Question

A BOD test is run with 150 mL of wastewater mixed with 150 mL of deionized...

A BOD test is run with 150 mL of wastewater mixed with 150 mL of deionized water. The initial DO = 8.5 mg/L, the DO after 5 days = 4.2 mg/L, and the DO ultimately levels off at 1.7 mg/L. Assume all nitrification has been inhibited and the temperature remains constant.

a) Calculate the CBOD5 (y5, mg/L)

b) Calculate the ultimate CBOD (Lo, mg/L)

c) Calculate the CBOD remaining after 5 days (L5, mg/L)

d) Calculate the deoxygenation rate constant (kL, day^-1)

PS: I think this is a Determonation of Sturation DO Concentration question. I am not sure 100%

0 0
Add a comment Improve this question Transcribed image text
Answer #1

CBOD = 8.5-4.2 = 4.3mg/L (because all nitrification has been inhibited and temperature is also constant).

Add a comment
Know the answer?
Add Answer to:
A BOD test is run with 150 mL of wastewater mixed with 150 mL of deionized...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 7. A BOD test is run using 100 mL of treated wastewater effluent mixed with 200...

    7. A BOD test is run using 100 mL of treated wastewater effluent mixed with 200 mL of dilution water (containing no BOD). The initial DO of the mix is 9.0 mg/L After 5 days, the DO is 4.0 mg/L. After a long period of time, the DO is 2.0 mg/L, and it no longer seems to be dropping. Assume nitrification has been inhibited so that only carbonaceous BOD is measured (a) What is the 5-day BOD of the wastewater?...

  • 4. A 20 mL wastewater sample was mixed with 280 mL deionized (DI) water in a...

    4. A 20 mL wastewater sample was mixed with 280 mL deionized (DI) water in a standard BOD bottle. The 5- day BOD test provided an initial dissolved oxygen concentration (DO intial) = 9 mg/L and a final dissolved oxygen concentration (DO final) = 3.5 mg/L. The BOD experiment was carried out at 20°C and the BOD rate constant at 20°C is 0.22/day. a. Find the dilution factor (P) and BODs value at 20°CY b. Find the ultimate BOD of...

  • The East Hartford WWTP discharges 150 MGD of treated wastewater into the CT river. The treated was tested in the lab and found to have aBOD5 of 17.5 mg/L and an ultimate BOD (Lo) of 92 mg/L at 20oC....

    The East Hartford WWTP discharges 150 MGD of treated wastewater into the CT river. The treated was tested in the lab and found to have aBOD5 of 17.5 mg/L and an ultimate BOD (Lo) of 92 mg/L at 20oC. Find the reaction rate K, for the treated wastewater. The CT river has a flow of 250 m2/s and an ultimate BOD (Lo) of 6 mg/L. the DO of the river is 4.5 mg/L and the DO of the wastewater is0.2...

  • 2. To determine the BOD in an industrial wastewater sample, a seeded BOD analysis was conducted;...

    2. To determine the BOD in an industrial wastewater sample, a seeded BOD analysis was conducted; data are summarized in the table below. Ten mL of wastewater was added per 300 mL bottle to determine the dissolved oxygen demand of the aged, settled wastewater seed (test A). The seeded test bottles (test B) contained 2.5 mL of industrial wastewater and 1.2 mL of seed wastewater a) What is the five-day BOD in this industrial wastewater? What is the k-value using...

  • Problem 4 (10 Points). You collect a sample of wastewater and run a BOD test. To...

    Problem 4 (10 Points). You collect a sample of wastewater and run a BOD test. To run this test, you mix 25 mL of the wastewater sample with clean (unseeded) dilution water to bring the total volume in the bottle to 300 ml, After measuring the initial dissolved oxygen concentration, you cap the BOD bottle and monitor the dissolved oxygen concentration of the diluted sample in the BOD bottle as a function of time (see Figure below). Previous tests have...

  • O The town of Gangnam in Seoul, Korea discharges 17360 m3/day of treated wastewater in the Han River. The treated w...

    O The town of Gangnam in Seoul, Korea discharges 17360 m3/day of treated wastewater in the Han River. The treated wastewater has a BOD5 of 12 mg/L and a BOD decay constant, k, of 0.12 day at 20 °C. The river has a flow rate of 0.43 m/s and an ultimate BOD, Lo, of 5 mg/L. The DO of the river is 6.5 mg/L and the D0 of the wastewater is 1.0 mg/L (a) Calculate the ultimate BOD of the...

  • 0. What is the BODs of the wastewater sample if the DO values for the blank and diluted samples a...

    0. What is the BODs of the wastewater sample if the DO values for the blank and diluted samples afher 5 days are R1 mg/L and 3.5 mg/L, respectively, and the wastewater is diluted from 2 miL to 200 milL? (10 points) If the 20 day BOD (assume that this is the ultimate BOD) is 700.0 mg/L determine the BOD rate constant k (in base e). (10 points) 0. What is the BODs of the wastewater sample if the DO...

  • (20)1. A wastewater has a BOD5 of 192 mg/L, a reaction rate k of 0.12/day, and...

    (20)1. A wastewater has a BOD5 of 192 mg/L, a reaction rate k of 0.12/day, and a total Kjeldahl nitrogen content (TKN) of 35 mg/L. (a) (b) Find the ultimate carbonaceous oxygen demand (CBOD). Find the ultimate nitrogenous oxygen demand (NBOD). Find the total ultimate (or theoretical) oxygen demand (TBOD). Find the remaining BOD (nitrogenous plus carbonaceous) after five days have elapsed.

  • Seawater+Wastewater mixing (Water pollution)

    Example 7.15: According to the following information, determine whether a fish, which requires 6 \(\mathrm{mg} / \mathrm{L}\) dissolved oxygen, would survive in the river.Given:Waste water flow rate \(\left(\mathrm{Q}_{w}\right)=1.2 \mathrm{~m}^{3} / \mathrm{s}\)River flow rate \(\left(\mathrm{Q}_{r}\right)=10 \mathrm{~m}^{3} / \mathrm{s}\)River temperature \(\left(T_{r}\right)=15^{\circ} \mathrm{C}\)River ultimate \(\mathrm{BOD}\left(\mathrm{BOD}_{r}\right)=2.3 \mathrm{mg} / 1\)River reaeration rate \(=0.4\) per day for BOD testBOD concentration was initially \(9 \mathrm{mg} / \mathrm{l}\), after 5 days was \(6 \mathrm{mg} / 1\) and after an indefinitely long period of time was \(2.5 \mathrm{mg} /...

  • Two samples are brought to the laboratory for the BOD test. Please calculate BOD5 for both,...

    Two samples are brought to the laboratory for the BOD test. Please calculate BOD5 for both, using laboratory DO data. a. For sample A, seeded BOD test was performed. The initial DO concentration in the BOD bottle, which was filled with 2 mL of sample and 298 mL of seeded dilution water, was 8.9 mg/L just after preparation. After an incubation period of five days, the final DO concentration was reported as 1.8 mg/L. Meanwhile, DO concentration in the seed...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT