> Almost all textbooks assume certain parameters to be constant, eg. terminal voltage, excitation,
> shaft speed, synchronous reactance, etc. in order to represent phasor diagrams.... but in
> the real world this is not so.

Most phasor diagrams use per unit values, in which everything is expressed in terms of the terminal voltage as 1. The synchronous reactance is the voltage drop in the windings at rated current.
In your case, the reactance can be treated as approximately proportional to speed or frequency. So if speed increases by 5 %, the synchronous reactance will also increase by 5 %.

I would suggest you start with the machine at rated conditions, and find the internal emf in per unit terms assuming a terminal voltage of 1 pu. Now, multiply this value by 1 factor to give the effect of speed (1.05 if speed is raised by 5 %) and another to give the effect of excitation current (1.05 if excitation is increased by 5%). Change the reactance in proportion to the change in speed.

This will give a simple voltage - reactance - resistance circuit that can be handled in the usual way to find the load current.

For a PM generator, you can forget (to a 1st approximation) the effect of field current.

 

From what I have understood so far on this topic, in my opinion, the load angle of the machine with lower emf remains the same. This is because firstly, the magnetic coupling strength remains the same & secondly, load angle is a function of driving torque (which is constant), increasing excitation changes the VARs & Vt but not the load on gen-2. The power factor is 'leading' as mentioned.

Thanks for your help

> The magnetic coupling strength for the machine with the
> lower emf stays the same as previous, therefore the load
> angle increases and as a result the machine increases the
> power factor towards "leading".

 

Shahvirb,I would like to raise a point to this response of urs

>the load angle of the machine with lower emf remains the same. This is because firstly, the magnetic coupling strength remains the same & secondly, load angle is a function of driving torque (which is constant), increasing excitation changes the VARs & Vt but not the load on gen-2. The power factor is 'leading' as mentioned.<

If someone decreases excitation or the dc field current, then obviously the magnetic strength is going to decrease and so the alternator has to take more armature current to offset this field drop as per the constant flux linkage theory. and this current will manifest itself as more Ia.Xs drop.vb. If terminal voltage is maintained constant, then this is only possible if machine takes a leading armature component so that armature mmf is magnetising in nature. So just as the power factor turns leading Ia.cos.theta is maintained at its previous constant value->this also indicates Ef. sin. delta is constant at its previous value, that's why real power doesn't change. For constant real power, Ef.sin.del = Ia.cos.theta has to be constant.Now if Ef.sin.del is constant and Ef decrease so del has to increase. i.e., load angle has to change.

Now, the job of maintaining constant terminal voltage goes to AVR. as if terminal voltage is not maintained constant, it will also decrease with excitation emf. and in this situation if both E and V decrease smilarly, then armature drop Ia.Xs will be similarly inclined from V axis in the phasor diagram. thus direction of Ia relative to V doesn't change. thus power factor will remain same, but the fact load angle will also remain same. if alternator was previously feeding resistive load, the power factor will still be unity, but Ef and Vt decrease will obviously lead to both less active and reactive power generated from armature terminals.

 

> In case of 2 loaded alternators in parallel, what will be
> the effect of increasing field excitation of Alt-1 if only
> stator winding resistance is considered neglecting all kinds
> of winding reactances? Will the system operate stably or
> become unstable resulting in pole slipping, etc.?

Coming back to my above question, this is what I have inferred.
Considering only internal stator resistances, if excitation of Alt-1 is increased, load current of Alt-1 will increase, shaft speed of Alt-1 will drop. since stator reactances have been neglected, auto-synchronism between the 2 machines will be lost. As a result Alt-1 will be running slower than Alt-2 which will result in unstable operation and pole slipping in both machines.

Please correct me if I am wrong.

 

shahvirb,

You are wrong.

When two synchronous generators are operated in parallel--that is, when they are <b>SYNCHRONIZED</b>, the speeds of the two must be the same. That's what synchronism means--the speeds of all synchronous generators <b>synchronized</b> together are the "same" (based on the formula N = (120 * F)/P).

I'm sure you would <i>never</i> say that two two-pole synchronous motors connected to a grid running at 50 Hz would operate at different speeds--regardless of the load being driven by either of the motors at any time while connected to the grid. The laws of physics and magnetism dictate they cannot operate at different speeds. (By the way, the only difference between motors and generators is the direction of current flowing in the stator windings. If it's flowing out of the stator windings then the machine is converting torque to amperes and is acting as a generator. If it's flowing in to the stator windings then the machine is converting amperes to torque and is acting as a motor. True story!)

Synchronous generators operating in parallel--regardless of the number of them synchronized together--<b>all operate at speeds dictated by the number of poles (synchronous speed)</b>, per the formula above. (And some people say I don't like maths formulas! HAH!)

If there are two synchronous generators synchronized together (operating in parallel) to supply a load, if the load is stable and the energy flow-rates into the two prime movers are such that the frequency is, say 50.0 Hz, and some well-meaning but ill-trained operator increases the energy flow-rate into one of the two generator prime movers then what will happen is that the load (the total number of lights and motors and televisions and computers and computer monitors) will stay the same for both machines. But the extra energy being input to the one generator prime mover will cause the speed of it's prime mover and generator to increase--which will also cause the speed of the other generator and its prime mover to increase. So, the frequency of the system (the generators and the loads being driven by the generators) will increase. The extra power--over and above that required to keep the system supplying the constant load (see the statement above) will not increase the load--that's the total umber of motors and lights and TVs and such--which doesn't change when the energy flow-rates into the generator prime movers changes. But the extra torque will increase the speed of BOTH generators. Because they are locked in synchronism--<b>synchronized</b> by the magnetic forces inside the generator between the seemingly rotating field of the stator which is keeping the magnetic field of the rotor--and hence the rotor speed--in check, locked in step (speed) with each other. If the speed of one increases, the speed of both must increase.

The above example presumes the two prime mover governors are NOT in speed control (neither Droop nor Isochronous). They are just methods for stabling controlling fuel as a function of the operator's commands. Once the operator has adjusted the governors such that the frequency is at rated then if someone turns off a motor or turns off some lights then if the load (the total of all the lights and motors and TVs, etc.) has changed--but if the operator doesn't do something very quickly then the frequency of the entire system will go up, with the speeds of ALL the generators increasing.

It's really all about magnetism and that one little formula.

As for slipping a pole, that has to do with decreasing the strength of the rotating generator field to the point that the magnetic forces between the stator and rotor magnetic fields can no longer keep the rotor in synchronism--so the torque being applied by the prime mover causes the rotor to accelerate VERY fast for a very short period time--and the load coupling between the prime mover and generator rotor breaks, or worse--as the generator suddenly stops again as the magnetic forces inside the generator try to stop the rotor from accelerating.

That's why loss of field (excitation) relays are so important for synchronous machines.

Alternating current systems are bound by F = (P * N)/120. Synchronous machines (generators or generators) are bound by the same formula when connected to the same grid. Even induction motors are bound by the same formula, though by definition they must have some slip (slight speed "droop" below synchronous speed) to operate. But they don't operate at wildly different speeds (unless driven by adjustable speed drives).

It's magnetism, plain and simple. Attraction and repulsion--though hopefully ALL attraction and no repulsion. And when the total energy input to the generators--all of the generators synchronized to a grid, even if the energy of only a single generator out of many is increased--exceeds the load then the frequency of the system will increase and the speeds of ALL the generators and their prime movers will increase.

Because they are <b>synchronized</b> together via the forces of magnetic attraction at work in the generators.

 

Dear CSA,

I appreciate your enthusiasm for the topic and help towards me. But you have missed a very crucial point in my query. I have neglected all stator reactances and considered only internal resistances! Auto-synchronism between 2 machines in parallel is possible only due to synchronous reactances, in the absence of which the system will become unstable.

The synchronizing voltage Es & resultant current Is are in phase (superimposed on load current if the machines are loaded) as only stator internal resistance is considered. As a result, the synchronizing current does not contribute towards speeding up or down the other machine as both the machines are now in generating mode. The auto-synchronizing feature is absent in this case.

 

Actually, if you leave the emf the same on machine 1, but increase the emf on machine 2. and they are not connected to the grid (both are synchronised, but islanded from the grid) the following will happen.

- Machine 2 will increasing it's kvar output
- Machine 1 will absorb the increase in kvar from machine 2

The energy needs to go somewhere

 

A few clarifications need to be made here, generator speed is only related to the prime mover, not the excitation. The excitation affects the generator voltage only.

In saying that, if you wish to consider the imaginary machine which has no inductance, then you need to appreciate that there is no magnetic field in the machine to couple the rotor and stator. Therefore, if there is no magnetic field, then the machine will not work. If you wish to investigate a power source that doesn't require the magnetic field, then a battery powered system is the most appropriate.

So, if the terminal voltage for each battery system is exactly equal where the load connects, then both battery systems supply the load equally. If the terminal voltage differs, then the battery with the higher terminal voltage supplies the load and charges the other battery until they are balanced.

 

>Actually, if you leave the emf the same on machine 1, but
>increase the emf on machine 2. and they are not connected to
>the grid (both are synchronised, but islanded from the grid)
>the following will happen.
>
>- Machine 2 will increasing it's kvar output
>- Machine 1 will absorb the increase in kvar from machine 2
>
>The energy needs to go somewhere


These two articles below seem to contradict each other;

https://canteach.candu.org/Content Library/20050819.pdf, 4.2, Case-1 (Vt decreases if steam input of G1 is increased at constant excitation, AVR on manual)

https://canteach.candu.org/Content Library/20050821.pdf, Example 3, (b) (Vt increases if steam input of G1 increases at constant excitation, AVR on manual)

Even more surprising is this article below;

https://canteach.candu.org/Content Library/20050821.pdf, Example 2, (b) (If excitation of G1 is increased at constant steam input, Vt increases which causes both P (MW) and Q (Mvar) to rise in proportion.The load will now be consuming slightly more MW (due to the voltage increase) and slightly more Mvar (again due to the voltage increase).

This contradicts the concept that increasing excitation with constant input causes only KVAR to increase, the KW remaining constant!)

 

Well, this opens up another topic on various load sharing methods used for generators. The cases you have presented above are completely different as follows.

In 4.2 case 1 described in "20050819.pdf", the AVR for the generator control system is set to manual and as the load increases, the voltage drop across Xs (synchronous reactance) increases. Since the AVR is not in auto, then the voltage will not auto correct to the set point voltage level and will drop.

In example 3 described in "20050821.pdf", the AVR for the generator control system is set to manual (i.e. excitation level not adjusting to maintain the voltage), then the fuel to the G2 is increased. This increase in fuel will result in G2 taking on more load, and G1 unloading. As "droop" is used for load sharing, this results in and increase in speed / frequency of the generators. Since there is an increase in speed, then the voltage rise is due to the increase in speed (E=4.44.N.MagFlux.f).

In example 2 described in "20050821.pdf", the AVR for both generators are set to manual, and excitation is increased for 1 generator. The resulting rise in Vt causes a proportional increase in power (P=V^2/R and Q=V^2/X).

> This contradicts the concept that increasing excitation with constant input causes
> only KVAR to increase, the KW remaining constant!)

Well, yes you are right if the machines are operating in droop load sharing mode, but no you are not if they are operating in an isochronous (isoch) load sharing mode.

If you are not aware of droop and isoch, in summary, "droop" is when the voltage is decreased as the kvar increases, and the frequency is decreased as the kW increases. "Isoch" is when the voltage remains the same for the full range of kvar capability of the generator, and the frequency remains the same for the full range of kW capability of the generator.

I will not elaborate further on the how droop and isochronous load sharing modes work, as this has been done to death on this and various other forums. One such example for a great droop explanation is below as follows:

http://control.com/thread/1421810245#1421861327

A quick interweb search should also find literature for isochronous load sharing methodology.

Hope this helps

 

Dear Peter,

I greatly appreciate your help & effort in responding to my queries;

>In 4.2 case 1 described in "20050819.pdf", the AVR for the
>generator control system is set to manual and as the load
>increases, the voltage drop across Xs (synchronous
>reactance) increases. Since the AVR is not in auto, then the
>voltage will not auto correct to the set point voltage level
>and will drop.

>In example 3 described in "20050821.pdf", the AVR for the
>generator control system is set to manual (i.e. excitation
>level not adjusting to maintain the voltage), then the fuel
>to the G2 is increased. This increase in fuel will result in
>G2 taking on more load, and G1 unloading. As "droop" is used
>for load sharing, this results in and increase in speed /
>frequency of the generators. Since there is an increase in
>speed, then the voltage rise is due to the increase in speed
>(E=4.44.N.MagFlux.f).

Oh OK I got it. In the first case, the figure indicated 'steam input increased'. I guess this was a normal response by the governor to increase steam input to the machine as MW load had increased. I confused it with the case of 'increase in driving torque'. my bad


>In example 2 described in "20050821.pdf", the AVR for both
>generators are set to manual, and excitation is increased
>for 1 generator. The resulting rise in Vt causes a
>proportional increase in power (P=V^2/R and Q=V^2/X).

>Well, yes you are right if the machines are operating in
>droop load sharing mode, but no you are not if they are
>operating in an isochronous (isoch) load sharing mode.

I do not understand this, especially when both the machines are in droop mode & isoch operation is nowhere considered in the examples. I guess the reason why MW also increases is maybe because the machines are connected to finite busbars & so Vt increases due to increase in excitation of G1. In case of infinite busbars, since Vt & f remain constant, only MVAR increases due to increase in excitation of G1, MW remaining constant. But this is in my opinion.

 

> I do not understand this, especially when both the machines are in droop mode &
> isoch operation is nowhere considered in the examples.

I only brought up droop and isoch due so I can point out that outside of these examples, the generator response depends on the load sharing methods used.

> I guess the reason why MW also increases is maybe because the machines are
> connected to finite busbars & so Vt increases due to increase in excitation of G1.

That is correct.

>In case of infinite busbars, since Vt & f remain constant,
>only MVAR increases due to increase in excitation of G1,
>MW remaining constant. But this is in my opinion.

That is generally correct, however while frequency is quite solid in most stable networks, voltage will make small changes depending on things like:

- Load
- Transformer droop
- Cable voltage drop
- etc

 

Dear Peter,

Thanks very much for help

regards,
Shahvir