5.5.
Do2 = 1. 7 · 10-4m 2 /d
Dcoo = 0.4 · 1 0-4m 2 /d
VQ2,COD = 1. 7 g COD/g 02
Controlling for the removal rate applies:
So2 > Dcoo
Scoo ~ Do2 -v-o-2 . .;_c_o_o
0.4. 10-4
- - - - 4 - - = 0.14 g 02/g COD
1·7·10 ·1.7
Biofilters
If the oxygen concentration is less than 0.14 Scoo. it is limiting for the removal. At the
influent, the oxygen concentration is less than 0.14 g 0 2/g COD · 150 g COD/m 3 = 21
g 0 2/m 3 because it is greater than the saturation with the air: 1 0 g 0 2/m 3 at 1s•c.
If the oxygen concentration in the filter just ahead of the outlet is assumed to be 2 g
0 2/m 3 , the COD-concentration should be less than S0~0.14 = 2/0.14 = 14 g COD/m 3
in order to be limiting.
The conclusion is that in a filter treating dissolved organic matter with oxygen as the
only oxidant, the oxygen will be limiting for the removal in the major part of the filter.
Filter kinetics
For a given medium in a filter, the analysis of the filter processes can be made in two
stages as shown in Fig 5.9:
Fig 5.9
Removal in the biofilm itself is here assumed to be of zero order
I stage I
potentially limiting factor •
Expression (5.32) complied with?
______ _.,/ '- .... ____ _
•
•
Yes, that is
oxygen potentially
limiting
•
I stage II Biofilm utilization? •
Expression (5.19)
p a r t i a r ~~ly
lkl
~1
half order
zero order
process for
process
oxygen
No, that is
organic matter
potentially limiting
•
Biofilm utilization?
•
Expression (5.19)
partially J L, fully
. ...
P J!:!l
half order
process for
org. mat.
zero order
process
Biofilm kinetic user guide. The calculation procedure is shown which is used to
determine the kinetics in relation to the wastewater (bulk phase).
159