Max endurance speed is not the same for gliders, propeller planes (also for turbo prop, right?) and jet planes.

While for for jet planes max endurance is at the minimum drag speed ($V_{md}$), for propeller planes and gliders it is the speed where minimum power ($V_{mp}$) is required. While I found hundreds of posts explaining that the the speeds are different because the formulas mentioned above are the way they are, I could not quite figure out why they are different (I am not an engineer and could not quite follow some of the more lengthy, mathematical explanations).

In this question the author states:

In case of propeller aircraft, the fuel flow rate is proportional to the power produced. Hence, the maximum endurance occurs at a point where the power is minimum. For (turbo)jets, the minimum fuel flow occurs when the thrust is minimum. Hence the maximum endurance occurs when the L/D is maximum.

Of course this explains why max endurance is different, but it creates the new question why propeller engines behave differently from jet engines.

At the end of the day both engines have the same task: Convert chemical energy to kinetic energy.

If I imagine a jet plane and a propeller plane with parking brakes set and a given throttle input they will both produce a lot of thrust, but zero power. Advancing the throttles will increase fuel flow for both planes, but power remains at zero, thus it seems that fuel flow depends on thrust, not on power.

I'd think to maintain straight and level flight we need a fixed amount of lift, but the drag is just disturbing us. So we are looking for all combination of $C_L$ and speed so that the resulting lift is as much as we require to maintain straight and level flight. Then from those values we select the value that has minimum drag. This is the speed we need to fly, irrespective of the type of engine. Then we burn as much fuel as we need to generate the the same amount of thrust as drag. This thinking of mine is obviously wrong, but why? What am I missing?

EDIT: While the answers I got seem to be perfectly logical to me and technically correct, it seems to me that they don't quite answer the question (or I am too stupid to make the connection). I guess I was not clear enough, so let me try to outline my mental picture so far...:

In an Otto engine (for simplicity let's assume a single cylinder engine) a fuel-air mixture is injected into a cylinder, where it is compressed and then it is ignited. The fuel burns and thus the air becomes hot and wants to expand, but there is a piston in the way. Thus the pressure rises and exerts a force on the piston. This force (measured in Newton) causes the piston to move from TDC (top dead center) to BDC (bottom dead center). Thus a force is applied to over a distance (the distance between TDC and BDC), resulting in torque, measured in Newton-meters, equivalent to energy measured in Joule. In a perfect engine the amount of energy released that way would be the same as the amount of chemical energy contained in the fuel.

Increasing the amount of fuel that is injected into the cylinder will cause the the air in the cylinder to become hotter and expand more forcefully. The distance between TDC and BDC remain the same, but the force increases, resulting in increased torque.

But for power we need to know at which rate we are releasing the energy, so we divide the energy released during one stroke by the time it takes from the ignition to the next ignition. Thus we will get the power measured in Newton-meters per second or Joule per second or simply Watt.

From the above it quite clear that fuel consumption for a piston engine is directly connected to the power it produces.

What will happen if the propeller is feathered? In this case the power absorption of the propeller is zero and if it were not for friction the propeller would indefinitely continue to rotate with zero fuel flow. No force is needed to keep the rotation is progress. The piston would still move up and down, but no force is there and thus power is also zero.

Now let's unfeather the propeller. The rotating propeller will now move through the air like a wing. This creates an aerodynamic force pulling the plane forward, which we call thrust. The amount of thrust it develops depends on the angle of attack and the speed of the propeller blade (and air density, but we'll assume that this is constant). The speed of the propeller is obviously different at the root and the tip, just like the AoA, but we can work with an average AoA and an average speed.

Here my mental model starts to get confused. How is the energy transferred? I know must have something to do with out TAS, but I don't get a picture how that happens.

To simplify things I picture a plane on the ground. We use a rope to attach the tail of the plane to a force meter and attach the force meter to a wall. Now we can startup the engine and read the thrust on the force meter. I think we will all agree that if we increase fuel flow, the force meter will show an increase in thrust. But what happens if we now add some head- or tailwind, while keeping fuel flow constant? In my mental picture the force meter will still display the same thrust (as long as we adjust prop speed or prop pitch so that fuel flow remains constant). But that seems to be wrong. Why?

  • $\begingroup$ I might come back tomorrow with a more detailed answer, but very roughly with a propeller you're delivering constant power so thrust drops with speed as F=P/V, but with a turbojet you're delivering constant thrust so effective power increases with speed for the same throttle setting as P=FV. $\endgroup$ – Talisker Apr 13 '17 at 22:12
  • $\begingroup$ "In case of propeller aircraft, the fuel flow rate is proportional to the power produced. Hence, the maximum endurance occurs at a point where the power is minimum." ... could this explain why Coast Guard C-130s would conduct searches with 2 engines shut down? $\endgroup$ – radarbob Apr 14 '17 at 1:20

Difference between engine types

That different engine types cause the optimum loiter speed to be different is due to the amount of air used for thrust creation. Propellers accelerate a lot of air by a little, while jets accelerate a little air by a lot.

Thrust is the difference between the exit impulse and the entry impulse of the air flowing through the engine rsp. propeller minus the installation drag (spill drag, cooling drag). If $v_0$ is the air speed ahead of the engine and $v_1$ that behind it, $\dot m_{air}$ the air mass flow and $\dot m_{fuel}$ the fuel flow, thrust $T$ is $$T = (\dot m_{air} + \dot m_{fuel})\cdot v_1 - \dot m \cdot v_0 - D_{installation}$$ As long as $v_0$ is low, the speed difference $\Delta v = v_1 - v_0$ can be low and still thrust is high if enough air is involved. However, in order to fly fast, $\Delta v$ must be high enough so maximum thrust is achieved with a small air mass flow - using a bigger air mass would drive up engine mass and installation drag.

As a consequence, propellers produce declining thrust with increasing flight speed (and, consequently, increasing entry impulse) while in jets with their high exit speed thrust is approximately constant over speed in the subsonic region - it is high enough so that subtracting the entry impulse doesn't make much of a difference. Jet engines are also better in making good use of that entry impulse: By precompressing the air ahead of and in the intake, this ram effect raises the pressure level and mass flow inside the engine.

A piston engine consumes fuel in proportion to its speed (RPM) and air density, but independent of air speed. Thus, its power output is independent of flight speed. Thrust is power divided by speed, so this is another way to show why thrust must go down with increasing flight speed. Contrast this with jets, where precompression in the intake increases mass flow over flight speed, so fuel consumption will go up with speed. A typical high-bypass turbofan will consume twice the fuel flow per unit of thrust at cruise than it will at zero speed.

Maximum endurance speed

Why the optimum speed for maximum endurance differs with the thrust-over-speed behavior of the engine is explained in this answer. But what about gliders?

To stay airborne for the longest time requires to fly at the minimum sink speed. The "fuel" which keeps the glider flying is its potential energy $E_{pot}$ which is continuously converted into kinetic energy to compensate for what is used up by drag. So we can equate: $$\frac{dE_{pot}}{dt} = m\cdot g\cdot \frac{dh}{dt} = m\cdot g\cdot v_z = D\cdot v$$ with $m$ = mass, $g$ = gravitational acceleration and $h$ = height. $v_z$ is the vertical speed which we aim to minimize.

If we call the flight path angle $\gamma$, we can write: $$m\cdot g = -\frac{L}{cos \gamma}\; \text{and}\; v_z = v\cdot sin\gamma$$ For small values of $\gamma$, the approximations $cos \gamma = 1$ and $sin \gamma = tan \gamma$ are valid. Together with the parabolic drag equation we can now write: $$v_z = v\cdot\frac{c_{D0}}{c_L} + v\cdot\frac{c_L}{\pi\cdot AR\cdot\epsilon} = v^3\cdot\frac{c_{D0}\cdot\rho\cdot S_{ref}}{2\cdot m\cdot g} + \frac{1}{v}\cdot\frac{2\cdot m\cdot g}{\rho\cdot S_{ref}\cdot\pi\cdot AR\cdot\epsilon}$$ and set the derivative of $v_z$ over flight speed $v$ to zero: $$\frac{dv_z}{dv} = 3\cdot v^2\cdot\frac{c_{D0}\cdot\rho\cdot S_{ref}}{2\cdot m\cdot g} - \frac{1}{v^2}\cdot\frac{2\cdot m\cdot g}{\rho\cdot S_{ref}\cdot\pi\cdot AR\cdot\epsilon}$$ $$= 3\cdot\frac{c_{D0}}{c_L} - \frac{c_L}{\pi\cdot AR\cdot\epsilon} \,\overset{!}{=}\, 0$$ This means that at the speed of minimum sink rate the induced drag must be three times as big as the zero-lift drag, which is different from the point of best glide speed, at which both are of the same size.

$c_{D0} \:$ zero-lift drag coefficient
$c_L \:\:\:$ lift coefficient
$S_{ref} \:$ reference area (wing area in most cases)
$v \:\:\:\:\:$ airspeed
$\rho \:\:\:\:\:$ air density
$\pi \:\:\:\:\:$ 3.14159$\dots$
$AR \:\:$ aspect ratio of the wing
$\epsilon \:\:\:\:\:$ the wing's Oswald factor
$m \:\:\:\:$ the aircraft's mass
$g \:\:\:\:\:$ gravitational acceleration

EDIT: Your expanded question now asks what will happen to a stationary, piston-engine-powered aircraft in varying wind. Details now depend on the type of propeller which is fitted to the aircraft. Let's look at both possible cases and assume we change from no wind to a strong headwind:

  1. Fixed-pitch propeller: With increasing headwind, the load on the propeller will go down, allowing the engine to run faster. If you keep the fuel flow constant, torque will go down, so the engine will settle at a higher speed where the load on the propeller blades is lower. Remember, power is also torque times angular velocity. Thrust will go down.
  2. Constant speed propeller: Now the propeller will adjust pitch such that the engine speed is held constant. The blades will have a steeper angle with the propeller disc, and the resulting forces on them will point more sideways than before. This requires an overall reduction in the load on the propeller blades to keep the circumferential force component constant, and the force component perpendicular to the propeller disc will have to go down. Thrust will go down.

Either way, flying faster will result in less thrust. Please adjust your mental picture accordingly.

  • $\begingroup$ I'll get my coat :) $\endgroup$ – ymb1 Apr 14 '17 at 10:33
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    $\begingroup$ @ymb1 Hey - I upvoted your answer, don't delete it! $\endgroup$ – Peter Kämpf Apr 14 '17 at 10:34
  • $\begingroup$ Thank you for your answer. While I believe that I understand what you are saying, I feel that the answer does not quite fit the question. I understand that propellers will produce less thrust at high TAS (on full throttle), but I think fuel flow will also reduce in this situation and thus this is no problem for endurance... I expanded my question... $\endgroup$ – yankee Apr 14 '17 at 13:51
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    $\begingroup$ @yankee: No, the fuel flow of piston engines is independent of speed. It only depends on throttle setting and air density. With jets this is different; their flow goes up with speed because precompression produces a higher mass flow. $\endgroup$ – Peter Kämpf Apr 14 '17 at 15:27

At the end of the day both engines have the same task: convert chemical energy to kinetic energy.

Not quite. The answer you linked gave away a golden hint.

In a piston engine aircraft (I am excluding turboprops now) chemical energy is transferred to rotational kinetic energy, which drives a propeller that transfers part of this rotational energy to the air in the form of kinetic energy (air being shoved backwards).

In a pure turbojet (no fan), most of the chemical energy is transferred directly to the air, pushing it out the back. See here: Which portion of combustion energy is used to spin the compressor of a pure jet engine?

A turbofan is part turbojet, part propeller, that's why the author's paragraph ends by saying:

[...] for turbofans, it is somewhere in-between.

Having a middle-man (propeller) for the reciprocating engine is the key—

enter image description here
(Source) Original graph is meant for propellers. I added the text that relates to turbojets.

Endurance (jet v prop)

In a turbojet it's easy to note the minimum thrust (= drag) on an L/D graph (bottom graph).

But since the output of a reciprocating engine is power, converting that to thrust horsepower (thp) requires multiplying a given power by propeller efficiency. The net propeller thrust varies with forward speed and RPM. There is no easy direct conversion to match an L/D graph.

Propellers lose thrust (waste power in terms of endurance) with forward speed. See here: Why are the top speeds for jet engines higher than for propellers?

But, converting the other way around—thrust to power—is easier. You multiply the thrust (y-axis) by the speed (x-axis) on an L/D graph. Thus, a new graph is born—the top one—with its endurance (lowest point) shifted to the left.

Note: A full throttled standstill prop or jet will be displacing air at a massive rate (velocity), thus power is not zero.

To expand on the different outputs, imagine this, if we remove the propelling nozzle of a turbojet (the back end), it's now become a gas turbine. Extend the shaft and connect it to a generator, and now the direct output we can measure is no longer a propelling force. It's now a shaft power that we can't plot directly on an L/D graph.

Endurance is better with an engine (or more) shutdown

Could this explain why Coast Guard C-130s would conduct searches with 2 engines shut down?

The C-130 is driven by four turboprops (gas turbines turning propellers). Shutting down two engines means the other two will need to run at higher torque levels.

Gas turbines consume the least amount of fuel per unit time per unit power when they are running at their cruise power. It's not a linear relation.

Conducting a search at low altitude and speed would not require this cruise power from all four engines. Hence the economy in shutting down two engines and running the other two at their maximum continuous power setting.

The graph below is for a turbofan, it illustrates the relation. At a given speed, the more thrust/torque a gas turbine produces, the more bang for your buck it gives back.

enter image description here
(Source) From a Boeing manual, yellow highlight is mine.

Note: A jet engine running faster is not like a jet plane flying faster, the latter will cause a drop in efficiency—higher TSFC. See here: How does the efficiency of a turbofan compare between take off speeds and optimal conditions?

For the 777-200, the fuel flow is 8694 lbs/hr @ FL190 with one engine out, and 9434 lbs/hr (2 engines) at the optimum FL of 430, both for the same weight (300,000 lbs) and KIAS of 232 (KTAS will be 307 and 469, respectively); that engine out 777 can stay for longer in the air. (The figures are from the 777 FCOM.)

In a piston engine it's different, the fuel / RPM / power relation is almost a linear one (see below—red line).

enter image description here
(Source) Piston engine.

  • $\begingroup$ “…power, converting that to thrust requires multiplying a given power by propeller efficiency”: Converting power to thrust requires, first and foremost, dividing it with velocity. The efficiency (efficiency is dimension-less ratio of output power to input power) of a propeller is actually fairly constant. It is efficiency of a jet that varies significantly with speed. I general I think you should explain the difference between power and thrust as it is kind of important here. $\endgroup$ – Jan Hudec Apr 14 '17 at 8:52
  • $\begingroup$ I still think it's more confusing than anything else. Normally, efficiency is defined in terms of power. And with that, efficiency of a piston engine with a propeller is fairly constant over the speed range (thrust is thus inversely proportional to speed). The odd behaviour is that of the jet, because it's efficiency increases with speed, and that is the key difference. Note that here I mean the total propulsive efficiency, that is ratio of thrust times speed (= output power) to fuel specific heat value times fuel flow (= input power). $\endgroup$ – Jan Hudec Apr 14 '17 at 9:03
  • $\begingroup$ @JanHudec - I appreciate the feedback, I've chosen better words now for the prop efficiency. $\endgroup$ – ymb1 Apr 14 '17 at 11:19
  • $\begingroup$ I think I understand what you described, but you also said "But since the output of a reciprocating engine is power...". I think all engines output power AND thrust, the question is more like how this relates to fuel flow. I expended my question with my current mental picture, to make it clearer. $\endgroup$ – yankee Apr 14 '17 at 13:54
  • $\begingroup$ @yankee: Power and thrust (as physical parameters) are related: Power is the work done by a force along a distance per unit of time. Thrust is a force, and the distance per time is the flight speed of the aircraft (or, more precise, the speed of the air flowing through the propeller disk). Thus, power is thrust times speed. $\endgroup$ – Peter Kämpf Apr 14 '17 at 18:02

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