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# p0226 - 158 CHAPTER 2 DIMENSIONAL ANALYSIS 2.26 Chapter 2...

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158 CHAPTER 2. DIMENSIONAL ANALYSIS 2.26 Chapter 2, Problem 26 Problem: Once transients have settled out, the radial pressure gradient, dp/dr , in a rotating tank of liquid is a function of the density, ρ , angular velocity, , and radial distance from the center of the tank, r . Deduce a formula for dp/dr as a function of ρ , and r using dimensional analysis. Solution: The dimensional quantities and their dimensions are [ dp/dr ] = F L 2 · 1 L = MLT 2 L 3 = M L 2 T 2 [ ρ ] = M L 3 , [ ] = 1 T , [ r ] = L There are 4 dimensional quantities and 3 independent dimensions ( M, L, T ) , so that the number of dimensionless groupings is 1. The appropriate dimensional equation is [ dp/dr ] = [ ρ ] a 1 [ ] a 2 [ r ] a 3 Substituting the dimensions for each quantity yields ML 2 T 2 = M a 1 L 3 a 1 T a 2 L a 3 = M a 1 L 3 a 1 + a 3 T a 2 Thus, equating exponents, we arrive at the following three equations. 1 = a 1 2 = 3 a 1 + a 3 2 = a 2 We can solve the first and third equations immediately for a 1 and a 2 , viz., a 1 = 1 and a 2 = 2 Substituting into the second equation, we find

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