Mass balances, denitrifying treatment plants
240
(
)
%
,..,
rA.Me = 2 Do2 kovi.MefvMe,02
· So2 2
(
-4
3)%,.,
= 2 . 1.7 . 10 . 90 . 10 /1.1
. 2 2
The oxygen penetration of the biofilm is:
%
L = (2 Do2 · So2J
~
kov102
%
= [2. 1.7. ;o-4. 2]
90. 10 . 1.1
= 83j.lm
For nitrate removal applies:
,..,
rA,N03,max = k%A,N03 · SNQ3 2
= 3 · 5% = 7 g N03-N/(m 2 · d)
kovi,N03 · Do2 · So2
kov1.02 · DN03 · SN03
30 ·1.7. 10-4.2
___;::..:;_..;,;,;___;_:._-=-- = 0.29
90. 1.1. 0.7. 10-4.5
It appears from Fig 7.5 that the denitrification rate is approx. 60 per cent of maximum.
rA,N03 = 0.6 · 7 = 4.2 g N03-N m- 2 d- 1
In spite of the aerobic conditions in the water, the denitrification rate is considerable.
The full penetration of oxygen and nitrate is
%
~L = (2 Do2 · So2 + 2 DN03 · SN03J
kov1.02
kovi,N03
= [2. 1.7. 10-4. 2 2. 0.7. 10- 4 . 5]%
3
+
3
90·10 ·1.1
30·10
= 174j.lm
The enthusiastic reader is welcome to solve the equations as regards the diffusion of
methanol to find out whether the concentration of methanol is big enough to penetrate
so deeply.