Slip-Ring Recovery System vs New Synchronous Motor

[jimmynora]

We have an asynchronous slip-ring induction Motor and we are currently evaluating ways to improve our energy efficiency. We have two concrete suggestions so far

Motor-Generator specs
6KV, 25MW
Pf= 0.9
R= 2975/min
f=50Hz

i) Slip-Ring Recovery System
Saving the slip power during the run-down phase of the generator and feeding it back into the grid

ii) Synchronous Motor
Replacing the old motor with a new synchronous Motor

My questions are :

Q1) Which would be the better suggestion (cost effective) and why ?

Q2) Which option would be cheaper ?

Q3) If we decide to install a new synchronous motor which specific type should we opt for ?

 

[LionelHutz]

Last time the motor was 2.5MW????

What do you mean by "run down" phase. Before you wrote about allowing it to coast. Then, you wrote about using the resistors for speed control. I believe you also posted that you have to do these tests at 3000rpm but I'm not even sure that is correct anymore. Frankly, no-one here has any clue how you operate this thing.

You should first optimize your existing system by ensuring the motor accelerates promptly to full-speed and by never running it at part speed. Then, measure the power while the machine is running idle at full speed. That is the only operating condition where you should be concerned about saving energy.

The slip recovery system would only save some starting energy compared to using resistors. At 3000rpm, all the resistors will be shorted with no losses meaning the full-speed system efficiency with a slip recovery system will be worse compared to the resistors. Idling the motor at any speed will use more energy than just removing power and letting it coast between tests.

The synchronous motor could be a few percent more efficient while running. You would have to determine the idle losses and efficiency of your existing motor before you could justify any kind of savings with a synchronous motor. However, you might need a VFD to operate it and that would kill any possible savings achieved at the motor.

My conclusion is that I highly doubt either method will show any reasonable payback period. The #1 rule in energy savings is to optimize what you have before spending a bunch of cash chasing the newest fad that is supposed to save energy.

 

[electricpete]

I take it you were looking at sync motor over induction motor based on efficiency considerations. The sync motor tends to have more advantage over induction motor at high-pole numbers than low-pole numbers. But since you have 2-pole, I don’t think sync motor buys you much. Plus, starting a 2-pole sync motor driving a generator and flywheel (if I remember correctly) is something that will not be trivial.

As Lionel suggested, maybe you can remind us of the complete application requirements facing this motor. What does it drive, what is the duty cycle...

=====================================
(2B)+(2B)' ?

 

[Drivesrock]

Whilst waiting for the complete application.

There are slip-ring energy recovery systems for large slip ring motors - like a regenerative ac drive. But I have seen only low voltage systems as close to synchronous speed the rotor voltage is low enough. This 'drive' would be used to control the speed of the motor within a small speed range of synchronous speed but if the application is a fan this would result in a larger air flow variation range. Some of these drives are also able to inject energy in to the rotor to take it super-synchronous thus increasing the speed range and possibly benefiting the process. Between the drive and the mains will be a step-up transformer.

The power rating of the 'Rotor-Drive' would be perhaps 25% of the power rating of the motor making it much cheaper than an HV variable speed drive connected to the motor stator.

The motor would probably be started via the rotor resistors which will be rated for HV - perhaps a liquid resistor and at close to synchronous speed the rotor-drive could be switched in.

The efficiency savings would be that you now control the motor speed - if working below synchronous - instead of perhaps using dampers.

These 'rotor-drives' will not be suitable to extract power from the motor slowing down.

 

[LionelHutz]

FYI, I just happen to be looking at a 12MW, 900rpm motor and fan application. The rotating energy is 194MJ and if I did the calculation right that is 54kWh of energy. At $0.20/kWh it would be possible to save $11 by recovering all of the rotating energy when stopping this motor. Of course, actually recovering all of the energy is impossible. Just giving an example, since it seems you are talking about installing equipment worth many, many thousands of dollars so you can slow the motor and recover this energy instead of just using the stop button.

 

[jimmynora]

Correction: The Motor is 2.5 MW

@ LionelHutz
The 'run down' phase refers to time when we switch off our motor and let the generator run down from its peak 2975 rot/min to about 1500 and maintaining this low speed by manually turn on/off the motor.

I am aware that during our operation we can only save energy at two points:
i) Improving the efficiency of the motor
ii) Recovering the energy during braking of the generator during the run-down phase

My suggestion was to recover the slip energy and put it back into the grid.

Another colleague suggested it will be cheaper to just put a new synchronous motor. I do no understand how a new motor would reduce the energy demand. Perhaps it draw less energy from the grid to power the generator. Plus there are so many different types of synchronous generator I do not know which one would be the 'better option'.

@Drivesrock

How does the regenerative AC drive concept differ from the conventional slip recovery system ?

Keep in mind we have a 6KV/280A/2500KW Motor-Generator

 

[jimmynora]

Question: Will the motor draw a lesser current to power the generator at a lower constant speed ?

 

[davidbeach]

Just let it run at full speed, unloaded, when not in use. Try it, measure the power consumed during full speed, no load operation. You will probably be surprised how low it really is. Notice, I said measure power, not amps. Ignore the amps, you don't pay for amps, you pay for power.

 

[jimmynora]

Unloaded motor ? Do you mean disconnect it from the Generator ?

 

[davidbeach]

No, if there's no load on the generator there's no load on the motor other than windage and other minor losses.

 

[jimmynora]

The motor runs at an average 900KW when keeping the generator up to speed.

During a test we did today the energy demand was 1650KWh. The test lasted two hours in which the motor was turned off once during the procedure for 15mins (which resulted in the generator running down its speed).

As you can see we are talking about huge energy demands over the month and year.

Slip recovery vs New Synchronous Motor remains the question. Perhaps there is a solution that allows us to combine both (having a feedback to grid system when generator speed runs down) ?


Feedback appreciated

 

[rasevskii]

for jimmynora:

Back to this thresd agin which we thought had been beaten to death on the previous old thread.

I cannot believe that the motor is taking 900Kw just to run the unit at full speed. Are all of the resistance steps in the resistor bank actually being cut out? Ie there a hanging contactor somewhere or other fault? Is there someone there that understands this equipment and how to maintain it?

How do you measure 900KW? Is there actually a wattmeter on the 6KV panel? Or a kWh meter? Or only an ammeter? Has the metering ever been tested recently?

Are you getting your information second hand from other people who may not like you, or are you looking at all this actually yourself?

WE ask again: Does the unit have a flywheel? If so the losses would be higher than if not.

rasevskii

 

[LionelHutz]

I guess it's time to be blunt.

Once you figure out that a motor using no power is always cheaper to run than a motor using some power then you will be much further ahead. Leaving that motor running idle for even 1 or 2 minutes will cost more than hitting the stop button. So my advice is learn how to use the damn stop button.

I also don't believe your power measurements. In your other thread concerning the power metering it appeared you were using phase-phase voltage and phase current to calculate the phase power, which will give you a phase power that is a factor of sqrt(3) too high. After seeing this error, I would have a hard time trusting the numbers not knowing what other errors you might have made.

I posted before that you must figure out the efficiency of that current motor before you can determine if another motor will be more efficient. Otherwise, replacing the current motor is nothing more than a science experiment that has a good chance of going badly wrong for you. Changing to a synchronous motor "on a guess" is an experiment worth at least a few hundred thousand dollars. You'll need a new job when the "energy savings" don't show any chance of payback.

The same as above applies to the current motor. Reduce the use of the resistors as much as possible by having the motor either off on at full speed. Then, that is the most efficient that motor can be. I'd bet a slip recovery system will then not show a payback either. That'd likely be a $50k to $100k project so you might keep your job on that one. Of course, you could just keep running the motor at part speeds to "prove" you found some energy savings which would really be complete hogwash. Well, fellow engineers would consider that solution complete hogwash but I'm sure it would be easy to fool many management types into making it appear you are a hero.

No-one here can answer the question you posed. There is not enough info to make an proper evaluation.

 

[waross]

Time to be even more blunt.
900 kW is about 36% of 2.5 MW. The first reaction of most of us seeing an MG set running unloaded at 36% of full load demand is disbelief. Under one HP we see that kind of inefficiency but once we get into integral HPs the efficiency is generally above 80% or 90%.
There is no way a unit that takes 900 kW at idle (about 1200 HP) will coast for 15 minutes.
Have you ever heard of apparent power or reactive power, or power factor?
Energy demand is not measured in KWHrs. (energy demand was 1650KWh)
Your numbers may make sense if you are measuring apparent power, or through some other error, some multiple of apparent power.
But to make one more try at helping have you looked at the revenue meter? If the unit is taking 900 kW at no load, you may be able to see it on the meter. Check the demand on the meter over 5 or 10 seconds with the kH number with the unit running at idle. Than push the stop button and recheck the meter consumption.


Bill
--------------------
"Why not the best?"
Jimmy Carter