Propeller torque & engine torque
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An engine in a test bed does NOTHING except churn up air thereby producing heat. If the object an engine is attached to does not increase in the total of GPE +KE as a result of the efforts of the engine then no *useful* work is being done. Likewise a person straining on a wrench is just producing heat. In both cases energy is certainly used, but not usefully.
Yet the dyno showed 327 rear wheel horsepower at 7200 RPM.
Lots of heat and noise, used 27 litres of fuel.
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oggers,
for once we agree!
That does make the point clearly. Power used, heat and noise generated, car remaining completely motionless - hence, no work done on the car. Efficiency zero (0).
Just as the aforementioned engine on a test stand, or non-moving aircraft during an engine runup. Not achieving anything, other than generating noise and heat.
Obviously not the point you've been failing for make for six pages of this thread, but I'm happy that you've seen the light at long last!
for once we agree!
That does make the point clearly. Power used, heat and noise generated, car remaining completely motionless - hence, no work done on the car. Efficiency zero (0).
Just as the aforementioned engine on a test stand, or non-moving aircraft during an engine runup. Not achieving anything, other than generating noise and heat.
Obviously not the point you've been failing for make for six pages of this thread, but I'm happy that you've seen the light at long last!
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That does make the point clearly. Power used, heat and noise generated, car remaining completely motionless - hence, no work done on the car
Work is being done because the rear wheels are turning, as I tried to explain it is "relative".
The same as any stationary engine, test bed or otherwise.
The book "The Jet Engine" by Rolls Royce gives the following version of the equation for propulsive efficiency for a jet engine.
Propulsive Efficiency = (Work done on the aircraft) / (Work done on the aircraft + Work done on the exhaust)
We can modify this to apply for a propeller aircraft as follows.
Propulsive Efficiency = (Work done on the aircraft ) / (Work done on the aircraft + Work done on the propwash)
We can also modify it to apply to your car on the dynamometer as follows:
Propulsive Efficiency = (work done on the car) / (Work done on the car + Work done on the dynamometer)
The work done on the jet engine exhaust, the propwash and the dynomometer are all wasted work.
In each case if the aircraft or the car are not moving then the work done on them is zero and the propulsive efficiency is zero.
If we have zero propulsive efficiency this means that we are wasting all of the power that is being generated by our propulsion system. This means that we have no thrust horsepower.
Propulsive Efficiency = (Work done on the aircraft) / (Work done on the aircraft + Work done on the exhaust)
We can modify this to apply for a propeller aircraft as follows.
Propulsive Efficiency = (Work done on the aircraft ) / (Work done on the aircraft + Work done on the propwash)
We can also modify it to apply to your car on the dynamometer as follows:
Propulsive Efficiency = (work done on the car) / (Work done on the car + Work done on the dynamometer)
The work done on the jet engine exhaust, the propwash and the dynomometer are all wasted work.
In each case if the aircraft or the car are not moving then the work done on them is zero and the propulsive efficiency is zero.
If we have zero propulsive efficiency this means that we are wasting all of the power that is being generated by our propulsion system. This means that we have no thrust horsepower.
Last edited by keith williams; 3rd Apr 2012 at 20:04.
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Keith, all reasonable contributions are welcome.
The assumption is we have thrust:
I disagree with your conclusion. For a helo in the hover propulsive efficiency is zero. But “the power being generated by our propulsion system” is not being wasted.
Also, propulsive efficiency is not the same as propeller efficiency. If there is thrust, propeller efficiency absolutely cannot be zero. Whereas propulsive efficiency must be zero when the aircraft is stationary, regardless of thrust.
For that reason, any attempt to use propulsive efficiency to prove that THP must be zero in the static thrust scenario, is a logical fallacy. All that is proved is that propulsive efficiency is zero.
THP = BHP x Propeller Efficiency.
If there is thrust PE cannot be zero therefore THP cannot be zero.
“THP is the propeller output or power that is converted to useable thrust by the propeller”. That statement and the equation above are both taken from the document cited by italia458 after he floated this canard.
Before anyone jumps on the word “useable” to suggest the thrust is not useable unless it’s moving the aircraft, consider the hover again. The thrust is not only useable, it is being used.
The assumption is we have thrust:
If we have zero propulsive efficiency this means that we are wasting all of the power that is being generated by our propulsion system. This means that we have no thrust horsepower.
Also, propulsive efficiency is not the same as propeller efficiency. If there is thrust, propeller efficiency absolutely cannot be zero. Whereas propulsive efficiency must be zero when the aircraft is stationary, regardless of thrust.
For that reason, any attempt to use propulsive efficiency to prove that THP must be zero in the static thrust scenario, is a logical fallacy. All that is proved is that propulsive efficiency is zero.
THP = BHP x Propeller Efficiency.
If there is thrust PE cannot be zero therefore THP cannot be zero.
“THP is the propeller output or power that is converted to useable thrust by the propeller”. That statement and the equation above are both taken from the document cited by italia458 after he floated this canard.
Before anyone jumps on the word “useable” to suggest the thrust is not useable unless it’s moving the aircraft, consider the hover again. The thrust is not only useable, it is being used.
Oggers,
My only purpose in contributing to this thread was an attempt to assist blackhand to understand why his car was not producing any THP. I can assure you that it was not in any way meant to convince you of your errors.
The past 6 pages of this thread have shown the following two things quite clearly:
1. You have a long held misunderstanding of the subjects being discussed in this thread.
2. No amount of logical argument by others will convince you that you are wrong.
For these reasons I (and clearly many other members) do not intend to engage in further pointless debate with you.
I suspect that you will come to recognise your mistakes only if you manage to discover them for yourself.
My only purpose in contributing to this thread was an attempt to assist blackhand to understand why his car was not producing any THP. I can assure you that it was not in any way meant to convince you of your errors.
The past 6 pages of this thread have shown the following two things quite clearly:
1. You have a long held misunderstanding of the subjects being discussed in this thread.
2. No amount of logical argument by others will convince you that you are wrong.
For these reasons I (and clearly many other members) do not intend to engage in further pointless debate with you.
I suspect that you will come to recognise your mistakes only if you manage to discover them for yourself.
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you have to split two things here : propulsive efficiency/ power / useful work of the pure engine vs the aircraft as a whole package ( or a car or whatever elese) .
since all we know that any useful work needs a motion by physics a tied up plane or a car on a dyno does not any useful work since it does not move- efficency zero.
BUT
the engine of the car or the plane moves ( spins) , produces a given torque at a given rpm, the prop translates this torque to thrust etc. so the engine by itself of course makes useful work , produces power and has a given efficiency.
the useful work of the pure engine is used to run the dyno,to stress the brakes when they hold the aircraft, to keep our helicopter in hover against gravity.
long story short conclusion : the aircraft as the whole package does not provide any useful work, its engines do with the moment as they start to spin.
cheers !
since all we know that any useful work needs a motion by physics a tied up plane or a car on a dyno does not any useful work since it does not move- efficency zero.
BUT
the engine of the car or the plane moves ( spins) , produces a given torque at a given rpm, the prop translates this torque to thrust etc. so the engine by itself of course makes useful work , produces power and has a given efficiency.
the useful work of the pure engine is used to run the dyno,to stress the brakes when they hold the aircraft, to keep our helicopter in hover against gravity.
long story short conclusion : the aircraft as the whole package does not provide any useful work, its engines do with the moment as they start to spin.
cheers !
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Keith, I'm not interested in your purpose for contributing. You did contribute.
If you believe that propeller efficiency cannot be other than zero when the aircraft is stationary just ask yourself next time you start the take off run where the aircraft finds the thrust and power to even begin moving.
If you believe that propeller efficiency cannot be other than zero when the aircraft is stationary just ask yourself next time you start the take off run where the aircraft finds the thrust and power to even begin moving.
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long story short conclusion : the aircraft as the whole package does not provide any useful work, its engines do with the moment as they start to spin.
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Induced Power
Hi oggers,
Is this what you are searching for?
"Induced Power
Induced Power is that portion of the power required to produce lift. It is the power required to overcome the portion of rotor drag which is caused by the induced flow tilting the total reaction rearwards.
Induced power is the force required to move a mass of air through the disk at the induced velocity.
If T is the rotor thrust (in a hover equal to weight (Mass * G)), which is a force, and this force moves the air at a velocity Vi , Pi = TVi .
Helicopter Power Required
Is this what you are searching for?
"Induced Power
Induced Power is that portion of the power required to produce lift. It is the power required to overcome the portion of rotor drag which is caused by the induced flow tilting the total reaction rearwards.
Induced power is the force required to move a mass of air through the disk at the induced velocity.
If T is the rotor thrust (in a hover equal to weight (Mass * G)), which is a force, and this force moves the air at a velocity Vi , Pi = TVi .
Helicopter Power Required
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A few thoughts:
At zero airspeed, how much induced drag does the aeroplane see? How much profile drag?
How much horsepower does it then take to (ahem) "move" it at zero airspeed?
There may be a clue here...
At zero airspeed, how much induced drag does the aeroplane see? How much profile drag?
How much horsepower does it then take to (ahem) "move" it at zero airspeed?
There may be a clue here...
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Perhaps it would help to remember that aircraft propulsion systems are not "lossless". Not all the power goes into moving the aircraft so you can't simply say the power generated by the engine is
Power = force x velocity
You have to write something like...
Power = (force x velocity) + "losses"
You can define:
force = thrust
velocity = aircraft airspeed
but best leave the definition of "losses" vague for the moment as there are too many things to list and they are NOT constant. Some of the losses are also dependant on velocity.
Now it's clear from the modified equation that the power output is not necessarily zero when the aircraft is stationary. In such a case "Losses" includes throwing a lot of air backwards for no purpose.
In the case of the car on the dyno, "losses" include the load the dyno places on the wheels (eg heating up the brake unit).
Interesting to consider what happens at take off. Lets say you have brakes on, run up the engines and then release the brakes... as the aircraft starts moving through the air the sum of all the "losses" must reduce to balance the equation. (I'll ignore the fact that some engines produce more power when moving).
Power = force x velocity
You have to write something like...
Power = (force x velocity) + "losses"
You can define:
force = thrust
velocity = aircraft airspeed
but best leave the definition of "losses" vague for the moment as there are too many things to list and they are NOT constant. Some of the losses are also dependant on velocity.
Now it's clear from the modified equation that the power output is not necessarily zero when the aircraft is stationary. In such a case "Losses" includes throwing a lot of air backwards for no purpose.
In the case of the car on the dyno, "losses" include the load the dyno places on the wheels (eg heating up the brake unit).
Interesting to consider what happens at take off. Lets say you have brakes on, run up the engines and then release the brakes... as the aircraft starts moving through the air the sum of all the "losses" must reduce to balance the equation. (I'll ignore the fact that some engines produce more power when moving).
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Hi barit1,
That depends on the acceleration which the power unit is giving to the aircraft at that very instant.
How much horsepower does it then take to (ahem) "move" it at zero airspeed?
There may be a clue here...
There may be a clue here...
Last edited by rudderrudderrat; 7th Apr 2012 at 13:20. Reason: typo
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The conditions I propose:
o Parked, brakes locked (zero GS)
o Zero wind (thus zero IAS)
The questions: How much induced AND profile drag?
and...
From a THP standpoint, how is this any different from running the engine on a test bench?
Keeping in mind the definition of one horsepower (i.e. 33,000 foot-pounds of work per minute), is there any useful work being done?
o Parked, brakes locked (zero GS)
o Zero wind (thus zero IAS)
The questions: How much induced AND profile drag?
and...
From a THP standpoint, how is this any different from running the engine on a test bench?
Keeping in mind the definition of one horsepower (i.e. 33,000 foot-pounds of work per minute), is there any useful work being done?
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A few thoughts:
At zero airspeed, how much induced drag does the aeroplane see? How much profile drag?
How much horsepower does it then take to (ahem) "move" it at zero airspeed?
There may be a clue here...
At zero airspeed, how much induced drag does the aeroplane see? How much profile drag?
How much horsepower does it then take to (ahem) "move" it at zero airspeed?
There may be a clue here...
The question is: if the engine is producing 200 BHP and the prop is providing thrust can we say that the prop is providing THP, even whilst the aircraft is stationary?
THP = BHP x PE
If there is thrust PE is not zero. If PE is not zero THP cannot be zero. THP is not the same as power required, though it may have the same value.
-------------------
RRT,
As an aside, the power required curve in the link you provided should be very familiar to helicopter pilots: the high power required to hover and the ‘hump’ near zero IAS as one transitions to forward flight are important characteristics for rotorheads to consider.
Last edited by oggers; 11th Apr 2012 at 11:43.
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THP is the power available.
THP is not the same as power required, though it may have the same value.