Help with a temperature question
 

Hi guys,

I need some help answering this question "what is the temperature at flight level 390 with a tas of 499 and mach .84", also if somebody could post the formula, I'd appreciate it. Thank you.

ok
 

Probably not good enough for an exam answer, but 'wet finger' calcs as follows:

Mach1 at -56C is 570kts appx

.84x570=479kts

TAS increases by appx 1kt/degree warmer than ISA

499 - 479 = 20kts

Therefore temp is roughly 20 degrees warmer than isa = -36C

Mmm - not far off Mathy's more accurate version!!

Cheers
mcdhu

Thank you for the answers. I came across a question on the JAA ATPL that says "what is the ISA deviation at fl390 with 499 tas and M.84". Any help would be much appreciated.

The JAA expect you to use

LSS = 38.94 time the square root of the Absolute Temperature.

And

Mach No = TAS / LSS

So working backwards

LSS = TAS / Mach = 499 / .84 = 594.047619

Dividing by 38.94 give Square root of absolute temp = 15.25546017

Squaring this gives absolute temp = 232.729065

subtracting 273 gives -40.27093497 degrees C

hi guys just for info

Mach 0.80 is 480 kt - TAT
so if you use .84 add 6 kts per 0.01 of mach

ex: mach .80 and TAT -25 degres is 480 - 25 = 455 kts

Cheers

.. or, if you want to do it the easy way -

M = V/a gives a = V/M

a/a0 = sqrt(T/T0) gives T = T0 * (a/a0)^2

so, plugging in the constants -

T = 288.15 * [(499/0.84)/661.5]^2 - 273.15

= - 40.8 C

or somewhere nearby, depending on the exact constant values you choose to use.

Going back to basics avoids carrying round offs etc through the calculations. In general, you will get a "better" answer by holding off using the constants until you get to the final equation.

what is the ISA deviation at fl390 with 499 tas and M.84

Normally we presume -56.5 C above 11km so deviation = 15.7 C degrees

I love it - the purists alongside the the pragmatists!
I seriously wish I was clever enough to be a purist.

Cheers,
mcdhu

Never mind .... some of us wish we could remember a few of the ripper rules of thumb which others come up with.

Thank you guys, it finally makes sense .........:O

sorry I ve made a small mistake
it is 487 and not 480

487 - TAT = TAS at Mach 0.80 and add or subtract 6 kts per 0.01 of Mach

Cheers again

Sebxl

Just for interest sake, I used a PC-based aviation calculator to arrive at the following:

At FL390 with a TAS of 499 one would expect the Mach No. to be 0.87 with an ISA atmosphere, resulting in a TAT of minus 23.7 C. (M 0.84 at FL 390 with ISA would produce a TAS of 482 kts, and TAT minus 25.9 C))

To achieve M 0.84 with the same conditions appears to require ISA + 16 C, thereby giving you a TAT of minus 7.7 C.

deeceethree probably not overly relevant to the first question but, nonetheless, so that others can see where the numbers are coming from ... again, for interest ..

(a) At FL390 with a TAS of 499 one would expect the Mach No. to be 0.87 with an ISA atmosphere, resulting in a TAT of minus 23.7 C.

Developing the sums from the earlier post ..

OAT = -40.8C = 232.4K

ISA = -56.5C = 216.7K

speed of sound = 661.5 SQRT (216.7/288.15) = 573.6 kt

M = 499/573.6 = 0.87

TAT = 216.7 * (1 + 0.87^2/5) - 273.15 = -23.7C

(b) M 0.84 at FL 390 with ISA would produce a TAS of 482 kts, and TAT minus 25.9 C

TAS = 573.6 * 0.84 = 481.8 kt (near enough to 482)

TAT = 216.7 * (1 + 0.84^2/5) - 273.15 = -25.9C

(c) To achieve M 0.84 with the same conditions appears to require ISA + 16 C, thereby giving you a TAT of minus 7.7 C.

We came up with ISA + 15.7C before - near enough to ISA + 16C. The equation for TAT comes up with about - 8.0C, which is close enough to - 7.7C, with the difference probably being tied up with some round offs.

capt_rs,

I seem to recall that in your original post, you asked for the applicable formula. Here it is :-

TAS = 38.975 X Mach No. X SQR (Absolute Temperature)..... (Where Absolute Temperature = SAT + 273.15)

The respondants, whilst providing sound replies, have made a rather simple question into a bleedin' dialogue!!!!!!!!!!!

By swapping around the variables to find what you want, here we go:-

Absolute Temperature = (TAS / 38.975 / Mach No.)^2

BY Substitution :-

Ta = (499 / 38.975 / .84)^2 = 232.31 Degrees Absolute - 273.15 = -40.84 Degrees C

Simple isn't it, and I thought that I was the last of the pedants. The F/L 390 is a "red herring", only necessary if you require the deviation from ISA. (In which case it is ISA + 15.66 C).

Regards,

Old Smokey (Reformed Pedant)

Thanks to all of you for taking the time to help me out. I appreciate that.