Torque Equation of the Induction Motor

The torque developed by the motor at the instant of starting is called the starting torque. In some cases it is greater than the normal running torque, and in others somewhat less.

At standstill \(s = 1\), so the rotor quantities take their standstill values:

This second form is the useful one for Section 61-5, because it isolates the dependence on rotor resistance — the one quantity a designer can choose freely.

So far the torque has been a proportionality. To obtain an actual value in newton metres, the constant must be evaluated. It turns out to be

where \(n_s\) is the synchronous speed in revolutions per second. The factor 3 is for the three phases, and \(2\pi n_s = \omega_s\) is the synchronous speed in radians per second.

• The resistance of a squirrel-cage rotor is fixed and small compared with its reactance, which is very large at start — because at standstill the frequency of the rotor currents equals the supply frequency.

• Hence the starting current \(I_2\), though very large in magnitude, lags by a very large angle behind \(E_{20}\), with the result that the starting torque per ampere is very poor.

• It is roughly 1.5 times the full-load torque, although the starting current is 5 to 7 times the full-load current.

• Such motors are therefore not very useful where the motor has to start against heavy loads.

• The starting torque of such a motor is increased by improving its power factor, by adding external resistance in the rotor circuit from a star-connected rheostat.

• The rheostat resistance is progressively cut out as the motor gathers speed.

Differentiate the starting torque with respect to the rotor resistance and set the result to zero.

Thus \(T_{st}\) is maximum when \(R_2 = X_{20}\). Substituting back:

The rotor EMF is induced by the flux, which is set by the applied voltage, so \(E_{20} \propto V\). Hence

• At standstill, \(s = 1\) and \(f_r = f\).

• \(E_{20}\) at standstill is maximum, because the relative speed between the rotor and the revolving stator flux is maximum.

• In fact, an induction motor is equivalent to a three-phase transformer with a short-circuited rotating secondary.

• Under running conditions the relative speed decreases, and since the rotor EMF is proportional to relative speed, it decreases too.

• Because the frequency of the rotor quantities also decreases, the rotor reactance decreases with it.

That asymmetry between \(R_2\) and \(X_2\) is the whole source of the machine's characteristic. Had both scaled with slip, the rotor power factor would be constant and the torque would be a plain straight line.

The derivation is the same as for starting torque, but with the running values substituted.

The torque under running conditions is

To make the calculation simple, take \(Y = 1/T_r\). Maximising \(T_r\) is the same as minimising \(Y\), and \(Y\) is much easier to differentiate because the slip appears in the numerator rather than buried in a quotient.

• Starting torque follows from \(T = K_1E_{20}I_2\cos\phi_2\) with the standstill values: \(T_{st} = K_1E_{20}^{2}R_2/(R_2^{2}+X_{20}^{2})\).

• \(K_1 = 3/2\pi n_s\), with \(n_s\) in rev/s. It arises because torque is air-gap power divided by synchronous angular speed.

• A cage motor has small \(R_2\) and large \(X_{20}\) at start, so the current is huge but badly lagging: about 1.5 × full-load torque for 5 to 7 × full-load current.

• A slip-ring motor adds rotor resistance. At first the improved power factor beats the reduced current and torque rises; past the optimum the impedance wins and torque falls.

• Maximum starting torque when \(R_2 = X_{20}\), giving \(\phi_2 = 45^{\circ}\).

• \(T \propto V^{2}\): a 5 % voltage change gives about a 10 % torque change.

• Under running conditions \(f_r = sf\), \(E_2 = sE_{20}\), \(X_2 = sX_{20}\) — but \(R_2\) is unchanged.

• The torque equation: \(T_r = K_1sE_{20}^{2}R_2/[R_2^{2}+(sX_{20})^{2}]\). Putting \(s = 1\) recovers \(T_{st}\).

• Maximum torque at \(s_{max} = R_2/X_{20}\), found by minimising \(Y = 1/T_r\).

• \(T_{max} = K_1E_{20}^{2}/2X_{20}\) — independent of \(R_2\). Rotor resistance moves the peak without changing its height.

• Using \(N_s\) in rev/min in \(K_1 = 3/2\pi n_s\). It must be rev/s, or the torque comes out 60 times too small.

• Dividing air-gap power by the rotor speed. Torque is \(P_g/\omega_s\), using the synchronous speed.

• Thinking maximum torque rises when rotor resistance is added. Only its position moves; \(T_{max}\) is unchanged.

• Believing more rotor resistance always means more starting torque. Beyond \(R_2 = X_{20}\) it falls again.

• Leaving the starting rheostat in circuit while running. It wastes power and drops the speed.

• Taking torque as proportional to \(V\). It goes as \(V^{2}\).

• Confusing \(X_{20}\) with \(X_2\). The running reactance is \(sX_{20}\), smaller by a factor of 25 at 4 % slip.

• Treating \(R_2\) as slip-dependent. It is not, and the whole characteristic depends on that.

• Applying \(T \propto s\) at large slip. It holds only while \(sX_{20} \ll R_2\).

• Assuming large starting current means large starting torque. At standstill the rotor power factor is about 0.2.

Convert \(N_s\) to rev/s before using \(K_1\). For maximum-torque questions, \(s_{max} = R_2/X_{20}\) and \(T_{max} = K_1E_{20}^{2}/2X_{20}\) are all that is needed.

1. P61.1 A 6-pole, 50 Hz motor has \(E_{20} = 100\) V, \(R_2 = 0.30~\Omega\) and \(X_{20} = 1.00~\Omega\) per phase. Find \(K_1\) and the starting torque.

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   \[n_s = \frac{2(50)}{6} = 16.667~\mathrm{rev/s}, \qquad K_1 = \frac{3}{2\pi(16.667)} = 0.028648\]

   \[T_{st} = (0.028648)\frac{(10000)(0.30)}{0.09+1.00} = (0.028648)\left(\frac{3000}{1.09}\right) = 78.85~\mathrm{N\,m}\]

2. P61.2 For P61.1, find \(s_{max}\) and \(T_{max}\), and the ratio \(T_{st}/T_{max}\).

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   \[s_{max} = \frac{0.30}{1.00} = 0.300\]

   \[T_{max} = \frac{(0.028648)(10000)}{(2)(1.00)} = 143.24~\mathrm{N\,m}\]

   \[\frac{T_{st}}{T_{max}} = \frac{78.85}{143.24} = 0.5505 = 55.05\,\%\]

3. P61.3 For P61.1, find the external rotor resistance for maximum starting torque, and confirm the torque then obtained.

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   \[R_{ext} = X_{20} - R_2 = 1.00 - 0.30 = 0.70~\Omega\ \text{per phase}\]

   \[T_{st} = (0.028648)\frac{(10000)(1.00)}{1.00+1.00} = (0.028648)(5000) = 143.24~\mathrm{N\,m}\]

   Exactly \(T_{max}\) from P61.2, as it must be.

4. P61.4 For P61.1, find the torque at 4 % slip and the ratio \(T_{max}/T_{fl}\).

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   \[sX_{20} = 0.0400, \qquad R_2^{2}+\left(sX_{20}\right)^{2} = 0.09 + 0.0016 = 0.0916\]

   \[T_r = (0.028648)\frac{(0.04)(10000)(0.30)}{0.0916} = (0.028648)\left(\frac{120}{0.0916}\right) = 37.53~\mathrm{N\,m}\]

   \[\frac{T_{max}}{T_{fl}} = \frac{143.24}{37.53} = 3.816\]

5. P61.5 A motor develops 60 N m of starting torque at rated voltage. Find the starting torque at 80 % voltage.

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   \[T = (60)(0.80)^{2} = (60)(0.64) = 38.4~\mathrm{N\,m}\]

   A 20 % voltage loss costs 36 % of the torque.

6. P61.6 A 4-pole, 50 Hz motor has \(R_2 = 0.20~\Omega\) and \(X_{20} = 1.00~\Omega\). What external resistance places maximum torque at a slip of 0.50?

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   \[s_{max} = \frac{R_2 + R_{ext}}{X_{20}} = 0.50 \quad\Longrightarrow\quad R_2 + R_{ext} = (0.50)(1.00) = 0.50~\Omega\]

   \[R_{ext} = 0.50 - 0.20 = 0.30~\Omega\ \text{per phase}\]

7. P61.7 Derive the torque equation under running conditions and show that \(s = 1\) recovers the starting torque.

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   Torque depends on the flux, the rotor current, and the phase angle between rotor EMF and rotor current:

   \[T_r \propto \Phi I_2\cos\phi_2\]

   Since \(E_{20} \propto \Phi\), and using the running values \(E_2 = sE_{20}\) and \(X_2 = sX_{20}\) from Chapter 60:

   \[T_r \propto \left(E_{20}\right)\left(\frac{sE_{20}}{\sqrt{R_2^{2}+\left(sX_{20}\right)^{2}}}\right)\left(\frac{R_2}{\sqrt{R_2^{2}+\left(sX_{20}\right)^{2}}}\right) = \frac{sE_{20}^{2}R_2}{R_2^{2}+\left(sX_{20}\right)^{2}}\]

   With the constant evaluated,

   \[T_r = \frac{3}{2\pi n_s}\left[\frac{sE_{20}^{2}R_2}{R_2^{2}+\left(sX_{20}\right)^{2}}\right]\]

   Setting \(s = 1\):

   \[T_r = \frac{3}{2\pi n_s}\cdot\frac{E_{20}^{2}R_2}{R_2^{2}+X_{20}^{2}} = T_{st}\]

   The starting torque is not a separate formula but simply this one at \(s = 1\), which is what standstill means. One equation covers the whole range from standstill to synchronous speed.

8. P61.8 Derive \(s_{max}\) and \(T_{max}\), and state the significance of the result.

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   Maximising \(T_r\) is the same as minimising \(Y = 1/T_r\), which is easier because the slip separates:

   \[Y = \frac{R_2^{2}+s^{2}X_{20}^{2}}{K_1sE_{20}^{2}R_2} = \frac{1}{K_1E_{20}^{2}R_2}\left[\frac{R_2^{2}}{s}+sX_{20}^{2}\right]\]

   \[\frac{dY}{ds} = \frac{1}{K_1E_{20}^{2}R_2}\left[-\frac{R_2^{2}}{s^{2}}+X_{20}^{2}\right] = 0 \quad\Longrightarrow\quad s_{max} = \frac{R_2}{X_{20}}\]

   Substituting back, with \(s_{max}X_{20} = R_2\) so the denominator is \(2R_2^{2}\):

   \[T_{max} = K_1\frac{\left(R_2/X_{20}\right)E_{20}^{2}R_2}{2R_2^{2}} = \frac{K_1E_{20}^{2}}{2X_{20}}\]

   Significance. \(R_2\) has cancelled entirely. Rotor resistance fixes where the peak occurs but not how high it is.

   A slip-ring machine can therefore slide its peak torque anywhere between full-load slip and standstill by adding resistance, losing nothing in peak value. That is the whole justification for the arrangement — and it also means a cage machine's \(T_{max}\) is fixed once \(E_{20}\) and \(X_{20}\) are settled, no matter what the bars are made of.

9. P61.9 Why does a cage motor give only about 1.5 times full-load torque while drawing 6 times full-load current?

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   Because torque depends on \(I_2\cos\phi_2\), and at standstill the rotor power factor is very poor.

   Why the power factor is poor. At \(s = 1\) the rotor frequency equals the supply frequency, so the rotor reactance takes its full value \(X_{20}\). A cage rotor's resistance is deliberately small, so \(X_{20} \gg R_2\) and

   \[\cos\phi_2 = \frac{R_2}{\sqrt{R_2^{2}+X_{20}^{2}}} \approx 0.2\]

   In the reference machine the reactance contributes 1.44 of the 1.5025 in \(Z_{20}^{2}\) — 95.8 % of it.

   Roughly four fifths of that huge starting current is therefore in quadrature and produces no torque at all; it merely heats the rotor and burdens the supply.

   The remedies all attack the same thing — raising the effective \(R_2\) at start: external resistance in a slip-ring machine, or a deep-bar or double-cage rotor whose skin effect raises the bar resistance at the high standstill frequency and lowers it again as the frequency falls to \(sf\).

10. P61.10 Explain why torque varies as \(V^{2}\), and what follows for starting large machines.

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    The applied voltage sets the air-gap flux, and the rotor EMF is induced by that flux, so \(E_{20} \propto V\). The torque expression contains \(E_{20}^{2}\):

    \[T_{st} = \frac{K_1E_{20}^{2}R_2}{R_2^{2}+X_{20}^{2}} = \frac{K_3V^{2}R_2}{Z_{20}^{2}} \quad\Longrightarrow\quad T \propto V^{2}\]

    The square arises because the flux acts twice — once in inducing the rotor EMF and hence the current, and again in producing the force on that current.

    Numerically: a 5 % change gives \((1.05)^{2} = 1.1025\) or about 10 %; a 10 % dip gives \((0.90)^{2} = 0.81\), costing 19 % of the torque.

    Consequence for starting. A large motor draws 5 to 7 times rated current at start, which depresses the supply voltage through the feeder impedance — so the very current drawn to start the machine reduces the voltage that produces its torque.

    If the remaining torque falls below the load torque, the machine cannot accelerate: the slip stays high, the current stays high, the voltage stays depressed, and the motor stalls. Starting methods that reduce current in a controlled way — star-delta, autotransformer, soft starters — are preferable to letting the supply sag uncontrolled.