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It's my understanding that hydraulics are merely a means of gear reduction to move a given object. So then I'm wondering why I can't find a formula or way to determine the following issue:

I want to run a gear pump (specifically a hydraulic motor as a pump; I know it will be more inefficient) with a DC electric motor. The system is meant to run intermittently so will not be subject to constant load like one would find in industry. I can't utilize a large bulky large HP motor so I am trying to run it with a below-spec HP motor that has a large amount of torque. Everything I've managed to find though seems to read as torque is insignificant, the HP rating is the determining factor. Is this correct? If horsepower=torque * RPM, wouldn't a small HP/large torque motor work, just at a lower RPM/speed?

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  • $\begingroup$ What do you mean by "below-spec"? As specified by the gear pump? $\endgroup$
    – jko
    Feb 25, 2021 at 16:48
  • $\begingroup$ hard to understand your question fully without numbers and reference to specific parts and documentation. But very generally, yes you can probably use gear reduction and sacrifice speed for reduced power requirement. But it will depend on the pump mechanism to a degree. You would not have the equivalent performance in the hydraulics, obviously. $\endgroup$
    – Pete W
    Feb 25, 2021 at 19:31
  • $\begingroup$ Let me be more specific: I am looking to utilize a small 2.2 kW BLDC electric motor to drive a hydraulic motor as a less-efficient gear pump to drive a cylinder. The motor has a 0.25 cu. in. displacement and runs 4000 RPM while at 3000 psi. I am calculating 10 hp required to drive the pump, but the motor is only capable of 2.9 hp. I'm wondering if gearing the motor down is a workaround, because I see torque being a more determining factor here than horsepower. Am I wrong in thinking as long as the driver motor's RPM stays above a threshold it can run the system? $\endgroup$
    – Ryan Curtis
    Feb 25, 2021 at 22:23
  • $\begingroup$ The pump will need both a minimum torque and a minimum speed to produce the pressure you need, which determines whether the cylinder can produce the force your system presumably requires, assuming it's more important than speed. I do not have experience working at this scale, but in principle you should look thru the torque-vs-RPM curve for the DC motor, account for the gear reduction, and again the pressure-vs-RPM curve for the pump head. You can probably get away with running the pump head in the 1000 RPM ballpark (wild guess), so see what gear reduction maximizes the torque at that speed. $\endgroup$
    – Pete W
    Feb 26, 2021 at 0:15

1 Answer 1

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The short answer is - always verify power AND torque for any system. Then you never have any surprises.

Pete's guess on speed (1000 rpm) is in the ballpark. Most hydraulic gear pumps/motors have a minimum operating speed around 300-500 rpm. I'll use 800 rpm as a conservative example. At that speed your flow rate would be

Q = (0.25 in^3/rev) * (800 rev/min) / (231 in^3/gal) = 0.86 gal/min = 3.26 L/min

Power at 3000 psi (20.7 MPa) would be

P = (3.26 L/min) * (20.7 MPa) / (1000 L/m^3) / (60 sec/min) = 1.13 kW = 1.52 hp

Your 2.9 hp electric motor has plenty of capacity there. Even allowing the typical extra 15% power for hydraulic efficiency loss, you're still fine. Next you need to check pump torque.

T_hydraulic = (3000 psi) * (0.25 in^3/rev) / (2 * 3.1415) = 119.4 in * lb = 13.5 N*m

The next step is to find the right gearbox ratio. Lets assume you picked a nominal 2000 rpm electric motor. You would need a reduction ratio of 2000/800 = 2.5

Assuming you find one with that exact ratio, your electric motor torque would be

T_electric = 47.8 in * lb = 5.4 N*m

Take that torque value and compare with the speed/torque curve in your electric motor catalog. Then you're good to go.

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