Synchronous motor rotor excitation effect on current draw

[Simon]

I have come across this latest question in my assignment.

A 30 kW star-connected synchronous motor is supplied from a
415 volt 50 Hz source. When fully loaded and running at unity power
factor the input power is 35 kW. The excitation current under these
conditions is 10 A.
What will be the value of input power if the load remains constant and the
excitation
(i) is reduced to 5 A
(ii) is increased to 20 A?

I am baffled. Firstly why state both 30KW and 35KW. What the motor is designed for is irrelevant as the question is not about will it break. It can't be a KW versus VA thing because this is at a power factor of 1.

While I find a lot of reference around about generically under and over exciting the rotor I have not found any explanation on how I would calculate the stator currents from the information given. Obviously the course material was it's usual vacuous confusing rubbish with 1/4 just waffling and 1/4 showing examples totally different to this question.

I would think that I would need a decent understanding of the magnetic's in this case and know the change in load angle.

Can I assume that with a change in excitation current there will be a proportionate change in back EMF that will in turn change the stator currents so that IXs lineV and back e.m.f V phasors all add up again but this assumes no movements in those phasors which I am sure will happen and I can't see how I would calculate the angles.

[jbb]

Hmm

Looking at the question, I think there isn’t enough information for a detailed analysis.

I therefore suggest that you make (and state!) an assumption or two.  And I also note that the powers they gave you were in kW.

[Simon]

Oh they never talk VA, I have to work out from the subtle wording if they mean W or VA. 70% the challenge with this course is dealing with it's own failings rather than trying to learn the subject. I have got to the point of being too scared to pick a new module fearful of the disasters it may hold. I want to be judged on my own mistakes, not theirs.

What assumption can i make? is my idea of back emf proportional to excitation correct? ish?

[T3sl4co1l]

Hmm, guessing:
1. The "30kW" figure is the mechanical rating and load, so, "40HP" if it were over here.
2. 35kW is the electrical consumption, so, 35kVA = 35kW (PF = 1) is the input and presumably the added 5kW is losses.
3. The asked-for "input power" is Apparent Power (VA), and PF < 1 under the specified conditions.

Phase angle varies with rotor current but I don't remember by how much, and also what that does for maximum torque output, if half field at full load is likely to stall the motor or what.  That point may also simply be ignored here.

In any case, I think the expected result is VAs goes up, capacitive or inductive respectively (or the other way around, I forget), while the real power remains ~same (because the load is assumed constant).

If you're curious, I had: Electrical Power and Controls, Skvarenina and DeWitt, Pearson Prentice Hall, 2nd ed. (2004).  I remember some problems like this, but not the formulas for them.  I don't remember them being nearly as nonsensical as your assignments have been...

Tim

[Simon]

Yes at power factor 1 the current in the stator is minimum. So either scenario will produce an increase in current.

Ah hang on they say "input power" this sounds like a trick question. If by power input they mean watts not VA then presumably that is unchanged as the VA will be what changes with the lower power factor.

[Simon]

Right done a bit more digging. I can calculate the load angle as at unity power factor the supply voltage vector and the IXs vector are at right angles to each other. This allows calculation of the back emf and the angle between that phasor and the supply voltage is the load angle. I would have to assume that changing the excitation of the rotor would not change the load angle (unlikely) meaning that if I can make an assumption about the back EMF change (linear to current change?) i can work out the new IXs vector and therefore the new I and P. What 30KW and 35 KW have to do with each other is a mystery unless as Tim say's it could be mechanical power and electrical power but that was a bit sly if they did that.....

[Benta]

When you go somewhere to buy a motor, you specify the power needed at the output shaft. What the motor draws is irrelevant (to begin with).
That's the background of the 30/35 kW confusion. Tricky, but a situation you'll meet in real life. Unfortunately, EEs often ignore the non-electrical aspects

[Simon]

Yes but we are talking about a very basic course here.

[Benta]

Yes but we are talking about a very basic course here.

True, but it's also a very basic fact. The term 30 kW motor always refers to mechanical power. Perhaps it's easier to understand if you talk about a "40 HP motor".

[Simon]

The language in the material is always vague. They do not establish standards and I suspect don't always use the language of industry.

[duak]

I had a different reply here originally but wasn't happy with it.   I googled "synchronous motor line current vs excitation" and got a bunch of interesting results.  One of which is "synchronous motor V curves".  I didn't specialize in power but it did ring a bell once I saw the shape of the curves.  I looked that up: https://en.wikipedia.org/wiki/V_curve

Bottom line, line current increases, power factor changes and in some cases the motor may become unstable and probably stalls.  It looks like these curves are determined empirically and one would need much more detailed information to get anything more than a rough number.

As an aside, my sons are in high school and sometimes have trouble with math and physics.  I swear that sometimes the questions are written in authentic frontier gibberish because they're so hard to figure out.  Once that's done, the problem itself is usually easy.

Cheers,

[Simon]

My problem with this question is that they have not furnished me with the knowledge to answer it properly. So i have to make some assumptions and provide an answer of little practical value. I really wonder with this course if the questions are clever or just lazily put.

[Simon]

I am confused, If i use trig as per below on my V (239.6) and E (338.85) on my calculator I get 294V instead of 239.4, I did this to test the formula before using it on new values and getting a wrong answer.

Yet if i use this online calculator the answer is correct: https://www.calculator.net/triangle-calculator.html?vc=45&vx=239.6&vy=338.85&va=&vz=&vb=&angleunits=d&x=78&y=26

c = √b^2 + a^2 - 2ba·cos(C) = 239.60313

[Simon]

And the answer is that I had my calculator on radians not degrees.

[RudiB]

Hello all, apologies for digging up an old thread but I am stumped also with this question. I can't see a way forward with it and my tutor has basically just told me that it's a trick question ().
Anyway do I actually need to do any calcs for this or just provide a statement, eg. leading PF and lagging PF and that E would change also.
Really appreciate any help as I'm keen to put this question to bed as been stuck on it for a good few hours now.

Thanks

[jbb]

Well, it doesn’t look like you’ve got enough information to do detailed calculations.

So maybe they’re looking for a couple of reasonable assumptions (eg does an excitation current of 10A mean anything special?) and a qualitative answer.

[duak]

I remembered that synchronous motors can be used to provide an adjustable leading or lagging power factor by varying the excitation current.  See https://en.wikipedia.org/wiki/Synchronous_condenser

I don't think the AC losses change much as the excitation is varied so, to conserve energy, the input power should be constant.  After all, it is a synchronous machine that runs at a constant RPM.  The VARs (reactive power) change as per the V-curve.  The stator current is at a minimum when the PF is 1.  I'll bet that the rotor will pull out of lock with the rotating field and stall if the excitation is dropped too low, especially if it's 100% loaded.
 

[RudiB]

Am I right in thinking that the PF will become leading if the excitation is reduced to 5a, and that when reduced to 10a it will be lagging and to the point of becoming unstable?

[duak]

RudiB,

no, see attached V curve graph.  Nominal excitation (10 A) = unity PF.

Over excitation = leading power factor = capacitive reactance.

This graph also shows the stability limit for excitation and load.  Underexcitation reduces the magnetic flux developed and thus reduces the output torque leading to a stall.

[schmitt trigger]

FWIW, I scanned a page of the book Electric Machinery by Fitzgerald, Kingsley and Umans, related to steady state synchronous machine operation.

[RudiB]

thanks for your advice guys, and for the scanned pages Schmitt Trigger, large power electrics really isn't my strongest subject I'm more of a digital electronics person.
Anyway, looking at the V curves, and having them explained, as my course notes make no attempt to explain V curves, would I now be right in saying that at 5a the power factor would be lagging at around 0.8 and the load would be inductive and that the motor would be under-excited, and that at 20a the power factor would be leading by over 0.8 and the motor would therefore be over-excited?

[schmitt trigger]

Correct