Pancyprian TS Marine Mech. II study materials
Pancyprian TS Marine Mech. II study resources

Pancyprian TS Marine Mech. II: Ship electrical installations, automation, instrumentation and alarm systems

Pancyprian TS Marine Mech. II · Work through your exam, one topic at a time.

All guide sections

Ship electrical installations, automation, instrumentation and alarm systems

This domain tests safe operation of the ship's electrical plant, from generation and distribution through protection and emergency power, plus the automation, instrumentation and alarm systems that allow unattended machinery spaces. Expect numerical questions on current, load sharing and synchronising.

Generation and distribution

  • Three-phase 380 V/50 Hz or 440 V/60 Hz; alternator voltage is held by the AVR and frequency by the prime mover governor.
  • Parallel operation needs equal voltage, equal frequency, the same phase sequence and a small phase difference; close the breaker with the synchroscope rotating slowly in the fast direction and passing 12 o'clock.
  • kW sharing is set by governor droop, kvar sharing by excitation. Reverse-power protection trips a generator that starts motoring; preferential tripping sheds non-essential load.
  • Insulation resistance is measured with a 500 V tester, minimum 1 MΩ; earth fault alarms indicate an insulated system losing its insulation.

Emergency power

  • The emergency generator must start and connect automatically within 45 s and feed the emergency switchboard; a battery-backed transitional supply covers the gap; capacity for cargo ships is 18 hours.

Instrumentation and alarms

  • 4-20 mA two-wire transmitters use a live zero, so a broken wire reads as a fault rather than as zero.
  • PID control: proportional responds to error, integral removes offset, derivative anticipates; too much proportional gain causes cycling.
  • Alarms need a set point, a dead band and a delay. UMS requires alarm transfer to the bridge and engineer's call, a deadman alarm in the control room and periodic rounds.

Worked example

A 640 kW generator at 440 V and power factor 0.8: I = P/(√3 × U × cos φ) = 640,000/(1.732 × 440 × 0.8) = 1,050 A. At 500 kW and 0.8 pf the current is 820 A; at 500 kW and 0.9 pf it falls to 729 A. The reasoning shows why voltage, power factor and apparent power must all be stated before a current can be trusted.

Traps

Using 440 V as phase voltage; synchronising with the synchroscope rotating in the wrong direction; forgetting reverse-power protection; assuming the emergency generator supplies power instantly; treating an alarm dead band as a faulty sensor.