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6 Pages

Elementary Principles 193

Course: CHEMICAL E 312, Fall 2011
School: The University of Akron
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Word Count: 209

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Basis: 6.38 1 mol outlet gas/min n0 ( mol / min) y 0 ( mol CH 4 / mol) (1 - y 0 ( mol C 2 H 6 / mol) n1 (mol O 2 / min) 3.76n1 (mol N 2 / min) 1 mol / min @ 573K, 105 kPa y1 (mol CO 2 / mol) y 2 (mol H 2 O / mol) (1 - y1 - y 2 ) mol N 2 / mol CH 4 + 2O 2 CO 2 + 2H 2 O C2 H 6 + 7 O 2 2CO 2 + 3H 2 O 2 pCO 2 = 80 mmHg y1 = 80 mmHg 101325 Pa = 01016 mol CO 2 / mol . 105000 Pa 760 mmHg 100% O2 conversion : 2no...

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Basis: 6.38 1 mol outlet gas/min n0 ( mol / min) y 0 ( mol CH 4 / mol) (1 - y 0 ( mol C 2 H 6 / mol) n1 (mol O 2 / min) 3.76n1 (mol N 2 / min) 1 mol / min @ 573K, 105 kPa y1 (mol CO 2 / mol) y 2 (mol H 2 O / mol) (1 - y1 - y 2 ) mol N 2 / mol CH 4 + 2O 2 CO 2 + 2H 2 O C2 H 6 + 7 O 2 2CO 2 + 3H 2 O 2 pCO 2 = 80 mmHg y1 = 80 mmHg 101325 Pa = 01016 mol CO 2 / mol . 105000 Pa 760 mmHg 100% O2 conversion : 2no yo + 7 no 1 - yo = n1 . C balance: no yo + 2no 1 - yo = 01016 b g 2 b g (1) (2) (3) (4) . N2 balance: 376n1 = 1 - y1 - y2 H balance: 4no yo + 6no 1 - yo = 2 y2 b g Rn 0.0770 = mol | y = 0.6924 mol CH / mol | Solve equations 1 to 4 S . |n = 01912 mol O |y = 01793 mol H O / mol T . Dew point: 01793b105000g Pa 760 mmHg . p dT i = = 141.2 mmHg T o o 4 1 2 2 2 * H2 O dp 101325 Pa dp = 58.8 o C Table B.3 b g 6.39 Basis: 100 mol dry stack gas n P (mol C 3 H 8) n B (mol C 4H10 ) n out (mol) 0.21 O2 0.79 N2 P = 780 mm Hg Stack gas: Tdp = 46.5C 100 mol dry gas 0.000527 mol C 3 H 8/mol 0.000527 mol C 4H 10/mol 0.0148 mol CO/mol 0.0712 mol CO 2/mol + O2 N 2 , nw (mol H2O) C 3 H 8 + 5O 2 3CO 2 + 4H 2 O C 4 H 10 + 13 O 2 4CO 2 + 5H 2 O 2 6-27
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The University of Akron - CHEMICAL E - 312
6.36 (cont'd)Circulation rate of dry nitrogen = 5.875 (1 - 0.1324) = = 0182 lb m / h . 5.097 lb - mole h lb - mole 28.02 lb mVinlet =5.387 lb - moles 359 ft 3 STP hr 1 lb - moleb g(200 + 460) R 492 R= 2590 ft 3 h6.37Basis: 100 mol C 6 H 14100 mol
The University of Akron - CHEMICAL E - 312
6.36 (cont'd)Circulation rate of dry nitrogen = 5.875 (1 - 0.1324) = = 0182 lb m / h . 5.097 lb - mole h lb - mole 28.02 lb mVinlet =5.387 lb - moles 359 ft 3 STP hr 1 lb - moleb g(200 + 460) R 492 R= 2590 ft 3 h6.37Basis: 100 mol C 6 H 14100 mol
The University of Akron - CHEMICAL E - 312
6.36 b.300 lbm/h wet product 0.2 = 0167 lb m T(l) / lb m . 1 + 0.2 0.833 lb m D / lb mn1 (lb-mole/h)m1 (lbm/h)Dryer0.02 / (102) = 0.0196 lb m T(l) / . 0.9804 lb m D / lb m @ 200OF, y3 (lb-mole T/lb-mole) (1-y3)( lb-mole N2/lb-mole)n3 (lb-mole/h)y1
The University of Akron - CHEMICAL E - 312
6.36 b.300 lbm/h wet product 0.2 = 0167 lb m T(l) / lb m . 1 + 0.2 0.833 lb m D / lb mn1 (lb-mole/h)m1 (lbm/h)Dryer0.02 / (102) = 0.0196 lb m T(l) / . 0.9804 lb m D / lb m @ 200OF, y3 (lb-mole T/lb-mole) (1-y3)( lb-mole N2/lb-mole)n3 (lb-mole/h)y1
The University of Akron - CHEMICAL E - 312
6.35 (cont'd)n2 =8.47 10 -3 kmol Hex1 kmol = 0.0086 kmol 0.984 kmol HexN 2 balance on dryer: n1 = 1 - 0.984 0.0086 = 1376 10 -4 kmol .Antoine Eq.bgSaturation at outlet: y3 =p hex ( 28C ) P=106.88555 - 1175.817 ( 28 + 224.867 )5 ( 760 )= 0.04
The University of Akron - CHEMICAL E - 312
6.35 (cont'd)n2 =8.47 10 -3 kmol Hex1 kmol = 0.0086 kmol 0.984 kmol HexN 2 balance on dryer: n1 = 1 - 0.984 0.0086 = 1376 10 -4 kmol .Antoine Eq.bgSaturation at outlet: y3 =p hex ( 28C ) P=106.88555 - 1175.817 ( 28 + 224.867 )5 ( 760 )= 0.04
The University of Akron - CHEMICAL E - 312
6.34Basis: 500 lb m hr dried leather (L)n1 (lb - moles / h)@130o F, 1 atm n0 (lb - moles dry air / h)@140o F, 1 atm y1 (lb - moles H2 O / lb - mole) (1- y1 )(lb - moles dry air / lb - mole) 500 lb m / h 0.06 lb m H2 O(l) / lbm 0.94 lb m L / lbm0 (lb m
The University of Akron - CHEMICAL E - 312
6.34Basis: 500 lb m hr dried leather (L)n1 (lb - moles / h)@130o F, 1 atm n0 (lb - moles dry air / h)@140o F, 1 atm y1 (lb - moles H2 O / lb - mole) (1- y1 )(lb - moles dry air / lb - mole) 500 lb m / h 0.06 lb m H2 O(l) / lbm 0.94 lb m L / lbm0 (lb m
The University of Akron - CHEMICAL E - 312
6.33n0 (kmol/min wet air) @ 28C, 760 mmHg n1 (kmol/min wet air) @ 80C, 770y1 (mol H2O/mol) (1-y1) (mol dry air/mol) 50% rel. sat. 1500 kg/min wet pulp 0.75 /(1 + 0.75) kg H2O/kg 1/1.75 kg dry pulp/kgy2 (mol H2O/mol) (1-y2) (mol dry air/mol) Tdew point
The University of Akron - CHEMICAL E - 312
6.33n0 (kmol/min wet air) @ 28C, 760 mmHg n1 (kmol/min wet air) @ 80C, 770y1 (mol H2O/mol) (1-y1) (mol dry air/mol) 50% rel. sat. 1500 kg/min wet pulp 0.75 /(1 + 0.75) kg H2O/kg 1/1.75 kg dry pulp/kgy2 (mol H2O/mol) (1-y2) (mol dry air/mol) Tdew point
The University of Akron - CHEMICAL E - 312
6.32 (cont'd)1.2(39.5) kmol/min 39.5 kmol/min, 283K, 35 atm 0.40 mol H2/mol 0.35 mol CO/mol 0.20 mol CO2/mol 0.05 mol CH4/moln1 (kmol/min), 261 K, 35 atmyMeOH sat'd yH2 yCH4 (2% of feed) yCO n2 (kmol/min), liquid xMeOH xCO2 xCH4 (98% of feed) p MeOH 26
The University of Akron - CHEMICAL E - 312
6.32 (cont'd)1.2(39.5) kmol/min 39.5 kmol/min, 283K, 35 atm 0.40 mol H2/mol 0.35 mol CO/mol 0.20 mol CO2/mol 0.05 mol CH4/moln1 (kmol/min), 261 K, 35 atmyMeOH sat'd yH2 yCH4 (2% of feed) yCO n2 (kmol/min), liquid xMeOH xCO2 xCH4 (98% of feed) p MeOH 26
The University of Akron - CHEMICAL E - 312
6.31 (cont'd) When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both to c. increase the fractional condensation. When you decrease Tf, less compression is required to achieve a specified fractional condensation. d.A lower Tf
The University of Akron - CHEMICAL E - 312
6.31 (cont'd) When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both to c. increase the fractional condensation. When you decrease Tf, less compression is required to achieve a specified fractional condensation. d.A lower Tf
The University of Akron - CHEMICAL E - 312
6.31 a.Basis: n0 mol feed gas . S = solvent , G = solvent - free gasn1 (mol) @ Tf (C), P4 (mm Hg) n0 (mol) @ T0 (C), P0 (mm Hg) y0 (mol S/mol) (1-y0) (mol G/mol) Td0 (C) (dew point) y1 [mol S(v)/mol] (sat'd) (1y1) (mol G/mol) n2 (mol S (l)Inlet dew poi
The University of Akron - CHEMICAL E - 312
6.31 a.Basis: n0 mol feed gas . S = solvent , G = solvent - free gasn1 (mol) @ Tf (C), P4 (mm Hg) n0 (mol) @ T0 (C), P0 (mm Hg) y0 (mol S/mol) (1-y0) (mol G/mol) Td0 (C) (dew point) y1 [mol S(v)/mol] (sat'd) (1y1) (mol G/mol) n2 (mol S (l)Inlet dew poi
The University of Akron - CHEMICAL E - 312
6.30 a.Room air - T = 22 C , P = 1 atm , hr = 40% :y1 P = 0.40 p H2O ( 22C ) y1 =( 0.40 )19.827 mm Hg = 0.01044 mol H O760 mm Hg2molSecond sample - T = 50 C , P = 839 mm Hg , saturated:y2 P = p H2 O ( 50C ) y2 =92.51 mm Hg = 0.1103 mol H 2 O mol
The University of Akron - CHEMICAL E - 312
6.30 a.Room air - T = 22 C , P = 1 atm , hr = 40% :y1 P = 0.40 p H2O ( 22C ) y1 =( 0.40 )19.827 mm Hg = 0.01044 mol H O760 mm Hg2molSecond sample - T = 50 C , P = 839 mm Hg , saturated:y2 P = p H2 O ( 50C ) y2 =92.51 mm Hg = 0.1103 mol H 2 O mol
The University of Akron - CHEMICAL E - 312
6.28 (cont'd)Condensation rate:0.84 lb - mol 18.02 lb m 1 ft 3 7.48 gal = 181 gal / min . min lb - mol 62.4 lb m 1 ft 3 23.98 lb - mol 359 ft 3 (STP) 525o R 29.92 in Hg = 9223 ft 3 / min o min 1 lb - mol 492 R 29.8 in HgAir delivered @ 65oF:6.29 Basis
The University of Akron - CHEMICAL E - 312
6.28 (cont'd)Condensation rate:0.84 lb - mol 18.02 lb m 1 ft 3 7.48 gal = 181 gal / min . min lb - mol 62.4 lb m 1 ft 3 23.98 lb - mol 359 ft 3 (STP) 525o R 29.92 in Hg = 9223 ft 3 / min o min 1 lb - mol 492 R 29.8 in HgAir delivered @ 65oF:6.29 Basis
The University of Akron - CHEMICAL E - 312
6.27Basis:12500 L 1 mol 273 K 103000 Pa = 528.5 mol / h h 22.4 L(STP) 293 K 101325 Pa528.5 (mo l/h) @ 20o C, 103 KPa y1 [mol H2O(v)/mol] y1 mol (mol DA/mol) 1 y1 H2 O(v)/ mol (sat'd) (1-y1 ) mol DA/mo ln o (mol/h) @ 35o C, 103 KPa y0 [mol H2O(v)/mol]
The University of Akron - CHEMICAL E - 312
6.26 a. To increase profits and reduce pollution. b. Assume condensation occurs. A=acetonen 1 mol @ To C, 1 at m 1 mo l @ 90o C, 1 atm 0.20 mol A(v)/ mol 0.80 mol N2 /mo l y1 mol A(v)/ mol (sat'd) (1-y1 ) mol N2 /mo l n 2 mol A(l)For cooling water at 20
The University of Akron - CHEMICAL E - 312
6.24 (cont'd)ln(40.0 / 5.00) 5269 5269 A = 5269, B = ln(5.00) + = 19.23 p * = exp(19.23 - ) 1 1 T ( K) 299 - 339 299 At lower explosion limit, y = 0.008 kmol C 9 H 20 / kmol p * ( T ) = yP = (0.008)( 760 mm Hg) -A = = 6.08 mm HgFormula for p*T = 302 K
The University of Akron - CHEMICAL E - 312
6.24 (cont'd)ln(40.0 / 5.00) 5269 5269 A = 5269, B = ln(5.00) + = 19.23 p * = exp(19.23 - ) 1 1 T ( K) 299 - 339 299 At lower explosion limit, y = 0.008 kmol C 9 H 20 / kmol p * ( T ) = yP = (0.008)( 760 mm Hg) -A = = 6.08 mm HgFormula for p*T = 302 K
The University of Akron - CHEMICAL E - 312
6.23 (cont'd) b.Assume no condensation occurs during the compressionV1 (m3 / min) 0.682 kmol/min 0.05 H(v), sat'd 0.95 N2 T1 (oC), 10 atm2.18 kmol/min 0.703 H(v) 0.297 N2 80oC, 1 atmCompressorV0 ( m 3 / min) 2.18 kmol/min 0.703 H(v) 0.297 N2 T0 (oC),
The University of Akron - CHEMICAL E - 312
6.23 (cont'd) b.Assume no condensation occurs during the compressionV1 (m3 / min) 0.682 kmol/min 0.05 H(v), sat'd 0.95 N2 T1 (oC), 10 atm2.18 kmol/min 0.703 H(v) 0.297 N2 80oC, 1 atmCompressorV0 ( m 3 / min) 2.18 kmol/min 0.703 H(v) 0.297 N2 T0 (oC),
The University of Akron - CHEMICAL E - 312
6.22 (cont'd) c.6.53 kmol/h 0.15 C6H14 (v) 0.85 N2 nV (kmol/h) y (kmol C6H14 (v)/kmol), sat'd (1-y) (kmol N2/kmol) T (oC), 2 atmnL (kmol C6H14 (l)/h) 80% of C6H14 in feed80% condensation: Mole balance: Hexane balance: Raoult's law: Antoine equation:6.
The University of Akron - CHEMICAL E - 312
6.22 (cont'd) c.6.53 kmol/h 0.15 C6H14 (v) 0.85 N2 nV (kmol/h) y (kmol C6H14 (v)/kmol), sat'd (1-y) (kmol N2/kmol) T (oC), 2 atmnL (kmol C6H14 (l)/h) 80% of C6H14 in feed80% condensation: Mole balance: Hexane balance: Raoult's law: Antoine equation:6.
The University of Akron - CHEMICAL E - 312
* * 6.21 a. Antoine equation ptol (85o F) = ptol (29.44 o C) = 35.63 mmHg = ptolMole fraction of toluene in gas: y = Toluene displaced = yntotal = =ptol 35.63 mmHg = = 0.0469 lb - mole toluene / lb - mole 760 mmHg PyPV RT 0.0469 lb - mole tollb - mole
The University of Akron - CHEMICAL E - 312
* * 6.21 a. Antoine equation ptol (85o F) = ptol (29.44 o C) = 35.63 mmHg = ptolMole fraction of toluene in gas: y = Toluene displaced = yntotal = =ptol 35.63 mmHg = = 0.0469 lb - mole toluene / lb - mole 760 mmHg PyPV RT 0.0469 lb - mole tollb - mole
The University of Akron - CHEMICAL E - 312
6.19 Liquid H 2 O initially present:25 L 100 kg .1 kmol 18.02 kgSaturation at outlet: y H 2 O = * p H 2 O 25 CbPg=L= 1387 kmol H 2 O l .bg23.76 mm Hg = 0.0208 mol H 2 O mol air 15 760 mm Hg .0.0208 = 0.0212 mol H 2 O mol dry air 1 - 0.020815
The University of Akron - CHEMICAL E - 312
6.19 Liquid H 2 O initially present:25 L 100 kg .1 kmol 18.02 kgSaturation at outlet: y H 2 O = * p H 2 O 25 CbPg=L= 1387 kmol H 2 O l .bg23.76 mm Hg = 0.0208 mol H 2 O mol air 15 760 mm Hg .0.0208 = 0.0212 mol H 2 O mol dry air 1 - 0.020815
The University of Akron - CHEMICAL E - 312
6.17 (cont'd)n1 mol @ 15C, 383.1 mm Hg y1 (mol H2O (v)/mol) (sat'd) (1-y1) (mol dry air/mol) n2 mol H2O (l)0.0258 mols @ 200C, 760 mm Hg 0.200 H2O mol /mol 0.800 mol air/molSaturation Condition: y1 =c.* p H 2 O 15 CbPg = 12.79 mm Hg = 0.03339 mol
The University of Akron - CHEMICAL E - 312
6.17 (cont'd)n1 mol @ 15C, 383.1 mm Hg y1 (mol H2O (v)/mol) (sat'd) (1-y1) (mol dry air/mol) n2 mol H2O (l)0.0258 mols @ 200C, 760 mm Hg 0.200 H2O mol /mol 0.800 mol air/molSaturation Condition: y1 =c.* p H 2 O 15 CbPg = 12.79 mm Hg = 0.03339 mol
The University of Akron - CHEMICAL E - 312
6.16 T = 90 F = 32.2 C , p = 29.7 in Hg = 754.4 mm Hg , hr = 95%Basis: 10 gal water condensed/min ncondensed = 10 gal H 2 O min 1 ft 3 7.4805 gal 62.43 lb m ft31 lb-mol 18.02 lb m= 4.631 lb-mole/minV1 (ft 3 / m in) n1 (lb - m oles / m in)n2 (lb - mo
The University of Akron - CHEMICAL E - 312
6.16 T = 90 F = 32.2 C , p = 29.7 in Hg = 754.4 mm Hg , hr = 95%Basis: 10 gal water condensed/min ncondensed = 10 gal H 2 O min 1 ft 3 7.4805 gal 62.43 lb m ft31 lb-mol 18.02 lb m= 4.631 lb-mole/minV1 (ft 3 / m in) n1 (lb - m oles / m in)n2 (lb - mo
The University of Akron - CHEMICAL E - 312
6.14 (cont'd)Mass of air:0.038 mol H 2 O 18.02 g 0.962 mol dry air 29.0 g + = 28.58 g 1 mol 1 mol 1 mol 22.4 L STP 1 molVolume of air: Density =b g b273.2 + 32.2gK = 25.04 L273.2K28.58 g = 1141 g L . 25.04 LIncrease in T increase in V decrease in d
The University of Akron - CHEMICAL E - 312
6.13 (cont'd)hm = ha =0.0281 = 0.0289 mol H 2 O mol dry air 1 - 0.0281 0.0289 mol H 2 O 18.02 g H 2 O mol dry air = 0.0180 g H 2 O g dry air mol dry air mol H 2 O 29.0 g dry airhp =hm 0.0289 100% = 100% = 86.5% 24.559 [ 759.5 - 24.559] p ( 25.56C ) P
The University of Akron - CHEMICAL E - 312
6.12 a.T1 = 58.3 C , p1 = 755 mm Hg - 747 - 52 mm Hg = 60 mm HgT2b g = 110 C , p = 755 mm Hg - b577 - 222gmm Hg = 400 mm Hg2ln p = a=a +b T Kln p2 p1 1 T2bb g-1 T1g=ln 400 601 110 + 273.2bg-1 58.3+ 273.2= -46614 .b = ln p1 - ln p =a
The University of Akron - CHEMICAL E - 312
6.9a.m=2 =2 F =2+2-2=2.Two intensive variable values (e.g., T & P) must bespecified to determine the state of the system. 1209.6 b. log p MEK = 6.97421 - = 2.5107 p MEK = 10 2.5107 = 324 mm Hg 55 + 216. Since vapor & liquid are in equilibrium p MEK =
The University of Akron - CHEMICAL E - 312
6.6a.p*(mm Hg) =758.9 + hright -hleft -3 42.7 3.1710 34.9 58.9 3.0110-3 78.9 -3 68.3 2.9310 122.9 77.9 2.8510-3 184.9 88.6 2.7610-3 282.9 -3 98.3 2.6910 404.9 105.8 2.6410-3 524.9 H v -51438 K . b.Plot is linear, ln p = - + B ln p = + 19.855 RT TT(C)1
The University of Akron - CHEMICAL E - 312
6.3 (cont'd)ln p (45o C) = -b4151 + 18.49 p = 2310 mm Hg . 45 + 273.2g. 2310 - 234.5 100% = -15% error . 234.5c.p =FG 118.3 - 760IJ b45 - 29.5g + 118.3 = 327.7 mm Hg H 29.5 - 77 Kg327.7 - 234.5 100% = 39.7% error 234.56.4Plot p log scale vs1
The University of Akron - CHEMICAL E - 312
CHAPTER SIX6.1 a.AB: Heat liquid - -V constantBC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T = 100 o C.CD: Heat vapor - -T increases, V increases .b. Point B: Negl
The University of Akron - CHEMICAL E - 312
5.83 Basis: 10 mL C5 H 10 l charged to reactorbgC5 H 10 +15 O 2 5CO 2 + 5H 2 O 210 mL C5 H 10 l n1 (mol C5 H 10 )bgn 3 (mol CO 2 ) n 4 mol H 2 O(v) n 5 (mol N 2 ) 75.3 bar (gauge), Tadn 2 (mol air) 0.21 O 2 0.79 N 2 27 o C, 11.2 L, Po (bar)bgd C
The University of Akron - CHEMICAL E - 312
5.81 (cont'd)Final: Tc ' =c. FG 0104 IJ 986.4 R + FG 0.027 IJ 227.16 R = 830 R H 0131K H 0131 K . . . F 0104 IJ 47.7 atm + FG 0.027 IJ 33.5 atm = 44.8 atm P '= G H 0131K H 0131 K . .o o oT=550 FoTr ' = 122 .dV irideal=Fig. 5.4-2 VPc ' 2.2 ft 3
The University of Akron - CHEMICAL E - 312
5.79H 2 : Tc = (33.3 + 8) K = 41.3 K Pc = (12.8 + 8) atm = 20.8 atm1 - butene: Tc = 419.6 K Pc = 39.7 atm Tr ' = 0.89rTc ' = 0.15(413 K) + 0.85(419.6 K) = 362.8 K . Pc ' = 0.15(20.8 atm) + 0.85(39.7 atm) = 36.9 atm V=U V P ' = 0.27 WFig. 5.4-2 z =
The University of Akron - CHEMICAL E - 312
5.78 Basis: 54.5 kmol CH 3OH hCO + 2H 2 CH 3OHn1 (kmol CO / h) 2n1 (kmol H 2 / h) 644 K 34.5 MPaCatalyst BedCondenserCO, H 254.5 kmol CH 3OH (l ) / ha.n1 =54.5 kmol CH 3OH 1 kmol CO react 1 kmol CO fed = 218 kmol h CO h 1 kmol CH 3OH 0.25 kmol CO
The University of Akron - CHEMICAL E - 312
5.76 CO: Tc = 133.0 K, Pc = 34.5 atm Tc = 0.60 133.0 + 0.40 33 + 8 = 96.2 K H 2 : Tc = 33 K, Pc = 12.8 atm Pc = 0.60 34.5 + 0.40 12.8 + 8 = 29.0 atmU V Wb g b gbbg gTurbine inlet:Tr = 150 + 273.2 96.2 = 4.4 Pr = 2000 psi 29.0 atmbTurbine exit: Tr
The University of Akron - CHEMICAL E - 312
5.74 CH 4 : Tc = 190.7 K , Pc = 458 atm .C 2 H 6 : Tc = 305.4 K , Pc = 48.2 atm C 2 H 4 : Tc = 2831 K , Pc = 50.5 atm .b gb g b gb g b gb g Pseudocritical pressure: P = b0.20gb45.8g + b0.30gb48.2g + b0.50gb50.5g = 48.9 atm U b90 + 273.2gK = 134 Reduced
The University of Akron - CHEMICAL E - 312
5.72 a.For N 2 : Tc = 126.20 K = 227.16o R, Pc = 335 atm .After heater: Tr = 609.7 o R = 2.68 227.16o R z = 1.02 600 psia 1 atm = 1.2 Pr = . 335 atm 14.7 psiaU | | V | | Wn=150 SCFM = 0.418 lb - mole / min 359 SCF / lb - mole zRTn 102 0.418 lb - mole
The University of Akron - CHEMICAL E - 312
5.70 (cont'd) c.143 kmol N 2 .n initial = 0.204 kmol y O = 0.21 kmol O 2 / kmol2143 kmol N 2 .143 kmol N 2 . y1143 kmol N 2 . y2N 2 at 700 kPa gauge = 7.91 atm abs. Pr = 0.236, Tr = 2.36 => z = 0.99Fig 5.4-2n2 = y1 = y2 =P2 V 7.91 atm 5000 L . =
The University of Akron - CHEMICAL E - 312
5.69 a.V 50.0 mL 44.01 g = = 4401 mL / mol . n 5.00 g mol RT 82.06 mL atm 1000 K P= = = 186 atm mol K 440.1 mL / mol V V=b. For CO 2 : Tc = 304.2 K, Pc = 72.9 atm T 1000 K Tr = = = 3.2873 Tc 304.2 KVr ideal =VPc 4401 mL 72.9 atm . mol K = = 1.28 RTc m
The University of Akron - CHEMICAL E - 312
5.65 a. kg / m3 ==dim (kg) (MW)P = RT V (m3 )30 kg kmol 9.0 MPa 10 atm = 69.8 kg m3 m3 atm 1.013 MPa 465 K 0.08206 kmolKFig. 5.4-3b.. Tr = 465 310 = 15rU V P = 9.0 4.5 = 2.0 W z = 0.84=(MW)P 69.8 kg m3 = = 831 kg m3 . zRT 0.845.66 Moles of
The University of Akron - CHEMICAL E - 312
5.63 (cont'd)READ (5, *) T, P IF (T.LT.Q.) STOP R = 0.42747 *R*R/PC*TC*TC B = 0.08664 *R*TC/PC . . M = 0.48508 + W = 155171 - W015613 ALP = 1.+ M 1 - T / TC 0.5 2 .dc bbghigVP = R T / P DO 20 I = 7, 15 V = VP F = R * T/(V B) ALP * A/V/(V + B) P FP
The University of Akron - CHEMICAL E - 312
5.62 (cont'd)mO 2 =32.0 lb m V 2.5 ft 3 MW = = 37.4 lb m 3 2.139 ft / lb - mole lb - mole VIdeal gas gives a conservative estimate. It calls for charging less O2 than the tank can safely hold.c. 1. 2. 3. 4.Pressure gauge is faulty The room temperatur
The University of Akron - CHEMICAL E - 312
5.61 (cont'd) c.W = 53,900 NVhn o (kmol) Vo (m3 ) 1 atm, 25o Cho=> n (kmol) P (atm), 25o Cadd 3.63 kg CO 2ho d(m)d(m)Given T, Vo , h, find d V Initial: n o = o Po = 1 RT V d 2 h 3.63 (kg) Final: V = Vo + , n = no + = o + 0.0825 4 44 (kg / kmol)
The University of Akron - CHEMICAL E - 312
5.60 O 2 : TC = 154.4 K ; PC = 49.7 atm ; = 0.021 ; T = 208.2 K 65 C ; P = 8.3 atm ;bgm = 250 kg h ; R = 0.08206 L atm mol K. SRK constants: a = 138 L2 atm mol 2 ; b = 0.0221 L mol ; m = 0.517 ; = 0.840 SRK equation: f V = a d i dVRTbi - VdV+ bi - P =
The University of Akron - CHEMICAL E - 312
5.58 C 3 H 8 : TC = 369.9 KPC = 42.0 atm 4.26 106 Pa 44.09 kg 1 kmol 1 kmol 103 moldi = 0152 .Specific Volume5.0 m3 75 kg= 2.93 10 -3 m3 molCalculate constantsa= b= 0.42747 0.08664d8.314 m Pa mol Ki b369.9 Kg3 22d8.314 m Pa mol Ki b369.9 Kg =
The University of Akron - CHEMICAL E - 312
5.57 a.van der Waals equation: P =Multiply both sides by V 2 V - b PV 3 - PV 2 b = RTV 2 - aV + ab PV 3 + -Pb - RT V 2 + aV - ab = 0 c 3 = P = 50.0 atm c 2 = -Pb - RT = -50.0 atm 0.0366 L / mol - 0.08206 . c 1 = - a = 133 atm L / mol2 2bgdid V - b
The University of Akron - CHEMICAL E - 312
5.56b g T bC H g = 369.9 K, P = 42.0 atm From Table 5.3 -1 bCH OH g = 0.559, bC H g = 0.152From Table B.1 Tc CH 3OH = 513.2 K, Pc = 78.50 atmc 3 8 c 3 3 8RTc PV B = 1+ B = Bo + B1 RT Pc Vbg0.422 0.422 = 0.083 - = -0.619 1.6 1.6 Tr 373.2K 513.2K 0.4
The University of Akron - CHEMICAL E - 312
5.54 (cont'd)237 m (STP) gFGH 25.3 kmol / hh IJK FGH 22.4 kmol IJK = 18,700 SCMH kmol F 237 kmol / h IJ = 444 kmol / h Reactor effluent flow rate: b 49.4 kmol / hgG H 25.3 kmol / sK F kmol IJ FG 22.4 m (STP) IJ = 9946 SCMH V G 444 H h K H kmol K Process