The induction motor in depth

Lesson A4 introduced the induction motor. This lesson turns it into numbers you can use to choose one. All the figures below come from the same example motor as Lab I: 230 V per phase, $R_1 = 0.5$ Ω, $R_2' = 0.4$ Ω, $X_1 + X_2' = 2$ Ω, four poles, 50 Hz.

The equivalent circuit

Per phase, the motor behaves like a transformer whose secondary load is the resistance $R_2'/s$ . Neglecting the magnetising branch, the rotor current is

I_2' = \frac{V_1}{\sqrt{(R_1 + R_2'/s)^2 + (X_1 + X_2')^2}}

The resistance $R_2'/s$ is the clever part: it is large when the slip is small, so a lightly loaded motor draws little current, and it falls to $R_2'$ at standstill, so a starting motor draws a lot.

FoundationStart here if this is new to you

You do not have to love circuits to use this. Remember one thing: the motor looks like a resistance that gets smaller as the motor slows down. Smaller resistance, bigger current, bigger torque, until the motor slows too much and the curve turns over.

Where the power goes

The power that crosses the air gap to the rotor, $P_\text{ag} = 3 I_2'^2 R_2'/s$ , splits in a fixed way:

P_\text{cu,rotor} = s\,P_\text{ag}, \qquad P_\text{mech} = (1 - s)\,P_\text{ag}, \qquad T = \frac{P_\text{ag}}{\omega_s}

For the example motor carrying 80 N·m, the slip is 3.56 %. Then $P_\text{ag} = 80 \times 157.1 = 12.57$ kW, of which 447 W heats the rotor bars and 12.12 kW reaches the shaft (less friction and windage). Add the stator copper and iron losses and a motor of this size is typically 88 to 92 % efficient.

The torque-speed curve

Torque follows from the circuit:

T(s) = \frac{3 V_1^2\, (R_2'/s)}{\omega_s\left[(R_1 + R_2'/s)^2 + (X_1 + X_2')^2\right]}

Try it: induction motor against a load

Load torque80 N·m

Supply voltage100 %

Rotor resistance R₂′0.40 Ω

The motor starts and settles where the two curves cross.

Starting torque: 84.0 N·m
Breakdown torque: 197.2 N·m
Breakdown slip: 0.194
Operating speed: 1,447 rpm
Operating slip: 3.56 %

Three points of that curve matter when you choose a motor:

Point	Where	Example motor	Why it matters
Starting (locked-rotor) torque	$s = 1$	84 N·m	must beat the load's breakaway torque
Breakdown (pull-out) torque	$s_\text{bd} = R_2' / \sqrt{R_1^2 + X^2} = 0.194$	197 N·m	the largest load peak it can carry
Rated point	$s \approx 2$ –5 %	rated torque	continuous, thermally allowed

Catalogues give the first two as a percentage of rated torque, for example " $T_\text{start}$ 220 %, $T_\text{bd}$ 300 %".

Predict first

The supply voltage sags to 90 %. By how much does the motor's torque at a given slip fall?

To 90 %To 81 %It does not change

Starting current

At standstill the impedance is at its smallest. The example motor draws

I_\text{start} = \frac{230}{\sqrt{0.9^2 + 2^2}} = 105\ \text{A}

against 19.3 A at its 80 N·m operating point: 5.4 times more. Real motors draw 5 to 8 times their rated current when started direct-on-line. The supply, the cables and the protection must survive it, and it causes a voltage dip for everyone else on the line. Chapter 2 compares the ways of reducing it.

Changing the curve

Voltage scales the whole curve by $V^2$ and leaves the breakdown slip where it is.
Rotor resistance moves the breakdown point to a higher slip without changing its height. A wound-rotor motor uses external resistors to start with high torque and low current, then shorts them.
Rotor bar design does the same in a cage motor: deep or double-cage bars have a high resistance at standstill (current crowds to the surface) and a low one when running. IEC 60034-12 groups cage motors into designs such as N (normal starting torque) and H (high starting torque).
Frequency moves the synchronous speed. With V/f kept constant, the curve slides along the speed axis with almost the same shape: that is how a VFD controls speed (lesson A5, Chapter 2).

ExplorerGo deeper: derivations and open questions

Kloss's formula. Neglect $R_1$ and the curve becomes

\frac{T}{T_\text{max}} = \frac{2}{s/s_\text{bd} + s_\text{bd}/s}

With only two catalogue numbers, the breakdown torque and its slip, you can sketch the whole curve. How far is it from the exact curve of the example motor at $s = 1$ ? Where does the error come from?

Magnetising current. The circuit above neglects the magnetising branch $X_m$ . Put it back and the no-load current is no longer zero: typically 25 to 50 % of rated current, almost all reactive. How does that explain the poor power factor of a lightly loaded motor?