Example Solved Problems on Trickling Filters

PROBLEMS

5. Design suitable dimensions of a circular trickling filter unit for treating 5 million litres of sewage per day. The BOD of sewage is 150 mg/l.

Solution:

Total BOD of sewage = 5 MLD × 150 mg/l

= 5 x 10⁶ × 150 mg/l

10=750 kg/d.

Note: After primary treatment 30% of BOD is removed, so only 70% BOD enters the trickling filter.

Assume organic loading = 1500 kg/ha-m/day. (range is 900 to 2200 kg/ha-m/d)

Volume of filter media = Total BOD of sewage / Organic loading rate

= 750kg/d /1500kg/ha-m/d

= 0.5ha-m.

=5000m³

Assume effective depth of filter as 2 m.

Surface area of filter required = Volume of filter media / Effective depth of filter

Check for Hydraulic Loading (22 to 44 ML/ha/day).

Surface area required = Total sewage treated/day)/Hydraulic loading/day

5 MLD / 22 ML/ha/d = 0.2 ha

S.A =0.2 ha = 2000 m².

Efficiency of Filter (if required)

6. AHRTF of 15 m ϕ operated with primary effluent of 1500 m3/d, recirculated effluent flow 1000 m3/d. Calculate recirculation factor and hydraulic loading rate on filter. 

Solution:

In high rate trickling filter, rate of hydraulic loading ranges between 10 to 30 m³/d/m².

7. The sewage is flowing at 4.5 ML per day from a primary clarifier to a standard rate T.F. The 5-day BOD of influent is 160 mg/L. The value of adopted organic loading is 160 g/m³/d. Surface loading is 2000 l/m2/d. Determine volume of filter and depth. Calculate efficiency of filter unit.

Solution:

8. A single stage filter is designed for an organic loading of 1000 kg of BOD in raw sewage per hectare metre per day with a recirculation ratio of 1.2. This filter treats a flow of 4 MLD of raw sewage with a BOD of 220 mg/l. Using NRC formula, determine strength of effluent.

Solution:

Flow = 4 MLD

BOD = 220 mg/l.

Total BOD-raw sewage = Flow x BOD concentration

= (4 × 10⁶) (220 × 10^-6)

= 880 kg/d.

Total BOD-Raw Sewage/ Organic loading rate

880 kg/d / 1000kg/ha-m/d = 880 m³.

Assume 35% BOD removal in primary clarifier.

BOD of influent applied to filter = 0.65 x 880 = 572 kg.

Efficiency of filter, E = 100 /1+0.44√U

U - kg/m³/d.

Note: In case the previous design was two stage trickling filter, then what would be the BOD of effluent?

Total volume of filter remains the same.

Filter volume of single stage = Filter volume of two stage (two filters)

Recirculation ratio of each filter = 1.2.

R = 1.2

F = 1.754.

First Stage:

Second Stage:

9. Design a HRTF for treating domestic sewage with a BOD of 285 mg/l and average flow of 40 MLD. The effluent BOD desired is 10 mg/l. Use NRC equation.

Solution:

10. Determine the size of a high rate trickling filter for the following data:

(i) Sewage Flow = 4.5 MLD

(ii) Recirculator Ratio = 1.5

(iii) BOD of raw sewage = 250 mg/l

(iv) BOD removal in primary tank = 30%

(v) Final effluent BOD desired = 30 mg/l

Solution:

Quantity of sewage flowing into the filter per day = 4.5 MLD

BOD concentration in raw sewage = 250 mg/l

.. Total BOD present in raw sewage = 4.5 Ml x 250 mg/l

1125 kg.

BOD removed in primary tank = 30%.

BOD left in sewage entering per day in the filter unit = = (1125) × 0.7 = 787.5 kg.

BOD concentration desired in final effluent = 30 mg/l

.. Total BOD left in the effluent per day = 4.5 x 30 kg

BOD removed by the filter = 135 kg.

Efficiency of the filter = 787.5 - 135 = 652.5 kg.