>how can the reactive power affect the
>i'm confused about this plz help me
Reactive power and fuel consumption
the straight and simple answer under all practical consideration is NO. reactive power will not have any effect on fuel consumption. to check this out (if you are in a site ). go to the GCP, decrease the MVAR from the AVR control to a min value, raise the MVAR to a max value there will no change in the fuel input.
theoretically the exciter which is connected to the turbine shaft must produce more power for a greater excitation this should increase the fuel consumption, but it is so small that it will not reflect in the fuel flow much.
first, THANKS for your response
so, you say that the exciter of 21 KW rated (one of our plant generators data) will produce a rated reactive power of 2250 KVAR is that right or logical.
and if it is right how can you explain that Q=1.73VI Sin (phi)
as the reactive power will vary with the imaginary part of the current and the torque will response for this current
"so, you say that the exciter of 21 KW rated (one of our plant generators data) will produce a rated reactive power of 2250 KVAR is that right or logical. and if it is right how can you explain that Q=1.73VI Sin (phi) as the reactive power will vary with the imaginary part of the current and the torque will response for this current"
well i cannot really understand what you are trying to say. is 21 KW the power rating of the exciter?? are you trying to equate between the "exciter" power and the "generator" reactive power ??. it is really hard to equate between the two without knowing
1. Generator excitation data , excitation curves
2. Excitation type used in generator
3. AVR details
"if it is right how can you explain that Q=1.73VI Sin (phi)" well this is a very generic equation , generator active and reactive power equation is different
E - The generator Excitation voltage , which is governed by the generator excitation system
V - The terminal voltage of the generator
I - The stator current
Xd - The generator Reactance
The Active power output from the generator is given by
P = E V / Xd Sin delta where , delta is the load angle , the angle between the voltage vectors E and V .
The reactive power output from the generator is given by
Q = EV/Xd cos delta - V^2/Xd where , delta is the load angle , the angle between the voltage vectors E and V .
in every text book the derivation of the above formula will be there. basically the generator is modeled as a power transfer through a inductive circuit.
"as the reactive power will vary with the imaginary part of the current and the torque will response for this current"
no as the reactive power varies the torque will not vary. have a go through of the generator real and reactive power equations , generator operating diagram to get a clear view. there is just too much to explain and with no diagrams it is more or less a lost cause.
don't get me wrong here, but in my opinion a read through of a book in electrical machines, especially chapters like ac machine fundamentals and synchronous machines will be helpful. you need to get the basics right :) and a textbook will be ideal for that :). of course if you have doubts you can always ask here.
I agree with you partially but not fully with your comment "this should increase the fuel consumption, but it is so small that it will not reflect in the fuel flow much." Actually this amount depends on the Alternator Efficiency Change. The Alternator efficiency changes @ of change of power factor. Suppose for p.f 0.95 the efficiency = 97% but for pf 0.80 it is 96% (can check from the alternator test results). So, this change in efficiency needs more power from the prime mover (Engine or turbine). Hence it requires more fuel consumption accordingly. So, you cannot say this is not so much. it is eventually a lot for large scale generation.
Oh, you want to be really, really careful when using the words 'reactive' and 'power' together in a sentence on control.com.
As for how "it" affects fuel consumption, it doesn't--for all intents and purposes. Fuel produces torque, and torque is used to produce real power (watts). That reactive thing you are referring to is apparent power, which is necessary for the magnetization of induction electric motors but which doesn't produce any real work (power; watts; torque).
I suggest you use your preferred Internet search engine and look up the definition of (and I'm only repeating the original poster's usage here!) reactive power. While I didn't find one relevant search result that had a really good definition, I did find several that were, together, very good.
But, as for any effect on fuel consumption, again, for all intents and purposes there is no effect on fuel consumption. Efficiency, yes, but fuel consumption, virtually none.
If you have more questions after doing your Internet research, please be more specific about the nature of your confusion. Tell us how you think fuel consumption would be impacted by (and, again, I'm just using the original poster's term!) reactive power.
the simple answer is that the generator efficiency drops so you require more fuel for the same energy delivered
Mohamed Ragab... paraphrasing your specific question, "How does kVAr affect fuel consumption?", then 'd-' provided the right answer, "... more fuel... required... !"
Now for the "How" part! Any increase in kVAr, either lagging or leading, will increase stator-current (ac) and rotor current (dc), thus increasing generator stator-winding and rotor-winding losses, and in-turn, decrease generator efficiency. The magnitude of the loss-increase will negatively impact the generator's rated efficiency, being less of an influence on a generator with a rated efficiency of 80 percent, then one rated at 95%.
CSA is right, of course, when he stated, "... for all intents and purposes there is no effect on fuel consumption." That is true because the GT has a much, much lower efficiency than the alternator!
Mohamed... if you want a quantitative value of loss increase, contact me. Or, in the interim, you can refer to my paper, "The Physics of... Armature Reaction" which illustrates the impact of lagging kVAr or leading kVAr on stator-current and excitation-current magnitudes.
Phil Corso (cepsicon[at]AOL[dot]com)
thanks for your short answer
but if it is true why you are paying only for active power while you are using both of them
Mohamed Ragab... because "Reactive Power" is not power. To learn why read the Control.Com Thread:
http://www.control.com/thread/1026236550 "The Physics of... Electrical Power!"
thank you very much, and this is the answer i was searching for
i have already finished reading abook of electrical power system that explain how that reactive power is not a power.
I'm very grateful to all of you and happy to find people like you who care with others even if for easy Questions.
again thank you and all who waste time in posting my question
Because for the average consumer the power utility factors into the price per KWH the price of the reactive power.
For very large industrial plants with lots of inductive loads (induction motors, primarily) power utilities do install VAr-hour meters in addition to the Watt-hour meters, and make those users pay for the reactive power in addition to the real power they consume. So, many of these large users will install various forms of power factor correction to reduce the amount of VAr-hours they "consume".
(We're really treading on thin ice here talking about reactive power and its production or consumption. There are those who usually chide us for using those terms on control.com, even though the rest of the world uses them frequently and they are found in texts and references everywhere. So, please be very careful!)
Ragab... for the record I would appreciate knowing the title of the text that explains, "... reactive power is not a power."
CSA... Congratulations for your slow, albeit sure, continuing use of the term, VAr, instead of the term, reactive power! Hopefully, you have also noted it has spread in use in the Control.Com forum! Keeep up the good work!
Special regards, Phil Corso (cepsicon[at]AOL[dot]com)
Power Value... please explain your derivation of the P and Q formulas which involve the product of E and V?
Phil i am uploading a small pic here. it is taken from a electrical engineering text book i have. this will give you the derivation of the formula.
this is one of the simplest derivation i have seen. the actual rigorous derivation without the assumption of R = 0 is in another text book, but i do not have a soft copy. if i get it i will upload it too.
Process V... my error! You identified E as Excitation Voltage when you really meant internal Air-gap voltage, Egp!
My experience is concurs my theoretical understanding of the subject in above matter. The generator is energy device & it generates electrical energy which is in KVA. The KW & KVAR part depends upon load. If demand of any of this component is increased, so will be increase in KVA generated.
For a short circuit fault, which is basically inductive in nature, will draw current & we calculate fault KVAr at different points & not KW. So if there is short circuit fault near generator, KW will be almost zero & KVA= KVAr. Generator will generate high KVA & turbine will need more torque (i.e. fuel) to feed this fault till it is cleared by protection system.
Coming to our normal operation, the change in power factor is generally between 0.02 to 0.12 max. Simply saying, we aim to improve our power factor from 0.83 (on an average) to 0.92 at the most. The difference is 0.09 between these 2 power factors. So difference in KVA for these 2 loads for same KW will be KW/0.83- KW/0.92= KW[(0.92-083)/(0.92*0.83)]= kw* 0.118.
Thus, for above case KVA generated will be 11.*8% more for operating at 0.83 power factor instead of 0.92. This will be reflected in generator current & subsequent heating in stator winding temperatures.
Now coming to your question, for this additional 11.8% KVA, additional fuel will be required. However the increase in fuel is seen only marginal. This is because, fuel consumption vs KVA is non linear characteristic. At no load turbine draws more 50% of fuel compared to full load. As load increase from zero to full load, the increase in fuel consumption slows down. Thus, 11.8% increase in KVA would not be reflected as 11.8% increase in fuel. The increase in fuel may not be significant but it is there.
You can check this in your plant before taking planned outage of any generator. Bring down the load of generator to nearly 20% of its capacity at some higher power factor ( 0.92 to 0.97). Note down KW, Amp & winding temperatures & fuel consumption of all running machines. Now raise excitation of above generator & check that its power factor is dropping. Ensure that generator current is less than overload setting. Try to bring power factor as low as possible while KW remaining constant. The load should be however less than generator capability curve provided by OEM at this reduced power factor. Now note the readings again. You will notice significant change in amps & winding temperature in all generators. There will be noticeably rise in fuel consumption in machine under test. The change in fuel consumption in other generators will be not as noticeable as the machine under test but still you can see there will be change in fuel consumption.
Thus, in case we are forced to run generator at lower load, care should be taken that it is running at better power factor to save additional fuel cost,
Please can you tell me how does the generator exciter current affect Power Factor? I have increased the exciter current and the Power factor has dropped but this does not seem right according to theory. The generator is rated at 0.85 PF.
This topic has been covered many times before on control.com. There is a 'Search' field at the far right of the Menu bar of every control.com page (I recommend using the 'Search' Help before searching for the best results!).
Basically, Power Factor is a measure of the "efficiency" of the generator at converting the total power being input to the generator into "real" power: watts (or KW or MW). When ALL of the torque being applied to the generator rotor by the prime mover (turbine or engine) is converted to real power, watts (which implies a resistive load with no reactive current), the Power Factor is 1.0--which is effectively the same as saying the generator is 100% efficient at converting the torque being applied to the generator rotor into useful power, watts (or KW or MW).
When the excitation is increased or decreased without changing anything else (the fuel or steam being admitted to the generator's prime mover) while the Power Factor is 1.0, then the efficiency of the generator decreases, and the Power Factor DECREASES below 1.0. So, when the Power Factor is 0.93, the generator is approximately 93% efficient at converting the torque being applied to the generator rotor into useful power, watts (or KW or MW). When the Power Factor is 0.85, it's 85% efficient.
And that's true if the Power Factor (and VAr) reading is Leading or Lagging when it's less than 1.0. The Power Factor is never greater than 1.0, and can only be 1.0 or less, regardless of whether or not the Power Factor is Leading or Lagging.
(Sometimes the Power Factor is displayed as being a positive or negative number, and when that's done usually positive means Lagging and negative means Leading. So, if your looking at a display that is digital, remember: the polarity indicates which "direction" the Reactive Current is flowing but still: the number is never more than 1.0 and can only be 1.0 or something less than 1.0.)
Lastly, decreasing excitation without changing anything else (the fuel or steam being admitted to the generator's prime mover) when the Power Factor is 1.0 will cause the Power Factor to DECREASE from 1.0--again, because efficiency of the generator is decreasing. The more reactive current (VArs) that are flowing in the generator stator the less watts that can be produced.
To see the effects of reactive current, and a lower Power Factor, on real power output of a generator, operate the generator at a low power output, say 10-15% of rated, adjust the excitation until the Power Factor is 1.0 (Unity) and record the watts (or KW or MW). Then increase the excitation until the Power Factor reaches, say, 0.85 or less (lower won't hurt the turbine for a few minutes)--without changing anything else--and record the watts. The watts will decrease slightly as the Power Factor decreases (from 1.0) and the reactive current (VArs) increases.
Do the same by decreasing the excitation when the power factor is 1.0--without changing anything else--and record the watts when the power factor reaches, say, 0.85 (don't go too low when decreasing the excitation for this test!). The watts will decrease slightly as the Power Factor decreases (from 1.0) and the reactive current (VArs) increases.
The Power Factor rating of a generator just refers to the total mount of power (real and reactive) that can be produced by the generator. It doesn't mean the generator has to be operated at that power factor, only that rated power can be produced for extended periods of time at that Power Factor.
I would recommend researching 'Power Factor' on sites like www.wikipedia.org (if available in your part of the world).
Hope this helps!
This is correct as per theory. The power factor of generator is proportional to ratio MW/MVAr. So any increase in numerator with denominator remaining constant will result in improved power factor. similarly any increase in denominator alone (i.e. MVAr) will result in dropping of power factor.
In a grid, the MW is directly proportional to speed reference. Any change in it will not affect MVAr. Similarly MVAr is directly proportional to magnitude of terminal voltage which is directly proportional to excitation.
So by increasing field excitation, you are increasing terminal voltage of generator. This added terminal voltage will result in generator supplying more MVAr & thus resulting in dropped power factor
Prasad... your explanation is in error because even though kVA, hence armature-current, increases, the amp-to-torque analogy doesn't apply!
In fact, for a 3-ph near-terminal fault the generator will speed up because the load kW is suddenly lost!
Hopefully you will conduct a study to determine if the generator and prime-move are in danger of overspeeding
Regards, Phil Corso
Yes Phil, during a fault near generator terminal will certainly reduce load MW to almost zero depending upon fault impedance. But at the same time the fault will draw large current from generator. This fault current will be inductive in nature & thus this component will contain only MVAr & no MW. In this case generator MVA will be equal to fault MVAr. This fault MVA will be far too higher than rated MVA capacity of generator. So obviously turbine has to provide far too mechanical/ kinetic energy to generator beyond its capability. So in such case obviously turbine/ prime mover speed is going to drop.
Also now that MW is almost zero during fault, fuel required by turbine before tripping will not drop down to no load value. In fact it will increase to cater to fault MVA. And this is what I wanted to prove (i.e. reactive power will affect fuel consumption of turbine.)
Keep in mind that whatever heat is generated during a fault is real power and real MW's. I have seen an off line fault of a bad lightning arrestor that caused the turbine speed control system to call for more steam to maintain speed. I agree that most of the current flow is reactive, but not all.
Prasad: remember, current does not "contain" MW.
Watts are produced when current flows thru a resistance and all generators, bus work, transformers etc have some unless they are superconducting. So when current flows thru a resistance, watts are produced in the form of heat in an amount equal to I-squared-R. Those watts dissipate energy which must have come from somewhere. In this case, "somewhere" is your generator which converted mechanical energy into electrical energy. Your generator received the mechanical energy from the conversion of thermal energy in your turbine, therefore the turbine must have burned more fuel. So increased reactive power generation increases fuel consumption.
The law of conservation of energy always applies.
> how can the reactive power affect the fuel consumption
> i'm confused about this plz help me
Reactive power can't affect fuel consumption. Because if reactive power(say lagging) increases then the frequency of excitation voltage decreases, but that doesn't mean that the angular velocity of rotor changes.
So to get back the actual frequency we use power factor correction and after that automatically the reactive power decreases and the frequency gets back to its actual value.
Reactive power is the power which is given to the system in the positive half of an AC cycle and received back in the negative half. overall average power through one complete cycle is zero. thus theoretically speaking there is no increase in fuel when you decrease your PF of simple increase MVAr.
But then why do we try to maintain high PF? It is because any system is not ideal. All system will have resistive losses which are in turn directly proportional to square of current through the system. even though as mentioned reactive power over a complete cycle is zero, reactive current will still cause a resistive loss in cables, transformers, etc. thus decreasing efficiency. but as mentioned in earlier posts, the effect is seen over a period of time. when you calculate you will have a substantial saving overall in an annual period.