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ABE 463 Electrohydraulic Systems

ABE 463 Electrohydraulic Systems. Control of Hydraulic Pumps. Goals of this Lecture. Basic Principles of Hydraulic Pump Controls Pressure limiting Load sensing Torque limiting Basic Concept of Pump Efficiency Pump efficiency definitions Key parameters. v o , F o. P 1 Q 1. P 4 Q 4.

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ABE 463 Electrohydraulic Systems

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  1. ABE 463 Electrohydraulic Systems Control of Hydraulic Pumps

  2. Goals of this Lecture • Basic Principles of Hydraulic Pump Controls • Pressure limiting • Load sensing • Torque limiting • Basic Concept of Pump Efficiency • Pump efficiency definitions • Key parameters

  3. vo, Fo • P1 Q1 • P4 Q4 • ni, Ti • M • no, To • EnergyLevel Energy Flow in a Fluid Power System

  4. Power Wasted in a Fixed-Displacement Pump

  5. M 1,000 psi P R H T M Solution to Reduce Wasted Power By controlling the displacement of the pump, we can adjust the amount of fluid discharged from the pump. It will significantly reduce the power loss due to the excess flow.

  6. Pressure Limiting for Displacement Control

  7. Operation of Pressure Limiting

  8. Power Saved through Pressure Limiting

  9. Load Compensation Displacement Control

  10. Operation of Load Compensation

  11. Power Saved through Load Compensation

  12. Torque Compensation Displacement Control

  13. Operation of Torque Compensation

  14. Power Saved via Torque Compensation

  15. Pump Power Output HPout =Q·P/1714 Q = flow rate, in gpm P = pressure, in psi Theoretical Pump Power Output In theory, pump power output can be determined based on its output flow rate and outlet pressure. Power can be measured by: horsepower (hp) Kilowatt (kW)

  16. Pump Input Power HPin =T·n/63025 T = Torque load on the power, in lb.-in n = engine speed, in rpm Theoretical Pump Power Consumption As any other machinery, the hydraulic pump needs an input power to drive the pump to produce an output flow and to overcome the load (outlet pressure). Input power is also measured by: horsepower (hp) Kilowatt (kW)

  17. Pump efficiency Effoa = (HPout/ Hpin)x100% Effoa = EffvolumexEffmech Effoa = overall efficiency, in % Effvolume = volumetric efficiency, in % Effmech = mechanical efficiency, in % Pump Efficiency Although modern hydraulic pumps are well designed and efficient, they are not 100% efficient in converting all of the mechanical input power into hydraulic output power. Some of the power is consumed by the pump. We can use pump efficiency to measured the power loss.

  18. Pump Power Output HPout = Q·P/1714 = 8x3000/1714 = 14 (Hp) Volumetric efficiency Mechanical efficiency Overall efficiency Effoa = (HPout/ Hpin)x100% = 14/20X100 = 70% Effvol = (Qcap/Qact)x100% = 8/10X100 = 80% Effmech = (Effoa/Effvol)x100% = 70/80X100 = 87.5% Example: Determination of Pump Efficiency One pump takes 20 input horsepower to delivery 8 gpm fluid at 3000 psi. The pump capacity is 10 gpm. Try to determine pump efficiencies.

  19. Effmech=85% Effvol=80% Effoverall=85x80=68% Exercise: Estimating Pump Efficiency

  20. Lecture Summary • Discussed the basic principles of three commonly used pump control methods, as well as efficiencies: • Pressure limit • Load sensing • Torque limit • Pump efficiency

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