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► A jet aeroplane has a cruising fuel consumption of 4060 kg/h, and 3690 kg/h during holding. If the destination is an isolated airfield, the aeroplane must carry, in addition to contingency reserves, additional fuel of:
Answer: 8120 kg
For an isolated airfield procedure, a Jet needs 2 hours at cruise consumption as additional fuel
► The flight crew of a turbojet aeroplane prepares a flight using the following data:
Flight leg distance= 3 500NM
Flight level: FL310
True airspeed: 450 kts
Headwind component at this level: 5 kts
Initially planned takeoff mass (without extra fuel on board): 180.000 kg
Fuel price: 0.35 Euro/l at departure, 0.315 Euro/l at destination
To maximize savings, the commander chooses to carry extra fuel in addition to that which is necessary. Using the appended annex, the optimum quantity of fuel which should be carried in addition to the prescribed quantity is:
Answer: The fuel transport operation is not recommended in this case
Before starting with Questions like these, first have a look at the fuel prices because the Fuel tankering procedure is only financially interesting when the price at destination is higher than that at departure.
Price destination = 0.315
Price departure = 0.35
In this case the procedure is not recommended. This saves time during the exam!
► The flight crew of a turbojet aeroplane prepares a flight using the following data:
Flight leg distance= 4 000NM
Flight level: FL310
True airspeed: 450 kts
Headwind component at this level: 50 kts
Initially planned takeoff mass (without extra fuel on board): 170.000 kg
Fuel price: 0.27 Euro/l at departure, 0.30 Euro/l at destination
To maximize savings, the commander chooses to carry extra fuel in addition to that which is necessary. Using the appended annex, the optimum quantity of fuel which should be carried in addition to the prescribed quantity is:
Answer: 8.000 kg
Step 1) Is the fuel price at destination higher than at departure? In this case "yes" so we can continue the calculations
Step 2) Enter the given graph at the right with NAM and at the left with the fuel/price ratio
NAM = (NGM / GS) x TAS
NAM = (4000 / 400) x 450
NAM = 4500
Fuel/price ratio = Price dep/Price dest
Fuel/price ratio = 0.27/0.30
Fuel/price ratio = 0.90
At the intersection on the graph we go vertically down and read an Optimum aircraft weight of 178.000 kg
Optimum = 178.000 kg
Our initially planned = 170.000 kg
We can therefore take 8.000 kg extra
► A turbojet aeroplane is prepared for a 1 300 NM flight at FL350, with a true airspeed of 460 kts and a headwind of 160 kts. The takeoff runway limitation is 174.000 kg, the planned departure mass is 160.000 kg. The departure fuel price is equal to 0.92 times the arrival fuel price (fuel price ratio = 0.92). In order for the airline to optimize its savings, the additional fuel quantity that must be loaded on board is:
Answer: 14.000 kg
Enter the graph by the same way we did in the previous Question and find an optimum aircraft mass of approx 210.000kg. However, this time we have a runway limitation of 174.000 kg which we cannot exceed for any take off. If we would load up until the optimum we will exceed this limit.
The extra fuel we can take on will therefore be 174.000 - 160.000 = 14.000
► The flight crew of a turbojet aeroplane prepares a flight using the following data:
Flight leg distance= 3 500NM
Flight level: FL310
True airspeed: 450 kts
Headwind component at this level: 55 kts
Initially planned takeoff mass (without extra fuel on board): 180.000 kg
Fuel price: 0.27 $/l at destination
The commander may carry on board 8.000kg more fuel than that which is necessary. For this fuel transport operation to be cost effective, the maximum fuel price at departure must be:
Answer: 0.24 $/l
Step 1) Enter the graph on the right with the NAM and on the bottom with the optimum weight.
NAM = (NGM / GS) x TAS
NAM = (3500 / 395) x 450
NAM = 3987 (approx 4000 to make it easier in the graph)
Optimum aircraft weight = 180.000 + 8.000 = 188.000 kg
Step 2) Find the Fuel/Price ratio
Fuel/price ratio = 0.90
0.90 = Price at dep/0.27
Price at dep = 0.27 x 0.90
Price at dep = 0.24
► Assuming a departure/destination fuel price ratio of 0.92, the commander decides to optimize fuel tankering by using the following data:
Cruiseflight level: FL350
Air distance to be covered: 1.830 NM
Planned take off mass: 190.000 kg
(With a minimum prescribed fuel quantity of 30.000 kg that includes a trip fuel of 22.000 kg)
Maximum landing mass: 180.000kg
Maximum takeoff mass: 205.000kg
Maximum tank capacity: 40.000kg
The additional fuel quantity will be:
Answer: 10.000kg
Step 1) Enter the graph like before with the NAM and the fuel price ratio (both already given this time)
Find the optimum aircraft weight to be approx 210.000kg
Planned takeoff mass = 190.000kg
However, we are given limitations so we should consider them!
Step 2) What are the limitations
1- max takeoff mass 205.000kg
-> Our planned take off mass was 190.000, we have a margin of 15.000kg
2- max landing mass + trip fuel = 202.000kg
-> Our landing mass will be (190.000 - trip) 168.000kg, we have a margin of 34.000kg
3- max tank capacity = 40.000kg
-> Our fuel is 30.000 which gives us a margin of 10.000kg
Now take the lowest value so that we are sure we do not exceed any of the 3 limitations. The lowest value is 10.000kg
Answer: 8120 kg
For an isolated airfield procedure, a Jet needs 2 hours at cruise consumption as additional fuel
► The flight crew of a turbojet aeroplane prepares a flight using the following data:
Flight leg distance= 3 500NM
Flight level: FL310
True airspeed: 450 kts
Headwind component at this level: 5 kts
Initially planned takeoff mass (without extra fuel on board): 180.000 kg
Fuel price: 0.35 Euro/l at departure, 0.315 Euro/l at destination
To maximize savings, the commander chooses to carry extra fuel in addition to that which is necessary. Using the appended annex, the optimum quantity of fuel which should be carried in addition to the prescribed quantity is:
Answer: The fuel transport operation is not recommended in this case
Before starting with Questions like these, first have a look at the fuel prices because the Fuel tankering procedure is only financially interesting when the price at destination is higher than that at departure.
Price destination = 0.315
Price departure = 0.35
In this case the procedure is not recommended. This saves time during the exam!
► The flight crew of a turbojet aeroplane prepares a flight using the following data:
Flight leg distance= 4 000NM
Flight level: FL310
True airspeed: 450 kts
Headwind component at this level: 50 kts
Initially planned takeoff mass (without extra fuel on board): 170.000 kg
Fuel price: 0.27 Euro/l at departure, 0.30 Euro/l at destination
To maximize savings, the commander chooses to carry extra fuel in addition to that which is necessary. Using the appended annex, the optimum quantity of fuel which should be carried in addition to the prescribed quantity is:
Answer: 8.000 kg
Step 1) Is the fuel price at destination higher than at departure? In this case "yes" so we can continue the calculations
Step 2) Enter the given graph at the right with NAM and at the left with the fuel/price ratio
NAM = (NGM / GS) x TAS
NAM = (4000 / 400) x 450
NAM = 4500
Fuel/price ratio = Price dep/Price dest
Fuel/price ratio = 0.27/0.30
Fuel/price ratio = 0.90
At the intersection on the graph we go vertically down and read an Optimum aircraft weight of 178.000 kg
Optimum = 178.000 kg
Our initially planned = 170.000 kg
We can therefore take 8.000 kg extra
► A turbojet aeroplane is prepared for a 1 300 NM flight at FL350, with a true airspeed of 460 kts and a headwind of 160 kts. The takeoff runway limitation is 174.000 kg, the planned departure mass is 160.000 kg. The departure fuel price is equal to 0.92 times the arrival fuel price (fuel price ratio = 0.92). In order for the airline to optimize its savings, the additional fuel quantity that must be loaded on board is:
Answer: 14.000 kg
Enter the graph by the same way we did in the previous Question and find an optimum aircraft mass of approx 210.000kg. However, this time we have a runway limitation of 174.000 kg which we cannot exceed for any take off. If we would load up until the optimum we will exceed this limit.
The extra fuel we can take on will therefore be 174.000 - 160.000 = 14.000
► The flight crew of a turbojet aeroplane prepares a flight using the following data:
Flight leg distance= 3 500NM
Flight level: FL310
True airspeed: 450 kts
Headwind component at this level: 55 kts
Initially planned takeoff mass (without extra fuel on board): 180.000 kg
Fuel price: 0.27 $/l at destination
The commander may carry on board 8.000kg more fuel than that which is necessary. For this fuel transport operation to be cost effective, the maximum fuel price at departure must be:
Answer: 0.24 $/l
Step 1) Enter the graph on the right with the NAM and on the bottom with the optimum weight.
NAM = (NGM / GS) x TAS
NAM = (3500 / 395) x 450
NAM = 3987 (approx 4000 to make it easier in the graph)
Optimum aircraft weight = 180.000 + 8.000 = 188.000 kg
Step 2) Find the Fuel/Price ratio
Fuel/price ratio = 0.90
0.90 = Price at dep/0.27
Price at dep = 0.27 x 0.90
Price at dep = 0.24
► Assuming a departure/destination fuel price ratio of 0.92, the commander decides to optimize fuel tankering by using the following data:
Cruiseflight level: FL350
Air distance to be covered: 1.830 NM
Planned take off mass: 190.000 kg
(With a minimum prescribed fuel quantity of 30.000 kg that includes a trip fuel of 22.000 kg)
Maximum landing mass: 180.000kg
Maximum takeoff mass: 205.000kg
Maximum tank capacity: 40.000kg
The additional fuel quantity will be:
Answer: 10.000kg
Step 1) Enter the graph like before with the NAM and the fuel price ratio (both already given this time)
Find the optimum aircraft weight to be approx 210.000kg
Planned takeoff mass = 190.000kg
However, we are given limitations so we should consider them!
Step 2) What are the limitations
1- max takeoff mass 205.000kg
-> Our planned take off mass was 190.000, we have a margin of 15.000kg
2- max landing mass + trip fuel = 202.000kg
-> Our landing mass will be (190.000 - trip) 168.000kg, we have a margin of 34.000kg
3- max tank capacity = 40.000kg
-> Our fuel is 30.000 which gives us a margin of 10.000kg
Now take the lowest value so that we are sure we do not exceed any of the 3 limitations. The lowest value is 10.000kg
