# How efficient would an airliner be if it did not have to carry its own fuel?

Hypothetically, if all the fuel was supplied externally or pulled from the atmosphere to where the plane had never to carry any fuel on board, how much more efficient would an airliner be? The plane would not fluctuate in weight.

• By "externally", do you mean aerial refueling? If so, the logistics of it would void (a hundred times over) any benefit. – Digital Dracula Feb 14 '20 at 5:53
• What exactly do you mean by "supplied externally"? – Pondlife Feb 14 '20 at 6:00
• IMO this is the wrong sort of hypothetical question for the site. See: aviation.meta.stackexchange.com/questions/447/… and aviation.meta.stackexchange.com/questions/3594/… – Jamiec Feb 14 '20 at 8:48
• You are certainly not asking about the max. range of such an aircraft as that would be infinite. Maybe you are asking about the hypothetical fuel burn rate at MZFW (maximum zero fuel weight), i.e. just the plane, pax, luggage, and maybe a bit of cargo? Which makes me assume that somebody would then redesignate the fuel tanks and put more cargo in there... – PerlDuck Feb 14 '20 at 10:04
• I think you need to define how you measure efficiency. Cost per mile, or fuel burned per mile? Where you get your hypothetical fuel from will make a huge difference in these parameters. – Michael Hall Feb 15 '20 at 3:21

Short answer is about twice as efficient for a mid range route,(3000 nautical miles) when based on fuel used per ton of cargo per mile traveled.

Explanation: I will use the 737-400 as an example because I have an operational manual for it. First we need to define terms. I assume we are talking about fuel burn per something: per mile, per hour, or per payload ton-mile. Gross weight is the total weight if you set the plane on a scale in some flying condition. Max gross weight is the maximum weight at which the aircraft is considered safe to fly. Operating empty weight is the lowest gross weight achievable in a flying condition, no cargo, no extra crew or passengers, and no usable fuel. Useful load is the difference between max gross weight and operating empty weight. Payload is useful load minus fuel load and extra crew.(that is load that pays, aka cargo and passengers)

The thrust required(and thus fuel consumption) for level flight is determined by aerodynamic drag and the majority of drag is broken into parasitic drag and induced drag; parasitic drag comes from shape or texture and dynamic pressure,(basically speed if you keep the density constant) go faster get more parasite drag, this does not change with gross weight.

Induced drag is caused by the lift created by practical wing designs, for a set wing design and speed the induced drag increases with weight. So high gross weights have higher induced drag and so require more thrust.

Now for the example. A 737-400 has a maximum gross weight of 143500 pounds and a typical operating empty weight of 77100 pounds(typical airline equipment with seats and such) giving a useful load of 66400 pounds. This weight can be split up however needed between fuel and payload depending on the length of the flight.(In practice other factors can limit gross weight such as available runway length, and payload as some engineering optimizations assume a lower landing weight than takeoff weight to save on structural weight.) At least up to the limit of installed fuel tank size, this model can hold 37708 pounds of fuel.

Some reserve of extra fuel is required for safety, diversions, holding and such, so this is usually extra fuel at landing. This is typically 4000 to 9000 pounds depending on specifics of the flight.

Assuming flight altitude of 33000 feet and mach .74 cruise(430 knots). I am using nautical miles and short tons(2000 pounds) for all of this.

A the most exaggerated case will be maximum range with reserve fuel of 9000 pounds.(maximum fuel at start, reduced payload) Payload will be at most 28692 pounds or 14.3 tons. Fuel burn at the start will be 6234 pounds per hour for 1.014 pounds fuel per ton-mile of payload (6234/(14.3*430), and at the end of flight 5066 pounds per hour for 0.824 pounds fuel per ton-mile of payload (5066/(14.3*430). (geometric mean of .914 pounds fuel per payload ton-mile. Change in rate is non-linear.)

If all of the useful load could be used for payload as in the case of an external fuel supply, or an infinitely short flight with no reserve fuel, you get 66400 pounds or 33.2 tons of payload. The fuel burn will still be 6234 pounds per hour but you get .437 pounds fuel per payload ton-mile. 6234÷(33.2×430)

So cruise fuel consumption with an external fuel supply will be approximately .437/.914 = 47.8% or about half of the fuel consumed per unit of cargo. Actual results will depend on variables if I must state the obvious. Longer flights would see more improvement, shorter flights less, payloads less than maximum allowable will also see less improvement, flights requiring less reserve fuel will also see less improvement.

Airliners can typically carry about 40-45% of their maximum total weight as fuel. If they didn't need to, they could theoretically carry that much extra payload instead, or they could fly lighter and burn 40-45% less fuel (because fuel burn rate is closely related to the total weight).

Obviously it's not quite that simple. Planes usually only carry as much fuel as they need (plus a safety margin), and they burn fuel as they fly, so the fuel weight is almost zero at the end of a flight. That would at least halve the improvement, so say 20%.

This assumes that you magic fuel either weighs nothing, or you're somehow teleporting it to the engines from the ground, or air-to-air refuelling where you don't count the cost of the tanker. And that you're only interested in fuel burned, not cost, because there's no way the estimate the cost of hypothetical magic :-)

If all current aviation engines could run by burning external fuel (let's say it's able to magically extract sufficient amount of hydrogen from atmospheric water vapor, even when relative humidity is very low) and mix it with atmospheric oxygen, then these engines would have unlimited autonomy, or limited by maintenance cycles only.

If these engines exhibit the same thrust/power to weight ratio as their kerosene burning twins,

then there wouldn't be airliners anymore, since VTOL pods, individual or family sized, or sized to any payload, could become the norm. Just put the required amount of GE90 around your house, and move it where you want.

Basically, typologies of flying devices would broaden spectacularly, ground and water transportation and most roads would become useless except for cars/boats nostalgics and maybe air-sick people.

As a side effect this would also solve global energetic crisis. But it wouldn't change our current inability to reach other stellar systems.

• It might be reasonable to assume this technology would be orders of magnitude more expensive than "traditional" power sources. – Jpe61 Feb 14 '20 at 9:49
• People have a hard enough time commanding their land-yachts on the 2-D roads. I'd hate to see what would happen with GE90-powered houses zipping around in 3-D space. There are multiple reasons the Jetson's flying cars aren't real yet. – FreeMan Feb 14 '20 at 14:12
• @FreeMan main reason among those multiple reasons being : There is no magical engine. Navigation & traffic control is solvable using current technology. – qq jkztd Feb 15 '20 at 9:11
• @qqjkztd that was intended to be tongue-in-cheek and goes along with the "Where are the flying cars they promised us?" meme that floats around the interwebz these days. – FreeMan Feb 17 '20 at 12:20

There are several answers, depending on what precisely your idea of zero fuel means.

The first is the simplest: We use a regular aircraft and simply assume that it takes off with zero fuel and still flies as if its tanks are full. Now the answer depends on the range: Longer flights need disproportionally more fuel. The linked answer gives 3.88% of landing mass per 1000 km for 2000 km range but 4.16% per 1000 km for a flight over 5700 km. The benefit is obviously highest for intercontinental traffic. If you now consider that ultra-long range flights are only possible with a fraction of the potential full passenger load, the increase in efficiency would be up to a factor of 4.

This splendid answer contains a table which lists how much more fuel is used to carry fuel along for a trip exceeding 2800 nautical miles. As you can see, extending the flight to 8000 miles needs 12% of the fuel load only to carry the extra fuel.

A more refined answer would also include the mass and volume requirements for the tanks. If the airliner is designed without fuel mass and volume in mind, it could be smaller and lighter (not counting that magical fuel-catching device, of course). Wings could have a higher aspect ratio and systems could be simpler. Parametric formulas for the tank mass estimate fuel tanks to weigh about 6.4% of the fuel they contain and wing tanks 1.8% on a basis of 20 tons (the percentage gets smaller with bigger tanks) plus a contribution in kg that is proportional to 23% of static thrust given in kN. Leaving out the tanks will increase payload by about 0.8% of total mass and allow for a slightly better L/D. Relative to payload mass which is between 30% on short and 10% on long range flights the mass savings alone add another 2.4 to 8% to efficiency.

In total, the gain in efficiency is likely to be about 5% for very short range and 15% to 25% for long range flights as long as take-off mass does not limit the payload which can be carried and the tank mass is included in the calculation.

...how much more efficient would an airliner be?

Answer: $$\frac {\text{fuel weight percentage at TO - fuel weight percentage at touchdown}}{2}$$

This answer computes the saving in fuel consumption due to a saving in empty weight - 10% less empty weight results in a fuel saving of 10%.

That is empty weight. In a traditional, fuel burning aeroplane, total weight decreases linearly over time due to using fuel. So fuel saving due to used fuel weight saving decreases linearly over the duration of the flight as well.

Let's take a B777-300ER for example:

• OEW = 167,829 kg
• MTOW = 351,533 kg
• Fuel capacity = 145,538 kg

If the plane takes off at MTOW with maximum fuel on board, the fraction of fuel weight = 145,538/351,533 = 41.4%. If at touchdown there is no more fuel left, fuel weight fraction = 0%. Total average fuel savings over the duration of the flight = (41.4 -0)/2 = 20.7%.

A normal B777-300ER would burn 145,538kg of fuel. A magic fuel accumulating B777ER would burn (1 - 0.207) * 145,538 = 115,410 kg of fuel. It is 20.7% more efficient.