### Re: How can I calculate the cooling of nitrogen as it expands through an orific

Date: Tue Nov 29 18:32:46 2005
Posted By: David Coit, Aerospace Engineer, Naval Air Warfare Center - Weapons Division
Area of science: Physics
ID: 1132770709.Ph
Message:
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Dave,

Yes it is possible, but first we need to know the original temperature of
the nitrogen in the bottle, and if we want the problem to be easy we have
to assume that the flow coming out of the bottle is isentropic, or has a
constant entropy. An isentropic flow is one that is both adiabatic and
reversible. "Adiabatic" means that there is no heat transfer from the flow
to the surroundings or vice versa. "Reversible" in this context means that
you could reverse the process and recover the fluid's original state, i.e.
there are no losses such as friction which would take some of the usable
energy in the flow and turn it into unusable heat. It also means that there
are no large gradients in the flow, i.e. that there are no large changes in
pressure, temperature, velocity, etc. such as shock waves. Strictly
speaking, there are no isentropic flows in the real world since there is
always some kind of friction or gradient in the flow, however in this case
the flow will be very close to isentropic so we could expect to see the
real result be off by only a few percent from the isentropic result. The
extra work to do a more detailed calculation would be immense by
comparison, and in a real-world application there is probably more error in
your pressure regulator than there is in assuming the flow is isentropic.
All of that being said, just keep in mind that this is a key assumption
that we've made to solve the problem, so if you were to do something quirky
like put lots of drag inducers or vortex generators in the flow, or expand
the flow so slowly that a significant amount of heat is transferred in from
the surroundings, or something like that, you will see a different
resulting temperature. For a simple expansion through an orifice or nozzle,
this should be a pretty good approximation.

I'll give references below, but the isentropic flow equation we're
interested in for this problem is:

P/Pt = (T/Tt)^(k/(k-1))

P = Static pressure, i.e. the pressure that you've expanded the flow to. In
this case this is atmospheric pressure.
Pt = Total pressure or stagnation pressure. This is the pressure the flow
would have if you isentropically slowed the flow to zero velocity. Since we
are assuming an isentropic process that starts from rest, this is simply
the pressure at the exit of the regulator, 50 psi.
T = Static temperature. This is the answer you're looking for. It's the
temperature of the moving flow after it's been expanded.
Tt = Total temperature or stagnation temperature. This is similar to the
total pressure and again it will merely be the temperature of the flow at
the exit of the regulator, which will be the temperature of the gas in the
tank.
k = Ratio of specific heats of the gas. This is a physical property of the
gas and for most gasses it is relatively constant for a wide range of
temperatures and pressures. We can simply look this up in a textbook or
some other reference. For nitrogen it is about 1.4. Interestingly, if you
ask a mechanical engineer what the symbol for the ratio of specific heats
is, they will likely say "k" while an aerospace engineer will typically
recognize the lowercase greek gamma as the ratio of specific heats. Don't
look at me, I didn't decide to teach it differently to different people.

So we simply plug in the known values of P, Pt, Tt, and k and then solve
for T. If we go from 50 psi to atmospheric pressure (about 14.7 psi) and if
the temperature in the tank starts out at 540 Rankine (about 80 degrees
Fahrenheit), we'd get a final temperature of about 380.6 Rankine, or about
-70 degrees Fahrenheit.

A couple of other notes while I'm at it:
As you allow nitrogen to escape from the bottle and the remaining nitrogen
expands, its temperature will be reduced according to the Ideal Gas Law
(Pv=nRT). Keep that in mind if you're very concerned about the resulting
temperature being uniform, as this will make a noticable difference if you
leave the orifice open for very long.

Note that I have not even mentioned the rate at which nitrogen escapes from
the bottle. For given values of upstream and downstream pressure, the flow
rate of gas through the orifice will depend only on the size of the
orifice. You can either pick the size and calculate the resulting flow
rate, or pick the flow rate and calculate the required orifice size. You
can do these calculations using the same isentropic flow equations found on
the NASA webpage I've listed in the references, but it might not be as
straightforward as the calculation above. Keep in mind that the mass flow
rate will be the product of the density, velocity, and cross-sectional area
at a given location, and the volume flow rate will be the product of the
velocity and cross-sectional area. If you run into problems, feel free to
follow up with another question and I'd be glad to go into more detail.

I hope this helps,
David Coit

References: http://www.grc.nasa.gov/WWW/K-12/airplane/isentrop.html
Introduction to Flight, 3rd ed., John D. Anderson Jr.

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