Two stirred tank reactors are available at a chemical plant, one of volume 100 m^{3} and the other of volume 30 m^{3}. It is suggested that these tanks be used as a two stage CSTR for carrying out an irreversible liquid phase reaction, A + B →
Products. The two reactants only will be present in the feed stream in equimolar amounts, C_{Ao} = C_{Bo} = 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).

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

Calculations:

-r_{A }= -dC_{A}/dt = k C_{A}C_{B}

i.e.,

-r_{A} = k (C_{Ao} - C_{Ao}X_{A}) (C_{Bo} - C_{Bo}X_{B})

The amounts of A and B which have reacted at any time t are equal and given by C_{Ao}X_{A} = C_{Bo}X_{B}

And for the given problem C_{Ao} = C_{Bo}

Therefore,

-r_{A,i} = k C_{Ao}^{2}(1 - X_{A,i})^{2}

Taking smaller reactor as the first reactor, volume (V_{1})= 30 m^{3}

The design equation for CSTRs in series is,

τ_{i}/C_{Ao} = (X_{A,i} - X_{A,i-1}) / (-r_{A})_{i}

where C_{Ao} = Initial concentration of A into the first reactor = 1.5 gmol/lit

Volumetric flow rate (v) = 20 lit/min

τ_{1} = V_{1}/v = 30 x 1000 /20 = 1500 min

1500/ C_{Ao} = (X_{A,i} - X_{A,i-1})/ k C_{Ao}^{2}(1 - X_{A,i})^{2}

1500 = (X_{A,1} - 0)/ (0.011 C_{Ao}(1 - X_{A,1})^{2})

1500 = X_{A,1}/(0.011 x 1.5 x (1 - X_{A,1})^{2
})

24.75 = X_{A,1}/(1 - X_{A,1})^{2}

24.75(1 - 2X_{A,1} + X_{A,1}^{2}) = X_{A,1}

24.75 - 50.5 X_{A,1} + 24.75 X_{A,1}^{2} = 0

X_{A,1} = 0.82

For the second reactor, τ_{2} = V_{2}/v = 100 x 1000 /20 = 5000 min

5000/ C_{Ao} = (X_{A,2} - 0.82)/ k C_{Ao}^{2}(1 - X_{A,2})^{2}

5000 x 0.011 x 1.5 = (X_{A,2} - 0.82)/ (1 - X_{A,2})^{2}

82.5 (1 - 2X_{A,2} + X_{A,2}^{2}) = X_{A,2} - 0.82

83.32 - 166 X_{A,2} + 82.5 X_{A,2}^{2} = 0

X_{A,2} = 0.96

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

For the first reactor:

τ_{1} = V_{1}/v = 1000 x 1000 /20 = 5000 min

5000/ C_{Ao} = (X_{A,1} - 0)/ k C_{Ao}^{2}(1 - X_{A,1})^{2}

5000 = X_{A,1}/(0.011 x 1.5 x (1 - X_{A,1})^{2}

82.5 = X_{A,1}/(1 - X_{A,1})^{2}

82.5(1 - 2X_{A,1} + X_{A,1}^{2}) = X_{A,1}

82.5 - 166 X_{A,1} + 82.5 X_{A,1}^{2} = 0

X_{A,1} = 0.90

For the second reactor, τ_{2} = V_{2}/v = 30 x 1000 /20 = 1500 min

1500/ C_{Ao} = (X_{A,2} - 0.90)/ k C_{Ao}^{2}(1 - X_{A,2})^{2}

1500 x 0.011 x 1.5 = (X_{A,2} - 0.90)/ (1 - X_{A,2})^{2}

24.75 (1 - 2X_{A,2} + X_{A,2}^{2}) = X_{A,2} - 0.90

25.65 - 50.5 X_{A,2} + 24.75 X_{A,2}^{2} = 0

X_{A,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%.**

Last Modified on: 30-Apr-2024

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