Unformatted text preview: xercise 4.12 (continued) Analysis: (a) (continued)
Molecular weights are MA = 78 and MB = 92
For a given temperature, compute saturated liquidphase 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 yx 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 yx diagram to obtain compositions of vapor and liquid for 50 mol% vaporized.
From the equation above Eq. (46), the slope of the qline is [(V/F)1]/(V/F) = (0.51.0)/0.5 = 1.
The construction is shown on the yx 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 hxy 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/kgoC, 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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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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