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

# The bottoms flow rate from the stripper b 993 lbmolh

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Unformatted text preview: tes and compositions of the exiting streams by the group (Kremser) method. Analysis: Apply Kremser's method, with stripping factors, S = KV/L , computed for L = entering feed = 1000 kmol/h, V = entering steam rate = 100 kmol/h, and K-values from Fig. 2.8 at the average of the two entering temperatures, (250 + 300)/2 = 275oF and a pressure of 50 psia. The feed gas composition and the resulting K-values and values of S are as follows: Component C1 C2 C3 nC 4 nC 5 nC10 Total: f, kmol/h 0.3 2.2 18.2 44.7 85.9 848.7 1000.0 K-value 67 26 12.2 6.0 2.85 S = KV/L 6.7 2.6 1.22 0.60 0.285 The fraction of a gas component not stripped, φS, is given by the Kremser Eq. (5-50), and by material balance the amounts stripped and not stripped follow: S −1 φ S = N +1 and l1 = φ S f and v3 = f − l1 S −1 The results are as follows: Component C1 C2 C3 nC 4 nC 5 Total: φS 0.0028 0.036 0.181 0.46 0.72 l1, kmol/h 0.0 0.1 3.3 20.6 61.8 85.8 The exiting gas is 65.5 + 100 = 165.5 kmol/h. The exiting liquid is 85.8 + 848.7 = 934.5 kmol/h. v3, kmol/h 0.3 2.1 14.9 24.1 24.1 65.5 Exercise 9.26 Subject: Liquid-liquid extraction of a hydrocarbon mixture by diethylene glycol (DEG) by the group (Kremser) method. Given: 100 kmol/h of an equimolar mixture of benzene (B), toluene (T), n-hexane (C6), and n-heptane (C7). Extraction at 150oC with 300 kmol/h of DEG. Extractor has 5 equilibrium stages. Distribution coefficients given below. Find: Flow rates and compositions of extract and raffinate. Analysis: Treat the problem like a stripper. Apply Kremser's method, with extraction factors, Eq. (9-39), E = KDV/L , computed for L = entering feed = 100 kmol/h, V = entering DEG solvent rate = 300 kmol/h, and given values of KD = mole fraction in extract/mole fraction in raffinate. The extract is like the vapor in stripping and the raffinate is like the liquid in stripping. The feed composition and the K-values and values of S are as follows: Component Benzene Toluene n-Hexane n-Heptane Total: f, kmol/h 25 25 25 25 100 KD 0.33 0.29 0.050 0.043 E = KDV/L 0.99 0.87 0.15 0.129 The fraction of a feed component not extracted, φE , is given by the Kremser Eq. (9-44), and by material balance the amounts stripped and not stripped follow: E −1 φ E = N +1 and l1 = φ E f and v3 = f − l1 E −1 The results are as follows: Component Benzene Toluene n-Hexane n-Heptane Total: φE 0.177 0.230 0.850 0.871 l1, kmol/h 4.3 5.8 21.2 21.8 53.1 v3, kmol/h 20.7 19.2 3.8 3.2 46.9 Now compute the transfer of DEG to the raffinate using Eqs. (9-40) and (9-43), using KD = 30 U = L/KDV = 100/[30(300)] = 0.0111. φU = 0.9889. Therefore 0.9889(300) = 296.7 kmol/h of DEG leaves in the extract and 300 - 296.7 = 3.3 kmol/h in the raffinate. The exiting extract flow rate = 46.9 + 296.7 = 343.6 kmol/h The exiting raffinate flow rate = 53.1 + 3.3 = 56.4 kmol/h. Exercise 9.27 Subject: Reboiled stripping of a normal paraffin hydrocarbon feed liquid by the group method. Given: Feed liquid at 39oF and 300 psia and of composition giv...
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