FLUID MECHANICS 8TH EDITION WHITE SOLUTIONS MANUAL CHAPTER 2 • PRESSURE DISTRIBUTION IN A FLUID |QUESTIONS AND CORRECT ANSWERS|

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FLUID MECHANICS 8TH EDITION WHITE SOLUTIONS MANUAL
CHAPTER 2 • PRESSURE DISTRIBUTION IN A FLUID |QUESTIONS
AND CORRECT ANSWERS|2023-2024

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 gives the freebody shown at right. Fig. P2.1
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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2-2 SoluC
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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? Bourdon
P2.4 Pressure gages, such as the Bourdon gage W 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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2-3 SoluC
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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/ρg, 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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2-4 SoluC
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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
slug ft lbf lbf
p = ρ g h = (1.73 )(32.2 )(36 ft) = 2005 = 13.9 gage Ans.(a)
bottom oil 3 2 2
ft s2 ft in

(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