Reaction engineering
Question -3 -the first order reaction A-(B is carried out in tubular reactor in which volumetric flow rate is constant . derive an equation relating to the reactor volume to the entering and exiting concentration of A . the rate constant = Kc and the volumetric flow rate Vs. what is the reactor volume necessary to reduce the exiting concentration to 10% of the entering concentration when volumetric flow rate = 10 d m3/min and specific reaction rate K= 0.23/min. Solution - For the tubular section - The molar balance on species A ( J=A) was shown to be- dFa/dV= ra
( 1) For , the first order- -ra = k.Ca
Since volumetric flow rate = Vo is constant- dFa/Dv= dFa/Dv = -dCa.Vo/Dv = Vo.dCa/Dv = ra-vo,dCa/Dv = -ra = K.Ca Rearranging -(-vo/K)(dCa/Ca) = Dv
(3) Using the condition , the entrance of the reactor when V=0, then Ca = Cao ( -vo/K)*?( limit (Cao to Ca) dCa/Ca =? (limit o to V) Dv This gives- V= vo/K .ln Cao/Ca= (10 dm3/min/0.23 /min)* ln Cao/0.1*Cao = 100 dm3 ?
( limit (Cao to Ca) dCa/Ca =?
(limit o to V) Dv
This gives- V= vo/K .ln Cao/Ca= (10 dm3/min/0.23 /min)* ln Cao/0.1*Cao = 100 dm3 onstant .
derive an equation relating to the reactor volume to the entering and exiting concentration of A .
the rate constant = Kc and the volumetric flow rate Vs. what is the reactor volume necessary to reduce the exiting concentration to 10% of the entering concentration when volumetric flow rate = 10 d m3/min and specific reaction rate K= 0.23/min.
Solution - For the tubular section - The molar balance on species A ( J=A) was shown to be- dFa/dV= ra -------( 1) For , the first order- -ra = k.Ca Since volumetric flow rate = Vo is constant- dFa/Dv= dFa/Dv = -dCa.Vo/Dv = Vo.dCa/Dv = ra