So imagine a cube in some fluid, and the fluid is flowing in some arbitrary direction and
flows lets say in this case, generally from left to right, but it also has some vertical
components.
It is hard draw on the screen, but it also has some component of velocity into, and out
of the screen.
The cube that we are imagining is a control volume.
We can imagine a control volume of any arbitrary shape, and more specifically we can imagine
a control volume that has any arbitrary volume, and that includes the volume that is infinitesimal
small, and what I mean by that is I can imagine a cube that is this big.
A cube that is this big.
A cube that is a little bit smaller, and a cube that is small as I can possibly imagine.
So lets think about this cube.
It is a tinny little cube and in an ocean of fluid, and I am going to zoom in on it.
To know what we are dealing with.
So again we have our fluid here, and the streamlines again are generally flowing from some arbitrary
direction.
So here is our cube again.
To give you a sense of the scale we are talking about, and lets define a coordinate axis.
We will say right hand axis x,y, and z in Cartesian coordinates.
Lets put the origin into that axis right at that corner of the cube, and I will put back
so we can see the direction, but I am going to draw a red dot, and at the red dot, x,y,
and z all equal 0.
The cube is three infinitesimal lengths, associated.
Here is dx, dy, and dz.
All the continuity equation is a statement of conservation of mass.
So we can say that the mass flow into the cube minus the mass flow out of the cube is
equal to the rate at which the mass within the cube changes with respect to time.
So another words if mass flow into the cube and no mass flowed out.
Then the mass within the cube would increase with u is a function of time.
Or I can say if mass leaves the cube more quickly then when it flows in then the amount
of mass in the cube decreases with time.
To keep it simple lets say that mass will flow in from the left side.
We will deal with that first, and out through the right side.
So here is my x-component of velocity.
So lets look at the mass flow in on the left side of the cube.
Just dealing with that.
What it is, is the density of the fluid times its volumetric flow rate.
It would give the mass flow rate in at that location.
So lets say that the density at some arbitrary position x times its velocity at some arbitrary
position x times the cross-sectional area.
In this case the cross sectional area would be dydz.
So relatively what I am look at is the density times the volumetric flow rate and that gives
me a mass flow rate coming in from this surface.
I will also have fluid traveling upwards in the positive y direction.
So that is fluid crossing this face of the cube.
There is also fluid flowing out of the screen or in the positive z direction, and that is
flowing from this back surface of the cube.
So here is the second term with dealing with mass flow in from the bottom, and the analogies
term to account for fluid coming in from the back face, and I can do something similar
to account for the mass flowing out of the cube.
So to keep it simple lets deal with mass leaving the right side of the cube.
Mass leaving the top of the cube.
Plus the mass leaving the front side of this cube.
So I have cleaned up those equations and put them up into the corner to get them out of
the way.
Lets deal with mass in the cube.
The mass is simply the density of the fluid times some differential volume.
So that will equal the density of the fluid times dxdydz.
If I take the derivative of the mass with respect to time.
dm/dt is equal to drho/dt times dxdydz.
So now if I get those out of the way.
What I am going to do is make the substitution for m dot in, mdot out, and dm/dt.
So here is my mass entering minus my mass leaving.
It is equal to the rate in which mass is accumulating within the control volume.
So lets take this equation.
We will divided both sides by dxdydz.
So naturally on the right hand side the all three differential terms drop out.
We are left with 1 differential length for each of the remaining terms.
So for example for this first term all we have left is 1 over dx term.
So after canceling out all the terms and rearranging I got rho times the x component of velocity
evaluated at x, and I have subtracted off this term the x plus dx.
I have done the same thing for the y and the z component.
Finally since we are dealing with differential lengths, and the limits for dx, dy, dz are
all approaching 0.
What I come up with is the 3 partial derivatives on the left, and if I simplify them and move
them over I get the continuity equation in its traditional form.
There are a couple of common simplifications that can be made to the continuity equation.
The first would be if the flow is steady then the density wont change with respect to time,
and I am left with the sum of those 3 terms are equal to 0 or if I said that if the fluid
is in-compressible then the density is not a function of x, y,or z, and that comes out.
If it is in-compressible the density cannot change with respect to time, it is also steady.Pulling
out density I am left with a simplified form of du/dx plus the partial of v with respect
to y, plus the partial of w with respect to z, and the sum of those three have got to
equal 0.
To think about this equation lets assume that the flow 2-d.
So there is no flow in the z direction, and w is equal to 0.
So what we are left with is the sum of these 2 terms.
We got that they equal 0, and lets say that the fluids are in-compressible, 2-d, and lets
say that the velocity on the right side of the cube is bigger then the x component entering
in the left side.
What that would say then is that if that was true du/dx would have to be greater then 0.
Partial of u with respect to x will have to be bigger then 0.
If that is bigger then 0, that means dv/dy has to be less then 0.
So what this is saying is that whether some fluid flowing into the cube at the bottom
as some high rate.
It means that fluid flowing upward leaving the cube from the top direction got to be
less fluid coming into it, In essence it is saying that some of the fluid entering the
bottom of it has to leave to the right of it, and that is what is increasing the velocity
leaving in the right side of the cube.
So the continuity equation in essence is the conservation of mass, and for in-compressible
flow the mass within that cube has to remain constant.