384 Solutions Manual x Fluid Mechanics, Fifth Edition (b) If we swing the pendulum on the moon at the same 20q, we may use similarity: 1/2 §g · T1 ¨ 1 ¸ © L1 ¹ 1/2 § 9.81 m/s2 · (2.04 s) ¨ ¸ © 1.0 m ¹ or: T2 1/2 6.39 2.75 s § 1.62 m/s2 · T2 ¨ ¸ , 0.3 m © ¹ Ans. (b) 5.27 In studying sand transport by ocean waves, A. Shields in 1936 postulated that the bottom shear stress W required to move particles depends upon gravity g, particle size d and density Up, and water density U and viscosity P. Rewrite this in terms of dimensionless groups (which led to the Shields Diagram in 1936). Solution: There are six variables (W, g, d, Up, U, P) and three dimensions (M, L, T), hence we expect n j 6 3 3 Pi groups. The author used (U, g, d) as repeating variables: W U gd § U g1/2 d 3/2 Up · , ¸ fcn ¨ P U¹ © Ans. The shear parameter used by Shields himself was based on net weight: W /[(Up U)gd]. 5.28 A simply supported beam of diameter D, length L, and modulus of elasticity E is subjected to a fluid crossflow of velocity V, density U, and viscosity P. Its center deflection G is assumed to be a function of all these variables. (a) Rewrite this proposed function in dimensionless form. (b) Suppose it is known that G is independent of P, inversely proportional to E, and dependent only upon UV2, not U and V separately. Simplify the dimensionless function accordingly. Solution: Establish the variables and their dimensions: G {L} fcn( U , D , L , E , V , P ) {M/L3} {L} {L} {M/LT2} {L/T} {M/LT} Then n 7 and j 3, hence we expect n j and selected by myself, as follows (a): Well-posed final result: 73 G L 4 Pi groups, capable of various arrangements § L U VD E · fcn ¨ , , Ans. (a) 2 ¸ D P V U © ¹ (b) If P is unimportant, then the Reynolds number (UVD/P) drops out, and we have already cleverly combined E with UV2, which we can now slip out upside down: 385 Chapter 5 x Dimensional Analysis and Similarity2 If P drops out and G v GE U V2L or: 1 G , then E L §L· fcn ¨ ¸ ©D¹ UV §L· fcn ¨ ¸ , E ©D¹ Ans. (b) 5.29 When fluid in a pipe is accelerated linearly from rest, it begins as laminar flow and then undergoes transition to turbulence at a time ttr which depends upon the pipe diameter D, fluid acceleration a, density U, and viscosity P. Arrange this into a dimensionless relation between ttr and D. Solution: Establish the variables and their dimensions: ttr {T} U , D , fcn( a , P ) {M/L3} {L} {L/T2} {M/LT} Then n 5 and j 3, hence we expect n j 5 3 2 Pi groups, capable of various arrangements and selected by myself, as required, to isolate ttr versus D: § U a2 t tr ¨ ¨ P © 1/3 · ¸ ¸ ¹ ª § U 2 a ·1/3 º ¸ » « © P2 ¹ » ¬ ¼ fcn « D ¨ Ans. 5.30 The wall shear stress Ww for flow in a narrow annular gap between a fixed and a rotating cylinder is a function of density U, viscosity P, angular velocity :, outer radius R, and gap width 'r. Using (U, :, R) as repeating variables, rewrite this relation in dimensionless form. Solution: The relevant dimensions are {Ww} {ML1T2}, {U} {ML3}, {P} {ML1T1}, {:} {T1}, {R} {L}, and {'r} {L}. With n 6 and j 3, we expect n j k 3 pi groups. They are found, as specified, using (U, :, R) as repeating variables: U a : b R cW w 31 U : RP 3 U a : b R c 'r b c M ½ a 1 ½b c M ½ ® 3 ¾ ® ¾ {L} ® ¾ ¯ L ¿ ¯T ¿ ¯ LT ¿ 1 32 a M ½ a 1 ½b c M ½ ® 3 ¾ ® ¾ {L} ® 2 ¾ ¯ L ¿ ¯T ¿ ¯ LT ¿ 1, b M 0 L0T 0 , solve a 1 M 0 L0T 0 , solve a 1, b 1, c M ½ a 1 ½b c 0 0 0 ® 3 ¾ ® ¾ {L} {L} M L T , solve a ¯ L ¿ ¯T ¿ 0, b 0, c 1 The final dimensionless function has the form: 31 fcn(3 2 , 33 ), or: 2, c W wall U: 2 R 2 § U:R2 'r · fcn ¨ , ¸ R¹ © P Ans. 2 2

Download
# W U P