fluid chap.2

# fluid chap.2 - Chapter 2 Pressure Distribution in a Fluid...

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Chapter 2 Pressure Distribution in a Fluid 2.1 For the two-dimensional stress field in Fig. P2.1, let xx yy 3000 psf 2000 psf σ σ = = xy 500 psf σ = Find the shear and normal stresses on plane AA cutting through at 30 ° . Solution: Make cut “AA” so that it just hits the bottom right corner of the element. This gives the freebody shown at right. Now sum forces normal and tangential to side AA. Denote side length AA as “L.” AA n,AA L F 0 (3000sin30 500cos30)Lsin30 (2000cos30 500sin30)Lcos30 σ = = + + Fig. P2.1 AA Solve for . (a) Ans σ 2 2683 lbf/ft t,AA AA F 0 L (3000cos30 500sin30)Lsin30 (500cos30 2000sin30)Lcos30 τ = = AA Solve for . (b) Ans τ 2 683 lbf/ft 2.2 For the stress field of Fig. P2.1, change the known data to σ xx = 2000 psf, σ yy = 3000 psf, and σ n (AA) = 2500 psf. Compute σ xy and the shear stress on plane AA. Solution: Sum forces normal to and tangential to AA in the element freebody above, with σ n (AA) known and σ xy unknown: n,AA xy xy F 2500L ( cos30 2000sin30 )Lsin30 ( sin30 3000cos30 )Lcos30 0 σ σ = ° + ° ° ° + ° ° = xy Solve for (2500 500 2250)/0.866 (a) Ans. σ = ≈ − 2 289 lbf/ft

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62 Solutions Manual Fluid Mechanics, Fifth Edition In like manner, solve for the shear stress on plane AA, using our result for σ xy : t,AA AA F L (2000cos30 289sin30 )Lsin30 (289cos30 3000sin30 )Lcos30 0 τ = °+ ° ° + ° + ° ° = AA Solve for 938 1515 (b) Ans. τ = ≈ − 2 577 lbf/ft This problem and Prob. 2.1 can also be solved using Mohr’s circle. 2.3 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/m 3 . The capillary rise in the tube, from Example 1.9 of the text, is 3 2Ycos 2(0.073 / )cos(0 ) 0.030 m (9790 / )(0.0005 ) cap N m h R N m m θ γ ° = = = Then the rise due to applied pressure is less by that amount: h press = 0.25 m 0.03 m = 0.22 m. The applied pressure is estimated to be p = γ h press = (9790 N/m 3 )(0.22 m) 2160 Pa Ans. 2.4 Given a flow pattern with isobars p o Bz + Cx 2 = constant. Find an expression x = fcn(z) for the family of lines everywhere parallel to the local pressure gradient p. Solution: Find the slope (dx/dz) of the isobars and take the negative inverse and integrate: 2 o p const gradient gradient d dx dx B 1 (p Bz Cx ) B 2Cx 0, or: dz dz dz 2Cx (dx/dz) dx 2Cx dx 2Cdz Thus , integrate , . dz B x B Ans = + = − + = = = = − = | | 2Cz/B x const e = 2.5 Atlanta, Georgia, has an average altitude of 1100 ft. On a U.S. standard day, pres- sure gage A reads 93 kPa and gage B reads 105 kPa. Express these readings in gage or vacuum pressure, whichever is appropriate.
Chapter 2 Pressure Distribution in a Fluid 63 Solution: We can find atmospheric pressure by either interpolating in Appendix Table A.6 or, more accurately, evaluate Eq. (2.27) at 1100 ft 335 m: g/RB 5.26 a o o Bz (0.0065 K/m)(335 m) p p 1 (101.35 kPa) 1 97.4 kPa T 288.16 K = = Therefore: Gage A 93 kPa 97.4 kPa 4.4 kPa (gage) Gage B 105 kPa 97.4 kPa . Ans = = − = = = + + 4.4 kPa (vacuum) 7.6 kPa (gage) 2.6 Express standard atmospheric pressure as a head, h = p/ ρ g, in (a) feet of ethylene glycol; (b) inches of mercury; (c) meters of water; and (d) mm of methanol.

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