: ,
n = =
700 m K 10 mol
min 295 K 22.4 m STP
mol min
3 3
3
273
28 920
b g
SO balance
2 ( t = 0 is the instant after the SO2 is released into the room):
N x
mol mol SO mol mol SO in room
2 2
b g b g=
Accumulation = –output.
d
dt
Nx nx
dx
dt
x
N
n
b g= − ⇒ = −
=
=
.
,
,
45 440
28 920
0 6364
t x
= = = × −
0
15
45 440
330 10 5
,
.
,
.
mol SO
mol
mol SO mol
2
2
b. The plot of x vs. t begins at (t=0, x=3.30×10-5
). When t=0, the slope (dx/dt) is
− × × = − ×
− −
0 6364 330 10 210 10
5 5
. . . . As t increases, x decreases.⇒
dx dt x
= −0 6364
. becomes less negative, approaches zero as t → ∞ . The curve
is therefore concave up.

11-7
c. Separate variables and integrate the balance equation:
dx
x
dt
x
t x e
x t
t
330 10 0
5
5 0 6364
5
0 6364
330 10
0 6364 330 10
.
.
. ln
.
. .
×
−
− −
−
z z
= − ⇒
×
= − ⇒ = ×
( ) /
. . .
.
1
0 6364 330 10 0 6364
5 0 6364
t = 0, x = 3.30 10 mol SO mol satisfies the initial condition;
(2)
dx
dt
reproduces the mass balance.
-5
2
× ⇒
= − × × = − ⇒
− −
e x
t
d. C
x
x e t
SO
2
3
3 3 2
2
moles mol SO 1 m
1100 m mol 10 L
mol SO L
= = × = ×
− − −
45 440
4131 10 13632 10
2 6 0 6364
,
. . /
.
i) t C
= ⇒ = × −
2 382 10 7
min
mol SO
liter
SO
2
2
.
ii) x t
= ⇒ =
×
−
=
−
− −
10
10 330 10
0 6364
55
6
6 5
ln .
.
.
e j min
e. The room air composition may not be uniform, so the actual concentration of the SO2
in parts of the room may still be higher than the safe level. Also, “safe” is on the average;
someone would be particularly sensitive to SO2.
0
t
x
11.8 (cont’d)

11-8
11.9 a. Balance on CO: Accumulation=-output
N x
n
P
RT
n x
P
RT
x
d Nx
dt
P
RT
x
dx
dt
P
NRT
x
PV NRT
dx
dt V
x
t x
p
p
p
p
p
( ) (
)
)
( )
, .
mol mol CO / mol) = total moles of CO in the laboratory
Molar flow rate of entering and leaving gas: (
kmol
h
Rate at which CO leaves: (
kmol
h
kmol CO
kmol
=
CO balance: Accumulation = -output
kmol CO
kmol
=
F
HG I
KJ
= − ⇒ = −
F
HG I
KJ
E =
= −
= =
ν
ν
ν
ν
ν
0 0 01
b.
dx
x V
dt t
V
x
x
p
t
r
p
r
0 01 0
100
.
ln
z z
= − ⇒ = −
ν
ν
b g
c. V = 350 m3
tr = − × × =
−
350
700
100 35 10 283
6
ln .
e j hrs
d. The room air composition may not be uniform, so the actual concentration of CO
in parts of the room may still be higher than the safe level. Also, “safe” is on the
average; someone could be particularly sensitive to CO.
Precautionary steps:
Purge the laboratory longer than the calculated purge time. Use a CO detector
to measure the real concentration of CO in the laboratory and make sure it is
lower than the safe level everywhere in the laboratory.
11.10 a. Total mass balance: Accumulation = input – output
dM
dt
m m M
= − = ⇒∴ =
kg min is a constant kg
b g 0 200
b. Sodium nitrate balance: Accumulation = - output
x = mass fraction of NaNO3
d xM
dt
xm
dx
dt
m
M
x
m
x
t x
b g b g
= −
E
= − = −
= = =
min
, .
kg
200
0 90 200 0 45

11-9
dx
dt
, x decreases when t increases
dx
dt
becomes less negative until x reaches 0;
Each curve is concave up and approaches x = 0 as t ;
increases
dx
dt
becomes more negative x decreases faster.
= −
→ ∞
⇒ ⇒
m
x
m
200
0
d.
dx
x
m
M
dt
x t
0 45 0
.
z z
= − ⇒ ln
.
. exp
x m
t x
mt
0 45 200
0 45
200
= − ⇒ = −
F
HG I
KJ
Check the solution:
( )
. exp( )
1
0 45
200 200 200
t = 0, x = 0.45 satisfies the initial condition;
(2)
dx
dt
satisfies the mass balance.
⇒
= − × − = − ⇒
m mt m
x
e. ln .
m t x f
= ⇒ = −
100 2 0 45
kg min d i
90% ⇒ = ⇒ =
x t
f 0 045 4 6
. . min
99% ⇒ = ⇒ =
x t
f 0 0045 9 2
. . min
99.9% ⇒ = ⇒ =
x t
f 0 00045 138
. . min
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0 5 10 15 20 25
t(min)
x
0
0.45
t(min)
x
11.10 (cont’d)
c.
m
m
m
=
=
=
50
100
200
kg / min
kg / min
kg / min
m
m
m
=
=
=
50
100
200
kg / min
kg / min
kg / min

11-10
11.11 a. Mass of tracer in tank: V C
m kg m
3 3
e j e j
Tracer balance: Accumulation = –output. If perfectly mixed, C C C
out tank
= =
d VC
dt
C
b g b g
= −
ν kg min
dC
dt V
C
t C
m
V
= −
= =

,
ν
0 0
b.
dC
C V
dt
C
m V
t
V
C
m
V
t
V
m V
C t
0 0 0
0
z z
= − ⇒
F
HG I
KJ = − ⇒ = −
F
HG I
KJ

ln

exp

ν ν ν
c. Plot C (log scale) vs t (rect. scale) on semilog paper: Data lie on straight line (verifying assumption
of perfect mixing) through t C
= = × −
1 0 223 10 3
, .
e j t C
= = × −
2 0 050 10 3
, .
e j.
− =
−
= −
E
= =
−
−
ln . .
.
min . .
ν
V
V
0 050 0 223
2 1
1495
30 1495 201
1
1 3
b g
e j e j
min
m min m
3
11.12 a. In tent at any time, P=14.7 psia, V=40.0 ft3
, T=68°F=528°R
⇒ = = = ⋅
⋅
=
N
PV
RT
m(liquid)
14.7 psia 40.0 ft
10.73
ft psia
lb - mole R
528 R
lb - mole
3
3
o
o 01038
.
b. Molar throughout rate:

min
.
n n n
in out
= = =
°
°
=
60 ft 492 R 16.0 psia 1 lb - mole
528 R 14.7 psia 359 ft STP
lb - mole min
3
3
b g 01695
Moles of O2 in tank= N(lb - mole)
lb - mole O
lb - mole
2
×
F
HG I
KJ
Balance on O2: Accumulation = input – output
d Nx
dt
n xn
dx
dt
x
dx
dt
x
t x
b g b g b g
= − ⇒ = − ⇒
= −
= =
0 35 01038 01695 0 35
163 0 35
0 0 21
. . . .
. .
, .
c.
dx
x
dt
x
t
x t
0 35
163
0 35
0 35 0 21
163
0 21 0
.
. ln
.
. .
.
. −
= ⇒ −
−
−
=
z z b g
b g
⇒
−
= ⇒ = −
− −
0 35
014
0 35 014
1 63 1 63
.
.
. .
. .
x
e x e
t t
x t
= ⇒ = −
−
−
F
HG I
KJ
L
NM O
QP=
0 27
1
163
0 35 0 27
0 35 0 21
0 343
.
.
ln
. .
. .
. ( )
min or 20.6 s
V is constant

11-11
11.13 a. Mass of isotope at any time = V C
liters mg isotope liter
b g b g
Balance on isotope: Accumulation = –consumption
d
dt
VC kC V
b g b g
= −
⋅
F
HG I
KJ
mg
L s
L
dC
dt
kC
t C C
= −
= =
0 0
,
Separate variables and integrate
dC
C
kdt
C
C
kt t
C C
k
C
C t
0 0 0
0
z z
= − ⇒
F
HG I
KJ = − ⇒ =
−
ln
lnb g
C C t
k
t
k
= ⇒ =
−
⇒ =
05
05 2
0 1 2
.
ln . ln
b g 1 2
b. t k
1 2
1
2 6
2
2 6
0 267
= ⇒ = = −
.
ln
.
.
hr
hr
hr
C C
= 0 01 0
. t =
−
=
ln .
.
0 01
0 267
b g 17.2 hr
11.14 A → products
a. Mole balance on A: Accumulation = –consumption
d C V
dt
kC V
A
A
b g = − V constant; cancels
b g
t C C
dC
C
kdt
C
C
kt C C kt
A A
A
A
C
C t
A
A
A A
A
A
= =
⇒ = − ⇒
F
HG I
KJ = − ⇒ = −
z z
0 0
0 0
0
0
,
ln expb g
b. Plot CA (log scale) vs. t (rect. scale) on semilog paper. The data fall on a straight line (verifies
assumption of first-order) through t CA
= =
213 0 0262
. , .
b g t CA
= =
120 0 0 0185
. , .
b g.
ln ln
ln . .
. .
. .
C kt C
k k
A A
= − +
− =
−
= − × ⇒ = ×
− − − −
0
3 1 3 1
0 0185 0 0262
120 0 213
353 10 35 10
b g min min
11.15 2 2
A B C
→ +
a. Mole balance on A: Accumulation = –consumption
d C V
dt
kC V
A
A
b g = − 2
V constant; cancels
b g
t C C
dC
C
kdt
C C
kt C
C
kt
A A
A
A
C
C t
A A
A
A
A
A
= =
⇒ = − ⇒ − + = − ⇒ = +
L
NM O
QP
z z
−
0
1 1 1
0
2 0 0 0
1
0
,
t=-ln(C/C0)/k
Cancel V

11-12
b. C C
C C
kt t
kC
A A
A A A
= ⇒ − + = − ⇒ =
05
1
05
1 1
0
0 0
1 2 1 2
0
.
.
; but C
n
V
P
RT
t
RT
kP
A
A
0
0 0
1 2
0
= = ⇒ =
n n
A A
= 05 0
.
n n A B A n
B A A
= =
05 2 2 05
0 0
. .
mol react. mol mol react.
b gb g
n n A C A n
C A A
= =
05 1 2 0 25
0 0
. .
mol react. mol mol react.
b gb g
total moles = ⇒ = =
125 125 125
0 1 2
0
0
. . .
n P
n RT
V
P
A
A
c. Plot t1 2 vs. 1 0
P on rectangular paper. Data fall on straight line (verifying 2nd order
decomposition) through t P
1 2 0
1060 1 1 0135
= =
, .
d i t P
1 2 0
209 1 1 0 683
= =
, .
d i
Slope: s atm
K L atm mol K
143.2 s atm
L mol s
RT
k
k
=
−
−
= ⋅
⇒ =
⋅ ⋅
⋅
= ⋅
1060 209
1 0135 1 0 683
1432
1015 0 08206
0582
. .
.
.
.
b gb g
d. t
RT
k P
E
RT
t P
RT k
E
R T
1 2
0 0
1 2 0
0
1 1
=
F
HG I
KJ ⇒
F
HG
I
KJ = +
exp ln ln
Plot t P RT
1 2 0 (log scale) vs. 1 T (rect. scale) on semilog paper.
t P R
1 2 0 1 0 08206
s atm, L atm / (mol K) T K
b g b g
, . ,
= = ⋅ ⋅
Data fall on straight line through t P RT T
1 2 0 74 0 1 1 900
= =
. ,
d i
t P RT T
1 2 0 0 6383 1 1 1050
= =
. ,
d i
E
R
=
−
=
ln . .
,
0 6383 74 0
1 1050 1 900
29 940
b g K E = ×
2 49 105
. J mol
ln ln .
,
. .
1
0 6383
29 940
1050
28 96 379 10
0
0
12
k
k
= − = − ⇒ = × ⋅
b g L (mol s)
e. T k k
E
RT
= ⇒ = −
F
HG I
KJ = ⋅
980 0 204
0
K L (mol s)
exp .
CA0
2
0 70 120
0 08206 980
1045 10
=
⋅ ⋅
= × −
. .
.
.
atm
L atm mol K K
mol L
b g
b gb g
90% conversion
C C t
k C C
A A
A A
= ⇒ = −
L
NM O
QP=
×
−
×
L
NM O
QP
= =
− −
010
1 1 1 1
0 204
1
1045 10
1
1045 10
4222 70 4
0
0
3 2
.
. . .
.
s min
R=8.314 J/ (mol ·K)
11.15 (cont’d)

11-13
11.16 A B
→
a. Mole balance on A: Accumulation = –consumption(V constant)
dC
dt
k C
k C
t C C
k C
k C
dC dt
k
C
C
k
k
C C t t
k
k
C C
k
C
C
A A
A
A A
A
A
A
C
C t
A
A
A A A A
A
A
A
A
= −
+
= =
+
= − ⇒ + − = − ⇒ = − −
z z
1
2
0
2
1 0 1 0
2
1
0
2
1
0
1 0
1
0
1 1 1
0
,
ln ln
b g b g
b. Plot t C C
A A
− 0
b g vs. ln /
C C C C
A A A A
0 0
b g b g
− on rectangular paper:
t
C C k
C C
C C
k
k
A A
y
A A
A A
x
0 1
0
0
2
1
1
−
= −
−
+
b g ;
b g
1

slope intercept
ln
Data fall on straight line through 116 28 0 2111
1 1
. , .
y x
−
F
HG I
KJ 130 01 0 2496
2 2
. , .
y x
−
F
HG I
KJ
− =
−
− − −
= − ⇒ = × ⋅
−
1 130 01 116 28
0 2496 0 2111
356 62 2 80 10
1
1
3
k
k
. .
. .
. .
b g L (mol s)
k
k
k
2
1
2
130 01 356 62 0 2496 4100 0115
= + − = ⇒ =
. . . . .
b g L mol
11.17 CO Cl COCl
2 2
+ ⇒
a.
3.00 L 273 K 1 mol
303.8 K 22.4 L STP
mol gas
b g= 012035
.
C
C
i
i
CO
Cl 2
mol 3.00 L mol L CO
mol 3.00 L mol L Cl
initial concentrations
2
b g b g
d i b g
= =
= =
U
V
|
W
|
0 60 012035 0 02407
0 40 012035 0 01605
. . .
. . .
C t C t
C t C t
p
p
CO
Cl
2
2
Since 1 mol COCl formed requires 1 mol of each reactant
b g b g
b g b g
= −
= −
U
V
|
W
|
0 02407
0 01605
.
.
b. Mole balance on Phosgene: Accumulation = generation
d VC
dt
C C
C C
p
p
d i
d i
=
+ +
8 75
1 58 6 34 3
2
.
. .
CO Cl
Cl
2
2
dC
dt
C C
C
t C
p p p
p
p
=
− −
−
= =
2 92 0 02407 0 01605
1941 24 3
0 0
2
. . .
. .
,
d id i
d i
c. Cl2 limiting; 75% conversion ⇒ = =
Cp 0 75 0 01605 0 01204
. . .
b g mol L
t
C
C C
dC
p
p p
p
=
−
− −
z
1
2 92
1941 24 3
0 02407 0 01605
2
0
0 01204
.
. .
. .
. d i
d id i
V=3.00 L

11-14
d. REAL F(51), SUM1, SUM2, SIMP
INTEGER I, J, NPD(3), N, NM1, NM2
DATA NPD/5, 21, 51/
FN(C) = (1.441 – 24.3 * C) ** 2/(0.02407 – C)/(0.01605 – C)
DO 10 I = 1, 3
N = NPD(I)
NM1 = N – 1
NM2 = N – 2
DO 20 J = 1, N
C = 0.01204 * FLOAT(J – 1)/FLOAT(NM1)
F(J) = FN(C)
20 CONTINUE
SUM1 = 0.
DO 30 J = 2, NM1, 2
SUM = SUM1 + F(S)
30 CONTINUE
SUM2 = 0.
DO 40 J = 3, NM2, 2
SUM2 = SUM2 + F(J)
40 CONTINUE
SIMP = 0.01204/FLOAT(NM1)/3.0 * (F(1) + F(N) + 4.0 * SUM1 + 2.0 * SUM2)
T = SIMP/2.92
WRITE (6, 1) N, T
10 CONTINUE
1 FORMAT (I4, 'POINTS —', 2X, F7.1, 'MINUTES')
END
RESULTS
5 POINTS — 91.0 MINUTES
t = 90 4
. minutes
11.18 a. Moles of CO in liquid phase at any time cm mols cm
2
3 3
= V CA
e j e j
Balance on CO in liquid phase:
2 Accumulation = input
d
dt
VC kS C C
dC
dt
kS
V
C C
t C
A A A
V
A
A A
A
b g e j e j
= −
F
HG I
KJ⇒
= −
= =
÷
*
*
,
mols
s
0 0
Separate variables and integrate. Since p y P
A A
= is constant, C p H
A A
*
= is also a constant.
dC
C C
kS
V
dt C C
kS
V
t
C C
C
kS
V
t
C
C
e C C e
A
A A
C t
A A
C
C
C C
A A
A
A
A
kSt V
A A
kSt V
A
A
A
A A
*
*
*
*
exp
*
*
ln
ln
*
−
= ⇒ − − =
⇒
−
= − ⇒ − = ⇒ = −
z z =
−
− −
0 0 0
1
1 1
e j
e j
b g
11.17 (cont’d)

11-15
11.18 (cont’d)
b. t
V
kS
C
C
A
A
= − −
L
N
MM
O
Q
PP
ln *
1
V k S CA
= = = = = ×
−
5 5000 0 020 785 0 62 10
3
L cm cm s cm mol / cm
3 2 3
, . , . , .
C y P H
A A
*
. .
= = ⋅ = ×
−
0 30 20 9230 0 65 10
3
a fa f d i
atm atm cm mol mol cm
3 3
t = − −
×
×
F
HG
I
KJ = ⇒
−
−
5000
0 02 785
1
0 62 10
0 65 10
9800
3
3
cm
cm s cm
s 2.7 hr
3
2
e j
b ge j
. .
ln
.
.
(We assume, in the absence of more information, that the gas-liquid interfacial surface area equals
the cross sectional area of the tank. If the liquid is well agitated, S may in fact be much greater than
this value, leading to a significantly lower t than that to be calculated)
11.19 A B
→
a. Total Mass Balance: Accumulation = input
dM
dt
d V
dt
v
= =
E
( )

ρ
ρ
dV
dt
v
=
t V
= =
0 0
,
A Balance: Accumulation = input – consumption
dN
dt
C v kC V
A
A A
= −
0 ( )
dN
dt
C v kN
A
Ao A
= −

t N A
= =
0 0
,
b. Steady State:
dN
dt
N
C v
k
A
A
A
= ⇒ =
0 0
c. dV vdt V vt
V t
0 0
z z
= ⇒ =

dN
C v kN
dt
A
A A
N t
A
0
0 0
−
=
z z
⇒ −
−
F
HG I
KJ = ⇒
−
= −
1 0
0
0
0
k
C v kN
C v
t
C v kN
C v
e
A A
A
A A
A
kt
ln

⇒ = − − → ∞ ⇒ =
N
C v
k
kt t N
C v
k
A
A
A
A
0 0
1

exp

b g
C
N
V
C kt
kt
A
A A
= =
− −
0 1
[ exp( )]
CA=NA/V

11-16
When the feed rate of A equals the rate at which A reacts, NA reaches a steady value.
NA would never reach the steady value in a real reactor. The reasons are:
( )
1 In our calculation, V = t , V .
But in a real reactor, the volume is limited by the reactor volume;
(2) The steady value can only be reached at t . In a real reactor, the reaction time is finite.
vt ⇒ → ∞ → ∞
→ ∞
d. lim lim
[ exp( )]
lim
t
A
t
A
t
A
C
C kt
kt
C
kt
→∞ →∞ →∞
=
− −
= =
0 0
1
0
From part c, t
N
V
A
→ ∞ → → ∞ ⇒ = →
, N a finite number, V C
A A 0
11.20 a. MC
dT
dt
Q W
v = −

M
C C
W
v p
=
= = ⋅ ⋅
=
( . ( . .
( .

300 100 300
0 0754 1
0
L) kg / L) = kg
kJ / mol C)( mol / 0.018 kg) = 4.184 kJ / kg C
o o
dT
dt
Q
t T
= 0 0797
. (kJ / s)
= 0, = 18 C
o
b. dT .
kJ
s
4.29 kW
18 C
C
0
s
o
o
100 240
0 0797
100 18
240 0 0797
4 287
z z
= ⇒ =
−
×
= =

.
.
Q dt Q
c. Stove output is much greater.
Only a small fraction of energy goes to heat the water.
Some energy heats the kettle.
Some energy is lost to the surroundings (air).
11.21 a. Energy balance: MC
dT
dt
Q W
v = −

M
C C
Q
W
v p
=
≈ = ⋅ ⋅
= =
=
20 0
0 0754 1
0 97 2 50 2 425
0
.
( .
. ( . ) .

kg
kJ / mol C)( mol / 0.0180 kg) = 4.184 kJ / (kg C)
kJ s
o o
a f
dT
dt
= °
0 0290
. C s
b g , t T
= = °
0 25
, C
The other 3% of the energy is used to heat the vessel or is lost to the surroundings.
b. dT dt T t s
T t
25 0
0 0290 25 0 0290
o
C
C
z z
= ⇒ = ° +
. . b g
c. T t
= ° ⇒ = − = ⇒
100 100 25 0 0290 2585 43.1 min
C s
b g .
No, since the vessel is closed, the pressure will be greater than 1 atm (the pressure at the normal
boiling point).
11.19 (cont’d)

11-17
11.22 a.Energy balance on the bar
MC
dT
dt
Q W UA T T
v
b
b w
= − = − −
b g
M
C T
U
A
v w
= =
= ⋅° = °
= ⋅ ⋅°
= + + =
B
60 7 7 462
0 46 25
0 050
2 2 3 2 10 3 10 112
cm g cm g
kJ (kg C), C
J (min cm C)
cm cm
3 3
2
2 2
Table B.1
e je j
a fa f a fa f a fa f
.
.
.
dT
dt
T
b
b
= − − °
0 02635 25
. b gb g
C min
t Tb
= = °
0 95
, C
b.
dT
dt
T T
b
bf bf
= = − − ⇒ = °
0 0 02635 25 25
. d i C
c.
dT
T
dt
b
b
t
Tb
−
= −
z
z 25
0 02635
0
95
.
⇒
−
−
F
HG I
KJ = −
ln .
T
t
b 25
95 25
0 02635
⇒ = + −
T t t
bb g b g
25 70 0 02635
exp .
Check the solution in three ways:
( )
. . ( )
.
1 25 70 95
70 0 02635 0 02635 25
25
0 02635
t = 0, T satisfies the initial condition;
(2)
dT
dt
reproduces the mass balance;
(3) t , T confirms the steady state condition.
b
o
b
b
o
= + = ⇒
= − × = − − ⇒
→ ∞ = ⇒
−
C
e T
C
t
b
T t
b = ° ⇒ =
30 100
C min
5
15
25
35
45
55
65
75
85
95
0
t
Tb(
o
C)

11-18
11.23
a. Energy Balance: MC
dT
dt
mC T UA T T
v p
= − + −
25
b g b g
steam
M
m
C C
UA
T
dT dt T t T
v p
=
=
≈ = ⋅°
= ⋅°
= °
= − = =
760
12 0
2 30
115
167 8
150 00224 0 25
kg
kg min
o o
kJ (min C)
kJ (min C)
sat'd; 7.5bars C
C C
steam
.
.
.
.
/ . . ( min), ,
a f
b. Steady State:
dT
dt
T T
s s
= = − ⇒ = °
0 150 0 0224 67
. . C
c.
dT
T
dt t
T
T
t
t
Tf
150 0 0224
1
0 0224
150 0 0224
094
150 094 0 0224
0 0224
0
25
. . .
ln
. .
.
. . exp( . )
.
−
= ⇒ = −
−
F
HG I
KJ ⇒ =
− −
z
z
t T
= ⇒ = °
40 498
min. C
.
d. U changed. Let x UA new
= ( ) . The differential equation becomes:
dT
dt
x x T
= + − +
0 3947 0 096 0 01579 5721
. . ( . . )
kJ / (min C)
o
dT
x x T
dt
x
x x
x x
x
0 3947 0 096 0 01579 5721 10
1
0 01579 5721 10
0 3947 0 096 0 01579 5721 10 55
0 3947 0 096 0 01579 5721 10 25
40
14 27
4
0
40
25
55
4
4
4
. . ( . . )
. .
ln
. . . .
. . . .
.
+ − + ×
=
⇒ −
+ ×
+ − + × ×
+ − + × ×
L
N
MMM
O
Q
PPP
=
⇒ = ⋅
−
−
−
−
z
z
e j
e j
Δ Δ
U
U
UA
UA
initial initial
= =
−
× =
( )
( )
. .
.
.
14 27 115
115
100% 241%
25
67
0
t
T(
o
C)
12.0 kg/min
T (o
C)
12.0 kg/min
25o
C
Q (kJ/min) = UA (Tsteam-T)

11-19
11.24 a. Energy balance: MC
dT
dt
Q W
v = −

, .
. .
W C
M Q W
v
= = ⋅°
= = =
0 177
350 40 2 40 2
J g C
g, J s
dT
dt
t T
T t
T t
= °
= = °
U
V
|
W
|
⇒
= +
= ° ⇒ = ⇒
0 0649
0 20
20 0 0649
40 308 51
.
,
.
. min
C s
C
s
C s
b g b g
b. The benzene temperature will continue to rise until it reaches Tb = °
801
. C ; thereafter the heat
input will serve to vaporize benzene isothermally.
Time to reach neglect evaporation s
T t
b b g:
.
.
=
−
=
801 20
0 0649
926
Time remaining: 40 minutes 60 s min s s
b g− =
926 1474
Evaporation: Δ .
Hv = =
30 765 1000 393
kJ mol 1 mol 78.11 g J kJ J g
b gb gb g
Evaporation rate = =
40.2 J s 393 J g g s
b g b g
/ .
0102
Benzene remaining = − =
350 0102 1474 200
g g s s g
.
b gb g
c. 1. Used a dirty flask. Chemicals remaining in the flask could react with benzene. Use a clean flask.
2. Put an open flask on the burner. Benzene vaporizes⇒ toxicity, fire hazard.
Use a covered container or work under a hood.
3. Left the burner unattended.
4. Looked down into the flask with the boiling chemicals. Damage eyes. Wear goggles.
5. Rubbed his eyes with his hand. Wash with water.
6. Picked up flask with bare hands. Use lab gloves.
7. Put hot flask on partner’s homework. Fire hazard.
11.25 a. Moles of air in room: n = =
60 m 273 K 1 kg- mole
283 K 22.4 m STP
kg- moles
3
3
b g 258
.
Energy balance on room air: nC
dT
dt
Q W
v = −

.

Q m H T T
W
s v
= − −
=
Δ H O, 3bars, sat'd
2
b g b g
30 0
0
0
nC
dT
dt
m H T T
v s v
= − −
.
Δ 30 0 0
b g
N
C
H
T
v
v
=
= ⋅°
=
= °
258
208
2163
0
0
.
.

kg- moles
kJ (kg- mole C)
kJ kg from Table B.6
C
Δ b g
dT
dt
m T
s
= − °
40 3 0559
. . C hr
b g
t T
= = °
0 10
, C
(Note: a real process of this type would involve air escaping from the room and a constant pressure
being maintained. We simplify the analysis by assuming n is constant.)

11-20
b. At steady-state, dT dt m T m
T
s s
= ⇒ − = ⇒ =
0 40 3 0559 0
0559
40 3
. .
.
.
T ms
= ° ⇒ =
24 0 333
C kg hr
.
c. Separate variables and integrate the balance equation:
dT
m T
dt
s
T t
f
40 3 0559
10 0
. .
−
=
z z dT
T
t
134 0559
10
23
. .
−
E
=
z
t = −
−
−
L
NMM
O
QPP=
1
0559
134 0559 23
134 0559 10
48
.
ln
. .
. .
.
b g
b g hr
11.26 a. Integral energy balance t t
= =
0 20
to min
b g
Q U MC T
v
= = =
− °
⋅°
= ×
Δ Δ
250 kg 4.00 kJ 60 C
kg C
kJ
20
4 00 104
b g .
Required power input: .
Q =
×
=
4.00 10 kJ 1 min 1 kW
20 min s 1 kJ s
kW
4
60
333
b. Differential energy balance: MC
dT
dt
Q
v = dT
dt
Q t
= 0 001
. b g
t T
= = °
0 20
, C
Integrate: dT Q dT T Qdt
T t t
20 0 0
0 001 20
o
C
o
C
z z z
= ⇒ = +
.
Evaluate the integral by Simpson's Rule (Appendix A.3)
kJ
s

Qdt
0
600
30
3
33 4 33 35 39 44 50 58 66 75 85 95
2 34 37 41 47 54 62 70 80 90 100 34830
z = + + + + + + + + + +
+ + + + + + + + + + =
b g
b g
⇒ = = °
T 600 s 20 0001 34830 548
b g e jb g
o o
C+ C/ kJ kJ C
. .
c. Past 600 s,
Q t t
= + − =
100
10
600 6
kW
60 s
s
b g
T Qdt Qdt
t
dt
t t
= + = + +
L
N
MMMMM
O
Q
PPPPP
z z z
20 0001 20 0001
6
0 0
600
34830
600
. .

⇒ T
t
t T
= + −
F
HG
I
KJ ⇒ = −
548
0 001
6 6
600
2
12000 248
2 2
.
.
.
s
b g b g
T t
= ° ⇒ = = ⇒ +
85 850 14 10
C s min, 10 s explosion at 10:14 s
.
m
T
s
f
=
= °
0 333
23 C
M
Cv
=
= ⋅°
250
4 00
kg
kJ kg C
.
11.25 (cont’d)

11-21
11.27 a.Total Mass Balance:
Accumulation=Input– Output
dM
dt
d( V)
dt
tot
E
= − ⇒ = −
. .
m m
i o
ρ
ρ ρ
8 00 4 00
dV
dt
t
=
= =
4 00
0 400
.
,
L / s
V L
0
KCl Balance:
Accumulation=Input-Output⇒ = − ⇒ = × −
dM
dt
d( )
dt
KCl . . .
, ,
m m
CV
C
i KCl o KCl 100 8 00 4 00
⇒ + = −
V
dC
dt
C
dV
dt
C
8 4
dC
dt
C
V
t
=
−
= =
8 8
0 0
, C g / L
0
b. (i)The plot of V vs. t begins at (t=0, V=400). The slope (=dV/dt) is 4 (a positive constant).
V increases linearly with increasing t until V reaches 2000. Then the tank begins to overflow
and V stays constant at 2000.
(ii) The plot of C vs. t begins at (t=0, C=0). When t=0, the slope (=dC/dt) is (8-0)/400=0.02.
As t increases, C increases and V increases (or stays constant)⇒ dC/dt=(8-8C)/V becomes
less positive, approaches zero as t→ ∞. The curve is therefore concave down.
c.
dV
dt
dV dt V t
V t
= ⇒ = ⇒ = +
z z
4 4 400 4
400 0
0
2000
t
V
400
0
1
t
C
ρ=constant
dV dt = 4

11-22
dC
dt
C
V
=
−
8 8 dC
dt
C
t
=
−
+
1
50 05
.
dC
C
C t
t
t
t C
t
C t C t
1
1 2 50 05
2
50 05
50
1 0 01
1 0 01 1
1
1 0 01
0 0 0 0
2
2
2
−
= ⇒ − − = +
⇒ =
+
= +
⇒ = + ⇒ = −
+
z z dt
50+ 0.5t
ln(1- C)
1
1- C
-1
ln( ) ln( . )
ln
.
ln( . )
( . )
( . )
When the tank overflows, V t t
= + = ⇒ =
400 4 2000 400 s
C = 1-
1
1+ 0.01 400
g / L
×
=
b g2
0 96
.
11.28 a.Salt Balance on the 1st
tank:
Accumulation=-Output
d(C
dt
dC
dt
g / L
S1 S1
E
= − ⇒ = − = −
= =
V
C v C
v
V
C
C
S S S
S
1
1 1
1
1
1
0 08
0 1500 500 3
)

.
( )
Salt Balance on the 2nd tank:
Accumulation=Input-Output
d(C dC
dt
g / L
S2 S2
E
= − ⇒ = − = −
=
V
dt
C v C v C C
v
V
C C
C
S S S S S S
S
2
1 2 1 2
2
1 2
2
0 08
0 0
)
( )

. ( )
( )
Salt Balance on the 3rd tank:
Accumulation=Input-Output
d(C dC
dt
g / L
S3 S3
E
= − ⇒ = − = −
=
V
dt
C v C v C C
v
V
C C
C
S S S S S S
S
3
2 3 2 3
3
2 3
3
0 04
0 0
)
( )

. ( )
( )
b.
0
3
t
C
S1
,
C
S2
,
C
S3
CS1
CS2
CS3
V t
= +
400 4
11.27 (cont’d)

11-23
The plot of CS1 vs. t begins at (t=0, CS1=3). When t=0, the slope (=dCS1/dt) is − × = −
0 08 3 0 24
. . .
As t increases, CS1 decreases ⇒ dCS1/dt=-0.08CS1 becomes less negative, approaches zero as
t→ ∞. The curve is therefore concave up.
The plot of CS2 vs. t begins at (t=0, CS2=0). When t=0, the slope (=dCS2/dt) is 0 08 3 0 0 24
. ( ) .
− = .
As t increases, CS2 increases, CS1 decreases (CS2 CS1)⇒ dCS2/dt =0.08(CS1-CS2) becomes less
positive until dCS2/dt changes to negative (CS2 CS1). Then CS2 decreases with increasing t as well
as CS1. Finally dCS2/dt approaches zero as t→∞. Therefore, CS2 increases until it reaches a
maximum value, then it decreases.
The plot of CS3 vs. t begins at (t=0, CS3=0). When t=0, the slope (=dCS3/dt) is 0 04 0 0 0
. ( )
− = .
As t increases, CS2 increases (CS3 CS2)⇒ dCS3/dt =0.04(CS2-CS3) becomes positive ⇒ CS2
increases with increasing t until dCS3/dt changes to negative (CS3 CS1). Finally dCS3/dt
approaches zero as t→∞. Therefore, CS3 increases until it reaches a maximum value then it
decreases.
c.
11.29 a.(i) Rate of generation of B in the 1st
reaction: r r C
B A
1 1
2 0 2
= = .
(ii) Rate of consumption of B in the 2nd
reaction: − = =
r r C
B B
2 2
2
0 2
.
b.Mole Balance on A:
Accumulation=-Consumption
(
mol / L
E
= − ⇒ = −
= =
d C V
dt
C V
dC
dt
C
t C
A
A
A
A
A
)
. .
, .
01 01
0 100
0
Mole Balance on B:
Accumulation= Generation-Consumption
(
mol / L
E
= − ⇒ = −
= =
d C V
dt
C V C V
dC
dt
C C
t C
B
A B
B
A B
B
)
. . . .
,
0 2 0 2 0 2 0 2
0 0
2 2
0
0
0.5
1
1.5
2
2.5
3
0 20 40 60 80 100 120 140 160
t (s)
C
S1
,
C
S2
,
C
S3
(g/L)
CS1
CS2
CS3
11.28 (cont’d)

11-24
c.
The plot of CA vs. t begins at (t=0, CA=1). When t=0, the slope (=dCA/dt) is − × = −
01 1 01
. . .
As t increases, CA decreases ⇒ dCA/dt=-0.1CA becomes less negative, approaches zero as
t→∞. CA→0 as t→∞. The curve is therefore concave up.
The plot of CB vs. t begins at (t=0, CB=0). When t=0, the slope (=dCB/dt) is 02 1 0 02
. ( ) .
− = .
As t increases, CB increases, CA decreases ( CB
2
CA)⇒ dCB/dt =0.2(CA- CB
2
) becomes less positive
until dCB/dt changes to negative ( CB
2
CA). Then CB decreases with increasing t as well as CA.
Finally dCB/dt approaches zero as t→∞. Therefore, CB increases first until it reaches a maximum
value, then it decreases. CB→0 as t→∞.
The plot of CC vs. t begins at (t=0, CC=0). When t=0, the slope (=dCC/dt) is 0 2 0 0
. ( ) = . As t
increases, CB increases ⇒ dCc/dt =0.2 CB
2
becomes positive also increases with increasing t
⇒ CC increases faster until CB decreases with increasing t ⇒ dCc/dt =0.2 CB
2
becomes less positive,
approaches zero as t→∞ so CC increases more slowly. Finally CC→2 as t→∞. The curve is therefore
S-shaped.
d.
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
0 10 20 30 40 50
t (s)
C
A
,
C
B
,
C
C
(mol/L)
CA
CB
CC
0
1
2
t
C
A
,
C
B
,
C
C
CA
CB
CC
11.29 (cont’d)

11-25
11.30 a. When x =1, y =1.
y
ax
x b
a
b
a b
x y
=
+
⇒ =
+
⇒ = +
= =
1 1
1
1
1
,
b.Raoult’s Law: p yP xp y
xp
P
C H C H
C H
5 12 5 12
5 12
46
46
= = ⇒ =
* ( )
* ( )
o
o
C
C
Antoine Equation: 5 12
1060.793
(6.84471 )
o 46 231.541
* (46 C) 10 1053 mm Hg
C H
p
−
+
= =
⇒ = =
×
=
y
xp
P
C H
* ( ) .
.
5 12
46 0 7 1053
760
0970
o
C
y
ax
x b
a
b
a
b
=
+
=
+
R
S
|
T
|
⇒
=
=
R
S
|
T
|
From part (a), a =1+ b
0970
0 70
0 70
1
2
1078
0 078
.
.
.
( )
( )
.
.

c. Mole Balance on Residual Liquid:
mol
E
= −
= =
dN
dt
n
t N
L
V
L

,
0 100
Balance on Pentane:
dx
dt
x = 0.70
E
= − ⇒ + = −
+
E = −
= −
+
−
F
HG I
KJ
=
d N x
dt
n y x
dN
dt
N
dx
dt
n
ax
x b
dN dt n
n
N
ax
x b
x
t
L
V
L
L V
L V
V
L
( )

/

,
0
d.Energy Balance: Consumption=Input
kJ / mol
E
= =

.
n H Q n
Q
V vap V
Δ

27 0
b g
From part (c),
dN
dt
n N n t
Qt
L
V L V
= − = − = −

.
100 100
27 0
.

.
n
N
Q
Qt
V
L
=
27 0
100
27 0
-
Substitute this expression into the equation for dx/dt from part (c):
x=0.70, y=0.970
Δ

Hvap=27 0
. kJ/mol
t N L
= =
0 100
, mol

11-26
dx
dt
n
N
ax
x b
x
Q
Qt
ax
x b
x
V
L
= −
+
−
F
HG I
KJ = −
+
−
F
HG I
KJ
.

.
270
100
270
-
x(0) = 0.70
e.
f. The mole fractions of pentane in the vapor product and residual liquid continuously decrease over a
run. The initial and final mole fraction of pentane in the vapor are 0.970 and 0, respectively. The
higher the heating rate, the faster x and y decrease.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 200 400 600 800 1000 1200 1400 1600 1800
t(s)
x,
y
y (Q=1.5 kJ/s)
x (Q=1.5 kJ/s)
y (Q=3 kJ/s)
x (Q=3 kJ/s)
11.30 (cont’d)

Solucionario_Felder.pdf

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Solucionario_Felder.pdf