Question

A river with a flow rate of 7 m3/s discharges into a lake with a volume of 9,000,000 m3. The concentration of a particular VOC in the river is 3 mg/L. The lake has a decay coefficient for the VOC of 0.3 per day. What is the concentration of the VOC in the river downstream?

Answer 1

Answer:

The concentration downstream reduces to 0.03463mg/L

Explanation:

Initially let us calculate the time it is required to fill the lake, since for that period of time the pollutant shall remain in the lake before being flushed out.

Thus the detention period is calculated as

$t=\frac{V}{Q}\\\\t=\frac{9000000}{7}=1.285\times 10^{6}seconds\\\\\therefore t = 14.872days$

Now the concentration of the pollutant after 14.872 days is calculated as

$N_{t}=N_{0}e^{-kt}$

where

$N_{o}$ is the initial concentration

't' is the time elapsed after which the remaining concentration is calculated

k is the dissociation constant.

Applying values we get

$N_{t}=3\times e^{-0.3\times 14.872}$

$N_{t}=0.03463mg/L$