Electrical Machines · Chapter 76

Salient-Pole Machines and Two-Reaction Theory

Part 5 · Synchronous Machines — when the air gap is not uniform, one reactance becomes two.

Prof. Mithun Mondal Engineering Devotion Digital Textbook
i Learning Objectives

By the end of this chapter you should be able to:

  1. Explain why a single synchronous reactance is inadequate for a salient-pole machine.

  2. Resolve armature current and MMF into direct-axis and quadrature-axis components (Blondel's two-reaction theory).

  3. Draw the phasor diagram and compute the excitation EMF and load angle.

  4. Derive and interpret the power–angle equation including reluctance power.

  5. Determine \(X_d\) and \(X_q\) from the slip test.

Section 76-1

Why One Reactance Is Not Enough

In a cylindrical-rotor machine the air gap is uniform, so armature MMF produces the same flux whatever its position relative to the poles. A salient-pole rotor has a short air gap under the pole faces and a long one between poles. The same MMF therefore produces more flux along the pole axis than across it.

Direct axis (d)

Along the pole centre line. Low reluctance → large reactance \(X_d\).

Quadrature axis (q)

90 electrical degrees away, between poles. High reluctance → smaller reactance \(X_q\).

Typical salient-pole values: \(X_d \approx 0.6\)–\(1.5\) pu, \(X_q \approx 0.4\)–\(1.0\) pu, with \(X_q / X_d \approx 0.6\)–\(0.7\).

Section 76-2

Blondel's Two-Reaction Theory

Blondel resolved the armature current \(I_a\) into two components, each acting on one axis where the reluctance is known:

\[I_d = I_a \sin\psi, \qquad I_q = I_a \cos\psi, \qquad \psi = \delta + \phi\]

where \(\psi\) is the angle between \(I_a\) and the excitation EMF \(E_f\) (the internal angle), \(\delta\) the load angle and \(\phi\) the power-factor angle (lagging, generator). Each component produces a voltage drop across its own reactance:

⟂
Generator voltage equation
Two reactances, two current components
\[\vec E_f = \vec V + R_a\vec I_a + jX_d\vec I_d + jX_q\vec I_q\]

Setting \(X_d = X_q = X_s\) recovers the cylindrical-rotor equation of Chapter 75.

Section 76-3

Phasor Diagram and the Load Angle

The difficulty is that \(\psi\) is unknown until \(E_f\) is known. The trick is to first draw \(\vec V + R_a\vec I_a + jX_q\vec I_a\): this phasor lies exactly along \(\vec E_f\), so it fixes the direction of the q-axis. Neglecting \(R_a\):

\[\tan\psi = \frac{V\sin\phi + I_aX_q}{V\cos\phi}, \qquad \delta = \psi - \phi\]
\[E_f = V\cos\delta + I_dX_d\]
Why the trick works. \(jX_q\vec I_a = jX_q\vec I_d + jX_q\vec I_q\). The term \(jX_q\vec I_d\) is along \(\vec E_f\) (since \(\vec I_d\) is perpendicular to it), so the tip of \(\vec V + jX_q\vec I_a\) differs from \(\vec E_f\) only by \(j(X_d - X_q)\vec I_d\), which is also along \(\vec E_f\).
Salient-pole generator, lagging power factor (Ra neglected) V Ia jXq·Iq jXd·Id Ef (q-axis) δ
Figure 76.1 — Phasor diagram of a salient-pole generator. The excitation EMF lies along the q-axis and leads the terminal voltage by the load angle δ.
Section 76-4

Power–Angle Equation and Reluctance Power

Neglecting \(R_a\), the per-phase power output of a salient-pole machine is

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

  • The first term is the excitation power, identical in form to the cylindrical machine.

  • The second is reluctance power: it exists even with zero field current, because the salient rotor tends to align its low-reluctance axis with the stator field. It is the basis of the reluctance motor (Chapter 91).

  • The reluctance term makes the curve steeper at small \(\delta\) and moves the maximum power to \(\delta < 90^\circ\).

1 Worked Example 76.1 — Excitation EMF and Load Angle

Problem. A salient-pole generator has \(X_d = 1.0\) pu and \(X_q = 0.6\) pu, with negligible \(R_a\). It delivers rated current at rated voltage and 0.8 power factor lagging. Find \(\delta\), \(I_d\), \(I_q\) and \(E_f\), and check the power.

\[\tan\psi = \frac{1(0.6) + 1(0.6)}{1(0.8)} = 1.5 \;\Rightarrow\; \psi = 56.31^\circ, \qquad \delta = 56.31^\circ - 36.87^\circ = 19.44^\circ\]
\[I_d = \sin 56.31^\circ = 0.832~\text{pu}, \qquad I_q = \cos 56.31^\circ = 0.555~\text{pu}\]
\[E_f = \cos 19.44^\circ + 0.832(1.0) = 0.943 + 0.832 = 1.775~\text{pu}\]

Check.

\[P = \frac{1.775}{1.0}\sin 19.44^\circ + \frac{1}{2}\left(\frac{1}{0.6} - 1\right)\sin 38.88^\circ = 0.591 + 0.209 = 0.800~\text{pu} \checkmark\]

Treating the machine as cylindrical with \(X_s = X_d\) would give \(E_f = 1.789\) pu — close, but the load angle would be quite different, and the 26 % of the power that is reluctance power would be missed.

Section 76-5

The Slip Test

The rotor is driven at a speed slightly different from synchronous with the field winding open, and a reduced balanced three-phase voltage is applied to the armature. As the stator field slips slowly past the poles, it alternately aligns with the d-axis and the q-axis.

  • When aligned with the d-axis, the impedance is high: current is minimum and terminal voltage maximum.

  • When aligned with the q-axis, impedance is low: current is maximum and voltage minimum.

\[X_d = \frac{V_\text{max}}{I_\text{min}}, \qquad X_q = \frac{V_\text{min}}{I_\text{max}} \qquad \text{(per-phase values)}\]

The slip must be very small, otherwise currents induced in the damper winding distort the readings. A voltmeter across the open field winding shows the slip frequency directly.

2 Worked Example 76.2 — Slip-Test Reduction

Problem. In a slip test, the per-phase readings swung between 96 V and 108 V, and between 9 A and 12 A. Find \(X_d\), \(X_q\) and their ratio.

\[X_d = \frac{108}{9} = 12.0~\Omega, \qquad X_q = \frac{96}{12} = 8.0~\Omega, \qquad \frac{X_q}{X_d} = 0.667\]
Section 76-6

Summary and Common Mistakes

Key Points

  • Saliency makes reluctance depend on axis: \(X_d > X_q\).

  • \(\vec V + jX_q\vec I_a\) gives the direction of \(\vec E_f\); then \(E_f = V\cos\delta + I_dX_d\).

  • Power has an excitation term in \(\sin\delta\) and a reluctance term in \(\sin 2\delta\).

  • Slip test: \(X_d = V_\text{max}/I_\text{min}\), \(X_q = V_\text{min}/I_\text{max}\).

Common Mistakes

  • Pairing \(V_\text{max}\) with \(I_\text{max}\) in the slip test.

  • Taking \(\psi\) as the power-factor angle — it is \(\delta + \phi\).

  • Assuming maximum power occurs at \(\delta = 90^\circ\) for a salient-pole machine.

Section 76-7

Chapter Review

Practice Problems

  1. P76.1 Repeat Example 76.1 at unity power factor.

    Show answer
    \(\tan\psi = 0.6/1 \Rightarrow \psi = \delta = 30.96^\circ\); \(I_d = 0.514\); \(E_f = \cos 30.96^\circ + 0.514 = 0.857 + 0.514 = 1.372\) pu.

  2. P76.2 With zero excitation, what is the maximum reluctance power of the machine of Example 76.1 at \(V = 1\) pu?

    Show answer
    \(P_\text{max} = \tfrac12(1/0.6 - 1) = 0.333\) pu, at \(\delta = 45^\circ\).
Multiple-Choice Questions

  1. MCQ 1. For a salient-pole machine:
    (a) \(X_d = X_q\)   (b) \(X_d > X_q\)   (c) \(X_d < X_q\)   (d) \(X_q = 0\)

    Show answer
    (b) — the d-axis path has the smaller air gap.

  2. MCQ 2. Reluctance power varies as:
    (a) \(\sin\delta\)   (b) \(\cos\delta\)   (c) \(\sin 2\delta\)   (d) \(\cos 2\delta\)

    Show answer
    (c)
Conceptual Questions

  1. Why does a salient-pole machine develop torque with no field current?

  2. Why must the slip be kept very small in the slip test?

  3. Under what condition does the two-reaction model reduce to the synchronous-reactance model?

Looking Ahead

Chapter 77 connects the alternator to a live busbar: the conditions for synchronising, and how load and reactive power are shared between machines in parallel.