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### Ch08

1. 1. CHAPTER EIGHT 8.1 a. U (T ) = 25.96T + 0.02134T 2 J / mol U (0 o C) = 0 J / mol U (100 o C) = 2809 J / mol Tref = 0 o C (since U(0 o C) = 0) b. We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to U (100 o C) . c. Q − W = ΔU + ΔE k + ΔE p ΔE k = 0, ΔE p = 0, W = 0 Q = ΔU = (3.0 mol)[(2809 − 0) J / mol] = 8428 J ⇒ 8400 J F ∂U I d. Cv = GH ∂T JK V = dU dT = [25.96 + 0.04268T ] J / (mol⋅ o C) F I z z OP T2 100 100 ΔU = Cv (T )dT = GG (25.96 + 0.04268T )dT = 25.96T + 0.04268 T2 JJ J / mol T1 0 H 2 QP 0 K ΔU = (3.0 mol) ⋅ ΔU (J / mol) = (3.0 mol) ⋅ [25.96(100 − 0) + 0.02134(100 2 − 0)] (J / mol) = 8428 J ⇒ 8400 J 8.2 a. b g Cv = C p − R ⇒ Cv = 35.3 + 0.0291T [J / (mol⋅° C)] − 8.314 [J / (mol ⋅ K)] 1 K 1° Cb gb g ⇒ Cv = 27.0 + 0.0291T [J / (mol⋅° C)] 100 T2 ⎤ 100 ΔH = ∫ C p dT = 35.3T ]25 + 0.0291 ⎥ = 2784 J mol 100 b. ˆ 25 2 ⎦ 25 z z z 100 100 100 c. ΔU = Cv dT = C p dT − b gb RdT = ΔH − RΔT = 2784 − 8.314 100 − 25 = 2160 J mol g 25 25 25 d. H is a state property 8.3 a. Cv [ kJ / (mol⋅ o C)] = 0.0252 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2 . PV (2.00 atm)(3.00 L) n= = = 0.245 mol RT (0.08206[atm ⋅ L / (mol ⋅ K)](298 K) z 1000 Q1 = nΔU 1 = (0.245 mol) ⋅ 0.0252 dT ( kJ / mol) = 6.02 kJ 25 z 1000 Q2 = nΔU 2 = (0.245) ⋅ [0.0252 + 1547 × 10 −5 T ] dT = 7.91 kJ . 25 z 1000 Q3 = nΔU 3 = (0.245) ⋅ [0.0252 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2 ] dT = 7.67 kJ . 25 6.02 - 7.67 % error in Q1 = × 100% = −215% . 7.67 7.91- 7.67 % error in Q2 = × 100% = 313% . 7.67 8-1
2. 2. 8.3 (cont’d) b. C p = Cv + R C p [ kJ / (mol⋅ o C)] = (0.0252 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2 ) + 0.008314 . = 0.0335 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2 . z T2 Q = ΔH = n C P dT T1 z 1000 = (0.245 mol) ⋅ [0.0335 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2 ] dT [kJ / (mol⋅ o C)] = 9.65 × 10 3 J . 25 Piston moves upward (gas expands). c. The difference is the work done on the piston by the gas in the constant pressure process. 8.4 a. (C ) p C H (l ) 6 6 ( 40 C ) = 0.1265 + 23.4 × 10 ( 40 ) = 0.1360 [kJ/(mol ⋅ K)] o −5 dC i b g b40° Cg = 0.07406 + 32.95 × 10 b40g − 25.20 × 10 b40g b g −5 −8 + 77.57 × 10 −12 40 2 3 b. p C H v 6 6 = 0.08684 [kJ / (mol⋅ o C)] c. dC i b g b313 Kg = 0.01118 + 1095 × 10 b313g − 4.891 × 10 b313g p C s . −5 2 −2 = 0.009615 [ kJ / (mol ⋅ K)] d. ΔHC6 H6 bv g = 0.07406T + 32.95 × 10 −5 2 2520 × 10 −8 3 77.57 × 10 −12 4 T − . T + OP 300 = 3171 kJ mol 2 3 4 T PQ 40 . e. ΔH C b sg = 0.01118T + 1095 × 10 −5 2 . T + 4.891 × 10 2 T −1 OP 573 = 3.459 kJ / mol 2 PQ 313 8.5 H 2 O (v, 100 o C, 1 atm) → H 2 O (v, 350 o C, 100 bar) a. H = 2926 kJ kg − 2676 kJ kg = 250 kJ kg z 350 b. H= 0.03346 + 0.6886 × 10 −5 T + 0.7604 × 10 −8 T 2 − 3593 × 10 −12 T 3 dT . 100 = 8.845 kJ mol ⇒ 491.4 kJ kg Difference results from assumption in (b) that H is independent of P. The numerical difference b g is ΔH for H 2 O v, 350° C, 1 atm → H 2 O v, 350° C, 100 bar b g z 80 8.6 b. dC ip n − C H (l) 6 14 = 0.2163 kJ / (mol⋅ o C) ⇒ ΔH = [0.2163] dT = 1190 kJ / mol . 25 The specific enthalpy of liquid n-hexane at 80oC relative to liquid n-hexane at 25oC is 11.90 kJ/mol c. dC ip n − C H (v) [ kJ 6 14 / (mol⋅ o C)] = 013744 + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10 −12 T 3 . z 0 ΔH = [013744 + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10 −12 T 3 ] dT = −110.7 kJ / mol . 500 The specific enthalpy of hexane vapor at 500oC relative to hexane vapor at 0oC is 110.7 kJ/mol. The specific enthalpy of hexane vapor at 0oC relative to hexane vapor at 500oC is –110.7 kJ/mol. 8-2
3. 3. 8.7 b g 18 T ′b° Fg − 32 = 0.5556T ′b° Fg − 17.78 T °C = 1 . C bcal mol⋅° Cg = 6.890 + 0.001436 0.5556T ′b° Fg − 17.78 = 6.864 + 0.0007978T ′b° Fg p 1° C C ′ b Btu lb - mole⋅° Fg = b100gC cal 453.6 mol 1 Btu = C . p E mol⋅° C 1 lb - mole 252 cal 1.8° F drop primes p p b g C p Btu lb - mole⋅° F = 6.864 + 0.0007978T ° F b g 8.8 dC i bT g = 01031 + b0158810001031g T = 01031 + 0.000557T [kJ / (mol⋅ p CH CH OH(l) 3 2 . . − . . o C)] 55.0 L 789 g 1 mol F GH 01031T + 0.000557 T OPQ 78.5 Q = ΔH = . 2 s 1 L 46.07 g 2 20 kJ mol = 941.9 × 7.636 kJ / s = 7193 kW 8.9 a. kJ mol z 200 b Q = ΔH = 5,000 mol s ⋅ g 0.03360 + 1367 × 10 −5 T − 1607 × 10 −8 T 2 + 6.473 × 10 −12 T 3 dT . . 100 = 17,650 kW b. b Q = ΔU = ΔH − ΔPV = ΔH − nRΔT = 17,650 kJ − 5.0 kmol ⋅ 8.314 [kJ / (kmol ⋅ K)] ⋅ 100 K gb gb g = 13,490 kJ The difference is the flow work done on the gas in the continuous system. c. Qadditional = heat needed to raise temperature of vessel wall + heat that escapes from wall to surroundings. 8.10 a. C p is a constant, i. e. C p is independent of T. Q b. Q = mC p ΔT ⇒ C p = mΔT Q 1 L 86.17 g 10 3 J (16.73 - 6.14) kJ Cp = = = 0.223 kJ / (mol ⋅ K) mΔT (2.00 L)(3.10 K) 659 g 1 mol 1 kJ Table B.2 ⇒ C p = 0.216 kJ / (mol⋅ o C) = 0.216 kJ / (mol ⋅ K) PV = RT FG ∂H IJ = FG ∂U IJ a∂ ∂T f P FG ∂U IJ 8.11 H = U + PV =====> H = U + RT =====> H ∂T K H ∂T K p p + R ⇒ Cp = H ∂T K p +R F ∂U I = dU = F ∂U I depends only on T, G But since U H ∂T JK dT GH ∂T JK p V ≡ Cv ⇒ C p = Cv + R 8-3
4. 4. 8.12 a. dC i p H O(l) 2 = 75.4 kJ / (kmol⋅ o C) =75.4 kJ/(kmol.oC) V = 1230 L , Vρ 1230 L 1 kg 1 kmol n= = = 68.3 kmol M 1 L 18 kg zd T2 n⋅ Cp i H 2 O(l) dT Q T` 68.3 kmol 75.4 kJ (40 − 29) o C 1 h Q= = = = 1967 kW . t t kmol⋅ o C 8h 3600 s b. Qtotal = Qto the surroundings + Qto water , Qto the surroundings = 1967 kW . z 40 n ⋅ C P ( H2 O ) dT Qto water 68.3 kmol 75.4 kJ / (kmol⋅ o C) 11 o C Qto water = = 29 = = 5.245 kW t t 3h 3600 s / h Qtotal = 7.212 kW ⇒ E total = 7.212 kW × 3 h = 21.64 kW ⋅ h c. Cost heating up from 29 o C to 40 o C = 21.64 kW ⋅ h × \$0.10 / (kW ⋅ h) = \$2.16 Costkeeping temperature constant for 13 h = 1.967 kW × 13 h × \$0.10/(kW ⋅ h)=\$2.56 Costtotal = \$2.16 + \$2.56 = \$4.72 d. If the lid is removed, more heat will be transferred into the surroundings and lost, resulting in higher cost. 8.13 a. ΔH N 2 (25 C) → N 2 (700 C) o o = HN 2 (700 o C) − HN o b 2 (25 C) g = 20.59 − 0 = 20.59 kJ mol b. ΔH H o F) → H 2 (77 o F) = HH o − HH = b0 − 5021g = −5021 Btu / lb - mol o 2 (800 2 (77 F) 2 (800 F) c. ΔH CO o C) → CO 2 (1250o C) = H CO o −H = b63.06 − 1158g = 5148 kJ mol CO 2 (300o C) . . 2 (300 2 (1250 C) d. ΔH O o F) → O 2 (0o F) = HO o − HO o = b−539 − 6774g = −7313 Btu / lb - mol 2 (970 2 (0 F) 2 (970 F) 300 kg 1 min 1000 g 1 mol 8.14 a. m = 300 kg / min n = = 178.5 mol / s min 60 s 1 kg 28.01 g Q = n ⋅ ΔH = n ⋅ zT1 T2 C p dT z 50 = (178.5 mol / s) ⋅ [0.02895 + 0.411 × 10 −5 T + 0.3548 × 10 −8 T 2 − 2.22 × 10 −12 T 3 ] dT [kJ / mol] 450 b = (178.5 mol / s) −12.076 [kJ / mol] = −2,156 kW g b. Q = n ⋅ ΔH = n ⋅ H (50o C) − H ( 450o C) = (178.5 mol / s)(0.73-12.815[kJ / mol]) = −2,157 kW 8.15 a. n = 250 mol / h 250 mol (2676 − 3697) kJ 1 kg 1 h 18.02 g i) Q = nΔH = = −1.278 kW h 1 kg 1000 g 3600 s 1 mol Q = nΔH = n ⋅ z T1 T2 C p dT z ii) 250 mol 1 h 100 = [003346 + 06880 × 10−5 T + 07604 × 10−8 T 2 − 3593 × 10−12 T 3 ] = −1.274 kW . . . . h 3600 s 600 8-4
5. 5. 8.15 (cont’d) iii) 250 mol Q= 3600 s b g ⋅ 2.54 − 20.91 [kJ / mol] = −1276 kW . b. Method (i) is most accurate since it takes into account the dependence of enthalpy on pressure and (ii) and (iii) do not. c. The enthalpy change for steam going from 10 bar to 1 atm at 600oC. 8.16 Assume ideal gas behavior, so that pressure changes do not affect ΔH . 200 ft 3 492 o R 12 atm 1 lb - mol . n= o = 0.6125 lb - mole / h h 537 R 1 atm 359 ft 3 (STP) Q = nΔH = (0.6125 lb - mole h b g ) ⋅ (2993 − 0) [Btu / lb - mole] = 1833 Btu / h 8.17 a. 50 kg 1.14 kJ b50 − 10g° C = 2280 kJ kg⋅° C b. (C ) p Na CO 2 3 ≈ 2 (C p ) Na + ( C p ) + 3 ( C p ) = 2 ( 0.026 ) + 0.0075 + 3 ( 0.017 ) = 0.1105 kJ mol ⋅ °C C O 50,000 g 0.1105 kJ 1 mol b50 − 10g° C = 2085 kJ mol⋅° C 105.99 g 2085 − 2280 % error = × 100% = −8.6% error 2280 8.18 dC i p C H O(l) 6 14 b g b g b g = 6 0.012 + 14 0.018 + 1 0.025 = 0.349 kJ / (mol⋅ o C) (Kopp’s Rule) dC i p CH COCH (l) 3 3 = 01230 + 18.6 × 10 −5 T kJ (mol⋅° C) . Assume ΔH mix ≅ 0 ↓ CH 3 COCH 3 ↓ C 6 H 14 O 0.30 ( 0.1230+18.6 × 10 T ) kJ 1 mol −5 0.70 ( 0.349 ) kJ 1 mol C pm = + mol ⋅ °C 58.08 g mol ⋅ °C 102.17 g = [0.003026 + 9.607 × 10−7 T] kJ (g ⋅ °C) ΔH = ∫ [0.003026 + 9.607 × 10−7 T] dT = −0.07643 kJ g 20 ˆ 45 8.19 Assume ideal gas behavior, ΔH mix ≅ 0 Mw = 1 3 b g b g 2 16.04 + 32.00 = 26.68 3 g mol ΔH = O2 z dC i dT = 10.08 kJ / mol, ΔH = dC i dT = 14.49 kJ / mol 350 25 p O 2 CH 4 z 25 350 p CH 4 L1 OF 1000 g IJ FG 1 mol IJ = 433 kJ kg H = M b14.49 kJ / molg + b10.08 kJ / molgPG 2 N3 3 QH 1 kg K H 26.68 g K 8-5
6. 6. 1000 m 3 1 min 273 K 1 kmol 8.20 n = min 60 s 303 K 22.4 m 3 STP b g = 0.6704 kmol s = 670.4 mol / s Energy balance on air: Table B.8 for ΔH 670.4 mol 0.73 kJ 1 kW Q = ΔH = nΔH Q= = 489.4 kW s mol 1 kJ s 489.4 kW heating 1 kW solar energy Solar energy required = = 1631 kW 0.3 kW heating 1631 kW 1000 W 1 m 2 Area required = = 1813 m 2 1 kW 900 W 8.21 C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O 135 × 10 5 SCFH 1 lb - mol . lb - mol n fuel = 3 = 376 h 359 ft h 376 lb − mol 5 lb - mol O 2 1 lb - mol air 115 . lb − mol n air = = 103 × 10 4 . h 1b - mol C 3 H 8 0.211b - mol O 2 h T2 Q =ΔH =n ⋅ ∫ C p dT T1 ⎛ lb − mol ⎞ 302 ∫ [0.02894 + 0.4147 × 10 −5 = ⎜1.03 × 104 ⎟⋅ T + 0.3191 × 10−8 T 2 − 1.965 × 10−12 T 3 ] dT ⎝ h ⎠ 0 1.03 × 10 lb-mol 8.954 kJ 453.593 mol 9.486 × 10-1 Btu 4 = =3.97 × 107 Btu/h h mol lb-mol kJ 8.22 a. Basis: 100 mol feed (95 mol CH4 and 5 mol C2H6) 7 CH 4 + 2O 2 → CO 2 + 2H 2 O C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2 n O 2 = 125 ⋅ LM 95 mol CH 4 2 mol O 2 + 5 mol C 2 H 6 35 mol O 2 . = 259.4 mol O 2 OP . MN 1 mol CH 4 1 mol C 2 H 6 PQ Product Gas: CO 2 : 95(1)+5(2)=105 mol CO 2 H 2 O : 95(2)+5(3)=205 mol H 2 O O 2 : 259.4-95(2)-5(3.5)=51.9 mol O 2 N 2 : 3.76(259.4)=975 mol N 2 Energy balance (enthalpies from Table B.8) ΔH CO 2 = H (CO , 450o C) − H (CO , 900o C) = 18.845 − 42.94 = −24.09 kJ / mol 2 2 ΔH H 2 O = H (H 450o C) − H (H 900o C) = 1512 − 33.32 = −18.20 kJ / mol . 2 O, 2 O, ΔH O 2 = H (O 450o C) − H (O 900o C) = 13.375 − 28.89 = −15.51 kJ / mol 2, 2, ΔH N 2 = H (N 450o C) − H (N 900o C) = 12.695 − 27.19 = −14.49 kJ / mol 2, 2, Q = ΔH = 105(-24.09) + 205(-18.20) + 51.9(-15.51) + 975(-14.49) Q = 21,200 kJ / 100 mol feed b. From Table B.5: H liq (40 o C) = 167.5 kJ / kg; H vap (50 bars) = 2794.2 kJ / kg; Q = n ⋅ ΔH = n(2794.2 -167.5) = 21200 ⇒ n = 8.07 kg / 100 mol feed 8-6
7. 7. 8.22 (cont’d) c. From part (b), 8.07 kg steam is produced per 100 mol feed 1250 kg steam 01 kmol feed 1 h . n feed = = 4.30 × 10 −3 kmol / s h 8.07 kg steam 3600 s 4.30 mol feed 1336.9 mol product gas 8.314 Pa ⋅ m 3 723 K V product gas = = 3.41 m 3 / s s 100 mol feed mol ⋅ K 1.01325 × 10 Pa 5 d. Steam produced from the waste heat boiler is used for heating, power generation, or process application. Without the waste heat boiler, the steam required will have to be produced with additional cost to the plant. 8.23 Assume ΔH mix ≅ 0 ⇒ ΔH = ΔH C10 H12 O2 + ΔH C6 H6 Kopp’s rule: C p d i C10 H12 O2 e j = 10(12) + 12(18) + 2(25) = 386 J mol⋅ o C = 2.35 J g⋅ o C e j 20.0 L 1021 g 1 kJ 2.35 J (71 − 25) o C ΔH C10 H12 O2 = = 2207 kJ L 10 3 J g⋅ o C ΔH C6 H6 = 15.0 L 879 g 1 mol L 78.11 g ⋅ LM N z298 348 OP [0.06255 + 23.4 × 10 −5 T] dT = 1166 kJ Q ΔH = 2207 + 1166 = 3373 kJ 8.24 a. 100 mol C3 H8 @ 40 o C, 250 kPa 100 mol C3 H8 @ 240 o C, 250 kPa VP 1 (m3 ) VP 2 (m3 ) mw kg H2 O(l, sat‘d) @ 5.0 bar mw kg H2 O(v) @ 300 o C, 5.0 bar Vw2 (m3 ) Vw1 (m3 ) b. References: H2O (l, 0.01 oC), C3H8 (gas, 40 oC) z 240 C3H8: Hin = 0 kJ / mol; Hout = CpC H dT =1936 kJ mol (Cp from Table B.2) . 3 8 40 H 2 O : H in = 3065 kJ/kg (Table B.7); H out = 640.1 kJ/kg (Table B.6) ˆ ˆ c. ΔH C3 H8 = 19.36 kJ/mol, ΔH w = (640.1 − 3065) kJ/kg = −2425 kJ/kg ˆ ˆ Q = ΔH = 100ΔH C3 H 8 + mw ΔH w = 0 ⇒ mw = 0.798 kg b From Table B.7: Vsteam 5.0 bar, 300° C = 0.522 m 3 kg g 0.008314 m ⋅ kPa (mol ⋅ K) 313 K b g 3 VC3 H8 40° C, 250 kPa = = 0.0104 m 3 mol C 3 H 8 250 kPa 0.798 kg steam 0.522 m3 steam 1 mol C3 H8 = 0.400 m 3 steam m 3 C3 H8 100 mol C3 H8 1 kg steam 0.0104 m3 C3 H8 d. 0.798 kg steam 2425 kJ 1 mol C3 H8 kJ Q = mw ΔH w = ˆ = 1860 100 mol C3 H8 kg steam 0.0104 m 3 C3 H 8 m 3 C3 H8 fed e. A lower outlet temperature for propane and a higher outlet temperature for steam. 8-7
8. 8. 8.25 a. 5500 L(ST P)/ min CH3 OH (v) 65o C n 2 mol/min CH3 O H (v) 260o C n 2 (mol/ min) mw kg/ min H2 O(l, s at‘d) @ 90o C mw kg/ min H2 O(v, s at‘d) @ 300o C Vw 2 (m3 /min ) Vw 1 (m3 /min ) 5500 L(STP) 1 mol n2 = = 245.5 mol CH 3OH(v)/min min 22.4 L(STP) An energy balance on the unit is then written, using Tables B.5 and B.6 for the specific enthalpies of the outlet and inlet water, respectively, and Table B.2 for the heat capacity of methanol vapor. The only unknown is the flow rate of water, which is calculated to be 1.13 kg H 2 O/min. b. ⎛ kg ⎞ ⎛ kJ ⎞ ⎛ 1 min ⎞ ⎛ 1 kW ⎞ Q = ⎜ 1.13 ⎟ ⎜ 2373.9 ⎟⎜ ⎟⎜ ⎟ = 44.7 kW ⎝ min ⎠ ⎝ kg ⎠ ⎝ 60 sec ⎠ ⎝ 1 kJ/s ⎠ 8.26 a. 100 mol/s (30o C) n 2 mol/s (30o C) 0.100 mol H2 O(v)/ mo l 0.020 mol H2 O(v)/ mo l 0.100 mol CO/ mol y 2 mol CO/s (mol CO/mol) 0.800 mol CO2 /mol (0.980-y 2 ) mol CO2 /s (mol CO2/mol) m3 kg humid air/s (50o C) H 2O(v) only m4 kg humid air/s (30ooC) (48 C) (0.002 /1.002 ) kg H2 O(v)/kg humid air y 4 kg H2 O(v)/kg humid air (1.000 /1.002 ) kg dry air/ kg humid air (1-y 4 ) kg dry air/kg humid air Basis: 100 mol gas mixture/s 5 unknowns: n2, m3, m4, y2, y4 – 4 independent material balances, H2O(v), CO, CO2 , dry air – 1 energy balance equation 0 degrees of freedom ( all unknowns may be determined) b. (1) CO balance: (100)(0.100) = n 2 y 2 U | V ⇒ n 2 = 9184 mol / s, x 2 = 01089 mol CO / mol (2) CO 2 balance: (100)(0.800) = n 2 (1 − y 2 ) | W . . 1000 . (3) Dry air balance: m3 = m4 (1 − y 4 ) 1002 . (100)(0.100)(18) 0.002 (0.020)(18) (4) H 2 O balance: + m3 = 9184 . + m4 y 4 1000 . 1002 1000 References: CO, CO2, H2O(v), air at 25oC ( H values from Table B.8 ) substance n in ( mol / s) H in (kJ / mol) n out ( mol / s) H out (kJ / mol) H2O(v) 10 0.169 91.84(0.020) 0.169 CO 10 0.146 10 0.146 CO2 80 0.193 80 0.193 H2O(v) m3(0.002/1.002)(1000/18) 0.847 m4y4(1000/18) 0.779 dry air m3(1.000/1.002) (1000/29) 0.727 m4(1-y4) (1000/29) 0.672 8-8
9. 9. 8.26 (cont’d) (5) Energy balance: 10(0169) + m3 FG 0.002 IJ FG 1000IJ (0.847) + m FG 1000 IJ FG 1000 IJ (0.727) . . H 1002 K H 18 K . H 1002 K H 29 K . 3 = 91.84(0.020)(0169) + m y (0.779)G F 1000IJ + m (1 − y )(0.672)FG 1000 IJ . H 18 K 4 4 4 H 29 K 4 Solve Eqs. (3)–(5) simultaneously ⇒ m3 = 2.55 kg/s, m4 = 2.70 kg/s, y4 = 0.0564 kg H2O/kg 2.55 kg humid air / s kg humid air = 0.0255 100 mol gas / s mol gas 00564 kg H 2 O . 29 kg DA 1 kmol H 2 O kmol H 2 O Mole fraction of water : =.0963 (1-.0564) kg dry air kmol DA 18 kg H 2 O kmol DA 0.0963 kmol H 2 O kmol H 2 O ⇒ = 00878 . (1 + 00963) kmol humid air . kmol humid air p H 2O (0.0878)(760 mm Hg) Relative humidity: = × 100% = 79.7% * p H 2O e48 Cj o 83.71 mm Hg c. The membrane must be permeable to water, impermeable to CO, CO2, O2, and N2, and both durable and leakproof at temperatures up to 50oC. 8.27 a. y H 2O = b p * 57° C g = 129.82 mm Hg = 0171 mol H O mol . 2 P 760 mm Hg ↓ 28.5 m 3 STP b g 1 mol = 1270 mol h ⇒ 217.2 mol H 2 O h b3.91 kg H O hg h 0.0224 m 3 STPb g 2 R 89.5 mol CO h |110.5 mol CO h 1270 − 217.2 = 1053 mol dry gas | =======> S given 2 h | 5.3 mol O h percentages |847.6 mol N h 2 T 2 1270 mol/h, 620°C 425°C m (kg H2 O( l )/h), 20°C References for enthalpy calculations: e CO, CO 2 , O 2 , N 2 at 25°C (Table B.8); H 2 O l , 0.01o C (steam tables) j substance n in H in n out H out CO 89.5 18.22 89.5 12.03 U n in mol h | CO 2 110.6 27.60 110.6 17.60 V H in kJ mol | O2 N2 5.3 847.6 19.10 18.03 5.3 847.6 12.54 11.92 W bg H 2O v 3.91 3749 3.91 + m 3330 U n in kg h V H in kJ kg H Obl g W m 83.9 -- -- 2 8-9
10. 10. 8.27 (cont’d) ΔH = ∑n H −∑n H out i i in i i = 0 ⇒ −8504 + 3246m = 0 ⇒ m = 2.62 kg h b. When cold water contacts hot gas, heat is transferred from the hot gas to the cold water lowering the temperature of the gas (the object of the process) and raising the temperature of the water. 8.28 2°C, 15% rel. humidity ⇒ p H 2 O = 015 5.294 mm Hg = 0.7941 mm Hg . b gb g dy i H 2O inhaled b0.7941g b760g = 1045 × 10 . −3 mol H 2 O mol inhaled air 5500 ml 273 K 1 liter 1 mol ninhaled = min 3 275 K 10 ml 22.4 liters STP b g = 0.2438 mol air inhaled min Saturation at 37 °C ⇒ y H 2 O = b p * 37° C g = 47.067 = 0.0619 mol H 2 O mol exhaled dry gas 760 mm Hg 760 0.2438 mol/min 2oC n2 kmol/min 37oC 1.045 x 10-3 H2O 0.0619 H2O 0.999 dry gas 0.9381 dry gas n1 mol H2O(l)/min 22o C Mass of dry gas inhaled (and exhaled) = b0.2438gb0.999gmol dry gas 29.0 g = 7.063 g min min mol Dry gas balance: b0.999gb0.2438g = 0.9381 n 2 ⇒ n2 = 0.2596 mols exhaled min H 2 O balance: b0.2438ge1045 × 10 j + n = b0.2596gb0.0619g ⇒ n = 0.0158 mol H O min . −3 1 1 2 References for enthalpy calculations: H Obl g at triple point, dry gas at 2 °C 2 substance min H in mout H out Dry gas 7.063 0 7.063 36.75 m in g min mH2 O = 18.02nH2 O bg H 2O v 0.00459 2505 0.290 2569 H in J g ˆ H Obl g 0.285 92.2 — — H H2 O from Table 8.4 2 H dry gas = 1.05 (T − 2 ) ˆ 966.8 J 60 min 24 hr Q = ΔH = ∑m H − ∑m H out i i in i i = min 1 hr 1 day = 1.39 × 10 6 J day 8-10
11. 11. 8.29 a. bg 75 liters C 2 H 5 OH l 789 g 1 mol = 1284 mol C 2 H 3 OH l bg liter 46.07 g e j (C p ) CH 3OH = 01031 + 0.557 × 10 −3 T kJ / (mol⋅ o C) (fitting the two values in Table B.2) . 55 L H 2 O l bg 1000 g 1 mol = 3054 mol H Obl g 2 b (C p ) H 2 O = 0.0754 kJ mol⋅° C g liter 18.01 g 1284 mol C2H5 OH(l) (70.0oC) 1284 mol C2H5 OH (l) (To C) 3054 mol H2O(l) (20.0oC) 3054 mol H2O(l) (To C) T T 0 = 1284 ∫ ( 0.1031 + 0.557 × 10−3 T ) dT + 3054 ∫ ( 0.0754 ) dT Q = ΔU ≅ ΔH ( liquids ) ⎫ 70 25 ⎪ ⎬ ⇒ ⇓ Integrate, solve quadratic equation Q = 0 ( adiabatic ) ⎪ ⎭ T=44.3 o C b. 1. Heat of mixing could affect the final temperature. 2. Heat loss to the outside (not adiabatic) 3. Heat absorbed by the flask wall & thermometer 4. Evaporation of the liquids will affect the final temperature. 5. Heat capacity of ethanol may not be linear; heat capacity of water may not be constant 6. Mistakes in measured volumes & initial temperatures of feed liquids 7. Thermometer is wrong 8.30 a. 1515 L/s air 500oC, 835 tor, Tdp=30o C 1515 L/s air , 1 atm 110 g/s H2O(v) 110 g/s H2O, T=25oC Let n1 (mol / s) be the molar flow rate of dry air in the air stream, and n 2 (mol / s) be the molar flow rate of H2O in the air stream. 1515 L 835 mm Hg mol ⋅ K n1 + n 2 = = 26.2 mol / s s 773 K 62.36 L ⋅ mm Hg n2 p * (30 o C) 31824 mmHg . =y= = = 0.0381 mol H 2 O / mol air n1 + n 2 Ptotal 835 mmHg ⇒ n1 = 25.2 mol dry air / s; n 2 = 10 mol H 2 O / s . 8-11
12. 12. 8.30 (cont’d) References: H2O (l, 25oC), Air (v, 25oC) substances nin (mol / s) H in (kJ / mol) n out (mol / s) H out (kJ / mol) dry air 25.2 14.37 25.2 zdT Cp i dT z zd 25 air H2O(v) 1.0 dC i 7.1 i 100 100 p H O ( l ) dT + H vap Cp dT + H vap H2 O ( l ) z zd 25 2 25 dC i i 500 T p H O ( v ) dT Cp H2 O ( v ) dT 100 2 100 H2O(l) 6.1 0 -- -- ΔH = 0 = n out ⋅ H out − nin ⋅ H in z25 T p air z b25.2gFGH dC i dT IJK + b7.1gFGH dC i dT + H + dC i dT IJK 25 100 p H O(l ) 2 vap z 100 T p H O(v ) 2 . H z F dC i dT + H + dC i dT IJ = 0 −b25.2gb14.37g − b100gG 25 100 p H O(l ) 2 K vap z100 500 p H O(v ) 2 Integrate, solve : T = 139 o C b. Q = − ( 25.2 ) ∫ (C ) dT − (1.00 ) ∫ (C ) 139 139 p air p H O(v) dT = −290 kW 500 500 2 This heat goes to vaporize the entering liquid water and bring it to the final temperature of 139oC. c. When cold water contacts hot air, heat is transferred from the air to the cold water mist, lowering the temperature of the gas and raising the temperature of the cooling water. 8-12
13. 13. 520 kg NH 3 103 g 1 mol 1h 8.31 Basis: = 8.48 mol NH 3 s h 1 kg 17.03 g 3600 s 8.48 mol NH 3 /s 25°C n 2 (mol/s) n 1 (mol air/s) 0.100 NH 3 T °C 0.900 air 600°C Q = –7 kW NH 3 balance: 8.48 = 0100n2 ⇒ n2 = 84.8 mol s . b gb g Air balance: n1 = 0.900 84.8 = 76.3 mol air s References for enthalphy calculations: NH 3 g , air at 25° C bg NH 3 Hin = 0.0 z C p from Hout = 25 600 dC i p NH dT 3 ⇒ Hout Table B.2 = 25.62 kJ mol Air: C p ( J mol ⋅ °C ) = 0.02894 + 0.4147 × 10−5 T + 0.3191 × 10−8 T 2 − 1.965 × 10−12 T 3 ˆ T H in = ∫ C p dT 25 = ( −0.4913 × 10−12 T 4 + 0.1064 × 10−8 T 3 + 0.20735 × 10−5 T 2 + 0.02894T − 0.7248 ) ( kJ mol ) 600 ˆ H out = ∫ C p dT = 17.55 kJ mol 25 Energy balance: Q = ΔH = ∑n H − ∑n H out ˆ i ˆ i in i i ⇓ −7 kJ s = ( 8.48 mols NH 3 s )( 25.62 kJ mol ) + ( 76.3 mols air s )(17.55 kJ mol ) − ( 8.48 )( 0.0 ) − ( 76.3) ( −0.4913 × 10−12 T 4 + 0.1064 × 10−8 T 3 + 0.20735 × 10−5 T 2 + 0.02894T − 0.7248 ) Solve for T by trial-and-error, E-Z Solve, or Excel/Goal Seek ⇒ T = 691o C 8.32 a. Basis: 100 mol/s of natural gas. Let M represent methane, and E for ethane Stack gas (900oC) Stack gas (ToC) n3 mol CO2/s n3 mol CO2/s n4 mol H2O/s n4 mol H2O/s 100 mol/s n5 mol O2/s n5 mol O2/s Heat n6 mol N2/s Exchanger n6 mol N2/s 0.95 mol M/mol Furnace 0.05 mol E/mol air (245oC) 20 % excess air (20oC) n1 mol O2/s n1 mol O2/s n2 mol N2/s n2 mol N2/s CH 4 + 2O 2 → CO 2 + 2H 2 O b g C 2 H 6 + 7 / 2 O 2 → 2CO 2 + 3H 2 O 8-13
14. 14. 8.32 (cont’d) ⎡ 95 mol M 2 mol O 2 4.76 mol air 5 mol E 3.5 mol O 2 4.76 mol air ⎤ nair = 1.2 ⎢ + ⎥ ⎣ s 1 mol M mol O 2 s 1 mol E mol O 2 ⎦ nair = 1185 mol air/s n1 = 0.21 × 1185 = 249 mol O 2 /s, n2 = 0.79 × 1185 = 936 mol N 2 /s 95 mol M 1 mol CO 2 5 mol E 2 mol CO 2 n3 = + = 105 mol CO 2 /s s 1 mol M s 1 mol E 95 mol M 2 mol H 2 O 5 mol E 3 mol H 2 O n4 = + = 205 mol H 2 O/s s 1 mol M s 1 mol E 95 mol M 2 mol O 2 5 mol E 3.5 mol O 2 n5 = 249 − + = 41.5 mol O 2 /s s 1 mol M s 1 mol E n6 = n2 = 936 mol N 2 /s Energy balance on air: 245 ⎛ mol air ⎞⎛ kJ ⎞ kJ Q = nair ∫ (C p ) air dT = ⎜1185 ⎟⎜ 6.649 ⎟ = 7879 ( = 7879 kW) 20 ⎝ s ⎠⎝ mol air ⎠ s Energy balance on stack gas: ( ( C ) dT ) 6 Q = −ΔH = −∑ ni ∫ T 900 p i i =3 −7879 = n3 ∫ (C ) (C ) (C ) (C ) T dT + n4 ∫ T dT + n5 ∫ T dT + n6 ∫ T p CO p H O (v ) p O p N dT 900 2 900 2 900 2 900 2 Substitute for the heat capacities (Table B.2), integrate, solve for T using E-Z Solve⇒ T = 732 o C b. 350 m 3 (STP) mol 1000 L 1 h = 4.34 mol / s h 22.4 L(STP) m 3 3600 s 4.34 mol / s Scale factor = = 0.0434 100 mol / s b g Q ′ = 0.0434 7851 = 341 kW 8.33 a. ΔH = z0 600 100 C p dT = 3 b g b 335 + 4 351 + 38.4 + 42.0 + 2 36.7 + 40.2 43.9 = 23100 J mol . . g 150 mol 23100 J 1 kW Q = ΔH = nΔH = = 3465 kW s mol 1000 J / s b. The method of least squares (Equations A1-4 and A1-5) yields (for X = T , y = C p ) b g Cp = 0.0334 + 1732 × 10−5 T ° C kJ (mol ⋅° C) ⇒ Q = 150 . z 0 600 0.0334 + 1732 × 10−5 T dT = 3474 kW . The estimates are exactly identical; in general, (a) would be more reliable, since a linear fit is forced in (b). 8.34 a. ln C p = bT 1 2 + ln a ⇒ C p = a exp bT 1 2 , e j T1 = 7.1 , C p1 = 0.329 , T2 = 17.3 , C p 2 = 0.533 b= ln C p 2 C p1 = 0.0473 U | T2 − T1 |⇒C V e = 0.235 exp 0.0473T 1 2 j = 0.235| p −1.4475 ln a = ln C p1 − b T1 = −14475 ⇒ a = e . | W 8-14
15. 15. 8.34 (cont’d) z b0.235gb2g Rexpe0.473T jLT S 1 OPU 150 e j QV 150 b. 1800 0.235 exp 0.0473T 1 2 dT = 0.0473 T 12 MN 12 − .0473 W 1800 = −1730 cal g DIMENSIONS CP(101), NPTS(2) WRITE (6, 1) 1 FORMAT (1H1, 20X'SOLUTION TO PROBLEM 8.37'/) NPTS(1) = 51 NPTS(2) = 101 DO 200K = 1, 2 N = NPTS (K) NM1 = N – 1 NM2 = N – 2 DT = (150.0 – 1800.0)/FLOAT (NM1) T = 1800.0 DO 20 J = 1, N CP (J) = 0.235*EXP(0.0473*SQRT(T)) 20 T = T + DT SUMI = 0.0 DO 30 J = 2, NM1, 2 30 SUMI = SUMI + CP(J) SUM2 = 0.0 DO 40 J = 3, NM2, 2 40 SUM2 = SUM2 + CP (J) DH = DT*(CP(1) + 4.0 = SUM1 + 2.0 = SUM2 + CP(N))/3.0 WRITE (6, 2) N, DH 2 FORMAT (1H0, 5XI3, 'bPOINT INTEGRATIONbbbDELTA(H)b= ', E11.4,'bCAL/G') 200 CONTINUE STOP END Solution: N = 11 ⇒ ΔH = −1731 cal g N = 101 ⇒ ΔH = −1731 cal g Simpson's rule with N = 11 thus provides an excellent approximation 8.35 a. m = 175 kg / min U | 175 kg 1000 g 1 mol 56.9 kJ 1 min M .W . = 62.07 g / mol V ⇒ Q = ΔH = = 2670 kW ΔH = 56.9 kJ / mol W v | min kg 62.07 g mol 60 s b. The product stream will be a mixture of vapor and liquid. c. The product stream will be a supercooled liquid. The stream goes from state A to state B as shown in the following phase diagram. P B A T 8-15
16. 16. 8.36 a. Table B.1 ⇒ Tb = 68.74 o C, ΔH v (Tb ) = 28.85 kJ / mol Assume: n - hexane vapor is an ideal gas, i.e. ΔH is not a function of pressure bC H g 6 14 l, 20o C ΔH ⎯ ⎯⎯→ ⎯ Total bC H g 6 14 v, 200o C B ΔH 1 A ΔH 2 bC H g b ⎯→ bC H g ⎯ ⎯⎯ g ΔH v Tb 6 14 l, 68.74 o C 6 14 v, 68.74 o C ΔH1 = z 20 68.74 0.2163 dT = 10.54 kJ / mol ΔH2 = z 200 68.74 013744 + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10 −9 T 3 dT . ΔH2 = 24.66 kJ / mol b g ΔHTotal = ΔH1 + ΔH 2 + ΔHv Tb = 10.54 + 24.66 + 28.85 = 64.05 kJ / mol b. ΔH = −64.05 kJ / mol c. e U 200 o C , 2 atm = H − PV j Assume ideal gas behavior ⇒ PV = RT = 3.93 kJ / mol U = 64.05 − 3.93 = 6012 kJ / mol . 8.37 Tb = 100.00° C b g ΔHv tb = 40.656 kJ mol e j e j H 2 O l, 50o C ΔH v 50o C ⎯ ⎯⎯⎯ → ⎯ e j H 2 O v, 50o C B ΔH 1 A ΔH 2 H O el, 100 Cj o e ΔH v 100 C ⎯ ⎯⎯⎯ → ⎯ o j H O e v, 100 Cj o 2 2 z 100 ΔH1 = C pH 2 Ob l g dT = 3.77 kJ mol 25 z 25 ΔH2 = C pH 2 Ob v g dT = −169 kJ mol . 100 Table B.1 B b g ΔHv 50° C = 3.77 + 40.656− 169 = 42.7 kJ mol . ( 2547.3 − 104.8) kJ 18.01 g 1 kg Steam table: = 44.0 kJ mol kg 1 mol 1000 g The first value uses physical properties of water at 1 atm (Tables B.1, B.2, and B.8), while the heat of vaporization at 50oC in Table B.5 is for a pressure of 0.1234 bar (0.12 atm). The difference is ΔH for liquid water going from 50oC and 0.1234 bar to 50oC and 1 atm plus ΔH for water vapor going from 50oC and 1 atm to 50oC and 0.1234 bar. 8.38 1.75 m3 879 kg 1 kmol 1000 mol 1 min = 164.1 mol/s 2.0 min m3 78.11 kg 1 kmol 60 s Tb = 801° C , ΔHv Tb = 30.765 kJ mol . b g 8-16
17. 17. 8.38 (cont’d) e j ⎯→ C 6 H 6 v, 580 o C⎯ e C 6 H 6 l, 25o C j B ΔH 1 A ΔH 2 −Δ H C H e v, 80.1 Cj ⎯ → el, 80.1 Cj o v o 6 6 ⎯ C6H 6 z 80.1 ΔH1 = C pC 6 H 6 b v g dT = −77.23 kJ mol 580 z 298 ΔH2 = C pC 6 H 6 b l g dT = −7.699 kJ mol 353.1 d i ΔH = ΔH1 − ΔH v 801o C + ΔH 2 = −115.7 kJ / mol . Q = ΔH = nΔH = b164.1 mol / sgb −115.7 kJ / molg = −190 x10 . −4 kW Antoine B 8.39 35° C ⇒ yCCl 4 = 015 V . U P ∗ 25° C V = 015 . 176.0 mm Hg b g = 0.0347 mol CCl 4 mol 15% relative saturation W 1 atm 760 mm Hg Table B.1 kJ 10 mol 0.0347 mol CCl 4 30.0 kJ ( ΔH v ) CCl 4 = 30.0 ⇒ Q = ΔH = = 10.4 kJ min mol min mol mol CCl 4 Time to Saturation 6 kg carbon 0.40 g CCl 4 1 mol CCl 4 1 mol gas 1 min = 45.0 min g carbon 153.84 g CCl 4 0.0347 mol CCl 4 10 mol gas 8.40 a. b g CO 2 g, 20° C → CO 2 s, − 78.4° C : ΔH = b g z −78.4 20 dC i p CO g 2 b gdT − ΔH sub b−78.4° Cg In the absence of better heat capacity data; we use the formula given in Table B.2 (which is strictly applicable only above 0° C ). ΔH ≈ 20 z −78.4 .03611 + 4.233 × 10 −5 T − 2.887 × 10 −8 T 2 + 7.464 × 10 −12 T 3 dT kJ mol FG IJ H K cal 4.184 × 10 −3 kJ − 6030 = −28.66 kJ mol mol 1 cal 300 kg CO 2 10 3 g 1 mol 28.66 kJ removed Q = ΔH = nΔH = = 195 × 10 5 k J h . h 1 kg 44.01 g mol CO 2 (or 6.23 × 10 7 cal hr or 72.4 kW ) b. According to Figure 6.1-1b, Tfusion=-56oC Q = ΔH = nΔH where, ΔH = z −56 20 dC ip CO (v) dT 2 e j z dC i +ΔHv −56o C + −78.4 −56 p CO (l) dT 2 Q=n LMz dC i −56 dT +ΔH e −56 Cj + z dC i dT OPQ o −78.4 N 20 p CO (v) 2 v −56 p CO (l) 2 8-17
18. 18. 8.41 a. C p = a + bT b= 53.94 − 50.41 = 0.01765 U | ⇒ C bJ mol ⋅ Kg = 4512 + 0.01765T bKg 500 − 300 V | . p a = 53.94 − b0.01765gb500g = 4512 | . W NaCl b s, 300 Kg → NaClb s, 1073 Kg → NaClbl , 1073 Kg ΔH = z C dT + ΔH b1073 Kg = M 300 1073 ps N . m z L b4512 + 0.01765T gdT OP J + 30.21 kJ Q mol mol 1073 300 10 3 J 1 kJ = 7.44 × 10 4 J mol b. Q = ΔU = n z 1073 300 Cv dT + ΔU m 1073 K b g Cv ≈ C p ΔU m ≅ ΔHm 200 kg 10 3 g 1 mol 74450 J Q ≈ ΔH = nΔH = = 2.55 × 10 8 J 1 kg 58.44 g mol 2.55 × 10 8 J s 1 kJ c. t= = 100 s 0.85 × 3000 kJ 10 3 J 8.42 ΔHv = 35.98 kJ mol , Tb = 136.2° C = 409.4 K , Pc = 37.0 atm , Tc = 619.7 K (from Table B.1) Trouton's rule: ΔHv ≈ 0.088Tb = 0.088 409.4 K = 36.0 kJ mol 01% error . b gb g b g Chen's rule: LM FG T IJ − 0.0327 + 0.0297 log P OP b ΔH ≈ HT K NM Tb 0.0331 c QP = 35.7 kJ mol (–0.7% error) 10 c v FT I 107 − G J b . HT K c Watson’s correlation : ΔH b100° Cg ≈ 35.98G F 619.7 − 373.2 IJ = 38.2 kJ mol 0.38 H 619.7 − 409.4 K v 8.43 b g b g C 7 H 2 N : Kopp's Rule ⇒ C p ≈ 7 0.012 + 12 0.018 + 0.033 = 0.333 k J (mol ⋅° C) Trouton's Rule ⇒ ΔH b200° Cg = 0.088b200 + 273.2g = 41.6 kJ mol v C H N bl , 25° Cg → C H Nbl , 200° Cg → C H Nbv , 200° Cg 7 12 7 12 7 12 200 kJ kJ ΔH = ˆ ∫ C dT + ΔH ( 200°C ) ≈ 0.333(200 − 25) mol 25 p ˆ v + 41.6 mol = 100 kJ mol 8-18
19. 19. 8.44 a. Antoine equation: Tb ° C = b g 1211033 b g . 6.90565 − log 100 − 220.790 = 261° C . F 562.6 − 299.3IJ ΔH b261° Cg = 30.765G 0.38 = 33.6 kJ mol Watson Correction: v. H 562.6 − 3531 K . b g b. Antoine equation: Tb 50 mm Hg = 118° C ; Tb 150 mm Hg = 35.2° C . b g Clausius-Clapeyron: ln p = − ΔH v + C ⇒ ΔH v = − R ln p 2 p1 b g RT 1 T2 − 1 T1 R | b g U | ΔH v = −0.008314 kJ S | ln 150 50 = 34.3 kJ mol V | T mol ⋅ K 1 308.4 K − 1 285.0 K W c. C 6 H 6 ( l , 26.1°C) C6 H 6 (v , 26.1°C) Δ H1 Δ H2 Δ Hv (80.1°C) C 6 H 6 ( l , 80.1°C) C6 H 6 (v , 80.1°C) zd 80.1 ΔH1 = i C p dT = 7.50 kJ mol l 26.1 zd 26.1 ΔH2 = i C p dT = −4.90 kJ mol v 80.1 b g ΔHv 261° C = 7.50 + 30.765 − 4.90 = 33.4 kJ mol . 8.45 a. Tout = 49.3oC. The only temperature at which a pure species can exist as both vapor and liquid at 1 atm is the normal boiling point, which from Table B.1 is 49.3oC for cyclopentane. b. Let n f , n v , and n l denote the molar flow rates of the feed, vapor product, and liquid product streams, respectively. Ideal gas equation of state 1550 L 273 K 1 mol nf = = 44.66 mol C 5 H 10 (v) / s s 423 K 22.4 L(STP) 55% condensation: n l = 0.550(44.66 mol / s) = 24.56 mol C 5 H 10 (l) / s Cyclopentane balance ⇒ n v = (44.66 − 24.56) mol C 5 H 10 / s = 20.10 mol C 5 H 10 (v) / s Reference: C5H10(l) at 49.3oC Substance n in H in n out H out (mol/s) (kJ/mol) (mol/s) (kJ/mol) C5H10 (l) — — 24.56 0 C5H10 (v) 44.66 Hf 20.10 Hv H i = ΔH v + z Ti 49.3o C C p dT 8-19