Meteorology (EXAM questions) Flashcards
List the values of the ISA:
Humidity:
Air pressure at MSL:
Air density at MSL:
Air temperature at MSL:
Vertical position of the tropopause:
Temperature at the tropopause:
Air pressure at the tropopause:
Temperature gradient below the tropopause:
Temperature gradient above the tropopause:
Humidity: 0.
Air pressure at MSL: 1013.25 hPa (29.92 inHg).
Air density at MSL: 1.225 kg/m³.
Air temperature at MSL: 15°C.
Vertical position of the tropopause: 36,000 ft MSL.
Temperature at the tropopause: -56.5°C.
Air pressure at the tropopause: 226 hPa.
Temperature gradient below the tropopause: -2°C/1,000 ft.
Temperature gradient above the tropopause: 0°C/1,000 ft (up to 20 km), 0.3°C/1,000 ft (above 20 km to 32 km).
What is the temperature in the ISA at:
a. 5,000 ft MSL?
b. 7,000 ft MSL?
c. 10,000 ft MSL?
d. 11,000 ft MSL?
a. 5°C.
b. 1°C.
c. -5°C.
d. -7°C.
QFE = 1,000 hPa, field elevation = 1410 ft. Calculate QNH and QNE.
QNH = QFE + (FE/30)
-> QNH = 1000 + (1410/30) = 1000 + 47 = 1047 hPa
QNE = (1013-QFE) x 30
-> QNE = (1013-1000) x 30 = 13 x 30 = 390ft
QNH = 1047.
QNE = 390 ft.
Fill in the proper terms or values:
QNH = 1,023 hPa, FE = 500 ft. Calculate the altitude and the height of FL100.
Altitude = FL + (QNH-1013) x 30
-> Altitude = 10 000ft + (1023-1013) x 30 = 10 300ft
Height = Altitude - FE
-> Height = 10 300ft - 500ft = 9800ft
Altitude = 10,300 ft.
Height = 9,800 ft.
QNH = 1,000 hPa, FE = 1,000 ft. Calculate the altitude and the height of FL070.
Altitude = FL + (QNH - 1013) x 30
-> Altitude = FL + (1000 - 1013) x 30 = 7000 ft + (-13 x 30) = 7000 -390 = 6610ft
Height = Altitude - FE
-> Height = 6610ft - 1000ft = 5610ft
Altitude = 6610 ft.
Height = 5610 ft.
Explain the formation of a surface inversion when there is no wind.
The surface is cooled by radiation and the air above is cooled by contact.
Explain the formation of a surface inversion when there is wind.
Warm air moves over colder ground and is cooled by contact.
Explain the formation of an upper inversion caused by horizontal movement of air.
Warm air moves over cold air.
Explain the formation of the upper inversion caused by vertical currents.
When air descends, it is warmed adiabatically.
When is air considered to be saturated and how can that be achieved?
Relative humidity is 100% – By cooling the air or adding water.
QFE = 913 hPa, temperature at field elevation (2,400 ft) = 35°C. Calculate the density altitude.
QFE: 913hPa
QNE = (1013-913) x 30ft = 3000ft
Temperature at field elevation (2400 ft) = 35°C
(Field elevation does not matter in this case)
DA = QNE + (ΔTemp x 120ft)
ΔTemp = (Current Temp - Temp ISA)
Temp ISA = 15º - 2º x 3 = 9º (-2º per 1000ft QNE)
ΔTemp = 35º - 9º = 26º
DA = QNE + (ΔTemp x 120ft)
DA = 3000ft + (26º x 120ft) = 3000ft + 3120ft = 6120ft
== 6120 ft
At which visibility and relative humidity is mist and haze reported in aviation weather reports?
Mist: 1 km up to 5 km / Relative humidity at least 80%.
Haze: 1 km up to 5 km. / Relative humidity is less than 80%.
How many octas of cloud cover can you expect when clouds are reported as:
a. Broken?
b. Scattered?
c. Overcast?
d. Few?
a. 5-7/8.
b. 3-4/8.
c. 8/8.
d. 1-2/8.
Define gusts.
Wind increases of at least 10 kt above the average wind during the last 10 minutes.
