To Whom It May Concern:
I am writing this reference at the request of Edward Berezhnoy. I have
known Edward for few months in my capacity as a teacher at Visions in
Education Charter School. Based on Edwards grades, attendance and
overall performance I am
SF State - Undergraduate
Student Tour
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Friday, November 6, 2015 at 2 pm
Room 106 of t
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
96
Solutions Manual Fluid Mechanics, Fifth Edition
This is the answer, since again it is negligible to go up to point A in low-density air.
(b) Given pA = 135 kPa, go down from point A to point B with negligible air-pressure
change, then jump across the m
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.
94
Solutions Manual Fluid Mechanics, Fifth Edition
2.46 In Fig. P2.46 both ends of the
manometer are open to the atmosphere.
Estimate the specific gravity of fluid X.
Solution: The pressure at the bottom of the
manometer must be the same regardless of
whi
Chapter 2 Pressure Distribution in a Fluid
93
2.44 Water flows downward in a pipe at
45, as shown in Fig. P2.44. The mercury
manometer reads a 6-in height. The pressure
drop p2 p1 is partly due to friction and
partly due to gravity. Determine the total
pr
92
Solutions Manual Fluid Mechanics, Fifth Edition
2.42 Small pressure differences can be
measured by the two-fluid manometer in
Fig. P2.42, where 2 is only slightly larger
than 1. Derive a formula for pA pB if
the reservoirs are very large.
Solution: App
Chapter 2 Pressure Distribution in a Fluid
91
2.40 In Fig. P2.40 the pressures at A and B are the same, 100 kPa. If water is
introduced at A to increase pA to 130 kPa, find and sketch the new positions of the
mercury menisci. The connecting tube is a unif
90
Solutions Manual Fluid Mechanics, Fifth Edition
Solution: The specific weight of the oil is (0.827)(62.4) = 51.6 lbf/ft3. If the reservoir
level does not change and L = 1 inch is the scale marking, then
lbf
lbf 1
p A (gage) = 1 2 = oil z = oil L sin =
Chapter 2 Pressure Distribution in a Fluid
89
2.35 Water flows upward in a pipe
slanted at 30, as in Fig. P2.35. The
mercury manometer reads h = 12 cm. What
is the pressure difference between points
(1) and (2) in the pipe?
Solution: The vertical distance
88
Solutions Manual Fluid Mechanics, Fifth Edition
2.34 To show the effect of manometer
dimensions, consider Fig. P2.34. The
containers (a) and (b) are cylindrical and
are such that pa = pb as shown. Suppose the
oil-water interface on the right moves up a
Chapter 2 Pressure Distribution in a Fluid
87
2.33 In Fig. P2.33 the pressure at point A
is 25 psi. All fluids are at 20C. What is the
air pressure in the closed chamber B?
Solution: Take = 9790 N/m3 for water,
8720 N/m3 for SAE 30 oil, and (1.45)(9790)
=