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Elementary Principles 68
Course: CHEMICAL E 312, Fall 2011School: The University of Akron
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Word Count: 266
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Extractor 4.36 (cont'd) S balance: 87.0 kg S = x2 m2 Extractor oil balance: 13.0 kg oil = y2 m2 Filter S balance: 87.0 kg S = 0.75m3 Filter mass balance: m2 kg = m3 + m4 Oil / hexane ratio in filter cake: b g y3 0.25 - y3 = y2 1 - x2 - y2 Filter oil balance: 13.0 kg oil = y3m3 + y4 m4 Evaporator hexane balance: 1 - y4 m4 = m5 Mixing pt. Hexane balance: m1 + m5 = 300 kg C6 H14 Evaporator oil balance: y4 m4 =...
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Extractor 4.36
(cont'd) S balance: 87.0 kg S = x2 m2
Extractor oil balance: 13.0 kg oil = y2 m2 Filter S balance: 87.0 kg S = 0.75m3 Filter mass balance: m2 kg = m3 + m4 Oil / hexane ratio in filter cake:
b g
y3 0.25 - y3
=
y2 1 - x2 - y2
Filter oil balance: 13.0 kg oil = y3m3 + y4 m4 Evaporator hexane balance: 1 - y4 m4 = m5 Mixing pt. Hexane balance: m1 + m5 = 300 kg C6 H14 Evaporator oil balance: y4 m4 = m6
b.
b
g
Yield =
m6 118 kg oil . = = 0118 kg oil / kg beans fed . 100 100 kg beans fed m 28 kg C6 H14 Fresh hexane feed = 1 = = 0.28 kg C 6 H14 / kg beans fed 100 100 kg beans fed m 272 kg C 6 H14 recycled Recycle ratio = 5 = = 9.71 kg C 6 H14 recycled / kg C 6 H14 fed 28 m1 kg C6 H14 fed
b
g
b b
g
g
c. 4.37
Lower heating cost for the evaporator and lower cooling cost for the condenser.
m lb m dirt 1 98 lb m dry shirts 3 lb m Whizzo
b
g
100 lbm 2 lbm dirt 98 lbm dry shirts Tub Filter
m lb m 4 013 lb m dirt / lb m . 0.87 lb m Whizzo / lb m
m lb m Whizzo 2
b
g
m lb m 3 0.03 lb m dirt / lb m 0.97 lb m Whizzo / lb m
b g
b g
m lb m 5 0.92 lb m dirt / lb m 0.08 lb m Whizzo / lb m
b g
m6 lbm 1 - x lb m dirt / lbm x
m
b g b g blb Whizzo/ lb g
m
Strategy 95% dirt removal m1 ( = 5% of the dirt entering) Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling the chart) m2 , m5 (solves Part (a))
4- 27
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The University of Akron - CHEMICAL E - 312
4.36(cont'd) Extractor S balance: 87.0 kg S = x2 m2Extractor oil balance: 13.0 kg oil = y2 m2 Filter S balance: 87.0 kg S = 0.75m3 Filter mass balance: m2 kg = m3 + m4 Oil / hexane ratio in filter cake:b gy3 0.25 - y3=y2 1 - x2 - y2Filter oil balan
The University of Akron - CHEMICAL E - 312
4.35 (cont'd)Total molar flow rate of liquid feed to stripper and mole fraction of CO2: n3 n3 + n4 = 680 mol / h, x3 = = 0.0478 mol CO 2 / mol n3 + n4e.Scale up to 1000 kg/h (=106 g/h) of product gas:. . bn g = d10 10 g / hib16.28 g / molg = 6142 10 m
The University of Akron - CHEMICAL E - 312
4.35 (cont'd)Total molar flow rate of liquid feed to stripper and mole fraction of CO2: n3 n3 + n4 = 680 mol / h, x3 = = 0.0478 mol CO 2 / mol n3 + n4e.Scale up to 1000 kg/h (=106 g/h) of product gas:. . bn g = d10 10 g / hib16.28 g / molg = 6142 10 m
The University of Akron - CHEMICAL E - 312
4.35a.Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge CO2 to the atmosphere. Absorber function: Separates CO2 from CH4. Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused. The top
The University of Akron - CHEMICAL E - 312
4.35a.Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge CO2 to the atmosphere. Absorber function: Separates CO2 from CH4. Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused. The top
The University of Akron - CHEMICAL E - 312
4.34 (cont'd)Overall mass balance: m1 = 175 + 10m2 + m2 Overall K balance: . 0196m1 = 10m2 + 0.400m2 Production rate of crystals = 10m2U | V | W45% evaporation: 175 kg evaporated min = 0.450m5 W balance around mixing point: 0.804m1 + 0.600m3 = m5 Mass
The University of Akron - CHEMICAL E - 312
4.34 (cont'd)Overall mass balance: m1 = 175 + 10m2 + m2 Overall K balance: . 0196m1 = 10m2 + 0.400m2 Production rate of crystals = 10m2U | V | W45% evaporation: 175 kg evaporated min = 0.450m5 W balance around mixing point: 0.804m1 + 0.600m3 = m5 Mass
The University of Akron - CHEMICAL E - 312
4.33 (cont'd)m 1 vs. x 60.03500 0.03000 0.02500 0.02000 0.01500 0.01000 0.00500 0.00000 0 2000 4000 6000 8000 10000 12000x 6 (kg Cr/kg)m 1 (kg/h)d.Cost of additional capacity installation and maintenance, revenue from additional recovered Cr, antici
The University of Akron - CHEMICAL E - 312
4.33 (cont'd)m 1 vs. x 60.03500 0.03000 0.02500 0.02000 0.01500 0.01000 0.00500 0.00000 0 2000 4000 6000 8000 10000 12000x 6 (kg Cr/kg)m 1 (kg/h)d.Cost of additional capacity installation and maintenance, revenue from additional recovered Cr, antici
The University of Akron - CHEMICAL E - 312
4.33a.m4 kg Cr / hbgm1 kg / hbgm2 kg / hbgTreatment Unit0.0515 kg Cr / kg 0.9485 kg W / kg0.0515 kg Cr / kg 0.9485 kg W / kgb g x b kg Cr / kgg 1 - x b kg W / kggm5 kg / h5 5b g x b kg Cr / kgg 1 - x b kg W / kggm6 kg / h6 6m3 kg / hb
The University of Akron - CHEMICAL E - 312
4.33a.m4 kg Cr / hbgm1 kg / hbgm2 kg / hbgTreatment Unit0.0515 kg Cr / kg 0.9485 kg W / kg0.0515 kg Cr / kg 0.9485 kg W / kgb g x b kg Cr / kgg 1 - x b kg W / kggm5 kg / h5 5b g x b kg Cr / kgg 1 - x b kg W / kggm6 kg / h6 6m3 kg / hb
The University of Akron - CHEMICAL E - 312
4.31 (cont'd) c. B fraction in bottoms: x B = 0100 mol B / mol .Moles of overhead: n1 = 46.0 mol Recovery of toluene:B 3. b1 - x gn 100% = b1 - 010gb54.02g 100% = 97% 0.50b100g 0.50b100gMoles of bottoms: n3 = 54.0 mol4.32a.m3 kg H 2O BypassBasis:
The University of Akron - CHEMICAL E - 312
4.31 (cont'd) c. B fraction in bottoms: x B = 0100 mol B / mol .Moles of overhead: n1 = 46.0 mol Recovery of toluene:B 3. b1 - x gn 100% = b1 - 010gb54.02g 100% = 97% 0.50b100g 0.50b100gMoles of bottoms: n3 = 54.0 mol4.32a.m3 kg H 2O BypassBasis:
The University of Akron - CHEMICAL E - 312
4.31a.2n1 molb gCondenser0.97 mol B / mol 0.03 mol T / moln1 mol100 mol 0.50 mol B / mol 0.50 mol T / molb gn1 mol (89.2% of Bin feed ) 0.97 mol B / mol 0.03 mol T / molb g0.97 mol B / mol 0.03 mol T / molStillb gb y b mol B / molg 1 - y b mo
The University of Akron - CHEMICAL E - 312
4.31a.2n1 molb gCondenser0.97 mol B / mol 0.03 mol T / moln1 mol100 mol 0.50 mol B / mol 0.50 mol T / molb gn1 mol (89.2% of Bin feed ) 0.97 mol B / mol 0.03 mol T / molb g0.97 mol B / mol 0.03 mol T / molStillb gb y b mol B / molg 1 - y b mo
The University of Akron - CHEMICAL E - 312
4.30a.100 kg / h 0.035 kg S / kg 0.965 kg H 2O / kgm3 kg / h1x3 kg S / kgb bg1 - x3 kg H 2O / kgbgg4b g x b kg S / kgg 1 - x b kg H O / kggm4 kg / h4 4 2m10 kg / hbg100.050 kg S / kg 0.950 kg H 2O / kgmw kg H 2O / hbg0100mw kg H 2O
The University of Akron - CHEMICAL E - 312
4.30a.100 kg / h 0.035 kg S / kg 0.965 kg H 2O / kgm3 kg / h1x3 kg S / kgb bg1 - x3 kg H 2O / kgbgg4b g x b kg S / kgg 1 - x b kg H O / kggm4 kg / h4 4 2m10 kg / hbg100.050 kg S / kg 0.950 kg H 2O / kgmw kg H 2O / hbg0100mw kg H 2O
The University of Akron - CHEMICAL E - 312
4.29a.100 mol / h 0.300 mol B / mol 0.250 mol T / mol 0.450 mol X / molb g Column 1 x b mol T / molg 1 - x - x b mol X / molg n b mol / h gx B 2 mol B / molT2 B2 T23n2 mol / hbgn4 mol / hbgColumn 20.940 mol B / mol 0.060 mol T / mol0.020 mo
The University of Akron - CHEMICAL E - 312
4.27 (cont'd) T1 c.P1 h1 R1 Trial 1 2 3 4 5 6 7 8 9 1075 F 150 psig 210 torr 82.4 x4 (kg SO2/kg) 0.10 0.10 0.10 0.10 0.10 0.20 0.20 0.20 0.20 0.20y1 V1 n10.07 kmol SO2/kmol 207 m3/h 95.26 kmol/hy3 (kmol SO2/kmol) V2 (m3/h) n3 (kmol/h) 0.050 0.025 0.0
The University of Akron - CHEMICAL E - 312
4.26 (cont'd) A balance: 1 - 0.0702 95.34 = 1 - 0.00100 n3 n3 = 88.7 kmol minSO2gb g b g balance: b0.0702gb9534g( 64.0 kg / kmol) = b0.00100gb88.7g( 64) + m x .b4 4(1)B balance: 1300 = m4 (1 - x4 ) (2) Solve (1) and (2) simultaneously m4 = 1723 kg /
The University of Akron - CHEMICAL E - 312
4.26 (cont'd)8 unknowns ( n1 , n3 , v1 , m2 , m4 , x4 , y1 , y3 ) 3 material balances 2 analyzer readings 1 meter reading 1 gas density formula 1 specific gravity 0 DFb.Orifice meter calibration: A log plot of V vs. h is a line through the points h1 =
The University of Akron - CHEMICAL E - 312
4.24a.n1 kmol / minbgn3 kmol / min20.0 kg CO 2 / min n2 kmol / minbgbg0.023 kmol CO 2 / kmol0.015 kmol CO 2 / kmol20.0 kg CO 2 kmol = 0.455 kmol CO 2 / min min 44.0 kg CO 2 Overall balance: 0.455 + n2 = n3 CO2 balance: 0.455 + 0.015n2 = 0.023
The University of Akron - CHEMICAL E - 312
4.24a.n1 kmol / minbgn3 kmol / min20.0 kg CO 2 / min n2 kmol / minbgbg0.023 kmol CO 2 / kmol0.015 kmol CO 2 / kmol20.0 kg CO 2 kmol = 0.455 kmol CO 2 / min min 44.0 kg CO 2 Overall balance: 0.455 + n2 = n3 CO2 balance: 0.455 + 0.015n2 = 0.023
The University of Akron - CHEMICAL E - 312
4.22 (cont'd) b.nP =mP 22.813Overall balance: n A + nB =mP 22.813 x m H2 balance: x A n A + x B nB = P P 22.813 nA = mP x B - x P 22.813 x B - x AXB 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50b bg gnB =mP x P - x A 22.813 x B - x
The University of Akron - CHEMICAL E - 312
4.22 (cont'd) b.nP =mP 22.813Overall balance: n A + nB =mP 22.813 x m H2 balance: x A n A + x B nB = P P 22.813 nA = mP x B - x P 22.813 x B - x AXB 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50b bg gnB =mP x P - x A 22.813 x B - x
The University of Akron - CHEMICAL E - 312
4.21 (cont'd) b.500 - 150 RA - 25 mA = 7.78 RA - 44.4 70 - 25 800 - 200 mB - 200 = RB - 20 mB = 15.0 RB - 100 60 - 20 ln 100 - ln 20 ln x - ln 20 = Rx - 4 ln x = 0.2682 Rx + 1923 x = 6.841e 0.2682 Rx . 10 - 4 300 + 44.4 400 + 100 mA = 300 RA = = 44.3, mB
The University of Akron - CHEMICAL E - 312
4.21 (cont'd) b.500 - 150 RA - 25 mA = 7.78 RA - 44.4 70 - 25 800 - 200 mB - 200 = RB - 20 mB = 15.0 RB - 100 60 - 20 ln 100 - ln 20 ln x - ln 20 = Rx - 4 ln x = 0.2682 Rx + 1923 x = 6.841e 0.2682 Rx . 10 - 4 300 + 44.4 400 + 100 mA = 300 RA = = 44.3, mB
The University of Akron - CHEMICAL E - 312
4.19 (cont'd)Assume volume additivity:m1 kgb gm kg m3 400 kg m 3 m3 + = 3 (2) 1000 kg 7440 kg 1480 kgb gSolve (1) and (2) simultaneously m1 = 668 kg H 2O, m3 = 1068 kg suspension v1 =b. c. 4.20 a.668 kgm3 = 0.668 m 3 water fed to tank 1000 kgSpe
The University of Akron - CHEMICAL E - 312
4.19 (cont'd)Assume volume additivity:m1 kgb gm kg m3 400 kg m 3 m3 + = 3 (2) 1000 kg 7440 kg 1480 kgb gSolve (1) and (2) simultaneously m1 = 668 kg H 2O, m3 = 1068 kg suspension v1 =b. c. 4.20 a.668 kgm3 = 0.668 m 3 water fed to tank 1000 kgSpe
The University of Akron - CHEMICAL E - 312
4.18a.100 kg 0.800 kgS / kg 0.200 kg H 2O / kgm1 kg H 2O 85% of entering waterbgbgb g m b kg H Ogm2 kgS3 2gb g Sugar balance: m = 0.800b100g = 80.0 kgS285% drying: m1 = 0.850 0.200 100 = 17.0 kg H 2ObOverall balance: 100 = 17 + 80 + m3 m3 =
The University of Akron - CHEMICAL E - 312
4.18a.100 kg 0.800 kgS / kg 0.200 kg H 2O / kgm1 kg H 2O 85% of entering waterbgbgb g m b kg H Ogm2 kgS3 2gb g Sugar balance: m = 0.800b100g = 80.0 kgS285% drying: m1 = 0.850 0.200 100 = 17.0 kg H 2ObOverall balance: 100 = 17 + 80 + m3 m3 =
The University of Akron - CHEMICAL E - 312
4.16a.4.00 mol H 2SO 4 0.098 kg H 2SO 4 L of solution = 0.323 kg H 2SO 4 / kg solution L of solution mol H 2SO 4 1213 kg solution .v1 Lbgb.bg100 kg 0.200 kg H 2SO 4 / kg 0.800 kg H 2 O / kg SG = 1139 .bg m b kg gv3 L35 unknowns ( v1 , v2 , v3
The University of Akron - CHEMICAL E - 312
4.16a.4.00 mol H 2SO 4 0.098 kg H 2SO 4 L of solution = 0.323 kg H 2SO 4 / kg solution L of solution mol H 2SO 4 1213 kg solution .v1 Lbgb.bg100 kg 0.200 kg H 2SO 4 / kg 0.800 kg H 2 O / kg SG = 1139 .bg m b kg gv3 L35 unknowns ( v1 , v2 , v3
The University of Akron - CHEMICAL E - 312
4.15a.100 kg / s 0.600 kg E / kg 0.050 kg S / kg 0.350 kg H 2 O / kgm kg / sbg3 unknowns ( m, xE , xS ) 3 balances 0 DF0.900 kg E / kg 0100 kg H 2 O / kg .b g x b kg S / kg g 1 - x - x b kg H O / kg gx E kg E / kgS E S 2m kg / sbgb.Overall
The University of Akron - CHEMICAL E - 312
4.15a.100 kg / s 0.600 kg E / kg 0.050 kg S / kg 0.350 kg H 2 O / kgm kg / sbg3 unknowns ( m, xE , xS ) 3 balances 0 DF0.900 kg E / kg 0100 kg H 2 O / kg .b g x b kg S / kg g 1 - x - x b kg H O / kg gx E kg E / kgS E S 2m kg / sbgb.Overall
The University of Akron - CHEMICAL E - 312
4.13 (cont'd) b. Effluent: x = 0.000145 388 1.3645 = 0.494 kg P / kg pb g Product: x = 0.000145b583g = 0.861 kg P / kg Waste: x = 0.000145b140g = 0123 kg P / kg . 0.861b1239g Efficiency = 100% = 95.8% 0.494b2253g1.3645p1.3645pc.Mass balance on puri
The University of Akron - CHEMICAL E - 312
4.13 (cont'd) b. Effluent: x = 0.000145 388 1.3645 = 0.494 kg P / kg pb g Product: x = 0.000145b583g = 0.861 kg P / kg Waste: x = 0.000145b140g = 0123 kg P / kg . 0.861b1239g Efficiency = 100% = 95.8% 0.494b2253g1.3645p1.3645pc.Mass balance on puri
The University of Akron - CHEMICAL E - 312
4.12a.1000 kg / h 0.500 kg CH 3OH / kg 0.500 kg H 2O / kgm kg / hbg2 unknowns ( m, x ) 2 balances 0 DF0.960 kg CH3OH / kg 0.040 kg H 2O / kg673 kg / hg 1 - x b kg H O / kggx kg CH 3OH / kg2bb.Overall balance: 1000 = m + 673 m = 327 kg / hMe
The University of Akron - CHEMICAL E - 312
4.12a.1000 kg / h 0.500 kg CH 3OH / kg 0.500 kg H 2O / kgm kg / hbg2 unknowns ( m, x ) 2 balances 0 DF0.960 kg CH3OH / kg 0.040 kg H 2O / kg673 kg / hg 1 - x b kg H O / kggx kg CH 3OH / kg2bb.Overall balance: 1000 = m + 673 m = 327 kg / hMe
The University of Akron - CHEMICAL E - 312
4.10a.300 gal m1 lb mb gm3 lb m0.750 lb m C 2 H 5OH / lb m 0.250 lb m H 2O / lb mb g0.600 lb m C 2 H 5OH / lb m 0.400 lb m H 2O / lb mV40 gal2 mb g m b lb g4 unknowns ( m1 , m2 ,V40 , m3 ) 2 balances 2 specific gravities 0 DF0.400 lb m C 2 H 5
The University of Akron - CHEMICAL E - 312
4.10a.300 gal m1 lb mb gm3 lb m0.750 lb m C 2 H 5OH / lb m 0.250 lb m H 2O / lb mb g0.600 lb m C 2 H 5OH / lb m 0.400 lb m H 2O / lb mV40 gal2 mb g m b lb g4 unknowns ( m1 , m2 ,V40 , m3 ) 2 balances 2 specific gravities 0 DF0.400 lb m C 2 H 5
The University of Akron - CHEMICAL E - 312
4.7 (cont'd) d.H 2O some CH3COOH CH3COOH H 2O C 4 H9OH C4 H9OH CH3COOHCH 3COOHExtractorDistillation ColumnC4 H 9OH4.8a.120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/eggX-large: 25 broken eggs/min 35 45 unbroken eggs/min Large: n 1 broken egg
The University of Akron - CHEMICAL E - 312
4.7 (cont'd) d.H 2O some CH3COOH CH3COOH H 2O C 4 H9OH C4 H9OH CH3COOHCH 3COOHExtractorDistillation ColumnC4 H 9OH4.8a.120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/eggX-large: 25 broken eggs/min 35 45 unbroken eggs/min Large: n 1 broken egg
The University of Akron - CHEMICAL E - 312
4.5 (cont'd) b. Overall objective: To produce C3H6 from C3H8. Preheater function: Raise temperature of the reactants to raise the reaction rate. Reactor function: Convert C3H8 to C3H6. Absorption tower function: Separate the C3H8 and C3H6 in the reactor e
The University of Akron - CHEMICAL E - 312
4.5 (cont'd) b. Overall objective: To produce C3H6 from C3H8. Preheater function: Raise temperature of the reactants to raise the reaction rate. Reactor function: Convert C3H8 to C3H6. Absorption tower function: Separate the C3H8 and C3H6 in the reactor e
The University of Akron - CHEMICAL E - 312
4.4b.n (mol) 0.500 mol N 2 mol 0.500 mol CH 4 mol100.0 g / s xP xB0.500 n mol N 2 28 g N 2 1 kg = 0.014 n kg N 2 mol N 2 1000 gnE =bgbgc.x E g C2 H 6 g3 8d.b g Rn blb - mole DA sg | 0.21 lb - moleO lb - mole DA U | S | 0.79 lb - mole N lb -
The University of Akron - CHEMICAL E - 312
4.4b.n (mol) 0.500 mol N 2 mol 0.500 mol CH 4 mol100.0 g / s xP xB0.500 n mol N 2 28 g N 2 1 kg = 0.014 n kg N 2 mol N 2 1000 gnE =bgbgc.x E g C2 H 6 g3 8d.b g Rn blb - mole DA sg | 0.21 lb - moleO lb - mole DA U | S | 0.79 lb - mole N lb -
The University of Akron - CHEMICAL E - 312
CHAPTER FOUR4.1 a. b. Continuous, Transient Input Output = Accumulation No reactions Generation = 0, Consumption = 06.00c.dn kg kg dn kg - 3.00 = = 3.00 dt dt s s st=100 m3 1000 kg 1 s . = 333 s 1 m3 3.00 kg4.2a. b. c.Continuous, Steady Statek =
The University of Akron - CHEMICAL E - 312
CHAPTER FOUR4.1 a. b. Continuous, Transient Input Output = Accumulation No reactions Generation = 0, Consumption = 06.00c.dn kg kg dn kg - 3.00 = = 3.00 dt dt s s st=100 m3 1000 kg 1 s . = 333 s 1 m3 3.00 kg4.2a. b. c.Continuous, Steady Statek =
The University of Akron - CHEMICAL E - 312
3.54 (cont'd)60.01.550.00 0.0277 60.00 0.0231K L / MOL MIN : 0.707110. 10.0 20.0 30.0 40.0 50.0 60.0 127. ETC 6 3.5 1.8 1.2 0.92 0.73 0.61 6bgbat 94CgTEMPERATURE (K): 383.15K L / MOL MIN : 1.758bgK0 L / MOL - MIN : 0.2329E + 10b g E bJ / MOL
The University of Akron - CHEMICAL E - 312
3.54 (cont'd)60.01.550.00 0.0277 60.00 0.0231K L / MOL MIN : 0.707110. 10.0 20.0 30.0 40.0 50.0 60.0 127. ETC 6 3.5 1.8 1.2 0.92 0.73 0.61 6bgbat 94CgTEMPERATURE (K): 383.15K L / MOL MIN : 1.758bgK0 L / MOL - MIN : 0.2329E + 10b g E bJ / MOL
The University of Akron - CHEMICAL E - 312
3.54 (cont'd)4 CONTINUE CALL LS (X, Y, NT, SLOPE, INTCPT) KO = EXP INTCPTbg23 510E = -8.314 = SLOPE WRITE (6, 5) KO, E FORMAT (' TEMPERATURE (K): ', F6.2, / * ' TIME CA', /, * ' (MIN) (MOLES)', / * 100 (IX, F5.2, 3X, F7.4, /) FORMAT (' K (L/MOL MIN
The University of Akron - CHEMICAL E - 312
3.54 (cont'd)4 CONTINUE CALL LS (X, Y, NT, SLOPE, INTCPT) KO = EXP INTCPTbg23 510E = -8.314 = SLOPE WRITE (6, 5) KO, E FORMAT (' TEMPERATURE (K): ', F6.2, / * ' TIME CA', /, * ' (MIN) (MOLES)', / * 100 (IX, F5.2, 3X, F7.4, /) FORMAT (' K (L/MOL MIN
The University of Akron - CHEMICAL E - 312
3.53 (cont'd) (d) n1 =26.93 + 27316 . n3 = n1 + n2 = 10.33 kmol min(e) V3 =n2b0.016034gb987.9gb947 60g = 0.8331 kmol CH 26.95 + 27316 . . b0.016034gb9119gb195g = 9.501 kmol air min = b g b b gb gb g g4minn3 T2 + 27316 . 10.33 2251 + 27316 . . = = 3
The University of Akron - CHEMICAL E - 312
3.53 (cont'd) (d) n1 =26.93 + 27316 . n3 = n1 + n2 = 10.33 kmol min(e) V3 =n2b0.016034gb987.9gb947 60g = 0.8331 kmol CH 26.95 + 27316 . . b0.016034gb9119gb195g = 9.501 kmol air min = b g b b gb gb g g4minn3 T2 + 27316 . 10.33 2251 + 27316 . . = = 3
The University of Akron - CHEMICAL E - 312
3.52 (cont'd) (b) ntot = . b500 + 14.696g 35 = 0.308 lb - mole 10.73 b85 + 459.7gmCO =0308 lb - mole 0.30 lb - mole CO 28 lb m CO . = 2.6 lbm CO lb - mole lb - mole CO(c) T =. b3000 + 14.696g 35 - 459.7 = 2733 F 10.73 0.3083.53 (a) T C = a r ohms +
The University of Akron - CHEMICAL E - 312
3.52 (cont'd) (b) ntot = . b500 + 14.696g 35 = 0.308 lb - mole 10.73 b85 + 459.7gmCO =0308 lb - mole 0.30 lb - mole CO 28 lb m CO . = 2.6 lbm CO lb - mole lb - mole CO(c) T =. b3000 + 14.696g 35 - 459.7 = 2733 F 10.73 0.3083.53 (a) T C = a r ohms +
The University of Akron - CHEMICAL E - 312
3.50bT g = 100 C bT g (a) V b mVg = aT b Cg + bb H 2Om AgCl= 455 C5.27 = 100a + b24.88 = 455a + b b = -0.2539 mV V mV = 0.05524T C - 0.2539a = 0.05524 mV Cb gb g T C = 1810V mV + 4.596 .b gb g. . . . (b) 100 mV 136 mV 1856 C 2508 C 3.51 (a)
The University of Akron - CHEMICAL E - 312
3.50bT g = 100 C bT g (a) V b mVg = aT b Cg + bb H 2Om AgCl= 455 C5.27 = 100a + b24.88 = 455a + b b = -0.2539 mV V mV = 0.05524T C - 0.2539a = 0.05524 mV Cb gb g T C = 1810V mV + 4.596 .b gb g. . . . (b) 100 mV 136 mV 1856 C 2508 C 3.51 (a)
The University of Akron - CHEMICAL E - 312
3.47 (cont'd) (c) h = 23 P = 0.02274 23 = 0.523 mm Hg V = 183 0.523132 mL 0.791 g s mL = 104 g s 104 g s 1 mol 58.08 g = 180 mol s .bgb gbg0.5= 132 mL s. . 3.48 (a) T = 85 F + 4597 = 544 R / 18 = 303 K - 273 = 30 C. (b) T = -10 C + 273 = 263 K 18
The University of Akron - CHEMICAL E - 312
3.47 (cont'd) (c) h = 23 P = 0.02274 23 = 0.523 mm Hg V = 183 0.523132 mL 0.791 g s mL = 104 g s 104 g s 1 mol 58.08 g = 180 mol s .bgb gbg0.5= 132 mL s. . 3.48 (a) T = 85 F + 4597 = 544 R / 18 = 303 K - 273 = 30 C. (b) T = -10 C + 273 = 263 K 18