# What is the Fuel-Oxygen ratio for a large turbofan at cruise conditions?

What is the Fuel-Oxygen ratio (by mass) for a large turbofan at cruise conditions?

In order to be concrete, I'll just pick an engine at random. I pick the GE90-115B. Okay it's not really random, it's just the largest turbofan I know of.

• for what its worth: the ge90-115b is the largest turbofan (it has the world record for most thrust) – Daniel K Sep 4 '17 at 11:41

The GE90-115B has a bypass ratio of 9:1. So if 1000 units [of air] enter the fan, only 100 will make it to the combustion chamber.

Kerosene burns efficiently at 15:1 (air to fuel). But not all the air in the combustion chamber is burned. Only around 22 (average) of the 100 will be burned, the rest will provide cooling, flame stabilization, and dilution.*

So of the 1000, 22 are burned at the 15:1 ratio. Total ratio will be 1000:1.5 (air to fuel). Since oxygen is ~20% of air, then the ratio is 130:1 (oxygen to fuel). (Note: bypass is included.)

To validate that figure:

If the fuel flow (per engine) is 2500 kg/hour (modest value in cruise), then that means the engine takes in ~1,667,000 kg of air per hour, or 460 kg per second. That agrees (give or take) with the numbers in this post (PDF does not confirm the GE90 version and the fuel flow).

According to The GE90 - An Introduction, the GE90 has a mass flow rate of 1,350 kg/s at take-off and 576 kg/s at cruise.

* The Jet Engine - Combustion Process.
It varies with the airplane weight, cruise altitude, speed, etc. See here (in lb/hr) for the GE90-90B from a 777-200 flight manual, it varies from 2140 to 4440 kg/hr.
Mass flow rate is for the air, as NASA puts it: "The total mass flow rate through the inlet is the sum of the core and fan flows."

FWIW, the -115B1 is the exclusive engine for the 777-300ER. Based on this official data from Boeing (last slide), at a 365-passenger load on a 6,000 NM journey, each engine burns an average of 3790 kg/hr (heavy load; takeoff and climb included), which is within the above -90B range, i.e., 2500 is a correct value to use as the -115B will also have a similar wide range. (Arriving at that figure requires the 777 cruise speed of 892 km/h.)

• Comments are not for extended discussion; this conversation has been moved to chat. – Farhan Sep 26 '17 at 18:53
• The 22% fuel use sounds incredibly fuel-rich, while the other answer claims an oxygen-rich burn. And considering that many planes take off with 1/3 of their weight in fuel, burning every pound of fuel is key to a reasonable efficiency. – MSalters Jun 5 '18 at 12:47
• @MSalters - It's 22% of the air that gets to the core. – ymb1 Jun 5 '18 at 13:40

I chose the GE90-115B because it is the largest turbofan in terms of thrust. Unfortunately, it was hard to come by specific fuel consumption figures for cruise conditions, maybe because the -115B is a relatively new version of the GE90.

But I finally found enough for a preliminary answer, so here it is.

From this pdf, I found that the GE90 at cruise consumes 1.079 kg/s of fuel. It also cites an air-flow of 576 kg/s. But I am not sure how trustworthy this pdf is, as it does not specify the exact model of the GE90 (there are many) and it had no in-line citation, and the only web link in its list of references was "http://www.ge.com/geae/ge90" which is a dead link. Nevertheless it is the closest thing I've found so far for a GE90-115B's fuel consumption at cruise.

(I did try to investigate GE's official site but it is hard to find specific data, which is a real shame.)

From Wikipedia's specs, I find that the 777 cruise speed is 892 kmh and service ceiling (which I'm assuming is cruise altitude) is 13.1 km.

From more Wikipedia specs, I find the fan diameter is 330 cm.

Therefore I can calculate the air intake via the cross section ($\pi \cdot 1.65^2$ = 8.553 sq meters) and the airspeed (892 kmh = 247.778 m/s) and the air density at 13.1 km above MSL from this chart (which I hate btw because it should put altitude on the x-axis, but instead it's on the y-axis, making it very hard to read), which I read as 0.2325 kg/m$^3$.

So air flow rate is 247.778 * 8.553 * 0.2325 = 492.725 kg/s.

The pdf says air flow is 576 kg/s. I'm not sure if this is a significant disagreement. In this case, I think it is reasonable to believe the pdf because the air intake cross section should be a little bit bigger than the fan diameter. The nacelle cowling is curved, so the diameter we want should be from the axis to the midpoint of the cowling. Judging from this photo, the extra radius looks like about 13.5% more, so 1.1335^2 = 1.2896 and multiplying that by 493 gives us 635.773 kg/s, only slightly closer to the pdf's value.

Anyway if we use 576 kg/s of airflow and 1.079 kg/s of fuel flow, we would have a fuel-air ratio of 1:534.

But that can't be right, seems way too high, and indeed there's more. We need to multiply by the bypass ratio...actually the inverse of it (and off by 1) because we only want airflow through the core. Then multiply by 0.2 because oxygen is only about 20% of air by mass.

The bypass ratio according to this is 9:1. But according to this pdf it's 7.08:1. Personally I think 9:1 is more believable because the GE90-115B is supposed to be a recent, advanced turbofan with the latest technology, but at the end of the day that's only my guess. If anyone can reconcile these disagreeing figures and show which one is wrong, please let me know.

Edit: Bypass ratio in cruise, according to the original pdf, is 8.1:1. Thanks again to Peter Kampf for clarifying that this figure looks realistic. I adjusted the math accordingly.

Therefore, 534*0.2/9.1... I get a fuel-oxygen ratio of 1:11.7 ! For simplification we could call this a 1:12 ratio of fuel to oxygen.

Interesting. This is not stoichiometric. Wikipedia says the stoichiometric for dodecane (C12H26) and O2 is 1 mole to 18.5 moles, so if we assume the mass of dodecane is 170 and the mass of O2 is 32, then the stoichiometric ratio by mass is 170:592 = 1:3.482 !

(Btw, dodecane is the simplified chemical for kerosene, which itself is the typically-assumed formula for jet fuel.)

So the 1:12 ratio is very oxygen-rich. I don't know for sure how to interpret this. I guess jet engines just cannot extract more power for some reason? I remember they are limited by core temperatures to prevent the blades from melting. That could be what's preventing them from burning stoichiometrically.

There is something else missing from the puzzle too. Is fuel combustion complete? That is, are all fuel molecules burned or are some just not burned for some reason as it goes through the engine? I know photos of older jet engines tend to have a lot of smoke, which means a lot of unburnt (wasted) fuel. Today I don't think I've ever seen smoke from a big turbofan, so I'm gonna guess that unburnt fuel is so low that we don't have to worry about it in our F-O ratio.

Edit: thanks to Peter Kampf for confirming, burn-rate is >99% and the less than stoichiometric ratio is indeed because temperatures would otherwise be too great for turbine materials of today.

That was a lot more work than it should have been, and if we're being honest here, it's not done yet because I would really like clarification on the 1.079 kg/s of fuel flow the pdf cited. It did not specify which version of the GE90 engine it was quoting. It's been pointed out that cruise conditions vary so there is no one authoritative figure. Too true. I guess the most authoritative figure would be what the manufacture designed it for at cruise conditions and maximum load right after it reached cruise.

If I had known how much trouble it would be to find this figure, I might not have chosen the GE90-115B. But it is the largest engine by thrust, so I think it's an important figure to know. I hope I did it justice.

• Bypass ratio is not constant, but changes with flight speed. Same goes for SFC (it doubles between SLS and cruise). You always need to make sure that the conditions for any figure are well defined. The reason for the oxygen-rich combustion is the maximum turbine temperature: If burning would be stoichiometric, the turbine would melt! Not only nitrogen but also oxygen are used as a process gas. The oxygen-rich, hot combustion is the reason that NOx emissions have gone up, like in modern diesel engines. – Peter Kämpf Sep 4 '17 at 18:14
• Advances in burner design result in combustion efficiency of >99% (that is also quoted in the PDF). So yes, combustion is nearly complete. – Peter Kämpf Sep 4 '17 at 18:21
• @PeterKämpf Ah, so the ratio is 9:1 at takeoff but 7.08:1 at cruise? And what does SLS stand for? Thanks for the stoichiometric and burn-rate clarifications too, I will make an edit. – DrZ214 Sep 5 '17 at 4:34
• SLS is sea level static. This is the performance on a test stand or on the ground. The change in the bypass ratio is not as wide as your numbers indicate. The figures in the PDF look more realistic (8.4 at take-off and 8.1 in cruise). – Peter Kämpf Sep 5 '17 at 6:36
• @PeterKämpf Thanks again. I edited and adjusted the math accordingly. New ratio is 1:11.7, which fortunately doesn't really change any conclusions. – DrZ214 Sep 5 '17 at 9:00