# What is the fuel burn of a 787-9 on a ferry flight?

I can't find any literature on 787-9 fuel burn for ferry flights, meaning no cargo or passengers, just crew.

An 787-9 burns about 5400 litres when on normal commercial service.

How does it change when on a ferry flight?

• The amount of missing information necessary to solve this is far greater than the amount of information provided, making the question entirely unanswerable in its present state. If you have "upcoming ops" involving a real aircraft, then whoever is operating it can provide much more specific information about your specific operation than a message board ever could (especially given the utter lack of necessary details here). VTC
– Ralph J
Oct 19, 2022 at 6:09

You can figure out the fuel burn using the Breguet equation (similar to how I did in this answer):

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

Here, TOW is the takeoff weight, LW is the landing weight and $$R$$ is the flight range. I grouped all the constants together into a factor $$f$$. The TOW consists of the OEW (Operating Empty Weight, 128,850 kg) plus the total fuel. The LW is then the OEW plus the fuel reserves (let's say 5000 kg).

To determine the unknown factor $$f$$, we look at the Payload-Range-Diagram, which can be found in the Boeing Airplane Characteristics for Airport Planning document: The last line on the right limits the range based on maximum fuel capacity. On a ferry flight, the payload is zero, so we can read off the maximum range at full fuel (around 9400 NM). Note that this is larger than the ranges you would typically find (e.g. 7645 NM on Wikipedia) because those ranges includes typical payload.

Now we can plug in the number into our equation and solve for $$f$$:

$$f = \frac{1}{R} \ln \left( \frac{\text{OEW} + \text{Fuel}}{\text{OEW} + \text{Reserve}} \right) \approx \frac{1}{9400 \, \text{NM}} \ln \left( \frac{230 \, 294 \, \text{kg}}{133 \, 850 \, \text{kg}} \right) \approx 5.773 \times 10^{-5} \, \text{NM}^{-1}$$

With the known $$f$$, we can now plot the fuel burn on a ferry flight as a function of range:

$$\text{Fuel Burn} = \text{Fuel} - \text{Reserve} = \left( \exp(f \cdot R) - 1 \right) \cdot (\text{OEW} + \text{Reserve})$$ This would be the total fuel burn on the ferry flight. To get the fuel rate, we divide by the flight time (assuming a typical cruise speed of 488 kt): The 787-9 therefore burns between ~3.8 and ~5.0 t of fuel per hour on a ferry flight.