How much chlorine will be in the fish tank when the tank is full?

Question

You are setting up a new aquarium and have purchased a large fish tank can hold 160 liters of water. The tank is partially filled with 80 liters of tap water in which 200 milligrams of chlorine is dissolved. You add tap water containing 4 mg of chlorine per liter into the tank at a rate of 8 L/min. At the same time, a filter is run to filter out the chlorine in the water. The tank water flows into the filter at a rate of 8 L/min. The filter then removes half of the chlorine in the water that passes through it before pumping the water back into the tank at a rate of 8 L/min. How much chlorine will be in the fish tank when the tank is full?

Answer 1

Hi... after looking at this problem a couple of times, it appears that the answer may not be correct using the concentration method? Using the mass method approach, I obtain a final mass of 417.2 mg and final concentration of 2.61 mg/l.

The concentration method given in the above answer yields a final concentration of 4.11 mg/l which would yield a mass of 657.6 mg?

I then placed the conditions in an EXCEL spreadsheet with 1 second intervals to obtain a mass of 417.. and concentration of 2.6...?

Thanks... just curious if there is a quirk to this problem? What answer did your professor provide?

Answer 2

160 liters = V_F, 80 liters = V₀ dV/dt = 8 l/min the recycle stream does not affect the volume. Therefore, the final time is (160 l - 80 l)/ (8 liters/min) = 10 minutes = t_F

C₀ = 200 mg/80 l = 2.5 mg/l

VdC/dt = (8 l/min) * 4mg/l - (.5C)(8 l/min)

Change of mass in the tank = rate of input of mass and rate mass is pulled out by filtration.

In numbers, m and t, the diff eqn becomes:

(80+8t)dC/dt = 32 - 4C where t is in minutes and C is in mg/l

dC/(32-4C) = dt/(80+8t) where C(0) = 2.5 mg/l

You can integrate both sides as definite integrals or as indefinite integrals and solve for the integration constant using the initial value. Mass of Cl is V_F*C(t_F)

Good luck!