Consider the set up consisting of two brine tanks where water cascades from tank
ID: 2987594 • Letter: C
Question
Consider the set up consisting of two brine tanks where water cascades from tank 1 down to tank 2, and then out of tank 2. Suppose that when the system is set into motion, the first tank contains 80 gallons of brine, while the tank below it contains 160 gallons of pure water. When the system starts, fresh water is pumped into tank 1 at a constant rate of 2 gal/min. at the same time the brine solution in tank 1 drains into tank 2 at a rate of 2 gal/min, and tank 2 drains at this same rate. Hence the volumes in both tanks stay constant. Assume that the solutions in each tank remain well-mixed, so that although the salt concentrations are changing in time, the concentration of salt leaving each tank equals the average concentration in that tank.
(a) if the first tank originally contains 60 lbs of salt, solve an initial value problem to find the amount x(t) of salt in tank 1 at time t.
(b) suppose that y(t) is the amount of salt in tank 2 at time t. Show first that dy/dt = x/40 - y/80. Then solve for y(t) using the function x(t) found in part (a)
Explanation / Answer
a) x(0) = 60 and dx/dt = 0*2-(x/80)*2 = -x/40. Solving this ODE yields x(t) = 60e^(-t/40).
b) We know y(0) = 0. dy/dt, the rate at which the salt in tank 2 is changing, is the difference between the rate at which salt flows in [(x(t)/80]*2 = x(t)/40 = 1.5e^(-t/40)] and the rate at which it is flowing out [(y(t)/160)*2 = y(t)/80].
So dy/dt = 1.5e^(-t/40) - y(t)/80
y' + (1/80)y = 1.5e^(-t/40)
[e^(t/80) * y ]' = 1.5e^(-t/80)
e^(t/80) * y' = -120e^(-t/80) + C
y = -120e^(-t/40) + Ce^(-t/80)
0 = -120 + C so C = 120
ANSWER: y = -120e^(-t/40) + 120e^(-t/80)