Does anyone out there know what the maximum speed is for aluminum skin as a function of altitude?
21 thoughts on “Aerothermal Bleg”
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Does anyone out there know what the maximum speed is for aluminum skin as a function of altitude?
Comments are closed.
In the back of Dan Raymer’s tome on aircraft design he used to have a graph of altitude/airspeed/temp. Sorry, am running out the door, otherwise I’d look for you.
Unfortunately, not being an aircraft designer, I don’t possess a copy of Raymer.
African or European?
I don’t… WAAAAAAAAAAAAAAAAAAAAAH!
it’s not speed, it’s temperature and then the question
is what degree of softening you can allow,
and then the induced heating is a function of drag,
profile and reynolds number and speed.
you can take optimum numbers, but it just gets worse from there.
The chart in Raymer is on page 356, basically he uses the total temperature equation:
Tt = T * (1+0.2 M^2). He also puts 250 F as the upper temperature limit for Al. That puts the upper Mach limit at Mach 2 or below up to about 100 kft.
I assume that he’s including wave drag in the heating assumptions?
It’s traditional (at least at the companies I have worked for) to use some sort of ablative insulation to protect the aluminum (or composite) skin of the rocket. Cork works pretty well, but I have also seen RTV and even ceramics used.
I vaguely recall the nominal aluminum temperature limit to be 160 deg C (320 deg F), but I could be wrong. The temperature limit was based on where the aluminum began to lose significant structural strength.
The question had nothing to do with rockets. It’s about aircraft design.
Is this aluminum laden, or unladen, with coconuts?
*sorry.. I know the joke is already taken, but I couldn’t resist*
T0=T*(1+0.2M^2) is simply total temperature, not the temperature behind the shock (note that this assumes gamma=1.4, which is correct for air) but perhaps Raymer correlates surface temperature to this number (I’m not familiar with his book). I agree that aluminum should probably be kept below 250F, maybe a little higher.
I’m sure however that surface temperature will not be the same from sea level to 30km since the temperature is 60C lower and density is 1% sea level meaning that heat transfer to the surface is about 4% that at sea level.
If your leading edge is blunt and your Mach number is low (1.5 you should use the oblique shock relationships.
The last part got cut off.
If your leading edge is blunt and your Mach number is low (1.5 you should use the oblique shock relationships.
Leading edge will be razor sharp, and there will be no shock.
Your web site doesn’t seem to like some symbols. last try.
If your leading edge is blunt and your Mach number is low (less than 1.5) you can assume a normal shock and use normal shock relations.
T2/T1=(1+(2gamma)/(gamma+1)(M1^2-1))((2+(gamma-1)M1^2)/(gamma+1)M1^2). (M=2, 440F at sea level) Then calculate the surface temperature as you would with flow over a flat plate, or just use T2 to be conservative (at sea level). If M greater than 1.5 you should use the oblique shock relationships.
“Leading edge sharp, and no shock”
Ok, so since it’s a subsonic vehicle, Al is OK for all speeds and all altitudes.
No, it’s supersonic. With no shock.
That’s a fascinating concept. Your aero surfaces look like an X-15’s then?
I’m not sure what you mean by “look like an X-15’s.” If you mean there’s no extreme sweep, then yes, but they can be much longer.
“No, it’s supersonic. With no shock.”
In a world without dimension, viscosity or angle of attack that might be possible, however in the real world… Or maybe this is a homework problem, not a real engineering question?
Anyway, I think I understand what you’re trying to approximate.
With a powerful enough magnifying glass all supersonic leading edges have infinite radius, so at the stagnation line the normal shock relationships are appropriate. Then assume infinite heat transfer to the surface, which means that the surface temperature is the temperature behind the normal shock. Al is good up to at least 175C, so at sea level M=1.9 is good, then at 30km (T static = 227K) so M=2.4 is ok. Both of these numbers are pretty conservative for at least 3 reasons I can think of, so you could probably justify M=2.1 at sea level and M=2.9 at 30km if you wanted to sharpen your pensile and do further calculations.
Rand,
Is this work related in any way to that company that has been doing research on shock-free supersonic craft you have been hinting about for years?
Rand,
A lot depends on how much loss of strength you can live with. If you look up the Mil-5-hdbk online, you should be able to find it. It has some temperature vs. strength curves for different aluminums. 2219 is pretty good. It actually maintains a decent amount of strength up to the 650-750F range, IIRC.
~Jon