Influence of form of –rA on reactor volume. The reaction A -> B is to be carried

Question

Influence of form of –rA on reactor volume. The reaction

A -> B is to be carried out isothermally in a continuous-flow reactor. Calculate both the CSTR and PFR reactor volumes necessary to consume 99% of A (i.e., CA = 0.01CA0) when the enteringmolar flowrate is 5 mol/h and volumetric flowrate is 10 dm3 h-1. Assume the reaction rate –rA has the following form:

a. –rA = k with k = 0.05 mol dm-3 h-1

b. –rA = k CA with k = 1×10-4 s-1

c. –rA= kCA2 with k = 1 dm3 mol-1 h-1

d. Repeat (a), (b), and (c) to calculate the time necessary to consume 99.9% of species A in a 1000 dm3 (1 dm3 = 1 L) constant volume batch reactor with CA0 = 0.5 mol dm-3. Explain the trends observed.

Explanation / Answer

For a CSTR, T= CAO*XA/(-rA)

T= space time = VCAO/FAO

V= Volume of reactor, CAO= initial concentration of A, XA= conversion and –rA= rate of reaction. FAO= molar flow rate of A

For case (a), T= VCAO/FAO= CAO*XA/ K , V/FAO= XA/K

V = XA*FAO/K =0.99*5/0.05 = 99dm3

For case (b), V/VO= CAOXA/KCAO*(1-XA) ( for 1st order reaction -rA= KCA= KCAO*(1-XA)

V/VO= XA/(K*(1-XA) , VO= 10dm3/hr= 10dm3/3600se

V*3600/10 = (0.99/1-0.99)/10-4, V= 2750 dm3

for case (c), T= V/VO= CAO*XA/KCAO2*(1-XA)2

(V/VO)*KCAO = XA/(1-XA)2 = 9900

Vo= FAO/CAO or CAO= FAO/Vo = 5/10 =0.5 mol/dm3

Hence (V/10)*1*0.5 = 9900, V= 9900*10/0.5= 19800dm3

For the case of PFR

T= CAO*

T= V/VO= VCAO/FAO

For zero order reaction the integral becomes XA/K

Hence for case (a), T= VCAO/FAO = CAOXA/K

Or V/FAO= XA/K

Or V/5= 0.99/0.05, V= 0.99*5/0.05 =99 dm3

Hence for zero order reactions, the volume is independent of type of reactor.

For 1st oder reaction, the PFR equation becomes

KT= -ln(1-XA)

Or T= V/VO , VO= 10/3600 dm3/sec

K= 10-4/sec

V*10-4*3600/10 = -ln(1-0.99), V = 0.165 dm3

For a 2nd order reaction in a PFR , KCAOT= XA/(1-XA) in a PFR

Hence 1*0.5*V/10 = 0.99/(1-0.99) =99

V= 99*10/0.5 = 1980 dm3

For a batch reactor, t= batch time = T for PFR

Hence t for zero order reaction t= (CAO-CA)/ K, but CA = CAO*(1-XA)

Hence batch time , t = CAOXA/K = 0.5*0.99/0.05= 99dm3

For a 1st order reaction , t= -ln(1-XA)/K =-ln(1-0.99)/ 10-4 =46052 sec= 12.79 hr

For a 2nd order reaction, KCAo*t= XA/(1-XA)= 99,

Hence batch time, t= 99/(KCAO) =99/1*0.5 = 198 hrs

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