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Mass Of Air 'Consumed' In Propulsion  
User currently offlinefaro From Egypt, joined Aug 2007, 1556 posts, RR: 0
Posted (2 years 10 months 4 weeks 1 day 18 hours ago) and read 4007 times:

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
18 replies: All unread, jump to last
 
User currently offlineJetlagged From United Kingdom, joined Jan 2005, 2565 posts, RR: 25
Reply 1, posted (2 years 10 months 4 weeks 2 hours ago) and read 3693 times:

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.
User currently offlineBMI727 From United States of America, joined Feb 2009, 15780 posts, RR: 27
Reply 2, posted (2 years 10 months 3 weeks 6 days 23 hours ago) and read 3660 times:

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?
User currently offlinefaro From Egypt, joined Aug 2007, 1556 posts, RR: 0
Reply 3, posted (2 years 10 months 3 weeks 6 days 20 hours ago) and read 3632 times:

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
User currently offlinerwessel From United States of America, joined Jan 2007, 2368 posts, RR: 2
Reply 4, posted (2 years 10 months 3 weeks 6 days 18 hours ago) and read 3608 times:
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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.


User currently offlineWingedMigrator From United States of America, joined Oct 2005, 2215 posts, RR: 56
Reply 5, posted (2 years 10 months 3 weeks 6 days 7 hours ago) and read 3476 times:

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!  


User currently offlineJetlagged From United Kingdom, joined Jan 2005, 2565 posts, RR: 25
Reply 6, posted (2 years 10 months 3 weeks 6 days 7 hours ago) and read 3460 times:

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.
User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 7, posted (2 years 10 months 3 weeks 5 days 22 hours ago) and read 3363 times:

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.


User currently offlineprebennorholm From Denmark, joined Mar 2000, 6481 posts, RR: 54
Reply 8, posted (2 years 10 months 3 weeks 5 days 3 hours ago) and read 3221 times:

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, Preben Norholm
User currently offlineJetlagged From United Kingdom, joined Jan 2005, 2565 posts, RR: 25
Reply 9, posted (2 years 10 months 3 weeks 4 days 13 hours ago) and read 3146 times:

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.
User currently offlinehal9213 From Germany, joined May 2009, 302 posts, RR: 0
Reply 10, posted (2 years 10 months 3 weeks 4 days 12 hours ago) and read 3123 times:

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.


User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 11, posted (2 years 10 months 3 weeks 4 days 1 hour ago) and read 3056 times:

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.


User currently offlinerwessel From United States of America, joined Jan 2007, 2368 posts, RR: 2
Reply 12, posted (2 years 10 months 3 weeks 3 days 21 hours ago) and read 3027 times:
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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.


User currently offlineflipdewaf From United Kingdom, joined Jul 2006, 1577 posts, RR: 0
Reply 13, posted (2 years 10 months 3 weeks 3 days 19 hours ago) and read 3015 times:
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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


User currently offlinefaro From Egypt, joined Aug 2007, 1556 posts, RR: 0
Reply 14, posted (2 years 10 months 3 weeks 3 days 16 hours ago) and read 3003 times:

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
User currently offlinetdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 15, posted (2 years 10 months 3 weeks 3 days 15 hours ago) and read 2991 times:

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.


User currently offlineBMI727 From United States of America, joined Feb 2009, 15780 posts, RR: 27
Reply 16, posted (2 years 10 months 3 weeks 3 days 2 hours ago) and read 2919 times:

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?
User currently offlinehal9213 From Germany, joined May 2009, 302 posts, RR: 0
Reply 17, posted (2 years 10 months 2 weeks 4 days 18 hours ago) and read 2659 times:

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!


User currently offlineBMI727 From United States of America, joined Feb 2009, 15780 posts, RR: 27
Reply 18, posted (2 years 10 months 2 weeks 4 days 7 hours ago) and read 2607 times:

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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