Today we are going to talk about nondimensionizing the basic differential equation.
An other words, how did we get from this.
This top thing here to this on the bottom, and if you go through these all of these are
non-dimensional, and the only thing that is in front of our non-dimensional variables
are these dimensionless parameters.
Here is 1 over the Re.
Here it is 1 over Fr number squared.
So it is very important when you are nondimensionlize differential equations that you go beyond
just the variables.
You have to collect everything else in a parameter that is also non-dimensional other wise it
wont be non-dimensional.
The first thing that we need to do.
Is define our non-dimensional variables.
We choose a reference variable and then take our variable and divided it by it.
Here we are going to take this as v0 or v free stream.
So our u* is u over v0/ free stream.
We do the same.
We can use the same reference for both u and v.
We do it for x.
So that is x over L. As well as y*, which is y over L. Some of the things though that
you have to be careful of are things like Tor p because those you can do it in terms
of tou or p0, but in order to be at your dimensionless parameter together need to put it in terms
of other variables.
So for example if we wanted a dimensionless time we would define T by L over v. So here
we are are going to define our p* or our dimensionless pressure as P divided by rho v free stream
squared.
Again this is a type of pressure term.
We have seen it before.
As you see this allows you to nondimensionlize your equations in terms of variables or things
that will be in our dimensionless.
So what I am going to do is I am going to take this top one, and show you how to go
through every single one of these variables, and then divide through.
So that we are able to have dimensionless variables.
So we will start with the.
Obviously we are not going to do anything with rho right now, but we will start with
u.
If you look above.
We can define u or just rewrite it as u star over v infinity.
Now lets look at du/dx.
Well this is the same as d(u*v infinity) we have replaced our u, and we have replaced
our dx with 1 over dx* over dx*/dx.
If you notice it is the same thing as dx*/dx is just 1.
So it is like multiplying it by 1.
So here we pull out the v infinity and then all we have is du*/dx* So that is the first
term there.
How about dx*/dx.
Well lets go back to our definition of x*, and if we differentiate x* with respect to
x.
What do we get?
We get 1/L. So du/dx can be written in terms of du*/dx* by putting v infinity divided by
L.
We do the exact same thing for dv/dy and because we are using the same references.
What you will see this comes out to be the exact same.
dv*/dy.
Also when we want to look at v.
We go up to our definition of v* is.
Then we now that v can just be rewritten as v* times v infinity.
Okay now lets do dp/dx.
Since that is a little bit different.
So we start with dp/dx, and again like we did before.
It is d(p*rho v infinity squared over dx*, dx*/dx.
Now we now that that is.
So we jut pull out our reference variable.
So that is rho v infinity squared over L. Where we got that L, again that is from dx*/dx
like we did before.
dp*/dx*, alright.
Now lets do the second derivative.
The terms that we find on the right hand side of the equation.
In fact lets go to to another page to be able to do this.
So what we are looking for is d^2u/dx^2.
So d^2u/dx^2 is exactly the same as d/dx of (du/dx), but we have already defined du/dx
with terms of dimensionless variables.
So that is d/dx of that (v infinity over L)(du*/dx*).
So now what do we do with our d/dx well we do exactly what we did before.
It is d/dx*, dx*/dx of the ( v infinity over L)(du*/dx*) So that becomes.
Remember this dx*/dx is 1/L. So this becomes v free stream over L^2 d^2u*/dx^2* and so
as you might have expected if what a derivative is.
Each time we take a derivative it is what we are taking the derivative with respect
to that increases, and that ids way du/dx had a 1/L d^2u/dx^2 has a 1/L^2, and when
we put that together, and when we do the same thing for the other second derivative.
What we are left with is this.
At rho (u* V infinity V infinity over L du*/dx*.
So that was our du/dx plus v* V infinity over L, dv*/dy*.
So that was the right hand side.
So now lets look at the left hand side.
That is equal to minus rho, v infinity squared, over L, dp*/dx* plus mu V infinity over L^2,
and that is what came out of those second derivatives.
d^2u*/dx*^2 plus d^2v*/dy*^2 and now we are almost done.
However our variables might be dimensionless, but the actual equation is not because you
can see by looking at it that there are still some dimension there that the dimensions with
rho, mu, and everything like that.
So we divided everything through by rho v^2 over L. Alright.
So my left side hand becomes u*du*/dx* plus v* dv*/dy*, and our right hand side becomes
minus dp*/dx* plus mu/rho v infinity L times d^2u*/dx^2*plus d^2v*/dy*^2.
Now we are dimensionless, and what we see here is 1 parameter, which is 1 over Re.
So I hope this helps you to figure out how to nondimesionlize any equation that you will
be given.