ch04.pdf

4- 1 CHAPTER FOUR 4.1 a. Continuous, Transient b. Input ? Output = Accumulation No reactions ? Generation = 0, Consumption = 0 600 300 300.. . kg s kg s kg s ?=?= dn dt dn dt c. t == 100 1 1 300 333 s . . m 1000kg m s kg 3 3 4.2 a. Continuous, Steady State b. kk=? = =?? =00CC C AA0 A c. Input ? Output ? Consumption = 0 Steady state ? Accumulation = 0 A is a reactant ? Generation = 0 VC VC kVC C C kV V AAAA A m s mol m m s mol m mol s 3 3 3 3 F H G I K J F H G I K J = F H G I K J F H G I K J + F H G I K J ?= + 0 0 1 4.3 a. 100 kg / h 0.550 kg B / kg 0.450 kg T / kg m v kg / h 0.850 kg B / kg 0.150 kg T / kg bg m l kg / h 0.106 kg B / kg 0.894 kg T / kg b g Input ? Output = 0 Steady state ? Accumulation = 0 No reaction ? Generation = 0, Consumption = 0 (1) Total Mass Balance: 100 0. kg / h = +mm vl (2) Benzene Balance: 0550 1000 0850 0106.. . . ×=+ kg B / h mm vl Solve (1) & (2) simultaneously ? . .mm vl = =59 7 40 3kg h, kg h b. The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg). The balance equations are also identical (initial input = final output). c. Possible explanations ? a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors. 4- 2 4.4 b. n(mol) mol N mol molCH mol 2 4 0 500 0 500 . . 0 500 28 1 0 014 . . n n mol N gN mol N kg 1000g kg N 2 2 2 2 bg bg= c. 100 0. g/s g C H g g C H g g C H g 26 38 410 x x x E P B bg bg bg . . n x x E E E = = 100 1 453593 30 3600 26 45 g C H s lb g lb - mole C H lb C H s h lb - mole C H / h 26 m26 m26 26 bg bg d. lb - mole H O s lb - mole DA s lb - moleO lb - mole DA lb - mole N lb - mole DA 2 2 2 . . n n 1 2 021 079 bg R S | T | U V | W | ( ) 2 2 2 O2 2 12 HO 12 22 O 12 0.21 lb-mole O /s lb-mole H O lb-mole 0.21 lb-mole O lb-mole nn n x nn n x nn = ?? = ?? + ?? ?? = ?? + ?? e. ( ) () () 2 2 NO 2 NO 2 4 mol 0.400mol NO mol mol NO mol 0.600 mol N O mol n y y? ( ) 24 2 NO NO 2 4 0.600 mol N Onn y??=? ?? 4.5 a. 1000 lb C H / h m38 Still . . n 7m m38 m m36 m lb / h lb C H / lb lb C H / lb bg 097 003 . . n 6m m38 m m36 m lb / h lb C H / lb lb C H / lb bg 002 098 n n 1m38 2m36 lb C H / h lb C H / h bg n n n n 1m38 2m36 3m 4 4m2 lb C H / h lb C H / h lb CH / h lb H / h bg bg bg n n 3m 4 4m2 lb CH / h lb H / h bg bg n 5m lb / hbg n n n 1m38 2m36 5m lb C H / h lb C H / h lb oil / h bg bg Reactor Absorber Stripper Compressor Basis: 1000 lb m C 3 H 8 / h fresh feed (Could also take 1 h operation as basis - flow chart would be as below except that all / h would be deleted.) Note: the compressor and the off gas from the absorber are not mentioned explicitly in the process description, but their presence should be inferred. 4- 3 4.5 (cont?d) b. Overall objective: To produce C 3 H 6 from C 3 H 8 . Preheater function: Raise temperature of the reactants to raise the reaction rate. Reactor function: Convert C 3 H 8 to C 3 H 6 . Absorption tower function: Separate the C 3 H 8 and C 3 H 6 in the reactor effluent from the other components. Stripping tower function: Recover the C 3 H 8 and C 3 H 6 from the solvent. Distillation column function: Separate the C 3 H 5 from the C 3 H 8 . 4.6 a. 3 independent balances (one for each species) b. 7 unknowns ( , , ,,,,mmmx y yz 13 52244 ) ? 3 balances ? 2 mole fraction summations 2 unknowns must be specified c. yx 22 1=? A Balance: 5300 1200 0 70 23 xm kg A h kg A h F H G I K J =+ F H G I K J . bgbg Overall Balance: mmm 135 5300 1200+ F H G I K J =+ + F H G I K J kg h kg h B Balance: 0 03 5300 1200 0 60 12 45 . . mx ym+ F H G I K J =+ F H G I K J kg B h kg B h zy 44 1070=? ?. 4.7 a. 3 independent balances (one for each species) b. Water Balance: 400 0 885 0 995 356 g mg m R R min . min . gH O g gH O g gmin 22 =?= bg b g Acetic Acid Balance: 400 0115 0 005 0 096bgb g.. . gCH OOH min g CH OOH min 33 F H G I K J =+ F H G I K J mm RE ? = m E 461g min Overall Balance: mmmm CREC + F H G I K J =+ F H G I K J ?=400 417 g min g min gmin c. 0115 400 0 005 356 0 096 461 44 44.. .bgbgbgbg bgbg? F H G I K J = F H G I K J ? g min g min gmin= gmin 4- 4 4.7 (cont?d) d. Extractor CH COOH HO 3 2 HO someCH COOH 2 3 CHOH CH COOH 49 3 CHOH 49 Distillation Column CHOH 49 CH COOH 3 4.8 a. 120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/egg X-large: 25 broken eggs/min 35 unbroken eggs/min Large: broken eggs/min unbroken eggs/min n 1 n 2 b. ( ) ()() 12 12 1 21 120 25 45 eggs min 50 11 390.30 120 25 nn nn n nn ?=+++ ?+= = ? ? ? ==+ ? ? c. nn 12 50+= large eggs min n 1 large eggs broken/50 large eggs ==11 50 0 22bg. d. 22% of the large eggs (right hand) and 25 70 36%bg? of the extra-large eggs (left hand) are broken. Since it does not require much strength to break an egg, the left hand is probably poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed. 4.9 a. m 1 015 085 lb strawberries lb S / lb lb W / lb m mm mm bg . . m 2 lb S sugar m ch m 3 lb W evaporated m bg 100 0667 0333 . . . lb jam lb S / lb lb W / lb m mm mm b. 3 unknowns (mmm 123 ,,) ? 2 balances ? 1 feed ratio 0 DF c. 12 12 1m 2m Feed ratio: / 45 / 55 (1) S balance: 0.15 0.667 (2) Solve simultaneously 0.49 lb strawberries, 0.59 lb sugar mm mm mm = += ?= = X-large: 25 broken eggs/min 45 unbroken eggs/min 4- 5 4.10 a. 300 0750 0250 1 gal lb lb C H OH / lb lb H O / lb m m25 m m2 m m bg . . V m 40 2 0400 0600 gal lb lb C H OH / lb lb H O / lb m m25 m m2 m bg bg . . m 3 0600 0400 lb lb C H OH / lb lb H O / lb m m25 m m2 m bg . . 4 unknowns (mmV m 12403 ,,,) ? 2 balances ? 2 specific gravities 0 DF b. m 1 300 1 74805 0877 624 2195= × = gal ft gal lb ft lb 3 m 3 m . .. Overall balance: mm m 12 3 + = (1) C 2 H 5 OH balance: 0 750 0 400 0 600 12 3 .. .mmm+ = (2) Solve (1) & (2) simultaneously ? = =mm 23 1646 3841lb lb m, m , V 40 1646 7 4805 1 207= × = lb ft 0.952 62.4lb gal ft gal m 3 m 3 . 4.11 a. . . n 1 0 0403 0 9597 mol / s mol C H / mol mol air / mol 38 bg . . n 2 021 079 mol air / s mol O / mol mol N / mol 2 2 bg . . n 3 0 0205 0 9795 mol / s mol C H / mol mol air / mol 38 bg 3 unknowns ( , , nnn 123 ) ? 2 balances 1 DF b. Propane feed rate: 0 0403 150 3722 11 . nn= ? = mol / sb g Propane balance: 0 0403 0 0205 7317 133 . . nnn= ? = mol/sb g Overall balance: 3722 7317 3600 22 + = ? = mol/sb g c. >. The dilution rate should be greater than the value calculated to ensure that ignition is not possible even if the fuel feed rate increases slightly. 4- 6 4.12 a. 1000 0500 0500 kg / h kgCH OH / kg kg H O / kg 3 2 . . . . m kg / h kgCH OH / kg kg H O / kg 3 2 bg 0 960 0 040 673 1 kg / h kgCH OH / kg kg H O / kg 3 2 x x bg ? 2 unknowns ( ,m x ) ? 2 balances 0 DF b. Overall balance: 1000 673 327= + ? = mmkg / h Methanol balance: 0 500 1000 0 960 327 673 0 276.. .bg bgbg=+?=xxkgCH OH / kg 3 Molar flow rates of methanol and water: 673 0 276 1000 32 0 580 10 673 0 724 1000 18 271 10 3 4 kg h kgCH OH kg g kg mol CH OH gCH OH mol CH OH / h kg h kg H O kg g kg mol H O gH O mol H O / h 33 3 3 22 2 2 . . . . . =× =× Mole fraction of Methanol: 580 10 580 10 271 10 0176 3 34 . .. . × ×+ × = mol CH OH / mol 3 c. Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the system is not at steady state. 4.13 a. Feed Reactor effluent Product Waste 2253kg 2253kg R=388 1239 kg R = 583 m w kg R = 140 bg Reactor Purifier Analyzer Calibration Data x p = 0.000145R 1.364546 0.01 0.1 1 100 1000 R x p 4- 7 4.13 (cont?d) b. Effluent: x p ==0 000145 388 0 494 1 3645 .. . bg kg P / kg Product: x p 0 000145 583 0 861 13645. bg kg P / kg Waste: x p ==0 000145 140 0123 13645 .. . bg kg P / kg Efficiency =×= 0 861 1239 0 494 2253 100% 958% . . . bg bg c. Mass balance on purifier: 2253 1239 1014= + ? =mm ww kg P balance on purifier: Input: 0 494 2253 1113. kg P / kg kg kg Pbgbg= Output: 0 861 1239 0123 1014 1192.. kg P / kg kg kg P / kg kg kg Pb bgb bg+= The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady state; additional reaction occurs in purifier; normal data scatter. 4.14 a. . . n 1 00100 09900 lb- mole / h lb- mole H O / lb - mole lb- mole DA / lb - mole 2 b g n v 2 2 lb- mole HO / h ft / h 2 3 b g d i . . n 3 0100 0900 lb - mole / h lb- mole H O / lb - mole lb- mole DA / lb- mole 2 b g 4 unknowns ( , , , nnnv 123 ) ? 2 balances ? 1 density ? 1 meter reading = 0 DF Assume linear relationship: v aR b= + Slope: a vv RR = ? ? = ? ? = .. . 21 21 96 9 40 0 50 15 1626 Intercept: bv aR a =? = ? = .. . 1 40 0 1626 15 15 61b g ..v 2 1626 95 15 61 170=+=bg c h ft / h 3 . n 2 170 62 4 589== ft h lb ft lb - mol 18.0lb lb - moles H O / h 3 m 3 m 2 bg DA balance: 0 9900 0 900 13 . . nn= (1) Overall balance: nn n 12 3 + = (2) Solve (1) & (2) simultaneously ? = = , nn 13 5890 6480 lb - moles / h lb - moles / h b. Bad calibration data, not at steady state, leaks, 7% value is wrong, v? R relationship is not linear, extrapolation of analyzer correlation leads to error. 4- 8 4.15 a. 100 0600 0050 0350 kg / s kg E / kg kgS / kg kg H O / kg 2 . . . . . m kg / s kg E / kg kg H O / kg 2 bg 0 900 0100 m x x xx E S ES kg / s kg E / kg kgS / kg kg H O / kg 2 bg bg bg bg1?? 3 unknowns ( ,,mx x ES ) ? 3 balances 0 DF b. Overall balance: 100 2 50 0=?= .mm kg / sb g S balance: 0 050 100 50 0100..bg bg b g=?=xx SS kgS / kg E balance: 0 600 100 0 900 50 50 0 300.. .b g bg bg=+?=xx EE kg E / kg kg Ein bottomstream kg Ein feed kg Ein bottomstream kg Ein feed == 0300 50 0600100 025 . . . bg bg c. xaR x ab R b xx RR axbR a xR R x a b b =? = + == = =? = ? =??=× =× = F H G I K J = × F H G I K J = ? ? ? ln ln ln ln / ln / ln . / . ln / . ln ln ln ln . . ln . . . . . . . . bg bg bg bg bg bg bg b g b g b g b g b g 21 21 11 3 3 1 491 1 3 1 1 491 0 400 0100 38 15 1491 0100 1491 15 6 340 1764 10 1764 10 0900 1764 10 655 d. Device not calibrated ? recalibrate. Calibration curve deviates from linearity at high mass fractions ? measure against known standard. Impurities in the stream ? analyze a sample. Mixture is not all liquid ? check sample. Calibration data are temperature dependent ? check calibration at various temperatures. System is not at steady state ? take more measurements. Scatter in data ? take more measurements. 4- 9 4.16 a. 400 0098 1213 0323 .. . . mol H SO L of solution kg H SO mol H SO L of solution kgsolution kg H SO / kgsolution 24 24 24 24 = bg b. v 1 100 0200 0800 1139 L kg kg H SO / kg kg H O / kg SG 24 2 bg . . .= v m 2 2 0600 0400 1498 L kg kg H SO / kg kg H O / kg SG 24 2 bg bg . . .= v m 3 3 0323 0677 1213 L kg kg H SO / kg kg H O / kg SG 24 2 bg bg . . .= 5 unknowns (vvvmm 123 2 3 ,,, , ) ? 2 balances ? 3 specific gravities 0 DF Overall mass balance: Water balance kg kg 100 0 800 100 0 400 0 677 44 4 144 23 23 2 3 += += U V W ? = = mm mm m m:. . . . bg v 1 100 1139 87 80== kg L kg L20%solution . . v 2 44 4 1498 29 64 60%== . . . kg L kg L solution v v 1 2 87 80 29 64 296 60% == . . . L 20%solution L solution c. 1250 44 4 144 1498 257 kg P h kg60%solution kg P L kgsolution L/h . . = 4.17 m 1 025 075 kg @$18 / kg kg P / kg kg H O / kg 2 bg . . m 2 012 088 kg @$10 / kg kg P / kg kg H O / kg 2 bg . . 100 017 083 . . . kg kg P/ kg kg H O / kg 2 Overall balance: mm 12 100+ = . (1) Pigment balance: 0 25 012 017 100 12 .. ..mm+ = b g (2) Solve (1) and (2) simultaneously ? = =mm 12 0 385 0 615.,.kg25% paint kg12% paint Cost of blend: 0 385 00 0 615 00 08. $18. . $10. $13.b g b g+ = per kg Selling price:110 08 39. $13. $14.b g= per kg 4- 10 4.18 a. 100 0800 0200 kg kgS / kg kg H O / kg 2 . . m m 2 3 kgS kg H O 2 b g b g m 1 kg H O 85%of entering water 2 bgb g 85% drying: m 1 0 850 0 200 100 17 0==.. .bgbg kg H O 2 Sugar balance: m 2 0800100 800bg kgS Overall balance: 100 17 80 3 33 = + + ? =mmkg H O 2 x w = + = 3 380 00361 kg H O kg kg H O / kg 2 2 bg . m mm 1 23 17 80 3 0205 + = + = kg H O kg kg H O / kg wetsugar 2 2 bg b. 1000 3 100 30 tonswet sugar day tonsH O tons wet sugar tons H O / day 2 2 = 1000 0 800 2000 15 365 810 7 tonsWS day tons DS ton WS lb ton lb days year per year m m .$. $8.=× c. xxxx xx x x wwww ww w w =+++= =?+?= =± == 1 10 0 0504 1 9 0 00181 0 0504 3 0 00181 0 0450 0 0558 12 10 1 2 10 2 ... . ... . .. ., . bg bgb g bg kg H O / kg SD kg H O / kg Endpoints Lower limit Upper limit 2 2 d. The evaporator is probably not working according to design specifications since x w =<0 0361 0 0450... 4.19 a. v m SG 1 1 100 m kg H O 3 2 ch bg = . v SG 2 400 744 m kg galena 3 di = . v m SG 3 3 148 m kg suspension 3 di bg = . 5 unknowns (vvvmm 123 1 3 ,,,, ) ? 1 mass balance ? 1 volume balance ? 3 specific gravities 0 DF Total mass balance: mm 13 400+ = (1) 4- 11 4.19 (cont?d) Assume volume additivity: mm 13 1000 400 7440 1480 kg m kg kg m kg kg m kg 33 3 bg bg += (2) Solve (1) and (2) simultaneously ? = =mm 13 668 1068kg H O kgsuspension 2 , v 1 668 1000 0668== kg m kg m water fed totank 3 3 . b. Specific gravity of coal < 1.48 < Specific gravity of slate c. The suspension begins to settle. Stir the suspension. 1.00 < Specific gravity of coal < 1.48 4.20 a. . . n 1 0040 0960 mol / h mol H O / mol mol DA / mol 2 bg n x x 2 1 mol / h mol H O / mol mol DA / mol 2 bg bg? n 3 97% mol H Oadsorbed / h of H O in feed 2 2 bg Adsorption rate: .. . .n 3 354 340 0 0180 1556= ? = bgkg 5 h mol H O kg H O mol H O / h 2 2 2 97% adsorbed: 156 097 004 401 11 ... .=?=nnbg mol / h Total mole balance: .. .nnn n 123 2 401 1556 38 54= + ? = ? = mol / h Water balance: () ( )() 3 2 0.040 40.1 1.556 38.54 1.2 10 mol H O/molxx ? =+ ?=× b. The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction will reach that of the inlet stream, i.e. 4%. 4.21 a. 300 055 045 lb / h lb H SO / lb lb H O / lb m m2 4 m m2 m . . . . m B lb / h lb H SO / lb lb H O / lb m m2 4 m m2 m bg 090 010 . . m C lb / h lb H SO / lb lb H O / lb m m2 4 m m2 m bg 075 025 Overall balance: 300+ = mm BC (1) H 2 SO 4 balance: 055 300 0 90 0 75.. . b g+ =mm BC (2) Solve (1) and (2) simultaneously ? = = , 400 700lb / h lb / h mm 4- 12 4.21 (cont?d) b. ..mRmR AAAA ?= ? ? ??= ?150 500 150 70 25 25 7 78 44 4bg . BBBB ?= ? ? ??= ?200 800 200 60 20 20 15 0 100bg ln ln ln ln ln . . .xRxRxe xx R x ?= ? ? ?? = + ?=20 100 20 10 4 4 0 2682 1923 6 841 0.2682 bg mR mR xR AA BB x =?= + ==?= + = =?= F H G I K J = 300 300 44 4 778 44 3 400 400 100 15 0 333 55% 1 0 268 55 6841 778 . . ., . ., . ln . . c. Overall balance: mmm ABC + = H 2 SO 4 balance: 001 090 075 075 075 001 015 . . . . .. . xm m m m m m xm ABC ABB A +== +?= ? b g bg ??= ?? ?= ? + ? 15 0 100 075 0016841 778 444 015 259 0236 135 813 0.2682 0.2682 0.2682 . ... . . . .. . . R eR ReRe B R A B R A R x xx dibg di Check: RR R e e Ax B ==?=? + ?=44 3 7 78 2 59 0 236 44 3 135 813 333 0.2682 7.78 0.2682 7.78 ., . . . . . . . bg bg ej 4.22 a. . . n A kmol / h kmol H / kmol kmol N / kmol 2 2 bg 010 090 . . n B kmol / h kmol H / kmol kmol N / kmol 2 2 bg 050 050 100 020 080 kg / h kmol / h kmol H / kmol kmol N / kmol 2 2 . . n P bg MW kg / kmol=+ =0 20 2 016 0 80 28 012 22 813.. . . .bg b g ?= = . .n P 100 22 813 438 kg h kmol kg kmol / h Overall balance: .nn AB + = 438 (1) H 2 balance: 010 0 50 0 20 4 38. . ..nn AB +=bg (2) Solve (1) and (2) simultaneously ? = = ., .nn AB 329 110kmol / h kmol / h 4- 13 4.22 (cont?d) b. . n m P P = 22 813 Overall balance: . nn m AB P += 22 813 H 2 balance: xn xn xm AA BB PP . += 22 813 ?= ? ? = ? ? . . n m xx xx n m xx xx A P BP BA B P PA BA 22 813 22 813 bg bg bg bg c. Trial X A X B X P m P n A n B 10.100.500.10 10 4.380.0 20.100.500.20 10 3.291.0 30.100.500.30 10 2.192.19 40.100.500.40 10 1.03.29 50.100.500.50 10 0.04.38 60.100.500.60 10 -1.05.48 7 0.10 0.50 0.10 250 10.96 0.00 80.100.500.20 2508.2 2.74 90.100.500.30 2505.485.48 10 0.10 0.50 0.40 250 2.74 8.22 11 0.10 0.50 0.50 250 0.00 10.96 12 0.10 0.50 0.60 250 -2.74 13.70 The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot blend a 10% H 2 mixture with a 50% H 2 mixture and obtain a 60% H 2 mixture. d. Results are the same as in part c. 4.23 Arterial blood ml / min mg urea / ml 200 0 190 . . Dialyzing fluid ml / min1500 Venous blood ml / min mg urea / ml 195 0 175 . . Dialysate ml / min mg urea / ml v c bg a. Water removal rate: 2000 1950 50...? = ml / min Urea removal rate: 190 200 0 175 195 0 38 8.....bg bg?= mg urea / min b. ./minv =+=1500 5 0 1505ml 38.8 mg urea/min 0.0258 mg urea/ml 1505 ml/min c == c. 27 11 206 .. min ? = bg mg removed 1 min 10 ml 5.0 L ml 38.8 mg removed 1 L (3.4 h) 3 4- 14 4.24 a. . n 1 20 0 kmol / min kgCO / min 2 bg . n 2 0015 kmol / min kmolCO / kmol 2 bg . n 3 0023 kmol / min kmolCO / kmol 2 bg . min . .n 1 20 0 44 0 0 455== kg CO kmol kg CO kmol CO / min 2 2 2 Overall balance: 0455 23 . + =nn (1) CO 2 balance: 0455 0015 0023 23 .. . + =nn (2) Solve (1) and (2) simultaneously ? = = ., .nn 23 55 6 561 kmol / min kmol / min b. u== 150 18 833 m s m/s. AD D== ?= 1 4 561 0123 1 60 8 33 108 2 ? . min . min . . kmol m kmol s s m m 3 4.25 Spectrophotometer calibration: CkA C A A C = ====> = = = 0.9 3 3 333 g/L?bg. Dye concentration: AC=?= =018 3333 018 0 600....bgbg g/L? Dye injected == 0.60 cm L 5.0 mg 10 g 10 cm L 1 mg g 33 33 1 1 30 ? ?. ?=?=30 0600 50.. . gL g/L L??bgbgVV 4.26 a. V n y y 1 1 1 1 1 m/min kmol / min kmol SO / kmol kmol A / kmol 3 2 di bg bg? 1000 2 LB/ min kg B / min m bg n y y 3 3 3 1 kmol / min kmol SO / kmol kmol A / kmol 2 bg bg? m x x 4 4 4 1 kg / min kg SO / kg kg B / kg 2 bg bg? 4- 15 4.26 (cont?d) 8 unknowns ( , , , , ,,,nnvmmxyy 131 2 4 4 13 ) ? 3 material balances ? 2 analyzer readings ? 1 meter reading ? 1 gas density formula ? 1 specific gravity 0 DF b. Orifice meter calibration: A log plot of vs. is a line through the points and , , .Vh h V h V 11 2 2 100 142 400 290== == did i ln ln ln ln ln ln ln . ln ln ln ln . ln . . . . Vbh aVah b VV hh aVbh ae V h b =+?= == = =? = ? =?==?= 21 21 11 258 0.515 290 142 400 100 0515 142 0 515 100 2 58 13 2 13 2 d h bg bg bg bg Analyzer calibration: ln lnybR a yae bR =+ ?= b yy RR aybR a ye R = ? = ? = =?= ? =? E =× U V | | | W | | | ?= × ? ? ln ln . . . ln ln ln . . . . . 21 21 11 4 4 0.0600 01107 0 00166 90 20 0 0600 0 00166 0 0600 20 7 60 500 10 500 10 bgb g bg bg c. hV 11 0.515 210 13 2 210 207 3=?= = mm m min 3 ..bg ? feed gas 3 3 3 atm K mol / L = 0.460 kmol / m m min kmol m kmol min = + + = E == 12 2 150 14 7 14 7 75 460 18 0460 207 3 0 460 9534 1 ... . . .. . bgb g bg bgbg n Ry Ry m 11 4 33 4 2 82 4 500 10 0 0600 82 4 0 0702 116 500 10 0 0600 116 0 00100 1000 130 1300 =?=× × = =?=× × = == ? ? ..exp... ..exp... . bg bg kmol SO kmol kmol SO kmol L B min kg L B kg / min 2 2 4- 16 4.26 (cont?d) A balance: 1 0 0702 9534 1 0 00100 88 7 33 ?=??=.. . .bgbgb gnn kmol min SO balance: kg / kmol) (1) B balance: 1300 = (2) Solve (1) and (2) simultaneously = 1723 kg / min, = 0.245 kg SO absorbed / kg SO removed = kg SO / min 2 2 22 00702 9534 640 000100 887 64 1 422 44 44 44 44 ..(. . .() () bgbg b gbg=+ ? ? = mx mx mx mx d. Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a higher rate of transfer of SO 2 from the gas to the liquid phase. 4.27 a. ,, , V n y y PT Rh 1 1 1 1 11 11 1 m/min kmol / min kmolSO / kmol kmol A / kmol 3 2 di bg bg? V m 2 2 m/min kg B / min 3 di bg n y y R 3 3 3 3 1 kmol / min kmolSO / kmol kmol A / kmol 2 bg bg? m x x 4 4 4 1 kg / min kgSO kg kg B / kg 2 bg bg bg? b. 14 unknowns ( , ,,,,,, , , ,,, ,nVyPTRhVmn y Rmx 11111 112 2 3 3 3 4 4 ) ? 3 material balances ? 3 analyzer and orifice meter readings ? 1 gas density formula (relates nV 11 and ) ? 1 specific gravity (relates mV 22 and ) 6 DF A balance: 11 11 33 ?=?yn ynbgbg (1) SO 2 balance: yn yn xm 11 33 44 64 =+ kgSO / kmol 2 (2) B balance: mxm 244 1=?bg (3) Calibration formulas: ye R 1 4 0.060 500 10 1 =× ? . (4) R 3 40.060 500 10 3 =× ? . (5) .Vh 11 0.515 132= (6) Gas density formula: ../. /. n P T V 1 1 1 1 12 2 14 7 14 7 460 18 = + + bg bg (7) Liquid specific gravity: SG V m =?=130 1300 2 2 . kg h m kg 3 bg (8) 4- 17 4.27 (cont?d) c. T 1 75 °F y 1 0.07 kmol SO 2 /kmol P 1 150 psig V 1 207 m3/h h 1 210 torr n 1 95.26 kmol/h R 1 82.4 Trial x 4 (kg SO 2 /kg) y 3 (kmol SO 2 /kmol) V 2 (m 3 /h) n 3 (kmol/h) m 4 (kg/h) m 2 (kg/h) 1 0.10 0.050 0.89 93.25 1283.45 1155.11 2 0.10 0.025 1.95 90.86 2813.72 2532.35 3 0.10 0.010 2.56 89.48 3694.78 3325.31 4 0.10 0.005 2.76 89.03 3982.57 3584.31 5 0.10 0.001 2.92 88.68 4210.72 3789.65 6 0.20 0.050 0.39 93.25 641.73 513.38 7 0.20 0.025 0.87 90.86 1406.86 1125.49 8 0.20 0.010 1.14 89.48 1847.39 1477.91 9 0.20 0.005 1.23 89.03 1991.28 1593.03 10 0.20 0.001 1.30 88.68 2105.36 1684.29 V 2 vs. y 3 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 0.000 0.020 0.040 0.060 y 3 (kmol SO 2 /kmol) V 2 (m 3 /h ) x4 = 0.10 x4 = 0.20 For a given SO 2 feed rate removing more SO 2 (lower y 3 ) requires a higher solvent feed rate ( V 2 ). For a given SO 2 removal rate (y 3 ), a higher solvent feed rate ( V 2 ) tends to a more dilute SO 2 solution at the outlet (lower x 4 ). d. Answers are the same as in part c. 4.28 Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3 Overall mass balance ? m 3 Mass balance - Unit 1 ? m 1 A balance - Unit 1 ? x 1 Mass balance - mixing point ? m 2 A balance - mixing point ? x 2 C balance - mixing point ? y 2 4- 18 4.29 a. 100 0 300 0 250 0 450 mol / h mol B / mol mol T / mol mol X / mol . . . n x x xx B T BT 2 2 2 22 1 mol / h mol B / mol mol T / mol mol X / mol bg bg bg bg?? . . n 4 0 940 0 060 mol / h mol B / mol mol T / mol bg . . n 3 0 020 0 980 mol / h mol T / mol mol X / mol bg n x x xx B T BT 5 5 5 55 1 mol / h mol B / mol mol T / mol mol X / mol bg bg bg bg?? Column 1 Column 2 Column 1 Column 2: 4 unknowns ( , ,,nnx x BT23 2 2 ) 4 unknowns ( , , ,nnny x345 ) ?3 balances ? 3 balances ? 1 recovery of X in bot. (96%) ? 1 recovery of B in top (97%) 0 DF 0 DF Column 1 96% X recovery: 096 0450 100 098 3 .. . bgbg= n (1) Total mole balance: 100 23 = + nn (2) B balance: 0 300 100 22 . bg= xn B (3) T balance: 0 250 100 0 020 22 3 . . bg=+xn n T (4) Column 2 97% B recovery: 0 97 0 940 22 4 . . xn n B = (5) Total mole balance: nnn 245 = + (6) B balance: xn n xn BB22 4 55 0940 . = + (7) T balance: xn n xn TT22 4 55 0 060 . = + (8) b. () .() . () . () . () . () . () . 1 441 2 559 3 0 536 4 0 431 53095 62496 7036 8082 32 45 55 ?= ?= ?= ?= ?= ?= ?= ?= nn xx BT mol / h mol / h mol B / mol mol T / mol mol / h mol / h mol B / mol mol T / mol Overall benzene recovery: 0 940 30 95 0300100 100% 97% .. . bg bg ×= Overall toluene recovery: 0 892 24 96 0250100 100 89% .. . bg bg ×= 4- 19 4.30 a. 100 0035 0965 kg / h kg S / kg kg H O / kg 2 . . m x x 3 3 3 1 kg / h kg S / kg kg H O / kg 2 bg bg ? m x x 4 4 4 1 kg / h kg S / kg kg H O / kg 2 bg bg ? . . m 10 0050 0 950 kg / h kg S / kg kg H O / kg 2 bg 0100. m w kg H O / h 2 bg 0100. m w kg H O / h 2 bg 0100. m w kg H O / h 2 bg m w kg H O / h 2 bg 1 41 b. Overall process 100 kg/h (m 10 kg / h) 0.035 kg S/kg 0.050 kg S/kg 0.965 kg H 2 O/kg 0.950 kg H 2 O/kg ()m w kg H O / h 2 Salt balance: 0 035 100 0 050 10 .. b g= m Overall balance: 100 10 = + mm w H 2 O yield: Y m w w = . kg H O recovered kg H Oin fresh feed 2 2 bg 96 5 First 4 evaporators 100 0035 0965 kg/h kg S/ kg kg HO/kg2 . . 4 0100× . mw kg HO/h2 bg m x x 4 4 4 1 kg /h kg S/ kg kg HO/kg2 bg bg ? Overall balance: 100 4 0100 4 = +. b gmm w Salt balance: 0 035 100 44 . b g= xm c. Y x w = = 031 0 0398 4 . . 4- 20 4.31 a. 100 050 050 mol mol B / mol mol T / mol . . 2 097 003 1 . . n mol mol B / mol mol T / mol bg . . n 1 097 003 mol mol B / mol mol T / mol bg () . . n 1 097 003 mol 89.2%of Bin feed mol B / mol mol T / mol bg n y y B B 4 1 mol 45%of feed toreboiler mol B / mol mol T / mol bgb g bg bg? n z z B B 2 1 mol mol B / mol mol T / mol bg bg bg? n x x B B 3 1 mol mol B / mol mol T / mol bg bg bg? Still Condenser Reboiler Overall process: 3 unknowns ( , ,nnx B13 ) Still: 5 unknowns ( , , ,,nnny z BB124 ) ? 2 balances ? 2 balances ? 1 relationship (89.2% recovery) 3 DF 0 DF Condenser: 1 unknown ( n 1 ) Reboiler: 6 unknowns ( , , ,,,nnnx yz BBB234 ) ? 0 balances ? 2 balances 1 DF ? 2 relationships (2.25 ratio & 45% vapor) 3 DF Begin with overall process. b. Overall process 89.2% recovery: 0 892 0 50 100 0 97 1 .. . bgbg= n Overall balance: 100 13 = + nn B balance: 0 50 100 0 97 13 .. bg= +nxn B Reboiler Composition relationship: yy xx BB BB / / . 1 1 225 ? ? = e j bg Percent vaporized: . nn 42 045= (1) Mole balance: nnn 234 = + (2) (Solve (1) and (2) simultaneously.) B balance: zn xn yn BBB 234 = + 4- 21 4.31 (cont?d) c. B fraction in bottoms: x B = 0100.molB/mol Moles of overhead: .n 1 46 0= mol Moles of bottoms: .n 3 54 0= mol Recovery of toluene: 1 050100 100% 1 010 54 02 0 50 100 100% 97% 3 ? ×= ? ×= xn B bg bg bgbg bg . .. . 4.32 a. 100 0 12 0 88 kg kg S / kg kg H O / kg 2 . . m 1 012 0 88 kg kg S / kg kg H O / kg 2 b g . . m 4 0 58 0 42 kg kg S / kg kg H O / kg 2 b g . . m 2 012 0 88 kg kg S / kg kg H O / kg 2 b g . . m 5 0 42 0 58 kg kg S / kg kg H O / kg 2 b g . . m 3 kg H O 2 b g Mixing point Bypass Evaporator Overall process: 2 unknowns (mm 35 , ) Bypass: 2 unknowns (mm 12 , ) ? 2 balances ? 1 independent balance 0 DF 1 DF Evaporator: 3 unknowns (mmm 134 ,,) Mixing point: 3 unknowns (mmm 245 ,,) ? 2 balances ? 2 balances 1 DF 1 DF Overall S balance: 012 100 042 5 ..bg= m Overall mass balance: 100 35 = +mm Mixing point mass balance: mmm 42 5 + = (1) Mixing point S balance: 058 012 042 42 5 ...mmm+ = (2) Solve (1) and (2) simultaneously Bypass mass balance: 100 12 = +mm b. mmmmm 12345 90 05 9 95 714 18 65 28 6=== = =., ., ., ., . kg kg kg kg kg product Bypass fraction: m 2 100 0 095= . c. Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a stream consisting of 90% solids could be hard to transport. Basis: 100 kg 4- 22 4.33 a. . . m 1 0 0515 09485 kg / h kgCr / kg kgW / kg bg . . m 2 00515 0 9485 kg / h kgCr / kg kgW / kg bg . . m 3 00515 0 9485 kg / h kgCr / kg kgW / kg bg m 4 kgCr / hbg m x x 5 5 5 1 kg / h kgCr / kg kgW / kg bg bg bg? m x x 6 6 6 1 kg / h kgCr / kg kgW / kg bg bg bg? Treatment Unit b. mm 12 6000 4500=?= kg / h kg / h maximum allowed valuebg Bypass point mass balance: m 3 6000 4500 1500= ? = kg / h 95% Cr removal: .. .m 4 0 95 0 0515 4500 220 2==bgbg kg Cr / h Mass balance on treatment unit: ..m 5 4500 220 2 4279 8= ? = kg / h Cr balance on treatment unit: x 5 0 0515 4500 220 2 4779 8 0 002707= ? = .. . . bg kgCr / kg Mixing point mass balance: ..m 6 1500 4279 8 5779 8= + = kg / h Mixing point Cr balance: x 6 0 0515 1500 0 0002707 4279 8 5779 8 0 0154= + = ... . . bg b g kgCr / kg c. m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h) x 5 m 6 (kg/h) x 6 1000 1000 0 48.9 951 0.00271 951 0.00271 2000 2000 0 97.9 1902 0.00271 1902 0.00271 3000 3000 0 147 2853 0.00271 2853 0.00271 4000 4000 0 196 3804 0.00271 3804 0.00271 5000 4500 500 220 4280 0.00271 4780 0.00781 6000 4500 1500 220 4280 0.00271 5780 0.0154 7000 4500 2500 220 4280 0.00271 6780 0.0207 8000 4500 3500 220 4280 0.00271 7780 0.0247 9000 4500 4500 220 4280 0.00271 8780 0.0277 10000 4500 5500 220 4280 0.00271 9780 0.0301 4- 23 4.33 (cont?d) m 1 vs. x 6 0.00000 0.00500 0.01000 0.01500 0.02000 0.02500 0.03000 0.03500 0 2000 4000 6000 8000 10000 12000 m 1 (kg/h) x 6 ( k g C r / k g) d. Cost of additional capacity ? installation and maintenance, revenue from additional recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon, regulatory limits on Cr emissions. 4.34 a. Evaporator . . m 1 0196 0 804 kg / s kg K SO / kg kg H O / kg 24 2 bg m m 4 5 kg K SO / s kg H O / s 24 2 bg bg Filtrate kg / s kg K SO / kg kg H O / kg 24 2 . . m 3 0 400 0 600 bg 175 kg H O / s 45% of water fed to evaporator 2 bg Filter cake kg K SO / s kgsoln / s kg K SO / kg kg H O / kg 24 24 2 10 0 400 0600 2 2 . . m m bg bg R S T U V W m m 6 7 kg K SO / s kg H O / s 24 2 bg bg Crystallizer Filter Let K = K 2 SO 4 , W = H 2 Basis: 175 kg W evaporated/s Overall process: 2 unknowns ( , mm 12 ) Mixing point: 4 unknowns ( , , , mmmm 1345 ) - 2 balances - 2 balances 0 DF 2 DF Evaporator: 4 unknowns ( , , , mmmm 4567 ) Crystallizer: 4 unknowns ( , , , mmmm 2367 ) ? 2 balances ? 2 balances ? 1 percent evaporation 2 DF 1 DF Strategy: Overall balances % evaporation Balances around mixing point Balances around evaporator verify that each chosen subsystem involves no more than two unknown variables ? ? ? ? U V | | W | | , , , mm m mm mm 12 5 34 67 4- 24 4.34 (cont?d) Overall mass balance: Overall K balance: . . mmm mm m 122 12 2 175 10 0196 10 0 400 =+ + =+ U V | W | Production rate of crystals=10 2 m 45% evaporation: 175 0 450 5 kg evaporated min = . m W balance around mixing point: 0 804 0 600 135 . . mmm+ = Mass balance around mixing point: mmmm 1345 + = + K balance around evaporator: mm 64 = W balance around evaporator: mm 57 175= + Mole fraction of K in stream entering evaporator = m mm 4 45 + b. Fresh feed rate: m 1 221= kg / s Production rate of crystals kg K s s==10 416 2 .m bg Recycle ratio: . . . m m 3 1 352 3 220 8 160 kg recycle s kg fresh feed s kg recycle kg fresh feed bg == c. Scale to 75% of capacity. Flow rate of stream entering evaporator = . ( kg / s) = kg / s . . 0 75 398 299 46 3% K, 537% W d. Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and the cooling cost for the crystallizer. 4- 25 4.35 a. Overall objective: Separate components of a CH 4 -CO 2 mixture, recover CH 4 , and discharge CO 2 to the atmosphere. Absorber function: Separates CO 2 from CH 4 . Stripper function: Removes dissolved CO 2 from CH 3 OH so that the latter can be reused. b. The top streams are liquids while the bottom streams are gases. The liquids are heavier than the gases so the liquids fall through the columns and the gases rise. c. 100 0 300 0 700 mol / h mol CO / mol molCH / mol 2 4 . . . . n 1 0010 0990 mol / h mol CO / mol mol CH / mol 2 4 bg . . n 2 0005 0995 mol / h mol CO / mol mol CH OH / mol 2 3 bg n n 3 4 mol CO / h mol CH OH / h 2 3 bg / n n 5 6 mol N h mol CO / h 2 2 bg /n 5 mol N h 2 bg Absorber Stripper Overall: 3 unknowns ( , , nnn 156 ) Absorber: 4 unknowns ( , , , nnnn 1234 ) ? 2 balances ? 3 balances 1 DF 1 DF Stripper: 4 unknowns ( , , , nnnn 2345 ) ? 2 balances ? 1 percent removal (90%) 1 DF Overall CH 4 balance: 0 700 100 0 990 1 .. bgbgb gmol CH / h 4 = n Overall mole balance: 100 16 mol / hbg=+ nn Percent CO 2 stripped: 090 36 . nn= Stripper CO 2 balance: . nn n 36 2 0005= + Stripper CH 3 OH balance: . nn 42 0 995= d. ., ., ., ., . nnn n n 123 4 6 70 71 6510 32 55 647 7 29 29 === = = mol / h mol / h mol CO / h mol CH OH / h mol CO / h 23 2 Fractional CO 2 absorption: f n CO 2 2 . mol CO absorbed / mol fed= ? = 30 0 0 010 30 0 0976 1 .. . 4- 26 4.35 (cont?d) Total molar flow rate of liquid feed to stripper and mole fraction of CO 2 : , .nn x n nn 34 3 3 34 680 0 0478+= = + =mol / h molCO / mol 2 e. Scale up to 1000 kg/h (=10 6 g/h) of product gas: MW g CO / mol g CH / mol g / mol g/h g/mol mol/h mol / h mol / h) mol / h) mol / h 24 1 feed 1 64 44 0 01 44 0 99 16 16 28 10 10 16 28 6142 10 100 6142 10 70 71 8 69 10 =+= =× = × =× .. . . (. /( . . bgbg bg dibg bg b g n n new new f. TT as < The higher temperature in the stripper will help drive off the gas. PP as > The higher pressure in the absorber will help dissolve the gas in the liquid. g. The methanol must have a high solubility for CO 2 , a low solubility for CH 4 , and a low volatility at the stripper temperature. 4.36 a. Basis: 100 kg beans fed m 1 kg C 6 H 14 ej 300 kg C H 614 130 87 0 . . kg oil kg S m x y xy 2 2 2 22 1 kg kg S / kg kg oil / kg kg C H / kg 614 bg bg bg bg?? m 5 kgC 6 H 14 ej m y y 3 3 3 075 025 kg kg S / kg kg oil / kg kg C H / kg 614 bg bg bg . . ? m y y 4 4 4 1 kg kg oil / kg kg C H / kg 614 bg bg ? m 6 kg oilbg Ex F Ev Condenser Overall: 4 unknowns (mmmy 1363 ,,,) Extractor: 3 unknowns (mxy 222 ,,) ? 3 balances ? 3 balances 1 DF 0 DF Mixing Pt: 2 unknowns (mm 15 , ) Evaporator: 4 unknowns (mmmy 4564 ,,,) ? 1 balance ? 2 balances 1 DF 2 DF Filter: 7 unknowns (mmmxyyy 2342234 ,,,,,,) ? 3 balances ? 1 oil/hexane ratio 3 DF Start with extractor (0 degrees of freedom) Extractor mass balance: 300 87 0 130 2 ++ =.. kgm 4- 27 4.36 (cont?d) Extractor S balance: 87 0 22 . kg S= xm Extractor oil balance: 130 22 . kg oil= ym Filter S balance: 87 0 0 75 3 .. kg S = m Filter mass balance: mmm 234 kgbg=+Oil / hexane ratio in filter cake: y y y xy 3 3 2 22 025 1. ? = ?? Filter oil balance: 130 33 44 . kg oil = +ym ym Evaporator hexane balance: 1 44 5 ?=ym mbg Mixing pt. Hexane balance: mm 15 300+ = kg C H 614 Evaporator oil balance: ym m 44 6 = b. Yield kg oil kg beans fed kg oil / kg beans fed== = m 6 100 118 100 0118 . . bg Fresh hexanefeed kg C H kg beans fed kg C H kg beans fed 614 614 == = m 1 100 28 100 028./bg Recycle ratio kg C H recycled kg C H fed kg C H recycled / kg C H fed 614 614 614 614 == = m m 5 1 272 28 971. c. Lower heating cost for the evaporator and lower cooling cost for the condenser. 4.37 100 2 98 lb m lb dirt lb dry shirts m m m 2 lb Whizzo m bgm 3 003 097 lb lb dirt / lb lb Whizzo / lb m mm bg . . m 4 013 087 lb lb dirt / lb lb Whizzo / lb m mm bg . . m 5 092 008 lb lb dirt / lb lb Whizzo / lb m mm bg . . m x x 6 1 lb lb dirt/lb lb Whizzo/ lb m mm mm bg bg? m 1 98 3 lb dirt lb dry shirts lb Whizzo m m m bg Tub Filter Strategy 95% dirt removal ? m 1 (= 5% of the dirt entering) Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling the chart) ?mm 25 , (solves Part (a)) 4- 28 4.37 (cont?d) Balances around the mixing point involve 3 unknowns mmx 36 ,,bg, as do balances around the filter mmx 46 ,,bg, but the tub only involves 2 mm 34 ,bg and 2 balances are allowed for each subsystem. Balances around tub ? mm 34 , Balances around mixing point ? mx 6 , (solves Part (b)) a. 95% dirt removal: m 1 005 20 010==.. .bgbg lb dirt m Overall dirt balance: 2 0 010 0 92 2 065 55 .. . .=+ ?=bgmm lb dirt m Overall Whizzo balance: m 2 3 0 08 2 065 317=+ =.. .bgb gb glb Whizzo lb Whizzo mm b. Tub dirt balance: 2 0 03 010 013 34 + = +...mm (1) Tub Whizzo balance: 0 97 3 0 87= + (2) Solve (1) & (2) simultaneously ? mm 34 20 4 19 3= =., . lb lb mm Mixing pt. mass balance: 317 20 4 17 3 66 ..+ = ? = lb lb Mixing pt. Whizzo balance: ()()( ) mm 3.17 17.3 0.97 20.4 0.961 lb Whizzo/lb 96% Whizzo, 4% dirtxx+= ?= ? 4.38 a. C mixer 3 Filter 3 Discarded 3 kg L L C 3 kg S S 2720 kg S F 3 kg L L F 3 kg S S C 2 kg L L C 2 kg S S mixer 2 Filter 2 mixer 1 Filter 1 C 1 kg L L C 1 kg S S 3300 kg S F 2 kg L L F 2 kg S S F 1 kg L L F 1 kg S S To holding tank 620 kg L mixer filter 1 kg L balance: kg L mixer filter 2 balance: mixer filter 3 kg L kg L kg L balance: 613.7 = 6.1+ C kg L 3L :. . .. : : .. . . . . . . 001620 62 620 6 2 6138 0016138 6138 001 62 6137 61 607 6 11 11 32 323 23 2 2 3 3 bg bg =? = =+? = += +=+ = U V | W | ? = = = ?= FF CC FF FFC CF F C F C LL LL LL LLL LL L L L L 4- 29 4.38 (cont?d) Solvent m f 1 kg S balance: kg S mf 2 balance: mf 3 balance: 2720 + C kg S kg S kg S kg S 2S :. : : . . . . . . 015 3300 495 3300 495 2805 015 495 495 015 2720 482 6 2734 6 480 4 2722 2 11 11 32 322 23 33 2 2 3 3 bg bg bg =? = =+? = += +=+ += =+ U V | | W | | ? = = = = CC FF FC FCF CC FC C F C F SS SS SS SSS SS SS S S S S Holding Tank Contents 62 62 124 2805 2734 6 5540 .. . . += += kg leaf kg solvent b. Extraction Unit Steam Stripper 5540 0165 0835 kgS kg E / kg kg W / kg . . Q Q D F kg D kg F bg bg Q Q Q E D F kg E kg D kg F bg bg bg Q R kg kg E / kg 0.15kg F / kg 0.855kgW / kg bg 013. Q 0 0200 kg kg E / kg 0.026kg F / kg 0.774kgW/ kg bg . Q B kg kg E / kg 0.987kgW/ kg bg 0013. Q 3 kg steambg Mass of D in Product: 1 kg D 620 kg leaf 1000 kg leaf kg D==062. Q D Water balance around extraction unit: 0835 5540 0855 5410..b g= ?=QQ RR kg Ethanol balance around extraction unit: 0165 5540 013 5410 211..bg bg b g=+?=QQ EE kg ethanol in extract c. F balance around stripper 0 015 5410 0 026 3121 00 ..bg b g=?=QQ kg mass of stripper overhead product E balance around stripper 013 5410 0 200 3121 0 013 6085.. .bg bg b g=+?=QQ BB kg mass of stripper bottom product W balance around stripper 0 855 5410 0 774 3121 0 987 6085 3796...bg bg bg+= + ? =QQ SS kg steam fed to stripper 4.39 a. CH 2H CH mol H react / mol C H react kmol C H formed / kmol H react 22 2 26 222 26 2 +? 2 05. 4- 30 4.39 (cont?d) b. n n H CH 2 222 2 22 2 22 H islimiting reactant mol H fed molC H fed molC H required (theoretical) excess C H mol fed mol required mol required = = = =?=? =?=?= =+= =+= ==? ? =?= 70 0 70 70 80 70 10 70 70 487 70 70 70 10 487 1706 0 0 00 0 0 0 0 0 ?? ? ? ? ? mol mol mol mol mol mol methanol fed... bgbg bgbg 4- 37 4.46 (cont?d) Product gas mol mol mol mol mol CH OH mol mol CH COOH mol mol CH COOCH mol mol H O mol mol 3 3 33 2 n n n n y y y y n A B C D A B C D total =?= = = = U V | | W | | ? = = = = = 170 6 70 100 6 10 70 70 0401 0040 0 279 0 279 250 6 .. . . . . . d. Cost of reactants, selling price for product, market for product, rate of reaction, need for heating or cooling, and many other items. 4.47 a. CO (A) HO (B) CO (C) H (D) 222 +??? + 100 020 010 040 030 . . . . . mol mol CO / mol mol CO / mol mol H O / mol mol I / mol 2 2 n n n n n A B C D I mol CO mol H O mol CO mol H mol I 2 2 2 bg bg bg bg bg Degree of freedom analysis: 6 unknowns ( nnnnn ABCDI ,,,,,? ) ? 4 expressions for n i ?bg ? 1 balance on I ? 1 equilibrium relationship 0 DF b. Since two moles are prodcued for every two moles that react, nn total out total in molbgbg bg==100. n A =?020. ? (1) n B =?040. ? (2) n C =+010. ? (3) n D =? (4) n I = 030. (5) n tot =100.mol At equilibrium: yy yy nn nn CD AB CD AB == + ?? = F H G I K J ?= 010 020 040 0 0247 4020 1123 0110 . .. .exp . ?? ?? ? bgbg bgbg mol yn DD ===? 0110.molH/mol 2 bg c. The reaction has not reached equilibrium yet. 4- 38 4.47 (cont?d) d. T (K) x (CO) x (H 2 O) x (CO 2 ) Keq Keq (Goal Seek) Extent of Reaction y (H 2 ) 1223 0.5 0.5 0 0.6610 0.6610 0.2242 0.224 1123 0.5 0.5 0 0.8858 0.8856 0.2424 0.242 1023 0.5 0.5 0 1.2569 1.2569 0.2643 0.264 923 0.5 0.5 0 1.9240 1.9242 0.2905 0.291 823 0.5 0.5 0 3.2662 3.2661 0.3219 0.322 723 0.5 0.5 0 6.4187 6.4188 0.3585 0.358 623 0.5 0.5 0 15.6692 15.6692 0.3992 0.399 673 0.5 0.5 0 9.7017 9.7011 0.3785 0.378 698 0.5 0.5 0 7.8331 7.8331 0.3684 0.368 688 0.5 0.5 0 8.5171 8.5177 0.3724 0.372 1123 0.2 0.4 0.1 0.8858 0.8863 0.1101 0.110 1123 0.4 0.2 0.1 0.8858 0.8857 0.1100 0.110 1123 0.3 0.3 0 0.8858 0.8856 0.1454 0.145 1123 0.5 0.4 0 0.8858 0.8867 0.2156 0.216 The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of carbon monoxide and water also maximizes the extent of reaction. 4.48 a. A2B C+? ln lnKAETK e =+ 0 bg E KK TT ee = ? = × ? = ? ln / ln . / . 12 12 4 11 10 5 2 316 10 1373 1573 11458 bgdi ln ln ln . . .AK T A e01 1 0 13 11458 10 5 11458 373 28 37 4 79 10=? =? =??=× ? KTKK=× ? = ?? 4 79 10 11458 450 0 0548 13 2 1 .exp ().bg ch atm atm b. nn nn nn nn yn n yn n yn n nnnn AA BB CC TT AA T BB T CC T TABC =? =? =+ =? U V | | W | | ? =? ? =? ? =+ ? =++ 0 0 0 0 00 00 00 0000 2 2 2 22 2 ? ? ? ? ?? ?? ?? bgb g bgbg bgb g At equilibrium, y yy P nn nn P KT C AB CeT e AeB e e 22 00 2 00 22 1 2 2 1 = +? ?? = ?? ?? bgb g bgb g bg (substitute for K e Tbg from Part a.) c. Basis: 1 mol A (CO) nnn n ABC T000 0 110 2=== ? = , P= 2atm , T = 423K ?? ?? ee ee e K 22 112 1 4 423 0 278 2 2 ? ?? == bg bgb g bg atm atm 2 -2 . ? ?? ee 2 01317 0?+ =. 4- 39 4.48 (cont?d) (For this particular set of initial conditions, we get a quadratic equation. In general, the equation will be cubic.) ? e = 0156. , 0.844 Reject the second solution, since it leads to a negative n B . yy AA BB CC =? ? ? = =? ? ? = =+ ? ? = 1 0156 2 2 0156 0 500 1 2 0156 2 2 0156 0 408 0 0156 2 2 0156 0 092 .. . .. . .. . bgbgch bgchbgch bgbgch Fractional Conversion of CO A nn nn AA AA AA bg= ? == 0 00 0156 ? . mol reacted / mol feed d. Use the equations from part b. i) Fractional conversion decreases with increasing fraction of CO. ii) Fractional conversion decreases with increasing fraction of CH 3 OH. iii) Fractional conversion decreases with increasing temperature. iv) Fractional conversion increases with increasing pressure. * REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI, FN, FDN, NT, CON, YA, YB, YC INTEGER NIT, INMAX TAU = 0.0001 INMAX = 10 A = 4.79E?13 E = 11458. READ (5, *) YA0, YB0, YC0, T, P KE = A * EXP(E/T) P2KE = P*P*KE C0 = YC0 ? P2KE * YA0 * YB0 * YB0 C1 = 1. ? 4. * YC0 + P2KE * YB0 * (YB0 + 4. * YA0) C2 = 4. * (YC0 ?1. ? P2KE * (YA0 + YB0)) C3 = 4. * (1. + P2KE) EK = 0.0 (Assume an initial value ? e = 00. ) NIT = 0 1 FN = C0 + EK * (C1 + EK * (C2 + EK * C3)) FDN = C1 + EK * (2. * C2 + EK * 3. * C3) EKPI = EK - FN/FDN NIT = NIT + 1 IF (NIT.EQ.INMAX) GOTO 4 IF (ABS((EKPI ? EK)/EKPI).LT.TAU) GOTO 2 EK = EKPI GOTO 1 2 NT = 1. ? 2. * EKPI YA = (YA0 ? EKPI)/NT YB = (YB0 ? 2. + EKPI)/NT YC = (YC0 + EKPI)/NT 4- 40 4.48 (cont?d) CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP 4 3 WRITE (6, 5) INMAX, EKPI FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X)) FORMAT ('DID NOT CONVERGE IN', I3, 'ITERATIONS',/, * 'CURRENT VALUE = ', F6.3) END $ DATA 0.5 0.5 0.0 423. 2. RESULTS: YA = 0.500, YB = 0.408, YC = 0.092, CON = 0.156 Note: This will only find one root ? there are two others that can only be found by choosing different initial values of ? a 4.49 a. CH O HCHO H O 42 2 +???+ (1) CH 2O CO 2H O 42 22 +???+ (2) 100 050 050 mol/s mol CH / mol mol O / mol 4 2 . . n n n n n 1 2 3 4 5 mol CH / s mol O / s mol HCHO / s mol H O / s mol CO 4 2 2 2 bg bg bg bg 7 unknowns ( , , , , , , nnnnn 12345 12 ??) ?5 12 equations for , n i ?? e j 2 DF b. n 112 50=???? (1) n 212 50 2=???? (2) n 31 =? (3) n 41 2 2=+?? (4) n 52 =? (5) c. Fractional conversion: 50 50 0 900 5 00 1 1 ? =?= . . n n bg molCH 4 /s Fractional yield: . . n n 3 3 50 0 855 42 75=?=molHCHO / s Equation 3 Equation 1 Equation 2 Equation 4 Equation 5 mol CH / mol mol O / mol mol HCHO / mol mol H O / mol mol CO / mol CH 4 4 O 2 2 HCHO H 2 O2 CO 2 2 ?= ?= ?= ?= ?= U V | | | W | | | ? = = = = = ? ? 1 2 2 4 5 42 75 225 275 47 25 225 0 0500 0 0275 04275 04725 0 0225 . . . . . . . . . . n n n y y y y y Selectivity: 22 [(42.75mol HCHO/s)/(2.25mol CO /s) 19.0 mol HCHO/mol CO= 52 (mol CO / s)n 4- 41 4.50 a. Design for low conversion and feed ethane in excess. Low conversion and excess ethane make the second reaction unlikely. b. C 2 H 6 + Cl 2 ? C 2 H 5 Cl + HCl, C 2 H 5 Cl + Cl 2 ? C 2 H 4 Cl 2 + HCl Basis: 100 mol C 2 H 5 Cl produced n 1 (mol C 2 H 6 ) 100 mol C 2 H 5 Cl 5 unknowns n 2 (mol Cl 2 ) n 3 (mol C 2 H 6 ) ?3 atomic balances n 4 (mol HCl) 2 D.F. n 5 (mol C 2 H 5 Cl 2 ) c. Selectivity: 100 14 5 mol C H Cl (mol C H Cl 25 24 2 = n ) ? =n 5 7143. mol CHCl 24 2 15% conversion: C balance: 1015 2210227143 13 13 ?= =++ U V | W | . . bg bg b g nn nn ? = = n n 1 3 714 3 114 3 . . mol C H in mol C H out 26 26 H balance: 6 714 3 5 100 6 114 3 4 7 143 4 ..bgbgbg bg=+ ++n ?=n 4 607 1. mol HCl Cl balance: 2 100 607 1 2 7 143 2 n =+ +..bg ? =n 2 114 3. mol Cl 2 Feed Ratio: 114 3 016 66 ./ . / mol Cl 714.3 mol C H mol Cl mol C H 22 2 = Maximum possible amount of C 2 H 5 Cl: n max == 114.3 mol Cl 1 mol C H Cl 1 mol Cl 114.3 mol C H Cl 225 2 25 Fractional yield of C 2 H 5 Cl: n n CHCl 25 100 114 3 0875 max . .== mol mol d. Some of the C 2 H 4 Cl 2 is further chlorinated in an undesired side reaction: C 2 H 5 Cl 2 + Cl 2 ? C 2 H 4 Cl 3 + HCl 4.51 a. C 2 H 4 + H 2 O ? C 2 H 5 OH, 2 C 2 H 5 OH ? (C 2 H 5 ) 2 O + H 2 O Basis: 100 mol effluent gas n n 1 2 (mol C 2 H 4 ) [mol H 2 O (v)] n 3 (mol I) 100 mol 0.433 mol C 2 H 4 /mol 0.025 mol C 2 H 5 OH / mol 0.0014 mol (C 2 H 5 ) 2 O/mol 0.093 mol I / mol 0.4476 mol H 2 O (v) / mol 3 unknowns -2 independent atomic balances -1 I balance 0 D. F. (1) C balance: 2 100 2 0 433 2 0 025 4 0 0014 1 n =?+?+?...bg (2) H balance: 4 2 100 4 0 433 6 0 025 10 0 0014 2 0 4476 12 nn+= ? +? +? +?.. . . (3) O balance: n 2 100 0 025 0 0014 0 4476=++.. .bg Note; Eq.()()() Eq. Eq. 12 32 2?+ ? = ?2 independent atomic balances (4) I balance: n 3 = 9.3 4- 42 4.51 (cont'd) b. (1) 46.08 mol C H (3) 47.4 mol H O (4) 9.3 mol I Reactor feed contains 44.8% C H , 46.1% H O, 9.1% I 126 22 3 26 2 ?= ?= ?= U V | W | ? n n n % conversion of C 2 H 4 : 46 08 43 3 46 08 100% 6 0% .. . . ? ×= If all C 2 H 4 were converted and the second reaction did not occur, n CHOH 25 46 08 di max .= mol ? Fractional Yield of C 2 H 5 OH: nn C H OH C H OH 25 25 2 5 46 08 0 054/./.. max dibg == Selectivity of C 2 H 5 OH to (C 2 H 5 ) 2 O: 2.5 mol C H OH 0.14 mol (C H ) O 17.9 mol C H OH / mol (C H ) O 25 252 25 252 = c. Keep conversion low to prevent C 2 H 5 OH from being in reactor long enough to form significant amounts of (C 2 H 5 ) 2 O. Separate and recycle unreacted C 2 H 4 . 4.52 CaF s H SO l CaSO s 2HF g 224 4 bg bg bg bg+?+ 1 metric ton acid 1000 kg acid 0.60 kg HF 1 metric ton acid 1 kg acid kg HF= 600 Basis: 100 kg Ore dissolved (not fed) (kg H O) n 1 (kg HF) n 2 (kg H SO ) 2 n 4 (kg CaSO ) n 5 4 n 3 (kg H SiF ) 4 6 100 kg Ore d issolved 0.96 kg Ca F /kg 2 0.04 kg SiO / kg 2 n 0.93 H SO kg/ kg 4 0.07 H O kg/kg 2 A (kg 93% H SO ) 2 4 2 2 4 Atomic balance - Si: ( ) 3262 326 262 (kg H SiF ) 28.1 kg Si0.04 100 kg SiO 28.1 kg Si 9.59 kg H SiF 144.1 kg H SiF60.1 kg SiO n n=?= Atomic balance - F: ( ) 22 2 26 2 26 (kg HF) 19.0 kg F0.96 100 kg CaF 38.0 kg F 20.0 kg HF78.1 kg CaF 9.59 kg H SiF 114.0 kg F 41.2 kg HF 144.1 kg H SiF n n = +?= 600 1533 kg or kg HF 100 kg ore diss. 1 kg ore feed 41.2 kg HF 0.95 kg ore diss. e= n 1 (kg CaSO 4 ) n 2 (kg HF) n 3 (kg H 2 SiF 6 ) n 4 (kg H 2 SO 4 ) n 5 (kg H 2 O) 4- 43 4.53 a. CH Cl CHCl HCl C H Cl Cl C H Cl HCl C H Cl Cl C H Cl HCl 66 2 65 65 2 64 2 64 2 2 63 3 +? + +? + +? + Convert output wt% to mol%: Basis 100 g output species g Mol. Wt. mol mol % CH 66 65.0 78.11 0.832 73.2 CHCl 65 32.0 112.56 0.284 25.0 CHCl 64 2 2.5 147.01 0.017 1.5 CHCl 63 3 0.5 181.46 0.003 0.3 total 1.136 Basis: 100 mol output n 1 (mol HCl(g)) n 4 (mol C H ) 6 n 3 (mol I) 6 n 2 (mol Cl ) 2 n 3 (mol I) 32.0 65.0 mo l C H 6 6 mo l C H Cl 6 5 0.5 2.5 mo l C H Cl 6 4 mo l C H Cl 6 3 2 3 4 unknowns -3 atomic balances -1 wt% Cl 2 in feed 0 D.F. b. C balance: () 1 6 6 73.2 25.0 1.5 0.3n =+++? =n 1 100 mol C H 66 H balance: ()( )( ) ( ) ( ) 4 6 100 6 73.2 5 25.0 4 1.5 3 0.3 n=++++ 4 28.9 mol HCln?= Cl balance: ( ) ( ) 2 2 28.9 25.0 2 1.5 3 0.3n =++ + 22 28.9 mol Cln?= Theoretical C H 66 ( ) 266 2 6 28.9 mol Cl 1 mol C H 1 mol Cl 28.9 mol C H== Excess C H 66 : () 66 100 28.9 28.9 100% 246% excess C H?×= Fractional Conversion: () 66 100 73.2 100 0.268 mol C H react/mol fed?= Yield: 65 65 (25.0 mol C H Cl) (28.9 mol C H Cl maximum)=0.865 () 22 22 66 66 66 28.9 mol Cl 70.91 g Cl 1 g gas Gas feed: 2091 g gas mole Cl 0.98 g Cl g gas 0.268 g liquid 78.11 g C H Liquid feed: 100 mol C H 7811 g liquid mol C H ? = ? ? ? ? ?? ? = ?? ? ?? ? c. Low conversion ? low residence time in reactor ? lower chance of 2nd and 3rd reactions occurring. Large excess of CH Cl 66 2 ? much more likely to encounter CH 66 than substituted C H 66 ? higher selectivity. d. Dissolve in water to produce hydrochloric acid. e. Reagent grade costs much more. Use only if impurities in technical grade mixture affect the reaction rate or desired product yield. 73.2 mol C 6 H 6 25.0 mol C 6 H 5 Cl 1.5 mol C 6 H 4 Cl 2 0.3 mol C 6 H 3 Cl 3 4- 44 4.54 a. 2CO 2CO O 2A 2B C ON 2NO CD2E 22 22 ?+ ? + +? +? nn nn nn nn nn yn n yn n yn n yn n yn n AA e BB e CC e e DD e EE e AA eTe BB eTe CCee T e DD e T e EE e Te =? =+ =+? =? =+ =? + =+ + ?= +? + =? + =+ + 01 02 012 02 02 0101 0101 012 01 0201 0201 2 2 2 2 2 1 2 ? ? ?? ? ? ?? ?? ? ?? bgbg bgbg bg bg n total = nnnnn Te TABCDE01 000000 +=++? Equilibrium at 3000K and 1 atm yy y nn BC A BeCee AeTe 2 2 01 2 012 01 2 01 2 2 01071= ++? ?+ = ??? ?? bgb g bgbg . () ()() 2 2 02 012 02 2 0.01493 Ee E CD A e e D e n y yy n n ? ?? ? + == +? ? E =?+?++?= =+???+= U V | W | fn n n n fn nn f f AeTeBeCee CeeDe E e 101 2 01 0 1 2 012 2 01202 0 2 2 11 2 21 2 01071 2 2 0 0 01493 2 0 . . , , ?? ??? ?? ? ? ?? ?? bgbgbgb g bg g bg bg Defines functions and b. Given all n io ?s, solve above equations for ? e1 and ? e2 ? n A , n B , n C , n D , n E ? y A , y B , y C , y D , y E c. n A0 = n C0 = n D0 = 0.333, n B0 = n E0 = 0 ? ? e1 =0.0593, ? e2 = 0.0208 ? y A = 0.2027, y B = 0.1120, y C = 0.3510, y D = 0.2950, y E = 0.0393 d. ad ad f ad ad f d af af aa aa d af af aa aa dd ee e e 11 1 12 2 1 21 1 22 2 2 1 12 2 22 1 11 22 12 21 2 21 1 11 2 11 22 12 21 111 212 +=? +=? = ? ? = ? ? =+ =+?? ??bg bg new new (Solution given following program listing.) . IMPLICIT REAL * 4(N) WRITE (6, 1) 1FORMAT('1', 30X, 'SOLUTION TO PROBLEM 4.57'///) 30 READ (5, *) NA0, NB0, NC0, ND0, NE0 IF (NA0.LT.0.0)STOP WRITE (6, 2) NA0, NB0, NC0, ND0, NE0 4- 45 2 FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/) NTO = NA0 + NB0 + NC0 + ND0 + NE0 NMAX = 10 X1 = 0.1 X2 = 0.1 DO 100 J = 1, NMAX NA = NA0 ? X1 ? X1 NB = NB0 + X1 + X1 NC = NC0 + X1 ? X2 ND = ND0 ? X2 NE = NE0 + X2 + X2 NAS = NA ** 2 NBS = NB ** 2 NES = NE ** 2 NT = NT0 + X1 F1 = 0.1071 * NAS * NT ? NBS * NC F2 = 0.01493 * NC * ND ? NES A11 = ?0.4284 * NA * NT * 0.1071 * NAS ? 4.0 * NB * NC ? NBS A12 = NBS A21 = 0.01493 * ND A22 = ?0.01493 * (NC + ND) ? 4.0 * NE DEN = A11 * A22 ? A12 * A21 D1 = (A12 * F2 ? A22 * F1)/DEN D2 = (A21 * F1 ? A11 * F2)/DEN X1C = X1 + D1 X2C = X2 + D2 WRITE (6, 3) J, X1, X2, X1C, X2C 3FORMAT(20X, 'ITER *', I3, 3X, 'X1A, X2A =', 2F10.5, 6X, 'X1C, X2C =', * 2F10.5) IF (ABS(D1/X1C).LT.1.0E?5.AND.ABS(D2/X2C).LT.1.0E?5) GOTO 120 X1 = X1C X2 = X2C 100 CONTINUE WRITE (6, 4) NMAX 4 120 5 FORMAT('0', 10X, 'PROGRAM DID NOT CONVERGE IN', I2, 'ITERATIONS'/) STOP YA = NA/NT YB = NB/NT YC = NC/NT YD = ND/NT YE = NE/NT WRITE (6, 5) YA, YB, YC, YD, YE FORMAT ('0', 15X, 'YA, YB, YC, YD, YE =', 1P5E14.4///) GOTO 30 END $DATA 0.3333 0.00 0.3333 0.3333 0.0 0.50 0.0 0.0 0.50 0.0 0.20 0.20 0.20 0.20 0.20 SOLUTION TO PROBLEM 4.54 NA0, NB0, NC0, ND0, NE0 = 0.33 0.00 0.33 0.33 0.00 ITER = 1 X1A, X2A = 0.10000 0.10000 X1C, X2C = 0.06418 0.05181 ITER = 2 X1A, X2A = 0.06418 0.05181 X1C, X2C = 0.05969 0.02986 ITER = 3 X1A, X2A = 0.05969 0.02486 X1C, X2C = 0.05937 0.02213 4.54 (cont?d) 4- 46 ITER = 4 X1A, X2A = 0.05437 0.02213 X1C, X2C = 0.05931 0.02086 ITER = 5 X1A, X2A = 0.05931 0.02086 X1C, X2C = 0.05930 0.02083 ITER = 6 X1A, X2A = 0.05930 0.02083 X1C, X2C = 0.05930 0.02083 YA, YB, YC, YD, YE= 2.0270E 1 E 3 E 2 E 3 E ? ? ? ?? 01 1197 01 5100 01 9501 01 9319 02 .. .. NA0, NB0, NC0, ND0, NE0 = 0.20 0.20 0.20 0.20 0.20 ITER = 1 X1A, X2A = 0.10000 0.10000 X1C, X2C = 0.00012 0.00037 ? ITER = 7 X1A, X2A = ?0.02244 ?0.08339 X1C, X2C = ?0.02244 ?0.08339 YA, YB, YC, YD, YE= 2.5051E 01 1.5868E 01 2.6693E 01 2.8989E 01 3.3991E 02 ? ?? ?? 4.55 a. 0 (kg/h) 0 (B) (1 ) R x fm = ? null 0 (A) (kg/h) (kg R/kg) RA m x null 0 (kg R/kg) (kg/h) (1 ) RA x fm? null 1 1 (kg/h) (kg R/kg) R m x null (P) (kg/h) 0.0075 kg R/kg P mnull 0 (kg/h) (kg R/kg) RA fm x null Mass balance on reactor: 01 2(1 ) (1)fm m? =nullnull 99% conversion of R: 11 0 0.01(1 ) (2) RRA mx f mx= ?nullnull Mass balance on mixing point: 10 (3) P mfmm+ =nullnullnull R balance on mixing point: 11 0 0.0075 (4) RRA P mx fmx m+ =nullnull null The system has 6 unknowns 011 (, ,,, , ) RA R P mx fmx mnullnullnull and four independent equations relating them, so there must be two degrees of freedom. b. 01 2(1 )f mm?=nullnull 11 0 0.01(1 ) RRA mx f mx=?nullnull 10P mfmm+=nullnullnull 11 0 0.0075 RRA P mx fmx m+=nullnull null 4850 0.0500 P RA m x = = null 4.54 (cont?d) Reactor: 99% conv. of R E-Z Solve ? ??? ? 0 2780 kg/h 0.254 kg bypassed/kg fresh feed m f = = null 4- 47 4.55 (cont?d) c. m P x RA m A0 m B0 f 4850 0.02 3327 1523 0.54 4850 0.03 3022 1828 0.40 4850 0.04 2870 1980 0.31 4850 0.05 2778 2072 0.25 4850 0.06 2717 2133 0.21 4850 0.07 2674 2176 0.19 4850 0.08 2641 2209 0.16 4850 0.09 2616 2234 0.15 4850 0.10 2596 2254 0.13 m P x RA m A0 m B0 f 2450 0.02 1663 762 0.54 2450 0.03 1511 914 0.40 2450 0.04 1435 990 0.31 2450 0.05 1389 1036 0.25 2450 0.06 1359 1066 0.22 2450 0.07 1337 1088 0.19 2450 0.08 1321 1104 0.16 2450 0.09 1308 1117 0.15 2450 0.10 1298 1127 0.13 f vs. x RA 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.00 0.02 0.04 0.06 0.08 0.10 0.12 x RA (kg R/kg A) f (kg bypass/kg fresh feed) 4- 48 4.56 a. 900 1 30 03 30 0 kg HCHO kmol HCHO kg HCHO kmol HCHO / h h . .= n 1 (kmol CH 3 OH / h) 30.0 kmol HCHO / h (kmol H 2 /h) (kmol CH 3 OH / h) 2 3 n n % conversion: 30 0 060 500 1 1 . .. n n=?= kmol CH OH / h 3 b. n 1 (kmol CH 3 OH / h) 30.0 kmol HCHO / h 2 (kmol H 2 /h) 3 (kmol CH 3 OH / h) n n 30.0 kmol HCHO / h 2 (kmol H 2 /h)n n 3 (kmol CH 3 OH / h) Overall C balance: n 1 (1) = 30.0 (1) ? n 1 = 30.0 kmol CH 3 OH/h (fresh feed) Single pass conversion: 30 0 13 060 3 20 0 . .. nn n + =?= kmol CH 3 OH / h n 1 + n 3 = 50.0 kmol CH 3 OH fed to reactor/h c. Increased x sp will (1) require a larger reactor and so will increase the cost of the reactor and (2) lower the quantities of unreacted methanol and so will decrease the cost of the separation. The plot would resemble a concave upward parabola with a minimum around x sp = 60%. 4.57 a. Convert effluent composition to molar basis. Basis: 100 g effluent: 22 2 2 33 3 3 10.6 g H 1 mol H 5.25 mol H 2.01 g H 64.0 g CO 1 mol CO 2.28 mol CO 28.01 g CO 25.4 g CH OH 1 mol CH OH 32.04 g CH OH 0.793 mol CH OH = = = ? H 2 : 0.631 mol H 2 /mol CO: 0.274 mol CO / mol CH 3 OH: 0.0953 mol CH 3 OH / mol 4- 49 4.57 (cont?d) 4 0.004 mol CH OH(v)/mol 3 (mol CO/mol) (0.896 - ) (mol H / mol) 2 (mol/min) x x nnull 1 2 (mol CO/min) (mol H / min) 2 n n null null CO H 2 CH 3 OH+? 350 mol/min 0.631 mol CH 3 OH(v)/mol 0.274 mol CO/mol 0.0953 mol H 2 mol/ 3 (mol CH OH(l)/min) 3 nnull Condenser Overall process 3 unknowns 34 (, ,)nnxnullnull 2 unknowns 12 (, )nnnullnull ?3 balances ?2 independent atomic balances 0 degrees of freedom 0 degrees of freedom Balances around condenser 33 4 44 34 32.1 mol CH OH(l)/min CO: 350 0.274 H : 350 0.631 (0.996 ) 318.7 mol recycle/min 2 .301 molCO/mol CH OH : 350 0.0953 0.004 3 n nx nxn x nn ? = ?=? ? ? ?=? ? ?= ? ? = ?=+? ? ? null null nullnull nullnull Overall balances 1 2 13 23 2 32.08 mol/min CO in feed 64.16 mol/min H in feed C: = H: 2 =4 n n nn nn = = ? ? ? ? null null nullnull nullnull Single pass conversion of CO: ( ) () %07.25%100 3009.072.31808.32 274.03503009.072.31808.32 =× ?+ ???+ Overall conversion of CO: %100%100 08.32 008.32 =× ? b. ? Reactor conditions or feed rates drifting. (Recalibrate measurement instruments.) ? Impurities in feed. (Re-analyze feed.) ? Leak in methanol outlet pipe before flowmeter. (Check for it.) Reactor Cond. 4- 50 4.58 a. Basis: 100 kmol reactor feed/hr Reactor Cond. Absorb Still n 1 (kmol CH 4 /h) n 2 (kmol Cl 2 /h) 100 kmol /h 80 kmol CH 4 /h 20 kmol Cl 2 /h n 3 (kmol CH 4 /h) n 4 (kmol HCl /h) 5n 5 (kmol CH 3 Cl /h) n 5 (kmol CH 2 Cl 2 /h) n 3 (kmol CH 4 /h) n 4 (kmol HCl /h) Solvent n 4 (kmol HCl /h) n 3 (kmol CH 4 /h) 5n 5 (kmol CH 3 Cl /h) n 5 (kmol CH 2 Cl 2 /h) n 5 (kmol CH 2 Cl 2 /h) 5n 5 (kmol CH 3 Cl /h) Overall process: 4 unknowns (n 1 , n 2 , n 4 , n 5 ) -3 balances = 1 D.F. Mixing Point: 3 unknowns (n 1 , n 2 , n 3 ) -2 balances = 1 D.F. Reactor: 3 unknowns (n 3 , n 4 , n 5 ) -3 balances = 0 D.F. Condenser: 3 unknowns (n 3 , n 4 , n 5 ) -0 balances = 3 D.F. Absorption column: 2 unknowns (n 3 , n 4 ) -0 balances = 2 D.F. Distillation Column: 2 unknowns (n 4 , n 5 ) -0 balances = 2 D.F. Atomic balances around reactor: 5 n, 4 n, 3 nfor Solve 5 2n 5 5n 4 n40 :balance Cl )3 5 2n 5 15n 4 n 3 4n320 :balance H )2 5 n 5 5n 3 n80 :balance C 1) ? ? ? ? ? ? ++= +++= ++= CH 4 balance around mixing point: n 1 = (80 ? n 3 ) Solve for n 1 Cl 2 balance: n 2 = 20 b. For a basis of 100 kmol/h into reactor n 1 = 17.1 kmol CH 4 /h n 2 = 20.0 kmol Cl 2 /h n 3 = 62.9 kmol CH 4 /h n 4 = 20.0 kmol HCl/h 5n 5 = 14.5 kmol CH 3 Cl/h c. (1000 kg CH 3 Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH 3 Cl/h Scale factor = 366.1 Cl/h 3 CH kmol 5.14 Cl/h 3 CH kmol 81.19 = Fresh feed: 2 Cl mole% 54.0 , 4 CH mol% 0.46 kmol/h 6.50 tot n /h 2 Cl kmol 3.27)366.1)(0.20( 2 n /h 4 CH kmol 3.23)366.1)(1.17( 1 n = ? ? ? ? == == Recycle: n 3 = (62.9)(1.366) = 85.9 kmol CH 4 recycled/h 4- 51 4.59 a. Basis: 100 mol fed to reactor/h ? 25 mol O 2 /h, 75 mol C 2 H 4 /h reactor Seperator n 1 (mol C 2 H 4 //h) n 2 (mol O 2 /h) n 1 (mol C 2 H 4 //h) n 2 (mol O 2 /h) n 3 (mol C 2 H 4 O /h) n 4 (mol CO 2 /h) n 5 (mol H 2 O /h) n 4 (mol CO 2 /h) n 5 (mol H 2 O /h) n 3 (mol C 2 H 4 O /h) 75 mol C 2 H 4 //h 25 mol O 2 /h n C2H4 ( mol C 2 H 4 /h) n O2 (mol O 2 /h) Reactor 5 unknowns (n 1 - n 5 ) -3 atomic balances -1 - % yield -1 - % conversion 0 D.F. Strategy: 1. Solve balances around reactor to find n 1 - n 5 2. Solve balances around mixing point to find n O2 , n C2H4 (1) % Conversion ? n 1 = .800 * 75 (2) % yield: )OHC of raten (productio n HC mol 100 OHC mol 90 HC mol )75)(200(. 423 42 42 42 =× (3) C balance (reactor): 150 = 2 n 1 + 2 n 3 + n 4 (4) H balance (reactor): 300 = 4 n 1 + 4 n 3 + 2 n 5 (5) O balance (reactor): 50 = 2 n 2 + n 3 + 2 n 4 + n 5 (6) O 2 balance (mix pt): n O2 = 25 ? n 2 (7) C 2 H 4 balance (mix pt): n C2H4 = 75 ? n 1 Overall conversion of C 2 H 4 : 100% b. n 1 = 60.0 mol C 2 H 4 /h n 2 = 13.75 mol O 2 /h n 3 = 13.5 mol C 2 H 4 O/h n 4 = 3.00 mol CO 2 /h n 5 = 3.00 mol H 2 O/h n O2 = 11.25 mol O 2 /h n C2H4 = 15.0 mol C 2 H 4 /h 100% conversion of C 2 H 4 c. Scale factor = h/mol h/mollb 363.3 OHC mol 5.13 h OHC lbm 05.44 OHC mole-lb 1 h OHC lbm 2000 4242 4242 ? = n C2H4 = (3.363)(15.0) = 50.4 lb-mol C 2 H 4 /h n O2 = (3.363)(11.25) = 37.8 lb-mol O 2 /h separator 4- 52 4.60 a. Basis: 100 mol feed/h. Put dots above all n?s in flow chart. reactor cond. 100 mol/h 32 mol CO/h 64 mol H 2 / h 4 mol N 2 / h 500 mol / h x 1 (mol N 2 /mol) x 2 (mol CO / mol) 1-x 1 -x 2 (mol H 2 / h) n 1 (mol /h) .13 mol N 2 /mol n 3 (mol / h) x 1 (mol N 2 /mol) x 2 (mol CO / mol) 1-x 1 -x 2 (mol H 2 / h) n 3 (mol CH 3 OH / h) Purge Mixing point balances: total: (100) + 500 = 1 nnull ? 1 nnull = 600 mol/h N 2 : 4 + x 1 * 500 = .13 * 600 ? x 1 = 0.148 mol N 2 /mol Overall system balances: N 2 : 4 = .148 * 3 nnull ? 3 nnull = 27 mol/h Atomic C: 32 = 2 nnull + x 2 *27 2 nnull = 24.3 mol CH 3 OH/h Atomic H: 2 * 64 = 4*24.3 + 2*(1-0.148-x 2 )*27 x 2 = 0.284 mol CO/mol Overall CO conversion: 100*[32-0.284(27)]/32 = 76% Single pass CO conversion: 24.3/ (32+.284*500) = 14% b. Recycle: To recover unconsumed CO and H 2 and get a better overall conversion. Purge: to prevent buildup of N 2 . 4.61 a. Reactor Condenser 1 mol (1-X I0 )/4 (mol N 2 / mol) 3/4 (1-X I0 ) (mol H 2 / mol) X I0 (mol I / mol) n r (mol) n 1 (mol N 2 ) 3n 1 (mol H 2 ) n 2 (mol I) n r (mol) (1-f sp ) n 1 (mol N 2 ) (1-f sp ) 3n 1 (mol H 2 ) n 2 (mol I) 2 f sp n 1 (mol NH 3 ) n p (mol) 2 f sp n 1 (mol NH 3 ) y p (1-f sp ) n 1 (mol N 2 ) y p (1-f sp ) 3n 1 (mol H 2 ) y p n 2 (mol I) (1-f sp ) n 1 (mol N 2 ) (1-f sp ) 3n 1 (mol H 2 ) n 2 (mol I) (1-y p ) (1-f sp ) n 1 (mol N 2 ) (1-y p ) (1-f sp ) 3n 1 (mol H 2 ) (1-y p ) n 2 (mol I) 2N 2 + 3H 2 -> NH 3 n 2 (mol CH 3 OH/h) ? N 2 + 3H 2 ? 2NH 3 4- 53 4.61 (cont?d) At mixing point: N 2 : (1-X I0 )/4 + (1-y p )(1-f sp ) n 1 = n 1 I: X I0 + (1-y p ) n 2 = n 2 Total moles fed to reactor: n r = 4n 1 + n 2 Moles of NH 3 produced: n p = 2f sp n 1 Overall N 2 conversion: %100 4/)X1( n)f1(y4/)X1( 0I 1spp0I × ? ??? b. X I0 = 0.01 f sp = 0.20 y p = 0.10 n 1 = 0.884 mol N 2 n 2 = 0.1 mol I n r = 3.636 mol fed n p = 0.3536 mol NH 3 produced N 2 conversion = 71.4% c. Recycle: recover and reuse unconsumed reactants. Purge: avoid accumulation of I in the system. d. Increasing X I0 results in increasing n r , decreasing n p , and has no effect on f ov . Increasing f sp results in decreasing n r , increasing n p , and increasing f ov . Increasing y p results in decreasing n r , decreasing n p , and decreasing f ov . Optimal values would result in a low value of n r and f sp , and a high value of n p , this would give the highest profit. X I0 f sp y p n r n p f ov 0.01 0.20 0.10 3.636 0.354 71.4% 0.05 0.20 0.10 3.893 0.339 71.4% 0.10 0.20 0.10 4.214 0.321 71.4% 0.01 0.30 0.10 2.776 0.401 81.1% 0.01 0.40 0.10 2.252 0.430 87.0% 0.01 0.50 0.10 1.900 0.450 90.9% 0.10 0.20 0.20 3.000 0.250 55.6% 0.10 0.20 0.30 2.379 0.205 45.5% 0.10 0.20 0.40 1.981 0.173 38.5% 4- 54 4.62 a. i -C H C H C H 410 48 818 + = Basis: 1-hour operation reactor (n-C H ) 4 n 510 (i-C H ) 4 n 610 (C H ) 8 n 718 (91% H SO ) 2 m 84 (n-C H ) 4 n 210 (i-C H ) 4 n 310 (C H ) 8 n 118 (91% H SO ) 2 m 44 decanter still (C H ) 8 n 118 (n-C H ) 4 n 210 (i-C H ) 4 n 310 (C H ) 8 n 118 (n-C H ) 4 n 210 F PD E C B A (kg 91% H SO ) 2 m 44 (i-C H ) 4 n 310 40000 kg kmoln 0 0.25 0.50 0.25 i-C H 4 10 n-C H 4 10 C H 48 Units of n: kmol Units of m: kg Calculate moles of feed MM M M=++=+ = ?? 0 25 0 50 0 25 0 75 5812 0 25 5610 57 6 ....... . LCH nCH CH 410 410 48 kg kmol bgb gbgb g n 0 40000 1 694== kg kmol 57.6 kg kmolbgb g Overall n - C H balance: 410 n 2 050 694 347==.bgbg kmol n - C H in product 410 C H balance: 818 n 1 694 1735== 0.25 kmol C H react 1 mol C H 1 mol C H kmol C H in product 48 818 48 88 bgbg . At (A), 5 mol - C H 1 mole C H n mol - C H kmol 410 48 410 A moles C H at A=173.5 -C H at A and B 48 410 ii i ?==bgbggbg bg bg 5 0 25 694 867 5.. nullnullnullnullnullnullnull Note: n mol C H 173.5 48 bg= at (A), (B) and (C) and in feed in ni - C H balance around first mixing point kmol - C H recycled from still 410 410 ?+= ?= 0 25 694 867 5 694 3 3 ..bgbg At C, 200 mol - C H mol C H n mol - C H kmol - C H 410 48 410 C 410 i ii?==bgbgbg200 1735 34 700., 4- 55 4.62 (cont?d) in n - C H balance around second mixing point l C H in recycle E 410 410 ? + = ?= 867 5 34 700 33,800 kmo 6 6 ., Recycle E: Since Streams (D) and (E) have the same composition, n n ni ni n 5 2 6 3 5 moles n - C H moles n - C H moles - C H moles - C H 16,900 kmol n - C H 410 E 410 D 410 E 410 D 410 bgbg =?= n n n n n 7 1 6 3 7 moles C H moles C H 8460 kmol C H 818 E 818 D 418 bg =?= Hydrocarbons entering reactor: 347 16900 5812+ F H G I K Jbgb g kmol n - C H kg kmol 410 . ++ F H G I K J + F H G I K J 867 5 33800 5812 1735 5610. ...bgb gkmol - C H kg kmol kmol C H kg kmol 410 48 i + F H G I K J =×8460 kmol 114.22 4 00 10 6 CH kg kmol kg 818 . . H SO solution entering reactor and leaving reactor 4.00 10 kg HC 2 kg H SO aq 1 kg HC kg H SO aq 24 6 24 24 bg bg bg = × =×800 10 6 . mn nn m 8 6 5 25 8 6 800 10 784 10 H SO in recycle H SO leaving reactor n - C H in recycle n - C H leaving reactor kg H SO aq in recycle E 24 24 410 410 24 bgbg bg . . × = + ?= × m 4 5 16 10 = ? =× H SO entering reactor H SO in E kg H SO aq recycled from decanter 24 24 24 . bg ? 16 10 091 1480 5 ..×= dibg b g kg H SO 1 kmol 98.08 kg kmol H SO in recycle 24 24 16 10 009 799 5 .. dibg b g kg H O 1 kmol 18.02 kg kmol H O from decanter 22 Summary: (Change amounts to flow rates) Product: 173.5 kmol C H h 347 kmol - C H h Recycle from still: 694 kmol - C H h Acid recycle: 1480 kmol H SO h 799 kmol H O h Recycle E: 16,900 kmol n - C H h 33,800 kmol L - C H h , 8460 kmol C H h, kg h 91% H SO 72,740 kmol H SO h , 39,150 kmol H O h 818 410 410 24 2 410 410 818 24 24 2 , , , . n i 784 10 6 ×? 4- 56 4.63 a. A balance on ith tank (input = output + consumption) nullnull , , null, null vC vCkCC V CCkCC A i Ai Ai Bi A i Ai Ai Bi vVv L min mol L mol liter min L note / bgbg b gbg ? ? =+ ? E =+ ÷= 1 1 ? ? B balance. By analogy, CCkCC B i Bi Ai Bi, ? = + 1 ? Subtract equations ? ? = ? = ?== ? ?? A ? ?? CCC C C C C C Bi Ai B i A i i Bi Ai B A,, , ,11 1 22 00 from balances on tank st bg ? b. CCC C CCC C Bi Ai B A Bi Ai B A ?=??= + ? 00 00 . Substitute in A balance from part (a). CCkCCCC A i Ai Ai Ai B A, ? =+ + ? 10 ? bg. Collect terms in C Ai 2 , C Ai 1 , C Ai 0 . Ck C kC C C CC k kCC C Ai AL B A A i AL AL B A A i 2 00 1 2 00 1 01 ?? ??? ???? ? ++ ??= ?++= ==+?=? ? ? bg bg , , ,, where Solution: C Ai = ?+ ??? ?? ? 2 4 2 (Only + rather than ±: since ?? is negative and the negative solution would yield a negative concentration.) (x min = 0.50, N = 1), (x min = 0.80, N = 3), (x min = 0.90, N = 4), (x min = 0.95, N = 6), (x min = 0.99, N = 9), (x min = 0.999, N = 13). As x min ? 1, the required number of tanks and hence the process cost becomes infinite. d. (i) k increases ? N decreases (faster reaction ? fewer tanks) ()nullii increases increases (faster throughput less time spent in reactor lower conversion per reactor) vN?? ? (iii) V increases ? N decreases (larger reactor ? more time spent in reactor ? higher conversion per reactor) c. k = 36.2 N gamma CA(N) xA(N) v = 5000 1 -5.670E-02 2.791E-02 0.5077 V = 2000 2 -2.791E-02 1.512E-02 0.7333 CA0 = 0.0567 3 -1.512E-02 8.631E-03 0.8478 CB0 = 0.1000 4 -8.631E-03 5.076E-03 0.9105 alpha = 14.48 5 -5.076E-03 3.038E-03 0.9464 beta = 1.6270 6 -3.038E-03 1.837E-03 0.9676 7 -1.837E-03 1.118E-03 0.9803 8 -1.118E-03 6.830E-04 0.9880 9 -6.830E-04 4.182E-04 0.9926 10 -4.182E-04 2.565E-04 0.9955 11 -2.565E-04 1.574E-04 0.9972 12 -1.574E-04 9.667E-05 0.9983 13 -9.667E-05 5.939E-05 0.9990 14 -5.939E-05 3.649E-05 0.9994 4- 57 4.64 a. Basis: 1000 g gas Species m (g) MW n (mol) mole % (wet) mole % (dry) C 3 H 8 800 44.09 18.145 77.2% 87.5% C 4 H 10 150 58.12 2.581 11.0% 12.5% H 2 O 50 18.02 2.775 11.8% Total 1000 23.501 100% 100% Total moles = 23.50 mol, Total moles (dry) = 20.74 mol Ratio: 2.775 / 20.726 = 0.134 mol H O / mol dry gas 2 b. C 3 H 8 + 5 O 2 ? 3 CO 2 + 4 H 2 O, C 4 H 10 + 13/2 O 2 ? 4 CO 2 + 5 H 2 O Theoretical O 2 : CH 100 kg gas h 80 kg C H 100 kg gas 1 kmol C H 44.09 kg C H 5 kmol O 1 kmol C H 9.07 kmol O / h 38 38 38 38 2 38 2 : = CH 100 kg gas h 15 kg C H 100 kg gas 1 kmol C H 58.12 kg C H 6.5 kmol O 1 kmol C H 1.68 kmol O / h 410 410 410 410 2 410 2 : = Total: (9.07 + 1.68) kmol O 2 /h = 10.75 kmol O 2 /h Air feed rate: 10.75 kmol O h 1 kmol Air .21 kmol O 1.3 kmol air fed 1 kmol air required 66.5 kmol air / h 2 2 = The answer does not change for incomplete combustion 4.65 5 L C H 0.659 kg C H L C H 1000 mol C H 86 kg C H 38.3 mol C H 614 614 614 614 614 614 = 4 L C H 0.684 kg C H L C H 1000 mol C H 100 kg C H 27.36 mol C H 716 716 716 716 716 716 = C 6 H 14 +19/2 O 2 ? 6 CO 2 + 7 H 2 O C 6 H 14 +13/2 O 2 ? 6 CO + 7 H 2 O C 7 H 16 + 11 O 2 ? 7 CO 2 + 8 H 2 O C 7 H 16 + 15/2 O 2 ? 7 CO + 8 H 2 O Theoretical oxygen: 38.3 mol C H 9.5 mol O mol C H 27.36 mol C H 11 mol O mol C H mol O required 614 2 614 716 2 716 2 +=665 O 2 fed: (4000 mol air )(.21 mol O 2 / mol air) = 840 mol O 2 fed Percent excess air: 840 665 665 100% 26 3% ? ×=. excess air 4- 58 4.66 CO 1 2 O CO H 1 2 OHO 22 222 +? +? 175 kmol/h 0.500 kmol N 2 /kmol x (kmol CO/mol) (0.500?x) (kmol H 2 /kmol) 20% excess air Note: Since CO and H 2 each require 0./5 mol O mol fuel 2 for complete combustion, we can calculate the air feed rate without determining x CO . We include its calculation for illustrative purposes. A plot of x vs. R on log paper is a straight line through the points Rx 11 10 0 0 05==., .bg and Rx 22 99 7 10==., .bg. ln ln ln ln . . ln . . . ln ln . . ln . . . . exp . . .. xbR a xaR b b axR a Rx =+ = == =? =??=× ? =?=× ? =?= @ 10 005 997 100 1303 10 1303 997 600 249 10 31303 6 00 2 49 10 3 38 3 0 288 bgb g bg b g bg moles CO mol Theoretical O 2 175 kmol kmol CO 0.5 kmol O 2 h kmol kmol CO 175 kmol kmol H 2 0.5 kmol O 2 h kmol kmol H 2 kmol O 2 h :. . . 0288 0 212 43 75+= Air fed: 43.75 kmol O 2 required 1 kmol air 1.2 kmol air fed h 0.21 kmol O 2 1 kmol air required kmol air h = 250 4.67 a. CH 2O CO 2H O CH 7 2 O2CO3HO CH 5O 3CO 4HO CH 13 2 O4CO5HO 42 22 26 2 2 2 38 2 2 2 410 2 2 2 +?+ +? + +? + +?+ 100 kmol/h 0.944 CH 4 0.0340 C 2 H 6 0.0060 C 3 H 8 0.0050 C 4 H 10 17% excess air n a (kmol air/h) 0.21 O 2 0.79 N 2 Theoretical O 0.944 100 kmol CH kmol O h 1 kmol CH 0.0340 100 kmol C H .5 kmol O h 1 kmol C H 0.0060 100 kmol C H kmol O h 1 kmol C H 0.0050 100 kmol C H .5 kmol O h 1 kmol C H kmol O h 2 42 4 26 2 26 38 2 33 410 2 410 2 : . bg bg bg bg 23 56 207 0 + ++ = 4- 59 4.67 (cont?d) Air feed rate: n f == 207.0 kmol O 1 kmol air 1.17 kmol air fed h 0.21 kmol O kmol air req. kmol air h 2 2 1153 b. nnx x x x P af xs =++++2 35 5 6 5 1 100 1 0 21 1234 .. .bgbgb c. ,( .) . ,( ) . naRn R n R nbRn R n R fff f f f aaa a a a == = ? = == == kmol / h, kmol / h, 75 0 60 125 550 25 22 0 xkA xk A k A x A A i ii i i i i i i =? = =?= ?= ?? ? ? , = CH C H C H C H ii i i 42438410 1 1 ,,, d. Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the flowmeter calibration formulas might not be linear, or the stack gas analysis could be incorrect. 4.68 a. C 4 H 10 + 13/2 O 2 ? 4 CO 2 + 5 H 2 O Basis: 100 mol C 4 H 10 n CO2 (mol CO 2 ) n H2O (mol H 2 O) n C4H10 (mol C 4 H 10 ) P xs (% excess air) n O2 (mol O 2 ) n air (mol air) n N2 (mol N 2 ) 0.21 O 2 0.79 N 2 D.F. analysis 6 unknowns (n, n 1 , n 2 , n 3 , n 4 , n 5 ) -3 atomic balances (C, H, O) -1 N 2 balance -1 % excess air -1 % conversion 0 D.F. Run P xs R f A 1 A 2 A 3 A 4 1 15% 62 248.7 19.74 6.35 1.48 2 15% 83 305.3 14.57 2.56 0.70 3 15% 108 294.2 16.61 4.78 2.11 Run n f x 1 x 2 x 3 x 4 n a R a 1 77.5 0.900 0.0715 0.0230 0.0054 934 42.4 2 103.8 0.945 0.0451 0.0079 0.0022 1194 54.3 3 135.0 0.926 0.0523 0.0150 0.0066 1592 72.4 4- 60 4.68 (cont?d) b. i) Theoretical oxygen = (100 mol C 4 H 10 )(6.5 mol O 2 /mol C 4 H 10 ) = 650 mol O 2 n mol O mol air / 0.21 mol O mol air air 2 2 = =()( )650 1 3095 100% conversion ? n C4H10 = 0, n O 2 = 0 n n n N2 CO2 4 10 2 4 10 2 H2O 4 10 2 4 10 2 2 2 2 mol 100 mol C H react 4 mol CO mol C H mol CO 100 mol C H react 5 mol H O mol C H mol H O 73.1% N 12.0% CO 14.9% H O == U V | W | 0 79 3095 mol 2445 400 500 .bgb g b g bg ii) 100% conversion ? n C4H10 = 0 20% excess ? n air = 1.2(3095) = 3714 mol (780 mol O 2 , 2934 mol N 2 ) Exit gas: 400 mol CO 2 500 mol H 2 O 130 mol O 2 2934 mol N 2 10.1% CO 2 12.6% H 2 O 3.3% O 2 74 0%. N 2 iii) 90% conversion ? n C4H10 = 10 mol C 4 H 10 (90 mol C 4 H 10 react, 585 mol O 2 consumed) 20% excess: n air = 1.2(3095) = 3714 mol (780 mol O 2 , 2483 mol N 2 ) Exit gas: 10 mol C 4 H 10 360 mol CO 2 450 mol H 2 O (v) 195 mol O 2 2934 mol N 2 0.3% C 4 H 10 9.1% CO 2 11.4% H 2 O 4.9% O 2 74 3%. N 2 4.69 a. C 3 H 8 + 5 O 2 ? 3 CO 2 + 4 H 2 O H 2 +1/2 O 2 ? H 2 O C 3 H 8 + 7/2 O 2 ? 3 CO + 4 H 2 O Basis: 100 mol feed gas 100 mol 0.75 mol C 3 H 8 n 1 (mol C 3 H 8 ) 0.25 mol H 2 n 2 (mol H 2 ) n 3 (mol CO 2 ) n 4 (mol CO) n 0 (mol air) n 5 (mol H 2 O) 0.21 mol O 2 /mol n 6 (mol O 2 ) 0.79 mol N 2 /mol n 7 (mol N 2 ) Theoretical oxygen 75 mol C H mol O mol C H mol H mol O mol H mol O 38 2 38 22 2 2 : . . 52505 387 5+= 4- 61 4.69 (cont?d) Air feed rate mol O h kmol air 0.21 kmol O kmol air fed 1 kmol air req'd. mol air 2 2 : . . .n 0 387 5 115 2306 5== 90% 0100 75 85% 0150 25 95% 095 675 3 192 4 5% 005 675 3 101 1 2 3 3 propane conversion mol C H ) = 7.5 mol C H (67.5 mol C H reacts) hydrogen conversion mol C H ) = 3.75 mol H CO selectivity mol C H react) mol CO generated mol C H react mol CO CO selectivity mol C H react) mol CO generated mol C H react mol CO 38 38 38 38 2 2 38 2 38 2 38 38 ? = ?= ?= = ?= = n n n n .( .( .(. . .(. . H balance (75 mol C H mol H mol C H mol H mol C H mol H mol H O)(2) mol H O 38 38 2 38 2 2 2 :) ()( (. )() (. )() ( . 825 7 5 8 3 75 2 2912 55 F H G I K J + =+ ?=nn O balance ( .21 2306.5 mol O )(2 mol O mol O ) 4 mol CO mol CO mol H O)(1) + 2 mol O mol O 2 2 2 22 2 :(. (. )()( . ( ) . 0 192 2 101 1 2912 1413 66 ×= ++ ?=nn N balance: mol N mol N 22 n 7 0 79 2306 5 1822= =.( .) Total moles of exit gas = (7.5 + 3.75 + 192.4 + 10.1 + 291.2 + 141.3 + 1822) mol = 2468 mol CO concentration in exit gas = 101 10 4090 6 . mol CO 2468 mol ppm×= b. If more air is fed to the furnace, (i) more gas must be compressed (pumped), leading to a higher cost (possibly a larger pump, and greater utility costs) (ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the product gas temperature decreases and less steam is produced. 4- 62 4.70 a. C 5 H 12 + 8 O 2 ? 5 CO 2 + 6 H 2 O Basis: 100 moles dry product gas n 1 (mol C 5 H 12 ) 100 mol dry product gas (DPG) 0.0027 mol C 5 H 12 /mol DPG Excess air 0.053 mol O 2 /mol DPG n 2 (mol O 2 ) 0.091 mol CO 2 /mol DPG 3.76n 2 (mol N 2 ) 0.853 mol N 2 /mol DPG n 3 (mol H 2 O) 3 unknowns (n 1 , n 2 , n 3 ) -3 atomic balances (O, C, H) -1 N2 balance -1 D.F. ? Problem is overspecified b. N 2 balance: 3.76 n 2 = 0.8533 (100) ? n 2 = 22.69 mol O 2 C balance: 5 n 1 = 5(0.0027)(100) + (0.091)(100) ? n 1 = 2.09 mol C 5 H 12 H balance: 12 n 1 = 12(0.0027)(100) + 2n 3 ? n 3 = 10.92 mol H 2 O O balance: 2n 2 = 100[(0.053)(2) + (0.091)(2)] + n 3 ? 45.38 mol O = 39.72 mol O Since the 4 th balance does not close, the given data cannot be correct. c. n 1 (mol C 5 H 12 ) 100 mol dry product gas (DPG) 0.00304 mol C 5 H 12 /mol DPG Excess air 0.059 mol O 2 /mol DPG n 2 (mol O 2 ) 0.102 mol CO 2 /mol DPG 3.76n 2 (mol N 2 ) 0.836 mol N 2 /mol DPG n 3 (mol H 2 O) N 2 balance: 3.76 n 2 = 0.836 (100) ? n 2 = 22.2 mol O 2 C balance: 5 n 1 = 100 (5*0.00304 + 0.102) ? n 1 = 2.34 mol C 5 H 12 H balance: 12 n 1 = 12(0.00304)(100) + 2n 3 ? n 3 = 12.2 mol H 2 O O balance: 2n 2 = 100[(0.0590)(2) + (0.102)(2)] + n 3 ? 44.4 mol O = 44.4 mol O ? Fractional conversion of C 5 H 12 : fed react/mol mol 870.0 344.2 00304.0100344.2 = ×? Theoretical O 2 required: 2.344 mol C 5 H 12 (8 mol O 2 /mol C 5 H 12 ) = 18.75 mol O 2 % excess air: air excess %6.18%100 required O mol 18.75 required O mol 18.75 - fed O mol 23.22 2 22 =× 4- 63 4.71 a. h/OHCH mol 6.296 g 04.32 mol ml g 792.0 L ml 1000 h OHCH L 12 3 3 = CH 3 OH + 3/2 O 2 ? CO 2 +2 H 2 O, CH 3 OH + O 2 ? CO +2 H 2 O 296.6 mol CH 3 OH(l)/h (n 2 mol dry gas / h) 0.0045 mol CH 3 OH(v)/mol DG 0.0903 mol CO 2 /mol DG /n 1 (mol O h) 2 0.0181 mol CO/mol DG 376 1 . /n (mol N h) 2 x (mol N 2 /mol DG) (0.8871?x) (mol O 2 /mol DG) (n 3 mol H O(v) / h) 2 4 unknowns ( , , ,)nnn x 123 ? 4 balances (C, H, O, N 2 ) = 0 D.F. b. Theoretical O 2 : 296.6 (1.5) = 444.9 mol O 2 / h C balance: 296.6 = n 2 (0.0045 + 0.0903 + 0.0181) ? n 2 = 2627 mol/h H balance: 4 (296.6) = n 2 (4*0.0045) + 2 n 3 ? n 3 = 569.6 mol H 2 O / h 69.65 x)]-2(0.8871.018102(0.0903) 627[0.00452 1 n2 296.6 :balance O ++++=+ N 2 balance: 3.76 n 1 = x ( 2627) Solving simultaneously ? ./ /nx 1 574 3= mol O h, = 0.822 mol N mol DG 22 Fractional conversion: fed react/mol OHCH mol 960.0 6.296 )0045.0(26276.296 3 = ? % excess air: %1.29%100 9.444 9.4443.574 =× ? Mole fraction of water: /molOH mol 178.0 mol )6.5692627( OH mol 6.569 2 2 = + c. Fire, CO toxicity. Vent gas to outside, install CO or hydrocarbon detector in room, trigger alarm if concentrations are too high 4.72 a. G.C. Say n s mols fuel gas constitute the sample injected into the G.C. If x CH 4 and x CH 26 are the mole fractions of methane and ethane in the fuel, then nx nx x x s s mol mol C H mol 2 mol C 1 mol C H mol mol CH mol 1 mol C 1 mol CH mol C H mol fuel mol CH mol fuel mole C H mole CH in fuel gas CH 2 2 2 6 CH 4 4 CH 2 6 CH 4 26 4 26 4 26 4 bg b gb g bg b gb g bg = E = 20 85 01176. 4- 64 4.72 (cont?d) Condensation measurement: 1.134 g H O 1 mol 18.02 g mol product gas mole H O mole product gas 2 2 bgb g 050 0126 . .= Basis: 100 mol product gas. Since we have the most information about the product stream composition, we choose this basis now, and would subsequently scale to the given fuel and air flow rates if it were necessary (which it is not). CH 2O CO 2H O CH 7 2 O2CO3HO 42 22 26 2 2 2 +?+ +? + n 1 (mol CH 4 ) 0.1176 n 1 (mol C 2 H 6 ) n 2 (mol CO 2 ) n 3 (mol O 2 / h) 376 n 3 (mol N 2 / h) 100 mol dry gas / h 0.126 mol H 2 O / mol 0..874 mol dry gas / mol 0.119 mol CO 2 / mol D.G. x (mol N 2 / mol) (0.881-x) (mol O 2 / mol D.G.) Strategy: H balance n? 1 ; C balance ? n 2 ; N balance O balance n 2 3 U V W ? , x H balance: mol CH in fuel 4 4 6 01176 100 0126 2 5 356 11 1 nn n+= ?=bgb g b gb gbg... ? 0.1176(5.356) = 0.630 mol C 2 H 6 in fuel C balance: 5 356 2 0 630 100 0 874 0119 3 784 22 .. .. .++= ?=bgb g b gb gb gnn mol CO in fuel 2 Composition of fuel: 5356. mol CH 4 , 0 630. mol CH 26 , 3 784. mols CO 2 ? 0 548. CH , 0.064 C H , 0.388 CO 426 2 N balance: 2 376 100 0874 3 ..nx=bgb g O balance: 2 3 784 2 100 0126 100 0 874 2 0119 0 881 3 bgb g bgb gbgb gbg b g...+= + + ? Solve simultaneously: n 3 18 86= . mols O fed 2 , x = 0813. Theoretical O 5.356 mol CH mol O 1 mol CH 0.630 mol C H .5 mol O 1 mol CH mol O required 2 42 4 26 2 4 2 : . 23 12 92 + = Desired O 2 fed: air mol O mol 0.21 fuel mol 1 air mol 7fuel mol 3.784) 0.630 (5.356 2 ++ = 14.36 mol O 2 Desired % excess air: %11%100 92.12 92.1236.14 =× ? b. Actual % excess air: %46%100 92.12 92.1286.18 =× ? Actual molar feed ratio of air to fuel: 1:9 feedmol77.9 air mol )21.0/86.18( = 4- 65 4.73 a. C 3 H 8 +5 O 2 ? 3 CO 2 + 4 H 2 O, C 4 H 10 + 13/2 O 2 ? 4 CO 2 + 5 H 2 O Basis 100: mol product gas n 1 (mol C 3 H 8 ) 100 mol n 2 (mol C 4 H 10 ) 0.474 mol H 2 O/mol x (mol CO 2 /mol) n 3 (mol O 2 ) (0.526?x) (mol O 2 /mol) Dry product gas contains 69.4% CO 2 ? /molCO mol 365.0x 6.30 4.69 x526.0 x 2 =?= ? 3 unknowns (n 1 , n 2 , n 3 ) ? 3 balances (C, H, O) = 0 D.F. O balance: 2 n 3 = 152.6 ? n 3 = 76.3 mol O 2 10 H 4 C 34.9% , 8 H 3 C %1.65 10 H 4 C mol 3.8 2 n 8 H 3 C mol 7.1 1 n 94.8 2 n 10 1 n 8 :balance H 36.5 2 n 4 1 n 3 :balance C ? = = ? ? ? ? =+ =+ b. n c =100 mol (0.365 mol CO 2 /mol)(1mol C/mol CO 2 ) = 365 mol C n h = 100 mol (0.474 mol H 2 O/mol)(2mol H/mol H 2 O)=94.8 mol H ? 27.8%C, 72.2% H From a: C %8.27%100 HC mol H)(C mol 14HC mol 3.80 HC mol H)(C mol 11HC mol 7.10 HC mol C mol 4HC mol 3.80 HC mol C mol 3HC mol 7.10 104 104 83 83 104 104 83 83 =× + + + + 4.74 Basis: 100 kg fuel oil Moles of C in fuel: C kmol 08.7 C kg 12.01 C kmol 1 kg C kg 0.85kg 100 = Moles of H in fuel: H kmol 0.12 H kg 1 H kmol 1 kg H kg 0.12kg 100 = Moles of S in fuel: S kmol 053.0 S kg 32.064 S kmol 1 kg S kg 0.017kg 100 = 1.3 kg non-combustible materials (NC) 4- 66 4.74 (cont?d) 100 kg fuel oil 7.08 kmol C n 2 (kmol N 2 ) 12.0 kmol H n 3 (kmol O 2 ) 0.053 kmol S C + O 2 ? CO 2 n 4 (kmol CO 2 ) 1.3 kg NC (s) C + 1/2 O 2 ? CO (8/92) n 4 (kmol CO) 2H + 1/2 O 2 ? H 2 O n 5 (kmol SO 2 ) 20% excess air S + O 2 ? SO 2 n 6 (kmol H 2 O) n 1 (kmol O 2 ) 3.76 n 1 (kmol N 2 ) Theoretical O 2 : 2 222 O kmol 133.10 S kmol 1 O kmol 1S kmol .0530 H kmol 2 O kmol 5.H kmol 21 C kmol 1 O kmol 1C kmol 08.7 =++ 20 % excess air: n 1 = 1.2(10.133) = 12.16 kmol O 2 fed O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n 3 ? n 3 = 2.3102 kmol O 2 C balance: 7.08 = n 4 +8n 4 /92 ? n 4 = 6.514 mol CO 2 ? 8 (6.514)/92 = 0.566 mol CO S balance: n 5 = 0.53 kmol SO 2 H balance: 12 = 2n 6 ? n 6 = 6.00 kmol H 2 O N 2 balance: n2 = 3.76(12.16) = 45.72 kmol N 2 Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol = 61.16 kmol ? 10.7% CO, 0.92% CO, 0.087% SO 9.8% H O, 3.8% O , 74.8% N 22 2 2 , 4.75 a. Basis: 5000 kg coal/h; 50 kmol air min kmol air h= 3000 C + 0 2 --> CO 2 2H + 1/2 O 2 -->H 2 O S + O 2 --> SO 2 C + 1/2 O 2 --> CO 5000 kg coal / h 0.75 kg C / kg 0.17 kg H / kg 0.02 kg S / kg 0.06 kg ash / kg 3000 kmol air / h 0.21 kmol O 2 / kmol 0.79 kmol N 2 / kmol n 1 (kmol O 2 / h) n 2 (kmol N 2 / h) n 3 (kmol CO 2 / h) 0.1 n 3 (kmol CO / h) n 4 (kmol SO 2 / h) n 5 (kmol H 2 O / h) mo kg slag / h Theoretical O 2 : C: 0.75 5000 kg C 1 kmol C 1 kmol O h kg C 1 kmol C 2 bg 12 01. = 312 2. kmol O h 2 4- 67 4.75 (cont?d) H: 0.17 5000 kg H 1 kmol H 1 kmol H O 1 kmol O h kg H 2 kmol H 2 kmol H O 22 2 bg 101. = 210 4. kmol O h 2 S: 0.02 5000 kg S 1 kmol S 1 kmol O h1 kmol S 2 bg 32.06 kg S = 3.1 kmol O 2 /h Total = (312.2+210.4 + 3.1) kmol O 2 /h = 525.7 kmol O h 2 O fed 2 = 0 21 3000 630. bg= kmol O h 2 Excess air: 630 525 7 525 7 100% 19 8% ? ×= . . . excess air b. Balances: C: 094 075 12 01 .. . bgbgb g5000 kg C react 1 kmol C h kg 0.1 33 =+nn ?= .n 32 kmol CO h266 8 , 01 267. .n 3 kmol CO h= H: 017 101 . . bgb g5000 kg H 1 kmol H 1 kmol H O h kg H 2 kmol H 2 5 = n ? =n 52 kmol H O h420 8. S: (from part a) 3.1 kmol O for SO 1 kmol SO h kmol O 22 2 2 4 bg 1 = n ? = .n 42 kmol SO h31 N 2 : 0 79 3000. bgb g kmol N h n 22 = ? = n 22 kmol N h2370 O: 0 21 3000 2 2 2 266 8 1 26 68 2 31 1 420 8.( ) .. . .bg bg bgbgbgbgbg=+ + + +n 1 ? ./n 1 136 4= kmol O h 2 Stack gas total = 3223 kmol h Mole fractions: x CO mol CO mol==× ? 26 7 3224 8 3 10 3 .. x SO 2 2 mol SO mol==× ? 31 3224 9 6 10 4 .. c. SO O SO SO H O H SO 22 3 32 24 +? +? 1 2 3.1 kmol SO 1 kmol SO 1 kmol H SO 98.08 kg H SO h kmol SO 1 kmol SO kmol H SO 304 kg H SO h 2324 24 23 24 1 = 4- 68 4.76 a. Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile matter 100 g c.a.r. 1.147 g a.d.c. 1.207 g c.a.r. g air - dried coal; 4.97 g H O lost by air drying 2 = 95 03. 95.03 g a.d.c 1.234 g H O 1.234 g a.d.c. g H O lost in second drying step 2 2 ? = 1204 231 . . bg Total H O g g g moisture 2 =+=497 231 728... 95.03 g a.d.c g v.m. H O 1.347 g a.d.c. g H O g volatile matter 2 2 1347 0 811 231 3550 .. .. ?+ ?= bgbg 95.03 g a.d.c .111 g ash 1.175 g a.d.c. .98 g ash 0 8= Fixed carbon =?? ? =100 7 28 3550 8 98 48 24... .bgg g fixed carbon 7.28 g moisture 48.24 g fixed carbon 35.50 g volatile matter 8.98 g ash 100 g coal as received 7.3% moisture 48.2% fixed carbon 35.5% volatile matter 9.0% ash ? b. Assume volatile matter is all carbon and hydrogen. CCO CO 22 +?: 1 mol O 1 mol C 10 g 1 mol air 1 mol C 12.01 g C 1 kg 0.21 mol O mol air kg C 2 3 2 = 396 5. 2H O H O 2 +? 1 2 2 : 0.5 mol O 1 mol H 10 g 1 mol air 2 mol H 1.01 g H 1 kg 0.21 mol O mol air kg H 2 3 2 =1179 Air required: 1000 kg coal 0.482 kg C 396.5 mol air kg coal kg C + 1000 kg 0.355 kg v.m. 6 kg C 396.5 mol air kg 7 kg v.m. kg C +=× 1000 kg 0.355 kg v.m. 1 kg H 1179 mol air kg 7 kg v.m. kg H mol air372 10 5 . 4- 69 4.77 a. Basis 100 mol dry fuel gas. Assume no solid or liquid products! C + 0 2 --> CO 2 C + 1/2 O 2 --> CO 2H + 1/2 O 2 -->H 2 O S + O 2 --> SO 2 n 1 (mol C) n 2 (mol H) n 3 (mol S) n 4 (mol O 2 ) (20% excess) 100 mol dry gas 0.720 mol CO 2 / mol 0.0257 mol CO / mol 0.000592 mol SO 2 / mol 0.254 mol O 2 / mol n 5 (mol H 2 O (v)) ? ? ? ? ? =++ ++++= = n ]n 0.25 0.0592 (74.57 (1.20) :O excess % 20 n (0.254)] 2 (0.000592) 2 0.0257 2(0.720) [ 100 n 2 :balance O n 2 n :balance H 422 54 52 ? n 2 = 183.6 mol H, n 4 = 144.6 mol O 2 , n 5 = 91.8 mol H 2 O Total moles in feed: 258.4 mol (C+H+S) ? 28.9% C, 71.1% H, 0.023% S 4.78 Basis: 100 g oil 0.87 g C/g mol On 12 (25% excess) 100 g oil 0.10 g H/g 0.03 g S/g mol Nn 12 3.76 furnace mol Nn 22 mol On 32 mol COn 42 mol SOn 52 mol H On 62 (1 ? )x mol SOn 52 (N , O , CO , H O) 2222 scrubber (1 ? )x mol SOn 52 0.90 Alkaline solution x mol SOn 32 (N , O , CO , H O) 2222 (1 ? )x mol SOn 52 0.10 (N , O , CO , H O) 2222 Stack N , O , CO , H O 2222 SO , 2 (612.5 ppm SO ) 2 CO 2 : 0.87 100 g C 1 mol C 1 mol CO g C 1 mol C mol CO mol O consumed 2 42 2 bg 12 01 7244 7244 . . . ?= F H G I K J n HO 2 : 0.10 100 g H 1 mol H 1 mol H O g H 2 mol H mol H O .475 mol O consumed 2 62 2 bg 101 495 2 . .?= F H G I K J n 4- 70 4.78 (cont?d) SO 2 : 0.03 100 g S 1 mol S 1 mol SO g S 1 mol S n mol SO .0956 mol O consumed 2 52 2 bg 32 06 0 0936 0 . .?= F H G I K J 25% excess O 2 : n 12 mol O=++?125 7 244 2 475 0 0936 12 27.. . . .bg O balance: 12.27 mol O fed mol O consumed mol O 23 2 2 2 n =?+ = 7 244 2 475 0 0936 246 ... . bg N balance: 2 n mol mol N 22 ==3761227 4614.. .bg SO in stack SO balance around mixing point 22 : xx x n 0 0936 010 1 0 0936 0 00936 0 0842 5 .. ... F H I K +? = +bgb g b gmol SO 2 Total dry gas in stack (Assume no CO 2 , O 2 , or N 2 is absorbed in the scrubber) 7 244 2 46 4614 0 00936 0 0842 5585 0 0842.... . .. CO O N SO 22 2 2 mol dry gas bgbgbg bg bgb++ + + = +xx 612 5. ppm SO dry basis in stack gas 2 bg 000936 00842 5585 0 0842 612 5 10 10 0 295 30% 6 .. .. . . . + + = × ?= ? x x x bypassed 4.79 Basis: 100 mol stack gas C + O 2 CO 2 ? 2H + O 2 1 2 H O 2 ? S + O 2 SO 2 ? (mol C)n 1 (mol H)n 2 (mol S)n 3 (mol O )n 42 (mol O )n 42 3.76 100 mol 0.7566 N 2 0.1024 CO 2 0.0827 H O 2 0.0575 O 2 0.000825 SO 2 a. b. C balance: mol C H balance: mol H mol C mol H mol C mol H 1 2 n n == ?= 100 01024 10 24 100 0 0827 2 16 54 10 24 16 54 062 bgb g bgb gbg .. . . . The C/H mole ratio of CH 4 is 0.25, and that of CH 26 is 0.333; no mixture of the two could have a C/H ratio of 0.62, so the fuel could not be the natural gas. S balance: n mol S 3 ==100 0 000825 0 0825bgb g.. 10 24 122 88 16 54 16 71 0265 122 88 16 71 735 265 142 24 100% 1 .. . . . . . . mol C 12.0 g 1 mol g C mol H 1.01 g 1 mol g H .0825 mol S 32.07 g 1 mol g S g C g H .9% S No. 4 fuel oil bgb g bgb g = = = U V | W | ? = ×= ? 4- 71 4.80 a. Basis: 1 mol C p H q O r C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O2 --> SO2 1 mol C p H q O r no (mol S) X s (kg s/ kg fuel) P (% excess air) n 1 (mol O 2 ) 3.76 n 1 (mol N 2 ) n 2 (mol CO 2 ) n 3 (mol SO 2 ) n 4 (mol O 2 ) 3.76 n 1 (mol N 2 ) n 5 (mol H 2 O (v)) Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C) q (mol H) (1 g / mol) = q (g H) ? (12 p + q + 16 r) g fuel r (mol O) (16 g / mol) = 16 r (g O) S in feed: n o = S) (mol )X-(107.32 r) 16 q p (12X S g 32.07 S mol 1 fuel) (g )X-(1 S) (g Xfuel g 16r) q p (12 s s s s ++ = ++ (1) Theoretical O 2 : O mol 2 O mol 1O) mol r( H mol 2 O mol 0.5H) (mol q C mol 1 O mol 1C) (mol p 222 ?+ requiredO mol r) /21q 1/4(p 2 ?+= % excess ? n 1 = (1 + P/100) (p +1/4 q ? ½ r) mol O 2 fed (2) C balance: n 2 = p (3) H balance: n 5 = q/2 (4) S balance: n 3 = n 0 (5) O balance: r + 2n 1 = 2n 2 + 2n 3 + 2n 4 + n 5 ? n 4 = ½ (r+2n 1 -2n 2 -2n 3 -n 5 ) (6) Given: p = 0.71, q= 1.1, r = 0.003, X s = 0.02 P = 18% excess air (1) ? n 0 = 0.00616 mol S (2) ? n 1 = 1.16 mol O 2 fed (3) ? n 2 = 0.71 mol CO 2 (5) ? n 3 = 0.00616 mol SO 2 (6) ? n 4 = 0.170 mol O 2 (4) ? n 5 = 0.55 mol H 2 O (3.76*1.16) mol N 2 = 4.36 mol N 2 Total moles of dry product gas = n 2 + n 3 + n 4 + 3.76 n 1 =5.246 mol dry product gas Dry basis composition y CO2 = (0.710 mol CO 2 / 5.246 mol dry gas) * 100% = 13.5% CO 2 y O2 = (0.170 / 5.246) * 100% = 3.2% O 2 y N2 = (4.36 / 5.246) * 100% = 83.1% N 2 y SO2 = (0.00616 / 5.246) * 10 6 = 1174 ppm SO 2 default Microsoft Word - chap04-1-30.doc