Running a motor on Low Voltage

[xJAyx]

Can anyone tell me what will happen if I try to run a motor on a lower than spec voltage?
My application is as follows:
Centrifugal pump
Delta/Y 50Hz 400/690 15kW motor. 1455 rpm
The power system i have is 400/230 3p/4w. The motor is presently running in delta on 400V. I need to slow the motor down and one of the electricians working with me has suggested running the motor in Y. This should work as the load will still be balanced, but what will the power output and new FLA be?
Thanks in Advance.

 

[xJAyx]

Oh I forgot to mention that the motor is able to run at a lower speed without cooling problems so that is not a concern, unless runnign at a low voltage will gnerate excessive heat.

 

[electricpete]

unless runnign at a low voltage will gnerate excessive heat

That seems like a good possibility to me. As a first approximation, the speed is constant with voltage change, so the driven load will demand the same torque, so the motor will demand the same power. So the load component of the current will increase by the ratio that the voltage decreased. If load is anywhere near nameplate horsepower, the motor might overheat or trip overloads.


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[sreid]

The important part of electricpete's reply is that the motor speed will not change going from Delta to Wye. The motor speed is determined by line frequency and number of motor poles.

 

[gepman]

There is a manufacturer that markets a device called an "evaporator fan controller" which reduces the speed of a three phase motor when no refrigerant liquid is flowing to the evaporator by reducing the voltage to the motor. I have been to his shop and observed testing for speed of the fan and power to the device. It does work in that it reduces the speed to about 20% of synchronous speed and reduces power by about that much (80%) also (if it was a VFD you would expect power reduction to be about 99% neglecting VFD efficiency). The manufacturer has a patent pending and claims that it does not harm the motor. Of course it is a variable torque load in a cool or cold environment. They would not let me measure the voltage at the motor terminals. Since the unit did not vary the frequency (my assumption since it I could not measure any properties at the motor terminals) I assume that the slip increased greatly. They have provided units for up to about 20hp fans I believe.

Small reductions in voltage just tends to increase the amperage to the motor.

 

[xJAyx]

Sorry, I didn't mean actually reducing the voltage supplied, but rather the coil voltage. In Delta, each coil gets 400V, in wye it gets 230V. It definately slows the motor down, and draws the exact same current under the same load. I'm trying to understand why it slows down, but it's been a long time since I had to deal with motor calcs. Can someone provide a good online resource for this or give an explaination?

 

[7anoter4]

Hi xJAyx
I think the centrifugal pump does not require a constant torque that means the same power for a certain speed. I agree with gepman VFD is the best solution. Also I agree with centripete if the torque would be constant.
I think the centrifugal pump torque is [approximative] proportional with the square of speed[n^2] .Reducing the voltage from 400 to 230 [delta to Y] the motor torque decrease with square of voltage [U^2]. Then the motor and the pump torque curve should meet at another speed.I am afraid this time on the instable region .That means the voltage is reduced too much. Attached jpg may illustrate the case I think.
Best Regards

 

[ScottyUK]

When you observe it slowing down it means that the slip has increased. Increasing the slip will increase the rotor losses, which will cause the rotor to get hotter. Rotors are tricky to protect because they obviously rotate and so direct measurement is pretty much impossible on small motors. Sometimes the first sign is the magic smoke getting out, or a blue oxide colour on the surface of the rotor.



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[jraef]

Running it connected in Y with the lower voltage will be exactly the same as starting with a Y-Delta starter, except without the transition to Delta. It will then accomplish the following:
1) The motor torque will be 33% of normal.
2) The speed will remain the same (or try to).
3) Do the math; the motor output kW will decrease proportionally. kW = Tq (N-m) x RPM / 9550

So you have a 15kW motor at 1455RPM, therefore we can determine that your Tq, is 15 x 9550 / 1455 = 99 N-m

With 33% torque, you have 99 x .33 = 33 N-m
So 33 x 1455 / 9550 = 5kW

If the load power requirement is decreased at that time to 5kW or less, then no problem. But if not, even though the motor current is 33% of normal, the motor will see excessive slip and overload.

 

[jraef]

Left out the emphasis that the speed remains the same. Sometimes people perceive that it runs slower, but that's just because the slip is increasing. But because the motor is in Y, the current is 33% of normal as well, so you can run it for a while at increased slip as long as the current remains below FLA. But it's a dangerous edge game.A slight increase in load has a disproportionate effect on slip and you can overload it much faster than you might think.

 

[aolalde]

The voltage reduction (57.73%) due to the wye reconnection will drop the torque similar to the sample lower curve. The load torque will produce high speed SLIP. If the load torque exceeds the new reduced breakdown torque, the motor shaft will stall and the current goes to the locked rotor condition ( around twice the normal full load current for the original voltage). The winding will burn if it is left for extended time under that condition.