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11-1
Solutions Manual
for
Fundamentals of Thermal Fluid Sciences
5th Edition
Yunus A. Çengel, John M. Cimbala, Robert H. Turner
McGraw-Hill, 2017
Chapter 11
FLUID STATICS
PROPRIETARY AND CONFIDENTIAL
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without the prior written permission of McGraw-Hill Education.
PROPRIETARY MATERIAL. © 2017 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
11-1C
Solution We are to define resultant force and center of pressure.
11-2C
Solution We are to examine a claim about hydrostatic force.
11-3C
Solution We are to consider the effect of plate rotation on the hydrostatic force on the plate surface.
11-4C
Solution We are to explain why dams are bigger at the bottom than at the top.
11-5C
Solution We are to explain how to determine the horizontal component of hydrostatic force on a curved surface.
11-3
11-6C
Solution We are to explain how to determine the vertical component of hydrostatic force on a curved surface.
11-7C
Solution We are to explain how to determine the line of action on a circular surface.
Analysis The resultant hydrostatic force acting on a circular surface always passes through the center of the circle
11-8
Solution A car is submerged in water. The hydrostatic force on the door and its line of action are to be determined for
the cases of the car containing atmospheric air and the car is filled with water.
Assumptions 1 The bottom surface of the lake is horizontal. 2 The door can be approximated as a vertical rectangular
11-4
11-9E
Solution The height of a water reservoir is controlled by a cylindrical gate hinged to the reservoir. The hydrostatic
force on the cylinder and the weight of the cylinder per ft length are to be determined.
Assumptions 1 The hinge is frictionless. 2 Atmospheric pressure acts on both sides of the gate, and thus it can be ignored
in calculations for convenience.
lbf 1747
ft/slbm 32.2
2
Vertical force on horizontal surface (upward):
avg bottom
32
1 lbf
yC
F P A gh A gh A
FH
R=2 ft
s = 13 ft
b=R
11-10
Solution An above the ground swimming pool is filled with water. The hydrostatic force on each wall and the
distance of the line of action from the ground are to be determined, and the effect of doubling the wall height on the
hydrostatic force is to be assessed.
Assumptions Atmospheric pressure acts on both sides of the wall of the pool, and thus it can be ignored in calculations
m 0.667 3
2
3
h
yP
(from the bottom)
11-6
11-11E
Solution A dam is filled to capacity. The total hydrostatic force on the dam, and the pressures at the top and the
bottom are to be determined.
Assumptions Atmospheric pressure acts on both sides of the dam, and thus it can be ignored in calculations for
convenience.
11-12
Solution A room in the lower level of a cruise ship is considered. The hydrostatic force acting on the window and the
pressure center are to be determined.
Assumptions Atmospheric pressure acts on both sides of the window, and thus it can be ignored in calculations for
convenience.
digit)t significan (three N 2843]4/m) 3.0()[N/m 221,40(
]4/[
22
2
2840
DPAPF aveaveR
The line of action of the force passes through the pressure center, whose vertical
FR
D=0.3 m
11-8
11-13
Solution The cross-section of a dam is a quarter-circle. The hydrostatic force on the dam and its line of action are to
be determined.
Assumptions Atmospheric pressure acts on both sides of the dam, and thus it can be ignored in calculations for
convenience.
N 10682.1
m/skg 1
7
2
N 10643.2
m/skg 1
N 1
/4]m) (7m) 70)[(m/s 81.9)(kg/m 1000(
]4/[
7
2
223
2
RwggWFV
V
Then the magnitude and direction of the hydrostatic force acting on the surface of the dam become
57.5
N 103.13 7
1.571
N 10682.1
N 10643.2
tan
N) 10643.2(N) 10682.1(
7
7
272722
H
V
VHR
F
F
FFF
R = 7 m
FH
W
11-14
Solution The force required to hold a gate at its location is to be determined.
Pap3.42185.0981086.0
Now, the question is how much fluid from the second one can make the same pressure.
cmm
p
hSG 3535.0
3.12066
3.4218
981023.1
23.1
2
010.8.0
m
Ay
I
yy
cg
xc
cgcp 1227.0
)29.0(55.0
129.02
55.0
3
NhAF1110124.035.080.0981023.1
2
Take moment about hinge will give
0
2
4.0
111011227.015.1152044.0 F
F = 17.8 kN
SG=1.23
Hinge
450
10 cm
A
C
SG=1.23
35 cm
F1
F2
11-10
11-15
Solution The resulting force acting on a triangular gate and its line of action are to be determined.
o
Sin 16.64,9.0
1
9.0
)(
N1658
7.07.0
2
1
)16.64(7.0
3
2
3.09810 SinAhF cgR
In order to locate FR on the gate xcp, and ycp must be found.
I
xyc
water
0.7 m
0.7 m
0.3 m
0.9 m
11-11
11-16
Solution A rectangular plate hinged about a horizontal axis along its upper edge blocks a fresh water channel. The
plate is restrained from opening by a fixed ridge at a point B. The force exerted to the plate by the ridge is to be determined.
Assumptions Atmospheric pressure acts on both sides of the plate, and thus it can be ignored in calculations for
m 1.613m) 5m 5)(kN/m 53.24(2 APF aveR
m )333.31(
PF
ys
FR
h = 5 m
11-17
g=9.81 "m/s2"
rho=1000 "kg/m3"
s=1"m"
11-18E
Solution The flow of water from a reservoir is controlled by an L-shaped gate hinged at a point A. The required
weight W for the gate to open at a specified water height is to be determined.
Assumptions 1 Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for
convenience. 2 The weight of the gate is negligible.
Analysis The average pressure on a surface is the pressure at the centroid
2
2
374 4 lbf/ft
.
Then the resultant hydrostatic force acting on the dam becomes
2
avg 374 4 lbf/ft 12 ft 5 ft 22,464 lbf
R
F P A .
The line of action of the force passes through the pressure center, which is 2h/3
from the free surface,
ft 8
3
ft) 12(2
3
2
h
yP
Taking the moment about point A and setting it equal to zero gives
ABWysFM PRA
)( 0
Solving for W and substituting, the required weight is determined to be
lbf 30,900
lbf) 464,22(
ft 8
ft )83(
R
PF
AB
ys
W
FR
h=12 ft
8 ft
11-14
11-19E
Solution The flow of water from a reservoir is controlled by an L-shaped gate hinged at a point A. The required
weight W for the gate to open at a specified water height is to be determined.
Assumptions 1 Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for
convenience. 2 The weight of the gate is negligible.
11-20
Solution Two parts of a water trough of semi-circular cross-section are held together by cables placed along the
length of the trough. The tension T in each cable when the trough is full is to be determined.
Assumptions 1 Atmospheric pressure acts on both sides of the trough wall, and thus it can be ignored in calculations for
N 5297
m/skg 1
2
The vertical force on the horizontal surface is zero, since it coincides with the
free surface of water. The weight of fluid block per 11-m length is
N 1
]4/[
223
2
RwggWFV
V
R = 0.6 m
FH
T
W
A
FR
11-16
11-21
Solution A cylindrical tank is fully filled by water. The hydrostatic force on the surface A is to be determined for
three different pressures on the water surface.
Assumptions Atmospheric pressure acts on both sides of the cylinder, and thus it can be ignored in calculations for
kN 1.97N1972 4
8.0
4.09810
2
R
F
m
Ay
I
yy
cg
xc
cgcp 5.0
48.04.0
648.0
4.0 2
4
m
Pa
p
h
water
air 58.30
9810
1035
11-22
Solution An open settling tank contains liquid suspension. The resultant force acting on the gate and its line of action
are to be determined.
Assumptions Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for
convenience.
0
0
0
Y
Y
Y
kN 140N140,428
3
0
2/523
5
2
3
10
8.98502
Y
Y
RYSinYF
m1.64 428,140
961,230
cp
y
(from bottom), and
0x cp
obviously.
11-23
Solution An open settling tank contains liquid suspension. The density of the suspension depends on liquid depth
linearly. The resultant force acting on the gate and its line of action are to be determined.
Assumptions Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for
convenience.
11-24
Solution A tank is filled by oil. The magnitude and the location of the line of action of the resultant force acting on
(a)
kN 615
NAghFcgABR087,61565.225.25.3981088.0
m4.873
65.275.4
125.26
75.4
3
Ay
I
yy
cg
xc
cgcp
MN2.52 NApF BDBD
6
10517.26)1.08(51797
The weight of the oil is
MNNW 049.110049.1981088.05.31.01.05.21.86 6
It is interesting that the weight of the oil is pretty less than the pressure force acting on the bottom surface of the tank.
However, if we calculate the force acting on top surface,
MNNApF ACAC 468.110468.16)1.08(5.3981088.0 6
(upward)
The difference between FBD and FAC would give the weight of the oil
MNW049.1468.1517.2
3.5 m
2.5 m
8 m
10 cm
A
B
C
D
11-20
11-25
Solution Two parts of a water trough of triangular cross-section are held together by cables placed along the length
of the trough. The tension T in each cable when the trough is filled to the rim is to be determined.
well as the weight of the liquid block are determined as follows:
Horizontal force on vertical surface:
avg
32
2
2
1 N
1000 kg/m 9 81 m/s 0 530 2 m (0.530 m 6 m) 1 kg m/s
8267 N
H x C
F F P A gh A g h / A
. . /
The vertical force on the horizontal surface is zero since it coincides with the free surface
of water. The weight of fluid block per 6-m length is
N 8267
m/skg 1
N 1
m)/2] m)(0.530 m)(0.530 6)[(m/s 81.9)(kg/m 1000(
]2/[
2
23
bhwggWFV
V
FH
W
0.75 m
45
A
