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  • Bioprocess Engineering Basic
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  • Bioprocess Engineering Basic

Solution Manual Contemporary Engineering Economics Park 7th edition - Updated 2024 Complete Solution Manual With Answers

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  • 21 de octubre de 2024
  • 19
  • 2024/2025
  • Examen
  • Preguntas y respuestas

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  • solution manual contemporary engineering economics
  • solution manual contemporary engineering

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  • Bioprocess Engineering Basic
  • Bioprocess Engineering Basic
  • Bioprocess Engineering Basic
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Chapter 3

Problem 3.1

S E k2
k1
 ES1 k3
k4
(ES) 2 
k5
P  E
rate  k5  ES2

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

r1 S E   r2  ES1 (1)
r3  ES1  r4  ES2 (2)
 E0    E    ES1   ES2 (3)
r
from (2)  ES1  4  ES2 (4)
r3
r2 r4  ES2
from (1) + (4)  E   (5)
r1r3 S

Substitute (4) and (5) into (3):

 r2 r4  r1 r4  S  r1 r3  S 
 E 0      ES2
 r1 r3  S 
r1 r3  E 0  S
  ES2 
r2 r4  r1 r4  S  r1 r3  S
r5 E 0 S r2 r
rate  where Km1  , Km 2  4
Km 2  Km1  S  S r1 r3

b.) Quasi–steady–state approach

d  ES1
 r1  E  S  r2  ES1  r4  ES 2  r3  ES 1  0 (6)
dt
d  ES2
 r3  ES1  r4  ES2  r5  ES2  0 (7)
dt
r4  r5
from (7)  ES1   ES2 (8)
r3

r r r r r r 
from (6) + (8):  E    2 4 2 5 3 5   ES2 (9)
 r1 r3  S 

substitute (8) and (9) into (3):



1

,  r2 r4  r2 r5  r3 r5  r1 r4 S   r1 r5 S   r1 r3 S  
 E 0      ES2
 r1 r3  S  
 r1 r3  E 0  S 
 ES2   
 r2 r4  r2 r5  r3 r5  r1 r4 S   r1 r5 S   r1 r3 S  
r5 E0 S
 rate 
 Km2  r5 r3  Km1  S  r5 r1  S
r r
where Km1  2 and Km 2  4
r1 r3

Problem 3.2
* r values are the same as k values all are rate constants
E S r1
r1
 ES r2
r2
EP
d  ES
 r1  E S  r2  E  P   r1  ES  r2  ES  0 (1)
dt
d P
rate   r2  ES  r2  E  P  (2)
dt
 E0    E    ES ,  E    E0    ES (3)

r1  r2
from (1):  E    ES (4)
r1 S  r2  P 

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

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

r1 r2 E 0  S  r2 r2 E 0  P 
rate   r  2  E 0   ES   P 
r1  r2  r1  S  r2  P 

Substituting (ES) in terms of E0:

r1 r2 E  S   r1 r2 E 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 
rate 
r1  r2  r1  S  r2  P 


2

, Divide top and bottom by (r-1 + r2) and let
1 r 1 r
 1 and  2
K m r1  r2
1
K P r1  r2


 rate 
V s K1m  S   VP K P  P
S P
1 1 
K m KP

Problem 3.3

k 1  k 2
a) Km   4.5 105 M
k1

b) E 0  106 M
S  103 M
V S
V  m1
Km  S
but Vm  r2 E0
 V  9.58 104 Msec1


Problem 3.4.

At low substrate concentrations So< 150 mg/l substrate inhibition is negligible.
V0 = Vm0 So / (Km + So) or 1/V0 = 1/ Vm + Km/Vm (1/S0)
For S0 < 150 mg/l Plot 1/V0 versus 1/S0
Slope = Km/Vm0 = 13.8 y-intercept = 1/ Vm = 0.023
Then, Vm = 43.5 mg/l-h and Km = 600 mg/l

At high substrate concentrations above 150 mg/l , substrate inhibition is significant.
V0 = Vm0/ (1+ S0/Ksı) or 1/V0 = 1/Vm0 + S0/Vm Ksı
For S0 > 150 mg/l plot 1/V0 versus S0
Slope = 1/ Vm0Ksı = 2.59x 10-3 Then, Ksı = 8.9 mg/l

Low Ksı indicates severe substrate inhibition.

Problem 3.5.

Plot 1/V versus 1/S at different inhibitor concentrations
Since the lines intercept at the same point on y-axis inhibition is competitive (Constant Vm,
increased Km).
For I = 0 , No inhibitor : From the intercept on y axis , 1/V m = 0.2 and Vm = 5 mM/h
And from the intercept on X-axis, - 1/Km = -1.2 and Km = 0.83 mM
From 1/V versus 1/S plot for I = 1.3 mM and S 0 = 0.50 mM V = 1.3 mM/h
Then, V = Vm S/ (Km(1+I/Ksı)) + S , 1.3 = 5(0.5)/ (0.83(1+1.3/Kı) + 0.5 )
Then Kı = 1.82 mM

3

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