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

Exercise 21 subject minimum work for separating a

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Unformatted text preview: ponent flow rates, ni , of feed and product 1, in kmol/h. Phase condition; temperature in K; enthalpy, h, in kJ/kmol; and entropy, s, in kJ/kmol-K for feed, product 1, and product 2. Infinite heat sink temperature = T0 = 298.15 K. Find: Minimum work of separation, Wmin , in kJ/h Analysis: From Eq. (4), Table 2.1, Wmin = nb − out From Eq. 2-1, nb in b = h − T0 s For the feed stream (in), n = 30 + 200 + 370 + 350 + 50 = 1,000 kmol/h b = 19,480 - (298.15)(36.64) = 8,556 kJ/kmol For product 1 (out), n = 30 + 192 + 4 + 0 + 0 = 226 kmol/h b = 25,040 - (298.15)(33.13) = 15,162 kJ/kmol For product 2 (out), n = nfeed - nproduct 1 = 1,000 - 226 = 774 kmol/h b = 25,640 - (298.15)(54.84) = 9,289 kJ/kmol From Eq. (4), Table 2.1, Wmin = 226(15,162) + 774(9,289) - 1,000(8,556) = 2,060 kJ/h Exercise 2.2 Subject: Minimum work for separating a mixture of ethylbenzene and xylene isomers. Given: Component flow rates, ni , of feed ,in lbmol/h. Component split fractions for three products, Phase condition; temperature in oF; enthalpy, h, in Btu/lbmol; and entropy, s, in Btu/lbmol-oR for feed and three products. Infinite heat sink temperature = T0 = 560oR. Find: Minimum work of separation, Wmin , in kJ/h Analysis: From Eq. (4), Table 2.1, Wmin = nb − out From Eq. 2-1, nb in b = h − T0 s For the feed stream (in), n = 150 +190 + 430 + 230 = 1,000 lbmol/h b = 29,290 - (560)(15.32) = 20,710 Btu/lbmol For product 1 (out), using Eq. (1-2), n = 150(0.96) + 190 (0.005) + 430(0.004) = 146.7 lbmol/h b = 29,750 - (560)(12.47) = 22,767 Btu/lbmol For product 2 (out), using Eq. (1-2), n = 150(0.04) + 190(0.99) + 430(0.99) + 230(0.015) = 623.3 lbmol/h b = 29,550 - (560)(13.60) = 21,934 Btu/lbmol For product 3 (out), by total material balance, n = 1,000 - 146.7 - 623.3 = 230 lbmol/h b = 28,320 - (560)(14.68) = 20,099 Btu/lbmol From Eq. (4), Table 2.1, Wmin = 146.7(22,767) + 623.3(21,934) + 230(20,099) - 1,000(20,710) = 924,200 Btu/h Exercise 2.3 Subject: Second-law analysis of a distillation column Given: Component flow rates, ni , from Table 1.5 for feed, distillate, and bottoms in kmol/h for column C3 in Figure 1.9. Condenser duty, QC ,= 27,300,00 kJ/h. Phase condition; temperature in K; enthalpy, h, in kJ/kmol; and entropy, s, in kJ/kmol-K for feed, distillate and bottoms. Infinite heat sink temperature = T0 = 298.15 K. Condenser cooling water at 25oC = 298.15 K and reboiler steam at 100oC = 373.15 K. Assumptions: Neglect shaft work associated with column reflux pump. Find: (a) (b) (c) (d) (e) Reboiler duty, QR , kJ/h Production of entropy, ∆Sirr , kJ/h-K Lost work, LW, kJ/h Minimum work of separation, Wmin , kJ/h Second-law efficiency, η Analysis: (a) From Eq. (1), the energy balance for column C3, QR = QC − nh + in nh out = 27,300,000 - 445.5(17,000) + 175.5(13,420) + 270(15,840) = 26,360,000 kJ/h (b) From Eq. (2), the entropy balance for column C3, ∆Sirr = ns + out Q − Ts ns + in Q Ts = 175.5(5.87) + 270(21.22) + 27,300,000 26, 360, 000 − 445.5(25.05) − 298.15 373.15...
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