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► The fuel burn off is 200 kg/h with a relative fuel density of 0,8. If the relative density is 0,75, the fuel burn will be:
A) 213 kg/h
B) 200 kg/h
C) 188 kg/h
D) 267 kg/h
A fuel burn in mass per hour will not change with the SG.
► The fuel burn of an turbine engine is 220 kg/h with a fuel density of 0,80. If the density is 0,75, the fuel burn will be:
A) 235 l/h
B) 206 l/h
C) 220 l/h
D) 176 l/h
A fuel burn in volume per hour will change with the SG.
220 l/h at 0,80 = 176kg/h
176 kg/h at 0,75 = 235 l/h
► (For this Question use Fuel Planning MRJT1) Finish the ENDURANCE/FUEL CALCULATION and determine ATC ENDURANCE for a twin jet aeroplane, with the help of the table provided. Contingency is 5% of the planned trip fuel and fuel flow for extra fuel is 2400 kg/h.
A) ATC ENDURANCE: 04:07
B) ATC ENDURANCE: 03:52
C) ATC ENDURANCE: 03:37
D) ATC ENDURANCE: 04:12
Table looks like this:
-----------------------Fuel---------Time
Trip------------------5800---------2:32
Cont 5% -------------290---------0:07
Alternate------------ 1800--------0:42
Final Reserve--------1325--------0:30
Min T/O--------------9215
Extra-----------------585---------0:14
Taxi-------------------200
Ramp----------------10000
- Trip fuel (5800kg) / time (2h32) = 2289kg/h fuel flow
- Contingency is 5% of 5800 = 290kg (at fuel flow of 2289kg/h this gives 7 minutes)
- Final reserve fuel for a jet is fuel for 30 minutes
- Add these values together (5800+290+1800+1325) = Min T/O fuel of 9215kg
- Ramp was 10.000kg - 200kg lost for taxi = 9800kg at start of take off which means the extra fuel is (9800 - 9215) 585kg. Fuel flow for extra fuel is given as 2400 which makes the time 0:14
- Add up all the times = 04:05 (Closest answer being 04:07 , you might find a more accurate answer if you also counted the seconds but the given answers are far enough apart to just use the hours and minutes)
► The planned flight is over a distance of 440 NM. Based on the wind charts at altitude the following components are found:
FL50: -30kt
FL100: -50kt
FL180: -70kt
The operations manual in appendix details the aircraft performances. Which of the following flight levels (FL) gives the best range performance?
A) FL050
B) Either FL050 or FL100
C) FL180
D) FL100
The first step is to open the appendix and find the TAS and fuel flow for the 3 flight levels, you will need to use interpolation.
you will find:
FL50 = 192 (TAS) 208 (FF)
FL100 = 201 (TAS) 192 (FF)
FL180 = 216 (TAS) 163 (FF)
Now use the wind components to transfer TAS into GS:
FL50 = 192 (TAS) 208 (FF) 162 (GS)
FL100 = 201 (TAS) 192 (FF) 151 (GS)
FL180 = 216 (TAS) 163 (FF) 146 (GS)
Now you can use 2 methods, either you divide the fuel flow by the GS and look which gives you the lowest number (gross fuel flow). Or you use T = D / V and then you multiply that time by the fuel flow to find which FL gives you the lowest number (fuel consumed for 440NM).
To save time on the exam, the first method will be faster (FF/GS):
FL50 = 1.28
FL100 = 1.27
FL180 = 1.11 <-- LOWEST
► Using the following information, calculate the range.
Given:
Aeroplane mass at start up: 3.663 lbs
Fuel load (density 6lbs/gal): 74 gal
Take off altitude: sea level
Headwind: 40kt
Cruise altitude: 8000ft
Power setting: fuel throttle
2.300 RPM
20°C lean of peak
A) 633 NM
B) 844 NM
C) 730 NM
D) 547.5 NM
This is an annoying Question since the answer you would like to choose probably will be 844 NM. However, the bottom of the graph states that these are Nautical AIR miles and (although they don't clearly specify it) the Question wants to know Nautical GROUND miles.
If you get the answer (NAM) from the graph using 8000ft and the full throttle 2300RPM line, use the TAS at that line (160) to find the groundspeed (120) and with this convert NAM into NGM = approx 633 NM
► A flight is to be made from one airport (elevation 3000ft) to another in a multi engine piston aeroplane. The cruising level will be FL110. The temperature at FL110 is ISA -10°C. The temperature at the departure aerodrome is -1°C. Calculate the fuel to climb with mixture rich.
A) 10 US Gallons
B) 6 US Gallons
C) 12 US Gallons
D) 3 US Gallons
Enter the graph with 11.000ft and -17°C and find approx 9.5 USG
If we were taking off from an airport at SL this would be our answer. However, we are taking off from an elevation of 3000ft (so we don't need the fuel to climb this 3000ft). Enter the graph a second time with the elevation and -1°C and find approx 3 USG
9.5 - 3 = 6.5 USG
► An aircraft cruising at FL350 in light and variable winds turns at waypoint "ALPHA" weighing 53.500 kg and later turns way-point "BRAVO" now weighing 50.200 kg. Assuming standard conditions what is the TAS, distance and specific fuel consumption between "ALPHA" and "BRAVO"?
A) 429 kt , 627 NM , 5.26 kg/NM
B) 426 kt , 631 NM , 5.22 kg/NM
C) 429 kt , 627 NM , 6.25 kg/NM
D) 429 kt , 573 NM , 5.24kg/NM
Go into the table for the 2 masses and find the TAS (no interpolation needed here):
ALPHA (53.500 kg) = 429 TAS
BRAVO (50.200 kg) 429 TAS
For the distance however, we do need interpolation:
ALPHA (53.500 kg) = 3931 NAM
BRAVO (50.200 kg) = 3304 NAM
3931 - 3304 = 627 NAM travelled
In these 627 NAM we lost (53.500-50.200) 3300 kg of fuel:
3300 / 627 = 5.26 kg/NM
► A twin jet aeroplane is flying at FL310 - below the optimum altitude (range loss = 6%). Using the following data find the specific range:
Cruise: Mach 0.74 at FL310
Gross mass: 50.000kg
Temperature: ISA
A) 2.807 NAM / 1000 kg
B) 187 NAM / 1000 kg
C) 2994 NAM / 1000kg
D) 176 NAM / 1000 kg
In the table, find 50.000kg and find 2994 NAM . We want to know the change per 1000kg so we can either look at 51.000 or 49.000 in the table:
50.000kg = 2994 NAM
51.000kg = 3179 NAM
3179 - 2994 = 185 NAM
6% loss = 185 - 11 = 174 NAM / 1000kg
► Find the fuel flow for the twin jet aeroplane. Given:
Cruise Mach 0,74 at FL 310
Mass: 50.000kg
Temperature: ISA
A) 2.560 kg / hr
B) 1.150 kg / hr
C) 2.994 kg / hr
D) 2.300 kg / hr
In the table;
50.000kg = 2994 NAM
TAS = 434 (meaning that we travel 434NM in 1 hour)
So after 1 hour we are now at (2994 NAM - 434) 2560 NAM
Find the closest value to 2560NAM in the table and find the corresponding mass = 47.700kg
So after 1 hour, the mass of fuel we lost = 50.000 - 47.700 = 2300 kg
A) 213 kg/h
B) 200 kg/h
C) 188 kg/h
D) 267 kg/h
A fuel burn in mass per hour will not change with the SG.
► The fuel burn of an turbine engine is 220 kg/h with a fuel density of 0,80. If the density is 0,75, the fuel burn will be:
A) 235 l/h
B) 206 l/h
C) 220 l/h
D) 176 l/h
A fuel burn in volume per hour will change with the SG.
220 l/h at 0,80 = 176kg/h
176 kg/h at 0,75 = 235 l/h
► (For this Question use Fuel Planning MRJT1) Finish the ENDURANCE/FUEL CALCULATION and determine ATC ENDURANCE for a twin jet aeroplane, with the help of the table provided. Contingency is 5% of the planned trip fuel and fuel flow for extra fuel is 2400 kg/h.
A) ATC ENDURANCE: 04:07
B) ATC ENDURANCE: 03:52
C) ATC ENDURANCE: 03:37
D) ATC ENDURANCE: 04:12
Table looks like this:
-----------------------Fuel---------Time
Trip------------------5800---------2:32
Cont 5% -------------290---------0:07
Alternate------------ 1800--------0:42
Final Reserve--------1325--------0:30
Min T/O--------------9215
Extra-----------------585---------0:14
Taxi-------------------200
Ramp----------------10000
- Trip fuel (5800kg) / time (2h32) = 2289kg/h fuel flow
- Contingency is 5% of 5800 = 290kg (at fuel flow of 2289kg/h this gives 7 minutes)
- Final reserve fuel for a jet is fuel for 30 minutes
- Add these values together (5800+290+1800+1325) = Min T/O fuel of 9215kg
- Ramp was 10.000kg - 200kg lost for taxi = 9800kg at start of take off which means the extra fuel is (9800 - 9215) 585kg. Fuel flow for extra fuel is given as 2400 which makes the time 0:14
- Add up all the times = 04:05 (Closest answer being 04:07 , you might find a more accurate answer if you also counted the seconds but the given answers are far enough apart to just use the hours and minutes)
► The planned flight is over a distance of 440 NM. Based on the wind charts at altitude the following components are found:
FL50: -30kt
FL100: -50kt
FL180: -70kt
The operations manual in appendix details the aircraft performances. Which of the following flight levels (FL) gives the best range performance?
A) FL050
B) Either FL050 or FL100
C) FL180
D) FL100
The first step is to open the appendix and find the TAS and fuel flow for the 3 flight levels, you will need to use interpolation.
you will find:
FL50 = 192 (TAS) 208 (FF)
FL100 = 201 (TAS) 192 (FF)
FL180 = 216 (TAS) 163 (FF)
Now use the wind components to transfer TAS into GS:
FL50 = 192 (TAS) 208 (FF) 162 (GS)
FL100 = 201 (TAS) 192 (FF) 151 (GS)
FL180 = 216 (TAS) 163 (FF) 146 (GS)
Now you can use 2 methods, either you divide the fuel flow by the GS and look which gives you the lowest number (gross fuel flow). Or you use T = D / V and then you multiply that time by the fuel flow to find which FL gives you the lowest number (fuel consumed for 440NM).
To save time on the exam, the first method will be faster (FF/GS):
FL50 = 1.28
FL100 = 1.27
FL180 = 1.11 <-- LOWEST
► Using the following information, calculate the range.
Given:
Aeroplane mass at start up: 3.663 lbs
Fuel load (density 6lbs/gal): 74 gal
Take off altitude: sea level
Headwind: 40kt
Cruise altitude: 8000ft
Power setting: fuel throttle
2.300 RPM
20°C lean of peak
A) 633 NM
B) 844 NM
C) 730 NM
D) 547.5 NM
This is an annoying Question since the answer you would like to choose probably will be 844 NM. However, the bottom of the graph states that these are Nautical AIR miles and (although they don't clearly specify it) the Question wants to know Nautical GROUND miles.
If you get the answer (NAM) from the graph using 8000ft and the full throttle 2300RPM line, use the TAS at that line (160) to find the groundspeed (120) and with this convert NAM into NGM = approx 633 NM
► A flight is to be made from one airport (elevation 3000ft) to another in a multi engine piston aeroplane. The cruising level will be FL110. The temperature at FL110 is ISA -10°C. The temperature at the departure aerodrome is -1°C. Calculate the fuel to climb with mixture rich.
A) 10 US Gallons
B) 6 US Gallons
C) 12 US Gallons
D) 3 US Gallons
Enter the graph with 11.000ft and -17°C and find approx 9.5 USG
If we were taking off from an airport at SL this would be our answer. However, we are taking off from an elevation of 3000ft (so we don't need the fuel to climb this 3000ft). Enter the graph a second time with the elevation and -1°C and find approx 3 USG
9.5 - 3 = 6.5 USG
► An aircraft cruising at FL350 in light and variable winds turns at waypoint "ALPHA" weighing 53.500 kg and later turns way-point "BRAVO" now weighing 50.200 kg. Assuming standard conditions what is the TAS, distance and specific fuel consumption between "ALPHA" and "BRAVO"?
A) 429 kt , 627 NM , 5.26 kg/NM
B) 426 kt , 631 NM , 5.22 kg/NM
C) 429 kt , 627 NM , 6.25 kg/NM
D) 429 kt , 573 NM , 5.24kg/NM
Go into the table for the 2 masses and find the TAS (no interpolation needed here):
ALPHA (53.500 kg) = 429 TAS
BRAVO (50.200 kg) 429 TAS
For the distance however, we do need interpolation:
ALPHA (53.500 kg) = 3931 NAM
BRAVO (50.200 kg) = 3304 NAM
3931 - 3304 = 627 NAM travelled
In these 627 NAM we lost (53.500-50.200) 3300 kg of fuel:
3300 / 627 = 5.26 kg/NM
► A twin jet aeroplane is flying at FL310 - below the optimum altitude (range loss = 6%). Using the following data find the specific range:
Cruise: Mach 0.74 at FL310
Gross mass: 50.000kg
Temperature: ISA
A) 2.807 NAM / 1000 kg
B) 187 NAM / 1000 kg
C) 2994 NAM / 1000kg
D) 176 NAM / 1000 kg
In the table, find 50.000kg and find 2994 NAM . We want to know the change per 1000kg so we can either look at 51.000 or 49.000 in the table:
50.000kg = 2994 NAM
51.000kg = 3179 NAM
3179 - 2994 = 185 NAM
6% loss = 185 - 11 = 174 NAM / 1000kg
► Find the fuel flow for the twin jet aeroplane. Given:
Cruise Mach 0,74 at FL 310
Mass: 50.000kg
Temperature: ISA
A) 2.560 kg / hr
B) 1.150 kg / hr
C) 2.994 kg / hr
D) 2.300 kg / hr
In the table;
50.000kg = 2994 NAM
TAS = 434 (meaning that we travel 434NM in 1 hour)
So after 1 hour we are now at (2994 NAM - 434) 2560 NAM
Find the closest value to 2560NAM in the table and find the corresponding mass = 47.700kg
So after 1 hour, the mass of fuel we lost = 50.000 - 47.700 = 2300 kg
