presentation on whats PONCHON-SAVARIT-METHOD

 For a non-ideal system, where the molar latent heat is
no longer constant and where there is a substantial
heat of mixing, the calculations become much more
tedious.
 For binary mixtures of this kind a graphical model has
been developed by RUHEMANN, PONCHON, and
SAVARIT, based on the use of an enthalpy-composition
chart.
 It is necessary to construct an enthalpy-composition
diagram for particular binary system over a
temperature range covering the two-phase vapor-
liquid region at the pressure of the distillation.

The following data are needed:
1. Heat capacity as a function of temperature,
composition and pressure.
2. Heat of mixing and dilution as a function of
temperature and composition.
3. Latent heats of vaporization as a function of
composition and pressure or temperature.
4. Bubble-point temperature as a function of
composition and pressure.



n
i
i
m h
h (1)
In “regular” / ideal mixtures:
o
i
i
i h
x
h 
(2)
For gaseous / vapor mixtures at normal T and P:

 


n
i
i
i
n
i
i
m y
h
H (3)
Enthalpy of liquid:

The equations used to calculate enthalpy of liquid and
vapor are:
sol
B
B
A
A
mix H
h
x
h
x
h 



    sol
ref
B
,
P
B
ref
A
,
P
A
mix H
T
T
C
x
T
T
C
x
h 





mix
mix
mix h
H 


B
B
A
A
mix x
x 




(3)
(4)
(5)
(6)

EXAMPLE 2
Devise an enthalpy-concentration diagram for the
heptane-ethyl benzene system at 760 mm Hg, using the
pure liquid at 0C as the reference state and assuming
zero heat of mixing.
SOLUTION
heptane ethyl benzene
TB (C) 136.2 98.5
CP (cal/mole K) 51.9 43.4
 (cal/mole) 7575 8600

t, C xH yH H EB
136.2 0.000 0.000 -- 1.00
129.5 0.080 0.233 1.23 1.00
122.9 0.185 0.428 1.19 1.02
119.7 0.251 0.514 1.14 1.03
116.0 0.335 0.608 1.12 1.05
110.8 0.487 0.729 1.06 1.09
106.2 0.651 0.834 1.03 1.15
103.0 0.788 0.904 1.00 1.22
100.2 0.914 0.963 1.00 1.27
98.5 1.000 1.000 1.00 --

t = 136.2C
xH = 0.0 xEB = 1.0
    sol
ref
EB
,
P
EB
ref
H
,
P
H H
t
t
C
x
t
t
C
x
h 





= 0 + 1.0 (43.4) (136.2 – 0) + 0 = 5,911 cal/mole
mix = xH H + xEB EB = 0 + 1.0 (8,600) = 8,600
H = h + mix = 5,920 + 8,600 = 14,511 cal/mole

t = 129.5C
xH = 0.08 xEB = 0.92
    sol
ref
EB
,
P
EB
ref
H
,
P
H H
t
t
C
x
t
t
C
x
h 





= 0.08 (51.9) (129.5) + 0.92 (43.4) (136.2)
= 5,708 cal/mole
mix = xH H + xEB EB = 0.08 (7,575) + 0.92 (8,600) = 8,518
H = h + mix = 5,708 + 8,518 = 14,226 cal/mole
The computations are continued until the last point
where xH = 1.0 and xEB = 0.0

t, C xH h
(cal/mole)
H
(cal/mole)
136.2 0.000 5,911 14,511
129.5 0.080 5,708 14,226
122.9 0.185 5,527 13,937
119.7 0.251 5,450 13,793
116.0 0.335 5,365 13,621
110.8 0.487 5,267 13,368
106.2 0.651 5,197 13,129
103.0 0.788 5,160 12,952
100.2 0.914 5,127 12,790
98.5 1.000 5,112 12,687

0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
16,000
18,000
20,000
0 0.2 0.4 0.6 0.8 1
x
H
Vapor
2 Phase
Liquid Saturated liquid
Saturated vapor

q
F
V
L
Over-all material balance:
F = V + L
F xF = V y + L x
F hF = V H + L h
(7)
(8)
(9)
The enthalpy-concentration diagram may be used to
evaluate graphically the enthalpy and composition of
streams added or separated.
Steady-state flow system
with phase separation
and heat added
Component material balance
Enthalpy balance:

For adiabatic process, q = 0:
V (H – hF) = L (hF – h)
V (y – xF) = L (xF – x)
h
h
h
H
V
L
F
F



x
x
x
y
V
L
F
F



(10)
(12)
(11)
(13)
Substituting eq. (7) to (9) gives:
  h
L
H
V
h
L
V F 



x
x
x
y
h
h
h
H
F
F
F
F





(14)
x
x
h
h
x
y
h
H
F
F
F
F





Eq. (14) can be rearranged:
(15)




h
hF
H
L
F
V
x xF
y
Enthalpy-concentration
lines – adiabatic, q = 0
:
is
VF
line
of
slope
The
___
F
F
x
y
h
H


:
___
is
FL
line
of
slope
The
x
x
h
h
F
F


According to eq. (15), the
slopes of both lines are the
same.
Since both lines go through
the same point (F), the lines
lie on the same straight line.

L
F
V
H
hF
h
x xF y
A
B
LEVER-ARM RULE PRINCIPLE
h
h
h
H
V
L
F
F



Consider triangle LBV
V
L
h
h
h
H
BA
AV
LF
FV
F
F
___
___
___
___





___
___
LF
FV
V
L

Similarly: ___
___
LV
LF
F
V
 ___
___
LV
FV
F
L


D, xD, HL
D
B, xB, HL
B
F
xF
HF
qD
qB
V1
L0
x0
HL
0
Over-all material balance:
F = D + B (16)
Component material balance:
F xF = D xD + B xB (17)
F xF = D xD + (F – D) xB (18)
 
B
D
B
F
x
x
x
x
F
D


 (19)
Enthalpy balance:
(20)
B
D
D
B
F h
B
q
h
D
q
h
F 




V1 = L0 + D (21)
V1 y1 = L0 x0 + D x0 (22)
Material balance around condenser:
Enthalpy balance:
A
V1
L1
L0
D
xD
qD
qD + V1 H1 = L0 h0 + D hD
(23)

Combining eqs. (21) and (24):
D
1
1
D
D
0
h
H
H
Q
h
D
L




Internal reflux is shown as:
0
D
D
1
D
D
1
0
h
Q
h
H
Q
h
V
L





(25)
(26)
Designating:
D
q
Q D
D 
V1 H1 = L0 h0 + D (hD – QD) (24)

Internal reflux between
each plate, until a point
in the column is reached
where a stream is added
or removed, can be
shown as:
m
D
D
m
D
D
m
m
h
Q
h
H
Q
h
V
L




 

1
1
(27)
A
Vm+1
Lm
L0
D
xD
qD
m


L0, D
h0, hD
V1
H1

(hD – QD), xD
hD – QD – H1
H1 – hD
y1, x0, xD
H
or
h


V1 = L0 + D (28)
y1 V1 = x0 L0 + yD D (29)
Material balance:
Enthalpy balance:
qD + V1 H1 = L0 h0 + D HD (30)
Designating:
D
q
Q D
D 
V1 H1 = L0 h0 + D (hD – QD) (31)
A
v1
v2
v3
vF
L1
L2
LF-1
L0
D, yD
F
L
F
xF
qD

0
1
1
D
D
0
h
H
H
Q
H
D
L




Internal reflux is shown as:
0
D
D
1
D
D
1
0
h
Q
H
H
Q
H
V
L





(32)
(33)
Internal reflux between each plate, until a point in the
column is reached where a stream is added or removed,
can be shown as:
m
D
D
1
m
D
D
1
m
m
h
Q
H
H
Q
H
V
L




 

(34)

h0, x0
V1
HD, yD

(HD – QD), yD
HD – QD – H1
H1 – h0
y1, x0, yD
H
or
h
D




V1
V3
Vn+1
L1
L2
Ln
L0
D
xD
F
L
qD
The material balance equation
maybe rearranged in the from
of difference:
L0 – V1 = L1 – V2 = L2 – V3
= . . . . = Lm – Vm+1
= – D = 

(35)
L0 – V1 = – D = 
n
V2

xD = x (37)
For the component material balance:
L0 x0 – V1 y1 = L1 x1 – V2 y2
= L2 x2 – V3 y3 = . . . .
= Lm xm – Vm+1 ym+1
= – D xD =  x
(36)
L0 x0 – V1 y1 = – D xD =  x

For the enthalpy balance:
L0 h0 – V1 H1 = L1 h1 –V2 H2 = L2 h2 –V3 H3 = . . . .
= Lm hm – Vm+1 Hm+1 = – D (hD – QD) =  h
(38)
• These 3 independent equations [eqs. (35), (36), and (37)] can be
written for rectifying section of the column between each plate.
• On the enthalpy scale and on the composition scale, the
differences in enthalpy and in composition always pass through
the same point,  ([xD, (hD – QD)]
• This is designated as point , the difference point, and all lines
corresponding to the combined material and enthalpy balance
equations (operating line equations) for the rectifying section of
the column pass through this intersection.
h = hD – QD
(39)

PLATE-TO-PLATE GRAPHICAL PROCEDURE FOR DETERMINING
THE NUMBER OF EQUILIBRIUM STAGES:
1. Use R, xD, HD or hD to establish the location of point  with x = xD
and h = hD – QD or h = HD – QD
2. Use Equilibrium data alone to establish the point L1 at (x1, h1).
Since L1 is assumed to be a saturated liquid, x1 must lie on the
saturated-liquid line.
3. Draw the operating line between L1 and . This line intersects the
saturated-vapor line at V2 (y2, H2).
4. Repeat steps 2 and 3 until the feed plate is reached.


x or y
H
or
h
V1
V2
V3
V4
D
L1
L2
L3
 (x, h)

B
xB
qB

m
N
m
L
1

m
V
1

M
V
M
V 1

M
L
M
L

The material balance equation maybe rearranged in
the from of difference:
M
M
M
M V
L
B
V
L 


 
 1
1
.
.
.
1
2 

 
 M
M L
L



 1
m
m V
L (40)

For the component material balance:
M
M
M
M
B
M
M
M
M y
V
x
L
x
B
y
V
x
L 


 


 1
1
1
1
.
.
.
1
1
2
2 

 


 M
M
M
M y
V
x
L


 


 x
y
V
x
L m
m
m
m 1
1 (41)
B
x
x 
 (42)

For the enthalpy balance:
(43)
  M
M
M
M
B
B
M
M
M
M H
V
h
L
Q
h
B
H
V
h
L 



 


 1
1
1
1
.
.
.
1
1
2
2 

 


 M
M
M
M H
V
h
L


 


 h
H
V
h
L m
m
m
m 1
1
B
B Q
h
h 

 (44)

• These 3 independent equations [eqs. (40), (41), and
(43)] can be written for stripping section of the column
between each plate.
• On the enthalpy scale and on the composition scale,
the differences in enthalpy and in composition always
pass through the same point, [xB, (hB – QB)].
• This is designated as point , the difference point, and
all lines corresponding to the combined material and
enthalpy balance equations (operating line equations)
for the stripping section of the column pass through
this intersection.



F = D + B (45)
Combining eq. (45) with eqs. (35) and (40) gives:
F



 (46)
Equation (46) implies that lies on the extension of the
straight line passing through F and .

QB is usually not known. It can be derived from over-all
material balance:

PLATE-TO-PLATE GRAPHICAL PROCEDURE FOR
DETERMINING THE NUMBER OF EQUILIBRIUM STAGES:
1. Draw a straight line passing through F and .
2. Draw a vertical straight line at xB all the way down until it
intersects the extension of line F in
3. Assuming the reboiler to be an equilibrium stage, the vapor
VM+1 is in equilibrium with the bottom stream.
4. Use equilibrium data alone to establish the value of ym+1 on
the saturated-vapor line.
5. Draw the operating line between Lm(xm, hm) and VM+1. This
line intersects the saturated-liquid line at




x or y
H
or
h
hB
1
M
V  M
V 1
M
V 
 
B
x
,

LM LM-1

D
B
F
qD
qB
V1
L0


• The construction may start from
either side of the diagram,
indicating either the condition at
the top or the bottom of the
column.
• Proceed as explained in previous
slides.
• In either case, when an equilibrium
tie line crosses the line connecting
the difference points through the
feed condition, the other
difference point is used to
complete the construction.

H
or
h
F
1
2
3
4
5
6
7
8
9

xF xD
xB



EXAMPLE 3
Using the enthalpy-concentration diagram from Example 2,
determine the following for the conditions in Example 1, assuming a
saturated liquid feed.
a. The number of theoretical stages for an operating reflux ratio of
R = L0/D = 2.5
b. Minimum reflux ratio L0/D.
c. Minimum equilibrium stages at total reflux.
d. Condenser duty feeding 10,000 lb of feed/hr, Btu/hr.
e. Reboiler duty, Btu/hr.
SOLUTION
(a) From the graph: hD = h0 = 5,117 cal/mole
H1 = 12,723 cal/mole
D
1
1
D
D
0
h
H
H
Q
h
D
L




117
,
5
723
,
12
723
,
12
Q
117
,
5
5
.
2 D



 QD = – 26,621 cal/mole

h = hD – QD = 5,117 – (– 26,621) = 31,738 cal/mole
The coordinate of point  is:
x = xD = 0.97
• Draw a straight line passing through  and F.
• Extend the line until it intersects a vertical line passing
through xB, at
• Draw operating lines and equilibrium lines in the whole
column using the method explained in the previous slides.

Number of stages = 11


0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.2 0.4 0.6 0.8 1
x
y

F
0
5,000
10,000
15,000
20,000
25,000
30,000
35,000
0 0.2 0.4 0.6 0.8 1
H

 = 21,700 cal/mole
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.2 0.4 0.6 0.8 1
x
y
0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
16,000
18,000
20,000
0 0.2 0.4 0.6 0.8 1
H

F
D
1
1
D
D
0
h
H
H
Q
h
D
L




D
1
1
h
H
H
h


 
117
,
5
723
,
12
723
,
12
700
,
21
D
L
min
0









= 1.18
(b)

0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.2 0.4 0.6 0.8 1
x
y

0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
16,000
18,000
20,000
0 0.2 0.4 0.6 0.8 1
H
F
1
2
3
4
5
6
7
N = 7
(c)

(d) hD – QD = h = 31,738 cal/mole
hD = 5,117 cal/mole
QD = – 26,621 cal/mole
F
mole
lb
103
hr
F
lb
000
,
10
F
mole
D
mole
426
.
0
mole
cal
mole
lb
Btu
8
.
1
mole
cal
621
,
26
QD 










= – 1,981,843 Btu/hr
(e) hB – QB = – 14,350 cal/mole
hB = 5,886 cal/mole
QB = 14,350 + 5,886 = 20,236 cal/mole
F
mole
lb
103
hr
F
lb
000
,
10
F
mole
B
mole
574
.
0
mole
cal
mole
lb
Btu
8
.
1
mole
cal
236
,
26
QD 









= 2,631,751 cal/mole

presentation on whats PONCHON-SAVARIT-METHOD

More Related Content

Similar to presentation on whats PONCHON-SAVARIT-METHOD (20)

More from ssuserb6ce69 (20)

Recently uploaded (20)

presentation on whats PONCHON-SAVARIT-METHOD