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HW4-Solutions_13107

HW4-Solutions_13107 - 4.7 4.8 3 independent balances(one...

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Unformatted text preview: 4.7 4.8 3 independent balances (one for each species) 400 0.885 H 0 ’5? (g 0.995 H 0 . Water Balance: . g g 2 : =R_ ) g 2 I) mix : 356g/rn1n min g (mm) g = Acetic Acid Balance: [(400)(0.1 131%] = [0.005% + 0.096%" [m] mm : [run :> 152E : 46lg/rnin Overall Balance: firc + 400][i] : [mx + mg ][i] :> me = 417 g/rnin — mm = — min [(0.115)(400)7(0.005)(356)][g] = [(0.096)(461)][g] 3 44 g/rnin =44 g [min min min H20 I CH3COOH someCHSCOOfi CH3COOH H20 I Extractor C4H90H Distillation —> C HQOH CHSCOOH Column 4. —> X-la e: 25 broken e 35 unbroken eggslrnin 0.30 broken egg/egg 0.70 unbroken eggfegg Large: n1 broken eggslmin n2 unbroken eggsfmin 120225+35+nl+n2 (eggs/min):>nl+n2=50 r21 :11 (0.30)(120):25+n1 n2 2 39 111 + n2 : 50 large eggs/min n1 large eggs broken150 large eggs : (1 1/5 0) : 0.22 22% of the large eggs (right hand) and (25/70) :> 36% 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. 4.11 a. 3 unknowns (221,222,223) 121(molls) 7 2 balances —> 0040313101 C3H8 fmol 1 DP 0.9 597 mol air / mo] fl3 (“101 / 5) 0.0205 mol C3H8 / mol n2 (molair / 5) 0.9795 molair / mol —> 0.21m0102 / mol 0.79 molN2 f mol 11- Propane feed rate: 0.04032?l : 150 :> 2'11 : 3722(11101/3) Propane balance: 0.0403521 = 0.0205223 :> n3 2 7317(11101/3) Overall balance: 3722 + 2'22 : 7317 3 222 : 3600(m01i s) c. 3. 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. 4.12 a. . Talks/h) 1000kg/h I 0950 kgCH3OH’ kg 2 unknowns (m, x) 0.500 kg CHSOH/ kg O0401951420 " kg — 2 balances 0.500 kg H20 r kg 0 DP 673kg/h -—> x(kg CH3OH I kg) 1—x(kgH20/kg) I). Overall balance: 1000 2 01+ 673 :> m = 327' kg/ h Methanol balance: 0.500(1000) = 0.960(327) + x(673) : x = 0.276 kg CH3OHK kg Molar flow rates of methanol and water: 673kg 0.276 kgCH3OH 1000g molCH30H h kg kg 32.0 gCH30H 673kg 0.724kgH20 1000 g molHZO h kg i kg iIBgHZO Mole fraction of Methanol: 5.80 x103 5.30 ><103 +2.7'1x104 _ 5.30 x103 molCH30Hf h —2.71><104molH20/h = 0.176 mol CH30H/ mol c. Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the system is not at steady state. 4.20 a. 1511 (mol l h) 132(mol/h] 0.0401310] H20 1' mol x(molH20 (11101) 0.960molDA fmol 17x(molDA/ mol] 14; (mol H20 adsorbed f h) 97% of H20 in feed (3.54 — 3.40) kg molH10 5 h 0.01301;ng 97% adsorbed: 1.56 = 097(00441) :5 a1 = 40.111101/11 Total mole balance: :41 : 142 + 1513 :> rig : 40.1 — 1.556 : 385411101! h Water balance: 0.040(401) = 1.566 + x(38.54) :5 x = 1.2 x10‘5(molH20 lmol) Adsoggtion rate: 143 = = 1.55611101H201’ h b. The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction will reach that of the inlet stream, i.e. 4%. 4-25 a- 1000 LB 1’ min 43 (mol; mm) p”; (m3 ,1 min] 17 y3 (101101 A / kmol] W’ 044(ng min] yl (kmol so2 r 166161) 41 (kg SO2 1kg) l—y1 (161161 A 1 kmol) 17x4 (kg B f kg) 8 unknowns (fil,n3,fil,ri12,m4,x4,y1,y3) — 3 material balances — 2 analyzer readings i 1 meter reading — 1 gas density formula — 1 specific gravity 0 DP I). Orifice meter calibration: A log plot of Vvs. h is a line through the points (F11 = 100, V1 2142) and (112 = 400, V2 2 290). anzblnh+ 111a :> Vzahb 1002 M) 1n(290/142) b 2 111012 #11) Z 111(400/100) Z 0515 lna=lI1V-1*blnhl=ln(142)70.5151n100 2.58 )8 a”8 13.2 >1? 13.28”” Analyzer eah'bration: lny=bR+lna=>y=ae b 111(y2/y1) 1n(0.1107/0.00166) 31R R2 4 R1 0 90 4 20 7 00600 Ina = 1n y1 — 6121 = 111(000166) — 0.0600(20) = —7.60 :5 y = 5.00 X 10‘1 e" ”5"“ ll : a = 5.00 x 10’4 9- h, = 210 rmn :5 V'l =13.2(210)°'515 = 207.3 m3/h (12.2)[(150 +14.7)/14.7](atm) pfm” _ [(75+460]/1.8](K] U 3 _ 0.460 mo]! L = 0.460 krnol / m5 . 207.3 n1 r11 = 0.460 kmol m3 = 95.34 kmol/ min min R1 = 82.4 :> yl = 5.00 x 10*t exp(0.0600 x 82.4) = 0.0702 kmol 802/161161 R3 = 11.6 : y3 = 5.00 ><10_4 exp(0.0600 x116): 0.00100 kmol SOZ/knlol 1000 LB 1.30 kg min LB A balance: (1 — 0.0702)(9534) = (1 —0.00100)n3 :> 713 = 88.7 kmol/min so2 balance: (0.071121953411540 kg 2‘ 1411101): (0.00100](88.7)(64)+m4x4 (1) B balance: 1300 = 1514(1—x4) (2) Solve (1) and (2) simultaneously: m4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg SO2 removed = m4x4 = 422 kg SO2 1’ min (1. Decreasing the bubble size increases the bubble surface—to—volurne ratio, which results in a higher rate of transfer of 502 from the gas to the h'quid phase. ...
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