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![1
CHAPTER 1
Basic Considerations
1.1 Conservation of mass — Mass — density
Newton’s second law — Momentum — velocity
The first law of thermodynamics — internal energy — temperature
1.2 a) density = mass/volume = M L
/
3
b) pressure = force/area = F L ML T L M LT
/ / /
2 2 2 2
= =
c) power = force × velocity = F L T ML T L T ML T
× = × =
/ / / /
2 2 3
d) energy = force × distance = ML T L ML T
/ /
2 2 2
× =
e) mass flux = ρAV = M/L3
× L2
× L/T = M/T
f) flow rate = AV = L2
× L/T = L3
/T
1.3 a) density =
M
L
FT L
L
FT L
3
2
3
2 4
/
/
=
b) pressure = F/L2
c) power = F × velocity = F × L/T = FL/T
d) energy = F × L = FL
e) mass flux =
M
T
FT L
T
FT L
= =
2
/
/
f) flow rate = AV = L2
× L/T = L3
/T
1.4 (C) m = F/a or kg = N/m/s2
= N.
s2
/m.
1.5 (B) [µ] = [τ/du/dy] = (F/L2
)/(L/T)/L = F.
T/L2
.
1.6 a) L = [C] T2
. ∴[C] = L/T2
b) F = [C]M. ∴[C] = F/M = ML/T2
M = L/T2
c) L3
/T = [C] L2
L2/3
. ∴[C] = L T L L L T
3 2 2 3 1 3
/
/ /
⋅ ⋅ =
Note: the slope S0 has no dimensions.
1.7 a) m = [C] s2
. ∴[C] = m/s2
b) N = [C] kg. ∴[C] = N/kg = kg ⋅ m/s2
⋅ kg = m/s2
c) m3
/s = [C] m2
m2/3
. ∴[C] = m3
/s⋅m2
⋅ m2/3
= m1/3
/s
1.8 a) pressure: N/m2
= kg ⋅ m/s2
/m2
= kg/m⋅ s2
b) energy: N⋅ m = kg ⋅ m/s2
× m = kg⋅ m2
/s2
c) power: N⋅ m/s = kg ⋅ m2
/s3
d) viscosity: N⋅ s/m2
=
kg m
s
s
1
m
kg / m s
2 2
⋅
⋅ = ⋅](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-2-2048.jpg)
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e) heat flux: J/s =
N m
s
kg m
s
m
s
kg m / s
2
2 3
⋅
=
⋅
⋅ = ⋅
f) specific heat:
J
kg K
N m
kg K
kg m
s
m
kg K
m / K s
2
2 2
⋅
=
⋅
⋅
=
⋅
⋅
⋅
= ⋅
1.9 kg
m
s
m
s
m
2
+ + =
c k f. Since all terms must have the same dimensions (units) we
require:
[c] = kg/s, [k] = kg/s2
= N s / m s N / m,
2 2
⋅ ⋅ = [f] = kg m / s N.
2
⋅ =
Note: we could express the units on c as [c] = kg / s N s / m s N s / m
2
= ⋅ ⋅ = ⋅
1.10 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm3
e) 1.2 cm2
f) 76 mm3
1.11 a) 1.25× 108
N b) 3.21× 10−5
s c) 6.7× 108
Pa
d) 5.6× 10−12
m3
e) 5.2 × 10−2
m2
f) 7.8 × 109
m3
1.12 (A) 8 9
2.36 10 23.6 10 23.6 nPa.
− −
× = × =
1.13 2 2 2
0.06854
0.225 0.738
0.00194 3.281
m m
d d
λ
ρ ρ
= =
×
where m is in slugs, ρ in slug/ft3
and d in feet. We used the conversions in the
front cover.
1.14 a) 20 cm/hr = 5
20
/3600 5.555 10 m/s
100
−
= × 5
20
/3600 5.555 10 m/s
100
−
= ×
b) 2000 rev/min = 2000 2
× π/60 = 209.4 rad/s
c) 50 Hp = 50 × 745.7 = 37 285 W
d) 100 ft3
/min = 100 × 0.02832/60 = 0.0472 m3
/s
e) 2000 kN/cm2
= 2× 106
N/cm2
× 1002
cm2
/m2
= 2× 1010
N/m2
f) 4 slug/min = 4 × 14.59/60 = 0.9727 kg/s
g) 500 g/L = 500 × 10−3
kg/10−3
m3
= 500 kg/m3
h) 500 kWh = 500 × 1000× 3600 = 1.8× 109
J
1.15 a) F = ma = 10 × 40 = 400 N.
b) F − W = ma. ∴ F = 10 × 40 + 10 × 9.81 = 498.1 N.
c) F − W sin 30° = ma. ∴ F = 10 × 40 + 9.81 × 0.5 = 449 N.
1.16 (C) The mass is the same on earth and the moon: [4(8 )] 32 .
du
r r
dr
τ µ µ µ
= = =
1.17 The mass is the same on the earth and the moon:](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-3-2048.jpg)
![5
1.34 τ= µ
du
dr
= 1.92 × 10−5
2
30(2 1/12)
(1/12)
×
= 0.014 lb/ft2
1.35 T = force × moment arm = τ2πRL × R = µ
du
dr
2πR2
L = µ
2
0.4
1000
R
+
2πR2
L.
∴µ=
2 2
2
0.0026
0.4 0.4
1000 2 1000 2 .01 0.2
12
T
R L
R
π π
=
+ + × ×
= 0.414 N.
s/m2
.
1.36 Use Eq.1.5.8: T =
3
2 R L
h
π ω µ
=
( )
2 .5 / 12
π
π
× ×
×
× ×
3 2000 2
60
4 006
01 12
.
. /
= 2.74 ft-lb.
Hp =
Tω
550
2 74 209
550
=
×
. .4
= 1.04 Hp
1.37 Fbelt = µ
du
dy
A = × −
131 10
10
002
3
.
.
(.6 × 4) = 15.7 N.
Hp =
F V
×
=
×
746
15 7 10
746
.
= 0.210 Hp
1.38 Assume a linear velocity so .
du r
dy h
ω
= Due to the area
element shown, dT = dF × r = τdA × r = µ
du
dy
2πr dr × r.
dr
r
τ
T = 3
0
2
R
r dr
h
µω π
∫ =
2
4
2 36 10
400 2
60
3 12
2 08 12
4
5 4
πµω
π
π
h
R
=
× × ×
×
×
×
−
. ( / )
. /
= 91 × 10−5
ft-lb.
1.39
2
30(2 1/12)
(1/12)
×
2 2
0 0
[32 / ] 32 / .
du
r r r r
dr
τ µ µ µ
= = = ∴τr = 0 = 0,
τr=0.25 = 32 × 1 × 10−3
× 2
.25/100
(.5/100)
= 3.2 Pa, τr=0.5 = 32 × 1 × 10−3
× 2
.5/100
(.5/100)
= 6.4 Pa
1.40 (A) 3
[10 5000 ] 10 10 5000 0.02 1 Pa.
du
r
dr
τ µ µ −
= = × = × × × =
1.41 The velocity at a radius r is rω. The shear stress is τ µ
=
∆
∆
u
y
.
The torque is dT = τrdA on a differential element. We have](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-6-2048.jpg)
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2.8 ∆p = ρgh = 0.0024 × 32.2 (–10,000) = –773 psf or –5.37 psi.
2.9 (D) 0 84000 1.00 9.81 4000 44760 Pa
p p gh
ρ
= − = − × × =
2.10
inside
100 9.81
3 13.51 Pa
.287 253
100 9.81
3 11.67 Pa
.287 293
outside o
o
base
i
i
pg
p g h h
RT
p
pg
p g h h
RT
ρ
ρ
×
∆ = ∆ = ∆ = × =
×
∴∆
×
∆ = ∆ = ∆ = × =
×
= 1.84 Pa
If no wind is present this ∆pbase would produce a small infiltration since the higher
pressure outside would force outside air into the bottom region (through cracks).
2.11 p = ρgdh where h = –z. From the given information S = 1.0 + h/100 since S(0) = 1 and
S(10) = 1.1. By definition ρ = 1000 S, where ρwater = 1000 kg/m3
.
Then dp = 1000 (1 + h/100) gdh. Integrate:
dp h gdh
p
= +
∫
∫ 1000 1 100
0
10
0
( / )
p = × +
×
1000 9 81 10
10
2 100
2
. ( ) = 103 000 Pa or 103 kPa
Note: we could have used an average S: Savg = 1.05, so that ρavg = 1050 kg/m3
.
2.12
v
∇ = + +
p
p
x
i
p
y
j
p
z
k
∂
∂
∂
∂
∂
∂
$ $ $
= – ρa i
x
$ – ρα y j
$ – ρα z k
$ – ρgk
$ = – ( )
ρ a i a j a k
x y z
$ $ $
+ + – ρgk
$
= – ρ
v
a – ρ
v
g
∴ ∇ = − +
v v v
p a g
ρ( )
2.13 /
0 0
[( ) / ]g R
atm
p p T z T α
α
= −
= 100 [(288 − 0.0065 × 300)/288]9.81/.0065 × 287
= 96.49 kPa
p p gh
atm
= − = −
×
× ×
ρ 100
100
287 288
9 81 300 1000
.
. / = 96.44 kPa
% error =
96 96
96
100
.44 .49
.49
−
× = −0.052%
The density variation can be ignored over heights of 300 m or less.](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-14-2048.jpg)
![18
2.41 F hA
= γ = 9810 × 10 × π × .32
/4 = 6934 N.
2.42
5 1 5 5
(2 ) (2 ) [9800 1 3 (2 )]. 32670 N
3 3 3 3
P P
× × = × × × × × × × ∴ = a) F = pc A = 9800 × 2
× 42
= 313 600 N or 313.6 kN
b)
2 2
9800 1 (2 4) 9800 2 9800 1 98000 N or 98.0 kN
3 3
c
F p A
= = × × × + × × + × × =
c) F = pc A = 9800 × 1 × 2 × 4 × 2 = 110 900 N or 110.9 kN
d) F = pc A = 9800 × 1 × 2 × 4/.866 = 90 500 N or 90.5 kN
2.43 For saturated ground, the force on the bottom tending to lift the vault is
F = pc A = 9800 × 1.5 × (2 × 1) = 29 400 N.
The weight of the vault is approximately
W gV
ρ
= 2400 9.81
walls = × [2(2×1.5×.1) + 2(2×1×.1) + 2(.8×1.3×.1)] = 28 400 N.
The vault will tend to rise out of the ground.
2.44 F = pc A = 6660 × 2 × π × 22
= 167 400 N or 167.4 kN
Find γ in Table B.5 in the Appendix.
2.45 a) F = pc A = 9800 (10 − 2.828/3) (2.828 × 2/2) = 251 000 N or 251 kN
where the height of the triangle is (32
− 12
)1/2
= 2.828 m.
b) F = pc A = 9800 × 10 (2.828 × 2/2) = 277 100 N or 277.1 kN
c) F = pc A = 9800 (10 − 2.828 × .866/3) (2.828 × 2/2) = 254 500 N or 254.5 kN
2.46 a) F hA
= = × × =
γ 62 27 33 24
.4 . 40,930 lb.
yp = +
×
×
27 33
6 8 36
27 33 24
3
.
/
.
= 27.46'. ∴y = 30 – 27.46 = 2.54'.
8/5.46 = 3/x. ∴x = 2.05’. (2.05, 2.54) ft.
b) F = 62.4 × 30 × 24 = 44,930 lb. The centroid is the center of pressure.
y = 2.667'. 8/5.333 = 3/x. ∴x = 2.000' (2.000, 2.667) ft.
c) F = 62.4 (30 – 2.667 × .707) × 24 = 42,100 lb.
yp = +
×
×
39 77
6 8 36
39 77 24
3
.
/
.
= 39.86'. y = 42.43 – 39.86 = 2.57'
8/5.43 = 3/x. ∴x = 2.04'. (2.04, 2.57) ft.
2.47 (B) The force acts 1/3 the distance from the hinge to the water line:
5 1 5 5
(2 ) (2 ) [9800 1 3 (2 )]. 32670 N
3 3 3 3
P P
× × = × × × × × × × ∴ =
(x, y)
y
x](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-19-2048.jpg)
 /
3 2 36
2 3
∴ =
+
+
+
+ +
=
+
+
+
=
+
y
h H h H
h H h H
h H h H h H
p
3
2 36
3 3 6 2
4
2
( ) /
( ) ( ) /
a) h
h H
=
+
2
. ∴h = H = 0.9 m
b) h = H = 1.2 m
c) h = H = 1.5 m
2.56 The gate is about to open when the center of pressure is at the hinge.
a) y H H
b
H b
p = + = + +
×
+
12 18 2
18 12
9 18
3
. ( . / )
. /
(. ) .
. ∴H = 0.](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-21-2048.jpg)
![26
2.81 From a force balance FB = W + pA.
a) The buoyant force is found as follows (h > 16'):
cos ,
θ =
− −
h R
R
15
Area = θR2
– (h – 15 – R) R sinθ
∴FB = 10 × 62.4[πR2
− θR2
+ (h – 15 – R) R sinθ].
FB = 1500 + γhA.
The h that makes the above 2 FB’s equal is found by trial-and-
error:
h = 16.5: 1859 ? 1577 h = 16.8: 1866 ? 1858
h = 17.0: 1870 ? 1960 ∴h = 16.82 ft.
b) Assume h > 16
1
3
ft. and use the above equations with R = 1.333':
h = 16.4: 1857 ? 1853 ∴h = 16.4 ft.
c) Assume h < 16
2
3
ft. With R = 1.667',
FB = 10 × 62.4[θR2
− (R – h + 15) R sinθ].
FB = 1500 + γhA. cos θ =
− +
R h
R
15
Trial-and-error for h:
h = 16: 1849 ? 1374 h = 16.2: 1853 ? 1765
h = 16.4: 1857 ? 2170 ∴h = 16.25 ft.
2.82 a) W FB
= . [ ]
0 01 136 1000 015 4 9 81 9810
2
. . . / . .
+ × × × × = −
h V
π
− =
×
× +
×
× = × −
V
π π
.
.
.
. . .
015
4
15
005
4
06 2 769 10
2 2
5 3
m ∴h = 7.361 × 10−3
m
∴ =
mHg
2
13.6 1000 .015 / 4
hπ
× × × = 0.01769 kg
b) (.01 + .01769) 9.81 = 9810
π π
×
× +
×
×
.
.
.
. .
015
4
15
005
4
12
2 2
Sx ∴Sx = 0.959.
c) (.01 + .01769) 9.81 = 9810
π×
×
.
.
015
4
15
2
Sx. ∴Sx = 1.045.
2.83 (. ) .
.
.
.
. .
01 9 81 9810
015
4
15
005
4
12
2 2
+ =
×
× +
×
×
mHg
π π
∴mHg = 0.01886.
a) (.01 + .01886) 9.81 = 9810
π×
×
.
.
015
4
15
2
Sx. ∴Sx = 1.089.
b) mHg = 0.01886 kg.
h − 15
θ R
pA
FB
W
h − 15
θ R](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-27-2048.jpg)
![33
c) For ω = 10, part of the bottom is bared.
π π π
× × = −
. . .
6 2
1
2
1
2
2
0
2
1
2
1
r h r h
Using Eq. 2.6.5:
ω2
0
2
2
r
g
h
= ,
ω2
1
2
2
r
g
h
= 1.
∴ = −
0144
2 2
2
2
2 1
2
.
g
h
g
h
ω ω
or
h h
2
1
2
2
0 144 10
2 981
− =
×
×
.
.
.
Also, h – h1 = 0.8. 1.6h – 0.64 = .7339. ∴h = 0.859 m, r1 = 0.108 m.
∴pA =
1
2
× 1000 × 102
(0.62
– 0.1082
) = 17 400 Pa.
d) Following part (c): h h
2
1
2
2
0144 20
2 9 81
− =
×
×
.
.
. 1.6h – .64 = 2.936.∴ h = 2.235 m.
∴pA =
1
2
× 1000 × 202
(0.62
– 0.2652
) = 57 900 Pa r1 = 0.265 m
2.103 The answers to Problem 2.102 are increased by 25 000 Pa.
a) 33 150 Pa b) 35 780 Pa c) 42 400 Pa d) 82 900 Pa
2.104 p r r g h
( ) [ (. )].
= − − −
1
2
0 8
2 2
ρω ρ
p r r h
( ) (. )
= + −
500 9810 8
2 2
ω if h < .8.
p r r r
( ) ( )
= −
500 2 2
1
2
ω if h > .8.
a) F p rdr r r dr
= = +
∫
∫ 2 2 12 500 3650
3
0
6
π π ( )
.
= 6670 N.
(We used h = .428 m)
b) F p rdr r r dr
= = +
∫
∫ 2 2 24 500 1962
3
0
6
π π ( )
.
= 7210 N. (We used h = 0.6 m)
c) F p rdr r r dr
= = −
−
∫
∫ 2 2 50 000 108
3
108
6
2
π π ( ( . )
.
.
= 9520 N. (We used r1 = 0.108 m)
d) F p rdr r r dr
= = −
∫
∫ 2 2 200 000 265
3
265
6
2
π π ( ( . )
.
.
= 26 400 N. (r1 = 0.265 m)
1
h
z
r
A
r0
h1
dr
dA = 2πrdr](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-34-2048.jpg)
![36
b) cos $ / / ( ) ( ) . .
α α
= ⋅ = − − + − = − ∴ =
v o
V i V 2 2 8 0 2425 104
2 2
v
V n i j n i n j
n n
n n
n n
n n
x y
x y
x y
x y
y y
⋅ = − − ⋅ + =
− − =
+ =
∴
= −
+ =
$ . ( $ $) ( $ $) .
0 2 8 0
2 8 0
1
4
16 1
2 2 2 2
∴ = = − = − +
n n n i j
y x
1
17
4
17
1
17
4
, $ ( $ $).
or
c) cos $ / / ( ) . . .
α α
= ⋅ = + − = ∴ = −
v o
V i V 5 5 8 0 6202 51 67
2 2
v
V n i j n i n j
n n
n n
n n
n n
x y
x y
x y
x y
y y
⋅ = − ⋅ + =
− =
+ =
∴
=
+ =
$ . ( $ $) ( $ $) .
0 5 8 0
5 8 0
1
8
5
64
25
1
2 2
2 2
∴ = = = +
n n n i j
y x
5
89
8
89
1
89
8 5
, $ ( $ $).
or
3.11 a) [ ]
v v
V dr x i xtj dxi dyj
× = + + × + =
0 2 0
. ( )$ $ ( $ $) .
∴ + − =
+
=
( ) .
x dy xtdx t
xdx
x
dy
2 0
2
or
Integrate: [ ]
t
xdx
x
dy t x n x y C
+
= − + = +
∫
∫ 2
2 2
. .
l
2 1 2 3 2
( ) .
− = − + ∴
ln C C = 0.8028.
[ ]
t x n x y
− + = +
2 2 0 8028
l .
b) [ ]
v v
V dr xyi y j dxi dyj
× = − × + =
0 2 0
2
. $ $ ( $ $) .
∴ + = = −
xydy y dx
dx
x
dy
y
2 0
2
2
or .
Integrate: 2 2 1 2
l l l l
nx n y C n n C
= − = − −
( / ). ( ) ( / ).
∴ = − = − − ∴ = −
C nx n y x y
2 2 2
2 2
. ( / ). .
l l
c) [ ]
v v
V dr x i y tj dxi dyj
× = + − × + =
0 4 0
2 2
. ( )$ $ ( $ $) .
( ) .
x dy y tdx
tdx
x
dy
y
2 2
2 2
4 0
4
+ + =
+
= −
or
Integrate:
t x
C
y
C
2 2
1 2
2
1
2
1
2
1 1
tan . tan .
− −
+
= +
= −
∴ = − −
=
−
C yt
x
0 9636
2
0 9636 2
1
. . tan .](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-37-2048.jpg)
![39
3.19
V V
a u
t x
∂ ∂
∂ ∂
= +
v v
v
v
+
V
w
y
∂
+
∂
v
ˆ.
V u
i
z t
∂
∂
∂
=
∂
v
For steady flow ∂ ∂
u t a
/ .
= =
0 0
so that
v
3.20 Assume u(r,x) and v(r,x) are not zero. Then, replacing z with x in the appropriate
equations of Table 3.1 and recognizing that vθ ∂ ∂θ
= =
0 0
and / :
r x
v v u u
a v u a v u
r x r x
∂ ∂ ∂ ∂
∂ ∂ ∂ ∂
= + = +
3.21 a) u e t
t
= − − = = ∞
−
2 1 0 1 10
( )( ) .
/
2 m / s at
( )
a
u
t
e t
x
t
= = −
= =
−
∂
∂
2 1 0
1
10
0 2 0
10
( ) . .
/
m / s at
2
b) u e t
t
= − − = = ∞
−
2 1 0 1
2 10
( .5 )( ) .
/
1.875 m / s at
a e t
x
t
= −
= =
−
2 1 0 2
1
10
0 0125 0
2 2 10
( .5 / ) . .
/
m / s at
2
c) u e t
t
= − − =
−
2 1 2 2 1
2 2 10
( / )( ) .
/
0 for all
a e t
x
t
= −
=
−
2 1 2 2
1
10
0
2 2 10
( / ) .
/
for all
3.22
DT T
u
Dt x
∂
∂
= v
+
T
w
y
∂
∂
+ 2
20(1 ) sin 0.5878
100 100 5
T T t
y
z t
∂ ∂ π π π
∂ ∂
+ = − − = − ×
= −0.3693 °C/s.
3.23
D
Dt
u
x
v
y
w
z t
e
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= + + + = − × − − × −
10 1 23 10 4 3000 10 4
( . ) = − ×
⋅
−
9 11 10 4
. .
kg
m s
3
3.24
D
Dt
u
x
v
y
w
z t
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= + + + = −
10
1000
4
= −
⋅
2500
kg
m s
3
.
3.25
D
Dt
u
x
ρ ∂ρ
∂
= = ×
4 01
(. ) = 0.04 kg/m3⋅s
3.26 (D)
2
2
10
[10(4 ) ]
(4 )
x
u u u u u
a u v w u x
t x y z x x
x
−
∂ ∂ ∂ ∂ ∂ ∂
= + + + = = −
∂ ∂ ∂ ∂ ∂ ∂
−
3 2
2
10 10 1
10( 2)( 1)(4 ) 20 6.25 m/s .
4 8
(4 )
x
x
−
= − − − = × × =
−
3.27
D
Dt
V
t
= ⋅∇ +
v v ∂
∂
observing that the dot product of two vectors
v
A A i A j A k
x y z
= + +
$ $ $
and
v v v
B B i B j B k A B A B A B A B
x y z x x y y z z
= + + ⋅ = + +
$ $ $ .
is](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-40-2048.jpg)
![40
3.28
a
u
t
V u
a
v
t
V v
a
w
t
V w
a
V
t
V V
x
y
z
= + ⋅∇
= + ⋅∇
= + ⋅∇
∴ = + ⋅∇
∂
∂
∂
∂
∂
∂
∂
∂
v v
v v
v v
v
v
v v v
( )
3.29 Using Eq. 3.2.12:
a)
v v
v v v v v v
v
v
A a
d s
dt
V r
d
dt
r
= + + × + × × + ×
2
2
2Ω Ω Ω
Ω
( )
= 2 20 4 20 20 1
( $ $) $ ( $ .5$)
k i k k i
× + × × = 160 600
$ $
j i
− m / s
2
b)
v v v v v v o
A V r k j k k i
= × + × × = × − + × ×
2 2 20 20 30 20 20 3
Ω Ω Ω
( ) ( $ cos $) $ ( $ $) = −507$
i
3.30
v
Ω =
× ×
= × −
2
24 60 60
7 272 10 5
π $ . $
k k rad/s.
v
v i k i k
= − − = − −
5 707 707 3 3
( . $ . $) .535$ .535$ m/s.
v v v v v v
A V r
= × + × ×
2Ω Ω Ω
( )
= 2 7 272 10 3 3 7 272 10
5 5
× × × − − + × ×
− −
. $ ( .535$ .535 $) . $
k i k k
[ . $ ( . $ . $)]
7 272 10 6 10 707 707
5 6
× × × − +
−
k i k = − × +
−
51 4 10 0 0224
5
. $ . $
j i m / s2
.
Note: We have neglected the acceleration of the earth relative to the sun since it is quite small
(it is d s dt
2 2
v
/ ). The component 5 ˆ
( 51.4 10 )
j
−
− × is the Coriolis acceleration and causes air
motions to move c.w. or c.c.w. in the two hemispheres.
3.31 a) two-dimensional (r, z) b) two-dimensional (x, y)
c) two-dimensional (r, z) d) two-dimensional (r, z)
e) three-dimensional (x, y, z) f) three-dimensional (x, y, z)
g) two-dimensional (r, z) h) one-dimensional (r)
3.32 Steady: a, c, e, f, h Unsteady: b, d, g
3.33 b. It is an unsteady plane flow.
3.34 a) d) e)
3.35 f, h](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-41-2048.jpg)
![43
3.53 (C)
2 2 2
1 2 1
. 0.200 0.600. 2 9.81 0.400 2.80 m/s.
2 2 2
V V V
p
V
g g g
γ
+ = + = ∴ = × × =
3.54 (B) The manometer reading h implies:
2 2
2
1 1 2 2
2 2
2
or (60 10.2). 9.39 m/s
2 2 1.13
V p V p
V V
ρ ρ
+ = + = − ∴ = The
temperature (the viscosity of the water) and the diameter of the pipe are
not needed.
3.55 a)
2
2
0
2 2
V
V p
ρ
+ =
2
2
( 10 )
. . 50
2
o o
o
p p
x p
p p x ρ
ρ ρ ρ
−
+ + = ∴ = −
b)
2
2
0
2 2
V
V p
ρ
+ =
2
2
(10 )
. . 50
2
o o
o
p p
y p
p p y ρ
ρ ρ ρ
+ + = ∴ = −
3.56
2
2
2 2
U p
V p
ρ ρ
∞ ∞
+ = + .
a) v v U r r
r c
θ θ
= = = − −
∞
0 180 1 1
2 2
and o
: ( / )( ).
( )
∴ = − = −
∞ ∞
p U v U
r
r
r
r
r
c c
ρ ρ
2 2
2
2 2 2
2
2
4
.
b) Let r r p U
c T
= = ∞
:
ρ
2
2
c) ( ) [ ]
v r r v U p v U
r c
= = − ∴ = − = −
∞ ∞ ∞
0 2
2 2
1 4
2 2 2
and = U 2
: sin . sin
θ θ
θ
ρ ρ
θ
d) Let θ ρ
= = − ∞
90
3
2
90
2
o
: p U
3.57
2
2
2 2
U p
V p
ρ ρ
∞ ∞
+ = + .
a) vθ θ
= =
0 180
and o
: ( )
p U v U
r
r
r
r
r
c c
= − =
−
∞ ∞
ρ ρ
2 2
2
2 2 2
3 6
.
b) Let r r p U
c T
= = ∞
:
1
2
2
ρ .
c) ( ) [ ]
v r r p v U
r c
= = − = −
∞ ∞
0
2 2
1 4
2 2 2
and = U 2
: sin
ρ ρ
θ
θ
d) Let θ ρ
= = − ∞
90
3
2
90
2
o
: p U](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-44-2048.jpg)
![51
4.6 a) The energy equation (the 1st law of Thermo).
b) The conservation of mass.
c) Newton’s 2nd law.
d) The energy equation.
e) The energy equation.
4.7
4.8
4.9
4.10 $ $ $ . ($ $)
n i j i j
1
1
2
1
2
0 707
= − − = − + . $ . $ .5$
n i j
2 0 866 0
= − . $ $
n j
3 = − .
1 1 1
ˆ ˆ ˆ
ˆ 10 [ 0.707( )] 7.07 fps
n
V V n i i j
= ⋅ = ⋅ − + = −
v
V V n i i j
n
2 2 2 10 0866 05 866
= ⋅ = ⋅ − =
v
$ $ ( . $ . $) . fps
3 3 2
ˆ ˆ
ˆ 10 ( ) 0
n
V V n i j
= ⋅ = ⋅ − =
v
4.11 flux = ηρ$
n VA
⋅
v
flux1 = ηρ ηρ
[ . ($ $)] $ / .
− + ⋅ = −
0 707 10 0 707 10
i j iA A
flux2 = ηρ ηρ
( . $ .5$) $ / .
0 866 0 10 0 866 10
i j iA A
− ⋅ =
flux3 = ηρ( $) $
− ⋅ =
j iA
10 0
3
n̂
v n̂
v
ω
n̂ v n̂
v
n̂
v
n̂
v n̂
n̂
n̂
v v
v
v
n̂
n̂
n̂
n̂
v
v
v
v
n̂ n̂ v
v](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-52-2048.jpg)
![70
To find &
m3 use continuity:
& & & . ( ) & .
m m m y dy m
1 2 3
2
3
0
2
2 4 10 32 2 28 10 2
= + × × = + +
∫
ρ ρ
∴ = − × +
=
& . .
m3 640 10 28 2
8
3
53 3
ρ ρ ρ
Rate of K.E. loss = & &
m
V
m
V u
dy
1
1
2
3
1
2 3
0
2
2
2
2
2
2
10
− − ∫
ρ
= − − +
∫
1280
32
2
53 3 32 10 28
2
2 2 3
0
2
ρ ρ ρ
. ( )
y dy
[655360 54579 507320] 115000 .
W
ρ
= − − =
4.98 The average velocity at section 2 is also 8 m/s. The kinetic-energy-
correction factor for a parabola is 2 (see Example 4.9). The energy
equation is:
V
g
p V
g
p
hL
1
2
1
2
2
2
2
2 2
+ = + +
γ
α
γ
.
8
2 9 81
150
2
8
2 9 81
110
2 2
×
+ =
×
+ +
. .
.
000
9810
000
9810
hL
∴ =
hL 0 815
. .
m
4.99 V
A
VdA y dy
= = + = × +
=
∫
∫
1 1
2
28
1
2
28 2
2
3
29 33
2
3
0
2
( ) . m / s
α = =
×
+
∫ ∫
1 1
2 29 33
28
3
3
3
2 3
0
2
AV
V dA y dy
.
( )
[ ]
=
×
× + × × + × × + =
1
2 29 33
28 2 3 28 2 3 3 28 2 5 2 7 1005
3
3 2 3 5 7
.
/ / / .
4.100 a)
2 2 4
0.01
2 2 2 2
0
1 1 20 0.01 0.01
10 1 2 5 m/s
2
0.01 0.01 0.01 4 0.01
r
V VdA rdr
A
π
π
= = − = − =
∫ ∫
× ×
α
π
π
= =
× ×
−
∫ ∫
1 1
0 01 5
10 1
0 01
2
3
3
2 3
3
2
2
3
0
0 01
AV
V dA
r
rdr
. .
.
=
×
−
×
×
+
×
×
−
×
=
2000
0 01 5
0 01
2
3 0 01
4 0 01
3 0 01
6 0 01
0 01
8 0 01
2 00
2 3
2 4
2
6
4
8
6
.
. .
.
.
.
.
.
.](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-71-2048.jpg)
![81
− = − = × × − − ∴ =
R m V V R
x x x x
&( ) . ( . cos cos ).
2 1
2
1000 015 100 19 32 60 100 30 6800
π o o
N .
∴ = = × × = ×
& . .
W R V
x B
12 12 6800 50 4 08 106
W
4.136 a) Refer to Fig. 4.16:
50 30
50 30
2500 86 6
1 1
1 1
1
2 2
sin sin
cos cos
.
o
o
=
− =
∴ = − +
V
V V
V V V
r
B r
r B B
α
α
2 2 2 2 2
2 1
2 2
30sin60 sin
900 30 .
30cos60 cos
r
r r B B
r B
V
V V V V
V V
α
α
=
∴ = = + +
− =
o
o
Combine the above: VB = 13 72
. m / s. Then, α α
1 2
59 4 42 1
= =
. , . .
o o
− = − = × × − − ∴ =
R m V V R
x x x x
&( ) . ( cos cos ).
2 1
2
1000 01 50 30 60 50 30 916
π o o
N.
∴ = = × × =
& . .
W V R
B x
15 15 13 72 916 188 500 W
b)
50 30
50 30
2500 86 6
1 1
1 1
1
2 2
sin sin
cos cos
.
o
o
=
− =
∴ = − +
V
V V
V V V
r
B r
r B B
α
α
∴ =
VB 14 94
. m / s.
2 2 2 2
2
2 2
30sin70 sin
900 20.52 .
30cos70 cos
r
r B B
r B
V
V V V
V V
α
α
=
∴ = + +
− =
o
o
α 1 41 4
= . o
,α 2 48 2
= . o
2
2 1
( ) 1000 .01 50( 30cos70 50cos30 ). 841 N.
x x x x
R m V V R
π
− = − = × × − − ∴ =
o o
&
∴ = = × × =
& . .
W V R
B x
15 15 14 94 841 188 500 W
c)
50 30
50 30
2500 86 6
1 1
1 1
1
2 2
sin sin
cos cos
.
o
o
=
− =
∴ = − +
V
V V
V V V
r
B r
r B B
α
α
∴ =
VB 16 49
. m / s
2 2 2 2
2
2 2
30sin80 sin
900 10.42 .
30cos80 cos
r
r B B
r B
V
V V V
V V
α
α
=
∴ = + +
− =
o
o
α 1 43
= o
, α 2 53 7
= . o
− = − = × × − − ∴ =
R m V V R
x x x x
&( ) . ( cos cos ).
2 1
2
1000 01 50 30 80 50 30 762
π o o
N.
∴ = = × × =
& . .
W V R
B x
15 15 16 49 762 188 500 W
4.137 To find F, sum forces normal to the plate: ( )
n out n
F m V
Σ = & 1 .
n
V
−
a) [ ]
∴ = × × × − − =
F 1000 02 4 40 40 60 11
. . ( sin ) .
o
080 N (We have neglected friction)
2 2 3 3 1
0 ( ) 40sin30 .
t
F m V m V m
Σ = = + − − × o
& & & Bernoulli: V V V
1 2 3
= = .
∴ = − −
= +
∴ = = × =
=
Continuity:
kg / s.
kg / s.
0 75 75 320 240
80
2 3 1
1 2 3
2 1
3
& & .5 &
& & &
& . & .
&
m m m
m m m
m m
m
b)
1 20
1.94 120( 120sin60 ) 3360 lb.
12 12
F
∴ = − × × × − =
o
(We have neglected friction)
ΣF m V m V m
t = = + − − ×
0 120 30
2 2 3 3 1
& & ( ) & sin .
o
Bernoulli: V V V
1 2 3
= = .](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-82-2048.jpg)
![82
2 1
2 3 1
1 2 3
3
20
.75 .75 1.94 120
0 0.5
144
Continuity:
22.6 slug/sec. and 9.7 slug/sec.
m m
m m m
m m m
m
∴ = = × × ×
∴ = − −
= + = =
& &
& & &
& & &
&
4.138 F m V
r r n
= = × × × + =
& ( ) . . ( ) sin
1
2
1000 02 4 40 20 60 24
o
940 N.
F W
x = = ∴ = × =
24 30 21 21 20 432
940 600 N. 600 000 W
cos & .
o
4.139 F m V V F V
r r n B x B
= = × × − = −
& ( ) . . ( ) sin . ( )sin .
1
2 2 2
1000 02 4 40 60 8 40 60
o o
& ( ) . ( ).
W V F V V V V V
B x B B B B B
= = − × = − +
8 40 75 6 1600 80
2 2 3
dW
dV
V V V
B
B B B
&
( ) . .
= − + = ∴ =
6 1600 160 3 0 13 33
2
m / s.
4.140 (A) Let the vehicle move to the right. The scoop then diverts the water to the
right. Then
2 1
( ) 1000 0.05 2 60 [60 ( 60)] 720000 N.
x x
F m V V
= − = × × × × − − =
&
4.141 1
(
r
F m V
= & 2
)(cos 1) 1000 .1 .6 ( )( 2) 120 .
B B B B
V V V V
α
− − = × × − − =
At
2
120 1000
0 : 120 133 300 N.
3600
t F
×
= = × =
2
133 300
1.33 m/s
100 000
o
a = =
−
= =
−
∴ − = ∫
∫
F
m
dV
dt
V dV
V
dt
B B B
B
t
120
100
0012
2
2
0
3333
16 67
000
. . .
.
.
∴ −
= ∴ =
1
16 67
1
33 33
0012 26 6
. .
. . .
t. sec
t
4.142 1
( )(cos 1) 90 .8 2.5 13.89 ( 13.89)( 1) 34700 N.
r B
F m V V α
= − − = × × × × − − =
&
50 1000
13.89 m/s 34700 13.89 482 000 W or 647 Hp.
3600
B
V W
×
= = ∴ = × =
&
4.143 See the figure in Problem 4.141.
F m V V V V
r B B B
= − − = × × × − − =
& ( )(cos ) . . ( )( )
1 1 1000 06 2 2 24
α V .
B
2
− = ∴− =
F mV
dV
dx
V V
dV
dx
B
B
B B
B
. .
24 5000
2
− = − = − ∴ =
∫
∫
24
5000
24
5000
27 78 250 458
250
27 78
0
m
dx dV
V
x n n x
B
B
x
. . . .
.
l l
V2
V1
= 0
F](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-83-2048.jpg)
![84
& .
. .
. .
m = × ×
+
=
194
10
12
29 33 58 67
2
186 2
2
π slug/ sec
4.148 Fix the reference frame to the boat: V V
1 2
10 20
= =
m / s, m / s.
∴Thrust = &( ) . ( ) .
m V V
2 1 1000 0 2 20 10 2000
− = × − = N
& .
W F V
= × = × =
1 2000 10 20 000 W or 26.8 Hp
4.149 0 2 2 10 1 2 20 0 1
1 1 1 1 1 1
. . . . ( ) ( . ).
= = × × ∴ = ∴ = ∴ = −
V A V V V V y y
m / s. m / s.
max
flux in = 2 2 1000 20 1 800
1
3
267
2 2 2
3
0
1
0
1
ρV dy y dy
= × − = =
∫
∫ (. )
.
.
.
000 N.
The slope at section 1 is −20. ∴ = − +
V y y A
2 20
( ) .
Continuity: A V A V V V
1 1 2 2 2 1
2 2
= ∴ = =
. m / s.
V A
V A
V A
2
2
2
0
05 1
1 2
( )
(. )
/ .
=
= −
∴ = −
2 1 2 2 2 20
2
= − ∴ = ∴ = −
A A V y y
/ . .5. ( ) .5 .
flux out = 2 1000 2 20 800
125
3
800
0 00153
2
3
0
05
0
05
( .5 )
( . )
[ . ]
.
.
− =
−
=
∫ y dy
y
000
000
3
= 408 3
. N. ∴change = 408 − 267 = 141 N.
4.150 a) β = =
−
× ×
= =
∫ ∫
V dA
V A
y dy
2
2
2 2
0
1
2
3
2 20 1
1 2 1 0
4000
1
3
4
3
(. )
. .
.
.
.
b) See Problem 4.149: V y y y V
2 2
20 0125 05 0 2
( ) ( . ), . .
= − ≥ ≥ = m / s.
β = =
−
× ×
=
−
=
∫ ∫
V dA
V A
y dy
y
2
2
2 2
0
05
2
3
0
05
2 20 125
2 1 10
2000
125
3
1021
( . )
. .
( . )
. .
.
.
4.151 From the c.v. shown: ( ) .
p p r r L
w o
1 2 0
2
2
− =
π τ π
∴ = = ∴ =
× ×
× × × −
τ µ
w
o
w w
p r
L
du
dr
du
dr
∆
2
0 03 144 75 12
2 30 2 36 10 5
.
. . /
.
= 191
ft / sec
ft
.
4.152 Write the equation of the parabola: V r V
r
r
( ) .
max
= −
1
2
0
2
p1
A1
p2
A2
τw2πro
L](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-85-2048.jpg)
![87
= − = −
∫
8 0 36
0
3
ρ ρ
AV k rdr AV k
Ω Ω
$ . $.
.
Σ
v v v v v v
M
d
dt
r V d V r V V n dA i V j V k V A
c v
e e e e
c s
= × − = × ⋅ = × +
∫ ∫
0 0 3 707 707
and ( ) . ( $) . $ (. $ . $) .
. . . .
ρ ρ ρ
The z-component of
v v v
r V V n dA V A
e e
c s
× ⋅ = ×
∫ ( $) . . .
. .
ρ ρ
3 707 2
Finally, − = = × × =
( ) . . . ,
M AV V A AV A V
I z e e e e
0 36 4 3 707 2
ρ ρ
Ω . Using
0 36 4 3 707 19 89
. . . . .
Ω = × × × ∴ =
Ω 46 9
. .
rad / s
4.160 A moment
v
M resists the motion thereby
producing power. One of the arms is shown.
v
M ri k Vi Adr AV k rdr AV k
I = × − × = − = −
∫
∫4 2 8 2 778
0
10 12
0
25
$ ( $ $) $ . $.
/
.
Ω Ω Ω
ρ ρ ρ
Σ
v v v v v v
M Mk
d
dt
r V d V r V V n dA V A k
c v
e e
c s
= × − = × ⋅ = × ×
∫ ∫
$, ( ) , ( $) $.
. . . .
and
0
10
12
4
2
ρ ρ
Thus, M + ×
× × = × ×
×
2 778 194
75
12
200
9
30 200
10
12
194
1 4
12
4
2
2
2
. .
.
.
/
.
π π
∴ =
M 309 ft - lb.
& .
W M
= = × =
Ω 309 30 9270 ft - lb /sec
4.161 & . . .
m AV V V
= = = × ∴ =
10 1000 01 31 8
2
0 0
ρ π m / s.
Continuity: V V V r
e
0
2 2
01 01 006 05
π π
× = × + × −
. . . ( . ).
0
2
.01 .006 .15. 11.1 m/s.
e e
V V V
π × = × × ∴ =
∴ = − − = −
V V r V r
e
0 191 05 42 4 212
. ( . ) . .
v
M ri k V i Adr ri k r i Adr
I = × + × + × + × −
∫
∫2 2 2 2 42 4 212
0
05
2
0
05
$ ( $ $) $ [ $ ( . )$]
.
.
.
Ω Ω
ρ ρ
= + −
∫
∫
4 4 42 4 212
0
2
05
2
0
05
Ω Ω
V Ak rdr Ak r r dr
ρ ρ
$ $ ( . )
.
.
.
= × × × × + × ×
4 31 8 1000 01
05
2
4 1000 01
2
2
2
Ω Ω
. .
. $ .
π π
k
42 4
2
2 05
212
3
2 05
2 2 3 3
.
(. . ) (. . ) $
− − −
k
x
y r
Ve
V
Ω](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-88-2048.jpg)
![93
5.16 From Table 5.1, spherical coordinates:
1 1
2
2
r r
r v
r
v
r
∂
∂ θ
∂
∂θ
θ
θ
( )
sin
( sin ).
= −
∴ = +
1 1
10
40
2
2
2
3
r r
r v
r r
r
∂
∂ θ
θ θ
( )
sin
sin cos .
∴ = +
+ = −
+
∫
r v r
r
dr f r
r
f
r
2
3
2
10
40
2 10
80
cos ( ) cos ( )
θ θ θ θ
4 2 10 2
80
2
0 0
2
v f f
r ( , ) cos ( ) . ( ) .
θ θ θ θ
= × −
+ = ∴ =
∴ = −
v
r
r 10
80
3
cos .
θ
5.17 Continuity:
∂
∂
ρ ρ
ρ
x
u
du
dx
u
d
dx
( ) . .
= ∴ + =
0 0
ρ = =
×
×
= =
−
×
=
p
RT
du
dx
18 144
1716 500
0 00302
526 453
2 2 12
219
3
. .
/
slug
ft
fps / ft.
∴ = − = − × = −
d
dx u
du
dx
ρ ρ .
. .
00302
486
219 0 00136 slug/ ft 4
5.18 [ ]
∂
∂
∂
∂
∂
∂
u
x
v
y x
e e
x x
+ = − − = −
− −
0 20 1 20
. ( )
Hence, in the vicinity of the x-axis:
∂
∂
v
y
e v ye C
x x
= = +
− −
20 20
and .
But v y C
= = ∴ =
0 0 0
if . .
v ye e
x
= = =
− −
20 20 0 2 0
2
( . ) .541 m / s
5.19 [ ]
1
0 20 1 20
r r
rv
v
z z
e e
r
z z z
∂
∂
∂
∂
∂
∂
( ) . ( )
+ = − − = −
− −
Hence, in the vicinity of the z-axis:
1
20
2
20
2
r r
rv e rv
r
e C
r
z
r
z
∂
∂
( ) .
= = +
− −
and
But v r C
r = = ∴ =
0 0 0
if . .
v re e
r
z
= = =
− −
10 10 0 2 0 271
2
( . ) . m / s
5.20 The velocity is zero at the stagnation point. Hence,
0 10
40
2
2
= − ∴ =
R
R
. m
The continuity equation for this plane flow is
∂
∂
∂
∂
u
x
v
y
+ = 0. Using
∂
∂
u
x
x
= −
80 3
,](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-94-2048.jpg)
![103
CHAPTER 6
Dimensional Analysis
and Similitude
6.1
g
V
V
g
p
g
z
g
V
V
g
p
g
z
1
2
1
2
1
1
1
2
2
2
2
2
2 2
+ +
= + +
ρ ρ
.
1
2
1
2
1
1
2
1
1
2
2
2
1
2
2
1
2
2
1
2
+ + = + +
p
V
gz
V
V
V
p
V
gz
V
ρ ρ
.
or
1
2
1
2
1
1
2
1
1
2
2
2
2
2
2
2
2
2
1
2
+ + = + +
p
V
gz
V
p
V
gz
V
V
V
ρ ρ
6.2 a) [ ]
& . .
m
FT
L
= =
⋅
⋅
=
⋅
∴
kg
s
N s
m s
N s
m
2
b) [ ]
p
F
L
= ∴
N
m2
. 2
c) [ ]
ρ = =
⋅
⋅
=
⋅
∴
kg
m
N s
m m
N s
m
3
2
3
2
4
. .
FT
L
2
4
d) [ ]
µ =
⋅
∴
N s
m2
.
FT
L2
e) [ ]
W FL
= ⋅ ∴
N m.
f) [ ]
& .
W
FL
T
=
⋅
∴
N m
s
g) [ ]
σ = ∴
N / m.
F
L
6.3 (A) The dimensions on the variables are as follows:
2 2
2 3 2 2
/
[ ] [ ] , [ ] , [ ] , [ ]
L L ML ML T M L
W F V M d L p V
T T
T T L LT
= × = × = = ∆ = = =
&
First, eliminate T by dividing W
& by ∆p. That leaves T in the denominator so
divide by V leaving L2
in the numerator. Then divide by d2
. That provides
2
W
pVd
π =
∆
&
6.4 1
2 5 2
, , , .
T e r
f
R R R
R R
µ
ρω ρω
∴ =
l](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-104-2048.jpg)
![104
6.5 (A) ( , , , , ). The units on the variables on the rhs are as follows:
V f d l g ω µ
=
1
2
[ ] , [ ] , [ ] , [ ] , [ ]
L ML
d L l L g T
T
T
ω µ
−
= = = = =
Because mass M occurs in only one term, it cannot enter the relationship.
6.6 [ ] [ ] [ ] [ ]
V f V
L
T
L
M
L
M
LT
= = = = =
( , , ). , , , .
l l
ρ µ ρ µ
3
∴There is one π − term: π
ρ
µ
1 =
Vl
.
∴ = = ∴ = =
π π ρ
µ
1 1 2
0
f
V
C
( ) , Re
Const. or Const.
l
6.7 [ ] [ ] [ ] [ ]
V f d V
L
T
M
T
M
L
d L
= = = = =
( , , ). , , , .
σ ρ σ ρ
2 3
∴ = ∴ = = ∴ =
π
σ
ρ
π π
σ
ρ
1 2 1 1 2
0
2
V d
f C
V d
C
. ( ) ,
onst. or We = Const.
6.8 [ ] [ ] [ ] [ ]
V f H g m V
L
T
g
L
T
m M H L
= = = = =
( , , ). , , , .
2
∴ = ∴ = ∴ =
π π
1
0
2 1
gHm
V
C V gH C
. . / .
6.9 [ ] [ ] [ ] [ ] [ ] [ ]
V f H g m V
L
T
H L g
L
T
m M
M
L
M
LT
= = = = = = =
( , , , , ). , , , , , .
ρ µ ρ µ
2 3
Choose repeating variables H g
, ,ρ (select ones with simple dimensions-we couldn’t
select V, H, and g since M is not contained in any of those terms):
π ρ π ρ π µ ρ
1 2 3
1 1 1 2 2 2 3 3 3
= = =
VH g mH g H g
a b c a b c a b c
, , .
∴ = = = = =
π
ρ
π
ρ
π
µ
ρ
µ
ρ
1
0
2 3 3 3 2 3
V
g H
V
gH
m
H gH gH
. . .
/
∴ =
V
gH
f
m
H gH
1 3 3
ρ
µ
ρ
, .
Note: The above dimensionless groups are formed by observation: simply combine the
dimensions so that the π − term is dimensionless. We could have set up equations similar
to those of Eq. 6.2.11 and solved for 1 1 1 2 2 2 3 3 3
, , and , , c and , , .
a b c a b a b c But the
method of observation is usually successful.
6.10 [ ] [ ] [ ] [ ] [ ]
F f d V F
ML
T
d L V
L
T
M
LT
M
L
D D
= = = = = =
( , , , , ). , , , , .
l µ ρ µ ρ
2 3](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-105-2048.jpg)
![105
π ρ π ρ π µ ρ
1 2 3
1 1 1 2 2 2 3 3 3
= = =
F V dV V
D
a b c b c a a b c
l l l
, , .
∴ = = =
π
ρ
π π
µ
ρ
1 2 2 2 3
F
V
d
V
D
l l l
, , .
∴ =
F
V
f
d
V
D
ρ
µ
ρ
l l l
2 2 1 , .
We could write
π
π π
π
π
1
2
2 2
2
3
2
1
=
f , or
F
d V
f
d dV
D
ρ
µ
ρ
2 2 2
=
l
, . This is equivalent
to the above. Either functional form must be determined by experimentation.
6.11 [ ] [ ] [ ] [ ] [ ]
F f d V F
ML
T
d L V
L
T
M
LT
M
L
D D
= = = = = =
( , , , , ). , , , , .
l µ ρ µ ρ
2 3
π µ π µ π ρ µ
1 2 3
1 1 1 2 2 2 3 3 3
= = =
F d V d V d V
D
a b c a b c a b c
, , .
l
By observation we have 1 2 3
, , .
D
F Vd
Vd d
ρ
π π π
µ µ
= = =
l
∴ =
F
Vd
f
d
Vd
D
µ
ρ
µ
1
l
, .
Rather than π π π
1 1 2 3
= f ( , ), we could write
π
π
π
π
1
3
2 2
3
1
=
f , , an acceptable form:
F
V d
f
d Vd
D
ρ
µ
ρ
2 2 2
=
l
, .
6.12 [ ] [ ] [ ] [ ] [ ] [ ]
h f d g h L
M
T
d L
M
L T
g
L
T
= = = = = = =
( , , , , ). , , , , , .
σ γ β σ γ β
2 2 2 2
1
Select d g
, ,
γ as repeating variables.
π γ π σ γ π β
1 2
1 1 1 2 2 2
= = =
hd g d g
a b c a b c
, , .
3
∴ = = =
π π
σ
γ
π β
1 2 2 3
h
d d
, , .
∴ =
h
d
f
d
1 2
σ
γ
β
, . Note: gravity does not enter the answer.
6.13 [ ] [ ] [ ] [ ]
F f m R F
ML
T
m M
T
R L
C C
= = = = =
( , , ). , , , .
ω ω
2
1
∴ = = ∴ = ∴ =
π ω
ω ω
ω
1 2 2
2
F m R
F
m R
F
m R
C F Cm R
C
a b c C C
C
. .
6.14 [ ] [ ] [ ] [ ]
σ σ
= = = = =
f M y I
M
LT
M
ML
T
y L I L
( , , ). , , , .
2
2
2
4
∴ =
π σ
1 M y I
a b c
.](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-106-2048.jpg)
![106
Given that 1
b = − , 1 Const. C .
I My
yM I
σ
π σ
= = ∴ =
6.15 [ ] [ ] [ ]
V f d
dp
dx
V
L
T
M
LT
d L
dp
dx
M
L T
=
= = =
=
( , , . , , , .
µ µ 2 2
∴ =
π µ
1 V d
dp
dx
a b
c
. Let’s start with the ratio
µ
dp dx
/
so that “M” is accounted for.
Then the π1 − term is
µV
dp dx d
/
.
2
Hence,
π
µ
µ
1 2
2
= = ∴ =
V
dp dx d
V
d dp dx
/
/
.
Const. Const
6.16 [ ] [ ] [ ] [ ]
V f H g V
L
T
H L g
L
T
M
L
= = = = =
( , , ). , , , .
ρ ρ
2 3
∴ = = = ∴ =
π ρ
ρ
1
0
VH g V
g H
V gH
a b c
Const. Const. .
Density does not enter the expression.
6.17 [ ] [ ] [ ] [ ] [ ] [ ]
3 2
( , , , , ). , , , , , .
L M M L
V f H g d V H L g d L
T LT L T
µ ρ µ ρ
= = = = = = =
π ρ π µ ρ π ρ
1 2
1 1 1 2 2 2 3 3 3
= = =
VH g H g dH g
a b c a b c a b c
, , .
3
Repeating
variables
H g
, , .
ρ
π π
µ
ρ
π
1 2 3 2 3
= = =
V
gH gH
d
H
, , .
/
∴ = =
π π π
µ
ρ
1 1 2 3 1 3
f
V
gH
f
gH
d
H
( , ), , .
or
6.18 ∆p f V d L
= ( , , , , , ).
ν ε ρ
[ ] [ ] [ ] [ ] [ ] [ ] [ ]
2
2 3
, , , , , , .
M L L M
p V d L L L L
T T
LT L
ν ε ρ
∆ = = = = = = =
Repeating variables: V d
, , .
ρ
π ρ π ν ρ π ρ π ε ρ
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =
∆pV d V d L V d V d
a b c a b c a b c a b c
, , , .
∴ = = = =
π
ρ
π
ν
π π
ε
1 2 3
∆p
V Vd
L
d d
, , , .
2 4
π π π π
ρ
ν ε
1 1 2 3 4 2 1
= ∴ =
f
p
V
f
Vd
L
d d
( , , ). , , .
∆](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-107-2048.jpg)
![107
6.19 F f V c h r w
D = ( , , , , , , , , )
ρ µ φ α where c is the chord length, h is the maximum thickness, r
is the nose radius, φ is the trailing edge angle, and α is the angle of attack. Repeating
variables: V c
, , .
ρ Theπ − terms are
π
ρ
π
ρ
µ
π π π φ π π α
1 2 2 2 3 4 5 7
= = = = = = =
F
V c
V c c
h
c
r
c
w
D
, , , , , , .
6
Then,
F
V c
f
V c c
h
c
r
c
w
D
ρ
ρ
µ
φ α
2 2 1
=
, , , , ,
6.20 [ ] [ ] [ ] [ ] [ ] [ ]
3
2
2
( , , , , ). , , , , 1, .
L L
Q f R A e S g Q R L A L e L s g
T T
= = = = = = =
There are only two basic dimensions. Choose two repeating variables, R and g. Then,
π π π π
1 2 3 4
1 1 2 2 3 3 4 4
= = = =
QR g AR g eR g sR g
a b a b a b a b
, , , .
∴ = = = =
π π π π
1 5 2 2 2 3 4
Q
gR
A
R
e
R
s
/
, , , .
∴ = ∴ =
π π π π
1 1 2 3 4 5 1 2
f
Q
gR
f
A
R
e
R
s
( , , ). , , .
6.21 [ ] [ ] [ ] [ ]
2 2 3
( , , , ). , , , , .
p p
L L M M
V f h g V h L g
T T T L
σ ρ σ ρ
= = = = = =
Repeating variables: h g V h g h g
p
a b c a b c
, , . , .
ρ π ρ π σ ρ
∴ = =
1 2
1 1 1 2 2 2
∴ = = ∴ =
π π
σ
ρ
σ
ρ
1 2 2 1 2
V
hg gh
V
gh
f
gh
p p
, . .
6.22 F f V e I d
D = ( , , , , , ).
µ ρ Repeating variables: V d
, , .
ρ
[ ] [ ] [ ] [ ] [ ] [ ] [ ]
2 3
, , , , , 1, .
D
ML L M M
F V e L I d L
T LT
T L
µ ρ
= = = = = = =
π ρ π µ ρ π ρ π ρ
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =
F V d V d e V d I V d
D
a b c a b c a b c a b c
, , , .
∴ = = = =
π
ρ
π
µ
ρ
π π
1 2 2 2 3 4
F
V d V d
e
d
I
D
, , , .
∴ =
F
V d
f
V d
e
d
I
D
ρ
µ
ρ
2 2 1 , , .
6.23 F f V D g
D s
= ( , , , , , ).
ρ ρ µ Repeating variables: V D
, , .
ρ
[ ] [ ] [ ] [ ] [ ] [ ] [ ]
2 3 3 2
, , , , , , .
D s
ML L M M M L
F V D L g
T LT
T L L T
ρ ρ µ
= = = = = = =](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-108-2048.jpg)
![108
π ρ π ρ ρ π µ ρ π ρ
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =
F V D V D V D gV D
D
a b c
s
a b c a b c a b c
, , , .
∴ = = = =
π
ρ
π
ρ
ρ
π
µ
ρ
π
1 2 2 2 3 4 2
F
V D VD
gD
V
D s
, , , .
∴ =
F
V D
f
VD
gD
V
D s
ρ
ρ
ρ
µ
ρ
2 2 1 2
, , .
6.24 F f V d e r c
D = ( , , , , , , ).
µ ρ Repeating variables: V d
, , .
ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
F
ML
T
V
L
T
M
LT
M
L
d L e L r L c
L
D = = = = = = = =
2 3 2
1
µ ρ
π ρ π µ ρ π ρ π ρ π ρ
1 2 3 4 5
1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
= = = = =
F V d V d eV d rV d cV d
D
a b c a b c a b c a b c a b c
, , , , .
∴ = = = = =
π
ρ
π
µ
ρ
π π π
1 2 2 2 3 4 5
2
F
V d Vd
e
d
r
d
cd
D
, , , , .
∴ =
F
V d
f
Vd
e
d
r
d
cd
D
ρ
µ
ρ
2 2 1
2
, , , .
6.25 f g V d f
T
M
LT
M
L
V
L
T
d L
= = = = = =
( , , , ). [ ] , [ ] , [ ] , [ ] , [ ] .
µ ρ µ ρ
1
3
Repeating variables, V d f V d V d
a b c a b c
, , . ,
ρ π ρ π µ ρ
1 2
1 1 1 2 2 2
= =
∴ = = ∴ =
π π
µ
ρ
µ
ρ
1 2 1
fd
V Vd
fd
V
g
Vd
, . .
6.26 F f V c t
L c
= ( , , , , ).
, ρ α
l Repeating variables: V c
, , .
ρ l
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
F
ML
T
V
L
T
c
L
T
M
L
L t L
L c
= = = = = = =
2 3
1
ρ α
l
π ρ π ρ π ρ π α ρ
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =
F V cV tV V
L
a b
c
c a b
c
c a b
c
c a b
c
c
l l l l
, , , .
∴ = = = =
π
ρ
π π π α
1 2 2 2 3 4
F
V
c
V
t
L
c c
l l
, , , .
∴ =
F
V
f
c
V
t
L
c c
ρ
α
2 2 1
l l
, , .
6.27
2
2 3
1
( , , , , ). [ ] , [ ]= ,[ ]= , [ ] , [ ] , [ ] .
ML M M
T f d t T d L t L
T LT
T L
ω ρ µ ω ρ µ
= = = = =
Repeating variables: d Td d t d
a b c a b c a b c
, , . , , .
3
ω ρ π ω ρ π µ ω ρ π ω ρ
1 2
1 1 1 2 2 2 3 3 3
= = =](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-109-2048.jpg)
![109
1 2 3
2 5 2
, , .
T t
d
d d
µ
π π π
ρω ρω
∴ = = =
3 5
1 1
2 5 2 2
, . , .
T t t
f W d f
d d
d d d
µ µ
ρω
ρω ρω ρω
∴ = =
&
6.28 F f V d L
D c
= ( , , , , , , )
ρ µ ρ ω where d is the cable diameter, L the cable length, ρc
the cable density, andω the vibration frequency.
Repeating variables: V d
, , .
ρ The π − terms are
π
ρ
π
ρ
µ
π π
ρ
ρ
π
ω
1 2 2 2 3 4 5
= = = = =
F
V d
Vd d
L
V
d
D
c
, , , ,
We then have
F
V d
f
Vd d
L
V
d
D
c
ρ
ρ
µ
ρ
ρ ω
2 2 1
=
, , ,
6.29 ∆p f D h d d
= ( , , , , , ).
ω ρ 1 0 Repeating variables: D, , .
ω ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ]
∆p
M
LT
D L h L
T
M
L
d L d L
= = = = = = =
2 3 1 0
1
ω ρ
π
ρω
π π π
1 2 2 2 3
1
4
0
= = = =
∆p
D
h
D
d
D
d
D
, , , .
∴ =
=
∆p
D
f
h
D
d
D
d
D
W
ρω 2 2 1
1 0
, , . & force × velocity = ∆pD D
2
×ω .
∴ =
& , , .
W D f
h
D
d
D
d
D
ρω3 5
1
1 0
6.30 T g f d H N h
= ( , , , , , , ).
,
ω ρ
l Repeating variables: ω ρ
, , .
d
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
T
ML
T
f
T T
d L H L L N h L
M
L
= = = = = = = = =
2
2 3
1 1
1
ω ρ
l
π
ρω
π
ω
π π π π
1 2 5 2 3 4 5 6
= = = = = =
T
d
f H
d d
N
h
d
, , , , , .
l
∴ =
T
d
g
f H
d
N
h
d
ρω ω
2 5 1 , , , , .
d
l
6.31 Q f H w g
= ( , , , , , ).
µ ρ σ Repeating variables: H g
, , .
ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
Q
L
T
H L w L g
L
T
M
LT
M
L
M
T
= = = = = = =
3
2 3 2
µ ρ σ
∴ = = = =
π π π
µ
ρ
π
σ
ρ
1 5 2 3 3 4 2
Q
gH
w
H gH gH
, , , .](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-110-2048.jpg)
![110
∴ =
Q
gH
f
w
H gH gH
5 1 3 2
, , .
µ
ρ
σ
ρ
6.32 d f V V D
j a
= ( , , , , , , ).
σ ρ µ ρ Repeating variables: V D
j , , .
ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
d L V
L
T
V
L
T
D L
M
T
M
L
M
LT
M
L
j a
= = = = = = = =
σ ρ µ ρ
2 3 3
π π π
σ
ρ
π
µ
ρ
π
ρ
ρ
1 2 3 2 4 5
= = = = =
d
D
V
V V D V D
j j j
a
, , , , .
∴ =
d
D
f
V
V V D V D
j j j
a
1 2
, , , .
σ
ρ
µ
ρ
ρ
ρ
6.33 T f H h R t
= ( , , , , ).
ω µ ρ
, , Repeating variables: ω ρ
, , .
h
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
T
ML
T T
H L h L R L t L
M
LT
M
L
= = = = = = = =
2
2 3
1
ω µ ρ
π
ρω
π π π π
µ
ρω
1 2 5 2 3 4 5 2
= = = = =
T
d
H
h
R
h
t
h h
, , , ,
∴ =
T
d
f
H
h
R
h
t
h h
ρω
µ
ρω
2 5 1 2
, , , .
6.34 µ ρ
= f D H g V
( , , , , , )
l . D = tube dia., H = head above outlet, l = tube length.
Repeating variables: D V
VD
H
D D
gD
V
, , . , , ,
ρ π
µ
ρ
π π π
1 2 3 4 2
= = = =
l
∴ =
µ
ρVD
f
H
D D
gD
V
1 2
, , .
l
6.35 T f R e r
= ( , , , , ).
, ,
ω ρ µ l Repeating variables: R, , .
ω ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
T
ML
T
R L
T
M
L
e L r L
M
LT
L
= = = = = = = =
2
2 3
1
ω ρ µ l
π
ρω
π π π π
µ
ρω
1 2 5 2 3 4 5 2
= = = = =
T
R
e
R
r
R R R
, , , ,
l
1
2 5 2
, , , .
T e r
f
R R R
R R
µ
ρω ρω
∴ =
l
6.36 y f V y g
2 1 1
= ( , , , ).
ρ Neglect viscous wall shear.
[ ] , [ ] , [ ] , [ ] , [ ] .
y L V
L
T
y L
M
L
g
L
T
2 1 1 3 2
= = = = =
ρ Repeating variables: V y
1 1
, , .
ρ](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-111-2048.jpg)
![111
π π
1
2
1
2
1
1
2
= =
y
y
gy
V
, . ( ρ does not enter the problem).
∴ =
y
y
f
gy
V
2
1
1
1
2
.
6.37 f g d V f
T
d L L
M
L
M
LT
V
L
T
= = = = = = =
( , , , , ). [ ] , [ ] , ] , [ ] , [ ] , [ ] .
[
l l
ρ µ ρ µ
1
3
Repeating variables: d V
, , .
ρ (l = length of cylinder).
π π π
µ
ρ
µ
ρ
1 2 3 1
. , , . , .
= = = ∴ =
fd
V d Vd
fd
V
f
d Vd
l l
6.38
Q
Q
V
V
p
p
V
V
F
F
V
V
m
p
m m
p p
m
p
m m
p p
p m
p p
m m m
p p p
= = =
l
l
l
l
2
2
2
2
2 2
2 2
, ,
( )
( )
∆
∆
ρ
ρ
ρ
ρ
τ
τ
ρ
ρ
ρ
ρ
ρ
ρ
m
p
m m
p p
m
p
m m m
p p p
m
p
m m m
p p p
V
V
T
T
V
V
Q
Q
V
V
= = =
2
2
2 3
2 3
3 2
3 2
, ,
&
&
l
l
l
l
( &
Q has same dimensions as .)
W
&
6.39 (A) Re Re . m m
m p
m
V L
ν
=
p p
p
V L
ν
= . 12 9 108 m/s.
p
m p
m
L
V V
L
∴ = = × =
6.40 A)
5
6
1.51 10
Re Re . . 4 10 461 m/s.
1.31 10
p p p
m m m
m p m p
m p m p
V L L
V L
V V
L
ν
ν ν ν
−
−
×
= = ∴ = = × =
×
6.41 a) Re Re . . .
m p
m m
m
p p
p
m
p
p
m
V d V d V
V
d
d
= = ∴ = =
ν ν
7
Q
Q
V
V
Q Q
V
V
m
p
m m
p p
m p
m
p
m
p
= ∴ = = × × =
l
l
l
l
2
2
2
2 2
15 7
1
7
0 214
. . . .
m / s
3
&
&
. & .
W
W
V
V
W
m
p
m m m
p p p
m
= = × = ∴ = × =
ρ
ρ
3 2
3 2
3
2
7
1
7
7 7 200 1400
l
l
kW
b) Re Re .
.
.
. .
m p
m
p
p
m
m
p
V
V
d
d
= ∴ = = × =
ν
ν
7
9
13
4 85
3
2
3
2
1
1.5 4.85 0.148 m /s.
7
1
4.85 200 466 kW
7
m
m
Q
W
= × × =
= × × =
&](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-112-2048.jpg)
![119
∂
∂
∂
∂
∂
∂
∂
∂
u
t
p
x
u
r r
u
r
*
*
*
*
*
*2 *
*
*
Re
= − + +
2
1
where Re =
Vd
ν
6.71 Assume w
z
= =
0 0
and
∂
∂
. The x-component Navier-Stokes equation is then
u
t
∂
∂
u
u v
x
∂
∂
+ +
u u
w
y z
∂ ∂
∂ ∂
+
1 p
x
∂
ρ ∂
= −
2 2 2
2 2 2
x
u u u
g
x y z
∂ ∂ ∂
ν
∂ ∂ ∂
+ + + +
With g g
x = the simplified equation is
u
u
x
g
u
x
u
y
∂
∂
ν
∂
∂
∂
∂
= + +
2
2
2
2
Let u u V x x h y y h
* * *
/ , / / .
= = =
and Then
V
h
u
u
x
g
V
h
u
x
u
y
2
2
2 2
*
*
*
*
*2
*
*2
∂
∂
ν
∂
∂
∂
∂
= + +
The normalized equation is
u
u
x
u
x
u
y
*
*
*
*
*
*
*
Re
∂
∂
∂
∂
∂
∂
= + +
1 1
2
2
2
2
2
Fr
where Fr and
= =
V
hg
Vh
Re
ν
6.72 * * * * * *2 2 2
, , , , , .
o
u v T x y
u v T x y
U U T
= = = = = ∇ = ∇
l
l l
ρ
∂
∂
∂
∂
c
UT T
x
UT T
y
K
T T
p
o o
o
l l l
*
*
*
*
* *
.
+
= ∇
2
2
Divide by ρc UT
p o / :
l
∂
∂
∂
∂ ρ
T
x
T
y
K
c U
T
p
*
*
*
*
* *
.
+ = ∇
l
2
Parameter =
K
c U
p
µ
µ
ρ l
=
1 1
Pr Re
.
6.73 ρ
ρ
ρ
* * * * * * *
, , , , , , .
= = = ∇ = ∇ ∇ = ∇ = =
o o o
V
V
U
t
tU
p
p
p
T
T
T
v
v
l
v
l
v
l
1 1
2
2
2
momentum: ρ ρ
µ µ
o
o
U DV
Dt
p
p
U
V
U
V
*
*
*
* * * * * * *
( ).
2
2
2
2
3
l
v
l
v
l
v
l
v v v
= − ∇ + ∇ + ∇ ∇ ⋅
Divide by ρoU 2
/ :
l
[ ]
ρ
ρ
µ
ρ
*
*
*
* * * * * * *
( ) .
DV
Dt
p
U
p
U
V V
o
o o
v
v
l
v v v v
= − ∇ + ∇ + ∇ ∇ ⋅
2
2
energy: ρ ρ
*
*
*
* * * * *
.
c T
U DT
Dt
K
T T p
U
p V
v o o o o
l l l
v v
= ∇ − ∇ ⋅
2
2
Divide by ρo v o
c T U / :
l
ρ
ρ ρ
*
*
*
* * * * *
.
DT
Dt
K
c U
T
p
c T
p V
o v
o
o v o
= ∇ − ∇ ⋅
l
v v
2](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-120-2048.jpg)
![165
8.17 The velocities associated with the two Re's are
V
D
1
1
5 5
3 10 1 5 10
0445
101
= =
× × ×
=
−
Re .
.
ν
m /s,
V
D
2
2
4 5
6 10 1 5 10
0445
20
= =
× × ×
=
−
Re .
.
ν
m /s.
The drag, between these two velocities, is reduced by a factor of 2.5
( ) ( )
[ ]
C C
D D
high low
and
= =
0 5 0 2
. . . Thus, between 20 m/s and 100 m/s the drag is
reduced by a factor of 2.5. This would significantly lengthen the flight of the
ball.
8.18 a) F V AC V C V C
D D D D
= ∴ = × ×
∴ =
1
2
0 5
1
2
00238
2
12
4810
2 2
2
2
ρ π .
. . . .
Re
/
.
. . : .
= =
×
×
= = = ×
−
VD V
V C V
D
ν
4 12
1 6 10
2080 5 98
4
Try fps, Re = 2 105
Try fps, Re= 2.3 105
C V
D = = ×
. : .
4 110
b) C V V
D = = × ×
× ∴ =
0 2
1
2
00238
2
12
2 155
2
2
. : . . .
0.5 fps.
π
8.19 4 2
1
2
1000 1 0 267
2
10
2 10
2 2 2
6
5
. . . . . Re
.
.
= × × ∴ = =
×
= ×
−
V C V C
V
V
D D
π
Try C V
D = ∴ = = × ∴
0 5 0 73 1 46 105
. : . Re . .
m /s. OK.
8.20 Re
.
. . .
= =
×
×
= × ∴ =
−
VD
CD
ν
40 2
15 10
53 10 07
5
6
. (This is extrapolated from
Fig. 8.8.) ∴ = × × × × × =
FD
1
2
122 40 2 60 7 81
2
. ( ) . .
900 N
M = 81 900 × 30 = 2 46 106
. .
× ⋅
N m
8.21 a) Re
.
.
. . Re . . Re . .
1 5
5
2
5
3
5
25 05
1 08 10
1 2 10 1 8 10 2 4 10
=
×
×
= × = × = ×
−
Assume a
rough cylinder (the air is highly turbulent).
( ) ( ) ( )
∴ = = =
C C C
D D D
1 2 3
0 7 0 8 0 9
. , . , . .
∴ = × × × × + × × + × × =
FD
1
2
1 45 25 05 10 7 075 15 8 1 20 9 1380
2
. (. . . . . . ) .
N
M = × × × × × + × × × + × × × = ⋅
1
2
1 45 25 05 10 7 40 075 15 8 27 5 1 20 9 10 25 700
2
. (. . . . . . . ) .
N m
b) Re
.
.
. . Re . , Re . .
1 5
4
2
5
3
5
25 05
1 65 10
7 6 10 1 14 10 1 5 10
=
×
×
= × = × = ×
−](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-125-2048.jpg)
![180
d) u
x
x
= = + ∴ = − ∴
0 0 10
10
1
: . . Stag pt: (−1, 0)
e) y
a v
=
v v
u
y x
∂ ∂
∂ ∂
+ 0 on axis. x
u
x a u v
x
∂
∂
= − = +
2
10 10
10 .
u
y x x
∂
∂
= + −
2
10
( 2,0) (10 5) 12.5 m/s .
4
x
a
∴ − = − − = −
8.78
2 3
2 5
( , ) 5 .
2 3
y y
u x y y y C
y
∂ψ
ψ
∂
= − = ∴ = − + 2 3
1
. (3 10 ).
6
y y
ψ
∴ = −
q udy y y dy
= = − = − = × −
∫
∫ ( )
. .
. .
.
.
5
02
2
5
0 2
3
6 667 10
2
2 3
3
0
2
0
2
m / s
2
ψ ψ
2 1
2 3 3
1
6
3 0 2 10 02 0 6667 10
− = × − × − = × −
( . . ) . .
m / s
2
ω
∂
∂
= − = − + ≠
u
y
y
1 10 0. ∴φ doesn’t exist.
8.79 ψ
π
π
θ θ θ
= + = +
30
5
2
30
5
2
y r sin .
a) v
r r
r = = + =
1
30
5
2
0
∂ψ
∂θ
θ
cos .
At θ π
= = ∴ =
, . . '.
5
2
30 0 0833
r
r
s
s
Stag. pt: ( " , ).
−1 0
b) At θ π ψ
π π π
= = = +
, sin .
=.0833,
r r
s
5
2
30
2
5
2 2
∴ = =
r yinter .0119 ft.
c) q U H H H
= × = ∴ = ∴
∆ψ
π π
. . .
=
5
60
30
5
2
Thickness = 2
5
30
H =
π
ft or 1.257".
d) v u
r ( , ) cos . . . . .
1 30
5
2
30 2 5 27 5 27 5
π π
= + = − + = − ∴ = fps
8.80 [ ] [ ] [ ]
φ
π
π
π
π
= + + − − + + = + +
2
1
2
1 10
1
4
1
2 1 2 2 2 1 2 2 2
l l l
n x y n x y x n x y
( ) ( ) ( )
/ /
[ ]
− − + +
1
4
1 10
2 2
ln x y x
( ) .
u
x
x
x
x
x x x
v y
y
= =
+
+
−
−
−
+ =
+
−
−
+ = =
=
∂φ
∂ 0
2 2
1
4
2 1
1
1
4
2 1
1
10
1
2 1
1
2 1
10 0 0
[ ( )]
( )
[ ( )]
( ) ( ) ( )
. .
if
30 fps
y
x
= 0
= 5π/2](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-140-2048.jpg)
![181
At the stagnation point, u
x x x
= ∴
+
−
−
+ = ∴
−
=
0
1
2 1
1
2 1
10 0
2
1
20
2
.
( ) ( )
. .
∴ = ∴ = ± ∴ ×
x x
2
1 1 1 049
. . . .
m. oval length = 2 1.049 = 2.098 m
All the flow from the source goes to the sink, i.e., π
π
m /s, or m / s for
2 2
2
0
y > .
u y
x y y y
x
( )
( ) ( )
.
= =
+
−
−
+
+ =
+
+
=
∂φ
∂ 0
2 2 2
1
4
2
1
1
4
2
1
10
1
1
10
1
2
0
1
10 . tan 10 .
2 2
1
h
q dy h h
y
π π
−
= + = ∴ + =
∫
+
h = 0.143 m so that thickness = 2h = 0.286 m.
The minimum pressure occurs on the oval surface at (0,h).
There u =
+
+ =
1
1 143
10 10 98
2
.
. m /s.
Bernoulli:
V p V p p
2 2 2 2
2 2
10 98
2 1000
10
2
+ = + + = +
∞ ∞
ρ ρ
.
.
.
10 000
1000
Pa
min
∴ = −
p 280 .
8.81 [ ] [ ] [ ]
φ
π
π
π
π
= + + +
−
− + + = + +
2
2
1
2
2
1 2
1
2
1
2 2 1 2 2 2 1 2 2 2
l l l
n x y n x y x n x y
( ) ( ) ( )
/ /
[ ]
− − + +
1
2
1 2
2 2
ln x y x
( ) .
u
x
x
x y
x
x y
v
y
x y
y
x y
= =
+
+ +
−
−
− +
+ =
− +
−
− +
∂φ
∂
1
2
2 1
1
1
2
2 1
1
2
1 1
2 2 2 3 2 2 2 2
( )
( )
( )
( )
.
( ) ( )
Along the x-axis (y = 0), v = 0 and u
x x
=
+
−
−
+
1
1
1
1
2.
Set u
x x
x x
=
−
−
+
= = ∴ = ±
0
1
1
1
1
2 2 2
2
: , . .
or
Stag. pts.: ( , ), ( , ).
2 0 2 0
−
u v
( , ) . ( , ) .
4 0
1
4 1
1
4 1
2 4 0 0
=
− +
−
− −
+ = − =
1.867 m / s
u v
( , ) . ( , ) .
0 4
1
1 4
1
1 4
2 0 4
4
1 4
4
1 4
0
2 2 2 2
=
+
−
−
+
+ = =
+
−
+
=
2.118 m / s
8.82 φ
π
π
π
π
= + − + + +
2
2
1
2
2
1
2 2 1 2 2 2 1 2
l l
n x y n x y
[ ( ) ] [ ( ) ]
/ /
= + − + + +
1
2
1
1
2
1
2 2 2 2
l l
n x y n x y
[ ( ) ] [ ( ) ].
(0, h)
y
x](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-141-2048.jpg)
![182
u
x
x
x y
x
x y
= =
+ −
+
+ +
∂φ
∂ 2 2 2 2
1 1
( ) ( )
.
v
y
y
x y
y
x y
= =
−
+ −
+
+
+ −
∂φ
∂
1
1
1
1
2 2 2 2
( ) ( )
.
At (0, 0) u = 0 and v = 0. At (1, 1)
2 2 2 2 2
2 2 1 1
0 0.4 m/s. 1.2 m/s.
5
2 1 1 2 1
v u
= + = = = + =
+ +
∴ = +
v
V i j
12 04
. $ . $ .
m/ s
8.83 φ
π
π
π
π
= − + + + + + ∞
2
2
1
2
2
1
2 2 1 2 2 2 1 2
l l
n y x n y x U x
[( ) ] [( ) ] .
/ /
= − + + + + + ∞
1
2
1
1
2
1
2 2 2 2
l l
n y x n y x U x
[( ) ] [( ) ] .
a) Stag. pts. May occur on x-axis, y =0.
u
x
x
x
x
x
y
= =
+
+
+
+
=
∂φ
∂ 0
2 2
1 1
10.
2
0.2 1 0.
x x
∴ + + = ∴no stagnation points exist on the x-axis.
(They do exist away from the x-axis.)
Along the y-axis: u y q udy
h
( ) . ( )
= = = =
∫
10
1
2
2
0
m / s.
2
π π
∴ = = ∴ =
∫
π 10 10 0314
0
dy h h
h
. m
. .
b) u
x
x
x x x
=
+
+ ∴ + + = ∴ = −
2
1
1 2 1 0 1
2
2
. . m.
Stag. pt.: (−1, 0)
Along the y-axis: u h h
= ∴ = × ∴ =
1 0 1 3 14
. . . .
. m
π
c) u
x
x
x x x
=
+
+ ∴ + + = ∴ = − −
2
1
02 10 1 0 9 90
2
2
. . . . , 0.10 m.
Stag. pts.: (−9.9, 0) , (−0.1, 0).
Along the y-axis: u h h
= ∴ = ∴ =
0 2 02 1571
. . . . .
. m
π
8.84 φ θ θ
= +
60
8
r
r
cos cos .
a) v
r r r
r = = − + = −
∂φ
∂
θ θ θ
60
8 8
60
2 2
cos cos cos .
At the cylinder surface vr = 0 for all θ. Hence,
x
y
x
y
x
y](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-142-2048.jpg)
![184
c) v v
r ( . , / ) / . . , ( . , / )
.
.
14 14 4 2 14 14 0 1414 14 14 4
10
14 14
0 707
π π
θ
= = = − = −
v v
r ( . , / ) / . , ( . , / ) / . .
0 1 2 2 0 1 20 0 1 2 10 0 1 100
π π
θ
= = = − = −
Bernoulli:
20 000
1.2
13 760 Pa
+
+
= +
+
∴ =
0 1414 0 707
2 1 2
20 100
2
2 2 2 2
. .
.
.
p
p
We used ρair
3
kg / m at standard conditions.
= 1 2
.
8.86 Along the y-axis v v
r
r = = − −
0 10
40
2
and θ .
We have set θ
π
=
2
in Eq. 8.5.27. rc = =
40
10
2.
b) v
r
r = − − ⇒
10
40
4 3 5 126 9
2
cos cos . ( , ) ( , . ).
θ θ o
v
r
v v
r
θ θ
θ θ
= − − ∴ = − = −
10
40
6 96 9 28
2
sin sin . . , . .
m /s m / s
c) Use Eq. 8.5.28: p p U
= − ∞
0
2 2
2ρ α
sin
Drag = p r d L p r L p p U
c c
cos . .
/
/
α α ρ
π
π
− × = − ∞
−
∫ 90 90 0
2
2
2
2 2
= − − ×
∞
∫
2 2 2
0
2 2
90
0
2
( sin )cos
/
p U r Ld p r L
c c
ρ α α α
π
[ ]
= −
− − =
∞ ∞ ∞
2 2
3
2 2
8
3
0
2
3
0
2
0
2 2
r L p U p U r L r L U
c c c
sin
sin
.
/
α ρ
α
ρ ρ
π
C
U A
r L U
U r L
D
c
c
= = = =
∞
∞
∞
Drag
1
2
8 3
1
2
2
8
3
2 667
2
2
2
ρ
ρ
ρ
( / )
. .
8.87 v U
r
U r U
r c
= − = = = × =
∞ ∞ ∞
cos cos . , .
θ
µ
θ µ
2
2 2
4 1 4 4
Let
For θ π
= = − +
, .
v
r
r 4
4
2
b) v
r
U
rc
θ
∂ψ
∂
θ
µ θ
θ θ
= − = − − = − −
= −
∞ sin
sin
sin sin .
2 2
4
4
1
8
c) p p
V v
c = + − = + × −
∞
∞
ρ ρ
θ
θ
2 2 2 2 2
2 2
1000
4
2
1000
8
2
50 000
sin
.
∴ = −
pc 58 32 2
sin θ kPa.
10 m/s
20 m/s
dα
p(α)
α
p90
x = −1
−x
−vr](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-144-2048.jpg)
![190
%error in low
τ 0
332 289
332
100 13%
( )
. .
.
.
x =
−
× =
8.106 τ ρ
δ δ
ρ
δ δ
δ
δ
δ
0
2 2
6
2
0
6
3 1 3
1
3
1
1
3
= −
+ +
− −
∞ ∞
∫
∫
d
dx
U
y y
dy U
y y
dy
/
/
/
+ +
− −
∞
∫ρ
δ δ
δ
δ
U
y y
dy
2
2
3
2
3
1
3
2
3
/
= = ∴ =
∞
∞
∞
d
dx
U
U d
dx
U
ρ δ µ
δ
δ
δ
µ ρ
2
0 1358
3
22 08
( . ) . . / .
Thus, δ τ ρ ρ
( ) . / , ( ) .
.
. Re .
/
x vx U x U
v
U x
U x
= =
=
∞ ∞
∞
∞
−
665 01358
665
2
0451
0
2 2 1 2
%error for δ =
−
× =
6 65 5
5
100 33%
.
. %error for τ 0
0 451 0 332
0 332
100 36%
=
−
× =
. .
.
8.107 Continuity from entrance to x: U H u y dy U x H
0
0
2 2
= + −
∫ ( ) ( )( ).
δ
δ
Write U x U x dy U x dy
( ) ( ) ( ) .
δ
δ
δ
= = ∫
∫ 0
0
Then, continuity provides
U H u U dy UH
0
0
2
= − +
∫( )
δ
= − −
∫
UH U u dy
2
0
( )
δ
= − ∴ =
−
UH U U x
U H
H
d
d
2
2
0
δ
δ
. ( ) .
If we were to move the walls out a distance δd x
( ), then U x
( ) would be constant
since ( )
[ ]
H d d
− +
2 2
δ δ would be constant; then U x U
( ) .
= 0 For a square wind
tunnel, displace one wall outward 4 0
δ d dp dx
for / .
=
8.108 The given velocity profile is that used in Example 8.13. There we found
δ ν
= = = = =
∞
−
5 48 5 48 10 10 0 00173 0 00173 3 0 003
6
. / . / . . .
x U x x m.
Assume the streamline is outside the b.l. Continuity is then
10 0 02 10
2
003 003
003 10
2
2
0
003
× = −
+ −
∫
.
. .
( . )
.
y y
dy h
= + − ∴ =
0 02 10 0 03 0 021
. . . .
h h m or 2.1 cm
[ ]
δ d
y y
dy
= − +
= − + =
∫
1
10
10
20
003
10
003
1
10
03 03 01 0 001
2
2
0
003
. .
. . . .
.
m
h − = − =
2 2 1 2 0 1
. . cm or 0.001 m.
The streamline moves away from the wall a distance δ d .](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-150-2048.jpg)
![192
8.111 a) δ
ν
= =
× ×
=
∞
−
4 65 4 65
1 6 10 20
12
0 0759
4
1 2
. .
.
. .
/
x
U
ft
b) τ ρ
ν
0
2 2
4
1 2
5
323 323 0024 12
1 6 10
20 12
9 11 10
= = × ×
×
×
= ×
∞
∞
−
−
. . .
.
. .
/
U
xU
psf
c) Drag =
2
1
20 15 1.29
2
U
LU
ν
ρ ∞
∞
× × ×
1 / 2
4
2
1 1.6 10
.0024 12 300 1.29 0.0546 lb.
2 20 12
−
×
= × × × × =
×
d) δ
∂
∂ δ δ
δ
x
u
x
U
y y d
dx
=
−
∞
=
× ×
= = − +
10
4
2
3
4
4 65
1 6 10 10
12
0 0416
3
2
3
2
.
.
. .
ft.
∴ = −
×
+
×
×
×
= − +
−
∂
∂
u
x
y y
y y
12
3
2 0416
3
2 0416
4 65
2
1 6 10
10 12
27 9 16140
2
3
4
4
3
. . . .
. .
. .
∴ = − = × − × =
∫
v
u
x
dy
∂
∂
δ
27 9
2
0416
16140
4
0416 0 0121
2 4
0
.
. . . .
fps
8.112 a) δ
ν
= =
× ×
=
∞
−
4 65 4 65
15 10 6
4
00221
5 1 2
. .
.
. .
/
x
U
m
b) τ ρ
ν
0
2 2
5 1 2
0323 323 122 4
15 10
6 4
0 00498
= = × ×
×
×
=
∞
∞
−
. . .
.
. .
/
U
xU
Pa
c) Drag =
1
2
129
1
2
122 4 6 5 129
15 10
6 4
0299
2 2
5 1 2
ρU Lw
v
LU
∞
∞
−
× = × × × × ×
×
×
=
. . .
.
. .
/
N
d)
3
2 4
3 3
2
2
u y y d
U
x dx
∂ δ
∂ δ δ
∞
= − +
2 3
2 5 4 5 2
3 4 3 4
4 .
2
2 4.65 1.5 10 3 4.65 (1.5 10 3)
y d
y
dx
δ
− −
×
= − +
× × × × × × ×
[ ]
∴ = − + ×
×
= − + ×
−
∂
∂
u
x
y y y y
4 6166 253 10
465
2
15 10
4
1
3
641 2 63 10
7 3
5
5 3
.
. .
. . .
∴ = − = × −
×
× =
∫
v
u
x
dy
∂
∂
δ
641
2
0156
2 63 10
4
0156 0 00391
2
5
4
0
.
.
.
. . ,
m / s
where δx=
−
=
× ×
=
3
5
465
15 10 3
4
01560
.
.
. m.](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-152-2048.jpg)
![194
b) Drag =
1
2
0024 20 12 15 074
1 58 10
20 12
1700
1 58 10
20 12
0 27
2
4 2 4
× × × ×
×
×
−
×
×
=
− −
. ( ) .
. .
. .
.
lb
c) Drag =
1
2
0024 20 12 15 074
1 58 10
20 12
2080
1 58 10
20 12
0 25
2
4 2 4
× × × ×
×
×
−
×
×
=
− −
. ( ) .
. .
. .
.
lb
8.117 a) Drag =
1
2
1000 1 2 1 2 074
10
1 2 1
1060
10
1 2 1
5 21
2
6 2 6
× × × ×
×
−
×
=
− −
. ( ) .
. .
. .
.
N
b) Drag =
1
2
1000 1 2 1 2 074
10
1 2 1
1700
10
1 2 1
4 44
2
6 2 6
× × × ×
×
−
×
=
− −
. ( ) .
. .
. .
.
N
c) Drag =
1
2
1000 1 2 1 2 074
10
1 2 1
2080
10
1 2 1
3 99
2
6 2 6
× × × ×
×
−
×
=
− −
. ( ) .
. .
. .
.
N
8.118 U∞ = = = ×
×
×
=
60
1000
3600
16 67 38 100
16 67 10
235
5
2
. .
.
.
.
m /s. 000
1.5 10
m
-5
δ
τ ρ
0
2 2
5
5
2
1
2
1
2
1 22 16 67 059
1 5 10
16 67 10
0 0618
= = × × ×
×
×
=
∞
−
U c f . . .
.
.
. .
.
Pa
b) τ ρ
0
2 2
5
5
2
1
2
1
2
1 22 16 67
455
06
16 67 10
1 5 10
0 151
= = × ×
×
×
=
∞
−
U c
n
f . .
.
.
.
.
. .
l
Pa
∴ = = ∴ =
×
+ ∴ =
−
u n
τ
δ
δ
.
.
.
.
.
.
.
.
. . .
151
1 22
351
16 67
351
2 44
351
1 5 10
7 4 585
5
m /s. m
l
Both (a) and (b) are in error, however, (b) is more accurate. 0.
p
x
∂
∂
<
8.119 a) 5 =
uτ ν
δ
ν
(See Fig. 8.24 b). ∴ =
× ×
= ×
−
−
δν
5 1 5 10
351
2 14 10
5
4
.
.
. .
m
b) δ
µ
δ δ
τ τ
δ
δ
δ
δ
δ
ν
d
U
U u dy
U
n
y
dy
u
U
n
y
dy
= − = −
+ −
∞
∞
∞ ∞
∫
∫
∫
1
2 5 2 44 3 74
15
15
0
( ) . . .
.
.
l l
( )
= − − −
− −
∞
= =
u
U
y n
y
y y n
y
y
v
τ
ν
δ
δ δ
δ δ
δ δ
2 5 15 2 44 3 74
15 87 8
87 8
585
. . . .
. .
.
l l
= + − + − =
.
.
[ . ] . .
351
16 67
219 620 008 2188 951 43 7 m
Note: We cannot use zero as a lower limit since the ln-profile does not go to the](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-154-2048.jpg)
![197
8.126 u
y
U x
dF
d y
U x F
U
x
U F
= = = =
∞ ∞
∞
∞
∂ψ
∂
ν
η
∂η
∂
ν η
ν
η
'( ) '( ).
We used Eq. 8.6.50 and Eqs. 8.6.48.
( )
v
x x
U x F
U
x
F U x
F
x
= − = − = − −
∞
∞
∞
∂ψ
∂
∂
∂
ν
ν
ν
∂
∂η
∂η
∂
1
2
= − − −
∞
∞
∞ −
1
2
1
2
3 2
U
x
F U x F y
U
x
ν
ν
ν
' /
= − + = −
∞ ∞ ∞ ∞
1
2 2
1
2
U
x
F
y U
x
U
x
F
U
x
F F
ν
ν
ν ν
η
' ( ' ).
8.127 The results are shown in Table 8.5.
8.128 a) τ 0
2
5
0 332 1 22 5
1 5 10
2 5
0 0124
= × ×
×
×
=
−
. .
.
. .
Pa
b) δ =
× ×
=
−
5
15 10 2
5
0 0122
5
.
. .
m
c) v
U
x
F F
max
max
m /s
= −
=
× ×
× =
∞
−
ν
η
1
2
1 5 10 5
2
8605 0 00527
5
( ' )
.
. . .
d) Q udy U
dF
d
dy U
dF
d
d
vx
U
= = =
∞ ∞
∞
∫ ∫
∫ 0 0
0
δ δ
δ
η η
η
= − =
× ×
× =
∞
∞
−
U
x
U
F F
ν
δ
[ ( ) ( )]
.
. . /
0 5
1 5 10 2
5
3 28 0 04
5
m s / m
2
8.129 a) τ 0
2
4
4
332 0024 15
1 6 10
6 15
2 39 10
= × ×
×
×
= ×
−
−
. .
.
. .
psf
b) δ =
× ×
=
−
5
1 6 10 6
15
0 04
4
.
. ft.
c) v
U
x
F F
max
max
fps
= −
=
× ×
× =
∞
−
ν
η
1
2
1 6 10 15
6
8605 0 0172
4
( ' )
.
. . .
d) Q udy U
x
U
F
= = =
× ×
× =
∞
∞
−
∫
ν
δ
δ
( )
.
. . / .
15
1 6 10 6
15
3 28 0 394
4
0
ft sec /ft
2
8.130 At x = 2m, Re = 5 × 2/10-6 = 107. ∴Assume turbulent from the leading edge.
a) τ ρ
0
2
2
1
2
0 455
0 06
= ∞
U
n x
.
( . Re )
l](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-157-2048.jpg)
![212
M2
833
1 4 1716 1191
0 492
=
× ×
=
.
. .
b) 1 2 2
M 3000/ 1.4 1716 500 2.74. M 0.493. 8.592 12 103.1 psia.
p
= × × = ∴ = = × =
T2 2
2 386 500 1193
103 1 144
1716 1193
0 00725
= × = ∴ =
×
×
=
. .
.
. .
o o
R or 733 F slug / ft3
ρ
9.46
[ ]
ρ
ρ
2
1
2
1
1
2
1
2 2
1
2
1
2
1
2
1
2
1
2
2 1
1
1
1
1
2
4 2 2
1
2 1
= =
− +
+
+
+
−
− +
=
+
+ −
p
p
T
T
k k
k
k
k
k k
k
k
M M
M M
M
M
( ) ( )
( )
.
M1
2 2
1
1
2
1
2
=
+
+
−
k
k
p
p
k
k
. (This is Eq. 9.4.12). Substitute into above:
ρ
ρ
2
1
2
1
2
1
2
1
2 2
1
1 1 1
4 1 1 1
1 1 1
1 1 1
=
+ + + −
+ − + + −
=
+ + + −
+ + − +
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )( )
.
k k
p
p
k
k k k
p
p
k
k k
p
p
k
k k k
p
p
=
− + +
+ + −
k k p p
k k p p
1 1
1 1
2 1
2 1
( ) /
( ) /
.
For a strong schock in which
p
p
k
k
2
1
2
1
1
1
1
>> =
+
−
, .
ρ
ρ
9.47 Assume standard conditions: T1 1
15 101
= =
o
C, kPa.
ρ
∴ = × × =
V1 2 1 4 287 288 680
. m / s.
1 2 2
M 2. M .5774. 1.688 288 486 K.
T
= ∴ = = × =
2 4.5 101 454 kPa.
p = × =
2 .5774 1.4 287 486 255 m/s.
V
∴ = × × =
induced 1 2 680 255 425 m/s.
V V V
∴ = − = − =
The high pressure and high induced velocity cause extreme damage.
9.48 If M then M m /s
2 1 1
5 2 645 2 645 1 4 287 293 908
= = ∴ = × × =
. , . . . . .
V
p2 2
8 00 200 1600
1600
287 2 285 293
8 33
= × = =
× ×
=
. .
. ( . )
. .
kPa abs kg /m3
ρ
9.49 If M then M fps
2 1 1
5 2 645 2 645 1 4 1716 520 1118
= = ∴ = × × =
. , . . . . .
V
p2 2
8 00 30 240
240 144
1716 2 285 520
0 01695
= × = =
×
× ×
=
. .
( . )
. .
psia slug /ft3
ρ
V1
V2
stationary
shock](https://image.slidesharecdn.com/mecnicadefluidosmerlec-221106122408-ffbf4f41/75/Mecanica-de-Fluidos_Merle-C-Potter-David-C-Wiggert_3ed-Solucionario-172-2048.jpg)
Uploaded byEdgar Ortiz Sánchez
Mecánica de Fluidos_Merle C. Potter, David C. Wiggert_3ed Solucionario
The document provides information about free engineering solucionarios (solution manuals) and textbooks that can be downloaded. It states that the solution manuals contain fully solved and clearly explained solutions to all the problems in the textbooks. Visitors are invited to download the files for free. The document then lists the contents of Chapter 1 from an engineering textbook, which includes concepts like mass, density, pressure, power, energy, viscosity, and dimensional analysis.
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Mecánica de Fluidos_Merle C. Potter, David C. Wiggert_3ed Solucionario
- 1. http://www.elsolucionario.net LIBROS UNIVERISTARIOS Y SOLUCIONARIOSDE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS.
- 2. 1 CHAPTER 1 Basic Considerations 1.1Conservation of mass — Mass — density Newton’s second law — Momentum — velocity The first law of thermodynamics — internal energy — temperature 1.2 a) density = mass/volume = M L / 3 b) pressure = force/area = F L ML T L M LT / / / 2 2 2 2 = = c) power = force × velocity = F L T ML T L T ML T × = × = / / / / 2 2 3 d) energy = force × distance = ML T L ML T / / 2 2 2 × = e) mass flux = ρAV = M/L3 × L2 × L/T = M/T f) flow rate = AV = L2 × L/T = L3 /T 1.3 a) density = M L FT L L FT L 3 2 3 2 4 / / = b) pressure = F/L2 c) power = F × velocity = F × L/T = FL/T d) energy = F × L = FL e) mass flux = M T FT L T FT L = = 2 / / f) flow rate = AV = L2 × L/T = L3 /T 1.4 (C) m = F/a or kg = N/m/s2 = N. s2 /m. 1.5 (B) [µ] = [τ/du/dy] = (F/L2 )/(L/T)/L = F. T/L2 . 1.6 a) L = [C] T2 . ∴[C] = L/T2 b) F = [C]M. ∴[C] = F/M = ML/T2 M = L/T2 c) L3 /T = [C] L2 L2/3 . ∴[C] = L T L L L T 3 2 2 3 1 3 / / / ⋅ ⋅ = Note: the slope S0 has no dimensions. 1.7 a) m = [C] s2 . ∴[C] = m/s2 b) N = [C] kg. ∴[C] = N/kg = kg ⋅ m/s2 ⋅ kg = m/s2 c) m3 /s = [C] m2 m2/3 . ∴[C] = m3 /s⋅m2 ⋅ m2/3 = m1/3 /s 1.8 a) pressure: N/m2 = kg ⋅ m/s2 /m2 = kg/m⋅ s2 b) energy: N⋅ m = kg ⋅ m/s2 × m = kg⋅ m2 /s2 c) power: N⋅ m/s = kg ⋅ m2 /s3 d) viscosity: N⋅ s/m2 = kg m s s 1 m kg / m s 2 2 ⋅ ⋅ = ⋅
- 3. 2 e) heat flux:J/s = N m s kg m s m s kg m / s 2 2 3 ⋅ = ⋅ ⋅ = ⋅ f) specific heat: J kg K N m kg K kg m s m kg K m / K s 2 2 2 ⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ 1.9 kg m s m s m 2 + + = c k f. Since all terms must have the same dimensions (units) we require: [c] = kg/s, [k] = kg/s2 = N s / m s N / m, 2 2 ⋅ ⋅ = [f] = kg m / s N. 2 ⋅ = Note: we could express the units on c as [c] = kg / s N s / m s N s / m 2 = ⋅ ⋅ = ⋅ 1.10 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm3 e) 1.2 cm2 f) 76 mm3 1.11 a) 1.25× 108 N b) 3.21× 10−5 s c) 6.7× 108 Pa d) 5.6× 10−12 m3 e) 5.2 × 10−2 m2 f) 7.8 × 109 m3 1.12 (A) 8 9 2.36 10 23.6 10 23.6 nPa. − − × = × = 1.13 2 2 2 0.06854 0.225 0.738 0.00194 3.281 m m d d λ ρ ρ = = × where m is in slugs, ρ in slug/ft3 and d in feet. We used the conversions in the front cover. 1.14 a) 20 cm/hr = 5 20 /3600 5.555 10 m/s 100 − = × 5 20 /3600 5.555 10 m/s 100 − = × b) 2000 rev/min = 2000 2 × π/60 = 209.4 rad/s c) 50 Hp = 50 × 745.7 = 37 285 W d) 100 ft3 /min = 100 × 0.02832/60 = 0.0472 m3 /s e) 2000 kN/cm2 = 2× 106 N/cm2 × 1002 cm2 /m2 = 2× 1010 N/m2 f) 4 slug/min = 4 × 14.59/60 = 0.9727 kg/s g) 500 g/L = 500 × 10−3 kg/10−3 m3 = 500 kg/m3 h) 500 kWh = 500 × 1000× 3600 = 1.8× 109 J 1.15 a) F = ma = 10 × 40 = 400 N. b) F − W = ma. ∴ F = 10 × 40 + 10 × 9.81 = 498.1 N. c) F − W sin 30° = ma. ∴ F = 10 × 40 + 9.81 × 0.5 = 449 N. 1.16 (C) The mass is the same on earth and the moon: [4(8 )] 32 . du r r dr τ µ µ µ = = = 1.17 The mass is the same on the earth and the moon:
- 4. 3 m = 60 32 2 1863 . .. = ∴ Wmoon = 1.863× 5.4 = 10.06 lb 1.18 (C) shear sin 4200sin30 2100 N. F F θ = = = o shear 4 2100 = 84 kPa 250 10 F A τ − = = × 1.19 a) λ ρ = = × × × = × − − − . . . . ( . ) .43 225 225 4 8 10 184 3 7 10 10 2 26 10 2 6 m d m or 0.00043 mm b) λ ρ = = × × × = × − − − . . . . ( . ) . 225 225 4 8 10 00103 3 7 10 7 7 10 2 26 10 2 5 m d m or 0.077 mm c) 26 2 10 2 4.8 10 .225 .225 .0039m .00002 (3.7 10 ) m d λ ρ − − × = = = × × or 3.9 mm 1.20 Use the values from Table B.3 in the Appendix. a) 52.3 + 101.3 = 153.6 kPa. b) 52.3 + 89.85 = 142.2 kPa. c) 52.3 + 54.4 = 106.7 kPa (use a straight-line interpolation). d) 52.3 + 26.49 = 78.8 kPa. e) 52.3 + 1.196 = 53.5 kPa. 1.21 a) 101 − 31 = 70 kPa abs. b) 760 − 31 101 × 760 = 527 mm of Hg abs. c) 14.7 − 31 101 × 14.7 = 10.2 psia. d) 34 − 31 101 × 34 = 23.6 ft of H2O abs. e) 30 − 31 101 × 30 = 20.8 in. of Hg abs. 1.22 p = po e−gz/RT = 101 e−9.81 × 4000/287 ×(15 + 273) = 62.8 kPa From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is % error = 62.8 61.6 61.6 − × 100 = 1.95 %. 1.23 a) p = 973 + 22,560 20,000 25,000 20,000 − − (785 - 973) = 877 psf T = −12.3 + 22,560 20,000 25,000 20,000 − − (−30.1 + 12.3) = −21.4°F b) p = 973 + .512 (785 − 973) + .512 2 (−.488) (628 − 2 × 785 + 973) = 873 psf T = −12.3 + .512 (−30.1 + 12.3) + .512 2 (−.488) (−48 + 2 × 30.1 − 12.3) = −21.4°F Note: The results in (b) are more accurate than the results in (a). When we use a linear interpolation, we lose significant digits in the result.
- 5. 4 1.24 T =−48 + 33,000 30,000 35,000 30,000 − − (−65.8 + 48) = −59°F or (−59 − 32) 5 9 = −50.6°C 1.25 (B) 1.26 p = n F A = 4 26.5 cos 42 152 10− × o = 1296 MN/m2 = 1296 MPa. 1.27 4 n 4 t (120000) .2 10 2.4N 20 .2 10 .0004N F F − − = × × = = × × = F = 2 2 n t F F + = 2.400 N. θ= tan−1 .0004 2.4 =.0095° 1.28 ρ= m V − = 0 2 180 1728 . / = 1.92 slug/ft3 . τ = ρg = 1.92 × 32.2 = 61.8 lb/ft3 . 1.29 ρ= 1000 − (T − 4)2 /180 = 1000 − (70 − 4)2 /180 = 976 kg/m3 γ = 9800 − (T − 4)2 /18 = 9800 − (70 − 4)2 /180 = 9560 N/m3 % error for ρ= 976 978 978 − × 100 = −.20% % error for γ = 9560 978 9.81 978 9.81 − × × × 100 = −.36% 1.30 S = 13.6 − .0024T = 13.6 − .0024 × 50 = 13.48. % error = 13.48 13.6 13.6 − × 100 = −.88% 1.31 a) m = W V g γ = 6 12 400 500 10 9.81 g − × × = = 0.632 kg b) m = 6 12 400 500 10 9.77 − × × = 0.635 kg c) m = 6 12 400 500 10 9.83 − × × = 0.631 kg 1.32 S = / water m V ρ ρ = 10/ . 1.2 water V ρ = . 1.94 ∴ V = 4.30 ft3 1.33 (D) 2 2 3 water ( 4) (80 4) 1000 1000 968 kg/m 180 180 T ρ − − = − = − =
- 6. 5 1.34 τ= µ du dr =1.92 × 10−5 2 30(2 1/12) (1/12) × = 0.014 lb/ft2 1.35 T = force × moment arm = τ2πRL × R = µ du dr 2πR2 L = µ 2 0.4 1000 R + 2πR2 L. ∴µ= 2 2 2 0.0026 0.4 0.4 1000 2 1000 2 .01 0.2 12 T R L R π π = + + × × = 0.414 N. s/m2 . 1.36 Use Eq.1.5.8: T = 3 2 R L h π ω µ = ( ) 2 .5 / 12 π π × × × × × 3 2000 2 60 4 006 01 12 . . / = 2.74 ft-lb. Hp = Tω 550 2 74 209 550 = × . .4 = 1.04 Hp 1.37 Fbelt = µ du dy A = × − 131 10 10 002 3 . . (.6 × 4) = 15.7 N. Hp = F V × = × 746 15 7 10 746 . = 0.210 Hp 1.38 Assume a linear velocity so . du r dy h ω = Due to the area element shown, dT = dF × r = τdA × r = µ du dy 2πr dr × r. dr r τ T = 3 0 2 R r dr h µω π ∫ = 2 4 2 36 10 400 2 60 3 12 2 08 12 4 5 4 πµω π π h R = × × × × × × − . ( / ) . / = 91 × 10−5 ft-lb. 1.39 2 30(2 1/12) (1/12) × 2 2 0 0 [32 / ] 32 / . du r r r r dr τ µ µ µ = = = ∴τr = 0 = 0, τr=0.25 = 32 × 1 × 10−3 × 2 .25/100 (.5/100) = 3.2 Pa, τr=0.5 = 32 × 1 × 10−3 × 2 .5/100 (.5/100) = 6.4 Pa 1.40 (A) 3 [10 5000 ] 10 10 5000 0.02 1 Pa. du r dr τ µ µ − = = × = × × × = 1.41 The velocity at a radius r is rω. The shear stress is τ µ = ∆ ∆ u y . The torque is dT = τrdA on a differential element. We have
- 7. 6 0.08 0 = = 2 0.0002 r TrdA rdx ω τ µ π ∫ ∫ , 2000 2 209.4 rad/s 60 π ω × = = where x is measured along the rotating surface. From the geometry 2 x = r, so that 0.08 0.08 2 3 0 0 209.4 / 2 329000 = 0.1 2 329000 (0.08 ) 0.0002 3 2 x x T dx x dx π × = = ∫ ∫ = 56.1 N . m 1.42 If τ µ = du dy = cons’t and µ= AeB/T = AeBy/K = AeCy , then AeCy du dy = cons’t. ∴ du dy = De−Cy . Finally, 0 0 y u Cy du De dy − = ∫ ∫ or u(y) = 0 y Cy D e C − − = E (e−Cy − 1) where A, B, C, D, E, and K are constants. 1.43 µ = = = Ae Ae Ae B T B B / / / . . 001 000357 293 353 ∴A = 2.334 × 10−6 , B = 1776. µ40 = 2.334 × 10−6 e1776/313 = 6.80 × 10−4 N. s/m2 1.44 m = ρV . Then dm = ρd V + V dρ. Assume mass to be constant in a volume subjected to a pressure increase; then dm = 0. ∴ρd V = − V dρ, or d V V . d ρ ρ = − 1.45 B = V − p V ∆ ∆ 2200 MPa. = V ∴∆ V − = 2 10 2200 p B ∆ − × = = −0.00909 m3 or −9090 cm3 1.46 Use c = 1450 m/s. L = c∆t = 1450 × 0.62 = 899 m 1.47 = B V p ∆ ∆ − V = −2100 −13 20 . = 136.5 MPa 1.48 a) 327,000 144/1.93 c = × = 4670 fps b) 327,000 144/1.93 c = × = 4940 fps c) 308,000 144/1.87 c = × = 4870 fps 1.49 V ∆ =3.8 × 10−4 × −20 × 1 = .0076 m3 . ∆p = −B V ∆ V .0076 2270 1 − = − = 17.25 MPa 1.50 p = 2 2 0741 5 10 6 σ R = × × − . = 2.96 × 104 Pa or 29.6 kPa. Bubbles: p = 4σ/R = 59.3 kPa
- 8. 7 1.51 Use TableB.1: σ= 0.00504 lb/ft. ∴p = 4 4 .00504 1/32 12 R σ × = × = 7.74 psf or 0.0538 psi 1.52 See Example 1.4: h = 4 cos 4 0.0736 0.866 0.130 m. 1000 9.81 0.0002 gD σ β ρ × × = = × × 1.53 (D) 6 4 cos 4 0.0736 1 3 m or 300 cm. 1000 9.81 10 10 h gD σ β ρ − × × = = = × × × 1.54 See Example 1.4: h = 4 cos 4 0.032cos130 1.94 13.6 32.2 0.8/12 gD σ β ρ × = × × × o = −0.00145 ft or −0.0174 in 1.55 force up = σ× L × 2 cosβ= force down = ρghtL. ∴h = 2σ β ρ cos . gt 1.56 Draw a free-body diagram: The force must balance: W = 2σL or π ρ σ d L g L 2 4 2 = . ∴ = d g 8σ πρ W σL σL needle 1.57 From the free-body diagram in No. 1.47, a force balance yields: Is π ρ d g 2 4 < 2σ? π(. ) . . 004 4 7850 9 81 2 0741 2 × < × 0.968 < 0.1482 ∴No 1.58 Each surface tension force = σ× πD. There is a force on the outside and one on the inside of the ring. ∴F = 2σπD neglecting the weight of the ring. F D 1.59 h(x) h dW σdl From the infinitesimal free-body shown: cos . d gh xdx σ θ ρ α = l cosθ= dx dl . / d d x d h g xdx g x σ σ ρ α ρ α ∴ = = l l We assumed small αso that the element thickness is αx.
- 9. 8 1.60 The absolutepressure is p = −80 + 92 = 12 kPa. At 50°C water has a vapor pressure of 12.2 kPa; so T = 50°C is a maximum temperature. The water would “boil” above this temperature. 1.61 The engineer knew that water boils near the vapor pressure. At 82°C the vapor pressure from Table B.1 is 50.8 (by interpolation). From Table B.3, the elevation that has a pressure of 50.8 kPa is interpolated to be 5500 m. 1.62 At 40°C the vapor pressure from Table B.1 is 7.4 kPa. This would be the minimum pressure that could be obtained since the water would vaporize below this pressure. 1.63 The absolute pressure is 14.5 − 11.5 = 3.0 psia. If bubbles were observed to form at 3.0 psia (this is boiling), the temperature from Table B.1 is interpolated, using vapor pressure, to be 141°F. 1.64 The inlet pressure to a pump cannot be less than 0 kPa absolute. Assuming atmospheric pressure to be 100 kPa, we have 10 000 + 100 = 600 x. ∴x = 16.83 km. 1.65 (C) 1.66 ρ= = × + = p RT 1013 0 287 273 15 . . ( ) 1.226 kg/m3 . γ = 1.226 × 9.81 = 12.03 N/m3 1.67 3 in 101.3 1.226 kg/m . 0.287 (15 273) p RT ρ = = = × + 3 out 85 1.19 kg/m . 0.287 248 ρ = = × Yes. The heavier air outside enters at the bottom and the lighter air inside exits at the top. A circulation is set up and the air moves from the outside in and the inside out: infiltration. This is the “chimney” effect. 1.68 3 750 44 0.1339 slug/ft . 1716 470 p RT ρ × = = = × m V ρ = 0.1339 15 2.01 slug. = × = 1.69 (C) pV m = 800 4 59.95 kg 0.1886 (10 273) RT × = = × + 1.70 p W V RT = 100 (10 20 4) 9.81 9333 N. 0.287 293 g = × × × × = ×
- 10. 9 1.71 Assume thatthe steel belts and tire rigidity result in a constant volume so that m1 = m2: V 1 V = 1 1 2 2 2 1 2 2 2 1 1 or . 150 460 (35 14.7) 67.4 psia or 52.7 psi gage. 10 460 m RT m RT p p T p p T = + ∴ = = + = − + 1.72 The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have 100000 1 9.81. 10200 kg. m m × = × ∴ = Hence, pV m = 3 100 4 / 3 10200. 12.6 m or 25.2 m. 0.287 288 r r d RT π × = = ∴ = = × 1.73 2 2 1 0 ( 10). 20 32.2. 25.4 fps. 2 KE PE mV mg V V = ∆ + ∆ = + − ∴ = × ∴ = 2 2 1 0 ( 20). 40 32.2. 35.9 fps. 2 mV mg V V = + − ∴ = × ∴ = 1.74 2 2 1-2 1 . a) 200 0 5( 10 ). 19.15 m/s. 2 f f W KE V V = ∆ × = × − ∴ = b) 10 2 2 0 1 20 15( 10 ). 2 f sds V = × − ∫ 2 2 2 10 1 20 15( 10 ). 15.27 m/s. 2 2 f f V V × = × − ∴ = c) 10 2 2 0 1 200cos 15( 10 ). 20 2 f s ds V π = × − ∫ 2 2 20 1 200sin 15( 10 ). 16.42 m/s. 2 2 f f V V π π × = × − ∴ = 1.75 2 1 2 1 2 2 1 1 . 10 40 0.2 0 . 40000. 2 E E u u u u = × × + = + ∴ − = % % % % 40000 . 55.8 C where comes from Table B.4. 717 v v u c T T c ∆ = ∆ ∴∆ = = o % The following shows that the units check: 2 2 2 2 2 car 2 2 2 air kg m / s m kg C m kg C C kg J/(kg C) N m s (kg m/s ) m s m V m c × ⋅ ⋅ ⋅ ⋅ ⋅ = = = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ o o o o where we used N = kg. m/s2 from Newton’s 2nd law.
- 11. 10 1.76 2 2 2 1H O 1 . . 2 E E mV m c T = = ∆ 2 6 1 100 1000 1500 1000 2000 10 4180 . 69.2 C. 2 3600 T T − × × × = × × × ∆ ∴∆ = o We used c = 4180 J/kg. C o from Table B.5. (See Problem 1.75 for a units check.) 1.77 water . 0.2 40000 100 4.18 . 19.1 C. f f m h m c T T T = ∆ × = × ∆ ∴∆ = o The specific heat c was found in Table B.5. Note: We used kJ on the left and kJ on the right. 1.78. (B) ice water ice water water . 320 . E E m m c T ∆ = ∆ × = × ∆ 6 3 5 (40 10 ) 1000 320 (2 10 ) 1000 4.18 . 7.66 C. T T − − × × × × = × × × ∆ ∴∆ = o We assumed the density of ice to be equal to that of water, namely 1000 kg/m3 . Ice is actually slightly lighter than water, but it is not necessary for such accuracy in this problem. 1.79. W pdV = mRT V = ∫ d V d V mRT = ∫ V ln V mRT = ∫ 2 V 2 1 1 ln p mRT p = since, for the T = const process, 1 p V 1 2 p V = 2. Finally, 1-2 4 1 1716 530ln 78,310 ft-lb. 32.2 2 W = × × = − The 1st law states that 0. 78,310 ft-lb or 101 Btu. v Q W u mc T Q W − = ∆ = ∆ = ∴ = = − − % 1.80 If the volume is fixed the reversible work is zero since the boundary does not move. Also, since V 1 2 1 2 , mRT T T p p p = = so the temperature doubles if the pressure doubles. Hence, using Table B.4 and Eq. 1.7.17, 200 2 a) (1.004 0.287)(2 293 293) 999 kJ 0.287 293 v Q mc T × = ∆ = − × − = × b) 200 2 (1.004 0.287)(2 373 373) 999 kJ 0.287 373 v Q mc T × = ∆ = − × − = × c) 200 2 (1.004 0.287)(2 473 473) 999 kJ 0.287 473 v Q mc T × = ∆ = − × − = × 1.81 W pdV = ( p V = 2 V − 1 1). If = const, T p V 2 1 T V = 2 1 2 so if 2 , T T = ∫ then V 2 2V = 1 and (2 W p V = 1 V − 1) pV = 1 1. mRT = a) 2 0.287 333 191 kJ W = × × = b) 2 0.287 423 243 kJ W = × × =
- 12. 11 c) 2 0.287473 272 kJ W = × × = 1.82 = 1.4 287 318 357 m/s. 357 8.32 2970 m. c kRT L c t = × × = = ∆ = × = 1/ 0.4/1.4 2 2 1 1 500 (20 273) 151.8 K or 121.2 C 5000 k k p T T p − = = + = − o 1.83 We assume an isentropic process for the maximum pressure: / 1 1.4/0.4 2 2 1 1 423 (150 100) 904 kPa abs or 804 kPa gage. 293 k k T p p T − = = + = Note: We assumed patm = 100 kPa since it was not given. Also, a measured pressure is a gage pressure. 1.84 / 1 1.4/0.4 2 2 1 1 473 100 534 kPa abs. 293 k k T p p T − = = = 2 1 ( ) (1.004 0.287)(473 293) 129 kJ/kg. v w u c T T =−∆ = − − = − − − = − We used Eq. 1.7.17 for cv. 1.85 a) 1.4 287 293 343.1 m/s c kRT = = × × = b) 1.4 188.9 293 266.9 m/s c kRT = = × × = c) 1.4 296.8 293 348.9 m/s c kRT = = × × = d) 1.4 4124 293 1301 m/s c kRT = = × × = e) 1.4 461.5 293 424.1 m/s c kRT = = × × = Note: We must use the units on R to be J/kg. K in the above equations. 1.86 (D) For this high-frequency wave, 287 323 304 m/s. c RT = = × = 1.87 At 10 000 m the speed of sound 1.4 287 223 299 m/s. c kRT = = × × = At sea level, 1.4 287 288 340 m/s. c kRT = = × × = 340 299 % decrease 100 12.06 %. 340 − = × = 1.88 a) = 1.4 287 253 319 m/s. 319 8.32 2654 m. c kRT L c t = × × = = ∆ = × = b) = 1.4 287 293 343 m/s. 343 8.32 2854 m. c kRT L c t = × × = = ∆ = × = c) = 1.4 287 318 357 m/s. 357 8.32 2970 m. c kRT L c t = × × = = ∆ = × =
- 13. 12 C HAPTER 2 FluidStatics 2.1 Σ ∆ ∆ ∆ ∆ F ma p z p s y z a y y y y = − = : sinα ρ 2 Σ ∆ ∆ ∆ ∆ ∆ ∆ F ma p y p s y z a g y z z z z z = − = + : cosα ρ ρ 2 2 Since ∆ ∆ s y cosα = and ∆ ∆ s z sin α = , we have p p y a y y − = ρ ∆ 2 and ( ) p p z a g z z − = + ρ ∆ 2 Let ∆y → 0 and ∆z → 0: p p p p y z − = − = 0 0 ∴ = = p p p y z . 2.2 p = γh. a) 9810 × 10 = 98 100 Pa or 98.1 kPa b) (0.8 × 9810) × 10 = 78 480 Pa or 78.5 kPa c) (13.6 × 9810) × 10 = 1 334 000 Pa or 1334 kPa d) (1.59 × 9810) × 10 = 155 980 Pa or 156.0 kPa e) (0.68 × 9810) × 10 = 66 710 Pa or 66.7 kPa 2.3 h = p/γ. a) h = 250 000/9810 = 25.5 m b) h = 250 000/(0.8 × 9810) = 31.9 m c) h = 250 000/(13.6 × 9810) = 1.874 m d) h = 250 000/(1.59 × 9810) = 16.0 m e) h = 250 000/(0.68 × 9810) = 37.5 m 2.4 (C) (13.6 9810) (28.5 0.0254) 96600 Pa Hg p h γ = = × × × = 2.5 S = 20 144 62.4 20 p h γ × = × = 2.31. ρ = 1.94 × 2.31 = 4.48 slug/ft3 . 2.6 a) p = γh = 0.76 × (13.6 × 9810) = 9810 h. ∴h = 10.34 m. b) (13.6 × 9810) × 0.75 = 9810 h. ∴h = 10.2 m. c) (13.6 × 9810) × 0.01 = 9810 h. ∴h = 0.136 m or 13.6 cm. 2.7 a) p = γ1h1 + γ2h2 = 9810 × 0.2 + (13.6 × 9810) × 0.02 = 4630 Pa or 4.63 kPa. b) 9810 × 0.052 + 15 630 × 0.026 = 916 Pa or 0.916 kPa. c) 9016 × 3 + 9810 × 2 + (13.6 × 9810) × 0.1 = 60 010 Pa or 60.0 kPa. y z pz∆y py∆z p∆s ∆s α ∆z ∆y ρg∆V
- 14. 13 2.8 ∆p =ρgh = 0.0024 × 32.2 (–10,000) = –773 psf or –5.37 psi. 2.9 (D) 0 84000 1.00 9.81 4000 44760 Pa p p gh ρ = − = − × × = 2.10 inside 100 9.81 3 13.51 Pa .287 253 100 9.81 3 11.67 Pa .287 293 outside o o base i i pg p g h h RT p pg p g h h RT ρ ρ × ∆ = ∆ = ∆ = × = × ∴∆ × ∆ = ∆ = ∆ = × = × = 1.84 Pa If no wind is present this ∆pbase would produce a small infiltration since the higher pressure outside would force outside air into the bottom region (through cracks). 2.11 p = ρgdh where h = –z. From the given information S = 1.0 + h/100 since S(0) = 1 and S(10) = 1.1. By definition ρ = 1000 S, where ρwater = 1000 kg/m3 . Then dp = 1000 (1 + h/100) gdh. Integrate: dp h gdh p = + ∫ ∫ 1000 1 100 0 10 0 ( / ) p = × + × 1000 9 81 10 10 2 100 2 . ( ) = 103 000 Pa or 103 kPa Note: we could have used an average S: Savg = 1.05, so that ρavg = 1050 kg/m3 . 2.12 v ∇ = + + p p x i p y j p z k ∂ ∂ ∂ ∂ ∂ ∂ $ $ $ = – ρa i x $ – ρα y j $ – ρα z k $ – ρgk $ = – ( ) ρ a i a j a k x y z $ $ $ + + – ρgk $ = – ρ v a – ρ v g ∴ ∇ = − + v v v p a g ρ( ) 2.13 / 0 0 [( ) / ]g R atm p p T z T α α = − = 100 [(288 − 0.0065 × 300)/288]9.81/.0065 × 287 = 96.49 kPa p p gh atm = − = − × × × ρ 100 100 287 288 9 81 300 1000 . . / = 96.44 kPa % error = 96 96 96 100 .44 .49 .49 − × = −0.052% The density variation can be ignored over heights of 300 m or less.
- 15. 14 2.14 / 0 0 0 g R atm atm Tz p p p p p T α α − ∆ = − = − = 100 288 0065 20 288 1 9 81 0065 287 − × − × . . /. = −0.237 Pa or −0.000237 kPa This change is very small and can most often be ignored. 2.15 Eq. 1.5.11 gives 310,000 144 . dp d ρ ρ × = But, dp = ρgdh. Therefore, 7 4.464 10 gdh d ρ ρ ρ × = or 2 7 32.2 4.464 10 d dh ρ ρ = × Integrate, using 0 ρ = 2.00 slug/ft3 : 2 7 2 0 32.2 4.464 10 h d dh ρ ρ ρ = ∫ ∫ × . ∴ 1 1 2 ρ − − = 7.21 × 10-7 h or 7 2 1 14.42 10 h ρ − = − × Now, 7 7 7 0 0 2 2 ln(1 14.42 10 ) 1 14.42 10 14.42 10 h h g g p gdh dh h h ρ − − − = = = − × ∫ ∫ − × − × Assume ρ = const: 2.0 32.2 64.4 p gh h h ρ = = × × = a) For h = 1500 ft: paccurate = 96,700 psf and pestimate = 96,600 psf. 96,600 96,700 % error 100 0.103 % 96,700 − = × = − b) For h = 5000 ft: paccurate = 323,200 psf and pestimate = 322,000 psf. 322,000 323,200 % error 100 0.371 % 323,200 − = × = − c) For h = 15,000 ft: paccurate = 976,600 psf and pestimate = 966,000 psf. 966,000 976,600 % error 100 1.085 % 976,600 − = × = − 2.16 Use the result of Example 2.2: p = 101 e−gz/RT . a) p = 101 e−9.81 ×10 000/287 ×273 = 28.9 kPa. b) p = 101 e−9.81 ×10 000/287 ×288 = 30.8 kPa. c) p = 101 e−9.81 ×10 000/287 ×258 = 26.9 kPa. 2.17 Use Eq. 2.4.8: p = 9.81 0.0065 287 101(1 0.0065 /288) . z × − a) z = 3000. ∴p = 69.9 kPa. b) z = 6000. ∴p = 47.0 kPa. c) z = 9000. ∴p = 30.6 kPa. d) z = 11 000. ∴p = 22.5 kPa.
- 16. 15 2.18 Use theresult of Example 2.2: / 0 = . gz RT p e p − 0 ln . p gz p RT = − 0.001 32.2 ln . 14.7 1716 455 z = − × ∴z = 232,700 ft. 2.19 p = γh = (13.6 × 9810) × 0.25 = 33 350 Pa or 33.35 kPa. 2.20 a) p = γh. 450 000 = (13.6 × 9810) h. ∴h = 3.373 m b) p + 11.78 × 1.5 = (13.6 × 9810) h. Use p = 450 000, then h = 3.373 m % error is 0.000 %. 2.21 Referring to Fig. 2.6a, the pressure in the pipe is p = ρgh. If p = 2400 Pa, then 2400 = ρgh = ρ × 9.81 h. a) ρ = × 2400 981 36 . . = 680 kg/m3 . ∴gasoline b) ρ = × 2400 981 272 . . = 899 kg/m3 . ∴benzene c) ρ = × 2400 981 245 . . = 999 kg/m3 . ∴water d) ρ = × 2400 9 81 154 . . = 1589 kg/m3 . ∴carbon tetrachloride 2.22 Referring to Fig. 2.6a, the pressure is p = ρwgh = 2 1 . 2 aV ρ Then V gh w a 2 2 = ρ ρ . a) V2 2 1000 9 81 06 123 = × × × . . . = 957. ∴V = 30.9 m/s b) V2 2 194 32 2 3 12 00238 = × × × . . / . = 13,124. ∴V = 115 ft/sec c) V2 2 1000 9 81 1 123 = × × × . . . = 1595. ∴V = 39.9 m/s d) V2 2 194 32 2 5 12 00238 = × × × . . / . = 21,870. ∴V = 148 ft/sec 2.23 (C) 0 30000 0.3 9810 0.1 8020 Pa w atm x x water w p p h h γ γ = + − = + × − × = 2.24 See Fig. 2.6b: p1 = –γ1h + γ2H. p1 = –0.86 × 62.4 × 5 12 + 13.6 × 62.4 × 9 5 12 . = 649.5 psf or 4.51 psi. 2.25 0 1 1 2 2 3 3 4 4 p p gh gh gh gh ρ ρ ρ ρ = + + + + = 3200 + 917×9.81×0.2 + 1000×9.81×0.1 + 1258×9.81×0.15 + 1593×9.81×0.18 = 10 640 Pa or 10.64 kPa
- 17. 16 2.26 ( )( ) ( ) p p p p p p p p 1 4 1 2 2 3 3 4 − = − + − + − (Use ∆ ∆ p g h = ρ ) 40 000 – 16 000 = 1000×9.81(–.2) + 13 600×9.81×H + 920×9.81×.3. ∴H = .1743 m or 17.43 cm 2.27 ( ) ( ) ( ) p p p p p p p p 1 4 1 2 2 3 3 4 − = − + − + − (Use ∆ ∆ p g h = ρ ) po – pw = 900×9.81(–.2) + 13 600×9.81(–.1) + 1000×9.81×.15 = –12 300Pa or –12.3 kPa 2.28 ( ) ( ) ( ) ( ) p p p p p p p p p p 1 5 1 2 2 3 3 4 4 5 − = − + − + − + − p1 = 9810(–.02) + 13 600×9.81(.–04) + 9810(–.02)+13 600×9.81×.16 = 15 620 Pa or 15.62 kPa 2.29 pw + 9810 × .15 – 13.6 × 9810 × .1 – .68 × 9810 × .2 + .86 × 9810 × .15 = po. ∴pw – po = 11 940 Pa or 11.94 kPa. 2.30 pw – 9810 × .12 – .68 × 9810 × .1 + .86 × 9810 × .1 = po. With pw = 15 000, po = 14 000 Pa or 14.0 kPa. 2.31 a) p + 9810 × 2 = 13.6 × 9810 × .1. ∴p = –6278 Pa or –6.28 kPa. b) p + 9810 × .8 = 13.6 × 9810 × .2. ∴p = 18 835 Pa or 18.84 kPa. c) p + 62.4 × 6 = 13.6 × 62.4 × 4/12. ∴p = –91.5 psf or –0.635 psi. d) p + 62.4 × 2 = 13.6 × 62.4 × 8/12. ∴p = 441 psf or 3.06 psi. 2.32 p – 9810 × 4 + 13.6 × 9810 × .16 = 0. ∴p = 17 890 Pa or 17.89 kPa. 2.33 (A) (13.6 9810) 0.16 21350 Pa. a p H γ = − = − × × = − , 21350 10000 11350 13.6 9810 . 0.0851 m aafter after after p H H = − + = − = × ∴ = 2.34 8200 + 9810 × .25 = 1.59 × 9810 × H. ∴H = 0.683 m Hnew = .683 + .273 = .956 m. ∆H = .273 2 = .1365. p + 9810 (.25 + .1365) = 1.59 × 9810 × .956. ∴p = 11 120 Pa or 11.12 kPa. 2.35 p + 9810 × .05 + 1.59 × 9810 × .07 – .8 × 9810 × .1 = 13.6 × 9810 × .05. ∴p = 5873 Pa or 5.87 kPa. Note: In our solutions we usually retain 3 significant digits in the answers (if a number starts with “1” then 4 digits are retained. In most problems a material property is used, i.e., S = 1.59. This is only 3 sig. digits! ∴ only 3 are usually retained in the answer! H ∆H ∆H
- 18. 17 2.36 Before pressureis applied the air column on the right is 48" high. After pressure is applied, it is (4 – H/2) ft high. For an isothermal process 1 p V 1 2 p V = 2 using absolute pressures. Thus, 14.7 × 144 × 4A = p2(4 – H / 2 )A or p2 = 8467 4 2 − H / . From a pressure balance on the manometer (pressures in psf): 30 × 144 + 14.7 × 144 = 13.6 × 62.4 H + 8467 4 2 − H / , or H2 – 15.59 H + 40.73 = 0. ∴H = 12.27 or 3.32 ft. 2.37 a) ( ) ( ) ( ) ( ) p p p p p p p p p p 1 5 1 2 2 3 3 4 4 5 − = − + − + − + − 4000 = 9800(0.16–0.22) + 15 600(0.10–0.16) + 133 400H + 15 600(0.07–H). ∴H = .0376 m or 3.76 cm b) 0.6×144 = 62.4(–2/12) + 99.5(–2/12) + 849H + 99.5(2.5/12 – H). ∴H = .1236 ft or 1.483 in. 2.38 a) 2 2 2 2 1 1 2 3 2 2 / 2 2( ) / H D d p D d γ γ γ γ ∆ = ∆ − + + − 2 2 2(.1/.005) 9800 2 15 600 2(133 400 15 600)(.1/.005) = − + × + − = × − 8 10 6 .487 H ∴ = × × − ∆H 8 10 400 6 .487 = 0.0034 m or 3.4 mm b) ∆H = − + × + − × 2 4 2 62 2 99 5 2 99 5 4 2 06 144 2 2 ( /. ) .4 . (849 . )( /. ) . = 0.01153 ft or 0.138 in. 2.39 ( ) ( ) ( ) p p p p p p p p 1 4 1 2 2 3 3 4 − = − + − + − (poil = 14.0 kPa from No. 2.30) 15 500 – 14 000 = 9800(0.12 + ∆z) + 680(0.1 – 2∆z) + 860(–0.1 – ∆z). ∴∆z = 0.0451 m or 4.51 cm 2.40 a) pair = –6250 + 625 = –5620 Pa. –5620 + 9800(2 + ∆z) – 13 600 × 9.81(0.1 + 2∆z) = 0. ∴∆z = 0.0025. ∴h = 0.1 + 2∆z = .15 m or 15 cm b) pair = 18 800 + 1880 = 20 680 Pa. 20 680 + 9800(0.8 + ∆z) – 13 600 × 9.81(0.2 + 2∆z) = 0. ∴∆z = 0.00715 m ∴h = .2+ 2∆z = .214 or 21.4 cm c) pair = –91.5 + 9.15 = –82.4 psf. –82.4 + 62.4(6 + ∆z) – 13.6 × 62.4(4/12 + 2∆z) = 0. ∴∆z = 0.00558 ft. ∴h = 4/12 + 2 (0.00558) = 0.3445 ft or 4.13 in. d) pair = 441 + 44.1 = 485 psf 485 + 62.4(2 + ∆z) – 13.6 × 62.4(8/12 + 2∆z) = 0. ∴∆z = 0.0267 ft. ∴h = 8/12 + 2 (0.0267) = 0.7205 ft or 8.65 in.
- 19. 18 2.41 F hA =γ = 9810 × 10 × π × .32 /4 = 6934 N. 2.42 5 1 5 5 (2 ) (2 ) [9800 1 3 (2 )]. 32670 N 3 3 3 3 P P × × = × × × × × × × ∴ = a) F = pc A = 9800 × 2 × 42 = 313 600 N or 313.6 kN b) 2 2 9800 1 (2 4) 9800 2 9800 1 98000 N or 98.0 kN 3 3 c F p A = = × × × + × × + × × = c) F = pc A = 9800 × 1 × 2 × 4 × 2 = 110 900 N or 110.9 kN d) F = pc A = 9800 × 1 × 2 × 4/.866 = 90 500 N or 90.5 kN 2.43 For saturated ground, the force on the bottom tending to lift the vault is F = pc A = 9800 × 1.5 × (2 × 1) = 29 400 N. The weight of the vault is approximately W gV ρ = 2400 9.81 walls = × [2(2×1.5×.1) + 2(2×1×.1) + 2(.8×1.3×.1)] = 28 400 N. The vault will tend to rise out of the ground. 2.44 F = pc A = 6660 × 2 × π × 22 = 167 400 N or 167.4 kN Find γ in Table B.5 in the Appendix. 2.45 a) F = pc A = 9800 (10 − 2.828/3) (2.828 × 2/2) = 251 000 N or 251 kN where the height of the triangle is (32 − 12 )1/2 = 2.828 m. b) F = pc A = 9800 × 10 (2.828 × 2/2) = 277 100 N or 277.1 kN c) F = pc A = 9800 (10 − 2.828 × .866/3) (2.828 × 2/2) = 254 500 N or 254.5 kN 2.46 a) F hA = = × × = γ 62 27 33 24 .4 . 40,930 lb. yp = + × × 27 33 6 8 36 27 33 24 3 . / . = 27.46'. ∴y = 30 – 27.46 = 2.54'. 8/5.46 = 3/x. ∴x = 2.05’. (2.05, 2.54) ft. b) F = 62.4 × 30 × 24 = 44,930 lb. The centroid is the center of pressure. y = 2.667'. 8/5.333 = 3/x. ∴x = 2.000' (2.000, 2.667) ft. c) F = 62.4 (30 – 2.667 × .707) × 24 = 42,100 lb. yp = + × × 39 77 6 8 36 39 77 24 3 . / . = 39.86'. y = 42.43 – 39.86 = 2.57' 8/5.43 = 3/x. ∴x = 2.04'. (2.04, 2.57) ft. 2.47 (B) The force acts 1/3 the distance from the hinge to the water line: 5 1 5 5 (2 ) (2 ) [9800 1 3 (2 )]. 32670 N 3 3 3 3 P P × × = × × × × × × × ∴ = (x, y) y x
- 20. 19 2.48 a) FhA = = × × γ π 9810 6 22 = 739 700 N or 739.7 kN. y y I Ay p = + = + × × 6 2 4 4 6 4 π π / = 6.167 m. ∴(x, y)p = (0, –0.167) m b) F hA = = × × γ π 9810 6 2 = 369 800 N or 369.8 kN. yp = + × × 6 2 8 2 6 4 π π / = 6.167 m. x2 + y2 = 4 x F x pdA x y xdy y y dy p = = − = − − ∫ ∫ ∫ − − 2 2 6 2 4 6 2 2 2 2 2 γ γ ( ) ( )( ) . ∴ × = − − + = − ∫ x y y y dy pγ π γ γ 6 2 2 24 4 6 32 2 2 2 3 ( ) . ∴xp = 0.8488 m ∴(x, y)p = (0.8488, –0.167) m c) F = 9810 × (4 + 4/3) × 6 = 313 900 N or 313.9 kN. yp = + × × 5333 3 4 36 5 333 6 3 . / . = 5.500 m. ∴y = –1.5 4/2.5 = 1.5/x. ∴x = 0.9375. ∴(x,y)p = (0.9375, –1.5) m d) F = × + × 9810 4 2 3 4 ( sin 36.9°) × 6 = 330 000 N yp = + × × 5 6 5 2 36 6 5 6 3 . .4 / . = 5.657 m. ∴y = 0.343 m 3 cos 53.13° = 1.8, 2.5 – 1.8 = 0.7, 2.4/2.057 = . / 7 1 x . ∴ x1 = 0.6. x = 1.8 + 0.6 = 2.4. ∴(x,y)p = (2.4, 0.343) m. 2.49 F hA = = × × × γ 62 11 6 10 .4 ( ) = 41,180 lb. y y I yA p = + = + × × 11 6 10 12 11 60 3 / = 11.758'. (16 – 11.758) 41,180 = 10P. ∴P = 17,470 lb. 2.50 F hA = = × × γ 9810 6 20 = 1.777 × 106 N, or 1177 kN. y y I Ay p = + = + × × 7 5 4 5 12 7 5 20 3 . / . = 7.778 m. (10 – 7.778) 1177 = 5 P. ∴P = 523 kN. y x (x, y) dA dy x y 3 4 53.13 o yp P F
- 21. 20 2.51 F hA == × × γ 9810 12 20 = 2.354 × 106 N, or 2354 kN. y y I Ay p = + = + × × 15 4 5 12 15 20 3 / = 15.139 m. (17.5 – 15.139) 2354 = 5 P. ∴P = 1112 kN. 2.52 y y I Ay H bH bH H H H H p = + = + × = + = 2 12 2 2 6 2 3 3 / / . yp is measured from the surface. ∴From the bottom, H y H H H p − = − = 2 3 1 3 . Note: this result is independent of the angle α, so it is true for a vertical area or a sloped area. 2.53 3 1 sin40 3 . ( 2) sin40 . 2( 2) . 2 3 l F l l F l P l l P γ γ = × × = + ∴ = + o o a) 9810× 23 = 2(2 + 2)P. ∴P = 9810 N b) 9810× 43 = 2(4 + 2)P. ∴P = 52 300 N c) 9810× 53 = 2(5 + 2)P. ∴P = 87 600 N 2.54 h = − 122 2 . .4 = 1.1314 m. A = 1.2 × 1.1314 + .4 × 1.1314 = 1.8102 m2 Use 2 forces: F h A c 1 1 9800 5657 1 2 11314 = = × × × γ . ( . . ) = 7527 N F h A c 2 2 9800 11314 3 11314 = = × × × γ . (.4 . ) = 1673 N yp1 2 3 11314 = ( . ). y y I A y p2 2 2 3 11314 3 11314 36 11314 2 11314 3 = + = + × × × . .4 . / .4 . / . / = 0.5657 m ΣMhinge = 0: 7527 11314 3 1673 11314 0 5657 11314 × + × − − . ( . . ) . P = 0. ∴P = 3346 N. 2.55 To open, the resultant force must be just above the hinge, i.e., yp must be just less than h. Let yp = h, the condition when the gate is about to open: y h H A h H I h H h H = + = + = + + ( ) / , ( ) , [ ( )]( ) / 3 2 36 2 3 ∴ = + + + + + = + + + = + y h H h H h H h H h H h H h H p 3 2 36 3 3 6 2 4 2 ( ) / ( ) ( ) / a) h h H = + 2 . ∴h = H = 0.9 m b) h = H = 1.2 m c) h = H = 1.5 m 2.56 The gate is about to open when the center of pressure is at the hinge. a) y H H b H b p = + = + + × + 12 18 2 18 12 9 18 3 . ( . / ) . / (. ) . . ∴H = 0.
- 22. 21 b) y HH b H b p = + = + + × + 12 2 0 2 2 12 1 2 3 . ( . / ) / ( ) . ∴H = 0.6667 m. c) y H H b H b p = + = + + × + 12 2 2 2 2 2 12 11 2 2 3 . ( . / ) . / ( . ) . . ∴H = 2.933 m. 2.57 (A) The gate opens when the center of pressure in at the hinge: 3 1.2 11.2 (1.2 ) /12 5. 5 1.2. 2 2 (1.2 ) (11.2 ) / 2 p h I h b h y y y Ay h b h + + + = + = + = + = + + + This can be solved by trial-and –error, or we can simply substitute one of the answers into the equation and check to see if it is correct. This yields h = 1.08 m. 2.58 F H bH bH 1 2 2 1 2 = × = γ γ F H b b H 2 = × = γ γ l l 1 2 3 2 2 γ γ bH H b H × = × l l . ∴ = H 3l a) H = × 3 2 = 3.464 m b) H = 1.732 m c) H = 10.39' d) H = 5.196' Assume 1 m deep 2.59 The dam will topple if the moment about “O” of F1 and F3 exceeds the restoring moment of W and F2. a) W = × × + × ( .4 )( / ) 2 9810 6 50 24 50 2 = 21.19 × 106 N dw = × + × + 300 27 600 16 300 600 = 19.67 m. F2 = 9810 × 5 × 11.09 = 0.544 × 106 N. d2 11 09 3 = . = 3.697 m. F1 9810 45 2 45 = × × = 9.933 × 106 N. d1 = 15 m. F3 9810 45 10 2 30 = × + × = 8.093 × 106 N. d3 2 943 15 5150 20 2 943 5150 = × + × + . . . . = 18.18 m. Wd F d Fd F d w + = × ⋅ + = × ⋅ 2 2 6 1 1 3 3 6 418 8 10 2961 10 . . N m N m ∴will not topple. b) W = (2.4 × 9810) (6 × 65 + 65 × 12) = 27.55 × 106 N. dw = 390 27 780 16 390 780 × + × + = 19.67 m. F2 6 0 54 10 ≅ × . N. d2 3 70 ≅ . m. F1 = 9810 × 30 × 60 = 17.66 × 106 N. d1 = 20 m. F3 9810 60 10 2 30 = × + × = 10.3 × 106 N. d3 2 943 15 7358 20 2 943 7 358 = × + × + . . . . = 18.57 m. F2 F1 H/3 l/2 F1 F2 F3 W O
- 23. 22 Wd F d FdF d w + = × ⋅ + = × ⋅ 2 2 6 1 1 3 3 6 543 9 10 544 5 10 . . N m N m ∴it will topple. c) Since it will topple for H = 60, it certainly will topple if H = 75 m. assume 1 m deep 2.60 The dam will topple if there is a net clockwise moment about “O.” a) W W W W = + = × × × × 1 2 1 6 43 1 62 2 . ( ) .4 .4 = 38,640 lb. W2 24 43 2 62 2 = × × × ( / ) .4 .4 = 77,280 lb. W3 40 22 33 2 62 = × × ( . / ) .4 = 27,870 lb @ 20.89 ft. F1 62 20 40 1 = × × × .4 ( ) = 49,920 lb @ 40/3 ft. F2 62 5 10 1 = × × × .4 ( ) = 3120 lb @ 3.33 ft 1 3 2 = 18,720 lb @ 15 ft = 28,080 lb @ 20 ft p p F F F = O M Σ : (49,920)(40/3) + (18,720)(15) + (28,080)(20) − (38,640)(3) − (77,280)(14) − (27,870)(20.89) − (3120)(3.33) < 0. ∴won’t tip. b) W1 = 6 × 63 × 62.4 × 2.4 = 56,610 lb. W2 = (24 × 63/2) × 62.4 × 2.4 = 113,220 lb. F1 62 30 60 = × × .4 = 112,300 lb. 3 (60 22.86/2) 62.4 W = × × = 42,790 lb. F2 62 5 10 = × × .4 = 3120 lb Fp1 62 10 30 = × × .4 = 18,720 lb. Fp2 62 50 30 2 = × × .4 / = 46,800 lb. O M Σ : (112,300)(20) + (18,720)(15) + (46,800)(20) − (56,610)(3) − (113,220)(14) − 42,790(21.24) = 799,000 > 0. ∴will tip. c) Since it will topple for H = 60 ft., it will also topple for H = 80 ft. 2.61 ΣMhinge = 0. 2.5P – dw × W – d1 × F1 = 0. ∴ P = × × × + × × × × × 1 2 5 2 3 9800 1 8 4 2 3 9800 2 4 4 2 . π π = 62 700 N Note: This calculation is simpler than that of Example 2.7. Actually, We could have moved the horizontal force FH and a vertical force FV (equal to W) simultaneously to the center of the circle and then 2.5P = 2FH.=2F1. This was outlined at the end of Example 2.7. 2.62 Since all infinitesimal pressure forces pass thru the center, we can place the resultant forces at the center. Since the vertical components pass thru the bottom point, they produce no moment about that point. Hence, consider only horizontal forces: ( ) 9810 2 (4 10) 784 800N ( ) 0.86 9810 1 20 168 700N water H H oil F F = × × × = = × × × = ΣM P : . . . 2 784 8 2 168 7 2 = × − × ∴P = 616.1 kN. F1 F2 F3 W O W3 F1 dw d1 P W
- 24. 23 2.63 Place theresultant force v v F F H V + at the center of the circular arc. v FH passes thru the hinge showing that P FV = . a) P FV = = × × + × = 9810 6 2 4 4 594 ( ) π 200 N or 594.2 kN. b) P = FV = 62.4 (20 × 6 × 12 + 9π × 12) = 111,000 lb. 2.64 (D) Place the force v v F F H V + at the center of the circular arc. FH passes through the hinge: 2 4 1.2 9800 ( 1.2 /4) 9800 300000. 5.16 m. V P F w w w π ∴ = = × × + × × = ∴ = 2.65 Place the resultant v v F F H V + at the circular arc center. v FH passes thru the hinge so that P FV = . Use the water that could be contained above the gate; it produces the same pressure distribution and hence the same FV . P FV = = 9810 (6 × 3 × 4 + 9π) = 983 700 N or 983.7 kN. 2.66 Place the resultant v v F F H V + at the center. v FV passes thru the hinge 2 × (9810 × 1 × 10) = 2.8 P. ∴P = 70 070 N or 70.07 kN. 2.67 The incremental pressure forces on the circular quarter arc pass through the hinge so that no moment is produced by such forces. Moments about the hinge gives: 3 P = 0.9 W = 0.9 × 400. ∴P = 120 N. 2.68 The resultant v v F F H V + of the unknown liquid acts thru the center of the circular arc. v FV passes thru the hinge. Thus we use only ( ) . FH oil Assume 1 m wide. a) ΣM R R R R S R R R R x : . 3 9810 2 4 3 9800 4 2 2 + = π π γ ∴γ x = 4580 N/m3 b) ΣM R R R R S R R R R x : . . . 3 62 4 2 4 3 62 4 4 2 2 + = π π γ ∴γ x = 29.1 lb/ft3 2.69 The force of the water is only vertical (FV)w, acting thru the center. The force of the oil can also be positioned at the center: a) P FH o = = × × × ( ) ( . ) . . 0 8 9810 0 3 3 6 = 8476 N. ΣF W F F y V o V w = = + − 0 ( ) ( ) 0 = S × 9810 π × .62 × 6 + . . 36 36 4 − π × 6 × (.8 × 9810) – 9810 × π × .18 × 6 − × × × − 9810 8 2 6 6 2 . . . ∴ = S 0 955 . . b) g V ρ . W = = 1996 lb. ΣF W F F y V o V w = = + − 0 ( ) ( )
- 25. 24 0 = S× 62.4 × π × 22 × 20 + 4 4 4 − π × 20 × .8 × 62.4 – 62.4 × π × 2 × 20 − × × × × 62 4 8 2 2 20 2 . . . ∴ = S 0 955 . . 2.70 The pressure in the dome is a) p = 60 000 – 9810 × 3 – 0.8 × 9810 × 2 = 14 870 Pa or 14.87 kPa. The force is F = pAprojected = (π × 32 ) × 14.87 = 420.4 kN. b) From a free-body diagram of the dome filled with oil: Fweld + W = pA Using the pressure from part (a): Fweld = 14 870 × π × 32 – (.8 × 9810) × 1 2 4 3 33 π × = –23 400 N or –23.4 kN 2.71 A free-body diagram of the gate and water is shown. H F d W H P w 3 + = × . a) H = 2 m. F = 9810 × 1 × 4 = 39 240 N. W xdy y dy = = = × ∫ ∫ 9810 2 9810 2 2 2 9810 2 2 3 2 1 2 0 2 0 2 3 2 / / / = 26 160 N. d x x xdy xdy x dx x dx w = = = = ∫ ∫ ∫ ∫ 2 1 2 4 4 1 2 1 4 1 3 3 0 1 2 0 1 / / = 0.375 m. ∴ = × + × P 1 3 39 0 375 2 26 160 240 . = 17 980 N or 17.98 kN. b) H = 8'. F = 62.4 × 4 × 32 = 7,987 lb. W xdy x dx = = × = × × ∫ ∫ 62 4 62 4 4 62 16 2 3 2 0 2 3 .4 .4 .4 / = 2,662 lb. d x x dx x dx w = = = ∫ ∫ 1 2 4 4 1 2 16 4 8 3 3 0 2 2 0 2 / / = 0.75'. 1 8 7,987 0.75 2,662 2910 lb 8 3 P = × + × = 2.72 (A) W V γ = 900 9.81 9810 0.01 15 . 6 m w w × = × × ∴ = W pA Fweld dA=xdy y x F h/3
- 26. 25 2.73 W =weight of displaced water. a) 20 000 + 250 000 = 9810 × 3 2 (6 /2). d d + ∴d2 + 12d – 18.35 = 0. ∴d = 1.372 m. b) 270 000 = 1.03 × 9810 × 3 2 (6 /2). d d + d2 + 12d – 17.81 = 0. ∴d = 1.336 m. 2.74 25 + FB = 100. ∴FB = 75 = 9810 − V . ∴ − V = 7.645 × 10−3 m3 γ × 7.645 × 10−3 = 100. or 7645 cm3 ∴γ = 13 080 N/m3 . 2.75 3000 × 60 = 25 × 300 ∆d × 62.4. ∴∆d = 0.3846' or 4.62". 2.76 100 000 × 9.81 + 6 000 000 = (12 × 30 + 8h × 30) 9810 ∴h = 1.465 m. ∴distance from top = 2 – 1.465 = 0.535 m 2.77 T + FB = W. (See Fig. 2.11 c.) T = 40 000 – 1.59 × 9810 × 2 = 8804 N or 8.804 kN. 2.78 The forces acting on the balloon are its weight W, the buoyant force FB, and the weight of the air in the balloon Fa. Sum forces: FB = W + Fa or 4 3 1000 4 3 3 3 π ρ π ρ R g R g a = + 4 3 5 100 9 81 287 293 1000 4 3 5 100 981 287 3 3 π π × × × = + × × . . . . . Ta ∴Ta = 350.4 K or 77.4°C 2.79 The forces acting on the blimp are the payload Fp, the weight of the blimp W, the buoyant force FB, and the weight of the helium Fh: FB = Fp + W + Fh 1500 150 100 9 81 287 288 2 π × × × × . . = Fp + 0.1 Fp + 1500 π × 1502 × 100 981 2 077 288 × × . . 4 o /64. I d π = . Npeople = 986 10 800 8 . × = 1.23 × 106 Of course equipment and other niceties such as gyms, pools, restaurants, etc., would add significant weight. 2.80 Neglect the bouyant force of air. A force balance yields FB = W + F = 50 + 10 = 60 = 9800 − V . ∴ − V = .006122 m3 Density: g V ρ . W = ρ × × 9 81 006122 . . = 50. ∴ρ = 832.5 kg/m3 Specific wt: γ = ρg = 832.5 × 9.81 = 8167 N/m3 Specific gravity: S = ρ/ρwater = 832.5/1000 = 0.8325
- 27. 26 2.81 From aforce balance FB = W + pA. a) The buoyant force is found as follows (h > 16'): cos , θ = − − h R R 15 Area = θR2 – (h – 15 – R) R sinθ ∴FB = 10 × 62.4[πR2 − θR2 + (h – 15 – R) R sinθ]. FB = 1500 + γhA. The h that makes the above 2 FB’s equal is found by trial-and- error: h = 16.5: 1859 ? 1577 h = 16.8: 1866 ? 1858 h = 17.0: 1870 ? 1960 ∴h = 16.82 ft. b) Assume h > 16 1 3 ft. and use the above equations with R = 1.333': h = 16.4: 1857 ? 1853 ∴h = 16.4 ft. c) Assume h < 16 2 3 ft. With R = 1.667', FB = 10 × 62.4[θR2 − (R – h + 15) R sinθ]. FB = 1500 + γhA. cos θ = − + R h R 15 Trial-and-error for h: h = 16: 1849 ? 1374 h = 16.2: 1853 ? 1765 h = 16.4: 1857 ? 2170 ∴h = 16.25 ft. 2.82 a) W FB = . [ ] 0 01 136 1000 015 4 9 81 9810 2 . . . / . . + × × × × = − h V π − = × × + × × = × − V π π . . . . . . 015 4 15 005 4 06 2 769 10 2 2 5 3 m ∴h = 7.361 × 10−3 m ∴ = mHg 2 13.6 1000 .015 / 4 hπ × × × = 0.01769 kg b) (.01 + .01769) 9.81 = 9810 π π × × + × × . . . . . 015 4 15 005 4 12 2 2 Sx ∴Sx = 0.959. c) (.01 + .01769) 9.81 = 9810 π× × . . 015 4 15 2 Sx. ∴Sx = 1.045. 2.83 (. ) . . . . . . 01 9 81 9810 015 4 15 005 4 12 2 2 + = × × + × × mHg π π ∴mHg = 0.01886. a) (.01 + .01886) 9.81 = 9810 π× × . . 015 4 15 2 Sx. ∴Sx = 1.089. b) mHg = 0.01886 kg. h − 15 θ R pA FB W h − 15 θ R
- 28. 27 2.84 a) 4 4 o (10/12) 6464 d I π π × = = = 0.02367 ft4 . − = = × × × × V W rH O 2 8 62 5 12 12 12 62 2 . .4 ( / ) / .4 π = 0.4363. depth = .4363 ( / ) π 5 12 2 = 0.8' ∴ = − − GM . /.4363 (. .4) 02367 5 = –0.0457'. ∴It will not float with ends horizontal. b) Io = 0.02367 ft4 , − V = 0.3636 ft3 , depth = 0.6667' GM = − − . /. ( ) / 02367 3636 5 4 12 = –0.01823'. ∴It will not float as given. c) − V = 0.2909, depth = 6.4", GM = . . . 02367 2909 4 3 2 12 − − = 0.0147. ∴It will float. 2.85 With ends horizontal 4 o /64. I d π = The displaced volume is − = × = × − V d h d x x γ π γ 2 5 3 4 9800 8 014 10 / . since h = d. The depth the cylinder will sink is depth = − = × = × − − V A d d d x x 8 014 10 4 10 20 10 5 3 2 5 . / / . γ π γ The distance CG is CG h d x = − × − 2 10 2 10 2 5 . / γ . Then GM d d d d x x = × − + × > − − π γ γ 4 5 3 5 64 8 014 10 2 10 2 10 2 0 / . . / . This gives (divide by d and multiply by γx): 612.5 – .5 γx + 5.1 × 10-5 γ x 2 > 0. Consequently, γx > 8369 N/m3 or γx < 1435 N/m3 2.86 3 3 . water water water S d W V S d γ γ γ − = = = 3 3 . water water water S d W V S d γ γ γ − = = = ∴h = Sd. GM d Sd d Sd d S S = − − = − + 4 3 12 2 2 1 12 1 2 2 / ( / / ) ( ). If GM = 0 the cube is neutral and 6S2 – 6S + 1 = 0. ∴ = ± − S 6 36 24 12 = 0.7887, 0.2113. The cube is unstable if 0.2113 < S < 0.7887. Note: Try S = 0.8 and S = 0.1 to see if GM > 0. This indicates stability. 2.87 As shown, y = × + × + 16 9 16 4 16 16 = 6.5 cm above the bottom edge. G S S A A = × + × + × × + × + × 4 9 5 16 8 5 16 4 5 8 2 8 16 γ γ γ γ γ γ . . . = 6.5 cm. C G h
- 29. 28 ∴130 + 104SA = 174 + 64 SA. ∴ SA = 1.1. 2.88 a) y = × + × + × + + 16 4 8 1 8 7 16 8 8 = 4. x = × + × + × + + 16 1 8 4 8 4 16 8 8 = 2.5. For G: y = × × + × × + × × × + × + × 12 16 4 5 8 1 1 5 8 7 1 2 16 5 8 15 8 . . . . . . = 4.682. x = × + × × + × × × + × + × 12 16 5 8 4 15 8 4 12 16 5 8 15 8 . . . . . . = 2.364. G must be directly under C. tan . . . θ = 136 682 ∴θ =11.3°. b) y = × + × + × + + 4 2 2 1 2 2 3 5 4 2 2 . = 2. x = × + × + × + + 4 1 2 2 2 2 2 4 2 2 = 1.25 For G:y = × × + × + × × + × + × 12 4 2 5 1 15 7 12 4 5 2 15 2 . . . . . . = 2.34. x = × + × + × × + × + × 12 2 5 4 15 4 1 2 4 5 2 15 2 . . . . . . = 1.182 ∆y = 0.34, ∆x = 0.068. tan . . . θ = 068 34 ∴θ = 11.3°. 2.89 The centroid C is 1.5 m below the water surface. ∴ CG = 1.5 m. Using Eq. 2.4.47: GM = × × × − = − = > l l 8 12 8 3 15 1777 15 0 277 0 3 / . . . . . ∴The barge is stable. 2.90 y = × + × + 8 3 16 97 1 8 16 97 .485 .414 . .485 . = 1.8 m. ∴CG = − 18 15 . . = 0.3 m. Using Eq. 2.4.47: GM = × − = − = l l 8 12 34 97 3 1 3 116 3 .485 / . . .46 . . . ∴Stable. 2.91 (A) 2 5 20000 20000 6660 (1.2 ) 24070 Pa 9.81 24070 0.02 30.25 N plug plug plug p h F p A γ π = + = + × × = = = × × = . 2.92 a) tan . . α = = 20 9 81 4 H ∴H = 8.155 m. pmax = 9810 (8.155 + 2) = 99 620 Pa b) pmax = ρ(g + az) h = 1000 (9.81 + 20) × 2 = 59 620 Pa c) pmax = 1.94 × 60 (–12) – 1.94 (32.2 + 60) (–6) = 2470 psf or 17.15 psi d) pmax = 1.94 (32.2 + 60) (–6) = 1073 psf or 7.45 psi 0.682 C 0.136 G
- 30. 29 2.93 The airvolume is the same before and after. ∴ 0.5 × 8 = hb/2. tan . . α = = 10 9 81 h b 4 9 81 10 = h h 2 . . ∴h = 2.856. ∴Use dotted line. 2 5 1 2 2 5 2 4 . . .452 . w + × × = ∴w = 0.374 m. a) pA = –1000 × 10 (0 – 7.626) – 1000 × 9.81 × 2.5 = 51 740 Pa or 51.74 kPa b) pB = –1000 × 10 (0 – 7.626) = 76 260 Pa or 76.26 kPa c) pC = 0. Air fills the space to the dotted line. 2.94 Use Eq. 2.5.2: Assume an air-water surface as shown in the above figure. a) 60 000 = –1000 ax (0–8) – 1000 × 9.81 0 2 5 8 981 − − . . ax 4 = h ax 2 9 81 2 × . 60 = 8 ax + 24.52 – 9.81 8 9 81 ax . . ax – 4.435 = 1.1074 ax . ax 2 – 10.1 ax + 19.67 = 0 ∴ax = 2.64, 7.46 m/s2 b) 60 000 = –1000 ax (–8) – 1000 (9.81 + 10) − + 2 5 8 9 81 . . . ax 60 = 8 ax + 49.52 – 19.81 8 19 81 ax . . ax – 1.31 = 1.574 ax . ax 2 – 5.1 ax + 1.44 = 0 ∴ax = 0.25, 4.8 m/s2 c) 60 000 = –1000 ax (–8) – 1000 (9.81 + 5) (–2.5 + 8 14 81 ax . ). 60 = 8 ax + 37.0 – 14.81 8 14 81 ax . . ax – 2.875 = 1.361 ax . ax 2 – 7.6 ax + 8.266 = 0 ∴ax = 1.32, 6.28 m/s2 2.95 a) ax = 20 × .866 = 17.32 m/s2 , az = 10 m/s2 . Use Eq. 2.5.2 with the peep hole as position 1. The x-axis is horizontal passing thru A. We have pA = –1000 × 17.32 (0 – 1.232) – 1000 (9.81 + 10) (0 – 1.866) = 58 290 Pa b) pA = –1000 × 8.66 (0 – 1.848) – 1000 (9.81 + 5) (0 – 2.799) = 57 460 Pa c) The peep hole is located at (3.696, 5.598). Use Eq. 2.5.2: pA = –1.94 × 51.96 (0 – 3.696) – 1.94 (32.2 + 30) (0 – 5.598) = 1048 psf d) The peep hole is located at (4.928, 7.464). Use Eq. 2.5.2: pA = –1.94 × 25.98 (–4.928) – 1.94 (32.2 + 15) (–7.464) = 932 psf 2.96 a) The pressure on the end AB (z is zero at B) is, using Eq. 2.5.2, p(z) = –1000 × 10 (–7.626) – 1000 × 9.81(z) = 76 260 – 9810 z α b h A B z 1 x w C
- 31. 30 ∴ = − ∫ Fz dz AB ( ) . 76 260 9810 4 0 2 5 = 640 000 N or 640 kN b) The pressure on the bottom BC is p(x) = –1000 × 10 (x – 7.626) = 76 260 – 10 000 x. ∴ = − ∫ F x dx BC ( ) . 76 260 10 000 4 0 7 626 = 1.163 × 106 N or 1163 kN c) On the top p(x) = –1000 × 10 (x – 5.174) where position 1 is on the top surface: ∴ = − ∫ F x dx top ( ) . 51 740 10 000 4 0 5 174 = 5.35 × 105 N or 535 kN 2.97 a) The pressure at A is 58.29 kPa. At B it is pB = –1000 × 17.32 (1.732–1.232) – 1000 (19.81) (1–1.866) = 8495 Pa. Since the pressure varies linearly over AB, we can use an average pressure times the area: FAB = + × × 58 290 8495 2 15 2 . = 100 200 N or 100.2 kN b) pD = 0. pC = –1000 × 17.32 (–.5–1.232) − 1000 × 19.81(.866–1.866) = 49 810 Pa. FCD = × × × 1 2 49 810 15 2 . = 74 720 N or 74.72 kN. c) pA = 58 290 Pa. pC = 49 810 Pa. ∴ = + × FAC 58 29 49 81 2 15 . . . = 81.08 kN. 2.98 Use Eq. 2.5.2 with position 1 at the open end: a) pA = 0 since z2 = z1. pB = 1000 × 19.81 × 0.6 = 11 890 Pa. pC = 11 890 Pa. b) pA = –1000 × 10 (.9–0) = –9000 Pa. pB = –000 × 10 (.9)–1000 × 9.81(-.6) = –3114 Pa pC = –1000 × 9.81 × (–.6) = 5886 Pa. c) pA = –1000×20 (0.9) = –18 000 Pa. pB = –1000 × 20 × 0.9–1000×19.81(−0.6) = –6110 Pa. pC = 11 890 Pa d) pA = 0. pB = 1.94 × (32.2-60) 25 12 = −112 psf. pC = –112 psf. e) pA = 1.94 × 60 − 37 5 12 . = −364 psf. pB = 1.94 × 60 − 37 5 12 . – 1.94 × 32.2 − 25 12 = –234 psf. x z A B C x z 1
- 32. 31 pC = –1.94× 32.2 − 25 12 = 130 psf. f) pA = 1.94 × 30 37 5 12 . = 182 psf. pB = –1.94(–30) 37 5 12 . – 1.94 × 62.2 − 25 12 = 433 psf. pC = –1.94 × 62.2 × − 25 12 = 251 psf. 2.99 Use Eq. 2.6.4 with position 1 at the open end: ω π = × 50 2 60 = 5.236 rad/s. a) pA = × × × 1000 5 236 2 6 15 2 2 . (. . ) = 11 100 Pa. pB = 1 2 × 1000 × 5.2362 × .92 + 9810 × .6 = 16 990 Pa. pC = 9810 × .6 = 5886 Pa. b) pA = 1 2 × 1000 × 5.2362 × 0.62 = 4935 Pa. pB = 1 2 × 1000 × 5.2362 × 0.62 + 9810 × 0.4 = 8859 Pa. pC = 9810 × 0.4 = 3924 Pa. c) pA = 1 2 × 1.94 × 5.2362 × 37 5 12 2 . = 259.7 psf. pB = 1 2 × 1.94 × 5.2362 × 37 5 12 62 25 12 2 . .4 + × = 389.7 psf. pC =62 25 12 .4 × = 130 psf. d) pA = 1 2 × 1.94 × 5.2362 × 22 5 12 2 . = 93.5 psf. pB = 1 2 × 1.94 × 5.2362 × 22 5 12 2 . + 62 15 12 .4 × = 171.5 psf. pC = 62 15 12 .4 × = 78 psf. A B C z 1 r ω
- 33. 32 2.100 Use Eq.2.6.4 with position 1 at the open end. a) pA = 1 2 × 1000 × 102 (0 – 0.92 ) = –40 500 Pa. pB = –40 500 + 9810 × 0.6 = –34 600 Pa. pC = 9810 × 0.6 = 5886 Pa. b) pA = 1 2 × 1000 × 102 (0 – 0.62 ) = –18 000 Pa. pB = –18 000 + 9810 × 0.4 = –14 080 Pa. pC = 9810 × 0.4 = 3924 Pa. c) pA = 1 2 × 1.94 × 102 0 37 5 144 2 − . = –947 psf. pB = -947 + 62.4 × 25 12 = –817 psf. pC = 62.4 × 25 12 = 130 psf. d) pA = 1 2 × 1.94 × 102 − 22 5 12 2 2 . = –341 psf. pB = –341 + 62.4 × 15 12 = –263 psf. pC = 62.4 × 15 12 = 78 psf. 2.101.1Use Eq. 2.6.4 with position 1 at the open end and position 2 at the origin. Given: p2 = 0. a) 0 = 1 2 × 1000 ω2 (0 – 0.452 ) – 9810 (0 – 0.6). ∴ω = 7.62 rad/s. b) 0 = 1 2 × 1000 ω2 (0 – 0.32 ) – 9810 (0 – 0.4). ∴ω = 9.34 rad/s. c) 0 = 1 2 × 1.94 ω2 0 18 75 12 2 2 − . – 62.4 − 25 12 . ∴ω = 7.41 rad/s. d) 0 = 1 2 × 1.94 ω2 − 11 25 12 2 2 . – 62.4 − 15 12 . ∴ω = 9.57 rad/s. 2.102 The air volume before and after is equal. ∴ = × × 1 2 6 2 0 2 2 π π r h . . . ∴ r h 0 2 = 0.144. a) Using Eq. 2.6.5: r0 2 2 5 2 × / = 9.81 h ∴h = 0.428 m ∴pA = 1 2 × 1000 × 52 × 0.62 – 9810 (–0.372) = 8149 Pa. b) r0 2 2 7 2 × / = 9.81 h. ∴h = 0.6 m. ∴pA = 1000 2 × 72 × 0.62 + 9810 × 0.2 = 10 780 Pa. A B C z 1 r ω z 1 r ω 2 1 2 h z r A r0
- 34. 33 c) For ω= 10, part of the bottom is bared. π π π × × = − . . . 6 2 1 2 1 2 2 0 2 1 2 1 r h r h Using Eq. 2.6.5: ω2 0 2 2 r g h = , ω2 1 2 2 r g h = 1. ∴ = − 0144 2 2 2 2 2 1 2 . g h g h ω ω or h h 2 1 2 2 0 144 10 2 981 − = × × . . . Also, h – h1 = 0.8. 1.6h – 0.64 = .7339. ∴h = 0.859 m, r1 = 0.108 m. ∴pA = 1 2 × 1000 × 102 (0.62 – 0.1082 ) = 17 400 Pa. d) Following part (c): h h 2 1 2 2 0144 20 2 9 81 − = × × . . . 1.6h – .64 = 2.936.∴ h = 2.235 m. ∴pA = 1 2 × 1000 × 202 (0.62 – 0.2652 ) = 57 900 Pa r1 = 0.265 m 2.103 The answers to Problem 2.102 are increased by 25 000 Pa. a) 33 150 Pa b) 35 780 Pa c) 42 400 Pa d) 82 900 Pa 2.104 p r r g h ( ) [ (. )]. = − − − 1 2 0 8 2 2 ρω ρ p r r h ( ) (. ) = + − 500 9810 8 2 2 ω if h < .8. p r r r ( ) ( ) = − 500 2 2 1 2 ω if h > .8. a) F p rdr r r dr = = + ∫ ∫ 2 2 12 500 3650 3 0 6 π π ( ) . = 6670 N. (We used h = .428 m) b) F p rdr r r dr = = + ∫ ∫ 2 2 24 500 1962 3 0 6 π π ( ) . = 7210 N. (We used h = 0.6 m) c) F p rdr r r dr = = − − ∫ ∫ 2 2 50 000 108 3 108 6 2 π π ( ( . ) . . = 9520 N. (We used r1 = 0.108 m) d) F p rdr r r dr = = − ∫ ∫ 2 2 200 000 265 3 265 6 2 π π ( ( . ) . . = 26 400 N. (r1 = 0.265 m) 1 h z r A r0 h1 dr dA = 2πrdr
- 35. 34 CHAPTER 3 Introduction toFluids in Motion 3.1 3.2 Pathlines: Release several at an instant in time and take a time exposure of the subsequent motions of the bulbs. Sreakline: Continue to release the devises at a given location and after the last one is released, take a snapshot of the “line” of bulbs. Repeat this for several different release locations for additional streaklines. 3.3 3.4 streakline pathline streamline streakline pathline hose boy time t t = 0 streamlines t = 2 hr pathline t = 2 hr streakline at t = 3 hr y x
- 36. 35 3.5 a)u dx dt t v dy dt t == + = = 2 2 2 x t t c y t c = + + = + 2 1 2 2 2 = + y y 2 ∴ − + = x xy y y 2 2 2 4 ∴parabola. b) x t t c c c = + + ∴ = − = − 2 1 1 2 2 8 4 . , . and = + + ± + − y y 4 2 4 8 ( ) ∴ − + + − = x xy y x y 2 2 2 8 12 0. ∴parabola. 3.6 ˆ ˆ ˆ ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ using , . z V dr udy vdx V ui vj wk dr dxi dy j dzk i j k j i k × = − = + + = + + × = × = − v v v v 3.7 Lagrangian: Several college students would be hired to ride bikes around the various roads, making notes of quantities of interest. Eulerian: Several college students would be positioned at each intersection and quantities would be recorded as a function of time. 3.8 a) At t V = = = 2 0 0 0 2 2 2 and m / s ( , , ) . At t V = − = + = 2 1 2 0 3 2 3 606 2 2 and m / s ( , , ) . . b) At t V = = 2 0 0 0 0 and ( , , ) . At t V = − = − + − = 2 1 2 0 2 8 8 246 2 2 and m/ s ( , , ) ( ) ( ) . . c) At t V = = − = 2 0 0 0 4 4 2 and m / s ( , , ) ( ) . At t V = − = + − + − = 2 1 2 0 2 4 4 6 2 2 2 and m / s ( , , ) ( ) ( ) . 3.9 (D) 5 ˆ ( 51.4 10 ) j − − × A simultaneous solution yields 4/5 and 3/5. x y n n = = (They must both have the same sign. 3.10 a) cos $ / ( )/ . . . α α = ⋅ = + + = ∴ = v o V i V 1 2 3 2 0 832 33 69 2 2 v V n i j n i n j n n n n n n n n x y x y x y y x x x ⋅ = + ⋅ + = + = + = ∴ = − + = $ . ( $ $) ( $ $) . 0 3 2 0 3 2 0 1 3 2 9 4 1 2 2 2 2 ∴ = = − = − n n n i j x y 2 13 3 13 1 13 2 3 , $ ( $ $). or 39.8o y x streamlines t = 5 s (27, 21) (35, 25)
- 37. 36 b) cos $/ / ( ) ( ) . . α α = ⋅ = − − + − = − ∴ = v o V i V 2 2 8 0 2425 104 2 2 v V n i j n i n j n n n n n n n n x y x y x y x y y y ⋅ = − − ⋅ + = − − = + = ∴ = − + = $ . ( $ $) ( $ $) . 0 2 8 0 2 8 0 1 4 16 1 2 2 2 2 ∴ = = − = − + n n n i j y x 1 17 4 17 1 17 4 , $ ( $ $). or c) cos $ / / ( ) . . . α α = ⋅ = + − = ∴ = − v o V i V 5 5 8 0 6202 51 67 2 2 v V n i j n i n j n n n n n n n n x y x y x y x y y y ⋅ = − ⋅ + = − = + = ∴ = + = $ . ( $ $) ( $ $) . 0 5 8 0 5 8 0 1 8 5 64 25 1 2 2 2 2 ∴ = = = + n n n i j y x 5 89 8 89 1 89 8 5 , $ ( $ $). or 3.11 a) [ ] v v V dr x i xtj dxi dyj × = + + × + = 0 2 0 . ( )$ $ ( $ $) . ∴ + − = + = ( ) . x dy xtdx t xdx x dy 2 0 2 or Integrate: [ ] t xdx x dy t x n x y C + = − + = + ∫ ∫ 2 2 2 . . l 2 1 2 3 2 ( ) . − = − + ∴ ln C C = 0.8028. [ ] t x n x y − + = + 2 2 0 8028 l . b) [ ] v v V dr xyi y j dxi dyj × = − × + = 0 2 0 2 . $ $ ( $ $) . ∴ + = = − xydy y dx dx x dy y 2 0 2 2 or . Integrate: 2 2 1 2 l l l l nx n y C n n C = − = − − ( / ). ( ) ( / ). ∴ = − = − − ∴ = − C nx n y x y 2 2 2 2 2 . ( / ). . l l c) [ ] v v V dr x i y tj dxi dyj × = + − × + = 0 4 0 2 2 . ( )$ $ ( $ $) . ( ) . x dy y tdx tdx x dy y 2 2 2 2 4 0 4 + + = + = − or Integrate: t x C y C 2 2 1 2 2 1 2 1 2 1 1 tan . tan . − − + = + = − ∴ = − − = − C yt x 0 9636 2 0 9636 2 1 . . tan .
- 38. 37 3.12 (C) 2 ˆˆ ˆ ˆ ˆ ˆ 2 (2 ) (2 2 ) 16 8 16 . V V V V a u v w xy yi y xi yj i i j t x y z ∂ ∂ ∂ ∂ = + + + = − − = − − + ∂ ∂ ∂ ∂ v v v v v 2 2 8 16 17.89 m/s a ∴ = + = 3.13 a) DV Dt u V x v V y w V z V t v v v v v = + + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ =0. b) u V x v V y w V z V t x i y j xi yj ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ v v v v + + + = + = + 2 2 2 2 4 4 ( $) ( $) $ $ = 8 4 $ $ i j − c) u V x v V y w V z V t x t xti ytj xyt xtj ztk x i xyj ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ v v v v + + + = + + + + + 2 2 2 2 2 2 2 2 ( $ $) ( $ $) $ + = − − 2 68 100 54 yzk i j k $ $ $ d) u V x v V y w V z V t x i yzj xyz xzj tz xyj tk zk ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ v v v v + + + = − − − + − + + ($ $) ( $) ( $ $) $ 2 2 2 2 = xi yz x yz xyzt j zt z k $ ( )$ ( ) $ − − + + + 2 4 2 2 2 2 =2 114 15 $ $ $ i j k − + 3.14 v Ω = − + − + − 1 2 1 2 1 2 ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ w y v z i u z w x j v x u y k $ $ $. a) v Ω = − = 1 2 20 ∂ ∂ u y k yk $ $ = −20 $ k b) v Ω = − + − + − 1 2 0 0 1 2 0 0 1 2 0 0 ( )$ ( )$ ( ) $ i j k = 0 c) v Ω = − + − + − 1 2 2 0 1 2 0 0 1 2 2 0 ( )$ ( )$ ( ) $ zt i j yt k = 6 2 $ $ i k − d) v Ω = + + − + − − 1 2 0 2 1 2 0 0 1 2 2 0 ( )$ ( )$ ( )$ xy i j yz k = − + 2 3 $ $ i k 3.15 The vorticity v v ω = 2Ω. Using the results of Problem 3.7: a) v ω = −40$ i b) v ω = 0 c) v ω = 12 4 $ $ i k − d) v ω = − + 4 6 $ $ i k 3.16 a) ε ∂ ∂ ε ∂ ∂ ε ∂ ∂ xx yy zz u x v y w z = = = = = = 0 0 0 , , . ε ∂ ∂ ∂ ∂ ε ∂ ∂ ∂ ∂ xy xz u y v x y u z w x = + = − = = + = 1 2 20 20 1 2 0 , , ε ∂ ∂ ∂ ∂ yz v z w y = + = ∴ = 1 2 0 0 20 0 20 0 0 0 0 0 . rate - of strain
- 39. 38 b) ε ε ε εε ε xx yy zz xy xz yz = = = = = = 2 2 0 0 0 0 , , . , , . rate-of strain = 2 0 0 0 2 0 0 0 0 c) ε ε ε xx yy zz xt xt yt = = = = = = − 2 8 2 8 2 4 , , . ε ε ε xy xz yz yt zt = = − = = = = 1 2 2 2 1 2 0 0 1 2 2 6 ( ) , ( ) , ( ) . rate-of strain = 8 2 0 2 8 6 0 6 4 − − − d) ε ε ε xx yy zz xz t = = − = − = = 1 2 12 2 , , . ε ε ε xy xz yz yz xy = − = = = = − = 1 2 2 3 1 2 0 0 1 2 2 2 ( ) , ( ) , ( ) . rate-of strain = 1 3 0 3 12 2 0 2 2 − 3.17 a) a r r r r r r = − − + − − 10 40 80 10 40 1 40 2 3 2 2 cos cos sin ( sin ) θ θ θ θ − + = − − − 1 10 40 10 2 1 125 1 2 2 2 r r sin ( .5)( ) . ( ) θ = 9.375 m/s2. a r r r r r θ θ θ θ θ = − + + + 10 40 80 10 40 10 40 2 3 2 2 cos sin sin cos − − 1 100 1600 4 r r sin cos θ θ = 0 since sin 180° = 0. aφ = 0. b) ω ω ω θ θ θ r z r r r r = = = − + − − − 0 0 1 10 40 1 10 40 2 2 , , sin ( sin ) = 0. At (4, 180°) v ω = 0 since v ω = 0 everywhere. 3.18 a) a r r r r r r = − − + − − 10 80 240 10 80 10 80 3 4 3 3 cos cos sin ( sin ) θ θ θ θ − + = − − 10 80 8 75 1 9375 1 3 2 2 r r sin . ( )(. )( ) θ = 8.203 m/s2 aθ = 0 since sin 180° = 0. aφ = 0 since vφ = 0. b) ω r = 0, ω θ = 0, ω φ = 0, since sin 180° = 0.
- 40. 39 3.19 V V a u tx ∂ ∂ ∂ ∂ = + v v v v + V w y ∂ + ∂ v ˆ. V u i z t ∂ ∂ ∂ = ∂ v For steady flow ∂ ∂ u t a / . = = 0 0 so that v 3.20 Assume u(r,x) and v(r,x) are not zero. Then, replacing z with x in the appropriate equations of Table 3.1 and recognizing that vθ ∂ ∂θ = = 0 0 and / : r x v v u u a v u a v u r x r x ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + = + 3.21 a) u e t t = − − = = ∞ − 2 1 0 1 10 ( )( ) . / 2 m / s at ( ) a u t e t x t = = − = = − ∂ ∂ 2 1 0 1 10 0 2 0 10 ( ) . . / m / s at 2 b) u e t t = − − = = ∞ − 2 1 0 1 2 10 ( .5 )( ) . / 1.875 m / s at a e t x t = − = = − 2 1 0 2 1 10 0 0125 0 2 2 10 ( .5 / ) . . / m / s at 2 c) u e t t = − − = − 2 1 2 2 1 2 2 10 ( / )( ) . / 0 for all a e t x t = − = − 2 1 2 2 1 10 0 2 2 10 ( / ) . / for all 3.22 DT T u Dt x ∂ ∂ = v + T w y ∂ ∂ + 2 20(1 ) sin 0.5878 100 100 5 T T t y z t ∂ ∂ π π π ∂ ∂ + = − − = − × = −0.3693 °C/s. 3.23 D Dt u x v y w z t e ρ ∂ρ ∂ ∂ρ ∂ ∂ρ ∂ ∂ρ ∂ = + + + = − × − − × − 10 1 23 10 4 3000 10 4 ( . ) = − × ⋅ − 9 11 10 4 . . kg m s 3 3.24 D Dt u x v y w z t ρ ∂ρ ∂ ∂ρ ∂ ∂ρ ∂ ∂ρ ∂ = + + + = − 10 1000 4 = − ⋅ 2500 kg m s 3 . 3.25 D Dt u x ρ ∂ρ ∂ = = × 4 01 (. ) = 0.04 kg/m3⋅s 3.26 (D) 2 2 10 [10(4 ) ] (4 ) x u u u u u a u v w u x t x y z x x x − ∂ ∂ ∂ ∂ ∂ ∂ = + + + = = − ∂ ∂ ∂ ∂ ∂ ∂ − 3 2 2 10 10 1 10( 2)( 1)(4 ) 20 6.25 m/s . 4 8 (4 ) x x − = − − − = × × = − 3.27 D Dt V t = ⋅∇ + v v ∂ ∂ observing that the dot product of two vectors v A A i A j A k x y z = + + $ $ $ and v v v B B i B j B k A B A B A B A B x y z x x y y z z = + + ⋅ = + + $ $ $ . is
- 41. 40 3.28 a u t V u a v t V v a w t Vw a V t V V x y z = + ⋅∇ = + ⋅∇ = + ⋅∇ ∴ = + ⋅∇ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ v v v v v v v v v v v ( ) 3.29 Using Eq. 3.2.12: a) v v v v v v v v v v A a d s dt V r d dt r = + + × + × × + × 2 2 2Ω Ω Ω Ω ( ) = 2 20 4 20 20 1 ( $ $) $ ( $ .5$) k i k k i × + × × = 160 600 $ $ j i − m / s 2 b) v v v v v v o A V r k j k k i = × + × × = × − + × × 2 2 20 20 30 20 20 3 Ω Ω Ω ( ) ( $ cos $) $ ( $ $) = −507$ i 3.30 v Ω = × × = × − 2 24 60 60 7 272 10 5 π $ . $ k k rad/s. v v i k i k = − − = − − 5 707 707 3 3 ( . $ . $) .535$ .535$ m/s. v v v v v v A V r = × + × × 2Ω Ω Ω ( ) = 2 7 272 10 3 3 7 272 10 5 5 × × × − − + × × − − . $ ( .535$ .535 $) . $ k i k k [ . $ ( . $ . $)] 7 272 10 6 10 707 707 5 6 × × × − + − k i k = − × + − 51 4 10 0 0224 5 . $ . $ j i m / s2 . Note: We have neglected the acceleration of the earth relative to the sun since it is quite small (it is d s dt 2 2 v / ). The component 5 ˆ ( 51.4 10 ) j − − × is the Coriolis acceleration and causes air motions to move c.w. or c.c.w. in the two hemispheres. 3.31 a) two-dimensional (r, z) b) two-dimensional (x, y) c) two-dimensional (r, z) d) two-dimensional (r, z) e) three-dimensional (x, y, z) f) three-dimensional (x, y, z) g) two-dimensional (r, z) h) one-dimensional (r) 3.32 Steady: a, c, e, f, h Unsteady: b, d, g 3.33 b. It is an unsteady plane flow. 3.34 a) d) e) 3.35 f, h
- 42. 41 3.36 a) inviscid.b) inviscid. c) inviscid. d) viscous inside the boundary layer. e) viscous inside the boundary layers and separated regions. f) viscous. g) viscous. h) viscous. 3.37 d and e. Each flow possesses a stagnation point. 3.38 3.39 (C) The only velocity component is u(x). We have neglected v(x) since it is quite small. If v(x) in not negligible, the flow would be two-dimensional. 3.40 Re = VL/ ν = 2 × .015/.77 × 10-6 = 39 000. ∴Turbulent. 3.41 Re = = VL ν .2 × .8/1.4 × 10-5 = 11 400. ∴Turbulent. 3.42 Re . . = = × × − VL ν 4 06 17 10 5 = 14 100. ∴Turbulent. Note: We used the smallest dimension to be safe! 3.43 a) Re . . .51 = = × × = − V D ν 1 2 0 01 1 10 795. 5 Always laminar. b) Re . .51 = = × × = − VD ν 12 1 1 10 79 5 500. May not be laminar. 3.44 Re = 3 × 105 = T Vx ν . ν µ ρ = / where µ µ = ( ). T a) T = 223 K or −50°C. ∴ = × ⋅ − µ 1 10 5 .5 N s / m . 2 ∴ = × × = × − − ν 1 5 10 3376 1 23 2 5 10 5 5 . . . . . m /s 2 3 10 900 1000 3600 2 10 5 5 × = × × × − xT .5 . ∴xT = 0.03 m or 3 cm b) T = −48°F. ∴µ = 3.3 × 10−7 lb-sec/ft2. ν = × = × − − 3 3 10 00089 3 7 10 7 4 . . . ft2/sec. 3 10 600 5280 3600 3 7 10 5 4 × = × × × − xT . . ∴xT = 0.13' or 1.5"
- 43. 42 3.45 Assume theflow is parallel to the leaf. Then 3 × 105 =VxT / . ν ∴ = × = × × × = − x V T 3 10 3 10 1 4 10 6 8 17 5 5 4 ν / .5 . / . m. The flow is expected to be laminar. 3.46 a) M V c = = × × = 100 14 287 236 0 325 . . . For accurate calculations the flow is compressible. Assume incompressible flow if an error of 4%, or so, is acceptable. b) M V c = = × × = 80 14 287 288 0 235 . . . ∴Assume incompressible. c) M V c = = × × = 100 14 287 373 0 258 . . . ∴Assume incompressible. 3.47 D Dt u x v y w z t ρ ∂ρ ∂ ∂ρ ∂ ∂ρ ∂ ∂ρ ∂ = + + + = 0. For a steady, plane flow ∂ρ ∂ / t = 0 and w = 0. Then u x v y ∂ρ ∂ ∂ρ ∂ + = 0. 3.48 D u Dt x ρ ∂ρ ∂ = v y ∂ρ ∂ + w + z t ∂ρ ∂ρ ∂ ∂ + 0. = ∴incompressible. 3.49 (B) 2 9810 0.800 . 113 m/s. 2 1.23 water air h V p V γ ρ ρ × = = = ∴ = 3.50 V p 2 2 = ρ . Use ρ = 0.0021 slug/ft3. a) v p = = × × 2 2 3 144 0021 / . / . ρ = 203 ft/sec. b) v p = = × × 2 2 9 144 0021 / . / . ρ = 351 ft/sec. c) v p = = × × 2 2 09 144 0021 / . / . ρ = 111 ft/sec. 3.51 p V = = × ρ 2 2 2 123 120 1000 3600 2 . / = 683 Pa. ∴F = pA = 683 π × 0.0752 = 12.1 N. 3.52 V p 2 2 0 + = ρ . ∴ = − = × V p 2 2 2000 123 ρ . = 57.0 m/s
- 44. 43 3.53 (C) 2 22 1 2 1 . 0.200 0.600. 2 9.81 0.400 2.80 m/s. 2 2 2 V V V p V g g g γ + = + = ∴ = × × = 3.54 (B) The manometer reading h implies: 2 2 2 1 1 2 2 2 2 2 or (60 10.2). 9.39 m/s 2 2 1.13 V p V p V V ρ ρ + = + = − ∴ = The temperature (the viscosity of the water) and the diameter of the pipe are not needed. 3.55 a) 2 2 0 2 2 V V p ρ + = 2 2 ( 10 ) . . 50 2 o o o p p x p p p x ρ ρ ρ ρ − + + = ∴ = − b) 2 2 0 2 2 V V p ρ + = 2 2 (10 ) . . 50 2 o o o p p y p p p y ρ ρ ρ ρ + + = ∴ = − 3.56 2 2 2 2 U p V p ρ ρ ∞ ∞ + = + . a) v v U r r r c θ θ = = = − − ∞ 0 180 1 1 2 2 and o : ( / )( ). ( ) ∴ = − = − ∞ ∞ p U v U r r r r r c c ρ ρ 2 2 2 2 2 2 2 2 4 . b) Let r r p U c T = = ∞ : ρ 2 2 c) ( ) [ ] v r r v U p v U r c = = − ∴ = − = − ∞ ∞ ∞ 0 2 2 2 1 4 2 2 2 and = U 2 : sin . sin θ θ θ ρ ρ θ d) Let θ ρ = = − ∞ 90 3 2 90 2 o : p U 3.57 2 2 2 2 U p V p ρ ρ ∞ ∞ + = + . a) vθ θ = = 0 180 and o : ( ) p U v U r r r r r c c = − = − ∞ ∞ ρ ρ 2 2 2 2 2 2 3 6 . b) Let r r p U c T = = ∞ : 1 2 2 ρ . c) ( ) [ ] v r r p v U r c = = − = − ∞ ∞ 0 2 2 1 4 2 2 2 and = U 2 : sin ρ ρ θ θ d) Let θ ρ = = − ∞ 90 3 2 90 2 o : p U
- 45. 44 3.58 2 2 2 2 U p Vp ρ ρ ∞ ∞ + = + . a) ( ) p U u x x = − = − + = − + ∞ ρ ρ π π ρ 2 2 10 10 20 2 50 1 1 1 2 2 2 2 2 = − + 50 2 1 2 ρ x x b) 1 0 when 1. 50 ( 2 1) 50 u x p ρ ρ − = = − = − − + = c) ( ) 2 2 2 2 2 60 1 30 30 450 1 1 2 2 2 p U u x x ρ ρ π ρ π ∞ = − = − + = − + = − + 450 2 1 2 ρ x x d) 1 0 when 1. 450 ( 2 1) 450 u x p ρ ρ − = = − = − − + = 3.59 V p V p V p p 1 2 1 2 2 2 1 1 2 2 2 0 20 + = + = − = ρ ρ . and kPa. ( ) V p p V 2 2 1 2 2 2 2 1000 20 6 32 = − = ∴ = ρ ( . 000) = 40. m / s 3.60 Assume the velocity in the plenum is zero. Then 2 2 2 1 1 2 2 2 2 2 or (60 10.2). 9.39 m/s 2 2 1.13 V p V p V V ρ ρ + = + = − ∴ = We found ρ = 113 . kg / m3 in Table B.2. 3.61 Bernoulli from the stream to the pitot probe: p V p T = + ρ 2 2 . Manometer: . T Hg p H H h p h γ γ γ γ + − − = − Then, 2 2 Hg V p H H p ρ γ γ + + − = . 2 (2 ) Hg V H γ γ ρ − ∴ = a) V V 2 13 6 1 9800 1000 2 0 04 3 14 = − × ∴ = ( . ) ( . ). . m / s b) V V 2 13 6 1 9800 1000 2 0 1 4 97 = − × ∴ = ( . ) ( . ). . m / s c) V V 2 13 6 1 62 4 194 2 2 12 1162 = − × ∴ = ( . ) . . ( / ). . fps d) V V 2 13 6 1 62 4 194 2 4 12 16 44 = − × ∴ = ( . ) . . ( / ). . fps
- 46. 45 3.62 The pressureat 90° from Problem 3.56 is 2 90 3 /2. p U ρ ∞ = − The pressure at the stagnation point is 2 /2. T p U ρ ∞ = The manometer provides: p H p T − = γ 90 2 2 1 3 1.204 9800 0.04 1.204 . 12.76 m/s 2 2 U U U ∞ ∞ ∞ × − × = − × ∴ = 3.63 The pressure at 90° from Problem 3.57 is 2 90 3 /2. p U ρ ∞ = − The pressure at the stagnation point is 2 /2. T p U ρ ∞ = The manometer provides: p H p T − = γ 90 2 2 1 3 1.204 9800 0.04 1.204 . 12.76 m/s 2 2 U U U ∞ ∞ ∞ × − × = − × ∴ = 3.64 Assume an incompressible flow with point 1 outside in the room where p1 0 = and v1 0 = . The Bernoulli’s equation gives, with p h w 2 2 = γ , 2 1 2 V 1 p ρ + 2 2 2 . 2 V p ρ = + a) 0 2 9800 0 02 1 204 18 04 2 2 2 = + − × ∴ = V V . . . . m / s b) 0 2 9800 0 08 1 204 36 1 2 2 2 = + − × ∴ = V V . . . . m / s c) 0 2 62 4 1 12 0 00233 66 8 2 2 2 = + − × ∴ = V V . / . . . fps d) 0 2 62 4 4 12 0 00233 133 6 2 2 2 = + − × ∴ = V V . / . . . fps 3.65 Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel where p V 1 1 0 0 = = and . Bernoulli’s equation gives 0 2 1 2 2 2 2 2 2 2 = + ∴ = − V p p V ρ ρ . a) ρ = = × = ∴ = − × × = − p RT p 90 0 287 253 1 239 1 2 1239 100 6195 2 2 . . . . kg / m Pa 3 b) ρ = = × = ∴ = − × × = − p RT p 95 0 287 273 1 212 1 2 1212 100 6060 2 2 . . . . kg / m Pa 3 c) ρ = = × = ∴ = − × × = − p RT p 92 0 287 293 1 094 1 2 1094 100 5470 2 2 . . . . kg / m Pa 3 d) ρ = = × = ∴ = − × × = − p RT p 100 0 287 313 1113 1 2 1113 100 5566 2 2 . . . . kg / m Pa 3 3.66 (A) 2 1 2 V g 2 1 2 2 2 p V p g γ γ + = + 2 2 2 800000 . . 40 m/s. 9810 2 9.81 V V = ∴ = ×
- 47. 46 3.67 a) ph V h h A A A = = × = = = γ 9800 4 39 0 2 200 Pa, Using . , 2 2 A V g A A p h γ + + 2 2 2 2 2 V p h g γ = + + 2 2 2 . 2 A V p p g γ = − = − × × = − 39 14 2 9 81 9800 58 2 200 700 Pa . b) 0 and 0. B B p V = = Bernoulli’s eq. gives, with the datum through the pipe, V g p h V g p h p B B B 2 2 2 2 2 2 2 2 2 4 14 2 9 81 9800 58 + + = + + = − × = − γ γ . . 700 Pa 3.68 Bernoulli: 2 2 2 2 V p g γ + 2 1 1 2 V p g γ = + Manometer: 2 2 1 2 2 Hg V p z H H z p g γ γ γ γ γ + + − − = + Substitute Bernoulli’s into the manometer equation: ( ) 2 1 1 1. 2 Hg V p H p g γ γ γ + − = + a) Use H = 0.01 m: V V 1 2 1 9800 2 9 81 13 6 1 9800 0 01 1 × × = − × ∴ = . ( . ) . .572 m / s Substitute into Bernoulli: p V V g 1 2 2 1 2 2 2 2 20 1 2 9 81 9800 198 = − = − × × = γ .572 . 600 Pa b) Use H = 0.05 m: V V 1 2 1 9800 2 9 81 13 6 1 9800 0 05 3 × × = − × ∴ = . ( . ) . .516 m / s Substitute into Bernoulli: p V V g 1 2 2 1 2 2 2 2 20 3 2 9 81 9800 193 = − = − × × = γ .516 . 600 Pa c) Use H = 0.1 m: V V 1 2 1 9800 2 9 81 13 6 1 9800 0 1 4 972 × × = − × ∴ = . ( . ) . . m / s Substitute into Bernoulli: p V V g 1 2 2 1 2 2 2 2 20 4 972 2 9 81 9800 187 = − = − × × = γ . . 400 Pa
- 48. 47 3.69 Bernoulli acrossnozzle: 2 1 2 V 2 1 2 2 2 p V p ρ ρ + = + 2 1 . 2 / V p ρ ∴ = Bernoulli to max. height: 2 1 2 V g 1 1 p h γ + + 2 2 2 V g = 2 p γ + 2 2 1 . / . h h p γ + ∴ = a) V p 2 1 2 2 700 1000 37 42 = = × = / / . ρ 000 m / s h p 2 1 700 = = / γ 000 / 9800 = 71.4 m b) V p 2 1 2 2 1 1000 52 92 = = × = / / . ρ 400 000 m / s h p 2 1 = = / γ 1 400 000/ 9800 = 142.9 m c) V p 2 1 2 2 100 194 1218 = = × × = / / . . ρ 144 fps h p 2 1 = = × / γ 100 144 / 62.4 = 231 ft d) V p 2 1 2 2 200 194 172 3 = = × × = / / . . ρ 144 fps h p 2 1 200 = = × / γ 144/ 62.4 = 462 ft 3.70 a) Apply Bernoulli’s eq. from the surface to a point on top of the downstream flow: 2 1 2 V g 1 p γ + 2 2 2 1 2 V p h g γ + = + 2 2 . 2 ( ) h V g H h + ∴ = − b) Apply Bernoulli’s eq. from a point near the bottom upstream to a point on the bottom of the downstream flow: 2 1 2 V g 2 1 2 2 1 2. 2 p V p h h g γ γ + + = + + Using p H p h h h V g H h 1 2 1 2 2 2 = = = = − γ γ , , ( ) and 3.71 2 1 2 V 2 1 2 2 . 2 p V p ρ ρ + = + p2 = −100 000 Pa, the lowest possible pressure. a) 600 2 2 2 000 1000 100 000 1000 = − V . ∴V2 = 37.4 m/s. b) 300 2 2 2 000 1000 100 000 1000 = − V . ∴V2 = 28.3 m/s.
- 49. 48 c) 80 144 1.94 14.7 144 1.94 × =− × V2 2 2 . ∴V2 = 118.6 ft/sec. d) 40 144 1.94 14.7 144 1.94 × = − × V2 2 2 . ∴V2 = 90.1 ft/sec. 3.72 A water system must never have a negative pressure, since a leak could ingest impurities. ∴ The least pressure is zero gage. V p gz V p gz 1 2 1 1 2 2 2 2 2 2 + + = + + ρ ρ . V V 1 2 = . Let z1 0 = , and p2 0 = . 500 000 1000 = 9 81 2 . . z ∴ z2 = 51.0 m. 3.73 a) ( ) ( ) p V V 1 2 2 1 2 2 2 2 1000 2 2 10 = − = − ρ = −48 000 Pa b) ( ) ( ) 2 2 2 2 1 2 1 902 2 10 43300 Pa 2 2 p V V ρ = − = − = − c) ( ) ( ) 2 2 2 2 1 2 1 680 2 10 32600 Pa 2 2 p V V ρ = − = − = − d) ( ) ( ) 2 2 2 2 1 2 1 1.23 2 10 59.0 Pa 2 2 p V V ρ = − = − = − 3.74 V p V p 1 2 1 2 2 2 2 2 + = + ρ ρ . ( ) ( ) 2 2 2 2 1 2 1 1.23 2 8 2 2 p V V ρ = − = − = −36.9 Pa 3.75 (D) ( ) ( ) 2 2 2 2 1 2 1 902 30 15 304400 Pa 2 2 p V V ρ = − = − = 3.76 Apply Bernoulli’s equation between the exit (point 2) where the radius is R and a point 1 in between the exit and the center of the tube at a radius r less than R: V p V p p V V 1 2 1 2 2 2 1 2 2 1 2 2 2 2 + = + ∴ = − ρ ρ ρ . . Since V V 2 1 < , we see that p1 is negative (a vacuum) so that the envelope would tend to rise due to the negative pressure over most of its area (except for a small area near the end of the tube). 3.77 Re . = V D ν For air ν ≅ × − 1 10 5 .5 . Use reasonable dimensions from your experience!
- 50. 49 a) Re . .5 . = × × = × − 200 03 1 10 4 10 5 4 ∴Separate b) Re . .5 . = × × = − 20 0 005 1 10 6700 5 ∴Separate c) Re .5 . . = × × = × − 20 2 1 10 2 7 10 5 6 ∴Separate d) Re . .5 . = × × = − 5 0 002 1 10 670 5 ∴Separate e) Re .5 . . = × × = × − 20 2 1 10 2 7 10 5 6 ∴Separate f) Re .5 . = × × = × − 100 3 1 10 2 10 5 7 ∴It will tend to separate, except streamlining the components eliminates separation. 3.78 A burr downstream of the opening will create a region that acts similar to a stagnation region thereby creating a high pressure since the velocity will be relatively low in that region. 3.79 ∆ ∆ p V R n = = × = ρ 2 2 1000 10 0 05 0 02 . . 40 000 Pa Along AB, we expect V V A B > < 10 10 m / s and m /s. 3.80 The higher pressure at B will force the fluid toward the lower pressure at A, especially in the wall region of slow moving fluid, thereby causing a secondary flow normal to the pipe’s axis. This results in a relatively high loss for an elbow. 3.81 Refer to Bernoulli’s equation: V p V p 1 2 1 2 2 2 2 2 + = + ρ ρ p p A B > since V V A B < p p C D < since V V C D > p p B D > since V V D B > stagnation region A B VA VB
- 51. 50 CHAPTER 4 The IntegralForms of the Fundamental Laws 4.1 a) No net force may act on the system: Σ v F = 0. b) The energy transferred to or from the system must be zero: Q - W = 0. c) If 3 3 2 ˆ ˆ ˆ 10 ( ) 0 n V V n i j = ⋅ = ⋅ − = v is the same for all volume elements then Σ v v F D Dt V dm = ∫ , or Σ v v F D Dt mV = ( ). Since mass is constant for a system Σ v v F m DV Dt = . Since DV Dt a F ma v v v v = = , . Σ 4.2 Extensive properties: Mass, m; Momentum, mV v ; kinetic energy, 1 2 mV 2 ; potential energy, mgh; enthalpy, H. Associated intensive properties (divide by the mass): unity, 1; velocity, v V;V2/2; gh; H/m = h (specific enthalpy). Intensive properties: Temperature, T; time, t; pressure, p; density, ρ; viscosity, µ. 4.3 (B) 4.4 System ( ) t V = 1 c.v.( ) t V = 1 System ( ) t t V + ∆ = 1 V + 2 c.v.( ) t t V + ∆ = 1 4.5 System ( ) t V = 1 V + 2 c.v.( ) t V = 1 V + 2 System ( ) t t V + ∆ = 2 V + 3 c.v.( ) t t V + ∆ = 1 V + 2 1 2 1 2 3 pump
- 52. 51 4.6 a) Theenergy equation (the 1st law of Thermo). b) The conservation of mass. c) Newton’s 2nd law. d) The energy equation. e) The energy equation. 4.7 4.8 4.9 4.10 $ $ $ . ($ $) n i j i j 1 1 2 1 2 0 707 = − − = − + . $ . $ .5$ n i j 2 0 866 0 = − . $ $ n j 3 = − . 1 1 1 ˆ ˆ ˆ ˆ 10 [ 0.707( )] 7.07 fps n V V n i i j = ⋅ = ⋅ − + = − v V V n i i j n 2 2 2 10 0866 05 866 = ⋅ = ⋅ − = v $ $ ( . $ . $) . fps 3 3 2 ˆ ˆ ˆ 10 ( ) 0 n V V n i j = ⋅ = ⋅ − = v 4.11 flux = ηρ$ n VA ⋅ v flux1 = ηρ ηρ [ . ($ $)] $ / . − + ⋅ = − 0 707 10 0 707 10 i j iA A flux2 = ηρ ηρ ( . $ .5$) $ / . 0 866 0 10 0 866 10 i j iA A − ⋅ = flux3 = ηρ( $) $ − ⋅ = j iA 10 0 3 n̂ v n̂ v ω n̂ v n̂ v n̂ v n̂ v n̂ n̂ n̂ v v v v n̂ n̂ n̂ n̂ v v v v n̂ n̂ v v
- 53. 52 4.12 ( $)( .5$ . $) $( ) v B n A i j j ⋅ = + ⋅ × 15 0 0 866 10 12 = × × = 15 0 866 120 1559 . cm 3 Volume = 15 60 10 12 1559 sin cm 3 o × × = 4.13 The control volume must be independent of time. Since all space coordinates are integrated out on the left, only time remains; thus, we use an ordinary derivative to differentiate a function of time. But, on the right, we note that ρ and η may be functions of (x, y, z, t); hence, the partial derivative is used. 4.14 4.15 4.16 4.17 If fluid crosses the control surface only on areas A1 and A2, ρ ρ ρ $ $ $ . . n VdA n VdA n VdA A A c s ⋅ = ⋅ + ⋅ = ∫ ∫ ∫ v v v 0 2 1 For uniform flow all quantities are constant over each area: ρ ρ 1 1 1 2 2 2 0 2 1 $ $ n V dA n V dA A A ⋅ + ⋅ = ∫ ∫ v v Let A1 be the inlet so $ n V V A 1 1 1 2 ⋅ = − v and be the outlet so $ . n V V 2 2 2 ⋅ = v Then − + = ρ ρ 1 1 1 2 2 2 0 V A V A or ρ ρ 2 2 2 1 1 1 A V A V = 1 2 1 system (∆t) is in volumes 1 and 2 c.v. (0) = c.v. (∆t) = volume 1 1 2 3 system (∆t) = V1 + V2 + V3 c.v. (∆t) = V1 + V2 system boundary at (t + ∆t)
- 54. 53 4.18 Use Eq.4.4.2 with mV representing the mass in the volume: 0 = + ⋅ ∫ dm dt n VdA V c s ρ$ . . v = + − dm dt A V A V V ρ ρ 2 2 1 1 = + − dm dt Q m V ρ &. Finally, dm dt m Q V = − & . ρ 4.19 Use Eq. 4.4.2 with mS representing the mass in the sponge: 0 = + ⋅ ∫ dm dt n VdA S ρ$ v = + + − dm dt A V A V A V S ρ ρ ρ 2 2 3 3 1 1 = + + − dm dt m A V Q S & . 2 3 3 1 ρ ρ Finally, dm dt Q m A V S = − − ρ ρ 1 2 3 3 & . 4.20 (D) 2 200 0.04 70 0.837 kg/s 0.287 293 p m AV AV RT ρ π = = = × × = × & . 4.21 A1V1 = A2V2. π × 125 144 2 . × 60 = π × 2 5 144 2 . V2. ∴V2 = 15 ft/sec. & . . m AV = = × ρ π 194 1 25 144 60 2 = 3.968 slug/sec. Q = AV = π 125 144 2 . × 60 = 2 045 . / . ft sec 3 4.22 A1V1 = A2V2.π × .0252 × 10 = (2π × .6 × .003)V2. ∴V2 = 1.736 m/s. & . m AV = = × × ρ π 1000 025 10 2 = 19.63 kg/s. Q = AV=π × .0252 × 10 =0 01963 . . m / s 3 4.23 & min = ρA1V1 + ρA2V2. 200 = 1000 π × .0252 × 25 + 1000 Q2. ∴Q2 =0 1509 . . m / s 3 4.24 ρ1 1 1 40 144 1716 520 = = × × p RT = .006455 slug/ft3. ρ2 7 144 1716 610 = × × = .000963 slug/ft3. & . & . ( / ). . m AV V m A = ∴ = = × ρ ρ π 1 1 1 2 2 2 144 006455 ∴V1 = 355 fps. & . . ( / ) . m V 2 2 0 2 000963 2 3 144 = = × × ∴V2 = 4984 fps.
- 55. 54 4.25 ρ ρρ ρ 1 1 1 2 2 2 1 1 2 500 287 393 4 1246 287 522 8 317 A V A V p RT = = = × = = × = . . .433 . . . kg m kg m 3 3 4.433 π × .052 × 600 = 8.317 π × .052 V2. ∴V2 = 319.8 m/s. & m A V = ρ1 1 1 = 20.89 kg/s. Q A V 1 1 1 = = 4 712 . . m / s 3 Q2 =2.512 . m / s 3 4.26 ρ ρ 1 1 1 2 2 2 A V A V = p RT A V p RT A V 1 1 1 1 2 2 2 2 = 200 293 0 05 40 120 0 03 120 2 2 2 π π × × = × × . . . T ∴ = − T2 189 9 83 . . K or C o 4.27 a) A V A V 1 1 2 2 = . (2 × 1.5 + 1.5 × 1.5) 3 =π d2 2 4 2 × . ∴d2 = 3.167 m b) (2 × 1.5 + 1.5 × 1.5) 3 =π d2 2 4 2 2 × . ∴d2 = 4.478 m c) (2 × 1.5 + 1.5 × 1.5) 3 = 1 3 2 866 2 2 πR R R − × × . . ∴R = 3.581 m. ∴d2 = 7.162 m 4.28 (A) Refer to the circle of Problem 4.27: 2 3 75.7 2 ( 0.4 0.10 0.40 sin75.5 ) 3 0.516 m /s. 360 Q AV π × = = × × − × × × = o 4.29 a) v r r r V vdA r r rdr r r r dr r r r = − = = − = − ∫ ∫ ∫ 10 1 10 1 2 20 0 0 2 0 0 0 2 0 0 0 0 0 . . π π π ∴ = − = V r r r 20 2 3 10 3 0 2 0 2 0 2 = 3.333 m/s. & . . m AV = = × × × ρ π 1000 04 3 33 2 = 16 75 . . kg / s Q = AV =0 01675 . . m / s 3 b) v r r r V r r rdr r r r = − = − = − ∫ 10 1 10 1 2 20 2 4 2 0 2 0 2 2 0 2 0 0 2 0 2 0 . . π π π ∴V = 5 m/s & . m AV = = × × × ρ π 1000 04 5 2 = 25 13 . kg / s. Q = AV =0 02513 . m / s. 3 c) v r r r V r r rdr r r r = − = − + ∫ 20 1 20 1 2 10 4 0 0 2 0 2 0 2 0 0 . / . / π π π ∴V = 5.833 m/s & . . m AV = = × × × ρ π 1000 04 5833 2 = 29 32 . kg / s. Q =0 02932 . . m / s 3 θ R cosθ = 1/2 θ = 60o
- 56. 55 4.30 a) Sincethe area is rectangular, V = 5 m/s. & . . m A V = = × × × ρ 1000 08 8 5 = 320 kg / s. Q = & m ρ = 0 32 . . m / s 3 b) v y h y h = − 40 2 2 with y = 0 at the lower wall. ∴ = − = × ∫ Vhw y h y h wdy h w h 40 40 6 2 2 0 . ∴V = 6.667 m/s. & . . . m A V = = × × × ρ 1000 08 8 6 667 = 426 7 . kg / s. Q = 0 4267 . . m / s 3 c) V × .08 = 10 × .04 + 5 × .02 + 5 × .02. ∴V = 7.5 m/s. & . . .5 m A V = = × × × ρ 1000 08 8 7 = 480 kg / s. & & Q m = ρ = 0 48 . . m / s 3 4.31 a) A V v dA 1 1 2 = ∫ . π π π × × = − = ∫ 1 24 6 1 2 2 4 2 0 2 0 2 0 2 0 v r r rdr v r r max max . With r0 1 24 = , vmax = 12 fps. ∴ = v r ( ) 12 1 576 2 ( ) . − r fps b) A V v dA 1 1 2 = ∫ . 1 12 6 1 4 3 2 2 × × = − = − ∫ w v y h wdy v w h h h max max . With h = 1 24 , vmax = 9 fps. ∴ = v y ( ) 9 1 576 2 ( ) . − y fps c) 1 1 2 . AV v dA = ∫ 0 2 2 2 0 max max 2 0 0 0.01 2 1 2 2 . 4 r r r v rdr v r π π π × × = − = ∫ With r0 = 0.01 m, max v = 4 m/s. ( ) v r ∴ = 2 4(1 10000 ) m/s. r − d) n̂ 2 max max 2 4 0.02 2 1 . 3 h h y h w v wdy v w h − × × = − = ∫ With h = 0.01 m, max v = 3 m/s. ( ) v y ∴ = 2 3(1 10000 ) m/s. y − 4.32 If dm dt / , = 0 then ρ ρ ρ 1 1 1 2 2 2 3 3 3 A V A V A V = + . In terms of & m Q 2 3 and this becomes, letting ρ ρ ρ 1 2 3 = = , 1000 0 02 12 1000 0 01 2 2 × × × = + × π . & . . m ∴ = & . . m2 5 08 kg / s 4.33 v dA A V r 1 0 2 2 1 ∫ = . v r r rdr r max . . 0 2 1 2 2 1 1 2 0025 2 ∫ − = × × π π ∴ = × × 2 005 4 0025 2 2 2 π π vmax . . . ∴vmax = 1 m/s. ∴ = v r ( ) 1 005 2 2 − r . . m / s
- 57. 56 4.34 & &&. . ( ) . &. . m m m y y dy m in out = + × × × = − + × × × + ∫ ρ ρ ρ 2 2 10 10 20 100 2 1 2 10 2 0 1 Note: We see that at y = 0.1 m the velocity u(.1) = 10 m/s. Thus we integrate to y = 0.1, and between y = 0.1 and 0.2 the velocity u = 10. 4 4 3 2 ρ ρ ρ = + + &. m ∴ = & . m 0 6667ρ = 0.82 kg/s. 4.35 V h u y dy h 1 1 0 = ∫ ( ) . 10 05 10 20 100 2 0 × = − ∫ . ( ) y y dy h = − 10 10 100 3 2 3 h h . ∴666.7 h3 − 200 h2 = −1. This can be solved by trial-and-error: h = .06: −.576? −1. h = .07: −.751? −1. h = .08: −.939? −1. h = .083: −.997? −1. h = .084: −1.016? −1. ∴h = 0.0832: or 8.32 cm. Note: Fluid does not cross a streamline so all the flow that enters on the left leaves on the right. The streamline simply moves further from the wall. 4.36 ( ) & . . ( ) / m VdA y y y dy = = − − × ∫ ∫ρ 2 2 1 3545 6 9 2 5 0 1 3 2 ( ) = − − + ∫ 22 6 2 127 9 3 19 2 2 3 0 1 3 y y y y dy . . / = 4.528 slug/sec. V u = = × = 2 3 2 3 2 4 3 max . fps (See Prob. 4.31b). ρ = + 2 2 194 2 . . = 2.07 slug/ft3. ∴ = × × × ρV A 2 07 4 3 5 1 3 . = 4.6 slug/sec. Thus, ρVA m ≠ & since ρ = ρ(y) and V = V(y) so that V V ρ ρ ≠ . 4.37 A V A V 1 1 2 2 = . π π × × = × × . ( . . ) . 01 8 2 2 04 2 2 V cos 30o ∴V2 = 0.05774 m/s. 4.38 2000 4 3 0015 9000 5 3 3 3 3 × × × × π . m of H O m of air m of air s 2 = 1.5 × (1.5h). ∴h = 0.565 m. 4.39 Use Eq. 4.3.3: 0 1 1 1 1 = − + ⋅ ∫ ∂ρ ∂ ρ t d V V n A v $ . v V n V 1 1 1 ⋅ = − $ . ∴ = − ρ ∂ρ ∂ 1 1 1 A V t Vtire . ( . ) . 37 14 7 144 1716 520 1 96 180 17 2 + × × × × = × π ∂ρ ∂t
- 58. 57 ∴ = ∂ρ ∂t 3 0110 5 . . × − − slug ft sec 3 4.40 & & & . m m m in = + 2 3 V1 = 20 m/s (see Prob. 4.31c). 20 1000 02 10 1000 02 2 2 3 × × = + × π π . . . V ∴ V3 = 12.04 m/s. 4.41 0 2 3 1 = + = + + − d dt m m d dt m m m m c v net c v . . . . & & & & ∴ = − − = × × × − − × × d dt m m m m c v . . & & & . . 1 2 3 2 2 1000 02 20 10 1000 02 10 π π = 2.57 kg/s. 4.42 The control surface is close to the interface at the instant shown. ∴Vi = interface velocity. ρ ρ e e e i i i A V A V = . 1 15 300 8000 287 673 1 2 2 .5 . . . × × × = × × π π Vi ∴Vi = 0.244 m/s. 4.43 Assume an incompressible flow: 4 1 2 2 Q A V = . 4 1500 60 2 4 2 × = × / ( ) . V ∴ = V2 12.5 . fps 4.44 For an incompressible flow (low speed air flow) udA A V A = ∫ 2 2 1 . 20 0 8 0 15 1 5 2 2 0 0 2 y dy V / . . . . × = × ∫ π 20 0 8 5 6 0 2 015 6 5 2 2 × = × . . . . / π V ∴ = V2 27 3 . . m / s 4.45 A V v dA A V e e 1 1 2 + = ∫ π π π ( . . ) . . . 01 0 025 4 200 1 0 025 2 0 1 2 2 2 2 2 0 0 025 − × + − = × ∫ r rdr Ve 0 1178 01963 0 0314 . . . . + = Ve ∴ = Ve 10 0 . . m / s 4.46 Draw a control volume around the entire set-up: 0 2 2 1 1 = + − dm dt V A V A tissue ρ ρ = + − − & & ( tan ) & m d d h h h tissue ρπ ρπ φ 2 2 2 2 1 2 1 4 Ve n̂ n̂ Vi
- 59. 58 or & & &tan . m d d h h h tissue = − + ρπ φ 2 2 2 2 1 2 1 2 4 4.47 The width w of the channel is constant throughout the flow. Then 0 2 2 1 1 = + − dm dt A V A V ρ ρ . 0 2 2 1 1 = + − d dt whL A V A V ( ) ρ ρ ρ 0 100 0 2 8 4 0 2 = × + × − × ρ ρ ρ dh dt w w w . . . ∴ = & . . h 0 008 m / s 4.48 0 2 2 1 1 = + − dm dt A V A V ρ ρ = + × × − × − & ( . . / ). m 1000 0 003 0 02 10 10 60 2 6 π ∴ = × − & . . m 3 99 10 4 kg / s 4.49 ρ ρ 1 1 1 2 2 2 A V A V = . & m A V 1 2 2 2 = ρ . 400 10 900 0 2 0 05 10 100 6 2 e Ve − − × × = × × / . . . π ∴ = Ve 207 m / s. 4.50 0 3 3 1 1 1 2 = + − − dm dt Q A V m ρ ρ & where m Ah = ρ . a) 0 1000 0 6 1000 0 6 60 1000 0 02 10 10 2 2 = × + × − × × − π π . & . / . . h ∴ = & . h 0 0111 m / s or 11.1 mm / s b) 0 1000 0 6 1000 0 01 0 20 2 = × + × − − π . & . . h ∴ = & . . h 0 00884 m / s or 8.84 mm/ s c) 2 2 0 1000 0.6 1000 1.0/60 1000 0.02 5 10. h π π = × + × − × × − & 0.000339 m/s or 0.339 mm/s. h ∴ = & 4.51 A V A V 1 1 2 2 = where A2 is an area just under the top surface. a) π π × × = × − 0 02 10 60 2 10 2 . ( tan ) / e h dh dt t o ∴ = − h dh e dt t 2 10 0 001333 . . / ∴ = − + − h e t 3 10 0 04 0 04 . . . / Finally, h t e t ( ) . ( ) . / / = − − 0 342 1 10 1 3 b) 0 04 10 10 60 10 10 . ( tan ) & / × × = × − e h h t o ∴ = − hdh e dt t 0 2309 10 . . / ∴ = − + − h e t 2 10 4 62 4 62 . . . / Finally, /10 1/2 ( ) 2.15(1 ) . t h t e− = −
- 60. 59 4.52 & W TpAV du dy Abelt = + + ω µ = × × + × × × + × × × × − 20 500 2 60 400 0 4 0 10 181 10 100 0 0 8 5 π / . .5 . .5 . = + + = 1047 800 0 000724 1847 . W 4.53 If the temperature is essentially constant, the internal energy of the c.v. does not change and the flux of internal energy into the pipe is the same as that leaving the pipe. Hence, the two integral terms are zero. The losses are equal to the heat transfer exiting the pipe. 4.54 80% of the power is used to increase the pressure while 20% increases the internal energy ( & Q = 0 because of the insulation). Hence, & ~ . & m u W ∆ = 0 2 1000 0 02 4 18 0 2 500 × × = × . . . . ∆T ∴ = ∆T 0 836 . . C o 4.55 (D) 2 2 2 1 2 P W V V Q g γ − = & 2 1 1200 200 . . 0.040 P p p W γ γ γ − − + = × & 40 40 kW and energy req'd = 47.1 kW. 0.85 P W ∴ = = & 4.56 & . W Q H P P p = γ η 5 746 9800 20 0 87 × = × × Q . . ∴ = Q 0 01656 . . m / s 3 4.57 − = − × & & . . W mg T 40 0 89 a) & . . WT = × × × = 40 0 89 200 9 81 69 850 W b) & . ( / ) . WT = × × × = 40 0 89 90 000 60 9 81 523 900 W c) & . ( / ) . WT = × × × × = 40 0 89 8 10 3600 9 81 776 6 100 W 4.58 10000000 . 0.89 50. 1.273 m/s 100 3 60 9.8 T T W z V AVg V η ρ − = ∆ = × ∴ = × × × × & 4.59 V g p z V g p z 1 2 1 1 2 2 2 2 2 2 + + = + + γ γ . 12 2 32 2 6 36 64 4 2 2 2 2 2 × + = + . . . h h 8 236 20 1 2 2 2 . . . = + h h Continuity: 3 × 12 = h2 V2. 3 ft V1 h2 V2
- 61. 60 This can besolved by trial-and-error. 2 8': 8.24 ? 8.31 h = 2 7.9': 8.24 ? 8.22 h = Qh2 = 7 93 . . ' 2 1.8': 8.24 ? 8.00 h = 2 1.75': 8.24 ? 8.31 h = Qh2 = 176 . '. 4.60 V g z V g z hL 1 2 1 2 2 2 2 2 + = + + . ∴ × + = × + + 4 2 9 81 2 16 2 9 81 0 2 2 2 2 2 . . . . h h ∴2 615 0 815 2 2 2 . . / . = + h h Trial-and-error provides the following: h2 2 2 615 2 63 = = .5: . ? . h2 2 45 615 2 = = . : . ? .59. 2 ∴ = h2 2 47 . m h2 0 65 2 615 2 = = . : . ? .58 h2 0 64 615 2 63 = = . : . ? . . 2 ∴ = h2 0 646 . m 4.61 Manometer: Position the datum at the top of the right mercury level. 9810 4 9810 2 1000 9810 13 6 4 9810 2 2 2 2 2 1 × + + + × = × × + × + . ( . ) . z p V p Divide by γ = 9810: . . . . 4 2 13 6 4 2 2 2 2 2 1 + + + = × + + z p V g p γ γ (1) Energy: V g p z V g p z 1 2 1 1 2 2 2 2 2 2 + + = + + γ γ . (2) Subtract (1) from (2): With z1 = 2 m, V g 1 2 2 12 6 4 = × . . . ∴V1 = 9.94 m/s 4.62 The manometer equation (see Prob. 4.61) is 0 4 2 13 6 4 2 2 2 2 2 1 . . . . + + + = × + + z p V g p γ γ (1) Energy: V g p z V g p z V g 1 2 1 1 2 2 2 2 2 2 2 2 0 05 2 + + = + + + γ γ . . (2) Subtract (1) from (2): With z1 = 2 m, and with V2 = 4V1 (continuity) 18 2 12 6 0 4 1 2 . . . . V g = × ∴V1 = 7.41 m/s. 4.63 (A) 2 2 0 V = 2 1 2 1 2 V p p g − − + 2 2 2 120 . 0 . 7 200000 Pa. 2 9.8 9810 p p γ − = + ∴ = × 4.64 Q = 120 × 0.002228 = π × 1 12 2 1 V . ∴V1 = 12.25 fps.
- 62. 61 Continuity: π π × = × 1 12 1 12 2 1 2 2 V V .5 . ∴V2 = 5.44 fps. Energy: V g p V g p V g 1 2 1 2 2 2 1 2 2 2 0 37 2 + = + + γ γ . . ∴ = × + − p2 2 2 60 144 62 4 0 63 12 25 64 4 5 44 64 4 . . . . . . = 8702.9 psf or 60.44 psi 4.65 Q = 600 × 10-3/60 = π × .022 V1. ∴V1 = 7.958 m/s. 3 2 0.02 3 3 3 3 2 0 1 1 10 1 0.02 6.67 0.02 y V dA wdy AV w α = = − ∫ ∫ × × V A V A 2 1 1 2 2 2 04 7 958 06 = = × . . . = 3.537 m/s. Energy: V g p V g p hL 1 2 1 2 2 2 2 2 + = + + γ γ . ∴ = − × + − hL 7 958 3 2 9 81 690 000 700 000 9810 2 2 . .537 . = 1.571 m 4.66 V Q A 1 1 2 0 08 03 = = × / . . π = 28.29 m/s. ∴V2 = 9V1 = 254.6 m/s. Energy: V g p V g p V g 1 2 1 2 2 2 1 2 2 2 2 2 + = + + γ γ . . ∴ = × − × p1 2 2 9810 254 6 2 9 81 0 8 28 29 2 9 81 . . . . . = 32 1 106 . . × Pa 4.67 a) Across the nozzle: 2 2 1 2 .07 .025 . V V π π × = × ∴V2 = 7.84 V1. Energy: V g p V g p 1 2 1 2 2 2 2 2 + = + γ γ . 2 2 1 1 7.84 1 9810 . 2 9.81 p V − ∴ = × For the contraction: π π × = × . . . 07 05 2 1 2 3 V V ∴V3 = 1.96 V1. Energy: V g p V g p 1 2 1 3 2 3 2 2 + = + γ γ . Manometer: γ γ × + = × + . . . . 15 136 15 1 3 p p ∴ = × + p p 1 3 12 6 15 γ γ . . . Subtract the above 2 eqns: V g V g V g 1 2 3 2 2 1 2 2 12 6 15 2 196 2 + × = = . . . . ∴ − = × × ( . ) . . . 196 1 12 6 15 2 2 1 2 V g ∴V1 = 3.612 m/s. ∴p1 = 394 400 Pa.
- 63. 62 From the reservoirsurface to section 1: V g p z V g p z 0 2 0 0 1 2 1 1 2 2 + + = + + γ γ H = + 3 612 19 62 2 . . 394 400 9810 = 40.0 m. b) Manometer: γ γ × + = × + . . . . 2 136 2 1 3 p p ∴ = × + p p 1 3 12 6 2 γ γ . . . Energy: V g p V g p 1 2 1 3 2 3 2 2 + = + γ γ . Also, V3 = 1.96 V1. ∴ + × = V g V g 1 2 2 1 2 2 12 6 2 196 2 . . . . ∴V1 = 4.171 m/s. The nozzle is the same as in part (a): ∴p1 = 534 700 Pa. From the reservoir surface to the nozzle exit: V g p z V g p z 0 2 0 0 2 2 2 2 2 2 + + = + + γ γ . ∴ = = × H V g 2 2 2 2 32 7 2 9 81 . . = 54.5 m. 4.68 a) Energy: V g p z V g p z 0 2 0 0 2 2 2 2 2 2 + + = + + γ γ . ∴ = = × × V gz 2 0 2 2 9 81 2 4 . . = 6.862 m/s. Q = AV = .8 × 1 × 6.862 = 5 49 . . m / s 3 For the second geometry the pressure on the surface is zero but it increases with depth. The elevation of the surface is 0.8 m. ∴ = + z V g h 0 2 2 2 . ∴ = − = × × V g z h 2 0 2 2 9 81 2 ( ) . = 6.264 m/s. ∴Q = .8 × 6.264 = 5 01 . . m / s 3 Note: z0 is measured from the channel bottom in the 2nd geometry. ∴z0 = H + h. b) V g p z V g p z 0 2 0 0 2 2 2 2 2 2 + + = + + γ γ . ∴ = = × × + = V gz 2 0 2 2 32 2 6 2 2 2123 . . fps. ∴Q = AV = (2 × 1) × 21.23 = 42.5 cfs. For the second geometry, the bottom is used as the datum: ∴ = + + z V g h 0 2 2 2 0 . ∴ = + − V g H h h 2 2 2 ( ) . ∴ = = × × V gH 2 2 2 32 2 6 . = 19.66 fps. ∴Q = 39.3 cfs.
- 64. 63 4.69 From thereservoir surface to the exit: Continuity: V g p z V g p z K V g 0 2 0 0 2 2 2 2 1 2 2 2 2 + + = + + + γ γ . V V 1 2 2 2 03 08 = . . = .1406 V2. 10 2 5 1406 2 2 2 2 2 2 = + × V g V g . ∴V2 = 13.36 m/s. ∴Q = 13.36 × π × .0152 =0 00944 . . m / s 3 The velocity in the pipe is V1 = 1.878 m/s. Energy 0 → A: 10 1878 2 9 81 9810 8 1878 2 9 81 3 2 2 = × + + × + . . . . . . pA ∴pA = 65 500 Pa. Energy 0 → B: 10 1878 2 9 81 9810 2 0 1878 2 9 81 10 2 2 = × + + × + . . . . . . pB ∴pB = −5290 Pa. Energy 0 → C: 10 1878 2 9 81 9810 12 2 8 1878 2 9 81 2 2 = × + + + × . . . . . . pC ∴pC = −26 300 Pa. Energy 0 → D: 10 1878 2 9 81 9810 0 5 1878 2 9 81 2 2 = × + + + × . . . . . pD ∴pD = 87 500 Pa. 4.70 V g p z V g p z 0 2 0 0 2 2 2 2 2 2 + + = + + γ γ . 80 000 9810 + = × 4 2 9 81 2 2 V . . ∴V2 = 19.04 m/s. a) Q A V = = × × 2 2 2 025 19 04 π . . = 0 0374 . . m / s 3 b) Q A V = = × × 2 2 2 09 19 04 π . . = 0 485 . . m / s 3 c) Q A V = = × × 2 2 2 05 19 04 π . . = 0 1495 . . m / s 3 4.71 a) p z V g V g 0 0 2 2 1 2 2 1 2 γ + = + .54 . 80 000 9810 + = + 4 16 2 1 2 1 2 1 2 V g V g .54 . ∴V1 = 3.687 m/s. Q A V = = × × 1 1 2 05 3 687 π . . = 0 0290 . . m / s 3 b) A V A V 1 1 2 2 = . V V V 1 2 2 2 2 09 05 3 24 = = . . . . 80 000 9810 + = + 4 2 2 3 3 24 2 2 2 2 2 2 V g V g . . . ∴V2 = 3.08 m/s. ∴Q A V = 2 2 = 0 0784 . . m / s 3 c) 80 000 9810 + = + 4 2 1 2 2 2 2 2 V g V g .5 . ∴V2 = 9.77 m/s. ∴Q A V = 2 2 = 0 0767 . . m / s 3 4.72 (C) Manometer: 2 2 1 2 2 V H p g p g γ ρ + = + or 2 2 1 9810 0.02 . 2 V p g g ρ × + = Energy: 2 100000 7.96 . 3.15. 2 9.81 9810 K K = ∴ = ×
- 65. 64 Combine the equations: 2 1 1 98100.02 1.2 . 18.1 m/s. 2 V V × = × ∴ = 4.73 Manometer: . 6 . 13 2 1 p z H p z H + + = + + γ γ γ γ ∴ = + p H p 1 2 12 6 γ γ . . Energy: p V g p V g 1 1 2 2 2 2 2 2 γ γ + = + . Combine energy and manometer: 12 6 2 2 2 1 2 . . H V V g = − Continuity: V d d V 2 1 2 2 2 1 = . ∴ = × − V H g d d 1 2 1 4 2 4 12 6 2 1 . / . ∴ = = × − Q V d H g d d d 1 1 2 1 4 2 4 1 2 1 2 4 4 12 6 2 1 π π . / / = 12 35 1 2 2 2 1 4 2 4 1 2 . / d d H d d − 4.74 Use the result of Problem 4.73: a) Q = 12.35 ×. . . . . / 16 08 2 16 08 2 2 4 4 1 2 × − = 0 0365 . . m / s 3 b) Q = 12.35 ×. . . . . / 24 08 4 24 08 2 2 4 4 1 2 × − = 0 0503 . . m / s 3 c) Using English units with g = 32.2: Q d d H d d = − 22 37 1 2 2 2 1 4 2 4 1 2 . . / Q = × − 22 37 1 2 1 4 10 12 25 2 2 4 4 1 2 . / .5 . / = 1.318 cfs. d) Q = 22.37 ×1 1 3 15 12 1 3333 2 2 4 4 1 2 × − / . / = 2.796 cfs. 4.75 (B) 2 2 0.040 . 7.96 m/s. 2 0.04 L V p Q h K V g A γ π ∆ = = = = = × 2 100000 7.96 . 3.15. 2 9.81 9810 K K = ∴ = × 4.76 a) Energy from surface to outlet: V g H 2 2 2 = . ∴ = V gH 2 2 2 . Energy from constriction to outlet: p V g p V g 1 1 2 2 2 2 2 2 γ γ + = + .
- 66. 65 Continuity: V V 12 4 = . With p1 = pv = 2450 Pa and p2 = 100 000 Pa, 2450 9810 16 2 9 81 2 100 000 9810 1 2 9 81 2 + × × = + × × . . . gH gH ∴H = 0.663 m. b) With p1 = 0.34 psia, p2 = 14.7 psia, . . . . . 34 144 62 4 16 2 2 14 7 144 62 4 1 2 2 × + = × + g gH g gH ∴H = 2.21 ft. 4.77 Continuity: V V 1 2 4 = . Energy surface to exit: V g H 2 2 2 = . Energy constriction to exit: p V g p V g v γ γ + = + 1 2 2 2 2 2 2 . ∴ = + − = − = − × × p p V V g p H v 2 2 2 2 2 2 16 2 15 100 000 15 65 9810 γ γ . = 4350 Pa. From Table B.1, T = 30°C. 4.78 Energy surface to surface: z z hL 0 2 = + . ∴ = + 30 20 2 2 2 2 V g . Continuity: V1 = 4V2. ∴V1 2 = 160 g. ∴V2 2 = 10 g. Energy surface to constriction: 30 160 2 94 000 9810 1 = + − + g g z ( ) ∴z1 = −40.4 m. ∴H = 40.4 + 20 = 60.4 m. 4.79 Continuity: V V 2 2 2 1 10 6 = = 2.778 V1. Energy: V g p V g p 1 2 1 2 2 2 2 2 + = + γ γ . V g V g 1 2 2 1 2 2 200 2 778 2 2450 9810 + = + 000 9810 . . ∴V1 = 7.67 m/s. ∴Q = π × .052 × 7.67 = 0 0602 . . m / s 3 4.80 Velocity at exit = Ve . Velocity in constriction = V1. Velocity in pipe = V2 . Energy — surface to exit: V g H e 2 2 = . ∴ = V gH e 2 2 . Continuity across nozzle: V D d Ve 2 2 2 = . Also, V V 1 2 4 = . Energy — surface to constriction: H V g pv = + 1 2 2 γ .
- 67. 66 a) 5 1 2 16 2 2 5 97 4 4 =× × × + − g D g . . 550 9810 ∴ = D 0131 . m b) 15 1 2 16 8 12 2 15 34 14 7 144 62 4 4 4 = × × + − g D g ( / ) (. . ) . . ∴ = ′ ′′ D 0 446 . or 5.35 4.81 Energy — surface to exit: 3 2 4 2 1177 2 2 2 2 2 2 = + ∴ = V g V g V . . . Energy — surface to “A”: 3 1177 2 9 81 1176 100 3 1 1177 2 9 81 2 = × + − + + + × . . ( ) .5 . . . 000 9810 H ∴ = H 8.57 . m 4.82 & . . m AV = = × × × = ρ π 1 94 1 12 120 5 079 2 slug / sec. & . . . . / . , . WP = × − × + × = 5 079 32 2 30 120 2 32 2 120 144 62 4 0 85 12 950 2 2 ft - lb sec or 23.5 Hp 4.83 & . . m AV = = × × × = ρ π 1000 02 40 50 27 2 kg / s. 20 40 2 9 81 9810 0 82 2 000 = 50.27 9.81 102 × − × + . / . . ∆p ∴ = × ∆p 1088 106 . . Pa 4.84 (C) 2 2 2 1 2 P W V V Q g γ − = & . p γ ∆ + 16 0.040 400 16 kW. 18.0 kW. 0.89 P P W W Q p η = ∆ = × = = = & & 4.85 − = × × − × + − × & . . . . . WT 2 1000 9 81 0 10 2 2 9 81 600 0 87 2 000 9810 ∴ = × & . . WT 1304 106 W We used V Q A 2 2 2 2 25 10 2 = = × = / . . π m / s. 4.86 V V 1 2 2 2 450 3 15 9 450 3 75 10 19 = × = = × = π π . . . . . fps fps − × = × × − × + − 10 000 1 746 550 450 194 32 2 1019 159 2 32 2 18 140 144 62 4 2 2 , . . . . . . ( ) . . ηT ∴ = ηT 0924 .
- 68. 67 4.87 a) && & ( ) . Q W mg V V g p p z z c g T T S v − = − + − + − + − 2 2 1 2 2 2 1 1 2 1 2 1 2 γ γ The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13. γ γ 1 1 1 2 2 2 85 9 81 287 293 9 92 600 9 81 287 20 = = × × = = × = p g RT T T . . . . . . . N / m 500 3 ∴− − × × + − + − ( . .5 . ( ) . 1 2 9 81 600 20 85 716 9 81 293 2 2 500 000) = 5 9.81 200 000 500 000 9.92 2 T T ∴ = T2 572 K or 299 C . o Be careful of units! p cv 2 600 716 = = ⋅ 000 Pa, J K kg .5 b) −60 000 + 1 500 000 = same as above. ∴ = T2 560 K or 287 C. o 4.88 γ γ 1 1 1 2 14 7 144 32 2 1716 520 0 0764 60 144 32 2 1716 760 0 213 = = × × × = = × × × = p g RT . . . . . . . lb ft lb ft 3 3 c mg AVg AV v = = = = × × × = 4296 213 1 24 600 697 2 ft - lb slug - R lb /sec. o . & . . ρ γ π Use Eq. 4.5.17 with Eqs. 4.5.18 and 1.7.13: & & & ( ) . Q W mg V V g p p c g T T z z c v + = − + − + − + − 2 2 1 2 2 2 1 1 2 1 2 1 2 γ γ − × × + = × + × − × + − 10 778 697 697 600 2 32 2 60 144 213 14 7 144 0764 4296 32 2 300 60 2 . & . . . . . . ( ) Wc ∴ = & . Wc 40 600 ft - lb sec or 73.8 Hp 4.89 Energy — surface to exit: − = − + & & . . W mg V g V g T T η 2 2 2 2 2 20 4 5 2 V mg Q 2 2 15 6 1326 15 9810 147 = × = = = × = π γ . . & m / s. 150 N / s. − × = × − + × ∴ = & . . . . . . & . W W T T 08 147 2 981 20 4 5 1326 2 9 81 2 150 13.26 5390 kW 2 4.90 (D) 2 2 4.58 7.16 36.0 15 3.2 . 416 000 Pa 2 9.81 9810 2 9.81 B B p p + = + + ∴ = × × In the above energy equation we used 2 2 0.2 with 4.42 m/s. 2 0.2 L V Q h K V g A π = = = = ×
- 69. 68 4.91 Energy —surface to “C”: & . & . . . . . W mg P × + × = × + + × 8 10 10 2 9 81 200 7 7 10 2 9 81 7705 2 2 000 9810 ( & . . .5 & . mg AVg WP = = × × × × = ∴ = ρ π 1000 05 10 9 81 770 52 2 N / s.) 700 W Energy — surface to “A”: 30 10 2 9 81 9810 1 10 2 9 81 169 2 2 = × + + × ∴ = . .5 . . . p p A A 300 Pa Energy — surface to “B”: 2 2 2 2 2 B O B O B P P B O V V p p V W mg z z K g g η γ − − = + + − + & & 52 2 9 81 9810 30 15 10 2 981 706 2 700 .8 = 770.5 10 100 Pa 2 × × + − + × ∴ = . . . . . p p B B 4.92 Manometer: γ γ γ γ ρ × + + = × + + + 20 12 13 6 20 12 2 1 1 2 2 2 2 z p z p V . . ∴ + + = × + + + 20 12 13 6 20 12 2 1 1 2 2 2 2 z p z p V g γ γ . . Energy: V g z p H z p V g T 1 2 1 1 2 2 2 2 2 2 + + = + + + γ γ . ∴ = × + − = × = 20 12 13 6 20 12 2 18 1 3 516 1 2 1 2 . . . V g H V T fps. π ∴ = × + × = = = × × × H W Q H T T T T 12 6 20 12 516 2 32 2 62 3 62 4 18 9 62 3 2 . . . . ' . & . . . γ η = 62 115 , . 980 ft - lb sec or Hp 4.93 Energy—across the nozzle: 2 2 2 1 1 2 2 2 1 1 2 5 . 6.25 . 2 2 2 p V p V V V V g g γ γ + = + = = 2 2 2 1 1 1 6.25 400 000 . 4.58 m/s 9810 2 9.81 2 9.81 V V V ∴ + = ∴ = × × , 7.16 m/s A V = , 2 28.6 m/s. V = Energy—surface to exit: 2 2 2 28.6 4.58 7.16 15 1.5 3.2 . 2 9.81 2 9.81 2 9.81 P H + = + + × × × 36.8 m. P H ∴ = 2 / 9810 ( .01 ) 28.6 36.8/.85 3820 W. P P P W QH γ η π ∴ = = × × × × = &
- 70. 69 Energy —surface to“A”: 2 2 7.16 7.16 15 3.2 . 39 400 Pa 2 9.81 9810 2 9.81 A A p p = + + ∴ = × × Energy —surface to “B”: 2 2 4.58 7.16 36.0 15 3.2 . 416 000 Pa 2 9.81 9810 2 9.81 B B p p + = + + ∴ = × × 4.94 (A) V Q A = = × = 0 1 04 19 89 2 . . . π m / s. Energy —surface to entrance: H V g p z K V g P = + + + 2 2 2 2 2 2 2 2 γ . ∴ = × + + + × = H P 19 89 2 9 81 180 50 5 6 19 89 2 9 81 201 4 2 2 . . . . . . 000 9810 m. ∴ = = × × = & / . . / . . W QH P P P γ η 9810 0 1 2014 0 75 263 000 W 4.95 Energy —surface to exit: 10 2 2 2 2 2 2 2 2 2 2 = + + + V g p z V g γ . . ∴ = = = × ∴ = V Q d d 2 2 2 2 7 83 0 02 7 83 4 0 0570 . . . / . . m / s. m. π 4.96 Depth on raised section = y2. Continuity: 3 3 2 2 × = V y . Energy (see Eq. 4.5.21): 3 2 3 2 0 4 2 2 2 2 g V g y + = + + ( . ). ∴ = + − + = 3 059 9 2 3 059 4 128 0 2 2 2 2 2 3 2 2 . , . . g y y y y or Trial-and-error: 2 2 2 2.0: .11 ? 0. 1.85 m. 1.8: .05 ? 0. y y y = − ∴ = = + 2 2 2 2.1: .1 ? 0. 2.22 m. 2.3: .1 ? 0. y y y = − ∴ = = + The depth that actually occurs depends on the downstream conditions. We cannot select a “correct” answer between the two. 4.97 Mass flux occurs as shown. The velocity of all fluid elements leaving the top and bottom is approximately 32 m/s. The distance where u y = = ± 32 m /s is 2 m. m1 m3 m2 m3 . . . .
- 71. 70 To find & m3use continuity: & & & . ( ) & . m m m y dy m 1 2 3 2 3 0 2 2 4 10 32 2 28 10 2 = + × × = + + ∫ ρ ρ ∴ = − × + = & . . m3 640 10 28 2 8 3 53 3 ρ ρ ρ Rate of K.E. loss = & & m V m V u dy 1 1 2 3 1 2 3 0 2 2 2 2 2 2 10 − − ∫ ρ = − − + ∫ 1280 32 2 53 3 32 10 28 2 2 2 3 0 2 ρ ρ ρ . ( ) y dy [655360 54579 507320] 115000 . W ρ = − − = 4.98 The average velocity at section 2 is also 8 m/s. The kinetic-energy- correction factor for a parabola is 2 (see Example 4.9). The energy equation is: V g p V g p hL 1 2 1 2 2 2 2 2 2 + = + + γ α γ . 8 2 9 81 150 2 8 2 9 81 110 2 2 × + = × + + . . . 000 9810 000 9810 hL ∴ = hL 0 815 . . m 4.99 V A VdA y dy = = + = × + = ∫ ∫ 1 1 2 28 1 2 28 2 2 3 29 33 2 3 0 2 ( ) . m / s α = = × + ∫ ∫ 1 1 2 29 33 28 3 3 3 2 3 0 2 AV V dA y dy . ( ) [ ] = × × + × × + × × + = 1 2 29 33 28 2 3 28 2 3 3 28 2 5 2 7 1005 3 3 2 3 5 7 . / / / . 4.100 a) 2 2 4 0.01 2 2 2 2 0 1 1 20 0.01 0.01 10 1 2 5 m/s 2 0.01 0.01 0.01 4 0.01 r V VdA rdr A π π = = − = − = ∫ ∫ × × α π π = = × × − ∫ ∫ 1 1 0 01 5 10 1 0 01 2 3 3 2 3 3 2 2 3 0 0 01 AV V dA r rdr . . . = × − × × + × × − × = 2000 0 01 5 0 01 2 3 0 01 4 0 01 3 0 01 6 0 01 0 01 8 0 01 2 00 2 3 2 4 2 6 4 8 6 . . . . . . . . .
- 72. 71 b) 2 3 0.02 2 2 0 11 10 0.02 10 1 0.02 6.67 m/s 0.02 0.02 0.02 3 0.02 y V VdA wdy A w = = − = − = ∫ ∫ × 3 2 0.02 3 3 3 3 2 0 1 1 10 1 0.02 6.67 0.02 y V dA wdy AV w α = = − ∫ ∫ × × = × − × × + × × − × = 1000 0 02 6 67 0 02 3 0 02 3 0 02 3 0 02 5 0 02 0 02 7 0 02 1 3 3 2 5 4 7 6 . . . . . . . . . .541 4.101 V A VdA R u r R rdr u n n n n n R = = − = − + − + ∫ ∫ 1 1 1 2 2 2 1 1 2 1 0 π π max / max K E V V dA u r R rdr u R n n n n n R . . max / max = = − = − + − + ∫ ∫ ρ ρ π ρπ 2 3 3 3 2 0 2 2 1 2 3 2 3 a) V u u = − − = 2 5 11 5 6 0 758 max max . K E R u R u . . . max max = − = ρπ ρπ 2 3 2 3 5 8 5 13 0 24 α ρ ρπ ρπ = = × = K E AV R u R u . . . . . max max 1 2 0 24 1 2 0 758 1102 3 2 3 2 3 3 b) V u u = − − = 2 7 15 7 8 0 817 max max . K E u R R u . . . max max = − = ρπ ρπ 3 2 2 3 7 10 7 17 0 288 α ρ ρπ ρπ = = × = K E AV V R u R u u . . . . . . max max max 2 2 3 2 2 2 2 0 288 0 817 0 817 2 1056 c) V u u = − − = 2 9 19 9 10 0 853 max max . K E R u R u . . . max max = − = ρπ ρπ 2 3 2 3 9 12 9 21 0 321 α ρ ρπ ρπ = = × = K E AV R u R u . . . . . max max 1 2 0 321 1 2 0 853 1034 3 2 3 2 3 3
- 73. 72 4.102 Engine power= F V m V V u u D × + − + − ∞ & ~ ~ 2 2 1 2 2 1 2 & & ( ) m g F V m V V c T T f f D v = + − + − ∞ 2 2 1 2 2 1 2 4.103 & W F V D η = × 10 5 930 100 3600 015 1340 1000 100 3 − × × × × = × m km kg m kJ kg km s 000 3600 3 3 qf . ∴ = qf 48 030 kJ / kg 4.104 0 2 2 32 2 2 2 2 2 1 2 1 1 2 = + + − − − + α γ γ ν V g p z V g p z LV gD 0 2 2 9 81 0 35 32 10 180 9 81 0 02 2 6 2 = × − + × × × − V V . . . . . V V V Q 2 5 14 4 3 434 0 0 235 7 37 10 + − = ∴ = = × − . . . . . m / s and m / s 3 4.105 Energy from surface to surface: H V g p z V g p z K V g P = + + − − − + 2 2 2 2 1 2 1 1 2 2 2 2 γ γ . a) H Q Q P = + × × × = + 40 5 0 04 2 9 81 40 50 7 2 2 2 π . . . Try Q H H P P = = = 0 25 43 2 58 . : . (energy). (curve) Try Q H H P P = = = 0 30 44 6 48 . : . (energy). (curve) Solution: Q = 0 32 . . m / s 3 b) H Q Q P = + × × × = + 40 20 0 04 2 9 81 40 203 2 2 2 π . . Try Q H H P P = = = 0 25 52 7 58 . : . (energy). (curve) Solution: Q = 0 27 . m / s 3 Note: The curve does not allow for significant accuracy. 4.106 Continuity: A V A V A V 1 1 2 2 3 3 = + π π π × × = × × + × ∴ = 0 06 5 0 02 20 0 03 1111 2 2 2 3 3 . . . . . V V m / s Energy: energy in + pump energy = energy out & & & & m V p W m V p m V p P P 1 1 2 1 2 2 2 2 3 3 2 3 2 2 2 + + × = + + + ρ η ρ ρ
- 74. 73 1000 0 065 5 2 120 0 85 1000 0 02 20 20 2 300 2 2 2 2 π π × × + + = × × + . . & . 000 1000 000 1000 W P + × × + 1000 0 03 1111 1111 2 500 2 2 π . . . 000 1000 ∴ = & WP 26 700 W 4.107 (A) After the pressure is found, that pressure is multiplied by the area of the window. The pressure is relatively constant over the area. 4.108 V g p V g p 1 2 1 2 2 2 2 2 + = + γ γ . 2 2 1 1 2 4 . ( /2) d V V V d = = a) V V 1 2 1 2 2 9 81 200 16 2 9 81 × + = × . . . 000 9810 ∴ = V1 5164 . m / s. ( ) p A F m V V 1 1 2 1 − = − & . 200 000 03 1000 03 5 164 4 5 164 5164 2 2 π π × − = × × × − . . . ( . . ). F ∴ = F 339 N. b) V V 1 2 1 2 2 9 81 400 16 2 9 81 × + = × . . . 000 9810 ∴ = V1 7 303 . m / s. 400 000 03 1000 03 7 303 4 7 303 7 303 2 2 π π × − = × × × − . . . ( . . ). F ∴ = F 679 N. c) 2 2 1 1 16 200 000 . 2 9.81 9810 2 9.81 V V + = × × 1 5.164 m/s. V ∴ = 2 2 200 000 .06 1000 .06 5.164(4 5.164 5.164). F π π × − = × × × − 1356 N. F ∴ = d) 2 2 1 1 1 16 30 144 . 17.24 fps. 2 32.2 62.4 2 32.2 V V V × + = ∴ = × × 2 2 2 30 1.5 1.94 (1.5/12) 17.24 (4 1). 127 lb. F F π π × × − = × × × − ∴ = e) 2 2 1 1 1 16 60 144 . 24.38 fps. 2 32.2 62.4 2 32.2 V V V × + = ∴ = × × 2 2 2 60 1.5 1.94 (1.5/12) 24.38 (4 1). 254 lb. F F π π × × − = × × × − ∴ = f) 2 2 1 1 1 16 30 144 . 17.24 fps. 2 32.2 62.4 2 32.2 V V V × + = ∴ = × × 2 2 2 30 3 1.94 (3/12) 17.24 (4 1). 509 lb. F F π π × × − = × × × − ∴ = 4.109 V g p V g p 1 2 1 2 2 2 2 2 + = + γ γ . V V V 2 2 2 1 1 9 3 9 = = . V1 2 2 9 81 2 81 2 9 81 × + = × . . . 000 000 9810 V1 2 ∴ = V1 2 50.
- 75. 74 p A Fm V V m V 1 1 2 1 1 8 − = − = &( ) & 2 045 1000 045 8 50 2 2 000 000π π × − = × × × . . F ∴ = F 10 180 N. 4.110 2 2 1 1 2 2 . 2 2 V p V p g g γ γ + = + 0 2 .01 .006 .15. 11.1 m/s. e e V V V π × = × × ∴ = ΣF m V V x x x = − &( ). 2 1 a) V V V 2 2 2 1 1 10 8 1 = = .562 . V V 1 2 1 2 2 9 81 400 2 441 2 9 81 × + = × . . . . 000 9810 ∴ = V1 23.56 . m / s ∴ − = − p A F m V V 1 1 2 1 &( ). 400 05 1000 05 23 23 2 2 000π π × − = × × × . . .56(.562 .56). F ∴ = F 692 N. b) V V V 2 2 2 1 1 10 6 2 778 = = . . V g V g 1 2 1 2 2 400 7 716 2 + = 000 9810 . . ∴ = V1 10 91 . . m / s 400 05 1000 05 10 91 1778 10 91 2 2 000π π × − = × × × . . . ( . . ). F ∴ = F 1479 N. c) V V V 2 2 2 1 1 10 4 6 25 = = . . V g V g 1 2 1 2 2 400 39 06 2 + = 000 9810 . . ∴ = V1 4.585 . m / s 400 05 1000 05 4 5 25 4 2 2 000π π × − = × × × . . .585( . .585). F ∴ = F 2275 N. d) V V V 2 2 2 1 1 10 2 25 = = . V g V g 1 2 1 2 2 400 625 2 + = 000 9810 . ∴ = V1 1132 . . m / s 400 05 1000 05 1132 24 1132 2 2 000π π × − = × × × . . . ( . ). F ∴ = F 2900 N. 4.111 (C) 2 2 1 1 2 2 2 2 V p V p g g γ γ + = + 2 2 1 (6.25 1) 12.73 . 9810 3085000 Pa. 2 9.81 p − × = × = × 2 1 1 2 1 ( ). 3085000 0.05 1000 0.1 12.73(6.25 1) p A F Q V V F ρ π − = − × × − = × × − 17500 N. F ∴ = 4.112 V V p V V g 2 1 1 2 2 1 2 2 2 4 120 2 62 4 120 30 2 32 2 13 = = = − = − × = fps. 080 psf. γ . . , 2 2 1 1 2 1 1.5 1.5 ( ) 13,080 1.94 30( 120 30) 1072 lb. 12 12 x x F p A m V V π π = − − = − × × − − = & 4.113 V V V g p V g p V g p 2 1 1 2 1 2 2 2 1 2 1 4 2 2 15 2 = + = + ∴ = . . . γ γ γ a) V1 2 2 9 81 15 9810 200 000 26 67 = × × × = . . . ∴ = = V V 1 2 516 20 7 . . m / s, m / s. p A F m V V x x x 1 1 2 1 − = − &( ). ∴ = × + × × = Fx 200 000 04 1000 04 5 16 1139 2 2 2 π π . . . . N
- 76. 75 F m VV y y y = − &( ). 2 1 ∴ = × × = Fy 1000 04 5 16 20 7 537 2 π . . ( . ) . N b) V1 2 2 9 81 15 9810 400 000 53 33 = × × × = . . . ∴ = = V V 1 2 7 30 29 2 . . m / s, m / s. p A F m V V x x x 1 1 2 1 − = − &( ). ∴ = × + × × = Fx 400 000 04 1000 04 7 3 2280 2 2 2 π π . . . . N F m V V y y y = − &( ) 2 1 =1000 04 7 3 29 2 1071 2 π× × × = . . ( . ) . N c) V1 2 2 9 81 15 9810 800 000 106 7 = × × × = . . . ∴ = = V V 1 2 10 33 41 3 . . m / s, m / s. 2 2 2 2 1 1 1 1 800 000 .04 1000 .04 10.33 4560 N. x F p A AV ρ π π = + = × + × × = F m V y y = &( ) 2 =1000 04 10 33 41 3 2140 2 π× × = . . ( . ) . N 4.114 V V 2 2 2 1 40 10 80 = = m / s. V g p V g p 1 2 1 2 2 2 2 2 + = + γ γ ∴ = × − × = × p1 2 2 6 9810 80 2 9 81 5 2 9 81 3 19 10 . . . Pa. p A F m V V x x 1 1 2 1 − = − & ( ). ∴ = × × − × × − = F 3 19 10 2 1000 2 5 80 5 353 6 2 2 . . . ( ) π π 000 N. 4.115 A V A V 1 1 2 2 = . π π × × = − . (. . ) . 025 4 025 02 2 2 2 2 V ∴ = V2 1111 . m / s. p V g p V g 1 1 2 2 2 2 2 2 γ γ + = + . 2 2 1 11.11 4 9810 53700 Pa. 2 9.81 p − = = × 1 1 2 1 ( ). p A F m V V − = − & 2 2 53 700 .025 1000 .025 4(11.11 4) 49.6 N. F π π ∴ = × − × × − = 4.116 Continuity: . . . . 7 1 7 1 2 2 1 V V V V = ∴ = Energy: V g p z V g p z 1 2 1 1 2 2 2 2 2 2 + + = + + γ γ V V V V 1 2 1 2 1 2 2 9 81 7 49 2 9 81 1 0 495 3 467 × + = × + ∴ = = . . . . . . , . m / s. Momentum: F F R m V V x 1 2 2 1 − − = − &( ) 9810 35 7 1 9810 05 0 1 1 1000 1 1 3 467 3 467 495 × × − × × − = × × × − . (. .5) . ( . .5) (. .5) . ( . . ) Rx ∴ = Rx 1986 N. ∴Rx acts to the left on the water, and to the right on the obstruction. F V2 p1 A1 F p1A1 F2 F1 Rx
- 77. 76 4.117 Continuity: 62 30 1 2 2 1 V V V V = ∴ = . . . Energy (along bottom streamline): V g p z V g p z 1 2 1 1 2 2 2 2 2 2 + + = + + γ γ 2 2 2 2 /900 6 0.2. 2 9.81 2 9.81 V V + = + × × 2 1 10.67, .36 m/s. V V ∴ = = Momentum: F F F m V V 1 2 2 1 − − = − &( ) 9810 3 6 4 9810 1 2 4 1000 2 4 10 67 10 67 36 × × − × × − = × × × − ( ) . (. ) (. ) . ( . . ) F ∴ = F 618 000 N. (F acts to the right on the gate.) 4.118 a) 8 6 2 2 × = . . V y F F m V V 1 2 2 1 − = − & ( ). γ γ ρ × × − = × × − . . . . . 3 6 2 6 8 8 6 8 2 2 2 w y y w w y γ ρ 2 36 4 8 8 6 6 4 8 8 2 9 81 2 2 2 2 2 2 (. ) . . . (. ) . . . − = × − ∴ + = × × y y y y y y y y 2 2 2 2 6 7 829 0 2 + − = ∴ = . . . .51 . m (See Example 4.12.) b) y y y g y V 2 1 1 2 1 1 2 2 2 1 2 8 1 2 4 4 8 9 81 4 12 3 23 = − + + = − + + × × = . . . . . . m c) y y y g y V 2 1 1 2 1 1 2 2 2 1 2 8 1 2 2 2 8 32 2 2 20 6 12 = − + + = − + + × × = . . . ft d) y y y g y V 2 1 1 2 1 1 2 2 2 1 2 8 1 2 3 3 8 32 2 3 30 11 = − + + = − + + × × = . .54 . ft 4.119 Continuity: V y V y V y y y 2 2 1 1 2 1 2 1 4 4 = = ∴ = . . Use the result of Example 4.12: y y y g y V 2 1 1 2 1 1 2 1 2 1 2 8 = − + + / a) y2 4 8 3 2 = × = . . . m 1/2 2 2 1 1 8 3.2 .8 .8 .8 . 2 9.81 V = − + + × × ∴ = V1 8 86 . m / s. b) y2 4 2 8 = × = ft. 8 1 2 2 2 8 32 2 2 2 1 2 1 2 = − + + × × . . / V ∴ = V1 25 4 . fps. F2 F1 F
- 78. 77 4.120 V = × = 9 33 1 m / s. 1 2 9 81 3 2 9 81 2 1 2 1 × + = × + . . . V y V y 1 1 1 3 = × . ∴ = + 3 05 19 62 3 1 2 1 . . . V V Trial-and-error: V V 1 1 7 3 05 2 93 7 2 3 05 3 06 = = = = : . ? . . : . ? . V1 7 19 417 = = . . m / s. y m. 1 1/2 2 2 2 1 8 .417 .417 .417 7.19 1.90 m. 2 9.81 y = − + + × × = V2 19 7 19 417 × = × . . . . V2 1 = .58 . m / s 4.121 Refer to Example 4.12: γ γ ρ y y w w w y V y 1 1 1 1 1 2 3 6 6 10 10 60 6 10 − × × = × × − = ⋅ . ( ). ∴ − = − γ ρ 2 36 600 6 1 2 1 1 ( ) . y y y ∴ + = = ∴ = = ( ) . . . . , . . y y y V 1 1 1 1 6 1200 32 2 37 27 3 8 15 8 ft fps 4.122 Continuity: 20 015 03 2 2 2 × × = × π π . . . V ∴ = V2 5 m / s. Momentum: p A p A m V V 1 1 2 2 2 1 − = − &( ). 60 03 03 1000 015 20 5 20 2 2 2 2 000π π π × − × = × × − . . . ( ). p ∴ = p2 135 kPa. 4.123 2 1 1 2 2 2 2 .05 2 . 15 30 m/s. 2 .025 V A V A V π π × = = = × p V g p V g p 1 1 2 2 2 2 1 2 2 2 2 9810 30 15 2 9 81 337 γ γ + = + ∴ = − × = . . 500 Pa. ( ) 2 1 1 1 1 . ( ). x x x F m V V p A F m V Σ = − − = − & & ∴ = + = × + × × = F p A mV 1 1 1 2 2 2 337 05 1000 05 15 4420 & . . . 500 N π π 4.124 & . . m1 2 1000 03 12 33 93 = × × = π kg / s. & . . m3 2 1000 02 8 10 05 = × × = π kg / s. ∴ = − = = × ∴ = & & & . . . . m m m V V 2 1 3 2 2 2 23 88 1000 03 8 446 π m / s. Energy from 1 → 2: V g p V g p p 1 2 1 2 2 2 2 2 2 2 500 8 446 2 9 81 9810 + = + ∴ = − × × γ γ . . . 000+ 122 = 536 300 Pa. p1A1 p2A2 p1A1 p2A2 p3A3 Ry Rx
- 79. 78 Energy from 1→ 3: V g p V g p 1 2 1 3 2 3 2 2 + = + γ γ . ∴ = + − × = p3 2 2 500 12 8 2 9 81 9810 540 000 000 Pa. . p A p A R m V m V m V x x x x 1 1 2 2 2 2 3 3 1 1 − − = + − & & & . ∴ = × − × + × − × = Rx 500 03 536 03 33 93 12 23 88 8 446 103 2 2 000 300 N π π . . . . . . p A R m V m V m V y y y y 3 3 3 3 2 2 1 1 − = + − & & & . ∴ = × − × − = Ry 540 10 05 8 759 000 .02 N 2 π . ( ) . 4.125 a) ( ) ΣF m V V F mV V m A x x x = − − = − = & . & . & 2 1 1 1 1 ρ = × 300 1000 052 π . = 38.2 m/s ∴ = × = F 300 38 2 11 . . 460 N b) − = − − F m V V r B & ( )(cos ). 1 1 α ∴ = × − = F 300 28 2 38 2 38 2 10 6250 . . ( . ) . N c) − = − − F m V V r B & ( )(cos ). 1 1 α ∴ = × − − = F 300 48 2 38 2 38 2 10 18 . . ( . ( )) . 250 N 4.126 a) − = − F m V V x x &( ). 2 1 200 1 94 1 25 12 2 1 2 = × . . . π V ∴ = V1 55 fps. b) − = − − F m V V r B & ( )(cos ). 1 1 α 200 1 94 125 12 30 2 1 2 = − . . ( ) . π V ∴ = V1 85 fps. c) − = − − F m V V r B & ( )(cos ). 1 1 α 200 1 94 125 12 30 2 1 2 = + . . ( ) . π V ∴ = V1 25 fps. 4.127 a) − = − F m V V x x &( ). 2 1 − = × × − 700 1000 04 30 2 1 1 1 π . ( cos ). V V V o ∴ = V1 32 24 . m / s. ∴ = = × × = & . . . . m A V ρ π 1 1 2 1000 04 32 24 162 1 kg / s b) − = − − F m V V r B & ( )(cos ). 1 1 α − = × − − 700 1000 04 8 866 1 2 1 2 π . ( ) (. ). V ∴ = V1 40 24 . m / s. ∴ = = × × = & . . . m A V ρ π 1 1 2 1000 04 40 24 202 kg / s c) − = − − F m V V r B & ( )(cos ). 1 1 α − = × + − 700 1000 04 8 866 1 2 1 2 π . ( ) (. ). V ∴ = V1 24 24 . m / s. ∴ = = × × = & . . . . m A V ρ π 1 1 2 1000 04 24 24 1218 kg / s 4.128 (D) 2 1 ( ) 1000 0.01 0.2 50(50cos60 50) 2500 N. x x x F m V V − = − = × × × − = − o & F V1 V2
- 80. 79 4.129 a) −= − = × × − R m V V x x x &( ) . ( cos ). 2 1 2 1 94 1 12 120 120 60 120 π o ∴ = Rx 305 lb. R m V V y y y = − = × × × & ( ) . ( . ). 2 1 2 1 94 1 12 120 120 866 π ∴ = Ry 528 lb. b) − = − − = × × × − R m V V x r B & ( )(cos ) . (.5 ). 1 2 1 194 1 12 60 60 1 α π ∴ = Rx 76 2 . . lb R m V V y r B = − = × × × & ( )sin . ( . ). 1 2 194 1 12 60 60 866 α π ∴ = Ry 132 lb. c) − = − − = × × × − R m V V x r B & ( )(cos ) . (.5 ). 1 2 1 1 94 1 12 180 180 1 α π ∴ = Rx 686 lb. R m V V y r B = − = × × × & ( )sin . ( . ). 1 2 194 1 12 180 180 866 α π ∴ = Ry 1188 lb. 4.130 V R B = = × = ω 0 30 15 .5 m / s. − = − − = × × × − R m V V x B &( )(cos ) . (.5 ). 1 2 1 1000 025 40 25 1 α π ∴ = Rx 982 N. ∴ = = × × = & . W R V x B 10 10 982 15 147 300 W 4.131 a) − = − = × − − R m V V x x x &( ) . ( cos ). 2 1 2 4 02 400 400 60 400 π o ∴ = Rx 1206 N. R m V V y y y = − = × × & ( ) . ( sin ). 2 1 2 4 02 400 400 60 π o ∴ = Ry 696 N. b) − = − − = × − − R m V V x r B & ( )(cos ) . ( . ). 1 2 2 120 1 4 02 300 5 1 o π ∴ = Rx 679 N. R m V V y r B = − = × × × & ( )sin . . . 1 2 2 4 02 300 866 α π ∴ = Ry 392 N. c) − = − − = × − − R m V V x r B & ( )(cos ) . ( . ). 1 2 2 120 1 4 02 500 5 1 o π ∴ = Rx 1885 N. R m V V y r B = − = × × × & ( )sin . . . 1 2 2 4 02 500 866 α π ∴ = Ry 1088 N. 4.132 − = − − = × × − − − F m V V x B & ( )(cos ) . ( ) ( .5 ). 1 2 2 120 1 4 02 400 180 1 o π ∴ = Rx 365 N. VB = × = 12 150 180 . m / s. & . W = × × = 15 365 180 986 000 W The y-component force does no work. 4.133 (A) 2 2 1 ( ) 1000 0.02 60 (40cos45 40) 884 N. x r r x x F m V V π − = − = × × × × − = o & Power 884 20 17700 W. x B F V = × = × = 4.134 a) Refer to Fig. 4.16: 1 1 1 2 1 1 750sin sin45 507 fps. 750cos 300 cos45 r r r r V V V V β β = = ∴ = − = o o Note: V V V V V V V x x r B r B r 2 1 2 2 1 1 1 2 1 − = − + − − = − + cos cos (cos cos ). α α α α ∴ = + = × × + = R mV x r & (cos cos ) . .5 (cos cos ) . 1 2 1 2 015 12 750 507 30 45 48 9 α α π o o lb.
- 81. 80 ∴ = =× × = & . , W R V x B 15 15 48 9 300 220 000 ft - lb sec or 400 Hp. b) 750 60 750 300 60 554 1 1 1 1 1 2 sin sin cos cos . β β = − = = V V V V r r r r o o fps = ∴ = + = × × × + = R mV x r & (cos cos ) . .5 (cos cos ) . 1 2 1 2 015 12 750 554 30 60 46 4 α α π o o lb. ∴ = = × × = & . , W R V x B 15 15 46 4 300 209 000 ft - lb sec or 380 Hp. c) 750 90 750 300 90 687 1 1 1 1 1 2 sin sin cos cos . β β = − = = V V V V r r r r o o fps = ∴ = + = × × × + = R mV x r & (cos cos ) . .5 (cos ) .5 1 2 1 2 015 12 750 687 30 0 36 α α π o lb. ∴ = = × × = & . , W R V x B 15 15 36 5 300 164 300 ft - lb sec or 299 Hp. 4.135 a) Refer to Fig. 4.16: 100 30 100 30 20 36 9 83 3 1 1 1 1 1 1 sin sin cos cos . , . o o o = − = ∴ = = V V V r r r α α α m / s. 2 2 2 2 2 2 sin60 83.3sin 71.5, 48 . cos60 83.3cos 20 V V V α α α = = = = − o o o − = − = × × − − ∴ = R m V V R x x x x &( ) . ( .5cos cos ). 2 1 2 1000 015 100 71 60 100 30 8650 π o o N. ∴ = = × × = × & . . W V R B x 12 12 20 8650 2 08 106 W b) 100 30 100 30 40 47 68 35 1 1 1 1 1 1 2 sin sin cos cos , . o o o = − = ∴ = = = V V V V r r r r α α α m / s. 2 2 2 2 2 2 sin60 68.35sin 38.9 m/s, 29.5 . cos60 68.35cos 40 V V V α α α = = = = − o o o − = − = × × − − ∴ = R m V V R x x x x &( ) . ( . cos cos ). 2 1 2 1000 015 100 38 9 60 100 30 7500 π o o N. ∴ = = × × = × & . . W V R B x 12 12 40 7500 3 60 106 W c) 100 30 100 30 50 53 8 6196 1 1 1 1 1 1 2 sin sin cos cos . , . o o o = − = ∴ = = = V V V V r r r r α α α m / s. 2 2 2 2 2 2 sin60 61.76sin 19.32 m/s, 15.66 . cos60 61.96cos 50 V V V α α α = = = = − o o o
- 82. 81 − = −= × × − − ∴ = R m V V R x x x x &( ) . ( . cos cos ). 2 1 2 1000 015 100 19 32 60 100 30 6800 π o o N . ∴ = = × × = × & . . W R V x B 12 12 6800 50 4 08 106 W 4.136 a) Refer to Fig. 4.16: 50 30 50 30 2500 86 6 1 1 1 1 1 2 2 sin sin cos cos . o o = − = ∴ = − + V V V V V V r B r r B B α α 2 2 2 2 2 2 1 2 2 30sin60 sin 900 30 . 30cos60 cos r r r B B r B V V V V V V V α α = ∴ = = + + − = o o Combine the above: VB = 13 72 . m / s. Then, α α 1 2 59 4 42 1 = = . , . . o o − = − = × × − − ∴ = R m V V R x x x x &( ) . ( cos cos ). 2 1 2 1000 01 50 30 60 50 30 916 π o o N. ∴ = = × × = & . . W V R B x 15 15 13 72 916 188 500 W b) 50 30 50 30 2500 86 6 1 1 1 1 1 2 2 sin sin cos cos . o o = − = ∴ = − + V V V V V V r B r r B B α α ∴ = VB 14 94 . m / s. 2 2 2 2 2 2 2 30sin70 sin 900 20.52 . 30cos70 cos r r B B r B V V V V V V α α = ∴ = + + − = o o α 1 41 4 = . o ,α 2 48 2 = . o 2 2 1 ( ) 1000 .01 50( 30cos70 50cos30 ). 841 N. x x x x R m V V R π − = − = × × − − ∴ = o o & ∴ = = × × = & . . W V R B x 15 15 14 94 841 188 500 W c) 50 30 50 30 2500 86 6 1 1 1 1 1 2 2 sin sin cos cos . o o = − = ∴ = − + V V V V V V r B r r B B α α ∴ = VB 16 49 . m / s 2 2 2 2 2 2 2 30sin80 sin 900 10.42 . 30cos80 cos r r B B r B V V V V V V α α = ∴ = + + − = o o α 1 43 = o , α 2 53 7 = . o − = − = × × − − ∴ = R m V V R x x x x &( ) . ( cos cos ). 2 1 2 1000 01 50 30 80 50 30 762 π o o N. ∴ = = × × = & . . W V R B x 15 15 16 49 762 188 500 W 4.137 To find F, sum forces normal to the plate: ( ) n out n F m V Σ = & 1 . n V − a) [ ] ∴ = × × × − − = F 1000 02 4 40 40 60 11 . . ( sin ) . o 080 N (We have neglected friction) 2 2 3 3 1 0 ( ) 40sin30 . t F m V m V m Σ = = + − − × o & & & Bernoulli: V V V 1 2 3 = = . ∴ = − − = + ∴ = = × = = Continuity: kg / s. kg / s. 0 75 75 320 240 80 2 3 1 1 2 3 2 1 3 & & .5 & & & & & . & . & m m m m m m m m m b) 1 20 1.94 120( 120sin60 ) 3360 lb. 12 12 F ∴ = − × × × − = o (We have neglected friction) ΣF m V m V m t = = + − − × 0 120 30 2 2 3 3 1 & & ( ) & sin . o Bernoulli: V V V 1 2 3 = = .
- 83. 82 2 1 2 31 1 2 3 3 20 .75 .75 1.94 120 0 0.5 144 Continuity: 22.6 slug/sec. and 9.7 slug/sec. m m m m m m m m m ∴ = = × × × ∴ = − − = + = = & & & & & & & & & 4.138 F m V r r n = = × × × + = & ( ) . . ( ) sin 1 2 1000 02 4 40 20 60 24 o 940 N. F W x = = ∴ = × = 24 30 21 21 20 432 940 600 N. 600 000 W cos & . o 4.139 F m V V F V r r n B x B = = × × − = − & ( ) . . ( ) sin . ( )sin . 1 2 2 2 1000 02 4 40 60 8 40 60 o o & ( ) . ( ). W V F V V V V V B x B B B B B = = − × = − + 8 40 75 6 1600 80 2 2 3 dW dV V V V B B B B & ( ) . . = − + = ∴ = 6 1600 160 3 0 13 33 2 m / s. 4.140 (A) Let the vehicle move to the right. The scoop then diverts the water to the right. Then 2 1 ( ) 1000 0.05 2 60 [60 ( 60)] 720000 N. x x F m V V = − = × × × × − − = & 4.141 1 ( r F m V = & 2 )(cos 1) 1000 .1 .6 ( )( 2) 120 . B B B B V V V V α − − = × × − − = At 2 120 1000 0 : 120 133 300 N. 3600 t F × = = × = 2 133 300 1.33 m/s 100 000 o a = = − = = − ∴ − = ∫ ∫ F m dV dt V dV V dt B B B B t 120 100 0012 2 2 0 3333 16 67 000 . . . . . ∴ − = ∴ = 1 16 67 1 33 33 0012 26 6 . . . . . t. sec t 4.142 1 ( )(cos 1) 90 .8 2.5 13.89 ( 13.89)( 1) 34700 N. r B F m V V α = − − = × × × × − − = & 50 1000 13.89 m/s 34700 13.89 482 000 W or 647 Hp. 3600 B V W × = = ∴ = × = & 4.143 See the figure in Problem 4.141. F m V V V V r B B B = − − = × × × − − = & ( )(cos ) . . ( )( ) 1 1 1000 06 2 2 24 α V . B 2 − = ∴− = F mV dV dx V V dV dx B B B B B . . 24 5000 2 − = − = − ∴ = ∫ ∫ 24 5000 24 5000 27 78 250 458 250 27 78 0 m dx dV V x n n x B B x . . . . . l l V2 V1 = 0 F
- 84. 83 4.144 − =− − = × − − F m V V V V r B B & ( )(cos ) . . ) ( ). 1 1 2 1 194 125 12 2 α π 2 ( ∴ = − = F V V dV dt B B 0 1323 20 1 2 . ( ) . At t V dV dt V B B = = = 0 0 0 1323 1 2 , . . . Then 20 With dV dt V B = = 6 30 1 1 , . . fps For t dV V dt V V B B B B VB > − = = − − ∴ = ∫ ∫ 0 30 1 0 006615 0 01323 1 30 1 1 30 1 8 57 2 0 2 0 , ( . ) . . . . . . . . fps 4.145 For this steady-state flow, we fix the boat and move the upstream air. This provides us with the steady-state flow of Fig. 4.17. This is the same as observing the flow while standing on the boat. & . . . ( . W FV F F V = × ∴ = = 1 1 50 1000 3600 1440 13 89 20 000 = N m / s) F m V V V V V = − = × + − ∴ = & ( ). . . ( . ). . 2 1 2 2 2 2 1440 123 1 13 89 2 13 89 30 6 m / s. π ∴ = = × + = Q A V 3 3 2 1 30 6 13 89 2 69 9 π . . . m / s. 3 η p V V = = = 1 3 13 89 22 24 0 625 . . . . or 62.5% 4.146 Fix the reference frame to the aircraft so that V1 200 1000 3600 55 = × = .56 m / s. V m 2 2 320 1000 3600 88 89 1 2 11 55 88 89 2 329 = × = ∴ = × × + = . & . . .56 . .5 m / s. kg / s. π F = − = 329 88 89 55 10 .5( . .56) 980 N. = × ∴ = ∆ ∆ p p π 11 2890 2 . . . Pa & . W F V = × = × 1 10 980 55.56 = 610 000 W or 818 Hp 4.147 Fix the reference frame to the boat so that V1 20 88 60 29 33 = × = . fps. V F m V V 2 2 1 2 40 88 60 58 67 1 94 10 12 29 33 58 67 2 58 67 29 33 = × = ∴ = − = × + − . &( ) . . . ( . . ) fps. π = 5460 lb. & . , . W F V = × = × = 1 5460 29 33 160 000 ft - lb sec or 291 Hp V2 F VB
- 85. 84 & . . . .. m = × × + = 194 10 12 29 33 58 67 2 186 2 2 π slug/ sec 4.148 Fix the reference frame to the boat: V V 1 2 10 20 = = m / s, m / s. ∴Thrust = &( ) . ( ) . m V V 2 1 1000 0 2 20 10 2000 − = × − = N & . W F V = × = × = 1 2000 10 20 000 W or 26.8 Hp 4.149 0 2 2 10 1 2 20 0 1 1 1 1 1 1 1 . . . . ( ) ( . ). = = × × ∴ = ∴ = ∴ = − V A V V V V y y m / s. m / s. max flux in = 2 2 1000 20 1 800 1 3 267 2 2 2 3 0 1 0 1 ρV dy y dy = × − = = ∫ ∫ (. ) . . . 000 N. The slope at section 1 is −20. ∴ = − + V y y A 2 20 ( ) . Continuity: A V A V V V 1 1 2 2 2 1 2 2 = ∴ = = . m / s. V A V A V A 2 2 2 0 05 1 1 2 ( ) (. ) / . = = − ∴ = − 2 1 2 2 2 20 2 = − ∴ = ∴ = − A A V y y / . .5. ( ) .5 . flux out = 2 1000 2 20 800 125 3 800 0 00153 2 3 0 05 0 05 ( .5 ) ( . ) [ . ] . . − = − = ∫ y dy y 000 000 3 = 408 3 . N. ∴change = 408 − 267 = 141 N. 4.150 a) β = = − × × = = ∫ ∫ V dA V A y dy 2 2 2 2 0 1 2 3 2 20 1 1 2 1 0 4000 1 3 4 3 (. ) . . . . . b) See Problem 4.149: V y y y V 2 2 20 0125 05 0 2 ( ) ( . ), . . = − ≥ ≥ = m / s. β = = − × × = − = ∫ ∫ V dA V A y dy y 2 2 2 2 0 05 2 3 0 05 2 20 125 2 1 10 2000 125 3 1021 ( . ) . . ( . ) . . . . 4.151 From the c.v. shown: ( ) . p p r r L w o 1 2 0 2 2 − = π τ π ∴ = = ∴ = × × × × × − τ µ w o w w p r L du dr du dr ∆ 2 0 03 144 75 12 2 30 2 36 10 5 . . . / . = 191 ft / sec ft . 4.152 Write the equation of the parabola: V r V r r ( ) . max = − 1 2 0 2 p1 A1 p2 A2 τw2πro L
- 86. 85 Continuity: π π ×× = − ∴ = ∫ . . . max max . 006 8 1 006 2 16 2 2 2 0 006 V r rdr V m / s. Momentum: p A p A F V dA mV 1 1 2 2 2 1 − − = − ∫ Drag ρ & . 40 006 1000 16 1 006 2 1000 006 8 8 2 2 2 2 2 2 0 006 000 Drag π π π × − = × − − × × × × ∫ . . . . F r rdr 4 9 651 7 238 .524 . . . − = − FDrag ∴ = FDrag N 2 11 . . 4.153 & ( ) . ( ) . m A V V y dA y dy top = − = × × − + = ∫ ∫ ρ ρ 1 1 2 2 0 2 123 2 10 32 28 10 65 6 kg / s. − = + − = + + × − × × ∫ ∫ F V dA m V m V y dy top 2 1 23 28 10 65 6 32 1 23 20 32 2 1 1 1 2 2 2 0 2 ρ & & . ( ) . . . ∴ = F 3780 N. 4.154 a) & & & ( ) . . ( ) . m m m A V u y dA y y dy top = − = − = × × − − × ∫ ∫ 1 2 1 1 2 0 1 123 1 2 8 20 100 8 2 ρ ρ = = = 0 656 0 1 8 . . ( ) ). kg / s. (Note: for y u y Momentum: − = − + × − × × × ∫ F y y dy Drag ρ ρ 64 20 100 2 656 8 1 2 8 2 2 2 0 1 ( ) . . . = × + − × ∴ = 1 23 6 83 5 25 123 12 8 2 1 . . . . . . . N Drag F b) To find h: 8 8 20 100 2 0 1 h y y dy = − ∫ ( ) . . ∴ = × − × = h 20 1 2 100 001 3 0 0667 2 . . . m. Momentum: − = − − × × × ∫ F y y dy Drag 1 23 64 20 100 2 123 0667 2 8 2 2 2 0 1 . ( ) . . . . = × − 1 23 6 83 10 . . .50. ∴ = FDrag N 2 1 . . 4.155 a) Energy: V g z V g z hL 1 2 1 2 2 2 2 2 + = + + . See Problem 4.118(a). 8 2 9 81 0 6 1912 2 9 81 2 2 2 × + = × + + . . . . .51 . hL ∴ = hL 1166 . m. ∴ = × × × × = losses = 900 W / m of width. γA V hL 1 1 9810 6 1 8 1166 54 (. ) . b) See Problem 4.120: V g z V g z hL 1 2 1 2 2 2 2 2 + = + + .
- 87. 86 7 19 2 981 417 1 2 9 81 1 9 2 2 . . . .58 . . . × + = × + + hL ∴ = hL 1025 . m. ∴ = × × × × = losses = 300 W γA V hL 1 1 9810 417 3 7 19 1025 90 . . . c) See Problem 4.121: 5 17 2 9 81 116 3 2 9 81 2 2 2 . . . . . × + = × + + hL ∴ = hL 0 0636 . m. ∴ = × × × = losses = / m of width. γA V hL 1 1 9810 116 5 17 0 0636 3740 W . . . 4.156 See Problem 4.122: V V p p 1 2 1 2 20 5 60 135 = = = = m / s, m / s, kPa, kPa. Then, V g p V g p h h L L 1 2 1 2 2 2 2 2 2 2 20 2 9 81 60 5 2 9 81 135 + = + + × + = × + + γ γ . . . . 000 9810 000 9810 ∴ = = × ∴ = h K g K K L 11 47 2 20 2 9 81 0 562 2 . . . . . m = V1 2 4.157 Continuity: V D Vd V d D V 1 2 2 1 2 2 = ∴ = . . Energy: V g H t V g V gH t 1 2 2 2 2 2 + = ∴ = ( ) . ( ). Momentum: ΣF F d dt V d V m V V d s dt a x I x x c v x x x x − = − + − = ∫ ( ) & ( ). . . . ρ 2 1 2 2 v ∴− = = − ∫ a m t d V V m t m d V t dt x o t ( ) ( ). ( ) ( ) . ρ π ρ π 2 2 0 4 4 But, V dH dt dH dt d D gH dH H d D gdt H gd D t Ho 1 2 2 1 2 2 2 1 2 2 2 2 2 2 2 = − ∴− = ∴− = ∴ = + . . . . / / ∴ = + + − ∫ a d g gd D t H d g gd D t H dt m x o t o o ρπ ρ π 2 2 2 2 2 0 2 2 4 2 2 2 4 2 2 2 4.158 This is a very difficult design problem. There is an optimum initial mass of water for a maximum height attained by the rocket. It will take a team of students many hours to work this problem. It involves continuity, energy, and momentum. 4.159 V m A V e e = = × × × = & . . . ρ π 4 1000 4 004 19 89 2 m / s. Velocity in arm = v v v M r V d V ri k Vi Adr I c v = × × − = × − × ∫ ∫ ( ) $ ( $ $) . . . 2 4 2 0 3 Ω Ω ρ ρ
- 88. 87 = − =− ∫ 8 0 36 0 3 ρ ρ AV k rdr AV k Ω Ω $ . $. . Σ v v v v v v M d dt r V d V r V V n dA i V j V k V A c v e e e e c s = × − = × ⋅ = × + ∫ ∫ 0 0 3 707 707 and ( ) . ( $) . $ (. $ . $) . . . . . ρ ρ ρ The z-component of v v v r V V n dA V A e e c s × ⋅ = × ∫ ( $) . . . . . ρ ρ 3 707 2 Finally, − = = × × = ( ) . . . , M AV V A AV A V I z e e e e 0 36 4 3 707 2 ρ ρ Ω . Using 0 36 4 3 707 19 89 . . . . . Ω = × × × ∴ = Ω 46 9 . . rad / s 4.160 A moment v M resists the motion thereby producing power. One of the arms is shown. v M ri k Vi Adr AV k rdr AV k I = × − × = − = − ∫ ∫4 2 8 2 778 0 10 12 0 25 $ ( $ $) $ . $. / . Ω Ω Ω ρ ρ ρ Σ v v v v v v M Mk d dt r V d V r V V n dA V A k c v e e c s = × − = × ⋅ = × × ∫ ∫ $, ( ) , ( $) $. . . . . and 0 10 12 4 2 ρ ρ Thus, M + × × × = × × × 2 778 194 75 12 200 9 30 200 10 12 194 1 4 12 4 2 2 2 . . . . / . π π ∴ = M 309 ft - lb. & . W M = = × = Ω 309 30 9270 ft - lb /sec 4.161 & . . . m AV V V = = = × ∴ = 10 1000 01 31 8 2 0 0 ρ π m / s. Continuity: V V V r e 0 2 2 01 01 006 05 π π × = × + × − . . . ( . ). 0 2 .01 .006 .15. 11.1 m/s. e e V V V π × = × × ∴ = ∴ = − − = − V V r V r e 0 191 05 42 4 212 . ( . ) . . v M ri k V i Adr ri k r i Adr I = × + × + × + × − ∫ ∫2 2 2 2 42 4 212 0 05 2 0 05 $ ( $ $) $ [ $ ( . )$] . . . Ω Ω ρ ρ = + − ∫ ∫ 4 4 42 4 212 0 2 05 2 0 05 Ω Ω V Ak rdr Ak r r dr ρ ρ $ $ ( . ) . . . = × × × × + × × 4 31 8 1000 01 05 2 4 1000 01 2 2 2 Ω Ω . . . $ . π π k 42 4 2 2 05 212 3 2 05 2 2 3 3 . (. . ) (. . ) $ − − − k x y r Ve V Ω
- 89. 88 = + = (. . )$ . $. 0 05 0 3 0 35 Ω Ω Ω k k ri V j V dr rdr k k e e $ ( $) . . . $ . $. . . . . × − × = − × × = − ∫ ∫ ρ 006 111 1000 006 13 86 2 05 2 05 2 ∴ − = − 0 35 13 86 . . . Ω ∴ = Ω 39 6 . . rad /s 4.162 1000 1000 500 2 = ∴ = = ⋅ M M Ω. N m. v M ri k V r i r dr I r r = × − × × ∫ $ ( $ ( )$ ) . 2 2 02 Ω ρ π = ∫ 0 08 2 0 . ( ) $. πΩ r V r drk R Continuity: V r r V R V r RV r r r ( ) . cos . . ( ) . / . 2 02 30 2 02 0 866 π π × = × ∴ = o v v v o o r V V n dA R R V V R k V V k r r r r c s × ⋅ = − + × = − + ∫ ( $) ( sin ) cos . $ . ( . )$. . . ρ ρ π Ω 30 30 2 02 00301 35 5 ∴− − = − + ∴ − − = ∫ 2 16 32 00301 35 52 1 1333 0 0 15 2 . . ( .5 ). . . . V r dr V V V V r r r r r ∴ = ± + × = Vr 1 2 52 1 52 1 4 1333 70 9 2 ( . . ) . m / s. The flow rate is Q A V e r = = × × × × = cos . . . . . . 30 2 15 02 70 9 866 116 o π m / s 3 4.163 See Problem 4.159. V V e = = × = 19 89 008 02 19 89 318 2 2 . . . . . m / s. m / s. v M ri k Vi d dt k ri Adr A A I e = × − × + − × = × = × ∫ 4 2 01 004 0 3 2 2 $ ( $ $) $ $ . . , . . . Ω Ω ρ π π = − − = − − ∫ ∫ 8 4 360 36 0 3 2 0 3 ρ ρ AV k rdr A d dt k r dr AV k A d dt k Ω Ω Ω Ω $ $ $ $. . . ( ) ( $) $. . . v v v r V V n dA V A k z e e c s × ⋅ = ∫ ρ 212 2 Thus, 360 36 212 31 8 373 2 AV A d dt V A d dt e e Ω Ω Ω Ω + = + = or . . The solution is Ω = + − Ce t 31 8 11 73 . . . The initial condition is Ω( ) . 0 0 = ∴ = − C 1173 . . Finally, Ω = − − 11 73 1 31 8 . ( ) . . e t rad /s 4.164 This design problem would be good for a team of students to do as a project. How large a horsepower blower could be handled by an average person?
- 90. 89 CHAPTER 5 The DifferentialForms of the Fundamental Laws 5.1 0 = − + ⋅ ∫ ∫ ∂ρ ∂ ρ t d V V ndA c v c s . . . . $ . v Using Gauss’ theorem: 0 = − + ∇ ⋅ − = + ∇ ⋅ − ∫ ∫ ∫ ∂ρ ∂ ρ ∂ρ ∂ ρ t d V V d V t V d V c v c v c v . . . . . . ( ) ( ) . v v v v Since this is true for all arbitrary control volumes (i.e., for all limits of integration), the integrand must be zero: ∂ρ ∂ ρ t V + ∇ ⋅ = v v ( ) . 0 This can be written in rectangular coordinates as − = + + ∂ρ ∂ ∂ ∂ ρ ∂ ∂ ρ ∂ ∂ ρ t x u y v z w ( ) ( ) ( ). This is Eq. 5.2.2. The other forms of the continuity equation follow. 5.2 & & . m m m t in out element − = ∂ ∂ ( ) ρ θ ρ ∂ ∂ ρ θ v rd dz v r v dr r dr d dz r r r ( ) ( ) − + + + − + ρ ρ ∂ ∂θ ρ θ θ θ θ v drdz v v d drdz ( ) + + − + + = + ρ θ ρ ∂ ∂ ρ θ ∂ ∂ ρ θ v r dr d dr v z v dz r dr d dr t r dr d drdz z z z 2 2 2 ( ) . Subtract terms and divide by rd drdz θ : − − + − − + = + ρ ∂ ∂ ρ ∂ ∂θ ρ ∂ ∂ ρ ∂ ∂ ρ θ v r r v r dr r v r z v r dr r t r dr r r r z ( ) ( ) ( ) / / . 1 2 2 Since dr is an infinitesimal, ( )/ ( / )/ . r dr r r dr r + = + = 1 2 1 and Hence, ∂ρ ∂ ∂ ∂ ρ ∂ ∂θ ρ ∂ ∂ ρ ρ θ t r v r v z v r v r z r + + + + = ( ) ( ) ( ) . 1 1 0 This can be put in various forms.
- 91. 90 5.3 & &. m m m t in out element − = ∂ ∂ ρ θ θ φ ρ ∂ ∂ ρ θ θ φ v rd r d v r v dr r dr d r dr d r r r ( ) sin ( ) ( ) ( )sin − + + + + + − + + ρ θ φ ρ ∂ ∂θ ρ θ θ φ θ θ θ v dr r dr d v v d dr r dr d 2 2 sin ( ) sin + + − + + ρ θ ρ ∂ ∂φ ρ φ θ φ φ φ v dr r dr d v v d dr r dr d 2 2 ( ) = + ∂ ∂ ρ θ θ φ t r dr drd d 2 2 sin Because some areas are not rectangular, we used an average length ( / ). r dr + 2 Now, subtract some terms and divide by rd d dr θ φ : − − − + − + ρ θ ρ θ ∂ ∂ ρ θ ∂ ∂θ ρ θ θ v v r v r dr r v r dr r r r r sin sin ( )sin ( ) ( ) sin 2 2 − + = + ∂ ∂φ ρ ∂ρ ∂ θ φ ( ) sin v r dr r t r dr r 2 2 2 Since dr is infinitesimal ( ) / ( / ) / . r dr r r r dr r + = + = 2 2 1 and Divide by r sinθ and there results ∂ρ ∂ ∂ ∂ ρ ∂ ∂θ ρ θ ∂ ∂φ ρ ρ θ φ t r v r v r v r v r r + + + + = ( ) ( ) sin ( ) 1 1 2 0 5.4 For a steady flow ∂ρ ∂t = 0. Then, with v w = = 0 Eq. 5.2.2 yields ∂ ∂ ρ ρ ρ x u du dx u d dx ( ) . = + = 0 0 or Partial derivatives are not used since there is only one independent variable. 5.5 Since the flow is incompressible D Dt ρ = 0. This gives 3 2 3 1 200 1 200 ˆ ˆ ˆ ˆ cos2 sin2 r r p p p i i i i r r r r r θ θ ∂ ∂ ρ ρ θ θ ∂ ∂θ ∴∇ = + = − − v or u x w z ∂ρ ∂ ∂ρ ∂ + = 0. Also, v v ∇ ⋅ = V 0, or ∂ ∂ ∂ ∂ u x w z + = 0.
- 92. 91 5.6 Given: ∂ ∂ ∂ρ ∂ t z =≠ 0 0 , . Since water can be considered to be incompressible, we demand that D Dt ρ = 0. ∴u x w z ∂ρ ∂ ∂ρ ∂ + = 0, assuming the x-direction to be in the direction of flow. Also, we demand that v v ∇ ⋅ = V 0, or ∂ ∂ ∂ ∂ u x w z + = 0. 5.7 We can use the ideal gas law, ρ = p RT . Then, the continuity equation D Dt V ρ ρ = − ∇ ⋅ v v becomes, assuming RT to be constant, 1 RT Dp Dt p RT V = − ∇ ⋅ v v or 1 p Dp Dt V = −∇ ⋅ v v . 5.8 a) Use cylindrical coordinates with v vz θ = = 0: 1 0 r r rvr ∂ ∂ ( ) = Integrate: rv C r = . ∴ = v C r r . b) Use spherical coordinates with v v θ φ = = 0: 1 0 2 2 r r r vr ∂ ∂ ( ) = Integrate: r v C r 2 = . ∴ = v C r r 2 . 5.9 D Dt V u x v y ρ ρ ρ ∂ ∂ ∂ ∂ = − ∇ ⋅ = − + = − × + × = − ⋅ v v 2 3 200 1 400 1 1380 . ( ) . kg m s 3 5.10 In a plane flow, u u x y v v x y = = ( , ) ( , ). and Continuity demands that ∂ ∂ ∂ ∂ u x v y + = 0. If u u x = = const, then ∂ ∂ 0 and hence ∂ ∂ v y = 0. Thus, v = const also.
- 93. 92 5.11 If uC v C = = 1 2 and , the continuity equation provides, for an incompressible flow, ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ u x v y w z w z w C + + = ∴ = = 0 0 3 . . and The z-component of velocity w is also constant. We also have D Dt t u x v y w z ρ ∂ρ ∂ ∂ρ ∂ ∂ρ ∂ ∂ρ ∂ = = + + + 0 The density may vary with x, y, z and t. It is not, necessarily, constant. 5.12 ∂ ∂ ∂ ∂ u x v y + = 0. ∴ + = A v y ∂ ∂ 0. ∴ = − + v x y Ay f x ( , ) ( ). But, v x o f x ( , ) ( ). = = 0 ∴ = − v Ay. 5.13 ∂ ∂ ∂ ∂ u x v y + = 0. ∴ = − = − + − + = − − + ∂ ∂ ∂ ∂ v y u x x y x x x y x y x y ( )5 ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 5 2 5 5 ∴ = − + + = + + = ∴ = + ∫ v x y y x x y dy f x y x y f x f x v y x y ( , ) ( ) ( ) ( ). ( ) . . 5 5 5 0 5 2 2 2 2 2 2 2 2 2 5.14 From Table 5.1: 1 1 1 10 4 2 r r rv r v r r r ∂ ∂ ∂ ∂θ θ θ ( ) . sin . = − = − + ∴ = + + = − + ∫ rv r dr f r r f r 10 4 10 4 2 . sin ( ) . sin ( ). θ θ θ θ . (. , ) . . . sin ( ) . ( ) . 2 2 10 2 4 2 0 0 v f f r θ θ θ θ = × − + = ∴ = ∴ = − v r r 10 0 4 2 . sin . θ 5.15 From Table 5.1: 1 1 20 1 1 2 r r rv r v r r r ∂ ∂ ∂ ∂θ θ θ ( ) cos . = − = − + ∴ = − + + = − − + ∫ rv r dr f r r f r 20 1 1 20 1 2 cos ( ) cos ( ). θ θ θ θ v f f r ( , ) ( )cos ( ) . ( ) . 1 20 1 1 0 0 θ θ θ θ = − − + = ∴ = ∴ = − − v r r 20 1 1 2 cos . θ
- 94. 93 5.16 From Table5.1, spherical coordinates: 1 1 2 2 r r r v r v r ∂ ∂ θ ∂ ∂θ θ θ ( ) sin ( sin ). = − ∴ = + 1 1 10 40 2 2 2 3 r r r v r r r ∂ ∂ θ θ θ ( ) sin sin cos . ∴ = + + = − + ∫ r v r r dr f r r f r 2 3 2 10 40 2 10 80 cos ( ) cos ( ) θ θ θ θ 4 2 10 2 80 2 0 0 2 v f f r ( , ) cos ( ) . ( ) . θ θ θ θ = × − + = ∴ = ∴ = − v r r 10 80 3 cos . θ 5.17 Continuity: ∂ ∂ ρ ρ ρ x u du dx u d dx ( ) . . = ∴ + = 0 0 ρ = = × × = = − × = p RT du dx 18 144 1716 500 0 00302 526 453 2 2 12 219 3 . . / slug ft fps / ft. ∴ = − = − × = − d dx u du dx ρ ρ . . . 00302 486 219 0 00136 slug/ ft 4 5.18 [ ] ∂ ∂ ∂ ∂ ∂ ∂ u x v y x e e x x + = − − = − − − 0 20 1 20 . ( ) Hence, in the vicinity of the x-axis: ∂ ∂ v y e v ye C x x = = + − − 20 20 and . But v y C = = ∴ = 0 0 0 if . . v ye e x = = = − − 20 20 0 2 0 2 ( . ) .541 m / s 5.19 [ ] 1 0 20 1 20 r r rv v z z e e r z z z ∂ ∂ ∂ ∂ ∂ ∂ ( ) . ( ) + = − − = − − − Hence, in the vicinity of the z-axis: 1 20 2 20 2 r r rv e rv r e C r z r z ∂ ∂ ( ) . = = + − − and But v r C r = = ∴ = 0 0 0 if . . v re e r z = = = − − 10 10 0 2 0 271 2 ( . ) . m / s 5.20 The velocity is zero at the stagnation point. Hence, 0 10 40 2 2 = − ∴ = R R . m The continuity equation for this plane flow is ∂ ∂ ∂ ∂ u x v y + = 0. Using ∂ ∂ u x x = − 80 3 ,
- 95. 94 we see that ∂ ∂ v y x =− − 80 3 near the x-axis. Consequently, for small ∆y, ∆ ∆ v x y = − − 80 3 so that v = − − = − 80 3 0 1 0 296 3 ( ) ( . ) . . m / s 5.21 The velocity is zero at the stagnation point. Hence 0 40 10 2 2 = − ∴ = R R . m ( ) 1 1 40 10 20 2 2 2 2 r r r v r r r r r ∂ ∂ ∂ ∂ = − = − ( ) . Near the negative x-axis continuity provides us with ( ) 1 20 r v r sin sin . θ ∂ ∂θ θ θ = Integrate, letting θ = 0 from the y-axis: v C θ θ θ sin cos = − + 20 Since vθ = 0 when θ = = 90 0 o , . C Then, with α = = − tan . . , 1 0 1 3 1909o vθ θ θ = − = − = − = 20 20 88 091 88 091 20 0 0333 0 999 0 667 cos sin cos . sin . . . . m / s 5.22 Continuity: ∂ ∂ ∂ ∂ u x v y v y u x + = ∴ = − = − − × = − 0 13 5 11 3 2 005 220 . . . . . m /s m ∆ ∆ ∆ ∆ ∴ = − = − ∴ = − × = − ∆ ∆ v v y v 0 220 220 004 0 88 . . . . m / s b) a u u x x = = × + = ∂ ∂ 12 6 220 2772 . ( ) . m /s2 5.23 ΣF ma y y = . For the fluid particle occupying the volume of Fig. 5.3: τ ∂τ ∂ τ ∂τ ∂ τ ∂τ ∂ yy yy zy zy xy xy y dy dxdz z dz dxdy x dx dydz + + + + + 2 2 2 − − − − − − τ ∂τ ∂ τ ∂τ ∂ τ ∂τ ∂ yy yy zy zy xy xy y dy dxdz z dz dxdy x dx dydz 2 2 2 + = ρ ρ g dx dy dz dx dy dz Dv Dt y Dividing by dx dy dz, and adding and subtracting terms: ∂τ ∂ ∂τ ∂ ∂τ ∂ ρ ρ xy yy zy y x y z g Dv Dt + + + = . 5.24 Check continuity: ∂ ∂ ∂ ∂ ∂ ∂ u x v y w z x y x x x y x y y y x y + + = + − + + + − + = ( )10 ( ) ( ) ( )10 ( ) ( ) . 2 2 2 2 2 2 2 2 2 2 10 2 10 2 0
- 96. 95 Thus, it isa possible flow. For a frictionless flow, Euler’s Eqs. 5.3.7 give, with g g x y = = 0: . u u p u v x y x ∂ ∂ ∂ ρ ρ ∂ ∂ ∂ + = − 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 10 10 10 10 20 100( ) ( ) ( ) ( ) p x y x y xy x y y x x y x y x y x y x y ∂ ρ ρ ρ ∂ − − + ∴ = − − = + + + + + . v v p u v x y y ∂ ∂ ∂ ρ ρ ∂ ∂ ∂ + = − 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 10 20 10 10 10 100( ) ( ) ( ) ( ) p x xy y x y x y y y x y x y x y x y x y ∂ ρ ρ ρ ∂ − − + ∴ = − − = + + + + + 2 2 2 2 2 2 2 2 2 100 100 100 ˆ ˆ ˆ ˆ ˆ ˆ ( ). ( ) ( ) ( ) p p x y p i j i j xi yj x y x y x y x y ∂ ∂ ρ ρ ρ ∂ ∂ ∴∇ = + = + = + + + + v 5.25 Check continuity (cylindrical coord from Table 5.1): 1 1 10 1 1 10 1 1 0 2 2 r r rv r v r r r r r ∂ ∂ ∂ ∂θ θ θ θ ( ) cos cos . + = + + − + = ∴It is a possible flow. For Euler’s Eqs. (let ν = 0 in the momentum eqns of Table 5.1) in cylindrical coord: 2 2 2 2 2 2 3 100 1 1 20 1 sin 10 1 cos r r r v v v v p v r r r r r r r r θ θ ∂ ∂ ∂ ρ ρ ρ ρ θ ρ θ ∂ ∂ ∂θ = − − = + − − − + − 10 1 1 10 10 2 2 2 ρ θ r r r sin . 4 1 100 1 1 sin cos r r v v v v v p v r r r r r r θ θ θ θ ∂ ∂ ∂ ρ ρ ρ ρ θ θ ∂θ ∂ ∂θ = − − − = − − − − + 10 1 1 20 100 1 1 2 3 2 2 ρ θ θ ρ θ θ r r r r cos sin sin cos . 3 2 3 1 200 1 200 ˆ ˆ ˆ ˆ cos2 sin2 r r p p p i i i i r r r r r θ θ ∂ ∂ ρ ρ θ θ ∂ ∂θ ∴∇ = + = − − v 5.26 This is an involved problem. Follow the steps of Problem 5.25. Good luck! ( ) ∂ ∂ ρ ρ ∂ ∂ ρ ∂ ∂θ θ φ θ p r v v r v v r v r v r r r = + − − 2 2 1 r p v v r v v r v r v r r ∂ ∂θ ρ ρ ∂ ∂ ρ ∂ ∂θ θ θ θ θ = − − − ( )
- 97. 96 5.27 2 2 . . 33 p p V p p V µ µ λ λ ∴ = − + ∇ ⋅ ∴ − = − + ∇ ⋅ v v v v ∂ ∂ $ $ $ $ . s s s s n R n R ≅ = − = − ∆ ∆ ∆α ∆α ∂ ∂ ∂θ ∂ $ $ $ $ . s t s t n t n t ≅ = = ∆ ∆ ∆θ ∆ 2 ˆ ˆ. DV V V V V s V n Dt t s t R ∂ ∂ ∂θ ∂ ∂ ∂ ∴ = + + − v For steady flow, the normal acc. is − V R 2 , the tangential acc. is V V s ∂ ∂ . 5.28 For a rotating reference frame (see Eq. 3.2.15), we must add the terms due to v Ω. Thus, Euler’s equation becomes 2 ( ) . DV d V r r p g Dt dt ρ ρ Ω + Ω × + Ω × Ω × + × =−∇ − v v v v v v v v v v 5.29 2 xx u p x ∂ τ µ ∂ = − + V λ + ∇ ⋅ v v 30 psi. = − τ τ yy zz p = = − = −30 psi. xy u v y x ∂ ∂ τ µ ∂ ∂ = + 5 5 .1 10 30 1440 18 10 psf. 12 − − = − × = × τ τ τ τ xz yz xx = = = × × = × − − 0 18 10 30 144 4 17 10 5 8 . . . xy 5.30 2 2 3 9/5 2 13/5 9/5 2 13/5 16 16 8 16 . ( , ) ( ). 3 v u y y y y v x y f x y x C x C x Cx C x ∂ ∂ ∂ ∂ = − = − ∴ = − + v x o f x C C ( , ) . ( ) . . . . / = ∴ = = ∴ = 0 0 8 1000 0 0318 4 5 ∴ = − − − u x y yx y x ( , ) . / 629 9890 4/ 5 2 8 5 v x y y x y x ( , ) . / / = − − − 252 5270 2 9 5 3 13 5 τ µ ∂ ∂ xx p u x = − + = − + = − 2 100 0 100 kPa. τ τ yy zz p = = − = −100 kPa. 5 4/5 5 2 10 629 1000 5.01 10 Pa. xy u v y x ∂ ∂ τ µ ∂ ∂ − − − = + = × × = × τ τ xz yz = = 0.
- 98. 97 5.31 Du u Dt t ∂ ∂ =( ) . u v w u V u x y z ∂ ∂ ∂ ∂ ∂ ∂ + + + = ⋅∇ v v Dv v Dt t ∂ ∂ = ( ) . u v w v V v x y z ∂ ∂ ∂ ∂ ∂ ∂ + + + = ⋅∇ v v Dw w Dt t ∂ ∂ = ( ) u v w w V w x y z ∂ ∂ ∂ ∂ ∂ ∂ + + + = ⋅∇ v v ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) . DV Du Dv Dw i j k V ui vj wk V V Dt Dt Dt Dt ∴ = + + = ⋅∇ + + = ⋅∇ v v v v v v 5.32 Follow the steps that lead to Eq. 5.3.17 and add the term due to compressible effects: 2 ˆ ˆ ˆ 3 3 3 DV p g V Vi Vj Vk Dt x y z µ ∂ µ ∂ µ ∂ ρ ρ µ ∂ ∂ ∂ =−∇ + + ∇ + ∇ ⋅ + ∇ ⋅ + ∇ ⋅ v v v v v v v v v v = −∇ + + + + + ∇ ⋅ v v v v v p g V x i y j z k V ρ µ∇ µ ∂ ∂ ∂ ∂ ∂ ∂ 2 3 $ $ $ ∴ = −∇ + + ∇ + ∇ ∇ ⋅ ρ ρ µ µ DV Dt p g V V v v v v v v v 2 3 ( ). 5.33 If u=u(y), then continuity demands that ∂ ∂ v y v C = ∴ = 0. . But, at y=0 (the lower plate) v=0. ∴ = = C v x y 0 0 , ( , ) . and 2 2 2 2 2 2 . x u u u Du u u u u p u v w g Dt t x y z x x y z ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ρ ρ ρ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∴ = + + + = − + + + + ∴ = − + 0 2 2 ∂ ∂ µ ∂ p x u ay . 0 . Dv p Dt y ∂ ρ ∂ = = − ρ ∂ ∂ ρ ∂ ∂ ρ Dw Dt p z g p z g = = − + − ∴ = − − 0 0 ( ). . 5.34 Continuity: ( ) 0. . r r rv rv C r ∂ ∂ = ∴ = At 0, . 0. r r v C = ≠ ∞ ∴ = 1 0 . r Dv p Dt r ∂ ρ ∂ = = − 1 0 . Dv p Dt r θ ∂ ρ ∂θ = = −
- 99. 98 z z Dv v Dtt ∂ ρ ρ ∂ = r v + z v v r θ ∂ ∂ + z z z v v v r z ∂ ∂ ∂θ ∂ + 2 2 2 2 2 1 1 z z z v v v p z r r r r ∂ ∂ ∂ ∂ µ ∂ ∂ ∂ ∂θ = − + + + 2 2 z v z ∂ ∂ + ∴ = − + + 0 1 2 2 ∂ ∂ µ ∂ ∂ ∂ ∂ p z v r r v r z z . 5.35 Continuity: 1 0 0 0 2 2 2 1 r r r v r v C r r v C r r r ∂ ∂ ( ) . . , . . = ∴ = = = ∴ = At 2 2 2 cot . v v p r r r θ θ ∂ ρ µ θ ∂ − = − + − 2 2 2 2 1 1 0 sin v v p r r r r r r θ θ ∂ ∂ ∂ µ ∂θ ∂ ∂ θ = − + − 0 1 = − r p sin . θ ∂ ∂φ 5.36 For an incompressible flow v v ∇⋅ = V 0. Substitute Eqs. 5.3.10 into Eq. 5.3.2 and 5.3.3: ρ ∂ ∂ µ ∂ ∂ ∂ ∂ µ ∂ ∂ ∂ ∂ ∂ ∂ µ ∂ ∂ ∂ ∂ ρ Du Dt x p u x y u y v x z u z w x gx = − + + + + + + 2 . = − + + + + + + + ∂ ∂ µ ∂ ∂ µ ∂ ∂ µ ∂ ∂ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ρ p x u x u y u z x u x v y w z gx 2 2 2 2 2 2 ∴ = − + + + + ρ ∂ ∂ µ ∂ ∂ ∂ ∂ ∂ ∂ ρ Du Dt p x u x u y u z gx 2 2 2 2 2 2 . 2 . y Dv u v v v w p g Dt x y x y y z z y ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ρ µ µ µ ρ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + − + + + + = − + + + + + + + ∂ ∂ µ ∂ ∂ µ ∂ ∂ µ ∂ ∂ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ρ p y v x v y v z y u x v y w z gy 2 2 2 2 2 2 ∴ = − + + + + ρ ∂ ∂ µ ∂ ∂ ∂ ∂ ∂ ∂ ρ Dv Dt p y v x v y v z gy 2 2 2 2 2 2 . 2 z Dw u w v w w p g Dt x z x y z y z z ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ρ µ µ µ ρ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + + + − + +
- 100. 99 = − ++ + + + + + ∂ ∂ µ ∂ ∂ µ ∂ ∂ µ ∂ ∂ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ρ p z w x w y w z z u x v y w z gz 2 2 2 2 2 2 ∴ = − + + + + ρ ∂ ∂ µ ∂ ∂ ∂ ∂ ∂ ∂ ρ Dw Dt p z w x w y w z gz 2 2 2 2 2 2 . 5.37 If we substitute the constitutive equations (5.3.10) into Eqs. 5.3.2 and 5.3.3., with µ µ = ( , , ) x y z we arrive at 2 2 2 2 2 2 2 x Du p u u u u u v u w g Dt x x x y y x z z x x y z ∂ ∂ ∂ ∂ ∂µ ∂ ∂µ ∂ ∂ ∂µ ∂ ∂ ρ ρ µ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = − + + + + + + + + + 5.38 If plane flow is only parallel to the plate, v w = = 0. Continuity then demands that ∂ ∂ u x / . = 0 The first equation of (5.3.14) simplifies to u u u t x ∂ ∂ ρ ∂ ∂ + v + u w y ∂ ∂ + u p z x ∂ ∂ ∂ ∂ = − x g ρ + 2 2 u x ∂ µ ∂ + 2 2 2 2 u u y z ∂ ∂ ∂ ∂ + + 2 2 u u t y ∂ ∂ ρ µ ∂ ∂ = We assumed g to be in the y-direction, and since no forcing occurs other than due to the motion of the plate, we let ∂ ∂ p x / . = 0 5.39 From Eqs. 5.3.10, − + + = − + + − ∇ ⋅ τ τ τ µ ∂ ∂ ∂ ∂ ∂ ∂ λ xx yy zz p u x v y w z V 3 2 3 v v . 2 2 . . 3 3 p p V p p V µ µ λ λ ∴ = − + ∇ ⋅ ∴ − = − + ∇ ⋅ v v v v 5.40 ˆ ˆ ˆ ( ) ( ) V V u v w ui vj wk x y z ∂ ∂ ∂ ⋅∇ = + + + + ∂ ∂ ∂ v v ˆ ( ) w w w v v v V V u v w u v w i y x y z z x y z ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∇ ⋅∇ = + + − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ × v v ˆ u u u w w w u v w u v w j z x y z x x y z ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ˆ v v v u u u u v w u v w k x x y z y x y z ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ Use the definition of vorticity: ˆ ˆ ˆ ( ) ( ) ( ) w v u w v u i j k y z z x x y ω ∂ ∂ ∂ ∂ ∂ ∂ = − + − + − ∂ ∂ ∂ ∂ ∂ ∂ v
- 101. 100 ˆ ˆ ˆ ( )( ) ( ) ( ) ( ) w v u w v x V ui vj wk y z x z x y x y z ω ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⋅∇ = − + − + − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ v v ˆ ˆ ˆ ( ) ( ) ( ) ( ) w v u w v u V u v w i j k x y z y z z x x y ω ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⋅∇ = + + − + − + − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ v v Expand the above, collect like terms, and compare coefficients of ˆ, i ˆ, j and ˆ. k 5.41 Studying the vorticity components of Eq. 3.2.21, we see that / z u y ω ∂ ∂ = − is the only vorticity component of interest. The third equation of Eq. 5.3.24 then simplifies to D Dt z z ω ν ω = ∇2 2 2 z y ∂ ω ν ∂ = since changes normal to the plate are much larger than changes along the plate, i.e., . z z y x ∂ω ∂ω ∂ ∂ >> 5.42 If viscous effects are negligible, as they are in a short section, Eq. 5.3.25 reduces to D Dt z ω = 0 that is, there is no change in vorticity (along a streamline) between sections 1 and 2. Since (see Eq. 3.2.21), at section 1, ω ∂ ∂ ∂ ∂ z v x u y = − = −10 we conclude that, for the lower half of the flow at section 2, . 10 = y u ∂ ∂ This means the velocity profile at section 2 is a straight line with the same slope of the profile at section 1. Since we are neglecting viscosity, the flow can slip at the wall with a slip velocity u0 ; hence, the velocity distribution at section 2 is u y u y 2 0 10 ( ) . = + Continuity then allows us to calculate the profile: V A V A 1 1 2 2 = 1 2 10 0 04 0 04 10 0 02 2 0 02 0 3 0 0 ( . )( . ) ( . / )( . ). . × = + × ∴ = w u w u m / s. Finally, u y y 2 0 3 10 ( ) . = +
- 102. 101 5.43 No. Thefirst of Eqs. 5.3.24 shows that, neglecting viscous effects, x x y z D u u u Dt x y z ω ∂ ∂ ∂ ω ω ω ∂ ∂ ∂ = + + so that ω y , which is nonzero near the snow surface, creates ω x through the term / , y u y ω ∂ ∂ since there would be a nonzero ∂ ∂ u y / near the tree. 5.44 k T ndA t V gz u d V V gz u p V ndA c v c s c s v v ∇ ⋅ = + + − + + + + ⋅ ∫ ∫ ∫ $ ~) ~ $ . . . . . . ∂ ∂ ρ ρ ρ 2 2 2 2 v v v v ∇⋅ ∇ − = + + − + ∇ ⋅ + + + − ∫ ∫ ∫ ( ) ~ ~ . . . . . . k T d V t V gz u d V V V gz u p d V c v c v c v ∂ ∂ ρ ρ ρ 2 2 2 2 ∴ − ∇ + + + + ∇ ⋅ + + + − = ∫ k T t V gz u V V gz u p d V c v 2 2 2 2 2 0 ∂ ∂ ρ ρ ρ ρ ρ ~ ~ . . . v v 2 2 2 2 2 2 V V p V V gz V t t ∂ ∂ρ ρ ρ ρ ∂ ρ ∂ + ∇ ⋅ + + = + ∇ ⋅ v v v v V p V V V g z t ∂ ρ ∂ ρ ∇ + ⋅ + ⋅∇ + + ∇ v v v v v v 0. = continuity momentum ∴− ∇ + + ⋅∇ = ∴ = ∇ k T t u V u Du Dt k T 2 2 0 ∂ ∂ ρ ρ ρ ~ ~ . ~ . v v 5.45 Divide each side by dxdydz and observe that 2 2 2 2 2 2 , , y dy y x dx x z dz z T T T T T T y y x x z z T T T dx dy dz x x z + + + ∂ ∂ ∂ ∂ ∂ ∂ − − − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = = = ∂ ∂ ∂ Eq. 5.4.5 follows. 5.46 ( / ) Du D h p Dh Dp p D Dh Dp p V Dt Dt Dt Dt Dt Dt Dt ρ ρ ρ ρ ρ ρ ρ ρ ρ − = = − + = − + − ∇ ⋅ v % where we used the continuity equation: / . D Dt V ρ ρ = − ∇ ⋅ v Then Eq. 5.4. 9 becomes 2 Dh Dp p V K T p V Dt Dt ρ ρ ρ − + − ∇ ⋅ = ∇ − ∇ ⋅ v v which is simplified to 2 Dh Dp K T Dt Dt ρ = ∇ +
- 103. 102 5.47 See Eq.5.4.9: ~ . . u cT c T t u T x v T y w T z k T = ∴ + + + = ∇ ρ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ 2 Neglect terms with velocity: ρ ∂ ∂ c T t k T = ∇2 . 5.48 The dissipation function Φ involves viscous effects. For flows with extremely large velocity gradients, it becomes quite large. Then ρc DT Dt p = Φ and DT Dt is large. This leads to very high temperatures on reentry vehicles. 5.49 u r u r r r y = − ∴ = − × 10 1 10 000 2 105 ( ). . ) ( takes the place of 2 ∂ ∂ From Eq. 5.4.17, Φ = = × 2 1 2 4 10 2 2 10 µ ∂ ∂ µ u y r . At the wall where r = = × × × × = ⋅ − 0 01 18 10 4 01 10 72 5 2 10 . . . . m, N / m s 2 Φ At the centerline ∂ ∂ u r = = 0 0 so Φ . At a point half-way: Φ = × × × × = ⋅ − 18 10 4 005 10 18 5 2 10 . . . N / m s 2 5.50 (a) Momentum: ∂ ∂ ν ∂ ∂ u t u y = 2 2 Energy: ρ ∂ ∂ ∂ ∂ µ ∂ ∂ c T t K T y u y = + 2 2 2 . (b) Momentum: ρ ∂ ∂ µ ∂ ∂ ∂µ ∂ ∂ ∂ u t u y y u y = + 2 2 Energy: ρ ∂ ∂ ∂ ∂ µ ∂ ∂ c T t K T y u y = + 2 2 2 .
- 104. 103 CHAPTER 6 Dimensional Analysis andSimilitude 6.1 g V V g p g z g V V g p g z 1 2 1 2 1 1 1 2 2 2 2 2 2 2 + + = + + ρ ρ . 1 2 1 2 1 1 2 1 1 2 2 2 1 2 2 1 2 2 1 2 + + = + + p V gz V V V p V gz V ρ ρ . or 1 2 1 2 1 1 2 1 1 2 2 2 2 2 2 2 2 2 1 2 + + = + + p V gz V p V gz V V V ρ ρ 6.2 a) [ ] & . . m FT L = = ⋅ ⋅ = ⋅ ∴ kg s N s m s N s m 2 b) [ ] p F L = ∴ N m2 . 2 c) [ ] ρ = = ⋅ ⋅ = ⋅ ∴ kg m N s m m N s m 3 2 3 2 4 . . FT L 2 4 d) [ ] µ = ⋅ ∴ N s m2 . FT L2 e) [ ] W FL = ⋅ ∴ N m. f) [ ] & . W FL T = ⋅ ∴ N m s g) [ ] σ = ∴ N / m. F L 6.3 (A) The dimensions on the variables are as follows: 2 2 2 3 2 2 / [ ] [ ] , [ ] , [ ] , [ ] L L ML ML T M L W F V M d L p V T T T T L LT = × = × = = ∆ = = = & First, eliminate T by dividing W & by ∆p. That leaves T in the denominator so divide by V leaving L2 in the numerator. Then divide by d2 . That provides 2 W pVd π = ∆ & 6.4 1 2 5 2 , , , . T e r f R R R R R µ ρω ρω ∴ = l
- 105. 104 6.5 (A) (, , , , ). The units on the variables on the rhs are as follows: V f d l g ω µ = 1 2 [ ] , [ ] , [ ] , [ ] , [ ] L ML d L l L g T T T ω µ − = = = = = Because mass M occurs in only one term, it cannot enter the relationship. 6.6 [ ] [ ] [ ] [ ] V f V L T L M L M LT = = = = = ( , , ). , , , . l l ρ µ ρ µ 3 ∴There is one π − term: π ρ µ 1 = Vl . ∴ = = ∴ = = π π ρ µ 1 1 2 0 f V C ( ) , Re Const. or Const. l 6.7 [ ] [ ] [ ] [ ] V f d V L T M T M L d L = = = = = ( , , ). , , , . σ ρ σ ρ 2 3 ∴ = ∴ = = ∴ = π σ ρ π π σ ρ 1 2 1 1 2 0 2 V d f C V d C . ( ) , onst. or We = Const. 6.8 [ ] [ ] [ ] [ ] V f H g m V L T g L T m M H L = = = = = ( , , ). , , , . 2 ∴ = ∴ = ∴ = π π 1 0 2 1 gHm V C V gH C . . / . 6.9 [ ] [ ] [ ] [ ] [ ] [ ] V f H g m V L T H L g L T m M M L M LT = = = = = = = ( , , , , ). , , , , , . ρ µ ρ µ 2 3 Choose repeating variables H g , ,ρ (select ones with simple dimensions-we couldn’t select V, H, and g since M is not contained in any of those terms): π ρ π ρ π µ ρ 1 2 3 1 1 1 2 2 2 3 3 3 = = = VH g mH g H g a b c a b c a b c , , . ∴ = = = = = π ρ π ρ π µ ρ µ ρ 1 0 2 3 3 3 2 3 V g H V gH m H gH gH . . . / ∴ = V gH f m H gH 1 3 3 ρ µ ρ , . Note: The above dimensionless groups are formed by observation: simply combine the dimensions so that the π − term is dimensionless. We could have set up equations similar to those of Eq. 6.2.11 and solved for 1 1 1 2 2 2 3 3 3 , , and , , c and , , . a b c a b a b c But the method of observation is usually successful. 6.10 [ ] [ ] [ ] [ ] [ ] F f d V F ML T d L V L T M LT M L D D = = = = = = ( , , , , ). , , , , . l µ ρ µ ρ 2 3
- 106. 105 π ρ πρ π µ ρ 1 2 3 1 1 1 2 2 2 3 3 3 = = = F V dV V D a b c b c a a b c l l l , , . ∴ = = = π ρ π π µ ρ 1 2 2 2 3 F V d V D l l l , , . ∴ = F V f d V D ρ µ ρ l l l 2 2 1 , . We could write π π π π π 1 2 2 2 2 3 2 1 = f , or F d V f d dV D ρ µ ρ 2 2 2 = l , . This is equivalent to the above. Either functional form must be determined by experimentation. 6.11 [ ] [ ] [ ] [ ] [ ] F f d V F ML T d L V L T M LT M L D D = = = = = = ( , , , , ). , , , , . l µ ρ µ ρ 2 3 π µ π µ π ρ µ 1 2 3 1 1 1 2 2 2 3 3 3 = = = F d V d V d V D a b c a b c a b c , , . l By observation we have 1 2 3 , , . D F Vd Vd d ρ π π π µ µ = = = l ∴ = F Vd f d Vd D µ ρ µ 1 l , . Rather than π π π 1 1 2 3 = f ( , ), we could write π π π π 1 3 2 2 3 1 = f , , an acceptable form: F V d f d Vd D ρ µ ρ 2 2 2 = l , . 6.12 [ ] [ ] [ ] [ ] [ ] [ ] h f d g h L M T d L M L T g L T = = = = = = = ( , , , , ). , , , , , . σ γ β σ γ β 2 2 2 2 1 Select d g , , γ as repeating variables. π γ π σ γ π β 1 2 1 1 1 2 2 2 = = = hd g d g a b c a b c , , . 3 ∴ = = = π π σ γ π β 1 2 2 3 h d d , , . ∴ = h d f d 1 2 σ γ β , . Note: gravity does not enter the answer. 6.13 [ ] [ ] [ ] [ ] F f m R F ML T m M T R L C C = = = = = ( , , ). , , , . ω ω 2 1 ∴ = = ∴ = ∴ = π ω ω ω ω 1 2 2 2 F m R F m R F m R C F Cm R C a b c C C C . . 6.14 [ ] [ ] [ ] [ ] σ σ = = = = = f M y I M LT M ML T y L I L ( , , ). , , , . 2 2 2 4 ∴ = π σ 1 M y I a b c .
- 107. 106 Given that 1 b= − , 1 Const. C . I My yM I σ π σ = = ∴ = 6.15 [ ] [ ] [ ] V f d dp dx V L T M LT d L dp dx M L T = = = = = ( , , . , , , . µ µ 2 2 ∴ = π µ 1 V d dp dx a b c . Let’s start with the ratio µ dp dx / so that “M” is accounted for. Then the π1 − term is µV dp dx d / . 2 Hence, π µ µ 1 2 2 = = ∴ = V dp dx d V d dp dx / / . Const. Const 6.16 [ ] [ ] [ ] [ ] V f H g V L T H L g L T M L = = = = = ( , , ). , , , . ρ ρ 2 3 ∴ = = = ∴ = π ρ ρ 1 0 VH g V g H V gH a b c Const. Const. . Density does not enter the expression. 6.17 [ ] [ ] [ ] [ ] [ ] [ ] 3 2 ( , , , , ). , , , , , . L M M L V f H g d V H L g d L T LT L T µ ρ µ ρ = = = = = = = π ρ π µ ρ π ρ 1 2 1 1 1 2 2 2 3 3 3 = = = VH g H g dH g a b c a b c a b c , , . 3 Repeating variables H g , , . ρ π π µ ρ π 1 2 3 2 3 = = = V gH gH d H , , . / ∴ = = π π π µ ρ 1 1 2 3 1 3 f V gH f gH d H ( , ), , . or 6.18 ∆p f V d L = ( , , , , , ). ν ε ρ [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2 2 3 , , , , , , . M L L M p V d L L L L T T LT L ν ε ρ ∆ = = = = = = = Repeating variables: V d , , . ρ π ρ π ν ρ π ρ π ε ρ 1 2 3 4 1 1 1 2 2 2 3 3 3 4 4 4 = = = = ∆pV d V d L V d V d a b c a b c a b c a b c , , , . ∴ = = = = π ρ π ν π π ε 1 2 3 ∆p V Vd L d d , , , . 2 4 π π π π ρ ν ε 1 1 2 3 4 2 1 = ∴ = f p V f Vd L d d ( , , ). , , . ∆
- 108. 107 6.19 F fV c h r w D = ( , , , , , , , , ) ρ µ φ α where c is the chord length, h is the maximum thickness, r is the nose radius, φ is the trailing edge angle, and α is the angle of attack. Repeating variables: V c , , . ρ Theπ − terms are π ρ π ρ µ π π π φ π π α 1 2 2 2 3 4 5 7 = = = = = = = F V c V c c h c r c w D , , , , , , . 6 Then, F V c f V c c h c r c w D ρ ρ µ φ α 2 2 1 = , , , , , 6.20 [ ] [ ] [ ] [ ] [ ] [ ] 3 2 2 ( , , , , ). , , , , 1, . L L Q f R A e S g Q R L A L e L s g T T = = = = = = = There are only two basic dimensions. Choose two repeating variables, R and g. Then, π π π π 1 2 3 4 1 1 2 2 3 3 4 4 = = = = QR g AR g eR g sR g a b a b a b a b , , , . ∴ = = = = π π π π 1 5 2 2 2 3 4 Q gR A R e R s / , , , . ∴ = ∴ = π π π π 1 1 2 3 4 5 1 2 f Q gR f A R e R s ( , , ). , , . 6.21 [ ] [ ] [ ] [ ] 2 2 3 ( , , , ). , , , , . p p L L M M V f h g V h L g T T T L σ ρ σ ρ = = = = = = Repeating variables: h g V h g h g p a b c a b c , , . , . ρ π ρ π σ ρ ∴ = = 1 2 1 1 1 2 2 2 ∴ = = ∴ = π π σ ρ σ ρ 1 2 2 1 2 V hg gh V gh f gh p p , . . 6.22 F f V e I d D = ( , , , , , ). µ ρ Repeating variables: V d , , . ρ [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2 3 , , , , , 1, . D ML L M M F V e L I d L T LT T L µ ρ = = = = = = = π ρ π µ ρ π ρ π ρ 1 2 3 4 1 1 1 2 2 2 3 3 3 4 4 4 = = = = F V d V d e V d I V d D a b c a b c a b c a b c , , , . ∴ = = = = π ρ π µ ρ π π 1 2 2 2 3 4 F V d V d e d I D , , , . ∴ = F V d f V d e d I D ρ µ ρ 2 2 1 , , . 6.23 F f V D g D s = ( , , , , , ). ρ ρ µ Repeating variables: V D , , . ρ [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2 3 3 2 , , , , , , . D s ML L M M M L F V D L g T LT T L L T ρ ρ µ = = = = = = =
- 109. 108 π ρ πρ ρ π µ ρ π ρ 1 2 3 4 1 1 1 2 2 2 3 3 3 4 4 4 = = = = F V D V D V D gV D D a b c s a b c a b c a b c , , , . ∴ = = = = π ρ π ρ ρ π µ ρ π 1 2 2 2 3 4 2 F V D VD gD V D s , , , . ∴ = F V D f VD gD V D s ρ ρ ρ µ ρ 2 2 1 2 , , . 6.24 F f V d e r c D = ( , , , , , , ). µ ρ Repeating variables: V d , , . ρ [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] . F ML T V L T M LT M L d L e L r L c L D = = = = = = = = 2 3 2 1 µ ρ π ρ π µ ρ π ρ π ρ π ρ 1 2 3 4 5 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 = = = = = F V d V d eV d rV d cV d D a b c a b c a b c a b c a b c , , , , . ∴ = = = = = π ρ π µ ρ π π π 1 2 2 2 3 4 5 2 F V d Vd e d r d cd D , , , , . ∴ = F V d f Vd e d r d cd D ρ µ ρ 2 2 1 2 , , , . 6.25 f g V d f T M LT M L V L T d L = = = = = = ( , , , ). [ ] , [ ] , [ ] , [ ] , [ ] . µ ρ µ ρ 1 3 Repeating variables, V d f V d V d a b c a b c , , . , ρ π ρ π µ ρ 1 2 1 1 1 2 2 2 = = ∴ = = ∴ = π π µ ρ µ ρ 1 2 1 fd V Vd fd V g Vd , . . 6.26 F f V c t L c = ( , , , , ). , ρ α l Repeating variables: V c , , . ρ l [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] . F ML T V L T c L T M L L t L L c = = = = = = = 2 3 1 ρ α l π ρ π ρ π ρ π α ρ 1 2 3 4 1 1 1 2 2 2 3 3 3 4 4 4 = = = = F V cV tV V L a b c c a b c c a b c c a b c c l l l l , , , . ∴ = = = = π ρ π π π α 1 2 2 2 3 4 F V c V t L c c l l , , , . ∴ = F V f c V t L c c ρ α 2 2 1 l l , , . 6.27 2 2 3 1 ( , , , , ). [ ] , [ ]= ,[ ]= , [ ] , [ ] , [ ] . ML M M T f d t T d L t L T LT T L ω ρ µ ω ρ µ = = = = = Repeating variables: d Td d t d a b c a b c a b c , , . , , . 3 ω ρ π ω ρ π µ ω ρ π ω ρ 1 2 1 1 1 2 2 2 3 3 3 = = =
- 110. 109 1 2 3 25 2 , , . T t d d d µ π π π ρω ρω ∴ = = = 3 5 1 1 2 5 2 2 , . , . T t t f W d f d d d d d µ µ ρω ρω ρω ρω ∴ = = & 6.28 F f V d L D c = ( , , , , , , ) ρ µ ρ ω where d is the cable diameter, L the cable length, ρc the cable density, andω the vibration frequency. Repeating variables: V d , , . ρ The π − terms are π ρ π ρ µ π π ρ ρ π ω 1 2 2 2 3 4 5 = = = = = F V d Vd d L V d D c , , , , We then have F V d f Vd d L V d D c ρ ρ µ ρ ρ ω 2 2 1 = , , , 6.29 ∆p f D h d d = ( , , , , , ). ω ρ 1 0 Repeating variables: D, , . ω ρ [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] ∆p M LT D L h L T M L d L d L = = = = = = = 2 3 1 0 1 ω ρ π ρω π π π 1 2 2 2 3 1 4 0 = = = = ∆p D h D d D d D , , , . ∴ = = ∆p D f h D d D d D W ρω 2 2 1 1 0 , , . & force × velocity = ∆pD D 2 ×ω . ∴ = & , , . W D f h D d D d D ρω3 5 1 1 0 6.30 T g f d H N h = ( , , , , , , ). , ω ρ l Repeating variables: ω ρ , , . d [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] . T ML T f T T d L H L L N h L M L = = = = = = = = = 2 2 3 1 1 1 ω ρ l π ρω π ω π π π π 1 2 5 2 3 4 5 6 = = = = = = T d f H d d N h d , , , , , . l ∴ = T d g f H d N h d ρω ω 2 5 1 , , , , . d l 6.31 Q f H w g = ( , , , , , ). µ ρ σ Repeating variables: H g , , . ρ [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] . Q L T H L w L g L T M LT M L M T = = = = = = = 3 2 3 2 µ ρ σ ∴ = = = = π π π µ ρ π σ ρ 1 5 2 3 3 4 2 Q gH w H gH gH , , , .
- 111. 110 ∴ = Q gH f w H gHgH 5 1 3 2 , , . µ ρ σ ρ 6.32 d f V V D j a = ( , , , , , , ). σ ρ µ ρ Repeating variables: V D j , , . ρ [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] . d L V L T V L T D L M T M L M LT M L j a = = = = = = = = σ ρ µ ρ 2 3 3 π π π σ ρ π µ ρ π ρ ρ 1 2 3 2 4 5 = = = = = d D V V V D V D j j j a , , , , . ∴ = d D f V V V D V D j j j a 1 2 , , , . σ ρ µ ρ ρ ρ 6.33 T f H h R t = ( , , , , ). ω µ ρ , , Repeating variables: ω ρ , , . h [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] . T ML T T H L h L R L t L M LT M L = = = = = = = = 2 2 3 1 ω µ ρ π ρω π π π π µ ρω 1 2 5 2 3 4 5 2 = = = = = T d H h R h t h h , , , , ∴ = T d f H h R h t h h ρω µ ρω 2 5 1 2 , , , . 6.34 µ ρ = f D H g V ( , , , , , ) l . D = tube dia., H = head above outlet, l = tube length. Repeating variables: D V VD H D D gD V , , . , , , ρ π µ ρ π π π 1 2 3 4 2 = = = = l ∴ = µ ρVD f H D D gD V 1 2 , , . l 6.35 T f R e r = ( , , , , ). , , ω ρ µ l Repeating variables: R, , . ω ρ [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] . T ML T R L T M L e L r L M LT L = = = = = = = = 2 2 3 1 ω ρ µ l π ρω π π π π µ ρω 1 2 5 2 3 4 5 2 = = = = = T R e R r R R R , , , , l 1 2 5 2 , , , . T e r f R R R R R µ ρω ρω ∴ = l 6.36 y f V y g 2 1 1 = ( , , , ). ρ Neglect viscous wall shear. [ ] , [ ] , [ ] , [ ] , [ ] . y L V L T y L M L g L T 2 1 1 3 2 = = = = = ρ Repeating variables: V y 1 1 , , . ρ
- 112. 111 π π 1 2 1 2 1 1 2 = = y y gy V ,. ( ρ does not enter the problem). ∴ = y y f gy V 2 1 1 1 2 . 6.37 f g d V f T d L L M L M LT V L T = = = = = = = ( , , , , ). [ ] , [ ] , ] , [ ] , [ ] , [ ] . [ l l ρ µ ρ µ 1 3 Repeating variables: d V , , . ρ (l = length of cylinder). π π π µ ρ µ ρ 1 2 3 1 . , , . , . = = = ∴ = fd V d Vd fd V f d Vd l l 6.38 Q Q V V p p V V F F V V m p m m p p m p m m p p p m p p m m m p p p = = = l l l l 2 2 2 2 2 2 2 2 , , ( ) ( ) ∆ ∆ ρ ρ ρ ρ τ τ ρ ρ ρ ρ ρ ρ m p m m p p m p m m m p p p m p m m m p p p V V T T V V Q Q V V = = = 2 2 2 3 2 3 3 2 3 2 , , & & l l l l ( & Q has same dimensions as .) W & 6.39 (A) Re Re . m m m p m V L ν = p p p V L ν = . 12 9 108 m/s. p m p m L V V L ∴ = = × = 6.40 A) 5 6 1.51 10 Re Re . . 4 10 461 m/s. 1.31 10 p p p m m m m p m p m p m p V L L V L V V L ν ν ν ν − − × = = ∴ = = × = × 6.41 a) Re Re . . . m p m m m p p p m p p m V d V d V V d d = = ∴ = = ν ν 7 Q Q V V Q Q V V m p m m p p m p m p m p = ∴ = = × × = l l l l 2 2 2 2 2 15 7 1 7 0 214 . . . . m / s 3 & & . & . W W V V W m p m m m p p p m = = × = ∴ = × = ρ ρ 3 2 3 2 3 2 7 1 7 7 7 200 1400 l l kW b) Re Re . . . . . m p m p p m m p V V d d = ∴ = = × = ν ν 7 9 13 4 85 3 2 3 2 1 1.5 4.85 0.148 m /s. 7 1 4.85 200 466 kW 7 m m Q W = × × = = × × = &
- 113. 112 6.42 a) ReRe . . . m p m m m p p p m p p m V d V d V V d d = = ∴ = = ν ν 5 2 2 2 1 1 5. 800/5 160 kg/s. 5 5 m m m m m p p p p p m V m m m V ρ ρ = = × ∴ = = = & l & & & l ∆ ∆ ∆ ∆ p p V V p p m p m m p p m p = = ∴ = = × = ρ ρ 2 2 2 5 25 25 600 15 . . 000 kPa b) Re Re . . . . . m p m p p m m p V V d d = ∴ = = × = ν ν 5 8 114 3 51 2 2 1 800 3.51 112 kg/s. 600 3.51 7390 kPa. 5 m m m p = × × = ∆ = × = & 6.43 a) Re Re . . . m p m m m p p p m p p m V d V d V V d d = = ∴ = = ν ν 10 F F V V F F m p m m m p p p m p = = × ∴ = = ρ ρ 2 2 2 2 2 10 10 l l 1 10 = 1. lb 2 . b) Re Re . . .41 . . m p m p p m m p V V d d = ∴ = = × = ν ν 10 1 06 1 7 52 p p m F F ρ = m ρ 2 2 2 2 2 2 1 10 10 17.68 lb. 7.52 p p m m V L V L = × × = 6.44 Re Re . . . m p m m m p p p m p p m m p m p V V V V = = ∴ = = = assuming l l l l ν ν ν ν ν ν 10 1 ∴ = = V V m p 10 1000 km / hr. This velocity is much too high for a model test; it is in the compressibility region. Thus, small-scale models of autos are not used. Full-scale wind tunnels are common. 6.45 Re Re . . . m p m m m p p p m p p m m p V V V V = ∴ = ∴ = l l l l ν ν ν ν Water: V V V V m p p m m p m p = = = ∴ = = l l 10 10 900 assuming km / hr. ν ν . Air: V V m p p m m p = = × × × = − − l l ν ν 90 10 15 10 1 10 13 5 6 . 500 km / hr. Neither a water channel or a wind tunnel is recommended. Full-scale testing in a water channel is suggested.
- 114. 113 6.46 Re Re. . / / . m p m m m p p p m p p m m p V V V V = = ∴ = = = if l l l l ν ν ν ν 10 ∴ = × = Vm 10 50 500 m / s. This is in the compressibility range so is not recommended. Try a water channel for the model study. Then 6 5 1 10 10 0.662. 1.5 10 p m m p m p V V ν ν − − × = = × = × l l 33.1 m/s. m V ∴ = This is a possibility, although 33.1 m/s is still quite large. 2 2 2 2 2 2 ( ) 1000 1 0.662 3.56. ( ) 1.23 10 D m m m m D p p p p F V F V ρ ρ = = × × = l l 6.47 5 3 1.06 10 Re Re . . 2.5 1 5.5 10 p p p m m m m p m p m p m p V d V V d d d V ν ν ν ν − − × = = ∴ = = × × × = 0.0048 ft. Find oil ν using Fig. B.2. Then 2 2 2 1.94 1 1.11. 1.94 0.9 m m m p p p p V p V ρ ρ ∆ = = × = ∆ × 6.48 Re Re . . . . . m p m m m p p p m p p m m p p m V V V V = = ∴ = = × × × − l l l l l l ν ν ν ν 0 1 025 10 3 If l l l p p m m V ≅ = = = 5 5 0025 2000 0 005 cm, then and m /s. . . We could try lp m V ≅ = 50 0 05 cm, but m /s. . Each of these Vm' s is quite small — too small for easy measurements. Let’s try a wind tunnel. Then, V V m p p m m p p m p = = × × × × × = = − − − l l l l l ν ν 01 025 10 1 10 18 10 0 28 5 3 3 5 . . . . m / s if cm. Or, if lp m V = = 50 2 8 cm, m / s. . This is a much better velocity to work with in the lab. Thus, choose a wind tunnel. 6.49 Re Re . . . . . m p m m m p p p m p m m m p p p m p V V V g V g V V = ∴ = = ∴ = ∴ = Fr Fr l l l l ν ν 2 2 1 30 V V m p p m m p m p m p m = = = ∴ = ∴ = × − l l ν ν ν ν ν ν ν 30 1 30 1 164 6 1 10 9 . . . . m / s Impossible! 2 6.50 (C) 2 Fr Fr . m m p m m V l g = 2 p p p V l g = 1 . 2 0.5 m/s. 4 m m p p l V V l ∴ = = × =
- 115. 114 6.51 (A) FromFroude’s number m m p p l V V l = . From the dimensionless force we have: 2 2 * * 2 2 2 2 2 2 2 or . 10 25 25 156000 N. p p p m m p p m m m m p p p m m F V l F F F F F V l V l V l ρ ρ = = ∴ = = × × = 6.52 Fr Fr m /s m p m m m p p p m p m p V g V g V V = ∴ = ∴ = = = . . . . 2 2 10 1 60 1 29 l l l l ( ) ( ) . ( ) ( ) . . F F V V F V V F D m D p m m m p p p D p p m p m D m = ∴ = × = × × = × ρ ρ 2 2 2 2 2 2 2 2 2 6 60 60 10 216 10 l l l l N 6.53 Fr Fr m p m m m p p p m p m p V g V g V V = = ∴ = . . 2 2 l l l l . a) Q Q V V Q Q V V m p m m p p m p m p m p = ∴ = = × × = l l l l 2 2 2 2 2 2 1 10 1 10 0 00632 . . . m / s 3 b) F F V V F F V V m p m m m p p p p m p m p m = ∴ = = × × = ρ ρ 2 2 2 2 2 2 2 2 2 12 10 10 l l l l . . 12 000 N 6.54 Neglect viscous effects. Fr Fr fps m p m p m p p V V V = ∴ = = ∴ = . . . . l l 1 10 63 2 F F V V F F V V m p m m m p p p p m p m p m = ∴ = = × × = ρ ρ 2 2 2 2 2 2 2 2 2 0 8 10 10 800 l l l l . . . lb 6.55 Neglect viscous effects, and account for wave (gravity) effects. Fr Fr m p m m m p p p m p m p m p m m p p V g V g V V V V = ∴ = ∴ = = . . . / / . 2 2 l l l l l l ω ω ∴ = = × × = ω ω m p m p p m V V l l 600 1 10 10 1897 rpm. T T V V T T V V m p m m m p p p p m p m p m = ∴ = = × × = ⋅ ρ ρ 2 3 2 3 2 2 3 3 3 1 2 10 10 120 l l l l . . . 000 N m 6.56 Fr Fr 6 100 m p m m m p p p m p m p m p p m V g V g V V = ∴ = ∴ = = ∴ = . . . . . 2 2 278 l l l l l l l l
- 116. 115 6.57 Check theReynolds number: Re . p p p p V d = = × = × − ν 15 2 10 30 10 6 6 This is a high-Reynolds-number flow. Re / . . m = × = × − 2 2 30 10 1 33 10 6 5 This may be sufficiently large for similarity. If so, & & . . W W V V m p m m m p p p = = × = × − ρ ρ 3 2 3 2 3 3 2 6 2 15 1 30 2 63 10 l l ∴ = × × = − & ( . ) / . . Wp 2 2 15 2 63 10 1633 6 kW 6.58 This is due to the separated flow downwind of the stacks, a viscous effect. ∴ Re is the significant parameter. Re . . . p = × × = × − 10 4 15 10 26 7 10 5 5 This is a high-Reynolds-number flow. Let’s assume the flow to be Reynolds number independent above Re = × 5 105 (see Fig. 6.4). Then 5 5 4/20 Re 5 10 . 37.5 m/s. 1.5 10 m m m V V − × = × = ∴ ≥ × 6.59 Re . . . p = × × = × − 20 10 15 10 13 3 10 5 6 This is a high-Reynolds-number flow. Let Re .4 . . . m m m V V = = × × ∴ ≥ − 10 15 10 3 75 5 5 m / s for the wind tunnel. Re . . . m m m V V = = × × ∴ ≥ − 10 1 1 10 1 0 5 6 m / s for the water channel. Either could be selected. The better facility would be chosen. F F V V F F m m m m m m m m m m 1 2 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 3 2 1000 1 23 2 15 1 416 = = ∴ = × = ρ ρ l l . . . . .4 . .4 . . N & & . . & ( . ) . . W W V V W m p m m m p p p p = = × × ∴ = × × = ρ ρ 3 2 3 2 3 2 3 2 3 3 2 2 15 4 20 10 15 3 2 20 15 10 4 71 l l 100 W 6.60 Re is the significant parameter. This is undoubtedly a high-Reynolds- number flow. If the model is 4' high then l l p m = 250, and the model’s diameter is 45/250 = 0.18'. For Re , m = × 3 105 we have Re . . . m m m V V = × = × × ∴ ≥ − 3 10 18 15 10 250 5 4 fps,and a study is possible.
- 117. 116 6.61 Mach No.is the significant parameter. M M m p = . a) M M . . 200 m/s. p m m p m p m p V V V V c c = ∴ = ∴ = = F F V V F m p m m m p p p p = ∴ = × × = ρ ρ 2 2 2 2 2 2 10 1 20 4000 l l . . N b) 255.7 200 186 m/s. 296 p p p m m m m c T V V V c T = = = = F F V V p m m p p m m m = = × × × = ρ ρ 2 2 2 2 2 2 2 10 601 186 200 20 2080 l l . . N c) 223.3 200 174 m/s. 296 p p p m m m m c T V V V c T = = = = F F V V p m p m p m p m = = × × × = ρ ρ 2 2 2 2 2 2 2 10 338 174 200 20 1023 l l . . N 6.62 273 M M . . 250 276 m/s. 223.3 p m m p m m p V V V c c = ∴ = ∴ = = 223.3 . 290 262 m/s. 273 m m m p p p p V c T V V c T = = ∴ = = p p V V p p V V m p m m p p p m p m p m o o = ∴ = = = ρ ρ ρ ρ ρ ρ 2 2 2 2 2 2 80 338 8 262 290 34 6 . . . . . kPa, abs α p = 5o for similarity. (Note: we use ρm at 2700 m where T = 0°C.) 6.63 a) Fr Fr m p m m m p p p m p m p V g V g V V = = ∴ = . . . 2 2 l l l l ω ω ω m p m p p m m V V = = × ∴ = × = l l 1 10 10 2000 10 10 6320 . . rpm b) Re Re . . . m p m m m p p p m p p m V V V V = = ∴ = = l l l l ν ν 10 ω ω ω m p m p p m m V V = = × = ∴ = l l 10 1 10 1 2000 . . rpm 6.64 There are no gravity effects nor compressibility effects. It is a high-Re flow. T T V V T T V V m p m m m p p m p m p m p m = ∴ = = × × = ⋅ ρ ρ 2 3 2 3 2 2 3 3 2 2 3 12 15 60 10 750 l l l l . . N m
- 118. 117 ω ω ω ω m p m p p m p m p m m p V V V V =∴ = = × × = l l l l . . . rpm 500 15 60 1 10 12 5 6.65 Re Re . . m p m m m p p p m p p m V V V V = ∴ = ∴ = = × = m / s. l l l l ν ν 10 10 100 This is too large for a water channel. Undoubtedly this is a high-Re flow. Select a speed of 5 m/s. For this speed, Re . , m = × × = × − 5 01 1 10 5 10 6 5 where we used 0.1 ( 1 m, m p = = l l i.e., the dia. of the porpoise). ω ω m p m p p m V V = = × × = l l 1 5 10 10 5 motions / second. 6.66 ρ ρ ρ * * * * * * , , , , , . = = = = = = o t tf u u V v v V x x y y l l Substitute in: * * * * * o * * * ( ) ( ) 0. o o V u V v f t x y ∂ρ ∂ ρ ∂ ρ ρ ρ ρ ∂ ∂ ∂ + + = l l Divide by ρoV / : l ∴ + + = f V t x u y v f V l l ∂ρ ∂ ∂ ∂ ρ ∂ ∂ ρ * * * * * * * * ( ) ( ) . . 0 parameter = 6.67 * * * * * * * * * 2 , , , , , , , , . V u v w x y z p V u v w x y z p t tf U U U U U ρ = = = = = = = = = v v l l l Substitute into Euler’s equation and obtain: Uf V t U u V x U v V y U w V z U p ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ρ ρ v l v l v l v l v * * * * * * * * * * * * * . + + + = − ∇ 2 2 2 2 Divide by U2 / : l f U V t u V x v V y w V z p l v v v v ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ * * * * * * * * * * * * * . + + + = −∇ Parameter = f U l 6.68 v v l v l v l V V U t tU p p U h h * * * * * , , , , . = = ∇ = ∇ = = ρ 2 Euler’s equation is then ρ ρ ρ U DV Dt U p g h 2 2 l v l v l l v * * * * * * . = − ∇ − ∇ Divide by ρU 2 / : l DV Dt p g U h v v l v * * * * * * . = −∇ − ∇ 2 Parameter = g U l 2 . 6.69 There is no y- or z-component velocity so continuity requires that∂ ∂ u x / . = 0 There is no initial pressure distribution tending to cause motion so∂ ∂ p x / . = 0 The
- 119. 118 x-component Navier-Stokes equationis then u u u t x ∂ ∂ ∂ ∂ + v + u w y ∂ ∂ + 1 u p z x ∂ ∂ ∂ ρ ∂ = − 2 2 x u g x ∂ ν ∂ + + 2 2 2 2 u u y z ∂ ∂ ∂ ∂ + + (wide plates) This simplifies to ∂ ∂ ν ∂ ∂ u t u y = 2 2 . a) Let u u U y y h t tU h * * * / , / / . = = = and Then U h u t U h u y 2 2 2 ∂ ∂ ν ∂ ∂ * * * *2 = The normalized equation is ∂ ∂ ∂ ∂ u t u y * * * *2 Re = 1 2 where Re = Uh ν b) Let u u U y y h t t h * * * / , / / . = = = and Then ν 2 ν ∂ ∂ ν ∂ ∂ U h u t U h u y 2 2 2 * * * *2 = The normalized equation is ∂ ∂ ∂ ∂ u t u y * * * *2 = 2 6.70 The only velocity component is u. Continuity then requires that ∂ ∂ u x / = 0 (replace z with x and vz with u in the equations written using cylindrical coordinates). The x-component Navier-Stokes equation is r u v t ∂ ∂ + v u r θ ∂ ∂ + u u u r x ∂ ∂ ∂θ ∂ + 1 x p g x ∂ ρ ∂ = − + 2 2 2 2 2 1 1 u u u r r r r ∂ ∂ ∂ ν ∂ ∂ ∂θ + + + 2 2 u x ∂ ∂ + This simplifies to ∂ ∂ ρ ∂ ∂ ν ∂ ∂ ∂ ∂ u t p x u r r u r = − + + 1 1 2 2 a) Let * * * * 2 * / , / , / , = / and / : u u V x x d t tV d p p V r r d ρ = = = = V d u t V d p x V d u r r u r 2 2 2 2 1 ∂ ∂ ρ ρ ∂ ∂ ν ∂ ∂ ∂ ∂ * * * * * *2 * * * = − + + The normalized equation is ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ u t p x u r r u r * * * * * *2 * * * Re = − + + 1 1 2 where Re = Vd ν b) Let * * * 2 * 2 * / , / , / , = / and / : u u V x x d t t d p p V r r d ν ρ = = = = ν ∂ ∂ ρ ρ ∂ ∂ ν ∂ ∂ ∂ ∂ V d u t V d p x V d u r r u r 2 2 2 2 1 * * * * * *2 * * * = − + + The normalized equation is
- 120. 119 ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ u t p x u r r u r * * * * * *2 * * * Re =− + + 2 1 where Re = Vd ν 6.71 Assume w z = = 0 0 and ∂ ∂ . The x-component Navier-Stokes equation is then u t ∂ ∂ u u v x ∂ ∂ + + u u w y z ∂ ∂ ∂ ∂ + 1 p x ∂ ρ ∂ = − 2 2 2 2 2 2 x u u u g x y z ∂ ∂ ∂ ν ∂ ∂ ∂ + + + + With g g x = the simplified equation is u u x g u x u y ∂ ∂ ν ∂ ∂ ∂ ∂ = + + 2 2 2 2 Let u u V x x h y y h * * * / , / / . = = = and Then V h u u x g V h u x u y 2 2 2 2 * * * * *2 * *2 ∂ ∂ ν ∂ ∂ ∂ ∂ = + + The normalized equation is u u x u x u y * * * * * * * Re ∂ ∂ ∂ ∂ ∂ ∂ = + + 1 1 2 2 2 2 2 Fr where Fr and = = V hg Vh Re ν 6.72 * * * * * *2 2 2 , , , , , . o u v T x y u v T x y U U T = = = = = ∇ = ∇ l l l ρ ∂ ∂ ∂ ∂ c UT T x UT T y K T T p o o o l l l * * * * * * . + = ∇ 2 2 Divide by ρc UT p o / : l ∂ ∂ ∂ ∂ ρ T x T y K c U T p * * * * * * . + = ∇ l 2 Parameter = K c U p µ µ ρ l = 1 1 Pr Re . 6.73 ρ ρ ρ * * * * * * * , , , , , , . = = = ∇ = ∇ ∇ = ∇ = = o o o V V U t tU p p p T T T v v l v l v l 1 1 2 2 2 momentum: ρ ρ µ µ o o U DV Dt p p U V U V * * * * * * * * * * ( ). 2 2 2 2 3 l v l v l v l v v v = − ∇ + ∇ + ∇ ∇ ⋅ Divide by ρoU 2 / : l [ ] ρ ρ µ ρ * * * * * * * * * * ( ) . DV Dt p U p U V V o o o v v l v v v v = − ∇ + ∇ + ∇ ∇ ⋅ 2 2 energy: ρ ρ * * * * * * * * . c T U DT Dt K T T p U p V v o o o o l l l v v = ∇ − ∇ ⋅ 2 2 Divide by ρo v o c T U / : l ρ ρ ρ * * * * * * * * . DT Dt K c U T p c T p V o v o o v o = ∇ − ∇ ⋅ l v v 2
- 121. 120 The parameters are: p U RT U kRT kU c kUk o o o o ρ 2 2 2 2 2 2 1 = = = = M . µ ρoUl = 1 Re . K c U K c c c U K o v p p v o ρ µ µ ρ l l = = Pr Re . p c T RT c T c c c K o o v o o v o p v v ρ = = − = −1. The significant parameters are K, , Re, Pr. M
- 122. 162 CHAPTER 8 External Flows 8.1 8.2Re . . . . = = ∴ = × × = × − − 5 5 1 51 10 20 3 78 10 5 5 VD D ν m 8.3 8.4 8.5 (C) 8.6 (C) A B C A-B: favorable B-C: unfavorable A-C: favorable separated region inviscid flow boundary layer near surface inviscid flow viscous flow near sphere no separation separation wake separation separated region building inviscid flow boundary layer wake A-B: favorable B-C: unfavorable A-D: favorable C-D: undefined A B C D separated flow
- 123. 163 8.7 ( B) 6 0.80.008 Re 4880. 1.31 10 VD ν − × = = = × 8.8 5 5 5 1 22 10 8 12 000915 5 = ∴ = = × × = − VD V D V ν ν a) fps. / . . / . . b) V = × × = − 5 388 10 8 12 000291 5 . / . . fps. c) V = × × = − 5 1 6 10 8 12 0 012 4 . / . . fps. 8.9 Re . . . = = × × = × − VD D D ν 20 1 51 10 13 25 10 5 5 a) Re . . . = × × = × 13 25 10 6 7 9 10 5 6 ∴Separated flow. b) Re . . . . = × × = × 13 25 10 06 7 9 10 5 4 ∴Separated flow. c) Re . . . = × × = 13 25 10 006 7950 5 ∴Separated flow. 8.10 F pdA p A p r rdr p p D A = − = − = − = ∫ ∫ back back front 0 2 0 0 0 1 1 2 2 1 2 1 4 1 2 ( ) π π π Bernoulli: p V p p ∞ ∞ + = ∴ = × × = 1 2 1 2 1 21 20 242 2 0 0 2 ρ . . Pa. ∴ = = FD 1 2 242 380 π( ) N C F V A D D = = × × × × = 1 2 2 2 2 2 380 1 21 20 1 0 5 ρ π . . 8.11 F F F total bottom top 000 .3 .3 +10 000 N. = + = × × × × = 20 3 3 2700 . . Flift cos 10 N = = 2700 2659 o Fdrag 10 N = = 2700 469 sin o C F V A L L = = × × × × = 1 2 2 2 2 2659 1000 5 3 3 2 36 ρ . . . C F V A D D = = × × × × = 1 2 2 2 2 469 1000 5 3 3 0 417 ρ . . . 8.12 F p A Lw F p A Lw Lw u u u l l l o = = × = = × = 26 8000 2 5 4015 000 . cos F F F Lw L u = − = l o o cos cos 5 10 21 950 F F F Lw D u = − = l o o sin sin 5 10 1569
- 124. 164 C F V A Lw Lw L L = = × × = 1 2 2 2 221 3119 750 0 25 ρ 950 . . C F V A Lw Lw D D = = × × = 1 2 2 2 2 1569 3119 750 0 0179 ρ . . 8.13 If CD = 1 0 . for a sphere, Re = 100 (see Fig. 8.8). ∴ × = V . , 1 100 ν ν V = 1000 . a) V FD = × × = ∴ = × × × × − 1000 1 46 10 0146 1 2 1 22 0146 05 1 0 5 2 2 . . . . . . m / s. π = × − 3 25 10 7 . . N b) V FD = × × × = ∴ = × × × × × − 1000 1 46 10 015 1 22 0 798 1 2 015 1 22 798 05 1 0 5 2 2 . . . . (. . ) . . . m /s. π = × − 4 58 10 5 . . N c) V FD = × × = ∴ = × × × × − 1000 1 31 10 00131 1 2 1000 00131 05 1 0 6 2 2 . . . . . m /s. π = × − 6 74 10 6 . . N 8.14 a) Re . . . . = = × × = × ∴ = − VD CD ν 6 5 1 5 10 2 10 0 45 5 5 from Fig. 8.8. ∴ = = × × × × × = F V AC D D 1 2 1 2 1 22 6 25 45 1 94 2 2 2 ρ π N. . . . . b) Re . . . . = × × = × ∴ = − 15 5 1 5 10 5 10 0 2 5 5 CD from Fig. 8.8. ∴ = = × × × × × = F V AC D D 1 2 1 2 1 22 15 25 2 5 4 2 2 2 ρ π N. . . . . 8.15 (B) Assume a large Reynolds number so that CD = 0.2. Then 2 2 2 1 1 80 1000 1.23 5 0.2 4770 N. 2 2 3600 D F V AC ρ π × = = × × × × × = 8.16 (D) Assume a Reynolds number of 105. Then CD = 1.2. 2 2 1 1 . 60 1.23 40 4 1.2. 0.0041 m. 2 2 D F V AC D D ρ = ∴ = × × × × × ∴ = 5 6 40 0.0041 Re 1.64 10 . 1.2. The assumption was OK. 10 D VD C ν − × = = = × ∴ =
- 125. 165 8.17 The velocitiesassociated with the two Re's are V D 1 1 5 5 3 10 1 5 10 0445 101 = = × × × = − Re . . ν m /s, V D 2 2 4 5 6 10 1 5 10 0445 20 = = × × × = − Re . . ν m /s. The drag, between these two velocities, is reduced by a factor of 2.5 ( ) ( ) [ ] C C D D high low and = = 0 5 0 2 . . . Thus, between 20 m/s and 100 m/s the drag is reduced by a factor of 2.5. This would significantly lengthen the flight of the ball. 8.18 a) F V AC V C V C D D D D = ∴ = × × ∴ = 1 2 0 5 1 2 00238 2 12 4810 2 2 2 2 ρ π . . . . . Re / . . . : . = = × × = = = × − VD V V C V D ν 4 12 1 6 10 2080 5 98 4 Try fps, Re = 2 105 Try fps, Re= 2.3 105 C V D = = × . : . 4 110 b) C V V D = = × × × ∴ = 0 2 1 2 00238 2 12 2 155 2 2 . : . . . 0.5 fps. π 8.19 4 2 1 2 1000 1 0 267 2 10 2 10 2 2 2 6 5 . . . . . Re . . = × × ∴ = = × = × − V C V C V V D D π Try C V D = ∴ = = × ∴ 0 5 0 73 1 46 105 . : . Re . . m /s. OK. 8.20 Re . . . . = = × × = × ∴ = − VD CD ν 40 2 15 10 53 10 07 5 6 . (This is extrapolated from Fig. 8.8.) ∴ = × × × × × = FD 1 2 122 40 2 60 7 81 2 . ( ) . . 900 N M = 81 900 × 30 = 2 46 106 . . × ⋅ N m 8.21 a) Re . . . . Re . . Re . . 1 5 5 2 5 3 5 25 05 1 08 10 1 2 10 1 8 10 2 4 10 = × × = × = × = × − Assume a rough cylinder (the air is highly turbulent). ( ) ( ) ( ) ∴ = = = C C C D D D 1 2 3 0 7 0 8 0 9 . , . , . . ∴ = × × × × + × × + × × = FD 1 2 1 45 25 05 10 7 075 15 8 1 20 9 1380 2 . (. . . . . . ) . N M = × × × × × + × × × + × × × = ⋅ 1 2 1 45 25 05 10 7 40 075 15 8 27 5 1 20 9 10 25 700 2 . (. . . . . . . ) . N m b) Re . . . . Re . , Re . . 1 5 4 2 5 3 5 25 05 1 65 10 7 6 10 1 14 10 1 5 10 = × × = × = × = × −
- 126. 166 ( ) () ( ) ∴ = = = = × = C C C D D D 1 2 3 8 7 8 101 287 308 1 17 . , . , . . . . . kg / m3 ρ ∴ = × × × × + × × + × × = FD 1 2 1 17 25 05 10 8 075 15 7 1 20 8 1020 2 . (. . . . . . ) . N M = × × × × × + × × × + × × × = ⋅ 1 2 1 17 25 05 10 8 40 075 15 7 27 5 1 20 8 10 19 2 . (. . . . . . . ) . 600 N m 8.22 Atmospheric air is turbulent. ∴Use the "rough" curve. ∴ = CD 0 7 . . F V D V D V V D = = × × × ∴ = × × − 10 1 2 00238 6 7 2000 10 2000 1 6 10 2 5 2 4 . . . . / . . = 2 2 2 2 2 min o 0.0024 = 30 104 11.8 psf. 2 2 p U v ρ ∞ ∴ = − − = − ∴ = ∴ = = V D V D 2 2370 148 0 108 . . . '. fps 8.23 Since the air cannot flow around the bottom, we imagine the structure to be mirrored as shown. Then L D C C D D / / . . . = = ∴ = ∞ 40 5 8 0 66 Re . . . . . . min min = = × × = × ∴ = × = − VD CD ν 30 2 1 5 10 4 10 1 0 66 66 5 6 ∴ = × × × + × × = FD 1 2 1 22 30 2 8 2 20 66 36 2 . . . 000 N 8.24 . B D W F F F + = 3 2 2 3 4 1 4 9810 1000 9810 7.82 . 3 2 3 D r V r C r π π π × + × = × × Re . = × = × − V r Vr 2 10 2 10 6 6 ∴ = V C r D 2 178 a) r V V CD = ∴ = . , . . 05 89 2 m. Re = 105 Assume a smooth sphere. Try C V D = ∴ = × . : . . 5 4 22 m / s. Re = 4.22 105 This is too large for Re. Try C V D = ∴ = × . : . . 2 667 m / s. Re = 6.67 105 OK. b) r V V CD = × = . , . . 025 4 45 2 m. Re = 5 104 Try C V D = = × . : . . 2 4 72 m / s. Re = 2.4 105 OK. c) r V V CD = = . , . . 005 089 2 m. Re = 104 Try C V D = = × . : . . 5 133 m / s. Re = 1.33 104 OK. d) r V V CD = × = . , . . 001 0178 2 m. Re = 2 103 Try C V D = = × . : . . 4 067 m / s. Re = 1.33 103 OK. W FB FD
- 127. 167 8.25 3 2 3 2 410 1 10 4 10 . .077 .00238 62.4 . 3 12 2 12 3 12 B D W D F F F V C S π π π + = × + × = 3 2 4 10/12 Re 5.2 10 . 1 .0139 810 1.6 10 D V V V C S − × = = × ∴ + = × a) S V CD = = . . . 005 219 2 Assume atmospheric turbulence, i.e., rough. Try C V C V D D = = = × ∴ = = . : . . Re . . . . . 4 23 4 1 2 10 3 27 5 fps fps b) S V C C V D D = = = = = × ∴ . . . . : . Re . . 02 1090 4 52 2 7 10 2 5 Try fps OK. c) S V C C V D D = = = = 1 0 58 4 381 2 . . . . : . 200 Try fps 8.26 Assume a 180 lb, 6' sky diver, with components as shown. If V is quite large, then Re > 2 × 105. F F D W = . 1 2 00238 2 3 1 2 1 0 7 2 5 1 2 1 0 7 18 12 2 5 1 0 4 12 4 180 2 × × × × × × × × × + × × + × × = . . . . . . . . . . V +2 π We used data from Table 8.1. ∴ = V 140 fps. 8.27 From Table 8.2 C F V V D D = = × × × = 0 35 1 2 1 22 3 2 0 35 683 2 2 . . . . . . . a) F W D = × × = ∴ = × = . & . 683 80 1000 3600 337 337 80 1000 3600 7500 2 N. W or 10 Hp b) V F W D = = × = ∴ = × = 25 683 25 427 427 25 10 2 m /s. N. 700 W or 14.3 Hp . & . c)V F W D = = × = ∴ = × = 27 8 683 27 8 527 527 27 8 14 2 . . . & . . m /s. N. 700 W or 19.6 Hp 8.28 1 2 1 1 400 1 2 1 1 2 . . . . . F F V AC C D D D D = × = = ρ 1 2 1 2 1 22 2 3 1 1 1 1 400 2 . . ( ) . . . × × × × × = × V ∴ = V 95 . . m / s 8.29 Re ( . . . = = × = × ∴ = VD CD ν 40 4 42 10 0 35 5 000 / 3600)0.6 1.51 10-5 from Fig. 8.8. a) F V AC D D = = × × × × × = 1 2 1 2 1 204 40 0 6 6 0 35 93 6 2 ρ . ( . . . 000 / 3600) N 2 b) F L D D = × = = = 93 6 0 68 63 7 6 0 6 10 . . . / / . . N where c) F L D D = × = = 93 6 0 76 71 1 20 . . . / N where we can use since only one end is free. The ground acts like the mid-section of a 12-m-long cylinder. 3 ft 6 in 8 in. dia. 2.5 ft 6 in 2.5 ft 18 in 1.1 m 1.2 m FW FD Fy Fx
- 128. 168 8.30 a) Curledup, she makes an approximate sphere of about 1.2 m in diameter (just a guess!). Assume a rough sphere at large Re. From Fig. 8.8, CD = 0 4 . : F V AC D D = 1 2 2 ρ 80 9 8 1 2 1 21 0 6 0 4 53 7 2 2 × = × × × × ∴ = . . . . . . . V V π m /s Check Re: Re . . . . . = × × = × ∴ − 53 7 1 2 1 51 10 4 27 10 5 6 OK. b) F V AC D D = 1 2 2 ρ . From Table 8.2, CD = 1.4: 80 98 1 2 121 4 14 4 29 2 2 × = × × × × ∴ = . . . . . . V V π m / s Check Re: Re . . . . = × × = × − 4 29 8 151 10 2 27 10 5 6 Should be larger but the velocity should be close. c) 2 1 2 D D F V AC ρ = 2 2 1 80 9.8 1.21 1 1.4. 17.2 m/s. 2 V V π × = × × × × ∴ = Check Re: Re . . . . = × × = × − 172 1 151 10 114 10 5 6 This should be greater than 107 for CD to be acceptable. Hence, the velocity is approximate. 8.31 With the deflector the drag coefficient is 0.76 rather than 0.96. The required power (directly related to fuel consumed) is reduced by the ratio of 0.76/0.96. The cost per year without the deflector is Cost = (200 000/1.2) × 0.25 = $41,667. With the deflector it is Cost = 41,667 × 0.76/0.96 = $32,986. The savings is $41.667 − 32,986 = $8,800. 8.32 2 2 1 1 .00238 88 (6 2) 1.1 122 lb. 2 2 D D F V AC ρ = = × × × × × = & , . W F V D = × = × = 122 88 10 700 ft - lb sec or 19.5 Hp 8.33 F V AC D D = = × × × × × × = 1 2 1 2 122 27 8 16 05 11 1043 2 2 2 ρ π . ( . . ) . . . N. & . ( . . ) . W F V D = × × = × × × = 2 1043 27 8 16 2 226 W or 1.24 Hp
- 129. 169 8.34 The projectedarea is ( . ) . . 2 0 3 2 4 4 6 + × = m2 F V AC D D = = × × × × = 1 2 1 2 1 18 20 4 6 0 4 434 2 2 ρ . . . N. Since there are two free ends, we use Table 8.1 with L D / / . . , = = 4 1 15 3 47 and approximate the force as FD = × = 434 0 62 269 . . N 8.35 The net force acting up is (use absolute pressure) 3 3 up 4 4 120 0.4 1.21 9.8 0.5 0.4 9.8 2.16 N 3 3 2.077 293 F π π = × × × − − × × = × From a force triangle (2.16 N up and FD to the right), we see that tan / . α = F FD up a) FD = = 2 16 80 0 381 . /tan . . o 0 381 1 2 1 21 0 4 0 2 2 50 2 2 . . . . . . = × × × ∴ = V V π m / s. Check Re: Re . . . . . = × × = × − 2 5 0 8 1 51 10 1 33 10 5 5 Too low. Use CD = 0 5 . : 0 381 1 2 1 21 0 4 0 5 1 58 2 2 . . . . . . = × × × ∴ = V V π m / s b) FD = = 2 16 70 0 786 . /tan . . o 0 786 1 2 1 21 0 4 0 2 3 60 2 2 . . . . . . = × × × ∴ = V V π m / s. Check Re: Re . . . . . = × × = × − 3 6 0 8 1 51 10 1 9 10 5 5 Too low. Use CD = 0 5 . : 0 786 1 2 1 21 0 4 0 5 2 27 2 2 . . . . . . = × × × ∴ = V V π m / s c) FD = = 2 16 60 1 25 . /tan . . o 1 25 1 2 1 21 0 4 0 5 2 2 . . . . . . = × × × ∴ = V V π 2.86 m / s Check Re: Re . . . . . = × × = × − 2 86 0 8 1 51 10 1 5 10 5 5 ∴OK. d) FD = = 2 16 50 1 81 . /tan . . o 1 81 1 2 1 21 0 4 0 5 2 2 . . . . . . = × × × ∴ = V V π 3.45 m / s Check Re: 5 5 3.45 0.8 Re 1.8 10 . 1.51 10− × = = × × Close, but OK. 8.36 Assume each section of the tree is a cylinder. The average diameter of the tree is 1 m. The top doesn't have a blunt end around which the air flows, however,
- 130. 170 the bottom does;so assume L D / ( / ) . = × = 5 2 2 5 So, use a factor of 0.62 from Table 8.1 to multiply the drag coefficient. The force acts near the centroid of the triangular area, one-third the way up. Finally, F d × = 5000 1 2 1 21 5 0 4 0 62 5 3 0 6 5000 2 × × × × + = = . ( ) . . . . . V V 54.2 m /s 8.37 Power to move the sign: F V V AC V D D = × 1 2 2 ρ = × × × × × = 1 2 1 21 11 11 0 72 1 1 11 11 657 2 . . . . . J /s. This power comes from the engine: 657 12 0 3 1 825 10 4 = × × ∴ = × − ( & . . & . 000 1000) kg /s. m m Assuming the density of gas to be 900 kg/m3, 1825 10 10 3600 6 52 1000 900 030 4 . . $683 × × × × × × × = − 8.38 The power expended is F V V D × = × = . ( / ) / . . m / s 25 88 60 3 281 11 18 1 2 121 1118 0 56 1 2 121 0 4 08 3 3 × × × × = × × × × × . . . . . . C V C D D ∴ = V 1347 . . m / s or 30.1 mph 8.39 & . W F V V AC V AC V D D D = × = × = × = 40 746 1 2 1 2 2 3 η ρ ρ ∴ × × = × × × ∴ = 40 746 9 1 2 122 3 035 34 7 3 . . . . . . V V m / s or 125 km / hr 8.40 (C) 5 4 0.02 0.02 Re 5000. St 0.21 . 4 1.6 10 VD fD f V ν − × × = = = ∴ = = = × 4 m/s 42 Hz (cycles/second). distance = 0.095 m/cycle. 42 cycles/s V f f ∴ = = = 8.41 5 .003 40 Re 10 000. 40< 10 000. 0.2< 50 m/s. 1.5 10 V V − × < < < ∴ < × low .003 St = 0.12 = . 8 Hz. .2 f f × ∴ = St =.21= .003 50 Hz. lhigh f f × ∴ = . 3500 The vortices could be heard over most of the range.
- 131. 171 8.42 5 5 6 40 .8.13 10 ft. 1.22 10 VD D D ν − − > = ∴ < × × 10 6 122 10 0020 5 000 < ft or 0.24" VD D D ν = × ∴ > − . . . . 8.43 From Fig. 8.9, Re is related to St. St = f D V V × = × 02 1 . . . Re . . . . = = × × = ∴ − VD V V ν 1 15 10 0 095 5 Try St =.21: m / s. Re = 630. This is acceptable. ∴ = V 0095 . . m / s 8.44 St = Re = Use Fig. 8.9. fD V V VD V = × = × − . . . 002 2 2 10 6 ν Try St = 0.21: m / s Re = 38 10 OK. 3 V = × ∴ . . . 0191 8.45 Let St = 0.21 for the wind imposed vorticies. When this frequency equals the natural frequency, or one of its odd harmonics, resonance occurs: f T L d = / ρ π 2 2 2 2 0.21 10 30 000/7850 0.016 . 0.525 m 0.016 L L π × = × × ∴ = Consider the third and fifth harmonics: f T L d = 3 2 2 / . ρ π ∴ = L 1.56 m. f T L d = 5 2 2 / . ρ π ∴ = L 2.62 m. 8.46 (C) By reducing the separated flow area, the pressure in that area increases thereby reducing that part of the drag due to pressure. Fig. 8.8 Table 8.1 8.47 Re / . . . . . . = × × = × = × × × × × × = − 88 6 12 1 6 10 2 8 10 1 2 00238 88 1 0 8 6 6 12 4 5 2 22 lb. FD The coefficient 1.0 comes from Fig. 8.8 and 0.8 from Table 8.1. & . W F V D = × = × = 22 88 1946 ft - lb /sec or 3.5 Hp ( ) C F W D D streamlined lb. ft - lb sec or 0.12 Hp = ∴ = = 0 035 0 77 67 8 . . . & . . 8.48 Re . . . = = × × = − VD ν 3 08 1 5 10 16 5 000 ∴ = × × × × × × = FD 1 2 122 3 008 2 12 78 0822 2 . ( . ) . . . N The coefficient 1.2 comes from Fig. 8.8 and 0.78 from Table 8.1. ( ) C F D D streamlined N. % reduction = = ∴ = ∴ − × = . . . . . . . 35 0 24 0822 0 24 0822 100 708%
- 132. 172 8.49 Re . . . == × = × ∴ = − VD CD ν 2 0 8 10 1 6 10 0 45 6 6 . from Fig. 8.8. L D CD = = ∴ = × = 4 0 8 5 0 62 0 45 0 28 . . . . . . Because only one end is free, we double the length. F V AC D D = = × × × × × = 1 2 1 2 1000 2 08 2 028 2 2 ρ . . . 900 N If streamlined, CD = × = 0 03 0 62 0 0186 . . . . ∴ = × × × × × = FD 1 2 1000 2 0 8 2 0 0186 2 . . . 60 N 8.50 V = × = 50 1000 3600 13 9 / . m /s. Assume the ends to not be free. ∴Use CD from Fig. 8.8. ( ) Re . . . . . . . . = × × = × ∴ = = − 13 9 0 02 1 5 10 1 85 10 1 2 0 3 5 4 streamlined C C D D & . . . . . W F V V AC D D = × = = × × × × × = 1 2 1 2 1 2 13 9 0 02 20 1 2 773 3 3 ρ W or 1.04 Hp & . . . . Wstreamlined W or 0.26 Hp = × × × × × = 1 2 1 2 13 9 0 02 20 0 3 193 3 8.51 V = × = 50 1000 3600 139 / . m / s. Re . . . . . . = × × = × ∴ = − 13 9 0 3 1 5 10 2 8 10 0 4 5 5 CD We assumed a head diameter of 0.3 m and used the rough sphere curve. FD = × × × × = 1 2 12 139 03 4 04 2 2 . . ( . / ) . . π 3.3 N FD = × × × × = 1 2 12 139 03 4 0035 2 2 . . ( . / ) . . π 0.29 N 8.52 σ ρ γ = − = − × = + = ∞ ∞ p p V V p h p v 1 2 0 7 150 1000 150 2 2 . . 000 1670 1 2 where 000 Pa. atm ∴ = V 20.6 m /s. 8.53 C F V A L L = = × × × × = ∴ ≅ 1 2 200 1000 12 4 10 0 69 3 2 2 ρ α 000 1 2 . . . . o C F F D D D = = × × × × ∴ = N . . . . 0165 1 2 1000 12 4 10 4800 2
- 133. 173 σcrit 000) 1670 1 2 = > ×+ − × × = . ? ( . . . 75 9810 4 101 1000 12 143 2 ∴No cavitation. 8.54 C F V A L L = = × × × × = ∴ = 000 1 2 1 2 50 194 35 16 12 30 105 7 3 2 2 ρ α . . . . . o C F F D D D = = × × × × ∴ = lb . . . . 027 1 2 194 35 16 12 30 1280 2 σcrit 1 2 = > × + − × × × = 16 62 4 16 12 2117 25 144 194 35 182 2 . ? . / . . . . ∴No cavitation. 8.55 p pv ∞ = × + = × = × 9810 5 101 1670 16 106 000 = 150 000 Pa. Pa. Re = 20 .8 10-6 . σ σ = − × × = ∴ = + = + = 150 1000 20 0 74 0 1 3 1 74 52 2 000 1670 1 2 . . ( )( ) . ( . ) . C C D D ∴ = = × × × × × = F V AC D D 1 2 1 2 1000 20 4 52 2 2 2 ρ π . . . 52 000 N Note: We retain 2 sig. figures since CD is known to only 2 sig. nos. 8.56 For a 6° angle of attack we find from Table 8.4 CL = 0 95 . . F V AC L L L = = × × × × × = × 1 2 1 2 1000 15 4 0 4 95 12 98 2 2 ρ . . . . 000 ∴ = L 069 . . m 8.57 ΣF ma a a = − × × = ∴ = . . . . . a) 1.75 m /s2 400 9810 4 3 2 400 9 81 3 π b) 400 9810 4 3 2 400 9 81 1 2 1000 4 3 2 3 3 − × × = + × × × ∴ = π π . . . . . a a 1.24 m /s 2 8.58 F ma V a a F V m V a = = × × − ∴ = − = × − 1 1 1 1000 1 2 1200 0 2 1000 . . . . . F m m a a F V V F V a a = + ∴ = − + − = − ( ) . . 2 2 2 1200 200 1400 is true acceleration. ∴ − × = − − − − × = % . . error = a a a F V F V F V 2 1 2 100 1400 1200 1400 100 16 7%
- 134. 174 8.59 (B) 1 2 2 FromFig. 8.12a 1.1. . L L L F C C V cL ρ = = 2 2 2 1200 9.81 1088. 33.0 m/s. 1.23 16 1.1 L W V V cLC ρ × × ∴ = = = ∴ = × × 8.60 C F V A C L L D = = × × × × = ∴ = = 1 2 1000 9 81 412 80 15 0 496 3 2 0065 2 2 ρ α . . . . . . . . 1 2 o & . . . W F V D = = × × × × × = 1 2 412 80 15 0065 80 10 2 300 W or 13.8 Hp 8.61 a) C V V L = = × + × × × ∴ = 122 1500 981 3000 1 2 122 20 2 . . . . . 34.5 m / s b) ( ) C V V L max 50 m /s = = × + × × × ∴ = 1 72 1500 9 81 3000 1 2 412 20 2 . . . . . (at 10 000 m) c) & . . W F V D = = × × × × × = 1 2 412 80 20 0065 80 13 2 700 W or 18.4 Hp where we found CD as follows: ( ) C C L D cruise = × + × × × = ∴ = 1500 9 81 3000 1 2 412 80 20 67 0065 2 . . . . . , from Fig. 8.12. ∴Power = 184 045 409 . . . . = Hp 8.62 C V V L = = × + × × × ∴ = 122 1500 9 81 3000 1 2 1007 20 2 . . . . . 38.0 m / s 8.63 ( ) C C L D cruise = × + × × × = ∴ = = 1500 9 81 3000 1 2 1 007 80 20 0 275 0 275 48 0 0057 2 . . . . . . . ∴ = = × × × × = & . . W F V D 1 2 1 007 80 20 0 0057 29 3 400 W or 39.4 Hp % change = 39 4 18 4 18 4 100 114% . . . − × = increase The increased power is due to the increase in air density.
- 135. 175 8.64 C V V L == × + × × × ∴ = 122 1500 981 9000 1 2 122 20 2 . . . . . 39.9 m / s 8.65 C V V L = = × × × × × ∴ = 172 250 981 1 2 122 60 8 2 . . . . . 000 69.8 m / s 8.66 a) C V V L = = × × × × × ∴ = 172 250 981 1 2 105 60 8 752 2 . . . . . . 000 m / s % change = 75 2 69 8 69 8 100 7 77% . . . . − × = increase b) C V V L = = × × × × ∴ = = × = 172 250 9 81 1 2 1515 60 8 62 6 1013 287 233 1515 2 . . . . . . . . 000 m / s kg / m3 ρ % change = 62 6 69 8 69 8 100 10 3% . . . . − × = − c) C V V L = = × × × × ∴ = = × = 172 250 9 81 1 2 1093 60 8 737 1013 287 323 1093 2 . . . . . . . . 000 m / s kg / m3 ρ % change = 73 7 69 8 69 8 100 5 63% . . . . − × = increase 8.67 For a conventional airfoil assume C C C L D L / . . . = = 47 6 0 3 at 0 3 9 81 1 2 0 526 222 200 30 2 38 10 2 6 . . . . . = × × × × × ∴ = × m m kg & . . . W F V D = = × × × × × = 1 2 0 526 222 200 30 0 3 47 6 490 3 000 W or 657 Hp 8.68 v v v v v v v ∇ × + ⋅∇ + ∇ − ∇ = ∂ ∂ ρ ν V t V V p V ( ) . 2 0 v v v v v v v v v ∇ × = ∇ × = ∇ × ∇ = ∇ × ∇ = ∂ ∂ ∂ ∂ ∂ω ∂ ρ ρ V t t V t p p ( ) . . 1 0 v v v v v ∇ × ∇ = ∇ ∇ × = ∇ ( ) ( ) 2 2 2 V V ω (we have interchanged derivatives) 2 2 1 1 ( ) ( ) ( ) 2 2 V V V V V V ∇ × ⋅∇ = ∇ × ∇ − × ∇× = ∇ × ∇ v v v v v v v v v v v ( ) V ω − ∇ × × v v v ( ) V ω = ∇ ⋅ v v v ( ) V ω − ∇ ⋅ v v v ( ) ( ) V V ω ω + ⋅∇ − ⋅∇ v v v v v v
- 136. 176 , . . xL u U U y ∂ψ ∂ = = ∴ = = ⋅ ∇ − ⋅∇ ∇ ⋅ = ∇ ⋅ ∇ × = ( ) ( ) ( ) . v v v v v v v v v v v V V V ω ω ω where 0 There results: ∂ω ∂ ω ω ν ω v v v v v v v v t V V + ⋅∇ − ⋅∇ − ∇ = ( ) ( ) . 2 0 This is written as D Dt V v v v v v ω ω ν ω = ⋅∇ + ∇ ( ) . 2 8.69 x-comp: ∂ω ∂ ∂ω ∂ ∂ω ∂ ∂ω ∂ ω ∂ ∂ ω ∂ ∂ ω ∂ ∂ ν ω x x x x x y z x t u x v y w z u x u y u z + + + = + + + ∇2 y-comp: ∂ω ∂ ∂ω ∂ ∂ω ∂ ∂ω ∂ ω ∂ ∂ ω ∂ ∂ ω ∂ ∂ ν ∂ ω ∂ ∂ ω ∂ ∂ ω ∂ y y y y x y z y y y t u x v y w z v x v y v z x y z + + + = + + + + + 2 2 2 2 2 2 z-comp: ∂ω ∂ ∂ω ∂ ∂ω ∂ ∂ω ∂ ω ∂ ∂ ω ∂ ∂ ω ∂ ∂ ν ∂ ω ∂ ∂ ω ∂ ∂ ω ∂ z z z z x y z z z z t u x v y w z w x w y w z x y z + + + = + + + + + 2 2 2 2 2 2 8.70 x w y ∂ ω ∂ = v z ∂ ∂ − 0. y u z ∂ ω ∂ = = w x ∂ ∂ − 0. 0. z v u x y ∂ ∂ ω ∂ ∂ = = − ≠ 2 2 z ( ) ; . z z z D D w DT Dt ω ω ω ν ω ν ω = ⋅∇ + ∇ ∴ = ∇ v v If viscous effects are negligible, then D Dt z ω = 0. Thus, for a planer flow, ω z = const if viscous effects are negligible. 8.71 a) v v ∇ × = − + − + − = V w y v i u z w x j v x u y k ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ z $ $ $ . 0 ∴irrotational ∂φ ∂ φ x x x f y = ∴ = + 10 5 2 . ( ) ∂φ ∂ ∂ ∂ y f y y f y C C = = ∴ = + = 20 10 0 2 . . . Let ∴ = + φ 5 10 2 2 x y b) v v ∇ × = + + − = V i j k 0 0 8 8 0 $ $ ( )$ . ∴irrotational ∂φ ∂ φ x y xy f y z = ∴ = + 8 8 . ( , ). ∂φ ∂ ∂ ∂ ∂ ∂ y x f y x f y f f z = + = ∴ = = 8 8 0 . ( ). and ∂φ ∂z df dz z f z C C = = − ∴ = − + = 6 3 0 2 . . . Let
- 137. 177 ∴ = − φ8 3 2 xy z c) v v ∇ × = + + − + + − − + + = − − V i j y x y x x y x x y y x y k 0 0 1 2 2 1 2 2 0 2 2 1 2 2 2 2 2 1 2 2 2 $ $ ( ) ( ) $ . / / ∴irrotational ∂φ ∂ φ x x x y x y f y = + ∴ = + + 2 2 2 2 . ( ) ∂φ ∂ ∂ ∂ ∂ ∂ y x y y f y y x y f y f C C = + + = + ∴ = ∴ = = − 1 2 2 0 0 2 2 1 2 2 2 ( ) . . . . / Let ∴ = + φ x y 2 2 d) v v ∇ × = + + − + − − + = V i j y x x y x y x y k 0 0 2 2 0 2 2 2 2 2 2 $ $ ( ) ( ) ( ) ( ) $ . ∴irrotational ∂φ ∂ φ x x x y n x y f y = + ∴ = + + 2 2 2 2 1 2 . ( ) ( ) l ∂φ ∂ ∂ ∂ ∂ ∂ y y x y y x y f y f y f C C = + = + + ∴ = ∴ = = 2 2 2 2 1 2 2 0 0 . . . . Let ∴ = + φ ln x y 2 2 8.72 ∂ ψ ∂ ∂ ψ ∂ 2 2 2 2 0 x y + = . This requires two conditions on x and two on y. At , . . x L u U U y ∂ψ ∂ = − = ∴ = At , . . x L u U U y ∂ψ ∂ = = ∴ = At y h = − , . = 0 ψ At , = h. y h U ψ = × (See Example 8.9). The boundary conditions are stated as: ∂ψ ∂ ∂ψ ∂ ψ ψ y L y U y L y U x h x h Uh ( , ) , ( , ) , ( , ) , ( , ) . − = = − = = 0 2 8.73 u y y f x v x df dx f x C = = ∴ = + = − = − = ∴ = − + ∂ψ ∂ ψ ∂ψ ∂ 100 100 50 50 . ( ). . . ∴ = − ψ( , ) . x y y x 100 50 (We usually let C = 0.) u x x f y v y df dy f y C = = ∴ = + = = = ∴ = + ∂φ ∂ φ ∂φ ∂ 100 100 50 50 . ( ). . . ∴ = + φ( , ) . x y x y 100 50 y = h y = 0 x = −L y x U
- 138. 178 8.74 a) ψθ = 40 . b) 1 1 1 40 1 0 0 2 r r r r r r r ∂ ∂ ∂ψ ∂θ ∂ ψ ∂θ∂ ∂ ∂ ∂ ∂θ + − = + − = ( ) ( ) . ∴It is incompressible since the above continuity equation is satisfied. Note: The continuity equation is found in Table 5.1. c) ∂φ ∂ ∂ψ ∂θ φ θ r r r nr f = = ∴ = + 1 40 40 . ( ) l ∂φ ∂θ ∂ ∂θ ∂ψ ∂ = = − = ∴ = = f r r f C C 0 0 . . . Let ∴ = φ 40ln r d) v r v a v v r r r r r r r = = = = − = − 40 0 40 40 10 2 , . . θ ∂ ∂ ∴ = r 5 43 . m 8.75 u y y x y x y x f y = = + = ∴ = − + − ∂ψ ∂ ∂φ ∂ φ 20 2 40 2 2 1 . tan ( ). v y x y x f y x x y f y x x y f C C = = − + + = − + + = − + ∴ = = ∂φ ∂ ∂ ∂ ∂ ∂ 40 1 40 20 2 0 2 2 2 2 2 2 / / . . . Let φ = − − 40 1 tan . y x 8.76 a) ∂ ψ ∂ ∂ ψ ∂ ∂ψ ∂ 2 2 2 2 2 2 2 0 10 2 x y x y x y x + = = + − . ( ) ( ). ∂ ψ ∂ 2 2 2 2 2 20 x y x y = − − ( ) − + − 80 2 2 2 3 x y x y ( ) ∂ψ ∂y x y y x y y = − + + + − − 10 10 10 2 2 2 1 2 2 2 ( ) ( ) ( ). ∂ ψ ∂ 2 2 2 2 2 20 y y x y = + − ( ) + + − + − − 40 80 2 2 2 3 2 2 3 y x y y x y ( ) ( ) . ∴ + = + − + + + − + ∂ ψ ∂ ∂ ψ ∂ 2 2 2 2 2 2 2 2 2 2 3 2 2 2 3 2 2 3 20 80 60 80 x y y x y x y x y y x y y x y ( ) ( ) ( ) ( ) = + + − + − + = + − − + = 80 80 80 80 80 80 80 0 2 2 2 2 3 2 2 2 3 3 2 2 3 2 3 2 3 2 2 3 y x y x y x y x y y x y x y y x y y x y ( ) ( ) ( ) ( ) ( ) . b) In polar coord: ψ θ θ θ θ θ ( , ) sin sin sin sin . r r r r r r = − = − 10 10 10 10 2 1 10 10 10 10 2 r r r r f ∂ψ ∂θ θ ∂φ ∂ φ θ θ = − = ∴ = + + cos . cos ( ).
- 139. 179 2 2 1 110 10 10 sin 10sin sin . 0. df df f C r r d r d r r ∂φ ∂ψ θ θ θ ∂θ θ ∂ θ = − + = − = − − = = . ∴ = + = + + φ θ φ 10 1 10 10 2 2 r r x y x x x y cos ( , ) , or where we let r x r x y cos . θ = = + and 2 2 2 c) Along the x-axis, v x = − = ∂ψ ∂ 0 where we let y = 0 in part (a) and u y x y y x y x y = = − + + + = − = ∂ψ ∂ 10 10 20 10 10 0 2 2 2 2 2 2 2 ( ) . with Euler’s Eq: ρ ∂ ∂ ∂ ∂ ρ ∂ ∂ u u x p x x x p x = − ∴ − = − . . 10 10 20 2 3 ∴ = − = − + + = ∫ p x x dx x x C C ρ ρ 200 200 50 100 50 5 3 4 2 . 000. = − + 1000 100 50 50 2 4 x x 000 Pa. (Could have used Bernoulli!) d) Let u x x = = − ∴ = ± ∴ 0 0 10 10 1 2 : . . Stag pts: (1, 0), (−1, 0) 8.77 a) ∂ φ ∂ ∂ φ ∂ ∂ ∂ ∂ ∂ 2 2 2 2 2 2 2 2 2 2 2 2 2 10 10 10 10 2 x y x x x y y y x y x y x x x y + = + + + + = + − + ( )10 ( ( ) + + − + = + − + + − + = ( )10 ( ) ( ) ( ) . x y y y x y x y x x y y x y 2 2 2 2 2 2 2 2 2 2 2 2 2 2 10 2 10 10 20 10 10 20 0 b) Polar coord: φ θ = + 10 5 2 r n r cos . l (See Eq. 8.5.14.) ∂φ ∂ θ ∂ψ ∂θ ψ θ θ r r r r r f r = + = ∴ = + + 10 10 1 10 10 2 cos . sin ( ) 1 10sin 10sin . df f C r r dr ∂φ ∂ψ θ θ ∂θ ∂ = − = − = − − ∴ = . 10 sin 10 . r ψ θ θ ∴ = + ∴ = + − ψ( , ) tan . x y y y x 10 10 1 c) v y y x y = = + ∂φ ∂ 10 2 2 . Along x-axis (y = 0) v = 0. u x x x y = = + + ∂φ ∂ 10 10 2 2 . Along x-axis u x = + 10 10 . Bernoulli: 2 2 V p gz ρ + + 2 2 V p gz ρ ∞ ∞ ∞ = + + (assume ) z z∞ = ( / ) . . 10 10 2 10 2 100 50 2 1 2 2 2 + + = + ∴ = − + x p p x x ρ ρ 100 000 kPa
- 140. 180 d) u x x = =+ ∴ = − ∴ 0 0 10 10 1 : . . Stag pt: (−1, 0) e) y a v = v v u y x ∂ ∂ ∂ ∂ + 0 on axis. x u x a u v x ∂ ∂ = − = + 2 10 10 10 . u y x x ∂ ∂ = + − 2 10 ( 2,0) (10 5) 12.5 m/s . 4 x a ∴ − = − − = − 8.78 2 3 2 5 ( , ) 5 . 2 3 y y u x y y y C y ∂ψ ψ ∂ = − = ∴ = − + 2 3 1 . (3 10 ). 6 y y ψ ∴ = − q udy y y dy = = − = − = × − ∫ ∫ ( ) . . . . . . 5 02 2 5 0 2 3 6 667 10 2 2 3 3 0 2 0 2 m / s 2 ψ ψ 2 1 2 3 3 1 6 3 0 2 10 02 0 6667 10 − = × − × − = × − ( . . ) . . m / s 2 ω ∂ ∂ = − = − + ≠ u y y 1 10 0. ∴φ doesn’t exist. 8.79 ψ π π θ θ θ = + = + 30 5 2 30 5 2 y r sin . a) v r r r = = + = 1 30 5 2 0 ∂ψ ∂θ θ cos . At θ π = = ∴ = , . . '. 5 2 30 0 0833 r r s s Stag. pt: ( " , ). −1 0 b) At θ π ψ π π π = = = + , sin . =.0833, r r s 5 2 30 2 5 2 2 ∴ = = r yinter .0119 ft. c) q U H H H = × = ∴ = ∴ ∆ψ π π . . . = 5 60 30 5 2 Thickness = 2 5 30 H = π ft or 1.257". d) v u r ( , ) cos . . . . . 1 30 5 2 30 2 5 27 5 27 5 π π = + = − + = − ∴ = fps 8.80 [ ] [ ] [ ] φ π π π π = + + − − + + = + + 2 1 2 1 10 1 4 1 2 1 2 2 2 1 2 2 2 l l l n x y n x y x n x y ( ) ( ) ( ) / / [ ] − − + + 1 4 1 10 2 2 ln x y x ( ) . u x x x x x x x v y y = = + + − − − + = + − − + = = = ∂φ ∂ 0 2 2 1 4 2 1 1 1 4 2 1 1 10 1 2 1 1 2 1 10 0 0 [ ( )] ( ) [ ( )] ( ) ( ) ( ) . . if 30 fps y x = 0 = 5π/2
- 141. 181 At the stagnationpoint, u x x x = ∴ + − − + = ∴ − = 0 1 2 1 1 2 1 10 0 2 1 20 2 . ( ) ( ) . . ∴ = ∴ = ± ∴ × x x 2 1 1 1 049 . . . . m. oval length = 2 1.049 = 2.098 m All the flow from the source goes to the sink, i.e., π π m /s, or m / s for 2 2 2 0 y > . u y x y y y x ( ) ( ) ( ) . = = + − − + + = + + = ∂φ ∂ 0 2 2 2 1 4 2 1 1 4 2 1 10 1 1 10 1 2 0 1 10 . tan 10 . 2 2 1 h q dy h h y π π − = + = ∴ + = ∫ + h = 0.143 m so that thickness = 2h = 0.286 m. The minimum pressure occurs on the oval surface at (0,h). There u = + + = 1 1 143 10 10 98 2 . . m /s. Bernoulli: V p V p p 2 2 2 2 2 2 10 98 2 1000 10 2 + = + + = + ∞ ∞ ρ ρ . . . 10 000 1000 Pa min ∴ = − p 280 . 8.81 [ ] [ ] [ ] φ π π π π = + + + − − + + = + + 2 2 1 2 2 1 2 1 2 1 2 2 1 2 2 2 1 2 2 2 l l l n x y n x y x n x y ( ) ( ) ( ) / / [ ] − − + + 1 2 1 2 2 2 ln x y x ( ) . u x x x y x x y v y x y y x y = = + + + − − − + + = − + − − + ∂φ ∂ 1 2 2 1 1 1 2 2 1 1 2 1 1 2 2 2 3 2 2 2 2 ( ) ( ) ( ) ( ) . ( ) ( ) Along the x-axis (y = 0), v = 0 and u x x = + − − + 1 1 1 1 2. Set u x x x x = − − + = = ∴ = ± 0 1 1 1 1 2 2 2 2 : , . . or Stag. pts.: ( , ), ( , ). 2 0 2 0 − u v ( , ) . ( , ) . 4 0 1 4 1 1 4 1 2 4 0 0 = − + − − − + = − = 1.867 m / s u v ( , ) . ( , ) . 0 4 1 1 4 1 1 4 2 0 4 4 1 4 4 1 4 0 2 2 2 2 = + − − + + = = + − + = 2.118 m / s 8.82 φ π π π π = + − + + + 2 2 1 2 2 1 2 2 1 2 2 2 1 2 l l n x y n x y [ ( ) ] [ ( ) ] / / = + − + + + 1 2 1 1 2 1 2 2 2 2 l l n x y n x y [ ( ) ] [ ( ) ]. (0, h) y x
- 142. 182 u x x x y x x y == + − + + + ∂φ ∂ 2 2 2 2 1 1 ( ) ( ) . v y y x y y x y = = − + − + + + − ∂φ ∂ 1 1 1 1 2 2 2 2 ( ) ( ) . At (0, 0) u = 0 and v = 0. At (1, 1) 2 2 2 2 2 2 2 1 1 0 0.4 m/s. 1.2 m/s. 5 2 1 1 2 1 v u = + = = = + = + + ∴ = + v V i j 12 04 . $ . $ . m/ s 8.83 φ π π π π = − + + + + + ∞ 2 2 1 2 2 1 2 2 1 2 2 2 1 2 l l n y x n y x U x [( ) ] [( ) ] . / / = − + + + + + ∞ 1 2 1 1 2 1 2 2 2 2 l l n y x n y x U x [( ) ] [( ) ] . a) Stag. pts. May occur on x-axis, y =0. u x x x x x y = = + + + + = ∂φ ∂ 0 2 2 1 1 10. 2 0.2 1 0. x x ∴ + + = ∴no stagnation points exist on the x-axis. (They do exist away from the x-axis.) Along the y-axis: u y q udy h ( ) . ( ) = = = = ∫ 10 1 2 2 0 m / s. 2 π π ∴ = = ∴ = ∫ π 10 10 0314 0 dy h h h . m . . b) u x x x x x = + + ∴ + + = ∴ = − 2 1 1 2 1 0 1 2 2 . . m. Stag. pt.: (−1, 0) Along the y-axis: u h h = ∴ = × ∴ = 1 0 1 3 14 . . . . . m π c) u x x x x x = + + ∴ + + = ∴ = − − 2 1 02 10 1 0 9 90 2 2 . . . . , 0.10 m. Stag. pts.: (−9.9, 0) , (−0.1, 0). Along the y-axis: u h h = ∴ = ∴ = 0 2 02 1571 . . . . . . m π 8.84 φ θ θ = + 60 8 r r cos cos . a) v r r r r = = − + = − ∂φ ∂ θ θ θ 60 8 8 60 2 2 cos cos cos . At the cylinder surface vr = 0 for all θ. Hence, x y x y x y
- 143. 183 60 8 2 739 2 r r c c =∴ = . . m b) Bernoulli: ∆p U = = = ∞ ρ 2 2 2 1000 8 2 32 000 Pa or 32 kPa c) v r r θ ∂φ ∂θ θ θ = = − − 1 60 8 2 sin sin . At r r v c = = − − = − , sin sin sin θ θ θ θ 8 8 16 d) ∆p v = = = ρ 90 2 2 2 1000 16 2 128 o 000 Pa or 128 kPa 8.85 ψ π π θ π π θ = + = + 4 2 20 2 2 10 l l n r n r At ( , ) ( , ), ( , ) ( , / ). x y r = = 0 1 1 2 θ π v r r ( , / ) ( ) . 1 2 1 1 1 2 2 π ∂ψ ∂θ = = = v r θ π ∂ψ ∂ ( , / ) . 1 2 10 1 10 = − = − = − vr ( . , / ) . . 1 7 4 2 1 7 1 18 π = = , vθ π ( . , / ) . . 1 7 4 10 1 7 5 88 = − = − vr ( . , ) . . 3 2 0 2 3 2 0 625 = = , vθ ( . , ) . . 3 2 0 10 3 2 3 125 = − = vr ( , / ) . 6 4 2 6 0 333 − = = π , vθ π ( , / ) . 6 4 10 6 1 67 − = − = − , etc. Note: We scaled the radius at each 45° increment to find r. b) v r v r r = = − 2 10 and θ . From Table 5.1 (use the l.h.s. of momentum) a Dv Dt v r v v r v r r r r r = + = − θ θ ∂ ∂ 2 2 = − − = − 2 2 100 104 1 2 3 r r r = −104 m /s2 a Dv Dt v v r v v r v v r r r r θ θ θ θ θ ∂ ∂ = + = + = + − = 2 10 2 10 0 2 3 r r r ( ) . ∴ = − v a( , ) ( , ) 0 1 104 0 m /s2 x y
- 144. 184 c) v v r( . , / ) / . . , ( . , / ) . . 14 14 4 2 14 14 0 1414 14 14 4 10 14 14 0 707 π π θ = = = − = − v v r ( . , / ) / . , ( . , / ) / . . 0 1 2 2 0 1 20 0 1 2 10 0 1 100 π π θ = = = − = − Bernoulli: 20 000 1.2 13 760 Pa + + = + + ∴ = 0 1414 0 707 2 1 2 20 100 2 2 2 2 2 . . . . p p We used ρair 3 kg / m at standard conditions. = 1 2 . 8.86 Along the y-axis v v r r = = − − 0 10 40 2 and θ . We have set θ π = 2 in Eq. 8.5.27. rc = = 40 10 2. b) v r r = − − ⇒ 10 40 4 3 5 126 9 2 cos cos . ( , ) ( , . ). θ θ o v r v v r θ θ θ θ = − − ∴ = − = − 10 40 6 96 9 28 2 sin sin . . , . . m /s m / s c) Use Eq. 8.5.28: p p U = − ∞ 0 2 2 2ρ α sin Drag = p r d L p r L p p U c c cos . . / / α α ρ π π − × = − ∞ − ∫ 90 90 0 2 2 2 2 2 = − − × ∞ ∫ 2 2 2 0 2 2 90 0 2 ( sin )cos / p U r Ld p r L c c ρ α α α π [ ] = − − − = ∞ ∞ ∞ 2 2 3 2 2 8 3 0 2 3 0 2 0 2 2 r L p U p U r L r L U c c c sin sin . / α ρ α ρ ρ π C U A r L U U r L D c c = = = = ∞ ∞ ∞ Drag 1 2 8 3 1 2 2 8 3 2 667 2 2 2 ρ ρ ρ ( / ) . . 8.87 v U r U r U r c = − = = = × = ∞ ∞ ∞ cos cos . , . θ µ θ µ 2 2 2 4 1 4 4 Let For θ π = = − + , . v r r 4 4 2 b) v r U rc θ ∂ψ ∂ θ µ θ θ θ = − = − − = − − = − ∞ sin sin sin sin . 2 2 4 4 1 8 c) p p V v c = + − = + × − ∞ ∞ ρ ρ θ θ 2 2 2 2 2 2 2 1000 4 2 1000 8 2 50 000 sin . ∴ = − pc 58 32 2 sin θ kPa. 10 m/s 20 m/s dα p(α) α p90 x = −1 −x −vr
- 145. 185 d) Drag =2 58 32 1 1 26 2 1 2 0 2 ( sin ) cos / − × × − × × ∫ α α α π d = − − = 2 58 32 1 3 52 42.7 kN. (See the figure in Problem 8.86c.) 8.88 On the cylinder 1000 2 sin 60sin , 2 2 3.651 c v U r θ θ θ π π ∞ Γ = − − = − − × where we have Used 400 3.651 ft. 30 c r U µ ∞ = = = If 2 2 2 2 2 2 2 2 6 6 6 6 ( , ) .0318 ( 6) ( 2) ( 6) ( 2) ( 6) ( 2) ( 6) ( 2) x x x x u x y x y x y x y x y − − + + = − + + + − + − − + + + + − + + + 227 , 313 . θ ∴ = o o Stag. pts.: (3.651 ft, 227°) , (3.651 ft, 313°). Max. pressure occurs on the cylinder at a stagnation pt.: 2 2 2 2 max o 0.0024 = 30 0 1.08 psf. 2 2 p U v ρ ∞ ∴ = − − = Min. pressure occurs at the top of the cylinder where θ = 90o and the velocity is: 90 o 1000 2 sin 2 30 104 fps 2 2 3.651 v U r θ π π ∞ Γ = − − = − × − = × 2 2 2 2 min o 0.0024 = 30 104 11.8 psf. 2 2 p U v ρ ∞ ∴ = − − = − 8.89 vθ θ π = − × − × 2 20 2 4 sin . . Γ For one stag. pt.: vθ θ = = 0 270 at o : 0 2 20 270 2 4 2 20 2 4 100 5 = − × − × ∴ = × × × = sin . . . . o Γ Γ π π m / s. 2 Γ Γ = ∴ = = × = 2 2 100 5 2 4 2 2 2 π ω ω π π r r c c . . . . 100 rad /s (See Example 8.12.) Min. pressure occurs where vθ is max, i.e., θ π = / . 2 There vθ π = − × × − × = 2 20 1 100 5 2 4 80 . . m /s. ∴ = + − = + × − × = − ∞ ∞ p p V v min Pa 2 2 2 2 2 2 0 20 2 1 22 80 2 1 22 3660 ρ ρ θ . . .
- 146. 186 8.90 Γ == × × × = = = × = 2 2 6 120 2 60 28 42 6 3 1 08 2 2 2 2 π ω π π µ r r U c c . / . . . m /s. m /s. 2 3 ∴ = − × − × ∴ = − vθ θ π θ 2 3 28 42 2 6 1256 sin . . sin . . . Impossible. ∴Stag. pt. is off the cylinder at θ = > 270o , . but r rc From Eq. 8.5.29, v r U r r r r θ ∂ψ ∂ θ µ θ π π = − = − − − = − − − − − = ∞ sin sin ( ) . ( ) . . 2 2 2 3 1 1 08 1 28 42 2 0 Γ ∴ + = ∴ − + = ∴ = 3 1 08 4 523 1 508 0 36 0 1 21 2 2 . . . . . . . r r r r r m. Stag. pt.: (1.21, 270°). ( ) . . . vθ π 90 2 3 2842 2 6 1354 o = − × − × = − m / s. Min. pressure occurs at θ = = = − = − 90 3 2 1354 2 122 106 2 2 o , : . . . at Pa min r r p c Max. pressure occurs at θ = = = − = − 270 3 2 154 2 122 4 04 2 2 o , : . . . . at Pa max r r p c 8.91 At 15,000 ft, ρ =. . 0015 slug / ft3 Lift = ρU L ∞ = × × × = Γ . , . 0015 350 15 000 60 472,000 lb 8.92 Place four sources as shown. Then, with q = 2π for each: u x y x x y x x y x x y ( , ) ( ) ( ) ( ) ( ) ( ) ( ) = − − + − + + + + − + − − + + 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 + + + + + x x y 2 2 2 2 2 ( ) ( ) v x y y x y y x y y x y ( , ) ( ) ( ) ( ) ( ) ( ) ( ) = − − + − + + − + + + − + + − 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 + + + + + y x y 2 2 2 2 2 ( ) ( ) 8.93 Place four sources with q = 0 2 . m / s, as shown. 2 u x y x x y x x y x x y x x y ( , ) . ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = − − − + − + − − + + + + + + − + + + + + 0318 6 6 2 6 6 2 6 6 2 6 6 2 2 2 2 2 2 2 2 2 v x y y x y y x y y x y y x y ( , ) . ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = − − − + − + − + + − + + − + + + + + + + 0318 2 6 2 2 6 2 2 6 2 2 6 2 2 2 2 2 2 2 2 2 where q 2 2 2 0318 π π = − = − . . . x y x y (6, 2)
- 147. 187 At (4,3) u(, ) . . . 4 3 0318 2 4 1 2 4 25 10 100 1 10 100 25 0 00922 = − − + + − + + + + + = m /s v( , ) . . . 4 3 0318 1 4 1 1 100 1 5 4 25 5 100 25 0 01343 = − + + + + + + + = − m /s 8.94 Re . / . crit = ∴ = × = ∞ U x x T T ν ν ν 6 10 300 2000 5 a) ν = × ∴ = × × = − − 1 56 10 2000 1 56 10 0 312 4 4 . . . ' ft /sec. or 3.74" 2 xT b) ν µ ρ = = × ∴ = × × = − − 2 1 10 2000 2 1 10 0 42 4 4 . . . ' ft /sec. or 5.04" 2 xT c) ν = × ∴ = × × = − − 3 47 10 2000 3 47 10 0 694 4 4 . . . ' ft /sec. or 8.33" 2 xT 8.95 a) Use Re / . . crit = × = × − 3 10 10 1 51 10 5 5 xT ∴ = xT 0 453 . . m b) Use Re / . . crit = = × − 10 10 1 51 10 6 5 xT ∴ = xT 1 51 . . m c) Use Re / . . crit = × = × − 3 10 10 1 51 10 5 5 xT ∴ = xT 0 453 . . m d) Use Re / . . crit = × = × − 3 10 10 1 51 10 5 5 xT ∴ = xT 0 453 . . m e) 4 5 growth growth Re 6 10 10 /1.51 10 . 0.091 m or 9.1 cm x x − = × = × ∴ = . Note: A rough plate, high free-stream disturbances, or a vibrated smooth plate all experience transition at the lower Re . crit 8.96 a) Use Re / . crit = × = − 3 10 10 10 5 6 xT ∴ = xT 0 03 3 . . m or cm b) Use Re / . crit = = − 10 10 10 6 6 xT ∴ = xT 0 1 10 . . m or cm c) Use Re / . crit = × = − 3 10 10 10 5 6 xT ∴ = xT 0 03 3 . . m or cm d) Use Re / . crit = × = − 3 10 10 10 5 6 xT ∴ = xT 0 03 3 . . m or cm e) 2 2 ( ) 20 000 2 1000 10 sin ( /2) p x x = − × × 8.97 Re . . . crit For a wind tunnel: = × = × × = × × ∞ ∞ − 6 10 2 6 10 2 1 5 10 5 5 5 U U ν m / s. ∴ = ∞ U 4 5 . For a water channel: 6 10 2 10 0 3 5 6 × = × ∴ = ∞ − ∞ U U . . m / s . 8.98 The x-coordinate is measured along the cylinder surface as shown in Fig. 8.19. The pressure distribution (see solution 8.86) on the surface is p p U r x c = − = ∞ 0 2 2 2ρ α α α sin ( where is zero at the stagnation point). Then 2 2 ( ) 20 000 2 1000 10 sin ( /2) p x x = − × × = − 20 200 2 2 sin ( / ) x kPa The velocity U(x) at the edge of the b.l. is U(x) on the cylinder wall:
- 148. 188 v r θ θθ π α α ( ) sin sin sin( ) sin = = − − = − − = 2 10 10 20 20 ∴ = U x x ( ) sin( / ) 20 2 8.99 U x v r v U x x x r c c ( ) . sin . ( ) sin . = = = ∴ = = θ θ α α at since 1 8 8 p x x ( ) sin sin = − = − 58 32 58 32 2 2 α kPa 8.100 The height h above the plate is h x mx m m ( ) . . . . . = + = × + ∴ = − 4 1 2 4 15 ∴ = − × = ∴ = − h x x U x h U x x ( ) . . . . ( ) . ( ) . . . 0 4 15 6 4 2 4 0 4 15 Continuity: or U x x ( ) . . = − 16 2 67 Euler’s Eqn: ρ ∂ ∂ ∂ ∂ ρ u u x p x dp dx x x = − ∴ = − − . . ( . ) 16 2 67 16 2 67 2 = − 256 2 67 3 ( . ) . x 8.101 a) top out in 0 0 0 0 m m m udy udydx udy udydx x x δ δ δ δ ρ ρ ρ ρ ∂ ∂ = − = + − = ∫ ∫ ∫ ∫ ∂ ∂ & & & b) 0 ( ) ( )( ) 2 x dp F p dx p d p dp d δ τ δ δ δ Σ = − + + − + + 0 higher order terms dx dp τ δ = − − + 2 2 2 out in top 0 0 0 0 2 0 0 ( ) ( ) mom mom mom u dy u dydx u dy U x udydx x x u dydx U x udydx x x δ δ δ δ δ δ ρ ρ ρ ρ ρ ρ ∂ ∂ − − = + − − ∫ ∫ ∫ ∫ ∂ ∂ ∂ ∂ = − ∫ ∫ ∂ ∂ & & & 2 3 2 5 4 5 2 3 4 3 4 4 . 2 2 4.65 1.5 10 3 4.65 (1.5 10 3) y d y dx δ − − × = − + × × × × × × × 8.102 τ δ ρ ρ δ δ 0 2 0 0 = − + − ∫ ∫ dp dx U x d dx udy d dx u dy ( ) = − + − − ∫ ∫ ∫ δ ρ ρ ρ δ δ δ dp dx d dx uUdy dU dx udy d dx u dy 0 2 0 0 where we have used g df dx dfg dx f dg dx U g f udy = − = = ∫ . , . Here ρ δ 0 ∴ = − + − − = ∫ ∫ τ δ ρ ρ ρ δ δ 0 0 0 dp dx d dx u U u dy dU dx udy ( ) . ( const.)
- 149. 189 8.103 dp dx d dx U U dU dx dU dx Udy UUdy = − = − = − = ∫ ∫ ρ ρ ρ δ δ δ δ 2 1 1 2 0 0 where . ∴ = − − + − ∫ ∫ τ δ ρ δ ρ θ ρ δ δ 0 0 2 0 1 dU dx Udy d dx U dU dx udy ( ) = + − = + ∫ ρ θ ρ ρ θ ρ δ δ d dx U dU dx U u dy d dx U dU dx U d ( ) ( ) ( ) . 2 2 0 8.104 If dp dx dU dx d dx u U u dy / ( ) . = = = − ∞ ∫ 0 0 0 0 then and τ ρ δ τ ρ π δ π δ ρ δ π π δ ρ δ π δ δ δ 0 2 0 2 0 2 2 1 2 2 2 2 2 2 = − = − − = − ∞ ∞ ∞ ∫ d dx U y y dy U d dx y y U d dx sin sin cos τ µ ∂ ∂ µ π δ 0 0 2 0 = = = ∞ u y U y cos . ∴ = ∴ = ∴ = ∞ ∞ ∞ ∞ µ π δ ρ δ δ δ ν δ ν U U d dx d U dx x U 2 137 11 5 4 79 2 . . . . . . b) τ µ π ν µ ν 0 2 1 4 79 0 328 = = ∞ ∞ ∞ ∞ U U x U U x . . . c) ∂ ∂ ∂ ∂ π ν ∂ ∂ ∂ ∂ u x U x y U x U x a x U ax a x v y = × = = − = − ∞ ∞ ∞ ∞ − sin . sin cos . / 2 4 79 2 3 2 ∴ = = ∞ ∞ ∞ ∞ ∞ ∞ ∫ ∫ v U y x U y U x dy U U U y dy . cos . . cos . . / 164 328 0316 189 3 2 0 0 ν ν ν ν δ δ 8.105 u U y d dx U y y dy = = − ∞ ∞ ∫ δ τ ρ δ δ δ . 0 2 0 1 = − = ∞ ∞ ρ δ δ ρ δ d dx U U d dx 2 2 2 3 1 6 . τ µ ∂ ∂ µ δ µ δ ρ δ δ δ ν 0 2 1 6 6 = = ∴ = ∴ = ∞ ∞ ∞ ∞ u y U U U d dx d U dx . . ∴ = = = ∞ ∞ ∞ ∞ δ ν δ ν τ µ ν 2 0 12 3 46 0 289 U x x x U U U x . ( ) . . . . %error in δ( ) . . x = − × = 5 3 46 5 100 30 8% low. U δ u = Uy/δ y
- 150. 190 %error in low τ0 332 289 332 100 13% ( ) . . . . x = − × = 8.106 τ ρ δ δ ρ δ δ δ δ δ 0 2 2 6 2 0 6 3 1 3 1 3 1 1 3 = − + + − − ∞ ∞ ∫ ∫ d dx U y y dy U y y dy / / / + + − − ∞ ∫ρ δ δ δ δ U y y dy 2 2 3 2 3 1 3 2 3 / = = ∴ = ∞ ∞ ∞ d dx U U d dx U ρ δ µ δ δ δ µ ρ 2 0 1358 3 22 08 ( . ) . . / . Thus, δ τ ρ ρ ( ) . / , ( ) . . . Re . / x vx U x U v U x U x = = = ∞ ∞ ∞ ∞ − 665 01358 665 2 0451 0 2 2 1 2 %error for δ = − × = 6 65 5 5 100 33% . . %error for τ 0 0 451 0 332 0 332 100 36% = − × = . . . 8.107 Continuity from entrance to x: U H u y dy U x H 0 0 2 2 = + − ∫ ( ) ( )( ). δ δ Write U x U x dy U x dy ( ) ( ) ( ) . δ δ δ = = ∫ ∫ 0 0 Then, continuity provides U H u U dy UH 0 0 2 = − + ∫( ) δ = − − ∫ UH U u dy 2 0 ( ) δ = − ∴ = − UH U U x U H H d d 2 2 0 δ δ . ( ) . If we were to move the walls out a distance δd x ( ), then U x ( ) would be constant since ( ) [ ] H d d − + 2 2 δ δ would be constant; then U x U ( ) . = 0 For a square wind tunnel, displace one wall outward 4 0 δ d dp dx for / . = 8.108 The given velocity profile is that used in Example 8.13. There we found δ ν = = = = = ∞ − 5 48 5 48 10 10 0 00173 0 00173 3 0 003 6 . / . / . . . x U x x m. Assume the streamline is outside the b.l. Continuity is then 10 0 02 10 2 003 003 003 10 2 2 0 003 × = − + − ∫ . . . ( . ) . y y dy h = + − ∴ = 0 02 10 0 03 0 021 . . . . h h m or 2.1 cm [ ] δ d y y dy = − + = − + = ∫ 1 10 10 20 003 10 003 1 10 03 03 01 0 001 2 2 0 003 . . . . . . . m h − = − = 2 2 1 2 0 1 . . cm or 0.001 m. The streamline moves away from the wall a distance δ d .
- 151. 191 8.109 From Prob.8.107 we found that we should displace the one wall outward 4δ d . From the definition of δ d : h x y y dy d ( ) = = − + = − + = ∫ 4 4 10 10 20 10 4 3 4 3 2 2 0 δ δ δ δ δ δ δ δ = × × = − 4 3 5 48 1 86 10 10 160 287 303 0 00735 5 . . / /(. ) . x x m We used δ( ) x found in Example 8.13, ρ ν µ ρ = = p RT / , / . and 8.110 a) u U y y U U y y dy d = − = − + = − + = ∞ ∞ ∞ ∫ 3 2 1 2 1 1 3 2 1 2 3 4 1 8 375 3 3 3 3 0 δ δ δ δ δ δ δ δ δ δ . . . From Eq. 8.6.16, δ ν ν d x U x U = × = ∞ ∞ . . . . 375 4 65 1 74 %error = 1.2%. θ δ δ δ δ δ δ = − − + = ∞ ∞ ∫ 1 3 2 1 2 1 3 2 1 2 0 139 2 2 3 3 3 3 0 U U y y y y dy . . ∴ = × = − × = ∞ ∞ θ ν ν . . . . . . . . 139 4 65 0 648 648 644 644 100 0 62% x U x U %error = b) u U y y y y dy d = − = − + = − + = ∞ ∫ 2 1 2 3 3 2 2 2 2 0 δ δ δ δ δ δ δ δ δ δ . / . See Example 8.13. ∴ = = − × = ∞ ∞ δ ν ν d x U x U 5 48 3 1 83 1 83 1 72 1 72 100 6 4% . . . . . . . %error = . θ δ δ δ δ δ δ δ δ δ δ δ δ = − − + = − − + + − = ∫ 2 1 2 1 3 4 3 2 4 2 4 1 5 1333 2 2 2 2 0 y y y y dy . . ∴ = × = − × = ∞ ∞ θ ν ν . . . . . . . . 1333 5 48 0 731 731 644 644 100 13 5% x U x U %error = . c) 0 2 1 sin 0.363 . See Problem 8.104. 4.79 . 2 d y x dy U δ π δ ν δ δ δ δ δ π ∞ = − = − = = ∫ ∴ = × = − × = ∞ ∞ δ ν ν d x U x U 0 363 4 79 1 74 1 74 1 72 1 72 100 1 2% . . . . . . . . %error = θ π δ π δ δ π π δ δ δ π δ δ δ = − = − − + − = − + = ∫sin sin cos sin . . y y dy y y 2 1 2 2 2 2 2 2 0 137 0 0 term ∴ = × = − × = ∞ ∞ θ ν ν . . . . . . . . . 137 4 79 0 654 654 644 644 100 1 6% x U x U %error =
- 152. 192 8.111 a) δ ν == × × = ∞ − 4 65 4 65 1 6 10 20 12 0 0759 4 1 2 . . . . . / x U ft b) τ ρ ν 0 2 2 4 1 2 5 323 323 0024 12 1 6 10 20 12 9 11 10 = = × × × × = × ∞ ∞ − − . . . . . . / U xU psf c) Drag = 2 1 20 15 1.29 2 U LU ν ρ ∞ ∞ × × × 1 / 2 4 2 1 1.6 10 .0024 12 300 1.29 0.0546 lb. 2 20 12 − × = × × × × = × d) δ ∂ ∂ δ δ δ x u x U y y d dx = − ∞ = × × = = − + 10 4 2 3 4 4 65 1 6 10 10 12 0 0416 3 2 3 2 . . . . ft. ∴ = − × + × × × = − + − ∂ ∂ u x y y y y 12 3 2 0416 3 2 0416 4 65 2 1 6 10 10 12 27 9 16140 2 3 4 4 3 . . . . . . . . ∴ = − = × − × = ∫ v u x dy ∂ ∂ δ 27 9 2 0416 16140 4 0416 0 0121 2 4 0 . . . . . fps 8.112 a) δ ν = = × × = ∞ − 4 65 4 65 15 10 6 4 00221 5 1 2 . . . . . / x U m b) τ ρ ν 0 2 2 5 1 2 0323 323 122 4 15 10 6 4 0 00498 = = × × × × = ∞ ∞ − . . . . . . / U xU Pa c) Drag = 1 2 129 1 2 122 4 6 5 129 15 10 6 4 0299 2 2 5 1 2 ρU Lw v LU ∞ ∞ − × = × × × × × × × = . . . . . . / N d) 3 2 4 3 3 2 2 u y y d U x dx ∂ δ ∂ δ δ ∞ = − + 2 3 2 5 4 5 2 3 4 3 4 4 . 2 2 4.65 1.5 10 3 4.65 (1.5 10 3) y d y dx δ − − × = − + × × × × × × × [ ] ∴ = − + × × = − + × − ∂ ∂ u x y y y y 4 6166 253 10 465 2 15 10 4 1 3 641 2 63 10 7 3 5 5 3 . . . . . . ∴ = − = × − × × = ∫ v u x dy ∂ ∂ δ 641 2 0156 2 63 10 4 0156 0 00391 2 5 4 0 . . . . . , m / s where δx= − = × × = 3 5 465 15 10 3 4 01560 . . . m.
- 153. 193 8.113 a) δ ν == × × = ∞ − 5 5 1 5 10 2 10 0 00866 5 1 2 x U . . . / m Use τ ρ ν 0 2 332 = ∞ ∞ . . U xU Drag = τ 0 2 5 0 332 1 22 10 1 5 10 10 2 1 2 4 0 561 wdx L = × × × × = − ∫ . . . / . . N b) δ = × × × = − . . . . . 38 2 1 5 10 10 2 0 0453 5 2 m Drag = 1 2 074 1 2 1 22 10 2 4 074 1 5 10 10 2 2 15 2 2 2 5 2 ρ ν U Lw U L ∞ ∞ − × = × × × × × × × = . . . . . . . . N 8.114 a) δ τ = × × × = = × × × × × − − . . . . . . . . . 38 6 1 5 10 20 6 0 0949 1 2 1 22 20 059 1 5 10 20 6 5 2 0 2 5 2 m = . . 6 Pa b) δ τ = × × = = × × × × − − . . . . . . 38 6 10 20 6 0 0552 1 2 1000 20 059 10 20 6 6 2 0 2 6 2 m = 286 Pa. 8.115 u y U u y U y u y U y ( ) . . / . / / = = = = ∞ ∞ − − = ∞ δ ∂ ∂ δ ∂ ∂ δ δ 1 7 1 7 6 7 1 7 ∂ ∂ δ u y y= should be zero. Thus, this condition is not satisfied. τ µ ∂ ∂ µ δ 0 0 1 7 1 7 1 0 = = = ∞ = ∞ − u y U y / . Thus, this is unacceptable and ∂ ∂ u y at, and near, the wall is not valid. u U y y = − ∞ 3 2 1 2 3 3 δ δ . u U y = ∞ δ 1 7 / . 8.116 a) Drag = 1 2 0024 20 12 15 074 1 58 10 20 12 1060 1 58 10 20 12 0 31 2 4 2 4 × × × × × × − × × = − − . ( ) . . . . . . lb cubic turb (power-law) U y u u
- 154. 194 b) Drag = 1 2 002420 12 15 074 1 58 10 20 12 1700 1 58 10 20 12 0 27 2 4 2 4 × × × × × × − × × = − − . ( ) . . . . . . lb c) Drag = 1 2 0024 20 12 15 074 1 58 10 20 12 2080 1 58 10 20 12 0 25 2 4 2 4 × × × × × × − × × = − − . ( ) . . . . . . lb 8.117 a) Drag = 1 2 1000 1 2 1 2 074 10 1 2 1 1060 10 1 2 1 5 21 2 6 2 6 × × × × × − × = − − . ( ) . . . . . . N b) Drag = 1 2 1000 1 2 1 2 074 10 1 2 1 1700 10 1 2 1 4 44 2 6 2 6 × × × × × − × = − − . ( ) . . . . . . N c) Drag = 1 2 1000 1 2 1 2 074 10 1 2 1 2080 10 1 2 1 3 99 2 6 2 6 × × × × × − × = − − . ( ) . . . . . . N 8.118 U∞ = = = × × × = 60 1000 3600 16 67 38 100 16 67 10 235 5 2 . . . . . m /s. 000 1.5 10 m -5 δ τ ρ 0 2 2 5 5 2 1 2 1 2 1 22 16 67 059 1 5 10 16 67 10 0 0618 = = × × × × × = ∞ − U c f . . . . . . . . Pa b) τ ρ 0 2 2 5 5 2 1 2 1 2 1 22 16 67 455 06 16 67 10 1 5 10 0 151 = = × × × × = ∞ − U c n f . . . . . . . . l Pa ∴ = = ∴ = × + ∴ = − u n τ δ δ . . . . . . . . . . . 151 1 22 351 16 67 351 2 44 351 1 5 10 7 4 585 5 m /s. m l Both (a) and (b) are in error, however, (b) is more accurate. 0. p x ∂ ∂ < 8.119 a) 5 = uτ ν δ ν (See Fig. 8.24 b). ∴ = × × = × − − δν 5 1 5 10 351 2 14 10 5 4 . . . . m b) δ µ δ δ τ τ δ δ δ δ δ ν d U U u dy U n y dy u U n y dy = − = − + − ∞ ∞ ∞ ∞ ∫ ∫ ∫ 1 2 5 2 44 3 74 15 15 0 ( ) . . . . . l l ( ) = − − − − − ∞ = = u U y n y y y n y y v τ ν δ δ δ δ δ δ δ 2 5 15 2 44 3 74 15 87 8 87 8 585 . . . . . . . l l = + − + − = . . [ . ] . . 351 16 67 219 620 008 2188 951 43 7 m Note: We cannot use zero as a lower limit since the ln-profile does not go to the
- 155. 195 wall. Hence, weuse δν ; the lower limit provides a negligible contribution to the integral. 8.120 a) Use Eq. 8.6.40: c n f = × × = − . . . . . 455 06 300 20 1 58 10 0 00212 4 2 l b) τ ρ 0 2 2 1 2 1 2 0024 300 00212 0 229 = = × × × = ∞ U c f . . . . psf uτ = = . . . 229 0024 9 77 fps. c) δ ν ν τ = = × × = × − − 5 5 1 58 10 9 77 8 09 10 4 5 u . / . . . ft d) 300 9 77 2 44 9 77 1 58 10 7 4 0 228 4 . . . . . . . . = × + ∴ = − ln δ δ ft 8.121 a) τ ρ 0 2 2 6 2 1 2 1 2 1000 10 455 06 10 3 10 110 = = × × × = ∞ − U c n f . . . l Pa ∴ = = ∴ = = × = × − − u u τ ν τ δ ν 110 1000 332 5 5 10 332 1 51 10 6 5 . . . . m /s. m b) u u = = × = 5 5 332 1 66 τ . . . m /s c) y =. . 15δ — Do part (d) first! ∴ = × = y . . . . 15 0333 0 005 m d) 10 332 2 44 332 10 7 4 0 0333 6 . . . . . . . = + ∴ = − ln δ δ m 8.122 Assume flat plates with dp dx n f / . . . . . = = × = − 0 523 06 10 100 10 00163 6 2 C l ∴Drag = 2 1 2 1000 10 10 100 00163 2 × × × × × × = . . 163 000 N To find δ τ max we need u . τ τ 0 2 6 2 1 2 1000 10 455 06 10 100 10 70 9 70 9 1000 0 266 = × × × = ∴ = = − . . . . . ln u Pa. m /s. 10 266 2 44 266 10 7 4 0 89 6 . . . . . . . = + ∴ = − ln δ δ m max 8.123 a) Assume a flat plate of width πD. 8 5 15 600 Re 6 10 . 1.5 10 UL ν − × = = = × ×
- 156. 196 2 8 1/52 1 1 drag 0.073(6 10 ) 1.2 15 600 100 32600 N 2 2 f C U L D ρ π π − = = × × × × × × × = power 32600 15 489000 W or 655 hp or 164 hp/engine D F U = × = × = . b) 3 helium 100 0.167 kg/m . 2.077 288 p RT ρ = = = × air helium B F W W V ρ = − = ∆ × 2 7 (1.2 0.167) 9.8 50 600/2 2.38 10 π = − × × × × = × payload = 6 6 6 23.8 10 9.8 1.2 10 12 10 N B F W − = × − × × = × 8.124 u y u x x y v x u y y u y y = = = − = = ∂ψ ∂ ∂ ∂ ∂ ψ ∂ ∂ ∂ψ ∂ ∂ ∂ ∂ ψ ∂ ∂ ∂ ∂ ψ ∂ , , , , . 2 2 2 2 2 3 3 Substitute into Eq. 8.6.45 (with dp dx / ): = 0 ∂ψ ∂ ∂ ψ ∂ ∂ ∂ψ ∂ ∂ ψ ∂ ν ∂ ψ ∂ y x y x y y 2 2 2 3 3 − = . 8.125 We also have ∂ψ ∂ ∂ψ ∂φ ∂φ ∂ ∂ψ ∂η ∂η ∂ ∂ ψ ∂ ∂ ∂ ∂ψ ∂ ∂φ ∂φ ∂ ∂ ∂ψ ∂ ∂η ∂η ∂ x x x x y y x y x = + = + , ( / ) ( / ) 2 Recognizing that ∂φ ∂ ∂φ ∂ ∂η ∂ ν / , / , / / , x y x y U x = = = − ∞ 1 0 2 3 and ∂η ∂ ν / / , y U x = ∞ ∂ψ ∂ νφ ∂ψ ∂η ∂ψ ∂ ∂ψ ∂φ ν ∂ψ ∂η y U x y U x = = − ∞ ∞ , 2 3 ∂ ψ ∂ ∂ νφ ∂ ψ ∂φ∂η νφ ∂ψ ∂η νφ ∂ ψ ∂η νφ 2 2 3 2 2 3 2 x y U U U y U = + − ∞ ∞ ∞ ∞ - 1 2 ∂ ψ ∂ νφ ∂ ψ ∂η νφ ∂ ψ ∂ νφ ∂ ψ ∂η νφ 2 2 2 2 3 3 3 3 y U U y U U = = ∞ ∞ ∞ ∞ , Equation 8.6.47 then becomes, using U y ∞ = / / , νφ η η ∂ψ ∂η η ∂ ψ ∂φ∂η η ∂ψ ∂η η ∂ ψ ∂η ∂ψ ∂φ η ∂ψ ∂η η ∂ ψ ∂η y y yx yx x y 2 2 2 2 2 2 2 2 2 2 2 − − − − = ∞ ν ν η ∂ ψ ∂η U x y 3 3 Multiply by y2 2 /η and Eq. 8.6.49 results: − + − = ∞ 1 2 2 2 2 2 3 3 φ ∂ψ ∂η ∂ ψ ∂φ∂η ∂ψ ∂η ∂ψ ∂φ ∂ ψ ∂η ν ∂ ψ ∂η νφ U
- 157. 197 8.126 u y U x dF dy U x F U x U F = = = = ∞ ∞ ∞ ∞ ∂ψ ∂ ν η ∂η ∂ ν η ν η '( ) '( ). We used Eq. 8.6.50 and Eqs. 8.6.48. ( ) v x x U x F U x F U x F x = − = − = − − ∞ ∞ ∞ ∂ψ ∂ ∂ ∂ ν ν ν ∂ ∂η ∂η ∂ 1 2 = − − − ∞ ∞ ∞ − 1 2 1 2 3 2 U x F U x F y U x ν ν ν ' / = − + = − ∞ ∞ ∞ ∞ 1 2 2 1 2 U x F y U x U x F U x F F ν ν ν ν η ' ( ' ). 8.127 The results are shown in Table 8.5. 8.128 a) τ 0 2 5 0 332 1 22 5 1 5 10 2 5 0 0124 = × × × × = − . . . . . Pa b) δ = × × = − 5 15 10 2 5 0 0122 5 . . . m c) v U x F F max max m /s = − = × × × = ∞ − ν η 1 2 1 5 10 5 2 8605 0 00527 5 ( ' ) . . . . d) Q udy U dF d dy U dF d d vx U = = = ∞ ∞ ∞ ∫ ∫ ∫ 0 0 0 δ δ δ η η η = − = × × × = ∞ ∞ − U x U F F ν δ [ ( ) ( )] . . . / 0 5 1 5 10 2 5 3 28 0 04 5 m s / m 2 8.129 a) τ 0 2 4 4 332 0024 15 1 6 10 6 15 2 39 10 = × × × × = × − − . . . . . psf b) δ = × × = − 5 1 6 10 6 15 0 04 4 . . ft. c) v U x F F max max fps = − = × × × = ∞ − ν η 1 2 1 6 10 15 6 8605 0 0172 4 ( ' ) . . . . d) Q udy U x U F = = = × × × = ∞ ∞ − ∫ ν δ δ ( ) . . . / . 15 1 6 10 6 15 3 28 0 394 4 0 ft sec /ft 2 8.130 At x = 2m, Re = 5 × 2/10-6 = 107. ∴Assume turbulent from the leading edge. a) τ ρ 0 2 2 1 2 0 455 0 06 = ∞ U n x . ( . Re ) l
- 158. 198 = × × × = 1 2 10005 0 455 0 06 10 32 1 2 7 2 . ( . ) . ln Pa b) uτ τ ρ = = = 0 32 1 1000 0 1792 / . / . m /s 5 0 1792 2 44 0 1792 10 7 4 0 0248 6 . . . . . . = + ∴ = − ln m or 24.8 mm δ δ c) Use the 1/7 the power-law equation: Q y dy = = ∫5 0 0248 0 109 1 7 0 0 0248 ( / . ) . / . m /s /m 3 8.131 From Table 8.5 we would select η = 6: a) 5 1.5 10 2 6 6 0.0147 m 5 x U ν δ − ∞ × × = = = b) 5 15.8 10 6 6 6 0.047 ft or 0.57 in. 15 x U ν δ − ∞ × × = = = 8.132 From Table 8.5 we interpolate for F' . = 0 5 to be η = − − − + = 0 5 0 3298 0 6298 0 3298 2 1 1 1 57 . . . . ( ) . = × × ∴ = − y y 5 1 5 10 2 0 00385 5 . . . m or 3.85 mm ( ) v U x F F = − ∞ ν η 1 2 ' = × × = − 1 5 10 5 2 0 207 0 00127 5 . ( . ) . m / s u v y x ∂ ∂ τ µ ∂ ∂ = + 2 " F U xU ν ρ ∞ ∞ = 5 2 1.5 10 0.291(1.2)5 0.011 Pa 2 5 − × = = × 8.133 If v y v y v y = = > = < 0 10 0 0 at and at then δ δ ∂ ∂ , / and continuity demands that / 0. The component, for must then be greater than , u x u y U ∂ ∂ δ > > as shown in (b); there should be a slight “overshoot”. Also, consider the control volume of (c) where the lower boundary is just above y v y = = δ. , If at large say 0 y = 10δ , then continuity demands that u out the right area be greater than : U an “overshoot”. It is not reasonable to assume that v = const as in (a); y y = δ v v y y = δ v = 0 v (a) (b) (c) U v v = 0 u > U
- 159. 199 reality would demanda profile such as that sketched in (b). The overshoot would be quite small and is neglected in boundary layer theory. 8.134 u U y y = − ∞ 3 2 1 2 3 3 δ δ For the Blasius profile: see Table 8.5. (This is only a sketch. The student is encouraged to draw the profiles to scale.) 8.135 8.136 A: 0. (favorable) p x ∂ ∂ < B: ∂ ∂ p x ≅ 0. C: 0. (unfavorable) p x ∂ ∂ > D: 0. p x ∂ ∂ > E: 0. p x ∂ ∂ < y U cubic Blasius A B C D y y y y 2U∞ zero velocity gradient separation streamline backflow inviscid profile low velocity outside b.l. y A B C D E δD δΒ δC δΑ δE
- 160. 200 CHAPTER 9 Compressible Flow 9.1 Btuft-lb lbm ft-lb 0.24 778 32.2 6012 Btu slug lbm- R slug- R p c = = o o c c R v p = − = − = = 6012 1716 4296 4296 1 778 1 32 2 ft -lb slug- R ft -lb slug- R Btu ft - lb slug lbm o o . Btu 0.171 lbm- R = o 9.2 c c R c kc c c k R c k R p v p v p p p = + = ∴ = + − = . . . or 1 1 ∴ = − c Rk k p /( ). 1 9.3 If ∆s = 0, Eq. 9.1.9 can be written as c n T T R n p p n T T n p p p c R p l l l l 2 1 2 1 2 1 2 1 = = or It follows that, using c c R c c k p v p v = + = and / , T T p p p p R c k p 2 1 2 1 2 1 1 1 = = − / . Using Eq. 9.1.7, T T p p p p p p k k 2 1 2 1 2 1 2 1 1 1 1 2 2 1 1 = = = − − ρ ρ ρ ρ or / . Finally, this can be written as p p k 2 1 2 1 = ρ ρ . 9.4 Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change: & & & ~ ~ . Q W m V V p p u u S − = − + − + − 2 2 1 2 2 2 1 1 2 1 2 ρ ρ
- 161. 201 Enthalpy is definedin Thermodynamics as h u pv u p = + = + ~ ~ / . ρ Therefore, & & & . Q W m V V h h S − = − + − 2 2 1 2 2 1 2 Assume the fluid is an ideal gas with constant specific heat so that ∆ ∆ h c T p = . Then ( ) & & & . Q W m V V c T T S p − = − + − 2 2 1 2 2 1 2 Next, let c c R k c c c R k k p v p v p = + = = − and so that / / ( ). 1 Then, with the ideal gas law T p R = / , ρ the first law takes the form & & & . Q W m V V k k p p S − = − + − − 2 2 1 2 2 2 1 1 2 1 ρ ρ 9.5 Differentiate p c d xy ydx xdy k ρ− = = + using ( ) : ρ ρ ρ − − − − = k k dp pk d 1 0. Rewrite: dp d k p ρ ρ = . 9.6 The speed of sound is given by c dp d = / . ρ For an isothermal process TR p K K = = / , ρ where is a constant. This can be differentiated: dp Kd RTd = = ρ ρ. Hence, the speed of sound is c RT = . 9.7 Eq. 9.1.4 with & & Q W V c T S p = = + = 0 2 2 is: cons't. V c T V V c T T V V V V c T c T p p p p 2 2 2 2 2 2 2 2 + = + + + = + + + + ( ) ( ) ( ) . ∆ ∆ ∆ ∆ ∆ 2 2 ( ) 0 2 2 V V V ∆ ∆ ∴ = + . . p p c T V V c T h + ∆ ∴− ∆ = ∆ = ∆ We neglected (∆V)2 . The velocity of a small wave is V c h c V = ∴ = − . . ∆ ∆ 9.8 For water ρ ρ dp d = × 2110 106 Pa Since ρ = 1000 kg / m we see that 3 ,
- 162. 202 c dp d =/ ρ = × = 2110 10 1000 1453 6 / m / s 9.9 For water c p dp d = ≅ = × = ∆ ∆ρ ρ 2110 10 1000 1453 6 m /s. L = × × velocity time = 1453 0.6 = 872 m. 9.10 Since c = 1450 m/s for the small wave, the time increment is ∆t d c = = = 10 1450 0 0069 . seconds 9.11 a) M = = × × = V c 200 1 4 287 288 0 588 . . . b) M 600/ 1.4 1716 466 0.567. = × × = c) M = × × = 200 1 4 287 223 0 668 / . . . d) M 600/ 1.4 1716 392 0.618. = × × = e) M = × × = 200 1 4 287 238 0 647 / . . . 9.12 c kRT d ct = = × × = ∴ = = × = 1 4 287 263 256 256 1 21 309 . . . m / s. m 9.13 a) Assume T = 20°C: c kRT = = × × = 1 4 287 293 343 . m /s. d c t = = × = ∆ 343 2 686 m b) Assume T = 70°F: c kRT = = × × = 1 4 1716 530 1130 . fps. d c t = = × = ∆ 1130 2 2260 ft. For every second that passes, the lightning flashed about 1000 ft away. Count 5 seconds and it is approximately one mile away. 9.14 c M c V = × × = = = 1 4 287 263 256 1 . sin . m / s. α 1000 sin 0.256. tan 0.2648 . 3776 m L L α α = ∴ = = ∴ = ∆t = = 3776 1000 3 776 . . s 1000 m V L
- 163. 203 9.15 Use Eq.9.2.13: a) c V V = = × × = sin . sin α or m /s 1 4 287 288 22 908 o b) c V V = = × × = sin . sin α or fps 1 4 1716 519 22 2980 o 9.16 Eq. 9.2.4: ∆ ∆ ∆ V p c p kRT = − = − = − × × = − ρ ρ 0 3 00237 1 4 1716 519 0 113 . . . . . fps Energy Eq: V c T V V c T T V V c T p p p 2 2 2 2 0 + = = + + ∴ = + ( ) ( ). . ∆ ∆ ∆ ∆ ∴ = − = − × × − × × = ∆ ∆ T c V cp / . ( . ) /( . . ) . . 1 4 1716 519 113 0 24 778 32 2 0 021o F Note: cp = × × = × × . . . . . 24 778 32 2 24 778 32 2 Btu lbm - F ft - lb Btu lbm slug ft - lb slug - F o o Then ft ft - lb /(slug - F) ft lb - sec F sec ft - lb - ft F. 2 2 2 2 /sec 2 o o o = − − − = (units can be a pain!) 9.17 a) ρ ρ ρ ρ ρ ρ ρ ρ ρ AV AV AdV AVd Ad dV VdA dAdV Vd dA d dAdV = + + + + + + + Keep only the first order terms (the higher order terms—those with more than one differential quantity—will be negligible): 0 = + + ρ ρ ρ AdV AVd VdA Divide by ρAV: dV V d dA A + + = ρ ρ 0 b) Expand the r.h.s. of Eq. 9.3.5 (keep only first order terms): V k k p V VdV k k p dp d 2 2 2 1 2 2 1 + − = + + − + + ρ ρ ρ . Hence, 0 2 2 1 = + − + + − VdV k k p dp d p ρ ρ ρ = + − + − − + VdV k k p dp p pd d 1 2 ρ ρ ρ ρ ρ ρ ρ = + − − VdV k k dp pd 1 2 ρ ρ ρ where we neglected ρ ρ ρ d compared to 2 . For an isentropic process Eq. 9.2.8 gives ρ ρ dp kpd = , so the above becomes
- 164. 204 0 1 2 = + − − VdV k k kpdpd ρ ρ ρ = + − − VdV k k k pd 1 1 2 ( ) ρ ρ = + VdV k p d ρ ρ 2 But d dV V dA A ρ ρ / / / = − − so that the above equation is 0 = + − − VdV k p dV V dA A ρ which can be written as dA A V kp dV V = − 2 1 ρ . Since c kp 2 = / , ρ and M = V/c, this is put in the form dA A V c dV V = − 2 2 1 or ( ) dA A dV V = − M2 1 c) Substituting in V c c kRT R c k k p = = = − M and we find , , / ( ) / , 2 1 T T V c T c c T kRT c T p p p 0 2 2 2 2 2 1 2 1 2 1 = + = + = + M M = − + = + − M M 2 2 1 2 1 1 1 2 k k k k ( ) . d) & /( ) / m p k TR A p k T k k R A k k = = + − + − − − M M M M 0 2 1 0 2 1 2 1 1 2 1 1 2 = + − + − p k RT A k k k 0 0 2 1 2 1 1 1 2 M M ( ) At the critical area A* * , . M = 1 Hence, & . * ( ) m p k RT A k k k = + + − 0 0 1 2 1 1 2 e) Since & m is constant throughout the nozzle, we can equate Eq. 9.3.17 to Eq. 9.3.18: p k RT A k p k RT A k k k k k 0 0 2 1 2 1 0 0 1 2 1 1 1 2 1 2 M M + − = + + − + − ( ) * ( ) or A A k k k k * ( ) ( ) = + − + + − 1 2 1 1 2 1 2 1 M M
- 165. 205 9.18 a) atm10 69.9 10 79.9 kPa abs. s p p = + = + = 1 69.9 kPa abs. p = From 1 → s : V p p p p s s s s k 1 2 1 1 1 1 1 1 1 4 2 906 79 9 69 9 0 997 + = = = = ρ ρ ρ ρ . . . . . . / / . kg /m3 ∴ + = ∴ = V V 1 2 1 2 69 79 77 3 900 .906 900 .997 m / s . . . b) p p s = + = = 26 4 10 36 4 26 4 1 . . . kPa abs. kPa abs. From 1 → s : V p p p p s s s s k 1 2 1 1 1 1 1 1 1 4 2 0 412 36 4 26 4 0 518 + = = = = ρ ρ ρ ρ . . . . . . / / . kg /m3 V V 1 2 1 2 26 36 111 + = ∴ = 400 .412 400 .518 m /s . . 9.19 a) 1/ 1/1.4 2 3 1 1 1 1 1 105 . 1.22 1.254 kg/m . 2 101 k s s s s p p V p p ρ ρ ρ ρ + = = = = 2 1 1 1.4101 000 105 0001.4 . 81.3 m/s. 2 .4 1.22 1.254 .4 V V + = ∴ = b) 2 1 1 4000 81.3-81 . 81.0 m/s. % error = 100 0.42%. 2 1.22 81.3 V V = ∴ = × = 9.20 Is p p r < × = kPa. . ? . . 5283 0 5283 200 105 7 0 a) p p V kRT p r e e e e < ∴ ∴ = ∴ = = choked flow. M kPa. . . . . . 5283 1 105 7 0 2 1000 298 1 4 287 2 1000 248 1 315 8 × = × + ∴ = = . . . . T T T V e e e e K, m / s. ρ π e m = × = ∴ = × × × = 105 7 287 248 1 1 484 1 484 01 315 8 0 1473 2 . . . . . & . . . . . kg / m kg / s 3 b) p p V r e e > ∴ < × + = M 1000 298 = 000 e 2 e . . . . . . . . 5283 1 2 1 4 4 130 130 200 2 338 0 1 4 ρ ρ ρ ρ 0 200 287 298 2 338 1 7187 257 9 = × = ∴ = ∴ = . . . . . . kg / m m / s. 3 e e V ∴ = × × × = & . . . . . m 1 7187 01 257 9 0 1393 2 π kg /s 9.21 Is p p r < × = psia. . ? . . 5283 0 5283 30 15 85 0 a) p p V kRT r e e e < ∴ = = = 15 choked flow and M psia. . . , . . 85 1 15 85 2 0 24 530 1 4 1716 2 778 32 2 0 24 441 7 1030 . . ( . ) . . . × = × × × + ∴ = = T T T V e e e e R, fps. o V s 1 Vs=0
- 166. 206 3 15.85 144 0.003011 slug/ft. 1716 441.7 e ρ × = = × 2 .5 .003011 1030 0.01692 slug/sec. 12 m π ∴ = × × = & b) 15.85. M 1, and 20 psia. r e e p p > ∴ < = 3 0 30 144 .00475 slug/ft . 1716 530 ρ × = = × 1/1.4 3 20 .00475 .003556 slug/ft . 30 e ρ ∴ = = 2 1.4 20 144 0.24 530(778 32.2) . 2 .4 .003556 e V × × × = + ∴ = ∴ = × × = V m e 838 9 003556 5 12 838 9 0 01627 2 . & . . . . . fps. slug /sec π (Note: ft-lb ft-lb 0.24 Btu/lbm- R=0.24 778 0.24 778 32.2 .) lbm- R slug- R cp = × = × × o o o 9.22 a) p p p T r e e e < ∴ = ∴ = × = = × = M kPa. K. . . . . . . . 5283 1 5283 200 105 7 8333 298 248 3 0 ρe e V = × = = × × = 105 7 287 248 3 1 483 1 4 287 248 3 315 9 . . . . . . . . kg / m m / s. 3 ∴ = × × × = & . . . . . m 1 483 01 315 9 0 1472 2 π kg /s b) p p p p p T T r e e e e > ∴ = = ∴ = = kPa, M . . . . . , . 5283 130 0 65 81 884 0 0 0 ρe e V = × = = × × = 130 287 263 4 1 719 81 1 4 287 263 4 263 5 . . . , . . . . kg / m m / s. 3 ∴ = × × × = & . . . . . m 1 719 01 263 5 0 1423 2 π kg /s 9.23 a) p p p r e e < ∴ = ∴ = × = M psia. . . . . . 5283 1 5283 30 15 85 0 Te = × = . . 8333 530 441 6o R. ∴ = × × = = × × = ρe e V 15 85 144 1716 441 6 003012 1 4 1716 441 6 1030 . . . . . . slug ft fps. 3 & . . . . m = × × = 003012 5 12 1030 0 01692 2 π slug /sec b) p p p p p T T r e e e e > ∴ = = = ∴ = = psia. M . . . . . . . . 5283 20 20 30 6667 785 0 890 0 0 0 ∴ = × × = = × × = ρ0 20 144 1716 472 00356 785 1 4 1716 472 836 . . . . fps. Ve ∴ = × = & . . . . m 00356 5 12 836 0 01664 2 π slug /sec
- 167. 207 9.24 p T ee = × = = × = . . . . . 5283 400 211 3 8333 303 252 5 kPa abs K. V m e = × × = ∴ = × × × = 1 4 287 252 5 318 5 211 3 287 252 5 05 318 5 7 29 2 . . . & . . . . . . . m / s. kg /s π 9.25 p p p T e e = = ∴ = = × = . . . . . . 5283 101 191 2 8333 283 235 8 0 0 kPa kPa abs K. V m e = × × = ∴ = × × × = 1 4 287 235 8 307 8 101 287 235 8 03 307 8 1 30 2 . . . & . . . . . . m / s. kg /s π p p p T e e 0 0 2 191 2 382 4 5283 202 0 235 8 = × = = = = . . . . . kPa abs. kPa abs. K. V m e e = = ∴ = × × × = 307 8 1 202 287 235 8 03 307 8 2 60 2 . . & . . . . . . m / s since M kg / s π 9.26 p p p T e e = = ∴ = = × = . . . . . . . 5283 14 7 27 83 8333 500 416 6 0 0 psia psia R. o 1.4 1716 416.6 1000 fps. e V = × × = 3 0.3203 kg/m and 199.4 kPa abs. e e p ρ ∴ = = 0 0 2 27.83. 0.5283 29.4 psia, 416.6 R, 1000 fps. e e e p p p T V = × = = = = o ∴ = & . . m 0 202 slug / sec 9.27 Treat the pipeline as a reservoir. Then, p p e = = . . 5283 264 5 0 kPa abs. M and m /s. e e V = = × × = 1 1 4 287 8333 283 307 8 . (. ) . & . . (. ) . . m = × × × × × = − 264 5 287 8333 283 30 10 307 8 3 61 4 kg /s. ∆ ∆ − = = × × × × = V m t & . . / (. . ) . ρ 361 6 60 2645 287 8333 283 333 m3 9.28 5193 300 1 667 2077 2 5193 225 200 225 300 1 667 667 × = × + ∴ = ∴ = . . . . K. T T T p e e e e =97.45 kPa. Next, T p V t t t = = ∴ = × × = 225 97 45 1 667 2077 225 882 6 K, kPa; m /s. . . . 3 2 2 97.45 = 0.2085 kg/m . 0.2085 × × .03 × 882.6 = 0.075 2.077 225 t e e V ρ π ρ π = × × 5193 300 2 1 667 667 200 200 2 077 300 1330 2 1 667 1 667 × = + = × = V p p e e e e e e . . . / . . . ρ ρ ρ kPa. = + × × − V V e e 2 3 667 2 3324 10 9 54 . . . or 3 116 10 63 420 10 91 8 6 2 3 667 . . . . × = + × = − V V V e e e Trial - and - error: m /s. 3 0.3203 kg/m and 199.4 kPa abs. e e p ρ ∴ = =
- 168. 208 9.29 ρ ρ 1 1 1 2 11 4 300 100 287 293 4 757 4 757 340 400 4 236 = = + × = = = p RT . . . . . . / . kg / m kg /m 3 3 V V V V 1 2 2 2 2 1 4 757 10 4 236 5 4 492 × × = × × ∴ = . . . . . V k k p V k k p V V 1 2 1 1 2 2 2 2 1 2 2 1 2 2 1 2 1 2 1 4 4 400 4 492 2 1 4 4 340 + − = + − + = + ρ ρ . . . . . . . 000 4.757 000 4.236 ∴ = V1 37 35 . . m s ∴ = = × × × = & . . . . . m A V ρ π 1 1 1 2 4 757 05 37 35 1 395 kg /s 9.30 ρ1 1 1 45 14 7 144 1716 520 0 009634 = = + × = p RT ( . ) . . slug ft3 slug /ft3 ρ2 1 1 4 009634 50 7 59 7 008573 = = . . . . . / . V V V V 1 2 2 2 2 1 009634 4 008573 2 4 495 × × = × × ∴ = . . . . . V V V 1 2 2 1 2 1 2 1 4 4 59 7 144 4 495 2 1 4 4 50 7 144 121 9 + × = + × ∴ = . . . . . . . . . .009634 .008573 fps. ∴ = × × = & . ( / ) . . . m 009634 2 12 121 9 0 1025 2 π slug /sec 9.31 Energy 0 → 2: 1000 303 2 1000 3 2 2 2 2 2 × = + = V T V kRT . 2 1.627 20 32.5 kPa. p ∴ = × = l 1.4 .4 3 2 2 107.9 5.39 200 5.390 kPa. 0.1740 kg/m . 303 .287 107.9 p ρ ∴ = = = = × Energy 0 → 1: 1000 303 2 1000 1 4 287 318 4 252 3 1 2 1 1 × = + × ∴ = = V V T V m /s, K. 1 2 . . . . p1 1 4 4 1 200 252 3 303 105 4 105 4 287 252 3 1 455 = = = × = . . . . . . . . . kPa. kg /m3 ρ Continuity: 1 455 05 318 4 174 4 3 1 4 287 107 9 0 2065 2 2 2 2 . . . . . . . . . π π × × = × × × ∴ = d d m 9.32 V kRT T T T V t t t t t t 2 1000 293 1 4 287 2 1000 244 0 313 1 = × = × + ∴ = = . . . . . K. m / s. ∴ = = ∴ = × = pt t 500 244 293 263 5 263 5 287 244 3 763 1 4 4 . . . . . . . kPa abs. kg /m3 ρ 0 1 2
- 169. 209 2 2 2 1.4 1.4 1.4263 500 1000 293 . 3.763 .025 313.1 .075 . 2 .4 3.763 e e e e e e e V p p V π ρ π ρ ρ × = + × × × = × = 2 6 .4 293 000= 1.014 10 . Trial-and-error: 22.2 m/s, 659 m/s. 2 e e e V V V − ∴ + × = ∴ = ∴ = ρe e p 5 897 0 1987 494 2 4 29 . , . . . , . . kg / m kPa kPa abs 3 9.33 * 0 9. 0.997 from Table D.1. 500 .997 498.5 kPa. e e e A p p p A = ∴ = ∴ = × = and 0 0.00855 from Table D.1. 4.28 kPa abs. e e p p p = ∴ = 9.34 M psia, R. t = ∴ = × = = × = 1 5283 120 63 4 8333 520 433 3 . . . . . p T t t o ∴ = ρt . . 01228 slug ft3 & . . . . . . m d d t t = = × × ∴ = 1 01228 4 1 4 1716 433 3 0 319 2 π ft p p T V e e 0 15 120 125 2 014 552 520 287 2 014 1 4 1716 287 = = ∴ = = × = = × × . . . , . . . M R, e o = 684 fps. A A d d e e * . . . . . . . = ∴ = × ∴ = 1 708 4 1 708 319 4 0 417 2 2 ft π π 9.35 * M 4. 10.72, .006586 2000 13.17 kPa, .2381 293 69.76 K. e e e A p T A = = = × = = × = For A A p p e * . , . . . . . . = = ∴ = = × = 10 72 0584 9976 9976 2000 1995 2 0 M kPa abs e 9.36 Let 1 150 M 1. Neglect viscous effects. M 0.430. 1.4 287 303 t = = = × × 2 2 1 * .05 1.5007. . 0.0816 m or 8.16 cm. 1.5007 1.5007 4 t t t d A A A d A π π × ∴ = ∴ = = = ∴ = 9.37 p T e es = × = = × = . . . . . 5283 400 211 3 8333 303 252 5 kPa abs. 303 .96 . 254.5 K. 1.4 287 254.5 319.8 m/s. 303 252.5 e e e T T V − = ∴ = ∴ = × × = − 2 211.3 .05 319.8 7.27 kg/s. .287 254.5 m π ∴ = × × = × &
- 170. 210 9.38 Isentropic flow.Since k = 1.4 for nitrogen, the isentropic table may be used. M = = 3 4 235 : . . * A A 3 100 3 1.4 297 373 1181 m/s. .9027 kg/m . .297 373 i i V ρ = × × = = = × ∴ = = × = ∴ = = A m V A i i i t & . . . . . . . ρ 10 9027 1181 0 00938 00938 4 235 0 00221 m m 2 2 At M = = = 3 3571 02722 0 0 , . , . . T T p p ∴ = = = = = = T T p p e e 0 0 373 3571 1044 100 02722 3670 . . . . K or 772 C kPa o 9.39 Isentropic flow. Since k = 1.4 for nitrogen, the isentropic table may be used. M = = 3 4 235 : . . * A A Vi i = × × = = × × = 3 1 4 1776 660 3840 15 144 1776 660 001843 . . . fps. slug ft3 ρ A A i t = × = ∴ = = . . . . . . . . 2 001843 3840 0283 0283 4 235 0 00667 ft ft 2 2 At M = = = 3 3571 02722 0 0 , . , . . T T p p ∴ = = = = = = T T p p e e 0 0 660 3571 1848 15 02722 551 . . . . o o R or 1388 F psia 9.40 Assume pe e = = × = 101 101 189 1273 4198 kPa. Then kg / m3 ρ . . . Momentum: 2 2 2 80 000 9.81 . .4198 .25 . 6 F mV AV V ρ π × = = = × & 1260 m/s. V ∴ = 9.41 F mV AV = = = × = & . . . . ρ ρ 2 101 287 873 403 kg / m (Assumegases are air.) 3 4 2 100 9.81 .403 200 10 . 349 m/s. V V − × = × × ∴ = 9.42 M M t e = = ∴ = = 1 4 2 94 02980 0 . ; . , . . * A A p p e e 0 .3665 .3665 300 109.95 K, e T T = = × = 0 0 100 .0298 . 3356 kPa abs. e p p p = = ∴ = i t e Ve = 0 Mt = 1 M > 1 M < 1 ~ i t e Ve = 0 Mt = 1 M > 1 M < 1 ~ Ve FB p0 A0
- 171. 211 ∴ = ×× = Ve 2 94 1 4 287 109 95 618 . . . m / s. ∴ = − × × × + × = FB 100 287 109 95 05 618 3 2 2 2 2 . . . . . π π 356 000 412 000 N 9.43 Assume an isentropic flow; Eq. 9.3.13 provides 103 1 1 2 2 1 1 . . p p k k = + − − M Using k = 1.4 this gives M or M 2 00424 0 206 = = . . . For standard conditions V c = = × × = M m / s 0206 14 287 288 70 . . . 9.44 a) 2 2 2 2 0.9850 1000 . 80 000 0.985 1000( 1000) V p V ρ × = − = × − V p 2 2 2 2 2 1 1000 2 1 4 4 287 283 0 80 287 283 9850 − + − × = = × = . . . . . . ρ ρ kg /m3 2 2 2 2 2 1000 1.4 ( 985 1 065 000) 284 300 = 0 2 2 .4 985 V V V − + − + − ∴ − + ∴ = = 3 3784 784 261 3 774 2 2 2 2 2 V V V 300 = 0. m /s. kg / m3 ρ . . Substitute in and find p2 808 = kPa. M K or 473 C 1 2 1000 1 4 287 283 2 966 808 287 3 774 746 = × × = = × = . . . . . . T o M2 261 1 4 287 746 0 477 = × × = . . . b) M M 0.477. kPa 1 2 2 1 1000 1 4 287 283 2 97 10 12 809 6 = × × = ∴ = = = / . . . . . . p p T2 2 2 644 283 748 809 6 287 748 3 771 = × = ∴ = × = . . . . . . K or 475 C kg / m3 o ρ 9.45 a) ρ ρ 1 2 2 12 144 1716 500 002014 002014 3000 = × × = × = . . . . slug ft3 V Momentum: 2 2 12 144 .002014 3000( 3000). p V × − = × − 2 2 2 2 2 3000 1.4 1716 500 0. 2 .4 V p ρ − + − × = V V V 2 2 2 2 2 6 3000 7 6 042 19 854 6 042 6 006 10 − + − − × . ( , . ) . = 0. ∴ − + × ∴ = = 6 23 15 10 833 0 00725 2 2 2 6 2 2 V V V , . . 000 =0. fps. slug ft3 ρ p2 102 9 = . . psia M R or 731 F 1 2 3000 1 4 1716 500 2 74 102 9 144 1716 00725 1191 = × × = = × × = . . . . . . T o o
- 172. 212 M2 833 1 4 17161191 0 492 = × × = . . . b) 1 2 2 M 3000/ 1.4 1716 500 2.74. M 0.493. 8.592 12 103.1 psia. p = × × = ∴ = = × = T2 2 2 386 500 1193 103 1 144 1716 1193 0 00725 = × = ∴ = × × = . . . . . o o R or 733 F slug / ft3 ρ 9.46 [ ] ρ ρ 2 1 2 1 1 2 1 2 2 1 2 1 2 1 2 1 2 1 2 2 1 1 1 1 1 2 4 2 2 1 2 1 = = − + + + + − − + = + + − p p T T k k k k k k k k k M M M M M M ( ) ( ) ( ) . M1 2 2 1 1 2 1 2 = + + − k k p p k k . (This is Eq. 9.4.12). Substitute into above: ρ ρ 2 1 2 1 2 1 2 1 2 2 1 1 1 1 4 1 1 1 1 1 1 1 1 1 = + + + − + − + + − = + + + − + + − + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) . k k p p k k k k p p k k k p p k k k k p p = − + + + + − k k p p k k p p 1 1 1 1 2 1 2 1 ( ) / ( ) / . For a strong schock in which p p k k 2 1 2 1 1 1 1 >> = + − , . ρ ρ 9.47 Assume standard conditions: T1 1 15 101 = = o C, kPa. ρ ∴ = × × = V1 2 1 4 287 288 680 . m / s. 1 2 2 M 2. M .5774. 1.688 288 486 K. T = ∴ = = × = 2 4.5 101 454 kPa. p = × = 2 .5774 1.4 287 486 255 m/s. V ∴ = × × = induced 1 2 680 255 425 m/s. V V V ∴ = − = − = The high pressure and high induced velocity cause extreme damage. 9.48 If M then M m /s 2 1 1 5 2 645 2 645 1 4 287 293 908 = = ∴ = × × = . , . . . . . V p2 2 8 00 200 1600 1600 287 2 285 293 8 33 = × = = × × = . . . ( . ) . . kPa abs kg /m3 ρ 9.49 If M then M fps 2 1 1 5 2 645 2 645 1 4 1716 520 1118 = = ∴ = × × = . , . . . . . V p2 2 8 00 30 240 240 144 1716 2 285 520 0 01695 = × = = × × × = . . ( . ) . . psia slug /ft3 ρ V1 V2 stationary shock
- 173. 213 9.50 p T 11 1 2615 101 26 4 223 3 1000 1 4 287 223 3 3 34 = × = = = × × = . . . / . . . . kPa. K. M ∴ = = × = = × = M kPa. K. 2 2 2 4578 12 85 26 4 339 3 101 223 3 692 5 . . . . . . . p T For isentropic flow from ‚ → €: For M = .458, p = .866 p0 and T T p T = ∴ = = = = . . /. . . /. . 960 339 866 391 692 5 960 721 0 0 0 kPa abs K or 448 C o 9.51 After the shock M kPa abs. 2 2 4752 10 33 800 8264 = = × = . , . p For isentropic flow from ‚ → €: For M = .475, p = .857 p0 . ∴ = = p0 8264 857 9640 /. . kPa abs 9.52 A A p p p e * . . . . /. . . = ∴ = = ∴ = = 4 147 985 101 985 102 5 0 0 M kPa abs e M kPa. K. t = = × = = × = 1 5283 102 5 54 15 8333 298 248 3 . . . . . . p T t t ∴ = × = = × × = ρt t V 5415 287 2483 7599 14 287 248 3 3159 . . . . . . . . kg / m m / s. 3 ∴ = × × × = & . . . . . m 7599 025 315 9 0 471 2 π kg /s If throat area is reduced, Mt remains at 1, ρ π t m = = × × × = . & . . . . . 7599 7599 02 315 9 0 302 2 kg /m and kg /s 3 9.53 p p A A p p e = = ∴ = = 101 4 2 94 9 918 2 1 2 1 kPa = M and . . . , / . . * ∴ = = = = p p p 1 1 0 101 9 918 10 18 2 94 0298 / . . . , / . . kPa. At M ∴ = = p0 10 18 0298 342 . /. . kPa abs M kPa abs K. t = = × = = × = 1 5283 342 181 8333 293 244 1 , . . . . p T t t ∴ = × × = Vt 1 4 287 244 1 313 . . . m /s M kPa abs K. 1 1 1 2 94 10 18 3665 293 107 4 = = = × = . , . . . . p T ∴ = × × = V1 2 94 1 4 287 107 4 611 . . . . m /s M kPa K. 2 2 4788 101 2 609 107 4 280 2 = = = = × = . , . . . . p T T e e ∴ = × × = V2 4788 1 4 287 280 2 161 . . . . m /s 9.54 p p A A p p e = = ∴ = = 14 7 4 2 94 9 918 2 1 2 1 . . . . , / . . * psia = M and ∴ = = = = p p p 1 1 0 14 7 9 918 1 482 2 94 0298 . / . . . , / . . psia. At M ∴ = = p0 1 482 0298 49 7 . /. . . psia M psia R. t = = × = = × = 1 5283 49 7 26 3 8333 520 433 3 , . . . . . . p T t t o ∴ = × × = Vt 1 4 1716 433 3 1020 . . . fps M psia R. 1 1 1 2 94 1 482 3665 520 190 6 = = = × = . , . . . . p T o ∴ = × × = V1 2 94 1 4 1716 190 6 1989 . . . . fps
- 174. 214 M psia R. 22 4788 14 7 2 609 190 6 497 3 = = = = × = . , . . . . . p T T e e o ∴ = × × = V2 4788 1 4 1716 497 3 523 . . . . fps 9.55 M kPa K. t t t p T = = × = = × = 1 5283 500 264 8333 298 248 3 . . . . . A A p 1 2 2 1 1 8 5 2 56 2 47 0613 500 30 65 * . . . , . . . = = ∴ = = × = M T V 1 1 451 298 134 4 2 47 1 4 287 134 4 574 = × = ∴ = × × = . . . . . . K. m /s M kPa K. 2 2 2 516 6 95 30 65 213 2 108 134 4 283 3 = = × = = × = . , . . . . . . p T After the shock it’s isentropic flow. At M = = . , . . * 516 1 314 A A p A 02 2 511 500 255 5 04 1 314 003825 = × = = × = . . . . . . * kPa. m2 π A A p p e e r * . . . . . . . . . . = × = ∴ = × = = π 05 003825 2 05 940 255 5 240 298 2 kPa abs = Me T V e e = = ∴ = × × = 283 3 213 240 273 8 298 1 4 287 273 8 99 2857 . . . . . . . K. m /s 9.56 p p T t t = = × = = = . . . / . 546 546 1200 655 673 655 1200 585 0 3 1 3 kPa. K. ∴ = × = = × × = = ρt t V 655 462 585 2 42 1 3 462 585 593 1 . . . . ( .) kg / m m /s. M 3 t & . . . . . m A V d d t t t t t = ∴ = × × × ∴ = ρ π m or 6 cm 4 2 42 4 593 0 060 2 Te e = = ∴ = × = 673 101 1200 380 2 101 462 380 2 575 3 1 3 . / . . . . . . K kg /m3 ρ Ve 2 2 1872 380 2 1872 673 + × = × . . (Energy from € → e .) (cp = ⋅ 1872 J / kg K) ∴ = ∴ = × ∴ = V d d e e e 1050 4 575 4 1050 0 092 2 m /s. m or 9.2 cm . . . . π 9.57 M kPa. e = = = × = 1 546 546 1000 546 0 . . . p p e Te e = = ∴ = × = 623 546 1000 542 546 462 542 2 18 3 1 3 . . . . . K. kg m3 ρ V d d e e e = × × = = × ∴ = 1 3 462 542 571 15 2 18 4 571 0 124 2 . . . . . m / s. m or 12.4 cm π
- 175. 215 9.58 M psia.R. e = = × = = = 1 546 150 81 9 1160 81 9 150 1009 3 1 3 . . . . . . p T e e o ∴ = × × = = × × = ρe e V 81 9 144 2762 1009 00423 1 3 2760 1009 1903 . . . . slug ft fps. 3 . . . . . 25 00423 4 1903 0 199 2 = × ∴ = πd d e e ft. or 2.39" 9.59 M kPa. K. t = = × = = = 1 546 1200 655 673 655 1200 585 3 1 3 . . . / . p T t t ∴ = × × = = × = Vt t 1 3 462 585 593 655 462 585 2 42 . . . . m / s. kg / m3 ρ ∴ = × × × = & . . . m 2 42 0075 593 0 254 2 π kg /s per nozzle Te = = 673 120 1200 396 3 1 3 . / . . K 9.60 M1 800 1 4 287 303 2 29 = × × = . . . From Fig. 9.15, β = 46 79 o o , . a) 1n 46 . M 2.29sin46 1.65. β = ∴ = = o o 2n 2 M .654 M sin(46 20 ). ∴ = = − o o ∴ = M2 1 49 . . p T 2 2 3 01 40 120 4 1 423 303 431 = × = = × = . . . . kPa abs K. V2 1 4 287 431 1 49 620 = × × × = . . . m /s b) β = ∴ = = ∴ = = − 79 2 29 79 2 25 541 79 20 2 o o o o . . sin . . . sin( ). M M M 1n 2n ∴ = M2 0 631 . . p T 2 2 5 74 40 230 1 90 303 576 = × = = × = . . . kPa abs K. V2 1 4 287 576 631 303 = × × × = . . . m /s c) V1 V2 = 20o V1 = 35o a detached shock
- 176. 216 9.61 β θ 140 10 = ∴ = o o . . M M M M =1.58. 1n 2n = = ∴ = = − ∴ 2 40 1 29 791 40 10 2 2 sin . . . sin( ). o o o If θ β 2 2 10 1 58 51 1 58 51 1 23 = = = = ∴ = o o o then, with M M M 2n 2n . , . . sin . . . ∴ = = − ∴ = = − = − = M M M 3n . sin( ). . . . 824 51 10 1 26 10 51 10 41 3 3 2 o o o β β 9.62 M M K. 1n 2n = = ∴ = = × = 3 5 35 2 01 576 1 696 303 514 2 . sin . . . . . o T M2 1 2 2 576 35 20 2 26 20 47 = − = = = ∴ = . sin( ) . . . . o o o o θ θ β M M M M 2n 3n 3 = = ∴ = = − ∴ = 2 26 47 1 65 654 47 20 1 44 3 . sin . . . sin( ). . . o o o T V kRT 3 3 3 3 1 423 514 731 1 44 1 4 287 731 780 = × = = = × × = . . . . K. M m /s 9.63 M M R. 1n 2n = = ∴ = = × = 3 5 35 2 01 576 1 696 490 831 2 . sin . . . . . o o T M2 1 2 2 576 35 20 2 26 20 47 = − = = = ∴ = . sin( ) . . . . o o o o θ θ β M M M M 2n 3n = = ∴ = = − ∴ = 2 26 47 1 65 654 47 20 1 44 3 3 . sin . . . sin( ). . . o o o T V kRT 3 3 3 3 1 423 831 1180 1 44 1 4 1716 1180 2420 = × = = = × × = . . . . o R. M fps 9.64 M M M 1n 2n 1 1 3 10 28 3 28 1 41 736 = = ∴ = = = ∴ = , . . sin . . . . θ β o o o ∴ = × = p2 2 153 40 86 1 . . kPa. M kPa 2 3 736 28 10 2 38 6 442 86 1 555 = − = ∴ = × = . sin( ) . . . . . o o p ( ) . . p3 10 33 40 413 normal kPa = × = 9.65 At M1 1 1 3 49 8 19 47 = = = , . , . . θ µ o o (See Fig. 9.18.) θ θ 1 2 2 49 8 25 74 8 4 78 + = + = ∴ = . . . . . o M From isentropic flow table: p p p p p p 2 1 0 1 2 0 20 1 02722 002452 1 80 = = × × = . . . . kPa 0 2 2 1 2 1 0 1 253 .1795 127K or 146 C. 12.08 . .3571 T T T T T T µ = = × × = − = o o V2 4 78 1 4 287 127 1080 90 25 70 53 12 08 32 4 = × × = = + − − = . . . . . . . m /s α o 9.66 θ θ 1 26 4 4 65 8 = = = . . , . . o o For M (See Fig. 9.18.) ∴ = − = θ 65 8 26 4 39 4 . . . . o T T T T T T V 2 1 0 1 2 0 2 273 1 5556 2381 117 4 1 4 287 117 867 = = × = ∴ = × × = . . . . K. m /s T2 156 = − o C.
- 177. 217 9.67 θ θ == = 26 4 4 65 8 . . , . . o o For M ∴ = − = θ 65 8 26 4 39 4 . . . . o 0 2 2 1 1 0 1 490 .2381 210 R or 250 F. .5556 T T T T T T = = × = − o o V2 4 1 4 1716 210 2840 = × × = . . fps 9.68 a) θ θ 1 2 2 39 1 39 1 5 44 1 2 72 20 1 0585 04165 = = + = ∴ = = × . . . . . . . . . o o Mu u p = 14.24 kPa. For θ β = = = = = ∴ = 5 2 5 27 2 5 27 1 13 889 o o o and M M M 1n 2n . , . . sin . . . . ∴ = × = p2 1 32 20 26 4 l . . . kPa M M l o o = = − = 2 889 27 5 2 37 . sin( ) . . b) M M M 1n 2n = = ∴ = = = = 2 72 5 25 2 72 25 1 15 875 . , . . . sin . , . . θ β o o o M2u = − = . sin( ) . . 875 25 5 2 56 o o For M = 36.0 For M2 = = + = = 2 37 36 5 41 2 58 . , . , . . θ θ o o l c) Force on plate = ( . . ) . 26 4 14 24 1000 − × × = A F C F V A A A L = = × × × × × = cos . . . . . . 5 1 2 12 2 996 1000 1 2 1 4 2 5 20 000 0 139 1 1 2 2 o ρ d) C F V A A A D = = × × × × × = sin . . . . . . 5 1 2 12 2 1000 0872 1 2 1 4 2 5 20 000 0 0122 1 1 2 2 o ρ 9.69 β = = = ∴ = × = = 19 4 19 1 30 1 805 20 36 1 786 2 o o . sin . . . . . . M kPa. M 1n 2n p M M 2 1 2 3 786 19 5 3 25 54 36 59 36 3 55 = − = = = ∴ = . sin( ) . . . . . . . . o o θ θ p p p p p p 3 2 0 2 3 0 36 1 1 0188 0122 23 4 = = × × = . . . . kPa. C A A V A D = − = × × × × = 36 1 2 23 4 2 5 1 2 6 35 0872 1 2 1 4 4 20 0 0025 1 2 2 . . sin . . . . . o ρ Lift Drag F Airfoil surface M1 M2 M3 p2 p3 shock
- 178. 218 9.70 If θβ = = → 5 4 1 o o with M then Fig. 9.15 =18 , . 1n 2n M 4sin18 1.24. M .818. = = ∴ = o 2 1.627 20 32.5 kPa. p ∴ = × = l ∴ = − = M2l o o . sin( ) . . 818 18 5 3 64 At M At M2u u 1 1 2 1 0 2 0 4 65 8 75 8 4 88 20 002177 006586 = = = = = , . . . . . . . θ o o p p p p p p = 6.61 kPa. C V A A A A A L = = − × − × × × × × = Lift 1 2 325 5 20 2 6 61 2 10 1 2 14 4 20 0 0854 1 2 2 ρ . cos / . / cos . . . o o C V A A A A D = = − × × × × × = Drag 1 2 32 5 5 6 61 2 10 1 2 1 4 4 20 0 010 1 2 2 ρ . sin . / sin . . . o o M1 M2u M2l shock shock