University Lake is a reservoir with a surface area of 8 acres and an average dep

Question

University Lake is a reservoir with a surface area of 8 acres and an average depth of 10 feet. The volume of the lakes is 222.2 acre-ft. University Lake receives water from stream inflow at a rate of 302 acre-ft/year with phosphorus (P) concentration of 0.045 mg P/L. Direct precipitation to the lake totals 80 cm/year and has a phosphorus (P) concentration of 0.011 mg/L. Estimated lake evaporation is 1888 m3/year. The stream outflow from the lake is 297 acre-ft/year. You can assume that the evaporation is pure water and carries no other chemicals with it. (There are 1233.5 m3 in 1 acre-ft) Show all of your work! (Hand write calculations and submit assignment in class. )

1) What is the concentration of phosphorus in the lake? (report in µg/L, assume that the concentration associated with the out flow equals the concentration in the lake)

2) How would increasing the stream inflow by 30% affect the concentration in the lake water? (repeat number 4 but multiply the inflow by 1.3)

3) How would increasing the outflow by 25% affect the concentration of the lake? (repeat number 4 but multiply the outflow by 1.25)

Explanation / Answer

1)

a. Phosphorus (P) concentration due to inflow = 302 acre-ft/year = 302 x1.233x106 litre/year(1 acre-ft = 1.233x106) = 372.366 x106 litre/year

phosphorus (P) concentration in 372.366 x106 litre = of 0.045 mg P/L x 372.366 x106 litre = 16.756 x106 mg

phosphorus (P) concentration in 372.366 x106 litre = 16.756 x103 kilogram

b. Phosphorus (P) concentration due to Direct precipitation to the lake:

Direct precipitation to the lake in an year =  80 cm/year = 0.0328084 x 80 ft/year = 2.62467 ft/year (1cm = 0.0328084 feets)

Surface area of lake = 8 acres

Volume of water due to Direct precipitation to the lake in an year = 2.62467 x  8 acres = 20.99736 acre.ft/ year

Volume of water due to Direct precipitation to the lake in an year = = 20.99736 x 1.233x106 = 25.89x106 litre

phosphorus (P) concentration = 25.89x106 litre x 0.011 mg/L = 284787mg or 284.787kg

concentration of phosphorus in the lake = (16.756 x103 + 284.787kg)/(372.366 x106 litre + 25.89x106 litre)

concentration of phosphorus in the lake = 4.2788 x 10-5 kg/ litre or 42.788 µg/L

2)

Phosphorus (P) concentration due to 30% increase in inflow = (130/100) x 16.756 x103 kilogram

Phosphorus (P) concentration due to 30% increase in inflow = 21.7828 x103 kilogram

New concentration of phosphorus in the lake = (21.7828 x103 + 284.787kg)/(1.3 x 372.366 x106 litre + 25.89x106 litre)

4.327268 x 10-5 kg/ litre or 43.272 µg/L

New concentration of phosphorus in the lake =43.272 µg/L while concentration of phosphorus in the lake was 42.788 µg/L.

3)

Phosphorus (P) concentration due to 25% increase in inflow = (125/100) x 16.756 x103 kilogram

Phosphorus (P) concentration due to 25% increase in inflow = 20.945 x103 kilogram

New concentration of phosphorus in the lake = (20.945 x103 + 284.787kg)/( 1.25 x 372.366 x106 litre + 25.89x106 litre)

4.320727 x 10-5 kg/ litre or 43.2072 µg/L

New concentration of phosphorus in the lake =43.2072 µg/L while concentration of phosphorus in the lake was 42.788 µg/L.

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