Faro
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### Mass Of Air 'Consumed' In Propulsion

Purely out of curiosity, what total mass of air would be sucked in and propulsed out of the engines on, say, a max-range-at-max-payload 77W flight?

Reply 7 of this thread: Speed Of Air Out Of The GE-90-115's Back End (by UAL747 Sep 10 2006 in Tech Ops) indicates a mass flow of 3,000 lbs/second. I assume that this is at max T/O thrust; would anyone be able to compute the actual mass flows applied to relevant durations of flight segments (T/O, climb, crz, etc) as a function of segment weight for such a trip?

I am curious as to what multiple of MTOW the propulsion air 'consumption' would represent on such a flight. IIRC, the more air mass one uses in propulsion, the more efficient it makes the flight.

Faro
The chalice not my son

Jetlagged
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### RE: Mass Of Air 'Consumed' In Propulsion

You could take the total fuel used and multiply by the typical air/fuel ratio, which I've seen estimated at anywhere between 45:1 and 150:1, and then multiply by the bypass ratio (9:1). The 77W can carry 320,000 lb of fuel, so assuming a fuel/air ratio of 45:1, the core mass of air used is 14,400,000 lb so the total mass of air passing through the engines is 129,600,000 lb. That's 167 times the MTOW.
The glass isn't half empty, or half full, it's twice as big as it needs to be.

BMI727
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting faro (Thread starter):I assume that this is at max T/O thrust; would anyone be able to compute the actual mass flows applied to relevant durations of flight segments (T/O, climb, crz, etc) as a function of segment weight for such a trip?

Yes, but it would be a lot of work. The mass flow is changing at every point throughout the flight where the conditions are changing.

It's not really a function of segment weight, it's more a function of altitude, mach number, and throttle setting. The mass flow calculations are actually a rather important portion of integrating the airframe and powerplant for drag purposes.
Why do Aerospace Engineering students have to turn things in on time?

Faro
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting Jetlagged (Reply 1): You could take the total fuel used and multiply by the typical air/fuel ratio, which I've seen estimated at anywhere between 45:1 and 150:1, and then multiply by the bypass ratio (9:1). The 77W can carry 320,000 lb of fuel, so assuming a fuel/air ratio of 45:1, the core mass of air used is 14,400,000 lb so the total mass of air passing through the engines is 129,600,000 lb. That's 167 times the MTOW.

Wow, that's quite a multiple; I saw it maxing out at 50 or so. I assume that a 45:1 air/fuel ratio corresponds to a typical cruise power setting?

Faro
The chalice not my son

rwessel
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting faro (Reply 3):Wow, that's quite a multiple; I saw it maxing out at 50 or so. I assume that a 45:1 air/fuel ratio corresponds to a typical cruise power setting?

No - even in the core, the substantial majority of the air does not get combusted.

WingedMigrator
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting Jetlagged (Reply 1):The 77W can carry 320,000 lb of fuel, so assuming a fuel/air ratio of 45:1, the core mass of air used is 14,400,000 lb so the total mass of air passing through the engines is 129,600,000 lb. That's 167 times the MTOW.

Another crude way to look at it: the 77W will fly 7800 nm at max fuel load. The volume of air swept by two 128-inch fan disks traveling over 7800 nm is 180 sq ft times 7800 nm = 1.8E2 ft2 * 4.74E7 ft = 8.5E9 ft3 of air. At cruise altitude that air weighs 0.0218 lb/ft3, for a total of 185,000,000 lb.

Pretty close if you ask me!

Jetlagged
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting faro (Reply 3):Wow, that's quite a multiple; I saw it maxing out at 50 or so. I assume that a 45:1 air/fuel ratio corresponds to a typical cruise power setting?

No, it applies to all stages of flight, any steady state power setting. The less dense the air the less fuel is needed. The higher the mass flow the more fuel is needed. For complete and efficient combustion you need a fuel air ratio of about 14.7:1 (by mass). A gas turbine uses a much weaker mixture than that as much of the air is used for cooling the combustion chamber. Incidentally this is the reason an afterburning turbojet works. If all the oxygen passing through the combustion chamber was used there'd be nothing else to aid combustion in the exhaust.
The glass isn't half empty, or half full, it's twice as big as it needs to be.

tdscanuck
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting Jetlagged (Reply 6):For complete and efficient combustion you need a fuel air ratio of about 14.7:1 (by mass).

I think that's backwards...14.7 kg of fuel to 1 kg of air would be *way* on the rich side. Fuel air ratio (by mass) is normally way less than 1. For almost all basic turbofan analysis problems it's so much less than one that you just assume it's zero and ignore it.

Tom.

prebennorholm
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting tdscanuck (Reply 7):Quoting Jetlagged (Reply 6): For complete and efficient combustion you need a fuel air ratio of about 14.7:1 (by mass). I think that's backwards...14.7 kg of fuel to 1 kg of air would be *way* on the rich side. Fuel air ratio (by mass) is normally way less than 1. For almost all basic turbofan analysis problems it's so much less than one that you just assume it's zero and ignore it.

Dear tdscanuck, Jetlagged just put the figures upside down. The optimal combustion fuel-air ratio isn't 14.7:1 but 1:14.7.

And it is correct when he states (reply #1) that a typical rate to the full core flow is between 1:45 to 1:150 depending on operating conditions, the lower figure probably being at take-off and the higher figure at flight idle.

It just means that roughly 70 to 90% of the air flow in the core does not contribute to the combustion, but is used for various cooling and mixing with the combusted air. Otherwise the HPT would melt immediately.

So assuming an average ratio of 1:75 and a bypass ratio of 6, then 75x(6+1) = 525 lbs air will pass the nacelle for every lb of fuel burned. (the "1" being the air passing through the core).

It is really an art as much as science to get fairly close to that 1:14.7 ratio in the combustion chamber at alll relevant operating conditions, but it is needed to get the most efficient combustion and a fairly clean exhaust. The funny shapes of the holes in the combustion chamber play a major role here.

Many modern engines have dual annular combustion chambers. One of them is sometimes shut off (no fuel injected) at some power settings, sometimes only half of the circle is shut off. It makes the engine more complicated, heavier, and makes the cooling process more challenging, but it gives a wider range of power settings with fairly optimal combustion with overall cleaner exhaust. And on short sectors, mainly, it will reduce overall fuel consumption. That's mainly because during ground taxi and descend the engines will typically always shut off one chamber.

It is almost like changing our V8 into an inline four cylinder when we are stuck in a motorway queue. Pretty smart if you ask me.
Always keep your number of landings equal to your number of take-offs

Jetlagged
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting tdscanuck (Reply 7):I think that's backwards...14.7 kg of fuel to 1 kg of air would be *way* on the rich side. Fuel air ratio (by mass) is normally way less than 1. For almost all basic turbofan analysis problems it's so much less than one that you just assume it's zero and ignore it.

I typed fuel/air when I meant air/fuel. If you look at my first post (reply 1) you'll see I got it the right way round that time. Never made a typo Tom?

[Edited 2011-12-01 07:41:00]
The glass isn't half empty, or half full, it's twice as big as it needs to be.

hal9213
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting WingedMigrator (Reply 5):Another crude way to look at it: the 77W will fly 7800 nm at max fuel load. The volume of air swept by two 128-inch fan disks traveling over 7800 nm is 180 sq ft times 7800 nm = 1.8E2 ft2 * 4.74E7 ft = 8.5E9 ft3 of air. At cruise altitude that air weighs 0.0218 lb/ft3, for a total of 185,000,000 lb.

Actually, you calculated the mass of air, that would travel "idled" through the engine housing, if there was no fan nor engine. However, the fan itself rotates, sucking and pushing much more air than the idle amount, which obviously is the purpose of the fan and engine.

tdscanuck
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting prebennorholm (Reply 8):Dear tdscanuck, Jetlagged just put the figures upside down.
 Quoting Jetlagged (Reply 9):Never made a typo Tom?

Whoa guys, wasn't trying to pick on Jetlagged, just pointing out a potential error. I've made more than my share of typos and I count on everyone else to catch them when I do.

 Quoting hal9213 (Reply 10):Actually, you calculated the mass of air, that would travel "idled" through the engine housing, if there was no fan nor engine.

True, although WingedMigrator's solution is a very nice and elegant "back-of-the-envelope" method. Also, since the whole point of high bypass turbofans is to move large mass at low deltaV, the higher your bypass ratio gets the closer the real and "idled" airflow become.

Tom.

rwessel
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting hal9213 (Reply 10):Actually, you calculated the mass of air, that would travel "idled" through the engine housing, if there was no fan nor engine. However, the fan itself rotates, sucking and pushing much more air than the idle amount, which obviously is the purpose of the fan and engine.

At high subsonic speed (IOW, cruise), there isn't (and can't be) much additional flow because of the suction from the engine (compared to the nominal cross section of the engine hitting the oncoming air stream). At supersonic speeds there can be none at all.

flipdewaf
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting tdscanuck (Reply 11):True, although WingedMigrator's solution is a very nice and elegant "back-of-the-envelope" method. Also, since the whole point of high bypass turbofans is to move large mass at low deltaV, the higher your bypass ratio gets the closer the real and "idled" airflow become.

For me as an engineer, the fact that the two "back of the envelope" calculations derived in completely different ways "from first principles" (almost) are both of the same magnitude is a good sign we are close to a reasonable solution to the problem.

Fred

Faro
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting flipdewaf (Reply 13):Quoting tdscanuck (Reply 11): True, although WingedMigrator's solution is a very nice and elegant "back-of-the-envelope" method. Also, since the whole point of high bypass turbofans is to move large mass at low deltaV, the higher your bypass ratio gets the closer the real and "idled" airflow become. For me as an engineer, the fact that the two "back of the envelope" calculations derived in completely different ways "from first principles" (almost) are both of the same magnitude is a good sign we are close to a reasonable solution to the problem.

Agreed; very elegant approximations indeed.

I must admit I'm still surprised by the result; +150 times MTOW in air consumption, that's no joke. Now there's an interesting pre-flight announcement that would get passengers' attention. It also puts into focus the incredible reliability of modern turbofan engines, man-handling over a hundred million pounds of matter per flight for +20 years on. Amazing when one thinks about it...

Faro
The chalice not my son

tdscanuck
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting faro (Reply 14):I must admit I'm still surprised by the result; +150 times MTOW in air consumption, that's no joke. Now there's an interesting pre-flight announcement that would get passengers' attention.

Airflow in aviation are often a lot higher than you'd think. The airflow that the wing is shoving down to create lift is even more impressive!

Tom.

BMI727
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting hal9213 (Reply 10):Actually, you calculated the mass of air, that would travel "idled" through the engine housing, if there was no fan nor engine.

For an airliner that rough calculation probably isn't so bad, since the engine is generally optimized for cruise. For a fighter, it wouldn't work so well.
Why do Aerospace Engineering students have to turn things in on time?

hal9213
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting rwessel (Reply 12):At high subsonic speed (IOW, cruise), there isn't (and can't be) much additional flow because of the suction from the engine (compared to the nominal cross section of the engine hitting the oncoming air stream).

True, havent thought about that, thanks!

BMI727
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### RE: Mass Of Air 'Consumed' In Propulsion

 Quoting rwessel (Reply 12):At high subsonic speed (IOW, cruise), there isn't (and can't be) much additional flow because of the suction from the engine (compared to the nominal cross section of the engine hitting the oncoming air stream)

In that condition you are likely faster than the design speed and are dumping air around the engine. In other words, the stream tube of air headed for the engine is smaller than the engine diameter.
Why do Aerospace Engineering students have to turn things in on time?

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