Separation Process Principles- 2n - Seader & Henley - Solutions Manual

Solve for the liquid l from eq 7 repeat this procedure

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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 . K-values 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 Rachford-Rice 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 Rachford-Rice equation. The calculations are summarized in the following table: Assumed P, psia K-values: 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. K-values 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 Rachford-Rice 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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