kt 中间产物收率Yp= CA(1+kr)1+k2) k k1(1-xA) kx(1-x4) (3.2-10) 1(+k,x11+k,x1-)4(-x1)+kx 将(3.2-9)式对τ求导: dek(1+k)1+k2)-k(+k2)+k(1+k) (+k1r)1+k2) ,可2=0,则:(1+k)1+k1)-k2(1+k2)+k(+2)=0 开整理得:k1k22=1即 (3.2-11) ∴P的最大收率YPm(1+kz)(1+k/ √k (3.2-12) P的最大浓度C (1+
中间产物收率 1 0 1 2 (1 )(1 ) P P A C k Y C k k = = + + (3.2-9) 或： 1 1 1 0 1 2 1 2 1 1 (1 ) (1 ) (1 ) (1 )(1 ) (1 ) (1 ) A P A A A P A A A A A A A x k C k x k x x Y C k x k x x x k k k x k x − − = = = − + + + − − (3.2-10) 将(3.2-9)式对τ求导: 2 1 1 2 1 2 1 2 1 2 1 2 (1 )(1 ) (1 ) (1 ) [(1 )(1 )] P dY k k k k k k k k d k k + + − + + + = + + 令： 0 P dY d = ，则： 1 2 1 2 2 1 (1 )(1 ) (1 ) (1 ) 0 + + − + + + = k k k k k k 展开整理得： 2 1 2 k k =1 即： 1 2 1 opt k k = (3.2-11) ∴P 的最大收率 1 ,max 1 2 (1 )(1 ) opt P opt opt k Y k k = + + 1 1 2 1 2 1 2 1 2 1 1 1 (1 )(1 ) k k k k k k k k k = + + 2 2 1 1 (1 ) k k = + (3.2-12) P 的最大浓度 0 ,max 2 2 1 (1 ) A P C C k k = + (3.2-13)