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6 Pages

### Elementary Principles 117

Course: CHEMICAL E 312, Fall 2011
School: The University of Akron
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Word Count: 204

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Assume 5.14 ideal gas behavior kg L = V2 cm s = V1 cm 3 d i d 3 b F I si G J H K 1 2 g nbkmolgM(kg / kmol) ====&gt; PM Vb Lg RT n P = V RT 12 12 = V1 P1M1T2 P2 M 2 T1 a. b. VH 2 = 350 cm3 758 mm Hg 28.02 g mol 323.2K s 1800 mm Hg 2.02 g mol 298.2K M = 0.25M CH 4 + 0.75M C3H 8 = 0.25 16.05 + 0.75 44.11 = 37.10 g mol LM N b gb g b gb OP Q 12 = 881 cm3 s g cm3 Vg = 350 s 5.15 a. Reactor LM...

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Assume 5.14 ideal gas behavior kg L = V2 cm s = V1 cm 3 d i d 3 b F I si G J H K 1 2 g nbkmolgM(kg / kmol) ====> PM Vb Lg RT n P = V RT 12 12 = V1 P1M1T2 P2 M 2 T1 a. b. VH 2 = 350 cm3 758 mm Hg 28.02 g mol 323.2K s 1800 mm Hg 2.02 g mol 298.2K M = 0.25M CH 4 + 0.75M C3H 8 = 0.25 16.05 + 0.75 44.11 = 37.10 g mol LM N b gb g b gb OP Q 12 = 881 cm3 s g cm3 Vg = 350 s 5.15 a. Reactor LM b758gb28.02gb323.2g OP N b1800gb37.10gb298.2g Q 12 = 205 cm3 s h soap 2 R 2 h 0.012 m PV = V= = 4 t RT b. n CO 2 d i 2 . 12 m 60 s = 11 10 -3 m3 min / . 7.4 s min n CO2 = 5.16 755 mm Hg 1 atm 1.1 10-3 m3 / min 1000 mol = 0.044 mol/min 3 m atm 300 K 1 kmol 0.08206 kmolK 760 mm Hg m air = 10.0 kg / h n air (kmol / h) n (kmol / h) y CO (kmol CO 2 / kmol) 2 VCO 2 = 20.0 m3 / h n CO (kmol / h) 2 150 o C, 1.5 bar Assume ideal gas behavior 10.0 kg 1 kmol n air = = 0.345 kmol air / h h 29.0 kg air n CO 2 = PV 15 bar . 100 kPa 20.0 m3 / h = = 0.853 kmol CO 2 / h 3 RT 8.314 m kPa 1 bar 423.2 K kmolK y CO 2 100% = b 0.853 kmol CO 2 / h 100% = 712% . 0.853 kmol CO 2 / h + 0.345 kmol air / h g 5-5
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The University of Akron - CHEMICAL E - 312
5.9P = 1 atm +10 cm H 2 O1m 1 atm = 101 atm . 2 10 cm 10.333 m H 2 OT = 25 C = 298.2 K , V =2.0 m3 = 0.40 m3 min = 400 L min 5 min m = n mol / min MW g / molbgbga.m=L PV 1.01 atm 400 min 28.02 MW = Latm RT 0.08206 molK 298.2 Kg mol= 458 g mi
The University of Akron - CHEMICAL E - 312
5.4 (cont'd) e.Basis: 1 kg slurry x c kg crystals , Vc m3 crystals =bg dix c kg crystalsbb1- x gbkg liquid g, V dmc l3liquid =liquid i b1-x dgbkg/ m i g kgc 3 l c kg / md3ig sl =bV + V gdm i3 c l1 kg=cxc+b1 1 - xclg5.5Assume
The University of Akron - CHEMICAL E - 312
5.4a.P1 = P0 + sl gh1 P2 = P0 + sl gh 2 P = P1 - P2 = sl h = h1 - h 2 1b. sl=cxc+b1 - x g check units!cU | V | WF IF I e jge jhbmgGH 11 N JK GH 11 Pa JK = ghkg m3 m s2 kgm s2 N m2 sll1 kg crystals / kg slurry kg liquid / kg slurry = + kg sl
The University of Akron - CHEMICAL E - 312
CHAPTER FIVE5.1Assume volume additivity Av. density (Eq. 5.1-1):1=0.400 0.600 + = 0.719 kg L 0.703 kg L 0.730 kg LAAOa.Dmmass of tank at time tA= mt + m0 m =mass of empty tankAb250 - 150gkg = 14.28 kg min bm = mass flow rate of liquidg b1
The University of Akron - CHEMICAL E - 312
4.80a.Basis: 1 mol CpHqOr1 mol CpHqOr no (mol S) Xs (kg s/ kg fuel)C + 02 -&gt; CO2 2H + 1/2 O2 -&gt;H2O S + O2 -&gt; SO2P (% excess air) n1 (mol O2) 3.76 n1 (mol N2)n2 (mol CO2) n3 (mol SO2) n4 (mol O2) 3.76 n1 (mol N2) n5 (mol H2O (v)Hydrocarbon mass: p (
The University of Akron - CHEMICAL E - 312
4.78 (cont'd)SO 2 :0.03 100 g Sb g0.0956 mol O 2 1 mol S 1 mol SO 2 n5 = 0.0936 mol SO 2 consumed 32.06 g S 1 mol S25% excess O 2 : n1 = 125 7.244 + 2.475 + 0.0936 12.27 mol O 2 . O 2 balance: n3 = 12.27 mol O 2 fed - 7.244 + 2.475 + 0.0936 mol O 2 c
The University of Akron - CHEMICAL E - 312
4.77a.Basis 100 mol dry fuel gas. Assume no solid or liquid products!n1 (mol C) n2 (mol H) n3 (mol S)100 mol dry gas C + 02 -&gt; CO2 C + 1/2 O2 -&gt; CO 2H + 1/2 O2 -&gt;H2O S + O2 -&gt; SO2 0.720 mol CO / mol 2 0.0257 mol CO / mol 0.000592 mol SO / mol 2 0.254
The University of Akron - CHEMICAL E - 312
4.76 a.Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile matter100 g c.a. r. 1.147 g a.d.c. 1.207 g c.a. r. = 95.03 g air - dried coal; 4.97 g H 2 O lost by air drying295.03 g a.d.c. b1.234 - 1204g g H O =
The University of Akron - CHEMICAL E - 312
4.75 (cont'd)H:0.17 5000 kg H 1 kmol H 1 kmol H 2 O h 0.02 5000 kg S h 101 kg H . 1 kmol S 32.06 kg S 2 kmol H 1 kmol O 2 1 kmol Sb g b gb1 kmol O 2 2 kmol H 2 O= 210.4 kmol O 2 hS:= 3.1 kmol O2/hTotal = (312.2+210.4 + 3.1) kmol O2/h = 525.7 kmol
The University of Akron - CHEMICAL E - 312
4.74 (cont'd)100 kg fuel oil 7.08 kmol C 12.0 kmol H 0.053 kmol S 1.3 kg NC (s) 20% excess air n1 (kmol O2) 3.76 n1 (kmol N2)C + O2 CO2 C + 1/2 O2 CO 2H + 1/2 O2 H2O S + O2 SO2n2 (kmol N2) n3 (kmol O2) n4 (kmol CO2) (8/92) n4 (kmol CO) n5 (kmol SO2) n6
The University of Akron - CHEMICAL E - 312
4.73a.C3H8 +5 O2 3 CO2 + 4 H2O, C4H10 + 13/2 O2 4 CO2 + 5 H2O Basis 100: mol product gasn1 (mol C3H8) n2 (mol C4H10) n3 (mol O2)100 mol 0.474 mol H2O/mol x (mol CO2/mol) (0.526x) (mol O2/mol)x 69.4 = x = 0.365 mol CO 2 /mol 0.526 - x 30.6Dry product
The University of Akron - CHEMICAL E - 312
4.72 (cont'd)mole H 2 O 0.50 mol product gas mole product gas Basis: 100 mol product gas. Since we have the most information about the product stream composition, we choose this basis now, and would subsequently scale to the given fuel and air flow rates
The University of Akron - CHEMICAL E - 312
4.71a.12 L CH 3 OH 1000 ml 0.792 g mol = 296.6 mol CH 3 OH / h h L ml 32.04 g CH3OH + 3/2 O2 CO2 +2 H2O, CH3OH + O2 CO +2 H2O 296.6 mol CH3OH(l)/hn1 (mol O 2 / h) 3.76n1 (mol N 2 / h)n2 ( mol dry gas / h) 0.0045 mol CH3OH(v)/mol DG 0.0903 mol CO2/mol
The University of Akron - CHEMICAL E - 312
4.70a.C5H12 + 8 O2 5 CO2 + 6 H2O Basis: 100 moles dry product gas n1 (mol C5H12) Excess air n2 (mol O2) 3.76n2 (mol N2) 100 mol dry product gas (DPG) 0.0027 mol C5H12/mol DPG 0.053 mol O2/mol DPG 0.091 mol CO2/mol DPG 0.853 mol N2/mol DPG n3 (mol H2O)3
The University of Akron - CHEMICAL E - 312
4.69 (cont'd)Air feed rate: n0 =387.5 mol O 2 1 kmol air 1.25 kmol air fed = 2306.5 mol air h 0.21 kmol O 2 1 kmol air req' d.. 90% propane conversion n1 = 0100(75 mol C 3 H 8 ) = 7.5 mol C 3 H 8 (67.5 mol C 3 H 8 reacts) . 85% hydrogen conversion n2 =
The University of Akron - CHEMICAL E - 312
4.68 (cont'd) b.i) Theoretical oxygen = (100 mol C4H10)(6.5 mol O2/mol C4H10) = 650 mol O2n air = (650 mol O 2 )(1 mol air / 0.21 mol O 2 ) = 3095 mol air100% conversion n C4H10 = 0 , nO 2 = 0n N2 = 0.79 3095 mol = 2445 mol nCO2 n H2O4 10 4 10U 73.1
The University of Akron - CHEMICAL E - 312
4.67 (cont'd)Air feed rate: n f =b.207.0 kmol O 2 h1 kmol air 0.21 kmol O 21.17 kmol air fed = 1153 kmol air h kmol air req.na = n f 2 x1 + 35x 2 + 5x 3 + 6.5x 4 1 + Pxs 100 1 0.21 .bgbgbgc.n f = aR f , (n f = 75.0 kmol / h, R f = 60) n f = 12
The University of Akron - CHEMICAL E - 312
4.66CO +1 O 2 CO 2 2H2 +1 O2 H 2O 2175 kmol/h 0.500 kmol N2/kmol x (kmol CO/mol) (0.500x) (kmol H2/kmol) 20% excess airNote: Since CO and H 2 each require 0 .5 mol O 2 / mol fuel for complete combustion, we can calculate the air feed rate without de
The University of Akron - CHEMICAL E - 312
4.64a.Basis: 1000 g gas Species C3H8 C4H10 H2O Total m (g) 800 150 50 1000 MW 44.09 58.12 18.02 n (mol) 18.145 2.581 2.775 23.501mole % (wet) 77.2% 11.0% 11.8% 100% 100% mole % (dry) 87.5% 12.5%Total moles = 23.50 mol, Total moles (dry) = 20.74 mol Ra
The University of Akron - CHEMICAL E - 312
4.63a.A balance on ith tank (input = output + consumption) v L min C A, i -1 mol L = vC Ai + kC Ai C Bi mol liter min V LbgbgE v, note V / v = bgbgC A , i -1 = C Ai + k C Ai C Bi B balance. By analogy, C B , i -1 = C Bi + k C Ai C Bi Subtract eq
The University of Akron - CHEMICAL E - 312
4.62 (cont'd)i - C 4 H 10 balance around second mixing point 867.5 + n6 = 34,700 n6 = 33,800 kmol C 4 H 10 in recycle E Recycle E: Since Streams (D) and (E) have the same composition, n5 moles n - C 4 H 10 n2 n7 n14b g = n bmoles i - C H g n bmoles n -
The University of Akron - CHEMICAL E - 312
4.62a.i - C 4 H 10 + C 4 H 8 = C 8 H 18Basis: 1-hour operationDn 2 (n-C 4 H10 ) n 3 (i-C 4 H 10) n 1 (C 8 H18 ) m4 (91% H 2 SO4 ) EP F decanter n 1 (C 8 H18 ) n 2 (n-C 4 H10 ) n 3 (i-C 4 H 10)n 1 (C 8 H18 ) n 2 (n-C 4 H10 ) stillUnits of n: kmol U
The University of Akron - CHEMICAL E - 312
4.61 (cont'd) At mixing point:N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1 I: XI0 + (1-yp) n2 = n2 Total moles fed to reactor: nr = 4n1 + n2 Moles of NH3 produced: np = 2fspn1 Overall N2 conversion: (1 - X I0 ) / 4 - y p (1 - f sp )n 1 (1 - X I0 ) / 4 100%b.X
The University of Akron - CHEMICAL E - 312
4.60a.Basis: 100 mol feed/h. Put dots above all n's in flow chart.100 mol/h 32 mol CO/h 64 mol H 2 / h 4 mol N 2 / h n1 (mol /h) .13 mol N 2 /mol reactor cond. n3 (mol CH 3 OH / h) n2 (mol CH3OH/h)500 mol / h x1 (mol N 2 /mol) x2 (mol CO / mol) 1-x1-x
The University of Akron - CHEMICAL E - 312
4.59a.Basis: 100 mol fed to reactor/h 25 mol O2/h, 75 mol C2H4/hn1 (mol C 2H 4 /h) n2 (mol O 2 /h) n3 (mol C 2H 4O /h)reactor nC2H4 ( mol C 2H 4 /h) nO2 (mol O 2 /h) 75 mol C 2H 4 /h 25 mol O 2 /h n1 (mol C 2H 4 /h) n2 (mol O 2 /h) n3 (mol C 2H 4O /h)
The University of Akron - CHEMICAL E - 312
4.58a.Basis: 100 kmol reactor feed/hrn3 (kmol CH 4 /h) 100 kmol /h Reactor n1 (kmol CH 4 /h) 80 kmol CH4 /h n2 (kmol Cl 2 /h) 20 kmol Cl2 /h n3 (kmol CH 4 /h) n4 (kmol HCl /h) 5n5 (kmol CH 3Cl /h) n5 (kmol CH 2Cl 2 /h) Cond. n3 (kmol CH 4 /h) n4 (kmol
The University of Akron - CHEMICAL E - 312
4.57 (cont'd)n4 (mol/min)0.004 mol CH3OH(v)/mol x (mol CO/mol) (0.896 - x ) (mol H 2 / mol)350 mol/ minReactorn1 (mol CO/min) n2 (mol H2 / min)Cond.n3 (mol CH 3OH(l)/min)0.631 mol CH 3OH(v)/ mol0.274 mol CO/ mol CO + H 2 CH 3OH 0.0953 mol H / mol
The University of Akron - CHEMICAL E - 312
4.57 (cont'd)n4 (mol/min)0.004 mol CH3OH(v)/mol x (mol CO/mol) (0.896 - x ) (mol H 2 / mol)350 mol/ minReactorn1 (mol CO/min) n2 (mol H2 / min)Cond.n3 (mol CH 3OH(l)/min)0.631 mol CH 3OH(v)/ mol0.274 mol CO/ mol CO + H 2 CH 3OH 0.0953 mol H / mol
The University of Akron - CHEMICAL E - 312
4.56a.900 kg HCHO 1 kmol HCHO = 30.0 kmol HCHO / h h 30.03 kg HCHOn (kmol CH OH / h) 1 330.0 kmol HCHO / h n2 (kmol H 2 / h) n3 (kmol CH 3OH / h)% conversion:b.30.0 = 0.60 n1 = 50.0 kmol CH 3 OH / h n1n (kmol CH OH / h) 1 330.0 kmol HCHO / h n2 (
The University of Akron - CHEMICAL E - 312
4.56a.900 kg HCHO 1 kmol HCHO = 30.0 kmol HCHO / h h 30.03 kg HCHOn (kmol CH OH / h) 1 330.0 kmol HCHO / h n2 (kmol H 2 / h) n3 (kmol CH 3OH / h)% conversion:b.30.0 = 0.60 n1 = 50.0 kmol CH 3 OH / h n1n (kmol CH OH / h) 1 330.0 kmol HCHO / h n2 (
The University of Akron - CHEMICAL E - 312
4.55 (cont'd) c.mP 4850 4850 4850 4850 4850 4850 4850 4850 4850 mP 2450 2450 2450 2450 2450 2450 2450 2450 2450xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10mA0 3327 3022 2870 2778 2717 2674 2641 2616
The University of Akron - CHEMICAL E - 312
4.55 (cont'd) c.mP 4850 4850 4850 4850 4850 4850 4850 4850 4850 mP 2450 2450 2450 2450 2450 2450 2450 2450 2450xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10mA0 3327 3022 2870 2778 2717 2674 2641 2616
The University of Akron - CHEMICAL E - 312
4.54 (cont'd)ITER = 4 X1A, X2A = 0.05437 ITER = 5 X1A, X2A = 0.05931 ITER = 6 X1A, X2A = 0.059300.02213 0.02086 0.02083YA, YB, YC, YD, YE =2.0270E - 01 2.9501E - 010.20 0.20X1C, X2C = 0.05931 X1C, X2C = 0.05930 X1C, X2C = 0.059301.1197 E - 01 3.931
The University of Akron - CHEMICAL E - 312
4.54 (cont'd)ITER = 4 X1A, X2A = 0.05437 ITER = 5 X1A, X2A = 0.05931 ITER = 6 X1A, X2A = 0.059300.02213 0.02086 0.02083YA, YB, YC, YD, YE =2.0270E - 01 2.9501E - 010.20 0.20X1C, X2C = 0.05931 X1C, X2C = 0.05930 X1C, X2C = 0.059301.1197 E - 01 3.931
The University of Akron - CHEMICAL E - 312
4.54 (cont'd)2 FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/) NTO = NA0 + NB0 + NC0 + ND0 + NE0 NMAX = 10 X1 = 0.1 X2 = 0.1 DO 100 J = 1, NMAX NA = NA0 X1 X1 NB = NB0 + X1 + X1 NC = NC0 + X1 X2 ND = ND0 X2 NE = NE0 + X2 + X2 NAS = NA * 2 NBS = NB
The University of Akron - CHEMICAL E - 312
4.54 (cont'd)2 FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/) NTO = NA0 + NB0 + NC0 + ND0 + NE0 NMAX = 10 X1 = 0.1 X2 = 0.1 DO 100 J = 1, NMAX NA = NA0 X1 X1 NB = NB0 + X1 + X1 NC = NC0 + X1 X2 ND = ND0 X2 NE = NE0 + X2 + X2 NAS = NA * 2 NBS = NB
The University of Akron - CHEMICAL E - 312
4.54a.2CO 2 2CO + O 2 O 2 + N 2 2NO2A 2B + C C + D 2En A = n A 0 - 2 e1 nB nC nD nEyA = = n B 0 + 2 e 2 yB = = nC 0 + e1 - e 2 y C = = n D0 - e2 yD = = n E 0 + 2 e 2 yE =bn bn bn bn bnA0 B0- 2 e1 + 2 e1C0D0E0+ e1 - e 2 nT 0 + e1 - 1 e 2 nT 0 +
The University of Akron - CHEMICAL E - 312
4.54a.2CO 2 2CO + O 2 O 2 + N 2 2NO2A 2B + C C + D 2En A = n A 0 - 2 e1 nB nC nD nEyA = = n B 0 + 2 e 2 yB = = nC 0 + e1 - e 2 y C = = n D0 - e2 yD = = n E 0 + 2 e 2 yE =bn bn bn bn bnA0 B0- 2 e1 + 2 e1C0D0E0+ e1 - e 2 nT 0 + e1 - 1 e 2 nT 0 +
The University of Akron - CHEMICAL E - 312
4.53a.C 6 H 6 + Cl 2 C 6 H 5 Cl + HCl C 6 H 5 Cl + Cl 2 C 6 H 4 Cl 2 + HCl C 6 H 4 Cl 2 + Cl 2 C 6 H 3 Cl 3 + HCl Convert output wt% to mol%: Basis 100 g output species C6 H 6 C 6 H 5 Cl C 6 H 4 Cl 2 C 6 H 3 Cl 3 g 65.0 32.0 2.5 0.5 Mol. Wt. 78.11 112.5
The University of Akron - CHEMICAL E - 312
4.53a.C 6 H 6 + Cl 2 C 6 H 5 Cl + HCl C 6 H 5 Cl + Cl 2 C 6 H 4 Cl 2 + HCl C 6 H 4 Cl 2 + Cl 2 C 6 H 3 Cl 3 + HCl Convert output wt% to mol%: Basis 100 g output species C6 H 6 C 6 H 5 Cl C 6 H 4 Cl 2 C 6 H 3 Cl 3 g 65.0 32.0 2.5 0.5 Mol. Wt. 78.11 112.5
The University of Akron - CHEMICAL E - 312
4.51 (cont'd) b. (1) n1 = 46.08 mol C 2 H 6(3) n2 = 47.4 mol H 2 O (4) n3 = 9.3 mol I % conversion of C2H4:U | Reactor feed contains 44.8% C H , 46.1% H O, 9.1% I V | W2 6 246.08 - 43.3 100% = 6.0% 46.08If all C2H4 were converted and the second react
The University of Akron - CHEMICAL E - 312
4.51 (cont'd) b. (1) n1 = 46.08 mol C 2 H 6(3) n2 = 47.4 mol H 2 O (4) n3 = 9.3 mol I % conversion of C2H4:U | Reactor feed contains 44.8% C H , 46.1% H O, 9.1% I V | W2 6 246.08 - 43.3 100% = 6.0% 46.08If all C2H4 were converted and the second react
The University of Akron - CHEMICAL E - 312
4.50a.Design for low conversion and feed ethane in excess. Low conversion and excess ethane make the second reaction unlikely. C2H6 + Cl2 C2H5Cl + HCl, C2H5Cl + Cl2 C2H4Cl2 + HCl Basis: 100 mol C2H5Cl producedn1 (mol C2H6) n2 (mol Cl2) 100 mol C2H5Cl n
The University of Akron - CHEMICAL E - 312
4.50a.Design for low conversion and feed ethane in excess. Low conversion and excess ethane make the second reaction unlikely. C2H6 + Cl2 C2H5Cl + HCl, C2H5Cl + Cl2 C2H4Cl2 + HCl Basis: 100 mol C2H5Cl producedn1 (mol C2H6) n2 (mol Cl2) 100 mol C2H5Cl n
The University of Akron - CHEMICAL E - 312
4.48 (cont'd)CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP WRITE (6, 5) INMAX, EKPI FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X) FORMAT ('DID NOT CONVERGE IN', I3, 'ITERATIONS',/, * 'CURRENT VALUE = ', F6.3) END \$ DATA 0.5 0.5 0.0 423. 2. RESULTS: YA = 0.5
The University of Akron - CHEMICAL E - 312
4.48 (cont'd)CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP WRITE (6, 5) INMAX, EKPI FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X) FORMAT ('DID NOT CONVERGE IN', I3, 'ITERATIONS',/, * 'CURRENT VALUE = ', F6.3) END \$ DATA 0.5 0.5 0.0 423. 2. RESULTS: YA = 0.5
The University of Akron - CHEMICAL E - 312
4.48 (cont'd)(For this particular set of initial conditions, we get a quadratic equation. In general, the equation will be cubic.). b g c2 - 2b0156gh y = 0.500 y = c1 - 2b0156gh c2 - 2b0156gh y = 0.408 . . y = b0 + 0156g c2 - 2b0156gh y = 0.092 . . n -n
The University of Akron - CHEMICAL E - 312
4.48 (cont'd)(For this particular set of initial conditions, we get a quadratic equation. In general, the equation will be cubic.). b g c2 - 2b0156gh y = 0.500 y = c1 - 2b0156gh c2 - 2b0156gh y = 0.408 . . y = b0 + 0156g c2 - 2b0156gh y = 0.092 . . n -n
The University of Akron - CHEMICAL E - 312
4.47(cont'd) d.T (K) 1223 1123 1023 923 823 723 623 673 698 688 1123 1123 1123 1123 x (CO) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.2 0.4 0.3 0.5 x (H2O) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.4 0.2 0.3 0.4 x (CO2) 0 0 0 0 0 0 0 0 0 0 0.1 0.1 0 0 Ke
The University of Akron - CHEMICAL E - 312
4.47(cont'd) d.T (K) 1223 1123 1023 923 823 723 623 673 698 688 1123 1123 1123 1123 x (CO) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.2 0.4 0.3 0.5 x (H2O) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.4 0.2 0.3 0.4 x (CO2) 0 0 0 0 0 0 0 0 0 0 0.1 0.1 0 0 Ke
The University of Akron - CHEMICAL E - 312
4.46 (cont'd) Product gas n A = 170.6 - 70 = 100.6 molnB = 10 mol nC = 70 mol nD = 70 mol ntotal = 250.6 mold.U y | y | V y | y | WA B C D= 0.401 mol CH3OH mol = 0.040 mol CH3COOH mol = 0.279 mol CH3COOCH 3 mol = 0.279 mol H 2O molCost of reactants,
The University of Akron - CHEMICAL E - 312
4.46 (cont'd) Product gas n A = 170.6 - 70 = 100.6 molnB = 10 mol nC = 70 mol nD = 70 mol ntotal = 250.6 mold.U y | y | V y | y | WA B C D= 0.401 mol CH3OH mol = 0.040 mol CH3COOH mol = 0.279 mol CH3COOCH 3 mol = 0.279 mol H 2O molCost of reactants,
The University of Akron - CHEMICAL E - 312
4.46a.A + B = C + D nA = nA - 0nB = nB - 0 0y A = n A - nT0nC = n C + nD = nD + 0yB yC yDnI = nI Total nT = niAt equilibrium:0e = en = en = enB0C0 + D0j - j n j n + j nTT TnC0 + c nD0 + c yC y D = = 4.87 ( nT 's cancel) y A yB n A0 - c
The University of Akron - CHEMICAL E - 312
4.46a.A + B = C + D nA = nA - 0nB = nB - 0 0y A = n A - nT0nC = n C + nD = nD + 0yB yC yDnI = nI Total nT = niAt equilibrium:0e = en = en = enB0C0 + D0j - j n j n + j nTT TnC0 + c nD0 + c yC y D = = 4.87 ( nT 's cancel) y A yB n A0 - c
The University of Akron - CHEMICAL E - 312
4.45a.Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line throughd R = 10, C11= 0.30 g m 3 andiFH R2= 48, C2 = 2.67 g m 3IKln C = bR + ln a C = ae br b= ln 2.67 0.30b48 - 10 C = 0169e 0.0575 R .g = 0.0575 , ln a = ln
The University of Akron - CHEMICAL E - 312
4.45a.Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line throughd R = 10, C11= 0.30 g m 3 andiFH R2= 48, C2 = 2.67 g m 3IKln C = bR + ln a C = ae br b= ln 2.67 0.30b48 - 10 C = 0169e 0.0575 R .g = 0.0575 , ln a = ln
The University of Akron - CHEMICAL E - 312
4.43 (cont'd)O 2 : n3 = 0.21 160.7 - N 2 : n41 b g 2 = 12.5 mol O = 0.79b160.7g = 127 mol N22Cl 2 : n5 = = 42.5 mol Cl 2 H 2 O: n6 = = 42.5 mol H 2 O These molar quantities are the same as in part (a), so the mole fractions would also be the same. Us
The University of Akron - CHEMICAL E - 312
4.43 (cont'd)O 2 : n3 = 0.21 160.7 - N 2 : n41 b g 2 = 12.5 mol O = 0.79b160.7g = 127 mol N22Cl 2 : n5 = = 42.5 mol Cl 2 H 2 O: n6 = = 42.5 mol H 2 O These molar quantities are the same as in part (a), so the mole fractions would also be the same. Us
The University of Akron - CHEMICAL E - 312
4.43a.2HCl +1 O 2 Cl 2 + H 2 O 2Basis: 100 mol HCl fed to reactor100 mol HCln1 mol airbg0.21 mol O 2 / mol 0.79 mol N 2 / mol 35% excess2b g n b mol O g n b mol N g n b mol Cl g n b mol H Ogn2 mol HCl3 2 4 2 5 2 6 2 2mol O bO gstoic = 100 mo
The University of Akron - CHEMICAL E - 312
4.41 (cont'd)e.10 5 45 0.0119 76.5 - 0.0605 - = 53.9 mV 7 7 7 5 nc = 53.9 + = 127.4 kmol / h 3 3 Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond range of calibration data, system had not reached steady state yet.
The University of Akron - CHEMICAL E - 312
4.41 (cont'd) b.nc =3.00 10 2 kmol 0.85 kmol H 2 S 1 kmol SO 2 = 127.5 kmol SO 2 / h h kmol 2 kmol H 2 Sc.C a lib r a t io n C u r v e1 .2 0 1 .0 0 X (mol H 2 S/mol) 0 .8 0 0 .6 0 0 .4 0 0 .2 0 0 .0 0 0 .0 2 0 .0 4 0 .0 6 0 .0 8 0 .0 1 0 0 .0R a (m