Efficiency of gears?

[Skogsgurra]

Hi all,

I will be measuring power consumed by rotary kiln drives the following days. The drives are in the 300 - 500 kW range and there are gears taking down motor speed from around 1500 RPM to something like 10 or 20 RPM. The gears are plain "cog-wheel" gears. I am sure there is a correct name for that type of gear, but it slipped my mind - if it ever was there...

Anyhow. What would the efficiency of such a gear be? How does it change with load and how does it change with speed?

I am guessing losses around 3 - 5 percent at rated speed and load. But would like to have a sanity check on that. Or, even better, a table or curve showing efficiency over the operating envelope.

Gunnar Englund

http://www.gke.org

 

[stardelta]

Skoggs?
Its a spur-gear that you describe but thats as far as I can go, this link might help you but my guess is the guys in the Gear and Pulley Engineering forum will answer the question.

http://www.mech.uwa.edu.au/DANotes/gears/intro/intro.html

 

[aolalde]

I guess you are right since for a reduction of 150 you will need at least 3 steps of reduction and assuming 1.5% losses per reduction step it should be around 95.5%total Efficiency. When the load is reduced, only part of the losses are reduced and the other part is fix and proportional to the speed.

 

[CJCPE]

"Spur gear" is the term for plain, cut straight across gears. If you are talking about a large gear on the outside of the kiln driven by a much smaller one, the large one is a "ring gear" or "girth gear" and the small one is a "pinion gear."

A quick search of "girth gear" and "efficiency" turned up something that indicates that such a gear set can be more than 99% efficient.

 

[Warpspeed]

Oh this is a toughie.

There are many types of gears, the straight cut spur gears are the most efficient, but helical gears with sliding tooth action run more silently.

There will be losses from oil shear between the teeth, and losses from windage and oil pumping between the gears, and both will change with speed and load, and type of lubricant. It all ends up as heat if that is any help.

The only way I have ever seen this done is to run the thing for many hours to thermal equilibrium and measure the oil temperature rise in the sump above ambient. Then you switch on the electrical resistance heater you cunningly inserted into the oil drain plug, and run it for many more hours. The total oil temperature rise will be proportional to the power being dissipated into the oil, so it is then a fairly simple matter to work out your gearbox losses if you know the wattage of your heater. Simple in theory, anyhow.

This sort of thing can be almost as much fun as electronics.

Good luck Gunnar...

 

[DickDV]

I'm out of that field now, but, when I was involved with gearing, we used the general rule of 3% losses per stage for spur and helical gears. For worm gears, we used 6% per stage. Actually, for worm gears, 8% would have been better.

While spur and helical gearing often worked out to be slightly more efficient, these "rules of thumb" always worked out quite well, especially for thermal estimates.

 

[blacksea]

Must be noted than losses (efficiency) for low power gear are significant higher (even for one stage) and non-linear. For example for maxon planetary gear up to 120Nm 1 stage gear efficiency is 80%(!), for 2 stages - 75%, for 3 stages - 70%.

 

[ozmosis]

Gunnar
There are four typical characteristic gear types:

Helical Gear Units
Bevel-helical Gear Units
Planetary Gear Units
Worm Gear Units

It depends on the ratio but I guess if you are looking at a 100:1 and possibly Helical Gear unit then the efficiency will be 95-92%. It is typical to work out the losses as the figure quoted by the gearbox manufacturer given for rated output torque. The rated output torque will include for efficiency losses so it is a case of subtracting this from the theoretical calculation of output torque to derive the efficiency. Speed control will not have too much effect on losses as long as it is not oversped.
Looking at the power in your particular application, I would possibly expect a helical bevel. These are slightly lower efficiency, but not much (90-95%)
A few useful sites to check on the gearbox losses using this method would be:

http://www.flender.com/content_manager/page.php?ID=36084&dbc=7853621a0a87c868f3514117bb2bfaa0

http://www.sew-eurodrive.co.uk/products/index.html

http://www.bonfiglioli.com/riduttori_uk.html

 

[ozmosis]

as an example. If you calculate the actual torque output of a prime mover with 500kW at 10rpm output, you get a torque value of 477500Nm. This excludes efficiency. If the gearbox manufacturer advises the torque output of their particular gearbox is 44120Nm, then the eff of the box is 92.4% eff

 

[ozmosis]

441200Nm I mean.
(if your're going to give an example, get it right....)