SAT TAT FREE AIR TEMP can someone explain

[eji]

Hello, I've been reading an advanced flight training manual and came across these terms;SAT, TAT, and FREE AIR TEMP GAUGE. Could somebody explain these terms so even I could understand them.It would be much appreciated. I also have another perplexing question that I just can't seem to grasp. Here goes... If a jet is travelling at max power at sea level and an identical jet is doing the same at 37,000 ft which jet is going faster ? (WHY) could someone please enlighten me. :confused:

[Denti]

SAT means static air temperature and is the temperature which you can measure if youre standing still in the surrounding air and measure it there. An airplane-fixed temperature probe can do that only parked on the ground (and in reality not even there, cause most of them are aspirated probes). So the temperature probes in airplanes measure another temperature, the so called total air temperature (TAT). That is basically the temperature at the leading edge. This temperature is higher than the SAT in flight cause its affected by the compression-effects you get when flying, the effect is notable above an airspeed of 180-200kts and/or at high altitudes. You can calculate your SAT from TAT with the mach-number, but i can't give you the formula right now, have to look it up.
The free airtemperature gauge is basically a thermometer just hold in the free airstream, like in most small airplanes (AFAIK correct me if i'm wrong).

[ 17 October 2001: Message edited by: Denti ]

[frigatebird1]

You need to think about airspeed, groundspeed and air density. In laymans terms, (for the pedants). Both aircraft will be doing the same speed through the air. As you climb through the Earth's atmosphere the air becomes less dense, that is to say, the molecules of air are further apart the higher up you go. The aircraft at 37000' will have the same speed through the air as the one at sea level but it will need to travel further to get the same amount of air molecules to pass over it's wings to give it this same airspeed. Therefore it will have a higher speed over the ground. Make any sense?

[heavy_landing]

It won't actually have the same speed through the air, but I know what you're trying to say.
Forgetting instrument error etc., SAT is the temperature of the outside air, disregarding the motion of the aircraft; it is calculated from the TAT.
TAT is the temperature of the outside air, measured directly, ie it takes account of the kinetic heating of the air - easier to measure than SAT.
TAT is of more use, if the TAT(for example) in cloud is below+11c=engine antice on.
Free air temp?no idea!

[Al Weaver]

I think the knowledge of the answers are there, I just can't sort them out yet.

The questions were posed as several, so the answers may be clearer if I knew which question is being answered. I think I got it that TAT includes kinetic energy (friction?) effects? while SAT doesn't.

I couldn't tell if "frigate" was answering a temperature question or a speed vs altitude question.

As I understand the speed vs thrust vs altitude question, one needs to sort out what level of thrust coresponds to max power at the two altitudes in question and then to sort out what expected aircraft performance would occur in speed for this thrust.

If the question were meant to answer the speed question for the same thrust at two different altitudes then the answer may be more straight forward.

OK I'll stand back and see what answers go with which question ;)

[411A]

In answer to your speed question, a jet transport aircraft could not cruise at max power at sea leval for very long because it would soon exceed Vmo.

[Checkboard]

If you are redifeining the speed question to "the same thrust at two different altitudes" then the answer is easy - the aircraft at the higher level will be travelling faster because it will experience less drag, as the air density is lower. The resistance of a body travelling throught the air can be stated in the simplest form as:

R = K þ S V²

Where K would be the drag coefficient of the aircraft determined by experiement, þ the desity of the air, S the frontal area and V the velocity.

Now, as you originally stated your question: If a jet is travelling at max power at sea level and an identical jet is doing the same at 37,000 ft which jet is going faster ?
Things become a little more difficult to answer, as the problem is complicated by other factors. A jet at sea level can produce up to ten times the thrust as a jet at altitude, so "flat chat" at sea level might mean 20,000 lbs of thrust per engine, while flat chat at the maximum altitude might mean 2,000 lbs of thrust per engine.

As 411A stated above, flat chat at sea level will quickly accelerate the jet past it's maximum speed - eventually bits may begin to fall off the airframe :eek: but if you ignore this, then I imagine the sea level jet may be able to sqeak over the line in the outright speed stakes - although it would be destroyed in the process.

[quid]

The speed questions have been answered. Now on to the temperatures.

For all practical purposes (for pilots, not engineers) SAT (static air temp), OAT (outside air temp) and Free Air Temp are all the same. TAT (total air temp) and RAT (ram air temp) are the same.

The difference is the "ram" effect. Let's say you are flying on an ISA standard day at FL 330. OAT will be -51C. If you're indicating M.80, the ram rise will be about 30C, so your RAT / TAT will be indicating around -21. Your TAS will be +/- 467. For each M.01 difference, your TAS will vary 5 kts. For each degree of RAT, your TAS will vary 1 kt.

[strobes_on]

This is from the distant past, but from memory I think it's correct .

RAM RISE = (TAS x 0.0115) squared

[Bellerophon]

Strobes_on

The formula is often quoted in the form: t°/T° = (1 + M²/5)

where t = stagnation temperature (°K), T = SAT (°K), M = Mach Number

Thus we get: Stagnation Temp (°K) = T° + 0.2T°M²

and it follows: Ram Rise (°C) = 0.2 T°M²

Substituting for Mach Number, using the formula TAS = 38.89M x Square Root of T°

we get: Ram Rise (°C) = (TAS x 0.0115)²

Well remembered!

[acegreaser]

To add to all these confusion:

If speed of sound travels FASTER in MORE DENSED air, then why is it slower if you are moving from a warmer to a colder place, assuming you are in level flight. Isn't colder air denser than warmer air. The question: Is speed of sound faster/slower in colder air.

I know the speed of sound is a function of temperature. So how does the density effect it if sound travels faster in denser air.

formula Mach 1= 39*square root of K
K=temperature in Kelvin

[john_tullamarine]

To add a couple of minor comments to Bellerophon's observations, it is usual to include the transducer loss (usually called something like the 'recovery factor') to arrive at

TAT = SAT(1 + kM^2/5)

So far as the alternative TAS equation is concerned, one should be aware that the coefficients are fairly sensitive to the constant values used in their derivation, so you might find some variation if you do the sums yourself. For aircraft where temperature rise is a problem which needs to be considered, you normally have a mach reading so there is little obvious point in playing with TAS.


Contemplating acegreaser's question, one needs to dig out the old physics textbook (dust exits stage left)...

Speed of sound in a medium (air, steel, whisky, ... whatever) is normally described in terms of

speed of sound = square root(bulk modulus/density)

where the bulk modulus (of elasticity) is a fairly simple measure of medium compressibility. Thus it is not just a matter of considering the medium density by itself. If the terms in the above equation are massaged a trifle, things substituted and the like, the equation ends up something like the more useful ratio expression

a1/a2 = square root (t1/t2)

where a is the speed of sound for two temperature states and t is the measure of the temperature state in absolute units (either Kelvin or Rankin, depending on your preference for centigrade, sorry, celcius, or farenheit)

Thus it follows that the speed of sound increases with increased temperature which is what we see - speed of sound is higher near sea level (higher OAT) and lower at cruise level (lower OAT)

[ 20 October 2001: Message edited by: john_tullamarine ]

[max motor]

I have written an air data conversion tool that will convert various air data parameters (including SAT, TAT TAS Mach etc) The calculations are based on those used in a typical air data computer. A Trial version is available for free download at:
http://www.avionic-cbt.co.uk/adc.asp

[ 20 October 2001: Message edited by: max motor ]

[Keith.Williams.]

The aircraft at high altitude will fly more slowly than at sea level, but this is not simply a function of power available. In order to understand the way in which increasing altitude affects airspeed we need to consider the following factors:


RELATIONSHIP BETWEEN INDICATED AIRSPEED AND DRAG
The airspeed indicator gives an output (IAS) that is proportional to dynamic pressure (1/2 Rho V squared), where Rho is air density and V is the true airspeed (TAS). If we ignore the complexities of compressibility, then any given value of (1/2 Rho V squared) will always give the same IAS at all altitudes. Airspeed indicators are calibrated to give IAS equal to TAS at sea level in the International Standard Atmosphere.

Drag is also proportional to (1/2 Rho V squared) so the drag force will be constant at any given IAS at all altitudes. So if our aircraft climbs at constant IAS it means that it climbs at constant (1/2 Rho V squared) and constant drag.


RELATIONSHIP BETWEEN INDICATED AIRSPEED AND TRUE AIRSPEED
As altitude increases, air density decreases. But climbing at constant IAS means climbing at constant (1/2 Rho V squared). This means that in order to maintain constant IAS and (1/2 Rho V squared), the value of V squared must increase to balance the rate of decrease in Rho.

At an altitude of 40000 feet in the International Standard Atmosphere, Rho is approximately ¼ of its sea level value. So if an aircraft climbs from sea level to 40000 feet, the decrease in Rho to ¼ of its starting value must be matched by an increase in V squared to 4 times its starting value. But V squared is TAS squared. If TAS squared increases by a factor of 4, then TAS must increase by a factor of 2. This means that at 40000 feet the TAS is approximately twice the IAS.


RELATIONSHIP BETWEEN ALTITUDE AND POWER REQUIRED
Power required is equal to drag multiplied by TAS. So when climbing at any given IAS, although the drag remains constant, the power required increases due to the increasing TAS. This means that at 40000 feet in the International Standard Atmosphere, the power required is greater than at the same IAS at sea level.

To find out how much the power required has increased we need to look again at the drag equation. Drag is proportional to (1/2 Rho V squared) where the V is TAS. When we multiply drag by TAS to get power required, we have TAS squared multiplied by TAS, which is TAS cubed. This means that power required is proportional to TAS cubed.

But at 40000 feet the TAS at any given IAS is twice its sea level value. The power required at 40000 feet must therefore be 2 cubed (or 8) times its sea level value. So flying at any given IAS requires 8 times as much power at 40000 feet as it does at sea level.


RELATIONSHIP BETWEEN ALTITUDE AND POWER AVAILABLE
The power output of any aircraft propulsion system is proportional to the mass flow of air passing through it. If we (incorrectly) assume that our engines are running at maximum RPM at sea level, then as we climb reducing air density will reduce both air mass flow and power output. At 40000 feet where the air is on ¼ of its sea level density we would have approximately ¼ of our maximum sea level power available. In reality the RPM would increase with increasing altitude. This would partly offset the reduction in air density, but would eventually become impracticable when the limiting RPM or EGT were reached. The overall effect would therefore be a reduction in power available as altitude increased.


BRINGING THESE FACTORS TOGETHER
The TAS and power required at any given IAS are much greater at 40000 feet than at sea level.
The power available is much less at 40000 feet.
Even if power available were to remain constant, the increasing power required would cause both TAS and IAS to decrease with increasing altitude.

The increase in power required is more significant than the decrease in power available, in causing airspeed to decrease as altitude increases.

[eji]

Thank you very much for your expertise in clarifying these terms for me. It's been of great help. :) :) :) :D