Three Phase Induction Machines

Given: 8 poles, 50 Hz, $T_{st} = 50$ kg-m, $Z_2 = 0.8 + j4\ \Omega$ per phase, so $R_2 = 0.8\ \Omega$, $X_2 = 4\ \Omega$.

$$N_s = \frac{120\times 50}{8} = 750\ \text{rpm}$$

Slip at maximum torque:

$$s_m = \frac{R_2}{X_2} = \frac{0.8}{4} = 0.2$$

Speed at maximum torque:

$$N_m = N_s(1 - s_m) = 750\times 0.8 = 600\ \text{rpm}$$

Maximum torque: using $\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1+s_m^2}$ (stator impedance neglected)

$$\frac{T_{st}}{T_{max}} = \frac{2\times 0.2}{1 + 0.2^2} = \frac{0.4}{1.04} = 0.3846$$ $$T_{max} = \frac{50}{0.3846} = 130\ \text{kg-m} \quad (= 130\times 9.81 = 1275.3\ \text{N-m})$$

Answer: the motor develops maximum torque at 600 rpm, and $T_{max} = 130$ kg-m ($\approx 1275$ N-m).

Given: 8 poles, 50 Hz, $T_{st} = 50$ N-m, rotor impedance $0.8 + j4\ \Omega$ per phase at standstill, so $R_2 = 0.8\ \Omega$, $X_2 = 4\ \Omega$.

Synchronous speed

$$N_s = \frac{120 f}{P} = \frac{120\times 50}{8} = 750\ \text{rpm}$$

Slip and speed at maximum torque

$$s_m = \frac{R_2}{X_2} = \frac{0.8}{4} = 0.2$$ $$N_m = N_s(1-s_m) = 750\times(1 - 0.2) = 600\ \text{rpm}$$

Maximum torque

The torque at slip $s$ compared to maximum torque is $\frac{T}{T_{max}} = \frac{2 s s_m}{s^2 + s_m^2}$. At starting $s = 1$:

$$\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1 + s_m^2} = \frac{2\times 0.2}{1 + 0.2^2} = 0.3846$$ $$T_{max} = \frac{T_{st}}{0.3846} = \frac{50}{0.3846} = 130.0\ \text{N-m}$$

Answer: maximum torque occurs at 600 rpm (slip 0.2), and $T_{max} = 130.0$ N-m.

Given: 8 poles, 50 Hz, $T_{st} = 50$ N-m, rotor impedance $0.8 + j2\ \Omega$ per phase at standstill, so $R_2 = 0.8\ \Omega$, $X_2 = 2\ \Omega$.

Synchronous speed

$$N_s = \frac{120 f}{P} = \frac{120\times 50}{8} = 750\ \text{rpm}$$

Slip and speed at maximum torque

$$s_m = \frac{R_2}{X_2} = \frac{0.8}{2} = 0.4$$ $$N_m = N_s(1-s_m) = 750\times(1 - 0.4) = 450\ \text{rpm}$$

Maximum torque

The torque at slip $s$ compared to maximum torque is $\frac{T}{T_{max}} = \frac{2 s s_m}{s^2 + s_m^2}$. At starting $s = 1$:

$$\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1 + s_m^2} = \frac{2\times 0.4}{1 + 0.4^2} = 0.6897$$ $$T_{max} = \frac{T_{st}}{0.6897} = \frac{50}{0.6897} = 72.5\ \text{N-m}$$

Answer: maximum torque occurs at 450 rpm (slip 0.4), and $T_{max} = 72.5$ N-m.

Given: 4 poles, 50 Hz, $T_{st} = 50$ N-m, rotor impedance $0.8 + j2\ \Omega$ per phase at standstill, so $R_2 = 0.8\ \Omega$, $X_2 = 2\ \Omega$.

Synchronous speed

$$N_s = \frac{120 f}{P} = \frac{120\times 50}{4} = 1500\ \text{rpm}$$

Slip and speed at maximum torque

$$s_m = \frac{R_2}{X_2} = \frac{0.8}{2} = 0.4$$ $$N_m = N_s(1-s_m) = 1500\times(1 - 0.4) = 900\ \text{rpm}$$

Maximum torque

The torque at slip $s$ compared to maximum torque is $\frac{T}{T_{max}} = \frac{2 s s_m}{s^2 + s_m^2}$. At starting $s = 1$:

$$\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1 + s_m^2} = \frac{2\times 0.4}{1 + 0.4^2} = 0.6897$$ $$T_{max} = \frac{T_{st}}{0.6897} = \frac{50}{0.6897} = 72.5\ \text{N-m}$$

Answer: maximum torque occurs at 900 rpm (slip 0.4), and $T_{max} = 72.5$ N-m.

The torque-slip characteristic shows how the torque developed by a 3-phase induction motor varies with slip (from $s = 1$ at standstill to $s = 0$ at synchronous speed), at rated voltage and frequency.

Torque of a 3-phase induction motor at slip $s$ (per-phase equivalent):

$$T = \frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2 + (sX_2)^2}\ \text{N·m}$$

where $E_2$ and $X_2$ are the rotor emf and reactance per phase at standstill and $\omega_s$ is the synchronous speed in rad/s.

 Torque T
   |          Tmax
   |         .--*--.
   |  Tst   /   |   \          Operating
   |    *--'    |    \          region
   |            |     \___ ___ (stable)
   |            |     
   +------------+-----------+----> slip s
   1 (standstill) sm       0 (sync)

Regions of the curve

1. Starting point ($s = 1$): the torque is the starting torque $T_{st}$. Here the rotor reactance $sX_2$ is large, so rotor power factor is low and the torque is less than the maximum.
2. Low-slip (operating) region ($0 < s < s_m$): the rotor reactance is small, so $sX_2\ll R_2$ and $T\approx \dfrac{3sE_2^2}{\omega_sR_2}\propto s$. The curve is almost a straight line. The motor works stably here: if the load rises, the speed falls slightly, slip increases and torque increases to meet it. The full-load slip is only about 2–5%.
3. Maximum torque point ($s = s_m$): the pull-out or breakdown torque $T_{max}$ occurs at $s_m = R_2/X_2$ and has the value

$$T_{max} = \frac{3E_2^2}{2\omega_sX_2}$$

4. High-slip region ($s_m < s < 1$): torque decreases as slip increases because $X_2$ dominates ($T\propto 1/s$). This region is unstable for normal loads; if the load torque exceeds $T_{max}$, the motor stalls.

Effect of rotor resistance

$T_{max}$ does not depend on $R_2$, but the slip at which it occurs does: $s_m = R_2/X_2$.

 T
 |      R2 low    R2 med    R2 high
 |      .-*-.     .-*-.      .-*-.
 |     /     \   /     \    /     \
 |    /       \ /       \  /       \
 +---+--------+---------+--+--------+--> s

• Increasing $R_2$ (wound-rotor motor with external resistance) moves the peak toward higher slip (toward $s = 1$). When $R_2 = X_2$, $s_m = 1$ and maximum torque occurs at starting.
• The starting torque increases and the starting current falls (better rotor power factor).
• The slope in the operating region becomes less steep, so the speed regulation worsens (more slip for the same torque), and the running efficiency falls due to $I_2^2R_2$ loss.
• Hence external rotor resistance is used for starting and speed control in slip-ring motors, and then cut out for running.

Torque-slip characteristics

The torque-slip curve shows the variation of the torque of a 3-phase induction motor with slip, at constant supply voltage and frequency.

Torque of a 3-phase induction motor at slip $s$ (per-phase equivalent):

$$T = \frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2 + (sX_2)^2}\ \text{N·m}$$

where $E_2$ and $X_2$ are the rotor emf and reactance per phase at standstill and $\omega_s$ is the synchronous speed in rad/s.

 Torque T
   |          Tmax
   |         .--*--.
   |  Tst   /   |   \          Operating
   |    *--'    |    \          region
   |            |     \___ ___ (stable)
   |            |     
   +------------+-----------+----> slip s
   1 (standstill) sm       0 (sync)

• At standstill ($s = 1$): starting torque $T_{st}$, with high rotor current but a poor power factor.
• At small slip: $sX_2\ll R_2$, so $T\propto s$. The curve is nearly a straight line, and this is the stable operating region (full-load slip 2–5%). Load increases give a small fall in speed.
• At $s = s_m$: the torque reaches its maximum value $T_{max}$.
• At large slip: $T\propto 1/s$, so torque decreases as $s$ rises. This part is unstable for normal loads.
• At synchronous speed ($s = 0$): torque is zero.

Condition for maximum torque

Let $K = \dfrac{3E_2^2}{\omega_s}$. Then

$$T = K\,\frac{sR_2}{R_2^2 + s^2X_2^2}$$

For the maximum value, $\dfrac{dT}{ds} = 0$:

$$\frac{dT}{ds} = K\,\frac{(R_2^2 + s^2X_2^2)R_2 - sR_2(2sX_2^2)}{(R_2^2 + s^2X_2^2)^2} = 0$$ $$R_2^2 + s^2X_2^2 - 2s^2X_2^2 = 0 \;\Rightarrow\; R_2^2 = s^2X_2^2$$ $$s_m = \frac{R_2}{X_2}$$

So the torque is maximum when the rotor resistance equals the rotor reactance at slip $s_m$, that is $R_2 = s_mX_2$ (the rotor power factor is then $0.707$ lagging).

Substituting $s = s_m$:

$$T_{max} = K\,\frac{s_mR_2}{2R_2^2} = \frac{3E_2^2}{2\omega_sX_2}$$

$T_{max}$ is independent of the rotor resistance, is proportional to $V^2$ and inversely proportional to $X_2$.

Also: $\dfrac{T}{T_{max}} = \dfrac{2ss_m}{s^2 + s_m^2}$.

The core loss (hysteresis plus eddy current) in a magnetic material depends on the frequency of the alternating flux in it:

$$P_h\propto f\,B_m^{1.6}, \qquad P_e\propto f^2B_m^2$$

Reason:

1. In an induction motor the stator field rotates at synchronous speed, and the rotor turns at nearly the same speed. The flux therefore moves relative to the rotor at the slip speed only, so the rotor core frequency is $f_r = sf$.
2. At normal running the slip is only 2–5%. For a 50 Hz supply, $f_r = 0.03\times 50\approx 1$–$2.5$ Hz.
3. The core loss is proportional to $f$ (hysteresis) and $f^2$ (eddy current), so at 1–2.5 Hz the rotor core loss is a tiny fraction of the stator core loss, which occurs at the full supply frequency of 50 Hz.

Example: at $s = 0.04$, $f_r = 2$ Hz. The rotor eddy-current loss is $(2/50)^2 = 0.0016$ of the loss at 50 Hz, i.e. 0.16%.

Hence in the equivalent circuit the core-loss branch is shown only on the stator side, and the rotor iron loss is neglected. The rotor copper loss $I_2^2R_2$ is the only significant rotor loss. (At starting, $s = 1$ and the rotor frequency is 50 Hz, but the rotor core loss is not important at that moment since the starting period is short.)

Synchronous speed:

$$N_s = \frac{120f}{P} = \frac{120\times 50}{4} = 1500\ \text{rpm}$$

Slip

$$s = \frac{N_s - N}{N_s} = \frac{1500 - 1440}{1500} = 0.04 \ (4\%)$$

Frequency of rotor emf

$$f_r = sf = 0.04\times 50 = 2\ \text{Hz}$$

Rotor induced emf per phase at 1440 rpm

The rotor emf per phase is proportional to slip: $E_{2r} = sE_{2}$, where $E_2 = 130$ V is the standstill value.

$$E_{2r} = 0.04\times 130 = 5.2\ \text{V}$$

Stator to rotor turns ratio

The stator is delta connected, so the stator phase voltage equals the line voltage, 440 V. At standstill the induction motor behaves like a transformer, so

$$\frac{N_1}{N_2} = \frac{E_1}{E_2} = \frac{440}{130} = 3.385$$

The turns ratio is independent of speed, so it is the same at 1440 rpm.

Answer: slip $= 0.04$ ($4\%$); rotor frequency $= 2\ \text{Hz}$; rotor emf per phase $= 5.2\ \text{V}$; turns ratio $N_1/N_2 = 3.385$ (stator phase turns : rotor phase turns, about 3.38:1).

An induction machine works as an induction generator when its rotor is driven by a prime mover above synchronous speed, so the slip becomes negative and the torque reverses. It then delivers active power to the supply.

Conditions for generator operation

1. Speed above synchronous speed: $N_r > N_s = \frac{120f}{P}$, so slip $s = \frac{N_s - N_r}{N_s} < 0$.
2. Source of reactive power: the machine has no separate field winding. It must take its magnetising current from the grid (grid-connected), or from a capacitor bank across the stator terminals (self-excited, stand-alone).
3. Frequency and voltage fixed by the grid: when grid-connected, the grid keeps the frequency and terminal voltage. For stand-alone use, residual magnetism and sufficient capacitance are needed for voltage build-up.
4. Prime mover with enough power: a turbine or engine must drive the rotor so that the mechanical input exceeds losses. The speed is only 2 to 5 % above $N_s$ at rated load.
5. Capacitor value must exceed a critical value so that the magnetising line is crossed (self-excited case).

Explanation with T-S curve

The torque-slip curve extends on both sides of $s = 0$.

• $0 < s < 1$: motoring region (torque positive, power flows from the supply to the shaft).
• $s < 0$: generating region. The curve is an inverted image of the motoring part, with torque negative. The maximum (pull-out) generating torque occurs at $s \approx -s_m$.
• $s > 1$: braking (plugging) region, rotor runs opposite to the field.

 T
 |        motoring
 |         _.-*-._
 |      _-'       `-._
 +-----+-------+-----------> s
 s=1       s=0   (N=Ns)
 ...  _.-'  generating  (s<0)
 |  *-'
 |

Beyond $N_s$ the slip is negative, the rotor emf and current reverse, and the electromagnetic torque opposes the prime mover. The air-gap power $P_g = \frac{P_{conv}}{1-s}$ becomes negative, meaning power is fed to the stator and then to the grid. If the driving torque exceeds the maximum generating torque, the machine runs away (loses stability), so the prime mover torque must stay below the pull-out value.

Induction motor as induction generator

If a three-phase induction motor connected to the supply is driven by a prime mover above synchronous speed, the slip becomes negative ($s<0$). The rotor conductors now cut the rotating field faster than the field itself rotates, so the rotor emf and current reverse. The torque opposes the driving torque, and the machine converts mechanical power into electrical power, which it delivers to the supply. It still draws lagging reactive power from the supply (or from capacitors) for magnetisation.

For an isolated (stand-alone) system the reactive power is supplied by capacitors connected across the stator terminals. This is the self-excited induction generator.

Procedure to find the excitation capacitor

1. Run the machine as a motor at the supply frequency and no load. Measure terminal (phase) voltage $V$ against magnetising current $I_m$ to get the no-load magnetisation curve (V against $I_m$). It is linear at first and then saturates.
2. A capacitor of capacitance $C$ per phase draws a current $I_c = \frac{V}{X_c}$, with $X_c = \frac{1}{2\pi f C}$. In the V-I plane it is a straight line through the origin with slope $X_c$.
3. For stable voltage, the capacitor line must cut the magnetisation curve. The intersection point gives the generated voltage.
4. For a required voltage $V_0$: read $I_m$ at $V_0$ from the curve. Then

$$X_c = \frac{V_0}{I_m}, \qquad C = \frac{I_m}{2\pi f V_0}$$

5. Minimum (critical) capacitance: the line tangent to the linear part of the curve has slope $X_m$ (unsaturated magnetising reactance), so

$$C_{min} = \frac{1}{\omega X_m} = \frac{1}{\omega^2 L_m}, \quad \omega = 2\pi f$$

If $C < C_{min}$, the line does not meet the curve and the voltage does not build up.

 V
 |            /  Xc line (larger C: flatter)
 |        _.-*---- saturation curve
 |     .-'  /
 |   .'   /
 |  / /
 +------------------> Im
   V0 found at crossing

The generated frequency depends on speed: $f \approx \frac{P N_r}{120}$ less the small slip, so frequency is a little different from that used in the test. Use the frequency of operation in $X_c$.

Voltage build-up in a self-excited induction generator is a regenerative process that starts from the residual magnetism of the rotor iron and is supported by a capacitor bank across the stator terminals.

Process

1. The rotor is driven above synchronous speed by a prime mover. The small residual flux cuts the stator conductors and induces a small emf $E_r$ (a few volts) at the terminals.
2. This emf drives a leading current $I_c = \frac{E_r}{X_c}$ through the capacitors.
3. The capacitor current flows in the stator winding as a magnetising current and increases the air-gap flux in the same direction as the residual flux.
4. The increased flux raises the terminal emf, which raises the capacitor current, which increases the flux further. The voltage thus builds up in steps.
5. The build-up stops at the point where the capacitor line ($V = I_c X_c$) intersects the magnetisation (saturation) curve. Beyond this point, saturation of the iron stops the rise of emf. This is the stable no-load voltage.

 V
 |           /  capacitor line
 |       _.-*   <- stable point
 |    _-' /
 |  .'  /
 | / ->/
 |/ /  Er
 +------------------> Im

Conditions for build-up

• Residual magnetism must exist in the rotor iron (if lost, flash the field with a battery).
• Capacitance must be greater than the critical value, so the capacitor line is flatter than the linear part of the magnetisation curve.
• Speed must be high enough, because the emf is proportional to speed.
• Load must not be too heavy, otherwise the voltage collapses.

With a higher capacitance the line is flatter and the final voltage is higher. A larger load needs a larger capacitance to hold the voltage.

Production of rotating magnetic field

A three-phase stator winding has three identical coils a, b, c placed $120^\circ$ apart in space. A balanced three-phase supply makes their fluxes alternate with a $120^\circ$ time displacement:

$$\phi_a = \phi_m \sin\omega t, \quad \phi_b = \phi_m \sin(\omega t - 120^\circ), \quad \phi_c = \phi_m \sin(\omega t - 240^\circ)$$

Each flux acts along the axis of its own coil. The resultant flux is the vector sum at each instant.

 Axes of coils:         Resultant at wt = 0, 60, 120:
        a                      R(0)  R(60)
        |                       \    /
        |                        \  /   -> rotates
       / \                        \/        clockwise
      /   \                   ----+----
     c     b                       |
 (120 deg apart)                 R(120)

The resultant has a constant magnitude of $1.5\phi_m$ and rotates by $60^\circ$ electrical for every $60^\circ$ of time, so it makes one revolution per cycle for a 2-pole winding. For $P$ poles, its speed is

$$N_s = \frac{120 f}{P}\ \text{rpm}$$

The direction of rotation follows the phase sequence a-b-c; swapping any two supply lines reverses it.

How the RMF makes the rotor rotate

1. The RMF sweeps across the stationary rotor conductors and, by Faraday's law, induces an emf in them.
2. The rotor bars are short-circuited, so a rotor current flows.
3. A current-carrying conductor in a magnetic field experiences a force ($F = BIl$). The forces on all conductors give a torque.
4. By Lenz's law, the rotor tries to reduce the relative speed between itself and the field, so it rotates in the direction of the RMF.
5. The rotor speed $N_r$ stays below $N_s$. If $N_r = N_s$, there would be no relative motion, no emf and no torque. The difference gives slip $s = \frac{N_s - N_r}{N_s}$.

Rotating magnetic field (RMF)

A rotating magnetic field is a magnetic field of constant magnitude ($1.5\phi_m$ for a three-phase winding) whose axis rotates in space at synchronous speed $N_s = \frac{120f}{P}$ rpm. It is produced by feeding balanced three-phase currents to three windings displaced $120^\circ$ in space.

Torque-speed (T-N) characteristics

The torque developed is

$$T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2}$$

 T
 |            Tmax
 |          _.-*-._
 |   Tst  /         \
 |    *--'            \
 |   /                  \
 |  /                     \
 +--+------+----------------+--> N
  N=0   N(Tmax)             Ns

Main points of the curve:

1. Starting (N = 0, s = 1): the torque is the starting torque $T_{st}$. For a squirrel-cage motor it is about 1.5 to 2 times full-load torque.
2. Between standstill and pull-out: the torque rises with speed, because the rotor reactance $sX_2$ falls and the power factor improves, until it reaches the maximum (breakdown) torque at $s_m = \frac{R_2}{X_2}$, where $T_{max} = \frac{3E_2^2}{2\omega_s X_2}$.
3. Stable region ($s < s_m$): from $T_{max}$ down to $N_s$ the torque falls almost linearly with speed. The motor works here at rated load with slip of 2 to 5 %.
4. At $N = N_s$ ($s=0$): the torque is zero.
5. Unstable region ($s > s_m$): if the load torque exceeds $T_{max}$, the speed falls and the torque falls too, so the motor stalls.

Increasing the rotor resistance moves $s_m$ toward standstill (the same $T_{max}$), and with $R_2 = X_2$ the maximum torque occurs at starting. A fall of supply voltage reduces torque in proportion to $V^2$.

The torque-slip characteristic shows how the electromagnetic torque of a three-phase induction motor varies with slip, from $s=1$ (standstill) to $s=0$ (synchronous speed).

$$T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2}$$

 T
 |          Tmax
 |        _.-*-._
 |  Tst _-'      `-._
 |   *-'              `-.   <- rated
 |  /                      `*  load T_fl
 +--+--------+-------------+--> s
  s=1       s_m           s=0
  (unstable)  (stable region)

(slip decreases to the right, so speed increases to the right)

Regions of the curve

1. Starting torque ($s = 1$):

$$T_{st} = \frac{3}{\omega_s}\cdot\frac{E_2^2 R_2}{R_2^2 + X_2^2}$$

It is the torque at standstill and must exceed the load torque for the motor to start. 2. Maximum torque ($s = s_m = R_2/X_2$):

$$T_{max} = \frac{3E_2^2}{2\omega_s X_2}$$

It is independent of $R_2$ and is also called pull-out or breakdown torque, typically 2 to 3 times full-load torque. 3. Normal operating region ($0 < s < s_m$): here $sX_2 \ll R_2$, so $T \approx \frac{3E_2^2}{\omega_s R_2}\,s$, which is a straight line. The torque is proportional to slip. A small drop in speed gives a large rise in torque, so the speed stays nearly constant (shunt-like behaviour). This region is stable and the motor runs here with slip of about 1 to 5 %. 4. Unstable region ($s_m < s < 1$): the torque falls as slip increases (reactance dominates, $T \propto 1/s$). If the load torque is more than the electromagnetic torque, the motor slows further and stalls.

For stability, the load torque line must cut the motor curve in the normal region.

Torque-slip characteristics

The torque of a three-phase induction motor is

$$T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2}$$

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

 T
 |    R2 small     R2 larger
 |      Tmax ______ Tmax
 |       .-*-.   .-*-.
 |      /     \ /     \
 |  Tst1       Tst2 (higher)
 +--+---------+--------------> s
   s=1      sm1 sm2       0

The curve rises from $T_{st}$ at $s=1$ to a maximum and then falls to zero at $s = 0$. It is nearly linear near $s=0$ (stable region) and the motor runs at low slip there.

Condition for maximum torque

Differentiate $T$ with respect to $s$ and equate to zero:

$$\frac{dT}{ds} = 0 \;\Rightarrow\; R_2^2 - s^2X_2^2 = 0 \;\Rightarrow\; s_m = \frac{R_2}{X_2}$$

So the torque is maximum when the rotor resistance equals the rotor reactance at that slip ($R_2 = s_mX_2$). Substituting:

$$T_{max} = \frac{3E_2^2}{2\omega_s X_2}$$

Effect of rotor resistance

• $T_{max}$ is independent of $R_2$; only its position changes.
• $s_m \propto R_2$: with higher $R_2$ the maximum torque occurs at higher slip (lower speed).
• Starting torque rises as $R_2$ increases (up to $R_2 = X_2$, where $T_{st} = T_{max}$).
• Running slip at full load increases, so speed falls and rotor copper loss rises, so efficiency falls.

This is why slip-ring motors insert external resistance for starting and speed control.

Torque-slip characteristics

The torque produced by a three-phase induction motor at slip $s$ is

$$T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2}$$

The curve starts at $T_{st}$ for $s=1$, rises to $T_{max}$ at $s_m = R_2/X_2$, then drops to zero at $s=0$.

 T
 |          Tmax
 |        _.-*-._
 |  Tst _-'      `-._
 |   *-'              `-.
 |                        `-.
 +--+--------+--------------*--> s
  s=1       s_m   linear   s=0
          (unstable)(stable)

• Between $s=1$ and $s_m$ the torque falls as slip rises ($T \propto 1/s$): unstable.
• Between $s_m$ and $0$ the torque is nearly proportional to slip ($T \propto s$): stable and almost linear.
• The maximum torque is $T_{max} = \frac{3E_2^2}{2\omega_s X_2}$.

Why the motor operates only in the linear portion

1. For small slip, $sX_2 \ll R_2$, so $T \approx \frac{3E_2^2}{\omega_s R_2}s \propto s$. The curve is a straight line.
2. In this region, if the load rises the speed falls slightly, slip rises, and the torque increases in proportion. A new equilibrium is reached quickly. The operation is stable.
3. Beyond $s_m$, a rise in load makes the speed fall, which reduces the torque, so the speed falls more and the motor stalls. The operation is unstable.
4. In the linear region the rotor current is large in comparison with the reactance effect, the rotor power factor is nearly unity, and slip is small (2 to 5 %), so rotor copper loss $= sP_g$ is small and efficiency is high.
5. Speed remains almost constant (close to $N_s$) from no load to full load, which is the required behaviour for most drives.

So the motor is run in the low-slip region of the curve, with full-load slip well below $s_m$.

Slip

Slip is the fractional difference between synchronous speed and rotor speed, expressed as a fraction or percentage of synchronous speed:

$$s = \frac{N_s - N_r}{N_s}$$

Here $N_s = \frac{120f}{P}$ is the speed of the rotating magnetic field and $N_r$ is the rotor speed. At standstill $s = 1$; at synchronous speed $s = 0$. At rated load it is about 2 to 5 %. The rotor frequency is $f_r = sf$.

Why only the linear portion is used

The torque-slip curve is nearly a straight line for small slip (from $s=0$ to the slip at about full load) because $sX_2 \ll R_2$, so

$$T \approx \frac{3E_2^2}{\omega_s R_2}\,s \;\propto\; s$$

• This region lies on the stable side of the maximum torque ($s < s_m$). An increase in load slows the rotor a little, slip and torque increase, and the motor settles at a new steady speed.
• Beyond the maximum torque ($s > s_m$) the torque falls as slip increases, so any extra load makes the motor stall.
• In the linear region the rotor copper loss ($sP_g$) is small, so efficiency is high, and the speed is nearly constant.

Torque-speed characteristics

The torque of an induction motor is

$$T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2}$$

 T
 |            Tmax
 |          _.-*-._
 |   Tst  /         \
 |    *--'            \
 |   /                  \
 +--+------+----------------+--> N
  N=0   N(Tmax)             Ns

• At $N=0$ the torque is the starting torque, which is moderate.
• The torque rises as the motor accelerates until the maximum (breakdown) torque occurs at $s_m = R_2/X_2$.
• From there to $N_s$ the torque falls nearly linearly to zero. Rated operation is on this falling part at 2 to 5 % slip, so speed changes very little with load.
• At $N_s$ the torque is zero.
• The part left of $T_{max}$ is unstable; the part right of it is stable.

Effect of rotor resistance

• $T_{max} = \frac{3E_2^2}{2\omega_s X_2}$ does not depend on $R_2$.
• The slip at maximum torque $s_m = R_2/X_2$ increases with $R_2$, so the peak shifts to lower speed.
• Starting torque rises with $R_2$ and is maximum when $R_2 = X_2$.
• Full-load speed falls and rotor copper loss rises (lower efficiency).
• Used in wound-rotor motors for high-torque starting and speed control.

 T   R2 low   R2 med   R2 = X2
 |   peak at high N     peak at N = 0

Effect of applied voltage

$E_2 \propto V_1$, so $T \propto V_1^2$ at any speed.

• Both $T_{st}$ and $T_{max}$ vary as $V^2$ (a 10 % voltage drop gives about 19 % less torque).
• $s_m$ does not change, so the peak stays at the same speed.
• At a fixed load torque, a lower voltage gives a higher slip, a higher rotor current and more heating.
• Reduced-voltage starting (star-delta, autotransformer) reduces starting current but also starting torque.

Derivation of torque

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

At slip $s$: rotor emf $= sE_2$, reactance $= sX_2$, so rotor current per phase

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

Rotor power factor $\cos\phi_2 = \frac{R_2}{\sqrt{R_2^2+(sX_2)^2}}$.

Torque is proportional to the rotor power input (air-gap power) $P_g$ divided by synchronous speed:

$$T = \frac{P_g}{\omega_s} = \frac{3I_2^2 R_2/s}{\omega_s}$$

Substituting $I_2$:

$$T = \frac{3}{\omega_s}\cdot\frac{s E_2^2 R_2}{R_2^2 + (sX_2)^2}\ \text{N-m}, \qquad \omega_s = \frac{2\pi N_s}{60}$$

Since $E_2 \propto V_1$, $T \propto V_1^2$.

• Starting torque ($s=1$): $T_{st} = \frac{3}{\omega_s}\cdot\frac{E_2^2R_2}{R_2^2+X_2^2}$.

Torque-slip characteristic

 T
 |          Tmax
 |        _.-*-._
 |  Tst _-'      `-._
 |   *-'              `-.
 +--+--------+-----------*--> s
  s=1       s_m         s=0

Condition for maximum torque

For maximum $T$, $\frac{dT}{ds} = 0$. Writing $T \propto \frac{s R_2}{R_2^2 + s^2X_2^2}$:

$$\frac{(R_2^2 + s^2X_2^2)R_2 - sR_2(2sX_2^2)}{(R_2^2+s^2X_2^2)^2} = 0 \;\Rightarrow\; R_2^2 = s^2X_2^2$$ $$s_m = \frac{R_2}{X_2}$$

Maximum torque occurs when rotor resistance equals the rotor reactance at running slip. Substituting:

$$T_{max} = \frac{3E_2^2}{2\omega_s X_2}$$

The ratio $\frac{T}{T_{max}} = \frac{2 s s_m}{s^2 + s_m^2}$ is used for numericals.

Torque developed is $T = \frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2+(sX_2)^2}$. The slip at which the torque is maximum is $s_m = R_2/X_2$, and the value $T_{max} = \frac{3E_2^2}{2\omega_s X_2}$.

 T
 |        R2 = X2 (peak at start)
 |   _.-*-._  R2 higher
 |  /  _-*-.  .
 | / /'    `-.   R2 low
 |*/ /        `.
 +--+--+--+-------> s
   1  .6  .3     0

Effects of increasing rotor resistance:

1. $T_{max}$ unchanged, since it is independent of $R_2$.
2. $s_m$ increases in proportion to $R_2$: the peak moves toward lower speed (higher slip).
3. Starting torque increases (as $R_2$ rises toward $X_2$); when $R_2 = X_2$, $s_m = 1$ and $T_{st} = T_{max}$. Beyond that it decreases again.
4. Speed at a given load falls, as slip $\propto R_2$ in the linear part, so speed regulation gets worse.
5. Rotor copper loss increases and efficiency decreases.
6. Starting current falls and starting power factor improves.

This is why a slip-ring motor uses external resistance at starting (high torque, low current) and cuts it out when running. Squirrel-cage motors with high-resistance rotors (double cage, deep bar) give high starting torque, at the expense of slip and efficiency.

Principle

In a slip-ring (wound-rotor) induction motor, extra resistance can be inserted in the rotor circuit through slip rings and brushes. The rotor resistance is $R_2 + R_{ext}$. This changes the slip at which a given load torque is produced, and therefore the speed. Speed can be lowered below normal, but not raised above it.

Circuit diagram

 3-phase supply
 R ---+----+----+
 Y ---+--+-|--+-|--+
 B ---+--|-|--|-|--+
      [ STATOR ]      [ ROTOR ]
        winding  ------ winding
                         | | |  slip rings
                        brushes
                         | | |
                        [Rext]  3-phase
                        rheostat (Y)

Torque-slip characteristics

 T
 |   TL ------------------------ load line
 |   _.-*-._  _.-*-._  _.-*-._
 |  /  R2    \   R2+r1    R2+r2
 +--+----+-----+---+----+------> s
  1   s3   s2     s1  (low speed to high)

Since $T_{max}$ is independent of the rotor resistance, increasing $R_2$ shifts the peak towards the high-slip side. For constant load torque $T_L$:

$$\frac{R_2}{s_1} = \frac{R_2 + R_{ext}}{s_2}$$

(the torque is constant, so $R_2/s$ is constant). So $s_2 = s_1\frac{R_2+R_{ext}}{R_2}$, and speed $N = N_s(1-s_2)$ decreases as $R_{ext}$ increases.

Operation

1. At $R_{ext} = 0$ the motor runs at normal speed with small slip $s_1$.
2. Insert $R_{ext}$ using a rheostat: the T-S curve flattens and the operating point on the load line moves to larger slip, so speed falls.
3. The more resistance, the lower the speed.

Merits and demerits

• Merits: simple, smooth control, good starting torque, same arrangement used for starting.
• Demerits: power lost in the rheostat ($sP_g$) so efficiency falls with speed; poor speed regulation (speed varies with load); only speed reduction; rheostat is bulky and expensive; not for light loads (little change in speed).

Synchronous speed

Synchronous speed is the speed of the rotating magnetic field produced by the stator winding of a three-phase induction motor:

$$N_s = \frac{120 f}{P}\ \text{rpm}$$

where $f$ is the supply frequency and $P$ the number of poles. For $f = 50$ Hz and $P=4$, $N_s = 1500$ rpm.

Why the rotor rotates in the direction of the field

1. The rotating field cuts the stationary rotor conductors and induces an emf (Faraday's law).
2. The closed rotor circuit carries a current, so each conductor in the field experiences a force.
3. By Lenz's law, the effect of the induced current opposes its cause, which is the relative motion between the field and the rotor conductors. The rotor can reduce this relative motion only by moving in the same direction as the field.
4. So the torque on the rotor acts in the direction of the field and the rotor follows it.

Why the rotor cannot reach synchronous speed

• If the rotor ran at $N_s$, the relative speed between field and rotor would be zero.
• No emf would be induced in the rotor, so no rotor current would flow.
• Without current there is no torque, and the rotor would slow down due to friction and windage losses and load.
• It therefore settles at a speed $N_r < N_s$ where the induced current gives the torque needed for the load. The difference $N_s - N_r$ is the slip speed, and slip $s = \frac{N_s - N_r}{N_s}$ is 2 to 5 % at full load.

For this reason the induction motor is also called an asynchronous motor.

The statement is true because an induction motor works by induction, and induction needs relative motion between the rotating magnetic field and the rotor conductors.

1. At synchronous speed $N_s = \frac{120f}{P}$, the rotor and the field rotate together. The slip is zero:

$$s = \frac{N_s - N_r}{N_s} = 0$$

2. The rotor conductors do not cut any flux, so the rotor emf is $E_{2s} = sE_2 = 0$.
3. With no emf, the rotor current is $I_2 = 0$, so there is no rotor mmf and no force on rotor conductors.
4. Torque is $T = \frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2+(sX_2)^2}$, which is zero at $s = 0$.
5. Without torque, friction and windage slow the rotor. Slip then appears, an emf is induced, and torque is produced again.

So the motor can only run at a speed slightly below synchronous speed, where the rotor current produces just enough torque to supply the load and the mechanical losses. At no load, the slip is very small (about 0.5 %) but not zero.

This also shows on the T-S curve, where the torque curve passes through zero at $s = 0$.

How the motor adjusts its current with load

An induction motor behaves like a transformer with a rotating secondary.

1. At no load, the rotor runs near $N_s$, the slip is tiny, the rotor emf $sE_2$ is small, so rotor current is very small. The stator takes only the no-load current $I_0$ (mostly magnetising, power factor 0.1 to 0.2).
2. When shaft load increases, the load torque exceeds the electromagnetic torque and the rotor slows down.
3. Slip $s$ increases, so the relative speed between field and rotor increases. The rotor emf $sE_2$ and its frequency $sf$ increase, so the rotor current $I_2 = \frac{sE_2}{\sqrt{R_2^2+(sX_2)^2}}$ increases. The rotor power factor also improves.
4. The torque therefore rises until it matches the load, and the motor settles at a new speed with slightly higher slip.
5. The rotor mmf opposes the stator flux (Lenz's law), tending to reduce it. To keep the flux constant ($\phi \propto V_1$), the stator draws an extra current $I_2'$:

$$\vec I_1 = \vec I_0 + \vec I_2'$$

So the stator current rises automatically with load, and the input power factor improves with load.

Effect of stator connection (slip-ring motor)

The stator can be connected in star or delta (the rotor is normally star). For the same line voltage $V_L$:

• A delta-connected winding is used for normal running; it takes more current but delivers full torque.
• Star connection at start reduces starting current to one-third and gives lower torque (star-delta starter for delta-rated motors).
• Rotor connection does not change stator current directly, but a star rotor has $E_{ph} = E_{L}/\sqrt3$ between slip rings, which sets rotor current and the external resistance values.

Given: 500 V, 50 Hz, 6-pole, $N_r = 975$ rpm, input $P_{in} = 40$ kW, stator loss $= 1$ kW, friction loss $= 2$ kW. (The question says "6-pole", so 6 poles are used.)

(a) Slip

$$N_s = \frac{120\times 50}{6} = 1000\ \text{rpm}, \qquad s = \frac{1000-975}{1000} = 0.025 = 2.5\%$$

(b) Rotor copper loss

$$P_g = 40 - 1 = 39\ \text{kW}$$ $$P_{cu2} = sP_g = 0.025\times 39 = 0.975\ \text{kW}$$

(c) Output HP

$$P_{m} = P_g - P_{cu2} = 39 - 0.975 = 38.025\ \text{kW}$$ $$P_{out} = 38.025 - 2 = 36.025\ \text{kW} = \frac{36025}{746} = 48.29\ \text{HP}$$

(d) Efficiency

$$\eta = \frac{36.025}{40}\times 100 = 90.06\%$$

Answer: (a) $s = 2.5\%$; (b) rotor copper loss $= 0.975$ kW; (c) output $= 36.03$ kW $\approx 48.29$ HP; (d) $\eta = 90.06\%$.

Given: 6-pole, 50 Hz, $T_{max} = 30$ N-m at 960 rpm, $R_2 = 0.6\ \Omega$, $s = 0.06$.

$$N_s = \frac{120\times 50}{6} = 1000\ \text{rpm}$$

Slip at maximum torque:

$$s_m = \frac{1000-960}{1000} = 0.04$$

(This gives $X_2 = R_2/s_m = 15\ \Omega$ at standstill, but the ratio method does not need it.)

Torque at any slip relative to maximum torque (neglecting stator impedance):

$$\frac{T}{T_{max}} = \frac{2\,s\,s_m}{s^2 + s_m^2}$$ $$\frac{T}{T_{max}} = \frac{2\times 0.06\times 0.04}{0.06^2 + 0.04^2} = \frac{0.0048}{0.0052} = 0.9231$$ $$T = 0.9231\times 30 = 27.69\ \text{N-m}$$

Answer: torque at 6 % slip $= 27.69$ N-m.

Given: 4-pole, 50 Hz, $R_2 = 0.4\ \Omega$, $X_2 = 4\ \Omega$ per phase (standstill).

Speed at maximum torque

$$N_s = \frac{120\times 50}{4} = 1500\ \text{rpm}, \qquad s_m = \frac{R_2}{X_2} = \frac{0.4}{4} = 0.1$$ $$N_m = N_s(1-s_m) = 1500\times 0.9 = 1350\ \text{rpm}$$

Ratio $T_{max}/T_{st}$

$$\frac{T_{st}}{T_{max}} = \frac{2s_m}{1+s_m^2} = \frac{0.2}{1.01} = 0.198 \;\Rightarrow\; \frac{T_{max}}{T_{st}} = \frac{1+s_m^2}{2s_m} = \frac{1.01}{0.2} = 5.05$$

Resistance for $T_{st} = \frac12 T_{max}$

Let the new total rotor resistance be $R_2'$, so $s_m' = R_2'/X_2$:

$$\frac{2s_m'}{1+s_m'^2} = 0.5 \;\Rightarrow\; s_m'^2 - 4s_m' + 1 = 0$$ $$s_m' = 2 - \sqrt3 = 0.2679 \quad (\text{the root} < 1)$$ $$R_2' = s_m' X_2 = 0.2679\times 4 = 1.0718\ \Omega$$

Additional resistance to be added per phase $= 1.0718 - 0.4 = 0.6718\ \Omega$.

Answer: speed at $T_{max}$ = 1350 rpm; $T_{max}/T_{st} = 5.05$; rotor resistance per phase $= 1.072\ \Omega$ (external $0.672\ \Omega$ added).