Calculating ram rise?
Joined: Apr 2003
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From: orbital
TAT/SAT=1+(g-1)/2 *M^2
g= gamma= cp/cv. The ratio of specific heats at constant P,V. (1.4 for air at STP.)
M= mach number.
The temp as measured by a probe (eg rosemount) is slightly different to the true TAT, due to a number of factors.
g= gamma= cp/cv. The ratio of specific heats at constant P,V. (1.4 for air at STP.)
M= mach number.
The temp as measured by a probe (eg rosemount) is slightly different to the true TAT, due to a number of factors.
Last edited by Re-entry; 14th September 2006 at 15:04.
Joined: Apr 2003
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From: orbital
I suppose I should add that gamma varies with temp. For dry air g=1.401 @0 deg C. g=1.405 @ -80 deg C.
Also note TAT and SAT are expressed in deg K (ie deg C+273.16).
The value of gamma is an experimentally derived number. The theoretical value for a diatomic gas is 7/5 ie 1.4 . Air is of course only about 99% diatomic O2 and N2. The other 1% being other gases (mainly Ar).
Also note TAT and SAT are expressed in deg K (ie deg C+273.16).
The value of gamma is an experimentally derived number. The theoretical value for a diatomic gas is 7/5 ie 1.4 . Air is of course only about 99% diatomic O2 and N2. The other 1% being other gases (mainly Ar).
Last edited by Re-entry; 14th September 2006 at 02:00.
Joined: Jun 2004
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From: Australia
TAT= SAT + (TAS/87.1)^2, which is exact
Or, if your entry variable is Mach Number -
TAT= SAT + (1+.2*M^2), which is also exact
Prop operators may prefer the former, jet jockeys the latter.
Regards,
Old Smokey
Joined: Apr 2003
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From: orbital
Old Smokey, I think for the second formula you meant
TAT= SAT*(1+ .2*M^2).
And the temps must be in deg K.
btw it is not exactly exact, due to the small variations in the value of gamma, as I mentioned before. But it is mighty close!
TAT= SAT*(1+ .2*M^2).
And the temps must be in deg K.
btw it is not exactly exact, due to the small variations in the value of gamma, as I mentioned before. But it is mighty close!
Joined: Jun 2004
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From: Australia
Thanks Re-entry, my speed typing skills leave a lot to be desired.
Agreed abiout the small variations in Gamma, but most ADCs / FMCs etc use the 'standard' 1.4, so unless you're actually converting test flying results into standard results, using the standard will keep the results in line with standard instrument calibrations. As you said earlier, the difference is very small in any case.
Just in case someone copied my mis-typing, I'd better say again
TAT= SAT X (1+.2*M^2),
The problem was a cut and paste from the formula derived for TAS, didn't edit it as fully as I should have
, Thanks again
.
Regards,
Old Smokey
Agreed abiout the small variations in Gamma, but most ADCs / FMCs etc use the 'standard' 1.4, so unless you're actually converting test flying results into standard results, using the standard will keep the results in line with standard instrument calibrations. As you said earlier, the difference is very small in any case.
Just in case someone copied my mis-typing, I'd better say again
TAT= SAT X (1+.2*M^2),
The problem was a cut and paste from the formula derived for TAS, didn't edit it as fully as I should have
, Thanks againRegards,
Old Smokey

Joined: Jun 2005
Posts: 954
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From: USA
Best regards,
Westhawk

Joined: Jun 2005
Posts: 954
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From: USA
Thank you.
So since SAT is displayed in degrees C, my cheat sheet ( for quick reference)should read like this:
TAT = SAT + 273 X (1+.2*M^2) - 273?
So if:
mach = .74
SAT = -54 deg C
Then:
-54 + 273 X (1 + .2 X .74^2) - 273 = - 30.015 deg C, or about 23.085 deg C of ram rise. Is that correct?
Sorry to drag this out. I just want to make sure I get it right.
Thanks for the clarification.
Westhawk
So since SAT is displayed in degrees C, my cheat sheet ( for quick reference)should read like this:
TAT = SAT + 273 X (1+.2*M^2) - 273?
So if:
mach = .74
SAT = -54 deg C
Then:
-54 + 273 X (1 + .2 X .74^2) - 273 = - 30.015 deg C, or about 23.085 deg C of ram rise. Is that correct?
Sorry to drag this out. I just want to make sure I get it right.
Thanks for the clarification.
Westhawk

Joined: Jun 2001
Aviation Qualifications: LAME
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From: Over The Hills And Far Away
It would be :
SAT(K)(1+0.2m^2)=TAT(K)
TAT(K)/(1+0.2m^2)=SAT(K)
As 411 says, the quickest way would be a AFM graph.
SAT(K)(1+0.2m^2)=TAT(K)
TAT(K)/(1+0.2m^2)=SAT(K)
As 411 says, the quickest way would be a AFM graph.
Last edited by Techman; 14th September 2006 at 22:55.

Joined: Jun 2005
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From: USA
Actually, I'll try both tomorrow. Then I'll compare the results to the SAT/TAS reading on the air data page. Gotta do something during cruise, and this seems both safer than, and preferable to, allowing any further discussions of religion or politics to commence!
Best regards,
Westhawk


Joined: Jan 2006
Posts: 2,480
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From: In a far better place
That's probably right captjns. But it will be ever so much more fun to give the co-pilot the formula and the calculator and ask him to find the SAT from our mach number and our OAT/RAT reading, then explain about the 10% of IAS thing after!
Actually, I'll try both tomorrow. Then I'll compare the results to the SAT/TAS reading on the air data page. Gotta do something during cruise, and this seems both safer than, and preferable to, allowing any further discussions of religion or politics to commence!
Best regards,
Westhawk
Actually, I'll try both tomorrow. Then I'll compare the results to the SAT/TAS reading on the air data page. Gotta do something during cruise, and this seems both safer than, and preferable to, allowing any further discussions of religion or politics to commence!
Best regards,
Westhawk




