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Unformatted text preview: equilibrium. Then by Dalton's law,
s
PPA
nPA =
[(8,000 − 67) + nPA ]
(1)
P
s
PPA
(7,933)
Solving for nPA , with P = 770 torr,
nPA = 770 s
(2)
PPA
1−
770
and the percent recovery = (97  nPA)/97 x 100%
Solving Eqs. (2) and (3),
Ps, torr nPA , lbmol/h % Recovery
in gas
of PA
0.7
7.2
89.3
0.4
4.1
93.9
0.1
1.03
98.5 (3) Exercise 4.66
Subject: Desublimation of anthraquinone (A) from nitrogen (N).
Given: Gas containing 10 mol% A and 90 mol% N at 760 torr and 300oC is cooled to 200oC.
Assumptions: Given vapor pressure data for solid A can be fitted to Antoine equation
Find: % desublimation of A assuming no change in total pressure.
Analysis: Using Polymath, the given vapor pressure data are fitted to the following Antoine
eq.:
PAs = 10
ss
A o 32 .8653− 49 , 762 .2
1327 .24 + T (1) where P is in torr and T is in C. From Eq. (1), P = 1.915 torr at 200oC. Assume 100 moles
of initial gas, containing 90 moles N and 10 moles A. Let nA = moles of A still in the gas at
equilibrium at 200oC. Then by Dalton's law, where pA = PAs ,
s
A pA
Ps
1.915
ntotal = A 90 + nA =
90 + nA
P
P
760
1.915
(90)
nA = 760
= 0.227 mole
Solving for nA ,
1.915
1−
770
and the percent recovery = (10  0.227)/10 x 100% = 97.73%
nA = (2) (3) Exercise 4.67
Subject: Separation of a gas mixture of propylene (A) and propane (P) by adsorption.
Given: T = 25oC, P = 101 kPa, gas containing 0.7 mole A and 1.3 mole P, and S = 0.1 kg silica
gel. Equilibrium adsorption data in Fig. 4.30.
Find: Equilibrium compositions and amounts of adsorbate and final gas.
Analysis: Let y and x be the mole fractions of P in the vapor and adsorbate, respectively. Let G
and W be the moles of equilibrium vapor and adsorbate, respectively. Then a propane material
balance is: 1.3 = Wx + Gy . A total material balance is: 2.0 = W + G. Combining balances,
y= 13 − Wx
.
2 −W (1) To solve, assume W, plot the equation on Fig. 4.30a and find intersection with the equilibrium
curve to find y and x. Read the mmole adsorbate/g adsorbent = moles adsorbate/kg adsorbent
from Fig. 4.30b. Calculate kg adsorbent. If it is not 0.1 kg, repeat with a new guess of W.
For W = 0.195 moles, G = 1.805 moles, y = 0.72  0.108x, y = 0.67, x = 0.41, S = 0.10 kg. Exercise 4.68
Subject: Separation of a gas mixture of propylene (A) and propane (P) by adsorption.
Given: Gas containing 50 mol% A and 50 mol% P, and S = 1,000 lb of less of silica gel/lbmol
of feed gas. Equilibrium adsorption data in Fig. 4.30 apply.
Find: If one equilibrium stage can produce a vapor with 90 mol% P and an adsorbate of 75
mol% A. If not, what separation can be achieved.
Analysis: With a basis of 1 lbmole of feed gas, let y and x be the mole fractions of P in the
equilibrium vapor and adsorbate, respectively. Then we desire, y = 0.9 and x = 0.25. Let G and
W be the lbmol of equilibrium vapor and adsorbate, respectively. Then a propane material
balance is: 0.5 = Wx + Gy . A total material balance is: 1.0 = W + G. Combining balances,
y= 0.5 − Wx
1−W (1) From the equilibr...
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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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