Chapter 2. Pressure Distribution in a Fluid P2.1 For the two-dimensional stress field in Fig. P2.1, let Find the shear and normal stresses on plane AA cutting through at 30°. Solution: Make cut “AA” so that it just hits the bottom right corner of the element. This gives the freebody shown at right. Now sum forces normal and tangential to side AA.

1. Fluid Mechanics White 8th Solutions
2. Manual Fluid Mechanics White Solution
3. Manual Fluid Mechanics White 8th

View Solution Manual for Fluid Mechanics 8th Edition by White from MIE 312 at University of Toronto. Chapter2PressureDistributioninaFluid P2. Daugherty, R. Franzini, and E. Finnemore, Fluid Mechanics with. Dickinson, C., Pumping Manual, 8th ed. M., Fluid Mechanics, 6th ed. Allison Transmission (NYSE: ALSN) is the world's largest manufacturer of fully automatic transmissions for medium- and heavy-duty commercial vehicles and is.

Denote side length AA as “L.” Fig. P2.1 P2.2 For the stress field of Fig. P2.1, change the known data to σxx = 2000 psf, σyy = 3000 psf, and σn(AA) = 2500 psf. Compute σxy and the shear stress on plane AA.

Solution: Sum forces normal to and tangential to AA in the element freebody above, with σn(AA) known and σxy unknown: 2-2 Solutions Manual. Fluid Mechanics, Eighth Edition In like manner, solve for the shear stress on plane AA, using our result for σxy: This problem and Prob. P2.1 can also be solved using Mohr’s circle.

P2.3 A vertical clean glass piezometer tube has an inside diameter of 1 mm. When a pressure is applied, water at 20°C rises into the tube to a height of 25 cm. After correcting for surface tension, estimate the applied pressure in Pa. Solution: For water, let Y = 0.073 N/m, contact angle θ = 0°, and γ = 9790 N/m3. The capillary rise in the tube, from Example 1.9 of the text, is Then the rise due to applied pressure is less by that amount: hpress = 0.25 m − 0.03 m = 0.22 m. The applied pressure is estimated to be p = γhpress = (9790 N/m3)(0.22 m) ≈ 2160 Pa Ans. P2.4 Pressure gages, such as the Bourdon gage W θ?

P2.4, are calibrated with a deadweight piston. If the Bourdon gage is designed to rotate the pointer 10 degrees for every 2 psig of internal pressure, how many degrees does the pointer rotate if the piston and 2 cm diameter Oil Fig. P2.4 weight together total 44 newtons?

Solution: The deadweight, divided by the piston area, should equal the pressure applied to the Bourdon gage. Stay in SI units for the moment: Bourdon gage 2-4 Solutions Manual. Fluid Mechanics, Eighth Edition From Table A.3, methanol has ρ = 791 kg/m3 and a large vapor pressure of 13,400 Pa. Then the manometer rise h is given by P2.8 Suppose, which is possible, that there is a half-mile deep lake of pure ethanol on the surface of Mars. Estimate the absolute pressure, in Pa, at the bottom of this speculative lake.

Solution: We need some data from the Internet: Mars gravity is 3.71 m/s2, surface pressure is 700 Pa, and surface temperature is -10ºF (above the freezing temperature of ethanol). Then the bottom pressure is given by the hydrostatic formula, with ethanol density equal to 789 kg/m3 from Table A.3. Convert ½ mile = ½(5280) ft = 2640 ft.

0.3048 m/ft = 804.7 m. Then P2.9 A storage tank, 26 ft in diameter and 36 ft high, is filled with SAE 30W oil at 20°C. (a) What is the gage pressure, in lbf/in2, at the bottom of the tank? (b) How does your result in (a) change if the tank diameter is reduced to 15 ft? (c) Repeat (a) if leakage has caused a layer of 5 ft of water to rest at the bottom of the (full) tank.

Solution: This is a straightforward problem in hydrostatic pressure. From Table A.3, the density of SAE 30W oil is 891 kg/m3 ÷ 515.38 = 1.73 slug/ft3.

(a) Thus the bottom pressure is pbottom = ρoil g h = (1.73 slug ft 3 )(32.2 ft s2 )(36 ft) = 2005 lbf ft 2 = 13.9 lbf in2 gage Ans.(a) (b) The tank diameter has nothing to do with it, just the depth: pbottom = 13.9 psig. Ans.(b) (c) If we have 31 ft of oil and 5 ft of water (ρ = 1.94 slug/ft3), the bottom pressure is Chapter 2. Pressure Distribution in a Fluid 2-5 P2.10 A large open tank is open to sea level atmosphere and filled with liquid, at 20ºC, to a depth of 50 ft. The absolute pressure at the bottom of the tank is approximately 221.5 kPa. From Table A.3, what might this liquid be? Solution: Convert 50 ft to 15.24 m. Use the hydrostatic formula to calculate the bottom pressure: P2.11 In Fig.

P2.11, sensor A reads 1.5 kPa (gage). All fluids are at 20°C. Determine the elevations Z in meters of the liquid levels in the open piezometer tubes B and C. Solution: (B) Let piezometer tube B be an arbitrary distance H above the gasolineglycerin interface. The specific weights are γair ≈ 12.0 N/m3, γgasoline = 6670 N/m3, and γglycerin = 12360 N/m3. Then apply the hydrostatic formula from point A to point B: Fig.

P2.11 Solve for ZB = 2.73 m (23 cm above the gasoline-air interface) Ans. (b) Solution (C): Let piezometer tube C be an arbitrary distance Y above the bottom. Then 1500 + 12.0(2.0) + 6670(1.5) + 12360(1.0 − Y) − 12360(ZC − Y) = pC = 0 (gage) Solve for ZC = 1.93 m (93 cm above the gasoline-glycerin interface) Ans. (c) Chapter 2. Pressure Distribution in a Fluid P2.14 2-7 oil, SG= 0.78 For the three-liquid system shown, compute h1 and h2. Water mercury Neglect the air density. P2.14 27 cm h1 Solution: The pressures at h2 8 cm 5 cm the three top surfaces must all be atmospheric, or zero gage pressure.

Compute γoil = (0.78)(9790) = 7636 N/m3. Also, from Table 2.1, γwater = 9790 N/m3 and γmercury = 133100 N/m3. The surface pressure equality is P2.15 In Fig. P2.15 all fluids are at 20°C. Gage A reads 15 lbf/in2 absolute and gage B reads 1.25 lbf/in2 less than gage C. Com-pute (a) the specific weight of the oil; and (b) the actual reading of gage C in lbf/in2 absolute. P2.15 Solution: First evaluate γair = (pA/RT)g = 15 × 144/(1717 × 528)(32.2) ≈ 0.0767 lbf/ft3.

Take γwater = 62.4 lbf/ft3. Then apply the hydrostatic formula from point B to point C: Solutions Manual. Fluid Mechanics, Eighth Edition 2-8 With the oil weight known, we can now apply hydrostatics from point A to point C: P2.16 If the absolute pressure at the interface between water and mercury in Fig. P2.16 is 93 kPa, what, in lbf/ft2, is (a) the pressure at the Water surface, and (b) the pressure at the bottom of the container? P2.16 75° 28 cm 75° Mercury 8 cm 32 cm Solution: Do the whole problem in SI units and then convert to BG at the end. The bottom width and the slanted 75-degree walls are irrelevant red herrings. Just go up and down: 2-10 Solutions Manual.

Fluid Mechanics, Eighth Edition P2.19 The U-tube at right has a 1-cm ID and contains mercury as shown. If 20 cm3 of water is poured into the right-hand leg, what will be the free surface height in each leg after the sloshing has died down? Solution: First figure the height of water added: Then, at equilibrium, the new system must have 25.46 cm of water on the right, and a 30-cm length of mercury is somewhat displaced so that “L” is on the right, 0.1 m on the bottom, and “0.2 − L” on the left side, as shown at right. The bottom pressure is constant: Thus right-leg-height = 9.06 + 25.46 = 34.52 cm Ans. Left-leg-height = 20.0 − 9.06 = 10.94 cm Ans. P2.20 The hydraulic jack in Fig.

P2.20 is filled with oil at 56 lbf/ft3. Neglecting piston weights, what force F on the handle is required to support the 2000-lbf weight shown?

P2.20 Solution: First sum moments clockwise about the hinge A of the handle: 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: Chapter 2. Pressure Distribution in a Fluid 2-11 Therefore the handle force required is F = P/16 = 222/16 ≈ 14 lbf Ans. P2.21 In Fig.

P2.21 all fluids are at 20°C. Gage A reads 350 kPa absolute. Determine (a) the height h in cm; and (b) the reading of gage B in kPa absolute. Solution: Apply the hydrostatic formula from the air to gage A: Fig. P2.21 Then, with h known, we can evaluate the pressure at gage B: P2.22 The fuel gage for an auto gas tank reads proportional to the bottom gage pressure as in Fig. If the tank accidentally contains 2 cm of water plus gasoline, how many centimeters “h” of air remain when the gage reads “full” in error?

P2.22 Solution: Given γgasoline = 0.68(9790) = 6657 N/m3, compute the gage pressure when “full”: Set this pressure equal to 2 cm of water plus “Y” centimeters of gasoline: Therefore the air gap h = 30 cm − 2 cm(water) − 27.06 cm(gasoline) ≈ 0.94 cm Ans. Chapter 2. Pressure Distribution in a Fluid 2-13.P2.25 As measured by NASA’s Viking landers, the atmosphere of Mars, where g = 2 3.71 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 ≈ 1.3E-5 m-1 and To ≈ 250 K. For example, at 20,000 m altitude, T ≈ 193 K.

Fluid Mechanics White 8th Solutions

(a) Find an analytic formula for the variation of pressure with altitude. (b) Find the altitude where pressure on Mars has dropped to 1 pascal.

Solution: (a) The analytic formula is found by integrating Eq. (2.17) of the text: (b) From Table A.4 for CO2, R = 189 m2/(s2-K). Substitute p = 1 Pa to find the altitude: P2.26 For gases over large changes in height, the linear approximation, Eq. (2.14), is inaccurate. Expand the troposphere power-law, Eq.

(2.20), into a power series and show that the linear approximation p ≈ pa - ρa g z is adequate when Solution: The power-law term in Eq. (2.20) can be expanded into a series: Multiply by pa, as in Eq.

Manual Fluid Mechanics White Solution

(2.20), and note that panB/To = (pa/RTo)gz = ρa gz. Then the series may be rewritten as follows: 2-14 Solutions Manual. Fluid Mechanics, Eighth Edition For the linear law to be accurate, the 2nd term in parentheses must be much less than unity. If the starting point is not at z = 0, then replace z by δz: P2.27 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! P2.28 A correlation of computational fluid dynamics results indicates that, all other things being equal, the distance traveled by a well-hit baseball varies inversely as the 0.36 power of the air density. If a home-run ball hit in NY Mets Citi Field Stadium travels 400 ft, estimate the distance it would travel in (a) Quito, Ecuador, and (b) Colorado Springs, CO. Solution: Citi Field is in the Borough of Queens, NY, essentially at sea level. Hence the standard pressure is po ≈ 101,350 Pa.

Look up the altitude of the other two cities and calculate the pressure: (0.0065)(2850) 5.26 = (101350)(0.705) = 71,500 Pa 288.16 (0.0065)(1835) 5.26 (b)Colorado Springs: z ≈1835m, pCC = po 1− = (101350)(0.801) = 81,100 Pa 288.16 (a)Quito: z ≈ 2850 m, pQ = po 1− Then the estimated home-run distances are: 101350 0.36 ) = 400(1.134) ≈ 454 ft Ans.(a) 50 0.36 (b) Colorado Springs: X = 400( ) = 400(1.084) ≈ 433 ft 81100 (a) Quito: X = 400( Ans.(b) The Colorado result is often confirmed by people who attend Rockies baseball games. 2-16 Solutions Manual. Fluid Mechanics, Eighth Edition P2.31 In Fig.

P2.31 determine Δp between points A and B. All fluids are at 20°C. P2.31 Solution: Take the specific weights to be Benzene: 8640 N/m3 Kerosene: 7885 N/m3 Mercury: 133100 N/m3 Water: 9790 N/m3 and γair will be small, probably around 12 N/m3. Work your way around from A to B: P2.32 For the manometer of Fig.

Manual Fluid Mechanics White 8th

P2.32, all fluids are at 20°C. If pB − pA = 97 kPa, determine the height H in centimeters. Solution: Gamma = 9790 N/m3 for water and 133100 N/m3 for mercury and (0.827)(9790) = 8096 N/m3 for Meriam red oil. Work your way around from point A to point B: Fig. P2.32 Chapter 2. Pressure Distribution in a Fluid 2-17 P2.33 In Fig. P2.33 the pressure at point A is 25 psi.

All fluids are at 20°C. What is the air pressure in the closed chamber B?

Solution: Take γ = 9790 N/m3 for water, 8720 N/m3 for SAE 30 oil, and (1.45)(9790) = 14196 N/m3 for the third fluid. Convert the pressure at A from 25 lbf/in2 to 172400 Pa. Compute hydrostatically from point A to point B: Fig.

P2.33 P2.34 To show the effect of manometer dimensions, consider Fig. 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 Δh.

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