Reaction engg problem NO5 -
QUESTION -5 -
For the consecutive reaction , A----K1(B-K2--(C, numerical values are K1 = 0.85 hr^-1, K2 = 0.13 hr^-1, Cao = 64 kg -mole /m3 and Cbo = Cco =0 . what is the maximum concentration attained by B when as 1).Batch reactor ,( 2 ) -single stage continuous stirred tank reactor , (3) - two stage continuous stirred reactor Solution -
The reaction - A---K1-(B-----K2--(C Rate of formation of B is- R =r = K1.Ca - K2.Cb For consecutive reaction- Ca =Cao .e^-E/R.T ( 1) in case of batch reactor Cb /Cao = (K1/K2-K1 )*(e^-K1*.t --e^-K2*t ) Where , t = residence time = V/vo Also ,( Cb)max/Cao = ( K1/K2)^K2/K2-K1 OR, ( Cb)max = Cao [k1/k2]^k2/k2-k1 = 64[0.35/0.13]^0.13/0.13-0.35 = 35.68 kg-mole /m3
Part (2) ---
In case of single stage CSTR- Ca1 =Cao/1+K1*t Cb1 = Cao *K1*t/(1+K1*t)*(1+K2*t) Differentiating Cb1 with respect to t for maximum concentration and equating to zero , we get- T =t = (1/k1*k2)^1/2 = 4.68 Cb = 64**0.35*4.68/(1+0.35*4.68)*(1+0.13*4.68) = 24.71 kg-mole/m3
Part - (3)-
In case of two stage CSTR - Ca2 = Cao/(1+k1*?)*(1+K2*?) Cb2 =2.Cao*K1*t/(1+K2*t)^2*(1+K1*t) Differentiating Cb2 with respect to t for maximum concentration and equating to zero, we get T = t =2.678 (Cb)max =2*64*0.35*2.678/(1+0.13*2.676)^2*(1+0.35*2.678) = 34.07 kg-mole/m3 centration and equating to zero , we get- T =t = (1/k1*k2)^1/2 = 4.68 Cb = 64**0.35*4.68/(1+0.35*4.68)*(1+0.13*4.68) = 24.71 kg-mole/m3
Part - (3)-
In case of two stage CSTR - Ca2 = Cao/(1+k1*?)*(1+K2*?) Cb2 =2.Cao*K1*t/(1+K2*t)^2*(1+K1*t) Differentiating Cb2 with respect to t for maximum concentration and equating to zero, we get T = t =2.678 (Cb)max =2*64*0.35*2.678/(1+0.13*2.676)^2*(1+0.35*2.678) ...