Methane recovery by most of the microbial biomass produced in aerobic growth (biogas), can be used as alternate fuel source ( green solution ).
Reduces production of landfill gas, which when broken down aerobically releases methane into atmosphere (a powerful greenhouse gas).
Sludge occupies less volume, easier to dry.
Lower operating costs.
Odours/flies typically removed from system.
Disadvantages of anaerobic digesters:
Accumulation of heavy metals and contaminants in sludge.
Narrow temperature control range.
Installing and managing an interrelated group of systems to safely handle heating of the tank, hydrogen sulphide reduction, methane transfer, heat production, electrical production, inter connection with the electrical grid and surplus heat management.
Mixing can be accomplished with a variety of gas mixers, mechanical mixers, and draft tubes with mechanical mixers or simply recirculation pumps. Most municipal digesters are intensely mixed to reduce the natural stratification that occurs in a low profile tank.
Figure 19.9 shows the various types of mixing flow regimes used in anaerobic digesters: (a) gas recycle and draft tube (b) gas recycle and injection (c) impeller and draft tube and (d) impeller.
The most efficient mixing device in terms of power consumed per gallon mixed is the mechanical mixer.
The volume of digesting sludge in a digester is a function of the volume of fresh sludge added daily, the volume of digested sludge produced daily and required digestion time (days).
Empirical evidence (batch experiments) has shown that if supernatant is removed from a batch of digesting sludge as it is produced, volume of remaining digested sludge vs. digestion time is a parabolic function.
For a parabolic function : average volume = initial volume – (2/3)(final volume – initial volume) therefore,
where: V avg = average volume of digesting sludge (m 3 / day)
V 1 = volume of fresh sludge added daily (m 3 / day)
V 2 = volume of digested sludge produced daily (m 3 / day)
Anaerobic Digestion Wastewater Engineering (Fair et al) (19.6)
Digesters may also be designed based on organic loading (kg/m 3 day), or mean cell residence time , θ c ,(days)
The mean cell residence time is often referred to as ‘ solids retention time ’ ( θ s ).
where: θ c = mean cell residence time (days)
X = kilograms of dry solids in digester
Δ X = kilograms of dry solids produced per day in digester
Since no. of cells in feed is negligible compared to no. of cells in digester, the mean cell residence time is equal to the hydraulic retention time, and we assume the biological reactor is completely mixed therefore: θ c = θ h = θ s
A heated, low-rate anaerobic digester is to be designed for an activated sludge plant treating the wastewater from 25,000 persons. The fresh sludge has 0.11 kg dry solids/cap-day, volatile solids are 70% of the dry solids, the dry solids are 5% of the sludge, and the wet specific gravity is 1.01. Sixty-five percent of the volatile solids are destroyed by digestion, and the fixed solids remain unchanged. The digested sludge has 7% dry solids and a wet specific gravity of 1.03. The operating temperature is 35 °C and the sludge storage time is 45 days. The sludge occupies the lower half of the tank depth, and the supernatant liquor and the gas occupy the upper half of the tank depth. Determine the digester volume.
A heated high-rate anaerobic digester is to be designed for an activated sludge plant treating the wastewater from 25,000 persons. The feed to the digester (primary and secondary sludge) is 56.1 m 3 /day and the operating temperature is 35 °C. Determine the digester volume.
From Table 19.3, minimum cell residence time, θ c min , for 35°C is 4 days.
Therefore the design cell residence time, θ c = (2.5) θ c min = 2.5(4) = 10 days.
Assume: completely mixed biological reactor without recycle, the θ c = θ h .
As a result we can use, the volume of a high rate digester formula,
V = Q • θ h = (56.1 m 3 /day)(10 days) = 561 m 3 .
Sludge pumping records in various plants may be used to determine amount of primary and secondary sludge required in plant design, however if this data is not available we may use the following equation, developed by (Eckenfelder and Weston, 1956),
kg solids /day = YS r - k e Х
where: Y = yield coefficient, kg of TSS or VSS per kg of BOD 5 or COD removed per day (kg/kg)
S r = kg of BOD 5 or COD removed per day (kg/day)
k e = endogenous coefficient (day -1 )
X = mass of TSS or VSS in the aeration tank (kg)
Table 19.4 shows some typical values of sludge quantities and solids concentrations.
A low-rate trickling filter plant treats the wastewater from 5000 persons. Short-term analyses using composite samples of the influent wastewater show the BOD 5 and suspended solids concentrations to be those expected from average values for municipal wastewaters in the US. The plant gets 95% BOD 5 removal, 33% BOD 5 removal by the primary clarifier, and 65% suspended solids removal by the primary clarifier. Assume that the primary sludge has 4% dry solids, primary sludge specific gravity = 1.01, secondary sludge has 5% dry solids, the specific gravity = 1.02, and the biological solids produced is 0.35kg/kg BOD 5 removed. Determine the primary and secondary sludge produced per day.
The primary solids = (5000) × (0.091)(0.65) = 295.8 kg/day.
An activated sludge treatment plant has an anaerobic digester and serves a population of 25,000 people. The mixture of primary and secondary sludge amounts to 0.11 kg/cap-day, the fresh sludge has 4.5% solids on a dry basis, and the digester operating temperature is 35ºC. During the coldest month, January, the sludge temperature is 12.8ºC. The specific heat constant is 4200 J/kgºC. Determine the heat required to raise fresh sludge temperature to that of the digester.
The number of kilograms of fresh dry sludge solids added per day is (25,000 capita)(0.11 kg/cap-day) = 2750 kg/day.
The heat required will be,
Q s = (2750 kg/day)(100 kg/4.5 kg) (35 ºC – 12.8ºC)(day / 24 hours)(4200 J/kgºC) =
The anaerobic digester in Example 4 is 22.9 m in diameter, wall height of 9.14 m, and has a floating cover. The digester is recessed 0.61 m in the earth and has a freeboard of 0.91 m. For insulation, the earth is mounded around the digester up to a height of 4.57 m. A wall height of 9.14 m – (0.61 m + 0.91 m + 4.57 m) = 3.05 m is exposed to air on the outside and to digester contents on the inside. The average temperature in January is 7.2ºC. The digester operating temperature is 35ºC, and the dry earth mounded around digester has temperature = avg. monthly temperature = digester operating temperature. Below the digester, earth temperature is 35ºC and temperature of earth around wall is 35ºC. Determine heat required to make up losses from digester.
The area of the floating cover is ( π /4)(22.9 m) 2 = 411.9 m 2 .
The area of the wall where air is on the outside and digester contents are on the inside is ( π )(22.9 m)(4.57 m) = 328.8 m 2 .
The temperature of the dry earth mounded around the digester is ( 35ºC + 7.2ºC)/2 = 21.1ºC
An aerobic digester is to be designed for an activated sludge plant treating WW from 10,000 persons. The flowsheet is shown. The primary and secondary sludge are to be blended in a blended tank and then thickened and then sent to digester. Pertinent data: BOD 5 = 200 mg/L, influent SS = 250 mg/L, average flow = 380 L/cap-day, 33% BOD 5 removal and 62% SS removal by primary clarifier, primary sludge has 5% solids, sludge density index = 6000 mg/L, secondary sludge = 0.031 kg/cap-day, volatile solids for primary and secondary sludge = 70%, and the specific gravity of primary, secondary, blended and thickened sludge = 1.01. Determine:
1. Primary, secondary and blended sludge flow in m 3 /day.
2. Solids concentration in blended flow.
3. The thickened sludge flow if thickener increases solids by 2.5 times and supernatant liquor has negligible SS.
4. The supernatant liquor flow.
5. Digester volume if minimum operating temperature is 18 °C.
6. Oxygen requirements of digester if 90% of primary BOD 5 is destroyed and 65% of volatile solids in WAS are destroyed.
7. The air required if air has 0.281 kg O 2 /m 3 and field transfer is 4%. Is mixing adequate?
Aerobic Digestion Wastewater Engineering
Example 7 (diagram) Aerobic Digestion Wastewater Engineering