Suppose that a large mixing tank initially holds 300 gallons of water in which 5

Question

Suppose that a large mixing tank initially holds 300 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gallons per minute, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gallons per minute. If the concentration of the solution entering is 2 pounds per gallon, determine a differential equation for the amount of salt A(t) in the tank at time t.

I have a ton of these problems to do so if I can just be walked through one I should be okay from here on out.

Explanation / Answer

Let a(t) be the amount of salt in the tank at time t. Then, the amount of salt in the tank at t=0 is 50 pounds, so the initial condiiton is: a(0) = 50 Now, salt is flowing into the tank at 3 gal/min at a concentration of 2 pounds per gallon. Therefore, it is entering at (3 gal/min)x(2 lb/gal) = 6 lb/min. This is the first term in the following differential equation. da/dt = 6 - [2a/(300+t)] But, brine is flowing out of the tank as well; this is the last term subtracted from the differential equation.. At any time, t, the tank holds (300 + t) gallons of fluid because a net increase of 1 gallon per minute is happening. 2a/(300+t) represents the pounds of salt leaving the tank at any time, t. Now that you have the equation and the initial condition, it can be solved. Since the question is only to formulate the differential equation, I will stop here

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