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p0230

# p0230 - 2.30 CHAPTER 2 PROBLEM 30 143 2.30 Chapter 2...

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2.30. CHAPTER 2, PROBLEM 30 143 2.30 Chapter 2, Problem 30 Figure 2.4 shows the geometry of the liquid-air interface. c σ ρ λ . . . ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...... . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... . . . . . . . . . . . . . . .... . . . . . . . . . . ......................................... . . . . . . . . . . . Figure 2.4: Geometry of a ripple. The dimensional quantities and their dimensions are [ σ ]= F L = MLT 2 L = M T 2 [ ρ M L 3 , [ c L T , [ λ 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 [ c ]=[ ρ ] a 1 [ σ ] a 2 [ λ ] a 3 Substituting the dimensions for each quantity yields LT 1 = M a 1 L 3 a 1 M a 2 T 2 a 2 L a 3 = M a 1 + a 2 L 3 a 1 + a 3 T 2 a 2 Thus, equating exponents, we arrive at the following three equations: 0= a 1 + a 2 1= 3 a 1 + a 3 2 a 2 We can solve the first and third equations immediately for a 1 and a 2 ,v iz . , a 1 = 1 / 2 and a 2 =1 / 2 Substituting into the second equation, we find 3 / 2+ a 3 = a 3 = 1 / 2

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144 CHAPTER 2. DIMENSIONAL ANALYSIS Substituting back into the dimensional equation, we have
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