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p0246

# p0246 - viz a 1 = 1 a 2 = a 4 − 2 a 3 = a 4 − 2...

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166 CHAPTER 2. DIMENSIONAL ANALYSIS 2.46 Chapter 2, Problem 46 The dimensional quantities and their dimensions are [ F ] = ML T 2 , [ ρ ] = M L 3 , [ μ o ] = T L , [ σ ] = 1 L , [ U ] = L T , [ L ] = L where denotes ohms. There are 6 dimensional quantities and 4 independent dimensions ( M, L, T, ) , so that the number of dimensionless groupings is 2. The appropriate dimensional equation is [ F ] = [ ρ ] a 1 [ μ o ] a 2 [ σ ] a 3 [ U ] a 4 [ L ] a 5 Substituting the dimensions for each quantity yields MLT 2 = M a 1 L 3 a 1 a 2 T a 2 L a 2 a 3 L a 3 L a 4 T a 4 L a 5 = M a 1 L 3 a 1 a 2 a 3 + a 4 + a 5 T a 2 a 4 a 2 a 3 Thus, equating exponents, we arrive at the following four equations: 1 = a 1 1 = 3 a 1 a 2 a 3 + a 4 + a 5 2 = a 2 a 4 0 = a 2 a 3 We can solve immediately for a 1 , a 2 and a 3 from the first, third and fourth equations,
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Unformatted text preview: viz., a 1 = 1 , a 2 = a 4 − 2 , a 3 = a 4 − 2 Substituting these values into the second equation yields 1 = − 3 − ( a 4 − 2) − ( a 4 − 2) + a 4 + a 5 = ⇒ a 4 = a 5 Therefore, the solution for a 1 , etc. is a 1 = 1 , a 2 = a 5 − 2 , a 3 = a 5 − 2 , a 4 = a 5 Substituting back into the dimensional equation, we have [ F ] = [ ρ ][ μ o ] a 5 − 2 [ σ ] a 5 − 2 [ U ] a 5 [ L ] a 5 = ^ ρ μ 2 o σ 2 ± [ μ o σ UL ] a 5 Therefore, the dimensionless groupings are: Fμ 2 o σ 2 ρ and μ o σ UL...
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