Question

26. A 2m3 CSTR processes an aqueous feed containing reactant A. This is a second order reaction occurring at a constant temperature of 80°C. Calculate the conversion of reactant A in a CSTR. Round your answer to three decimal places. Data: 2A C Qo = 100 L min-1 40 = 100 mol L-1 ra =-0.02 CA2 [mol L-1 min 1]

Answer 1

The performance equation of CSTR is

where V is the volume of the reactor.

q_o is the volumetric flow rate of reactant.

$X_{A_{exit}$ is the conversion at the exit of the reactor. or simply the total conversion.

$-r_A_{exit}$ is the reaction rate at the exit or the overall reaction rate.

we have

V=2 m³.

qo=100 Lmin-1

$-r_A_{exit}=.02*C_A^2$

we have $C_A=C_{A_o}(1-X_A)$

Thus $-r_A_{exit}=.02*C_A^2=.02*(C_{A_o}(1-X_A))^2$

Substitute this in Performance equation.

$\frac{2 m^3}{100 L min^{-1}}=\frac{X_A_{exit}}{.02*(C_A_o(1-X_A_{exit}))^2}$

$\frac{2 *10^3 L}{100 L min^{-1}}=\frac{X_A_{exit}}{.02*(100*(1-X_A_{exit}))^2}$

Now simplifying and solving for $X_{A_{exit}$ .

$\frac{X_A_{exit}}{(1-X_A_{exit})^2}=\frac{2 *10^3*.02*100^2 }{100}=4000$

I've solved this equation with trial and error.

and the answer is 0.984

Hope this is clear. Have a nice day.

Please ask if having any doubts or need any clarification.

Rate/Feedback is really appreciated.