Question

I am unsure how to solve this question.

The flow rate and NaCl concentration of a river are 2 million gallon per day and 32 ppm. A
food processing plant produces and releases a saline solution at a rate of 85 gpm into the
stream. The salts quickly become uniformly distributed in the stream. Downstream of the
discharge point is a fishing spot. Because the fish cannot tolerate the salt concentrations
above 200 mg/L, the NaCl concentration of effluent must be limited. Calculate the
maximum NaCl concentration of effluent releasing into the stream in mg/L.

Answer 1

Initial concentration of salt in river = 32ppm

1 ppm = 1mg/L

Now we can add 200-32 = 168 mg/L to maximum.

So It is the question of dilution

N1V1 = N2V2

For calculation of per minute.

Volume initial per minute before addition of Release = 2*10^6*3.78/24/60 Liter = 5250 L

Initial concetration = 32 mg/L

Final voume after addition = 5250+321.3

= 5571.3 L

Final concentration of liquid = 200mg/L

Increament in 1 liter = 200-32 = 168 mg.

Total increament per minute = 5571.3*168 = 935978.4 mg

this increase is led by 85 Gallon = 321.3 Liter

N1V1 = N2V2

==>935978.4/321.3

2913.09 ppm.