salt in a tank

Question

A 200-gallon tank is currently half full of water that contains 30 pounds of salt. A solution containing 2 pounds of salt per gallon enters the tank at a rate of 6 gallons per minute, and the well-stirred mixture is withdrawn from the tank at a rate of 6 gallons per minute. How many pounds of salt are in the tank 10 minutes later? Round your answer to 2 decimal place

Answer 1

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Answer 2

Hi Noah.

 

This is what we call a tank problem in differential equations and even though this scenario is simplified quite a bit (we assume a perfectly mixed tank) it has many important applications.

 

Tell you what, I'll get you to the part where we make the differential equation, and then you can solve it.  If you have trouble, I'll help you further.

 

First thing is first, we need a differential equation.  Let's let S(t) denote the amount of salt in the tank (in pounds) at some time t > 0.  Then S'(t) would represent the rate of change of salt in the tank.  We can say that S'(t) = (Flow of salt in) - (Flow of salt out).  I'll leave it to you to figure out why that makes sense.  It's not too bad.

 

Flow of salt in is simple.  We're told that solution enters the tank at 6gal/min.  For each gallon, there are 2 pounds of salt.  To figure out how many pounds per minute we're dealing with, multiply the two.  Flow of salt in = 6gal/min x 2lb/gal = 12lb/min of salt entering the tank.  (Note that this is amount per time, consistent with the units of S'(t))

 

So far we have S'(t) = 12 - (flow out).  This part is slightly harder.  We're told solution leaves at 6gal/min.  However, we don't know how concentrated these gallons of water are with salt.  With the flow of rate in, we did know the concentration: 2lbs/gal.  We need to know this for the solution flowing out.

 

Concentration is amount/gallon.  The number of gallons of the tank wont change since we're told that solution enters and leaves at 6gal/min.  So our concentration is (amount of salt)/200.  But we already have a variable for amount of salt; S(t)!  Therefore we do what we did before:  The amount flowing out is 6gal/min x S(t)lb/200gal.

 

So our differential equation we need to solve is:

 

S'(t) = 12 - 6S(t)/200