Stirred Tank Reactors of Unequal Volumes in Series

Acetic acid is hydrolysed in three stirred tank reactors operated in series. The feed flows to the first reactor (V = 1 lit) at a rate of 400 cm³/min. The second and third reactors have volumes of 2 and 1.5 litres respectively. The first order irreversible rate constant is 0.158 min^-1. Calculate the fraction hydrolysed in the effluent from the third reactor.

Calculations:

The design equation for series, steady flow mixed reactor is

V_i =F_Ao (X_A,i - X_A,i-1) / (-r_A)_i

Where V_i = volume of reactor i

F_Ao = molal flow rate of A into the first reactor

X_A,i = fractional conversion of A in the reactor i

X_A,i-1 = fractional conversion of A in the reactor i-1

For first order reaction, -r_A,i = kC_A,i = kC_Ao(1 - X_A,i)

v = volumetric flow rate of A = 400 cm³/min = 0.4 lit/min

(-r_A)₁ = (kC_A)₁ = k C_A,1 = k C_Ao ( 1- X_A,1)

C_Ao= F_Ao / v

i.e., F_Ao = v C_Ao

X_A,i-1 = X_A,0 = 0

Therefore,

V_i =F_Ao (X_A,i - X_A,i-1) / (-r_A)_i

1 = 0.4 (X_A,1 - 0) / (0.158 x ( 1 - X_A,1 ) )

X_A,1 = 0.283

(-r_A)₂ = (kC_A)₂ = k C_A,2 = k C_Ao ( 1- X_A,2)

Therefore,

(-r_A)₂ = k C_Ao ( 1- X_A,2)

X_A,1 = 0.283

F_Ao = v C_Ao

V_i =F_Ao (X_A,i - X_A,i-1) / (-r_A)_i

2 = 0.4 (X_A,2 - 0.283) / ( k ( 1- X_A,2) )

5 k = (X_A,2 - 0.283) / ( 1- X_A,2)

0.79 - 0.79 X_A,2 = X_A,2 - 0.283

1.073 = 1.79 X_A,2

X_A,2 = 0.60

(-r_A)₃= (kC_A)₃ = k C_A,3 = k C_Ao ( 1- X_A,3)

X_A,2 = 0.6

F_Ao = v C_Ao

V_i =F_Ao (X_A,i - X_A,i-1) / (-r_A)_i

1.5 = 0.4 (X_A,3 - 0.60) / ( k ( 1- X_A,3) )

0.5925 = (X_A,3 - 0.60) / ( 1- X_A,3)

0.5925 - 0.5925 X_A,3 = X_A,3 - 0.60

1.1925 = 1.5925 X_A,3

X_A,3 = 0.749