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 18th Jun 2000, 21:13 #1 (permalink) DECU Guest   Posts: n/a Why are jet engines more fuel efficient at high altitude? Hi all, Can anyone please tell me why jeyt engines are more fuel efficient at high altitude? Thanks, DECU
 19th Jun 2000, 03:48 #2 (permalink) Tinstaafl Guest   Posts: n/a My, admittedly simple, understanding: Jet engines are most efficient operating at max permitted RPM, subject to operating limits. Unfortunately, at lower altitudes Temp & pressure limits would be exceeded so, to produce a given amount of thrust eg maximum or some cruise setting, the RPM must be reduced from maximum. As altitude increases air density reduces so thrust output for a given RPM reduces. To maintain this thrust RPM must be increased. As RPM is increased (but not beyond limits) efficiency rises. Couple with this, the airframe is designed to be most efficient at a particular Mach No. This will cause a certain amount of drag which in turn will require a certain fixed amount of thrust (all other factors being equal). As altitude increases with the airframe maintaining an efficient airspeed, the thrust is produced by an engine that operates closer & closer to it's max RPM. Hence overall efficiency rises. NB: Rate of change of gain in thrust efficiency would have to be greater than rate of change of TAS reduction, with gain in altitude at this 'efficient' Mach No. Someone please jump in and correct my errors!!
 19th Jun 2000, 07:12 #3 (permalink) flite idol Guest   Posts: n/a Ok for starters TAS is Equivalent airspeed corrected for density and it is related inversly, in other words a decrease in density will give an increase in TAS. Its relationship to altitude requires no furthur explanation.
 19th Jun 2000, 23:54 #4 (permalink) Tinstaafl Guest   Posts: n/a Flite Idol, My comment regarding the reduction in TAS with increasing altitude was related to maintaining a constant Mach No. As temp reduces with altitude so does the TAS associated with a constant Mach No. I was trying to refer to flight where Mach No. is the limiting factor and not the relationship between EAS & TAS with changing density. T ------- Grammar! Do we really need it?
 20th Jun 2000, 00:55 #5 (permalink) mik Guest   Posts: n/a D. P. Davies's "Handling the Big Jets" has a section on why higher is better for jet transport aircraft. Its a book that is well worth getting hold of, although it is now a bit old. Mik p.s. the reason I'm not going to quote Mr. Davies's explanation is that a) it's fairly long, and b) it includes graphs. [This message has been edited by mik (edited 19 June 2000).]
 20th Jun 2000, 01:11 #6 (permalink) flite idol Guest   Posts: n/a Okeydokey T!!
 20th Jun 2000, 04:13 #7 (permalink) Tinstaafl Guest   Posts: n/a sigh I loved my copy of 'Handling the Big Jets' - but it disappeared years ago. sniff
 20th Jun 2000, 04:41 #8 (permalink) CaptainSquelch Guest   Posts: n/a Handling the big jets is nice but if you want an answer you just might start with "Gas turbine theory" by Cohen, Rogers and Saravanamuttoo ISBN 0 582 44927 8 If you have any questions after that there is a nice load of literature mentioned in it. BTW it has something to do with cold air and thin air. hence the optimum around the tropopause..... good luck and a lot of reading pleasure.
 20th Jun 2000, 14:35 #9 (permalink) Atlas Guest   Posts: n/a Hmmm. Thought one of the real engineering types would venture in here. In the meantime, just the musings of another pilot who tries to retain some of his 6th form physics, so take with a grain of salt (KISS as always) . As Tinstaafl and others points out, the jets' guts are optimized for operation at high altitude where they spend most of their time, therefore they will have somewhat poorer sfc lower down, but where the efficiency of the aircraft is concerned, that isn't the dramatic difference. Jets do their work by throwing masses of air out the pipe at a velocity, at a rate, which is pretty much force=massxacceleration. The forward velocity (TAS) of the A/C is largely secondary. Consider that the thrust (=drag) to maintain 250-300 CAS (about what jet airliners cruise at) will be similar at MSL as it is at FL310 and above. Thus, the fuel burn at 250-300 (CAS=TAS) at MSL is similar to that at about 440-480 TAS at the FL's (250-300 IAS!), actually a little greater due to the inefficiencies to which Mr. T. et al referred. I know I don't wish to read several long tomes to answer every little curiosity I have, but if you do want to read up a bit, in addition to the other books referred to, you might look at MECHANICS of FLIGHT by A.C. Kermode. If its in your nearest library, the index will direct you to the dozen or so pages that will explain it all quite nicely. Cheers, Atlas
 21st Jun 2000, 14:46 #10 (permalink) Join Date: Aug 1998 Location: Ex-pat Aussie in the UK Posts: 3,726 Ahh, the simplest questions always generate the most complex answers and vice-versa! The efficiency of a heat engine is largely governed by the compression ratio. A heat engine is one that converts heat in a gas (added by a fuel) to do work. All combustion engines, like the Otto cycle (four stroke) engine, Diesel engine and Brayton Cycle (gas turbine) engine are heat engines. How does the amount the air/fuel mix is compressed effect the efficiency? Imagine a piston travelling up a cylinder and compressing the gas in that cylinder (like pushing in the handle of a bike pump with your thumb over the exit hole). If released, the piston would now travel back down the cylinder until the pressure inside equalled the atmospheric pressure. The work done in compressing the gas is recovered as the piston moves back down the cylinder under the pressure of the compressed gas (less friction losses etc.) If the piston compresses the gas to half its original volume, then it has a compression ratio of 2:1, as the piston recovers it can do work based on the compression it started with. The more the air is compressed, the more work can be done as it recovers. Otto cycle piston engines have a compression ratio of around 8:1. Now lets introduce some combustion after the piston has compressed the air, as happens in an Otto cycle engine (a "suck, squeeze, bang, blow four stroke). If the charge was only compressed at a 2:1 ratio, then the effect of the combustion can only do work until the piston recovers to ambient pressure. The more the piston compresses the charge, the further away from ambient pressure it is and thus more work can be done by the piston as it recovers. So the higher the fuel/air mix is compressed, the more efficiently that fuel can expend its energy. Why not compress the charge more in a four stroke, and get even better efficiency? Because as the charge is compressed it heats up, until at higher compressions it heats enough to explode at the wrong time ("known as "pinging", "knocking" or detonation) which may create enough pressure to exceed the strength of the engine components. Typically you put holes in pistons or cylinder heads - not considered a good thing. Diesel engines actually take advantage of this effect and use "compression ignition" instead of a spark plug, so they can achieve double the compression of Otto engines and are consequentially more efficient. An aviation diesel engine, burning kerosene (Jet fuel) is being developed and should be available on the Socata Trinidad in a year or so. It is lighter, has less moving parts and burns less fuel with more power than your current Lycoming or Continental. All this can be proved mathematically from first principles through the laws of thermodynamics, a branch of Physics usually studied at second year degree level in Mechanical Engineering. I won't bore you with the proof, but if you are really interested you can click here. The same principle applies in gas turbine engines, although technically you don't talk about compression ratios, but rather cycle pressure ratios. Known as the thermal efficiency, or internal efficiency of the engine the higher the pressure ratios, the higher the efficiency of combustion. As the compressor is not a cylinder travelling a fixed distance, but a "fan blowing air into a small space" the pressure ratio is governed by the engine RPM. So gas turbine engines like to run at high RPM. Next you need to look at the propulsive efficiency, or external efficiency of the engine (the study in Physics known as mechanics). The amount of thrust provided by an engine (propeller or jet) is related to the amount of air they throw out the back, and the speed at which they throw that air. In symbols, if m kilograms is the mass of air affected per second, and if it is given an extra velocity of v meters per second by the propulsion device, then the momentum given to the air per second is mv, so Thrust = mv (per second). A propeller engine uses a large m and a small v, a gas turbine engine uses a small m and a large v. 10 kg of air given a velocity of 1 m/s has the same thrust as 1 kg/s of air given a velocity of 10 m/s. Which is most efficient? Well, the rate at which kinetic energy is given to the air (the work done) is ½mv² watts. So the first case requires 5 watts of work and the latter requires 50 watts of work. Clearly the piston is more efficient. Problems occur as the speed increases, and the propeller efficiency breaks down. An easier way to think of it is by considering the waste energy in the flow. Stand behind a propeller engine at takeoff, and it will knock your hat off. Stand behind a jet giving the same thrust and it will knock you off your feet! Waste energy dissipated in the jet wake, which represents a loss, can be expressed as [W(vj-V)²]÷2g (W is the mass flow, vj is the jet velocity, V is the aircraft velocity, so (vj-V) is the waste velocity). As the jet exhaust leaves the gas turbine at roughly the same speed whether it is standing still or moving, the faster the engine is moving the less waste energy lost. Assuming an aircraft speed of 375 mph and a jet velocity of 1,230 mph the efficiency of a turbo-jet is approx. 47%. At 600 mph the efficiency is approx. 66%. Propeller efficiency at these speeds is approx. 82% and 55% respectively. So gas turbine engines like to fly at high TAS. The problem is that at high RPM at sea level turbo-jets are sucking in very thick air, so when they add a heap of fuel and burn it they make tremendous amounts of thrust - good for take-off but way too much for level flight at low level (with the associated high IAS). Aircraft generate too much drag at high Indicated Air Speed and keeping the engine turning at an efficiently high RPM at sea level would quickly exceed the aircraft speed limitations. As you increase altitude, however, the amount of thrust the engine can produce reduces (as it is sucking in "thinner" air) even though it is still operating at high compression ratios (for good thermal efficiency). Also you can have a high TAS (for good propulsive efficiency) at a low IAS (for lower drag on the airframe). There are a few other advantages, like the cold temperature, which keeps the turbine temperatures down as well. So jet engines like to fly at high power and at high TAS, while aeroplanes like a moderate IAS, and the regime were all this can be achieved together is at high altitude. [This message has been edited by Checkboard (edited 04 July 2000).]
 22nd Jun 2000, 11:21 #11 (permalink) Atlas Guest   Posts: n/a Checkboard - When I was composing my own briefer response , I prefaced it by asking DECU whether he was asking about jets' efficiency in comparison to [1] itself at lo altitude, or [2] a propellor (at hi altitude, presumably). I used as a quick reference for the latter, the afore-mentrioned MECHANICS of FLIGHT, which covered the propellor comparison much as you did (p. 133-135). On rereading DECU's question, it seemed pretty clear to me he was asking about the hi-lo jet comparison, so I ditched it all, save the mention of the text, and then went on to address the efficiency that most pilots care about when they ask about the engine - nm/lb rather than sfc #'s. One small quibble - Thrust is a force (massxacceleration). When applying the T=mv formula, this yeids the momentum given to a particular mass of air, more properly stated in basic units of measurement as = MxL/T. The thrust/force would be derived from the rate at which momentum is given to masses air, or MxL/T/T(again), or mass x acceleration. N'est-ce pas? You are obviously much more deeply schooled in this stuff than I, hence my repeated retreat behind KISS principles. Cheers, Atlas [This message has been edited by Atlas (edited 22 June 2000).]
 22nd Jun 2000, 18:10 #12 (permalink) Mice Guest   Posts: n/a Jeeesus Checkboard! I think I might just give my engineer licence back to the authorities, I am obviously not worthy! As you said, complex answers for a simple question. As the aircraft frame is encountering less air resistance at higher altitude, there is less requirement for power to propell it at constant speed through the air. Therefore, as the air is less dense, the optimum fuel/air ratio is maintained by reducing the fuel input. Modern turbofans are gas path optimised for best SFC in CRZ also. The advantage they have over a piston engine at high altitude is the high mass airflow the turbine engine design can process, whereas the piston engine mass flow is a limit of the swept volume, and the ability of other equipment to compress intake air into the cylinder at pressures that allow the engine to still function. Remember, the turbine engine mass airflow capacity is variable, whereas this is not exactly so for the piston design.
 22nd Jun 2000, 21:10 #13 (permalink) DECU Guest   Posts: n/a Hi all, Wow! Thanks for all your inputs. It is very much appreciated. I asked this question because it was a question which I was asked during an interview recently, and without going into too much detail, I just told them the basics, which I myself do not know if it was the correct answer. In fact, I cant even really remember what I told the interviewers, except that it was due to a higher TAS and also the air being less dense. Once again, I really appreciate the comments and in depth explanations. I guess I'll know what to tell the interviewers the next time round. Cheers, DECU
25th Jun 2000, 16:44   #14 (permalink)

Join Date: Aug 1998
Location: Ex-pat Aussie in the UK
Posts: 3,726

Quote:
 then the momentum given to the air per second is mv, so Thrust = mv.
F=ma, a is in m/s/s, v = m/s, for momentum per second = ma = mv/s

[This message has been edited by Checkboard (edited 04 July 2000).]

 25th Jun 2000, 19:51 #15 (permalink) N2 Guest   Posts: n/a Checkboard, In an effort to reduce ripping through dusty boxes of college notes, some stub of my brain seems to recall the natural limiting fwd speed the engine may attain once the active (thrust) and reactive forces are in balance. Our savior here now being ram effect. Does ram effect now provide the bulk of compressive efficiency at altitude and crz speed?
 28th Jun 2000, 21:35 #16 (permalink) Elevation Guest   Posts: n/a If I remember correctly, it has something to do with the exhaust gas. The cooler the outside air temp, the more the exhaust gas expands as it exits the chamber hence creating more forward "push". Then again I better go back and dig for my old textbook on turbine
 1st Jul 2000, 11:06 #17 (permalink) twistedenginestarter Guest   Posts: n/a I've just found this at http://www.faa.gov/avr/afs/acs/61-107.txt (1) The efficiency of the jet engine at high altitudes is the primary reason for operating in the high-altitude environment. The specific fuel consumption of jet engines decreases as the outside air temperature decreases for constant revolutions per minute (RPM) and TAS. Thus, by flying at a high altitude, the pilot is able to operate at flight levels where fuel economy is best and with the most advantageous cruise speed. For efficiency, jet aircraft are typically operated at high altitudes where cruise is usually very close to RPM or exhaust gas temperature limits.
 1st Jul 2000, 14:32 #18 (permalink) ironbutt57 Guest   Posts: n/a so all the above gobbledygook translates to a simple answer... jet-powered AIRCRAFT operate more efficiently at optimum altitudes which tend to be higher than other aircraft types for many many reasons
 1st Jul 2000, 22:19 #19 (permalink) Flap 5 Guest   Posts: n/a DECU God help you if Checkboard was your interviewer!
4th Jul 2000, 12:19   #20 (permalink)

Join Date: Aug 1998
Location: Ex-pat Aussie in the UK
Posts: 3,726

N2,
Does ram effect now provide the bulk of compressive efficiency at altitude and crz speed?

At usual operating thrust the engine exhaust achieves supersonic velocity, referred to as being "choked". Thrust levels above this increase the mass flow through the engine, but the exhaust speed is fixed. As in my original explaination this means that the faster the engine is going (in terms of TAS), the less thrust it is producing (but at a greater efficiency).

There is also a loss through momentum drag - if you are an airmolecule hanging around in the sky, and are scooped up by an engine rushing past, then the engine has to accelerate you up to its speed as you pass into the engine, and this is known as momentum drag, which has a net effect of reducing the thrust.

On the upside is the "ram ratio effect". The higher dynamic pressure into the front of the engine increases the mass flow (Force = mass x acceleration, so the more mass the engine processes, the more thrust it produces). The higher mass flow requires a higher fuel flow to meet it, the ram effect covers the losses from momentum drag, but doesn't cover the losses associated with the higher airspeed.

Elevation:
Quote:
 it has something to do with the exhaust gas. The cooler the outside air temp, the more the exhaust gas expands as it exits the chamber hence creating more forward "push".
The jet engine is a reaction engine - it produces thrust by throwing mass out the back, what happens to the exhasut after it has passes the jet nozzle has no effect on thrust produced. See below for more on temperature, though.

This common misconception also has people thinking that the jet blast deflectors seen on carriers increase the takeoff thrust of the jets on a cat launch - they don't, they just stop people & equipment behind the jet being toasted

twistedenginestarter:
Quote:
 The specific fuel consumption of jet engines decreases as the outside air temperature decreases for constant revolutions per minute (RPM) and TAS.
The reason the statement refers to a constant RPM is that the RPM controls the pressure ratio of the engine, which is the dominating factor in engine efficiency.

Having said that, when the inlet air temperature is lowered, a given heat addition can provide greater changes in pressure or volume & greater efficiency. If the RPM was limited due to EGT considerations a higher RPM can be achieved as well.

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