An imposed displacement of 01 m acts upon the coupled

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Unformatted text preview: 5 0.1 ( 4)(1.0 × 10 ) ( VX)average = 0.1 1 ∫ (1.6 × 105 )2 0 2.0 (1 − (1.6 × 105 )r 2 ) r dr = 1m / sec so Re = Sc = (1.0)(1.0)(0.005) = 500 1× 10−5 1.0 × 10−5 1.43 × 10 −5 ≅ 0. 7 So, we can neglect the second term on the RHS. Substituting the dimensionless velocity profile in the species transport equation we obtain: ∂ 2θi ∂r +2 + 1 ∂θi ∂θ = (1 − r +2 ) ∂x + r + ∂r + And boundary conditions become: θi = 1at x + = 0 θi = 0 at r + = 1 ∂θi ∂r 1–476 + = 0 at r + = 0 ANSYS Verification Manual . ANSYS Release 9.0 . 002114 . © SAS IP, Inc. VM209 The final solution takes the form: ∞ 2 θi ( x + , r + ) = ∑ Cn Rn (r + ) exp( −λn x + ) n=0 where the λn are the eigenvalues, the Rn are the corresponding eigenfunctions, and Cn are constants. Next, averaged mass fractions (Θi)m can be evaluated as: (θi )m = ∞G ( Yi )w − ( Yi )m n exp( −λ 2 x + ) =8 ∑ n 2 ( Yi )w − ( Yi ) in n = 0 λn or in dimensional form: x D AB 2 exp −λn 2 n = 0 λn 2R2 ( VX)average ∞ Gn...
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