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

Analysis with two times the absorbent rate of example

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Unformatted text preview: 11.19 12.58 12.60 12.50 12.60 C3 26.09 34.28 35.26 35.30 35.30 nC4 51.46 77.04 87.94 93.34 98.64 nC5 74.29 93.08 95.24 95.33 95.33 Total Gas 36.26 40.32 40.98 41.15 41.33 oil 0.05 0.05 0.05 0.05 0.05 Additional stages have little effect on the absorption of C1, C2, and C3 because their absorption factors are less than one. Exercise 5.14 (continued) Exercise 5.15 Subject: Multistage absorption of a hydrocarbon gas with a high-molecular-weight oil. Given: Hydrocarbon feed gas and absorbent oil of Example 5.3 in Fig. 5.11. Absorber pressure of 400 psia and average temperature of 97.5oF, with K-values at these conditions. Find: Material balance for absorbent rate of 330 lbmol/h instead of the 165 lbmol/h and 3 stages instead of the 6 in Example 5.3. Compare results with Example 5.3 and discuss effect of trading stages for absorbent flow rate. Analysis: With two times the absorbent rate of Example 5.3, double the absorption factors and halve the stripping factors given in Example 5.3. Then, using the same procedure as in that example, the following results are obtained and compared. Component C1 C2 C3 nC 4 nC 5 Oil Total N = 3 and L0 = 330 lbmol/h: N = 6 and L0 = 165 lbmol/h: υ1 , lbmol/h l3 , lbmol/h υ1 , lbmol/h l6 , lbmol/h 150.1 9.9 155.0 5.0 278.1 91.9 323.5 46.5 93.8 146.2 155.4 84.6 1.5 23.6 3.0 22.0 0.3 6.3 0.3 5.5 0.1 328.3 0.1 164.1 523.9 606.2 637.3 327.7 The results show that considerably more absorption takes place when the absorbent rate is doubled and the number of stages is halved. Exercise 5.16 Subject: Multistage absorption of a hydrocarbon gas with a high-molecular-weight oil. Given: Hydrocarbon feed gas and absorbent oil of Example 5.3 in Fig. 5.11. Absorber pressure of 400 psia and average temperature of 97.5oF, with K-values at these conditions. Find: Minimum absorbent rate to reduce the propane flow rate to 155.4 lbmol/h in exit vapor. Analysis: The minimum absorbent rate corresponds to N = ∞ . Use Eq. (5-48) to obtain the absorption factor, A = L/KV, needed for propane. Because A for propane < 1 for N = 6, it will be < 1 for N = ∞ , since the absorbent rate, L, will be lower. From Eq. (5-48), Fraction of propane not absorbed = 155.4/240 = 0.648 = φΑ = (Α − 1)/(ΑΝ+1 −1) When A < 1, ΑΝ+1 = 0, therefore, Eq. (5-48) reduces to: A = 1 - φΑ = 1 - 0.648 = 0.352 From Eq. (5-38), L = AKV = 0.352(0.584)(800) = 165 lbmol/h = minimum rate This is the same L as in Example 5.3 because for φΑ > about 0.5, N above 5 has no effect on A. Exercise 5.17 Subject: Multistage absorption of a hydrocarbon gas with a high-molecular-weight oil. Given: Hydrocarbon feed gas and absorbent oil of Example 5.3 in Fig. 5.11. Absorber pressure of 400 psia and isothermal operation with N = 6 equilibrium stages at: (a) 125oF. (b) 150oF. Assumptions: Applicability of Kremser's method. Find: % absorption of components and % stripping of oil. Analysis: From Fig. 2.8, the following K-values are obtained, where for methane, the oil is as...
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