Deactivation kinetics……Half life….

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v  k e
3
v  k e
3
k  Ae
3






 Ea / RT







de a
dt
 k d ea
 e a  e 0 exp(  k d t )
k
d
 A exp(  E / RT )
d
d
Deactivation kinetics…………
• The value of vmax for enzymatic reaction
depends on the amount of active enzyme
present…..
v  k e
m
3
a
 k e exp(  k t )
3
0
d

v
m0
v v
m
m0
exp(  k t )
 k e  initial value of v
3
0
before
d
m
deactivati on
occurs
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
 t
1/ 2

ln 2
k
d
• For a batch reactor,
v s
v 
m
K
M
ds
 
s

dt
s
 
s
K
K ln
M
exp(  k t )
d
M
s
s
0
s
s
t
ds  v  exp(  k t ) dt
m0
s
0
m0
dt
K
s
M
v
 
ds
 (s  s)  
0
d
0
v
m0
k
d
exp(  k
d
t )  1
k 
s

exp(  k t )  1 
 (s  s)
 K ln
v 
s

d
0
d
M
0
m0

k 
s

 k t  ln 1 
 (s  s)
 K ln
v 
s


d
0
d
M
0
m0


k 
s

t 
ln 1 
 (s  s)
 K ln
k
v 
s


1
d
0
M
d
m0
0
Without deactivation………………
v s
v 
max
K
M
s
 
ds
dt
tb 

K
K M ln
s
M
s
so
s
ds  v
max


so
 (s0  s)
 K M ln
s


1
v max
 dt
 ( s 0  s )  v max t b
 tb 
1
v max

1
 K M ln
1 X

A

 s0 X A 

•
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 = (10.75)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 = (S0Sf)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.

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