Aircraft Energy Question

[QJB]

Hi guys,

I wanted to get some clarification on a few things that have been bothering me recently:

1/ I know that an efficient aircraft has a high L/D ratio. Thus thrust provided by the engine is significantly less than lift that acts upon the aircraft. However from an overall energy perspective where does the energy for lift come from. Is it that burning fuel provides a certain amount of energy, this energy is used to create a total aerodynamic reaction from the aircraft (the result of pressure and shear distribution), the total aerodynamic reaction is then resolved into two forces lift and drag. However we consider lift as not taxing the aircraft because it does not impede horizontal motion? I hope I am being clear. Surely the energy for lift has to come from somewhere!

2/ I know that work = force x distance. Therefore if I push on one side of an object with force A and move is distance B the work done will be AB. However! If someone else pushes with an equal force on the opposing side there will be no movement and thus no work done. Therefore my question is, in level flight all the forces on an aircraft are balanced, there is no net force and thus no acceleration. How is it then that the aircraft can be said to be doing work and using energy? Is this something to do with frame of reference?

Cheers

J

[TyroPicard]

From my simple brain.... in unaccelerated S+L flight...
2. Think of Work as Drag x Distance .. you have to do work to overcome the drag. In a space rocket in a vacuum with the engine off no work is being done?
1. Lift is related to the Kinetic Energy of the a/c which the fuel provides by overcoming drag...

[FlyingStone]

As I understand, the work is force times distance, but force in the equation isn't really (vector) sum of all external forces acting on the object in question, but only (again, vector) sum of all active forces acting on the object. As you probably know, for every action there is an opposite reaction (3rd Newton law), thus if act on an object with an active force, the reaction force formes in the opposite direction.

So basically, if you push an aircraft with force 1 kN for 100 meters, the work done is 100 kJ, even if the forces acting on the aircraft are in equilibrium (the resultant of all external forces acting on the aircraft is zero) and the aircraft is consequently moving at a constant speed.

As for the first question, it is very simple (or very complicated, but I'd rather use the simple explanation). To gain lift (or aerodynamic reaction, whatever you call it), you need difference between pressure on the top and on the bottom of the wing (Bernoulli's theorem) and usually this is created by exposing a wing profile to an airflow. In order to create an airflow, you have two choices: either fix the aircraft and move the air with a fan (for ex. wind tunnel), either move the aircraft through a still air (which is what we want to do, since aircraft are a means of travel). So if we want to move the aircraft through still air in order to create an airflow, we must create a force, which moves the aircraft in the desired direction (let's say forward) and that force is called thrust. And to create thrust force, we need to convert fuel into rotating force and create thrust either with turbojet engine either with a propeller (you could do a PhD in thermodynamics, but for the sake of argument it's really not worth going into details how fuel is burned).

So to summarize it: converting (burning) fuel creates thrust force, which moves aircraft through still air, therefor creating airflow to which wing is exposed, hence lift force (and drag of course) is created, and the aircraft flies.

[used2soar]

One way to look at it is what is the potential energy of the aircraft; this is mass * gravity * height above ground, or mgh. In order to get a 500 Kg glider 3000 meters in the air, 500 kg * 9.8 m/s^2 * 3000 m ~ 15,000,000 = 15 Mega Joule has to be expended. With a glider that has 30:1 L/D, an extra 3% has to be spent overcoming drag. Assuming a motor glider, 15.45 Mega Joules, which is 15.45 Mega Watt Seconds. One horse power is 750 watts. So with a 45 Hp engine, climb to 3000 meters should be 8 minutes. (((500 * 3 000 * 10.3) / 750) / 60) / 45 = 7.62962963. But not all 45 Hp goes to forward thrust due to prop efficiency considerations, reality is more like 25 minutes.

[Rivet gun]

So we are considering an aircraft in straight and level (un accelerated) flight.

Looking at the problem from the aircraft's frame of refrence, we resolve the forces into vertical and horizonal components.

In the vertical direction, weight is balanced by lift but there is no vertical motion so no work is done or energy consumed in providing lift per se.

In the horizonal direction however a certain amount of drag is an inevitable consequence of generating lift. To maintain equilibrium drag must be balanced by thrust. Thrust comes from a propellor or jet which works by causing air to accelerate (i.e. it adds momentum to a mass of air). Adding momentum to air requires energy and the energy comes from burning fuel.

[syedo]

try wikipedia...works wonder!! ;)

[QJB]

Thrust comes from a propellor or jet which works by causing air to accelerate (i.e. it adds momentum to a mass of air)

Ok so would this be more correct: The engine burns fuel and a percentage of the energy from this process is used by the propeller to give air momentum and as an equal and opposite reaction the aircraft moves fwd.

Theoretically with a wing of infinite span and non viscous incompressible flow there will be no drag due to there being no shear forces and pressure forces balancing out. In this way couldn't you say that lift is proportional to the energy imparted to the airflow as a consequence of the propeller.

Obviously for a real flow there is drag, but still doesn't the energy for lift still come from the engine? :confused:

[Wizofoz]

doesn't the energy for lift still come from the engine?

Yes it does.

I think your confusion comes from thinking that because the FORCE of thrust is less than the FORCE of lift, the ENERGY put into thrust can't produce the ENERGY needed to supply lift.

Not so- to give a very simple example of a device that puts less force in than it gets out, give me a lever, a fulcrum and a place to stand......

[jtt]

if you look for the energy you can't think about the aircraft alone - it's flying throught the air and what happens to the air must also be taken into consideration. You are correct that in level, non-accelerated flight the energy of the airplane doesn't change. As you wrote the work done is the product of force times displacement. And in level, non-accelerated flight both the vertical forces (gravity and lift) as well as the horizontal forces (drag and thrust) cancel exactly (otherwise the airplane would accelerate, i.e. it would start to rise or fall or get slower or faster). And since there's no net force no work is done and the energy of the airplane remains unchanged. This doesn't depend on the reference system, in each reference frame the net force on the airplane is zero.

Things are different for the air you're flying through. The air you hit must be shoved out of the way (and to get it to start moving it must be accelerated). The air going through the engines gets accelerated, thereby generating the thrust that balances the drag. And the air flowing around the wings gets pushed downwards, producing the lift that keeps you up against gravity. Whenever something is made to move energy is needed (if you accelerate something from zero speed to a speed of v the energy required to do so is half its mass times v squared, i.e. the kinetic energy it obtains). And this is where all the chemical energy from the fuel goes to (disregarding losses) - it ends up in accelerating the air you're flying through.

The drag the airplane experiences has two components. First there's the "normal" drag (also sometimes called "profile drag") that also would be there without wings, it's due to the air getting pushed out of the way and friction with the body of the airplane. And then there's the so called "induced drag" which results from the wings pushing air downwards. Normal drag roughly increases with the square of the speed of the airplane while the induced drag drops with speed (always assuming level flight at all speeds and adjusting the AoA to maintain level flight). At low speeds the induced drag is the major contribution to the total drag, at higher speeds it's the "normal" drag. And at one speed there's a "sweet spot" where the total drag has a minimum, and that's the speed for the best L/D ratio.

[HazelNuts39]

And the air flowing around the wings gets pushed downwards, producing the lift that keeps you up against gravity. Whenever something is made to move energy is needed (if you accelerate something from zero speed to a speed of v the energy required to do so is half its mass times v squared, i.e. the kinetic energy it obtains). (...) And then there's the so called "induced drag" which results from the wings pushing air downwards.
Hi jtt, excellent explanation!

It just reminds me of an analogy that I found enlightening at the time. A parcel of air with mass m that is pushed downwards at a speed v represents a momentum of m * v, which equals its contribution to the lift force and, as you say, that requires an energy of 0,5 * m * v-squared. The power ( = energy per unit time) required to produce the lift force is "induced drag" force times the airplane velocity. That energy-per-unit-time should equal the energy required for "pushing air downwards". An equation that is not easy to solve, because the air mass involved is infinitely large, the 'induced' velocity asymptotically approaching zero with increasing distance from the airplane. However, if the induced drag corresponds to an elliptical lift distribution over the wing span, and if one crudely assumes that the "air mass" (per unit time) is that passing through a circular area normal to the free stream with a diameter equal to the wing span, and that it is pushed downwards at a uniform 'induced' velocity 'v', then it can be shown that the two energies (per unit time) are equal.

regards,
HN39

[Checkboard]

Surely the energy for lift has to come from somewhere!
QJB - if you climb onto a table - where does the energy to hold you above the floor (the "lift") come from?

[bearfoil]

A chair, or static system, has a focal density that stabilizes the mass.
"Air" or in the case of flight, a dynamic system, does the same thing, exactly. Newton applies.

The fact that the "Air has no shape" is irrelevant. Besides, prove that the fluid is shapeless!! There is no mystery. The aircraft creates its own support (by the combustion of fuel) in an infinitely sequential manner.

To go to the extreme, one can say without fear that the chair is not fixed, it is moving!

airbear

[bookworm]

I know that work = force x distance

The key that you may be missing is that it's force x distance in the direction of the force.

In the case of lift, there is no movement in the direction of the lift, so no work is done.

In the case of drag, the aircraft moves in direction of the thrust, and therefore work is done by the thrust (against the drag). It's immaterial that the forces balance, you still do work when you move in the direction of the force.

[bearfoil]

Hmmm.

Resisting gravity with an equal force up is not work? None of the four arrows in the classic "image" is actually straight. They are all curved, this is true regardless the genesis of the diagram. In any dynamic system, there is no "fixed" image such that the forces are all "equal". Each vector is a curve, though the motive is to portray a "snapshot". I'm disappointed the "theoretical is eliminated" in the classic portrayal of the four horsemen.

That there is no "movement" means there is no "work"? Hold your fist at arm's length for twenty minutes and say there is no effort (work) involved.

If it helps, consider walking, which is technically simply falling and catching one's fall.
A hover is only slightly challenging, but is perhaps easier to envision.

just a little devil's advocacy, as it were.

[QJB]

Now I'm even more confused haha. I think I understand JTT's explanation though. So although no work is being done on the aircraft (ie. its kinetic and potential energy remains unchanged) work is being done on the air and therefore energy from the fuel is being consumed in proportion with the rate of change of momentum.

[bookworm]

So although no work is being done on the aircraft (ie. its kinetic and potential energy remains unchanged) work is being done on the air

It's not strictly true to say that no work is being done on the aircraft. The engine does work on the aircraft, and if thrust equals drag, the aircraft does work on the air at the same rate. Thus the energy of the aircraft doesn't change. If you were to consider the aircraft climbing instead, thrust exceeds drag and the engine does more work on the aircraft than the aircraft does on the air. Thus the aircraft gains potential energy.

and therefore energy from the fuel is being consumed in proportion with the rate of change of momentum.

Fuel is consumed in (approximate) proportion to the work the engine has to do on the aircraft in either of the above cases.

[jtt]

It's not strictly true to say that no work is being done on the aircraft. The engine does work on the aircraft, and if thrust equals drag, the aircraft does work on the air at the same rate. Thus the energy of the aircraft doesn't change.When the energy of the aircraft doesn't change then no work has been done on it per definition (at least in the language of physics). And there is neither change of kinetic nor potential energy of the aircraft in level, non-accelerated flight...

I think we don't have to discuss the "normal drag", i.e. the drag that also woul be there if the aircraft wouldn't have wings. A certain amount of the thrust is required to balance this drag.

The tricky part seems to be the relation between lift and "induced drag". To get a clearer picture let's forget for a moment about moving forward and consider a (hovering) helicopter.

Or better let's start with a helicopter that sits on a supporting platform. I guess you'll agree that the platform doesn't do work on the helicopter, despite keeping it a bit above the floor. The platform exerts an upward force on the helicopter (otherwise the helicopter would succumb to gravity), but the helicopter doesn't move up or down and there is no motor or anything similar in the platform that would use energy to do any work. The legs of the platform just rest on the ground, exerting a certain force on the ground that is balanced by a contrary force from the ground. The molecules in the legs of the platform experience a force from above that is balanced by an equal force by the molecules below them. Nothing moves, so no work is done anywhere and no energy is needed.

Now remove the platform. Unless you switch the helicopters motor on it will fall down - the molecules of the air aren't tightly bound to each other, so there are no strong forces to keep them in place - they easily can get out of the way to make room for the now dropping helicopter.

What does the rotor do? It accelerates air downwards. When running at the right speed the helicopter won't climb or fall, so its potential energy doesn't change and thus no work is done on it (same as with the aircraft in level, non-accelerated flight). All the energy the motor gets from the fuel (if we don't consider thermal losses, friction in the motor etc.) goes into pushing air downwards. Due to Newton's principle that actio equals reactio the force the air experiences from the rotor (which is the mass of the air times its acceleration) is offset by an equal but opposite force on the rotor (and thereby the helicopter). If this upward force on the rotor is equal to the gravitational pull the helicopter simply hovers.

Now let's get back to an aircraft in flight. The main mental stumbling block seems to be how vertical lift translates into horizontal induced drag and vice versa. It can be bit mysterious since one hardly ever sees the air moving over the wing getting pushed down. But, on the other hand, it's an everyday experience that applying a force in horizontal direction moves something vertically - just think of a simple pulley.

With a pulley you can pull down e.g. a helium balloon by exerting a horizontal force. Of course, then there's also a force on the pulley - the vertical force that pulls down the balloon results in an equal upward force on the pulley. That force is counteracted by the mounting of the pulley.

Now mentally replace the balloon by an air mass and the pulley by the wing of an aircraft. The situation is the same - pulling down some air mass results in an upward force on the wing. Now the wing isn't fixed to some mounting that keeps it from moving up but for that we've got gravity. And, of course, we still need a horizontal force to induce the pulling, and that comes from the thrust produced by the engines. So the "induced drag" is actually nothing else than the the force needed to "pull down" the air "around a corner", redirected by the wing like a pulley does, which in turn keeps the aircraft away from terra firma.

Admittedly, pulleys don't resemble wings much;-) And, of course, the way a pulley works is a bit different from what a wing does and there's no rope to be seen or pulled on easily. I just did bring in the pulley to illustrate that it's not unheard of that a horizontal force is "turned" into an downward force, which then has some repercussions on what (pulley/wing) does the "redirection" of the force.

Now, how a wing does the pulley's job on air is a different story. And it's immaterial to the discussion about where the energy of the fuel goes to, so I won't try to go into that here. But if you want to get a feeling for what a wing actually does with the air take a spoon, go to the kitchen or bathroom, open a tab and move the backside of the spoon slowly sideways into the water stream and see what happens;-)

Regards, Jens

PS: Concerning the argument that one has to use some effort to hold up something heavy with ones arms and thus work is being done: this doesn't coincide with the definition of "work" used in physics - but then concepts in physics are somethimes a bit counterintuitive. The main problem here is that human muscles aren't meant to exert a constant force for long times. A muscle isn't a steel scaffolding, muscles are dynamical systems that can keep up a contraction only for short times. Then they automatically relax. And when one notices that they relax one has to apply a conscious effort to make them contract again (and muscles don't like that for extended times, so it starts to hurt). So in the case of holding something up one actually may do some work, but just because when the muscles relax the load drops a bit and then gets lifted up again a bit...

[bearfoil]

Hello Jens

Are we introducing the "Coanda" effect on the spoon? The wing behaves more like an inclined plane (alpha) than a NOTAR. Redirecting airflow is not much more complex than this, although it is possible to introduce more complex considerations. I appreciate your introduction of the hovering helicopter; moving air around keeps us all in business. Likewise, a simplistic view of muscle kinesiology is perhaps helpful. The far end of the discussion involves the tax on all movements, entropy. A stable or "unaccelerated" airframe in flight resists the acceleration of gravity in its simple expression. Can we call the resistance against 16ft/sec/sec work? I would. Similarly, forward motion is put in the bank. This is the interesting part of the hover. Work is being "done" on the air. The third law doesn't need name tags, it all adds up in the end.

The "four vector" diagram is wanting a fifth, where is the "lift" vector pushing down on the horizontal Stabilizer? There are others. Thanks for your comments.

rgds, bill

[HazelNuts39]

The far end of the discussion involves the tax on all movements, entropy. Right, the thermal energy in the fuel ultimately ends up heating the atmosphere.

where is the "lift" vector pushing down on the horizontal Stabilizer?It is part of the resultant lift force generated by wing+fuselage+tail that balances the weight (and moments, but those include thrust and drag as well).

regards,
HN39

[DERG]

Second part: you mean no net accn due to gravity, stable level flight. So you have the mass of the aircraft in equilibrium with the lift of the wings. So LIFT L is balanced by mass M and gravity g.

so L=Mg

Where does the energy to enable the above to work. For most powered flight from chemical release of combustion. For a jet engine this goes SUCK SQUEEZE BURN BLOW
Thrust.

Other methods: jump of a cliff in a hang glider: gravitation potential..you worked to get up on the cliff LOL. Your legs. A tow rope. A pull from another aircraft. Rocket propulsion as in Fat Albert.

Main thing about energy is that it transfers. Never lost only traded from one sort to another.

Need to consider stress and strain on the airframe too, a bit like the fulcrum in your question.

[barit1]

Why not use a sailplane/hang glider etc. in a rising air mass as an example?

If you launch into the wind from a ridge, you can gain potential energy (i.e. climb) - just as if you were a powered aircraft. Ditto in a thermal, or wave soaring.

I even read an account of an old KC-97 with a pilot who was glider rated - he found a nice mountain wave, pulled power back to idle, and climbed! :ok:

Does the rising air mass affect the L/D? The answer is NO - even though your apparent L/D approaches infinity or beyond.

[glum]

As I picture it, an aircraft that weighs (for example) 200 tons, has to be creating 200 tons of lift every single instant to stay aloft. To climb it needs to create more than 200 tons, to descend less than 200.

It does this due to the effect of the wing angle of attack coupled with the shape*, which when combined force 200 tons of air downward every single instant, thus winning the battle against gravity, which is trying to pull the aircraft down at 9.81m/s at all times.

That's why work is being done even in straight and level flight.

Plus of course barging a plane shape's worth of air out the way, 'dragging' along a heap of air due to the boundary layer, creating all those wing vortices etc. Whole bunch of stuff going on when you get up close and personal with air at speed!

Same reason your arm aches - holding it out is actually resisting gravity which is trying to accelerate your arm downwards. Don't think there's a boundary layer or drag to consider here though.;)

*Ignoring body lift etc as it's easier to simply think of the wing as creating the lift.

[barit1]

Not quite right, glum. Anytime the ship's in steady-state flight, be it cruise, climb, or descent, net lift must be 200 tons. (Note: net lift is the summation of wing and horizontal tail lift, plus any vertical component of engine thrust).

It is during the transition to climb or descent that lift changes. I.E. in cruise, ROC = 0. To transition to climb of 1000 ft/min, we must accelerate the aircraft to this vertical velocity. To do that takes a few additional tons of net lift. Once stabilized in climb, net lift is again 200 tons. Opposite is true in descent.

This is much more obvious in a highly maneuverable airplane; the change in net lift is read out on a G-meter. :)

[DERG]

Barit have you actually ever flown an aircraft?

[barit1]

Check my profile. My father instructed in many types during WWII; I've instructed on SE light types.

And Newton's laws apply to all of them. :)

[glum]

I think that's what I meant! And from recent lectures (MSc Aircraft design), the tail isn't really creating lift, it's there to shift the centre of lift to a better place for stability taking the C of G into account, and generally has a negative angle of attack since the C of G is aft of the centre of pressure if you only had a wing - hence why flying wings are very very hard to make work...

[barit1]

It's very easy to make a flying wing stable, IF it's a swept wing. A bit of washout in the wing makes a stabilizing couple - the outboard wing is aft, and behaves as a horizontal stabilizer.

Relatively efficient too, because the "wingtip" vortex is now inboard, and countered by the negative lift at the real tip.

I learned this lesson building model gliders as a pre-teen.

[HazelNuts39]

And Newton's laws apply to all of them.Newton was a pilot?

[barit1]

You mean he wasn't? :cool:

[DERG]

Barit...God help the poor buggers you instructed.

The only machine they are going to motivate is the Space Shuttle or a friendly alien space craft.

Are you serious?

[bfisk]

No energy, as such, is required for stable, level flight. Since the aircraft is not changing it's height or speed, no work (in the physical sense) is being performed. So that kind of makes the whole discussion nonsensical, doesn't it?

The wings produce lift, used to counter gravity. This is no different from you sitting on your chair right now - the chair is acting on you with a force equally big but in the opposite direction of gravity. That force prevents you from falling. However, if your chair is not lifting you up or letting you down, there is no work being performed.

If you now also decide to scoot your (rolling-)chair along in the corridor, according to Newton, it would want to continue rolling along the corridor with unaltered direction and speed. However, friction works to slow it down, so you use force (your feet) to maintain said speed. Again, no net work is being performed as long as you maintain speed. You do work, and so does friction, but the physical resultant work is zero. There is no net change in energy.

If you could create a chair with frictionless, or low-friction wheels, that would make YOU perform less work to maintain speed, because the chair is doing less work the other way. So it would be beneficial to you that the chair drag was lower for the same chair-up-force (the one that keeps you sitting), and you could calculate a ratio for that, and call it.... :)

Edit: and Barit1: The net lift on a 200-ton airplane is not 200 tons in climb or descent. Lift is defined as the components of forces acting normally to the flight path, and so if the flight path is not normal to the force of gravity, lift and gravity is not 180 degrees apart, and for them to equal out, they must be different, and have a lift/drag component as well. This is easily provable by asking how much lift is required for a 90 degree (vertical) climb? None! But a whole lot of thrust. Again, in a vertical descent, all the lift in the world would not make you decrease your rate of descent, only drag would.

Edit: perhaps I didn't make myself clear (perhaps I don't have it figured all out, that's probably it, actually). I was starting the post out as a discussion on work being performed along the normal axis, ie vertically, and the thrust/drag component aside. And then I obviously went away from my original thoughts. I haven't edited the contents above thogh, because of the posts below.

[Golf-Sierra]

The wing as it moves through the air has two components of drag:

induced drag - is the drag which results from the fact that the wing produces lift. There will always be a component of induced drag. It is impossible (even theoretically) to design a wing which would have 0 induced drag.

parasitic drag - is the drag resulting from everything else - designers strive to reduce this.

Now as the plane moves through the air the power is simply P=f*v (force to overcome induced drag * velocity). The fact that neither the kinetic nor potential energy of the aircraft are changing makes no difference. (In reality - as the plane 'ploughs' through the air the air molecules will be accelerated - heated - hence their kinetic energy is increasing).

So what is the magnitude of this induced drag? I think, assuming some kind of 'ideal' wing, that the power expended to overcome induced drag will be be equal to the power required to overcome the constant accelerating force of gravity on the plane. I guess we could try to imagine this as the wing 'deflecting' a certain mass of air downwards. The mass of this air * the acceleration (i.e. force) will need to be equal to the force exerted on the aircraft by gravity - 9.8 N/kg.


Regards,

Gregory

[DERG]

Why the F... did I answer this????

"no work (in the physical sense) is being performed. So that kind of makes the whole discussion nonsensical, doesn't it?"

Nonsense? That means we need no fuel right? Free energy?
Another poster who drives a UFO!!

FGS sake get yourselves educated..PLEASE
You are embrrassing yourselves!:bored:

[barit1]

DERG seems to me to be a devoted student of Saul Alinsky. We haven't yet seen his theory of aviation, but rather plenty of his expert invective.

I'm willing to listen, however. Go ahead, sir.

[DERG]

The only reason you folks hang around here is to suck the blood outta people like me. If you think I am going to give you a 101 lecture on Newtonian mechanics you are wrong. As for the other WANNABE no comment.

[barit1]

That's what I thought; so be it. I defer to your obviously superior knowledge and tactful discourse.

[DERG]

No need to grovel Barit.:bored:

[Microburst2002]

When we talk about energy, we talk about energy transfers.

Steady level flight, constant speed:

The state of energy of the airplane remains constant (if we neglect the mass reducion due to fuel burn).

However, to maintain that state of energy you need to burn fuel, right? Otherwise you would slowdown or descend. This is because drag removes energy from the airplane, transfering it to the air in several forms (moving it and heating it, mainly). You replace that energy with the energy coming from the engines (much of which is wasted in the air as well, by the way).


There are many energy transfers, between the airplane and the air, and in both senses. The fuel burnt allows you to counter the drag caused by your airspeed. Your airspeed makes the lift that keeps your altitude constant.
The net result, in this case, I think it would be the fuel burnt. That is what keeps you up there for as long as you want.

In the case of a glider maintaining constant altitude and speed, the energy lost by drag has to come from the rising air, entirely.

nice thread

[FullWings]

I think the original question contains a better explanation than anything posted afterwards, if I could be so bold. ;)
1/ I know that an efficient aircraft has a high L/D ratio. Thus thrust provided by the engine is significantly less than lift that acts upon the aircraft. However from an overall energy perspective where does the energy for lift come from. Is it that burning fuel provides a certain amount of energy, this energy is used to create a total aerodynamic reaction from the aircraft (the result of pressure and shear distribution), the total aerodynamic reaction is then resolved into two forces lift and drag. However we consider lift as not taxing the aircraft because it does not impede horizontal motion? I hope I am being clear. Surely the energy for lift has to come from somewhere!

Except maybe for the last sentence. Lift is defined as being normal to the vector of flight, so cannot do "work" in this frame of reference. Energy is required to move the drag force along the flight path, so that's where the engines come in (and/or a change in mgh or 1/2mv^2).

2/ I know that work = force x distance. Therefore if I push on one side of an object with force A and move is distance B the work done will be AB. However! If someone else pushes with an equal force on the opposing side there will be no movement and thus no work done. Therefore my question is, in level flight all the forces on an aircraft are balanced, there is no net force and thus no acceleration. How is it then that the aircraft can be said to be doing work and using energy? Is this something to do with frame of reference?

See above. :ok:

[Microburst2002]

Hi full wings

As I mentioned, the state of energy of the airplane remains constant. This does not mean that there are no energy transfers ocurring between the airplane and the air.

They occur. The airplane gives mechanical energy to the air. The airplane moves the air about it, creating a circulation around the wings, the reaction of which is Lift, Drag and aerodyinamic moment. Lift helps to maintain the state of energy. Drag does not. It would reduce it, but the Thrust from the engines counteracting Drag helps to maintain airspeed and thus the energy state constant.

The air is not giving energy to the airplane in the form of Lift. Lift is just a force (a much more misterious magnitude than energy, in my opinion). It gives energy to the airplane in the form of heat, vibration and others, but not in the form of potential or kinetic energy. These remain constant.

In the case of the glider in an updraft flying at constant altitude and speed, the air isn't giving potential or kinetic energy either. But it supplies the energy wasted by drag.

My conclusion is that an external supply of energy is needed for sustained flight. It can be energy from fuel or energy from rising air, or from a towing airplane, or any other. The amount of such energy is proportional to Drag.

[Gillegan]

I think that some of the confusion here is a good illustration of why thinking of problems in terms of energy is not always the best way to solve a problem. Other times, it is much simpler than other "models". Energy models work best when you just think in terms of energy...PE=mgh and KE=1/2 m(v)squared.

If you want to think of the problem in terms of work, then work = force x distance. In level, unaccelerated flight, the "work" done by the "thrust" is exactly equal to the work done by the drag, thus no overall work to the system. There is still "energy" being put into the system by the engines but stepping back and looking at the whole thing, no excess work is being done anywhere.

The lift can be thought of the same way with the induced drag being what keeps the whole lift/drag thing related. Thinking in terms of energy for level, unaccelerated flight doesn't really help you with too much. Now, when you get into climbing flight is where energy models become really useful. In climbing, unaccelerated flight, you are changing the potential energy of the aircraft. As your lift still equals weight and thrust in the direction of flight still equals drag, the thrust over and above drag becomes the excess "energy" in the system that results in changing your energy...either potential (height) or kinetic (you speed up). If you draw all kinds of vectors (curved?) with angles and stuff and are able to keep track of it, you will come out with the same answers that you come up with by using the much simpler energy equations.

[DERG]

I like that..it's neat. Not in my mindset but I can see how that works. The other way to look at it: any vector change in any three dimensional flight path demands an energy change. The tougher the airframe the more work it can do transferring the changes and moreover the rate (accn/dccn) at which the transfer occurs.:)

[FullWings]

I think a major source of confusion may be the fact that lift and drag are mathematical constructs: there to make the concepts/sums easier. The "reaction force" vector equals the vector combination of gravity and thrust in steady (unaccelerated) flight; any imbalance between these forces will result in a velocity change.

In the lift/drag model, energy is expended by moving the "drag force" along the flightpath; this is lost to the aircraft as a system and expresses itself in heat and motion of the air. Energy is also traded w.r.t. gravity, but as this is a conservative field, returning to the same point will leave no net expenditure.

[Microburst2002]

There is work being done, but not by lift.

As long as there is thrust and the airplane moves, there is work being done.

An airplane with engines delivering 5,000 Kg of thrust that flies 100 km from A to B has made a work of 500,000,000 jules between those points.

In a steady climb, we have first to establish a few assumptions. In a constant EAS climb, for instance, TAS increases. So there is an acceleration, as well. The energy to increase both potential energy and kinetic energy comes from the fuel, too, along with the energy used to compensate drag and other energy losses.

If we switch to a constant mach climb, below tropopause (OAT decreasing with altitude) TAS will decrease, as well as EAS. So some kinetic energy will be traded for potential energy, knot by knot, and is felt inmediately in the VSI. Reducing EAS should help because of the reduced Drag. Above the tropopause, TAS will become constant again. Only EAS decreasing will help in the climb. On top of that, we actually climb at constant IAS, which at high mach numbers over reads with respect to EAS, which complicates calculations...

Depending on the particular problem is easier to use energy or forces. You can look at a steady climb from many points of view. From the point of view of energy, if the state of energy of the airplane increases, energy must come from outside of the system (fuel in the tanks is considered out, for this matter).

Similarly we can explain Lift using energy conservation (Bernoulli equation) or using forces (newton laws). Or such a thing as the koanda effect, which I will never understand.

Energy is like an accountability book. If everything is OK, the number in the left page must be the same as in the right page.

Forces explain in more detail what is going on. To me, Newton's point of view to explain Lift is better than Bernouill's.

[DERG]

"From the point of view of energy, if the state of energy of the airplane increases, energy must come from outside of the system (fuel in the tanks is considered out, for this matter)."

Just don't follow this at all..please explain. Thanks.

[Microburst2002]

Hi, let me try

If we consider the airplane as "the system", the law of energy conservation states that the energy at a given instant is: the initial energy plus any energy that came into the system and minus any energy that went out. Energy cannot disappear or be created. Energy is just transfered from one system to another.

In the case of the airplane flying (or a car running), there is energy going out of the system (to the air) due to Drag, friction, etc... In order to maintain the vehicle energy constant, energy from outside must replace the lost energy. This energy comes from the fuel. In this case, I consider fuel as external to the system. But we could consider fuel a part of the system too.

Fuel as external: Airplanes energy state (simplified): potential + kinetic. Energy remains constant as long as there is fuel.

Fuel as internal: Airplanes energy state (simplified): potential + kinetic + fuel energy. In this case, the energy state will decrease continuously until the fuel is depleted. And we have to take into account the engine efficiency, since a large part of the fuel energy will be wasted into the air, too.

[DERG]

Yes, thank you, thats clear. Of course the new generation engines are more efficient so we get farther for less fuel. I suppose the best illustration of energy exchange was was the "Gimli" incicident.