Can anyone explain theoretically how a synchronous motor can be changed from a lagging to leading power factor? I understand that it can be done by adjusting DC excitation, but I don't understand how a device consisting of resistance and inductance can draw a leading PF.
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synchronous motor leading pf 1
- Thread starter slushin
- Start date
Lets assume your synchronous motor is connected to the grid, is running at rated voltage and is delivering rated active power. Lets also assume that the excitation system is adjusted in such a way that zero reactive power is generated, i.e. the motor is running at 1.0 p.f. If you then increase the excitation current, the motor voltage cannot follow because the grid voltage remains stable. Instead, the motor will deliver overexcited reactive power to the system. If you decrease excitation, the motor voltage again doesn't change but underexcited reactive power will result.
Wolf
Wolf
Following wolf39 explanation I'll try to give some simply mathematics.
First of all the relationship:
E=V+R*I+j*X*I in complex numbers where:
E=EMF the voltage produces in armature [the induced voltage] by rotating poles [the inductor] proportionally with the magnetic flux variation in time E=-K*d(F)/dt.
R and X are the armature winding characteristics [considered linearly and including field losses and armature reaction] I=Ire+/-j*Iim the armature current [intensity] and V the grid/generator terminals voltage. From the attached sketch we could state:
E^2=(V+RI*cos(fi)+XI*sin(fi))^2+(XI*cos(fi)-RI*sin(fi)^2 and as R<<X we could neglect R then:
E^2=(V+XI*sin(fi))^2+(XI*cos(fi))^2 I*cos(fi)=Ire; I*sin(fi)=Iim
E^2=(V+X*Iim)^2+(X*Ire)^2 or E^2-V^2-2*V*X*Iim-X^2*Iim^2-X^2*Ire^2=0
The V is constant and if we'll take it as reference then Vre=V and Vim=0.
The apparent power S=P+j*Q S=sqrt(3)*V*(Ire+j*Iim) then V*Ire=P and V*Iim=Q.
If the primary driver [turbine, diesel or else] power remain constant P=constant so Ire=constant.
From equation a*x^2+2*b*x+c=0 x1,2=[-b+/-sqrt(b^2-a*c)]/a taking a=-X^2 b=-X c=E^2-V^2-X^2*Ire^2 we'll get:
Iim=-V/X+sqrt(E^2-X^2*Ire^2)/X .Let's say for Iim=0 Ecritic=sqrt(V^2+X^2*Ire^2)then:
If E<Ecritic V/X> sqrt(E^2-X^2*Ire^2)/X and Iim<0[underexcitation] ;if E>Ecritic Iim>0[superexcitation].
First of all the relationship:
E=V+R*I+j*X*I in complex numbers where:
E=EMF the voltage produces in armature [the induced voltage] by rotating poles [the inductor] proportionally with the magnetic flux variation in time E=-K*d(F)/dt.
R and X are the armature winding characteristics [considered linearly and including field losses and armature reaction] I=Ire+/-j*Iim the armature current [intensity] and V the grid/generator terminals voltage. From the attached sketch we could state:
E^2=(V+RI*cos(fi)+XI*sin(fi))^2+(XI*cos(fi)-RI*sin(fi)^2 and as R<<X we could neglect R then:
E^2=(V+XI*sin(fi))^2+(XI*cos(fi))^2 I*cos(fi)=Ire; I*sin(fi)=Iim
E^2=(V+X*Iim)^2+(X*Ire)^2 or E^2-V^2-2*V*X*Iim-X^2*Iim^2-X^2*Ire^2=0
The V is constant and if we'll take it as reference then Vre=V and Vim=0.
The apparent power S=P+j*Q S=sqrt(3)*V*(Ire+j*Iim) then V*Ire=P and V*Iim=Q.
If the primary driver [turbine, diesel or else] power remain constant P=constant so Ire=constant.
From equation a*x^2+2*b*x+c=0 x1,2=[-b+/-sqrt(b^2-a*c)]/a taking a=-X^2 b=-X c=E^2-V^2-X^2*Ire^2 we'll get:
Iim=-V/X+sqrt(E^2-X^2*Ire^2)/X .Let's say for Iim=0 Ecritic=sqrt(V^2+X^2*Ire^2)then:
If E<Ecritic V/X> sqrt(E^2-X^2*Ire^2)/X and Iim<0[underexcitation] ;if E>Ecritic Iim>0[superexcitation].
- Moderator
- #4
In words:
A motor must have the core magnetized in order to function.
An induction motor draws reactive current from the grid to magnetize the magnetic circuit.
When a synchronous motor is started as an induction motor, it draws magnetizing current or reactive current from the grid and the power factor is lagging.. When the field is applied, the field supplies the energy to magnetize the magnetic circuit. If the field does not supply enough energy to completely magnetize the magnetic circuit the remainder of the magnetizing force will be drawn from the grid as a reactive current. The power factor will be improved but still lagging.
As the field strength is increased, the motor will draw progressively less reactive current and the power factor will improve. When the field is at the proper level the power factor will be unity. If the field strength is increased further the power factor will go leading.
If this is an islanded installation over exciting the field may cause the supply voltage to rise.
Synchronous motors may be supplied without shaft extensions and with field windings suitable for higher than normal current levels. These may be run at full rated current with no mechanical load by increasing the field strength. The current will be reactive and leading. This is used to supply the VARs needed to correct the power factor at large industrial plants.
A synchronous motor used in this fashion is called a synchronous condenser.
Bill
--------------------
"Why not the best?"
Jimmy Carter
A motor must have the core magnetized in order to function.
An induction motor draws reactive current from the grid to magnetize the magnetic circuit.
When a synchronous motor is started as an induction motor, it draws magnetizing current or reactive current from the grid and the power factor is lagging.. When the field is applied, the field supplies the energy to magnetize the magnetic circuit. If the field does not supply enough energy to completely magnetize the magnetic circuit the remainder of the magnetizing force will be drawn from the grid as a reactive current. The power factor will be improved but still lagging.
As the field strength is increased, the motor will draw progressively less reactive current and the power factor will improve. When the field is at the proper level the power factor will be unity. If the field strength is increased further the power factor will go leading.
If this is an islanded installation over exciting the field may cause the supply voltage to rise.
Synchronous motors may be supplied without shaft extensions and with field windings suitable for higher than normal current levels. These may be run at full rated current with no mechanical load by increasing the field strength. The current will be reactive and leading. This is used to supply the VARs needed to correct the power factor at large industrial plants.
A synchronous motor used in this fashion is called a synchronous condenser.
Bill
--------------------
"Why not the best?"
Jimmy Carter
automatic2
Industrial
in yet other words;
stator current is a function of the sum of supply and Cemf. At unity, the summed voltage lies at 90 degrees and impresses a current 90 degrees lagging, which happens to fall at 0 degrees, our supply voltage reference. If we increase Cemf, the resultant voltage becomes greater than 90 degrees and the impressed current now leads the supply voltage.
stator current is a function of the sum of supply and Cemf. At unity, the summed voltage lies at 90 degrees and impresses a current 90 degrees lagging, which happens to fall at 0 degrees, our supply voltage reference. If we increase Cemf, the resultant voltage becomes greater than 90 degrees and the impressed current now leads the supply voltage.
It would be much easier to find a decent machines text and look at the internal vector diagram of the machine showing the relationship between internal EMF, terminal voltage, stator current, etc than try to describe it here in words or equations. There are a few references in FAQ238-1287.
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If we learn from our mistakes I'm getting a great education!
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If we learn from our mistakes I'm getting a great education!
You are right, Scotty. First of all I didn't think it will be so complicate at the beginning. Second I don't realize why I described the synchronous generator instead of the motor [it is not so big difference anyway].
Actually, I don't remember I saw anywhere an explanation of leading/lagging current phenomenon using only L, C and R parameters. But it is possible using La and Le instead of EMF to get such an explanation.
Actually, I don't remember I saw anywhere an explanation of leading/lagging current phenomenon using only L, C and R parameters. But it is possible using La and Le instead of EMF to get such an explanation.
-
1
- Moderator
- #8
May I try again?
A motor needs Vars. An induction motor draws VARs from the grid and the power factor is lagging.
A synchronous motor develops its own VARs depending on the field strength.
But if the field is not strong enough not enough VARs will be produced and the power factor will be lagging. At the optimum field strength the motor will generate just enough VARs and there will be no flow of VARs into or out of the machine. The power factor will be unity. As the field strength is increased the motor will generate more VARs than it needs and the surplus VARs will be exported to the grid. The power factor will be leading.
Bill
--------------------
"Why not the best?"
Jimmy Carter
A motor needs Vars. An induction motor draws VARs from the grid and the power factor is lagging.
A synchronous motor develops its own VARs depending on the field strength.
But if the field is not strong enough not enough VARs will be produced and the power factor will be lagging. At the optimum field strength the motor will generate just enough VARs and there will be no flow of VARs into or out of the machine. The power factor will be unity. As the field strength is increased the motor will generate more VARs than it needs and the surplus VARs will be exported to the grid. The power factor will be leading.
Bill
--------------------
"Why not the best?"
Jimmy Carter
- Moderator
- #10
Thanks 7aniter4. I hang out with a lot of very good people and it helps. You and Scotty are in the group as well as all the regulars here. I don't want to start listing for fear of missing good people, but you all know who you are and you know that you're good.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Bill
--------------------
"Why not the best?"
Jimmy Carter
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