Unformatted text preview: 3
98.90
98.50
98.35 Raoult’s law
xA
yA
0.018 0.026
0.276
0.350
0.459
0.547
0.589
0.672
0.729
0.793
0.876
0.910
0.954
0.967
0.992
0.995
1.007
1.005 Experimental
xA
yA
0.025
0.048
0.129
0.205
0.250
0.349
0.354
0.454
0.497
0.577
0.692
0.742
0.843
0.864
0.940
0.948
0.994
0.993 The Raoult’s law values are in very poor agreement with the experimental values. Analysis: (e) (continued) Exercise 4.9 (continued) Analysis: (e) (continued) Exercise 4.9 (continued) Comparison with Experimental Data Exercise 4.10
Subject: Continuous, single stage distillation of A and B to produce a distillate and bottoms.
Given: Saturated liquid feed of 50 mol% A and 50 mol% B fed to a still at 40 mol/h.
Relative volatility = αA,B = 2. Bottoms rate = 30 mol/h
(a) Total condenser with a reflux ratio = 1.
(b) No reflux.
Assumptions: Still is an equilibrium stage.
Find: (a) Composition of the two products.
(b) Composition of the two products.
Analysis: From the definition of the relative volatility,
α A,B = yA x B yA 1 − xA
=
=2
xA y B xA 1 − yA (1) (a) Distillate = D = F  W = 40  30 = 10 mol/h
Material balance for A: 0.5(40) = 20 = yA(10) + xA(30)
(2)
Solving Eqs. (1) and (2) simultaneously by eliminating yA , we obtain:
2
3 xA + 3 xA − 2 = 0 (3) Solving Eq. (3), a quadratic equation, get only one postive root:
xA = 0.4575, xB = 0.5425 for the bottoms
Substitution into Eq. (2), gives,
yA = 0.6275, yB = 0.3725 for the distillate
(b) Note that the solution to Part (a) was independent of the reflux ratio. Accordingly, the
solution to Part (b) is the as for Part (a) Exercise 4.11
Subject: Distillation of an acetone (A)  water (B) mixture that is partially vaporized.
Given: Feed is 57 mol% A and 43 mol% B as a liquid at 125oC and 687 kPa. It is flashed
across a valve to the column pressure of 101.3 kPa, with a resulting temperature of 60oC.
Vaporliquid equilibrium data at column pressure. Enthalpy data at column conditions.
Compositions of the distillate and bottoms.
Assumptions: Feed is at equilibrium downstream of the feed valve. Will have to check if feed
valve operates adiabatically. Given heat capacities are for the liquid and are constant. Heats of
vaporization are constant. No effect of pressure on enthalpy.
Find: Mole ratio of liquid to vapor in the feed downstream of the valve. Construct an Hxy
diagram.
Analysis: From the equilibrium data, at 60oC, xA = 0.50 and yA = 0.85 .
Take a basis of F = feed rate = 1 kmol/s.
Total material balance around feed valve: F = 1 = V + L
(1)
Acetone material balance around feed valve: 0.57(1) = 0.85V + 0.50L
Solving Eqs. (1) and (2) simultaneously,
V = 0.2 kmol/s and L = 0.8 kmol/s (2) Therefore, after the valve, moles L/moles V = 0.8/0.2 = 4
Now check whether valve is operating adiabatically.
Enthalpy of liquid entering valve = 0 (as given)
Enthalpy of feed after the valve, using given enthalpies =
27,200(0.2) + (5,270)(0.8) = 1224 kJ/s
Therefore, the enthalpy increases across the valve by 1224 kJ/s
To construct an enthalpy diagram for 1 atm pressure, take as an enthalpy datum, A and B as
liquids a...
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 Spring '11
 Levicky
 The Land

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