Current on rotor connected to rheostat

[philly]

I have working with a 4.16kV wound rotor motor with the rotor connected through slip rings to a rheostat during starting. Obviously after the motor has started the leads coming from the rotor are shorted together in the rheostat.

My question is after the rotor leads are shorted together is there still voltage and current on these leads. My question regarding the voltage is that yes there is a voltage (referenced to ground) on these leads because there is still a voltage on the rotor proportional to the stator voltage, slip, and transformation ratio "a" between the stator and rotor.

Now what about current. I know in a squierel cage induction motor there is current circulating in the rotor as a result of the induced voltage in the rotor and the fact that the rotor is shorted on both ends. I'm not exactly how this current interacts for the three phases because I guess the rotor cant be viewed simply as three different phases but rather as all three phase voltages and currents somehow interacting on a single rotor surface. How do the different phases interact with one another if it is one surface and not broken into the different phases?

So would the same apply on the cables between the rotor slip rings and rheostat when they are shorted. Can these be viewed as three different phases somehow having current circulating on them? For the slip ring connections tapped off of the rotor, are they somehow tapped off of seperate phases or sections of the rotor or are they all tapped off of essentially the same rotor surface but just termed the "a, b, and c" rotor phases?

 

[mcclary's electrical]

Yes, there is still voltage and current once the leads are shorted. When up tp full speed, and the rheostat has been fully compressed, the shorting contactor kicks in. Once that pulls in, the rheostat is bypassed, and the rheostat can be openend back up to it's strting position. All the shorting contactor is doing is bypassing the rheostat after starting. Is this a liquid rheostat? Salt water? And the slip ring are tapped off differant phases. How big is this motor?

 

[philly]

Yes, there is still voltage and current once the leads are shorted. When up tp full speed, and the rheostat has been fully compressed, the shorting contactor kicks in. Once that pulls in, the rheostat is bypassed, and the rheostat can be openend back up to it's strting position. All the shorting contactor is doing is bypassing the rheostat after starting. Is this a liquid rheostat? Salt water? And the slip ring are tapped off differant phases. How big is this motor?

This liquid Rheostat is a Salt Water Rheostat and is controlling the rotor on a 6500HP Ball Mill Motor.

I thought there was both voltage and current on the Rheostat when up to full speed. Thanks for the confirmation.

I am trying to visualize how the different voltages are established on the rotor, and how the different phases exist and are tapped as you mention. I'm trying to see how these different voltgaes are set up on the rotor in order for current to be established on the cables from the rotor both during starting and during full speed operation.

 

[philly]

I noticed when looking at this today that there appear to be cables coming from inside the rotor when looking at the end of the rotor/shaft. These cables then connect onto (3) small pieces of what look like bus which are then welded to the part where the (3) slip ring connections are. Are these cables simply an extension of the rotor through the ODE shaft?

 

[mcclary's electrical]

I noticed when looking at this today that there appear to be cables coming from inside the rotor when looking at the end of the rotor/shaft. These cables then connect onto (3) small pieces of what look like bus which are then welded to the part where the (3) slip ring connections are. Are these cables simply an extension of the rotor through the ODE shaft?

Yes, that's essentially what you have. Although I have never seen one done that way. This sounds like an old, low rpm motor. Around 300-400?

 

[philly]

Yes, that's essentially what you have. Although I have never seen one done that way. This sounds like an old, low rpm motor. Around 300-400?

The motor is actually a 1200rpm motor.

One of the reasons that sparked my origonal question was that the box that the slip rings are located in is getting extremely hot. Folks here were saying that it couldn't be getting hot because of rotor current because there was no voltage or current in the cables leaving the slip ring. I argued otherwise saying that there was voltage and current on these rotor cables (although I'm not saying it the cause of heating) and thus asked my O.P. After I asked the question however I became curious as to the physical nature of the 3 phases on the rotor?

 

[winnie]

As you know, the stator has a 'three phase' winding where conductors from each phase are located on specific regions, such that the successive current flow in each phase causes a 'rotating' magnetic field.

The rotor has essentially the same sort of three phase winding, where each phase is located in specific regions. At the terminals of the rotor winding, the output voltage, phase angle, and _frequency_ depend upon the position and speed of the rotor. Essentially the rotor winding 'sees' a changing magnetic field which is the _difference_ between the speed of the rotor and the speed of the magnetic field; if the rotor were at synchronous speed then the rotor winding would 'see' a DC magnetic field.

When the rotor circuit is shorted, the current flowing in the rotor will depend upon load. The voltage will be quite low; whatever is needed to push the current through the low resistance of the rotor circuit. The rotor circuit frequency will be quite low, the slip frequency.

-Jon

 

[Jraef]

Absolutely there is energy coming off of that rotor, and when shorted, all of that energy is passing through the shorting contacts. In fact there is so much energy coming off, there are companies that make energy recovery systems for WRIMs. But back to your situation, "getting extremely hot" is, however nebulous, something to be genuinely concerned with. You may have excessive resistance on the slip ring connections. Given that the owner doesn't seem to understand how it really works, his assumptions may have lead to his not maintaining them.

Sometimes, people ASS-u-me they understand things that they really don't. Bravo to you for wanting to be more thorough.

 

[philly]

Absolutely there is energy coming off of that rotor, and when shorted, all of that energy is passing through the shorting contacts. QUOTE]

On the datasheet for this motor the FLC is listed at 900A. The secondary rotor current is listed at 875A.

The motor typically runs at around 700A. With this motor running at 700A will we see most of the rated rotor current on the rotor cables and shorting block at this operating point? So lets say maybe 500-600A with minimal voltage?

I still would like to see how the rotor is physically configured to seperate the 3 different phaes at one time if anyone has a good visual. Maybe I'll try to sketch something and post to see if I get it.

 

[LarryFine]

I remember calling these wound-rotor motors, and to look at them like a rotary transformer. As the motor starts, the secondary resistors are slowly removed (bypassed) from the rotor circuit, which reduces the rotor impedance and increases the magnetic coupling.

Eventually, the rotor windings are shorted with no resistors in the loop, which makes the rotor behave as if it had shorting bars like an induction motor, and provides maximum torque. The advantage is the low initial starting current and high full-speed running power.

 

[zbang]

Does the "hot box" contain the resistance, and is all of the resistance being cut out/shorted? ISTR that some resistance grids are only designed for intermittant duty. If the last contacts don't complete short the rotor, it'll get mighty hot there.

BTW, the first wound-rotor motor I ran into is (still) turning a 1921 carousel. They used it for the incredibly soft start (10hp 900 rpm geared down to about 5rpm).

 

[Besoeker]

Absolutely there is energy coming off of that rotor, and when shorted, all of that energy is passing through the shorting contacts.

Current yes. Energy other that conductor losses, no.
When the rotor is shorted, slip is low and sE2 is just enough to maintain the required current through that low resistance circuit of which the rotor resistance is a major component.

In fact there is so much energy coming off, there are companies that make energy recovery systems for WRIMs.

Like the static Kramer?
This used as a means of speed control. At full speed with shorted rotor, there is no energy to recover.
If you are interested, here is a paper written by a guy who used to work for me. And I accommodated in my family home.
http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=35428
The Cortina connection is me.

Sometimes, people ASS-u-me they understand things that they really don't.

Quite so.

 

[Jraef]

Like the static Kramer?
This used as a means of speed control. At full speed with shorted rotor, there is no energy to recover.

I left that part out, but of course, it applies to when the WRIM is used in Variable Speed applications, such as pumping. My experience was with a company called Flowmatcher who makes Liquid Rheostat controllers for WRIMS used in the pumping industry. They bought another company called Marathon, who made a slip recovery system for WRIMs. They now incorporate the Marathon Slip Recovery system into a product they call the Wattmiser. My point was, there is enough energy coming back off of the rotor when the speed is reduced, to be worth recovering and putting back into the system to raise the overall efficiency. Where there is energy, there are losses, so for sure, it is possible that a connection box can get warm.

 

[Jraef]

I remember calling these wound-rotor motors, and to look at them like a rotary transformer. As the motor starts, the secondary resistors are slowly removed (bypassed) from the rotor circuit, which reduces the rotor impedance and increases the magnetic coupling.

Eventually, the rotor windings are shorted with no resistors in the loop, which makes the rotor behave as if it had shorting bars like an induction motor, and provides maximum torque. The advantage is the low initial starting current and high full-speed running power.

That's a fairly accurate description.

 

[LarryFine]

That's a fairly accurate description.

Thanks, I think. :grin:

I also remember installing the conductors from the huge drum switch to the grid of cast-iron resistors.

 

[mcclary's electrical]

I remember calling these wound-rotor motors, and to look at them like a rotary transformer. As the motor starts, the secondary resistors are slowly removed (bypassed) from the rotor circuit, which reduces the rotor impedance and increases the magnetic coupling.

Eventually, the rotor windings are shorted with no resistors in the loop, which makes the rotor behave as if it had shorting bars like an induction motor, and provides maximum torque. The advantage is the low initial starting current and high full-speed running power.

That's the way I understand it too Larry. And the advantage is definately the low starting current. Ther's a 2500 close to me that we've maintained for years. It never draws ovver 100 amps starting, that's pretty impressive since it can pull 800 running (short term)

 

[Besoeker]

My point was, there is enough energy coming back off of the rotor when the speed is reduced, to be worth recovering and putting back into the system to raise the overall efficiency.

I was referring to the your comment in post #8:

all of that energy is passing through the shorting contacts

If the shorting contactor is closed, the motor will run at full speed, not reduced speed.

You may have gathered from my previous post, that we design and manufacture slip recovery drives, mostly in the 1,000kW to 5,000kW range and usually for pumps, centrifugal compressors, and fans.
There are some merits in using SRDs for this sort of power range. Motors for LV (say. 400V or 480V) get a bit difficult to make much above about 600kW. For the SRD, the motor stator voltage can be anything convenient. Typically, they are rated for the plant incoming supply, usually 11kV in UK and 13.8kV elsewhere. Being powered directly from the plant incoming supply means that the cost and losses associated with step down transformers are eliminated. The total recovery kVA on a centrifugal load is around one seventh of what it would be with a variable frequency inverter so losses and harmonics are correspondingly reduced. And it's less complex than a VFI of comparable rating.

There are down sides too.
Slip rings and brushes require maintenance. Super-synchronous operation, while possible, adds a level of complication that negates the merit of simplicity that operation in the sub-synchronous confers.

It's horses for courses. I'm not plugging a product.
We do pretty much all types of variable speed drives.
So no particular axe to grind.

 

[philly]

Current yes. Energy other that conductor losses, no.
When the rotor is shorted, slip is low and sE2 is just enough to maintain the required current through that low resistance circuit of which the rotor resistance is a major component.

So then the voltage on the rotor is proportional to the voltage transformation "a" between the stator and rotor, and the slip "s". So there could be a small voltage on the rotor and still a very large current on the rotor and the cables leaving the rotor due to the low resistance of the rotor? Can we expect upwards of 500-600A on the rotor cables when operating near full load?

To make things clear, the shorting contactor is in the rheostat which is located remotely from the slip ring box mentioned above?

Is the rotor voltage listed on the datasheet of 3750V the voltage seen on the rotor when s=1? If so then I guess "a" could be calculated from these values?

Does anyone have a good diagram showing all these voltage and current interactions on a rotor?

 

[LarryFine]

So there could be a small voltage on the rotor and still a very large current on the rotor and the cables leaving the rotor due to the low resistance of the rotor?

To say it more accurately, the low voltage and high current is the (desired) result of the low resistance of the rotor circuit, external to the motor itself. The internal rotor wiring is always of low resistance.

Can we expect upwards of 500-600A on the rotor cables when operating near full load?

I would think so, possibly approaching that of the motor's line current. The starting resistors wouldn't be of such high power capacity if the current was small.

To make things clear, the shorting contactor is in the rheostat which is located remotely from the slip ring box mentioned above?

Yes, it's in the controller. Think of it as the mechanical bypass most dimmers now have when in the full-on position. The switch contacts act as shorting bars.

Is the rotor voltage listed on the datasheet of 3750V the voltage seen on the rotor when s=1? If so then I guess "a" could be calculated from these values?

If "s=1" means the maximum resistance, the starting position, then it's either that, or it's with an open rotor circuit, zero current.

Does anyone have a good diagram showing all these voltage and current interactions on a rotor?

I don't, but as I mentioned earlier, it's basically a transformer whose primary current varies inversely with the secondary's load resistance.

It doesn't overheat at maximum current because the primary and secondary are not mechanically fixed together, rotating instead of burning up.

Any stalled motor behaves very much like a transformer with a secondary short. This motor at the full-speed setting acts as an induction motor.

 

[Besoeker]

So then the voltage on the rotor is proportional to the voltage transformation "a" between the stator and rotor, and the slip "s". So there could be a small voltage on the rotor and still a very large current on the rotor and the cables leaving the rotor due to the low resistance of the rotor?
Can we expect upwards of 500-600A on the rotor cables when operating near full load?

At full load torque, the rotor will take full rated rotor current. This is often given on the motor nameplate or data sheet.
To a good approximation, you can get it from P = sqrt(3) x Vr x Ir/1000
where
P is the motor power in kW
Vr is the rotor open circuit voltage
Ir is the rotor current at full load.

For example, we have a drive on a 3,065 kW machine with a stator voltage of 11kV and a rotor voltage of 2,150V. The above calculation gives a rotor current of 823A. This is pretty close to the nameplate value of 814A

Is the rotor voltage listed on the datasheet of 3750V the voltage seen on the rotor when s=1? If so then I guess "a" could be calculated from these values?

Yes. I have more often seen the rotor voltage given as open circuit voltage at standstill (i.e. when s = 1)

Does anyone have a good diagram showing all these voltage and current interactions on a rotor?

This the starting sequence with a five stage starter.

The resistance bank is divided into five sections. At the start, all five sections are connected in series to the rotor. As the motor speeds accelerates and the rotor voltage drops, the sections are shorted out in sequence until finally, the rotor shorting contactor comes in.