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Elementary Principles 469
Course: CHEMICAL E 312, Fall 2011School: The University of Akron
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mol 10.13 37.5 C2H4O a. 50 mol C2H4 50 mol O2 208.3333 mol C2H4 50 mol O2 166.6667 18.75 37.5 8.333333 8.333333 Separator mol C2H4 mol O2 mol C2H4O mol CO2 mol H2O 8.333333 18.75 8.333333 8.333333 mol C2H4 mol O2 mol CO2 mol H2O Reactor Xsp = 0.2 Ysp = 0.9 158.3333 mol C2H4 (Ra) Rc-Ra = 158.3333 mol C2H4 (Rc) 0 Procedure: Assume Ra, perform balances on mixing point, then reactor, then separator. Rc is...
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mol 10.13
37.5 C2H4O a. 50 mol C2H4 50 mol O2 208.3333 mol C2H4 50 mol O2 166.6667 18.75 37.5 8.333333 8.333333 Separator mol C2H4 mol O2 mol C2H4O mol CO2 mol H2O 8.333333 18.75 8.333333 8.333333 mol C2H4 mol O2 mol CO2 mol H2O
Reactor Xsp = 0.2 Ysp = 0.9 158.3333 mol C2H4 (Ra) Rc-Ra =
158.3333 mol C2H4 (Rc) 0
Procedure: Assume Ra, perform balances on mixing point, then reactor, then separator. Rc is recalculated recycle rate. Use goalseek to find the value of Ra that (Rc-Ra) drives to zero. b. Xsp 0.2 0.2 0.3 0.3 Ysp 0.72 1 0.75333 1 Yo 0.6 0.833 0.674 0.896 no 158.33 158.33 99.25 99.25
The second reaction consumes six times more oxygen per mole of ethylene consumed. The lower the single pass ethylene oxide yield, the more oxygen is consumed in the second reaction. At a certain yield for a specified ethylene conversion, all the oxygen in the feed is consumed. A yield lower than this value would be physically impossible.
10-21 21
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The University of Akron - CHEMICAL E - 312
10.12 (cont'd)500SP(N + 1) = T RETURN ENDProgram Output: 0.884 mol/s carbon monoxide 0.642 mol/s oxygen 3.777 mol/s nitrogen 0.723 mol/s carbon dioxide T = 1560.43 C10- 20
The University of Akron - CHEMICAL E - 312
10.12 (cont'd)900CC C 100 C200 CFORMAT ('Product Stream', F7.3, ' mols/s carbon monoxide', /, *15X, F7.3, 'mols/s oxygen', /. *15X, F7.3, 'mols/s nitrogen', /. * *15X, F7.3, 'mols/s carbon dioxide', /, 15X, F7.2, 'C') END SUBROUTINE REACTAD (SF, SP,
The University of Akron - CHEMICAL E - 312
10.12 (cont'd)Solution to Problem 10.12 1-CO -2 1.607 0.45 0.36 0.88385 0.02895 4.11E-06 3.55E-09 -2.22E-12 0.1799 5.00E-05 -2.90E-11 -6.57E-12 650 -110.52 -566 1560 -4.7E-08 0 0 -393.5 0.642425 3.777 0.72315 0.0291 0.029 0.03611 1.16E-05 2.20E-06 4.23E-
The University of Akron - CHEMICAL E - 312
10.12 a. Extent of reaction equations: = - SF IX X NU IXb g b g SPb I g = SF b I g + NU b I g I = 1, NEnergy Balance: Reference states are molecular species at feed stream temperature.Q = H = H r + nout H out = 0 0 = NU I HF I + SP Ii =1 I =1Nbg bg
The University of Akron - CHEMICAL E - 312
10.11 (cont'd)C *CHAPTER 10 PROBLEM 11 DIMENSION SF(8), SP(8), CP(7), HF(7) REAL NU(7) DATA NU/1., 5, 0., 3., 4., 0., 0./ DATA CP/0.1431, 0.0330, 0.0308, 0.0495, 0.0375, 0., 0./ DATA HF/103.8, 0., 0., 393.5, 241.83, 0., 0./ COMMON CP, HF SF(1) = 3.348 SF
The University of Akron - CHEMICAL E - 312
10.11 1 A1 + 2 A2 + 3 A3 +. m Am = 0a. Extent of reaction equations: = -[ SF IX X ] NU IXb g b g SPb I g = SF b I g + NU b I g I = 1,2, . N g TP = SPb N + 1g = HF b I g NU b I gN I =1Energy Balance: Reference states are molecular species at 298K. TF
The University of Akron - CHEMICAL E - 312
10.10 (cont'd) e.C *CHAPTER 10 - PROBLEM 10 DIMENSION SF(3), SV(3), SL(2) COMMON A, B, C, CPL, HV, CPV, CPG DATA A, B, C / 7.87863, 1473.11, 230.0/ DATA CPL, HV, CPV, CPG,/ 0.078, 35.27, 0.050, 0.029/ FLOW = 1.0 SF(1) = 0.704*FLOW SF(2) = FLOW SF(1) YC =
The University of Akron - CHEMICAL E - 312
10.10 (cont'd)10 2 1 1 1 5b.variables (n F , x F , TF , P , nv , x v , T , nl , p *, , Q) A material balances Antoine equation Raoult's law energy balance degrees of freedom References: A(l), B(g) at 25oC Substance nin H in A(l) A(v) B(g) - nF x F - H1
The University of Akron - CHEMICAL E - 312
10.9 (cont'd) 600700800NL = NF NVP DO 700 I = 1, N XL(I) = XF(I)*NF/(NF + NV*(PV(I)/P 1) SL(I) = XL(I)*NL XV(I) = XL(I)*PV(I)/P SV(I) = SF(I) SL(I) Q1 = 0. Q2 = 0. DO 800 I = 1, N Q1 = Q1 + CP(I)*SF(I) Q2 = Q2 + HV(I)*XV(I) Q = Q1*(T TF) + Q2*NVP RETUR
The University of Akron - CHEMICAL E - 312
10.9 (cont'd)900901 CC100 200C 300 C500400 900 CFORMAT (A15, F7.4,' mols/s n-pentane', /, *15X, F7.4,' mols/s n-hexane', /, *15X, F7.4,' mols/s n-hephane', /, * 15X, F7.2,' K') WRITE (6, 901) Q FORMAT ('Heat Required', F7.2, 'kW') END SUBROUTINE F
The University of Akron - CHEMICAL E - 312
10.9 (cont'd)where: PV I = 10 A I - B Ibg d b g b g cCb I g + T hi I = 1,2,. N - 1 3 + 3b N - 1g + N + 4 variables b NF , NL, NV , XF ( I ), XL( I ), XV ( I ), PV ( I ), TF , T , P , Qg- N mass balance -1 energy balances - N equilibrium relations - N A
The University of Akron - CHEMICAL E - 312
10.8 (cont'd)901 CFORMAT ('Heat Required', F7.2,' kW') END SUBROUTINE FLASN2 (SF, SL, SV, T, P, Q) REAL NF, NL, NV DIMESION SF(3), SL(3), SV(3) COMMON A1, B1, C1, C2, CP1, CP2, NV1, NV2 Vapor Pressure PV1 = 10.*(A1 B1/(T 273.15 + C1) PV2 = 10.*(A2 B2/(T
The University of Akron - CHEMICAL E - 312
10.8 a. Let Bz = benzene, Tl = toluene * Antoine equations: pBz = 106.89272-1211.033/(T + 220.790) (=1350.491)* pTl = 106.95805-1346.773/(T + 219.693) (=556.3212) * * * * Raoult's law: xBz = (P - pTl )/(pBz -pTl ) (=0.307) , yBz = xBz pBz / P ( = 0.518)
The University of Akron - CHEMICAL E - 312
10.7 (cont'd) b. C*CHAPTER 10 - PROBLEM 7 DIMENSION SF(8), S1(8), S2(8) FLOW = 150. N=3 SF(1) = 0.35*FLOW SF(2) = 0.57*FLOW SF(3) = 0.08*FLOW SF(8) = 315. X1 = 0.60 CALL SPLIT (SF, S1, S2, X1, N) WRITE (6, 900)' STREAM 1', S1(1), S1(2), S1(3), S1(B) WRIT
The University of Akron - CHEMICAL E - 312
10.6 (cont'd) c.nr n2 = 100 + 0.85nr n3 = nr 1 - 0.85bgk =1 k = 2 k = 3 100.0 132.3 1515 . 185.0 212.5 228.8 15.0 120.25 79.75 n6 = 67.69 n r = 132.3 19.85 138.1 94.21 80.76 1515 . 22.73 148.7 102.8 88.55 163.0n4 = 0.65n2 n5 = n 2 + n 3 - n 4 n 4 + n
The University of Akron - CHEMICAL E - 312
10.6 n - C 4 H 10 i - C 4 H 10 n - B = i - Bn 1 (mol n-B) mixer n 2 (mol n-B) n 3 (mol i-B)bgn 4 (mol n-B) n 5 (mol i-B) still n 6 (mol) x 6 (mol n-B/mol) (1 x 6)(mol i-B/mol)reactorn r (mol) x r (mol n-B/mol) (1 x r)(mol i-B/mol)a. Mixer:5 variab
The University of Akron - CHEMICAL E - 312
10.5 a. n 1 (mol CO)n 2 (mol H2 ) reactor n 3 (mol C3 H6 )n 4 (mol C3 H6 ) n 5 (mol CO) n 6 (mol H2 ) n 7 (mol C7 H8 O) n 8 (mol C4 H7 OH) n 9 (kg catalyst)Flash tankn 16(mol C3 H6 ) n 11(mol C3 H ) 6 n 12(mol CO) n 17(mol CO) n 13(mol H ) Separation
The University of Akron - CHEMICAL E - 312
10.3 (cont'd) c. Design set: x B , x c , T0 , n2 , Ql lq qSpecifying x B determines n2 impossible design set.d. Design set: x B , x c , T0 , Pex , QCalculate n2 from x B , n3 from x B n0 from x B , x c and Pex n1 from C material balance, n4 from tota
The University of Akron - CHEMICAL E - 312
10.2 10 variables n1 , n2 , n3 , n4 , x1 , x 2 , x 3 , x 4 , T , Pbg g b g bg-2 material balances* * -2 equilibrium relations: [ x 3 P = x 4 PB T , 1 - x 3 P = 1 - x 4 PC T ]bgb16 degrees of freedoma. A straightforward set:ln , n , n , x , x , Tq
The University of Akron - CHEMICAL E - 312
CHAPTER TEN10.1 b. Assume no combustionn 1 (mol gas),T1 (C) x 1 (mol CH4 /mol) x 2 (mol C2 H6 /mol) 1 x 1 x 2 (mol C3 H8 /mol) n 2 (mol air), T2 (C) n 3 (mol), 200C y 1 (mol CH /mol) 4 y 2 (mol C2 H6 /mol) y 3 (mol C3 H8 /mol) 1 y 1 y 2 y 3 (mol air/mol
The University of Akron - CHEMICAL E - 312
9.70 (cont'd)Air fed (25% excess) = 1.25(4.76 kmol air kmol O2 kmol air )(338 ) = 2011 kmol O2 d d2011 kmol 29 kg 1 tonne E = 58.3 tonnes/ d (air to boiler) d kmol 103 kgEnergy balance on boiler air preheater:2011 kmol 103 mol 2.93 kJ kJ ^ Q0 = = 5.8
The University of Akron - CHEMICAL E - 312
9.70 (cont'd)Solids balance on dryer: 0.35 24,000 kg / d = 0.75n2 n2 = 11200 kg / d F11.2 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
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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