Design of biofilters
178
Hence at least 7 units must be arranged in a series. The first six will work identically
because the HAc-concentration is greater than the limit for being rate limiting in relation to oxygen. In the first six units 6 · 7.3 = 44 kg HAc/d are removed, and hence there are 6 kg HAc/d left corresponding to an influent concentration of 60 g HAc/m 3 .
In the seventh unit, HAc is expected to be rate limiting.
%
k%,HAc = ( 2 · DHAc · kovt,HAc)
%
k%.02 = ( 2002 · kovt,02)
kovt,HAc
- - - = VQ2 HAc = 2.1 g HAc/g02
kovt,02
·
1;2
%
k%.HAc __ ( 2 DHAc koVI,HAc) __ ( DHAc
)
k1,t 2.o2
2 Do2 kovt,02
Do2
· VQ2,HAc
DHAc
0.7 · 10- 4 m 2 /d
Do2
1.7 · 1 o- 4 m 2 /d, see Table 5.2
k%,HAc = k%A,02(~o:G VQ2,HAcJ%= 3.5 · (~:;· 2.1 f= 3.3 (g HAc)1; 2 m- %d- 1
The balance for HAc:
0 81 - rA,HAc · A2 = 0 83
6000 g HAc/d- 3.3 · 500 · 83% = 100 · 83
This gives the solution:
83 % = 3.1
(or 83 % =- 19)
83
= 9.6 g HAc/m 3
At this concentration HAc will be limiting for the removal:
Dred
= ~ = 0.20
Dox·Vox,red 1.7·2.1
Sox
8red
Notice: The reductant is thus limiting for the process, see Expression (5.32). HAc is limiting when 8HAc is less than 4/0.20 = 20 g HAc/m 3 .
An eighth unit is necessary as the concentration is still higher than 2 · Ks. In the eighth
unit, the removal follows a first order process as the concentration is less than a 2 . Ks
= 4 g HAc/m 3 .