Speed Control and Braking

The two operating slips:

Constant torque fixes the rotor current, and that is the key to the whole calculation. Torque is \(3I_2^2R/(s\omega_s)\), so if \(T\) and \(I_2\) are both unchanged then \(R/s\) is unchanged:

Slip is proportional to total rotor resistance at constant torque — five times the resistance, five times the slip. The relation is exact when \((sX_2)^2\) may be neglected against \(R^2\), and at \(s = 0.2\) with \(R = 0.25\) Ω that is still comfortably true.

The air-gap power is also unchanged, because \(P_g = T\omega_s\) and both factors are fixed:

The rotor current, from the copper loss at the original operating point:

Identical, as constant torque requires. The stator current is essentially unchanged too, so the machine is drawing the same input while delivering less — which is exactly where the loss appears.

The new power balance:

The rotor itself is no hotter than before — same current, same resistance — and all the extra loss is in the external resistors, where it can be got rid of. That is the one redeeming feature of the method.

Set the two conditions side by side:

A 17% speed reduction has cost 3.3 kW — a sixth of the machine's output — burnt in a resistor bank. Take the speed down to half and the rotor circuit would be 50% efficient.

Write the two characteristics that must intersect. On the working part of its curve the motor's torque is proportional to the slip and to the square of the applied voltage; the fan's torque is proportional to the square of the speed:

The first is the low-slip linear approximation of the torque–slip curve, valid while \((sX_2)^2 \ll R_2^2\) — which is why the machine must have a high-resistance rotor for this method to work over any range.

Set the two equal at each voltage and divide, so that \(k\), \(c\) and \(N_s\) all cancel:

The other root, 5.797, is outside the physical range and is discarded.

The speeds:

A 30% voltage reduction has bought an 8% speed reduction. That is the fundamental limitation of the method: slip is a poor lever because it only has the range 0 to \(s_{\max T}\) to work in.

The fan power falls as the cube of speed:

The rotor copper loss, before and after:

The rotor loss went up while the output went down — that is what slip control always does. But it went up by only 763 W. Had the load been a constant-torque one, the air-gap power would have stayed at 16.67 kW and the rotor loss would have been \(0.1725\times16667 = 2875\) W. The falling torque demand is what keeps the fan case tolerable.

The stalling limit for a constant-torque load. Breakdown torque scales as \(V^2\), so the motor can hold full-load torque only while

At 67.4% of rated voltage the peak of the torque curve has fallen to exactly the load torque, and the slightest further reduction makes the machine stall. And 67.4% of voltage on a constant-torque load buys a speed reduction of only a few per cent before the curve becomes unstable.

Why the fan is different. As the fan slows it asks for less torque, so the operating point moves down the load curve as well as down the motor curve, and the intersection remains on the stable side. A conveyor or a hoist demands the same torque at every speed, so the motor curve must be pushed down while the load line stays put — and the two lose contact at 67.4% voltage. Fans and centrifugal pumps are the only loads for which stator-voltage control gives a usable range, and even then the range is modest and the rotor loss real.

The V/f law and its consequences:

Both the synchronous speed and the reactance are proportional to frequency; the resistance is not, and that single fact is the whole of parts (c) and (d).

The ideal argument. With \(R_1\) neglected the maximum torque is

and every symbol on the right carries a factor of frequency: \(V \propto f\), \(\omega_s \propto f\), \(X_{01}\propto f\). So \(T_{\max} \propto f^2/(f\cdot f)\), a constant. Evaluating at 50 Hz:

This is the promise of constant V/f: constant flux, and therefore the machine's full torque capability available all the way down to zero speed.

Now include the stator resistance, which changes the denominator to \(R_1+\sqrt{R_1^2+X_{01}^2}\):

Read what has happened. The constant-flux promise fails progressively as the frequency falls:

The cause is visible in the second column. At 50 Hz the resistance is a sixth of the reactance and barely matters; at 10 Hz it is comparable with it, and a fixed 0.5 Ω is now taking a serious fraction of a phase voltage that has fallen to 46 V.

The boost. Solve the same expression for the voltage that restores 143.8 N·m at 10 Hz:

In practice the inverter's V/f characteristic is given a fixed offset at zero frequency rather than a curve — the "IR compensation" or "torque boost" setting — chosen to be roughly the stator resistance drop at rated current. Here \(I_{ph}R_1\) at a current of the order of 40 A would be about 20 V per phase, which is the right order.

Speed at 25 Hz on rated torque. At constant torque the machine holds the same slip speed, not the same slip, because the torque depends on the rotor frequency rather than on the slip alone:

The slip has doubled while the slip speed and the rotor frequency (2 Hz) are unchanged. This is the practical signature of V/f control: the speed error caused by loading is a fixed number of rev/min at every setting, which is why open-loop V/f drives hold speed poorly at low frequency and a slip-compensation term is added.

The two synchronous speeds:

Doubling the poles halves the speed exactly; the difference in slip makes the running speeds not quite a factor of two apart.

The 4-pole torque:

Constant-torque changeover. If the reconnection is arranged so the machine develops the same torque on both connections, the output follows the speed:

Half the speed, half the power, same torque. This is the connection used for conveyors, hoists and any load whose torque demand does not change with speed. In winding terms it is the series-delta to parallel-star change.

Constant-power changeover is the other standard arrangement (parallel-star to series-delta). It holds the output at 22 kW on both connections, so the torque on 8 poles becomes

Double the torque at half the speed. Machine tools use it, because the cutting power required is roughly the same at every spindle speed.

Four speeds from one machine. A single consequent-pole winding gives two speeds in the ratio 2 : 1 and nothing else. Two independent windings, each pole-changeable, give four:

The second winding sits in the same slots as the first and is idle whenever the other is energised, so a four-speed machine is physically larger and dearer than a single-speed one of the same rating. Pole-amplitude modulation is the refinement that obtains two speeds in a ratio other than 2 : 1 from a single winding.

What pole changing is and is not good for. The speeds are discrete, so no continuous adjustment is possible — but each of them is a synchronous speed, so the machine runs at a low slip and full efficiency at every setting. Nothing is wasted, which is precisely what rotor-resistance and voltage control cannot claim. It remains common on fans, pumps, lifts and machine tools where two or four fixed speeds are all that is wanted, and it is confined to cage machines, since a wound rotor would have to be re-poled as well.

Reversing two leads reverses the direction of the rotating field. The rotor, still turning forwards at 1440 rev/min, now runs against a field going backwards at 1500 rev/min, so the relative speed is the sum:

Nearly twice line frequency in the rotor bars, and it falls to 50 Hz only when the machine reaches standstill. Deep-bar effects are therefore pronounced during plugging.

The full-load torque, for reference:

Compare the two torques using \(T \propto sR_2'/\left[R_2'^2+(sX)^2\right]\) at constant applied voltage:

Only about half full-load torque, because at \(s' = 1.96\) the rotor reactance \(s'X = 7.84\) Ω swamps the 0.4 Ω of resistance and the rotor power factor collapses to 0.051. Plugging a cage motor gives a modest braking torque and a very large current.

The energy balance. With the field reversed, the air-gap power flows into the rotor from the supply while the kinetic energy also flows into it from the shaft; both end up as rotor copper loss. Writing \(J\,d\omega_m/dt = T\) and integrating from \(\omega_{m0}\) to zero:

Split it into its two sources:

The supply supplies twice as much energy as the load gives up, and every joule of it is dissipated in the rotor. For a machine plugged from very nearly synchronous speed the total is \(3\times\tfrac12J\omega_s^2\) — three times the stored kinetic energy, and three times the energy that a free run-up from rest would dissipate.

Why the supply must be opened at standstill. Nothing in the physics stops the machine at zero speed — the reversed field continues to produce torque and the machine simply accelerates the other way. A zero-speed relay or a plugging switch is therefore an essential part of the scheme.

Maximum braking torque. The condition is the same as for maximum motoring torque, \(R = sX\), applied at \(s' = 1.96\):

2.60 times full-load torque instead of 0.525 — a fivefold improvement, obtained by making the rotor circuit resistive at 98 Hz. It requires a slip-ring machine, which is why heavy plugging duty is a slip-ring application.

Above synchronous speed the slip is negative:

The rotor now cuts the field in the opposite sense, so the rotor emf, the rotor current and the torque all reverse: the machine develops a torque opposing rotation, and power flows from shaft to supply. It has become an induction generator.

The power balance, taken in the direction of flow. Mechanical input first:

What reaches the line:

Compare Problem 5: plugging dissipated 280 kJ to stop the same machine, while regeneration returns 91.5% of what it receives. That difference is the whole argument for regenerative braking wherever the duty allows it.

The machine still needs the supply. The magnetising current comes from the line, so the stator must stay connected and the line must be live: an induction machine cannot regenerate into a dead network without capacitors to supply the reactive power. It also draws lagging reactive power while exporting real power — the current reverses in phase but not in quadrature.

Regeneration by pole changing. Switching the machine of Problem 4 from 4 poles to 8 while it runs at 1440 rev/min drops the synchronous speed to 750, so

A very large negative slip, so a heavy braking torque and a large current — the machine regenerates strongly while decelerating from 1440 to about 712 rev/min, after which it motors again. This is the standard way of obtaining a fast, efficient deceleration between two speeds on a multi-speed drive, and it is why lift motors are often pole-changing machines.

The three modes together:

Dynamic braking is included for completeness: the stator is disconnected from the AC supply and fed with direct current, so the field is stationary and the machine behaves as a synchronous generator feeding its own rotor resistance. It is efficient in the sense that the supply contributes almost nothing, but all the kinetic energy still ends up in the rotor.