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MARINA OIC-EW Theoretical Exam study materials
MARINA OIC-EW Theoretical Exam study resources

MARINA OIC-EW Theoretical Exam: Marine Engineering Knowledge (Main Propulsion and Auxiliary Machinery)

MARINA OIC-EW Theoretical Exam · Work through your exam, one topic at a time.

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Marine Engineering Knowledge (Main Propulsion and Auxiliary Machinery)

Marine Engineering Knowledge (Main Propulsion and Auxiliary Machinery)

This domain tests whether you can explain what is happening inside the propulsion plant and auxiliaries from the readings you record, and whether you can put numbers on it.

Propulsion plant

  • Slow-speed two-stroke: uniflow scavenging through scavenge ports, exhaust valve, separate cylinder lubrication (high-BN oil), constant-pressure turbocharging, crosshead guides lubricated from system oil.
  • Parameters that reveal condition: scavenge air pressure, exhaust gas temperature after each unit, Pmax and compression pressure from pressure cards, fuel index, turbocharger speed, air cooler outlet temperature.
  • Rising exhaust temperature at constant load with falling Pmax points to late injection, a worn fuel valve or a leaking exhaust valve.

Auxiliaries

  • Boiler: water level, three-element feed control, water treatment, safety valve lifting test under steam.
  • Purifier: correct gravity disc for the density at the actual separation temperature, back pressure, sludge discharge interval.
  • Steering gear: two independent power units; SOLAS requires the main gear to put the rudder from 35 degrees on one side to 30 degrees on the other in not more than 28 seconds at rated speed.

Worked example

Six-cylinder two-stroke, bore 500 mm, stroke 1 200 mm, mean indicated pressure 19 bar, 120 rpm, mechanical efficiency 0.90.

  • Area = pi/4 x 0.5 squared = 0.1963 m2; swept volume = 0.1963 x 1.2 = 0.2356 m3.
  • Work per cycle = 1 900 kPa x 0.2356 m3 = 447.6 kJ.
  • Two-stroke: one power stroke per revolution, so 447.6 x (120/60) = 895 kW per cylinder.
  • Indicated power = 6 x 895 = 5 372 kW; brake power = 5 372 x 0.90 is about 4 835 kW.

Traps

Applying the four-stroke factor (one power stroke per two revolutions) to a two-stroke; quoting indicated power as brake power; mixing bar and kPa; using the bore instead of the radius; diagnosing a scavenge fire (smoke and high temperature at the scavenge space) as an exhaust valve fault.