/ -In(S 6 IS 1 ) \ i/n
™ (kw2s)D )Using values from above, the area and diameter of the trickling filter are
A = 694 gal/min ( ^^ft J^ =^6 "'^6 ff^29 '^9 ft (9>1 m)- Calculate the hydraulic and organic loading on the filter
The hydraulic loading (QIA ) is then calculated:
Hydraulic loading = 694 gal/min/699.6 ft
2- 0.99 gal/min-ft
2
(0.672 L/s-m
2
)
For plastic media trickling filters, the hydraulic loading ranges from 0.2 to 1.20
gal/min-ft
2
(0.14 to 0.82 L/s-m
2
).
The organic loading to the trickling filter is calculated by dividing the BOD 5 load to
the filter by the filter volume as follows:
(1.0 Mgd)(240 mg/L)(8.34 lb-L/mg-Mgal)
Organic loading = (699.6 ft
2
)(25 ft)(10
3
fWlOOO ft
3
)= 114
To^Fd(557kg/m2
'd)For plastic media trickling filters, the organic loading ranges from 30 to 200 Ib/10
3ft
3
-d (146.6 to 977.4 kg/m
2
-d).
- Determine the required dosing rate for the filter
To optimize the treatment performance of a trickling filter, there should be a continual
and uniform growth of biomass and sloughing of excess biomass. To achieve uniform
growth and sloughing, higher periodic dosing rates are required. The required dosing rate
in inches per pass of distributor arm may be approximated using the following:
Dosing rate = (organic loading, lb/10^3 ft^3 -d)(0.12)Using the organic loading calculated above, the dosing rate is:
Dosing rate = (114 lb/10^3 ft^3 -d)(0.12) = 13.7 in/pass (34.8 cm/pass)Typical dosing rates for trickling filters are listed in Table 2. To achieve the typical
dosing rates, the speed of the rotary distributor can be controlled by (1) reversing the lo-
cation of some of the existing orifices to the front of the distributor arm, (2) adding re-
versed deflectors to the existing orifice discharges, and (3) by operating the rotary distrib-
utor with a variable speed drive.
- Determine the required rotational speed of the distributor
The rotational speed of the distributor is a function of the instantaneous dosing rate and
may be determined using the following:
= L6(6r)
" (WPK)