ekv3

RTEaAek

/

3

ekv3

)exp(0 tkee

ekdt

de

da

ada

)/exp( RTEAkddd

Deactivation kinetics…………• The value of vmax for enzymatic reaction depends

on the amount of active enzyme present…..

amekv3

)exp(03

tkekd

)exp(0

tkvvdmm

occursondeactivatibefore

vofvalueinitialekvmm

030

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,

dt

ds

sK

svv

M

m

ssK

tkv

dt

ds

M

dm

)exp(

0

dttkvdss

sK ts

sdm

M

00

)exp(0

1)exp()(ln 0

0

0 tkk

vss

s

sK

d

d

m

M

)(ln1ln

0

0

0

sss

sK

v

ktk

M

m

d

d

)(ln1ln

10

0

0

sss

sK

v

k

kt

M

m

d

d

)(ln1)exp(

0

0

0

sss

sK

v

ktk

M

m

d

d

)(ln1

0max

sss

sK

vt o

Mb

dt

ds

sK

svv

M

max

dtvdss

sKM

max

bo

M tvsss

sK max0 )(ln

AA

Mb XsX

Kv

t 0max 1

1ln

1

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.