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Unformatted text preview: /5 NW. 2990: The ﬂow diagram is 30 CO CO
55 H2 A reactor H2
5 N2 MeOH 5N2 with the reactor outlet stream assumed at equilibrium. The material balance equations are hCOput = 30 ‘57
ﬁHZput = 65  ﬁMeOHput = E
ﬁN 2,0ut = 5
ﬂout g 100 " 2‘5 (with all ﬂows in kmol/h.) The equilibrium constant for this gasphase reaction, written in terms
of extent of reaction, is Ka = ML = M3.
ycq(YH2)2 P2 (305)(65 415)2 P2 To ﬁnd K“, we ﬁrst calculate A6; = —162.32 + 137.27 = ——25.05 kl/gmdl A AH; = 200.94 + 110.53 = —90.41 kJ/gmol At 298 K, K a = exp(——2—5——’0£) = 24605
8.314 x 298 From the van’t Hoff equation, at the reaction temperature of 473 K ana = _ 1 —25050 + 90410 _ 90410] : _3_39
, 8.314 298 473
 Ka =0.0337 4mm The reactor pressure is 4925 kPa, or49:6 atm, We need to convert to atm, because the standard
state for the Gibbs energy of reaction is ' k _ a; ‘r ‘ \ V, . . 2
KaPZ = 0033701336? 49.6; m
(30 — §)(65 — 2;)2 We know that the extent of reaction must be less than 30 and more than 0. Using an equation
solver, we ﬁnd the root of the equation as: 5:2558 Inserting this value into the material balance equations, we ﬁnd that the reactor outlet ﬂows are
(in kmol/h): flCOput = 4.2; a.“ '
ﬁHZput a 13 4' KR
ﬁMeOHput = 25's! A ﬁN 2,0ut = 5 ﬂout ‘ 48.41, “WW .. . W... U. WWW...» (6 c0 :: 3 o éé W l 30 092 \ From data in Appendix B, we calculate: = (—1)(110.5)+ (—1)(241.8) + (—3935) = —41.2 kJ/gmol
A191: = (—i)(—137.3) + (—l)(—228.6) + (—3944) = 28.5 kJ/gmol From the van’t Hoff equation as ana =_ l —28500+41200_41200 =_5.126+ 4955 8.314 298 s T Ka as a function of T was calculated in a spreadsheet T(°C) T(K) ’ K,
100 373 3491.83
200 473 210.54
300 573 33.83
400 673 9.36
500 773 3.61
600 873 1.73
700 973 0.97
800 1073 0.60
900 1173 0.41
1000 ‘ 1273 0.29
We know that = yCOZyHZ = E2 K
a yCOYHzO (l—E)2 assuming equimolar (1:1) feed of CO and H20. From this we can calculate the ektent of reaction at equilibrium as a function of temperature, and
the mole fractions of CO and H2, with r g z (from the quadratic formula). K0714? 1;;
2 and yCO‘
E
sz =5 The results are plotted. A reactor temperature of 488 °C gives a 2:1 mole ratio of H22CO. 0.9
0.‘ OJ Mole fraction 02 H [00 Male Ratio 0.l 0 200; 400 600 800 1000 0 200 400 600 800 1000 Temperature (’C) Temperature (°C) and at 30°C the mole fraction of ox
yozP a 0.21(1 atm)
H02 47,500 atm ygen dissolved in water is only = 4.4 x 10‘6 x02 ‘= There is more oxygen dissolved in the cooler waters, so that is where the ﬁsh go gal Use M for methane and E for ethane. Flows are given in gmol/min, and compositions as mol%.
’ Use f, d, and b to indicate feed, distillate (or overhead), and bottoms streams, respectively. The ﬂow diagram is 90% M
10% E
70 M
30 E distillation
2% M
98% E The 2 material balance equations are;
70 = ﬂMd + I'le == 0.90fld + 0.02f1}2 V 30 = nEd + an, =0.10r'zd +0.98% We combine and solve to ﬁnd
rid = 77.3 gmol/min rib = 22.7 gmol/min From the deﬁnitions we calculate the fractional recoveries of each component; a 0.90 77.3
fRMd = 44d— = A?) = 0.994 n Mf
fRE g 2.121 = 0.98(22.7) g 0.74
' nEf 30 ‘ The flow diagram is W
solvent 8 2 3200 gmol/h
72% M M
13& coz coz
12% H28 H25
3% cos 0.3% cos
HZS
cos
5 with M for methane and S for the solvent. We also know that the gas:solvent ratio is 3:1 and the
solvent absorbs 97.2% of the H28 fed. DOF anal sis Number of variables Stream variables S stem variables _ N 0 reaction, assumed stead state
Number ofconstraims —  _ _
Speciﬁed ﬂows 2 Stream 1 =.3200 gmol/h
Stream 2 = 1067 mol/h 4 in stream 1, 1 in stream 2, 4 in
stream 3, 3 in stream 4 S eciﬁed comositions 3 in stream 1, 1 in stream 3 System performance 97.2% recovery of H28 
s o  ciﬁcations Material balances — DOF:12—(2+4+1+5)=12—12=0. The material balances for methane and CO2 are simple because each appears in only one input
and one output stream.
it M1 = ﬁM3 = O.72(3200) = 2304 gmol/h I'lCOZ’l = ﬁc02’3 8 = ngl/h From the system performance speciﬁcation for H28 we know
ﬁstA a "HZSA
1': H2 SJ 0.12(3200) r'zHZSA = 373.25 gmol/h fRHZSA = 0.972 = Now from material balance on H28 we ﬁnd
1.111253 = ﬁHZS,1  ileSA =  = ngl/h From the speciﬁed stream composition of the exiting gas,
r“: it
0003 g cos,3 = cos,3 113 m
flCOS’3 = 8.2:_gmol/h From material balance on COS we ﬁnd
I‘lCOS’4 == ﬁCoSJ  IICOS’3 =  8.2 = 87.8 ngI/h The solvent in streams 2 and 4 is 1067 gmol/h. From these calculations we ﬁnd that the exit gas ﬂow rate is 2739 gmol/h and the gas contains
84.1 mol% CH4, 15.2 mol% C02, 0.4 mol% H28 and 0.3 mol% COS. The liquid stream ﬂow rate
is 1528 gmol/h; it contains 24.4 mol% H28 and 5.75 mol% COS in addition to the solvent. ...
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