Question

Two tanks are interconnected. Tank A contains 60 grams of salt in 50 liters of water, and Tank B contains 80 grams of salt in 40 liters of water.
A solution of 2 gram/L flows into Tank A at a rate of 5 L/min, while a solution of 3 grams/L flows into Tank B at a rate of 7 L/min. The tanks are well mixed.

The tanks are connected, so 8 L/min flows from Tank A to Tank B, while 3 L/min flows from Tank B to Tank A. An additional 12 L/min drains from Tank B.

Letting x represent the grams of salt in Tank A, and y represent the grams of salt in Tank B, set up the system of differential equations for these two tanks.

Answer 1

Let a(t) and b(t) denote the amounts of salt in tanks A and B, respectively.

The volume of liquid in tanks A and B after t minutes are

A: 50 L + (5 L/min + 3 L/min - 8 L/min)t = 50 L

B: 40 L + (7 L/min + 8 L/min - 3 L/min - 12 L/min)t = 40 L

so the amount of solution in the tanks stays constant.

Salt flows into tank A at a rate of

(2 g/L)*(5 L/min) + (b(t)/40 g/L)*(3 L/min) = (10 + 3/40 b(t)) g/min

and out at a rate of

(a(t)/50 g/L)*(8 L/min) = 4/25 a(t) g/min

so the net flow rate is given by the differential equation

$\dfrac{\mathrm da(t)}{\mathrm dt}=10+\dfrac{3b(t)}{40}-\dfrac{4a(t)}{25}$

We do the same for tank B: salt flows in at a rate of

(3 g/L)*(7 L/min) + (a(t)/50 g/L)*(8 L/min) = (21 + 4/25 a(t)) g/min

and out at a rate of

(b(t)/40 g/L)*(3 L/min + 12 L/min) = 3/8 b(t) g/min

and hence with a net rate of

$\dfrac{\mathrm db(t)}{\mathrm dt}=21+\dfrac{4a(t)}{25}-\dfrac{3b(t)}8$

Replace a(t) and b(t) with x and y. Then the system is (in matrix form)

$\dfrac{\mathrm d}{\mathrm dt}\begin{bmatrix}x\\y\end{bmatrix}=\dfrac1{200}\begin{bmatrix}-32&15\\32&-75\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}10\\21\end{bmatrix}$

with initial conditions x(0) = 60 g and y(0) = 80 g.