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assignment # 3 Solution

# assignment # 3 Solution - Ql Solution C I 14 00 kg& C...

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Unformatted text preview: Ql Solution: C I 14 00 kg/& C, W“ E2 98299 6 ' ’ w 2039! ® {mm mm 1% *Ww” 8979c a 50:4,; 93“ “1W‘k C. s @g% w x, @gVoS 5Y9 1’68? 5m. Mgg’s 90F :l‘cﬁw—RO f»? :5} TM W‘ \ » Mme-tr etc/2 we? 99% MM mﬁw a" , 727,5 = 0435' a? f Wool-0.? : 9w bay/i Whig :ﬂiI/é : 71115, : 0/1,,” : 955 4%.,55 72,6: mpg/0.; : wow/e95 : /9¢0 Lyn: ﬂ;th : 0:510:95» 9m, razjléd‘ﬁw : (0 32152334: m 2 905.;— 71:31:" my,“ mas—750w}? 1" 343 451/: 7):) s: a}: —-'7L :Zf‘jxeasawW-aooiaézo Ade/A .N-S .é 1,19 ___ may I 74w ;, (Mtge: ’Vﬁ-W'r’: é5¥"§é€9 “ T7 fig/:4. 1/1; : o’ajﬁ/ 0.2 ; age/4),; : 490g) my: orcpg} @941; ; 052/4905) : 15% Lg/A 5%”: 04391453537? .9 37%?) 499/51. 7W3: st "(Awe 1§“%E>~ 4939' : g3)” gag/A 0’45; ’ 719p- 9454,: W2 ~ w = mm @M» W33: 500%»01 : ﬂ; ' : Visa/0.3;: 500/03? {14‘26762 Wugﬂa,éf’wzi 926.9291“? 54> “Jr” : 6350 #1 - .%W,.; ' 1:— 41;”: 428,9 + 21561922347106, log/£1 915,12. 4451+ 745;: 500 + 23%; 1 [5m [Lg/A Q9 1. (E) ‘x LE3 ‘1 CE) 3 @ g 980 {a --» —7 L990 @907 4 1984:) — w “WM @2643 — 628.6 1mg ea‘ﬂrsn 1 H— 54; § I812 500 500 ﬁ (37%! 54312029; 243 o 92 Solution: DOF analysis: Variables Explanation Constraints Explanation Stream 32 5 in stream 1 3 in stream 2 3 in stream 3 l in stream 4 3 in stream 5 2 in stream 6 3 in stream 7 3 in stream 8 2 in stream 9 3 in stream 10 2 in stream 1 l 2 in stream 12 System 0 no chemical reactions; steady—state. so no accumulation Specified flows 4 Streams 1.4. 6 and 11 Specified 8 4 in stream 1 compositions l in stream 6 l in stream 8 l in stream 9 l in stream 12 Specified system 0 performance Material balance 20 5 for l”l separator 3 for 2”d separator 3 for mixer 3 for 3"] separator 3 for 4lh separator 3 for 5‘h separator total 32 32 DOF:32—32:{) Q3 Solution: ’1 ‘j 1“ w a I I H I I a _ 7 Lsc the integral mass balance. msﬂf nijﬂﬂ — J. interested in what happens over a defined time interval: Io Ill‘ (ff — 1H foam“.th because we are to 2 a 2 _ (a) mmr : £ (1 + 2:)m : t +1 0 : 6 kg 2 a 2 (b) nrw = f(3cxp(2r))m = iexpﬁll‘) = 81.9 — 1.5 = 80.4 kg 0 " 0 Q4 Solution: We choose differential mole balance equation. because we are interested in what is happening at any instance of time. Mixer is at steady state. We start with the total mole balance. which is easy because there is no reaction. no“, = n}- = 64 + 97 = 161 kgmolfh alljin 64 kgmol/h 8.3% EA 6.2% E W l EA mixer —|—} E AA 97 kgmol/b 3.7% EA 2.6% AA 5.4% E W All stream compositions given in mol‘7c. with EA : ethyl acetate. E : ethanol. AA : acetic acid. and W : water. DOF analysis: No. of variables No. of constraints Stream variables 1 l Specified flow 2 System variables {) Specified 5 composition Specified 0 performance- Material balances 4 Total 1 l l l DOF:ll—ll:{) Now the component mole balances: rim‘om = 285A),- = (0.083)64 + (0.037)97 = 8.9 kgmol/h alljin am“, = E a“ = (0.062)64 + (0.054)97 = 9.21 kgmol/h alljin iii/mm, : Eamj : (0.026)97 : 2.52 kgmolfh alljin 8W9“, = 2 am}- : (.855)64 + (0.883)97 =140.37 kgmoth alljin From these we calculate the mol% in the output stream: % x 100% : 5.5 mol% ethyl acetate 9.21 l x100% = 5.7 mol% ethanol 2.52 . . 161 x100% : 1.6 mol% acetic aetd 7, 140" 7 x1009? = 87.2 mol% water 95 Solution: leak t R an ’40 wt% nitric acid The tank initial contains 5000 L. or 6280 kg solution (calculated from the solution density). At t = 0. the leak rate is 5 Umin (or 6.28 kgfmin); the leak rate increases linearly such that at I = 10 min, the leak rate is 69.08 kgfmin. We write this as a linear equation: mm = 6.28 + 6.281: where t is in minutes. Since we are interested in what happens over a finite interval of time. we use an integral balance; 30 20 mf # m” = {7513mm = —f(6.28 + 6.281)dt = —[6.28(20) + (l 0 6.28 2 (202)]=—1381.6 kg This is the total loss of material from the tank, since the material is 40 wt% nitric acid, there have. been 552.6 kg nitric acid spilled onto the ﬂoor in 20 minutes. ...
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