66.-Merle C. Potter, David C. Wiggert, Bassem H. Ramadan - Mechanics of Fluids - Instructor’s solution manual-Cengage Learning (2012).pdf

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
© 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.
1
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 [τ du/dy] = (F/L2
)/(L/T)/L = F.
T/L2
.
1.3 (A) 8 9
2.36 10 23.6 10 23.6 nPa.
1.4 (C) The mass is the same on earth and the moon: [4(8 )] 32 .
du
r r
dr
1.5 (C) shear sin 4200sin30 2100 N.
F F
3
shear
4 2
2100 N
= 84 10 Pa or 84 kPa
250 10 m
F
A
1.6 (B)
1.7 (D)
2 2
3
water
( 4) (80 4)
1000 1000 968 kg/m
180 180
T
1.8 (A) 3
[10 5000 ] 10 10 5000 0.02 1 Pa.
du
r
dr
1.9 (D) 3 2 6
4 cos 4 0.0736 N/m 1
3 m or 300 cm.
1000 kg/m 9.81 m/s 10 10 m
h
gD
We used kg = N·s2
/m
1.10 (C)
1.11 (C)
pV
m
2 3
800 kN/m 4 m
59.95 kg
0.1886 kJ/(kg K) (10 273) K
RT

Chapter 1 / Basic Considerations
2
1.12 (B) ice water ice water water
. 320 .
E E m m c T
6 3
5 (40 10 ) 1000 320 (2 10 ) 1000 4.18 . 7.66 C.
T T
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, 287 323 304 m/s.
c RT
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 L
/ 3
b) pressure = force/area = F L ML T L M LT
/ / /
2 2 2 2
c) power = force velocity = F L T ML T L T ML T
/ / / /
2 2 3
d) energy = force distance = ML T L ML T
/ /
2 2 2
e) mass flux = ρAV = M/L3
× L2
× L/T = M/T
f) flow rate = AV = L2
× L/T = L3
/T
1.16 a) density =
M
L
FT L
L
FT L
3
2
3
2 4
/
/
b) pressure = F/L2
c) power = F × velocity = F L/T = FL/T
d) energy = F×L = FL
e) mass flux =
M
T
FT L
T
FT L
2
/
/
f) flow rate = AV = L2
L/T = L3
/T
1.17 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] = L T L L L T
3 2 2 3 1 3
/ / /
Note: the slope S0 has no dimensions.
1.18 a) m = [C] s2
. [C] = m/s2
b) N = [C] kg. [C] = N/kg = kg m/s2
kg = m/s2
c) m3
/s = [C] m2
m2/3
. [C] = m3
/s m2
m2/3
= m1/3
/s

3
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
d) viscosity: N s/m2
=
kg m
s
s
1
m
kg / m s
2 2
e) heat flux: J/s =
N m
s
kg m
s
m
s
kg m / s
2
2 3
f) specific heat:
J
kg K
N m
kg K
kg m
s
m
kg K
m / K s
2
2 2
1.20 kg
m
s
m
s
m
2
c k f. Since all terms must have the same dimensions (units) we require:
[c] = kg/s, [k] = kg/s2
= N s / m s N / m,
2 2
[f] =kg m / s N.
2
Note: we could express the units on c as [c] = kg / s N s / m s N s / m
2
1.21 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm3
e) 1.2 cm2
f) 76 mm3
1.22 a) 1.25 108
N b) 3.21 10 5
s c) 6.7 108
Pa
d) 5.6 m3
e) 5.2 10 2
m2
f) 7.8 109
m3
1.23 2 2 2
0.06854
0.225 0.738
0.00194 3.281
m m
d d
where m is in slugs, in slug/ft3
and d in feet. We used the conversions in the front cover.
1.24 a) 20 cm/hr = 5
20/100
5.555 10 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
g) 500 g/L = 500 10 3
kg/10 m 500 kg/m3
h) 500 kWh = 500 1000 3600 = 1.8 109
J
1.25 a) F = ma = 10 40 = 400 N.
b) F W = ma. F = 10 40 + 10 9.81 = 498.1 N.
c) F W sin 30 = ma. F = 10 40 + 9.81 0.5 = 449 N.
1.26 The mass is the same on the earth and the moon:
m =
60
32 2
1863
.
. . Wmoon = 1.863 5.4 = 10.06 lb

4
1.27 a)
26
6
2 10 2
4.8 10
0.225 0.225 0.43 10 m
0.184 (3.7 10 )
m
d
or 0.00043 mm
b)
26
5
2 10 2
4.8 10
0.225 0.225 7.7 10 m
0.00103 (3.7 10 )
m
d
or 0.077 mm
c)
26
2 10 2
4.8 10
0.225 0.225 0.0039 m
0.00002 (3.7 10 )
m
d
or 3.9 mm
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 31 = 70 kPa abs. b) 760
31
101
760 = 527 mm of Hg abs.
c) 14.7
31
101
14.7 = 10.2 psia. d) 34
31
101
34 = 23.6 ft of H2O abs.
e) 30
31
101
30 = 20.8 in. of Hg abs.
1.30 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
% error =
62.8 61.6
61.6
100 = 1.95 %.
1.31 a) p = 973 +
22,560 20,000
25,000 20,000
(785 973) = 877 psf
T = 12.3 +
22,560 20,000
25,000 20,000
( 30.1 + 12.3) = 21.4 F
b) p = 973 + 0.512 (785 973) +
0.512
2
( .488) (628 2 785 + 973) = 873 psf
T = 12.3 + 0.512 ( 30.1 + 12.3) +
0.512
2
( .488) ( 48 + 2 30.1 12.3) = 21.4 F
Note: The results in (b) are more accurate than the results in (a). When we use a linear
interpolation, we lose significant digits in the result.
1.32 T = 48 +
33,000 30,000
35,000 30,000
( 65.8 + 48) = 59 F or ( 59 32)
5
9
= 50.6 C

5
1.33 p = n
F
A
= 4
26.5 cos 42
152 10
= 1296 MN/m2
= 1296 MPa.
1.34
4
4
(120000) 0.2 10 2.4 N
20 0.2 10 0.0004 N
n
t
F
F
F = 2 2
n t
F F = 2.400 N.
= tan 1 0.0004
2.4
=0.0095
Density and Specific Weight
1.35 =
m
V
0 2
180 1728
.
/
= 1.92 slug/ft3
. = g = 1.92 32.2 = 61.8 lb/ft3
.
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
% error for =
976 978
978
100 = .20%
% error for =
9560 978 9.81
978 9.81
100 = .36%
1.37 S = 13.6 0.0024T = 13.6 0.0024 50 = 13.48.
% error =
13.48 13.6
13.6
100 = .88%
1.38 a) m =
W V
g
6
12 400 500 10
9.81
g
= 0.632 kg
b) m =
6
12 400 500 10
9.77
= 0.635 kg
c) m =
6
12 400 500 10
9.83
= 0.631 kg
1.39 S =
/
water
m V 10/
. 1.2
water
V
.
1.94
V = 4.30 ft3

6
Viscosity
1.40 Assume carbon dioxide is an ideal gas at the given conditions, then
3
3
200 kN/m
2.915 kg/m
0.189 kJ/kg K 90 273 K
p
RT
3 2 2 2 3
2.915 kg/m 9.81 m/s 28.6 kg/m s 28.6 N/m
W mg
g
V V
From Fig. B.1 at 90°C, 5 2
2 10 N s/m , so that the kinematic viscosity is
5 2
6 2
3
2 10 N s/m
6.861 10 m /s
2.915 kg/m
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
piston
W 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
0
/ 2
piston
cylinder piston
V
V
r D D
Using piston piston
W m g , we can write
/ 2
piston
piston
cylinder piston
V
m g DL
D D
Solve :
piston
V
2
2 3
2 2
2
0.350 kg 9.81 m/s 0.1205 0.120 m
0.91 kg m /N s 0.91 m/s
2 0.025 N s/m 0.12 0.10 m
piston cylinder piston
piston
m g D D
V
DL
where we used N = kg·m/s2
.

7
1.42 The shear stress can be calculated using / .
du dy From the given velocity
distribution,
2
( ) 120(0.05 )
u y y y 120(0.05 2 )
du
y
dy
From Table B.1 at 10 C, 3 2
1.308 10 N s/m so, at the lower plate where y = 0,
1 3 3 2
0
120(0.05 0) 6 s 1.308 10 6 7.848 10 N/m
y
du
dy
At the upper plate where y = 0.05 m,
1 3 2
0.05
120(0.05 2 0.05) 6 s 7.848 10 N/m
y
du
dy
1.43 =
du
dr
= 1.92 2
30(2 1/12)
(1/12)
= 0.014 lb/ft2
1.44 2
30(2 1/12)
(1/12)
2 2
0 0
[32 / ] 32 / .
du
r r r r
dr
r = 0 = 0,
r = 0.25 = 32 1 10 3
2
0.25/100
(0.5/100)
= 3.2 Pa,
r = 0.5 = 32 1 10 3
2
0.5/100
(0.5/100)
= 6.4 Pa
1.45 T = force moment arm = 2 RL R =
du
dr
2 R2
L = 2
0.4
1000
R
2 R2
L.
=
2 2
2
0.0026
0.4 0.4
1000 2 1000 2 .01 0.2
12
T
R L
R
= 0.414 N.
s/m2
.
1.46 Use Eq.1.5.8: T =
3
2 R L
h
=
3 2000 2
2 0.5/12 4 0.006
60
0.01/12
= 2.74 ft-lb.
power =
T
550
2 74 209
550
. .4
= 1.04 hp

8
1.47 Fbelt = 3 10
1.31 10
0.002
du
A
dy
(0.6 4) = 15.7 N.
power =
F V
746
15 7 10
746
.
= 0.210 hp
1.48 Assume a linear velocity so .
du r
dy h
Due to the area
element shown, dT = dF r = dA r =
du
dy
2 r dr r.
dr
r
T = 3
0
2
R
r dr
h
=
5 4
4
400 2
2.36 10 (3/12)
2 60
4 2 0.08/12
R
h
= 91 10 5
ft-lb.
1.49 The velocity at a radius r is r . The shear stress is
u
y
.
The torque is dT = rdA on a differential element. We have
0.08
0
= = 2
0.0002
r
T rdA rdx ,
2000 2
209.4 rad/s
60
where x is measured along the rotating surface. From the geometry 2
x r, so that
0.08 0.08
2 3
0 0
209.4 / 2 329 000
= 0.1 2 329 000 (0.08 )
0.0002 3
2
x x
T dx x dx = 56.1 N .
m
1.50 If
du
dy
= cons’t and = AeB/T
= AeBy/K
= AeCy
, then
AeCy du
dy
= cons’t.
du
dy
= De Cy
.
Finally, or u(y) = 0
y
Cy
D
e
C
= E (e Cy
1) where A, B, C, D, E, and K are constants.
1.51 /
B T
Ae
/293
/353
0.001
0.000357
B
B
Ae
Ae
A = 2.334 10 6
, B = 1776.
40 = 2.334 10 6
e1776/313
= 6.80 10 4
N.
s/m2

9
Compressibility
1.52 m = V . Then dm = d V + V d . Assume mass to be constant in a volume subjected
to a pressure increase; then dm = 0. d V = V d , or
d V
V
.
d
1.53 B =
V p
V
2200 MPa. V
V 2 10
2200
p
B
= 0.00909 m3
or 9090 cm3
1.54 Use c = 1450 m/s. L = c t = 1450 0.62 = 899 m
1.55 =
B V
p
V
= 2100
13
20
.
= 136.5 MPa
1.56 a) 327,000 144/1.93
c = 4670 fps b) 327,000 144/1.93
c = 4940 fps
c) 308,000 144/1.87
c = 4870 fps
1.57 V =3.8 10 4
20 1 = 0.0076 m3
. p = B
V
V
0.0076
2270
1
= 17.25 MPa
Surface Tension
1.58 p = 6
2 2 0.0741
5 10
R
= 2.96 104
Pa or 29.6 kPa. Bubbles: p = 4 /R = 59.3 kPa
1.59 Use Table B.1: = 0.00504 lb/ft. p =
4 4 0.00504
1/(32 12)
R
= 7.74 psf or 0.0538 psi
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:
6
2 2 0.025 N/m
10 kPa
5 10 m
inside outside
p p p
r
Hence,
10 kPa 8000 10 8010 kPa
inside outside
p p
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 4 0.0736 0.866
0.130 m.
1000 9.81 0.0002
gD

10
1.62 See Example 1.4: h =
4 cos 4 0.032cos130
1.94 13.6 32.2 0.8/12
gD
= 0.00145 ft or 0.0174 in
1.63 force up = L 2 cos = force down = ghtL. h =
2 cos
.
gt
1.64 Draw a free-body diagram:
The force must balance:
W = 2 L or
d
L g L
2
4
2 .
d
g
8
W
L L
needle
1.65 From the free-body diagram in No. 1.47, a force balance yields:
Is
d
g
2
4
< 2 ?
2
(0.004)
7850 9.81 2 0.0741
4
0.968 < 0.1482 No
1.66 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.
F
D
1.67
h(x)
h
dW
dl
From the infinitesimal free-body shown:
cos .
d gh x dx cos =
dx
d
.
/
d dx d
h
g xdx g x
We assumed small so that the element
thickness is x.
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.

11
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
10 000 + 100 = 600 x. x = 16.83 km.
Ideal Gas
1.73
p
RT
1013
0 287 273 15
.
. ( )
1.226 kg/m3
. = 1.226 9.81 = 12.03 N/m3
1.74 3
in
101.3
1.226 kg/m .
0.287 (15 273)
p
RT
3
out
85
1.19 kg/m .
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.
1.75 3
750 44
0.1339 slug/ft .
1716 470
p
RT
m V 0.1339 15 2.01 slug.
1.76
p
W V
RT
100
(10 20 4) 9.81 9333 N.
0.287 293
g
1.77 Assume that the steel belts and tire rigidity result in a constant volume so that m1 = m2:
V 1 V 1 1 2 2
2
1 2
2
2 1
1
or .
150 460
(35 14.7) 67.4 psia or 52.7 psi gage.
10 460
m RT m RT
p p
T
p p
T

12
1.78 The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have
100 000 1 9.81. 10 200 kg.
m m
Hence,
pV
m
3
100 4 /3
10 200. 12.6 m or 25.2 m.
0.287 288
r
r d
RT
The First Law
1.79 2 2
1
0 ( 10). 20 32.2. 25.4 fps.
2
KE PE mV mg V V
2 2
1
0 ( 20). 40 32.2. 35.9 fps.
2
mV mg V V
1.80 2 2
1-2
1
. a) 200 0 5( 10 ). 19.15 m/s.
2
f f
W KE V V
b)
10
2 2
0
1
20 15( 10 ).
2
f
sds V
2
2 2
10 1
20 15( 10 ). 15.27 m/s.
2 2
f f
V V
c)
10
2 2
0
1
200cos 15( 10 ).
20 2
f
s
ds V
2 2
20 1
200sin 15( 10 ). 16.42 m/s.
2 2
f f
V V
1.81 2
1 2 1 2 2 1
1
. 10 40 0.2 0 . 40 000.
2
E E u u u u
40 000
. 55.8 C where comes from Table B.4.
717
v v
u c T T c
The following shows that the units check:
2 2 2 2 2
car
2 2 2
air
kg m /s m kg C m kg C
C
kg J/(kg C) N m s (kg m/s ) m s
m V
m c
where we used N = kg.
m/s2
from Newton’s 2nd
law.
1.82 2
2
2 1 H O
1
. .
2
E E mV m c T
2
6
1 100 1000
1500 1000 2000 10 4180 . 69.2 C.
2 3600
T T
We used c = 4180 J/kg.
C from Table B.5. (See Problem 1.75 for a units check.)

13
1.83 water . 0.2 40 000 100 4.18 . 19.1 C.
f f
m h m c T T T
The specific heat c was found in Table B.5. Note: We used kJ on the left and kJ on the
right.
1.84 W pd V
mRT
V
d V
d V
mRT
V
ln
V
mRT
2
V
2
1 1
ln
p
mRT
p
since, for the T = const process, 1
p V 1 2
p V 2. Finally,
1-2
4 1
1716 530ln 78,310 ft-lb.
32.2 2
W
The 1st
law states: 0. 78,310 ft-lb or 101 Btu.
v
Q W u mc T Q W
1.85 If the volume is fixed the reversible work is zero since the boundary does not move. Also,
since V 1 2
1 2
,
mRT T T
p p p
the temperature doubles if the pressure doubles. Hence, using
Table B.4 and Eq. 1.7.17,
200 2
a) (1.004 0.287)(2 293 293) 999 kJ
0.287 293
v
Q mc T
b)
200 2
(1.004 0.287)(2 373 373) 999 kJ
0.287 373
v
Q mc T
c)
200 2
(1.004 0.287)(2 473 473) 999 kJ
0.287 473
v
Q mc T
1.86 W pd V (
p V 2 V 1
1). If = const,
T
p
V
2
1
T
V
2 1
2
so if 2 ,
T T
then V 2 2V 1 and (2
W p V 1 V 1) pV 1 1.
mRT
a) 2 0.287 333 191 kJ
W
b) 2 0.287 423 243 kJ
W
c) 2 0.287 473 272 kJ
W
1.87 = 1.4 287 318 357 m/s. 357 8.32 2970 m.
c kRT L c t
1/ 0.4/1.4
2
2 1
1
500
(20 273) 151.8 K or 121.2 C
5000
k k
p
T T
p
1.88 We assume an isentropic process for the maximum pressure:
/ 1 1.4/0.4
2
2 1
1
423
(150 100) 904 kPa abs or 804 kPa gage.
293
k k
T
p p
T
Note: We assumed patm = 100 kPa since it was not given. Also, a measured pressure is a
gage pressure.

14
1.89
/ 1 1.4/0.4
2 1 2 1
/ 100 473/ 293 534 kPa abs.
k k
p p T T
2 1
( ) (1.004 0.287)(473 293) 129 kJ/kg.
v
w u c T T
We used Eq. 1.7.17 for cv.
Speed of Sound
1.90 a) 1.4 287 293 343.1 m/s
c kRT
b) 1.4 188.9 293 266.9 m/s
c kRT
c) 1.4 296.8 293 348.9 m/s
c kRT
d) 1.4 4124 293 1301 m/s
c kRT
e) 1.4 461.5 293 424.1 m/s
c kRT
Note: We must use the units on R to be J/kg.
K in the above equations.
1.91 At 10 000 m the speed of sound 1.4 287 223 299 m/s.
c kRT
At sea level, 1.4 287 288 340 m/s.
c kRT
340 299
% decrease 100 12.06 %.
340
1.92 a) = 1.4 287 253 319 m/s. 319 8.32 2654 m.
c kRT L c t
b) = 1.4 287 293 343 m/s. 343 8.32 2854 m.
c kRT L c t
c) = 1.4 287 318 357 m/s. 357 8.32 2970 m.
c kRT L c t

Chapter 2 / Fluid Statics
© 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.
15
CHAPTER 2
Fluid Statics
FE-type Exam Review Problems: Problems 2-1 to 2-9
2.1 (C) (13.6 9810) (28.5 0.0254) 96 600 Pa
Hg
p h

     
2.2 (D) 0 84 000 1.00 9.81 4000 44 760 Pa
p p gh

      
2.3 (C) 0 30 000 0.3 9810 0.1 8020 Pa
w atm x x water w
p p h h
 
        
2.4 (A) (13.6 9810) 0.16 21350 Pa.
a
p H

       
, 21350 10 000 11350 13.6 9810 . 0.0851 m
a after after after
p H H
        
2.5 (B) The force acts 1/3 the distance from the hinge to the water line:
5 1 5 5
(2 ) (2 ) [9800 1 3 (2 )]. 32 670 N
3 3 3 3
P P
           
2.6 (A) The gate opens when the center of pressure in at the hinge:
3
1.2 11.2 (1.2 ) /12
5. 5 1.2.
2 2 (1.2 ) (11.2 ) / 2
p
h I h b h
y y y
Ay h b h
  
       
 
This can be solved by trial-and –error, or we can simply substitute one of the
answers into the equation and check to see if it is correct. This yields h = 1.08 m.
2.7 (D) Place the force
 
F F
H V
 at the center of the circular arc. FH passes through the
hinge:
2
4 1.2 9800 ( 1.2 / 4) 9800 300 000. 5.16 m.
V
P F w w w

          
2.8 (A) W V


900 9.81 9810 0.01 15 . 6 m
w w
     
2.9 (A)
5
20 000 20 000 6660 (1.2 ) 24 070 Pa
9.81

      
plug
p h
2
24 070 0.02 30.25 N

    
plug plug
F p A .

16
Chapter 2 Problems: Pressure
2.10   
 
F ma p z p s
y z
a
y y y y
  
: sin 
2
  
   
F ma p y p s
y z
a g
y z
z z z z
   
: cos  
2 2
Since  
s y
cos  and  
s z
sin   , we have
p p
y
a
y y
  

2
and  
p p
z
a g
z z
  


2
Let y  0 and z  0: then
p p
p p
y
z
 
 



0
0
  
p p p
y z .
2.11 p = h. 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
2.13 S =
20 144
62.4 20
p
h




= 2.31.  = 1.94  2.31 = 4.48 slug/ft3
.
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.
y
z
pzy
pyz
ps
s

z
y
gV

17
2.17
inside
100 9.81
3 13.51 Pa
0.287 253
100 9.81
3 11.67 Pa
0.287 293


 
        
 


 
       

 
outside o
o
base
i
i
pg
p g h h
RT
p
pg
p g h h
RT
= 1.84 Pa
If no wind is present this pbase would produce a small infiltration since the higher
pressure outside would force outside air into the bottom region (through cracks).
2.18 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:
dp h gdh
p
 

 1000 1 100
0
10
0
( / )
p   

1000 9 81 10
10
2 100
2
. ( ) = 103 000 Pa or 103 kPa
Note: we could have used an average S: Savg = 1.05, so that avg = 1050 kg/m3
.
2.19
  
  
i j k
p p p
p
x y z
   
=  
       
           
i j k k i j k k a g
x y z x y z
a g a a a g
( )
 a g
p
   
2.20 /
0 0
[( )/ ]g R
atm
p p T z T 

 
= 100 [(288  0.0065  300)/288]9.81/0.0065  287
= 96.49 kPa
100
100 9.81 300/1000
0.287 288
atm
p p gh

     

= 96.44 kPa
% error =
96 96
96
100
.44 .49
.49

 = 0.052%
The density variation can be ignored over heights of 300 m or less.
2.21
/
0
0
0
g R
atm atm
T z
p p p p p
T


 

    
 
 
= 100
9.81/0.0065 287
288 0.0065 20
1
288

 
 
 

 
 
 
 
 
= 0.237 Pa or 0.000237 kPa
This change is very small and can most often be ignored.

18
2.22 Eq. 1.5.11 gives 310,000 144 .
dp
d


  But, dp = gdh. Therefore,
7
4.464 10
gdh d
 


 or 2 7
32.2
4.464 10
d
dh




Integrate, using 0
 = 2.00 slug/ft3
:
2 7
2 0
32.2
4.464 10



h
d
dh


  . 
1 1
2

 
 
 
 
= 7.21  107
h or 7
2
1 14.42 10 h
 

 
Now,
7
7 7
0 0
2 2
ln(1 14.42 10 )
1 14.42 10 14.42 10

h h
g g
p gdh dh h
h

 
    
   
 
Assume  = const:
2.0 32.2 64.4
p gh h h

    
a) For h = 1500 ft: paccurate = 96,700 psf and pestimate = 96,600 psf.
96,600 96,700
% error 100 0.103 %
96,700

   
b) For h = 5000 ft: paccurate = 323,200 psf and pestimate = 322,000 psf.
322,000 323,200
% error 100 0.371 %
323,200

   
c) For h = 15,000 ft: paccurate = 976,600 psf and pestimate = 966,000 psf.
966,000 976,600
% error 100 1.085 %
976,600

   
2.23 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
0.0065 287
101(1 0.0065 / 288) .
z 

a) z = 3000. p = 69.9 kPa. b) z = 6000. p = 47.0 kPa.
c) z = 9000. p = 30.6 kPa. d) z = 11 000. p = 22.5 kPa.

19
2.25 Use the result of Example 2.2: /
0
= .
gz RT
p
e
p

0
ln .
p gz
p RT
 
0.001 32.2
ln .
14.7 1716 455
z
 

z = 232,700 ft.
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 = gh =   9.81h or
2400
.
9.81h
 
a)
2400
9.81 0.36
 

= 680 kg/m3
. gasoline
b)
2400
9.81 0.272
 

= 899 kg/m3
. benzene
c)
2400
9.81 0.245
 

= 999 kg/m3
. water
d)
2400
9.81 0.154
 

= 1589 kg/m3
. carbon tetrachloride
2.29 Referring to Fig. 2.6a, the pressure is p = wgh = 2
1
.
2
aV
 Then 2 2
.
w
a
gh
V



a) 2 2 1000 9.81 0.06
1.23
V
  
 = 957. V = 30.9 m/s
b) 2 2 1.94 32.2 3/12
0.00238
V
  
 = 13,124. V = 115 ft/sec
c) 2 2 1000 9.81 0.1
1.23
V
  
 = 1595. V = 39.9 m/s
d) 2 2 1.94 32.2 5/12
0.00238
V
  
 = 21,870. V = 148 ft/sec

20
2.30 See Fig. 2.6b: p1 = –1h + 2H.
p1 = –0.86  62.4 
5
12
+ 13.6  62.4 
9 5
12
.
= 649.5 psf or 4.51 psi.
2.31 0 1 1 2 2 3 3 4 4
p p gh gh gh gh
   
    
= 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      
p p p p p p p p
1 4 1 2 2 3 3 4
       (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      
p p p p p p p p
1 4 1 2 2 3 3 4
       (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        
p p p p p p p p p p
1 5 1 2 2 3 3 4 4 5
        
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 or 11.94 kPa.
2.36 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.
2.37 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.
2.38 p – 9810  4 + 13.6  9810  0.16 = 0. p = 17 890 Pa or 17.89 kPa.

21
2.39 8200 + 9810  0.25 = 1.59  9810  H. H = 0.683 m
Hnew = 0.683 + 0.273 = 0.956 m. H =
0.273
2
= 0.1365.
p + 9810 (0.25 + 0.1365) = 1.59  9810  0.956.
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
water oil HG
0.07 0.1 0.09sin 40
  
      
A B
p p
Solve for pB:
 
 
water HG oil
water water water
water
3
0.07 0.09sin 40 0.1
0.07 13.6 0.09sin 40 0.87 0.1
0.07 13.6 0.09sin 40 0.87 0.1
10 kPa 0.07 13.6 0.09sin 40 0.87 0.1 9.81 kN/m 2.11 kPa
  
  

      
      
      
       
B A
A
A
p p
p
p
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.
A
B
10 cm
7 cm
9 cm
Mercury
Water
Oil (S = 0.87)
40
o
h
h
From the previous problem we have
  water HG oil
1
0.07 0.09sin40 0.1
  
      
B A
p p (1)
For the new condition
     
water HG oil
2
0.07 0.11sin40 0.1 sin40
  
         
B A
p p h h (2)
where h in this case is calculated from the new manometer reading as
/ sin 40 11 9 0.783 cm
       
h h h
H
H
H

22
Subtracting Eq.(1) from Eq.(2) yields
       
2 1
0.02sin40 sin40
B B water HG oil
p p h h
  
        
Substituting the values of h and  1
B
p gives  2
B
p
     
2
2.11 0.00783 13.6 0.02sin 40 0.87 0.00783sin 40 9.81
0.52 kPa
 
       
 

B
p
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 1
p V 1 2
p V
 2 using absolute
pressures. Thus,
14.7  144  4A = p2(4 – H / 2 )A or p2 = 2 8467 / (4 / 2)
p H
 
From a pressure balance on the manometer (pressures in psf):
30  144 + 14.7  144 = 13.6  62.4 H +
8467
4 2
 H /
,
or H2
– 15.59 H + 40.73 = 0. H = 12.27 or 3.32 ft.
2.44 a)        
p p p p p p p p p p
1 5 1 2 2 3 3 4 4 5
        
4000 = 9800(0.16–0.22) + 15 600(0.10–0.16) + 133 400H + 15 600(0.07–H).
H = 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)
2 2
2 2
1 1 2 3 2
2 /
2 2( ) /
H D d
p D d
   


    
2
2
2(0.1/ 0.005)
9800 2 15 600 2(133 400 15 600)(0.1/ 0.005)

    
  
8 10 6
.487 H
   

H 8 10 400
6
.487 = 0.0034 m or 3.4 mm
b)
2
2
2(4 / 0.2)
0.06 144
62.4 2 99.5 2(849 99.5)(4 / 0.2)
H
  
    
= 0.01153 ft or 0.138 in.
2.46      
p p p p p p p p
1 4 1 2 2 3 3 4
       (poil = 14.0 kPa from No. 2.30)
15 500 – 14 000 = 9800(0.12 + z) + 680(0.1 – 2z) + 860(–0.1 – z).
z = 0.0451 m or 4.51 cm
2.47 a) pair = –6250 + 625 = –5620 Pa.
–5620 + 9800(2 + z) – 13 600  9.81(0.1 + 2z) = 0. z = 0.0025.
h = 0.1 + 2z = 0.15 m or 15 cm

23
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. z = 0.00558 ft.
h = 4/12 + 2 (0.00558) = 0.3445 ft or 4.13 in.
d) pair = 441 + 44.1 = 485 psf
485 + 62.4(2 + z) – 13.6  62.4(8/12 + 2z) = 0. z = 0.0267 ft.
h = 8/12 + 2 (0.0267) = 0.7205 ft or 8.65 in.
Forces on Plane Areas
2.48 F hA
  = 9810  10    0.32
/4 = 6934 N.
2.49
5 1 5 5
2 2 9800 1 3 2 . 32 670 N
3 3 3 3
 
     
           
     
 
     
 
P P
a) F = pc A = 9800  2  42
= 313 600 N or 313.6 kN
b)
2 2
9800 1 (2 4) 9800 2 9800 1 98 000 N or 98.0 kN
3 3
c
F p A
           
c) F = pc A = 9800  1  2  4  2 = 110 900 N or 110.9 kN
d) F = pc A = 9800  1  2  4/0.866 = 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

24
2.53 a) 62.4 27.33 24

    
F hA 40,930 lb.
3
6 8 /36
27.33
27.33 24
p
y

 

= 27.46 ft. y = 30 – 27.46 = 2.54 ft.
8/5.46 = 3/x. x = 2.05’. (2.05, 2.54) ft.
b) F = 62.4  30  24 = 44,930 lb. The centroid is the center of pressure.
y = 2.667 ft.
8 3
5.333 x
 . x = 2.000 ft (2.000, 2.667) ft.
c) F = 62.4 (30 – 2.667  0.707)  24 = 42,100 lb.
yp  


39 77
6 8 36
39 77 24
3
.
/
.
= 39.86 ft. y = 42.43 – 39.86 = 2.57 ft
8 3
5.46 x
 8/5.43 = 3/x. x = 2.04 ft. (2.04, 2.57) ft.
2.54 a) 2
9810 6 2
 
   
F hA = 739 700 N or 739.7 kN.
4
/ 6 2 /4(4 6)
 
     
p
y y I Ay = 6.167 m. (x, y)p = (0, –0.167) m
b) F hA
   
 
9810 6 2 = 369 800 N or 369.8 kN.
yp  


6
2 8
2 6
4


/
= 6.167 m. x2
+ y2
= 4
2 2
2
2 2
(6 ) (4 )(6 ) .
2 2 2
 
 
     
  
p
x
x F pdA x y xdy y y dy
2
2 3
2
6 2 (24 4 6 ) 32 .
2

  

      

p
x y y y dy xp = 0.8488 m
(x, y)p = (0.8488, –0.167) m
c) F = 9810  (4 + 4/3)  6 = 313 900 N or 313.9 kN.
yp  


5 333
3 4 36
5 333 6
3
.
/
.
= 5.500 m. y = –1.5
4/2.5 =
1.5
x
. x = 0.9375. (x, y)p = (0.9375, –1.5) m
d) F    
9810 4
2
3
4
( sin 36.9°)  6 = 330 000 N
3
5.6 5 2.4 /36(6 5.6)
p
y     = 5.657 m. y = 0.343 m
3 cos 53.13 = 1.8, 2.5 – 1.8 = 0.7, 2.4/2.057 = . /
7 1
x .  x1 = 0.6.
x = 1.8 + 0.6 = 2.4. (x, y)p = (2.4, 0.343) m.
(x, y)
y
x
y
x
(x, y)
dA
dy
x
y
3 4
53.13
o

25
2.55 F hA
    
 62 11 6 10
.4 ( ) = 41,180 lb.
y y
I
yA
p    


11
6 10 12
11 60
3
/
= 11.758 ft.
(16 – 11.758) 41,180 = 10P. P = 17,470 lb.
2.56 F hA
   
 9810 6 20 = 1.777  106
N, or 1177 kN.
y y
I
Ay
p    


7 5
4 5 12
7 5 20
3
.
/
.
= 7.778 m.
(10 – 7.778) 1177 = 5 P. P = 523 kN.
2.57 F hA
   
 9810 12 20 = 2.354  106
N, or 2354 kN.
y y
I
Ay
p    


15
4 5 12
15 20
3
/
= 15.139 m.
(17.5 – 15.139) 2354 = 5 P. P = 1112 kN.
2.58 y y
I
Ay
H bH
bH H
H H
H
p    

  
2
12
2 2 6
2
3
3
/
/
. yp is measured from the surface.
From the bottom, H y H H H
p
   
2
3
1
3
.
Note: This result is independent of the angle , so it is true for a vertical area or a sloped
area.
2.59 3
1
sin40 3 . ( 2) sin40 . 2( 2) .
2 3
l
F l l F l P l l P
 
       
a) 981023
= 2(2 + 2)P. P = 9810 N
b) 981043
= 2(4 + 2)P. P = 52 300 N
c) 981053
= 2(5 + 2)P. P = 87 600 N
2.60 2 2
1.2 0.4
h   = 1.1314 m.A = 1.2  1.1314 + 0.4  1.1314 = 1.8102 m2
Use 2 forces: 1 1 9800 0.5657 (1.2 1.1314)
c
F h A

     = 7527 N
2 2
1.1314
9800 (0.4 1.1314)
3
c
F h A

     = 1673 N
yp1
2
3
11314
 ( . ).
3
2
2
2
1.1314 0.4 1.1314 / 36
3 0.4 (1.1314 / 2) (1.1314 / 3)
p
I
y y
A y

   
 
= 0.5657 m
Mhinge  0: 7527 1.1314/3 1673 (1.1314 0.5657) 1.1314P
     = 0. P = 3346 N.
yp
P
F

26
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 A h H I h H h H
      
( ) / , ( ) , [ ( )]( ) /
3 2 36
2 3
 



 






y
h H h H
h H h H
h H h H h H
p
3
2 36
3 3 6 2
4
2
( ) /
( ) ( ) /
a) h
h H


2
. h = H = 0.9 m
b) h = H = 1.2 m
c) h = H = 1.5 m
2.62 The gate is about to open when the center of pressure is at the hinge.
a)
3
1.8 /12
1.2 (1.8/2 ) .
(0.9 )1.8
p
b
y H H
H b

    

H = 0.
b)
3
2 /12
1.2 (2.0/2 ) .
(1 )2
p
b
y H H
H b

    

H = 0.6667 m.
c)
3
2.2 /12
1.2 (2.2/2 ) .
(1.1 )2.2
p
b
y H H
H b

    

H = 2.933 m.
2.63 F
H
bH bH
1
2
2
1
2
  
 
F H b b H
2   
 
 
1
2 3 2
2
 
bH
H
b H
  


.  
H 3
a) H  
3 2 = 3.464 m b) H = 1.732 m c) H = 10.39' d) H = 5.196'
2.64 A free-body-diagram of the gate and block is sketched.
Sum forces on the block:
0
    
y B
F W T F
where FB is the buoyancy force which is given by
2
(3 )
B
F R H
 
 
 
 
Take moments about the hinge:
3.5 (3 )
H p
T F y
   
where FH is the hydrostatic force acting on the gate. It is,
using 2
1.5 m and 2 3 6 m ,
   
h A
Fstop
0
T
FH
Rx
Ry
FB
T
W
yp
  
3 2
9.81 kN/m 1.5 m 6 m 88.29 kN

   
H
F hA
F2
F1
H/3
l/2

27
From the given information,
 
3
2 3 /12
1.5 2 m
1.5 6
p
I
y y
yA
    

 
88.29 3 2
25.23 kN
3.5
 
  
T
 
2
70 25.23 44.77 kN. 3 44.77

       
B
F W T R H
   
2
3
44.77 kN
3 m 1.55 m
9.81 kN/m 1 m

  
H
Assume 1 m deep
2.65 The dam will topple if the moment about “O” of F1 and F3 exceeds
the restoring moment of W and F2.
a) W     
( .4 )( / )
2 9810 6 50 24 50 2 = 21.19  106
N
300 27 600 16
300 600
w
d
  


= 19.67 m. (dw is from O to W.)
F2 = 9810  5  11.09 = 0.544  106
N. d2
1109
3

.
= 3.697 m.
F1 9810
45
2
45
   = 9.933  106
N. d1 = 15 m. (d1 is from O to F1.)
F3 9810
45 10
2
30
 

 = 8.093  106
N. d3
2 943 15 5150 20
2 943 5150

  

. .
. .
= 18.18 m.
Wd F d
F d F d
w    
   



2 2
6
1 1 3 3
6
418 8 10
2961 10
.
.
N m
N m
will not topple.
b) W = (2.4  9810) (6  65 + 65  12) = 27.55  106
N.
dw =
390 27 780 16
390 780
  

= 19.67 m.
F2
6
0 54 10
 
. N. d2 3 70
 . m.
F1 = 9810  30  60 = 17.66  106
N. d1 = 20 m.
F3 9810
60 10
2
30
 

 = 10.3  106
N. d3
2 943 15 7 358 20
2 943 7 358

  

. .
. .
= 18.57 m.
Wd F d
F d F d
w    
   



2 2
6
1 1 3 3
6
543 9 10
544 5 10
.
.
N m
N m
it will topple.
c) Since it will topple for H = 60, it certainly will topple if H = 75 m.
F1
F2
F3
W
O

28
Assume 1 m deep
2.66 The dam will topple if there is a net clockwise moment about “O.”
a) W W W W
      
1 2 1 6 43 1 62 2
. ( ) .4 .4 = 38,640 lb.
W2 24 43 2 62 2
   
( / ) .4 .4 = 77,280 lb.
W3 40 22 33 2 62
  
( . / ) .4 = 27,870 lb @ 20.89 ft.
F1 62 20 40 1
   
.4 ( ) = 49,920 lb @ 40/3 ft.
F2 62 5 10 1
   
.4 ( ) = 3120 lb @ 3.33 ft
1
3
2
= 18,720 lb @ 15 ft
= 28,080 lb @ 20 ft
p
p
F
F
F


 


O
M
 : (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 30 60
  
.4 = 112,300 lb. 3 (60 22.86/2) 62.4
W    = 42,790 lb.
F2 62 5 10
  
.4 = 3120 lb
Fp1 62 10 30
  
.4 = 18,720 lb. Fp2 62 50 30 2
  
.4 / = 46,800 lb.
O
M
 : (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 Mhinge = 0. 2.5P – dw  W – d1  F1 = 0.
 P     

 








1
2 5
2
3
9800 1 8
4 2
3
9800
2
4
4
2
. 

= 62 700 N
Note: This calculation is simpler than that of Example 2.7. Actually, We could
have moved the horizontal force FH and a vertical force FV (equal to W)
simultaneously to the center of the circle and then 2.5P = 2FH.=2F1. This
was outlined at the end of Example 2.7.
2.68 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:
( ) 9810 2 (4 10) 784 800N
( ) 0.86 9810 1 20 168 700N
water
H
H oil
F
F
    
    
M P
: . . .
2 784 8 2 168 7 2
    P = 616.1 kN.
F1
F2
F3
W
O
W3
F1
dw
d1
P
W

29
2.69 Place the resultant force
 
F F
H V
 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:
2
1

 
H
V
F F
F W F
where and
H V
F F are the horizontal and vertical components
of the force acting on the water by the surface AB. Hence,
   
3
2 9.81 kN/m 8 1 2 4 706.3 kN
    
H
F F
F2
FH
FV
F1
W
.A
.
B
xV
The line of action of FH is the same as that of F2. Its distance from the surface is
 
3
4 2 12
9 9.037 m
9 8
p
I
y y
yA
    

To find FV we find W and F1:
   
3 2
9.81 kN/m 2 2 2 4 33.7 kN
4


 
     
 
 
W V
 
3
1 9.81 kN/m 8 2 4 628 kN
   
F
1 33.7 628 662 kN
     
V
F F W
To find the line of action of FV, we take moments at point A:
1 1 2
    
V V
F x F d W d
where
   
1 2
2 2 2
1 m, and 1.553 m:
3 4 3 4
 

   
 
R
d d
1 1 2 628 1 33.7 1.553
1.028 m
662
     
   
V
V
F d W d
x
F
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.

30
2.71 Place the resultant 
F F
H V 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 
F F
H V at the center. FV passes thru the hinge
2  (9810  1  10) = 2.8 P. P = 70 070 N or 70.07 kN.
2.73 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 
F F
H V of the unknown liquid acts thru the center of the circular arc. FV
passes thru thehinge. Thus, we use only ( ) .
FH oil Assume 1 m wide:
a) M
R R
R
R
S
R
R
R
R
x
: .
3
9810
2
4
3
9800
4 2
2





 





 








  x  4580 N/m3
b) M
R R
R
R
S
R
R
R
R
x
: . . .
3
62 4
2
4
3
62 4
4 2
2





 





 








  x  29.1 lb/ft3
2.75 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 36 = 8476 N.
0 ( ) ( )
y V o V w
F W F F
    
0 = S  9810  0.62
 6 +
0.36
0.36
4

 

 
 
 6  (0.8  9810) – 9810  0.18  6
2
9810 0.8 2 0.6 6
      
S 0 955
. .
b) g V
 .
W
 = 1996 lb.
0 ( ) ( )
y V o V w
F W F F
    
0 = S  62.4    22
 20 + 4
4
4








 20  0.8  62.4 – 62.4    2  20
    
62 4 8 2 2 20
2
. . .  
S 0 955
. .

31
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
Using the pressure from part (a):
Fweld = 14 870    32
– (0.8  9810) 
1
2
4
3
33
 





 = –23 400 N
or –23.4 kN
2.77 A free-body diagram of the gate and water is shown.
H
F d W H P
w
3
   .
a) H = 2 m. F = 9810  1  4 = 39 240 N.
W xdy
y
dy
  



9810 2 9810 2
2
2 9810
2
2
3 2
1 2
0
2
0
2 3 2
/ /
/
= 26 160 N.
d x
x
xdy
xdy
x dx
x dx
w    










2
1
2
4
4
1
2
1 4
1 3
3
0
1
2
0
1
/
/
= 0.375 m.
    
P
1
3
39
0 375
2
26 160
240
.
= 17 980 N or 17.98 kN.
b) H = 8 ft. F = 62.4  4  32 = 7987 lb
W xdy x dx
     


62 4 62 4 4 62 16 2 3
2
0
2
3
.4 .4 .4 / = 2662 lb.
2
3
0
2
2
0
1
4
2 1 16/4
2 8/3
4
w
x dx
d x
x dx
 
    
 


= 0.75 ft. 
1 8
7987 0.75 2662 2910 lb
8 3
 
    
 
 
P
W
pA
Fweld
dA=xdy
y
x
F
h/3

32
Buoyancy
2.78 W = weight of displaced water.
a) 20 000 + 250 000 = 9810  3 2
(6 /2).
d d
 d2
+ 12d – 18.35 = 0. d = 1.372 m.
b) 270 000 = 1.03  9810  3 2
(6 /2).
d d
 d2
+ 12d – 17.81 = 0. d = 1.336 m.
2.79 25 + FB = 100. FB = 75 = 9810 
V .  
V = 7.645  
m3
  7.645  
= 100. or 7645 cm3
 = 13 080 N/m3
.
2.80 3000  60 = 25  300 d  62.4. d = 0.3846' or 4.62".
2.81 100 000  9.81 + 6 000 000 = (12  30 + 8h  30) 9810
h = 1.465 m. distance from top = 2 – 1.465 = 0.535 m
2.82 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:
FB = W + Fa or
4
3
1000
4
3
3 3
   
R g R g
a
 
3 3
4 100 9.81 4 100 9.81
5 1000 5 .
3 0.287 293 3 0.287 a
T
 
 
   

Ta = 350.4 K or 77.4C
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
2 100 9.81
1500 150
0.287 288


 

= Fp + 0.1 Fp + 1500   1502

100 9 81
2 077 288


.
.
FP = 9.86 × 108
and Npeople =
9 86 10
800
8
. 
= 1.23  106
Of course equipment and other niceties such as gyms, pools, restaurants, etc., would add
significant weight.

33
2.85 Neglect the bouyant force of air. A force balance yields
FB = W + F
= 50 + 10 = 60 = 9800 
V .  
V = 0.006122 m3
Density: g V
 .
W

9.81 0.006122
   = 50.  = 832.5 kg/m3
Specific wt:  = g = 832.5  9.81 = 8167 N/m3
Specific gravity:
water
832.5
1000
S


  S = 0.8325
2.86 From a force balance FB = W + pA.
a) The buoyant force is found as follows (h > 16'):
cos ,
 
 
h R
R
15
Area = R2
– (h – 15 – R) R sin
FB = 10  62.4[R2
 R2
+ (h – 15 – R) R sin].
FB = 1500 + hA.
The h that makes the above 2 FB’s equal is found by trial-and-
error:
h = 16.5: 1859 ? 1577 h = 16.8: 1866 ? 1858
h = 17.0: 1870 ? 1960 h = 16.8 ft.
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  
 
R h
R
15
Trial-and-error for h:
h = 16: 1849 ? 1374 h = 16.2: 1853 ? 1765
h = 16.4: 1857 ? 2170 h = 16.25 ft.
2.87 a) W FB
 .  
0 01 13 6 1000 015 4 9 81 9810
2
. . . / . .
      
h V

 

 

   
V
 
.
.
.
. . .
015
4
15
005
4
06 2 769 10
2 2
5 3
m h = 7.361  
m
 
mHg
2
13.6 1000 .015 / 4
h
   = 0.01769 kg
h  15
 R
pA
FB
W
h  15
 R

34
b) (0.01 +0 .01769) 9.81 = 9810
2 2
0.015 0.005
0.15 0.12 .
4 4
x
S
 
 
 
  
 
 
 
Sx = 0.959.
c) (0.01 + 0.01769) 9.81 = 9810
2
0.015
0.15
4
 
 Sx. Sx = 1.045.
2.88
2 2
0.015 0.005
(0.01 )9.81 9810 0.15 0.12 .
4 4
Hg
m
 
 
 
    
 
 
 
mHg = 0.01886.
a) (0.01 +0 .01886) 9.81 = 9810
2
0.015
0.15
4
 
 Sx. Sx = 1.089.
b) mHg = 0.01886 kg.
Stability
2.89 a)
4 4
o
(10/12)
64 64
d
I
  
  = 0.02367 ft4
.
  
   
V
W
rH O
2
8 62 5 12 12 12
62
2
. .4 ( / ) /
.4

= 0.4363. depth = 2
0.4363
(5 /12)

= 0.8 ft
0.02367/ 0.4363 (0.5 0.4)
GM
    = –0.0457'. It will not float with ends horizontal.
b) Io = 0.02367 ft4
, 
V = 0.3636 ft3
, depth = 0.6667 ft
0.02367/ 0.3636 (5 4) /12
GM    = –0.01823 ft. It will not float as given.
c) 
V = 0.2909, depth = 6.4", GM =
0.02367 4 3.2
0.2909 12

 = 0.0147 ft. It will float.
2.90 With ends horizontal 4
o / 64.
I d

 The displaced volume is
     
V d h d
x x
  
2 5 3
4 9800 8014 10
/ . since h = d. The depth the cylinder will
sink is
depth =

   
 
V
A
d d d
x x
8 014 10 4 10 20 10
5 3 2 5
. / / .
  
The distance CG is CG
h
d
x
   
2
10 2 10 2
5
. /
 . Then
GM
d
d
d
d
x
x


   





4
5 3
5
64
8 014 10 2
10 2 10 2 0
/
.
. / .
This gives (divide by d and multiply by x): 612.5 – 0.5 x + 5.1  105
 x
2
> 0.
Consequently, x > 8369 N/m3
or x < 1435 N/m3

35
2.91
3
3
.
water
water water
S d
W
V S d

 
   
3
3
.
water
water water
S d
W
V S d

 
    h = Sd.
4
3
/12 1 1
( / 2 / 2) .
12 2 2
d S
GM d Sd d
S
Sd
 
     
 
 
If GM = 0 the cube is neutral and 6S2
– 6S + 1 = 0.
 
 
S
6 36 24
12
= 0.7887, 0.2113.
The cube is unstable if 0.2113 < S < 0.7887.
Note: Try S = 0.8 and S = 0.1 to see if GM  0. This indicates stability.
2.92 As shown, 16 9 16 4/(16+16)
y     = 6.5 cm above the bottom edge.
4 9.5 16 8.5 16 4
0.5 8 2 8 16
A
A
S
G
S
  
  
    

    
= 6.5 cm.
130 + 104 SA = 174 + 64 SA.  SA = 1.1.
2.93 a) y 
    
 
16 4 8 1 8 7
16 8 8
= 4. x 
    
 
16 1 8 4 8 4
16 8 8
= 2.5.
For G:
1.2 16 4 0.5 8 1 1.5 8 7
1.2 16 0.5 8 1.5 8
y
       

    
= 4.682.
1.2 16 0.5 8 4 1.5 8 4
1.2 16 0.5 8 1.5 8
x
      

    
= 2.364.
G must be directly under C.
0.136
tan .
0.682
   =11.3.
b) y 
    
 
4 2 2
1
2
2 3 5
4 2 2
.
= 2. x 
    
 
4
1
2
2 2 2 2
4 2 2
= 1.25
For G:
1.2 4 2 0.5 1 1.5 7
1.2 4 0.5 2 1.5 2
y
     

    
= 2.34.
1.2 2 0.5 4 1.5 4
1.2 4 0.5 2 1.5 2
x
    

    
= 1.182
y = 0.34, x = 0.068.
0.068
tan .
0.34
   = 11.3.
2.94 The centroid C is 1.5 m below the water surface.  CG = 1.5 m.
Using Eq. 2.4.47: GM 

 
    


8 12
8 3
15 1777 15 0 277 0
3
/
. . . . .
The barge is stable.
0.682
C
0.136
G
C
G
h

36
2.95 y 
  

8 3 16 97 1
8 16 97
.485 .414 .
.485 .
= 1.8 m. CG  
18 15
. . = 0.3 m.
Using Eq. 2.4.47:
3
8.485 /12
0.3 1.46 0.3 1.16.
34.97
GM

     Stable.
Linearly Accelerating Containers
2.96 a) tan
.
.
  
20
9 81 4
H
H = 8.155 m. 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 The air volume is the same before and after.
 0.5  8 = hb/2. tan
.
.
  
10
9 81
h
b
4
9 81
10

h
h
2
.
.
h = 2.856. Use dotted line.
1
2.5 2.5 2.452 4.
2
w     w = 0.374 m.
a) pA = –1000  10 (0 – 7.626) – 1000  9.81  2.5 = 51 740 Pa or 51.74 kPa
b) pB = –1000  10 (0 – 7.626) = 76 260 Pa or 76.26 kPa
c) pC = 0. Air fills the space to the dotted line.
2.98 Use Eq. 2.5.2: Assume an air-water surface as shown in the above figure.
a) 60 000 = –1000 ax (0–8) – 1000  9.81 0 2 5
8
9 81
 














.
.
ax
4 =
h
ax
2
9 81
2
 .
60 = 8 ax + 24.52 – 9.81
8
9.81
x
a
or ax – 4.435 = 1.1074 ax .
ax
2
– 10.1 ax + 19.67 = 0 ax = 2.64, 7.46 m/s2
b) 60 000 = –1000 ax (–8) – 1000 (9.81 + 10)  






2 5
8
9 81
.
.
.
ax
60 = 8 ax + 49.52 – 19.81
8
19.81
x
a
or ax – 1.31 = 1.574 ax .
ax
2
– 5.1 ax + 1.44 = 0 ax = 0.25, 4.8 m/s2

b
h
A
B
z
1
x
w
C

37
c) 60 000 = –1000 ax (–8) – 1000 (9.81 + 5) (–2.5 +
8
14 81
ax
.
).
60 = 8 ax + 37.0 – 14.81
8
14.81
x
a
or ax – 2.875 = 1.361 ax .
ax
2
– 7.6 ax + 8.266 = 0 ax = 1.32, 6.28 m/s2
2.99 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
  

F z dz
AB ( )
.
76 260 9810 4
0
2 5
= 640 000 N or 640 kN
b) The pressure on the bottom BC is
p(x) = –1000  10 (x – 7.626) = 76 260 – 10 000 x.
  

F x dx
BC ( )
.
76 260 10 000 4
0
7 626
= 1.163  106
N or 1163 kN
c) On the top p(x) = –1000  10 (x – 5.174) where position 1 is on the top surface:
  

F x dx
top ( )
.
51 740 10 000 4
0
5 174
= 5.35  105
N or 535 kN
2.101 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
2
15 2
. = 100 200 N or 100.2 kN
b) pD = 0. pC = –1000  17.32 (–0.5–1.232)  1000  19.81(0.866–1.866) = 49 810 Pa.
FCD    
1
2
49 810 15 2
. = 74 720 N or 74.72 kN.
c) pA = 58 290 Pa. pC = 49 810 Pa.  


FAC
58 29 49 81
2
15
. .
. = 81.08 kN.
x
z

38
2.102 Use Eq. 2.5.2 with position 1 at the open end:
a) pA = 0 since z2 = z1.
pB = 1000  19.81  0.6 = 11 890 Pa.
pC = 11 890 Pa.
b) pA = –1000  10 (0.9–0) = –9000 Pa.
pB = –000  10 (0.9)–1000  9.81(0.6) = –3114 Pa
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
d) pA = 0. pB = 1.94  (32.2-60)
25
12





 = 112 psf. pC = –112 psf.
e) pA = 1.94  60 






37 5
12
.
= 364 psf.
pB = 1.94  60 






37 5
12
.
– 1.94  32.2 






25
12
= –234 psf.
pC = –1.94  32.2 






25
12
= 130 psf.
f) pA = 1.94  30
37 5
12
.





 = 182 psf.
pB = –1.94(–30)
37 5
12
.





 – 1.94  62.2 






25
12
= 433 psf.
pC = –1.94  62.2  






25
12
= 251 psf.
Rotating Containers
2.103 Use Eq. 2.6.4 with position 1 at the open end:




50 2
60
= 5.236 rad/s.
a) 2 2
(1000 5.236 /2) (0.6 1.5)
A
p     = 11 100 Pa.
pB 
1
2
 1000  5.2362
 0.92
+ 9810  0.6 = 16 990 Pa.
pC = 9810  0.6 = 5886 Pa.
b) pA 
1
2
 1000  5.2362
 0.62
= 4935 Pa.
pB 
1
2
 1000  5.2362
 0.62
+ 9810  0.4 = 8859 Pa.
pC = 9810  0.4 = 3924 Pa.
A
B
C
x
z
1
A
B
C
z
1
r


39
c) pA 
1
2
 1.94  5.2362

37 5
12
2
.





 = 259.7 psf.
pB 
1
2
 1.94  5.2362

37 5
12
62
25
12
2
.
.4





   = 389.7 psf.
pC = 62.4 25/12
 = 130 psf.
d) pA 
1
2
 1.94  5.2362

22 5
12
2
.





 = 93.5 psf.
pB 
1
2
 1.94  5.2362

22 5
12
2
.





 + 62
15
12
.4  = 171.5 psf.
pC = 62.4 15/12
 = 78 psf.
2.104 Use Eq. 2.6.4 with position 1 at the open end.
a) pA 
1
2
 1000  102
(0 – 0.92
) = –40 500 Pa.
pB = –40 500 + 9810  0.6 = –34 600 Pa.
pC = 9810  0.6 = 5886 Pa.
b) pA 
1
2
 1000  102
(0 – 0.62
) = –18 000 Pa.
pB = –18 000 + 9810  0.4 = –14 080 Pa.
pC = 9810  0.4 = 3924 Pa.
c) pA 
1
2
 1.94  102
0
37 5
144
2







.
= –947 psf.
pB = 947 + 62.4  25/12 = –817 psf. pC = 62.4  25/12 = 130 psf.
d) pA 
1
2
 1.94  102







22 5
12
2
2
.
= –341 psf.
pB = –341 + 62.4  15/12 = –263 psf. pC = 62.4  15/12 = 78 psf.
2.105 Use Eq. 2.6.4 with position 1 at the open end and position 2 at the origin. Given: p2 = 0.
a) 0 =
1
2
 1000 2
(0 – 0.452
) – 9810 (0 – 0.6).  = 7.62 rad/s.
b) 0 =
1
2
 1000 2
(0 – 0.32
) – 9810 (0 – 0.4).  = 9.34 rad/s.
c) 0 =
1
2
 1.94 2
0
18 75
12
2
2







.
– 62.4 






25
12
.  = 7.41 rad/s.
d) 0 =
1
2
 1.94 2







1125
12
2
2
.
– 62.4 






15
12
.  = 9.57 rad/s
A
B
C
z
1
r

z
1
r



40
2.106 The air volume before and after is equal.
   
1
2
6 2
0
2 2
 
r h . . .  r h
0
2
= 0.144.
a) Using Eq. 2.6.5: r0
2 2
5 2
 / = 9.81 h
h = 0.428 m
pA =
1
2
 1000  52
 0.62
– 9810 (–0.372)
= 8149 Pa.
b) r0
2 2
7 2
 / = 9.81 h. h = 0.6 m.
pA =
1000
2
 72
 0.62
+ 9810  0.2 = 10 780 Pa.
c) For  = 10, part of the bottom is bared.
  
   
. . .
6 2
1
2
1
2
2
0
2
1
2
1
r h r h
Using Eq. 2.6.5:
2
0
2
2
r
g
h
 ,
 2
1
2
2
r
g
h
 1.
  
0144
2 2
2
2
2 1
2
.
g
h
g
h
 
or
h h
2
1
2
2
0144 10
2 9 81
 


.
.
.
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.
d) Following part (c): h h
2
1
2
2
0144 20
2 9 81
 


.
.
. 1.6h – 0.64 = 2.936. h = 2.235 m.
pA = 1000  202
(0.62
– 0.2652
)/2 = 57 900 Pa r1 = 0.265 m
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 d) 82 900 Pa


h
z
r
A
r0

h
z
r
A
r0
h1

41
2.108 2 2
1
( ) [0 (0.8 )].
2
p r r g h
 
   
2 2
( ) 500 9810(0.8 )
p r r h

   if h < 0.8.
p r r r
( ) ( )
 
500 2 2
1
2
 if h > 0.8.
a)
0.6
3
0
2 2 (12 500 3650 )
F p rdr r r dr
 
  
  = 6670 N.
(We used h = 0.428 m)
b)
0.6
3
0
2 2 (24 500 1962 )
F p rdr r r dr
 
  
  = 7210 N. (We used h = 0.6 m)
c)
0.6
3 2
0.108
2 2 (50 000( 0.108 )
F p rdr r r dr
 

  
  = 9520 N. (We used r1 = 0.108 m)
d)
0.6
3 2
0.265
2 2 (200 000( 0.265 )
F p rdr r r dr
 
  
  = 26 400 N. (We used r1 = 0.265 m)
dr
dA = 2rdr

42

Chapter 3 / Introduction to Fluid Motion
43
CHAPTER 3
Introduction to Fluids in Motion
3.1 (D) ˆ ˆ ˆ ˆ
ˆ 0. ( ) (3 4 ) 0 or 3 4 0
x y x y
n n n n
n V i j i j
Also 2 2
1
x y
n n since n̂ is a unit vector. A simultaneous solution yields
4/ 5 and 3/ 5.
x y
n n (Each with a negative sign would also be OK.)
3.2 (C) 2
ˆ ˆ ˆ ˆ ˆ ˆ
2 (2 ) (2 2 ) 16 8 16
u v w xy y y x y
t x y z
V V V V
a i i j i i j
2 2
( 8) 16 17.89 m/s
a
3.3 (D) 2
2
10
10(4 )
(4 )
x
u u u u u
a u v w u x
t x y z x x
x
3 2
2
10 10 1
10( 2)( 1)(4 ) 20 6.25 m/s .
4 8
(4 )
x
x
3.4 (C) 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.
3.5 (B)
2
9810 0.800
. 113 m/s.
2 1.23
water
air
h
V p
V
3.6 (C)
2 2 2
1 2 1
. 0.200 0.600. 2 9.81 0.400 2.80 m/s.
2 2 2
V V V
p
V
g g g
3.7 (B) The manometer reading h implies:
2 2
2
1 1 2 2
2 2
2
or (60 10.2). 9.39 m/s
2 2 1.13
V p V p
V V
The temperature (the viscosity of the water) and the diameter of the pipe are not
needed.
3.8 (A)
2
1
2
V
g
2
1 2 2
2
p V p
g
2
2
2
800 000
. . 40 m/s.
9810 2 9.81
V
V

44
3.9 (D) 2 2 2 2
1 2 1
902
30 15 304 400 Pa
2 2
p V V
Chapter 3 Problems: Flow Fields
3.10
3.11 Pathline: Release several at an instant in time and take a time exposure of the subse-
quent 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
3.13
3.14 a) 2 2 2
dy
dx
u t v t
dt dt
x t t c y t c
2
1
2
2
2
y y
2
x xy y y
2 2
2 4 parabola.
streakline
pathline
streamline
streakline
pathline hose
boy
time t
t = 0
streamlines
t = 2 hr
pathline
t = 2 hr
streakline at t = 3 hr
y
x
39.8o
y
x
streamlines
t = 5 s
(27, 21)
(35, 25)

45
b) x t t c c c
2
1 1 2
2 8 4
. , .
and
y y
4 2 4 8
( )
x xy y x y
2 2
2 8 12 0. parabola.
3.15
ˆ ˆ ˆ ( )
ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ
using , .
z
d udy vdx
u v w
d dx dy dz
V r
V i j k
r i j k i j k j i k
3.16 Lagrangian: Several college students would be hired to ride bikes around the
various roads, making notes of quantities of interest.
Eulerian: Several college students would be positioned at each intersection
and quantities would be recorded as a function of time.
3.17 a) At 2
2 and (0,0,0) 2 2 fps.
t V
At 2 2
2 and (1, 2,0) 3 2 3.606 fps.
t V
b) At t V
2 0 0 0 0
and ( , , ) .
At 2 2
2 and (1, 2,0) ( 2) ( 8) 8.246 fps.
t V
c) At 2
2 and (0,0,0) ( 4) 4 fps.
t V
At 2 2 2
2 and (1, 2,0) 2 ( 4) ( 4) 6 fps.
t V
3.18 a)
2 2
ˆ 1 2
cos 0.832. 33.69
3 2
V
V i
2 2
2 2
3
3 2 0
2
ˆ ˆ ˆ ˆ
ˆ 0. (3 2 ) ( ) 0.
9
1
1
4
y x
x y
x y
x y
x x
n n
n n
n n
n n
n n
V n i j i j
2 3 1 ˆ ˆ
ˆ
, or (2 3 ).
13 13 13
x y
n n n i j
b)
2 2
ˆ 2
cos 0.2425. 104
( 2) ( 8)
V
V i
2 2 2 2
2 8 0 4
ˆ ˆ ˆ ˆ
ˆ 0. ( 2 8 ) ( ) 0.
1 16 1
x y x y
x y
x y y y
n n n n
n n
n n n n
V n i j i j
1 4 1 ˆ ˆ
ˆ
, or ( 4 ).
17 17 17
y x
n n n i j

46
c)
2 2
ˆ 5
cos 0.6202. 51.67
5 ( 8)
V
V i
2 2
2 2
8
5 8 0
5
ˆ ˆ ˆ ˆ
ˆ 0. (5 8 ) ( ) 0.
64
1
1
25
x y
x y
x y
x y
y y
n n
n n
n n
n n
n n
V n i j i j
5 8 1 ˆ ˆ
ˆ
, or (8 5 ).
89 89 89
y x
n n n i j
3.19 a) ˆ ˆ ˆ ˆ
0. ( 2) ( ) 0.
d x xt dx dy
V r i j i j
( ) .
x dy xtdx t
xdx
x
dy
2 0
2
or
Integrate: . 2ln 2 .
2
xdx
t dy t x x y C
x
2(1 2ln3) 2 . 0.8028.
C C
2ln 2 0.8028
t x x y
b) 2
ˆ ˆ ˆ ˆ
0. 2 ( ) 0.
d xy y dx dy
V r i j i j
xydy y dx
dx
x
dy
y
2 0
2
2
or .
Integrate: 2 ln( / ). 2ln(1) ln( 2/ ).
nx y C C
2 2
2. ln ln( / 2). 2.
C x y x y
c) 2 2
ˆ ˆ ˆ ˆ
0. ( 4) ( ) 0.
d x y t dx dy
V r i j i j
( ) .
x dy y tdx
tdx
x
dy
y
2 2
2 2
4 0
4
or
Integrate: 1 1
1 2 1 1
tan . tan .
2 2 2 2 2
t x
C C
y
1
0.9636. tan 0.9636 2
2
x
C yt
3.20 a) 0
D
u v w
Dt x y z t
V V V V V
b) ˆ ˆ ˆ ˆ
2 (2 ) 2 (2 ) 4 4
u v w x y x y
x y z t
V V V V
i j i j= ˆ ˆ
8 4
i j

47
c) 2 2
ˆ ˆ ˆ ˆ ˆ
ˆ ˆ
(2 2 ) 2 (2 2 ) 2 ) 2
u v w x t xt yt xyt xt zt x xy yz
x y z t
V V V V
i j j k i j k
ˆ ˆ ˆ
68 100 54
i j k
d) ˆ ˆ ˆ ˆ ˆ ˆ
( 2 ) 2 ( 2 ) ( 2 )
u v w x yz xyz xz tz xy t z
x y z t
V V V V
i j j j k k
= 2 2 2
ˆ ˆ ˆ
(2 4 2 ) ( )
x yz x yz xyzt zt z
i j k
= ˆ ˆ ˆ
2 114 15
i j k
3.21
1 1 1
ˆ ˆ ˆ
2 2 2
w v u w v u
y z z x x y
Ω i j k
a)
1 ˆ ˆ
20
2
u
y
y
Ω k k = ˆ
20k
b)
1 1 1
ˆ ˆ ˆ
(0 0) (0 0) (0 0)
2 2 2
Ω i j k = 0
c)
1 1 1
ˆ ˆ ˆ
ˆ ˆ
(2 0) (0 0) (2 0) 6 2
2 2 2
zt yt
Ω i j k i k
d)
1 1 1
ˆ ˆ ˆ
ˆ ˆ
(0 2 ) (0 0) ( 2 0) 2 3
2 2 2
xy yz
Ω i j k i k
3.22 The vorticity 2 .
ω Ω Using the results of Problem 3.21:
a) ˆ
40
ω i b) ω 0 c) ˆ ˆ
12 4
ω i k d) ˆ ˆ
4 6
ω i k
3.23 a) 0, 0, 0.
xx yy zz
u v w
x y z
1 1
20 20, 0,
2 2
xy xz
u v u w
y
y x z x
0 20 0
1
0. rate-of strain 20 0 0
2
0 0 0
yz
v w
z y
b)
xx yy zz
xy xz yz
2 2 0
0 0 0
, , .
, , .
rate-of strain =
2 0 0
0 2 0
0 0 0

48
c) xx yy zz
xt xt yt
2 8 2 8 2 4
, , .
xy xz yz
yt zt
1
2
2 2
1
2
0 0
1
2
2 6
( ) , ( ) , ( ) .
rate-of strain =
8 2 0
2 8 6
0 6 4
d) xx yy zz
xz t
1 2 12 2
, , .
xy xz yz
yz xy
1
2
2 3
1
2
0 0
1
2
2 2
( ) , ( ) , ( ) .
rate-of strain =
1 3 0
3 12 2
0 2 2
3.24 a) a
r r r r r
r 10
40 80
10
40
1
40
2 3 2 2
cos cos
sin
( sin )
1
10
40
10 2 1 125 1
2
2
2
r r
sin ( .5)( ) . ( ) = 9.375 m/s2
.
a
r r r r r
10
40 80
10
40
10
40
2 3 2 2
cos sin
sin
cos
1
100
1600
4
r r
sin cos = 0 since sin 180 = 0.
a = 0.
b) r z
r r r r
0 0
1
10
40 1
10
40
2 2
, , sin ( sin ) = 0.
At (4, 180 ) ω= 0 since ω = 0 everywhere.
3.25 a) a
r r r r r
r 10
80 240
10
80
10
80
3 4 3 3
cos cos
sin
( sin )
10
80
8 75 1 9375 1
3
2 2
r r
sin
. ( )(. )( ) = 8.203 m/s2
a = 0 since sin 180 = 0. a = 0 since 0.
v
b) r = 0, = 0, = 0, since sin 180 = 0.
3.26 u
t x
V V
a v w
y
V ˆ.
u
z t
V
i For steady flow / 0 so that 0.
u t a

49
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 0 and / 0:
v
r x
v v u u
a v u a v u
r x r x
3.28 a) /10
2(1 0)(1 ) 2 m/s at .
t
u e t
/10 2
1
2(1 0) 0.2 m/s 0.
10
t
x
u
a e at t
t
b) 2 /10
2(1 0.5 )(1 ) 1.875 m/s at .
t
u e t
2 2 /10 2
1
2(1 0.5 / 2 ) 0.0125 m/s at 0.
10
t
x
a e t
c) u e t
t
2 1 2 2 1
2 2 10
( / )( ) .
/
0 for all
a e t
x
t
2 1 2 2
1
10
0
2 2 10
( / ) .
/
for all
3.29
DT T
u
Dt x
v
T
w
y
2
20(1 ) sin 0.5878
100 100 5
T T t
y
z t
= 0.3693 C/ s
3.30
4
4 3000 10
10( 1.23 10 )
D
u v w e
Dt x y z t
= 9 11 10 4
. .
kg
m s
3
3.31
1000
10
4
D
u v w
Dt x y z t
= 2500
kg
m s
3
.
3.32 4 (.01)
D
u
Dt x
= 0.04 kg/m3
s
3.33
D
Dt
V
t
observing that the dot product of two vectors ˆ ˆ ˆ
x y z
A A A
A i j k
and ˆ ˆ ˆ is
x y z x x y y z z
B B B A B A B A B
B i j k A B .
3.34 ( )
x
y
z
u
a u
t
v
a v
t t
w
a w
t
V
V
V a V V
V

50
3.35 Using Eq. 3.2.12:
a)
2
2
2 ( )
d d
dt
dt
S Ω
A a Ω V Ω Ω r r
= 2
ˆ ˆ ˆ ˆ
ˆ ˆ ˆ
2(20 4 ) 20 (20 1.5 ) 160 600 m /s
k i k k i j i
b) ˆ ˆ ˆ
ˆ ˆ ˆ
2 ( ) 2(20 20cos30 ) 20 (20 3 ) 507
A Ω V Ω Ω r k j k k i i = ˆ
507i
3.36 5
2 ˆ ˆ
7.272 10
24 60 60
Ω k k rad/s.
ˆ ˆ
ˆ ˆ
5( 0.707 0.707 ) 3.535 3.535
V i k i k m/s.
2 ( )
A Ω V Ω Ω r
= 5 5
ˆ
ˆ ˆ ˆ
2 7.272 10 ( 3.535 3.535 ) 7.272 10
k i k k
5 6 ˆ
ˆ ˆ
7.272 10 (6.4 10 )( 0.707 0.707 )
k i k = 5 2
ˆ ˆ
52 10 0.0224 m/s
j i .
Note: We have neglected the acceleration of the earth relative to the sun since it is quite
small (it is 2 2
/ ).
d dt
S The component 5ˆ
( 51.4 10 )
j 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) d) e)
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.

51
3.44
3.45 Re VL/ = 2 0.015/0.77 10 6
= 39 000. Turbulent.
3.46 Re
VL
0.2 0.8/1.4 10 5
3.47 Re
.
.
VL 4 06
17 10 5
Note: We used the smallest dimension to be safe!
3.48 a) 5
1.2 0.01
Re 795.
1.51 10
VD
Always laminar.
b) Re
.
.51
VD 12 1
1 10
79
5
500. May not be laminar.
3.49 Re = 3 105
= T
Vx
. / where ( ).
T
a) T = 223 K or 50 C. 5 2
1.5 10 N s/m ,
5
5 2
1.5 10
2.5 10 m /s.
0.3376 1.23
3 10
900 1000
3600 2 10
5
5
xT
.5
. xT = 0.03 m or 3 cm
b) T = 48 F = 3.3 10 7
lb-sec/ft2
.
7
4
3.3 10
3.7 10
0.00089
ft2
/sec.
5
4
600 5280
3 10 .
3600 3.7 10
T
x
xT = 0.13' or 1.5"
3.50 Assume the flow is parallel to the leaf. Then 3 105
= / .
T
Vx
5 5 4
3 10 / 3.5 10 1.4 10 / 6 8.17 m.
T
x V
The flow is expected to be laminar.
3.51 a)
100
M 0.325.
1.4 287 236
V
c
For accurate calculations the flow is
compressible. Assume incompressible flow if an error of 4%, or so, is acceptable.
b)
80
M 0.235.
1.4 287 288
V
c
Assume incompressible.
c)
100
M 0.258.
1.4 287 373
V
c
Assume incompressible.

52
3.52 0.
D
u v w
Dt x y z t
For a steady, plane flow / 0
t and 0.
w
Then 0.
u v
x y
3.53
D
u
Dt x
v
y
w
z t
0. incompressible.
Bernoulli’s Equation
3.54
2
.
2
p
V
Use = 0.0021 slug/ft3
.
a) 2 / 2 0.3 144 / 0.0021
v p = 203 ft/sec
b) 2 / 2 0.9 144 / 0.0021
v p = 351 ft/sec
c) 2 / 2 0.09 144 / 0.0021
v p = 111 ft/sec
3.55
2
2
1.23 120 1000
2 2 3600
V
p = 683 Pa F = pA = 683 0.0752
= 12.1 N.
3.56
2
0.
2
p
V 2 2 2000
1.23
p
V = 57.0 m/s
3.57 a)
2
2
0
2 2
p V
V 2
2
0 0
0
( 10 )
. . 50
2
p p p
x
p p x
b)
2
2
0
2 2
p V
V 2
2
0 0
0
(10 )
. . 50
2
p y p p
p p y
3.58
2
2
2 2
p p
U
V
. a) 2 2
0 and 180 , (1 / )( 1).
r c
v v U r r
4
2
2 2 2
2
2 .
2 2
c c
r
r r
p U v U
r
r
b) Let r r p U
c T
:
2
2
c) 2 2 2 2
0 and , 2sin . U 1 4sin
2 2
r c
v r r v U p v U
d) Let 90
3
2
90
2

: p U

53
3.59
2
2
2 2
U p
V p
.
a) 0 and 180 :
v
3 6
2 2 2
2 .
2 2
c c
r
r r
p U v U
r r
b) Let r r p U
c T
:
1
2
2
.
c) 2 2 2 2
0 and : 1 4sin
2 2
r c
v r r p U v U
d) Let 2
90
3
90 :
2
p U
3.60
2
2
2 2
U p
V p
.
a)
2 2
2 2 2 20 1
( ) 10 10 50 1 1
2 2 2
p U u
x x
50
2 1
2
x x
b) 1
0 when 1. 50 ( 2 1) 50
u x p
c)
2 2
2 2 2 60 1
( ) 30 30 450 1 1
2 2 2
p U u
x x
450
2 1
2
x x
d) 1
0 when 1. 450 ( 2 1) 450
u x p
3.61
V p V p
V p p
1
2
1 2
2
2
1 1 2
2 2
0 20
. and kPa.
2
2 1 2 2
2 2
(20 000) 40. 6.32 m/s
1000
V p p V
3.62 Assume the velocity in the plenum is zero. Then
2 2
2
1 1 2 2
2 2
2
or (60 10.2). 9.39 m/s
2 2 1.13
V p V p
V V
We found 113
. kg / m3
in Table B.2.

54
3.63 Applying the Bernoulli equation between points 1 and 2 along the streamline on the
centerline of the flow:
2 2
1 2
1 1 2 2
2 2
V V
p z p z
From the manometer reading we find
2
2
1 1 2 2 Hg
( )
2
V
p z p z H H
where we have used Eq. 3.4.11. Subtract the manometer equation from Bernoulli’s
equation and we have
2
1
Hg
( )
2
V
H
Substitute in the given information and there results
3
1 3
2 (13.6 1) 9810 N/m 0.12 m
5.45 m/s
1000 kg/m
V
3.64 Bernoulli from the stream to the pitot probe: p
V
p
T
2
2
.
Manometer: .
T Hg
p H H h p h
Then,
2
2
Hg
V
p H H p . 2
(2 )
Hg
V H
a) 2 (13.6 1)9800
(2 0.04). 3.14 m/s
1000
V V
b) 2 (13.6 1)9800
(2 0.1). 4.97 m/s
1000
V V
c) 2 (13.6 1)62.4
(2 2 /12). 11.62 fps
1.94
V V
d) 2 (13.6 1)62.4
(2 4 /12). 16.44 fps
1.94
V V
3.65 Applying Bernoulli’s equation between the two sections connected by the manometer we
write
2 2
1 2
1 1 2 2
2 2
V V
p z p z

55
The manometer equation is
2
1
1 air 1 2 air 2 water
( )
2
V
p z p z H H
Subtract the manometer equation from Bernoulli’s equation and obtain
2
2
air water
0 ( )
2
V
H
Since air is an ideal gas we calculate the density as follows:
2
3
120 N/m
1.38 kg/m
0.287 N m/kg K 30 273 K
p
RT
Substitute in the given information:
3
water air
2 3
2( ) 2 (9810 1.38 9.81) N/m 0.05 m
26.6 m/s
1.38 kg/m
H
V
The air column could have been neglected.
3.66 The pressure at 90 from Problem 3.58 is 2
90 3 / 2.
p U The pressure at the
stagnation point is 2
/ 2.
T
p U The manometer provides: p H p
T 90
2 2
1 3
1.204 9800 0.04 1.204 . 12.76 m/s
2 2
U U U
3.67 The pressure at 90 from Problem 3.59 is 2
90 3 / 2.
p U The pressure at the
stagnation point is 2
/ 2.
T
p U The manometer provides: p H p
T 90
2 2
1 3
1.204 9800 0.04 1.204 . 12.76 m/s
2 2
U U U
3.68 Bernoulli:
2
2 2
2
V p
g
2
1 1
2
V p
g
Manometer:
2
2
1 2
2
Hg
V
p z H H z p
g
Substitute Bernoulli’s into the manometer equation:
2
1
1 1.
2
Hg
V
p H p
g

56
a) Use H = 0.01 m:
2
1
1
9800
(13.6 1)9800 0.01 1.572 m/s
2 9.81
V
V
Substitute into Bernoulli:
p
V V
g
1
2
2
1
2 2 2
2
20 1
2 9 81
9800 198
.572
.
600 Pa
b) Use H = 0.05 m:
2
1
1
9800
(13.6 1)9800 0.05 3.516 m/s
2 9.81
V
V
p
V V
g
1
2
2
1
2 2 2
2
20 3
2 9 81
9800 193
.516
.
600 Pa
c) Use H = 0.1 m:
2
1
1
9800
(13.6 1)9800 0.1 4.972 m/s
2 9.81
V
V
p
V V
g
1
2
2
1
2 2 2
2
20 4 972
2 9 81
9800 187
.
.
400 Pa
3.69 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:
2 2
1 2
1 1 2 2
2 2
V V
p z p z
Since the flow is horizontal 1 2
z z , p1 = (120 + 100) kPa absolute, and p2 = 1.7 kPa
absolute so Bernoulli’s equation takes the form
2 2
1 1
1
1000 1000 (4 )
220 000 1700 5.40 m/s
2 2
V V
V
Substitute in the units to make sure they check.
3.70 Write Bernoulli’s equation between points 1 and 2 along the center streamline:
2 2
1 2
1 1 2 2
2 2
V V
p z p z
Since the flow is horizontal, 1 2
z z and Bernoulli’s equation becomes
2 2
1 2
0.5 1.125
1000 1000
2 2
p p

57
From fluid statics, the pressure at 1 is 1 9810 0.25 2452 Pa
p h and at 2, using
p2 = H, Bernoulli’s equation predicts
2 2
0.5 1.125
2452 1000 9810 1000 0.1982 m or 19.82 cm
2 2
H H
3.71 Assume an incompressible flow with point 1 outside in the room where p1 0 and
1 0.
v The Bernoulli’s equation gives, with p h
w
2 2 ,
2
1
2
V 1
p 2
2 2
.
2
V p
a)
2
2
2
9800 0.02
0 . 18.04 m/s
2 1.204
V
V
b)
2
2
2
9800 0.08
0 . 36.1 m/s
2 1.204
V
V
c) 0
2
62 4 1 12
0 00233
66 8
2
2
2
V
V
. /
.
. . fps
d) 0
2
62 4 4 12
0 00233
133 6
2
2
2
V
V
. /
.
. . fps
3.72 Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel where
p V
1 1
0 0
and . Bernoulli’s equation gives
2
2
2
2
2 2
1
0 .
2 2
p
V
p V
a) 3 2
2
90 1
1.239 kg/m . 1.239 100 6195 Pa
0.287 253 2
p
p
RT
b) 3 2
2
95 1
1.212 kg/m . 1.212 100 6060 Pa
0.287 273 2
p
p
RT
c) 3 2
2
92 1
1.094 kg/m . 1.094 100 5470 Pa
0.287 293 2
p
p
RT
d) 3 2
2
100 1
1.113 kg/m . 1.113 100 5566 Pa
0.287 313 2
p
p
RT
3.73 a) p h V h h
A A A
9800 4 39 0 2
200 Pa, Using
. ,
2
2
A
V
g
A
A
p
h
2
2 2
2
2
V p
h
g
2
2
2
.
2
A
V
p p
g
39
14
2 9 81
9800 58
2
200 700 Pa
.

58
b) 0 and 0.
B B
p V Bernoulli’s eqn gives, with the datum through the pipe,
V
g
p
h
V
g
p
h p
B B
B
2
2
2
2
2 2
2
2 2
4
14
2 9 81
9800 58
.
.
700 Pa
3.74 Bernoulli across nozzle:
2
1
2
V 2
1 2 2
2
p V p
2 1
. 2 /
V p
Bernoulli to max. height:
2
1
2
V
g
1
1
p
h
2
2
2
V
g
2
p
2 2 1
. / .
h h p
a) 2 1
2 / 2 700 000 /1000 37.42 m/s
V p
h p
2 1 700
/ 000/ 9800 = 71.4 m
b) 2 1
2 / 2 1 400 000 /1000 52.92 m/s
V p
h p
2 1 / 1 400 000/ 9800 = 142.9 m
c) 2 1
2 / 2 100 144 /1.94 121.8 fps
V p
h p
2 1 / 100 144/ 62.4 = 231 ft
d) 2 1
2 / 2 200 144 /1.94 172.3 fps
V p
h p
2 1 200
/ 144/ 62.4 = 462 ft
3.75 a) Apply Bernoulli’s eq. from the surface to a point on top of the downstream flow:
2
1
2
V
g
1
p 2
2 2
1
2
V p
h
g
2 2
. 2 ( )
h V g H h
b) Apply Bernoulli’s eq. from a point near the bottom upstream to a point on the bottom
of the downstream flow:
2
1
2
V
g
2
1 2 2
1 2.
2
p V p
h h
g
Using p H p h h h V g H h
1 2 1 2 2 2
, , ( )
and
3.76
2
1
2
V 2
1 2 2
.
2
p V p
p2 = 100 000 Pa, the lowest possible pressure.
a)
600
2
2
2
000
1000
100 000
1000
V
. V2 = 37.4 m/s.
b)
300
2
2
2
000
1000
100 000
1000
V
. V2 = 28.3 m/s.

59
c)
80 144
1.94
14.7 144
1.94
V2
2
2
. V2 = 118.6 ft/sec.
d)
40 144
1.94
14.7 144
1.94
V2
2
2
. V2 = 90.1 ft/sec.
3.77 A water system must never have a negative pressure, since a leak could ingest impurities.
The least pressure is zero gage:
V p
gz
V p
gz
1
2
1
1
2
2
2
2
2 2
. V V
1 2 . Let z1 0, and p2 0.
500 000
1000
9 81 2
. .
z z2 = 51.0 m.
3.78 a) 2 2 2 2
1 2 1
1000
( ) (2 10 )
2 2
p V V = 48 000 Pa
b) 2 2 2 2
1 2 1
902
( ) (2 10 ) 43300 Pa
2 2
p V V
c) 2 2 2 2
1 2 1
680
( ) (2 10 ) 32 600 Pa
2 2
p V V
d) 2 2 2 2
1 2 1
1.23
( ) (2 10 ) 59.0 Pa
2 2
p V V
3.79
V p V p
1
2
1 2
2
2
2 2
. 2 2 2 2
1 2 1
1.23
2 30
2 2
p V V = 551 Pa
3.80 Apply Bernoulli’s equation between the exit (point 2) where the radius is R and a point 1
in between the exit and the center of the tube at a radius r less than R:
V p V p
p
V V
1
2
1 2
2
2
1
2
2
1
2
2 2 2
. .
Since V V
2 1 , we see that p1 is negative (a vacuum) so that the envelope would tend to
rise due to the negative pressure over most of its area (except for a small area near the
end of the tube).
3.81 Re .
VD
For air 5
1.5 10 m2
/s. Use reasonable dimensions from your experience!
a) Re
.
.5
.
20 0 03
1 10
4 10
5
4
Separate
b) Re
.
.5
.
20 0 005
1 10
6700
5
Separate
c) Re
.5
. .
20 2
1 10
2 7 10
5
6
Separate

60
d) Re
.
.5
.
5 0 002
1 10
670
5
Separate
e) Re
.5
. .
20 2
1 10
2 7 10
5
6
Separate
f) Re
.5
.
100 3
1 10
2 10
5
7
It will tend to separate, except
streamlining the components eliminates separation.
3.82 A burr downstream of the opening will create a region that acts
similar to a stagnation region thereby creating a high pressure since
the velocity will be relatively low in that region.
3.83 p
V
R
n
2 2
1000
10
0 05
0 02
.
. 40 000 Pa Along AB, we
expect 10 m/s and 10 m/s.
A B
V V
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:
V p V p
1
2
1 2
2
2
2 2
p p
A B since V V
A B
p p
C D since V V
C D
p p
B D since V V
D B
stagnation
region
A
B
VA
VB

Chapter 4 / The Integral Forms of the Fundamental Laws
61
CHAPTER 4
The Integral Forms of the
Fundamental Laws
4.1 (B)
4.2 (D) 2
200
0.04 70 0.837 kg/s
0.287 293
p
m AV AV
RT
 
     

.
4.3 (A) Refer to the circle of Problem 4.27:
2 3
75.7 2
( 0.4 0.10 0.40 sin75.5 ) 3 0.516 m /s.
360
Q AV 

        
4.4 (D)
2 2
2 1
2
P
W V V
Q g


 2 1 1200 200
. .
0.040
P
p p W
  
 
 

40
40 kW and energy req'd = 47.1 kW.
0.85
P
W
  
4.5 (A)
2
2
0
V

2
1 2 1
2
V p p
g
 

2
2
2
120
. 0 . 7 200 000 Pa.
2 9.8 9810
p
p


   

4.6 (C) Manometer:
2
2
1 2
2
V
H p g p
g
 
   or
2
2
1
9810 0.02 .
2
V
p g
g

  
Energy:
2
100 000
7.96
. 3.15.
2 9.81 9810
K K
  

Combine the equations:
2
1
1
9810 0.02 1.2 . 18.1 m/s.
2
V
V
    
4.7 (B)
2
2
0.040
. 7.96 m/s.
2 0.04
L
V p Q
h K V
g A
 

    

2
100 000
7.96
. 3.15.
2 9.81 9810
K K
  


62
4.8 (C)
2 2
2 1
2
P
W V V
Q g


 .
p



16
0.040 400 16 kW. 18.0 kW.
0.89
P
P
W
W Q p

      
4.9 (D)
2 2
4.58 7.16
36.0 15 3.2 . 416 000 Pa
2 9.81 9810 2 9.81
B
B
p
p
     
 
In the above energy equation we used
2
2
0.2
with 4.42 m/s.
2 0.2
L
V Q
h K V
g A 
   

4.10 (A) V
Q
A
 


0 1
04
19 89
2
.
.
.

m / s.
Energy —surface to entrance: H
V
g
p
z K
V
g
P    
2
2
2
2
2
2
2 2

.
 

  


HP
19 89
2 9 81
180
50 5 6
19 89
2 9 81
201 4
2 2
.
.
.
.
.
.
000
9810
m.
     
 / . . / . .
W QH
P P P
  9810 0 1 2014 0 75 263 000 W
4.11 (A) After the pressure is found, that pressure is multiplied by the area of the
window. The pressure is relatively constant over the area.
4.12 (C)
2 2
1 1 2 2
2 2
V p V p
g g
 
  
2 2
1
(6.25 1) 12.73
. 9810 3 085 000 Pa.
2 9.81
p
 
  

2
1 1 2 1
( ). 3 085 000 0.05 1000 0.1 12.73(6.25 1)
p A F Q V V F
 
         
17 500 N.
F
 
4.13 (D) 2 1
( ) 1000 0.01 0.2 50(50cos60 50) 2500 N.
x x x
F m V V
         
4.14 (A) 2
2 1
( ) 1000 0.02 60 (40cos45 40) 884 N.
x r r
x x
F m V V 
         
Power 884 20 17 700 W.
x B
F V
    
4.15 (A) Let the vehicle move to the right. The scoop then diverts the water to the
right. Then
2 1
( ) 1000 0.05 2 60 [60 ( 60)] 720 000 N.
x x
F m V V
         

63
Chapter 4 Problems: Basic Laws
4.16 a) No net force may act on the system: 0.
 
F
b) The energy transferred to or from the system must be zero: Q  W = 0.
c) If 3 3 2
ˆ ˆ
ˆ 10 ( ) 0
n
V      
V n i j is the same for all volume elements then
,
D
dm
Dt
  
F V or ( ).
D
m
Dt
 
F V Since mass is constant for a system .
D
m
Dt
 
V
F
Since , .
D
m
Dt
  
V
a F a
4.17 Extensive properties: Mass, m; Momentum, ;
mV kinetic energy,
1
2
mV2
;
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, .
4.18
System ( )
t V
 1
c.v.( )
t V
 1
System ( )
t t V
   1 V
 2
c.v.( )
t t V
   1
4.19
System ( )
t V
 1 V
 2
c.v.( )
t V
 1 V
 2
System ( )
t t V
   2 V
 3
c.v.( )
t t V
   1 V
 2
4.20 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
v
ˆ
a)
v
n
ˆ
b)
n
v
ˆ
c)
n
v
ˆ
d)
n
v

ˆ
e)
1 2
1
2
3
pump

64
4.22
n
v
n
n n
n
v
v
v
v
ˆ
ˆ
ˆ
ˆ
4.23
n
n
n
n n
n
v
v
v
v
v
v
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
System-to-Control Volume Transformation
4.24 1
1 1
ˆ ˆ ˆ ˆ
ˆ 0.707( )
2 2
    
n i j = i j . 2
ˆ ˆ
ˆ 0.866 0.5
 
n i j. 3
ˆ
ˆ  
n j .
1 1 1
ˆ ˆ ˆ
ˆ 10 0.707( ) 7.07 fps
n
V  
       
 
V n i i j
2 2 2
ˆ ˆ ˆ
ˆ 10 (0.866 0.5 ) 8.66 fps
n
V      
V n i i j . 3 3 2
ˆ ˆ
ˆ 10 ( ) 0
n
V      
V n i j
4.25 flux = ˆ A
 
n V flux1 =
ˆ ˆ ˆ
0.707( ) 10
10
0.707


 
  
   
i j iA
A
flux2 =
ˆ ˆ ˆ
(0.866 0.5 ) 10
10
0.866


 

i j iA
A flux3 = 3
ˆ ˆ
( ) 10 0
A
   
j i
4.26 ˆ ˆ ˆ
ˆ
( ) 15(0.5 0.866 ) (10 12)
A
    
B n i j j    
15 0 866 120 1559
. cm 3
Volume = 3
15 sin 60 10 12 1559 cm
  
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
1
2
1
system (t) is in
volumes 1 and 2
c.v. (0) = c.v. (t)
= volume 1

65
4.29
4.30
Conservation of Mass
4.31 If fluid crosses the control surface only on areas A1 and A2,
1 2
. .
ˆ ˆ ˆ 0
c s A A
dA dA dA
  
     
  
n V n V n V
For uniform flow all quantities are constant over each area:
1 2
1 1 1 2 2 2
ˆ ˆ 0
A A
dA dA
 
   
 
n V n V
Let A1 be the inlet so 1 1 1 2
ˆ and
V A
  
n V be the outlet so 2 2 2
ˆ .
V
 
n V Then
  
 
1 1 1 2 2 2 0
V A V A
or
 
2 2 2 1 1 1
A V A V

4.32 Use Eq. 4.4.2 with mV representing the mass in the volume:
. .
ˆ
0 V
c s
dm
dA
dt

  
 n V 2 2 1 1
V
dm
A V A V
dt
 
  
.
V
dm
Q m
dt

  
Finally,
.
V
dm
m Q
dt

 
4.33 Use Eq. 4.4.2 with mS representing the mass in the sponge:
ˆ
0 S
dm
dA
dt

  
 n V    
dm
dt
A V A V A V
S
  
2 2 3 3 1 1    
dm
dt
m A V Q
S
 .
2 3 3 1
 
Finally,
dm
dt
Q m A V
S
  
 
1 2 3 3
 .
1
2
3
system (t) = V1
+ V2
+ V3
c.v. (t) = V1
+ V2
system boundary
at (t + t)

66
4.34 A1V1 = A2V2.  
125
144
2
.
 60 =  
2 5
144
2
.
V2. V2 = 15 ft/sec.
 .
.
m AV
  
 
194
1 25
144
60
2
= 3.968 slug/sec. Q = AV = 
125
144
2
.
 60 = 3
2.045 ft /sec.
4.35 A1V1 = A2V2.   .0252
 10 = (2  0.6  0.003)V2. V2 = 1.736 m/s.
2
1000 0.025 10
m AV
 
    = 19.63 kg/s. Q = AV=  0.0252
 10 = 3
0.1509 m /s.
4.36 
min = A1V1 + A2V2. 200 = 1000   0.0252
 25 + 1000 Q2. Q2 = 3
0.1509 m /s.
4.37 1
1
1
40 144
1716 520
 


p
RT
= 0.006455 slug/ft3
. 2
7 144
1716 610



= 0.000963 slug/ft3
.
1 2
1 1
0.2
. .
( 2 /144) 0.006455
m
m AV V
A

 
   
 
V1 = 355 fps.
2 2
0.2 0.000963 (2 3/144) .
m V
    V2 = 4984 fps.
4.38 1
1 1 1 2 2 2 1 2
3 3
500 kg 1246 kg
. 4.433 . 8.317
0.287 393 0.287 522
m m
p
A V A V
RT
   
     
 
4.433   0.052
 600 = 8.317   0.052
V2. V2 = 319.8 m/s.

m A V
 1 1 1 = 20.89 kg/s. Q A V
1 1 1
 = 3
4.712 m /s. Q2 = 3
2.512 m /s.
4.39  
1 1 1 2 2 2
A V A V
 .
p
RT
A V
p
RT
A V
1
1
1 1
2
2
2 2

200
293
0 05 40
120
0 03 120
2
2
2
 
    
. . .
T
  
T2 189 9 83
. .
K or C

4.40 a) A V A V
1 1 2 2
 . (2  1.5 + 1.5  1.5) 3 =
d2
2
4
2
 . d2 = 3.167 m
b) (2  1.5 + 1.5  1.5) 3 =
d2
2
4
2
2
 . d2 = 4.478 m
c) (2  1.5 + 1.5  1.5) 3 =
1
3 2
866 2
2
R
R
R
 





 
. .
R = 3.581 m. and d2 = 7.162 m
 R
cos = 1/2
 = 60o

67
4.41 a)
0 0 0 2
2
0
0 0 0
0 0 0
10 1 . 10 1 2 20 .
r r r
r r r
v r V vdA rdr r dr
r r r
  
 
   
      
 
     
     
  
  





 
V
r
r r
20
2 3
10
3
0
2
0
2
0
2
= 3.333 m/s.
2
1000 0.04 3.33
m AV
 
     = 16 75
. .
kg / s Q = AV = 3
0.01675 m /s.
b)
0 2 2
2 2
2 0 0
0
2 2
0 0
0
10 1 . 10 1 2 20 .
2 4
r
r r
r r
v r V rdr
r r
  
     
     
     
     
     
 V = 5 m/s
2
1000 0.04 5
m AV
 
     =25.13 kg/s. Q = AV = 3
0.02513 m /s.
c)
0
0
2 2
0 0
0 0
/2
20 1 . 20 1 2 10 / 4.
r
r
r r
v r V rdr r
r r
  
   
    
   
   
 V = 5.833 m/s
2
1000 0.04 5.833
m AV
 
     =29.32 kg/s. Q = 3
0.02932 m /s.
4.42 a) Since the area is rectangular, V = 5 m/s.
1000 0.08 0.8 5
m AV

     =320 kg/s. Q =

m

= 3
0.32 m /s.
b)
2
2
40
y y
v
h h
 
 
 
 
 
with y = 0 at the lower wall.
2
2
0
40 40 .
6
h
y y h
Vhw wdy w
h h
 
    
 
 
 
 V = 6.667 m/s.
1000 0.08 0.8 6.667
m AV

     =426.7 kg/s. Q = 3
0.4267 m /s.
c) V  0.08 = 10  0.04 + 5  0.02 + 5  0.02. V = 7.5 m/s.
1000 0.08 0.8 7.5
m AV

     =480 kg/s.  
Q
m


= 3
0.48 m /s.
4.43 a) 1 1 2 .
A V v dA
 
0
2 2
2
0
max max
2
0
0
1
6 1 2 2 .
24 4
r
r
r
v rdr v
r
  
 
 
    
 
   
   

With r0
1
24
 , max
v = 12 fps. ( )
v r
  12 1 576 2
( ) .
 r fps
b) 1 1 2 .
A V v dA
 
2
max max
2
1 4
6 1 .
12 3
h
h
y h
w v wdy v w
h

 
 
    
 
   
   

With h =
1
24
, max
v = 9 fps. ( )
v y
  9 1 576 2
( ) .
 y fps

66.-Merle C. Potter, David C. Wiggert, Bassem H. Ramadan - Mechanics of Fluids - Instructor’s solution manual-Cengage Learning (2012).pdf

66.-Merle C. Potter, David C. Wiggert, Bassem H. Ramadan - Mechanics of Fluids - Instructor’s solution manual-Cengage Learning (2012).pdf

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