Increasing dc motor speed by armature control 1

[edison123]

My client has a 550 V, 1500 HP, 1000 RPM DC motor for a rubber mill. Since the motor is not working well in field weakening mode (sparking issues), he wants to increase the motor speed to 1200 RPM by increasing the armature voltage to 660 V. The DC drive is rated for 700 V max.

Is this advisable ?

 

[aolalde]

Edison123:

That will be 20% over speed (in the border limits), the motor can work with that, but the problem is; increasing the speed the power demand of the load will increase too. For a rubber mill probably at constant torque the HP demand will be 1200HP.
The combination over-speed and over temperature due to overload could soften the mechanical support of the armature winding, leading to a catastrophic failure.

Whoever, if the cooling volume can be improved, or the motor design and construction are conservative and the temperature rise is not over the insulation safety operation limits, it could work.

 

[edison123]

Thx aoloade.

I forgot to mention that the motor was rated for 1500 RPM in the field weakened mode. But due to serious sparking in this mode, the motor could not run beyond the base speed of 1000 RPM. Inadequate field stength seems to be the problem with field weakening.

BTW, what about bar to bar voltage if the armature voltage is increased to 660 V ? The motor has 6 poles, 63 slots and 189 segments. I calculated the bar to bar voltage as 17.5 V at 550 V and 21 V at 660 V. Will this create sparking at 660 V?

 

[aolalde]

Edison123:

For a commutator with 1mm thick micas the maximum recommended voltage per bar is 40V.
The volts per bar you calculated are affected by the ratio of the pole arc to the pole pitch (around 0.7) assuming your machine has pole face compensating windings.

At 660 V armature: Volts/bar = 660*6/(189*0.7) = 29.93 Volts.

 

[edison123]

Thx aolalde. This motor has pole face comp. windings.

Is there any reason why volts/bar is affected by pole arc/pole pitch ratio ?

I had thought it was just a straight forward calculation of armature volts / Segments between two brushes since the voltage is applied between these brushes.

 

[aolalde]

Edison123; you are right, since this is a motor, the voltage between brush arms is the line voltage V and then the maximum volts per bar are same as the average volts per bar.
Vave = Vline*p / comm. bars
When the machine is a generator, only those conductors under the pole faces generate voltage and the peak voltage between those active bars is higher than the average.