Consider the Concorde supersonic transport
flying at twice the speed of sound at an altitude
of 14 km.
At a point on the wing, the metal surface
temperature is 347 K.
The intermediate layer of air in contact with
the wing at that point has the same temperature
and is at a pressure of 1.04 x 10^4 N/m^2.
Calculate the air density at this point.
This is one of those tricky questions because
if you read the first sentence and you're
rushing because you're in a test you're going
to say, “oh, twice the speed of sound, I
see an altitude here, I'm going to go look
at my standard atmosphere tables to figure
out my values at that height, then I'm going
to maybe try and figure out do I need to use
Bernoulli's to try and take those parameters
over to find that point on the wing,” but,
in reality, it's not that hard.
Let's just start by writing out the “knowns”
and the “unknowns” and you'll notice that
it's not as hard as it seems.
So, the first known: twice the speed of sound.
Oh, let me write “knowns”.
So, the velocity of the free stream is equal
to two times the speed of sound.
The altitude is equal to 14 kilometers.
Let's see, the temperature…
Now, remember, this is the temperature at
the point on the wing it's not the free stream
temperature.
So, it's going to be 347 K. And then, of course,
pressure of 1.04 x 10^4 Newton meters squared.
Now, a quick note about this temperature,
you'll notice that they gave a temperature
of the metal surface at a point on the wing.
They didn't give the temperature of the air
at that same point.
However, what they did was they specified
that the air in contact has the same temperature.
Now this is actually really nice of them because
sometimes this is one of things that is hidden
or that you have to kind of know when you're
working a problem.
Sometimes, instead of saying same temperature,
it might say that the metal on the wing and
the air at that point are in thermal equilibrium
or in a steady state or steady flight.
It means the same thing.
Let's look at our unknowns.
For a problem with so much wording, the only
unknown that it is asking for is the density
at that point.
Now this could be a tricky problem because
you're seeing these values here.
However, those are actually not needed.
These are sort of a red herring.
If you notice, we have the pressure of the
air at that point, and we have the temperature
of the air at that point, and we're looking
for the density.
Pressure, temperature, density that screams
Ideal Gas Law.
Pressure equals density times the gas constant
times temperature.
Now since the problem up here does not specify
whether it is a perfect gas, or an ideal gas,
or whether it's a real gas…
So, if we just come in here and we'll say,
“assumes perfect gas.”
I think that's reasonable.
Arranging that, of course, algebraically gives
us the density is equal to the pressure over
the gas constant times temperature.
Throwing that over here…
Putting the values into the equation, we get
pressure is 1.04 x 10^4 Newton's per meter squared.
Hey, look at that.
That's the correct units.
Gas constant is 287 Newton meters per kilogram
Kelvin.
Do you have a memorized yet, lol?
And then, of course, the temperature which
is going to be 347 Kelvin.
You're going to go ahead and just get the
standard units out, right, kilograms per cubic
meter.
So, let's go ahead and plug and chug into
our calculator.
We are going to get an answer of 0.104 kilograms
per cubic meter.
