Chapter 16 — The induction machine

The synchronous machine of Chapter 15 is the smaller half of the rotating-machine problem. There are a few thousand large synchronous generators on a continental grid; there are tens of millions of induction motors, and electric motors between them consume something like 40 % of the world's electricity, the great majority of it through induction machines. When a fault depresses voltage across a substation, the load that reacts most violently, and the load most likely to decide whether the system recovers, is induction motors.

This chapter is shorter than the last, because most of the machinery is already built. Park's transformation, the swing equation, and the voltage-behind-reactance interface all carry over unchanged. What changes is the rotor.

Learning objectives

By the end of this chapter you should be able to:

• Define slip and explain why an induction machine develops torque only when it is not synchronous.
• Write the $dq$ model of a cage rotor and say what a second cage is for.
• Construct the torque-speed characteristic from the equivalent circuit and locate the operating point, breakdown torque and starting torque.
• Predict how a load's torque-speed characteristic changes the settling point and the starting duty.
• Explain why induction motors dominate dynamic load behaviour and what stalling does to a recovering system.

16.1 What changes

Strip the field winding off a synchronous rotor and short the dampers on themselves, and you have an induction machine. That single change propagates through everything:

• No excitation. There is nowhere to inject field current, so the machine has no independent source of flux. All of it comes across the air gap from the stator, which means an induction machine always absorbs reactive power, whether it is motoring or generating.
• No synchronism. With no field, the rotor carries no flux of its own to lock onto the rotating stator field. It therefore has no preferred speed, no rotor angle, and no pull-out in the synchronous sense.
• Torque requires relative motion. Rotor current exists only because the stator field sweeps past the rotor bars and induces it. If the rotor turned at exactly synchronous speed the bars would see a constant flux, no emf would be induced, no current would flow, and no torque would be produced.

That last point defines the machine's one essential variable, slip:

$$s=\frac{{\omega }_s-{\omega }_r}{{\omega }_s}=1-{\omega }_r\text{(pu)}.$$

At standstill $s=1$; at synchronous speed $s=0$; a motor under load runs at a small positive slip, typically 0.5 to 3 %. Drive the shaft above synchronous speed and $s$ goes negative and the machine generates — which is how a fixed-speed wind turbine works, and why the same component serves both roles.

The frequency of the rotor currents is $s{f}_0$. At 1 % slip on a 60 Hz system, the bars carry 0.6 Hz. This is why rotor resistance matters so much and rotor reactance so little at normal running speeds, and why the picture inverts completely at standstill.

16.2 The $dq$ model

The transformation is the same one Park wrote for the synchronous machine, with one change of frame. There is no rotor angle to reference, so the natural choice is a frame rotating at synchronous speed rather than rotor speed. In that frame the stator quantities are constant in the steady state and the rotor equations pick up a slip-speed coupling:

$$0=R_r{i}_{dr}+\frac{d{\psi }_{dr}}{dt}-s{\omega }_0{\psi }_{qr},0=R_r{i}_{qr}+\frac{d{\psi }_{qr}}{dt}+s{\omega }_0{\psi }_{dr}.$$

The left-hand zeros are the whole story of a cage: the bars are shorted by the end rings, so there is no applied rotor voltage. NumaSim keeps these two flux linkages (four for a double cage) as state variables and integrates them trapezoidally, retaining the slip-speed coupling, in exactly the way the synchronous model integrates its field and damper fluxes.

The parameters are the classical equivalent-circuit set rather than the derived reactances of §15.4, because for an induction machine the circuit is the datasheet: stator $R_s$ and $X_{ls}$, magnetizing $X_m$, and rotor $R_{r1}$, $X_{lr1}$. A typical set for a 1 MVA low-voltage motor is $R_s=0.03$, $X_{ls}=0.1$, $X_m=3.0$, $R_{r1}=0.025$, $X_{lr1}=0.1$ pu.

Note how much smaller $X_m$ is than a transformer's. A transformer has no air gap and a magnetizing reactance of 50 to 200 pu; an induction machine must push its flux across a physical gap, so $X_m$ lands around 3 pu and the magnetizing current is a third of rated rather than one per cent. That gap is also why the machine's power factor is unimpressive even at full load.

Two cages

A single cage forces an uncomfortable compromise. Torque near synchronous speed is roughly proportional to $s/R_r$, so a low rotor resistance gives efficient running. Torque at standstill depends on $R_r$ competing against the leakage reactance, so a high rotor resistance gives good starting. One number cannot do both.

The engineering answer is to make the rotor resistance depend on frequency. Put a small high-resistance cage near the rotor surface and a large low-resistance cage deep in the iron; at standstill the rotor frequency is 60 Hz, skin effect confines current to the outer cage, and the machine sees high resistance. At 1 % slip the rotor frequency is 0.6 Hz, current spreads into the deep bars, and the machine sees low resistance. A deep-bar rotor achieves the same thing with one physically tall bar. NumaSim's rotor_type selects single or double cage, adding $R_{r2}$ and $X_{lr2}$ for the second.

16.3 Steady state and the torque-speed curve

Set the derivatives to zero and the $dq$ model collapses to the single-phase equivalent circuit every motor course begins with: a series $R_s+j{X}_{ls}$, a shunt $j{X}_m$, and a rotor branch $j{X}_{lr}+R_r/s$. The $R_r/s$ is not a modelling trick — it is what falls out algebraically when you refer a rotor quantity at slip frequency to the stator, and its excess over $R_r$ is exactly the mechanical power.

Reduce the stator side to a Thévenin equivalent $V_{th}$, $R_{th}+j{X}_{th}$ and the developed torque follows:

$$T_e(s)=\frac{|V_{th}{|}^2(R_r/s)}{(R_{th}+R_r/s{)}^2+(X_{th}+X_{lr}{)}^2}\text{pu}.$$

Three features of this expression carry the whole subject.

Torque-speed characteristic computed from the equivalent circuit of the lab machine. Doubling the rotor resistance leaves the breakdown torque untouched and moves it to twice the slip, nearly doubling the starting torque and doubling the running slip. The two load curves show why a fan is easy to start and a constant-torque load is not.

Near synchronous speed the $R_r/s$ term dominates the denominator, so $T_e\approx |V_{th}{|}^{2}s/R_r$ — torque is proportional to slip and inversely proportional to rotor resistance. This is the linear region where a motor actually runs, and it is why the running slip is small: a few per cent of speed buys full torque.

Breakdown torque occurs where $R_r/s$ matches the magnitude of the rest of the impedance:

$$s_{max}=\frac{R_r}{\sqrt{R_{th}^2+(X_{th}+X_{lr}{)}^2}},T_{max}=\frac{|V_{th}{|}^2}{2[R_{th}+\sqrt{R_{th}^2+(X_{th}+X_{lr}{)}^2}]}.$$

Look at what $R_r$ does and does not appear in. It sets where the peak occurs but not how high it is. Doubling the rotor resistance slides the whole peak to twice the slip while leaving its height at 2.06 pu — which is precisely the freedom a double-cage rotor exploits.

Breakdown torque scales as $|V_{th}{|}^2$, so a motor that develops 2.06 pu at rated voltage develops 1.32 pu at 0.8 pu voltage and only 0.51 pu at half voltage. Hold that thought for §16.5.

Starting torque, at $s=1$, is 0.56 pu for this machine and 1.04 pu with the doubled rotor resistance. Starting current is the other half of the story: at $s=1$ the $R_r/s$ term is at its smallest and the machine looks like a short circuit behind the leakage reactances, $X_{ls}+X_{lr}=0.2$ pu, so it draws about five times rated current. That is the direct-on-line starting problem in one sentence — poor torque and enormous current, both because the rotor is turning at the wrong frequency.

16.4 The mechanical side

The shaft obeys the same swing equation as Chapter 15, with the rotor angle simply dropped since an induction machine has none:

$$2H\frac{d{\omega }_r}{dt}=T_m-T_e-D({\omega }_r-1).$$

Inertia constants are smaller than a generator's — 0.5 to 3 s for a motor and its driven equipment against 3 to 6 s for a turbine set — so motors respond faster and stall sooner.

What is worth adding is that the load torque is rarely constant. NumaSim's load_curve option supplies the standard polynomial

$$T_{\text{load}}=T_0(A+B{\omega }_r+C{\omega }_r^2),$$

with load_a, load_b, load_c selecting the mixture. The three terms are not arbitrary: $A$ is Coulomb friction and any constant-torque duty such as a conveyor or a positive-displacement pump; $B$ is viscous friction; $C$ is the aerodynamic or hydraulic term that dominates fans, centrifugal pumps and compressors. The default $C=1$ gives the fan law, torque proportional to speed squared, which is why fans start easily — at standstill they demand no torque at all — and why a small speed reduction saves so much energy.

The intersection of the machine curve and the load curve is the operating point, and it is stable if the machine curve is steeper. On the high-speed side of breakdown it always is: a small slowdown increases slip, which increases torque, which restores speed. On the low-speed side the slope reverses and the equilibrium is unstable — a machine pushed past breakdown decelerates all the way to standstill.

16.5 Motors as loads

The reason a system study cares is that motors do not behave like the constant-power or constant-impedance loads a load flow assumes.

Consider a fault that depresses voltage at a substation to 0.6 pu. Breakdown torque falls with the square of voltage, to 0.36 of its rated value. If the motors were carrying more than that, they cannot, so they decelerate. Deceleration increases slip, which — past the breakdown point — reduces torque further and increases current draw, which depresses the voltage further. The process runs away, and the motors stall.

Stalled motors are close to short circuits. They draw five or six times rated current at very poor power factor, and that reactive draw holds the voltage down even after the fault is cleared. The result is fault-induced delayed voltage recovery: the fault clears in five cycles but the voltage takes seconds to come back, because thousands of stalled air-conditioner compressors are holding it there. Thermal protection eventually trips them, and the voltage recovers as load is lost. This is a real and much-studied phenomenon on summer-peaking distribution systems, and it is invisible to any model that represents air conditioning as a constant-power load.

The same mechanism sets a slower limit. Because an induction motor's absorbed reactive power rises steeply as slip increases, a heavily motor-loaded area can reach a point where more load produces less delivered power — the voltage-stability limit. Motors are the reason that limit is where it is.

Load flow versus EMT

NumaSim's induction machine exports to load flow as a constant-PQ row (the pf_p_mw / pf_q_mvar parameters), which is the standard and correct treatment for finding an initial operating point. But that row is a snapshot at one voltage. Everything in this section happens because the real machine is not a constant-PQ device once the voltage moves, which is exactly what the dynamic model represents and the load-flow row does not.

16.6 Lab: motoring against a load torque

Open the induction motoring sample in simulator →

A 1 MVA, 480 V single-cage machine is connected directly across an ideal source with a constant 0.4 pu braking torque on the shaft. NumaSim uses the generator convention throughout, so a load is a negative drive torque: tm0 = -0.4 pu is what makes this a motor. init_slip = 0 releases the rotor at synchronous speed with only its magnetizing flux seeded, so the run starts with the machine unloaded-and-turning and watches it pull in.

1. Read the operating point. By about 1 s the rotor settles at w = 0.9890, giving slip = 0.0110, and Te = -0.400 pu exactly balancing the load. Reactive draw is 0.349 MVAr against 0.409 MW of real power — a power factor of 0.76, which is what a 3 pu magnetizing reactance costs you.
2. Check the slip against theory. Solve the torque expression of §16.3 for the slip that develops 0.4 pu with this machine's parameters and it returns $s=0.01103$. The run settles at $0.01103$. A full EMT model integrating four flux linkages and a swing equation, and a phasor equivalent circuit from a first course in machines, land on the same number — which is exactly what should happen once the transients are over.
3. Move the rotor resistance. Double Rr1 to 0.05 pu. The settled slip goes to 0.0220 — almost exactly double, as the linear region predicts. Note that the machine is now measurably less efficient at the same torque, because rotor loss is $s$ times air-gap power.
4. Watch the connection transient. The first three cycles are the price of switching a machine on directly. Full voltage lands on windings whose working flux has yet to build, so the machine draws whatever the leakage reactances allow. Phase current peaks at 16.5 kA around 6.9 ms — 9.7 times the 1.70 kA rated peak once the DC offset is counted, against 0.92 kA in the settled state — and Te spikes to +9.6 pu, dragging the rotor down to w = 0.976 before useful torque develops. This is a reduced version of the direct-on-line problem, since the rotor was already at speed.
5. Change the inertia. Raise H and the pull-in takes proportionally longer, exactly as $2Hd\omega /dt=T_m-T_e$ says. The settling point does not move, because it is set by the torque balance and not by inertia.

Starting from standstill

A genuine direct-on-line start from $s=1$ is not a supported initialization path for this model today. Use init_slip to place the rotor where you want it and study the pull-in from there, or use step 4 above for the switching inrush. The physics of a full DOL start — 0.56 pu of torque against five times rated current, for however many seconds the inertia demands — is in §16.3, and the torque-speed figure tells you what it would look like.

16.7 Summary

• An induction machine is a synchronous machine with no field winding and a shorted rotor. It develops torque only at nonzero slip, and always absorbs reactive power.
• Rotor currents run at slip frequency, $s{f}_0$, which is why rotor resistance dominates at running speed and leakage reactance dominates at standstill.
• The $dq$ model is written in a synchronously rotating frame; the rotor equations have zero on the left and a slip-speed coupling term.
• A double cage or deep bar makes the effective rotor resistance frequency-dependent, buying starting torque without spoiling running efficiency.
• The torque-speed curve has a linear region near synchronous speed ($T\propto s/R_r$), a breakdown peak whose height scales as $V^2$ and whose slip scales as $R_r$, and a low starting torque at $s=1$.
• Load torque is usually speed-dependent; the fan law $T\propto {\omega }^2$ is the common case and makes starting easy.
• Motors dominate dynamic load behaviour. Depressed voltage can stall them, and stalled motors hold the voltage down — the mechanism behind delayed voltage recovery and a major factor in voltage stability.

16.8 Problems

Problem 16.1. A 4-pole 60 Hz induction motor runs at 1,764 rpm. What is its slip, and what is the frequency of its rotor currents?

Solution 16.1

Synchronous speed for a 4-pole machine at 60 Hz is $120\times 60/4=\mathrm{1,800}$ rpm. So

$$s=\frac{1800-1764}{1800}=0.02,$$

two per cent, and the rotor currents run at $s{f}_0=0.02\times 60=1.2$ Hz.

Problem 16.2. Show that doubling the rotor resistance doubles the slip at which breakdown occurs but leaves the breakdown torque unchanged.

Solution 16.2

From §16.3,

$$s_{max}=\frac{R_r}{\sqrt{R_{th}^2+(X_{th}+X_{lr}{)}^2}},T_{max}=\frac{|V_{th}{|}^2}{2[R_{th}+\sqrt{R_{th}^2+(X_{th}+X_{lr}{)}^2}]}.$$

$R_r$ appears linearly in the numerator of $s_{max}$ and nowhere at all in $T_{max}$. Doubling it therefore doubles the breakdown slip and leaves the peak height alone.

Physically: at breakdown the rotor branch resistance $R_r/s$ has grown to match the reactive part of the circuit. Doubling $R_r$ means that match now happens at twice the slip, but the impedance magnitude at the match is the same, so the same current flows and the same torque is developed.

Problem 16.3. A motor with a breakdown torque of 2.06 pu is carrying 0.8 pu. A nearby fault depresses its terminal voltage to 0.55 pu for 150 ms. Will it stall?

Solution 16.3

Breakdown torque scales with the square of voltage: $2.06\times {0.55}^2=0.62$ pu. That is below the 0.8 pu load, so throughout the fault the machine cannot develop enough torque at any slip, and it decelerates continuously.

Whether it actually stalls depends on the inertia and the duration. With a net decelerating torque of at least $0.8-0.62=0.18$ pu and, say, $H=1$ s, the speed drop over 150 ms is at least $0.18\times 0.15/(2\times 1)=1.4\mathrm{\%}$ — which is comparable to the running slip, so the machine ends the fault near or past its breakdown point. It is a marginal case, which is exactly why it needs simulating rather than estimating. Note that the real decelerating torque is larger than 0.18 pu, since developed torque falls further as slip grows past breakdown.

Problem 16.4. A fan-driven motor and a conveyor-driven motor are identical and both carry 0.6 pu at rated speed. Which is harder to start direct-on-line, and by how much?

Solution 16.4

The conveyor, by a wide margin. Its load is constant-torque ($A=1$, $C=0$), so it demands 0.6 pu from standstill — but the machine only develops 0.56 pu at $s=1$, so it never accelerates at all. The motor sits at zero speed drawing locked-rotor current until protection trips it.

The fan follows $T=0.6{\omega }_r^2$, so it demands nothing at standstill and the full 0.56 pu of starting torque is available to accelerate the inertia. It runs up cleanly, with the accelerating margin shrinking only as it approaches the operating point.

This is why constant-torque loads are the ones that get soft starters, and why the torque-speed curve is always drawn with a load curve on it.

Problem 16.5. Why does an induction machine absorb reactive power when generating, unlike a synchronous machine?

Solution 16.5

A synchronous machine makes its own flux with a DC field winding, and by over-exciting that winding it can produce more flux than it needs and export the surplus as reactive power.

An induction machine has no field winding. Every bit of its air-gap flux has to be magnetizing current drawn from the terminals, and the air gap makes that current large — $X_m\approx 3$ pu instead of a transformer's 100. Reversing the direction of real power by driving the shaft above synchronous speed does not change where the flux comes from. This is why fixed-speed induction generators need capacitor banks or a nearby synchronous source, and why modern wind turbines use converters instead.

Problem 16.6. During a system study you notice the voltage at a distribution substation takes 4 seconds to recover from a fault that was cleared in 5 cycles. The load model is constant-power. What should you suspect, and what would you change?

Solution 16.6

Suspect that the slow recovery is not being produced by the load model at all, because a constant-power load has no dynamics and cannot delay a recovery — so the 4 seconds must be coming from something else in the case, and is worth tracing before it is believed.

The physical mechanism that does produce this signature is stalled induction motors (§16.5). To represent it, replace the aggregate constant-power load with a mixture that includes explicit induction machine models, sized to the motor fraction of the load — typically 50 to 70 % on a summer-peaking distribution feeder, much of it single-phase air-conditioner compressors with low inertia and no undervoltage protection.

16.9 References

• P. Kundur, Power System Stability and Control, McGraw-Hill — Section 7.1 for load modelling in general; Section 7.2 for the induction motor model, its equivalent circuit, double-cage rotors and the torque-slip relation; Section 7.2.6 for motor stalling and its system consequences.
• J. Arrillaga and N. R. Watson, Power Systems Electromagnetic Transients Simulation, IET Power and Energy Series 39 — Chapter 7 for the EMT treatment of rotating machines and their network interface.
• IEEE Task Force on Load Representation for Dynamic Performance, "Load representation for dynamic performance analysis," IEEE Transactions on Power Systems, 1993 — the standard reference for aggregate load composition.
• NERC, Fault-Induced Delayed Voltage Recovery technical reference — the industry account of the stalling phenomenon in §16.5.

---

Previous: Chapter 15 — The synchronous machine · Next: Chapter 17 — Modeling control systems.