Design og modellering af metanolanlæg til VEnzin-visionen Bilag

VEnzin.for
c:/dna/source/
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100 CONTINUE
KOMTY = ’MIXING_TANK’
GOTO 9999
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C Component characteristics
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200 CONTINUE
KOMTY = ’MIXING_TANK’
ANTKN = 5
ANTPK = 2
ANTLK = 3
ANTEX = 2
ANTM1 = 2
ANTM2 = 2
MEDIE(1) = REALFL$
MEDIE(2) = REALFL$
MEDIE(3) = REALFL$
MEDIE(4) = REALFL$
MEDIE(5) = HEAT$
ANTME = 4
VARME(1) = NODE1$
VARME(2) = NODE1$
VARME(3) = NODE3$
VARME(4) = NODE3$
PARNAM(1) = ’Pressure loss side 1’
PARNAM(2) = ’Pressure loss side 2’
ZANAM(1) = ’Flu 2 mass−%’
ZANAM(2) = ’Flu 2 vol−%’
IF (FKOMP.EQ.6) GOTO 600
** FKOMP = 3
GOTO 9999
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C Component equations. All in residual form.
C Do not include the conservation laws. These are treated automatically
C by DNA.
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400 CONTINUE
C
DPA = PAR(1)
DPB = PAR(2)
C
C Pressure losses
C
RES(1) = P(1) − P(2) − DPA
RES(2) = P(3) − P(4) − DPB
C
CALL STATES(P(2),H(2),T2,V,S,X,U,1,2,MEDIE(2))
CALL STATES(P(4),H(4),T4,V,S,X,U,1,2,MEDIE(4))
RES(3) = T2−T4
C
RES(4) = ZA(1) − MDOT(3)/(MDOT(3)+MDOT(1))
RES(5) = ZA(2) − (MDOT(3)/RM_MOL(MEDIE(3)))/(MDOT(3)
$ /RM_MOL(MEDIE(3))+MDOT(1)/RM_MOL(MEDIE(1)))
C
IF (FKOMP.EQ.5) GOTO 500
GOTO 9999
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C Solution check
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500 CONTINUE
IF (MDOT(1).LT.−1D−10) GOTO 550
IF (MDOT(2).GT.1D−10) GOTO 550
IF (MDOT(3).LT.−1D−10) GOTO 550
IF (MDOT(4).GT.1D−10) GOTO 550
IF (Q.GT.1D−10) GOTO 550
GOTO 9999
550 FBETI = .FALSE.
GOTO 9999
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C Write component information
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600 CONTINUE
KOMDSC = ’Heat exchanger used as a model for a mixing tank. This
$ means that the outlet temperatures are set equal.’
K_PAR(1) = ’Pressure loss side 1, $\\Delta p_{12}$ [bar]’
K_PAR(2) = ’Pressure loss side 2, $\\Delta p_{34}$ [bar]’
K_BET =
37/67
19−03−2007