Reference no: EM13297370

1. You add 100 mg/L of glucose to pure water, then decide that this was a mistake and decide to remove it biochemically before you serve the water to your friends at a cocktail party. To remove the glucose, you seed the water with a harmless bacterial culture. No problem. You purchase the seed culture for a BOD test and then run the test, adding nitrogen and phosphorus for bacterial growth.
Assume that all the glucose is biodegradable. To get things going, you measure the initial rate of O2 utilization in the 100 mg/L glucose solution in a BOD bottle seeded in exactly the same way as a BOD measurement. In 5 hours, 2 mg/L of O2 is gone (initial rate only because you will soon run out of dissolved oxygen).
a. Find the ultimate BOD, k (the first order rate constant for BOD removal) and the 5-day BOD of the water in the bottle. Note, there is no dilution in this test so you will have to rely on the initial rate data provided above.
b. You are exceptionally clever and find a way to recreate the bacterial and culture conditions from the BOD test in a CSTR under steady conditions. That is, Cin = 100 mg/L glucose, cell number, nutrient concentrations and so forth are unchanged. If the flow rate is 1 MGD, how large should the CSTR be to provide an effluent concentration of 1 mg/L glucose? How long would it take in a PFR to achieve the same degree of treatment under the same conditions?
2. A wastewater has a BOD5 of 200 mg/L at 25oC, and under these conditions, k = 0.25 day-1. Find,
a. the ultimate BOD
b. the BOD5 at 20oC.

3. A wastewater treatment plant discharges 0.5 m3/s of effluent with Lo = 80 mg/L into a stream flowing at 10.5 m3/s. Just above the discharge point, L in the stream is 3.0 mg/L. Assume that kd = 0.22 day-1.
a. What is the ultimate BOD of the completely mixed wastewater/water mixture immediately downstream from the outfall?
b. If the cross-sectional area of the stream is 55 m2 (a constant), find the ultimate BOD at a point 10,000 m downstream.
c. At some points, the stream width is reduced by half, but the depth is unchanged. Would this tend to reduce or to increase the ultimate BOD at the 10,000m point? How about the dissolved O2 level at that point? Explain.