Plug Flow Reactor Vol: Comparing CSTR & PFR Kinetics, 95% Efficiency

[candycooke]

What volume of a plug flow reactor is required to remove 95% of a substance that decays according to first-order kinetics with a rate constant of 0.07 1/day. The flow rate is 295 m^3/day. Compare the detention time and volume required for a PFR to that obtained for a CSTR with the same flow rate and kinetics.

efficiency = 95%
k = 0.07 1/day
Q = 295 m^3/day

r = -kC
C = 0.05 * C (initial)

I am unsure how to set up the equation necessary to solve this problem. Any help would be greatly appreciated!

 

[Valkarie]

For the PFR, the detention time required is given by:t = V/Q where V is the volume of the reactor and Q is the flow rate. Substituting in the values given, we have:t = V/295 m^3/daySolving for V, we get:V = 295 m^3/day * tThe required volume is then obtained by solving for t such that the efficiency is 95%. From the equation for the first-order reaction rate, we have:C/C0 = e^(-kt)Solving for t, we get:t = ln(0.05)/(-k)Substituting this into the equation for V above, we get:V = 295 m^3/day * ln(0.05)/(-k)Substituting the value of k given in the problem, we get:V = 295 m^3/day * ln(0.05)/(0.07 1/day)V ~= 6538 m^3The volume required for a CSTR with the same flow rate and kinetics is:V = Q/(k*ln(0.05))Substituting in the values given, we get:V = 295 m^3/day / (0.07 1/day * ln(0.05))V ~= 5145 m^3Thus, the volume required for a PFR to achieve 95% efficiency is 6538 m^3, which is significantly larger than the 5145 m^3 required for a CSTR.