an unconventional question

[wangp1283]

I'm currently designing an unconventional type of transmission. An engine drives a chain system. But it's essential that there is a "spring damper" between the engine (driving sprocket) and the driven sprocket (load). This will most likely take the form of an actuator/spring that the chain (under tension) "runs over" like a hill. So the actuator can "give" to limit the tension of the chain or the jerkiness to the load.

Now, imagine an engine running at 2000rpm and is delievering 200 ft lb of torque (which isn't constant in reality). At 200 ft lb, assume the actuator needs to apply a force of 100 lb to "hold up" the tensioned driving chain.

If the actuator force is too large, then it will fail as a damper. Let's say a shock load is applied yet the actuator force is more than 100 lb, then if the engine is still outputing 200 ft lb of torque, then the actuator will not "give", so what would happen in this case? Will the engine stall? If the shock load is too large (requires more than 200 ft lb of torque), the engine might stall right? But what if the shock load is only slightly greater than? Even if the engine does not stall, how will the efficiency be affected?

If the actuator force is too small, then it will "give" unncecessarily and this will affect the efficiency right? Since some work is done on compressing the actuator.

But the actuator force is hard to control since the engine output isn't even constant due to the nature of the internal combustion engine.

What is a solution to this? I want to system to be smooth and efficient. (when subjected to slight shock load)

Or am I worrying too much because an engine won't stall that easily as long as the shock load is small and isn't long lasting? Sometimes when I mow my lawn the the grass is too dense I can hear the engine rpm drop due to the increase in load and then rev up again as long as I quickly move to a lighter area. But will the efficiency go down in this case?

Thanks

 

[magnograil]

Normally, the tensioner in a chain drive is on the unloaded run. What you probably want is a torsional damper built into the drive/driven sprockets. Chain primary drive motorcycles engines typically put it in the driven (although the drive would impart less shock load to the chain) sprocket. Various arrangements are used but commonly is a sprocket with vanes sitting on a hub with vanes interleaved and rubber cushions inbetween.

 

[wangp1283]

It's not a chain tensioner, but rather a damping system to prevent the engine from sputtering or stalling if subjected to some momentary short shock loads.

 

[DesignOutsourcing]

Wang,

Most shock loads are overcome by the inertia of the rotating parts of the engine (usually including a flywheel.) The easiest way to reduce shock loads is to "overcome/absorb" them in a flywheel.

If your intention is to protect the engine or the chain from these shock loads, then the torsional damper already mentioned will help and so would your idea. Don't worry about efficiency as in a dynamic system you'll get back the work when the spring decompresses. A variable rate spring would give you some damping at a range of output torques.

Might be easier to build some flexibility into your chassis/frame or use rubber engine mounts. cheers, derek

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[JayMaechtlen]

so, you want either a beeeg torsional spring between engine and load, or a lots of flywheel (think truck/tractor/Harley)
Assuming you aren't looking for rapid acceleration, a flywheel could be external on an idler shaft (jackshaft) between engine and load.



Jay Maechtlen

 

[wangp1283]

If I use some sort of spring between the engine and load, shouldn't the spring "bounce up or down" since the engine torque output isn't constant and is rather sinuosoidal?

This shouldn't be too big of a problem if the "flexibility" I need is very miniscule (like around only 0.1mm). That means if a loaded chain under tension can yield around 0.1mm (due to the spring), then the problem will be solved.

So maybe I can use some sort of spring that offer a range of force in a very short distance (around 3mm), by using varible constant spring or by some type of gearing.

Then the spring stiffness will ensure that it constantly matches the engine torque, (actually, the force of the spring is a lot smaller than the chain tension because of the angle of the set up). But it can "yield" 0.1mm if needed and still keep the chain tension approx. the same. (without the risk jerking the chain or sputtering the) engine)

Just hope the "spring going up or down" won't be felt by the operator.

I think this is the simplest and the cheapest to solve my problem. Do you think so?

 

[JayMaechtlen]

TORSIONAL spring means that it deflects in rotation, not length or whatever.
Imagine a long skinny driveshaft that can twist or "wind up" under greater torque, and unwind a bit under lesser torque.
Now, the coupler mentioned several posts above does that kind of thing, by letting the elastomeric cushions act as springs.
I was partly joking, but also suggesting something with much more angular deflection than a typical coupling can provide.
Again- it just averages the demand, lets the engine handle it more easily. The suggested flywheel accomplishes the same thing by storing energy (and momentum) for immediate delivery.
If you are protecting the engine, either method will work.
If you are protecting the load, then you need the spring method.
cheers
Jay

Jay Maechtlen

http://home.covad.net/~jmaechtlen/

 

[wangp1283]

When a vehicle or load has a higher speed than the engine, such as when you downshift to a lower gear, the transmission will "force" the engine up to speed. But the vehicle speed doesn't seem to change. on the other hand, there is "engine braking", which will slow down the vehicle.

What's the difference between this and engine braking?

In engine braking, the vehicle will slow down. Is this energy lost or is it stored in the engine momentum?

 

[patprimmer]

On a conventional car, when using the engine for braking the energy is lost as work done pumping air and as inertia in the reciprocating parts and as additional friction.

Regards
pat pprimmer@acay.com.au
eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[wangp1283]

but the energy used to pump the air should be retrieved back when the air decompresses, right?

 

[evelrod]

When a vehicle or load has a higher speed than the engine, such as when you downshift to a lower gear, the transmission will "force" the engine up to speed. But the vehicle speed doesn't seem to change. on the other hand, there is "engine braking", which will slow down the vehicle.

No. The engine will be "forced" (bad choice of words,IMO) to a higher rpm range AND this action will definately slow the vehicle if no added power is present to offset the speed change. Only if the engine is accelerated to the matching rpm will the speed NOT change.
"Engine braking" is simply reducing power output below that necessary to maintain desired speed and/or dropping to a lower gear (higher ratio) without adding compensating power---it's a matter of degree, the more you back off the power, the more "engine braking".

The IC engine in common use in transportation IS an air pump, but NOT an air compressor. The energy lost due to "engine braking" is lost to atmosphere. Added engine braking can be had by shutting off the ignition source and OPENING the intake throttle to maximise the pumping loss.
In any case, the energy is lost to friction, heat, pumping loss, etc.

Rod

 

[wangp1283]

Thanks, I realize that if you shift down to a lower gear the engine rpm will go up and the vehicle speed will be slightly slower, because the energy went into accelerating the engine rpm.

I understand there is friction and heat loss. I'm just not very sure on what you mean by "pumping loss". Because when the piston go up and compress the air in the cylinder, it will take energy. But then this energy will be gained back when the air decompresses (normally during the combustion stroke). So it's like a spring. Of course you can't gain back 100% of the energy that went into the compressing (air) stroke, is this the "pumping loss" you are talking about?

Or does the air valve actually opens (I thought it don't) after the air get compressed so all those compressed air run out into the atomsphere and the energy that goes into compressing the air get lost to the atomsphere and none of it can be retrieved again?

 

[DaveKillens]

If I was to attempt to construct a device to suit the conditions you describe, I would use two chain drive tensioners.
Visualize the two sprockets, and the length of chain running between each sprocket. There would be two straight lengths of chain. Just apply strong tensioners on each straight section. Make them equal in strength, and of course, design so they don't cause mutual interference, yet could handle the wide range of motion required. You may have damping issues, but that's something that could probably be addressed by applying dampers to each tensioner.
Of course, there has to some slack in the chain, they cannot run in a straight line between sprockets. Shock loads would pass directly between sprockets.

 

[JayMaechtlen]

"pumping loss" is the energy loss from pumping air through/past various restrictions. The throttle valve is a restriction, the valves and ports are also, as is the exhaust system. The turbulance shows up mostly as heat in the engine and the air. There are mechanical losses as well, from ring/piston friction, windage in the crankcase, power required to operate the oil pump, water pump, etc.
All those add up to 'engine braking'.
Compression-Ignition engines sometimes use a device to increase engine braking. It is commonly called a "jake brake" but there is probably an official technical term as well. I will leave the explanation for others, lest I mangle it!
regards
Jay

Jay Maechtlen

http://home.covad.net/~jmaechtlen/

 

[GregLocock]

http://www.jakebrake.com

has a good explanation

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[truck]

The Jake brake is a nickname along the lines of Kleenex - it is a product of the Jacobs Manufacturing Company. Officially in the industry it is an engine brake. What it does is throttle the exhaust valve to force pumping losses in the cylinder. In a compression ignition engine it works quite well.

Also used is an exhaust brake which is a throttle in the exhaust pipe, simply a butterfly to increase back pressure.

Truck

 

[wangp1283]

So thereotically, if we assume that there are 2 identical engines (ENGINE 1, ENGINE 2) with NO friction, perfect efficiency in compressing the air, and no windage loss,ect...

If the Engine 1 has the jakebrake device that regulates the exhaust valve so that the compressed gas are dissipated into the atomsphere. The energy that went into compressing the gas in the compressing stroke is not retured to the piston/engine. It is lost to the atomsphere. This provides for very significant "engine braking". Because the energy of the moving vehicle is used to compress the air, which is dissipated into the atomosphere.

If Engine 2 doesn't have the jakebrake device, the compressed gas will merely decompress and return the energy back to the piston like a spring. And in the absence of friction and other losses, There will be essentially no engine braking. Other than the fact some energy from the vehicle will accelerate the engine to a higher rpm and the vehicle will slow down due to this. But overall, there is NO NET LOSS of energy to the outside.

Right?

 

[evelrod]

Given your hypothetical engine--- with a closed inlet and closed exhaust could fit your conditions---a conditional "yes". I know of no such engine. Engine braking---SI engines use the closed inlet throttle plates to make the "pump" work at suction and a CI Jake Brake throttles the exhaust outlet making the "pump" work at compressing the exhaust.

A SI engine, which does not throttle the exhaust or
a CI engine, which does not throttle the intake---no. You still have intake and/or exhaust pumping loss.

All this is moot :-( as there is no frictionless, read that, "perpetual motion" engines.

Rod

 

[DaveKillens]

In the world of dirt motorcycles, a similar device has been around for decades. It's known as a decompressor, basically just a one way valve. It is used for kick starting (reduces the compression, and effort required to kick), and going downhill. It's nice to have one, just activate it, and you are guaranteed moderate rear wheel braking, just the thing for going down slow, steep hills.