ATPL theory questions

[keith williams]

Subtracting 1100 kg climb fuel from the 56000 kg take-off mass gives a mass of 56000 kg – 1100 kg = 54900 kg at the start of the cruise.


In Figure 4.5.3.1 the second cells in the 55000 kg row indicates that the TAS = 437 kts.

Note 2D at the base of the page states “decrease TAS by 1 kt for each 1 deg C below ISA.

So for the existing ISA deviation of -5 deg C the required correction is -5 kts.

So the corrected TAS = 437 – 5 = 432 kts.

The question states that the leg time is 48 min.

In 48 minutes at 432 kts TAS the distance flown = 432 x 48 / 60 = 345.6 NAM.

In Figure 4.5.3.1 the cell at the intersection of the 54000 kg row and the 900 kg column (for the mass of 54900 kg) indicates an initial cruise range of 3736 NAM.

Subtracting the 345.6 NAM flown in this leg gives 3736 – 345.6 = 3390.4 NAM remaining.

The closest figure to this in the table is 3395, which appears at the intersection of the 52000 kg row and the 900 kg column. This indicates a final mass of 52900 kg.

Subtracting this final mass from the initial mass gives a mass change of (54900 – 52900) = 2000 kg caused by the burning of fuel.

Note 2B at the base of the page states “decrease fuel by 0.6% for each 10 deg C below ISA.

So for the existing -5 deg C ISA deviation, the fuel must be decreased by 0.6% x 5/10 = 0.3%.

0.3% of 2000 kg = 6 kg. So the corrected fuel = 2000 kg – 6 kg = 1994 kg.

So the fuel consumed in the cruise is 1994 kg.

So for the cruise leg of this flight the distance flown is approximately 345 Nm and the fuel burned is approximately 1994 kg.