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 =1543.41**d**m^3

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

=>X_{1} =2.0733-2.0733*X_{1}
=**X _{1}** =2.0733/(1+2.0733)
=

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
=>**X _{2}**=2.7479/3.0733
=

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