I have a 36 inch snow auger that needs to be run at 900-950rpms. What size hydraulic pump and motor would i need to run it. I have a 2 cylinder 13hp honda gas motor that i am going to use to power it.
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EXAMPLE - Guideline Only:
Torque(in-lb) X Speed(RPM)/63025=HP
Torque(using 950)= 862 in-lbs
Assume 2000 PSI system PSI
Torque=PSI X Motor Disp (cu in)/6.28
Motor Disp = 2.7 cu in/Rev
Flow(GPM)=RPM X DISP/231
Flow = 11 GPM
I have a pump with the specs of- cc/rev
So would i have to find a motor with matching #'s or would this pump not work. Sorry i don't know much about this stuff thx
Sorry, I have no idea what "6-36" means. cc/rev, as I have seen it, is usually stated in a decimal format. Does that mean it is a variable displacement pump with 6 to 36 cc/rev?
Any help out there?
Ok i found a motor and a pump here are the specs let me know if this sounds like it will work to power a 36"blower (needs to run 900-950rpm's)
DISP - 6.36
MAX BAR - 110
MAX RPM - 4000
MAX RPM - 969
FLOW GPM - 12
TORQUE lbin - 650
PRESSURE psi - 1800 Thanks
If your motor needs 12 GPM to turn at 969 RPM, this won't work. Your pump at 4000 RPM will put out 6.7 GPM. 6.7/12 X 969 = 542 RPM.
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