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Fluid Mechanics 8th Edition White SOLUTIONS MANUAL
Chapter 2 • Pressure Distribution in a Fluid
P2.1 For the two-dimensional stress field in
Fig. P2.1, let
Find the shear and normal stresses on plane AA
cutting through at 30°.
Solution: Make cut ―AA‖ so that it just hits
the bottom right corner of the element. This Fig. P2.1
gives the freebody shown at right.
Now sum forces normal and tangential to
side AA. Denote side length AA as ―L.‖
P2.2 For the stress field of Fig. P2.1, change the known data to σxx = 2000 psf, σyy = 3000 psf, and
σn(AA) = 2500 psf. Compute σxy and the shear stress on plane AA.
Solution: Sum forces normal to and tangential to AA in the element freebody above,
with σn(AA) known and σxy unknown:
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In like manner, solve for the shear stress on plane AA, using our result for σxy:
This problem and Prob. P2.1 can also be solved using Mohr‘s circle.
P2.3 A vertical clean glass piezometer tube has an inside diameter of 1 mm. When a pressure
is applied, water at 20°C rises into the tube to a height of 25 cm. After correcting for surface
tension, estimate the applied pressure in Pa.
Solution: For water, let Y = 0.073 N/m, contact angle θ = 0°, and γ = 9790 N/m3. The capillary
rise in the tube, from Example 1.9 of the text, is
Then the rise due to applied pressure is less by that amount: hpress = 0.25 m − 0.03 m = 0.22 m. The
applied pressure is estimated to be p = γhpress = (9790 N/m3)(0.22 m) ≈ 2160 Pa Ans.
θ?
P2.4 Pressure gages, such as the Bourdon gage W Bourdon
gage
in Fig. P2.4, are calibrated with a deadweight piston.
If the Bourdon gage is designed to rotate the pointer
2 cm Oil
diameter
10 degrees for every 2 psig of internal pressure, how
many degrees does the pointer rotate if the piston and
Fig. P2.4
weight together total 44 newtons?
Solution: The deadweight, divided by the piston area, should equal the pressure applied to the
Bourdon gage. Stay in SI units for the moment:
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At 10 degrees for every 2 psig, the pointer should move approximately 100 degrees. Ans.
P2.5 Quito, Ecuador has an average altitude of 9,350 ft. On a standard day, pressure gage A in
a laboratory experiment reads 63 kPa and gage B reads 105 kPa. Express these readings in gage
pressure or vacuum pressure, whichever is appropriate.
Solution: Convert 9,350 ft x 0.3048 = 2,850 m. We can interpolate in the Standard Altitude
Table A.6 to a pressure of about 71.5 kPa. Or we could use Eq. (2.20):
Good interpolating! Then pA = 71500-63000 = 8500 Pa (vacuum pressure) Ans.(A), and pB
= 105000 - 71500 = 33500 Pa (gage pressure) Ans.(B)
P2.6 Express standard atmospheric pressure as a head, h = p/ρ g, in (a) feet of glycerin;
(b) inches of mercury; (c) meters of water; and (d) mm of ethanol.
Solution: Take the specific weights, γ = ρg, from Table A.3, divide patm by γ :
(a) ) Glycerin: h = (2116 lbf/ft2)/(78.7 lbf/ft3) ≈ 26.9 ft Ans. (a)
(b) Mercury: h = (2116 lbf/ft2)/(846 lbf/ft3) = 2.50 ft ≈ 30.0 inches Ans. (b) (c)
Water: h = (101350 N/m2)/(9790 N/m3) ≈ 10.35 m Ans. (c)
(d) Ethanol: h = (101350 N/m2)/(7740 N/m3) = 13.1 m ≈ 13100 mm Ans. (d)
P2.7 La Paz, Bolivia is at an altitude of approximately 12,000 ft. Assume a standard
atmosphere. How high would the liquid rise in a methanol barometer, assumed at 20°C?
[HINT: Don‘t forget the vapor pressure.]
Solution: Convert 12,000 ft to 3658 meters, and Table A.6, or Eq. (2.20), give
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From Table A.3, methanol has ρ = 791 kg/m3 and a large vapor pressure of 13,400 Pa. Then
the manometer rise h is given by
__ _____ _____ _____ _____ _____ _____ __ _____ _____ _____ ___
P2.8 Suppose, which is possible, that there is a half-mile deep lake of pure ethanol on the
surface of Mars. Estimate the absolute pressure, in Pa, at the bottom of this speculative lake.
Solution: We need some data from the Internet: Mars gravity is 3.71 m/s2, surface pressure is
700 Pa, and surface temperature is -10ºF (above the freezing temperature of ethanol). Then
the bottom pressure is given by the hydrostatic formula, with ethanol density equal to 789
kg/m3 from Table A.3. Convert ½ mile = ½(5280) ft = 2640 ft * 0.3048 m/ft = 804.7 m. Then
P2.9 A storage tank, 26 ft in diameter and 36 ft high, is filled with SAE 30W oil at 20°C. (a) What
is the gage pressure, in lbf/in2, at the bottom of the tank? (b) How does your result in (a) change
if the tank diameter is reduced to 15 ft? (c) Repeat (a) if leakage has caused a layer of 5 ft of
water to rest at the bottom of the (full) tank.
Solution: This is a straightforward problem in hydrostatic pressure. From Table A.3, the density of
SAE 30W oil is 891 kg/m3 ÷ 515.38 = 1.73 slug/ft3. (a) Thus the bottom pressure is
pbottom = ρoil g h = (1.73
slug ft )(36 ft) = 2005 lbf = 13.9 lbf gage Ans.(a)
)(32.2
ft3 s2 ft2 in2
(b) The tank diameter has nothing to do with it, just the depth: pbottom = 13.9 psig. Ans.(b)
(c) If we have 31 ft of oil and 5 ft of water (ρ = 1.94 slug/ft3), the bottom pressure is
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P2.10 A large open tank is open to sea level atmosphere and filled with liquid, at 20ºC, to a
depth of 50 ft. The absolute pressure at the bottom of the tank is approximately
221.5 kPa. From Table A.3, what might this liquid be?
Solution: Convert 50 ft to 15.24 m. Use the hydrostatic formula to calculate the bottom
pressure:
P2.11 In Fig. P2.11, sensor A reads 1.5 kPa
(gage). All fluids are at 20°C. Determine the
elevations Z in meters of the liquid levels in the
open piezometer tubes B and C.
Solution: (B) Let piezometer tube B be an
arbitrary distance H above the gasoline-
glycerin interface. The specific weights are γair ≈
12.0 N/m3, γgasoline = 6670 N/m3, and γglycerin =
12360 N/m3. Then apply the hydrostatic formula
from point A to point B: Fig. P2.11
Solve for ZB = 2.73 m (23 cm above the gasoline-air interface) Ans. (b) Solution (C):
Let piezometer tube C be an arbitrary distance Y above the bottom. Then
1500 + 12.0(2.0) + 6670(1.5) + 12360(1.0 − Y) − 12360(ZC − Y) = pC = 0 (gage)
Solve for ZC = 1.93 m (93 cm above the gasoline-glycerin interface) Ans. (c)
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P2.12 In Fig. P2.12 the tank contains water
and immiscible oil at 20°C. What is h in
centimeters if the density of the oil is 898
kg/m3?
Solution: For water take the density = 998
kg/m3. Apply the hydrostatic relation from the
oil surface to the water surface, skipping the
8-cm part:
Fig. P2.12
P2.13 In Fig. P2.13 the 20°C 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 bottom pressure must be
the same whether we move down through the water or through the gasoline into the third fluid:
Fig. P2.13
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Fliughidth Edition 2-7
P2.14 For the three-liquid system SG=
0.78
shown, compute h1 and h2. water
mercury
h2
27 cm
8 cm
Solution: The pressures at h1 5 cm
the three top surfaces must all be
atmospheric, or zero gage pressure. Compute γ oil = (0.78)(9790) = 7636 N/m3. Also,
3
from Table 2.1, γ water = 9790 N/m and γ mercury = 133100 N/m . The surface pressure
3
equality is
P2.15 In Fig. P2.15 all fluids are at 20°C. Gage A reads 15 lbf/in2 absolute and gage B reads
1.25 lbf/in2 less than gage C. Com-pute (a) the specific weight of the oil; and (b) the actual
reading of gage C in lbf/in2 absolute.
Fig. P2.15
Solution: First evaluate γair = (pA/RT)g = [15 × 144/(1717 × 528)](32.2) ≈ 0.0767 lbf/ft3.
Take γwater = 62.4 lbf/ft3. Then apply the hydrostatic formula from point B to point C:
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Fliughidth Edition 2-8
With the oil weight known, we can now apply hydrostatics from point A to point C:
P2.16 If the absolute pressure at the interface
between water and mercury in Fig. P2.16 is 93 kPa,
what, in lbf/ft2, is (a) the pressure at the Water 28 cm
surface, and (b) the pressure at the bottom of
75° 75°
the container?
Fig. P2.16 Mercury 8 cm
32 cm
Solution: Do the whole problem in SI units and then convert to BG at the end. The bottom width
and the slanted 75-degree walls are irrelevant red herrings. Just go up and down:
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Fluid Mechanics 8th Edition White SOLUTIONS MANUAL
Chapter 2 • Pressure Distribution in a Fluid
P2.1 For the two-dimensional stress field in
Fig. P2.1, let
Find the shear and normal stresses on plane AA
cutting through at 30°.
Solution: Make cut ―AA‖ so that it just hits
the bottom right corner of the element. This Fig. P2.1
gives the freebody shown at right.
Now sum forces normal and tangential to
side AA. Denote side length AA as ―L.‖
P2.2 For the stress field of Fig. P2.1, change the known data to σxx = 2000 psf, σyy = 3000 psf, and
σn(AA) = 2500 psf. Compute σxy and the shear stress on plane AA.
Solution: Sum forces normal to and tangential to AA in the element freebody above,
with σn(AA) known and σxy unknown:
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t iohnaps teMra2n ua•l P•r esFsluride M
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inigahtFhluEidddition 2-2
In like manner, solve for the shear stress on plane AA, using our result for σxy:
This problem and Prob. P2.1 can also be solved using Mohr‘s circle.
P2.3 A vertical clean glass piezometer tube has an inside diameter of 1 mm. When a pressure
is applied, water at 20°C rises into the tube to a height of 25 cm. After correcting for surface
tension, estimate the applied pressure in Pa.
Solution: For water, let Y = 0.073 N/m, contact angle θ = 0°, and γ = 9790 N/m3. The capillary
rise in the tube, from Example 1.9 of the text, is
Then the rise due to applied pressure is less by that amount: hpress = 0.25 m − 0.03 m = 0.22 m. The
applied pressure is estimated to be p = γhpress = (9790 N/m3)(0.22 m) ≈ 2160 Pa Ans.
θ?
P2.4 Pressure gages, such as the Bourdon gage W Bourdon
gage
in Fig. P2.4, are calibrated with a deadweight piston.
If the Bourdon gage is designed to rotate the pointer
2 cm Oil
diameter
10 degrees for every 2 psig of internal pressure, how
many degrees does the pointer rotate if the piston and
Fig. P2.4
weight together total 44 newtons?
Solution: The deadweight, divided by the piston area, should equal the pressure applied to the
Bourdon gage. Stay in SI units for the moment:
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At 10 degrees for every 2 psig, the pointer should move approximately 100 degrees. Ans.
P2.5 Quito, Ecuador has an average altitude of 9,350 ft. On a standard day, pressure gage A in
a laboratory experiment reads 63 kPa and gage B reads 105 kPa. Express these readings in gage
pressure or vacuum pressure, whichever is appropriate.
Solution: Convert 9,350 ft x 0.3048 = 2,850 m. We can interpolate in the Standard Altitude
Table A.6 to a pressure of about 71.5 kPa. Or we could use Eq. (2.20):
Good interpolating! Then pA = 71500-63000 = 8500 Pa (vacuum pressure) Ans.(A), and pB
= 105000 - 71500 = 33500 Pa (gage pressure) Ans.(B)
P2.6 Express standard atmospheric pressure as a head, h = p/ρ g, in (a) feet of glycerin;
(b) inches of mercury; (c) meters of water; and (d) mm of ethanol.
Solution: Take the specific weights, γ = ρg, from Table A.3, divide patm by γ :
(a) ) Glycerin: h = (2116 lbf/ft2)/(78.7 lbf/ft3) ≈ 26.9 ft Ans. (a)
(b) Mercury: h = (2116 lbf/ft2)/(846 lbf/ft3) = 2.50 ft ≈ 30.0 inches Ans. (b) (c)
Water: h = (101350 N/m2)/(9790 N/m3) ≈ 10.35 m Ans. (c)
(d) Ethanol: h = (101350 N/m2)/(7740 N/m3) = 13.1 m ≈ 13100 mm Ans. (d)
P2.7 La Paz, Bolivia is at an altitude of approximately 12,000 ft. Assume a standard
atmosphere. How high would the liquid rise in a methanol barometer, assumed at 20°C?
[HINT: Don‘t forget the vapor pressure.]
Solution: Convert 12,000 ft to 3658 meters, and Table A.6, or Eq. (2.20), give
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From Table A.3, methanol has ρ = 791 kg/m3 and a large vapor pressure of 13,400 Pa. Then
the manometer rise h is given by
__ _____ _____ _____ _____ _____ _____ __ _____ _____ _____ ___
P2.8 Suppose, which is possible, that there is a half-mile deep lake of pure ethanol on the
surface of Mars. Estimate the absolute pressure, in Pa, at the bottom of this speculative lake.
Solution: We need some data from the Internet: Mars gravity is 3.71 m/s2, surface pressure is
700 Pa, and surface temperature is -10ºF (above the freezing temperature of ethanol). Then
the bottom pressure is given by the hydrostatic formula, with ethanol density equal to 789
kg/m3 from Table A.3. Convert ½ mile = ½(5280) ft = 2640 ft * 0.3048 m/ft = 804.7 m. Then
P2.9 A storage tank, 26 ft in diameter and 36 ft high, is filled with SAE 30W oil at 20°C. (a) What
is the gage pressure, in lbf/in2, at the bottom of the tank? (b) How does your result in (a) change
if the tank diameter is reduced to 15 ft? (c) Repeat (a) if leakage has caused a layer of 5 ft of
water to rest at the bottom of the (full) tank.
Solution: This is a straightforward problem in hydrostatic pressure. From Table A.3, the density of
SAE 30W oil is 891 kg/m3 ÷ 515.38 = 1.73 slug/ft3. (a) Thus the bottom pressure is
pbottom = ρoil g h = (1.73
slug ft )(36 ft) = 2005 lbf = 13.9 lbf gage Ans.(a)
)(32.2
ft3 s2 ft2 in2
(b) The tank diameter has nothing to do with it, just the depth: pbottom = 13.9 psig. Ans.(b)
(c) If we have 31 ft of oil and 5 ft of water (ρ = 1.94 slug/ft3), the bottom pressure is
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P2.10 A large open tank is open to sea level atmosphere and filled with liquid, at 20ºC, to a
depth of 50 ft. The absolute pressure at the bottom of the tank is approximately
221.5 kPa. From Table A.3, what might this liquid be?
Solution: Convert 50 ft to 15.24 m. Use the hydrostatic formula to calculate the bottom
pressure:
P2.11 In Fig. P2.11, sensor A reads 1.5 kPa
(gage). All fluids are at 20°C. Determine the
elevations Z in meters of the liquid levels in the
open piezometer tubes B and C.
Solution: (B) Let piezometer tube B be an
arbitrary distance H above the gasoline-
glycerin interface. The specific weights are γair ≈
12.0 N/m3, γgasoline = 6670 N/m3, and γglycerin =
12360 N/m3. Then apply the hydrostatic formula
from point A to point B: Fig. P2.11
Solve for ZB = 2.73 m (23 cm above the gasoline-air interface) Ans. (b) Solution (C):
Let piezometer tube C be an arbitrary distance Y above the bottom. Then
1500 + 12.0(2.0) + 6670(1.5) + 12360(1.0 − Y) − 12360(ZC − Y) = pC = 0 (gage)
Solve for ZC = 1.93 m (93 cm above the gasoline-glycerin interface) Ans. (c)
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P2.12 In Fig. P2.12 the tank contains water
and immiscible oil at 20°C. What is h in
centimeters if the density of the oil is 898
kg/m3?
Solution: For water take the density = 998
kg/m3. Apply the hydrostatic relation from the
oil surface to the water surface, skipping the
8-cm part:
Fig. P2.12
P2.13 In Fig. P2.13 the 20°C 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 bottom pressure must be
the same whether we move down through the water or through the gasoline into the third fluid:
Fig. P2.13
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Fliughidth Edition 2-7
P2.14 For the three-liquid system SG=
0.78
shown, compute h1 and h2. water
mercury
h2
27 cm
8 cm
Solution: The pressures at h1 5 cm
the three top surfaces must all be
atmospheric, or zero gage pressure. Compute γ oil = (0.78)(9790) = 7636 N/m3. Also,
3
from Table 2.1, γ water = 9790 N/m and γ mercury = 133100 N/m . The surface pressure
3
equality is
P2.15 In Fig. P2.15 all fluids are at 20°C. Gage A reads 15 lbf/in2 absolute and gage B reads
1.25 lbf/in2 less than gage C. Com-pute (a) the specific weight of the oil; and (b) the actual
reading of gage C in lbf/in2 absolute.
Fig. P2.15
Solution: First evaluate γair = (pA/RT)g = [15 × 144/(1717 × 528)](32.2) ≈ 0.0767 lbf/ft3.
Take γwater = 62.4 lbf/ft3. Then apply the hydrostatic formula from point B to point C:
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Fliughidth Edition 2-8
With the oil weight known, we can now apply hydrostatics from point A to point C:
P2.16 If the absolute pressure at the interface
between water and mercury in Fig. P2.16 is 93 kPa,
what, in lbf/ft2, is (a) the pressure at the Water 28 cm
surface, and (b) the pressure at the bottom of
75° 75°
the container?
Fig. P2.16 Mercury 8 cm
32 cm
Solution: Do the whole problem in SI units and then convert to BG at the end. The bottom width
and the slanted 75-degree walls are irrelevant red herrings. Just go up and down:
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