Meteorology (EXAM questions)

Question 1

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:

Answer 1

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).

Question 2

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?

Answer 2

a. 5°C.
b. 1°C.
c. -5°C.
d. -7°C.

Question 3

QFE = 1,000 hPa, field elevation = 1410 ft. Calculate QNH and QNE.

Answer 3

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.

Question 4

Fill in the proper terms or values:

Answer 4

Question 5

QNH = 1,023 hPa, FE = 500 ft. Calculate the altitude and the height of FL100.

Answer 5

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.

Question 6

QNH = 1,000 hPa, FE = 1,000 ft. Calculate the altitude and the height of FL070.

Answer 6

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.

Question 7

Explain the formation of a surface inversion when there is no wind.

Answer 7

The surface is cooled by radiation and the air above is cooled by contact.

Question 8

Explain the formation of a surface inversion when there is wind.

Answer 8

Warm air moves over colder ground and is cooled by contact.

Question 9

Explain the formation of an upper inversion caused by horizontal movement of air.

Answer 9

Warm air moves over cold air.

Question 10

Explain the formation of the upper inversion caused by vertical currents.

Answer 10

When air descends, it is warmed adiabatically.

Question 11

When is air considered to be saturated and how can that be achieved?

Answer 11

Relative humidity is 100% – By cooling the air or adding water.

Question 12

QFE = 913 hPa, temperature at field elevation (2,400 ft) = 35°C. Calculate the density altitude.

Answer 12

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

Question 13

At which visibility and relative humidity is mist and haze reported in aviation weather reports?

Answer 13

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%.

Question 14

How many octas of cloud cover can you expect when clouds are reported as:

a. Broken?
b. Scattered?
c. Overcast?
d. Few?

Answer 14

a. 5-7/8.
b. 3-4/8.
c. 8/8.
d. 1-2/8.

Question 15

Define gusts.

Answer 15

Wind increases of at least 10 kt above the average wind during the last 10 minutes.

Question 16

Which visibility, weather phenomena, and cloud conditions can you expect when CAVOK is reported?

Answer 16

Visibility at least 10 km

no weather phenomena

no clouds below 5,000 ft

no CB (Cumulonimbus)

no TCU (Towering Cumulus).

Question 17

112. The following weather reports are displayed:

ETSL 081320Z 30018G23KT 1900 R21/1800D SHRA SCT018 BKN028TCU BKN050 17/10 Q1008
YLO BLU TEMPO YLO=
ETSN 081320Z 26015KT 9000 FEW008 BKN014TCU BKN030 15/11 Q1007
GRN BLU TEMPO GRN=
EDDM 081250Z 23013KT 5000 VCSH BKN014CB BKN020 17/10 Q1006 RERA NOSIG=

a. Determine the ceiling at ETSL.
b. Determine the visibility at ETSL.
c. Determine the spread at ETSL.
d. Determine the QNH at EDDM.
e. Determine the wind direction at ETSN.
f. Determine the weather phenomenon (in plain language) at ETSL.
g. Determine the weather phenomenon (in plain language) at EDDM.

Answer 17

a. Determine the ceiling at ETSL. = 2800ft
b. Determine the visibility at ETSL. = 1900m
c. Determine the spread at ETSL. = 7°
d. Determine the QNH at EDDM. = 1006 hPa
e. Determine the wind direction at ETSN. = 260°
f. Determine the weather phenomenon (in plain language) at ETSL. = SHRA
g. Determine the weather phenomenon (in plain language) at EDDM. = VCSH

Question 18

113. The following weather reports are displayed:

ETHC 151020Z VRB03KT 1000 BR M00/M00 Q1019=
AMB 152300Z 25005KT 4000 MIFG FEW010 00/00 Q1018=
GRN=
ETNW 151020Z 26007KT 3000 FG SCT002 01/01 Q1018=
YLO YLO TEMPO BCFG=

a. Determine the ceiling at ETHC.
b. Determine the ceiling at ETHB.
c. At which base, ETNW or EDDV, do you find the higher relative humidity?
d. Determine the weather phenomenon (in plain language) at ETHC, at ETHB, at ETNW, and at EDDV.

Answer 18

a. Determine the ceiling at ETHC. = NO CEILING
b. Determine the ceiling at ETHB. = NO CEILING
c. At which base, ETNW or EDDV, do you find the higher relative humidity? = EDDV
d. Determine the weather phenomenon (in plain language) at ETHC, at ETHB, at ETNW, and at EDDV. = MIST - SHALLOW FOG - FOG PATCHES - FOG IN THE VICINITY

Question 19

114. The following TAF is displayed:

ETHB 200520Z 2006/2015 VRB03KT 2000 BR SKC
BECMG 2007/2009 09007KT 9000 NSW SCT200=

a. For which time period is this TAF valid?
b. Which cloud conditions can be expected at 0630Z?
c. Which weather phenomenon (in plain language) can be expected at 1200Z?

Answer 19

a. For which time period is this TAF valid? = 06z-15z
b. Which cloud conditions can be expected at 0630Z? = SKY CLEAR
c. Which weather phenomenon (in plain language) can be expected at 1200Z? = NSW

Question 20

115. The following TAF is displayed:

ETHF 020820Z 0209/0218 22012KT 9000 NSW BKN020 BKN050
TEMPO 0212/0215 24015G25KT 5000 SHRA SCT012 BKN020 BKN050=

a. For which time period is this TAF valid?
b. Which visibility can be expected at 1400Z?
c. Which visibility can be expected at 1800Z?

Answer 20

a. For which time period is this TAF valid? = 09z-18z
b. Which visibility can be expected at 1400Z? = Prevailing 9000m, tempo 5000m
c. Which visibility can be expected at 1800Z? = 9000m

Question 21

116. The following TAF is displayed:

ETMN 091120Z 0912/0921 22015KT 9000 -RA BKN030 OVC050
BECMG 0914/0916 25020KT 3000 RA BKN007 OVC012
BECMG 0917/0918 27015KT 6000 -RA BKN018 BKN050 BKN080=

a. Which ceiling can be expected at 1300Z?
b. Which visibility can be expected at 1600Z?
c. Which ceiling can be expected at 2000Z?

Answer 21

a. Which ceiling can be expected at 1300Z? = 3000ft
b. Which visibility can be expected at 1600Z? = 3000m
c. Which ceiling can be expected at 2000Z? = 1800ft

Question 22

117. The following TAF is displayed:

ETHB 060520Z 0606/0615 19008KT 4000 HZ SKC
BECMG 0607/0609 25012KT 9000 NSW BKN025 BKN050=
TEMPO 0612/0614 29015G25KT 3000 SHRA BKN012 BKN090=

a. Which visibility can be expected at 0630Z?
b. Which ceiling can be expected at 1100Z?
c. Which ceiling can be expected at 1430Z?

Answer 22

a. Which visibility can be expected at 0630Z? = 4000m
b. Which ceiling can be expected at 1100Z? = 2500ft
c. Which ceiling can be expected at 1430Z? = 2500ft