SAMPLE CIVL 456 Exam #3 QUESTIONS

1. Suppose that A.E. Staley was considering the use of an innovative ‘Biofor’ system for treating their raw wastewater (average daily flow of 1.8 MGD; raw soluble starch (i.e., C₆H₁₂O₆) of 6800 mg/L; raw nitrogen (organic plus ammonia plus nitrate plus nitrite) nitrogen of 25 mg/L); raw total phosphate of 100 mg/L). Clearly this waste is going to pose a problem in terms of biological treatment, in that it is nutrient deficient. Estimate the daily pounds of ammonia-nitrogen and phosphate (as sodium phosphate) which would have to be added to this waste stream to insure a balanced nutrient availability. You can assume all other nutrients are sufficiently available such that no other nutrient addition is required.

• Based on a balanced stoichiometric oxidation (C₆H₁₂O₆ to CO₂) and reduction (O₂ to H₂O), one mole of glucose will produce 24 electrons, and will in turn require 6 moles of oxygen gas to consume these electrons. Therefore, the molar ration would be 6 M oxygen per 1 M glucose, or on a weight basis, 192 gms oxygen required to fully oxidize 180 grams glucose. Overall, the apparent COD (and since glucose is highly biodegradable, the COD will be about equal to the BOD) would be (192/180)x(6,800) = 7,250 mg/L
• If you then estimate that ~80% of this COD is going to be removed in the first anaerobic reactor presently used at Staley’s, and that this system will maintain a corresponding yield of perhaps 0.05, then the daily anaerobic biomass production rate would be (0.80)0.05x7250 = 270 mg cells/L. Following this anaerobic system, the remaining 20% of the raw COD would then likely be fully oxidized (optimistically!), and with an expected observed yield of (0.5/(1+0.05x10) [or about 0.33], this subsequent aerobic tank would produce biomass at about a rate of 0.33x0.20*7250 mg/L.
• As a result, the system’s expected overall biomass production will be about 358 + 360 mg/L biomass, or 718 mg/L, or which 100 mg/ L would be ‘N’ and about 21 mg/L would be ‘P.’
• With an influent ‘N’ value of only 25 mg/L, they will consequently need 100-25 mg/L, which at 1.8 MGD corresponds to a daily mass of 1.8x75x8.34 = 1,125 pounds as ‘N’ required for supplementation.
• With an influent ‘P’ value of 100 mg/L, and only a requirement of 21 mg/L, the raw waste should have enough influent ‘P.’

2. Suppose that A.E. Staley was considering the use of an innovative ‘Biofor’ system for treating their raw wastewater’s organic content, with an expected design loading of 5 kg COD per cubic meter of tank volume per day. Estimate the necessary tank volume required for this application.

• From Problem #1 we know the incoming waste has a COD of 7250 mg/L, which with 1.8 MGD of flow would then correspond to a daily mass of 1.8x8.34x7250 = 108,840 pounds COD.
• Converting this latter figure to kilograms, we obtain 108,840/2.2 = 49,470 kg
• Now we should divide this daily load in kg by the desired loading to obtain our necessary reactor volume, 49,470/5 = 9,900 cubic meters.

3. Suppose that the City of West Lafayette were to renovate their existing anaerobic sludge digestors, such that they were being loaded at a rate of 0.2 lbs VSS per day per cubic foot of tank volume (you should assume the system is a two-tank in series, high-rate operation, with this loading being based on their total volume). How big would these two tanks have to be (assuming equal volume per tank)? You should also assume the WWTP has an average daily flow of 7.8 MGD, and a solids concentration entering the digestors of 3% (NOTE: they are using a gravity belt thickener with polymer addition to preconcentrate the primary and secondary sludges).

• The West Lafayette plant’s daily sludge production can be estimated at about 7.8 tons, and with 75% being volatile (estimated!), the daily VSS would be about 5.9 tons, or 11,700 pounds.
• Dividing this latter load by the desired reactor loading, we obtain 58,500 cubic feet for the total system volume, or about 30,000 cubic feet each.
• If each reactor were 15 feet deep, they’d have to have diameters of about 50 feet.

4. Suppose that an alternative ‘aerobic’ digestor system were to be used at the latter application in lieu of anaerobic digestion. If this system were assumed to operate at a temperature of 20oC, and based on a desired compliance with the ‘503’ reg stipulations for PSRP of this sludge, what would the necessary VSS loading (in lbs per cubic foot per day) have to be for these conditions?

• The ‘503’ reg indicates that PSRP is obtained in an aerobic digestor with a 40 day SRT (which is also the HRT) at a temperature of 20oC.
• To solve this problem, you also would have to determine the incoming sludge flow.
• If it contains 30 gms TSS per liter (i.e., 3% TSS), it would have 30x0.75 = 22.5 gms VSS per liter.
• If the incoming VSS load is 11,700 pounds, this would equate to 11,700x454/22.5 = 236,100 liters of sludge per day, or 236,100/[3.78x7.48] = 8,350 cubic feet per day.
• In order to obtain an HRT of 40 days, the requisite tank volume would then have to be 40x8,350 = 334,000 cubic feet, which is a very large tank!

5) Suppose that a new suburban housing development in southern Florida were to employ a reed bed FWS treatment scheme for their 1.5 MGD wastewater flow. Approximately how many acres would this system require for conventional treatment?

• You’d have to make an assumption about the incoming strength of this wastewater, and a BOD of about 150 would likely be in the ballpark.
• Therefore, the daily BOD would be 1.5x8.34x150 = 1900 lbs BOD.
• From the book, we can obtain a desired loading of 25 lbs BOD per acre per day.
• By dividing the incoming load by this latter value (1900/25) we obtain a required land area of 76 acres.