Double-Cage Motors and the Induction Generator

i Learning Objectives

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

Explain the conflict between high starting torque and high running efficiency in a cage rotor.
Describe the double-cage rotor and the deep-bar rotor, and explain how slip frequency shares current between the cages.
Compute the torque contributed by each cage from its impedance.
Explain how an induction machine generates at negative slip, and where its magnetising power comes from.
Calculate the power flow of a grid-connected induction generator.

Section 70-1

The Rotor-Resistance Dilemma

Chapter 62 showed that maximum torque occurs at slip \(s_m = R_2'/X_2'\). A high rotor resistance moves the peak towards standstill and raises starting torque; a low rotor resistance gives small running slip, low rotor copper loss and high efficiency. A slip-ring motor resolves this with external resistance, but a cage rotor has no terminals.

⇅

The key observation

Rotor frequency changes from \(f\) at start to \(sf\) when running

At standstill the rotor current alternates at 50 Hz; at full load only at 1–3 Hz. If the rotor's effective resistance can be made to depend on frequency, it will be high at start and low when running — automatically.

Section 70-2

Construction of the Double-Cage Rotor

Outer cage

Small cross-section bars of high-resistivity material (brass, bronze) near the surface. High resistance, low leakage reactance — little iron surrounds them.

Inner cage

Large copper bars deep in the iron, joined to the outer slot by a narrow slit. Low resistance, high leakage reactance — they are encircled by iron and link much leakage flux.

At starting (\(f_r = f\)): the inner cage's reactance is large, so current is forced into the outer cage. Its high resistance gives high starting torque.
When running (\(f_r = sf\), small): reactances are negligible, current divides according to resistance, and the low-resistance inner cage carries most of it — high efficiency.

Deep-bar rotor. A single tall, narrow bar does the same thing more simply. At high rotor frequency, current crowds to the top of the bar (skin effect) so the effective resistance is high; at slip frequency the current spreads over the full depth. NEMA Design B and C motors use deep bars or double cages.

Section 70-3

Equivalent Circuit and Torque Sharing

Neglecting stator impedance and the mutual leakage between the cages, both cages see the same rotor EMF \(E_2\) and are in parallel. With standstill reactances \(X_o\), \(X_i\), each cage's torque is proportional to its air-gap power:

\[T_o \propto E_2^2\,\frac{R_o/s}{(R_o/s)^2 + X_o^2}, \qquad T_i \propto E_2^2\,\frac{R_i/s}{(R_i/s)^2 + X_i^2}\]

The total torque–slip curve is the sum of two curves: one peaking near standstill (outer cage) and one peaking near synchronous speed (inner cage).

Figure 70.1 — The outer cage supplies starting torque, the inner cage running torque; the sum is high over the whole speed range.

1 Worked Example 70.1 — Torque Sharing Between the Cages

Problem. The standstill impedances per phase of a double-cage rotor are: outer \(1.0 + j1.0~\Omega\), inner \(0.25 + j3.0~\Omega\). Neglecting stator impedance, find the ratio of the torques of the two cages (a) at standstill, (b) at 5 % slip.

(a) Standstill, \(s = 1\).

\[\frac{T_o}{E_2^2} \propto \frac{1.0}{1.0^2 + 1.0^2} = 0.500, \qquad \frac{T_i}{E_2^2} \propto \frac{0.25}{0.25^2 + 3.0^2} = 0.0276\]

\[T_o : T_i = 18.1 : 1\]

(b) Running, \(s = 0.05\). Now \(R_o/s = 20~\Omega\) and \(R_i/s = 5~\Omega\):

\[\frac{20}{20^2 + 1^2} = 0.0499, \qquad \frac{5}{5^2 + 3^2} = 0.147 \;\Rightarrow\; T_i : T_o = 2.95 : 1\]

The roles swap: the outer cage produces almost all the starting torque, the inner cage three-quarters of the running torque.

Section 70-4

The Induction Generator

If a prime mover drives an induction machine connected to a live supply above synchronous speed, the slip is negative. The term \(R_2'/s\) becomes a negative resistance: the rotor branch now delivers real power and the machine generates.

\[s = \frac{N_s - N}{N_s} < 0 \quad\Rightarrow\quad P_\text{ag} = 3I_2'^2\,\frac{R_2'}{s} < 0\]

The machine has no field winding, so it cannot supply its own magnetising reactive power. It takes it from the grid (grid-connected) or from capacitors across its terminals (self-excited, isolated).
It always operates at a leading power factor as seen from the network: it delivers real power but absorbs reactive power.
No synchronising is needed — the frequency is fixed by the grid regardless of shaft speed. This makes it ideal for small hydro and fixed-speed wind turbines.
A self-excited generator needs residual magnetism and enough capacitance; its voltage and frequency both vary with load and speed.

2 Worked Example 70.2 — Grid-Connected Induction Generator

Problem. The 400 V, 4-pole, star-connected motor of Example 65.1 (\(R_1 = 0.5\), \(R_2' = 0.3\), \(X_1 + X_2' = 2.166~\Omega\)) is driven at 1530 rev/min on a 50 Hz grid. Using the approximate circuit and neglecting the magnetising branch, find the current, the mechanical input and the electrical output.

\[s = \frac{1500 - 1530}{1500} = -0.02, \qquad \frac{R_2'}{s} = -15~\Omega\]

\[Z = (0.5 - 15) + j2.166 = 14.66~\Omega\ \text{(magnitude)}, \qquad I = \frac{230.9}{14.66} = 15.75~\text{A}\]

\[P_\text{ag} = 3(15.75)^2(-15) = -11.17~\text{kW}, \qquad P_\text{mech} = P_\text{ag}(1 - s) = -11.39~\text{kW}\]

\[P_\text{elec} = 3I^2R_1 + P_\text{ag} = 0.37 - 11.17 = -10.79~\text{kW}\]

The negative signs mean power flows in the generating direction: 11.39 kW enters at the shaft, 0.22 kW is lost in the rotor, 0.37 kW in the stator, and 10.79 kW reaches the grid. Friction and core losses would reduce this further.

Section 70-5

Summary and Common Mistakes

Key Points

Double-cage and deep-bar rotors exploit the change of rotor frequency from \(f\) to \(sf\) to give high starting torque and low running loss.
Outer cage: high \(R\), low \(X\); inner cage: low \(R\), high \(X\).
Above \(N_s\) the slip is negative and the machine generates, but it always draws magnetising VArs.

Common Mistakes

Swapping the properties of the cages — the outer cage has the high resistance.
Forgetting that rotor reactance is \(sX_2\) at slip \(s\) when working with the rotor at rotor frequency.
Assuming an induction generator can run isolated without capacitors.

Section 70-6

Chapter Review

Practice Problems

P70.1 A 6-pole, 50 Hz induction generator delivers power at a slip of −2.5 %. Find its shaft speed.

Show answer
\(N_s = 1000\) rev/min; \(N = N_s(1 - s) = 1000(1.025) = 1025\) rev/min.
P70.2 For the cages of Example 70.1, find the torque ratio \(T_o : T_i\) at \(s = 0.5\).

Show answer
Outer: \(2/(4 + 1) = 0.400\); inner: \(0.5/(0.25 + 9) = 0.0541\). Ratio \(T_o : T_i = 7.4 : 1\).

Multiple-Choice Questions

MCQ 1. In a double-cage rotor at starting, most current flows in the:
(a) inner cage (b) outer cage (c) end rings only (d) shaft

Show answer
(b) — the inner cage's high leakage reactance blocks current at supply frequency.
MCQ 2. An induction generator connected to the grid:
(a) supplies reactive power (b) absorbs reactive power (c) needs synchronising (d) runs below \(N_s\)

Show answer
(b) — it has no field winding and must be magnetised from the grid.

Conceptual Questions

Why does the inner cage have higher leakage reactance than the outer cage?
Explain the operation of a deep-bar rotor using skin effect.
Why is the induction generator popular for small hydro plants?

Looking Ahead

Part 5 turns to the machine that does carry a field winding and therefore runs at exactly synchronous speed: the synchronous machine. Chapter 71 introduces the alternator, the generator behind almost all of the world's electricity.