Three Phase Synchronous Motor

A synchronous motor runs at constant speed whatever its excitation. For a constant load (constant input power $VI\cos\phi$), changing the field current changes the back emf $E_b$, and hence the magnitude and power factor of the armature current. So the same motor can behave as an inductive (lagging) or a capacitive (leading) load.

Motor equation (per phase, $R_a$ neglected):

$$\bar V = \bar E_b + jX_s\bar I \qquad\text{or}\qquad \bar I = \frac{\bar V - \bar E_b}{jX_s}$$

The resultant voltage $\bar E_r = \bar V - \bar E_b$ drives the current, which lags $E_r$ by 90°.

1. Under-excitation ($E_b < V$): lagging pf (inductive)

           V
   O-------------->
    \`-.        /
     \  `-.   / Er = jIXs
      \    `/
   I   \   Eb (short)
  (lags V)

• $E_b\cos\delta < V$; the motor draws a lagging current.
• The armature current supplies the missing magnetisation (demagnetising requirement), like an induction motor or inductor.

2. Normal excitation ($E_b \approx V$): unity pf

• Field current is just enough for the required flux; current is in phase with $V$ and is minimum for that load.

   O--------------> V
   O-------->  I (in phase)
     \       /
      Eb    / Er

3. Over-excitation ($E_b > V$): leading pf (capacitive)

     I (leads V)
      ^       V
      |  O---------->
      |    \        .
      |      \    . Er
      |   Eb (long)

• $E_b > V$; the resultant $E_r$ makes the current lead $V$.
• The motor supplies reactive power to the system and acts as a capacitor. Running at no load like this, it is called a synchronous condenser, used for power-factor improvement and voltage control.

Summary

These variations are shown by the V-curves (armature current vs field current at constant load), whose lowest point is at unity pf.

In a synchronous motor with constant load (constant power $P = 3VI\cos\phi$), varying the DC field current changes the back emf $E_b$. Since $V$ and $P$ are fixed, $I\cos\phi$ stays constant, so the tip of the current phasor moves along a line perpendicular to $V$, while $E_b\sin\delta$ also stays constant.

(a) Normal excitation

• Field current gives $E_b$ such that the current is in phase with $V$: unity pf.
• Armature current is minimum for that load.

(b) Under-excitation

• $E_b$ is smaller; the resultant voltage $E_r = V - E_b$ swings so that the current lags $V$.
• Armature current is larger than at normal excitation; the motor draws lagging reactive power (magnetising current) from the supply.

(c) Over-excitation

• $E_b$ is larger; the current leads $V$.
• Armature current again increases; the motor supplies reactive power to the supply (synchronous condenser action).

 Phasors at constant load (V reference, Rs = 0)

   I3 (lead, over-excited)
     \
      \     I1 (upf, normal)
  O----+------> V      I tip moves on the
      /                vertical line I cos(phi)
     /                 = constant
   I2 (lag, under-excited)

V-curves and inverted V-curves

  Ia                         pf
  |  \       /  full load    |      _
  |   \     /                |    /   \  (inverted V)
  |  \ \   / /               |   /     \
  |   \_\_/_/  no load       |  /       \
  +------------> If          +--------------> If
  lag | upf | lead          lag  upf  lead

• V-curves: $I_a$ vs $I_f$ at constant load, shaped like a "V". The minimum point of each curve is at unity pf. The left side (under-excited) is lagging, the right side (over-excited) is leading. Higher load gives a higher curve.
• Inverted V-curves: pf vs $I_f$; the peak is unity pf. The locus of minimum points (unity pf) shifts to the right as load increases, since more excitation is needed at higher load.

This property lets an over-excited synchronous motor drive its load and also correct the plant power factor.

Why a synchronous motor is not self-starting

1. When balanced 3-phase supply is given to the stator, a magnetic field rotating at synchronous speed $N_s = 120f/P$ is set up.
2. The rotor poles are excited by DC and are stationary at start.
3. At one instant, a stator N pole is in front of a rotor S pole: attraction tends to turn the rotor in the direction of the field (say clockwise).
4. Half a cycle later (1/100 s at 50 Hz), the stator poles have moved by one pole pitch; now a stator N is in front of the rotor N: repulsion tends to turn the rotor anticlockwise.
5. Because of its inertia, the rotor cannot move in such a short time; the torque reverses every half cycle, so the average starting torque is zero. The rotor only vibrates.

 t = 0:   stator N  -> attracts rotor S  -> torque CW
 t = T/2: stator S  -> repels  rotor S   -> torque CCW
          average torque = 0  (rotor stays at rest)

Hence the rotor must first be brought close to synchronous speed by some other means.

Damper winding method (starting as induction motor)

Construction: copper or brass bars are placed in slots in the rotor pole faces and short-circuited at both ends by end rings, forming a squirrel-cage (damper / amortisseur) winding.

Procedure

1. Keep the DC field winding open or short-circuited through a discharge resistance (to avoid very high induced voltage in the many-turn field winding).
2. Switch on the 3-phase supply (often at reduced voltage through an auto-transformer or star–delta starter to limit current).
3. The rotating field induces currents in the damper bars, and the motor starts as an induction motor, accelerating to about 95–98% of synchronous speed.
4. Now apply DC excitation to the field. The rotor poles lock with the stator field and are pulled into synchronism.
5. At synchronous speed there is no relative motion, so no current flows in the damper bars; they carry current only during speed changes, when they also damp hunting.

Advantages: simple, no separate starting motor, also prevents hunting. Limitation: low starting torque, so it suits motors that start at light load.

Hunting is the periodic oscillation (swinging) of the rotor of a synchronous machine about its steady equilibrium position (load angle $\delta$) when the load or supply changes suddenly. The rotor speed swings above and below synchronous speed while the average speed stays synchronous. It is also called phase swinging.

How it occurs

1. Suppose the motor runs at load angle $\delta_1$ with load torque $T_1$.
2. The load is suddenly increased to $T_2$. The rotor slows down momentarily, and $\delta$ increases to give more torque.
3. Due to its inertia, the rotor does not stop at the new angle $\delta_2$; it overshoots to $\delta_3 > \delta_2$, where the developed torque is more than the load torque.
4. The excess torque accelerates the rotor; it swings back past $\delta_2$ towards a smaller angle; then it is retarded again.
5. The rotor thus oscillates about $\delta_2$ at the natural frequency of the machine–system. Small oscillations die out if there is enough damping.

   delta
    |     .-.
 d3 |    /   \     .-.
 d2 |---/-----\---/---\---.--------  new steady value
    |  /       \_/     '-'
 d1 |_/
    +--------------------------------> time
       load step

Causes

• Sudden change in load (especially periodic loads such as compressors and pumps).
• Sudden change in supply voltage or frequency, or faults in the system.
• Sudden change in field current.
• Pulsating torque of the prime mover (generators driven by diesel engines).
• Resonance between the natural frequency of the rotor and the frequency of load pulsations.

Effects

• Large mechanical stresses and fatigue in the shaft.
• Large current and power swings, voltage fluctuations and increased losses.
• If the swing exceeds the stability limit, the machine falls out of step (loses synchronism).
• Can cause resonance and damage.

Methods to minimise hunting

1. Damper (amortisseur) windings: short-circuited copper bars in the pole faces. When the rotor swings, relative motion induces currents in them which produce a torque opposing the swing (like an induction motor), damping the oscillation. This is the main method.
2. Flywheel: adds inertia and smooths the torque of reciprocating prime movers or loads; it changes the natural frequency away from the forcing frequency.
3. Design the machine with suitable synchronising power (proper $X_s$ and excitation), so the natural frequency is not near the frequency of load pulsations.
4. Avoid sudden large changes in load, and use fast-acting excitation systems with power system stabilisers.

Given: star, 5 kVA, 208 V, 50 Hz, $X_s = 8\ \Omega$, $R_a \approx 0$; unity pf, input 3 kW.

(i) Back emf and power angle

$$\begin{aligned} V &= \frac{208}{\sqrt3} = 120.09\ \text{V} \\ I &= \frac{3000}{\sqrt3\times208\times1} = 8.327\ \text{A}\quad(\text{in phase with } V) \\ \bar E_b &= \bar V - jX_s\bar I = 120.09 - j(8\times8.327) = 120.09 - j66.62 \\ E_b &= \sqrt{120.09^2 + 66.62^2} = 137.33\ \text{V per phase} \\ \delta &= \tan^{-1}\frac{66.62}{120.09} = 29.02^\circ\ (E_b \text{ lags } V) \end{aligned}$$

Line value of back emf $= \sqrt3\times137.33 = 237.9$ V.

   O-------------------> V = 120.09
   O--------> I (upf)  |
     `-.               | jIXs = 66.62
        `-.            |
           `-. Eb = 137.33, delta = 29.02 deg

(ii) Power angle increased by 20%, excitation constant

$$\begin{aligned} \delta' &= 1.2\times29.02^\circ = 34.82^\circ \\ \bar E_b' &= 137.33\angle -34.82^\circ = 112.74 - j78.42 \\ \bar I' &= \frac{\bar V - \bar E_b'}{jX_s} = \frac{(120.09 - 112.74) + j78.42}{j8} = \frac{7.35 + j78.42}{j8} \\ &= 9.802 - j0.919\ \text{A} \\ I' &= 9.845\ \text{A},\quad \angle -5.36^\circ \\ \text{pf} &= \cos 5.36^\circ = 0.996\ \text{lagging} \end{aligned}$$

New input power $= 3\times120.09\times9.802 = 3531$ W.

Answer: (i) E_b = 137.33 V/phase (237.9 V line), δ = 29.02°; (ii) I = 9.85 A, pf = 0.996 lagging (input ≈ 3.53 kW).

A synchronous motor always runs at synchronous speed $N_s = 120f/P$. When the shaft load changes it cannot slow down (as an induction motor does); instead the rotor shifts backwards in space by a larger load angle $\delta$, and the armature current changes to supply the new power.

Adjustment to change in load

1. At no load the rotor poles are almost in line with the stator field ($\delta \approx 0$), and the motor draws only enough current to meet losses.
2. When load torque increases, the rotor momentarily slows down. The rotor poles fall back with respect to the rotating stator field, so $\delta$ increases.
3. Back emf $E_f$ stays the same in magnitude (excitation unchanged) but moves further behind $V$. The resultant voltage $E_r = V - E_f$ grows, so armature current $I_a = E_r / jX_s$ increases.
4. Power drawn $P = \dfrac{3VE_f}{X_s}\sin\delta$ rises until it equals the new load plus losses. The rotor then runs again at $N_s$ with a larger fixed $\delta$.
5. If load exceeds $P_{max}$ (at $\delta = 90^\circ$ for a cylindrical rotor), the motor pulls out of step and stops.

          V (fixed)
   ------------------>
     \  delta1
      \--------> Ef (light load)
       \
        \ delta2 > delta1
         \------> Ef (heavy load, same length)

Effect of change in excitation (constant load)

With load (and hence $P = VI_a\cos\phi$) constant, $I_a\cos\phi$ is constant, so the tip of $I_a$ moves on a line perpendicular to $V$, and $E_f\sin\delta$ is also constant.

           I(lead)   I(unity)
             \         |
              \        |
  ------------ +-------+--------> V
                \      |
             I(lag)  locus of I tip
                      (I cos(phi) const)

• Raising the field current from a low value first reduces $I_a$ (lagging pf improves) until $I_a$ is minimum at unity pf.
• Further increase makes $I_a$ rise again with a leading pf.
• Plotting $I_a$ against field current gives the V-curves; the pf vs field current curve is an inverted V.

This is why an over-excited synchronous motor can act as a synchronous condenser to improve plant power factor.

Given: $P_{out} = 75$ kW, $V_L = 400$ V (star), $R_a = 0.04\ \Omega$, $X_s = 0.4\ \Omega$, pf = 0.8 leading, $\eta = 92.5\%$.

Input power and current

$$\begin{aligned} P_{in} &= \frac{P_{out}}{\eta} = \frac{75}{0.925} = 81.081\ \text{kW} \\ I_a &= \frac{P_{in}}{\sqrt3\,V_L\cos\phi} = \frac{81081}{\sqrt3 \times 400 \times 0.8} = 146.29\ \text{A} \\ V &= \frac{400}{\sqrt3} = 230.94\ \text{V per phase} \end{aligned}$$

Open-circuit (excitation) emf

For a motor $\vec V = \vec E_f + \vec I_a Z_s$, so $\vec E_f = \vec V - \vec I_a Z_s$. Take $V$ as reference; leading current is at $+36.87^\circ$.

$$\begin{aligned} Z_s &= 0.04 + j0.4 = 0.402\angle 84.29^\circ\ \Omega \\ \vec I_a &= 146.29\angle 36.87^\circ = 117.03 + j87.77\ \text{A} \\ \vec I_a Z_s &= 58.81\angle 121.16^\circ = -30.43 + j50.32\ \text{V} \\ \vec E_f &= 230.94 - (-30.43 + j50.32) \\ &= 261.37 - j50.32 \\ &= 266.17\angle -10.90^\circ\ \text{V} \end{aligned}$$

Line value: $\sqrt3 \times 266.17 = 461.0$ V. As expected for leading pf, $E_f > V$ (over-excited), and the load angle is $\delta = 10.90^\circ$.

Mechanical power developed

$$\begin{aligned} P_{cu} &= 3I_a^2R_a = 3 \times 146.29^2 \times 0.04 = 2568\ \text{W} \\ P_{mech} &= P_{in} - P_{cu} = 81081 - 2568 = 78513\ \text{W} \end{aligned}$$

Check: $3\,\text{Re}(\vec E_f \vec I_a^{*}) = 78.51$ kW, the same value.

(The difference between 78.51 kW and the 75 kW output is the friction, windage, core and stray loss.)

Answer: $E_f = 266.2$ V per phase (461.0 V line), load angle $10.9^\circ$; mechanical power developed $= 78.51$ kW.

The synchronous motor is over-excited so it supplies leading kVAR, which cancels part of the lagging kVAR of the factory load.

Factory load

$$\begin{aligned} P_1 &= 240\ \text{kW},\quad \cos\phi_1 = 0.8\ \text{lag},\quad \tan\phi_1 = 0.75 \\ Q_1 &= 240 \times 0.75 = 180\ \text{kVAR (lagging)} \end{aligned}$$

Combined load

$$\begin{aligned} P &= 240 + 60 = 300\ \text{kW} \\ \cos\phi &= 0.9 \ \Rightarrow\ \tan\phi = \tan(25.84^\circ) = 0.4843 \\ Q &= 300 \times 0.4843 = 145.30\ \text{kVAR (lagging)} \end{aligned}$$

kVAR from the motor

$$\begin{aligned} Q_m &= Q_1 - Q = 180 - 145.30 = 34.70\ \text{kVAR (leading)} \\ S_m &= \sqrt{60^2 + 34.70^2} = 69.31\ \text{kVA} \\ \cos\phi_m &= \frac{60}{69.31} = 0.866\ \text{leading} \end{aligned}$$

     kVAR (lag +)
      |
 180  +  factory load
      |
145.3 +  combined (pf 0.9)
      |   motor gives -34.7 kVAR
      +--------------------------- kW
         60   240   300

Answer: The motor supplies 34.70 kVAR leading and works at a power factor of 0.866 leading (input 69.31 kVA).

The power angle characteristic gives the power of a salient pole machine as a function of the load angle $\delta$. Because the air gap is not uniform, the armature current is split into a direct-axis part $I_d$ (sees $X_d$) and a quadrature-axis part $I_q$ (sees $X_q$), with $X_d > X_q$ (two-reaction theory).

Derivation (generator, $R_a$ neglected)

             E (q-axis)
            /|
  jXq Iq  /  |
         /   |  jXd Id
        /    |
  V    /delta|
 -----+------+------> d-axis
   I lags V by phi; psi = delta + phi

Take $E$ along the q-axis and $V$ lagging $E$ by $\delta$. From the phasor diagram, resolving $V$ along the two axes:

$$\begin{aligned} V\cos\delta &= E - X_d I_d \ \Rightarrow\ I_d = \frac{E - V\cos\delta}{X_d} \\ V\sin\delta &= X_q I_q \ \Rightarrow\ I_q = \frac{V\sin\delta}{X_q} \end{aligned}$$

Power per phase is $P = VI\cos\phi$. Projecting $I_d$ and $I_q$ on $V$ (the angle between $V$ and the q-axis is $\delta$):

$$\begin{aligned} P &= V\left(I_q\cos\delta + I_d\sin\delta\right) \\ &= V\left[\frac{V\sin\delta}{X_q}\cos\delta + \frac{E - V\cos\delta}{X_d}\sin\delta\right] \\ &= \frac{EV}{X_d}\sin\delta + V^2\sin\delta\cos\delta\left(\frac{1}{X_q} - \frac{1}{X_d}\right) \end{aligned}$$

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

$$P = \frac{EV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin 2\delta \quad \text{(per phase)}$$

For three phases multiply by 3. The same expression holds for a motor (with $\delta$ the angle by which $E$ lags $V$).

• First term: excitation power, depends on field current.
• Second term: reluctance power, due to saliency; it exists even when $E = 0$ and vanishes for a cylindrical rotor ($X_d = X_q$).

 P |      .--.  resultant
   |    .'    '.
   |  .'  exc.  '.
   | /  .-.reluct.\
   |/.-'   '-.     \
   +-----------+----+---- delta
   0   ~60-75  90  180 deg

Condition for maximum power

$$\frac{dP}{d\delta} = \frac{EV}{X_d}\cos\delta + V^2\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\cos2\delta = 0$$

Let $a = \dfrac{EV}{X_d}$ and $b = V^2\left(\dfrac{1}{X_q}-\dfrac{1}{X_d}\right)$. With $\cos2\delta = 2\cos^2\delta - 1$:

$$2b\cos^2\delta + a\cos\delta - b = 0 \ \Rightarrow\ \cos\delta_{m} = \frac{-a + \sqrt{a^2 + 8b^2}}{4b}$$

So maximum power occurs at $\delta_m < 90^\circ$ (unlike the cylindrical machine, where $P_{max}$ is at $90^\circ$). The salient pole machine is therefore "stiffer": it gives more synchronising power for a given $\delta$.

Assumption: star connection (not stated); $R_a$ neglected.

Given: $V_L = 320$ V, $X_s = 7.2\ \Omega$, $P = 4.6$ kW at unity pf.

(i) Back emf and power angle

$$\begin{aligned} V &= \frac{320}{\sqrt3} = 184.75\ \text{V per phase} \\ I_a &= \frac{4600}{\sqrt3 \times 320 \times 1} = 8.299\ \text{A} \\ I_aX_s &= 8.299 \times 7.2 = 59.76\ \text{V} \end{aligned}$$

Motor: $\vec E_f = \vec V - j X_s \vec I_a$, with $I_a$ in phase with $V$:

$$\begin{aligned} \vec E_f &= 184.75 - j59.76 \\ E_f &= \sqrt{184.75^2 + 59.76^2} = 194.18\ \text{V per phase} \\ \delta &= \tan^{-1}\frac{59.76}{184.75} = 17.92^\circ \end{aligned}$$

Line value of $E_f = \sqrt3 \times 194.18 = 336.3$ V.

  V = 184.75 -------------->
   \ delta = 17.92 deg      |
    \                       | jXs Ia = 59.76
     \--------------------->|
          Ef = 194.18
  Ia in phase with V

(ii) Load angle increased by 25 %

$$\delta' = 1.25 \times 17.92^\circ = 22.40^\circ, \qquad \vec E_f = 194.18\angle -22.40^\circ = 179.52 - j74.01\ \text{V}$$ $$\begin{aligned} \vec I_a' &= \frac{\vec V - \vec E_f}{jX_s} = \frac{(184.75 - 179.52) + j74.01}{j7.2} \\ &= \frac{5.23 + j74.01}{j7.2} = 10.279 - j0.727 \\ &= 10.30\angle -4.04^\circ\ \text{A} \end{aligned}$$

Power factor $= \cos 4.04^\circ = 0.9975$ lagging.

Check: new input $= 3 \times 184.75 \times 10.30 \times 0.9975 = 5.70$ kW, equal to $\dfrac{3VE_f}{X_s}\sin22.40^\circ$.

Answer: (i) $E_f = 194.2$ V/phase (336.3 V line), $\delta = 17.92^\circ$. (ii) $I_a = 10.30$ A at 0.9975 lagging pf.

Work in per unit: $V = 1\angle0^\circ$ pu, rated current $I_a = 1$ pu at 0.8 lagging, so $\vec I_a = 1\angle -36.87^\circ = 0.8 - j0.6$ pu.

Method (two-reaction theory, motor, $R_a = 0$): $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

a) Excitation voltage

$$\begin{aligned} \vec E' &= 1 - j0.4(0.8 - j0.6) = 1 - 0.24 - j0.32 \\ &= 0.76 - j0.32 = 0.8246\angle -22.83^\circ\ \text{pu} \\ \delta &= 22.83^\circ,\quad \psi = 22.83^\circ - 36.87^\circ = -14.04^\circ \\ I_d &= 1 \times \sin(-14.04^\circ) = -0.2425\ \text{pu},\quad I_q = 0.9701\ \text{pu} \\ E_f &= 0.8246 + (0.8 - 0.4)(-0.2425) = 0.7276\ \text{pu} \end{aligned}$$

$E_f = 0.7276 \times 11 = 8.00$ kV (line), i.e. 4.62 kV per phase. $E_f < V$, as expected for an under-excited (lagging) motor.

b) Excitation power and reluctance power

$$P = \underbrace{\frac{E_fV}{X_d}\sin\delta}_{\text{excitation}} + \underbrace{\frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta}_{\text{saliency}}$$ $$\begin{aligned} P_{exc} &= \frac{0.7276 \times 1}{0.8}\sin22.83^\circ = 0.3529\ \text{pu} = 17.65\ \text{MW} \\ P_{rel} &= \frac{1}{2}(2.5 - 1.25)\sin45.67^\circ = 0.4471\ \text{pu} = 22.35\ \text{MW} \\ P &= 0.3529 + 0.4471 = 0.8\ \text{pu} = 40\ \text{MW} \end{aligned}$$

Check: input $= VI\cos\phi = 1 \times 1 \times 0.8 = 0.8$ pu. Shaft output $= 0.8 - 0.15 = 0.65$ pu $= 32.5$ MW.

c) Field current reduced to zero

With $E_f = 0$ only reluctance power remains:

$$P_{rel,max} = \frac{V^2}{2}\left(\frac{1}{X_q} - \frac{1}{X_d}\right) = \frac12(2.5 - 1.25) = 0.625\ \text{pu} \ (\text{at } \delta = 45^\circ)$$

The motor must take $0.8$ pu (0.65 pu load + 0.15 pu rotational loss). Since $0.625 < 0.8$, the reluctance torque cannot carry the load, so the machine falls out of synchronism. It would stay in step as a reluctance motor only if the shaft load were reduced so that total input is below 0.625 pu (load below 0.475 pu).

Answer: (a) $E_f = 0.728$ pu (8.0 kV line), $\delta = 22.8^\circ$; (b) excitation power 0.353 pu (17.65 MW), saliency power 0.447 pu (22.35 MW); (c) No, it loses synchronism.

A synchronous motor has no net starting torque when its field is excited and the stator is switched on, so it must be brought near synchronous speed by some other means.

Why it is not self-starting

• The stator 3-phase supply produces a field rotating at $N_s = 120f/P$ (1500 rpm for 4-pole, 50 Hz) instantly.
• The rotor poles are fixed by DC excitation, and the heavy rotor is at rest.
• Suppose a stator N pole is just ahead of a rotor S pole: attraction pulls the rotor one way. Half a cycle (0.01 s) later a stator S pole sits there and pushes the rotor the opposite way.
• Due to rotor inertia it cannot follow these rapid reversals, so the average torque is zero; the rotor only vibrates.

 t = 0      : stator N over rotor S -> torque CW
 t = T/2    : stator S over rotor S -> torque CCW
 average over one cycle             -> 0

Starting methods

1. Damper (amortisseur) winding / induction motor starting - most common.
2. Pony (auxiliary) motor - a small induction motor or DC machine on the same shaft brings the rotor near $N_s$, then the field is excited and the machine is synchronised.
3. Slip-ring induction motor method - rotor has a 3-phase winding connected through slip rings and resistances; starts as a slip-ring induction motor, then DC is fed to the rotor.
4. Variable frequency supply (inverter) - frequency is raised slowly from zero so the rotor follows the field from rest.
5. DC exciter on shaft used as a DC motor to run the machine up.

Damper winding starting method

• Copper or brass bars are placed in slots in the pole faces and shorted at both ends by end rings, forming a partial squirrel cage.
• Step 1: Field winding is not connected to DC; it is shorted through a discharge resistance (to avoid very high induced voltage in it). Reduced voltage (autotransformer or star-delta) is applied to the stator.
• Step 2: The rotating field induces currents in the damper bars, producing induction-motor torque. The rotor accelerates to about 95-97 % of $N_s$.
• Step 3: DC excitation is switched on. The rotor poles lock with the stator poles (synchronising or "pull-in" torque) and the rotor runs at $N_s$.
• Step 4: At synchronous speed there is no relative motion, so no current flows in the damper bars; they carry current only when speed changes, which also damps hunting.

 3-ph AC --> [Autotransformer] --> Stator
                                     |
 Rotor: pole faces with damper bars  |
 Field --[discharge R]-- (start)     |
 Field --[DC exciter ]-- (near Ns)   |

The motor starts with low torque because the cage is small; hence this method suits pumps, fans and compressors started at light load.

Assumptions: the motor operates at rated voltage and rated current ($V = 1$ pu, $I_a = 1$ pu) with losses and $R_a$ neglected; bases are the machine ratings (11 kV line, 6.351 kV phase).

$\cos\phi = 0.98$ leading $\Rightarrow \phi = 11.48^\circ$, $\vec I_a = 1\angle 11.48^\circ = 0.98 + j0.199$ pu.

Method (two-reaction theory, motor, $R_a = 0$): $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

Load angle

$$\begin{aligned} \vec E' &= 1 - j0.8(0.98 + j0.199) = 1 + 0.159 - j0.784 \\ &= 1.1592 - j0.784 = 1.3994\angle -34.07^\circ\ \text{pu} \end{aligned}$$

So the load angle is $\delta = 34.07^\circ$.

Internal emf

$$\begin{aligned} \psi &= \delta + 11.48^\circ = 45.55^\circ \ (\text{leading current}) \\ I_d &= 1 \times \sin45.55^\circ = 0.7139\ \text{pu},\quad I_q = \cos45.55^\circ = 0.7003\ \text{pu} \\ E_f &= 1.3994 + (1.2 - 0.8)(0.7139) = 1.3994 + 0.2856 = 1.685\ \text{pu} \end{aligned}$$

In volts: $E_f = 1.685 \times 6.351 = 10.70$ kV per phase $= 18.53$ kV line.

 V = 1.0 ---------------------->
   \                  Ia (leads V by 11.5 deg)
    \ delta = 34.07
     \
      '--> Ef = 1.685 pu  (Ef > V: over-excited)

Check: $P = \dfrac{E_fV}{X_d}\sin\delta + \dfrac{V^2}{2}\left(\dfrac{1}{X_q}-\dfrac{1}{X_d}\right)\sin2\delta = 0.787 + 0.193 = 0.98$ pu $= V I\cos\phi$.

Answer: Internal emf $E_f = 1.685$ pu (10.70 kV/phase, 18.53 kV line); load angle $\delta = 34.07^\circ$.

A synchronous motor is an AC motor whose rotor, excited by DC, locks with the stator rotating field and runs at exactly synchronous speed $N_s = 120f/P$.

Characteristic features

1. Runs only at synchronous speed; speed does not change with load (zero speed regulation) as long as supply frequency is constant.
2. Not self-starting; needs damper winding, pony motor or variable frequency supply to start.
3. Power factor can be controlled by field current: lagging, unity or leading.
4. When over-excited it supplies reactive power, so it can work as a synchronous condenser.
5. If load torque exceeds the pull-out torque, it falls out of step and stops.
6. Needs a DC source for excitation, so it is costlier and more complex than an induction motor.
7. Efficiency is high, especially in large and low-speed ratings.
8. May hunt (oscillate about $N_s$) under sudden load changes; damper windings reduce this.

Applications

• Power factor correction in industries and substations (synchronous condenser).
• Constant speed drives: compressors, large fans, blowers, centrifugal pumps, rolling mills, cement mills.
• Low-speed high-power drives: reciprocating compressors, ball mills.
• Motor-generator sets, frequency changers; small versions (reluctance, hysteresis) in clocks and recorders.
• Voltage regulation at the end of long transmission lines.

Effect of excitation on armature current and pf

At constant load the input $3VI_a\cos\phi$ is fixed, so the active component $I_a\cos\phi$ is constant. Changing the field current changes $E_f$ and only the reactive component of $I_a$.

 Ia (A)
   |\                 /
   | \  lagging      / leading
   |  \             /
   |   \___________/  <- min Ia, upf
   +----------------------- If
    under   normal   over-excited

• Under-excitation ($E_f < V$): resultant $E_r = V - E_f$ makes $I_a$ lag $V$; large current, lagging pf.
• Normal excitation ($E_f$ roughly equal to $V$): $I_a$ in phase with $V$; current minimum, unity pf.
• Over-excitation ($E_f > V$): $I_a$ leads $V$; current rises again, leading pf.

         Ia(lead)
           \      Ia(upf)
            \       |
  -----------\------+------> V
              \     |
            Ia(lag) |
     tip of Ia moves on vertical line

The plot of $I_a$ against $I_f$ is the V-curve; joining the minimum points gives the unity pf compounding curve. Curves for higher load lie above, and the pf curve has an inverted-V shape.

Given: 4 kVA, 110 V, star, $X_d = 3\ \Omega$, $X_q = 2\ \Omega$, full load at 0.8 lagging, $R_a$ neglected.

$$\begin{aligned} V &= \frac{110}{\sqrt3} = 63.51\ \text{V},\quad I_a = \frac{4000}{\sqrt3 \times 110} = 20.99\ \text{A} \\ \vec I_a &= 20.99\angle -36.87^\circ = 16.80 - j12.60\ \text{A} \end{aligned}$$

Method (two-reaction theory, motor, $R_a = 0$): $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

Load angle

$$\begin{aligned} jX_q\vec I_a &= j2(16.80 - j12.60) = 25.19 + j33.59 \\ \vec E' &= 63.51 - 25.19 - j33.59 = 38.32 - j33.59 = 50.96\angle -41.24^\circ\ \text{V} \end{aligned}$$

Load angle $\delta = 41.24^\circ$.

Excitation emf

$$\begin{aligned} \psi &= \delta - \phi = 41.24^\circ - 36.87^\circ = 4.37^\circ \\ I_d &= 20.99\sin4.37^\circ = 1.600\ \text{A},\quad I_q = 20.99\cos4.37^\circ = 20.93\ \text{A} \\ E_f &= 50.96 + (3 - 2)(1.600) = 52.56\ \text{V per phase} \end{aligned}$$

Line value $= \sqrt3 \times 52.56 = 91.03$ V.

Maximum power for this excitation

$$P = 3\left[\frac{E_fV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\right] = 3\left[1112.57\sin\delta + 336.11\sin2\delta\right]$$

Check at $\delta = 41.24^\circ$: $3(733.5 + 334.2) = 3200$ W $= 4000 \times 0.8$. Correct.

Setting $dP/d\delta = 0$ with $a = 1112.57$, $b = V^2(1/X_q - 1/X_d) = 672.22$:

$$\begin{aligned} \cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = \frac{-1112.57 + \sqrt{1112.57^2 + 8(672.22)^2}}{4(672.22)} = 0.4055 \\ \delta_m &= 66.08^\circ \\ P_{max} &= 3[1112.57\sin66.08^\circ + 336.11\sin132.15^\circ] \\ &= 3[1017.0 + 249.2] = 3798\ \text{W} \end{aligned}$$

Answer: $E_f = 52.56$ V per phase (91.0 V line), load angle $= 41.24^\circ$, maximum power $\approx 3.80$ kW at $\delta = 66.1^\circ$.

A damper (amortisseur) winding is a set of copper or brass bars embedded in slots on the rotor pole faces and short-circuited at both ends by end rings. It forms a partial squirrel cage, so the synchronous motor can start as an induction motor and then pull into synchronism.

Construction

     end ring                     end ring
        ||==========bar============||
        ||==========bar============||   pole face
        ||==========bar============||
   (bars of all poles joined -> squirrel cage)

• Bars are placed in semi-closed slots on each pole shoe.
• End rings (or segments) join the bars of all poles, giving a closed cage.
• Bar resistance is chosen to give reasonable starting torque.

Starting procedure

1. Field kept unexcited: the DC field winding is disconnected from the exciter and shorted through a discharge resistance. This avoids a dangerously high voltage being induced in the many-turn field winding at standstill, and the induced field current adds some starting torque.
2. Supply applied: 3-phase supply is given to the stator, usually at reduced voltage (autotransformer or star-delta) to limit starting current.
3. Induction motor action: the rotating field cuts the damper bars, induces emf and current in them, and produces torque exactly as in a squirrel-cage induction motor. The rotor accelerates.
4. Near synchronous speed: the rotor reaches about 95-98 % of $N_s$ (slip of 2-5 %). It cannot reach $N_s$ by induction action alone.
5. Field excited: the discharge resistor is removed and DC is applied. The rotor poles are now magnetised; because slip is small, the stator poles drag the rotor poles into step (pull-in torque), and the motor runs at $N_s$.
6. Full voltage is then applied to the stator.

Behaviour after synchronising

• At $N_s$ there is no relative motion between the stator field and the damper bars, so no emf, no current and no torque in them; they do not affect normal running.
• If the rotor speed swings above or below $N_s$ (hunting), currents are induced that oppose the swing, so the same winding damps hunting.

Limitations

• Starting torque is low (cage is small and fits only on pole faces), so the motor must be started at light load.
• Pull-in is possible only if load and inertia are not too large.

A synchronous motor runs at constant synchronous speed $N_s = 120f/P$ and, unlike an induction motor, its power factor can be set by the DC field current.

Characteristic features

1. Speed is constant at $N_s$ from no load to full load; it depends only on supply frequency and number of poles.
2. Not self-starting; started by damper winding, pony motor or variable frequency.
3. Operates at lagging, unity or leading pf depending on excitation.
4. Over-excited motor on no load acts as a synchronous condenser.
5. Breaks down (pulls out of step) if load exceeds the maximum (pull-out) torque.
6. Requires a separate DC excitation source.
7. Prone to hunting on sudden load changes.
8. High efficiency, especially for large, low-speed drives.

How it draws lagging or leading current

For a motor (with $R_a$ neglected): $\vec V = \vec E_f + jX_s\vec I_a$, so

$$\vec I_a = \frac{\vec V - \vec E_f}{jX_s} = \frac{\vec E_r}{jX_s}$$

The current lags the resultant voltage $E_r$ by $90^\circ$. Keeping the load (and so $I_a\cos\phi$ and $E_f\sin\delta$) constant, only $E_f$ is changed by the field current.

1. Under-excitation, $E_f < V$ - lagging current

   V ---------------------->
    \  Er (from Ef tip to V tip)
     \ delta
      '-----> Ef (short)
   Ia = Er/jXs lags V   -> lagging pf

The resultant $E_r$ is nearly in phase with $V$, so $I_a$ (90° behind $E_r$) lags $V$. The motor takes magnetising (lagging) reactive power from the supply to make up for its weak field.

2. Normal excitation - unity pf

$E_f$ is such that $E_r$ is $90^\circ$ ahead of $V$; then $I_a$ is in phase with $V$, and current is minimum.

3. Over-excitation, $E_f > V$ - leading current

   V ---------------------->
    \          Er points above V
     \ delta
      '-----------------> Ef (long)
   Ia = Er/jXs leads V  -> leading pf

$E_r$ swings ahead of $V$ by more than $90^\circ$, so $I_a$ leads $V$. The excess field means the motor supplies lagging reactive power to the network, behaving like a capacitor.

This control is the basis of the V-curves and of using synchronous motors for power factor correction.

Assumption: the power factor is not stated as lagging or leading; the usual textbook version of this problem takes 0.8 leading (over-excited motor). The lagging result is given at the end. Star connection, all losses neglected, so input = 2 MW.

$$\begin{aligned} V &= \frac{6600}{\sqrt3} = 3810.51\ \text{V per phase} \\ I_a &= \frac{2\times10^6}{\sqrt3 \times 6600 \times 0.8} = 218.69\ \text{A} \\ \vec I_a &= 218.69\angle 36.87^\circ = 174.95 + j131.22\ \text{A} \end{aligned}$$

Method (two-reaction theory, motor, $R_a = 0$): $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

Load angle

$$\begin{aligned} jX_q\vec I_a &= j3.1(174.95 + j131.22) = -406.77 + j542.36 \\ \vec E' &= 3810.51 - (-406.77 + j542.36) = 4217.28 - j542.36 \\ &= 4252.0\angle -7.33^\circ\ \text{V} \end{aligned}$$

So $\delta = 7.33^\circ$.

Excitation emf

$$\begin{aligned} \psi &= \delta + \phi = 7.33^\circ + 36.87^\circ = 44.20^\circ \\ I_d &= 218.69\sin44.20^\circ = 152.46\ \text{A},\quad I_q = 156.79\ \text{A} \\ E_f &= 4252.0 + (5 - 3.1)(152.46) = 4541.7\ \text{V per phase} \end{aligned}$$

Line value $= \sqrt3 \times 4541.7 = 7866$ V.

Check: $3\left[\dfrac{E_fV}{X_d}\sin\delta + \dfrac{V^2}{2}\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right)\sin2\delta\right] = 2.000$ MW. Correct.

          Ia (leads V by 36.87 deg)
         /
        /
  -----+-------------------> V = 3811 V
        \  delta = 7.33 deg
         '----------> Ef = 4542 V

If the pf is 0.8 lagging: $\vec E' = 3446.7\angle -9.05^\circ$, $\psi = -27.82^\circ$, $I_d = -102.05$ A, $E_f = 3446.7 - 1.9 \times 102.05 = 3252.8$ V per phase (5634 V line).

Answer (0.8 leading): $E_f = 4542$ V per phase $= 7.87$ kV line, load angle $\delta = 7.33^\circ$.

Given: 25 MVA, 11 kV, star, $X_d = 48\ \Omega$, $X_q = 3.5\ \Omega$, $R_a = 0$, rated load at unity pf and rated voltage.

$$\begin{aligned} V &= \frac{11000}{\sqrt3} = 6350.85\ \text{V per phase} \\ I_a &= \frac{25\times10^6}{\sqrt3 \times 11000} = 1312.16\ \text{A}\ (\text{in phase with } V) \end{aligned}$$

Method (two-reaction theory, motor, $R_a = 0$): $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

(i) Excitation voltage

$$\begin{aligned} \vec E' &= V - jX_qI_a = 6350.85 - j(3.5 \times 1312.16) = 6350.85 - j4592.56 \\ &= 7837.4\angle -35.87^\circ\ \text{V} \quad \Rightarrow\ \delta = 35.87^\circ \\ \psi &= \delta - 0 = 35.87^\circ \ (\text{upf}) \\ I_d &= 1312.16\sin35.87^\circ = 768.90\ \text{A},\quad I_q = 1063.28\ \text{A} \\ E_f &= E' + (X_d - X_q)I_d = 7837.4 + 44.5 \times 768.90 \\ &= 42053\ \text{V per phase} = 72.84\ \text{kV line} \end{aligned}$$

Check: $P$ at $\delta = 35.87^\circ$ from the power-angle equation $= 25.00$ MW $=$ rated input.

(ii) Maximum power and its power angle

$$P = 3\left[\frac{E_fV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\right]$$

Let $a = \dfrac{E_fV}{X_d} = 5.5641\times10^6$ and $b = V^2\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right) = 10.6835\times10^6$ (per phase).

$\dfrac{dP}{d\delta} = 0 \Rightarrow 2b\cos^2\delta + a\cos\delta - b = 0$:

$$\begin{aligned} \cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = 0.5888 \\ \delta_m &= 53.93^\circ \\ P_{max} &= 3a\sin\delta_m + \tfrac{3b}{2}\sin2\delta_m \\ &= 13.49 + 15.25 = 28.75\ \text{MW} \end{aligned}$$

The maximum occurs below $90^\circ$ because of the reluctance term.

Answer: (i) $E_f = 42053$ V per phase (72.84 kV line), load angle $35.87^\circ$; (ii) maximum power angle $\delta_m = 53.93^\circ$, $P_{max} = 28.75$ MW.