Unformatted text preview: ) Exercise 4.22
Subject: Equilibrium flash of a hydrocarbon mixture.
Given: 100 kmoles of 25 mo1% nC4 , 40 mol% nC5, and 35 mol% nC6 . Kvalues in Fig. 2.8
Assumptions: Amounts are per hour.
Find: Pressure and liquid and vapor compositions for equilibrium at 240oF to recover, in the
liquid phase, 80% of the nC6 in the feed.
Analysis: For 80% recovery of nC6 , the liquid product must contain (0.35)(0.80)(100) = 28
kmole/h of nC6 . Must solve by trial and error by assuming values of pressure to obtain the Kvalues from Fig. 2.8. Then solve the RachfordRice equation (Eq. (3), Table 4.4),
f {Ψ} = zi 1 − Ki
=0
i =1 1 + Ψ Ki − 1
C for Ψ = V/F by a nonlinear solver, such as Newton's method. Compute V from Eq. (4), Table
4.4. Then solve Eqs. (5) and (6), Table 4.4 for the equililbrium vapor and liquid compositions.
Solve for the liquid, L, from Eq. (7). Repeat this procedure until 28 kmol/h of nC6 are found in
the equilibrium liquid as computed from nnC 6 = xnC 6 L , noting that each assumed pressure
L requires an iterative procedure to solve for Ψ from the RachfordRice equation. The calculations
are summarized in the following table:
Assumed P, psia
Kvalues: nC4
nC 5
nC6
V/F
L, kmol/h
x of nC4
nL of nC4 , kmol/h 100
2.40
1.20
0.53
0.706
29.4
0.524
15 110
2.30
1.08
0.50
0.482
51.8
0.461
24 117
2.25
1.00
0.48
0.335
66.5
0.424
28 Therefore, the converged pressure is 117 psia. The equilibrium vapor and liquid compositions in
terms of amounts are:
Component
nC 4
nC 5
nC6
. Vapor flows, kmol/h
13
13
7 Liquid flows, kmol/h
12
27
28 Exercise 4.23
Subject: Equilibrium flash vaporization of a hydrocarbon mixture.
Given: Equimolar mixture of C2, C3, nC4, and nC5. Kvalues from Fig. 2.8 and 2.9
Find: Amounts and compositions of equilibrium liquid and vapor at 150oF and 205 psia.
Conditions of T and P where 70% of C2 and no more than 5% of nC4 is in the vapor.
Analysis: Take as a basis, a feed of 100 lbmol/h. From Fig. 2.8, at 150oF and 205 psia, the Kvalues are as given in the table below. Then solve the RachfordRice equation (Eq. (3), Table
4.4),
f {Ψ} = zi 1 − Ki
=0
i =1 1 + Ψ Ki − 1
C for Ψ = V/F by a nonlinear solver, such as Newton's method. Compute V from Eq. (4), Table
4.4. Then solve Eqs. (5) and (6), Table 4.4 for the equililbrium vapor and liquid compositions.
The calculations are summarized in table below, which also includes other conditions of
T and P to obtain 70% of C2 and no more than 5% of nC4 in the vapor. Thus, we desire
(0.7)(25) = 17.5 lbmol/h of C2 in the vapor and 25  17.5 = 7.5 lbmol/hr of C2 in the liquid. At
the same time we desire no more than (0.05)(25) = 1.25 lbmol/h of nC4 in the vapor,
corresponding to 25  1.25 = 23.75 lbmol/h of nC4 in the liquid. In searching for these
other conditions, we note that at the base conditions, 75.8% of the C2 goes to the vapor, which is
very close to the desired 70%. But 30% of the nC4 also goes to the vapor, which is much higher
than the desired 5%. The relative volatility of C2 to nC4 at th...
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 Spring '11
 Levicky
 The Land

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