p0219 - there must be 1 additional dimensionless grouping...

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128 CHAPTER 2. DIMENSIONAL ANALYSIS 2.19 Chapter 2, Problem 19 The dimensional quantities and their dimensions are [ u τ ] = L T , [ ρ ] = M L 3 , [ τ w ] = F L 2 = MLT 2 L 2 = M LT 2 There are 3 dimensional quantities and 3 independent dimensions ( M, L, T ) , so that the number of dimensionless groupings is 0. The appropriate dimensional equation is [ u τ ] = [ τ w ] a 1 [ ρ ] a 2 Substituting the dimensions for each quantity yields LT 1 = M a 1 L a 1 T 2 a 1 M a 2 L 3 a 2 = M a 1 + a 2 L a 1 3 a 2 T 2 a 1 Thus, equating exponents, we arrive at the following three equations: 0 = a 1 + a 2 1 = a 1 3 a 2 1 = 2 a 1 If we add the second and third equations, we find 0 = 3( a 1 + a 2 ) which is 3 times the first equation. Thus, the equations are linearly dependent so that
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Unformatted text preview: there must be 1 additional dimensionless grouping. In the present case, this means there is one such grouping. We can solve the first and third equations immediately for a 1 and a 2 , i.e., a 1 = 1 2 and a 2 = − 1 2 Substituting the values of a 1 and a 2 into the dimensional equation, we have [ u τ ] = [ τ w ] 1 / 2 [ ρ ] − 1 / 2 Therefore, we expect u τ to vary according to u τ = constant · ± τ w ρ Selecting a value of one for the constant, we conclude that u τ = ± τ w ρ...
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