Three-Phase Induction Machines

The torque–slip characteristic of a 3-phase induction motor is the curve of developed torque against slip (or speed) at constant supply voltage and frequency.

Torque equation

$$T = \frac{K\, s\, E_2^2\, R_2}{R_2^2 + (s X_2)^2}$$

where $E_2$ and $X_2$ are rotor emf and reactance per phase at standstill, $R_2$ is rotor resistance, $s$ is slip and $K = 3/(2\pi n_s)$.

Shape of the curve

 T
 |          Tmax
 |         .--.
 |       .'    '.
 |      /        '.
 |     /           '-.
 |    /               '-. Tst
 |   /  stable  |  unstable
 |  /           |
 | /            |
 |/_____________|____________ s
 0             sm           1
(N=Ns)                    (N=0)

1. Low-slip region ($s \ll s_m$, normal running): $sX_2 \ll R_2$, so $T \approx K s E_2^2/R_2$, i.e. $T \propto s$. The curve is nearly a straight line through the origin. This is the stable operating region; speed falls only slightly as load rises.
2. Maximum torque: occurs at $s_m = R_2/X_2$, with $T_{max} = K E_2^2/(2X_2)$. It is also called pull-out or breakdown torque.
3. High-slip region ($s > s_m$): $sX_2 \gg R_2$, so $T \approx K E_2^2 R_2/(s X_2^2)$, i.e. $T \propto 1/s$. The curve is a rectangular hyperbola. Operation here is unstable: if load increases, speed falls, torque falls further, and the motor stalls.
4. At $s = 1$ the torque is the starting torque $T_{st} = K E_2^2 R_2/(R_2^2 + X_2^2)$.

Effect of rotor circuit resistance

When $R_2$ is increased, the whole curve keeps the same peak height but the peak moves to the right (towards $s = 1$).

Using $T_{st}/T_{max} = 2a/(1 + a^2)$ with $a = R_2/X_2$, for $X_2 = 1\ \Omega$:

• Maximum torque is unchanged: $T_{max} = KE_2^2/(2X_2)$ does not contain $R_2$.
• Slip at maximum torque increases: $s_m = R_2/X_2$, so higher $R_2$ shifts the peak towards $s = 1$ (lower speed).
• Starting torque increases up to $R_2 = X_2$, when $s_m = 1$ and $T_{st} = T_{max}$. Beyond that, starting torque falls again.
• Starting current decreases and starting power factor improves.
• Full-load slip increases, so running efficiency falls (rotor copper loss $= s P_{ag}$) and speed regulation becomes poorer.

This is why slip-ring motors use external rotor resistance only during starting (or for speed control) and short it out in normal running. Double-cage and deep-bar motors get the same effect automatically.

An induction generator is an ordinary induction machine driven by a prime mover above synchronous speed, so that slip becomes negative and the machine converts mechanical power into electrical power.

Motor to generator action

Slip $s = (N_s - N)/N_s$.

• $N < N_s$: $s$ positive, rotor conductors cut the field in one direction, machine works as a motor.
• $N = N_s$: $s = 0$, no emf in rotor, no torque.
• $N > N_s$: $s$ negative. The rotor now cuts the rotating field in the opposite direction, so rotor emf, rotor current and torque reverse. The torque opposes the prime mover, and the air-gap power $P_{ag} = 3I_2^2R_2/s$ becomes negative: power flows from rotor to stator to the supply. The machine is a generator.

 T
 |   motoring
 |      .--.
 |    .'    '.
 |___/________'.______ N
 |  Ns   (s<0)
 | '.  .'  generating
 |   ''    (T negative)

The induction generator cannot produce its own magnetizing current. It always needs reactive power (lagging VAr) to set up the rotating field. Where this comes from decides the two modes.

Grid-connected mode

 Prime    +--------+     3-phase grid
 mover ===| IG     |=====(fixes V and f,
 N > Ns   +--------+      supplies VAr)

1. The stator is connected to the grid (infinite bus). The grid fixes voltage and frequency and supplies the magnetizing (reactive) current.
2. The machine is first run up near $N_s$ (or started as a motor), then driven above $N_s$ by the turbine.
3. Active power delivered rises with negative slip (typically 1–5% above $N_s$). Frequency stays equal to grid frequency even if turbine speed varies a little, so no synchronizing is needed.
4. Used in wind farms and small hydro plants connected to the grid.

Isolated (self-excited) mode

 Prime    +--------+
 mover ===| IG     |===+=====> Load
 N > Ns   +--------+   |
                     ===== C (delta/star
                           capacitor bank)

1. There is no grid, so a capacitor bank across the stator terminals supplies the lagging VAr (capacitors take leading current = machine's lagging magnetizing current).
2. Voltage build-up: residual magnetism in the rotor gives a small emf when it is driven. This drives a small current through the capacitors, which increases the flux, which raises the emf further. The process continues until the magnetizing curve of the machine meets the capacitor line $V = I_c X_c$ — the operating voltage.
3. Voltage and frequency depend on speed, capacitance and load. Higher speed or larger C raises voltage; a load needing VAr lowers it. Regulation is poor, so voltage/frequency controllers or electronic load controllers are used.
4. Used in micro-hydro and stand-alone wind systems in remote areas.

Given: $V_L = 400$ V (star), 4 poles, 50 Hz, slip-ring voltage at standstill $= 200$ V (star), $Z_1 = 0.8 + j3\ \Omega$, $R_2 = 0.08\ \Omega$, $X_2 = 0.25\ \Omega$, $T_{st} = 10$ N·m.

Method: Refer the stator impedance to the rotor side and use torque ratios (the supply voltage cancels). The 10 N·m starting torque is used as the reference value.

Step 1: Transformation ratio and referred impedance

$$\begin{aligned} K &= \frac{E_2}{E_1} = \frac{200}{400} = 0.5 \\ R_1' &= K^2 R_1 = 0.25 \times 0.8 = 0.2\ \Omega \\ X_1' &= K^2 X_1 = 0.25 \times 3 = 0.75\ \Omega \\ X &= X_1' + X_2 = 0.75 + 0.25 = 1.0\ \Omega \end{aligned}$$

Torque (with $V'$ = stator voltage referred to rotor):

$$T = \frac{3}{\omega_s}\cdot\frac{V'^2 (R_2/s)}{(R_1' + R_2/s)^2 + X^2}$$

$N_s = 120 \times 50/4 = 1500$ rpm.

(i) Maximum torque

Maximum torque occurs when $R_2/s_m = \sqrt{R_1'^2 + X^2}$:

$$\begin{aligned} s_m &= \frac{R_2}{\sqrt{R_1'^2 + X^2}} = \frac{0.08}{\sqrt{0.04 + 1}} = \frac{0.08}{1.0198} = 0.0784 \end{aligned}$$ $$T_{max} = \frac{3}{\omega_s}\cdot\frac{V'^2}{2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)}$$

Ratio with starting torque ($s = 1$):

$$\begin{aligned} \frac{T_{max}}{T_{st}} &= \frac{(R_1' + R_2)^2 + X^2}{2R_2\left(R_1' + \sqrt{R_1'^2+X^2}\right)} \\ &= \frac{(0.28)^2 + 1^2}{2 \times 0.08 \times (0.2 + 1.0198)} \\ &= \frac{1.0784}{0.19517} = 5.525 \end{aligned}$$ $$T_{max} = 5.525 \times 10 = 55.25\ \text{N·m}$$

(It occurs at $N = 1500(1 - 0.0784) \approx 1382$ rpm.)

(ii) Torque at 1430 rpm

$$s = \frac{1500 - 1430}{1500} = 0.04667, \qquad \frac{R_2}{s} = \frac{0.08}{0.04667} = 1.714\ \Omega$$ $$\begin{aligned} \frac{T}{T_{st}} &= \frac{R_2/s}{R_2}\cdot\frac{(R_1'+R_2)^2 + X^2}{(R_1' + R_2/s)^2 + X^2} \\ &= \frac{1}{0.04667}\times\frac{1.0784}{(0.2 + 1.714)^2 + 1} \\ &= 21.43 \times \frac{1.0784}{4.6645} = 4.954 \end{aligned}$$ $$T = 4.954 \times 10 = 49.54\ \text{N·m}$$

Note: if the stator impedance were neglected, $T_{max}/T_{st} = (R_2^2+X_2^2)/(2R_2X_2) = 1.72$ only; including $Z_1$ (as given) changes the result a lot, so it has been included.

Answer: (i) $T_{max} \approx 55.25$ N·m (at slip 0.0784); (ii) torque at 1430 rpm $\approx 49.5$ N·m.

The torque of an induction motor is proportional to the product of the stator flux, the rotor current and the rotor power factor:

$$T \propto \phi\, I_2 \cos\phi_2$$

Derivation of running torque

Let at standstill: rotor emf per phase $= E_2$, rotor resistance $= R_2$, rotor reactance $= X_2$. At slip $s$:

• rotor frequency $f_2 = s f$
• rotor emf $= sE_2$
• rotor reactance $= sX_2$

Rotor current and power factor:

$$I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}, \qquad \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}}$$

Since $E_2 \propto \phi$, write $\phi = E_2/k_1$:

$$\begin{aligned} T &= k\,\phi\, I_2 \cos\phi_2 \\ &= k\,\frac{E_2}{k_1}\cdot\frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}\cdot\frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} \\ &= \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2} \end{aligned}$$

The same result follows from power: air-gap power $P_{ag} = 3I_2^2 R_2/s$ and $T = P_{ag}/\omega_s$, giving $K = 3/\omega_s = 3/(2\pi n_s)$ with $n_s$ in rps.

Torque–slip characteristic

 T
 |        Tmax
 |       .--.
 |     .'    '.
 |    /        '.
 |   /           '-._
 |  /                '-- Tst
 | / T ~ s   |  T ~ 1/s
 |/__________|___________ s
 0          sm          1
 N=Ns                 N=0

• Small $s$: $T \propto s$ (straight line) — stable running region.
• Large $s$: $T \propto 1/s$ (hyperbola) — unstable region.
• At $s = 1$: starting torque.

Condition for maximum torque

$E_2$, $R_2$, $X_2$ are constant, so $T$ is maximum when $\dfrac{dT}{ds} = 0$. Write

$$T = \frac{K E_2^2 R_2}{\dfrac{R_2^2}{s} + sX_2^2}$$

$T$ is maximum when the denominator $y = R_2^2/s + sX_2^2$ is minimum:

$$\begin{aligned} \frac{dy}{ds} &= -\frac{R_2^2}{s^2} + X_2^2 = 0 \\ \Rightarrow\ R_2 &= s X_2 \\ \Rightarrow\ s_m &= \frac{R_2}{X_2} \end{aligned}$$

($d^2y/ds^2 = 2R_2^2/s^3 > 0$, so it is a minimum of $y$, i.e. maximum of $T$.)

Condition: torque is maximum when rotor resistance equals rotor reactance at that slip, $R_2 = s X_2$.

Substituting $s_m$:

$$T_{max} = \frac{K (R_2/X_2) E_2^2 R_2}{R_2^2 + R_2^2} = \frac{K E_2^2}{2X_2}$$

Results:

1. $T_{max}$ does not depend on $R_2$; only the slip at which it occurs does.
2. $T_{max} \propto E_2^2 \propto V^2$ and $\propto 1/X_2$.
3. For maximum starting torque, $s_m = 1$, i.e. $R_2 = X_2$ (done with external rotor resistance in slip-ring motors).

A 3-phase induction motor is an AC motor in which the rotor current is produced by electromagnetic induction from the stator's rotating field; no electrical connection to the rotor is needed.

Construction

1. Stator (stationary part)

• Frame of cast iron or steel for support and protection.
• Core of thin (0.4–0.5 mm) silicon-steel laminations with slots on the inner surface, to reduce eddy-current and hysteresis loss.
• A 3-phase distributed winding in the slots, wound for a definite number of poles, connected in star or delta and fed from the 3-phase supply.

2. Rotor (rotating part) — laminated cylindrical core with slots, mounted on the shaft. Two types:

• Squirrel-cage rotor: copper or aluminium bars in the slots, short-circuited at both ends by end rings. Simple, robust, cheap. Bars are usually skewed to reduce noise and cogging.
• Slip-ring (wound) rotor: 3-phase winding (usually star) with the same number of poles as the stator. The three ends are brought to three slip rings on the shaft; brushes allow external resistance to be added for starting and speed control.

3. Other parts: a small air gap (0.4–2 mm) between stator and rotor, shaft, bearings, end covers and cooling fan.

   +-------------------------+
   |  Stator core + 3-ph wdg |
   |   +-----------------+   |
   |   |  air gap        |   |
   |   |   +---------+   |   |
   |   |   | Rotor   |===|===|== shaft
   |   |   +---------+   |   |
   |   +-----------------+   |
   +-------------------------+

Working principle

1. Rotating magnetic field: a balanced 3-phase supply to the 3-phase stator winding produces a magnetic field of constant magnitude rotating at synchronous speed $N_s = 120f/P$.
2. Induced emf: the field cuts the stationary rotor conductors, inducing emf in them (Faraday's law).
3. Rotor current: the rotor circuit is closed (end rings or external resistors), so current flows.
4. Torque: the current-carrying rotor conductors in the magnetic field experience a force. By Lenz's law, the rotor turns in the same direction as the field to reduce the relative speed.
5. Slip: the rotor can never reach $N_s$; if it did, there would be no relative motion, no emf and no torque. It runs at speed $N$ slightly below $N_s$, with slip $s = (N_s - N)/N_s$, typically 2–5% at full load.

Example: a 4-pole, 50 Hz motor has $N_s = 1500$ rpm; at 4% slip it runs at 1440 rpm and the rotor frequency is $sf = 2$ Hz.

Given: $P = 4$, $f = 50$ Hz, $N = 1440$ rpm, input $P_{in} = 35$ kW, stator losses $= 1$ kW, friction (mechanical) losses $= 1.5$ kW.

Power flow: input → (minus stator losses) → air-gap power → (minus rotor Cu loss) → gross mechanical power → (minus friction) → output.

Synchronous speed and slip

$$\begin{aligned} N_s &= \frac{120 f}{P} = \frac{120 \times 50}{4} = 1500\ \text{rpm} \\ s &= \frac{N_s - N}{N_s} = \frac{1500 - 1440}{1500} = 0.04 \end{aligned}$$

(i) Rotor copper loss

$$\begin{aligned} P_{ag} &= P_{in} - \text{stator losses} = 35 - 1 = 34\ \text{kW} \\ P_{cu2} &= s \, P_{ag} = 0.04 \times 34 = 1.36\ \text{kW} \end{aligned}$$

(ii) Efficiency

$$\begin{aligned} P_{m} &= P_{ag} - P_{cu2} = 34 - 1.36 = 32.64\ \text{kW} \\ P_{out} &= P_m - \text{friction} = 32.64 - 1.5 = 31.14\ \text{kW} \\ \eta &= \frac{P_{out}}{P_{in}} = \frac{31.14}{35} = 0.8897 = 88.97\% \end{aligned}$$

Answer: (i) Rotor copper loss = 1.36 kW; (ii) efficiency = 88.97%.

A starter is a device that limits the starting current of an induction motor and gives overload and no-volt protection.

Why a starter is needed

At standstill, slip $s = 1$, so the rotor behaves like a short-circuited transformer secondary. The rotor emf is maximum and the only opposition is the small leakage impedance:

$$I_{2,st} = \frac{E_2}{\sqrt{R_2^2 + X_2^2}}$$

The stator draws 5–7 times full-load current at rated voltage. This causes:

• a large voltage dip in the supply line, disturbing other loads;
• heating and mechanical stress on the windings;
• tripping of protective devices.

Small motors (below about 5 kW) can be started direct-on-line; larger ones need a starter that reduces the voltage or adds rotor resistance during starting.

Starting methods

1. Direct-on-line (DOL) starter — full voltage applied through a contactor with overload relay. Simple and cheap; starting current 5–7 × FL. Used for small motors.

2. Stator resistance (reactor) starting — resistors in series with each stator phase reduce the voltage to $xV$. Current falls to $x$ times, but torque falls to $x^2$ times. Resistors are shorted after run-up. Wasteful; used rarely.

3. Auto-transformer starting — a 3-phase auto-transformer applies a tapped voltage $xV$. Motor current falls to $x$ times, line current to $x^2$ times, starting torque to $x^2$ times. Better than resistance starting; used for large motors.

4. Star–delta starting — for motors designed to run in delta. At start the winding is connected in star, so each phase gets $V/\sqrt{3}$; after speed builds up it is switched to delta.

$$I_{line,\,Y} = \tfrac{1}{3} I_{line,\,\Delta}, \qquad T_{st,\,Y} = \tfrac{1}{3} T_{st,\,\Delta}$$

Cheap and common for medium squirrel-cage motors.

 Supply ---[TPDT switch]--- Stator
          Start: Y          Run: Delta

5. Rotor resistance starting (slip-ring motors only) — external resistance in the rotor via slip rings reduces starting current and, by increasing $s_m$ towards 1, increases the starting torque. Resistance is cut out in steps as the motor speeds up.

6. Soft starter / VFD — thyristor (or inverter) control raises voltage (or frequency) smoothly. Low current, smooth acceleration; used in modern installations.

Both are 3-phase induction motors with the same stator; they differ only in the rotor. A squirrel-cage rotor has short-circuited bars, while a slip-ring rotor has a 3-phase winding brought out through slip rings.

Basic differences

Characteristics

• Squirrel cage: low rotor resistance gives small full-load slip (2–5%), so it runs at nearly constant speed with good efficiency, but has low starting torque and high starting current. Double-cage and deep-bar designs improve starting torque.
• Slip ring: adding rotor resistance shifts maximum torque towards starting ($s_m = R_2/X_2$), giving high starting torque with low starting current. In running the rings are shorted and it behaves like a cage motor. Speed can be reduced (with loss) by keeping resistance in.

Applications

• Squirrel cage: fans, blowers, pumps, lathes, drilling machines, compressors, conveyors — general constant-speed drives (about 90% of industrial motors).
• Slip ring: cranes, hoists, lifts, elevators, crushers, large compressors, mills — loads needing high starting torque or some speed control.

Given: $V_L = 440$ V (Y), $f = 50$ Hz, $P = 4$, $N = 1425$ rpm, rotor impedance at standstill $= 0.4 + j4\ \Omega$/phase, rotor/stator turns ratio $K = 0.8$. Stator impedance is neglected.

Slip and rotor emf

$$\begin{aligned} N_s &= \frac{120 \times 50}{4} = 1500\ \text{rpm}, \quad s = \frac{1500 - 1425}{1500} = 0.05 \\ E_1 &= \frac{440}{\sqrt{3}} = 254.03\ \text{V/phase} \\ E_2 &= K E_1 = 0.8 \times 254.03 = 203.23\ \text{V/phase (standstill)} \end{aligned}$$

(ii) Rotor current

$$\begin{aligned} I_2 &= \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}} = \frac{0.05 \times 203.23}{\sqrt{0.4^2 + (0.05 \times 4)^2}} \\ &= \frac{10.161}{\sqrt{0.16 + 0.04}} = \frac{10.161}{0.4472} = 22.72\ \text{A} \end{aligned}$$

(iii) Full-load rotor copper loss

$$P_{cu2} = 3 I_2^2 R_2 = 3 \times 22.72^2 \times 0.4 = 619.5\ \text{W}$$

(i) Full-load torque

$$\begin{aligned} P_{ag} &= \frac{P_{cu2}}{s} = \frac{619.5}{0.05} = 12\,390\ \text{W} \\ \omega_s &= \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} \\ T &= \frac{P_{ag}}{\omega_s} = \frac{12\,390}{157.08} = 78.88\ \text{N·m} \end{aligned}$$

Check: $P_m = (1-s)P_{ag} = 11\,771$ W and $T = P_m/\omega = 11\,771/149.23 = 78.88$ N·m. (checks) (This is the gross developed torque; mechanical losses are not given.)

Answer: (i) Full-load torque = 78.88 N·m; (ii) rotor current = 22.72 A; (iii) rotor copper loss = 619.5 W.

An isolated (self-excited) induction generator is a 3-phase induction machine driven above synchronous speed by a prime mover, with a capacitor bank across its stator terminals to supply the reactive power it needs, feeding a local load with no grid.

Operation in isolated mode

 Turbine   +---------+
 (N > Ns)==| Ind.    |===+=========> Local load
           | machine |   |
           +---------+  === C (3-phase bank)

1. Need for capacitors: an induction machine cannot make its own magnetizing current. With no grid, the capacitors supply the leading current that acts as the machine's lagging magnetizing current.
2. Voltage build-up: the rotor has a little residual magnetism. When driven, it induces a small emf in the stator. This sends a small leading current through the capacitors, which strengthens the air-gap flux, which increases the emf. This cumulative process continues until the magnetization curve ($V$ vs $I_m$) meets the capacitor line $V = I_c/(\omega C)$.

 V |          capacitor line
   |        /   V = Ic.Xc
   |      / .----- magnetization
   |    /.'          curve
   |  .'/  <- operating point
   |.' /      (intersection)
   |__/_____________ I

3. Generation: slip is negative ($N > N_s$), so power flows from the shaft to the stator and load.
4. Voltage and frequency depend on speed, capacitance and load. Too small C (line steeper than the curve's air-gap line) means no build-up. A lagging load takes extra VAr and lowers voltage; so an electronic load controller or extra capacitors are used for regulation.

Role in renewable energy

• Wind energy: squirrel-cage (fixed speed) and doubly-fed (slip-ring) induction generators are widely used in wind turbines; they tolerate speed variation and need no synchronizing.
• Micro and small hydro: self-excited induction generators (often motors used as generators) supply villages in remote hilly areas, e.g. many micro-hydro schemes in Nepal.
• Advantages: rugged brushless cage rotor, low cost, little maintenance, easily available as standard motors, inherent short-circuit protection (voltage collapses on fault).
• Limitations: need for reactive power, poor voltage and frequency regulation in isolated mode.

A 3-phase induction motor works on electromagnetic induction: the stator's rotating field induces current in the rotor, and the interaction produces torque.

Working

1. A balanced 3-phase supply to the 3-phase stator winding produces a rotating magnetic field of constant magnitude moving at $N_s = 120f/P$.
2. The field cuts the rotor conductors and induces emf in them.
3. The rotor circuit is closed, so rotor current flows.
4. Current-carrying conductors in the field experience force; by Lenz's law the rotor turns in the direction of the field to reduce relative motion.
5. The rotor settles at speed $N < N_s$; slip $s = (N_s - N)/N_s$. If $N = N_s$, no emf, no torque.

Torque equation

At slip $s$: rotor emf $= sE_2$, rotor reactance $= sX_2$.

$$I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}, \qquad \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}}$$

Torque $T \propto \phi I_2\cos\phi_2$ and $E_2 \propto \phi$:

$$\begin{aligned} T &= k\,\phi\,I_2\cos\phi_2 \\ &= k' E_2 \cdot \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}\cdot\frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} \\ &= \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2} \end{aligned}$$

Using air-gap power, $T = \dfrac{P_{ag}}{\omega_s} = \dfrac{3I_2^2R_2/s}{2\pi N_s/60}$, so $K = \dfrac{3}{2\pi N_s/60}$.

Starting torque ($s = 1$): $T_{st} = \dfrac{K E_2^2 R_2}{R_2^2 + X_2^2}$.

Maximum torque: $s_m = R_2/X_2$, $T_{max} = \dfrac{KE_2^2}{2X_2}$.

Relation between speed and torque

Put $s = (N_s - N)/N_s$:

• Near synchronous speed (small $s$): $sX_2 \ll R_2$, so $T \approx \dfrac{K E_2^2}{R_2}\, s = \dfrac{K E_2^2}{R_2}\cdot\dfrac{N_s - N}{N_s}$. Torque rises linearly as speed falls — stable region.
• Low speeds (large $s$): $T \approx \dfrac{K E_2^2 R_2}{s X_2^2}$, i.e. $T \propto 1/s$. Torque falls as speed falls — unstable region.

 N
 Ns|'-._
   |    '-.  normal running
   |       '.
 Nm|--------: Tmax (pull-out)
   |       .'
   |     .'   unstable
   |   .'
  0|_.'___________ T
     Tst

So the motor runs at almost constant speed from no load to full load, like a DC shunt motor.

Use the exact per-phase equivalent circuit (magnetizing reactance in parallel with the rotor branch; core loss is included in the 405 W constant loss).

 I1   R1=0.29  jX1=0.503       jX2=0.209
 o--[====]---[~~~~~]---+-----[~~~~~]----+
                       |                |
 V1=230.94 V       jXm=13.25      R2/s = 7.0
                       |                |
 o---------------------+----------------+

Given: $V_L = 400$ V (Y), $f = 50$ Hz, $P = 6$, $s = 0.02$.

Speed

$$\begin{aligned} N_s &= \frac{120 \times 50}{6} = 1000\ \text{rpm} \\ N &= (1 - s)N_s = 0.98 \times 1000 = 980\ \text{rpm} \end{aligned}$$

Impedances

$$\begin{aligned} V_1 &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ Z_2 &= \frac{R_2}{s} + jX_2 = \frac{0.14}{0.02} + j0.209 = 7 + j0.209\ \Omega \\ Z_p &= \frac{jX_m\, Z_2}{jX_m + Z_2} = \frac{j13.25 (7 + j0.209)}{7 + j13.459} = 5.340 + j2.983\ \Omega \\ Z_{in} &= (0.29 + j0.503) + Z_p = 5.630 + j3.486 = 6.622\angle 31.77^\circ\ \Omega \end{aligned}$$

Stator current and power factor

$$\begin{aligned} I_1 &= \frac{V_1}{Z_{in}} = \frac{230.94}{6.622} = 34.88\ \text{A}\ (\angle -31.77^\circ) \\ \text{pf} &= \cos 31.77^\circ = 0.850\ \text{lagging} \end{aligned}$$

Powers

Air-gap power (all power into $Z_p$ goes to $R_2/s$):

$$\begin{aligned} P_{ag} &= 3 I_1^2 R_p = 3 \times 34.876^2 \times 5.340 = 19\,485\ \text{W} \\ P_{m} &= (1 - s)P_{ag} = 0.98 \times 19\,485 = 19\,096\ \text{W} \\ P_{out} &= P_m - 405 = 18\,691\ \text{W} \approx 18.69\ \text{kW} \end{aligned}$$

Output torque

$$\begin{aligned} \omega &= \frac{2\pi \times 980}{60} = 102.63\ \text{rad/s} \\ T_{out} &= \frac{P_{out}}{\omega} = \frac{18\,691}{102.63} = 182.1\ \text{N·m} \end{aligned}$$

Efficiency

$$\begin{aligned} P_{in} &= 3V_1 I_1\cos\phi = 3 \times 230.94 \times 34.876 \times 0.8502 = 20\,543\ \text{W} \\ \eta &= \frac{18\,691}{20\,543} = 0.9098 = 90.98\% \end{aligned}$$

Check of losses: stator Cu $= 3 \times 34.876^2 \times 0.29 = 1058$ W; rotor Cu $= sP_{ag} = 390$ W; fixed $= 405$ W; total $= 1853$ W $= 20\,543 - 18\,691$. (checks)

Note: at 400 V the computed output at 2% slip exceeds the 7.5 kW rating; these are the values the given circuit data produce.

Answer: N = 980 rpm, $T_{out}$ = 182.1 N·m, $P_{out}$ = 18.69 kW, $I_1$ = 34.88 A, pf = 0.850 lagging, η = 90.98%.

The T–s characteristic shows how the torque developed by a 3-phase induction motor varies with slip (speed) at constant voltage and frequency. It follows from

$$T = \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2}$$

 T
 |          Tmax
 |         .--.
 |       .'    '.
 |      /  B     '.
 |     /           '-._
 |    / A              '-- Tst (C)
 |   /     |
 |  /stable|  unstable
 | /       |
 |/________|______________ s
 0        sm              1
 (Ns)                    (N=0)

Regions of the curve

1. Low-slip region ($0 < s < s_m$): $sX_2$ is small compared with $R_2$, so

$$T \approx \frac{K E_2^2}{R_2}\, s \quad\Rightarrow\quad T \propto s$$

The curve is almost a straight line through the origin (region OA–B). 2. Maximum torque (B): at $s_m = R_2/X_2$, $T_{max} = KE_2^2/(2X_2)$ — pull-out torque, usually 2–3 times full-load torque. 3. High-slip region ($s_m < s \le 1$): $sX_2 \gg R_2$, so

$$T \approx \frac{K E_2^2 R_2}{s X_2^2} \quad\Rightarrow\quad T \propto \frac{1}{s}$$

The curve is a rectangular hyperbola, ending at starting torque $T_{st}$ at $s = 1$ (C). 4. For $s < 0$ the machine generates; for $s > 1$ it brakes (plugging).

Why the motor operates only in the linear portion

1. Stability: in the linear region, if load torque increases, the speed falls slightly, slip rises and the motor torque rises to meet the new load; a new steady point is reached. So $dT/ds > 0$ gives stable operation. In the region beyond $s_m$, an increase in load reduces speed, slip increases and the motor torque falls; the speed keeps falling until the motor stalls. Operation there is unstable.
2. Near-constant speed: in the linear region slip is only 2–5%, so speed changes very little from no load to full load, which most drives need.
3. Efficiency: rotor copper loss $= s \times P_{ag}$. At low slip this loss is small; at high slip most air-gap power is wasted as heat in the rotor.
4. Current and heating: rotor current rises sharply with slip; at high slip the current is near starting value (5–7 × FL), which would overheat the machine.
5. Power factor: rotor power factor $R_2/\sqrt{R_2^2 + (sX_2)^2}$ is high at low slip and poor at high slip.

Hence the motor is designed so that full-load torque lies on the linear part, well below $T_{max}$, leaving a margin for temporary overloads.

The no-load test gives the shunt (magnetizing) branch; the blocked-rotor test gives the series (leakage) branch. Star connection: phase voltage $= V_L/\sqrt{3}$, phase current $=$ line current.

No-load test (400 V, 10 A, 1400 W)

$$\begin{aligned} V_{ph} &= \frac{400}{\sqrt{3}} = 230.94\ \text{V} \\ \cos\phi_0 &= \frac{P_0}{\sqrt{3}V_L I_0} = \frac{1400}{\sqrt{3}\times 400 \times 10} = 0.2021 \\ \sin\phi_0 &= 0.9794 \\ I_w &= I_0\cos\phi_0 = 10 \times 0.2021 = 2.021\ \text{A} \\ I_\mu &= I_0\sin\phi_0 = 10 \times 0.9794 = 9.794\ \text{A} \\ R_0 &= \frac{V_{ph}}{I_w} = \frac{230.94}{2.021} = 114.29\ \Omega \\ X_0 &= \frac{V_{ph}}{I_\mu} = \frac{230.94}{9.794} = 23.58\ \Omega \end{aligned}$$

($R_0$ represents core plus friction and windage loss; the small stator Cu loss at no load is neglected.)

Blocked-rotor test (200 V, 55 A, 7000 W)

$$\begin{aligned} Z_{eq} &= \frac{200/\sqrt{3}}{55} = \frac{115.47}{55} = 2.0995\ \Omega \\ R_{eq} &= \frac{P_{sc}}{3 I_{sc}^2} = \frac{7000}{3 \times 55^2} = 0.7713\ \Omega \\ X_{eq} &= \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{2.0995^2 - 0.7713^2} = 1.9526\ \Omega \end{aligned}$$

Separating stator and rotor values

$$\begin{aligned} R_1 &= 1.2 \times 0.6 = 0.72\ \Omega \\ R_2' &= R_{eq} - R_1 = 0.7713 - 0.72 = 0.0513\ \Omega \\ X_1 &= X_2' = \frac{X_{eq}}{2} = \frac{1.9526}{2} = 0.976\ \Omega \end{aligned}$$

(Equal split of leakage reactance is the usual assumption.)

Equivalent circuit referred to stator (per phase)

 I1  R1=0.72   jX1=0.976    jX2'=0.976  R2'=0.0513
 o--[====]----[~~~~]---+---+---[~~~~]---[====]--+
                       |   |                    |
 V=230.94 V      R0=114.3  jX0=23.58     R2'(1-s)/s
                       |   |              (load)
 o---------------------+---+--------------------+

Answer: $R_0 = 114.29\ \Omega$, $X_0 = 23.58\ \Omega$, $R_1 = 0.72\ \Omega$, $R_2' = 0.0513\ \Omega$, $X_1 = X_2' = 0.976\ \Omega$ per phase.

The developed torque of a 3-phase induction motor at slip $s$ is

$$T = \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2}, \qquad K = \frac{3}{2\pi n_s}$$

where $E_2$, $X_2$ are rotor emf and reactance per phase at standstill and $R_2$ is rotor resistance per phase.

Condition for maximum torque

For a given motor and supply, $E_2$, $R_2$ and $X_2$ are constant; only $s$ varies. Divide numerator and denominator by $s$:

$$T = \frac{K E_2^2 R_2}{\dfrac{R_2^2}{s} + s X_2^2}$$

$T$ is maximum when the denominator $y = \dfrac{R_2^2}{s} + sX_2^2$ is minimum:

$$\begin{aligned} \frac{dy}{ds} &= -\frac{R_2^2}{s^2} + X_2^2 = 0 \\ s^2 X_2^2 &= R_2^2 \\ s_m &= \frac{R_2}{X_2} \end{aligned}$$

Since $\dfrac{d^2y}{ds^2} = \dfrac{2R_2^2}{s^3} > 0$, $y$ is minimum and $T$ is maximum.

Condition: torque is maximum when $R_2 = s X_2$, i.e. rotor resistance equals rotor reactance under running conditions.

Maximum torque:

$$\begin{aligned} T_{max} &= \frac{K (R_2/X_2) E_2^2 R_2}{R_2^2 + R_2^2} = \frac{K E_2^2}{2 X_2} \end{aligned}$$

Conclusions:

1. $T_{max}$ is independent of $R_2$; $R_2$ only changes the slip at which it occurs.
2. $T_{max} \propto 1/X_2$.
3. $T_{max} \propto E_2^2 \propto V^2$.
4. For maximum starting torque, $R_2 = X_2$ ($s_m = 1$).

(With stator impedance included: $s_m = R_2'/\sqrt{R_1^2 + (X_1 + X_2')^2}$.)

T–s characteristics for different supply voltages

Since $E_2 \propto V$, at any slip $T \propto V^2$, while $s_m = R_2/X_2$ does not depend on $V$.

 T            V1 > V2 > V3
 |        .--.   V1
 |      .'    '.
 |     /  .-.   '.
 |    / .'   '.   '-.   V2
 |   / /  .-.  '-.   '-.
 |  / / .'   '-.  '-.   '-
 | / / /        '-. '-.  V3
 |/_/_/____:_________'-.-'--- s
 0         sm               1
           (same for all V)

Example: a 10% fall in voltage reduces $T_{max}$ and $T_{st}$ to $0.9^2 = 0.81$ of their values, i.e. by 19%. This is why induction motors are sensitive to supply voltage dips, and why reduced-voltage starters (star–delta, auto-transformer) reduce starting torque.

With negligible stator impedance, torque depends on $R_2/s$:

$$T = \frac{K s E_2^2 R_2}{R_2^2 + (sX_2)^2} = \frac{K E_2^2 (R_2/s)}{(R_2/s)^2 + X_2^2}$$

So for the same torque, $R_2/s$ stays constant, i.e. $s \propto R_2$. Also $s_m = R_2/X_2 \propto R_2$.

Given: rated output 40 kW at $N = 0.96 N_s$, so $s_1 = 0.04$. $s_{m1} = 4 \times 0.04 = 0.16$. New rotor resistance $R_2' = 5R_2$.

Assumption: 40 kW is the mechanical power developed at rated torque (mechanical losses neglected).

Original condition

$$\begin{aligned} P_{ag} &= \frac{P_m}{1 - s_1} = \frac{40}{0.96} = 41.667\ \text{kW} \\ P_{cu2} &= s_1 P_{ag} = 0.04 \times 41.667 = 1.667\ \text{kW} \end{aligned}$$

(i) Rated torque with $5R_2$

$$\begin{aligned} s_2 &= 5 s_1 = 5 \times 0.04 = 0.20 \\ N_2 &= (1 - 0.20) N_s = 0.80\, N_s \end{aligned}$$

Torque and synchronous speed are unchanged, so air-gap power $P_{ag} = T\omega_s$ is unchanged $= 41.667$ kW.

$$\begin{aligned} P_{out} &= (1 - s_2) P_{ag} = 0.8 \times 41.667 = 33.33\ \text{kW} \\ P_{cu2} &= s_2 P_{ag} = 0.2 \times 41.667 = 8.333\ \text{kW} \end{aligned}$$

(ii) Speed at maximum torque

$$\begin{aligned} s_{m2} &= 5 s_{m1} = 5 \times 0.16 = 0.80 \\ N_{m2} &= (1 - 0.80) N_s = 0.20\, N_s \end{aligned}$$

(The maximum torque value itself is unchanged.)

Answer: (i) speed $= 0.8N_s$, output $= 33.33$ kW, rotor Cu loss $= 8.333$ kW; (ii) speed at maximum torque $= 0.2N_s$ (slip 0.8). For example, for a 4-pole 50 Hz machine ($N_s = 1500$ rpm) these speeds would be 1200 rpm and 300 rpm.

Given: $V_L = 400$ V, stator in delta (so stator phase voltage $= 400$ V), rotor in star with 190 V between slip rings at standstill, $Z_1 = 0.5 + j2.5\ \Omega$, $R_2 = 0.06\ \Omega$, $X_2 = 0.3\ \Omega$, $T_{max} = 150$ N·m, 4 poles, 50 Hz, full-load slip $= 0.04$.

Step 1: Turns ratio and stator impedance referred to rotor

$$\begin{aligned} E_{2,ph} &= \frac{190}{\sqrt{3}} = 109.70\ \text{V}, \qquad E_{1,ph} = 400\ \text{V} \\ K &= \frac{109.70}{400} = 0.27424, \qquad K^2 = 0.075208 \\ R_1' &= K^2 R_1 = 0.075208 \times 0.5 = 0.0376\ \Omega \\ X_1' &= K^2 X_1 = 0.075208 \times 2.5 = 0.1880\ \Omega \\ X &= X_1' + X_2 = 0.1880 + 0.3 = 0.4880\ \Omega \end{aligned}$$

(i) Slip at maximum torque

$$\begin{aligned} s_m &= \frac{R_2}{\sqrt{R_1'^2 + X^2}} = \frac{0.06}{\sqrt{0.0376^2 + 0.4880^2}} = \frac{0.06}{0.48947} = 0.1226 \end{aligned}$$

(Speed at maximum torque $= 1500(1 - 0.1226) \approx 1316$ rpm.)

(ii) Full-load torque and power

$$T = \frac{3}{\omega_s}\frac{V'^2 (R_2/s)}{(R_1' + R_2/s)^2 + X^2}, \qquad T_{max} = \frac{3}{\omega_s}\frac{V'^2}{2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)}$$

At $s = 0.04$: $R_2/s = 0.06/0.04 = 1.5\ \Omega$.

$$\begin{aligned} \frac{T_{fl}}{T_{max}} &= \frac{(R_2/s)\cdot 2\left(R_1' + \sqrt{R_1'^2 + X^2}\right)}{(R_1' + R_2/s)^2 + X^2} \\ &= \frac{1.5 \times 2(0.0376 + 0.48947)}{(1.5376)^2 + 0.4880^2} \\ &= \frac{1.5 \times 1.05414}{2.6024} = 0.6076 \end{aligned}$$ $$T_{fl} = 0.6076 \times 150 = 91.14\ \text{N·m}$$

Full-load speed $N = (1 - 0.04) \times 1500 = 1440$ rpm:

$$\begin{aligned} \omega &= \frac{2\pi \times 1440}{60} = 150.80\ \text{rad/s} \\ P_{out} &= T_{fl}\,\omega = 91.14 \times 150.80 = 13\,744\ \text{W} \approx 13.74\ \text{kW} \end{aligned}$$

(This is the gross mechanical power developed; mechanical losses are not given.)

Answer: (i) $s_m = 0.1226$; (ii) full-load torque $= 91.14$ N·m, power output $\approx 13.74$ kW.