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6 Pages

Elementary Principles 244

Course: CHEMICAL E 312, Fall 2011
School: The University of Akron
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Word Count: 198

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S=sucrose, 6.101 Let I=trace impurities, A=activated carbon mS (kg S) m I0 (kg I) R0 (color units / kg S) V (L) Add m A (kg A) mS (kg S) m I (kg I) R (color units / kg S) V (L) Come to equilibrium m A (kg A) m IA (kg I adsorbed) no sucrose is adsorbed solution volume (V) is not affected by addition of the carbon m a. R(color units/kg S) = kCi (kg I / L) = k I (1) V Assume k R = k (Ci 0 - Ci ) = (m I 0...

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S=sucrose, 6.101 Let I=trace impurities, A=activated carbon mS (kg S) m I0 (kg I) R0 (color units / kg S) V (L) Add m A (kg A) mS (kg S) m I (kg I) R (color units / kg S) V (L) Come to equilibrium m A (kg A) m IA (kg I adsorbed) no sucrose is adsorbed solution volume (V) is not affected by addition of the carbon m a. R(color units/kg S) = kCi (kg I / L) = k I (1) V Assume k R = k (Ci 0 - Ci ) = (m I 0 - mI ) V mIA = mI 0 - mI R = kmIA V (2) (3) (4) % removal of color = km IA / V m R x100% x100 = = 100 IA mI 0 R0 kmI 0 / V m Equilibrium adsorption ratio: X i* = IA mA Normalized percentage color removal: = m m % removal ( 3) 100 m IA / m I 0 = 100 IA S = m A / mS m A / mS m A mI 0 m mI 0 = 100X * S X i* = i 100mS mI 0 (1),( 5) (5) Freundlich isotherm X i* = K F Ci mI 0 R = KF ( ) 100mS k = 100mS K F mI 0 k ' R = KF R A plot of ln vs. ln R should be linear: slope = ; 9.500 9.000 y = 0.4504x + 8.0718 intercept = lnK 'F ln v 8.500 8.000 0.000 1.000 2.000 3.000 ln R 6-78
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The University of Akron - CHEMICAL E - 312
6.100 (cont'd)ln X * = 1406 ln p NO2 - 1.965 X * = e -1.965 p 1.406 = 0140 p 1.406 . . NO2 NO2K F = 0.140 (kg NO 2 / 100 kg gel)(mm Hg) -1.406 ; = 1406 .b. Mass of silica gel : mg = * (0.05 m) 2 (1m) 10 3 L 0.75 kg gel1m 3 L= 5.89 kg gelMaximum NO2
The University of Akron - CHEMICAL E - 312
6.100 (cont'd)ln X * = 1406 ln p NO2 - 1.965 X * = e -1.965 p 1.406 = 0140 p 1.406 . . NO2 NO2K F = 0.140 (kg NO 2 / 100 kg gel)(mm Hg) -1.406 ; = 1406 .b. Mass of silica gel : mg = * (0.05 m) 2 (1m) 10 3 L 0.75 kg gel1m 3 L= 5.89 kg gelMaximum NO2
The University of Akron - CHEMICAL E - 312
6.99 (cont'd)50% CCl4 adsorbed nc = 0.500nc 0 = 0662 mol CCl 4 (= nads). Total moles in tank: n tot = n0 - nads = (1985 - 0.0662) mol = 1.919 molPressure in tank. Assume T = T0 and V = V0.P=n tot RT0 (1919)(0.08206)(307) . = atm V0 50.0FG HIJ FG 76
The University of Akron - CHEMICAL E - 312
6.99 (cont'd)50% CCl4 adsorbed nc = 0.500nc 0 = 0662 mol CCl 4 (= nads). Total moles in tank: n tot = n0 - nads = (1985 - 0.0662) mol = 1.919 molPressure in tank. Assume T = T0 and V = V0.P=n tot RT0 (1919)(0.08206)(307) . = atm V0 50.0FG HIJ FG 76
The University of Akron - CHEMICAL E - 312
6.98. a.1.50 L / min 25o C, 1atm, rh = 25% n0 (mol / min) y0 (mol H2 O / mol) (1- y0 ) (mol dry air / mol)n0 =M (g gel) Ma (g H2O)(1 atm)(1.50 L / min) PV = = 0.06134 mol / min RT (0.08206 L atm / mol K)(298 K)r.h.=25%pH 2 O* pH2O (25o C)= 0.25X
The University of Akron - CHEMICAL E - 312
6.98. a.1.50 L / min 25o C, 1atm, rh = 25% n0 (mol / min) y0 (mol H2 O / mol) (1- y0 ) (mol dry air / mol)n0 =M (g gel) Ma (g H2O)(1 atm)(1.50 L / min) PV = = 0.06134 mol / min RT (0.08206 L atm / mol K)(298 K)r.h.=25%pH 2 O* pH2O (25o C)= 0.25X
The University of Akron - CHEMICAL E - 312
6.97(cont'd)Overall composition of feed to Stage 1:b200gb0.30g = 60 kg A h U 500 kg h | 200 - 60 = 140 kg M h V 12% A, 28% M, 60% W 300 kg W h | WFigure 6.6-1 Mass balance Acetone balance: Extract: x1A = 0.095, x1W = 0.880, x1M = 0.025 Raffinate: y1A
The University of Akron - CHEMICAL E - 312
6.97(cont'd)Overall composition of feed to Stage 1:b200gb0.30g = 60 kg A h U 500 kg h | 200 - 60 = 140 kg M h V 12% A, 28% M, 60% W 300 kg W h | WFigure 6.6-1 Mass balance Acetone balance: Extract: x1A = 0.095, x1W = 0.880, x1M = 0.025 Raffinate: y1A
The University of Akron - CHEMICAL E - 312
6.96 a. Basis: 100 kg; A=acetone, W=water, M=MIBKSystem 1: x a,org = 0.375 mol A, x m,org = 0.550 mol M, x w,org = 0.075 mol W x a,aq = 0.275 mol A, x m,aq = 0.050 mol M, x w,aq = 0.675 mol Wmaq,1 = 417 kg . Mass balance: maq ,1 + morg ,1 = 100 Acetone
The University of Akron - CHEMICAL E - 312
6.96 a. Basis: 100 kg; A=acetone, W=water, M=MIBKSystem 1: x a,org = 0.375 mol A, x m,org = 0.550 mol M, x w,org = 0.075 mol W x a,aq = 0.275 mol A, x m,aq = 0.050 mol M, x w,aq = 0.675 mol Wmaq,1 = 417 kg . Mass balance: maq ,1 + morg ,1 = 100 Acetone
The University of Akron - CHEMICAL E - 312
6.93W = water, A = acetone, M = methyl isobutyl ketonex W = 0.20 x A = 0.33 x M = 0.47U | V | WFigure 6.6-1Phase 1: x W = 0.07, x A = 0.35, x M = 0.58 Phase 2: x W = 0.71, x A = 0.25, x M = 0.04Basis: 1.2 kg of original mixture, m1=total mass in pha
The University of Akron - CHEMICAL E - 312
6.93W = water, A = acetone, M = methyl isobutyl ketonex W = 0.20 x A = 0.33 x M = 0.47U | V | WFigure 6.6-1Phase 1: x W = 0.07, x A = 0.35, x M = 0.58 Phase 2: x W = 0.71, x A = 0.25, x M = 0.04Basis: 1.2 kg of original mixture, m1=total mass in pha
The University of Akron - CHEMICAL E - 312
6.92a. P-penicillin; Ac-acid solution; BA-butyl acetate; Alk-alkaline solutionBroth Mixing tank Acid100 kg 0.015 P 0.985 Acm1 (kg BA) Extraction Unit ID.F. analysis: Extraction Unit I 3 unknown (m1, m2p, m3p) 1 balance (P) 1 distribution coefficient
The University of Akron - CHEMICAL E - 312
6.92a. P-penicillin; Ac-acid solution; BA-butyl acetate; Alk-alkaline solutionBroth Mixing tank Acid100 kg 0.015 P 0.985 Acm1 (kg BA) Extraction Unit ID.F. analysis: Extraction Unit I 3 unknown (m1, m2p, m3p) 1 balance (P) 1 distribution coefficient
The University of Akron - CHEMICAL E - 312
6.91a. Basis: 100 g feed 40 g acetone, 60 g H 2 O. A = acetone, H = n - C 6 H 14 , W = water40 g A 60 g W 25C e 1 (g A) 60 g W 25C e 2 (g A) 60 g W100 g H100 g H r 1 (g A)75 g H75 g H r 2 (g A)xA in H phase / xA in W phase = 0.343 x = mass fraction
The University of Akron - CHEMICAL E - 312
6.91a. Basis: 100 g feed 40 g acetone, 60 g H 2 O. A = acetone, H = n - C 6 H 14 , W = water40 g A 60 g W 25C e 1 (g A) 60 g W 25C e 2 (g A) 60 g W100 g H100 g H r 1 (g A)75 g H75 g H r 2 (g A)xA in H phase / xA in W phase = 0.343 x = mass fraction
The University of Akron - CHEMICAL E - 312
6.89 Basis: 100 kg/h.A=oleic acid; C=condensed oil; P=propane95.0 kg C / h m 2 kg A / h m 3 kg A / h100 kg / h 0.05 kg A / kg 0.95 kg C / kgm1 kg P / hm1 kg P / ha. 90% extraction: m3 = (0.09)(0.05)(100 kg / h) = 4.5 kg A / hBalance on oleic acid:
The University of Akron - CHEMICAL E - 312
6.89 Basis: 100 kg/h.A=oleic acid; C=condensed oil; P=propane95.0 kg C / h m 2 kg A / h m 3 kg A / h100 kg / h 0.05 kg A / kg 0.95 kg C / kgm1 kg P / hm1 kg P / ha. 90% extraction: m3 = (0.09)(0.05)(100 kg / h) = 4.5 kg A / hBalance on oleic acid:
The University of Akron - CHEMICAL E - 312
I 6.86 x a =. b0.5150 gg b1101 g molg b0.5150 gg b110.1 g molg+ b100.0 gg b94.10 g molg = 0.00438 mol solute mol2 RTm0Tm =b1 - 0.00523g mol solvent0.00523 mol soluteH mxs I Tm II Tm=I xs II xsII xs=I xsII Tm I Tm= 0.00438mol solute 0.49 C
The University of Akron - CHEMICAL E - 312
I 6.86 x a =. b0.5150 gg b1101 g molg b0.5150 gg b110.1 g molg+ b100.0 gg b94.10 g molg = 0.00438 mol solute mol2 RTm0Tm =b1 - 0.00523g mol solvent0.00523 mol soluteH mxs I Tm II Tm=I xs II xsII xs=I xsII Tm I Tm= 0.00438mol solute 0.49 C
The University of Akron - CHEMICAL E - 312
6.84Moles of diphenyl (DP): Moles of benzene (B): x DP =56.0 g = 0.363 mol 154.2 g mol 550.0 ml 0.879 g mlU | | V 1 mol = 6.19 mol | | 78.11 g W0.363 = 0.0544 mol DP mol 6.19 + 0.363* * p B T = (1 - x DP ) p B T = 0.945 120.67 mm Hg = 114.0 mm Hgbg
The University of Akron - CHEMICAL E - 312
6.84Moles of diphenyl (DP): Moles of benzene (B): x DP =56.0 g = 0.363 mol 154.2 g mol 550.0 ml 0.879 g mlU | | V 1 mol = 6.19 mol | | 78.11 g W0.363 = 0.0544 mol DP mol 6.19 + 0.363* * p B T = (1 - x DP ) p B T = 0.945 120.67 mm Hg = 114.0 mm Hgbg
The University of Akron - CHEMICAL E - 312
6.83 (cont'd)Overall S balance:1000 kg H2SO4 h 32 kg S 98 kg H2SO4 + = m5 (kg / h) (0.96 + (0.04)(0.00079) (kg CaSO4 ) kg 32 kg S 136 kg CaSO4m8 (kg / h) 0.00079 (kg CaSO4 ) 32 kg S kg 136 kg CaSO4 (1') 326.5 = 0.226m5 + 0.000186m8Overall N balance:
The University of Akron - CHEMICAL E - 312
6.83 (cont'd)Overall S balance:1000 kg H2SO4 h 32 kg S 98 kg H2SO4 + = m5 (kg / h) (0.96 + (0.04)(0.00079) (kg CaSO4 ) kg 32 kg S 136 kg CaSO4m8 (kg / h) 0.00079 (kg CaSO4 ) 32 kg S kg 136 kg CaSO4 (1') 326.5 = 0.226m5 + 0.000186m8Overall N balance:
The University of Akron - CHEMICAL E - 312
6.83 (cont'd)Overall H balance :1000 (kg H 2SO4 ) h = 2 kg H 98 kg H 2SO4 kg + 1000 kg HNO3 h 2 kg H 18 kg H 2 O (5) + 1kg H 63 kg HNO3 + mw (kg / h) 2 kg H 18 kg H 2 O 2 kg H 18 kg H 2 O kg0.04m5 (kg / h) 500 X a (kg H 2 O)m8 (kg / h) 500 X a (kg H 2
The University of Akron - CHEMICAL E - 312
6.83 (cont'd)Overall H balance :1000 (kg H 2SO4 ) h = 2 kg H 98 kg H 2SO4 kg + 1000 kg HNO3 h 2 kg H 18 kg H 2 O (5) + 1kg H 63 kg HNO3 + mw (kg / h) 2 kg H 18 kg H 2 O 2 kg H 18 kg H 2 O kg0.04m5 (kg / h) 500 X a (kg H 2 O)m8 (kg / h) 500 X a (kg H 2
The University of Akron - CHEMICAL E - 312
6.83 (cont'd) b. Acid is corrosive to pipes and other equipment in waste water treatment plant. c. Acid feed:1000 kg H 2 SO 4 / h = 0.10 mw = 8000 kg H 2 O / h (2000 + mw ) kg / hOverall S balance:1000 kg H 2 SO 4 h + 32 kg S 98 kg H 2 SO 4 = m5 (kg /
The University of Akron - CHEMICAL E - 312
6.83 (cont'd) b. Acid is corrosive to pipes and other equipment in waste water treatment plant. c. Acid feed:1000 kg H 2 SO 4 / h = 0.10 mw = 8000 kg H 2 O / h (2000 + mw ) kg / hOverall S balance:1000 kg H 2 SO 4 h + 32 kg S 98 kg H 2 SO 4 = m5 (kg /
The University of Akron - CHEMICAL E - 312
6.82(cont'd)Overall mass balance Overall MgSO 4U m ,m V balance W1Diss. tank overall mass balance Diss. tank MgSO 4 balance4U m ,m V W26( MW) MgSO4 = (24.31 + 32.06 + 64.00) = 120.37, ( MW) MgSO4 7H2O = (120.37 + 7 * 18.01) = 246.44Overall MgSO4
The University of Akron - CHEMICAL E - 312
6.82(cont'd)Overall mass balance Overall MgSO 4U m ,m V balance W1Diss. tank overall mass balance Diss. tank MgSO 4 balance4U m ,m V W26( MW) MgSO4 = (24.31 + 32.06 + 64.00) = 120.37, ( MW) MgSO4 7H2O = (120.37 + 7 * 18.01) = 246.44Overall MgSO4
The University of Akron - CHEMICAL E - 312
6.81(cont'd) (c) Liquid feed: (100)(56.39) = 5640 kg / hTo calculate V , we would need to know the density of a 7 wt% aqueous Na2CO3 solution.(d) If T dropped in the filter, more solid NaHCO3 would be recovered and the residual solution would contain le
The University of Akron - CHEMICAL E - 312
6.81(cont'd)O balance (not counting O in the air): 48 16 n1 (0.700)(932) + 100 (0.07)( ) + 100 (0.93)( ) 106 18 48 16 = (n2 w )(16) + n2 c (32) + (m3 + 0.024m4 )( ) + 0.976m4 ( ) 84 18 22.4n1 + 8584 = 16n2 w + 32n2 c + 0.5714(m3 + 0.024m4 ) + 0.8676m4 .
The University of Akron - CHEMICAL E - 312
6.81. Basis : 100 kg liquid feed. Assume Patm=1 atm100 kg Feed 0.07 kg Na 2 CO 3 / kg 0.93 kg H 2 O / kgn 2w (kmol H 2 O )(sat' d) n 2c (kmol CO 2 ) n 2a (kmol Air) 70 o C, 3 atm(absolute)Reactor Reactoren1 (kmol) 0.70 kmol CO 2 / kmol 0.30 kmol Air
The University of Akron - CHEMICAL E - 312
6.81. Basis : 100 kg liquid feed. Assume Patm=1 atm100 kg Feed 0.07 kg Na 2 CO 3 / kg 0.93 kg H 2 O / kgn 2w (kmol H 2 O )(sat' d) n 2c (kmol CO 2 ) n 2a (kmol Air) 70 o C, 3 atm(absolute)Reactor Reactoren1 (kmol) 0.70 kmol CO 2 / kmol 0.30 kmol Air
The University of Akron - CHEMICAL E - 312
6.80 (cont'd). . K balance on dryer: 0.90m1 + 01667 010m1 = 1000 kg h m1 = 1090 kg hMass balance around crystallizer-centrifugebgbgm f + mr = me + m1 + mr me = 4760 - 1090 = 3670 kg h water evaporated95% solution recycled mr =b0.10 1090g kgh not
The University of Akron - CHEMICAL E - 312
6.80 (cont'd). . K balance on dryer: 0.90m1 + 01667 010m1 = 1000 kg h m1 = 1090 kg hMass balance around crystallizer-centrifugebgbgm f + mr = me + m1 + mr me = 4760 - 1090 = 3670 kg h water evaporated95% solution recycled mr =b0.10 1090g kgh not
The University of Akron - CHEMICAL E - 312
6.79 (cont'd)0100(m0 ) min = 1000 kg NaCl / h (m0 ) min = 10,000 kg / min . Evaporation rate: m2 = 9000 kg H 2 O / h Exit solution flow rate: m1 = 0b.m2 [kg H 2 O(v) / h]m0 (kg/h) solution0.100 kg NaCl/kg 0.900 kg H2O/kgm1 (kg/h) sat'd solution @ 80
The University of Akron - CHEMICAL E - 312
6.79 (cont'd)0100(m0 ) min = 1000 kg NaCl / h (m0 ) min = 10,000 kg / min . Evaporation rate: m2 = 9000 kg H 2 O / h Exit solution flow rate: m1 = 0b.m2 [kg H 2 O(v) / h]m0 (kg/h) solution0.100 kg NaCl/kg 0.900 kg H2O/kgm1 (kg/h) sat'd solution @ 80
The University of Akron - CHEMICAL E - 312
6.77 (cont'd)m0 = 2150 kg feed / h Mass balance: m0 = m1 + 1000 kg / h MgSO 4 balance: 0.35m0 = 0.23m1 + 0.488(1000) kg MgSO 4 / h m = 1150 kg soln / h 1 The crystals would yield 0.488 1000 kg / h = 4886.78kg anhydrous MgSO 4 hBasis: 1 lbm feed soluti
The University of Akron - CHEMICAL E - 312
6.77 (cont'd)m0 = 2150 kg feed / h Mass balance: m0 = m1 + 1000 kg / h MgSO 4 balance: 0.35m0 = 0.23m1 + 0.488(1000) kg MgSO 4 / h m = 1150 kg soln / h 1 The crystals would yield 0.488 1000 kg / h = 4886.78kg anhydrous MgSO 4 hBasis: 1 lbm feed soluti
The University of Akron - CHEMICAL E - 312
6.75 (cont'd)Analysis of feed: 2KOH + H 2 SO 4 K 2 SO 4 + 2H 2 Ox0 = 22.4 mL H 2 SO 4 l 5 g feed solnbg1L 10 mL30.85 mol H 2 SO 4 L2 mol KOH 1 mol H 2 SO 456.11 g KOH 1 mol KOH= 0.427 g KOH g feed 60% recovery: 875 ( 0.427 )( 0.60 ) = 224.2 kg KO
The University of Akron - CHEMICAL E - 312
6.75 (cont'd)Analysis of feed: 2KOH + H 2 SO 4 K 2 SO 4 + 2H 2 Ox0 = 22.4 mL H 2 SO 4 l 5 g feed solnbg1L 10 mL30.85 mol H 2 SO 4 L2 mol KOH 1 mol H 2 SO 456.11 g KOH 1 mol KOH= 0.427 g KOH g feed 60% recovery: 875 ( 0.427 )( 0.60 ) = 224.2 kg KO
The University of Akron - CHEMICAL E - 312
6.73 (cont'd)Equilibrium condition: At G1, p H 2S = 0.04 18 atm = 0.072 atm . x3 = p H 2S H H 2S = 0.072 atm = 2.67 10 -3 mole H 2 S mole 27 atm mol fractionb gbgStrategy: Overall H 2 and H 2 S balances n1 , n2 n2 + air flow rate volumetric flow rate
The University of Akron - CHEMICAL E - 312
6.73 (cont'd)Equilibrium condition: At G1, p H 2S = 0.04 18 atm = 0.072 atm . x3 = p H 2S H H 2S = 0.072 atm = 2.67 10 -3 mole H 2 S mole 27 atm mol fractionb gbgStrategy: Overall H 2 and H 2 S balances n1 , n2 n2 + air flow rate volumetric flow rate
The University of Akron - CHEMICAL E - 312
6.72 (cont'd) b. Mole fraction of water in dried gas =yw =n4 2.218 lb - moles W / d lb - moles W(v) = = 1.99 10 -4 4 lb - mole n3 + n4 (2.218 + 1.112 10 ) lb - moles / dHenry's law: ywP = Hwxw ( x w ) max =(199 10 -4 )(500 psia)(1 atm / 14.7 psia) .
The University of Akron - CHEMICAL E - 312
6.72 (cont'd) b. Mole fraction of water in dried gas =yw =n4 2.218 lb - moles W / d lb - moles W(v) = = 1.99 10 -4 4 lb - mole n3 + n4 (2.218 + 1.112 10 ) lb - moles / dHenry's law: ywP = Hwxw ( x w ) max =(199 10 -4 )(500 psia)(1 atm / 14.7 psia) .
The University of Akron - CHEMICAL E - 312
6.72 a. G = dry natural gas, W = watern 3 (lb - mole G / d) n 4 (lb - mole W / d) 10 lb m W / 10 6 SCF gas 90 o F, 500 psiaAbsorbern 7 (lb - mole W / d)4.0 10 6 SCF / d 4 80 = 320 lb m W / d n1 (lb - mole G / d) n 2 [lb - mole W(v) / d]FG lb - mole T
The University of Akron - CHEMICAL E - 312
6.72 a. G = dry natural gas, W = watern 3 (lb - mole G / d) n 4 (lb - mole W / d) 10 lb m W / 10 6 SCF gas 90 o F, 500 psiaAbsorbern 7 (lb - mole W / d)4.0 10 6 SCF / d 4 80 = 320 lb m W / d n1 (lb - mole G / d) n 2 [lb - mole W(v) / d]FG lb - mole T
The University of Akron - CHEMICAL E - 312
6.71 (cont'd)yE3 =* 0.450 pE ( -40 C )yH30.450(0.360) = 2.13 10-4 kmol E kmol P 760 = 1 - yA3 - yE3 = 0.9674 kmol H 2 kmol =Mole balance about still: nc = n p + nr nc = 22.67 + nr A balance about still: 0.550nc = 0.97(22.67) + 0.05nr A balance about
The University of Akron - CHEMICAL E - 312
6.71 (cont'd)yE3 =* 0.450 pE ( -40 C )yH30.450(0.360) = 2.13 10-4 kmol E kmol P 760 = 1 - yA3 - yE3 = 0.9674 kmol H 2 kmol =Mole balance about still: nc = n p + nr nc = 22.67 + nr A balance about still: 0.550nc = 0.97(22.67) + 0.05nr A balance about
The University of Akron - CHEMICAL E - 312
6.71 Basis: 1000 kg/h productnH4 (mol H 2 /h) E = C2 H5 OH ( M = 46.05) A = CH 3 CHO ( M = 44.05) P = 760 mm Hg Fresh feed n0 (mol E/h) nA1 (mol A/h) nE1 (mol E/h) 280C reactor nA2 (mol A/h) nE2 (mol E/h) nH2 (mol H 2/h) nr (mol/h) 0.05 A 0.95 E scrubber
The University of Akron - CHEMICAL E - 312
6.71 Basis: 1000 kg/h productnH4 (mol H 2 /h) E = C2 H5 OH ( M = 46.05) A = CH 3 CHO ( M = 44.05) P = 760 mm Hg Fresh feed n0 (mol E/h) nA1 (mol A/h) nE1 (mol E/h) 280C reactor nA2 (mol A/h) nE2 (mol E/h) nH2 (mol H 2/h) nr (mol/h) 0.05 A 0.95 E scrubber
The University of Akron - CHEMICAL E - 312
6.69 (cont'd) d.Txy diagram (P=1 atm) 85 80 T(oC) 75 70 x 65 60 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Mole fraction of choloroform 1 y yc xcRaoult's law: Tbp = 71o C, y = 0.58 71 - 75.3 T = 100% = -5.7% error in Tbp 75.3 Tactual y y actual = 0.58 - 0.6
The University of Akron - CHEMICAL E - 312
6.69 a. B = benzene, C = chloroform. At 1 atm, (Tbp)B = 80.1oC, (Tbp)C = 61.0oCThe Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1oC when xC = 0 and at 61.0oC when xC = 1. (See solution to part c.)b.Txy Diagram for an Ideal
The University of Akron - CHEMICAL E - 312
6.68 (cont'd) c. (i) x A = 0.34; y A = 0.55(ii) Mole bal.: 1 = nV + nL A bal.: nV = 0.762 mol vapor, nL = 0.238 mol liquid 0.50 = 0.55nV + 0.34nL 76.2 mole% vapor(iii) A( l ) = 0.791 g/cm3 , E(l) = 0.789 g/cm3 l 0.790 g/cm3 (To be more precise, we coul
The University of Akron - CHEMICAL E - 312
6.68 (cont'd) c. (i) x A = 0.34; y A = 0.55(ii) Mole bal.: 1 = nV + nL A bal.: nV = 0.762 mol vapor, nL = 0.238 mol liquid 0.50 = 0.55nV + 0.34nL 76.2 mole% vapor(iii) A( l ) = 0.791 g/cm3 , E(l) = 0.789 g/cm3 l 0.790 g/cm3 (To be more precise, we coul
The University of Akron - CHEMICAL E - 312
6.66 (cont'd)V=3.16 mol vapor 0.08206 L atm 373 K 760 mm Hg - 0.6 L = 97.4 L mol K 750 mm Hg 1 atm (Liquid volume is about 0.6 L)iii) 600 mm Hgv=10 mol vapor 0.08206 L atm 373K 760 mm Hg = 388 L mol K 600 mm Hg 1 atm6.67 a. M = methanoln f (mol) x
The University of Akron - CHEMICAL E - 312
6.66 (cont'd)V=3.16 mol vapor 0.08206 L atm 373 K 760 mm Hg - 0.6 L = 97.4 L mol K 750 mm Hg 1 atm (Liquid volume is about 0.6 L)iii) 600 mm Hgv=10 mol vapor 0.08206 L atm 373K 760 mm Hg = 388 L mol K 600 mm Hg 1 atm6.67 a. M = methanoln f (mol) x
The University of Akron - CHEMICAL E - 312
6.64 (cont'd) e. If the column is long enough, the liquid flowing down eventually approaches equilibrium with the entering gas. At 70oC, the mole fraction of hexane in the exiting liquid in equilibrium with the mole fraction in the entering gas is 4.56x10
The University of Akron - CHEMICAL E - 312
6.64 (cont'd) e. If the column is long enough, the liquid flowing down eventually approaches equilibrium with the entering gas. At 70oC, the mole fraction of hexane in the exiting liquid in equilibrium with the mole fraction in the entering gas is 4.56x10
The University of Akron - CHEMICAL E - 312
6.64 Basis: 100 mol/s gas feed. H=hexane.200 mol oil/s nGN (mol/s) yN (mol H/mol) (1yN) (mol N2/mol) 100 mol/s 0.05 mol H/mol 0.95 mol N2/mol nL1 (mol/s) x1 (mol H/mol) (99.5% of H in feed) (1x1) (mol oil/mol) nL (mol/s) xi1 (mol H/mol) nG (mol/s) yi (mo