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Unformatted text preview: ME 309 Fall 2008 Section 2 (Merkle) Homework # 18 Due Fri 10 Oct 2008 Problem 18-1 . 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 pi-groups Primary dimensions associated with each parameter: D u , ρ l , ρ l , μ , g L Lt-1 ML-3 ML-3 ML-1 t -1 Lt-2 Choose three base parameters for each pi group: ( D, u , ρ l ) Use viscosity for first pi-group: D a u ∞ b ρ l c μ = Π 1 Insert units to appropriate powers: L a ( Lt-1 ) b ( ML-3 ) c ( ML-1 t-1 ) = 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 pi-group is then: Π 1 = D-1 V ∞-1 ρ l-1 μ which is again just the inverse of the Reynolds number Final result: D u l ∞ = Π ρ μ 1 For the second pi-group , 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 ( Lt-1 ) b ( ML-3 ) 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 pi-group is then: l s ρ ρ = Π 2 This pi-group 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