Question

Suppose a large mixing tank initially holds 200 gallons of water in which 100 pounds of salt have been dissolved. Another brine solution with 2 Ibs/gal is pumped into the tanked at a rate of 3 gal/min, and after the solution is well stirred, the tank is drain at a rate of 5 gal/min. determine the amount of salt at any time t > 0.

Answer 1

The amount of salt in a mixing tank over time where brine is added and then drained can be determined by setting up and solving a differential equation that accounts for the rates of salt entering and leaving the tank.

Step-by-step explanation:

A student asked about determining the amount of salt at any time t > 0 in a mixing tank scenario. The tank starts with 200 gallons of water and 100 pounds of salt. Additional brine solution with 2 lbs/gal concentration is added at 3 gal/min, and the mixture is drained at 5 gal/min. To solve this, we set up a differential equation considering the rates of salt coming in and going out, and the fact that the solution is well stirred, meaning the concentration of salt in the water leaving the tank is the same as that in the tank at any given time.

We let S(t) be the amount of salt in the tank at any time t. The rate of salt entering is 2 lbs/gal × 3 gal/min = 6 lbs/min. The rate of salt leaving is (S(t) lbs)/(V(t) gal) × 5 gal/min, where V(t) = 200 - 2t gallons, since the tank is being drained faster than it is being filled. The differential equation is thus dS/dt = incoming rate - outgoing rate = 6 - (5S(t))/(200 - 2t). Solving this differential equation will give the function S(t) which represents the amount of salt at any time t.