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View Full Version : ISA Temperature & ISA Deviation


jlowell
02-06-2018, 02:21 AM
I was browsing so site and youtube to understand the PA, DA or ISA Temps Deviation.

Then I got this formula:
Pressure Alt= Elevation + (1013 - QNH) x 30
ISA Temp = 15 - (2 x Elevation/Altitude/1000)

After than i have browsed some website for online test to do some practice and solve some question and I got all good but saw a some question on a forum that i try to solve with question with answer but when i tried to do my answer is not same as their answer.

Example:
What is the ISA Temperature & ISA Deviation if the height is 37,000 feet & is -45 degrees ?
(Ans: -57 degrees for ISA temperature & +12 for ISA deviation)

I used the formula : ISA Temp = 15 - (2 x Elevation/Altitude/1000)
ISA Temp = 15 - (2 x 37000/1000)
ISA Temp = 15 - (2 x 37) = -59 degress
ISA deviation= -45 - -59
ISA deviation= -45 + 59 = +14 degrees

How he get -57 for Isa Temp and +12 for deviation ?

Example 2:
What is the ISA Temperature & ISA Deviation if the height is 57,000 feet & is -67 degrees ?
(His Ans: -57 degrees for ISA temperature & -10 for ISA deviation)

ISA Temp = 15 - (2 x 57000/1000)
ISA Temp = 15 - (2 x 57) = -99degrees
ISA deviation= -67 - -99
ISA deviation= -67 + 99 = +32degrees

How he get -57 for Isa Temp and -10 for ISA deviation ?

The explanation there for Example 1 is:

In the ISA the tropopause is at 36000 feet, where the temperature is –56.25 degrees Celsius (or approximately –57). Above the tropopause the temperature remains constant.

So at 37000 feet the ISA temperature is approximately –57 degrees Celsius.

ISA deviation = Actual Temp – ISA temp

So if actual temp = -45 at 37000 feet then:

ISA deviation = -45 – (-57) = +12

The explanation there for Example 2 is:

(Ans: -57 degrees for ISA temperature & -10 for ISA deviation)

In the ISA the tropopause is at 36000 feet, where the temperature is –56.25 degrees Celsius (or approximately –57). Above the tropopause the temperature remains constant.

So at 37000 feet the ISA temperature is approximately –57 degrees Celsius.

ISA deviation = Actual Temp – ISA temp

So if actual temp = -67 at 36000 feet then:

ISA deviation = -67 – (-57) = -10


But still confuse since the explanation is based on 2 altitude that i don't know how they get it.

Thank for help.

BTW am learning the theory for PPL just got an ideas before I go to do my PPL since I will have to go to south africa and time is money so I start to understand before going :)


jlowell
02-06-2018, 05:15 AM
My first question how do he get -57 for Isa Temp and +12 for deviation ? following their explaination above


From what i had understand on the web the temperation from 38 000 ft to 65,500 feet is constant.

My second question is that mean that from 38 000 ft to 65 000ft we don't need to calculate Isa temperation since it is the same -57 degree ?

galaxy flyer
02-15-2018, 01:21 PM
The “standard” atmosphere is -56.5C at 36,000’, then the lapse rate is 0, as, by definition, one is in the stratosphere where the temp remains constant.

GF


Allegheny
02-21-2018, 03:46 AM
One of the things that might be confusing the OP is that the Tropopause is that it does change height. The STANDARD ATMOSPHERE MODEL shows the Trop at around 36000ft. In actuality it changes:

The tropopause height does not gradually drop from low to high latitudes. Rather, it drops rapidly in the area of the subtropical and polar front jets (STJ and PFJ respectively in the Figure on the left), as shown in the Palmen-Newton model of the general circulation (Fig 12.16 or Fig on left). Especially when the jet is strong and the associated front at low levels intense, then the tropopause height drops suddenly across the jet stream. Sometimes the tropopause actually folds down to 500 hPa (5.5 km) and even lower, just behind a well-defined cold front. The subsided stratospheric air within such a tropopause fold (or in the less pronounced tropopause dip) is much warmer than the tropospheric air it replaces, at the same level, and this warm advection aloft (around 300 hPa) largely explains the movement of the frontal low (at the surface) into the cold airmass, a process called occlusion.

The height of the tropopause (http://www-das.uwyo.edu/~geerts/cwx/notes/chap01/tropo.html)

So, the apparent problem of the Trop layer being static should not be assumed. Use the formula and forget the standard trop level. It's like the 15 Deg. C / 59 F value. An assumed standard only, used for the basis of reference, not an in-fact condition most of the time.

rickair7777
02-21-2018, 05:55 AM
For academic problems, unless otherwise stated, assume the lapse rate stops lapsing at 36,000 and -57* C.

In the real world, that will rarely be the case. But any airliner that can fly that high will do all the math for you.

HercDriver130
02-21-2018, 11:41 AM
For academic problems, unless otherwise stated, assume the lapse rate stops lapsing at 36,000 and -57* C.

In the real world, that will rarely be the case. But any airliner that can fly that high will do all the math for you.

ya tell my jet that its all academic when Im at 35000 over Siberia and the temp is -68.... lol... when it gets so cold for so long you get fuel freeze caution messages.... lol