ch02 - Exercise 2.1 Subject Minimum work for separating a...

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Exercise 2.1 Subject: Minimum work for separating a hydrocarbon stream. Given: Component flow rates, n i , 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 = T 0 = 298.15 K. Find: Minimum work of separation, W min , in kJ/h Analysis: From Eq. (4), Table 2.1, W nb nb min out in = - ° ° From Eq. 2-1, b h T s = - 0 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 = n feed - n product 1 = 1,000 - 226 = 774 kmol/h b = 25,640 - (298.15)(54.84) = 9,289 kJ/kmol From Eq. (4), Table 2.1, W min = 226(15,162) + 774(9,289) - 1,000(8,556) = 2,060 kJ/h
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Exercise 2.2 Subject: Minimum work for separating a mixture of ethylbenzene and xylene isomers. Given: Component flow rates, n i , of feed ,in lbmol/h. Component split fractions for three products, Phase condition; temperature in o F; enthalpy, h , in Btu/lbmol; and entropy, s , in Btu/lbmol- o R for feed and three products. Infinite heat sink temperature = T 0 = 560 o R. Find: Minimum work of separation, W min , in kJ/h Analysis: From Eq. (4), Table 2.1, W nb nb min out in = - ° ° From Eq. 2-1, b h T s = - 0 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, W min = 146.7(22,767) + 623.3(21,934) + 230(20,099) - 1,000(20,710) = 924,200 Btu/h
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Exercise 2.3 Subject: Second-law analysis of a distillation column Given: Component flow rates, n i , from Table 1.5 for feed, distillate, and bottoms in kmol/h for column C3 in Figure 1.9. Condenser duty, Q C ,= 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 = T 0 = 298.15 K. Condenser cooling water at 25 o C = 298.15 K and reboiler steam at 100 o C = 373.15 K. Assumptions: Neglect shaft work associated with column reflux pump. Find: (a) Reboiler duty, Q R , kJ/h (b) Production of entropy, Δ S irr , kJ/h-K (c) Lost work, LW, kJ/h (d) Minimum work of separation, W min , kJ/h (e) Second-law efficiency, η Analysis: (a) From Eq. (1), the energy balance for column C3, in out 27,300,000 - 445.5(17,000) + 175.5(13,420) + 270(15,840) 26,36 0,000 kJ/h C R Q nh nh Q = - + = = ° ° (b) From Eq. (2), the entropy balance for column C3, out i rr n i 27,300,000 26,360,000 175.5(5.87) + 270(21.22) + 445.5(25.05 16,520 kJ/h-K ) 298.15 373.15 s s Q Q ns ns T T S ± ² ± ² = + - + ³ ´ ³ ´ µ µ - Δ = = - ° ° (c) From Eq. (2-2), LW = T 0 Δ S irr = 298.15(16,520) = 4,925,000 kJ/h (d) Combining Eqs. (3) and (4) of Table 2.1, 0 0 mi st am cw n e 1 1 LW 298.15 298.15 26,360,000 1 27,300,000 1 4,925,000 373.15 298.15 373,000 kJ/h R C T T Q Q T W T ± ² ± ² = - - - - ³ ´ ³ ´ µ µ ± ² ± ² = - - - - ³ ´ ³ ´ µ µ = (e) From Eq. (5), Table 2.1, η = 373,000/(4,925,000 + 373,000) = 0.0704 = 7.04%
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Exercise 2.4 Subject: Second-law analysis for a membrane separation of a gas mixture Given: Component flow rates in lbmol/h for the feed. Permeate of 95 mol% H 2 and 5 mol% CH 4 . Separation factor, SP, for H 2 relative to CH 4 , of 47. Phase condition; temperature in o F; enthalpy, h , in Btu/lbmol; and entropy, s , in Btu/lbmol- o R for the feed, permeate, and retentate. Infinite heat sink temperature = T 0 = 539.7 o R.
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