Consider the cascade of two tanks of brine shown in the figure above, with V1=10
ID: 2944984 • Letter: C
Question
Consider the cascade of two tanks of brine shown in the figure above, with V1=105 gallons and V2=210 gallons being the volumes of brine in the two tanks. Each tank also initially contains 55 lb of salt. The three flow rates indicated in the figure are each 4.5 gal/min, with pure water flowing into tank 1.
(a) Find the amount x(t) of salt in tank 1 at time t.
x(t)= ?
(b) Suppose that y(t), the amount of salt in tank 2 at time t, is given by the differential equation
dy/dt = (4.5x/105)-(4.5y/210)
Solve for y(t) using the function x(t) found in part (a).
y(t) = ?
(c) Finally, find the maximum amount of salt ever in tank 2.
Maximum amount of salt = ? lbs.
Explanation / Answer
A)
first, set up your standard tank equation
x' = rate in - rate out
rate in = 4.5gal/min*0lb/gal = 0 lb/min
rate out = 4.5gal/min*xlb/105gal
x' = -3/70*x
this becomes a first order linear ODE of the form
x' +3/70*x = 0
which you can easily solve and obtain
x(t) = Ce^(-3t/70)
using our initial condition x(0) = 55 you obtain C = 55, so
x(t) = 55e^(-3t/70)
B)
Plugging in x(t) you obtain
y' = 33/14e^(-3t/70)-3y/140
You can rearrange this to obtain a first order ODE
y' + 3y/140 = 33/14e^(-3t/70)
This can be solved using the method
= eintegral(3/140)dt= e3t/140
g = 33/14e^(-3t/140)
integral(g) = -110 e^(-3 t/140)+C
y(t) = (-110 e^(-3 t/140)+C)/(e3t/140)
y(t) = -110e-3t/70+Ce-3t/140
y(0) = 55 so C = 165
y(t) = -110e-3t/70+165e-3t/140
we know that the maximum simply occurs when the derivative = 0, so derive your equation for tank 2:
y' = -33/28 e^(-3t/70)(3e^(3t/140)-4)
setting this equal to 0 you obtain t = 140/3 (2 log(2)-log(3))
plugging this back into your equation for y(t) you obtain a maximum value of 495/8 or 61.875 lbs.