Salt water pumped into a tank problem.

Suppose that a large mixing tank initially holds 400 gallons of water in which 65 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 6 gal/min, and when the solution is well stirred, it is pumped out at a faster rate of 8 gal/min. If the concentration of the solution entering is 3 lb/gal, determine a differential equation for the amount https://live.wilkes.edu/content/enfo...o4Mphov0RLzYEy of salt in the tank at time https://live.wilkes.edu/content/enfo...o4Mphov0RLzYEy

Re: Salt water pumped into a tank problem.

Hey derail.

What is the rate of change of salt? Is the density of salt the same for all forms of brine?

Re: Salt water pumped into a tank problem.

Since the tank initially "holds 400 gallons of water" and "brine solution is pumped into the tank at a rate of 6 gal/min" and "it is pumped out at a faster rate of 8 gal/min" so after t minutes it will hold 400+ 6t- 8t= 400- 2t gallons of water. The density of the brine is the amount of salt, A(t), divided by the volume of water, 400- 2t, is $\displaystyle \frac{A(t)}{400- 2t}$. The point of that is that 8 gallons of water being pumped out each minute will contain $\displaystyle \frac{8A(t)}{400- 2t}$ pounds of salt. At the same time, a brine solution with 3 lbs/gal is coming in at 6 gal/min and so salt is coming in at (3 lbs/gal)(6 gal/min)= 18 gal/min.

dA/dt is the rate at which salt is coming into the tank (positive for inflow, negative for outflow) so $\displaystyle dA/dt= 18- \frac{8A}{400- 2t}$.