Answer to Question #155928 in Civil and Environmental Engineering for Kevin O. Vargas

Question

A tank initially contains 60 gal of brine in which there is dissolved 8 lb of salt. Brine containing 2 lb of dissolved salt per gallon flows into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring, and the stirred mixture simultaneously flows out at the slower rate of 3 gal/min. How much salt is in the tank at any time?

Answer 1

Let x(t) represent the amount of salt in the tank at time t, where t is given in minutes.

The flow rate in is = 5 gal/min and the flow rate out is = 3 gal/min. So the Volume in the tank at the time t is

V(t) = 60 + (5 - 3)t = 60 + 2t

in gallons.

The concentration of salt in the tank (and hence of the solution flowing out of the tank) is then

c₀(t) = x(t) / V(t) = x(t) / (60 + 2t)

Since the incoming concentration is c_i = 2 lb/gal, the net rate of change of the salt is

dx/dt = r_ic_i - r₀c₀

dx/dt = 2 • 5 - (3 • x(t)) / (60 + 2t)

10 = x'(t) + (3 • x(t)) / (60 + 2t)

which is linear differential equation, where p(t) = 3 / (60 + 2t) and q(t) = 10, so that an integration factor is

m(t) = e^(-$\smallint$p(t)dt) = e^(-3$\smallint$ dt/(60 + 2t)) = (60 + 2t)^-3

Hence we have

x(t) / (60 + 2t)³ = $\smallint$ 10dt / (60 + 2t)³

x(t) / (60 + 2t)³ = 5 / (60 + 2t)² - C

x(t) = 60 + 2t - C(60 + 2t)³

Since x(0) = 8 then

8 = 60 + 0 - C(60 + 0)³

C = 13/54 000

Hence

x(t) = 60 + 2t - 13(60 + 2t)³/54 000