Arranging Stirred Tank Reactors in Series

Arranging CSTRs

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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 → 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_AC_B

i.e.,

-r_A = k (C_Ao - C_AoX_A) (C_Bo - C_BoX_B)

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

And for the given problem C_Ao = C_Bo

Therefore,

-r_A,i = k C_Ao²(1 - X_A,i)²

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

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

τ₁ = V₁/v = 30 x 1000 /20 = 1500 min

1500/ C_Ao = (X_A,i - X_A,i-1)/ k C_Ao²(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,1²) = X_A,1

24.75 - 50.5 X_A,1 + 24.75 X_A,1² = 0

X_A,1 = 0.82

For the second reactor, τ₂ = V₂/v = 100 x 1000 /20 = 5000 min

5000/ C_Ao = (X_A,2 - 0.82)/ k C_Ao²(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,2²) = X_A,2 - 0.82

83.32 - 166 X_A,2 + 82.5 X_A,2² = 0

X_A,2 = 0.96
If the reactor is arranged with bigger reactor as the first one:

For the first reactor:

τ₁ = V₁/v = 1000 x 1000 /20 = 5000 min

5000/ C_Ao = (X_A,1 - 0)/ k C_Ao²(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,1²) = X_A,1

82.5 - 166 X_A,1 + 82.5 X_A,1² = 0

X_A,1 = 0.90

For the second reactor, τ₂ = V₂/v = 30 x 1000 /20 = 1500 min

1500/ C_Ao = (X_A,2 - 0.90)/ k C_Ao²(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,2²) = X_A,2 - 0.90

25.65 - 50.5 X_A,2 + 24.75 X_A,2² = 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%.

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Last Modified on: 30-Apr-2024

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