Earlier in the course we derived the isentropic flow relations,
which govern how changes in the thermodynamic state of a gas
its temperature and pressure, are related to each other
But what about the velocity of the flow?
We know that a flow speed
is one of its most important properties,
but the isentropic flow relations tell
us nothing about it.
In this module we'll look at how the thermodynamic state
of the flow and its speed are related
for adiabatic flows. This means a flow where there's
no heat transfer to or from it.
To achieve this, we're going to consider flow
through what's known as a stream tube. This
is an imaginary tube of fluid where the sides are
made up of streamlines so there's no flow across them.
We're going to apply
conservation of energy in its control volume form
to this stream tube. Let's examine
the first term. The internal energy (U)
plus a 1/2 V^2, plus gz
is the total energy per unit volume
of the fluid. If we multiply this by density
and integrate over the control volume, this gives us
the total energy within that stream tube.
If we take
the time derivative of this, then what we end up with
is the rate of energy accumulation within our stream tube.
The second term, is the net flux
of that energy, through the control volume surface.
Next we have the work that's done by the pressure
and the net heat transfer to, and
work output from, the control volume. For this
stream tube, we can make a number of
simplifications to this equation. First, let's
assume we have steady state flow, so the time derivative
makes the first term zero. Second, let's
assume we have no shaft work, so that
work term is equal to zero. Next we'll assume the
flow is adiabatic,
seeing as we are trying to derive some
adiabatic flow relations,
and that means that Q dot is also
going to be equal to zero.
The next thing we can do is to combine
the pressure work term, with the internal energy
in that flux term, using the fact that the enthaply is equal
to the internal energy plus p on ρ.
We then obtain much a simpler equation
which is telling us that the net flux of the enthalpy
plus a 1/2 v^2, plus gravity times the height of the fluid
across the control surface is equal to zero. 
But we can simplify this expression
even further. Let's assume that our stream tube is very thin
so that the variations across the inflow plane A1
and the outflow plane A2 are zero. So we've got uniform flow
across those planes. Second for compressible flows of gases
we can usually neglect that gravity times density term.
So let's also do that. So then let's look how we can simplify
our integral of the flux across the control surface
We're only going to get contributions to this
flux from the ends of our stream tube
This is because there is no flow, across the top and the bottom
Now because we've assumed that we've got
uniform flow across those planes
basically all we need to do is multiply
h plus a half V squared at
A1 and A2, and then multiply them by
rho times V dot n, times those areas.
So let's do that. So we see we've got a
a simple expression with no integrals, but we
can simplify this even further, by
recognizing
that density times normal velocity times area
is equal to the mass flow rate. Ok, so now 
we've just got
h plus a half V squared at each end multiplied by
the mass flow rates across those planes. 
Now because we got steady
flow through our stream tube, the mass flow
rate coming into the stream tube
must be equal to the mass flow rate going out.
We're not dealing with any nuclear
reactions
so there can't be any mass created or
destroyed as we go through that stream tube
Now if you make this substitution in the
equation
that m dot 1 is equal to m dot 2,
which is just equal to the mass flow rate
then what we're going to get is that the enthalpy
plus a half the velocity squared is equal to a constant
and this is known as the stagnation enthalpy.
So if we have a fluid flowing with velocity V
and it has enthalpy h, and then we bring it to rest
then the resulting enthalpy when it's stationary is going to be
equal to this stagnation enthalpy , h naught.
So you can see that we've successfully related the velocity
to the thermodynamic state of the gas, at least the enthalpy.
In the next module we're going to look at
how the remaining thermodynamic properties change.
