Aircraft Energy Question
Join Date: Jun 2001
Location: In the State of Perpetual Confusion
Posts: 185
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
Last edited by Gillegan; 16th Nov 2010 at 10:07.
Join Date: Oct 2009
Location: Durham
Posts: 483
Likes: 0
Received 0 Likes
on
0 Posts
Yeah..
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.
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.
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.
Join Date: Mar 2005
Location: Uh... Where was I?
Posts: 1,338
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
Join Date: Oct 2009
Location: Durham
Posts: 483
Likes: 0
Received 0 Likes
on
0 Posts
Request clarification please
"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.
Just don't follow this at all..please explain. Thanks.
Join Date: Mar 2005
Location: Uh... Where was I?
Posts: 1,338
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
Join Date: Oct 2009
Location: Durham
Posts: 483
Likes: 0
Received 0 Likes
on
0 Posts
Hiya
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.
