INSTRUCTOR'S SOLUTIONS MANUAL
TO ACCOMPANY
MECHANICS of FLUIDS
FOURTH EDITION
MERLE C. POTTER
Michigan State University
DAVID C. WIGGERT
Michigan State University
BASSEM RAMADAN
Kettering University
Contents
Chapter 1
Basic Considerations
1
Chapter 2
Fluid Statics
15
Chapter 3
Introduction to Fluids in Motion
43
Chapter 4
The Integral Forms of the Fundamental Laws
61
Chapter 5
The Differential Forms of the Fundamental Laws
107
Chapter 6
Dimensional Analysis and Similitude
125
Chapter 7
Internal Flows
145
Chapter 8
External Flows
193
Chapter 9
Compressible Flow
237
Chapter 10
Flow in Open Channels
259
Chapter 11
Flows in Piping Systems
303
Chapter 12
Turbomachinery
345
Chapter 13
Measurements in Fluid Mechanics
369
Chapter 14
Computational Fluid Dynamics
375
Chapter 1/ Basic Considerations
CHAPTER 1
Basic Considerations
FE-type Exam Review Problems: Problems 1-1 to 1-14.
1.1
(C)
m = F/a or kg = N/m/s2 = N.s2/m.
1.2
(B)
[μ
1.3
(A)
2.36 10
1.4
(C)
The mass is the same on earth and the moon:
1.5
(C)
Fshear
1.6
(B)
1.7
(D)
1.8
(A)
1.9
(D)
[τ du/dy] = (F/L2)/(L/T)/L = F.T/L2.
8
F sin
F
= shear
A
1000
water
du
dr
h
23.6 10
4 cos
gD
9
23.6 nPa.
4200sin 30
2100 N
250 10
(T 4)2
180
4
m
du
dr
[4(8r )] 32 r.
2100 N.
84 103 Pa or 84 kPa
2
(80 4)2
180
1000
[10 5000r ] 10
3
968 kg/m3
10 5000 0.02 1 Pa.
4 0.0736 N/m 1
1000 kg/m3 9.81 m/s2 10 10
6
m
3 m or 300 cm.
We used kg = N·s2/m
1.10
1.11
(C)
(C)
m
pV
RT
800 kN/m 2 4 m3
0.1886 kJ/(kg K) (10 273) K
1
59.95 kg
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Chapter 1 / Basic Considerations
1.12
(B)
Eice
Ewater . mice 320 mwater cwater T .
5 (40 10 6 ) 1000 320 (2 10 3 ) 1000 4.18 T .
T 7.66 C.
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.
1.13
(D)
For this high-frequency wave, c
RT
287 323 304 m/s.
Chapter 1 Problems: Dimensions, Units, and Physical Quantities
1.14
Conservation of mass — Mass — density
Newton’s second law — Momentum — velocity
The first law of thermodynamics — internal energy — temperature
1.15
a) density = mass/volume = M / L3
b) pressure = force/area = F / L2 ML / T 2 L2 M / LT 2
c) power = force velocity = F L / T ML / T 2 L / T
d) energy = force distance = ML / T 2 L ML2 / T 2
e) mass flux = ρAV = M/L3 × L2 × L/T = M/T
f) flow rate = AV = L2 × L/T = L3/T
1.16
M FT 2 / L
a) density = 3
L
L3
b) pressure = F/L2
ML2 / T 3
FT 2 / L4
c) power = F × velocity = F
L/T = FL/T
d) energy = F×L = FL
M FT 2 / L
e) mass flux =
FT / L
T
T
f) flow rate = AV = L2 L/T = L3/T
1.17
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] = L3 / T L2 L2 / 3 L1/3 T
Note: the slope S0 has no dimensions.
1.18
a) m = [C] s2.
b) N = [C] kg.
c) m3/s = [C] m2 m2/3.
[C] = m/s2
[C] = N/kg = kg m/s2 kg = m/s2
[C] = m3/s m2 m2/3 = m1/3/s
2
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Chapter 1/ Basic Considerations
1.19
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
kg m
1
s 2 kg / m s
d) viscosity: N s/m2 = 2
s
m
N m kg m m
kg m 2 / s 3
e) heat flux: J/s =
2
s
s
s
kg
m m
J
N m
m 2 / K s2
f) specific heat:
2
kg K kg K
s
kg K
1.20
kg
m
s2
m
km f . Since all terms must have the same dimensions (units) we require:
s
[c] = kg/s, [k] = kg/s2 = N s 2 / m s 2 N / m, [f] = kg m / s 2 N.
c
Note: we could express the units on c as [c] = kg / s
1.21
a) 250 kN
e) 1.2 cm2
1.22
a) 1.25 108 N
d) 5.6
m3
1.23
1.24
0.225
b) 572 GPa
f) 76 mm3
2 2
0.738
N s/m
d) 17.6 cm3
c) 42 nPa
b) 3.21 10 5 s
e) 5.2 10 2 m2
0.06854m
N s2 / m s
c) 6.7 108 Pa
f) 7.8 109 m3
m
0.00194 3.281 d
d2
where m is in slugs, in slug/ft3 and d in feet. We used the conversions in the front cover.
20/100
5.555 10 5 m/s
3600
b) 2000 rev/min = 2000 2 /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
500 kg/m3
g) 500 g/L = 500 10 3 kg/10 m
h) 500 kWh = 500 1000 3600 = 1.8 109 J
a) 20 cm/hr =
1.25
a) F = ma = 10 40 = 400 N.
b) F W = ma.
c) F W sin 30 = ma.
1.26
The mass is the same on the earth and the moon:
60
1.863.
m=
Wmoon = 1.863 5.4 = 10.06 lb
32.2
F = 10
F = 10
40 + 10 9.81 = 498.1 N.
40 + 9.81 0.5 = 449 N.
3
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Chapter 1 / Basic Considerations
1.27
a)
0.225
b)
0.225
c)
0.225
m
d
2
m
d
2
m
d2
0.225
0.225
0.225
4.8 10
26
0.184 (3.7 10
4.8 10
)
26
0.00103 (3.7 10
4.8 10
0.43 10 6 m or 0.00043 mm
10 2
10 2
7.7 10 5 m or 0.077 mm
10 2
0.0039 m or 3.9 mm
)
26
0.00002 (3.7 10
)
Pressure and Temperature
1.28
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.
1.29
a) 101
c) 14.7
e) 30
31 = 70 kPa abs.
b) 760
31
14.7 = 10.2 psia.
d) 34
101
31
30 = 20.8 in. of Hg abs.
101
31
760 = 527 mm of Hg abs.
101
31
34 = 23.6 ft of H2O abs.
101
1.30
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
62.8 61.6
100 = 1.95 %.
% error =
61.6
1.31
a) p = 973 +
1.32
T = 48 +
22,560 20,000
(785 973) = 877 psf
25,000 20,000
22,560 20,000
T = 12.3 +
( 30.1 + 12.3) = 21.4 F
25,000 20,000
0.512
( .488) (628 2 785 + 973) = 873 psf
b) p = 973 + 0.512 (785 973) +
2
0.512
T = 12.3 + 0.512 ( 30.1 + 12.3) +
( .488) ( 48 + 2 30.1 12.3) = 21.4 F
2
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.
33,000
35,000
30,000
( 65.8 + 48) = 59 F or ( 59
30,000
4
32)
5
= 50.6 C
9
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Chapter 1/ Basic Considerations
1.33
1.34
p=
Fn
Ft
26.5 cos 42
Fn
=
= 1296 MN/m2 = 1296 MPa.
4
A
152 10
(120 000) 0.2 10
20 0.2 10
4
4
2.4 N
0.0004 N
Fn2
F=
= tan
1
Ft2 = 2.400 N.
0.0004
=0.0095
2.4
Density and Specific Weight
m
V
0.2
= 1.92 slug/ft3.
180 / 1728
1.35
=
1.36
= 1000 (T 4)2/180 = 1000 (70 4)2/180 = 976 kg/m3
= 9800 (T 4)2/18 = 9800 (70 4)2/180 = 9560 N/m3
976 978
% error for =
100 = .20%
978
9560 978 9.81
% error for =
100 = .36%
978 9.81
1.37
S = 13.6
1.38
W
a) m =
g
32.2 = 61.8 lb/ft3.
0.0024T = 13.6 0.0024 50 = 13.48.
13.48 13.6
100 = .88%
% error =
13.6
V
g
12 400 500 10
9.81
12 400 500 10
b) m =
9.77
12 400 500 10
c) m =
9.83
1.39
= g = 1.92
S=
m/ V
water
water
6
6
. 1.2
6
= 0.632 kg
= 0.635 kg
= 0.631 kg
10/ V
.
1.94
V = 4.30 ft3
5
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Chapter 1 / Basic Considerations
Viscosity
1.40
Assume carbon dioxide is an ideal gas at the given conditions, then
200 kN/m3
0.189 kJ/kg K 90 273 K
p
RT
W
V
mg
V
2.915 kg/m3
2.915 kg/m3 9.81 m/s2
g
From Fig. B.1 at 90°C,
28.6 kg/m2 s2
28.6 N/m3
2 10 5 N s/m2 , so that the kinematic viscosity is
2 10 5 N s/m2
2.915 kg/m3
6.861 10 6 m2 /s
The kinematic viscosity cannot be read from Fig. B.2; the pressure is not 100 kPa.
1.41
At equilibrium the weight of the piston is balanced by the resistive force in the oil due to
wall shear stress. This is represented by
Wpiston
DL
where D is the diameter of the piston and L is the piston length. Since the gap between
the piston and cylinder is small, assume a linear velocity distribution in the oil due to the
piston motion. That is, the shear stress is
Vpiston 0
V
r
Using Wpiston
Dcylinder
Dpiston / 2
mpiston g , we can write
Vpiston
mpiston g
Dcylinder
Dpiston / 2
DL
Solve Vpiston :
Vpiston
mpiston g Dcylinder
2
Dpiston
DL
0.350 kg 9.81 m/s2 0.1205 0.120 m
2
2 0.025 N s/m
2
0.12 0.10 m
0.91 kg m2 /N s3
0.91 m/s
where we used N = kg·m/s2.
6
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Chapter 1/ Basic Considerations
1.42
du /dy . From the given velocity
The shear stress can be calculated using
distribution,
120(0.05 2 y )
1.308 10 3 N s/m2 so, at the lower plate where y = 0,
From Table B.1 at 10 C,
du
dy
du
dy
y2 )
u ( y ) 120(0.05 y
120(0.05 0) 6 s 1
1.308 10 3
6 7.848 10 3 N/m2
y 0
At the upper plate where y = 0.05 m,
du
dy
1.43
1.44
120(0.05 2 0.05)
2
r = 0.25
r = 0.5
1.45
T = force
=
30(2 1/12)
(1/12)
du
dr
30(2 1/12)
(1/12)
7.848 10 3 N/m2
y 0.05
du
= 1.92
dr
=
1
6s
= 32
= 32
[32r / r02 ] 32 r / r02 .
1
1
= 0.014 lb/ft2
2
10
0.25 /100
3
10
3
(0.5 /100) 2
0.5 /100
(0.5 /100) 2
moment arm = 2 RL
T
2 R3 L
1.46 Use Eq.1.5.8: T =
h
power =
= 6.4 Pa
0.0026
0.4
1000 2 R 2 L
2
R
0.4
1000 2
12
=
T
550
2
0.5/12
= 0,
= 3.2 Pa,
du
2 R2L =
dr
R=
r=0
.012 0.2
3
2000 2
60
0.01/12
0.4
R
2
1000 2 R2L.
= 0.414 N.s/m2.
4 0.006
= 2.74 ft-lb.
2.74 209.4
= 1.04 hp
550
7
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Chapter 1 / Basic Considerations
1.47
Fbelt =
du
A 1.31 10
dy
power =
1.48
F V
746
3
10
(0.6
0.002
15.7 10
= 0.210 hp
746
r
. Due to the area
h
du
r = dA r =
2 r dr r.
dy
Assume a linear velocity so
element shown, dT = dF
T=
R
2
2 3
r dr =
h
h
0
1.49
4) = 15.7 N.
du
dy
400 2
60
2 0.08/12
2.36 10
R4
4
5
dr
r
(3/12) 4
= 91
10 5 ft-lb.
u
.
y
The velocity at a radius r is r . The shear stress is
The torque is dT = rdA on a differential element. We have
T=
rdA=
0.08
0
2000 2
60
r
2 rdx ,
0.0002
209.4 rad/s
where x is measured along the rotating surface. From the geometry x
T=
0.08
0.1
0
1.50
209.4 x / 2
2
0.0002
x
dx 329 000
2
0.08
x 2dx
0
2 r, so that
329 000
(0.083 ) = 56.1 N . m
3
du
= cons’t and = AeB/T = AeBy/K = AeCy, then
dy
du
du
AeCy
= cons’t.
= De Cy.
dy
dy
D Cy y
= E (e Cy 1) where A, B, C, D, E, and K are constants.
Finally, or u(y) =
e
0
C
If
1.51
Ae
40
B/T
= 2.334
0.001 Ae B/293
0.000357
Ae
A = 2.334
B /353
10 6 e1776/313 = 6.80
10 6, B = 1776.
10 4 N.s/m2
8
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Chapter 1/ Basic Considerations
Compressibility
1.52
m=
1.53
B=
1.54
Use c = 1450 m/s. L = c t = 1450
V . Then dm = d V + V d . Assume mass to be constant in a volume subjected
d
dV
.
d V = V d , or
to a pressure increase; then dm = 0.
V
V p
V
V
2200 MPa.
B V
=
V
V p
B
0.62 = 899 m
1.3
= 136.5 MPa
20
1.55
p=
1.56
a) c
327,000 144 /1.93 = 4670 fps
c) c
308,000 144 /1.87 = 4870 fps
1.57
V =3.8
10
2100
4
20
2 10
= 0.00909 m3 or 9090 cm3
2200
b) c
1 = 0.0076 m3.
p= B
327,000 144 /1.93 = 4940 fps
V
V
2270
0.0076
= 17.25 MPa
1
Surface Tension
2
R
2 0.0741
104 Pa or 29.6 kPa.
1.58
p=
1.59
Use Table B.1:
1.60
The droplet is assumed to be spherical. The pressure inside the droplet is greater than the
outside pressure of 8000 kPa. The difference is given by Eq. 1.5.13:
5 10
p
Hence,
6
= 2.96
= 0.00504 lb/ft.
2
r
pinside
pinside
p=
poutside
poutside 10 kPa
4
R
Bubbles: p = 4 /R = 59.3 kPa
4 0.00504
= 7.74 psf or 0.0538 psi
1/(32 12)
2 0.025 N/m
10 kPa
5 10 6 m
8000 10 8010 kPa
In order to achieve this high pressure in the droplet, diesel fuel is usually pumped to a
pressure of about 20 000 kPa before it is injected into the engine.
1.61
See Example 1.4:
h=
4 cos
gD
4 0.0736 0.866
1000 9.81 0.0002
9
0.130 m.
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Chapter 1 / Basic Considerations
1.62
See Example 1.4: h =
4 cos
gD
4 0.032cos130
1.94 13.6 32.2 0.8/12
= 0.00145 ft or
L
2 cos = force down = ghtL.
1.63
force up =
1.64
Draw a free-body diagram:
The force must balance:
d2
W = 2 L or
L g
4
d
1.65
1.66
0.0174 in
h=
L
2 cos
.
gt
L
needle
2 L.
W
8
g
From the free-body diagram in No. 1.47, a force balance yields:
d2
(0.004)2
Is
g< 2 ?
7850 9.81 2 0.0741
4
4
0.968 < 0.1482
No
Each surface tension force =
D. There is a force on the
outside and one on the inside of the ring.
F=2
D neglecting the weight of the ring.
1.67
D
From the infinitesimal free-body shown:
dx
d cos
gh x dx.
.
cos =
d
d dx/d
h
g xdx
g x
We assumed small so that the element
thickness is x.
dl
h
h(x)
F
dW
Vapor Pressure
1.68
The absolute pressure 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.
10
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Chapter 1/ Basic Considerations
1.69
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.
1.70
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.
1.71
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.
1.72
The inlet pressure to a pump cannot be less than 0 kPa absolute. Assuming atmospheric
pressure to be 100 kPa, we have
x = 16.83 km.
10 000 + 100 = 600 x.
Ideal Gas
p
RT
1.73
1.74
in
101.3
0.287 (273 15)
p
RT
1.226 kg/m3.
= 1.226
101.3
1.226 kg/m3 .
0.287 (15 273)
out
9.81 = 12.03 N/m3
85
1.19 kg/m3.
0.287 248
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.
p
RT
1.75
750 44
1716 470
p
Vg
RT
0.1339 slug/ft 3.
m
V
0.1339 15 2.01 slug.
100
(10 20 4) 9.81 9333 N.
0.287 293
1.76
W
1.77
Assume that the steel belts and tire rigidity result in a constant volume so that m1 = m2:
V1
p2
V
2
p1
or
T2
T1
m1RT1
p1
m2 RT2
.
p2
(35 14.7)
150 460
10 460
11
67.4 psia or 52.7 psi gage.
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Chapter 1 / Basic Considerations
1.78
The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have
100 000 1 m 9.81.
m 10 200 kg.
Hence,
m
pV
RT
100 4 r 3 / 3
10 200.
0.287 288
r 12.6 m or d
PE
1
mV 2 mg ( 10).
2
20 32.2.
25.2 m.
The First Law
1.79
KE
0
1
mV 2 mg ( 20).
2
0
1.80
W1-2
b)
KE. a) 200 0
10
c)
0
1.81
102
20
2
200 cos
20
40 32.2.
1
5(V f2 102 ).
2
Vf
V
V
25.4 fps.
35.9 fps.
19.15 m/s.
1
15(V f2 102 ).
2
20sds
0
10
V2
V2
1
15(V f2 102 ).
2
s
ds
20
200sin
Vf
15.27 m/s.
1
15(V f2 102 ).
2
1
15(V f2 102 ).
2
2
Vf
16.42 m/s.
1
10 402 0.2u1 0 u2 .
u2 u1 40 000.
2
40 000
55.8 C where cv comes from Table B.4.
u cv T.
T
717
The following shows that the units check:
E1
E2 .
mcar V 2
mair c
kg m2 / s 2
m2 kg C
m2 kg C
kg J/(kg C)
N m s2
(kg m/s2 ) m s 2
where we used N = kg.m/s2 from Newton’s 2nd law.
1.82
E2
E1.
1
1500
2
1
mV 2
2
C
mH2Oc T .
100 1000
3600
2
1000 2000 10
6
4180 T .
T
69.2 C.
We used c = 4180 J/kg. C from Table B.5. (See Problem 1.75 for a units check.)
12
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Chapter 1/ Basic Considerations
1.83
m f h f mwater c T . 0.2 40 000 100 4.18 T .
T 19.1 C.
The specific heat c was found in Table B.5. Note: We used kJ on the left and kJ on the
right.
1.84
p
V2
mRT
dV
d V mRT
mRT ln
mRT ln 2
p1
V
V
V1
since, for the T = const process, p1 V 1 p2 V 2 . Finally,
4
1
W1-2
1716 530ln
78,310 ft-lb.
32.2
2
W
pd V
The 1st law states:
Q W
0.
u mcv T
Q W
78,310 ft-lb or
101 Btu.
1.85
If the volume is fixed the reversible work is zero since the boundary does not move. Also,
mRT T1 T2
since V
the temperature doubles if the pressure doubles. Hence, using
,
p
p1 p2
Table B.4 and Eq. 1.7.17,
200 2
a) Q mcv T
(1.004 0.287)(2 293 293) 999 kJ
0.287 293
200 2
b) Q mcv T
(1.004 0.287)(2 373 373) 999 kJ
0.287 373
200 2
c) Q mcv T
(1.004 0.287)(2 473 473) 999 kJ
0.287 473
1.86
W
pd V
then V
1.87
1.88
T1
2
V 1 ). If p = const,
2V 1 and W
a) W 2 0.287
b) W 2 0.287
c) W 2 0.287
2
c= kRT
T2
p( V
p(2 V 1 V 1)
333 191 kJ
423 243 kJ
473 272 kJ
T1
V1
pV 1
T2
so if T2
V2
mRT1.
2T1,
1.4 287 318 357 m/s. L c t 357 8.32 2970 m.
p2
p1
k 1/ k
(20 273)
500
5000
0.4 /1.4
151.8 K or
121.2 C
We assume an isentropic process for the maximum pressure:
k /k 1
1.4 / 0.4
T
423
p2 p1 2
(150 100)
904 kPa abs or 804 kPa gage.
T1
293
Note: We assumed patm = 100 kPa since it was not given. Also, a measured pressure is a
gage pressure.
13
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Chapter 1 / Basic Considerations
1.89
p2
w
p1 T2 / T1
u
k /k 1
100 473 / 293
cv (T2 T1)
1.4/0.4
534 kPa abs.
(1.004 0.287)(473 293)
129 kJ/kg.
We used Eq. 1.7.17 for cv.
Speed of Sound
1.90
a) c
kRT
1.4 287 293 343.1 m/s
b) c
kRT
1.4 188.9 293 266.9 m/s
c) c
kRT
1.4 296.8 293 348.9 m/s
d) c
kRT
1.4 4124 293 1301 m/s
e) c
kRT
1.4 461.5 293 424.1 m/s
Note: We must use the units on R to be J/kg.K in the above equations.
1.91
kRT
1.4 287 223
At 10 000 m the speed of sound c
kRT
1.4 287 288 340 m/s.
At sea level, c
340 299
% decrease
100 12.06 %.
340
1.92
a) c= kRT
1.4 287 253 319 m/s. L c t 319 8.32 2654 m.
b) c= kRT
1.4 287 293 343 m/s. L c t 343 8.32 2854 m.
c) c= kRT
1.4 287 318 357 m/s. L c t 357 8.32 2970 m.
14
299 m/s.
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Chapter 2 / Fluid Statics
CHAPTER 2
Fluid Statics
FE-type Exam Review Problems: Problems 2-1 to 2-9
2.1
(C)
p   Hg h  (13.6  9810)  (28.5  0.0254)  96 600 Pa
2.2
(D)
p  p0   gh  84 000  1.00  9.81  4000  44 760 Pa
2.3
(C)
pw  patm   x hx   water hw  0  30 000  0.3  9810  0.1  8020 Pa
2.4
(A)
pa   H  (13.6  9810)  0.16  21 350 Pa.
pa,after  21350  10 000  11350  13.6  9810Hafter .  Hafter  0.0851 m
2.5
(B)
2.6
(A)
2.7
(D)
The force acts 1/3 the distance from the hinge to the water line:
5
1
5
5
(2  )  P   (2  )  [9800  1 3  (2  )].  P  32 670 N
3
3
3
3
The gate opens when the center of pressure in at the hinge:
1.2  h
I
11.2  h
b(1.2  h)3 /12
y
 5. y p  y 


 5  1.2.
2
Ay
2
(1.2  h)b(11.2  h) / 2
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.


Place the force FH  FV at the center of the circular arc. FH passes through the
hinge:
 P  FV  4 1.2w  9800  ( 1.22 / 4)w  9800  300 000.  w  5.16 m.
2.8
(A)
W  V
900  9.81  9810  0.0115w.  w  6 m
2.9
(A)
p plug  20 000   h  20 000  6660  (1.2 
5
)  24 070 Pa
9.81
Fplug  p plug A  24 070    0.022  30.25 N .
15
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Chapter 2 / Fluid Statics
Chapter 2 Problems: Pressure
2.10
yz
ay
2
yz
yz
Fz  ma z : pz y  ps cos  
a z  g
2
2
Since scos  y and s sin   z, we have
Fy  ma y : p y z  ps sin  
py  p  
y
ay
2
Let y  0 and z  0:
2.11
p = h.
z
ps
pyz
z
s
y
gV
z
 a z  g
2
py  p  0

pz  p  0 

pzy
pz  p  
and
then
 p y  pz  p.
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
2.12
h = p/.
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
p
20  144

= 2.31.
 h 62.4  20
 = 1.94  2.31 = 4.48 slug/ft3.
2.13
S=
2.14
a) p = h = 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.
2.15
a) p = 1h1 + 2h2 = 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.
2.16
p = gh = 0.0024  32.2 (–10,000) = –773 psf or –5.37 psi.
16
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y
Chapter 2 / Fluid Statics
100  9.81
pg

h 
 3  13.51 Pa 
0.287  253
RTo

 pbase = 1.84 Pa
100  9.81
pg
pinside  i g h 
h 
 3  11.67 Pa 

0.287  293
RTi
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).
poutside  o g h 
2.17
2.18
p = gdh where h = –z. From the given information S = 1.0 + h/100 since S(0) = 1 and
S(10) = 1.1. By definition  = 1000 S, where water = 1000 kg/m3. Then
dp = 1000 (1 + h/100) gdh. Integrate:
p
10
0
0
 dp   1000(1  h / 100)gdh
10 2
p  1000  9.81(10 
) = 103 000 Pa or 103 kPa
2  100
Note: we could have used an average S: Savg = 1.05, so that  avg = 1050 kg/m3.
2.19
p 
p
p
p
i
j k
x
y
z


=   ax i   y j   z k   gk    ax i  a y j  az k   gk    a   g
p    (a  g )
2.20
2.21
p  patm [(T0   z ) / T0 ]g /  R
= 100 [(288  0.0065  300)/288]9.81/0.0065  287 = 96.49 kPa
100
 9.81 300 /1000 = 96.44 kPa
p  patm   gh  100 
0.287  288
96.44  96.49
 100 = 0.052%
% error =
96.49
The density variation can be ignored over heights of 300 m or less.
 T  z 
p  p  p0  patm  0

 T0 
g / R
 patm
 288  0.0065  20 9.81/0.0065287 
= 100 
 1 = 0.237 Pa or

288



This change is very small and can most often be ignored.
17
0.000237 kPa
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 2 / Fluid Statics
2.22
Eq. 1.5.11 gives 310, 000  144  
 gdh 
4.464  107

dp
. But, dp = gdh. Therefore,
d
d  or
Integrate, using 0 = 2.00 slug/ft 3:

d
h
32.2
  2  4.464 107  dh .
2
0
Now,
h

p   gdh 
0
Assume  = const:
h
d
2

32.2
4.464  107
 1 1
     = 7.21  107 h
  2
2g
dh
or  
2
1  14.42  107 h
2g
 1  14.42 107 hdh  14.42 107 ln(1  14.42 10
7
h)
0
p   gh  2.0  32.2  h  64.4h
a) For h = 1500 ft: paccurate = 96,700 psf and pestimate = 96,600 psf.
% error 
96, 600  96, 700
 100  0.103 %
96, 700
b) For h = 5000 ft: paccurate = 323,200 psf and pestimate = 322,000 psf.
% error 
322, 000  323, 200
 100  0.371 %
323, 200
c) For h = 15,000 ft: paccurate = 976,600 psf and pestimate = 966,000 psf.
% error 
2.23
966, 000  976, 600
 100  1.085 %
976, 600
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.
2.24
Use Eq. 2.4.8:
p=
9.81
287 .
0.0065
101(1  0.0065 z / 288)
a) z = 3000.
p = 69.9 kPa.
b) z = 6000.
c) z = 9000.
p = 30.6 kPa.
d) z = 11 000. p = 22.5 kPa.
18
p = 47.0 kPa.
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Chapter 2 / Fluid Statics
2.25
p  gz / RT
=e
.
p0
0.001
32.2 z
ln
.

14.7
1716  455
Use the result of Example 2.2:
ln
p
gz

.
p0
RT
z = 232,700 ft.
Manometers
2.26
p = h = (13.6  9810)  0.25 = 33 350 Pa or 33.35 kPa.
2.27
a) p = h.
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
The % error is 0.000 %.
2.28
Referring to Fig. 2.6a, the pressure in the pipe is p = gh. If p = 2400 Pa, then
2400
.
9.81h
2400 = gh =   9.81h or  
2.29
a)  
2400
= 680 kg/m3.
9.81 0.36
gasoline
b)  
2400
= 899 kg/m3.
9.81 0.272
benzene
c)  
2400
= 999 kg/m3.
9.81 0.245
water
d)  
2400
= 1589 kg/m3.
9.81 0.154
carbon tetrachloride
Referring to Fig. 2.6a, the pressure is p = wgh =
2  gh
1
 aV 2 . Then V 2  w .
a
2
a) V 2 
2 1000  9.81 0.06
= 957.
1.23
b) V 2 
2 1.94  32.2  3 /12
= 13,124.
0.00238
V = 115 ft/sec
c) V 2 
2 1000  9.81 0.1
= 1595.
1.23
V = 39.9 m/s
d) V 2 
2 1.94  32.2  5 /12
= 21,870.
0.00238
V = 148 ft/sec
V = 30.9 m/s
19
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Chapter 2 / Fluid Statics
p1 = –1h + 2H.
5
9.5
p1 = –0.86  62.4 
+ 13.6  62.4 
= 649.5 psf or 4.51 psi.
12
12
2.30
See Fig. 2.6b:
2.31
p  p0  1 gh1  2 gh2  3 gh3  4 gh4
= 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
2.32
p1  p4   p1  p2    p2  p3    p3  p4 
(Use p  gh)
40 000 – 16 000 = 10009.81(–0.2) + 13 6009.81H + 9209.810.3.
H = 0.1743 m or 17.43 cm
2.33
p1  p4   p1  p2    p2  p3    p3  p4 
(Use p  gh)
po – pw = 9009.81(–0.2) + 13 6009.81(–0.1) + 10009.810.15
= –12 300Pa or –12.3 kPa
2.34
p1  p5   p1  p2    p2  p3    p3  p4    p4  p5 
p1 = 9810(–0.02) + 13 6009.81(–0.04) + 9810(–0.02) + 13 6009.810.16
= 15 620 Pa or 15.62 kPa
2.35
pw + 9810  0.15 – 13.6  9810  0.1 – 0.68  9810  0.2 + 0.86  9810  0.15 = po.
pw – po = 11 940 Pa
2.36
2.37
2.38
or
11.94 kPa.
pw – 9810  0.12 – 0.68  9810  0.1 + 0.86  9810  0.1 = po.
With pw = 15 000, po = 14 000 Pa or
14.0 kPa.
a) p + 9810  2 = 13.6  9810  0.1.
p = –6278 Pa
or
–6.28 kPa.
b) p + 9810  0.8 = 13.6  9810  0.2.
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.
p – 9810  4 + 13.6  9810  0.16 = 0.
p = 17 890 Pa
or
17.89 kPa.
20
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Chapter 2 / Fluid Statics
2.39
8200 + 9810  0.25 = 1.59  9810  H.
H = 0.683 m
H
0.273
Hnew = 0.683 + 0.273 = 0.956 m.
H =
= 0.1365.
2
p + 9810 (0.25 + 0.1365) = 1.59  9810  0.956.
H
H
p = 11 120 Pa or 11.12 kPa.
2.40
p + 9810  0.05 + 1.59  9810  0.07 – 0.8  9810  0.1 = 13.6  9810  0.05.
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 significant digits!  only 3 are usually retained in the answer!
2.41 The equation for the manometer is
pA   water  0.07  pB   oil  0.1   HG  0.09sin 40
Solve for pB:
pB  p A   water  0.07   HG  0.09sin 40   oil  0.1
 p A   water  0.07  13.6 water  0.09sin 40  0.87 water  0.1
 p A   0.07  13.6  0.09sin 40  0.87  0.1   water
 10 kPa   0.07  13.6  0.09sin 40  0.87  0.1  9.81 kN/m3  2.11 kPa
2.42 The distance the mercury drops on the left equals the distance along the tube that the
mercury rises on the right. This is shown in the sketch.
Oil (S = 0.87)
B
10 cm
Water
h
A
9 cm
7 cm
h
Mercury
40
o
From the previous problem we have
 pB 1  pA   water  0.07   HG  0.09sin 40  oil  0.1
(1)
For the new condition
 pB 2  pA   water   0.07  h    HG  0.11sin 40   oil   0.1  h sin 40
(2)
where h in this case is calculated from the new manometer reading as
h  h / sin 40  11  9  h  0.783 cm
21
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Chapter 2 / Fluid Statics
Subtracting Eq.(1) from Eq.(2) yields
 pB 2   pB 1   water   h    HG  0.02sin 40  oil   h sin 40
Substituting the values of h and  pB 1 gives  pB 2
 pB 2  2.11   0.00783  13.6  0.02sin 40  0.87   0.00783sin 40   9.81
 0.52 kPa
2.43 Before pressure is 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 p1 V 1  p2 V 2 using absolute
pressures. Thus,
p2 = p2  8467 / (4  H / 2)
14.7  144  4A = p2(4 – H / 2 )A or
From a pressure balance on the manometer (pressures in psf):
8467
30  144 + 14.7  144 = 13.6  62.4 H +
,
4H /2
or
H2 – 15.59 H + 40.73 = 0.
H = 12.27 or 3.32 ft.
2.44 a) p1  p5   p1  p2    p2  p3    p3  p4    p4  p5 
4000 = 9800(0.16–0.22) + 15 600(0.10–0.16) + 133 400H + 15 600(0.07–H).
H = 0.0376 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 = 0.1236 ft or 1.483 in.
2.45
a)
2D2 / d 2
H

p1  1  2 2  2( 3   2 ) D 2 / d 2

2(0.1/ 0.005)2
9800  2 15 600  2(133 400  15 600)(0.1/ 0.005)
 H  8.487  10 6  400
b) H 
2.46
2
 8.487  10 H 6
= 0.0034 m or 3.4 mm
2(4 / 0.2)2
62.4  2  99.5  2(849  99.5)(4 / 0.2) 2
0.06 144 = 0.01153 ft or 0.138 in.
p1  p4   p1  p2    p2  p3    p3  p4  (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
2.47
a) pair = –6250 + 625 = –5620 Pa.
–5620 + 9800(2 + z) – 13 600  9.81(0.1 + 2z) = 0.
h = 0.1 + 2z = 0.15 m or 15 cm
22
z = 0.0025.
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Chapter 2 / Fluid Statics
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 = 0.2+ 2z = 0.214 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.
h = 4/12 + 2 (0.00558) = 0.3445 ft or 4.13 in.
z = 0.00558 ft.
d) pair = 441 + 44.1 = 485 psf
485 + 62.4(2 + z) – 13.6  62.4(8/12 + 2z) = 0.
h = 8/12 + 2 (0.0267) = 0.7205 ft or 8.65 in.
z = 0.0267 ft.
Forces on Plane Areas
2.48 F  h A = 9810  10    0.32/4 = 6934 N.
1  5 
 5
 5 
2.49  2    P    2    9800  1 3   2    .  P  32 670 N
3  3 
 3
 3 
2
a) F = pc A = 9800  2  4 = 313 600 N or 313.6 kN
2
2
b) F  pc A  9800  1 (2  4)  9800   2  9800   1  98 000 N or 98.0 kN
3
3
c) F = pc A = 9800  1  2  4  2 = 110 900 N or 110.9 kN
d) F = pc A = 9800  1  2  4/0.866 = 90 500 N or 90.5 kN
2.50
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  g V
walls
 2400  9.81 [2(21.50.1) + 2(210.1) + 20(.81.30.1)] = 28 400 N.
The vault will tend to rise out of the ground.
2.51
F = pc A = 6660  2    22 = 167 400 N or 167.4 kN
Find  in Table B.5 in the Appendix.
2.52
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  0.866/3) (2.828  2/2) = 254 500 N or 254.5 kN
23
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Chapter 2 / Fluid Statics
2.53
a) F   hA  62.4  27.33  24  40,930 lb.
6  83 /36
= 27.46 ft.
yp  27.33 
27.33  24
8/5.46 = 3/x.
x = 2.05’.
y
(x, y)
y = 30 – 27.46 = 2.54 ft.
x
(2.05, 2.54) ft.
b) F = 62.4  30  24 = 44,930 lb. The centroid is the center of pressure.
8
3
(2.000, 2.667) ft.
y = 2.667 ft.
 . x = 2.000 ft
5.333 x
c) F = 62.4 (30 – 2.667  0.707)  24 = 42,100 lb.
6  83 / 36
= 39.86 ft.
39.77  24
y p  39.77 
8
3
 8/5.43 = 3/x.
5.46 x
2.54
y = 42.43 – 39.86 = 2.57 ft
x = 2.04 ft.
(2.04, 2.57) ft.
a) F   hA  9810  6   22 = 739 700 N or 739.7 kN.
y p  y  I /Ay  6    24 /4(4  6) = 6.167 m.
(x, y)p = (0, –0.167) m
b) F  hA  9810  6  2 = 369 800 N or 369.8 kN.
  24 / 8
= 6.167 m. x2 + y2 = 4
yp  6 
2  6
x

xp F 
pdA 
2
2

 x p 6  2 
2

2

x(6  y ) xdy 
2
(24  4 y  6 y
2
2

2
2

y
2
(4  y )(6  y )dy.
dA
dy
(x, y)
x
2
 y 3 )dy  32 . xp = 0.8488 m
2
(x, y)p = (0.8488, –0.167) m
c) F = 9810  (4 + 4/3)  6 = 313 900 N or 313.9 kN.
y p  5.333 
4/2.5 =
1.5
.
x
3  4 3 / 36
= 5.500 m.
5.333  6
y
y = –1.5
x
(x, y)p = (0.9375, –1.5) m
x = 0.9375.
2
 4 sin 36.9°)  6 = 330 000 N
3
yp  5.6  5  2.43 /36(6  5.6) = 5.657 m. y = 0.343 m
3
d) F  9810  ( 4 
3 cos 53.13 = 1.8,
2.5 – 1.8 = 0.7,
x = 1.8 + 0.6 = 2.4.
2.4/2.057 = .7 / x1 .
53.13
4
o
 x1 = 0.6.
(x, y)p = (2.4, 0.343) m.
24
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Chapter 2 / Fluid Statics
2.55
F  h A  62.4  11  (6  10) = 41,180 lb.
6  10 3 / 12
I
= 11.758 ft.
yp  y 
 11 
11  60
yA
(16 – 11.758) 41,180 = 10P.
2.56
2.59
2.60
F
4  53 / 12
I
= 7.778 m.
 7.5 
7.5  20
Ay
(10 – 7.778) 1177 = 5 P.
2.58
P
F  h A  9810  6  20 = 1.777  106 N, or 1177 kN.
yp  y 
2.57
P = 17,470 lb.
yp
P = 523 kN.
F  h A  9810  12  20 = 2.354  106 N, or 2354 kN.
4  53 / 12
I
= 15.139 m.
yp  y 
 15 
15  20
Ay
(17.5 – 15.139) 2354 = 5 P. P = 1112 kN.
I
H
bH 3 / 12
H H 2
 
   H. y p is measured from the surface.
Ay 2 bH  H / 2 2 6 3
2
1
From the bottom, H  y p  H  H  H.
3
3
Note: This result is independent of the angle , so it is true for a vertical area or a sloped
area.
yp  y 
1
l
l sin 40  3l. F   (l  2) P sin 40 .   l 3  2(l  2) P.
2
3
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
F 
h  1.22  0.42 = 1.1314 m. A = 1.2  1.1314 + 0.4  1.1314 = 1.8102 m2
Use 2 forces: F1   hc A1  9800  0.5657  (1.2 1.1314) = 7527 N
1.1314
 (0.4 1.1314) = 1673 N
F2   hc A2  9800 
3
2
I
1.1314
0.4 1.13143 / 36
y p1  (11314
.
).
= 0.5657 m

yp2  y  2 
3
3
0.4  (1.1314 / 2)  (1.1314 / 3)
A2 y
M hinge  0: 7527 1.1314/3  1673  (1.1314  0.5657)  1.1314P = 0. P = 3346 N.
25
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Chapter 2 / Fluid Statics
2.61
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 ) / 3, A  (h  H ) 2 , I  [2(h  H )](h  H ) 3 / 36
hH
hH hH hH
2(h  H ) 4 / 36



 yp 

2
3
(h  H ) (h  H ) / 3
3
6
2
hH
.
2
b) h = H = 1.2 m
a) h 
h = H = 0.9 m
c) h = H = 1.5 m
2.62
The gate is about to open when the center of pressure is at the hinge.
b 1.83 /12
.
a) y p  1.2  H  (1.8/2  H ) 
H = 0.
(0.9  H )1.8b
b) y p  1.2  H  (2.0/2  H ) 
b  23 /12
.
(1  H )2b
H = 0.6667 m.
c) y p  1.2  H  (2.2/2  H ) 
b  2.23 /12
.
(1.1  H )2.2b
H = 2.933 m.
1
H
 bH  bH 2
2
2
F2  H  b  bH
1
H

bH 2   bH  .
 H  3
2
3
2
a) H  3  2 = 3.464 m b) H = 1.732 m c) H = 10.39' d) H = 5.196'
2.63
F1  
F1
H/3
l/2
F2
2.64 A free-body-diagram of the gate and block is sketched.
Sum forces on the block:
Fy  0
T
W  T  FB
where FB is the buoyancy force which is given by
T
0
F stop
FB
FB    R2 (3  H )
Take moments about the hinge:
yp
FH
T  3.5  FH  (3  yp )
where FH is the hydrostatic force acting on the gate. It is,
using h  1.5 m and A  2  3  6 m2 ,
W
Rx
Ry
FH   hA   9.81 kN/m3 1.5 m  6 m2   88.29 kN
26
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Chapter 2 / Fluid Statics
From the given information,
 
2 33 /12
I
yp  y 
 1.5 
2m
1.5  6
yA
T 
88.29   3  2 
3.5
 25.23 kN
FB  W  T  70  25.23  44.77 kN.
H 3 m
2.65
44.77 kN
 9.81 kN/m   1 m 
3
2
 R2  3  H   44.77
 1.55 m
The dam will topple if the moment about “O” of F1 and F3 exceeds
the restoring moment of W and F2.
Assume 1 m deep
F1
F2
W
a) W  (2.4  9810)(6  50  24  50 / 2) = 21.19  106 N
O
300  27  600 16
F3
= 19.67 m. (dw is from O to W.)
dw 
300  600
11.09
F2 = 9810  5  11.09 = 0.544  106 N. d 2 
= 3.697 m.
3
45
 45 = 9.933  106 N.
F1  9810 
d1 = 15 m. (d1 is from O to F1.)
2
45  10
2.943  15  5150
.
 20
F3  9810 
 30 = 8.093  106 N. d 3 
= 18.18 m.
2
2.943  5150
.
Wd w  F2 d 2  418.8  10 6 N  m 
 will not topple.
F1 d1  F3 d3  296.1  10 6 N  m 
b) W = (2.4  9810) (6  65 + 65  12) = 27.55  106 N.
390  27  780  16
dw =
= 19.67 m.
390  780
d 2  3.70 m.
F2  0.54  10 6 N.
6
F1 = 9810  30  60 = 17.66  10 N.
d1 = 20 m.
60  10
2.943  15  7.358  20
F3  9810 
d3 
 30 = 10.3  106 N.
= 18.57 m.
2
2.943  7.358
Wd w  F2 d 2  543.9  10 6 N  m 
 it will topple.
F1 d1  F3 d3  544.5  10 6 N  m 
c) Since it will topple for H = 60, it certainly will topple if H = 75 m.
27
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Chapter 2 / Fluid Statics
2.66
The dam will topple if there is a net clockwise moment about “O.”
a) W  W1  W2 . W1  (6  43  1)  62.4  2.4 = 38,640 lb.
W2  (24  43 / 2)  62.4  2.4 = 77,280 lb.
Assume 1 m deep
W3
F1
F2
W
O
W3  (40  22.33 / 2)  62.4 = 27,870 lb @ 20.89 ft.
F3
F1  62.4  20  ( 40  1) = 49,920 lb @ 40/3 ft.
F2  62.4  5  (10  1) = 3120 lb @ 3.33 ft
 Fp1 = 18,720 lb @ 15 ft
F3  
 Fp 2 = 28,080 lb @ 20 ft
M O : (49,920)(40/3) + (18,720)(15) + (28,080)(20) 38,640)(3)
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  62.4  30  60 = 112,300 lb.
W3  (60  22.86/2)  62.4 = 42,790 lb.
F2  62.4  5  10 = 3120 lb
Fp1  62.4  10  30 = 18,720 lb.
Fp2  62.4  50  30 / 2 = 46,800 lb.
M O : (112,300)(20) + (18,720)(15) + (46,800)(20)
will tip.
c) Since it will topple for H = 60 ft., it will also topple for H = 80 ft.
Forces on Curved Surfaces
2.67
M hinge = 0.
2.5P – dw  W – d1  F1 = 0.
dw

  22
42
1 2
 9800  1  8 
 9800 
 4  = 62 700 N
P 

4
3
2.5  3

W
P
F1
d1
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.
2.68
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:
( FH ) water  9810  2  (4  10)  784 800N
(FH )oil  0.86  9810  1  20  168 700N
M: 2 P  784.8  2  168.7  2.
P = 616.1 kN.
28
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Chapter 2 / Fluid Statics
2.69



Place the resultant force FH  FV at the center of the circular arc. FH passes thru the
hinge showing that P  FV .
a) P  FV  9810(6  2  4    4)  594 200 N or 594.2 kN.
b) P = FV = 62.4 (20  6  12 + 9  12) = 111,000 lb.
2.70
a) A free-body-diagram of the volume of water in the vicinity of the surface is shown.
Force balances in the horizontal and vertical directions give:
F1
.
FH  F2
A
FV  W  F1
where FH and FV are the horizontal and vertical components
of the force acting on the water by the surface AB. Hence,
FH  F2   9.81 kN/m
3
 8 1 2  4  706.3 kN
FH
F2
.
W
FV
xV
B
The line of action of FH is the same as that of F2. Its distance from the surface is
 
4 23 12
I
yp  y 
9
 9.037 m
98
yA
To find FV we find W and F1:



W   V  9.81 kN/m3 2  2  22   4  33.7 kN
4




 
F1  9.81 kN/m3 8  2  4   628 kN
 FV  F1  W  33.7  628  662 kN
To find the line of action of FV, we take moments at point A:
FV  xV  F1  d1  W  d2
where d1  1 m, and d2 
 xV 
2R
2 2

 1.553 m:
3 4    3 4   
F1  d1  W  d2 628 1  33.7 1.553

 1.028 m
FV
662
Finally, the forces FH and FV that act on the surface AB are equal and opposite to those
calculated above. So, on the surface, FH acts to the right and FV acts downward.
b) If the water exists on the opposite side of the surface AB, the pressure distribution would
be identical to that of Part (a). Consequently, the forces due to that pressure distribution
would have the same magnitudes. The vertical force FV = 662 N would act upward and
the horizontal force FH = 706.3 N would act to the left.
29
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 2 / Fluid Statics
2.71
Place the resultant FH  FV at the circular arc center. 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 . We have
P  FV = 9810 (6  3  4 + 9) = 983 700 N or 983.7 kN.
2.72
Place the resultant FH  FV at the center. FV passes thru the hinge
2  (9810  1  10) = 2.8 P.
2.73
P = 70 070 N or 70.07 kN.
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.
2.74 The resultant FH  FV of the unknown liquid acts thru the center of the circular arc. FV
passes thru the hinge. Thus, we use only ( FH ) oil . Assume 1 m wide:
2.75
a) M :
R 2 
R
R  4R 
 R 
 9810 R  
  R  x R .
 9800S

3
2  3 
4 
2 
  x  4580 N/m3
b) M :
R
R  4R 
R 2 
 R 
 62.4 R  
  R   x R .
 62.4S
 2 
3
2  3 
4 
  x  29.1 lb/ft3
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.
Fy  0  W  ( FV )o  (FV )w
0.36 

0 = S  9810  0.62  6 +  0.36 
  6  (0.8  9810) – 9810  0.18  6
4 

9810  0.8  2  0.62  6
 S  0.955.
b)  g V  W . = 1996 lb.
Fy  0  W  (FV )o  (FV )w
4 
0 = S  62.4    22  20 +  4 
  20  0.8  62.4 – 62.4    2  20

4 
62.4 .8  2  2 2  20.
 S  0.955.
30
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Chapter 2 / Fluid Statics
2.76
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 = (  32)  14.87 = 420.4 kN.
b) From a free-body diagram of the dome filled with oil:
Fweld + W = pA
W
Using the pressure from part (a):
Fweld = 14 870    32 – (0.8  9810) 
Fweld
pA
14

   33  = –23 400 N


2 3
or –23.4 kN
2.77
A free-body diagram of the gate and water is shown.
H
F  d w W  H  P.
3
a) H = 2 m. F = 9810  1  4 = 39 240 N.
2
2
0
0
W  9810  2 xdy  9810  2
y
F
h/3
dA=xdy
x
y1/ 2
2  9810 2 3/ 2
= 26 160 N.
dy 
2
2 3/ 2
1
1
x
4 x 3 dx
xdy

2
2
1  1 / 4
 10
 
dw  x 
 = 0.375 m.
 1 / 3
2
xdy
2

 4 x dx
0
P 
1
0.375
 39 240 
 26 160 = 17 980 N or 17.98 kN.
3
2
b) H = 8 ft.
F = 62.4  4  32 = 7987 lb
2
W  62.4 4 xdy  62.4  4 4 x 2 dx  62.4  16  2 3 / 3 = 2662 lb.
0
2
dw  x 
1
4x3dx
2 0
2
2
 4x dx
18
1  16/4 

 
 = 0.75 ft.  P    7987  0.75  2662   2910 lb
83
2  8/3 

0
31
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Chapter 2 / Fluid Statics
Buoyancy
2.78
W = weight of displaced water.
a) 20 000 + 250 000 = 9810  3 (6d  d 2 /2). d2 + 12d – 18.35 = 0. d = 1.372 m.
b) 270 000 = 1.03  9810  3 (6d  d 2 /2).
2.79
25 + FB = 100.
V.
FB = 75 = 9810 
d2 + 12d – 17.81 = 0.
d = 1.336 m.
V = 7.645   m3
 
or 7645 cm3
  7.645   = 100.
 = 13 080 N/m3.
2.80
3000  60 = 25  300 d  62.4.
2.81
100 000  9.81 + 6 000 000 = (12  30 + 8h  30) 9810
h = 1.465 m.
2.82
d = 0.3846' or 4.62".
distance from top = 2 – 1.465 = 0.535 m
T + FB = W. (See Fig. 2.11 c.)
T = 40 000 – 1.59  9810  2 = 8804 N or 8.804 kN.
2.83 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:
4 3
4
FB = W + Fa or
R g  1000  R 3  a g
3
3
4
100  9.81
4
100  9.81
Ta = 350.4 K or 77.4C
 1000    53
  53
.
3
0.287  293
3
0.287Ta
2.84
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
100  9.81
100  9.81
= Fp + 0.1 Fp + 1500   1502 
2.077  288
0.287  288
8
9.86  10
= 1.23  106
FP = 9.86 × 108
and
Npeople =
800
Of course equipment and other niceties such as gyms, pools, restaurants, etc., would add
significant weight.
1500 1502 
32
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Chapter 2 / Fluid Statics
2.85
Neglect the bouyant force of air. A force balance yields
FB = W + F
V.
= 50 + 10 = 60 = 9800 
 g V  W.
Density:
 = 832.5 kg/m3
  9.81 0.006122 = 50.
2.86
V = 0.006122 m3

Specific wt:
 = g = 832.5  9.81 = 8167 N/m3
Specific gravity:
S

water

832.5
S = 0.8325
1000
From a force balance FB = W + pA.
FB
a) The buoyant force is found as follows (h > 16'):
h  15  R
cos  
, Area = R2 – (h – 15 – R) R sin
R
FB = 10  62.4[R2  R2 + (h – 15 – R) R sin].
W
pA
FB = 1500 + hA.

The h that makes the above 2 FB’s equal is found by trial-anderror:
h = 16.8: 1866 ? 1858
h = 16.5: 1859 ? 1577
h = 17.0: 1870 ? 1960 h = 16.8 ft.
R
h  15
b) Assume h > 16.333 ft and use the above equations with R = 1.333 ft:
h = 16.4:
1857 ? 1853
h = 16.4 ft.
c) Assume h < 16.667ft. With R = 1.667 ft,
FB = 10  62.4[R2  (R – h + 15) R sin]
FB = 1500 + hA.
cos  
Trial-and-error for h:
h = 16:
h = 16.4:
2.87 a) W  FB .

V 
R  h  15
R
1849 ? 1374
1857 ? 2170
h = 16.2:
4
.15 
 .005 2
4
h  15
1853 ? 1765
h = 16.25 ft.
0.01  13.6  1000  h .015
 .015 2
 R
2

/ 4  9.81  9810 
V .
.06  2.769  10 5 m 3 .
h = 7.361   m
 m Hg  13.6  1000  h  .0152 / 4 = 0.01769 kg
33
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Chapter 2 / Fluid Statics
   0.0152

  0.0052
b) (0.01 +0 .01769) 9.81 = 9810 
 0.15 
 0.12 Sx . Sx = 0.959.
4
4


c) (0.01 + 0.01769) 9.81 = 9810
2.88
  0.0152
4
 0.15 Sx.
Sx = 1.045.
   0.0152

  0.0052
(0.01  mHg )9.81  9810 
 0.15 
 0.12  .
4
4


a) (0.01 +0 .01886) 9.81 = 9810
  0.0152
4
 0.15 Sx.
mHg = 0.01886.
Sx = 1.089.
b) mHg = 0.01886 kg.
Stability
2.89
d4
  (10 /12)4
= 0.02367 ft4.
64
64
.8  62.4    (5 / 12) 2  12 / 12
0.4363
W
= 0.4363. depth =
= 0.8 ft

V 

62.4
rH2O
 (5 /12) 2
a) I o 

GM  0.02367 / 0.4363  (0.5  0.4) = –0.0457'. It will not float with ends horizontal.
V = 0.3636 ft3, depth = 0.6667 ft
b) Io = 0.02367 ft4, 
2.90
It will not float as given.
GM  0.02367 / 0.3636  (5  4) /12 = –0.01823 ft.
0.02367 4  3.2
V = 0.2909, depth = 6.4", GM =
c) 
= 0.0147 ft. It will float.

0.2909
12
With ends horizontal I o   d 4 / 64. The displaced volume is

V   x d 2 h / 4  9800  8.014  10 5  x d 3 since h = d. The depth the cylinder will
sink is

V
 8.014  10 5  x d 3 / d 2 / 4  10.20  10 5  x d
depth =
A
h
The distance CG is CG   10.2  10 5  x d / 2 . Then
2
d 4 / 64
d
GM 
  10.2  10 5  x d / 2  0.
3
5
8.014  10  x d
2
This gives (divide by d and multiply by x):
Consequently, x > 8369 N/m3
612.5 – 0.5 x + 5.1  105  2x > 0.
x < 1435 N/m3
or
34
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Chapter 2 / Fluid Statics
2.91
V
 
W

 water
S  water d 3
 water
3
Sd .
V
 
W
 water

S  water d 3
 water
 S d 3 . h = Sd.
d 4 /12
 1 1 S
 (d / 2  Sd / 2)  d 
  .
Sd
 12S 2 2 
If GM = 0 the cube is neutral and 6S2 – 6S + 1 = 0.
6  36  24
S 
= 0.7887, 0.2113.
12
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.
GM 
3
2.92
As shown, y  16  9  16  4/(16+16) = 6.5 cm above the bottom edge.
4  9.5  16  8.5  16SA  4
= 6.5 cm.
G
0.5  8  2  8  SA 16
130 + 104 SA = 174 + 64 SA.
 SA = 1.1.
2.93
a) y 
16  4  8  1  8  7
= 4.
16  8  8
For G: y 
x
x
1.2 16  4  0.5  8 1  1.5  8  7
= 4.682.
1.2 16  0.5  8  1.5  8
1.2 16  0.5  8  4  1.5  8  4
1.2 16  0.5  8  1.5  8
1
 2  3.5
2
= 2.
422
422
For G: y 
= 2.364.
0.136
C
G
x
4
0.682
1
2222
2
= 1.25
422
1.2  4  2  0.5 1  1.5  7
1.2  2  0.5  4  1.5  4
= 2.34. x 
= 1.182
1.2  4  0.5  2  1.5  2
1.2  4  0.5  2  1.5  2
y = 0.34, x = 0.068. tan  
2.94
h
C
16  1  8  4  8  4
= 2.5.
16  8  8
G must be directly under C.
0.136
 =11.3.
tan  
.
0.682
b) y 
G
0.068
.  = 11.3.
0.34
The centroid C is 1.5 m below the water surface.
 CG = 1.5 m.
  8 3 / 12
 1.5  1.777  1.5  0.277  0.
Using Eq. 2.4.47: GM 
83
The barge is stable.
35
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Chapter 2 / Fluid Statics
2.95
8.485  3.414  16.97  1
= 1.8 m.
 CG  1.8  1.5 = 0.3 m.
8.485  16.97
 8.4853 /12
Using Eq. 2.4.47: GM 
 0.3  1.46  0.3  1.16. Stable.
34.97
y
Linearly Accelerating Containers
2.96
a) tan  
20
H
 . H = 8.155 m.
9.81 4
pmax = 9810 (8.155 + 2) = 99 620 Pa
b) pmax = (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
2.97
z
The air volume is the same before and after.
A
10
h
 .
 0.5  8 = hb/2. tan  
9.81 b
h 9.81
h. h = 2.856. Use dotted line.
4
B
2 10
1
1
2.5w   2.5  2.452  4. w = 0.374 m.
2
a) pA = –1000  10 (0 – 7.626) – 1000  9.81  2.5 = 51 740 Pa or 51.74 kPa
b

w
h
C
x
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.
2.98
Use Eq. 2.5.2: Assume an air-water surface as shown in the above figure.
 
8a x  
a) 60 000 = –1000 ax (0–8) – 1000  9.81 0   2.5 

9.81  
 
8a x
h 2  9.81
60 = 8 ax + 24.52 – 9.81
or ax – 4.435 = 1.1074
4=
9.81
2a x
a x2 – 10.1 ax + 19.67 = 0
ax .
ax = 2.64, 7.46 m/s2

8a x 
b) 60 000 = –1000 ax (–8) – 1000 (9.81 + 10)  2.5 
.
9.81 

60 = 8 ax + 49.52 – 19.81
a x2 – 5.1 ax + 1.44 = 0
8ax
or ax – 1.31 = 1.574
19.81
ax = 0.25, 4.8 m/s2
36
ax .
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Chapter 2 / Fluid Statics
c) 60 000 = –1000 ax (–8) – 1000 (9.81 + 5) (–2.5 +
60 = 8 ax + 37.0 – 14.81
a x2 – 7.6 ax + 8.266 = 0
2.99
8a x
).
14.81
8ax
or ax – 2.875 = 1.361
14.81
ax = 1.32, 6.28 m/s2
ax .
a) ax = 20  0.866 = 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
2.100 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
2.5
 FAB   (76 260  9810 z)4dz = 640 000 N or 640 kN
0
b) The pressure on the bottom BC is
p(x) = –1000  10 (x – 7.626) = 76 260 – 10 000 x.
 FBC 
7.626
 (76 260  10 000 x)4dx
= 1.163  106 N or 1163 kN
0
c) On the top p(x) = –1000  10 (x – 5.174) where position 1 is on the top surface:
 Ftop 
5.174
 (51 740  10 000 x)4dx = 5.35  10
5
N or 535 kN
0
2.101 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
 1.5  2 = 100 200 N or 100.2 kN
2
z
x
b) pD = 0. pC = –1000  17.32 (–0.5–1.232)  1000  19.81(0.866–1.866) = 49 810 Pa.
1
FCD   49 810  1.5  2 = 74 720 N or 74.72 kN.
2
58.29  49.81
 1.5 = 81.08 kN.
c) pA = 58 290 Pa.
pC = 49 810 Pa.  FAC 
2
37
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 2 / Fluid Statics
2.102 Use Eq. 2.5.2 with position 1 at the open end:
a) pA = 0 since z2 = z1.
z
1
A
pB = 1000  19.81  0.6 = 11 890 Pa.
pC = 11 890 Pa.
b) pA = –1000  10 (0.9–0) = –9000 Pa.
pB = –000  10 (0.9)–1000  9.81(0.6) = –3114 Pa
C
B
x
pC = –1000  9.81  (–0.6) = 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
25
pB = 1.94  (32.2-60)   = 112 psf.
pC = –112 psf.
d) pA = 0.
 12 
37.5 
e) pA = 1.94  60  
 = 364 psf.
 12 
37.5 
 25 
pB = 1.94  60  
 – 1.94  32.2    = –234 psf.
 12 
 12 
25
pC = –1.94  32.2    = 130 psf.
 12 
37.5 
f) pA = 1.94  30 
 = 182 psf.
 12 
37.5 
 25 
pB = –1.94(–30) 
 – 1.94  62.2    = 433 psf.
 12 
 12 
25
pC = –1.94  62.2     = 251 psf.
 12 
Rotating Containers
2.103 Use Eq. 2.6.4 with position 1 at the open end:
50  2

= 5.236 rad/s.
60
a) p A  (1000  5.2362 /2)  (0.6 1.5) 2 = 11 100 Pa.
1
p B   1000  5.2362  0.92 + 9810  0.6 = 16 990 Pa.
2
pC = 9810  0.6 = 5886 Pa.
1
b) p A   1000  5.2362  0.62 = 4935 Pa.
2
1
p B   1000  5.2362  0.62 + 9810  0.4 = 8859 Pa.
2
pC = 9810  0.4 = 3924 Pa.
38
z
1
A

C
B
r
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 2 / Fluid Statics
2
1
37.5 
c) p A   1.94  5.2362  
 = 259.7 psf.
 12 
2
2
1
37.5
25
p B   1.94  5.2362  
= 389.7 psf.
  62.4 


2
12
12
pC = 62.4  25/12 = 130 psf.
2
1
22.5
d) p A   1.94  5.2362  
 = 93.5 psf.
 12 
2
2
1
15
22.5

2
p B   1.94  5.236  
= 171.5 psf.
 + 62.4 
 12 
2
12
pC = 62.4  15/12 = 78 psf.
2.104 Use Eq. 2.6.4 with position 1 at the open end.
1
a) p A   1000  102 (0 – 0.92) = –40 500 Pa.
2
pB = –40 500 + 9810  0.6 = –34 600 Pa.
pC = 9810  0.6 = 5886 Pa.
1
b) p A   1000  102 (0 – 0.62) = –18 000 Pa.
2
pB = –18 000 + 9810  0.4 = –14 080 Pa.
pC = 9810  0.4 = 3924 Pa.
1
37.5 2 
2 
p

c) A
 1.94  10  0 
 = –947 psf.

2
144 
pB = 947 + 62.4  25/12 = –817 psf.
1
 22.5 2 
d) p A   1.94  102  
 = –341 psf.
 12 2 
2
pB = –341 + 62.4  15/12 = –263 psf.
z
A
1

r
C
B
pC = 62.4  25/12 = 130 psf.
pC = 62.4  15/12 = 78 psf.
2.105 Use Eq. 2.6.4 with position 1 at the open end and position 2 at the origin. Given: p2 = 0.
1
1
a) 0 =  1000 2 (0 – 0.452) – 9810 (0 – 0.6).  = 7.62 rad/s.
2
z
1
2
2
b) 0 =  1000  (0 – 0.3 ) – 9810 (0 – 0.4).  = 9.34 rad/s.
2

1
25 
18.75 2 

2 
r
c) 0 =  1.94   0 
 – 62.4    .  = 7.41 rad/s.

2
12 2 
12
d) 0 =

2
1
 11.25 
15
 1.94 2  
– 62.4    .  = 9.57 rad/s
2 
 12 
 12 
2
39
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 2 / Fluid Statics
2.106 The air volume before and after is equal.
1
 r02 h = 0.144.
 r02 h   .6 2 .2.
2
a) Using Eq. 2.6.5: r02  5 2 / 2 = 9.81 h
h = 0.428 m
1
pA =  1000  52  0.62 – 9810 (–0.372)
2
= 8149 Pa.
z

r0
h

A
r
b) r02  7 2 / 2 = 9.81 h. h = 0.6 m.
pA =
1000
 72  0.62 + 9810  0.2 = 10 780 Pa.
2
z
c) For  = 10, part of the bottom is bared.
1
2
1
2
r0
 .6 2 .2  r02 h  r12 h1 .
Using Eq. 2.6.5:
 2 r02
 h,
2g
 0.144 
 2 r12
2g
2g
h2 
2g
h
 h1 .
h12
2
h1 
or


0.144  10 2
2
2
.
h  h1 
2  9.81
2
A
Also, h – h1 = 0.8. 1.6h – 0.64 = 0.7339. h = 0.859 m, r1 = 0.108 m.
pA = 1000  102 (0.62 – 0.1082)/2 = 17 400 Pa.
0.144  20 2
. 1.6h – 0.64 = 2.936. h = 2.235 m.
2  9.81
pA = 1000  202 (0.62 – 0.2652)/2 = 57 900 Pa
r1 = 0.265 m
d) Following part (c): h 2  h12 
2.107 The answers to Problem 2.105 are increased by 25 000 Pa.
a) 33 150 Pa
b) 35 780 Pa
c) 42 400 Pa
40
d) 82 900 Pa
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
r
Chapter 2 / Fluid Statics
2.108 p(r ) 
1
 2r 2   g[0  (0.8  h)].
2
dA = 2rdr
p(r )  500 2r 2  9810(0.8  h)
p(r )  500 2 (r 2  r12 )
a) F   p 2 rdr  2
b) F   p 2 rdr  2
c) F   p 2 rdr  2
d) F   p 2 rdr  2
0.6
 (12 500r
0
if h < 0.8.
dr
if h > 0.8.
3
 3650r )dr = 6670 N.
(We used h = 0.428 m)
0.6
 (24 500r
3
 1962r )dr = 7210 N. (We used h = 0.6 m)
0
0.6

(50 000(r 3  0.1082 r )dr = 9520 N. (We used r1 = 0.108 m)
0.108
0.6

(200 000(r 3  0.2652 r )dr = 26 400 N. (We used r1 = 0.265 m)
0.265
41
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 2 / Fluid Statics
42
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3 / Introduction to Fluid Motion
CHAPTER 3
Introduction to Fluids in Motion
FE-type Exam Review Problems: Problems 3-1 to 3-10
3.1
(D)
(nxˆi ny ˆj) (3ˆi 4ˆj) 0
nˆ V 0.
Also nx2 ny2
3.2
(C)
(D)
nx
4 / 5 and n y
a
V
V
V
V
u
v
w
t
x
y
z
ax
2xy(2yˆi ) y 2 (2xˆi 2yˆj)
(4 x)
16ˆi 8ˆi 16ˆj
17.89 m/s
u
u
u
u
u
v
w
t
x
y
z
2
0
3 / 5. (Each with a negative sign would also be OK.)
( 8)2 162
10
3nx 4ny
1 since n̂ is a unit vector. A simultaneous solution yields
a
3.3
or
u
x
u
10( 2)( 1)(4 x)
3
10
10(4 x)
2
2
(4 x) x
10
1
20
6.25 m/s 2 .
4
8
The only velocity component is u(x). We have neglected v(x) since it is quite
small. If v(x) were not negligible, the flow would be two-dimensional.
3.4
(C)
3.5
(B)
V2
2
p
3.6
(C)
V12
2g
p
3.7
(B)
The manometer reading h implies:
3.8
(A)
water h
9810 0.800
.
1.23
air
V22
.
2g
V12
2g
0.200
0.600.
V
113 m/s.
V
2 9.81 0.400
2.80 m/s.
V12 p1 V22 p2
2
or V22
(60 10.2).
V2 9.39 m/s
2
2
1.13
The temperature (the viscosity of the water) and the diameter of the pipe are not
needed.
V12
2g
p1
V22
2g
p2
.
800 000
9810
43
V22
.
2 9.81
V2
40 m/s.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 3 / Introduction to Fluid Motion
3.9
p1
(D)
2
902
302 152
2
V22 V12
304 400 Pa
Chapter 3 Problems: Flow Fields
3.10
pathline
streamline
3.11
streakline
Pathline: Release several at an instant in time and take a time exposure of the subsequent motions of the bulbs.
Streakline: 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.
3.12
streakline
pathline
t=0
hose
time t
boy
3.13
y
streakline at t = 3 hr
pathline
t = 2 hr
streamlines
t = 2 hr
x
3.14
dx
2t 2
dt
t 2 2t c1
a) u
x
y
x
2
v
dy
2t
dt
y t 2 c2
y
(27, 21)
2 y
2xy
y
2
4y
streamlines
t=5s
parabola.
44
(35, 25)
39.8o
x
© 2021 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3 / Introduction to Fluid Motion
b) x
t2
y
2t c 1 .
x
3.15
3.16
3.17
V
dr
2
4.
4) 8
2xy y
2
8x 12y
uˆi vˆj wkˆ
dx ˆi dy ˆj dz kˆ
parabola.
0.
(V dr ) z
udy vdx
using ˆi ˆj kˆ ,
ˆj ˆi
kˆ .
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.
a) At t
2 and (0, 0, 0)
V
At t
2 and (1, 2, 0)
V
b) At t
2 and (0,0,0)
V
At t
2 and (1, 2, 0)
V
c) At t
At t
3.18
y
4 2(
8 , and c 2
c1
a) cos
b) cos
2 and (0, 0, 0)
V
2 and (1, 2, 0)
V ˆi
V
V
1 2
32
22
0.
nx
2
, ny
13
ny
32 2 2
3.606 fps.
0.
( 2) 2 ( 8) 2
( 4) 2
4 fps.
0.832.
6 fps.
33.69
3
or nˆ
13
( 2)2 ( 8)2
0.2425.
4
or nˆ
17
45
3nx
2ny
nx2 ny2
0
1
ny
nx2
1
(2ˆi 3ˆj).
13
( 2ˆi 8ˆj) (nxˆi ny ˆj) 0.
1
, nx
17
8.246 fps.
22 ( 4) 2 ( 4) 2
2
V nˆ 0.
2 fps.
(3ˆi 2ˆj) (nx ˆi ny ˆj) 0.
V nˆ
V ˆi
V
22
3
nx
2
9 2
nx 1
4
104
2nx 8ny
nx2 ny2 1
0
nx
4ny
16ny2 ny2 1
1
( 4ˆi ˆj).
17
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 3 / Introduction to Fluid Motion
V ˆi
V
c) cos
V nˆ
a) V dr
0.6202.
52 ( 8)2
5
, nx
89
8
or nˆ
89
( x 2)ˆi xtˆj
0.
(x
51.67
5nx 8ny
(5ˆi 8ˆj) (nx ˆi ny ˆj) 0.
0.
ny
3.19
5
2)dy
xtdx
nx2 ny2
xdx
x 2
Integrate: 2 nx
2.
C
c) V dr
(x 2
4)dy
Integrate:
C
3.20
0 or
lnx2
y 2 tdx
t
tan
2
0.9636.
y C.
(dxˆi dyˆj) 0.
0 or
1
x
2
1
2.
(dxˆi dyˆj) 0.
dy
.
y2
tdx
1
.
y
x
C
2
yt tan
x2 y
ln( y / 2).
( x2 4)ˆi y 2tˆj
0.
1
dy
2dx
.
x
y
ln( y / C). 2ln(1)
ln( 2 / C).
2 y 2 dx
xydy
64 2
ny ny2
25
y 0.8028
xyˆi 2 y 2ˆj
0.
8
ny
5
dy .
xdx
dy. t x 2ln x 2
x 2
2(1 2 ln 3)
2 C.
C 0.8028.
b) V dr
nx
0.
Integrate: t
t x 2 ln x 2
1
1
(8ˆi 5ˆj).
89
(dxˆi dyˆj)
0 or t
0
4
2
tan
2
x
0.9636
2
1
1
C
2
1
.
2
2
V
V
V
V
DV
u
v
w
0
Dt
x
y
z
t
V
V
V
V
b) u
v
w
2x(2ˆi ) 2 y(2ˆj) 4xˆi 4 yˆj = 8ˆi 4ˆj
x
y
z
t
a)
46
© 2021 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3 / Introduction to Fluid Motion
V
V
V
v
w
x
y
z
c) u
V
t
x2t (2xtˆi 2ytˆj) 2xyt (2xtˆj 2ztkˆ ) x 2ˆi 2xyˆj) 2yzkˆ
68ˆi 100ˆj 54kˆ
V
V
V
v
w
x
y
z
d) u
V
t
x(ˆi 2yzˆj) 2xyz( 2xzˆj) tz( 2xyˆj tkˆ ) zkˆ
= xˆi (2 yz 4x2 yz 2 2xyzt )ˆj ( zt 2 z)kˆ
= 2ˆi 114ˆj 15kˆ
3.21
1
2
Ω
a) Ω
3.22
3.23
v ˆ 1
i
z
2
w
y
1 uˆ
k
2 y
w ˆ 1
j
x
2
u
z
v
x
u ˆ
k
y
20 y kˆ = 20 kˆ
b) Ω
1
1
1
(0 0)ˆi
(0 0)ˆj
(0 0)kˆ = 0
2
2
2
c) Ω
1
1
1
(2zt 0)ˆi
(0 0)ˆj
(2 yt 0)kˆ
2
2
2
d) Ω
1
1
1
(0 2xy)ˆi
(0 0)ˆj
( 2 yz 0)kˆ
2
2
2
6ˆi 2kˆ
2ˆi 3kˆ
The vorticity ω 2Ω. Using the results of Problem 3.21:
b) ω 0
c) ω 12ˆi 4kˆ
d) ω
a) ω
40ˆi
a)
b)
u
x
xx
0,
v
y
yy
xy
1
2
u
y
v
x
yz
1
2
v
z
w
y
20 y
0.
0,
zz
20,
xz
w
z
1
2
rate-of strain
0.
u
z
w
x
2,
yy
2,
zz
0.
xy
0,
xz
0,
yz
0.
0,
0 20 0
20 0 0
0
xx
4ˆi 6kˆ
0
0
2 0 0
rate-of strain = 0 2 0
0 0 0
47
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Chapter 3 / Introduction to Fluid Motion
c)
xx
2xt
8,
1
(2 yt )
2
xy
2xt
yy
2,
xz
rate-of strain =
d)
xx
1,
2xz
yy
1
( 2 yz )
2
xy
3,
1
(0)
2
8
2
2 8
0
6
12, zz
a) ar
10
1
40
cos
r2
1
10
r
40
a
10
r2
1
100
r
a = 0.
40
r2
0,
0,
b)
r
80
cos
r3
2
3.25
a) ar
10
80
sin
r3
1
r
ω= 0
0
6.
6
4
t
2.
yz
1
( 2xy )
2
40 sin
r2
r
40 sin
r2
r
10
10
z
80
cos
r3
1
(2zt )
2
yz
10
1600
sin cos
r4
At (4, 180 )
0,
4.
1
2.
40
( sin )
r2
(10 2.5)( 1)1.25( 1) = 9.375 m/s2.
sin 2
cos
2 yt
zz
1
(0) 0,
2
3 0
12 2
2 2
xz
rate-of strain = 3
0
3.24
8,
240
cos
r4
=0
10
40
cos
r2
since sin 180 = 0.
40
1
40
sin
( sin ) = 0.
10
2
r
r
r2
since ω = 0 everywhere.
10
80 sin
( sin ) 10
r3
r
80
r3
2
80 sin 2
10
8.75( 1)(.9375)( 1) = 8.203 m/s2
r3
r
a = 0 since v 0.
a = 0 since sin 180 = 0.
b)
3.26
a
r
= 0,
V
V
u
t
x
= 0,
v
= 0,
V
y
w
V
z
since sin 180 = 0.
uˆ
i. For steady flow u / t
t
48
0 so that a 0.
© 2021 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3 / Introduction to Fluid Motion
3.27
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 v
0 and /
0:
v
v
u
u
ar v
u
ax v
u
r
x
r
x
3.28
a) u 2(1 0)(1 e t /10 )
u
1
ax
2(1 0)
t
10
b) u 2(1 0.52 )(1 e
ax
c) u
ax
2 m/s at t
e
t /10
)
t /10
0.2 m/s2 at t
1.875 m/s at t
1 t /10
e
10
2(1 2 2 / 2 2 )(1 e t / 10 )
1 t/ 10
2(1 2 2 / 2 2 )
e
10
2(1 0.52 / 22 )
T
z
.
0.0125 m/s 2 at t 0.
0 for all t .
0 for all t .
T
t
t
100
DT
Dt
u
3.30
D
Dt
u
3.31
D
Dt
u
3.32
D
Dt
u
3.33
D
observing that the dot product of two vectors A
V
Dt
t
and B Bxˆi By ˆj Bzkˆ is A B Ax Bx Ay By Az Bz .
3.34
ay
az
x
x
x
u
V
t
v
V
t
w
V
t
T
y
0.
3.29
ax
T
x
.
v
v
y
v
y
w
w
w
z
z
t
t
20(1 y 2 )
100
10( 1.23 10 4 e
10
1000
4
sin
3000 10 4
= 2500
0.5878
5
= 0.3693 C / s
) = 9.11 10
4
kg
.
m3 s
kg
.
m3 s
4 (.01) = 0.04 kg/m3 s
Axˆi Ay ˆj Azkˆ
u
v
a
V
(V
t
)V
w
49
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Chapter 3 / Introduction to Fluid Motion
3.35
3.36
Using Eq. 3.2.12:
d 2S
dΩ
a) A a
r
2Ω V Ω (Ω r )
2
dt
dt
= 2(20kˆ 4ˆi) 20kˆ (20kˆ 1.5iˆ) 160 ˆj 600 iˆ m 2/s
b) A 2Ω V Ω (Ω r) 2(20kˆ 20cos30 ˆj) 20kˆ (20kˆ 3ˆi)
507ˆi = 507iˆ
2
kˆ 7.272 10 5 kˆ rad/s.
24 60 60
3.535ˆi 3.535kˆ m/s.
V 5( 0.707ˆi 0.707kˆ )
Ω
A
2Ω V Ω ( Ω r )
= 2 7.272 10 5 kˆ ( 3.535ˆi 3.535kˆ ) 7.272 10 5 kˆ
7.272 10 5 kˆ (6.4 106 )( 0.707ˆi 0.707kˆ ) = 52 10 5 ˆj 0.0224ˆi m/s2 .
Note: We have neglected the acceleration of the earth relative to the sun since it is quite
small (it is d 2S /dt 2 ). The component ( 51.4 10 5 ˆj) is the Coriolis acceleration and causes
air motions to move c.w. or c.c.w. in the two hemispheres.
Classification of Fluid Flows
3.37
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)
3.38
Steady: a, c, e, f, h
Unsteady: b, d, g
3.39
b. It is an unsteady plane flow.
3.40
a)
3.41
f, h
3.42
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.
3.43
d and e. Each flow possesses a stagnation point.
d)
e)
50
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Chapter 3 / Introduction to Fluid Motion
3.44
3.45
Re
3.46
Re
3.47
3.48
3.49
VL /
VL
=2
0.2
0.8/1.4
Turbulent.
VL
4 .06
= 14 100.
1.7 10 5
Note: We used the smallest dimension to be safe!
Re
a)
Re
b)
Re
Re = 3
3 10 5
1.2 0.01
VD
VD
105 =
1.2 1
1.51 10
VxT
3 105
5
.
600 5280 xT
3600 3.7 10
Always laminar.
May not be laminar.
79 500.
/
1.5 10
900 1000x T
.
3600 2.5 10 5
= 3.3
.
(T ).
where
5
2
N s/m ,
1.5 10 5
0.3376 1.23
xT = 0.03 m or
xT = 0.13' or
Assume the flow is parallel to the leaf. Then 3
xT 3 105 / V 3.5 105 1.4 10
The flow is expected to be laminar.
4
2.5 10
5
m2 /s.
3 cm
3.3 10 7
0.00089
10 7 lb-sec/ft2.
4
Turbulent.
795.
5
1.51 10
b) T = 48 F
3.51
Turbulent.
10 5 = 11 400.
a) T = 223 K or 50 C.
3.50
10 6 = 39 000.
0.015/0.77
3.7 10
4
ft2/sec.
1.5"
105 = VxT / .
/ 6 8.17 m.
100
V
0.325. For accurate calculations the flow is
c
1.4 287 236
compressible. Assume incompressible flow if an error of 4%, or so, is acceptable.
80
V
0.235.
Assume incompressible.
b) M
c
1.4 287 288
100
V
0.258.
c) M
Assume incompressible.
c
1.4 287 373
a) M
51
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Chapter 3 / Introduction to Fluid Motion
3.52
D
Dt
u
Then
3.53
u
D
Dt
v
x
u
v
x
v
x
w
y
z
For a steady, plane flow
0.
t
0 and w 0.
/ t
0.
y
w
y
z
incompressible.
0.
t
Bernoulli’s Equation
3.54
V2
2
.
Use
= 0.0021 slug/ft3.
a) v
2p /
2 0.3 144 / 0.0021 = 203 ft/sec
b) v
2p /
2 0.9 144 / 0.0021 = 351 ft/sec
c) v
2p /
2 0.09 144 / 0.0021 = 111 ft/sec
3.55
p
3.56
V2
2
3.57
a)
3.58
p
V2
2
p
1.23 120 1000
2
3600
0.
p
V02
2
p0
V2
b)
2
p
V02
2
p0
V2
2
p
U2
2
p
.
( 10x)2
2
.
(10 y)2
2
a) v
.
2 2000
= 57.0 m/s
1.23
p
c) vr
d) Let
rc :
pT
2
0 and r
rc , v
90  :
p 90
p0
p
p0
0 and
p
b) Let r
0.0752 = 12.1 N.
F = pA = 683
= 683 Pa
2p
V
V2
2
2
.
.
p
p
180 , vr
U
2
2
vr2
p0 50x2
p0 50 y 2
U (1 rc2 / r 2 )( 1).
2
U
2
2
rc2
r2
rc
r
4
.
U2
U 2sin .
p
2
U2
v2
2
U 2 1 4sin 2
3
U2
2
52
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Chapter 3 / Introduction to Fluid Motion
3.59
V2
2
U2
2
p
0 and
a) v
b) Let r
rc :
V2
2
a) p
rc : p
2
p90
U2
2
p
p
(U 2
U2
2
vr2
2
U2 2
U2
2
3
U2
2
v2
c) p
2
3.61
102
2
u )
0 when x
V 12
2
p1
2
2
1.
p2
V 22
2
p1 p2
p
1.
2
d) u
V22
3.62
0 when x
(U
.
2
.
u2 )
2
6
rc
r
U 2 1 4sin 2
10
2
20
2 x
50
1
50
b) u
3
rc
r
1
U2 .
2
pT
90 :
d) Let
.
180 : p
0 and r
c) vr
3.60
p
.
30
V1
450
450 ( 2 1)
1
0 and p1
2
(20 000)
1000
p2
40.
2
2
x
1
x2
1
1
1
x
50
2
60
30
2 x
2
p
50 ( 2 1)
1
1
x
1
2
450
2
x
1
x2
450
20 kPa.
V2
6.32 m/s
Assume the velocity in the plenum is zero. Then
V12
2
p1
We found
V22
2
p2
or V22
2
(60 10.2).
1.13
V2
9.39 m/s
113
. kg / m 3 in Table B.2.
53
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Chapter 3 / Introduction to Fluid Motion
3.63 Applying the Bernoulli equation between points 1 and 2 along the streamline on the
centerline of the flow:
V12
V22
z1 p2
z2
2
2
From the manometer reading we find
p1
p1
z1
V22
2
p2
( z2
H)
H
Hg
where we have used Eq. 3.4.11. Subtract the manometer equation from Bernoulli’s
equation and we have
V12
2
(
)H
Hg
Substitute in the given information and there results
V1
3.64
2 (13.6 1) 9810 N/m3 0.12 m
1000 kg/m3
Then,
3.65
V2
2
p
H
Hg H
h
H
Hg H
p.
p
V2
2
pT
Bernoulli from the stream to the pitot probe:
Manometer: pT
5.45 m/s
p.
h.
Hg
V2
(2 H )
a) V 2
(13.6 1)9800
(2 0.04).
1000
V
3.14 m/s
b) V 2
(13.6 1)9800
(2 0.1).
1000
V
4.97 m/s
c) V 2
(13.6 1)62.4
(2 2 /12).
1.94
V
d) V 2
(13.6 1)62.4
(2 4 /12).
1.94
V 16.44 fps
11.62 fps
Applying Bernoulli’s equation between the two sections connected by the manometer we
write
p1
V12
2
z1
p2
V22
2
54
z2
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Chapter 3 / Introduction to Fluid Motion
The manometer equation is
p1
V12
2
z
p2
air 1
( z2
air
H)
water
H
Subtract the manometer equation from Bernoulli’s equation and obtain
V22
2
0
(
air
water
)H
Since air is an ideal gas we calculate the density as follows:
120 N/m2
0.287 N m/kg K 30 273 K
p
RT
1.38 kg/m3
Substitute in the given information:
V2
2(
water
air
2 (9810 1.38 9.81) N/m3 0.05 m
1.38 kg/m3
)H
26.6 m/s
The air column could have been neglected.
3.66
U 2 / 2. The manometer provides: pT
stagnation point is pT
1
1.204U 2
2
3.67
3
1.204U 2 .
2
9800 0.04
U
1
1.204U 2
2
U 2 / 2. The manometer provides: pT
Bernoulli:
Manometer:
V22
2g
p1
3
1.204U 2 .
2
9800 0.04
p2
z
V12
2g
Hg H
U
The pressure at the
H
p90
12.76 m/s
3 U 2 / 2.
The pressure at 90 from Problem 3.59 is p90
stagnation point is pT
3.68
3 U 2 / 2.
The pressure at 90 from Problem 3.58 is p90
The pressure at the
H
p90
12.76 m/s
p1
H
z
V22
2g
p2
Substitute Bernoulli’s into the manometer equation:
V12
p1
H
p1.
Hg
2g
55
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Chapter 3 / Introduction to Fluid Motion
a) Use H = 0.01 m:
V12 9800
2 9.81
Substitute into Bernoulli:
V 22 V 12
p1
2g
(13.6 1)9800 0.01
20 2 1.572 2
2 9.81
V12 9800
b) Use H = 0.05 m:
2 9.81
Substitute into Bernoulli:
V 22 V 12
p1
2g
Substitute into Bernoulli:
V 22 V 12
p1
2g
3.69
198 600 Pa
(13.6 1)9800 0.05
20 2 3.516 2
2 9.81
V12 9800
c) Use H = 0.1 m:
2 9.81
9800
V1 1.572 m/s
9800
(13.6 1)9800 0.1
20 2 4.972 2
2 9.81
9800
V1 3.516 m/s
193 600 Pa
V1
4.972 m/s
187 400 Pa
Cavitation will occur when the pressure in the liquid becomes equal to the vapor
pressure. For water at 15°C the vapor pressure is 1.7 kPa absolute (consult the
Appendix). The minimum pressure in the flow will occur at the minimum flow area.
Apply Bernoulli’s equation between points 1 and 2 which lie on the centerline:
p1
V12
2
z1
p2
V22
2
z2
Since the flow is horizontal z1 z2 , p1 = (120 + 100) kPa absolute, and p2 = 1.7 kPa
absolute so Bernoulli’s equation takes the form
1000 V12
220 000
2
1000 (4V1 )2
1700
2
V1
5.40 m/s
Substitute in the units to make sure they check.
3.70
Write Bernoulli’s equation between points 1 and 2 along the center streamline:
p1
V12
2
z1
Since the flow is horizontal, z1
p1 1000
0.52
2
p2
V22
2
z2
z2 and Bernoulli’s equation becomes
p2 1000
56
1.1252
2
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Chapter 3 / Introduction to Fluid Motion
From fluid statics, the pressure at 1 is p1
p2 = H, Bernoulli’s equation predicts
2452 1000
3.71
0.52
2
H 0.1982 m or 19.82 cm
0. The Bernoulli’s equation gives, with p 2
V12
2
a) 0
b) 0
c) 0
d) 0
V22
2
V22
2
V 22
2
V 22
2
p1
V22
2
p2
9800 0.02
.
1.204
9800 0.08
.
1.204
62.4 1 / 12
.
0.00233
62.4 4 / 12
.
0.00233
w
0 and
h2 ,
.
V2
18.04 m/s
V2
36.1 m/s
V2
66.8 fps
V2
133.6 fps
Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel where
p1 0 and V1 0. Bernoulli’s equation gives
V22 p2
1
p2
0
.
2
2
p
90
1.239 kg/m3 .
RT 0.287 253
p
95
1.212 kg/m3 .
RT 0.287 273
p
92
1.094 kg/m3 .
RT 0.287 293
p
100
1.113 kg/m3 .
RT 0.287 313
a)
b)
c)
d)
3.73
1.1252
2
9810H 1000
Assume an incompressible flow with point 1 outside in the room where p1
v1
3.72
h 9810 0.25 2452 Pa and at 2, using
a) p A
VA2
2g
h
pA
9800 4
hA
V22
2g
39 200 Pa, V A
p2
h2 .
p2
V22
p2
p2
p2
p2
0.
pA
1
1.239 100 2
2
1
1.212 100 2
2
1
1.094 100 2
2
1
1.113 1002
2
Using hA
6060 Pa
5470 Pa
5566 Pa
h2 ,
V22
2g
39 200
57
6195 Pa
14 2
9800
2 9.81
58 700 Pa
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 3 / Introduction to Fluid Motion
b) pB
2
B
V
2g
3.74
0 and VB
pB
0. Bernoulli’s eqn gives, with the datum through the pipe,
h2 .
p2
4
Bernoulli across nozzle:
V12
2
p1
V22
2
Bernoulli to max. height:
V12
2g
p1
2 p1 /
a) V2
h2
p1 /
h2
h2 .
h2
p1 / .
37.42 m/s
52.92 m/s
2 100 144 /1.94 121.8 fps
p1 /
h2
p2
2 p1 /
1 400 000 / 9800 = 142.9 m
100 144 / 62.4 = 231 ft
2 p1 /
d) V2
V2
.
V22
2g
h1
2 1 400 000 /1000
p1 /
h2
p2
58 700 Pa
700 000 / 9800 = 71.4 m
2 p1 /
c) V2
14 2
9800
2 9.81
2 700 000 /1000
2 p1 /
b) V2
3.75
p2
V 22
2g
hB
2 200 144 /1.94 172.3 fps
p1 /
200 144 / 62.4 = 462 ft
a) Apply Bernoulli’s eq. from the surface to a point on top of the downstream flow:
V12
2g
p1
h1
V22
2g
p2
h2 .
V2
2g (H
h)
b) Apply Bernoulli’s eq. from a point near the bottom upstream to a point on the bottom
of the downstream flow:
V12
2g
p1
Using p1
3.76
V12
2
p1
V22
2g
h1
H , p2
V22
2
p2
p2
h and h1
.
h2 .
h2 ,
V2
2 g( H
h)
p2 = 100 000 Pa, the lowest possible pressure.
600 000
a)
1000
V 22
2
100 000
.
1000
V 2 = 37.4 m/s.
300 000
b)
1000
V 22
2
100 000
.
1000
V 2 = 28.3 m/s.
58
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Chapter 3 / Introduction to Fluid Motion
3.77
80 144
c)
1.94
V 22
2
14.7 144
.
1.94
V 2 = 118.6 ft/sec.
40 144
d)
1.94
V22
2
14.7 144
.
1.94
V 2 = 90.1 ft/sec.
A water system must never have a negative pressure, since a leak could ingest impurities.
The least pressure is zero gage:
V 12
2
p1
500 000
1000
3.78
a) p1
b) p1
c) p1
d) p1
3.79
3.80
V 12
2
2
2
2
2
p1
p2
V 22
2
gz 1
(V22
(V22
V 22
2
1000 2
(2
2
902 2
V12 )
(2
2
680 2
V12 )
(2
2
1.23 2
V12 )
(2
2
p2
.
V2 .
Let z 1
0, and p 2
0.
z 2 = 51.0 m.
9.81 z 2 .
(V22 V12 )
(V22
V1
gz 2 .
102 ) = 48 000 Pa
102 )
43 300 Pa
102 )
32 600 Pa
102 )
59.0 Pa
p1
2
V22 V12
1.23 2
2 302 = 551 Pa
2
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 22 V 12
V 12 p1 V 22 p 2
.
.
p1
2
2
2
Since V2 V1 , 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).
3.81
Re
VD
. For air
a) Re
b) Re
c) Re
1.5 10 5 m2/s. Use reasonable dimensions from your experience!
20 0.03
4 10 4 .
5
1.5 10
20 0.005
6700.
1.5 10 5
20 2
2.7 10 6 .
1.5 10 5
Separate
Separate
Separate
59
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Chapter 3 / Introduction to Fluid Motion
5 0.002
Separate
670.
1.5 10 5
20 2
e) Re
Separate
2.7 10 6 .
1.5 10 5
100 3
It will tend to separate, except
f) Re
2 10 7 .
5
1.5 10
streamlining the components eliminates separation.
d) Re
3.82
3.83
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.
10 2
V2
0.02 40 000 Pa
n 1000
0.05
R
expect VA 10 m/s and VB 10 m/s.
p
stagnation
region
B
Along AB, we
VB
A
VA
3.84
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.
3.85
Refer to Bernoulli’s equation:
V12
2
p1
pA
pB
since
VA
VB
pC
pD
since
VC
VD
pB
pD
since
VD
VB
V 22
2
60
p2
© 2021 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
CHAPTER 4
The Integral Forms of the
Fundamental Laws
FE-type Exam Review Problems: Problems 4-1 to 4-16
4.1
(B)
4.2
(D)
m   AV 
4.3
(A)
Refer to the circle of Problem 4.27:
75.7  2
Q  AV  (  0.42 
 0.10  0.40  sin 75.5 )  3  0.516 m3 /s.
360
4.4
(D)
200
p
  0.042  70  0.837 kg/s .
AV 
0.287  293
RT
WP V22  V12
p  p1

 2
.
Q
2g

WP  40 kW
V22  V12
and energy req'd =
p2  p1
. 0
4.5
(A)
0
4.6
(C)
Manometer:  H  p1   g
2g
Energy: K

WP
1200  200

.
  0.040


40
 47.1 kW.
0.85
p
1202
 2 .  p2  7 200 000 Pa.
2  9.8 9810
V22
V2
 p2 or 9810  0.02  p1   g 2 .
2g
2g
100 000
7.962

.
2  9.81
9810
 K  3.15.
Combine the equations: 9810  0.02  1.2 
4.7
(B)
hL  K
V 2 p

.

2g
V12
.
2
V1  18.1 m/s.
0.040
Q

 7.96 m/s.
A   0.042
100 000
7.962
K

.
 K  3.15.
2  9.81
9810
V 
61
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.8
(C)
WP V22  V12 p

.

Q
2g

WP  Qp  0.040  400  16 kW.
4.9
(D)
4.10
(A)
WP


16
 18.0 kW.
0.89
pB
4.582
7.162
36.0  15 

 3.2
.  pB  416 000 Pa
2  9.81 9810
2  9.81
In the above energy equation we used
V2
Q
0.2
hL  K
with V  
 4.42 m/s.
2g
A   0.22
V 
Q
0.1

 19.89 m / s.
A  .04 2
V 22 p 2
V 22

 z2  K
.
Energy —surface to entrance: H P 
2g 
2g
19.89 2 180 000
19.89 2
HP 

 50  5.6
 201.4 m.
9810
2  9.81
2  9.81
W P  QH P /  P  9810  0.1  201.4 / 0.75  263 000 W.
4.11
(A)
4.12
(C)
After the pressure is found, that pressure is multiplied by the area of the
window. The pressure is relatively constant over the area.
V12 p1 V22
p


 2.

2g 
2g
p1  9810 
(6.252  1)  12.732
 3 085 000 Pa.
2  9.81
p1 A1  F   Q(V2  V1 ). 3 085 000    0.052  F  1000  0.1 12.73(6.25  1)
 F  17 500 N.
4.13
(D)
 Fx  m(V2 x  V1x )  1000  0.01  0.2  50(50cos 60  50)  2500 N.
4.14
(A)
Fx  m(Vr 2 x  Vr1x )  1000    0.022  60  (40cos 45  40)  884 N.
Power  Fx  VB  884  20  17 700 W.
4.15
(A)
Let the vehicle move to the right. The scoop then diverts the water to the
right. Then
F  m(V2 x  V1x )  1000  0.05  2  60  [60  (60)]  720 000 N.
62
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Chapter 4 / The Integral Forms of the Fundamental Laws
Chapter 4 Problems: Basic Laws
4.16
4.17
4.18
a) No net force may act on the system: F  0.
b) The energy transferred to or from the system must be zero: Q  W = 0.
c) If V3n  V3  nˆ 2  10ˆi  (ˆj)  0 is the same for all volume elements then
D
D
DV
F 
V  dm, or F 
(mV). Since mass is constant for a system F  m
.
Dt
Dt
Dt
DV
Since
 a,
 F  ma.
Dt
1
Extensive properties: Mass, m; Momentum, mV ; kinetic energy, mV 2 ;
2
potential energy, mgh; enthalpy, H.
Associated intensive properties (divide by the mass): unity, 1; velocity, V; V2/2;
gh; H/m = h (specific enthalpy).
Intensive properties: Temperature, T; time, t; pressure, p; density, ; viscosity, .
System (t )  V 1
c.v.(t )  V 1
System (t  t )  V 1  V
c.v.(t  t )  V 1
4.19
System (t )  V 1  V
c.v.(t )  V 1  V
2
2
2
2
System (t  t )  V
V
c.v.(t  t )  V 1  V
4.20
1
2
2
1
3
3
pump
2
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.
4.21
n
ˆ
a)
v
ˆn
v
n
ˆ
v
b)
v
c)
63
v

ˆn
ˆn
d)
e)
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.22
ˆn
n
ˆ
n
ˆ
v
v
ˆn
v
v
v
n
4.23
ˆn
n
ˆ
v
v
ˆn
ˆn
v
v
n
ˆ
v
ˆn
v
System-to-Control Volume Transformation
1 ˆ 1 ˆ
i
j = 0.707(ˆi  ˆj) .
nˆ 2  0.866ˆi  0.5ˆj . nˆ 3  ˆj .
2
2
V1n  V1  nˆ 1  10ˆi   0.707(ˆi  ˆj)   7.07 fps
V2n  V2  nˆ 2  10ˆi  (0.866ˆi  0.5ˆj)  8.66 fps .
V3n  V3  nˆ 2  10ˆi  (ˆj)  0
4.24
nˆ 1  
4.25
flux = nˆ  VA
flux2 =
4.26
flux1 =
 (0.866ˆi  0.5ˆj) 10ˆiA
0.866
 10 A
  0.707(ˆi  ˆj)  10ˆiA
0.707
 10 A
flux3 =  (ˆj) 10ˆiA3  0
(B  nˆ ) A  15(0.5ˆi  0.866ˆj)  ˆj (10 12)  15  0.866  120  1559 cm 3
Volume = 15 sin 60 10 12  1559 cm3
4.27 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  and  may be functions of
(x, y, z, t); hence, the partial derivative is used.
4.28
2
1
1
system (t) is in
volumes 1 and 2
64
c.v. (0) = c.v. (t)
= volume 1
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.29
2
system (t) = V1 + V2 + V3
3
c.v. (t) = V1 + V2
1
4.30
system boundary
at (t + t)
Conservation of Mass
4.31
If fluid crosses the control surface only on areas A1 and A2,
 nˆ  VdA   nˆ  VdA   nˆ  VdA  0
c.s.
A1
A2
For uniform flow all quantities are constant over each area:
1nˆ 1  V1  dA  2nˆ 2  V2  dA  0
A1
A2
Let A 1 be the inlet so nˆ 1  V1  V1 and A2 be the outlet so nˆ 2  V2  V2 . Then
  1V1 A 1   2V2 A 2  0
or
 2 A 2V2   1 A 1V1
4.32
Use Eq. 4.4.2 with mV representing the mass in the volume:
dmV
dmV
0
   nˆ  VdA 
  A2V2   A1V1
dt c.s.
dt

Finally,
dmV
 Q  m.
dt
dmV
 m   Q.
dt
4.33
Use Eq. 4.4.2 with m S representing the mass in the sponge:
dm S
dm S
dmS
0
   nˆ  VdA 
 A 2 V 2  A 3V 3  A 1V1 
 m 2  A 3V 3  Q 1 .
dt
dt
dt
Finally,
dm S
 Q 1  m 2  A 3 V 3 .
dt
65
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Chapter 4 / The Integral Forms of the Fundamental Laws
1.25 2
2.5 2
 60 =  
V2.
V2 = 15 ft/sec.
144
144
1.25 2
1.25 2
m  A V  1.94
 60 = 3.968 slug/sec. Q = AV = 
 60 = 2.045 ft 3 /sec.
144
144
4.34
A1V1 = A2V2.  
4.35
A1V1 = A2V2.   .0252  10 = (2  0.6  0.003)V2.
V2 = 1.736 m/s.
m   AV  1000  0.0252  10 = 19.63 kg/s. Q = AV=  0.0252  10 = 0.1509 m3 /s.
4.36
4.37
m in = A1V1 + A2V2. 200 = 1000   0.0252  25 + 1000 Q2. Q2 = 0.1509 m3 /s.
p1
7  144
40  144
= 0.006455 slug/ft3.  2 
= 0.000963 slug/ft 3.

1716  610
RT1 1716  520
0.2
m
V1 = 355 fps.
.

m   AV .  V1 
2
1A1 (  2 /144)  0.006455
1 
V2 = 4984 fps.
m2  0.2  0.000963  (2  3 /144)V2 .
4.38
1A1V1   2 A2V2 . 1 
p1
500
kg
1246
kg

 4.433 3 .  2 
 8.317 3
0.287  522
RT 0.287  393
m
m
4.433   0.052  600 = 8.317   0.052 V2. V2 = 319.8 m/s.
m   1 A1V1 = 20.89 kg/s.
4.39
Q 1  A1V1 = 4.712 m3 /s.
p
p1
A 1V1  2 A 2 V 2
RT2
RT1
 1 A1V1   2 A2V2 .
120
200
  0.05 2  40 
  0.03 2  120.
T2
293
4.40
a) A 1V1  A 2 V 2 .
 T2  189.9 K
(2  1.5 + 1.5  1.5) 3 = 
d 22
 2.
4
d 22 2
 . d2 = 4.478 m
4 2
1
R
c) (2  1.5 + 1.5  1.5) 3 =  R 2  .866 R  2.
3

2
b) (2  1.5 + 1.5  1.5) 3 = 
R = 3.581 m.
Q2 = 2.512 m3 /s.
and
or
83 C.
d2 = 3.167 m
cos = 1/2
 = 60o

R
d2 = 7.162 m
66
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.41

r
a) v  10 1   .
 r0 
r0
r0

 r02V   vdA   10 1 
0
V 

0
2
0
r0

r
r2 


2
20
rdr
r



  r0 dr.
r0 

0
2
0
r  10
20  r
= 3.333 m/s.
 
2 
r0  2
3
3
m   AV  1000    0.042  3.33 = 16.75kg / s.
 r2 
b) v  10 1  2  .
 r 
0 

 r02V
Q = AV = 0.01675 m3 /s.
 r2 
 r2 r2 
  10 1  2 2 rdr  20  0  0  . V = 5 m/s
 r 
 2 4
0 



0
r0
m   AV  1000    0.042  5 = 25.13 kg/s.

r
c) v  20 1   .
 r0 
 r02V

Q = AV = 0.02513 m3 /s.
r0

r
20 1  2 rdr  10 r02 / 4.
 r0 
r0 /2

m   AV  1000    0.042  5.833 = 29.32 kg/s.
4.42
a) Since the area is rectangular, V = 5 m/s.
m   AV  1000  0.08  0.8  5 = 320 kg/s.
Q=
V = 5.833 m/s
Q = 0.02932 m3 /s.

m

= 0.32 m 3 /s.
 y y2 
b) v  40   2  with y = 0 at the lower wall.
h h 


 y y2 
h
 Vhw   40   2 wdy  40  w. V = 6.667 m/s.
h h 
6


0
m   AV  1000  0.08  0.8  6.667 = 426.7 kg/s.
Q = 0.4267 m3 /s.
h
c) V  0.08 = 10  0.04 + 5  0.02 + 5  0.02.
V = 7.5 m/s.
m
m   AV  1000  0.08  0.8  7.5 = 480 kg/s. Q 
= 0.48 m3 /s.

4.43
a) A1V1   v2dA.
With r0 
1
,
24
b) A1V1   v2dA.
With h =
1
,
24
r0
2
 r2 
r2
 1 
     6   vmax 1  2  2 rdr  2 vmax 0 .
 r 
4
 24 
0 

0
vmax = 12 fps.
 v(r )  12(1  576r 2 ) fps.
h
 y2 
4h
1




w
v
6
1  2 wdy  vmax w .
max 



 h 
3
 12



h
vmax = 9 fps.
67
 v( y)  9(1  576 y 2 ) fps.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible websit e, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
c) A1V1   v2dA.
0
With r0 = 0.01 m,
d) n̂
0.02  w  2 
r02
r2 
rdr
v

2

2

.

max
4
r02 
 v(r )  4(1  10 000r 2 ) m/s.
vmax = 4 m/s.
 y2 
4h
v
 max 1  h2 wdy  vmax w 3 .


h
h
With h = 0.01 m,
4.44



r0
  0.012  2   vmax 1 
vmax = 3 m/s.
 v( y)  3(1  10 000 y 2 ) m/s.
If dm / dt  0, then  1 A 1V1   2 A 2V2   3 A 3V3 . In terms of m 2 and Q 3 this
becomes, letting  1   2   3 ,
1000    0.02 2  12  m 2  1000  0.01.
4.45
r1
 v dA  A V .
1
2
2
0
 2 vmax
4.46
 r2 
2
 vmax 1  r 2  2 rdr    0.0025  2.
1 

0
r1
0.0052
   0.00252  2.  vmax = 1 m/s.
4

r2 
 v (r )   1 
m/s.
 0.0052 


 0.1

min  mout  m.   0.2  2 10     10(20 y  100 y 2 )2dy    0.1 2 10   m.
 0

Note: We see that at y = 0.1 m the velocity u(0.1) = 10 m/s. Thus, we integrate to y = 0.1,
and between y = 0.1 and 0.2 the velocity u = 10:
4

4      2    m .
3


4.47
 m2  5.08 kg/s.
h
V1 h1   u( y )dy .
0
 m  0.6667  = 0.82 kg/s.
h
10.05   10(20 y  100 y 2 )dy
0
100 3 

 10 10h 2 
h .


3
666.7 h3  200 h2 = 1. This can be solved by trial-and-error:
?
?
h = 0.07:
0.751 1
h = 0.06:
0.576 1
?
?
h = 0.08: 0.939 1
h = 0.083: 0.997 1
?
h = 0.084: 1.016 1
h = 0.0832 m: 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.
68
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.48
m   VdA 
1/ 3
 2.21.3545y (6 y  9y
2
)2  5dy
0
1/ 3


 22  6 y  2.127 y 2  9 y 2  3.19 y 3 dy = 4.528 slug/sec.
0
2
2
4
u max   2  fps. (See Prob. 4.43b).
3
3
3
1
4 
2.2  1.94
= 2.07 slug/ft3.  V A  2.07    5   = 4.6 slug/sec.

3
3 
2
Thus, V A  m since  = (y) and V = V(y) so that V  V .
V 
  0.012  8  (2  0.2  0.04)V2cos 30 .
4.49
A 1V1  A 2V 2 .
4.50
m3 of H 2O
4
m3 of air
 9000  5
2000    0.00153 3
= 1.5  (1.5h).
3
s
m of air
4.51
Use Eq. 4.4.3:
0

d V  1V1  nˆ A1
t
V2 = 0.05774 m/s.
h = 0.565 m.
V1  nˆ 1  V1.
2
(37  14.7)144


 1 
     180 
17.
 1A1V1 
V tire
1716  520
t
t
 96 
slug

.

 3.01  10 5 3
ft  sec
t
4.52
 in  m
2 m
 3.
m
V1 = 20 m/s (see Prob. 4.43c).
20 1000  0.022  10  1000  0.022  V3.
 V3 = 12.04 m/s.
d
d
 net  mc .v.  m
2 m
3 m
1
mc .v.  m
dt
dt
d
 mc.v.  m1  m2  m3  1000    0.022  20  10  1000  0.022 10
dt
= 2.57 kg/s.
4.53
0
4.54
The control surface is close to the
interface at the instant shown.
Vi = interface velocity.
 e A eV e   i A i V i .
1.5    0.152  300 
Ve
8000
 12  Vi .
0.287  673
ˆ
n
Vi
ˆ
n
Vi = 0.244 m/s.
69
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.55
Assume an incompressible flow:
4Q1  A 2V2 .
4.56
4  1500 / 60  (2  4)V2 .
For an incompressible flow (low speed air flow)
 udA  A 2V 2 .
A1
0. 2
 20 y
1/ 5
 0.8dy    0.15 2 V 2 .
0
5
20  0.8 0.2 6/ 5    0.15 2 V 2 .
6
4.57
 V 2  12.5 fp s.
 V2  27.3 m/s
A1V1   v2dA  Ae Ve
 (0.12  0.025 2 )  4 
0. 025

0

r2 
200 1 
2rdr    0.12 V e
2
0.025 

0.1178  0.1963  0.0314Ve .
Ve  10.0 m/s
4.58
Draw a control volume around the entire set-up:
dm tissue
0
 V 2 A 2  V1 A 1
dt
 d22  d 2  
2
 m tissue   
 h2   ( h1 tan  ) h1
 4 
or

 d 2  d 22 
 tissue   
m
h2  h12 h 1 tan 2  .

 4
4.59
The width w of the channel is constant throughout the flow. Then
dm
d
 A 2V 2  A 1V1 .
0  ( whL)  A 2 V 2  A 1V1
dt
dt
dh
0   w  100  0.2w  8   4w  0.2.
h  0.008 m/s
dt
0
4.60
4.61
dm
 A 2V 2  A 1V1
dt
  1000(  0.003 2  0.02  10  10 6 / 60).
m
0
 1 A 1V1   2 A 2V2 .
m  3.99 104 kg/s
m 1   2 A 2 V 2 .
400e10/ 100  10 6  900  0.2    0.05 2 Ve .
70
Ve  207 m/s
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.62
0
dm
  3 Q 3   1 A 1V1  m 2 where m  A h.
dt
a) 0  1000  0.6 2 h  1000  0.6 / 60  1000  0.02 2  10  10.
h  0.0111 m/s
or
11.1 mm/s
b) 0  1000  0.6 2 h  1000  0.01  0  20.
h  0.00884 m/s
or
8.84 mm/s.
c) 0  1000  0.62 h  1000  1.0 / 60  1000  0.022  5  10.
 h  0.000339 m/s
4.63
or
0.339 mm/s.
Choose the control volume to consist of the volume of the liquid in the tank. So the
control surface will move with the liquid surface. Apply the conservation of mass:
0
d
  dV  c.s.  V  ndA
dt c.v.
Since the density of the fluid is constant and there is no flow into the tank, the above
equation becomes
0
dV
 Ve Ae
dt
where Ve and Ae are the velocity and area at the exit. From Bernoulli’s equation we
determine Ve  2 gh where h is the height of liquid at any time. Note that h will vary as
the fluid flows out of the tank.
The volume of liquid in the tank is given by V  hA , where A is the cross-sectional area
of the tank. Substituting in the conservation of mass we get
A
dh
  Ae 2 gh
dt
where we divided by the constant density. Rearranging, we write
dt  
A h1/2 dh
Ae 2 g
tf
A
Integrating:  dt  
Ae
0
tf  

2g
At
Ae

0

hf
h1/2 dh
. Then, where hf = 1.5 m,
2g
2h1f 2  2
  0.52
  0.012
1.5
 1383 s
2  9.81
71
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.64
Choose the control volume to consist of the air volume inside the tank. The conservation
of mass equation is
0
d
  dV  CS  V  n dA
dt CV
Since the volume of the tank is constant, and for no flow into the tank, the equation is
0V
d
 eVe Ae
dt
p
d
1 dp
. At the instant of interest,
.

RT
dt RT dt
Substituting in the conservation of mass equation, we get
Assuming air behaves as an ideal gas,  


1.8 kg/m3  200 m/s    0.015 m 
eVe Ae
dp
( RT )  

1.5 m3
dt
V
dp
  14.5 kPa/s
dt
4.65
2


kJ
0.287 kg  K  298 K 


A 1V1  A 2V 2 where A 2 is an area just under the top surface.
dh
a)   0.02 2  10e  t/ 10    (h tan 60  ) 2
dt
2
 t / 10
 h dh  0.001333 e
dt .
 h 3  0.04 e  t / 10  0.04.
Finally,
h(t )  0.342(1  et/ 10 )1/ 3 .
b) 0.04  10  10e  t/ 10  (h tan 60  )  10h
 hdh  0.2309e  t / 10 dt .
 h 2  4.62 e  t / 10  4.62.
Finally,
h(t )  2.15(1  et /10 )1/ 2 .
Energy Equation
4.66
W  T  pAV  
du
Abelt
dy
 20  500  2 /60  400  0.4  0.5 10  1.81105 100  0.5  0.8
 1047  800  0.000724  1847 W
4.67
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.
72
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.68
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  0.2W
1000  0.02  4.18T  0.2  500.
4.69
  Q H P .
W
P
4.70

p
5  746 

W
T
 40  0.89.

mg
T  0.836C
Q  9800  20
.
0.87
Q  0.01656 m3 /s.
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 106 /3600)  9.81  776 100 W
WT
10 000 000
 T z.
 0.89  50.  V  1.273 m/s
 AVg
100  3  60  V  9.8
4.71

4.72
V 12 p1
V2 p

 z1  2  2  z 2 .
2g 
2g 
3 ft
h2
V1
36 2
12 2
6 
 h2 .
64.4 h22
2  32.2
20.1
8.236  2  h2 .
h2
V2
Continuity: 3  12 = h2 V2.
This can be solved by trial-and-error.
h2  7.9 ' : 8.24 ? 8.22
h2  7.93' .
h2  1.8' : 8.24 ? 8.00
h2  1.75' : 8.24 ? 8.31
h2  1.76 '.
V 12
V2
 z 1  2  z 2  hL .
2g
2g

h2  8' :
4.73
8.24 ? 8.31
16
42
2
 h2  0.2.
2  9.81h22
2  9.81
 2.615  0.815 / h22  h2 . Trial-and-error provides the following:
h2  2.5: 2.615 ? 2.63
? 2.59.
h2  2.45: 2.615 
 h2  2.47 m
h2  0.65: 2.615 ? 2.58
h2  0.64: 2.615 ? 2.63.
 h2  0.646 m
73
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.74
Manometer: Position the datum at the top of the right mercury level.
9810.4  9810 z 2  p 2 
Divide by   9810:
V 22
 1000  (9810  13.6).4  9810  2  p1
2
.4  z 2 
p2


p
V 22
 13.6.4  2  1 .
2g

V 12 p1
V2 p

 z1  2  2  z 2 .
2g 
2g 
Energy:
4.75
(2)
V 12
 12.6 .4.
2g
Subtract (1) from (2): With z1 = 2 m,
(1)
V1 = 9.94 m/s
The manometer equation (see Prob. 4.74) is
Energy:
p2
p
V 22
 13.6 .4  2  1 .
2g

(1)
V 12 p1
V 22 p 2
V 22
.

 z1 

 z 2  0.05
2g 
2g 
2g
(2)
0.4  z 2 


Subtract (1) from (2): With z1 = 2 m, and with V2 = 4V1 (continuity)
1.8V 12
 12.6  0.4.
2g
V1 = 7.41 m/s.
2
4.76
 1
Q = 120  0.002228 =     V 1 .
 12 
2
V1 = 12.25 fps.
2
1
 1.5 
 V1      V 2 .
 12 
 12 
Continuity:
  
Energy:
V 12 p1 V 22 p 2
V 12
.



 0.37
2g 
2g 
2g
V2 = 5.44 fps.

12.25 2 5.44 2 
= 8702.9 psf or 60.44 psi
 p 2  60  144  62.4 0.63

64.4
64.4 

74
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.77
Q = 600  10-3/60 =   0.022 V1.

1
AV 3
V2 
0.02  w  6.673

0
3

y2 
wdy
10 1 
 0.022 


3
A1V1 0.04  7.958
= 3.537 m/s.

A2
0.062
V 12 p1 V 22 p 2



 hL .
2g 
2g 
7.958 2  3.537 2 690 000  700 000

= 1.571 m
2  9.81
9810
 hL 
V1  Q / A1 
dA 
0.02
1
2
Energy:
4.78
V
3
V1 = 7.958 m/s.
0.08
= 28.29 m/s.
 .03 2
V2 = 9V1 = 254.6 m/s.
V12 p1 V22 p2
V2



.2 1 .
2g 
2g 
2g
Energy:
 254.6 2
28.29 2 
 p1  9810 
 0.8
= 32.1  10 6 Pa.

2  9.81 
 2  9.81
4.79
a) Across the nozzle:   0.072  V1    0.0252  V2 .
V12 p1 V22 p2
.



2g 
2g 
Energy:
 p1  9810
For the contraction:   0.072  V1    0.052  V3.
Energy:
V12 p1 V32 p3


 .
2g 
2g 
Manometer:
  0.15  p1  13.6  0.15  p3.
Subtract the above 2 eqns:
V2 = 7.84 V1.
7.842  1 2
V1 .
2  9.81
V3 = 1.96 V1.

p1

 12.6 .15 
p3

.
V2
V12
V2
 12.6  0.15  3  1.962 1 .
2g
2g
2g
 (1.962  1)V12  12.6  0.15  2 g.
V1 = 3.612 m/s.
p1 = 394 400 Pa.
From the reservoir surface to section 1:
V02 p0
V2 p
  z0  1  1  z1
2g 
2g 
H
3.612 2 394 400

= 40.0 m.
19.62
9810
75
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Chapter 4 / The Integral Forms of the Fundamental Laws
b) Manometer:   0.2  p1  13.6  0.2  p3.
V12 p1 V32 p3


 .
2g 
2g 
1.96 2 V12
V2
.
 1  12.6.2 
2g
2g
The nozzle is the same as in part (a):

p1

 12.6  0.2 
p3

.
Also, V3 = 1.96 V1.
Energy:
V1 = 4.171 m/s.
p1 = 534 700 Pa.
From the reservoir surface to the nozzle exit:
V02 p0
V2 p

 z0  2  2  z2 .
2g 
2g 
4.80 a) Energy:
H 
V22
32.7 2

= 54.5 m.
2 g 2  9.81
V 02 p 0
V2 p

 z 0  2  2  z 2 .  V 2  2 gz 0  2  9.81  2.4 = 6.862 m/s.
2g 
2g 
Q = AV = 0.8  1  6.862 = 5.49 m3 /s.
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:
 z0 
V22
 h.
 V2  2g( z 0  h)  2  9.81  2 = 6.264 m/s.
2g
Q = 0.8  6.264 = 5.01 m3 /s.
Note: z0 is measured from the channel bottom in the 2nd geometry.
z0 = H + h.
V02 p0
V22 p2

 z0 

 z2 .
b)
2g 
2g 
2

 V2  2 gz0  2  32.2   6    21.23 fps.
2

Q = AV = (2  1)  21.23 = 42.5 cfs.
For the second geometry, the bottom is used as the datum:
V22
V22
 z0 
 0  h.

 ( H  h)  h.
2g
2g
 V2  2gH  2  32.2  6 = 19.66 fps. Q = 39.3 cfs.
4.81
0.032
= 0.1406 V2 .
Continuity: V1  V2
0.082
From the reservoir surface to the exit:
V02 p0
V2 p
V2

 z0  2  2  z2  K 1 .
2g 
2g 
2g
10 
0.14062 V22
V22
 5
2g
2g
V2 = 13.36 m/s.
Q = 13.36    .0152 = 0.00944 m3 /s.
76
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
The velocity in the pipe is V1 = 1.878 m/s.
pA
1.8782
1.8782
Energy 0  A: 10 

 0.8
 3.
2  9.81 9810
2  9.81
p
1.878 2
1.878 2
Energy 0  B: 10 
 B  2.0
 10.
2  9.81
2  9.81 9810
p
1.878 2
1.878 2
Energy 0  C: 10 
 C  12  2.8
.
2  9.81 9810
2  9.81
p
1.878 2
1.878 2
Energy 0  D: 10 
 D 05
.
2  9.81 9810
2  9.81
4.82
V 22
80 000
4
.
9810
2  9.81
V 02 p 0
V2 p

 z0  2  2  z2.
2g 
2g 
pA = 65 500 Pa.
pB = 5290 Pa.
pC = 26 300 Pa.
pD = 87 500 Pa.
V2 = 19.04 m/s.
a) Q  A2V2    0.0252 19.04 = 0.0374 m 3 /s.
b) Q  A2V2    0.092 19.04 = 0.485 m 3 /s.
c) Q  A2V2    0.052 19.04 = 0.1495 m 3 /s.
4.83
a)
p0

 z0 
V 22
V2
 1.54 1 .
2g
2g
16V 12
V2
80 000
4
 1.54 1 .
9810
2g
2g
V1 = 3.687 m/s.
Q  A1V1    0.052  3.687 = 0.0290 m 3 /s.
b) A 1V1  A 2V 2 .
0.092
 V2  3.24V2 .
0.052
V2
3.24 2 V 22
80 000
. V2 = 3.08 m/s.  Q  A 2V2 = 0.0784 m 3 /s.
 4  2  2.3
9810
2g
2g
c)
4.84
V1 
V2
V2
80 000
 4  2  1.5 2 .
9810
2g
2g
Manometer:
Energy:
 H   z  p1  13.6 H   z  p2 .
p1


V2 

p1

 12.6 H 
p2

.
V 12 p 2 V 22


.

2g
2g
12.6 H 
Combine energy and manometer:
Continuity:
 Q  A 2V2 = 0.0767 m 3 /s.
V2 = 9.77 m/s.
d12
d22
V1 .
 V12
 d14 
 12.6H  2 g  4  1 .
d

 2 
1/2
d 2   12.6H  2 g 
 Q  V1 1   4 4

4 4  d1 / d2  1 
77
V 22  V 12
.
2g
d12
 H
 12.35d12d22  4 4
 d d
 1
2
1/2



© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible websit e, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.85
Use the result of Problem 4.84:
2
0.2
1/2
2
0.4
1/2

a) Q = 12.35  0.16  0.08 
4
4
 0.16  0.08 
2
= 0.0365 m3 / s .

b) Q = 12.35  0.24  0.08 
4
4
 0.24  0.08 
2
 H 
Q  22.37 d d  4

 d1  d24 
2
1
c) Using English units with g = 32.2:
2
= 0.0503 m3 / s .
1/2
2
 1   1   10 /12 
Q  22.37       4

 2   4   0.5  0.254 
1/2
2
 1   15 /12 
d) Q = 22.37  1     4

 3   1  0.33334 
2
4.86
Energy from constriction to outlet:
Continuity:
V1  4V2 .
p1


.
= 1.318 cfs.
= 2.796 cfs.
V 22
 H.
2g
a) Energy from surface to outlet:
1/ 2
2
2
 V 22  2 gH .
V 12 p 2 V 22


.

2g
2g
With p1 = pv = 2450 Pa and p2 = 100 000 Pa,
2450
16
100 000
1

 2 gH 

 2 gH .
9810 2  9.81
9810
2  9.81
H = 0.663 m.
b) With p1 = 0.34 psia, p2 = 14.7 psia,
0.34 144 16
14.7 144 1
2 gH 
2gH.


62.4
2g
62.4
2g
4.87
Continuity:
V1  4V2 . Energy: surface to exit:
Energy: constriction to exit:
 pv  p2 
V22
 16V22
2g
pv


H = 2.21 ft.
V 22
 H.
2g
V12 p2 V22


.
2g 
2g
  p2  15H  100 000  15  0.65  9810 = 4350 Pa.
From Table B.1, T = 30C.
78
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.88
V 22
Energy: surface to surface: z 0  z 2  hL .  30  20  2
.
2g
 V 22 = 10 g.
V1 = 4V2.  V 12 = 160 g.
160 g (94 000)
Energy: surface to constriction: 30 

 z1
2g
9810
z1 = 40.4 m. H = 40.4 + 20 = 60.4 m.
Continuity:
4.89
Continuity:
Energy:
10 2
V 1 = 2.778 V1.
62
V 12 p1 V 22 p 2
V 12 200 000 2.778 2 V 12 2450


 .
.



2g 
2g 
2g
9810
2g
9810
V2 
V1 = 7.67 m/s.
4.90
Q =   0.052  7.67 = 0.0602 m3 /s.
Velocity at exit = V e . Velocity in constriction = V1 . Velocity in pipe = V2 .
Energy: surface to exit:
V e2
 H.
2g
 V e2  2 gH .
D2
V e . Also, V1  4V2 .
d2
V12 pv

.
Energy: surface to constriction: H 
2g 
Continuity across nozzle: V 2 
a) 5 
 97 550
D4
1 
.
 16  4  2 g  5 
.2
9810
2g 

 D  0.131 m
 (.34  14.7)144
D4
1 
b) 15 
.
 2 g  15 
 16
4
2 g  (8 / 12)
62.4

4.91
Energy: surface to exit:
3
Energy: surface to “A”:
3
V 22
V2
4 2 .
2g
2g
 D  0.446  or 5.35 
 V 22  11.77.
11.77
1176  100 000
11.77 2
.

 ( H  3)  1.5
9810
2  9.81
2  9.81
 H  8.57 m .
2
4.92
 1
m  A V  1.94       120  5.079 slug / sec.
 12 
 302  1202 120 144 
ft-lb
WP  5.079  32.2 
or 23.5 hp.

 / 0.85  12,950
62.4 
sec
 2  32.2
79
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.93
m  AV  1000   .02 2  40  50.27 kg / s.
p 
 10 2  40 2
20 000 = 50.27  9.81

 / 0.82.
 2  9.81 9810 
4.94
 0  10.2 2 600 000 
W T  2  1000  9.81

 0.87.
9810 
 2  9.81
We used V2 
4.95
V1 
 p  1.088  10 6 Pa.
W T  1.304  10 6 W.
Q
2

 10.2 m/s
A2   0.252
450
 15.9 fps.
  32
V2 
450
 10.19 fps.
  3.75 2
10.192  15.92 (18  140)144 
 1 
10, 000 

T .
  550  450 1.94  32.2 
62.4
 0.746 
 2  32.2

 T  0.924
4.96
 V 2  V12 p2 p1

c
a) Q  WS  mg  2
   z2  z1  v (T2  T1 )  .
 2 1
g
 2 g

The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13.
1 
p1g
85  9.81

 9.92 N/m3.
RT1 0.287  293
2 
600  9.81 20 500

.
0.287 T2
T2
600 000T2 85 000 716.5

 200 2
 (1 500 000) = 5  9.81



(T2  293).
20 500
9.81
9.92

 2  9.81
 T2  572 K
or
299 C .
Be careful of units. p2  600 000 Pa,
b) 60 000 + 1 500 000 = same as above.
80
cv  716.5 J/kg K
 T2  560 K
or
287  C.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.97
1 
p1 g 14.7  144  32.2
lb
lb
60  144  32.2
 0.213 3 .
 0.0764 3 .  2 

ft
RT1
ft
1716  760
1716  520
ft - lb
c v  4296
.
slug - R
2
1
  AVg  AV .213       600 .697 lb /sec.
mg
 24 
Use Eq. 4.5.17 with Eqs. 4.5.18 and 1.7.13:
 V22  V12 p2 p1 cv

   (T2  T1 )  z2  z1  .
Q  WC  mg 
 2 1 g
 2 g

 6002

60 144 14.7 144 4296
10  778  0.697  WC  .697 



(300  60) 
0.0764
32.2
 2  32.2 0.213

WC  40 600
4.98
ft-lb
sec
or
73.8 hp.
V22
V22 
   mg

Energy: surface to exit: W
20
4
5
.



.
T T
2g 
 2g
V2 
15
 13.26 m / s.
 .62
  Q  15  9810  147 150 N / s.
mg
 13.26 2
13.26 2 

.
 WT  0.8  147 150
 20  4.5
2  9.81
 2  9.81
4.99
 WT  5390 kW.
Energy: surface to “C”:
 10 2
200 000
10 2 

  10  
WP .8  mg

 7.7
770.5.
9810
2  9.81
 2  9.81
(mg   AVg  1000    0.052 10  9.81  770.5 N/s.)
 WP  52 700 W.
Energy: surface to “A”:
pA
10 2
10 2
30 

 1.5
.
2  9.81
2  9.81 9810
Energy: surface to “B”:
V 2  VO2 pB  pO
V2 

 z B  zO  K B 
WPP  mg  B
2 g 

 2 g
 10 2
pB
10 2 
.
.
52 700.8 = 770.5

 30  15
2  9.81
 2  9.81 9810
81
 p A  169 300 Pa .
 p B  706 100 Pa.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible websit e, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.100 Choose a control volume that consists of the entire system and apply conservation of
energy:
V12
p2 V22
HP  
 z  HT  
z h
 2g 1
 2g 2 L
p1
Carburetor
0.5 m
Section (1)
Section (2)
Pump
We recognize that HT = 0, V1 is negligible and hL = 210 V 2/2g where V = Q/Aand
V2  Q / A2 . Rearranging we get:
HP 

p2  p1

V22

 z2  z1  hL
2g

6.3 106 m3 /s
Q
(0.321) 2
V 
 0.321 m/s  hL  210
 1.1 m
A  2.5 103 m 2
2  9.81


Q 6.3 106 m3 /s

 12.53 m/s
V2 
A2  4 104 m 2


Substituting the given values we get:
95  100  kN/m2

12.53 m/s 

 0.5m 
2
HP

6.660 kN/m3
2 9.81 m/s2

 1.1 m  8.85 m
The power input to the pump is:



6660N/m3 6.3 106 m3 /s 8.85m 
 QH P
WP 

 0.5 W
P
0.75
82
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.101 Manometer:
Energy:
V2
20
20
  z1  p1  13.6    z2  p2  2  .
12
12
2
2
p
p
V
20
20

 z 1  1  13.6 
 z2  2  2 .
12
2g
12



p
p
V 12
V2
 z1  1  H T  z 2  2  2 .
2g


2g
20
20 V12
  13.6  
 HT .
12
12 2 g
 H T  12.6 
V1 
18
  1/ 3
20
51.6 2

 62.3' .
12 2  32.2
2
 51.6 fps.
W T  Q  T H T  62.4  18.9  62.3
 62,980
4.102 Energy: across the nozzle:
V12 p2 V22
.



2g
 2g 
p1
6.252 V12
V12
400 000
.



9810
2  9.81 2  9.81
V2 
52
22
ft-lb
sec
or
115 hp
V1  6.25V1.
V1  4.58 m/s , VA  7.16 m/s , V2  28.6 m/s.
Energy: surface to exit:
H P  15 
 WP 
28.62
4.582
7.162
 1.5
 3.2
.
2  9.81
2  9.81
2  9.81
 H P  36.8 m.
 QHP 9810  (  0.012 )  28.6  36.8

 3820 W.
P
0.85
Energy : surface to “A”:
15 
p
7.162
7.162
 A  3.2
.
2  9.81 9810
2  9.81
 p A  39 400 Pa
Energy: surface to “B”:
pB
4.582
7.162
36.0  15 

 3.2
.
2  9.81 9810
2  9.81
4.103 Energy: surface to exit:
10 
 V2  7.83 m/s.
 pB  416 000 Pa
V 22 p 2
V2

 z 2  2.2 2 .
2g 
2g
Q  0.02  7.83   d22 / 4.
83
 d2  0.0570 m.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible websit e, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.104 Depth on raised section = y 2 . Continuity: 3  3  V 2 y 2 .
V2
32
 3  2  (0.4  y 2 ).
2g
2g
Energy (see Eq. 4.5.21):
 3.059 
Trial-and-error:
92
 y2 ,
2 g y 22
y2  2.0 :
y2  1.8 :
y2  2.1:
y2  2.3 :
or
y 23  3.059 y 22  4.128  0
 0.11 ? 0 

 0.05 ? 0 
 y2  1.85 m.
 0.1 ? 0 

 0.1 ? 0 
 y2  2.22 m.
The depth that actually occurs depends on the downstream conditions. We cannot select a
“correct” answer between the two.
.
m3
4.105 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  32 m /s is y   2 m.
.
m2
.
m1
.
m3
To find m 3 use continuity:
m 1  m 2  2m 3 .
2
 4  10  32   2  (28  y 2 )10dy  2m 3 .
0
8

 m 3  640  10 28  2    53.3.
3

2
2
2
V
V
u3
Rate of K.E. loss = m 1 1  2m 3 1   2  10dy
2
2
2
0
2
32 2
 1280 
 53.3  32 2  10   (28  y 2 ) 3 dy
2
0
 1.23  [655 360  54 579  507 320]  115 000 W.
4.106 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 12 p1
V2 p

  2 2  2  hL .
2g 
2g 
82
150 000
82
110 000

2

 hL .
9810
9810
2  9.81
2  9.81
 hL  0.815 m .
84
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
2
1
1
1
23 
4.107 V   VdA   (28  y 2 )dy   28  2    29.33 m/s
A
20
2 
3 
1

AV
2
1
2 3
 V dA  2  29.33 3 0 (28  y ) dy
3
3
3
 3
25 27 
2 2

 28  2  3  28   3  28     1.005
3
5 7 
2  29.333 
1
4.108
a) V 
1
1
VdA 

A
  0.012

1
AV 3
V
3
dA 
0.01

0

r2 
20
10 1 
2 rdr 
 0.012 
0.012


0.01
1

  0.012  53
2
0
 0.012
0.014 


  5 m/s
4  0.012 
 2
3
r2 
10 1 
2 rdr
 0.012 


3
2000  0.01
3  0.01
3  0.016
0.018 




  2.00

0.012  5 3  2
4  0.012 6  0.014 8  0.016 
1
1
b) V   VdA 
0.02w
A

1
AV 3
V
3
dA 

4.109 V 
0.02

0
4

y2 
10 
0.023 
10 1 
0.02
wdy




  6.67 m/s
2

0.02 
3  0.022 
 0.02 
0.02
1
0.02  w  6.673
1000
0.02  6.67 3
0

3  0.02 3 3  0.02 5
0.02 7 



0
02
.
  1.541

3  0.02 2 5  0.02 4 7  0.02 6 

1/ n
R

3

y2 
wdy
10 1 
 0.022 


3
1
1
r


VdA 
u
1
max




A
 R
 R2 0
n 
 n

2 rdr  2umax 

 2n  1 n  1 
R
3/ n
V2 
 2 n
n 
 3  r
3
R 

K.E.    V 
dA   umax 1   2 rdr   umax


 2 
20
 R
 3  2n 3  n  



 5 5
a) V  2umax     0.758 umax
 11 6 
5
3 5
2 3
K.E.   R2umax
    0.24  R umax
 8 13 

K.E.
1
2
 AV 3

1
2
3
0.24  R2umax
3
 R2  0.7583 umax
85
 1.102
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible websit e, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
 7 7
b) V  2umax     0.817 umax
 15 8 
7 7
3
3
K.E.   umax
R2     0.288  R2umax
 10 17 

3
0.288 R2umax
K.E.

2
V2 
 0.8172 umax
2
 AV 
  R  0.817umax 
2
 2 




 1.056
9 9
c) V  2umax     0.853 umax
 19 10 
9
3  9
2 3
K.E.   R2umax
    0.321  R umax
12
21



K.E.
1
2
 AV 3

1
2
3
0.321 R2umax
3
 R2  0.8533 umax
 1.034
 V 2  V12

4.110 Engine power = FD  V  m  2
 u2  u1 


2


 V 2  V12

m f q f  FDV  m  2
 cv (T2  T1 ) 
2


4.111
W   FD  V
 kJ  100  km
10 3  m3 
1340 100 000
 kg 

. 
  015


  930 3   q f   
m 
5  km
1000
3600
 kg  3600  s 
 q f  48 030 kJ/kg
4.112 0   2
V22 p2
V2 p
 LV

 z2  1  1  z1  32
2g 
2g 
gD 2
02
106 180V
V2
 0.35  32 
.
2  9.81
9.81 0.022
V 2  14.4V  3.434  0.
 V  0.235 m/s
86
and
Q  7.37 105 m3 /s
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.113 Choose a control volume that consists of the entire system and apply conservation of
energy as follows:
HP 
p1


p V2
V12
 z1  HT  2  2  z2  hL
2g
 2g
Note that sections 1 and 2 in this case are on the water surface. Hence, p2  p1 , V2  V1 ,
and z2  z1 . The energy equation simplifies to
V2
H P  hL  51
2g
Since V  Q A , the equation is re-written as
H P  51
Q2
2 gA
2

51Q 2
2  9.81 (  0.1)
2
 2634Q 2
Since the pump characteristic curve is given, the operating point is at the intersection
between the pump curve and system curve.
HP
Operating
Point
Pump
Curve
HPD
System
Curve
Q
QD
We determine the operating point and the flow rate by setting the system curve equation
equal to the pump curve equation as follows:
2634Q2  15  11Q  150Q2
Rearranging, we get the quadratic equation: 2784Q2  11Q  15  0. The roots of this
equation are determined using the binomial theorem:
Q
11 
 11
2
 4  2784  15 
2  2784

11  409
5568
We choose the positive root Q = 0.075 m3/s and we calculate
2
HP  2634   0.075  14.8 m.
Hence, the power input to the pump is
WP   QH P  9.81 kN/m3  0.075 m3 /s 14.8 m  10.9 kW
87
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.114 Energy from surface to surface:
V12 p 1
V22 p 2
V2
HP 

 z2 

 z1  K
.
2g 
2g 
2g
Q2
 40  50.7 Q 2
2
  0.04  2  9.81
a) H P  40  5
Try Q  0.25:
Try Q  0.30:
H P  43.2 (energy).
H P  44.6 (energy).
H P  58 (curve)
H P  48 (curve)
Solution: Q  0.32 m3 /s
b) H P  40 
20 Q 2
 40  203 Q 2
  0.04 2  2  9.81
Try Q  0.25:
Solution:
H P  52.7 (energy).
H P  58 (curve)
Q  0.27 m3 /s
Note: The curve does not allow for significant accuracy.
4.115 Continuity:
A1V1  A2V2  A3V3
  0.062  5    0.022  20    0.032 V3.
Energy:
energy in + pump energy = energy out
 V3  11.11 m/s
V2 p 
V2 p 
V2 p 
m1  1  1   WP P  m2  2  2   m3  3  3 
 2
 2
 2
 
 
 



 52 120 000 
 202 300 000 
 0.85WP  1000  0.022  20 

1000  0.062  5  




1000 
1000 
 2
 2
 1111
. 2 500 000 
2
1000  0.03  1111
. 


1000 
 2
W  26 700 W
P
Momentum Equation
4.116
V 12 p1 V 22 p 2


 .
2g 
2g 
a)
V12
200 000 16 V12


.
2  9.81
9810
2  9.81
p1 A 1  F  m V 2  V1 .
V2 
d2
(d / 2)2
V1  4 V1.
 V1  5.164 m/s.
200 000 .03 2  F  1000 .03 2  5.164(4  5.164  5.164).  F  339 N .
b)
V12
400 000 16 V12
 V1  7.303 m/s.


.
2  9.81
9810
2  9.81
400 000 .03 2  F  1000 .03 2  7.303(4  7.303  7.303).  F  679 N .
88
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
c)
V12
30 144 16 V12
d)


 V1  17.24 fps.
.
2  32.2
62.4
2  32.2
30    1.52  F  1.94    (1.5 /12)2  17.242 (4  1).
 F  127 lb.
V12
60 144 16 V12


 V1  24.38 fps.
.
2  32.2
62.4
2  32.2
60    1.52  F  1.94    (1.5 /12)2  24.382 (4  1).
 F  254 lb.
e)
f)
4.117
V12
200 000 16 V12
V1  5.164 m/s.


.
2  9.81
9810
2  9.81
200 000  .062  F  1000  .062  5.164(4  5.164  5.164).  F  1356 N.
V12
V12
30  144 16
.

V1  17.24 fps.
2  32.2
62.4
2  32.2
30    32  F  1.94    (3/12)2  17.242 (4  1).

V 12 p1 V 22 p 2


 .
2g 
2g 
V2 
 F  509 lb.
92
V 1  9V 1 .
32
V 12
81 V12
2 000 000


.
2  9.81
9810
2  9.81
V12  50.
p1 A1  F  m(V2  V1 )  m  8V1
  A1V1  8V1
2 000 000  0.0452  F  1000(  0.0452 )  8  50
 F  10 180 N .
V12 p1 V22 p2
4.118



.
2g 
2g 
V0  0.012  Ve  0.006  0.15.
 Ve  11.1 m/s.
Fx  m(V2 x  V1x )
a) V 2 
V 12
400 000 2.441 V 12


.  V1  23.56 m/s.
2  9.81
9810
2  9.81
 p1 A 1  F  m (V 2  V1 ).
10 2
V 1  1.562 V 1 .
82
400 000  0.052  F  1000  0.052  23.56(0.562  23.56) .  F  692 N .
b) V 2 
10 2
V 1  2.778 V 1 .
62
V 12 400 000 7.716 V 12
.  V1  10.91 m/s.


2g
9810
2g
400 000  0.052  F  1000  0.052 10.91(1.778 10.91).
89
 F  1479 N.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible websit e, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
10 2
c) V 2  2 V 1  6.25 V 1 .
4
V 12 400 000 39.06 V 12
.  V1  4.585 m / s.


2g
9810
2g
400 000  0.052  F  1000  0.052  4.585(5.25  4.585).
10 2
V1  25 V1 .
22
d) V 2 
V 12 400 000 625 V 12
.  V1  1132


.
m / s.
2g
9810
2g
400 000  0.052  F  1000  0.052 1.132(24 1.132).
4.119 V2  4V1  120 fps.
 F  2275 N .
 F  2900 N .
V 2  V12 
 120 2  30 2 

p1    2
62
.
4

 2  32.2   13 ,080 psf.
g
2




2
2
 1.5 
 1.5 
F  p1 A1  m(V2 x  V1x )  13,080  
  1.94   
  30(120  30)  1072 lb.
 12 
 12 
15 V 12 p1
V 12 p1 V 22 p 2


 .

 .
4.120 V 2  4 V 1 .
2g 
2g 
2g

2  9.81
a) V12 
 V1  5.16 m / s, V2  20.7 m / s.
 200 000  26.67.
15  9810
p1 A 1  Fx  m (V2 x  V1x ).  Fx  200 000  0.042  1000  0.042  5.162  1139 N.
 Fy  1000  0.042  5.16(20.7)  537 N.
Fy  m (V 2 y  V1y ).
2  9.81
 400 000  53.33.  V1  7.30 m/s, V2  29.2 m/s.
15  9810
p1 A 1  Fx  m (V2 x  V1x ).  Fx  400 000 .04 2  1000 .04 2  7.3 2  2280 N .
Fy  m (V 2 y  V1y ) = 1000 .04 2  7.3  (29.2)  1071 N .
b) V12 
2  9.81
 800 000  106.7.  V1  10.33 m/s, V2  41.3 m/s.
15  9810
Fx  p1A1   A1V12  800 000  0.042  1000  0.042 10.332  4560 N.
c) V12 
Fy  m (V 2 y ) = 1000  0.042 10.33(41.3)  2140 N.
4.121 V2 
402
10
2
V1  80 m/s.
V 12 p 1 V 22 p 2



2g 
2g 
p 1A 1
 80 2
52 
6
 p1  9810

  3.19  10 Pa.
2
9
.
81
2
9
.
81




F
V2
p1A1  F  m(V2 x  V1x ).
 F  3.19 106   0.22  1000  0.22  5(80  5)  353 000 N.
90
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
  0.0252  4   (0.0252  0.022 )V2 .
4.122 A 1V1  A 2V2 .
 V2  11.11 m/s.
p1

2

V 12 p 2 V 22


.

2g
2g
2
 11.11  4
p1  9810 
 2  9.81

p1 A1  F  m(V2  V1 ).
p1A1
F

  53 700 Pa.

 F  53 700  0.0252  1000  0.0252  4(11.11  4)
 49.6 N.
4.123 Continuity:
Energy:
0.7 V1  0.1 V2 .
 V2  7 V1.
p
p
V
V
 1  z1 
 2  z2
2g 
2g 
2
1
2
2
V12
49V12
 0.7 
 0.1.
2  9.81
2  9.81
Momentum:
F1
F2
Rx
 V1  0.495, V2  3.467 m/s.
F1  F2  Rx  m(V2  V1 )
9810  0.35(0.7 1.5)  9810  0.05(0.11.5)  Rx
 1000  (0.11.5)  3.467(3.467  0.495)
 R x  1986 N.
Rx acts to the left on the water, and to the right on the obstruction.
4.124 Continuity:
6 V1  0.2 V2 .
 V2  30 V1.
Energy (along bottom streamline):
F1
V 12 p1
V2 p

 z1  2  2  z 2
2g 
2g 
F
F2
V22 / 900
V22
6
 0.2.
2  9.81
2  9.81
 V2  10.67, V1  0.36 m/s.
Momentum:
F1  F2  F  m (V2  V1 )
9810  3(6  4)  9810  0.1(0.2  4)  F  1000  ( o.2  4) 10.67(10.67  0.36)
 F  618 000 N .
91
(F acts to the right on the gate.)
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible websit e, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.125 a) 8  0.6  V2 y2 .
  0.3  0.6w  

2
F1  F2  m(V2  V1 ).
 8  0.6 
y2
y2 w    0.6w  8 
 8 .
2
 y2

(0.36  y22 )  4.8  8
y22  0.6 y2  7.829  0.
0.6  y2
.
y2
4.8  8  2
.
9.81
 (0.6  y2 ) y2 
 y2  2.51 m.
(See Example 4.12.)
b) y2 

1
8
2 8
2 1
2
 0.4 122   3.23 m.
  y1  y1  y1V1    0.4  0.4 
2 
g
9.81
 2 

c) y2 

1
8
2 8
2 1
2
 2  202   6.12 ft.
  y1  y1  y1V1   2  2 
2 
g
32.2

 2 
d) y2 

1
8
2 8
2 1
2
 3  302   11.54 ft.
  y1  y1  y1V1   3  3 
2 
g
32.2

 2 
4.126 Continuity:
V2 y2  V1y1  4V2 y1.
Use the result of Example 4.12:
 y2  4 y1.
1/2 


1
8
y2    y1   y12  y1V12 
2
g






a) y2  4  0.8  3.2 m.
3.2 
1/2 
1
8


 0.8  V12 
 0.8   0.82 
2 
9.81


8
b) y2  4  2  8 ft.
4.127 V 
 3.05 
y2 
19.62
1/2 
1
8


 2  V12 
 2   22 
2 
32.2


V12
12
3
 y1.
2  9.81
2  9.81
9
 1 m/s.
3 3
V12
 V1  8.86 m/s.
.


3
. Trial-and-error:
V1
.

 V1  25.4 fps.
V1y1  1 3.

3.05 ? 2.93 

 V1  7.19 m/s.

y  0.417 m.
?
V 1  7.2: 3.05  3.06 1

V 1  7:
1/2 
1
8


 0.417  7.192 
 0.417   0.4172 
2 
9.81


  1.90 m.

V2  1.58 m/s.
V2 1.9  7.19  0.417.
92
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.128 Refer to Example 4.12:

y
60 
 1 y1w    3  6w    6w 10 10   .
y1 
2

(V1y1  6 10).
 y 6

1200
 ( y12  36)  600   1
 37.27.  y1  3.8 ft, V1  15.8 fps.
 or ( y1  6) y1 
2
y
32.2
 1 
4.129 Continuity: 20    0.0152  V2    0.032.
 V2  5 m/s.
Momentum:
p2A2
p1A1
p1A1  p2 A2  m(V2  V1 ).
60 000  0.032  p2  0.032  1000  0.0152  20(5  20).
4.130 V1A1  2V2 A2 .
p1


 p2  135 kPa.
  0.052
V2  15
 30 m/s.
2  0.0252
V12 p2 V22


.
2g 
2g
Fx  m  V2x  V1x  .
 p1  9810
302  152
 337 500 Pa.
2  9.81
p1A1  F  m(V1 ).
 F  p1A1  mV1
 337 500  0.052  1000  0.052 152  4420 N.
4.131 By choosing a control volume around the elbow and drawing a free-body-diagram as
shown we have:
P2A2
y
Rx
P1A1
x
Ry
P3A3
93
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Chapter 4 / The Integral Forms of the Fundamental Laws
First, applying conservation of momentum in the x-direction we write:
p1A1  p3 A3 cos 40  Rx  m3V3 cos 40  m1V1
Rx  p1A1  m1V1  p3 A3 cos 40  m3V3 cos 40
Next we calculate the mass flow rates.
2
m1   V1 A1  1000 kg/m3  15 m/s    0.1 m   471 kg/s
2
m2   V2 A2  1000 kg/m3  5 m/s    0.225 m   795 kg/s
m3  m1  m2  471  795  1266 kg/s
2
V3  m3 / A3  1266 / 1000    0.125    25.8 m/s


Substitute in the conservation of momentum equation:
2
2
Rx  250000     0.1  47115  170000     0.125  1266  25.8  cos 40  16500 N


Now, apply conservation of momentum in the y-direction to write:


Ry  p2 A2  p3 A3 sin 40  m3 V3 sin 40  m2  V2 
Substituting the given values we get:
2
2
Ry  30000     0.225  159  5  170000     0.125  1266  25.8 sin 40  20150 N


The minus signs indicate that the direction of Rx is to the right and Ry is downwards.
4.132 a)  Fx  m(V2 x  V1x ),
V1 
 F  mV1.
V2
300
m

 38.2 m/s
 A1 1000    0.052
F
V1
 F  300  38.2  11 460 N .
b)  F  m r (V1  VB )(cos   1).
28.2
(38.2  10)  6250 N .
 F  300 
38.2
c)  F  mr (V1  Vb )(cos  1)
4.133 a)  F  m (V 2 x  V1x ).
 F  300 
48.2
(38.2  (10))  18 250 N .
38.2
2
 1.25 
2
200  1.94 
  V1 .
 12 
V1  55 fps.
2
 1.25 
2
b)  F  m r (V 1  V B )(cos   1). 200  1.94 
 (V 1  30) . V1  85 fps.
 12 
2
 1.25 
2
c)  F  m r (V1  V B )(cos   1). 200  1.94 
 (V 1  30) . V1  25 fps.
 12 
94
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.134 a)  F  m (V 2 x  V1x ).
700  1000 .04 2 V1 (V1  cos 30  V1 ).
V1  32.24 m/s
 m  A 1V1  1000 .04 2  32.24  162.1 kg / s.
b)  F  m r (V 1  V B )(cos   1). 700  1000 .04 2 (V1  8) 2 (.866  1). V1  40.24 m/s
 m  A 1V1  1000 .04 2  40.24  202 kg / s.
c)  F  m r (V1  V B )(cos   1). 700  1000 .04 2 (V1  8) 2 (.866  1). V1  24.24 m/s
 m  A 1V1  1000 .04 2  24.24  121.8 kg / s.
4.135 a)  R x  m (V 2 x
2
 1
 V 1x )  1.94     120(120 cos 60   120).  R x  305 lb.
 12 
R y  m (V 2 y
2
 1
 V 1y )  1.94    120  (120.866).
 12 
 R y  528 lb.
2
 1
b)  R x  m r (V 1  V B )(cos   1)  1.94     60  60(.5  1).  R x  76.2 lb.
 12 
2
 1
R y  m r (V 1  V B )sin   1.94    60  (60.866).
 12 
 R y  132 lb.
2
 1
c)  R x  m r (V 1  V B )(cos   1)  1.94     180  180(.5  1).  R x  686 lb.
 12 
2
 1
R y  m r (V 1  V B )sin   1.94    180  (180.866).  R y  1188 lb.
 12 
4.136
VB  R  0.5  30  15 m / s.
R x  m (V1  VB )(cos   1)  1000 .025 2  40  25(.5  1).  R x  982 N.
W  10R x VB  10  982  15  147 300 W.
4.137 a) R x  m (V2 x  V1x )  4 .02 2  400(400 cos 60  400).
R y  m (V 2 y  V1y )  4 .02 2  400(400 sin 60  ).
 R x  1206 N.
 R y  696 N.
b)  Rx  mr (V1  VB )(cos120  1)  4  0.022  3002 (0.5  1).  R x  679 N.
R y  m r (V1  VB )sin   4 .02 2  300 2 .866.
 R y  392 N.
c)  Rx  mr (V1  VB )(cos120  1)  4  0.022  5002 (0.5  1).  R x  1885 N.
R y  m r (V1  VB )sin   4 .02 2  500 2 .866.
 R y  1088 N.
4.138  Fx  m (V1  VB )(cos 120  1)  4 .02 2  (400  180)2 (.5  1).  R x  365 N.
V  1.2  150  180 m / s.
W  15  365  180  986 000 W.
B
The y-component force does no work.
95
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.139 a) Refer to Fig. 4.16:
750sin 1  Vr1 sin 45 
  Vr1  507 fps.

 Vr 2
750 cos 1  300  Vr1 cos 45 

Note: V2 x  V1x  Vr 2 cos  2  VB  Vr1 cos 1  VB
 Vr1(cos  2  cos 1)
2
 0.5 
 Rx  mVr1 (cos  2  cos 1 )  0.015 
  750  507(cos 30  cos 45 )  48.9 lb.
 12 
ft-lb
or 400 hp.
W  15RxVB  15  48.9  300  220,000
sec
b)
750sin 1  Vr1 sin 60 

 Vr1  554 fps = Vr 2
750 cos 1  300  Vr1 cos 60 

2
 0.5 
 Rx  mVr1 (cos  2  cos 1 )  0.015  
  750  554(cos 30  cos 60 )  46.4 lb.
 12 
ft-lb
or 380 hp
W  15RxVB  15  46.4  300  209,000
sec
c)
750sin 1  Vr1 sin 90 

 Vr1  687 fps = Vr 2
750 cos 1  300  Vr1 cos 90 

2
 0.5 
 Rx  mVr1 (cos  2  cos 1 )  0.015  
  750  687(cos 30  0)  36.5 lb.
 12 
W  15RxVB  15  36.5  300  164,300
4.140 a) Refer to Fig. 4.16:
ft-lb
or 299 hp
sec
100sin 30  Vr1 sin 1 

 1  36.9 , Vr1  83.3 m/s
100 cos 30  20  Vr1 cos 1 



 V2  71.5,  2  48
V2 cos 60  83.3cos  2  20 

V2 sin 60  83.3sin  2
 Rx  m(V2 x  V1x )  1000  0.0152 100(71.5cos 60  100cos 30 ).  Rx  8650 N
 W  12VB Rx  12  20  8650  2.08  106 W
96
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Chapter 4 / The Integral Forms of the Fundamental Laws
b)
100sin 30  Vr1 sin 1 

 1  47 , Vr1  Vr 2  68.35 m/s
100 cos 30  40  Vr1 cos 1 



 V2  38.9 m/s,  2  29.5
V2 cos 60  68.35cos  2  40 

V2 sin 60  68.35sin  2
 Rx  m(V2 x  V1x )  1000  0.0152 100(38.9cos 60  100cos 30 ).  Rx  7500 N
 W  12VB Rx  12  40  7500  3.60  106 W
c)
100sin 30  Vr1 sin 1 

 1  53.8 , Vr1  Vr 2  61.96 m/s
100 cos 30  50  Vr1 cos 1 



 V2  19.32 m/s,  2  15.66
V2 cos 60  61.96 cos  2  50 

V2 sin 60  61.76sin  2
 Rx  m(V2 x  V1x )  1000  0.0152 100(19.32cos 60  100cos 30 ).  Rx  6800 N
 W  12RxVB  12  6800  50  4.08 106 W
4.141 a) Refer to Fig. 4.16:
50 sin 30   V r 1 sin  1 
2
2
  V r 1  2500  86.6V B  V B

50 cos 30  V B  V r 1 cos  1 



30 cos 60  Vr 2 cos  2  VB 

30sin 60  Vr 2 sin  2
Vr22  Vr21  900  30VB  VB2 .
Combine the above: VB  13.72 m / s. Then,  1  59.4  ,  2  42.1 .
R x  m (V2x  V1x )  1000 .012  50( 30 cos 60  50 cos 30 ).
 R x  916 N.
W  15VB R x  15  13.72  916  188 500 W.
50 sin 30   V r 1 sin  1 
2
2
b)
  V r 1  2500  86.6V B  V B

50 cos 30  V B  V r 1 cos  1 



30 cos 70  Vr 2 cos  2  VB 

30sin 70  Vr 2 sin  2
 VB  14.94 m/s.
Vr22  900  20.52VB  VB2 .  1  41.4  ,  2  48.2 
 Rx  m(V2 x  V1x )  1000  0.012  50(30cos 70  50cos 30 ).
W  15V R  15  14.94  841  188 500 W.
B
 Rx  841 N.
x
97
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Chapter 4 / The Integral Forms of the Fundamental Laws
c)
50 sin 30   V r 1 sin  1 
2
2
  V r 1  2500  86.6V B  V B  VB  16.49 m/s
50 cos 30   V B  V r 1 cos  1 




2
2
 Vr 2  900  10.42VB  VB .  1  43 ,  2  53.7
30 cos80  Vr 2 cos  2  VB 

R x  m (V2x  V1x )  1000 .012  50(30 cos 80  50 cos 30 ).  R x  762 N.
W  15V R  15  16.49  762  188 500 W.
30sin 80  Vr 2 sin  2
B
x
4.142 To find F, sum forces normal to the plate: Fn  m  Vout n  V1n  .




a)  F  1000.02.4  40 (40 sin 60 )  11 080 N . (We have neglected friction)
Ft  0  m2V2  m3 (V3 )  m1  40sin 30 .
Bernoulli: V1  V 2  V 3 .
0  m2  m3  0.5m1   m2  .75m1  0.75  320  240 kg/s.

Continuity: m1  m2  m3 
m3  80 kg/s.

1 20
  120(120sin 60 ) 3360 lb. (We have neglected friction)
12 12
Ft  0  m2V2  m3 (V3 )  m1 120sin 30 . Bernoulli: V1  V2  V3.
b) F  1.94 
20
120
0  m2  m3  0.5m1   m2  0.75m1  0.75 1.94 
144

Continuity: m1  m2  m3 
 22.6 slug/sec and m3  9.7 slug/sec

4.143 F  mr (V1r )n  1000  0.02  0.4  (40  20) 2 sin 60  24 940 N.
Fx  24 940cos 30  21 600 N.
 W  21 600  20  432 000 W.
4.144 F  mr (V1r )n  1000  0.02  0.4(40  VB ) 2 sin 60 .
Fx  8(40  VB2 )sin 2 60 .
W  VB Fx  8VB (40  VB )2  0.75  6(1600VB  80VB2  VB3 ).
dW
 6(1600  160VB  3VB2 )  0.
dVB
 VB  13.33 m/s.
98
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.145 F  mr ( V1  VB )(cos   1)  1000  .1  .6VB (VB )(2)  120VB2 .
V2
2
 120  1000 
At t  0 : F  120  
  133 300 N.
 3600 
133 300
ao 
 1.33 m/s2
100 000
F
V1 = 0
t
16.67
dV
 F dVB  120VB2
.    2B .0012  dt .


100 000
m
dt
VB
33.33
0
1 
 1
.0012 t.


 16.67 33.33 
 t  26.6 sec.
4.146 F  mr (V1  VB )(cos  1)  90  .8  2.5  13.89  (13.89)(1)  34 700 N.
50 1000


 13.89 m/s   W  34 700 13.89  482 000 W or 647 hp
 VB 
3600


4.147 See the figure in Solution 4.145.
F  mr (V1  VB )(cos   1)  1000  0.06  0.2  VB (VB )(2)  24VB2 .
 F  mVB
x
24 dx

5000
0

dVB
.
dx
27.78

250
 24VB2  5000VB
dVB
.
VB

dVB
.
dx
100 1000
 27.78 m/s
3600
24
x  ln 27.78  ln 250.
5000
 x  458 m
4.148 To solve this problem, choose a control volume attached to the reverse thruster vanes, as
shown below. The momentum equation is applied to a free body diagram:
Vr2
Momentum:
 Rx  0.5m  (Vr 2 ) x  (Vr 3 ) x   m   Vr1 x
Assume the pressure in the gases equals the
atmospheric pressure and that Vr1 = Vr2 = Vr3.
Hence,
Rx
Vr1
(Vr1 ) x  Vr1  800 m/s , and
(Vr 2 ) x  (Vr 3 ) x  Vr1  sin 
where = 20 . Then, momentum is
 Rx  0.5m  2Vr1 sin    m  Vr1
Vr3
 mVr1  sin   1
99
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Chapter 4 / The Integral Forms of the Fundamental Laws
Momentum:
Rx  0.5m  2Vr1 sin    m  Vr1  mVr1  sin   1
The mass flow rate of the exhaust gases is
m  mair  mfuel  mair 1  1 40   100 1.025  102.5 kg/s
Substituting the given values we calculate the reverse thrust:


Rx  102.5 kg/s 800 m/s  1  sin 20  110 kN
Note that the thrust acting on the engine is in the opposite direction to Rx, and hence it is
referred to as a reverse thrust; its purpose is to decelerate the airplane.
2
. 
 125
.  
 (V  VB ) 2 ( 2).
4.149  F  m r (V1  VB )(cos  1)  194
 12  1
dV
 F  0.1323(V1  VB ) 2  20 B .
dt
dV
At t  0, V B  0. Then 20 B  0.1323V12 .
dt
dVB
With
 6, V1  30.1 fps.
dt
VB
V2
F
VB
2
dVB
1
1
For t  0 , 
. VB  8.57 fps.
 0.006615  dt. 0.01323 

2
( 30.1  VB )
30.1  VB 30.1
0
0
4.150 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  FV1.
20 000  F 
F  m(V2  V1 ).
1440  1.23 12 
Q  A3V3   12 
p 
50 1000
.  F  1440 N.
3600
(V1  13.89 m/s)
V2  13.89
(V2  13.89).
2
 V2  30.6 m/s.
30.6  13.89
 69.9 m3 /s
2
V1 13.89

 0.625 or
V3 22.24
100
62.5%
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.151 Fix the reference frame to the aircraft so that V1 
V2 
320 1000
 88.89 m/s.
3600
200 1000
 55.56 m/s.
3600
 m  1.2   1.12 
55.56  88.89
 329.5 kg/s.
2
F  329.5(88.89  55.56)  10 980 N  p  11
. 2.
W  F  V1  10 980  55.56  610 000 W or
4.152 Fix the reference frame to the boat so that V1  20 
V2  40 
88
 58.67 fps.
60
 p  2890 Pa.
818 hp.
88
 29.33 fps.
60
2
 10  29.33  58.67
(58.67  29.33)  5460 lb.
 F  m(V2  V1 )  1.94    
2
 12 
W  F  V1  5460  29.33  160,000
ft-lb
or 291 hp.
sec
2
 10  29.33  58.67
m  1.94      
 186.2 slug/sec
2
 12 
4.153 Fix the reference frame to the boat: V1  10 m/s, V2  20 m/s.
Thrust = m(V2  V1 )  1000  0.2  (20  10)  2000 N.
W  F  V1  2000 10  20 000 W or
26.8 hp.
4.154 0.2  V1A1  V1  .2 1.0. V1  1 m/s. V1 max  2 m/s.
0.1
0.1
0
0
V1( y)  20(0.1  y).
flux in = 2   V 2dy  2  1000  202 (0.1  y) 2 dy  800 000
0.13
 267 N.
3
The slope at section 1 is 20.  V2 ( y)  20 y  A.
Continuity: A1V1  A2V2. V2  2V1  2 m/s.

  V2  A  1/ 2.
V2 (0.05)  A  1
1
2  A  .  A  2.5.  V2 ( y)  2.5  20 y.
2
0.05
flux out = 2

0
V2 (0)  A
0.05
 ( y  0.125)3 
800 000

1000(2.5  20 y) dy  800 000 
[0.00153]

3
3

 0
change = 408  267 = 141 N.
 408.3 N.
2
101
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Chapter 4 / The Integral Forms of the Fundamental Laws
0.1
4.155 a)  
V
2
dA

V 2A
2  202 (0.1  y)2 dy
0
12  0.2 1.0
 4000 
0.13 4
 .
3
3
b) See Problem 4.155: V2 ( y)  20(0.125  y), 0.05  y  0. V2  2 m/s.

V
2
dA
2
V A
2

0.05

202 ( y  0.125) 2 dy
0
22  0.11.0
( y  0.125)3
 2000
3
0.05
 1.021
0
4.156 From the c.v. shown: ( p1  p2 ) r02   w 2 r0 L.
 w 
p ro
du

.
dr w
2L

du
dr

w
w2roL
0.03  144.75 / 12
2  30  2.36  10 5
p 1A 1
p 2A 2
 191 ft/sec/ft
 r2 
4.157 Write the equation of the parabola: V (r )  Vmax 1  2  .
 r 
0 

0.006

r2 
2 rdr.
Continuity:   0.0062  8   Vmax 1 
 0.0062 


0
 Vmax  16 m/s.
Momentum: p1A1  p2 A2  FDrag    V 2dA  mV1.
2
40 000  0.006  FDrag 
0.006

0
r2
2
2

1000 16 1 
2 rdr  1000    0.006 2  8  8
 0.0062 


4.524  FDrag  9.651  7.238.
 FDrag  2.11 N.
4.158 mtop   A1V1    V2 ( y)dA
2


 1.23  2 10  32   (28  y 2 )10dy   65.6 kg/s


0
2
F
    V 2dA  mtop V1  m1V1  1.23 (28  y 2 ) 210dy  65.6  32  1.23  20  32 2
2
0
 F  3780 N .
102
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.159 a) mtop
0.1


 m1  m2   A1V1    u( y )dA  1.23 0.1 2  8   (20 y  100 y 2 )8  2dy 


0
 0.656 kg/s. (Note: y  0.1 for u( y)  8).
Momentum:  FDrag  
0.1
 64(20 y  100 y
2 2
) 2dy  0.656  8    0.1 2  82
0
 1.23 6.83  5.25  1.23 12.8.
b) To find h:
8h 
0.1
 8(20 y  100 y
2
FDrag  2.09 N
)dy.
0
20  0.12 100  0.001
h 

 0.0667 m.
2
3
0.1
Momentum:  FDrag  1.23  64(20 y  100 y 2 ) 2 2dy  1.23  0.0667  2  82
0
 FDrag  2.10 N
 1.23  6.83  10.50.
Momentum and Energy
V 12
V 22
 z1 
 z 2  hL . See Problem 4.125(a).
4.160 a) Energy:
2g
2g
1.912 2
82
 0.6 
 2.51  hL .
 hL  1166
.
m.
2  9.81
2  9.81
 losses =  A1V1hL  9810  (0.6 1)  8 1.166  54 900 W/m of width.
b) See Problem 4.127:
V 12
V2
 z 1  2  z 2  hL .
2g
2g
1.58 2
7.19 2
.417 
 1.9  hL .
 hL  1.025 m.
2  9.81
2  9.81
 losses = A 1V1 hL  9810.417  3  7.19  1.025  90 300 W
32
5.17 2
 1.16 
 2  hL .  hL  0.0636 m.
c) See Problem 4.128:
2  9.81
2  9.81
 losses =  A1V1hL  9810 1.16  5.17  0.0636  3740 W/m of width.
4.161 See Problem 4.129: V1  20 m/s, V2  5 m/s, p1  60 kPa, p2  135 kPa.
V12 p1 V22 p2
Then,
 
  hL .
2g 
2g 
202
60 000
52
135 000



 hL .
2  9.81 9810
2  9.81
9810
 hL  11.47 m  K  V12 /2g  K  202 /2  9.81.
103
 K  0.562.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible websit e, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
V1D2  Vd2 .
4.162 Continuity:
Energy:
V12
V2
 H (t ) 
.
2g
2g
Momentum:
Fx  ( FI ) x 
ax m(t )  
But, V1  
dH
.
dt
 ax 
d 2
4
d2
 V1 
D2
V.
 V  2 gH (t ).

 d2s 
d




V
d
V
m
V
V

(
).
 2   ax .
2x
1x
x
dt c.v .
 dt  x
 d2
4
t
m(t )  mo   
V (V ).
0
 d2
4
V (t )dt.
2 gd2
dH d
dH
d
 2 2 gH .  1/2  2 2gdt.  H1/2 
t  Ho
dt D
2D2
H
D
2

2
2

 2 gd 2
2 g 
t  H o 
2

 2D
 d 2

 4
t

0


 2 gd 2
2 g 
t  H o  dt  m o 
2


 2D
4.163 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 find
a solution to this problem. It involves continuity, energy, and momentum, resulting in a
set of non-linear differential equations.
Moment of Momentum
4.164 Ve 
4
m

 19.89 m/s.
 Ae 1000  (4    0.0042 )
MI 
 r  (2Ω  V)  d V
c.v.
 8  AV kˆ
Velocity in arm = V .
0.3
 4  rˆi  (2kˆ  V ˆi )  Adr
0
0.3
 rdr  0.36 AV kˆ
0
M  0
and
d
 (r  V)  d V  0
dt c.v.
 (r  V)V  nˆ  dA  0.3ˆi   0.707Ve ˆj  0.707Vekˆ  Ve  Ae
c.s.
The z-component of
2
 r  V(V  nˆ )  dA  0.3  0.707Ve Ae 
c.s.
Finally, ( MI ) z  0.36  AV   4  0.3  0.707Ve2 Ae  . Using AV  Ae Ve ,
   46.9 rad / s.
0.36  4  0.3 0.707 19.89.
104
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Chapter 4 / The Integral Forms of the Fundamental Laws
y
V
4.165 A moment M resists the motion thereby
producing power. One of the arms is shown.
MI 
0.25

r
x

Ve
4rˆi  (2kˆ  V ˆi )  Adr  8 AV kˆ
10/12

0
rdr  2.778 AV kˆ .
0
d
 M  Mkˆ and dt  (r  V) d V  0 and
c.v.
10
 (r  V)V  nˆ  dA  12  Ve  Ae (4kˆ )
2
c.s.
2
2
10
 0.75  200
 1/4 
 30  2002  1.94 
Thus, M  2.778 1.94 
 
 4
9
12
 12 
 12 
W  M  309  30  9270 ft-lb/sec
 M  309 ft-lb
4.166 m  10   AV  1000  0.012 V0 .
 V0  31.8 m/s.
V0  0.012  V   0.012  Ve  0.006(r  0.05).
Continuity:
V0  0.012  Ve  0.006  .15.
 Ve  11.1 m/s.
 V  V0  19.1(r  0.05)Ve  42.4  212r.
MI 
0.05

2rˆi  (2kˆ  V0ˆi )  Adr 
0
 4V0  Akˆ
0.2

2rˆi  [2kˆ  (42.4  212r )ˆi ] Adr
0.05
0.05

0.2
rdr  4 Akˆ
0

(42.4r  212r 2 )dr
0.05
212
 42.4

= 
(0.22  0.052 ) 
(0.23  0.053)  kˆ
3
 2

 (0.05  0.3)kˆ  0.35kˆ .
0.2

rˆi  (Ve ˆj)Ve   0.006dr  11.12  1000  0.006
0.05
0.2

rdr kˆ  13.86 kˆ
0.05
 0.35  13.86.
105
  39.6 rad/s.
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Chapter 4 / The Integral Forms of the Fundamental Laws
1000
 2 N  m.
500
MI   rˆir  (2kˆ  V (r)ˆir )  2 r  0.02dr
4.167 1000  M.
M 
R
 0.08  r 2V (r )drkˆ
0
Continuity: V (r )2 r  0.02  Vr cos30 2 R  0.02.
 V (r )  0.866R
Vr
r
 r  V(V  nˆ )  dA   R(R  Vr sin 30 )Vr cos 30
 2 R  0.02kˆ  0.00301Vr (35  0.5Vr )kˆ
 r  V(V  nˆ )  dA   R(R  Vr sin 30 )Vr cos 30
 2 R  0.02 kˆ  0.00301Vr (35  0.5Vr )kˆ
c.s.
c.s.
 2  16.32V r
.15

r dr  .00301V r (35.5V r ).
 V r2  52.1V r  1333  0.
0
1
 V r  (52.1  52.12  4  1333 )  70.9 m / s.
2
The flow rate is Q  Ae Vr cos 30  2  0.15  0.02  70.9  0.866  1.16 m3 /s
4.168 See Problem 4.165. Ve  19.89 m/s.
V
0.3
0.0082
0.022
19.89  3.18 m/s.

 d ˆ  ˆ 
M I  4  rˆi  (2kˆ  V ˆi )   
k   ri  Adr. A    0.012 , Ae    0.0042
 dt 


0
 8  AV kˆ
0.3

0
d ˆ
rdr  4  A
k
dt
 360 AV kˆ  36 A
0.3
r
2
dr
0
d ˆ
k
dt
 (r  V)z (V  nˆ )  dA  212Ve Aekˆ
2
c.s.
Thus, 360 AV   36 A
d
d
 212Ve2 Ae or
 31.8  373.
dt
dt
The solution is   Ce 31.8t  11.73. The initial condition is (0)  0.  C  1173
. .
Finally,
  11.73(1  e31.8t ) rad/s.
4.169 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?
106
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Chapter 5 / Differential Forms of the Fundamental Laws
CHAPTER 5
The Differential Forms
of the
Fundamental Laws
Differential Continuity Equation
5.1
0

d V    V  nˆ dA . Using Gauss’ theorem, this can be written as

t
c.v.
c.s.

0

dV 

t
c.v.

 

   ( V)d V    t    ( V)d V
c.v.
c.v.
Since this is true for all arbitrary control volumes (i.e., for all limits of integration), the
integrand must be zero:

   (  V)  0.
t
This can be written in rectangular coordinates as
 


 ( u)  (  v)  (  w).
t x
y
z
This is Eq. 5.2.2. The other forms of the continuity equation follow.

5.2
melement
. This is expressed as
t






 vr (rd dz)    vr   vr  dr  (r  dr )d dz   v drdz    v  (  v )d  drdz
r





dr 
dr 
dr 

  




  vz  r   d dr    vz  (  vz )dz   r   d dr     r   d drdz  .
2
2
2
z
t  




min  mout 
Subtract terms and divide by rddrdz:

 vr
r

r  dr 
r  dr / 2  r  dr / 2
1 

(  vr ) 
(  v )  (  vz )
 
.
r r
r 
r
r
z
t
Since dr is an infinitesimal, (r  dr )/r  1 and (r  dr /2)/r  1. Hence,
 
1 

1
 (  vr ) 
(  v )  (  vz )   vr  0.
t r
z
r 
r
This can be put in various forms. If  = const, it divides out.
107
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Chapter 5 / Differential Forms of the Fundamental Laws
5.3
melement
. This takes the form
t



 vr (rd )r sin  d   vr  ( vr )dr  (r  dr )d (r  dr ) sin d
r



 dr 

  dr 
  v dr  r   sin  d    v 
(  v )d  dr  r   sin  d

2
2


 

 

dr 
dr 

(  v )d  dr  r   d
  v dr  r   d    v 
2
2



 
min  mout 
2

dr 
  
    r   drd sin  d 
t  
2

Because some areas are not rectangular, we used an average length (r  dr /2). Now,
subtract some terms and divide by rdddr:
(r  dr )2 

r  dr /2
(  vr )sin 
(  v )
sin 

r

r
r
  vr sin    vr sin  
2
r  dr /2   r  dr /2 

 (  v )

sin 
r
r

t
Since dr is infinitesimal (r  dr )2 /r  r and (r  dr /2) / r  1. Divide by r sin  and there
results
 
1 
1

2
 (  vr ) 
(  v ) 
(  v )   vr  0
t r
r 
r sin  
r
5.4

 0. Then, with v  w  0, Eq. 5.2.2 yields
t

du
d
(  u)  0
 0.
 u
or
x
dx
dx
For a steady flow
Partial derivatives are not used since there is only one independent variable.
5.5
Since the flow is incompressible
D
 0. This gives
Dt
2-D
steady


 
D
 0.
u
v
w

x
y
z
t
Dt
u


w
 0.
x
z
Also, since the flow is incompressible,
  V  0, or
u w

 0.
x z
108
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Chapter 5 / Differential Forms of the Fundamental Laws


 0,
 0. Since water can be considered to be incompressible, we demand
t
z


D
w
 0, assuming the x-direction
that
 0. Equation (5.2.8) then provides u
x
z
Dt
to be in the direction of flow. There is no variation with y. Also, we demand that
  V  0, or
u w

 0.
x z
5.6
Given:
5.7
We can use the ideal gas law,  
p
. Then, the continuity equation (5.2.7)
RT
D
    V becomes, assuming RT to be constant,
Dt
p
1 Dp

  V or
RT Dt
RT
5.8
1 Dp
   V.
p Dt
a) Use cylindrical coordinates with v  vz  0 :
1 
( rv r )  0
r r
Integrate:
rvr  C.
 vr 
C
.
r
1  2
(r vr )  0 .
r 2 r
b) Use spherical coordinates with v  v  0 :
Integrate:
r 2 vr  C.
5.9
 vr 
C
r2
.
(a) Since the flow is steady and incompressible then VA = constant, where the constant is
determined by using the conditions at the inlet that is,  VA inlet  40 1  40 m3 /s. And,
since the flow is inviscid, the velocity is uniform in the channel, so u  V . Hence, at any x
position within the channel the velocity u can be calculated using u  V  40/A. Since the
flow area is not constant it is given by A  2hw, where the vertical distance h is a
function of x and can be determined as, h  0.15x  0.5H . Substituting, we obtain the
following expression for the velocity:
u( x) 
40
20

m/s
2(0.15x  0.5) 0.15x  0.5
109
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Chapter 5 / Differential Forms of the Fundamental Laws
(b) To determine the acceleration in the x-direction, we use (see Eq. 3.2.9) ax  u
where
du
dx
du
3

dx
(0.15x  0.5) 2
Hence, the expression for acceleration is
ax 
20
3
60
m/s2


2
3
0.15x  0.5 (0.15x  0.5)
(0.15x  0.5)
Note that the minus sign indicates deceleration of the fluid in the x-direction.
5.10
u v
 u 

 0, we write v    dy  C. With the
x y
 x 
3
u
v
result from Problem 5.9:

  , we integrate to find
2
x
y
(0.15x  0.5)
3y
v(x, y) 
(0.15x  0.5)2

(a) Using the continuity equation
and since v  0 at y  0, then C  0.
(b) To determine the acceleration in the y-direction, we use (see Eq. 3.2.9)
v
v
ay  u  v .
x
y
From part (a) we have
0.9 y
v

x
(0.15x  0.5)3
Substituting in the expression for acceleration we get
ay 
0.9y
9 y
3y
20
3




0.15x  0.5 (0.15x  0.5)3 (0.15x  0.5)2 (0.15x  0.5)2
(0.15x  0.5)4
5.11
 u v 
D
kg
    V.        2.3(200 1  400  1)  1380 3
Dt
m s
 x y 
5.12
In a plane flow, u  u(x, y) and v  v(x, y). Continuity demands that
If u  const, then
u v

 0.
x y
u
v
 0 and hence
 0. Thus, we also have v  const and
x
y
D/Dt = 0.
110
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Chapter 5 / Differential Forms of the Fundamental Laws
5.13
If u  C1 and v  C2 , the continuity equation provides, for an incompressible flow,
u v w
 
 0.
x y z

w
 0 so w  C3.
z
The z-component of velocity w is also constant. We also have




D
0
u
v
w
t
x
y
z
Dt
The density may vary with x, y, z and t. It is not, necessarily, constant.
5.14
u v

 0.
x y
 A
v
 0.
y
But, v(x,0)  0  f (x).
5.15
5.16
 v   Ay.
5x 2  5y 2
( x 2  y 2 )5  5 x(2 x)
v
u
 2



y
x
(x 2  y 2 )2
(x  y 2 )2
u v

 0.
x y
 v( x, y) 
 v(x, y)   Ay  f (x).
5 y 2  5x 2
 (x
2
From Table 5.1:
 y2 )
dy  f ( x) 
2
5y
x2  y 2
 f ( x). f ( x)  0.
v 
5y
x2  y 2
.
1 
1 v
1
0.4 
  10  2  sin  .
(rvr )  
r r
r 
r
r 
0.4 
0.4 


 rvr  10  2  sin  dr  f ( )  10r 
 sin   f ( ).
r 

r 


0.4 

0.2vr (0.2, )  10  0.2 
 sin   f ( )  0.
0.2 

 f ( )  0.
0.4 

 vr  10  2  sin  .
r 

5.17
From Table 5.1:
1 
1 v 20 
1
( rv r )  

 1  2  cos .

r r
r 
r
r 
1

 1
 rvr  20 1  2  cos  dr  f ( )  20  r   cos   f ( ).
 r
 r 

vr (1, )  20(1  1) cos   f ( )  0.
 f ( )  0.
1

 vr  20 1  2  cos  .
 r 
111
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Chapter 5 / Differential Forms of the Fundamental Laws
5.18
From Table 5.1, spherical coordinates:

1  2
1

(r vr )  
( v sin ).
2
r r
r sin  
1  2
1 
40 
(r vr ) 
 10  3  2 sin  cos .
2
r r
r sin  
r 
40 
80 


 r 2vr  r 10  3  2cos  dr  f ( )  10r 2   cos   f ( )
r 

r 


80 

4vr (2, )  10  22   cos   f ( )  0.
2

 f ( )  0.
80 

 vr  10  3  cos  .
r 

5.19
Continuity:


5.20

( u)  0.
x

du
d
u
 0.
dx
dx
p
18 144
slug

 0.00302 3 .
RT 1716  500
ft
du 526  453

 219 fps/ft
dx 2  2 /12
d
 du
0.00302


 219  0.00136 slug/ft 4 .
dx
u dx
486
u v


 0.
20(1  e  x )  20e  x
x y
x
Hence, in the vicinity of the x-axis:

v
 20e  x
y
But v  0 if y  0.

and v  20ye  x  C.
C  0.
v  20 ye  x  20(0.2)e 2  0.541 m/s
5.21
From Table 5.1,
v
1 
(rvr )  z  0.
z
r r
 
20(1  e  z )   20e  z


z
Hence, in the vicinity of the z-axis:
r2
1 
z
(rvr )  20e
and rvr  20e  z  C.
r r
2
But vr  0 if r  0.
C  0.
vr  10re  z  10(0.2)e 2  0.271 m/s
112
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Chapter 5 / Differential Forms of the Fundamental Laws
5.22
The velocity is zero at the stagnation point. Hence, 0  10 
R2
.  R  2 m.
u
u v
 80 x 3 ,

 0. Using
x
x y
The continuity equation for this plane flow is
we see that
40
v
 80x 3 near the x-axis. Consequently, for small y,
y
v  80x 3y so that v  80(3)3 (0.1)  0.296 m/s.
5.23
The velocity is zero at the stagnation point. Hence, 0  (40/R2 ) 10.  R  2 m.
1  2
1 
20
Use continuity from Table 5.1:
r v r  2 ( 40  10r 2 )   .
2
r r
r r
r


Near the negative x-axis continuity provides us with
1

20
v sin   .

r sin  
r
v sin   20 cos  C
0.1
Since v  0 when   90  , C  0. Then, with   tan 1
 1.909  ,
3
cos 
cos88.091
0.0333
v  20
 20
 20
 0.667 m/s
sin 
sin 88.091
0.999
Integrate, letting   0 from the y-axis:
5.24
13.5  11.3
m/s
u v
v
u
.

 0.



 220
2  0.005
m
x y
y
x
v  v  0  220y.
 v  220  0.004  0.88 m/s.
u
b) ax  u  12.6  (220)  2772 m/s 2.
x
Continuity:
Differential Momentum equation
5.25
Fy  may . For the fluid particle occupying the volume of Fig. 5.3:
 yy dy 
 zy dz 
 xy dx 



 yy 
 dxdz   zy 
 dxdy   xy 
 dydz
y 2 
z 2 
x 2 



 yy dy 
 zy dz 
 xy dx 



  yy 
 dxdz   zy 
 dxdy   xy 
 dydz
y 2 
z 2 
x 2 



  gy dx dy dz  dx dy dz
Dividing by dx dy dz , and adding and subtracting terms:
 xy
x

 yy
y

 zy
z
  gy  
113
Dv
Dt
Dv
.
Dt
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Chapter 5 / Differential Forms of the Fundamental Laws
5.26
Check continuity:
2
2
2
2
u v w ( x  y )10  10 x(2 x) ( x  y )10  10 y(2 y )
 0.




(x 2  y 2 )2
(x 2  y 2 )2
x y z
Thus, it is a possible flow. For a frictionless flow, Euler’s Eqs. 5.3.7 give, with
g x  g y  0:
u
p
u
u
 v
 .
x
y
x
p
10 y
20xy
100( x2  y 2 ) y
10x 10 y 2  10x2
   2
 2

x
( x2  y 2 )3
x  y 2 ( x2  y 2 )2
x  y 2 ( x2  y 2 )2
u

p
v
v
 v
 .
x
y
y
p
20xy
10 y 10x2  10 y 2
100(x2  y 2 ) y
10x


  2


y
(x 2  y 2 )3
x  y 2 ( x 2  y 2 )2
x 2  y 2 (x 2  y 2 )2
p 
5.27
p ˆ p ˆ
100 y  ˆ
100 x ˆ
100 
( xˆi  yˆj)
i
j 2
i 2
j 2
2
2
2
2
2 2
x
y
(x  y )
(x  y )
(x  y )
Check continuity (cylindrical coord from Table 5.1):
1 
1 v 10 
1
1
10 
( rv r ) 

 1  2  cos  
 1  2  cos   0. It is a possible flow.



r r
r 
r
r
r
r 
For Euler’s Eqs. (let v = 0 in the momentum eqns of Table 5.1) in cylindrical coord:
2
p
v
v2
v v 100  
1
1

 20 
     vr r    r 
1  2  sin 2   10  1  2  cos 2   3 

r
r
r
r 
r  r 
 r 
r 
10  
1
10 
2 

 1  2  sin   10  2  .



r
r
r 
vv
v
v v
1 p
100  
1
1  4  sin  cos 
   r    vr      

r
r 
r
r 
r  r 
2
1
1
 20  100 

10   1  2  cos  sin   3  
 1  2  sin  cos  .
r 

r 
r 
r 
p 
5.28
p ˆ 1 p ˆ
200   1
200 

ir 
i  3  2  cos 2  ˆir  3 sin 2 ˆi
r 
r
r r
r

Follow the steps of Problem 5.27. The components of the pressure gradient are
2
2
v  v
p
v
v v

  vr r    r
r
r
r
r 
vv
v v
v
1 p
   r    vr     
r 
r
r 
r
114
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Chapter 5 / Differential Forms of the Fundamental Laws
5.29
 2

 2

p  p 
     V.
p  p  
     V.
 3

 3

 sˆ sˆ
 nˆ
nˆ


 .
R
R
 s s
 sˆ sˆ nˆ 

.


 nˆ
t t
t
t

DV  V
V    V 2 

V

 nˆ
 sˆ   V
Dt  t
R 
s   t
 V2 
V
For steady flow, the normal acc. is  
, the tangential acc. is V
.
 R 
s


5.30
For a rotating reference frame (see Eq. 3.2.15), we must add the terms due to Ω.
Thus, Euler’s equation becomes
dΩ 
 DV
 2Ω  V  Ω  (Ω  r) 
 r   p   g

dt
 Dt

5.31
 xx   p  2
 yy
u
    V  30 psi
x
  zz   p  30 psi.
 u
 xy   
 y

v
x
 xz   yz  0.
5.32

0.1 
5 
5
  10 30  1440    18 10 psf
12




5
 xy 18 10

 4.17 108.
 xx 30 144
16 y
16 y 2
v
u



.
y
x C x9/5 C 2 x13/5
v( x, o)  0.
 f ( x)  0.
 v( x, y) 
8y 2
Cx9/5
8  C 10004/5.

16 y3
3C 2 x13/5
 f ( x).
C  0.0318.
 u( x, y)  629 yx4/5  9890 y 2 x 8/5 .
v( x, y)  252y2 x9/5  5270y3 x13/5 .
u
 100  0  100 kPa.
x
  p  100 kPa.
 xx   p  2
 yy   zz
 u
 xy   
 y

v 
5
4/5 
 5.01105 Pa.
  2 10 629 1000

x 
 xz   yz  0.
115
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Chapter 5 / Differential Forms of the Fundamental Laws
5.33
Du u  

 

  u  v  w  u  ( V   )u.
Dt
t  x
y
z 
Dv v  

 

  u  v  w  v  ( V   ) v.
Dt t  x
y
z 
Dw w  

 

  u  v  w  w  (V   )w
Dt
t  x
y
z 

5.34
DV Du ˆ Dv ˆ Dw ˆ

i
j
k  V   (uˆi  vˆj  wkˆ )  (V   )V.
Dt
Dt
Dt
Dt
Follow the steps that lead to Eq. 5.3.17 and add the term due to compressible effects:

 
 
 
DV
 p   g  2V 
  V ˆi 
  V ˆj 
  V kˆ
Dt
3 x
3 y
3 z
 p   g   2 V 

5.35
  ˆ  ˆ  ˆ
j  k   V
 i
3  x
y
z 
DV

 p   g  2 V  (   V ).
Dt
3
If u = u(y), then continuity demands that
plate), v = 0.
v
 0.  v  C. But, at y = 0 (the lower
y
C  0, and v (x , y )  0.
  2u  2u  2u 
 u
p
u
u
u 
Du

    u  v  w      gx    2  2  2  .
 x
x
y
z 
x
Dt
y
z 
 t

0  
p
 2u
  2.
x
ay

p
Dv
0 .
y
Dt

p
Dw
 0     ( g ).
z
Dt
116

p
  g .
z
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 5 / Differential Forms of the Fundamental Laws
5.36
The x-component Navier-Stokes equation can be written as
 u

 t
u
  2u  2u  2u 
p
u
u
u 
 v  w      gx    2  2  2 

x
y
z 
x
y
z 
 x
Based on the given conditions the following assumptions can be made:
 u

One-dimensional  v  w  0 
Steady state 
 0
 t

 dp

Zero pressure-gradient 
 0
 dx

 u

A wide channel 
 0
 z

Incompressible    constant 
 u

Fully-developed flow 
 0
 x

The Navier-Stokes equation takes the simplified form 
 2u
 0 or
 2u
 0. Integrating
y 2
y 2
twice yields, u( y)  ay  b . To determine a and b we apply the following boundary
conditions: u  V1 at y = 0, and u  V2 at y = h. This gives b  V1 and
a  (V1  V2 ) / h. The velocity distribution between the plates is then
 V  V2 
u( y)    1
 y  V1
 h 
5.37
Using the x-component Navier-Stokes equation with x being vertical and the following
assumptions:
 u

Steady state 
One-dimensional  v  w  0 
 0
 t

 u

Fully-developed flow 
Incompressible    constant 
 0
 x

The x-component Navier-Stokes equation reduces to
0
p
2u
 g   2
x
y
Integrate the above differential equation twice (see Problem 5.36):
u( y) 
1  dp

  g  y 2  ay  b

2  dx

Applying the no-slip boundary condition at both plates (see Problem 5.36) we get
u( y) 

1  dp

  g  y 2  hy

2  dx

117

© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5 / Differential Forms of the Fundamental Laws
5.38
 v

Steady state  z  0 
 t

Horizontal ( gz  0)
Assumptions: One-dimensional  vr  v  0 
Incompressible    constant 
 v

Fully-developed flow  z  0 
 z

Dvr
Dv
1 p
1 p
.
.
0
0
Dt
Dt
 r
 r 

5.39
  2 v 1 v
 v
p
vz v vz
v 
Dvz
1  2 vz  2 vz
z
   z  vr

 vz z       2z 

 2
2
 t

 r



Dt
r
r
z
z
r
r



z 2
r





  2 vz 1 vz 
p
0      2 
.

r r 
z
 r




 v

Assumptions: One-dimensional flow  v  vr  0  Steady state  z  0 
 t

Incompressible    constant 
Horizontal ( gz  0)
 v

Fully-developed flow  z  0 
 z

The Navier-Stokes equation in cylindrical form provides the following equation:
0
  2 v 1 vz 
p
   2z 

r r 
z
 r
Rearrange the above equation and integrate:
0
p    vz 

,
r
z r r  r 
Integrating again yields: vz (r ) 
vz 1 p  r  C1

 
r  z  2  r
1 p  r 2 
   C ln r  C2
 z  4  1
C1 and C2 are determined using the boundary conditions: vz  0, at r  ro , and vz  Vc
at r  ri . Hence,
Vc 
1 p  ri2 
1 p  ro2 
ln
and
0
C
r
C



 
   C ln r  C2
2
 z  4  1 i
 z  4  1 o
118
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Chapter 5 / Differential Forms of the Fundamental Laws
Subtracting the second equation from the first yields,
C1 
Vc 
1 p 2 2
ri  ro
4 z
ln  ri ro 


The drag force on the inner cylinder is zero when the shear stress  rz on the inner cylinder
 v 
 v v 
is zero, i.e.,  rz    r  z 
 0 . Since vr  0 , then  rz    z 
 0 . From
r  r ri
 z
 r  r ri
the above expression for vz we find
vz
r

r ri
1 p 2
C
1 p
ri  1  0. Then C1  
ri .
ri
2 z
2 z
Combining with the above expressions for C1 we solve for Vc . The result is:
Vc 
5.40

1  2
(r vr )  0.
 r 2 vr  C. At r  r1, vr  0. C  0.
2 r
r
2
p
p
 1   2 v 
v 
v
 2v



0
r
         2 cot  .



r  r sin 2  
r
r
 r

 r r 
Continuity:
0
5.41

1 p  2 2
ri  ro  2ri2 ln  ri ro 

4 z 
1 p
.
r sin  
 v

Steady state  z  0 
 t

Vertical ( gr  g  0)
Assumptions: One-dimensional flow  vz  vr  0
Incompressible    constant 
 v

Developed flow  r  0 
 

The simplified differential equation from Table 5.1 is
which can be re-written as
Integrating we get:
2
 v
r
2

v v

 C1
r
r
r
2

1 v v

0
r r r 2
  v 
 0.
r  r 
The above equation can be re-written as
Integrating again yields
 2 v
rv  C1
1 
 rv   C1.
r r
r2
r C
 C2  v  C1  2
2
2 r
119
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Chapter 5 / Differential Forms of the Fundamental Laws
Apply the boundary conditions v  0 at r  ri , and v  ro at r  ro . We have
0  C1
ri C2

2 ri
and
ro  C1
Solving for C1 and C2 yields C1  2
ro C2

2 ro
 ro2
ri2  ro2
and C2 
 ri2ro2
ri2  ro2
.
  r 2    r 2r 2  1
Finally, v    2 o 2  r   2 i o2 
 r r   r r  r
o 
o 
 i
 i
5.42
For an incompressible flow   V  0. Substitute Eqs. 5.3.10 into Eq. 5.3.2 and 5.3.3:
Du
u    u v    u w 
 


   


  p  2  
  g x .
Dt x 
x  y  y x  z  z x 



 2u 2u 2u
p
Du
    2  2  2   g x .
x
y
z 
Dt
 x
Dv   u v   
v    v w 
       p  2     

   gy .
Dt x  y x  y 
y  z  z y 



p
  u v w 
2u
2u
2u
 2  2  2  


  g x
x  x y z 
z
y
x
x
p
 2v
 2v
 2v
  u  v  w 
 2  2  2   

   gy
y
y  x y z 
x
y
z
  2v  2v  2v 
p
Dv
     2  2  2    gy .

Dt
y
y
z 
 x
Dw   u w    v w   
w 
 

    p  2
  
   gz
Dt x  z x  y  z y  z 
z 
p
 2w
 2w
 2w
  u v w 
  2  2  2   

   gz
z
z  x y z 
x
y
z

5.43
  2w  2w  2w 
p
Dw
     2  2  2    gz .
 x
Dt
z
y
z 

If we substitute the constitutive equations (5.3.10) into Eqs. 5.3.2 and 5.3.3., with
  (x , y , z ) we arrive at

  2u  2u  2u 
p
 u    u  v     u  w 
Du
    gx    2  2  2   2

  
 

 x

x






z x 
Dt
x
x
y
y
x
z



y
z




120
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Chapter 5 / Differential Forms of the Fundamental Laws
5.44
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

 t
u
  2u  2u  2u
p
u 
u
u



w



v


g


x
 x 2 y 2 z 2
z 
x
y
x

2
u
 u

 2
t
y




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.
5.45
From Eqs. 5.3.10, 
 xx   yy   zz
3
 p
2  u v w 

 
     V.
3  x y z 
 2

 2

p  p 
     V.  p  p   
     V.
 3

 3

Vorticity
5.46
 

 
( V   ) V   u  v  w  (uˆi  vˆj  wkˆ )
y
z 
 x
   w
w
w    v
v
v  
 × (V   )V    u
v
w
   u  v  w   ˆi
y
z  z  x
y
z  
 y  x
   u
u
u     w
w
w  ˆ
  u  v  w    u
v
w
 j
y
z   x   x
y
 z  
 z  x
   v
v
v    u
u
u 
   u  v  w    u  v  w   kˆ
y
z   y   x
y
 z  
 x  x
Use the definition of vorticity: ω  (
w v ˆ u w ˆ v u ˆ
 )i  ( 
) j  (  )k
y z
z x
x y
 w v 
u w 
v x  
(ω   )V  (
 ) ( 
)  (  )  (uˆi  vˆj  wkˆ )
 y z x z x y x y z 
 

   w v ˆ u w ˆ v u ˆ 
(V   )ω  u  v  w  (
) j  (  )k 
 )i  ( 
y
z   y z
z x
x y 
 x
Expand the above, collect like terms, and compare coefficients of ˆi, ˆj, and kˆ .
121
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Chapter 5 / Differential Forms of the Fundamental Laws
5.47
Studying the vorticity components of Eq. 3.2.21, we see that z  u/y is the only
vorticity component of interest. The third equation of Eq. 5.3.24 then simplifies to
Dz
  2z
Dt
 2z

y 2
since changes normal to the plate are much larger than changes along the plate, i.e.,
z

 z .
y
x
5.48
If viscous effects are negligible, as they are in a short section, Eq. 5.3.25 reduces to
D z
0
Dt
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,
z 
v u

 10
x y
u
 10.
y
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 2 ( y )  u 0  10y . Continuity then allows us to calculate the profile:
we conclude that, for the lower half of the flow at section 2,
V1A1  V2 A2
1
(10  0.04)(0.04w)  (u0  10  0.02 / 2)(0.02w).
2
Finally,
5.49
u0  0.3 m/s.
u2 ( y)  0.3  10y
No. The first of Eqs. 5.3.24 shows that, neglecting viscous effects,
Dx
u
u
u
 x
 y
 z
x
y
z
Dt
so that  y , which is nonzero near the snow surface, creates  x through the term
y u/y, since there would be a nonzero u/y near the tree.
122
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Chapter 5 / Differential Forms of the Fundamental Laws
Differential Energy Equation
5.50

KT  nˆ dA 
c.s.
c.v.


  (KT )d V 

 V2
 gz  u  d V 


t  2


c.v.
c.v.
V2
p
 gz  u    V  nˆ dA

 
2
c.s. 


  V2
  gz   u d V 
 

t  2

V2
p
  V 
 gz  u   d V
 2
 

c.v.



V2
p 
  V2
2
   K T   
  gz   u      V 
 gz  u   d V  0

 2
t  2
  



c.v. 

 V2

p  V 2  
p
 V

 V2
   V 
 gz   
   V   V  

 V  V 
 gz   0.


t
  2  t

2
 t


 2
continuity
K2T 
5.51
Du
 K 2T .
Dt

or
Divide each side by dxdy dz and observe that
T
x

x  dx
T
x
x
dx
Eq. 5.4.5 follows.
5.52

u   V  u  0.
t
momentum


2
 T
x 2
,
T
y

y  dy
T
y
y
dy

2
 T
x 2
,
T
z

z dz
T
z
dz
z

 2T
z 2
D(h  p / )
Du
Dh Dp p D
Dh Dp p






     V 
Dt
Dt
Dt Dt  Dt
Dt Dt 
where we used the continuity equation: D  /Dt     V. Then Eq. 5.4. 9 becomes

Dh Dp p

     V   K2T  p  V
Dt Dt 
which is simplified to

5.53
See Eq. 5.4.9: u  cT.
Dp
Dh
 K 2T 
Dt
Dt
 T
T
T
T 
2
 c 
u
v
w
  K T.
x
y
z 
 t
Neglect terms with velocity:
c
T
 K 2T.
t
123
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Chapter 5 / Differential Forms of the Fundamental Laws
5.54
The dissipation function  involves viscous effects. For flows with extremely large
velocity gradients, it becomes quite large. Then
 cp
and
5.55
DT

Dt
DT
is large. This leads to very high temperatures on reentry vehicles.
Dt
u  10(1  10 000 r 2 ).

u
 2r 105.
r
(r takes the place of y)
 1  u 2 
From Eq. 5.4.17,   2         4r 2 1010.
 2  y  


At the wall where r  0.01 m,
At the centerline
  1.8 105  4  0.012 1010  72 N/m2  s.
u
 0 so   0.
r
At a point half-way:   1.8 105  4  0.0052 1010  18 N/m2  s.
5.56
(a) Momentum:
u
 2u
 2
t
y
 u 
T
 2T
Energy:  c
 K 2  
t
y
 y 
(b) Momentum: 
2
u
 2u  u
 2 
t
y y
y
 u 
T
 2T
Energy:  c
 K 2  
t
y
 y 
2
124
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Chapter 6 / Dimensional Analysis and Similitude
CHAPTER 6
Dimensional Analysis
and Similitude
FE-type Exam Review Problems: Problems 6-1 to 6-7
6.1
(A)
The dimensions on the variables are as follows:
L L ML2
ML / T 2
M
[W ] [ F V ] M 2
,
[
d
]
L
,
[
p
]
, [V ]
T
T
T3
L2
LT 2
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
W
L
T
pVd 2
6.2
(A)
6.3
(A)
6.4
A)
V
f (d , l , g ,
). The units on the variables on the rhs are as follows:
L
ML
[d ] L, [l ] L, [ g ]
, [ ] T 1, [ ]
2
T
T
Because mass M occurs in only one term, it cannot enter the relationship.
Re m
Re m
Vm Lm
Re p .
Re p .
,
Vp Lp
p
m
Vm Lm
Vp Lp
p
Vm2
V p2
lm g m
lp g p
(C)
Frm
6.6
(A)
From Froude’s number Vm
Fp* or
Fm
2 2
mVm lm
Vm
.
m
6.5
Fm*
Frp .
Vm
.
Lp
Vp
Vp
12 9 108 m/s.
Lm
Lp
m
Lm
p
lm
lp
1
4
1.51 10
5
1.31 10
6
461 m/s.
.
Vm
Vp
lm
. From the dimensionless force we have:
lp
Fp
.
2
2
pV p l p
125
Vp
4 10
Fp
Fm
2
V p2 l 2p
Vm2 lm2
0.5 m/s.
10 25 252 156 000 N
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
Chapter 6 Problems: Introduction
6.7
g V12
V12 2 g
or
6.8
p1
g
g V 22
V12 2 g
z1
1
2
p1
V12
gz1
V12
1 V22
2 V12
1
2
p1
V12
gz1
V12
1
2
c) [ ]
e) [W ]
N m.
z2 .
p2
V12
gz 2
.
V12
gz 2 V22
V22 V12
p2
V22
N s2 N s
.
m s
m
N s2 N s2
.
m m3
m4
kg
s
kg
m3
a) [m]
p2
g
N
.
m2
N s
d) [ ]
.
m2
N m
f) [W ]
.
s
FT
.
L
FT 2
.
L4
b) [ p]
FL
F
L2
FT
L2
FL
T
F
L
g) [ ] N/m.
Dimensional Analysis
T
6.9
6.10
R
V
e r
, , ,
R R R
f1
2 5
V
f1 (
f ( , , d).
V
2
)
V d
f (H, g, m).
1
0
2
V
Const.
[V ]
1
6.12
M
,[ ]
L3
[V ]
There is one
6.11
.
L
, [ ] L, [ ]
T
V
.
term: 1
f ( , , ).
1
R2
.
[V ]
gHm 0
.
V2
L
M
, [ ]
, [ ]
T
T2
1
f1 (
0
2
C,
M
.
LT
or Re
M
, [d] L.
L3
Const.
)
Const.
V 2d
C , or We = Const.
L
L
, [ g]
, [m] M, [ H] L.
T
T2
1
C.
V
126
gH / C .
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.13
L
L
M
M
, [ H] L, [ g]
, [m] M, [ ]
, [ ]
.
2
3
T
LT
T
L
Choose repeating variables H , g , (select ones with simple dimensions-we couldn’t
select V, H, and g since M is not contained in any of those terms):
V
f (H, g, m, , ). [V ]
VH a1 g b1
V 0
g H
1
1
c1
V
gH
mH a2 g b2 c2 ,
m
.
2
H3
,
2
V
.
gH
m
,
H3
f1
H a3 g b3
3
3
c3
.
gH 3/ 2
gH 3
.
.
gH 3
Note: The above dimensionless groups are formed by observation, simply combine the
dimensions so that the
term is dimensionless. We could have set up equations similar
to those of Eq. 6.2.11 and solved for a1 , b1 , c1 and a2 , b2 , c 2 and a3 , b3 , c3 . But the
method of observation is usually successful.
6.14
FD
ML
L
,
[
]
,
[
]
,[ ]
d
L
V
T
T2
f (d, , V , , ). [ FD ]
1
1
FD  a1 V b1
FD
,
2
V 2
c1
,
d
,

2
FD
2V 2
dV b2
2
c2
3
 a2 ,
V
3
M
,[ ]
LT
 a3 V b3
c3
M
.
L3
.
.
d
,
.
 V
f1
FD

f2 ,
. This is equivalent
2 2
d V
d dV
2
2
to the above. Either functional form must be determined by experimentation.
We could write
6.15
FD
1
f2
1
2
2
f (d, , V , , ). [ FD ]
1
FD d a1
b1
V c1 ,
By observation we have
1
FD
Vd
Rather than
1
3
1
f2
2
,
2
1
,
3
or
ML
L
M
, [d] L, [V ]
,[ ]
,[ ]
2
T
LT
T
d a2 b2 V c2 ,
d a3 b3 V c3 .
2
3
FD
Vd
,
, 3
.
2
Vd
d
M
.
L3
 Vd
,
.
d
f1
f1 (
,
3
), we could write
, an acceptable form:
3
127
FD
V 2d 2
f2

,
.
d Vd
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.16
M
, [d] L, [ ]
T2
Select d , , g as repeating variables:
h
hd a1
1
b1
1
h
d
6.17
FC
f1
d2
1
d2
V
d
3
3
.
,
.
Note: gravity does not enter the answer.
Rc
m
FC
2
R
dp
L
. [V ]
,[ ]
dx
T
a
g c2 ,
1
ML
, [m] M, [ ]
, [R] L.
2
T
T
.
m
FC
2
R
V
(dp /dx) d 2
dp
M
, [d] L, [ ]
LT
dx
6.20
V
FC
Cm
2
R
dp / dx
so that “M” is accounted for. Then the
1
term is, by
.
V
Const.
(dp /dx) d2
f (H, g, ). [V ]
M
.
L T2
2
c
Hence,
1
C.
dp
.
dx
b
Let’s start with the ratio
inspection
2
b2
2
M
ML2
,
[
M
]
, [ y] L, [ I ] L4 .
2
2
LT
T
I
My
C
Const.
.
1
yM
I
1,
f ( , d,
d a2
2
.
b
f ( M, y, I ). [ ]
V
,
FC m a
Given that b
6.19
g c1 ,
h
,
d
f (m, , R). [FC ]
1
6.18
M
L
, [ ] 1, [ g]
.
2
LT
T2
f ( , d, , , g). [h] L, [ ]
Const
V
L
L
, [H ] L, [ g]
,[ ]
T
T2
(dp /dx)d2
.
M
.
L3
0
1
VH a g b
c
V
g H
Const.
V
Const.
gH .
Density does not enter the expression.
128
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Chapter 6 / Dimensional Analysis and Similitude
6.21
V
VH a1
1
By inspection:
1
6.22
L
, [ H ] L, [ ]
T
f ( H , , , g , d ). [V ]
p
b1
g c1 ,
V
,
gH
1
f1(
H a2
2
2
,
),
3
b2
2
or
M
,[ ]
LT
g c2 ,
gH 3/ 2
V
gH
3
f1
,
gH 3
M
L2
, [ L]
T
L
, [d ] L, [ ]
T
LT
Repeating variables: V , d , .
, [V ]
2
pV a1 db1
1
By inspection:
1
6.24
,
d
.
H
f (V , d, , L, , ).
p
6.23
L
,
[
]
, [ d ] L.
g
L3
T2
Repeating
H , , g.
dH a3 b3 g c3 .
variables
d
.
H
3
M
1
f1 (
c1
,
2
p
,
2
V
2
2 , 3 , 4 ).
V a2 db2
Vd
,
p
V
2
,
3
L V a3 db3
L
,
d
4
e
.
d
c2
3
f1
L, [e]
L, [ ]
c3
,
4
M
L3
.
e V a4 db4
c4
.
L e
, , .
Vd d d
FD f (V , , , c, h, r, , w, ) where c is the chord length, h is the maximum thickness, r
is the nose radius, is the trailing edge angle, and
is the angle of attack. Repeating
variables: V , c, . The
terms are
FD
V c
c
c
c
, 2
, 3
, 4
, 5
, 6
, 7
.
1
2 2
h
r
w
V c
Then,
FD
V c c c
c
, , , , ,
f1
2 2
h r
w
V c
L3
L
, [ R] L, [ A] L2 , [e] L, [ s] 1, [ g ]
.
T
T2
There are only two basic dimensions. Choose two repeating variables, R and g. Then,
Q
f ( R, A, e, S , g ). [Q]
1
1
1
ARa2 g b2 , 3 eRa3 g b3 , 4
A
e
, 3
, 4 s.
2
2
R
R
A e
Q
f1 2 , , s .
f1 ( 2 , 3 , 4 ).
5
R R
gR
QRa1 g b1 ,
Q
,
gR 5 / 2
2
129
sRa4 g b4 .
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.25
Vp
L
L
M
, [h] L, [ g]
, [ ]
, [ ]
2
T
T
T2
f (h, g, , ). [Vp ]
Repeating variables: h, , g.
Vp
1
6.26
FD
hg
,
2
Vp h a1
1
gh
f1
gh
h a2
2
FDV a1 b1 d c1 , 2
V a2 b2 d c2 ,
FD
e
, 2
, 3
, 4
2 2
V d
V d
d
1
FD
V 2d 2
f1
M
2
T
a
3
L
M
LT
b
L
T
d c3 ,
4
I V a4
b4
d c4 .
c
d
L
T
2
a b c
g V dle
FD
e
L
1 a b
a 1 b
T:
2
b 2c d
d
L:
1
3a b c d e
1
2 b 2c
3(1 b) b c (2 b 2c) e
2 b c
1 b
FD
f ( , , V , D)
b c
gV
a b
2 b 2c
1 a b
T:
2
b c
L:
1
3a b c d
2 b. T
2 b c
l
or
V c Dd
M:
d
b3
L.
I.
or
M:
Hence, e
g c2 .
e
, ,I .
V d d
6.27 Using the exponent method we write:
FD f ( , , g, V , l)
ML
.
L, [ I ] 1, [d]
e V a3
3
b2
L3
.
gh 2
f (V , , , e, I , d). Repeating variables: V , , d .
ML
L
M
M
[ FD ]
, [V ]
, [ ]
, [ ]
, [ e]
2
T
LT
T
L3
1
6.28 T
g c1 ,
Vp
.
2
b1
M
b
2 2
V l
ML
T
2
gl
Vl
M
a
M
LT
3
L
V
b
c
FD
V 2l 2
2
L
T
c
L
Re, Fr
d
a 1 b
c
1 b
2 b
1
b
V
3(1 b) b (2 b) d
2 b
D
2 b
130
2
V D
b
2
VD
and
T
V 2 D2
Re
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
FD
6.29
f (V ,
s
[ FD
, , , D, g). Repeating variables: V , , D.
ML
L
M
M
M
L
, [V ]
,[ s]
,[ ]
,[ ]
, [ D] L, [ g ]
.
2
3
3
T
LT
T
L
L
T2
FDV a1 b1 Dc1 ,
FD
, 2
V 2 D2
1
1
s
2
s
,
FD
V 2 D2
FD
6.30
b1
d c1 ,
V a2
2
b2
2
1
FL
s
V a3
3
,
fd
,
V
gD
.
V2
4
1
1
ML
T
2
, [V ]
FLV a1
b1
gV a4
4
b4
Dc4 .
,
3
b3
d c3 ,
e
,
d
rV a4
4
r
,
d
4
1
.
L2
L, [e] L, [r] L, [c]
5
b4
d c4 ,
5
cV a5
b5
d c5 .
cd 2 .
e r
, , , cd 2 .
Vd d d
f1
M
, [ ]
LT
f V a1 db1
1
2
Dc3 ,
Vd
M
3
, [V ]
,
2
L
c1
fd
V
.
L
, [d] L.
T
V a2 db2
g1
Vd
c2
.
). Repeating variables: V , ,  c .
f (V , c , , c , t ,
[ FL ]
,
b3
gD
.
VD V 2
,
eV a3
3
Vd
1
, [ ]
T
g( , , V , d). [ f ]
Repeating variables, V , d, .
6.32
Dc2 ,
VD
f1
FD
V 2d 2
f
3
d c2 ,
FD
,
V 2d 2
1
6.31
b2
f (V , , , d , e, r, c). Repeating variables: V , , d .
ML
L
M
M
[FD ]
, [V ]
,[ ]
,[ ]
, [d]
2
T
LT
T
L3
FDV a1
1
V a2
L
L
, [c]
, [ ]
T
T
 c c1 ,
FL
,
V 2c2
2
2
c
,
V
FL
V 2c2
cV a2
3
f1
b2
L3
, [ c]
 c c2 ,
t
,
c
c t
, ,
V c
131
M
L, [t]
tV a3
3
b3
 c c3 ,
L, [ ] 1.
4
V a4
b4
 c c4 .
.
4
.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.33
T
f (d ,
Repeating variables: d,
T
T
2 5
d
T
f1
2 5
d
, [d ]=L, [ ]=
2
, .
1
6.34
ML2
, t ). [T ]
, ,
Td a1
1
,
2
d
,
2
b1
d
t
.
d
2
c1
,
M
1
, [ ]
T
,
3
L
d a2
2
b2
c2
,
3
t d a3
L.
b3
c3
.
t
.
d
3
3 5
W
d f1
d
,
2
t
.
d
FD f (V , , , d , L, c , ) where d is the cable diameter, L the cable length, c the
the vibration frequency. Repeating variables: V , d , . The
cable density, and
terms are
1
FD
2 2
V d
,
2
Vd
,
3
d
,
L
FD
V 2d2
p
6.35
f ( D, h,
1
p
D2
2
f1
W
g( f ,
[T ]
Vd
,
d
,
L
,
c
5
c
V
d
V
d
, , d1 , d0 ). Repeating variables: D, , .
M
1
M
, [ D] L, [h] L, [ ]
, [ ]
, [d1] L, [d0 ] L
2
T
LT
L3
p
d0
d1
h
, 2
, 3
, 4
.
2 2
D
D
D
D
[ p]
T
f1
,
4
We then have
6.36
M
, [t ]
LT
, [ ]
h d1 d 0
,
,
. W force
D D D
h d1 d 0
3
D 5 f1
,
,
.
D D D
velocity =
pD 2
D.
, d , H , , N , h, ). Repeating variables: , d , .
1
1
ML2
, [f]
,[ ]
, [d ] L, [ H ] L, [] L, [ N ] 1, [h]
2
T
T
T

h
T
f
H
, 2
, 3
, 4
, 5 N, 6
.
1
2 5
d
d
d
d
T
2 5
d
g1
f
,
L, [ ]
M
.
L3
H
h
, , N,
.
d d
d
132
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.37
Q
f (H, w, g,
[Q]
). Repeating variables: H, g, .
, ,
L3
L
, [ H ] L, [w] L, [ g]
,[ ]
T
T2
Q
1
,
gH 5
w
,
H
2
Q
6.38
d
f (V , Vj , D,
[d ]
1
, , ,
d
,
D
V
,
Vj
2
d
D
6.39
T
1
M L2
, [ ]
T2
T
2
d
5
,
V
Vj
6.40
V D
,
H
,
h
2
d
2
j
2
j
V D
1
, [H ]
T
T
2
4
gH 2
f1
5
3
L
M
,[ ]
T2
.
.
.
gH 2
M
,[ ]
T2
L, [ ]
,
4
,
Vj D
Vj D
,
L, [h]
R
,
h
3
H R t
,
, ,
h h h
a
]
M
.
L3
.
, h, .
t
,
h
h2
a
M
,[
LT
.
L , [R ]
4
M
,[ ]
L3
5
a
,
, ). Repeating variables:
f ( , H , h, R, t ,
[T ]
L
, [ D]
T
3
f1
,
gH 3
,
M
). Repeating variables: Vj , D, .
a
L
, [Vj ]
T
L, [V ]
gH 3
w
,
H
f1
gH 5
3
M
,[ ]
LT
L, [t ]
5
L, [ ]
M
, [ ]
LT
M
.
L3
h2
.
f ( D, H , , g, , V ) . D = tube dia., H = head above outlet,  = tube length.
Repeating variables: D, V , .
VD
f1
1
VD
,
2
H
,
D
3

,
D
4
gD
V2
H  gD
,
,
.
D D V2
133
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Chapter 6 / Dimensional Analysis and Similitude
6.41
T
[T ]
f ( R,
, ). Repeating variables: R, , .
, , e, r ,
ML2
, [ R] L, [ ]
T2
T
, 2
1
2 5
R
T
R

,
R
L, [ ]
5
M
, []
LT
L.
R2
.
R2
y2
f (V1 , y1 , , g). Neglect viscous wall shear.
L
M
L
[ y2 ] L, [V1 ]
, [ y1 ] L, [ ]
, [ g]
. Repeating variables: V1 , y1 , .
3
T
L
T2
y2
gy1
, 2
. ( does not enter the problem).
1
y1
V12
y2
y1
6.43
L, [r ]
e r
, , ,
R R R
f1
2 5
6.42
M
, [e]
L3
r
, 4
R
1
,[ ]
T
e
, 3
R
gy1
.
V12
f
1
M
, [d ] L, [] L, [ ]
,[ ]
T
L3
Repeating variables: d , , V . (  = length of cylinder).
f
M
, [V ]
LT
g( d , , , , V ). [ f ]
1.
fd
,
V

,
d
2
3
Vd
fd
V
.
f1

,
d
Vd
L
.
T
.
Similitude
6.44
Qm
Qp
Vm  2m
,
Vp  2p
m
p
Vm2 ( Fp ) m
,
2
( Fp ) p
p Vp
pm
pp
Vm2  2m
2 2
p Vp  p
m
m
Vm2  3m Q m
,
2 3
Q p
p Vp  p
Vm2 Tm
,
2
Tp
p Vp
m
Vm3  2m
3 2
p Vp  p
m
m
(Q has same dimensions as W .)
6.45
a) Re m
Re p .
Qm
Qp
W m
W p
Vm d m
m
Vm
Vp
2
m
2
p
.
dp
Vm
Vp
.
p
Vm3  2m
3 2
p Vp  p
m
Vp d p
dm
2
m
2
p
Qm
Qp
Vm
Vp
73
1
72
7.
134
7.
1.5 7
W m
1
72
0.214 m3 /s.
7 200
1400 kW.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
b) Re m
6.46
a) Re m
1.5 4.85
Wm
4.853
Vm2
2
p Vp
b) Re m
Re m
Re p .
Vm
dp
Vm
Vp
dm
m
10Vp
10
L2m
Vp  p
p
.
Fm
7.52
Vm
Vp
2
7390 kPa.
10.
dm
1
600 3.512
pm
1.06
1.41
10
p
V p2 L2p
2
m Vm
Vm  m
m
15 000 kPa.
3.51.
1
= 1.
10 2
10 2
800 / 5 160 kg/s.
25 600
dp
Vm
Vp
.
p
Vm2  2m
2 2
p Vp  p
Fm
1
5
25 p p
.8
114
.
5
p
Vp d p
m
p
m
mp
3.51 112 kg/s.
2
m
Re p .
mm
5.
5.
dm
pm
dm
Vm d m
Re p .
Fp
6.48
5
5
2
dp
1
800
1
dp
Vm
Vp
.
52 .
Vm
Vp
Re p .
466 kW
p
m
Fm
Fp
200
Vp d p
2
m mVm
2
p pV p
mm
a) Re m
72
4.85.
0.148 m3 /s.
2
m
pm
pp
6.47
1
7
.9
1.3
7
p
1
Vm d m
Re p .
m
dm
Qm
mm
mp
b) Re m
dp
Vm
Vp
Re p .
Fp
10 lb.
7.52.
102
p
m
m
p
17.68 lb.
10 assuming
m
1.
p
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.
135
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Chapter 6 / Dimensional Analysis and Similitude
6.49
Re m
m
p
Vm
Vp
Water:
Vp  p
Vm  m
Re p .
Vm
Vp
.
p
10 assuming
m
p
m
p
p
m
m
p
.
.
Vm
10Vp
900 km / hr.
1.5 10 5
13 500 km / hr.
1 10 6
m p
Neither a water channel nor a wind tunnel is recommended. Full-scale testing in a water
channel is suggested.
Vm
Air:
6.50
Vp
m
90 10
Properties of the atmosphere at 8 km altitude: T = 37ºC +273 = 236 K and pressure =
35.7 kPa, density = 0.526 kg/m3, and viscosity = 1.527 10 N·s/m2.
Properties of air at standard atmosphere: T = 20ºC, p = 101.325 kPa, density = 1.204
kg/m3, dynamic viscosity = 1.82 10 N·s/m2.
VD
VD
Use the Reynolds number to achieve dynamic similarity,
m
Then Vm
D
Vp
D
p
and Vm
Vp
Dp
p m
6.51
Re m
Tm
Re p .
Vm
Vm  m
p
10 50
2
V D
1.82 10
.
5
1.527 10
5
10
1
1041 km/hr
T
2
p
V 2 D2
0.526 10412 102
10
1.204 2002 12
Vp  p
m
0.526
1.204
T
To calculate the thrust apply:
2 2
p V p Dp
2 2
m Vm Dm
m
200
Dm
m p
Then Tp
p
Vm / Vp
 p / m
m
1184N
10 if
m
p
.
500 m / s.
This is in the compressibility range so is not recommended. Try a water channel for the
model study. Then
Vm
1 10 6
p m
10
0.662.
Vm 33.1 m/s.
Vp
1.5 10 5
m p
This is a possibility, although 33.1 m/s is still quite large.
( FD ) m
( FD ) p
2
mVm
2
pV p
2
m
2
p
1000
1
0.6622
1.23
102
136
3.56.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.52
Re m
Find
6.53
Re m
oil
Vp d p
Vm d m
.
dm
using Fig. B.2. Then
pm
pp
Re p .
m
p
Vp  p
Vm  m
Re p .
m
If
p
m
Vm
Vm
Vp
2.5 1
p
2
mVm
2
pV p
p
1.06 10
5
5.5 10
3
1.94
12
1.94 0.9
p
m
m
p
1.11.
0.1 .025 10
p
3
m
p
p
5 cm.
2.8 m/s. This is a much better velocity to work with in the lab.
50 cm, Vm
p
.
5
2000 and Vm 0.005 m/s.
0.0025
m
50 cm, but Vm 0.05 m/s. Each of these Vm 's is quite small, too
small for easy measurements. Let’s try a wind tunnel. Then,
1 10 3
p m
p
Vm Vp
0.1 .025 10 3
0.28 m/s if
5
m p
m 1.8 10
Or, if
= 0.0048 ft.
p
5 cm, then
We could try
.
Vp
dp
Thus, choose a wind tunnel.
6.54
Re m
Vp  p
Vm  m
Re p .
m
Vm
Vp
6.55
Frm
p
m
m
Frp .
6.56
Frm
a)
Qm
Qp
F
b) m
Fp
Vm
Vp
2
m
2
p
.
Vm2  2m
.
2 2
p Vp  p
m
1
.
30
m
p
Vp2
p gp
.
 pgp
m
Vp
Vp2
 2p
Vm2
 2m
Vm
Vp
.
Vm
Vp
Fm
pgp
1
.
164
10
p
Qp
Fp
p
( FD ) p
V p2
Qm
m
Vm
m
Vm2
 m gm
Fr p .
30
Vm2  2m
.
2 2
p Vp  p
( FD ) m
( FD ) p
Frp .
p
p
Vm2
m gm
. Frm
Vp2
Vm2
m gm
m
1
60
( FD ) m
.
Vm
Vp
6.1 10
1
.
30
9
m 2 /s. Impossible!
1.29 m/s.
60 60 2
10
2.16 10 6 N.
m
.
p
2
m
2
p
2
Vp2  2p
Vm2  2m
137
1
1
10 102
0.00632 m3 /s.
12 10 10 2
12 000 N.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.57
Neglect viscous effects: Frm
Vm2  2m
.
2 2
p Vp  p
Fm
Fp
6.58
Fp
m
m
p
Vm
Vp
Frp .
Fm
Vp2  2p
1
.
10
0.8 10 10 2
Vm2  2m
Vp
63.2 fps.
800 lb.
Neglect viscous effects, and account for wave (gravity) effects.
Vp2
Vm2
Vm
Vm /  m
m
m
Frm Frp .
.
.
.
Vp
Vp /  p
m gm  p g p
p
p
m
Tm
Tp
p
Vm  p
Vp  m
Vm2  3m
.
2 3
p Vp  p
Tp
m
Vm2
m gm
Vp2
6.59
Frm
6.60
Check the Reynolds number:
Frp .
1
10
600
pgp
.
Tm
10 1897 rpm.
Vp2  3p
1.2 10 10 3
Vm2  3m
m
.
p
Vm
Vp
6
100
120 000 N m.
m
.
p
p
m
278.
Vp d p
15 2
30 10 6 .
6
10
p
This is a high-Reynolds-number flow.
Re p
2 2 / 30
1.33 10 5 .
6
10
This may be sufficiently large for similarity. If so,
Rem

W
m

Wp

W
p
6.61
Vm3  m2
3 2
pVp  p
m
23
15 3
1
30 2
(2 2.15) / 2.63 10
2.63 10 6 .
6
1633 kW.
This is due to the separated flow downwind of the stacks, a viscous effect. Re is the
10 4
26.7 10 5 . This is a high-Reynolds-number
significant parameter. Re p
5
1.5 10
flow. Let’s assume the flow to be Reynolds number independent, above Re 5 10 5
(see Fig. 6.4). Then
Rem
5 105
Vm 4 / 20
1.5 10
5
.
138
Vm
37.5 m/s.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.62
Re p
20 10
1.5 10 5
13.3 10 6 . This is a high-Reynolds-number flow.
Vm 0.4
105
For the wind tunnel, let Rem
.
Vm 3.75 m/s
1.5 10 5
Vm 0.1
For the water channel, let Rem 105
.
Vm 1.0 m/s
1 10 6
Either could be selected. The more convenient facility would be chosen.
Fm2
2
m1 Vm1
2
m2 Vm2
Wm
Wp
3
mVm
3
p Vp
Fm1
6.63
2
m
2
p
3.2
.
Fm2
3.2
Fm2
153 0.42
.
203 102
1000 2.42
1.23 152
0.12
0.42
203 102
(15 3.2) 3
15 0.42
Wp
4.16 N.
71 100 W or 95 hp
Re is the significant parameter. This is undoubtedly a high-Reynolds-number flow. If the
p
model is 4 ft high then
250, and the model’s diameter is 45/250 = 0.18 ft. For
m
Re m
3 10 5 , we have
Vm .18
.
1.5 10 4
3 10 5
Re m
6.64
2
m1
2
m2
Vm
250 fps, and a study is possible.
Mach No. is the significant parameter: M m
a) M m
Vm
Fp
c) V p
Vm
Fp
cp
Vm2  2m
.
2 2
p Vp  p
Fm
Fp
b) V p
Vp
Vm
cm
M p.
Fp
m
cp
cm
Fm
cp
cm
Fm
Vm
Tp
Tm
2
m Vp
2
mVm
Vm
p
Vm
.
200
2
p
2
m
Tp
Tm
Vp2
2
m Vm
200
2
p
2
m
Vp
10 12
255.7
296
10 0.601
Mp .
200 m/s.
20 2
4000 N.
186 m/s.
1862
200
223.3
296
2
202
2080 N.
174 m/s.
10 0.338
139
1742
200
2
202 1023 N.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.65
Mm
Vm
Vp
6.66
Vp
290
223.3
273
Vp2
m
Vm2
5 for similarity. (Note: we use
m
a) Frm
b) Re m
.
pp
pm

p
pgp
Vm  p
Vp  m
10.
Vp  p
m
Vm  p
Vp  m
1
10
10
2902
34.6 kPa, abs.
m
.
p
10
10
2000
p
Vm
Vp
p
2622
0.338 o
0.8 o
at 2700 m where T = 0 C.)
m
.
276 m/s.
262 m/s.
80
Vm
Vp
.
1
10
Vm  m
Re p .
m
Vp2
Vm2
m gm
Frp .
273
223.3
250
p
p
10.
m
1.
m
6320 rpm.
2000 rpm.
There are no gravity effects or compressibility effects. It is a high-Re flow.
Vm2  3m
.
2 3
p Vp  m
Tm
Tp
m
p
6.68
Tm
.
Tp
2
mVm
2
p Vp
m
6.67
Vm
.
cp
cm
cp
pm
pp
p
Vp
Vm
cm
M p.
Re m
Tp
m
Vm  p
.
Vp  m
Vm
Re p .
Tm
p
Vp
m
m
p
m
.
Vp2  3p
2
m
V 
3
m
Vp  m
Vm  p
Vm
Vp
p
12
500
p
15 2
60 2
10 3
15
60
1
10
750 N m.
12.5 rpm.
10 10 100 m/s.
m
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
where we used
m
m
5 0.1
1 10 6
0.1 (
p
p
Vm  p
Vp  m
5 10 5 ,
1 m, i.e., the dia. of the porpoise).
1
5
10
10
140
5 motions / second.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
Normalized Differential Equations
6.70
*
, t*
u *
,v
V
tf , u*
0
*
f
0
Divide by
f
V
6.71
t
t
V ( *u* )
0
x*
*
0V /
*
*
*
x
v *
,x
V
0
x
y
, y*
V (
* *
v )
y
0.
*
:
* *
(
u )
y
*
(
* *
v ) 0.
v
w * x
V * u
, u
, v*
, w*
, x
, y*
U
U
U
U
Substitute into Euler’s equation and obtain:
V*
U2
t*
V*
u*
U2
v*
x*
f
.
V
parameter
y
V*
Uf
. Substitute in:
V*
U2
y*
w*
, z*
z
U2
V*
p
, p*
U
* *
p
z*
2
, t*
tf .
.
Divide by U 2 / :
f V*
U t*
6.72
V*
V *
, t
U
V*
u*
tU
*
x
*
,
U 2 DV*
Dt
v*
V*
y
*
* *
g
* *
p.
z
U
p
*
* *
*
p
, p*
U2
V*
w*
2
h
, h*
Parameter =
f
U
. Euler’s equation is then
* *
h.
g
Divide by U 2 / :
DV*
Dt*
6.73
* *
p
h.
U2
Parameter =
g
.
U2
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 x-component
Navier-Stokes equation is then
u
u
u
t
x
v
u
y
w
1
u
z
p
x
2
gx
u
x
2
2
u
y
2
2
u
z2
(wide plates)
This simplifies to
u
t
2
u
y2
.
141
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Chapter 6 / Dimensional Analysis and Similitude
a) Let u*
u/U, y*
y /h and t*
tU /h. Then
U 2 u*
U 2u*
h t* h 2 y*2
The normalized equation is
u*
1 2u*
t* Re y*2
b) Let u*
u/U, y*
y /h and t*
U u*
2
t /h2 . Then
2 *
U
*
Uh
where Re
u
2
h t
h y*2
The normalized equation is
2 *
u*
u
t*
6.74
y*2
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
u
t
u
r
vr
v
u
u
r
2
1 p
x
u
x
gx
u 1 u
r2 r r
1
r
2
2
u
2
2
u
x2
This simplifies to
u
t
a) Let u*
2
1 p
x
u/V , x*
u 1 u
r2 r r
x /d, t*
tV /d, p*
V 2 u*
V 2 p*
d t*
d x*
The normalized equation is
b) Let u*
u*
p*
t*
x*
u/V , x*
1
Re
x /d, t*
V
d2
t*
u
1 u*
r*2
r* r*
u
1 u*
r*2
r* r*
t /d 2 , p*
V 2 p*
d x*
d 2 t*
The normalized equation is
Re
2 *
2 *
V u*
u*
p / V 2 and r*
V
d2
2 *
u
1 u*
r*2
r* r*
2 *
u
1 u*
x*
r*2
r* r*
142
Vd
where Re
p / V 2 and r*
p*
r /d :
r /d :
where Re
Vd
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
6.75
Assume w 0 and
u
t
With g x
1
u
u
w
y
z
v
2
p
x
gx
2
u
x2
2
u
y2
u
z2
g the simplified equation is
2
u
u
x
Let u*
u
x
u
0. The x-component Navier-Stokes equation is then
z
g
x
u/V , x*
2
u
2
u
y2
x /h and y*
V 2 * u*
u
h
x*
V
g
h
y /h. Then
2 *
2 *
*2
y*2
u
2
u
x
The normalized equation is
u*
u*
6.76
u*
u
,
U
*
1
2
x
Fr
v*
v
,
U
cp
Divide by
1
Re
T*
UT0 T *
*
2 *
2 *
*2
*2
u
u
x
T
,
T0
y
x
x
x*
UT0 T *
y
*
where Fr
K
2
,
T0
y*
y
,
V
and Re
hg
*2
2
2
Vh
.
*2 *
T .
cpUT0 / :
T*
T*
x*
y*
K
c pU
*2 *
T .
Parameter =
143
K
cp
U
1 1
.
Pr Re
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Chapter 6 / Dimensional Analysis and Similitude
6.77
*
, V*
0
V *
,t
U
tU
p
, T*
p0
, p*
T
,
T0
1
*2
,
*
1
U
*
*
2
2
.
Momentum:
*U
0
Divide by
0U
2
2
DV*
Dt*
U
* *
p
*2
2
V*
3
2
(
V* ).
/ :
*
DV*
Dt
Energy:
*
cv 0T0
Divide by
p0
0cvT0U /
*
p0
*
0U
U DT *
K
*
2
Dt
* *
T0
*2
p
2
*2 *
T
0U
p0
U
p*
V*
*
*
(
*
V* ) .
V* .
:
DT *
Dt
K
0cvU
*
p0
p*
0cvT0
*2 *
T
*
V*.
The parameters are:
p0
0U
2
K
0cvU
RT0
kRT0
c2
U
kU
kU
2
2
1
.
Re
0U
K cp
c p cv 0U
p0
0cvT0
RT0
cvT0
1
2
kM 2
.
K
.
Pr Re
c p cv
cv
K 1.
The significant parameters are K, M, Re, Pr.
144
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Chapter 7 / Internal Flows
CHAPTER 7
Internal Flows
FE-type Exam Review Problems: Problems 7-1 to 7-13
7.1
(D)
7.2
(A)
7.3
(D) The flow in a pipe may be laminar at any Reynolds number between 2000 and
perhaps 40 000 depending on the character of the flow.
7.4
(D) The friction factor f depends on the velocity (the Reynolds number).
7.5
(A)
7.6
(B)
Δp = f
L V2
e .15
.
=
= .0075. ∴ Moody's diagram gives, assuming Re > 105
D 2 g D 20
15 V 2
. ∴V = 6.79 m/s and
0.02 2 × 9.81
Q = AV = π × 0.012 × 6.79 = 0.00214 m3 /s.
6.79 × 0.02
= 1.36 × 105. ∴ OK .
Check the Reynolds number: Re =
−6
10
f = 0.034. Then 60 000 = 0.034 ×
7.7
(D)
e 0.26
=
= 0.00325.
80
D
hL
1 V2
= sin θ = f
.
L
D 2g
Check Re: Re =
7.8
(B)
VD
ν
Assume Re > 3 × 105 . Then f = 0.026.
sin 30D = 0.026
=
5.49 × 0.08
10
−6
1
V2
.
0.08 2 × 9.81
= 4.39 × 105.
∴V = 5.49 m/s.
∴ OK.
e 0.26
4 × 0.06
=
= 0.0043. Re =
= 3 × 104 ∴ f = 0.033 from Moody's diagram.
−6
D
60
8 × 10
Δp = −γ f
L V2
20
42
+ γ Δh = −9810 × 0.033 ×
+ 9810 × 20 = 108 000 Pa
D 2g
0.06 2 × 9.81
145
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Chapter 7 / Internal Flows
7.9
(A)
0.02
A 4× 4
Q
=12.5 m/s.
=
= 1 cm. V = =
P 4× 4
A 0.04 × 0.04
R=
Re =
4RV
Δp = f
ν
4 × 0.01× 12.5
=
10
−6
0.046
e
=
= 0.00115. ∴ f = 0.021.
4 R 4 × 10
= 5 × 105.
2
40
12.52
L V
= 0.021×
×
= 167 Pa
4R 2 g
4 × 0.01 2 × 9.81
7.10
(B)
Viscous effects (losses) are important.
7.11
(A)
A negative pressure must not exist anywhere in a water system for a community
since a leak would suck into the system possible impurities.
7.12
(C)
1
1
AR 2 / 3 S1/ 2 =
0.8 × 2.4 × 0.482 / 3 × 0.0021/ 2 = 4.39 m3 /s
n
0.012
0.8 × 2.4
A
where we have used R =
=
= 0.48 m.
Pwetted 0.8 + 0.8 + 2.4
Q=
Chapter 7 Problems: Laminar or Turbulent Flow
7.13
7.14
VD
V ×2
.
a) 2000 =
.
−6
ν
1 × 10
1× 10−6
V × 0.02
b) 2000 =
∴ V = 0.1 m/s.
.
1× 10−6
V × 0.002
c) 2000 =
∴ V = 1.0 m/s.
.
1× 10−6
Re =
VD
=
Re =
Vh
=
ν
b) 1500 =
7.15
Re =
Vh
7.16
Re =
Vh
ν
ν
Vh
.
1 × 10 −6
a) 1500 =
V ×4
1× 10−6
V ×1
1 × 10
. ∴V = 0.0015 m/s.
−6
∴ V = 0.001 m/s.
. ∴V = 0.000375 m/s.
c) 1500 =
V × 0.3
1 × 10−6
. ∴V = 0.005 m/s.
=
1.5×.2 / 12
= 1790. Using Recrit = 1500, the flow is turbulent.
1.4 × 10 −5
=
(1 / 2) × 1.4
= 700 000.
10 −6
∴Very turbulent
146
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Chapter 7 / Internal Flows
7.17
Re =
VD
ν
.
a) V =
Re×ν 2000 ×10−6
=
= 0.1 m/s.
D
0.02
b) V =
Re×ν 40 000 ×10−6
=
= 2 m/s.
D
0.02
Entrance and Developed Flow
7.18
LE
= 0.065 Re
D
Re =
a) LE = 0.065 ×
b) LE = 0.065 ×
c) LE = 0.065 ×
ν
. V=
0.1592 × 0.04
1.31× 10−6
0.1592 × 0.04
1.007 ×10−6
0.1592 × 0.04
d) LE = 0.065 ×
7.19
VD
0.661×10−6
0.1592 × 0.04
0.367 × 10−6
0.0002
π × 0.022
= 0.1592 m/s.
× 0.04 = 12.6 m.
× 0.04 = 16.4 m.
× 0.04 = 25.0 m.
× 0.04 = 45.1 m.
Re×ν 1000 ×1.51×10−5
=
= 0.378 m/s.
a) V =
D
0.04
LE = 0.065 Re × D = 0.065 × 1000 × 0.04 = 2.6 m .
L
2.6
= 0.65 m.
Li ≅ E =
4
4
b) V =
Re×ν 80 000 ×1.51× 10−5
=
= 30.2 m/s.
D
0.04
LE ≅ 120D = 120 × 0.04 = 4.8 m .
Li ≅ 10D = 10 × 0.04 = 0.4 m .
7.20
V =
0.025
π × 0.03
2
= 8.84. Re =
8.84 × 0.06
1.007 × 10
∴ LE = 120 × 0.06 = 7.2 m.
7.21
−6
= 5.3 × 105. ∴Turbulent.
∴Developed.
Q (18 L/1000 L/m3 )/(2 hr × 3600 s/hr)
V= =
= 0.796 m/s
A
π × 0.0012 m 2
Re =
VD
ν
=
0.796 × 0.002
= 139.6.
1.14 × 10 −5
∴ laminar.
LE = 0.065 Re × D = 0.065 × 139.6 × 0.002 = 0.0181 m.
∴ negligible
147
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Chapter 7 / Internal Flows
7.22
LE = 0.04 Re× h = 0.04 × 7700 × .012 = 3.7 m.
( LE )min = 0.04 ×1500 × 0.012 = 0.72 m.
7.23
( LE )turb
7.24
5 × .06
× 0.06 = 75.5 m.
1.55 × 10−5
= 120 × 0.06 = 7.2 m.
(Re = 32 300)
( LE )lam = 0.065 Re D = 0.065
0.2 × 0.04
= 8000.
10 −6
ν
a) If laminar, LE = 0.065 × Re × D = 0.065 × 8000 × 0.04 = 20.8 m
Li = LE / 4 = 20.8 / 4 = 5.2 m
Re =
VD
=
b) This is a low Reynolds number turbulent flow. A minimum entrance length
would be L E = 120D = 120 × 0.04 = 4.8 m with a minimum inviscid core
length of Li = 10D = 10 × 0.04 = 0.4 m .
7.25
out − mom
in .
ΣFx = Δpπ r02 − τ 0 2π r0 Δx = mom
.
.
momin
momout
Δp 2τ 0 Δ mom
∴
=
+
pA
(p + Δp)A
Δx
r0
Δx
Δp 2τ 0
τ 0A0
= Const. From the
since mom
=
For developed flow
Δx r0
velocity distribution in an entrance (see Fig. 7.1) it is obvious that
Δ mom
∂u
(τ 0 ) entrance > (τ 0 ) d evelop ed since
is greater in the entrance. Also,
> 0 since
Δx
∂y wall
the momentum flux increases from the inlet to the developed flow. Hence,
⎛ Δp ⎞
⎛ Δp ⎞
>⎜ ⎟
.
⎜ ⎟
⎝ Δx ⎠ entrance ⎝ Δx ⎠ d eveloped
7.26
a) For a high Re flow transition to turbulence occurs near the origin. In the entrance
region the velocity gradient ∂u/∂y at the wall is very large resulting in a large wall
shear. This large wall shear requires a large pressure gradient. In addition, the
momentum flux is increasing in the x-direction, also requiring an increased pressure
gradient (see the solution to 7.13 for more detail).
b) For a low Re turbulent flow, the flow is laminar through much of the entrance region,
up to about Ld (see Fig. 7.2). The laminar flow results in a much smaller velocity
gradient at the wall compared with that of the turbulent flow of part (a) requiring a
much smaller pressure gradient. This results in the lower distribution of Fig. 7.3.
148
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Chapter 7 / Internal Flows
c) The pressure distribution must move from the lower distribution to the higher distribution of Fig. 7.3 as the Re increases. This occurs at an intermediate Re when
transition occurs near Li . Research that gives accurate data for such low turbulent Re
transition does not exist.
Laminar Flow in a Pipe
7.27
1 dpk 2
1 Δpk 2
(r − r02 ) =
(r − r02 ).
4μ dx
4μ L
u (r ) =
7.28 In a developed flow, dh/dx = slope of the pipe and p is a linear function of x so that
dp/dx = const. Therefore, d(p + γh)/dx = const and it can be moved outside the integral.
Then,
r
d( p + γ h) 0 2 2
1 d( p + γ h) ⎛ r04 r02 2 ⎞ r02 d( p + γ h)
−
=
r
r
rdr
(
)
⎜⎜ − × r0 ⎟⎟ =
0
∫
2
dx
dx
dx
μ
r
4 μ r02
2
0
⎝ 4 2
⎠ 8μ
0
2
7.29
VD
1600 =
ν
=
1.06 ×10−5
∴ V = 0.254 fps.
.
Using Eq. 7.3.14:
Δp r02 0.07 ×144 × 0.42 /144
=
= 268 ft.
L=
8μ V 8 × 2.06 ×10−5 × 0.254
Using Eq. 7.3.18:
τ0 =
∴f =
7.30
V × 0.8/12
VD
1500 =
ν
τ0
1
ρV 2
8
8ν V
r02 g
=
6.27 ×10−4
1
× 1.94 × 0.2542
8
V × 0.01
=
6.61×10−7
.
= 0.04.
∴ V = 0.0992 m/s.
Δp
Δh 8μV
+γ
= 2 .
L
L
r0
Eq. 7.3.14:
∴α =
=
r0 Δp 0.4 /12 × 0.07 ×144
=
= 6.27 ×10−4 psf .
2L
2 × 268
∴
8 × 6.61×10−7 × 0.0992
0.005 × 9.81
2
Δh 8μV
.
=
L r02 ρ g
α
Δh
L
α = Δh/L
= 0.00214 rad or 0.123D
Q = AV = π × 0.0052 × 0.0992 = 7.79 ×10−6 m3 /s.
7.31
V = Q /A = 0.0002/(π × 0.012 ) = 0.637 m/s.
a) Δp =
8μ VL
r02
=
8 × 0.1× 0.637 ×10
0.012
Use Eq. 7.3.14.
= 51 000 Pa.
149
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Chapter 7 / Internal Flows
b) Δp =
c) Δp =
8μ VL
r02
8μ VL
r02
=
=
8 ×1×10−3 × 0.637 × 10
8 × 1.5 × 0.637 ×10
0.012
0.01
7.32
∫
Q=
Eq. 7.3.14:
= 510 Pa.
0.012
9810(0.00015)
2 ×10
0
= 764 000 Pa.
−3
(0.02 y − y 2 )100dy
= 73 600 × (0.01× 0.012 −
0.013
) = 0.049 m3 /s
3
For a vertical pipe Δh = L. Thus,
ρgr02 gr02 9.81×.012 1.226 × 10 −4
.
V =
=
=
=
8μ
8ν
8ν
ν
a) V =
1.226 × 10−4
1.52 × 10
−6
= 80.7 m/s. ∴ Q = π × 0.012 × 80.7 = 0.0254 m3 /s.
Re =
80.7 × 0.02
1.52 × 10
−6
= 1.06 × 106. ∴not laminar
1.226 ×10−4
b) V =
= 0.33 m/s. ∴ Q = 1.04 × 10−4 m3 /s. Re = 17.8. ∴ laminar.
0.34 / 917
1.226 ×10−4
= 0.103 m/s. ∴ Q = 3.24 × 10−5 m3 /s. Re = 1.73. ∴ laminar.
c) V =
1.5 /1258
7.33
2000 =
VD
ν
=
Δp =
7.34
4QD ρ
πD μ
2
8μVL
r02
=
=
4 × 0.12 × 1.78
π × 2 × 10
−4
×D
8 × 2 × 10−4 × 30
0.342
=
1360
.
D
∴ D = 0.680 ft.
= 0.415 psf.
Neglect the effects of the entrance region and assume developed flow for the whole
length. Also, assume p inlet = γh (neglect V 2 /2 g compared to 4 m).
∴Δp = γ h − 0.
∴ ρ gh =
8μ VL
r02
. ∴V =
ghr02 9.81× 4 × 0.00252
=
= 0.766 m/s.
8ν L
8 ×1×10−6 × 40
∴ Q = AV = π × 0.00252 × 0.766 = 1.5 × 10−5 m3s.
150
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Chapter 7 / Internal Flows
LE = 0.065 ×
7.35
0.766 × 0.005
× 0.005 = 1.2 m.
1×10−6
Neglect the effects of the entrance region and assume developed flow for the whole
length. Also, assume p inlet = γh (neglect V 2 /2 g compared to 4 m):
8μ VL
∴Δp = γ h =
r02
r02
and
8V μ L 8(0.0034 / 60 × 60) × 10−3 × 4
=
=
γh
(π r02 ) 9800 × 4
where V = Q /A = Q /π r02 . This gives r0 = 7.04 × 10 −4 m or 0.704 mm .
The velocity is then V =
0.0034/60 × 60
π × (0.000704) 2
= 0.6066 m/s.
V 2 0.6066 2
=
= 0.0188 m. This is negligible compared to 4 m.
2 g 2 × 9.81
VD
LE = 0.065 Re D = 0.065 ×
ν
× D = 0.065
0.6066 × 0.001408
10−6
× 0.001408 = 0.0782 m
This entrance region is insignificant; to include its effect would be quite difficult.
7.36
Re = 2000 =
Δp =
7.37
Δp + γΔh =
VD
ν
=
8μ VL
r02
8μ VL
r02
V × 0.8 /12
1.6 × 10−4
=
7.38
Δp =
VD
ν
(0.4 /12) 2
= 0.396 psf .
− 6000 + 9810 × 10sin10 =
.
Re =
2τ 0 L
.
r0
Re = 40 000=
8 × 3.82 ×10−7 × 4.8 × 30
D
∴ V = 0.552 m/s.
Δp + γΔh =
∴ V = 4.8 fps.
.
VD
ν
=
0.552 × 0.004
1×10−6
V × 0.1
1.51×10−5
0.0022
.
= 2210.
− 6000 + 9810 × 10sin10D =
=
8 ×1×10−3 V ×10
2 ×τ 0 ×10
.
0.002
∴τ 0 = 1.1 Pa.
. ∴ V = 6.04 m/s. ∴ Vmax = 2V = 12.1 m/s
8μ VL 8 ×1.81×10−5 × 6.04 ×10
=
= 3.5 Pa. LE = 0.065 × 40 000 × 0.1 = 260 m.
r02
0.052
151
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Chapter 7 / Internal Flows
7.39
If Q = πR 2 2 gH , then V = 2 gH . The parabolic velocity profile is
⎛
⎛
⎛
r2 ⎞
r2 ⎞
r2 ⎞
u(r ) = u max ⎜ 1 − 2 ⎟ = 2V ⎜ 1 − 2 ⎟ = 2 2 gH ⎜ 1 − 2 ⎟ .
R ⎠
R ⎠
R ⎠
⎝
⎝
⎝
But, the manometer requires
u2
u2
p+
ρ − p = γH .
.
∴H =
2
2g
Substituting yields
⎛
r2 ⎞
u = 2u ⎜ 1 − 2 ⎟ .
⎜ R ⎟
⎝
⎠
7.40
V =−
∴r2 =
R2
or r = R / 2.
2
R2 Δp + γΔh
R2 γ (− L) γ R2 9800 × 0.0012
=−
=
=
= 1.225 m/s
8μ
8μ L
8μ
L
8 ×10−3
Q = AV = π × 0.0012 × 1.225 = 3.85 ×10−6 m3 /s
Re =
VD
ν
=
122.5 × 00.02
= 2450
10 − 6
It probably is not a laminar flow. Since Re > 2000 , it would most likely be turbulent. If the pipe
were smooth, disturbance and vibration free, with a well-rounded entrance it could be laminar.
7.41
a) First, determine the average velocity of the flow
⎡ liter
m3 min ⎤
2
× 10−3
×
π ( 0.02m ) = 0.053 m/s
V = Q A = ⎢4
⎥
liter 60s ⎦
⎣ min
Then determine if the flow is laminar or turbulent:
Re =
ρVD 103 × 0.053 × 0.04
=
= 1860
1.14 ×10−3
μ
This is less than 2000 and hence the flow is laminar. The average velocity is given in Eq.
(7.3.13) for laminar flow as
r02 d
r02 ⎡ dp
dh
V =−
( p + γ h ) = − ⎢ + γ ⎤⎥
dx ⎦
8μ dx
8μ ⎣ dx
Since the flow is vertically downward then dh/dx = −1, hence the pressure gradient is
dp
8μV
8 ×1.14 × 10−3 N ⋅ s/m2 × 0.053 m/s
= γ − 2 = 9810 −
= 9808 Pa/m
dx
r0
0.022 m2
Note that in this case the pressure drop due to viscous effects is negligible compared to
the hydrostatic pressure.
152
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Chapter 7 / Internal Flows
Next, determine the pressure drop over a distance of 10 m using dp/dx = Δp/L:
dp
Δp = L
= 10 × 9808 = 98 080 Pa
dx
(b) The friction head loss is determined from Eq.(7.3.22) as
−3
32μVL 32 1.14 × 10 ( 0.053)(10 )
hL =
=
= 0.00123 m
γ D2
9810 × 0.042
(c) The shear stress at the wall can be determined from Eq.(7.3.17) rearranged as
r d ( p + γ h)
r ⎛ dp
⎞
= − 0 ⎜ − γ ⎟ = −0.01 m ( 9808 − 9810 N/m3 ) = 0.02 N/m2
τ0 = − 0
dx
2
2 ⎝ dx
⎠
(
7.42
a) Combine Eq. 7.3.12 and 7.3.15:
⎛ r2 ⎞
u(r ) = V = 2V ⎜1 − 2 ⎟ .
⎜ r ⎟
0 ⎠
⎝
)
⎛ r2 ⎞
u = umax ⎜1 − 2 ⎟ .
⎝ r0 ⎠
∴r2 =
r02
and r = 0.707 r0
2
b) From Eq. 7.3.17: τ = Cr where C is a constant. Then τ w = Cr0 .
If τ = τ w /2, then τ w /2 = Cr and r = τ w /2C = r0 /2 .
7.43
Qpipe
Qannulus
=
π r04 Δp
8μ L
2
2 2⎫
⎧
4
π Δp ⎪ 4 ⎛ r0 ⎞ ⎡⎣r0 − (r0 / 2) ⎤⎦ ⎪
⎨r0 − ⎜ ⎟ −
⎬
8μ L ⎪
ln(2r0 / r0 ) ⎪
⎝2⎠
⎩
⎭
r04
=
r04
r04 9r04 /16
−
−
16
ln2
= 7.938
V × 2/12
. ∴ V = 1.44 fps.
ν
1.2 ×10−5
Δp 8 μVL 8 × 2.36 × 10 −5 × 1.44 × 30
= 0.0188 ft.
hL =
=
=
γ
γ r02
62.4 × (1 / 12) 2
r Δp (1/12) × 0.0188 × 62.4
2
= 0.00163 psf . LE = 0.065 × 20, 000 × = 217 ft.
τ0 = 0 =
12
2L
2 × 30
VD
7.44
Re = 20 000 =
7.45
π (−Δp) ⎡ 4 4 (r22 − r12 )2 ⎤
See Example 7.2: Q = −
⎢ r2 − r1 −
⎥
8μ L ⎣⎢
ln(r2 / r1 ) ⎦⎥
∴Q = −
=
π
8 ×1×10−3
2
2 2
(−100) ⎡
4
4 (0.03 − 0.02 ) ⎤
3
−4
⎢ 0.03 − 0.02 −
⎥ = 1.31× 10 m /s.
10 ⎣⎢
ln(0.03/0.02) ⎦⎥
r22 − r12 1 ⎤
∂u
1 (−Δp) ⎡
=
τ r1 = μ
⎢ 2r1 −
⎥
ln(r2 / r1 ) r1 ⎦⎥
∂r r =r1 4 L ⎣⎢
153
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Chapter 7 / Internal Flows
=−
7.46
100 ⎡
0.032 − 0.022
1 ⎤
2
0.02
×
−
×
⎢
⎥ = 0.054 Pa.
4 × 10 ⎢⎣
ln(0.03 / 0.02) 0.02 ⎥⎦
π (−Δp) ⎡ 4 4 (r22 − r12 )2 ⎤
Q=−
⎢ r2 − r1 −
⎥
8μ L ⎣⎢
ln(r2 / r1 ) ⎦⎥
See Example 7.2:
2
2 2
⎡
4
4 (0.03 − 0.02 ) ⎤
3
−4
∴Q =
⎢0.03 − 0.02 −
⎥ = 7.25 ×10 m /s.
−5
ln(0.03/0.02) ⎦⎥
8 ×1.81×10 ×10 ⎣⎢
π ×10
∴V =
7.25 ×10−4
Q
=
= 0.462 m/s.
A π (0.032 − 0.022 )
τ r1 = μ
=−
7.47
Δp ⎡
r22 − r12 1 ⎤
∂u
=−
−
r
2
⎢ 1
⎥
4 L ⎢⎣
ln(r2 / r1 ) r1 ⎥⎦
∂r r =r1
10 ⎡
0.032 − 0.022
1 ⎤
×
−
×
2
0.02
⎢
⎥ = 0.0054 Pa.
4 × 10 ⎢⎣
ln(0.03/0.02) 0.02 ⎥⎦
See Example 7.2:
dp μ (T ) d ⎛ du ⎞
=
⎜ r ⎟ . The function μ (T ) requires that T (r ) be known.
dx
r dr ⎝ dr ⎠
The energy equation is needed to find T (r ). But, as Eq. 5.5.1 shows, we must know
u(r ) to find T (r ). Hence, the above momentum equation and the energy equation are
coupled. A simultaneous solution (a numerical approach is needed) would provide
u(r ) and T (r ).
7.48
1 dp ⎡ 2 2 r22 − r12
r⎤
From Example 7.2 u(r ) =
ln ⎥
⎢r − r2 +
4μ dx ⎣⎢
ln(r1 / r2 ) r2 ⎦⎥
As r1 → 0, ln(r1 / r2 ) → −∞ so that
(
)
r22 − r12
ln(r / r2 ) → 0.
ln(r1 / r2 )
Thus, u(r ) =
1 dp 2
r − r02
4 μ dx
As r1 → r2 ,
r22 − r12
0
= . ∴Differentiate w.r.t r1:
ln(r1 / r2 ) 0
w here r2 = r0 . See Eq. 7.3.11.
Also, ln
⎛
y⎞
y
r
= ln ⎜ 1 − ⎟ ≅ − , where y = r2 − r.
r2
r2
⎝ r2 ⎠
∴ u(r ) =
1 dp ⎛ 2 2 2r12
⎜ r − r2 +
4 μ dx ⎜⎝
r2
−2r1
= −2r12 .
1 / r1
⎞ 1 dp ⎡ 2
2r12
2
−
+
y⎟ =
y
r
y
2
⎟ 4μ dx ⎢⎢
r2
⎠
⎣
⎤ 1 dp 2
( y − ay ).
y⎥ =
⎥⎦ 4 μ dx
154
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Chapter 7 / Internal Flows
Laminar Flow between Parallel Plates
7.49
a) 2000 =
Vh
ν
. ∴V =
2000 ×1.2 × 10−5
= 0.576 fps. ∴ Q = AV = 0.04 ft 3 /sec.
1/ 24
1 20
2000 ×1.6 × 10−5
b) V =
×
×.768 = 0.053 cfs.
= 0.768 fps. ∴ Q =
24 12
1/ 24
7.50
V
y.
a
There is no pressure gradient. ∴Eq. 7.4.13 gives u =
The friction balances the weight component.
∂u
0.2
V
.
τ A = W sin θ .
τ =μ
=μ =μ
∂y
0.0004
a
0.2
× 1× 1 = 40sin 20D. ∴ μ = 0.0274 N ⋅ s/m 2 .
a) μ
0.0004
b) μ
7.51
0.2
× 1× 1 = 40sin 30D.
0.0004
b)
7.52
U
y. Thus, τ A = W sin θ .
a
1×10−3 V
a)
×1× 1 = 40sin 20D.
0.0004
1×10−3 V
×1× 1 = 40sin 30D.
0.0004
W
∴ μ = 0.04 N ⋅ s/m 2 .
With the pressure gradient zero, Eq. 7.4.13 gives u =
∂u
V 1× 10−3
τ =μ
=μ =
V.
∂y
a 0.0004
τA
∴ V = 5.47 m/s.
∴ V = 8 m/s.
The depth of water is a/2 with the maximum velocity at the
surface:
−
dh
= sin θ .
dx
Q=−
a /2
∫
0
=
H ence, u( y ) = −
γ sin θ 2
( y − ay ).
2μ
y
a = 0.012 m
γ sin θ 2
50γ sin θ ⎛ a3 a3 ⎞ 25γ a3
( y − ay )50dy = −
sin θ
⎜ − ⎟=
2μ
2μ ⎜⎝ 24 8 ⎟⎠ 12μ
25
12 × 10
−3
× 9810 × sin 20D × 0.0123 = 12.1 m3 /s. ∴ V =
∴ Re =
Va /2
40.3 × 0.006
= 241 000.
10−6
The assumption of laminar flow was not a good one, but we shall stay with it to answer
the remaining parts.
ν
=
12.1
= 40.3 m/s.
0.006 × 50
155
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Chapter 7 / Internal Flows
umax =
9810 × sin 20D ⎛ 0.0122 ⎞
⎜
⎟ = 60.4 m/s.
2 ×10−3 ⎜⎝ 4 ⎟⎠
τ0 = μ
γ sin θ
∂u
9810sin 20D
( − a) =
=−
× 0.012 = 20.1 Pa.
2
2
∂y y =0
Obviously the flow would be turbulent and the above analysis would have to be modified
substantially for an actual flow.
7.53
u( y ) =
γ dh 2
9810
( y − ay ) =
× (−0.00015)( y 2 − 0.02 y)
−3
2μ dx
2 ×10
0.01
Q=
∫
9810(0.00015)
2 × 10
0
−3
y
( 0.02y − y2 )100dy
a = 0.02 m
⎛
0.013 ⎞
3
= 73 600 ⎜ 0.01× 0.012 −
⎟⎟ = 0.049 m /s
⎜
3 ⎠
⎝
du
0.049
9810 × 0.00015 × 0.02
V=
= 0.049 m/s. τ 0 = μ
=
= 0.015 Pa.
dy 0
100 × 0.01
2
f=
7.54
Eq. 7.4.17:
8τ 0
ρV 2
Δp =
=
8 × 0.015
1000 × 0.0492
12 μ VL
a2
. ∴V =
Re = 490.
= 0.050
50 × 0.022
12 × 1.81× 10−5 × 60
= 1.53 m/s.
∴ Q = AV = 0.02 × 0.9 × 1.53 = 0.028 m3 /s.
This is maximum since laminar flow is assumed. Check the Reynolds number:
Re =
Va
1.53 × 0.02
= 2030.
1.51× 10−5
This is marginally high. Care should be taken to eliminate vibrations, disturbances, or
rough walls.
7.55
ν
=
( p A − p B ) static = γh = 9810 × 20 sin 30 D = 98 100 or 98.1 kPa.
Eq. 7.4.17:
Δp + γΔh =
∴ V = 0.56 m/s.
∴f =
8τ 0
ρV
2
=
12 μ LV
a2
τ0 =
.
∴−96 000 + 98 100 =
∴ flow is d ow n.
12 × 10−3 × 20V
0.0082
.
aΔp aγΔh 0.008
(−96 000 + 98 100) = 0.42 Pa.
+
=
2L
2L
2 × 20
8 × 0.42
1000 × 0.562
= 0.0107.
156
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Chapter 7 / Internal Flows
7.56
Assume laminar flow:
Δp =
12 μ VL
a2
. ∴ 600 × 144=
12 ×1.2 × 10−3V × 2/12
(0.02 /12) 2
. ∴ V = 100 fps.
∴ Q = AV = (0.02 × 4 /144) ×100 = 0.0556 ft 3 /sec or 0.0556 cfs
7.57
u( y) =
1 dp 2
U
( y − ay ) + y.
a
2μ dx
a) τ y =a
U
dp
1 dp
2 μU
2 ×1.81× 10−5 × 6
=0=
= −13.6 Pa/m.
(2a − a ) + μ . ∴ = − 2 = −
2 dx
a
dx
0.0042
a
b) τ y =0 = 0 = −
0.004
c) Q =
∫
0
=
du 1 dp
U
(2 y − a ) + μ .
=
dy 2 dx
a
a dp
U
dp 2 μU 2 × 1.81× 10−5 × 6
+μ . ∴
= 2 =
= 13.6 Pa/m.
2 dx
a
dx
0.0042
a
⎡ 1 dp 2
U
⎢ 2μ dx ( y − ay) + a
⎣
1
2 ×1.81×10−5
d) u(0.002) = 4 =
⎤
y ⎥dy
⎦
dp ⎛ 0.0043 0.0043 ⎞
6
0.0042
−
+
×
= 0.
⎜
⎟
2 ⎟⎠ 0.004
2
dx ⎜⎝ 3
1
2 × 1.81× 10
∴
7.58
τ =μ
−5
∴
dp
= 40.7 Pa/m.
dx
dp
6
(0.0022 − 0.004 × 0.002) +
× 0.002.
dx
0.004
dp
= 9.05 Pa/m
dx
1 dp 2
U
du 1 dp
U
( y − ay ) + y. τ = μ
(2 y − a) + μ .
=
a
dy 2 dx
a
2 μ dx
U
1
a) τ y =0.006 = (−20)(0.006) + 1.95 ×10−5
= 0. ∴U = 18.5 m/s.
2
0.006
1
U
b) τ y =0 = (−20)(0.006) + 1.95 × 10−5
= 0. ∴U = −18.5 m/s.
2
0.006
u=
0.006
c) Q =
∫
0
U
1
⎡
⎤
(−20)( y 2 − 0.006 y ) +
y ⎥dy ∴U = −6.15 m/s.
⎢
−5
0.006 ⎦
⎣ 2 × 1.95 × 10
d) u(0.002) =
1
2 × 1.95 × 10
(−20)(0.0022 − 0.006 × 0.002) +
−5
U
× 0.002.
0.006
∴U = −12.3 m/s.
157
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Chapter 7 / Internal Flows
7.59
(i) The solution for laminar flow between two parallel plates is given in Eq.(7.4.13) as
u( y) =
(
)
1 d
U
( p + γ h ) y2 − ay + y
a
2μ dx
For a horizontal flow dh/dx = 0, and hence the velocity profile is given by
u( y) =
1 dp 2
U
y − ay ) + y
(
a
2 μ dx
(ii) To determine the pressure gradient we set u = 0 at y = a/2, that is
0=
1 dp ⎛ a2 a2 ⎞ U
⎜ − ⎟+
2 μ dx ⎝ 4 2 ⎠ a
⎛a⎞
⎜ ⎟
⎝2⎠
Simplifying and solving for the pressure gradient we get
dp
2
= 4μU a2 = 4 ( 0.4 N ⋅ s/m2 ) ( 2 m/s ) ( 0.01m ) = 32 kPa/m
dx
Note that since the pressure gradient is positive in the flow direction, it is considered
to be an adverse pressure gradient.
7.60
The velocity profile for laminar flow between two inclined parallel plates is given by Eq.
7.4.11 as
u( y) =
λ
y2 + Ay + B .
2
We determine A and B by applying the two boundary conditions: u = 0 at y = 0 , and
u = −U at y = h . The first boundary condition results in B = 0 . The second condition
λ
⎛U λ ⎞
gives −U = h2 + Ah , which yields A = − ⎜ + h ⎟ , and hence we have
2
⎝h 2 ⎠
u( y) =
7.61
u=
U
1 d ( p + γ h) 2
y − hy − y
2μ
dx
h
(
)
45
U
du
y=
y = 67,500y. τ = μ
= 10−4 × 67,500 = 6.75 psf.
0.008 /12
a
dy
∴ F = τ A = 6.75 × (2π × 10/144) = 2.95 lb.
7.62
vθ =
U
y.
a
∴τ = μ
dvθ
0.2 × 30
= 0.1×
= 750 Pa.
0.0008
dy
T = F × R = τπ D × L × R = 750π × 0.4 × 0.8 × 0.2 = 151 N ⋅ m.
158
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Chapter 7 / Internal Flows
7.63
Assume a linear velocity profile between the rotating disc and the wall:
∴ vθ =
U
rω
y=
y.
a
a
∴τ = μ
dvθ μω
0.01× 60
r=
r = 500 r.
=
0.0012
dy
a
.2
.2
0
0
T = ∫ τ dA × r = ∫ 500r × r 2π rdr = 1000π ∫ r 3dr =
The largest Re occurs at r = 0.2 m: (Re) max =
dA = 2πrdr
dr
r
τ
1000π × .24
= 1.26 N ⋅ m.
4
Rω × a
ν
=
.02 × 60 × 0.0012
= 1240.
0.01/(0.86 ×1000)
The laminar flow assumption is valid.
7.64
Neglect the shear on the cylinder bottom; assume a linear velocity profile:
vθ =
0.1× 30
U
y=
y = 3000 y.
0.001
a
∴τ = μ
dvθ
= 0.42 × 3000 = 1260 Pa.
dy
∴ T = F × R = π DL × τ × R = π × 0.2 × 0.1× 1260 × 0.1 = 7.9 N ⋅ m.
7.65
Assume a linear velocity profile: vθ =
U
rω
50r
y=
y=
y.
0.002
a
a
dv
τ = μ θ = 25 000 r. T = ∫ τ × dA × r =
dy
A
2π
∴T = 25000 ×
0.707
7.66
0.0707
∫
0
0.0707
∫
0
(25 000 r )(2π r
dr
) × r.
0.707
50 000π 0.0707 4
×
= 1.388 N ⋅ m.
r dr =
0.707
4
3
Assume that all losses occur in the 8-m-long channel. The velocity through the straws
and screens is so low that the associated losses will be neglected. Assume a developed
flow in the channel:
V=
Re×ν 7000 ×1.5 × 10−5
=
= 8.75 m/s
0.012
h
12 × 1.8 × 10 −5 × 8.75 × 8
Δp =
= 105 Pa
0.012 2
Energy:
Δp
ΔpAV
= Δp m
=
W
( ρ AV ) =
fan
ρη
=
ρη
η
105 ×1.2 × 0.012 × 8.75
= 18.9 W
0.7
159
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
Laminar Flow Between Rotating Cylinders
7.67
vθ =
The solution for vθ (r ) is (see Eq. 7.5.15)
A
B
r+ .
r
2
B
. Also, vθ = Rω at r = R.
r
B
∴T = τ1 2π RL × R = 4πμω R2 L. ∴ Rω = .
∴ B = ω R2 . ∴ vθ = ω R2 /r.
R
The shear stress at r = R is
If vθ = 0 as r → ∞, A = 0.
⎡
⎣
τ1 = − ⎢ μ r
∴ vθ =
dvθ /r ⎤
= 2μω. ∴ T = τ 1 2πRL × R = 4πμωR 2 L.
⎥
dr ⎦
⎛ 1000 × 2π
∴ T = 4π × 2.36 ×10 × ⎜
60
⎝
−5
2
⎞ ⎛ 1 ⎞ 40
= 7.19 × 10−4 ft-lb.
⎟×⎜ ⎟ ×
12
12
⎠ ⎝ ⎠
Use units on the variables of lb, ft, rad, and sec and the units will work out. You should
check to make sure.
7.68
4πμ r12 r22 Lω1
Use Eq. 7.5.19: T =
r22 − r12
∴ T = 0.040 N ⋅ m
Re =
∴ μ = 0.0134
0.032 − 0.022
= Tω = 0.04 × (3000 × 2π /60) = 12.6 W.
∴W
4 μπ r12 r22 Lω1
N ⋅s
m2
r22 − r12
. Re =
= 0.015 =
4πμ × 0.042 × 0.052 × 0.5 × 40
0.052 − 0.042
40 × 0.04(0.05 − 0.04) × 1000 × 0.9
= 1070. Eq. 7.5.17 is OK.
0.0134
With ω 1 = 0 , Eq. 7.5.15 is vθ =
∴ T2 = τ 2 A 2 r2 =
7.71
4π × 0.035 × 0.022 × 0.032 × 0.4 × (3000 × 2π /60)
ω r1 (r2 − r1 ) ρ (3000 × 2π /60) × 0.02 × (0.01) × 917
=
= 1650. ∴Eq. 7.5.15 is OK.
0.035
μ
7.69 Use Eq. 7.5.19: T =
7.70
=
r22ω 2 ⎛
r12 ⎞
d ⎛ vθ ⎞ 2 μr12ω 2
−
.
.
=
τ
μ
r
r
⎜
⎟
⎜ ⎟=
2
r22 − r12 ⎝
r⎠
dr ⎝ r ⎠ r22 − r12
2 μr12ω 2
4πμr12 r22 Lω 2
2
π
=
.
r
Lr
2
2
r22 − r12
r22 − r12
4πμr12 r22 Lω 1 4π × 0.1 × 0.2 2 × 0.2008 2 × 0.8 × 30
= 152 N ⋅ m
T=
=
0.2008 2 − 0.2 2
r22 − r12
% error =
152 − 151
× 100 = 0.66%
151
160
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Chapter 7 / Internal Flows
Turbulent Flow
7.72
Let u = u + u′, v = v + v′, w = w + w′. The continuity equation becomes
∂
∂
∂
∂u ∂v ∂w ∂u′ ∂v′ ∂w′
+
+
+
+
+
(u + u′) + (v + v′) + (w + w′) =
.
∂x
∂y
∂z
∂x ∂y ∂z ∂x ∂y ∂z
Now, time-average the above equation recognizing that
Then,
∂u ∂u
∂u′ ∂
=
= u′ = 0.
and
∂x ∂x
∂x ∂x
∂u ∂v ∂w
+
+
= 0. Substitute this back into the continuity equation, so that
∂x ∂y ∂z
∂u′ ∂v′ ∂w′
+
+
= 0.
∂x ∂y ∂z
7.73
∂
∂
∂
∂
Du
= (u + u′) (u + u′) + (v + v′) (u + u′) + (w + w′) (u + u′) + (u + u′)
∂x
∂y
∂z
∂t
Dt
=u
∂u
∂u′
∂u
∂u′
∂u
∂u′
∂u
∂u′
∂u
∂u′
+u
+ u′
+v +v
+ v′
+w
+w
+ u′
+ v′
∂x
∂x
∂x
∂x
∂y
∂y
∂y
∂y
∂z
∂z
+ w′
=u
∂u
∂u′ ∂u
+ w′
+ .
∂z
∂z ∂t
∂u
∂u
∂u ∂u
∂u′
∂u′
∂u′
+v +w
+
+ u′
+ v′
+ w′ .
∂x
∂y
∂t ∂t
∂x
∂y
∂z
∂u
∂u
∂u ∂u
Du
=u +v +w
+ .
∂x
∂z
∂z ∂t
Dt
∴
∂
Du D u ∂ 2 ∂
−
= u′ + u′v′ + u′w′.
Dt Dt ∂x
∂y
∂z
7.74 Use the fact that
(We used continuity.)
∂
∂
u′v′ = u′v′. See Eq. 7.6.2. This is equivalent to
∂y
∂y
T
⎞ 1T ∂
∂ ⎛1
′
′
⎜
u v dt ⎟ = ∫ (u′v′)dt ,
⎟ T ∂y
∂y ⎜⎝ T ∫0
0
⎠
which is obviously correct. Also,
⎛ ∂u′ ∂v′ ∂w′
∂
∂
∂
u′u′ + u′v′ + u′w′ = u′ ⎜
+
+
⎜
∂x
∂y
∂z
⎝ ∂x ∂y ∂z
⎞
∂u′
∂u′
∂u′
+ v′
+ w′
.
⎟ + u′
⎟
∂x
∂y
∂z
⎠
Time average both sides and obtain the result.
161
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Chapter 7 / Internal Flows
7.75
The x-component Navier-Stokes equation for a horizontal channel is
⎛ ∂ 2u ∂ 2u ∂ 2 u ⎞
⎛ ∂u
∂p
∂u
∂u
∂u ⎞
+u +v +w ⎟ = − + μ⎜ 2 + 2 + 2 ⎟
⎜
∂x
∂y
∂z ⎠
∂x
∂y
∂z ⎠⎟
⎝ ∂t
⎝ ∂x
ρ⎜
Substitute u = u + u′, v = v′, w = w′ into the N-S eq and time-average:
⎡ ∂u
ρ⎢
+
⎛ ∂2 u ∂2 u ∂2 u ⎞
⎤
∂ p ∂ p′
∂u′ ∂ 2 ∂
∂
+ u′ + u′v′ + u′w′⎥ = −
−
+μ⎜ 2 + 2 + 2 ⎟
⎜ ∂x
∂t ∂x
∂y
∂z
∂x ∂x
∂y
∂z ⎟⎠
⎦
⎝
⎣ ∂t
where we used the result written in Problem 7.71. Then
∂p
∂
∂2 u
ρ u′v′ = − + μ 2
∂y
∂x
∂y
In terms of stresses
−
∂p ∂τ
∂τ turb
= − + lam
∂y
∂x
∂y
∂p ∂
= (τ turb + τ lam )
∂x ∂y
or
v′ = v − v .
u = Σui /11 = 16.2 m/s.
t
u′
u′2
v′
v′2
u′v′
0
-.1
.01
9.5
.02
-5.6
.03
1.1
.04
-11
.05
-6
.06
0.9
.07
12.4
.08
-9.5
.09
3
.1
5.4
.01
90.2
31.4
1.2
121
36
.81
153.8
90.2
9
3.2
-3.8
-7
5.1
5.7
-4.4
0.2
8.3
-3.6
-6.6
29.2 u′2 = 51.2 m 2 /s 2
3.1
10.2
14.4
49
26
32.5
19.4
.04
68.9
13.0
43.6
9.6
-.32
-36.1
39.2
5.6
-62.7
26.4
.2
102.9
34.2
-19.8
16.7 u′v′ = 9.7 m2 /s 2
7.77
u′v′ = ν
7.78
η=−
v = Σvi /11 = −1.6 m/s.
u′ = u − u ,
7.76
v′2 = 26.1 m 2 /s 2
∂u τ
∂u r Δp
− =ν
−
∂y ρ
∂y 2 Lρ
76.9 − 60.7 0.69
8
= 1.6 ×10−4
−
= −26.3 ft 2 / sec 2
0.09
2 30 × .0035
u′v′
−26.3
=−
= 0.146 ft 2 /s.
(76.9 − 60.7) / 0.09
du / dy
Kuv =
u′v′
u′2 v′2
=
−26.3
= −0.118
316 156
A m = η / ∂u / ∂y = 0.146 / (76.9 − 60.7) / 0.09 = 0.0285 ft or 0.342 in.
162
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Chapter 7 / Internal Flows
7.79
1
2π
v′ = sin
t.
2
0.2
u′v′ =
=
η=−
1
0.2
1
0.8
0.2
∫
0
0.2
1
⎛1
⎞⎛ 1
⎞
⎜ sin10π t ⎟⎜ sin10π t ⎟ dt =
0.8
⎝2
⎠⎝ 2
⎠
⎛1
⎞
1
∫ sin
2
10π t dt
0
1 ⎛ 0.2 ⎞
2 2
⎟ = 0.125 m /s .
2 ⎠
∫ ⎝⎜ 2 + 2 cos 20π t ⎠⎟ dt = 0.8 ⎝⎜
0
0.125
u′v′
=−
= 0.0125 m 2 /s.
du / dy
−10
η
Am =
∂u / ∂y
=
0.0125
= 0.0354 m or 3.54 cm
10
1
u′2 = sin 2 10π t = v′2 ∴ u′2 = v′2 = 0.125.
4
7.80
0.2
Kuv =
u′v′
u′2 v′2
=
0.125
= 1.0
0.125 × 0.125
e 0.26
0.02 × 0.2
= 4000,
=
= 0.0013.
−6
ν
D 200
10
From the Moody diagram, this is effectively a “smooth” pipe so we conclude that
δ ν > e.
0.2 × 0.2
e
b) Re =
= 40 000,
= 0.0013.
−6
D
10
From the Moody diagram, this is in the transition zone where δ ν may be near, in
magnitude, to e. The pipe is rough.
e
c) Re = 400 000,
= 0.0013 and the Moody diagram indicates a rough pipe.
D
a) Re =
VD
=
6 × 0.1
e
= 0.001, then the pipe is
= 5455. From the Moody diagram, if
−4
ν
D
1.1 × 10
“smooth.” Thus, e = 0.001 D = 0.001× 100 = 0.1 mm .
VD
7.81
Re =
=
7.82
a) Using Re = 4000 and e /D = 0.0013, the Moody diagram provides f = 0.04. Then
f
0.04
τ 0 = ρV 2 =
1000 × 0.02 2 = 0.002 Pa
8
8
τ
0.002
∴ uτ = 0 =
= 0.001414 m/s
1000
ρ
Eq. 7.6.16 gives
ur
⎛
⎞
umax = uτ ⎜ 2.44 ln τ o + 5.7 ⎟
ν
⎝
⎠
0.001414 × 0.1
⎛
⎞
= 0.001414 ⎜ 2.44 ln
+ 5.7 ⎟ = 0.0251 m/s
10
⎝
⎠
163
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
b) With Re = 40 000, e /D = 0.0013 the Moody diagram gives f = 0.026. Then
τ0 =
0.026
1000 × 0.2 2 = 0.130 Pa
8
uτ = 0.130/1000 = 0.01140 m/s
Eq. 7.6.17 gives
r
0.1
⎛
⎞
⎛
⎞
+ 8.5 ⎟ = 0.262 m/s
umax = uτ ⎜ 2.44 ln o + 8.5 ⎟ = 0.0114 ⎜ 2.44 ln
0.00026
e
⎝
⎠
⎝
⎠
c) With Re = 400 000, e /D = 0.0013 we find f = 0.022:
τ0 =
0.022
1000 × 2 2 = 11 Pa ,
8
uτ = 11/1000 = 0.105 m/s
r
0.1
⎛
⎞
⎛
⎞
+ 8.5 ⎟ = 2.41 m/s
umax = uτ ⎜ 2.44 ln 0 + 8.5 ⎟ = 0.105 ⎜ 2.44 ln
0.00026
e
⎝
⎠
⎝
⎠
7.83
Here n = 7 so that, from Eq. 7.6.21, f =
1
n
2
=
1
= 0.0204. And from Eq. 7.6.20,
49
V = 4.9 m/s.
1
1
a) ∴τ 0 = ρ V 2 f = × 1000 × 4.92 × 0.0204 = 61.2 Pa.
8
8
b)
1
du
= 9.2 × y −6/7
7
dy
c)
∂p
Δp
2τ
2 × 61.2
=−
=− 0 =−
= −2450 Pa/m.
∂x
0.05
L
ro
y =0
= ∞. (The profile is not too good near the wall!)
1
d) τ varies linearly with r. ∴τ r =2.5 cm = τ 0 = 30.6 Pa.
2
τ = ρη
7.84
V=
(r0 = 5 cm)
du
1
⎛
⎞
. ∴ 30.6 = 1000η ⎜ 9.2 × × 0.025−6/7 ⎟ . ∴η = 9.85 × 10−4 m 2 /s.
7
dy
⎝
⎠
2.5
Q
=
= 18.33 fps.
A π × (2.5/12) 2
Re =
VD
ν
=
18.33 × 5/12
= 7.09 × 105.
−5
1.08 ×10
From Table 7.1, n ≅ 8.5. From Eq. 7.6.20
umax =
(n + 1)(2n + 1)
9.5 × 18
×V =
× 18.33 = 21.7 fps.
2
2n
2 × 8.52
164
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Chapter 7 / Internal Flows
7.85
n = 5: V =
2n2
2 × 52
umax =
umax = 0.758umax .
(n + 1)(2n + 1)
6 ×11
ro
u dA ∫0
α=∫
=
3
3
V A
=
4.59
ro2+3/ n
3
umax
y3/ n 2π (ro − y )(dy )
3
π ro2ro3/ n
0.7583 umax
=
2
0.7583 ro2+3/ n
3/ n
1+3/ n
∫0 ( ro y − y ) dy
ro
⎡ ro2+3/ n
ro2+3/ n ⎤
1 ⎞
⎛ 1
−
−
⎢
⎥ = 4.59 ⎜
⎟ = 1.10
1
3/
2
3
/
1.6
2.6
n
n
+
+
⎝
⎠
⎣
⎦
2 × 102
umax = 0.866 umax .
With n = 10, V =
11× 21
α=
7.86
2 ⎛ 1
1 ⎞
−
⎟ = 1.03.
3⎜
0.866 ⎝ 1.3 2.3 ⎠
With n = 7, Eq. 7.6.21 gives f =
1
1
=
= 0.0204.
2
n
49
1
1
∴τ 0 = ρ V 2 f = ×1000 × 102 × 0.0204 = 1020 Pa.
8
2
Since τ varies linearly with r and is zero at r = 0,
τ = τ0
τ lam
y
τturb
C (r = 0)
τlam
τ
r 1020
=
r = 20 400 r.
r0 0.05
1 1 −6/7 ⎤ 10−3 ×12.24 −6/7
∂u
−3 ⎡
y
=μ
= 10 ⎢umax 1/7 y ⎥ =
= 0.00268y −6/7 .
1/7
7 r0
∂y
⎣
⎦ 7 × 0.05
τ turb = τ − τ lam = 20 400 r − 0.00268 y −6/7 where y + r = 0.05.
τ lam ( y ) is good away from y = 0 (the wall).
τ lam (.00625) = .21, τ lam (.003125) = .38, τ lam (.00156) = .68, τ lam (.00078) = 1.24
dp
2τ
2 × 1020
=− 0 =−
= −40 800 Pa/m.
0.05
dx
r0
7.87
19.63 × 0.8
= 71 400.
ν
2.2 × 10−4
e
1
= 0. ∴ From Fig. 7.13 f = 0.019. τ 0 = × 917 × 19.632 × 0.019 = 840 Pa.
b)
D
8
2
2n
19.63 × 8 × 15
c) V =
umax . ∴ umax =
= 24.0 m/s
(n + 1)(2n + 1)
2 × 49
a) V =
1.2
Q
=
= 19.63 m/s.
A π × 0.42
Re =
VD
=
165
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
d) δν =
5ν 5 × 2.2 × 10−4
=
= 0.0015 m where uτ = 840/917 = 0.957 m/s.
0.957
uτ
0.957 × 0.4
⎡
⎤
+ 5.7 ⎥ = 22.9 m/s.
e) umax = 0.957 ⎢ 2.44 ln
−4
2.2 ×10
⎣
⎦
Note that u max is nearly the same using either form of the velocity profile.
7.88
We must find uτ :
V=
1.2
Q
=
= 19.63 m/s.
A π × 0.42
e 0.26
=
= 0.000325.
D 800
τ0 =
Re =
19.63 × 0.8
= 71 400
2.2 × 10−4
∴ Moody diagram ⇒ f = 0.021
1
1
fρV 2 = × 0.021 × 917 × 19.63 2 = 928 Pa
8
8
uτ = τ 0 /ρ = 928/917 = 1.006 m/s.
1.006 × 0.4
⎡
⎤
∴ umax = 1.006 ⎢ 2.44 ln
+ 5.7 ⎥ = 24.2 m/s
−4
2.2 ×10
⎣
⎦
7.89
a)
Δp 1.5 × 144
2τ
2τ 0
=
= 14.4 = 0 =
. ∴ τ 0 = 1.2 psf. ∴ uτ =
L
r0
2 / 12
15
1.2
= 0.786 fps.
1.94
ur
0.786 × 2 /12
⎡
⎤
⎡
⎤
+ 5.7 ⎥ = 23.2 fps.
b) umax = uτ ⎢ 2.44 ln τ 0 + 5.7 ⎥ = 0.786 ⎢ 2.44 ln
−5
ν
0.736 ×10
⎣
⎦
⎣
⎦
2
2n
2 × 64
umax =
× 23.2 = 19.41 fps.
c) Assume n = 8 : V =
(n + 1)(2n + 1)
9 ×17
d) Check Re :
Re =
VD
ν
=
19.41× 4 /12
= 8.79 ×105 . ∴ n ≅ 8 is OK.
−5
0.736 × 10
e) Q = AV = π × (2/12) 2 ×19.41 = 1.69 cfs.
7.90
a) From a control volume of the 10-m section of pipe
πD 2
0.12 × 5000
= 15 Pa
4
4 × 10
1
1
= 0.0204 .
Assume n = 7. Then Eq. 7.6.21 gives f = 2 =
n
49
From Eq. 7.3.19,
Δp
V2 =
= τ 0 πDL.
∴τ 0 =
8τ 0
8 × 15
=
= 6.41 or V = 2.53 m/s
ρ f 917 × 0.0204
166
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Chapter 7 / Internal Flows
2.53 × 0.12
= 1381. This suggests laminar flow. Use Eq. 7.3.14:
2.2 × 10 −4
r02 Δp
0.062 × 5000
=
= 1.12 m/s.
V=
8μ L 8 × (917 × 2.2 × 10−4 ) × 10
Check: Re =
1.12 × 0.12
= 611. OK.
2.2 × 10 −4
Finally, Q = AV = π × 0.062 × 1.12 = 0.0127 m3 /s
Check: Re =
b) From a control volume or Eq. 7.6.18:
Assume n = 6. Then Eq. 7.6.21 gives
and
V2 =
Check: Re =
0.06(20 000)
= 60 Pa
2 × 10
1
f=
= 0.0278
36
τ0 =
8τ 0
8 × 60
=
= 18.83 or V = 4.34 m/s.
ρ f 917 × 0.0278
4.34 × 012
.
= 2370. OK.
2.2 × 10 − 4
Q = π × 0.062 × 4.34 = 0.0491 m3 /s
c) τ 0 =
0.06(200 000)
= 600 Pa
2 ×10
Assume n = 7. Then f =
V=
Check: Re =
1
= 0.0204 . And
49
8τ 0
8 × 600
=
= 16.0 m/s
917 × 0.0204
ρf
16.0 × 012
.
= 8740. OK.
2.2 × 10 − 4
Q = π × 0.062 × 16 = 0.181 m3 /s
7.91
u
u max
⎛y⎞
=⎜ ⎟
⎝ r0 ⎠
1/ 8
23.2 fps
n=8
Eq. 7.6.15
167
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Chapter 7 / Internal Flows
Turbulent Flow in Pipes and Conduits
7.92
a) Re =
VD
ν
=
(0.020 / π × 0.042 ) × 0.08
10
−6
= 3.18 × 105.
e
= 0. ∴ f = 0.0143.
D
b) Eq. 7.6.26 provides f by trial-and-error. Try f = 0.0143 from Moody’s diagram:
(
)
1
= 0.86 ln 3.18 ×105 0.0143 − 0.8. ∴ f = 0.0146.
f
Another iteration may be recommended but this is quite close. The value for f is
essentially the same using either method. The equations could, however, be
programmed on a computer.
7.93
a) Re =
VD
ν
=
(0.03 / π × 0.052 ) × 0.1
1.14 × 10
−6
= 3.35 × 105.
e 0.26
=
= 0.0026. ∴ f = 0.026.
D 100
b) Eq. 7.6.28 provides f by trial-and-error. Try f = 0.026 from the Moody diagram:
⎛ 0.26
⎞
1
2.51
= −0.86 ln ⎜
+
⎟ = 6.189. ∴ f = 0.0261
5
f
⎝ 3.7 × 100 3.35 × 10 × 0.026 ⎠
The value for f is essentially the same using either method. The equations could,
however, be programmed on a computer.
7.94
7.95
0.025 × 0.04
= 1000. ∴ laminar. ∴ f =
10 −6
e 0.26
0.25 × 0.04
= 10 000 ,
=
= 0.0065.
b) Re =
−6
D
40
10
e 0.26
2.5 × 0.04
c) Re =
= 100 000 ,
=
= 0.0065.
−6
D
40
10
e
= 0.0065. ∴ f = 0.033
d) Re = 10 6 ,
D
a) Re =
V=
0.02
Q
=
= 2.55 m/s.
A π × 0.052
a) Re =
2.55 × 0.1
= 2.55 ×105.
−6
10
Δp = γ f
64ν
64 × 10 −6
=
= 0.064
V D 0.025 × 0.04
∴ f = 0.04
∴ f = 0.034
100 2.552
L V2
=ρf
×
= 3251ρ f .
0.10
2
D 2g
e 0.046
=
= 0.00046.
D 100
∴ f = 0.0185
∴Δp = 1000 × 0.0185 × 3251 = 60 100 Pa.
0.9
⎧
⎛ 10−6 × 0.1 ⎞ ⎤ ⎪⎫
0.022 ×100 ⎪ ⎡ 0.046
Δp = γ hL = 9800 × 1.07
+ 4.62 ⎜
⎨ln ⎢
⎟ ⎥⎬
0.02 ⎠ ⎥ ⎪
9.8 × 0.15 ⎪ ⎢ 3.7 ×100
⎝
⎦⎭
⎩ ⎣
−2
= 59 200 Pa
168
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
b) Re =
2.55 × 0.1
= 3.1× 103.
−5
8.1×10
e
= 0.00046.
D
∴ f = 0.042.
∴Δp = 1260 × 0.042 × 3251 = 172 000 Pa.
Δp = γ hL
.9
⎧
⎛ 8.1×10−5 × 0.1 ⎞ ⎤ ⎫⎪
0.022 ×100 ⎪ ⎡ 0.046
= 12 360 ×1.07
+ 4.62 ⎜
⎨ln ⎢
⎟ ⎥⎬
0.02
9.8 × 0.15 ⎪ ⎢ 3.7 × 100
⎝
⎠ ⎦⎥ ⎪⎭
⎩ ⎣
c) Re =
2.55 × 0.1
= 1.06 × 103.
−4
2.4 ×10
∴ laminar.
∴f =
−2
= 180 400 Pa
64
= 0.06.
Re
∴Δp = 917 × 0.06 × 3251 = 179 000 Pa.
Eq. 7.6.29 is not applicable if Re < 3000.
d) Re =
2.55 × 0.1
= 1.02 ×105.
−6
2.5 ×10
e
= 0.00046.
D
∴ f = 0.02.
∴Δp = 809 × 0.02 × 3251 = 52 600 Pa.
.9
⎧
⎛ 2.5 ×10−6 × 0.1 ⎞ ⎤ ⎪⎫
0.02 ×100 ⎪ ⎡ 0.046
Δp = 7940 ×1.07
+ 4.62 ⎜
⎨ln ⎢
⎟ ⎥⎬
0.02
9.8 × 0.15 ⎪ ⎢ 3.7 × 100
⎝
⎠ ⎦⎥ ⎪⎭
⎩ ⎣
−2
2
7.96
V=
Q
0.06
=
= 4.89 fps.
A π (0.75 /12) 2
hL = f
Re =
VD
ν
=
= 52 800 Pa
4.89 × 1.5 /12
= 5 × 104.
−5
1.22 ×10
L V2
600 4.892
= f
= 1782 f .
1.5 /12 2 × 32.2
D 2g
a)
e 0.00085
=
= 0.0068
D 1.5/12
b)
e 0.0005
=
= 0.004
D 1.5/12
c)
e 0.00015
=
= 0.0012
D 1.5/12
d)
e
=0
D
∴ f = 0.0205.
∴ f = 0.035.
∴ f = 0.033.
∴ hL = 1782 × 0.035 = 62 ft.
∴ hL = 1782 × 0.033 = 59 ft.
∴ f = 0.0245.
∴ hL = 1782 × 0.0245 = 42 ft.
∴ hL = 1782 × 0.0205 = 36.5 ft.
169
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Chapter 7 / Internal Flows
7.97
a) ρ =
p
500
=
= 5.95 kg/m3 .
RT 0.287 × 293
∴V =
ν=
m
1.2
=
= 25.7 m/s.
ρ A 5.95π × 0.052
∴ f = 0.032. Δp = γ f
μ 1.81×10−5
=
= 3.04 × 10−4 m 2 /s.
ρ
5.95
Re =
VD
ν
=
25.7 × 0.1
= 8450.
3.04 × 10−4
L V2
10 25.7 2
= (5.95 × 9.81) × 0.032
×
= 6290 Pa.
D 2g
0.1 2 × 9.81
0.9
⎧
⎛ 3.04 ×10−4 × 0.1 ⎞ ⎤ ⎫⎪
0.202 × 10 ⎪ ⎡
⎢
Δp = γΔh = 5.95 × 9.8 × 1.07
⎨ln 0 + 4.62 ⎜
⎟ ⎥⎬
0.202
9.8 × 0.15 ⎪ ⎢
⎝
⎠ ⎥⎦ ⎪⎭
⎩ ⎣
−2
2
b) ρ =
p
500
=
= 9.03 kg/m3 .
RT 0.189 × 293
∴V =
ν=
1.2
m
=
= 16.92 m/s.
ρ A 9.03π × 0.052
= 6360 Pa
μ 1.2 ×10−5
=
= 1.329 × 10−4 m 2 /s.
ρ
9.03
Re =
VD
ν
=
16.92 × 0.1
= 12 730.
1.329 × 10−4
L V2
10 16.922
∴ f = 0.029. Δp = γ f
= (9.03 × 9.81) × 0.029
×
= 3750 Pa.
D 2g
0.1 2 × 9.81
0.9
⎧
⎛ 1.33 × 10−4 × 0.1 ⎞ ⎤ ⎫⎪
0.133 ×10 ⎪ ⎡
Δp = 9.03 × 1.07
⎨ln ⎢0 + 4.62 ⎜
⎟ ⎥⎬
0.133
0.15
⎢
⎝
⎠ ⎥⎦ ⎪⎭
⎪⎩ ⎣
2
−2
= 3740 Pa
p
500
μ 7 ×10−6
=
= 0.414 kg/m3 . ν = =
= 17 × 10−6 m 2 /s.
ρ
RT 4.124 × 293
0.414
1.2
369 × 0.1
V=
= 369 m/s. Re =
= 2.2 × 106.
2
0.414π × 0.05
17 × 10−6
L V2
10 3692
∴ f = 0.01. Δp = γ f
= (0.414 × 9.81) × .01
= 28 000 Pa.
D 2g
0.1 2 × 9.81
c) ρ =
0.9
⎧
⎛ 17 × 10−6 × 0.1 ⎞ ⎤ ⎫⎪
2.92 ×10 ⎪ ⎡
Δp = 0.414 ×1.07
⎨ln ⎢ 0 + 4.62 ⎜
⎟ ⎥⎬
2.9
0.15 ⎪ ⎢
⎥⎪
⎝
⎠
⎦⎭
⎩ ⎣
7.98
V=
−2
= 28 700 Pa
0.08
Q
e 0.15
=
= 4.73 m/s.
=
= 0.001.
2
A π × 0.075
D 150
L V2
10
4.732
Δp = γ f
= (917 × 9.81) f
×
= 6.84 × 105 f .
D 2g
0.15 2 × 9.81
170
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Chapter 7 / Internal Flows
4.73 × 0.15
64
= 322. ∴ laminar and Δp = 6.84 × 105 ×
= 1.36 × 105 Pa.
0.0022
Re
Eq. 7.6.29 is not applicable to a laminar flow.
a) Re =
b) Re =
4.73 × 0.15
= 3220.
0.00022
∴ f = 0.042.
0.082 × 10
Δp = 9.8 × 917 ×1.07
9.8 × 0.155
Δp = 6.84 × 105 × 0.042 = 29 400 Pa.
0.9
⎧⎪ ⎡ 0.15
⎛ 0.00022 × 0.15 ⎞ ⎤ ⎫⎪
+
ln
4.62
⎨ ⎢
⎜
⎟ ⎥⎬
0.08
⎝
⎠ ⎥⎦ ⎪⎭
⎪⎩ ⎢⎣ 3.7 × 50
−2
= 29 400 Pa.
c) Re =
4.73 × 0.15
= 16 300.
0.000044
∴ f = 0.029. Δp = 6.84 × 105 × 0.029 = 20 000 Pa.
0.9
⎧
⎛ 4.4 × 10−5 × 0.15 ⎞ ⎤ ⎫⎪
0.08 × 10 ⎪ ⎡ 0.15
Δp = 9.8 × 917 ×1.07
+ 4.62 ⎜
⎨ln ⎢
⎟ ⎥⎬
0.08
9.8 × 0.155 ⎪ ⎢ 3.7 × 50
⎝
⎠ ⎥⎦ ⎭⎪
⎩ ⎣
= 20 700 Pa.
−2
2
d) Re =
4.73 × 0.15
= 46 000.
1.5 × 10−5
∴ f = 0.0245. Δp = 6.84 × 105 × 0.0245 = 17 000 Pa.
0.9
⎧
⎛ 1.5 × 10−5 × 0.15 ⎞ ⎤ ⎫⎪
0.08 × 10 ⎪ ⎡ 0.15
Δp = 9.8 × 917 ×1.07
+ 4.62 ⎜
⎨ln ⎢
⎟ ⎥⎬
0.08
9.8 × 0.152 ⎪ ⎢ 3.7 × 50
⎝
⎠ ⎥⎦ ⎭⎪
⎩ ⎣
= 16 800 Pa.
−2
2
7.99
V=
0.4/60
Q
=
= 3.395 m/s.
A π × 0.0252
Re =
3.395 × 0.05
= 1.7 × 105.
10−6
L V2
10 3.3952
= 9800 f
×
= 36 000. ∴ f = 0.031.
a) Δp = γ f
D 2g
0.05 2 × 9.8
e
∴ = 0.005 and e = 0.005 × 50 = 0.25 mm. ∴ Cast iron
D
b) Δp = 9800 f
∴
10 3.3952
×
= 24 000.
0.05 2 × 9.8
∴ f = 0.021.
e
=.0009 and e =.0009 × 50 =.045 mm .
D
∴ Wrought iron
171
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
c) Δp = 9800 f
∴
7.100 V =
10 3.3952
×
= 19 000.
0.05 2 × 9.8
∴ f = 0.0165.
e
= 0.0003 and e = 0.0015.
D
Q
0.3
=
= 8.8 fps.
A π (1.25/12) 2
Re =
∴ Drawn tubing
8.8 × 2.5/12
= 1.3 × 105.
−5
1.41× 10
e
= 0.
D
L V2
∴ f = 0.0167. ∴Δp − γΔh = γ f
. (Recall: Δp = p1 − p 2 , Δh = h2 − h1 )
D 2g
∴Δp = 62.4 × 0.0167
7.101 V =
For
300
8.82
×
+ 62.4 × 300sin 30D = 11,200 psf .
2.5/12 2 × 32.2
0.02
Q
=
= 10.2 m/s.
A π × 0.0252
e
= 0 : f = 0.013.
D
For
Re =
10.2 × 0.05
= 7.7 × 105.
−6
0.661× 10
e 0.046
=
= 0.00092.
50
D
e
= 0.00092 : f = 0.0197.
D
This is a significant difference. ∴ The roughness is significant.
7.102 V =
5
Q
=
= 9.95 m/s.
A π × 0.42
Re =
9.95 × 0.8
= 8 × 106.
−6
10
e 1.6
=
= 0.002 (using an average “e” value). ∴ f = 0.0237.
D 800
∴Δp = γ f
L V2
100 9.952
= 9810 × 0.0237 ×
×
= 147 000 Pa.
D 2g
0.8 2 × 9.81
⎡ D 5 hL ⎤
7.103 Use Eq. 7.6.30: Q = −0.965 ⎢ g
⎥
L ⎦
⎣
a) hL =
Δp
γ
=
500 000
= 51.0 m.
9810
Δp
γ
=
500 000
= 40.5 m.
12 350
0.5
⎡ e
⎛ 3.17ν 2 L ⎞ ⎤
+⎜
ln ⎢
⎟ ⎥.
⎢ 3.7 D ⎜⎝ gD3hL ⎟⎠ ⎥
⎣
⎦
e 0.26
=
= 0.0026. ν = 10−6 m 2 /s.
D 100
⎡ 9.81× 0.15 × 51 ⎤
∴ Q = −0.965 ⎢
⎥
200
⎣
⎦
b) hL =
0.5
0.5
⎡ 0.0026 ⎛ 3.17 × 10−12 × 200 ⎞0.5 ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.035 m /s.
3
⎢ 3.7
⎝ 9.81× 0.1 × 51 ⎠ ⎥⎦
⎣
e
=0.0026. ν = 0.0012 m 2 /s.
D
172
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Chapter 7 / Internal Flows
⎡ 9.81× 0.15 × 40.5 ⎤
∴ Q = −0.965 ⎢
⎥
200
⎣
⎦
c) hL =
Δp
γ
=
500 000
= 55.6 m.
9000
1/2
⎡ 9.81× .15 × 55.6 ⎤
∴ Q = −.965 ⎢
⎥
200
⎣
⎦
d) hL =
Δp
γ
=
500 000
= 63.1 m.
7930
0.5
⎡ 0.0026 ⎛ 3.17 × .00122 × 200 ⎞0.5 ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.013 m /s.
3
3.7
⎢
⎝ 9.81× 0.1 × 40.5 ⎠ ⎥⎦
⎣
e
=0.0026. ν = 1.1×10−4 m 2 /s.
D
⎡ .0026 ⎛ 3.17 × 1.122 × 10−8 × 200 ⎞.5 ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.027 m /s.
3
9.81× .1 × 55.6
⎢ 3.7 ⎝
⎠ ⎥⎦
⎣
e
=0.0026. ν = 1.85 × 10−6 m 2 /s.
D
.5
⎡ .0026 ⎛ 3.17 × 1.852 × 10−12 × 200 ⎞.5 ⎤
⎡ 9.81× .15 × 63.1 ⎤
3
∴ Q = −.965 ⎢
+⎜
⎟ ⎥ = 0.039 m /s.
⎥ ln ⎢
3
200
3.7
×
×
9.81
.1
63.1
⎢
⎥
⎣
⎦
⎝
⎠ ⎦
⎣
7.104 Use Eq. 7.6.30: hL =
Δp
γ
=
200 000
= 20.4 m. ν = 10−6 m 2 /s.
9810
⎡ 9.81× 0.045 × 20.4 ⎤
a) Q = −0.965 ⎢
⎥
100
⎣
⎦
0.5
0.5
⎡ 0.26
⎛ 3.17 ×10−12 × 100 ⎞ ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.0027 m /s.
3
⎢ 3.7 × 40 ⎝ 9.81× 0.04 × 20.4 ⎠ ⎥
⎣
⎦
0.5
⎡ 0.046 ⎛ 3.17 ×10−12 ×100 ⎞0.5 ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.0033 m /s.
⎢ 3.7 × 40 ⎝ 9.81× 0.043 × 20.4 ⎠ ⎥
⎣
⎦
0.5
⎡ ⎛ 3.17 × 10−12 × 100 ⎞0.5 ⎤
3
ln ⎢0 + ⎜
⎟ ⎥ = 0.0038 m /s.
3
⎢ ⎝ 9.81× 0.04 × 20.4 ⎠ ⎥
⎣
⎦
⎡ 9.81× 0.045 × 20.4 ⎤
b) Q = −0.965 ⎢
⎥
100
⎣
⎦
⎡ 9.81× 0.045 × 20.4 ⎤
c) Q = −0.965 ⎢
⎥
100
⎣
⎦
7.105 Use Eq. 7.6.30:
hL = 40 − 10 = 30 m, ν = 1.31× 10−6 m 2 /s,
3.17 × (1.31×10−6 ) 2 × 200
= 3.7 × 10−12.
9.8 × 30
⎡ 9.8 × 0.045 × 30 ⎤
a) Q = −0.965 ⎢
⎥
200
⎣
⎦
0.5
0.5
⎡ 0.15
⎛ 3.7 ×10−12 ⎞ ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.0025 m /s.
3
⎢ 3.7 × 40 ⎝ 0.04
⎠ ⎥⎦
⎣
⎡ 9.8 × 0.085 × 30 ⎤
b) Q = −0.965 ⎢
⎥
200
⎣
⎦
0.5
0.5
⎡ 0.15
⎛ 3.7 ×10−12 ⎞ ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.0157 m /s.
⎢ 3.7 × 80 ⎝ 0.083 ⎠ ⎥
⎣
⎦
173
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
⎡ 9.8 × 0.125 × 30 ⎤
c) Q = −0.965 ⎢
⎥
200
⎣
⎦
0.5
0.5
⎡ 0.15
⎛ 3.7 × 10−12 ⎞ ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.0459 m /s.
3
×
3.7
120
⎢
⎝ 0.12
⎠ ⎦⎥
⎣
0.5
0.5
⎡ 0.15
⎛ 3.7 × 10−12 ⎞ ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.0979 m /s.
3
⎢ 3.7 ×160 ⎝ 0.16
⎠ ⎥⎦
⎣
⎡ 9.8 × 0.165 × 30 ⎤
d) Q = −0.965 ⎢
⎥
200
⎣
⎦
e
0.046
=
= 0.000104.
3.7 D 3.7 ×120
p
Δp
200
400
a) ρ =
=
= 2.23 kg/m3 . ∴ hL =
=
= 18.3 m.
γ
9.81× 2.23
RT 0.287 × 313
7.106 Use Eq. 7.6.30:
ν=
= ρQ.
m
2 ×10−5
= 9 × 10−6 m 2 /s.
2.23
⎡ 9.81× .125 ×18.3 ⎤
Q = −.965 ⎢
⎥
400
⎣
⎦
0.5
0.5
⎡
⎛ 3.17 × 92 ×10−12 × 400 ⎞ ⎤
ln ⎢.000104 + ⎜
⎟ ⎥ = 0.0235.
3
×
×
9.81
0.12
18.3
⎢
⎝
⎠ ⎥⎦
⎣
= 0.052 kg/s.
∴m
b) ρ =
p
200
=
= 3.38 kg/m3 .
RT 0.189 × 313
ν=
∴ hL =
400
= 12.1 m.
9.81× 3.38
μ 1.3 ×10−5
=
= 3.8 ×10−6 m 2 /s.
ρ
3.38
⎡ 9.81× .125 ×12.1 ⎤
Q = −.965 ⎢
⎥
400
⎣
⎦
0.5
0.5
⎡
⎛ 3.17 × 3.82 ×10−12 × 400 ⎞ ⎤
ln ⎢.000104 + ⎜
⎟ ⎥ = 0.0205.
9.81× 0.123 ×12.1 ⎠ ⎥
⎢
⎝
⎣
⎦
= 0.069 kg/s.
∴m
c) ρ =
p
200
400
=
= 0.155 kg/m3 . ∴ hL =
= 263 m.
9.81× 0.155
RT 4.12 × 313
μ 7.2 ×10−6
ν= =
= 46 × 10−6 m 2 /s.
ρ
0.155
⎡ 9.81× .125 × 263 ⎤
Q = −.965 ⎢
⎥
400
⎣
⎦
0.5
0.5
⎡
⎛ 3.17 × 462 × 10−12 × 400 ⎞ ⎤
ln ⎢.000104 + ⎜
⎟ ⎥ = 0.086.
3
×
×
9.81
0.12
263
⎢
⎝
⎠ ⎥⎦
⎣
= 0.0133 kg/s.
∴m
174
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
7.107 a) Use an intermediate value for e = 0.005 ft.
Q2
30 × 144
600
L Q2
f
or
or fQ 2 = 4694.
=
D 2A 2 g
62.4
4 2(π × 2 2 ) 2 × 32.2
hL = f
Guess: f = 0.02. Then Q = 484 ft3/sec.
Check: V =
484
= 38.6 fps,
π × 22
Re =
38.6 × 4
= 12.6 × 106 ,
1.22 ×10−5
e 0.005
=
= 0.00125.
4
D
∴ f = 0.021.
This is good. Use f = 0.021 and Q = 473 ft 3 /sec.
e
30 ×144
0.005
= 69.2 ft.
=
= 0.00034 using an average
62.4
3.7 D 3.7 × 4
value for e. ν = 1.22 × 10−5 ft 2 /sec.
b) Use Eq. 7.6.30: hL =
⎡ 32.2 × 45 × 69.2 ⎤
Q = −0.965 ⎢
⎥
600
⎣
⎦
0.5
0.5
⎡
⎛ 3.17 ×1.222 ×10−10 × 600 ⎞ ⎤
ln ⎢.00034 + ⎜
⎟ ⎥ = 475 cfs.
3
×
×
32.2
4
69.2
⎢
⎝
⎠ ⎥⎦
⎣
5.2 ⎤
2 4.75
⎡
1.25 ⎛ LQ ⎞
9.4 ⎛ L ⎞
7.108 Use Eq. 7.6.31: e = 0 D = 0.66 ⎢ e ⎜
⎟ +ν Q ⎜
⎟ ⎥
gh
gh
⎢
L
L
⎝
⎠ ⎦⎥
⎝
⎠
⎣
Δp
a) hL =
γ
=
200 000
= 20.4 m.
9810
Δp
γ
=
200 000
= 16.2 m.
12 350
0.04
= 0.032 m.
ν = 7.9 × 10−5 m 2 /s.
5.2
⎡
100
⎞ ⎤
9.4 ⎛
−5
D = 0.66 ⎢7.9 ×10 × 0.002 ⎜
⎟ ⎥
⎝ 9.81×16.2 ⎠ ⎦⎥
⎣⎢
c) hL =
Δp
γ
=
200 000
= 25.2 m.
7930
.
ν = 10−6 m 2 /s.
5.2
⎡ −6
100
⎞ ⎤
9.4 ⎛
D = 0.66 ⎢10 × 0.002 ⎜
⎟ ⎥
⎝ 9.81× 20.4 ⎠ ⎥⎦
⎢⎣
b) hL =
0.04
0.04
= 0.040 m.
ν = 2.1× 10−6 m 2 /s.
5.2
⎡
100
⎞ ⎤
9.4 ⎛
−6
D = 0.66 ⎢ 2.1× 10 × 0.002 ⎜
⎟ ⎥
⎝ 9.81× 25.2 ⎠ ⎥⎦
⎢⎣
0.04
= 0.031 m.
175
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
d) hL =
Δp
γ
=
200 000
= 22.2 m.
9020
ν = 3.8 × 10−5 m 2 /s.
5.2
⎡
⎛ 100
⎞ ⎤
D = 0.66 ⎢3.8 × 10−5 × 0.0029.4 ⎜
⎟ ⎥
⎝ 9.81× 22.2 ⎠ ⎦⎥
⎣⎢
0.04
= 0.036 m.
7.109 a) Use an intermediate value for e: e = 1.6 mm.
hL = f
300
52
L V2
or 20 = f
or 31.0 f = D5 .
D 2g
D (π D 2 /4) 2 × 2 × 9.8
Guess: f = 0.02. Then D = 0.91 m .
Check: V =
e 1.6
5
7.69 × 0.91
= 7.69 m/s, Re =
= 7.0 × 106 ,
=
= 0.00176.
−6
D 910
π × 0.455
10
∴ f = 0.0225.
This is good. Use f = 0.0225 and D = 0.93 m. Choose a standard diameter, i.e., D = 1.0 m
(It cannot be smaller than 0.93 m).
b) Use Eq. 7.6.31: Select an average “e”: e = 1.6 mm. hL = 20 m .
5.2
2 4.75
⎡
300 ⎞ ⎤
1.25 ⎛ 300 × 5 ⎞
9.4 ⎛
−6
⎢
D = 0.66 0.0016 ⎜
⎟ + 10 × 5 ⎜
⎟ ⎥
9.81
20
9.81
20
×
×
⎢
⎝
⎠ ⎥
⎝
⎠
⎣
⎦
A diameter of 1 m would be selected.
0.04
= 0.96 m.
L V2
0.4/60
1200 (0.00849/D 2 ) 2
2
=
=
=
D
h
f
f
0.00849
/
,
or
3
or
L
D 2g
D
2 × 9.8
π D 2 /4
f = 680 D 5 .
Guess: f = 0.02. Then the above equation gives D = 0 .124 m.
7.110 a) V =
Check: V = 0.551 m/s, Re =
0.551× 0.25
e 0.0015
= 1.05 × 105 ,
=
= 1.2 × 10−5.
−6
124
D
1.31× 10
∴ f = 0.0175
Use f = 0.0175 and D = 0.121 m.
b) Use Eq. 7.6.31:
9.4
5.2
2 4.75
⎡
1200 ⎞ ⎤
⎛
−5 1.25 ⎛ 1200 × (0.4/60) ⎞
−6 ⎛ 0.4 ⎞
D = 0.66 ⎢(1.5 ×10 ) ⎜
⎟ + 1.31× 10 × ⎜
⎟ ⎜
⎟ ⎥
9.8 × 3
⎢
⎝ 60 ⎠ ⎝ 9.8 × 3 ⎠ ⎥
⎝
⎠
⎣
⎦
0.04
= 0.127 m.
176
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
V 2 0.892 2
=
= 0.041 m.
Exiting K.E. =
2 g 2 × 9.8
∴ negligible.
w2
= w /4. But, R = D /4.
4w
Sheet metal is relatively smooth like drawn tubing. ∴Use e = 0.001 mm.
7.111 hL = 10 m. For a square duct, R =
5.2
2 4.75
⎡
200 ⎞ ⎤
−6 1.25 ⎛ 200 × 4 ⎞
−5
9.4 ⎛
∴ D = 0.66 ⎢(10 ) ⎜
+
×
×
1.6
10
4
⎟
⎜
⎟ ⎥
×
×
9.81
10
9.81
10
⎢
⎝
⎠ ⎥
⎝
⎠
⎣
⎦
0.829
∴R =
= 0.207 m.
∴ w = 4R = 0.83 m.
4
0.04
= 0.829 m.
80
0.02 × 0.04
, ν = 10−6 m 2 /s, use D = 4 R = 4
= 0.027 m.
9810
2(0.02 + 0.04)
0.5
⎡ 0.0015 ⎛ 3.17 × 10−12 × 2 × 9810 ⎞0.5 ⎤
⎡ 9.81× 0.0275 80 ⎤
Q = −0.965 ⎢
×
+⎜
⎟ ⎥
⎥ ln ⎢
3
2
9810
3.7
27
×
9.81
0.027
80
×
×
⎢
⎣
⎦
⎝
⎠ ⎥⎦
⎣
7.112 Use Eq. 7.6.30: hL =
= 0.000143 m3 /s.
0.04 × 0.1
= 0.057 m. Use Eq. 7.6.30 with e = 0. ν = 10−6 m 2 /s.
2(0.04 + 0.1)
0.5
⎡ ⎛ 3.17 × 10−12 × 5 × 9810 ⎞0.5 ⎤
⎡ 9.81× 0.0575 100 ⎤
3
×
Q = −0.965 ⎢
⎟ ⎥ = 0.00074 m /s.
⎥ ln ⎢0 + ⎜
3
5
9810 ⎦
⎢ ⎝ 9.81× 0.057 × 100 ⎠ ⎥
⎣
⎣
⎦
7.113 D = 4R = 4
7.114 a) R =
A
0.3 × 1.2
=
= 0.2 m , hL = 10 000 × 0.0015 = 15 m (from energy eq.)
Pw et
1.8
⎡ 9.8(4 × .2)5 ×15 ⎤
Q = −0.965 ⎢
⎥
10 000
⎣
⎦
b) R =
0.5
⎡ 1.5
⎛ 3.17 ×10−12 × 10 000 ⎞ ⎤
3
ln ⎢
+⎜
⎟ ⎥ = 0.0351 m /s.
3
⎢ 3.7 × 800 ⎝ 9.8 × (4 × .2) ×15 ⎠ ⎥
⎣
⎦
A
0.6 × 1.2
=
= 0.3 m , hL = 10 000 × 0.0015 = 15 m (from energy eq.)
Pw et
2.4
⎡ 9.8 × 1.25 ×15 ⎤
Q = −0.965 ⎢
⎥
⎣ 10 000 ⎦
c) R =
0.5
0.5
0.5
⎡ 1.5
⎛ 3.17 ×10−12 × 10 000 ⎞ ⎤
3
ln ⎢
+⎜
⎟ ⎥ = 0.257 m /s.
3
9.8 × 1.2 × 15
⎢ 3.7 × 1200 ⎝
⎠ ⎥⎦
⎣
A
0.9 × 1.2
=
= 0.36 m, hL = 10 000 × 0.0015 = 15 m (from energy eq.)
Pw et
3.0
177
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Chapter 7 / Internal Flows
⎡ 9.8 × 1.445 ×15 ⎤
Q = −0.965 ⎢
⎥
10 000
⎣
⎦
0.5
0.5
⎡ 1.5
⎛ 3.17 ×10−12 × 10 000 ⎞ ⎤
3
ln ⎢
+⎜
⎟ ⎥ = 2.37 m /s.
3
3.7
1440
×
⎢
⎝ 9.8 × 1.44 × 15 ⎠ ⎥⎦
⎣
Minor Losses
7.115 Energy: 0 =
V22 − V12 p2 − p1
+
+ hL .
γ
2g
p1A2
V1
V2
Note: The area at the sudden
expansion is A2 but the pressure right
at the expansion is p1. It then increases
to p2.
Momentum: ( p1 − p2 ) A2 = ρ A2V2 (V2 − V1 )
Using hL = K
p2A2
V22
, along with A1V1 = A2V2 ,
2g
the energy and momentum equations are combined to provide
⎞ V22
V 2 V 2 − V22 p1 − p2 V22 ⎛ A22
1
+
=
−
K 2 = 1
⎜
⎟⎟ +
γ
2g
2g
2 g ⎜⎝ A12
⎠ g
⎛
⎛ A2
⎞
A2 ⎞
− 1⎟
⎜1 −
⎟. ∴ K = ⎜
A1 ⎠
⎝
⎝ A1
⎠
V2
Δn (n is in the direction of the
7.116 Referring to the equation Δp = − ρ
R
center of curvature), we observe that Δp is negative all along the
line CB from C to B in Fig. P7.115. Hence, the pressure decreases
from C to B with pC > pB. Using Bernoulli’s equation we see that VB
> VC. Fluid moves from the high pressure region at the outside of the
bend toward the low pressure region at the outside of the bend
creating a secondary flow.
2
pC
pB
V
0.02
100 + 50
= 63.7, V2 = 1 = 15.9 m/s. ρ1 =
= 1.78 kg/m3 .
2
4
0.287 × 293
π × 0.01
2
2
2
p 2 − 50 000 ⎛
15.9 − 63.7
1 ⎞ 63.7 2
. ∴ p 2 = 51 400 Pa .
+ ⎜1 − ⎟
+
Energy: 0 =
1.78 × 9.81 ⎝
2 × 9.81
4 ⎠ 2 × 9.81
7.117 a) V1 =
b) V1 = 63.7, V2 =
V1
= 7.08 m/s. ρ1 = 1.78 kg/m3 .
9
1 ⎞ 63.7 2
7.08 2 − 63.7 2 p 2 − 50 000 ⎛
. ∴ p 2 = 50 700 Pa .
+ ⎜1 − ⎟
+
Energy: 0 =
1.78 × 9.81 ⎝
2 × 9.81
9 ⎠ 2 × 9.81
2
7.118 a) V1 = 63.7, V2 = 15.9,
Energy: 0 =
ρ1 = 1.78 kg/m3 .
15.92 − 63.7 2 p2 − 50 000
(63.7 − 15.9) 2
.
+
+ 0.4
2 × 9.81
1.78 × 9.81
2 × 9.81
∴ p2 = 52 600 Pa.
178
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Chapter 7 / Internal Flows
b) V1 = 63.7, V2 = 7.08,
Energy: 0 =
ρ1 = 1.78 kg/m3 .
7.082 − 63.7 2 p2 − 50 000
(63.7 − 7.08) 2
.
+
+ 0.4
2 × 9.81
1.78 × 9.81
2 × 9.81
∴ p2 = 52 400 Pa.
3
⎡
⎛1⎞ ⎤
7.119 a) Ac = ⎢0.62 + 0.38 ⎜ ⎟ ⎥ A2 = 0.626 A2 . Neglect losses from 1 → c (see Fig. 7.16).
⎝ 4 ⎠ ⎥⎦
⎢⎣
2
2
2
⎛ Ac ⎞ Vc2
2 ⎛ A2 ⎞ V2
From c → 2: hL = ⎜1 −
= (1 − 0.626) ⎜
⎟
⎟
⎝ A2 ⎠ 2 g
⎝ Ac ⎠ 2 g
V22
1
0.14
.
= 0.140 ×
×
∴K =
= 0.36.
2
0.626 2 g
0.6262
2
⎡
⎛1⎞ ⎤
b) Ac = ⎢0.6 + 0.4 ⎜ ⎟ ⎥ A0 = 0.625 A0 . Neglect losses from c → 2 (see Fig. 7.16).
⎝ 4 ⎠ ⎥⎦
⎢⎣
2
2
2
⎛ Ac ⎞ Vc2 ⎛ Ac A0 ⎞ ⎛ A2 ⎞ V22
From c →2: hL = ⎜1 −
= ⎜1 −
.
⎟ ⎜
⎟
⎟
⎝ A2 ⎠ 2 g ⎝ A0 A2 ⎠ ⎝ Ac ⎠ 2 g
2
1 ⎞ A2 A2
1
⎛
∴ K = ⎜1 − 0.625 × ⎟ 22 × 02 = 0.712 × 42 ×
= 29.
4 ⎠ A0 Ac
0.6252
⎝
2
⎡
⎛1⎞ ⎤
c) Ac = ⎢0.6 + 0.4 ⎜ ⎟ ⎥ A0 = 0.625 A0 . Neglect losses from 1 → c (see Fig. 7.16).
⎝ 4 ⎠ ⎦⎥
⎣⎢
2
2
⎛
A ⎞ V2 ⎛
A A ⎞ A 22 V 22
From c → 2: hL = ⎜ 1 − c ⎟ c = ⎜ 1 − c 0 ⎟
.
A 2 ⎠ 2g ⎝
A 0 A 2 ⎠ A c2 2 g
⎝
2
1 ⎞ A22 A02
1
⎛
∴ K = ⎜1 − 0.625 × ⎟ 2 2 = 0.866 × 92 ×
= 180.
9 ⎠ A0 Ac
0.6252
⎝
7.120 V =
L V2
Q
0.12
≅ 0).
=
=
Neglect
pipe
friction
(i.e.,
f
5.5
fps.
D 2g
A π (1/12) 2
5.52 − 02
5.52
Energy: 0 =
.
+ 0 − 6 + (K + 0.03)
2 × 32.2
2 × 32.2
∴ K = 11.7.
7.121 a) manometer: 0.04 × 9810 + p1 = 0.04 × (9810 × 13.6) + p2 .
Energy:
0=
V=
0.006
= 4.77 m/s.
π × 0.022
V22 − V12
p − p1
V2
.
+ 2
+K
γ
2g
2g
179
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Chapter 7 / Internal Flows
∴K =
b) K =
p1 − p2
γ
×
0.04 × 9810 ×12.6 × 2 × 9.81
= 0.435.
9810 × 4.77 2
2 g 0.08 × 9810 ×12.6 × 2 × 9.81
=
= 0.869.
V2
9810 × 4.77 2
Simple Piping Systems
7.122 Assume completely turbulent regime:
e 0.046
=
= 0.0012.
D
40
Energy:
0=
∴ f = 0.0205, Re > 106.
2
V2
⎡ L
⎤V
− 40 + ⎢ f + 2Kelbow + Kentrance ⎥
2g
⎣ D
⎦ 2g
2
110
⎡
⎤ V
40 = ⎢1 + 0.0205 ×
. (Assume a screwed elbow.)
+ 2 × 1.0 + 0.5⎥
0.04
⎣
⎦ 2 × 9.81
∴ V = 3.62 m/s. Re =
∴ V = 3.50 m/s.
3.62 × 0.04
= 1.4 ×105.
−6
10
∴ Try f = 0.022.
∴ Q =AV =π × 0.022 × 3.5 = 0.0044 m3 /s.
The EGL and HGL have sudden drops at the elbows, and a gradual slope over
the pipe length.
7.123 Energy: 0 =
2
V2
⎛ L
⎞V
.
− H + ⎜ f + Kentrance + K valve ⎟
2g
⎝ D
⎠ 2g
e 0.00085
=
= 0.0026.
D
4/12
a) Assume f = 0.027 (slightly larger than the value of the completely turbulent flow):
1200 ⎞ V 2
⎛
15 = ⎜1 + 0.5 + 15 + 0.027 ×
. ∴ V = 2.91 ft/sec.
⎟
4 /12 ⎠ 2 × 32.2
⎝
2.91× 4/12
Re =
= 9.2 ×104.
−5
1.06 ×10
b) Assume f = 0.026:
2
∴ f is OK.
⎛ 2⎞
∴ Q = π × ⎜ ⎟ × 2.91 = 0.25 cfs.
⎝ 12 ⎠
1200 ⎞ V 2
⎛
30 = ⎜16.5 + 0.026 ×
.
⎟
4/12 ⎠ 64.4
⎝
∴ V = 4.19 fps.
2
4.19 × 4/12
⎛ 2⎞
Re =
= 1.3 × 105. ∴ f is OK. ∴ Q = π × ⎜ ⎟ × 4.19 = 0.37 cfs.
−5
1.06 × 10
⎝ 12 ⎠
1200 ⎞ V 2
⎛
.
b) Assume f = 0.026: 60 = ⎜ 16.5+.026
⎟
⎝
4 / 12 ⎠ 2 g
∴ V = 5.92 fps.
180
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Chapter 7 / Internal Flows
2
Re = 1.8 × 10 . ∴ f is OK.
5
7.124 Assume screwed elbows: 0 =
⎛ 2⎞
∴ Q = π × ⎜ ⎟ × 5.92 = 0.52 cfs.
⎝ 12 ⎠
V2
L ⎞V2
⎛
.
− 2 + ⎜ Kentrance + 2Kelbows + f ⎟
D ⎠ 2g
2g
⎝
26 ⎞ V 2
⎛
.
a) Assume f = 0.022 : 2 = ⎜ 0.8 + 1 + 2 × 1 + 0.022 ×
⎟
0.04 ⎠ 2 × 9.81
⎝
1.47 × 0.04
∴ Re =
= 5.2 ×104 and f = 0.021.
1.14 × 10−6
∴ V = 1.50, Q = π × 0.022 × 1.50 = 0.0019 m3 /s.
∴ V = 1.47 m/s.
26 ⎞ V 2
⎛
b) Assume f = 0.018 : 2 = ⎜ 0.8 + 1 + 2 × 0.8 + 0.018
. ∴ V = 2.06 m/s.
⎟
0.08 ⎠ 2 × 9.81
⎝
2.06×.08
. × 105 . f =.0164. ∴V = 2.12, Q = π ×.04 2 × 2.12 = 0.011 m 3 / s .
= 15
∴ Re =
114
. × 10 − 6
26 ⎞ V 2
⎛
c) Assume f = 0.013 : 2 = ⎜ 0.8 + 1 + 2 × 0.6 + 0.01
. ∴ V = 2.76 m/s.
⎟
0.12 ⎠ 2 × 9.81
⎝
2.76 × 0.12
∴ Re =
= 3 × 105 and f = 0.014.
−6
1.14 ×10
∴ V = 2.55, Q = π × 0.062 × 2.55 = 0.029 m3 /s.
7.125 a) e / D1 = 0.26 / 20 = 0.013. ∴ f1 = 0.041.
e / D2 = 0.26 / 40 = 0.0065. ∴ f2 = 0.033.
2
⎞ V1
EGL
2
⎞ V2
20
40
⎛
⎛
0 = −10 + ⎜ .5 + .56 + 5 + .041 ⎟
.
+ ⎜ 1 + .033
⎟
.02 ⎠ 2 g ⎝
.04 ⎠ 2 g
⎝
HGL
V2
300 ⎞ V 2
⎛
But, H P = 20 + ⎜ 0.5 + 1.0 + 0.021
= 20 + 1.68V 2 = 20 + 1700Q 2 .
⎟
0.2 ⎠ 2 g
⎝
∴ Q = A1V1 = π × 0.012 × 2.00 = 6.28 × 10−4 m3 /s.
Lowest pressure occurs in the small pipe. This occurs at the enlargement:
p
V2
20 ⎞ V12
⎛
0 = 1 + 1 − H + ⎜ 0.5 + 5 + 0.041
.
⎟
2 g 9810
0.02 ⎠ 2 g
⎝
With H = 12 m and V1 = 2.00 m/s, then p1 = 22 720 Pa.
20 ⎞ V12 ⎛
40 ⎞ V22
⎛
1
0.033
.
+
+
b) 0 = −20 + ⎜ 0.5 + 0.56 + 5 + 0.041
⎟
⎜
⎟
0.02 ⎠ 2 g ⎝
0.04 ⎠ 2 g
⎝
But, V1 = 4V2 . ∴ V2 = 0.706, V1 = 2.82 m/s.
181
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Chapter 7 / Internal Flows
Lowest pressure occurs in the small pipe. This occurs at the enlargement.
0=
p
V12
20 ⎞ V12
⎛
.
+ 1 − H + ⎜ 0.5 + 5 + 0.041
⎟
2 g 9810
0.02 ⎠ 2 g
⎝
With H = 22 m and V1 = 2.82 m/s, then p1 = 26 950 Pa.
20 ⎞ V12 ⎛
40 ⎞ V22
⎛
.
+ ⎜1 + 0.033
c) 0 = −30 + ⎜ 0.5 + 0.56 + 5 + 0.041
⎟
⎟
0.02 ⎠ 2 g ⎝
0.04 ⎠ 2 g
⎝
But, V1 = 4V2 . ∴ V2 = 0.865, V1 = 3.56 m/s.
∴ Q = A1V1 = π × 0.012 × 3.56 = 0.00109 m3 /s.
Lowest pressure occurs in the small pipe. This occurs at the enlargement.
0=
p
V12
20 ⎞ V12
⎛
.
+ 1 − H + ⎜ 0.5 + 5 + 0.041
⎟
2 g 9810
0.02 ⎠ 2 g
⎝
With H = 32 m and V1 = 3.56 m/s, then p1 = 12 920 Pa.
7.126 Assume constant water level, neglect the velocity head at the pipe exit and
let p exit = 0. Then the energy equation says hL = 0.8 m:
0.8 = f
10
V2
V2
+ 2 × 1.5
0.025 2 × 9.8
2 × 9.8
Try f = 0.02. Then V = 0.319 m/s, Re =
or
0.8 = 400 fV 2 + 0.153V 2
0.313 × 0.025
= 1.1× 104.
−6
0.73 × 10
∴ f = 0.029.
Try f = 0.029. Then V = 0.262 m/s.
∴ Q = AV = π × 0.01252 × 0.262 = 1.28 × 10−4 m3 /s.
Finally, Δt =
V
−
10
=
= 78.1 sec or 1.3 min.
Q
.128
7.127 The siphon operates unless p min = p vapor = 1.176 kPa abs or −98.8 kPa. Since pressure
decreases along the pipe, we will let the pressure after the second elbow be −98.8 kPa.
From there, the pressure increases to zero at the exit. If there are no losses,
γ H = − pelbow or H =
98 800
= 10.1 m.
9800
182
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Chapter 7 / Internal Flows
0=
Energy:
2
V2
⎛ 8 + 2H
⎞V
.
−6+⎜ f
+ 0.8 + 2 × 2.0 ⎟
2g
⎝ 0.03
⎠ 2g
Let H = 6 m and assume f = 0.02. Then energy provides V = 2.48 m/s.
Check: Re =
2.48 × 0.03
−6
= 5.7 × 104. From the Moody diagram, f = 0.02. ∴Good.
1.31× 10
Now, check the pressure after the elbow:
0 − pelbow
12 2.482
0=
. ∴ pelbow = −93 000 Pa.
− (6 + 6) + 0.02
×
9800
0.03 2 × 9.8
Try H = 6.5 m and the pressure after the second elbow is close to pv .
V32 p0
V2
1.2 V12
1.2 V22
− − 1.2 + 0.8 1 + f1
+ f2
2g γ
2g
0.008 2 g
0.005 2 g
25
64
V3 =
V2 , V2 =
V1 .
Continuity:
4
25
Assume
f1 = f2 = 0.02. Then energy says V1 = 2.09 m/s.
2.09 × 0.008
= 16 700. ∴ f = 0.026. Then V1 = 2.88 m/s.
Check: Re1 =
10−6
Finally,
V3 = 16V1 = 46.1 m/s
7.128 Energy:
0=
e 1.65
=
= 0.0021. ∴ f = 0.024.
D 800
L⎞ V 2
+ K valve + K exit + f ⎟
.
D⎠ 2g
7.129 Use an average value of e:
⎛
0 = Δz + ⎜ K entrance
⎝
2000 ⎞ V 2
⎛
−Δz = ⎜ 0.8 + 1 + 1 + 0.024
⎟
0.8 ⎠ 2 × 9.81
⎝
a) Δz = −80. ∴ V = 5.0, Q = π × 0.42 × 5 = 2.5 m3 /s.
b) Δz = −150. ∴ V = 6.85, Q = π × 0.42 × 6.85 = 3.44 m3 /s.
c) Δz = −200. ∴ V = 7.9, Q = π × 0.42 × 7.9 = 3.97 m3 /s.
5 × 0.8
Check: Re =
This is sufficiently large.
= 4 ×106.
−6
10
300 V 2
7.130 Energy:
0 = −3 + f
.
0.0094 2 × 9.8
Assume turbulent:
Check:
Re =
f = 0.04. Then V = 0.215 m/s.
0.215 × 0.0094
10−6
= 2020. ∴ marginally laminar.
183
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Chapter 7 / Internal Flows
Assume laminar flow with f = 64/ Re. Then
3=
64 × 10−6
300
V2
.
×
×
V × 0.0094 0.0094 2 × 9.8
∴ V = 0.271 m/s.
0.271× 0.0094
= 2540. ∴turbulent.
10−6
The flow is neither laminar nor turbulent but may oscillate between the two. The wall
friction is too low in the laminar state so it speeds up and becomes turbulent. The wall
friction is too high in the turbulent state so it slows down and becomes laminar, etc., etc.
Re =
Check:
7.131 a) V =
0.6
π × (0.75 /12)
2
= 48.9. Re =
48.9 ×1.5 /12
1.22 × 10
−5
= 5 × 105.
e 0.00085
=
= 0.0068
D 1.5/12
∴ f = 0.033.
48.92 30 ×144
900 ⎞ 48.92
⎛
+
+ 30 + ⎜ 0.5 + 2 × 1.2 + 0.033
= 9066 ft.
⎟
32.2 × 2
62.4
1.5/12 ⎠ 2 × 32.2
⎝
= γ QH P = 62.4 × 0.6 × 9066 = 424, 000 ft-lb/sec or 771 hp.
∴W
P
ηP
0.8
0.6
12.2 × 3 /12
e
= 12.2 fps. Re =
= 2.5 ×105.
= 0.0034.
b) V =
2
−5
D
1.22 ×10
π × (1.5 /12)
∴ f = 0.027.
HP =
12.22 30 ×144
900 ⎞ 12.22
⎛
+
+ 30 + ⎜ 0.5 + 2 × 0.8 + 0.027
= 331 ft.
⎟
64.4
62.4
3 /12 ⎠ 64.4
⎝
= γ QH P = 62.4 × 0.6 × 331 = 15,500 ft-lb/sec or 28.2 hp.
∴W
P
ηP
0.8
0.6
5.43 × 4.5/12
e
c) V =
= 5.43 fps. Re =
= 1.67 ×105.
= 0.0023.
−5
2
D
1.22 × 10
π × (2.25/12)
∴ f = 0.026.
∴ HP =
5.432 30 ×144
900 ⎞ 5.432
⎛
+
+ 30 + ⎜ 0.5 + 2 × 0.6 + 0.026
= 129 ft.
⎟
64.4
62.4
4.5/12 ⎠ 64.4
⎝
= γ QH P = 62.4 × 0.6 × 129 = 6040 ft-lb/sec or 11.0 hp.
∴W
P
ηP
0.8
∴ HP =
7.132 V =
0.01
7.96 × 0.04
Q
e 0.0015
=
= 7.96 m/s. Re =
= 2.8 × 105.
=
= 3.8 × 10−5
−
2
6
40
A π × 0.02
D
1.14 × 10
∴ f = 0.0145.
800 ⎞ 7.962
⎛
H P = 80 − 10 + ⎜ 0.5 + 1.0 + 0.0145
= 1010 m.
⎟
0.04 ⎠ 2 × 9.81
⎝
184
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Chapter 7 / Internal Flows
= γ QH P = 9810 × 0.01×1010 = 117 000 W.
∴W
P
ηP
0.85
7.962 1670 − 100 000
L ⎞ 7.962
⎛
0=
. ∴ L = 13.0 m.
+
+ 0 − 10 + ⎜ 0.5 + 0.0145
⎟
2 × 9.81
9810
0.04 ⎠ 2 × 9.81
⎝
7.133 V =
2
3.14 × 0.9
Q
e 0.26
6
3.14
m/s.
Re
2.5
10
.
=
=
=
=
×
=
= 2.9 × 10−5
−
2
6
A π × 0.45
D 900
1.14 ×10
∴ f = 0.015.
400 ⎞ 3.142
⎛
a) − HT = −20 + ⎜ 0.5 + 1 + 0.015
= −15.9 m.
⎟
0.9 ⎠ 2 × 9.81
⎝
= γ H Qη = 9810 × 2 ×15.9 × 0.85 = 265 000 W.
∴W
T
T
T
400 ⎞ 3.142
⎛
b) − HT = −60 + ⎜ 0.5 + 1 + 0.015
= −55.9 m.
⎟
0.9 ⎠ 2 × 9.81
⎝
= γ QH η = 9810 × 2 × 55.9 × 0.85 = 932 000 W.
∴W
T
T T
400 ⎞ 3.142
⎛
c) − HT = −100 + ⎜ 0.5 + 1 + 0.015
= −95.9 m.
⎟
0.9 ⎠ 2 × 9.81
⎝
= γ QH η = 9810 × 2 × 95.9 × 0.85 = 1 600 000 W or 1.6 MW.
∴W
T
T T
7.134
Energy across nozzle (neglect losses):
∴ V1 = 31.5 fps. Re =
HP =
31.5 × 2/12
1.06 × 10−5
V 12 100 × 144 V 22 ( 4V 1 ) 2
+
=
=
.
2g
62.4
2g
2g
= 4.95 ×105.
e 0.00015
=
= 0.0009. ∴ f = 0.02.
2/12
D
(4 × 31.5)2
1200 ⎞ 31.52
⎛
− 60 + ⎜ 0.5 + 0.02
= 2410 ft.
⎟
64.4
2/12 ⎠ 64.4
⎝
= γ QH P = 62.4 × (π × (1/12 ) × 31.5) × 2410 = 138,000 ft-lb/sec or 251 hp.
∴W
P
ηP
0.75
2
0=
7.135
31.52 (0.34 − 14.7)144
L ⎞ 31.52
⎛
. ∴ L = 37.9 ft.
+
− 60 + ⎜ 0.5 + 0.02
⎟
64.4
62.4
2/12 ⎠ 64.4
⎝
e 0.26
=
= 0.0013. ∴ f = 0.021. Q = AV = π × 0.12 V = 0.0314V
D 200
300 ⎞ V 2
⎛
H P = 20 + ⎜ 0.5 + 1.0 + 0.021
= 20 + 1.68V 2 = 20 + 1700Q 2 .
⎟
0.2 ⎠ 2 g
⎝
185
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Chapter 7 / Internal Flows
Try Q = 0.2 m3 /s: ( H P )energy = 88 m.
Try Q = 0.18 m3 /s: ( H P )energy = 75 m.
( H P )char. = 74 m.
EGL
( H P )char. = 77 m.
HGL
∴ Q = 0.182 m3 /s.
∴ H P = 20 + 1700 × 0.1822 = 76 m.
= 9810 × 0.182 × 76 / 0.7 = 194 000 W.
∴W
P
7.136
e
= 0.0013. ∴ f = 0.021. Q = 0.0314V .
D
300 ⎞ V 2
⎛
H P = −20 + ⎜ 0.5 + 1.0 + 0.021
= −20 + 1700Q 2 .
⎟
0.2 ⎠ 2 g
⎝
Try Q = 0.23 m3 /s: ( H P ) E = 70 m.
( H P )C = 70 m.
= 9810 × 0.23 × 70 = 190 000 W.
∴W
P
0.83
EGL
HGL
Re = 1.5 × 106. ∴OK.
7.137 Energy:
300 ⎞ V 2
⎛
− HT = −20 + ⎜ 0.5 + 1.0 + 0.021
⎟
0.2 ⎠ 2 g
⎝
800Q = −20 + 1706Q 2 . ∴ Q = 0.493 m3 /s.
= H γ Qη = 800 × 0.493 × 9800 × 0.88 = 3.4 × 106 W or 3.4 MW.
W
T
T
7.138
e 0.046
=
= 0.00029. ∴ f = 0.015. Q = π × 0.082 V = 0.0201V .
D 160
Q2
50 ⎞
⎛
= 25 + 882Q 2 .
H P = 25 + ⎜ 0.5 + 2 × 0.4 + 1 + 0.015
⎟
2
0.16
⎝
⎠ 2 g × 0.0201
Try Q = 0.23 m3 /s. ( H P ) E = 72 m. ( H P )C. = 70 m.
= 9810 × 0.23 × 72 = 195 000 W.
a) ∴ W
P
0.83
b) 0 =
p
11.442
10 ⎞ 11.442
⎛
+ 2 − 8 + ⎜ 0.5 + 0.015
. ∴ pin = −81 000 Pa.
⎟
2 × 9.81 9810
0.16 ⎠ 2 × 9.81
⎝
c) 72 =
p
11.442
10 ⎞ 11.442
⎛
+ 3 − 8 + ⎜ 0.5 + 0.015
. ∴ pout = 625 000 Pa.
⎟
2 × 9.81 9810
0.16 ⎠ 2 × 9.81
⎝
186
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Chapter 7 / Internal Flows
EGL
d)
HGL
7.139
e
= 0.00029. ∴ f = 0.015. Q = 0.0201 V .
H P = −25 + 882Q 2 .
D
Try Q = 0.31 m3 /s. ( H P ) E = 60 m. ( H P )C. = 60 m.
= 9810 × 0.31× 60 = 223 000 W.
a) ∴ W
P
0.82
2
p
15.42
40 ⎞ 15.422
⎛
+ 2 − 33 + ⎜ 0.5 + 0.015
.
b) 0 =
⎟
2 × 9.81 9810
0.16 ⎠ 2 × 9.81
⎝
∴ p in = −300 000 Pa. This is below absolute zero and the pump would not function
according to the curves of Example 7.16. Part (a) is also not correct. The pump must
be repositioned for reverse flow.
c) Since p in is too low, the problem is not workable as posed.
d) The EGL and HGL curves would
appear as sketched.
7.140
EGL
HGL
e 1.65
=
= 0.0014. ∴ f = 0.021. Q = π × 0.62 V = 1.13V .
D 1200
Q2
1000 ⎞
⎛
− HT = −60 + ⎜ 0.8 + 1 + 0.021
. ∴ HT = 60 − 0.77Q 2 .
⎟
2
1.2 ⎠ 1.13 × 2 × 9.81
⎝
= γ QH η = 9810 × (60Q − 0.77Q3 ) × 0.9 = 530 000Q − 6800Q3.
W
T
T T
= 500 000 Q (power in watts).
But, the characteristic curve is W
T
Hence, 500 000Q = 530 000Q − 6800Q3. ∴Q = 2.1 m3/s.
= 1.05 MW.
∴W
T
187
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Chapter 7 / Internal Flows
7.141 The conservation of energy equation when applied to the entire system gives
⎛V2 p
⎞
⎛V2 p
⎞
HP + ⎜
+ + z⎟ = ⎜
+ + z ⎟ + hL
⎜ 2g γ
⎟
⎜
⎟
⎝
⎠in ⎝ 2 g γ
⎠out
Since the inlet and outlet sections lie on a reservoir surface, the energy equation reduces to
H P = hL where hL represents the total losses in the system. It is given by
2
⎛ L
⎞V
hL = ⎜ f + Kinlet + Koutlet + K filter + Kvalve + Kbends ⎟
⎝ D
⎠ 2g
Using f = 0.04, Kinlet = 0.8, Koutlet = 1.0, Kfilter = 12.0, Kvalve = 6.0, Kbend = 1.1, we have
2
60
⎛
⎞ Q
+ 1.0 + 0.8 + 12 + 6 + 5 ×1.1⎟
= 40 735Q 2
H P = ⎜ 0.04
2
0.1
⎝
⎠ 2 gA
Equating the above equation to the pump characteristic equation we have
10 + 12Q − 150Q 2 = 40735Q 2
Solving the above equation gives Q = 0.016 m3 /s and the mass flow rate is
[
(
)(
)
= ρQ = 103 kg/m3 0.016 m3 /s = 16 kg/s
m
The pump head is calculated as follows
H P = 10 + 12Q − 150Q 2 = 10 + 12 × 0.016 − 150 × 0.0162 = 10.15 m
And the power input to the pump is
(
)
2
= mgH
W
P
P = (16 kg/s ) 9.81 m/s × 10.15 m = 1593 W
7.142 τ w Aw = W sin θ . sin θ = 0.001.
51.68
∴ A1 =
π × 32 − 1.308 × 2.7 = 0.5285 ft 2 .
360
τw = 0.170 psf.
7.143 τ w Aw = W × 0.0016 (see fig. of Prob. 7.141).
10
cosα = . ∴α = 60D.
20
120
120
⎛
⎞
− 0.1× 0.1732 ⎟ × 0.0016 L.
τ w 2π × 0.2 ×
L = 9810 ⎜ π × 0.22
360
360
⎝
⎠
∴τ w = 0.921 Pa.
L
τwAw
W
θ
10
α
20
188
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Chapter 7 / Internal Flows
1.0
1.0
⎛ 2 × 0.6 ⎞
7.144 a) Q =
AR2/3S1/2 =
× (2 × 0.6) ⎜
⎟
n
0.012
⎝ 2 + 2 × 0.6 ⎠
2/3
× 0.0011/2 = 1.64 m3 /s.
b) Planed wood and finished concrete have the same “n” (i.e., roughness).
∴Use a value for “e” which is relatively small, say e = 0.6 mm. The solution
is not too sensitive to this choice. Then, with R = 0.375 m,
e
e
0.6
=
=
= 0.0004 ∴ f = 0.016
D 4R 4 × 375
2 × 9.81× 0.001× (4 × 0.375)
L V2
= LS. ∴ V 2 =
. ∴ V = 1.36 m/s.
hL = f
0.016
D 2g
1.36 × (4 × 0.375)
⎛
⎞
∴ Q = 1.36 × 2 × 0.6 = 1.63 m3 /s. ⎜ Check: Re =
= 2.0 ×106. ∴ OK ⎟
−6
10
⎝
⎠
7.145 a) R =
1.49
π × 32
= 1.5 ft. ∴ Q =
× π × 32 × 1.52/3 × 0.00121/2 = 159 cfs.
6π
0.012
2.7
. α = 25.84D.
3
51.68
∴ A1 =
π × 32 − 1.308 × 2.7 = 0.5285 ft 2 .
360
27.75 × 360
∴ A = 9π − 0.5285 = 27.75 ft 2 . R =
= 1.719 ft.
6π × 308.3
1.49
∴Q =
× 27.75 ×1.7192/3 × 0.00121/2 = 171 cfs.
0.012
A1
b) cos α =
c) R =
α
3
9π /2
1.49 9π
= 1.5 ft. ∴ Q =
×
× 1.52/3 × 0.00121/2 = 79.7 cfs.
3π
0.012 2
d) α = 60D. A = π × 9 ×
120
5.52
− 1.5 × 2.6 = 5.52. R =
= 0.879 ft.
360
2π × 3 × 120/360
1.49
× 5.52 × 0.8792/3 × 0.00121/2 = 21.8 cfs.
0.012
2.5
67.11
e) cos α =
. ∴α = 33.56D. A = 9π ×
− 2.5 × 1.658 = 1.125 ft 2 .
3
360
1.125
1.49
R=
= 0.32 ft. ∴ Q =
× 1.125 × 0.322/3 × 0.00121/2 = 2.26 cfs.
6π × 67.11/360
0.012
Q=
7.146 a) R =
1.2 × 0.5 + 0.5 × 0.5
1.0
= 0.325 m ∴ Q =
× 0.85 × 0.3252/3 × 0.0011/2 = 0.794 m3 /s.
1.2 + 2 × 0.5/0.707
.016
V=
Q 0.794
=
= 0.934 m/s
A 0.85
189
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Chapter 7 / Internal Flows
e
3
=
= 0.0023. ∴ f = 0.025.
D 4 × 325
4 × 0.325 × 2 × 9.81× 0.001
∴V 2 =
and V = 1.01 m/s.
0.025
b) Use an “e” for rough concrete, e = 3 mm. ∴
hL = LS = f
L V2
.
4R 2 g
∴ Q = 1.01× 0.85 = 0.86 m3 /s.
⎛ y ⎞
y
2y
1.0
=
. 5=
2y ⎜
7.147 a) R =
⎟
2 + 2y 1 + y
0.016 ⎝ 1 + y ⎠
2/3
× 0.001 .
1/2
⎛ y ⎞
∴y⎜
⎟
⎝ 1+ y ⎠
2/3
= 1.265
? 1.265
y = 1.5: 1.07 =
Trial-and-error:
⎫
y = 1.8: 1.34 =? 1.265⎪
⎪
⎬ ∴ y = 1.71 m.
?
y = 1.7: 1.25 = 1.265 ⎪
⎪⎭
3(1 + y )
e
.
b) Use an “e” value of rough concrete: e = 3 mm. ∴
=
4 R 4 y × 1000
4 y × 0.001× 2 × 9.81
y
L V2
= 0.0785
. ∴ V 2 = 4 RS × 2 g / f =
.
hL = LS = f
4R 2 g
( y + 1) f
(1 + y ) f
e
⎫
= 0.0012. ∴ f = 0.02, V = 1.57. Q = 5.35 m3 /s. ⎪
⎪
4R
⎬ ∴ y = 1.67 m.
e
3 ⎪
Try y = 1.6 :
= 0.0012. ∴ f = 0.022, V = 1.48. Q = 4.74 m /s.
⎪⎭
4R
Try y = 1.7 :
7.148 Assume y > 3. R =
30 + 20( y − 3) 20 y − 30 10 y − 15
.
=
=
26 + 2( y − 3)
2 y + 20
y + 10
⎛ 10 y − 15 ⎞
1.0
100 =
(20 y − 30) ⎜
⎟
0.022
⎝ y + 10 ⎠
Or
2/3
× 0.0011/2
⎛ 10 y − 15 ⎞
6.96 = (2 y − 3) ⎜
⎟
⎝ y + 10 ⎠
y
2/3
.
Try y = 5: 6.96 =? 12.3 . y = 4: 6.96 =? 7.36 .
⎫
y = 3.8: 6.96 =? 6.47 ⎪
⎪
⎬ ∴ y = 3.91 m
?
y = 3.9: 6.96 = 6.91⎪
⎪⎭
190
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Chapter 7 / Internal Flows
⎛ 2 y + y2 ⎞
2 y + y2
1.0
. 10 =
(2 y + y 2 ) ⎜
7.149 R =
⎜ 2 + 2.83 y ⎟⎟
2 + 2.83 y
0.012
⎝
⎠
Try y = 1.2: 3.06 =? 3
Try y = 1.5: 4.68 =? 4
∴ y > 2 ft. R =
Q=
4π × α /180
,
2/3
A = 4π
2/3
(2 y + y 2 )5/3
(2 + 2.83 y ) 2/3
=3
∴ y = 1.19 m .
(2 y + y 2 )5/3
(0.0016)1/2 or
(2 + 2.83 y ) 2/3
Try y = 1.4: 4.10 =? 4
1.49
⎛ 2π ⎞
× 2π ⎜
⎟
0.013
⎝ 2π ⎠
A
(0.0016)1/2 or
Try y = 1.15: 2.83 =? 3
⎛ 2 y + y2 ⎞
1.0
(2 y + y 2 ) ⎜
b) 10 =
⎟⎟
⎜
0.016
⎝ 2 + 2.83 y ⎠
7.150 Try y = 2 ft : Q =
2/3
=4
∴ y = 1.38 m .
× 0.0011/2 = 22.8. cos α =
y−2
2
180 − α
+ ( y − 2)2sin α .
180
α
0.6
1.49
AR2/3 × 0.0011/2 = 24. ∴ AR2/3 = 6.62.
0.013
y
⎫
Try y = 2.1' : α = 87.13 D , A = 6.683 ft 2 , R = 1.099' . 7.12 =? 6.62 ⎪
⎪
⎬y = 2.04'.
D
2
?
Try y = 2.04' : α = 88.85 , A = 6.443 ft , R = 1.039' . 6.61 = 6.62⎪
⎪⎭
1.0 π × 0.42 ⎛ 0.08π
⎜
0.013
2
⎝ 0.4π
A
0.4 − y
cos α =
, R=
,
0.4
0.8π × α /180
1.0
Q=
AR2/3 × 0.0011/2 = 0.2.
0.013
7.151 Try y = 0.4 m : Q =
⎞
⎟
⎠
2/3
× 0.0011/2 = 0.209 m3 /s. ∴ y < 0.4 m.
A = 0.42 π
α
180
− (0.4 − y ) × 0.4sin α .
∴ AR2/3 = 0.0822.
Try y =.396 m : α = 89.43 D , A =.2481 m 2 , R =.1987 m.
.0845 =? .0822
Try y =.392 m : α = 88.85 D , A =.2449 m 2 , R =.1974 m .
.0830 =? .0822
Try y =.391 m: α = 88.71D , A =.2441 m 2 , R =.1971 m.
.0827 =? .0822
∴ y = 0.389 m .
191
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Chapter 7 / Internal Flows
192
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Chapter 8 / External Flows
CHAPTER 8
External Flows
FE-type Exam Review Problems: Problems 8-1 to 8-9
8.1
(C)
8.2
(C)
8.3
(B)
Re 
8.4
(B)
Assume a large Reynolds number so that CD = 0.2. Then
F
8.5
(D)
0.8  0.008
1.31  106
 4880.
2
1
1
 80  1000 
2
V 2 ACD   1.23  
    5  0.2  4770 N.
2
2
 3600 
1
1
V 2 ACD .  60   1.23  402  4  D  1.2.  D  0.0041 m.
2
2
Re 
(C)


Assume a Reynolds number of 105. Then CD = 1.2.
F
8.6
VD
Re 
VD

VD



40  0.0041
10
6
4  0.02
1.6  105
 1.64  105.  CD  1.2. The assumption was OK.
 5000.  St  0.21 
 f  42 Hz (cycles/second). distance =
fD f  0.02

.
V
4
4 m/s
V

 0.095 m/cycle.
f 42 cycles/s
8.7
(C)
By reducing the separated flow area, the pressure in that area increases thereby
reducing that part of the drag due to pressure.
8.8
(B)
From Fig. 8.12a, C L  1.1. C L 
V 2 
1
2
FL
.
 V 2cL
2W
2 1200  9.81

 1088 and V  33.0 m/s.
 cLCL 1.23 16 1.1
193
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Chapter 8 / External Flows
Chapter 8 Problems: Separated Flows
8.9
B
A
B C
A
C
A-B: favorable
B-C: unfavorable
A-D: favorable
C-D: undefined
A-B: favorable
B-C: unfavorable
A-C: favorable
8.10
Re  5 
VD

. D 
inviscid
flow
D
5  1.51  10 5
 3.78  10 5 m.
20
no separation
inviscid
flow
viscous flow
near sphere
8.11
separated
flow
separated
region
boundary layer
near surface
separation
separation
wake
8.12
separated
region
boundary
layer
building
wake
inviscid
flow
8.13
5
VD

5
5 1.22 105
 V  . a) V 
 9.15 104 fps.
D
0.8/12
5  0.388 10 5
b) V 
 2.91 10 4 fps.
0.8/12
5 1.6 104
c) V 
 0.012 fps.
0.8/12
194
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Chapter 8 / External Flows
8.14
8.15
Re 
VD


20  D
1.5110
 13.25 105 D.
a) Re  13.25 105  6  7.9 106 .
Separated flow.
b) Re  13.25 105  0.06  7.9 104 .
Separated flow.
c) Re  13.25 105  0.006  7950.
Separated flow.
FD 

Afront
1
1 1 1
pdA  pback Aback  p0  (1  r 2 )2 rdr  p0 2      p0
2 4 2
0
Bernoulli: p 
1
1
V2  p 0 .  p 0   1.21  20 2  242 Pa.
2
2
1
 ( 242)  380 N
2
 FD 
CD 
FD
1
2
8.16
5
V 2 A

2  380
 0.5
1.21  20 2    1 2
Ftotal  Fbottom  Ftop  20 000  0.3  0.3+10 000  0.3  0.3  2700 N.
Flift  2700 cos 10   2659 N
Fdrag  2700 sin 10   469 N
CL 
CD 
8.17
1
2
1
2
FL
V 2 A
FD
2
V A


2659
1
2
 1000  5  0.3  0.3
2
469
1
2
 1000  5  0.3  0.3
2
 2.36
 0.417
F  p  A  26 000  Lw. Fu  pu Au  8000 
Lw
 4015 Lw
2 cos 5 
FL  F cos 5  Fu cos10  21 950Lw
FD  F sin 5  Fu sin10  1569 Lw
F
21 950 Lw
1
 0.25
CL  1 L
2
V 2 A
 0.3119  750 Lw
2
CD 
1
2
2
FD
2
V A

1569 Lw
1
2
 0.3119  750 Lw
2
195
 0.0179
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 8 / External Flows
8.18
If C D  1.0 for a sphere, Re = 100 (see Fig. 8.9). 
V  0.1

 100,
V =1000 .
1
a) V  1000 1.46 105  0.0146 m/s.  FD  1.22  0.01462   0.052 1.0
2
 3.25  10 7 N.
b) V  1000 
1.46 105
1
 0.798 m/s.  FD   (0.015 1.22)  0.7982   0.052 1.0
0.015 1.22
2
 4 .58  10 5 N.
1
c) V  1000 1.31106  0.00131 m/s.  FD  1000  0.001312   0.052 1.0
2
 6.74  10 6 N.
8.19
a) Re 
VD


6  0.5
1.5 10
 FD 
b) Re 
8.20
 2 105.
CD  0.45 from Fig. 8.9.
1
1
 V 2 ACD  1.22  62    0.252  0.45  1.94 N.
2
2
15  0.5
1.5 10
 FD 
5
5
 5 105.
 CD  0.2 from Fig. 8.8.
1
1
 V 2 ACD  1.22 152    0.252  0.2  5.4 N.
2
2
The velocities associated with the two Reynolds numbers are
V1 
Re1 3 105 1.5 105

 101 m/s,
D
0.0445
Re2  6 104 1.5 105
V2 

 20 m/s.
D
0.0445
The drag, between these two velocities, is reduced by a factor of 2.5
C 
D high

 0.5 and CD  low  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.
8.21
2
1
1
 2
a) FD   V 2 ACD .  0.5   0.00238V 2    CD .  V 2CD  4810.
2
2
 12 
Re  VD /  (V  4/12)/1.6 104  2080V . Try CD  0.5 :
Try CD  0.4 :
V  98 fps, Re  2 105 .
V  110 fps, Re  2.3 105.
2
1
 2
b) CD  0.2 : 0.5   0.00238V 2     0.2.  V  155 fps. (Check those units.)
2
 12 
196
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Chapter 8 / External Flows
8.22
V  0.2
1
 2 105 V .
4.2  1000V 2  0.12 CD .  V 2CD  0.267. Re 

6
2
10
 V  0.73 m/s. Re  1.46 105.  OK.
Try CD  0.5 :
8.23
Re 
VD


40  2
5
 5.3 106. CD  0.7 . (This is extrapolated from Fig. 8.9.)
1.5 10
1
 FD  1.22  402  (2  60)  0.7  81 900 N.
2
M = 81 900  30 = 2.46  10 6 N  m.
8.24
a) Re1 
25  0.05
1.08 10
5
 1.2 105.
Re 2  1.8 105.
Re3  2.4 105.
Assume a rough cylinder (the air is highly turbulent).
  CD 1  0.7,  CD 2  0.8,  CD 3  0.9.
1
 FD  1.45  252 (0.05 10  0.7  0.075 15  0.8  0.1 20  0.9)  1380 N.
2
1
M   1.45  252 (0.05 10  0.7  40  0.075 15  0.8  27.5  0.1 20  0.9  10)
2
 25 700 N  m.
b) Re1 
25  0.05
1.65 10
5
 7.6 104. Re2  1.14 105 , Re3  1.5 105.
CD 1  0.8, CD 2  0.7, CD 3  0.8.  
101
 1.17 kg/m3.
0.287  308
1
 FD  1.17  252 (0.05 10  0.8  0.075 15  0.7  0.1 20  0.8)  1020 N.
2
1
M   1.17  252 (0.05  10  0.8  40  0.075  15  0.7  27.5  0.1 20  0.8  10)
2
 19 600 N  m.
8.25
Atmospheric air is turbulent. Use the "rough" curve: CD  0.7.
1
FD  10   0.00238V 2  6D  0.7.
2
 pmin 
2
 2000=V D.
5
10 
V  2000/V 2
1.6 104
.

0.0024  2
30  1042   11.8 psf.
U2  vo2  =




2
2
V 2 D  2370.
V  148 fps.
197
D  0.108 ft
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 8 / External Flows
8.26
Since the air cannot flow around the bottom, we imagine the structure
to be mirrored as shown. Then
L / D  40/5  8.
Remin 
VDmin


30  2
1.5 10
5
CD  0.66CD .
 4 106.
CD  1.0  0.66  0.66.
1
 28

 FD  1.22  302  
 20   0.66  36 000 N.
2
 2

8.27
FB  FD  FW .
FB
4
1
4
9810   r 3   1000V 2 r 2CD  9810  7.82   r 3 .
3
2
3
Re 
V  2r
106
 2 106 Vr.  V 2CD  178r
FD
W
a) r  0.05 m.  Re  105V , V 2CD  8.9. Assume a smooth sphere.
Try CD  0.5 :  V  4.22 m/s. Re  4.22 105. This is too large for Re.
Try CD  0.2 :  V  6.67 m/s. Re  6.67 105. OK.
b) r  0.025 m. Re  5 104V , V 2CD  4.45.
Try CD  0.2 : V  4.72 m/s. Re  2.4 105 . OK.
c) r  0.005 m. Re  104V , V 2CD  0.89.
Try CD  0.5 : V  1.33 m/s. Re  1.33 104 . OK.
d) r  0.001 m. Re  2 103V , V 2CD  0.178.
Try CD  0.4 : V  0.67 m/s. Re  1.33 103. OK.
8.28
3
2
3
4  10  1
4  10 
 10 
FB  FD  FW . 0.077       0.00238V 2   CD  62.4S    .
3  12  2
3  12 
 12 
Re 
V 10/12
1.6 10
4
 5.2 103 V .
1  0.0139V 2CD  810S
a) S  0.005. V 2CD  219. Assume atmospheric turbulence, i.e., rough.
Try CD  0.4 : V  23.4 fps. Re  1.2 105. CD  0.3 and V  27 fps.
b) S  0.02. V 2CD  1090. Try CD  0.4 : V  52 fps. Re  2.7 105. OK.
c) S  1.0. V 2CD  58, 200. Try CD  0.4 : V  381 fps.
198
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Chapter 8 / External Flows
8.29
6 in
Assume a 180 lb, 6-ft sky diver, with components
as shown. If V is quite large, then Re > 2  105 and
FD  FW .
6 in
8 in. dia.
18 in
3 ft
2.5 ft
2.5 ft
1
1
1
18


 4
 0.00238V 2  2  3  1.0  0.7+2  2.5  1.0  0.7   2.5 1.0       0.4   180.
2
2
2
12
 12 


We used data from Table 8.1.
8.30
From Table 8.2 CD  0.35.
2
V  140 fps.
1
FD  1.22V 2  3.2  0.35  0.683V 2 .
2
 80 1000 
a) FD  0.683  
  337 N.
 3600 
 W  337
80 1000
 7500 W or 10 hp.
3600
b) V  25 m/s. FD  0.683 252  427 N. W  427  25  10 700 W or 14.3 hp.
c) V  27.8 m/s. FD  0.683 27.82  527 N. W  527  27.8  14 700 W or 19.6 hp.
8.31
1.2FD  1.1  400.
1.2 
FD 
1
V 2 ACD . CD  1.1
2
1
 1.22V 2  (2  3)  1.1  1.1  400.
2
 V  9.5 m/s.
FD
1.1 m
1.2 m
FW
Fx
Fy
8.32
( 40 000 / 3600)0.6
 4.42  10 5 .  CD  0.35 from Fig. 8.9.
1.51  10 -5

1
1
a) FD   V 2 ACD  1.204  (40 000/3600)2  0.6  6  0.35  93.6 N
2
2
b) FD  93.6  0.68  63.7 N where L/D  6/0.6  10.
c) FD  93.6  0.76  71.1 N where we can use L/D  20 since only one end is free. The
ground acts like the mid-section of a 12-m-long cylinder.
8.33
a) Curled up, she makes an approximate sphere of about 1.2 m in diameter (just a guess!).
Assume a rough sphere at large Re. From Fig. 8.9, C D  0.4 :
1
1
FD   V 2 ACD 80  9.8  1.21 V 2  0.62  0.4.  V  53.7 m/s.
2
2
53.7 1.2
 4.27 106.  OK.
Check Re: Re 
5
1.5110
Re 
VD

199
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Chapter 8 / External Flows
b) FD 
1
 V 2 ACD . From Table 8.2, CD = 1.4:
2
1
80  9.8  1.21 V 2  42 1.4.
2
Check Re: Re 
close.
c) FD 
4.29  8
5
1.51 10
 V  4.29 m/s.
 2.27  106. Should be larger but the velocity should be
1
V 2 ACD
2
1
80  9.8  1.21 V 2 12 1.4.
2
Check Re: Re 
17.2 1
5
 V  17.2 m/s.
 1.14 106. This should be greater than 107 for C D to
1.5110
be acceptable. Hence, the velocity is approximate.
8.34
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.
1
1
 V 2 ACD   0.00238  882  (6  2) 1.1  122 lb.
2
2
W  FD  V  122  88  10, 700 ft-lb/sec or 19.5 hp.
8.35
FD 
8.36
FD 
8.37
The projected area is
1
1
 V 2 ACD  1.22  (27.8 1.6)2    0.052 1.1  10.43 N.
2
2
W  FD  V  2  10.43  (27.8 1.6)  2  226 W or 1.24 hp.
(2  0.3)
 4  4.6 m 2 .
2
1
1
FD   V 2 ACD  1.18  202  4.6  0.4  434 N.
2
2
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.
200
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Chapter 8 / External Flows
8.38
The net force acting up is (use absolute pressure)
4
4
120
Fup    0.43  1.21 9.8  0.5    0.43
 9.8  2.16 N
3
3
2.077  293
From a force triangle (2.16 N up and FD to the right), we see that
tan   Fup / FD .
a) FD  2.16 / tan 80   0.381. Assume CD = 0.2:
1
0.381  1.21V 2  0.42  0.2.  V  2.50 m/s.
2
2.5  0.8
Check Re: Re 
 1.33  10 5 . Too low. Use C D  0.5:
1.51  10 5
1
0.381  1.21V 2  0.42  0.5.  V  1.58 m/s
2
b) FD  2.16 / tan 70   0.786. Assume CD = 0.2:
1
0.786  1.21V 2  0.42  0.2.  V  3.60 m/s.
2
3.6  0.8
Check Re: Re 
 1.9  10 5 . Too low. Use C D  0.5:
5
1.51  10
1
0.786  1.21V 2  0.42  0.5.  V  2.27 m/s
2
c) FD  2.16 / tan 60   1.25. Assume CD = 0.5:
1
1.25  1.21V 2  0.42  0.5.  V  2.86 m/s.
2
2.86  0.8
OK.
Check Re: Re 
 1.5  10 5 .
5
1.51  10
d) FD  2.16 / tan 50   1.81. Assume CD = 0.5:
1
1.81  1.21V 2  0.42  0.5.  V  3.45 m/s.
2
3.45  0.8
 1.8  105.
Check Re: Re 
Close, but OK.
5
1.51  10
8.39
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, 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
 5

2
 2 1.21V  5  0.4  0.62   3  0.6   5000. V  54.2 m/s.
201
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Chapter 8 / External Flows
8.40
Power to move the sign:
1
V 2 ACD  V
2
1
  1.21  11.11 2  0.72  1.1  11.11  657 J / s.
2
FDV 
This power comes from the engine:
  0.3.
657  (12 000  1000)m
Assuming the density of gas to be 900 kg/m3,
.
 10 4  10  3600  6  52 
1825
  1.825  10 4 kg /s.
m
1000
 0.30  $683
900
8.41
The power expended is FD  V . V  (25  88 / 60) / 3.281  11.18 m/s
1
1
1.2111.183  0.56  CD  1.21 V 3  0.4  CD  0.8
2
2
V  13.47 m/s or 30.1 mph.
8.42
W  40  746  FD  V 
1
1
 V 2 ACD  V   ACDV 3.
2
2
1
 40  746  .9  1.22  3  0.35V 3.  V  34.7 m/s or 125 km/hr.
2
Vortex Shedding
8.43
40  Re  10 000. 40 
V  0.003
1.5 105
 10 000.  0.2  V  50 m/s.
f  0.003
.  flow  8 Hz.
0.2
St = 0.12 =
St  0.21=
f  0.003
.  fhigh  3500 Hz.
50
The vortices could be heard over most of the range.
8.44
40 
VD


6D
1.22  10
10 000 <
8.45
5
VD


.  D  8.13  105 ft.
6D
.  D  0.020 ft or 0.24".
.  10 5
122
From Fig. 8.10, Re is related to St: St 
Re 
VD


This is acceptable.
V  0.1
1.5 105
f  D 0.2  0.1
.

V
V
. Try St  0.21: V  0.095 m/s.  Re  630.
 V  0.095 m/s.
202
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Chapter 8 / External Flows
8.46
St 
fD 0.002  2

.
V
V
Re 
VD

Try St  0.21: V  0.0191 m/s.
8.47
V 2
. Use Fig. 8.9.
106
Re  38 103.  OK.

Let St = 0.21 for the wind imposed vorticies. When this frequency equals the natural
frequency, or one of its odd harmonics, resonance occurs:
T
f
2 2
L d 
0.21 10
 30 000/7850L2  0.0162   .
0.016
Consider the third and fifth harmonics:
f  3 T / L2 d 2 .
 L  1.56 m.
 L  0.525 m
f  5 T / L2 d 2 .
 L  2.62 m.
Streamlining
8.48
Re 
88  6/12
1.6 10
4
1
6

 2.8 105. FD   0.00238  882 1.0  0.8   6    22 lb.
2
 12 
The coefficient 1.0 comes from Fig. 8.9 and 0.8 from Table 8.1. We have
W  FD  V  22  88  1946 ft-lb/sec or
CD streamlined  0.035.
8.49
Re 
VD


3  0.08
1.5 10
5
3.5 hp.
 FD  0.77 lb. W  67.8 ft-lb/sec or 0.12 hp.
1
 16 000.  FD  1.22  32  (0.08  2) 1.2 .78  0.822 N
2
The coefficient 1.2 comes from Fig. 8.9 and 0.78 from Table 8.1.
CD streamlined  0.35.
8.50
Re 
VD
2  0.8
 FD  0.24 N.
 % reduction =
0.822  0.24
100  70.8%
0.822
 1.6 106.
CD  0.45 from Fig. 8.9.
10
4
L

 5. CD  0.62  0.45  0.28.
D 0.8
Because only one end is free, we double the length.
1
1
FD   V 2 ACD  1000  22  0.8  2  0.28  900 N.
2
2


6
If streamlined, CD  0.03  0.62  0.0186.
1
 FD  1000  22  0.8  2  0.0186  60 N.
2
203
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Chapter 8 / External Flows
8.51
V  50 1000 / 3600  13.9 m/s.
Assume the ends to not be free. Use C D from Fig. 8.9.
Re 
13.9  0.02
1.5 10
5
W  FD  V 
 1.85 104. CD  1.2.
CD streamlined  0.3
1
1
 V 3 ACD  1.2 13.93  0.02  20 1.2  773 W or 1.04 hp.
2
2
1
Wstreamlined  1.2 13.93  0.02  20  0.3  193 W or 0.26 hp
2
8.52
Re 
V  50 1000/3600  13.9 m/s.
13.9  0.3
 2.8  10 5 .
1.5  10 5
 CD  0.4
We assumed a head diameter of 0.3 m and used the rough sphere curve.
FD 
1
1
 V 2 ACD  1.2 13.92 (  0.32 /4)  0.4  3.3 N.
2
2
FD 
1
1
 V 2 ACD  1.2 13.92 (  0.32 /4)  0.035  0.29 N.
2
2
Cavitation
8.53

p  pv
1
2
V 2
.
0.7 
150 000  1670
1
1000V 2
2
where p   h  patm  150 000 Pa.
 V  20.6 m/s.
8.54
CL 
1
2
FL
V 2 A

200 000
1
1000  122  0.4 10
2
C D  0.0165 
?
 crit  0.75 
 0.69.
FD
1
1000 122  0.4 10
2
.
(9810  0.4  101 000)  1670
1
1000 122
2
204
  3 .
 FD  4800 N.
 1.43.
 no cavitation
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Chapter 8 / External Flows
8.55
CL 
FL
1
V 2 A
2
50 000

16
1
.  352   30
 194
12
2
C D  0.027 
?
 crit  1.6 
8.56
. .
 105
FD
.
1
1.94  352  (16/12)  30
2
(62.4 16/12  2117)  0.25 144
1
1.94  352
2
   7.3 .
 FD  1280 lb.
 1.82.
p  9810  5  101 000  150 000 Pa. pv  1670 Pa. Re 

150 000  1670
1
1000  202
2
 0.74.
 no cavitation
20  0.8
10
6
 16 106.
 C D  C D (0)(1   )  0.3(1  0.74)  0.52
1
1
 V 2 ACD  1000  202    0.42  0.52  52 000 N.
2
2
Note: We retain 2 sig. figures since C D is known to only 2 sig. figures.
 FD 
8.57
For a 6 angle of attack we find from Table 8.4 C L  0.95.
FL 
1
1
V 2 ACL   1000  152  4  0.4 L.95  12 000  9.8.
2
2
 L  0.69 m.
Added Mass
8.58
4
400
F  ma. a) 400  9810    0.23 
a.
3
9.81
 a  1.75 m/s 2.
4
4
 400 1

 1000    0.23  a.
b) 400  9810    0.23  
3
3
 9.81 2

8.59
F  ma1  1000  1.2  V
 a 1.
F  (m  ma )a2 .
 a2 
 a1 
 a  1.24 m/s2 .
F
. ma  0.2  1000 
V.
1200 
V
F
F

.
1200 
V  200 V
1400 
V
F
F

a2  a1
1200 V
 % error =
 100  1400 V
F
a2
1400 V
205
a2 is true acceleration.
 100  16.7%.
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Chapter 8 / External Flows
Lift and Drag on Airfoils
8.60
The total aerodynamic drag consists of both lift and drag that is:
FTotal  FL  FD  FTotal  FL2  FD2  18 kN  FL2  FD2  18 kN 
2
which can be combined with FL  3FD to yield
9FD2  FD2  324  FD  324 10  5.69 kN
and hence, FL  3  5.69  17.1 kN so
CL 
8.61
1
2
FL
 V 2cL

17.1103
1 1.2  61.12 1.3 10
2
 0.587
From the measured force we can calculate the lift coefficient as follows
CL 
1
2
FL
2
 V cL

13.7
 0.70
1  0.0233 150  6 18 144
2
Where the velocity V was calculated from the given value of Re as
V
Re
4.586 105

 150 ft/sec
 c  0.00233   6 12  / 3.8110 7
The angle of attack is calculated from the given expression for CL
  sin1  CL 2   sin1  0.7 2   6.4
8.62
CL 
1
2
FL
2
V A

1000  9.81
1
 0.412  802 15
2
 0.496.
  3.2 . C D  0.0065.
1

W  FDV    0.412  802 15  0.0065   80  10 300 W or 13.8 hp.
2

8.63
a) C L  1.22 
1500  9.81  3000
1
1.22  V 2  20
2
b)  C L max  1.72 
.
 V  34.5 m/s.
1500  9.81  3000
1
 0.412  V 2  20
2
.
 V  50 m/s.
(at 10 000 m) c)
1

W  FDV    0.412  802  20  0.0065   80  13 700 W or 18.4 hp
2

where we found C D as follows:
1500  9.81  3000
 0.67.  C D  0.0065, from Fig. 8.13.
C L cruise  1
 0.412  802  20
2
Power = 18.4/0.45  40.9 hp.
206
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Chapter 8 / External Flows
1500  9.81  3000
8.64
C L  1.22 
8.65
C L cruise  1
1
1.007  V 2  20
2
.
1500  9.81  3000
2
1.007  802  20
 V  38.0 m/s.
 0.275.
 CD 
0.275
 0.0057.
48
1
W  FDV  1.007  803  20  0.0057  29 400 W or 39.4 hp
2
39.4  18.4
% change =
 100  114% increase
18.4
The increased power is due to the increase in air density.
1500  9.81  9000
8.66
C L  1.22 
8.67
C L  1.72 
8.68
a) C L  1.72 
1
1.22  V 2  20
2
250 000  9.81
1
1.22  V 2  60  8
2
b) CL  1.72 
.
 V  69.8 m/s.
250 000  9.81
250 000  9.81
 V  75.2 m/s.
101.3


 1.515 kg/m3 
.  V  62.6 m/s   
0.287  233


62.6  69.8
 100  10.3%
69.8
250 000  9.81
1
1.093V 2  60  8
2
% change =
.
75.2  69.8
 100  7.77% increase
69.8
1
1.515V 2  60  8
2
% change =
c) CL  1.72 
 V  39.9 m/s.
1
1.05  V 2  60  8
2
% change =
8.69
.
.
101.3


 V  73.7 m/s   
 1.093 kg/m3 
0.287  323


73.7  69.8
 100  5.63% increase
69.8
For a conventional airfoil assume C L /CD  47.6 at C L  0.3.
0.3 
m  9.81
1
 0.526  2222  200  30
2
.
 m  2.38  106 kg
1
0.3
W  FDV   0.526  2223  200  30 
 490 000 W or 657 hp
2
47.6
207
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Chapter 8 / External Flows
Vorticity, Velocity Potential, and Stream Function
8.70
p
 V

 
 (V )V 
   2 V   0.

 t


V 

.
 ( V ) 
t t
t

p


1

p  0.
 (2V)  2 ( V)  2ω (we have interchanged derivatives)
1
 1
 (V )V     V 2  V  ( V)   (V 2 )   (V  ω)
2
 2
 V (  ω)  ω (  V )  (V )ω  (ω )V
 (V )ω  (ω )V
since   ω    (  V )  0 and   V  0.
There results:
ω
 (V )ω  (ω )V 2ω  0.
t
This is written as
Dω
 (ω )V 2ω.
Dt
8.71
Starting with the vorticity equation, Eq. (8.5.3), we write
D
      V  2 where,   xˆi  y ˆj  zkˆ
Dt
Since initially y-vorticity exists in the flow then, x  z  0 .
To explain the existence of x-vorticity, write the vorticity equation in the x-direction:
Dx
u
 y
  2x . Initially, x  0, so 2x  0 .
Dt
y
Dx
u
u
 0  y
 0, and hence
0
Downstream of the obstruction,
Dt
y
y
which indicates that x-vorticity is being generated in the flow due to the re-orientation of
the y-vorticity tube in the x-direction.
8.72
x-comp:
y-comp:
z-comp:
x



u
u
u
 u x  v x  w x  x
 y
 z
2x
t
x
y
z
x
y
z
y
t
u
y
x
v
y
y
w
y
z
 x
v
v
v
 y
 z
2y
x
y
z
z



w
w
w
 u z  v z  w z  x
 y
 z
2z
t
x
y
z
x
y
z
208
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Chapter 8 / External Flows
8.73
x 
w v
u w

 0. y 

 0.
y
z
z
x
z 
v u

 0.
x y
Dz
Dz
 (ω )w  2z ; 
  2z .
DT
Dt
D z
 0.
Dt
Thus, for a planer flow,  z  const if viscous effects are negligible.
If viscous effects are negligible, then
8.74
 w v  ˆ  u w  ˆ  v u  ˆ
a)   V  
 i  
 j     k  0.
 y z   z x   x y 
irrotational

 10x.   5x2  f ( y)
x
 f

 20 y.
y y
 f  10 y 2  C. Let C  0.
   5 x 2  10 y 2
b)  V  0ˆi  0ˆj  (8  8)kˆ  0.
irrotational

 8y.   8xy  f ( y, z) .
x
 df

 6z.
z dz
f

 8x 
 8x.
y
y

f
 0 and f  f ( z).
y
 f  3z 2  C. Let C  0.
   8 xy  3z 2
  y 1 ( x 2  y 2 ) 1/2 2x  x 1 ( x 2  y 2 ) 1/2 2 y 
2
2
 kˆ  0. irrotational

c)   V  0ˆi  0ˆj  
2
2


x y
x2  y 2



x

.   x 2  y 2  f ( y )
2
2
x
x y
y
f
 1 2
.
 ( x  y 2 )1/2 2 y 

y 2
y
x2  y 2

f
 0.  f  C. Let C  0.
y
  x2  y 2
209
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Chapter 8 / External Flows
  y (2 x)
 x(2 y )  ˆ

d)   V  0ˆi  0ˆj   2
 k  0.
2 2
( x 2  y 2 ) 2 
 ( x  y )

1
x
 2
.   ln( x 2  y 2 )  f ( y )
2
x x  y
2
irrotational
y
f
f
1 2y

.
 2



 0.  f  C  0.
y x  y 2 2 x2  y 2 y
y
8.75
 2
x 2

 2
  ln x2  y 2
 0. This requires two conditions on x and two on y.
y 2
At x   L, u  U .
At x  L, u  U .


At y   h,  = 0.
At y  h,   U  h.

 U.
y
y=h
y
U

 U.
y
x = L
x
y=0
(See Example 8.9).
The boundary conditions are stated as:


( L, y)  U ,
( L, y )  U ,  ( x, h)  0,  ( x, h)  2Uh.
y
y
8.76
u
df


 100.   100 y  f ( x). v  

 50.  f  50x  C.
dx
y
x
 ( x, y)  100y  50x.
u
(We usually let C = 0.)

 df
 100.   100x  f ( y). v 

 50.  f  50 y  C.
x
y dy
  ( x , y )  100 x  50 y.
8.77
a)   40 .
b)
1   

r r  
2
1 
 1    1 

(40)   
(0)   0.



 

r  
 r  r  r r

It is incompressible since the above continuity equation is satisfied.
Note: The continuity equation is found in Table 5.1.
210
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Chapter 8 / External Flows
c)
 1  40

 .   40ln r  f ( )
r r 
r
 f


 r
 0.
 
r
 f  C. Let C  0.
  40 ln r
40
, v  0.
r
d) vr 
ar  vr
vr 40  40 
   2   10.
r
r  r 
 r  5.43 m
8.78
u
2y


1 y
 20 2




 f ( y).

.
40
tan
y
x
x  y 2 x
v
f
f
40 / x
40x
2x

.  f  C. Let C  0.


 2

 20 2
2
2
2
y
y
y
1 y / x
x y
x  y2
y
x
  40 tan 1 .
8.79
a)
 2
x
2
 2
x
2

 2
y
2
 0.

 10 y ( x 2  y 2 ) 2 (2 x).
x
 20 y( x 2  y 2 )2 80x 2 y( x 2  y 2 ) 3

 10  10(x2  y 2 )1  10y(x2  y 2 )2 (2y).
y
 2
y

2
 20 y( x 2  y 2 ) 2 40 y( x2  y 2 )2  80 y3 ( x2  y 2 )3.
 2
x2


 2
y 2

20 y
( x2  y 2 )2
80 y( x 2  y 2 )
(x 2  y 2 ) 3


80x2 y
60 y
80 y3


(x2  y 2 )3 (x2  y 2 )2 (x2  y 2 )3
80 x 2 y
(x 2  y 2 ) 3

80y 3
(x 2  y 2 ) 3
211

80x 2 y  80y 3  80x 2 y  80y 3
(x 2  y 2 ) 3
 0.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 8 / External Flows
b) In polar co-ord:  ( r ,  )  10r sin  
10r sin 
10
 10r sin   sin  .
2
r
r
1  
10 

10 

 10  2  cos   .   10   cos   f ( ).
r
r  
r 

r 
1  1 df 
10 
10
df


 10  2  sin   
 10sin   2 sin  .
 0. f  C .
r  r d 
d
r
r 
r

   10 r 

1
10 x
,
 cos  or  ( x , y )  10 x  2

r
x  y2
where we let r cos   x and r 2  x 2  y 2 .

 0 where we let y = 0 in part (a) and
x
20 y 2

10
10
u
 10  2
 2
 10  2 with y  0.
2
2
2
y
(x  y )
x y
x
c) Along the x-axis, v  
p
p
10  20 
u

  .   10  2  3    .
x
x
x
x  x 

 200 200 
 50 100 
 p    5  3  dx    4  2   C. C  50 000.
x 
x 
 x
 x
 100 50 
 1000  2  4   50 000 Pa. (Could have used Bernoulli!)
x 
x
10
d) Let u  0: 0  10  2 .  x  1.  Stag pts: (1, 0), (1, 0)
x
Euler’s Eq: u

8.80
a)
 2
x 2

 2
y 2


 
10 x    10 y  ( x 2  y 2 )10  10 x(2x

10

 

x 
(x 2  y 2 )2
x 2  y 2  y  x 2  y 2 
( x 2  y 2 )10  10 y(2 y )
(x 2  y 2 )2

10 x 2  10 y 2  20 x 2  10 x 2  10 y 2  20 y 2
(x 2  y 2 )2
 0.
b) Polar co-ord:   10r cos   5ln r 2 .
(See Eq. 8.5.14.)

10r 1 
 10cos   2 
.   10r sin   10  f (r )
r
r 
r
1 

df
 10sin   
 10sin   .  f  C .   10r sin   10 .
r
r 
dr
y
 ( x , y )  10 y  10 tan 1 .
x
212
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8 / External Flows
10 y

 2
. Along x-axis (y = 0) v = 0.
y x  y 2

10 x
10
u
 10  2
.
Along x-axis u  10  .
2
x
x
x y
c) v 
Bernoulli:
V2 p
V2 p
  gz      gz

2 
2
(assume z  z )
(10  10/x) 2 p 102 100 000
2 1 
 

.  p  100  50   2  kPa.


2
2
x x 
d) u  0:
0  10 
10
.
x
 x  1.
Stag pt: (1, 0)


e) ay  v v /y  u v /x  0 on x-axis. ax  uu/x  v u/y  10  10/x  10/x2 .
 ax (2, 0)  (10  5)  10/4   12.5 m/s2.
8.81
u ( x, y )  y  5 y 2 
q
0.2

0

y 2 5 y3
1
.  

 C .   (3 y 2  10 y 3 ).
y
2
3
6
udy 
1
6
0.2

( y  5y 2 )dy 
0
0.22
0.23
5
 6.67 103 m 2 /s.
2
3
 2 1  (3  0.22  10  0.23 )  0  6.67 103 m2 /s.

u
 1  10 y  0.
y
  doesn’t exist.
Superposition of Simple Flows
8.82
5
5
  30r sin    .
2
2
1 
5
a) vr 
 30cos    0.
2r
r 
5
At    ,
 30.  rs  0.0833' .
2rs
( 1" ,0).
Stag. pt:
  30y 
b) At    , r =.0833,  s 
 r  yinter  0.0119 ft.
30 fps
y
= 5/2
=0
x
5
 5
 30r sin 
.
2
2 22
213
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 8 / External Flows
c) q  U  H   .  30 H 
d) vr (1,  )  30cos  
8.83

5
5
5
. H =
. Thickness = 2 H 
ft or 1.257".
2
60
30
5
 30  2.5  27.5.  u  27.5 fps.
2
1/2
1/2
 
 
ln (x  1)2  y  
ln (x  1)2  y 2   10x


2 
2 
1
1
 ln (x  1)2  y 2   ln ( x  1)2  y 2   10x.



4
4 
u

x
y 0
1
1
[2( x  1)]
[2( x  1)]
1
1
4
4

 10 

 10. v  0 if y  0.
2
2
2( x  1) 2( x  1)
( x  1)
( x  1)
At the stagnation point, u  0. 
 x 2  1.1.
2
1
1

 10  0.  2
 20.
2( x  1) 2( x  1)
x 1
 x  1.049 m.
 oval length = 2  1.049 = 2.098 m.
All the flow from the source goes to the sink, i.e.,  m2 /s, or

u( y) 
x
x 0

2
1
1
(2)
(2)
1
 4 2  4 2  10 
 10.
1 y
1 y
1  y2
m2 /s for y  0.
y
(0, h)
x
h

1


 10 dy  .  tan 1 h  10h  .
q  
2


2
2

0  1 y
h = 0.143 m so that thickness = 2h = 0.286 m.
The minimum pressure occurs on the oval surface at (0, h).
There u 
Bernoulli:
1
1  0.1432
 10  10.98 m/s.
p
10 2 10 000
V 2 p V2 p  10.98 2
.
.
 




1000
2

2

2
1000
2
 p min  280 Pa.
214
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Chapter 8 / External Flows
8.84

1/2 2
1/2
2 
ln (x  1)2  y 2  
ln (x  1)2  y 2   2x



2 
2
1
1
 ln (x  1)2  y 2   ln (x  1)2  y 2   2x.
 2 

2 
1
1
2( x  1)
2( x  1)
y
y

2
2.
u
v
 2 2




x ( x  1)  y 2 ( x  1)2  y3
( x  1)2  y 2 ( x  1)2  y 2
Along the x-axis (y = 0), v = 0 and u 
Set u  0:
Stag. pts.:
1
1

 2, or x 2  2.  x   2 .
x1 x1
( 2 ,0), (  2 ,0).
u(4,0) 
u(0, 4) 
8.85
1
1

 2.
x1 x1
1
1

 2  1.867 m/s.
4  1 4  1
1
1 4
2

1
1 4
2
v(4,0)  0.
 2  2.118 m/s.
v(0, 4) 
1/2 2
1/2
2  2
ln x  ( y  1)2  
ln  x2  ( y  1)2 


2 
2 
1
1
 ln  x2  ( y  1)2   ln  x2  ( y  1)2  .
 2 

2 
x
x

u
.
 2
 2
2
x x  ( y  1)
x  ( y  1) 2
4
1 4
2

4
1  42
 0.

v
y
y 1
y 1

.
 2
 2
2
y x  ( y  1)
x  ( y  1)2
x
At (0, 0) u = 0 and v = 0.
At (1, 1) v  0 
8.86
2
2
2
2 1
 0.4 m/s. u 
1
2
1

1
2
2
2 1
 1.2 m/s.  V  1.2ˆi  0.4ˆj m/s.
1/2 2
1/2
2 
ln ( y  1)2  x2  
ln ( y  1)2  x2   U x.


2 
2 
1
1
 ln ( y  1)2  x2   ln ( y  1)2  x2   U x.



2
2 

215
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Chapter 8 / External Flows
a) Stag. pts. May occur on x-axis, y = 0.

x
x


 10.
u
2
x y 0 1  x 1  x2
y
x
 x 2  0.2 x  1  0. no stagnation points exist on the x-axis.
(They do exist away from the x-axis.)
h
1
Along the y-axis: u( y )  10. q   udy  (2 )   m 2 /s.
2
0
h
   10dy  10 h.
 h  0.314 m.
0
2x
 x 2  2 x  1  0.
2  1.
1 x
Stag. Pt: (1, 0)
b) u 
2x
 0.2.  x 2  10x  1  0.
1 x2
Stag. pts: (9.9, 0) , (.1, 0).
c) u 
8.87

y
x
   1  h.
Along the y-axis: u  1.0.
Along the y-axis: u  0.2.
 x  1 m.
 h  3.14 m.
 x  9.90,  0.10 m.
   0.2 h.
 h  15.71 m.
60
cos   8r cos  .
r
a) vr 

60
 60 
  2 cos   8cos    8  2  cos  .
r
r
r 

At the cylinder surface v r  0 for all  . Hence,
60
 8.
rc2
b) Bernoulli:
c) v 
p  
 rc  2.739 m
U 2
82
 1000
 32 000 Pa or 32 kPa
2
2
1 
60
  2 sin   8sin  .
r 
r
At r  rc , v  8 sin   8 sin   16 sin 
2
v 90

16 2
 1000
 128 000 Pa or 128 kPa
d) p  
2
2
216
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Chapter 8 / External Flows
8.88

4
20

ln r  2  10ln r at (x, y)  (0,1) which is r( , ) (1, /2).
2
2
a) vr (1,  /2) 
1  1
 (2)  2.
r  1

10
   10.
r
1
2
 1.18.
1.7
v (1.7,  /4)  
2
 0.625.
3.2
v (3.2,0)  
2
 0.333.
6
v (6,  /4)  
vr (1.7,  /4) 
vr (3.2,0) 
v (1,  / 2)  
vr (6,  /4) 
10
 5.88
1.7
10
 3.125
3.2
10
 1.67, etc.
6
2
10
and v   . From Table 5.1 (use the l.h.s. of momentum):
r
r
2
v v 2 2  2  100
Dvr v

 vr r      2   3  104 m/s 2
ar 
r
Dt
r
r
r r  r
Dv vr v
vv
v
2  10  2(10)
a 

 vr   r    2  
0
Dt
r
r
r r 
r
r3
b) vr 
a(0,1)  104i r or (ax , ay )  (0, 104) m/s2
2
10
 0.1414, v (14.14,  /4)  
 0.707 m/s
14.14
14.14
2
10
 20, v (0.1,  /2) 
 100 m/s
vr (0.1,  /2) 
0.1
.1
p 20 2  100 2
20 000 0.1414 2  0.707 2



.  p  13 760 Pa
Bernoulli:
2
1.2
2
1.2
We used air  1.2 kg/m3 at standard conditions.
c) vr (14.14,  /4) 
8.89
40
.
r2
40
rc 
 2.
10
Along the y-axis v r  0 and v  10 
We have set  

2
in Eq. 8.5.27.
40
cos  . ( 4 ,3)  (5,126.9  ).
2
r
40
v  10sin   2 sin  .  vr  6.96 m/s, v  9.28 m/s.
r
a)
10 m/s
b) v r  10 cos  
217
20 m/s
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Chapter 8 / External Flows
c) Use Eq. 8.5.28: p  p 0  2 U 2 sin 2 
Drag =
 /2

 /2
p cos  rc Ld  p90  2rc L.

0
p
p90
d

 /2
2
p90 
p0  2 U 2 .
 p0  2U2 sin 2   cos  rc Ld  p90  2rc L
 /2

sin 3  
 2rc L  p0 sin   2 U 2

3 

0
  p0  2 U 2  2rc L


8
 rc LU2 .
3
(8 / 3)rc LU2 8
Drag
CD  1
 1
  2.667.
3
U2 A
U2 2rc L
2
8.90
a) v r  U  cos  
2

r2
4
.
r2

 sin  
4
 U sin  
  4  2  sin   8sin  .
b) v  
2
r
1 

rc
For    , v r  4 
c) pc  p   
x
cos  . Let U   4 ,   rc2U   12  4  4.
x = 1
vr
v2
V2
8 2 sin 2 
42
    50 000  1000 
 1000
.
2
2
2
2
 pc  58  32sin 2  kPa.
 /2
d) Drag = 2

(58  32sin 2  ) cos  11d  26  2 1
0

 1 
 2 58  32     52  42.7 kN. (See the figure in Problem 8.89c.)
 3 

218
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Chapter 8 / External Flows
8.91
On the cylinder v  2U sin  
Used rc 

U

1000
 60sin  
, where we have
2 rc
2  3.651
400
 3.651 ft.
30


x6
x6
If u( x, y)  0.0318 

2
2
2
( x  6)  ( y  2) 2
 ( x  6)  ( y  2)

  227 , 313 .
x6
2
( x  6)  ( y  2)
2

x6
2

2
( x  6)  ( y  2) 
Stag. pts.: (3.651 ft, 227) , (3.651 ft, 313).
Max. pressure occurs on the cylinder at a stagnation pt.:
v  2U sin  
 pmax 

1000
 60sin  
,
2 rc
2  3.651

0.0024  2 2 
U2  vo2  =
30  0  1.08 psf.



2
2 
Min. pressure occurs at the top of the cylinder where   90  and the velocity is:
v90  2U sin  
 pmin 
8.92

1000
 2  30 
 104 fps
2 ro
2  3.651

0.0024  2
U2  vo2  =
30  1042   11.8 psf.


2
2 
v  2  20sin  

. For one stag. pt.: v  0 at   270 :
2  0.4
0  2  20sin 270 
  2 rc2.
 

.
2  0.4

2 rc2

  2  20  2  0.4  100.5 m2 /s.
100.5
2  0.42
 100 rad/s. (See Example 8.12.)
Min. pressure occurs where v is max, i.e.,    / 2. There
v  2  20 1 
 pmin
100.5
 80 m/s.
2  0.4
v2
V2
202
802
    0
 p 
1.22 
1.22  3660 Pa.
2
2
2
2
219
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Chapter 8 / External Flows
8.93
  2 rc2  2  0.62 120  2 /60  28.42 m 2 /s.   rc2U  0.62  3  1.08 m3 /s.
28.42
 v  2  3sin  
. sin   1.256. Impossible.
2  0.6
Stag. pt. is off the cylinder at   270  , but r  rc . From Eq. 8.5.29,



1.08
28.42
v  
 U sin   2 sin  
 3(1)  2 (1) 
 0.
r
2 r
2 r
r
r
1.08 4.523
3  2 
.  r 2  1.508r  0.36  0.  r  1.21 m.
r
r
28.42
Stag. pt: (1.21, 270). (v )90  2  3 
 13.54 m/s.
2  0.6
 32 13.542 
Min. pressure occurs at   90 , at r  rc : pmin   
1.22  106 Pa.
 2
2 

 32 1.542 
Max. pressure occurs at   270 , at r  rc : pmax   
1.22  4.04 Pa.
 2
2


8.94
At 15,000 ft,   0.0015 slug/ft 3.
Lift = UL  0.0015  350 15,000  60  472,000 lb.
8.95
For a flat plate, transition to turbulent flow occurs when Rex  3 105 for flow on rough
plates or when Rex  5 105 for flow on smooth plates. Assuming the wing has a rough
surface, or the flow is disturbed, we determine the distance from the leading edge at
which transition to turbulence occurs as follows:
U xT
3  105
 3  105  xT 

U 
The density of air is determined using the ideal gas law

p
628

 8.88  104 slug/ft 3
RT 1716 48  460 R


At an altitude of 30,000 ft the dynamic viscosity is   3.11107 lb-sec/ft 2 .
Solving for xT yields:
3 105
xT 
 0.143 ft
ft/sec 
3 
2
4
7
8.88 10 slugs/ft  500 mph 1.466
3.1110 lb-sec/ft
mph 




220

© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8 / External Flows
8.96
Place four sources as shown. Then, with q  2 for each:
x2
x2

u( x, y) 
2
2
( x  2)  ( y  2)
( x  2) 2  ( y  2) 2
x2
x2


2
2
2
( x  2)  ( y  2)
( x  2)  ( y  2) 2
y2
v(x, y) 
y2

( x  2)2  ( y  2)2 ( x  2)2  ( y  2)2
v(4, 3) = (0.729, 0.481) m/s

y2
( x  2)2  ( y  2) 2
y
x

y2
( x  2) 2  ( y  2) 2
y
8.97
2
Place four sources with q  0.2 m /s, as shown.
(6, 2)
x


x6
x6
x6
x6



u( x , y )  .0318 
2
2
2
2
2
2
2
2 
( x  6)  ( y  2)
( x  6)  ( y  2)
( x  6)  ( y  2) 
 ( x  6)  ( y  2)


y2
y2
y2
y2



v( x , y )  .0318 
2
2
2
2
2
2
2
2 
( x  6)  ( y  2 )
( x  6)  ( y  2)
( x  6)  ( y  2) 
 ( x  6)  ( y  2 )
q
0.2

 0.0318.
2
2
2
10
10 
 2
Then u(4,3)  0.0318 



 0.00922 m/s.
 4  1 4  25 100  1 100  25 
1
5
5 
 1
v(4,3)  0.0318 



 0.01343 m/s.
 4  1 100  1 4  25 100  25 
where
Boundary Layers
8.98
Recrit 
U xT

.
 xT 
6 105
 2000 .
300
a)   1.56 104 ft 2 /sec.  xT  2000 1.56 104  0.312' or 3.74"
b)  

 2.1104 ft 2 /sec.  xT  2000  2.110 4  0.42 ' or 5.04"

c)   3.47 104 ft 2 /sec.  xT  2000  3.47 104  0.694' or 8.33"
221
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Chapter 8 / External Flows
8.99
a) Use Recrit  3 105 
10xT
.  x T  0.453 m.
1.51105
10xT
 x T  1.51 m.
b) Use Recrit  106 
.
1.51105
10xT
c) Use Recrit  3 105 
.  x T  0.453 m.
1.51105
10xT
d) Use Recrit  3 105 
.  x T  0.453 m.
1.51105
10xgrowth
e) Re  6 104 
 xgrowth  0.091 m or 9.1 cm .
.
1.51105
Note: A rough plate, high free-stream disturbances, or a vibrated smooth plate all
experience transition at the lower Re crit .
8.100 a) Use Recrit  3 105  10xT /106.
 x T  0.03 m or 3 cm.
b) Use Recrit  106  10xT /106 .
 x T  0.1 m or 10 cm.
c) Use Recrit  3 105  10xT /106 .  x T  0.03 m or 3 cm.
d) Use Recrit  3 105  10xT /106 .  x T  0.03 m or 3 cm.
e) Re  6 104  10xgrowth /106 .
8.101 Recrit  6 105 
U  2

 xgrowth  0.006 m or
. For a wind tunnel: 6 105 
For a water channel: 6 105 
6 mm
U  2
.
1.5 105
U  4.5 m/s.
U  2
106
. U  0.3 m/s .
8.102 The x-coordinate is measured along the cylinder surface as shown in Fig. 8.19. The
pressure distribution (see solution 8.89) on the surface is
p  p0  2U 2 sin 2  where rc  x ( is zero at the stagnation point).
Then
p( x)  20 000  2 1000 102 sin 2 ( x/2)
 20  200sin 2 (x/2) kPa
The velocity U(x) at the edge of the b.l. is U(x) on the cylinder wall:
v ( r  2)  10 sin   10 sin   20 sin(   )  20 sin 
U(x)  20sin(x/2)
222
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Chapter 8 / External Flows
8.103 U ( x)  v at rc  1. v  8sin  . U ( x)  8sin x since x   rc .
p( x)  58  32sin 2   58  32sin 2 x kPa
8.104 The height h above the plate is h( x)  mx  0.4.
0.1  m  2  0.4
 m  0.15
2.4
 h(x)  0.4  0.15x. Continuity: 6  0.4  U( x)h. U( x) 
or
0.4  0.15x
16
U ( x) 
.
2.67  x
dp
p
256
u
16
16

.
 

u   .
Euler’s Eqn:
dx
x
x
2.67  x (2.67  x )2 (2.67  x) 3
Von Karman Integral Equation
8.105 Refer to Fig. 8.25 to respond to this problem.

a) mtop  mout  min    udy 
0





 udy dx    udy    udy dx

x 0
x 0
0
dp
)d  ( p  dp)(  d )
2
  0dx   dp  higher order terms  momout  momin  momtop
b) Fx  p   0dx  ( p 

momout  momin  momtop    u2dy 
0




  


2
2
u
dy
dx
u
dy
U
(
x
)
udy
dx








 x 

x 0
0
 0


  


2
(
)
u
dy
dx
U
x
udy
dx





 x 

x 0
 0


dp
d
d
8.106  0  
 U ( x)  udy   u 2 dy
dx
dx 0
dx 0
 

 d 
dp d 
dU 
udy
uUdy




u 2 dy





dx  0
dx dx 0
 dx 0
where we have used g


dg 
df dfg

f
.  Here U  g , f   udy.
dx 
dx dx

0

dp d 
dU
  0  

u(U  u)dy  
udy.
dx 0
dx dx 0
223
(   const. )
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Chapter 8 / External Flows
8.107



dp
 d 2
1
dU
dU  1

 
U   U
  Udy where U   Udy.
2 dx
 0
dx
dx   0
dx


 dU  1 

dU
d
2
 0    
udy
Udy    (U )  

dx 0
dx
dx   0
 



d
dU
d
dU
(U 2 )  
(U  u)dy   (U 2 )  
U d .

dx
dx 0
dx
dx

dU
d
 0 and  0  
u(U  u)dy.
8.108 a) If dp /dx  0 then
dx
dx 0


y
y
y y
d
d  2
d  2  

   U2
 
 0    U2 sin
1  sin
cos
dy  U2




2 
2 
2 2  0
dx 0
dx  
dx   2 
Also, we have  0  
 U
b)  0  U
c)
u

cos 0.
 U
2
y y 0
d


 0.137 U2
.  d  11.5
dx.
2
dx
U
  4.79
x
U
.
 1
U
U
.
 0.328U
x
2 4.79  x
 ayx 3/2 
 y
ay
U 
u


v
 ay 
 U sin 


 .
sin
U
U

 cos


 


x
x
x
y
2 
x
 x
 2  4.79  x 



 v  U
0

U 
U3
dy
cos  0.328y
0.0316



 x 


0.164 y U
x
3/2
224


U  
y cos  0.189
 y  dy.



 

0

© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8 / External Flows
8.109 u  U
y


.
y
y y
d
0  
U2 1   dy
 
dx

U
0

0  
U
U
d
1
u
.
   .     U2


dx
6
y
 2  12

x.  ( x)  3.46
U

d 2    1
d
.
U     U2
dx
dx
2 3 6
% error in  (x) 
x
U
 d  6
.  0  0.289 U

U
u = Uy/
dx
U
.
x
5  3.46
100  30.8% low.
5
% error in  0 (x) 
0.332  0.289
100  13% low.
0.332
 /2
 /6
y
y 1
d 
2 y
2  y 1 
8.110  0 
  3U 1  3  dy  U   1    dy
dx 


  3   3 
 /6
0




y 2 
y 2 

 1 
  dy 
 3 3  3 3  
 /2

3U

d
d
U2 (0.1358 )    . 

 22.08
.

U
dx
dx


U2 
 6.65  
2
1/2
,  0 ( x)  0.1358 U2 
  0.451U Re x .
U
 2 U x 
6.65  5
0.451  0.332
% error for  
 100  33%. % error for  0 
 100  36%
5
0.332
Thus,  ( x)  6.65
x

8.111 Continuity from entrance to x: U 0 H  2 u( y )dy  U ( x )( H  2 ).
0


0
0
Write U ( x )  U ( x ) dy   U ( x )dy . Then, continuity provides


U0 H
.
H


2
d
0
0
If we were to move the walls out a distance  d ( x ), then U ( x ) would be constant since
U 0 H  2  ( u  U )dy  UH  UH  2  (U  u)dy  UH  2U d .
 H  2   2  would be constant; then U (x )  U
d
d
0
U ( x ) 
. For a square wind tunnel,
displace one wall outward 4 d for dp / dx  0.
225
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Chapter 8 / External Flows
8.112 The given velocity profile is that used in Example 8.13. There we found
  5.48  x /U  5.48 106 x /10  0.00173 x  0.00173 3  0.003 m.
Assume the streamline is outside the b.l. Continuity is then
0.003
10  0.02 

0
 2y
y2 
10 

dy  (h  0.003)10
 0.003 0.0032 


 0.02  10 h  0.03.
1
d 
10
0.003

0
 h  0.021 m or 2.1 cm

20 y
10 y 2 
1


dy
 0.03  0.03  .01  0.001 m
10 

2
0.003
10
0.003


h  2  2.1  2  0.1 cm or 0.001 m.
The streamline moves away from the wall a distance  d .
8.113 From Prob. 8.111 we found that we should displace the one wall outward 4 d .
From the definition of  d :
4
h( x)  4 d 
10


20 y 10 y 2 
 4

 2  dy  4        
10 
3 3


 
0

4
1.86 105 x /10 
  0.00735 x m
  5.48
3
160 / (0.287  303) 


We used  ( x ) found in Example 8.13,   p/RT, and    /.

 3 y 1 y3 
 3 y 1 y3 
1
3
1



  0.375 .
.
U
1
dy







8.114 a) u  U  



d

 2  2 3 
 2  2 3 
U
4
8





0

From Eq. 8.6.16,  d  0.375  4.65

1
U2


0
x
U
 1.74
x
U
.
% error = 1.2%.
1 y3   3 y 1 y3 


1
dy  0.139 .
3
3
 

 2  2   2  2  

3
U 2 

  0.139  4.65
y
x
U
 0.648
x
U
226
.
% error =
0.648  0.644
100  0.62%
0.644
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8 / External Flows

 y y2 
 2y y2 
 
b) u  U  2  2  . See Example 8.13.  d  1 
 2  dy       .
   


3 3
  


0

 d 


5.48  x
x
 1.83
.
U
3 U
 y
y2  
% error =
y2 
y
1.83  1.72
100  6.4% .
1.72
1
4
2
2
1
   2  2  1  2  2  dy              0.1333 .
   
  
3
3
4
4
5

0
  0.1333  5.48
x
U
 0.731
x
.
U
0.731  0.644
100  13.5% .
0.644
% error =

y
x
2

.
dy   
 0.363 . See Problem 8.108.   4.79
c)  d  1  sin


2 
U


0
 d  0.363  4.79
x
U
 1.74
x
U
.
1.74  1.72
100  1.2%
1.72
% error =


y
y
y y
 2
 2

  sin
1  sin
cos
dy   
  sin -term    
 0.137 .


2 
2 
2 2
2 
 
0

0
  0.137  4.79
8.115 a)   4.65
b)
x
U
x
U
x
 0.654
U
% error =
0.654  0.644
100  1.6%.
0.644
1/2
 1.6 104  20 
 4.65 


12


 0  0.323U2
.

xU
 0.0759 ft.
2  1.6 10
4 1/2
 0.323  0.0024 12 
 20 12 


 9.1110 5 psf .
1

c) Drag = U2  20 15 1.29
2
LU
1/2
 1.6  104 
1
  0.0024  122  300 1.29
 20  12 
2


227
 0.0546 lb.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 8 / External Flows
d)  x10
 3y 3y3  d
1.6 104 10
u
 4.65
 0.0416 ft.
 U   2  4  .
12
x
2  dx
 2

 4.65 1.6 104
3y
3y 3
u
  12 

 27.9 y  16140 y3.

2
4
10 12
x
0.2  0.0416  2
 0.2  0.0416


v  
0
8.116 a)   4.65
b)
x
U
27.9
16140
u
dy 
 0.04162 
 0.04164  0.0121 fps.
2
4
x
1/2
 1.5 105  6 
 4.65 


4


 0  0.323U2
 0.0221 m.
2  1.5  10

xU
 0.323 1.22  4 


6 4
5 1/2


 0.00498 Pa.
1/2
 1.5 105 
1
1

c) Drag = U2 Lw 1.29
 1.22  42  6  5 1.29 
 6  4 
2
LU 2


d)
 0.299 N.
 3y 3 y3  d
u
 U   2 

x
2  4  dx
 2

 d
42 y3
3 4
3
 4 


y
 .
2
4
5
2
5

2 4.65  (1.5 10  3)  dx
 2  4.65 1.5 10  3
5
u
1
7 3  4.65 1.5  10

  4 6166 y  2.53  10 y
 64.1y  2.63  105 y 3 .


x
2
4
3

5
64.1
u
2 2.63  10
 v   dy 
 0.0156 
 0.01564  0.00391 m/s,
2
4
x

0
where  x3  4.65
1.5 105  3
 0.01560 m.
4
228
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Chapter 8 / External Flows
Laminar and Turbulent Boundary Layers
8.117 a)   5
x
U
1/2
 1.5 10 5  2 
 5


10


 0.00866 m. Use  0 .332U 2
2
Drag =   0 wdx  0.332 1.22 102
0
 1.5 105 
b)   0.38  2 
 10  2 



xU 
.
1.5 105 2
 4  0.561 N.
10
1/ 2
0.2
 0.0453 m.
  
1
Drag = U2 Lw  0.074 

2
 U L 
0.2
 1.5 105 
1
 1.22 102  2  4  0.074 
 10  2 
2


0.2
 2.15 N.
 1.5 105 
8.118 a)   0.38  6 


 20  6 
0.2
 1.5 105 
1
 0.0949 m.  0  1.22  202  0.059 


2
 20  6 
0.2
 0.6 Pa.
 106 
b)   0.38  6 


 20  6 
0  
u
y
 106 
1
 0.0552 m.  0  1000  202  0.059 


2
 20  6 
u 1
 U y 6/7 1/7 .
y 7
8.119 uy   U .
u
y
0.2
u
y

y 
0.2
 286 Pa.
U
.
7
should be zero. Thus, this condition is not satisfied.
y 
y 0
1
1
u
  U  1/7  . Thus, this is unacceptable and
at and near the
7
0
y
wall is not valid.
We sketch
y
 3 y 1 y3 

u  U 
and
 2  2  3 


229
1/7
y
u  U  
 
U
.
u
u
cubic
turb (power-law)
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 8 / External Flows
0.2

 1.58 104 
 1.58 104  
1
2

8.120 a) Drag =  0.0024  20  (12 15) 0.074 
 1060 
 20 12 
 20 12  

2





 0.31 lb.
0.2

 1.58 104 
 1.58 104  
1
2


b) Drag =  0.0024  20  (12 15) 0.074 
 1700 
 20 12 
 20 12  

2





 0.27 lb.
0.2

 1.58 104 
 1.58 104  
1
2

c) Drag =  0.0024  20  (12 15) 0.074 
 2080 

 20 12 
 20 12  

2





 0.25 lb.
0.2

 106 
 106  
1
2
8.121 a) Drag = 1000 1.2  (1 2) 0.074 
1060


   5.21 N.
 1.2 1 

2
1.2
1




 

0.2

 106 
 106  
1
2

  4.44 N.
b) Drag = 1000 1.2  (1 2) 0.074 
 1700 
 1.2 1 
 1.2 1  

2





0.2

 106 
 106  
1
2
  3.99 N.
c) Drag = 1000 1.2  (1 2) 0.074 
 2080 
 1.2 1 
 1.2 1  

2





 1.5 105 
60 1000
 16.67 m/s.   0.38 100 000 
8.122 a) U 
 16.67 105 
3600


0.2
 235 m.
0.2

 1.5 105  
1
1
2
2 
  0.0618 Pa.
 0  U c f  1.22 16.67  0.059 
5


2
2
 16.67 10  

b)  0 
1
1
0.455
U2 c f  1.22 16.672
 0.151 Pa.
2
2
2
5 
 
16.67 10
ln  0.06

1.5 105  
 
 u 
0.151
16.67
0.351
 0.351 m/s. 
 2.44ln
 7.4.   585 m.
1.22
0.351
1.5 105
Both (a) and (b) are in error, however, (b) is more accurate.
230
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8 / External Flows
8.123 a) 5 
u  
(See Fig. 8.24 b).  

1
b)  d 
U


(U  u )dy 
0

U
0.15


5 1.5 105
 2.14 104 m.
0.351
y
u

 2.5  2.44 ln   dy  U





3.74 ln
0.15
y

dy

0.15 87.8
585 
y
y
u 




 2.5(0.15   )  2.44  y ln  y 

 3.74  y ln  y 
U 







87.8 



0.351
[219  620  0.008  2188  951]  43.7 m.
16.67
Note: We cannot use zero as a lower limit since the ln-profile does not go to the wall.
Hence, we use   ; the lower limit provides a negligible contribution to the integral.
8.124 a) Use Eq. 8.6.40: c f 
0.455
 
300  20  

ln  0.06
1.58 104  
 
2
 0.00212.
b)  0 
1
1
U2 c f   0.0024  3002  0.00212  0.229 psf.
2
2
c)  
5 5 1.58 104

 8.09 105 ft.
u
9.77
d)
300
9.77
 2.44ln
 7.4.
9.77
1.58 104
8.125 a)  0 
u 
0.229
 9.77 fps.
0.0024
  0.228 ft.
1
1
0.455
U2 c f  1000 102
 110 Pa.
2
2
2
 
10  3  
ln  0.06 6  
10  
 
 u 
110
 0.332 m/s.
1000
 
5 5 106

 1.51105 m.
u
0.332
b) u  5u  5  0.332  1.66 m/s.
c) y  0.15 .
d)
 y  0.15  0.0333  0.005 m.
Do part (d) first!
10
0.332
 2.44ln
 7.4.
0.332
106
  0.0333 m.
231
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 8 / External Flows
8.126 Assume flat plates with dp /dx  0. C f 
0.523
 
10 100  

ln  0.06
106  
 
2
 0.00163.
1
Drag = 2  1000 102 10 100  0.00163  163 000 N.
2
To find  max we need u :
1
2
 0  1000 102
0.455
 
10 100  

ln  0.06
106  
 
2
10
0.266
 2.44ln
 7.4.
0.266
106
8.127 a) Assume a flat plate of width D. Re 
drag  C f
UL

 70.9 Pa.  u 
70.9
 0.266 m/s.
1000
 max  0.89 m.

15  600
5
1.5  10
 6  108.
1
1
U 2 L D  0.073(6  108 )1/ 5   1.2  152  600   100  32 600 N
2
2
power  FD  U  32 600 15  489 000 W or 655 hp or 164 hp/engine .
b) helium 
100
p

 0.167 kg/m3 .
RT 2.077  288
FB  Wair  Whelium    V  (1.2  0.167)  9.8    502  600 / 2  2.38  107
payload = FB  W  23.8  106  9.8  1.2  106  12  106 N
Laminar Boundary Layer Equations
8.128 u 
 u  2


,
, v
,
y
x xy
x
u  2

,
y y 2
 2u
y 2

 3
y3
.
Substitute into Eq. 8.6.45 (with dp/dx  0) :
  2   2
 3
.



y xy x y 2
y3
232
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Chapter 8 / External Flows
 2 ( / y )   ( / y ) 


xy

x

x
y
Recognizing that  /x  1,  /y  0,  /x  
U  / x3 ,
2
U 
U


  y U 
and
,
,




x
 
y
y
x  2  x3 
    
,


x  x  x
8.129 We also have
U  2 1 U 
U  2  y U 
 2




xy
  2  3 
  2  2  3 
 2
U  2  U 
3 U 3 U
,



  2   
y 2
y3   3 
Equation 8.6.47 then becomes, using U /   /y,

 
y 
   2
   2  2     





2
   2x 



y
2
yx
2
yx







   2   2 U  3
 2  2 
x y  3
  y  
Multiply by y 2 / 2 and Eq. 8.6.49 results:
2
1   
 2    2
 3 U

 




2        2
 3 
8.130 u 
U
dF 

 U x
 U x F '( )   U F '( ).
x
d y
y
We used Eq. 8.6.50 and Eqs. 8.6.48.
v



x
x


U  x F  
1 U
F 
F  U  x
2
x
 x

U  1 3/2 
1 U
 x
F  U x F ' y

  2
2
x


y U U
1 U
1 U
F
F'
( F ' F ).
x
x
x
2
2 x
2
8.131 The results are shown in Table 8.5.
233
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Chapter 8 / External Flows
8.132 a)  0  0.332  1.22  5 2
1.5  10 5
 0.0124 Pa.
25
15
.  10 5  2
 0.0122 m.
5
b)   5
1.5 105  5

(
'
)
F
F


 0.8605  0.00527 m/s.


2
x  2
max
 U  1
c) vmax 



dF
dF
vx
dy   U 
d
d) Q   u(1 dy )   U 
d
d
U
0
0
0
 U
x
U
[ F ( )  F (0)]  5
8.133 a)  0  0.332  0.0024 152
1.5 105  2
 3.28  0.04 m3 /s/m
5
1.6 104
 2.39 104 psf.
6 15
1.6  10 4  6
 0.04 ft.
15
b)   5
1.6 104 15




 0.8605  0.0172 fps.
(
'
)
F
F

6
x  2
max
 U  1
c) vmax 

d) Q   udy  U
0
x
F ( )  15
U
1.6 104  6
 3.28  0.394 ft 2 /sec/ft.
15
8.134 At x = 2 m, Re = 5  2/106 = 107. Assume turbulent from the leading edge.
a)  0 
1
0.455
U2
2
ln(0.06 Rex )2
1
0.455
 1000  52
 32.1 Pa
2
2
(ln 0.06 107 ) 


b) u 
0
32.1

 0.1792 m/s.
1000

5
0.1792
 2.44ln
 7.4.
0.1792
106
  0.0248 m
c) Use the 1/7 the power-law equation:
Q
0.0248

0
1/7
 y 
5

 0.0248 
dy  0.109 m3 /s/m
234
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8 / External Flows
8.135 From Table 8.5 we would select   6:
a)   6
x
U
x
b)   6
U
6
1.5  105  2
 0.0147 m
5
6
15.8  105  6
 0.047 ft or 0.57 in.
15
8.136 From Table 8.5 we interpolate for F'  0.5 to be

0.5  0.3298
( 2  1)  1  1.57
0.6298  0.3298
1.57 
v
U
5
.
y
x
1.5 105  2
U   1 
  F ' F 
x  2
 y  0.00385 m or 3.85 mm
1.5 105  5
(0.207)  0.00127 m/s
2

5
 u v 

2 1.5  10
2
 0.291(1.2)5
 0.011 Pa
   
  F " U
25
xU
 y x 
8.137
y
v
y
v=0
v=0
U
y=
v
u>U
v
v
y=
(a)
(b)
(c)
If v  0 at y  10 and v  0 at y   , then v/y  0 and continuity demands that
u/x  0. The u component, for y   must then be greater than U, 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   . If v  0 at large y, say y  10 , 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); reality would demand a profile such as that sketched in (b). The
overshoot would be quite small and is neglected in boundary layer theory.
 3 y 1 y3 

8.138 u  U  
 2  2  3 


For the Blasius profile: see Table 8.5.
(This is only a sketch. The student is encouraged
to draw the profiles to scale.)
235
y
U
cubic
Blasius
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 8 / External Flows
8.139
y
y
low velocity
outside b.l.
A
8.140 A:
B:
C:
D:
E:
inviscid
profile
y
y
backflow
2U
B
D
C
zero velocity
gradient
p
 0. (favorable)
x
p
 0.
x
p
 0. (unfavorable)
x
p
 0.
x
p
 0.
x
separation
streamline
y
D
C
D 
C
236
  
E 
A
E
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
B
Chapter 9 / Compressible Flow
CHAPTER 9
Compressible Flow
Introduction
9.1
Btu
ft-lb
lbm
32.2
Btu
slug
cp
0.24
cv
cp R 6012 1716 4296
778
lbm- R
6012
ft-lb
slug- R
ft-lb
slug- R
4296
ft-lb
1 Btu 1 slug
slug- R 778 ft-lb 32.2 lbm
0.171
9.2
cp
cv
R.
kcv .
cp
If
s
cp
lbm- R
R or c p 1
k
1
k
R
Rk
k 1
cp
9.3
cp
Btu
0, Eq. 9.1.9 can be written as
p
R n 2
p1
T
cp n 2
T1
It follows that, using c p
T2
T1
R/ cp
p2
p1
or
cv
p2
p1
cp
T
n 2
T1
R
p
n 2
p1
R and c p / cv
k,
k 1/ k
Using Eq. 9.1.7,
T2
T1
p2
p2
p1
1
2 p1
k 1/ k
or
p2
p1
1
2
1/ k
.
Finally, this can be written as
p2
p1
k
2
.
1
237
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Chapter 9 / Compressible Flow
Speed of Sound
9.4
Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change:

Q W
V22 V12 p 2 p 1 ~ ~
S
u2 u1 .

2
m
2
1
~ pv
Enthalpy is defined in Thermodynamics as h u
~ p / . Therefore,
u

Q W
V22 V12
S
h2 h1 .

m
2
Assume the fluid is an ideal gas with constant specific heat so that
h
c p T . Then
Q WS V22 V12
c p (T2 T1 ).
m
2
Next, let cp cv R and k cp /cv so that cp / R k /(k 1). Then, with the ideal gas
RT , the first law takes the form
law p

Q W
S

m
9.5
V22
dp
k
2
Differentiate p
k
V12
k
pk
k
p2
p1
2
1
1
c using d( xy)
k 1
d
.
ydx xdy :
0.
Rewrite:
dp
d
9.6
k
p
.
The speed of sound is given by
dp /d .
c
For an isothermal process TR p/
K, where K is a constant. This can be
differentiated: dp Kd
RTd . Hence, the speed of sound is
RT .
c
9.7

Eq. 9.1.4 with Q W
S
V2
2
0
cpT
2V V
2
0 is:
V)2
(V
2
( V )2
2
V2
2
c pT
c p (T
cons' t.
T)
c p T.
V2
V V
2V V
2
( V)2
cp T
h.
We neglected ( V)2. The velocity of a small wave is V
238
c.
h
cpT
cp T.
c V.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 9 / Compressible Flow
9.8
For water
Bulk modulus =
p
For water c
dp
d
velocity
L
1453 m/s
1
2110 106
1000
dp
d
time = 1453
1453 m/s.
0.6 = 872 m.
Since c = 1450 m/s for the small wave, the time increment is
t
9.11
2110 106
1000
dp
d
c
9.10
10 6 Pa
2110
1000 kg/m3 , we see that
Since
9.9
dp
d
d
c
10
1450
0.0069 seconds
200
1.4 287
a) M
V
c
b) M
600 / 1.4 1716 466
288
c) M 200 / 1.4 287 223
d) M
600 / 1.4 1716 392
e) M 200 / 1.4 287 238
9.12
c
9.13
a) Assume T = 20 C:
kRT
c
d
0.567.
0.668.
0.618.
0.647.
1.4 287 263 325 m/s.
1.4 287 293
kRT
c t
0.588.
343
2
d ct
256 1.21 393 m.
343 m/s.
686 m
b) Assume T = 70 F:
c
d
kRT
c t
1.4
1130
1716 530
2
1130 fps.
2260 ft.
For every second that passes, the lightning flashed about 1000 ft away. Count 5 seconds
and it is approximately one mile away.
239
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Chapter 9 / Compressible Flow
9.14
1.4 287 263
c
sin
0.256.
9.16
tan
3776
1000
t
9.15
256 m/s.
1
M
sin
1000
.
L
0.2648
V
c
.
V
L
1000 m
3776 m
L
3.776 s.
Use Eq. 9.2.13:
1.4 287 288
a)
c
V
sin
or V
b)
c
V
sin
or V
Eq. 9.2.4:
1.4
1716 519
sin 22 
p
c
V
V2
2
Energy Eq:
kRT
c pT
V )2
(V
2980 fps
0.3
0.00237 1.4 1716 519
p
c p (T
2
T ).
0.113 fps.
0 V V
1.4 1716 519 ft/sec ( 0.113 ft/sec)
6012 ft-lb/slug- R
c V
cp
T
908 m/s
sin 22
( V )2
2
c p T.
0.021 R or 0.021 F
Note: Use slug = lb-sec2/ft (m = F/a). (Units can be a pain!)
Isentropic Flow
9.17
a) 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
AdV
AVd
VdA
Divide by AV :
d
dV
V
dA
A
0
b) Expand the r.h.s. of Eq. 9.3.5 (keep only first order terms):
V2
2
p
k
k
1
V2
2VdV
2
p
k
k
240
1
dp
.
d
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 9 / Compressible Flow
Hence,
2VdV
2
0
p
k
1
k
p
k
VdV
2
dp
k
p
d
pd
pd
2
1
k
p
dp
1
k
VdV
dp
d
where we neglected d compared to 2 . For an isentropic process Eq. 9.2.8
gives dp kpd , so the above becomes
pd
p
k kpd
k ( k 1)pd
0 VdV
VdV
VdV
k
d
2
2
2
k 1
k 1
dV /V dA /A so that the above equation is
But d /
p
dV dA
0 VdV k
V
A
which can be written as
V2
dA
dV
.
1
A
kp
V
Since c 2
kp / , and M = V/c, this is put in the form
V2
c2
dA
A
c) Substituting in V
T0
T

d) m
p
p0
1
kRT , and R / c p
M2 c 2
2c p T
1
2
k
T0 1
2
At the critical area A , M
*
1
(M2 1)
dV
V
( k 1) / k , we find
M 2 kRT
2c p T
1
M 2 k( k
2k
1)
1 1
1
k
2
M2 .
k /( 1 k )
M2
1
2
1
k
*
1
k
p0 1
k
MA 1
RT0
dA
A
or
Mc, c 2
V2
2c p T
k
AM
TR
dV
V
M2
1/2
M2
k
AM
R
k 1
2( 1 k )

1. Hence, m
241
p0
k 1
k
A*
RT0
2
k 1
2( 1 k )
.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9 / Compressible Flow
 is constant throughout the nozzle, we can equate Eq. 9.3.17 to Eq. 9.3.18:
e) Since m
or
9.18
p0
k
MA 1
RT0
A
A*
1 2 (k
M
k
a) ps
patm 10
p1
b) p s
26.4
V12
2
9.20
69.9 10
k 1
k
A*
RT0
2
p0
10
k 1
2( k 1)
79.9 kPa abs.
ps
1
s
p1
ps
1
s
.
1
s
79 900
.
0.997
V1
36.4 kPa abs. p 1
p1
ps
1
s
26 400
0.412
p1
.
V
.
s
1.4 101 000
0.4
1.22
V12
2
4000
.
1.22
1
1/ k
ps
p1
79.9
0.906
69.9
1/1.4
Vs=0
0.997 kg/m3 .
77.3 m/s.
ps
p1
1/ k
36.4
26.4
0.412
1/1.4
0.518 kg/m3 .
V1 111 m/s.
ps
p1
1/ k
1.22
105 000 1.4
.
1.254 0.4
V1 81.0 m/s.
105
101
% error =
0.5283p0 ?
0.5283 200 105.7 kPa.
a) pr
0.5283p0 .
choked flow.
1.4 287Te
1000Te .
2
105.7
1.484 kg/m3.
0.287 248.1
1/1.4
1.254 kg/m3 .
V1 81.3 m/s.
Is pr
e
s
26.4 kPa abs.
1
s
36 400
.
0.518
V12
2
1000 298
k 1
2( 1 k )
1
69 900
0.906
V2
s: 1
2
From 1
b)
M2
1)M 2
1
V2
s: 1
2
V12
2
9.19
2
k 1
2( 1 k )
69.9 kPa abs.
From 1
V2
a) 1
2
1
k
Me
Te
81.3 81
100 0.42%.
81.3
1.
248.1 K, Ve
m 1.484
242
Ve2
kRTe .
pe
105.7 kPa.
315.8 m/s.
0.012 315.8 0.1473 kg/s.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 9 / Compressible Flow
b) pr
0.5283p0 .
0
Me
200
0.287 298
Is pr
2.338.
Ve2
2
1.4 130 000
.
0.4
e
1.7187 kg/m3.
e
130
200
1.4
e
2.338
257.9 m/s.
Ve
0.012 257.9 0.1393 kg/s.
m 1.7187
9.21
1. 1000 298=
0.5283p0 ?
0.5283 30 15.85 psia.
a) p r
15.85.
choked flow and Me 1, p e 15.85 psia. Ve2 kRT.
1.4 1716 Te
Te 441.7 R, Ve 1030 fps.
6012 530
6012 Te .
2
15.85 144
0.003011 slug/ft 3 .
e
1716 441.7
0.5
12
m 0.003011
b) pr
e
Me 1, and pe 20 psia.
30 144
.00475 slug/ft 3 .
1716 530
Ve
0.24
0.003556 slug/ft 3 .
1.4 20 144
.
0.4 0.003556
m 0.003556
Btu
lbm- R
= 0.24 778
ft-lb
lbm- R
0.5
12
2
6012
838.9 0.01627 slug/sec.
ft-lb
slug- R
.
0.5283 p0 . M e 1. pe 0.5283 200 105.7 kPa. Te 0.8333 298 248.3 K.
105.7
1.483 kg/m3. Ve
1.4 287 248.3 315.9 m/s.
0.287 248.3
m 1.483
0.5283 p0 .
e
Ve2
2
838.9 fps.
Note: c p
b) pr
1/1.4
20
0.00475
30
6012 530
e
1030 0.01692 slug/sec.
15.85.
0
9.22 a) pr
2
0.012 315.9 0.1472 kg/s.
pe
130 kPa,
pe
p0
130
1.719 kg/m3 , Ve
0.287 263.4
m 1.719
0.65.
Me
0.81, Te
0.81 1.4 287 263.4
0.884T0
263.5 m/s.
0.012 263.5 0.1423 kg/s.
243
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Chapter 9 / Compressible Flow
9.23
a) pr
0.5283 p0 .
Me
1.
0.8333 530 441.6 R.
Te
15.85 144
1716 441.6
e
0.5283 p0 .
0
20 psia.
pe
20 144
1716 472
0.5
12
2
pe
p0
20
30
0.5
12
0.5283 p0 101 kPa.
pe
1.4 287 235.8
Ve
9.26
pe
. Ve
1.4 1716 441.6 1030 fps.
1030 0.01692 slug/sec.
0.6667.
Me
0.785 1.4 1716 472
m
211.3
0.287 252.5
p0 191.2 kPa abs. Te
307.8 m/s.
m
0.052 318.5 7.29 kg/ s.
101
0.287 235.8
pe
0.5283 p0
Ve
307.8 m/s since Me 1.
m
202
0.287 235.8
Ve
836 fps.
0.8333 283 235.8 K.
2 191.2 382.4 kPa abs.
p0
0.890T0 .
0.8333 303 252.5 K.
p0
0.5283 p0 14.7 psia.
0.785. Te
836 0.01664 slug/sec.
Te
1.4 287 252.5 318.5 m/s.
Ve
9.25
2
0.5283 400 211.3 kPa abs.
pe
ft 3
0.00356. Ve
m 0.00356
9.24
slug
0.003012
m 0.003012
b) pr
0.5283 30 15.85 psia.
pe
27.83 psia. Te
0.032 307.8 1.30 kg/s.
202.0 kPa abs. Te
235.8 K.
0.032 307.8 2.60 kg/s.
0.8333 500 416.6 R.
1.4 1716 416.6 1000 fps.
m
14.7 144
1716 416.6
p0
2 27.83.
m
29.4 144
1716 416.6
(1.25 /12)2 1000 0.101 slug/sec.
pe
0.5283 p0
29.4 psia,
Te
416.6 R, Ve 1000 fps.
(1.25 /12)2 1000 0.202 slug/sec.
244
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Chapter 9 / Compressible Flow
9.27
Treat the pipeline as a reservoir. Then, pe
Me
1 and Ve
1.4 287(0.8333 283)
264.5
30 10
0.287 (0.8333 283)
m
1.667 2077 Te
2
5193 300
4
5193 Te .
Te
264.5 kPa abs.
307.8 m/s.
307.8 3.61 kg/s.
3.61 6 60
264.5 / (0.287 0.8333 283)
m t
V
9.28
0.5283 p0
333 m3
225 K.
pe
200
225
300
1.667/0.667
= 97.45 kPa abs.
Next, Tt
225 K, pt
5193 300
V1
V12
2
m
1.667
1.667 pe
. pe
0.667 e
Ve2
2
3324 103 9.54Ve 0.667 .
0.3203 kg/m3 and pe
300 100
0.287 293
4.757
200
10 2
200 / 2.077 300
k p1
k 1 1
V22
2
4.757 kg/m .
52 .
k p2
.
k 1 2
V12
2
V1
37.35 m/s.
1A1V1
4.757
0.0752 Ve
e
1.667
e
kPa.
91.8 m/s.
199.4 kPa abs.
3
4.236
V2
1330
e
Ve2 63 420 103 Ve 0.667 . Trial-and-error: Ve
e
1
0.2085× ×0 .032×882.6 =
Ve2
2
or 3.116 106
p1
RT1
1.667 2077 225 882.6 m/s.
Vt
97.45
= 0.2085 kg/m3.
2.077 225
t
9.29
97.45 kPa;
2
340
4.757
400
V2
4.492 V1 .
1.4 400 000
0.4 4.757
1/1.4
4.236 kg/m 3.
4.4922 V12
2
1.4 340 000
.
0.4 4.236
0.052 37.35 1.395 kg/s.
245
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Chapter 9 / Compressible Flow
9.30 We need to determine the Mach number at the exit. Since the M = 1 at the throat,
then A* Athroat 9.7 cm 2 . Hence, the area ratio at the exit is Ae A* 13 9.7 1.34 .
Using the air tables, we find two possible solutions, one for subsonic flow, and the other for
supersonic flow in the diverging section of the nozzle. At the exit:
Subsonic Flow:
Me
0.5, Te T0
0.9524, and pe p0
0.8430 .
Hence,
Ve
M e ce
Supersonic Flow:
Me kRTe
Me
0.5 1.4 287 0.9524 295
1.76, Te T0
168 m/s
0.1850 .
0.6175, and pe p0
Hence,
Ve
M e ce
Me kRTe
1.76 1.4 287 0.6175 295
476 m/s
9.31 Since the flow is subsonic at the throat, the flow is also subsonic at the exit. Hence, for M =
0.72 at the throat
Athroat A* 1.0806
A*
9.7 1.0806 8.976 cm2
At the exit the area ratio is
Ae A* 13 / 8.976 1.448
From compressible flow tables for air we determine
Me
9.32
1
0.45 and Te T0
(45 14.7)144
1716 520
p1
RT1
2
0.009634
Ve
0.45 1.4 287 0.961 295
152 m/s
0.009634 slug/ft 3 .
50.7
59.7
V1 0.009634 42
V12
2
0.961
1/1.4
0.008573 slug/ft 3.
V2 0.008573 22.
1.4 59.7 144
0.4 0.009634
m 0.009634
4.4952 V12
2
V2
4.495 V1.
1.4 50.7 144
.
0.4 0.008573
V1 121.9 fps.
(2/12)2 121.9 0.1025 slug/sec.
246
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Chapter 9 / Compressible Flow
9.33
Energy 0
2: 1000
303
V22
2
1000 T2 . V2
3 kRT2
107.9
200
303
p2
Energy 0
T1
Vt2
5.39 kPa.
1 (M1 = 1): 1000 303
252.3 K. p1
252.3
200
303
2
Continuity: 1.455
9.34
1.4/0.4
0.05
1.4/0.4
244
500
293
Ve2
1000 293
2
5.39
0.287 107.9
V12
V12
1000
.
2
1.4 287
1.4/0.4
105.4 kPa.
1
263.5 kPa abs.
1.4 pe
. 3.763
0.4 e
2
0.1740 kg/m3.
V1 318.4 m/s
kRT1
105.4
1.455 kg/m3.
0.287 252.3
d22
318.4 0.174
3 1.4 287 107.9.
4
1.4 287 Tt
1000 Tt .
2
kRTt . 1000 293
pt
2
1
0
Substitute V2 into the energy equation and find T2 = 107.9 K.
d2
0.2065 m.
Tt
244.0 K. Vt
313.1 m/s.
t
263.5
0.287 244
3.763 kg/m3.
0.0252 313.1
e
0.0752Ve .
pe
1.4
e
263 500
3.7631.4
Ve2
1.014 106 Ve 0.4 . Trial-and-error: Ve 22.2 m/s, 659 m/s.
2
5.897, 0.1987 kg /m3 .
pe 494.2 kPa, 4.29 kPa abs.
293 000=
e
9.35
9.36
Ae
pe
p0
p
and e
p0
9.
A*
Mt
1.
t
p
p0
pt
.01228
15
120
0.00855 from Table D.1.
0.5283 120 63.4 psia, Tt
slug
.
ft 3
0.125.
A
*
A
pe
0.997 from Table D.1.
1.708.
m 1 0.01228
Me
2.014, Te
de2
4
1.708
500 0.997 498.5 kPa.
pe
4.28 kPa abs.
0.8333 520 433.3 R.
dt2
1.4 1716 433.3.
4
0.552 520 287 R, Ve
247
0.319 ft.
2.014 1.4 1716 287
= 684 fps.
2
0.319
.
4
dt
de
0.417 ft.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9 / Compressible Flow
9.37
Me
4. A/A* 10.72, pe
For A /A* 10.72, Me
9.38
0.0584.
0.9976 p0
pe
0.2381 293 69.76 K.
0.9976 2000 1995.2 kPa abs.
Using compressible flow tables for air, we determine the pressure ratio and temperature
ratio for M = 2.8 to be:
p
p0
and T
9.39
0.006586 2000 13.17 kPa, Te
0.03685, and
0.3894 T0
T
T0
V
125 K
p
0.3894 .
0.03685 p0
Mc 2.8 kRT
129 kPa abs
2.8 1.4 287 320 1004 m/s .
At the given section we have
M
A/A* 1.3398
0.5
A* 12.4/1.3398 9.255 cm 2
(a) At the throat: At /A* 10/9.255 1.0805 . Using the isentropic flow table, at the throat
Mt
0.72,
p p0
0.7080,
pt
0.708 600 424.8 kPa, Tt
Vt
Mt ct
M kRTt
0.9061
T T0
0.9061 303 275 K
0.72 1.4 287 275
239 m/s
(b) The area ratio Ae/At = 20/8 = 2.5. There are two entries in the table for Ae/A* = 20/8 =
2.5. The one at M = 0.24 is for subsonic flow throughout:
pe /p0
0.9607.
0.9607 600 576.4 kPa abs.
pe
The one at M = 2.44 is for supersonic flow throughout the diverging section:
pe /p0
(c) For Me
2.0
0.06426.
pe p0
pe
0.06426 600 38.56 kPa abs.
pe
0.1278, Te T0
0.5556,
0.1278 600 76.68 kPa and Te
Since the flow is supersonic at the exit
Ae A* 1.688
0.5556 303 168.3 K
Mth 1 and A*
9.255 cm2
Hence, Ae 1.688 9.255 15.62 cm2
The mass flow is calculated using:
pe
Me kRTe Ae
RTe
Substituting the given values we get:
m
m
e
Ve Ae
76.68
0.287 168.3
2 1.4 287 168.3 15.62 10 4 m2 1.29 kg/s
248
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Chapter 9 / Compressible Flow
9.40
Let Mt
A
pe
0.5283 400 211.3 kPa abs. Tes
303 Te
.
303 252.5
0.96
Te
9.42
A
m
i Vi
Ai
9.43
Te
A
A*
M<1
Mt = 1
Ve ~= 0
e
100
0.9027 kg/m3.
0.297 373
i
0.00938
4.235
At
373
1044 K or 772 C. p0
0.3571
100
0.02722
t
i
M>1
0.00221 m2 .
pe
M<1
Mt = 1
3670 kPa abs.
Ve ~= 0
e
4.235.
3 1.4 1776 660
3840 fps.
Ai
0.2
0.001843 3840
0.0283 ft 2 .
T0
t
M>1
Vi
At M 3, T
319.8 m/s.
0.02722 p0 .
Isentropic flow. Since k = 1.4 for
nitrogen, the isentropic flow table
may be used.
M 3:
1.4 287 254.5
10
0.00938 m2 .
0.9027 1181
0.3571 T0 , p
T0
Ve
i
3 1.4 297 373 1181 m/s.
At M 3, T
0.0816 m or 8.16 cm.
4.235.
A*
Vi
dt
0.052 319.8 7.27 kg/s.
Isentropic flow. Since k = 1.4 for
nitrogen, the isentropic flow table may
be used.
At M 3,
dt2
.
4
0.8333 303 252.5 R.
254.5 K.
m 211.3/(0.287 254.5)
0.430.
1.4 287 303
0.052
1.5007
A1
1.5007
At
1.5007.
A*
9.41
150
1. Neglect viscous effects. M1
i
At
15 144
1776 660
0.001843 slug/ft 3.
0.0283
0.00667 ft 2 .
4.235
0.3571 T0 , p 0.02722 p0 .
Te
660
1848 R or 1388 F. p0
0.3571
249
pe
15
0.02722
551 psia.
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Chapter 9 / Compressible Flow
9.44
9.45
Assume pe
101 kPa. Then
F
mV
AV 2 .
F
mV
AV 2 .
101
0.189 1273
e
80 000 9.81
0.4198
6
101
0.287 873
0.252 V 2 .
Mt
Ae
1.
A*
Te 0.3665 T0
pe
Ve
9.47
4;
Me
2.94, pe
V
p0
1260 m/s.
349 m/s.
FB
0.02980 p0 .
Ve
p0A0
0.3665 300 110.0 K,
100 0.0298 p0 .
V
0.403 kg/m3. (Assume gases are air.)
100 9.81 0.403 200 10 4 V 2.
9.46
0.4198 kg/m3.
3356 kPa abs.
2.94 1.4 287 109.95 618 m/s.
100
0.052 6182
FB
0.287 109.95
0.22
3 356 000
412 000 N.
Assume an isentropic flow; Eq. 9.3.13 provides
. p
103
p
1
k 1
k 1 2
M
1
2
Using k = 1.4 this gives M 2
.
0.0424 or M 0.206.
Mc
For standard conditions V
0.206 1.4 287 288
70 m/s or 157 mph
Normal Shock
9.48
a) 0.9850 1000
V22
1000
2
2
2V2 .
80 000
1.4 p2
0.4 2
p2
0.985 1000(V2 1000)
287 283
0.
1
80
0.287 283
0.9850 kg/m3.
V22 10002 1.4 V2
( 985V2 1 065 000) 284 300 = 0
2
2
0.4 985
3V22 3784V2 784 300 0.
V2 261 m/s. 2 3.774 kg/m3.
Substitute in and find p2
M1
M2
808 kPa abs.
1000
2.966. T2
1.4 287 283
261
0.477.
1.4 287 746
250
808
0.287 3.774
746 K or 473 C.
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Chapter 9 / Compressible Flow
1000
1.4 287 283
b) M1
a)
M2
0.477. p2 10.12 p1 809.6 kPa abs.
2.644 283 748 K or 475 C.
T2
9.49
2.97.
12 144
0.002014 slug/ft 3. 0.002014 3000
1716 500
Momentum: 12 144 p2 0.002014 3000(V2 3000).
2V2 .
1
V22 30002
2
V22
1.4 p2
1716 500
0.4 2
3000 2
7
V2
(19 ,854
6.042
p2
2
3000
1.4 1716 500
1
p2 T1
p1 T2
M 21
2.74.
M2
2kM12 k 1
k 1
k
1 p2
2k p 1
2
0.00725 slug/ft3.
1
1
2k
1)
2
4 k (k
103.1 144
0.00725 slug/ft 3.
1716 1193
1) ( k
k 1 2
M1 [4kM12 2k 2]
2
( k 1)M12
.
2 ( k 1)M12
. (This is Eq. 9.4.12). Substitute into above:
p2
p1
(k
p
1) 2
p1
1)p 2 / p 1
.
1)p 2 / p 1
p
For a strong shock in which 2
1,
p1
k
k
2
8.592 12 103.1 psia.
( k 1) 2 M12
k
1) ( k
p2
0.493.
2.386 500 1193 R or 733 F.
(k
1
833 fps.
3000
102.9 144
2.74. T2
1191 R or 731 F.
1716 0.00725
1.4 1716 500
833
0.492.
1.4 1716 1191
M2
T2
V2
102.9 psia.
M1
b) M1
0.
6.042V2 ) 6.006 10 6 = 0.
6V22 23,000V2 15 106 =0.
9.50
809.6
3.771 kg/m3.
0.287 748
2
1)
(k
(k
1)
(k
1) ( k
1)
2
1)
(k
p2
p1
1)( k
k
1
p
1) 2
p1
.
1 (k
1 (k
2
1
251
k
k
1
.
1
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Chapter 9 / Compressible Flow
9.51
Assume standard conditions: T1
15  C,
101 kPa.
1
V1
2 1.4 287 288
M1
2.
p2
4.5 101 454 kPa.
V2
0.5774 1.4 287 486 255 m/s.
V1 V2 680 255 425 m/s.
Vinduced
M2
stationary
shock
680 m/s.
.5774. T2
1.688 288
486 K.
V2
V1
The high pressure and high induced velocity cause extreme damage.
9.52
If M2
0.5, then M1
p2
9.53
If M 2
p2
9.54
p1
2.645 1.4 287 293 908 m/s.
1600
8.00 200 1600 kPa abs.
8.33 kg/m3.
2
0.287 (2.285 293)
0.5, then M1
2.645.
V1
2.645.
V1
8.00 30 240 psia.
0.2615 101 26.4 kPa.
M2
9.55
9.56
240 144
1716 (2.285 520)
2
223.3 K.
M1
0.960 T0 .
0.01695 slug/ft 3.
1000
1.4 287 223.3
0.4578. p2 12.85 26.4 339 kPa. T2
3.34.
3.101 223.3 692.5 K.
0: For M = 0.458, p = 0.866p0 and
For isentropic flow from 2
T
T1
2.645 1.4 1716 520 1118 fps.
p0
339
0.866
391 kPa abs.
T0
692.5
0.960
721 K or 448 C.
After the shock M 2 0.4752, p2 10.33 800 8264 kPa abs.
0: For M = 0.475, p = 0.857p0
For isentropic flow from 2
8264
p0
9640 kPa abs.
0.857
A
A*
101
102.5 kPa abs.
0.985
M t 1. pt 0.5283 102.5 54.15 kPa. Tt 0.8333 298 248.3 K.
54.15
0.7599 kg/m3. Vt
1.4 287 248.3 315.9 m/s.
t
0.287 248.3
m 0.7599
0.0252 315.9 0.471 kg/s. If throat area is reduced, Mt
4.
Me
remains at 1,
t
0.147. pe
0.985 p0
p0
0.7599 kg/m3 and m 0.7599
252
0.022 315.9 0.302 kg/s.
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Chapter 9 / Compressible Flow
9.57
pe
p1
A
A*
101 kPa = p 2 .
1, pt
M1
2.94, p1 10.18 kPa abs.
0.4788, pe
14.7 psia
p2 .
10.18
0.0298
342 kPa abs.
0.8333 293 244.1 K.
101 kPa .
Te
611 m/s.
2.609 107.4 280.2 K.
T2
A
A*
4.
M1
2.94, and p2 / p1
9.918.
14.7
1.482
1.482 psia. At M1 2.94, p / p0 0.0298.
p0
49.7 psia.
9.918
0.0298
Mt 1, pt 0.5283 49.7 26.3 psia. Tt 0.8333 520 433.3 R.
M1
1.4 1716 433.3
M2
2.94 1.4 1716 190.6
0.4788, pe 14.7 psia . Te
V2
1, pt
M2
T2
1989 fps.
2.609 190.6 497.3 R.
0.4788 1.4 1716 497.3
0.5283 500 264 kPa. Tt
A1 /A*
T1
1020 fps.
2.94, p1 1.482 psia. T1 0.3665 520 190.6 R.
V1
Mt
p0
0.4788 1.4 287 280.2 161 m/s.
Vt
9.59
0.0298.
0.3665 293 107.4 K.
T1
2.94 1.4 287 107.4
V2
p1
2.94, p / p0
9.918.
1.4 287 244.1 313 m/s.
V1
M2
2.94 , and p 2 / p 1
0.5283 342 181 kPa abs. Tt
Vt
pe
M1
101
10.18 kPa. At M1
9.918
Mt
9.58
4.
82 /52
2.56.
M1
0.451 298 134.4 K.
0.516, p2
0.8333 298 248.3 K.
2.47, p1
V1
523 fps.
0.0613 500 30.65.
2.47 1.4 287 134.4
574 m/s.
6.95 30.65 213 kPa.
After the shock it’s isentropic flow. At M
p02
0.511 500 255.5 kPa. A*
Ae
0.052
0.003825
T2 2.108 134.4 283.3 K.
A
0.516, * 1.314.
A
2
0.04
0.003825 m2 .
1.314
2.05.
pe 0.940 255.5 240 kPa abs = pr . Me 0.298.
A*
Te 283.3(213/240)0.2857 273.8 K.
Ve 0.298 1.4 287 273.8 99 m/s.
253
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Chapter 9 / Compressible Flow
Vapor Flow
9.60
0.546 p0
pt
0.546 1200 655 kPa. Tt
655
0.462 585
t

m
t
Te
673
Ve2
2
Ve
9.61
Me
1, pe
Me
1, pe
4
Mt
Vt
1, pt
0.3/1.3
546
1000
0.060 m or 6 cm.
dt
101
0.462 380.2
e
542 K.
e
de
1160
0.575 kg/m3.
e.) (cp 1872 J/kg K)
0.092 m or 9.2 cm.
593 m/s.
0.0075
81.9
150
0.124 m or 12.4 cm.
de
0.3/1.3
1009 R.
1.3 2760 1009 1903 fps.
0.199 ft. or 2.39".
de
t
2.18 kg/m3.
de2
571.
4
0.00423 slug/ft 3. Ve
0.546 1200 655 kPa. Tt
2
546
0.462 542
571 m/s. 15 2.18
81.9 144
2762 1009
1.3 462 585
m 2.42
593 m/s. (Mt 1.)
0.546 1000 546 kPa.
de2
0.25 0.00423
1903.
4
9.63
593.
380.2 K
0.546 150 81.9 psia. Te
e
585 K.
1.3 462 585
4 0.575( de2 /4) 1050.
1050 m/s.
623
d t2
0.3/1.3
1872 673. (Energy from 0
380.2
1.3 462 542
Ve
9.62
2.42
0.3/1.3
101
1200
1872
2.42 kg/m3. Vt
4
At Vt .
0.546 p0
Te
655
673
1200
673 655 /1200
655
0.462 585
0.3/1.3
2.42 kg/m3.
593 0.254 kg/s per nozzle , Te
254
585 K.
120
673
1200
0.3/1.3
396 K.
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Chapter 9 / Compressible Flow
Oblique Shock Wave
9.64
800
1.4 287
From Fig. 9.15,
M1
a)
46 .
M 2n
M1n
p2
20 ).
2.29sin 79
2.25.
5.74 40 230 kPa abs. T2
0.541 M 2 sin(79
20 ).
1.90 303 576 K.
1.4 287 576 0.631 303 m/s.
a detached
shock
40  .
1
2
2sin 40
M1n
1.29.
M 2n
10  then, with M 1.58 ,
M3n
M1n
= 35o
10  .
M1n
9.67
M 2n
0.631.
V1
9.66
1.423 303 431 K.
M1n
c)
If
1.49.
1.4 287 431 1.49 620 m/s.
V2
9.65
M2
V2
M2
= 20o
1.65.
3.01 40 120.4 kPa abs. T2
79 .
V2
V1
2.29sin 46
0.654 M 2 sin(46
p2
b)
2.29.
303
46  , 79  .
0.824 M3 sin(51
3.5sin 35
2.01.
M2
0.576/ sin(35
M 2n
2.26sin 47
M 2n
2
0.791 M 2 sin(40
51  . 1.58 sin 51 
10 ).
10 ).
M 2 =1.58.
M 2n .
M3 1.26.
2
10 51 10 41 .
0.576. T2 1.696 303 514 K.
20 )
2.26.
1.65.
M3n
1
20
2.
2
0.654 M3 sin(47
20 ).
V3
3.5sin 35
0.576. T2 1.696 490 831 R.
M2
0.576/ sin(35
M 2n
2.26sin 47
M 2n
M3 kRT3
47 .
T3 1.423 514 731 K.
2.01.
1.23.
M 2n
20 )
2.26.
1.65.
M3n
T3 1.423 831 1180 R. V3
1
20
255
1.44 1.4 287 731 780 m/s.
2.
0.654 M3 sin(47
M3 kRT3
M 3 1.44.
2
47 .
20 ).
M 3 1.44.
1.44 1.4 1716 1180
2420 fps.
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Chapter 9 / Compressible Flow
9.68
M1
3,
10 .
28 . M1n
1
p2
1.41.
M 2n
0.736.
2.153 40 86.1 kPa abs.
0.736
M2
3sin 28
sin(28
2.38.
10 )
p3
6.442 86.1 555 kPa abs.
( p3 )normal 10.33 40 413 kPa abs.
Expansion Waves
9.69
At M 1
1
3,
49.8  ,
1
49.8
2
25
19.47  . (See Fig. 9.18.)
1
74.8  .
From isentropic flow table: p2
9.70
9.71
1
T0 T2
T1 T0
M 2 4.78.
p p
1
0.002452 1.80 kPa abs.
p1 0 2 20
0.02722
p1 p0
1
0.1795 127K or
0.3571
T2
T1
V2
4.78 1.4 287 127 1080 m/s.
26.4  .
253
4,
65.8
26.4
273
90 25 70.53 12.08 32.4 .
1
0.2381 117 K.
0.5556
T1
V2
4 1.4 287 117
For M 4 ,
T0 T2
T1 T0
12.08 .
39.4  .
T2
26.4  .
2
65.8  . (See Fig. 9.18.)
For M
T0 T2
T1 T0
146 C.
65.8  .
T2
65.8
26.4
156  C.
39.4  .
1
0.2381 210 R or 250 F.
0.5556
T2
T1
V2
4 1.4 1716 210
490
867 m/s.
2840 fps.
256
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Chapter 9 / Compressible Flow
9.72
a)
39.1 .
1
39.1 5 44.1 .
2
Mu
2.72. p2u
(20/0.0585) 0.04165
= 14.24 kPa abs.
5 and M
For
2.5,
b) M 2.72 ,
5 .
25  . M 1n
.875
sin(25  5  )
M 2u
For M 2.37 ,
= 36.0  . For
c) Force on plate = ( 26.4
F cos 5 
1
2
1V1 A
2
F sin 5 
1
2
1V1 A
2
d) C D
19 . M1n
9.73
M2
p3
p2
p0 p3
p2 p0
36.1
CD
1
2
1.4
12.2
1
2
1.4
5)
36.1
2.5
2.5 2
1
0.889/ sin(27
0.889.
5)
2.37.
1.15 , M 2n .875.
36 5 41 , M2
2.58.
F.
A
1000A
0.139.
F
20 000A
Lift
Airfoil
surface
Drag
0.0122.
20 000 A
p2 1.805 20 36.1 kPa. M 2n
54.36.
2
59.36.
1
0.0122 23.4 kPa.
0.0188
A
A
23.4
sin 5
2
2
1
V12 A
2
M 2n
2.56.
1000 A .0872
1.30.
3.25.
2
M2
1.13.
2.72 sin 25 
1000
12.2 .996
4sin19
0.786
sin(19
14.24 )
2.5sin 27
M
1.32 20 26.4 kPa abs.
p2
CL
27 . M1n
6.35 0.0872
1
1.4 42 20
2
257
M3
0.786.
3.55.
shock
M1
p2
M2
p3
M3
0.0025.
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Chapter 9 / Compressible Flow
9.74
If
5 with M1
M1n
4sin18
p2
1.24.
1.627 20
0.818
M2
sin(18
At M1
4,
1
= 18 .
4, then Fig. 9.15
5)
M 2n
M1
0.818.
shock
32.5 kPa.
M2u
shock
M2l
3.64.
65.8 . At 75.8 , M 2u
4.88. p2u
p1
p0 p2
p p0
20
0.002177
0.006586
= 6.61 kPa.
CL
CD
Lift
32.5 A cos 5
Drag
32.5 A sin 5 6.61 ( A/2) sin10
1
1.4 42 20 A
2
1
V12 A
2
1
V12 A
2
20 A/2 6.61 ( A/2) cos10
1
1.4 42 20 A
2
258
0.0854.
0.010.
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Chapter 10 / Flow in Open Channels
CHAPTER 10
Flow in Open Channels
10.1
10.2
Type
Open Channel
(a) Steady,
uniform
Pumping between two
reservoirs in constant
diameter pipe
Terminal flow in a very
long prismatic channel
(b) Unsteady,
nonuniform
Water hammer
Flood wave in river, or
hydraulic bore
(c) Steady,
nonuniform
Constant discharge
in diffuser
Backwater conditions
behind a dam
(d) Unsteady,
uniform
Gradual deceleration
of flow in a pipe of
constant diameter
“Practically impossible
situation” (Chow); however, the kinematic wave
concept assumes this type
(Henderson)
  cos1 (1  2  0.3)  1159
.
rad or 66.4  ,
Q 2B
Q 2 d sin 

gA 3 gd 2 / 4 (  sin  cos  )  3

10.3
Closed Conduit
4 2 sin( 1.159)
 1.
9.81  (d 5 / 64 )  1.159  sin 1.159 cos 1.159) 3
This equation reduces to:
192.1  d5 .
(a) Steady, nonuniform
(c) Unsteady, nonuniform
(b) Unsteady, nonuniform
(d) steady, nonuniform
259
d  2.86 m
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Chapter 10 / Flow in Open Channels
Uniform Flow
10.4
260
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Chapter 10 / Flow in Open Channels
10.5
A  by  my 2 ; solve for b : b  A /y  my; then
P  b  2y 1  m2  A/y  my  2 y 1  m2 .
Set dP / dm  0, solve for m :
dP
 y 
dm
2my
1  m2
2m  1  m 2 ,
 0,
3m 2  1,
 m  3/3.
Set dP / dy  0, noting that dA / dy  0 likewise:
dP 1 dA A

 2  m  2 1  m2  0
dy y dy y
 A   my 2  2y 2 1  m 2

3
1
 y2
2 1 
3
 3

3
3
 y2
4
 A  3y 2 .
,
3 
 3
2
One can also show that b 
3y and P  2 3y  3b.
3
10.6
c1 A5/ 3
S0 .
Use Chezy-Manning equation in the form Q 
n P 2/ 3
Substitute in appropriate expressions for A and P, and solve for y 0 by trial and error.
(a)
35 

1 2


4.5y 0  2 y 0 ( 2.5  3.5)
5 /3

0.015 4.5  y 0 ( 1  2.5 2  1  3.5 2 )
2 /3
0.00035 .
(4.5 y0  3y02 )5/3
This reduces to 28.06 
(4.5  6.33y0 ) 2/3
Solving, y0  2.15 m, A  4.5  2.15  3  2.152  23.5 m 2 ,
P  4.5  6.33  2.15  18.1 m.
261
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Chapter 10 / Flow in Open Channels
2.1
(b) 425 
2

/ 4 (  sin  cos  )
5 /3
0.001.
0.012( 2.1 ) 2 /3
This reduces to 2.248 
(  sin  cos  ) 5 /3
 2 /3
Solving,   1.825 rad, y0 
A
,
2.1
(1  cos1.825)  1.31 m,
2
2.1 2
(1.825  sin 1.825 cos 1.825)  2.28 m 2 ,
4
P  2.1  1.825  3.83 m.
(c) 120 


1.49 6 2 / 4 (  sin  cos  )
5 /3
0.012( 6 ) 2/3
This reduces to 2.593 
0.001.
(  sin  cos  ) 5 /3
 2 /3
,
6
Solving,   1.965 rad, y0  (1  cos1.965)  4.15 ft,
2
36
( 1.965  sin 1.965 cos 1.965)  20.9 ft 2 ,
A
4
P  6  1.965  11.8 ft.
(d) From Problem. 10.5, m  3/3, A  3 y02 , P  2 3y0 .
Substitute into Chezy-Manning equation:
( 3 y 02 ) 5/3
(2 3 y 0 )
2 /3

Qn 15  0.011

 4.576.
s0
0.0013
This reduces to y08/3  4.195.
 y 0  4.195 3/8  1.71 m, A  3  1.71 2  5.08 m 2 ,
P  2 3  1.71  5.93 m.
1.49( 25y 0 ) 5 /3
(e) 1200 
0.02( 25  2y 0 ) 2 /3
0.0004 .
y 05 /3
,
This reduces to 3.768 
( 25  2y 0 ) 2 /3
Solving, y 0  10.2 ft , A  25  10.2  255 ft 2 , P  25  2  10.2  45.4 ft.
262
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Chapter 10 / Flow in Open Channels
10.7
Q 4

 2.857 m2
A 1.4
2/3
1 A
V  
S0 . Solve for P.
n P 
A
 S 
P  A 0 


 Vn 
3/2
 0.001 
 2.857 

 1.4  0.015 
3/2
 5.280 m.
 P  b  2y 0 1  m 2 ; 5.280  b  2y 0 1  1.75 2 ; b  5.280  4.031y 0 .
Substitute into area function:
A  by 0  my 02 ; 2.857  (5.280  4.031y 0 )  1.75y 02 .
This relation reduces to y02  2.315y0  1.253  0.
y0 
2.315 1
2.315 2  4  1.253  1.452 or 0.863 m.

2
2
Use lower value of y 0 , since y 0  1.452 results in a negative b:
 y 0  0.86 m, b  5.280  4.031  0.863  1.80 m.
10.8
b = 0, m1 = 8, m2 = 0
1
1
8
A  by  y 2 (m1  m2 )  m1y 2  y 2  4y 2
2
2
2
P  b y
Q  AR

2/3
 
1  m12  1  m22  y
 4y2 
S0
 4y2 

n
 9.062 y 
2/3
 
1  m12  1  y
0.0005
 3.456 y8/3
0.015
(a)
y  0.12 m, Q  3.956(0.12)8/3  0.0121 m3 /s
(b)
 0.08 
Q  0.08 m /s, y  

 5.456 
10.9 Q 
3
3/8
 0.244 m.
1 A5/3
1 d 2 (  sin  cos  )5/3

S
S0 ,
0
(d) 2/3
n P 2/3
n 4
Q 

65  1  9.062y


2
1 d    2/3

S0  d 4/3 S0
 d
8n
n 4 2
2
263
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Chapter 10 / Flow in Open Channels
(a) Q  1.75 m3 /s, S0  0.0003, n  0.014
 8nQ
d  
 S
0

3/4




3/4
 8  0.014 1.75 


   0.0003 
(b) S0  0.00005, d  1.30. Q 

8  0.014
 2.62 m
1.34/3  0.00005  0.281 m3 /s
2
 8nQ   8  0.014 0.45 2
(c) d  0.75 m, Q  0.45 m /s. S0   4/3   
 0.000554
 d     0.754/3 


3
Energy Concepts
10.10
q  2gy 2 (E  y), E  constant,
dq 1 4 gy ( E  y)  2 gy 2 2 gy ( E  y)  gy 2
.


dy 2
q
2 gy 2 ( E  y)
Setting dq / dy  0 and noting that q  0, so that the numerator is zero, one can
solve for y at the condition q  qmax (note: y  0 is a trivial solution):
gy[ 2(E  y)  y]  0,
10.11
2E  3y  0,
 y  2E/3  yc .
q2
Ey
,
2gy 2
y
q2
E


y c y c 2gy c y 2
y
q2
.


y c 2gy c3 ( y / y c ) 2
But q / gy c3  1, 
1
E y 1
.
 
yc yc 2 ( y / yc )2
1
d  E
0
   1
dy  y c 
(y / y c ) 3
 y / y c  1 and E / y c  1  1 / 2  3 / 2
264
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Chapter 10 / Flow in Open Channels
10.12
q  V1y1  3  2.5  7.5 m 2 /s,
E1  y 1 
q2
2 gy 12
 2.5 
32
 2.96 m ,
2  9.81
y c  3 q 2 / g  7.5 2 / 9.81  1.79 m, Ec 
(a) E2  E1  h  2.96  0.2  2.76 m, E2  Ec .
3
3
y c   1.79  2.68 m.
2
2
 y 2 is subcritical.
q2
7.5 2
2.87
,
or
2
.
76


 y2  2
y
2
2
2
2gy 2
2  9.81y 2
y2
 E2  y 2 
Solving, y2  2.13 m, and y2  h  y1  2.13  0.2  2.5  0.17 m.
(b) y1  (1  Kc )
q2
2 gy12
 y2  (1  Kc )
 2.5  (1  0.1)
q2
2 gy22
 h; Kc  0.1, h  0.15 m.
32
7.52
 y2  (1  0.1)
 0.15.
2  9.81
2  9.81y22
The equation reduces to 2.855  y 2 
3.154
.  Solving, y 2  2.20 m,
y 22
y2  h  y1  2.2  0.15  2.5   0.15 m
(c) Set E2  Ec  2.68 m and  maximum h is
hmax  E1  E2  E1  Ec  2.96  2.68  0.28 m.
 yc  hmax  y1  1.79  0.28  2.5  0.43 m
(d) y1  ( 1  K e )
q2
q2
1
y
(
K
)



 h; K e  0.2, h  0.2 m
2
e
2 gy12
2 gy 22
7.5 2
32
 y 2  ( 1  0.2 )
 0.2
2  9.81
2  9.81 y 22
2.294
The equation reduces to 3.067  y 2 
.  Solving, y 2  2.77 m,
y 22
 2.5  ( 1  0.2 )
y2  h  y1  2.77  0.2  2.5  0.07 m.
10.13
q1 
q
5
Q 25

 5 m2 / s, V1  1   2.5 m/s,
b1 5
y1 2
3
yc1  3 q12 /g  52 /9.81  1.37 m,
265
E1  y1 
q12
2 gy12
 2
2.52
 2.32 m.
2  9.81
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Chapter 10 / Flow in Open Channels
Q 25
3
3 4.542

 4.54 m2 /s, Ec2  3 q22 /g  3
 1.92 m.
b2 5.5
2
2 9.81
Set E1  E2 and solve for y2 , noting that y2 is subcritical, since Ec2  E2:
(a) q2 
4.54 2
1.051
2.32  y 2 
 y2 
.
2
y 22
2  9.81y 2
25
Q
(b) q2  
 5.263 m2 /s, Kc  0.1.
b2 4.75
y1  (1  Kc )
q12
2 gy12
 y2  (1  Kc )
q22
2 gy22
 Solving, y 2  2.07 m
,
52
5.2632

y

(1

0.1)
.
2
2  9.81 22
2  9.81y22
1.553
The equation reduces to 2.35  y2  2 .
y2
Solving (assuming subcritical conditions at location 2), y 2  1.94 m.
 2  (1  0.1)
10.14
(a) q1  V1y1  3  3  9 m2 /s, Q  b1q1  3  9  27 m3 /s,
E1  y1 
q12
2 gy12
 3
32
3
 3.46 m, yc1  92 /9.81  2.02 m,
2  9.81
3
Ec1   2.02  3.03 m.
2
Without change in width at loc. 2, E2  E1  h  3.46  0.7  2.76 m.
Since E2  Ec1 , width must change to prevent choking. Set E c 2  2.76 m.
 y c2 
2
 2.76  1.84 m, q 2  gy c 23  9.81  1.84 3  7.81 m 2 / s.
3
b2  Q / q 2  27 / 7.81  3.46 m.
(b) q1  V1y1  10 10  100 ft 2 /sec, Q  b1q1  10 100  1000 ft 3 /sec,
E1  y1 
q12
2 gy12
 10 
102
3
 11.55 ft, yc1  1002 /32.2  6.76 ft,
2  32.2
3
Ec1   6.76  10.14 ft.
2
Without change in width at loc. 2, E2  E1  h  11.55  2.3  9.25 ft.
Since E2  Ec1 , width must change to prevent choking. Set Ec 2  9.25 ft.
2
 yc2   9.25  6.17 ft, q2  g( yc2 )3  32.2  6.173  86.9 ft 2 /sec.
3
 b2  Q /q2  1000/86.9  11.51 ft.
266
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Chapter 10 / Flow in Open Channels
10.15
q  Q / b1  4.8 / 2  2.4 m2 / s , y c1  3 2.4 2 / 9.81  0.84 m.
(a) Desire y 2  y 1  h  1.22  0.1  1.32 m. Write energy eqn. from upstream loc. (1)
into transition loc. (2):
q2
q2
y1  1 2  h  y2  2 2 ,
2 gy1
2 gy2
and solve for the unknown q2 :
 1.22 
q 22
2.4 2
.
.
.
.
0
1
1
52
1
32




2  9.81  1.22 2
2  9.81  1.32 2
Reducing, q22  6.84,  q2  6.84  2.62 m 2 /s.  b2  Q / q2  4.8 / 2.62  1.84 m.
(b) Let E2  Ec2  E1  h  1.52 m. Then,
yc2 
2
2
Ec2   1.52  1.01 m, q2  1.013  9.81  3.195 m,
3
3
 b2  Q / q2  4.8 / 3.195  1.50 m.
10.16
q2
5.52
E1  y1 
 2.15 
 2.484 m
2 gy12
2  9.81 2.152
Fr1 
q 2 3 5.52

 0.557. yc 

 1.456  1.46 m
g
9.81
gy13
9.81 2.153
5.5
q
3
3
3
yc  1.456  2.183  2.18 m
2
2
(a) The maximum height of the raised bottom at location 2 will be one for which the
energy is a minimum:
Ec 
E 1  Ec  h , 2.484  2.183  h ,  h = 2.484 - 2.183 = 0.30 m
(b)
(c)
(d) Since Fr1 < 1, if h > 0.30 m, subcritical nonuniform flow will occur upstream of the
transition.
267
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Chapter 10 / Flow in Open Channels
10.17 Refer to Fig. 10.8. Since a steep channel exists downstream of the entrance, critical
conditions occur at the entrance. The two eqns. to be satisfied are
y1  z2  yc 
Ac
and Q  gAc3 /Bc .
2Bc
The first eqn is solved independently for y c , and then y c is substituted into the second eqn
to find Q. Note: Ac and Bc refer to critical conditions.
(a) Rectangular channel: Ac  by c  4y c , Bc  b  4.
y 1  z2  y c 
by c 3
2
2
 y c ,  y c  ( y 1  z 2 )   2.5  1.67 m,
2b 2
3
3
 Q  gb3 yc3 / b  9.81 42 1.673  27.0 m3 /s.
(b) Trapezoidal channel: Ac  byc  myc2 , Bc  b  2myc .
y1  z2  yc 
byc  myc2
,
2(b  2myc )
2.5  yc 
3yc  2.5yc2
,
2(3  2  2.5yc )
reduces to 12.5yc2  16yc  15  0
 yc 


1
16  16 2  4  12.5  15  1.91 m (use positive root).
2  12.5
Ac  3  1.91  2.5  1.91 2  14.85 m 2 , Bc  3  5  1.91  12.55 m.
Q  9.8114.853 /12.55  50.6 m3 /s
d2
d
(  sin  cos  ), Bc  d sin  , yc  (1  cos  ),
4
2
2
(d /4)(  sin  cos  )
y1  z2  yc 
,
2d sin 
(3.52 /4)(  sin cos )
2.5  yc 
, reduces to
2  3.5sin
(c) Circular channel: Ac 
2.5  1.75(1  cos )  0.4375(  sin  cos ) / sin ,   cos 1 (1 .5714 yc ).
 Solving, yc  179
. m (  1594
.
rad),
Ac  (3.52 /4)(1.594  sin1.594cos1.594)  4.95 m 2 ,
Bc  3.5  sin 1.594  3.50 m.
Q  9.81 4.953 /3.50  18.4 m3 /s.
268
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Chapter 10 / Flow in Open Channels
10.18
At the canal entrance, the condition of critical flow is Fr 2  Q2 B / gA3  1, to be solved
for the unknown width b.
(a) Rectangular channel: B  b, A  by, y  y c .

(b)
Q
Q 2b
Q2
18

 1. b 

 5.75 m.
3 3
2 3
gb y
gb y
9.81  1
gy 3
Trapezoidal channel: B  b  2my  b  2  3  1  b  6,
A  by  my 2  b  1  3  12  b  3.

Q 2B
18 2 (b  6)
(b  3) 3


 33.0.
1
,
or
(b  6 )
gA 3 9.81(b  3) 3
Solving, b  3.89 m.
(c)
Rectangular channel: B  b, A  by, y  y c .
Q 2B
Q2
 3  2 3  1.  b 
gA
gb y
(d)
Q
gy
3

635
32.2  3 3
 21.5 ft.
Trapezoidal channel: B  b  2my  b  2  3  3  b  18,
A  by  my 2  b  3  3  3 2  3b  27.

635 2 (b  18)
Q 2B
(b  9) 3


1
 463.8.
,
or
(b  18)
gA 3 32.2( 3b  27 ) 3
Solving, b  16.1 ft.
10.19
(a) Write energy eqn. from reservoir (section R) to section C, where critical conditions
exist:
y R  z R  Ec  zc , Ec  y R  z R  zc  103.6  102  1.6 m.
2
2
 y c  Ec   1.6  1.067 m.
3
3
3
 q  gy c  9.81  1.067 3  3.45 m2 / s.
(b) Energy eqn. from reservoir to section A:
3.45 2
0.6067
103.6  y A 
 100, or 3.6  y A 
.
2
2  9.81y A
y A2
Solving, yA  3.55 m.
Energy eqn. from reservoir to section B:
3.45 2
0.6067
103.6  y B 
 101, or 2.6  y B 
.
2
2  9.81y B
y B2
Solving, yB  2.50 m.
269
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Chapter 10 / Flow in Open Channels
10.20 Objective is to determine the zero of the function f ( y)  Q2 B( y) / g[ A( y)]3 1, in which
eqns. for B(y) and A(y) representing either a circular or trapezoidal section area can be
substituted. The false position algorithm is explained in Example 10.10; this was used to
determine the roots. The solutions are:
(a) yc  1.00 m, (b) yc  1.50 f t, (c) yc  1.81 m, (d) yc  3.48 f t.
10.21
A3 Q 2

, or
B
g
for critical flow is
A trial-and-error solution yields
10.22
3 2
y and B  10  3y c . The criterion
2 c
3
3 2

( y c  1.5)  10  2 y c 
16.5 2

 27.75
10  3y c
9.81
Assume yc > 1.5 m; then A  ( y c  1.5)  10 
yc  1.78 m .
In the lower section ( y  1 m ), the area is
y
y
0
0
A   bd   3  d  2 y 3/ 2 . Assume y  1 m; then Fr 2 
Q2 B Q2 3 y
3Q 2


gA3
g 8 y 9/ 2
8 gy 4
3Q 2
 1, or Q  8  9.81 / 3  5.1 m 3 / s
8  9.81  1 4
If y c  1 m , then
 when Q  5.1 m 3 / s, critical depth will be > 1 m. Cross-sectional area
including upper region (y > 1 m) is
A  2(1) 3/2  23( y c  1)  23y c  21, and B  23 m.
(a)
Q 2B
55 2  23

 1, ( 23y c  21) 3  7092,
3
3
gA
9.81( 23y c  21)
 yc 
(b)
10.23
(7092) 1/3  21
 1.75 m.
23
3  3.5 2
Q2B
 1, y c4  0.468,  y c  0.83 m.

3
4
8  9.81 y c
gA
y
y
0
0
x  10 , A   2  d   2 10 
A  4.216y
3/2
.
4
10y 3 /2 , or
3
25 2
1.792
Q2
.
Ey
y
y
2
2 3
2gA
2  9.81  4.216 y
y3
Determine depth alternate to y = 2.0 m:
2 
1.792
1.792
 2.224  y alt  3 ,
3
2
y alt
270
 Solving, y alt  1.21 m.
© 2012 Cengage learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 10 / Flow in Open Channels
Q2
, in which E is known. This
g[ A( y )] 3
relation can be solved using Excel Solver, or the “Find” function in Mathcad.
10.24
Determine the zero of the function f ( y )  E  y 
10.25
v  2 g , tan

2

w/2
,
Y 
Y
 
Q   vwd   2 g 2 tan (Y   )d
2

0
0
Y


 2 2 g  tan    Y 1/ 2   3/ 2  d
20

2
 
 8
   2
2 g  tan Y 5 / 2 .
 2 2 g  tan  Y 5 / 2  Y 5 / 2  
2  3
5
2

 15

Y
cd :
Multiply by the discharge coefficient
 Q  cd
8


2 g  tan Y 5 / 2 .
15
2

1.52
22.5
2
3
y

 0.612 m
 1.5 m /s, c
9.81
15
1.52
3
 1.28 m  E3
Ec   0.612  0.919 m, E0  1.2 
2  9.811.22
2
E2  E0  h  1.28  0.2  1.08 m > E3  no choking.
10.26 (a) q 
Since losses are neglected throughout the transition, y1 = y0, and y2 can be computed
by writing the energy equation between locations 1 and 2:
1.28  y2 
1.52
2  9.811.2
2
 h  y2 
0.115
y22
+0.2 ,  y2  0.95 m .
(b)
271
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Chapter 10 / Flow in Open Channels
10.27
v  2g ,
w / 2  x  y 2  (Y  ) 2
Y
Y
0
0
Y
Q   vwd   2 g  2(Y  )2 d  2 2 g  1 / 2 (Y 2  2Y  2 )d
0
Y
4
2
2

 2 2 g   Y 21/ 2  2Y 3/ 2  5/2  d  2 2 g  Y 7/ 2  Y 7/ 2  Y 7/ 2 


3
5
7
0
32
 70  84  30  7 / 2
 2 2g 
2 gY 7 / 2 .
Y , or  Q 
105
105


10.28 a) B = 3 ft, H = 1.2 ft
0.026
Q  4BH1.522B
0.026
 4  3 1.21.5223
 16 ft3 /sec
b) H (ft)
0.5 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5
Q (ft3/sec) 4.05 8.46 10.2 12.0 13.9 16.0 18.1 20.3 22.6
10.29 Let location 1 be upstream of the transition section, and location 2 be in the transition. For
both (a) and (b), b = 3.5 m, and h = 0.6 m.
(a) y1 = 1.5 m, Q = 4 m3/s.
q
4
 1.143 m 2 /s ,
3.5
yc  3
1.1432
 0.511 m ,
9.81
3
1.142
Ec   0.511  0.767 m ,
E1  1.5 
 1.529 m ,
2
2  9.811.52
Check: E2  E1  h  1.529  0.6  0.929  Ec
 0.929  y2 
1.142
2  9,81 y22
 y2 
272
0.0662
y22
 y2  0.835 m (T&E)
© 2012 Cengage learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 10 / Flow in Open Channels
(b) y1 = 1.0 m, Q = 4 m3/s.
q
3
 0.857 m2 /s ,
3.5
yc  3
3
Ec   0.422  0.633 m ,
2
0.8572
 0.422 m ,
9.81
E1  1.0
0.8572
2  9.81 1.02
 1.037 m ,
Check: E2  E1  h  1.037  0.6  0.437  Ec
 Choking occurs at location 2. Set E2  Ec , hence E1  Ec  h, or
y1 
0.857 2
2  9.81 y12
which reduces to
y1 
 0.633  0.6,
0.0374
y12
 1.233  y1  1.21 m (T&E)
(c) For situation (b), the raised section acts as a broad-crested weir, since critical flow
occurs over it.
2 2 
2 2 
10.30 Q 1  b
g  ( y 1  h) 3 /2 , Q2  b
g ( y2  h) 3 /2 ;
3 3 
3 3 
given Q1 , y 1 , Q2 , y 2 , solve for b and h.
y 1  h  Q1 
 
y 2  h  Q2 
(a) h 
2 /3
y 1  y 2 (Q 1 / Q 2 ) 2/3
.
, or h 
1  (Q 1 / Q 2 ) 2/3
1.05  1.75(0.15 / 30) 2 / 3
 1.03 m,
1  (0.15 / 30) 2 / 3
b 
(b) h 
Q2
2 2 
g  ( y 2  h) 3/2

3 3 

30
 28.8 m.

2 2
3 /2
 9.81  (1.75  1.03)

3 3

3.45  5.75(5 / 1000) 2 / 3
 3.38 ft ,
1  (5 / 1000) 2 / 3
b 
1000
 88.7 ft.

2 2
3 /2
 32.2  ( 5.75  3.38)

3 3

273
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Chapter 10 / Flow in Open Channels
10.31
Assume uniform, subcritical flow throughout the canal; then,
Elev. A  Elev. B
(by)5/3
S0 
, Q
S0 .
L
n(b  2 y ) 2/3
Write energy eqn. from reservoir to canal entrance:
Q2
Q2
hy
y
. Unknowns are Q and y.


2gA 2
2gb 2 y 2
Eliminate Q in energy eqn. and solve for y:
h
 b 4 /3 S 0 
y 4 /3
y 10 /3
1  b 10 /3 S 0 
y


.




2
4 /3
2gb 2 y 2  n 2  (b  2y ) 4 /3
 2gn  (b  2y )
Substitute given data into above relations, and compute S 0 , y, and Q .
(a) S0 
501.8  500.2
 0.00107,
1500
0.944 y 4 /3
y 4 /3
 2.5 4 /3  0.00107 
2y


y
,

 2  9.81  0.014 2  ( 2.5  2y ) 4 /3
( 2.5  2y ) 4 /3
 Solving, y  1.82 m,
Q 
(2.5 1.82)5/3 0.00107
0.014(2.5  2 1.82)
Check for y  y c :
2/3
 8.70 m3 /s.
yc  3 q 2 /g  3 (8.70 / 2.5) 2 /9.81  1.07 m  1.82 m.
 Assumption of subcritical flow is valid.
(b) S0 
1646  1641
 0.00102,
4920
y 4 /3
2.868y 4 /3
 1.49 2  8 4 /3  0.00102 
y
,
6.5  y  



 2  32.2  0.014 2  ( 8  2y ) 4 /3
( 8  2y ) 4 /3
 Solving, y  5.93 ft ,
Q 
1.49(8  5.93)5/3 0.00102
yc  3
0.014(8  2  5.93)
2/3
 288 ft 3 /sec.
(288/8) 2
 3.42 ft  5.93 ft.
32.2
274
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Chapter 10 / Flow in Open Channels
Momentum Concepts
10.32
 y2 q2 
 
M  b
 2 gy 
2
q2
M 1 y 
    2
by c2 2  y c 
gy c y
2
q2
1 y 
.
    3
2  yc 
gy c ( y / y c )
But q 2 /gyc3  1,
2
1 y 
1
M
 2    
(y / yc )
by c 2  y c 
10.33
Q2
F
 M1  ,
Determine the zero of the function f ( y )  A2 ( y )y ( y ) 
gA2 ( y )

in which M 1 and F are known. This relation can be solved using Excel Solver, or the
“Find” function in Mathcad.
10.34 (a) Conditions at location 1:
q2
1.5 2
E1  y1  1 2  1.8 
 1.835 m,
2 gy1
2  9.81  1.8 2
d/2
q1
b2
b1
3
yc  3 q12 /g  1.52 /9.81  0.612 m.
q2
d/2
The smallest constriction at location 2 is one that establishes critical flow, i.e., where
minimum energy exists:
 E 2  E c , or since E 1  E 2 ,
2
2
2
y c2  Ec2  E 1   1.835  1.22 m.
3
3
3
 q2  g( yc2 )3  9.811.223  4.22 m2 /s,
b2 
q1b1 1.5  6

 2.13 m.
4.22
q2
Hence the maximum diameter is d  b 1  b 2  6  2.13  3.87 m
275
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Chapter 10 / Flow in Open Channels
The momentum eqn. cannot be used to determine the drag since no information is given
with regard to location downstream of cofferdam. But we do have available the drag
relation, which will provide the required drag force:
Frontal area:
A  y 1 d  1.8  3.87  6.97 m2 ,
Approach velocity: V1 
q1 1.5

 0.833 m/s.
y1 1.8
1
1
 F  CD A V12   0.15  6.97 1000  0.8332  363 N,
2
2
A rather insignificant drag force!
(b) E 1  y 1 
q 12
16 2
6


 6.11 ft ,
2gy 12
2  32.2  6 2
y c  3 q12 / g  3 16 2 / 32.2  1.99 ft.
2
2
2
y c2  Ec2  E 1   6.11  4.07 ft,
3
3
3
3
 q 2  ( gy c2 )  32.2  4.07 3  46.6 ft 2 / sec,
b2  q1b1 / q2  16  20 / 46.6  6.87 ft ,
 d  b1  b2  20  6.87  13.1 ft .
Frontal area:
A  y 1 d  6  13.1  78.6 ft 2 ,
Approach velocity: V1  q 1 / y 1  16 / 6  2.67 ft / sec.
 F  C D AV12 / 2  0.15  78.6  1.94  2.67 2 / 2  81.5 lb.
10.35
Check flow conditions at location 2:
q  V2 y2  3  2.5  7.5 m 2 /s, yc  3 q 2 /g  3 7.5/9.81  1.79 m.
Flow is subcritical, and provided pipeline does not “choke” flow at location 3,
conditions will remain subcritical from locs. 1 to 2.
E 3  E 2  d  2.5  7.5 2 / (2  9.81  2.5 2 )  0.2  2.76 m,
Ec 
3
3
yc   179
.  2.68 m.  Ec  E2  d , and no choking occurs.
2
2
276
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Chapter 10 / Flow in Open Channels
(a)
Write the momentum eqn between locations 1 and 2:
2
y 12
q2
q2
V12  C D bq 2
F y2
F
1




. Now,
  C D A
.

2 gy 1 b 2 gy 2
2 
b b 
2gy 12
Substitute this along with known data into momentum eqn:
y 12
7.5 2
0.3  0.2  7.5 2 2.5 2
7.5 2




,
2 9.81y 1
2
9.81  2.5
2  9.81y 12
or
y 12 5.734 0.172


 5.419.
2
y1
y 12
Solving, y 1  2.52 m,
 V1  q /y1  7.5/2.52  2.98 m/s.
(b)
F  C D bd V12 /2  0.3 100  0.2 1000  2.982 /2  2.66 104 N.
10.36
1
wh
2
1
  2  3  0.3
2
 0.27 m 2
Asill 
y
1
y
3
A  my 2  3y 2
M 1  A1 y1 
1
Q2
Q2
Q2
0
.
125
,
 3  0.5 2   0.5 


3
7.358
gA1
9.81  3  0.5 2
M 2  A2 y 2 
Q2
1
Q2
Q2
 3  1.8 2   1.8 

5
.
832

,
gA2
3
95.35
9.81  3  1.8 2
Q2
0.4  0.27Q 2
F C D Asill Q 2



.

2gA12
2  9.81  ( 3  0.5 2 ) 2 102.2
Substitute into momentum eqn: M1  M2  F / .
Q2
Q2
Q2
,
 5.832 

7.358
95.35 102.2
1
1
1 
 1
2
. )
Q  (5.832  0125


  49.37,
 7.58 95.35 102.2 
 0.125 
Q  7.03 m3 /s.
277
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Chapter 10 / Flow in Open Channels
10.37
Continuity eqn:
y 2 V1  w

.
y1
w

y2 1 
(V1  w) 2
Momentum eqn:
  1 8
 1.
y 1 2 
gy 1

Eliminate y 2 / y 1 , substitute in known data, and solve for w:

(V  w) 2
V1  w 1 
  1 8 1
 1 ,
2 
w
gy 1


( 0.85  w ) 2
0.85  w 1 
 1 ,
  1 8
w
9.81  1.6
2 

0.85
1
 1
w
2


1  0.51( 0.85  w) 2  1 ,
3  1  0.51( 0.85  w) 2 
1.70
.
w
 V  w
 0.85  3.8 
Solving, w  3.8 m/s, and y 2  y 1  1
  1.96 m.
  1.6

 w 
3.8 
10.38
Given condition: V2 y 2  0.4V1 y 1 ,
Continuity eqn: y 1 (V1  w)  y 2 (V2  w),

y2 1 
(V  w) 2
  1 8 1
 1.
y 1 2 
gy 1

Unknowns are V2 , y2 and w. Eliminate V2 and w, and solve for y2 .
Momentum eqn:
278
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Chapter 10 / Flow in Open Channels
(a) V2 y 2  0.4  1  1.5  0.6,
1.5( 1  w)  y 2V2  y 2 w  0.6  y2 w, or w  0.9 /(y 2  1.5).
2


y 2 1 
0.9 


1  8 1 
 / ( 9.81  1.5)  1 , or

1.5 2 
y 2  1.5 



2

0.9 
1  1  0.5437 1 
  1.333y 2  0.
y 2  1.5 

Solving, y 2  1.77 m,
 V2  0.6 / y2  0.6 /1.77  0.339 m/s,
 w  0.9 / ( y2  1.5)  0.9 / (1.77  1.5)  3.33 m/s.
(b)
V2 y 2  0.4  3  5  6,
5( 3  w)  y 2V2  y 2 w  0.6  y2 w, or
2


y 2 1 
9 


1  8 3 
 / ( 32.2  5)  1 , or

y 2  5
5 2



2

9 
1  1  0.0497 3 
  0.4 y 2  0.
y 2  5

Solving, y 2  5.80 ft ,
 V2  0.6/y2  0.6/5.80  1.03 ft/sec,
w 
9
9

 11.2 ft/sec.
y2  5 5.80  5
10.39 Since a hydraulic jump is to occur between locs. 1 and 2, apply momentum eqn. to find
y 2 . Use energy eqn. to determine h, with critical conditions on sill, loc. 3:
q  Q / b  20 / 7.5  2.667 m2 /s,
Fr1  q / gy13  2.667 / 9.81  0.53  2.408  1.




y1
0.5
1  8Fr12  1 
1  8  2.4082  1  1.47 m,
2
2
 y 3  y c  3 q 2 / g  3 2.667 2 / 9.81  0.898 m.
(a)  y2 
279
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Chapter 10 / Flow in Open Channels
q2
3
(b) h  E 2  E 3  E 2  E c  y 2 
 yc ,
2
2gy 2 2
 h  1.47 
2.667 2
3
  0.898  0.29 m.
2
2
2  9.81  1.47
(c) Downstream of the sill the depth y4 is the depth alternate to y2:
2.667 2
E2  1.638  y 4 
,  Solving, y4 =0.59 m.
2  9.81  y 42
F    M 2  M4 
 y 22
q2
y 42
q2 

  b



2 gy 4 
 2 gy 2
 1.47 2
2.667 2
0.59 2
2.667 2 
4
 9810  7.5  



  1.26  10 N


2
9
.
81
1
.
47
2
9
.
81
0
.
59


(d)
(e) Fr1 = 2.4, hence from Table 10.2 the jump is characterized as weak.
10.40 (a) To compute the discharge, write the energy equation between locations 1 and 2, and
solve for q:
q
2 g( y2  y1 )

y12
 y22


2  9.81(0.10  2.5)
 2.5
2
 0.10
2

 0.687 m2 /s
 Q  bq  5  0.687  3.43 m3 /s
(b) To compute depth downstream, first compute the Froude number at 2:
q
0.687
Fr2 

 6.934
3
9.81  0.10 3
gy 2
 y3 
y2
2


1  8Fr22  1 
0.10
2
280


1  8  6.934 2  1  0.932 m
© 2012 Cengage learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 10 / Flow in Open Channels
(c) To calculate power lost, first compute the head loss in the jump:
( y 3  y 2 )3
(0.932  0.10)3
hj 

 1.544 m
4y 3 y 2
4  0.932  0.10
   Qh  9810  3.43  1.544  5.20  10 4 watt, or 52.0 kW
W
j
10.41 q  Q / b 
555
.
 185
. m 2 / s, Fr3  q / gy33 ,
3
2
(a) y3 
y
c
3
1
1  185
. 
( q / Fr3 )2  3

  0.853 m.
g
9.81  0.75

 3 q 2 / g  3 1.85 3 / 9.81  0.704 m.
q2
1.852
0
853

.

 1.093 m ,
2  9.81  0.8532
2 gy32
1
E 2  E 3  h3 , but h3 
2  0.707.
2
q2
1.852
or
 y2 


E
h
,
y

 1.093  0.707
3
3
2
2  9.81y22
2gy 22
E3  y 3 
Equation reduces to y2  0.1744 / y22  1.80
 y 2  1.74 m.
At loc. 1, the water surface elev. relative to the datum is E 3  h 3  1.80 m.
 y 22
q2
y 32
q2 
(b) F   ( M 2  M 3 )  b



,
2 gy 3 
 2 gy 2
 1.742

1.852
0.8532
1.852
 9810  3




9.81 1.74
2
9.81 0.853
 2
 F  2.77  10 4 N.
281
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Chapter 10 / Flow in Open Channels
10.42 Use Eqn. 10.5.16 with F = 0:
y2
Q2
M 1  1 ( 2my 1  3b) 
6
g(by 1  my 12 )
1.1 2
60 2
 44.55 m 3 .
(2  3  1.1  3  5) 
2
6
9.81(5  1.1  3  1.1 )
 M 2  M 1 , or
y 22
Q2
( 2my 2  3b) 
 44.55,
6
g(by 2  my 22 )

y 22
60 2
( 2  3y 2  3  5) 
 44.55,
6
9.81( 5y 2  3y 22 )
367
y 23  2.5y 22 
 44.55. Solving, y 2  2.55 m.
5y 2  3y 22
Energy loss across jump is h j :
hj  E1  E2  y 1 
Q2
Q2
y


,
2
2gA12
2gA22
A1  5  11
.  3  11
. 2  913
. m 2 , A2  5  2.55  3  2.552  32.26 m 2 ,
60 2
60 2
 h j  1.1 
 2.55 
 0.575 m.
2  9.81  32.26 2
2  9.81  9.13 2
 : W
  Qh  9810  60  0.575  3.38  10 5 W.
Power dissipated is W
j
j
j
10.43 (a) Compute the height of the step: q1  V1  y1  8  0.5  4 m2 /s
 y 2 q 2 y22 q 2 
V12
  b 1 
 

2
 2 gy1 2 gy2 
9.81 0.52 / 2  42 / (9.81 0.5   2 2 / 2  4 2 / (9.81 2))
F = (M1  M2 ), or CD hb
h 
1.2  82 / 2
 0.146  0.15 m
(b) Compute the downstream depth if step were not present:
Fr1 
 y2 
q
gy
y1
2
3
1


4
9.81  0.5 3

1  8Fr12  1 
 3.613
0.5
2


1  8  3.613 2  1  2.317 m, or 2.32 m
Note that the width b was not used in the problem.
282
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Chapter 10 / Flow in Open Channels
10.44 First find Q, then compute y 1 :
V2  Fr2 gy2  0.4 9.81 1.5  1.534 m/s,
 Q  V2 A2  1.534  5 1.5  11.51 m3 /s.
q  Q / b  11.51/ 5  2.30 m 2 /s,
yc  3 q 2 / g  3 2.302 / 9.81  0.814 m
To compute y 2 , use momentum eqn, M1  F /   M2 , or
 y 12
q 2  C D AV12 / 2  y 22
q2 







.
b
 2 gy 1 
 2 gy 2 
Substitute in known data, plus the relation V1  q / y 1 :
y 12
2.30 2
0.35  5  0.17  1000  2.30 2 1.5 2
2.30 2




,
2 9.81y 1
2
5  9810y 12
9.81  1.5 2
which reduces to y 12 
1.078 0.0321

 2.97.
y1
y 12
Solving, y 1  0.346 m.
Nonuniform Gradually Varied Flow
10.45 Assume steep channel slope, with critical flow at the entrance. Then, from the minimum
energy concept,
2
y c  (103.6  101.5)  1.40 m,
3
(a)  Q  b gyc3  8 9.811.403  41.5 m3 /s.
(b) Compute y 0 using Chezy-Manning eqn:
(8 y 0 )
A 5/ 3
Qn
415
.  0.014
 9186
,
. ,
2/ 3 
2/ 3 
P
c1 S 0 (8  2 y 0 )
1  0.004
Solving, y 0  1.21 m.
Since y c  y 0 , slope is steep, and assumption of critical flow at entrance is valid.
(c) A possible jump is located at B. Note that its location must be determined using
numerical analysis (see the figure that follows).
283
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Chapter 10 / Flow in Open Channels
3
10.46 (a) q  Q / b  0.35 /1.8  0.194 m 2 /s,
yc  3 q 2 /g  0.1942 /9.81  0.157 m.
Use Chezy-Manning eqn. to compute y 0 :
(1.8y0 )5/3
0.35  0.012
Qn

 0.0329 
(1.8  2 y0 )2/3
c1 S0 1 0.0163
Solving, y 0  0.094 m.
3
3
Ec  h  y c  h   0.157  0.1  0.335 m,
2
2
2
q
0.194 2
0
094

.

 0.311 m.
E0  y 0 
2 gy 02
2  9.81  0.094 2
Since E 0  E c  h , normal conditions
cannot exist at loc. 1, and choking will
occur at loc. 2. Compute alternate depths
at locs. 1 and 3: E 1  E 3  E c  h.
y
0.1942
2  9.81y
2
 y
0.00192
y2
 0.335.
Solving, y 1  0.316 m, y 3  0.088 m. (Note: loc. 2 is a critical control, with
subcritical flow upstream and supercritical flow downstream.) A jump is located
upstream of loc. 1. Find the depth conjugate to y0:
Fr03  q2 /gyo3  0.1942 /(9.81 0.0943 )  4.62,
 ycj 
y0
2
 1  8  4.62 1  0.243 m
 1 8Fr 1  0.094
2
2
0
(b) q  Q /b  12.5/6  2.08 ft 2 /s,
3
yc  3 q 2 /g  2.082 /32.2  0.51 ft.
Use Chezy-Manning eqn. to compute y 0 :
(6y 0 )5/3
Qn
12.5  0.012
,

 0.7885 
(6  2y 0 )2/3
c1 S0 1.49  0.0163
Solving, y0  0.31 ft.
3
3
Ec  h  yc  h   0.51  0.33  110
. ft ,
2
2
q2
2.082
0
31
E0  y0 

.

 1.01 ft.
2 gy02
2  32.2  0.312
y
2.082
0.0672
. ,
 110
2  y
2  32.2y
y2
284
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Chapter 10 / Flow in Open Channels
Solving, y1  1.04 ft , y3  0.29 ft.
Fr03  q 2 /gyo3  2.082 /(32.2  0.313 )  4.51,
 ycj 
10.47
y0
2
 1 8Fr 1  0.312  1 8 4.51 1  0.79 ft
2
0
From Problem 10.46a, q  0.194 m2 /s,
yc  0157
.
m, and Ec  h  0.335 m.
Use Chezy-Manning eqn. to compute y 0 :
(1.8y0 )5/3
Qn
0.35  0.012

 0.1165 
(1.8  2 y0 ) 2/3
c1 S0 1 0.0013
Solving, y 0  0.21 m.
E0  y 0 
q2
2 gy 02
 0.21 
0.194 2
 0.254 m.
2  9.81  0.21 2
Since E 0  E c  h , normal conditions
cannot exist at loc. 1, and choking will
occur at loc. 2. Compute alternate
depths at locs. 1 and 3: E 1  E 3  E c  h.
y
0.194 2
0.00192
y
 0.335,
2
y2
2  9.81y
Solving, y 1  0.316 m, y 3  0.088 m. (Note: loc. 2 is a critical control, with
subcritical flow upstream & supercritical flow downstream.) A jump is located
downstream of location 3 (see figure, next page). Find depth conjugate to y 0 :
Fr02  q 2 /gy03  0.1942 /(9.81 0.213 )  0.414,
y
0.21
1  8  0.414  1  0.113 m.
 y cj  0 1  8 Fr02  1 
2
2



285

© 2012 Cengage learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 10 / Flow in Open Channels
3
10.48 q  Q / b  33 / 4  8.25 m 2 /s, yc  3 q 2 /g  8.252 /9.81  1.91 m.
Use Chezy-Manning eqn. to compute y 0 :
(4y 0 )5/3
Qn
33  0.012
,

 13.43 
(4  2y 0 )2/3
c1 S0 1  0.00087
Solving, y 0  2.98 m.
Since y 0  y c , mild slope conditions prevail. With yentrance  yc , an M3 profile exists
downstream of the entrance. For free outfall conditions, y exit  y c , and an M 2 profile
exists upstream of the exit. The M 3 and M 2 profiles are separated by a hydraulic jump
located approximately 260 m downstream of the entrance (determined by numerical
analysis).
10.49 q1  Q / b1  8.5 / 3  2.83 m 2 /s,
3
yc1  3 q12 /g  2.832 /9.81  0.935 m,
q2  Q / b2  8.5 /1.8  4.72 m 2 /s,
3
yc2  3 q22 /g  4.722 /9.81  1.314 m,
Ec 2 
3
3
y c2   1.314  1.97 m,
2
2
E0  y 0 
q 12
2 gy 02
 1.54 
2.83 2
 1.71 m.
2  9.81  1.54 2
(a) Since E 0  Ec2 , choking occurs at loc. 2, and the depth at loc. 1 is greater than y 0 .
To compute y 1 , set E 1  E c 2 :
y1 
q 12
2.83 2
0.4082


 y1 
 1.97 ,
y
1
2
2
2gy 1
2  9.81y 1
y 12
Solving, y 1  1.85 m.
(b) Since y 0  y c 1 , a mild slope condition exists, and an M1 curve occurs upstream of the
constriction.
286
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Chapter 10 / Flow in Open Channels
10.50 A spreadsheet solution is shown below. An M3 profile is situated upstream, with a
hydraulic jump at approximately 150 m followed by an M1 profile.
Q=
b=
20
0
yc =
y0 =
n=
m1 =
m2 =
0.014
3.5
2.5
S0 =
yu =
0.001
0.50
L=
yd =
300 g =
2.50 c1 =
9.81
1.00
ycj
3.769
3.324
2.837
2.578
2.365
2.184
2.028
1.893
1.778
1.684
Depth Residual
1.554
0.00
1.809 -1.8E-09
Station
1
2
3
4
5
6
7
8
9
10
y
0.500
0.600
0.700
0.800
0.900
1.000
1.100
1.200
1.300
1.400
A
0.750
1.080
1.470
1.920
2.430
3.000
3.630
4.320
5.070
5.880
V
26.667
18.519
13.605
10.417
8.230
6.667
5.510
4.630
3.945
3.401
E
36.744
18.079
10.135
6.330
4.353
3.265
2.647
2.292
2.093
1.990
ym
S(ym)
0.550
0.650
0.750
0.850
0.950
1.050
1.150
1.250
1.350
5.720E-01
2.347E-01
1.094E-01
5.612E-02
3.101E-02
1.818E-02
1.119E-02
7.175E-03
4.760E-03
33
34
35
36
36
36
35
32
28
x
0
33
67
102
138
174
210
245
277
304
11
12
13
14
15
16
17
2.500
2.450
2.400
2.350
2.300
2.250
2.200
18.750
18.008
17.280
16.568
15.870
15.188
14.520
1.067
1.111
1.157
1.207
1.260
1.317
1.377
2.558
2.513
2.468
2.424
2.381
2.338
2.297
2.475
2.425
2.375
2.325
2.275
2.225
1.878E-04
2.094E-04
2.340E-04
2.621E-04
2.943E-04
3.313E-04
-56
-56
-57
-59
-60
-62
300
244
188
131
72
12
-51
FM
54.49
37.97
28.08
21.75
17.51
14.59
12.56
11.17
10.24
9.68
Residual
2.77E-04
-5.67E-05
3.64E-06
-5.48E-04
-1.19E-04
-2.52E-05
-4.53E-06
-1.48E-04
-6.60E-05
-3.90E-05
10.51
First, compute y 0 1 , y 0 2 and y c . This can be done by trial & error, or by use of Excel
Solver (see Problem 10.57). The results are:
y 0 1  1.82 m, y 0 2  0.89 m, y c  1.19 m.
Since y 0 1  y c and y 0 2  y c , the slope upstream of loc. A is mild, and downstream of A
the slope is steep. Hence, loc. A acts as a critical control, provided no backwater from the
reservoir influences the depth at A. (This is verified by the varied flow calculations shown
in Problem 10.57.) Upstream of loc. A there is an M 2 curve, with critical flow at A,
followed by an S 2 curve terminating in a hydraulic jump. An S1 curve exists behind the
jump.
287
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Chapter 10 / Flow in Open Channels
10.52
q  Q /b  15/4  3.75 m 2 /s,
3
yc  3 q 2 /g  3.752 /9.81  1.13 m.
y01  0.93 m, yc  y01
upstream reach has a steep slope.
y02  1.42 m, yc  y02
downstream reach has a mild slope.
Compute depth conjugate to y 0 1 :
Fr012  q 2 /( gy013 )  3.752 /(9.81 0.933 )  1.782
y0
0.93
1  8  1.782  1  135
. m
 y cj  1 1  8Fr012  1 
2
2
Hydraulic jump occurs upstream of transition, followed by an S 1 curve up to the
transition.




q 2 3 1.672
Q 5
  1.67 m3 /s , and yc  3

 0.657 m
9.81
b 3
g
Since y0 < yc, the channel upstream of A is steep. Compute the depth conjugate to y0:
q2
1.67 2
2
 4.43
Fr0  3 
gy0 9.81 0.43
10.53 (a) q 
ycj 
y0
2


1  8Fr02  1 
0.4
2


1  8  4.43  1  1.01 m
The depth at the outfall is 1.6  y c  2.26 m . There is an H2 profile upstream of the
weir up to location A; upstream of A an S1 profile exists. Between location A and the
outfall the water depth is always greater than 2.26 m, therefore the jump will be
upstream of A, where the depth on the S1 curve is equal to ycj.
(b)
288
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Chapter 10 / Flow in Open Channels
10.54 (a)
yc  3
q 2 3 5.75 2

 1.50 m . Since y01 < yc and y02 > yc, there will be a hydraulic
9.81
g
jump between A and C.
(b) Either an M3 curve exists downstream of B, with the depth equal to y01 upstream be,
or an S1 curve exists upstream of B, with the depth equal to y02 downstream of B.
(c) Compute the depth conjugate to y01:
5.752
1.0
Fr 
 3.37 ; y cj 
3
9..81  1.0
2
2
1


. m
1  8  3.37  1  214
Since ycj > y02, the hydraulic jump must be downstream of B. Hence, an M1 curve
exists downstream of B, increasing from y01 at B to ycj, followed by a jump to y02.
q 2 3 42

 1.18 m . Since y01 > yc and y02 < yc, an M2 curve is located
9.81
g
10.55 (a) y c  3
upstream of C and an S2 curve is downstream of C. Note that the depth at C is
critical.
(b) Extend the downstream channel slope from C to B, with an abrupt drop of height h
immediately upstream of B. Write the energy equation from B to C to determine the
magnitude of h:
h  y 01 
q2
42
3
3
.
1
6
y



 1.18  0.15 m
c
2
2 
2  9.81  1.6
2
2 gy 01 2
(c) Just as in part (a), an S2 profile exists downstream of location B.
10.56 (a) Compute y c using Fr 2  1:
Q2 B
gA3

Q 2d sin 
2.52  2.5sin 

3
3
9.81 (2.52 /4)(  sin  cos  ) 
g (d 2 /4)(  sin  cos  ) 




sin 
 0.4175
 1,
(  sin  cos  ) 3
Solving,
2.5
d
  1.115 rad, and  y c  (1  cos ) 
(1  cos 1.115)  0.70 m.
2
2
Compute y 0 using
(d 2 /4)(  sin  cos  ) 
Qn


2/3
c1 S0
( d)
5/3
,
2


2.5  0.015 (2.5 /4)(  sin cos )

1 0.001
(  2.5)2/3
5/3
(  sin  cos  ) 5 /3
.
which reduces to 1.039 
 2 /3
289
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Chapter 10 / Flow in Open Channels
d
2.5
Solving,   1.361 rad, and  y 0  (1  cos ) 
(1  cos 1.361)  0.99 m.
2
2
Since y 0  y c , a mild slope condition exists. The water surface consists of an M 3
curve beginning at the inlet, followed by a hydraulic jump to an M 2 curve, which
terminates at critical depth at the outlet. The numerically-predicted water surface and
energy grade line are provided in the following table.
x
(m)
0
7
14
20*
20*
230
350
(b)
y
(m)
0.40
0.43
0.46
0.49*
0.97*
0.96
0.93
E
(m)
1.64
1.44
1.29
1.18*
1.07*
1.065
1.05
x
y
(m)
(m)
406
0.90
442
0.87
465
0.84
480
0.82
490
0.79
500
0.70
*
hydraulic jump
E
(m)
1.03
1.01
1.00
0.98
0.97
0.95
x
y
(ft)
(ft)
1285
2.99
1405
2.89
1482
2.795
1566
2.60
1586
2.505
1600
2.31
*
hydraulic jump
E
(ft)
3.40
3.34
3.29
3.20
3.17
3.14
y c  2.31 ft , y 0  3.28 ft ,
x
(ft)
0
23
46
69*
69*
697
1087
y
(ft)
1.30
1.40
1.50
1.61*
3.21*
3.18
3.085
E
(ft)
5.57
4.84
4.32
3.92*
3.55*
3.53
3.46
10.57 A spreadsheet solution is shown below. An M2 profile is located upstream of location A.
Q=
b=
17.5
3
yc =
y0 =
Station
1
2
3
4
5
6
7
y
1.188
1.300
1.400
1.500
1.600
1.700
1.750
n=
m1 =
m2 =
0.012
1.8
1.8
S0 = 0.0003
yu = 0.00
L=
yd =
1000 g = 9.81
1.19 c1 = 1.00
E
1.607
1.624
1.661
1.714
1.776
1.847
1.885
ym
S(ym)
x
1.244
1.350
1.450
1.550
1.650
1.725
1.392E-03
1.009E-03
7.594E-04
5.810E-04
4.511E-04
3.763E-04
-16
-53
-114
-224
-468
-494
Depth Residual
1.188
0.00
1.823 -5.1E-05
A
6.104
6.942
7.728
8.550
9.408
10.302
10.763
V
2.867
2.521
2.264
2.047
1.860
1.699
1.626
290
x
1000
984
932
818
594
126
-368
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Chapter 10 / Flow in Open Channels
The profile downstream of location A is calculated next. Note that along the S2 curve,
normal flow conditions are reached approximately 210 m from the upstream end. An S2
curve exists downstream of the jump, which is located approximately 230 m downstream
from A.
Q = 17.5
b= 3
n=
m1 =
m2 =
0.012
1.8
1.8
S0 = 0.005
yu = 1.19
L = 500
yd = 3.00
g=
c1 =
9.81
1.00
Depth Residual
yc = 1.188
0.00
y0 = 0.892 -0.00061
Station
1
2
3
4
5
6
y
1.188
1.150
1.100
1.050
0.950
0.892
A
6.104
5.831
5.478
5.135
4.475
4.108
V
2.867
3.001
3.195
3.408
3.911
4.260
E
1.607
1.609
1.620
1.642
1.730
1.817
ym
S(ym)
1.169
1.125
1.075
1.000
0.921
1.775E-03
2.059E-03
2.454E-03
3.237E-03
4.423E-03
x
0
1
1
4
4
9
13
50 63
151 214
7
8
9
10
11
12
13
14
15
3.000
2.750
2.500
2.250
2.000
1.800
1.600
1.400
1.200
25.200
21.863
18.750
15.863
13.200
11.232
9.408
7.728
6.192
0.694
0.800
0.933
1.103
1.326
1.558
1.860
2.264
2.826
3.025
2.783
2.544
2.312
2.090
1.924
1.776
1.661
1.607
2.875
2.625
2.375
2.125
1.900
1.700
1.500
1.300
4.325E-05
6.426E-05
9.883E-05
1.584E-04
2.528E-04
3.994E-04
6.630E-04
1.171E-03
-49
-48
-47
-46
-35
-32
-27
-14
x
ycj
1.188
1.226
1.279
1.335
1.456
1.456
500
451
403
356
310
275
243
216
202
62
30
2
3
y
=
=1.54 m
,
 6 m /s
c
9.81
5
62
E1  2 
 2.459 m
2  9.81 22
10.58 (a) q 
E2  1.8 
62
2  9.811.82
 2.366 m
ym  1.9 m, Am  5 1.9  9.5 m 2 , Pm  5  2 1.9  8.8 m, Rm  9.5 / 8.8  1.08 m
2
Q 2 n2
1
 30  0.012 
S( ym )  2 4/3  
 0.0036

4/3
Am Rm
9.5

 1.08
E2  E1
2..366  2.459
x 

 20.2 m
S0  S( ym )1 0.001  0.0036
(b) An A2 profile is contained within the reach (which has an adverse slope with y2 < y1).
291
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Chapter 10 / Flow in Open Channels
10.59 (a) Use Eq. 10.4.27 to compute Q:
Q  0.58 
8
2  9.81 tan 60  0.375/2  0.198 m3 /s
15
(b) Compute q, yc, and y0:
q
0.198
0.1982
 1.98 m2 /s, yc  3
 0.16 m,
1.0
9.81
y05/3
Q n 0.198  0.017

 0.238 
, and by trial and errory0  0.57 m
S0
0.0002
(1  2y0 )2/3
Since y0 > yc, the slope is mild. The depth upstream of the weir is 0.5 + Y = 0.87 m.
Hence, an M1 curve is situated upstream of the weir.
10.60
Evaluate Q using x 
E2  E1
.
S0  S( ym )
E1  y 1 
Q2
Q2
Q2

1
.
05


1
.
05

,
135.2
2gA12
2  9.81  ( 2.5  1.05) 2
E2  y 2 
Q2
Q2
Q2

1
.
2


1
.
2

,
176.6
2gA22
2  9.81  ( 2.5  1.2) 2
1
1
y m  ( y 1  y 2 )  (1.05  1.2)  1.125 m,
2
2
2
 Qn  (b  2y m ) 4 /3
S( y m )  

 c 1  (by m ) 10 /3
Q 2  0.013 2  ( 2.5  2  1.125) 4 /3

1 2  ( 2.5  1.125) 10 /3
 4.296  10 5 Q 2 .
 xS0  S( y m )  E 2  E 1 ,
50( 0.005  4.296  10 5 Q 2 )  1.2 
Q2
Q2
 1.05 
,
176.6
135.2
which reduces to 4.141  10 4 Q 2  0.10. Solving, Q  15.5 m3 /s.
3
Q
yc   
b
2
15.5 
g 

 2.5 
3
2
9.81  1.58 m,
The profile is an S3 curve.
292
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Chapter 10 / Flow in Open Channels
10.61 (a) Consider location 1, immediately upstream of transition:
q1  Q / b1  5.5 / 3  1.83 m 2 /s,
3
yc1  3 q12 /g  1.832 /9.81  0.70 m.
 y c 1  y 0 , and steep slope condition prevails. Compute specific
energy at normal depth:
q 12
1.83 2
 0.5 
 1.18 m.
E0  y 0 
2 gy 02
2  9.81  0.5 2
At location 2, within the constriction:
q2  Q / b2  5.5 /1.5  3.67 m2 /s,
Ec 2 
3
yc2  3 q22 /g  3.672 /9.81  1.11 m,
3
3
y c2   1.11  1.67 m.
2
2
Since E c2  E 0 , uniform flow cannot exist at loc. 1, hence choking occurs at
loc. 2. Compute y 1 by setting E 1  E c2 :
y1 
q 12
1.83 2
0.71


 y 1  2  1.67 ,  y1  1.60 m.
y
1
2
2
2gy 1
2  9.81y 1
y1
Since y 1 is subcritical, there will be a hydraulic jump some distance upstream of the
transition, followed b an S 1 curve. The depth behind the jump is that depth conjugate
to y 0 :
Fr02  q12 /gy03  1.832 /(9.81 0.53 )  2.73
y
0.5
 ycj  0 1  8Fr02  1 
1  8  2.73  1  0.94 m.
2
2



293

© 2012 Cengage learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 10 / Flow in Open Channels
(b) See part (a) for development and figures.
q1  Q / b1  200 /10  20 ft 2 /sec,
3
yc1  3 q12 /g  202 /32.2  3.67 ft,
E0  y 0 
q 12
20 2
1
65

.

 3.93 ft.
2gy 02
2  32.2  1.65 2
q2  Q /b2  200/5  40 ft 2 /sec,
3
yc2  3 q22 /g  402 /32.2  3.67 ft,
3
3
y c2   3.67  5.505 ft.
2
2
2
q
202
6.211
y1  1 2  y1 
 y1  2  5.505.
2
y1
2 gy1
2  32.2 y1
Ec2 
 Solving, y 1  5.28 ft.
Fr02  q12 /gy03  202 /(32.2 1.653 )  2.76,
y
1.65
1  8  2.76  1  314
. ft.
 y cj  0 1  8Fr02  1 
2
2




10.62 Write energy eqn. across the gate to compute the discharge:
E 1  E 2 , or y 1 
q2
q2
y


.
2
2gy 12
2gy 22
 1
1
 q  2g( y 1  y 2 ) 2  2   1
 y2 y1 
1 
 1
  1  193
 2  9.81  (185
.  0.35)
. m 2 /s ,
2 
 0.35 185
. 2
Q  bq  4  1.93  7.72 m 3 / s.
y c  3 q 2 / g  1.93 2 / 9.81  0.72 m.
Compute y 0 using Chezy-Manning relation:
(4 y0 )5/3
7.72  0.014
Qn
.

 3.821 
(4  2 y0 ) 2/3
c1 S0 1 0.0008
 Solving, y 0  1.17 m. Since y 0  y c , mild channel conditions exist.
294
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Chapter 10 / Flow in Open Channels
Upstream of the gate there will be an M1 profile, with an M 3 downstream of the gate,
terminating in a hydraulic jump to normal flow conditions.
Compute the depth upstream of the jump, conjugate to y 0 :
Fr02  q 2 /gy03  1.932 /(9.811.173 )  0.237.
y0
2
 y cj 


1  8Fr02  1 
117
.
2


1  8  0.237  1  0.41 m.
Use the step method to compute water surface and energy grade line:
Ey
S( y ) 
q2
1.93 2
0.190

y

y
2
2
2gy
2  9.81y
y2
Q 2 n 2 ( 4  2y ) 4 /3 7.72 2  0.014 2 ( 4  2y ) 4 /3

( 4 y ) 10 /3
4 10 /3 y 10 /3
 1.15  10 4 (4  2y) 4 /3 y 10/3
M1 curve upstream of gate
y
E
(m)
1.85
(m)
1.906
1.7
1.766
1.5
1.584
1.3
1.412
1.2
1.332
ym
S( y m )
x
x
(m)
(m)
0
(m)
1.775
2.514104
255
1.6
3.336104
390
1.4
4.82410
4
542
1.25
6.627104
583
255
645
1187
1770
M 3 curve downstream of gate
y
E
ym
(m)
0.35
(m)
1.901
(m)
0.37
1.758
0.39
1.639
0.41
1.540
S( y m )
x
x
(m)
(m)
0
0.36
0.02741
5
0.38
0.02315
5
0.40
0.01973
5
5
10
15
Hydraulic jump occurs ~5 m from the gate.
295
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Chapter 10 / Flow in Open Channels
10.63
10.64
10.65 (a) First find the width of the channel using the Manning equation.
By trial, b = 20 m.
q
125
6.252
 6.25 m 2 /s , yc  3
 1.58 m .
20
9.81
Since yc < y0, an M1 profile exists upstream of the dam.
6.252
6.252
12.014
m,
11

E


 11.016 m,
A
2  9.81122
2  9.81112
0.0252  1252
n2Q 2
 1.973 105.
S( ym )  2 4/3 
2
4/3
230  5.35
Am Rm
(b) EB  12 
x 
12.014  11.016
0.0004  1.973 105
 2624 m
(c)
296
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Chapter 10 / Flow in Open Channels
10.66 Compute normal depth using Chezy-Manning eqn:
(3.66 y02 )5/3
Qn
15.38  0.017

 5.476 
(3.66  2 y02 )2/3
c1 S0 1 0.00228
 Solving, y 0 2  1.65 m.
Compute critical depth:
q  Q/b  15.38/3.66  4.20 m2 /s,
3
 yc  3 q 2 /g  4.202 /9.81  1.22 m.
Upstream of slope change the channel is steep, and downstream, the channel is
mild. The hydraulic jump will terminate at normal depth.
(a) Find depth conjugate to y 0 2 :
Fr022  q 2 /gy023  4.202 /(9.811.653 )  0.400,
y0
1.65
 y cj  2 1  8Fr0 22  1 
1  8  0.4  1  0.87 m.
2
2




(b) Step method:
Ey
S( y ) 
q2
4.20 2
0.900
,

y

y
2
2
2gy
2  9.81y
y2
Q 2 n 2 (b  2y ) 4 /3 15.38 2  0.017 2 ( 3.66  2y ) 4 /3

(by ) 10 /3
( 3.66) 10 /3 y 10 /3
 9.052  10 4 (3.66  2y) 4 /3 y 10/3 .
y
E
ym
(m)
0.6
(m)
3.100
(m)
0.7
2.537
0.8
2.206
0.87
2.059
S( y m )
x
x
(m)
(m)
0
0.65
0.03216
18.8
0.75
0.02104
17.6
0.835
0.01536
11.2
18.8
36.4
47.6  L j
(c) An M 3 curve exists downstream of the change in channel slope, terminating in a
hydraulic jump, approximately 50 m from the slope change.
297
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Chapter 10 / Flow in Open Channels
10.67
10.68
This is a design analysis problem. Some extensive calculations are required.
(a)
Find the depth of flow immediately upstream of the jump (call it location 1):
V2 
19
19
Q
=0.95 m 2 /s,

 1.267 m/s, q 
20
A2 0.75  20
Fr2 
1.267
0.75
 0.467, y1 
2
9.81 0.75
 1  8  0.467 1  0.246 m
2
Compute the loss across the jump, and subsequently the dissipated power:
0.95 2
0.952


 3.87 m
0
75
.
2  9.81  0.10 2
2  9.81  0.752
   Qh  9810  19  3.87  7.21  10 5 watt, or 721 kW
W
j
h j  E1  E2  0.10 
(b) The length of the apron is the distance from the toe to downstream of the jump.
First, compute the distance from ytoe to y1 using a single increment of length:
0.952
0.952
4.700
m,
0.246
Etoe  0.10 

E


 1.006 m,
1
2  9.81 0.102
2  9.81 0.2462
1
ym  (0.10  0.246)  0.173 m, Am  20  0.173  3.46 m 2 ,
2
 19  0.014 
Pm  20  2  0.173  20.35 m, Sm  

 3.46 
 x1  xtoe 
2
1
 3.46 / 20.354/3
 0.06275,
E1  Etoe
1.006  4.700

 59.3 m
S0  Sm 0.0005  0.06275
298
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Chapter 10 / Flow in Open Channels
The length of the jump is six times the downstream depth, or 6  0.75 = 4.5 m. Hence the
required length of the apron is L = 59.3 + 4.5 = 63.8 m, or approximately 65 m. Normal
design would require additional length as a safety factor.
(c) The normal depth is computed by trial and error to be y0  0.755 m. The critical depth
is yc  3 0.95 2/9.81 0.452 m . Since y0 > yc, the channel exhibits mild slope conditions.
The depth ahead of the dam is h + yc = 6 + 0.45 = 6.45 m, and is greater than y0. Hence an
M1 profile is upstream of the dam. On the apron, between the toe and the hydraulic jump,
an M3 curve exists.
 q 2n2 
10.69 (a) y0  

 S0 
3/10
 1.22  0.0172 


 0.0005 
1/3
3/10
 0.946 m,
1/3
 q2 
 1.22 
yc     
  0.528 m
 9.81 
 g
Since y0 > yc, the channel possesses a mild slope. With yA = 0.65 m and yB = 0.90 m,
we see that yc < yA < y0, and yc < yB < y0, so that an M2 curve exists. Because an M2
profile decreases with increasing x, location B is upstream of location A.
(b) From Ex. 10.16 we find, for a wide rectangular channel, J = 2.5, M = 3, and
N = 3.33. Using Eq. 10.7.13,
3

0.946 
 0.528  2.5
x
u
F
u
F (v, 2.5)
(
,
3.33)





0.005 
 0.946  3.33

 1892u  F (u , 3.33)  0.1305F (v , 2.5) ,
N 3.33

 1.33
J
2.5
Compute xA:
uA 
0.65
 0.687,
0.946
vA  0.6871.33  0.607,
and from Table 10.4, F (0.687,3.33)  0.742,
F (0.607, 2.5)  0.668,
so that xA  1892(0.687  0.742  0.1305  0.668)  61 m
Compute xB:
0.90
 0.951,
vB  0.9511.33  0.935,
0.946
F (0.951,3.33)  1.615,
F (0.935, 2.5)  1.488.
uB 
 xB  1892(0.951 1.615 0.1305 1.488)  889 m
Therefore the distance between A and B is xA  xB = 61  (889) = 950 m.
299
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Chapter 10 / Flow in Open Channels
10.70 Use the varied flow function to compute the water surface profile.
 q 2 n2 
y0  

 S0 
3/10
 1.5 2  0.015 2 

  1.627 m,
 0.0001 
1/3
 q2 
 1.5 2 
yc     
  0.612 m
 9.81 
 g
Therefore the estuary has a mild slope and an M1 curve exists upstream of the outlet. For a
wide rectangular channel (Ex. 10.16), J = 2.5, N = 3.33, M = 3, and N/J = 1.33. Use Eq.
10.7.13 to compute x:
3

1.627 
 0.612  2.5
(
,
2.5)
x
F
v
u  F (u,3.33)  


0.0001 
 1.627  3.33

 16 270 u  F (u,3.33)  0.04 F (v, 2.5) 
The results of the calculations are as follows:
y (m)
7
6
5
4
2
10.71
u
4.30
3.67
3.07
2.46
1.23
F(u,3.33)
0.015
0.021
0.032
0.052
0.348
v
6.96
5.67
4.45
3.31
1.32
F(v,2.5)
0.038
0.053
0.074
0.117
0.572
x (m)
69 800
59 700
49 500
39 500
14 700
x – 69 800 (m)
0
10 100
20 300
30 300
55 100
Let subscript 1 denote the side channel, and subscript 2 the main channel.
(a) A1  1150  150 m2 , P1  1 150  151 m ,
A1R12/3 150  0.9932/3
150
 0.993 m , K1 

 4977 ,
R1 
n1
151
0.03
A2  5  5  25 m 2 , P2  4  5  5  14 m ,
A2 R22/3 25 1.7862/3
25
 1.786 m , K2 

 1840
R2 
n2
14
0.02
Compute  using Eq. 10.7.9:

( A1  A 2 ) 2  K 13 K 23 
(150  25 ) 2  4977 3 1840 3 





  1.493  1.49

25 2 
(K 1  K 2 ) 3  A12 A 23  ( 4977  1840 ) 3  150 2
(b) Q  (K1  K2 ) S  (4977  1840) 0.0005 =152 m3 /s
300
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Chapter 10 / Flow in Open Channels
10.72 This is a design analysis problem, and requires extensive calculations that would best be
performed on a spreadsheet. Let location A be 400 m upstream of location B. Conditions
at B are known. The energy equation between A and B is predicted using total energy
H  y  z   V 2 / 2g :
L
H*A  HB  hL  HB  (SA  SB )
2
In the equation, HB and SB are known, and HA and SA are unknown. A trial and error
solution is required. With the table of given data, location A is associated with x = 0, and
location B with x = 400 m. The following parameters are computed at location B, where
yB = 3.0 m:
Location
Side channel
Main channel
AB (m2)
60
273
PB (m)
149
97
RB (m)
0.40
2.83
KB (Eq. 10.7.8)
11
18 210
Now B, VB, HB, and SB can be computed:
B 
(60  273) 2  113 182103 


  1.48
(11  18210)3  602
2732 
VB 
Q
280

 0.84 m/s
AB 60  273
SB 
2802
 0.000236
(11  18210)3
(Eq. 10.7.9)
(Eq. 10.7.7)
0.842
 18.153 m
2  9.81
With L = 400 m, the energy equation between A and B is predicted:
H B  3.0  15.1  1.48 
400
(SA  0.000236)  18.20  200SA
2
The total energy at location A is
H*A  18.153 
H A  yA  zA   A
V A2
V2
 y A  15.0   A A
2g
2g
The trial and error solution proceeds by assuming values of yA, computing the
corresponding A, VA, SA, HA, and H *A until the latter two are in close agreement. For
yA = 3.2 m, the hydraulic parameters at location A are provided in this table:
Location
Side channel
Main channel
AA (m2)
172
342
PA (m)
116
112
301
RA (m)
1.48
3.05
KA (Eq. 10.7.8)
4470
24 000
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Chapter 10 / Flow in Open Channels
The corresponding values of HA and H A* are
A 
 44703 240003 


  1.39
(4470  24000)3  1722
3422 
VA 
280
Q

 0.54 m/s
AA 172  342
SA 
(172  342)2
2802
(4470  24000)
3
 9.67 105
H A  3.2  15.0  1.39 
(Eq. 10.7.9)
(Eq. 10.7.7)
0.542
 18.22 m
2  9.81
H*A  18.200  200  0.0000967  18.22 m
Hence the depth at the upstream location is yA = 3.2 m.
302
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Chapter 11 / Flows in Piping Systems
CHAPTER 11
Flows in Piping Systems
11.1
V12
0.12
Q2


 0.52 m,
2 g 2 gA12 2  9.81(  0.22 /4) 2
V22
Q2
0.12


 0.21 m.
2 g 2 gA22 2  9.81(  0.252 /4) 2
Location
EGL
(m)
HGL
(m)
Loss
(m)
A
10.00
10.00
entrance
9.74
9.22
 H P  45.2 ( 1)
d/s pump
54.94
54.42
Kv V12 /2 g  2  0.52  1.04
d/s valve
53.90
53.38
R1Q 2  53.4  0.1 2  0.53
u/s bend 1
53.37
52.85
d/s bend 1
53.32
53.11
u/s bend 2
53.14
52.93
d/s bend 2
53.09
52.88
u/s exit
B
44.23
44.02
44.02
44.02
Ke V22 /2 g  0.25 0.21  0.05(2)
10
10
R2 
 Q 2  904 
 0.1 2  0.18 ( 3)
500
500
KeV22 / 2g  0.05
490
490
R2 
 Q 2  904 
 0.1 2  8.86
500
500
2
K2V2 / 2g  1  0.21  0.21
K1V12 / 2g  0.5  0.52  0.26
(1)
The pump adds energy, hence the minus sign.
Note change in kinetic energy term.
(3)
Note the 10 m length between elbows.
(2)
According to Ex. 11.2, the EGL an HGL at loc. B should be:

p
200  10 3
 20  44.0 m.


z


 B 9810  0.85

Some round-off error is present.
303
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Chapter 11 / Flows in Piping Systems

W
Q
f
11.2 Substitute V 
and HP 
into the energy equation:
Q
A

W
p1
p2
Q2
Q2
f




 hL
 2 gA12  Q  2 gA22
(a) Note that the kinetic energy terms can be neglected:
A1 

 0.052  0.00196 m2 , A2 

4
0.095
Q
 0.00158 m3 /s
60
4
 0.082  0.00503 m2 ,
Wf
350 103
0.001582
760 103
0.001582




 6.6
9810
9810
2  9.81 0.001962 9810  0.00158
2  9.81 0.005032
35.7  0.03  0.0645Wf  77.5 0.005 6.6.
 W f  750 watt
(b) Neglect the kinetic energy terms:
Q
25
 0.0557 ft 3 /sec
7.48  60
Wf
50 114
110 144
0

 0  20
62.4
62.4  0.0557
62.4
115.4  0.288W f  253.8  20
 W f  550 ft-lb/sec, or 1 hp
304
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Chapter 11 / Flows in Piping Systems
11.3
(a) Use a numerical solution. The system demand curve is
2
 Q
 L
H P  z   f   K 
 2 gA 2
 D
Q2
2440


 32   0.02
 12.5
 2  9.81  (0.7854  0.20 2 ) 2

0.20
 32  13250Q 2
The pump data can be approximated by the quadratic equation (for example, using a
least squares fit):
H P  54.5  38.6Q  2330Q 2
Eliminating HP from the two relations and solving yields
Q  0.039 m3/s and HP  52.3 m.
(b) A numerical approximation to the efficiency curve is
  29.4Q  349.2Q 2  1045Q 3
 29.4  0.039  349.2  0.039 2  1045  0.039 3
 0.68
Hence the required pump power is
   QH P  0.82  9810  0.039  52.3  2.42  10 4 watt, or 24.2 kW
W
P

0.68
11.4
Write energy eqn. between locs. 1 and 2:
H P  z1  z2 
Solving for D 1 ,
f
(L  Le )
D
Q2


2 g D 2 

4
2
.


Q 2 ( L  Le ) f
D 

2
 2 g(  / 4 ) ( H P  ( z2  z1 )) 
1/5
.
Substitute the relations
e
 

f  1325
.  ln 0.27  5.74 Re  0.9  

D
 
2
and Re 
4Q
kD
, Le 
,
f
 D
and solve for D by successive substitution.
305
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Chapter 11 / Flows in Piping Systems
(a) Compute H P from pump data using linear interpolation:
H P  453  ( 453  423)
Q  15 ,000 gpm 
(12 ,000  15 ,000)
 430.5 ft.
(12 ,000  16,000)
1 ft 3
1 min

 33.4 ft 3 / sec,
7.48 gal 60 sec
1/5


33.42 (1500  Le ) f
D  

2
 2  32.2( / 4) (430.5  120) 
D (ft)
e/D
1
1.30
1.29
1.29
0.003
0.0023
0.0023
Re 
 0.6184[(1500  Le ) f ]1/5.
3.016  10 6
D
f
3.02  10 6
2.32  10 6
2.34  10 6
Le 
0.026
0.024
0.024
2.50
(ft)
f
96
135
134
 D  1.29 ft.
(b) Compute H P from pump data using linear interpolation:
HP  154  (154  148)
(0.8  11
.)
 149.5 m.
(0.8  1.2)
1/5


1.12 (500  Le ) f
D  

2
 2  9.81( / 4) (149.5  40) 
D (m)
0.3
0.42
0.42
e/D
0.0033
0.0024
Re 
 0.2467[(500  Le ) f ]1/5 .
1.401  10 6
D
4.67  10 6
3.34  10 6
f
0.027
0.025
Le 
2.50
(m)
f
28
42
 D  420 mm.
306
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Chapter 11 / Flows in Piping Systems
2
 L
 Q
11.5 (a) z A  H P  z B   f  K 
, (see the following figure)
 D
 2 gA 2
Q2

2 gA 2
0.4 2
 0.2115 m,
2

2
2  9.81    0.5 

4
4Q
4  0.4
Re 

 2.39  10 6 ,
7
D   0.5  4.26  10
e
 

f  1.325 ln 0.27  5.74 Re 0.9  

D
 
2
0 . 9
 

1
6 
 1.325  ln  0.27 
 5.74(2.39  10  
 
500
 
2
 L
 Q
 H P  z B  z A   f  K 
 D
 2 gA 2
2
 0.0235.

 0.0235  5000
 220  100  
 2  2  0.2115  170.5 m;


0.5
  QH  0.81  9810  0.4  170.5  542 000 W, or W
  542 kW.
W
f
P
f
(b) Find the elevation at C when the pressure at that location is equal to the vapor
pressure (see the figure on the next page):
pc  pv  patm  55.2  100  44.8 kPa,
2
pc
Q2
 L
 Q
z c  z1  H P   f AC  K 


 D
 2 gA 2 
2 gA 2
44.8  103
 0.0235  4000 
 100  170.5  
 2  0.2115 
 0.2115


0.5
0.81  9810
 235.7 m
(c)
Location
HGL (m)
A
d/s pump
100.0
270.5
d/s valve
270.1
C
230.3
u/s valve
B
220.4
220.0
Loss (m)
170.5
2  0.2115 = 0.4
0.0235  4000  0.2115
 39.8
0.5
0.0235  1000  0.2115
 9.94
0.5
2  0.2115 = 0.4
307
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Chapter 11 / Flows in Piping Systems
11.6
(a) Energy eqn. from A to B:
 K1
p

p

K 2  2
  z    z  (R1  R 2  R3 )Q1.85  
2 
2 Q .

A 
B
 2 gA1 2 gA2 
Use Hazen-Williams resistance formula for R:
R
10.59L
.
C 1.85 D 4 .87
 R1 
K1
2
1
2 gA
K 2
2
2
2 gA
10.59  200
 1071,
100 1.85  0.2 4 .87

2
 103.3,
2  9.81  ( / 4) 2  (0.2) 4
R2 
10.59  150
 193.4 ,
120 1.85  0.25 4 .87

3
 63.5,
2  9.81  ( / 4) 2  (0.3) 4
R3 
10.59  300
 271.1.
90 1.85  0.3 4.87
Substitute known data into energy eqn:
250  107  (1071  193  271)Q 1.85  (103.3  63.5)Q 2 ,
143  1535Q 1.85  167Q 2 ,
F(Q)  1535Q 1.85  167Q 2  143,
Iter.
1
2
3
*
Q(m3 /s)
F
F
0.277*
0.265
0.265
12.60
0.284
1046
1007
1/1.85
 143 
Q

 1535 
F (Q)  2840Q 0.85  334Q
Q   F / F 
0.01205
0.00028
Q  0.265 m3 /s
(1st iteration)
308
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Chapter 11 / Flows in Piping Systems
(b) See part (a) for development.
4.72  600
R1 
100
K 1
1. 85
 8
 
 12 
4 . 87
 4.071,
2
 0.255,
2  32.2  ( / 4) 2  (8 /12) 4
4.72  300
R2 
 0.490 ,
4 . 87
 10 
 
120
 12 
K 2
3
4.72  900

 0157
. , R3  1.85
 1.030.
2
2
4
2 gA2 2  32.2  ( / 4)  (10 /12)
 (1) 4 .87
90
2
1
2 gA

1. 85
Substitute known data into energy eqn:
820  351  ( 4.971  0.49  1.03)Q 1.85  (0.255  0.157 )Q 2 ,
469  5.591Q 1.85  0.412Q 2 ,
F(Q)  5.591Q 1.85  0.412Q 2  469 , F (Q)  10.343Q 0.85  0.824 Q
Iter.
Q(ft 3 /sec)
F
F
Q   F /F 
1
2
3
4
10.96*
10.40
10.39
10.39
49.45
1.161
0.319
88.19
84.27
84.20
0.561
0.038
0.0038
Q  10.39 ft 3 /sec.
*
11.7
 469 
Q

 5.591 
1/1. 85
(1st iteration)
(a) Let H A  piezometric head at pipe inlet, and H B  piezometric head at
pipe outlet. Write energy eqn. across pipeline:



W
W
W
f1
f2
f3
HA 


 ( R1  R2  R 3 )Q 2  H B ,
Q
Q
Q
3
3
1
H A  HB 
 W f1  Q 2  R i  0.
i 1
Q i1
 /(R )]1/3 .
Q  [ W
But H  H ,  solving for Q,
A
fi
B
i
3
  200  103  200  103  250  103  650  103 ,
(b)  W
f1
i 1
3
 Ri  4  104  3  104  2  105  2.7  105 ,
i 1
Q  [650 103 /(8330  2.7 105 )]1/3  0.0663 m3 /s.
309
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Chapter 11 / Flows in Piping Systems
11.8 Write the energy equation from the liquid surface (location 1) to the valve exit (location 2):
p1

 z1 

W
f
L  Q2

  K  f  1
 z2
D  2 gA 2
Q 
(a) Rearrange and substitute in known data:
Wf 1 
fL  Q 2
P1
 z1  z2 
  K   1
0
D  2 gA2

 Q 
110 103
1104 1
 24  18 
0.68  9810
0.68  9810 Q
(0.5  3  0.26  2  0.015  450 / 0.30  1) 2
Q 0

2  9.81 (0.7854  0.302 ) 2
1.50
22.4 
 273.2Q 2  0.
By trial and error, Q  0.32 m3/s.
Q
b) Substitute in known data:
15 144
15  550 1 (0.5  3  0.26  2  0.015 1500 / 0.667  1) 2
Q 0
 75  60 

0.68  62.4
0.68  62.4 Q
2  32.2  (0.7854  0.667 2 ) 2
1
65.9  194.4  4.8Q 2  0.
By trial and error, Q  4.7 ft3/sec.
Q
11.9
Write energy eqn. between reservoirs:

2
10  2
H P   R1 
 R2  R3 
 Q  50;
2
2gA1
2gA32 

HP 
  1920  10 3  0.82 160.7
W
P
,


Q
Q
9800 Q
2
2

 0.03264 ,
2
2gA1 2  9.81  (  / 4 ) 2  1.5 4
10
10

 0.3985.
2
2gA3 2  9.81  (  / 4 ) 2  1.2 4
Compute the resistance coefficients assuming high Re-number flows:
L
R1  1.07 1 5
gD1
 
e1  
 ln 0.27  
D1  
 
2
200
 1.07
. 5
9.81  15
1 
 
 ln 0.27  1500  


2
 0.03864,
2
300  
1 
R2  1.07

 0.4846,
5  ln 0.27 
9.81  1  
1000  
2
120  
1 
ln 0.27 
R3  1.07

 0.07455.
9.81  1.25  
1200  
310
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Chapter 11 / Flows in Piping Systems
Substitute the data into the energy eqn:
160.7
 (0.03864  0.03264  0.4846  0.07455  0.3985)Q 2  50 ,
Q
160
 1.029Q 2  50
Q
Use Newton’s method: F (Q )  1.029Q 3  50Q  160.7 , F (Q )  3.087Q 2  50
Q(m3 /s)
2.0
2.842
2.775
2.774
Iter.
1
2
3
4
F
62.35
74.93
73.77
F
52.47
5.020
0.0389
Q   F / F 
+0.842
0.067
0.00053
Q  2.77 m3 /s.
Recalculating the resistance coefficients based on the current value of Q yields
R 1  0.03916, R 2  0.4878, R 3  0.07522. The new estimate of Q remains
2.77 m3 /s.
11.10 Initially work between locs. B and C:
Le 
8 f ( L  Le )
Dk
, R
f
g 2 D5
Le1 
1.2  2
8  0.015  260
 160, R1 
 0.1295,
0.015
9.81  2 1.25
Le 2 
1 3
8  0.02 1150
 150, R2 
 1.900,
0.02
9.81  2 15
Le 3 
0.5  2
8  0.018 1556
 56, R3 
 74.05,
0.018
9.81  2  0.55
0.75  4
 143,
0.021
Q2
W 
2 
 4 1 


 i 2 R 

i 
Le 4 
8  0.021 943
 6.895.
9.81  2  0.755
32
2  6.022 m,
1
1 
 1




 19
74.05
6.895 
.
R4 
Q2  W / R2  6.022 /1.9  1.780 m3 /s,
Q3  W / R3  6.022 / 74.05  0.285 m3 /s,
Q4  W / R4  6.022 / 6.895  0.935 m3 /s.
311
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Chapter 11 / Flows in Piping Systems
Check continuity: Q2  Q3  Q4  1.78  0.285  0.935  3  Q1,  OK.
To find W P , write energy equation from loc. A to B:

p
H P  R 1Q 12    z  R 1Q 12  W  20,
B

 H P  0.1295  3 2  6.022  20  27.2 m.
  Q1H P  9810  3  27.2  1.07  10 6 W, or W
  1.07 MW.
W
P
P
0.75

11.11 (a) Compute resistance coefficients:
Le1  D1K / f1  0.24  2 / 0.02  24,
R1 
8 f1 ( L1  Le1 )
8  0.02  54

 112.1,
2 5
g D1
9.81  2  0.245
R2 
8 f 2 L2
8  0.015  60

 232.4 ,
2
5
g D2 9.81   2  0.2 5
R3 
8 f 3 L3
8  0.025  90

 1773.
2
5
g D3 9.81   2  0.16 5
Let H = hydraulic grade line at the junction. Write energy equations for each branch:
Line 1:
10  H P  R1Q 12  H , or
H  10  ( 20  30Q 12 )  112.1Q 12  30  142.1Q 12 .
Line 2:
H  R2Q 22  20, or Q 2  ( H  20) / 232.4 .
Line 3:
H  R3Q 32  18, or Q 3  ( H  18) / 1773 .
Continuity at junction: Q  Q1  Q2  Q3  0.
Tabulate solution (assume Q 1 for each iteration, then solve for
H1 Q2 , Q3 and Q):
Q1
3
(m /s)
0.15
0.20
H
(m)
26.80
24.32
Q2
Q3
3
3
(m /s)
0.1711
0.363
(m /s)
0.0705
0.0597
Q
(m3 /s)
0.0916
+0.004
 Q1  0.20 m3 /s, Q2  0.14 m3 /s, Q3  0.06 m3 /s.
312
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Chapter 11 / Flows in Piping Systems
R1 
(b) Le1  10 /12  2 / 0.02  83.3,
R2 
Line 1:
8  0.02  183.3
 0.2296,
32.2   2  ( 10 / 12) 5
8  0.015  200
 0.5735,
32.2   2  ( 8 / 12) 5
8  0.025  300
 6.042.
32.2   2  ( 6 / 12) 5
R3 
H  20  ( 55  0.1Q 12 )  0.2296Q 12
 75  0.33Q 12 ,
Line 2:
Q 2  ( H  50) / 0.5735 ,
Line 3:
Q 3  ( H  45) / 6.042 .
H
Q1
3
(ft /sec)
6
6.5
6.26*
*
(ft)
63.12
61.06
62.07
Q2
Q3
3
3
(ft /sec)
4.783
4.391
4.587
Q
3
(ft /sec)
0.5148
+0.4788
-0.008
(ft /sec)
1.732
1.630
1.681
Estimated by linear interpolation between Q = 6 and Q = 6.5.
 Q1  6.26 ft3 /sec, Q2  4.59 ft 3 /sec, Q3  1.68 ft 3 /sec.
11.12 Use SI data from Pbm 11.11 (a):
Q  Q1  Q2  Q3  0, or
w( H ) 
30  H

142.1
Use false position method of solution: H r 
Iter
1
2
3
H
25
24.57
24.46
Hu
20
20
20
w( H  )
0.02193
0.00562
0.00144
H  20

232.4
H  18
.
1773
H  w( H u )  H u w( H  )
w( H u )  w( H  )
w( H u )
Hr
w( H r )
0.2317
0.2317
0.2317
24.57 0.00562
24.46 0.00144
24.43 0.00030
Sign
w( H  )w( H r )
+
+
+


0.0045
0.0012
Q1  (30  24.43) /142.1  0.198 m3 /s,
Q2  (24.43  20) / 232.4  0.138 m3 /s,
Q3  (24.43  18) / 1773  0.060 m3 /s.
313
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Chapter 11 / Flows in Piping Systems
2
e 
 
11.13 To compute a constant friction factor, use f  1.325ln  0.27   . Le  kD / f and
D 
 
R  8 f ( L  Le ) / ( g 2 D5 ) give the equivalent length and resistance coefficient. Finally the
discharge in each pipe is Q  W / R , in which W  ( p /   z) A  ( p /   z) B .
2
Pipe 1:
0.1  
 
Le  2 1/ 0.012  167,
f  1.325ln 0.27 
 0.012,


1000  

R  8  0.012  767 / ( 9.81   2  1 5 )  0.7605,
 Q1  50 / 0.7605  8.81 m3 /s.
2
Pipe 2:
0.15  
 
f  1.325ln 0.27 
 0.0125, Le  0

1200  
 
R  8  0.0125  1000 / ( 9.81   2  1.2 5 )  0.4151,
Q2  50 / 0.4151  10.98 m3 /s.
2
Pipe 3:
0.2  
 
Le  4  0.85 / 0.0142  239,
f  1.325ln 0.27 
 0.0142,

850  
 
R  8  0.0142  789 / ( 9.81   2  0.85 5 )  2.086,
 Q3  50 / 2.086  4.90 m3 /s.
2
Pipe 4:
0.1  
 
Le  11/ 0.012  83,
f  1.325ln 0.27 
 0.012,

1000  
 
R  8  0.012  883 / ( 9.81   2  1 5 )  0.8755,
 Q4  50 / 0.8755  7.56 m3 /s.
The total discharge through the piping is Q  Q1  Q2  Q3  Q4  32.25 m3 /s.
11.14 Continuity at junction: Q 1  (Q 2  Q 3  Q 4 )  0.
p0
 R1Q 12  H , or Q 1  ( p 0 /   H ) / R1

in which H = piezometric head at junction. For each branch (i= 2, 3, 4):
Energy eqn. across pipe 1:
H  R i Q 12 , or Q i  H / R i .
Substitute into continuity eqn:
 p0 /  H 


R1


314
1/2
4
H
  
i 2 R
 i

1/2
4
 H

i 2
1
Ri
.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 11 / Flows in Piping Systems
Solving for H,
H
p 0

1  R1 


1 

R i 
2

300  10 3 / 9810

1
1  1.6  10 

5
5
3
10

.

4

1
1  10 6



1.8  10 6 
1
2
30.58
 26.46 m.
1.556
Q1  (30.58  26.46) /1.6 104  0.01605 m3 /s,
Q2  26.46 / 5.3 105  0.00707 m3 /s,
Q3  26.46 /1106  0.00514 m3 /s,
Q4  26.46 /1.8 106  0.00383 m3 /s.
fL V 2
flV 2

11.15 W 
, and since W is constant for all pipes, V 
D 2g
D
V1 
D
.
fL
450
150
300
 4.33 , V2 
 3.40 , V3 
 3.48
0.012  2000
0.02  650
0.015 1650
Hence, pipe 1 has the largest velocity.
2
 4 1 
2
2
 Q12 ,
11.16 (a) Write energy eqn. across system: H P  a 0  a1Q1  R1Q1  
i
2


R i 
with Q i  H / R i , i  2, 3, 4, and H  H P  R1Q 12 , in which H = piezometric
head at the junction. The unknowns are Q 1 , Q 2 , Q 3 , Q 4 and H . The system can be
treated as a parallel piping system because the hydraulic grade line at the exit of pipes
2, 3 and 4 are the same.

(b) As opposed to a branching system, a parallel system requires no trial-and-error
solution.
(c) Substitute known data into energy equation, and solve for Q 1 :
2



1   2

 Q1 ,
a 0  a1  R1  

R i  




1

 45  10 4  34 650 + 


 82 500

315
1

127 900
1

115 500 
2
 2
2
Q 1  56 410Q 1 ,

© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 11 / Flows in Piping Systems
Q1  45 / 56 410  0.0282 m3 /s.
 H  45  (10 4  34 650)  0.0282 2  9.49 m ,
Q2  9.49 / 82 500  0.0107 m3 /s,
Q3  9.49 /127 900  0.0086 m3 /s,
Q4  9.49 / 115 500  0.0091 m3 /s.
(d) By increasing the diameter of pipe 1 by approximately 50%,
R1  34 650/1.55  4563, Q1  0.041 m3 /s, and H  20.5 m.
11.17 Let H = piezometric head at the junction. Write energy equation for line 2 to find H:
H  z 2  R2 Q 22  15  2000  0.035 2  17.45 m.
The discharges Q 3 and Q 4 can now be computed directly:
Q3 
H  z3
17.45  12

 0.0603 m3 /s,
R3
1500
Q4 
H  z4
17.45  10

 0.0863 m3 /s.
R4
1000
4
Q1   Qi  0.035  0.0603  0.0863  0.182 m3 /s,
i 2
and H P  H  R1Q 12  z 1  17.45  1400  0.182 2  3  60.82 m.
11.18 (a) For each pipe compute equivalent length and resistance coefficient:
Pipe 1:
Le  DK / f  0.05  3 / 0.02  7.5,
R1 
Pipe 2:
8 f ( L  Le )
8  0.02  37.5

 1.983 105 ,
2 5
2
5
g D
9.81   0.05
Le  0.075  5 / 0.025  15,
R2 
8  0.025  55
 4.788  10 4 ,
2
5
9.81    0.075
8  0.022  62.7
 1.466  10 5 .
2
5
9.81    0.06
2
(0.6 / 60) 2
Q
.

 1125
m.
W
2
[(1983
.
 105 ) 1/2  (4.788  10 4 ) 1/2  (1.466  105 ) 1/2 ]2
 3

 ( Ri ) 1/2 
 i 1



Pipe 3:
Le  0.06  1 / 0.022  2.7 ,
316
R3 
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 11 / Flows in Piping Systems
 Q1  W / R1  1.125 /1.983 105  0.00238 m3 /s, or 143 L/min
Q2  W / R2  1.125 / 4.788 104  0.00485 m3 /s, or 290 L/min
Q3  W / R3  1.125 /1.466 105  0.00277 m3 /s, or 166 L/min
(b) See part (a) for development.
Pipe 1:
Le 
2
 3 / 0.02  25,
12
R1 
8  0.02  115
 450.2,
32.2   2  ( 2 / 12) 5
Pipe 2:
Le 
3
 5 / 0.025  50,
12
R2 
8  0.025  170
 109.6,
32.2   2  ( 3 / 12) 5
Pipe 3:
Le 
R3 
8  0.022  189.5
 267.4.
32.2   2  ( 2.5 / 12) 5
W 
2.5
1/ 0.022  9.5,
12
[ 450.2  /2
0.352
 2.949 ft,
 109.6 1/2  267.4 1/2 ]2
Q1  2.949 / 450.2  0.081 ft3 /sec,
Q2  2.949 / 109.6  0.164 ft3 /sec,
Q3  2.949 / 267.4  0.105 ft3 /sec.
11.19 Write energy eqn. across system, excluding pipe 3 with loc. A at lower reservoir,
and loc. B at upper reservoir:
z A  H P  R 1Q 12  R2 Q 22  zB .
But,
R2 Q 22  R3 Q 32 , so that Q 3  ( R2 / R3 ) 1/2 Q 2 .
Write continuity eqn. at junction: Q 1  Q 2  Q 3  Q 2 [ 1  ( R2 / R3 ) 1/2 ], or
Q 2  Q 1 [ 1  ( R2 / R3 ) 1/2 ] 1 .
Substitute this expression along with
H P  150  5Q 12 into energy eqn. and solve for Q 1 :
z A  150  5Q 12  R1Q 12  R2 Q 12 [ 1  ( R2 / R3 ) 1/2 ] 2  zB ,
2

  1000  1/2  

 2
10  150  5Q  400  10001  
  Q 1  40,


1500

 



2
1
which reduces to 120  708.1Q 12 ,
317
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Chapter 11 / Flows in Piping Systems
3
Q1  120 / 708.1  0.412 m / s,
1/2
 1000 
Q3  

 1500 
  1000 1/2 
Q2  0.412 1  
 
  1500  
 0.227  0.185 m3 /s.
1
 0.227 m3 /s,
H P  150  5  0.412 2  149 m,
  QH P  9810  0.412  149  8.03  10 5 W, or W
  803 kW
W
P
P
0.75

11.20 Write energy eqn. from junction at A to suction side of pump at B:

K  2 p B
H PA   R1 
 zB ,
Q 

2 gA12 

where H PA  pump head and pB  pressure. The only unknown in the relation is Q, since
H PA
 
W
PA
1  10 6  0.76
95.6



.
Q
0.81  9810  Q
Q
Evaluate the constants:
R1 
K
2
8 f 1 L1
8  0.23  5000
 0.52,

 40.0,
2 
2
5
2
5
2 gA1 2  9.81  ( / 4) 2  0.754
g D1 9.81    0.75
pB

 zB 
150  103
 27  45.9 m.
0.81  9810
Substituting into the energy eqn.,
95.6
 40.5Q 2  45.9.
Q
Solve using Newton’s method: Q  1.053 m3 /s, and HPA  95.6/1.053  90.8 m.
The second reach is now analyzed. Write energy eqn. from suction side of pump at B to
the downstream reservoir:

K  2
 z B  H PB   R2 
 Q  50.

2 gA22 

Evaluate the constants:
pB
R2 
8 f 2 L2
8  0.023  7500

 60.1,
2
5
g D2 9.81   2  0.75 5
K
2
1
2 gA

10
 2.61,
2  9.81  ( / 4) 2  0.754
 45.9  H PB  ( 60.1  2.6)  1.053 2  50, or H PB  73.6 m.
W PB 
QH P

B

0.81  9810  1.053  73.6
  810 kW.
 8.10  10 5 W, or W
PB
0.76
318
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Chapter 11 / Flows in Piping Systems
11.21 Write energy equation from loc. A to loc. D:

p

p
  z  W AB  W BC  W CD    z ,
D

A

W AB  R1Q12 , W BC
2
 5 1 
 Q2 , W  R Q2 .
 
CD
1
6 1
i 2
R i 


Since the friction factors are unknown, an iterative solution is required. For the
2
e

first iteration assume flow in the fully rough zone, so that f  1.325 ln 0.27  .
D

Then the equivalent lengths and resistance coefficients are tabulated:
Pipe
1
2
3
4
5
6
e
D
0.0005
0.004
0.004
0.0033
0.0025
0.00057
f
Le 
0.0167
0.0284
0.0284
0.0269
0.0249
0.0172
 5 1 


 i 2 R 

i 

DK
f
8 f ( L  Le)
g 2 D 5
4.47  103
4.81  107
6.01  107
3.13  107
8.44  106
4.77  103
R
36
0
0
2
0
51
2
 1.578  10 6 ,
The energy eqn. becomes ( 4.47  10 3  1.578  10 6  4.77  10 3 )Q 12  70  10,
Q12  3.78 105 , and Q1  0.00615 m3 /s.
W BC  1.578  10 6 ( 3.78  10 5 )  59.6 m,
Q2  WBC / R2  59.65 / 4.81107  0.00111 m3 /s,
Q3  WBC / R3  59.65 / 6.01107  0.0010 m3 /s,
Q4  WBC / R4  59.65 / 3.13 107  0.00138 m3 /s,
Q5  WBC / R5  59.65 / 8.44 106  0.00266 m3 /s.
319
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Chapter 11 / Flows in Piping Systems
For the second iteration, assume water temperature of 20C, so that  = 1  10-6 m2/s:
4Q
QD
Q
Re 

 1.27  10 6 ,
A D
D
2
e
 

f  1.325ln 0.27  5.74 Re 0.9   ,

D
 
and the updated resistance coefficients are tabulated:
Pipe
1
2
3
4
5
6
f
0.0236
0.0306
0.0308
0.0292
0.0268
0.0233
Re
3.90  104
5.64  104
5.08  104
5.84  104
8.45  104
4.46  104
 5 1 


 i 2 R 

i 

Le
25
0
0
2
0
38
R
6.25  103
5.18  107
6.52  107
3.40  107
9.08  106
6.31  103
2
 1.704  10 6 ,
The energy eqn. becomes ( 6.25  10 3  1.704  10 6  6.31  10 3 )Q 12  60 ,
Q12  3.494 105 , and Q1  0.00591 m3 /s.
W BC  1.704  10 6 (3.494  10 5 )  59.54 m,
Q2  59.54 / 5.18 107  0.00107 m3 /s,
Q3  59.54 / 6.12 107  0.00095 m3 /s,
Q4  59.54 / 3.4 107  0.00132 m3 /s,
Q5  59.54 / 9.08 106  0.00256 m3 /s.
Calculation of the friction factors with the current estimates of the discharges reveals that
they change only in the third significant figure, hence new estimates of Q i will not change
significantly.
Q1  Q6  5.91 L/s, Q2  1.07 L/s,
Q3  0.95 L/s, Q4  1.32 L/s, Q5  2.56 L/s.
320
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Chapter 11 / Flows in Piping Systems
11.22 Compute resistance coefficients:
10.59 L
10.59  200
 7.6,

1. 85
4 . 87
C D
130 1.85  0.5 4 .87
10.59  1500
R3 
 686.5,
130 1.85  0.3 4 .87
R1 
10.59  600
 274.6,
130 1.85  0.3 4 .87
10.59  1500
R4 
 169.1.
130 1.85  0.4 4 .87
R2 
(a) Let H J  piezometric head at junction J. Assume Q1  2 m3 /s (out of J).
Then H J  30  R1Q 12  30  7.6  2 2  60.4 m,
Q2  (250  60.4) / 274.6  0.831 m3 /s (into J ),
Q3  (300  60.4) / 686.5  0.591 m3 /s (into J ),
Q4  (200  60.4) /169.1  0.909 m3 /s (into J ),
 Q  2  0.831  0.591  0.909  0.331.
Assume Q1  2.5 m3 /s. Then:
H J  30  7.6  2.5 2  77.5 m,
Q2  ( 250  77.5) / 274.6  0.793 m3 / s,
Q3  ( 300  77.5) / 686.5  0.569 m3 / s,
Q4  ( 200  77.5) / 169.1  0.851 m3 / s,
 Q  2.5  0.793  0.569  0.851  0.287.
Linear interpolation to find the next estimate of Q 1 yields
0.287  0.331 0.287

,
2.5  2
2.5  Q 1
Q1  2.27 m 3 / s.
Then: H J  30  7.6  2.27 2  69.2 m,
Q 2  ( 250  69.2) / 274.6  0.811 m 3 / s,
Q 3  ( 300  69.2) / 686.5  0.580 m 3 / s ,
Q 4  (200  69.2) / 169.1  0.880 m 3 / s ,
Q  2.27  0.811  0.580  0.880  0.001.  OK.
(b) Assume Q1  1 m3 /s (into J ). Then:
H J  30  (250  0.4Q1  0.1Q12 )  R1Q12  280  0.4  1  7.7  1 2  271.9 m,
Q2  (279.1  250) / 274.6  0.282 m3 /s (out of J ),
Q3  (300  279.1) / 686.5  0.202 m3 /s (into J ),
Q4  (279.1  200) /169.1  0.652 m3 /s (out of J ),
 Q  1  0.282  0.202  0.652  0.268.
321
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Chapter 11 / Flows in Piping Systems
Assume Q1  0.5 m3 /s (into J ). Then:
H J  280  0.4  0.5  7.7  0.5 2  277.9 m,
Q2  (277.9  250) / 274.6  0.319 m 3 / s,
Q3  ( 300  277.9) / 686.5  0.179 m 3 / s,
Q4  (279.9  250) / 169.1  0.679 m 3 / s,
 Q  0.5  0.319  0.179  0.679  0.319.
Use linear interpolation to find the new estimate of Q 1 :
0.268  ( 0.319) 0.268

,
1  0.5
1  Q1
Q1  0.772 m 3 / s.
Then: H J  280  0.4  0.772  7.7  0.772 2  275.1 m,
Q 2  ( 275.1  250) / 274.6  0.302 m 3 / s,
Q 3  ( 300  275.1) / 686.5  0.190 m 3 / s,
Q 4  ( 275.1  200) / 169.1  0.666 m 3 / s,
Q  0.772  0.302  0.19  0.666  0.006.  OK.
11.23 Compute equivalent lengths and
resistance coefficients using
Le 
8 f ( L  Le )
DK
and R 
:
f
g 2 D5
Le1 
0.35  2
 35,
0.02
Le 2  0,
R2
Le 3  0,
R3
0.20  2
 20,
0.02
0.25  2
Le5 
 25,
0.02
Le 4 
R4
R5
8  0.02 135
 42.5,
9.81  2  0.355
8  0.02  750

 3873,
9.81   2  0.2 5
8  0.02  850

 4390,
9.81   2  0.2 5
8  0.02  520

 2685,
9.81   2  0.2 5
8  0.02  375

 635.
9.81   2  0.2 5
R1 
Write the energy equation from A to E:
2
 1
p

1 
2

z
H
R
Q
Q 12  H E ,







P
1 1



A
R3 
 R2
in which H E  hyd. gr. line at loc. E. The following strategy is employed.
322
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Chapter 11 / Flows in Piping Systems
Assume Q1, compute H E , then compute Q4 and Q5 using
Q4 
H E  HC
R4
and Q5 
HE  HB

p

p
, where H C    z , H B    z .
R5
B

C

Finally check continuity at E with the relation Q  Q 1  Q 4  Q 5 .
Substitute known data into the above relations:
2
1 
 1
2
2
H E  0  60  10Q  42.5Q  

 Q1  60  1083Q1 ,
 3873

4390
2
1
2
1
Q4  ( H E  50) / 2685, Q 5  ( H E  48) / 635 .
The results are tabulated below. The third estimate of Q 1 is found by interpolation.
HE
Q1
(m3 /s)
0.1
0.09
0.0904
(m)
49.17
51.23
51.15
Q4
Q5
Q
(m3 /s)
0.01758
+0.02140
+0.02070
(m3 /s)
+0.04292
+0.07132
+0.07043
(m3 /s)
+0.07466
0.00272
0.00073
Compute hydraulic grade line at D:
H D  H P  R1Q 12  60  ( 10  42.5)  0.0904 2  59.57 m.
Q 2 
HD  HE

R2
59.57  51.15
 0.04663 m 3 / s,
3873
Q3 
HD  HE

R3
59.57  51.15
 0.04379 m 3 / s.
4390
Q1  90 L/s, Q2  46 L/s, Q3  44 L/s, Q4  20 L/s, and Q5  70 L/s.
11.24 Compute the R-values for each pipe:
1 0.333
 17 ft ,
0.02
2  0.333
Pipe 2: Le 
 33 ft ,
0.02
2  0.333
Pipe 3: Le 
 33 ft ,
0.02
4  0.333
Pipe 4: Le 
 67 ft ,
0.02
Pipe 1: Le 
8  0.02  27
3
32.2   2  0.3335
8  0.02  533
R2 
 66
32.2   2  0.3335
8  0.02  2, 033
R3 
 250
32.2   2  0.3335
8  0.02 1,817
 224
R4 
32.2   2  0.3335
R1 
323
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Chapter 11 / Flows in Piping Systems
(a) Compute the discharge and head loss in pipe 2:
Q 2  5,000
gal
h

1 ft 3
1 hr
ft 3

 0.186
,
7.48 gal 3600 sec
sec
 ( h L ) 2  R 2 Q 22  66  0.186 2  2.28 ft
(b) The flow path is pipe 1  pipe 2  pipe 3; there is no flow in pipe 4. The demand
curve from A to B is
H P  zB  z A  (hL )1  (hL )2  (hL )3  430  (3  66  250)Q2  430  319Q2
A trial and error solution is employed: assume Q, compute HP from the demand curve,
and compare it with HP from the pump characteristic curve.
Trial 1: Q = 10,000 gal/h = 0.371 ft 3/sec,
H P (demand )  430  319  0. 3712  474 ft , HP (pump) = 465 ft
Trial 2: Q = 9,000 gal/h = 0.334 ft 3/sec,
H P (demand )  430  319  0. 334 2  467 ft ,
HP (pump) = 466 ft
 The discharge is approximately 0.33 ft3/sec, or 9,000 gal/h.
(c) Q2  11,000 gal /min  0.409 ft 3 /sec, and from the pump curve, HP = 460 ft.
Write the energy equation from A to C:
HC  H P  (R1  R2 )Q22  460  69  0.409 2  449 ft
The energy equation from C to D is HC  zD  R4Q42 , where HC is the hydraulic grade
line at location C. Therefore the discharge in pipe 4 is
Q4 
11.25
HC  zD

R4
449  445
 0134
.
ft 3 /sec , or 3,600 gal/h.
224
(a) The discharge through the pump is Q  12,000 /(7.48  3600)  0.446 ft 3 /sec , and
from the characteristic curve the pump head is HP = 454 ft. The efficiency is P =
0.86. Hence the required power is
   QHP  62.4  0.446  454  1.47  104 ft - lb , or 26.7 hp
W
P
0.86
sec
P
(c) The energy equation from A to C is HP  zC  pC /  z A  (hL )1  (hL )2 .
pC

 454  300  (3  66)  0.4462  140 ft , and pC 
324
62.4  140
 61 lb /in 2
144
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Chapter 11 / Flows in Piping Systems
(d) Write the energy equation from C to D and solve for the pressure at location D:
pC /  zC  pD /  zD . (Since there is no flow in pipe 4, the energy equation
reduces to the law of hydrostatics.) From part (b), the pressure at location C is 61
lb/in2.
 pD  61  62.4 
11.26
300  445
  183
. lb /in 2
144
(a) From the pump curve, for HP = 460 ft, Q = 11,000 gal/h, or 0.409 ft 3/sec. Write the
energy equation from A to C and evaluate the hydraulic grade line:
HC  H P  ( hL )1  ( hL )2  460  69  0.409 2  449 ft
Thus the discharge in pipe 3, between location C and location B is
Q3 
HC  H B

R3
449  430
 0.276 ft 3 / sec, or 7,400 gal /h
250
(b) The flow path is pipe 1  pipe 2  pipe 4; there is no flow in pipe 3. Write the energy
equation from location A to location D:
H P  zD  z A  ( hL )1  ( hL )2  ( hL )4  445  (3  66  224 )Q 2  445  293Q 2
A trial and error solution is employed: assume Q, compute HP from the demand
curve, and compare it with HP from the pump characteristic curve.
Trial 1: Q = 10,000 gal/h = 0.371 ft 3/sec
HP (demand) = 445 + 293  0.3712 = 485 ft
HP (pump) = 465 ft
Trial 2: Q = 7,500 gal/h = 0.278 ft 3/sec
HP (demand) = 445 + 293  0.2782 = 468 ft
HP (pump) = 475 ft
Trial 3: Q = 8,000 gal/h = 0.297 ft 3/sec
HP (demand) = 445 + 293  0.2972 = 471 ft
HP (pump) = 472 ft
Hence, the head across the pump is HP  470 ft.
325
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Chapter 11 / Flows in Piping Systems
11.27 (a) Unknowns: Q 1 , Q 2 , Q 3 , Q 4 , Q 5 and H J .
Eqns:
H A  R1Q 12  H J , H J  R2Q 22  H B ,
H J  R3Q 32  H B , H J  R4 Q 42  H B ,
H J  R5Q52  H C , Q 1  Q2  Q3  Q4  Q5 .
Reduce to one equation in the one unknown H J :
H J  H B  R2Q 22  R3Q 32  R4 Q 42 ,
HA  HJ
Q1 
Furthermore,
R1
 Q3  Q2
, Q2 
R2
R2
and Q4  Q2
.
R4
R3
HJ  HB
HJ  HC
, Q5 
R2
R5
.
Substitute into continuity relation:

HA  HJ
R1

HJ  HB 
1 
R2

R2

R4
R2 

R4 
H J  HC
R5
(b)
 R1Q 12  R2 Q 22  H A  H B  0, R2 Q 22  R3 Q 32  0,
R3 Q 32  R4 Q 42  0, R4 Q 42  R5 Q 52  H C  H B  0
11.28 Q I 
( W 2  W 3 )  15
( W 1  W 2 )  10
, Q II 
,
G2  G3
G1  G2
Wi  RQi Qi , Gi  2 R Qi .
Loop Pipe
I
1
2
Q
3.5
0
W
24.50
0
 24.50
QI 
G
Q
W
14.00 2.46
12.10
0
+0.36
+0.26
 11.84
14.00
 24.50  10
 1.04
14.00
326
QI 
G
Q
9.84
1.44
11.28
2.30
+0.43
 1184
.  10
.
 016
11.28
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 11 / Flows in Piping Systems
II
2
3
0
3.5
0
24.5
 24.50
QII 
0
14.00
14.00
 24.50  15
 0.68
14.00
0.36
2.82
-0.26
15.90
 16.16
QII 
1.44
11.28
12.72
0.43
+2.73
 1616
.  15
 0.09
12.72
H Junction  50  R1Q 12  50  2  2.3 2  39.4
Q1  2.30 (into J ),
Q2  0.43 (into J ),
Q3  2.73 (out of J )
11.29 Consider the flow correction in a clockwise sense through the system:
Q 
 (W1  H P  W 2 )  H  30Q1 Q1  20.4 / Q1  20Q2 Q2  20

G1  G 2
60 Q1  20.4 / Q12  40 Q2
Q1 (m3/s)
1
0.655
0.631
0.632
Q2 (m3/s)
0.5
0.155
0.131
0.132
Q (m3/s)
0.345
0.024
+0.000155
+7.89  10
Therefore, Q1  0.63 m3/s and Q2  0.113 m3/s. The hydraulic grade line at location J is:
H J  5  H P  R1Q12  5 
20.4
 30  0.632 2  25.3 m
0.632
11.30 First, continuity is satisfied as shown in the figure:
(a) The hydraulic grade lines at the nodes (i.e., the nodal piezometric heads can now be
computed:
H A  100  R1 Q12  100  20  0.45 2  95.95 m,
H B  H A  R2 Q22  95.95  51  0.28 2  91.95 m,
H C  H A  R3 Q 32  95.95  280  0.17 2  87 .86 m,
H D  H B  R5 Q52  91.95  310  0.18 2  81.91 m.
327
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Chapter 11 / Flows in Piping Systems
(b)
pA  (H A  zA )  9810(95.95  10)  8.43  10 5 Pa,
pB  (HB  zB )  9810(91.95  20)  7.06  10 5 Pa,
pC  (HC  zC )  9810(87.86  5)  8.13  10 5 Pa,
pD  (H D  zD )  9810(81.91  0)  8.04  10 5 Pa.
11.31 Compute equivalent lengths and resistance coefficients using
Le 
8 f ( L  Le )
Dk
& R
:
f
g 2 D5
R1 
Le1  0,
2  3 /12
 25,
0.02
2  3 /12
Le3 
 25,
0.02
3  3 /12
Le 4 
 30,
0.025
3  4 /12
Le5 
 67,
0.015
R2 
Le 2 
R3 
R4 
R5 
8  0.015  800
32.2   2  ( 8 / 12) 5
8  0.02  625
32.2   2  ( 3 / 12) 5
8  0.02  675
32.2   2  ( 3 / 12) 5
8  0.025  455
32.2   2  ( 3 / 12) 5
8  0.015  1067
32.2   2  ( 4 / 12) 5
 2.29 ,
 322.2,
 348.0,
 293.2,
 97 .9.
Solve for single unknown H J using eqn. derived in Pbm. 11.27(a):
HA  HJ
R1
650  H J
2.29
R2 

R4 

HJ  HB 
1 

R2

R2

R3

H J  575 
1 
322.2 
322.2

348.0
H J  HC
R5
322.2 

293.2 
,
H J  180
97 .9
,
which reduces to F(H J )  650  H J  0.2538 H J  575  0.1529 H J  180 .
Take first derivative and employ Newton’s method of solution:
F ( H J )  
0.5

650  H J
0.1269
0.07645
.

H J  575
H J  180
HJ
F
F
H J   F / F 
600
626.74
624.86
624.84
2.669
0.2345
0.003064
0.0006379
0.09982
0.1249
0.1213
0.1213
+26.74
1.877
0.025
0.0026
 H J  624.8 ft ,
328
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Chapter 11 / Flows in Piping Systems
Q1 
Q2 
Q3 
HA  HJ
R1
HJ  HB
R2
HJ  HB
R3
HJ  HB
Q4 
Q5 
R4
HJ  HC
R5

650  624.8
 3.32 ft 3 / sec,
2.29

624.8  575
 0.39 ft 3 / sec,
322.2

624.8  575
 0.38 ft 3 / sec,
348.0

624.8  575
 0.41 ft 3 / sec,
293.2

624.8  180
 2.13 ft 3 / sec.
97 .9
11.32
Assume flow directions as shown. Then: Q 

2 R Q
 R 1Q 12  R 2 Q 22  R 3 Q 32
1
1
 R2Q 2  R3Q 3


Initial flow assumption: Q 1  25, Q 2  10, Q 3  25.
1st iteration: Q 

 3  25 2  5  10 2  2  25 2
2 3  25  5  10  2  25
  3.21,
 Q 1  25  3.21  21.79 , Q 2  10  3.21  6.79 ,
nd
2 iteration: Q 

 3  21.79 2  5  6.79 2  2  28.21 2
2 3  21.79  5  6.79  2  28.21
 Q 1  21.79  0.20  21.59 , Q 2  6.79  0.20  6.59 ,
rd
3 iteration: Q 

 3  21.59 2  5  6.59 2  2  28.41 2
2 3  21.59  5  6.59  2  28.41
Q 3  25  3.21  28.21.
  0.20,
Q 3  28.21  0. 20  28. 41.
  0.0041,
Q1  21.6, Q2  6.6, Q3  28.4. Units are L/s.
329
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Chapter 11 / Flows in Piping Systems
11.33 The solution was obtained using EPANET, Version 2.0.
Link - Node Table:
---------------------------------------------------------------------Link
Start
End
Length Diameter
ID
Node
Node
m
mm
---------------------------------------------------------------------1
1
3
500
300
2
2
4
600
250
3
3
4
50
150
4
3
5
200
250
5
4
5
200
300
Node Results:
---------------------------------------------------------------------Node
Demand
Head Pressure
Quality
ID
LPS
m
m
---------------------------------------------------------------------3
0.00
9.36
5.36
0.00
4
0.00
5.68
4.68
0.00
5
75.00
5.78
5.78
0.00
1
-137.01
15.00
0.00
0.00 Reservoir
2
62.01
2.00
0.00
0.00 Reservoir
Link Results:
---------------------------------------------------------------------Link
Flow Velocity Headloss
Status
ID
LPS
m/s
m/km
---------------------------------------------------------------------1
137.01
1.94
11.29
Open
2
-62.01
1.26
6.14
Open
3
36.48
2.07
73.51
Open
4
100.54
2.05
17.86
Open
5
-25.54
0.36
0.52
Open
330
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Chapter 11 / Flows in Piping Systems
11.34 A Mathcad solution is provided below. The solution converges after 3 iterations.
Given:
ORIGIN  1
 200 
L   300   m
 
 120 
 1500 
D   1000   mm


 1200 
Z  50 m
e  1 mm
Resistance coefficients:
f  1.325  ln  0.27
i
2
K   0 
 
 10 
 
newton
3
m
W P  1920 kW
  0.82
i  1  3
e

D 
i 
2
 0.018 
f   0.02 


 0.019 
D  K 
 
i i 
8 f  L 
 i i
f 
i 
 
R 
i
2
  9810
 0.071  2
s
R   0.487 

 5
 0.473  m
 i5
g   D
3
Initial flow estimate (note that the discharge is common to all three lines):
Q  2
1
m
s
Hardy Cross iteration:

3
 R  R  R  Q Q 
Q( Q) 
1
2

W P 

2 R  R  R  Q 
N  4
Solution:
j  1  N
1
Q
j 1
2
3
Q
 Z
W P 
2
Q
 j
 Q  Q Q
j
 j 
Q Q
 2 
 2.59 

 m3
Q   2.762 
 2.771  s


 2.771 
Residual:
0.59 m3
0.172
8.456·10 -3
s
1.762·10 -5
331
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.35 (a) H B  H P  H A  ( R1  R2 )Q 2 ,
10  100  826Q 2  35  ( 5000  300)Q 2 ,
6126 Q 2  75,
Q  0.111 m 3 / s.
(b)
 (R1Q12  R2 Q22  HP )  ( H A  HB )
Q 
2R1Q1  2R2Q2  2a1Q2
Let Q 1  Q 2  Q , and substituting in known data:
6126Q 2  75
( 5000  300)Q 2  100  826Q 2  25
Q 

12252Q
2( 5000  300  826)Q
Iteration
1
2
3
4
Q
0.20
0.1306
0.1122
0.1107
Q
-0.0694
-0.0184
-0.00152
-0.0000104
Q  0.111 m 3 / s.
332
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.36 The solution was obtained using EPANET, Version 2.0.
Link - Node Table:
---------------------------------------------------------------------Link
Start
End
Length Diameter
ID
Node
Node
m
mm
---------------------------------------------------------------------1
1
5
200
100
2
5
3
150
50
3
5
6
500
100
4
6
4
35
50
5
6
2
120
100
Node Results:
---------------------------------------------------------------------Node
Demand
Head Pressure
Quality
ID
LPS
m
m
---------------------------------------------------------------------5
0.00
121.72
121.72
0.00
6
0.00
116.23
116.23
0.00
1
-9.28
125.00
0.00
0.00 Reservoir
2
7.15
115.00
0.00
0.00 Reservoir
3
1.58
118.00
0.00
0.00 Reservoir
4
0.55
116.00
0.00
0.00 Reservoir
Link Results:
---------------------------------------------------------------------Link
Flow Velocity Headloss
Status
ID
LPS
m/s
m/km
---------------------------------------------------------------------1
9.28
1.18
16.40
Open
2
1.58
0.81
24.80
Open
3
7.69
0.98
10.98
Open
4
0.55
0.28
6.55
Open
5
7.15
0.91
10.24
Open
333
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.37
The solution was obtained using EPANET, Version 2.0.
Link - Node Table:
---------------------------------------------------------------------Link
Start
End
Length Diameter
ID
Node
Node
m
mm
---------------------------------------------------------------------1
5
6
200
100
2
6
3
150
50
3
6
7
500
100
4
7
4
35
50
5
7
2
120
100
7
1
5
#N/A
#N/A Pump
Energy Usage:
---------------------------------------------------------------------Usage
Avg.
Kw-hr
Avg.
Peak
Cost
Pump
Factor Effic.
/m3
Kw
Kw
/day
---------------------------------------------------------------------7
100.00 75.00
0.16
13.33
13.33
0.00
---------------------------------------------------------------------Demand Charge:
0.00
Total Cost:
0.00
Node Results:
---------------------------------------------------------------------Node
Demand
Head Pressure
Quality
ID
LPS
m
m
---------------------------------------------------------------------5
0.00
169.11
44.11
0.00
6
0.00
149.86
149.86
0.00
7
0.00
120.64
4.64
0.00
1
-23.12
125.00
0.00
0.00 Reservoir
2
15.77
115.00
0.00
0.00 Reservoir
3
4.78
118.00
0.00
0.00 Reservoir
4
2.57
116.00
0.00
0.00 Reservoir
Link Results:
---------------------------------------------------------------------Link
Flow Velocity Headloss
Status
ID
LPS
m/s
m/km
---------------------------------------------------------------------1
23.12
2.94
96.25
Open
2
4.78
2.44
212.39
Open
3
18.33
2.34
58.43
Open
4
2.57
1.31
132.65
Open
5
15.77
2.01
47.02
Open
7
23.12
0.00
-44.11
Open Pump
334
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.38 (a) Let Rj be the resistance coefficient for a single tube in the condenser. Then
N
Rj
Q   W / R j  N W / Rj . Solving for W, we have W  2 Q 2  R 2 Q 2 .
N
j1
Hence, R2 
Rj
N2
.
(b) H P  z B  z A  (R1  R2  R3 )Q 2 HP and Q are the two unknowns.
(c) Q 
(d) R1 
 (R1  R2  R3) ) QQ  (a 0  a1Q  a 2Q 2  a 3Q 3 )  z A  z B
2(R1  R2  R3) ) Q  (a1  2a 2Q  3a 3Q 2 )
8 f L1
8 f L2
8 f L3
 0.0103
2
5  0.00516 , R2 
2
2 5  2.538 , R3 
g 2 D35
g N D2
g D1
Using the equation in part (c), Q = 2.061 m3/s after 5 iterations, beginning
initial estimate Q = 3 m3/s.
Iteration
1
2
3
4
5
Q
3
2.389
2.113
2.062
2.061
with an
Q
0.611
0.277
0.050
0.0014
5.1  1013
(e)
HC  z A   a0  a1Q  a2Q 2  a3Q 32   R1Q 2
 2  30.4  31.8  2.06  18.6  2.062  4.0  2.063   0.00516  0.60 2
 10.83 m
pC   ( HC  zC )  9810  (10.83  6)  47380 Pa, or 47.4 kPa
HC'  HC  R2Q 2  10.83  2.538  2.062  0.060 m
pC'   ( HC'  zC )  9810  (0.060  6)  5830 Pa, or  58.3 kPa
335
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.39
The solution was obtained using EPANET, Version 2.0
Link - Node Table:
---------------------------------------------------------------------Link
Start
End
Length Diameter
ID
Node
Node
ft
in
---------------------------------------------------------------------1
1
3
1500
10
2
3
4
400
10
3
4
5
600
10
4
5
6
800
8
5
6
10
700
8
6
10
9
1000
10
7
9
8
750
10
8
8
7
450
10
9
7
6
550
10
10
7
4
800
10
11
8
3
800
10
12
2
9
1450
10
Node Results:
---------------------------------------------------------------------Node
Demand
Head Pressure
Quality
ID
CFS
ft
psi
---------------------------------------------------------------------3
0.00
246.28
39.55
0.00
4
2.00
242.23
39.96
0.00
5
1.00
241.55
39.67
0.00
6
1.00
241.71
41.90
0.00
7
1.00
242.26
39.98
0.00
8
0.00
244.65
41.01
0.00
9
1.00
245.40
43.50
0.00
10
1.00
242.35
44.35
0.00
1
-3.82
275.40
0.00
0.00 Reservoir
2
-3.18
265.40
0.00
0.00 Reservoir
Link Results:
---------------------------------------------------------------------Link
Flow Velocity Headloss
Status
ID
CFS
fps
ft/Kft
---------------------------------------------------------------------1
3.82
7.01
19.41
Open
2
2.69
4.93
10.14
Open
3
0.82
1.51
1.13
Open
4
-0.18
0.51
0.20
Open
5
-0.41
1.17
0.91
Open
6
-1.41
2.58
3.05
Open
7
0.77
1.41
1.00
Open
8
1.90
3.49
5.32
Open
9
0.77
1.41
1.00
Open
10
0.13
0.24
0.04
Open
11
-1.13
2.07
2.04
Open
336
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.40 The solution was obtained using
EPANET, Version 2.0
Link - Node Table:
---------------------------------------------------------------------Link
Start
End
Length Diameter
ID
Node
Node
ft
in
---------------------------------------------------------------------1
1
3
1500
10
2
3
4
400
10
3
4
5
600
10
4
5
6
800
8
5
6
10
700
8
6
10
9
1000
10
7
9
8
750
10
8
8
7
450
10
9
7
6
550
10
10
7
4
800
10
11
8
3
800
10
12
2
9
1450
10
Node Results:
---------------------------------------------------------------------Node
Demand
Head Pressure
Quality
ID
CFS
ft
psi
---------------------------------------------------------------------3
0.00
246.28
39.55
0.00
4
2.00
242.23
39.96
0.00
5
1.00
241.55
39.67
0.00
6
1.00
241.71
41.90
0.00
7
1.00
242.26
39.98
0.00
8
0.00
244.65
41.01
0.00
9
1.00
245.40
43.50
0.00
10
1.00
242.35
44.35
0.00
1
-3.82
275.40
0.00
0.00 Reservoir
2
-3.18
265.40
0.00
0.00 Reservoir
Link Results:
---------------------------------------------------------------------Link
Flow Velocity Headloss
Status
ID
CFS
fps
ft/Kft
---------------------------------------------------------------------1
3.82
7.01
19.41
Open
2
2.69
4.93
10.14
Open
3
0.82
1.51
1.13
Open
4
-0.18
0.51
0.20
Open
5
-0.41
1.17
0.91
Open
6
-1.41
2.58
3.05
Open
7
0.77
1.41
1.00
Open
8
1.90
3.49
5.32
Open
9
0.77
1.41
1.00
Open
10
0.13
0.24
0.04
Open
11
-1.13
2.07
2.04
Open
5.83
13.79
Open
337
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.41
The solution was obtained
using EPANET, Version 2.0.
Link - Node Table:
---------------------------------------------------------------------Link
Start
End
Length Diameter
ID
Node
Node
m
mm
---------------------------------------------------------------------1
6
8
610
350
2
8
9
914
350
3
9
10
760
350
4
10
13
610
350
6
11
12
457
300
7
11
10
610
350
8
7
11
914
350
9
6
7
760
350
10
8
11
610
350
11
10
2
30
200
12
4
12
61
150
13
5
6
1500
400
14
7
3
975
300
16
12
13
610
350
18
1
5
#N/A
#N/A Pump
Energy Usage:
---------------------------------------------------------------------Usage
Avg.
Kw-hr
Avg.
Peak
Cost
Pump
Factor Effic.
/m3
Kw
Kw
/day
---------------------------------------------------------------------18
100.00 75.00
0.52
971.98
971.98
0.00
---------------------------------------------------------------------Demand Charge:
0.00
Total Cost:
0.00
Node Results:
---------------------------------------------------------------------Node
Demand
Head Pressure
Quality
ID
LPS
m
m
---------------------------------------------------------------------5
0.00
146.93
143.93
0.00
6
0.00
61.82
49.82
0.00
7
60.00
41.24
26.24
0.00
8
0.00
40.32
28.32
0.00
9
110.00
31.32
13.32
0.00
10
110.00
30.78
15.78
0.00
11
0.00
36.07
24.07
0.00
12
60.00
31.94
25.94
0.00
13
60.00
30.77
18.77
0.00
1
-516.40
3.00
0.00
0.00 Reservoir
2
56.12
30.00
0.00
0.00 Reservoir
3
83.11
34.00
0.00
0.00 Reservoir
4
-22.83
34.00
0.00
0.00 Reservoir
Link Results:
338
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 11 / Flows in Piping Systems
---------------------------------------------------------------------Link
Flow Velocity Headloss
Status
ID
LPS
m/s
m/km
---------------------------------------------------------------------1
265.67
2.76
35.25
Open
2
145.21
1.51
9.85
Open
3
35.21
0.37
0.71
Open
4
4.72
0.05
0.02
Open
6
92.45
1.31
9.04
Open
7
135.63
1.41
8.68
Open
8
107.62
1.12
5.65
Open
9
250.73
2.61
27.08
Open
10
120.46
1.25
6.97
Open
11
56.12
1.79
25.86
Open
12
22.83
1.29
33.81
Open
13
516.40
4.11
56.74
Open
14
83.11
1.18
7.42
Open
16
55.28
0.57
1.92
Open
18
516.40
0.00
-143.93
Open Pump
11.42 The solution, determined by trying different pipe diameters, was obtained using
EPANET, Version 2.0. A useful pump power of 30 hp and a Hazen-Williams
coefficient of 100 for all pipes were employed in the program.
Link - Node Table:
---------------------------------------------------------------------Link
Start
End
Length Diameter
ID
Node
Node
ft
in
---------------------------------------------------------------------1
5
6
550
6
2
6
7
625
6
3
7
8
1000
6
4
8
9
725
6
5
9
10
725
6
6
10
3
800
6
7
3
11
675
6
8
3
4
950
6
9
4
5
725
6
10
2
3
475
8
11
11
7
825
6
12
1
2
#N/A
#N/A Pump
339
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 11 / Flows in Piping Systems
Energy Usage:
---------------------------------------------------------------------Usage
Avg.
Kw-hr
Avg.
Peak
Cost
Pump
Factor Effic.
/Mgal
Kw
Kw
/day
---------------------------------------------------------------------12
100.00 75.00
1035.69
29.83
29.83
0.00
---------------------------------------------------------------------Demand Charge:
0.00
Total Cost:
0.00
Node Results:
---------------------------------------------------------------------Node
Demand
Head Pressure
Quality
ID
GPM
ft
psi
---------------------------------------------------------------------2
0.00
747.25
107.13
0.00
3
0.00
743.63
101.23
0.00
4
25.00
740.98
100.09
0.00
5
50.00
739.62
90.83
0.00
6
80.00
739.31
86.36
0.00
7
0.00
739.39
82.06
0.00
8
50.00
738.84
81.82
0.00
9
100.00
738.84
88.32
0.00
10
75.00
739.97
95.31
0.00
11
100.00
740.28
91.11
0.00
1
-480.00
500.00
0.00
0.00 Reservoir
Link Results:
---------------------------------------------------------------------Link
Flow Velocity Headloss
Status
ID
GPM
fps
ft/Kft
---------------------------------------------------------------------1
55.66
0.63
0.57
Open
2
-24.34
0.28
0.12
Open
3
54.20
0.61
0.55
Open
4
4.20
0.05
0.00
Open
5
-95.80
1.09
1.57
Open
6
-170.80
1.94
4.57
Open
7
178.54
2.03
4.96
Open
8
130.66
1.48
2.78
Open
9
105.66
1.20
1.88
Open
10
480.00
3.06
7.63
Open
11
78.54
0.89
1.08
Open
12
480.00
0.00
-247.25
Open Pump
11.43 Using the EPANET program, the system can be designed by treating several groups of
greens/fairways independently, each group branching from a main feeder pipe. To simplify
the analysis, the four sprinklers on each green can be treated as a single sprinkler, and any
sprinkler can be handled either using a known flow demand, or acting as a valve emptying
into a reservoir (use a large K-value). Since all sprinklers are not run simultaneously, a
watering strategy must be developed.
340
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.44 Vss 
2 g( H1  H3 )
2  9.81 3

 0.307 m/s
0.025  2500 / 0.1  0.15
fL / D  K
V  0.99Vss  0.99  0.307  0.304 m/s
t

 (V  V )(Vss  V0 ) 
Vss L
ln  ss

2 g( H1  H3 )  (Vss  V )(Vss  V0 ) 
0.307  2500  (0.307  0.304)(0.307  0) 
ln 
  69.0 s
2  9.81 3
 (0.307  0.304)(0.307  0) 
11.45
0.4
0.3
Velocity (m/s)
0.2
0.1
0
0
20
40
60
80
Time (sec)
11.46 First compute the initial velocity V0, with H1-H3 = 8 m, and K0 = 275:
V0 
2  9.81  8
 0.552 m /s
0.015  800
 275
0.05
The final steady-state velocity is, with Kss = 5:
2  9.81  8
 0.800 m /s
0.015  800
5
0.05
The steady-state discharge is Q  0.7854  0.052  0.800  0.00157 m 3 / s , and the time
VSS 
to reach 95% of that value is
V  0.95Vss  0.95  0.800  0.760 m /s
t 
0.800  800  (0.800  0.760)(0.800  0.552) 
 8.03 s
ln
2  9.81  8  (0.800  0.760)(0.800  0.552) 
341
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible we bsite, in whole or in part.
Chapter 11 / Flows in Piping Systems
11.47 Compute the acoustic wave speed:
a
220  10 7
B
 1330 m /s

DB


200  220  10 7 
 1 

1000 1 


eE 
12  150  10 9 

(a) Wave travel time from valve to reservoir is
t  L/a  800/1330  0.602 s .
(b) Pressure change at valve due to doubling of discharge:
V 
Q
2

0.05
 1.592 m/s
 D /4 0.7854  0.22
p    aV  1000  1330  1.592  2.12  106 Pa, or  2120 kPa
Note the large pressure reduction due to the water hammer effect. The original
pressure at the valve must be sufficiently large so that cavitation will not occur.
Cavitation at the valve could be avoided by opening the valve slowly.
(c) Pressure change at valve due to halving of the discharge:
Q
1
1
0.05
V   

 0.796 m/s
2
2 D / 4
2  0.7854  0.22
p =  1000 1330  (  0.796)=1.06 106 Pa, or 1060 kPa
11.48 (a) Compute the acoustic wave speed:
a
B

DB

 1 


eE 
1.05  10 9
 1090 m /s.

50  1.05  10 9 
680   1 

2.5  70  10 9 

(b) The change in velocity due to rapid valve closure is V  VSS  0.800 m /s
(Problem 11.45). Hence the pressure rise is
p  aV  680  1090  (0.800)  5.93  10 5 Pa, or 593 kPa.
(c) From Problem 11.45, H  8 m, f  0.015 . Then the initial steady-state pressure at
the valve is

fL Vss2 

0.015  800 0.8 2 

pss   gH 
  680 9.81  8 
  1120 Pa
D 2 
0.05
2 


Hence the instantaneous pressure is
p  pss  p  1120  5.93  105  5.94  105 Pa, or 594 kPa.
The instantaneous pressure due to rapid valve closure is about 590 kPa, which is over
two times the allowable pressure in the pipe. It is possible that the pipe could rupture.
This could be avoided by closing the valve more slowly, thereby reducing the
maximum pressure rise due to water hammer activity.
342
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Chapter 11 / Flows in Piping Systems
11.49 Compute the acoustic wave speed:
a
217  10 3  144
 3610 ft /sec

20  217  10 3 
0.9  1.94   1 

0.40  29  10 6 

(a) The allowable changes in velocity and discharge due to pressure constraint are:
V 
Q 
p
90 144

 2.06 ft/s,
 a 0.9 1.94  3610

4
D2 V 
  20 
2
    (2.06)  4.49 ft 3 / sec
4  12 
Hence the tolerable flow rate decrease is (1  4.49/20) × 100 = 77%.
(b) Wave travel time from the downstream to the upstream end of pipe:
t
L 13000

 3.60 s
a 3610
343
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Chapter 11 / Flows in Piping Systems
344
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Chapter 12 / Turbomachinery
CHAPTER 12
Turbomachinery
12.1
(a) r1
Vt 1
T
0, Vt2
V2 cos
Q(r2Vt2
1000

W
H
(b) r1
3 m /s,
0.15 m, V1
90 , r2
1
6 cos 30
2
5.2
0)
r1
u2
r2
Vt2
u2
7.5 5.3cos 45
2
Q(r2Vt2

W
H
3.66 m /s,
Vt1
V1 cos
Vt2
V2 cos
H
2
64.1 N m,
3.75 5.3 cos 45
0,
machine is a pump.
2.86 m.
80 , r2
0.15 m, V2
0.64 m /s ,
61
. cos 30
5.28 m /s ,
0.057( 0.3
1
45 .
2
3.75 m /s ,
3.66 cos 80
Q(r1Vt1 r2Vt2 )
1000

W
1
1
5.3 m /s,
v2
v1 cos
u1
T 25 641
. 1600 W,
 / Q 1600 /(9810 0.057)
W
0.3 m, V1
T
0.3 m,
(assume pump)
r1Vt1 )
1000 0.057(0.3 3.75 0)
(d) r1
3.97 m.
5.3 m /s, 1 45 , r2
25 0.15 3.75 m / s, Vt1
25 0.3 7.5 m /s,
v2 cos
machine is a pump.
88.9 N m,
0.15 m, v1
T
(c) r1
5.20 m /s,
T 25 88.9 2220 W,

W / Q 2220 /(9810 0.057)
u1
30 .
2
(assume pump)
r1Vt1 )
0.057( 0.3
6 m /s,
0.3 m, V2
6.1 m /s,
2
30 .
(assume turbine)
0.64
0.15
5.28)
34.2 N m ,
a pump.
T 25 34.2 855 W,

W / Q 855 /(9810 0.057) 153
. m.
0.15 m, v1 3 m /s, 1 90 , r2 0.3 m, v2 8.7 m /s, 2 30 .
r2 25 0. 3 7.5 m / s,
u1
r1 25 015
.
3.75 m /s = Vt1 , u 2
Vt 2
T
u2
Q(r2Vt2
1000

W
H
v 2 cos
2
7.5
r1Vt1 )
0.057( 0
8.7 cos 30 
0,
(assume pump)
0.15
3.75)
32.1 N m,
machine is a turbine.
T 25 321
. 802 W,

W / Q 802 /(9810 0.057) 1.43 m.
345
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Chapter 12 / Turbomachinery
(e) r1
6 m /s,
0.3 m, V1
Vt1
V1 cos
1000
H
(f) r1
6cos30
1
0.057(0.3
3 m /s,
0.15 m, V2
5.20 m/s, Vt2
V2 cos
2
80 .
2
3cos80
0.52 m/s.
(assume turbine)
Q(r1Vt1 r2Vt2 )
T

W
30 , r2
1
5.2
0.15
0.52)
84.5 N m ,
a turbine.
T 25 84.5 2110 W,
 / Q 2110 /(9810 0.057) 3.77 m.
W
0.3 m, v1 3 m /s, 1 90 , r2 0.15 m, v2 4.33 m /s, 2 30 .
r2 25 0.15 3.75 m /s ,
u1
r1 25 0.3 7.5 m /s = Vt1 , u2
Vt2
v2 cos
u2
2
3.75 4.33 cos 30
(assume turbine)
Q(r1Vt1 r2Vt2 )
T
1000

W
0.057(0.3
25 128
T
0,
7.5
0)
machine is a turbine.
128 N m ,
3200 W,
H
 / Q
W
u1
u2
800
/ 30 83.8 rad /s,
r1 83.8 0.04 3.35 m /s,
r2 83.8 0.125 10.48 m /s,
Q
2 r1b1Vn1 , but Vn1
3200 /(9810 0.057)
5.72 m.
12.2
u1 since
1
45 ,
0.04 0.05 3.35 0.0421 m 3 / s.
Q
0.0421
2.14 m /s,
2 r2b2 2
0.125 0.025
Vn2
.
214
u2
10.48
6.77 m /s,
tan 2
tan 30
Q 2
Vn2
Vt2
Vt1
T

W
P
Ht
0
(
1
Q(r2Vt2
90 under ideal conditions).
r1Vt1 )
1000
0.0421(0.125
T 838
. 35.6 2980 W,
T / Q 2980 /(9810 0.0421)
346
6.77
0)
35.6 N m.
7.22 m.
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Chapter 12 / Turbomachinery
12.3
D1
200 mm, D2
150 mm,
0.81, S 2
S1
13.6,
H = 600 mm, Q = 115 L/s.
Manometer relation: p1
p2
p1
z2
1
z
1 1
2
z1
2
H
H
1 2
1 H
1
S2
S1
1 H
p1
Energy eqn. from loc. 1 to loc. 2:
1
Q2 1
2 g A22
HP
1
A12
p2
p1
1
0.115 2
1
2
2 9.81 ( / 4)
0.15 4
12.4
Q
2 gal / sec
1 ft 3
7.48 gal
2000
30
z
1
Q2
2 gA12
z2
p2 ,
13.6
0.81
z1
HP
1 0.6
p2
1
9.47 m
Q2
2 gA22
z2 ,
z1
1
( / 4) 2
0.2 4
9.47
10.95 m.
0.267 ft 3 / sec,
209 rad /sec,
0.267
6.12 ft /sec,
2.5 0.4
2
12 12
2.5
43.54 ft /sec,
r2 209
12
u2
43.54
(u2 Vn2 cot 2 )
( 43.54 6.12 cot 60  ) 54.1 ft.
g
32.2
Q
2 r2b2
Vn2
u2
Ht

W
W
QH t
62.4 0.8 0.267 541
.
721 ft - lb /sec , or
721
1.31 horsepower.
550
347
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Chapter 12 / Turbomachinery
12.5
r0
0.285 m, ri
0.135 m,
0.57
(0.2852 0.1352 )
Q
Vn
(r02 ri2 )
2.88 m/s,
.
0.285 0135
30
2
Use the theoretical head relation
u
u2
g
Ht
2.85
cot
12.6
1500
ravg
2
uVn
(cot
g
332
9.81
1
cot
2
); substitute in known data and solve for
33 2.88
(cot 60
9.81
10.58 ,
cot
2
)
111.0 9.69(0.5774 cot
2
2
:
), or
5.4 .
2
Compute loss in suction pipe:
Q2
L
K
hL
f
2 gA2
D
0.015
11
0.1
2
0.19
z
patm
pv
0.05 2
0.8
hL
5.85 m.
2
0.1 4
4
2340 Pa. Substitute known data into NPSH
2
Water at 20 C:
9792 N /m3 , pv
relation, solving for z:
12.7
33.0 m /s.
9.81
101 103 2340
9792
NPSH
5.85 3
1.23 m.
Neglecting losses, write energy
eqn from loc. 1 to loc. 2 to
determine the magnitude of z.
For water at 80 F,
(a)
z
0 ( 9.9 144)
621
.
NPSH =
(b)
z
0.51 lb/in 2 .
621
. lb / ft 3 and pv
(13 / 7.48)2
2
2 32.2
(14.7 - 0.51) 144
62.2
4
16.8
(12 0.51) 144
. 10.5 ft.
161
62.2
348
4
12
4
16.1 ft , and
16.8 ft.
16.1 / 115 0.14.
Change in el. is 16.8
10.5 = 6.3 ft lower.
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 12 / Turbomachinery
12.8
1
Q1
2
970
1200
N1
Q
N2 2
Q2
2

W
2
12.9
2
1
3
2
1
0.65 m 3 / s,

11.5 m and W
1
from Fig. 12.9, H1
H2
0.8
H1
1200
970

W
1
1200
970
91 kW.
2
11.5 17.6 m ,
3
91 172 kW.
From Fig. 12.9, at best efficiency, NPSH 1
2
3
2
0.8 m /s, NPSH2
For water at 50 C,
z
12.10 CQ
CH
Q
D3
gH
2
D2
patm
1200
970
NPSH1
1
5 m. For the new conditions 1200 rpm,
9693 N /m3 and pv
pv
NPSH
Q / 3600
304 0.2053
9.81H
3042 0.2052
2
5
7.65 m.
12.3 kPa.
101 103 12.3 103
9693
1.061 10 4 Q
. m.
7.65 151
(Q in m3 /h)
2.526 10 3 H (H in m)
Tabulate CQ and CH using selected values of Q and H from Fig. 12.6:
Q (m3/h)
0
50
100
150
200
250
300
CQ×10
0
0.53
1.06
1.59
2.12
2.65
3.18
3
H (m)
54
53
52
50
47
41
33
CH ×10
1.36
1.34
1.31
1.26
1.19
1.04
0.83
1
The dimensionless curve shown in Fig. 12.12 is for the 240-mm impeller. Since the
impellers are not the same (240 mm versus 205 mm) dynamic similitude does not exist and
thus, the curves are not the same.
349
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Chapter 12 / Turbomachinery
12.11 Given: N1 = 2900 rpm, N2 = 3300 rpm, Q2 = 200 m2/h. To find H2, we must first find the
corresponding Q1 and H1:
2
H2
H1
2
Q2
3300
1.295 ,
2900
Q1
200
176 m3 / h
1.138
2
1
Q1
Q2
1.138
From Fig. 12.6, H1
58 m.
12.12 Compute the specific speed:
1
75.1 m .
1.295 58
H2
Q
( gH P )3/4
P
3300
1.138
2900
2
1800
0.15
30
(9.81 22)3/4
hence use a mixed flow pump. As an alternate, since
pump could be employed.
12.13 Compute the specific speed:
Q
( gH P )3/4
P
hence use a radial flow pump.
0.75 (best eff. ), CQ
12.14 Fig. 12.13: At
CH
0.018, C W
750
45
0.049 78.5
(a) D
30
P
1.30,
is close to unity, a radial flow
2000
0.17
30
(9.81 104)3/4
0.033,
0.048,
0.0011, C NPSH
0.023.
78.5 rad /s,
1/3
2.27 ft ,
78.5 2 2.27 2
17.8 ft ,
H
32.2
0.023 78.5 2 2.27 2
H NPSH
22.7 ft ,
32.2
 0.0011 1.94 78.5 3 2.27 5 6.22 10 4 ft - lb / sec,
W
0.018

or W
(b) Q
6.22
10 4 / 550
78.5 2.27 3 CQ
113 horsepower.
918.2CQ
H ( 78.5 2 2.27 2 / 32.2)C H 986.1C H
 ( 1.94 78.5 3 2.27 5 / 550)C  1.028
W
W
350
10 5 CW
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Chapter 12 / Turbomachinery
A table presenting the head and power vs discharge follows.
Q(ft3 /sec)
CQ
0.015
0.02
0.03
0.04
0.049
0.055
12.15 N 1
13.8
18.4
27.6
36.7
45.0
50.5
400 rpm, D2 /D1
N2
Q2
N 2 D2
N 1 D1
H2
g1 N2
g2 N1
D1
D2
1
2

W
2
12.16
600

W
1

W
2
2
2
D2
D1
(1
H1
1
3
2
2
2
3
7.6
6.76 m,
1/4
(1 0.7)
0.67,
9.81
0.0252 6.76 / 0.67
2
1000
Q
0.0252 m 3 / s,
0.085
3
2
) 1
2
7.6 m,
400
2
400
1/4
Q
( gH P )3/4
Q1H1 /
Q2
Q1
134
113
62
2/3, g2 /g1 1/2
400 2
400 3
62.8 rad /s,
22.7 / 60
62.8 0.378
(9.81 19.5)3/4
0.378 m3 / s, N 1
19.5 m, Q1
N2
3
g 2Q2 H 2 /
30
P
12.17 H 1
35.5
30.6
23.7
21.7
17.8
5.9
0.085 m3 / s, H 1
400 rpm, Q1
W (hp)
H(ft)
0.378 m 3 / s,
0.751,
Use a radial flow pump.
600 rpm,
9810 0.378 19.5 / 0.7 103 103 W, H 2
H 2 D2
N1
H 1 D1
N 2 D2
N 1 D1
N2
N1
3
3
Q1
D2
D1
5
30.5
( 1)600
19.5
750 3
( 1) 0.378
600

W
1
1247 W.
750
600
351
30.5 m.
750 rpm,
0.472 m 3 / s,
3
( 1) 5 103
10 3
201 10 3 W
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Chapter 12 / Turbomachinery
H2
12.18
H1
2
2
D2
D1
2
1
2
2
1
Q2
Q1
2,
2
D2
D1
1
1
D2
D1
2
1
2
3
D2
D1
2
2,
3
,
D2
.
D1
2
or
1
4
2
2,
1
12.19
1400
30
146.6 rad /sec,
D3
Q / CQ
D
gH P
CH
and
227
D
119
.
Q
1
6.46 / 0.0165
32.2 200
0.125
0.52 ,
227,
P
D
227
173.3 rad/sec, or N
1.31
 /( Q)
W
f
Q
( gH P )3/4
2 D1 1.19 D1.
6.46 ft 3 / sec,
use radial flow pump.
3915
. ,
12.20 Assume a pump speed N = 2000 rpm, or
HP
4
, and D2
2900 / 449
146.6 6.46
(32.2 200)3/4
Q
( gH P )3/4
P
4
2
391.5
D
173.3 30
2000
30
200 103 /(8830 0.66)
209 0.66
(9.81 34.3)3/4
391.5
227
1.31 ft,
1655 rpm.
209 rad /s.
34.3 m,
2.16.
The specific speed suggests a mixed-flow pump. However, if N = 1000 rpm, a radial-flow
pump may be appropriate. Consider both possibilities.
Mixed flow: from Fig. 12.14, at best
C W
0.0117 , CQ
,:
0.148, C H
0.067.
Use
CQ
Q
D3
and CH
gH P
2
D2
352
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Chapter 12 / Turbomachinery
Combining and solving for D and
Q / CQ
D
Q
.
CQ D3
and
gH P / C H
0.66
g

W
P
3
D5
CW
282 30 /
0.251 m ,
2674 rpm.
900 kg/m3 ,
0.0117 900 2823
Radial flow: from Fig. 12.12, at best
0.2515
, : C W
0.66
0.0165 0.8783
0.0027
900
10 5 W , or 235 kW.
0.0165, C H
0.125.
0.878 m ,
59.1 rad/s, or N
59.1 3
2.35
0.027 , CQ
0.66 / 0.0165
9.81 34.3 / 0.125
D

W
P
0.66 / 0148
.
9.81 34.3 / 0.067
D
282 rad/s, or N
0.148 0.2513
8830
9.81
,
0.878 5
59.1 30
2.62
564 rpm.
10 5 W , or 262 kW.
Hence, a mixed-flow pump is preferred.
12.21 Compute required pump head using energy eqn.:
HP
z
f
30
15
600
30
L
D
K
Q2
2 gA2
0.019
500
0.75
62.8 rad/s,
12
3
19.1 m,
2
2
P
9.81
Q
( gHP )3/4
4
0.75
4
62.8 1
(9.81 191
. )3/4
1.24.
Use either a mixed-flow or radial-flow pump.
353
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Chapter 12 / Turbomachinery
12.22 Compute required pump head using energy eqn.:
HP
z
K
f L Q2
D 2 gA2
122
1.5
0.01 61
0.075
From Fig. 12.12: At best efficiency (
CH
0.125. Use CQ
0.0142 2
17.3 m.
2
2
9.81
4
0.76), CQ
P
Q
and CH
D3
0.075
4
0.0165 and
gH P
.
2
D2
By combining and solving for D and :
Q / CQ
D
gH P / C H
Q /(CQD3 )
or
N

W
P
0.0142 /(0.0165 0153
. 3)
240 30 /
QHP /
0.0142 / 0.0165
.
9.81 17.3 / 0125
P
0.153 m ,
240 rad / s,
2290 rpm.
9810 0.85 0.0142 17.3 / 0.76
2695 W.
Expected pump efficiency would be lower because of smaller scale and since oil has a
higher viscosity than water.
12.23
The intersection of the system demand curve with the head-discharge curve yields

Q 2.75 m 3 / min , H P 12.6 m, W
7.2 kW.
P
354
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Chapter 12 / Turbomachinery
12.24 N 1

W
1
2.75 m3 / min, H 1
1350 rpm, Q1
7.2 kW, N 2
Q2
H2

W
2
1200 rpm, D2
N2
N1
Q1
D2
D1
H1
N2
N1

W
1
N2
N1
2
3
D1 .
3
2.75
1200
1350
2
D2
D1
12.6
D2
D1
12.6 m,
5
2.44 m3 /min,
2
1200
1350
1200
7 .2
1350
9.96 m,
3
5.06 kW.
Efficiency will remain approximately the same.
12.25
1350
141.4 rad/s,
30
CQ ( D3 ) 1Q (141.4 0.2163 ) 1Q 0.702Q
CH
( D) 2 gHP
CW
(
At best

D5 ) 1 W
P
3
P
, CQ
(141.4 0.216) 2 HP
(1000 141.43
0.032, C H
0.134,
(Q in m3 /s)
0.0105HP

0.2165 )W
P
P
(HP in m)
 (W
 in W)
10 7 W
P
P
7.53
CQ1/2CH3/4
(0.032 / 2)1/2 (0134
. )
3/4
0.57
(Note: Use CQ /2 in place of CQ because of double entry.)
355
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Chapter 12 / Turbomachinery
12.26 Compute system demand:
L
f
D
V2
K
2g
Q VA 3
0.32
HP
P
4
Q
( gH P )3/4
12.27 From Fig. P12.27, at best
Q
CQ D3
HP
CH
2
0.01
31.4 0.212
(9.81 0.4)3/4
P , CQ
g
32
0.1
2 9.81
4
300
0.212 m 2 ,
0.212
0.049 0.33
D2
14
0.3
519
. ,
0.049, and CH
/ 30
0.40 m,
31.4 rad /s.
Axial pump is appropriate.
0.019.
160 rad /s , or N 160 30 /
0.019 160 2
9.81
0.32
1530 rpm,
4.46 m.
Since the head rise is too large for the demand, the pump is not well suited for the desired
application. In an attempt to reduce H P , try operating the pump at the extreme end of
the curve, i.e., CQ 0.055 and C H 0.005. Then:
0.212
0.055 0.33
0.005
and H P
143 rad /s (N 1370 rpm),
1143 2
9.81
0.3 2
0.94 m.
12.28 (a)
The intersection of the pump curve with the demand curve yields H P 64 m

From Fig. 12.6, W
64 kW and NPSH
8.3 m.
and Q 280 m 3 / h.
P
356
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Chapter 12 / Turbomachinery
(b)
The intersection of the two curves gives H P
From Fig. 12.6; with Q
7.6 m.
and NPSH
12.29 Q
67 m and Q 510m3 / h.

255 m 3 / h for one pump, W
2 60 120 kW,
P
p /( g ) 1000 103 /(9.81 607) 168 m,
600 / 3600 0.167 m 3 / s, H P
2500
/ 30 262 rad /s.
P
Q
( gH P )3/4
262 0167
.
(9.81 168)3/4
From Fig. 12.12, at best efficiency, CQ
Q
CQ
D
H
CH
2
g
D2
1/3
0.41,
use a radial-flow pump.
0.0165 and C H
0167
.
0.0165 262
0.125,
11/3
0125
262 2 0.338 2
.
9.81
0.338 m ,
100 m.
Since required pumping head is 168 m, use two stages.
357
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Chapter 12 / Turbomachinery
12.30 Use energy eqn. to establish system demand:
1.325 ln 0.27
f
HP
z
f
2
e
D
L Q2
D 2gA 2
From Fig. 12.6, at best
P
,Q
640
2
0.017 ,
640
0.017 3 5280
2
2
5Q
2 32.2 ( / 4) 15
.
1100
449
2.45 ft 3 / sec, and H P
Assume three pumps in series, so that H P
demand head is
HP
0.00085
1.5
1.325 ln 0.27
0.893
3
2.45 2
215
640 0.893Q 2 .
215 ft.
645 ft. Then the
645 ft.
Hence three pumps in series are appropriate. The required power is
W P
W P
or
12.31
QH P /
Given:
Q
62.4 0.86 2.45 645 / 0.75 1.13 10 5
P
1.13 10 5 / 550
500
7.48 60
206 hp.
1114
.
ft 3 / s, p 2
10
0.833 ft,
12
The required pumping head is
L 150 ft, D =
p2
HP
z
1
ft - lb
,
sec
80 144 11520 lb /ft 2 ,
z
fL V 2
D 2g
75 ft, and V2
11520
62.4
75
1
1114
.
0.7854 0.8332
2.044 ft /s
0.02 150 2.044 2
0.833
2 32.2
259.9 ft
The performance data for the mixed flow pump (Fig. P12.31b) is
CQ
0.0165, C H
0.75 .
0.124,
Determine the speed and head delivered by a one-stage impeller:
d
8
12
N
0.667 ft,
30
Q
1.114
CQ d3
0.0165 0.6673
=2180 rpm, and H
2 2
CH
d
g
358
228 rad/s
0.124 2282 0.667 2
32.2
89 ft.
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Chapter 12 / Turbomachinery
Number of stages required:
HP
H
260
89
2.92 (use three stages)
Power requirement:
62.4 1.114 3 89
0.75

W
P
2.48 10 4 ft - lb /sec , or 45 hp .
12.32 (a) For water at 80 C, pv = 46.4 103 Pa, and = 9553 kg/m3. Write the energy
equation from the inlet (section i) to the location of cavitation in the pump:
Vi2
2g
pi
pv
NPSH ,
pi
NPSH
pv
Vi2
2g
(83 46.4) 10 3
9533
62
2 9.81
5.67 m.
(b) NPSH1 = 5.67 m, N1 = 2400 rpm, N2 = 1000 rpm, D2/D1 = 4.
NPSH2
NPSH1
12.33 Compute HP and
HP
z2
z1
= 600
P
K
1.5
30
D2
D1
2
1000
5.67
2400
2
(4) 2 15.8 m.
fL Q 2
D 2 gA 2
0.02 60
1.25 2
0.75
2 9.81 (0.7854 0.75 2 ) 2
(b) From Fig. 12.13,
P
QHP
P
4.265 m
62.8 rad / s
62.8 1.25
(9.81 4.265)3/4
Q
( gH P )3/4

W
P
2
:
23 20
(a)
N2
N1
4.27 . Hence use an axial pump.
0.75.
9810 1.25 4.265
0.75
359
6.97 10 4 W, or approx. 70 kW.
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Chapter 12 / Turbomachinery
12.34 Given:
200 m, D
L
0.05 m, z
1593 kg /m 3 , p v
z2
z1
3 m, V
86.2 10 3 Pa, p a
3 m /s,
6 10
7
m 2 / s,
101 10 3 Pa.
Compute the pump head:
3 0.05
Re
2.5 105
7
6 10
2
1.325 ln 5.74Re-0.9
f
HP
1
z
fL V 2
D 2g
3
1.325 ln 5.74 (2.5 105 )
0.015 200
32
0.05
2 9.81
1
0.9
2
0.015
25.0 m
(a) Choose a radial-flow pump. Use Fig. P12.35 to select the size and speed:
0.124, CQ
CH
3
Q
D
CHQ 2
CQ2 gH P
Q
CQ D 3
0.75,
0.00589 m 2 ,
0.05 2
4
4
0.0165,
4
0.124 0.00589 2
0.0165 2 9.81 25.0
0.00589
0.0165 0.090 3
0.090 m
490 rad / s, or N
490
30
4680 rpm.
(b) Available net positive suction head:
NPSH
pa
pv
g
z
101 10 3 86.2 10 3
1593 9.81
3
3.95 m.
(c)
360
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Chapter 12 / Turbomachinery
1.2 kg/m 3 , U
12.35 Given: D 1.5 m,
20 m/s, A 2m2 , N
2500 rpm.
Compute speed, discharge, and CQ:
2500
262 rad /s, Q
30
UA
20 2
40 m 3 / s,
40
Q
0.045
3
262 1.53
D
Hence, from Fig. 12.13, CW = 0.0012, and CH = 0.02.
CQ
(a) Evaluate the power:

W
3
CW
D5
0.0012 1.2 262 3
1.5 5
1.97 10 5 W, or 197 kW.
(b) Compute the head rise across the fan, and the corresponding pressure rise:
HP
p
12.36
120
CH
2
D2
g
gH P
. 2
0.02 262 2 15
315 m,
9.81
1.2 9.81 315 3710 Pa.
/ 30 12.6 rad /s,
1
cot 1 (2 r12b1 / Q cot
cot 1 (2
Vt1
u1 Vn1 cot
12.6 4.5
Vt2
u2

W
T
Q( rV
1 t1
2
r2
2
2
r2Vt2 )
T
Q
cot
2 r2b 2
150
cot100
2.5 0.85
2
29.52 m /s,
1000 150(4.5 58.37 2.5 29.52)
T 12.6 2.83 107
Under ideal conditions
HT
0.85 12.6 /150 cot 75 ) 6.1 .
Q
r1
cot 1
2 r1b1
150
cot 75 58.37 m /s,
4.5 0.85
1
Vn2 cot
12.6 2.5
T
4.52
1)
2.83 10 7 N m.
.
357
108 W, or 357 MW.

1, and W
T
 , hence
W
f
 / Q 357
.
W
108 /(9810 150) 243 m.
T
361
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Chapter 12 / Turbomachinery
12.37
D1
D2
1 
, W1
5
3 kW, N 1

W
2
From the similarity rules

W
3
N2
N1
1
5.8 m.
1.8 m, H 2
360 rpm, H 1
5
D2
D1
H
and 2
H1
N2
N1
2
D2
D1
2
.
Eliminate (N2 /N1 ) and solve for W2 . From the second relation,
N2
N1
H2
H1
1/2
D1
,
D2
and substituting into the first relation,
3 /2
2
D2
 H2
W
1
H1
D1
Subsequently N 2 is determined:

W
2
N2
12.38
N2
N1
H2
H1
1/2

240 rpm, W
2
D1
D2
5.8
3
1.8
3 /2
5.8
360
1.8
N2
N1
1
1/2
1
5

3 ft, W
1
2200 kW, D2

W
2
From the similarity rules,

W
( 5) 2
3
5
D2
D1
434 kW.
129 rpm.
9kW, H1
H
and 2
H1
25 ft.
N2
N1
2
D2
D1
2
.
Substitute second eqn. into the first to eliminate ( N 2 / N 1 ), and solve for D 1 :
D1
N2
D2
N1
W1
W2
H1
H2
12.39 From Fig. P12.39,
1/2
1/2
H2
H1
3/4
D2
D1
25
240
150
0.91, C H
T
9
3
2200
1/2
1/2
150
25
3/4
3
0.736
0.23, and CW
0.736 ft,
399 rpm.
0.027. Use definitions of
dimensionless coefficients to determine D and :
gHT
1/2
1/3
WT
CH D2
CW D 5
From given data compute W T : WT
D
and
2.14
0.027
10 6
1000
9.81 80
0.23 0.6312
1/2
, and
QH T
T
0.23
9.81 80
D
WT
CW
1/2
362
3/4
9810 3 80 0.91 2.14 106 W,
3 /4
0.631 m,
1/2
92.6 rad /s ,
CH
gHT
or N
92.6
30
884 rpm.
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Chapter 12 / Turbomachinery
12.40 Write energy eqn. from upper reservoir (loc. 1) to surge tank (loc. 2) and solve for Q:
D
(z z )
2 gA
fL 1 2
1/2
2
Q
1/2
2 9.81 ( / 4) 2 0.855
(650 648.5)
0.025 2000
0.401 m 3 / s.
Apply energy eqn. from loc. 2 to lower reservoir (loc. 3) and determine H T :
HT
z2
z3
L
D
f
QHT
Q2
2 gA 2
0.4012
0.02 100
1
0.85
648.5 595

W
T
Kv
53.4 m.
2
5
0.85
4
.
9810 0.401 53.4 0.9 189
105 W, or 189 kW.
T
2 9.81
From Fig. 12.32, use a Francis turbine.
A representative value of the specific speed is 2 (Fig. 12.20):
( gHT )5/4
 / )1/2
(W
T
2(9.81 53.3)5/4
(2.67 105 /1000)1/2
T
or
N
306 30 /
12.41 Prototype: N 1
Model: N 2
2920 rpm.
420 rpm, H 1
D2
D1
2000 rpm;
W1
Q1 H1
306 rad /s ,
0.312 m3 / s,
3 m, Q1
H1
9810 0.312 3 0.9 8260 W.
1
Q1

W
2
 N2
W
1
N1
1 (1
3
2
D2
D1
D2
D1
Q2
2
2
N2
N1
N2
N1
0.9;
1
.
6
 and
Use similarity rules to compute H 2 , Q 2 , W
2
H2
1
2000
420
0.312
2000
420
3
D2
D1
D1
1)
D2
3
5
8260
2
2000
420
2
:
2
1
6
1.89 m,
1
6
3
3
1
6
0.0688 m 3 / s,
5
115 W ,
1/4
1 (1 0.9)(6)1/4
363
0.84.
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Chapter 12 / Turbomachinery
12.42
200
/ 30
20.94 rad /s, and from Fig. P12.42 at best efficiency (
T
= 0.8),
0.42, cv = 0.94.
c v 2 gHT
V1
WT
HT
Q
0.94 2 9.81 120
4.5 10 6
9810 120 0.8
T
45.6 m /s,
4.78 m3 / s
This is the discharge from all of the jets.
2 gHT
Determine the wheel radius r : r
Hence, the diameter of the wheel is 2 r
Compute diameter of one jet:
Let N j
2
Dj
0.42 2 9.81 120
20.94
0.973
2r / 8
0.973 m,
1.95 m.
1.95 / 8
0.244 m, or 244 mm.
no. of jets. Then each jet has a discharge of Q / N j and an area
Q /N j
4
V1
D 2j .
Solving for N j :
Use three jets.
T
Nj
 / )1/2
(W
T
( gHT )5/4
Q
V1
1
4
D
4.78
2
j
45.6
20.94(4.5 10 6 /1000)1/2
(9.81 120)5/4
12.43
Assume patm
p1
0.244
4
V1
Q
r12
V2
Q
r22
85
52
2
2.24,
0.204.
1.08 m /s,
85
2.52
4.33 m /s.
101 kPa
(a) Write energy eqn. from loc. 2 to loc. 1 and solve for p 2 (absolute):
p2
p1
2
(V12 V22 )
101 10 3
(b)
z
patm
pv
z
998
(1.08 2
2
HT
4.33 2 ) 9800 2.5
101 10 3 2340
9800
364
0.14
6.77
31.8
10 4 Pa, or 68 kPa.
5.62 m.
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Chapter 12 / Turbomachinery
12.44 Q
73,530 / 6 12,260 ft 3 / sec (one unit),
WT
Q T
HT
427,300 550
62.4 12 260 0.85
361.4 ft
Write energy eqn. from upper reservoir (loc. 1) to lake (loc. 2):
Q2
K
2 gA 2
L
f
D
z1
0.01 1300
D
1030
12,260 2
0.5
361.4 660,
2
2 32.2
4.919 10 7
D5
which reduces to 8.6

W
T
z2 ,
HT
D4
4
1.892 10 6
D4
0.
Solving, D 25.8 ft
361.4 0.3048 110 m.
437,000 0.746 326,000 kW, HT
From Fig. 12.32, a Francis or pump/turbine unit is indicated.
12.45 From Fig. P12.45,
480
D
/ 30
1/2
gH T
CH 2
Q CQ D3
WT
0.91, CQ
T
QH T
0.13, CH
50.3 rad/s,
1/2
9.81 9.5
0.23 50.32
013
.
50.3 0.43
T
0.23.
0.400 m ,
0.418 m 3 / s ,
9.81 0.418 9.5 0.9
3.55
10 4 W , or 35.5 kW.
12.46 (a) Let H be the total head and Q the discharge delivered to the turbine; then
HT
and Q
0.95H

W
T
HT T
0.95 305 289.8 m
10.4 10 6
9810 289.8 0.85
365
4.30 m 3 / s.
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Chapter 12 / Turbomachinery
Write energy eqn. from reservoir to turbine outlet:
H
Q2
,
K
2 gA 2
L
f
D
HT
0.02 3000
4.32
305 289.8
2
,
D
2 9.81 ( / 4) 2 D4
91.67 3.06
Solving D = 1.45 m.
which reduces to 15.2
0.
D5
D4
c v 2 gHT
(b) Compute jet velocity: V1
0.98 2 9.81 289.8
73.9 m /s
The flow through one nozzle is Q/4, and the jet area is D 2j / 4. Hence
D2j
Q /4
V1
4
4
Dj
4.3 / 4
73.9
0.0146
0.0146 m 2 ,
0.136 m.
12.47 Determine power available from each turbine:

W
T
T
( gHT )5/4
2
2.42 (9.81 3.7)5/4
1000
50
/ 30
2
1.69 10 6 W
The total power developed by all turbines is
QHT
T
9810 282 3.7 0.9 9.21 10 6 W
Hence, the required number of units is 9.21/1.69 = 5.4,

12.48 W
T
1000
4.15
(9.81 3.7 ) 5 /4
50
/ 30
use six turbines.
2
4.99
10 6 W (one unit) .
Total power developed is 9.21 10 6 W. Hence, required number of units is
9.21/ 4.99 1.8. Use two turbines.
12.49 (a)
Q
HT
Wf
1200
60 1000
0.02 m3 /s ,
Q2
z1 z2
fL
D
70 47
0.02 105
0.022
2
0.10
2 9.81 (0.7854 0.102 ) 2
QHT
K
2 gA2
15.4 m ,
9810 0.02 15.4 3020 W.
366
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Chapter 12 / Turbomachinery
(b) From Fig. P12.49a, C H
2
(c)
0.02
0.132
Q
CQ
D3
0.23, CW
0.152 , or
0.027 , CQ
2
D6
0.132 , and
0.91 .
T
0.0231
gH T 9.81 15.4
0.0231
657 ,
0.077 m ,
D 4
0.23
657
CH
0.152
30
0.152
=
333 rad /s, or N 333
3180 rpm.
3
3
0.077
D
D2
2
D1
1)
D2
1 (1
WT
12.50 (a) HT
1/4
1 (1 0.91)
1/4
1000
77
3020 0.83 2510 W
Wf
fL Q 2
D 2 gA 2
z1 z2
0.015 350 0.252
915 892
0.3 2 9.81 (0.7854 0.32 ) 2
WT
0.83
QHT
11.8 m ,
9810 0.25 11.8 0.85 2.46 104 W , or 24.6 kW.
T
(b) Compute the specific speed of the turbine:
N
1200
30
30
WT /
T
gHT
126 rad/s ,
126 2.46 104 /1000
5/4
9.81 11.8
1.65
5/4
Hence, from Fig. 12.20, a Francis turbine is appropriate.
(c) From Fig. 12.24, the turbine with T = 1.063 is chosen: CH = 0.23, CQ = 0.13, and
T = 0.91.
C Q2
0.23 0.25 2
4
0.293 m, or approximately 0.30 m ;
D 4 2H
0.13 2 9.81 11.8
CQ gHT
0.25
0.13 0.30 3
Q
CQ D 3

W
T
QHT
T
71.2 rad / s, or N
9810 0.25 11.8 0.91
71.2
30
680 rpm ;
2.63 10 4 W, or 26.3 kW .
Calculate a new specific speed based on the final design data:
71.2
T
2.63 104 / 1000
(9.81 11.8)5/4
0.96
an acceptable value according to Fig. 12.20.
367
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Chapter 12 / Turbomachinery
368
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Chapter 13 / Measurements in Fluid Mechanics
CHAPTER 13
Measurements in Fluid Mechanics
13.1
A reading of 4 cm provides a vertical measurement of 4 sin 20 
1.368 cm.
p 0.01368 (9810 13.6) 1825 Pa.
13.2
V2
a)
2
p
V
b)
13.3
Hg h
100
;
V2
h
,
1.22 (9810 13.6)
2
100
46.8 h and C
46.8 .
V2
h
(0.8217 1.22) (9810 13.6)
.
2
100
1
V2
2
h.
51.6 h , and C 51.6.
V
1
1.22 82 / 9810 0.00398 m or 3.98 mm.
2
h
The reading is too small for accurate measurements.
13.4
Q
V
t
0.5 1/7.481
1.114 10
10 60
4
ft 3 /sec
Q 1.94 1.114 2.16 10
m
Q
A
V
1.114 10
2
4
0.05 /144
4
2.04 fps.
slug/sec.
Re
2.04 0.1/12
10
5
1702.
The flow is laminar.
13.5
Q
0.5 2.6 0.5 6.65 1 8.65 9.5 9.9 10
V
13.6
Q [
12 10
Q
A
(32 22 ) 9.5
(4.52 42 ) 6.65
Q
A
4
2 0.854 m3 /s.
0.854
8.54 m/s.
0.1 1
(22 12 ) 9.9
V
100 10
0.0592
0.052
(42 32 ) 8.65
(52 4.52 ) 2.6] 10
4
0.0592 m3 /s.
7.54 m/s.
369
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Chapter 13 / Measurements in Fluid Mechanics
13.7
Hg h.
p
h
Re
VD
2g
p1
9.55 105.
6
p2
7.96 m/s. V0
0.062
7.96 0.12
10
31.8 2
1.02 2
0.09
V1
.
Manometer: p 1
p1
H
p2
Hg
Q
A0
K 1.02.
31.8 2
1.02 2
p2
H.
D0
D
31.8 m/s.
0.5.
K 2g(h1 h2 ).
V0
9810
486 000 Pa.
2 9.81
p1 p2
H.
Hg
486 000
9810(13.6 1)
H
13.8
p1 p2
See the sketch above:
H
3.93 m.
Hg
H.
p1 p2
14 830
9810
.
m.
1512
9810(13.6 1) 0.12 14 830 Pa.
Hg
h1
p1 p2
9810
h2
a) Assume Re 105.
K 0.68.
Q
1.45 m/s.
A
Check: V
Re
D0
D
15
24
0.625.
Q 0.68
0.0752 2 9.81 1.512
1.45 0.24
3.5 105.
10
6
0.0654 m3 /s.
K 0.67 and Q 0.064 m3 /s.
b) Assume Re 105.
Check: V
13.9
Q
A
K 1.05.
2.23 m/s.
2.23 0.24
Re
10
a) Re 105.
K 1.0.
Check: V
K
14 830
9810
h2
Q
A
1.01.
b) Assume Re 105.
Check: V
Q
A
5.4 105.
6
p1 9810 0.12 p2 13.6 9810 0.12.
h1
0.0752 2 9.81 1.512
Q 1.05
0.101 m3 s/.
OK.
p1 p2 14 830 Pa.
D0
D
6
12
0.5
1.512 m.
Q 1.0
0.0154
0.06
2
0.032 2 9.81 1.512
=1.36 m/s.
Re
0.0154 m3 /s.
1.36 0.12
0.661 10
6
2.5 105.
OK.
K
0.63.
0.0097
0.06
2
Q 0.63
0.858 m/s.
370
0.032 2 9.81 1.512
Re
0.858 0.12
0.661 10
6
0.0097 m3 /s.
=1.6 105.
OK.
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Chapter 13 / Measurements in Fluid Mechanics
13.10 Q
KA
R p
where K
D
a) Assume Re 105. Then K
Check V
12.4 m/s.
2
0.05
12.4 0.1
Re
10
0.035 4
Check: Re
0.05 10
c) Assume Re 106. Then K
(
4/12) 10
d) Assume Re 106. Then K
v1( , pT , p1) =
v1( +
5
1.3 106.
9
2
12
0.99
0.99
OK
10 5 .
2
(8/12) 10 144
(4/12) 1.94
3.33 cfs.
OK.
2
1
12
(8/12) 10 144
(2/12) 1.94
1.18 cfs.
OK.
2(102 95) 10 3 / 680
, pT , p1) = 2(102 95) 10 3 / 730
v1( , pT + pT , p1) =
2(103 95) 10 3 / 680
v1( , pT, p1 + p1) =
2(102 96) 103 / 680
v1
b)
5
K
OK.
0.99. Q 0.99
1.18 4
2 / 12 10
Check: Re
13.11 a)
0.99. Q
3.33 4
Check: Re
=1.2 106 .
0.035 m3 /s.
9 105.
6
6
0.0974 m3 /s.
Q 0.098 m3 /s.
0.2 80 000
0.05 1000
0.0252
0.2 80 000
0.1 1000
0.052
b) Assume Re 106. Then K 0.99.
Q 0.99
Q4
.
D
VD
Re
0.98. Q 0.98
0.0974
Q
A
6.5
.
Re
1
4.537 m/s,
4.379 m/s,
4.581 m/s,
4.201 m/s,
(4.379 4.537)2 (4.851 4.537)2 (4.201 4.537) 2
0.486 m/s
v1( , pT p1) = 4.537 m/s
v1( + , pT p1) = 4.379 m/s
v1[ , pT p1 + (pT p1)] = 4.851 m/s
v1
(4.379 4.537)2 (4.851 4.537)2
0.3515 m/s
c) Arrangement (b) is preferable, since v1 is smaller (one less reading, resulting in one
less uncertainty measurement).
371
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Chapter 13 / Measurements in Fluid Mechanics
2
13.12 A0
K
33.3
8.709 10 4 m 2 ,
33.3 / 54
4
1000
Q
( h in meters of mercury)
A0 2 g(S 1) h
8.709
K
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
10
[ K(Q , h)
K (Q, h)
1.076
1.024
1.017
1.026
1.031
1.041
1.027
1.025
1.005
1.025
1.041
1.027
1.043
1.040
0.9935
4
Q
2 9.81 ( 13.6
K(Q
Q , h)] 2
1) h
73.03
[ K(Q , h)
K (Q Q, h)
1.147
1.093
1.082
1.095
1.100
1.112
1.096
1.084
1.063
1.083
1.099
1.085
1.103
1.109
1.050
0.617
Q
h
K(Q, h)
K(Q , h
K(Q, h ( h))
1.002
0.9805
0.9862
1.001
1.012
1.025
1.012
1.013
0.9938
1.015
1.032
1.018
1.036
1.033
0.9864
( h))] 2
K
0.1026
0.0816
0.0719
0.0734
0.0716
0.0728
0.0706
0.0602
0.0591
0.0589
0.0587
0.0587
0.0604
0.0694
0.0569
Compare the above with Fig. 13.10, curve labeled “Venturi meters and nozzles,”
372
0.6.
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Chapter 13 / Measurements in Fluid Mechanics
13.13 From Example 13.1:
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
13.14
K /K
0.1173
0.0859
0.0789
0.0703
0.0633
0.0628
0.0622
0.0614
0.0507
0.0507
0.0455
0.0449
0.0455
0.0448
0.0441
i
(S
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
(m water)
0.164
0.277
0.403
0.504
0.680
0.781
0.882
1.033
1.147
1.260
1.424
1.550
1.688
1.764
1.764
(1)
m
Relative uncertainty
xi
1) h
x i( 1)
K /K decreases with increasing Re.
y i( 2)
-1.8079
-1.2837
-0.9088
-0.6852
-0.3857
-0.2472
-0.1256
0.03247
0.1371
0.2311
0.3535
0.4383
0.5235
0.5676
0.5676
-2.5929
-6.3890
-6.1754
-5.9955
-5.8746
-5.7199
-5.6408
-5.5940
-5.5165
-5.4846
-5.4171
-5.3412
-5.3124
-5.2533
-5.2344
-5.2805
-84.2292
ln[(S 1) h], S 13.6 ( h in meters mercury)
18.1376 ( 2.5929)( 84.2292) / 15
7.7464 ( 2.5929) 2 / 15
xi y i
11.5507
7.9274
5.4487
4.0253
2.2062
1.3944
0.7026
-0.1791
-0.7519
-1.2519
-1.8881
-2.3284
-2.7501
-2.9710
-2.9972
+18.1376
(2)
yi
x i2
3.2685
1.6479
0.8259
0.4695
0.1488
0.06111
0.01578
0.001054
0.01880
0.05341
0.1250
0.1921
0.2741
0.3222
0.3222
+7.7464
ln Q (Q in m3 /s)
0.4902 ,
0.4902( 2.5929)
C exp( 5.5305) 0.00396 .
5.5305 ,
15
From Problem 13.12, A0 8.709 10 4 m 2 and K avg 1.029 . Hence in Eq. 13.3.8,
b
84.2292
Kavg A0 2 g 1.029 (8.709 10 4 ) 2 9.81 0.00397 , and the exponent is 0.500.
373
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Chapter 13 / Measurements in Fluid Mechanics
13.15
xi
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
-0.9493
-0.9729
-0.9835
-1.0300
-1.0906
-1.1394
-1.1648
-1.1842
-1.1973
-1.2140
-1.3356
-1.4106
-1.4355
-1.5799
-16.6876
yi
ln Q
-1.9733
-2.0402
-2.0557
-2.1716
-2.3403
-2.3677
-2.5145
-2.5498
-2.5889
-2.6636
-2.9604
-3.1350
-3.1749
-3.4451
-35.9810
x i2
xi y i
1.8733
1.9849
2.0218
2.2367
2.5523
2.6978
2.9289
3.0195
3.0997
3.2336
3.9539
4.4222
4.5576
5.4429
+44.0251
0.9012
0.9465
0.9673
1.0609
1.1894
1.2982
1.3568
1.4023
1.4335
1.4738
1.7838
1.9898
2.0549
2.4961
+20.3545
m
44.0251 ( 16.6876)( 35.9810) / 14
20.3545 ( 16.6876) 2 / 14
b
35.9810 2.4533( 16.6876)
14
C
exp(0.3542) 1.425 and Q 1.425 Y 2.45
In Eq. 10.4.27, use
Q
ln Y
60  , C d
2.4533 ,
0.3542 ,
0.58; then
60  5 / 2
8
2 32.2 tan
0.58
Y
2
15
374
1.433Y 5 / 2
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Chapter 14 / Computational Fluid Dynamics
CHAPTER 14
Computational Fluid Dynamics
14.3
In order to derive the backward difference operator we use Eqs.14.2.13 and 14.2.14. We
eliminate the third term involving ( 2u / y2 ) j in both equations by multiplying Eq.14.2.13
by ( 4) and adding it to Eq.14.2.14:
4u j
uj
Then u j
4u j
1
u
y
uj
2
4u j
2
u
y2
j
u
y2
u
y
2
j
j
j
2
4u j
uj
1
uj
2
O
y4
3
3
u
y3
2
j
y2
3!
O
y3 .
y2
O
2 y
y4
y4 .
O
2
O
3!
j
2 y
1
j
y3
3!
j
y3
3!
2 y
u
y3
u
y3
j
4u j
j
u
y3
3
3
y 4
j
3u j
3
4
2!
3u j
u
y
y2
2!
2 y
2
j
we get
u
y
2
y 4
2 y
3u j
1
u
y
Solving for
u
y
4
To derive the forward difference operator we use the following Taylor series equations
uj
uj
u
y
uj
1
u
y2
j
u
y
uj
2
2
y
y2
2!
j
u
y2
j
u
y3
2 y
2
2 y
3
j
2
3
u
y3
2!
j
y3
3!
O
y4
2 y
3
3!
j
O
y4
The first equation is multiplied by ( 4) and added to the second equation:
4u j
uj
4u j
1
2
Then u j
2
4
u
y
uj
4u j
u
y
1
2
u
y2
y 4
j
2 y
j
3u j
2 y
2
u
y2
2
u
y
j
y2
2!
2
y 4
j
375
u
y3
4
u
y3
3
u
y3
j
j
y3
3!
2 y
3
2!
j
3
j
y3
3!
3!
O
O
y4
O
y4
3
y4
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Chapter 14 / Computational Fluid Dynamics
u
y
Solving for
we get:
j
3u j
u
y
4u j
1
uj
2
2 y
j
u
y
3u j
3
u
y3
2
4u j
uj
1
j
2
O
2 y
j
y2
3!
O
y3
y2
14.5 The system is written in matrix form as Ax = b such that
2
2
0
0
x1
16
4
0
1
5
2
3
0
6
x2
x3
3
10
0
0
4
3
x4
6
Since matrix A is a tri-diagonal matrix the Gauss-Elimination method can be used to
eliminate elements a21, a32, and a43 to transform A into an upper triangular matrix as
2
2
0
0
x1
16
0
0
5
0
2
1
0
6
x2
x3
29
39
0
0
0 27
x4
162
Using backward substitution we solve the system to get
14.6
x4
162 27 6
x2
29 2 x3
5
x3
39 6 x4
39 36 3
x1
16 2 x2
2
1
7
Start by writing Eq.14.2.26:
unj
unj
1
unj 11 2unj
1
y2
t
Multiplying the above equation by
unj 11
1 2
unj 11
unj
1
unj 11
1
t , using
unj 1
1
unj 1 2unj unj 1
y2
t
y2 , and rearranging yields
1 21
unj
1
unj 1
The above system of equations can be written in matrix form as
376
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Chapter 14 / Computational Fluid Dynamics
1 2
1 2
A
1 2
,x
1 2
b
1
u3n
1 21
u2n
1
V0
1
n
4
u
1 21
n
3
u
1
u2n
1
u5n
1 21
u4n
1
u3n
n
uJL
1
n
uJL
1 21
2
u2n
u3n
u4n
1
1
1
n 1
uJL
1
1
14.10 To determine the consistency of Eq.14.2.21, which is given by,
unj
unj
1
unj 11 2unj
unj 11
1
O
y2
t
t , y2
we expand unj 11 , unj 11 and unj about unj 1 using Taylor series as follows
u
n 1
j 1
u
unj 11
unj
n 1
j
unj
unj
1
u
y
n 1
u
y
n 1
u
t
1
2
n 1
2
n 1
u
y2
y
j
u
y2
y
j
n 1
2
u
t2
t
j
j
j
n 1
j
y2
2!
n 1
3
n 1
u
y3
y2
2!
t2
2!
3
u
y3
3
u
t3
y3
3!
j
j
n 1
j
y3
3!
t3
3!
Substituting the above equations in Eq.14.2.21 yields
u
t
Now, as
n 1
j
n 1
2
u
t2
j
3
t
2!
u
t3
n 1
j
t2
3!
2
u
y2
n 1
2
j
4
u
y4
n 1
j
y2
4!
y approach zero the above equation reduces to
t and
u
t
n 1
j
2
u
y2
n 1
j
Which is identical to the original PDE, and hence the FDE given by Eq.14.2.21 is
consistent.
377
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Chapter 14 / Computational Fluid Dynamics
14.11 To determine the consistency of the following FDE
unj
unj
1
unj 1 unj
1
unj
1
unj 1
1
y2
2 t
t4
y2
t , y2 ,
O
We expand all the terms about unj using Taylor series as follows
u
n 1
j
unj
u
u
unj
1
n
j 1
u
unj 1
n
u
t
n
j
n
j
unj
j
n
u
y
n
u
y
n
n
2
n
2
n
2
n
u
t2
t
u
t
2
j
u
t2
t
j
j
u
y2
y
j
j
u
y2
y
j
j
t2
2!
3
n
3
n
u
t3
t2
2!
j
u
t3
y2
2!
t3
3!
j
3
n
3
n
u
y3
y2
2!
t3
3!
u
y3
j
j
y3
3!
y3
3!
Substituting the above equations into the FDE we get
u
t
n
3
u
t3
j
n
j
t2
3!
5
u
t5
n
2
u
y2
j
n
j
2
u
t2
2
u
t2
t2
Rearranging and letting
u
t
j
t4
5!
t and
Simplifying, and letting
u
t
n
2u
y
n
j
2
u
y2
2
2
2u
n
j
2
2
u
t2
n
y2
2!
j
n
j
t2
2!
2
2
4
u
y4
4
u
t4
n
y4
4!
j
n
j
t4
4!
y approach zero the above equation reduces to
n
j
t2
y2
y2 yields
2
u
y2
Hence, the given FDE is consistent.
378
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Chapter 14 / Computational Fluid Dynamics
14.14 To determine the numerical stability of Eq.14.2.21, we determine if the condition given
by Eq.14.3.21 is always satisfied, that is, Gm 1 where Gm Amn 1 Amn . For the FDE
as follows:
given by Eq.14.2.21 we write the evolution equation for the error
n 1
j
n
j
n 1
j 1
2
t
n 1
j
2
n 1
j 1
y
Next, the error is represented in terms of a Fourier series (see Eq.14.3.17),
n
j
m
Amn exp Ikm xj
Substituting the above in to the error equation we get
Amn 1
Ikm y j
Amn e
Ikm y j
2
t
m
y
y
Amn 1e
Ikm y j
2 Amn 1e
Ikm y j
Amn 1e
y
0
Since there are no interactions between the Fourier components, the above equation
requires every component to be zero, that is
Ikm y j
Amn 1
Amn e
t
Ikm y j
Divide by e
Ikm y j
y
2
Ikm y j
2 Amn 1e
Ikm y j
Amn 1e
y
0
and simplify:
t n 1 Ikm y
Am e
2 Amn 1
2
y
Using the relationship exp Ikm y
yields
Amn 1
Amn 1e
Ikm y
cos km y
t n1
Am 2 cos km y
y2
Amn 1
y
Amn 1e
Amn
I sin km y in the above equation
Amn
2
which can be further simplified to
Amn 1 2
t n1
A
2sin2 km y 2
2 m
y
Amn
And hence from the above equation the stability condition is
Gm
1
Amn 1 Amn
1 4
t 2
sin km y 2
y2
which is always less than one, and hence the FDE is unconditionally stable.
379
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Chapter 14 / Computational Fluid Dynamics
unj
14.15 (a) For the FDE given by
unj
1
unj 1 2unj unj 1
1
2 t
y
n 1
j
the corresponding disturbance equation is
t2 , y2 ,
O
2
n 1
j
n
j 1
2
2 t
n
j
2
n
j 1
y
which can be represented using a Fourier series as
Ikm y j
Amn 1
Amn 1 e
Ikm y j
2 t
m
Amn e
2
y
y
Ikm y j
Ikm y j
2 Amn e
Amn e
y
0
Since there are no interactions between the Fourier components, the above equation
requires every component to be zero, that is
Amn 1
Ikm y j
Amn 1 e
Ikm y j
2 t
y
y
Amn e
2
Ikm y j
2 Amn e
Ikm y j
Amn e
y
0
t n Ikm y
2 Amn Amn e Ikm y
Am e
2
y
Note that in the above equation we can substitute Amn 1 Gm Amn to give
Ikm y j
Divide by e
and simplify:
t n Ikm
Am e
y2
Gm Amn 2
y
2 Amn
Amn Gm 2
t Ikm
e
y2
Amn Gm
t
2 cos km y
y2
2
y
2 e
Amn 1 2
Amn e
Ikm y
Ikm y
Amn 1
Amn 1
Amn 1
Amn 1
2
Further simplification yields
Amn Gm 8
Now, using Amn
Gm Gm 8
k y
t
sin2 m
2
2
y
Amn 1
Gm Amn 1 and substituting in the above equation we get
k y
t
sin2 m
2
2
y
1
which can be written as Gm2 Gm 8
k y
t
sin2 m
2
y
2
1 0
whose roots indicate that Gm 1 and hence the method is unconditionally unstable
and it cannot be used. This method is known as Richardson’s method and usually is
presented for historic purposes only.
380
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Chapter 14 / Computational Fluid Dynamics
(b) For the FDE given by:
unj
1
unj
unj 1 unj
1
unj
1
unj 1
1
y2
2 t
t , y2 ,
O
t4
y2
The corresponding disturbance equation is
n 1
j
n 1
j
n
j 1
n 1
j
n 1
j
n
j 1
y2
2 t
Following a similar procedure for part (a) we write
Amn 1 2
t n Ikm
Am e
y2
y
Amn 1
Amn 1
Amn e
t
2 Amn cos km y Amn 1
2
y
Amn 1
Ikm y
Amn 1
Further simplification yields
Amn 1 2
For convenience let r
Amn 1 1 2r
Now substitute Amn
y2 and
t
Amn 4r cos
km y and substitute in the above:
Amn 1 1 2r
Gm Amn to get
1
Amn Gm 1 2r
Amn 1
Amn 4r cos
Amn 1 1 2r
which can be written as
Amn Gm 1 2r
4r cos
Next, we substitute Amn
Gm Gm 1 2r
Gm Amn 1 in the above equation:
4r cos
Gm2 1 2r
Its roots are Gm
2r cos
unconditionally stable.
Amn 1 1 2r
4r cos
1 2r
Gm
1 2r
1 4r2 sin2
1 2r
381
0
0
or Gm
1 and hence the method is
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Chapter 14 / Computational Fluid Dynamics
14.16 Start by writing Eq.14.2.26:
unj
unj
1
unj 11 2unj
unj 11
1
y2
t
unj
1 2
unj 11
1
y2
t
t , using
Multiplying the above equation by
unj 11
unj 1 2unj unj 1
1
unj 1
1
y2 , and rearranging yields
unj
1 21
1
unj 1
The corresponding disturbance equation is
n 1
j 1
n 1
j
1 2
n 1
j 1
n
j 1
1
n
j
1 21
1
n
j 1
which can be represented using a Fourier series as
eIkm
Amn 1
y
e
1 2
eIkm
Amn 1
y
Ikm y
1 21
1
e
Ikm y
It can be further simplified:
Amn 1
cos km y
2
Amn 2 1
1 2
cos km y
1 21
Hence, the amplification factor Gm is given by
Gm
cos km y
21
2
Gm
1 21
cos km y
sin2 km y 2
1 41
1 4
1 2
sin2 km y 2
The above equation shows that if 0.5
1 then Gm 1 for all values of and hence the
method is unconditionally stable. However, if 0
0.5 the method is stable only if
1
1 41
1 4
sin2 km y 2
sin2 km y 2
1
which yields the stability criterion
1
2 4
382
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Chapter 14 / Computational Fluid Dynamics
14.19 For transient one-dimensional fluid flow the governing differential equation is
2
u
t
u
y2
(a) The finite difference equation in explicit form is
unj
1
Runj 1
1 2R unj
Runj 1
Applying the above equation for node 3 we write
u3(2)
0.3u4(1)
1 2 0.3 u3(1) 0.3u2(1)
0.3 27 58
0.4 34 39.1 m/s
(b) The finite difference equation in implicit form is (see Eq. 14.2.22)
unj
unj 11
R
Note that
u31
unj
1 2
unj 11
1
0.3 . The finite difference equation for node 3 is written as
0.3u42
1 2 0.3 u32
0.3u22
0.3 30 1.6u32
34
43 1.6u32
0.3u22
0.3u22 ................………........(I)
In order to determine u32 we need to write the FDE for node 2 as
u21
0.3u32
1 2 0.3 u22
0.3u1 2
58
0.3u32
1.6u22
94
0.3u32
1.6u22 ……………………...(II)
0.3 120
Solving Eqs. (I) and (II) simultaneously yields
u3
39.3 m/s
(c) The Crank-Nicolson finite difference equation is given by Eq.14.2.26 with
written as
unj 11 2 1
unj
1
unj 11
unj 1 2 1
unj
1 2 and is
unj 1
Applying the above equation to nodes 2 and 3 yields the following two FDEs
0.3u32
2 1 0.3 u22
0.3u1 2
0.3u32
2.6u22
2
3
2
2
0.3u
2.6u
0.3u31
2 1 0.3 u21
0.3u11
0.3 120 0.3 34 1.4 58 0.3 100
157.4 …………………………(III)
383
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Chapter 14 / Computational Fluid Dynamics
0.3u42
2 1 0.3 u32
0.3u22
0.3u41
0.3 30 2.6 u32
0.3u22
0.3 27 1.4 34 0.3 58
2.6 u32
0.3u22
2 1 0.3 u31
0.3u21
82.1 ………………………….(IV)
Solving Eqs. (III) and (IV) simultaneously yields u3
39.1 m/s
14.20 For steady two-dimensional potential flow the governing differential equation is:
2
2
x2
y2
0
(a) The corresponding finite differences equation is
2
i 1, j
i, j
2
i 1, j
2
i, j 1
i, j
2
x
Since
i, j 1
y
0
y , the above equation is written as
x
i, j
1
4
i 1, j
i, j 1
i 1, j
i, j 1
(b) Applying the above equation for nodes 1, 2, and 3 we write the finite difference equations
for 1, 2, and 3 as
1
1.73 2.25 1.38 1.37 1.68
4
1
1.29 0.46 2 1.04 0.958
4
1
0.46 0.67 2 1.04 0.803
4
1
2
3
14.21 (a) The finite difference equation in explicit form is
uin, j 1
t
x
2
t
x2
If x = y, and using
uin, j 1
t
uin 1, j 2uin, j uin 1, j
uin 1, j uin, j
1
y
t
y2
uin 1, j uin, j
2
uin, j
1
2uin, j uin, j
1
, the above equation can be re-written as
1
384
1 4
uin, j
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Chapter 14 / Computational Fluid Dynamics
(b) The finite difference equation in implicit form is
t
t
uin, j 1 uin, j
uin 1,1 j 2uin, j 1 uin 1,1 j
uin, j 11 2uin, j 1 uin, j 11
2
2
x
y
t
x2
If x = y, and using
uin, j 1
1 4
t
, the above equation can be re-written as
y2
uin 1,1 j uin, j 11 uin 1,1 j uin, j 11
uin, j
(c) Using the Crank-Nicolson method the FDE is
t n1
ui 1, j 2uin, j 1 uin 1,1 j uin 1, j 2uin, j uin 1, j
2
2 x
t n1
ui , j 1 2uin, j 1 uin, j 11 uin, j 1 2uin, j uin, j 1
2
2 y
uin, j 1 uin, j
t
x2
If x = y, and using
uin, j 1
21 2
t
, the above equation can be re-written as
y2
uin 1,1 j uin 1,1 j uin, j 11 uin, j 11
uin, j
21 2
uin 1, j uin 1, j uin, j
1
uin, j
1
14.22 For steady two-dimensional potential flow the governing differential equation is:
2
2
x2
y2
0
The corresponding finite differences equation is
i 1, j
2
i, j
2
i 1, j
i, j 1
2
x
Since
x
i, j
2
i, j 1
y
0
y , the above equation is written as
i, j
1
4
i 1, j
i, j 1
i 1, j
i, j 1
When the above equation is written for node 5 we have
1
2
4
6
8
4
Using the Gauss-Seidel iterative method we write
5
k 1
5
1
4
k 1
2
k 1
4
2
5
k 1
5
1
4
k
6
2
2
k
8
2
4
1
6
1
8
0.445 0.434 0.387 0.307 / 4 0.393
385
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Chapter 14 / Computational Fluid Dynamics
386
© 2012 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.