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# 42 chapter 2 problem 42 191 the first equation tells

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Unformatted text preview: a1 + a3 + a4 1 = −a1 + a2 − 2a4 0 = −a1 − a2 − 2a3 − 2a4 2.42. CHAPTER 2, PROBLEM 42 191 The first equation tells us that a1 = −a3 − a4 Then, eliminating a1 from the second equation, (a3 + a4 ) + a2 − 2a4 = 1 =⇒ a2 = 1 − a3 + a4 Finally, substituting for a1 and a2 in third equation, there follows (a3 + a4 ) − (1 − a3 + a4 ) − 2a3 − 2a4 = 0 =⇒ a4 = − Finally, for this value of a4 , we have 1 1 − a3 and a2 = − a3 2 2 Substituting back into the dimensional equation, we have a1 = [h] = [µ] 1 −a3 2 [U ] 1 −a3 2 a3 [σ ] [ρg ] −1 2 µU = ρg 1 2 σ µU a3 Therefore, the dimensionless groupings are h ρg µU and σ µU Using E. S. Taylor’s method, we have the following. M L h 0 1 µ 1 −1 U 0 1 σ 1 0 ρg 1 −2 T M L 0 h 0 1 −1 µ 1 −1 → −1 U 0 1 −2 σ /µ 0 1 −2 ρg/µ 0 −1 T LT 0 h 1 0 −1 →U 1 −1 −1 σ /µ 1 −1 −1 ρg/µ −1 −1 −1 LT T T h/U 0 1 h/U 1 h ρg/(µU ) 0 →U → 1 −1 → σ /(µU ) 0 σ /(µU ) 0 σ /(µU ) 0 0 ρgU/µ −2 ρgU/µ −2 ρgU/µ 0 −2 Therefore, as with the indicial method, the dimensionless groupings are h ρg µU and σ µU 1 2...
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