Assignment 3 Solutions - C1451? QLH A %-— S 014/77ng W...

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Unformatted text preview: C1451? QLH A %-— S 014/77ng W C‘) Co+3H2-C#v~/¥a0:0 g2/2M 3/; 1 3 —1 —/ Ima Yea/+4 = Vm‘% : 4% = - ll W/WTL M20 1' Vida 2 4+2 :- ’12 (MM/mu : Vcw (g r. : [2 414% /4m“( *"% . (V111 % :; : 3742; W/Wq 5’" Mm'L/W : zéxglz): dag/Wig («#1: MIfq~Lleg : XX z: /(M/t [5H2 : ~r '3 r 7 ,m,“ gun—up 7%: €511.71», $10.54;»!le = 59 Mme/4 M5 = Mm ’ (4’94 +972): [00—- 27'5? 1/9kwd‘f/4 m WWW hi W W); WW‘I vimm 2 meg/i“ + ‘(zm : ammo-gawk é'lwfl/A \ Mama! : hay/Mr kw: a 1L 5 : EWC/flv } l %W,M c 4/I/u/l4jacv‘zw : 0,8;X/00f 5: 30 waif/4 flaw: a+5+§+ 90 = we W/& W 9; i , ~ -.— ’7'-.5‘>(:’£:W75‘ 22¢ (4) , may , . h '0) :9 ' I vu’ 3' 5 #au #63 I 3- ffOxOA ( M 0 ' ‘1‘” 5b.; WW : ajjm 15-?- Mw 2/9 a,“ My“ (“3424! I W ’M 35W wk Mt WW’V MM A semibatch system, with a single component of interest: glucose. 0.2 L/h 100 G glucose/L Reactor Reactor consumption = Reactor 25 g/h t=0 0<t<6 t_6 mg,sys=GOg - ‘ DOFanal sis: — No. of variables _ ugwcose m Specified flow System variables 2 (reaction, Specified 1 (initial glucose) accumulation) comositions Specified - rformance - _ _ ‘ DOF=3-3=O. We use an integral material balance, with units in g and h. tf tf tf "‘3;st ’ "’ng9 g f mgind‘ ' fmgputd’ + f Rgd’ to to to 6 6 "18,3st —60 = f20dt-0+ f—zsdt =120—150 0 O mg’sys’f = 30 g glucose To calculate the concentration, we need to do a total mass balance, recalling that 1 mL broth = 1 g. If t f If msysj ' msysfl = f ’h-gindt ‘ f ’hgputd’ + f Rgd‘ t0 t0 t0 6 msys’f-600=f200dt=1200 0 msys’f=1800g=1.8L Final concentration = 30 g glucose/ 1.8 L or 16.67 g/L. , ...
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This note was uploaded on 04/05/2010 for the course CHEBE 241 taught by Professor Ali during the Winter '08 term at The University of British Columbia.

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Assignment 3 Solutions - C1451? QLH A %-— S 014/77ng W...

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