HW 3 -- Chapter 4 Q _ A corrected

HW 3 -- Chapter 4 Q _ A corrected - CHEMS 201 MATERIAL...

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Unformatted text preview: CHEMS 201 MATERIAL & ENERGY BALANCES HOMEWORK 3 ASSIGNMENT Q1 Q2 Q3 Q4 4.7 (cont’d) d. H 2O some CH3COOH CH3COOH H 2O C 4 H9OH C4 H9OH CH3COOH CH 3COOH Extractor Distillation Column C4 H 9OH 4.8 a. 120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/egg X-large: 25 broken eggs/min X-large: 35 45 unbroken eggs/min Large: n 1 broken eggs/min n 2 unbroken eggs/min b. 120 25  45  n1  n2 eggs min Ÿ n1  n2 0.30 120 c. 25  n1 50 ½ n1 11 ° Ÿ ¾ ° n2 39 ¿ n1  n2 50 large eggs min d. b11 50g 0.22 22% of the large eggs (right hand) and b25 70g Ÿ 36% of the extra-large eggs (left hand) n1 large eggs broken/50 large eggs 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. m1 lb m strawberries 015 lb m S / lb m . 0.85 lb m W / lb m m2 lb m S sugar 4.9 a. b g m3 lb m W evaporated b g c h 1.00 lb m jam 0.667 lb m S / lb m 0.333 lb m W / lb m b. c. 3 unknowns ( m1 , m2 , m3 ) – 2 balances – 1 feed ratio 0 DF Feed ratio: m1 / m 2 45 / 55 (1) S balance: 0.15 m1  m 2 0.667 (2) Solve simultaneously Ÿ m1 0.49 lb m strawberries, m 2 0.59 lb m sugar 4-4 4.10 a. 300 gal m1 lb m bg m3 lb m 0.750 lb m C 2 H 5OH / lb m 0.250 lb m H 2O / lb m bg 0.600 lb m C 2 H 5OH / lb m 0.400 lb m H 2O / lb m V40 gal 2 m bg m b lb g 1ft 3 0.877 u 62.4 lb m 7.4805 gal ft 3 4 unknowns ( m1 , m2 ,V40 , m3 ) – 2 balances – 2 specific gravities 0 DF 0.400 lb m C 2 H 5OH / lb m 0.600 lb m H 2 O / lb m b. m1 300 gal 2195 lb m (1) (2) Overall balance: m1  m2 m3 C2H5OH balance: 0.750m1  0.400m2 Solve (1) & (2) simultaneously Ÿ m2 V40 4.11 a. 0.600m3 1646 lb m, , m3 207 gal 3841lb m 1646 lb m ft 3 7.4805 gal 0.952 u 62.4lb m 1ft 3 n1 mol / s b g n3 mol / s 0.0403 mol C3H 8 / mol 0.9597 mol air / mol n2 mol air / s b g 3 unknowns ( n1 , n2 , n3 ) – 2 balances 1 DF 0.0205 mol C3H 8 / mol b g 0.9795 mol air / mol 0.21 mol O 2 / mol 0.79 mol N 2 / mol b. Propane feed rate: 0.0403n1 Propane balance: 0.0403n1 Overall balance: 3722  n2 150 Ÿ n1 3722 mol / s 0.0205n3 Ÿ n3 7317 mol / s 7317 Ÿ n2 3600 mol / s b g b g b 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. 4-5 4.12 a. m kg / h 1000 kg / h 0.500 kg CH 3OH / kg 0.500 kg H 2O / kg b g 2 unknowns ( m, x ) – 2 balances 0 DF 0.960 kg CH3OH / kg 0.040 kg H 2O / kg 673 kg / h g 1  x b kg H O / kgg x kg CH 3OH / kg 2 b b. Overall balance: 1000 m  673 Ÿ m 327 kg / h Methanol balance: 0.500 1000 bg 0.960 327  x 673 Ÿ x bgbg 0.276 kg CH 3OH / kg Molar flow rates of methanol and water: 673 kg 0.276 kg CH3OH 1000 g mol CH3OH h kg kg 32.0 g CH3OH 5.80 u 103 mol CH3OH / h 673 kg 0.724 kg H 2O 1000 g mol H 2O 2.71 u 104 mol H 2O / h h kg kg 18 g H 2O Mole fraction of Methanol: 5.80 u 103 0176 mol CH 3OH / mol . 5.80 u 103  2.71 u 104 c. Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the system is not at steady state. Product F eed 2253 kg 4.13 a. Reactor Reactor effluent 2253 kg R = 388 1239 kg Purifier R = 583 W aste m w kg bg R = 140 Analyzer Calibration Data 1 xp x p = 0.000145R 0.1 1.364546 0.01 100 R 1000 4-6 4.13 (cont’d) b. Effluent: x p Product: x p Waste: x p Efficiency b g 0.494 kg P / kg 0.000145b583g 0.861 kg P / kg 0.000145b140g 0123 kg P / kg . 0.861b1239g u 100% 95.8% 0.494b2253g 0.000145 388 1.3645 1.3645 1.3645 c. Mass balance on purifier: 2253 1239  mw Ÿ mw P balance on purifier: Input: 0.494 kg P / kg 2253 kg 1113 kg P 1014 kg gb g Output: b0.861 kg P / kggb1239 kgg  b0123 kg P / kggb1014 kgg . b 1192 kg P 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. n1 lb- mole/ h b g n3 lb- mole/ h . 00100 lb- mole H2O/ lb- mole . 09900 lb- mole DA/ lb- mole b g . 0100 lb- mole H2O/ lb- mole . 0900 lb- mole DA/ lb- mole b v dft / hi 3 2 n2 lb- mole HO/ h 2 g 4 unknowns ( n1 , n2 , n3 , v ) – 2 balances – 1 density – 1 meter reading = 0 DF Assume linear relationship: v Slope: a Intercept: b v2  v1 R2  R1 aR  b 96.9  40.0 1.626 50  15 40.0  1.626 15 v a  aR1 bg b 15.61 v2 n2 1.626 95  15.61 170 ft 3 / h 170 f t 3 62 .4 lb m lb - mol h ft 3 18.0 lb m bg c h g (1) (2) 6480 lb - moles / h 589 lb - moles H 2 O / h DA balance: 0.9900n1 0.900n3 Overall balance: n1  n2 n3 Solve (1) & (2) simultaneously Ÿ n1 b. 5890 lb - moles / h, n3 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. 4-7 R 105 160 245 360 Xp 0.08 0.16 0.25 0.45 ln R 4.65396 5.07517 5.50126 5.8861 ln Xp -2.5257 -1.8326 -1.3863 -0.7985 R ln R lnXp Xp 388 5.96101 -0.7056 0.49381 Effluent 583 6.36819 -0.15 0.8607 Product 140 4.94164 -2.0965 0.12288 Waste From Chart 3 plot of LN R vs LN Xp 0 0 1 2 3 4 5 6 7 -0.5 y = 1.3645x - 8.8394 R2 = 0.9931 -1 ln Xp -1.5 -2 -2.5 -3 ln R ...
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