uranium exists naturally in groundwater,however if the concentration is too high

Question

uranium exists naturally in groundwater,however if the concentration is too high then the water is unsafe to drink suppose a tank containing 2000 liters of groundwater has 100 mg of dissolved uranium in it this is above the safety limit of 0.03mg/l. to correct this, water with a uranium concertation of 0.01mg/l is pumped into the tank at rate of 12 l per minute.the water in the tank is well stirred and is pumped out of the tank at the rate of 12 litters per minute. solve the differential equation for the amount of uranium in the thank at t?

Explanation / Answer

let S(t) be the amount of uranium in the water at time 't' minutes

=> dS/dt = uranium solution in the tank - urainum solution out of the tank

dS/dt = [.01 mg/l * 12 l/min] - [S/2000 * 12 l/min]

dS/dt = .12 - 3S/500

dS/dt = (60 - 3S)/500

dS/(60-3S) = dt/500

integrating both sides

=> -ln(60-3S)/3 = t/500 + C

ln(60-3S) = -3t/500 - D

(60-3S) = e^(-3t/500 -D) = e^(-3t/500)*e^(-D) = F*e^(-3t/500)

S = 20 - F/3*e^(-3t/500)

S = 20 - G*e^(-3t/500) , -------(1)

now the initial condition is S(t=0) = 100 mg

plug S(t=0) = 100 in (1)

=> 100 = 20 - G*e^(-3(0)/500)

100 = 20 -G , => G= -80

=> S = 20 + 80*e^(-3t/500) , ---------> the amount of uranium in the thank at t

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