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.
