Snow Auger

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Anonymous said…
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.

5 replies
  1. JDK
    JDK says:

    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

  2. Anonymous
    Anonymous says:

    I have a pump with the specs of- cc/rev
    6-36
    max bar
    110
    max rpm
    4000
    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

  3. JDK
    JDK says:

    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?

    Thank you.

  4. Anonymous
    Anonymous says:

    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)
    PUMP
    DISP – 6.36
    MAX BAR – 110
    MAX RPM – 4000
    MOTOR
    MAX RPM – 969
    FLOW GPM – 12
    TORQUE lbin – 650
    PRESSURE psi – 1800 Thanks

  5. JDK
    JDK says:

    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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