What are Trickling Filters?
Trickling filter is an attached growth process, i.e., a process in which microorganisms responsible for treatment are attached to an inert (non-reactive) packing material such as sand. Packing material used includes rock, gravel, slag, sand, redwood, and a wide range of plastic and other synthetic materials.
How do they work?
1: Process Description
The wastewater in trickling filter is distributed over the top area of the filter containing non-submerged packing material. Air circulation in the void space, by either natural draft or blowers, provides oxygen for the microorganisms growing as an attached bio-film. During operation, the organic componentsin the wastewater is metabolised by the biomass attached to the medium. The biological mucus thickens as the organic matter removed from the flowing sewage is synthesized into new cellular material.
The thickness of the aerobic layer is limited by the depth of oxygen penetration depth of the microbial layer. As a result of the increased thickness of the mucus layer, the microorganisms near the surface of the medium enter the endogenous phase when the substrate is metabolized before the substrate reaches the microorganisms near the surface of the medium and loses its ability to adhere to the surface of the medium.
The liquid then flushes the mucus out of the medium and a new layer of mucus begins to grow. This phenomenon of mucus peeling is called shedding. The stripped film and washed wastewater are collected by the sub-drain, which also allows air to circulate through the filter. The recovered liquid is sent to the settling tank used for solid-liquid separation.
B: Filter type
The hydraulic load factor is the total flow rate including the recirculation applied to the unit area of ??the filter in one day, while the organic load factor is 5 at 20 ° C excluding the recirculation BOD applied per unit volume per day.
Recirculation is generally not used with slow filters. A properly operated slow trickling filter in combination with a secondary purification device, can remove 75-90% BOD and produce highly cleaned wastewater. It is suitable for treating low to medium load domestic wastewater.
One-stage or two-stage high-speed sprinkler beds are recommended for moderate to relatively heavy domestic and industrial wastewater. The BOD removal efficiency is about 75-90%, but the wastewater is only partially nitrified. The one-stage system consists of a primary purification device, a filter, a secondary purification device, and equipment for recycling wastewater.
The two-stage filter consists of a primary clarifier, an intermediate clarifier that can be omitted in some cases, and two filters connected in series with the final clarifier. Trickling filters are classified as high rate or low rate, on the basis of organic and hydraulic loading applied to the unit.
How are Trickling Filters designed?
Trickli?ng filter design is based on empirical relationships to find the required filter volume.
Types of empirical relationships:
· NRC equations (National Research Council of USA)
· Rankine’s equation
· Eckenfelder equation
· Galler and Gotaas equation
NRC and Rankine’s equations are commonly used.
NRC equations give satisfactory values when there is no recirculation, the seasonal variations in temperature are not large and fluctuations with high organic loading.
Rankine's equation is used for high rate filters. This equation has been successfully used over a wide range of temperatures.
For single-stage filters, the following conditions must be met:
1: The BOD of raw household sediment wastewater applied to the filter should not exceed 1.2 kg BOD-5/day/m3 filter capacity.
2: The hydraulic load (including circulation) should not exceed the filter area of ??30 m3/m2 per day.
3: The recirculation rate (R/Q) should ensure that the BOD (including recirculation) entering the filter does not exceed 3 times the expected BOD for wastewater.
This means that as long as the above conditions are met, efficiency is only a function of recirculation and is given by the following equation:
E = {(R / Q) + 1)}/ {(R / Q) + 1.5)}
Let’s understand it with the help of an example:
Q: Design a low rate filter to treat 6.0 MLD of sewage of BOD of 210 mg/l. The final effluent should be 30 mg/l and organic loading rate is 320 g/m3/d. The recirculation ratio is 1.50.
Solution:
Assume 30% of BOD load removed in primary sedimentation,
= 210 x 0.30 = 63 mg/l.
Left over BOD
= 210 - 63 = 147 mg/l.
Required Percent of BOD removed
= (147-30) x 100/147 = 80%
BOD load applied to the filter
= flow x conc. of sewage (kg/d)
= 6 x 106 x 147/106 = 882 kg/d
To find out filter volume, using Rankine’sequation,
E = {(R / Q) + 1)}/ {(R / Q) + 1.5)}
= (1.5+1)/ (1.5+1.5) = 0.833
Efficiency of the filter = 83.3%
E2 = 100/ {1+ 0.44 (Y / V*F)}
80 = 100 (Rf1= 1.5)
1+0.44(882/V1*1.5)1/2
V1= 1821 m3
Depth of filter = 1.5 m,Filter area = 1821/1.5 = 1214.25 m2, andDiameter = 39.3 m < 60 m
Hydraulic loading rate = 6 x 106/103 x 1/1214.25= 4.94 m3/d/m2 > 4, hence not o.k.
Let us increase the volume to 2500 m3
Then,
Depth of filter = 1.5 m, Filter area = 2500/1.5 = 1666.66m2, and Diameter = 46.6 m < 60 m
Hydraulic loading rate = 6 x 106/103 x 1/1666.66= 3.600 m3/d/m2 < 4, hence o.k.
Organic loading rate = 882 x 1000/1821 = 326.18 g/d/m3 which is approx. equal to 320.
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