Expert: Paul Klarreich - 3/4/2007

Question
Hi Paul,

Below is my question, hope you can solve it.

A 100-litres tank is intially half filled with pure water. Water containing
10gm/litre of a pollutant is added at a rate of 5 litres/min and the water flows
out of the tank at a rate of 3 litres/min. When the water in the tank is full,
it will overflow.

(a) Show that  dx/dt+(3/2)(x/(25+t))=50

Where x is the amount of pollutant in the tank after t mins, before the tank
overflow. State any suitable assumptions you have made?

(b) How long will it take for the tank to overflow?

(c) Solve the above differential equation problem and state the values of t for
the above equation be valid.

(d) Hence find the amount and concentration of pollutants in the tank just
before it overflows.

Regards
James

Answer
Questioner:   James
Category:  Advanced Math
 
Subject:  Modeling differential equation
Question:  Hi Paul,

Below is my question, hope you can solve it.

A 100-liter tank is initially half full of pure water. Water containing
10gm/liter of salt is added at a rate of 5 liters/min and the water flows
out of the tank at a rate of 3 liter/min.

>> AND THIS LITTLE GUY STANDS THERE STIRRING LIKE CRAZY.

When the water in the tank is full, it will overflow.

(a) Show that  dx/dt+(3/2)(x/(25+t))=50

Where x is the amount of salt in the tank after t mins, before the tank overflow. State any suitable assumptions you have made?

(b) How long will it take for the tank to overflow?

(c) Solve the above differential equation problem and state the values of t for the above equation be valid.

(d) Hence find the amount and concentration of salt in the tank just before it overflows.

Regards
James

...................................
Hi, James,

I don't have a complete solution for you now.  I will give it some more thought, but I can send you this much.  If I come up with more, I'll send a followup.

Try this sequence:[I made 'salt' the pollutant. Less typing.]

1. The rate at which salt is being added to the tank is 50 g/min.

2. Let the Concentration of salt at time t be C(t) g/liter.

3. The rate at which salt leaves the tank is 3C(t), because 3 liters are leaking out.

4. The amount of solution in the tank at time t is 50 + 2t, because the net effect of adding 5 liters and losing 3 liters is an increase of 2 liters per minute.

5. The amount of salt in the tank at time t is  x(t) = C(t)(50 + 2t), i.e. the concentration times the amount.

6. The rate of change of the amount of salt is equal to the rate at which it enters minus the rate at which it leaves:

dx/dt = 50 - 3C(t)

But C(t) = x(t)/(50 + 2t)  

dx/dt = 50 - 3x/(50 + 2t)

That's basically your equation.

(b) I think it overflows when it reaches 100 liters, which will take 25 minutes at 2 liters increase /min.