Starting, Starters and Braking

Separate the field current from the armature current. A shunt motor splits the line current between the two:

Fix the machine constant now and reuse it all set. At 1000 rpm,

It is also convenient to note \(E_a/N = 240/1000 = 0.24\) V per rpm, since the field is held constant throughout this set.

Direct-on-line starting. At standstill the armature is turning at zero speed, so it generates nothing and the only limit is its own resistance:

Twenty-five times full load, and with it twenty-five times full-load torque applied to a stationary shaft. The commutator would flash over, the fuses would open, and if they did not, the coupling might not survive. The rule is general: the direct-on-line multiple is \(V_t/(I_{a,fl}R_a)\), and for any efficient machine \(I_{a,fl}R_a\) is a small percentage of \(V_t\), so the multiple is always large.

Size the starter. Permitting \(I_{\max} = 2.5\times50 = 125\) A:

Nine tenths of the starting resistance is in the starter and one tenth in the machine. The starter must dissipate \(125^2\times1.8 = 28.1\) kW at the instant of switching — briefly, which is why starter grids are made of bare wire on porcelain and are rated for seconds, not hours.

When may it all be removed? With \(R_{\text{ext}} = 0\) the current is \((250-E_a)/0.2\), and this must not exceed 125 A:

The motor must reach 93.75% of full-load speed before the last of the resistance can go. Removing it all at once at, say, half speed would give \((250-120)/0.2 = 650\) A. That is why a starter has intermediate studs, which is the next problem.

Why the resistances form a geometric progression. At the moment the starter moves from stud \(m\) to stud \(m+1\) the speed, and therefore the back emf, cannot change instantaneously. So the same \(E_a\) appears in both equations:

The ratio is the same at every stud, which is what makes the design a one-line calculation instead of a step-by-step simulation.

Count the studs. The progression starts at \(R_1 = V_t/I_{\max} = 2\ \Omega\) and must finish at the bare armature resistance, since the last stud has no external resistance left:

A starter cannot have half a stud, so take \(n = 4\). Rounding up is compulsory: with three studs the ratio would have to be larger, letting the current fall further than 50 A between steps — a slower, weaker start — or, if \(I_{\max}\) were kept, exceeding it.

Recompute the ratio for four studs. The upper limit of 125 A is the hard constraint, so hold \(R_1 = 2\ \Omega\) and let \(k\) adjust:

The current now swings between 125 A and 58 A instead of 125 A and 50 A. That is on the safe side of the specification: the starter is advanced a little earlier, the mean accelerating torque is higher, and the run-up is quicker.

The resistance at each stud is the progression \(R_1,\ kR_1,\ k^2R_1,\ k^3R_1\), and the sections of the starter are the differences between consecutive studs:

The sections check against Problem 1's total, and their sizes fall in the same ratio \(k\) — the first section is more than half the whole starter, because at standstill there is no back emf to help.

Follow the run-up stud by stud. The stud is left when the current has decayed to 58.0 A, at which moment \(E_a = V_t - I_{\min}R_m\) and \(N = E_a/0.24\):

Check the third row against Problem 1: the last stud is entered at 225 V, that is 937.5 rpm — precisely the speed at which the whole resistance could safely have been removed. The two calculations meet, as they must.

Reversing the supply does not reverse the back emf. The armature keeps turning the same way and its field is unchanged, so \(E_a\) keeps both its size and its direction — but the supply now opposes it no longer, it assists it. The two act together round the armature loop:

Nearly twice the supply voltage, which is why plugging always needs more resistance than starting does.

Size the braking resistance:

Compare Problem 1: starting the same machine at the same current limit needed only 1.8 Ω. Plugging asks for more than twice as much, because the machine's own emf is now helping the current instead of opposing it.

The braking torque. The flux is unchanged, so torque is simply proportional to current — and the current now flows the other way through the armature, so the torque opposes rotation:

Two and a half times full-load torque, applied against the direction of motion. Plugging is the fiercest of the three methods for exactly this reason: it is the only one that can develop full braking torque all the way down to zero speed.

Account for the power. Three quantities, and they must balance:

This is the damning feature of plugging: the supply pays 31 kW to stop the machine, and every watt of it, together with the 30 kW of kinetic energy being extracted, ends up as heat in the resistor. Nothing is recovered, and the resistor must be rated for 61 kW — for a few seconds, but 61 kW nonetheless.

What happens at zero speed. As the machine slows, \(E_a\) falls and the current falls with it, reaching \(250/3.92 = 63.8\) A at standstill — still 1.28 times full-load current, and still producing 1.28 times full-load torque. The supply must therefore be opened at the moment the shaft stops, or the motor will accelerate smoothly away in the opposite direction. A zero-speed relay or a centrifugal switch does this in practice.

Once the supply is removed the machine is a generator. Its field is still excited and its armature still turning, so it still generates 240 V — but now that emf drives current round a closed loop of \(R_a\) and \(R_b\) instead of opposing a supply:

Less resistance than plugging needs (3.72 Ω) and more than starting needs (1.8 Ω external), because the driving emf here is \(E_a\) alone rather than \(V_t+E_a\) or \(V_t\).

Initial torque and power. The current is reversed relative to motoring, so the torque opposes rotation:

The same braking torque as plugging, at the same current — unsurprising, since torque depends only on flux and current — but the supply contributes nothing. All 30 kW comes from the rotating mass, which is precisely what is being removed.

The torque decays with the speed, because the emf that produces the current does. With the flux held constant, \(E_a = 0.24N\):

Everything is proportional to speed, so the braking effort fades exactly when it is most wanted. Dynamic braking cannot hold a load at rest and cannot bring a machine to a definite stop; the last few revolutions must be taken by a mechanical brake, or by switching to plugging.

Compare the three methods at the same 125 A:

The emf follows the speed. With the flux unchanged, \(E_a = 0.24N\):

And 264 V is greater than the 250 V supply. That single inequality is the whole of regenerative braking.

The current reverses of its own accord. Nothing was switched; the armature equation simply changed sign:

While motoring, \(I_a = (V_t-E_a)/R_a = (250-240)/0.2 = 50\) A into the machine. The 10 V of net emf that used to drive current one way is now 14 V driving it the other. Note how violent the change is: a 10% rise in speed swings the armature current from \(+50\) A to \(-70\) A, because the whole 120 A swing appears across 0.2 Ω.

Torque and power. Reversed current means reversed torque, so the machine now opposes the descending load:

Close the energy balance. The descending load supplies the mechanical power, and the only loss on the way is in the armature copper:

Ninety-five per cent of the energy the load gives up is put back on the mains. Set that against Problem 3, where 61 kW was turned into hot air, and the attraction of regeneration is obvious — as is the reason every modern lift, crane and electric train uses it.

The lower limit. Regeneration needs \(E_a > V_t\), which at rated flux means

That is the ideal no-load speed of the motor. Below it the machine motors, above it it generates, and it passes through the boundary without any switching whatever — only the current changes sign.

To regenerate at 1000 rpm the flux must be raised instead. Since \(E_a \propto \phi N\), obtaining 264 V at 1000 rpm requires

A 10% increase in field current, obtained by cutting out field rheostat resistance, and the same 70 A flows back — at a slightly larger braking torque, because the flux is larger. This is why regenerative braking on a hoist is arranged with a strengthened field: the load can then be lowered at, rather than above, normal speed.

Write Newton's law for the shaft. The only torque acting is the electrical braking torque, and it opposes rotation:

The speed appears on both sides, because the emf that produces the braking current is itself proportional to the speed. The result is a first-order equation with no forcing term — the source-free response of Set 18 of the circuits book, with mechanical quantities in place of electrical ones.

Identify the time constant by comparing with \(d\omega_m/dt = -\omega_m/\tau\):

A useful sanity check: the initial deceleration is \(T_b/J = 286.5/5 = 57.3\) rad/s², and \(\omega_{m0}/57.3 = 104.72/57.3 = 1.83\) s — the time the machine would take to stop if the initial torque were maintained, which is the standard interpretation of a time constant.

The time to reach 10% of speed:

And exponentially, never quite zero: at 100 rpm the braking torque is down to a quarter of full-load torque, at 10 rpm to 2.5%, and friction alone finishes the job. A definite stop needs a mechanical brake, which is why hoists and lifts carry one regardless of the electrical scheme.

Energy: all of it comes from the rotating mass. With no supply connected and no load torque, the kinetic energy is the entire budget:

It divides between the two resistances in proportion to them, because they carry the same current at every instant:

The 2.9 kJ that lands in the armature is the reason repeated braking overheats a machine even though the rheostat takes nine tenths of the heat.

Check the energy against the power. The power decays as the square of the speed, so it has half the time constant of the speed:

The two routes to 27.4 kJ — mechanical kinetic energy and the integral of electrical power — agree exactly, which is the strongest possible check that the model is consistent.