why alternator cannot be run in leading power factor

 

The ability to run an alternator with a leading power factor is a function of the design and construction of the alternator. Some alternators are actually designed to be run with "large" leading power factors--they are called synchronous condensers.

Building an alternator to run with a "large" leading power factor is very costly. Also, most power systems are inductive in nature and running a generator with a leading power factor increases the lagging VAR load on the system.

markvguy

 

Physics.

 

That´s not true, all generation can run in leading power factor whitout damage, same like a sincronous capacitor or the older sistems whit a motor conected to the Utility line, and contributing whit capacitive charge only.

What is your real problem?? can you send me an e-mail maybe I can help you.

 

Most synchronous generators have a Reactive Capability Curve which is produced by the generator manufacturer that defines how much reactive power (leading and lagging) the generator can produce without damage. The best reference is to obtain and use the Reactive Capability Curve for your generator.

One of the more serious consequences of running a generator in a lagging power factor condition is that to do so requires reducing the excitation to the generator--i.e., running it in an under-excited condition. If the excitation is reduced too much, the generator can "slip a pole" and serious damage can result to the generator, the load coupling, and even the prime mover.

There are generally protective relays (ANSI Device Numbers 40 and/or 41) to detect an "extremely" underexcited condition, or, a loss of excitation, which will trip the generator breaker at a minimum to prevent "slipping a pole" or damaging the generator internals from running in an underexcited condition for an extended period of time.

markvguy

 

reactive power sharing require for variable load in parallel alternator? in our sustem lot of time trip in active reverse power?

 

My question is, what is the meaning of "slip a pole"?

 

My question is, when we operate synchronous motor in overexitation condition it works in leading power factor condition, whereas when we operate generator in overexitation condition it works in lagging power factor condition. what is the reason behind it?

 

Here is one source located by typing "slip+a+pole" into Google (there were many others):

http://www.kilowattclassroom.com/Archive/SyncMotors.pdf#search=""slip+a+pole""

In a synchronous machine, the magnetic forces of the rotor are "locked in step" with the magnetic forces present as a result of the AC flowing in the armature. (Synchronous motors can also slip a pole.) This is one reason why the machine is referred to as a synchronous machine--it's speed is fixed by the frequency of the AC flowing in the armature (as opposed to induction machines which require that the rotor speed be slightly less than synchronous speed to operate).

"Pull-out" is another term to describe what happens when the torque being applied to the rotor exceeds the magnetic force that is keeping the rotor in synchronism with the AC which is flowing in the armature windings. In a two-pole machine, the rotor will suddenly rotate as much as 180 degrees "ahead" of itself before being "caught" by the interaction of the magnetic forces at work in the machine.

This places tremendous forces on the rotor and load coupling and prime mover or driven device shafts--and usually results in serious damage to all the components.

In other words, "it's a bad, bad thing." There are usually relays and sensors to detect a loss of exciitation or a rotor undercurrent condition which might result in slipping a pole and trip (open) the generator or main breaker (at a minimum) to try to prevent slipping a pole or pulling the rotor of synchronism.

markvguy

 

It's the convention chosen to express the conditions of GENERATOR operation, which is different than the conventions chosen to express what happens on an electrical grid such as the VAR flow through a substation. So it seems confusing, but one way to remember it is: Lagging VARs feed a lagging load (from the generator perspective).

Mr. Corso, who frequents this forum, could probably best describe the exact details of the conventions and why they were chosen as they were. How about it, Mr. Corso?

markvguy

 

Responding to Anonymous, Aug 24, 1:12am query regarding pf readings:

You are probably aware that the lag/lead values are based on the displacement angle between stator (or line) current and terminal voltage. An example might be useful.

Consider the case of a single synchronous motor supplied by a single generator. When the motor's field excitation is increased its 'back emf' tends to increase. But, because terminal voltage is constant the increase is slight. Then, a counteracting demagnetizing current is produced in the stator. This current is 'wattless' because the load is constant. And, it must 'lag' the 'back emf' of the motor. The line current will lead the motor's terminal voltage.

Conversely, the opposite responses are observed when the motor's excitation is decreased.

This now brings us to the question "How were your observations made... From a common pf meter? Or, from separate meters, one measuring the motor's pf, the other the generator's?

Responding to Anonymous' Aug 24, 12:56am query:

In its simplest form a motor's rotor is a large rotating magnet. Its rotation is caused by the magnetic attraction between the pole faces with the rotating magnetic field produced in the motor's stator. If the load torque is excessive, or terminal voltage too low, or excitation insufficient, then the rotor will pull-out-of-step or "slip-a-pole!"

Protection schemes/devices are available that will trip the motor off-line to prevent major damage. Re-syncrhonization is possible, but that's a subject for another day (nay, more like many days!)

For additional information see thread:

http://www.control.com/1026221273 "Alternator Runnig in Leading kVAR"

Regards, Phil Corso, PE {Boca Raton, FL, USA} [[email protected]] ([email protected])

 

My question is, two machines are connected in parallel, equally sharing a common inductive load. Now suppose we increase the excitation of one m/c say A and decresing the excitation of other machine say B. At that time also power o/p and terminal voltage shared equally. Beacause A is overexcited its armature current lags, opposing its field. And B is underexcited so its armature current, leads, aiding its field flux. each machine produces a new reactive current which increses A's armature current, but reduces B's. so my question is that, what is the reason that what is the reason due to which, when we over exit the m/c it oppse the field and reverse when under exit.

 

Responding to markvguy's Aug 24, 10:10pm comment... thanks for the accolade (I presume). Now two comments:

1) The Google reference cited by 'markvguy' is an excellent one. It jogged my memory. The most common error in protective relaying is the inattention paid to instrument transformer wiring connections and polarities.

2) Anyone heard of a "Supersynchronous" motor?

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Mr. Corso, thanks for your quick reply. And you should take it as the complement it was intended to be; your replies are always concise and to the point and you're extremely knowledgeable about electrical principles and properties; this author always learns from your replies.

About the supersynchronous motor, it's not a familiar term.

markvguy

 

My question is, that due to which principle lagging current oppose field flux and leading current aids the field flux?

From shvetang dave

 

Responding to Anonymous' Aug 25, 9:55pm query....

The question was answered as Case 1 in the cited thread # 1026221273
"Alternator Running in Leading kVAR!"

Are you interested in a variation of that description?

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

To markgvuy... ask the older engineers at GE to describe the supersynchronous motor! They should know... GE "invented" it.

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

yes i want description about regarding to following question.

My question is, two machines are connected in parallel, equally sharing a common inductive load. Now suppose we increase the excitation of one m/c say A and decresing the excitation of other machine say B. At that time also power o/p and terminal voltage shared equally. Because A is overexcited its armature current lags, opposing its field. And B is underexcited so its armature current, leads, aiding its field flux. Each machine produces a new reactive current which increses A's armature current, but reduces B's. so my question is what is the reason, when we overexcite the m/c it oppose the field and reverse when underexcited?

Reply From karan ray and shvetang dave.

 

Responding to Shevtang's Aug 26, 2:58 and Ray-Shevtang' Aug 29, 12:01am
queries... it ain't easy but I'll try again!

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])

 

Responding to Shvetang's Aug 26, 2:58pm and Karen-Shevtang's Aug 29, 12:01pm queries.... the principle is called Armature Reaction.

It's easy to say but not so easy to explain, but I will give it another shot. First, let's identify key variables. Neglecting resistances, then:

Vt = terminal or grid voltage.
Eg= emf induced in the armature (stator) winding by the field (rotor) flux.
Xs= synchronous reactance of the armature.
Ia = armature current.
Vs= voltage-drop caused by IaXs.

Now, the concept of displacement angle <d is introduced. <d represents the electrical displacement between the internal voltage Eg, and
terminal voltage Vt. The displacement <d, is a measure of the power developed in a synchronous machine.

1) No-Load to On-load Operation
Consider a generator connected to the grid, but delivering zero load. Vt and Eg are in phase (<d is 0). That is, the Vt voltage is syncronized to
the grid.

Now increase the prime mover's input to increase its speed. Since the rpm is fixed at grid frequency, the increased input, as seen on a scope
synch'ed to the grid, advances the rotor's position, causing Eg, to lead Vt. A current Ia, causes an internal voltage-drop Vs. The resultant
triangle consisting of Vt, Eg, and Vs is formed. An important fact is that advance of Eg wit respect to Vt, forces the generator to deliver the current, thereby delivering power.

(I know it's heady stuff but you asked for it, and remember w/o diagrams)

2) Excitation's Role on Reactive Current.
Now we introduce the concept of power factor or, putting it simply, Cos <t, where <t represents the angle between terminal voltage Vt, and armature current Ia.

Consider the earlier case, that is, two machines in parallel (w/o grid):
V1=V2, E1=E2, and Ia1 = Ia2, <d1=<d2, and the load is inductive. Now, increase the excitation in one, while decreasing the other. Voltage balance dictates that the overexcited machine deliver a lagging current that opposes the field-flux. Conversely, the underexcited machine delivers a leading current that aids its field flux. In each machine a new reactive current Ir, is added vectorially to each machines' armature current. Thus machine 1 and 2 current's become Ia1' and Ia2', respectively. However, because the load was taken as inductive, the load's lagging current is supplied by machine 1 only. The leading current of Machine 2 produces a circulating current, Ic, that flows in both machines.

Dave and Ray... any clearer?

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])