Three Phase Synchronous Generator

Armature reaction is the effect of the magnetic field set up by the armature (stator) current on the main field flux produced by the rotor poles. When the alternator is loaded, the three-phase armature currents produce a rotating mmf that rotates at synchronous speed along with the rotor. This mmf distorts, weakens or strengthens the main flux depending on the power factor of the load, and so changes the terminal voltage.

In the equivalent circuit, armature reaction is represented by a fictitious reactance $X_a$; together with the leakage reactance $X_l$ it forms the synchronous reactance $X_s = X_a + X_l$, and

$$\bar E = \bar V + \bar I (R_a + jX_s)$$

(i) Unity power factor (resistive) load

• The armature current is in phase with the emf, so its maximum occurs when the pole axis is midway between the conductors; the armature flux is at 90° (in quadrature) to the main flux.
• The effect is cross-magnetising: flux is distorted (crowded at one pole tip, weakened at the other) but its average value changes little.
• The terminal voltage drops only slightly (mainly due to $IR_a$ and the small quadrature drop). $E$ is only a little greater than $V$.

 (a) Unity pf load
                    E
                  / |
                /   |  j I Xs
              /     |
            O-------+  V
            O------->  I  (in phase with V)

(ii) Lagging power factor (inductive) load

• For zero pf lagging, the current lags the emf by 90°; armature flux acts directly opposite to the main flux along the pole axis.
• The effect is demagnetising; the resultant air-gap flux and generated voltage fall.
• For a practical lagging pf (e.g. 0.8) the effect is partly cross-magnetising and partly demagnetising.
• The terminal voltage falls considerably with load; $V < E$, and regulation is large and positive. More field current is needed to keep $V$ constant.

 (b) Lagging pf load
                      E
                     /|
                   /  |  j I Xs
                 /    |
               O------+ V
                `-.
                   `-. I  (lags V by phi)

(iii) Leading power factor (capacitive) load

• For zero pf leading, the current leads the emf by 90°; the armature flux acts along the main flux.
• The effect is magnetising (strengthening); the air-gap flux increases.
• The terminal voltage rises with load; $V$ can be greater than $E$, so regulation can be negative. Less field current is needed.

 (c) Leading pf load
             I      E
              \    /'.
               \  /   '.  j I Xs
                \/      '.
                O----------+ V
      I leads V by phi;  E < V

Summary

Parallel operation means connecting two or more alternators to the same bus-bars so that they share the load. Connecting an incoming alternator to live bus-bars is called synchronising.

Conditions for parallel operation

1. Equal voltage: the rms terminal voltage of the incoming machine must equal the bus-bar voltage.
2. Equal frequency: the frequency of the incoming machine must equal the bus-bar frequency.
3. Same phase sequence: the phase sequence (R-Y-B) of the incoming machine must be the same as that of the bus-bars.
4. In phase: the incoming voltage must be in phase with the bus voltage at the instant of closing the switch (zero phase difference).
5. The waveforms should be of similar shape (both nearly sinusoidal).

If these are not met, large circulating currents and heavy mechanical shocks occur at the instant of switching.

Dark lamp method

Three lamps are connected across the three poles of the open synchronising switch, i.e. each lamp is connected between a bus-bar phase and the same phase of the incoming alternator (R–R', Y–Y', B–B').

 Bus-bars      R         Y         B
               |         |         |
              (L1)      (L2)      (L3)   lamps across
               |         |         |     open switch
 Incoming      R'        Y'        B'
 alternator    |_________|_________|
                  G2 + prime mover

The voltage across each lamp is the difference of the bus voltage and the incoming voltage of the same phase.

Procedure

1. Start the incoming alternator with its prime mover and bring it near synchronous speed.
2. Switch on its field and adjust the excitation until its voltmeter reads the bus-bar voltage.
3. Watch the lamps:
   • If all three lamps become bright and dark together, the phase sequence is correct. If they flicker one after another (rotating effect), the phase sequence is wrong; interchange any two leads of the incoming machine.
   • The rate of flickering equals the frequency difference $(f_1 - f_2)$. Adjust the prime-mover speed until the flickering is very slow.
4. When all three lamps are dark (at the middle of the dark period), the voltages are equal and in phase. Close the synchronising switch at this instant.
5. After closing, the incoming machine is "floating" on the bus; it takes load when its prime-mover input (steam/water) is increased. The reactive load is shared by adjusting the excitation.

Merits and limitations

• Simple and cheap; it checks phase sequence as well.
• Lamps go dark at about one-third of rated voltage, so the exact instant of synchronism is uncertain; a voltmeter across a lamp or a synchroscope is used for accuracy in large stations.
• Filament failure may be mistaken for darkness.

Pitch factor (coil span factor), $k_p$

The pitch factor is the ratio of the emf induced in a short-pitched (chorded) coil to the emf that would be induced in a full-pitched coil.

If the coil is short of full pitch (180° electrical) by an angle $\alpha$, the emfs in the two coil sides are not in phase and add as phasors:

$$k_p = \frac{\text{phasor sum}}{\text{arithmetic sum}} = \cos\frac{\alpha}{2}$$

Significance: short pitching slightly reduces the emf ($k_p < 1$) but it eliminates or reduces harmonics (e.g. $\alpha = 36^\circ$ removes the 5th harmonic, since $k_{p5} = \cos(5\alpha/2) = 0$), saves copper in end connections and reduces leakage reactance.

Distribution (breadth) factor, $k_d$

The distribution factor is the ratio of the phasor sum to the arithmetic sum of the emfs of coils distributed in several slots of a phase group.

With $m$ slots per pole per phase and slot angle $\beta = \dfrac{180^\circ}{\text{slots per pole}}$:

$$k_d = \frac{\sin(m\beta/2)}{m\sin(\beta/2)}$$

Significance: distributing the winding gives a more sinusoidal emf, better use of the core and better cooling; $k_d < 1$ only slightly reduces the emf.

EMF equation of an alternator

Let

• $P$ = number of poles, $N_s$ = speed in rpm,
• $\Phi$ = flux per pole (Wb), $Z$ = conductors in series per phase, $T = Z/2$ turns per phase.

In one revolution, each conductor cuts flux $P\Phi$ in time $60/N_s$ seconds.

$$\begin{aligned} \text{Average emf per conductor} &= \frac{d\Phi}{dt} = \frac{P\Phi}{60/N_s} = \frac{P\Phi N_s}{60} \\ \text{Since } f = \frac{PN_s}{120},\quad \frac{PN_s}{60} &= 2f \\ \text{Average emf per conductor} &= 2f\Phi\ \text{V} \\ \text{Average emf per turn} &= 2 \times 2f\Phi = 4f\Phi\ \text{V} \\ \text{Average emf per phase} &= 4f\Phi T \\ \text{RMS emf per phase} &= 1.11 \times 4f\Phi T = 4.44\, f\Phi T\ \text{V} \end{aligned}$$

(form factor of a sine wave = 1.11). Including the winding factors for a short-pitched, distributed winding:

$$E_{ph} = 4.44\, k_p\, k_d\, f\, \Phi\, T\ \text{volts}$$

For a star-connected machine, line voltage $E_L = \sqrt{3}\,E_{ph}$; for delta, $E_L = E_{ph}$. The product $k_w = k_p k_d$ is called the winding factor.

Connecting an incoming alternator to live bus-bars (synchronising) is safe only when the following conditions are met:

1. Equal voltage: rms terminal voltage of the incoming alternator = bus-bar voltage.
2. Equal frequency: frequency of the incoming alternator = bus-bar frequency.
3. Same phase sequence: R-Y-B of the machine must match R-Y-B of the bus.
4. Phase agreement: at the instant of closing, corresponding voltages must be in phase.

Otherwise heavy circulating current and mechanical shock occur.

Three lamp (two bright, one dark) method

One lamp $L_1$ is connected between corresponding phases (R–R'), while the other two are cross-connected: $L_2$ between Y and B', and $L_3$ between B and Y'.

 Bus-bars     R          Y          B
              |           \        /
            (L1)          (L2)  (L3)
              |              \  /
              |               \/
              |               /\
 Incoming     R'          B'      Y'
 alternator   |___________|_______|

Procedure

1. Run the incoming machine near synchronous speed and adjust its field so its voltage equals the bus voltage.
2. The lamps brighten and darken one after another in a rotating sequence. The direction of rotation shows whether the machine is fast or slow, and the rate shows the frequency difference.
3. Adjust the prime-mover speed until the sequence is very slow.
4. Close the switch at the instant when $L_1$ is dark and $L_2$, $L_3$ are equally bright.

Justification (phasors)

Let the bus phase voltages be $V_R, V_Y, V_B$ (each $V$) and the machine voltages $V_{R'}, V_{Y'}, V_{B'}$.

          VR = VR'
           ^
           |          lamp L1 : VR - VR' = 0  (dark)
           |
          O
         / \          lamp L2 : VY - VB' = sqrt3 V
        /   \         lamp L3 : VB - VY' = sqrt3 V
     VB=VB'  VY=VY'   (equal brightness)

• When the two systems are in phase, $V_R - V_{R'} = 0$, so $L_1$ is dark.
• $L_2$ sees $V_Y - V_{B'} = V_Y - V_B$, which is a line voltage of magnitude $\sqrt{3}V$; similarly $L_3$ sees $\sqrt{3}V$. Both are equally and fully bright.
• When the machine phasors rotate relative to the bus (frequency difference), the voltages across $L_1, L_2, L_3$ reach their maxima at different instants (120° apart), so the lamps glow in sequence. This gives the "rotating light" effect that tells fast/slow.
• If the phase sequence were wrong, all lamps would not follow this pattern (they would brighten/darken together), so the method also checks phase sequence.

Advantage over dark-lamp method: the exact instant is easier to judge, because the moment of one dark lamp with two equally bright lamps is sharper than "all dark", and the direction of rotation tells whether to speed up or slow down.

For a synchronous generator,

$$\bar E = \bar V + \bar I R_a + j\bar I X_l + j\bar I X_a = \bar V + \bar I(R_a + jX_s)$$

where $X_a$ represents armature reaction and $X_l$ the leakage reactance. With a capacitive (leading pf) load, $V$ becomes greater than $E$ for two reasons.

1. Armature reaction is magnetising

• With a purely capacitive load, the armature current leads the emf by 90°. The current in a conductor is maximum before the pole axis reaches it, so the armature mmf acts along the field axis and aids the main flux.
• The resultant air-gap flux increases, so the voltage actually generated in the air gap rises.
• For a practical leading pf (e.g. 0.8 lead), one component of armature mmf is magnetising and the other cross-magnetising; the net effect still raises the voltage.

2. Reactance drop adds to the voltage (phasor view)

The drop $jIX_s$ is perpendicular to $I$. When $I$ leads $V$, $jIX_s$ points backwards against $V$, so the vector sum makes $E$ shorter than $V$:

             I      E
              \    /'.
               \  /   '.  j I Xs
                \/      '.
                O----------+ V
                   I R  (small, along I)

From the components (taking current as reference):

$$E = \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi - IX_s)^2}$$

Because the reactive term is $V\sin\phi - IX_s$ (minus sign for leading pf), $E$ can be less than $V$, i.e. $V > E$.

Result

• Terminal voltage rises as the capacitive load increases; the external characteristic rises.
• Voltage regulation $=\dfrac{E - V}{V}\times 100\%$ is negative.
• A smaller field current is needed to keep rated voltage. Example: an alternator feeding a long unloaded cable or a capacitor bank (Ferranti-like rise).

The power angle characteristic is the curve of electrical power ($P$) against the power (load) angle $\delta$, the angle between the excitation emf $E$ and the terminal voltage $V$.

Derivation (armature resistance neglected)

For a cylindrical-rotor generator with $\bar E = E\angle\delta$, $\bar V = V\angle 0$:

$$\begin{aligned} \bar I &= \frac{E\angle\delta - V\angle 0}{jX_s} \\ \bar S &= 3\bar V\bar I^{*} \\ P &= \frac{3EV}{X_s}\sin\delta \\ Q &= \frac{3V}{X_s}\left(E\cos\delta - V\right) \end{aligned}$$

(per-phase values; for a motor, $\delta$ is negative and power flows into the machine).

The curve

  P
  |  generator       Pmax
  |  region       .-"""""-.
  |            .'           '.
  |          .'               '.
  |        .'                   '.
 -+------+------------+-----------+-----> delta
 -180   .0           90          180 deg
     .'
  .'   motor region (delta < 0)

Features

1. Power varies sinusoidally with $\delta$.
2. Maximum power occurs at $\delta = 90^\circ$: $$P_{max} = \frac{3EV}{X_s}$$ This is the steady-state stability limit. Beyond 90°, more $\delta$ gives less power, the machine cannot hold synchronism and falls out of step (pole slipping).
3. Positive $\delta$ (E leads V): generator action; negative $\delta$: motor action.
4. Normal operation is at $\delta$ = 20°–35°, leaving a margin for sudden load changes.
5. $P_{max}$ increases with excitation $E$ and with lower $X_s$; strong excitation improves stability.
6. Synchronising power coefficient $P_{syn} = \dfrac{dP}{d\delta} = \dfrac{3EV}{X_s}\cos\delta$ is the slope of the curve; it is largest at $\delta = 0$ and zero at $90^\circ$.

Given: star connected, $V_L$ = 12000 V, $R_a = 2\ \Omega$, $X_s = 25\ \Omega$ per phase, load 1250 kW at 0.75 pf.

$$\begin{aligned} V_{ph} &= \frac{12000}{\sqrt3} = 6928.2\ \text{V} \\ I &= \frac{1250\times10^3}{\sqrt3\times12000\times0.75} = 80.19\ \text{A} \\ \cos\phi &= 0.75,\quad \sin\phi = 0.6614 \\ V\cos\phi + IR_a &= 5196.15 + 80.19\times2 = 5356.53\ \text{V} \\ V\sin\phi &= 4582.58\ \text{V},\quad IX_s = 80.19\times25 = 2004.69\ \text{V} \end{aligned}$$

(i) 0.75 pf leading

$$\begin{aligned} E &= \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi - IX_s)^2} \\ &= \sqrt{5356.53^2 + (4582.58 - 2004.69)^2} \\ &= \sqrt{5356.53^2 + 2577.89^2} = 5944.57\ \text{V per phase} \\ E_L &= \sqrt3 \times 5944.57 = 10296\ \text{V} \\ \%\text{Reg} &= \frac{5944.57 - 6928.2}{6928.2}\times100 = -14.20\% \end{aligned}$$

(ii) 0.75 pf lagging

$$\begin{aligned} E &= \sqrt{5356.53^2 + (4582.58 + 2004.69)^2} \\ &= \sqrt{5356.53^2 + 6587.26^2} = 8490.26\ \text{V per phase} \\ E_L &= \sqrt3 \times 8490.26 = 14706\ \text{V} \\ \%\text{Reg} &= \frac{8490.26 - 6928.2}{6928.2}\times100 = 22.55\% \end{aligned}$$

Answer: (i) 0.75 lead: E = 5944.6 V/phase (10.30 kV line), regulation = −14.20%; (ii) 0.75 lag: E = 8490.3 V/phase (14.71 kV line), regulation = +22.55%.

A synchronous machine has a stationary armature (stator) and a rotating field system (rotor) excited by DC; the rotor runs at synchronous speed $N_s = 120f/P$.

        +-------------------------+
        |  stator frame           |
        |  +-------------------+  |
        |  | stator core with  |  |
        |  | 3-ph winding slots|  |
        |  |     +-------+     |  |
        |  |     | rotor | <-- DC via slip rings
        |  |     | (N/S) |     |  |
        |  |     +-------+     |  |
        |  +-------------------+  |
        +-------------------------+

Stator (armature)

• Frame of cast iron or welded steel supports the core.
• Core of thin silicon-steel laminations (to reduce eddy and hysteresis losses) with slots on the inner surface.
• Three-phase winding, usually double-layer, short-pitched and distributed, placed in the slots; usually star connected.

Rotor and field

The field winding is fed with DC through slip rings and brushes (or a brushless exciter). There are two types of rotor:

Need for parallel operation

Generating stations use several alternators connected in parallel to common bus-bars instead of one large machine because:

1. Continuity of supply: if one machine fails or needs maintenance, others keep supplying the load.
2. Efficiency: machines are most efficient near full load; units are switched in or out as load varies, so each runs near full load.
3. Maintenance and repair can be done one unit at a time without shutting down the station.
4. Future expansion: new units can be added as demand grows.
5. Reserve capacity: only a small spare unit is needed instead of a spare of the full station capacity.
6. Interconnected grids require all generators to run in parallel.

Dark lamp method of synchronising

Three lamps are connected across the open synchronising switch, each between a bus-bar phase and the same phase of the incoming machine (R–R', Y–Y', B–B').

 Bus-bars      R         Y         B
               |         |         |
              (L1)      (L2)      (L3)
               |         |         |
 Incoming      R'        Y'        B'
 alternator    |_________|_________|

Procedure

1. Run the incoming alternator up to about synchronous speed by its prime mover.
2. Adjust its field current until its voltage equals the bus voltage (voltmeters).
3. Observe the lamps. If they flicker together (all bright, all dark at the same time), the phase sequence is correct; if they glow one after another, interchange two leads of the incoming machine.
4. The flicker rate equals the frequency difference; adjust the speed until flickering is very slow.
5. Close the switch in the middle of the dark period of all three lamps; then the voltages are equal and in phase.
6. Increase prime-mover input to make the machine take active load; adjust excitation to share reactive load.

Limitation: lamps are dark over a range of voltage (up to about 1/3 of rated), so the exact moment is uncertain; a voltmeter across a lamp or a synchroscope gives better accuracy.

Effective resistance

DC resistance between two terminals = 0.16 Ω, which is two phases in series (star).

$$\begin{aligned} R_{dc,ph} &= \frac{0.16}{2} = 0.08\ \Omega \\ R_a &= 1.5 \times 0.08 = 0.12\ \Omega \end{aligned}$$

Synchronous impedance and reactance

The same field current (70 A) gives 700 V (line) on open circuit and 376 A on short circuit:

$$\begin{aligned} Z_s &= \frac{700/\sqrt3}{376} = \frac{404.15}{376} = 1.0749\ \Omega \\ X_s &= \sqrt{Z_s^2 - R_a^2} = \sqrt{1.0749^2 - 0.12^2} = 1.068\ \Omega \end{aligned}$$

Regulation at full load, 0.8 pf leading

$$\begin{aligned} V_{ph} &= \frac{2300}{\sqrt3} = 1327.91\ \text{V},\quad I = 376\ \text{A} \\ V\cos\phi + IR_a &= 1062.32 + 376\times0.12 = 1107.44\ \text{V} \\ V\sin\phi - IX_s &= 796.74 - 376\times1.068 = 796.74 - 401.62 = 395.12\ \text{V} \\ E &= \sqrt{1107.44^2 + 395.12^2} = 1175.82\ \text{V per phase} \\ \%\text{Reg} &= \frac{1175.82 - 1327.91}{1327.91}\times100 = -11.45\% \end{aligned}$$

Answer: $X_s$ = 1.068 Ω per phase; full-load regulation at 0.8 pf leading = −11.45% (E = 1175.8 V/phase, 2036.6 V line).

Yes. The terminal voltage $V$ of an alternator can be greater than the internally generated emf $E$ when it supplies a leading power factor (capacitive) load.

Why

The voltage equation is

$$\bar E = \bar V + \bar I(R_a + jX_s) \quad\Rightarrow\quad E = \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi \pm IX_s)^2}$$

with $+$ for lagging and $-$ for leading current. With leading current the term $V\sin\phi - IX_s$ is small, so $E < V$. Physically, the armature reaction of a leading current is magnetising: it strengthens the main field flux, so the voltage rises with load.

 Leading pf load (E < V)
             I      E
              \    /'.
               \  /   '.  j I Xs
                \/      '.
                O----------+ V

Effect of armature reaction: resistive (unity pf) load

• Current is in phase with emf; armature mmf is at 90° to the field mmf.
• Effect is cross-magnetising: flux is distorted (stronger at one pole tip, weaker at the other) with little change in its total value.
• Terminal voltage falls slightly with load; $E$ is a little more than $V$.

 Unity pf
                    E
                  / |
                /   |  j I Xs
              /     |
            O-------+  V
            O------->  I

Effect of armature reaction: inductive (lagging pf) load

• For zero pf lagging, current lags emf by 90°, the armature mmf directly opposes the field mmf.
• Effect is demagnetising; air-gap flux falls and the terminal voltage drops sharply with load.
• For practical lagging pf (e.g. 0.8), it is partly demagnetising and partly cross-magnetising. More field current is needed to hold rated voltage; regulation is large and positive.

 Lagging pf (E > V)
                      E
                     /|
                   /  |  j I Xs
                 /    |
               O------+ V
                `-.
                   `-. I

Given: delta connected, so $V_{ph} = V_L = 13.5$ kV; $R_a = 1.5\ \Omega$, $X_s = 3\ \Omega$ per phase; load 1.28 MW at 0.8 pf, i.e. $1.28/0.8 = 1.6$ MVA (full load).

$$\begin{aligned} I_{ph} &= \frac{1.6\times10^6}{3\times13500} = 39.51\ \text{A} \\ V\cos\phi + IR_a &= 13500\times0.8 + 39.51\times1.5 = 10800 + 59.26 = 10859.26\ \text{V} \\ V\sin\phi &= 8100\ \text{V},\quad IX_s = 39.51\times3 = 118.52\ \text{V} \end{aligned}$$

(i) 0.8 pf lagging

$$\begin{aligned} E &= \sqrt{10859.26^2 + (8100 + 118.52)^2} = \sqrt{10859.26^2 + 8218.52^2} \\ &= 13618.65\ \text{V} \\ \%\text{Reg} &= \frac{13618.65 - 13500}{13500}\times100 = 0.879\% \end{aligned}$$

(ii) 0.8 pf leading

$$\begin{aligned} E &= \sqrt{10859.26^2 + (8100 - 118.52)^2} = \sqrt{10859.26^2 + 7981.48^2} \\ &= 13476.93\ \text{V} \\ \%\text{Reg} &= \frac{13476.93 - 13500}{13500}\times100 = -0.171\% \end{aligned}$$

The regulation is small because the given impedance (3 Ω) is very small compared with the phase voltage.

Answer: (i) +0.88% at 0.8 pf lagging; (ii) −0.17% at 0.8 pf leading.

Given: star, 10 kVA, 230 V, $R_a = 0.5\ \Omega$, $X_s = 1.2\ \Omega$ per phase.

$$\begin{aligned} V_{ph} &= \frac{230}{\sqrt3} = 132.79\ \text{V},\quad I = \frac{10000}{\sqrt3\times230} = 25.10\ \text{A} \\ V\cos\phi + IR_a &= 106.23 + 12.55 = 118.78\ \text{V} \\ V\sin\phi &= 79.67\ \text{V},\quad IX_s = 30.12\ \text{V} \end{aligned}$$

(a) 0.8 pf lagging

$$\begin{aligned} E &= \sqrt{118.78^2 + (79.67 + 30.12)^2} = 161.76\ \text{V} \\ \%\text{Reg} &= \frac{161.76 - 132.79}{132.79}\times100 = 21.81\% \end{aligned}$$

(b) 0.8 pf leading

$$\begin{aligned} E &= \sqrt{118.78^2 + (79.67 - 30.12)^2} = 128.70\ \text{V} \\ \%\text{Reg} &= \frac{128.70 - 132.79}{132.79}\times100 = -3.08\% \end{aligned}$$

(c) Power factor for zero regulation

Zero regulation means $E = V$. With $Z_s = R_a + jX_s = 1.3\angle 67.38^\circ\ \Omega$ and current $I\angle\phi$ ($\phi$ positive for leading):

$$\begin{aligned} E^2 &= V^2 + 2VIZ_s\cos(\theta_z + \phi) + I^2Z_s^2 = V^2 \\ \cos(\theta_z + \phi) &= -\frac{IZ_s}{2V} = -\frac{25.10\times1.3}{2\times132.79} = -0.1229 \\ \theta_z + \phi &= 97.06^\circ \\ \phi &= 97.06^\circ - 67.38^\circ = 29.68^\circ\ \text{(leading)} \\ \text{pf} &= \cos 29.68^\circ = 0.869\ \text{leading} \end{aligned}$$

Answer: (a) 21.81%; (b) −3.08%; (c) zero regulation at pf = 0.869 leading.

Given: star, 1650 kVA, 13.8 kV, $R_a = 1.6\ \Omega$, $X_s = 35\ \Omega$; load 1320 kW at 0.8 pf (= 1650 kVA, full load).

$$\begin{aligned} V_{ph} &= \frac{13800}{\sqrt3} = 7967.43\ \text{V} \\ I &= \frac{1320\times10^3}{\sqrt3\times13800\times0.8} = 69.03\ \text{A} \\ V\cos\phi + IR_a &= 6373.95 + 110.45 = 6484.40\ \text{V} \\ V\sin\phi &= 4780.46\ \text{V},\quad IX_s = 2416.09\ \text{V} \end{aligned}$$

The power angle $\delta$ is the angle between $E$ and $V$. If $\theta$ is the angle of $E$ from the current phasor, $\theta = \tan^{-1}\dfrac{V\sin\phi \pm IX_s}{V\cos\phi + IR_a}$, then $\delta = \theta - \phi$ for lagging pf and $\delta = \phi - \theta$ for leading pf.

(a) 0.8 pf lagging

$$\begin{aligned} E &= \sqrt{6484.40^2 + (4780.46 + 2416.09)^2} = \sqrt{6484.40^2 + 7196.55^2} = 9686.98\ \text{V} \\ \theta &= \tan^{-1}\frac{7196.55}{6484.40} = 47.98^\circ \\ \delta &= 47.98^\circ - 36.87^\circ = 11.11^\circ \\ \%\text{Reg} &= \frac{9686.98 - 7967.43}{7967.43}\times100 = 21.58\% \end{aligned}$$

(b) 0.8 pf leading

$$\begin{aligned} E &= \sqrt{6484.40^2 + (4780.46 - 2416.09)^2} = \sqrt{6484.40^2 + 2364.37^2} = 6902.00\ \text{V} \\ \theta &= \tan^{-1}\frac{2364.37}{6484.40} = 20.03^\circ \\ \delta &= 36.87^\circ - 20.03^\circ = 16.84^\circ \\ \%\text{Reg} &= \frac{6902.00 - 7967.43}{7967.43}\times100 = -13.37\% \end{aligned}$$

(In both cases $E$ leads $V$, as it must for a generator.)

Answer: (a) regulation = 21.58%, power angle = 11.11°; (b) regulation = −13.37%, power angle = 16.84°.

Regulation is found by the synchronous impedance (EMF) method.

Armature resistance

DC test between two lines (star: two phases in series):

$$\begin{aligned} R_{dc,ph} &= \frac{1}{2}\times\frac{6}{6} = 0.5\ \Omega \\ R_a &= 1.667\times0.5 = 0.8335\ \Omega \end{aligned}$$

Synchronous impedance (same field current 12.5 A)

$$\begin{aligned} I_{rated} &= \frac{100\times10^3}{\sqrt3\times1100} = 52.49\ \text{A} \\ E_{oc,ph} &= \frac{420}{\sqrt3} = 242.49\ \text{V} \\ Z_s &= \frac{242.49}{52.49} = 4.620\ \Omega \\ X_s &= \sqrt{4.620^2 - 0.8335^2} = 4.544\ \Omega \end{aligned}$$

Regulation at full load, 0.8 pf lagging

$$\begin{aligned} V_{ph} &= \frac{1100}{\sqrt3} = 635.09\ \text{V} \\ V\cos\phi + IR_a &= 508.07 + 52.49\times0.8335 = 551.82\ \text{V} \\ V\sin\phi + IX_s &= 381.05 + 52.49\times4.544 = 619.56\ \text{V} \\ E &= \sqrt{551.82^2 + 619.56^2} = 829.67\ \text{V per phase} \\ \%\text{Reg} &= \frac{829.67 - 635.09}{635.09}\times100 = 30.64\% \end{aligned}$$

Answer: $Z_s$ = 4.62 Ω, $X_s$ = 4.544 Ω, regulation at 0.8 pf lagging = 30.64% (full load assumed).

A three-phase synchronous generator (alternator) converts mechanical energy into three-phase AC electrical energy at a frequency fixed by its speed, $f = \dfrac{PN_s}{120}$.

Construction

           stator frame
     +------------------------+
     |  laminated stator core |
     |  with 3-ph armature    |
     |  winding in slots      |
     |      +----------+      |
     |      |  rotor   |      |  DC field via
     |      |  N    S  |------+--- slip rings
     |      +----------+      |  and brushes
     +------------------------+
          shaft <- prime mover

Stator (armature) – stationary part

• Frame (cast iron/welded steel) for support and ventilation.
• Core of 0.35–0.5 mm silicon-steel laminations with slots, to reduce iron losses.
• Three-phase, double-layer, distributed, short-pitched winding, displaced 120° electrical, usually star connected.

Rotor (field) – rotating part, excited by DC from an exciter through slip rings (or brushless exciter):

1. Salient-pole rotor: projecting poles on a spider, large diameter, short length, many poles; used for low-speed hydro turbines. Damper bars in pole faces.
2. Cylindrical (non-salient) rotor: smooth forged-steel cylinder, small diameter, long axial length, 2 or 4 poles; used for high-speed steam/gas turbines.

Why stationary armature, rotating field? High voltage (11 kV and above) armature is easier to insulate when stationary; output is taken directly without slip rings; only low-voltage DC (100–400 V) is fed through slip rings; the rotor is lighter and better for high speed; the stator can be cooled better.

Principle of operation

1. The field winding is excited with DC, producing alternate N and S poles on the rotor.
2. The prime mover drives the rotor at synchronous speed $N_s$.
3. The rotating magnetic field cuts the stationary armature conductors. By Faraday's law of electromagnetic induction, an emf is induced in them: $e = Blv$.
4. As N and S poles pass a conductor alternately, the emf alternates; one cycle is produced per pair of poles passing. Hence $f = PN_s/120$.
5. The three phase windings are 120° electrical apart in space, so three emfs equal in magnitude and 120° apart in time are produced.
6. The rms emf per phase is $E = 4.44\,k_p k_d f\Phi T$.
7. When a load is connected, current flows; the armature mmf (armature reaction) and the impedance drop change the terminal voltage, which is corrected by adjusting the field current.

Using the two-reaction (Blondel) theory: $X_d$ = 0.6 pu, $X_q$ = 0.45 pu, $R_a$ = 0.015 pu, $V$ = 1 pu, $I$ = 1 pu, pf = 0.8 lagging ($\phi = 36.87^\circ$).

                      E (on q-axis)
                     /:
                    / : j Id (Xd - Xq)
                   /  :
                  / .-+ E' = V + I(Ra + jXq)
                 /.' delta
                O-------------- V
                 \  phi
                  \ I (psi = phi + delta from q-axis)

(i) Load angle $\delta$

$$\begin{aligned} \tan\psi &= \frac{V\sin\phi + IX_q}{V\cos\phi + IR_a} = \frac{0.6 + 0.45}{0.8 + 0.015} = \frac{1.05}{0.815} = 1.2883 \\ \psi &= 52.18^\circ \\ \delta &= \psi - \phi = 52.18^\circ - 36.87^\circ = 15.31^\circ \end{aligned}$$

(ii) $I_d$ and $I_q$

$$\begin{aligned} I_d &= I\sin\psi = \sin 52.18^\circ = 0.790\ \text{pu} \\ I_q &= I\cos\psi = \cos 52.18^\circ = 0.613\ \text{pu} \end{aligned}$$

(iii) Open-circuit voltage (excitation emf)

$$\begin{aligned} E &= V\cos\delta + I_qR_a + I_dX_d \\ &= \cos 15.31^\circ + 0.613\times0.015 + 0.790\times0.6 \\ &= 0.9645 + 0.0092 + 0.4740 = 1.448\ \text{pu} \end{aligned}$$

(iv) Voltage regulation

$$\%\text{Reg} = \frac{E - V}{V}\times100 = \frac{1.448 - 1}{1}\times100 = 44.8\%$$

Answer: δ = 15.31°, I_d = 0.790 pu, I_q = 0.613 pu, E = 1.448 pu, regulation = 44.8%.

Given: star, 1500 kVA, 13 kV, $R_a = 0.9\ \Omega$, $X_s = 8\ \Omega$ per phase; rated current at rated voltage.

$$\begin{aligned} V_{ph} &= \frac{13000}{\sqrt3} = 7505.55\ \text{V} \\ I &= \frac{1500\times10^3}{\sqrt3\times13000} = 66.62\ \text{A} \\ IR_a &= 66.62\times0.9 = 59.96\ \text{V},\quad IX_s = 66.62\times8 = 532.94\ \text{V} \end{aligned}$$

(i) Unity power factor

$$\begin{aligned} E &= \sqrt{(V + IR_a)^2 + (IX_s)^2} = \sqrt{7565.51^2 + 532.94^2} \\ &= 7584.26\ \text{V per phase}\quad (E_L = 13136\ \text{V}) \\ \%\text{Reg} &= \frac{7584.26 - 7505.55}{7505.55}\times100 = 1.05\% \end{aligned}$$

(ii) 0.8 pf lagging

$$\begin{aligned} V\cos\phi + IR_a &= 6004.44 + 59.96 = 6064.40\ \text{V} \\ V\sin\phi + IX_s &= 4503.33 + 532.94 = 5036.27\ \text{V} \\ E &= \sqrt{6064.40^2 + 5036.27^2} = 7882.95\ \text{V per phase}\quad (E_L = 13654\ \text{V}) \\ \%\text{Reg} &= \frac{7882.95 - 7505.55}{7505.55}\times100 = 5.03\% \end{aligned}$$

Answer: unity pf: E = 7584.3 V/phase, regulation 1.05%; 0.8 lag: E = 7883.0 V/phase, regulation 5.03%.

For a synchronous generator, per phase,

$$\bar E = \bar V + \bar I R_a + j\bar I X_s$$

Whether $V$ is less or more than $E$ depends on the power factor of the load, through the armature reaction and the synchronous reactance drop.

          +--- jXs ---- Ra ---+------+
          |                   |      |
         (E)                  V    Load
          |                   |      |
          +-------------------+------+
     per-phase equivalent circuit

Case 1: $V < E$ (unity and lagging pf loads)

• Unity pf (resistive): armature reaction is cross-magnetising; $E$ is slightly greater than $V$ due to $IR_a$ and the quadrature drop $IX_s$.
• Lagging pf (inductive): armature reaction is demagnetising; the drop $jIX_s$ is nearly in phase with $V$, so $E$ is much greater than $V$.

$$E = \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi + IX_s)^2} > V$$

 Lagging pf: E > V
                      E
                     /|
                   /  |  j I Xs
                 /    |
               O------+ V
                `-.
                   `-. I

Case 2: $V > E$ (leading pf, capacitive load)

• Armature reaction is magnetising, aiding the main flux.
• $jIX_s$ has a component opposite to $V$, so

$$E = \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi - IX_s)^2}$$

which becomes less than $V$ when $IX_s$ is large enough.

 Leading pf: E < V
             I      E
              \    /'.
               \  /   '.  j I Xs
                \/      '.
                O----------+ V

Summary

At one particular leading power factor, $E = V$ and regulation is zero.

Given: star, 10 kVA, 400 V, 50 Hz, $R_a = 0.5\ \Omega$, $X_s = 10\ \Omega$; rated load at 0.8 pf lagging.

Phase voltage and current

$$\begin{aligned} V_{ph} &= \frac{400}{\sqrt3} = 230.94\ \text{V} \\ I &= \frac{10\times10^3}{\sqrt3\times400} = 14.43\ \text{A} \\ \cos\phi &= 0.8,\quad \sin\phi = 0.6,\quad \phi = 36.87^\circ \end{aligned}$$

Excitation emf

$$\begin{aligned} V\cos\phi + IR_a &= 184.75 + 14.43\times0.5 = 191.97\ \text{V} \\ V\sin\phi + IX_s &= 138.56 + 14.43\times10 = 282.90\ \text{V} \\ E &= \sqrt{191.97^2 + 282.90^2} = 341.89\ \text{V per phase}\quad(592.2\ \text{V line}) \end{aligned}$$

Power angle

The angle of $E$ from the current phasor is

$$\begin{aligned} \theta &= \tan^{-1}\frac{282.90}{191.97} = 55.84^\circ \\ \delta &= \theta - \phi = 55.84^\circ - 36.87^\circ = 18.97^\circ \end{aligned}$$

                       E = 341.9 V
                      /
                     /  delta = 18.97 deg
                    O-------------- V = 230.9 V
                     \  phi = 36.87 deg
                      \ I = 14.43 A

Voltage regulation

$$\%\text{Reg} = \frac{E - V}{V}\times100 = \frac{341.89 - 230.94}{230.94}\times100 = 48.04\%$$

(The regulation is high because $X_s = 10\ \Omega$ is large compared with the rated impedance $V/I = 16\ \Omega$, i.e. $X_s \approx 0.625$ pu.)

Answer: power angle δ = 18.97°, voltage regulation = 48.04% (E = 341.9 V per phase).

Load (external) characteristics

The load or external characteristic of an alternator is the curve of terminal voltage $V$ against load (armature) current $I_a$, at constant speed and constant field current, for a given load power factor.

  Vt
  |                 ___.-- leading pf
  |          ___.--'
  |E ------'----------.___   unity pf
  |                       `--.
  |    `--.__                 lagging pf
  |          `--.__
  +--------------------------------> Ia
  0                  rated Ia

• All curves start at $V = E$ (no load).
• Lagging pf: voltage falls sharply because of the demagnetising armature reaction and the large $IX_s$ drop in phase with $V$.
• Unity pf: voltage falls slightly (cross-magnetising effect, $IR_a$ and small quadrature drop).
• Leading pf: voltage rises with load because armature reaction is magnetising.
• Hence a lower pf (lagging) gives poorer regulation; an automatic voltage regulator changes the field current to hold $V$ constant.

Why $V > E$ for capacitive load

Per phase, $\bar E = \bar V + \bar I(R_a + jX_s)$ and

$$E = \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi - IX_s)^2}\quad\text{(leading pf)}$$

The reactive term $V\sin\phi - IX_s$ is reduced by the reactance drop, so $E$ can be smaller than $V$.

Armature reaction view: with a capacitive load the current leads the emf (by 90° for a pure capacitor). The armature mmf then acts along the axis of the field poles and aids the field flux (magnetising effect). The air-gap flux increases, so the voltage at the terminals is higher than the open-circuit emf $E$ produced by the field current alone.

 Leading pf phasor diagram
             I      E
              \    /'.
               \  /   '.  j I Xs
                \/      '.
                O----------+ V      (E < V)

Consequence: voltage regulation is negative; less excitation is needed at leading pf, and over-voltage can occur when a lightly loaded long line or capacitor bank is fed.

Given: star, 1.5 MVA, 11 kV, $R_a = 1.2\ \Omega$, $X_s = 25\ \Omega$; load 1.4375 MVA at 0.8 pf.

$$\begin{aligned} V_{ph} &= \frac{11000}{\sqrt3} = 6350.85\ \text{V} \\ I &= \frac{1.4375\times10^6}{\sqrt3\times11000} = 75.45\ \text{A} \\ V\cos\phi + IR_a &= 5080.68 + 90.54 = 5171.22\ \text{V} \\ V\sin\phi &= 3810.51\ \text{V},\quad IX_s = 1886.23\ \text{V} \end{aligned}$$

(i) 0.8 pf lagging

$$\begin{aligned} E &= \sqrt{5171.22^2 + (3810.51 + 1886.23)^2} = 7693.79\ \text{V} \\ \%\text{Reg} &= \frac{7693.79 - 6350.85}{6350.85}\times100 = 21.15\% \end{aligned}$$

(ii) 0.8 pf leading

$$\begin{aligned} E &= \sqrt{5171.22^2 + (3810.51 - 1886.23)^2} = 5517.64\ \text{V} \\ \%\text{Reg} &= \frac{5517.64 - 6350.85}{6350.85}\times100 = -13.12\% \end{aligned}$$

Power factor for zero regulation

For $E = V$, with $Z_s = 1.2 + j25 = 25.029\angle 87.25^\circ\ \Omega$ and current angle $\phi$ (leading positive):

$$\begin{aligned} \cos(\theta_z + \phi) &= -\frac{IZ_s}{2V} = -\frac{75.45\times25.029}{2\times6350.85} = -0.1487 \\ \theta_z + \phi &= 98.55^\circ \\ \phi &= 98.55^\circ - 87.25^\circ = 11.30^\circ \\ \text{pf} &= \cos 11.30^\circ = 0.981\ \text{leading} \end{aligned}$$

Answer: (i) 21.15%; (ii) −13.12%; zero regulation at pf = 0.981 leading.

Given: star, 50 kVA, 440 V, $R_a = 0.25\ \Omega$, $X_s = 3.2\ \Omega$, leakage reactance $X_l = 0.5\ \Omega$ per phase; rated load at unity pf.

• The internal (air-gap) emf $E_i$ is found using only the leakage impedance $R_a + jX_l$.
• The no-load emf $E_0$ (excitation emf) uses the full synchronous impedance $R_a + jX_s$, since $X_s = X_l + X_a$ includes armature reaction.

$$\begin{aligned} V_{ph} &= \frac{440}{\sqrt3} = 254.03\ \text{V} \\ I &= \frac{50\times10^3}{\sqrt3\times440} = 65.61\ \text{A}\quad(\text{in phase with } V) \end{aligned}$$

(i) Internal emf

$$\begin{aligned} \bar E_i &= V + I(R_a + jX_l) = (254.03 + 65.61\times0.25) + j(65.61\times0.5) \\ &= 270.44 + j32.80 \\ E_i &= 272.42\ \text{V per phase}\quad(471.8\ \text{V line}) \end{aligned}$$

(ii) No-load emf

$$\begin{aligned} \bar E_0 &= V + I(R_a + jX_s) = 270.44 + j(65.61\times3.2) \\ &= 270.44 + j209.95 \\ E_0 &= 342.36\ \text{V per phase}\quad(593.0\ \text{V line}) \end{aligned}$$

(iii) Voltage regulation

$$\%\text{Reg} = \frac{E_0 - V}{V}\times100 = \frac{342.36 - 254.03}{254.03}\times100 = 34.77\%$$

              E0
             /|
            / |  I Xa (armature reaction)
           /  |
          / Ei+
         / /  |  I Xl
        O-----+  V + I Ra
        O-------> I  (unity pf)

Answer: (i) $E_i$ = 272.4 V/phase (471.8 V line); (ii) $E_0$ = 342.4 V/phase (593.0 V line); (iii) regulation = 34.77%.

Voltage regulation of an alternator is the change in terminal voltage from no load to full load (speed and field current constant), expressed as a percentage of the full-load voltage:

$$\%\text{Reg} = \frac{E_0 - V}{V}\times100$$

where $E_0 = \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi \pm IX_s)^2}$. The factors that affect it are:

1. Armature resistance drop ($IR_a$)

The resistance of the armature winding causes a drop in phase with the current. It is small in large machines but increases with temperature.

2. Leakage reactance drop ($IX_l$)

Part of the armature flux links only the armature conductors (slot, end-winding and tooth-tip leakage). It causes a reactive drop $IX_l$.

3. Armature reaction ($IX_a$)

The armature mmf changes the main flux: cross-magnetising at unity pf, demagnetising at lagging pf, magnetising at leading pf. It is represented by $X_a$; usually it is the biggest factor. $X_s = X_l + X_a$.

4. Power factor of the load

• Lagging pf: large positive regulation (voltage falls).
• Unity pf: small positive regulation.
• Leading pf: low or negative regulation (voltage rises).

5. Magnitude of load current

All drops are proportional to the current; regulation rises with load.

6. Saturation and field excitation

With saturation, the actual change in voltage is less than predicted by unsaturated $X_s$; this is why the EMF method gives pessimistic (higher) and the MMF method optimistic (lower) values, while ZPF (Potier) gives accurate values.

7. Speed and frequency

A change in prime-mover speed changes $E = 4.44fk_w\Phi T$ and $X_s = 2\pi fL$; constant speed is assumed.

Given: star, 1500 kVA, 11 kV, $R_a = 0.8\ \Omega$, $X_s = 4\ \Omega$; load $Z_L = 80 + j60\ \Omega$ per phase at rated voltage.

Load current

$$\begin{aligned} V_{ph} &= \frac{11000}{\sqrt3} = 6350.85\ \text{V} \\ Z_L &= 80 + j60 = 100\angle 36.87^\circ\ \Omega \\ I &= \frac{6350.85}{100} = 63.51\ \text{A},\quad \text{pf} = \frac{80}{100} = 0.8\ \text{lagging} \end{aligned}$$

EMF generated

$$\begin{aligned} V\cos\phi + IR_a &= 5080.68 + 63.51\times0.8 = 5131.49\ \text{V} \\ V\sin\phi + IX_s &= 3810.51 + 63.51\times4 = 4064.55\ \text{V} \\ E &= \sqrt{5131.49^2 + 4064.55^2} = 6546.20\ \text{V per phase} \\ E_L &= \sqrt3\times6546.20 = 11338\ \text{V} \end{aligned}$$

Voltage regulation

$$\%\text{Reg} = \frac{6546.20 - 6350.85}{6350.85}\times100 = 3.08\%$$

Overloaded or under-loaded?

$$\begin{aligned} I_{rated} &= \frac{1500\times10^3}{\sqrt3\times11000} = 78.73\ \text{A} \\ \text{Load kVA} &= 3\times6350.85\times63.51\times10^{-3} = 1210\ \text{kVA} \\ \text{Under-load} &= \frac{78.73 - 63.51}{78.73}\times100 = \frac{1500 - 1210}{1500}\times100 = 19.33\% \end{aligned}$$

Answer: E = 6546.2 V/phase (11.34 kV line); regulation = 3.08%; the generator is under-loaded by 19.33% (supplies 1210 kVA, 968 kW).

The statement is true. The power-angle equation of a non-salient (cylindrical-rotor) machine is

$$P = \frac{3EV}{X_s}\sin\delta$$

It assumes a uniform air gap, so the same reactance $X_s$ acts on the armature mmf in every direction.

Why it fails for salient-pole machines

1. A salient-pole rotor has a non-uniform air gap: small under the pole (direct axis, d-axis) and large between poles (quadrature axis, q-axis).
2. Reluctance along the d-axis is low, so $X_d$ is large; along the q-axis reluctance is high, so $X_q$ is smaller ($X_d > X_q$). A single $X_s$ cannot represent the machine.
3. The armature current must be split into $I_d$ and $I_q$ (two-reaction theory), and the power becomes

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

4. The second term is the reluctance power; it exists even with zero excitation ($E = 0$), because the rotor tends to align with the stator field along its minimum-reluctance path. A cylindrical rotor has no such term ($X_d = X_q$).

Effect on the curve

 P
 |        resultant (salient)
 |      .-"-.
 |    .'  .--'-.  excitation term
 |  .'  .'      '.
 |.' .-'-.        '.  reluctance term (sin 2d)
 +--------'----------'----> delta
 0     45   70  90     180

• Maximum power occurs at $\delta < 90^\circ$ (typically 60°–75°), not at 90°.
• $P_{max}$ is larger than for a cylindrical machine with the same $E$ and $X_d$; the curve is steeper at small $\delta$ (higher synchronising power).

Hence the cylindrical-rotor formula would give a wrong $P_{max}$ and wrong stability limit for salient-pole machines.