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

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Presentation on theme: "Deactivation kinetics………… The value of v max for enzymatic reaction depends on the amount of active enzyme present….."— Presentation transcript:

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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,

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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 = ( )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… g product 4320 g pdt is produced in 4.81 hrs……..therefore for ONE day i.e 24hrs… g of product is produced Upstream cost = 4800 $ /day Downstream cost, C=155-(0.33X) = C= $ /kg pdt Downstream cost = ( $ /kg pdt)(21.55kg pdt / day) = $ /day

12 Total expenditure = = $ /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 = hrs Given that 1 g substrate produces 1.2g of product Therefore, 4320 g of substrate produces… g product 5184 g pdt is produced in hrs……..therefore for ONE day i.e 24hrs… g of product is produced

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


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