p0258 - 2.58. CHAPTER 2, PROBLEM 58 185 2.58 Chapter 2,...

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2.58. CHAPTER 2, PROBLEM 58 185 2.58 Chapter 2, Problem 58 The dimensional quantities and their dimensions are [ σ ]= F L = ML/T 2 L = M T 2 [ D 1 L, [ D 2 L, [ ρ M L 3 , [ U L T , [ μ M LT There are 6 dimensional quantities and 3 independent dimensions ( M,L,T ) , so that the number of dimensionless groupings is 3. Using the indicial method, the appropriate dimensional equation is [ D 2 ]=[ D 1 ] a 1 [ ρ ] a 2 [ U ] a 3 [ μ ] a 4 [ σ ] a 5 Substituting the dimensions for each quantity yields L = L a 1 M a 2 L 3 a 2 L a 3 T a 3 M a 4 L a 4 T a 4 M a 5 T 2 a 5 = M a 2 + a 4 + a 5 L a 1 3 a 2 + a 3 a 4 T a 3 a 4 2 a 5 Thus, equating exponents, we arrive at the following three equations: 0= a 2 + a 4 + a 5 1= a 1 3 a 2 + a 3 a 4 a 3 a 4 2 a 5 We can solve immediately for a 2 and a 3 from the first and third equations, viz., a 2 = a 4 a 5 and a 3 = a 4 2 a 5 Substituting into the second equation yields a 1 =1+3 a 2 a 3 + a 4 =1+3( a 4 a 5 ) ( a 4 2 a 5 )+ a 4 =1 a 4 a 5 Substituting back into the dimensional equation, we have [ D 2 D 1 ] 1 a 4 a 5 [ ρ ] a 4 a 5 [ U ] a 4 2 a 5 [ μ ] a 4 [ σ ] a 5 =[ D 1 ] ^ μ ρ U D 1 ± a 4 ^ σ ρ U 2 D 1 ± a 5 Therefore, the dimensionless groupings are
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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p0258 - 2.58. CHAPTER 2, PROBLEM 58 185 2.58 Chapter 2,...

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