50/60 Hz Motor Selection 8

[roydm]

I wonder if you motor gurus could give me a few pointers.
We have a project in China (50 Hz) where we are installing a 200 kW pump. We want to run the pump on a VFD somewhere between 3000 and 3600 rpm.
The pump will draw ~ 200 kW right at the point where the client specify changing from 380 V to 10 kV
Q1) Should we be selecting a 60 Hz motor and underspeeding it or a 50 Hz motor and overspeeding?
Q2) Would the 50 Hz option be the same HP motor as the 60 Hz option?
Q3) If you overspeed a 50 Hz motor does the Voltage go over as well or would we set the VFD for 380 Volts at the new top speed?
Q4)Would a 10 kV 200 kW motor be the same frame size as a 380 V motor?
Q5) What is the relevent cost of a 10 kV motor v/s a 380 Volt one?
Thanks in advance for the help.
Regards
Roy

 

[ScottyUK]

Iron losses occur each time the magnetic flux reverses. If the frequency increases then the number of flux reversals per unit time increases linearly with frequency, and the iron losses therefore increase linearly with frequency assuming flux density remains constant. The V/Hz flux density question is related to core saturation, which is a slightly different topic.


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If we learn from our mistakes I'm getting a great education!

 

[Masbibe]

ScottyUK,

Your story is true for hysteresis losses, but eddy current losses are quadratic depend with frequency. So totaly you have something between linear and quadratic.

Milovan Milosevic.

 

[edison123]

Milovan - Yes, yo do have a point about iron loss increasing with the frequency (from linear up to power of 1.4, IIRR).

But that equation is valid for up to core saturation point only. Beyond saturation, the iron loss shoots up due to flux harmonics. And hence, the requirement that V/Hz be constant. Since the iron loss is a small component (up to the saturation point) in the motor losses, an increase of iron loss due to frequency will be off-set due to inherent increased cooling air flow/turbulence at the higher rotor speed.

Muthu

http://www.edison.co.in

 

[Masbibe]

Edison123,

I agree with you that iron losses are small part of total motor losses. I just wanted to hear from some expert for motor design that this is no big deal increasing iron losses for 20-30 %.

Thanks,

Milovan Milosevic

 

[LionelHutz]

roydm;

The NEMA specs really should not be hard to meet. I was thinking part 31, but both can apply - 30 is for general purpose motors that can be used with VFD's and 31 is for definite purpose inverter duty motors. 31 has some extra requirements which can be important. In any case, if the supplier feels his motors are not capable of meeting these specs then either he doesn't know what his motors are capable of or he knows he's building a very poor motor. Both cases are bad from a users point of view.

On the other hand, I have seen cases where motors from manufacturers that have a decent reputation fail on a VFD too...


Milivan;

I'd say you have a valid point if the motor is being operated right at it's limit. The iron losses do increase and the power required to turn the fan will increase (or it should increase depending on fan design??). So, a motor that can output 200hp at 60Hz probably is not capable of outputting 233Hp at 70Hz.

That is why I recommended just picking a 60Hz motor - it eliminates possible issues like this.

 

[roydm]

Gentlemen
A few final questions

We feel fairly confident but would just like to pass it by you gurus in case we missed something.

Attached is a graph showing how we think the Horsepower, Voltage and Torque relate with the curve (A) not (B)

We assume the horsepower keeps going up as long as there is enough voltage to maintain the V/Hz relationship or does it flatten off at 50 Hz?

Since the line voltage is 400 and the motor is 380 will the voltage at the VFD terminals continue to rise past 50 Hz or does entering the motor nameplate data (380 V @ 50 Hz) mean that at 50 Hz the terminal voltage is already 400?

As our desired operating speed is 3213 rpm with 363 ft lb of torque, will we have enough torque at the frequency required to give us 3213 rpm. How does the slip frequency factor into this?

Provided the HP doesn’t drop after 50 Hz we should be OK, is it reasonable to expect constant HP between 50 & 60 Hz?

Note: As I pointed out earlier, the motor ramps up to speed (3213) and stays there 24/7

Thanks in Advance
Roy Matson
Noram Engineering

 

[LionelHutz]

When you run the motor above 50Hz the curves will not continue as you have graphed. In theory, your graphs are correct.

The available shaft HP = HP produced by rotor minus the bearing, windage and fan losses. So, spinning the motor faster in the "constant HP region" will result in the available HP actually going down and not remaining constant.

Even running with a constant V/Hz I doubt the available HP will increase at the same rate as the over frequency.

We played with running it in the so-called constant HP region on a 5HP motor with an open shaft. We could not run the 60Hz motor above about 75Hz because the motor would "stall" and quit accelerating. So, theory may say one thing but in reality it's not necessarily true. In that experiment, there was 0HP available at 75Hz, not the 5HP that theory suggests.

I thought there was a thread a while back which addressed this in more detail but I can't find it now.

 

[elmotor]

Yes you are correct and as in your graph up to so-called field weakening point (380V@50Hz), motor will work in the constant torque area and the power will increase in proportion of the speed. After this point, motor will work in constant power area and produced maximum torque will decrease in square of the frequency (e.g. (50/70)^2 x Tmax).

Entering the motor nameplate 380V@50Hz will set the field weakening point, meaning that the voltage on the motor terminals is in principle 380V at 50 Hz, but the modulation and voltage losses in the chokes etc. inside the converter will decrease the voltage of the fundamental wave ie. producing the torque roughly saying appr. 5...8%. Therefore many motor manufacturers will give some 90 % of the nominal torque at the field weakening point. In your case 250 HP (185 kW) close to 165 kW, what is the same as in your application, 165 kW@3213 rpm.

Yes you should have enough torque. I suppose that the maximum torque of your motor is somewhere 2.5 x Tn (nominal torque) and if taking into account the freq. and assumed voltage drop, the torque would like ((360/380)^2 x (50/54)^2) x 2.5 = 0.77 x 2.5 = 1.9xTn. Even with the maximum torque of 2xTn, the maximum would be 1.5 x Tn. As your line voltage is 400V, I would set the field weakening point like this, 400V@53Hz, giving 380V@50Hz (minus the voltage losses)and 400V@52.6Hz and above that.

Yes, it is reasonable to expect to have constant HP between 50 to 60 Hz.

LionelHutz, sorry but I don't follow you. Why the 60 Hz motor should not run at no-load with 75 Hz ?

 

[LionelHutz]

Why the 60 Hz motor should not run at no-load with 75 Hz ?

You'd have to ask the motor. It refused to spin faster than about 2250rpm when increasing only the frequency.

I don't follow why you are calculating anything using 2.5 x Tn. This is the torque the motor could produce before stalling but while it is overloaded. You can not expect to run the motor continually above Tn.

I also don't follow your voltage arguement. The voltage drops that occur between the line voltage and the motor terminals is the reason the line voltage is 400VAC and the motor is rated for 380VAC. There is 400VAC available to ensure voltage drops in equipment and wires do not drop the voltage below 380VAC.

That squared ratio jogged my memory and I figured out which thread I was thinking about before.

thread237-247210

There is a frequency ratio squared theory as posted by jraef. This theory goes contrary to the constant HP theory but personally, my instincts tell me it is correct.

There is also a nice curve posted by ScottyUK which shows the torque is actually not reduced linearly with the overspeed but rather follows an exponential type curve. The chart using simple speed ratios to calculate the torque yet the torque is not a straight line. This curve was produced using the constant HP theory.

In the end, this is the calculation I would have done using the squared theory...
Rated torque is 438 ft-lbs.
The overspeed requires 53.5Hz.
(50/53.5)^2 x 438 = 382 ft-lbs

Using the constant HP theory gives 409 ft-lbs instead.

Of course, neither of the above account for the fan requiring more torque at the higher speed leading to annother reduction in the motor torque output capability. I expect the real torque capability is less than both of the above numbers...

I see another interesting (or maybe not so interesting) thing that may be an issue. The motor is listed as Class 1 Div 2 on that pdf - I don't believe you can install a motor in a Class 1 Div 2 enviroment and then apply a VFD. There might be a rule that it is allowed if the motor has been tested with the VFD. Hopefully, someone with more knowledge can comment on this.

 

[DickDV]

Lionel, I don't know why your motor acted the way it did but I assure you that the vast majority of motors will go to at least 25% and most to 50% overspeed with very constant horsepower. In fact, I routinely design machine power trains so the motor runs at 75-90Hz at full speed and have never been unsuccessful doing so.

And, that is to be expected if you can believe the published data from the motor manufacturers. Most publish torque-speed curves that go to 90hz and constant hp is right there to see.

Many inverter-duty motors have nameplate data that spec's constant hp up to 120hz. And, yes, that can be trusted too.

 

[LionelHutz]

Do you have some literature and sample curves from a motor manufacturer which address this?

 

[elmotor]

Well, running a motor test laboratory more than 25 years in one of the biggest motor manufacturers in Europe, this is my experience where theory and practise are daily met. Regarding the voltage, I understood so that the line voltage is 400V and the motor is rated to 380V, so if there 400V available on VFD terminals, why not to use as there is voltage drop in the VFD itself ?

 

[DickDV]

I use the curves from the Reliance/Baldor motor catalog. I don't have a current version in electronic form to send you. I suspect your local Baldor distributor would be able to email you one.

 

[LionelHutz]

Thanks elmotor. Could you clarify something. I believe your calculation was to show the new breakdown torque or peak torque, correct? How would you calculate the continous available running torque at the higher speed?

The theory I know and have proven to myself says that the available torque approximately reduces by the square of the voltage reduction. At 2x rated speed or base speed, the motor is effectively receiving 1/2 of its rated voltage. Squaring 50% means the available torque is 25% of rated. A 380VAC, 50Hz motor overspeed to 100Hz is expecting 760VAC, but still only has 380VAC applied. I have never read an explanation of why a VFD can maintain constant HP above the base speed, only "because it does".

I have seen some typial Baldor special VFD motor curves showing over speed but they didn't seem correct to me. The main issue I had was their curve showed the current was constant from 0Hz to 2x+ the base or rated frequency. That doesn't make any sense to me since the current should drop above the base frequeny. This data was shown for the special purpose Baldor blower cooled or non vented VFD motors. I have never seen data like that for standard induction motors.

 

[DickDV]

Lionel, when the motor reaches base speed, the voltage would normally be right at the nameplate value. The frequency would also be right at the nameplate value (the definition of base speed). At its nameplate hp load, the current also should be at nameplate amps.

When the motor is driven overspeed into the field-weakened range, the voltage can go no higher so it stays the same. The frequency continues up to whatever higher limit is set.

Since the torque has fallen off by the inverse ratio of overspeed, and hp = T x rpm/5250, the hp will remain constant. If the output power is constant, then the input power must remain at least constant too. Since the voltage hasn't changed, at the nameplate hp load the input amps would also not be changed.

Now, we both know that the motor is largely an inductor so it seems wrong that the current doesn't fall off as the frequency rises. I'm sure someone else can offer a more technical explanation than I but, I believe that the increased frequency causes the current vector that represents the magnetization current to drop as expected. It must be then that the torque-producing current vector increases to keep the total current the same. I can't say that I understand why but it must be true.

Someone want to explain why it takes more current to produce less torque as the motor goes over its base frequency? (Keep it basic enough for us ordinary folk, please!)

At some point in the overspeed range, the torque starts falling faster than the overspeed ratio. I suppose this is due to all kinds of losses including windage but mostly magnetic losses. When that happens, the hp also starts to fall and the motor probably isn't very useful above that point.

The motor's peak torque (the breakdown torque) drops as the square of the overspeed ratio. So, if a motor has 200% peak torque at base speed, at double speed it would have none since the continuous torque has fallen to 50% nameplate and so has the overload torque. This fact has caused a lot of disappointment when using inverter-duty motors deep into the overspeed range since it is often overlooked.

 

[itsmoked]

Someone want to explain why it takes more current to produce less torque as the motor goes over its base frequency? (Keep it basic enough for us ordinary folk, please!)

I don't believe it does. You've reached constant hp, it does not require more energy coming in than going out. You said this yourself just above.

As the motor spins faster the windage and fan loads will go up causing some loss in shaft output. Perhaps this is what you're seeing?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[roydm]

DickDV wrote
"Now, we both know that the motor is largely an inductor so it seems wrong that the current doesn't fall off as the frequency rises"

Would be true for motor under load, I believe the inductive reactance drops resulting in a PF change.
Interesting discussion.
Roy

 

[DickDV]

Keith, I know it doesn't take more current overall but, the torque-producing vector must be getting larger if the magnetizing vector is getting smaller and the total is holding constant.

What is happening to make the torque-producing vector get larger as the frequency goes up?

 

[roydm]

This may be a silly question but if you know you want to run a motor slightly overspeed why would you enter the nameplate voltage so it maxes out at synchronous, why not enter nameplate + 10% if you want to run it 10% over.
I know you will run out of voltage perhaps before you reach 10% over but why artificially limit the voltage at nameplate.

 

[ScottyUK]

Dick,

Torque current rises as slip increases in the weak field area? Just thinking aloud, brain ain't fully awake yet.


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If we learn from our mistakes I'm getting a great education!