View Full Version : Electric Motor burnout

SteveF

06-22-2011, 01:38 PM

Here's one for the resident motor experts.

We know that lower than rated voltage causes a motor to run hot. Hot enough and it can be destroyed. I presume it is because the motor is trying to pull more current to make up for the lower voltage but exactly why does this happen?

Thanks.

Steve

winchman

06-22-2011, 01:42 PM

I think it runs with so much less efficiency at slow speeds that more of the power is converted to heat. That's compounded by the fact that cooling is reduced by the low speed, too.

MaxHeadRoom

06-22-2011, 01:50 PM

An AC induction motor attempts to run at synchronism, i.e. the rotor attempts to run at the same rpm as the revolving field, the closer it gets, less slip difference, the lower the current, any load causes the rotor slip to increase, and with it the current.

Therefore anything that causes the slip to increase such as lower voltage will increase the current.

BTW, the rotor can never catch up to the revolving field as the rotor current would be zero.

This is why even after inrush, a stalled or locked rotor is like a shorted turn secondary to the incoming supply.

Max.

ironnut

06-22-2011, 02:29 PM

You are correct in that with a lower voltage the amount of current increases to maintain the same power output. AC circuits actually require more complex equations to calculate the power being used (power factor, etc.) but the DC equivalents are close enough. Power = Voltage x Current (P= VxI). If the Power stays the same then as the Voltage decreases the Current has to increase. Ohms law for DC says that Voltage = Current * Resistance ( V= IxR). If you substitute IxR for V in the Power equation you get Power = Current x Current x Resistance or I squared x R. If you have an electrical cord that has .01 ohms of resistance and you chase 15 amps of current through it then you have 22.5 watts (15 x 15 x .01) of heat coming out of that electrical cord. Bump that up to 20 amps and it is 40 watts. The wires in the motor acts the same way as an electrical cord. Small changes in current result in much larger changes in the power (heat) being dissipated by the windings. If the motor starts to slow down the slip frequency increases which increases the amount of power that the motor will draw as the back emf (electro-motive-force e.g voltage) decreases. The difference between the back emf of the motor and the applied voltage determines how much current the motor will draw. At start up the back emf is essentially zero so the inrush current is very high. As the motor comes up to speed the back emf increases and the current draw goes way down.

Way more than you probably wanted to know.

MaxHeadRoom

06-22-2011, 02:50 PM

Although with an induction motor, rotor current approaches zero as rpm increases or approaches synchronism.

The current in the shorted turns on the rotor decreases as the induced frequency approaches the stator frequency.

Many years ago the rotor conductor bars were made of copper and copper plates on the end of the rotor were sweated on, if the motor overheated and flung the solder, the motor turned in to a large inductor, the rotor did not turn and the current went down to zero.

Max.

J Tiers

06-22-2011, 09:37 PM

The motor may OR MAY NOT heat up.

The motor will tend to run hotter at any given power output if voltage is lower.

Ironnut started the explanation.

However....... for simplicity assume a "constant power" load. (they are less common, but simplify the explanation)

At a given power output, at full voltage, the back EMF is a given amount, because the speed is at nominal, or close to it. That nominal speed is the speed at which the generated voltage (back EMF) due to spinning of the rotor is just enough that the difference between back EMF and line voltage allows enough current flow to provide the required torque. The net power produced is related to the current x voltage x power factor.

If the voltage decreases, the motor must slow down in order to allow enough current flow to provide the required torque.

But that torque would be at a slower speed, and corresponds to less power.

So the torque must be LARGER at the slower speed to provide the same power... and torque tends to speed up the motor.

A balance is found where the speed (and back EMF) allows enough torque that the power output is restored, despite the slower speed.

but the torque will be higher, and higher torque means more current, and so more heating of the windings.

Most loads are not truly constant power, and so they don't act quite the same, but the idea is similar.... because the load is constant enough, and the overall speed change is relatively small.

If the net load on the motor is smaller, the increased current may remain below normal full-load current, and there is no problem.

SteveF

06-23-2011, 06:32 AM

Got it.

Thanks guys.

Steve

Metalmelter

06-23-2011, 07:41 AM

Here's one for the resident motor experts.

I presume it is because the motor is trying to pull more current to make up for the lower voltage but exactly why does this happen?

Thanks.

Steve

Simple >>> Ohm's Law.

The relationship of E(voltage)= I(current) X R(resistance) will never change. You can move the variables around via simple algebra and you can see why increasing one variable will decrease another to maintain the equality in the relationship. So in motor terms - if you cut the voltage in half you will make the motor attempt to draw twice the current (and run real hot) because the load (resistance in the windings) has not changed. Think of the simple starter in your car. It's only a mere 12 volts, yet draws hundreds of amps for a short amount of time before you literally cook the windings ;) It's a bit more complicated in AC versus DC motors but hopefully you will understand the concept here.