Reference no: EM13844721

1. Provide brief answers to the following:

a. What is tertiary treatment? Give two examples of tertiary treatment techniques.

b. What is a maturation pond and what is its primary purpose in wastewatertreatment?

c. What are the advantages and disadvantages to the use of decentralized wastewatertreatment?

d. Describe the five steps typically used in SBR treatment.

e. What is the purpose of an equalization tank? Where in the treatment process/trainwould it typically be placed and why?

f. What are the major nutrients required for growth of the microbial population inwastewater plants?

g. What is the difference between autotrophic and heterotrophic bacteria? What arethe main purposes of each in wastewater treatment?

h. Explain the purpose of the F/M ratio. How is F and M defined in terms of BOD andmixed liquor volatile suspended solids?

i. Define SVI and explain its use in the design and operation of an activated sludgeplant.

j. In a wastewater treatment plant having complete treatment facilities, what are thesources of sludge generation? How are these sludges handled and disposed?

2.

a. A Wastewater treatment facility being designed for a regional town requiresan equalisation basin to even out flow and BOD variations. The average dailyflow is approximately 0.307 m^3/s. The following flows and BOD5 have beenfound to be typical of the average variation over a day. What volumeequalisation basin is required to provide for a uniform outflow equal to theaverage daily flow? Assume the flows are hourly averages and that anaddition of 25% to the estimated volume will be provided to account forcontingencies.

TIME	Flow, m3/s	BODs, mg/L	TIME	Flow, m3/s	BODs, mg/L
00.00	0.275	150	12.00	0.425	220
01.00	0.221	115	13.00	0.405	210
02.00	0.164	75	14.00	0.385	200
03.00	0.130	50	15.00	0.352	190
04.00	0.105	45	16.00	0.326	180
05.00	0.099	60	17.00	0.326	170
06.00	0.119	90	18.00	0.328	175
07.00	0.204	130	19.00	0.365	210
08.00	0.354	175	20.00	0.399	280
09.00	0.411	200	21.00	0.399	305
10.00	0.425	220	22.00	0.379	245
11.00	0.430	220	23.00	0.345	180

b. Compute and plot the unequalised and the equalised hourly hydraulicloadings to the wastewater treatment facility for the maximum day. Assumethat the ratio of the maximum day flow rate to the average day flow rate is1.8. Determine the following ratios for the hydraulic loading: peak/average,minimum / average, peak/minimum. (Hint: use Excel spread sheet)

c. Compute and plot the unequalised and the equalised hourly BOD5 massloadings to the wastewater treatment facility for an average day. Determinethe following ratios for BOD5 mass loading: peak/average, minimum/average,peak/minimum.

3.

a. Wastewater from a winery contains approximately 1.8 g/L glucose (C₆H₁₂O₆), 1.8 g/L fructose (C₆H₁₂O₆) and 250 mg/L acetic acid (CH₃COOH).

What wouldbe the BOD5 and ultimate BOD assuming the first order rate constant is 0.23d^-1 (base e)?

b. Determine the number of completely mixed chlorine contact chambers eachhaving a detention time of 30 min that would be required in a seriesarrangement to reduce the bacterial count of a treated effluent from 10^6 to14.5 organisms/mL if the first order removal rate constant is equal to 6.1 h^-1.

If a plug flow chlorine contact chamber were used with the same detentiontime as the series of completely mixed chambers, what would the bacterialcount be after treatment?

c. A town of 25,000 inhabitants is to discharge treated domestic sewage to the neighbouring stream which has a minimum flow of 0.25 m³/s and a BOD5 of 3mg/L. Wastewater flow and BOD5 contribution from the town are 175L/(person.day), and 60 g /(person.day) respectively.

If the BOD5 in the streambelow the sewage outfall should not exceed 4 mg/L, determine thepercentage removal of BOD5 from the town wastewater before beingdischarged into the stream.

4. A completely mixed activated sludge plant is to be designed for a population of 85,000people. A treatability study was made using a pilot plant and the following data wasobtained:

• Influent flow = 360 L/(p.d)
• Influent BOD5 = 205 mg/L
• 34% of the influent BOD5 is removed by the primary clarifier
• μmax = 2.8 d^-1.
• Ks = 65 mg/L
• Y = 0.62 mg MLVSS/mg BOD5
• kd = 0.05 d^-1
• MLSS = 4000 mg/L
• MLVSS is 72% of the MLSS
• Effluent suspended solids have a BOD5 of 0.5 mg per mg of SS.
• Discharge license is 10 mg/L BOD5 and 15 mg/L SS.

Determine:

a. The required sludge age
b. The volume of the aeration basin
c. The flow rate of waste sludge
d. The flow rate of returned sludge
e. The food to microorganism ratio

5. A coastal community consists of a 10,000 population that increases to 30,000 in the summermonths, plus a frozen seafood processing plant. The local beach is the main reason forsummer tourism to the town.

a. What would be the key water quality parameters that would need to bemonitored to ensure human and environmental health objectives can be metif cliff-side discharge is utilized.Provide reasoning for your answer.

b. Sketch a flow diagram for a treatment system that would produce a suitablewater quality for cliff-side discharge. Explain the purpose for each step andwhy you would need it in this case.

c. The community is considering the provision of recycled water. Two optionshave been proposed, one for irrigating green spaces and one for providingnon-potable water to industry. What extra treatment would you recommendin each case? Which option would you recommend to the local authoritiesand why?