Question

A tank contains 100 gal of brine made by dissolving 60 lb of salt in water. Salt water containing 1 lb of slat per gallon runs in at the rate of 2 gal/min and the well stirred mixture runs out at the same rate of 3 gal/min. Find the amount of salt in the tank after 30 minutes.

Answer 1

Initially there are 100 gal in the tank. There are 60 lbs of salt. The concentration of salt in the tank is 60/100 or 0.6 lbs per gal.
Water flowing in has concentration of 1lb/gal.
Water flowing out has unknown concentration because mixture is constantly changing.
Tank is losing 1 gal per minute. (3-2=1)
Use this info to set up 3 functions, V(t), S(t), C(t) for volume, concentration, and salt content.
\[V(t) = 100-t\]
\[s(t) =60+2t-3t*C(t) \]
\[C(t) = \frac{s(t)}{V(t)} = \frac{60+2t-3t*C(t)}{100-t}\]
Using a little algebra solve for C(t)
\[\rightarrow C(t) = \frac{60+2t}{100+2t}\]
Now we know the salt concentration in the tank at any time t.
Evaluate at t=30.
\[C(30) = \frac{60+2(30)}{100+2(30)} = \frac{120}{160} = \frac{3}{4}\]
\[s(30) = 60+2(30)-3(30)(3/4) = 120-90(3/4) = 52.5\]
Therefore the amount of salt after 30 minutes is 52.5 lbs