Question

A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of 2lb of salt/gallon is admitted into the tank at a rate of 6 gallons per minute. The well-stirred solution is drained at the same rate. How long will it take for the tank to have 100 lb. of dissolved salt

Answer 1

Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min

so that the net rate at which salt is exchanged through the tank is

ds(t)/dt = 12 - s(t)/150 … (lb/min)

Solve for s(t). The DE is separable, so we have

ds/dt = 12 - s/150

150 ds/dt = 1800 - s

150/(1800 - s) ds = dt

Integrate both sides to get

-150 ln|1800 - s| = t + C

Solve for s :

ln|1800 - s| = -t/150 + C

1800 - s = exp(-t/150 + C )

1800 - s = C exp(-t/150)

s = 1800 - C exp(-t/150)

Now given that s(0) = 50, we solve for C :

50 = 1800 - C exp(-0/150)

50 = 1800 - C

C = 1750

Then the amount of salt in the tank at any time t ≥ 0 is

s(t) = 1800 - 1750 exp(-t/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for t in

100 = 1800 - 1750 exp(-t/150)

1700 = 1750 exp(-t/150)

34/35 = exp(-t/150)

ln(34/35) = -t/150

t = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for s(t), we could have instead stopped with

-150 ln|1800 - s| = t + C

Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so

-150 ln|1800 - 50| = 0 + C

C = -150 ln|1750|

==>   t = 150 ln(1750) - 150 ln|1800 - s|

Then s(t) = 100 lb when

t = 150 ln(1750) - 150 ln(1700)

t = 150 ln(1750/1700)

t = 150 ln(35/34)

t ≈ 4.348