Unformatted text preview: 0.814 0.0133 1.3376 0.801 73.76 75.10 18.625 0.0178 17.19 0.0388 0.0160 0.231 0.794 0.0127 1.2722 0.781 71.94 73.21 20.677 0.0174 19.08 0.0384 0.0156 0.257 0.773 0.0121 1.2084 0.761 70.11 71.32 22.725 0.0169 20.97 0.0379 0.0152 0.284 0.753 0.0114 1.1463 0.741 68.28 69.43 24.771 0.0165 22.86 0.0375 0.0148 0.312 0.732 0.0108 1.0858 0.721 66.46 67.55 26.812 0.0161 24.75 0.0371 0.0144 0.340 0.712 0.0103 1.0271 0.702 64.64 65.66 28.851 0.0156 26.63 0.0367 0.0140 0.369 0.692 0.0097 0.9700 0.682 62.82 63.79 30.886 0.0152 28.50 0.0363 0.0136 0.399 0.671 0.0091 0.9147 0.662 61.00 61.91 32.917 0.0148 30.38 0.0359 0.0132 0.429 0.651 0.0086 0.8610 0.642 59.18 60.04 34.945 0.0143 32.25 0.0354 0.0128 0.461 0.631 0.0081 0.8089 0.623 57.36 58.17 36.970 0.0139 34.12 0.0350 0.0124 0.493 0.610 0.0076 0.7585 0.603 55.55 56.31 38.991 0.0135 35.99 0.0346 0.0120 0.527 0.590 0.0071 0.7098 0.583 53.73 54.44 41.009 0.0130 37.85 0.0342 0.0116 0.562 0.570 0.0066 0.6627 0.564 51.92 52.58 43.024 0.0126 39.71 0.0338 0.0112 0.598 0.550 0.0062 0.6173 0.544 50.11 50.73 45.035 0.0122 41.56 0.0334 0.0108 0.635 0.530 0.0057 0.5735 0.524 48.30 48.88 47.042 0.0117 43.42 0.0329 0.0104 0.673 0.510 0.0053 0.5314 0.505 46.49 47.03 49.047 0.0113 45.27 0.0325 0.0100 0.713 0.490 0.0049 0.4909 0.485 44.69 45.18 51.048 0.0109 47.11 0.0321 Exercise 13.2
Subject: Simple differential batch (Rayleigh) distillation of a mixture of isopropanol and water.
Given: Charge of 40 mol% isopropanol (P) and 60 mol% water (W). Distillation at 1 atm until
70 mol% of the charge is distilled. Vaporliquid equilibrium data in Exercise 7.33.
Assumptions: Perfect mixing in the still. Exiting vapor in equilibrium with the liquid.
Find: Compositions of residue in still and cumulative distillate.
Analysis: Because the vaporliquid equilibrium data indicate that P is the more volatile
component below 40 mol% P in the liquid, base the calculations on P. Use Eq. (133), with x0 =
0.4 and W/W0 = 0.3 (70 mol% vaporized). Thus, Eq. (133) becomes:
x
0.4 dx
= ln 0.3 = −1.204 or
y−x 0.4
x dx
= 1204
.
y−x (1) where x is the mole fraction of P in the residue at 70 mol% distilled and y is in equilibrium with
x.
Equilibrium data from Exercise 7.33 that are in the potential range of this exercise are:
y
x 0.4620
0.0841 0.5242
0.1978 0.5686
0.3496 0.5926
0.4525 A perfect fit of these data to a cubic equation gives:
y = 0.38858 + 1.04439x  2.20358x2 + 1.97113x3 (2) Substitute Eq. (2) into Eq. (1) and, using a spreadsheet, integrate the resulting equation using the
trapezoidal rule with a ∆x increment of 0.01 starting from x = 0.4 with decreasing values of x
until the integral equals 1.204. As an example, using this method for the first four increments,
the integral is equal to ∆x[ 0.5fx=0.40 + fx=0.39 + fx=0.38 + fx=0.37 + 0.5fx=0.36 ], where f = 1/(yx).
The complete spreadsheet is given on the following page. From it, the following result is
obtained: Mole fraction of P in the residue at 70 mol% distilled = 0.0575.
For this value of x, a m...
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 Spring '11
 Levicky
 The Land

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