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Unformatted text preview: Calculate, using the numerical finite difference procedure for one dimensional conduction, the steady state temperature in a long flat plate of 1m thickness (and also 1m width) with a uniform energy generation of 4 x 104W/m3 The surface at x = 0 is at a constant temperature of 120C, and the surface at x = L = 1m loses heat by convection to air at 40C with a heat transfer coefficient h = 100W/m2K. The plate thermal conductivity is k = 100W/mk.. There is no heat transfer in the y or z directions. Take 5 points (nodes), T0, T1, T3, and T4 with T0 = 120C = 393K. Solve numerically for T1, T2, T3 and T4. T0 q, k L x T, h 0 1 x=L/4 2 3 4 Doing an energy balance around node m=1, 2, or 3 qL q m x E E + g= s i -o n u E E t t & qL - qR + qV = Est = 0 for steady state (Tm - Tm-1 ) (Tm +1 - Tm ) - - - +& kAc kAc q ( dxAc ) = 0 x x Tm+1 + Tm-1 - 2Tm = & q 2 ( x ) k qR Doing an energy balance around node m=4 qcond q x/2 E E + g= s i -o n u E E t t & qcond - qconv + qV = Est = 0 for steady state T dx & - - = kAc ( hAc (Ts - T ) ) + q Ac 0 x 2 k x & (T4 - T3 ) + h(T4 - T ) - q 0 = x 2 The resulting equations are: T2 + T0 - 2T1 = - T3 + T1 - 2T2 = - T4 + T2 - 2T3 = - T4 - T3 + & q 2 ( x ) k & q 2 ( x ) k & q 2 ( x ) k 4 qconv & hx q 2 (T4 - T ) = ( x ) k 2k & 2 q - k ( x ) - T0 & - q ( x ) 2 k & - q ( x ) 2 k & q h x 2 ( x ) + T 2 k k A matrix can be used to solve: - 0 2 1 0 T 1 -2 1 1 0 T2 = 0 1 -2 1 T 3 hx T 0 0 -1 + 1 4 k T1 ~ 172.5 C, T2 ~ 200 C, T3 ~ 202.5 C, T4 ~ 180 C ...
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This note was uploaded on 08/01/2008 for the course ME 109 taught by Professor Greif during the Spring '08 term at University of California, Berkeley.
- Spring '08
- Heat Transfer