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hw2solution - ® ® VF—Im-ymum\Mo 111(0.0/fi-val 13 5...

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Unformatted text preview: ® ® VF—Im-ymum \Mo); 111(0):.0 {/fi-val «(+13 \ 5 “(3)=~1Y.(.\\—Y‘(s)+c_"5l1(a§ g; v.5): “QT .‘I 3 WM- “(swam— ghqo): ‘33 Yd”? L4 “(\— 4H4 .1) (31'st --\3 (k: :71 (“\BC (4:? 5+1 L- 3&1 e S +3‘H ~3+‘ 1) L_..5+l :15 5+1 ML Q1 '77 (1 943+] H: M 41:30? Elk): b£>J=O=$1+5+\-=? 5=£~~———-:L-—""= ET..— Quoa; “(33:0 :—S+| 7-3-7 32—” .6) L. 603 K ~11;— ——— »- ..,.. ._.._....— _ - . -Mu . rm...”- - "CJH UNA] “'3 iii} HG) '= “[4“?- 1+ xi 3cm 4‘ng 01(3) @3351 _ q L «I» L “33—6qu0) m= H} + 5+1 4 $44.31 q ‘1 ch 7555‘" 5 :5 (—3.9—0‘ 9 w \1 ”£3 {¢*55’5}*1 14:3 q {I Jr-QUM‘ [+3 1+1. rim-P a . fl"..- __:1‘{_ 9434‘” ‘1 4' 3 J“ a 1“ bOM-l- ‘37-?371‘1 —‘q q q TU) “SE 1177+ Co-f‘L)‘ YU’) ‘ie’t‘l' el++§+£1+ ® “a yC+J=O 5m“ \{+s‘1J'—:ro Icy L HornU‘z- Al Jam LA (4-) l' YUJ— [. V = C) 1/ r Y 15" _. "1' ( flL '1’?“ 5*?!) 34° (.95) 5+ Problem 3: Part 1 Fe p, V const Fi, 01, 02 const I.C.: F40) = 0 Xi(0) = Xo(0) = 0 Mass balance d(pV)/dt = F1 — F0 Fi = F0 = F 3of6 Catal st balance d(pVx0)/dt = Fixi — Foxo = F(xi — x0) pV(dx0’/dt) = F(xi’ — xo’) X0'=Xo(t)-Xo Xi'=Xi(t)';i PVSXe’(S) = F(X1’(S) — Xo’(S)) Catalyst delay FXi(t) = Fc(t — 9i) mm) = Fga — 6i) Where Fc ' = FC (1‘) —Fc —0-s 1 —0-s ' in’(s) = e ’ Fc’(s) where xi’(s) = Fe ’ Fc (s) Reactor delay ya) : Xo(t _ 60) _ )"(0 = Xo’(t - 90) Where 3" = W) - y Y’(s) = e‘gvs X0’(s) where Xo’(s) = eBOSY’(s) stegvs Y’(s) = -FY’(s) + e_(9‘+00)317c'(s) —6k (st/F + 1)Y’(s) =e?Fc'(s) where e = ei + 90 where K = UP and I = pV/F 4of6 Problem 3: Part 2 Fc(t) = U10) + U20) + U30) FcM -tS FcM -t S Fc* -t S Fc(s) = U1(s) + U2(s) + U3(s) = e ‘ — e 2 — e 2‘ s s s KF M K F M _ F M Y,(S) = G(S)U(S) = c (”“1” — M e-(9+t2)s s(zs +1) s(zs +1) —(t—0—tl) —(t—t9—t2) y’<t)=KFcM[1—e T ]S(r—6—tl)—K(FCM—F,..*)[1—e , 150—641) Problem 3: Part 3 y(t*) = y* = Fc*/F where t* = t2 + 6 y’(t*> = y(t*) —§> = y(t*) = PM: = 0 =0 =1 —(t2—tl) —0 y’<t*)=KF,..M[1—e r ]S(t2§'I)-F(FCM-Fc*)[1—§;]S(0) y’(t*)= KFM[1—e 7 ] where At=t2—t1 At = t2 —t1 = 4 1n (1 —Fc*/FcM t1=t2+Tln(1—Fc*chM) I—l this term will be negative if F:chM is >0 and <1. 50f6 ...
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