Page 340 - Pressure Swing Adsorption
P. 340

PRESSURE SWING ADSORPTION   APPENDIX B                                317
 316

 Eouatlon  B.4 now  becomes:
 (B.l I)


 clx 8   .   [   /3n[l  - YA(i)]   • ]
 dT  (J)  = aB  I+ f3AyA(j)  +  !3.[,  - yA(i)]  - Xe(/)  '

 1  = 2, ... ,M + 1     1:;,  ~1~·
 The  following  set of linear algebraic equations is  obtained when  the dimen-
 sionless overall material  balance  eauatlon  (Ea.  B.7) and  the velocity  bound-
                              ~
                            0
 ary conditions (Eq. B.8) are combined and wntten m collocatmn form:   .Q   0
                            "  •
                         0    "'  .5
 M~• (A (.  ")  _ Ax(M + 2,i)Ax(J,M+ 2))-(")   (B.12)   ,.   "  "'
 ;_,   XJ,l   Ax(M+2M  2\   VI   '   0  •  '5
 •=2  -  '   •   +  ;       :;:
 j3AyA(i)   ·  )
 (
 =  -,fr  [  aA  1 + j3AyA(i)  + J3.[1  - YA(i)]  - xA(i)   "- .
                      %!~   I   ';,            0       0
                            C  0.
 13,.[1  - y,,(i)]   (  )]   -l<c   "- "I
 + Ys<r 11  -,---'--=~-'-''"[r'-''---~-)~] - x n  j)
 (
 + /JAy,,(j)  + /311  ] - y,,(j
                        ·-  N                  N
 -[Ax(_  )  _  Ax(J, M  + 2)Ax(M -r  2, 1) )-< ) _ l_  ap
 I
 J,l   Ax(M+ 2,M+ 2)   v   Par·
                              ~ t'
 J =  2, ... , M + 1        '5   -.  ;;
                         0   0   - O   0  0   .
 The  followmg ·eauat10ns,  which  are  denved  from  the  boundary  conditions,   .5   .Q  _g.:::  Q
                                ~
                            ~
                            N
 give the values  at  J =  l  and  M  +  2:   ~   ·c   N  .s  ~  •
                         n
                                    E
                        0     ~  1!'  E  ·o
 M+I                        ~  Js   .a  e  .5
 YA(!)=  -A, L  [A,Ax(M + 2,i) -Ax(l,i)]YA(i)   (B.13)   ~   0   -
 M+I
 -A L  Ax(M+2,i)yA(i)  +A,Peii(l)YAlz-o-  "  e
 4                      ~
 !-2
 M+i
 YA(M+2)  =A,  L  (A,Ax(M+2,i)-Ax(l,i)]yA(i)   (B.14)
 -A, Pe v(l)YAlz-o-
 v(M + 2)  =   Ax(M + 2, 1)   (1)   (B.15)
 Ax( M + 2, M + 2) v
 \  I
 M+ I
 -  ~  Ax(M+2,i)v(i))/Ax(M+2,M+2)
 (
 1 2
 Note:  Here J refers to the axiai  locatton m the bed and is  different from  tile
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