Aspen adsorption 自带例子丁烯脱氢生成丁二烯Fortran语句
subroutine pUgCtSl(Tg,p,y,ny,cs,nsol,: Rgas,ngas,Rtot,nreac,Rsol,nrsol,ifail)
C-----------------------------------------------------------
CReaction Rates for Butene dehydrogenation example
Creference: Froment, Bischoff: Chemical Reactor Analysis
Cand Design
CT.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
粗看了下，很适合我这种初学者 chenhaijunsjy 发表于 2015-6-17 08:54
粗看了下，很适合我这种初学者
能看懂吗？最后一步Rgas（1）、Rgas（2）、Rgas（3）、Rgas（4）是什么？
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