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Propeller torque & engine torque

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Propeller torque & engine torque

Old 31st Mar 2012, 14:32
  #81 (permalink)  
 
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And so when a PW4000 runs up to max on a terrestrial test bed, it is actually creating ~70,000 thrust HP? Is that what we surmise?
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Old 31st Mar 2012, 14:37
  #82 (permalink)  
 
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Old Fella:
...on the T56A-11 it was common to reach max permissable T.I.T before achieving limiting torque.
Hmmm. Since the aircraft performance is dependent on the props being driven at some torque rating, I must ask how you determine your performance-limited TOGW??
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Old 31st Mar 2012, 14:42
  #83 (permalink)  
 
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barit1:

And so when a PW4000 runs up to max on a terrestrial test bed, it is actually creating ~70,000 thrust HP? Is that what we surmise?
No. THP has to do with aircraft performance. When it's on a testbed, you wouldn't be analyzing THP. The engine would be producing thrust and you could compute how efficient the engine is in making thrust (ie: accelerating air).

EDIT: However, if you were to apply the concept of THP to the testbed scenario, the THP would be zero because the testbed/engine wouldn't be moving.
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Old 31st Mar 2012, 16:33
  #84 (permalink)  
 
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italia:

So, has this discussion changed your answer now, or do you still believe that the THP will equal the BHP because it's 'accelerating a mass of air rearwards'?
Not changed my pov. BHP x PE = THP. [notwithstanding Tx losses obviously]

After 4 pages you have now done a ma-housiv u-turn and decided that the aircraft can stay still after all and still output power from the prop.
Where did I say that the prop does not do work on the air when the aircraft is not moving? I never said that! So, to put it bluntly, you are lying.
Well, what I wrote and what you wrote underneath it aren't the same. But taking your little statement at face value, you are clearly accepting there that there is a power output from the prop even if the plane is stationary.

An equation for PE (that you have used yourself) is power output divided by power input. An equation for THP is BHP x PE. If you acknowledge there is a power output from the prop you must acknowledge there is THP.

I look forward to reading the wriggling you will do to mitigate this conclusion. Not that I've entirely finished myself.

No. THP has to do with aircraft performance. When it's on a testbed, you wouldn't be analyzing THP
I think most people here, being aviation professionals, would feel the putting of an engine on a tested had quite a lot to do with aircraft performance. Unless you are a glider pilot.
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Old 31st Mar 2012, 16:45
  #85 (permalink)  
 
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oggers:

You're wrong. End of story.

There is no point explaining it anymore.

Good day!
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Old 1st Apr 2012, 02:08
  #86 (permalink)  
 
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Limiting MTOW

Barit1, I probably did not articulate my point as well as I might. All I was trying to convey was that the 19600"/lbs limiting torque was often not achievable before reaching the Max permissable Turbine Inlet Temperature, especially on the T56A-11 which from memory was limited to 971 degrees C at Take-off power and 927 degrees C Max Continuous. The T56A-15 however with a higher T.I.T limit could readily reach Max Torque below the limiting T.I.T of 1083 degrees C for Take-off. Of course, achievable Torque for the given ambient conditions and Runway length available, surface type etc all are considered to determine Max Take-off weight allowed. Hope this is clearer.
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Old 1st Apr 2012, 08:24
  #87 (permalink)  
 
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Prop Torque

Hi,

In my reply (post #24), On RR Tyne with 4 bladed variable pitch props, we had to have both Prop RPM and torque within limits before we called power was set. The engine and gearbox were bolted together so it was not possible to measure their individual torques separately.
Hence the torque sensor (which was bolted between an engine mount and the airframe) simply measured the reaction to the propeller torque - so it indicated actual Prop Torque.

In order to keep passenger comfort during stepped climbs, we kept the prop RPM constant, but varied the fuel flow by "trimming" the (guarded) HP fuel levers. Power changes were called for ("50% trim" etc) and measured by observing the torque.

Since we measured both torque (force) and prop RPM (velocity) one could calculate the propeller horse power being generated (with a known p.e.) - even when stationary on the runway

Since there is no torque meter fitted on a bypass jet engine, the most useful parameters are N1 / EPR which only give an indication of thrust, not power.

Edit for Capt Pit Bull's comments below.
Rolls-Royce Tyne - Wikipedia, the free encyclopedia
"RTy.1 Mk 506
3,259 kW or 4,985 e.s.h.p"

Last edited by rudderrudderrat; 1st Apr 2012 at 19:14. Reason: power output for Mk 506
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Old 1st Apr 2012, 10:03
  #88 (permalink)  
 
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RPM & Torque

rudderrudderrat, the Allison T56 is essentially a constant speed engine/gearbox/propeller installation with propeller blade angle selectable only in the 'ground' range of throttle lever travel (except feathering or unfeathering of course). Once in the 'flight' range (flight idle to 90 degrees throttle angle) the throttle simply schedules fuel flow and the propeller is governed to 100% RPM (1021 RPM) + or - 2% in basic hydraulic governing mode. The obvious advantage in having such an engine is almost instantaneous response to throttle movement (limited to not moving the throttle from Flight Idle to Max in less than 1 second) Without getting the purists throwing stones the torque value indicated on the C130 is the load imposed on the engine torque shaft ( and thus the engine itself) by the reduction gearbox, accessories and propeller. This whole thread seems to have become the site for "ego boosting" by some, surely not what the original poster was looking for. I am sure you, barit1 and I each would understand that as a crew member our concern would be "Have I got the torque or power I expect within the RPM and T.I.T limitations applicable and, if not, why not?" I am familiar with the use of N1 and EPR (or IEPR on the RB211 engine) as a measure of thrust. Whether we call the motive force Horespower, Thrust, Torque or any other name doesn't really matter. Every aircraft I have operated has had a specific set of parameters which are to be met before and during flight.

Last edited by Old Fella; 1st Apr 2012 at 11:11.
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Old 1st Apr 2012, 10:26
  #89 (permalink)  
 
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Sorry I'm late.

Italia is correct (give or take the odd minor slip / assumed simplification e.g. zero wind etc).

Oggers, you don't understand efficiency.

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.

This is my one and only post on the topic. Oggers, you have a long ingrained 'alternative framework'. Unless you are at least prepared to consider that something you think you understand could be wrong I, like italia, see no point in spending any time over it.

pb

Last edited by Capt Pit Bull; 1st Apr 2012 at 10:27. Reason: typo
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Old 1st Apr 2012, 11:04
  #90 (permalink)  
 
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Oh yeah nice one Pitbull. I'll chalk that up to being 1st Apr
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Old 2nd Apr 2012, 13:54
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Capt Pit Bull:

The issue I’m disputing is the assertion by italia458 (note, he raised it, not me) that a plane at standstill can knock out 200 BHP and generate static thrust, yet there is zero THP.

“italia is correct.”
That’s a very definite conclusion to draw from the rather subjective argument you offer.

“Oggers, you don't understand efficiency.”
‘Power out over power in’ is pretty uncontroversial.

Unless you are at least prepared to consider that something you think you understand could be wrong I, like italia, see no point in spending any time over it.
THP = BHP x PE. If the thrust is not zero, PE cannot be zero, therefore THP is not zero. There is another equation out there based on airspeed for an arcraft in flight but it does not invalidate this one or vice versa. The trouble is Pit Bull, that italia458 is insisting it does, and you assert that he is correct. I would call that dogma and therefore I have to chuckle at the irony of your post.

“An engine in a test bed does NOTHING except churn up air thereby producing heat."

It develops thrust.


"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..”
Well, hopefullly the engine won't "increase in the total of GPE + KE" during the test because that wouldn't be terribly "useful work". It is on the test-bed for a reason. It is developing thrust and fulfilling a purpose other than moving an aircraft. You are entitled to your opinion - in a certain context I would agree with it - but it does not lead me to the conclusion that the propeller efficiency is zero, or that static thrust = zero THP.

Applying this to the helo scenario, if I want to maintain a steady hover, the work done on the air which provides the thrust to keep me there is 100% useful work. In no sense is it wasted.
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Old 2nd Apr 2012, 16:11
  #92 (permalink)  
 
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oggers ´n italia : do you not have more important things to sort out in your life?
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Old 2nd Apr 2012, 19:20
  #93 (permalink)  
 
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I squeeze it in between the pilates and AA meetings.
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Old 2nd Apr 2012, 20:25
  #94 (permalink)  
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If you're not doing any useful work, you can't claim to be efficient. End of story.

As for your uncontroversial proof of understanding, it dictates that if the Joules leaving the system per time unit equals the Joules entering the system per time unit, efficiency is 100%. Any less than 100% efficiency, and less Joules leave the system than enter it.



That's pretty amazing, and proves Pitbull's point rather nicely.
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Old 3rd Apr 2012, 12:11
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Blackhand: rolling road dyno - good example. Makes the point clearly.
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Old 3rd Apr 2012, 18:39
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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!
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Old 3rd Apr 2012, 19:51
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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.

Last edited by keith williams; 3rd Apr 2012 at 20:04.
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Old 5th Apr 2012, 12:45
  #98 (permalink)  
 
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Keith, all reasonable contributions are welcome.

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.
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.
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Old 5th Apr 2012, 15:26
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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.
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Old 5th Apr 2012, 16:49
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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 !
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