Biofilters
which is such a large concentration of glucose that oxygen is limiting for the removal.
Empirical constants for the volumetric enzyme production of biofilm and hydrolysis:
Enzyme prod. rv,E = 300 · 10 6 1U/(m 3 biofilm ·d)
Hydrolysis
kE = 0.05 m 3 /(IU ·d)
where IU is the "International Unit" for enzymes.
rA,E
= PL · rv,E = 41 · 10- 6 · 300 · 10 6 = 12,300 IU/(m 2 ·d)
kE · rA,E = 0.05 · 12,300 = 615 m/d
2
(These quantities are still very uncertain.)
Substitution in the formula for the degree of hydrolysis (5.67) gives:
DH=------~--~~
d1
1+_1 _ _ _ _ _
kE rA,E A2• · V2
1
= 0.86
1+615·100
where the "combined load" is 01/A2· · V2 = 500 2 /500 · 5 = 100 m/d 2 •
Starch concentration in the effluent is
XR,3 = C1(1- DH) = 100 · 0.14 = 14 g/m 3
Glucose concentration in the effluent is
500
3
s3 = 100 · 0.86-20 500 = 86-20 = 66 g/m
If the filter is made more efficient, for example by using a fluidized filter with a carrier
of fine particles and by increasing the specific volume to 1000 m 2 /m 3 , the volume is reduced to 0.5 m 3 .
at
5oo 2
A2.v2- 5oo. o.5 1000 mid
1
1000 = 0 · 38
1 + 615
XR,3 = 100 · 0.62 = 62 g!m 3
S3
100 · 0.38-20 ~~~ = 38-20 = 18 g/m 3
It is the hydrolysis which is the rate limiting step for the process.
5.12. Detailed model
So far the description of the processes in the biofilm has been influenced by such
simplifying assumptions that the equations can be solved analytically and that the
interaction of parameters is relatively easily ignored. Such a description has its
advantage in that it facilitates the understanding and gives a general view; but the
more complicated the interaction of processes is, the more difficult it is to find
suitable simplifications and analytical solutions. This interaction can, on the other
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