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Torque...it's Power


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#151 Magoo

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Posted 07 April 2012 - 12:59

There are at least three common torque wrenches around. Leaf spring who does not need to be tuned back to zero after use, and normal stretching spring. (what is the official name of it?)


All types of fastener installation measurement (wrenches of various types, angle meter, torque-to-yield, whatever) require the proper sealant and/or lubricant on the fastener threads (or completely dry if specified, say if the fastener has a specific treatment) to obtain an accurate result because friction in the fastener threads is such a large variable.

The torsion or beam type is by far the cheapest and most reliable torque wrench, and when used properly, at least as accurate as any. However, if you walk through any pit or garage area, even in the pros, you will find that maybe 70 percent of the users are doing it wrong. In the handle you will find a pivot pin, the purpose of which is to fix the effective length of the wrench's lever arm. You must apply all force through the pin. If you allow the handle to rock in or out, the lever arm is now shorter or longer than intended and the result is a faulty torque reading. Small torque wrenches use a ball handle for the same purpose.


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#152 gruntguru

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Posted 08 April 2012 - 00:05

GruntGuru- I'm not sure you're entirely right... I'd say Dmitriy's interpretation would be about right- on a return leg work is produced as well when the car is accelerating, which also reflects on your energy balance- fuel consumed is not only wasted in heat but to do the work on accelerating and moving at constant speed on the whole trip. Another point I'd like to make is that you've lumped all forces working against the direction of movement (negative) together as 'being done on the car'- there would be a case to argue that braking should be an exception (while the effect is negative on the car, the positive work is needed to achieve that, on part of driver or car itself)...

When you burn fuel to complete a round trip, it all ends up as heat. Even the air that was stirred up slows down due to friction, creating heat in the process. One exception is a bit of energy (also not recoverable) that goes into abrasion of things like tyres, road surface, bearing surfaces etc

#153 gruntguru

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Posted 08 April 2012 - 00:12

The torsion or beam type is by far the cheapest and most reliable torque wrench, and when used properly, at least as accurate as any. However, if you walk through any pit or garage area, even in the pros, you will find that maybe 70 percent of the users are doing it wrong. In the handle you will find a pivot pin, the purpose of which is to fix the effective length of the wrench's lever arm. You must apply all force through the pin. If you allow the handle to rock in or out, the lever arm is now shorter or longer than intended and the result is a faulty torque reading. Small torque wrenches use a ball handle for the same purpose.

Interesting point. Only applies to torque wrenches where the torque is inferred by measuring distorrtion at a point not on the axis of the thread being tightened. For example, if the torque is measured by torsional displacement (twist) of a shaft along the thread axis, the lever arm for application of the force can be any length.

#154 Magoo

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Posted 08 April 2012 - 11:55

Interesting point. Only applies to torque wrenches where the torque is inferred by measuring distorrtion at a point not on the axis of the thread being tightened. For example, if the torque is measured by torsional displacement (twist) of a shaft along the thread axis, the lever arm for application of the force can be any length.


Since no one has ever made such a torque wrench (torsion shaft with pivot handle), is going to make one, or has any cause to make one now or in the future, I am unable to share your concern on this point.

#155 bigleagueslider

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Posted 09 April 2012 - 02:33

All types of fastener installation measurement (wrenches of various types, angle meter, torque-to-yield, whatever) require the proper sealant and/or lubricant on the fastener threads (or completely dry if specified, say if the fastener has a specific treatment) to obtain an accurate result because friction in the fastener threads is such a large variable.

The torsion or beam type is by far the cheapest and most reliable torque wrench, and when used properly, at least as accurate as any. However, if you walk through any pit or garage area, even in the pros, you will find that maybe 70 percent of the users are doing it wrong. In the handle you will find a pivot pin, the purpose of which is to fix the effective length of the wrench's lever arm. You must apply all force through the pin. If you allow the handle to rock in or out, the lever arm is now shorter or longer than intended and the result is a faulty torque reading. Small torque wrenches use a ball handle for the same purpose.


I like the new subject matter of this thread.

When installing threaded fasteners, the most important factor is usually preload. A screw thread is basically a mechanical device that converts rotary motion to linear motion. Since its efficiency relies on sliding contact between metal surfaces, friction has a significant effect. While it's true that using beam or click type torque wrenches is the cheapest method for installing fasteners, it's also the least accurate. Since the critical factor is preload force, the most accurate results are obtained using methods that accurately measure fastener preload. These methods would include torque angle wrenches, bolt stretch measurement, hydraulic pre-tensioning, torque-to-yield, preload indicating washers, electronic strain measurement, etc.

Fastener installation processes that rely on strain are mostly immune from variations in friction. So thread lubrication is not so important. However, fastener stress due to installation preload must take into account the total combined stress levels. It must include preload axial tension, torsion due to installation torque, and bending due to misalignment.

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#156 gruntguru

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Posted 09 April 2012 - 03:39

Since no one has ever made such a torque wrench (torsion shaft with pivot handle), is going to make one, or has any cause to make one now or in the future, I am unable to share your concern on this point.

I wouldn't be too confident no one has EVER made such a torque wrench.

I wouldn't be at all confident no one will EVER make such a torque wrench.

As to cause - well for a start it would eliminate the issue of "70% of users doing it wrong".

As to "concern" well - I'm not even concerned about the 70% of threads you say are being incorrectly torqued - most of them are probably not too far off.

#157 saudoso

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Posted 09 April 2012 - 11:36

Really? I think it takes a lot of *work* to rotate a turbine through ~10W hydraulic oil at several thousand rpm. Imagine that instead of an engine, you are turning the converter by hand with a crank. You don't think you'd be performing work?

The absorption unit on a Superflow dyno is a water brake, i.e. hydraulic turbine. Do we mean to say that an engine on a dyno is not really performing work? What is the engine producing? Some kind of fake or simulated work?



The thing is it doesn't matter what gets in, just what comes out. That's the whole point of the argument.

If in the end a spring is loaded, the speed increased or it's sitting higher above the floor work was done.

If not you just released energy. For work to be done, you must have increased kinetic or potential energy in the system you are evaluating (the plane, not the air).

And in the case of the dyno, the engine is releasing heat into the environment. Gas burns->heats air->pushes piston->rotates crank->rotates brake->stirs water that can't escape and gets hot while damping the energy thrown in.

#158 Magoo

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Posted 10 April 2012 - 10:24

I like the new subject matter of this thread.

When installing threaded fasteners, the most important factor is usually preload. A screw thread is basically a mechanical device that converts rotary motion to linear motion. Since its efficiency relies on sliding contact between metal surfaces, friction has a significant effect. While it's true that using beam or click type torque wrenches is the cheapest method for installing fasteners, it's also the least accurate. Since the critical factor is preload force, the most accurate results are obtained using methods that accurately measure fastener preload. These methods would include torque angle wrenches, bolt stretch measurement, hydraulic pre-tensioning, torque-to-yield, preload indicating washers, electronic strain measurement, etc.

Fastener installation processes that rely on strain are mostly immune from variations in friction. So thread lubrication is not so important. However, fastener stress due to installation preload must take into account the total combined stress levels. It must include preload axial tension, torsion due to installation torque, and bending due to misalignment.

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If you recall the old empirical formula for fastener pretensioning (F = T/KD or suchlike) the greatest factor in torque coefficient is friction and the greatest factor in friction is lubrication. If a torque specification calls for lubricant/no lubricant and you do the opposite, you'll be well off the mark.

In "wet" applications (fastener screws into coolant or oil gallery) the sealant serves as the lubricant -- best example being the ubiquitous SB Chevy. Since time began the torque specification (60-65 ft-lbs) included sealant on the lower row (short) row of screws and since time began, mechanics have been ignoring it and installing them dry. When the fastener weeps or the gasket seeps, the mechanic simply puts another round on the fasteners. And often as not it works, displaying once again the significant tolerance for abuse among well-engineered machines.




#159 carlt

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Posted 10 April 2012 - 11:11

I find hitting the spanner with a hammer , using the correct force for each strike , stops the nuts and bolts coming undone

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#160 pugfan

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Posted 12 April 2012 - 01:32

Hi Dmitriy. Its the old frame-of-reference thing again.

If a car does a round trip back to its starting location, the nett work on the car from ALL sources is zero.

The work done on the car by the drivetrain will be some positive number but this will be balanced by the work done on the car by the braking system, air resistance, tyre resistance and other parasitic losses (all negative numbers). This has to be the case if the car has returned to its original energy state.

I have ignored heat energy and chemical energy stored in the car. Clearly the hot parts of the car will cool over time and the lower fuel level in the tank is an indication of how much heat energy was transfered to the environment as a result of the round trip.


Eh?

If we take the car as the boundary of the system (different to frame-of-reference) then when it is back at it's starting point and stationary (or at some other point with identical gravitational potential, Energy is not a vector) then the system has lost energy (fuel has been converted to heat) and has therefore done work on it's environment.

#161 Greg Locock

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Posted 12 April 2012 - 06:09

Eh?

If we take the car as the boundary of the system (different to frame-of-reference) then when it is back at it's starting point and stationary (or at some other point with identical gravitational potential, Energy is not a vector) then the system has lost energy (fuel has been converted to heat) and has therefore done work on it's environment.


Or it has just heated the environment. If it had sat there idling the entire time the same net work would have been done. None. The same fuel would be gone, the environment would be thathe same amount warmer.

#162 Magoo

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Posted 12 April 2012 - 10:44

One odd thing about this discussion has been the refusal to recognize the concept of thermodynamic work.

I thought this was settled in the 1840s, but people are really adamant about it, like they have a religious objection.


#163 gruntguru

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Posted 12 April 2012 - 11:17

If we take the car as the boundary of the system (different to frame-of-reference) then when it is back at it's starting point and stationary (or at some other point with identical gravitational potential, Energy is not a vector) then the system has lost energy (fuel has been converted to heat) and has therefore done work on it's environment.

I did say that - last paragraph you quoted. My point was - nett work done on the car is zero.

#164 pugfan

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Posted 13 April 2012 - 00:56

I did say that - last paragraph you quoted. My point was - nett work done on the car is zero.


Still not with you there. The car started off at a given energy level and lost a bunch of chemical energy which got thrown out the exhaust pipe as heated air or got turned into kinetic energy of air molecules etc. It therefore ended up at a lower energy state.

How does the nett work end up as zero?

Edit: Ah, I think when you say nett work you mean the car and it's environment as the boundary of the system.

Edited by pugfan, 13 April 2012 - 01:04.


#165 gruntguru

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Posted 13 April 2012 - 01:59

Yes its all in the wording. Nett work done "on the car" is zero. All the work and heat resulting from burning fuel ended up being done "on the environment".

#166 pugfan

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Posted 13 April 2012 - 02:35

Yes its all in the wording. Nett work done "on the car" is zero. All the work and heat resulting from burning fuel ended up being done "on the environment".


Agree with you about it's all in the wording and that's why I still disagree. Nett work done on the system whose boundary encompasses the car and it's environment is zero. Nett work done on the system whose boundary encompasses the car only is not zero.

#167 gruntguru

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Posted 13 April 2012 - 06:19

Nett work done on the system whose boundary encompasses the car only is not zero.

Lets say the fuel tank has lost 100MJ worth of fuel. Probably about 20MJ of work was produced and 80MJ of heat within the system. The 80MJ of heat mostly exits the system immediately and all of it has exited within a few hours after the trip. The 20MJ of work is used increasing the kinetic and possibly potential energy of the car and (mostly) overcoming parasitic losses. Once the car completes its round trip the kinetic and potential energies return to their initial level so all the work ends up going into parasitic losses (heat). The nett work done on the car is zero.

Similar to the aircraft in a level cruise, where the work done on the aircraft by the propulsion system is exactly offset by the negative work done on the aircraft by the environment (drag).

#168 Magoo

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Posted 13 April 2012 - 10:06

The notion that airplanes in flight produce no net work is rooted in the theory of equal transit times aka Bernoulli depiction of how airplanes fly. This explanation is popular, easy to understand, and alas, completely erroneous. In reality, it takes considerable work to generate lift.

#169 saudoso

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Posted 13 April 2012 - 10:28

The notion that airplanes in flight produces work depends solely on the system boundaries selected. That's the concept you miss here.

You're sitting there typing and travelling near 1000mph and you feel nothing, your energy state is table. Are you subject to work like that?

#170 Greg Locock

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Posted 13 April 2012 - 11:05

The notion that airplanes in flight produce no net work is rooted in the theory of equal transit times aka Bernoulli depiction of how airplanes fly. This explanation is popular, easy to understand, and alas, completely erroneous. In reality, it takes considerable work to generate lift.


...Unless your"height maintaining" system is a stepladder, or a hydrogen balloon, or many other examples. The really boring banal fact is that W=f.x is a good enough explanation, when applied properly, and all your counter examples are just badly expressed or misunderstood.

If a propeller were sufficiently large in diameter, and a wing were sufficiently large in area, and the weight was finite, then the power required from the engine of a propeller driven aircraft to support that weight tends towards zero. It doesn't matter what your theory is about circulation and lift and so on. Worrying about that is interesting but irrelevant.

#171 gruntguru

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Posted 14 April 2012 - 03:54

The notion that airplanes in flight produce no net work is rooted in the theory of equal transit times aka Bernoulli depiction of how airplanes fly. This explanation is popular, easy to understand, and alas, completely erroneous. In reality, it takes considerable work to generate lift.

You are confusing the wording. Yes an aircraft in steady level flight is producing work. It is doing work "on the environment", stirring the air with its propulsion system and its lift system.The amount of work done on the environment is in fact exactly equal to the amount of work done by the propusion system and the useful/wasted ratio improves.

Different question - what is the nett work done "on the aircraft". Well some percentage of the work done by the propulsion system is useful - ie does work "on the aircraft" pushing it along. The remainder is wasted, stirring the air. As Greg has suggested, the greater the mas of air accelerated (pushed backwards) by the propulsion system and the lower the velocity of that air, the greater the efficiency of the propulsion system.

Let's get back to the useful work done. This is the force exerted on the airframe by the propulsion system times the forward displacement of the aircraft. So there is some work done "on the aircraft". At the same time theree is an equal and opposite force acting on the aircraft - drag. If it were not equal and opposite the aircraft would be accelerating and not in "steady" flight. The drag force times the displacement of the aircraft represents a negative work being done "on the aircraft". So the nett work done "on the aircraft" is zero.

Incidentally all this is unrelated to which lift theory one chooses. One of the sources of the drag mentioned in the previous paragraph is a necessary side effect of the lift generation process (unless Greg manages to build that infinite area wing of course).

Edited by gruntguru, 14 April 2012 - 03:56.


#172 Catalina Park

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Posted 14 April 2012 - 03:55

Sorry Greg, but I always start to get worried when people start bringing propellers into the Technical Forum.

#173 Magoo

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Posted 14 April 2012 - 10:15

You may draw a dotted line at given points around a mechanical operation, call it a physical system, and note that the net work is zero, energy is conserved, whatever. Fine, knock yourself out. That is entirely valid and in the odd case might even be useful.

However, it is a completely different matter to claim that therefore, the machine or machines within the system are not performing work. That does not follow. That is not what it means. That is not supported by known mechanical principles.

That is how people confuse themselves into the notion that an engine on a dyno is not performing work.

#174 Magoo

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Posted 14 April 2012 - 10:35

The notion that airplanes in flight produces work depends solely on the system boundaries selected. That's the concept you miss here.


No, an aircraft in flight performs work -- or it wouldn't be in flight. Flying is not a word game or trick of semantics. If you want your airplane to fly, it's going to need to perform work.

#175 gruntguru

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Posted 14 April 2012 - 10:36

However, it is a completely different matter to claim that therefore, the machine or machines within the system are not performing work.

That is how people confuse themselves into the notion that an engine on a dyno is not performing work.

Perhaps someone has made such a claim somewhere in this thread. If I noticed at the time I probably corrected them.

#176 Magoo

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Posted 14 April 2012 - 11:45

Perhaps someone has made such a claim somewhere in this thread. If I noticed at the time I probably corrected them.


The proposition that an airplane in flight performs no work is a similar misstatement and/or delusion. An aircraft that does not perform work is an aircraft that does not fly. There is no set of boundaries or frame of reference within which an aircraft in flight is not performing work. That is rubbish. That never happened. The net work or energy in an entire physical system might be zero, but that is an entirely different statement -- a different kettle of fish altogether.

It is the ability to perform work that allows the aircraft to fly. If an aircraft cannot perform work, you have no physical system of an aircraft in flight to study. You have an aircraft sitting on the ground. Its net energy is also zero, note.



#177 saudoso

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Posted 14 April 2012 - 14:37

Semantics that is. Let's put this way: if the airplane's energy state does not change ( same speed same altitude) no work is being done *ON* the plane. All forces acting on it net to ZERO.

The bad dyno example ( if the barb is directed at me) was trying to support the direct correlation between power and torque curves in an ICE. What is also true. One torque curve will be paired to one and only power curve and having one you can plot the other. There is no arguing around this.

Edited by saudoso, 14 April 2012 - 14:38.


#178 Magoo

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Posted 14 April 2012 - 22:23

The bad dyno example ( if the barb is directed at me) was trying to support the direct correlation between power and torque curves in an ICE. What is also true. One torque curve will be paired to one and only power curve and having one you can plot the other. There is no arguing around this.


I wouldn't dream of arguing that point. It's absolutely true. However, that doesn't make power and torque the same information, let alone the same property. Power and torque are two different physical entities. There is no arguing that either.

This brings us back to the narrative in the RAM Truck commerical: "Torque...Isn't complicated...Torque is Power..." which does sort of echo your stated position that torque and power are "essentially the same information." So what about the commercial? Close enough for you?

#179 saudoso

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Posted 14 April 2012 - 22:58

If you know the speed and torque you know power. If you know the speed and power, you know torque.

If you know power and current, you know voltage. If you know current and voltage, you know power.

Are those the same? No.

Are they stuck like siamese brothers? Yes.

So they are not the same thing, but if you know one AND the speed involved you know the other.

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#180 bigleagueslider

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Posted 15 April 2012 - 00:09

If you know the speed and torque you know power. If you know the speed and power, you know torque.

If you know power and current, you know voltage. If you know current and voltage, you know power.

Are those the same? No.

Are they stuck like siamese brothers? Yes.

So they are not the same thing, but if you know one AND the speed involved you know the other.


saudoso,

Just to make the discussion even more fun, let's consider the case of a rocket engine propelling a spaceship vertically away from a launch pad. The rocket engine obviously produces power since the spaceship has mass and velocity. In this particular case there is no torque involved. Instead the rocket engine produces thrust by imparting momentum to the mass of rocket fuel being ejected out of the nozzle. The spaceship goes one way and the spent fuel mass goes the other.

Regards,
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#181 Magoo

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Posted 15 April 2012 - 00:23

If you know the speed and torque you know power. If you know the speed and power, you know torque.

If you know power and current, you know voltage. If you know current and voltage, you know power.

Are those the same? No.

Are they stuck like siamese brothers? Yes.

So they are not the same thing, but if you know one AND the speed involved you know the other.


So for you, the properties of torque and power themselves aren't significant, only the relationship between them?




#182 saudoso

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Posted 15 April 2012 - 02:11

But a "Torque Curve" is as good as a "power Curve".



They represent the same information and can be directly calculated one from the other, so I guess they are pretty much the same.



Torque and power are not the same information, and one cannot be determined from the other without knowing an additional variable, which is time.

Torque is a force applied at a radius about a shaft. Power is that torque applied over time.

The best explanation I have heard for engine torque and power is this: Torque is an engine's ability to do work. Power is the rate at which that work gets done.



For God's sake I give up.

Force X Speed = Power. Any set of units you pick. And we were talking about the effin power and torque curves on ICEs and some thick skulls fail to see how they are related.

I give up.







#183 gruntguru

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Posted 15 April 2012 - 04:19

The proposition that an airplane in flight performs no work is a similar misstatement and/or delusion.

I don't think anybody has put that proposition.

I will take yours one step further. Any engine operating with an efficiency greater than zero is doing work.

In the case of the aircraft, car, ship etc at constant speed and constant altitude, 100% of the work is going into the environment.

#184 bigleagueslider

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Posted 15 April 2012 - 05:09

saudoso,

Please, don't give up! If you do, who will I have to argue with?

If you want to limit the discussion to recip piston engines, consider these two cases:

First we have an engine with a flat "torque" curve over its rpm range. This engine would also have a power curve that increases linearly with engine rpm.

Second we have an engine that has a peak "torque" hump in the middle of its rpm range equivalent to the torque of the first engine. This engine would have a power curve that has a peak power equivalent to the first engine only at one engine speed.

If the first engine is used in a car with a one-speed transmission, and the second engine is used in a car equipped with a transmission that has an infinite number of gears, which car would be faster around a road course?

Regards,
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#185 saudoso

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Posted 15 April 2012 - 06:39

:wave:

#186 Canuck

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Posted 15 April 2012 - 16:52

2nd.

If all the work is going into the environment, does the same apply to the rocket? Both are using thrust to create motion, one via direct chemical reaction and one via chemical to mechanical conversion. If no work is being done, how do we have a location displacement? (or is my envelope slipping again)

#187 Greg Locock

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Posted 15 April 2012 - 22:11

If the rocket is not accelerating, and is in horizontal flight, then no net work is being done on the mass of the rocket. If it is accelerating it is gaining kinetic enegry, and if it moves up it is gaining potential energy, that is, work is being done on it.

Work is being done on the surrounding atmosphere in all 3 cases, or even in the case where the rocket is fastened to the ground before it launches, to get back to the clown and the chair. Late edit - although, confusingly, a rocket does not need a surrounding atmosphere if it is to do work.


Edited by Greg Locock, 16 April 2012 - 03:40.


#188 gruntguru

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Posted 16 April 2012 - 01:38

First we have an engine with a flat "torque" curve over its rpm range. This engine would also have a power curve that increases linearly with engine rpm.

Second we have an engine that has a peak "torque" hump in the middle of its rpm range equivalent to the torque of the first engine. This engine would have a power curve that has a peak power equivalent to the first engine only at one engine speed.

If the first engine is used in a car with a one-speed transmission, and the second engine is used in a car equipped with a transmission that has an infinite number of gears, which car would be faster around a road course?

2nd.
Note:
- Both engines have peak power at only one engine speed.
- Engine 2 has a higher rpm limit and makes peak power at higher rpm. VP would select engine 1 for his Jag every time. :lol:

#189 pugfan

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Posted 16 April 2012 - 04:42

The 80MJ of heat mostly exits the system immediately and all of it has exited within a few hours after the trip .... The nett work done on the car is zero.


:confused:

#190 Magoo

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Posted 16 April 2012 - 09:53

Magoo,
Technically, it does not take "work" to drive that turbine through the oil, it takes "energy". The hydraulic turbine has kinetic energy, and that kinetic energy is converted to thermal energy in the fluid mass via the mechanism of viscous shear.



Traditionally, when you have a machine that produces energy without performing work, you'll want to form a company and sell stock. You can get rich, go to jail, or both. Australia has been a popular venue in recent years.


The approach is rather amusing. How did the kinetic energy get into the turbine, one might well ask. Reminds me of the little old lady who accosted William James after a lecture to inform him he had it all wrong, that the earth and its environs were in truth supported by a giant turtle. Taken aback, he asked her what was supporting the turtle. "Very clever, young man," she answered, "but it's turtles all the way down."

#191 bigleagueslider

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Posted 17 April 2012 - 04:34

Traditionally, when you have a machine that produces energy without performing work, you'll want to form a company and sell stock. You can get rich, go to jail, or both. Australia has been a popular venue in recent years.

The approach is rather amusing. How did the kinetic energy get into the turbine, one might well ask. Reminds me of the little old lady who accosted William James after a lecture to inform him he had it all wrong, that the earth and its environs were in truth supported by a giant turtle. Taken aback, he asked her what was supporting the turtle. "Very clever, young man," she answered, "but it's turtles all the way down."


Magoo-

This thread is indeed very amusing.

Besides your example of William James and the pragmatic little old lady, I'd bring up a famous quote from Archimedes: "Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. While Archimedes was definitely one of history's greatest thinkers, and his quote sounds impressive on its surface, the proposition actually disregards many basic principles of physics. And since it also involves the ideas of force, torque and work it is also quite relevant to this thread.

Needless to say, Archimedes' proposition does considered the effects of gravity, friction, inertia, material properties, etc. In reality, Archimedes could never really accomplish such a feat regardless of how long the lever was. Just consider the mass of a lever long enough to allow Archimedes to move the earth. Even if the lever was made from the best structural materials known to mankind, the long lever could never be made strong/stiff enough to even support its own mass, let alone the additional mass of the earth.

To make a long story short, the physical principles of work, power, efficiency, torque, inertia, etc. are all very relevant and inter-related.

Regards,
slider

#192 manolis

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Posted 17 April 2012 - 05:18

If you want to limit the discussion to recip piston engines, consider these two cases:
First we have an engine with a flat "torque" curve over its rpm range. This engine would also have a power curve that increases linearly with engine rpm.
Second we have an engine that has a peak "torque" hump in the middle of its rpm range equivalent to the torque of the first engine. This engine would have a power curve that has a peak power equivalent to the first engine only at one engine speed.
If the first engine is used in a car with a one-speed transmission, and the second engine is used in a car equipped with a transmission that has an infinite number of gears, which car would be faster around a road course?
Regards,
slider


Bigleagueslider,

The answer depends on the road course, on the exact meaning of the “infinite number of gears”, on the rev limit of the first engine.

Provided you mean that:
1. The rev limit of the first engine is higher than the revs of the peak torque of the 2nd engine (which means the 1st engine peak power is higher then the peak power of the 2nd engine),
2. The transmission of the 2nd car has not just “infinite number of gears” (which could mean transmission ratios anywhere between 3.1:1 and 3.2:1) but an unlimited range of continuously variable transmission ratios (like, say, from 1000000:1 to 0.000001:1),
then the answer depends on the road course and on the single transmission ratio of the 1st car.

If the “road course” is slow enough (say it does not allow speeds over 150 Km/h) or short enough (say 400m from the start to the end), the 2nd car can be faster because from the beginning of the motion to the end of the motion all the peak power of the 2nd car is used to move the car. In comparison, at low speeds only a small percentage of the peak power of the 1st car can be used for its motion.
If the “road course” is fast enough and long enough, the 1st car can be the winner.

For instance, if the 1st engine has a peak power of 200 bhp at 10000 rpm, corresponding to, say, a speed of 200 Km/h (with its single gear ratio), while the 2nd car has a peak power of only 150 bhp at 7500 rpm, then:
At 50 Km the 1st car has only 50 bhp to accelerate it, while the 2nd car accelerates with all the 150 bhp of its engine.
At 100 Km/h the 1st car has only 100 bhp to move it, while the 2nd car has 150 bhp.
At 150 Km/h the 1st car and the 2nd car have both 150 bhp to push them forwards.
At 200 Km/h the 1st car has 200 bhp to push it forwards, while the 2nd car has only 150 bhp.

If the two cars move at speeds permanently above 150 Km/h, the 1st is faster.
If the two cars move at speeds lower than 150 Km/h, the 2nd is faster.


Way more interesting is the problem of the optimization of the gear-box ratios, especially for engines with peaky torque curve like the old 2-strokes:

Posted Image

Take a look at http://www.pattakon....attakonEduc.htm , at the RoadLoad DOS program (near the bottom of the page).

Thanks
Manolis Pattakos


#193 gruntguru

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Posted 17 April 2012 - 06:35

Manolis
Slider stated that both engines have the same peak power and the same peak torque. This makes the problem much easier to solve. Engine 2 can be operated constantly at its power peak and so the car fitted with engine 2 will have superior performance at all speeds except at the speed where engine one is at peak power. For that instant only, the performance of the two engines will be identical.

#194 Slumberer

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Posted 17 April 2012 - 06:49

Is the change between the infinite number of gears instantaneous?

(Ducks and runs for cover)

#195 gruntguru

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Posted 17 April 2012 - 06:52

The 80MJ of heat mostly exits the system immediately it is burned and all of it has exited within a few hours after the trip.


:confused:

Not sure what has you perplexed but I hope the red text above fixes the problem.

#196 Dmitriy_Guller

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Posted 17 April 2012 - 06:53

Bigleagueslider,

The answer depends on the road course, on the exact meaning of the “infinite number of gears”, on the rev limit of the first engine.

Provided you mean that:
1. The rev limit of the first engine is higher than the revs of the peak torque of the 2nd engine (which means the 1st engine peak power is higher then the peak power of the 2nd engine),
2. The transmission of the 2nd car has not just “infinite number of gears” (which could mean transmission ratios anywhere between 3.1:1 and 3.2:1) but an unlimited range of continuously variable transmission ratios (like, say, from 1000000:1 to 0.000001:1),
then the answer depends on the road course and on the single transmission ratio of the 1st car.

If the “road course” is slow enough (say it does not allow speeds over 150 Km/h) or short enough (say 400m from the start to the end), the 2nd car can be faster because from the beginning of the motion to the end of the motion all the peak power of the 2nd car is used to move the car. In comparison, at low speeds only a small percentage of the peak power of the 1st car can be used for its motion.
If the “road course” is fast enough and long enough, the 1st car can be the winner.

For instance, if the 1st engine has a peak power of 200 bhp at 10000 rpm, corresponding to, say, a speed of 200 Km/h (with its single gear ratio), while the 2nd car has a peak power of only 150 bhp at 7500 rpm, then:
At 50 Km the 1st car has only 50 bhp to accelerate it, while the 2nd car accelerates with all the 150 bhp of its engine.
At 100 Km/h the 1st car has only 100 bhp to move it, while the 2nd car has 150 bhp.
At 150 Km/h the 1st car and the 2nd car have both 150 bhp to push them forwards.
At 200 Km/h the 1st car has 200 bhp to push it forwards, while the 2nd car has only 150 bhp.

If the two cars move at speeds permanently above 150 Km/h, the 1st is faster.
If the two cars move at speeds lower than 150 Km/h, the 2nd is faster.


Way more interesting is the problem of the optimization of the gear-box ratios, especially for engines with peaky torque curve like the old 2-strokes:

Posted Image

Take a look at http://www.pattakon....attakonEduc.htm , at the RoadLoad DOS program (near the bottom of the page).

Thanks
Manolis Pattakos

You are ignoring the setup of the problem. Both engines have the same peak power.


#197 gruntguru

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Posted 17 April 2012 - 06:54

Is the change between the infinite number of gears instantaneous?

(Ducks and runs for cover)

:lol: If not, it would be unwise to use them all. :lol:

#198 manolis

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Posted 17 April 2012 - 07:49

Manolis
Slider stated that both engines have the same peak power and the same peak torque. This makes the problem much easier to solve. Engine 2 can be operated constantly at its power peak and so the car fitted with engine 2 will have superior performance at all speeds except at the speed where engine one is at peak power. For that instant only, the performance of the two engines will be identical.


Guntguru,

unless I miss something, Bigleagueslider says nothing about same peak power.

He writes:
"First we have an engine with a flat "torque" curve over its rpm range. This engine would also have a power curve that increases linearly with engine rpm.
Second we have an engine that has a peak "torque" hump in the middle of its rpm range equivalent to the torque of the first engine. This engine would have a power curve that has a peak power equivalent to the first engine only at one engine speed."

The two engines have "at a specific engine speed", in the middle of their rpm range, the same (equivalent) torque (which means that at the revs of the torque bump of the 2nd engine it makes its peak power, which is the same (equivalent) with the power the 1st engine makes at the same revs, but smaller than the peak power of the 1st engine).
The 1st engine keeps its torque to the red line, while the torque of the 2nd engine drops steeply after the "torque" bump revs.
I.e. the 1st engine has more peak power than the 2nd.

If the second engine is to have the same peak power with the first engine, then the "torque" bump of the 2nd engine needs to be not in the middle of its rpm range, but at the red line of their rpm range.

Bigleagueslider?

Thanks
Manolis Pattakos

#199 Dmitriy_Guller

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Posted 17 April 2012 - 12:37

Guntguru,

unless I miss something, Bigleagueslider says nothing about same peak power.

He writes:
"First we have an engine with a flat "torque" curve over its rpm range. This engine would also have a power curve that increases linearly with engine rpm.
Second we have an engine that has a peak "torque" hump in the middle of its rpm range equivalent to the torque of the first engine. This engine would have a power curve that has a peak power equivalent to the first engine only at one engine speed."

The two engines have "at a specific engine speed", in the middle of their rpm range, the same (equivalent) torque (which means that at the revs of the torque bump of the 2nd engine it makes its peak power, which is the same (equivalent) with the power the 1st engine makes at the same revs, but smaller than the peak power of the 1st engine).
The 1st engine keeps its torque to the red line, while the torque of the 2nd engine drops steeply after the "torque" bump revs.
I.e. the 1st engine has more peak power than the 2nd.

If the second engine is to have the same peak power with the first engine, then the "torque" bump of the 2nd engine needs to be not in the middle of its rpm range, but at the red line of their rpm range.

Bigleagueslider?

Thanks
Manolis Pattakos

You're assuming equal RPM range for both engines, which wasn't stated.


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#200 munks

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Posted 17 April 2012 - 14:31

You're assuming equal RPM range for both engines, which wasn't stated.


Good point. And I think the last line is ambiguous. The peak powers are equivalent? Or are the powers equivalent at one engine speed, which happens also to be the 2nd engine's peak power? The latter doesn't seem possible if we assume equal RPM range.