Post on 26-Mar-2015
transcript
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Deactivation kinetics…………• The value of vmax for enzymatic reaction depends
on the amount of active enzyme present…..
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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
dk
t2ln
2/1
• For a batch reactor,
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Without deactivation………………
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:
• 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
• 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
• 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 $
• 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
• For 90% conversion……the profit is only 6556 $/day
• Therefore…..first method is more economical.