How do I size a hydraulic motor for real world conditions?

jgibbens · Feb 2, 2012

I'd like to try running a small 3600 rpm generator with a hydraulic motor on a power beyond loop. I need to figure out the right displacement size for the hydraulic motor but from what I've been reading it isn't going to be as simple as I thought... The tractor is a kubota l3200. The utility output is supposed to be 6.5gpm and I *think* that's rated at the same rpm where the pto will spin at 540 rpm and also at no load. What I don't know is how much the flow rate will change under a load. (volumetric efficiency?) I'm assuming that as the pressure increases, there will be some loss in flow rate. I need to find the flow-rate/psi/tractor rpm tradeoff that will get the most HP out of the pump and match the hydraulic motor displacement so that at that flow rate, the generator will be spinning at the right rpm. If I assume no loss, I think it would be 6.5gpm X 2250PSI X 0.000583 = 8.5HP and then take 6.5gpm X 231 in^3/gal / 3600rpm = 0.417 Cubic inch displacement for the motor *if* the tractor can really put out that combination of flow and pressure and the motor is also has no loss.

So how do I adjust that for real-world conditions?

Sorry that was kinda long-winded.

oldnslo · Feb 2, 2012

jgibbens said:
I'd like to try running a small 3600 rpm generator with a hydraulic motor on a power beyond loop. I need to figure out the right displacement size for the hydraulic motor but from what I've been reading it isn't going to be as simple as I thought... The tractor is a kubota l3200. The utility output is supposed to be 6.5gpm and I *think* that's rated at the same rpm where the pto will spin at 540 rpm and also at no load. What I don't know is how much the flow rate will change under a load. (volumetric efficiency?) I'm assuming that as the pressure increases, there will be some loss in flow rate. I need to find the flow-rate/psi/tractor rpm tradeoff that will get the most HP out of the pump and match the hydraulic motor displacement so that at that flow rate, the generator will be spinning at the right rpm. If I assume no loss, I think it would be 6.5gpm X 2250PSI X 0.000583 = 8.5HP and then take 6.5gpm X 231 in^3/gal / 3600rpm = 0.417 Cubic inch displacement for the motor *if* the tractor can really put out that combination of flow and pressure and the motor is also has no loss.

So how do I adjust that for real-world conditions?

Sorry that was kinda long-winded.

Simple answer is yes there are inefficiency losses in hydraulics. Depending on the pump and motor style you can be looking at 10 - 40 % loss. This will also cause an equal amount of RPM variations in the generator.

Cheap gear pumps and motors are around 80 - 85% each.

Dumb question but I have to ask. Why not use the PTO it will be much more efficient and also maintain the Generator RPM closer to desired speed.

Roy

jgibbens · Feb 2, 2012

Mainly cost and complexity. I already have an old direct drive generator with engine problems and i have most of the hydraulic parts. I looked for a gearbox to go from 540 to 3600 and havent found any that are separate from the generator. It also looks like a belt drive would require 2 steps and be costly component wise. With hydraulics I figured all i would need to do is pull the engine out of the generator and install a hydraulic motor. The application would be intermittent use for my well pump when the power goes or power tools where an extension cord won't reach. What do you all think?

oldnslo · Feb 2, 2012

As a starting point you will need to know torque or HP requirements to drive the generator.
Then you can use the formulae: (torque in-lbs x 6.28) / PSI = CIR
CIR = hydraulic motor displacement in cubic inches per revolution.

Once you know the theoretical motor displacement you can figure input flow required by taking the (CIR x RPM)/ 231 = GPM

Based on this GPM required will your tractor provide enough flow.

NOTES:
1) I did NOT figure any efficiency in these equations. Normally would use 85% for the motor so you need to increase input flow by 15%.
2) Does your L3200 have an oil cooler for the hydraulics? I am not sure if this size tractor is designed for the continuos use of the hydraulics.

Attaching a page with some hydraulic formulas for your reference.

jgibbens · Feb 7, 2012

I looked into this some more and I think the volumetric effeciency will make it very hard to regulate the speed. This might be ok for restive loads but I wanted to drive a well motor.. Thanks for all the help.

oldnslo · Feb 8, 2012

jgibbens said:
I looked into this some more and I think the volumetric effeciency will make it very hard to regulate the speed. This might be ok for restive loads but I wanted to drive a well motor.. Thanks for all the help.

I suspect you may be correct on that. you may also find that the generator has some funky shaft configuration to mate to the gas engine and would be a real challenge to make an adapter. I have seen where one person gutted the gas engine and used the existing crack shaft for his input. The gas engine had the bearings to take the side load from belts. May not have been pretty but it worked.