1 / 14

# - PowerPoint PPT Presentation

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about '' - shani

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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

• 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

• 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

• 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