This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: ME 608 Numerical Methods in Heat, Mass, and Momentum Transfer Mid-Term Examination Solutions Date: March 5, 2008 6:00 – 7:30 PM Instructor: J. Murthy Open Book, Open Notes Total: 50 points 1. We are given the following problem statement. The governing equation is: ∂ ∂ x ( ρ C p u ( y ) T ) = ∂ ∂ y parenleftbigg k ∂ T ∂ y parenrightbigg + S The velocity profile is given by: u ( y ) u m = 3 2 parenleftbigg 1- parenleftBig y D parenrightBig 2 parenrightbigg Here u m is the mean velocity at any x location. Constant properties are assumed. Furthermore, ρ = 1 kg / m 3 , C p = 1000 J / kgK , u m = . 1 m / s , k = . 5 W / mK , D = . 1 m and S = 10 3 W / m 3 . u(y) x y Insulated Insulated 2D 1 2 3 Figure 1: Computational Domain for Problem 1 (a) Determine analytically the gradient ∂ T ∂ x , explaining your logic clearly. Compute its numerical value. Considering a control volume across the cross section of the channel, with height 2D and width dx , we obtain the following overall heat balance: ( ρ u m C p T b D )+ SDdx = ( ρ u m C p T b D )+( ρ u m C p D ) dT b dx dT b dx = S ρ u m C p = 10 K / m 1 Just as for given-flux boundary conditions, the presence of a heat source means that for thermally fully-developed flow, all temperatures rise at the same rate, so that ∂ T ∂ x = ∂ T b ∂ x (b) For cell 1, the discrete energy balance yields- k s parenleftbigg ∂ T ∂ y parenrightbigg s + S ∆ y- (...
View Full Document
This note was uploaded on 12/29/2011 for the course ME 608 taught by Professor Na during the Fall '10 term at Purdue.
- Fall '10