Aspen adsorption 自带例子丁烯脱氢生成丁二烯Fortran语 ...？ subroutine pUgCtSl(Tg,p,y,ny,cs,nsol, : Rgas,ngas,Rtot,nreac,Rsol,nrsol,ifail) C----------------------------------------------------------- C Reaction Rates for Butene dehydrogenation example C reference: Froment, Bischoff: Chemical Reactor Analysis C and Design C T.B. 06.09.98 C------------------------------------------------------------ implicit none integer ncomp, nsol, ngas, nreac, nrsol, ifail, itype, i, j, : ncg, ncs, nctot, nr, ny parameter (ncg=4, ncs=1, nctot=5, nr=3) double precision Tg, p, c(ncg), cs(nsol), stage, Rgas(ngas), : Rtot(nreac), Rsol(nrsol), R, Ea(nr), k0(nr), : kactive(nr), stoichg(ncg, nr), stoichs(ncs, nr), : F, ctot, pp(ncg), K, M_C, effect(nr), y(ny) data R /8.31441d0/ data K /0.0018d0/ data M_C /12.011d0/ data Ea /29236.0d0, 32860.0d0, 21042.0d0/ data k0 /5072.22d0, 43300.0d0, 141.89d0/ data effect /1d0, 1d0, 1d0/ data ((stoichg(i,j), i=1,1), j=1,nr) /0.0d0, 0.0d0, 1.0d0/ data ((stoichg(i,j), i=2,2), j=1,nr) /0.0d0, -1.0d0, 0.0d0/ data ((stoichg(i,j), i=3,3), j=1,nr) /1.0d0, 0.0d0, 0.0d0/ data ((stoichg(i,j), i=4,4), j=1,nr) /0.0d0, 0.0d0, 0.0d0/ data ((stoichs(i,j), i=1,1), j=1,nr) /0.0d0, 0.0d0, 0.0d0/ C------------------------------------------------------------------- C set active rate constants to be k0 or 0, depending on stage number C------------------------------------------------------------------- kactive(1) = k0(1) kactive(2) = k0(2) kactive(3) = k0(3) C------------------------------------------------------------------ C Calculate the reaction rates of each reaction C------------------------------------------------------------------ do i=1, nr rtot(i) = kactive(i)*dexp(-Ea(i)/(R*Tg))*effect(i) enddo do i=1,ncg c(i) = y(i)*p/(R*1.0d-2*Tg) enddo ctot = 0.0d0 do i=1, ncg ctot = ctot + c(i) enddo do i=1, ncg pp(i) = p*y(i) enddo C------------------------------------------------------------------ C The following are only dummies. To be used until rate eq is known c Substitute/Replace 1.0d-3 below with the right composition function F C------------------------------------------------------------------ 从这开始，我感觉是在定义反应速率，为什么还对反应速率进行比较？ F = 1.0D-3 Rtot(1) = rtot(1)*(pp(2)-pp(3)*pp(1)/K)/ : ((1.0d0+1.727d0*pp(2)+3.593d0*pp(3)+38.028d0*pp(1))**2.0d0)* : dexp(-42.12d0*cs(1)*M_C) if(pp(2) .lt. 1.0d-20) then Rtot(2) = 1.0d-20 else Rtot(2) = rtot(2)*pp(2)**0.743d0/ : ((1.0d0+1.695d0*dsqrt(pp(3)))**2.0d0)* : dexp(-45.53d0*M_C*cs(1))/M_C endif if(pp(1) .lt. 1.0d-20) then Rtot(3) = 1.0d-20 else Rtot(3) = rtot(2)*pp(1)**0.853d0/ : ((1.0d0+1.695d0*dsqrt(pp(3)))**2.0d0)* : dexp(-45.53d0*M_C*cs(1))/M_C endif C------------------------------------------------------------------ C Calculate the reaction rates for each component C------------------------------------------------------------------ Rgas(1) = Rtot(1)-Rtot(3)/(4.0d0) Rgas(2) = -(Rtot(1)+Rtot(2)/(4.0d0)) Rgas(3) = Rtot(1) Rgas(4) = 1.0d-20 C if(y(2) .gt. 0.99999d0)then C Rsol(1) = 0.0d0 C else Rsol(1) = Rtot(2)+Rtot(3) C endif return end 查看更多2个回答 . 4人已关注
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