Five million gallons per day (MGD) of wastewater, with a concentration od 10.0 m

Question

Five million gallons per day (MGD) of wastewater, with a concentration od 10.0 mg/ of a conservative pollutant, is released into a stream having an upstream flow of 10 MGD and pollutant concentration of 2.0 mg/L.
a. What's the concentration in ppm just downstream?

A lake receives water from a stream, a paper mill and a wastewater treatment plant (WWTP) and all three inputs contain the CONSERVATIVE pollutant "B". One river leaves the lake. The flow rates for these inputs are 0.2 L/min (incoming stream), 0.15 L/min (paper mill) and 0.5 L/min (WWTP). The concentrations of "B" entering the lake are 11.2 g/L (incoming stream), 302 ng/L (paper mill) and 507 ng/L (WWTP). Assume i) STEADY STATE conditions and ii) that the flow rate of the river leaving the lake is the sum of the three inputs of water entering the lake. Be careful about units a) What is the concentration (Hg/L) of "B" in the river leaving the lake? (3 points) b) Is this below the maximum contaminant level (MCL) for this chemical of 0.002 ppb? (1 point)

Explanation / Answer

Here two streams of different MGD's are mixed with different concentrations into a stream.

So now the net conentration of pollutant can be calculated by

((5x10) + (10x 2))/ (5+10) = 70/15 = 4.66 mg/L

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