Question

A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible isomerization. The reactor is 3 m long, with a cross- sectional area of 25 dm^2. After being built, a pulse tracer test on the reactor gave the following data: tm = 10 s and σ2 = 65 s2. What conversion can be expected in the real reactor?

Answer 1

Answer:

The conversion in the real reactor is = 88%

Explanation:

conversion = 98% = 0.98

process rate = 0.03 m^3/s

length of reactor = 3 m

cross sectional area of reactor = 25 dm^2

pulse tracer test results on the reactor :

mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2

note:  space time (t) =

t = $\frac{A*L}{Vo}$   Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor

therefore (t) = $\frac{25*3*10^{-2} }{0.03}$ = 25 s

since the reaction is in first order

X = 1 - $e^{-kt}$

$e^{-kt}$ = 1 - X

kt = In $\frac{1}{1-X}$

k = In $\frac{1}{1-X}$ / t  

X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then  

K = 0.156 $s^{-1}$

Calculating Da for a closed vessel

; Da = tk

      = 25 * 0.156 = 3.9

calculate Peclet number Per using this equation

0.65 = $\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})$

therefore

$\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0$

solving the Non-linear equation above( Per = 1.5 )

Attached is the Remaining part of the solution