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 Qo = 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 Qo = 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
