Design of biofilters
composite sample will show concentrations in the range 5- 20 g BOD I m 3 ; for Bv =
1000 g BODI(m 3 ·d) in the range 20-50 g BOD 1m 3 .
It is recommended to carry out the design on the basis of Bv = 400 g BOD I (m 3 · d),
0.5 mlh < BA,v < 1.0 mlh and R < 1.0. For trickling filters with plastics media it is
recommended that BA = 4 g BODI(m 3 ·d), 0.8 < BA,V < 1.8 mlh in the range 100 <
ro <200m 2 1m 3 and R < 1. For these loads no nitrification will take place.
The experience values in Table 5.3 only apply to ordinary municipal wastewater
which is considered to have a relatively homogenous composition. All the curves of
treatment results (whether expressed as treatment efficiency or as concentration in
the effluent), which can be found in great numbers in the literature, are characterized by a very large spread. This expresses that these simple loading rules cannot
take into account the numerous special circumstances which occur in practice. In the
above, for example, such basic conditions as: Filter medium, surface area, influent
concentration and the special property of the organic matter have not been taken
into consideration. We have to be very cautious about using the above loading rules
uncritically- although these rules have formed the basis of the successful design of
numerous full-scale trickling filters.
174
Example 5.7
A trickling filter with a diameter of 10 m and a height of 2 m is loaded with 235 m 3 /d of
mixed domestic and industrial wastewaters. The concentration of BOD is 500g/m 3 .
Find the volumetric loading rate and the necessary recycle.
According to Expression (5.54), the volumetric loading rate is:
Bv =a, · C,N2 = Q · C,!(rr · r2 ·h)
Substitution gives:
Bv = (235 m 3 /d) · (500 g BOD/m 3 )/(rr · (5m) 2 · 2 m)
Bv = 748 g BOD/(m 3 . d)
Hence the filter is interfacing normal and high loading.
(5.59)
The necessary recycle is controlled by the requirement of the hydraulic surface loading rate which, based on Table 5.3, in this case is estimated at 1.2 m/h.
From Expression (5.57) the necessary recycle, Q 6, can be determined:
BA,v = (Q, + Oe)/A2
Substitution gives:
1.2 m/h · 24 h/d = (235 m 3 /d + Q6)/(rr · (5m) 2 )
0 6 = 2,027 m 3 /d
that is, a recycle ratio R = Oe/01 = (2,027 m 3 /d)(235 m 3 tdr 1 = 8.6
(5.60)
This is a high recycle. It appears from Table 5.3 that the requirement of the hydraulic
surface loading rate, and hence the recycle, increases by increasing organic volume-
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