3 phase motor VFD frequency / current draw help needed?

Question

I am planning to covert a 1920 Drummond B type lathe to 3 phase motor / VFD combo and am trying to understand the math involved with VFDs mainly relating to frequency and current draw. How do they achieve constant torque? Here's the nameplate from a 3 phase motor I'm thinking of buying-

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So what I think I can deduce from this nameplate:

• Speed is 2860 RPM.
• So torque is 2.75 ft-lb / 3.72 Nm
• Can be wired either STAR or DELTA.
• If speed = 120 x frequency / #poles = 3000. So after losses this must be a 2 pole motor.

My target is 50 - 2500 RPM. I'm thinking out loud now, say I run the motor to max 10,000 RPM (not sure if mechanical limit will be met) by setting the VFD to 166.66 Hz. If I use a 4:1 pulley I will be back to down to 2500 RPM but with four times the torque. What's going on with the amperage?

The motor is cheap and just for kicks here's the Chinesey VFD I plan to use with it. Capable of 8 amps, motor is capable of 4.19 amps @ 230 V.

Thank you.

Answer 1

1. The motor datasheet says 3.4 Nm @2850 RP/M 50Hz. The torque will decrease with speed.

2. With no load you can spin it as fast as the VFD can generate. I did this with a cheapy motor recently and found out about the problems of motor imbalance. The no load speed is of little interest as if there is not enough torque, then it is useless to you.

Answer 2

How do they achieve constant torque?

A VFD provides the motor with the capability of producing rated torque at reduced speed by providing the optimum ratio of voltage to frequency. That provides the magnetic field for which the motor is designed without excessive current. There are a number of techniques for doing that. Many VFDs do that quite well without using shaft speed/position feedback. The technique is called "sensorless vector," "direct torque control," and perhaps other terms that are unique to the various manufacturers. The term "V/F open loop" used in the eBay listing generally describes a VFD that is not designed for constant torque but best suited to drive fans and centrifugal pumps.

My target is 50 - 2500 RPM. I'm thinking out loud now, say I run the motor to max 10,000 RPM (not sure if this is a mechanical limit) by setting the VFD to 166.66 Hz. If I use a 4:1 pulley bringing I will be back to down to 2500 RPM but with four times the torque. What's going on with the amperage?

There is a mechanical limit. The motor rotor is not likely to be well enough balanced to run at 10,000 RPM without excessive vibration. I believe that an IP65 motor is totally enclosed with an external fan attached to the shaft, At 10,000 RPM, that fan will require more than 30 times the power that it uses at 2880 RPM if vibration does not break it apart. The motor bearings may not be rated for that speed even if the rotor is properly balanced for 10,000 RPM.

The motor current will be pretty much proportional to torque if the V/Hz is optimized for constant torque. However with no load, the current only drops to something like 30 to 50 percent of rated current.

If speed is increased above the rated motor frequency with constant V/Hz (increasing the voltage above the nameplate voltage) the current mains constant for constant torque and the motor gives a "free" increase in horsepower. If the speed is increased by increasing only the frequency. The torque capability decreases roughly following a constant power curve. The current remains relatively constant. At some point, generally about 1.5 X rated frequency, the motor will be operating at peak torque (breakdown or pull-out torque). At that point the motor torque capability is below the constant power curve.