4
$\begingroup$

Here is my assumption:

In straight and level flight of a jet turbine engine:

One IAS = One drag = One thrust = One fuel flow

Therefore, the fuel flow will be approximately the same for a given IAS, regardless of altitude, temperature, or RPM necessary to maintain this IAS.

Is that correct? If not, where is the fallacy?

EDIT : Here is a screenshot from ATPL course material which supports this idea and led me to ask the question as it seems doubtful to me. Propulsive efficiency vs RPM at high altitude at the end is not clear as well. enter image description here

$\endgroup$
  • 1
    $\begingroup$ This is very much dependant on altitude, air pressure, temperature... Thinner air creates less drag, has less oxygen, etc. Not too mention that "IAS" changes with altitude all by itself... $\endgroup$ – Ron Beyer Aug 30 at 12:35
  • $\begingroup$ Related: Why do jet engines get better fuel efficiency at high altitudes? $\endgroup$ – Bianfable Aug 30 at 12:48
  • $\begingroup$ @Ron Beyer. It's true that IAS increases (in TAS terms) with altitude, but the OP presumes a fixed IAS... $\endgroup$ – xxavier Aug 30 at 13:26
  • $\begingroup$ @Ron Beyer as IAS is fixed, thinner air will not create less drag. IAS reflects dynamic pressure, so if maintained constant, dynamic pressure will be the same, and so will drag. It's true though that air gets thinner with altitude, but it is compensated by the increased TAS (more air, so overall dynamic pressure is constant). This part by itself is counter intuitive $\endgroup$ – Arnaud PROST Aug 30 at 13:34
  • $\begingroup$ @Bianfable Thanks for the link, I'm familiar with this topic. Engines get better fuel efficiency at high altitudes because for a same IAS, so a same fuel flow from my point of view, the TAS is much greater: therefore the consumption per unit of distance is much less $\endgroup$ – Arnaud PROST Aug 30 at 13:38
5
$\begingroup$

The most obvious difference is due to the temperature of the air.

Both turbine and piston engines are heat engines. They work by converting thermal power into mechanical power. The theoretical absolute maximum efficiency you can achieve is called the Carnot efficiency, $$\eta = 1-\dfrac{T_C}{T_H}$$

This is the efficiency of an ideal engine using the Carnot cycle, that works by transferring heat from a hot reservoir with temperature $T_H$ to a cold reservoir with temperature $T_C$. A typical jet engine is approximated by the Brayton cycle, and a piston engine the Otto or Diesel cycle, but neither can ever by more efficient than the efficiency noted above. The efficiency of a Brayton cycle is $$\eta = 1-\dfrac{T_C}{T_E}$$ with $T_E$ the EGT.

When flying higher, the temperature of the cold reservoir (the atmosphere) drops lower. You can see that the maximum efficiency of the engine will also increase (even if $T_H$ or $T_E$ respectively drop simultaneously with $T_C$). This means that, even if the required power would stay constant for constant thrust, the fuel flow changes, because a single unit of energy from a drop of fuel can be converted into more mechanical power.

$\endgroup$
  • $\begingroup$ thanks, I'm almost convinced. Here are my doubts: 1/ even though the Carnot cycle efficiency is very much dependent on temperature difference between hot and cold reservoir, is it the case for the Brayton cycle ? The maximum (Carnot efficiency) could vary with this temperature while the Brayton, staying below, would not be affected. 2/ Why wouldn't the hot reservoir just become colder (EGT lower) ? 3/ If the temperature is colder, density higher therefore TAS will be lower to fly at same IAS. Doesn't this compensate the extra energy from cold air ? $\endgroup$ – Arnaud PROST Aug 31 at 12:21
  • $\begingroup$ 1/ I quoted Carnot as to avoid the inevitable question "what if we used another cycle". I'll add it to the answer. 2/ If the temperature rise is approximately constant (which makes sense because $c_p$ is approximately constant as the atmosphere is approximately an ideal gas) you still get a rise in efficiency $1-\frac{T_C}{T_h}<1-\frac{T_C-\Delta T}{T_H-\Delta T}$. 3/ That effect is already accounted for if we only look at the IAS. $\endgroup$ – Sanchises Sep 1 at 6:15
  • $\begingroup$ Could you edit your answer to elaborate on 3/ ? It's not clear to me, and the whole difficulty of the problem. If we had fixed TAS, the efficiency difference would be sufficient to explain different fuel flows. But we fixed IAS, so even though more energy is produced out of a unit of fuel in colder air, in this colder air TAS will be lower to maintain the IAS. So less work will be needed, and less fuel. $\endgroup$ – Arnaud PROST Sep 1 at 8:24
  • $\begingroup$ @ArnaudPROST You're thinking the wrong way around. IAS is dynamic pressure, so what the aircraft actually 'feels' - it doesn't matter if that's due to temperature or altitude that the density drops. But after establishing that constant IAS leads to roughly constant thrust, you can then wonder how efficient thrust is generated. Density doesn't really come into the whole thermodynamic efficiency picture. $\endgroup$ – Sanchises Sep 1 at 22:07
  • $\begingroup$ I agree, but thermodynamic is not the only parameter influencing the efficiency of thrust generation, TAS is one as well. Thrust is approximately: $T=\dot{m_{air}}(V_{jet}-TAS)$ Thermo tells you how hard it is to accelerate the airflow to $V_{jet}$, but for a same thrust, you will need more or less of it depending on TAS $\endgroup$ – Arnaud PROST Sep 1 at 22:31
1
$\begingroup$

The fallacy is

One thrust = One fuel flow

The thrust-specific fuel consumption (TSFC) of a jet engine (the mass of fuel consumed per unit of thrust) isn't constant, so that equality is violated. TSFC will vary with all the usual factors: ambient temperature and pressure, airspeed, etc.

$\endgroup$
  • 1
    $\begingroup$ Could you perhaps elaborate why this is the case? $\endgroup$ – Sanchises Aug 31 at 7:53
  • $\begingroup$ Honestly, come to think of it, you can have multiple level flights at the same IAS with different AoAs or configurations, and thus different drags, too, so that part's also trivially fallacious, but the question also seemed to be about engines instead. Where it's also trivially fallacious because there's a whole concept devoted to the idea that one thrust doesn't equal one fuel flow. $\endgroup$ – Erin Anne Aug 31 at 9:49
  • $\begingroup$ @Erinn Anne this is true about configurations, but the question of course considers a constant configuration. Could you elaborate on different drags and AoAs for the same IAS on level flight ? It sounds wrong to me, from ATPL at least. For your answer, I also disagree: the difficulty is that ambient temperature and pressure changes, as well as airspeed, are included in IAS parameter (from density). IAS being fixed, it does not matter if one is changed, the others will compensate to maintain the same value of IAS and it could therefore be the same from the engine point of view. $\endgroup$ – Arnaud PROST Aug 31 at 12:11
  • $\begingroup$ On configuration: you could deploy some drag, like flaps or spoilers, add an equal amount of thrust, and now you're at the same IAS (because thrust - drag is still 0 and haven't changed altitude). So one IAS absolutely doesn't correspond to one drag (hence what we call the "drag coefficient", which changes with configuration, or with airframe...) On thrust: TSFC is an empirically observed fact. Look at the tables in the wikipedia article I linked. You can thought experiment at it all you like, but the reality is what it is. $\endgroup$ – Erin Anne Aug 31 at 21:55
  • $\begingroup$ @Erin Anne: configuration is constant. When you said "you can have multiple level flights at the same IAS with different AoAs OR configuration" it's false. For a level flight and an IAS you have only one AoA. Your reference to TSFC is not clear enough, and not linked to the statements of the question. Please elaborate your reasoning: "reality is what it is" does sound like an authority argument which does not help to understand underlying phenomena $\endgroup$ – Arnaud PROST Sep 1 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.