Originally Posted by ShoeDLG

The canted jets example is just a construction to illustrate that you can have the same rate of kinetic energy addition to the air, and the same power consumption, but different net thrust. I think most people will be able to generalize this example, and recognize that it highlights a serious issue with your definition of efficiency.

Shoe, this is very basic trigonometry. The thrust in the direction of motion will be motor thrust times Cosine 45 degrees.
Two motors, so the net thrust in the direction of motion is 1,414 times the combined output thrust of both units.

That is very inefficient but it has nothing to do with the efficiency of the propulsion unit.

I will get back to your previous post later.

Originally Posted by Montag DP

Mark Drela already showed you the math. It comes from a very basic application of Newton's 1st and 2nd laws to a control volume surrounding the propeller. If you refuse to accept that analysis, why should I waste my time trying to convince you?

Montag, first law, the velocity of a body is constant if there is no net force acting on it.

A body (toy aeroplane) moving through the air has a force acting on it called aerodynamic drag. In order for the net force to be zero there must be a force of the same magnitude acting in opposition.That force is thrust. That is Newton's Law, not mine.

Second law, the acceleration of a body is proportional to the net force acting on it.

If there were no thrust, the only force (therefore the net force) acting on the body would be drag. What keeps the model moving at a constant speed?

Those are the basic applications of Newton's 1st. and 2nd. Laws.

You can not tell anyone who has a basic understanding of physics that a model moving at a constant maximum speed is producing no thrust and if the maths tells you that is so, then the maths are wrong.

Dr Drela gave you a figure for thrust. He never told you the speed the plane is moving at. He told you that if the plane were to move at the same speed as the efflux there would be no thrust from the motor. How the plane got to that speed is irrelevant, we can assume it didn't get there because of the motor's thrust, because the maximum speed the plane can reach is when the thrust matches its drag. And that's well before the plane matches the efflux speed.

Originally Posted by TangoKilo

Montag, first law, the velocity of a body is constant if there is no net force acting on it.

A body (toy aeroplane) moving through the air has a force acting on it called aerodynamic drag. In order for the net force to be zero there must be a force of the same magnitude acting in opposition.That force is thrust. That is Newton's Law, not mine.

Second law, the acceleration of a body is proportional to the net force acting on it.

If there were no thrust, the only force (therefore the net force) acting on the body would be drag. What keeps the model moving at a constant speed?

Those are the basic applications of Newton's 1st. and 2nd. Laws.

You can not tell anyone who has a basic understanding of physics that a model moving at a constant maximum speed is producing no thrust and if the maths tells you that is so, then the maths are wrong.

You're arguing a strawman. I never said the plane was moving at constant speed, only that if the model's speed is equal to the efflux velocity, no net thrust is produced.

The fact that it is not moving at a constant speed in this situation should have been obvious when I said that the plane would slow down until thrust increases enough to counter drag.

[QUOTE=ShoeDLG;23004660]By referring to the airplane’s True Airspeed, the frame of reference remains the same in level flight or a steady hover: that being the frame of reference where the undisturbed air is at rest.

I was thinking in terms of that part of the free body diagram which is considered fixed. To which all motion is relative and on which all forces act.

You’re defining the efficiency of the propulsive system as:

Power supplied to increase the kinetic energy of the air / Power supplied to the propulsion system

That’s a way to define efficiency, but that’s not the standard definition for steady flight which is:

Power supplied to keep the airplane moving (Drag*TAS) / Power supplied to the propulsion system

Yes, but isn't the Power supplied to keep the airplane moving equal to the Power supplied to increase the kinetic energy of the air ?

It is entirely possible to have two such systems that, despite having the same kinetic energy transfer rate and power draw, produce different amounts of thrust.

I will have to think about that.

TangoKilo, why do you delete the [/QUOTE] in all your posts? It makes them very confusing to read.

Originally Posted by TangoKilo

Shoe, this is very basic trigonometry. The thrust in the direction of motion will be motor thrust times Cosine 45 degrees.
Two motors, so the net thrust in the direction of motion is 1,414 times the combined output thrust of both units.

That is very inefficient but it has nothing to do with the efficiency of the propulsion unit.

I will get back to your previous post later.

It has nothing to do with the efficiency of the propulsion system if you define efficiency in terms of the ability to blow air around. If you define efficiency in terms of ability to provide useful power to the vehicle, then it has everything to do with efficiency.

Originally Posted by TangoKilo

Yes, but isn't the Power supplied to keep the airplane moving equal to the Power supplied to increase the kinetic energy of the air ?

Absolutely not.

Originally Posted by ShoeDLG

It has nothing to do with the efficiency of the propulsion system if you define efficiency in terms of the ability to blow air around. If you define efficiency in terms of ability to provide useful power to the vehicle, then it has everything to do with efficiency.

Yes, I do define effiency as the "ability to blow air around", but if you blow it in the wrong direction it is not adding as much as it should to the propulsion of the vehicle.

Originally Posted by Brandano

Dr Drela gave you a figure for thrust. No he didn't. He never told you the speed the plane is moving at. Irrelevant. He told you that if the plane were to move at the same speed as the efflux there would be no thrust from the motor.So what is the motor doing with all those Watts? How the plane got to that speed is irrelevant,What has that got to do with the discussion? we can assume it didn't get there because of the motor's thrust, No we can't, what else got it there? because the maximum speed the plane can reach is when the thrust matches its drag. At last, something we can agree on. And that's well before the plane matches the efflux speed.What is the connection between efflux velocity and thrust?

How is thrust derived? What is your theory on the mechanics?

Originally Posted by Montag DP

You're arguing a strawman. I never said the plane was moving at constant speed, only that if the model's speed is equal to the efflux velocity, no net thrust is produced.

The fact that it is not moving at a constant speed in this situation should have been obvious when I said that the plane would slow down until thrust increases enough to counter drag.

Sorry, I have no idea what a strawman is.

Are we not talking about a model moving at its maximum speed which would be (more or less) constant?

Of course there is no net thrust, the sum of drag and thrust is zero.

Originally Posted by ShoeDLG

Absolutely not.

As you have previously said, there is a certain power required to move a body at a certain speed. Where does that power come from?

The only place it can come from is a fan moving air along a tube. So the power required to move a body through the air must be equal to the power required to move an amount of air through a tube. Where else can power come from?

Originally Posted by TangoKilo

Sorry, I have no idea what a strawman is.

Are we not talking about a model moving at its maximum speed which would be (more or less) constant?

Of course there is no net thrust, the sum of drag and thrust is zero.

Arguing a strawman means you are arguing something you have constructed yourself, instead of the actual point made by the other person in the debate.

We are talking about what happens when flight speed = efflux velocity. The flight speed is not constant in this case, if the ducted fan jet is the only thing producing thrust. My argument is that the maximum speed attainable by the aircraft under its own power must be less than the efflux velocity, because in that condition the propeller is not producing any thrust, which, as you pointed out, would be required to overcome drag. If the aircraft has reached the same speed as the efflux velocity (how that occurred is not important; we could just say it got dropped from a full-scale aircraft at that speed, for example), then it will quickly decelerate until thrust = drag.

The maximum velocity of the aircraft will be when drag equals thrust, as I'm sure you agree with. However, this maximum speed is not necessarily easy to predict because:
1) Thrust produced by the motor depends on forward flight speed
2) Drag depends on forward flight speed

Based on your posts, you seem to agree with #2 but take issue with #1, for some reason that I can't figure out. Is that correct? If so, could you explain why you do not agree with #1?

Originally Posted by TangoKilo

As you have previously said, there is a certain power required to move a body at a certain speed. Where does that power come from?

The only place it can come from is a fan moving air along a tube. So the power required to move a body through the air must be equal to the power required to move an amount of air through a tube. Where else can power come from?

Yes the thrust (and power) must come from moving the air. That does not mean that there must be equality between the rate of kinetic energy addition to the air and the power supplied to move the airplane.

Consider two airplanes moving at the same speed with the same drag. The propulsion systems of both airplanes impart horizontal momentum to the air at the same rate. In any given second, the propulsion system of Airplane A imparts a velocity change of V m/s to a volume of air with mass M kg. In the same second, the propulsion system of Airplane B imparts a velocity change of V/2 m/s to a volume of of air with mass 2M kg. The thrust, drag and momentum transfer rates are the same. Because the airplanes are moving at the same speed, the power supplied to keep them moving is the same (Drag*TAS). I think you will find that the rate at which the propulsion system of Airplane A changes the air's kinetic energy is different from the rate at which the propulsion system of Airplane B change's the air's kinetic energy.

The extra kinetic energy Airplane A "dumps" into the air has to come from somewhere, so the fuel flow (or battery drain rate) of Airplane A must be higher. So both propulsion systems provide the same power to move the airplane through the air, but one of the systems does it with less power consumption. You want to say that the net propulsive efficiencies of these systems are the same. That seems hard to justify.

Originally Posted by TangoKilo

Yes, I do define effiency as the "ability to blow air around".

It takes power to blow air around. In the context of airplane performance, what is the return on investment for blowing air around?

If I can generate the same thrust by blowing less air around (transferring less kinetic energy to the air), then I can generate that thrust with less power consumption. Most people would see this as more efficient. Again, in the context of airplane performance, there is no inherent goodness in dumping kinetic energy into the air. Designers go to great lengths (high aspect ratio wings, winglets, elliptical planforms, high bypass turbofan engines, bent wings to allow bigger propellers) just to reduce kinetic energy transfer to the air.