JOHN: I'm gonna briefly describe
how to use the interactive simulation
that models what's often referred to
as a Hilsch tube or a vortex tube,
and the idea is that we can feed in air,
feed at higher pressure within the atmospheric pressure,
and room temperature around room temperature,
and we can split the air into a cold gas,
so the cold gas is leaving here,
and the blue is used to indicate lower temperature.
Hot gas is leaving here,
so red is the other end of the temperature scale
in this example, and we can change
how much is cold gas and how much is hot gas
by changing the slider up here.
We also can change the feed pressure,
and both of these things will affect
how cold the cold gas is, how hot the hot gas is.
Now, for this particular settings
of a feed fraction, and this is cut off,
but this is 0.5,
a feed faction of 0.5 in the cold stream,
and of course, 0.5 in the hot stream,
we can see easily that for a constant heat capacity,
which we're using here, we satisfy the first law.
Half the gas undergoes a temperature decrease
of 42 degrees Kelvin,
half the gas undergoes a temperature increase
of 42 degrees Kelvin, so the first law is satisfied.
And then here, we're calculating the entropy change.
The entropy change is positive,
meaning the second law is also satisfied,
and the main reason the second law is satisfied
is that we undergo this pressure drop,
four bar, to we have one bar,
leaving for both cold gas and hot gas.
Now, the actual values shown here
are from experimental data,
and we can calculate, then, a comparison,
might call this isentropic efficiency.
If we were to take a gas at four bar and 298
and have it undergo adiabatic reversible expansion,
then we would get a largeR temperature decrease.
So that's what the denominator is here.
The numerator is the actual temperature decrease,
in this case, 42 degrees Kelvin.
The way the coefficient performs is
the product of the isentropic efficiency,
it's this multiplied by the fraction of the cold gas,
in this case, .5, and as soon as I get figures,
only given two here, that's why it's not just shown
as 0.15 instead of 0.145.
So the coefficient performance basically says,
if I had one mole of gas
or one kilogram of gas coming in,
and it went through isentropic expansion,
reversible adiabatic expansion,
we could cool that kilogram down to a certain temperature,
and if we compare, in this case,
basically, it's how much cooling we can obtain.
We only have half as much gas that gets cooled down,
so the net cooling available
would only 0.15,
so that's why we refer
this is coefficient of performance.
So let's look at the actual simulation.
Well, here is this simulation,
and if we decrease the fraction of cold gas,
you'll notice the temperature goes down,
try, and we can get a low temperature, 248 Kelvin.
Now, if we already increase the feed pressure,
we can get an even colder temperature.
So as you can see, here, we're getting
a 71-degrees-Kelvin drop in temperature.
If 20% of the gas is leaving as cold gas,
and we have a feed pressure, 7.8 bar.
If I increase the fraction of cold gas
and decrease the fraction of hot gas,
then the hot gas gets hotter,
and let's to extreme here.
Now, the hot gas is more than 100 degrees Kelvin hotter
than the feed gas, but only 20% of the gas is hot gas.
Likewise, you notice the coefficient performance
increases and then decreases,
so there's some maximum coefficient performance,
it's a function of the feed fraction.
So the idea of a Hilsch tube is
that it allows us to obtain cold temperatures
without refrigeration system
sometimes used to cool down cutting tools, for example,
and blowing that cold air onto the tool.