Talk:Total air temperature

Total Air Temperature is a term used generally in aviation. It is the temperature measured approximately by a temperature probe mounted on the surface of the aircraft by or on the pitot intakes. It is higher than the static air temperature due to the translational energy of the molecules hitting the proble while the aircraft is in motion.

In summary, an aircraft in motion with a true airspeed will have a TAT greater (hotter) than the static air temperature by itself.

-- Perhaps this article should be merged with the article for stagnation temperature. —Preceding unsigned comment added by 204.52.215.1 (talk) 00:20, 31 March 2008 (UTC)

Static temperature redirect

This is simply a wrong redirect. Total air temperature is defined correctly here and is completely separate from static temperature. If anything, based on the encyclopedia article, it would be best suited redirecting to the main temperature article, perhaps to the section on theoretical definition. 128.6.73.200 (talk) 18:22, 3 July 2009 (UTC)

I agree. Thanks for pointing it out. I have amended the redirect so that it now goes to Temperature. Dolphin51 (talk) 00:42, 4 July 2009 (UTC)

Remove kinetic heating

I propose to remove the reference in this article to kinetic heating. If you follow the link to stagnation temperature it is apparant that "total air temperature " is synonymous with "stagnation temperature" for isentropic compressible flow, which is adiabatic and reversible by definition. As used in subsonic aviation, the term "total air temperature" implies isentropic flow. Isentropic flow has no friction (kinetic) heating by definition.

Kinetic heating may be a factor in the ram rise of a direct reading thermometer, but it is not a component of total air temperature, as defined, (and should not affect a TAT probe).

The subsequent equations for ram rise are correct for isentropic temperature rise, and take no account of friction heating. The preceding references to kinetic heating are therefore misleading.

I also propose to remove the reference to molar mass (which is incorrect anyway). This is an aviation article. In aviation engineering it is conventional to define R as the specific (per kg) gas constant for air (287.05 J/kg.K). This eliminates the need for molar mass (m) in the equation which is required when the universal (per mol) gas constant is used.

Also note that (gamma - 1)/(gamma R) simplifies to 1/Cp (Cp = specific heat at constant pressure). Therefore the ram rise equation simplifies to RR = V^2/(2*Cp)

I also propose to change T0 to Ts for static temperature. This is to avoid confusion with subscript 0 used for ISA sea level values in other air data articles (i.e T0 = 288.15 K).

Benbow (talk) 17:04, 15 March 2011 (UTC)

@Benbow: You have written "total air temperature" is synonymous with "stagnation temperature" for isentropic compressible flow. That is what is currently stated by Wikipedia (see Stagnation temperature#Adiabatic) but it is not quite as simple as that. Stagnation temperature is the temperature of the fluid at a stagnation point. If the fluid is not stagnant anywhere in the flow field then there is no stagnation point and no stagnation temperature. However, total temperature is defined for every point in the flow field. For example, if we pick a point in the flow field, and the static temperature at that point is T_s, and the speed of the flow at that point is V, then the total temperature T_t at that point is:

$T_{t}=T_{s}+{\frac {V^{2}}{2C_{p}}}\,$ where C_p is the specific heat at constant pressure

Total temperature is the same throughout the flow field. Where the speed of the flow is high, the static temperature is low; and where the speed is low, the static temperature is high. If there is a stagnation point somewhere in the flow field, the temperature at that point will be equal to the total temperature.

So the two are numerically equal, although they are different things. There are many examples of this in physics. For example, Newton's Second Law states that force is equal to mass times acceleration. Mass times acceleration is numerically equal to the resultant force, but "mass times acceleration" is not synonymous with "force" - they are different things. If "mass times acceleration" was synonymous with "force" there would be no need for Newton's Second Law, and no-one would credit Newton with the momentous nature of his discovery. Dolphin (t) 01:42, 16 March 2011 (UTC)

I agree with you Benbow and i think you would agree that there is a lot needed to be done in this article! Seems that you have a lot more knowledge in this subject than i have. I am just a pilot student with maths background who searched in wiki for Ram rise and only found a couple of lines of it written. My intention was to expand the article from my own experience and notes for the time being, and return later on when more time would be available and more information gathered (i m right in the middle of my ATPL exams right now!). You are more than welcome to make any qualitative changes you think will improve the article.

Jetpipe (talk) 00:12, 16 March 2011 (UTC)

Hi Dolphin. I understand your point that total air temerature is not synomymous with stagnation temperature. I think whilst Total air temperature may be defined mathematically for any piont in the flow field it is only realised physically when the air is brought to rest at stagnation (e.g in TAT probe). Same arguement applies to total pressure and static pressure.

Corrected my typo for Cp.

However the point I was really making was about friction (kinetic) heating. As far as TAT theory is concerned, all the equations in the article are derived on the assumption of isentropic compressible flow. Therefore the reference to friction heating is misleading since this would violate the isentropic assumption underlying the equations. For real measurement with a TAT probe I think any non isentropic effects (friction, heat transfer etc.) are likely to be small and would be accounted for in calibation by use of the empirical recovery factor (e). This point is made in the first part ofthe article. When TAT is displayed in cockpit instruments, I think it the the true (isentropic) TAT which is displayed. Any known calibration errors of the TAT probe are adjusted for by the air data computer.

Jetpipe, good luck with your exams. We are often cautioned that Wikipedia is not to be taken unverified as a reliable source of knowledge. I think the same caveat should sometimes be applied to what they teach on ATPL courses.

As Dolphin states, the equation for ram rise in its simplest form is:

$T_{t}-T_{s}={\frac {V^{2}}{2C_{p}}}\,$

which you can derive on the basis that R=Cp-Cv and gamma = Cp/Cv, or from first principles on the basis of conservation of energy and the definitions of enthalpy and Cp, assuming isentropic flow.

For temperature in celsius or kelvin and speed in knots this evaluates to Tt-Ts=(V/87)^2

(use Cp = 1005 J/kg.K to evaluate in SI units, then convert 1 m/s = 1.944 knots)

Benbow (talk) 11:26, 16 March 2011 (UTC)

According to my ATPL book (Oxford Aviation Academy, Aircraft general knowledge 4 - Instrumentation, 5th Edition):

"...The laws of thermodynamics are such that the combination of kinetic and adiabatic heating will always add up to a figure known as the Total Ram Rise. Unfortunately, any measurement process always has leaks and inefficiencies, so we do not measure the full ram rise. The amount of ram rise actually sensed is called the Measured Ram Rise. If we could measure the Total Ram Rise, we would measure the Total Air Temperature. However, in practise, what we measure is the Ram Air Temperature which is lower..."

So, as i understand, it sais that by using the equations we are able to calculate the Total Ram Rise, but the instruments can't measure the total value and this value is called Measured ram rise.

TAT > RAT

and

TotalRR > MeasuredRR

If I am right, your point Benbow is, that allthough kinetic friction exists, the equations in the article does not take account for it. But at the same time you say it is being accounted in the recovery factor $e$? Air as you say is compressible but is the airflow always isentropic? A quick thought is that at Low Mach the air does not compress but the airflow is isentropic(?) and at High Mach numbers the air compresses but the airflow is not isentropic. Am I wrong? If an aircraft is flying at a constant presure altitude with increasing Mach speed, isn't the system gaining energy that could be transformed into heat? Just trying to sort things up in my head!

Jetpipe (talk) 18:13, 16 March 2011 (UTC)

Your flight school says "...The laws of thermodynamics are such that the combination of kinetic and adiabatic heating will always add up to a figure known as the Total Ram Rise "

How do they define their terms "kinetic heating" and "aidabatic heating"? How are they different? To me, "heating" implies transfer of heat energy across the system boundary, so the phrase "adiabatic heating" is a contradiction in terms. I prefer to talk of "adiabatic temperature change" to avoid possible misunderstanding.

On reflection, I think we can derive the ram rise as a functon of TAS (V) on the basis of conservation of energy (conversion of kinetic energy to enthalpy). So the isentropic assumption would not be essential to this derivation. However in practice I think any friction (viscous shear) heating component would be negligible in a TAT probe and therefore somewhat misleading in the TAT article.

The TAT probe is similar to a pitot tube but has a slight leak rate. In the TAT probe the air is brought almost to rest as it passes the sensing element. Therefore I suggest that any viscous shear heating of the sensing element heating would be quite negligible. Ideally, a TAT probe would be like a pitot probe and bring the air completely to rest. However if it did this the air in the probe would be stagnant and heat would transfer in from the (de iced) casing causing erroneous measurement. Therefore there has to be a slight flow through the TAT probe to minimise heat transfer error. Because of this the air is not completely brought to rest in the probe and therefore not not 100% of the kinetic energy of the air is recovered. Therefore the measured temperature is a little less that the true TAT. To account for this we apply an empirical recovery factor (e), to the measured temperature when calculating SAT. For modern TAT probes the revcovery factor is only a little less than 1. try this NASA link and scroll to temperature.

http://www.nasa.gov/centers/dryden/pdf/88377main_H-2044.pdf

No mention here of friction heating. They do mention the possible error due to heat transfer from the casing and possible self heating of the sensing element (which must have an electric current in it).

Returning to the theory however, we also have the equations which express the ratio Tt/Ts as a function Mach number. These are the most important to the air data computer because the ADC does not know TAS a priori. It calulates Mach number on the basis of Pt and Ps. Having done so, it can use the above equations to calculate Ts from (measured) Tt, and hence calculate TAS. Now in this process the isentropic assumption is more important since the equations relating Mach number, Pt, and Ps, Tt and Ts are based on isentropic flow theory.

As to your second point: Air is always compressible. However at low flight speeds the compression is low enough to be neglected for practical purposes. We can therefore develop very simple mathematical models (such as Bernoulli's equation) based on the assumption of incompressible (constant density) flow.

At higher subsonic speeds we can no longer neglect compressibility, and we have to develop more complex mathematical models to account for the compression. The "next most simple" mathematical model assumes that air is compressible and that the compression process is isentropic. The model also assumes that air obeys the Ideal Gas Equation of State (P = R Rho T).

Isentropic compression is both adiabatic and potentially reversible. Think of a thermally insulated frictionless bicycle pump sealed at the end which you use as a pneumatic accumulator: in this theoretical case you could recover 100% of the mechanical energy.

The "isentropic compressible flow" model holds good up to high subsonic speeds (but cannot account for flow across a supersonic shock wave). This model is the basis for the calibration of airspeed indicators and Mach meters in subsonic aircraft, as well as the TAT - SAT conversion. Above Mach 3 even more complex models are needed.

One way into the ram rise equation (perhaps not rigorous) goes like this:

Start with the incompressible pitot tube equation (in what follows, subscript t means total, s means static)

Pt-Ps = 1/2 rho V^2

This is a pressure equation. Pressure is energy per unit volume.

(Have you ever heard a controller say "the QNH is 1010 hectojoules per cubic meter"? Perhaps not. I would consider him a bit eccentric, but would congratulate him on his understanding of science: he realises that a pascal is equivalent to a joule per cubic meter)

If I devide through by rho, I get the equation in terms of energy per unit mass, which we call specific energy.

Pt/rho - Ps/rho = 1/2 V^2

This is a specific energy eqation, but it neglects the internal energy (which is related to temperature). For incompressible flow this omission does not matter since the internal energy is constant so it would cancel out. But not so once we have to consider compressible flow. So, moving to compressible flow, let's add in the specific internal energy which I will call u. Furthermore rho is no longer constant so I need to distingusih between rhot and rhos

(Pt/rhot + ut) - (Ps/rhos + us) = 1/2 V^2

Now I can make a substitution P/rho + u = h where h is called specific enthalpy

ht-hs = 1/2 V^2

The specific kinetic energy of the air is converted into specific enthalpy when the air is brought to rest.

For an ideal gas, a change in specific enthalpy is proportional to change in temperature, the constant of proportionality being the specific heat at constant pressure (Cp)

ht-hs = Cp(Tt-Ts)

So

Tt-Ts = V^2 /(2*Cp)

As an exercise, try deriving an expression for the Dry Adiabatic lapse Rate using similar reasoning. (hint: use specific potential energy instead of specific kinetic energy)

(I normally prefer to use symbol e for specific internal energy, but in the present discussion we have used e for recovery factor, so I'll use u for specific internal energy) Benbow (talk) 17:58, 18 March 2011 (UTC)

First of all, I want to thank you for your valuable help. You have helped me come a step closer in understanding the principles of the air. And I did my homework...(But I cheated a little bit and looked at the equation for DALR because I needed to see what i was looking for). So:

Potential Energy is

$E_{pot}=mgH$

But as we changed the Kinetic Energy for airflow we must change Potential energy too. We assume a Presure 2 at Height 2 and a P1 at H1.

P2 - P1 = - Rho g (H2 - H1)

P2/Rho2 - P1/Rho1 = - g (H2 - H1)

This is, as you say, a specific energy equation and it neglects the internal energy. Now i think I have to add the specific internal energy u as the equation is potentialy reversible?

(P2/Rho2 + u2) - (P1/Rho1 + u1) = - g (H2 - H1)

and I think the differential expresion should be something like this

d(P/Rho) + du = - g dH

but

P/Rho + u = h , which is the specific enthalpy, so next

dh = - g dH but the change of specific enthalpy is given by dh= Cp dT

From these two last equations we get:

g/Cp = -dT/dH

which is the DALR.

Jetpipe (talk) 02:44, 19 March 2011 (UTC)

Great, but we are getting off topic. If you wish, go to Pprune and send me a PM under the username Rivet gun Benbow (talk) 14:17, 19 March 2011 (UTC)

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