Mitchell21
Mechanical
I'm sizing a hydraulic pump and DC motor and I'm surprised about the electrical power in v's hydraulic power out I get in my calculation. I'd apperciate if someone could confirm if this is right.
I have been given motor data at 24V DC:
4950 rpm using 6 Amps when 0 Nm torque
3150 rpm using 90 Amps when 3 Nm torque
The pump displaces 0.306 cm^3 per rev. Assuming perfect efficiencies to keep things simple, when the hydraulic pressure is 210 bar, the pump will need 0.306 cc/rev x 210 bar = 1.024 Nm of torque to turn the shaft. Linearly interpolating the motor data gives a speed of 4336 rpm at this torque, and with the given pump displacement, the flow rate at this speed would be 1.33 liter/min. Linearly interpolating the motor data gives a current of 35 Amps at this torque.
The hydraulic power is pressure x flow => 1.33 liter/min x 210 bar = 465 W
The mechanical power is torque x rot. speed => 1.024 Nm x 4336 rpm = 465 W
The electrical power is voltage x current => 24 V x 35 Amp = 840 W
So only 55% of the electrical power is converted to hydraulic power. I know that the motor gets hot due to losses. (That's another issue). Considering losses in the pump, the actual efficiency would be even less. Is that sort of efficiency typical in DC motor systems? Or is there something wrong with my calculation or the motor data?
Appreciate any comments!
I have been given motor data at 24V DC:
4950 rpm using 6 Amps when 0 Nm torque
3150 rpm using 90 Amps when 3 Nm torque
The pump displaces 0.306 cm^3 per rev. Assuming perfect efficiencies to keep things simple, when the hydraulic pressure is 210 bar, the pump will need 0.306 cc/rev x 210 bar = 1.024 Nm of torque to turn the shaft. Linearly interpolating the motor data gives a speed of 4336 rpm at this torque, and with the given pump displacement, the flow rate at this speed would be 1.33 liter/min. Linearly interpolating the motor data gives a current of 35 Amps at this torque.
The hydraulic power is pressure x flow => 1.33 liter/min x 210 bar = 465 W
The mechanical power is torque x rot. speed => 1.024 Nm x 4336 rpm = 465 W
The electrical power is voltage x current => 24 V x 35 Amp = 840 W
So only 55% of the electrical power is converted to hydraulic power. I know that the motor gets hot due to losses. (That's another issue). Considering losses in the pump, the actual efficiency would be even less. Is that sort of efficiency typical in DC motor systems? Or is there something wrong with my calculation or the motor data?
Appreciate any comments!