Solution manual for bioprocess engineering 3rd edition by shuler

9/18/24, 12:49 Solution manual for bioprocess engineering 3rd edition
PM by shuler

Full file at https://testbanku.eu/


Solution Manual for Bioprocess Engineering 3rd Edition by Shuler Check TOC
for included chapters

S +E
k2
( )
k4   P + E
k5


rate = k (ES)
5


a.) Equilibrium approach
* r values are the same as k values all are rate constants

r1 (S)(E ) = r2 (ES)
1
(1)
r3 (ES) = r4 (ES)
1
(2)
(E ) = (E) + (ES) + (ES) (3)
0 1 r
from (2) (ES) = 4 (ES)
(4)
1 2
r3
r r (ES) 2
from (1) + (4) (E) = 2 4 (5)
r1r3 (S)

Substitute (4) and (5) into (3):
 r r + r r (S) + r r (S) 
(E ) =  2 4 1 4 1 3  (ES)
0  r1 r3 (S)  2
 
(ES) = r1 r3 ( E0 )(S)
2 r2 r4 + r1 r4 (S) + r1 r3
(S)
r2 r4
rate = r 5 E0 S where Km = , Km =
Km ( Km + S) 1 2
+ S r r
2 1 1 3



b.) Quasi–steady–state approach
d( ES)
1
= r ( E)(S) r (ES) + r (ES)  r (ES)  0 (6)
1 2 4 2 3 1
1
dt
d( ES)
2
=r ( ES)  r (ES)  r (ES)  0 (7)
3 4 2 5 2
1
dt r4 + r5
from (7) (ES) =
(8)
(ES) 1
2
3
r
r r + r r + r r
from (6) + (8): (E) = 2 4 2 5 3 5


( ES) (9)
  r1 r3 (S)  
2
 
substitute (8) and (9) into (3):




1

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,9/18/24, 12:49 Solution manual for bioprocess engineering 3rd edition
PM by shuler


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 r2 r4 + r2 r5 + r3 r5 + r1 r4 (S ) + r1 r5 ( S) + r1 r3 (S) 
( E )0 = r1 r3 ( S)  ( ES)
2
(ES) =  
r r (E )(S) 
2  r r + r r + r r + 1 3
r r (S) +r r (S) + r r ( S) 
0


24 2 5 3 5 1 4 1 5 1 3 
r 5 E0 S
 rate =
(Km2 + r5 r3 )(Km1 +S) +r5
r1 +S
r r
where Km = 2 and Km = 4
1 1 2
r 3
r
Problem 3.2
* r values are the same as k values all are rate constants
E +S (
d( ES )
r 2

) r1

= r1 ( E)(S) +r2 ( E)(P) r1 (ES) r2 (1)
(ES)  0
dt d P
()
rate = =r (ES) (E)(P) (2)
r
2 2
dt
(E ) = (E) +(ES), (E) = (E )  (ES) (3)
0 0




from (1): (E) r1 + r2
r1 (S) + r2
(ES) (4)
=
(P)
 r1 ( S) + r 2 ( P) + r 1 + r2 
Combine (3) + (4): ( E ) =

( ES)
r (S) +r (P ) 
1 2
0
 
(ES) = E 0 (r1 (S) + r2
(5)
(P))
r1 (S) + r2 (P) +
r1 + r2

Substitute (3) and (5) into
(2):

r rE (S) + r r E r 2 (E  (ES))(P )
(P)
rate = r 1 +
2 0 2 2 0
r + r (S) + r ( P)
0

1 2 1 2



Substituting (ES) in terms of E0:

r1 r2E(S)  r1 r2E 0 ( P)
rate = r1 +r2 +r1 S + r2 P
( ) ( )

LetVS = r2 E0 and VP = r-1 E0

r1 Vs (S)  r2 VP (P)

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, 9/18/24, 12:49 Solution manual for bioprocess engineering 3rd edition
PM rate by shuler
= r1 +r2 +r1 (S) + r2 (P)



2




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