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

# We must decrease the temperature to obtain the

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Unformatted text preview: e base conditions is: α C 2 , nC4 = KC 2 KnC4 = yC 2 / xC 2 ynC4 / xnC 4 = (nC 2 )V / (nC2 ) L (nnC 4 )V / (nnC4 ) L = (19.0 / 6.0) = 7.39 (7.5 / 17.5) But we need a relative volatility of: α C 2 , nC4 = (nC2 )V / (nC 2 ) L (nnC4 )V / (nnC 4 ) L = (17.5 / 7.5) = 44.3 (1.25 / 23.75) For ideal solutions, where the Raoult's law K-value applies, Eq. (2-21) combined with Eq. (3 in Table 2.3, gives relative volatility as independent of pressure and equal to the ratio of vapor pressures, which depend only on temperature. In general, as the temperature is reduced, the relative volatility increases. Assume that the hydrocarbon mixture, although not an ideal solution, Exercise 4.23 (continued) Analysis: (continued) follows these same trends. Thus, to increase the relative volatility, the pressure has little effect. We must decrease the temperature to obtain the desired relative volatility, and then adjust the pressure to obtain the required compositions. The base case and calculations leading to the desired separation are summarized in the following table: T, oF P, psia K-values: C2 C3 nC4 nC5 α of C2 to nC4 % C2 to vapor % nC4 to vapor Base Case 150 205 4.1 1.5 0.56 0.215 7.3 75.8 30 -70 14.7 4.0 0.46 0.055 0.077 73 -40 14.7 7.5 1.11 0.165 0.028 45 Desired Case -40 16.4 6.7 1.00 0.148 0.025 45 70 4.9 Thus, at -40oF and 16.4 psia, the desired 70% of the ethane is found in the vapor product, with only 5% of the n-butane. The compositions of the vapor and liquid products for the base case and the desired case are as follows: Base Case: Desired Case: y l, lbmol/h x y l, lbmol/h x Component υ, lbmol/h υ, lbmol/h C2 19.0 0.44 6.0 0.11 17.5 0.69 7.5 0.10 C3 13.4 0.31 11.6 0.20 6.4 0.25 18.6 0.25 7.5 0.17 17.5 0.31 1.22 0.05 23.78 0.32 nC4 nC5 3.5 0.08 21.5 0.38 0.22 0.01 24.78 0.33 Total: 43.4 1.00 56.6 1.00 25.34 1.00 74.66 1.00 Exercise 4.24 Subject: Cooling of a reactor effluent with recycle liquid from a partial condensation. Given: Reactor effluent temperature of 1000oF and composition in lbmol/h of 2000 H2, 2000 CH4, 500 benzene, and 100 toluene. Partial condensation conditions or 100oF and 500 psia, and component K-values at these conditions. Two heat exchangers in a recycle loop. Find: (a) Composition and flow rate of vapor leaving flash drum in Fig. 4.38. (b) Proof that vapor flow rate is independent of quench rate. Analysis: (a) Assume that vapor rate is independent of quench rate. Therefore, conduct the flash calculation on just the reactor effluent at the flash drum conditions of temperature and pressure. Use the Rachford-Rice equations (Eqs. (3) and (6), Table 4.4): f {Ψ} = yi = zi 1 − Ki =0 i =1 1 + Ψ Ki − 1 C (1) zi Ki 1 + Ψ Ki − 1 (2) Nonlinear Eq. (1) is solved for Ψ = V/F , followed by calculation of V = ΨF, and then calculations of vapor mole fractions from Eq. (2). The given input for Eq. (1) is: Component Hydrogen Methane Benzene Toluene f, lbmol/h 2,000 2,000 500 100 zi 0.4348 0.4348 0.1087 0.0217 Ki 80 10 0.010 0.004 One method f...
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