Electrical Machines · Chapter 84

Double-Revolving-Field and Cross-Field Theories

Part 6 · Single-Phase and Special Machines — two theories of how a field that only pulsates can still drive a rotor.

Prof. Mithun Mondal Engineering Devotion Digital Textbook
i Learning Objectives

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

  1. Resolve a pulsating field into two counter-rotating fields (double-revolving-field theory).

  2. Compute forward and backward slips and rotor frequencies.

  3. Explain why a single-phase motor has no starting torque yet runs in either direction once started.

  4. Use the forward/backward equivalent circuit to compute current, torque and power.

  5. Describe the cross-field theory as an alternative explanation.

Section 84-1

The Pulsating Field

A single stator winding carrying \(i = I_m\cos\omega t\) produces an air-gap flux that is fixed in space and pulsates in magnitude: \(\phi(\theta, t) = \Phi_m\cos\omega t\cos\theta\). Unlike the three-phase field of Chapter 59, it does not rotate — so how can it turn a rotor?

Section 84-2

Double-Revolving-Field Theory

A trigonometric identity resolves the pulsating field into two fields of equal, half amplitude, rotating in opposite directions at synchronous speed:

⇄
Ferraris' resolution
One pulsating field = two rotating fields
\[\Phi_m\cos\omega t\cos\theta = \underbrace{\tfrac{\Phi_m}{2}\cos(\theta - \omega t)}_{\text{forward}} + \underbrace{\tfrac{\Phi_m}{2}\cos(\theta + \omega t)}_{\text{backward}}\]

Each rotating field acts on the rotor exactly as in a three-phase motor. If the rotor turns forward at slip \(s\):

\[s_f = s, \qquad s_b = \frac{N_s - (-N)}{N_s} = 2 - s\]

  • At standstill \(s_f = s_b = 1\): the two torques are equal and opposite — zero starting torque.

  • Once turning forward, the forward field's torque exceeds the backward field's, and the motor accelerates in whichever direction it was started.

  • The backward field induces rotor currents of frequency \((2 - s)f\), close to \(2f\). These cause extra rotor copper loss, so a single-phase motor is less efficient than a three-phase motor of the same size, and its torque pulsates at \(2f\) (the source of its characteristic hum).

Forward, backward and resultant torque forward backward resultant +Ns −Ns 0
Figure 84.1 — The resultant torque is zero at standstill and has the same shape in both directions: the motor runs whichever way it is started.
Section 84-3

Equivalent Circuit

Each half-field is represented by half the magnetising reactance and half the rotor impedance. Neglecting core loss, the forward and backward impedances are

\[Z_f = \frac{j\frac{X_m}{2}\left(\frac{R_2'}{2s} + j\frac{X_2'}{2}\right)}{\frac{R_2'}{2s} + j\frac{X_2' + X_m}{2}}, \qquad Z_b = \text{same with } s \to 2 - s\]
\[\vec I_1 = \frac{\vec V}{R_1 + jX_1 + Z_f + Z_b}, \qquad P_{\text{ag},f} = I_1^2R_f, \quad P_{\text{ag},b} = I_1^2R_b\]
\[T = \frac{P_{\text{ag},f} - P_{\text{ag},b}}{\omega_s}, \qquad P_\text{mech} = (P_{\text{ag},f} - P_{\text{ag},b})(1 - s)\]

\(R_f\) and \(R_b\) are the real parts of \(Z_f\) and \(Z_b\). Near rated speed \(R_b\) is small, so the backward field absorbs little power.

1 Worked Example 84.1 — Performance from the Equivalent Circuit

Problem. A 230 V, 50 Hz, 4-pole single-phase motor has \(R_1 = 2.0\), \(X_1 = 3.0\), \(R_2' = 4.0\), \(X_2' = 2.0\), \(X_m = 60~\Omega\). At 1440 rev/min find the slips, rotor frequencies, input current, power factor, torque and gross mechanical power (neglect core loss).

\[s = \frac{1500 - 1440}{1500} = 0.04, \quad s_b = 1.96; \qquad f_{rf} = 2~\text{Hz}, \quad f_{rb} = 98~\text{Hz}\]
\[Z_f = 13.00 + j21.94~\Omega, \qquad Z_b = 0.955 + j0.999~\Omega\]
\[Z_\text{in} = 15.96 + j25.94 = 30.45\angle 58.4^\circ~\Omega \;\Rightarrow\; I_1 = 7.55~\text{A}, \quad \text{pf} = 0.524~\text{lag}\]
\[P_{\text{ag},f} = 7.55^2(13.00) = 741.7~\text{W}, \qquad P_{\text{ag},b} = 7.55^2(0.955) = 54.5~\text{W}\]
\[T = \frac{741.7 - 54.5}{157.08} = 4.37~\text{N·m}, \qquad P_\text{mech} = 687.2(0.96) = 659.7~\text{W}\]

The backward field takes about 7 % of the forward air-gap power. Its rotor copper loss is \(54.5(2 - s) \approx 107\) W — heat that a three-phase motor would not produce.

Section 84-4

Cross-Field Theory

The cross-field theory explains the same behaviour without decomposing the field.

  1. At standstill the pulsating stator flux \(\phi_d\) induces transformer EMFs in the rotor. The resulting rotor currents produce MMF on the same axis; the force on the conductors cancels, so there is no torque.

  2. Once the rotor turns, its conductors also cut \(\phi_d\), generating speed (rotational) EMFs. These drive currents whose MMF lies along the axis 90° away — the cross field \(\phi_q\).

  3. Because of the high rotor reactance, \(\phi_q\) lags \(\phi_d\) by nearly 90° in time. Two fluxes displaced 90° in space and in time constitute a rotating field — nearly uniform near synchronous speed.

Two theories, one machine. The double-revolving-field theory is preferred for quantitative work because it leads directly to the equivalent circuit; the cross-field theory gives clearer physical insight into why rotation itself creates the second field.
Section 84-5

Summary and Common Mistakes

Key Points

  • A pulsating field = two half-amplitude fields rotating in opposite directions.

  • \(s_f = s\), \(s_b = 2 - s\); equal and opposite torques at standstill.

  • Torque \(= (P_{\text{ag},f} - P_{\text{ag},b})/\omega_s\); torque pulsates at \(2f\).

  • Cross-field theory: rotation generates a quadrature flux that turns the field into a rotating one.

Common Mistakes

  • Using full \(X_m\) and \(R_2'\) in each half of the circuit instead of half-values.

  • Writing backward slip as \(1 + s\) instead of \(2 - s\).

  • Forgetting to subtract the backward air-gap power when computing torque.

Section 84-6

Chapter Review

Practice Problems

  1. P84.1 A 2-pole, 50 Hz single-phase motor runs at 2850 rev/min. Find \(s_f\), \(s_b\) and both rotor frequencies.

    Show answer
    \(s = 0.05\), \(s_b = 1.95\); \(f_{rf} = 2.5\) Hz, \(f_{rb} = 97.5\) Hz.

  2. P84.2 In a single-phase motor at 1440 rev/min (4-pole, 50 Hz), \(P_{\text{ag},f} = 500\) W and \(P_{\text{ag},b} = 40\) W. Find the torque and the total rotor copper loss.

    Show answer
    \(T = 460/157.08 = 2.93\) N·m. Rotor Cu loss \(= 0.04(500) + 1.96(40) = 20 + 78.4 = 98.4\) W.
Multiple-Choice Questions

  1. MCQ 1. The backward slip of a single-phase motor running at slip \(s\) is:
    (a) \(s\)   (b) \(1 - s\)   (c) \(2 - s\)   (d) \(1 + s\)

    Show answer
    (c)

  2. MCQ 2. Torque in a single-phase induction motor pulsates at:
    (a) \(f\)   (b) \(2f\)   (c) \(sf\)   (d) zero frequency

    Show answer
    (b) — the interaction of the forward and backward fields.
Conceptual Questions

  1. Why is the efficiency of a single-phase motor lower than that of a three-phase motor of equal rating?

  2. Explain, using cross-field theory, why the cross field is nearly 90° behind the main field in time.

  3. Why is the backward-field impedance small near rated speed?

Looking Ahead

Since a single-phase motor has no starting torque, something must give it a push. Chapter 85 shows how an auxiliary winding with a resistor or capacitor creates a starting rotating field — giving the split-phase, capacitor-start and permanent-split-capacitor motors.