Three tank mixture problem. Tanks A,B, and C can hold 200, 150, and 100 gallons

Question

Three tank mixture problem. Tanks A,B, and C can hold 200, 150, and 100 gallons respectively. 4 gallons of pure water runs into tank A every minute. A brine solution runs into tanks B and C at a rate of 4 gallons per minute. A brine solution also runs out of each tanks at a rate of 4 gallons per minute. Tank A initially contains 15 lbs of salt, Tank B initially contains 10 lbs of salt, and tank C initially contains 5 lbs of salt.

A) Set up a system of differential equations that model the amount of salt in each tank x1(t), x2(t), x3(t) at time t in tank A, B, and C respectively.

B) Predict the limiting values of x1(t), x2(t), x3(t) as t approach infinity without solving the system

C)Solve the system you found in part A and give functions for x1(t), x2(t), x3(t).

D) Use the initial conditions to find an explicit solution

E) Use maple or a graphing calculator to solve for when the amount of salt will be cumulatively less than or equal to .5 pounds

F) How much salt will be left in the tank as t approaches infinity? Prove your answer

Explanation / Answer

It is easiest to think about this the following way.

The salt in tank A at time t is x1(t)
The salt in tank A slightly later is x1(t +delta t)
The concentration of the salt in tank A is:
x(t)/200

The amount leaving the tank is the concentration of salt multiplied by 4 gallons per minute

So we get:

x1(t + delta t) = x(t) - 4*(x1(t) / 200) * (delta t)

Now if we move x(t) to the other side and divide by delta t we get:

[x1(t + delta t) - x(t)] / (delta t) = - 4 * (x1(t) / 200)

In the limit as delta t approaches zero we get:

dx1 / dt = -0.02 x1


Tank B and C can be done in a similar fashion, however keep in mind that there will be salt from tank A going into Tank B and salt leaving tank B. The same will be true for tank C

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