Ch. 5 - CHAPTER 5 Rotary Actuators Fluid Power Circuits and...

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1 CHAPTER 5 Rotary Actuators Fluid Power Circuits and Controls, John S.Cundiff, 2001 INTRODUCTION ± Concepts developed for pumps are applicable to hydraulic motors. ± Motors convert fluid energy back into mechanical energy and thus are the mirror image of pumps ± Typical motor designs are gear, vane and piston.

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2 INTRODUCTION ± Motor performance is a function of pressure. ± As Pressure increases, ² leakage increases ² speed decreases ² quantity of mechanical energy delivered to the load decreases. INTRODUCTION ± Motor volumetric efficiency is e vm = Actual motor speed Theoretical motor speed ± Pump volumetric efficiency is e vp = Actual flow Theoretical flow ± Purpose of the pump is to produce flow. Load sets the pressure. ± Purpose of the motor is to receive this flow and reproduce rotary motion.
3 INTRODUCTION ± Simple example : ² Suppose a motor has a displacement of 3.9 in 3 /rev. Measured flow is 10 GPM. The theoretical output speed is N mth = 231Q V mth where N mth = theoretical motor speed(rpm) Q = flow (GPM) V mth = motor displacement (in 3 /rev) INTRODUCTION ² Substituting, N mth = 231(10) = 592 rpm 3.9 Assuming the measured speed is 536 rpm , the motor volumetric efficiency is e vm = 536 x 100 = 90.5% 592

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4 INTRODUCTION ± The overall efficiency of a hydraulic motor is e om = Actual output power Input power ± Input power is hydraulic power measured at motor inlet port and the output power is mechanical power delivered by the motor output shaft. INTRODUCTION ± The motor in previous example (V mth =3.9in 3 /rev) operates at a 2000 psi pressure drop across the ports. Measured flow to the motor is 10 GPM, then hydraulic power input is P in = PQ 1714 Where P in = input hydraulic power (hp) P = pressure drop (psi) Q = flow (GPM)
5 INTRODUCTION ± Substituting, P in = 2000(10) = 11.67 hp 1714 Measured torque is 1080 lb f -in at 536 rpm. ± Output mechanical power is P out = TN / 63025 Where P out = mechanical power (hp) T = measured torque ( lb f -in) N = measured speed (rpm) INTRODUCTION ² Substituting, P out = 1080(536) 63025 = 9.18 hp M otor overall efficiency is e om = ( P out / P in ) x 100 = (9.18 / 11.67) X 100 = 78.7 %

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6 INTRODUCTION ± Torque efficiency describes hydraulic motor performance. e tm = Actual output torque Theoretical torque Theoretical output torque is T mth = P V mth / 2 π where P= pressure drop across motor V mth = displacement (in 3 /rev) INTRODUCTION ± Substituting, T mth = 2000(3.9) = 1241 lb f -in 2 π Α ctual output torque is 1080 lb f -in, torque efficiency is e tm = (1080/1241) x 100 = 87%
7 INTRODUCTION ± Overall efficiency is product of volumetric and torque efficiencies. Volumetric efficiency is e vm = N/ N mth Where N = actual output speed (rpm) N mth = theoretical output speed (rpm) INTRODUCTION Substitution of N mth = 231 (Q/V mth ) Into previous equation gives e vm = N V mth / 231Q ± Torque Efficiency defined by e tm = T/ T mth where T = measured output torque (lb f -in) T mth = theoretical output torque (lb f -in)

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This note was uploaded on 11/22/2011 for the course ABE 5152 taught by Professor Burks during the Fall '08 term at University of Florida.

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Ch. 5 - CHAPTER 5 Rotary Actuators Fluid Power Circuits and...

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