Question

Mainly a Chem Eng Q. Trying to study for my exams but the solution is not up on the slides: A liquid phase reaction: A+B→C the reaction is first order with respect to component A with k = 0.311 min-1 . The total entering volumetric flow rate of the reactants (A and B) is 2 dm3 s -1 . (a) If 80% is to be achieved, determine the necessary CSTR volume. (b)If two-800 dm3 reactors were arranged in series, what would be the corresponding conversion?

Answer 1

rate of reaction -r_A =k*C_A =k*C_Ao*(1-x)

=>volume of CSTR V=F_Ao*X/(-r_A)

where FAo =C_Ao*vo =molar flow rate & vo =total volumetric flow rate

X=fractionall conversion

=>V=C_Ao*vo*X / (k*C_Ao*(1-X)) = vo*X/(k*(1-X)) {CAo cancels out}

=>V =2*10^-3 m^3*0.8/(0.311/60*0.2) ( k=0.311 /60 since 0.311 min^-1 has to be converted to s^-1)

=>V =1543.41dm^3

b) conversion for 1st reactor X₁/(1-X₁)=V*k/(vo) =800*10^-3*0.311/(60*2*10^-3) =2.0733

=>X₁ =2.0733-2.0733*X₁ =X₁ =2.0733/(1+2.0733) =0.6746

V2 =vo*(X2-X1)/(k*(1-X2)) =>(X2-X1)/(1-X2)= =800*10^-3*0.311/(60*2*10^-3) =2.0733

=>(X2-0.6746)/(1-X2)=2.0733

=>X2-0.6746 =2.0733 - 2.0733 *X2

=>3.0733*X2 =2.7479 =>X₂=2.7479/3.0733 =0.8941