Chapter 2 Pressure Distribution in a Fluid

Transcription

1 Chapter Pressure Distribution in a Fluid.1 For the two-dimensional stress field in Fig. P.1, let xx 000 psf 000 psf xy yy 500 psf Find the shear and normal stresses on plane AA cutting through at 0. 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. F 0 n,aa AA L (000sin0 500cos0)Lsin0 (000 cos0 500 sin 0)L cos0 Fig. P.1 Solve for t,aa AA AA 68 lbf/ft Ans. (a) F 0 L (000cos0 500sin0)Lsin0 (500cos0 000sin0)L cos0 Solve for AA 68 lbf/ft Ans. (b). For the stress field of Fig. P.1, change the known data to xx 000 psf, yy 000 psf, and n(aa) 500 psf. Compute xy and the shear stress on plane AA. Solution: Sum forces normal to and tangential to AA in the element freebody above, with n(aa) known and xy unknown: F 500L ( cos0 000sin 0 )L sin 0 n,aa xy xy ( sin cos0 )L cos0 0

2 7 Solutions Manual Fluid Mechanics, Seventh Edition Solve for ( )/ lbf/ft Ans. (a) xy In like manner, solve for the shear stress on plane AA, using our result for xy: F L (000 cos0 89sin 0 )L sin 0 t,aa AA (89 cos0 000 sin 0 )L cos0 0 Solve for lbf/ft Ans. (b) AA This problem and Prob..1 can also be solved using Mohr s circle.. A vertical clean glass piezometer tube has an inside diameter of 1 mm. When a pressure is applied, water at 0C rises into the tube to a height of 5 cm. After correcting for surface tension, estimate the applied pressure in Pa. Solution: For water, let Y 0.07 N/m, contact angle 0, and 9790 N/m. The capillary rise in the tube, from Example 1.9 of the text, is h cap Y cos (0.07 Nm / ) cos(0 ) 0.00 m R (9790 N/ m )( m) Then the rise due to applied pressure is less by that amount: hpress 0.5 m 0.0 m 0. m. The applied pressure is estimated to be p hpress (9790 N/m )(0. m) 160 Pa Ans. P.4 Pressure gages, such as the Bourdon gage W? Bourdon gage in Fig. P.4, are calibrated with a deadweight piston. If the Bourdon gage is designed to rotate the pointer 10 degrees for every psig of internal pressure, how cm diameter Oil many degrees does the pointer rotate if the piston and weight together total 44 newtons? Fig. P.4 Solution: The deadweight, divided by the piston area, should equal the pressure applied to the Bourdon gage. Stay in SI units for the moment: p Bourdon F 44 N lbf 140, 060 Pa A piston ( / 4)(0.0 m) in

3 Chapter Pressure Distribution in a Fluid 7 At 10 degrees for every psig, the pointer should move approximately 100 degrees. Ans..5 Denver, Colorado, has an average altitude of 500 ft. On a U.S. standard day, pressure gage A reads 8 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. (.7) at 500 ft 1615 m: g/rb 5.6 Bz ( K/m)(1615 m) pa po1 (101.5 kpa) kpa T o K Therefore: Gage A 8 kpa 8.4 kpa 0.4 kpa (gage).4 kpa (vacuum) Gage B 105 kpa 8.4 kpa 1.6 kpa (gage) Ans..6 Express standard atmospheric pressure as a head, h p/g, in (a) feet of glycerin; (b) inches of mercury; (c) meters of water; and (d) mm of ethanol. Solution: Take the specific weights, g, from Table A., divide patm by : (a) Glycerin: h (116 lbf/ft )/(78.7 lbf/ft ) 6.9 ft Ans. (a) (b) Mercury: h (116 lbf/ft )/(846 lbf/ft ).50 ft 0.0 inches Ans. (b) (c) Water: h (10150 N/m )/(9790 N/m ) 10.5 m Ans. (c) (d) Ethanol: h (10150 N/m )/(7740 N/m ) 1.1 m 1100 mm Ans. (d) P.7 La Paz, Bolivia is at an altitude of approximately 1,000 ft. Assume a standard atmosphere. How high would the liquid rise in a methanol barometer, assumed at 0C? [HINT: Don t forget the vapor pressure.] Solution: Convert 1,000 ft to 658 meters, and Table A.6, or Eq. (.0), give Bz g /( RB) (0.0065)(658) 5.6 plapaz po (1 ) 10150[1 ] 64, 400 Pa T o

4 74 Solutions Manual Fluid Mechanics, Seventh Edition From Table A., methanol has = 791 kg/m and a large vapor pressure of 1,400 Pa. Then the manometer rise h is given by p p g h (791)(9.81) h LaPaz vap methanol Solve for h 6.57 m Ans. methanol.8 A diamond mine is miles 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) Convert miles 19 m and use a linear-pressure-variation estimate: a Then p p h 101,50 Pa (1 N/m )(19 m) 140,000 Pa 140 kpa Ans. (a) Alternately, the troposphere formula, Eq. (.7), predicts a slightly higher pressure: a o p p (1 Bz/T ) (101. kpa)[1 ( K/m)( 19 m)/88.16 K] 147 kpa Ans. (a) (b) The gage pressure at this depth is approximately 40,0001, m Hg or 00 mm Hg 1 mm Hg or 0. error. Thus the error in the actual depth is 0. of 0 m or about 10 m if all other parameters are accurate. Ans. (b) P.9 A storage tank, 6 ft in diameter and 6 ft high, is filled with SAE 0W oil at 0C. (a) What is the gage pressure, in lbf/in, 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., the density of SAE 0W oil is 891 kg/m = 1.7 slug/ft. (a) Thus the bottom pressure is slug ft lbf lbf pbottom oil g h (1.7 )(. )(6 ft) gage Ans.( a) ft s ft in (b) The tank diameter has nothing to do with it, just the depth: pbottom = 1.9 psig. Ans.(b)

5 Chapter Pressure Distribution in a Fluid 75 (c) If we have 1 ft of oil and 5 ft of water ( = 1.94 slug/ft ), the bottom pressure is p gh gh (1.7)(.)(1) (1.94)(.)(5) b oil oil water water.10 A closed tank contains 1.5 m of SAE 0 oil, 1 m of water, 0 cm of mercury, and an air space on top, all at 0C. If pbottom 60 kpa, what is the pressure in the air space? Solution: Apply the hydrostatic formula down through the three layers of fluid: p p h h h bottom air oil oil water water mercury mercury air lbf lbf Ans.( c) ft in or: Pa p (870 N/m )(1.5 m) (9790)(1.0 m) (1100)(0. m) Solve for the pressure in the air space: pair Pa Ans..11 In Fig. P.11, sensor A reads 1.5 kpa (gage). All fluids are at 0C. 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 air 1.0 N/m, gasoline 6670 N/m, and glycerin 160 N/m. Then apply the hydrostatic formula from point A to point B: Fig. P N/m (1.0 N/m )(.0 m) 6670(1.5 H) 6670(Z H 1.0) p 0 (gage) Solve for ZB.7 m ( cm above the gasoline-air interface) Ans. (b) Solution (C): Let piezometer tube C be an arbitrary distance Y above the bottom. Then (.0) 6670(1.5) 160(1.0 Y) 160(ZC Y) pc 0 (gage) Solve for ZC 1.9 m (9 cm above the gasoline-glycerin interface) Ans. (c) B B

6 76 Solutions Manual Fluid Mechanics, Seventh Edition.1 In Fig. P.1 the tank contains water and immiscible oil at 0C. What is h in centimeters if the density of the oil is 898 kg/m? Solution: For water take the density 998 kg/m. Apply the hydrostatic relation from the oil surface to the water surface, skipping the 8-cm part: p (898)(g)(h 0.1) atm (998)(g)( ) p, atm Solve for h 0.08 m 8.0 cm Ans. Fig. P.1.1 In Fig. P.1 the 0C 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/m and gasoline 6670 N/m. The bottom pressure must be the same whether we move down through the water or through the gasoline into the third fluid: bottom Fig. P.1 p (9790 N/m )(1.5 m) 1.60(9790)(1.0) 1.60(9790)h 6670(.5 h) Solve for h 1.5 m Ans.

7 Chapter Pressure Distribution in a Fluid 77 P.14 For the three-liquid system shown, compute h1 and h. Neglect the air density. water oil, SG= 0.78 mercury h Fig. P.14 7 cm Solution: The pressures at h1 8 cm 5 cm the three top surfaces must all be atmospheric, or zero gage pressure. Compute oil = (0.78)(9790) = 766 N/m. Also, from Table.1, water = 9790 N/m and mercury = 1100 N/m. The surface pressure equality is N N N N N (9790 )(0.7 m) (1100 ) h 1 (1100 )(0.08 m) (766 ) h (1100 )(0.05 m) m m m m m or : h Pa 786h Solve for h m 6.0 cm, h 0.5 m 5. cm Ans In Fig. P.15 all fluids are at 0C. Gage A reads 15 lbf/in absolute and gage B reads 1.5 lbf/in less than gage C. Com-pute (a) the specific weight of the oil; and (b) the actual reading of gage C in lbf/in absolute. Fig. P.15

8 78 Solutions Manual Fluid Mechanics, Seventh Edition Solution: First evaluate air (pa/rt)g [15 144/( )](.) lbf/ft. Take water 6.4 lbf/ft. Then apply the hydrostatic formula from point B to point C: p (1.0 ft) (6.4)(.0 ft) p p (1.5)(144) psf B oil C B Solve for 55. lbf/ft Ans. (a) oil With the oil weight known, we can now apply hydrostatics from point A to point C: p p gh (15)(144) (0.0767)(.0) (55.)(.0) (6.4)(.0) C A C or: p 95 lbf/ft 16.6 psi Ans. (b) P.16 If the absolute pressure at the interface between water and mercury in Fig. P.16 is 9 kpa, what, in lbf/ft, is (a) the pressure at the Water 8 cm surface, and (b) the pressure at the bottom of the container? Fig. P Mercury 75 8 cm 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: p p h 9000 Pa (9790 N / m )(0.8 m) surface interface water 9060 Pa lbf / ft Ans.( a) p p h 9000 Pa (1100 N / m )(0.08 m) bottom interface mercury Pa lbf / ft Ans.( b)

9 .17 All fluids in Fig. P.17 are at 0C. If p 1900 psf at point A, determine the pressures at B, C, and D in psf. Solution: Using a specific weight of 6.4 lbf/ft for water, we first compute pb and pd: B A water B A Fig. P.17 p p (z z ) (1.0 ft) 188 lbf ft Ans. (pt. B) p p (z z ) (5.0 ft) 1 lbf/ft Ans. (pt. D) D A water A D Finally, moving up from D to C, we can neglect the air specific weight to good accuracy: p p (z z ) 1 6.4(.0 ft) 087 lbf/ft Ans. (pt. C) C D water C D The air near C has lbf/ft times 6 ft yields less than 0.5 psf correction at C..18 All fluids in Fig. P.18 are at 0C. If atmospheric pressure 101. kpa and the bottom pressure is 4 kpa absolute, what is the specific gravity of fluid X? Solution: Simply apply the hydrostatic formula from top to bottom: Fig. P.18 or: p p h, bottom top (870)(1.0) (9790)(.0) (.0) (1100)(0.5) X X Solve for 157 N/m, or: SG 157/ Ans. X

10 80 Solutions Manual Fluid Mechanics, Seventh Edition.19 The U-tube at right has a 1-cm ID and contains mercury as shown. If 0 cm 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: 0 cm (1 cm) h, 4 or h 5.46 cm Then, at equilibrium, the new system must have 5.46 cm of water on the right, and a 0-cm length of mercury is somewhat displaced so that L is on the right, 0.1 m on the bottom, and 0. L on the left side, as shown at right. The bottom pressure is constant: p 1100(0. L) p 9790(0.546) 1100(L), or: L m atm atm Thus right-leg-height cm Ans. left-leg-height cm Ans..0 The hydraulic jack in Fig. P.0 is filled with oil at 56 lbf/ft. Neglecting piston weights, what force F on the handle is required to support the 000-lbf weight shown? Fig. P.0 Solution: First sum moments clockwise about the hinge A of the handle: MA 0 F(15 1) P(1), or: F P/16, where P is the force in the small (1 in) piston. Meanwhile figure the pressure in the oil from the weight on the large piston: p W 000 lbf psf, oil A -in ( /4)(/1 ft) 1 Hence P poila small (40744) lbf 4 1

11 Chapter Pressure Distribution in a Fluid 81 Therefore the handle force required is F P/16 /16 14 lbf Ans..1 In Fig. P.1 all fluids are at 0C. Gage A reads 50 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: p p h Fig. P.1 A air (9790)h 1100(0.8) Pa, Solve for h 6.49 m Ans. (a) Then, with h known, we can evaluate the pressure at gage B: p ( ) = Pa 51 kpa Ans. (b) B. The fuel gage for an auto gas tank reads proportional to the bottom gage pressure as in Fig. P.. If the tank accidentally contains cm of water plus gasoline, how many centimeters h of air remain when the gage reads full in error? Fig. P.

12 8 Solutions Manual Fluid Mechanics, Fifth Edition Solution: Given gasoline 0.68(9790) 6657 N/m, compute the gage pressure when full : p (full height) (6657 N/m )(0.0 m) 1997 Pa full gasoline Set this pressure equal to cm of water plus Y centimeters of gasoline: pfull (0.0 m) 6657Y, or Y m 7.06 cm Therefore the air gap h 0 cm cm(water) 7.06 cm(gasoline) 0.94 cm Ans.. In Fig. P. both fluids are at 0C. If surface tension effects are negligible, what is the density of the oil, in kg/m? Solution: Move around the U-tube from left atmosphere to right atmosphere: p (9790 N/m )(0.06 m) a oil (0.08 m) p, a oil Fig. P. solve for 74 N/m, or: 74/ kg m Ans. oil.4 In Prob. 1. 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: p W m g (6.0E18 kg)(9.81 m/s ) Pa air air sea-level Aearth 4Rearth 4 (6.77E6 m) This is a little off, thus our mass estimate must have been a little off. If global average sea-level pressure is actually Pa, then the mass of atmospheric air must be more nearly Ans. m earth sea-level air A p 4 (6.77E6 m) (10150 Pa) 5.8E18 kg g 9.81 m/s Ans.

13 Chapter Pressure Distribution in a Fluid 8 *P.5 As measured by NASA s Viking landers, the atmosphere of Mars, where g =.71 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.E-5 m -1 and To 50 K. For example, at 0,000 m altitude, T 19 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. (.17) of the text: p g zdz g z dz g Cz ln( ) ( e 1) p 0 0 o R T R Cz Te RT o oc g Cz or, finally, p po exp[ ( e 1)] Ans.( a) RT C o (b) From Table A.4 for CO, R = 189 m /(s -K). Substitute p = 1 Pa to find the altitude: g Cz.71 m/ s (1.E5) z p 1Pa po exp[ ( e 1)] (700 Pa) exp[ { e 1}] RTo C (189)(50)(1.E 5) 1 (1.E5) z or : ln( ) { e 1}, Solve for z 56,500 m Ans.( b) 700 P.6 For gases over large changes in height, the linear approximation, Eq. (.14), is inaccurate. Expand the troposphere power-law, Eq. (.0), into a power series and show that the linear approximation p pa - a g z is adequate when To g z, where n ( n 1) B RB Solution: The power-law term in Eq. (.0) can be expanded into a series: Bz (1 ) T o n 1 n Bz T o n( n 1) Bz ( )! T o... where n g RB

14 84 Solutions Manual Fluid Mechanics, Fifth Edition Multiply by pa, as in Eq. (.0), and note that panb/to = (pa/rto)gz = a gz. Then the series may be rewritten as follows: p p a n 1 Bz a gz( 1 T o For the linear law to be accurate, the nd term in parentheses must be much less than unity. If the starting point is not at z = 0, then replace z by z: n 1 B z To 1, or : z Ans. To ( n 1) B... ).7 This is an experimental problem: Put a card or thick sheet over a glass of water, hold it tight, and turn it over without leaking (a glossy postcard works best). Let go of the card. Will the card stay attached when the glass is upside down? Yes: This is essentially a water barometer and, in principle, could hold a column of water up to 10 ft high! P.8 A correlation of numerical results indicates that, all other things being equal, the horizontal distance traveled by a well-hit baseball varies inversely as the cube root of the air density. If a home-run ball hit in New York City travels 400 ft, estimate the distance it would travel in (a) Denver, Colorado; and (b) La Paz, Bolivia. Solution: New York City is approximately at sea level, so use the Standard Atmosphere, Table A.6, and take air = 1.55 kg/m. Modify Eq. (.0) for density instead of pressure: Bz ( g / RB ) z 4.6 (1 ) (1 ) a To Using nominal altitudes from almanacs, apply this formula to Denver and La Paz: ( a) Denver, Colorado : z 580 ft 1609m ; kg / m ( b) La Paz, Bolivia : z 1000 ft Finally apply this to the 400-ft home-run ball: 660m ; kg / m

15 Chapter Pressure Distribution in a Fluid 85 ( a) ( b) Denver : La Paz : Distance traveled Distance traveled 1.55 (400 ft)( ) (400 ft)( ) / 1/ 41ft 45ft Ans.( a) Ans.( b) In Denver, balls go 5% further, as attested to by many teams visiting Coors Field. P.9 An airplane flies at a Mach number of 0.8 at a standard altitude of 4,000 ft. (a) What is the plane s velocity, in mi/h? (b) What is the standard density at that altitude? Solution: (a) Convert 4,000 ft to 715 m. Find the standard temperature from Eq. (.19): T T B z K ( m)(715 m) 40.6 K o From the (absolute) temperature, we compute the speed of sound and hence the velocity: a krt 1.4(87 m / s K)(40.6 K) 11 m / s V ( Ma) a (0.8)(11 m / s) 55 m / s mi / h Ans.( a) (b) Given o = 1.55 kg/m, the power-law density formula is evaluated at T = 40.6 K: g 1 RB T o ( ) (1.55)( ) kg / m Ans.( b) T o P.0 For the traditional equal-level manometer measurement in Fig. E., water at 0C flows through the plug device from a to b. The manometer fluid is mercury. If L = 1 cm and h = 4 cm, (a) what is the pressure drop through the device? (b) If the water flows through the pipe at a velocity V = 18 ft/s, what is the dimensionless loss coefficient of the device, defined by K = p/( V )? We will study loss coefficients in Chap. 6.

16 86 Solutions Manual Fluid Mechanics, Fifth Edition Solution: Gather density data: mercury = 1550 kg/m, water = 998 kg/m. Example., by going down from (a) to the mercury level, jumping across, and going up to (b), found the very important formula for this type of equal-leg manometer: p p p ( ) g h ( kg / m )(9.81 m / s )(0.4 m) a b merc water or : p 9,600 Pa Ans.( a) (b) The loss coefficient calculation is straightforward, but we check the units to make sure. Convert the velocity from 18 ft/s to 5.49 m/s. Then K p 9600 N / m 9600 N / m 0.98 V (998 kg / m )(5.49 m / s) 0080 N / m Ans.( b).1 In Fig. P.1 determine p between points A and B. All fluids are at 0C. Fig. P.1 Solution: Take the specific weights to be Benzene: 8640 N/m Mercury: 1100 N/m Kerosene: 7885 N/m Water: 9790 N/m and air will be small, probably around 1 N/m. Work your way around from A to B: p (8640)(0.0 m) (1100)(0.08) (7885)(0.) (9790)(0.6) ( 1 )(0.09) A p, or, after cleaning up, p p 8900 Pa Ans. B A B

17 Chapter Pressure Distribution in a Fluid 87. For the manometer of Fig. P., all fluids are at 0C. If pb pa 97 kpa, determine the height H in centimeters. Solution: Gamma 9790 N/m for water and 1100 N/m for mercury and (0.87)(9790) 8096 N/m for Meriam red oil. Work your way around from point A to point B: A p (9790 N/m )(H meters) 8096(0.18) 1100(0.18 H 0.5) p p Solve for H 0.6 m.6 cm Ans. B A Fig. P.. In Fig. P. the pressure at point A is 5 psi. All fluids are at 0C. What is the air pressure in the closed chamber B?

18 88 Solutions Manual Fluid Mechanics, Fifth Edition Solution: Take 9790 N/m for water, 870 N/m for SAE 0 oil, and (1.45)(9790) N/m for the third fluid. Convert the pressure at A from 5 lbf/in to Pa. Compute hydrostatically from point A to point B: A Fig. P. p h (9790 N/m )(0.04 m) (870)(0.06) (14196)(0.10) pb Pa psi Ans.

19 Chapter Pressure Distribution in a Fluid 89.4 To show the effect of manometer dimensions, consider Fig. P.4. 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 distance h h. Derive a formula for the difference pa pb when (a) d << D; and (b) d 0.15D. What is the % difference? Fig. P.4 Solution: Take 9790 N/m for water and 870 N/m for SAE 0 oil. Let H be the height of the oil in reservoir (b). For the condition shown, pa pb, therefore (L h) (H h), or: H ( / )(L h) h (1) water oil water oil Case (a), d << D: When the meniscus rises h, there will be no significant change in reservoir levels. Therefore we can write a simple hydrostatic relation from (a) to (b): p (L h h) (H h h) p, a water oil b or: p p h Ans. (a) a b water where we have used Eq. (1) above to eliminate H and L. Putting in numbers to compare later with part (b), we have p h( ) 1070 h, with h in meters. Case (b), d 0.15D. Here we must account for reservoir volume changes. For a rise h h, a volume (/4)d h of water leaves reservoir (a), decreasing L by h(d/d), and an identical volume of oil enters reservoir (b), increasing H by the same amount h(d/d). The hydrostatic relation between (a) and (b) becomes, for this case, oil a water oil b p [L h(d/d) h h] [H h(d/d) h h] p, or: p p h 1 d D 1 d D Ans. (b) a b water oil where again we have used Eq. (1) to eliminate H and L. If d is not small, this is a considerable difference, with surprisingly large error. For the case d 0.15 D, with water and oil, we obtain p h[1.05(9790) (870)] 1486 h or 9% more than (a).

20 90 Solutions Manual Fluid Mechanics, Fifth Edition.5 Water flows upward in a pipe slanted at 0, as in Fig. P.5. The mercury manometer reads h 1 cm. What is the pressure difference between points (1) and () in the pipe? Solution: The vertical distance between points 1 and equals (.0 m)tan 0 or m. Go around the U-tube hydrostatically from point 1 to point : p 9790h1100h (1.155 m) p, 1 Fig. P.5 or: p p ( )(0.1) Pa Ans..6 In Fig. P.6 both the tank and the slanted tube are open to the atmosphere. If L.1 m, what is the angle of tilt of the tube? Fig. P.6 Solution: Proceed hydrostatically from the oil surface to the slanted tube surface: pa 0.8(9790)(0.5) 9790(0.5) 9790(.1sin ) p a, or: sin 0.45, solve 5 Ans..7 The inclined manometer in Fig. P.7 contains Meriam red oil, SG Assume the reservoir is very large. If the inclined arm has graduations 1 inch apart, what should be if each graduation represents 1 psf of the pressure pa? Fig. P.7

21 Chapter Pressure Distribution in a Fluid 91 Solution: The specific weight of the oil is (0.87)(6.4) 51.6 lbf/ft. If the reservoir level does not change and L 1 inch is the scale marking, then lbf lbf 1 p A(gage) 1 oil z oil L sin 51.6 ft sin, ft ft 1 or: sin 0.5 or: 1.45 Ans. P.8 If the pressure in container A is 150 kpa, compute the pressure in Fig. P.8 B container B. A Water 18 cm Solution: The specific weights are oil = (0.8)(9790) = 78 N/m, mercury = 1,100 N/m, and water = 9790 N/m. 16 cm 8 cm Oil, SG = 0.8 Mercury cm Go down 16 cm from A to the mercury interface, jump across and go up 14 cm ( cm - 8 cm) to the right-side mercury interface, and then up 18 cm of water to point B. Of course, you could also go straight down to the bottom of the tube and then across and up. Calculate p (78 N / m )(0.16 m) (1100)(0.14) (9790)(0.18) p A Given p 150, 000 Pa, solve for p 10,900 Pa 11 kpa Ans. A B B

22 .9 In Fig. P.9 the right leg of the manometer is open to the atmosphere. Find the gage pressure, in Pa, in the air gap in the tank. Neglect surface tension. Solution: The two 8-cm legs of air are negligible (only Pa). Begin at the right mercury interface and go to the air gap: 0 Pa-gage (1100 N/m )( m) airgap ( N/m )( m) p or: p 7951 Pa 71 Pa 5700 Pa-gage Ans. airgap Fig. P.9.40 In Fig. P.40 the pressures at A and B are the same, 100 kpa. If water is introduced at A to increase pa to 10 kpa, find and sketch the new positions of the

23 Chapter Pressure Distribution in a Fluid 9 mercury menisci. The connecting tube is a uniform 1-cm in diameter. Assume no change in the liquid densities. Fig. P.40 Solution: Since the tube diameter is constant, the volume of mercury will displace a distance h down the left side, equal to the volume increase on the right side; h L. Apply the hydrostatic relation to the pressure change, beginning at the right (air/mercury) interface: p B Hg( Lsin h) W( h Lsin p with h L or: 100, ( h)(1 sin15 ) 9790( h)(1 sin15 ) p 10,000 Pa Solve for h (0,000 Pa)/[( N/m )(1 sin15 )] 0.19 m Ans. The mercury in the left (vertical) leg will drop 19. cm, the mercury in the right (slanted) leg will rise 19. cm along the slant and 5 cm in vertical elevation. A.41 The system in Fig. P.41 is at 0C. Determine the pressure at point A in pounds per square foot. Solution: Take the specific weights of water and mercury from Table.1. Write the hydrostatic formula from point A to the water surface: Fig. P lbf p A (0.85)(6.4 lbf/ft ) ft (846) (6.4) p atm (14.7)(144) ft Solve for p 770 lbf/ft Ans. A

24 .4 Small pressure differences can be measured by the two-fluid manometer in Fig. P.4, where is only slightly larger than 1. Derive a formula for pa pb if the reservoirs are very large. Solution: Apply the hydrostatic formula from A to B: Fig. P.4 p gh gh g(h h) p A B Solve for p p gh Ans. A B 1 If ( 1) is very small, h will be very large for a given p (a sensitive manometer)..4 The traditional method of measuring blood pressure uses a sphygmomanometer, first recording the highest (systolic) and then the lowest (diastolic) pressure from which flowing Korotkoff sounds can be heard. Patients with dangerous hypertension can exhibit systolic pressures as high as 5 lbf/in. Normal levels, however, are.7 and 1.7 lbf/in, respectively, for systolic and diastolic pressures. The manometer uses mercury and air as fluids. (a) How high should the manometer tube be? (b) Express normal systolic and diastolic blood pressure in millimeters of mercury. Solution: (a) The manometer height must be at least large enough to accommodate the largest systolic pressure expected. Thus apply the hydrostatic relation using 5 lbf/in as the pressure, B h p / g (5 lbf/in )(6895 Pa/lbf/in )/(1100 N/m ) 0.6 m So make the height about 0cm Ans a (b) Convert the systolic and diastolic pressures by dividing them by mercury s specific weight. systolic diastolic h (.7 lbf/in )(144 in /ft )/(846 lbf/ft ) 0.46 ft Hg 140 mm Hg h (1.7 lbf/in )(144 in /ft )/(846 lbf/ft ) 0.89 ft Hg 88 mm Hg

25 Chapter Pressure Distribution in a Fluid 95 The systolic/diastolic pressures are thus 140/88 mm Hg. Ans. (b).44 Water flows downward in a pipe at 45, as shown in Fig. P.44. The mercury manometer reads a 6-in height. The pressure drop p p1 is partly due to friction and partly due to gravity. Determine the total pressure drop and also the part due to friction only. Which part does the manometer read? Why? Fig. P.44 Solution: Let h be the distance down from point to the mercury-water interface in the right leg. Write the hydrostatic formula from 1 to : 6 6 p sin 45 h h p, 1 1 p1 p ( )(6/1) 6.4(5sin 45 ) friction loss..... gravity head.. lbf 171 ft Ans. The manometer reads only the friction loss of 9 lbfft, not the gravity head of 1 psf.

26 .45 Determine the gage pressure at point A in Fig. P.45, in pascals. Is it higher or lower than Patmosphere? Solution: Take 9790 Nm for water and 1100 Nm for mercury. Write the hydrostatic formula between the atmosphere and point A: p (0.85)(9790)(0.4 m) atm (1100)(0.15 m) (1)(0.0 m) (9790)(0.45 m) p, A A atm Fig. P.45 or: p p 100 Pa 100 Pa (vacuum) Ans.

27 Chapter Pressure Distribution in a Fluid In Fig. P.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 which leg we approach through, left or right: p (870)(0.1) (9790)(0.07) atm (0.04) (left leg) X atm X Fig. P.46 p (870)(0.09) (9790)(0.05) (0.06) (right leg) or: X N/m, SG X 1.45 Ans The cylindrical tank in Fig. P.47 is being filled with 0C water by a pump developing an exit pressure of 175 kpa. At the instant shown, the air pressure is 110 kpa and H 5 cm. The pump stops when it can no longer raise the water pressure. Estimate H at that time. Fig. P.47 Solution: At the end of pumping, the bottom water pressure must be 175 kpa: pair 9790H Meanwhile, assuming isothermal air compression, the final air pressure is such that pair Volold R (0.75 m) Volnew R (1.1 m H) 1.1 H where R is the tank radius. Combining these two gives a quadratic equation for H: 0.75(110000) 9790H , or H 18.98H H The two roots are H 18.7 m (ridiculous) or, properly, H 0.61 m Ans.

28 98 Solutions Manual Fluid Mechanics, Fifth Edition P.48 The system in Fig. P.49 is open to 1 atm on the right side. (a) If L = 10 cm, what is the air pressure in container A? (b) Conversely, if pa = 15 kpa, what is the length L? 15 cm A Air cm B Mercury C D 5 18 cm Water L z = 0 Fig. P.48 Solution: (a) The vertical elevation of the water surface in the slanted tube is (1.m)(sin55) = 0.98 m. Then the pressure at the 18-cm level of the water, point D, is N pd patm water z 10150Pa (9790 )( m) 10900Pa m Going up from D to C in air is negligible, less than Pa. Thus pc pd = Pa. Going down from point C to the level of point B increases the pressure in mercury: p B p C mercury z CB N (1100 m )( m) This is the answer, since again it is negligible to go up to point A in low-density air. (b) Given pa = 15 kpa, go down from point A to point B with negligible air-pressure change, then jump across the mercury U-tube and go up to point C with a decrease: Once again, pc pd Pa, jump across the water and then go up to the surface: 11800Pa pc pb mercury zbc (1100)( ) 11400Pa Ans.( a)

29 Chapter Pressure Distribution in a Fluid 99 p atm p D water z Solve for ( z Then the slanted distance L z surface surface 1.16m 1.16m / sin m) 10150Pa 1.75m Ans.( b).48 Conduct an experiment: Place a thin wooden ruler on a table with a 40% overhang, as shown. Cover it with full-size sheets of newspaper. (a) Estimate the total force on top 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 7 in (0.686 m) by.5 in (0.57 m). Thus the force is: F pa (1015 Pa)(0.686 m)(0.57 m) 9700 N! Ans. Fig. P.48 (b) The newspaper will hold the ruler, which will probably break due to the chop. Ans. (c) Chop is fast, air does not have time to rush in, partial vacuum under newspaper. Ans.

30 100 Solutions Manual Fluid Mechanics, Fifth Edition P.50 A small submarine, with a hatch door 0 inches in diameter, is submerged in seawater. (a) If the water hydrostatic force on the hatch is 69,000 lbf, how deep is the sub? (b) If the sub is 50 ft deep, what is the hydrostatic force on the hatch? Solution: In either case, the force is pcgahatch. Stay with BG units. Convert 0 inches =.5 ft. For seawater, = 105 kg/m = 1.99 slug/ft, hence = (1.99)(.) = 64.0 lbf/ft. lbf ( a) F pcg A ( h) A 69,000 lbf (64 ) h (.5 ft) ; h 0 ft Ans.( a) ft 4 lbf ( b) F pcg A ( h) A (64 )(50 ft) (.5 ft) 110, 000 lbf Ans.( b) ft 4.51 Gate AB in Fig. P.51 is 1. m long and 0.8 m into the paper. Neglecting atmospheric-pressure effects, compute the force F on the gate and its center of pressure position X. Solution: The centroidal depth of the gate is AB oil CG gate hcg 4.0 ( )sin m, Fig. P.51 hence F h A ( )(5.08)(1. 0.8) 8750 N Ans. The line of action of F is slightly below the centroid by the amount y Ixx sin (1/1)(0.8)(1.) sin 40 CP hcga (5.08)(1. 0.8) m Thus the position of the center of pressure is at X m Ans.

31 Chapter Pressure Distribution in a Fluid 101 P.5 Example.5 calculated the force on plate AB and its line of action, using the moment-of-inertia approach. Some teachers p() A say it is more instructive to calculate these 6 ft by direct integration of the pressure forces. Using Figs. P.5 and E.5a, (a) find an expression for the pressure variation p() along the plate; B 8 ft Fig. P.5 (b) integrate this pressure to find the total force F; (c) integrate the moments about point A to find the position of the center of pressure. Solution: (a) Point A is 9 ft deep, and point B is 15 ft deep, and = 64 lbf/ft. Thus pa = (64lbf/ft )(9ft) = 576 lbf/ft and pb = (64lbf/ft )(15ft) = 960 lbf/ft. Along the 10-ft length, pressure increases by ( )/10ft = 8.4 lbf/ft /ft. Thus the pressure is p( ) ( lbf / ft ) Ans.( a) (b) Given that the plate width b = 5 ft. Integrate for the total force on the plate: F 10 p da pb d ( )(5 ft) d plate (5)( / ) ,400 lbf Ans.( b) 0 (c) Find the moment of the pressure forces about point A and divide by the force: The center of pressure is ft down the plate from Point A. M A plate p b da (5)(576 Then 10 0 ( )(5 ft) d / 8.4 CP M F / ) A ft lbf 8400lbf 08,000 ft lbf 5.4 ft Ans.( c)

32 10 Solutions Manual Fluid Mechanics, Fifth Edition.5 Panel ABC in the slanted side of a water tank (shown at right) is an isoceles triangle with vertex at A and base BC m. Find the water force on the panel and its line of action. Solution: (a) The centroid of ABC is of the depth down, or 8 m from the surface. The panel area is (1)( m)(5 m) 5 m. The water force is F h A (9790)(.67 m)(5 m ) 11,000 N Ans. (a) ABC CG panel (b) The moment of inertia of ABC is (16)( m)(5 m) 6.94 m 4. From Eq. (.44), y I sin /(h A ) 6.94sin(5 )/[.67(5)] m Ans. (b) CP xx CG panel The centroid is 5(/) =. m down from A along the panel. The center of pressure is thus ( ) =.75 m down from A, or 1.5 m up from BC..54 In Fig. P.54, the hydrostatic force F is the same on the bottom of all three containers, even though the weights of liquid above are quite different. The three bottom shapes and the fluids are the same. This is called the hydrostatic paradox. Explain why it is true and sketch a freebody of each of the liquid columns. Fig. P.54

33 Chapter Pressure Distribution in a Fluid 10 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 horizontal. In (c) upward side pressure helps reduce a heavy W..55 Gate AB in Fig. P.55 is 5 ft wide into the paper, hinged at A, and restrained by a stop at B. Compute (a) the force on stop B; and (b) the reactions at A if h 9.5 ft. Solution: The centroid of AB is.0 ft below A, hence the centroidal depth is h ft. Then the total hydrostatic force on the gate is Fig. P.55 CG gate F h A (6.4 lbf/ft )(7.5 ft)(0 ft ) 960 lbf The C.P. is below the centroid by the amount y CP Ixx sin (1/1)(5)(4) sin 90 hcga (7.5)(0) ft This is shown on the freebody of the gate at right. We find force Bx with moments about A: M B (4.0) (960)(.178) 0, A x x or: B 5100 lbf (to left) Ans. (a) The reaction forces at A then follow from equilibrium of forces (with zero gate weight): F A, or: A 460 lbf (to left) x x x F 0 A W A, or: A 0 lbf Ans. (b) z z gate z z

34 104 Solutions Manual Fluid Mechanics, Fifth Edition.56 For the gate of Prob..55 above, stop B breaks if the force on it equals 900 lbf. For what water depth h is this condition reached? Solution: The formulas must be written in terms of the unknown centroidal depth hcg: h h F h A (6.4)h (0) 148h y CG CG CG CG IXX sin (1/1)(5)(4) sin CP hcga h CG(0) hcg Then moments about A for the freebody in Prob..155 above will yield the answer: 1. MA 0 900(4) (148h CG), or hcg ft, h Ans. h ft CG.57 The tank in Fig. P.57 is m wide into the paper. Neglecting atmospheric pressure, find the resultant hydrostatic force on panel BC, (a) from a single formula; (b) by computing horizontal and vertical forces separately, in the spirit of curved surfaces. Fig. P.57 Solution: (a) The resultant force F, may be found by simply applying the hydrostatic relation F h A (9790 N/m )( 1.5 m)(5 m m) 440,550 N 441 kn Ans. (a) CG (b) The horizontal force acts as though BC were vertical, thus hcg is halfway down from C and acts on the projected area of BC. F (9790)(4.5)( ) 64,0 N 64 kn Ans. (b) H The vertical force is equal to the weight of fluid above BC, F (9790)[()(4) (1/)(4)()]() 5,440 5 kn Ans. (b) V The resultant is the same as part (a): F [(64) (5) ] 1/ 441 kn.

35 .58 In Fig. P.58, weightless cover gate AB closes a circular opening 80 cm in diameter when weighed down by the 00-kg mass shown. What water level h will dislodge the gate? Solution: The centroidal depth is exactly Fig. P.58 equal to h and force F will be upward on the gate. Dislodging occurs when F equals the weight: F hcga gate (9790 N/m ) h (0.8 m) W (00)(9.81) N 4 Solve for h 0.40 m Ans..59 Gate AB has length L, width b into the paper, is hinged at B, and has negligible weight. The liquid level h remains at the top of the gate for any angle. Find an analytic expression for the force P, perpendicular to AB, required to keep the gate in equilibrium. Solution: The centroid of the gate remains at distance L from A and depth h below the surface. For any, then, the hydrostatic force is F (h)lb. The moment of inertia of the gate is (11)bL, hence ycp (11)bL sin[(h)lb], and the center of pressure is (L ycp) from point B. Summing moments about hinge B yields PL F(L/ y ), or: P = (γhb/4)[l - L sin θ/(h)] Ans. CP

36 106 Solutions Manual Fluid Mechanics, Fifth Edition P.60 Determine the water hydrostatic force on one side of the vertical equilateral triangle panel BCD in Fig.. P.60. Neglect atmospheric pressure. Solution: Pythagoras says that the lengths of DC, Fig. P.60 B CG F C 7 0 cm 40 cm BC, and BD are all 50 cm. The centroid of the panel would lie along the bisector BF, whose length is 50 sin60 = 4. cm. D 5 0 cm The CG would be 1/ of the way from F to B, or 14.4 cm away from F. The CG would be 14.4sin7 = 8.66 cm vertically above F, and F is 0 cm above the bottom. Thus the CG is 60 cm 0 cm 8.66 cm = 1.4 cm below the surface of the water. The hydrostatic force is A (1/ )(50 cm)(4. cm) 108cm m triangle p h (9790 N / m )(0.14 m) 068 Pa CG water CG Finally, F p A (068 Pa)(0.108 m ) N Ans. panel CG.61 Gate AB in Fig. P.61 is a homo-geneous mass of 180 kg, 1. m wide into the paper, resting on smooth bottom B. All fluids are at 0C. For what water depth h will the force at point B be zero?

37 Chapter Pressure Distribution in a Fluid 107 Solution: Let 160 Nm for glycerin and 9790 Nm for water. The centroid of AB is 0.4 m vertically below A, so hcg m, and we may compute the glycerin force and its line of action: F ha (160)(1.567)(1.) 4 N y g CP,g (1/1)(1.)(1) sin m (1.567)(1.) These are shown on the freebody at right. The water force and its line of action are shown without numbers, because they depend upon the centroidal depth on the water side: w Fig. P.61 F (9790)h (1.) CG y CP CG (1/1)(1.)(1) sin h (1.) h CG The weight of the gate, W 180(9.81) 1766 N, acts at the centroid, as shown above. Since the force at B equals zero, we may sum moments counterclockwise about A to find the water depth: M 0 (4)(0.5461) (1766)(0.5cos60 ) A CG,water (9790)h (1.)( /h ) CG Solve for h.09 m, or: h h m Ans. CG CG

38 .6 Gate AB in Fig. P.6 is 15 ft long and 8 ft wide into the paper, hinged at B with a stop at A. The gate is 1-in-thick steel, SG Compute the 0 C water level h for which the gate will start to fall. Solution: Only the length (h csc 60) of the gate lies below the water. Only this part Fig. P.6 contributes to the hydrostatic force shown in the freebody at right: y h F hcg A (6.4) (8h csc 60 ) CP 88.h (lbf) (1/1)(8)(h csc60 ) sin60 (h/)(8h csc60 ) h csc60 6 The weight of the gate is (7.85)(6.4 lbf/ft )(15 ft)(1/1 ft)(8 ft) 4898 lbf. This weight acts downward at the CG of the full gate as shown (not the CG of the submerged portion). Thus, W is 7.5 ft above point B and has moment arm (7.5 cos 60 ft) about B. We are now in a position to find h by summing moments about the hinge line B:

39 Chapter Pressure Distribution in a Fluid 109 B M (10000)(15) (88.h )[(h/)csc60 (h/6)csc60 ] 4898(7.5cos60 ) 0, 1/ or: 110.9h , h (1161/110.9) 10.6 ft Ans..6 The tank in Fig. P.6 has a 4-cmdiameter plug which will pop out if the hydrostatic force on it reaches 5 N. For 0C fluids, what will be the reading h on the manometer when this happens? Solution: The water depth when the plug pops out is (0.04) CG CG Fig. P.6 F 5 N h A (9790)h or hcg.0 m 4 It makes little numerical difference, but the mercury-water interface is a little deeper than this, by the amount (0.0 sin 50) of plug-depth, plus cm of tube length. Thus p (9790)(.0 0.0sin ) (1100)h p, atm or: h 0.15 m Ans. atm

40 110 Solutions Manual Fluid Mechanics, Fifth Edition.64 Gate ABC in Fig. P.64 has a fixed hinge at B and is 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 meter) be the depth down to the level AB. The forces on AB and BC are shown in the freebody at right. The moments of these forces about B are equal when the gate opens: MB 0 H(0.)b(0.1) H H (Hb) or: H 0.46 m, h H m Ans. This solution is independent of both the water density and the gate width b into the paper. Fig. P Gate AB in Fig. P.65 is semicircular, hinged at B, and held by a horizontal force P at point A. Determine the required force P for equilibrium. Solution: The centroid of a semi-circle is at 4R/ 1.7 m off the bottom, as shown in the sketch at right. Thus it is m down from the force P. The water force F is F hcga (9790)( ) () N The line of action of F lies below the CG: y 4 Ixx sin ( )() sin 90 CP hcga (5 1.77)( /)() Fig. P m Then summing moments about B yields the proper support force P: M 0 (91000)( ) P, or: P N Ans. B

41 Chapter Pressure Distribution in a Fluid Dam ABC in Fig. P.66 is 0 m wide into the paper and is concrete (SG.40). Find the hydrostatic force on surface AB and its moment about C. Could this force tip the dam over? Would fluid seepage under the dam change your argument? Solution: The centroid of surface AB is 40 m deep, and the total force on AB is Fig. P.66 F h A (9790)(40)(100 0) CG 1.175E9 N The line of action of this force is two-thirds of the way down along AB, or m from A. This is seen either by inspection (A is at the surface) or by the usual formula: y Ixx sin (1/1)(0)(100) sin(5.1 ) CP hcga (40)(0 100) m to be added to the 50-m distance from A to the centroid, or m. As shown in the figure, the line of action of F is.67 m to the left of a line up from C normal to AB. The moment of F about C is thus M FL (1.175E9)( ).1E9 N m Ans. C This moment is counterclockwise, hence it cannot tip over the dam. If there were seepage under the dam, the main support force at the bottom of the dam would shift to the left of point C and might indeed cause the dam to tip over..67 Generalize Prob..66 with length AB as H, length BC as L, and angle ABC as, with width b into the paper. If the dam material has specific gravity SG, with no seepage, find the critical angle c for which the dam will just tip over to the right. Evaluate this expression for SG.40. Solution: By geometry, L Hcos and the vertical height of the dam is Hsin. The Fig. P.67

42 11 Solutions Manual Fluid Mechanics, Fifth Edition force F on surface AB is (H/)(sin)Hb, and its position is at H/ down from point A, as shown in the figure. Its moment arm about C is thus (H/ Lcos). Meanwhile the weight of the dam is W (SG) (L/)H(sin)b, with a moment arm L/ as shown. Then summation of clockwise moments about C gives, for critical tip-over conditions, H H L L MC 0 sin Hb L cos SG( ) H sin b with L H cos. Solve for cos c Ans. SG Any angle greater than c will cause tip-over to the right. For the particular case of concrete, SG.40, cosc 0.40, or c 64.5, which is greater than the given angle 5.1 in Prob..66, hence there was no tipping in that problem..68 Isosceles triangle gate AB in Fig. P.68 is hinged at A and weighs 1500 N. What horizontal force P is required at point B for equilibrium? Solution: The gate is.0/sin m long from A to B and its area is m. Its centroid is 1/ of the way down from A, so the centroidal depth is m. The force on the gate is F h A (0.8)(9790)(.667)(1.054) CG 8894 N The position of this force is below the centroid: y CP Ixx sin h A CG Fig. P.68 (1/ 6)(1.0)(.611) sin m (.667)(1.054) The force and its position are shown in the freebody at upper right. The gate weight of 1500 N is assumed at the centroid of the plate, with moment arm meters about point A. Summing moments about point A gives the required force P: MA 0 P(.0) 1500(0.559) 8894( ), Solve for P N Ans.

43 Chapter Pressure Distribution in a Fluid 11 P.69 Consider the slanted plate AB of length L in Fig. P.69. (a) Is the hydrostatic force F on the plate equal to the weight of the missing water above the plate? If not, correct this hypothesis. Neglect the atmosphere. A F Water, specific weight B Fig. P.69 (b) Can a missing water approach be generalized to curved plates of this type? Solution: (a) The actual force F equals the pressure at the centroid times the plate area: But the weight of the missing water is F p CG A plate h CG Lb Lsin Lb sin 1 Wmissing missing [ ( Lsin)( Lcos ) b] L bsin cos Why the discrepancy? Because the actual plate force is not vertical. Its vertical component is F cos = Wmissing. The missing-water weight equals the vertical component of the force. Ans.(a) This same approach applies to curved plates with missing water. Ans.(b) L b P.70 The swing-check valve in Fig. P.70 covers a.86-cm diameter Air opening in the slanted wall. The hinge is 15 cm from the centerline, as shown. 15 cm hinge h The valve will open when the hinge moment is 50 N-m. Find the value of 60 Water at 0C Fig. P.70

44 114 Solutions Manual Fluid Mechanics, Fifth Edition h for the water to cause this condition. Solution: For water, take = 9790 N/m. The hydrostatic force on the valve is N F pcg A h ( ) R (9790 ) h ( )(0.114 m) h m The center of pressure is slightly below the centerline by an amount sin I xx (9790)sin(0 )( / 4)(0.114) ycp F h h The 60 angle in the figure is a red herring we need the 0 angle with the horizontal. Then the moment about the hinge is Mhinge F l (401.8 h)(0.15 ) 50 N m h Solve for h 0.79 m Ans. Since ycp is so small ( mm), you don t really need EES. Just iterate once or twice. 4

45 Chapter Pressure Distribution in a Fluid In Fig. P.71 gate AB is m wide into the paper and is connected by a rod and pulley to a concrete sphere (SG.40). What sphere diameter is just right to close the gate? Solution: The centroid of AB is 10 m down from the surface, hence the hydrostatic force is F h A (9790)(10)(4 ) CG 1.175E6 N Fig. P.71 The line of action is slightly below the centroid: y CP (1/1)()(4) sin m (10)(1) Sum moments about B in the freebody at right to find the pulley force or weight W: MB 0 W(6 8 4 m) (1.175E6)( m), or W N Set this value equal to the weight of a solid concrete sphere: W N concrete D (.4)(9790) D, or: D sphere.15 m Ans Gate B is 0 cm high and 60 cm wide into the paper and hinged at the top. What is the water depth h which will first cause the gate to open? Solution: The minimum height needed to open the gate can be assessed by calculating the hydrostatic force on each side of the gate and equating moments about the hinge. The air pressure causes a force, Fair, which acts on the gate at 0.15 m above point D. Fig. P.7 F air (10,000 Pa)(0. m)(0.6 m) 1800 N

46 116 Solutions Manual Fluid Mechanics, Fifth Edition Since the air pressure is uniform, Fair acts at the centroid of the gate, or 15 cm below the hinge. The force imparted by the water is simply the hydrostatic force, w CG w F ( h A) (9790 N/m )(h 0.15 m)(0. m)(0.6 m) 176.h 64. This force has a center of pressure at, o (1/1)(0.6)(0.) (sin 90 ) ycp with h in meters ( h0.15)(0.)(0.6) h0.15 Sum moments about the hinge and set equal to zero to find the minimum height: Mhinge 0 (176.h 64.)[0.15 (0.0075/(h 0.15))] (1800)(0.15) This is quadratic in h, but let s simply solve by iteration: h 1.1 m Ans..7 Weightless gate AB is 5 ft wide into the paper and opens to let fresh water out when the ocean tide is falling. The hinge at A is ft above the freshwater level. Find h when the gate opens. Solution: There are two different hydrostatic forces and two different lines of action. On the water side, Fig. P.7 F h A (6.4)(5)(10 5) lbf w CG positioned at. ft above point B. In the seawater, h F s ( ) (5h) 159.9h (lbf) positioned at h/ above point B. Summing moments about hinge point A gives the desired seawater depth h: A M 0 (159.9h )(1 h/) (15600)(1.), or 5.h h , solve for h 9.85 ft Ans.

47 Chapter Pressure Distribution in a Fluid Find the height H in Fig. P.74 for which the hydrostatic force on the rectangular panel is the same as the force on the semicircular panel below. Solution: Find the force on each panel and set them equal: F h A (H/)[(R)(H)] RH rect CG rect F h A (H 4R/ )[( /)R ] semi CG semi Fig. P.74 Set them equal, cancel RH (/)R H R /, or: H (/)RH R / 0 1/ Finally, H R[π/4 +{(π/4) + /} ] 1.9R Ans. P.75 The cap at point B on the 5-cm-diameter tube in Fig. P.75 will be dislodged when the hydrostatic force on its base reaches lbf. Oil, SG = 0.8 B 1 m water h For what water depth h does this occur? m Fig. P.75 Solution: Convert the cap force to SI units: lbf x = 97.9 N. Then the dislodging: pressure just under cap B will be F 97.9 N pb 49,800 Pa ( gage) A ( / 4)(0.05 m) tube Begin at point B, go down and around the two fluids to the surface of the tank: N N N Pa (0.8)(9790 )(1 m) (9790 )( m) (9790 )( h) p 0 ( ) surface gage m m m 7750 Pa Solve for h 7.89 m Ans N/ m

48 118 Solutions Manual Fluid Mechanics, Fifth Edition.76 Panel BC in Fig. P.76 is circular. Compute (a) the hydrostatic force of the water on the panel; (b) its center of pressure; and (c) the moment of this force about point B. Solution: (a) The hydrostatic force on the gate is: F h A CG (9790 N/m )(4.5 m)sin 50 ( )(1.5 m) 9kN Ans. (a) (b) The center of pressure of the force is: y CP 4 r sin I xxsin 4 hcg A hcg A 4 4 (1.5) sin m Ans (4.5sin50 )( )(1.5 ). (b) Fig. P.76 Thus y is 1.65 m down along the panel from B (or 0.15 m down from the center of the circle). (c) The moment about B due to the hydrostatic force is, M (8550 N)(1.65 m) 87,600 N m 88 kn m Ans. (c) B.77 Circular gate ABC is hinged at B. Compute the force just sufficient to keep the gate from opening when h 8 m. Neglect atmospheric pressure. Solution: The hydrostatic force on the gate is F h A (9790)(8 m)( m ) CG N Fig. P.77

49 This force acts below point B by the distance y 4 Ixxsin ( /4)(1) sin 90 CP hcga (8)( ) m Summing moments about B gives P(1 m) (46050)(0.015 m), or P 7690 N Ans. P.78 Panels AB and CD are each 10 cm wide into the paper. (a) Can 0 cm you deduce, by inspection, which 40 cm A water D panel has the larger water force? (b) Even if your deduction is brilliant, 40 cm 40 B C cm calculate the panel forces anyway. Fig. P.78 Solution: (a) The writer is unable to deduce by inspection which panel force is larger. CD is longer than AB, but its centroid is not as deep. If you have a great insight, let me know. (b) The length of AB is (40cm)/sin40 = 6. cm. The centroid of AB is 40+0 = 60 cm below the surface. The length of CD is (50cm)/sin50 = 65.7 cm. The centroid of AB is 0+5 = 55 cm below the surface. Calculate the two forces: N FAB hab AAB (9790 )(0.6 m)(0.6 m)(1. m) 490 N m N FCD hcd ACD (9790 )(0.55 m)(0.657 m)(1. m) 40 N Ans.( b) m It turns out that panel AB has the larger force, but it is only 4 percent larger.

50 10 Solutions Manual Fluid Mechanics, Fifth Edition.79 Gate ABC in Fig. P.79 is 1-msquare and hinged at B. It opens automatically when the water level is high enough. Neglecting atmospheric pressure, determine the lowest level h for which the gate will open. Is your result independent of the liquid density? Fig. P.79 Solution: The gate will open when the hydrostatic force F on the gate is above B, that is, when y CP Ixxsin h A CG (1/1)(1 m)(1 m) sin m, (h 0.5 m)(1 m ) or: h m, or: h 0. m Ans. Indeed, this result is independent of the liquid density.

51 Chapter Pressure Distribution in a Fluid 11 *P.80 A concrete dam (SG =.5) is made in the shape of an isosceles triangle, as in Fig. P.80. Analyze this geometry to find the range of angles for which the h F W L hydrostatic force will tend to tip the dam B over at point B. The width into the paper is b. l Solution: The critical angle is when the hydrostatic force F causes a clockwise moment equal to the counterclockwise moment of the dam weight W. The length L of the slanted side of the dam is L = h/sin. The force F is two-thirds of the way down this face. The moment arm of the weight about point B is l = h/tan The moment arm of F about point B is quite difficult, and you should check this: Moment arm of F about B is L l cos 1 h sin h cos tan Evaluate the two forces and then their moments: F M B h h b ; W sin h b h hcos ( ) sin sin tan h SG dam SG h b tan SG h b h ( ) clockwise tan tan When the moment is negative (small, the dam is stable, it will not tip over. The moment is zero, for SG =.5, at = Thus tipping is possible in the range > Ans. NOTE: This answer is independent of the numerical values of h, g, or b but requires SG =.5. P.81 For the semicircular cylinder CDE in Ex..9, find the vertical hydrostatic force by integrating the vertical component of pressure around the surface from = 0 to =.

52 1 Solutions Manual Fluid Mechanics, Fifth Edition Solution: A sketch is repeated here. At any position, as in Fig. P.81, the vertical component of pressure is A p cos. The depth down to this point is h+r(1- cos), h C and the local pressure is times this depth. Thus F p cos da 0 [ h R(1 cos )] (cos )[ b R d ] R D br( h R) cos d br cos d 0 br 0 0 E Rewrite : Fdown R b Ans. The negative sign occurs because the sign convention for df was a downward force. p Fig. P.81.8 The dam in Fig. P.8 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 resultant strikes the dam. Solution: The horizontal force acts as if the dam were vertical and 0 m high: F h A H CG vert (9790 N/m )(10 m)(050 m ) 97.9 MN Ans. Fig. P.8

53 Chapter Pressure Distribution in a Fluid 1 This force acts / of the way down or 1. m from the surface, as in the figure. The vertical force is the weight of the fluid above the dam: F V (Vol) dam (9790 N/m ) (0 m) (50 m) 15.8 MN Ans. 4 This vertical component acts through the centroid of the water above the dam, or 4R/ 4(0 m)/ 8.49 m to the right of point A, as shown in the figure. The resultant hydrostatic force is F [(97.9 MN) (15.8 MN) ] 1/ 18. MN acting down at an angle of.5 from the vertical. The line of action of F strikes the circular-arc dam AB at the center of pressure CP, which is m to the right and.1 m up from point A, as shown in the figure. Ans.

54 14 Solutions Manual Fluid Mechanics, Fifth Edition.8 Gate AB is a quarter-circle 10 ft wide and hinged at B. Find the force F just sufficient to keep the gate from opening. The gate is uniform and weighs 000 lbf. Solution: The horizontal force is computed as if AB were vertical: Fig. P.8 F h A (6.4)(4 ft)(8 10 ft ) H CG vert lbf acting 5. ft below A The vertical force equals the weight of the missing piece of water above the gate, as shown below. F (6.4)(8)(8 10) (6.4)( /4)(8) (10) V lbf The line of action x for this 8570-lbf force is found by summing moments from above: M (of F ) 8570x 996(4.0) 166(4.605), or x ft B V Finally, there is the 000-lbf gate weight W, whose centroid is R/ 5.09 ft from force F, or ft from point B. Then we may sum moments about hinge B to find the force F, using the freebody of the gate as sketched at the top-right of this page: M B(clockwise) 0 F(8.0) (000)(.907) (8570)(1.787) (19968)(.667), or F 7480 lbf Ans. 8.0

55 Chapter Pressure Distribution in a Fluid 15 P.84 Panel AB is a parabola with its maximum at point A. It is 150 cm wide into the paper. Neglect atmospheric pressure. Find (a) the vertical 5 cm A water C and (b) horizontal water forces on the panel. 75 cm parabola Fig. P cm B Solution: (b) The horizontal force is calculated from the vertical projection of the panel (from point A down to the bottom). This is a rectangle, 75 cm by 150 cm, and its centroid is 7.5 cm below A, or ( ) = 6.5 cm below the surface. Thus N FH pcg, H A projected [9790 (0.65 m)][0.75 m(1.50 m)] 6880 N Ans.( b) m (a) The vertical force is the weight of water above the panel. This is in two parts (1) the weight of the rectangular portion above the line AC; and () the little curvy piece above the parabola and below line AC. Recall from Ex..8 that the area under a parabola is two-thirds of the enclosed rectangle, so that little curvy piece is one-third of the rectangle. Thus, finally, F V 1 (9790)(0.5)(0.4)(1.5) (9790)( )(0.75)(0.4)(1.5) 1469 N 1469 N 940 N Ans.( a).85 Compute the horizontal and vertical components of the hydrostatic force on the quarter-circle panel at the bottom of the water tank in Fig. P.85.

56 16 Solutions Manual Fluid Mechanics, Fifth Edition Solution: The horizontal component is F h A (9790)(6)( 6) H CG vert N Ans. (a) Fig. P.85 The vertical component is the weight of the fluid above the quarter-circle panel: F W( by 7 rectangle) W(quarter-circle) V (9790)( 76) (9790)( /4)() (6) N Ans. (b).86 The quarter circle gate BC in Fig. P.86 is hinged at C. Find the horizontal force P required to hold the gate stationary. The width b into the paper is m. Neglect the weight of the gate. Solution: The horizontal component of water force is Fig. P.86 H CG F h A (9790 N/m )(1 m)[( m)( m)] 58,740 N This force acts / of the way down or 1. m down from the surface (0.667 m up from C). The vertical force is the weight of the quarter-circle of water above gate BC: V water F (Vol) (9790 N/m )[( /4)( m) ( m)] 9,70 N FV acts down at (4R/) m to the left of C. Sum moments clockwise about point C:

57 Chapter Pressure Distribution in a Fluid 17 MC 0 ( m)p (58740 N)(0.667 m) (970 N)(0.849 m) P Solve for P 58,700 N 58 7 kn Ans..87 The bottle of champagne (SG 0.96) in Fig. P.87 is under pressure as shown by the mercury manometer reading. Compute the net force on the -in-radius hemispherical end cap at the bottom of the bottle. Solution: First, from the manometer, compute the gage pressure at section AA in the champagne 6 inches above the bottom: Fig. P.87 4 p AA ( ) ft ( ) ft patmosphere 0 (gage), 1 1 or: P 7 lbf/ft (gage) AA Then the force on the bottom end cap is vertical only (due to symmetry) and equals the force at section AA plus the weight of the champagne below AA: F F p (Area) W W V AA AA 6-in cylinder -in hemisphere (7) (4/1) ( ) (/1) (6/1) ( )( /)(/1) lbf Ans..88 Circular-arc Tainter gate ABC pivots about point O. For the position shown, determine (a) the hydrostatic force on the gate (per meter of width into the paper); and (b) its line of action. Does the force pass through point O? Solution: The horizontal hydrostatic force is based on vertical projection: Fig. P.88

58 F h A (9790)()(6 1) 1760 N at 4 m below C H CG vert The vertical force is upward and equal to the weight of the missing water in the segment ABC shown shaded below. Reference to a good handbook will give you the geometric properties of a circular segment, and you may compute that the segment area is.61 m and its centroid is m from point O, or 0.5 m from vertical line AC, as shown in the figure. The vertical (upward) hydrostatic force on gate ABC is thus F A (unit width) (9790)(.611) V ABC 196 N at m from B 1/ The net force is thus F [FH F V] N per meter of width, acting upward to the right at an angle of 10.7 and passing through a point 1.0 m below and m to the right of point B. This force passes, as expected, right through point O..89 The tank in the figure contains benzene and is pressurized to 00 kpa (gage) in the air gap. Determine the vertical hydrostatic force on circular-arc section AB and its line of action. Solution: Assume unit depth into the paper. The vertical force is the weight of benzene plus the force due to the air pressure: Fig. P.89 FV N (0.6) (1.0)(881)(9.81) (00,000)(0.6)(1.0) 1400 Ans. 4 m Most of this (10,000 N/m) is due to the air pressure, whose line of action is in the middle of the horizontal line through B. The vertical benzene force is 400 N/m and has a line of action (see Fig..1 of the text) at 4R/() 5.5 cm to the right or A.

59 Chapter Pressure Distribution in a Fluid 19 The moment of these two forces about A must equal to moment of the combined (1,400 N/m) force times a distance X to the right of A: (10000)(0 cm) (400)(5.5 cm) 1400( X), solve for X = 9.9 cm Ans. The vertical force is 1400 N/m (down), acting at 9.9 cm to the right of A. P.90 The tank in Fig. P.90 is 10 cm long into the paper. Determine the horizontal and vertical hydrostatic 150 Missing water forces on the quarter-circle panel AB. A The fluid is water at 0C. 75 cm Neglect atmospheric pressure. B 40 cm Solution: For water at 0C, take = 9790 N/m. Fig. P.90 The vertical force on AB is the weight of the missing water above AB see the dashed lines in Fig. P.90. Calculate this as a rectangle plus a square-minus-a-quarter-circle: Missing water (1.5m)(0.75m)(1.m) (1 / 4)(0.75m) m F V (9790 N / m )(.05m ),600N ( vertical force) The horizontal force is calculated from the vertical projection of panel AB: F H p CG h A projection N (9790 m 0.75 )(1.5 m)(0.75m)(1.m) 16,500N ( horizontal force).91 The hemispherical dome in Fig. P.91 weighs 0 kn and is filled with water and attached to the floor by six equally-spaced bolts. What is the force in each bolt required to hold the dome down?

60 10 Solutions Manual Fluid Mechanics, Fifth Edition Solution: Assuming no leakage, the hydrostatic force required equals the weight of missing water, that is, the water in a 4-m-diameter cylinder, 6 m high, minus the hemisphere and the small pipe: Fig. P.91 F W W W total -m-cylinder -m-hemisphere -cm-pipe (9790) () (6) (9790)( /)() (9790)( /4)(0.0) (4) N The dome material helps with 0 kn of weight, thus the bolts must supply or N. The force in each of 6 bolts is /6 or Fbolt N Ans..9 A 4-m-diameter water tank consists of two half-cylinders, each weighing 4.5 kn/m, bolted together as in Fig. P.9. If the end caps are neglected, compute the force in each bolt. Solution: Consider a 5-cm width of upper cylinder, as seen below. The water pressure in the bolt plane is p h (9790)(4) 9160 Pa 1 Fig. P.9

61 Chapter Pressure Distribution in a Fluid 11 Then summation of vertical forces on this 5-cm-wide freebody gives Fz 0 p1a1 Wwater Wtank Fbolt (9160)(4 0.5) (9790)( /)() (0.5) (4500)/4 F, bolt Solve for F 1100 N Ans. one bolt.9 In Fig. P.9 a one-quadrant spherical shell of radius R is submerged in liquid of specific weight and depth h R. Derive an analytic expression for the hydrodynamic force F on the shell and its line of action. Solution: The two horizontal components are identical in magnitude and equal to the force on the quarter-circle side panels, whose centroids are (4R/) above the bottom: Fig. P.9 4R Horizontal components: Fx Fy hcgavert h R 4 Similarly, the vertical component is the weight of the fluid above the spherical surface: 1 4 R Fz Wcylinder Wsphere R h R R h There is no need to find the (complicated) centers of pressure for these three components, for we know that the resultant on a spherical surface must pass through the center. Thus 1/ 1/ F Fx Fy F z R (h R/) (h 4R/ ) Ans. 4

62 1 Solutions Manual Fluid Mechanics, Fifth Edition P.94 Find an analytic formula for the vertical and horizontal forces on each of the semi-circular panels AB in Fig. P.94. The width into the paper is b. Which force is larger? Why? h d A d/ h A + + d B Fig. P.94 B Solution: It looks deceiving, since the bulging panel on the right has more water nearby, but these two forces are the same, except for their direction. The left-side figure is the same as Example.9, and its vertical force is up. The right-side figure has the same vertical force, but it is down. Both vertical forces equal the weight of water inside, or displaced by, the half-cylinder AB. Their horizontal forces equal the force on the projected plane AB. d FH pcg, AB Aprojected [ g ( h )]( b d) Ans. d FV ghalf cylinder g [ ( ) b] Ans.

63 Chapter Pressure Distribution in a Fluid 1.95 The uniform body A in the figure has width b into the paper and is in static equilibrium when pivoted about hinge O. What is the specific gravity of this body when (a) h 0; and (b) h R? Solution: The water causes a horizontal and a vertical force on the body, as shown: R R FH Rb at above O, 4R FV R b at to the left of O 4 These must balance the moment of the body weight W about O: R b R R b 4R s Rb4R R MO srhb s h Solve for: SGbody Ans. R 1 For h 0, SG / Ans. (a). For h R, SG /5 Ans. (b).

64 14 Solutions Manual Fluid Mechanics, Fifth Edition.96 Curved panel BC is a 60 arc, perpendicular to the bottom at C. If the panel is 4 m wide into the paper, estimate the resultant hydrostatic force of the water on the panel. Solution: The horizontal force is, F h A H CG h (9790 N/m )[ 0.5(sin 60 ) m] [(sin60 )m(4 m)] 5,650 N Fig. P.96 The vertical component equals the weight of water above the gate, which is the sum of the rectangular piece above BC, and the curvy triangular piece of water just above arc BC see figure at right. (The curvytriangle calculation is messy and is not shown here.) V above BC F (Vol) (9790 N/m )[( m )(4 m)] 161,860 N The resultant force is thus, R 1/ F [(5,650) (161,860) ] 7,65 N 7 kn Ans. This resultant force acts along a line which passes through point O at 1 tan (161,860/5,650) 5.7

65 Chapter Pressure Distribution in a Fluid 15 P.97 The contractor ran out of gunite mixture and finished the deep corner, of a 5-m-wide swimming pool, with a quarter-circle piece of PVC pipe, labeled AB in m A water Fig. P.97 Fig. P.97. Compute the (a) horizontal and (b) vertical water forces on the curved panel AB. 1 m B Solution: For water take = 9790 N/m. (a) The horizontal force relates to the vertical projection of the curved panel AB: N FH, AB hcg Aprojected (9790 )(.5 m)[(1 m)(5 m)] 1,000 N Ans.( a) m (b) The vertical force is the weight of water above panel AB: N FV (9790 )[( m)(1 m) (1 m) ](5 m) 16,000 N Ans.( b) m 4

66 16 Solutions Manual Fluid Mechanics, Seventh Edition.98 Gate ABC in Fig. P.98 is a quarter circle 8 ft wide into the paper. Compute the horizontal and vertical hydrostatic forces on the gate and the line of action of the resultant force. Solution: The horizontal force is F h A (6.4)(.88)( ) h CG h 7987 lbf located at y cp (1/1)(8)(5.657) 0.94 ft (.88)( ) Area ABC ( /4)(4) (4 sin 45 ) ft The resultant is found to be R v ABC Fig. P.98 Thus F Vol (6.4)(8)(4.566) 80 lbf 1/ F [(7987) (80) ] 800 lbf acting at 15.9 through the center O. Ans. P.99 The mega-magnum cylinder in Fig. P.99 has a hemispherical bottom and is pressurized with air to 75 kpa (gage). Determine (a) the horizontal and (b) the vertical hydrostatic forces on the hemisphere, in lbf. Air Water 0 ft Solution: Since the problem asks for BG units, convert the air pressure to BG: 75,000 Pa = 1566 lbf/ft. 1 ft Fig. P.99

67 Chapter Pressure Distribution in a Fluid 17 (a) By symmetry, the net horizontal force on the hemisphere is zero. Ans.(a) (b) The vertical force is the sum of the air pressure term plus the weight of the water above: F p A V air surface water water lbf lbf 1 4 (1566 ) (6 ft) (6.4 )[ (6 ft) (0 ft) ( )(6 ft) ] ft ft 177, 000lbf 11, 000 lbf 90,000 lbf Ans.( b).100 Pressurized water fills the tank in Fig. P.100. Compute the hydrostatic force on the conical surface ABC. Solution: The gage pressure is equivalent to a fictitious water level h p/ / m above the gage or 8. m above AC. Then the vertical force on the cone equals the weight of fictitious water above ABC: Fig. P.100 F V Vol above 1 (9790) () (8.) () (4) ,000 N Ans.

68 18 Solutions Manual Fluid Mechanics, Fifth Edition P.101 The closed layered box in Fig. P cm Air has square horizontal cross-sections everywhere. All fluids are at 0C. Estimate the 80 cm SAE 0W oil gage pressure of the air if (a) the 90 cm Water 0 cm A hydrostatic force on panel AB is 48 kn; C 160 cm B or if (b) the hydrostatic force on the Fig. P.101 bottom panel BC is 97 kn. Solution: At 0C, take oil = 891 kg/m and water = 998 kg/m. The wedding-cake shape of the box has nothing to do with the problem. (a) the force on panel AB equals the pressure at the panel centroid (45 cm down from A) times the panel area: F AB p CG 48000N A AB [ p ( p air air oil gh oil water (891)(9.81)(0.8m) (998)(9.81)(0.45m)][(0.9m)(1.6m)] ( pair Pa)(1.44m ) ; Solve pair 000Pa Ans.( a) (b) The force on the bottom is handled similarly, except we go all the way to the bottom: F BC p BC 97000N A AB [ p ( p air air oil gh oil water ( pair Pa)(.56m ) ; Solve pair 000Pa Ans.( b) gh watercg ), ), or : or : (891)(9.81)(0.8m) (998)(9.81)(0.9m)][(1.6 m)(1.6m)] gh water

69 Chapter Pressure Distribution in a Fluid A cubical tank is m and is layered with 1 meter of fluid of specific gravity 1.0, 1 meter of fluid with SG 0.9, and 1 meter of fluid with SG 0.8. Neglect atmospheric pressure. Find (a) the hydrostatic force on the bottom; and (b) the force on a side panel. Solution: (a) The force on the bottom is the bottom pressure times the bottom area: F bot bot bot p A (9790 N/m )[(08 1 m) (09 1 m) (10 1 m)]( m) 8,000 N Ans. (a) (b) The hydrostatic force on the side panel is the sum of the forces due to each layer: F side CG side h A ( N/m )(0.5 m)( m ) ( N/m )(1.5 m)( m ) (9790 N/m )(.5 m)( m ) 15,000 kn Ans. (b).10 A solid block, of specific gravity 0.9, floats such that 75% of its volume is in water and 5% of its volume is in fluid X, which is layered above the water. What is the specific gravity of fluid X? Solution: The block is sketched below. A force balance is W = B, or 0.9 (HbL) (0.75HbL) SG (0.5HbL) SG, SG X 0.6 Ans. X X

70 140 Solutions Manual Fluid Mechanics, Fifth Edition.104 The can in Fig. P.104 floats in the position shown. What is its weight in newtons? Solution: The can weight simply equals the weight of the displaced water (neglecting the air above): Fig. P.104 W displaced (9790) (0.09 m) (0.08 m) 5.0 N Ans Archimedes, when asked by King Hiero if the new crown was pure gold (SG 19.), found the crown weight in air to be 11.8 N and in water to be 10.9 N. Was it gold? Solution: The buoyancy is the difference between air weight and underwater weight: B W W N air water water crown But also W (SG), so W B(SG 1) air water crown in water Solve for SG 1 W /B 110.9/ (not pure gold) Ans. crown in water.106 A spherical helium balloon is.5 m in diameter and has a total mass of 6.7 kg. When released into the U. S. Standard Atmosphere, at what altitude will it settle?

71 Chapter Pressure Distribution in a Fluid 141 Solution: The altitude can be determined by calculating the air density to provide the proper buoyancy and then using Table A. to find the altitude associated with this density: air balloon sphere m /Vol (6.7 kg)/[ (.5 m )/6] kg/m From Table A., atmospheric air has kg/m at an altitude of about 4000 m. Ans..107 Repeat Prob..6 assuming that the 10,000 lbf weight is aluminum (SG.71) and is hanging submerged in the water. Solution: Refer back to Prob..6 for details. The only difference is that the force applied to gate AB by the weight is less due to buoyancy: (SG 1).71 F 1 net body (10000) 610 lbf SG.71 This force replaces in the gate moment relation (see Prob..6): h h MB 0 610(15) (88.h ) csc 60 csc (7.5cos60 ) 6 or: h 7680/ , or: h 8.8 ft Ans..108 A 7-cm-diameter solid aluminum ball (SG =.7) and a solid brass ball (SG = 8.5) balance nicely when submerged in a liquid, as in Fig. P.108. (a) If the fluid is water at 0C, what is the diameter of the brass ball? (b) If the aluminum D = 7 cm pulleys + + brass brass ball has a diameter of.8 cm, what is the density of the fluid? Fig. P.108

72 14 Solutions Manual Fluid Mechanics, Fifth Edition Solution: For water, take = 9790 N/m. If they balance, net weights are equal: ( SGalum SG fluid ) water Dalum ( SGbrass SG fluid ) water Dbrass We can cancel water and (/6). (a) For water, SGfluid = 1, and we obtain 6 6 (.7 1)(0.07m) (8.5 1) Dbrass ; Solve Dbrass 0.047m (b) For this part, the fluid density (or specific gravity) is unknown: Ans.( a) (.7 SG fluid )(0.07m) Thus (8.5 SG )(0.08m) 1.595(998) According to Table A, this fluid is probably carbon tetrachloride. fluid fluid ; Solve 159kg/m SG fluid Ans.( b) 1.595

73 Chapter Pressure Distribution in a Fluid The float level h of a hydrometer is a measure of the specific gravity of the liquid. For stem diameter D and total weight W, if h 0 represents SG 1.0, derive a formula for h as a function of W, D, SG, and o for water. Solution: Let submerged volume be o when SG 1. Let A D /4 be the area of the stem. Then oo o o o Fig. P.109 W (SG) ( Ah), or: h= W(SG 1) Ans. SG ( D /4) P.110 A solid sphere, of diameter 18 cm, floats in 0C water with 1,57 cubic centimeters exposed above the surface. (a) What are the weight and specific gravity of this sphere? (b) Will it float in 0C gasoline? If so, how many cubic centimeters will be exposed? Solution: The total volume of the sphere is (/6)(18 cm) = 054 cm. Subtract the exposed portion to find the submerged volume = = 157 cm. Therefore the sphere is floating exactly half in and half out of the water. (a) Its weight and specific gravity are kg m Wsphere water gsubmerged (998 )(9.81 )(157 E 6 m ) N Ans.( a) m s Wsphere kg 499 sphere 499, SG.( ) (9.81)(054 6) sphere 0.50 Ans a g E m 1000 sphere (b) From Table A., gasoline = 680 kg/m > sphere. Therefore it floats in gasoline. Ans.(b) (c) Neglecting air buoyancy on the exposed part, we compute the fraction of sphere volume that is exposed to be ( kg/m )/(680 kg/m ) = 0.66 or 6.6%. The volume exposed is exp osed 0.66sphere 0.66(054 cm ) 81 cm Ans.( c) Check buoyancy: the submerged volume, 41 cm, times gasoline specific weight = N.

74 .111 A hot-air balloon must support its own weight plus a person for a total weight of 100 N. The balloon material has a mass of 60 g/m. Ambient air is at 5C and 1 atm. The hot air inside the balloon is at 70C and 1 atm. What diameter spherical balloon will just support the weight? Neglect the size of the hot-air inlet vent. Solution: The buoyancy is due to the difference between hot and cold air density: cold p kg kg ; 1.00 RT (87)(7 5) m 87(7 70) m cold hot The buoyant force must balance the known payload of 100 N: W 100 N g Vol ( )(9.81) D, 6 Solve for D 168 or D 11.8 m Ans. balloon Check to make sure the balloon material is not excessively heavy: W(balloon) (0.06 kg/m )(9.81 m/s )( )(11.8 m) 56 N OK, only 0% of W. total

75 Chapter Pressure Distribution in a Fluid The uniform 5-m-long wooden rod in the figure is tied to the bottom by a string. Determine (a) the string tension; and (b) the specific gravity of the wood. Is it also possible to determine the inclination angle? Fig. P.11 Solution: The rod weight acts at the middle,.5 m from point C, while the buoyancy is m from C. Summing moments about C gives M 0 W(.5sin ) B(.0sin ), or W 0.8B C But B (9790)( /4)(0.08 m) (4 m) N. Thus W 0.8B N SG(9790)( /4)(0.08) (5 m), or: SG 0.64 Ans. (b) Summation of vertical forces yields String tension T B W N Ans. (a) These results are independent of the angle, which cancels out of the moment balance.

76 146 Solutions Manual Fluid Mechanics, Fifth Edition.11 A spar buoy is a rod weighted to float vertically, as in Fig. P.11. Let the buoy be maple wood (SG 0.6), in by in by 10 ft, floating in seawater (SG 1.05). How many pounds of steel (SG 7.85) should be added at the bottom so that h 18 in? Fig. P.11 Solution: The relevant volumes needed are Wsteel Spar volume (10) 0.78 ft ; Steel volume (6.4) Immersed spar volume (8.5) 0.6 ft 1 1 The vertical force balance is: buoyancy B Wwood Wsteel, Wsteel or: 1.05(6.4) (6.4)(0.78) W 7.85(6.4) or: W W, solve for W 5.4 lbf Ans. steel steel steel steel.114 The uniform rod in the figure is hinged at B and in static equilibrium when kg of lead (SG 11.4) are attached at its end. What is the specific gravity of the rod material? What is peculiar about the rest angle 0? Solution: First compute buoyancies: Brod 9790(/4)(0.04) (8) 98.4 N, and Wlead (9.81) 19.6 N, Blead 19.6/ N. Sum moments about B: M 0 ( SG 1)(98.4)(4 cos0 ) ( )(8cos0 ) 0 B Solve for SGrod 0.66 Ans. (a) The angle drops out! The rod is neutrally stable for any tilt angle! Ans. (b)

77 Chapter Pressure Distribution in a Fluid The inch by inch by 1 ft spar buoy from Fig. P.11 has 5 lbm of steel attached and has gone aground on a rock. If the rock exerts no moments on the spar, compute the angle of inclination. Solution: Let be the submerged length of spar. The relevant forces are: Wwood (0.6)(64.0) (1) 1.8 lbf at distance 6 sin to the right of A 1 1 Buoyancy (64.0) at distance sin to the right of A 1 1 The steel force acts right through A. Take moments about A: M A 0 1.8(6 sin) sin Solve for 86.4, or 9.95 ft ( submerged length) Thus the angle of inclination cos 1 (8.0/9.95) 0.6 Ans. P.116 The deep submersible vehicle ALVIN, in the chapter-opener photo, has a titanium (SG = 4.50) spherical passenger compartment with an inside diameter of in and a wall thickness of 1.9 in. (a) Would the empty sphere float in seawater? (b) Would it float if it contained 1000 lbm of people and equipment inside? (c) What wall thickness would cause the empty sphere to be neutrally buoyant? Solution: First convert the data to metric: Di = inches = 1.98 m, Do = (1.9) = inches =.081 m, titanium = 4.50(1000) = 4500 kg/m, seawater = 105 kg/m. (a) Compute the weight of the sphere and compare to the buoyancy.

78 148 Solutions Manual Fluid Mechanics, Fifth Edition kg m Weight (4500 )(9.81 ) [(.081 m) (1.98 m) ] 8, 084 N m s 6 kg m Buoyancy (105 )(9.81 ) (.081 m) 47, 466 N m s 6 You could also add in the weight of the air inside the sphere, but that is only 49 N. We see that the buoyancy exceeds the weight by 19 kn, or more than tons. The sphere would float nicely. Ans.(a) (b) If we add 1000 lbm of people and equipment, that would weigh 1000 lbf and be another (1000)(4.448) = 4448 N of weight, making the total sphere weight = 581 N. The excess buoyancy is = N, it will still float nicely. Ans. (b) NOTE: The entire ALVIN, not just the sphere, floats, which is a good safety feature. (c) For neutral buoyancy, equate W (empty) and B and solve for the outside diameter: W (4500)(9.81) [ Do (1.98) ] 49 N (105)(9.81) Do 6 6 Solve for D.1615 m, thickness ( ) / m.51in Ans.( c) o Slightly less than doubling the wall thickness would create neutral buoyancy for the sphere.

79 Chapter Pressure Distribution in a Fluid The balloon in the figure is filled with helium and pressurized to 15 kpa and 0C. The balloon material has a mass of 85 g/m. Estimate (a) the tension in the mooring line, and (b) the height in the standard atmosphere to which the balloon will rise if the mooring line is cut. Fig. P.117 Solution: (a) For helium, from Table A-4, R 077 m /s /K, hence its weight is Whelium Hegballoon (9.81) (10) 119 N 077(9) 6 Meanwhile, the total weight of the balloon material is W balloon kg m [ (10 m) ] 6 N m s Finally, the balloon buoyancy is the weight of displaced air: Bair airgballoon (9.81) (10) 6108 N 87(9) 6 The difference between these is the tension in the mooring line: T B W W N Ans. (a) line air helium balloon (b) If released, and the balloon remains at 15 kpa and 0C, equilibrium occurs when the balloon air buoyancy exactly equals the total weight of N: kg Bair 1401 N air (9.81) (10), or : air m From Table A-6, this standard density occurs at approximately Z 1,800 m Ans. (b) P.118 An intrepid treasure-salvage group has discovered a steel box, containing gold doubloons and other valuables, resting in 80 ft of seawater. They estimate the weight of the box and treasure (in air) at 7000 lbf. Their plan is to attach the box to a sturdy balloon, inflated with air to atm pressure. The empty balloon weighs 50 lbf. The box is ft wide, 5 ft long, and 18 in high. What is the proper diameter of the balloon to ensure an upward lift force on the box that is 0% more than required?

80 150 Solutions Manual Fluid Mechanics, Fifth Edition Solution: The specific weight of seawater is approximately 64 lbf/ft. The box volume is (ft)(5ft)(1.5ft) = 1 ft, hence the buoyant force on the box is (64)(1) = 768 lbf. Thus the balloon must develop a net upward force of 1.( lbf) = 7478 lbf. The air weight in the balloon is negligible, but we can compute it anyway. The air density is: p (116 lbf / ft ) slug At p atm, air RT (1716 ft / s R)(50 R) ft Hence the air specific weight is (0.0071)(.) = 0. lbf/ft, much less than the water. Accounting for balloon weight, the desired net buoyant force on the balloon is F net (640.lbf / ft Solve for D )( / 6) D 1.4 lbf balloon, 50lbf 7478lbf D balloon 6.14ft Ans..119 With a 5-lbf-weight placed at one end, the uniform wooden beam in the figure floats at an angle with its upper right corner at the surface. Determine (a) ; (b) wood. Fig. P.119 Solution: The total wood volume is (4/1) (9) 1 ft. The exposed distance h 9tan. The vertical forces are Fz 0 (6.4)(1.0) (6.4)(h/)(9)(4/1) (SG)(6.4)(1.0) 5 lbf The moments of these forces about point C at the right corner are: M 0 (1)(4.5) (1.5h)(6 ft) (SG)( )(1)(4.5 ft) (5 lbf)(0 ft) C where 6.4 lbf/ft is the specific weight of water. Clean these two equations up: 1.5h 1 SG 5/ (forces).0h 1 SG (moments) Solve simultaneously for SG 0.68 Ans. (b); h 0.16 ft; 1.0 Ans. (a)

81 Chapter Pressure Distribution in a Fluid A uniform wooden beam (SG 0.65) is 10 cm by 10 cm by m and hinged at A. At what angle will the beam float in 0C water? Solution: The total beam volume is (.1) 0.0 m, and therefore its weight is W (0.65)(9790)(0.0) N, acting at the centroid, 1.5 m down from point A. Meanwhile, if the submerged length is H, the buoyancy is B (9790)(0.1) H 97.9H newtons, acting at H/ from the lower end. Sum moments about point A: A Fig. P.10 M 0 (97.9H)(.0 H/)cos190.9(1.5cos ), or: H( H/).95, solve for H 1.5 m Geometry: H m is out of the water, or: sin 1.0/1.775, or 4. Ans..11 The uniform beam in the figure is of size L by h by b, with b,h << L. A uniform heavy sphere tied to the left corner causes the beam to float exactly on its diagonal. Show that this condition requires (a) b /; and (b) D [Lhb/{(SG 1)}] 1/. Solution: The beam weight W blhb and acts in the center, at L/ from the left corner, while the buoyancy, being a perfect triangle of displaced water, equals B Lhb/ and acts at L/ from the left corner. Sum moments about the left corner, point C: Fig. P.11 M 0 ( Lhb)(L/) ( Lhb/)(L/), or: / Ans. (a) C b b

82 15 Solutions Manual Fluid Mechanics, Fifth Edition Then summing vertical forces gives the required string tension T on the left corner: F 0 Lbh/ Lbh T, or T Lbh/6 since / z b b 1/ Lhb sphere But also T (W B) (SG 1) D, so that D Ans. (b) 6 π(sg 1).1 A uniform block of steel (SG 7.85) will float at a mercury-water interface as in the figure. What is the ratio of the distances a and b for this condition? Solution: Let w be the block width into the paper and let be the water specific weight. Then the vertical force balance on the block is Fig. P (a b)lw 1.0aLw 1.56 blw, a or: 7.85a 7.85b a 1.56b, solve for 0.84 Ans. b P.1 A barge has the trapezoidal shape shown in Fig. P.1 and is m long into the paper. If the total weight of barge and 60 H? 60.5 m cargo is 50 tons, what is the draft H of the barge when floating in seawater? 8 m Fig. P.1 Solution: For seawater, let = 105 kg/m. The top of the barge has length [8m+(.5)tan60] = = m. Thus the total volume of the barge is [( m)/](.5m)(m) = m. In terms of seawater, this total volume would be equivalent to (519.4m )(105kg/m )(9.81m/s ) = 5.E6N 4.448lbf/N 000lbf/ton =

83 Chapter Pressure Distribution in a Fluid tons. Thus a cargo of 50 tons = 700,000 lbf would fill the barge a bit more than halfway. Thus we solve the following equation for the draft to give W = 50 tons: H kg m 1 (m)( H )(8 m)(105 )(9.81 )( tan 60 m s 4.448lbf Solve by iteration or EES : H ) 700,000lbf / N 1.58m Ans..14 A balloon weighing.5 lbf is 6 ft in diameter. If filled with hydrogen at 18 psia and 60F and released, at what U.S. standard altitude will it be neutral? Solution: Assume that it remains at 18 psia and 60F. For hydrogen, from Table A-4, R 4650 ft /(s R). The density of the hydrogen in the balloon is thus p 18(144) H slug/ft RT (4650)(460 60) In the vertical force balance for neutral buoyancy, only the outside air density is unknown: Fz Bair WH W balloon air (.) (6) (0.0000)(.) (6).5 lbf 6 6 air Solve for slug/ft kg/m From Table A-6, this density occurs at a standard altitude of 6850 m 500 ft. Ans. P.15 A solid sphere, of diameter 0 cm, has a specific gravity of 0.7. (a) Will this sphere float in 0C SAE 10W oil? If so, (b) how many cubic centimeters are exposed, and (c) how high will a spherical cap protrude above the surface? NOTE: If your knowledge of offbeat sphere formulas is lacking, you can Ask Dr. Math at Drexel University, EES is recommended for the solution.

84 154 Solutions Manual Fluid Mechanics, Fifth Edition Solution: From Table A., the density of SAE 10W oil is 870 kg/m > 700 kg/m. The sphere floats in oil. Ans.(a) The volume of the sphere is (/6)(0 cm) = 4,189 cm. The fraction exposed is ( )/870 = or 19.5 % The volume exposed is (0.195)(4189) = 818 cm. Ans.(b) h R - h r R spherical cap From Dr. Math, or a good math book, the sphere formulas relate to the distances (r, h, R) shown in the figure at right. h r R 10 cm ; cap ( r h ) h 818 cm h 6 Knowing that h is small, of order 5 cm, you could guess your way to the answer. Or you could use EES and get the answer directly. In either case, the result is h = 5.67 cm Ans.(c) The sphere would poke out of the water to a height of 5.67 centimeters..16 A block of wood (SG 0.6) floats in fluid X in Fig. P.16 such that 75% of its volume is submerged in fluid X. Estimate the gage pressure of the air in the tank. Solution: In order to apply the hydrostatic relation for the air pressure calculation, the density of Fluid X must be found. The buoyancy principle is thus first applied. Let the block have volume V. Neglect the buoyancy of the air on the upper part of the block. Then Fig. P V (0.75V) air (0.5V) ; N /m water X X water

85 Chapter Pressure Distribution in a Fluid 155 The air gage pressure may then be calculated by jumping from the left interface into fluid X: 0 Pa-gage (78 N/m )(0.4 m) p 10 Pa-gage 10 Pa-vacuum Ans. air.17* Consider a cylinder of specific gravity S 1 floating vertically in water (S 1), as in Fig. P.17. Derive a formula for the stable values of D/L as a function of S and apply it to the case D/L 1.. Solution: A vertical force balance provides a relation for h as a function of S and L, D h/4 SD L/4, thus h SL Fig. P.17 To compute stability, we turn Eq. (.5), centroid G, metacenter M, center of buoyancy B: 4 ( D/) D MB I o/vsub MG GB and substituting h SL, MG GB 16 SL Dh where GB L/ h/ L/ SL/ L(1 S)/. For neutral stability, MG 0. Substituting, D L D 0 (1 S) solving for D/ L, S S Ans. 16SL L 8 (1 ) If D/L 1., S S , or 0 S 0.5 and S 1 for stability Ans.

86 156 Solutions Manual Fluid Mechanics, Fifth Edition.18 The iceberg of Fig..0 can be idealized as a cube of side length L as shown. If seawater is denoted as S 1, the iceberg has S Is it stable? Solution: The distance h is determined by w w hl S L, or: h SL Fig. P.18 The center of gravity is at L/ above the bottom, and B is at h/ above the bottom. The metacenter position is determined by Eq. (.5): 4 L /1 L L MB I o/ sub MG GB Lh 1h 1S Noting that GB L/ h/ L(1 S)/, we may solve for the metacentric height: L L 1 MG (1 S) 0 if S S 0, or: S 0.11 or S 6 Instability: 0.11 S Since the iceberg has S , it is stable. Ans..19 The iceberg of Prob..18 may become unstable if its width decreases. Suppose that the height is L and the depth into the paper is L but the width decreases to H L. Again with S 0.88 for the iceberg, determine the ratio H/L for which the iceberg becomes unstable. Solution: As in Prob..18, the submerged distance h SL 0.88L, with G at L/ above the bottom and B at h/ above the bottom. From Eq. (.5), the distance MB is sub Io LH /1 H L SL MB MG GB MG HL(SL) 1SL

87 Chapter Pressure Distribution in a Fluid 157 Then neutral stability occurs when MG 0, or H L H 1/ 1/ (1 S), or [6S(1 S)] [6(0.88)(1 0.88)] Ans. 1SL L.10 Consider a wooden cylinder (SG 0.6) 1 m in diameter and 0.8 m long. Would this cylinder be stable if placed to float with its axis vertical in oil (SG 0.85)? Solution: A vertical force balance gives 0.85R h 0.6R (0.8 m), or: h m The point B is at h/ 0.8 m above the bottom. Use Eq. (.5) to predict the metacenter location: o sub 4 MB I / [ (0.5) /4]/[ (0.5) (0.565)] m MG GB Now GB 0.4 m 0.8 m m, hence MG m. This float position is thus slightly unstable. The cylinder would turn over. Ans..11 A barge is 15 ft wide and floats with a draft of 4 ft. It is piled so high with gravel that its center of gravity is ft above the waterline, as shown. Is it stable? Solution: Example.10 applies to this case, with L 7.5 ft and H 4 ft: L H (7.5 ft) 4 ft MA.69 ft, where A is the waterline H (4 ft) Since G is ft above the waterline, MG ft, unstable. Ans.

88 158 Solutions Manual Fluid Mechanics, Fifth Edition.1 A solid right circular cone has SG 0.99 and floats vertically as shown. Is this a stable position? Solution: Let r be the radius at the surface and let z be the exposed height. Then Fig. P.1 z r Fz 0 w (R h r z) 0.99 w R h, with. h R z 1/ Thus (0.01) h The cone floats at a draft h z h. The centroid G is at 0.5h above the bottom. The center of buoyancy B is at the centroid of a frustrum of a (submerged) cone: h R Rr r 4 R Rr r Then Eq. (.5) predicts the position of the metacenter: sub 0.441h above the bottom 4 I o (0.154R) /4 R MB MG GB 0.99 R h h MG (0.5h 0.441h) MG h Thus MG 0 (stability) if (R/h) 10.9 or R/h.1 Ans..1 Consider a uniform right circular cone of specific gravity S 1, floating with its vertex down in water, S 1.0. The base radius is R and the cone height is H, as shown. Calculate and plot the stability parameter MG of this cone, in dimensionless form, versus H/R for a range of cone specific gravities S 1.

89 Solution: The cone floats at height h and radius r such that B W, or: h r r h(1.0) R H( S), or: S 1 H R Thus r/r h/h S 1/ for short. Now use the stability relation: 4 H h Io r R 4 4 sub rh/ 4H /4 MG GB MG 1/ Non-dimensionalize in the final form: MG = R 1+, S Ans. H 4 H This is plotted below. Floating cones pointing down are stable unless slender, R << H.

90 160 Solutions Manual Fluid Mechanics, Fifth Edition.14 When floating in water (SG 1), an equilateral triangular body (SG 0.9) might take two positions, as shown at right. Which position is more stable? Assume large body width into the paper. Fig. P.14 Solution: The calculations are similar to the floating cone of Prob..1. Let the triangle be L by L by L. List the basic results. (a) Floating with point up: Centroid G is 0.89L above the bottom line, center of buoyancy B is 0.45L above the bottom, hence GB ( )L 0.044L. Equation (.5) gives MB I / L MG GB MG 0.044L o sub Hence MG 0.07L Unstable Ans. (a) (b) Floating with point down: Centroid G is 0.577L above the bottom point, center of buoyancy B is 0.548L above the bottom point, hence GB ( )L 0.096L. Equation (.5) gives MB I / 0.186L MG GB MG 0.096L o sub Hence MG 0.15L Stable Ans. (b).15 Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG, floating in water (SG 1) with its axis vertical. Show that the body is stable if R/L [SG(1 SG)] Solution: For a given SG, the body floats with a draft equal to (SG)L, as shown. Its center of gravity G is at L/ above the bottom. Its center of buoyancy B is at (SG)L/ above the bottom. Then Eq. (.5) predicts the metacenter location: 1/ 4 R /4 R L L MB I o/ sub MG GB MG SG R (SG)L 4(SG)L Thus MG 0 (stability) if R /L > SG(1 SG) Ans. For example, if SG 0.8, stability requires that R/L

91 Chapter Pressure Distribution in a Fluid Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG 0.5, floating in water (SG 1) with its axis horizontal. Show that the body is stable if L/R.0. Solution: For the given SG 0.5, the body floats centrally with a draft equal to R, as shown. Its center of gravity G is exactly at the surface. Its center of buoyancy B is at the centroid of the immersed semicircle: 4R/() below the surface. Equation (.5) predicts the metacenter location: (1/1)(R)L L 4R MB I o/ sub MG GB MG (R /)L R L 4R or: MG 0 (stability) if Ans. R L/R >.17 A tank of water 4 m deep receives a constant upward acceleration az. Determine (a) the gage pressure at the tank bottom if az 5 m /s; and (b) the value of az which causes the gage pressure at the tank bottom to be 1 atm. Solution: Equation (.5) states that p (g a) (kg kaz) for this case. Then, for part (a), z p (g a ) S (998 kg/m )( m /s)(4 m) Pa (gage) Ans. (a) For part (b), we know p 1 atm but we don t know the acceleration: m p (g a ) S (998)(9.81 a )(4.0) Pa if a = 15.6 Ans. (b) s z z z

92 16 Solutions Manual Fluid Mechanics, Fifth Edition.18 A 1 fluid ounce glass, inches in diameter, sits on the edge of a merry-go-round 8 ft in diameter, rotating at 1 r/min. How full can the glass be before it spills? Solution: First, how high is the container? Well, 1 fluid oz in, hence 1 fl. oz in (1.5 in) h, or h.06 in It is a fat, nearly square little glass. Second, determine the acceleration toward the center of the merry-go-round, noting that the angular velocity is (1 rev/min)(1 min/60 s)( rad/rev) 1.6 rad/s. Then, for r 4 ft, x a r (1.6 rad/s) (4 ft) 6. ft/s Then, for steady rotation, the water surface in the glass will slope at the angle a 6. x tan 0.196, or: h left to center (0.196)(1.5 in) 0.94 in g a z. 0 Thus the glass should be filled to no more than inches This amount of liquid is (1.5 in) (.77 in) 19.6 in 10.8 fluid oz. Ans..19 The tank of liquid in the figure P.19 accelerates to the right with the fluid in rigid-body motion. (a) Compute ax in m/s. (b) Why doesn t the solution to part (a) depend upon fluid density? (c) Compute gage pressure at point A if the fluid is glycerin at 0C. Fig. P.19 Solution: (a) The slope of the liquid gives us the acceleration: a x 8 15 cm tan 0.1, or: 7.4 g 100 cm thus a 0.1g 0.1(9.81) 1.8 m/s Ans. (a) x (b) Clearly, the solution to (a) is purely geometric and does not involve fluid density. Ans. (b)

93 Chapter Pressure Distribution in a Fluid 16 (c) From Table A- for glycerin, 160 kg/m. There are many ways to compute pa. For example, we can go straight down on the left side, using only gravity: p gz (160 kg/m )(9.81 m/s )(0.8 m) 460 Pa (gage) Ans. (c) A Or we can start on the right side, go down 15 cm with g and across 100 cm with ax: p gz a x (160)(9.81)(0.15) (160)(1.8)(1.00) A x Pa Ans. (c).140 An elliptical-end fuel tank is 10 m long, with -m horizontal and -m vertical major axes, and filled completely with fuel oil ( 890 kg/m ). Let the tank be pulled along a horizontal road in rigid-body motion. Find the acceleration and direction for which (a) a constant-pressure surface extends from the top of the front end to the bottom of the back end; and (b) the top of the back end is at a pressure 0.5 atm lower than the top of the front end. Solution: (a) We are given that the isobar or constant-pressure line reaches from point C to point B in the figure above, is negative, hence the tank is decelerating. The elliptical shape is immaterial, only the -m height. The isobar slope gives the acceleration: m ax tanc B 0., hence ax 0.(9.81) 1.96 m/s Ans. (a) 10 m g (b) We are now given that pa (back end top) is lower than pb (front end top) see the figure above. Thus, again, the isobar must slope upward through B but not necessarily pass through point C. The pressure difference along line AB gives the correct deceleration: kg pab 0.5( 1015 Pa) oilaxxab 890 ax (10 m) m solve for a 5.69 m/s Ans. (b) x

94 164 Solutions Manual Fluid Mechanics, Fifth Edition This is more than part (a), so the isobar angle must be steeper: 5.69 tan 0.580, hence isobar The isobar in part (a), line CB, has the angle (a) tan 1 (0.) The same tank from Prob..19 is now accelerating while rolling up a 0 inclined plane, as shown. Assuming rigidbody motion, compute (a) the acceleration a, (b) whether the acceleration is up or down, and (c) the pressure at point A if the fluid is mercury at 0C. Fig. P.141 Solution: The free surface is tilted at the angle This angle must satisfy Eq. (.55): tan tan(.59 ) a /(g a ) But the 0 incline constrains the acceleration such that ax 0.866a, az 0.5a. Thus 0.866a m tan 0.416, solve for a.80 (down) Ans. (a, b) a s The cartesian components are ax.9 m/s and az 1.90 m/s. (c) The distance S normal from the surface down to point A is (8 cos) cm. Thus 1/ 1/ A x z p [a (g a ) ] (1550)[(.9) ( ) ] (0.8 cos 7.41 ) 00 Pa (gage) x Ans. (c) z.14 The tank of water in Fig. P.14 is 1 cm wide into the paper. If the tank is accelerated to the right in rigid-body motion at 6 m/s, compute (a) the water depth at AB, and (b) the water force on panel AB.

95 Chapter Pressure Distribution in a Fluid 165 Fig. P.14 Solution: From Eq. (.55), 6.0 tan a x/g 0.61, or Then surface point B on the left rises an additional z = 1 tan 7.4 cm, or: water depth AB cm Ans. (a) The water pressure on AB varies linearly due to gravity only, thus the water force is 0.16 FAB pcga AB (9790) m (0.16 m)(0.1 m) 15.7 N Ans. (b)

96 166 Solutions Manual Fluid Mechanics, Fifth Edition.14 The tank of water in Fig. P.14 is full and open to the atmosphere (patm 15 psi 160 psf) at point A, as shown. For what acceleration ax, in ft/s, will the pressure at point B in the figure be (a) atmospheric; and (b) zero absolute (neglecting cavitation)? Solution: (a) For pa pb, the imaginary free surface isobar should join points A and B: Fig. P.14 tan tan a / g, hence a g. ft/s Ans. (a) AB x x (b) For pb 0, the free-surface isobar must tilt even more than 45, so that p 0 p gz a x (.)() 1.94 a (), B A x x solve a 589 ft/s Ans. (b) x This is a very high acceleration (18 g s) and a very steep angle, tan 1 (589/.) Consider a hollow cube of side length cm, full of water at 0C, and open to patm 1 atm at top corner A. The top surface is horizontal. Determine the rigidbody accelerations for which the water at opposite top corner B will cavitate, for (a) horizontal, and (b) vertical motion. Solution: From Table A-5 the vapor pressure of the water is 7 Pa. (a) Thus cavitation occurs first when accelerating horizontally along the diagonal AB: p p a L ( 998) a ( 0. ), A B x, AB AB x, AB solve ax, AB 19 m/s Ans. (a) If we moved along the y axis shown in the figure, we would need ay m/s. (b) For vertical acceleration, nothing would happen, both points A and B would continue to be atmospheric, although the pressure at deeper points would change. Ans.

97 Chapter Pressure Distribution in a Fluid A fish tank 16-in by 7-in by 14-inch deep is carried in a car which may experience accelerations as high as 6 m/s. Assuming rigid-body motion, estimate the maximum water depth to avoid spilling. Which is the best way to align the tank? Solution: The best way is to align the 16-inch width with the car s direction of motion, to minimize the vertical surface change z. From Eq. (.55) the free surface angle will be tanmax a x/g 0.61, thus z tan 4.9 inches ( 1.5 ) 9.81 Thus the tank should contain no more than inches of water. Ans..146 The tank in Fig. P.146 is filled with water and has a vent hole at point A. It is 1 m wide into the paper. Inside is a 10-cm balloon filled with helium at 10 kpa. If the tank accelerates to the right at 5 m/s/s, at what angle will the balloon lean? Will it lean to the left or to the right? Solution: The acceleration sets up pressure isobars which slant down and to the right, in both the water and in the helium. This means there will be a buoyancy force on the balloon up and to the right, as shown at right. It must be balanced by a string tension down and to the left. If we neglect balloon material weight, the balloon leans up and to the right at angle Fig. P.146 1ax tan tan Ans. g measured from the vertical. This acceleration-buoyancy effect may seem counter-intuitive.

98 168 Solutions Manual Fluid Mechanics, Fifth Edition.147 The tank of water in Fig. P.147 accelerates uniformly by rolling without friction down the 0 inclined plane. What is the angle of the free surface? Can you explain this interesting result? Solution: If frictionless, F W sin ma along the incline and thus a g sin 0 0.5g. Fig. P.147 a 0.5g cos0 x Thus tan ; solve for 0! Ans. g az g 0.5g sin 0 The free surface aligns itself exactly parallel with the 0 incline. P.148 A child is holding a string onto which is attached a helium-filled balloon. (a) The child is standing still and suddenly accelerates forward. In a frame of reference moving with the child, which way will the balloon tilt, forward or backward? Explain. (b) The child is now sitting in a car that is stopped at a red light. The helium-filled balloon is not in contact with any part of the car (seats, ceiling, etc.) but is held in place by the string, which is held by the child. All the windows in the car are closed. When the traffic light turns green, the car accelerates forward. In a frame of reference moving with the car and child, which way will the balloon tilt, forward or backward? Explain. (c) Purchase or borrow a helium-filled balloon. Conduct a scientific experiment to see if your predictions in parts (a) and (b) are correct. If not, explain. Solution: (a) Only the child and balloon accelerate, not the surrounding air. This is not rigid-body fluid motion. The balloon will tilt backward due to air drag. Ans.(a) (b) Inside the car, the trapped air will accelerate with the car and the child, etc. This is rigid-body motion. The balloon will tilt forward, as in Prob. P.146. Ans.(b) (c) A student in the writer s class actually tried this experimentally. Our predictions were correct.

99 Chapter Pressure Distribution in a Fluid The waterwheel in Fig. P.149 lifts water with 1-ft-diameter half-cylinder blades. The wheel rotates at 10 r/min. What is the water surface angle at pt. A? Solution: Convert 10 r/min 1.05 rad/s. Use an average radius R 6.5 ft. Then x x Fig. P.149 a R (1.05) (6.5) 7.1 ft/s toward the center Thus tan a /g 7.1 /., or: 1.5 Ans..150 A cheap accelerometer can be made from the U-tube at right. If L 18 cm and D 5 mm, what will h be if ax 6 m/s? Solution: We assume that the diameter is so small, D L, that the free surface is a point. Then Eq. (.55) applies, and 6.0 tan a x/g 0.61, or Fig. P.150 Then h (L/) tan (9 cm)(0.61) 5.5 cm Ans. Since h (9 cm)ax/g, the scale readings are indeed linear in ax, but I don t recommend it as an actual accelerometer, there are too many inaccuracies and disadvantages..151 The U-tube in Fig. P.151 is open at A and closed at D. What uniform acceleration ax will cause the pressure at point C to be atmospheric? The fluid is water. Solution: If pressures at A and C are the same, the free surface must join these points: 45, a gtan g. ft/s Ans. x Fig. P.151