Set 26 — Torque and Torque–Slip Relations
Set 25 left the rotor described but the torque only implied. Writing the torque out as a function of slip closes that gap, and the function that appears has a single, well-behaved maximum: torque rises almost linearly with slip near synchronism, peaks where the rotor resistance equals the rotor reactance at that slip, and falls away beyond it. Two numbers therefore describe the whole curve — the height of the peak and the slip at which it occurs.
The most useful result of the set is that the peak height does not depend on the rotor resistance at all, while its position is directly proportional to it. That single fact turns most torque questions into ratio arithmetic: \(T/T_{max} = 2ss_b/(s^2+s_b^2)\) needs no voltage, no reactance and no machine constant, only two slips. Three of the four problems here are solved with it, and the fourth shows what it means to move the peak deliberately by changing the rotor circuit.
Torque is air-gap power divided by synchronous angular velocity, not by the actual shaft speed:
\[ T = \frac{P_g}{\omega_s} = \frac{3I_2^2R_2/s}{\omega_s}, \qquad \omega_s = \frac{2\pi N_s}{60}\ \text{rad/s} \]In terms of slip and the standstill rotor quantities this becomes the torque-slip relation:
\[ T = \frac{3}{\omega_s}\cdot\frac{sE_2^2R_2}{R_2^2+(sX_2)^2} \]Put \(s = 1\) for the starting torque. The same expression in stator-referred quantities replaces \(E_2\) by the stator phase voltage and \(R_2,X_2\) by \(R_2',X_2'\).
Referring the rotor to the stator uses the per-phase turns ratio \(K = N_2/N_1\):
\[ R_2' = \frac{R_2}{K^2}, \qquad X_2' = \frac{X_2}{K^2} \]Maximum torque occurs where the two terms of the denominator are equal:
\[ R_2 = s_bX_2 \;\Longrightarrow\; s_b = \frac{R_2}{X_2}, \qquad T_{max} = \frac{3E_2^2}{2\omega_sX_2} \]\(T_{max}\) contains no \(R_2\). Rotor resistance moves the peak along the slip axis but cannot raise or lower it.
The ratio form removes every machine constant and is the fastest route through most problems:
\[ \frac{T}{T_{max}} = \frac{2ss_b}{s^2+s_b^2}, \qquad \frac{T_{st}}{T_{max}} = \frac{2s_b}{1+s_b^2} \]At a fixed torque the slip is proportional to the rotor resistance. Doubling \(R_2\) doubles the slip at which a given load is carried — the basis of rotor-resistance speed control and of Problem 4.
Torque may be quoted in kg-m in older problems. Ratios are unaffected; convert with \(1\ \text{kg-m} = 9.81\ \text{N·m}\) only if newton metres are asked for.
A 150 kW, 3000 V, 50 Hz, 6-pole star-connected induction motor has a star-connected slip-ring rotor with a transformation ratio of 3.6 (stator/rotor). The rotor resistance is 0.1 Ω per phase and its per-phase leakage inductance is 3.61 mH. The stator impedance may be neglected. At rated voltage with the slip rings short-circuited, determine
- the starting current
- the starting torque.
Turn the rotor inductance into a reactance at supply frequency, since at standstill the rotor frequency is the supply frequency:
Refer both rotor quantities to the stator. With \(K = N_2/N_1 = 1/3.6\), impedances scale by \(1/K^2 = 3.6^2 = 12.96\):
Referring lets the stator phase voltage drive the rotor branch directly, which is legitimate here only because the stator impedance is to be neglected.
The starting current is the stator phase voltage across the referred rotor impedance at \(s = 1\):
The stator is in star, so this is also the line current drawn from the supply at start.
Synchronous angular velocity:
The starting torque is the air-gap power at \(s = 1\) divided by \(\omega_s\):
The same number comes from \(T = 3I_{st}^2R_2'/\omega_s = 3(117.4)^2(1.296)/104.7 = 512\ \text{N·m}\), since at \(s = 1\) the whole air-gap power is dissipated in the rotor resistance.
How poor a start that is. The peak of this machine's torque-slip curve sits at
So \(T_{st}/T_{max} = 2s_b/(1+s_b^2) = 0.175\) — the machine starts with only 17.5% of the torque it is capable of. Rotor resistance in the slip-ring circuit is what fixes this, by moving \(s_b\) towards 1.
Calculate the torque exerted by an 8-pole, 50 Hz, 3-phase induction motor operating with a 4% slip, given that it develops a maximum torque of 150 kg-m at a speed of 660 rpm. The resistance per phase of the rotor is 0.5 Ω.
Find the slip at which the maximum occurs, which is what the 660 rpm is really telling us:
The maximum-torque condition then gives the standstill rotor reactance:
This is the only use the rotor resistance is put to. Its value never enters the final ratio, because \(T_{max}\) does not depend on it.
Form the torque ratio. Writing both torques from the torque-slip relation and cancelling \(3E_2^2/\omega_s\):
Hence the torque at 4% slip:
Alternative method. The standard ratio form gets there without computing \(X_2\) at all — only the two slips are needed:
The rotor resistance was a red herring: it is needed only if the reactance itself is asked for.
A 3-phase, 50 Hz, 8-pole induction motor has a full-load slip of 2%. The rotor resistance and standstill rotor reactance per phase are 0.001 Ω and 0.005 Ω respectively. Find the ratio of the maximum torque to the full-load torque, and the speed at which the maximum torque occurs.
Synchronous speed:
The slip for maximum torque follows straight from the rotor constants:
The speed at which the maximum appears:
Well below the full-load speed of \(750(1-0.02) = 735\) rpm, so the machine normally runs on the steep, stable part of the curve and only reaches its peak torque during a heavy overload.
The torque ratio with \(s_{FL} = 0.02\) and \(s_b = 0.2\):
Invert it, because the question asks for maximum over full load:
A pull-out torque of five times full load is unusually large, and it is the price of a very small \(s_{FL}/s_b\) ratio: this machine reaches full-load torque at only a tenth of its breakdown slip.
For a 3-phase slip-ring induction motor the maximum torque is 2.5 times the full-load torque and the starting torque is 1.5 times the full-load torque. Determine the percentage reduction in rotor circuit resistance required to give a full-load slip of 3%. Neglect the stator impedance.
Use the starting-torque ratio to locate the present peak. Putting \(s = 1\) in the ratio relation:
A slip of 3 describes a machine driven backwards, so the physical root is \(s_b = 1/3\).
Translate that into the present rotor resistance:
Impose the new operating condition. The load is unchanged, so the machine must still develop \(T_{fl} = T_{max}/2.5\) — and \(T_{max}\) is untouched by a change of rotor resistance — but now at \(s = 0.03\). Writing \(a\) for the new value of \(R_2/X_2\):
The smaller root would put the breakdown slip below the operating slip, leaving the machine on the unstable side of its own curve, so \(a = 0.1437\) is the design value: \(R_{2,\text{new}} = 0.1437X_2\).
The percentage reduction — and \(X_2\) cancels, as it must, since only the resistance is being altered:
Alternative method. At a fixed torque the slip is exactly proportional to the rotor resistance, so the percentage reduction in resistance equals the percentage reduction in full-load slip. The original full-load slip follows from the same ratio with \(s_b = 1/3\):
The machine was carrying full load at 6.96% slip; cutting the rotor resistance by 56.9% brings that down to 3%, and the shaft speed up by the same 4% of synchronous speed.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Synchronous speed | \(N_s = 120f/P\), \(\omega_s = 2\pi N_s/60\) | Torque uses \(\omega_s\), not \(\omega_m\) |
| Torque | \(T = P_g/\omega_s = 3I_2^2R_2/(s\omega_s)\) | Problem 1 |
| Torque-slip relation | \(T = \dfrac{3}{\omega_s}\dfrac{sE_2^2R_2}{R_2^2+(sX_2)^2}\) | Problems 1, 2 |
| Starting torque | \(T_{st} = \dfrac{3}{\omega_s}\dfrac{E_2^2R_2}{R_2^2+X_2^2}\) | \(s = 1\) — Problem 1 |
| Referred rotor values | \(R_2' = R_2/K^2,\; X_2' = X_2/K^2\) | \(K = N_2/N_1\) — Problem 1 |
| Standstill reactance | \(X_2 = 2\pi fL_2\) | From inductance — Problem 1 |
| Slip for maximum torque | \(s_b = R_2/X_2 = R_2'/X_2'\) | Problems 1, 3 |
| Maximum torque | \(T_{max} = \dfrac{3E_2^2}{2\omega_sX_2}\) | Independent of \(R_2\) — Problem 4 |
| Torque ratio | \(\dfrac{T}{T_{max}} = \dfrac{2ss_b}{s^2+s_b^2}\) | Problems 2, 3, 4 |
| Starting-torque ratio | \(\dfrac{T_{st}}{T_{max}} = \dfrac{2s_b}{1+s_b^2}\) | Problems 1, 4 |
| Universal form | \(T/T_{max} = 2u/(1+u^2),\; u = s/s_b\) | One curve for every machine — Problem 2 |
| Overload capacity | \(T_{max}/T_{FL} \approx s_b/2s_{FL}\) | Valid when \(s_{FL}\ll s_b\) — Problem 3 |
| Resistance and slip | \(s \propto R_2\) at constant torque | Problem 4 |
| Torque units | \(1\ \text{kg-m} = 9.81\ \text{N·m}\) | Problem 2 |
Common Mistakes
Dividing the air-gap power by the rotor speed. Torque is \(P_g/\omega_s\); the shaft speed appears only when mechanical power is wanted — Problem 1.
Using rotor quantities with the stator voltage, or referred quantities with the rotor e.m.f. Either pair is correct, but mixing them is not: Problem 1 uses \(V_{1,ph}\) with \(R_2'\) and \(X_2'\) throughout.
Referring with \(K\) instead of \(K^2\). Impedances scale with the square of the turns ratio, so 0.1 Ω becomes 1.296 Ω, not 0.36 Ω — Problem 1.
Believing that rotor resistance raises the maximum torque. \(T_{max}\) contains only \(E_2\), \(\omega_s\) and \(X_2\); resistance only moves the peak — Problem 4.
Quoting \(T_{FL}/T_{max}\) when the question asked for \(T_{max}/T_{FL}\). The ratio formula naturally produces 0.198; the answer wanted is its reciprocal, 5.05 — Problem 3.
Taking the speed of maximum torque as a slip. 660 rpm has to be converted to \(s_b = (750-660)/750 = 0.12\) before it can be used — Problem 2.
Keeping the wrong root of the quadratic. \(s_b = 3\) and \(a = 0.00626\) are mathematically valid and physically wrong; check each root against the operating slip — Problem 4.
Mixing kg-m and N·m inside a ratio. Ratios are dimensionless, so 0.6 applies directly to the 150 kg-m; convert only at the end, if at all — Problem 2.
Forgetting that the torque-slip curve is symmetric in \(s/s_b\). Slips of \(s_b/3\) and \(3s_b\) give the same torque, which is why a ratio equation always has two roots — Problems 2 and 4.
The torque-slip relation now covers the whole speed range with two numbers, and the ratio form makes most questions independent of voltage, reactance and machine constants. What it does not yet say is where the power goes: the expression \(T = P_g/\omega_s\) was used as a definition, and the air-gap power was never split into the part that turns the shaft and the part that heats the rotor bars.
That split is fixed entirely by the slip, and it is the reason a large slip is expensive. A machine running at 5% slip loses 5% of its air-gap power as rotor copper loss no matter how well it is built — which sets a ceiling on efficiency that no amount of good design can lift.
Next: Set 27 — Power Flow and Efficiency, where \(P_g : P_{cu2} : P_m = 1 : s : (1-s)\) is established and used to work through the losses from stator input to shaft output.