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Propeller torque & engine torque

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Propeller torque & engine torque

Old 17th Apr 2012, 03:14
  #141 (permalink)  
 
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oggers, please re-read post #144. I made it pretty clear that there are two kinds of efficiency, one for static thrust, and one for propulsive efficiency. Please - when you refer to "efficiency" - make it clear which you are talking about.

According to Hubert C Smith, Associate Professor Emeritus, Penn State, in his “Illustrated Guide to Aerodynamics”:

“actual power available for thrust is obtained by multiplying BHP times propeller efficiency”

“The [amount of power] available is the thp, which is the bhp times propeller efficiency”
Smith here is referring to propulsive efficiency - where the plane is being propelled. Don't use this for the stationary example.
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Old 17th Apr 2012, 10:33
  #142 (permalink)  
 
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You say THP is thrust x velocity. I agree. But I say that only works for a specific set of assumptions. Static thrust is not the only scenario that doesn’t conform to the assumptions. Here is another:

Aircraft flying straight and level at speed Va, pilot sets new power to accelerate to Vb. Assume constant propeller efficiency between Va and Vb due variable prop. Final THP = thrust x Vb, we agree that much. Using the hypothesis you outlined above we can draw some conclusions:

The new power is set, the BHP has gone to the prop but THP doesn’t catch up until the plane reaches Vb?!

Thrust horsepower during the acceleration is less than final THP when you get to Vb?!

Thrust must have increased, but this didn’t reflect an increase in THP, this was because power was wasted in the propwash?!


OK lets’ look at your scenario and see if it really proves your point.

To simplify the situation and to avoid any ambiguities we will assume the following:


1. The aircraft is initially in steady non-accelerated flight. This means that the power required is equal to the THP being generated.


2. The initial speed Va is greater than Vmd so any increase in speed will produce an increase in drag and an increase in power required.


3. There is no reduction gearbox between the engine and the propeller, so the full engine torque is applied to the propeller.


4. The propeller is of the constant speed –variable pitch type.


5. To increase the power, the pilot simply pushes the throttle forward (selecting a higher RPM would just complicate the argument, but not really change the outcome).


6. The pilot has decided to increases airspeed simply by increasing the engine power output (you have already intimated this, but it is best to avoid unnecessary arguments about how best to increase airspeed).

With the aircraft in steady flight at constant RPM the engine torque (causing the propeller to rotate) will be equal and opposite of the propeller torque (the component of propeller total reaction that acts in the plane of rotation to oppose rotation in powered flight). This balance of two opposing torques will maintain a constant propeller RPM.


When the pilot opens the throttle the power output (BHP) of the engine will increase. The increased power will initially take the form of an increase in engine torque. This will cause engine torque to become greater than propeller torque, so the propeller RPM will start to increase.


When the propeller constant speed unit senses this increase in RPM, it will command the pitch change unit to increase the pitch of the propeller blades. This will increase the angle of attack of the blades, thereby increasing propeller total reaction. This in turn will increase both the propeller torque and the thrust. The increased propeller torque will cause the RPM to return to its initial (selected) value. So the overall effects of the pilot opening the throttle will be an increase in thrust, which will cause an immediate increase in THP.


Prior to the pilot opening the throttle, the aircraft was in steady balanced flight, so the thrust was equal to the drag and the power required (drag x TAS) was equal to the THP (thrust x TAS).

The increased thrust caused by the pilot opening the throttle would have the following effects:


1. The thrust would be greater than the drag.


2. The THP being generated would be greater than the power required.


The combination of thrust greater than drag and THP greater than power required would cause the airspeed to increase. As the airspeed increases, it would cause the drag and hence the power required to increase. The acceleration would continue until the increasing drag was once again equal to the thrust and the increasing power required was equal to the THP being generated. The aircraft would then settle at the new into balanced flight at the new higher speed.

Now let’s look again at your comments:


The new power is set, the BHP has gone to the prop but THP doesn’t catch up until the plane reaches Vb?!

Thrust horsepower during the acceleration is less than final THP when you get to Vb?!
The situation is actually one in which the THP increase occurred before the airspeed increase and contributed to it. The only thing that you have proved by proposing this scenario, is that you have given insufficient thought to the subject.


Static thrust is not the only scenario that doesn’t conform to the assumptions. Here is another:
I presume that this is a reference to your earlier suggestion that an aircraft attempting to accelerate from brake release was another “impossible chicken and egg” situation. I am quite happy to explain to you why this is not the impossibility that you suggest. But you will probably gain more by working it out for yourself.


work done on the propwash isn’t wasted if it’s giving thrust.

As an aircraft flies through the air it is continuously losing energy due to the drag force. If the airspeed and altitude are to be maintained, this lost energy must be replenished at a rate that is equal to the loss rate. Looked at in this context the purpose of the propulsion system is to take energy from the fuel, convert it into kinetic energy and transfer it to the aircraft. Any energy that is lost to the propwash (or jet efflux) represents a waste of energy.


The enormous improvements in fuel efficiency that have been achieved in turbojets / turbofans over the past few decades have been brought about by reducing the amount of kinetic energy that is wasted in the jet efflux. The old turbojets were quite able to produce thrust, but their very high exhaust velocities meant that they were throwing away enormous amounts of kinetic energy. The fact that modern engines are able produce the same (or even more) thrust, while throwing away much less energy, shows that the energy loss is wasteful.


I am quite happy to respond to the remainder of your post when you have stated the equation that you consider to be correct for propeller efficiency.
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Old 17th Apr 2012, 15:54
  #143 (permalink)  
 
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Propeller Efficiency

Hi oggers,

Have a look at:Propeller Aircraft Performance and The Bootstrap Approach: Background

"Of the four forces acting on the airplane – thrust, drag, lift, and weight – thrust is the most difficult to measure or predict. That is why most books about aircraft performance simply assume that propeller efficiency h is some constant. Commonly cited values are h = 80% and h = 85%. Then thrust T = h P, where P is the engine power. Unfortunately, propeller efficiency is in fact not constant; it varies with air speed and RPM or, more precisely, with the dimensionless ratio of those two variables:
J=V/nd
where J is the "propeller advance ratio." As the propeller rotates through one circle the airplane advances a distance V/n. J is then the ratio of that advance distance to the propeller’s diameter.
Figure 1 is an example of how propeller efficiency varies with advance ratio."

If there is zero advance, then efficiency = 0
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Old 17th Apr 2012, 16:57
  #144 (permalink)  
 
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Rudderrudderrat,

If Oggers had wanted to accept the fact that propeller efficiency is zero when forward speed is zero, he just needed to read the Essay to which he referred on page 4 of this thread.

It includes a graph that is pretty much like yours, but it shows propeller efficiency against forward speed.

And it starts at zero-zero!
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Old 15th Jun 2012, 16:58
  #145 (permalink)  
 
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keith Williams:

I have invited oggers to go back to basics and explain why he thinks that an aeroplane that is not moving (brakes on or hovering model) is generating THP. I think that this exercise will be more productive than thinking up imaginary scenarious about circling aircraft and rotating wings. You might like to try it
.

<<Thrust Horsepower: The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill Dictionary of Aviation

Zero thrust horsepower ≡ zero thrust. The model requires thrust to hover therefore the THP is not zero.

Power is required for thrust additional to losses. It is the induced power of the prop/rotor. In the case of the model ‘prop hanging’, that induced power is the thrust times the velocity induced at the prop, along the thrust axis. Induced power is therefore equivalent to thrust horsepower.

The position you are sticking to is that all power going into the slipstream is wasted power, which is true in the context of aircraft velocity but not in the context of thrust. Hence, propulsive efficiency for velocity, and FOM for thrust:

“Since it is the induced power which relates to the useful function of the rotor, the ratio of induced power to total power is a measure of rotor efficiency in the hover. This ratio is called the Figure of Merit” Sedddon: Basic Helicopter Aerodynamics

For a given prop the thrust depends on both induced velocity Vi and the mass flow through the prop: T = ρA(Va + Vi) x ΔV. Va may go to zero but if Vi goes to zero you have no thrust. Statically, the “power converted to thrust” is T x Vi. You will find this is in accordance with simple momentum theory for a prop as well as the definition at the top. It is therefore THP.

Non-statically this is T x (Va + Vi) which eventually becomes ≈ T x Va only. You have said as much but the point that T x Va is actually an approximation (though it is definitive for propulsive power) of ‘power converted to thrust’ seems to have escaped you. In the latter case it is fair to say that THP = T Va and swallow the approximation. But this becomes invalid at low speed as I have been stressing all along. This graph shows where the approximation holds good as the curves run along the top near 100% beginning at about 20m/s for lightly loaded props:





The efficiency increases as mass flow through the prop picks up at speed.

If Oggers had wanted to accept the fact that propeller efficiency is zero when forward speed is zero, he just needed to read the Essay to which he referred on page 4 of this thread. It includes a graph that is pretty much like yours, but it shows propeller efficiency against forward speed. And it starts at zero-zero!
The graph in the essay is the same as the one above. These are Froude efficiency curves. It does not show that THP goes to zero with Va for a given BHP which is what you think. The speedier end of the curve shows Froude efficiency ≈ 100% of induced efficiency whereas the slower end of the curve shows the opposite. Ergo TVa is a good proxy for induced power at one end of the curve but completely invalid at the other end.

When you add BET you get a graph like the one RRT posted for overall propulsive efficiency. Personally, when I compare the two graphs I see they are “pretty much alike” in the same way that r is pretty much like n (ie different). The r shaped curve is showing that Froude efficiency increases as a proportion of overall efficiency and stays there even after overall efficiency drops off again so TVa as a proportion of “power converted to thrust” stays high. OTOH, as speed goes down the approximation fails. That is the whole point of the essay, the very reason the guy felt the need:

"you should see a problem in that as your velocity goes to zero, no matter how much thrust you're producing, your efficiency goes to zero. So how do you know how good your prop is doing at low speeds or statically?"

..and he spends most of the essay working around the problem.

Looking at the model again, the position that you and barit1 have adopted is that the model has zero THP whilst prop hanging but some THP if it climbs. Therefore it has negative THP if it descends and the prop will drive the engine. In that case you are in autorotation as soon as you start to descend (or shortly after when -THP is sufficient to offset profile power). That would be nice, but it’s not the case. Autorotation does not begin until after RoD exceeds Vi. Any helicopter pilot will tell you that auto requires a healthy rate of descent, not merely a slight descent.

By contrast, if you take THP = induced power = T (Va + Vi) then you have THP in the hover, more THP in the climb and less THP in the descent until RoD exceeds Vi and autorotation begins. You will find this is in agreement with momentum theory for a helicopter in descent.

“A number of contributions to this thread have included attempts to draw comparisons between rotary wing pof and fixed wing pof. This is unlikely to be very helpful because the two are very different in many ways. The fact is that few of us know enough about rotary wing pof to make the comparisons very illuminating (I certainly don't)”
I find the comparison to be very helpful because as far as momentum theory goes the rotor and the prop are precisely the same. And we don’t need to go beyond that to determine the induced power required to provide a given thrust.

THP and BHP both have very specific definitions, which means that propeller efficiency also has a very specific definition.

THP = Thrust x Aircraft Velocity

Well, here are 3 credible definitions of THP:

<<The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill dictionary of aviation

<<THP: The horsepower equivalent of the thrust produced by a turbojet or turbofan engine>> FAA Handbook of Aeronautical Knowledge.

<<The amount of power that gets converted into thrust is referred to as thrust horsepower or thp>> Hubert C Smith Ph.D. Associate Professor Emeritus, Penn State

There is absolutely no doubt that the model is developing induced power T x (Va + Vi). That is a given. Induced power is the power converted to thrust for the model or lift for the helicopter. It fits all three of those definitions, whereas T x Va is an approximation that only works when the Froude efficiency is high. Ergo it is invalid statically.
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Old 15th Jun 2012, 18:17
  #146 (permalink)  
 
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Well, here are 3 credible definitions of THP:

<<The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill dictionary of aviation

<<THP: The horsepower equivalent of the thrust produced by a turbojet or turbofan engine>> FAA Handbook of Aeronautical Knowledge.

<<The amount of power that gets converted into thrust is referred to as thrust horsepower or thp>> Hubert C Smith Ph.D. Associate Professor Emeritus, Penn State
Well as far as I'm concerned none of them are precisely scientifically credible; words like 'equivalent' can be read as meaning 'exactly equal to' but this is not the case, the dimensions of the units involved are wrong. The discrepancy in all 3 cases, to make the units correct, is a velocity term. i.e. TAS. Which takes us right back to square 1.

really, none of those definitions is any more precise than saying 'well, its kind of like this.....'

I bet if you asked Hubert C Smith to give you a formula relating THP and thrust he would say: THP = Thrust x TAS.

pb
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Old 15th Jun 2012, 20:01
  #147 (permalink)  
 
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Oggers!... You're still arguing the same point I clearly described at the beginning of this thread, eh!
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Old 18th Jun 2012, 13:29
  #148 (permalink)  
 
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The author of the essay to which you have repeatedly referred did not actually state that the equation is invalid as speed approaches zero. What he actually said was:

While working on a project designing a propeller at work, I wanted to know just how good I was doing. Efficiency is one measure of how well a propeller is performing, but it's not necessarily a good indication of how well the design is performing up to its potential. In aviation, propulsive efficiency is defined as:
Efficiency = T x V / Pavail
where η is efficiency, T is Thrust, V is Velocity, and Pavail is Power Available, or power going into the propeller. Basically, power out divided by power in. This equation is very useful for many cases, but you should see a problem in that as your velocity goes to zero, no matter how much thrust you're producing, your efficiency goes to zero.

The author of the essay had a problem because he was attempting to use the propeller efficiency equation to calculate thrust. He was assuming that the purpose of the propeller is to produce thrust, but this is not true.

The purpose of the propeller is to propel the aircraft forward against the drag force, by producing thrust. The rate at which the propeller is achieving this purpose is equal to the thrust multiplied by the aircraft velocity. This is our old friend the THP.

Because the primary purpose of the propeller is to propel the aircraft forward, it is entirely logical that the overall efficiency of the propeller should be measured in terms of its propulsive efficiency.

Propeller efficiency = (Thrust x Aircraft Velocity/ Brake Horsepower

This can also be stated as (Thrust x TAS) / BHP provided we take care to note that the TAS is velocity of the aircraft relative to the air and is not the propwash velocity.

When the aircraft is running at full throttle on the ground, with the wheel brakes preventing it from moving forward, the propeller is absorbing power but is not achieving any success in propelling the aircraft forward. So the propeller efficiency is zero, which is exactly what the equation yields. The equation is in fact producing the correct result at all airspeeds including zero.

To understand why the author of the essay could not use the equation to calculate thrust when airspeed was zero we need to look at what he was attempting to do.

He was attempting to take the product of (Thrust x zero airspeed) and then divide it by zero airspeed to reveal the thrust.

For any number other than zero and infinity this would be a perfectly reasonable thing to do.

But the product of any number multiplied by zero is zero, and dividing this product by zero simply gives zero / zero = zero.

The simple fact is that the processes of multiplication and division are not reversible when one of the numbers is zero. This fact is not a function of any inaccuracies in the propeller efficiency equation. It is simply the result of one of the properties of the number zero. Any equation involving multiplication and division will encounter exactly the same problem when one of the values goes to zero. But this does not mean that all such equations are invalid.

Looking at the model again, the position that you and barit1 have adopted is that the model has zero THP whilst prop hanging but some THP if it climbs. Therefore it has negative THP if it descends and the prop will drive the engine. In that case you are in autorotation as soon as you start to descend (or shortly after when -THP is sufficient to offset profile power). That would be nice, but it’s not the case. Autorotation does not begin until after RoD exceeds Vi. Any helicopter pilot will tell you that auto requires a healthy rate of descent, not merely a slight descent.



Entering autorotation in a helicopter is not simply a matter of throttling back the power until the aircraft starts to descend. If you do this, the blades will stall, the rotor speed will decrease, the blades will fold upwards and the aircraft to crash. To enter autorotation you must lower the collective to get negative pitch (or at least a very low pitch) at which the ROD airflow produces an angle of attack, such that part of the total reaction drives the blades forward. This keeps the rotors turning and keeps generating lift. None of these good things will happen if we simply reduce power in a propeller driven aircraft, because the mechanisms are physically very different. Your use of this scenario simply illustrates your lack of knowledge regarding helicopter POF.


Well, here are 3 credible definitions of THP:

<<The amount of horsepower the engine-propeller combination transforms into thrust.>> McGraw-Hill dictionary of aviation

<<THP: The horsepower equivalent of the thrust produced by a turbojet or turbofan engine>> FAA Handbook of Aeronautical Knowledge.

<<The amount of power that gets converted into thrust is referred to as thrust horsepower or thp>> Hubert C Smith Ph.D. Associate Professor Emeritus, Penn State




In the second of your definitions the term “horsepower equivalent of the thrust” implies that it is the thrust x TAS. So this definition is quite correct.

In dealing with your first and third definitions I could quite easily find numerous sources of the definition THP = Thrust x TAS. I have at least one example on my office bookshelf. But trading references will prove nothing. Instead let’s see what your definitions produce in terms of a propeller efficiency equation.

Efficiency is simply a ratio of output divided by input, and standard efficiency equations are of the form Efficiency = Output / Input

In your first and third definitions the output is thrust and the input is power.

So we have Efficiency = Thrust output / Power input

If we use imperial units of pounds force for thrust and foot pounds force per minute for power we get:

Efficiency = (lbf) / (ft lbf / min)

Cancelling out the lbf on the top and bottom of this equation we get

Efficiency = 1 / ft/min which is Efficiency = min/ft.

The equation has yielded an efficiency figure that is in units of minutes per foot. What exactly does this mean?

Efficiency is just a ratio of output divided by input, so it has no units.

To get rid of the min/ft in our result we need to multiply by ft/min. But ft/mins are the units of velocity, so to get a proper value for efficiency we need to multiply by velocity. This converts our efficiency equation into Efficiency = (Thrust x TAS) / Power input.

This is of course the propulsive efficiency equation.

The above sequence proves something that we should already have known. That is the fact that power, which is the rate of doing work or expending energy, cannot be converted into a force. Power and force are two totally different things and one cannot be converted into another. So whatever the sources of your first and second definitions, they cannot be correct. It is of course possible that the authors did not intend their definitions to be interpreted literally.

Finally I think that you need to look again at your argument that induced power is the same things at THP. I have drawn the following quotes from various internet sources.

Induced power is the power required to maintain enough lift to overcome the force of gravity.
www.math.usu.edu/~powell/ornlab-html/node7.html

Induced Power is that portion of the power required to produce lift.

www.griffin-helicopters.co.uk/note/helicopterpower.htm - 13k


The minimum engine power required to hover is called the "induced power."

scienceworld.wolfram.com/physics/Helicopter.html - 15k


Induced power is what people are referring to when they say helicopters "beat the air into submission." Newton's 2nd law concerning action-reaction applies in this regime where we must force air down to keep the aircraft aloft.

www.navair.navy.mil/safety/documents/Power_Available_vs.doc



In each case they show that induced power is exclusively concerned with the process of generating lift. THP is concerned with propelling aircraft through the air, so they are not the same thing. By all means argue that your model aeroplane is generating induced power. But it is most certainly not generating any THP.
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Old 18th Jun 2012, 19:01
  #149 (permalink)  
 
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keith williams:
...The author of the essay had a problem because he was attempting to use the propeller efficiency equation to calculate thrust. He was assuming that the purpose of the propeller is to produce thrust, but this is not true.

The purpose of the propeller is to propel the aircraft forward against the drag force, by producing thrust. The rate at which the propeller is achieving this purpose is equal to the thrust multiplied by the aircraft velocity. This is our old friend the THP.
Well done!
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Old 21st Jun 2012, 10:49
  #150 (permalink)  
 
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Entering autorotation in a helicopter is not simply a matter of throttling back the power until the aircraft starts to descend. If you do this, the blades will stall, the rotor speed will decrease, the blades will fold upwards and the aircraft to crash. To enter autorotation you must lower the collective to get negative pitch (or at least a very low pitch) at which the ROD airflow produces an angle of attack, such that part of the total reaction drives the blades forward. This keeps the rotors turning and keeps generating lift. None of these good things will happen if we simply reduce power in a propeller driven aircraft, because the mechanisms are physically very different. Your use of this scenario simply illustrates your lack of knowledge regarding helicopter POF.
Keith, that is a big strawman, and not your first. Nobody has said that entering auto is a simple matter of throttling back the throttle What I wrote was that your description of what constitutes THP implies you would be in auto as soon as Thrust x RoD (equivalent to T x TAS) exceeded profile power, when what actually happens is that auto begins when RoD exceeds induced velocity by enough to offset profile power. This is a fact.

the mechanisms are physically very different
The actual mechanical arrangement is superfluous, this is about the over-arching physics. Simple momentum theory considers only that a 'disc' accelerates air. This does not tell you if the model is mechanically capable of going into auto, it tells you the theoretical relationship between induced velocity, RoD and thrust.
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Old 21st Jun 2012, 12:33
  #151 (permalink)  
 
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when what actually happens is that auto begins when RoD exceeds induced velocity by enough to offset profile power.
As the collective is lowered the aircraft starts to descend and the induced velocity and the induced power both fall to zero. As the aircraft accelerates downwards the direction of the total reaction is tilted in the direction of rotation at some points along the blades and is tilted in the opposite direction at other points. It is only when a certain rate of descent has been achieved that the sum total of the different parts of the total reaction is sufficient to keep the blades rotating in the correct direction and to produce sufficient lift to support the weight of the aircraft. When the sum of lift plus vertical component of drag equals the weight of the aircarft the ROD becomes constant. The actual rate of descent depends upon a number of things include blade pitch angle and aircraft weight.

So even with a helicopter, the autorotative force does not become significant until a certain ROD has been achieved. But none of this confirms your argument that induced power is the same thing as THP

The actual mechanical arrangement is superfluous, this is about the over-arching physics. Simple momentum theory considers only that a 'disc' accelerates air. This does not tell you if the model is mechanically capable of going into auto, it tells you the theoretical relationship between induced velocity, RoD and thrust.
And

Any helicopter pilot will tell you that auto requires a healthy rate of descent, not merely a slight descent.
I have known several hundred helicopter pilots over the years, but none of them have flown theoretical discs. They have all flown real helicopters with real rotor mechanisms. You are using the behaviour of real systems to support arguments about theoretical physics, then you are saying that the actual mechanical arrangement of the systems is superfluous. That line of argument is clearly nonsensical because the behaviour of the real systems is determined as much by their physical characteristics as it is by the theoretical physics.


Throughout this thread you have made various assertions, including the following:

1. Propellers continue produce THP when the aircraft velocity is zero.

2. When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.

3. If propellers produced no THP when the aeroplane is standing still on the ground, it will be impossible for the aeroplane to accelerate when the brakes are released.

4. If THP is Thrust x TAS it will be impossible for aircraft to accelerate in flight.

5. THP is the same thing as Induced Power.

6. The propulsive efficiency equation is not the correct equation for propeller efficiency (though interestingly you have never actually stated what the correct equation is).

7. As long as a propeller is producing thrust its efficiency cannot be zero.

In each case your assertions have been incorrect. You have certainly demonstrated tenacity in pursuing these arguments for so long, but you have also demonstrated a very poor understanding of these subjects.

Last edited by keith williams; 21st Jun 2012 at 15:40.
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Old 21st Jun 2012, 18:29
  #152 (permalink)  
 
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OP:

Torque is a measurement of power from the engine, at the gear box of the prop/rotor.

The reason for torque indication is the gear box has limitations on how much power it can handle, as it's being resisted on the other side by the prop/rotor, which is being resisted by the drag from the air.

Certainly you can measure torque of the engine, torque to the prop, but usually the gear box is the weakest link, as it's translating the power from the engine to the airfoil...so it usually gets the indicator, with your job keeping it out of the red.
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Old 22nd Jun 2012, 11:56
  #153 (permalink)  
 
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Keith:

As the collective is lowered the aircraft starts to descend and the induced velocity and the induced power both fall to zero. As the aircraft accelerates downwards the direction of the total reaction is tilted in the direction of rotation at some points along the blades and is tilted in the opposite direction at other points. It is only when a certain rate of descent has been achieved that the sum total of the different parts of the total reaction is sufficient to keep the blades rotating in the correct direction and to produce sufficient lift to support the weight of the aircraft. When the sum of lift plus vertical component of drag equals the weight of the aircarft the ROD becomes constant. The actual rate of descent depends upon a number of things include blade pitch angle and aircraft weight.
Yeah, but irrelevent.

I have known several hundred helicopter pilots over the years, but none of them have flown theoretical discs. They have all flown real helicopters with real rotor mechanisms.
..and they all imagine their thrust in terms of a man rowing a boat down a canal a la your little analogy?

You are using the behaviour of real systems to support arguments about theoretical physics, then you are saying that the actual mechanical arrangement of the systems is superfluous. That line of argument is clearly nonsensical because the behaviour of the real systems is determined as much by their physical characteristics as it is by the theoretical physics
I referred to actuator disc theory because it is the right one to tackle this meme of thrust without THP, and I prefer it to your rowing boat story.

When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.
I don't recall ever saying that. Feel free to paste a quote though, if you can find one.
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Old 22nd Jun 2012, 12:27
  #154 (permalink)  
 
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I thought you might wish to deny one or more of your errors, so I did a search.

1. Propellers continue produce THP when the aircraft velocity is zero.


Page 1 post 8 “Not that myth again. It's disappointing to have to point out to one who calls himself an instructor that in your 'brakes on' scenario the aircraft is still producing 200 THP as well as 200 BHP because it is accelerating a mass of air rearwards in a futile attempt to turn the earth and the atmosphere in opposite directions.”

2. When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.


Page 1 post 14 “But that does not mean a stationary aircraft can do no work and therefore produce no THP!!! The work is done by moving the mass of air one way and the mass of the Earth a tiny imperceptible amount the other way.

3. If propellers produced no THP when the aeroplane is standing still on the ground, it will be impossible for the aeroplane to accelerate when the brakes are released.


Page 6 post 120 “Sure, no power is required if the aircraft is at standstill but that does not mean no power is available. OTOH power is required to accelerate the mass of the vehicle from standstill with or without aerodynamic drag. No power available = no acceleration. THP is the power available. If the aircraft has to move before you get THP then you are in a chicken and egg situation - you can't accelerate the aircraft without power but you can't have power until you move the aircraft.”



4. If THP is Thrust x TAS it will be impossible for aircraft to accelerate in flight.


Page 7 post 148 “Aircraft flying straight and level at speed Va, pilot sets new power to accelerate to Vb. Assume constant propeller efficiency between Va and Vb due variable prop. Final THP = thrust x Vb, we agree that much. Using the hypothesis you outlined above we can draw some conclusions:

The new power is set, the BHP has gone to the prop but THP doesn’t catch up until the plane reaches Vb?!

Thrust horsepower during the acceleration is less than final THP when you get to Vb?!

Thrust must have increased, but this didn’t reflect an increase in THP, this was because power was wasted in the propwash?!”



If you look at item number 2 above you will see that

Page 1 post 14 “But that does not mean a stationary aircraft can do no work and therefore produce no THP!!! The work is done by moving the mass of air one way and the mass of the Earth a tiny imperceptible amount the other way.


implies that

When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.



It really is time for you to face up to the fact that virtually everything that you have written in this thread has been nonsense.

keith williams is offline  
Old 24th Jun 2012, 11:54
  #155 (permalink)  
 
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Keith Williams:

If you look at item number 2 above you will see that:

Page 1 post 14 “But that does not mean a stationary aircraft can do no work and therefore produce no THP!!! The work is done by moving the mass of air one way and the mass of the Earth a tiny imperceptible amount the other way. “

implies that

When an aeroplane is standing still on the ground with wheel brakes applied the THP is equal to Thrust x propwash velocity.
Actually it doesn’t imply that. From the essay again:

It's pretty well known that at the propeller, the air has accelerated one half of what it will do downstream (VI = 1/2 ΔV). So, at the propeller, the velocity is VA + 1/2 ΔV.
…which is a basic tenet of the fluid dynamics here. Vi is half ΔV. Those for whom this is not ‘pretty well known’ are liable to draw the wrong conclusion – as you have above. Yes, I say induced power is equivalent to THP but that wouldn’t make it “equal to Thrust x propwash velocity”. Statically it is T x Vi as I stated in a previous post.
oggers is offline  
Old 24th Jun 2012, 12:40
  #156 (permalink)  
 
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If you feel the need to hang on to that figleaf ogger then do so.

Life's too short for me to waste any more time conversing with you oggers.

I'm sure most of the people who have followed this thread have come to the same conclusion.

Last edited by keith williams; 24th Jun 2012 at 12:41.
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Old 24th Jun 2012, 13:59
  #157 (permalink)  
 
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Keith that's the second time you've scurried off. However, since your last reappearance you have spouted plenty of waffle in my direction [including explicitly "inviting" me to comment]. So whether you care to read it or not I will be commenting further on the many spurious points you addressed to me

Last edited by oggers; 24th Jun 2012 at 14:20.
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