Fluid Dynamics Sol ch5

# Fluid Dynamics Sol ch5 - Chapter 5 Dimensional Analysis and...

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Chapter 5 Dimensional Analysis and Similarity 5.1 For axial flow through a circular tube, the Reynolds number for transition to turbulence is approximately 2300 [see Eq. (6.2)], based upon the diameter and average velocity. If d = 5 cm and the fluid is kerosene at 20 ° C, find the volume flow rate in m 3 /h which causes transition. Solution: For kerosene at 20 ° C, take ρ = 804 kg/m 3 and µ = 0.00192 kg/m s. The only unknown in the transition Reynolds number is the fluid velocity: tr Vd (804)V(0.05) Re 2300 , solve for V 0.11m/s 0.00192 ≈== = 3 2 m Then Q VA (0.11) (0.05) 2.16E 4 3600 4s Ans. π == = × 3 m 0.78 hr 5.2 In flow past a thin flat body such as an airfoil, transition to turbulence occurs at about Re = 1E6, based on the distance x from the leading edge of the wing. If an airplane flies at 450 mi/h at 8-km standard altitude and undergoes transition at the 12 percent chord position, how long is its chord (wing length from leading to trailing edge)? Solution: From Table A-6 at z = 8000 m, 0.525 kg/m 3 , T 236 ° K, hence 1.53E 5 kg/m s. Convert U = 450 mi/hr 201 m/s. Then the transition Reynolds no. is 6 tr x,tr Ux (0.525)(201)(0.12C) Re 10 , solve for C 1.53E 5 Ans. = 1.21 m 5.3 An airplane has a chord length L = 1.2 m and flies at a Mach number of 0.7 in the standard atmosphere. If its Reynolds number, based on chord length, is 7E6, how high is it flying? Solution: This is harder than Prob. 5.2 above, for we have to search in the U.S. Stan- dard Atmosphere (Table A-6) to find the altitude with the right density and viscosity

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Chapter 5 Dimensional Analysis and Similarity 309 and speed of sound. We can make a first guess of T 230 K, a (kRT) 304 m/s, U = 0.7a 213 m/s, and µ 1.51E 5 kg/m s. Then our first estimate for density is 3 C UC (213)(1.2) Re 7E6 , or 0.44 kg/m and Z 9500 m (Table A-6) 1.51E 5 ρρ ρ == Repeat and the process converges to 0.41 kg/m 3 or Z 10100 m Ans . 5.4 When tested in water at 20 ° C flowing at 2 m/s, an 8-cm-diameter sphere has a measured drag of 5 N. What will be the velocity and drag force on a 1.5-m-diameter weather balloon moored in sea-level standard air under dynamically similar conditions? Solution: For water at 20 ° C take 998 kg/m 3 and 0.001 kg/m s. For sea-level standard air take 1.2255 kg/m 3 and 1.78E 5 kg/m s. The balloon velocity follows from dynamic similarity , which requires identical Reynolds numbers: balloon model model proto 1.2255V (1.5) VD 998(2.0)(0.08) Re 1.6E5 Re 0.001 1.78E 5 = = = | ρ or V balloon 1.55 m/s. Then the two spheres will have identical drag coefficients: balloon D,model D,proto 22 2 2 2 2 F F5 N C 0.196 C V D 998(2.0) (0.08) 1.2255(1.55) (1.5) = = = Solve for . Ans F 1.3 N balloon 5.5 An automobile has a characteristic length and area of 8 ft and 60 ft 2 , respectively. When tested in sea-level standard air, it has the following measured drag force versus speed: V , mi/h: 20 40 60 Drag, lbf: 31 115 249 The same car travels in Colorado at 65 mi/h at an altitude of 3500 m. Using dimensional analysis, estimate (a) its drag force and (b) the horsepower required to overcome air drag.
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## This note was uploaded on 05/07/2008 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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Fluid Dynamics Sol ch5 - Chapter 5 Dimensional Analysis and...

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