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(For this question use annex 033-046 rev. 05.11.2004) A turbojet aeroplane is flying using the following data:
Optimum flight level, Mach 0.80, mass of 190 000 kg
Temperature: ISA
Tailwind component: 100 kts
The fuel mileage and the fuel consumption per hour are:
  • A
    105 NM / 1 000 kg; 6 515 kg/hr.
  • B
    105 NM / 1 000 kg; 5 330 kg/hr.
  • C
    86 NM / 1 000 kg; 6 515 kg/hr.
  • D
    71 NM / 1 000 kg; 5 330 kg/hr.

Refer to figure.

1. Enter the table on the first column and find the value "190" (which corresponds to 190 000 kg) and intersect with "0" (0 hundreds of kg) - read the corresponding value for Distance and Time:

  • For 190 000 kg => 6515 NAM and 852 min

2. Find the fuel consumption per hour (kg/h):
To find consumption per hour, we need to fly for 1 h. Therefore, deduct 60 minutes from 852 min = 792 min. And find this value in the table - and read the corresponding distance and mass.
- There is no 792 minutes on the table. Therefore, we must interpolate between 791 min (184 600 kg) and 794 min (184 800 kg)
  • By interpolation, find a value of approx. 184 670 kg for 792 min.
=> The difference between 190 000 kg and 184 667 kg - will be the fuel consumption per hour:
  • 190 000 kg - 184 667 kg = 5330 kg (5330 kg/h)

3. Find fuel mileage (NM / 1 000 kg):
Find TAS for 190 000 kg => Continue on the same line until the last column of the table and read the value of 459 kts TAS.
Tailwind 100 kts - Therefore, GS = 459 + 100 = 559 kt - which means that we will cover 559 NM in 1 h.
  • At a GS of 559 kt = we cover 559 NM in 1 h
  • At a Fuel consumption of 5330 kg/h:
  • We can fly 559 x 1000/5330 = 105 NM per 1000 kg

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