suppose a 30L tank has two pipes feeding into it. One carries a salt water solut

Question

suppose a 30L tank has two pipes feeding into it. One carries a salt water solution with 2 kg of salt per liter and it comes in at a rate of 4 liters per minute. the second pipe has a salt water solution with 1 kg of salt per liter and it comes in at a rate of 2 liters per minute. the well mixed water from the tank exits the tank through pipe at the bottom that has the solution leaving at a rate of 4 liters per minute. suppose the tank initially contained 10 liters of solution that had 3 kg of salt in it. a) Give a sketch of this situation, making sure to label important pieces of imformation b) Let S(t) be the amount of salt in the tank after t minutes. Give the differential equation and initial value that models this situation. c) Find the solution to the differential equation above. Make sure to show all of your work in a clear fashion. d) How much salt will be in the tank the instant it becomes full?

Explanation / Answer

s(t) = kgs of salt in the tank at time t minutes.

and we know that

s'(t) = the rate at which the amount of salt in the tank is changing

s'(t) = (rate of salt going in) - (rate of salt going out)

The things we have to determine are the ate in" and ate out" quantities.

rate in = (2 kg/liter * 4 l/min) + (1 kg/liter * 2 l/min) = 8+2 = 10 kg of salt / minute

rate out = s(t) kg of salt / 10liter of solution * 4 liter/minute = .4s(t) kr of salt / minute

now ds/dt = rate in- rate out

ds/dt = 10 - .4s(t)

ds/[10 - .4s] =dt

integrating both sides

=> -ln(10-.4s)/.4 = t + C

ln(10-.4s) = -.4t - .4C

10 - .4s = e^(-.4t - .4C)

.4s = 10 - De^(-.4t)

s = 25 - 2.5De^(-.4t)

or s(t) = 25 - Fe^(-.4t) , F is the integration constant      ---------> (1)

the initial value is s(t=0) =3 kg of salt

now plug the initial value in

=> 3 = 25 - F , F = 22

hence the solution is :

s(t) = 25 - 22e^(-.4t)

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