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# assignchap5-95 - PROBLEM 5.95 KNOWN One-dimensional wall...

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Unformatted text preview: PROBLEM 5.95 KNOWN: One-dimensional wall suddenly subjected to uniform volumetric heating and convective surface conditions. FIND: Finite-difference equation for node at the surface, it = —L. SCHEMATIC: for Z' A 0,? 75ml»? T i Tum "L Lyx+L ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) Uniform q. ANALYSIS: There are two types of finite—difference equations for the explicit and implicit methods of solution. Using the energy balance approach, both types will be derived. Explicit Method. Perform an energy balance on the surface node shown above, . . . . ' T§+1_T(I)3 Ein ‘EoutJrEg :Est qconv+QCond +qV=p CVT (L) Tp—Tp Ax Ax TpH—Tp h 1.1 TweTp)+kI-1#+' 1-1-— = c1-1-— 0—1, 2 ( )i 0 ( ) Ax q 2 p 2 m ( ) For the explicit method, the temperatures on the LHS are evaluated at the previous time (p). The RHS provides a forward-difference approximation to the time derivative. Divide Eq. (2) by pch/2At and solve for T3 +1. hAt p ch km ,0 CAXZ- (Tf—Tglm—Aing. p c T§+1=2 (Tm—T§)+2 Introducing the Fourier and Biot numbers, F0 2 (k/p c) At/sz Bi 2 hAx/k 61sz 2k Tg+1:2Fo [TP+Bi-Too+ ]+(1—2Fo—2Fo-Bi)rg’. (3) The stability criterion requires that the coefficient of T}; be positive. That is, (1—2Fo—2Fo-Bi)20 or FoSl/2(1+Bi). (4)< Impiieit Method. Begin as above with an energy balance. In Eq. (2), however, the temperatures on the LHS are evaluated at the new (p+1) time. The RHS provides a backward— difference approximation to the time derivative. Tp-t-1_Tp+1 A p+1_ p l 1 o . X Ax To To h(T,_Tp+)+kW+ _= (3.2— 5 0 Ax q 2 p 2 At H . 2 (1+2 Fo(Bi +1))rg+1 —2 .Fo-TIPH :Tg +2Bi Pct-Tm +Fo qux . (6) < COMMENTS: Compare these results (Eqs. 3, 4 and 6) with the appropriate expression in Table 5.2. ...
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assignchap5-95 - PROBLEM 5.95 KNOWN One-dimensional wall...

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