Question

A liquid reaction A+B->C, r=kC_B*C_A^2 takes place in a CSTR of volume V_R in the presence of a large excess of reactants B. Assume the reactor achieves 50% conversion of A at steady state.
Material Balance for only one reactor
hint: solve for VR1 for XA=0.5
(a) what is the steady state conversion if the original reactor is replaced by two CSTR'S of volume V_R/2 in series?
hint: 0= Q1CA1-Q2CA2-kCA2^2VR2
hint: CA1/Caf=0.618
Xa2=0.568
(b) what is the conversion if the original all reactor is replaced by ghrrr CSTR'S of volume V_R/3?
hint: VR3=(1/3)VR1
XA3=0.597

Answer 1

for a CFSTR , under steady state, the mass balance equation can be written as

Rate of mass in = Rate of mass out+Rate of chemical reaction

FAO= FA+KC_BCA²V, FAo= initial molar flow rate of A, FA= Molar flow rate of A at any residence time, K= rate constant, V= Volume of reactor

FA=FAO*(1-XA), XA= conversion

FAO= FAO*(1-XA)= KC_BC_A²V

FAOXA= KC_BCA²V

since large excess B is used compared to A, KCB= K', K'= Rate constant

FAOXA= K'CA²V

but T= Residence time= V/VO, VO= Volumetric flow rate

T= VCAO/FAO, CAO= initial concentration of A

hence T= CAOXA/K'CA²

CA= CAO*(1-XA)

K'CAOT= XA/ (1-XA)²

given XA=0.5

K'CAOT= 0.5/0.25= 2

when the reactor is replaced by two reactors each having 50% of the original volume (V/2). New space time

T' has to be defined, XA1= new conversion from 1st reactor

K'CAOT'= XA₁/(1-XA₁)²

T'=T/2

KCAOT/2= XA₁/(1-XA1)²

XA1/(1-XA1)² =2/2=1, when solved using excel, XA1=0.382

at the inlet to second reactor, CA=CAO*(1-XA1)= CAO*(1-0.382)=0.618CAO

for the second reactor, XA2= conversion from 2nd reactor

KCAOT' becomes K*0.682CAOT/2= XA₂²/(1-XA₂)²= 0.682, XA1=0.3712

concentration at the end of second reactor, CA2= 0.682CAO*(1-0.3712)= 0.43CAO

The overall conversion is 1-CA2/CAO= 1-0.43=0.57

2, when the reactor is replaced by reactors volume of each is V/3, T'= T/3

hence K'CAOT'= XA1/(1-XA1)² ,XA1= conversion at the end of 1st reactor

2/3 = XA1/(1-XA1)², when solved using excel, XA1=0.313, CA1= CAO*(1-0.313)=0.687CAO

for the second reactor, XA2= conversion at the end of second reactor

K'CAOT' = K0.687CAOT/3= XA₂/(1-XA2)²= 0.458, when solved, XA2=0.255, CA=0.687CAO*(1-0.255)= 0.511CAO

for the 3rd reactor, K'CAOT' = K'0.511CAOT/3 = XA₃(1-XA₃)²= 0.34, when solved ,XA3=0.212

CA3= 0.511CAO*(1-0.212)= 0.402CAO

Overall conversion = 1-CA3/CAO= 1-0.402=0.598