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3. SYNCHRONOUS MACHINES
Work done by:
Diogo Mariano
Fernando Arantes
André Freitas
3.1. Construction of synchronous machines
Synchronous machines are AC machines that have a field circuit supplied by an external DC source. In a synchronous generator, a DC current is applied to the rotor winding producing a rotor magnetic field. The rotor is then turned by external means producing a rotating magnetic field, which induces a 3-phase voltage within the stator winding. In a synchronous motor, a 3-phase set of stator currents produces a rotating magnetic field causing the rotor magnetic field to align with it. The rotor magnetic field is produced by a DC current applied to the rotor winding. Field windings are the windings producing the main magnetic field (rotor windings for synchronous machines); armature windings are the windings where the main voltage is induced (stator windings for synchronous machines).
The rotor of a synchronous machine is a large electromagnet. The magnetic poles can be either salient (sticking out of rotor surface) or non-salient construction.
Alternator and Motor
These AC generators can be divided in many ways but we will discuss now two main types of alternator categorized according to their design. These are
Salient pole type
 It is used as low and medium speed alternator. It has a large number of projecting poles having their cores bolted or dovetailed onto a heavy magnetic wheel of cast iron or steel of good magnetic quality. Such generators are characterized by their large diameters and short axial lengths. These generators are look like big wheel. These are mainly used for low speed turbine such as in hydra power plant.
Non-Salient pole type
 It is used for steam turbine driven alternator. The rotor of this generator rotates in very high speed. The rotor consists of a smooth solid forged steel cylinder having a number of slots milled out at intervals along the outer periphery for accommodation of field coils. These rotors are designed mostly for 2 pole or 4 pole turbo generator running at 36000 rpm or 1800 rpm respectively.
3.2. 	Principle of operation on the synchronous machines
Synchronous motor is a doubly excited machine i.e two electrical inputs are provided to it. It’s stator winding which consists of a 3 phase winding is provided with 3 phase supply and rotor is provided with DC supply. The 3 phase stator winding carrying 3 phase currents produces 3 phase rotating magnetic flux. The rotor carrying DC supply also produces a constant flux. Considering the frequency to be 50 Hz, from the above relation we can see that the 3 phase rotating flux rotates about 3000 revolution in 1 min or 50 revolutions in 1 sec. At a particular instant rotor and stator poles might be of same polarity (N-N or S-S) causing repulsive force on rotor and the very next second it will be N-S causing attractive force. But due to inertia of the rotor, it is unable to rotate in any direction due to attractive or repulsive force and remain in standstill condition. Hence it is not self starting.
To overcome this inertia, rotor is initially fed some mechanical input which rotates it in same direction as magnetic field to a speed very close to synchronous speed. After some time magnetic locking occurs and the synchronous motor rotates in synchronism with the frequency.
3.2.1 Rotating Magnetic field
In order to understand the operation of 3-phase AC machines, it is essential to understand the magnetic field produced by currents in the three phases of the stator winding. We will study the mmf pattern of a 3-phase winding such as those found on the stator of 3-phase induction and synchronous machines. We will consider here a 2-pole machine with six slots so that each phase winding is concentrated in slots 180° electrical apart. Since it is a 2-pole machine, six slots will be considered i.e., 2-slots per pole as shown in Fig. 1.
Now consider the situation at t = 0. Fig. 1(a) the moment when current in phase a is at its maximum value.
The mmf of phase a then has its maximum value Fmax. perpendicular to the plane of the coil aa' as shown in Fig. 1 (a) and is drawn along the magnetic axis of phase 'a'.
 Illustration the production of a rotating magnetic field by means of 3-phase currents
At this moment the currents ib and ic are both Im/2 in the negative direction as shown in the phasor diagram which is the actual instantaneous direction. The corresponding mmfs of phase b and c are shown by the phasers Fb and Fe both equal to Fmax/2 drawn in the negative direction of respective magnetic axes. The resultant is obtained by taking the phasor sum of the three phasers which comes to 3/2Fmax centred along the magnetic axis of phase a.
It represents a sinusoidal space wave with its positive half wave centred on the axis of phase a and having an amplitude 3/2 times that of the phase a. 
Consider the instant t1 such that ωt1 = π/3, as shown in Fig. 1 (b) the currents in phase a and b are a positive half-maximum and that in phase c is a negative maximum as shown in corresponding phasor diagram. The individual mmf components and their resultant are shown in Fig. 1 (b). 
The resultant has the same magnitude as at t = 0 but it has now rotated anticlockwise 60° electrical in space. Similarly at ωt2 = 2π/t when phase b current is a positive maximum and 3 the phase a and c currents are a negative half maximum (see phasor diagram), the same resultant mmf distribution is obtained but it has rotated another 60° in anticlockwise direction and now it is along magnetic field axis of phase b. With passage of time the resultant mmf wave retains its sinusoidal form and amplitude but shifts progressively around the air gap. This shift corresponds to a field rotating uniformly around the circumference of the air gap.
In one cycle the resultant mmf comes back in the position of Fig. 1 (a). The mmf wave, therefore, makes one revolutions per cycle in a 2-pole machine and in one cycle the rotor has moved one complete revolution in a 2-pole machine and hence in a P-pole machine the wave travels one wavelength or 2/P revolutions per cycle. Therefore, the mmf wave rotates at synchronous speed in space and its constant magnitude equals 1.5 times the maximum mmf of one phase.
Analytical method : Consider Fig. 1 (d) where the phase a magnetic axis is taken as reference and consider the instant when the mmf of phase has moved through an angle θ when the magnitude of mmf of phase a is Fapeak, those of phase b and c are Fbpeak and Fcpeak respectively. 
Taking the components of mmf due to all the three phases along the reference axis, the total mmf along the axis is:
*(1)…	F (θ) = Fapeak.cosθ+ Fbpeak.cos(θ -120) +Fcpeak.cos(θ-240)
However, the amplitudes of the mmf vary with time in accordance with the variation of current. Thus with the time origin arbitrarily taken as the instant when the phase a current is a positive maximum. Hence:
Fapeak = Famax.cosωt
Fbpeak = Fbmax.cos(ωt-120°)
	Fcpeak = Fcmax.cos(ωt-240°)			
Here 120° phase displacement is due to displacement of current in the three phases and since the currents are balanced IIaI = IlbI = IlcI and hence:
Famax = Fbmax = Fcmax = Fmax
Hence equation *(1) can be rewritten as 
F(θ) =Fm.cosθ.cos.ωt+Fm.cos(θ-120).cos(ωt-120)+Fmcos(θ-240)cos(θt-240)
Now using the trigonometric relation:
cosα.cosβ = (1/2).[cos(α-β)+cos(α+β)]
we have
F(θ) = (Fm/2).[cos(θ -ωt)+cos(θ+ωt)] 
 +cos(θ-120-ωt+120)+cos(θ-120+ωt-120)
 +cos(θ-240-ωt+240)+cos(θ-240+ωt-240)] 
*(2)…	=(Fm/2).[cos(θ-ωt)+cos(θ+ωt)+cos(θ-ωt)+cos(θ+ωt-240)+cos(θ-ωt)+cos(θ+ωt-480)
Now the three terms in the above equation involving the angle θ+ωt, (θ+ωt-240) and (θ+ωt-480°) sum to zero as these are three phaors displaced in phase by 120°. Note that the term θ+ω t -480° is equivalent to (θ+ωt-120°).
 	Therefore, the equation *(2) reduces to 
*(3)…	F(θ) = (3Fm/2)cos.(θ-ωt)
 Which is a desired expression for theresultant mmf wave.
The wave described by equation *(3) is a sinusoidal function of the space angle θ. It has a constant amplitude and space-phase angle ωt which is a linear function of time. The angle ωt provides the rotation of the entire wave around the air gap at the constant angular velocity ω rad/ sec. 
3.3.	Mathematical model of synchronous machines
3.3.1.	Synchronous motor equivalent circuit
The following circuit diagrams illustrate the per phase equivalent circuits of a round rotor synchronous machine in the motor and generator mode respectively
Synchronous machine per phase equivalent circuits in (a) generator, and (b) motor reference directions
In the per phase equivalent circuit model illustrated above, there are three parameters need to be determined: winding resistance Ra, synchronous reactance Xs, and induced emf in the phase winding Ea. The phase winding resistance Ra can be determined by measuring DC resistance of the winding using volt-ampere method, while the synchronous reactance and the induced emf can be determined by the open circuit and short circuit tests.
3.3.2.	Phasor Diagrams
3.4.	Characteristics of synchronous motor
3.4.1.	Motor-torque
If the machine is being used as a generator, torque is being supplied by a prime-mover. The generator load current produces a rotating magnetic field in the stator (armature) which tries to oppose the rotation of the rotor. This action produces a torque which is transmitted to the generator foundation. Fig.4 shows the torque versus torque angle  characteristic in both the motor and generator modes. 
Fig.4 Steady-state torque-angle Characteristic.
3.4.2.	‘V’ Curve for the synchronous machine
The magnitude of the armature current reflects the transfer of reactive energy into or out of the machine. If the field is providing just the right amount of flux to produce the torque, the armature current will transmit only real power. At this condition, the armature current magnitude is minimized for this particular load and the terminal power factor is unity. Figure 11 shows the armature current versus field current curves for different real power conditions. These are called the V-curves. 
	
3.5.Synchronous Machine Operated as a Motor
Electromagnetic Power and Torque
When a synchronous machine is operated as a motor to drive a mechanical load, in steady state, the mechanical torque of the motor should balance the load torque and the mechanical loss torque due to friction and windage, that is
	
T Tload
Tloss
Multiplying the synchronous speed to both sides of the torque equation, we have the power balance equation as
Pem
Pload
Ploss
where Pem=Twsyn the electromagnetic power of the motor, Pload=Tloadwsyn is the mechanical power delivered to the mechanical load, and Ploss=Tlosswsyn the mechanical power loss of the system. Similar to the case of a generator, the electromagnetic power is the amount of power being converted from the electrical into the mechanical power. That is
Pem
3Ea I a cosjE Ia a
Twsyn
where EaIa is the angle between phasors Ea and Ia.
wsyn T
jXs Ra Ia
Mech. Ea
Load
Motor Va
Tload Tloss
A synchronous machine operated as motor
When the stator winding resistance is ignored, the per phase circuit equation can be approximately written
as d
Va
jXs I a
Va E a
jX s I a
j Ea
I a
The corresponding phasor diagram is shown on the
right hand side. From the phasor diagram, we can readily obtain
Motor phasor diagra
Therefore,
Pem
3E aVa 
X
sin d
and
T Pem 
wsyn
s
 3E aVa 
wsyn X s
sin d
where d is the load angle. When the stator winding resistance is ignored, d can also be regarded as the angle between the rotor and stator
Electromagnetic torque
vs. load angle
rotating magnetic fields. In motor mode, the stator field is ahead of the rotor. The electromagnetic torque of a synchronous machine is proportional to the sine function of the load angle, as plotted in the diagram above, where the curve in the third quadrant is for the situation when the machine is operated as a generator, where the electromagnetic torque is negative because the armature current direction is reversed.
3.6. Synchronous Motor Power Factor
Assume that a synchronous motor is driving a constant torque load. The active power converted by the machine is constant, no matter what the value of the field current is, since the motor speed is a constant. Thus,
 3Va E a 
T
w syn X s
sin d
constant
or Ea sin d constant
and
Pem
3Va I a cosj constant
or I a cosj constant
Using the phasor diagram below, we analyze the variation of the power factor angle of a synchronous motor when the rotor field excitation is varied. For a small rotor field current the induced emf in the stator winding is also small, as shown by the phasor Ea1. This yields a lagging power factor angle 1 > 0. As the excitation current increases, the lagging power factor angle is reduced. At a certain rotor current, the induced emf phasor Ea2 is perpendicular to the terminal voltage phasor, and hence the stator current phasor is aligned with the terminal voltage, that is a zero power factor angle 2 = 0. When the rotor current
further increases, the stator current leads the terminal voltage, or a leading power factor angle 3 < 0. In the phasor diagram, the above two conditions on Ea and Ia mean that they will only be able to vary along the horizontal and the vertical dotted lines, respectively, as
shown below.
I a cosj
Ia1
j3 Ia2 Va
Ea sind
j1 d 1
Ia3
d 2 d 3
jX s Ia1
jX s Ia3
jX s Ia2
Ea1 Ea2 Ea3
Phasor diagram of a synchronous motor in under excitation, unit power factor, and over excitation mode
For conversion of a certain amount of active electrical power into mechanical power, a certain amount of magnetic flux is required. In the case of a lagging power factor, the rotor field current is so small that some reactive power is required from the stator power supply, and hence the stator current lags the terminal voltage. This state is known as under excitation. When the rotor field current is just enough to produce the required magnetic flux, a unit power factor is obtained. If the rotor field current is more than required the spurious reactive power is to be exported to the power lines of the power supply. This state is known as over excitation.
In practice, because of this feature, synchronous motors are often run at no active load as synchronous condensers for the purpose of power factor correction. The diagram underneath the phasor diagram 
illustrates schematically the power factor compensation for an inductive load, which is common for factories using large induction motor drives, using a synchronous condenser. By controlling the rotor excitation current such that the synchronous condenser draws a line current of leading phase angle, whose imaginary component cancels that of the load current, the total line current would have a minimum imaginary component. Therefore, the overall power factor of the inductive load and the synchronous condenser would be close to one and the magnitude of the overall line current would be the minimum.
It can also be seen that only when the power factor is unit or the stator current is aligned with the terminal voltage, the magnitude of thestator current is minimum. By plotting the magnitude of the stator current against the rotor excitation current, a family of “V” curves can be obtained. It is shown that a larger rotor field current is required for a larger active
load to operate at unit power
Three PhaseI
Power Supply s
Inductive
Load
I cmp
Synchronous
Condenser
I load
Power factor compensation for an inductive load using a synchronous condenser
3.7. Synchronous Motor Drives
A synchronous motor cannot start in synchronous mode since the inertia and the mechanical load prevent the rotor to catch up with the rotating magnetic field at the synchronous speed. A common practice is to embed a few copper or aluminum bars short circuited by end rings in the rotor and to start the motor as an induction motor (the principle of induction motors is discussed in another chapter). When the rotor speed is close to the synchronous speed, the rotor is energized with a DC power supply and it will catch up or synchronize with the rotating magnetic field. This, however, is not a problem for power electronic inverter controlled synchronous motors because the inverter can ramp up the excitation frequency.
Since the rotor speed is proportional to the stator excitation frequency, the speed of a synchronous motor can only be controlled by varying the stator frequency. A common speed control strategy is the variable voltage variable frequency (VVVF) speed control, in which the ratio between the stator voltage and frequency is kept a constant. Below is the block diagram of an open loop synchronous motor drive. For rotor speeds below the rated speed, VVVF strategy is employed, and the maximum torque that the motor can produce is a constant. When the rotor speed required is higher than the rated speed, the stator voltage is capped to the rated voltage while the frequency is increased. The maximum torque is then reduced as the speed increases. As illustrated by the torque-speed curves in the diagram below, the motor drive is suitable for a constant torque load when the speed is below the rated speed, and would be suitable for a constant power load when the speed is higher than the rated speed.
r
max
rated
0 Tmax T
Torque speed curves of a synchronous motor with VVVF control
In the closed loop control, the stator excitation can be controlled according to the rotor position such that stator magnetic field is perpendicular to the rotor field and hence the electromagnetic torque the motor produces is always maximum under any load conditions. The torque speed curve of the motor in this case is essentially same as that of a DC motor. This type of motor drive is known as the brushless DC motors, which will discussed in another chapter. The diagrams below illustrate an optic position sensor and the block diagram of the closed loop synchronous motor drive.