This is a post purely to help anyone struggling with understand this. Most of this I'm sure is
self-bloody-evident to most but I know even the intellectual among us still work rote without thinking about *why* this is the case. And in any event, a lot of people struggle with this at my school so I thought if that is the case, might as well help
everyone.
Standard question: A tank of water has 10 L with 6kg of salt dissolved in it. A salt solution of concentration 5kg/L is added at a rate of 6L/min. There is a hole at the bottom of the tank, which allows water to leak out at a rate of 5 L/min. Find the change in the mass of salt over time, in terms of x kg of salt and t time in minutes.
So the first sentence relates to the initial condition, that is: when t = 0, x = 6... in English -> when we start there is 6 grams of salt. (IGNORE THIS WHEN SETTING UP THE DE)
The next thing starts off our first rate. Now usually we mechanically go 5*6 =30. That is fine for the machines among us. Why this is the case? Well simple dimensional analysis tells us that when we are multiplying those two things we cancel the "L"s because they are the same on the numerator and denominator. That is:
=
The other, intuitive way is to stop and think for a moment. You have say a 1L bottle. In that there is 5kg of salt. Now you are pouring that into the tank at a rate of 6L per minute, that is 6 bottle-fulls per minute, which means you are chucking in 6 lots of 5 kg = 30 kg every minute. This is the real reason why we mechanically multiply.
Ok so now you know partially the DE. You know the RATE IN is 5*6 = 30.
Now we need to find the RATE OUT. Why? Well so we can describe
which is in words: The change in salt over time. If we only have the rate in this does not give us the full picture, like foolishly using a magnifying glass to observe the Bayeux Tapestry in full.
So the rate out. This is the interesting part. We know that there is already some salt in the tank and salt is also pouring in. This means that we cannot easily determine how much salt is going to be taken out at any given instant. Which means we have:
xkg of salt in 10L.
BUT how is the 10L changing? Well we are going 6 L/min in and 5 L/min out so overall we are gaining 1 L every minute.
So really Volume = 10 + 1t
In effect:
of salt in
L
So then I ask: how many kgs per litre?
Well simple. Divide through by
This is kg/L and doesn't describe the rate as in how much is going in over time. So we need to multiply it by the rate going out for reasons already explained.
Now overall we have to take this rate away from the rate of inflow as it is being taken out of the picture, or the tank if you don't like to talk figuratively :knuppel2:
So now we have:
AT SPECH LEVEL YOU WILL NOT BE ABLE TO FIND THE GENERAL EQUATION OF THIS. This requires integration factors, something requiring knowledge of multi-variable calculus. You only need to know how to set one like this up.
I
think this covers the main ideas of these questions.
Sometimes they will be nasty and make the rate of inflow = 0 as they are just putting in water with no salt/serum or whatever they choose.
eg: 10L tank has 6kg of salt. Water is poured in at a rate of 6 L/min and the solution is drawn off at 6 L/min.
Following my ideas you get:
(the t disappears because the rate of inflow = rate of outflow meaning
no net gain of water over time)
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