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as2-sol - ME 608 Numerical Methods in Heat Mass and...

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ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer Solution to Assignment No: 2 Due Date: February 7, 2011 Instructor: J. Murthy 1. No solution attached. 2. For thermally fully developed flow with flux boundary conditions, temperature increases linearly in the axial direction and the shape of the temperature profile becomes constant. Thus, the wall temperature T w increases linearly in x, as does the bulk temperature T b ( x ) . Also dT b dx = T x = constant If axial diffusion is neglected, it is possible to non-dimensionalize the energy equation using the dimensionless temper- ature θ = ( T T b ) / ( qD / k )) and demonstrate that the energy equation and its boundary conditions do not generate any dimensionless groups. Thus, we may choose any q, D, k, U, ρ and C p and obtain the same dimensionless θ profile and the Nusselt number. The dimensional temperature would of course depend on the choice of properties. (a) Consider a macroscopic control volume spanning the duct cross section and of with x in the axial (x) direction. The heat balance yields: dT b dx = ( 4 D ) q ρ UC p D 2 where U is the prescribed mean velocity (=U for plug flow). Thus we see that dT b / dx is a constant; so is T / x . (b) For the control volumes shown in Fig. 1, the discrete equations are: Point 1 a E = k z / y ; a W = k z / y ; a N = k y / z ; a S = k y / z ; a P = Σ nb a nb ; b = ρ UC P ( T / x ) V Point 2 a E = k z / y ; a W = k z / y ; a N = k y / z ; a S = 0; a P = a E + a W + a N ; b = ρ UC P ( T / x ) V + q y Point 3 a E = k z / y ; a W = 0; a N = k y / z ; a S = 0; a P = a E + a N ; b = ρ UC P ( T / x ) V + q y + q z Point 4 a E = k z / y ; a W = k z / y ; a N = 0; a S = k y / z ; a P = a E + a W + a S ; b = ρ UC P ( T / x ) V Point 5 a E = k z / y ; a W = 0; a N = 0; a S = k y / z ; a P = a E + a S ; b = ρ UC P ( T / x ) V + q z Point 6 a E = 0; a W = k z / y ; a N = 0; a S = k y / z ; a P = a W + a S ; b = ρ UC P ( T / x ) V (c) The dimensionless temperature on the centerline is shown in Fig. 2. The plot is for a 20 × 20 mesh in one-quarter of the duct. The corresponding contours of θ are shown in Fig. 3. (d) Table 1 shows the Nusselt number computed for two different mesh sizes. The 20 × 20 result is within 0.06% of the 40 × 40 result. (e) The problem has all Neumann boundary conditions, and the governing equation admits T and T+C as solutions.
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