Hi everyone, first post on here In a bit of strife though and looking for a bit of guidance...
A 150 litre tank contains 40 litres of water and 20 litres of alcohol mixed together. Pure alcohol is run in at 5 litres/minute, and the mixture is stirred to keep it perfectly uniform. Simultaneously, the mixture is withdrawn at 3 litres/minute.
Derive a differential equation for the amount of alcohol, litres, in the tank at time minutes.
My attempt so far:
This is my first time dealing with two liquids in a mixture (L/L) as opposed to a solid (such as a pollutant in a solution which is measured in g/L)
Let be the amount of alcohol in the tank at time .
= rate alcohol flows in - rate alcohol flows out
Rate in = Concentration X Flow rate in = 1 L/L X 5
This is where I've hit a stumbling block as I'm having trouble calculating what rate out comes to.
Rate out = Concentration X Flow rate out = x/(60+2t) X 3
The reason I've gone with as the volume is that there's 60L at the start, and the difference between the two rates of flow is . Am I right in doing so?
And then the final differential equation I get to is:
= 5 - x/(20 + t) (simplified from above)
A premeditated thanks for the help!