We are synchronizing 12MW TG with grid in power factor mode. Rated exciter field current is 3.5Adc. We synchronized and at 9MW field current is 3.9Adc. We increased power factor from 0.85 to 0.95 so field current got decreased and came at 2.8Adc. What is the reason?
Power factor (at the generator terminals!) can be leading (positive) or lagging (negative). Since the Maximum value of power factor is 1.0, leading or lagging VArs/power factor is always less than 1.0. It would be necessary to know if the power factor was leading or lagging 0.85-0.95.
Having said that here's the general rule: To decrease power factor in the lagging direction (positive) from 1.0, increase the excitation applied to the synchronous generator. To decrease power factor in the leading direction (negative) from 1.0, decrease the excitation applied to the synchronous generator.
So, in your case if the unit was operating at 0.85 lagging and you wanted to increase the power factor to 0.95 lagging you would decrease the excitation.
When the excitation is exactly equal to the amount required to make the generator terminal exactly equal to grid voltage, the VAr flow will be zero, and the power factor will be 1.0 (unity). (Power factor is the measure of the efficiency of the generator at producing real power (KW) for the total amount of power 'being applied' (KVa). When the power factor is 1.0, the generator is 100% efficient at producing real power (KW).)
When the excitation is increased above the amount required to make the generator terminal voltage exactly equal to the grid voltage (in effect, you are trying to "boost" the grid voltage) then lagging VArs will flow in the synchronous generator stator windings, and the power factor will be less than 1.0 in the lagging direction. If you want to return the power factor to 1.0 or closer to 1.0 from the current condition, you would need to decrease the excitation.
When the excitation is decreased below the amount required to make the generator terminal voltage exactly equal to the grid voltage (in effect, you are trying to "buck" the grid voltage) then leading VArs will flow in the synchronous generator stator windings, and the power factor will be less than 1.0 in the leading direction. If you want to return the power factor to 1.0 or closer to 1.0 from the current condition, you would need to increase the excitation.
So, from your post I would presume the unit was operating at 0.85 lagging, and to increase the power factor (increase the efficiency of the generator) you would need to decrease the excitation to get the power factor closer to 1.0 (0.95 in your case).
Hope this helps!
Thanks for your response.
Our problem is that why exciter current cross the limit of its rated exciter current value (3.5Adc) at 9.0MW while its rated load is 12.0MW?
We have to increase power factor from 0.85 (lagging) to unity to decrease the value of exciter current. Rated pf is 0.8 lagging.
We already synchronized this unit with grid and it is working fine. At this stage we are facing problem that we can not go beyond 9.0 MW load.
From other posts here, I don't think we have the entire saga, and it would seem that configurations and settings have been changed and then changed. You haven't told us what kinds of units you are working on; if there is any transformer impedance between the units or if they are synchronized together to a common bus. There's just too much we dont' know.
It's also possible that the grid voltage (or, the voltage on the bus the generator is being synchronized to) is higher than nominal, and higher than rated generator terminal voltage. From your description it seems that is most likely the case, and that the generator is being "over-excited" and that is the cause of the excess exciter current. You haven't told us what voltage the generator nameplate lists, and what voltage it's being operated at when it's synchronized.
Generators are rarely operated at nameplate rating, especially at the power factor listed on the nameplate. Operation at rated nameplate power output (KW and KVAr; KVA) means a lot of heat is being produced by current flowing in the generator field and in the generator stator windings--by both amperes and reactive current. Nameplate is the maximum continuous operation that the generator can withstand--meaning it's ability to dissipate the heat by whatever method is used to cool the generator.
Operation at rated nameplate values also means the generator set is not producing power (real power, KW) at 100% efficiency--which means it's expending some of the energy in reactive power which is generally not producing any revenue.
Lastly, most synchronous generators are rated for more power output than the prime mover (steam turbine; gas turbine; reciprocating engine; etc.) can produce. This is to protect the generator and to allow the prime mover to produce as much power as it's capable of as ambient and machine conditions change. So, if you're trying to make generator nameplate rating you need to consider what the prime driving the generator is capable of producing because it's very likely that the prime mover isn't rated as high as the generator nameplate.
Again, based on your other posts and this thread, it would seem there is a lot we don't know about the conditions and equipment and the history of operation (start-up; commissioning; etc.) at your site. It would take a LOT of back-and-forth (us asking you questions, and you providing a lot of information (some of which you will deem unnecessary--but isn't) for us to be of any further help.
Lastly, you haven't said why you have only been able to reach 9.0 MW. Is it because the prime mover can't produce any more torque to drive the generator? Is it because you are unwilling or unable to increase the load because the exciter current is above rated? Synchronous generator power output is primarily a function of the energy flow-rate into the prime mover. More energy, more power; less energy, less power. Exciter current primarily controls voltage--especially at higher loads; and when synchronized to a bus or grid that is operating at approximately generator nameplate rated voltage, changing excitation changes reactive current flow--not real power flow, especially at higher loads.
If you want to pursue this thread further you will need to provide all the answers to the following questions:
1) What is the voltage of the generator/bus/grid when the generator is synchronized?
2) What is rated nameplate voltage of the generator?
3) What kind of prime mover is driving the generator?
4) Is there a transformer between this "12 MW" generator and other generators with which this generator is synchronized?
5) What is the frequency of the generator when it is synchronized and at 9.0 MW?
If you want more help, we need answers--to ALL of the above questions.
Looking forward to hearing the answers to ALL of the above questions.
Jeet... it will be easier to explain if you would provide what the corresponding armature-currents were for each of the conditions you cited?
Let's go little back to the classroom ok:
We know that
1. when current flows through a conductor, magnetic field is
2. When a conductor is moved in the vicinity of that magnetic field or when the magnetic field is moved in the vicinity of the conductor, voltage is generated into that conductor.
3. When a closed loop is formed by connecting a load to the conductor being moved through the magnetic field, current flows through the load.
Dc current (field current) is sent through the generator rotor windings and that produces magnetic field (refer to item (1) above).
In bigger generators, the magnetic field is being moved/ rotated (Generator rotor) by the turbine and the conductor (Generator stator windings) is kept stationary and by so doing voltage is generated/induced in the stator windings (refer to item (2) above).
When the generator is synchronized to the grid, current flows from the generator stator windings into the grid (refer to item (3) above).
The voltage (e) produced/induced in the conductor is basically given by:
E = blv (volts)
'E' is the induced voltage which is approximately equal to the generator
'b' is the flux density of the magnetic field produced (The strength of
the magnetic field produced in the generator rotor)
'l' is the length of the conductor (stator windings)
'v' is the speed at which the conductor or the field is being moved (speed
at which the generator rotor is being rotated by the turbine).
Due to inductive and capacity loads, when the voltage and current starts from the zero point in the alternating current cycle, one reaches its maximum value before the other and that makes one lags behind the other by a certain angle in the cycle and if you find the cosine of that angle between the current and voltage, it gives you power factor (p.f).
The higher the field current,(the stronger the generator rotor magnetic field strength) the higher the generator terminal voltage, and therefore the higher the reactive power generated to supply these inductive loads and hence the bigger the angle of lag between the current and voltage. If you find the cosine of a bigger angle, you will get a lower value which gives you your p.f (cosine inverse of 0.85 = 31.8 degrees angle) which is why at 0.85 p.f your field current is higher.
The lower the field current, (the weaker the generator rotor magnetic field strength) the lower the generator voltage, and therefore the lower the reactive power generated to supply these inductive loads and hence the smaller the angle of lag between the current and voltage. if you find the cosine of a smaller angle, you will get a higher value which gives you your p.f (cosine inverse of 0.95 = 18.2 degrees angle), which is why at 0.95 p.f, your field current is smaller.
Now, most generators I have seen has a p.f rating of 0.85 lagging.However operating such generators at this p.f means you are operating the generator at the edge of its reactive power capability curve. This heats up the generator stator windings and reduces the lifespan of the generator so please be cautious.
You can draw two power right angled triangles with the same base; one with a longer height (bigger VAr) and the other with a shorter height (smaller VAr) and find the cosine of both angles, that would help you to understand better.I hope this helps!!.