ATPL Q's using Temps at flight levels
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Join Date: Jul 2010
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ATPL Q's using Temps at flight levels
Hope someone can help here:
Having trouble with questions asking for Temperatures at flight levels, usually using Local Speed of Sound and often asking to add ISA + and -.
e.g 1
"Given the following, calculate TAS in ISA conditions.
Mach 0.83, Temp at FL300 -39°C"
I used LSS=38.94 x √Temp (kelvin), but didn't get the answer of 489kts
e.g 2
"Given the following, calculate the corresponding flight level in ISA.
Mach 0.767, TAS 477 kts"
So the answer is FL167, but how???
Is there a formula to work out temperature at altitude, like what is the temperature at FL33 in ISA conditions?
If any one smarter than me can help out I'd be very thankful.
Cheers!
Having trouble with questions asking for Temperatures at flight levels, usually using Local Speed of Sound and often asking to add ISA + and -.
e.g 1
"Given the following, calculate TAS in ISA conditions.
Mach 0.83, Temp at FL300 -39°C"
I used LSS=38.94 x √Temp (kelvin), but didn't get the answer of 489kts
e.g 2
"Given the following, calculate the corresponding flight level in ISA.
Mach 0.767, TAS 477 kts"
So the answer is FL167, but how???
Is there a formula to work out temperature at altitude, like what is the temperature at FL33 in ISA conditions?
If any one smarter than me can help out I'd be very thankful.
Cheers!
Join Date: Dec 2011
Location: UK
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The first one seems to contain a piece of useless/misleading information. The -39 degrees is irrelevant/wrong if you are using ISA conditions. In ISA conditions at FL300 the temp is -45, using the formula that you correctly stated I came out with 489KT. MN=TAS/LSS LSS=38.94*sqrt(-45+273)
0.83*588=489
The second one involves a bit more thinking. I'm not sure if I have the most efficient way of doing it but I got there.
Firstly you'll need to know the temperature, you can find this by TAS/MN=38.94*sqrt(temp K)
so (TAS/MN*38.94)^2=temp = 255K
Subtract 273 from your temp to get degrees centigrade, you end up with -17.9, then use the fact that per 1000 feet the temp decreases by 2 degrees, apply that and you'll get roughly 17000FT. I assume that FL167 is the closest of the answers?
Hope this helps
0.83*588=489
The second one involves a bit more thinking. I'm not sure if I have the most efficient way of doing it but I got there.
Firstly you'll need to know the temperature, you can find this by TAS/MN=38.94*sqrt(temp K)
so (TAS/MN*38.94)^2=temp = 255K
Subtract 273 from your temp to get degrees centigrade, you end up with -17.9, then use the fact that per 1000 feet the temp decreases by 2 degrees, apply that and you'll get roughly 17000FT. I assume that FL167 is the closest of the answers?
Hope this helps
