If a Reaction: e.g., ν g G->ν f F Then the individual balances, which include the reactor with "f" fraction overall conversion, become: G in - G out - f G in (ν g /ν g ) = 0 F in - F out - f (ν f /ν g ) G in = 0 e.g., ν g G +ν h H ->ν f F + ν d D with G limiting, f conversion. G in - G out - f G in (ν g /ν g ) = 0 F in - F out - f (ν f /ν g ) G in = 0 last term >0 H in - H out - f (ν h /ν g ) G in = 0 D in - D out - f (ν d /ν g ) G in = 0 last term >0 Recycle Ratio, RR = Recycle/Feed R= Feed*RR Overall: Total: Mass in = Mass out Individual species N Balances Mixing Point 1: Total: Mass Feed + Mass Recycle = Mass into Reactor Individual Species N Balances
Reactor with a limiting reactant (l) and a conversion f For each of the species Mi(Xi)) in - Mi(Xi)) out - f ( νi/ νl ) Ml(Xl) = 0 Separator Mass in = Mass out Individual species 3 species Mi(Xi)) in = Mi(Xi)) out Simplification: if the concentrations out (products and recycle) are equal. Otherwise, must solve. Lets take two examples: 1. Glucose (G) reacts to form Fructose (F) in a reactor with recycle. 40%w Glucose in Water (W) is fed into the system and blended with the recycle to result in a feed containing 4%w F into the reactor. The ratio of recycle to feed is 1/8.33 and both streams out of the separator (recycle and product) have the same composition. What is the composition of the product stream and what is the fractional conversion in the reactor? 2. Butane (C4H10-B) is being dehydrogenated to form butadiene (C4H6-D) in a reactor with recycle of the unreacted butane and butadiene. The hydrogen (H) product is not recycled. The overall conversion is 95% of butane to butadiene. The separator results in two streams: the recycle, R, containing only butane and butadiene (5% of that in the product stream, P) and the product stream which contains 0.5% of the butane leaving the reactor. What are the product composition, recycle ratio and single pass conversion of butane in the reactor?
Example 1: reaction G->F Step 1: Draw and label the diagram for all the unknown feeds and concentrations. Unknows : f, R,S,P,T, Wp,Wr,Wt,Ws, Gp,Gr,Gt,Gs, Fp,Fr,Ft, Fs, Concentrations of p, s, r are the same for each species: leaving f, R,S,P,T, Wt,Wrps, Gt,Grps, Ft,Frps, minus 2 (W+G+F=1 for t,rps); and S=T Next: pick a basis: 100kg feed weight (mass) basis, because mass ratio given 3. Overall Balances: Total : P = 100kg R= 100/8.33 = 12 Total: 100 + 12 = T = 112 Water: 60kg in -> 60kg out Wp = 0.6 Note for Wr,t,p always = 0.6 Mixing Point 1: G: 100(0.4) + 12(Gr) = 112 (Gt) F: 0+ 12 (Fr) = 112(0.04) Fr = 0.373 thus, since Wr = 0.6, Gr = 0.027 Reactor: T (Gt) - (S) (Gr=p=s) -f T (Gt) = 0 Reactor + Separator: T (Gt) - (R+P) (Gr=p) -f T (Gt) = 0 T (Ft) - (R+P) (Fr=p) + f T (Gt) = 0 112 (0.36) - (100+12) (0.027) -f 112 (0.36) = 0 f = 0.93
Example 2: reaction B->D + 2 H Step 1: Draw and label the diagram for all the unknown feeds and concentrations. Unknows : f, RR, R,S,P,T, Bp,Br,Bt,Bs, Dp,Dr,Dt,Ds, Hp,Hr,Ht, Hs, Hs = 0 minus 4 (B+D+H=1 for each stream); and S=T Choose a basis: 100 moles of feed molar basis (two phases- reasonable MW) pure feed Degrees of freedom (where to start) Overall:4 unknowns: P, Bp, Dp, Hp with Bp + Dp +Hp = 1 [-1] Independent Equations: each species plus 95% overall conversion specified. 0 degrees of freedom, thus all of these can be specified Mixing Point 1: 6 unknowns: R,T, Br, Bt, Dr, Dt with Br+Dr=1; Bt+Dt=1, R+Feed(assumed)=T, Balances on D and B 1 degrees of freedom, solve later Reactor: 7(6) unknown: T, S (note molar vs mass), Bt, Dt, Bs, Ts, Hs with Bs+Ds+Hs=1; Bt+Dt=1 plus Balances on D + H and B several degrees of freedom Separator: unknown S, R, P, Bs, Bp, Br, Ds, Dp, Dr, Hs, Hp, Hr, with Bs + Ds=1, Bp+Dp+Hp=1, Br+Tr=1, Hr=0, R+P=S, plus balances on D + H and B [3], specified: Dr = 0.5Dp and Bp=0.05Ds. [2] and the variables solved from the overall balance. This can be solved after the overall balance! Overall:4 unknowns: P, Bp, Dp, Hp with Bp + Dp +Hp = 1 [-1] Independent Equations: each species plus 95% overall conversion specified. Choose basis: 100 moles(/hr) of B into process P = 100 moles(/hr) + 2(100)*f= PBp + PDp + PHp = 290 m(hr) Given PBp = 5 and PDp = 95 Balance on C
100*4= PBp*4+ PDp*4 Balance on H 100*10= PBp*10+ PDp*6+ PHp*2 (1000-50-570)/2=190= PHp Balance on H2 PHp = 2f(100)= 190 Balance on B 100-PBp - f(100)= 0 PBp = 5 Balance on D -PDp + f(100)= 0 PDp = 95 65.5% H2 32.8%D 1.7% B Given: RDr = 0.05 PDp = 4.75 SBs*0.05 = PBp ; SBs=100 at separations on B SBs=PBp+RBr RBr=95 RDr= 4.95 R=99.95 Recycle ratio =.9995 Mixing Point 1: 6 unknowns: R,T, Br, Bt, Dr, Dt with Br+Dr=1; Bt+Dt=1, R+Feed(assumed)=T, Balances on D and B Overall T=R+100 = 199.95 Dt+Bt = 1= Dr+Br Balance on D RDr= TDt = 199.95*Dt= 4.95 Dt= 0.025 -> Bt =.975
Reactor: 7(6) unknown: T, S (note molar vs mass), Bt, Dt, Bs, Ts, Hs with Bs+Ds+Hs=1; Bt+Dt=1 plus Balances on D + H and B Balance on B: TBt-(PBp+RBr)-f TBt= 0 f= 1-(PBp+RBr)/TBt = 1-(5+95)/(199.95*.95)= 0.472 Separator: unknown S, R, P, Bs, Bp, Br, Ds, Dp, Dr, Hs, Hp, Hr, with Bs + Ds=1, Bp+Dp+Hp=1, Br+Tr=1, Hr=0, R+P=S, plus balances on D + H and B [3], specified: Dr = 0.5Dp and Bp=0.05Ds. [2] and the variables solved from the overall balance. This can be solved after the overall balance!
Mixing Point 1: reaction G->F Pw = 100= Fw R= 100/RR = Rw Total: 100 + Rw = Tw = Note for Wr,t,p always = 0.6 Mixing Point 1 G: 100(0.3) + Rw(Gr) = Tw (Gt) = (Rw+100) (Gin) F: 0+ Rw(Fr) = Tw(0.05) Fr = thus, since Wr= 0.6, Gr = Reactor: T (Gin) - (T) (Gr=p=s) -0.9 T (Gt) = 0 Reactor + Separator: T (Gin) - (R+P) (Gr=p) -0.9 T (Gin) = 0 Gr=p = 0.1Gin 100(0.3) + Rw(Gr) = Tw (Gin) = (Rw+100) (Gin) 30 + Rw(0.1Gin) = (Rw+100) (Gin) 30=0.9*Rw*Gin + 100*Gin for F: T (Fin) - (R+P) (Fr=p) +0.9 T (Gin) = 0 0.9 (Gin) = (Fr) - (Fin) Rw(Fr) = Tw(0.05) Rw(Fr) = 5+Rw(0.05) 5= Rw(Fr-0.05)