Unformatted text preview: the given vapor pressure data are fitted to Antoine equations, we obtain:
xA = Exercise 4.15 (continued)
Analysis: (e) continued PAs = exp 18.4854 − 392196
.
T + 230.91 (7) PBs = exp 25.0173 − 8010.6
T + 353.238 (8) Where vapor pressure is in torr and temperature is in oC. Solving, Eqs. (1) to (8),
T, oC
82.5
84.0
86.0
88.0
90.0
92.0
94.0
96.0
98.0
100.0 Ps o f B Ps o f A
KB
KA
xB
yB
αBA
760.0 392.1 1.0000 0.5159 1.000 1.000 1.938
809.5 416.2 1.0651 0.5476 0.874 0.931 1.945
879.9 450.2 1.1578 0.5924 0.721 0.835 1.954
955.7 486.6 1.2575 0.6402 0.583 0.733 1.964
1037.3 525.3 1.3649 0.6912 0.458 0.626 1.975
1125.0 566.6 1.4803 0.7456 0.346 0.513 1.985
1219.3 610.6 1.6043 0.8035 0.245 0.394 1.997
1320.5 657.4 1.7375 0.8650 0.155 0.269 2.009
1429.1 707.2 1.8804 0.9305 0.073 0.138 2.021
1545.6 760.0 2.0336 1.0000 0.000 0.000 2.034 These results are plotted below. Raoult’s law is badly in error when compared to the
experimental data.
For part (b), Raoult's law predicts a bubblepoint vapor with an isopropanol mole fraction of
0.56.
By coincidence, this compares well with the result determined with the experimental data.
For part (c), however, Raoult's law predicts isopropanol mole fractions of 0.28 for the liquid and
0.43 for the vapor. These are drastically different from the values of 0.14 and 0.50, respectively
from the experimental data. Raoult's law can not be used for the isopropanolwater system, for
which it also fails to predict an azeotrope. Exercise 4.15 (continued)
Analysis: (e) continued Exercise 4.16
Subject: Vaporization of mixtures of nhexane (H) and noctane (C) at 1 atm
Given: Txy diagram in Fig. 4.3, and yx diagram in Fig. 4.4. 100 kmol mixture.
Find: Temperature, kmol of vapor, mole fractions of H in liquid and vapor at equilibrium for
various flash conditions.
Analysis: Let zH = mole fraction of nhexane in the feed and Ψ = V/F.
Use inverse leverarm rule as displayed by Line DEF in Fig. 4.3.
The results for parts (a) through (f) are as follows:
Given
(a) zH = 0.5, Ψ = 0.2
(b) zH = 0.4, yH = 0.6
(c) zH = 0.6, xC = 0.7
(d) zH = 0.5, Ψ = 0.0
(e) zH = 0.5, Ψ = 1.0
(f) zH = 0.5, T = 200oF T, o F
196
220
210
188
230
200 V, kmol
20
48.6
73.7
0.0
100
31 yH
0.80
0.60
0.70
0.84
0.50
0.77 xH
0.43
0.21
0.30
0.50
0.14
0.38 Exercise 4.17
Subject: Derivation of equilibrium flash equations for a binary mixture (1, 2).
Given: Eqs. (5), (6), and (3) of Table 4.4.
Find: Derive given equations for x1, x2, y1, y2, and Ψ = V/F.
Analysis: First derive the equation for Ψ = V/F. From Eq. (3), Table 4.4,
z1 1 − K1
1 − z1 1 − K2
+
=0
1 + Ψ K1 − 1
1 + Ψ K2 − 1 (1) Solving Eq. (1) for Ψ, and simplifying,
Ψ= z1 K1 − K2 / 1 − K2 − 1
− z1 1 − K1 − 1 − z1 1 − K2
=
z1 1 − K1 K2 − 1 + 1 − z1 1 − K2 K1 − 1
K1 − 1 (3) Substituting Eq. (3) into Eq. (5) of Table 4.4 and simplifying gives the required equation for x1.
Then use y1 = K1x1 and simplify, followed by x2 = 1  x1 and y2 = 1 y1 . Exercise 4.18
Subject: Conditions for R...
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 Spring '11
 Levicky
 The Land

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