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 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 = 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 ² a 3 ^ μ ρ Ud ² a 5 Hence, noting that the reciprocal of μ/ ( ρ ) is the Reynolds number, the dimensionless
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This note was uploaded on 10/12/2009 for the course AME 309 at USC.

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

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