TWO PM ALTERNATORS IN PARALLEL QUERY

studiot

Joined Nov 9, 2007
4,998
It is not sensible to use either Ea or Eb as the reference vector here.

Second you have completely ignored my pointer about the reactance of the machines, so your phasor diagrams do not show these.

Thre is one current I, circulating in the loop and one Voltage, V common to both branches, each containing a machine and its associated impedance. The impedance, Z, is made up of the resistance R and the reactance X

Thus Z = R + jX:

Both I and V are comples quantities and I have shown them bold to emphasise this. They have a fixed phase relationship, \(\phi\). Since we are talking rotating machinery, the current will be lagging by this angle and X will never be insignificant.

I have shown the combined phasor diagram for this situation and also split it into machine A and machine B. This presentation makes things easier to see.
Note
that the vectors V and I are common to all three diagrams,
that Ea>V>Eb
that Ea leads V and Eb lags V
I have chosen the circulating current as the common baseline
 

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b.shahvir

Joined Jan 6, 2009
457
Dear Studiot, :)

Thanks very much. I guess these are the practical phasor diagrams I was after!

Humble request! could you please send me, if its not much trouble, similar phasor diagrams for the paralleled alternators supplying a purely resistive load (on-load); depicting Ea, Eb and terminal voltage across the resistive load Vt.

Second you have completely ignored my pointer about the reactance of the machines, so your phasor diagrams do not show these.

Thre is one current I, circulating in the loop and one Voltage, V common to both branches, each containing a machine and its associated impedance. The impedance, Z, is made up of the resistance R and the reactance X

Thus Z = R + jX:

Both I and V are comples quantities and I have shown them bold to emphasise this. They have a fixed phase relationship, \(\phi\). Since we are talking rotating machinery, the current will be lagging by this angle and X will never be insignificant.

In your 2nd attachment, you have considered stator resistances Ra and Rb alone. There is no mention whatsoever of stator reactances Xa and Xb or impedances Za and Zb for that matter! Hence, I had considered circulating current ‘I’ to be in-phase with the resultant voltage Ec.

Kind Regards,
Shahvir
 

studiot

Joined Nov 9, 2007
4,998
Do you understand the complex equation

V = E ± RI ±jXI

and can you apply it to the circuit in attachment 2?
Sorry if I didn't make it plain enough, I was not trying to catch you out.

paralleled alternators supplying a purely resistive load
Alternators have a huge inductance. You can't ignore it. Real world power generators have multiphase and balancing ( and computers to control it all these days).

It is the inductance which allows the synchronisation process. Energy is stored in the reactive components (don't forget this is impossible in resistive ones) and given back at appropriate times.

All rotating machines, of any number of phases, spend part of any one cycle generating and part motoring. The relative percentage of the cycle determines whether the machine is overall a generator or motor.
When you have coupled machines this mechanism allows transfer of energy from one machine to another by suitable changes to the phase angle.

Have you seen graphs showing this?
 

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b.shahvir

Joined Jan 6, 2009
457
Dear Studiot, :)

I am aware of the effects of synchronous reactance (Xs) on parallel operation of alternators.
‘Xs’ is responsible for maintaining synchronism between 2 paralleled alternators. I have seen graphs showing this.

But I do not intend to venture deep into these aspects as I am presently interested only in practical phasor diagrams like the ones you had sent earlier. I humbly request you, again, to send me practical phasor diagrams for 2 paralleled alternators on-load, wherein Alt ‘A’ takes up more load as compared to Alt ‘B’ (unequal load sharing).

You can easily neglect stator resistances Ra and Rb (as they are practically insignificant, anyways) replacing them instead by pure synchronous reactances Xa and Xb, only.

Attached is a set of phasor diagrams depicting my interpretation of unequal load sharing between two PM alternators in parallel. Please note:- I have neglected stator resisances Ra and Rb as they are practically insignificant.

Fig (a) depicts both the alternators sharing load equally, where, Ea = Eb; jIaXs = jIbXs; Ia + Ib = IL (load current).
Also, Ea > Eb > Vt.

Fig (b) depicts both the alternators sharing load unequally, where, Ea > Eb; jIaXs > jIbXs; Ia > Ib = IL

Input torque to Alt ‘A’ is increased intentionally causing Alt ‘A’ to take up more load as compared to Alt ‘B’. Magnitudes of voltages Ea, Eb and Vt would increase due to increase in both the alternator speeds (and hence busbar frequency). This is what I have attempted to show thru the phasor diagrams.
I humbly request you to comment on the same and correct me if my interpretation is wrong.

Thanks & Regards,
Shahvir
 

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studiot

Joined Nov 9, 2007
4,998
But you already have such phasor diagrams.

Can you not simply add your loads to Ra and/or Rb as appropriate and put some actual numbers in?

Of course you have to get some actual numbers for Xa and Xb. This is where the proceedure for onload/offload measurements comes in, similar to with a transformer. I presume you are familiar with these?
 

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b.shahvir

Joined Jan 6, 2009
457
Dear Studiot, :)

But is my phasor interpretation correct? please comment.

At present i cannot say that the phasors i have attached are accurate or not, as i myself am unable to gauge the practicallity of my phasor diagrams for on-load conditions.

If possible, please do modify or send me correct set of phasors for on-load conditions as you had sent earlier for no-load. I would be very much grateful, since it would give me an insight into the practical happenings in the paralleled alternators on load.

P.S. Please refer to my 'CT Magnetics' thread as i need to understand a bit more about the logic behind the representation of 'magnetic hysteresis' as a resistance. It would help me in visualizing the concept of hysteresis loss.
Sorry for trouble!

Thanks & Regards,
Shahvir
 

studiot

Joined Nov 9, 2007
4,998
If you re-read my recent posts carefully you will see that they contain the answer to your question.

There never has been a generator that does not posess inductance and there never will be. This is what makes them work. So you cannot do the analysis without this.

I suggested you add your load resistance to one or both of the series resistances in my circuit and my diagram.

The actual diagram will depend upon the phase angle which in turn depends upon the inductances. You cannot even say for certain what is leading or lagging without specific information.

You could also add the load resistance in parallel with both generators.

Either way you will need to do some circuit (network) analysis.
You have repeatedly declined to answer whether you are capable of this analysis and I do not intend to say the same thing for a third time without cooperation.

So post your details and we may be able to help further.
 
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Thread Starter

b.shahvir

Joined Jan 6, 2009
457
You could also add the load resistance in parallel with both generators.

Either way you will need to do some circuit (network) analysis.
You have repeatedly declined to answer whether you are capable of this analysis and I do not intend to say the same thing for a third time without cooperation.

So post your details and we may be able to help further.
:) Thanks, but presently would be difficult for me to do rigorous circuit analysis (you might consider it as my incapability). Hence i had expected tentative on-load phasor diagrams like the no-load ones you had sent earlier.

I am also aware the role of inductances in alternator operation, keeping it in mind i had sent you my interpretation of on-load phasor diagrams.

The diagrams you send me may not be exact as a result of rigorous mathematical calculations but resembling the ones you had sent me earlier for no-load.

P.S. While you are at it.... could you please also guide me for my query on Magnetic Hysteresis being represented as a resistance (please refer my earlier posts on the same)
Sorry for trouble!
 
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Wendy

Joined Mar 24, 2008
23,415
We generally try to teach. Reactance isn't a linear function, it uses trigonometry. For understanding electronics, there is no substitute for math.
 

studiot

Joined Nov 9, 2007
4,998
This is my last effort to help, if you don't want to follow what I am saying.

I am redoing the parallel generators from my post#66, but simplifying. The entire circuit (DC if you wish) is resistive to develop a point. This is a graphical procedure to allocate the load distribution between unequal generators.

In Fig1 I show a single supply (generator, battery, whatever) in series with its internal resistance. It has an EMF, E and the arrangement produces a terminal voltage V. E does not vary but V varies with load due to the internal resistance.

In Fig2 I have shown this variation as a graph ( as straight line ) and an equation. Note that negative currents mean that the machine is functioning as a motor, positive currents mean a generator. Note also that the load line crosses the voltage axis at the generator EMF, E - when I is zero.

In Fig3 I have redrawn the parallel arrangement of two generators showing values of E and internal resistance. I have also added a switched load.

When the switch is open there is a complete circuit around the parallel generators and their internal resistances. The total circuit resistance is 3 ohms and the difference in EMF is 10 volts so there is 10/3 =3.3 amps flowing. Remember this is positive in the direction of Ia, the greater EMF and negative in the direction of Ib. This tells us that a is generating and b is motoring, as we expect.

If we now close the switch and apply the (variable) load the terminal voltage V varies with the load (and current) just as with the simpler single generator example above.

The principle of superposition allows us to consider the effects of each generator as if it were the only one in circuit and add the results.
This means I(load) = Ia + Ib (either may be negative)

In Fig4 I have started this process and drawn up the load line (as Fig2) for generator b. The line is easy to draw as we only require two points for a straight line. We already know the voltage intercept (I=0) as 95volts. So if we substitute say 5 amps into the equation V = E - IR we can calculate V = 95 - (5) volts. So a second point is 90 volts and 5 amps.

In Fig5 I have added a second independent load line for generator a. Note these lines cross the voltage axis at their respective EMFs and have different slopes, given by their internal resistances.
Since for parallel operation the terminal voltages must be equal we can either solve

V = 105 - 2Ia or V = 95 - Ib and Ia = Ib simultaneously
or substitute Ia = 3.3 or Ib = -3.3 into either equation. Any of these yield the terminal voltage of 98.3 volts.

In Fig6 I have shown this voltage (98.3) as a dashed horizontal line. Note that it intersects the 'machine a' load line at +3.3 amps (generating) and 'machine b' load line at -3.3 amps (motoring) as previously calculated, as shown by the vertical dashed lines.

This gives us the load parallel combination intercept on the voltage axis, i.e. one point on the output terminal voltage load line. Note I(load) = Ia + Ib = (3.3 + (-3.3)) =0

We need one more point to draw a straight line.

So choose a voltage, say 90 volts.
Draw a horizontal line on our graph, at 90 volts, and note the current intercepts with 'machine a' at +7.5 amps and 'machine' b at +5 amps. Thus I(load) = (7.5 + 5) = 12.5 amps. We could also get this from our formulae.

So the second load line point is at 90 volts and 12.5 amps. Plot this point and join up to the 98.3 on the voltage axis.

This is the load line for the parallel combination of these two machines and I have shown it as a line of crosses in Fig7
 

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studiot

Joined Nov 9, 2007
4,998
It would be worth your while working through my procedure and drawing the graphs for yourself, making sure you understand how they work.

Then change over the internal resistances so that machine a has 1 ohm and machine b has 2 ohms and do the graphs again. It doesn't take long and the result is very instructive.

This procedure can be extended to any number of parallel generators. I have only used straight lines, but you can also use real curves if you know the voltage/current characteristics of your machines and they are not linear.
 

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b.shahvir

Joined Jan 6, 2009
457
Dear Studiot,

Attached is an article depicting decrease in terminal voltage Vt when input torque (steam supply) to one of the alternators is increased (and vice-versa). This is due to ‘Armature Reaction’. :confused:

Now how do I relate this phenomenon to your explanation…….. wherein Vt rises if input torque (steam supply) is increased? Please comment.

Kind Regards,
Shahvir
 

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studiot

Joined Nov 9, 2007
4,998
My first thoughts are that an array of 12 or more 800 megawatt generators make rather more than the usual AAA or CR2032 batteries we see at AAC. This is definitely power station engineering.

My second thoughts are that here is a prime example of why you should read things thoroughly.

My originally posted extract from Hughes referred to an Infinite Bus. This is the sort of material apprentices and technicians studying for day release and night school would use.
It is a very good text because it contains not only introductory level explanations but many practical procedures for such its target audience. In particular the infinite bus is much simpler to deal with and understand so is used as a starting point.

Your article is in much greater depth and specifically refers to a non-infinite bus. It too is very practical, but I would say would be targeted at students who had already finished a two year course of Hughes.

That said, rephrasing your specific question to
Why does the terminal voltage rise in the Hughes example and fall in the Cowling article?

Hughes is making the point that the rest of the bus will 'pull up' the laggard generator if the bus can be considered of sufficient capacity (infinite).

Cowling is basically making the point that if the bus has not sufficient capacity (because the generator is so large) then the voltage will fall. He also specifies particular conditions when this will happen. Note that this is only for a generator, connected but unloaded i.e. zero current.
If you are studying this you should be aware that the armature reaction will increase with current and what Cowling is saying is that if you start to draw current from a supply with droop the voltage will fall.
 

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b.shahvir

Joined Jan 6, 2009
457
Dear Studiot, :)

Okay!….. I have come to understand that there is a thin line between the generators operating with increased Vt or vice-versa due to increase in input torque or effects of ‘Armature Reaction’. However, my understanding of this is a bit different from what you have stated and is supplemented by what I have been reading.

1) If the generator is small, i.e. below 5% of the total bus capacity, then the system bus is considered to be ‘Infinite’. In this case, the bus voltage Vt is considered to be constant as the generator in question cannot influence changes in Vt (since it forms a very small part of the total power system)…….the generator speed equals Infinite bus frequency (which also remains constant).

2) If the generator is large, i.e. above 5% of the total bus capacity, then the system bus is considered to be ‘Finite’. In this case, the bus voltage Vt can vary if the input torque of the generator in question is increased. The bus voltage will then change……but whether it rises due to increased input torque, or falls due to ‘Armature Reaction’ is something which is subject to discussion. Now, the generator speed will rise and the Finite bus frequency will also increase as a by effect.

:confused: Also, one point eludes me in all this!……….during ‘load sharing’ process between two generators, if input torque to say, Gen A is increased, Vt rises but the total load current ‘I(load)’ (and hence output MW) remains constant.
So, my question is……… How will a voltmeter connected in parallel to the load (which now becomes an integral part of the connected load) indicate increase/decrease in Vt if the load current (and also voltmeter current) remains constant ?
I will be extremely grateful if you would be kind enough to comment on the same.
Sorry for trouble!

P.S. Attached are two articles by Cowling for your reference………. one on ‘Infinite bus’ while the other on ‘Alternator Loading’ [ Pg 6; Example 3; Part (b)] depicting increment in Vt (albeit by 1%) due to increase in input torque of generator, excitation considered to be constant. Please comment.

Thanks & Regards,
Shahvir
 

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studiot

Joined Nov 9, 2007
4,998
I have come to understand that there is a thin line between the generators operating with increased Vt or vice-versa due to increase in input torque or effects of ‘Armature Reaction’.
My second thoughts are that here is a prime example of why you should read things thoroughly.............................
....................................Note that this is only for a generator, connected but unloaded i.e. zero current.
Read Cowling's article again.

The situation you keep quoting is only one of three cases quoted and only with the regulator (AVR) disconnected. He also carefully states that the excitation is kept constant as changing this markedly changes the voltage.

If you think about Cowling's example he is describing opposite ends of a range (between no load and full load). Reality is somewhere in between.
 

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b.shahvir

Joined Jan 6, 2009
457
If you think about Cowling's example he is describing opposite ends of a range (between no load and full load). Reality is somewhere in between.

:rolleyes: Guess I’ll then have to go with the fact that at lighter loads, terminal voltage Vt rises if input torque of any one generator is increased and Vt would fall at full load (at heavy load currents) due to ‘Armature Reaction’, as is applicable to a Finite (weak) bus.
Correct me if I’m wrong, but I think that’s what Cowling might be trying to convey……. going by your explanation!

However, I still didn’t get a response to this peculiar doubt of mine……… so I am quoting it again below, albeit re-posited differently;

:confused: During ‘load sharing’ process between two power sources A and B (they might be DC batteries or Generators/Alternators) connected in parallel and sharing a common load…… if the EMF of say, source A is increased, Vt rises but the total load current ‘I(load)’ and hence output ‘Watts’ remains constant (since current shared by source B would tend to decrease proportionately).

So, my question is……… How will a voltmeter connected in parallel to the load (which now becomes an integral part of the connected load) indicate increase/decrease in Vt if the load current (and in the process, voltmeter current) remains constant ? The voltmeter can also be replaced by an additionally connected load, etc.
Awaiting a suitable reply. :)

Thanks & regards,
Shahvir
 

studiot

Joined Nov 9, 2007
4,998
However, I still didn’t get a response to this peculiar doubt of mine
All the good textbooks (including Hughes) on power engineering that I have seen describe the 3-Voltmeter or 3-Lamp technique for bringing a synchronous generator on line. You mention that you have read several.

Read one of these for your answer. However I am concerned that you still haven't understood the point Cowling was making. The statement

The voltmeter can also be considered akin to an additionally connected load, etc.
shows a very basic misconception about voltmeters, especially in conjunction with 800MW generators.
 
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