hw6_soln

# hw6_soln - 11.11 Notation Subscripts Benzene = bz Xyl.ene =...

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11.11 Notation: Subscripts Benzene = bz, Xyl.ene = xy, Solids = s W = mass !Taction, F' = liquid stream, S' = solids stream s" - ro bz- 1.0 ros:y =0 ros:~ 1.0 SO s' - s' ro bz-(1-roxy) s' ro xy ros' ~ 0 '- 1.0 s' - s' ro bz-(l-roxy) s' ro xy ros' =0 '~ '" ~ ~ S' ~ 2 I. S2 . F2 F' - F' ro bz- (l.ro xy) F' ro 'y "l: = 0.9 ~ ~ F' F'- F' ro bz- (l-ro ,y) F ' ro xy roF~=0,9 ~ roF;"'=O ro<y=O,l roF;= 0.9 1.0 I~: Basis is I hr (Fo= 2000 kg and SO= I.000 kg) Steps 6 and 7: Unknowns: . F' s' F' , , ' } Umt I: co." COxy' co." S, F ,F- Net=8 Unit 2: CO~y, oo;~,co~:'S', F', S2 Material Balances: Unit I: bz, xy, s Unit 2: bz, xy, s dditional Equations: 00;; = oo~~and oo:~ = ro~; (the related equations for benzene al redundant). Total equations =8. The solution is simplified if two balances are made first (not an essential ste All material balances are in kg. Solids balance on Unit 2 and also Unit I.: 2000 (0,9) = F' (0.9) 2000(0.9)= F2(0.9) F' =2000 F2=2000 Total balance on Unit I.and also Unit z: 2000+ S' = 2000+ S2 1000 + 2000 = 2000 + S I SI=S2 SI = 1000 S2= 1000 The other balances are Unit2: Benzene: Xylene: I000(1- oo~;) = 2000( 1- oo;~)+ I.000( 1- co~;) 2000(OJ) + IOOOoo~; = 2000oo;~ + IOOOoo~; lOoo~ = ro~; Unit I: Benzene: Xylene: 1.0(1000)+ 20oo(l-ooF') = 2000(1-00;;)+ IOOO(I-oo~;) 0 + 2000oo~y = 2oo0oo~ + I000( oo~;) lOco;; = ro;y Solve the equations to get the compositions ofthe streams: (continu

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b) Xylene in Feed = 2000 x 0.1= 200kg Xylene in Product = 1000x 0.18= 180kg 1%Recoveryl=~xloo=190%1 200 .. 11.14 Sten.U..J., 3 and 4: mol% 64.29H, 14.29SiCI, 21.42H,SiCl, C tD' 100% .-j Separator IE H"SiCI, ,HSiCI3 H,SiCl, ..HSiCI3 100%Si B SteD5: Basis: 100kgB 100kg Si l lkg mol Si = 3.560 kgmol Si - 28.09kgSi SteDs6 and 7: Unknowns: A,D,E Balances: Stem;JLa.nd9: Balances to be solved: System: overall process Si overall mole balance kgmolSi = 3.560 Dk I rO.l429kgmOlSiCl' l lkgmOISi 0.2142kgmOIH,SiCI' l
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