Page 751 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

706 Fundamentals of Water Treatment Unit Processes: Physical, Chemical, and Biological

Expressed symbolically, Equation 22.B.1 is 7. Substitute Equation 22.B.8 in Equation 22.B.7:
[E t ] [S]
k 1
S þ R þ E ! E S ! X þ P þ E (22:B:2) [ES] ¼ (22:B:9)
k 2
k 1 K s þ [S]
where 8. Express dX=dt, based upon [ES] Equation 22.B.2, as
R is the reactants; for example, O 2
dX
E is the enzyme concentration ¼ k 2 [ES] (22:B:10)
E S is enzyme substrate complex dt
X is the cells synthesized
9. Substituting Equation 22.B.9 in Equation 22.B.10
P is the end products; for example, CO 2
gives
k 1 , k 1 , k 2 are the kinetic constants
dX [E t ] [S]
Since the reactions depicted are hypothetical, there is no ¼ k 2 (22:B:11)
dt K m þ [S]
need to consider balancing them stoichiometrically. The fol-
lowing steps are based upon Equation 22.B.2 in terms of
10. Divide both sides of Equation 22.B.11 by [E t ]:
the associated kinetic expressions:
dX
=[E t ] m (22:B:12)
1. The rate of appearance of [ES] is dt
11. Now also in Equation 22.B.11, let [S] K s , which
d[ES]
¼ k 1 [E] [S] k 1 [ES] k 2 [ES] (22:B:3) gives
dt

dX
2. The total enzyme in the system E t is the sum of both dt ¼ k 2 [E t ] V max (22:B:13)
combined and free enzyme: [S]>>K m
12. Substituting (22.B.13) in Equation 22.B.11 gives
[E t ] ¼ [ES] þ [E] (22:B:4)
dX S
The total enzyme concentration, E t , represents the ¼ V max (22:B:14)
dt K m þ S
total catalytic activity of the system, which is not a
measurable quantity. Since viable cells provide the 13. Divide both sides of Equation 22.B.14 by [E t ], and let
catalytic activity, then E t should be proportional to
the concentration of viable cells X.
V max
3. Replace [E] in Equation 22.B.3 by Equation 22.B.4: b m ¼ (22:B:15)
[E t ]
d[ES]
¼ k 1 [E t ] [ES]g[S] k 1 [ES] k 2 [ES] (22:B:5) 14. The result is the Michaelis–Menten equation, which
f
dt
is the same as the Monod equation,
4. Apply the ‘‘steady-state approximation’’ and get [S]
m ¼ bm (22:B:16)
K m þ [S]
0 ¼ k 1 [E t ] [ES]g[S] k 1 [ES] k 2 [ES] (22:B:6)
f
It is easy to see, plotting Equation 22.B.9 as Figure 22.B.1, and
5. Solve for [ES] to give
Equation 22.B.14 as Figure 22.B.2, that the enzyme becomes
[E t ] [S]
(22:B:7)
[ES] ¼ [E ] t
(k 1 þ k 2 )=k 1 g þ [S]
f
6. Let

[ES]
k 1 þ k 2
(22:B:8)
K s ¼
k 1
In other words, K s is an equilibrium constant, and
[S]
expresses the net effect of forward rates and back-
ward rates of reaction. FIGURE 22.B.1 Enzyme isotherm plotted by Equation 22.B.9.