Concentration response in continuous stirred tank

Concentration response in continuous stirred tank
Example: Concentration response in continuous stirred tank
A stirred tank without chemical reaction is shown in Figure below having a steady state concentration of component A of 0.9 Kmol /m3. perfect mixing in the tank is assumed. Suddenly, the concentration of the component A is changed to 1 Kmol /m3 (step change). The outlet concentration will also reach 1 Kmol /m3. We assume constant liquid density ? and constant volume V of 5m 3 and constant inlet and outlet flow rates of the liquid is 1 m3/h. Determine:
a- The time constant, and what outlet concentration of component A at t=??
b- The outlet concentration of component A after time period of 3 hour?

Solution: With constant density and constant volume, the mass balance gives that the volumetric inlet and outlet flow rates are equal, qin = qout = q. We further assume perfect mixing in the tank such that cA,out = cA. The component balance for A in the tank is then:
(1)
With constant volume V this gives
(2)
With Y = cA and X = cA,in we see that this is in standard form (1st ODE) with

Here, the time constant in this case is V /q [s]. ,
In other words, it will take ? = 5h
the outlet concentration of component A at t= ? can be calculated by:
?Y(t)= 0.63 x A = 0.63 x 0.1 = 0.063oC
?Y(t) = y(t) – y(0) = 0.063oC
Y(t) = 0.063+ 0.9 = 0.963oC
b) Equation (2) becomes:

(3)

Taken Laplace for the equation

(the step response) is given by:
(4)
When solve (ao) and (a1) and from inverse Laplace, the solution of eq. (4) is given by:
( 5)
At s.s. the xo = yo = 0.9 Kmol /m3
A= 1 – 0.9 = 0.1 Kmol /m3
Y(t) = A(1- e-t/?)
Y(t) = 0.1(1- e-3/5) = 0.1*0.45
?Y(t)=y(t)-y(0)=0.045
Y(t)=0.9+0.045= 0.945 Kmol /m3

Example: Mixing tank with chemical reaction



2A B + C
Reaction Rate of component A = rA = kCA
CA: concentration of component A (Kmole/m3)
K: constant rate of reaction (1/h)
V: volume of tank (m3) constant
F: volume flow rate (m3 /h) constant
Material balance
In – out – rate of reaction = accumulation
,

the time constant for the system is:
Let R (the residence time) is:
The rearranged model of the 1st ODE system is then:
This can be shifted to Laplace transformed:

The step response is given by:
The solution can be found using the transfer function model:

Take Laplace inverse:

Impulse Input


Ramp Input