# Machinery Rated Power Calculation

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1. Jul 15, 2015

### Moteor

Hi everyone,

Thanks for the help in advance. I have a project with a piece of machinery that is rated to 105 HP @ 540 RPM.

I was wondering what the torque rating for the machine would be, whilst its being driven from a diesel engine?

I believe there is a direct relationship between Torque and Horspower from Engines;
TORQUE = HP x 5252 ÷ RPM​

Therefore Torque = 105 HPx 2525 /540 RPM= 490.97 lb ft = 665.665938072 Nm

1) So I have done the math, but is that right? Is that really what the torque the machine can handle?

I also have the torque and HP stats of the engine as below:

I need to find the torque of the engine related by the rated torque capacity of the wood chipper = 665.7 Nm. @540 RPM?

Also I need to find out what engine speed can provide the max torque rating for the machine as the engine has to be reduced by a ratio through pulleys to match the max RPM of the machine. My second question is:

2) Does reducing a gear ratio, say 3.7 : 1 multiply the torque by that much? So if;

Torque x gear ratio = torque at the wheel
Then @ 2000 RPM the engine would provide 169.1 Nm , therefore multiply by the 3.7037 times the motor is rotating faster than 540 RPM and you have the figure 625.92 which is under the figure we have for the rated capacity of the wood chipper.

My third question is

3) Am I right?

or is it all in my head?

Thanks all . ..

2. Jul 15, 2015

### dean barry

I would be tempted to stick with power ratings, as long as your deisel engine has 105 HP at working speed
And the deisel at working speed matches the chipper at 540 rpm
(youll have to gear it inbetween the two to suit)

3. Jul 15, 2015

### Moteor

Thanks Dean Barry, I am trying to work out the power ratings to be sure the machine works at the proper power.

To do this I need to reduce the engine speed by 4 from 2000 rpm to 540 to match the max speed of the the machine?

Also the engine has a power curve of HP so it's not always at 95 hp but that depends on the rpm.

4. Jul 15, 2015

### dean barry

Yes, so that when your engine is at 2000 rpm, the chipper is at 540 rpm, so whatever power your engine is producing at 2,000 rpm this will be reflected at the chipper regardless of the chipper speed, so make sure the chipper is at 540 rpm when youre engine is at 2,000 rpm, so yes the speed reduction ratio between your engine and the chipper is 2,000 : 540, so if it involved two sprockets and a chain, your engine sprocket would have say 25 teeth, and the chipper would have 25 * ( 2,000 / 540 ) = 92.59 teeth (say 93 teeth)
The torque increase from engine to chipper would be proportional to this ratio also.
signing off, back tommorow

5. Jul 15, 2015

### Moteor

That's what I thought would happen. The thing I was thinking is if the torque is increased then don't that effectively increase the HP in proportion?

What is the consequence of putting too much torque through the machine? Snapped belts and twisted shafts I think?

6. Jul 15, 2015

### SteamKing

Staff Emeritus
Since Power = Torque * RPM * constant
and
Torque out = Torque in * Gear Ratio
but
RPM out = RPM in / Gear Ratio

So

Power in = Torque in * RPM in * constant
and
Power out = Torque out * RPM out * constant
but
Power out = Torque in * Gear Ratio * (RPM in / Gear Ratio) * constant

The gear ratio cancels out of the power equation, so the power transmitted remains constant, except for some minor losses in the gearbox.

Among other things.

7. Jul 15, 2015

### Moteor

Thanks Steam King :)

So if the engine was running at 2000 RPM the power through the engine would be 170 Nm and the HP would be about 40?

So to ramp up the horsepower would have what effect on the machine (wood chipper)?

And if it gives the power, whats the best way to reduce the speed coming out of the engine?

Last edited: Jul 15, 2015
8. Jul 15, 2015

### jack action

Your wood chipper is meant to work at a constant 540 rpm. At this rpm, if the power input is 105 hp then the torque input is 1021 lb.ft (You made a mistake in your first post). These are unalterable facts. If you change one, you need to change at least another one. Since you have two of them fixed, the third one is also fixed by default.

It doesn't matter if the power source produces the 105 hp @ 54 rpm (i.e. with 10212 lb.ft of torque) or @ 5400 rpm (i.e. with 102 lb.ft of torque). Because you will have to use some sort of gear ratio to set it to the required 540 rpm (0.1:1 or 10:1, depending on the chosen engine) and in any case - once the correct 540 rpm is achieved - the torque will be 1021 lb.ft ( = 10212 * 0.1 = 102 * 10).

That is the beauty of power: It takes into account the energy delivery rate, no matter whether it comes from a large force recurring at a slow rate (high torque, low rpm) or a small force recurring at a fast rate (low torque, high rpm). With gear ratios, you can easily transfer the input energy rate from force to speed or vice-versa; but the total energy rate (a.k.a. power) will always stay the same. Energy cannot be created nor destroyed, but its «form» can be altered (force or speed, but can also be converted to heat, electrical current & voltage, potential energy, etc.).

9. Jul 15, 2015

### Moteor

So what would your recommendation be for the gear ratio to give as much power as possible?

I am thinking at peak HP @ around 3500 RPM would do it with a reduction of 6.48 to 1?

Thanks for the help

10. Jul 15, 2015

### insightful

Call a local vendor for a V-belt drive. Your 6.8 to 1 looks good.

11. Jul 15, 2015

### jack action

The maximum power of your engine is 62.6 kW (84 hp) @ 3789 rpm. To drop it to 540 rpm, you need a 7:1 gear ratio (=3789 / 540). The power at the chipper will be 84 hp (minus some losses in your gear/pulley/chain reducer, probably something like 5% or about 4 hp). So you will always be below the 105 hp, because you cannot increase or reduce the power produced by the engine.

For example, if you put a 3.7:1 gear ratio, you would have to run the engine at 2000 rpm to get the required 540 rpm (3.7 * 540 = 2000). At that rpm your engine produces a maximum power of about 35 kW (47 hp). Subtracting a 5% loss (2.4 hp), you end up with 44.6 hp at the chipper's input.

What it means is that your engine will not be able to perform as much work as it was intended to do, no matter the rpm you choose (and that is assuming you run your engine full throttle, which might not be recommended for a long period of time). Either you will have to feed it slower or you will have to put smaller branches. But if you want more from you chipper (or any machine for that matter), you need more power.

Forget about torque numbers, what is important is power. If the specs would have been 1021 lb.ft @ 540 rpm, you should have said to yourself: «I wonder how much power that is?» You do the math, find 105 hp and then you would have said: «Ah! I need an engine that can produce 105 hp and - no matter at what rpm it produces that 105 hp - I need to bring its speeds down to 540 rpm.»

Normally you would want to use the engine at a rpm where it has the best fuel consumption (which is usually somewhere around maximum torque; it depends on the load your putting on the engine). If you choose an ideal engine, it should be able to produce the 105 hp at that rpm.

So your engine is probably build to perform best at around 2500 rpm (that could include performance of the cooling and lubrication systems as well), where you won't be able to get more than 45 kW (60 hp), and that is full throttle. The engine would usually be used at 84 hp @ 3789 rpm for short period of time only (maybe that is your intended use?). Only racing engines are usually prepped to perform at maximum power "all the time" (where "all the time" could mean one season - or even only one race - before an overhaul is due).

12. Jul 16, 2015

### dean barry

Says it all really, so, to summarise:
if you had a more powerful engine and ran it (comfortably) below its peak power revs, then calculate the inbetween ratio to suit the chipper speed.
An issue may be the large size of the chipper sprocket (or pulley wheel) required, a more powerful lesser rpm engine might be the best option overall
Example:
Say you had a engine which produced a maximum 200 hp @ 4,000 rpm.
Reduce the running speed to say 2,000 rpm, which is 100 bhp or thereabouts (depending on which engine you use)
Your speed reduction ratio = 2,000 / 540 = 3.7 : 1
So your sprockets ( chipper : engine ) would be say: 74 : 20
The torque rating of the engine at 2,000 rpm multiplied by 3.7 would = the torque at the chipper at 540 rpm

13. Jul 16, 2015

### Nidum

An important consideration when matching engines to loads is speed regulation .

The rated power of your wood chipper only applies at max loading - ie biggest logs and highest feed rate . Actual horsepower demand will vary greatly with load and could drop to a very small value when no chipping is being done .

If engine is unregulated it could speed up considerably at lower loadings .

Ideally your engine should have built in speed regulation .

14. Jul 16, 2015

### Moteor

Thanks for all the replies, I have spent considerable time working this out. At the moment I am working with this:

My plan is to have a compound pulley system with 3 groove vbelt pulleys rated to at least +20% over 105 HP.

I am looking at the 2 pulley sizes of 117.5 and 305.5 and being in a drive chain to have 2 of each.

they give a ratio of 2.6 : 1 so 2 of them is 2.6 x 2.6 = 6.76.

so at 3,650 RPM it will be reduced by 6.76 times giving 539.4 rpm. So this gives peak power at peak revs.

Please can you confirm if this is right or REALLY IS ALL IN MY HEAD!!

15. Jul 18, 2015