Friday, June 9, 2023 7:22:46 AM

Running a hydraulic winch

3 years ago
#18 Quote
A member asked... I am designing a simple hydraulic system for running a winch.  I have designed the system and selected components.  I have also determined the theoretical horsepower and displacement requirements.  Can anyone suggest the best way to calculate actual efficiency for the system?  I am worried that my system may be inadequate if I don’t take into account inefficiencies and losses.
3 years ago
#19 Quote
A typical Hydraulic system is 70-75% efficient.
3 years ago
#20 Quote
Component suppliers should be able to provide efficiency numbers for their products if it is not in their literature.
3 years ago
#21 Quote
Efficiency or Overall efficiency is the combination of taking into account Mechanical losses resulting due to friction and Volumetric losses due to leakage. Mathematically overall efficiency = (Mechanical efficiency or M.E) X (Volumetric efficiency or V.E). Now a winch in its simplest form may have a pump running a hydraulic motor which maybe coupled to a gearbox. So the overall efficiency of this system will be = (M.E of pump)X(V.E of pump)X(M.E of hyd. motor)X(V.E of hyd motor)X(M.E of gearbox). These efficiency numbers can be obtained from respective suppliers

Another way of calculating overall efficiency would be to divide the output power at the load by the input power at the source

Overall efficiency = (Load X velocity) / (Input power from prime mover)
3 years ago
#22 Quote
If you do not have the efficiency of the mechanical components of the winch, you could apply a torque wrench to the input shaft with the winch under a load to see what the actual input torque will be.  With that information, you know the output torque requirements of the motor.  The manufacturer's published information should tell you what the input pressure will be to provide that torque.  That will provide the necessary information for the pressure setting of the relief or for the compensator. That info along with the desired speed (flow rate) will give the required power input.