Department of Civil Engineering-I.I.T. Delhi nd CEL 212: Environmental Engineering (2 Semester 2010-2011) Biological Processes Courtesy: Dr. Arvind K. Nema Example 1. Design of the aeration basis based on solids retention time. You are provided the following information about a municipal wastewater treatment plant. This plant will use the traditional activated sludge process. Population = 150,000 people, flow rate of 33.75x106 L/day (equals 225L/person/day) and influent BOD5 concentration of 444 mg/L (note this is high strength wastewater). Assume that the regulatory agency enforces an effluent standard of BOD5 = 20 mg/L and suspended solids standard of 20 mg/L in the treated wastewater. A wastewater sample is collected from the biological reactor and is found to contain a suspended solids concentration of 4,300 mg/L. The suspended solids concentration in the secondary sludge is 15,000 mg/L and the concentration in the secondary sludge is 5,000 mg/L. The concentration of suspended solids in the plant influent is 200 mg/L and that t hat which leaves the primary clarifier is 100 mg/L. The microorganisms in the activated sludge process can convert 100 grams of BOD5 into 55 grams of biomass. They have a maximum growth rate of 0.1/day, a first-order death rate constant of 0.05/day, and they reach ½ of their maximum growth rate when the BOD5 concentration is 10 mg/L. The mean cell retention time of the solids is 4 days and sludge is processed on the belt filter press every 5 days.
A. What is the design design volume of the aeration aeration basin? Solution: Assuming that 30% o f the plant influent BOD5 is removed during primary sedimentation, this means that So = 444 mg/L × 0.70 = 310 mg/L. Thus,
1/SRT = [(Qo × Y)/(V × X)) × (So-S)] – kd 6
1 / 4 days=[33.75x10 L/day (0.55 gm SS/gm BOD5) /V 4,300 mg SS/L (310 mg/L – 20 mg/L)] – 0.05/day Solve for V = 5,000,000 Liters B. What is the plant’s aeration aeration period? Solution: The plant’s aeration period is the number of hours that the wastewater is aerated during the activated sludge process. This equals the hydraulic detention time of the biological reactor.
HRT= V/Q = 5,000,000 L / 33.75 x 106 L/day L/day = 0.15 days = 3.6 hours C. How many kg of primary and secondary dry solids need to be processed daily from the treatment plant? Solution: The amount of solids processed from the primary sedimentation tanks equals the difference in suspended solids concentrations measured across the sedimentation tanks multiplied by the plant flow rate. 6
33.75x10 L/day (200 mg SS/L – 100 mgSS/L) × kg/1,00,000 mg = 2,275 kg primary solids per day
We have not provided with the concentration difference of suspended solids across the secondary sedimentation tanks so we can determine the amount of secondary solids produced daily in the same manner that we used for primary solids. However, careful examination of the expression of solids retention time shows that that the term QwXw equals the answer. 4 days = V × X / Qw × Xw = 5,000,000 L (4,300 mg SS/L) / Qw × Xw Solve for QwXw which equals 5,400 kg secondary dry solids per day. D. Determine the F/M ratio (in units of lbs BOD5/lb MLSS-day) using data provided in the above example problem. Solution.
By definition, F/M = Q So / X V = [33.75x106 L/day x 310 mg/L) / [4,300 mgSS/L × 5,000,000 L] = 0.48 lbs BOD5/lb MLSS-day Various reactor configurations are available, each with its own set of advantages and disadvantages. The two basic types are plug flow (PF) and completely mixed flow (CMF) reactors. PF reactors offer a higher treatment efficiency than CMF reactors, but are less able to handle spikes in the BOD load. Other modifications of the process are based on the manner in which waste and oxygen are introduced to the system. ++++++++++
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Problem 1: Design a complete mix activated sludge process to treat 0.25 m /s of wastewater with BOD5 of 250 mg/L. The effluent is to have BOD 5 of 20mg/L or less. Assume the o temperature is 20 C and the following conditions are applicable. - The influent and effluent microorganism concentrations are negligible. Food and microorganisms are completely mixed in the aeration basin. Wastewater contains adequate nitrogen, phosphorus and other trace nutrients for biological growth. - Ratio of MLVSS to MLSS is 0.8. (MLSS and MLVSS: Mixed liquor suspended solids (represents total solids) and mixed liquor volatile suspended solids (represents biological solids). - MLVSS concentration in the reactor = 3500 mg/L; Return sludge concentration = 10,000 mg/L; - Design mean cell residence time (θ C) is 10 days - Effluent contains 22 mg/L of biosolids, of which 65% is biodegradable -1 -1 - Kinetic coefficients: Ks = 50 mg/L; µ m = 5.0 d ; k d = 0.06 d and Y = 0.50 Suggested Steps: 1. Estimate the concentration of soluble BOD5 in the effluent (eff) BOD5 in the eff = Soluble BOD 5 in the eff + BOD 5 in the eff suspended solids BOD5 = 0.68 x BOD L 2. Determine the treatment efficiency 3. Compute the reactor volume
S =
K s (1 + k d θ c )
θ c ( µ m
−
X =
k d ) − 1
θ cY (S o
−
S )
θ =
θ (1 + k d θ c )
V Q
θ c =
VX Qw X r
S = BOD concentration in activated sludge (S = BOD 5 allowed – BOD 5 in SS) [mg/L] X = microorganism concentration in activated sludge [mg/L of MLVSS] X r = microorganism concentration in recycle [mg/L of VSS] S O = influent BOD [mg/L] θ c = mean cell resident time in the aeration tank [d] θ = hydraulic detention time [d] 3 V = aeration tank volume [m ] 3 Q = flow rate [m /d] 3 Qw = flow rate of waste sludge [m /d] -1 µ m = maximum specific substrate utilization rate [d ] K S = half-maximum rate concentration [mg/L] -1
k d = endogenous-decay rate coefficient [d ] Y = yield coefficient [mg/L MLVSS/mg/L] 4. Food to Microorganism Ratio: To keep the microorganisms efficient, the Food to Microorganism Ratio (F/M) must be keep low (around 0.10 to 1.0 mg/L-d). F = QS o To achieve a low F/M ratio, use a low sludge wasting rate ( Qw) creating a M VX
long cell detention time (θ c) 5. Waste Sludge Production: Excess sludge is produced during the activated sludge process that must be treated and disposed of. To estimate the excess sludge production, use the following equation: Y P x = Q(S o − S ) 10 − 3 kg g 1 + k d θ c
(
)
P x = waste activated sludge produced [kg VSS/d] 6. Oxygen Requirements: Activated sludge uses large volumes of oxygen in the production of sludge and the consumption of BOD. However, oxygen is produced during cell formation by moving from right to left per Equation 5-44: C5H7O2N + 5O2 <-----> 5CO2 + 2H2O + NH3 + energy The ratio of oxygen usage to cell formation is 5(32)/113 = 1.42. Subtracting cell formation from the oxygen consumed in the reduction of BOD ( S O-S ), the oxygen requirements of activated sludge can be estimated: Q(S o − S )(10 3 kg g ) − 1.42 P x ; f = conversion factor to convert BOD 5 to BOD L. M O2 = f Calculate air requirement −
