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A350-900 Range/Payload

When the TOW (takeoff weight) of an A350-900 is 250,606 kg (because of runway length limitation), with maximum payload, what is the maximum range it can reach using the chart above?

It is easy to figure out for a Boeing 787 since Boeing provides different brake release slopes, as shown below, unlike the A350 chart.

B787-10 Range/Payload

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    $\begingroup$ Sorry but whats not clear in the chart? $\endgroup$ – vasin1987 Nov 28 '19 at 6:31
  • $\begingroup$ @vasin1987 see if it's clearer now $\endgroup$ – ymb1 Nov 28 '19 at 9:07
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Since Airbus does not publish this in the charts, we have to calculate it ourselves. You are asking about an Airbus A350-900 with a TOW (TakeOff Weight) of 250,606 kg. This weight would then consist of:

$$ \text{TOW} = \text{OEW} + \text{Payload} + \text{Fuel} = 142 \, 400 \, \text{kg} + 53 \, 300 \, \text{kg} + 54 \, 906 \, \text{kg} $$

based on typical OEW (Operating Empty Weight) and max. payload from Wikipedia. This leaves us with about 54.9 tons of total fuel. Not all of this fuel will be usable however, since the aircraft still needs some fuel reserves: $ \text{Fuel} = \text{Trip Fuel} + \text{Reserves} $.


To calculate the range of the aircraft, we can use the Breguet range equation (see e.g. this answer):

$$ \frac{\text{TOW}}{\text{LW}} = \exp \left( \frac{R \cdot g \cdot b_f}{v \cdot L/D} \right) = \exp(f \cdot R) $$

We are only interested in the TOW, LW (Landing Weight) and the range $R$, so I combined all the other aircraft specific factors into one factor $f$. The landing weight can be calculated via the fuel use:

$$ \text{LW} = \text{TOW} - \text{Trip Fuel} = \text{TOW} - (\text{Fuel} - \text{Reserves}) $$

We now have to solve for the two unknowns $f$ and the reserve fuel. For this, I read off two values from the plot you provided:

values read from chart

We now have two given ranges for known masses:

  • At MTOW and max. payload (resulting in 84.3t of total fuel) the range is about 5800NM.
  • At MTOW and max. fuel (about 112.6t based on about 25t of payload) the range is 8600NM.

By plugging these values into the Breguet equation, one can numerically solve for the unknowns:

$$ f \approx 5.2711 \times 10^{-5} \; \text{NM}^{-1} \;\;\; \text{and} \;\;\; \text{Reserves} \approx 10\,545 \, \text{kg} $$

We can test, if these parameters make sense, by plugging in no payload at maximum fuel:

$$ R = \ln \left( \frac{\text{TOW}}{\text{LW}} \right) / f \approx 9698 \, \text{NM} $$

This range is consistent with the value where the magenta line crosses the range axis.


Now moving back to your numbers. We can now plug in a TOW of 250,606 kg at maximum payload. The trip fuel will then be:

$$ \text{Trip Fuel} = 54 \, 906 \, \text{kg} - 10 \, 545 \, \text{kg} = 44 \, 361 \, \text{kg} $$

This gives the final answer:

$$ R = \ln \left( \frac{\text{TOW}}{\text{TOW} - \text{Trip Fuel}} \right) / f \approx 3696 \, \text{NM} $$

final result

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