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

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Unformatted text preview: sumed to have a MW of 300. The K-values for the absorbent oil are rough guesses. The values are compared to those in Example 5.3. Component C1 C2 C3 nC 4 nC 5 Oil K-values at 400 psia: T = 97.5F T = 125oF 6.65 8.0 1.64 2.0 0.584 0.73 0.195 0.26 0.0713 0.098 0.0001 0.0002 T = 150oF 8.8 2.4 0.90 0.34 0.135 0.0003 Absorption and stripping factors are obtained with Eqs. (5-38) and (5-51), respectively, using the above K-values with the entering vapor and liquid flow rates. Using the same procedure as in Example 5.3 and letting: The % absorption of a component in the entering gas = l N / l0 + υ N +1 × 100% The % stripping of the oil in the absorbent = υ 1 oil / l0 oil × 100% The following results are obtained and compared with those of Example 5.3. % absorption for N = 6 stages at 400 psia: Component T = 97.5F T = 125oF T = 150oF C1 3.10 2.58 2.34 C2 12.60 10.31 8.59 C3 35.26 28.24 22.91 nC 4 87.94 74.09 59.32 nC 5 95.24 93.08 88.93 Oil 0.05 0.10 0.15 These results show that increasing temperature decreases the % absorption of gaseous components and increases the % stripping of the absorbent. Exercise 5.18 Subject: Multistage absorption of a light-hydrocarbon gas mixture with n-heptane. Given: VN+1 = 1,000,000 lbmol/day of feed gas containing in mole %, C1 94.9, C2 4.2, C3 0.7, nC4 0.1, nC5 0.1. Absorption at 550 psia and -30oF with N = 10 equilibrium stages and K-values below. Assumptions: Applicability of Kremser's method. Find: Absorbent flow rate, L0 , and component distribution for 50% absorption of C2. Analysis: From Eq. (5-48), φ A C2 = fraction of C 2 not absorbed = 0.50 = Solving and using Eq. (5-38), AC 2 = 0.50 = AC2 − 1 N AC2+1 − 1 L0 L0 = KC 2VN +1 (0.36)(1,000,000) Therefore, L0 = 0.50(0.36)(1,000,000) = 180,000 lbmol/day To compute the distribution of the other components,i, between products V1 and L10 , compute Ai = L0/KiVN+1 and φA for i from Eq. (5-48). Then Eq. (5-54) is used to compute the component flow rates in the exit vapor. Then an overall material balance is used to compute the component flow rates in the exit liquid: l N = l0 + υ N +1 − υ 1 The results are as follows, using a spreadsheet: Flow rates, lbmol/day: Component K-value A l10 φA υn+1 = υ11 υ1 C1 2.85 0.063 0.937 949,000 889,200 59,800 C2 0.36 0.50 0.500 42,000 21,000 21,000 C3 0.066 2.73 0.0000276 7,000 0.2 6,999.8 nC 4 0.017 10.59 0 1,000 0 1,000 nC 5 0.004 45.0 0 1,000 0 1,000 Exercise 5.19 Subject: Stripping of a hydrocarbon liquid with steam at 50 psia and 300oF. Given: In Fig. 5.8(b), LN+1 = 1,000 kmol/h of 0.03% C1, 0.22% C2, 1.82% C3, 4.47% nC4, 8.59 nC5, and 84.87% nC10 , in mol%. V0 = 1,000 kmol/h of superheated steam. Stripper with N = 3 equilibrium stages. K-value for nC10 = 0.20. Other K-values from Fig. 2.8. Assumptions: Applicability of Kremser's equation. Negligible absorption of steam. Find: Compositions and flow rates of exiting stripped liquid and rich gas. Dew point. Analysis: For a stripper, Eq. (5-55) applies, which for no hydro...
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