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Air is sent to a compressor-condenser then a heater

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Air is sent to a compressor-condenser then a heater. The air contains 10.0% water. Process conditions are shown in the flow diagram below. Air, T = 15.6°C T = 100°C 10.0% HO P-3.00 am P-3.00 atm Compressor Heater 12 Condenser T = 90°C V P=1.00 atm , H.O(1) The mole fraction of water in the inlet air is 0.10. . >;=0.10 In the Compressor Condenser, water is condensed and removed as liquid water. The remaining air is sent to the heater. Look up the vapor pressure of water at 15.6°C in a Vapor Pressure of Water table in reference books. PAO (15.6°C) -13.290 mm Hg Since the outlet of the Compressor Condenser is saturated, calculate the mole fraction of water in air, Y P (T) P 13.290 mm Hg 760 mm Hg 3.00 am atu };=0.00583 mol H.O 100 moi-air x 0.90 mol BDA mol air Assumen, is 100 moles of feed. Do a water balance around the Compressor/Condenser 0.10 mol H.O 100 mot mir -0.00583 mol H,0 (W) x, mohair mohair mot air +, mol HOI) Do a dry air balance around the Compressor/Condenser. =(1.00 -0.00583) mol BDA xn, molair mol air mol-air 90 moles BDA=0.99417m, moles BDA 7. = 90.53 moles air 0.10 mol H2O 100 mohair 100 0.00583 mol H/O (V)x90.53 motair mol air mot air +*, mol H,00) 10 mol 1.0 -0.528 mol H, 0+, mol H,O (1) » = 9.47 moles H.0 (1) Calculate the mole fraction of water condensed. fraction condensed" ? 9.47 moles liquid fraction condensed 100 moles feed fraction condensed 0.0947 At the outlet of the heater, the temperature is 100°C. Look up the vapor pressure of water at 100°C in Vapor Pressure of Water tables in reference books. P100°C) = 760.00 mm Hg The concentration and flow rate of water and air at the outlet of the heater is the same as the outlet of the Compressor Condenser since there is no reaction in the heater. Therefore the partial pressure of water at the outlet is: P. =P 760 mm Hg P.=0.00583x3 atau am P. = 13.92 mm Hg Calculate the relative humidity at 100°C, at the outlet of the heater. h P: -X100% p'(T) 13.92 mm Hg ht - -100% 760.00 mm Hg h = 1.749% The ratio of the outlet air volume and inlet air volume can be expressed by a ratio of the ideal gas law. At the outlet, the number of moles is equivalent to fly, or 90.53 moles air. At the inlet, it is 100 moles of air. Adjust the temperatures from "C to absolute, K XIX VTP, V. 90.53 motes air (273.2+100°C) K 1 atma V V 100 moles air (273.2 +90°C) K3 na V = 0.310 X 60 mm Hg P@ 130°C 731. 3092 metly Pet @ 58.3°C Preah = 101.3 kPa Protal = 760 mm Hg Y moles of Chlorobenzene moles of air psat Pr - peat @saturated Condition, (1200) = 25.48988 Y, = (721.3098 760-731.3098 -67 0.0257142 Y @ 583°C; Y = 450 760-60 (Y,-? Ý & Fraction of ? Chlorobenzene condensed 25-48988 - 0.0857142 25.48988 = 99.663% a Pure Chlorobenzene, From Clausius-Clapeyron equation's Inpsat = - AH / + , C = Constant at psat 60 mm Hg T= 58.3°C = 331.45 K. To = 110°C = 383.15 K. at psat = 400 mm m Hg from equis In p sat +G = AH I R Ti AH R Tz sat In Post = t - Alt Alt 1 - 4 In psat Psat AH R In 60 AH V 406 E 4660.055716 1 383.15 331.45 So, In/psat = 4660.055716 osstilt t ) psat at if we assume psat T=331.45K cond p. sat@ 130°C @ 403.15K = ? 10 160 => 4660.055716 403.15 psat ? pasat = 731.3098 mm trg -> @(130°C) = 339085.65 2 ? Page NO DATE = 339 1 kJ - Ailuring thel to beat u eoinly sfeel the amount heat recieved by 20% of the infurt = 0.2X 339085065 I = 67817.12 > Also exit temp. of the stream will be 220°C as to heat transfer would be perfect .. Assuming knit temp a 30°c, from & we have A7= (90-30) 6781712 = mº X/25-5 X. 60 L? Bapy autore ini mo=9.006 x 9 kg the molessbr. 9/20X100 mturing pretend becovered b) Collant @ - 35°C Aturning enit Temp. y qu = '0°C >* Q = 100*48440x%29+35) 60550oJ Portion og kost recipered by autor) z 12.100) 121100 = m0 x 128.5 X. 80 % mo = 12 - 1 kq /kr. recovery 60% NO 2 og Page We are given that 20 moly of infula o condensez is acetone. Atuning we have 100 moles of input fer kouz. mo Cinput) - 200 miles hre 20mobs acetone +80 moles N? o Cooling towar water lou mutes the Inpur - Aulent & Ne Meat lost by the input sheem - Q - mºCp AT At (021: Tect -20°c Contended, mistund in question! Q = 100 X Cp X (90-20) Cpacetone 188 /mol ki ?? = 29-176 Il molk Nitrogen Cp mixture (29-176X0-8) + (125-5X02) J/molk , = 48 44 5/ml K. Q 100 x 48.44 X 70 mot 5 XK hr molk xx [change in temps would units be same in korco Vapor pressure of ethyl alcohol, P, = 20 mmHg Temperature, T, = 8°C = 8+273K = 281K Normal boiling point = 78.4°C = 78.4 +273 K =351.4K Normal pressure, P2 = 760 mmHg Temperature, Tz = 45°C = 45+273 K =318 K Let the vapor pressure at 318 K be = P2 PART A According to the Antoine equation: B logo P=A B C+T or P=10 Here, A, B, and Care Antoine coefficients. For ethanol (between -57 to 80°C), A = 8.20417, B =1642.89, C = 230.3 Air is sent to a compressor-condenser then a heater. The air contains 10.0% water. Process conditions are shown in the flow diagram below. Air, T = 15.6°C T = 100°C 10.0% HO P-3.00 am P-3.00 atm Compressor Heater 12 Condenser T = 90°C V P=1.00 atm , H.O(1) The mole fraction of water in the inlet air is 0.10. . >;=0.10 In the Compressor Condenser, water is condensed and removed as liquid water. The remaining air is sent to the heater. Look up the vapor pressure of water at 15.6°C in a Vapor Pressure of Water table in reference books. PAO (15.6°C) -13.290 mm Hg Since the outlet of the Compressor Condenser is saturated, calculate the mole fraction of water in air, Y P (T) P 13.290 mm Hg 760 mm Hg 3.00 am atu };=0.00583 mol H.O 100 moi-air x 0.90 mol BDA mol air Assumen, is 100 moles of feed. Do a water balance around the Compressor/Condenser 0.10 mol H.O 100 mot mir -0.00583 mol H,0 (W) x, mohair mohair mot air +, mol HOI) Do a dry air balance around the Compressor/Condenser. =(1.00 -0.00583) mol BDA xn, molair mol air mol-air 90 moles BDA=0.99417m, moles BDA 7. = 90.53 moles air 0.10 mol H2O 100 mohair 100 0.00583 mol H/O (V)x90.53 motair mol air mot air +*, mol H,00) 10 mol 1.0 -0.528 mol H, 0+, mol H,O (1) » = 9.47 moles H.0 (1) Calculate the mole fraction of water condensed. fraction condensed" ? 9.47 moles liquid fraction condensed 100 moles feed fraction condensed 0.0947 At the outlet of the heater, the temperature is 100°C. Look up the vapor pressure of water at 100°C in Vapor Pressure of Water tables in reference books. P100°C) = 760.00 mm Hg The concentration and flow rate of water and air at the outlet of the heater is the same as the outlet of the Compressor Condenser since there is no reaction in the heater. Therefore the partial pressure of water at the outlet is: P. =P 760 mm Hg P.=0.00583x3 atau am P. = 13.92 mm Hg Calculate the relative humidity at 100°C, at the outlet of the heater. h P: -X100% p'(T) 13.92 mm Hg ht - -100% 760.00 mm Hg h = 1.749% The ratio of the outlet air volume and inlet air volume can be expressed by a ratio of the ideal gas law. At the outlet, the number of moles is equivalent to fly, or 90.53 moles air. At the inlet, it is 100 moles of air. Adjust the temperatures from "C to absolute, K XIX VTP, V. 90.53 motes air (273.2+100°C) K 1 atma V V 100 moles air (273.2 +90°C) K3 na V = 0.310 X 60 mm Hg P@ 130°C 731. 3092 metly Pet @ 58.3°C Preah = 101.3 kPa Protal = 760 mm Hg Y moles of Chlorobenzene moles of air psat Pr - peat @saturated Condition, (1200) = 25.48988 Y, = (721.3098 760-731.3098 -67 0.0257142 Y @ 583°C; Y = 450 760-60 (Y,-? Ý & Fraction of ? Chlorobenzene condensed 25-48988 - 0.0857142 25.48988 = 99.663% a Pure Chlorobenzene, From Clausius-Clapeyron equation's Inpsat = - AH / + , C = Constant at psat 60 mm Hg T= 58.3°C = 331.45 K. To = 110°C = 383.15 K. at psat = 400 mm m Hg from equis In p sat +G = AH I R Ti AH R Tz sat In Post = t - Alt Alt 1 - 4 In psat Psat AH R In 60 AH V 406 E 4660.055716 1 383.15 331.45 So, In/psat = 4660.055716 osstilt t ) psat at if we assume psat T=331.45K cond p. sat@ 130°C @ 403.15K = ? 10 160 => 4660.055716 403.15 psat ? pasat = 731.3098 mm trg -> @(130°C) = 339085.65 2 ? Page NO DATE = 339 1 kJ - Ailuring thel to beat u eoinly sfeel the amount heat recieved by 20% of the infurt = 0.2X 339085065 I = 67817.12 > Also exit temp. of the stream will be 220°C as to heat transfer would be perfect .. Assuming knit temp a 30°c, from & we have A7= (90-30) 6781712 = mº X/25-5 X. 60 L? Bapy autore ini mo=9.006 x 9 kg the molessbr. 9/20X100 mturing pretend becovered b) Collant @ - 35°C Aturning enit Temp. y qu = '0°C >* Q = 100*48440x%29+35) 60550oJ Portion og kost recipered by autor) z 12.100) 121100 = m0 x 128.5 X. 80 % mo = 12 - 1 kq /kr. recovery 60% NO 2 og Page We are given that 20 moly of infula o condensez is acetone. Atuning we have 100 moles of input fer kouz. mo Cinput) - 200 miles hre 20mobs acetone +80 moles N? o Cooling towar water lou mutes the Inpur - Aulent & Ne Meat lost by the input sheem - Q - mºCp AT At (021: Tect -20°c Contended, mistund in question! Q = 100 X Cp X (90-20) Cpacetone 188 /mol ki ?? = 29-176 Il molk Nitrogen Cp mixture (29-176X0-8) + (125-5X02) J/molk , = 48 44 5/ml K. Q 100 x 48.44 X 70 mot 5 XK hr molk xx [change in temps would units be same in korco Vapor pressure of ethyl alcohol, P, = 20 mmHg Temperature, T, = 8°C = 8+273K = 281K Normal boiling point = 78.4°C = 78.4 +273 K =351.4K Normal pressure, P2 = 760 mmHg Temperature, Tz = 45°C = 45+273 K =318 K Let the vapor pressure at 318 K be = P2 PART A According to the Antoine equation: B logo P=A B C+T or P=10 Here, A, B, and Care Antoine coefficients. For ethanol (between -57 to 80°C), A = 8.20417, B =1642.89, C = 230.3