Hydraulic Pump and Motor for Replacing 16HP Gas Engine

[lugnut]

need a hydraulic pump (gpm) ? to drive a hydraulic motor (torque)? - that has to replace a 16 horsepower gas engine - to drive a snowblower at 1000 rpm,s what is the formula for this? pump would be driven at arpox 3,000 rpm

 

[jack action]

The power from the pump, the hydraulic motor, the snowblower and the driving motor/engine must all be the same (taking into account their respective efficiencies). Power is conserve throughout the system.

Equations and calculators available http://www.surpluscenter.com/Hydraulic.htm" .

Torque needed by the 16 hp snowblower @ 1000 rpm is 1008 in.lb (Power = Torque X RPM / 63025).

Assuming 0.85 efficiency, the characteristics of the motor are (you need to know the operating pressure or the GPM):

GPM = Torque X RPM / PSI X 2 X pi / 231 / 0.85

GPM and PSI will be the same for the pump. Power and torque for the pump are (assuming 0.85 efficiency):

Power = PSI X GPM / 1714 / 0.85 --> according to the website: «HP is for electric motors, double this for gas engines.» Multiply by 746 to convert into Watt (electric motor).

Torque (ft.lb) = 5252 X Power / RPM

Displacement of the pump and motor (with their respective RPM):

Displacement (in³/rev) = GPM X 231 / RPM