Arranging CSTRs

Two stirred tank reactors are available at a chemical plant, one of volume 100 m3 and the other of volume 30 m3. It is suggested that these tanks be used as a two stage CSTR for carrying out an irreversible liquid phase reaction, A + B à Products. The two reactants only will be present in the feed stream in equimolar amounts, CAo = CBo = 1.5 gmol/lit. The volumetric feed rate will be 20 lit/min. The reaction is first order with respect to each of the reactants A and B i.e., second order overall. The rate constant is 0.011 lit/(gmol.min).

    1. Which tank should be used as the first stage for higher overall conversion?
    2. With this arrangement, calculate the overall conversion obtainable under steady state conditions.

Calculations:

-rA = -dCA/dt = k CACB

i.e.,

-rA = k (CAo - CAoXA) (CBo - CBoXB)

The amounts of A and B which have reacted at any time t are equal and given by CAoXA = CBoXB

And for the given problem CAo = CBo

Therefore,

-rA,i = k CAo2(1 - XA,i)2

Taking smaller reactor as the first reactor, volume (V1)= 30 m3

The design equation for CSTRs in series is,

ti/CAo = (XA,i - XA,i-1) / (-rA)i

where CAo = Initial concentration of A into the first reactor = 1.5 gmol/lit

Volumetric flow rate (v) = 20 lit/min

t1 = V1/v = 30 x 1000 /20 = 1500 min

1500/ CAo = (XA,i - XA,i-1)/ k CAo2(1 - XA,i)2

1500 = (XA,1 - 0)/ (0.011 CAo(1 - XA,1)2)

1500 = XA,1/(0.011 x 1.5 x (1 - XA,1)2

24.75 = XA,1/(1 - XA,1)2

24.75(1 - 2XA,1 + XA,12) = XA,1

24.75 - 50.5 XA,1 + 24.75 XA,12 = 0

XA,1 = 0.82

For the second reactor, t2 = V2/v = 100 x 1000 /20 = 5000 min

5000/ CAo = (XA,2 - 0.82)/ k CAo2(1 - XA,2)2

5000 x 0.011 x 1.5 = (XA,2 - 0.82)/ (1 - XA,2)2

82.5 (1 - 2XA,2 + XA,22) = XA,2 - 0.82

83.32 - 166 XA,2 + 82.5 XA,22 = 0

XA,2 = 0.96

If the reactor is arranged with bigger reactor as the first one:

For the first reactor:

t1 = V1/v = 1000 x 1000 /20 = 5000 min

5000/ CAo = (XA,1 - 0)/ k CAo2(1 - XA,1)2

5000 = XA,1/(0.011 x 1.5 x (1 - XA,1)2

82.5 = XA,1/(1 - XA,1)2

82.5(1 - 2XA,1 + XA,12) = XA,1

82.5 - 166 XA,1 + 82.5 XA,12 = 0

XA,1 = 0.90

For the second reactor, t2 = V2/v = 30 x 1000 /20 = 1500 min

1500/ CAo = (XA,2 - 0.90)/ k CAo2(1 - XA,2)2

1500 x 0.011 x 1.5 = (XA,2 - 0.90)/ (1 - XA,2)2

24.75 (1 - 2XA,2 + XA,22) = XA,2 - 0.90

25.65 - 50.5 XA,2 + 24.75 XA,22 = 0

XA,2 = 0.92

From the above calculations, it is seen that the reactor with the smallest volume should be the first one. And overall conversion for this configuration is 96%.