Jump to content

Motor Heating Effect Of Different Starting Methods


Recommended Posts

I was recently in another forum discussion on the use of soft starters, where someone made reference to the (I believe) myth of soft starters reducing motor heating during startup. Some years ago I was in a discussion of the exact opposite claim, that soft starters WILL cause more heating in the motor, and I was directed to a fantastic article written by an Allen Bradley engineer on the various methods of starting high inertia loads. In there, he made a wonderful case for the fact that any fixed frequency starting method has essentially the same net effect on raising the motor temperature, regardless of current limit etc. His main point was that the net slip losses from stand-still to full speed are always the same, only the amount of time can be varied. We all know that it takes a specific amount of energy to accelerate a motor, regardless of how that energy is delivered. So since heat represents energy and energy is a product of power and time, the net heat gain in a motor is going to be the same.


I happen to subscribe to this view, but I cannot find any corroborating information. Marke's paper on starting high inertia loads is good, but does not specifically address this issue, being something rather unique to soft starter applications. Unfortunately, that A-B engineer's paper appears to no longer be available.


Anyone else have any thoughts on this and/or documents I can point to on this subject?

"He's not dead, he's just pinin' for the fjords!"
Link to comment
Share on other sites

Hi Jraef


The contention is correct for the inertial component or the load and is not correct for the work component of the load.


Long start times are associated with high inertias so the load is primarliy inertial. For low inertia loads, the power dissipaqted could be higher, but the extended time allow for some of the heat to be dissipated away from the source.


For the inertial component of the load, the energy dissipated in the rotor is equal to the full speed kinetic energy of the driven load.


At any speed, the power dissipated in the rotor, is equal to the slip losses which are equal to the shaft torque times the slip speed.

In the case of the inertial component of the torque, the torque is "charging" up the energy stored in the load inertia. At half speed the energy dissipated in the rotor is equal to the energy transferred into the load as the torque is equal and the slip speed equals the load speed.

If we integrate the product of torque times slip speed and integrate the product of torque time load speed, we find that the two are equal and independent of the torque being used to accelerate the load.

Hence, the heating in the rotor is a function of the load intertia and not the starting method.


If we now add some work torque, then the rotor heating is increased due to the torque that is driving the load (doing work) rather than being stored in the load inertia.

This adds additional heat dissipation in the rotor. Provided that the work torque is having little influence over the start time, then the rotor power is very close to the full speed kinetic energy of the driven load.

If the work torque at any speed, is significant, then it will extend the start time for a given torque, and thereby increase the power dissipated in the rotor.

If we consider a load that has both an inertial component and a work component, then under full voltage start conditions, the additional heating due to the work component will be minimised. As the start torque is reduced by the application of a reduced voltage starter, the acceleration torque will be reduced by greater than the voltage reduction squared due to the presence of the work torque component.

For example, if we assume that the full voltage torque is 100% and the work torque is 20% (linear across the speed range for the purposes of example only) then at 100% voltage, the torque is 100% and the acceleration torque is 80%. If we now reduce the start voltage to 50%, then the start torque is reduced to 25% and the acceleration torque becomes 5% (25% - 5%). In this case, the acceleration time is going to become 16 times longer rather than 4 times longer and the rotor power will be 4 times as much.


If the start current was now set to 70.71%, then the start torque would be 50% and the accelleration torque woulde be 30%. The accelleration time would now be 2.66 times as long relative to the inertial load component only. The rotor heating would increase by 33% relative to a purely inertial load of the same inertia.


So the application of any reduced voltage starter will not reduce the heating in the motor relative to full voltage starting. If the work torque component is significant relative to the inertial component, the start time will be further extended and the rotor heating will actually be increased, not reduced.


Most loads exhibit a non linear torque curve and so the influence of the work torque component is restricted to a small speed range close to full speed, so provided that the motor and load accelerate freely over the whole speed range, the additional heating due to the work component can be small.


Best regards,

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
  • Create New...