Generator Power Factor

Generator Power Factor - cashmanequipment.com

Power factor is determined by the load, and it is what it is. A low power factor means that there is extra current that is not doing any useful work, and a generator rated at a lower power factor for a given power output can handle extra current.

Power factor = Real Power / Apparent Power

What it boils down to is that for a given amount of Real Power (MW), the total current is proportional to the inverse of the power factor. (1/pf)

The OP's talk would sell to an uneducated owner perhaps.

Compare the KVA ratings of the two generators. (This is usually given, but if not can be easily be computed, per previous post.)

rated MVA is 27.8 for 0.8 lag and 31.3 for 0.9 lag.i am actually concerned about stability of generator when synched with utility system.

Are you sure you have lag number correct vs MVA? 31.3 x 0.8 = 25 MW: 27.8 x 0.9 = 25 MW.

Are you sure both generators are same insulation classes and temperature rise? The apparently bigger one may just be hotter and less efficient.

Generators do not have highest efficiency at rated load, but at somewhat below. It may be that the bigger generator is same/more efficient at 27.8 MVA 0.9 p.f. This in addition to lower temperature rise in rotor and stator with massive effect on life expectancy.

You need to look very carefully at the loss claims for the two & adjust bigger for reduced current.

Bigger generator should have more "moment of inertia" kg- m² /inertia constant MWsecond/MVA, that usually aids stability.

the insulation class /temperature rise quoted is F/B . that means a 10% overload,right?

i am more concerned about the stability problem. with bigger size , will the generator respond quickly to transient disturbances /fluctuations fast enough when triggered by excitation system and governor? from your comment it seems that bigger generator has more moment of inertia and more stability but what about quick response of generator to changing loads?

I do not for your specific case, but most specification sheets come with a generator slew rate and a generator ramp rate (do not confuse the two). Slew is the incremental allowable change rate from control circuits. Ramp is the loading curve in load value / time.

An insulation class has a maximum temperature, [ambient + rise], but that does not mean it runs at that. It is common to use class F insulation on a machine designed & proven for B rises because it increases the insulation & hence machine life a lot - there is a theoretical doubling of rate of chemical reactions for 8'C rise, deterioration of insulation is largely a chemical process. Machine standards usually relate to 40 Celsius ambient, but you might have one quote based on {your ambient (<40'C) + rise} = maximum temperature for class (total temperature).

But your F/B does not imply any overload capacity - the rotor is often F with stator B.

Speed/power response (governor) depends mostly on the prime mover (and how much life you want to batter out of it) electronics & fuel/steam valves can do whatever it takes - do they meet your response requirement? More inertia means set goes off normal frequency slower for any given power change, prime mover MW/second can be less for given frequency change. "Bigger" generator rating means little - you can have short fat rotors or long thin rotors for same rating, look at kg-m2 or MJoule/MVA figures for inertia.

Voltage response depends on machine type, brushed rotors can have quickest response because external exciter can have more volts to change field current quick, despite rotor inductance.

Ask the suppliers for full load reject frequency & voltage response graphs & similar for step load applications.

For the size change you give, there will be little difference for similar machines, except that 10 MVA change for bigger machine is ..% less as proportion of rating and proportion of reactance voltage drop.

You have not indicated if the generator is connected to Grid or isolated.

And, as I wrote before, the "bigger" machine could be the same machine running hotter.

Have you taken the correct step and contacted the generator supplier, of is this just another homework question?

Do your own homework. CR4 is not a homework cheat site; however, if you have questions about understanding concepts or how a portion of a problem is derived, these types of questions will be accepted.

You are either a very lazy student or an incompetent worker.

It's 5 years now that you've come to the forum asking these test questions. Do your own work!

ah, so now you are the measuring scale of incompatency.so for now your output to my questions is a big zero.by the way ,why u even go through my questions if you are such a guru.just enjoy yourself man and chill.i am not being paid by you so why bother about my incompetency

Your attitude and lack of information about your problems will assign you to the "Do Not Help" list.

Consider the venomous attitude he was shown. Some of us here are generally lacking in manners, courtesy, and the Golden Rule.

Enough said? Or shall I continue?

Then why did you two respond to his further questions? Who is the "greater fool" here?

Generator Power Factor - cashmanequipment.com was the fist reply.#1.

Did you not see it???

I an't help it if you didn't bother to read it.

I have offered you help many times in the past and in this thread as well!

And now begins the time of unraveling of the entire purpose of our being here.

You'd be better served by doing some real learning than asking for a quick answer for your immediate problem.

Typically generator ratings are listed as xxMVA @ 0.x inductive, not MW, which is usually how the prime mover is rated. The thing you should be looking at is the GCC (Generator Capability Curve), that will tell you everything about the available real power for a given reactive power output and/or power factor.

You are getting a bigger machine at the 0.8 PF, so you would expect to pay more. The larger machine, run at the same kVA load, will run cooler and longer, with perhaps some windage load (slightly lower efficiency) turning the larger machine.

The ability to stabilize your power system, start large motors across the line, and trade some of that extra reactive capability for kilowatt power generation if you can squeeze more power from your prime mover might be worth the extra cost.

Small machines like this at 0.8PF were popular in the 60s and 70s, the standard I'm familiar with started at 50MW, and were typically purchased at 0.85PF, enough to be assured of taking care of a typical industrial induction motor load complement while keeping an eye on the budget.

Check what power actor your existing plant is running at and what the anticipated power factor will be when the new plant for which you are installing extra capacity is installed. As stated the 0.8 rated generator is likely to cost more. Investigate if you can purchase power factor correction equipment with the money saved by buying the cheaper machine. This will benefit both your existing and new capacity and save on your long term energy costs. It should give you a better ROI

our system expected load power factor is about 0.9 lag.

Dear Mr.coolyar,

My view is as follows.

For the same power i.e., MW, the generator with 0.8 PF will have more line current than the Alternator with PF 0.9 and hence the voltage drop will be more in the Alternator with a PF 0.8. This decides the DROOP CHARACTER of the Alternator, which is linked to the Load Sharing by the Alternator.

In this case the Alternator with 0.9 PF will share larger proportion of the Load than the Alternator with 0.8 PF. This is due to the sets are working with a common frequency and and common voltage in the system.

DHAYANANDHAN.S

Can you elaborate your point more openly. Actually we are building a captive power plant with 25MW steam turbine system based on coal. the unit will be running in parallel with national electrical system . our net demand is about 18MW or so. auxiliary plant consumption about 2.5MW. so rest will be injected into national electrical grid. the supplier has quoted two different generators : one of 25MW with 0.8lag and another 25MW with 0.9 lag. Now we need to select one of them. so i was wondering what will be the pros and cons of both when operated in parallel with national grid or even in island mode.

Mr. dhayanandhan has paralleled your two choices. The answer you seek has been presented at least 5 times in this thread, within the rather lean detail that you have provided.

A transient stability analysis is likely outside the scope of a thread in a forum, and you have yet to present even a remote target for your analysis, like being able to start a 1500hp I.D. fan motor across the line while running full load, or what complement of field regulation equipment might be supplied with the generator, saturable reactors, field forcing, brushless rotating exciter, all kinds of options possible.

How can you possibly be allowed on the site?

Please tell us where this site is located.

Whose national grid? Canada? Finalland? Semifinalland? Please tell me you are not the sole power engineer in Poptestiland?

For God sakes man, pick the one with the highest reactive loading. It will last better and serve you longer. I can only assume since you are adding capacity to your local grid (that can be islanded), this is a small country with many islands, or land with no inhabitants, and no power utilization for the most part, that you must purchase a package unit with turbine and generator. Is it LM 2500 GE? What is it?

As was already said by pcchatur, if the Diesel [or turbine] is able to an extra loading and if the price of generators is the same, then 0.8 lead [in my opinion the generator delivers reactive power] is better. If this is more expensive and the driver is not able to more load then you don't need this.

On the other hand in connection with steady state, dynamic and transient stability it depends on rotor inertia moment and not directly on rated power factor.

Usually smaller generator will be of less stability, indeed.

Also for record: static stability depends on excitation current limit [possibility to more overexciting] and less on the inertia moment as dynamic stability.

For record only: I did here a confusion indeed. In my opinion lagging or leading is a confusing notion for synchronous motor and generator .It will be better to say overexcited or underexcited and it seems that “overexcited” is what “lagging” does mean. So, of course, the overexciting for 0.8 power factor is better than 0.9 since the excitation current limit is superior and the static stability is then better.

You have just defined a synchronous condenser. Overexcitation.

Correct. However a synchronous condenser does not deliver active power only reactive.

A synchronous condenser is underexcited [stator e.m.f. less than line voltage].

And going back to #31, both a synchronous motor and a synchronous generator can be over-excited (lagging VAR output, to maintain inductive loads) or under-excited (leading VAR output).

Some generators have a drive clutch so they can operate [motoring, under-excited] as "synchronous condensers" with prime mover stationary or even make lagging output, as in usual generator operation.

It appears we all need a refresher course in synchronous generators.

In my opinion, it is not our knowledge but what lacks us is logic.

To understand the connection between lagging, leading, overexcitation and underexcitation it is simply.

For a generator Eg=V+i*I*xd [turbogenerator xd=xq] is the equation and for a motor Em=V-i*I*xd.

If cos(fi)=1[ fi=0] then Eg=Em fi=the angle between voltage at terminals and the current flowing through the terminals.

If fi<0 the current lag the voltage- then generator Eg is more [overexcited] and the motor Em is less [underexcited] since the e.m.f. is direct proportional [approximate] with the excitation current.

If fi=lead Eg is less than for cos(fi)=1 [generator underexcited] and Em is more [overexcited].

The problem is the reactive power. Here, since it is not an actual one as the active power is, we have a convention: Q>0 is capacitive power[power out] and Q<0 is inductive power[power in].

If we connect a synchronous motor at the terminals of a generator the active power delivered by

generator is absorbed by motor but the reactive power delivered has to be positive in order to

compensate the negative required by motor.

S[apparent power]=V*I” where I”=I conjugate

If ig=I*[cos(fi)+i*sin(fi)] in generator and im=I*[cos(fi)-i*sin(fi)]

Ig”= I*[cos(fi)-i*sin(fi)] and I”m= I*[cos(fi)+i*sin(fi)]

Sg=V*I*cos(fi)-i*V*I*sin(fi)

fi<0 Q>0

Sm=V*I*cos(fi)+i*V*I*sin(fi)

fi<0

motor Q<0

So the generator has to be overexcited and the motor underexcited.

You have drawn case for two machines connected together, with no other load

There can only be one voltage vector between Eg and Em, across inductance 2* Xd, where "motor" and "generator" have same reactance (and just one current between two machines) .

You have two equal voltage vectors from V, in opposite directions, whose sum is zero, implying no voltage difference between Eg and Em.

You have a current through two inductors in series which decides to change its direction 180 degrees when it passes point V.

Logical problems, inconsistent with physics.

The voltage vector between Eg and Em can be "swung round the clock" by changing Eg, Em magnitude and angle between them, consequently current angle can go round the clock too, lagging Xd voltage by 90 degrees.

Sorry 67- by the way: if I am Double O 7, so you are 6 [?!]. What’s news on MI6? -this was for a theoretical explanation only. Nobody intends to work so, of course. As I am a humble electrical machine constructor- I am not a operating engineer-which knows all this from the day when he was born -I have to explain all phenomena theoretically to my self.

Now I really really have to go read up on this again. You have thoroughly confused me. Never introduce mathematical symbols please without designating what they are. I too old and tired to figure our your notation.

Sorry for the delay James [“Double O 7” ???]. I am also retired-my daughter says I am not old-and I am a full time job grand-father-you know.

I thought the notations are international, then sorry again.

Let's say Eg=generator e.m.f.[the voltage produced in the generator stator in rotation and excited but not loaded still]. Em is the counter e.m.f from the synchronous motor in rotation and excited.

xd and xq are leakage reactance of the stator: the first when the rotor pole is under this winding and xq when it is between poles. However, only for salient poles xd and xq are different.

I conjugate [I* symbol usually but here I put it as I" as in excell language * it symbols "multiplied by"] .If I(t)=sqrt(2)*I*[cos(wt+fi)-j*I*sin(wt+fi) ] where w=2*pi()*freq. t[sec] fi[radians] I[rms] then conjugate will be I(t)*=sqrt(2)*I*[cos(wt+fi)+j*I*sin(wt+fi) ].

If I missed something don't hesitate to bring it to my attention.

Whatever happened to I = e^Iθ

I need to take a nap now, sorry.

Doesn't make much sense to run it, that way....

Normal is real power in, reactive power out, so over excited...