Treatment Plants for Denitrification
ScoD,3 = 20 = 6 _ 7 g COD < DN03 . VN0 3 COD= 8 _ 8 9 COD
SN03,3
3
g N03-N Dcoo
'
g N03-N
Just as the influent, nitrate is not limiting and the acetic acid is limiting.
By using Expression (5.13), p is determined in the filter, first in the influent, as
kovi,COD is measured at
260 kg COD/(m 3 - d)/(6.5 kg COD/kg NOs- N) = 40 kg NOs- N/(m 3 ·d)
L is set at 2 mm and Dcoo at 0.5 · 1 o· 4 m 2 /d.
That is,
p = ( 2 · Dcoo · Scoo.1)% = [ 2 · 0.5 · 10-4m 2 /d. 130 g COD/m 3 ]%
koVI,COD · L 2
260 · 10 3 g COD/(m 3 ·d)· (2 · 10- 3 m) 2
= 0.013< 1
that is, the film is only partially penetrated, and hence it is a half order reaction.
Correspondingly it is found for the effluent: p = 0.002 <1.
The whole filter therefore has half order kinetics in respect of the really limiting component: acetic acid.
The mass balance for the removal of acetic acid or nitrate can now be set up.
For acetic acid we have
QS+rA · 00· A· dy=Q(S+ ~~ · dyJ
rA.coo = k112A,coo · Scoo 1A!
According to Expression (5.37) and (5.42), the solution is:
~
~
A~
ScOD3 2 =Scoo1 2 -0,5·k1/2ACOD-Q ·
•
•
•
1
As
k1,j!A,COD = (2. Dcoo.2. kovt,COD r
A2• can be found:
~
~
(Scoo.3 2 - Scoo.1 2 ) · 01
A2• = ....:._:::..:::.:=---=::.!..:...--'--.:...
0.5 · (2 · DcoD,2 · koV!,COD)%
_
(1301; 2 -20%)(gCOD/m 3 )%
4 ,goom2
- 0.5 ( 2 ·0.5 · 10-4 m 2 /d · 260 · 10 3 g COD/(m 3 ·d))%
267
ScoD,3 = 20 = 6 _ 7 g COD < DN03 . VN0 3 COD= 8 _ 8 9 COD
SN03,3
3
g N03-N Dcoo
'
g N03-N
Just as the influent, nitrate is not limiting and the acetic acid is limiting.
By using Expression (5.13), p is determined in the filter, first in the influent, as
kovi,COD is measured at
260 kg COD/(m 3 - d)/(6.5 kg COD/kg NOs- N) = 40 kg NOs- N/(m 3 ·d)
L is set at 2 mm and Dcoo at 0.5 · 1 o· 4 m 2 /d.
That is,
p = ( 2 · Dcoo · Scoo.1)% = [ 2 · 0.5 · 10-4m 2 /d. 130 g COD/m 3 ]%
koVI,COD · L 2
260 · 10 3 g COD/(m 3 ·d)· (2 · 10- 3 m) 2
= 0.013< 1
that is, the film is only partially penetrated, and hence it is a half order reaction.
Correspondingly it is found for the effluent: p = 0.002 <1.
The whole filter therefore has half order kinetics in respect of the really limiting component: acetic acid.
The mass balance for the removal of acetic acid or nitrate can now be set up.
For acetic acid we have
QS+rA · 00· A· dy=Q(S+ ~~ · dyJ
rA.coo = k112A,coo · Scoo 1A!
According to Expression (5.37) and (5.42), the solution is:
~
~
A~
ScOD3 2 =Scoo1 2 -0,5·k1/2ACOD-Q ·
•
•
•
1
As
k1,j!A,COD = (2. Dcoo.2. kovt,COD r
A2• can be found:
~
~
(Scoo.3 2 - Scoo.1 2 ) · 01
A2• = ....:._:::..:::.:=---=::.!..:...--'--.:...
0.5 · (2 · DcoD,2 · koV!,COD)%
_
(1301; 2 -20%)(gCOD/m 3 )%
4 ,goom2
- 0.5 ( 2 ·0.5 · 10-4 m 2 /d · 260 · 10 3 g COD/(m 3 ·d))%
267