Differential Equations

[nic_pham]

A tank contains 100 litres of pure water. 2 brine solutions are pumped into the tank. The first 1 contains 0.1kg of salt per litre of water and enters the tank the tank at a rate of 5 litres per min. The second solution contains 0.05kg of salt per litre of water and enters the tank at 10 litre per min. The solution mixed and drains at 15 litre per min.
a) How much salt is left after t mins?
b) how much salt is left after 1 mour?

 

[galactus]

Let's see if we can't struggle through this. It's been a while since I used

diffeq, so bear with me.


For one pipe, the solution is entering the tank at
(5 L/min)(.1 kg/L)=.5 kg/min.
For the other pipe, the solution is entering at
(10L/min)(.05kg/L)=.5 kg/min.

Adding them we get 1 kg/min.

It is leaving the tank at (15L/min)(A/100 kg/L), so we subtract it.

So we put it together to get our system:

1 kg/min-3A/20 kg/min

Therefore, dA/dt=1-3A/20

dA/dt+3A/20=1, A(0)=0 (since there is initially no salt in the tank).

The integrating factor is e^(3t/20)

we have:

d/dt[Ae^(3t/20)]=e^(3t/20)

Integrating both sides, we get:

Ae^(3t/20)=(20e^(3t/20))/3+C

A=20/3+ce(-3t/20)

Since A=0 when t=0, we have:

c=-20/3

So, the amount of salt at any time t is given by:

If you want to find how much is in the tank at one hour, enter 60 in for t.

Do you see what is going on here?.

Good luck. I hope this helps.