synchronous motor leading pf
synchronous motor leading pf
(OP)
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.





RE: synchronous motor leading pf
Wolf
RE: synchronous motor leading pf
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].
RE: synchronous motor leading pf
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
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"Why not the best?"
Jimmy Carter
RE: synchronous motor leading pf
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.
RE: synchronous motor leading pf
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If we learn from our mistakes I'm getting a great education!
RE: synchronous motor leading pf
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.
RE: synchronous motor leading pf
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
RE: synchronous motor leading pf
RE: synchronous motor leading pf
Bill
--------------------
"Why not the best?"
Jimmy Carter