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# p0264 - 194 CHAPTER 2 DIMENSIONAL ANALYSIS 2.64 Chapter 2...

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194 CHAPTER 2. DIMENSIONAL ANALYSIS 2.64 Chapter 2, Problem 64 2.64(a): The dimensional quantities and their dimensions are [ F ] = ML T 2 , [ U ] = L T , [ d ] = L, [ ] = 1 T , [ ρ ] = M L 3 , [ μ ] = M LT There are 6 dimensional quantities and 3 independent dimensions ( M, L, T ) , so that the number of dimensionless groupings is 3. 2.64(b): The appropriate dimensional equation is [ F ] = [ U ] a 1 [ d ] a 2 [ ] a 3 [ ρ ] a 4 [ μ ] a 5 Substituting the dimensions for each quantity yields MLT 2 = L a 1 T a 1 L a 2 T a 3 M a 4 L 3 a 4 M a 5 L a 5 T a 5 = M a 4 + a 5 L a 1 + a 2 3 a 4 a 5 T a 1 a 3 a 5 Thus, equating exponents, we arrive at the following three equations: 1 = a 4 + a 5 1 = a 1 + a 2 3 a 4 a 5 2 = a 1 a 3 a 5 We can solve a 4 from the first equation, viz., a 4 = 1 a 5 and for a 1 from the third equation, a 1 = 2 a 3 a 5 Then, substituting for a 1 and a 4 in the second equation, there follows 2 a 3 a 5 + a 2 3(1 a 5 ) a 5 = 1 = a 2 = 2 + a 3 a 5 Substituting back into the dimensional equation, we have [ F ] = [ U ] 2 a 3 a 5 [ d ] 2+ a 3 a 5 [ ] a 3 [ ρ ] 1 a 5 [ μ ] a 5 = ± ρ U 2 d 2 = ^ d U a 3 ^ μ ρ Ud a 5 Hence, noting that the reciprocal of μ/ ( ρ Ud ) is the Reynolds number, the dimensionless

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p0264 - 194 CHAPTER 2 DIMENSIONAL ANALYSIS 2.64 Chapter 2...

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