Question

QUESTION 5: 10 marks An elementary aqueous phase reaction A --> B+C is to be carried out in 10 point a non-ideal CSTR which has both bypassing and a dead zone in it. The measured liquid volume in the reactor is 500 dm3 and the volumetric flow rate to the reactor is 10 dm3/min with a concentration of A as Imol/dm3. The reaction rate constant is 0.5 min-1. a) Calculate the conversion that can be expected in this reactor if the model parameters of a and B were determined from a tracer test and were found to be 0.8 and 0.3 respectively. (6 marks) b) Calculate the conversion if the reactor were an ideal CSTR. compare both and comment (4 marks)

advanced CRE subject

step by step solution

Answer 1

From tracer test model

$\alpha = 0.8$

$\beta = 0.3$

From tracer model

CA = BCA + (1 - BCAs

$C_{As}= \frac{C_{Ao}(1-\beta ) V_{o}}{(1-\beta )V_{o}+ \alpha Vk}$

C_Ao = 1 mol/dm3

Vo = 10 dm3/min

V = 500 dm3

k = 0.5/min

$C_{As}= \frac{1(1-0.3)10}{(1-0.3)10+ (0.8)(500)(0.5)}$

C_As = 0.03381 mol/dm3

C_A = 0.32367 mol/dm3

$X_{A}= 1-\frac{C_{A}}{C_{Ao}}$

X_A = 1-(0.32367/1) = 0.6763

Conversion from non ideal reactor = 0.6763 = 67.63%

B)

If CSTR was ideal then

Design equation is given by

$\frac{\tau }{C_{Ao}}= \frac{X_{A}}{kC_{Ao}(1-X_{A})}$



$\tau = \frac{V}{V_{o}} = \frac{500}{10}$ = 50 min

k = 0.5 /min

$50= \frac{X_{A}}{0.5(1-X_{A})}$

X_A = 0.9615

Conversion from ideal CSTR = 96.15%

Conversion from non ideal = 67.63%

Comparing both we can see that there is significant cchange im conversion and hence there is significant stagnation and bypassing in reactor

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