Synchronous Machine

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Transcript Synchronous Machine

Synchronous Machine
•The stator is similar in construction that of a induction motor
•The rotor can be Salient or Non-Salient (cylindrical rotor)
•Field excitation is provided on the rotor by either permanent or
electromagnets with number of poles equal to the poles of the
RMF caused by stator
•Non-excited rotors are also possible as in case of reluctance motors
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Synchronous Machine (2)
•The rotor gets locked to the RMF and rotates unlike induction
motor at synchronous speed under all load condition
•All conventional power plants use synchronous generators for
converting power to electrical form
•They operate at a better power factor and higher efficiency than
equivalent induction machines
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Synchronous Machine Construction
(a)CRSM
(b) SPSM
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Concept of synchronous reactance (1)
• Like dc machines synchronous machines will also have armature
reaction. However unlike dc machine we do not like to eliminate
it, but try to use it to our benefit.
•Essentially, this armature reaction will determine how much
power can be transferred to or from the synchronous machine and
limits the current that is flowing in the synchronous machine and
hence provides inherent short-circuit protection: a great boon
when we are talking about zillions of megawatts of power flow!
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Concept of synchronous reactance (2)
•Suppose we short-circuit a synchronous generator with the field
circuit excited. By Faraday’s law an emf will be induced in the
stator (armature) which by Lenz’s law has to oppose the original
field on the rotor. It means the resulting armature reaction will
induce an opposing emf to the one produced by the main field.
•One way to represent this is the following circuit where Xar
conjures up the effect of armature reaction. This can be proved
as follows: Suppose the original field flux is f= m cost.
By Farday’s and Lenz’s law this would produce a voltage
Ef= Em sint. This voltage produce a current and hence flux that
opposes f, under short-circuit. This current then has to of the form
Isc=-Im cost. Clearly Ef/jXar= Em sint/jXar would give such a current.
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Equivalent circuit of CRSM (1)
Machine
Generator (Appx.)
Machine
Motor(Appx.)
Machine
Machine
Generator (Exact)
Motor(Exact)
•Only difference is in current direction; in a generator it flows
out of it, in case of a motor
it flows into it.
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Equivalent circuit of CRSM (2)
Machine
Machine
Motor(Exact)
Generator (Exact)
Xs=Xar+Xal (Synchronous reactance)
Zs= Ra+jXs (Synchronous impedance)
Xal is leakage Reactance
Ra is armature resistance
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Phasor diagram of CRSM
Note:  is +ve for (a) generator and –ve for (b) motor
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Derivation of power equation for CRSM
on the green board
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Effect of Load Change (Field constant)
Note: Er same as Ef
Va same as Vt
Ra has been neglected
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Effect of Field Change (Load constant)
Note: Er same as Ef
Va same as Vt
Ra has been neglected
Question: 1)Why is the loci of stator current and excitation voltage
moves on a straight line?
2) What is happening to power factor as field is changed?
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V curves
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Effect of Field Change (Load constant)
for a generator
a Power
Ia2
Ef2
jIa2Xs
Ef1
jIa1Xs
Ia1
Vt
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a Power
Conclusion for effect for field change with
constant load on power factor
•For motor with increased (decreased)excitation power factor becomes
leading (lagging)
•For generator with increased (decreased) excitation power factor
becomes lagging (leading)
•Unloaded overexcited synchronous motors are sometimes used
to improve power factor. They are known as synchronous condensers
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Torque versus Electrical Load Angle
Normalized Torque, Power
1
0.5
Generator
0
Motor
-0.5
-1
Tmax,Pmax
-3
-2
-1
0
1
Delta(Radians)
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2
3
Torque versus Speed
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Example 1
A six-pole 60 Hz synchronous motor is operating with a developed
power of 5 hp and a torque angle of 5o. Find the speed and developed
torque. Suppose that the load increases such that the developed torque
doubles. Find the new torque angle. Find the pull-out torque and
maximum developed power for this machine.
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Example 2
An eight-pole, 240 V-rms, 60 Hz, delta connected synchronous motor
operates with a constant developed power of 50 hp and a torque angle
of 15o and unity power factor. Suppose the field current is increased
by 20%. Find the new torque angle and power factor. Is the new power
factor lagging or leading? Assume linear magnetic characteristics.
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SPSM and the concept of Direct and
Quadrature Axes
 Since in the salient pole machine the reluctance of the machine
varies with the position of the pole, flux due to armature reaction
varies with power factor. Thus Xar alone is no longer sufficient
for the equivalent circuit.
 Reluctance is minimum along polar (direct) axis. Hence component
of the armature reaction acting along this axis produce maximum flux.
Let this component be ad.
Reluctance is maximum along the inter-polar (quadrature )axis.
Hence the component of the armature reaction acting along this axis
produce minimum flux. Let this component be aq.
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SPSM and the concept of Direct and
Quadrature Axes (2)
Xd=Xad+Xal=(d)irect axis synchronous reactance)
Xq=Xaq+Xal= (q)uadrature axis synchronous reactance)
 Xad= d(irect) axis armature reactance =Lad
 Xaq = (q)uadrature axis armature reactance=Laq
Xal = leakage reactance
ad=LadId
aq=LaqIq
 Id= d(irect) axis component of the armature current
 Iq = (q)uadrature axis component of the armature current
Ia=Iq±jId
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Explaining d-q axes using diagrams
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Equivalent circuits of SPSM
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Power Angle Characteristics of SPSM
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Examples on SPSM
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