# Synchronous generators

Good evening

I have problem to understand the operation of Synchronous generators. When we want to connect it to the line network, we have to regulate the excitation and angle-speed in order to satisfy correct line frequency and voltage value. This regulation is generally done in open-circuit mode.

Why is also important to accelerate the rotor before connect it to the line?

Acceleration is done to obtain correct angle between generated and terminal voltages. It’s linked to the torque applied to the rotor axis. It corresponds to the electrical power delivered to the load.

If we don’t accelerate rotor, does the terminal voltage fall far short than line voltage?

I have attached the schematic rappresentation of this kind of regulation.

Thanks a lot.

If you don't synchronize massive currents flow and trip everything out.

Consider two heavy pendulums connected by a thread - so long as they swing in time the
thread can allow one pendulum to impart mechnical force to the other (within limits).

If out of time the thread snaps immediately.

The current is sensitive to the moment-by-moment difference in voltage between the network
and generator terminals open-circuit. You want that current to be within the ratings of the
windings, and the voltage difference is small (much smaller than the line voltage) at all times.
Unless you synchronize frequency and match phase you can get full mains peak-peak voltage
across the winding resistance, causing tripping (perhaps 30 times normal load currents or more).

Furthermore you want the phases matched to make the generator a non-reactive power source
(which will automatically be the least voltage difference).

Thank so much.

I try to deepen your clear and appreciated explanation.

1. If I don’t accelerate the rotor after open-circuit regulation, can the Potier diagram appear like in the figure attached?

I have another question.

1. If the active electrical power ( prop. to BC line in the figure ) is linked to the torque applied to the rotor (and so the mechanical power), what is the origin (or conterpart) of the reactive power ( prop. to AC line )?

Well you'd have to explain each of the phasors and angles in the diagram first, and hope I am
willing to do your homework for you...

Sorry, if I have understand your reply I try to describe more in depth.

• Eo is the generated voltage in stator windings

• U* is the correct value of external voltage obtained if the rotor is accelerated (according to standard procedure). This is the same value obtained during regulation in open circuit condition.

• jXsI is the voltage in the inductances associated to synchronous reactance (in Behn Eschemburg model). This is oriented perpendicular to the current.

• U' is the external voltage produced without change rotor speed.

• The first diagram rappresents the regulation in open circuit mode.

• The secound one represents the correction obtained accelering the rotor (and so the external voltage appears the same in open circuit case). The Eo voltage is incremented by the major angolar speed; "delta" is load-angle (between Eo and U*).

• In third case I describe what I suppose. Generator is connected without accelering and it's supplied by the same excitation. I think that the current phase not change in this case, but we have minor module (because the generated voltage is the same as in open circuit mode).
Load angle is reduced (due to current module). So the U' (external voltage) is reduced respect the desired value.

I think you know more than me about this particular setup - I would have expected reactances to
be minimal in an efficient generator and all the phasors to pretty much lineup...

I think you are making the problem more complicated than it really is.

You have a power system running at 50 or 60 Hz and a synchronous machine which you want to connect to it.

First requirement— Bring the incoming generator up to synchronous speed, ie the speed at which it runs when connected to the power system. This will have to be done by adjusting the speed of the prime mover to which the generator is connected (engine, waterwheel, whatever)

Second requirement— adjust the incoming generator voltage to equal the power system voltage. This is not as critical as speed; I would be happy within a few percent and after you are synchronized trim it to get the desired kvar.

Third requirement— You will need a synchroscope or oscilloscope to do this so I will assume a synchroscope. this instrument has dial and clock type hand which indicates the phase difference between the incoming generator and the power system; when the difference is 0 the hand is at 12 o’clock. So bump the speed a bit up or down to get the phase difference to 0 then close the breaker and you’re done!

Now adjust the generator excitation to set the required kvar and the prime mover regulator to get the required kw.

It is very important to get the phase difference to zero before you close the breaker, otherwise there is a big bang and if you’re lucky it trips off and you start again. If you’re not lucky you do some mechanical damage to the machinery.

Many years ago I saw it done improperly manually synchronizing a waterwheel generator. It shook the building and got a cascade of dust from the rafters.

Phoxx

Nowdays the big PV generators and wind generators all do it electronically.

Paul

Paul:
Yes they use a "synchronizing relay". This gives a string of pulses on one output if the speed is low and on another output if the speed is high. The pulses bump the engine throttle open or closed and get lower and lower frequency as the synchronous speed is approached and finally a "phase check relay" says it's ok to closed the breaker. Actually this scheme comes from the day when the electronics was pretty primitive and not universally trusted so some operators preferred to do it manually.

In the PV example the inverter frequency would be adjusted to get synchronism. For a wind generator I expect the blade pitch would be adjusted to adjust speed. I know of one experimental wind generator which ran at variable speed then used an inverter to get line frequency. But the wind blew it down.

Phoxx