Applications of Differential Equations

Concentration of a Brine Solution

Salt is dissolved in a tank filled with 100 gallons of water. Originally 25 lb of salt was dissolved in the tank. Salt water containing 2 lb of salt per gallon is poured in at the rate of 4 gal/min. The solution is well stirred. The mixture is poured out at a rate of 4 gal/min. Find the amount of salt as a function of time Q(t). How much salt remains after 3 hours?

Start by summarizing what is known.

t = 0 Q_{o} = 25 lb

NEED SKETCH

We need an equation. Here’s a hint: the rate of change of Q in the tank is the difference between what enters the tank and what exits the tank. See if you can write this a differential equation.

dQ/dt = (4gal/min)2 lb/gal – (4gal/min)Q/100gal

dQ/dt = 8 – 0.04Q

Now separate the variables and integrate:

dQ/(8-0.04Q) = dt

Integrating:

-25ln(8-0.04Q) = t + C

ln(8-0.04Q) = -0.04t + C

8 – 0.04Q = Ce^{-0.04}

Then solve for the constant, C and finish the problem..

Use t = 0 Q_{o} = 25 lb

8 – 0.04(25) = Ce^{-0.04(0)}

C = 7

The equation is then Q = 200 – 175e^{-0.04t}

And after 3 hours (180min):

Q = 200 – 175e^{-0.04(180)} = 199.9 lb