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6 Pages

Elementary Principles 447

Course: CHEMICAL E 312, Fall 2011
School: The University of Akron
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Word Count: 209

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(cont'd) Solids 9.70 balance on dryer: 0.35 24,000 kg / d = 0.75n2 n2 = 11200 kg / d F 11.2 tonnes / d (conc. sludge) Mass Balance on dryer: 24,000 = n1 + 11200 n1 = 12,800 kg / d Energy balance on sludge side of dryer: References : H 2 O(l,22 C), Solids(22 C) Substance Solids H 2 O(l) H 2 O(v) nin (kg d) 8400 15600 - ^ H in 0 0 - nout 8400 2800 12800 ^ H out (kJ kg) ^ H 1 (kJ kg) (kg d) ^ H2 ^ H 3...

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(cont'd) Solids 9.70 balance on dryer: 0.35 24,000 kg / d = 0.75n2 n2 = 11200 kg / d F 11.2 tonnes / d (conc. sludge) Mass Balance on dryer: 24,000 = n1 + 11200 n1 = 12,800 kg / d Energy balance on sludge side of dryer: References : H 2 O(l,22 C), Solids(22 C) Substance Solids H 2 O(l) H 2 O(v) nin (kg d) 8400 15600 - ^ H in 0 0 - nout 8400 2800 12800 ^ H out (kJ kg) ^ H 1 (kJ kg) (kg d) ^ H2 ^ H 3 ^ H1 = 2.5(100 - 22) = 195.0 kJ/kg ^ H = (419.1 - 92.2) = 326.9 kJ/kg 2 ^ H 3 = (2676 - 92.2) = 2584 kJ/kg ^ ( H water from Table B.5) Q2 = m H - m H Q i i i i out in 7 2 = 356 107 kJ day . Qsteam = . 356 10 = 6.47 107 kJ / d Q3 = 2.91 107 kJ / d 0.55 H v for 2 H O(sat'd, ) Energy balance on steam side of dryer: 6.47 107 B kJ kg = n3 2133 d d FG IJ H K FG kJ IJ F 1 tonne I n H kg K GH 10 kg JK 3 3 = 30.3 tonnes / d (boiler feedwater) Energy balance on steam side of boiler: Q1 = (30300 kg kJ )(2737.6 - 83.9) = 8.04 107 kJ / d d kg 8.04 107 = 13 108 kJ / d . 0.62 62% efficiency Fuel heating value needed = n4 = 130 108 kJ / d . 3.75 104 kJ / kg = 3458 kg / d D = 3.5 tonnes / day (fuel oil) S + O 2 SO 2 Air feed to boiler furnace: C + O 2 CO 2 , 4H + O 2 2H 2 O, (nO2 ) theo = 3458 kg kgC 1 kmol C 1 kmol O 2 1 1 1 1 (0.87 kg )( 12 kg )( 1 kmol C )+(0.10)( 1)( 4 ) + (0.0084)( 32 )(1) d = 338 kmol O 2 /d 9-95
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The University of Akron - CHEMICAL E - 312
9.70 a. W = H2Om 1 [ k g W (v)/d ] 100oCF24,000 kg sludge / d, 22 C 0.35 solids, 0.65 W(l)Waste gasoDRYERQ2m 2 (kg conc. sludge/d), 100 o C 0.75 solids, 0.25 W (l)INCINERATORm3 [kg W(v)/d] 4B, sat'dC m3 [kg W(l)/d] BOILER 20 o CQ (kJ / d) 3m3
The University of Akron - CHEMICAL E - 312
9.69 (cont'd)Hi =T 0 + C dT H i pi kJ mol 25 kJ molCzn Hii= -1575 10 6 kJ h .= -9.888 10 6 kJ h +inn Hi outiLM5595dC i + 1261dC i + 7931dC i N OP 1 kJ dT +52816dC i + 23311dC i +3458dC i + 3013dC i b g Q 10 J 1 kJ + dC i b g 10 J dTTout 25 p
The University of Akron - CHEMICAL E - 312
9.69 (cont'd)Stripper off-gas 279.7 mol CO h 279.7 mol CH 4 h 30308 mol H 2 O h 34.3 kmol h, 0.82% CO, 0.82% CH 4 , 88.5% H 2 O, 9.9% CO 2 3392 mol CO h3.4259 10 4c.d.mol I F 1 kmol I FG JG J = 466 kmol DMF h H h K H 10 mol K b0.991gb7955g mol C H in
The University of Akron - CHEMICAL E - 312
9.69 (cont'd)ge j CO: n = 0.988n + b0.00055ge5.086 10 j n = 23311 mol CO h CO : n = b0.0068ge5.086 10 j = 3458 mol CO h H O: n = b0.0596ge5.086 10 j = 30313 mol H O h n = b0.0567gn (mol CH ) 1 mol C Soot formation: n = 0.0567n b1g h 1 mol CHC 2 H 2 : n7
The University of Akron - CHEMICAL E - 312
9.69Absorber off-gas n 8 (mol H 2/h) n 12 (mol N 2/h) 0.988 n 9 (mol CO/h) 0.950 n 6 (mol CH4 /h) 0.006 n 7 (mol C 2H 2/h) Basis: 5000 kg/h Product gas n 1 (mol/h) 0.991 mol C2 H2 ( g)/mol 0.00059 mol H2 O/mol 0.00841 mol CO2 /mol 0.917 n 1 (mol DMF/h) L
The University of Akron - CHEMICAL E - 312
9.68 (cont'd) d.Run 1 yCH4 0.75 yC2H6 0.21 yC3H8 0.04 Tf 40 Ta 150 Pxs 25 ywo 0.0306 nO2i 3.04 nN2 11.44 nH2Oi 0.46 HCH4 -74.3 HC2H6 -83.9 HC3H8 -102.7 HO2i 3.6 HN2i 3.8 HH2Oi -237.6 nCO2 1.29 nH2O 2.75 nO2 0.61 nN2 11.44 Tad 1743.1 alph0 -1052 alph1 0.4
The University of Akron - CHEMICAL E - 312
9.68 (cont'd) c. References : C(s), H 2 ( g) at 25o CH CH 4 (T) = ( H o ) CH 4 + (C p ) CH 4 dT f25zTUsing ( H fo )CH 4 from Table B.1 and (C p )CH 4 from Table B.2HCH (T) = -7485 kJ / mol + .4F GG H25zT(0.03431+5.469 10-5 T + 03661 10-8 T 2 -
The University of Akron - CHEMICAL E - 312
9.68b.1 mol natural gas yCH 4 (mol CH 4 / mol)yC 2 H 6 (mol C 2 H 6 / mol) yC 3H 8 (mol C 3 H 8 / mol)nCO 2 (mol CO 2 ) nH 2 O (mol H 2 O) nN 2 (mol N 2 ) nO 2 mol O 2 )Humid air na (mol air) ywo (mol H20(v)/mol) (1-ywo) (mol dry air/mol) 0.21 mol O2
The University of Akron - CHEMICAL E - 312
9.67 (cont'd) b.Reference states: CH 4 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o Cbg bg bgnin mol 100 . 2.60 9.78 - -bg0 (We will use the values of Hc given in Table B.1, which are based on H 2 O l as abgcombustion product, and so must choose t
The University of Akron - CHEMICAL E - 312
9.66 (cont'd)H b = ( H v ) H2 O(25C)+ (C p ) H 2 O(v) dT25zTa.Energy Balance ^ ^ ^ H = (Hco )CH4 + nout Hout - nin Hin = 0^ Table B.2 for C pi (T ), ( H v ) H O = 44.01 kJ/mol 20.247(-890.36) + 0.494(44.01) + 0.0963(T - 25) + 1.02 10-5 (T 2 - 2
The University of Akron - CHEMICAL E - 312
9.65 (cont'd) c.T 7252 5634 4680 4426 4414f(T) 6.05E+03 1.73E+03 3.10E+02 1.41E+01 3.11E-02f'(T) Tnew 3.74 5634 1.82 4680 1.22 4426 1.11 4414 1.11 4414d. 9.66The polynomial formulas are only applicable for T 1500C 5.5 L/s at 25C, 1.1 atm n 1(mol CH4/
The University of Akron - CHEMICAL E - 312
9.65 (cont'd)substance C5 H 12 O2 CO 2 H2O nin Hin mol kJ mol 100 0 . 10.40 H1 - - - - nout Hout mol kJ mol - - 2.40 H2 5.00 H3 6.00 H4i = 2,3Hi =25zT(C p ) i dT= H v 25 C + (C p ) H 2 O(v) dT for H 2 O(v)o 25Table B.8ejzTH1 = H O 2 (75 o C
The University of Akron - CHEMICAL E - 312
9.64 (cont'd) c.Preheating step: References: C 3 H 6 g , O 2 , N 2 at 25 CSubstance C3H 6O O2 N2bgn in H in n out H out ( mol / s) (kJ / mol) (mol / s) ( kJ / mol) (562 C) (50 C) 52.34 315 52.34 67.66 . 209.4 0.826 209.4 17.72 787.3 0.775 787.3 16.65
The University of Akron - CHEMICAL E - 312
9.64C 3 H 6 O g + 4O 2 g 3CO 2 g + 3H 2 O l , Hio = -1821.4 kJ mol Basis :1410 m3 STP feed gas minbgbgbgafb g103 mol322.4 m STPb g1 min 60 s= 1049 mol s feed gasStochiometric proportion:1 mol C 3 H 6 O 4 mol O 2 4 3.76 = 15.04 mol N 2 1 + 4
The University of Akron - CHEMICAL E - 312
9.63 (cont'd)Theoretical oxygen = 2 mol O 2 1 mol CH 4 1 mol air 0.21 mol O 2 82 mol CH 4 + 3.5 mol O 2 1 mol C 2 H 6 18 mol C 2 H 6 = 227 mol O 2Air fed : n1 =1.2 227 mol O 2= 1297.14 mol airC balance : n2 = 82.00 1 + 18.00 2 n2 = 118.00 mol CO 2 H
The University of Akron - CHEMICAL E - 312
9.62Basis : 1 mol CO burned.1 mol CO n 0 mol O2 3.76n 0 mol N2 25Co CO + O 2 CO 2 , H c = -282.99 kJ mol121 mol CO2 (n 0 0.5) mol O2 3.76n 0 mol N2 1400Ca.Oxygen in product gas: n1 = n0 mol O 2 fed -1 mol CO react 0.5 mol O 2 = n0 - 0.5 1 mol CO
The University of Akron - CHEMICAL E - 312
9.61 (cont'd) 101 10 8 = 8.31 10 5 (Ta - 30) + 59.92(Ta2 - 30 2 ) + 0.031(Ta3 - 30 3 ) - 142 10 -5 (Ta4 - 30 4 ) . . Ta = 150 Cc.4400 kg ash/h at 450C40,000 kg coal/h 25C 5.95 10 6 mo l O /h 2 2.24 10 7 mo l N /h 2 3.61 10 5 mo l H O( v )/h 2 150C(= 1
The University of Akron - CHEMICAL E - 312
9.61 (cont'd)30% relative humidity (inlet air):Table B.3y H 2O P = 0.30 p H 2 Ob30 Cg 5.95 10n26+ 2.24 10 7 + n 2. b760 mm Hgg = 0.300 b31824 mm HggB n 2 = 3.61 10 5 mol H 2 O hVolumetric flow rate of inlet air:V=e5.95 106+ 224 10 7 + 3.61
The University of Akron - CHEMICAL E - 312
9.60 (cont'd)o H = ( H c ) CH 4 +FG 450 kmol IJ FG1000 mol IJ FG -890.36 kJ IJ + 2.99 10 kJ H h K H kmol K H mol K h L F kg I OL kJ O + Mm G J P Mb2914 - 105g P = 0 m = 1.32 10 kg steam / h H h K QN kg Q N + d H i = 0 Energy balance on preheater: H = d
The University of Akron - CHEMICAL E - 312
9.60 (cont'd)H p = n2 ( H v ) Ho 2 O(25 C)+ nstack gas (C p ) stack gas (Tstack gas - 25o C) 0.0315 kJ (300 - 25) o C mol o C3 5590 kmol 10 3 mol = 180 kmol H 2 O 10 mol 44.01 kJ + h 1 kmol mol h 1 kmol 7 = 5.63 10 kJ / ho Q = H = ( H c ) CH 4 +=FG
The University of Akron - CHEMICAL E - 312
9.60a.Basis: 450 kmol CH 4 fed hn a ( kmol air / h)@25 o C 0.21 kmol O 2 / kmol 0.79 kmol N 2 / kmol 450 kmol CH 4 / h @ 25 o C m w [kg H 2 O(l) / h] 25 CoCH 4 + 2O2 CO2 + 2H 2OStack gas@300 C n1 (kmol CO 2 / h) n 2 (kmol H 2 O / h) Q( kJ / h) n 3 (
The University of Akron - CHEMICAL E - 312
9.59 (cont'd)CH 4 + 2O 2 CO 2 + 2H 2 O , C 6 H 14 +Theoretical O 2 : 9.30 mol CH 4 sfed19 2O 2 6CO 2 + 7H 2 O + 0.70 mol C 6 H 14 s 9.5 mol O 2 1 mol C 6 H 14 = 25.3 mol O 2 s2 mol O 2 1 mol CH 42 theor. 4100% excess O 2N2b g = 2 bO g balance: 0
The University of Akron - CHEMICAL E - 312
9.59a.Basis:207.4 liters 273.2 K 1.1 atm 1 mol = 10.0 mols s fuel gas to furnace s 278.2 K 1.0 atm 22.4 liters STPb gH = C 6 H 14 ; M = CH 4Qc (kW) n0 (mol/s) y 0 (mol C 6 14/mol) H (1 y 0 (mol CH 4/mol) ) 60C, 1.2 atm T = 55C dp m (kg H 2O( l)/s) w
The University of Akron - CHEMICAL E - 312
9.58 (cont'd) d.XaPxsr 10 10 10 10 10 10 10 10 10 10 10Ta 150 150 150 150 150 150 150 150 150 150 150Ts 700 700 700 700 700 700 700 700 700 700 700Q00.0 5 0.0 5 1 0.0 5 1 0.0 5 1 0.0 5 1 0.0 5 1 0.1 5 0.1 10 0.1 20 0.1 50 0.1 100996 905 813 722 6
The University of Akron - CHEMICAL E - 312
9.58b.Basis : 1 mol fuel gas/sn0 (mol O 2 s) 3.76n0 (mol N 2 s) Ta ( C) 1 mol / s @ 25o C x m (mol CH 4 / mol) x a (mol Ar / mol) (1 - x m - x a ) (mol C 2 H 6 / mol)oStack gas, Ts ( o C) nO2 (mol O 2 / s) 3.76n0 (mol N 2 / s) nCO (mol CO / s) rnCO (
The University of Akron - CHEMICAL E - 312
9.57 (cont'd)Summary of component mass flow ratesStack gas at 600 F, 1 atm 2978 lb - moles CO 2 h 131000 lb m CO 2 h1352.5 lb - moles H 2 O h 24400 lb m H 2 O h 56.4 lb - moles SO 2 h 943 lb - moles O 2 h 17150 lb - moles N 2 h 3620 lb m SO 2 h 30200 l
The University of Akron - CHEMICAL E - 312
9.57Basis : 50000 lb m coal fed h b0.730gb50000glbhmC1b - mole C 12.01 lb m= 3039 1b - mole C h. b0.047gb50000g 101 = 2327 lb - moles H h (does not include H in water) b0.037gb50000g 32.07 = 57.7 lb - moles S h b0.068gb50000g 18.02 = 189 lb - moles
The University of Akron - CHEMICAL E - 312
9.56 (cont'd)Table B.1H = H fo +BzTable B.2, for CH 4 , C2 H 6T25C p dTBB = H fo + Hi (T) for O 2 , N 2 , CO, CO 2 , H 2 O vTable B.8bgEnergy balance:Q = H =n H - n Hi i ii=-2764 kJ 0.130 m 3 fuel= -2.13 104 kJ m3 fueloutin9-67
The University of Akron - CHEMICAL E - 312
9.56CH 4 + 2O 2 CO 2 + 2H 2 O, C 2 H 6 +7 O 2 2 CO 2 + 3H 2 O 2Basis : 100 mol stack gas. Assume ideal gas behavior.n1 (mol CH4 ) n2 (mol C2 H6 ) Vf (m3 at 25C, 1 atm) n3 (mol O2 ) 3.76n3 (mol N2 ) 200C, 1 atm100 mol at 800C, 1 atm 0.0532 mol CO 2 /m
The University of Akron - CHEMICAL E - 312
9.55 (cont'd) 10% excess O2 O 2 fed=1.1(2000 mol/s)=2200 mol/sC balance:(1000 mol CH 4 s )(1 mol Cmol CH 4 ) = (100 )(1) + n3 (1) + 10n3 (1) n3 = 81.8 mol CO s 10n3 = 818 mol CO 2 sH balance: (1000 )( 4 ) = (100 )( 4 ) + 2n4 n4 = 1800 mol H 2 O s O b
The University of Akron - CHEMICAL E - 312
9.54 (cont'd) b.Energy balance on vaporizer:LM Q = H = nH = 1 mol M MN1OP kJ C dT + H + C dT PP mol = 40.33 kJ A A A Q References : CH OHa vf, N (g), O (g), CO (g), H Oalf at 25 Cz64.725plvz10064.7pvTable B.2Table B.1Table B.232222Su
The University of Akron - CHEMICAL E - 312
9.53a.Energy balance: U = 0 n lb m fuel burnedbgo U c (Btu)o 0.00215 U c + 4.62 lb m 0.900 Btu lb m F 87.06 F - 77.00 F = 0 o U c = -19500 Btu lb mafbgbgalb m+ mCv Tout - 77 F = 0bgfb.o The reaction for which we determined U c is1 lb m
The University of Akron - CHEMICAL E - 312
9.51 (cont'd)Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o Cbgbg bg bgbgFor a constant volume process the heat released or absorbed is determined by the internal energy of reaction. nin nout U in U out Substance mol
The University of Akron - CHEMICAL E - 312
9.51 (cont'd)1 mol / s fuel gas 0.85 mol CH 4 / s , 0.15 mol C 2 H 6 / sTheoretical oxygen =2 mol O 2 1 mol CH 40.85 mol CH 4 s+3.5 mol O 2 1 mol C 2 H 60.15 mol C 2 H 6 s= 2.225 mol O 2 / sAssume 10% excess O 2 O 2 fed = 1.1 2.225 = 2.448 mol O
The University of Akron - CHEMICAL E - 312
9.49 (cont'd) a.x0 (kg O) 2 kg H H available for combustion = total H H in H 2 O ; latter is kg coal 16 kg OEq. (9.6-3) HHV = 32,764C + 141,790 H -FG HO + 9261S 8IJ Kin waterAThis formula does not take into account the heats of formation of the ch
The University of Akron - CHEMICAL E - 312
9.47 (cont'd)Energy balance: Q = H = H ro +Q= +n H - n Hi i i out in N 2 outi135 mol h 6360-47.3 kJeffluent solutiona fa f a fa f mol -22,000 kJ - a50,000fa0.003fa1.01f - a 49,850fa0.73f = h+ 15 0.40 + 49,850 0.292SO 2 out1h 1 kW = -6.11 kW 36
The University of Akron - CHEMICAL E - 312
9.47Basis : 50,000 mol flue gas/h50,000 mol/h 0.00300 SO2 0.997 N 2 50C n1 (mol solution/h) 0.100 (NH 4 ) 2 SO 3 0.900 H 2O( l ) 25Cn4 (mol SO2 /h) n5 (mol N 2 /h) 35C n2 (mol NH4 HSO3 /h) 1.5n2 (mol (NH 4)2 SO3 /h) n3 (mol H 2 O(l )/h) 35C90% SO2 rem
The University of Akron - CHEMICAL E - 312
9.46a.H 2SO 4 aq + 2NaOH aq Na 2SO 4 aq + 2H 2 O lb gb gb gbgH 2SO4 solution: : 75 ml of 4M H 2SO 4 solution 4 mol H 2SO 4 1 L acid soln2. b75 mLgb123 g mLg = 92.25 g, (0.3 mol H SO )b98.08 g molg = 29.42 g H SO b92.25 - 29.42g g H O b62.83 g H O
The University of Akron - CHEMICAL E - 312
9.45a.H 2 SO 4 aq + 2 NaOH aq Na 2 SO 4 aq + 2 H 2 O aq1 mol H 2SO 4 49 mol H 2O 25C 2 mol NaOH 38 mol H 2O 25Cb gb gb gb gBasis : 1 mol H 2 SO 4 fed1 mol Na 2SO 4 89 mol H 2O 40CReference states : Na s , H 2 g , S s , O 2 g at 25C2 4bg b g bg
The University of Akron - CHEMICAL E - 312
9.43a.CaCl 2 s + 10H 2 O l CaCl 2 aq , r = 10bgb.(3) b1g - b2g CaCl bsg + 6H Oblg CaCl H = H - H b Hess' s law g = -97.26 kJ mol From (1), H = e H j b g - eH j b g e H j b g = b-64.85 - 794.96g kJ mol = -859.81 kJ mol2 2 2 2 o r3 o r1 o r2 o r1 o f
The University of Akron - CHEMICAL E - 312
9.41 (cont'd)Volume ratio = 66.67 cm 3 NaOH(aq) 20.49 cm H 2 SO 4 (aq)3= 3.25 cm 3 caustic solution / cm 3 acid solutionb.H 2 SO 4 aq : r = 9 mol H 2 O / 1 mol H 2 SO 4b geH jo fsoln= H foejH 2SO 4 l. b g + eH f j H SO baq., r = 9g = b-81132
The University of Akron - CHEMICAL E - 312
9.39Mass of H 2 SO 4 = Mass of solution =3 m3 10 3 L 1 mol H 2 SO 4 1 m3 L 3 m3 10 3 L 10 3 mL 1 m3 L 1.064 g 1 mL= 3000 mol H 2 SO 4FG 98.02 g IJ = 2.941 10 H 1 mol K5g H 2 SO 4= 3192 10 6 g solution .1 mol ) = 161 10 5 mol H 2 O . 18.02 g2 4 M
The University of Akron - CHEMICAL E - 312
9.38 (cont'd)c.1 kg coal contains 45.23 mol C and 35.77 mol H 1 kg coal + nO 2 45.23 CO 2 + (35.77 / 2) mol H 2 O (l) H r = -21,400 kJ = 45.23( H fo ) CO 2 + (35.77 / 2)( H fo ) H 2 O(l) - ( H fo ) coal ( H fo ) coal = -1510 kJ / kgRe ferences : C(s),
The University of Akron - CHEMICAL E - 312
9.38a.Any C consumed in reaction (2) is lost to reaction (1). Without the energy released by reaction (2) to compensate for the energy consumed by reaction (1), the temperature in the adiabatic reactor and hence the reaction rate would drop. Basis : 1.0
The University of Akron - CHEMICAL E - 312
9.37bg bg bg bg CObgg + H Ob vg CO bgg + H bggC 3 H 8 g + 3H 2 O v 3CO g + 7H 2 g2 2 2Basis : 1 mol C 3 H 8 fedHeating gas 4.94 m3 at 1400C, 1 atm n g (mol) 1 mol C 3H 8(g ) 6 mol H 2 O( g ) 125C a n g (mol), 900C Product gas, 800C n 1 (mol C 3H 8 =
The University of Akron - CHEMICAL E - 312
9.362CH 4 C 2 H 2 + 3H 2 C 2 H 2 2C(s) + H 2Basis: 10.0 mol CH 4 (g)/s 1500 Con1 (mol CH 4 / s) n2 (mol C 2 H 2 / s) n3 (mol H 2 (s)/s) n4 (mol C(s)/s)975 kW1500o Ca.60% conversion n1 = 10 1 - 0.600 = 4.00 mol CH 4 sC balance: 10 1 = 4 1 + 2n 2 +
The University of Akron - CHEMICAL E - 312
9.35 (cont'd)C *DECREMENT BY 50 DEG. AND LOOK FOR A SIGN IN PSI DO 10I =1, 20 CALL PSICAL (T, PHI, PSI) IF (PSIL*PSI).LT.0.0) GO TO 40 TLAST = T PSIL = PSI T = T 50. 10 CONTINUE 40 IF (T.GE.0.0) GO TO 45 WRITE (3, 2) 2 FORMAT (1X, 'T LESS THAN ZERO - ERR
The University of Akron - CHEMICAL E - 312
9.35 (cont'd)T =bg145600 -zbT12739.419 + 01147T dT - .z1273Te26.90 + 4.167 10 T jdT . . b1135 + 01392T gdT1273-3gzT T =bg3052 + 36.2T + 0.05943T 2 127240 - 113T - 0.0696T 2 .c.F K I = 1 F K I - 1 = bT g = 0 GH 1 + K JK 1 + bT g GH 1 +
The University of Akron - CHEMICAL E - 312
9.35Basis : 1 mol C 2 H 6 fed to reactor1 mol C H 6 2 1273 K, P atm n (mols) @ T (K), P atm ) n C 2H6 (mol C 2H 6 ) n C 2H4 (mol C 2H 4 n H 2 (mol H 2)a.C2 H 6 C2 H 4 + H 2 , K p =x C2 H 4 x H 2 x C2 H 6P = 7.28 10 6 exp[ -17,000 / T ( K )](1)Frac
The University of Akron - CHEMICAL E - 312
9.34 (cont'd)b g CH bg, 800 Cg: H = 42.84 kJ / mol Sbl, 150 Cg: H = - 1.47 kJ / mol Sbg, 800 Cg: H = 103.83 kJ / mol4 2 3 4CH 4 g, 150 C : H1 = - 3.57 kJ / molCS2 g, 800 C : H5 = 19.08 kJ / mol2 6b g H Sbg, 800 Cg: H = 26.88 kJ / mol CH bg, 700 Cg:
The University of Akron - CHEMICAL E - 312
9.34 (cont'd)Energy balance on reactor:Q = H = H r + =n H -n Hi i iout ini= 41b0.20 f -274.0gb b1gg + 0.20b1 - f gb7.140g + 0.80b1 - f gb3.640g + 0.20 f b3180g + 0.40 f b4.480g .kJ s f = 0.800b.0.04 mol CH4 0.16 mol S(l ) 0.16 mol CS2 0.32
The University of Akron - CHEMICAL E - 312
9.33 (cont'd)n2 n3n1 = 1166 0.651 = 759.1 mol hb g = 1166b1 - 0.651g = 406.9 mol h = 2332b1 - 0.651g = 813.9 mol h Entot = 1980mol 1980 mol 22.4 L STP Vout = h h 1 molb g400 K 1.00 atm1 m3273 K 5.00 atm 103 L= 13.0 m 3 h9.34a.CH 4 g + 4S g CS
The University of Akron - CHEMICAL E - 312
9.33Basis :17.1 m 3 10 3 L 273 K 5.00 atm31 molh 1m 298 K 1.00 atm 22.4 L STP CO g + 2 H 2 g CH 3OH g ,bgH robg =e jH foCH 3OH gbg b g - e H jo fa f = 3497 mol h feedCO(g)= -90.68 kJ mol3497 mol/h n 1 (mol CH 3 OH /h) 0.333 mol CO/mol n 2
The University of Akron - CHEMICAL E - 312
9.32 (cont'd) 3130 Btu lb - mole b g CO bg,1830 Fg: H = H (1830 F) 20,880 Btu lb - mole CObg,1830 Fg: H = H (1830 F) 13,280 Btu lb - mole 0.24 Btu b1830 - 77g F Solid b1830 Fg: H = = 420 Btu lb lb F CO 2 g,400 F : H = HCO 2 (400 F)2CO 2 Table B.9 Table
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9-31 (cont'd)^ ^ ^ ^ ^ ^ H = H ro + (1.00) H 3 + (4.00) H 4 + (6.00) H 5 - (4.00) H1 - (6.00) H 2o ^ ^ ^ Substitute for , H r , and H1 through H 6H = (0.3479 Tout + 4.28 10-5 Tout 2 + 0.9285 10-8 Tout 3 - 4.697 10-12 Tout 4 ) - 972.24 kJ/mol = 0 E-Z So
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9.31 (cont'd)4 mol / s NH 3 6 mol / s O 2 Tin = 200o C n3 (mol O 2 ) n4 (mol NO) n5 (mol H 2 O) ToutO 2 consumed :5 mol O 2 4 mol NH 34 mol NH 3 fed s= 5 mol / s n 3 = (6 - 1) mol O 2 / s = 1 mol O 2 / = 4 mol NO / sNO produced : n 4 =4 mol NO prod
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9.30 (cont'd) c.References : C 2 H 4 g , C 2 H 6 g , C 2 H 5 Cl g , HCl g at 0 CbgC 2 H 4 g, 50 C : H = C p dT0ejzbgbgbg50Table B.22.181 kJ mol50 ^ C2 H 6 ( g, 50 C ) : H = C p dT 2.512 kJ mol Table B.2HCl g, 50 C : H = C p dT0ejz050
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9.29 (cont'd)Energy Balance : H =n H - n Hi i iouti=0 mb 27261 - 125.7 - 4.92 10 6 = 0 . mb = 1892 kg steam hbing9.30a.C 2 H 4 + HCl C 2 H 5ClBasis:1600 kg C 2 H 5Cl l hbg103 g 1 mol = 24800 mol h C 2 H 5Cl 1 kg 64.52 gC n 3 (mol HCl(g)/
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9.29 (cont'd)substance CH 3 OH O2 N2 H 2O HCHO H2 n in kmol / h 75.43 . 2188 82.29 ns - - H in kJ / kmol -195220 3620 3510 -237740 - - n out kmol / h 22.63 n2 82.29 n6 52.80 n5 H out kJ / kmol -163200 18410 17390 -220920 -88800 16810Energy Balance :H =
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9.29 (cont'd) ym P = pm Tm 0.42 760 mmHg = 319.2 mmHg = pm Tmb g b gbgb g Antoine equation p m = 319.2 mmHg Tm = 44.1 C c.Moles HCHO formed :=36 106 kg solution 0.37 kg HCHO 350 days 1 kg solution 1 kmol 30.03 kg HCHO 1 day 24 h= 52.80kmol HCHO
The University of Akron - CHEMICAL E - 312
9.28 (cont'd)QF = H =19400 mol H 2 O 18.0 g 1 kg h 1 mol 10 3 g6a3928 - 104.8fkJkg= 1.34 10Reactor C :kJ h steam generatorbgbg jReferences: C8 H 8 v , C8 H 10 v , H 2 g , H 2 O v at 600 CHi 560 C =ezd560 600bgbg bgbgC pv i dT for C8 H