Design og modellering af metanolanlæg til VEnzin-visionen Bilag
Design og modellering af metanolanlæg til VEnzin-visionen Bilag Design og modellering af metanolanlæg til VEnzin-visionen Bilag
VEnzin.for c:/dna/source/ C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− C Component name C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 100 CONTINUE KOMTY = ’MEOH_CONVERTER’ MMVAR(1) = 0 GOTO 9999 C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− C Component characteristics C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 200 CONTINUE KOMTY = ’MEOH_CONVERTER’ ANTKN = 3 ANTPK = 1 ANTEX = 6 ANTM1 = 2 ZANAM(1) = ’MULTIPLIER H’ ZANAM(2) = ’MULTIPLIER C’ ZANAM(3) = ’MULTIPLIER O’ ZANAM(4) = ’GIBBS ENERGY’ ZANAM(5) = ’T−COND’ MEDIE(1) = ANYGAS$ MEDIE(2) = ANYGAS$ MEDIE(3) = HEAT$ ANTME = 2 VARME(1) = NODE1$ VARME(2) = NODE2$ C 204 ANTEL(2) = 37 ANTLK = 1+ANTEL(2) VAREL(1,2) = H2$ VAREL(2,2) = CO$ VAREL(3,2) = CO2$ VAREL(4,2) = H2O_G$ VAREL(5,2) = CH3OH$ VAREL(6,2) = 2 VAREL(7,2) = 3 VAREL(8,2) = 5 DO I=8,36 VAREL(I+1,2) = I ENDDO C IF (FKOMP.EQ.6) GOTO 600 IF (CON) GOTO 404 *** FKOMP = 3 GOTO 9999 C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− C Component equations. All in residual form. C Do not include the conservation laws, since these are treated C automatically by DNA. C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 400 CONTINUE C CON=.true. GOTO 204 404 Continue C C Pressure C ANTGIBBS=5 RES(1) = P(1) − P(2) − PAR(1) RES_NR=1 C C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− C Find the composition of the equilibrium gas C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− C C Calculate mole flow of each species in C M_BL(1)=0.D0 DO I=1,ANTST M_BL(1)=X_J(MEDIE(1),I)*M_MOL(I)+M_BL(1) ENDDO DO I=1,ANTST NIN(I)= MDOT(1)/M_BL(1)*X_J(MEDIE(1),I) ENDDO C C Calculate mole flow of each species out 25/67 19−03−2007
VEnzin.for c:/dna/source/ C C C M_BL(2)=0.D0 DO I=1,ANTST M_BL(2)=X_J(MEDIE(2),I)*M_MOL(I)+M_BL(2) ENDDO NOUT(ANTST+1)=0.D0 DO I=1,ANTST NOUT(I)=(−MDOT(2))*X_J(MEDIE(2),I)/M_BL(2) NOUT(ANTST+1)=NOUT(ANTST+1)+NOUT(I) ENDDO DO J=ANTGIBBS+1,ANTEL(2) RES(RES_NR+J−ANTGIBBS)=NOUT(VAREL(J,2))−NIN(VAREL(J,2)) ENDDO RES_NR=RES_NR+ANTEL(2)−ANTGIBBS C C Molar balance for each atom (H,C,O) C DO J=1,2 RES(RES_NR+J) = 0.D0 DO I=1,ANTST RES(RES_NR+J)=RES(RES_NR+J)−(NIN(I)−NOUT(I))*EL(I,J) ENDDO ENDDO RES_NR=RES_NR+3 C RES(RES_NR)=1.D0 DO I=1,ANTST RES(RES_NR)= RES(RES_NR)−X_J(MEDIE(2),I) ENDDO RES_NR=RES_NR+1 C C Gibbs’ free energy of each compound C CALL STATES(P(2),H(2),TGAS,V,S,X,U,1,2,MEDIE(2)) TGAS = TGAS+273.15D0 C DO I=1,ANTGIBBS CALL GIBBS(VAREL(I,2),TGAS,P(2)*X_J(MEDIE(2),VAREL(I,2)),G(I)) ENDDO C C Gibbs free energy of the mixture C RES(RES_NR)=ZA(4) DO I=1,ANTGIBBS RES(RES_NR)=RES(RES_NR)−X_J(MEDIE(2),VAREL(I,2))*G(I) ENDDO C C Partial derivatives of the function to be minimized with respect to C each species molar fraction C DO I=1,ANTGIBBS RES(I+RES_NR)=G(I) DO J=1,2 RES(I+RES_NR) = RES(I+RES_NR) + ZA(J)*EL(VAREL(I,2),J) ENDDO RES(I+RES_NR) = RES(I+RES_NR) + ZA(3)*EL(VAREL(I,2),4) ENDDO RES_NR=RES_NR+ANTGIBBS+1 C C Check if condensation of water and methanol occurs at equllibrium C b_2_1=−1062.945621D0 b_1_2=3538.709318D0 c J/mol alpha=0.2994D0 R_u=8.314D0 c J/(mol*K) c T=ZA(5) X0 = 0.D0 CALL STATES(P_SAT_ST,HD,T,V,S,X0,U,3,6,WATHF$) X0 = 0.D0 CALL STATES(P_SAT_ME,HD,T,V,S,X0,U,3,6,MEOH$) T_K=T+273.15D0 c tau_2_1=b_2_1/(R_u*T_K) 26/67 19−03−2007
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<strong>VEnzin</strong>.for<br />
c:/dna/source/<br />
C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−<br />
C Component name<br />
C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−<br />
100 CONTINUE<br />
KOMTY = ’MEOH_CONVERTER’<br />
MMVAR(1) = 0<br />
GOTO 9999<br />
C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−<br />
C Component characteristics<br />
C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−<br />
200 CONTINUE<br />
KOMTY = ’MEOH_CONVERTER’<br />
ANTKN = 3<br />
ANTPK = 1<br />
ANTEX = 6<br />
ANTM1 = 2<br />
ZANAM(1) = ’MULTIPLIER H’<br />
ZANAM(2) = ’MULTIPLIER C’<br />
ZANAM(3) = ’MULTIPLIER O’<br />
ZANAM(4) = ’GIBBS ENERGY’<br />
ZANAM(5) = ’T−COND’<br />
MEDIE(1) = ANYGAS$<br />
MEDIE(2) = ANYGAS$<br />
MEDIE(3) = HEAT$<br />
ANTME = 2<br />
VARME(1) = NODE1$<br />
VARME(2) = NODE2$<br />
C<br />
204 ANTEL(2) = 37<br />
ANTLK = 1+ANTEL(2)<br />
VAREL(1,2) = H2$<br />
VAREL(2,2) = CO$<br />
VAREL(3,2) = CO2$<br />
VAREL(4,2) = H2O_G$<br />
VAREL(5,2) = CH3OH$<br />
VAREL(6,2) = 2<br />
VAREL(7,2) = 3<br />
VAREL(8,2) = 5<br />
DO I=8,36<br />
VAREL(I+1,2) = I<br />
ENDDO<br />
C<br />
IF (FKOMP.EQ.6) GOTO 600<br />
IF (CON) GOTO 404<br />
*** FKOMP = 3<br />
GOTO 9999<br />
C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−<br />
C Component equations. All in residual form.<br />
C Do not include the conservation laws, since these are treated<br />
C automatically by DNA.<br />
C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−<br />
400 CONTINUE<br />
C<br />
CON=.true.<br />
GOTO 204<br />
404 Continue<br />
C<br />
C Pressure<br />
C<br />
ANTGIBBS=5<br />
RES(1) = P(1) − P(2) − PAR(1)<br />
RES_NR=1<br />
C<br />
C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−<br />
C Find the composition of the equilibrium gas<br />
C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−<br />
C<br />
C Calculate mole flow of each species in<br />
C<br />
M_BL(1)=0.D0<br />
DO I=1,ANTST<br />
M_BL(1)=X_J(MEDIE(1),I)*M_MOL(I)+M_BL(1)<br />
ENDDO<br />
DO I=1,ANTST<br />
NIN(I)= MDOT(1)/M_BL(1)*X_J(MEDIE(1),I)<br />
ENDDO<br />
C<br />
C Calculate mole flow of each species out<br />
25/67<br />
19−03−2007