Three Phase Synchronous Machines

Principle

A synchronous motor runs at constant speed. For constant load and constant supply voltage $V$, the power input $P = \sqrt3 V_L I\cos\phi$ stays constant, so the active component $I\cos\phi$ of the armature current stays constant. When field excitation $I_f$ is changed, the back emf $E_f$ changes, and the current adjusts so that the reactive component changes. This changes the power factor. (Armature resistance is neglected.)

The phasor relation is $\vec V = \vec E_f + j\vec I X_s$ for a motor.

Three conditions

1. Normal excitation: $E_f \cos\delta = V$ approximately. The current is minimum, in phase with $V$, and the power factor is unity.
2. Under-excitation ($E_f$ smaller): the current lags the voltage. The motor draws reactive power from the supply like an inductive load: lagging power factor. Current is larger than at unity pf.
3. Over-excitation ($E_f$ larger): the current leads the voltage. The motor supplies reactive power to the supply: leading power factor. Current is again larger.

 Under-excited     Normal          Over-excited
     V                V                 V
     |                |                 |
  Ef |  jIXs       Ef |  jIXs        Ef |     jIXs
  <--+--->         <--+--->          <--+--------->
      \              |                 /
      I (lag)        I (in phase)     I (lead)

In each case the end of the $I\cos\phi$ line stays on a horizontal line (constant power), and the locus of $E_f$ is a horizontal line (constant power condition $E_f \sin\delta$ = constant).

Application

An over-excited, no-load synchronous motor is a synchronous condenser used for power factor correction. The V-curve plots armature current against field current for each load; the minimum current of each curve corresponds to unity power factor.

Why a synchronous motor is not self-starting

When three-phase supply is applied to the stator, a rotating field at synchronous speed $N_s = \frac{120f}{P}$ is produced. The rotor poles, excited by DC, are stationary. The rotating stator pole is attracted by one rotor pole, then in the next half cycle (in 1/100 s at 50 Hz) it has moved on and the force reverses. The rotor, because of its inertia, cannot accelerate to $N_s$ in that short time. The torque reverses every half cycle, so the average starting torque is zero. The rotor must first be brought close to synchronous speed by other means.

Starting methods

1. Using a pony (auxiliary) motor: a small induction or DC motor drives the rotor to near synchronous speed. The field is then excited, synchronised with the supply like an alternator, and the pony motor is disconnected. It is used for large motors that can start without load.
2. Using damper windings (induction motor method): squirrel-cage type bars are placed in the pole faces and short-circuited by end rings.
   • With the field winding short-circuited through a resistance (to avoid high induced voltages), the stator is supplied with reduced voltage (by autotransformer or star-delta starter).
   • The damper winding develops torque like a squirrel-cage rotor, and the motor accelerates to about 95 % of $N_s$.
   • DC excitation is then applied. The rotor pulls into step (synchronises) and the damper winding carries no current.
3. Using a variable-frequency supply: the frequency is raised gradually from a very low value, so the rotor locks to the field.
4. As a slip-ring motor (rotor resistance): special cases of large machines.

  Supply
    |
 [Autotransformer] -> stator
                     rotor: damper bars shorted
                     field winding --[R]-- (shorted)
    after ~95% Ns: remove R, apply DC field

The motor is started on no load or light load because pull-in torque is limited.

A synchronous motor has no starting torque because the stator field rotates at $N_s = \frac{120f}{P}$ while the rotor is stationary. The rotor poles are attracted and then repelled in each half cycle, so the average torque is zero. The motor must be brought near synchronous speed before DC field excitation is applied. The common methods are given below.

1. Using a pony (auxiliary) motor

A small induction or DC motor, mechanically coupled to the shaft, runs the rotor close to $N_s$ without load. The rotor is then excited and synchronised with the supply in the same manner as an alternator (lamp or synchroscope method), the pony motor is disconnected, and the load is applied. It needs an extra motor and is used for large machines.

2. Using damper (amortisseur) winding

Short-circuited copper bars are placed in slots in the pole faces and joined by end rings, similar to a squirrel-cage rotor.

• Reduced voltage is applied to the stator. The field winding is closed through a resistance (about 10 times the field resistance) to avoid dangerous high voltage in the many-turn field.
• The motor starts as an induction motor and rises to about 95 % of $N_s$.
• DC field is applied, and the rotor locks into synchronism.
• Once running at $N_s$ there is no relative motion, so the damper winding carries no current. It also reduces hunting.

  stator ---(autotransformer / star-delta)--- supply
  rotor: pole-face damper bars + end rings
  field winding --[discharge resistor]--

3. Using variable frequency supply

An inverter supplies the stator at very low frequency at first, so the rotating field is slow and the rotor locks to it. The frequency is gradually raised to the rated value. This is used in modern drives.

4. Using the DC motor method

The exciter or a DC machine on the same shaft runs the machine as a DC motor to bring it to speed.

Of these, the damper-winding method is the most common.

Given: 1200 kVA, 6600 V (line), star, $R_a = 0.4\ \Omega$, $X_s = 6\ \Omega$ per phase, full-load current, 0.8 pf lagging at rated voltage. Find the terminal voltage for the same excitation and current at 0.8 pf leading.

Rated quantities

$$V_{ph} = \frac{6600}{\sqrt3} = 3810.5\ \text{V}, \qquad I = \frac{1200\times10^3}{\sqrt3\times6600} = 104.97\ \text{A}$$ $$Z_s = 0.4 + j6\ \Omega$$

Excitation emf (0.8 lagging)

Taking $V_{ph}$ as reference, $\vec I = 104.97\angle -36.87^\circ$ A:

$$\vec E = \vec V + \vec I Z_s = 3810.5 + 104.97\angle-36.87^\circ(0.4 + j6)$$ $$IR = 41.99\ \text{V},\quad IX_s = 629.84\ \text{V}$$ $$E = \sqrt{(3810.5 + 41.99\times0.8 + 629.84\times0.6)^2 + (629.84\times0.8 - 41.99\times0.6)^2}$$ $$E_{ph} = 4249.05\ \text{V} \;(\text{line } 7359.6\ \text{V})$$

Terminal voltage at 0.8 leading (same $E$ and $I$)

Now $\vec I = 104.97\angle+36.87^\circ$ A. The drop $\vec I Z_s$ has magnitude $104.97\times 6.0133 = 631.23$ V at an angle $36.87^\circ + 86.19^\circ = 123.06^\circ$ with respect to $V$. For the same excitation, $|\vec V + \vec I Z_s| = E$, so

$$V^2 + 2V(631.23)\cos 123.06^\circ + 631.23^2 = 4249.05^2$$ $$V^2 - 688.62\,V - 17.656\times10^{6} = 0$$ $$V_{ph} = \frac{688.62 + \sqrt{688.62^2 + 4\times17.656\times10^{6}}}{2} = 4560.3\ \text{V}$$ $$V_L = \sqrt3\times 4560.3 = 7898.7\ \text{V}$$

Answer: terminal voltage at 0.8 leading $\approx 7899$ V (line), i.e. 4560 V per phase, higher than rated because of the leading current (Ferranti-type effect; armature reaction aids the field).

Given: 3.3 kV, 3-phase star motor, $Z_s = 0.2 + j2.2\ \Omega$ per phase, 0.5 pf leading, $I_L = 100$ A.

$$V_{ph} = \frac{3300}{\sqrt3} = 1905.26\ \text{V}, \qquad I = 100\ \text{A}$$

$\cos\phi = 0.5$ leading, so $\phi = 60^\circ$ and $\vec I = 100\angle +60^\circ$ A with $V_{ph}$ as reference.

For a motor, the back emf is

$$\vec E_b = \vec V - \vec I Z_s$$

Impedance drop:

$$\vec I Z_s = 100\angle60^\circ\times(0.2 + j2.2) = 100\angle 60^\circ\times 2.209\angle 84.8^\circ = 220.9\angle144.8^\circ = -180.53 + j127.32\ \text{V}$$ $$\vec E_b = 1905.26 - (-180.53 + j127.32) = 2085.79 - j127.32\ \text{V}$$ $$E_b = \sqrt{2085.79^2 + 127.32^2} = 2089.66\ \text{V}, \qquad \delta = \tan^{-1}\!\left(\frac{-127.32}{2085.79}\right) = -3.49^\circ$$

The line value is $\sqrt3\times 2089.66 = 3619.4$ V.

Phasor diagram

              V (reference)
  ------------>--------------------
     \   \            
      \    \  jIXs + IR (I Zs)
  Eb   \     \
        v      ^ I (leads V by 60 deg)

(Eb lags V by 3.49 deg; I leads V by 60 deg; the motor is over-excited because $E_b > V$.)

Answer: back emf per phase $= 2089.7$ V (line 3619 V), load angle $3.49^\circ$ (E lagging V).

Load characteristics of a synchronous generator

The load characteristic (external characteristic) is the plot of terminal voltage $V$ against load current $I$ at constant speed, constant field current and constant power factor.

 V
 |            leading pf (rises)
 |         _.-'
 |  ______-----   unity pf (small fall)
 |        `-._
 |            `-._  lagging pf (falls)
 +----------------------> I

• Lagging pf (inductive): the armature reaction is demagnetising and the drop $I Z_s$ is in the same direction, so $V$ falls considerably with load.
• Unity pf: the armature reaction is mainly cross-magnetising; $V$ falls slightly.
• Leading pf (capacitive): the armature reaction is magnetising, so $V$ rises with load.

Why $V > E$ with capacitive load

The phasor equation for a generator is $\vec E = \vec V + \vec I(R_a + jX_s)$, where $X_s$ includes armature reaction reactance ($X_s = X_{ar} + X_l$).

• With leading current, the armature flux aids the main field flux (magnetising armature reaction). Equivalent to a drop of $jIX_s$ that is directed opposite to the lagging case.
• For a leading current, the reactance drop $j\vec I X_s$ points opposite to $\vec V$ (it leads $\vec I$ by $90^\circ$, and $\vec I$ itself leads $\vec V$). In $\vec E = \vec V + j\vec I X_s$ the second term therefore reduces the magnitude, so $E < V$.

        jIXs
     <----------.         I leads V
      \          \        E = V + jIXs
       \  E       \       is shorter than V
        `-----------> V

Neglecting $R_a$, with $I$ leading by angle $\phi$ (and $\phi$ small compared to the voltage):

$$E \approx V - I X_s \sin\phi \quad\Rightarrow\quad V \approx E + IX_s\sin\phi > E$$

The same result is seen physically: the capacitive load current produces a field that adds to the main field, so the net flux (and the terminal voltage) is greater than that due to the field current alone. This is the Ferranti-like voltage rise, and it is why voltage regulation is negative for leading pf.

Working of a three-phase synchronous generator (alternator)

1. The rotor carries a DC-excited field winding (the poles). The prime mover turns the rotor at constant speed $N_s$.
2. The rotating flux cuts the three-phase armature (stator) windings, placed $120^\circ$ apart. By Faraday's law, an emf is induced in each phase:

$$E = 4.44\,f\,\phi\,T_{ph}K_wK_d \quad\text{with}\quad f = \frac{PN_s}{120}$$

3. The three emfs are equal in magnitude and displaced by $120^\circ$ in time, giving a balanced three-phase supply.
4. When a load is connected, a three-phase armature current flows, and the terminal voltage is $V = E - I(R_a + jX_s)$.

Armature reaction

Armature reaction is the effect of the armature (stator) mmf, produced by the load current, on the main field flux (its magnitude and position). Armature current produces a rotating field at synchronous speed which is stationary relative to the rotor, and it either opposes, aids or distorts the main flux.

Effect of load power factor

 Lagging: Fa opposes Ff     Leading: Fa aids Ff
 N <---Ff  <---Fa           N --->Ff  --->Fa

The armature reaction is represented by a reactance $X_a$, which together with leakage reactance gives the synchronous reactance $X_s = X_a + X_l$.

Armature reaction in a synchronous generator is the effect of the armature mmf $F_a$ (produced by load current) on the main field mmf $F_f$. Armature current in the three-phase winding produces a field that rotates at synchronous speed, so it is stationary with respect to the rotor poles. The result depends on the angle $\psi$ between the armature current and the induced emf $E$.

1. Resistive load (unity pf)

Current $I$ is in phase with $E$. The armature mmf $F_a$ lags $F_f$ by $90^\circ$, so it acts across the pole axis.

• Cross-magnetising armature reaction.
• The flux is distorted: the trailing pole tip is strengthened, the leading tip weakened; the net flux is roughly the same (slightly reduced by saturation).
• Terminal voltage drops slightly.

  Ff -->   Fa (down, 90 deg)   Result: Fr = Ff + Fa
   N-----S      |                 (shifted)

2. Inductive load (lagging pf)

Current $I$ lags $E$ by up to $90^\circ$. For zero-pf lagging, $F_a$ is directly opposite to $F_f$.

• Demagnetising (partly cross-magnetising for general lagging).
• Net flux is reduced, so the induced emf falls and $V$ drops significantly.
• To keep $V$ constant the field current must be increased.

  Ff -->        <-- Fa (zero pf lag)
  Fr = Ff - Fa

3. Capacitive load (leading pf)

Current $I$ leads $E$ by up to $90^\circ$. For zero-pf leading, $F_a$ is in the same direction as $F_f$.

• Magnetising (partly cross-magnetising for general leading).
• Net flux increases, the induced emf rises, so $V$ may exceed $E$.
• The field current must be decreased to hold voltage.

  Ff -->   --> Fa (zero pf lead)
  Fr = Ff + Fa

Derivation of the emf equation

Let

• $P$ = number of poles, $\phi$ = flux per pole (Wb)
• $N_s$ = speed in rpm, $f = \frac{PN_s}{120}$ = frequency
• $Z$ = total conductors in series per phase, $T_{ph} = Z/2$ turns per phase

Average emf per conductor. In one revolution a conductor cuts $P\phi$ weber in time $\frac{60}{N_s}$ s. So the average emf per conductor is

$$e_{avg} = \frac{P\phi}{60/N_s} = \frac{P\phi N_s}{60} = 2f\phi\ \text{V}$$

since $\frac{PN_s}{60} = 2f$.

Per turn (two conductors): $e_{turn} = 4f\phi$.

Per phase with $T_{ph}$ turns in series:

$$E_{avg} = 4 f \phi T_{ph}$$

Form factor for a sinusoidal emf $= \frac{\text{rms}}{\text{avg}} = 1.11$:

$$E_{ph} = 1.11\times 4f\phi T_{ph} = 4.44\, f\,\phi\, T_{ph}\ \text{V}$$

Winding factors. In a practical winding, the coils are short-pitched and the phase belt is distributed over several slots. This reduces the emf by two factors:

• Pitch factor $K_p = \cos(\alpha/2)$, where $\alpha$ is the angle by which the coil is short-pitched (electrical degrees).
• Distribution factor $K_d = \dfrac{\sin(m\gamma/2)}{m\sin(\gamma/2)}$, where $m$ is slots per pole per phase and $\gamma$ the slot angle.

$$\boxed{E_{ph} = 4.44\,K_pK_d\, f\,\phi\, T_{ph}\ \text{V}}$$

For a star connection, line emf $E_L = \sqrt3E_{ph}$.

The frequency of an alternator is related to speed and poles by

$$f = \frac{PN_s}{120} \quad\Rightarrow\quad N_s = \frac{120f}{P}$$

For $f = 50$ Hz:

• 4-pole: $N_s = \frac{120\times50}{4} = 1500$ rpm
• 6-pole: $N_s = \frac{120\times50}{6} = 1000$ rpm

Answer: 1500 rpm for the 4-pole alternator and 1000 rpm for the 6-pole alternator.

V-curve

A V-curve of a synchronous motor is the graph of armature current $I_a$ against field current $I_f$, at constant supply voltage and constant load (mechanical output). The curve has the shape of the letter V.

• The lowest point of each curve is at unity power factor, where the armature current is minimum.
• To the left (under-excited) the current lags the voltage; the pf is lagging, and $I_a$ increases as $I_f$ is reduced.
• To the right (over-excited) the current leads; the pf is leading, and $I_a$ increases as $I_f$ is increased.
• Curves are drawn for no load, half load, full load, etc. A higher load gives a curve higher up, and the minimum point moves to a larger field current.
• The dotted line joining the minima is the unity power factor line, while the region to the right of it is leading pf.

  Ia
   |\                         /
   | \   full load          /
   |  \                    /
   |   \     half load    /
   |    \    \          / /
   |     \    \  no   / /
   |      \    \ load/ /
   +-------\----*---/------> If
   lagging   unity pf  leading
   (under)  (min Ia)  (over)

Inverted V-curve

The inverted V-curve is the plot of power factor against field current at constant load. Power factor rises as the field current increases from the under-excited region, reaches a maximum of unity at the same field current at which the V-curve is at its minimum, then falls again (to leading) as the field current is raised further. The shape is like an inverted V.

  pf
  1.0|        /\
     |      /    \
     |    /        \
     |  /            \
     +-----------------> If
     lag    unity   lead

These curves show that a synchronous motor can be operated at any power factor by changing excitation, so it is used for power factor correction (a synchronous condenser).

Hunting

Hunting is the oscillation of the rotor of a synchronous machine about its steady-state synchronous position (the load angle $\delta$ swings above and below its mean value) following a sudden change in load or supply.

Cause: when load changes suddenly, the rotor does not take its new load angle immediately. Because of its inertia it overshoots, and the restoring (synchronising) torque pulls it back, so it oscillates about the new position. The oscillations are slowly damped.

Effects: variation in current, power and voltage, possible loss of synchronism, mechanical stress and noise.

Prevention: damper (amortisseur) windings on pole faces damp the oscillations by developing a torque that opposes relative motion. A flywheel (to increase inertia), and proper design of the governor (prime mover control) also help.

Loaded operation

At no load, the rotor poles lie nearly in line with the stator rotating field. The back emf $E_b$ and the supply voltage $V$ are almost in opposition, and the net voltage $E_r$ across the armature is tiny; the current is small.

When mechanical load is applied:

1. The rotor momentarily slows and falls back by load angle $\delta$ (the rotor stays at synchronous speed; only its position shifts).
2. $E_b$ shifts by angle $\delta$ with respect to $V$, so the resultant voltage $E_r = V - E_b$ increases.
3. Armature current $I_a = \frac{E_r}{Z_s}$ increases and lags $E_r$ by $\theta = \tan^{-1}(X_s/R_a)$.
4. The motor torque rises until it equals the load torque.
5. Speed remains constant at $N_s$ for all loads up to the pull-out torque.

$$P = \frac{3VE_b}{X_s}\sin\delta, \qquad T = \frac{P}{\omega_s}$$

Maximum power occurs at $\delta = 90^\circ$ (pull-out). If the load exceeds it, the motor loses synchronism and stops.

 Phasors (motor):
       V
   ---------->
        \  delta
     Eb  \     Er = V - Eb  -> Ia

Synchronising (paralleling) is the process of connecting an incoming alternator to a live bus or another running alternator. If this is done incorrectly, large circulating currents and mechanical shocks occur.

Conditions (criteria) to be satisfied

1. Equal terminal voltage: the rms voltage of the incoming machine must equal the bus voltage. Adjust it by the field current (excitation).
2. Equal frequency: the incoming alternator's frequency must equal the bus frequency. Adjust the prime-mover speed (governor) until the difference is nearly zero.
3. Same phase sequence: the phase order (R-Y-B) must be the same as the bus. Wrong sequence causes short-circuit.
4. Same phase angle (in phase): corresponding phase voltages must be in phase at the instant of closing the switch.
5. Same waveform (sinusoidal for both) is also desirable.

Detailed explanation

• Voltage: If voltages differ, a circulating current $I_c = \frac{E_1 - E_2}{Z_1 + Z_2}$ flows, mostly reactive. It is eliminated by matching the voltage with a voltmeter.
• Frequency: If frequencies differ, the phase difference changes continually and the machines pass in and out of phase at the beat frequency. A synchroscope or lamps show the beat. The closing is done when the beat is very slow.
• Phase sequence: Checked once at installation with a phase sequence indicator, or by lamps (all lamps should light and dark together in the dark-lamp method). Reversal of any two lines corrects it.
• Phase angle: The switch is closed when the synchroscope pointer stops at the vertical (12 o'clock) position, or when the dark lamps are fully dark.

Methods

• Three dark lamps method
• Two bright, one dark lamp method
• Synchroscope (most common)

 Bus ----+----+----+----
         |    |    |
       [S]--(Lamps / Synchroscope)--
              |
    Incoming alternator

Dark lamp method of synchronising

The dark lamp method uses three lamps to check voltage, frequency, phase sequence and phase angle between the incoming alternator and the bus-bars.

Connections

Three lamps are connected across the switch contacts, one in each phase: lamp $L_1$ between R and R', $L_2$ between Y and Y', $L_3$ between B and B'.

 Bus      R ----+--------[S]-------- R'
          Y ----|--+-----[S]-------- Y'   Incoming
          B ----|--|--+--[S]-------- B'   alternator
                L1 L2 L3 (across switch)

Procedure

1. Start the incoming alternator with its prime mover and bring it to near rated speed.
2. Adjust its field current so that its terminal voltage equals the bus voltage (check with a voltmeter).
3. Observe the lamps:
   • If the phase sequence is correct, all three lamps flicker (grow bright and dark) together.
   • If the sequence is wrong, the lamps become bright and dark one after another (rotating pattern). Interchange any two lines of the incoming alternator and repeat.
4. Adjust the speed of the prime mover (governor) until the flicker becomes very slow, which shows that the frequencies are almost equal.
5. When the lamps are dark (voltage across the lamps is zero) at the middle of a dark period, the incoming voltage is in phase with the bus voltage and equal in magnitude. Close the switch at that instant.
6. After closing, adjust the governor and field to share the load.

Principle

The voltage across each lamp is the phasor difference of bus and incoming voltage $|\vec V_{bus} - \vec E_{inc}|$. It becomes zero only if both are equal and in phase, so a dark lamp indicates the correct instant.

Disadvantage

Lamps are dark even when the voltage is considerably different (they need a minimum voltage to glow), so the method is not accurate. A synchroscope or the two bright, one dark method gives better results.

Two-reaction model

In a salient-pole machine the air gap is not uniform: it is small along the pole axis (direct axis, d-axis) and large between the poles (quadrature axis, q-axis). So the same mmf produces different flux along the two axes, and the single synchronous reactance of a cylindrical rotor cannot be used. Blondel's two-reaction theory resolves the armature mmf (and the current) into two components, one along each axis, and treats each separately.

• $I_d = I\sin\psi$: direct-axis component, in quadrature with the emf $E$, and acting along the pole axis (magnetising or demagnetising).
• $I_q = I\cos\psi$: quadrature-axis component, in phase with $E$, acting along the inter-polar axis (cross-magnetising).

Here $\psi$ is the angle between $I$ and $E$. Each component produces an armature reaction flux, and so a reactance:

$$X_d = X_{ad} + X_l, \qquad X_q = X_{aq} + X_l$$

with $X_d > X_q$ because the d-axis has the lower reluctance (typically $X_q \approx 0.6\text{--}0.7X_d$).

Phasor equation (generator)

$$\vec E = \vec V + \vec I R_a + j\vec I_d X_d + j\vec I_q X_q$$

          E
         /|
   jIdXd/ |     V at angle delta from E
       /  |
      /   |jIqXq
  ---+----+-----> V
   Iq in phase with E
   Id 90 deg behind/ahead

Load angle and components

From the phasor diagram (neglecting $R_a$), with $\delta$ the angle between $E$ and $V$:

$$I_q = \frac{V\sin\delta}{X_q}, \qquad I_d = \frac{E - V\cos\delta}{X_d}$$

These are used to find the power developed:

$$P = \frac{VE}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta$$

The second term is the reluctance power; it exists even without excitation.

For a cylindrical rotor, $X_d = X_q = X_s$, and the model reduces to the simple synchronous reactance model.

Two-reaction theory

In a salient-pole machine the air gap is not uniform: it is small along the pole axis (direct axis, d-axis) and large between the poles (quadrature axis, q-axis). So the same mmf produces different flux along the two axes, and the single synchronous reactance of a cylindrical rotor cannot be used. Blondel's two-reaction theory resolves the armature mmf (and the current) into two components, one along each axis, and treats each separately.

• $I_d = I\sin\psi$: direct-axis component, in quadrature with the emf $E$, and acting along the pole axis (magnetising or demagnetising).
• $I_q = I\cos\psi$: quadrature-axis component, in phase with $E$, acting along the inter-polar axis (cross-magnetising).

Here $\psi$ is the angle between $I$ and $E$. Each component produces an armature reaction flux, and so a reactance:

$$X_d = X_{ad} + X_l, \qquad X_q = X_{aq} + X_l$$

with $X_d > X_q$ because the d-axis has the lower reluctance (typically $X_q \approx 0.6\text{--}0.7X_d$).

Phasor equation (generator)

$$\vec E = \vec V + \vec I R_a + j\vec I_d X_d + j\vec I_q X_q$$

          E
         /|
   jIdXd/ |     V at angle delta from E
       /  |
      /   |jIqXq
  ---+----+-----> V
   Iq in phase with E
   Id 90 deg behind/ahead

Determination of $X_d$ and $X_q$: slip test

1. Drive the rotor by a prime mover at slightly less than synchronous speed (slip below 1 %), with the field winding open.
2. Apply a reduced balanced three-phase voltage (about 25 % of rated, with the same phase sequence) to the stator terminals.
3. The stator rotating field slowly slides past the rotor poles. The effective reluctance varies between the d-axis and q-axis values, so the armature current and terminal voltage oscillate slowly.
4. Record with an oscillograph (or ammeter and voltmeter readings) the maximum and minimum values:
   • When the field is aligned with the d-axis, the reactance is largest, so the current is minimum and the voltage is maximum.
   • When aligned with the q-axis, the current is maximum and the voltage is minimum.
5. Then

$$X_d = \frac{V_{max}}{I_{min}}, \qquad X_q = \frac{V_{min}}{I_{max}}$$

(per phase values; $R_a$ is small and neglected).

 Test connections
  3-phase reduced V --> stator
  field winding: OPEN (voltmeter across it
                 shows induced emf pulsating)
  prime mover: speed just below Ns

$X_d$ is also found from the open-circuit and short-circuit characteristics ($X_d = \frac{V_{oc}}{I_{sc}}$ at the same field current, unsaturated). $X_q$ by this test is typically 60 to 70 % of $X_d$.

Per phase, resistance neglected, a generator has terminal voltage $V$, excitation emf $E$, and load angle $\delta$ (angle between $E$ and $V$). Using the two-reaction model $\vec E = \vec V + j\vec I_d X_d + j\vec I_q X_q$.

Derivation

Components of current:

$$I_q = \frac{V\sin\delta}{X_q}, \qquad I_d = \frac{E - V\cos\delta}{X_d}$$

The output power per phase is the sum of $V I$ components in phase with $V$. The component $I_q$ makes angle $\delta$ with $V$ (in phase with $E$) and $I_d$ has a component along $V$:

$$P_{ph} = V I\cos\phi = V I_q\cos\delta + V I_d\sin\delta$$ $$P_{ph} = \frac{V^2\sin\delta\cos\delta}{X_q} + \frac{V\sin\delta\,(E - V\cos\delta)}{X_d}$$ $$P_{ph} = \frac{VE}{X_d}\sin\delta + V^2\sin\delta\cos\delta\left(\frac{1}{X_q} - \frac{1}{X_d}\right)$$

Using $\sin\delta\cos\delta = \tfrac12\sin2\delta$:

$$\boxed{P_{ph} = \frac{VE}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin 2\delta}$$

Three-phase power is $3P_{ph}$ (with phase values). For a cylindrical rotor, $X_d = X_q$ and the second term vanishes.

• First term: excitation power (fundamental, $\sin\delta$).
• Second term: reluctance power (double frequency, $\sin2\delta$). It exists even if $E=0$.

Power-angle characteristic

 P
 |        total
 |     _.-'-._
 |   .'  excit.  `-.
 | .' ____---___     `-.
 |/.-'   reluct. `-._    `-.
 +--+----+----+----+----+----> delta
 0       45   ~75-80  90   180
 (max power below 90 deg)

• The total curve is the sum of a sine curve and a $\sin2\delta$ curve.
• Maximum power occurs at $\delta < 90^\circ$ (usually 60 to 80°) for a salient pole machine, and at $90^\circ$ for a cylindrical machine.
• Stable operation is on the rising part ($0 < \delta < \delta_{max}$), where $dP/d\delta > 0$. Beyond it the machine loses synchronism.

In large alternators, the armature winding is on the stator and the field system rotates, instead of the opposite arrangement. The advantages are:

1. Easy collection of output: the high-power armature output is taken directly from stationary terminals, with no slip rings and brushes in the main power circuit. Only a small DC field current (a few amperes at 125 to 500 V) goes through two slip rings.
2. Easier insulation for high voltage: a stationary armature can be insulated well for 11 kV to 33 kV, because it is not subject to centrifugal forces. Sparking and insulation breakdown at slip rings would occur with such voltages in a rotating armature.
3. Better cooling: the large armature winding and core, where most of the heat is generated, are stationary and can be cooled by ducts, air or hydrogen, or water in the stator.
4. Light rotor: the field winding is simpler and lighter, so the rotor has lower inertia and can be built stronger against centrifugal stress.
5. Smaller slip-ring losses and cost: two small rings carry only the DC field current, so there is less wear, sparking and maintenance.
6. Better mechanical balance and simpler construction of the rotor, and a heavier, rigid frame for the stator.

The frequency of an alternator is $f = \frac{PN_s}{120}$. For $f = 50$ Hz the speed is $N_s = \frac{6000}{P}$. A low speed therefore needs many poles, and a high speed needs few poles.

Salient-pole alternator (low speed)

• Used with hydraulic turbines (100 to 400 rpm), engines, and wind turbines, which need many poles (e.g. $P = 24$ at 250 rpm).
• Many poles must fit on the rotor, so it has a large diameter and short axial length, with projecting poles.
• At low speed, the centrifugal force and windage loss are small, so the projecting poles are mechanically safe.

Cylindrical (non-salient) alternator (high speed)

• Used with steam or gas turbines at 1500 or 3000 rpm, with 2 or 4 poles.
• At high speed, the large-diameter projecting poles would suffer large centrifugal force and high windage loss and noise.
• A smooth cylindrical rotor of small diameter and long axial length (slotted forged steel with distributed winding) withstands the centrifugal stresses and gives low windage loss and a better sinusoidal flux distribution.

So the number of poles needed and the mechanical limits decide the type: salient poles for low-speed, many-pole machines, and cylindrical rotors for high-speed, 2- or 4-pole turbo-alternators.

Given: 1500 kVA, 6600 V, star, $R_a = 0.4\ \Omega$, $X_s = 6\ \Omega$ per phase, full-load at 0.8 pf lagging, rated voltage.

$$V_{ph} = \frac{6600}{\sqrt3} = 3810.5\ \text{V}, \qquad I = \frac{1500\times10^3}{\sqrt3\times6600} = 131.22\ \text{A}$$

(i) Excitation emf (0.8 lagging)

Taking $V_{ph}$ as reference, $\vec I = 131.22\angle-36.87^\circ$, $Z_s = 0.4 + j6$:

$$IR_a = 52.49\ \text{V}, \quad IX_s = 787.30\ \text{V}$$ $$\vec E = 3810.5 + 131.22\angle-36.87^\circ(0.4+j6)$$ $$E = \sqrt{(3810.5 + 52.49\times0.8 + 787.30\times0.6)^2 + (787.30\times0.8 - 52.49\times0.6)^2}$$ $$E_{ph} = 4366.07\ \text{V}, \qquad E_L = \sqrt3\times4366.07 = 7562.3\ \text{V}, \qquad \delta = 7.88^\circ$$

(ii) Terminal voltage at 0.8 leading (same $E$ and $I$)

$\vec I = 131.22\angle+36.87^\circ$. The drop $\vec IZ_s$ has magnitude $131.22\times6.0133 = 789.1$ V at $36.87^\circ + 86.19^\circ = 123.06^\circ$ from $V$. With $|\vec V + \vec IZ_s| = E$:

$$V^2 + 2V(789.1)\cos123.06^\circ + 789.1^2 = 4366.07^2$$

Solving this quadratic gives (taking the positive root)

$$V_{ph} = 4746.1\ \text{V}, \qquad V_L = \sqrt3\times4746.1 = 8220.5\ \text{V}$$

Answer: (i) $E_{ph} = 4366$ V (line 7562 V); (ii) terminal voltage $\approx 8220$ V line (4746 V per phase). The voltage rises because the leading current's armature reaction is magnetising.

Given: 1600 kVA, 13.5 kV, star, $R_a = 1.5\ \Omega$, $X_s = 30\ \Omega$ per phase. Load $P = 1.28$ MW at rated voltage.

$$V_{ph} = \frac{13500}{\sqrt3} = 7794.2\ \text{V}, \qquad I = \frac{1.28\times10^6}{\sqrt3\times13500\times\cos\phi}$$ $$\vec E = \vec V + \vec I(R_a + jX_s), \qquad \text{Regulation} = \frac{E - V}{V}\times100$$

(i) 0.8 pf lagging

$$I = \frac{1.28\times10^6}{\sqrt3\times13500\times0.8} = 68.43\ \text{A}$$ $$IR_a = 102.64\ \text{V}, \quad IX_s = 2052.80\ \text{V}$$ $$E = \sqrt{(7794.2 + 102.64\times0.8 + 2052.80\times0.6)^2 + (2052.80\times0.8 - 102.64\times0.6)^2}$$ $$E_{ph} = 9244.2\ \text{V}, \qquad E_L = 16011\ \text{V}$$ $$\text{Regulation} = \frac{9244.2 - 7794.2}{7794.2}\times100 = 18.60\%$$ $$\delta = \tan^{-1}\frac{1642.2-61.6}{7794.2 + 82.1 + 1231.7}\approx 9.85^\circ$$

(ii) Unity pf

$$I = \frac{1.28\times10^6}{\sqrt3\times13500} = 54.74\ \text{A}, \quad IR_a = 82.11\ \text{V}, \quad IX_s = 1642.24\ \text{V}$$ $$E = \sqrt{(7794.2 + 82.11)^2 + 1642.24^2}$$ $$E_{ph} = 8045.7\ \text{V}, \qquad E_L = 13935.6\ \text{V}$$ $$\text{Regulation} = \frac{8045.7-7794.2}{7794.2}\times100 = 3.23\%, \qquad \delta = \tan^{-1}\frac{1642.24}{7876.3} = 11.78^\circ$$

Answer: as in the table (emf is line value; power angle is the angle between $E$ and $V$).

Given: 5 MVA, 11 kV, star, $R_a = 0.12\ \Omega$, $X_s = 2\ \Omega$ per phase, full load at 0.9 pf lagging.

$$V_{ph} = \frac{11000}{\sqrt3} = 6350.85\ \text{V}, \qquad I = \frac{5\times10^6}{\sqrt3\times11000} = 262.43\ \text{A}$$ $$IR_a = 31.49\ \text{V}, \qquad IX_s = 524.86\ \text{V}$$

Voltage regulation at 0.9 lagging

$\cos\phi = 0.9$, $\sin\phi = 0.4359$:

$$E = \sqrt{(V + IR_a\cos\phi + IX_s\sin\phi)^2 + (IX_s\cos\phi - IR_a\sin\phi)^2}$$ $$= \sqrt{(6350.85 + 28.34 + 228.78)^2 + (472.38 - 13.73)^2} = \sqrt{6607.97^2 + 458.65^2}$$ $$E_{ph} = 6623.9\ \text{V}$$ $$\text{Regulation} = \frac{6623.9-6350.85}{6350.85}\times100 = 4.30\%$$

Power factor for zero regulation

For zero regulation $E = V$ at full load current. The current must lead. Let $\vec I = I\angle\theta$ with $\theta > 0$ leading, $Z_s = 2.003\angle86.57^\circ$ ($|Z_s| = \sqrt{0.12^2+2^2}= 2.0036\ \Omega$):

$$|\vec V + \vec IZ_s|^2 = V^2 \;\Rightarrow\; 2VI|Z_s|\cos(\theta + 86.57^\circ) + I^2|Z_s|^2 = 0$$ $$\cos(\theta+86.57^\circ) = -\frac{I|Z_s|}{2V} = -\frac{262.43\times2.0036}{2\times6350.85} = -0.0414$$ $$\theta + 86.57^\circ = 92.37^\circ \;\Rightarrow\; \theta = 5.81^\circ \text{ (leading)}$$ $$\cos\theta = 0.9949$$

Answer: regulation at 0.9 lagging $= 4.30\%$ (E = 6623.9 V per phase). Regulation is zero at 0.995 leading power factor.

Given: 6.6 kV, star, $X_s = 20\ \Omega$ per phase (resistance neglected), constant $V$ and constant excitation. Condition 1: input 100 kW at 0.8 leading. Find the power factor at 1500 kW input.

$$V_{ph} = \frac{6600}{\sqrt3} = 3810.5\ \text{V}$$

Step 1: excitation emf at 100 kW, 0.8 leading

$$I_1 = \frac{100\times10^3}{\sqrt3\times6600\times0.8} = 10.93\ \text{A}, \qquad \vec I_1 = 10.93\angle+36.87^\circ$$

For a motor, $\vec E = \vec V - j\vec I X_s$:

$$\vec E = 3810.5 - j20\,(8.748 + j6.561) = (3810.5 + 131.2) - j174.95 = 3941.7 - j174.95\ \text{V}$$ $$E = 3945.6\ \text{V per phase}\quad (\delta_1 = 2.54^\circ)$$

Step 2: at 1500 kW (E unchanged)

Power per phase $= 500$ kW:

$$P_{ph} = \frac{VE}{X_s}\sin\delta \;\Rightarrow\; \sin\delta = \frac{500\times10^3\times20}{3810.5\times3945.6} = 0.6651, \quad \delta = 41.69^\circ$$

(maximum power $= 3VE/X_s = 2255$ kW, so 1500 kW is below the pull-out limit.)

Current:

$$\vec I = \frac{\vec V - \vec E}{jX_s} = \frac{3810.5 - 3945.6\angle-41.69^\circ}{j20}$$ $$|I| = 138.15\ \text{A}, \qquad \angle I = -18.23^\circ$$ $$\cos\phi = \cos 18.23^\circ = 0.950 \ \text{(lagging)}$$

Check: $\sqrt3\times6600\times138.15\times0.9498 = 1500$ kW.

Answer: power factor at 1500 kW input $\approx 0.95$ lagging (current 138.15 A). The motor moved from leading to lagging because the excitation was fixed while the load increased.