Question

The reaction A→B is to be carried out isothermally in a continuous-flow reactor. Calculate the PFR volume to consume 99% of A (CA = 0.01CA0) when the entering molar flow rate is 5 mol/h (assume pure A), the volumetric flow rate is constant at 10 dm3 /h and the rate is: -rA=3CA 2 [ dm3 /mol•h]

Answer 1

Answer:

Explanation:

From the information above:

The equation for the reactor volume of CSTR is given by the formula:

$V = \dfrac{C_{Ao}V_o - C_A V_o }{-rA}$

$V = \dfrac{C_{Ao}V_o (1 - 0.01)}{KC_A^2}$

The reaction rate of A is:

$-r_A = kC_A^2$ and k = 3 dm³/mol.h

replacing the values in the above equation to solve for v;

∴

$V = \dfrac{(0.5 \ mol/dm^3) (10 \ dm^3/h) (1 - 0.01)}{ ( 3 \ dm^3/mol.h) (0.01 \times 0.5 \ mol/dm^3)^2}$

V = 66,000 dm³

Thus, the reactor volume of CSTR is 66000 dm³

Taking the differential mole balance equation for PFR is as follows:

$\dfrac{d(C_AV_o)}{dV} = -r_A$

$\dfrac{d(C_AV_o)}{dV} = Kc^2A$

Taking the integral between initial and final concentrations;

$\dfrac{v_o}{k} \int ^{C_A}_{C_{Ao}} \dfrac{dC_A}{C^2A}= - \int ^V_o \ dV$

$V = \dfrac{V_o}{K} \Big [\dfrac{1}{C_A}- \dfrac{1}{C_{Ao}} \Big ]$

replace the values in the above equation;

$V= \dfrac{(10 \ dm^3/h) }{(3 \ dm^3/mol.h)}\Big [ \dfrac{1}{0.01(0.5 \ mol/dm^3)} - \dfrac{1}{0.5 \ mol/dm^3} \Big ]$

$\mathbf{V = 66 0 \ dm^3}$

Thus, the reactor volume of PFR is 660 dm³