Large 8-pole induction motor for slow speed application 7

[HenryOhm]

All,

I am investigating the upgrade of an old 5,000Hp 150rpm DC motor to AC for a pump application. Am curious about an "off-the-shelf" induction machine operating off PWM versus synchronous due to up-front price. Appears that an 8-pole induction machine is not stretching this too far but would like to avoid a reduction gear if possible. I assume a PWM could be operated only in the 0-10Hz range to effectively operate an 8-pole 900rpm machine as a 150rpm motor? But, what will be the downsides? Low end speed regulation, how the torque curve may be affected, upstream harmonics...? Does an AFE affect the equation?

Any information including reference materials or whitepapers that might help would be great appreciated.

Thanks!

 

[electricpete]

At very low speeds, the lack of self-cooling of the motor may require derating below rated torque (or else external fan).

Sleeve bearings may require higher viscosity to maintain oil film at the low speeds (rolling element bearings won't care).

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(2B)+(2B)' ?

 

[waross]

As I understand it you want to run a 1200 RPM motor at 150 RPM and develop 5000 HP. To get the torque needed start with a 40,000 HP motor. If you feed the motor double voltage you may get by with a 20,000 HP motor.
You may want to re-think the case for a reduction gear.

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

 

[electricpete]

Yes, good point - would certainly be cost prohibitive.

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(2B)+(2B)' ?

 

[HenryOhm]

Sorry, first, the system would have separate forced air or TEWAC cooling so no issues related to actual RPM and cooling available.

Maybe I need to go back and hit my old textbooks, but I thought that a 5000Hp motor fed from a PWM VSI with the V/Hz curve adjusted to give full voltage at 10 Hz rather than 60 Hz, and with a direct torque type control that would output rated current at 10Hz, could acheive 5000Hp at the reduced 10Hz PWM output?

 

[jpts]

Mechanical power of the motor is angular speed multiplied by torque. Basically if you decrease speed from 900 rpm to 150 rpm, your output power will decrease to one sixth of the rated (at 900 rpm). To get the rated power at 150 rpm, you would have to use 6-fold torque and consequently 6-fold current, which is clearly not possible (for instance because you would have about 36 times higher copper losses in the winding. And industrial induction machines cannot provide 6-fold overloading torque). You could get a bit higher torque by boosting V/Hz curve (torque is proportional to V/Hz ratio squared), but this will easily saturate your core, decreasing your power factor.

One more issue relates to slip, if you have 1% of slip in 900 RPM and rated power, which means 9 RPM, then at 120 RPM you would have 7.5% slip if you apply rated torque. This is because slip in rpm stays constant regardless of speed, if the torque remains constant. In practice this means, that 7.5% of your input power turns into heat in the rotor winding instead of 1%. I would guess that this will cause problems with heating.

 

[LionelHutz]

You must maintain the V/Hz ratio. An induction motor operated with a constant V/Hz is a constant torque machine, not a constant hp machine. HP is directly proportional to the speed and you're thinking about running at 1/6th speed which means 1/6th of rated hp. The good news is you really only need a 30,000hp motor, not 40,000hp.

You need to re-think the gear drive or the motor selection.

 

[jraef]

Maybe I need to go back and hit my old textbooks, but I thought that a 5000Hp motor fed from a PWM VSI with the V/Hz curve adjusted to give full voltage at 10 Hz rather than 60 Hz, and with a direct torque type control that would output rated current at 10Hz, could acheive 5000Hp at the reduced 10Hz PWM output?

It's the part in bold that doesn't fit. The motor is going to provide full torque at any speed as long as you stay within 10% of the designed V/Hz ratio. So if you had a 4000V 60Hz motor, it has a ratio of 66.7:1. If you set the drive to provide full voltage at 10Hz, that is a ratio of 400:1, you would be SERIOUSLY saturating that motor.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[jraef]

Dangit, forgot I can't append posts here...

So to further that thought and tie it to Bill's observation, if you want full torque of a 5000HP 900RPM motor at 10Hz, no problem. But if what you want is the torque of a 5000Hp 150RPM motor at 10 Hz, then you have to start off with that much higher torque value at the motor design point, or a lot more poles in the motor (in this case, 48 poles). Remember, HP is a shorthand expression of "x torque at y speed".

Do your torque calcs first, then determine how you want to get there.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[electricpete]

Mechanical power of the motor is angular speed multiplied by torque. Basically if you decrease speed from 900 rpm to 150 rpm, your output power will decrease to one sixth of the rated (at 900 rpm). To get the rated power at 150 rpm, you would have to use 6-fold torque and consequently 6-fold current, which is clearly not possible (for instance because you would have about 36 times higher copper losses in the winding. And industrial induction machines cannot provide 6-fold overloading torque). You could get a bit higher torque by boosting V/Hz curve (torque is proportional to V/Hz ratio squared), but this will easily saturate your core, decreasing your power factor.

Agreed.

One more issue relates to slip, if you have 1% of slip in 900 RPM and rated power, which means 9 RPM, then at 120 RPM you would have 7.5% slip if you apply rated torque. This is because slip in rpm stays constant regardless of speed, if the torque remains constant. In practice this means, that 7.5% of your input power turns into heat in the rotor winding instead of 1%. I would guess that this will cause problems with heating.

I agree with the constant slip in rpm as we decrease frequency.
I don’t think this results in increased rotor heating. It would be a higher fraction of input power, but input power is decreasing (for constant torque with decreased supply frequency and proportionately decreased voltage). At constant torque, rotor I^2*R should stay relatively constant when decreasing frequency and voltage proportionately.


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(2B)+(2B)' ?

 

[dpc]

I'll just add that you can buy slower induction motors. 8-poles is not the limit.

But this seems like an obvious case where a synchronous machine is a better choice, at least this would have been the case in the past.

 

[jpts]

"I agree with the constant slip in rpm as we decrease frequency.
I don't think this results in increased rotor heating. It would be a higher fraction of input power, but input power is decreasing (for constant torque with decreased supply frequency and proportionately decreased voltage). At constant torque, rotor I^2*R should stay relatively constant when decreasing frequency and voltage proportionately."

Yes, you are right of course on the heating, rotor copper losses are still about the same as in rated point despite of the high (relative) slip. Well, at least efficiency will be quite poor due to this.

 

[waross]

Losses, or percentage losses?
The same losses expressed as a percentage of a 30,000 HP motor will not be as large as when expressed as a percentage of a 5000 HP motor.

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

 

[electricpete]

jpts and I are on the same page. See 2nd excerpt quoted 4 Apr 12 8:34. That particular sub-discussion was examining effect of change in speed~frequency of a given motor (with constant volts/hz and torque).

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[HenryOhm]

Thanks for all your help!

 

[gsimson]

Did you think of replacing the Pump-Motor as a unit? Whether any specific need for running the Pump at 150 RPM? If not we can consider replacing the Pump- Motor operating at 1500 RPM. The size of the Pump will be reduced to 1/10th. The system will operate at better efficiency. The spares for maintenance will be available 'off-the-shelf'.

 

[rasevskii]

Is this a centrifugal type pump? Vertical shaft or horizontal? Is the unit started under load or at no-load (pump dewatered?)...

Why do you need to replace the DC machine? Is a variable speed required? Have you considered a thyristor type drive for the existing machine (or is it already?)

Is this is actually a compressor (not a pump)?

A lot more info is needed for anyone on the forum to give any sensible advice.

If it is in fact a centrifugal pump, a custom built synchronous machine is the best solution, assuming that the pump is started up dewatered. There would be also a power factor advantage as the synchronous machine can supply VARS to the system if needed.

Also what is the operating environment? Is a brushless machine desired? Are there problems with the brushes/commutator of the existing DC unit?

rasevskii

 

[Barry1961]

I agree with jraef. Figure out what the actual torque is. You may have a 5000 hp motor doing a 1000 hp job.

Barry

 

[rasevskii]

@HenryOhm:

I think that after some net investigation that this has to be a floating dredge or dredge ship slurry pump. A very strange piece of machinery indeed due the slow speed and DC drive. Therefore a possible requirement for a heavy start and variable speed.

The suggestions above by myself and others are in that case possibly totally unusable.

Possibly it is some sort of Ward-Leonard drive system from an onboard DC generator...

How about it, HenryOhm, don't keep us guessing. What is this exotic piece of equipment...?

rasevskii

 

[electricuwe]

A typical application where you can find such kind of equipment is pumped storage hydropower. Typically these designs are vertical shaft low speed sychronous motors with a large number of pole-pairs running direcly on line.

However for some special applictions also direct drive systems with inverter are used. But this is usually not done to save moeny, but to fulfill special requirements. You may find such designs for example in direct driven wind turbines - of course working as generator is this case.