How to power hyd driven gen.

My suggestions:

1. You have to start from the generator and work backwards to theoretically match up everything or you may be very disappointed with the result.

2. Get label plate info from both the generator and hyd motor. You'll be flying blind if you don't have that.

3. You need to provide about 2 shaft hp per kw to the generator to get the full capacity. The generator must also spin at 3600 rpm (if that is on the plate) to obtain the proper frequency. Make sure it's 60 hz. and not some oddball. Voltage control is probably automatic within the generator itself.

4. Assuming no gearing between the hyd motor and generator, the motor must also spin at 3600 rpm. You need to determine what fluid flow in gpm is needed to obtain that speed. That should be on the nameplate. It may be given in cubic inches per revolution. If it is, you can convert to the gpm necessary to get 3600 rpm at the shaft.

5. The next step is to determine the hydraulic hp needed to be provided to the motor. Take the total generator shaft hp and divide by, oh, say 0.8 to account for some inefficiency to get hyd hp.

6. From the hyd hp and the gpm you can determine the needed hydraulic pressure to properly run that generator. To get the pressure use the equation psi = hp x 1714/gpm.

7. You now need to determine the appropriate hydraulic pump to provide the above flow and pressure at the speed you are going to operate it at...whatever your pto speed is if that is your choice. Be careful here. If you can't get a speed match, you may have to consider speed up gearing...something you want to avoid, at least from a cost standpoint.

8. Once you have the pump selected, take the hydraulic hp and divide by, say, 0.8 to get the shaft hp necessary to properly drive the pump. You will need to provide this much hp to the pump drive for everything to work properly.

9. If you can easily match all these characteristics up, you should be good to go. If you can't, you may have to rethink the thing as it could get to be a costly and frustrating experience trying to do it on a trial and error basis.

10. Good luck

Rod

Good Luck,
Dusty

Don't forget that gearing up removes torque multiplition just as gearing down adds torque. So if the requirement was 2hp at 540rpm you might need 6hp at 3600rpm. I didn't do the calculations, just an example.

T_Bone

If you have 6:1 pulleys you get 6x the rpm at 1/6 of the torque which is the same hp.

HP= torque x RPM / 5252. 5252= the rotational force in radians per second

example: needed 12hp@3600rpm with 17.5ft/lbs torque(t) with a 6:1 gear reduction torque multiplier, then:
t= hp / rpm x 5252 = 17.5ft/lbs torque
math check 17.5 x 3600 / 5252= 12hp

Loss of gear torque:
17.5ft/lbs torque / 6 = 2.92ft/lbs the new torque

HP= t x rpm / 5252
2.92 x 3600 / 5252= 2hp the new hp less gear reduction torque.

OR 10hp less than orginal required.

The new HP requirement would be 24hp without the gear reduction. New required engine spec's:
24hp @ 35ft/lbs @ 3600rpm

Torque is the most important number on engine spec's.

T_Bone

I certainly agree with your statement about torque being an important number in engine specs. I do, however have to basically agree with Charles about power transfer across a gear train.

I am confused by your "loss of gear torque" calculation. The loss in power transmitted across a gear train is determined by the train efficiency. This is, in effect, also the loss in torque transfer since there is no loss in speed (no slippage). I recognize that torque is either increased or decreased depending on whether it is a speed-down or speed-up drive...I am talking about the LOST torque in the process (and also, by the relationship, lost hp). That loss is caused by frictional effects at teeth, bearings, seals, etc. and by the churning and heat generation in any lubrication bath. For your old standard spur gear trains, that total efficiency should be about 95% or better. While some specialized trains such as worm drives can be worse than this because of high frictional losses, I think that using 95% of hp in to get hp out would be realistic for most "normal" gear trains with either up or down ratios.

Using the above, your example of a need for 12 hp would require 12/0.95 = 12.6 hp to be input to the reduction unit.

Using your calculation, only half of an engine's hp would ever get to the rear wheels in a typical car or truck in the lower gears :o).

Rod

Ahh but were "loosing" the gear torque multiplication from 540rpm to 3600rpm on output or rufffly 6:1 gear reduction torque what Charles used for his example. The acutal torque loss would be 6.6666:1

We would gain gear torque going from 3600rpm to 540rpm.

If we loose gear torque then we have to add engine torque to over come that loss at the "NEW" rpm.

I was trying to show that staying with a 6:1 gear reduction the HP requirement would be 12hp in the example at 540pm "output". This example assumes 12hp is the correct HP that is needed WITH the gear reduction at 540rpm output.

If we remove the gear reduction torque then the new HP requirement would be 24hp @ 3600rpm for this example.

If we were going from 3600rpm to 540rpm then there would be less HP requirement.

I ran into this problem alot on large HVAC system fans. A service tech would increase fan RPM and did not condsider the new HP required to drive the new higher rpm thus overload the smaller HP motor.

T_Bone

The service tech you referred to needs to go back to basic training in this area. He'll overload motors every time that way, assuming they are not oversized to begin with :o).

I think we also are using the terminology a bit differently with respect to "lost". The fact that the torque is decreased in a speed-up drive by nearly the same ratio as the speed is increased is not considered a "loss" by me. You are considering it as a "loss". That's OK as long as we accept it as a difference in terms. I see the gear drive as nothing more than an energy (hp) transfer from one state of speed and torque to a different state of speed and torque. The only "losses" that occur in such an energy transfer are those due to the irreversiblility of the process. In other words, If the drive were 100% efficient, the HP in would be the same as the HP out even though the torques (and speeds) are quite different. This would be true in either direction. A drive with a speed reduction of, say 6:1, would definitely NOT reduce the HP required to be transmitted, just the same as a speed-up drive of 6:1 would not increase it. By my way of thinking torque is not "lost" by simply being reduced. It can, in fact, be recovered (at least most of it) by an additional downstream gear reduction set. Think of a 6:1 speed-up drive immediately followed by a 6:1 speed down (reduction) drive. In such a case, the torque into the first drive would be the same as the torque out the final drive assuming 100% efficiency of both drives. In this instance there is no "lost" torque even though the intermediate torque available between the drives is 1/6 of the first input!

I don't want to turn this into some kind of contest. I just think that you are incorrect on this one.

Rod

I think the point with the genset is that they are set to a specific kv oputput.

So if it takes 2 hp to make 1 kv, it ain't going to matter if it's spinning at 540 or 3600 rpm. Both are going to output 1 kv. We are not taking a 540 genset & trying to speed it up like you do with a fan. We are just trying to increase the shaft speed to get to 3600 rpm, where we will get the 1 kv.

1kv = 2 hp. Shaft speeds are irrelevent.

To use the fan example, if we speed up the shaft, we would also be trimmimng down the fan size so as not to change airflow - and thus remain at the same motor size.

I think. But I'm just a simple dirt farmer, this starts to hurt my head. :)

--->Paul

For some reason I had it in my head that the HP requirement was BEFORE the gear reducer when in fact it's after the gear reducer.

Sorry about that guys!

T_Bone