4. Deactivation kinetics………… The value of v max for enzymatic reaction depends on the amount of active enzyme present…..

5. Deactivation kinetics……Half life…. Half life is the time required for half the enzyme activity to be lost as a result of deactivation i.e. if e a = e 0 /2 ==> t=t 1/2

6. For a batch reactor,

8. Without deactivation………………

9. 13.1 Economics of batch enzyme conversion An enzyme is used to convert substrate to a commercial product in a 1600L batch reactor. Vmax for the enzyme is 0.9g/L.h; Km is 1.5g/L. Substrate concentration at the start of the reaction is 3g/L; according to the stoichiometry of the reaction, conversion of 1g substrate produces 1.2g product. The cost of operating the reactor including labor, maintenance, energy and other utilities is estimated as $4800/day. The cost of recovering the product depends on the extent of substrate conversion and the resulting concentration of product in the final reaction mixture. For conversions between 70% and 100%, the cost of Down Stream Processing can be approximated as C=155-(0.33X) where C is the cost in $ per kg of pdt treated and X is the % substrate conversion. The market price for the product is $750/kg. Currently the enzyme reactor is operated with 75% substrate conversion; however it is proposed to increase this to 90%. Estimate the effect this will have ion the economics of the process:

10. Total Expenditure = Upstream cost + Downstream cost Profit = Price of product - Expenditure Given S 0 = 3 g/L For 75% conversion…. Since 75% is converted…….S f = (1- 0.75)3=0.75 g/L Amount of substrate consumed per batch = (S 0 -S f )x(volume of rector) = (2.25 g/L) 1600L = 3600 g Therefore, t b = 4.81hrs

11. Given that 1 g substrate produces 1.2g of product Therefore, 3600g of substrate produces…..4320 g product 4320 g pdt is produced in 4.81 hrs……..therefore for ONE day i.e 24hrs….. 21555 g of product is produced Upstream cost = 4800 $ /day Downstream cost, C=155-(0.33X) = C= 130.25 $ /kg pdt Downstream cost = (130.25 $ /kg pdt)(21.55kg pdt / day) = 2806.88 $ /day

12. Total expenditure = 4800+2806.88= 7606.88 $ /day Price of product = 750 $ / kg = (750 $ / kg)(21.55 kg/day) = 16,163 $ /day Therefore, Profit = Price of pdt - Expenditure Profit = 8556 $

13. For 90% conversion…. Since 90% is converted…….S f = (1-0.9)3=0.3 g/L Amount of substrate consumed per batch = (S 0 - S f )x(volume of rector) = (2.7 g/L) 1600L = 4320 g Therefore, t b = 6.837 hrs Given that 1 g substrate produces 1.2g of product Therefore, 4320 g of substrate produces…..5184 g product 5184 g pdt is produced in 6.837 hrs……..therefore for ONE day i.e 24hrs….. 18197.455 g of product is produced

14. For 90% conversion……the profit is only 6556 $/day Therefore…..first method is more economical.