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p0212

# p0212 - ω 2 = ω 2 ^ d dx X σν t d ω dx ~± = 1 L σ L...

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2.12. CHAPTER 2, PROBLEM 12 141 2.12 Chapter 2, Problem 12 Problem: The following empirical differential equation determines a quantity ω known as the specific dissipation rate in a one-dimensional, steady flow. u d ω dx = α X du dx ~ 2 βω 2 + d dx X σν t d ω dx ~ The quantity u denotes velocity, x is spatial distance and ν t is kinematic viscosity. Also, β and σ are dimensionless coefficients. Determine the dimensions of ω and of α required for this equation to be dimensionally homogeneous. Solution: We begin with the following equation. u d ω dx = α X du dx ~ 2 βω 2 + d dx X σν t d ω dx ~ First, we determine the dimensions of each term. ^ u d ω dx = L T [ ω ] L = [ ω ] T α X du dx ~ 2 = [ α ] L 2 T 2 L 2 = [ α ] T 2 ± βω 2 = = [ β
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Unformatted text preview: ][ ω ] 2 = [ ω ] 2 ^ d dx X σν t d ω dx ~± = 1 L [ σ ] L 2 T [ ω ] L = [ ω ] T Hence, three of the terms in the equation have dimension [ ω ] /T and the other two terms have dimension [ α ] /T 2 and [ ω ] 2 . So, in order for this equation to be dimensionally homogeneous, two conditions must be satisfied. First, we must have [ ω ] 2 = [ ω ] T = ⇒ [ ω ] = 1 T Second, since [ ω ] /T = 1 /T 2 , necessarily [ α ] T 2 = 1 T 2 = ⇒ [ α ] = 1 Therefore, the quantity ω has dimensions 1 /T and the coefficient α is dimensionless....
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