Question : Two tanks are connected by pipes. Initially, tank A contains 100 L of

Question

Question : Two tanks are connected by pipes. Initially, tank A contains 100 L of water in which 65g of salt is dissolved. Tank B contains 100 L of pure water.

Pure water is pumped into tank A at a rate of 10L per minute. Mixed liquid is pumped from tank A to tank B at a rate of 15L per minute, from tank B back to tank A at a rate of 5L per minute, and out of tank B at a rate of 10L per minute.

Assume that all liquids are mixed instantaneoulsy, and let y1 and y2 represent the grams of salt in tanks A and b, respectively. Find the vector function

that describes the amount of salt in each tank at any time t.

Explanation / Answer

Note that the net flow for tank A is 10-15+5 = 0, and the net flow for tank B is 15 - 5 - 10 = 0

This simplifies the equations substantially. We thus have, for the amount of salt, A(0) = 65 and B(0) = 0. Also, only pure water is being pumped in from outside, so eventually we should expect to have a value of salt equal to 0 in each tank.

Then, dy1(t)/dt = -15 y1(t)/100 + 5 y2(t)/100

dy2(t)/dt = 15 y1(t)/100 - 15 y2(t)/100

Then, the matrix is

-15      5

15   -15

Solve for the eigenvalues and eigenvectors.

We get

(-15-L)^2 -5*15 = 0

(-15-L)^2 = 75

L = -15 +- 5 sqrt(3)

Eigenvectors may be derived

Let the eigenvector be (x y)

L = -15 + 5 sqrt(3)

-15x + 5y = (-15+5 sqrt(3))x

5y = 15x + (-15+5 sqrt(3))x

let x = 1

5y = 5 sqrt(3)

y = sqrt(3)

(1 sqrt(3))

L = -15 - 5 sqrt(3)

-15x + 5y = (-15-5 sqrt(3))x

5y = 15x + (-15-5 sqrt(3))x

let x = 1

5y = -5 sqrt(3)

y = -sqrt(3)

(1 -sqrt(3))

Y = c1(1 sqrt(3))e^(-15+5 sqrt(3))t + c2(1 -sqrt(3))e^(-15-5 sqrt(3))t

Now, we can use the initial conditions of (65 0) to solve for c1 and c2

c1 + c2 = 65

sqrt(3)c1 - sqrt(3)c2 = 0

c1 - c2 = 0

Thus, c1 = 65/2 and c2 = 65/2

Thus, the solution is

Y = 65/2(1 sqrt(3)) e^(-15+5 sqrt(3))t + 65/2(1 sqrt(3)) e^(-15-5 sqrt(3))t, or

y(t) =

65/2 e^(-15+5 sqrt(3))t + 65/2 e^(-15-5 sqrt(3))t

65 sqrt(3)/2 e^(-15+5 sqrt(3))t - 65/2 e^(-15-5 sqrt(3))t

Note the behavior of y1 and y2. y1(t) simply decreases to 0 as t decreases, as each of the 2 terms is positive and decreasing to 0. y2 has a more interesting behavior, as the first term is positive and the second term is negative.

At t = 0, as the exponentials are both 1, we get 65/2 - 65/2 = 0. Initially as t increases, the first term doesn't decrease in magnitude as quickly as the second term. Thus, y2(t) initially grows, then decreases.

The result for y1 and y2 is exactly what one should expect given the flows.

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