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Salt Tank Differential Equation

 
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Oct19-10, 11:44 PM   #1
 

Salt Tank Differential Equation


Ok so this is the question

A tank contains 100 kg of salt and 2000 L of water. A solution of a concentration 0.025 kg of salt per liter enters a tank at the rate 5 L/min. The solution is mixed and drains from the tank at the same rate.

A.What is the concentration of our solution in the tank initially

B.Find the amount of salt in the tank after 4.5 hours

C.Find the concentration of salt in the solution in the tank as time approaches infinity.

I found A to be .05 and C to be .025. But I'm not sure how to get B... I know I have to set up an equation but I'm not very good at these word problems and could use some help setting up the equation and answering B so any help would be great!

Thanks!
 
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Oct21-10, 01:28 AM   #2
 
Rate of change of salt in tank = amount of salt in/min - amount of salt out/min

For the amount of salt in, just plug in the values you are given

[tex]Concentration * flowrate = \frac{.025kg}{L}\left(\frac{5L}{min}\right)[/tex]

For salt out, the concentration will depend on t. More explicitly, since concentration is
kg/L, and the number of liters will never change (since the rate of flow in and the rate of flow out are the same), the amount of salt will change with time.
 
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