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Deactivation kinetics…………

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

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Deactivation kinetics…………

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  1. Deactivation kinetics………… • The value of vmax for enzymatic reaction depends on the amount of active enzyme present…..

  2. 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 ea = e0/2 ==> t=t1/2

  3. For a batch reactor,

  4. Without deactivation………………

  5. 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:

  6. Total Expenditure = Upstream cost + Downstream cost • Profit = Price of product - Expenditure • Given S0 = 3 g/L For 75% conversion…. • Since 75% is converted…….Sf = (1-0.75)3=0.75 g/L • Amount of substrate consumed per batch = (S0-Sf)x(volume of rector) • = (2.25 g/L) 1600L = 3600 g • Therefore, tb = 4.81hrs

  7. 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

  8. 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 $

  9. For 90% conversion…. • Since 90% is converted…….Sf = (1-0.9)3=0.3 g/L • Amount of substrate consumed per batch = (S0-Sf)x(volume of rector) • = (2.7 g/L) 1600L = 4320 g • Therefore, tb = 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

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

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