Need help on rotary broom hydraulic conversion

[locknut]

Need advice on system components and layout for a C-48 Two-Way Sweepster broom.The pump would be driven off the 2000rpm pto at the front of the tractor.The hyd motor would need to turn around 540rpm to the broom drive.Long story short,I didn't discover a busted gear in one of the gearboxes till late in the restoration process.For power requirements,stand alone sweepers this size are commonly powered with 6-8hp gas engines and this particular broom was sold new to use use on 10-12hp garden tractors back in the 70's.The sweeper will be used primarily in cold weather for snow.

This is where the pump needs to be mounted,aprox 5" x 5" space.Driveshaft yoke is 3/4" w/3/16" keyway.



And the location of the motor.Currently has 7/8 hex shaft for sprocket drive but I could change that to 1" shaft,or mount motor with higher offset using another hex sprocket and chain loop to the motor.



Here's a pic of a conversion that someone done years ago.Apparently the stacked Durst boxes were not reliable.I did find a specialty shop in Ohio that had three of these and every one had the same busted gear as the one I have.

 

[oldnslo]

locknut,
Here is how I would figure component sizes required.
NOTE: These calculations are based on 12 HP direct drive to the sweeper with no gear reduction. If there was gear reduction the input torque would have to be recalculated.

Torque = HP x 5252 / RPM or 12 x 5252 / 540 = 117 ft lbs torque required to drive the sweeper based on 12 HP mentioned above.

Hydraulic motor size: torque (ft lbs) = motor CIR x pressure x efficency / (24 pi)
transpose this formula to solve for motor CIR (cubic inch/rev)

example: 117 x 24pi / ( pressure drop across the motor x efficency) or 117 x 24pi / (2000 x .85) = 5.2 CIR motor

Pump size required:
example: 5.2 CIR motor turning at 540 rpm requires 2808 CIM (cubic inches of oil per minute) @ 100% volumetric efficiency. 2808 CIM/ 2000 RPM = 1.4 CIR pump @ 100% efficient

 

[locknut]

locknut,
Here is how I would figure component sizes required.
NOTE: These calculations are based on 12 HP direct drive to the sweeper with no gear reduction. If there was gear reduction the input torque would have to be recalculated.

Torque = HP x 5252 / RPM or 12 x 5252 / 540 = 117 ft lbs torque required to drive the sweeper based on 12 HP mentioned above.

Hydraulic motor size: torque (ft lbs) = motor CIR x pressure x efficency / (24 pi)
transpose this formula to solve for motor CIR (cubic inch/rev)

example: 117 x 24pi / ( pressure drop across the motor x efficency) or 117 x 24pi / (2000 x .85) = 5.2 CIR motor

Pump size required:
example: 5.2 CIR motor turning at 540 rpm requires 2808 CIM (cubic inches of oil per minute) @ 100% volumetric efficiency. 2808 CIM/ 2000 RPM = 1.4 CIR pump @ 100% efficient

Thanks for the calcs.I called the surplus center today.The tech wanted to know the torque in/lb requirement before he would recommend a pump/motor.I have no way of knowing that.What he did throw out there is not far from what you figured. A 4.5 cu in motor and a 1.69 cu in pump.Now here's the kicker.He said no way this will work as it takes 28HP to run the combo! Is hydraulic power this inefficient and power hungry?

1.69 cu in CESSNA 26011LZC HYD PUMP

4.5 cu in CHAR-LYNN 101-1002 HYD MOTOR

 

[J_J]

22 GPM at 3000 psi will require about 38 HP.

The pump is not matched to the motor.

The motor needs about 15 GPM continuous

The motor is also 1800 psi continious

 

[Leejohn]

I have a motor that came off of a broom and I think it's a char-lynn. I can take a look and see what size it is. seems like the last #'s were 1008. I do know it was chain drive stright from motor to broom.

 

[MossRoad]

I don't understand how anyone came up with any numbers at all based on the OPs information.

All we know is the following:
An original stand alone unit ran on 6-8HP.
It it turned the broom at 540RPM.
The OP wants to run the hydraulic pump at 2000RPM.

The 12HP figure came from a garden tractor that has to propel itself AND the broom. Therefore, the numbers should be run for 8HP to the broom, not 12.

 

[oldnslo]

Moss,
I was just trying to provide the OP some examples of formulas on how to size a hydraulic drive.

Locknut,
You can figure 20% power loss at a minimum going hydraulic drive Vs direct mechanical drive. Everything attachments shows a 4' x 26" diameter requires 10 - 15 GPM @ 2000 - 2500 PSI. Does not state operating RPM's though.

 

[J_J]

I don't understand how anyone came up with any numbers at all based on the OPs information.

All we know is the following:
An original stand alone unit ran on 6-8HP.
It it turned the broom at 540RPM.
The OP wants to run the hydraulic pump at 2000RPM.

The 12HP figure came from a garden tractor that has to propel itself AND the broom. Therefore, the numbers should be run for 8HP to the broom, not 12.

It is possible to reverse engineer things.

My post was about the recommended parts from the Surplus guys which required to much HP.

If 12 HP was the original HP, then you could reverse engineer the requirement as to the force required to make something work.

12 HP can power a 3 cu in pump at 2500 psi and 7 GPM.

He wants a 540 motor so a hyd motor with a cu in displacement of 3 cu in, would turn at 540

This us at no loss, so if you figured the loss from the pump to motor, the you would need more pump GPM.

Torque would be 3 cu in at 2500 psi = 1194 in lbs.

If you think 8 HP is enough, then 12HP is even better .

8 HP can do 6 GPM at 2000 psi.

This is not .000 accuracy, but close enough.

 

[J_J]

I have a motor that came off of a broom and I think it's a char-lynn. I can take a look and see what size it is. seems like the last #'s were 1008. I do know it was chain drive stright from motor to broom.

Charlynn Hydraulic Motor, Model 101-1008-009

This Charlynn motor is a 22 cu in at 1800 psi. Takes 12 GPM for a speed of 123 rpm.

 

[locknut]

Charlynn Hydraulic Motor, Model 101-1008-009

This Charlynn motor is a 22 cu in at 1800 psi. Takes 12 GPM for a speed of 123 rpm.

That speed would be correct for a motor to drive the brush core directly,which all hydraulic powered brooms use.I don't want to drive the core bar directly as it would put the motor sticking out on the side of the broom.I want to mount it close to the lower sprocket drive shaft where it's better protected from being damaged.The original driveshaft at this location run at around 500rpm or so via the 4:1 reduction gearbox unit,which was driven by the 2000rpm front pto.The chain/sprocket drive on this broom reduces the 540 rpm to right around 135rpm(max) to the brush core shaft.

I received an email back from sweepster today.They had no idea what the in lb torque requirement would be to run the chain drive loop.Just for fun,I put a 7/8 socket on my little 12v master mechanic cordless drill,set it on the hex shaft and lowered the broom to the proper brush pattern.Don't know what the rpm was ,but was enough power to probably sweep the garage floor.Then I hooked up my honkin cheap harbor freight 1/2" low speed drill to it.I think I could lower it in the back yard and throw turf over the fence and dig a hole with it.Neither one of these drills have in/lbs torque ratings.

So,I have a pretty $800 boat anchor at this point. Thanks to all for trying to help.Maybe I can find a solution.

FYI,the Char Lynn 1002 motor that was used in the calcs that the surplus center tech used was given by me,as that is the motor used in the small picture above.I lifted that pic from a thread at WFM and that was the model number stated.He said there were no numbers on the pump.