Chapter 2 Pressure Distribution in a Fluid
95
of the newspaper due to air pressure.
(b) With everyone out of the way, perform
a karate chop on the outer end of the ruler.
(c) Explain the results in b.
Results: (a) Newsprint is about 27 in (0.686 m)
by 22.
Chapter 2 Pressure Distribution in a Fluid
2.29
85
Show that, for an adiabatic atmosphere, p = C()k, where C is constant, that
(k 1)gz
p/p o = 1
kRTo
k/(k 1)
, where k = c p /c v
Compare this formula for air at 5 km altitude with the U.S. standard atm
Solutions Manual Fluid Mechanics, Fifth Edition
84
Using p/p o from Ans. (a), we obtain
T (m 1)gz
= 1
To
mRTo
Ans. (b)
Note that, in using Ans. (a) to obtain Ans. (b), we have substituted po/o = RTo.
(c) Comparing Ans. (b) with the text, Eq. (2.27), w
Chapter 2 Pressure Distribution in a Fluid
83
2.25 Venus has a mass of 4.90E24 kg and a radius of 6050 km. Assume that its atmosphere is 100% CO2 (actually it is about 96%). Its surface temperature is 730 K, decreasing to 250 K at about z = 70 km. Average
Solutions Manual Fluid Mechanics, Fifth Edition
82
Solution: Given gasoline = 0.68(9790) = 6657 N/m3, compute the pressure when full:
pfull = gasoline (full height) = (6657 N/m 3 )(0.30 m) = 1997 Pa
Set this pressure equal to 2 cm of water plus Y centimet
Chapter 2 Pressure Distribution in a Fluid
Therefore the handle force required is F = P/16 = 222/16 14 lbf
81
Ans.
2.21 In Fig. P2.21 all fluids are at 20C.
Gage A reads 350 kPa absolute. Determine
(a) the height h in cm; and (b) the reading
of gage B in
80
Solutions Manual Fluid Mechanics, Fifth Edition
2.19 The U-tube at right has a 1-cm ID
and contains mercury as shown. If 20 cm3
of water is poured into the right-hand leg,
what will be the free surface height in each
leg after the sloshing has died dow
Chapter 2 Pressure Distribution in a Fluid
79
2.17 All fluids in Fig. P2.17 are at 20C.
If p = 1900 psf at point A, determine the
pressures at B, C, and D in psf.
Solution: Using a specific weight of
62.4 lbf/ft3 for water, we first compute pB
and pD:
Fig
Solutions Manual Fluid Mechanics, Fifth Edition
78
2.15 In Fig. P2.15 all fluids are at 20C.
Gage A reads 15 lbf/in2 absolute and gage
B reads 1.25 lbf/in2 less than gage C. Compute (a) the specific weight of the oil; and
(b) the actual reading of gage C
Chapter 2 Pressure Distribution in a Fluid
77
2.13 In Fig. P2.13 the 20C water and
gasoline are open to the atmosphere and
are at the same elevation. What is the
height h in the third liquid?
Solution: Take water = 9790 N/m3 and
gasoline = 6670 N/m3. The
Solutions Manual Fluid Mechanics, Fifth Edition
76
2.11 In Fig. P2.11, sensor A reads 1.5 kPa
(gage). All fluids are at 20C. Determine
the elevations Z in meters of the liquid
levels in the open piezometer tubes B
and C.
Solution: (B) Let piezometer tube
Chapter 2 Pressure Distribution in a Fluid
75
Solution: (a) Convert 2 miles = 3219 m and use a linear-pressure-variation estimate:
Then p pa + h = 101,350 Pa + (12 N/m3 )(3219 m) = 140,000 Pa 140 kPa
Ans. (a)
Alternately, the troposphere formula, Eq. (2.2
74
Solutions Manual Fluid Mechanics, Fifth Edition
Solution: Seawater specific weight at the surface (Table 2.1) is 10050 N/m3. It seems
quite reasonable to average the surface and bottom weights to predict the bottom
pressure:
10050 + 10520
p bottom po
Chapter 2 Pressure Distribution in a Fluid
73
For the linear law to be accurate, the 2nd term in parentheses must be much less than unity. If
the starting point is not at z = 0, then replace z by z:
2 To
n 1 B z
or :
z <
Ans.
< 1 ,
2 To
(n 1) B
_
2.5 Den
Solutions Manual Fluid Mechanics, Fifth Edition
72
Solve for xy = (2500 500 2250)/0.866 289 lbf/ft 2
Ans. (a)
In like manner, solve for the shear stress on plane AA, using our result for xy:
Ft,AA = AA L (2000 cos30 + 289sin 30)L sin 30
+ (289 cos30 + 30
Chapter 2 Pressure Distribution
in a Fluid
2.1 For the two-dimensional stress field
in Fig. P2.1, let
xx = 3000 psf yy = 2000 psf
xy = 500 psf
Find the shear and normal stresses on plane
AA cutting through at 30.
Solution: Make cut AA so that it just
hi
110
Solutions Manual Fluid Mechanics, Fifth Edition
P2.69 Consider the slanted plate AB of
A
length L in Fig. P2.69. (a) Is the hydrostatic
F
B
force F on the plate equal to the weight
Water, specific weight
of the missing water above the plate? If not,
Chapter 2 Pressure Distribution in a Fluid
109
force F on surface AB is (H/2)(sin)Hb, and its position is at 2H/3 down from point A,
as shown in the figure. Its moment arm about C is thus (H/3 Lcos). Meanwhile the
weight of the dam is W = (SG) (L/2)H(sin)
Solutions Manual Fluid Mechanics, Fifth Edition
108
2.66 Dam ABC in Fig. P2.66 is 30 m wide
into the paper and is concrete (SG 2.40).
Find the hydrostatic force on surface AB
and its moment about C. Could this force tip
the dam over? Would fluid seepage u
Chapter 2 Pressure Distribution in a Fluid
107
2.64 Gate ABC in Fig. P2.64 has a fixed
hinge at B and is 2 m wide into the paper.
If the water level is high enough, the gate
will open. Compute the depth h for which
this happens.
Solution: Let H = (h 1 met
Solutions Manual Fluid Mechanics, Fifth Edition
106
contributes to the hydrostatic force shown
in the freebody at right:
h
F = h CG A = (62.4) (8h csc 60)
2
= 288.2h 2 (lbf)
(1/12)(8)(h csc 60)3sin 60
(h/2)(8h csc 60)
h
= csc 60
6
y CP =
The weight of
Chapter 2 Pressure Distribution in a Fluid
105
AB is 0.433 m vertically below A, so hCG
= 2.0 0.433 = 1.567 m, and we may
compute the glycerin force and its line of
action:
Fg = hA = (12360)(1.567)(1.2) = 23242 N
y CP,g =
(1/12)(1.2)(1)3sin 60
= 0.0461 m
104
Solutions Manual Fluid Mechanics, Fifth Edition
2.61 Gate AB in Fig. P2.61 is a homogeneous mass of 180 kg, 1.2 m wide into
the paper, resting on smooth bottom B. All
fluids are at 20C. For what water depth h
will the force at point B be zero?
Solutio
Chapter 2 Pressure Distribution in a Fluid
103
the surface. For any , then, the hydrostatic force is F = (h/2)Lb. The moment of inertia
of the gate is (1/12)bL3, hence yCP = (1/12)bL3sin/[(h/2)Lb], and the center of
pressure is (L/2 yCP) from point B. Sum
102
Solutions Manual Fluid Mechanics, Fifth Edition
Solution: (a) The resultant force F, may be found by simply applying the hydrostatic
relation
F = h CG A = (9790 N/m 3 )(3 + 1.5 m)(5 m 2 m) = 440,550 N = 441 kN
Ans. (a)
(b) The horizontal force acts as
Chapter 2 Pressure Distribution in a Fluid
101
2.56 For the gate of Prob. 2.55 above, stop B breaks if the force on it equals 9200 lbf.
For what water depth h is this condition reached?
Solution: The formulas must be written in terms of the unknown centro
Solutions Manual Fluid Mechanics, Fifth Edition
100
Fig. P2.54
Solution: The three freebodies are shown below. Pressure on the side-walls balances
the forces. In (a), downward side-pressure components help add to a light W. In (b) side
pressures are horiz
Chapter 2 Pressure Distribution in a Fluid
MA =
p b dA =
99
10
(576 + 38.4 )(5 ft )d
=
0
plate
= (5)(576 2 / 2 + 38.4 3 / 3) |10 = 144000 + 64000 = 208,000 ft lbf
0
Then
CP =
MA
208000 ft lbf
=
F
38400 lbf
= 5.42 ft
Ans.(c)
The center of pressure is 5.
Solutions Manual Fluid Mechanics, Fifth Edition
98
P2.52 Example 2.5 calculated the force on
plate AB and its line of action, using the
p()
moment-of-inertia approach. Some teachers
say it is more instructive to calculate these
6 ft
by direct integration
Chapter 2 Pressure Distribution in a Fluid
97
Solution: (a) The total volume of oil in the vat is (3 m)(7 m)(4 m + 2 m)/2 = 63 m3.
Therefore the weight of oil in the vat is
W = oil (Vol) = (0.85)(9790 N/m 3 )(63 m 3 ) = 524, 000 N
Ans. (a)
(b) The force o