Chapter 2: Conversion and Reactor Sizing

Old Exam Questions

The adiabatic exothermic irreversible gas phase reaction

is to be carried out in a flow reactor for a stoichiometric feed of A and B.

Additional Information

a) What PFR volume is necessary to achieve 50% conversion?

V1 =                            

$V_1 = (500 - 100) (0.5) = 200 dm^3$

$V_1 = (500 - 100) (0.5) + (100)(0.5) = 250 dm^3$

$V_1 = \dfrac{1}{2}(500 - 100) (0.5) + (100)(0.5) = 150 dm^3$

Solution Part a

b) What CSTR volume is necessary to achieve 50% conversion?

V1 =                            

$V_1 = \dfrac{1}{2}(500) (0.5) = 125 dm^3$

$V_1 = (500) (0.5) = 250 dm^3$

$V_1 = (500 - 100) (0.5) = 200 dm^3$

Solution Part b

c) What CSTR volume must be added to raise the conversion in Part (b) to 80%?

V2 =                                      

$V_2 = (100) (0.8) = 80 dm^3$

$V_2 = (500 - 100) (0.8) = 160 dm^3$

$V_2 = (100) (0.8 - 0.5) = 30 dm^3$

Solution Part c

d) What PFR volume must be added to raise the conversion in Part (b) to 80%?

V2 =                                      

$V_2 = \dfrac{1}{2}(500 - 100) (0.8 - 0.5) = 60 dm^3$

$V_2 = \dfrac{1}{2}(500 - 100) (0.8 - 0.5) + 100(0.8 - 0.5)= 90 dm^3$

$V_2 = (500-100) (0.8 - 0.5)+100(0.8 - 0.5) = 150 dm^3$

Solution Part d











































Solution

a) What PFR volume is necessary to achieve 50% conversion?      V1 = 150 dm3

Back to Part a











































b) What CSTR volume is necessary to achieve 50% conversion?     V1 = 250 dm3

Shaded Area:    

Back to Part b











































c) What CSTR volume must be added to raise the conversion in Part (b) to 80%?      V2 = 30 dm3

Shaded Area:     

Back to Part c











































d) What PFR volume must be added to raise the conversion in Part (b) to 80%?     V2 = 90 dm3

Shaded Area:     

Back to Part d



Back to Chapter 2