9-14 If the BOD, values for two livestock wastes having k values of 0.3800 day and 0.240 day- are 16230.0 mg L-', what would be the ultimate BOD for each?
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- (a) Calculate the ultimate BOD of a waste that has a measured 5-day BOD of 20 mg/L, assuming a BOD rate coefficient of 0.15/day measured at 20 degrees C. (b) Estimate the rate coefficient if the temperature of the waste is raised to 30 degrees C. (c) What would be the BOD5 at 30 deg. C?If the BOD3 of waste is 75 mg/L and the K is 0.150 d-1 , what is the ultimate BOD?If the BOD4 of a waste is 200.0 mg/L and the ultimate BOD is 270.0 mg/L, what is the rate constant at 10⁰C.
- Please Solve the Question fully ! If the BOD3 of a waste is 75 mg/L and the K is 0.150 d^-1 , what is the ultimate BOD? (Report your answer to two significant figures).If the five-day BOD of a waste is 220mg/L and the ultimate BOD is 320mg/L, what is the rate constant?A.0.142/day B. 0.178/day C. 0.210/day D. 0.233/dayA test bottle containing just seeded dilution water has its DO level drop by 1.0 mg/L in a five-day test. A 300-ml BOD bottle just filled with 15 ml of wastewater and the rest seeded dilution water (sometimes expressed as a dilution of 1:20) experiences a drop of 7.2 mg/L in the same period. What would be the five day BOD of the waste?
- If the BOD4 of a waste is 135 mg/L and K is 0.075 day^-1, the 5-day BOD (BOD5) and ultimate BOD (BODu or L0) of this waste are nearly? Use equations K = (2.303)K relationship, if necessary.A BOD test is run using 100 ml of treated waste H20 mixed with 200ml of pure water. The initial DO of the mixture is 9.0 mg/ml. After 5 days, Do id 4.0 mg/ml. What is the 5-day BOD of the waste water?A PFR is designed to treat incoming waste water containing 7.33 mM of H2S at a flow rate of 100 liters/min. The observed removal rate constant is 0.003 L/mg/hr. 90% removal is required. What is the order of the observed reaction rate? What is the necessary retention time? What is the necessay reaction volume in m3?
- Qy Waste water has a 5 - day BoD of 600 ppm Cingle) and the k value a 2006 is 0.2 per day. Find the Ultimate BoD of the waste. What would be the 5 day. BOD if the value of t dropped to al per day?7-5. Assuming that the quantities of solid wastes generated daily at a commercial facility are distributed normally, with a mean value of 10 yd3 and a standard deviation of 7 yd3, what size container would you recommend for this facility? What are the important tradeoffs in the selection of container size?Working for the local wastewater treatment plant, you are asked by the labtechnician to help her set up BOD5 tests. The issue here is what dilution to use to get the BOD5 test to give an acceptable result. Of course, you do not know the actual BOD5 value, but you are able to determine from past tests that the total ultimate BOD (NBOD + L0) of the wastewater is546 mg/L and the organic Nitrogen (TKN) concentration of the wastewater is 32 mg N/L. You also know that the first order rate constant for the BOD test is 0.25 day-1. When you ask, the technician tells you that the normal DO concentration at the start of the test is 8.3 mg/L.Based on all this information, tell the lab technician which of the following wastewater sample volumes added to a standard BOD test bottle (300 mL) is going to produce an acceptable final result: 1 mL/3 mL/9 mL/30 mL/90 mL NOTE: Acceptable BOD test should comply with the two essential criteria: DOfinal ≥ 2 mg/L and DOinitial – DOfinal ≥ 2 mg/L.