Separation Process Principles- 2n - Seader & Henley - Solutions Manual

3 1724 hlb kjkg 1142 1233 1325 1418 1510 1603 1704 hva

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Unformatted text preview: xercise 4.12 (continued) Analysis: (a) (continued) Molecular weights are MA = 78 and MB = 92 For a given temperature, compute saturated liquid-phase mixture enthalpies in kJ/kg of mixture from, xA M A hLA + (1 − xA ) M BhLB hL = (9) xA M A + (1 − xA ) M B hV = Similarly for the vapor, yA M A hVA + (1 − yA ) M BhVB (10) yA M A + (1 − yA ) M B Will have to interpolate and extrapolate given saturated enthalpy data. Liquid enthalpy data are linear with temperature, therefore, it is found that: hLA = 185T − 32 , . hLB = 185T − 34 . (11, 12) Vapor enthalpy data are not quite linear, but fit the following quadratic equations: hVA = 427 + 0.85T + 0.0025T 2 , T, oC xA yA 80.1 85.0 90.0 95.0 100.0 105.0 110.5 1.000 0.775 0.579 0.408 0.259 0.128 0.000 1.000 0.899 0.776 0.630 0.459 0.259 0.000 hVB = 411 + 0.85T + 0.0025T 2 (hL)A , kJ/kg 116.2 125.3 134.5 143.8 153.0 162.3 172.4 (hL)B , kJ/kg 114.2 123.3 132.5 141.8 151.0 160.3 170.4 (hV)A , kJ/kg 511.1 517.3 523.8 530.3 537.0 543.8 551.5 (hV)B , kJ/kg 495.1 501.3 507.8 514.3 521.0 527.8 535.5 (13, 14) hL , kJ/kg 116.2 124.7 133.6 142.5 151.5 160.5 170.4 hV , kJ/kg 511.1 515.4 519.7 523.8 527.7 531.5 535.5 Plots of h in kJ/kg mixture as a function of saturated vapor and liquid mole fractions, and y-x are given on the next page. (b) Take a basis of 1 kmol of 30 mol% A - 70 mol% B feed mixture. Then, kg A = (0.30)(78) = 23.4 kg and kg B = (0.70)(92) = 64.4 kg or 87.8 kg total feed. Use y-x diagram to obtain compositions of vapor and liquid for 50 mol% vaporized. From the equation above Eq. (4-6), the slope of the q-line is [(V/F)-1]/(V/F) = (0.5-1.0)/0.5 = -1. The construction is shown on the y-x diagram, where the intersection with the equilibrium curve gives xA = 0.22 and yA = 0.38. The mass of liquid = (0.22)(0.5)(78) + (0.78)(0.5)(92) = 44.5 kg. The mass of vapor = 87.8 - 44.5 = 43.3 kg. On the h-x-y diagram, Point A is the saturated liquid feed with hL = 150 kJ/kg of feed. Point C is the liquid remaining after 50 mol% vaporization, withhL, = 158 kJ/kg. Since 44.5/87.8 or 0.507 of the feed is left as liquid, this is equivalent to (0.507)(158) = 80 kJ/kg feed. Point D is the vapor, with hV = 540 kJ/kg vaporized. Since 0.493 Exercise 4.12 (continued) Analysis: (b) (continued) of the feed is vaporized, this is equivalent to (0.493)(540) = 266 kJ/kg feed. Therefore, the energy required for partial vaporization = 266 + 80 - 150 = 196 kJ/kg of feed. Point B is the combined vapor and liquid phases after partial vaporization. Point E is condensed vapor as saturated liquid, with an enthalpy of 145 kJ/kg. This is equivalent to (0.493)(145) = 71 kJ/kg of feed. Therefore, the condenser duty = 266 - 71 = 195 kJ/kg feed. Point F is 10oC subcooled condensate, where the enthalpy change from saturation, based on a liquid specific heat of 1.85 kJ/kg-oC, is 1.85(10)(0.493) = 9 kJ/kg feed. Therefore, the condenser duty is now 195 + 9 = 204 kJ/kg feed. Exercise 4.12 (continued) Analysis: (a) Enthalpy – Composition Diagram Exercise 4....
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