How much of an improvement would a 1% weight decrease on an airplane be to the industry?

Question

Say I thought of a very easy and inexpensive way to reduce the takeoff weight of a commercial airplane (body, fuel and passengers) by 1%. How much would this mean to the industry? Would it make much of a difference, or would the savings be insignificant enough to not really be worth anyone's time?

Answer 1

There is an easy way to get an idea of what magnitude the savings are: The Breguet equation can answer this. You first calculate a reference aircraft and then start all over with one that weighs 1% less. The difference in fuel consumption shows how much fuel is saved per flight.

Our reference design is the A320 which has an L/D of 18 and we let it fly a trip of R = 2,500,000 m. $m_1$ is take-off mass and $m_2$ landing mass. Allow for 3 t of fuel reserves and plug that into the Breguet equation, using a fuel burn of $b_f$ = 0.000018 kg/Ns and a speed of Mach 0.78, which equates to $v$ = 262 m/s in 11,000 m altitude: $$m_1 = m_2 \cdot e^{\frac{R\cdot g\cdot b_f}{v\cdot L/D}}$$ The landing mass $m_2$ is the operational empty mass OEW + 18 t payload + 3 t of fuel reserves = 63.6 t. The take-off mass $m_1$ is 10% higher, so the difference in fuel consumption is 1% of that trip fuel of 6,239 kg or 62 kg. That is not much for a single trip.

But now consider that this one A320 will fly 50,000 trips before it retires. Over its lifetime the 1% in mass reduction will save 3,119 metric tons of fuel. This fills 29.4 DOT-111 railway tank cars!

And there are about 12,000 airliners of 150 seats and more and easily twice that number if all regional jets are included, flying 3.8 billion passengers and 53 million tons of freight per year on 35.4 million flights. If we make the heroic assumption that our A320 flight is typical, this translates into global fuel savings of about 2.2 million tons of fuel per year if only 1% of mass could be saved on every airplane.

Another way to approach this is the average fuel efficiency of modern airliners which consume between 3 and 3.5 liters of fuel per 100 passenger kilometers. Using ICAO's number of 6.6 trillion passenger kilometers in 2015, this would equate to 174 million tons of aviation fuel used per year for revenue passenger transport, assuming a mean density of kerosene of 0.81 kg/l. The same source gives an average load factor of 80%, so one fifth of seats is empty; the efficiency number above is for a full airplane, however. Therefore, our estimate needs to be increased to 217 million tons.

Saving only 1% of this would reduce fuel consumption by 2.17 million tons per year, which comes pretty close to the first estimate. This is enough to fill more than 41 Panamax oil tankers (the maximum size for the old Panama canal) or 18 Neopanamax oil tankers.

Or, to look at this from a different angle, with the current price of kerosene of 1.76 USD per gallon, the saved fuel equates to 1.25 billion USD if everyone paid US prices for Jet A-1 fuel. Compared to the average net profit of all airlines per year between 2007 and 2017, which amounts to 13 billion USD, this is almost 10% of yearly profits.

Now this doesn’t look so insignificant any more, does it?

Answer 2

One % of takeoff weight is huge. On a B747 with MTOW of 300 tons, that is a weight saving of 3,000 kg, weight that needs not be lugged around for the couple of decades that the aircraft will be flying.

At the aircraft factory where I used to work, there were Idea Boxes everywhere where one could deposit suggestions for weight savings, every kg saved adds to improvement of fuel economy. How much depends on range and a whole lot of other factors, and can be quantified for specific aircraft.

1%. Yuuge.

If we make an estimate for a B747-400 flying maximum range of 12,700 km according to slide 7/14 out of this presentation.

MTOW is given as 400,000 kg, fuel weight as 175,000 kg, this makes $W_{final}$ = MTOW - $W_{fuel}$ = 225,000 kg. With 1% saving, MTOW is reduced to 396,000 kg. Range stays the same if $\frac{W_{init}}{W_{final}}$ stays the same. So the new $W_{final}$ becomes:

$$ W_{final} = \frac{396,000}{400,000}\cdot 225,000 = 222,750 kg$$

$W_{final}$ is again MTOW - $W_{fuel}$ =>

$$W_{fuel} = MTOW - W_{final} = 396,000 - 222,750 = 173,250 kg$$

Final fuel saving is 175,000 - 173,250 = 1,750 kg

For one long distance, maximum range flight, a reduction of 1% in TO weight saves 1,750 kg of fuel. Every percent of weight savings equates to a percent of fuel saved. Every trip, for 2 - 3 decades.