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Unformatted text preview: PROBLEM 5.95 KNOWN: Onedimensional wall suddenly subjected to uniform volumetric heating and
convective surface conditions. FIND: Finitedifference equation for node at the surface, it = —L.
SCHEMATIC:
for Z' A 0,? 75ml»? T i Tum "L Lyx+L ASSUMPTIONS: (1) Onedimensional 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)+kI1#+' 11— = c11— 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 forwarddifference 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+BiToo+ ]+(1—2Fo—2FoBi)rg’. (3) The stability criterion requires that the coefficient of T}; be positive. That is, (1—2Fo—2FoBi)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. Tpt1_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 .FoTIPH :Tg +2Bi PctTm +Fo qux . (6) < COMMENTS: Compare these results (Eqs. 3, 4 and 6) with the appropriate expression in
Table 5.2. ...
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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at Wisconsin.
 Fall '08
 Rothamer
 Heat Transfer

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