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Unformatted text preview: ME 309 Fall 2008 Section 2 (Merkle) Homework # 18 Due Fri 10 Oct 2008 Problem 181 . A sphere is submerged in a liquid and released. Depending upon the densities of the sphere and the liquid, the sphere either rises (and ultimately floats) or sinks to the bottom. Find the dimensionless parameters ( Π groups) associated with this problem. Solution: Identify: Important dimensional parameters: D, ρ l , ρ s , μ , u , g Analysis : There are 6 parameters ( n = 6), and 3 primary dimensions ( r = 3), M, L, t, So there will be m = n – r= 6 – 3 = 3 pigroups Primary dimensions associated with each parameter: D u , ρ l , ρ l , μ , g L Lt1 ML3 ML3 ML1 t 1 Lt2 Choose three base parameters for each pi group: ( D, u , ρ l ) Use viscosity for first pigroup: D a u ∞ b ρ l c μ = Π 1 Insert units to appropriate powers: L a ( Lt1 ) b ( ML3 ) c ( ML1 t1 ) = Π 1 Sum powers of all primary units : M: c + 1 = 0 L: a + b 3c  1 = 0 t: b 1 = 0 From time :  b – 1 = 0 or b =  1 From Mass: c + 1 = 0 or c =  1 And from L: a + b 3c  1 = a  1 + 3  1 = 0 or a = 1 The first pigroup is then: Π 1 = D1 V ∞1 ρ l1 μ which is again just the inverse of the Reynolds number Final result: D u l ∞ = Π ρ μ 1 For the second pigroup , choose the density of the sphere as the parameter: D a u ∞ b ρ l c ρ s = Π 2 Insert units to appropriate powers: L a ( Lt1 ) b ( ML3 ) c ( ML 3 ) = Π 2 Sum powers of all primary units : M: c + 1 = 0 L: a + b 3c  3 = 0 t: b = 0 From time : b = 0 From Mass: c + 1 = 0 or c =1 And from L: a + b 3c  3 = a + 0 + 3  3 = 0 or a = 0 The second pigroup is then: l s ρ ρ = Π 2 This pigroup could have been obtained by inspection....
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This note was uploaded on 03/09/2010 for the course ME 309 taught by Professor Merkle during the Spring '08 term at Purdue University.
 Spring '08
 MERKLE

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