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

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VEnzin.for c:/dna/source/ $ ,P_ST,T,P_SAT_ME,P_SAT_ST,x_ME_2,x_ME_3,x_ST_3,P_ME_IN,NOUTME $ ,NOUTST,P_ST_IN,HPL,P_ST_2 CHARACTER*100 K_PAR(3),K_STAT(1) CHARACTER*500 K_LIG(47), K_BET CHARACTER*1000 KOMDSC, K_INP CHARACTER*100 KMEDDS(7) logical EVAP,EVAPST,EVAPME EXTERNAL STATES INCLUDE ’THERPROP.INI’ C======================================================================= GOTO (100,200,1,400,400,200,1) FKOMP 1 RETURN C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− C Component name C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 100 CONTINUE KOMTY = ’GASCOOL4’ GOTO 9999 C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− C Component characteristics C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 200 CONTINUE KOMTY = ’GASCOOL4’ ANTKN = 7 ANTPK = 3 ANTLK = 44 ANTEX = 9 ANTM1 = 4 ANTM2 = 2 MEDIE(1) = ANYGAS$ MEDIE(2) = ANYGAS$ MEDIE(3) = WATHF$ MEDIE(4) = MEOH$ MEDIE(5) = ANYFLU$ MEDIE(6) = ANYFLU$ MEDIE(7) = CONTROL$ ANTME = 4 VARME(1) = NODE1$ VARME(2) = NODE2$ VARME(3) = NODE5$ VARME(4) = NODE5$ ANTEL(1) = 0 ANTEL(2) = 37 ANTEL(3) = 0 ANTEL(4) = 0 DO I=1,36 VAREL(I,2) = I ENDDO VAREL(37,2) = CH3OH$ ZANAM(1) = ’t−diff boil’ ZANAM(2) = ’t−diff cond’ ZANAM(3) = ’T−boil’ ZANAM(4) = ’T−condense’ ZANAM(5) = ’CH3OH mass−%’ ZANAM(6) = ’CH3OH vol−%’ ZANAM(7) = ’Transferred’ C IF (FKOMP.EQ.6) GOTO 600 ** FKOMP = 3 GOTO 9999 C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− C Component equations. All in residual form. C Do not include the conservation laws. These are treated automatically C by DNA. C−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 400 CONTINUE C RES(1) = P(1) − P(2) − PAR(1) c RES(2) = P(2) − P(3) c RES(3) = P(2) − P(4) RES(4) = P(5) − P(6) − PAR(2) C C Same temperature out C CALL STATES(P(2),H(2),T2,V,S,X,U,1,2,MEDIE(2)) CALL STATES(P(3),H3,T2,V,S,X,U,1,3,MEDIE(3)) CALL STATES(P(4),H4,T2,V,S,X,U,1,3,MEDIE(4)) C 31/67 19−03−2007

VEnzin.for c:/dna/source/ RES(5) = H3 − H(3) RES(6) = H4 − H(4) C C Calculate mole flow in C M_BL1 = 0.D0 DO I=1,ANTST M_BL1=M_BL1+X_J(MEDIE(1),I)*M_MOL(I) ENDDO NIN = MDOT(1)/M_BL1 NINST=NIN*X_J(MEDIE(1),7) NINME=NIN*X_J(MEDIE(1),CH3OH$) N=NIN−NINST−NINME C C Calculate mole flow out C M_BL2 = 0.D0 DO I=1,ANTST M_BL2=M_BL2+X_J(MEDIE(2),I)*M_MOL(I) ENDDO NOUT = (−MDOT(2))/M_BL2 C C Outlet composition of gas C DO I = 1,6 RES(I+6) = X_J(MEDIE(2),I)*NOUT − X_J(MEDIE(1),I)*NIN ENDDO DO I = 8,35 RES(I+5) = X_J(MEDIE(2),I)*NOUT − X_J(MEDIE(1),I)*NIN ENDDO C RES(41)=1.0D0 DO I = 1,ANTST RES(41) = RES(41)−X_J(MEDIE(2),I) ENDDO C C Check for condensing water or methanol 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(4) X0 = 0.D0 CALL STATES(P_SAT_ST,HD,T,V,S,X0,U,3,6,MEDIE(3)) X0 = 0.D0 CALL STATES(P_SAT_ME,HD,T,V,S,X0,U,3,6,MEDIE(4)) T_K=T+273.15D0 c tau_2_1=b_2_1/(R_u*T_K) tau_1_2=b_1_2/(R_u*T_K) c x_ME_3=ZA(8) x_ST_3=1D0−x_ME_3 gamma_2=exp((x_ST_3**2*(tau_1_2*((exp(−(alpha $ *tau_1_2))/(x_ME_3+x_ST_3*exp(−(alpha*tau_1_2)))) $ )**2+(tau_2_1*(exp(−(alpha*tau_2_1))/(x_ST_3+x_ME_3 $ *exp(−(alpha*tau_2_1)))**2))))) gamma_1=exp((x_ME_3**2*(tau_2_1*((exp(−(alpha $ *tau_2_1))/(x_ST_3+x_ME_3*exp(−(alpha*tau_2_1)))) $ )**2+(tau_1_2*(exp(−(alpha*tau_1_2))/(x_ME_3+x_ST_3 $ *exp(−(alpha*tau_1_2)))**2))))) c P_ST_IN=X_J(MEDIE(1),7)*P(1) P_ME_IN=X_J(MEDIE(1),CH3OH$)*P(1) c P_ME=P_ME_IN*0.999d0 x_ME=P_ME/(gamma_2*P_SAT_ME) x_ST=1D0−x_ME P_ST_2=gamma_1*P_SAT_ST*x_ST P_ST=(P_ME_IN−P_ME+P_ST_IN)−(P_ME_IN−P_ME)/x_ME if (T.gt.239.1) then RES(52)=T−238 else RES(52)=P_ST−P_ST_2 32/67 19−03−2007

<strong>VEnzin</strong>.for<br />

c:/dna/source/<br />

RES(5) = H3 − H(3)<br />

RES(6) = H4 − H(4)<br />

C<br />

C Calculate mole flow in<br />

C<br />

M_BL1 = 0.D0<br />

DO I=1,ANTST<br />

M_BL1=M_BL1+X_J(MEDIE(1),I)*M_MOL(I)<br />

ENDDO<br />

NIN = MDOT(1)/M_BL1<br />

NINST=NIN*X_J(MEDIE(1),7)<br />

NINME=NIN*X_J(MEDIE(1),CH3OH$)<br />

N=NIN−NINST−NINME<br />

C<br />

C Calculate mole flow out<br />

C<br />

M_BL2 = 0.D0<br />

DO I=1,ANTST<br />

M_BL2=M_BL2+X_J(MEDIE(2),I)*M_MOL(I)<br />

ENDDO<br />

NOUT = (−MDOT(2))/M_BL2<br />

C<br />

C Outlet composition of gas<br />

C<br />

DO I = 1,6<br />

RES(I+6) = X_J(MEDIE(2),I)*NOUT − X_J(MEDIE(1),I)*NIN<br />

ENDDO<br />

DO I = 8,35<br />

RES(I+5) = X_J(MEDIE(2),I)*NOUT − X_J(MEDIE(1),I)*NIN<br />

ENDDO<br />

C<br />

RES(41)=1.0D0<br />

DO I = 1,ANTST<br />

RES(41) = RES(41)−X_J(MEDIE(2),I)<br />

ENDDO<br />

C<br />

C Check for condensing water or methanol<br />

C<br />

b_2_1=−1062.945621D0<br />

b_1_2=3538.709318D0<br />

c J/mol<br />

alpha=0.2994D0<br />

R_u=8.314D0<br />

c J/(mol*K)<br />

c<br />

T=ZA(4)<br />

X0 = 0.D0<br />

CALL STATES(P_SAT_ST,HD,T,V,S,X0,U,3,6,MEDIE(3))<br />

X0 = 0.D0<br />

CALL STATES(P_SAT_ME,HD,T,V,S,X0,U,3,6,MEDIE(4))<br />

T_K=T+273.15D0<br />

c<br />

tau_2_1=b_2_1/(R_u*T_K)<br />

tau_1_2=b_1_2/(R_u*T_K)<br />

c<br />

x_ME_3=ZA(8)<br />

x_ST_3=1D0−x_ME_3<br />

gamma_2=exp((x_ST_3**2*(tau_1_2*((exp(−(alpha<br />

$ *tau_1_2))/(x_ME_3+x_ST_3*exp(−(alpha*tau_1_2))))<br />

$ )**2+(tau_2_1*(exp(−(alpha*tau_2_1))/(x_ST_3+x_ME_3<br />

$ *exp(−(alpha*tau_2_1)))**2)))))<br />

gamma_1=exp((x_ME_3**2*(tau_2_1*((exp(−(alpha<br />

$ *tau_2_1))/(x_ST_3+x_ME_3*exp(−(alpha*tau_2_1))))<br />

$ )**2+(tau_1_2*(exp(−(alpha*tau_1_2))/(x_ME_3+x_ST_3<br />

$ *exp(−(alpha*tau_1_2)))**2)))))<br />

c<br />

P_ST_IN=X_J(MEDIE(1),7)*P(1)<br />

P_ME_IN=X_J(MEDIE(1),CH3OH$)*P(1)<br />

c<br />

P_ME=P_ME_IN*0.999d0<br />

x_ME=P_ME/(gamma_2*P_SAT_ME)<br />

x_ST=1D0−x_ME<br />

P_ST_2=gamma_1*P_SAT_ST*x_ST<br />

P_ST=(P_ME_IN−P_ME+P_ST_IN)−(P_ME_IN−P_ME)/x_ME<br />

if (T.gt.239.1) then<br />

RES(52)=T−238<br />

else<br />

RES(52)=P_ST−P_ST_2<br />

32/67<br />

19−03−2007

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