I'm considering building a hydraulic power unit to mount to the back of my Kubota R410 wheel loader so that I can run higher flow hydraulic attachments than the 11.3 GPM flow from the machine would normally allow. If I do this, I'll be reconfiguring a 3 pt, pto driven brush hog to skid steer QA and hydraulic drive. Trying to figure out what size pump (and engine to drive it) and hydraulic motor I should be looking at. I'm finding that brush hogs are normally rated by HP requirement, while hydraulic motors are rated by torque. Not exactly sure how to go about converting those specs. The brush hog is 5 ft. wide and it'll be used to cut tall grass and light brush, nothing real thick. Any ideas?
Hydraulic Pump and Motor Torque Requirements
- Thread starter BlueKnife
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SSdoxie
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First step determine motor RPM needed. Then (RPM X torque) / 5252 = HP
Dave M7040
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The HRS30 unit made by Landpride contains valuable information for your design.
I draw to your attention the input horsepower requirement of 52 HP and the output of 43 HP.
There are significant losses occurring in converting the mechanical power of the pto shaft into hydraulic power.
The same will occur at the Bush Hog end where you are converting hydraulic power back into mechanical power.
Bush Hog's web site says minimum of 20 HP to operate a 5' rotary cutter.
I cannot find exact spec's for a R410 pto shaft HP but you need to be certain there is enough power at that point to do what you want.
The R410 does not seem to be a high powered machine.
I see an engine power of 27 KW which converts to 36 HP..
The pto shaft power will be less.
Using a hydraulic power pack in summer conditions will require extensive hydraulic fluid cooling.
Dave M7040
I draw to your attention the input horsepower requirement of 52 HP and the output of 43 HP.
There are significant losses occurring in converting the mechanical power of the pto shaft into hydraulic power.
The same will occur at the Bush Hog end where you are converting hydraulic power back into mechanical power.
Bush Hog's web site says minimum of 20 HP to operate a 5' rotary cutter.
I cannot find exact spec's for a R410 pto shaft HP but you need to be certain there is enough power at that point to do what you want.
The R410 does not seem to be a high powered machine.
I see an engine power of 27 KW which converts to 36 HP..
The pto shaft power will be less.
Using a hydraulic power pack in summer conditions will require extensive hydraulic fluid cooling.
Dave M7040
The R410 doesn't have a pto. It's a wheel loader, not a tractor. I wasn't intending to run this system off of the loader engine at all. It would have it's own independent engine dedicated exclusively to running the auxiliary hydraulics. After doing some more research though, I'm actually starting to think that it might make more sense from an expense and complication standpoint to skip the hydraulics entirely and use direct drive from an engine mounted on the brush hog itself. It would be fairly easy to rig up the wiring to start and stop the brush hog engine from the seat. Biggest challenge I think would be the throttle. Of course it could just be started and stopped like normal but it'd be safer I think if it could be controlled without getting off the machine.
Dave M7040
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I realized the R410 was a loader. I did not know it lacked a pto shaft.
The independent engine is a wise choice. The throttle will be the easy part.
Most small engines.... 20 to 40 HP want to run at 3,600 rpm unless diesel which are more expensive and will run around 2,000 rpm.
The Bush hog will be designed to run at 540 rpm.
Dave M7040
I must have been half asleep last night when I wrote that. Of course you're right. Can't direct drive it. Belt drive would be the way to do that I imagine. How would you do the throttle? I could rig something up with a couple of pieces of paracord pretty easily, but I'd prefer something more streamlined and "professional" than that.
pmsmechanic
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If you just want to go from idle to full throttle then a solenoid from a power door lock setup would work. If you want to vary the throttle then use a stepper motor.
Dave M7040
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Remember to make provision for shear bolt.
In reality, having considered and worked through a similar speed reduction problem, you would be better off looking for a bush hog like this one where all the work has been done
http://www.bushhog.com/uploads/documents/GT42-G48-SPEC.pdf
Dave M7040
IceStationZebra
Gold Member
Fluid Power Horsepower (hp) = [ pressure (psi) x pump flow (gpm) x efficiency ] / 1,714
Adding to Dave M7040 you have to consider efficiency losses every time you convert energy to another form. And certain types of motors/pumps are better than others in this regard. Gear pumps/motors are ~80-85% while piston pumps/motors will be in the low 90's. In practice the efficiency also varies with speed, pressure, oil viscosity and unit wear. If you want to whole picture you also need to consider flow restriction in the plumbing.
Taking his example and using gear pump/motor: 52hp x 85% (rotation-to-hydraulic) x 95% (line losses) x 85% (hydraulic-to-rotation) = 35.6hp (68% of prime mover)
with piston pump/motor: 52hp x 92% (rotation-to-hydraulic) x 97% (line losses) x 92% (hydraulic-to-rotation) = 42.7hp (85% of prime mover)
Also realize that most of the energy loss is going to heat, so the more inefficient the design the larger the cooling load. 1hp ~ 42.4btu/min
gear pump/motor example puts out 695 btu/min while the piston pump/motor example is only 394 btu/min.
I also like the direct drive version Dave suggested. cheaper, lighter, significantly less maintenance, less fuel used
ISZ
Adding to Dave M7040 you have to consider efficiency losses every time you convert energy to another form. And certain types of motors/pumps are better than others in this regard. Gear pumps/motors are ~80-85% while piston pumps/motors will be in the low 90's. In practice the efficiency also varies with speed, pressure, oil viscosity and unit wear. If you want to whole picture you also need to consider flow restriction in the plumbing.
Taking his example and using gear pump/motor: 52hp x 85% (rotation-to-hydraulic) x 95% (line losses) x 85% (hydraulic-to-rotation) = 35.6hp (68% of prime mover)
with piston pump/motor: 52hp x 92% (rotation-to-hydraulic) x 97% (line losses) x 92% (hydraulic-to-rotation) = 42.7hp (85% of prime mover)
Also realize that most of the energy loss is going to heat, so the more inefficient the design the larger the cooling load. 1hp ~ 42.4btu/min
gear pump/motor example puts out 695 btu/min while the piston pump/motor example is only 394 btu/min.
I also like the direct drive version Dave suggested. cheaper, lighter, significantly less maintenance, less fuel used
ISZ
Would a shear bolt be needed with a belt drive system?
Seems like that would be very easy to damage with a machine this size. Those brush hogs are designed more to be pulled by an ATV or small tractor rather than being attached to the front of a machine that can lift over 4000 lbs.