Chapter7_SM - Chapter 7 Flow Past Immersed Bodies P7.1 An...

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Chapter 7 Flow Past Immersed Bodies P7.1 An ideal gas, at 20 ° C and 1 atm, flows at 12 m/s past a thin flat plate. At a position 60 cm downstream of the leading edge, the boundary layer thickness is 5 mm. Which of the 13 gases in Table A.4 is this likely to be? Solution : We are looking for the kinematic viscosity. For a gas at low velocity and a short distance, we can guess laminar flow. Then we can begin by trying Eq. (7.1 a ): s m E for Solve m s m Vx m m x x / 5 0 . 2 ) 6 . 0 )( / 12 ( 0 . 5 / 0 . 5 Re 0 . 5 6 . 0 005 . 0 2 = = = = = ν δ The only gas in Table A.4 which matches this viscosity is the last one, CH 4 . Ans . But wait! Is it laminar? Check Re x = (12)(0.6)/(2.0E-5) = 360,000. Yes, OK. P7.2 Air, equivalent to a Standard Altitude of 4000 m, flows at 450 mi/h past a wing which has a thickness of 18 cm, a chord length of 1.5 m, and a wingspan of 12 m. What is the appropriate value of the Reynolds number for correlating the lift and drag of this wing? Explain your selection. Solution: Convert 450 mi/h = 201 m/s, at 4000 m, ρ = 0.819 kg/m s, T = 262 K, μ = 1.66E 5 kg/m s. The appropriate length is the chord , C = 1.5 m, and the best parameter to correlate with lift and drag is ReC = (0.819)(201)(1.5)/1.66E 5 = 1.5E7 Ans. P7.3 Equation (7.1 b ) assumes that the boundary layer on the plate is turbulent from the leading edge onward. Devise a scheme for determining the boundary-layer thickness more accurately when the flow is laminar up to a point Re x ,crit and turbulent thereafter. Apply this scheme to computation of the boundary-layer thickness at x = 1.5 m in 40 m/s
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Chapter 7 Flow Past Immersed Bodies 565 flow of air at 20 ° C and 1 atm past a flat plate. Compare your result with Eq. (7.1 b ). Assume Re x ,crit 1.2E6. Fig. P7.3 Solution: Given the transition point xcrit, Recrit, calculate the laminar boundary layer thick- ness δ c at that point, as shown above, c/xc 5.0/Recrit 1/2 . Then find the “apparent” distance upstream, Lc, which gives the same turbulent boundary layer thickness, c Then begin x 1/7 /L 0.16/Re . cc L effective at this “apparent origin” and calculate the remainder of the turbulent boundary layer as /xeff 0.16/Reeff 1/7 . Illustrate with a numerical example as requested. For air at 20 ° C, take ρ = 1.2 kg/m 3 and μ = 1.8E 5 kg/m s. c crit c c 1/2 1.2(40)x 5.0(0.45) Re 1.2E6 if x 0.45 m, then 0.00205 m 1.8E 5 (1.2E6) == = = 1/6 7/6 c c U 0.00205 1.2(40) Compute L 0.0731 m 0.16 0.16 1.8E 5 ⎛⎞ ⎜⎟ ⎢⎥ ⎝⎠ Finally, at x = 1.5 m, compute the effective distance and the effective Reynolds number: eff c c eff 1.2(40)(1.123) x x L x 1.5 0.0731 0.45 1.123 m, Re 2.995E6 =+ − = + = = eff 1.5 m eff 1.8E 5 0.16x 0.16(1.123) Re (2.995E6) Ans. ≈= | 0.0213 m Compare with a straight all-turbulent-flow calculation from Eq. (7.1 b ): x1 . 5 m 1.2(40)(1.5) 0.16(1.5) Re 4.0E6, whence (25% higher) . 1.8E 5 (4.0E6) Ans =≈ | 0.027 m P7.4 A smooth ceramic sphere (SG = 2.6) is immersed in a flow of water at 20 ° C and 25 cm/s. What is the sphere diameter if it is encountering (a) creeping motion, Re d = 1; or (b) transition to turbulence, Re d = 250,000?
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566 Solutions Manual Fluid Mechanics, Fifth Edition Solution: For water, take ρ = 998 kg/m 3 and μ = 0.001 kg/m s.
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This note was uploaded on 04/24/2011 for the course MECH 122 at San Jose State.

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Chapter7_SM - Chapter 7 Flow Past Immersed Bodies P7.1 An...

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