Does value meet the eas quality requirements

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Reference no: EM131377472

Question: The village of Pongelton is in the valley of the River Pong. The mean flow of the river is 3.0 m3/s. The population of Pongelton is 6,000. The daily average water consumption is 250 litre / head / day. Water utilisation (i.e. the water spent for watering gardens, watering animals, etc) is 20%. Sewers are built in the village and the construction of a sewage treatment works is planned.

The local government also wishes to construct a small recreational reservoir downstream of the site of the planned effluent discharge. The Environment Agency requires an assurance that the aquatic environment of the planned reservoir will not be impaired by the discharge of the final effluent from the sewage treatment works. Thus, the Environment Agency will require the sewage treatment works to be designed for phosphorus removal, in order to avoid eutrophication downstream of the discharge. To achieve this, the concentration of phosphorus in the inflow to the planned reservoir must be less than 40 mg/m3. The average phosphorus concentration of household sewage is 10 mg/l and the phosphorus concentration of the unpolluted river, upstream of Pongelton is 10 mg/m3.

You are the designer and must address the following issues:

1. If no treatment facility is provided, what will the phosphorus concentration of the river be downstream of the sewage discharge?

2. Does this value meet the EA's quality requirements?

3. If not, in percentage terms, what degree of treatment should be provided for phosphorus removal. Express your answer in percentage terms (i.e. percentage of phosphorus to be removed)"

Verified Expert

In the village of Pongelton, the valley of the River Pong, population of Pongelton is 6000. And the river flow at 3 m 3 /s. Here we have to find out the concentration of the phosphorus. Here we got the phosphorus concentration 37 mg/L which is slight more than that if the standards set by the environment agency. The extra required removal of phosphorus is 5.714 %.

Reference no: EM131377472

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inf1377472

2/3/2017 4:29:01 AM

I gave you question 2. Please, can you help me in question 1 as well?. It is same thin and there is clarification from my tutor for starting solution. use 250 L/head/day D_0=10.8-1.84=8.96 mg/L to solve this question, please read this question and take look at graph because this data from my tutor. Thanks.

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