Answer to Question #315536 in Differential Equations for Allen

Question

A tank contains 100 gal of brine made by dissolving 60 lb of salt in water. Salt water containing 1 lb of salt 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

Solution: 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 $C(t)$

$C(t)=\frac{60+2t}{100+2t}$

Now we know the salt concentration in the tank at any time t.

Evaluate at t=30.

$C(30)=\frac{60+2(30)}{100+2(30)}=\frac{120}{160}=\frac{3}{4}$

$\\s(30)=60+2(30)-3(30)(\frac{3}{4})=120-90 (\frac{3}{4})=52.5$

Therefore the amount of salt after 30 minutes is 52.5 lbs