Need help on rotary broom hydraulic conversion

[MossRoad]

Yes,that tree is still up there.It gets plenty of water from the spring in the clock....:laughing:

AAAhhh!!! That's a good one!!! I laughed hard. Thanks. :laughing:

 

[locknut]

One of the techs at sweepster suggested a motor in the 6-700 rpm range and sized between 4-5 cu in. Pump size 1.2 cu in at 17-1800 psi system pressure at 10-12 gpm. So,I went to the S-Center horsepower requirement calculator and the gpm and psi called for 24-28 HP.Something just doesn't seem right.

 

[MossRoad]

...Something just doesn't seem right.

No sir, it does not.

 

[J_J]

I thought we had already figured out what was needed.

To turn the broom at 554 rpm, you need a 5 cu in motor and 12 GPM.

To develop 12 GPM you need a 1.4 cu in pump, turning at 2000 shaft rpm.

12 GPM, 1800 psi = 15HP

You said something about 8 HP.

8 HP and 1800 psi, .8 cu in pump= 6.5 GPM

6.5 GPM into a 2.8cu in hyd motor = 536 rpm

So what HP engine do you have?

 

[oldnslo]

Locknut,
You have to determine either how much torque is required to drive this broom and figure component sizes based on torque required or how much HP you have available to drive this sweeper and size components based on available HP. One or the other or a combination of both must be determined before final size of components can be determined..

Could you just drive this straight from the front or rear PTO via gearbox or jack shaft, chain and sprocket arrangement. you would only have one direction of rotation this way unless you installed a reversable gear box.

 

[locknut]

No sir, it does not.

No,it doesn't,considering 16-18 HP Case-Ingersoll-John Deere garden tractors run hyd tillers that work well,and in my mind,would take much more torque to operate.The only difference is those tillers are driven right on the shaft at around 100-150 rpm,same as most broom shafts run.The only difference in my setup is running a higher speed motor to power the chain drive,instead of the broom core shaft directly.

 

[locknut]

I thought we had already figured out what was needed.

To turn the broom at 554 rpm, you need a 5 cu in motor and 12 GPM.

To develop 12 GPM you need a 1.4 cu in pump, turning at 2000 shaft rpm.

12 GPM, 1800 psi = 15HP

You said something about 8 HP.

8 HP and 1800 psi, .8 cu in pump= 6.5 GPM

6.5 GPM into a 2.8cu in hyd motor = 536 rpm

So what HP engine do you have?

I mentioned 8HP as a general power requirement,so,like you mentioned earlier,could estimate a system to have similar power available.

The tractor has a 20HP Kohler.I want to keep the power requirement at a level to operate the broom AND run a 24V alternator on the backside of the front PTO drive(normally runs the deck) to power a 24V heater that sucks up 1hp on low and 2hp on high.

Concerning pumps,would a vane pump work? In the confined space I'd like to place a pump, the top ports look better to plumb and the input shafts match the driveshaft size and key width.

Surplus Center

Specifically this one. V10 7-ring 1.39 cu in.Graphs are here;

http://www.go2hydraulics.com/docs/mfr/vickers/vickers-v10-v20-v2010-v2020-vane-pumps.pdf

 

[MossRoad]

We figured out the HP it takes to drive the hydraulic PTO pump on my Power Trac PT425. It puts out 8gpm at 2500psi. That requires about 11.6hp to drive that pump, as the formula at the Surplus Center is:
HP to drive a hydraulic pump=Psi x GPM / 1714.

So going by the numbers provided by sweepster,

1700psi X 12gpm / 1714 = 11.9hp required to turn the pump.
As long as your front PTO can provide 11.9 hp or better at 2000RPM, you would be fine.

 

[locknut]

Locknut,
You have to determine either how much torque is required to drive this broom and figure component sizes based on torque required or how much HP you have available to drive this sweeper and size components based on available HP. One or the other or a combination of both must be determined before final size of components can be determined..

Could you just drive this straight from the front or rear PTO via gearbox or jack shaft, chain and sprocket arrangement. you would only have one direction of rotation this way unless you installed a reversable gear box.

Well,that's the thing I asked the sweepster techs and they did not know the torque requirement at the 540rpm drive input to the chain loop.From what they suggested,the torque would be around 1100 in lbs. I've looked hard for another gear driven solution and so far empty handed,especially being able to turn the broom 30deg left or right.

 

[locknut]

We figured out the HP it takes to drive the hydraulic PTO pump on my Power Trac PT425. It puts out 8gpm at 2500psi. That requires about 11.6hp to drive that pump, as the formula at the Surplus Center is:
HP to drive a hydraulic pump=Psi x GPM / 1714.

So going by the numbers provided by sweepster,

1700psi X 12gpm / 1714 = 11.9hp required to turn the pump.
As long as your front PTO can provide 11.9 hp or better at 2000RPM, you would be fine.

Your formula works out to that but the calculator at SC says 14HP and to double that for a gas engine.

Surplus Center