MET pre-pt 3 Flashcards
80% of the surface pressure will be typically at
a.18000ft
b.6000ft
c.10000ft
d.36000ft
b.6000ft
Given that the Transition Altitude (TA) is 13000ft, Transition Level (TL) is FL155, QNH 983. What is the thickness of the transition layer? (assume 30ft or 9m/hpa)
a.3400 ft
b.1600 ft
c.1900 ft
d.2500 ft
b.1600 ft
(TA is 13000ft with QNH 983. TL is 15500 with SPS 1013.
Pressure difference is 1013 - 983 = 30hPa; 30 x 30ft = 900ft
ISA 15500ft (Indicated) is equivalent to 14600 (True) with QNH 983.
14600 - 13000 = 1600ft
Actual layer thickness will be 1600ft)
An aircraft is over Spain at FL280 temp -35 degrees. The ISA deviation is
[+6]
ISA temperature can be calculated by applying the ISA temp lapse rate of 2 degrees per 1000ft altitude. ISA surface temperature is +15.
ISA temp: +15 - ((28000/1000) x 2); +15 - 56 = -41 degrees.
If the OAT at 28000ft is -35 degrees, it is 6 degrees warmer than ISA at that level so the deviation is ISA +6
As an aircraft with local QNH 1028 climbs through the transition altitude TA, the indication on the altimeter will
(assume 30ft or 9m/hpa)
[drop by 450ft]
“Wind off pressure, wind off height”
QNH 1028. On climbing to the TA the altimeter setting will be reduced from 1028 to 1013. As 1013 is a lower pressure than 1028 it will lie above the surface and therefore closer to the aircraft in the air.
1028 - 1013 = 15hPa; 15 x 30ft = 450ft
The reading will reduce by 450ft as the altimeter setting is adjusted down from 1028 to 1013
Overflying mountainous terrain, the highest obstacle is at 12500ft which must be cleared by 2000ft. The air temperature is ISA -25 and local QNH 993. What minimum indicated altitude must be flown?
(assume 30ft or 9m/hpa)
a.13650 ft
b.15950 ft
c.16550 ft
d.13050 ft
b.15950 ft
MSA/True altitude 14500ft. It is colder than ISA so the Indicated altitude will be greater than the True.
TEC: 4 x (Dev) 25 x 14500/1000 (layer of air) = 1450ft
We need to find Indicated altitude so 14500 (True) + 1450 = 15950ft
You are approaching an airfield with an elevation of 500ft, QNH 996. Given that the pressure altitude is 3000ft, indicated altitude is 2500ft and temp is ISA – 21, what is your true altitude?
(assume 30ft or 9m/hpa)
a.3178ft
b.2332ft
c.2842ft
d.2668ft
b.2332ft
TEC: 4 x (Dev) 21 x 2000/1000 (layer of air) = 168ft
It is colder than ISA so the True altitude will be lower than Indicated/FL
True altitude: 2500 - 168 = 2332ft
Given the temp deviation is ISA -10 and the aircraft is at a true altitude of 2400ft, calculate the indicated altitude.
[2500ft]
If it is colder than ISA the True altitude will be lower than Indicated/FL.
TEC: 4 x (Dev)10 x 2400/1000 (layer of air) = 96ft
(True Alt) 2400 + 96 = 2496ft (indicated Alt)
An aircraft flying at FL050 with a true altitude of 4700ft is overflying a region with a local QNH of
(assume 30ft or 9m/hpa)
[1003hPa]
The difference between Indicated/FL 5000ft and True altitude 4700ft is 300ft (5000 - 4700 = 300).
The difference in pressure between SPS 1013 for FL and QNH is equivalent to 300ft; 300/30hPa = 10hPa.
As true altitude is less than FL the local QNH must be a lower number than 1013 (closer to the aircraft). 1013 - 10hPa = 1003hPa.
An aircraft departing airfield A with an elevation of 200ft has the local QNH 1010 set. The aircraft travels 60nm east with the same QNH set before landing at airfield B with an elevation of 800ft and a local QNH of 1020. What will be indication on the altimeter on landing?
a.Zero
b.800ft
c.500ft
d.300ft
c.500ft
When the aircraft lands at airfield B it will be physically 800ft amsl as that is the elevation of the airfield. QNH/msl pressure is 1020 however the aircraft altimeter has 1010 set as reference. The pressure is 1020 at msl so 1010 is above the surface and therefore closer to the aircraft so the altimeter will give a reading of less than 800ft.
1020 - 1010 = 10hPa; 10 x 30ft = 300ft difference.
800ft - 300ft = 500ft
An aircraft is crossing through TA at 8000ft with QNH 1033 set. When changing from QNH to FL the difference in true altitude will be
[zero]
When adjusting the altimeter setting the indicated altitude/FL will change however the true altitude will be the same
An aircraft departs Paris CDG with QNH 1018 set. On crossing through the TA at 8000ft, what will be the difference in altimeter indication when changing from QNH to FL?
a.600ft increase
b.zero
c.150ft decrease
d.150ft increase
c.150ft decrease
Changing from QNH 1018 at the TA to SPS 1013 the indicated altitude will decrease as 1013 is a lower pressure so will be above the surface and therefore closer to the aircraft in the air.
“Wind off pressure, wind off height”
The difference in pressure is 1018 - 1013 = 5hPa; 5 x 30ft = 150ft.
The reading on the altimeter will therefore drop by 150ft
Calculate the RH with temp 30 degrees and dew point 21 degrees
a.55%
b.45%
c.90%
d.9%
a.55%
RH = 100 - ((t - dp) x 5)
30-21 = 9
9 x 5 = 45
100 - 45 = 55% RH
When flying from low pressure to high pressure the FL displayed will
[not change]
How will true altitude be affected when flying from low pressure to high pressure maintaining a constant altitude reading?
[It will increase]
Given a transition altitude TA of 6000ft and a transition layer TL at FL055 with local QNH 1042. The thickness of the transition layer will be
(assume 30ft or 9m/hpa)
[370 ft]
TA is 6000ft above QNH 1042. TL is 5500ft above 1013.
Pressure difference: 1042 - 1013 = 29hPa; 29hPa x 30ft = 870ft
5500ft + 870ft = 6370ft; 6370ft True is equivalent to 5500ft Indicated using QNH 1042.
TL 6370 - TA 6000 = 370ft difference so the TL is 370ft deep
