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Jane-DoH
3rd Apr 2011, 05:09
Okay, stagnation temperatures are the areas on the plane where the air is slowed all the way down to zero right? How much of an airplane's nose and leading edge experiences these temperatures? I'm guessing it couldn't be too much considering most of the air would be slowed down to a lower mach number, but not all the way to zero.

kenparry
3rd Apr 2011, 10:30
Not quite there.

Stagnation points are the places where the airflow stops. That's where the stagnation temperature is reached.

Where?

A point on the nose;
A line down the leading edge of each flying surface;
A line round the intake lip of each engine;
(In the case of the second and third, only if the leading edges are perpendicular to the local airflow) for example.

And, yes, elsewhere there is a lesser but variable amount of kinetic heating.

How much heating? Rule of thumb, stagnation temperature increase above ambient in Celcius is TAS(knots)squared/100
example: 500 kt TAS, increase 25 deg C.

Chris Scott
4th Apr 2011, 21:52
kenparry,

That rule of thumb for estimating the increment to be applied to SAT to get TAT seems very neat and simple.

For the record, did you mean:

[TAS/100]^2

(where TAS is in knots) ?

Chris

Jetpipe.
6th Apr 2011, 14:03
First.
At stagnation point, the flow is at zero speed relative to the body for a very small layer of molecules at the surface of, f.ex., the leading edge. This means that this amount of still, warmer air is too small and will not affect the main airflow.

The air is a compressible mix of gases which changes characteristics as speed is increased. At subsonic speeds the air is thought to be isentropic and therefore the first law of thermodynamics can be applyed. That is that as the dynamic pressure decreases (velocity/IAS) the static pressure and temperature of air is increased and vice versa. All this happens adiabatically, meaning without any energy loss (heat) from the system. Because of all this, the stagnation temperature will be equall to Total Air Temperature.

TAT=SAT+Ram Rise

So Ram Rise is what we measure with the rule of thumb mentioned above..
(The exact formula is RR=TAS^2/2Cp , TAS in kt, and Cp is the specific heat for constant pressure, so a more exact rule of thumb would be
RR=[TAS/87]^2 ) These equations do NOT take account for Kinetic heating

Ofcourse there is always some friction from the air and this is what is called Kinetic Heating/Aerodynamic Heating but for subsonic speeds this is almost negligible. The modern TAT measuring probes have a recovery factor q=0.98, so the ADC of the aircraft multiplyes the above equation with q to get the correct TAT.

At supersonic and hypersonic speeds the equations used for measuring TAT are different as the air starts behaving as a non isentropic fluid.. Meaning that the Stagnation temperature will not be equal to TAT..

Jetpipe.

Jane-DoH
9th Apr 2011, 00:38
TAS is used because it roughly translates to the mach number right?

selfin
9th Apr 2011, 02:05
No, it is used in lieu of the Mach number owing to a rearrangement of the conventional form, T0/T1 = 1+[(γ-1)/2]*M^2, noting M^2 = u^2/(γ R T1) in a calorically perfect gas.

Cp can be written γR/(γ-1) and the expression u^2/2cp = T0-T1 is consistent with this. Rewriting as [u/sqrt(2cp)]^2 you'll see that expressing u (TAS) in knots requires you to scale sqrt(2cp) accordingly (i.e. by 3600/1852 for the conversion from m/s to kts) to achieve the same result.

Cp for air (at room temperature - i.e. vibrational modes frozen out) is about 1004.5 J kg^-1 K^-1 so you can play with the numbers to see where the approximate constant of 87 comes from.

Jetpipe.
11th Apr 2011, 10:30
Since equations link airspeeds to eachother

EAS=TAS sqrt(ρ/ρ0)
where ρ= actual density , and ρ0= standard density at sea level 1.225 kg/m^3

M = TAS/sqrt(γ R T)
where γ= the adiabatic indeex (1.4) , R= the specific gas constant (286.9 J/kg K) , and T= Static Air Temperature in Kelvin

The Ram Rise equations are linked in the same way. We choose for which airspeed we want the TAT-SAT equation to be solved for, TAS, Mach or EAS.

TAT - SAT = TAS^2 / 2Cp

TAT - SAT = 0.5 M^2 (γ-1) SAT
or for others more convinient TAT/SAT = 1 + [(γ-1)/2] M^2

TAT - SAT = (ρ0/ρ) EAS^2 / 2Cp
Jetpipe.

Jane-DoH
13th Apr 2011, 03:48
SAT = Stagnation Air Temperature?

D O Guerrero
13th Apr 2011, 07:27
Static Air Temperature?

Jetpipe.
13th Apr 2011, 10:01
TAT= Total Air Temperatrure (or T0 by selfin)

SAT= Static Air Temperature (or T1 by selfin)

jetpipe.

kenparry
13th Apr 2011, 18:32
Chris Scott:

Yes, exactly what I meant, but I garbled it, as you suspected. Thanks.

Jane-DoH
16th Apr 2011, 01:24
Okay, I think this would be more helpful if I saw numbers plugged into this equation. Say we had an outside air temperature of 220 kelvin and we were doing Mach 2?

Jetpipe.
18th Apr 2011, 19:27
Jane-DoH, as I wrote in previous post, these equations can only be applied for subsonic speeds to get an exact answer. I guess we could use them to answer your example with some kind of correcting factor ''q'' as the aerodynamic heating at M 2.0 is still low. But then we would be assuming laminar isentropic flow for supersonic airspeeds which is quite wrong, however as i said, we should end up with an answer close to what we would have if we used the correct (quite complex) equations for super/hypersonic speeds..

So what do we know, M=2.0 and SAT=220 Kelvin

TAT - SAT = 1/2 M^2 (γ-1) SAT q

(a reasonable value for q at M 2.0 would be i think 3-5%, q= 1.03-1.05)

TAT - SAT = 0.5*4*(1.4 - 1)*220*1.05

TAT - SAT = 184.6 degrees of Ramrise

TAT= 184.6 + SAT = 404.6 K = 131 Celsius, would be heating up the leading edges of the aircraft..

Hope this helps :ok:
Jetpipe.

Jane-DoH
19th Apr 2011, 03:37
Jetpipe

Jane-DoH, as I wrote in previous post, these equations can only be applied for subsonic speeds to get an exact answer.

I did not know that

I guess we could use them to answer your example with some kind of correcting factor ''q'' as the aerodynamic heating at M 2.0 is still low.

Where do you derive "q" from?

But then we would be assuming laminar isentropic flow for supersonic airspeeds which is quite wrong, however as i said, we should end up with an answer close to what we would have if we used the correct (quite complex) equations for super/hypersonic speeds..

Understood. What formula is that out of curiousity?

blackmail
19th Apr 2011, 10:51
JD,
it might be the formula of Raleigh for supersonic speeds :

Pt-p/p =[ 166,92158M^7/(7M^2-1)^2,5]:cool: -1

Pt = Total Air Pressure
p = Static Air Pressure
M^7 = Mach to the power 7
M^2 = Mach to the power 2(square)
^2,5 = power 2,5

/ = diveded by. ...

hope this helps, but don't try to commit this formula to memory.
bm.

Jane-DoH
20th Apr 2011, 18:28
blackmail

JD,
it might be the formula of Raleigh for supersonic speeds :

Pt-p/p =[ 166,92158M^7/(7M^2-1)^2,5]:cool: -1

Pt = Total Air Pressure
p = Static Air Pressure
M^7 = Mach to the power 7
M^2 = Mach to the power 2(square)
^2,5 = power 2,5

And this gives you the "q-factor" or does it give you the stagnation temperature? I don't see any temperature figures in there...

BTW: How do you determine total air-pressure? Can I just set it as an "x-value" and then fill in the rest, then solve for it?

selfin
20th Apr 2011, 22:59
these equations can only be applied for subsonic speeds to get an exact answer

Unless a real gas model is constructed, including quantum-mechanical effects, the solution will be an approximation regardless of the regime. Within the static temperature limits of the troposphere/stratosphere (in ISA) the assumption of a calorically perfect gas is justified up to, shall we say, around Mach 2 or 3 or higher.

And this gives you the "q-factor" or does it give you the stagnation temperature? I don't see any temperature figures in there...

A correction term, in whatever form is chosen, can be included to account for discrepancies between expected values and those measured in practice. The Rayleigh supersonic pitot equation given above by Blackmail does not include any.

BTW: How do you determine total air-pressure? Can I just set it as an "x-value" and then fill in the rest, then solve for it?

Leave the Mach number as the independent variable to avoid iteration. Use the post-shock total pressure and pre-shock static pressure in the Rayleigh equation. A full derivation is given in Anderson (Fundamentals of Aerodynamics) which PBL has previously advised you to consult. Alternatively look up P. Balachandran (Fundamentals Of Compressible Fluid Dynamics), chapter 9, on Google Books. For the hypersonic regime refer to Anderson, J. (2000). Hypersonic and High Temperature Gas Dynamics. AIAA, or Fletcher, D.G. (2004). Fundamentals of Hypersonic Flow-Aerothermodynamics. NATO RTO AVT Lecture Series: RTO-EN-AVT-116 (http://www.rta.nato.int/Pubs/RDP.asp?RDP=RTO-EN-AVT-116).

Jane-DoH
1st May 2011, 04:50
selfin

I don't remember seeing the Rayleigh equation in there...