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mech7220-hw2-solns

# mech7220-hw2-solns - 3 Laminar Internal Flow Exercises 1...

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3 Laminar Internal Flow Exercises 1. Consider laminar flow between two parallel plates. The flow is incompressible, has constant properties, and is fully developed. (a) Derive the velocity profile and the mean velocity. (b) Derive the Nusselt number for the constant surface heat flux case (both surfaces are heated) (c) Derive the Nusselt number for the case in which one surface is insulated, and the other is main- tained at a constant temperature. Your should perform at least two iterations for the temperature profile. Let x and y denote the flow and normal directions. The momentum equation will be u ∂u ∂x + v ∂u ∂y = - 1 ρ dP dx + ν 2 u ∂y 2 (1) The FDF assumption means that u = u ( y ), and the convection terms drop out. The velocity is zero at y = 0 and L , and the solution is u = - 1 2 ρν dP dx ( y L - y 2 ) (2) The mean velocity is u m = 1 L Z L 0 u dy = - L 2 12 ρν dP dx (3) and u ( y ) = u u m = 6 y (1 - y ) , y = y L (4) represents the dimensionless velocity profile. For constant q 00 s on both surfaces, ρ u m C ∂T ∂x = ρ u m C dT m dx = q 00 s P A = 2 h ( T s - T m ) L (5) and the energy equation becomes - 2 u Nu L = d 2 T d y 2 , T = T - T s T m - T s (6) with zero temperature at the surfaces. The solution is T ( y ) = Nu L ( y - 2 y 3 + y 4 ) The mean temperature condition is 1 = Z 1 0 u T d y = 17 Nu L 140 Nu L = 70 17 = 4 . 12 The hydraulic diameter is twice L , and the Nusselt number based on this length is 8.24 which agrees with the book.

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mech7220-hw2-solns - 3 Laminar Internal Flow Exercises 1...

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