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Solutions Manual forFluid MechanicsSeventh Edition in SI UnitsFrank M. WhiteChapter 2Pressure Distribution in a FluidPROPRIETARY AND CONFIDENTIALThis Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By ...

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Solutions Manual for

Fluid Mechanics Seventh Edition in SI Units

Frank M. White Chapter 2 Pressure Distribution in a Fluid

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of the McGraw-Hill. © 2011 by The McGraw-Hill Companies, Inc. Limited distribution only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

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2.1 For the two-dimensional stress field in Fig. P2.1, let

σ xx = 150 kPa σ yy = 100 kPa σ xy = 25 kPa 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. Now sum forces normal and tangential to side AA. Denote side length AA as “L.”

Fig. P2.1

∑Fn,AA = 0 = σ AAL − (150 sin 30 + 25 cos 30)L sin 30 − (100 cos 30 + 25 sin 30)L cos 30

Solve for σ AA ≈ 134 kPa Ans. (a) ∑F t,AA = 0 = τ AA L − (150 cos 30 − 25 sin 30)L sin 30 − (25 cos 30 − 100 sin 30)L cos 30

Solve for τ AA ≈ 34 kPa

Ans. (b)

2.2 For the stress field of Fig. P2.1, change the known data to σxx = 100 kPa, σyy = 150 kPa, and σn(AA) = 125 kPa. Compute σ xy 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: ∑Fn,AA = 125L − ( σ xy cos 30° +100 sin 30°)L sin 30° − (σ xy sin 30° +150 cos 30°)L cos 30° = 0

Solve for σ xy = (125 − 25 −112.5)/0.866 ≈ −14.4 kPa

Ans. (a)

In like manner, solve for the shear stress on plane AA, using our result for σ xy:

∑F t,AA = τ AAL − (100 cos 30° +14.4 sin 30°)L sin 30° + (14.4 cos 30°+150sin 30°)L sin 30° = 0

Solve for τ AA = 46.9 − 75.8 ≈ −28.9 kPa

Ans. (b)

This problem and Prob. 2.1 can also be solved using Mohr’s circle.

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2.3 Pressure is independent of orientation. Is the pressure gradient also independent of orientation? Show that pressure gradient is equal in dimensions to force per unit volume. Determine the pressure gradient if the pressure is given by p(x, y, z) = –(x, y, gz), where g is the acceleration due to gravity. Solution: Pressure is independent of orientation, but pressure gradient is orientation dependent. pressure force/area = Pressure gradient = distance distance force = area × distance force = volume p =− (x, y, gz)  ∂p ∂p ∂p  , ∇p =  ,  =− (1, 1, g )  ∂x ∂y ∂z  2.4 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.

P2.5 Pressure gages, such as the Bourdon gage in Fig. P2.5, are calibrated with a deadweight piston. If the Bourdon gage is designed to rotate the pointer 10 degrees for every 10 kPa of internal pressure, how many degrees does the pointer rotate if the piston and weight together total 44 newtons?

Solution: The deadweight, divided by the piston area, should equal the pressure applied to the Bourdon gage. p Bourdon =

F 44 N = = 140,060 Pa ≈ 140 kPa A piston (π / 4)(0.02m)2

At 10 degrees for every 10 kPa, the pointer should move approximately 140 degrees. Ans.

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2.6 Denver, Colorado, has an average altitude of 1600 m. On a U.S. standard day, pressure gage A reads 83 kPa and gage B reads 105 kPa. Express these readings in gage or vacuum pressure, whichever is appropriate. Solution: We can find atmospheric pressure by either interpolating in Appendix Table A.6 or, more accurately, evaluate Eq. (2.27) at 1600 m:  Bz  pa = po 1−   To 

g/RB

5.26  (0.0065 K/m)(1600 m)  = (101.35 kPa) 1 −  ≈ 83.5 kPa 288.16 K  

Therefore: Gage A = 83 kPa − 83.5 kPa = −0.5 kPa (gage) = +0.4 kPa (vacuum) Gage B = 105 kPa − 83.5 kPa = 21.5 kPa (gage)

Ans.

2.7 Express standard atmospheric pressure as a head, h = p/ ρ g, in (a) m of glycerin; (b) cm of mercury; (c) meters of water; and (d) mm of ethanol. Solution: Take the specific weights, γ = ρ g, from Table A.3, divide patm by γ : (a) Glycerin: h = (101350 N/m2)/(12360 N/m3) ≈ 8.2 m Ans. (a) (b) Mercury: h = (101350 N/m2)/(132926 N/m3) = 0.76 m ≈ 76 cm 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)

P2.8 La Paz, Bolivia is at an altitude of approximately 3600 m. 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: Table A.6, or Eq. (2.20), give p LaPaz = po (1 −

B z g /( RB) (0.0065)(3600) 5.26 ] ≈ 64,916 Pa ) = 101350[1− To 288.16

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 p LaPaz − pvap = 64916 − 13400 = ρ methanol g h = (791)(9.81) h Solve for

hmethanol = 6.64 m

Ans.

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2.9 A diamond mine is 3000 m below sea level. (a) Estimate the air pressure at this depth. (b) If a barometer, accurate to 1 mm of mercury, is carried into this mine, how accurately can it estimate the depth of the mine? Solution: (a) Use a linear-pressure-variation estimate: 3 p ≈ p a + γ h = 101,350 Pa + (12 N/m )(3000 m) = 137,350 Pa ≈ 137.4 kPa Ans. (a)

Alternately, the troposphere formula, Eq. (2.27), predicts a slightly higher pressure: p ≈ p a (1 − Bz/To )5.26 = (101.35 kPa)[1 − (0.0065 K/m)( −3000 m)/288.16 K]5.26 = 143 kPa Ans. (a)

(b) The gage pressure at this depth is approximately 36,000/132,926 ≈ 0.27 m Hg or 270 mm Hg ±1 mm Hg or ±0.37% error. Thus the error in the actual depth is 0.37% of 3000 m or about ± 10 m if all other parameters are accurate. Ans. (b) P2.10 Find the gage pressure distribution in a saline solution with a variable density given by ρ(z) = α + βz, where z is the depth below the free surface and α , β are positive constants. Solution: For the downward z,

dp = ρg dz

ρ = α + βz dp ∴ = (α + β z)g dz ⇒ p=

z

∫ 0 (α + β z)g dz

= ( α z + β z 2 2 )g = ( α + 21 β z) zg P2.11 A storage tank, 8 m in diameter and 10 m high, is filled with SAE 30W oil at 20°C. (a) What is the gage pressure, in kPa, at the bottom of the tank? (b) How does your result in (a) change if the tank diameter is reduced to 4 m? (c) Repeat (a) if leakage has caused a layer of 1.5 m 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. (a) Thus the bottom pressure is 3 2 pbottom = ρ oil g h = (891 kg/m )(9.81 m/s )(10 m) = 87.4 kPa gage

Ans.( a )

(b) The tank diameter has nothing to do with it, just the depth: pbottom = 87.4 kPa.

Ans.(b)

(c) If we have 8.5 m of oil and 1.5 m of water ( ρ = 998 kg/m3), the bottom pressure is pb = ρ oil ghoil + ρ water ghwater = (891)(9.81)(8.5) + (998)(9.81)(1.5) = 74296 + 14686 = 89.0 kPa

Ans.(c)

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2.12 A closed tank contains 1.5 m of SAE 30 oil, 1 m of water, 20 cm of mercury, and an air space on top, all at 20°C. If pbottom = 60 kPa, what is the pressure in the air space? Solution: Apply the hydrostatic formula down through the three layers of fluid:

Solve for the pressure in the air space: pair ≈ 10500 Pa Ans. P2.13 If the mass density of water in the sea is given by ρ = (1000 + 0.005z3/2) kg/m3, where z in meters is the depth below the free surface, determine the depth at which the gage pressure is 1000 kPa. Solution: 3 dp = (1000 + 0.005z 2 )g dz

⇒ p=

z

∫ 0 (1000 + 0.005z

3

2

)g dz

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=(1000z + 0.002z 2 )g For pressure

1000 ×103 =(1000z + 0.002z 2 ) × 9.81 5

Solving by iteration (first guess by ignoring the z5/2 term) get z = 101.73 m

2.14 In Fig. P2.14, 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 gasolineglycerin 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. P2.14

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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2.15 In Fig. P2.15 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. P2.15

2.16 In Fig. P2.16 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. P2.16

P2.17

For the three-liquid system

shown, compute h1 and h2. Neglect the air density.

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Solution: The pressures at the three top surfaces must all beatmospheric, or zero gage pressure. Compute γoil = (0.78)(9790) = 7636 N/m3. Also, from Table 2.1, γwater = 9790 N/m3 and γ mercury = 133100 N/m3 . The surface pressure equality is

2.18 In Fig. P2.18 all fluids are at 20°C. Gage A reads 101 kPa absolute and gage B reads 8 kPa less than gage C. Compute (a) the specific weight of the oil; and (b) the actual reading of gage C in kPa absolute.

Fig. P2.18

Solution: First evaluate γair = (pA/RT)g = [101 × 103/(287 × 293)](9.81) ≈ 11.8 N/m3. Take γwater = 9800 N/m3. Then apply the hydrostatic formula from point B to point C: 3 p B + γ oil (0.3 m) + (9800)(0.6 m) = p C = p B + 8 × 10 Pa

Solve for γ oil ≈ 7067 N / m3

Ans. (a)

With the oil weight known, we can now apply hydrostatics from point A to point C: p C = pA + ∑ρ gh = (101 × 10 3 ) + (11.8)(0.6) + (7067)(0.6) + (9800)(0.6) or: p C = 111 kPa Ans. (b)

P2.19

If the absolute pressure at the interface

between water and mercury in Fig. P2.19 is 93 kPa, what, in kPa, is (a) the pressure at the

Water

surface, and (b) the pressure at the bottom of the container?

Fig. P2.19

75°

28 cm 75°

Mercury

32 cm

8 cm

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Solution: The bottom width and the slanted 75-degree walls are irrelevant red herrings. Just go up and down: psurface = pinterface − γ water Δh = 93000Pa − (9790 N / m3 )(0.28m) = 90260 Pa = 90.3 kPa

Ans.( a )

3 pbottom = pinterface + γmercury Δh = 93000Pa + (133100 N / m )(0.08m)

= 103650 Pa = 103.7 kPa

Ans.(b)

___________________________________________________________________________ 2.20 All fluids in Fig. P2.20 are at 20°C. If p = 90 kPa at point A, determine the pressures at B, C, and D in kPa. Solution: Using a specific weight of 9800 N/m3 for water, we first compute pB and pD:

Fig. P2.20 3 p B = p A − γ water (z B − z A ) = 90 ×10 − 9800(0.4 m) = 86.1 kPa 3 p D = p A + γ water (z A − z D ) = 90 ×10 + 9800(1.4 m) = 103.7 kPa

Ans. (pt. B) Ans. (pt. D)

Finally, moving up from D to C, we can neglect the air specific weight to good accuracy: 3 p C = p D − γ water (zC − z D ) = 103.7 ×10 − 9800(0.6 m) = 97.8 kPa

The air near C has γ ≈ 0.074

lbf/ft3

Ans. (pt. C)

times 6 ft yields less than 0.5 psf correction at C.

2.21 All fluids in Fig. P2.21 are at 20°C. If atmospheric pressure = 101.33 kPa and the bottom pressure is 242 kPa absolute, what is the specific gravity of fluid X? Solution: Simply apply the hydrostatic formula from top to bottom:

Fig. P2.21

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2.22 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 down? Solution: First figure the height of water added:

Then, at equilibrium, the new system must have 25.46 cm of water on the right, and a 30-cm length of mercury is somewhat displaced so that “L” is on the right, 0.1 m on the bottom, and “0.2 − L” on the left side, as shown at right. The bottom pressure is constant: Thus right-leg-height = 9.06 + 25.46 = 34.52 cm Ans. left-leg-height = 20.0 − 9.06 = 10.94 cm Ans.

2.23 The hydraulic jack in Fig. P2.23 is filled with oil at 8.8 kN/m3. Neglecting piston weights, what force F on the handle is required to support the 10 kN weight shown?

Fig. P2.23

Solution: First sum moments clockwise about the hinge A of the handle:

∑M A = 0 = F(40 + 2) − P(2), or: F = P/21, where P is the force in the small (2 cm) piston. Meanwhile figure the pressure in the oil from the weight on the large piston: p oil =

W 10 ×10 3 N = 1989 kN/m 2 = 2 A 3-in ( π /4)(0.08 m)

π Hence P = p oilAsmall = (1989 ×10 3 ) (0.02)2 = 625 N 4 Therefore the handle force required is F = P/21 = 625/21 ≈ 30 N Ans.

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2.24 In Fig. P2.24 all fluids are at 20°C. Gage A reads 350 kPa absolute. Determine (a) the height h in cm; and (b) the reading of gage B in kPa absolute. Solution: Apply the hydrostatic formula from the air to gage A: Fig. P2.24

Then, with h known, we can evaluate the pressure at gage B:

2.25 The fuel gage for an auto gas tank reads proportional to the bottom gage pressure as in Fig. P2.25. If the tank accidentally contains 2 cm of water plus gasoline, how many centimeters “h” of air remain when the gage reads “full” in error? Fig. P2.25

Solution: Given γgasoline = 0.68(9790) = 6657 N/m3, compute the gage pressure when “full”:

Set this pressure equal to 2 cm of water plus “Y” centimeters of gasoline:

Therefore the air gap h = 30 cm − 2 cm(water) − 27.06 cm(gasoline) ≈ 0.94 cm Ans. 2.26 In Fig. P2.26 both fluids are at 20°C. If surface tension effects are negligible, what is the density of the oil, in kg/m3? Solution: Move around the U-tube from left atmosphere to right atmosphere:

Fig. P2.26

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2.27 In Prob. 1.2 we made a crude integration of atmospheric density from Table A.6 and found that the atmospheric mass is approximately m ≈ 6E18 kg. Can this result be used to estimate sea-level pressure? Can sea-level pressure be used to estimate m? Solution: Yes, atmospheric pressure is essentially a result of the weight of the air above. Therefore the air weight divided by the surface area of the earth equals sea-level pressure:

This is a little off, thus our mass estimate must have been a little off. If global average sealevel pressure is actually 101350 Pa, then the mass of atmospheric air must be more nearly

*P2.28 As measured by NASA’s Viking landers, the atmosphere of Mars, where g = 3.71 2 m/s , is almost entirely carbon dioxide, and the surface pressure averages 700 Pa. The temperature is cold and drops off exponentially: T ≈ To e Cz, where C ≈ 1.3E-5 m-1 and To ≈ 250 K. For example, at 20,000 m altitude, T ≈ 193 K. (a) Find an analytic formula for the variation of pressure with altitude. (b) Find the altitude where pressure on Mars has dropped to 1 pascal. Solution: (a) The analytic formula is found by integrating Eq. (2.17) of the text:

(b) From Table A.4 for CO2, R = 189 m2/(s2-K). Substitute p = 1 Pa to find the altitude:

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P2.29 For gases over large changes in height, the linear approximation, Eq. (2.14), is inaccurate. Expand the troposphere power-law, Eq. (2.20), into a power series and show that the linear approximation p ≈ pa - ρa g z is adequate when

δ z 77.4°° . Ans. NOTE: This answer is independent of the numerical values of h, g, or b but requires SG = 2.5.

2.90 In Fig. P2.90, the 2-m-by-2-m square gate AB, which is made of uniform material, is hinged at B but free-ended at A. (a) Draw a labeled diagram to show the hydrostatic pressure distribution on the gate. Indicate the values of pressure at A and B. (b) Calculate the magnitude of the hydrostatic resultant force. (c) Find the location of the center of pressure. (d) Determine the minimum weight of the gate that is required to keep it shut for the water level shown in the figure.

Fig. P2.90

Solution:

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B is 1 m below free surface A is 1 – 2 sin 20° = 0.316 m below free surface. PB = 103 × 9.81 × 1 = 9.81 kPa, PA = 103 × 9.81 × 0.316 = 3.10 kPa

Hydrostatic force on AB = 1 (PA + PB ) × Area 2

2 F = 12 (9.81+ 3.10) × 2

= 25.82 kN Distance of center of pressure from B d=

2PA + PB 2 × 3.1 + 9.81 ×2= × 2 = 0.827 m 3(PA + PB ) 3(3.1+ 9.81)

Taking moment about B W ×1× cos20° = F × d ⇒W=

25.82 × 0.827 = 22.72 kN cos20°

P2.91 For the semicircular cylinder CDE in Ex. 2.9, find the vertical hydrostatic force by integrating the vertical component of pressure around the surface from θ = 0 to θ = π .

Solution: A sketch is repeated here. At any position θ, A

as in Fig. P2.91, the vertical component of pressure is p cosθ. The depth down to this point is h+R(1- cos θ), and the local pressure is γ times this depth. Thus

h

p

C θ R

D

E

Fig. P2.91

The negative sign occurs because the sign convention for dF was a downward force. __________________________________________________________________________

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2.92 The dam in Fig. P2.92 is a quarter-circle 50 m wide into the paper. Determine the horizontal and vertical components of hydrostatic force against the dam and the point CP where the r...