May 20, 201214 yr Hi Everyone! Can you tell me why TAT temperature always around at (~ -28C) at FL360, I think it must be around at -55C at this flight level. I got this problem in the begining. Thank You!
May 20, 201214 yr TAT is a temperature that is dependent on mach speed. SAT (as seen on the progress page of the FMS) is the temperature (uncorrected) at the current flight level. Kenny Lee"Keep climbing"
May 20, 201214 yr Commercial Member TAT is the temperature of the probe, as the aircraft moves through the air friction generates heat. The faster you fly the more heat is generated. SAT is the static air temperature. Rob Prest
May 20, 201214 yr TAT is a temperature that is dependent on mach speed. SAT (as seen on the progress page of the FMS) is the temperature (uncorrected) at the current flight level. SAT doesn't need to be corrected, it is what is measured: the Static Air Temperature. TAT is the temperature of the probe, as the aircraft moves through the air friction generates heat. The faster you fly the more heat is generated. SAT is the static air temperature. Strictly speaking it's kinetic energy which adds to SAT to make TAT. The air which hits the leading edge, or the TAT probe, doesn't have friction with the aircraft, but it has extra energy due to speed. </pedantry>
May 20, 201214 yr Kids, kids, you're both right. Sorta. TAT is SAT plus Ram Rise. Ram Rise is Kinetic (friction) and Adiabatic (compression). Unless the 737 is having a really bad, no-wingy dive type of day, it's going to see mostly Adiabatic RR. <waitingforthenextcorrectionbecauseiforgotmostofthisstuffalongtimeago/> Matt Cee
May 20, 201214 yr Matt, ram rise is adiabatic compression. Full stop (pardon the pun). Any skin friction heating will be in addition to that, but TAT doesn't include it. The TAT probe certainly won't record any frictional temperature rise. TAT = SAT * ( 1.0 + 0.2*M2) (absolute units)
May 21, 201214 yr Hi Everyone! Can you tell me why TAT temperature always around at (~ -28C) at FL360, I think it must be around at -55C at this flight level. I got this problem in the begining. Thank You! TAT (Total air temperature) and OAT (Outside air temperature) are not the same thing when flying in the air. http://en.wikipedia....air_temperature When you're on the ground on the other hand TAT is the same as OAT. You use the TAT temperature for when or not to use de-iceing Kind regards Peter
May 21, 201214 yr Thanks for perfect answers, so I see, I meant SAT temperature. SAT (Static air temperature) is the same as OAT (Outside air temperature) Kind regards Peter
May 21, 201214 yr It is due to the Joule/ Thomson effect which also explains how fridges work. In a fridge the coolant gas is allowed to expand quickly and in doing so uses energy and therefore the gas temperature drops. The reverse happens when air is compressed as it hits the airframe and energy is put into the system and therefore the temperature rises. One of the reasons I love aviation is that nothing is quite what it seems ie as well as temperature there are all the different measurements of air speed and all have a use. Flight levels are also normally not your exact height above sea level. DME distance is measured on the slant and therefore when you get near the beacon it becomes less meaningful etc. Regards Nixon Thomas
May 21, 201214 yr When you're on the ground on the other hand TAT is the same as OAT. <More pedantry> Only if there's no wind! </More pedantry> John-Alan Pascoe
May 22, 201214 yr <More pedantry> Only if there's no wind! </More pedantry> Yes and No. The formula to calculate TAT has nothing to do with wind but it uses the Mach number which in this case is so little a factor that it is insignificant. The formula is: where: static air temperature total air temperature Mach number ratio of specific heats = approx 1.400 for dry air So if you put figures into that formula you will get exactly two figures alike for Ttotal (TAT) and Ts (OAT) as the Mach number for a plane on the ground is almost 0 even if it moves. And yes I know the mach number not will be exactly zero, but it will be so close to zero, that it will be insignificant. And for the sake of it, the temperature is calculated in Kelvin degrees, but it does not matter what kind of degrees you use (be it Kelvin, Fahrenheit og Celsius). Let's take an example. The temperature outside (Ts) is 25 degrees Celsius. We will put that into the formula: So Ttotal equals 1+(((1400-1)/2)*0)*25 (notice all the parentheses). Kind regards Peter
May 23, 201214 yr Something is wrong either with that formula or some of your points. They dont fit... disregarding all the parentheses are wrong (you missed one to wrap the right side before multiplying by Ts) it is definitely not true that the kind of degrees dont matter. It would be true, if the zero was at the same point. By that you postulate that for a given mach number, where say Ts = 25°C and Ttotal = 30°C, following applies: 30/25 = 303,15/298,15 = 86/77 which is arguably not true. meanwhile the size of the unit (equal to degree Celsius or degree Fahrenheit, or different) matters not, but the zero point must be reffered to absolute zero (leaving us with using Kelvins or deg Rankine). In your case formula works, but only because you use a M0 movement. Substitute M1 and story changes. --Peter Fabian
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