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6 Pages

### Elementary Principles 420

Course: CHEMICAL E 312, Fall 2011
School: The University of Akron
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Word Count: 359

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: 9.57 Basis 50000 lb m coal fed h b0.730gb50000glb h mC 1b - 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 H O h . b0118gb50000g = 5900 lb ash h 2 m 50,000 lb m coal/h 3039 lb-moles C/h 2327 lb-moles H/h 57.7 lb-moles S/h 189 lb-moles H 2O/h 5900 lb m ash/h 77F,...

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: 9.57 Basis 50000 lb m coal fed h b0.730gb50000glb h mC 1b - 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 H O h . b0118gb50000g = 5900 lb ash h 2 m 50,000 lb m coal/h 3039 lb-moles C/h 2327 lb-moles H/h 57.7 lb-moles S/h 189 lb-moles H 2O/h 5900 lb m ash/h 77F, 1 atm (assume) n 1 (lb-moles air/h) 0.210 O 2 0.790 N 2 77F, 1 atm (assume) Stack gas at 600F, 1 atm (assume) n 2 (lb-moles CO2 /h) n 3 (lb-moles H2 O/h) n 4 (lb-moles SO 2 /h) n 5 (lb-moles O2 /h) n 6 (lb-moles N2 /h) m 7 (lbm fly ash/h) m 8 (lbm slag/h) at 600F 0.287 lb mC/lb m 0.016 lb mS/lb m 0.697 lb mash/lb m a. Feed rate of air : O 2 required to oxidize carbon C + O 2 CO 2 = b g 3039 lb - moles C 1 lb - mole O 2 h 1 lb - mole C = 3039 lb - moles O 2 h Air fed: n1 = 1.5 3039 lb - moles O 2 h 1 fed mole air 0.210 mole O 2 = 21710 lb - moles air h 30% ash in coal emerges in slag 0.697m8 = 0.30 5900 lb m h m8 = 2540 lb m slag / h m7 = 0.700 5900 = 4130 lb m fly ash h C balance: 3039 lb - moles C h = n 2 + 0.287 2540 12.01 n 2 = 2978 lb - moles CO 2 h M CO 2 = 44.01 b g b g b g b gb g 131 10 5 lb m CO 2 h . H balance: 2327 lb - moles H h + 189 2 = 2n 3 n 3 = 1352.5 lb - moles H 2 O h M H 2 O =18.02 b g b gb g 2.44 10 4 lb m H 2 O h M N 2 = 28.02 N 2 balance: n6 = 0.790 21710 lb - moles h = 17150 lb - moles N 2 h b g 4.81 105 lb m N 2 h S balance: 57.7 lb - moles S h = 1 n 4 + 0.016 2540 32.06 n 4 = 56.4 lb - moles SO 2 h M SO 2 = 64.2 b g bg bair g b g 3620 lb m SO 2 h O balance: 189 1 + 0.21 21710 2 = 2978 2 + 1352.5 1 + 56.4 2 + 2n5 b gb g b gb bcoalg gb g b bCO g 2 gb g b eH O j 2 gb g b gb g bSO g 2 bO g 2 n5 = 943 lb - moles O 2 h 30200 lb m O 2 h 9- 68
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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
The University of Akron - CHEMICAL E - 312
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
The University of Akron - CHEMICAL E - 312
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
The University of Akron - CHEMICAL E - 312
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
The University of Akron - CHEMICAL E - 312
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)/
The University of Akron - CHEMICAL E - 312
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 =
The University of Akron - CHEMICAL E - 312
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
The University of Akron - CHEMICAL E - 312
9.28 (cont'd)Reactor :.n 3 (mol C8H10 /h) n 4 (mol H2O( v )/h) 600C Qc (kJ/h)n5 (mol C8H10 /h) n 4 (mol H2O(v)/s) v 960 (mol C8H8 /s) 960 (mol H2 /s) 560C35% 1-pass conversion 0.35n3 mol C8 H 10 react h n3 = 2740 mol C 8 H10 h fed to reactorbg1 m
The University of Akron - CHEMICAL E - 312
9.27 (cont'd)Energy balance: H = 0 (Assume adiabatic) n eHii , out- Hi , in =j n C bTi piout- Tingi=0LM10 lb - moles C H h N912B 120 lb m 0.40 Btu 200 F - 400 F + 2.50 42.08 0.57 200 F - 400 1 lb - mole 1b m FOPe Qj b gbC 3H 6gb ge+
The University of Akron - CHEMICAL E - 312
9.27a.Basis:1200 lb m C 9 H 12 1 lb - mole = 10.0 lb - moles cumene produced h h 120 lb m Overall process :n 1 (lb-moles/h) 0.75 C 3H 6 0.25 C 4H 10 n 2 (lb-moles C 6 H 6 /h) n 3 (lb-moles C 3 H 6 /h) n 4 (lb-moles C 4 H10 /h)10.0 lb-moles C H12 /h 9
The University of Akron - CHEMICAL E - 312
9.26 (cont'd)Overall Process Substance C2 H4 O2 C2 H4 O CO2 H2 O l ReactorninHinnoutHout(mol) (kJ / mol) (mol) (kJ / mol) - - 0.500 52.28 . 0625 0substance C2 H 4 O2 C2 H 4 O CO2 H2O gi iHout (mol) (kJ / mol) (mol) ( kJ / mol) . 2 79.26 150 79.26
The University of Akron - CHEMICAL E - 312
9.26a.Basis: 2 mol C 2 H 4 fed to reactorC2 H 4 g +1 O 2 (g) C 2 H 4 O g 2 C 2 H 4 + 3O 2 2CO 2 + 2H 2 ObgbgQr (kJ) heat n1 (mol C 2H 4) n2 (mol O 2 ) 25C 2 mol C2 H4 1 mol O2 450C reactorn 6 (mol CO2 ) n 7 (mol H 2 O(l ) 25C separation n 3 (mol C
The University of Akron - CHEMICAL E - 312
9.25 (cont'd)Q=100 J 85 s s1 kJ 1000 J= 8.5 kJTable B.1Hro = (Hfo ) HCHO + (Hfo ) H2O - (Hfo ) CH4 = -282.88 kJ / molU ro = H ro - RT (=B. . b(-11590) + (-24183) - (-74.85)g kJ / molgaseous productsi -gaseous reactantsi)1 kJ 10 3 J = -282.
The University of Akron - CHEMICAL E - 312
9.25a.CH 4 ( g) + O 2 ( g) HCHO(g) + H 2 O(g)10 L, 200 kPa n0 (mol feed gas) at 25C 0.851 mol CH4/mol 0.15 mol O2 /mol Q (kJ)Basis : n0 = 200 kPa 1000 Pa 10 L 10 -3 m 3 1 kPa 1Ln3 (mol HCHO) n4 (mol H2O) n5 (mol CH4) T (C), P(kPa), 10L1 mol K 8.314
The University of Akron - CHEMICAL E - 312
9.24a.A+B C 2C D + B(1) (2)Basis: 1 molx AO (mol A / mol) x BO (mol B / mol) x IO (mol I / mol) n A (mol A) n B (mol B) nC (mol C) n D (mol D) n I (mol I) T ( C)Fractional conversion:fA =mol A consumed x AO - n A = n A = x AO (1 - f A ) mol A feed
The University of Akron - CHEMICAL E - 312
9.23 (cont'd) b. Basis : 1000 kg CaCO3 fed 10.0 kmol CaCO3 CaCO 3 (s) CaO(s) + CO 2 (g) 2CO + O 2 2CO 210 kmol CaCO3 25 oC 200 kmol at 900oC 0.75 N2 0.020 O2 0.090 CO 0.14 CO2Product gas at 900oC n2 (kmol CO2 ) n3 (kmol N2 ) n4 (kmol CO)n1 [kmol CaO(s)
The University of Akron - CHEMICAL E - 312
9.22(cont'd)Scale factor:a.bn g bn g55 actual 5 basis=29.3 lb m H 2 O 4h1b - mole H 2 O118.016 lb m H 2 O 15.0 lb - moles H 2 O= 0.02711 h -1V0 = 6.218 10 5 ft 3 0.02711 h -1 = 169 10 4 ft 3 h feed .Vpb.d id i = d6.443 10 ft id0.02711 h i =
The University of Akron - CHEMICAL E - 312
9.22 (cont'd)Energy balance on reactor (excluding cooling jacket)References : C s , H 2 g , O 2 g , N 2 g at 25 C 77 Fsubstance C 6 H 5CH 3 O2 N2 C 6 H 5CHO CO 2 H 2O nin Hinbg b g b g b g100 200 752 - - -ejnout Hout H4 H5 H6 H7 H8 H9 86.5 182.5 7
The University of Akron - CHEMICAL E - 312
9.22 (cont'd)Plan of attack: % excess air n0 0.5% CO 2 formation n4 C balance n1 H balance n5 O balance n2100% excess air:n0 = 100 lb - moles C 6 H 5CH 3 1 mol O 2 reqd 1 mole C 6 H 5CH 3Ideal gas equation of state V0 E.B. on reactor Q E.B. on jacket
The University of Akron - CHEMICAL E - 312
9.21 (cont'd)References: C ( s ) , H 2 ( g ) , O2 ( g ) at 25 C, N 2 ( g ) at 310 Csubstance C2 H 4 H2O N2 C2 H 5 OH nin ^ H in nout (mol) 0.510 0.3414 0.096 0.02417 1.415 10-3^ H out (kJ/mol) ^ H1(mol) (kJ/mol) ^ 0.537 H10.367 0.096 - -^ H2 0 -
Duke - DIVINITY - XianTheo32
Notes Christian Theology 32 September 28, 2007This isn't a creation like a top that YHWH winds up and then lets spin out by itself. The people of Israel was called into being. It has being only in its calling from YHWH. And he stays with them. Even in th
Duke - DIVINITY - XianTheo32
Lecture 1 The Theological Enterprise; or, the Music of the Christian Life August 29, 2007 A class in Contemporary Theology: &quot;When we look back, we're always looking at the present.&quot; A. The Problem of Modern Theology 1. Theology as a &quot;WhiteMale&quot; Enterprise
The University of Akron - CHEMICAL E - 312
9.20 (cont'd)Air g, 150 C : H2bgTable B.8=B3.67 kJ molNO g, 700 C : H3 = 90.37 + H 2 O g, 700 C : H 4 N 2 g, 700 C : H5bgbgTable B.1, Table B.8zTable B.1,Table B.2 70025C p dT=B111.97 kJ mol= =B-216.91 kJ mol 20.59 kJ molbgTable
Duke - DIVINITY - XianTheo32
Friday: August 31, 2007The Theological Enterprise; or the Music of the Christian Life, Part 2 Contextualists: Both modes of theology represent modes of abstraction whether it is whiteness or black power. Don't get lost in the language of whiteness, and o
The University of Akron - CHEMICAL E - 312
9.19 (cont'd)Makeup water required : 495,000 gal - 46,360 gal C 2 H 5OH 25 gal mash 2.6 gal C 2 H 5OH = 4.9 104 galb.46,360 gal C2 H5OH 1 bu 1 acre 1 batch 24 h 330 days acres = 175 105 . Acres reqd. : 1 batch 2.6 gal C2 H5OH 101 bu 8 h 1 day 1 year ye
Duke - DIVINITY - XianTheo32
Wednesday, September 05, 2007 Notes John Coltrane, Naima Live, 1965 (orgasmistic climax) Michael Brecker Naima (soloist on UTube) &quot;Brecker tries to blow his reality through Naima. It's *his* Naima. The most intimate engagement with the Naima that came bef
Duke - DIVINITY - XianTheo32
Notes Christian Theology 32 September 7, 2007The Name of God in respect to the name Father (of the Trinity) Next week, the Name of Son. The people of election. During the song together on Wednesday we're charged with singing faithfulness and fidelity to
The University of Akron - CHEMICAL E - 312
9.18 (cont'd)1.2 1.4 1.6 1.8 2.0 400 400 400 400 400 0.5 0.5 0.5 0.5 0.5 400 400 400 400 400 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.7 0.9 1.1 1.3 1.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 2.995 2.995 2.995 2.995 2.995 5.335 5.335 5.335 5.335 5.335 2
Duke - DIVINITY - XianTheo32
Notes Christian Theology 32 September 12, 2007The process he is taking us through is part of what he it trying to teach us. Theologians typically do their work by downloading the information to the listener. He is trying a different way: he believes one
The University of Akron - CHEMICAL E - 312
9.18 (cont'd) c. n0 = 2.0 mol CO, T0 = 350 K, T = 550 K, and X = 0.700 mol FeO reacted/mol FeO fed n1 = 1 - 0.7 = 0.3, n2 = 2 - 0.7 = 1.3, n3 = 0.7, n4 = 0.7, = 0.7 ^ ^ ^ Summary : H 0 = 1.520 kJ/mol, H1 = 13.48 kJ/mol, H 2 = 7.494 kJ/mol, ^ ^ H 3 = 7.20
Duke - DIVINITY - XianTheo32
Notes Christian Theology 32 September 14, 2007Questions he received. One: Knowledge. How do we know what we know? 1. Theological knowledge, epistemology how do we know something. 2. Election getting started when we fill it out with Christology. 3. Provid
Duke - DIVINITY - XianTheo32
Notes Christian Theology 32 September 19, 2007By notetaker: Stephanie Anne Jones (&lt;stephanie.a.jones@duke.edu&gt; Wednesday, September 19, 2007: The Formation of the World in Christ: Jesus of Nazareth, the Israel of God To this point in the course, the firs
The University of Akron - CHEMICAL E - 312
9.18b.References : FeO(s), CO(g), Fe(s), CO 2 (g) at 25o C Substance FeO CO Fe CO 2 nin nout Hin Hout ( mol) ( kJ / mol) ( mol) ( kJ / mol) 1.00 0 n1 H1 n0 H0 n2 H2 - - - -in H inn3 n4H3 H4Q = H ro +nout H out-n Q = H ro + n1 H1 + n2 H 2 + n3 H
Duke - DIVINITY - XianTheo32
Notes Christian Theology 32 September 21, 2007Christology: Burning love of God expressed. But what is the perfect response to the call of YHWH? And at the same time, it is YHWH's call. It is the articulation of YHWH's call to humanity. It is the perfect