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4.4  Finite-Difference  Methods  for  Parabolic  Equations            103


         Table  E4.3.  Comparison  of FDS  and  AS at  x  =  0.30
         for  r  =  0.5
          t       FDS       AS        Diff    %Error
         .000     .6000    .6004    -.0004    -.0006
         .005     .6000    .5966     .0034     .0057
         .010     .6000    .5799     .0201     .0347
         .020     .5500    .5334     .0166     .0312
         .100     .2484    .2444     .0040     .0165


         Table  E 4 . 4 .  FDS  at  several ^/-locations  for  r  =  1
                 x=0.1    x=0.2    x=0.3    x=0.4    x=0.5    x=0.6
         +->
         .0000    .200     .400     .600     .800    1.000     .800
         .0100    .200     .400     .600     .800     .600     .800
         .0200    .200     .400     .600     .400    1.000     .400
         .0300    .200     .400     .200    1.200    -.200    1.200
         .0400    .200     .000    1.400   -1.200    2.600   -1.200


         In  some problems the  boundary  conditions  are  expressed  in terms  of  derivatives
         rather  than  in  terms  of  T,  as  in  Eq.  (4.4.1).  In  those  cases  the  solution  of  Eq.
         (4.2.4)  by  an  explicit  or  implicit  method  requires  additional  work.  To  illustrate
         this,  consider  Example  4.1  with  the  boundary  conditions  of the  form
                                             dT
                                    i  =  0,  —  =  T                      (4.4.4a)
                                             ox


                                    i  =  I,  9 £  =  -T                  (4.4.4b)

         They  may  be  expressed  in  terms  of  central  differences,  Eq.  (4.3.7),  or  by  a  for-
         ward  difference,  Eq.  (4.3.8). With  the  choice  of  central  differences.  Eq.  (4.4.4a)
         can  be  written  as
                                      rpn   rpn
                                                   n
                                                 T 0                       (4.4.5)
                                        2Ax
        At  i  — 0,  Eq.  (4.4.3b)  becomes

                            n+1     n
                          T 0   =  T 0  +  aj^(Tf  -  2T 0" +  7 ^ )        (4.4.6)
                                         [ZAX)
         Eliminating  T ^  between  Eqs.  (4.4.5)  and  (4.4.6)  gives


                         r n + i  = n  +  2a-^[T?  -  (1  + Ax)TZ\         (4.4.7)
                                 T
         At  i  = ,  Eq  (4.4.3b)  becomes
               /
                           1
                         T^   =  Tf  +  a ^ a  (27+1  -  23?  +  7?-i)
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