Chapter 02 of 11 · ETO

Generation & Distribution

Get the mechanism behind excitation, synchronising and droop right and the plant stops feeling like a set of memorised rules. This chapter works through the reasoning and the numbers behind each control loop, not just the headline facts.

Worked examples3, fully stepped
Read time≈ 16 min
PrerequisiteNone

1. How the alternator actually makes its power

Every marine alternator you will sit an exam question on is a salient-pole synchronous machine: a rotating field on the shaft, a stationary three-phase winding in the stator, and nothing touching except a bearing. The rotor carries a DC field winding wound round projecting poles; feed that winding with direct current and it becomes a rotating electromagnet. As the prime mover turns it, the rotating field sweeps past the stator conductors and induces a three-phase EMF in them — that induced EMF is the machine's entire output, and everything else in this chapter is about controlling it.

The field current itself has to come from somewhere, and on a ship it almost always comes through a brushless exciter rather than slip rings and brushes. A small exciter armature is mounted on the same shaft, wound to produce AC; that AC is rectified by a ring of diodes also spinning on the shaft, and the resulting DC is fed straight into the main rotor winding — no sliding contact, no carbon dust in an engine room that does not need any more of it. The exciter's own field is what the AVR actually controls: turn up the exciter's field current a little and, a fraction of a second later, the main rotor's field current follows, and with it the machine's terminal voltage.

The one relationship worth fixing in your head before anything else is how frequency is set. It is not a control loop at all — it falls straight out of the machine's geometry and the speed the governor holds it at.

f = pN / 120 output frequency from pole count p and speed N (rpm)
The key idea

Once the number of poles is cast in iron, frequency is purely a function of shaft speed. The governor sets frequency by holding speed; the AVR has no route to frequency at all — it only ever touches the field, and through the field, voltage.

2. The AVR loop and why voltage sags under load

The AVR is a closed loop, and it is worth thinking of it as exactly that: it samples the generator's terminal voltage, compares it against a reference, and drives the exciter field up or down to null out the difference. Left alone, a synchronous machine's terminal voltage does not stay put as load is applied — armature reaction sees to that. Current flowing in the stator windings sets up its own magnetic field, and depending on the power factor of the load, that field either opposes or reinforces the rotor's field. A lagging (inductive) load — the normal case, with motors and lighting ballasts on the board — demagnetises the machine and terminal voltage sags. A leading load magnetises it further and voltage rises. The AVR's job is to counteract exactly this, raising excitation as lagging load comes on and trimming it back as it comes off, fast enough that the lamps do not visibly flicker.

Where this gets interesting is with two or more generators in parallel. Each machine's AVR is trying to hold its own idea of what the busbar voltage should be, and if their references do not agree exactly — and in practice they never agree to the last millivolt — one machine ends up trying to push more excitation onto the shared bus than the other. The result is a reactive circulating current flowing between the machines that does no useful work, only heats the windings and eats into the reactive capacity available for the actual load. Modern AVRs counter it with reactive droop, sometimes called cross-current compensation: each AVR is made to accept a small voltage droop as reactive current increases, which stabilises reactive sharing the same way governor droop stabilises real power sharing. The two mechanisms are cousins, and mixing them up under pressure — reaching for the governor to cure a reactive problem — is the single most common way marks are lost in this area.

3. Synchronising: why each of the four conditions matters

The examiner will accept "voltage, frequency, phase sequence, phase angle" as the list, but the mark is really in explaining what each one protects against, because that is what tells you how to check it. Equal voltage matters because, at the instant of closing, any difference appears directly across the breaker contacts and drives a circulating current limited only by the machines' own reactance — small in magnitude compared to a fault, but a needless stress on a healthy machine. Equal frequency matters because a persistent mismatch means the two machines' phasors are continuously rotating relative to each other; close on a bad moment and you close well away from the in-phase point.

Phase sequence is the one that catches people out after maintenance work — if a set of terminals has been reconnected, swapped conductors on an alternator or a shore connection can reverse two phases, and the synchroscope will still appear to behave sensibly even though the incoming machine's phase rotation no longer matches the bus. It is checked once, formally, with a phase-sequence indicator or phasing lamps after any rewiring, precisely because the synchroscope alone cannot be trusted to reveal it. Phase angle at the instant of closing is the condition the synchroscope exists to manage moment to moment: even with voltage, frequency and sequence all correct, closing while the two machines' voltage waveforms are out of step drives an instantaneous torque transient through the coupling, proportional to the sine of the angle between them. Close near 180° out of phase and the transient can be severe enough to shear a coupling or trip every protective device on the board simultaneously.

In practice the synchroscope pointer is watched creeping slowly in the "fast" direction — the incoming machine deliberately run a shade quick — and the breaker is closed just ahead of the in-phase mark to allow for its own closing time, as worked through numerically below. A check-synchronising relay is normally fitted as a back-stop, refusing to let the breaker close outside a narrow angle window regardless of what the operator does — a safeguard, not a replacement for watching the instrument.

4. Real and reactive load sharing, and why droop is stable

Once two machines are paralleled, the split of load between them is set by two entirely separate controls, and keeping that separation straight is most of what this topic examines. The governor, acting on the fuel rack or throttle, sets how much real power (kW) a machine delivers. The AVR, acting on excitation, sets how much reactive power (kVAr) it delivers. A circulating current with a reactive character — one that shows up mainly as a difference in the machines' power factors rather than a difference in shaft loading — is an excitation problem, and adjusting a governor to chase it does nothing but shift kilowatts around while the reactive imbalance sits there untouched.

Droop is what makes kW sharing stable rather than a tug-of-war. An isochronous governor — one that tries to hold exactly one fixed speed regardless of load — works perfectly for a single machine on its own, but put two of them in parallel and both will fight to hold the bus at their own reference frequency; the one with the marginally higher setting will try to take all the load, driving its own speed down slightly, which its isochronous governor reads as an error and corrects for by opening the fuel rack further, taking still more load. Droop breaks that fight by deliberately letting frequency fall a little as load rises — a governor with 4% droop, for instance, runs 4% faster at no load than at full load. With that slope built in, any machine that starts to take more than its share sees its own frequency sag, which the shared droop characteristic reads as "ease off" for every machine on the bus at once, settling naturally into a stable split roughly proportional to each machine's rating, as worked through below.

Vessels with a common load-sharing bus — typically DP-capable ships — often run their governors isochronous instead, but keep the same stability by exchanging a load-sharing signal electronically between control units so that each machine's fuel demand is trimmed in proportion to the others, doing in software what droop does mechanically.

5. Protection, discrimination and preferential trip

A ship's electrical protection has to do two jobs that can pull in different directions: clear a fault fast enough to prevent damage, and clear only the faulty section, leaving everything else running. The second job is discrimination, and it is achieved by grading protective devices two ways at once. Current grading sizes fuses and instantaneous trip settings so that a device closer to the fault sees a larger fault current than one further back and is set to operate on it; time grading adds a deliberate, increasing delay to devices further up the distribution tree, so that even where fault current is similar at two points, the device nearest the fault is given the shortest time to clear it before anything upstream is allowed to act. Get the grading wrong — settings too close together, or a device replaced with one of different characteristics — and a fault on one circuit can trip a main switchboard breaker instead of the final sub-circuit fuse, blacking out far more of the ship than the fault justifies.

Generators carry protection beyond simple overcurrent. Reverse-power protection watches for a paralleled machine's real power flow going negative — its prime mover has stopped contributing and the machine is now being driven as a motor by the others on the bus, which for a diesel engine risks damage from being spun without fuel and lubrication, so the relay trips it off promptly. Over- and under-frequency and over- and under-voltage protection guard the limits the AVR and governor are meant to hold; earth-fault protection catches insulation failure before it develops into something worse.

Preferential (load-shedding) trip sits ahead of all of this as a first line of defence against overload rather than a fault. As load climbs towards a running set's capacity, non-essential circuits — air conditioning and ventilation first, typically — are shed automatically in short, staged delays, buying the system time to shed enough load that the generator's own overload protection never has cause to operate. It is not a fault condition and not a system failure; it is the protection scheme doing exactly the job it was designed for, and treating it as an alarm to investigate rather than routine operation is a common misreading in this topic.

6. Blackout recovery, in the order it actually happens

A total blackout is the scenario every part of this chapter exists to avoid, but the exam still expects you to know the recovery sequence cold, because a slow or disordered recovery turns a survivable incident into a much longer one. The emergency generator is designed to answer first: on loss of the main switchboard it autostarts and comes on load at the emergency switchboard within a short, defined time, energising steering gear, navigation lights, emergency lighting, and fire and bilge pumps required for immediate safety — services chosen specifically because they cannot wait for the main plant to be re-started.

Restoring the main supply is then a sequence, not a single action, and rushing it tends to produce a second blackout rather than avoiding the first one for longer. A main generator is started, run up and synchronised onto its own switchboard section exactly as covered above — the blackout does not suspend the need for correct synchronising conditions. Once it is established and carrying its own housekeeping load, the bus tie to the emergency switchboard, or to other sections of the main board, is closed to consolidate the supply. From there, auxiliary machinery needed to support the main engine and further generation — cooling pumps, fuel and lube-oil systems, compressed air — is restored before large loads such as cargo pumps or bow thrusters are reconnected, and even then typically one at a time with a pause to confirm the system is stable before the next is brought on. Restoring everything at once simply reproduces the overload that may have caused the blackout in the first place, against a plant that has not yet built back its spinning reserve.

7. Worked examples

Three worked problems follow: synchronising an incoming machine against a breaker's own operating time, resolving how load actually splits between two generators on droop and correcting frequency without disturbing that split, and checking a preferential-trip stage against a generator's overload timer.

Worked example 1

Synchronising an incoming alternator onto a live bus

No.2 generator is running up to close onto the bus, which No.1 already supplies at a steady 60 Hz. No.2 is an 8-pole alternator turning at 906 r/min, and its breaker takes 150 ms to close from the moment the synchronising switch is operated. At what point on the synchroscope should the switchboard operator close the breaker?

Given

Running bus frequency, f₁ = 60 Hz Incoming alternator: p = 8 poles, N = 906 r/min Breaker closing (operating) time = 150 ms

  1. Find the incoming machine's frequency.

    f₂=pN / 120 =8 × 906 / 120 =60.4 Hz

    Frequency is fixed by pole count and shaft speed alone — the AVR has no influence over it at all.

  2. The slip between the two frequencies is what drives the synchroscope round.

    Δf=f₂ − f₁ =60.4 − 60 =0.4 Hz (incoming fast — correct direction)

    Check it is small, and that the incoming machine is running fast, before going any further.

  3. Convert that slip into the time for one full sweep of the pointer.

    T=1 / Δf =1 / 0.4 =2.5 s per revolution

    So the closing instant can be judged by eye.

  4. Allow for the breaker's own closing time.

    ω=360° / T =360 / 2.5 =144 °/s lead angle=ω × t_close =144 × 0.15 =21.6°

    The close command has to be given before the in-phase point, not at it, so the contacts actually meet while the phases coincide.

AnswerOperate the breaker when the synchroscope pointer is about 21.6° before the 12 o'clock (in-phase) mark — roughly the 11 o'clock position — moving slowly in the fast direction, so the contacts actually touch as the phases coincide.

The trap: waiting for the pointer to reach 12 o'clock exactly and closing then. By the time the breaker mechanism finishes moving, the phase has already slipped past the in-phase point and the sets close with an angle between them.

Worked example 2

Splitting load between two generators on matched droop

No.1 (600 kW) and No.2 (400 kW) generators are paralleled on the switchboard, both governors set with the standard 4% droop from the same 60 Hz rated point. Together they are carrying a 750 kW load. Work out how that load is actually splitting between them, and how to bring system frequency back to exactly 60 Hz without upsetting the split.

Given

No.1 rated 600 kW, No.2 rated 400 kW, both on 4% droop Droop reference (no-load) frequency for both = 60 × 1.04 = 62.4 Hz Combined load = 750 kW

  1. Set up the droop line for each machine.

    Droop band=62.4 − 60.0 = 2.4 Hz (same for both — droop is set in %, not in Hz per kW)

    Output falls linearly from rated kW at 60 Hz to zero at the no-load frequency. Because both machines carry the same percentage droop, both have the identical 2.4 Hz-wide band.

  2. Write each generator's output as its fraction of that band.

    P₁=600 × (62.4 − f)/2.4 P₂=400 × (62.4 − f)/2.4 P₁ + P₂=750

    And set the two outputs to sum to the known combined load — that pins down the one unknown, the running frequency f.

  3. Solve for f.

    1000 × (62.4 − f)/2.4=750 62.4 − f=750 × 2.4/1000 = 1.8 f=62.4 − 1.8 = 60.6 Hz
  4. Feed f back into each droop line to see the actual split.

    P₁=600 × 1.8/2.4 = 450 kW (75% of rating) P₂=400 × 1.8/2.4 = 300 kW (75% of rating)
  5. Correct the frequency without touching the split.

    New no-load reference=62.4 − 0.6 = 61.8 Hz (both machines) New f=61.8 − 1.8 = 60.0 Hz P₁=600 × 1.8/2.4 = 450 kW, P₂ = 400 × 1.8/2.4 = 300 kW — unchanged

    Because both governors sit on the same percentage droop, winding both speed-setting motors down by the same amount shifts both droop lines down together — the 1.8 Hz gap between reference and actual frequency, and so the 450/300 split, stays put; only the absolute frequency moves.

AnswerSystem frequency settles at 60.6 Hz with No.1 carrying 450 kW and No.2 300 kW — both at 75% of rating, correctly proportioned by size. Trimming both governors down by 0.6 Hz together brings frequency to exactly 60.0 Hz while leaving the 450/300 split untouched.

The trap: nudging only one governor to fix the frequency. That changes the gap on one droop line only, so it changes the split as well as the frequency, and the two machines no longer share load in proportion to their rating.

Worked example 3

Will preferential trip save the generator from an overload?

No.1 generator (600 kW rated) is running at 550 kW when a large cargo pump motor starts and adds 150 kW to the board almost instantly. The generator's overload protection is set to trip at 110% of rating if the overload persists for 10 s. The preferential trip system sheds Stage 1 (HVAC, 80 kW) after a 5 s delay and Stage 2 (a further 40 kW) after 10 s. Does the generator trip?

Given

Generator rating = 600 kW; overload trip = 110% sustained for 10 s Load before the motor starts = 550 kW; motor adds 150 kW Preferential Stage 1: −80 kW at 5 s. Stage 2: a further −40 kW at 10 s

  1. Find the overload trip threshold in kW,.

    Trip threshold=600 × 1.10 = 660 kW

    The actual loading can be compared against it directly instead of working in percentages.

  2. Work out the loading the instant the motor starts.

    Load=550 + 150 = 700 kW 700 / 600=116.7% of rating — above the 660 kW threshold

    Before anything has been shed.

  3. The overload relay only trips once the overload has persisted for its full 10 s delay, so check whether Stage 1 clears the condition before that timer runs out.

    At t=5 s, Stage 1 sheds 80 kW: Load=700 − 80 = 620 kW 620 kW<660 kW threshold — overload condition has ended
  4. Compare the two timers.

    The overload only existed for the first 5 s of the relay's 10 s delay, so the relay's timer never reaches its setting and resets.

AnswerNo — the generator does not trip. Stage 1 alone brings the load to 620 kW, under the 660 kW threshold, five seconds before the overload relay's 10 s timer would have expired. Stage 2 never needs to operate.

The trap: assuming Stage 2 is what saves the generator because it produces the more comfortable final load. Check the overload relay's timer against Stage 1 alone — by the time Stage 2 would fire, the generator was never actually going to trip.

Reference sheet
60-second recall
  1. Lagging loads sag voltage by armature reaction — the AVR raises excitation to hold it, not the governor.
  2. Reverse-power protection trips a paralleled generator before its prime mover is damaged by being driven as a motor.
  3. Discrimination means the device nearest the fault opens first, so only the faulty section is lost.
  4. Trim both governors' speed references by the same amount to shift system frequency without disturbing the kW split.
  5. Blackout recovery restores essential services before cargo or hotel load, one stage at a time, not all at once.