Question

In: Chemistry

I do not know how to solve this problem. A manufacture in Ontario has been fined...

I do not know how to solve this problem.

A manufacture in Ontario has been fined because it has been releasing 5 L/s effluent having
zinc concentration of 0.1 mg/L into a river. Upstream of the factory, the stream water rate is
100 L/s with zinc concentration of approximately zero. The factory has been forced to
reduce the zinc concentration of the effluent to below 20 μg/L. The engineer that is an
employee of the factory recommends that they divert parts of the stream into the site and
dilute the effluent to bring its zinc concentration to the required level.
a) Calculate the present concentration of zinc several miles downstream of the plant
where the zinc is diluted in stream completely.
b) Compute the amount of water needed to be diverted to the site to achieve the required
zinc concentration of effluent.
c) Determine the concentration of zinc downstream of the plant where the zinc is diluted
in stream completely if the engineer’s plan is put into operation.

Solutions

Expert Solution

a) When Zinc is diluted with in stream. Zinc entering from the plant= 5*0.1 mg/L= 0.5 mg/S

The upstream water is at 100 L/s and hence total flow rate of mixed stream is 100+5=105 L/s

The Zinc concentration after mixing several miles downstream of the plant= 0.5/105 mg/L=0.004762 mg/L (1 mg= 1000 ug) =4.762 ug/L

b) Let the water required to be added = V ( Zinc concentration=0)

Zinc concentration in the effluent= 5*0.1 =0.5 mg/S

The mixed stream volume = 5+V1

The zinc concentratino in the resulting stream need to be 20 ug/L =20*1000mg/L

1000*0.5/(5+V1)}= 20

0.5= 500/ (5+V1)=20

500= 100+20V1

400= 20V1

V1= 20 L/s

c) Once the water is diverteted 20L/ s is required to be mixed with effluents to get 20 L/S. Then balance water =80L/S

Volume of 20 ug/L stream =20+5=25 L/S

amount of zinc =25*20*1000 mg/S=500000mg/S

This is mixed with balance water having no zinc and its flow rate =80L/s

total flow rate= 105 L/S

and zinc concentration =500000/105=4761.905 mg/L= 4.761 ug/S


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