Environmental Engineering I Jagadish Torlapati, PhD Fall 2017 MODULE 3 WASTEWATER TREATMENT CONTROL PARAMETERS QS = VX F M

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1 F:M Ratio MODULE 3 WASTEWATER TREATMENT CONTROL PARAMETERS One commonly used parameter in the activated sludge process is the food to microorganism ratio (F/M or F:M) Food (F) Organic matter in the wastewater is called Food or Substrate (S). We use indirect measure of organic matter as oxygen demand to measure the amount of organic matter that is in the wastewater. Typically, we use 5-day BOD (BOD 5 ) as a measure of food in the wastewater since microorganisms are used to remove the dissolved BOD. Microorganism (M) This is the bacteria presented in the secondary treatment tank. The bacteria are seeded once, and the population dynamics are controlled by the process of wasting and recycling. The concentration of bacteria in the tank is measured using the solids experiment. The amount of volatile suspended solids (VSS) measured in the experiment are designed as the concentration of microorganisms. Typically, the microorganisms are referred to as Mixed Liquor Volatile Suspended Solids (MLVSS) F M QS = VX o If the F/M is high, what does that imply? If the F/M is low, what does that imply? What happens to the activated sludge process if the mean cell retention time (θ c ) is high? 1

2 Problem 1 Compute the F/M ratio for the new activated sludge plant where the flow rate is 0.15 m 3 /s, initial BOD is 84 mg/l. The volume of the tank is 970 m 3 and assume that the concentration of MLVSS is 2000 mg/l. Problem 2 How many lbs of MLVSS should be maintained in an aeration tank with a volume of MG receiving primary effluent BOD of 603 lbs/d? The desired F:M ratio is 0.3. Also find the concentration of MLVSS in mg/l 2

3 Problem 3 How many lbs of MLVSS should be maintained in an aeration tank with a volume of MG receiving primary effluent BOD of 2502 lbs/d? The desired F:M is 0.3. Find the concentration of MLVSS in mg/l Yield Coefficient (Y) This parameter tells us how many lbs of cells are produced for a lb of BOD removed. This is also called growth rate. At the facility, the yield coefficient is measured by diving the average solids wasted with the average pounds of BOD removed. Problem 4 The influent and effluent BOD concentrations in an aeration tank are 120 and 5 mg/l respectively. The influent flowrate is 2 MGD (a) What is the BOD removed in mg/l? (b) What is the amount of BOD removed in lbs/day? (c) If the yield coefficient is 0.7, what is the amount of biomass produced? 3

4 Problem 5 How would the amount of biomass produced change in the Problem 4, if the Y value is 0.5? What is the difference in biomass produced per day in both problems? What does this tell you about the Y value? Sludge Production The observed yield is given by Y Yobs = 1+ kdqc And the net activated sludge produced each day is given by -3 Px = YobsQ( So - S)(10 kg/g) Problem 6 Estimate the net activated sludge produced each day from a new activated sludge plant whose flowrate is 0.15 m 3 /s, yield coefficient is 0.5 kg VSS/kg BOD5 removed, the decay constant is 0.05 d -1 and the detention time is 5 days. The influent BOD is 84 mg/l and the effluent BOD is 11.1 mg/l 4

5 Oxygen Demand Oxygen is used in those reactions required to degrade the substrate to produce the high energy compounds required for cell synthesis. The mass of oxygen required may be estimated as ( - S) -3 Q So (10 kg/g) MO = P 2 x f Where Q is the wastewater flowrate into the aeration tank (m 3 /d), So in the influent soluble BOD 5 (mg/l), S is the effluent soluble BOD 5 (mg/l), f is the conversion factor for converting BOD 5 to ultimate BOD L and P x is the waste activated sludge produced. Problem 7 Estimate the volume of air to be supplied (m3/d) for the new activated-sludge plant at Gatesville. Assume that BOD5 is 68 percent of the ultimate BOD and the oxygen transfer efficiency is 8 %. 5