final-sol

# final-sol - ME 608 Numerical Methods in Heat Mass and...

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Unformatted text preview: ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer Final Examination Date: April 30, 2008 8:00-10:00 AM Instructor: J. Murthy Open Book, Open Notes Total: 50 points Use the finite volume method in all problems. 1. The flow domain is shown in Fig. 1. The flow is described by the following governing equations: ∂ u ∂ x = = − μ u κ − ∂ p ∂ x + ∂ ∂ x parenleftbigg μ ∂ u ∂ x parenrightbigg We are given μ = 10 − 3 kg / ( m − s ) , κ = 10 − 2 m 2 , L = . 3 m , u 1 = . 5 m / s and u 4 = . 5 m / s . Furthermore, we wish to start with an initial guess of p = Pa everywhere. We use under-relaxation factors of α u = 1 . 0 and α p = 1 . 0. B A C 1 2 3 4 L u 1 u 4 x Figure 1: Computational Domain for Problem 1 (a) Discrete momentum equation for cell 2: Integrate the governing equation over the staggered control volume around 2 to obtain: = − μ u 2 κ ∆ x +( p A − p B ) + parenleftbigg μ ∂ u ∂ x | e − μ ∂ u ∂ x | w parenrightbigg = − μ u 2 κ ∆ x +( p A − p B ) + parenleftbigg μ u 3 − u 2 ∆ x − μ u 2 − u 1 ∆ x parenrightbigg so that a P 2 = μ ∆ x κ + 2 μ ∆ x a E 2 = μ ∆ x a W 2 = b 2 = μ ∆ x u 1 1 and the discrete equation is given by: a P 2 u 2 = a E 2 u 3 + b 2 +( p A − p B ) For cell 3, similarly, = − μ u 3 κ ∆ x +( p B − p C ) + parenleftbigg μ ∂ u ∂ x | e − μ ∂ u ∂ x | w parenrightbigg = − μ u 3 κ ∆ x +( p B − p C ) + parenleftbigg μ u 4 − u 3 ∆ x − μ u 3 − u 2 ∆ x parenrightbigg so that a P 3 = μ ∆ x κ + 2 μ ∆ x a E 3 = a W 3 = μ ∆ x b 3 = μ ∆ x u 4 and the discrete equation is given by: a P 3 u 3 = a W 3 u 2 + b 3 +( p B − p C ) The numerical values of the coefficients for the two discrete equations are given below. a P 2 = . 03; a E 2 = . 01; a W 2 = . 0; b 2 = . 005 a P 3 = . 03; a E 3 = . 0; a W 3 = . 01; b 3 = . 005 (b) To develop the pressure-correction equations, we first find the corrections. Thus u 2 = u ∗ 2 + d 2 parenleftBig p ′ A − p ′ B parenrightBig d 2 = 1 a P 2 = 33 . 3333 u 3 = u ∗ 3 + d 3 parenleftBig p ′ B − p ′ C parenrightBig d 3 = 1 a P 3 = 33 . 3333 We note that, in deriving the correction equations above, we have ignored the corrections due to...
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final-sol - ME 608 Numerical Methods in Heat Mass and...

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