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Question 1Diesel propulsion mechanisms and supplied performance models

A 6-cylinder, 2-stroke slow-speed marine propulsion diesel engine has a cylinder bore of 700 mm and a piston stroke of 2,800 mm. At 90 rev/min the mean indicated pressure taken from the indicator cards is 18.5 bar. What is the total indicated power developed by the engine?

  1. 8,971 kW
  2. 35,883 kW
  3. 17,942 kW Correct answer.
  4. 2,990 kW

Explanation

The supplied two-stroke model gives power as mean indicated pressure times total swept volume times revolutions per second. Using SI units gives about 17941.48 kW, so the listed 17942 kW is the nearest supplied approximation. This is indicated power, not brake output, and does not establish actual propulsion performance.

Question 2Diesel propulsion mechanisms and supplied performance models

A main engine develops a steady brake power of 15,000 kW. Over a 24-hour sea passage the vessel burns 61.2 tonnes of fuel with a lower calorific value of 42.7 MJ/kg. What are the specific fuel oil consumption and the brake thermal efficiency of the plant?

  1. SFOC 170.0 g/kWh; brake thermal efficiency 49.6% Correct answer.
  2. SFOC 204.0 g/kWh; brake thermal efficiency 41.3%
  3. SFOC 145.0 g/kWh; brake thermal efficiency 58.1%
  4. SFOC 255.0 g/kWh; brake thermal efficiency 33.0%

Explanation

Work delivered in 24 hours = 15,000 × 24 = 360,000 kWh. Fuel burned = 61.2 t = 61,200,000 g, so SFOC = 61,200,000 / 360,000 = 170.0 g/kWh. The heat supplied per kilowatt-hour is 0.170 kg × 42,700 kJ/kg = 7,259 kJ, and since one kilowatt-hour of work is 3,600 kJ, the brake thermal efficiency is 3,600 / 7,259 = 0.496, that is 49.6% - a figure typical of a modern large two-stroke at its economical load.

Question 3Diesel propulsion mechanisms and supplied performance models

An out-of-phase (draw) indicator card taken from a 2-stroke diesel cylinder shows an abnormally high maximum combustion pressure, with the pressure rise beginning noticeably before top dead centre, and a sharp metallic knock is audible. What is the most probable fault?

  1. Retarded fuel injection timing, causing late burning during the expansion stroke.
  2. Advanced fuel injection timing, so that combustion begins too early while the piston is still rising on compression. Correct answer.
  3. Severely choked scavenge ports, giving insufficient combustion air.
  4. A leaking exhaust valve seat allowing blow-by into the exhaust receiver.

Explanation

Injecting too early gives the fuel a long period in a hot, still-compressing charge, so a large proportion burns before top dead centre. The result is a very steep rate of pressure rise, an excessive maximum pressure superimposed on the compression pressure, and the characteristic knock, along with high mechanical loading of the crosshead, crankpin, and running gear. Retarded timing produces the mirror image: low maximum pressure and high exhaust temperature as combustion trails into the expansion stroke.

Question 4Pumps, separation, freshwater, refrigeration and heat exchange

Under a supplied geometrically similar pump model with constant efficiency and the affinity-law assumptions, a pump at 1200 rpm gives 500 m3/h, 25 m head and 45 kW. At 960 rpm, what are the modeled flow, head and power rather than an actual plant-duty approval?

  1. 450 m³/h; 20.0 metres; 36.0 kW
  2. 320 m³/h; 12.8 metres; 18.4 kW
  3. 400 m³/h; 16.0 metres; 23.0 kW Correct answer.
  4. 400 m³/h; 20.0 metres; 30.0 kW

Explanation

The affinity laws relate performance to speed. The speed ratio is 960 / 1,200 = 0.80. Flow varies directly with speed: 500 × 0.80 = 400 m³/h. Head varies with the square of speed: 25 × 0.80² = 25 × 0.64 = 16.0 metres. Power varies with the cube of speed: 45 × 0.80³ = 45 × 0.512 = 23.0 kW. The cube relationship is why speed control saves so much more energy than throttling a valve, which leaves the pump absorbing nearly full power.

Question 5Pumps, separation, freshwater, refrigeration and heat exchange

What is the mechanism of cavitation in a centrifugal sea water cooling pump, and what condition must be maintained to prevent it?

  1. It is caused by stray electrical currents ionising the water; prevented by bonding the pump casing to the shaft earthing ring.
  2. It occurs when the local static pressure falls below the vapour pressure of the liquid, so vapour bubbles form and then collapse violently against the impeller; prevented by keeping the net positive suction head available above the net positive suction head required. Correct answer.
  3. It occurs when sea water density exceeds 1,025 kg/m³; prevented by adding fresh water to the sea chest.
  4. It is mechanical vibration from an unbalanced rotor; prevented by dynamic balancing.

Explanation

Pressure is lowest at the impeller eye, where velocity is highest. If the absolute pressure there falls below the saturation vapour pressure of the water at its temperature, the water flashes to vapour. The bubbles are swept outward into the higher pressure along the vanes and implode, producing intense local shock loading that pits the metal, along with the characteristic noise of gravel in the pump, vibration, and a fall in head and efficiency. The remedy is on the suction side: keep the net positive suction head available comfortably above the value the pump requires, by maintaining suction head, keeping strainers and sea chests clear, and avoiding excessive suction line losses.

Question 6Pumps, separation, freshwater, refrigeration and heat exchange

A disc-stack centrifugal fuel oil purifier is fitted with a gravity disc (regulating ring) whose bore is too large for the density and separating temperature of the oil being treated. What happens?

  1. The oil-water interface moves outward, past the outer edge of the top disc, so the water seal is broken and clean oil is discharged through the water outlet into the sludge system. Correct answer.
  2. The oil-water interface moves inward and water is carried over with the clean oil to the service tank.
  3. The driving motor trips on overcurrent from excessive centrifugal torque.
  4. The operating water seals the bowl permanently shut so that it cannot desludge.

Explanation

The radius of the interface is set by the balance between the head of oil from the light-phase outlet and the head of water from the gravity disc bore. Increasing that bore lets water leave at a larger radius, which pushes the interface outward towards the bowl periphery. Once the interface passes the outer edge of the top disc there is oil rather than water at the boundary between the two discharge paths, the water seal is lost, and oil runs out with the water - visible immediately as oil in the sludge tank. This is why the rule of thumb is to use the largest disc that does not break the seal, and why a denser oil calls for a smaller disc.

Question 7Hull form, free surfaces and bounded stability calculations

In a supplied small-angle model with an unrestricted rectangular tank surface, displacement is 14000 t, KM 8.40 m, KG 7.65 m, and the slack tank is 20 m by 12 m with oil density 0.95 t/m3. Using correction=rho×L×B^3/(12×displacement), which listed GM values follow?

  1. GM solid 0.95 m; correction 0.350 m; GM fluid 0.600 m
  2. GM solid 0.75 m; correction 0.050 m; GM fluid 0.700 m
  3. GM solid 1.15 m; correction 0.450 m; GM fluid 0.700 m
  4. GM solid 0.75 m; correction 0.195 m; GM fluid 0.555 m Correct answer.

Explanation

Solid GM is 8.40 minus 7.65 = 0.75 m. The supplied surface moment is 20 × 12^3 / 12 = 2880 m4, giving correction 2880 × 0.95 / 14000 = 0.19543 m and effective GM about 0.55457 m. These values use the stipulated unrestricted small-angle geometry; do not generalize identical correction to clipped free surfaces, large angles or every fill condition.

Question 8Hull form, free surfaces and bounded stability calculations

A vessel of 10,000 tonnes displacement enters drydock trimmed 1.00 metre by the stern. The longitudinal centre of flotation is 50 metres forward of the after keel blocks, and the moment to change trim one centimetre is 150 t·m/cm. What upthrust acts on the after blocks at the critical instant?

  1. 300 tonnes Correct answer.
  2. 150 tonnes
  3. 600 tonnes
  4. 75 tonnes

Explanation

As the water level falls the stern takes the blocks first, and the block reaction grows until the vessel lands over her whole length, by which point the trim has been taken out. The force required to remove that trim is P = (change of trim in centimetres × MCTC) / distance from the point of contact to the centre of flotation = (100 × 150) / 50 = 300 tonnes. That force acts upward at the keel and produces a virtual loss of metacentric height of approximately P × KM / displacement, which is why the trim on entering dock is kept as small as practicable.

Question 9Hull form, free surfaces and bounded stability calculations

What defines a hogging condition of the hull girder, and how are the deck and bottom structure stressed?

  1. Weight is concentrated amidships; the deck is in compression and the keel in tension.
  2. The vessel rolls beyond 45° and the stresses are purely torsional.
  3. Buoyancy exceeds weight amidships while weight exceeds buoyancy towards the ends, so the hull arches upward and the strength deck is in tension while the bottom shell and keel are in compression. Correct answer.
  4. Wind pressure on the deckhouse bends the hull.

Explanation

The hull behaves as a box girder loaded by the difference between the weight distribution and the buoyancy distribution along its length. When the net upward force is concentrated amidships - a wave crest amidships, or heavy cargo and ballast at the ends of a lightly loaded midbody - the girder bends convex upward. The upper flange of that girder, the strength deck, is stretched and so is in tension, while the lower flange, the bottom shell and keel, is compressed. Sagging is the reverse case and reverses the stresses. The distinction matters in practice because compression governs buckling, so the deck matters most in sagging and the bottom in hogging.

Question 10Programme scope, watchkeeping and accountable ship systems

Under section 10.3 of the ISM Code, what is critical equipment and what must be done about it?

  1. Equipment and technical systems whose sudden operational failure may result in a hazardous situation - steering gear, the emergency generator, the emergency fire pump, and engine safety trips among them; the safety management system must provide specific measures to promote their reliability, including regular testing of standby arrangements and of equipment not in continuous use. Correct answer.
  2. Any galley or mess room appliance costing more than a set sum.
  3. Equipment that only a classification surveyor may operate.
  4. Machinery exempt from routine inspection and oil sampling.

Explanation

Section 10.3 requires the company to identify equipment and technical systems the sudden operational failure of which may result in hazardous situations, and to provide in the safety management system for specific measures aimed at promoting their reliability. It singles out one measure in particular: the regular testing of standby arrangements and of equipment or technical systems that are not in continuous use, because a standby pump or an emergency generator that is never run is not a standby at all. In practice this drives a documented list of critical equipment, testing intervals in the planned maintenance system, a defined stock of critical spares, and priority given to defects on those items.

Question 11Electrical circuits, machines and digital logic

A 440 V, three-phase, 60 Hz auxiliary alternator supplies a balanced motor load of 450 kW at a lagging power factor of 0.82. What are the apparent power and the line current?

  1. 369.0 kVA and 484 A
  2. 548.8 kVA and 720 A Correct answer.
  3. 548.8 kVA and 1,247 A
  4. 450.0 kVA and 590 A

Explanation

Apparent power follows from the real power and the power factor: S = P / cos φ = 450 / 0.82 = 548.8 kVA. Line current then follows from the three-phase relationship P = √3 × V × I × cos φ, giving I = 450,000 / (1.732 × 440 × 0.82) = 450,000 / 624.9 = 720 A. The same answer comes from I = S / (√3 × V) = 548,800 / (1.732 × 440) = 720 A. Note how poor power factor inflates the current: at unity power factor the same 450 kW would draw only about 591 A.

Question 12Electrical circuits, machines and digital logic

Comparing star-delta starting with direct-on-line starting of a three-phase squirrel-cage induction motor, by what factor are the starting line current and the starting torque reduced while the motor is in star?

  1. Current halves while torque is unchanged.
  2. Current falls to 1/√3 while torque falls to one third.
  3. Both the starting line current and the starting torque fall to one third of their direct-on-line values. Correct answer.
  4. Both current and torque increase threefold.

Explanation

In star the voltage across each phase winding is the line voltage divided by √3, so the phase current is 1/√3 of what it would be in delta. In star the line current equals the phase current, whereas in delta the line current is √3 times the phase current, so the ratio of line currents is (1/√3) ÷ √3 = 1/3. Torque is proportional to the square of the applied phase voltage, and (1/√3)² is also 1/3. That symmetry is the whole trade-off of star-delta starting: the reduction in starting current comes at exactly the same cost in starting torque, so the method suits only machines that can run up against a light load.

Question 13Electrical circuits, machines and digital logic

In the supplied brushless alternator fault comparison, other causes have been checked and the rotating-diode hypothesis is under examination. Which listed description matches the excitation path and the stated hypothesis?

  1. Excitation comes from permanent magnets; the symptom indicates demagnetisation by sea water.
  2. The automatic voltage regulator feeds a stationary exciter field; the exciter armature on the rotor generates alternating current which a rotating diode bridge on the shaft rectifies and delivers to the main field. The symptom points to one of those rotating diodes having failed open circuit. Correct answer.
  3. Excitation is transferred by microwave transmission; the symptom indicates radio interference.
  4. Excitation comes from carbon brushes on a commutator; the symptom indicates worn brush springs.

Explanation

A brushless machine is really two machines on one shaft. The automatic voltage regulator controls the current in a stationary exciter field winding; the exciter armature, which rotates, generates three-phase alternating current; and a diode bridge mounted on the shaft rectifies that current and feeds it directly into the main rotor field through internal connections, so no brushes or slip rings are needed. If one diode in the bridge opens, the remaining diodes can still pass enough current to excite the field at no load, but they cannot supply the much larger field current that a loaded machine demands, so the voltage collapses as load is applied. The condition is confirmed by an exciter current waveform check or a rotating diode test.

Question 14Measurement, feedback, PLC and automation

In a supplied suitably tuned stable PID loop, proportional-only control has a steady-state offset, integral action can remove it and derivative action is used for appropriate damping. Which listed description matches those stated model roles?

  1. Proportional action produces an immediate output proportional to the present error but leaves a steady-state offset; integral action accumulates the error over time and drives that offset to zero; derivative action responds to the rate of change of error and damps overshoot. Correct answer.
  2. Proportional controls cooling water, integral controls lubricating oil, and derivative controls starting air.
  3. Proportional eliminates error, integral introduces a deliberate delay, and derivative amplifies noise.
  4. Proportional acts only during starting, integral only at sea, and derivative only during a blackout.

Explanation

Proportional action moves the final control element in proportion to how far the measured value is from the set point. It responds immediately, but because it needs a residual error to hold any given output it always leaves an offset. Integral action integrates that residual error with time and keeps adjusting the output until the error is zero, at the cost of a slower response and a tendency to overshoot or to wind up if the valve saturates. Derivative action looks at how fast the error is changing and applies a counteracting output in anticipation, which damps overshoot and oscillation but amplifies measurement noise, so it is used sparingly on noisy signals.

Question 15Measurement, feedback, PLC and automation

In a supplied functioning current loop within transmitter compliance limits, why do the series-current relationship and 4 mA live zero help remote instrumentation?

  1. Current loops generate magnetic fields that assist automatic steering.
  2. Digital equipment cannot measure a 0 to 10 V signal.
  3. The current is the same at every point in a series loop, so cable resistance and terminal contact resistance do not corrupt the reading over long shipboard runs, and the live zero of 4 mA means that a broken wire or dead transmitter reads as zero current and can be distinguished from a genuine zero measurement. Correct answer.
  4. Current signals need no power supply of any kind.

Explanation

Two properties account for the preference. First, in a series current loop the same current flows everywhere regardless of the resistance in the path, so voltage drops along a hundred metres of cable, in terminal blocks, or through changes in temperature do not affect the value the receiver sees, whereas a voltage signal is attenuated by exactly those effects. Second, the live zero at 4 mA turns a wiring failure into a detectable condition: a reading below about 3.6 mA or above about 21 mA is outside the measuring range and is used by the receiving system to raise a transmitter fault alarm rather than being mistaken for a valid measurement.

Question 16Measurement, feedback, PLC and automation

What is the operating principle of a Pt100 resistance temperature detector, and what are its nominal resistance and temperature coefficient?

  1. A bimetallic strip that snaps open at 100°C, with infinite resistance at 0°C.
  2. The electrical resistance of pure platinum rises predictably with temperature; the element has a resistance of 100.00 Ω at 0°C and a temperature coefficient of 0.00385 per °C, giving about 0.385 Ω per degree and 138.5 Ω at 100°C. Correct answer.
  3. It generates 100 mV at 0°C by the thermoelectric effect.
  4. It is an optical fibre that changes the colour of transmitted light with pressure.

Explanation

A Pt100 is a passive resistive sensor: platinum is used because its resistance-temperature relationship is stable, repeatable, and very nearly linear over the range met in machinery spaces. The standard element has a resistance of exactly 100 Ω at 0°C and a mean temperature coefficient of 0.00385 Ω per ohm per degree Celsius, so it reads 138.5 Ω at 100°C. Because the element resistance is small, lead resistance matters, which is why Pt100 sensors are wired in three-wire or four-wire configurations to cancel it out.

Question 17Thermal Physics

In a provision refrigeration plant working on the vapour-compression cycle, refrigerant enters the compressor as saturated vapour at 360 kJ/kg and leaves at 400 kJ/kg. It leaves the condenser as subcooled liquid at 240 kJ/kg and expands isenthalpically through the expansion valve. What is the theoretical coefficient of performance?

  1. 3.0 Correct answer.
  2. 4.5
  3. 2.0
  4. 1.5

Explanation

Expansion through the valve is isenthalpic, so the enthalpy entering the evaporator is the same 240 kJ/kg that left the condenser. The refrigerating effect is therefore 360 − 240 = 120 kJ/kg, and the work of compression is 400 − 360 = 40 kJ/kg. The coefficient of performance is the useful effect divided by the work input: 120 / 40 = 3.0, meaning three units of heat are removed from the cold space for every unit of work put in at the compressor.

Question 18Thermal Physics

A counter-flow plate heat exchanger cools jacket water from 85°C to 70°C using sea water entering at 25°C and leaving at 35°C. What is the logarithmic mean temperature difference?

  1. 35.0°C
  2. 55.0°C
  3. 47.4°C Correct answer.
  4. 22.5°C

Explanation

The supplied counter-flow terminal differences are 85 minus 35 = 50 degrees and 70 minus 25 = 45 degrees. Their logarithmic mean is (50 minus 45)/ln(50/45), about 47.456 degrees. The listed 47.4 degrees is the closest supplied approximation. This model assumes the stated flow arrangement and is not complete exchanger design approval.

Question 19Maintenance diagnosis, condition evidence and controlled restoration

During routine laboratory testing of a marine auxiliary diesel engine crankcase lubricating oil sample, the results reveal a significant decrease in kinematic viscosity accompanied by a drop in closed-cup flashpoint from 225°C down to 165°C. What defect does this indicate?

  1. Fuel oil dilution of the crankcase lubricating oil caused by leaking fuel pump seals, defective injector nozzles, or severe unburned fuel blow-by Correct answer.
  2. Severe soot loading and carbon accumulation caused by late exhaust valve closing
  3. Thermal oxidation and oil polymerization resulting from prolonged high-temperature operation
  4. Seawater contamination leaking from the charge air cooler or lube oil plate cooler

Explanation

A sharp drop in lube oil flashpoint (especially dropping below the SOLAS safe boundary of 180°C / 200°C) coupled with a drop in kinematic viscosity is a definitive diagnostic indicator of fuel oil dilution. Unburned marine diesel oil (flashpoint ~60°C) or distillate leaking past fuel pump plunger barrels, defective injector O-rings, or worn piston rings enters the crankcase, thinning the lube oil film and drastically increasing crankcase explosion risks.

Question 20Maintenance diagnosis, condition evidence and controlled restoration

In a controlled comparison with other causes excluded by supporting evidence, a pump/motor vibration pattern has dominant1X/2X components and the supplied alignment hypothesis is being tested. Which listed mechanical issue best matches that hypothesis?

  1. Angular and parallel shaft misalignment between the electric motor and pump shafts Correct answer.
  2. Severe dynamic unbalance of the pump impeller
  3. Cavitation erosion on the impeller suction eye
  4. Electrical phase unbalance in the motor stator windings

Explanation

Shaft misalignment across flexible couplings generates strong forces at once and twice shaft rotational frequency. When parallel and angular misalignment is present, the flexible coupling flexes twice per revolution, producing a characteristic high 2X running speed peak in both radial and axial directions that frequently equals or exceeds the 1X amplitude, along with 180° phase shifts across the coupling.

Question 21Maintenance diagnosis, condition evidence and controlled restoration

In condition-based vibration monitoring of a shipboard centrifugal pump, what specific mechanical failure is indicated by the emergence of non-synchronous peak frequencies corresponding to the Ball Pass Frequency Outer race (BPFO) accompanied by sidebands spaced at running speed (1X)?

  1. Localized fatigue spalling or a surface micro-crack on the outer raceway of the rolling-element bearing Correct answer.
  2. A broken impeller vane creating hydraulic blade-pass pressure pulses
  3. Excessive lubricating oil viscosity causing hydrodynamic boundary friction
  4. Severe mechanical unbalance of the pump shaft coupling

Explanation

The bearing-related characteristic frequency can support an outer-race defect hypothesis in the supplied analysis, but sensor setup, speed, geometry and other causes must be checked. A spectrum alone is not proof of a unique failure or permission to continue operation. Use comparable trends and the actual competent maintenance process.

Question 22Diesel propulsion mechanisms and supplied performance models

What sequence of events leads to a scavenge space fire in a 2-stroke crosshead engine, and what is the first indication in the engine room?

  1. Water leaking from the charge air cooler reacting with scavenge air; indicated by falling scavenge manifold pressure.
  2. Failure of the turbocharger thrust bearing flooding the compressor with oil; indicated by sudden engine overspeed.
  3. Unburnt cylinder oil and carbon residues accumulating in the scavenge trunk because of worn piston rings or leaking injectors, ignited by hot combustion gas blowing past the piston; indicated by a rise in scavenge and exhaust temperature on the affected unit. Correct answer.
  4. Excessive cylinder oil alkalinity reacting with fuel vanadium; indicated by rising jacket water expansion tank pressure.

Explanation

Drainings of unburnt cylinder lubricant, carbon, and fuel from a defective injector collect on the scavenge floor and diaphragm plate. When piston rings are worn, broken, or stuck, hot gas blows past the piston into the trunk and ignites those deposits. The first signs are a rise in scavenge air temperature and in exhaust temperature on that unit, followed by surging or loss of power, hot or blistering paint on the scavenge doors, and smoke from the scavenge drains. The response is to reduce to dead slow or stop, shut off the fuel and cylinder oil to that unit, close the scavenge drains, and let the fire burn out or smother it with the fixed extinguishing connection - never by opening the doors.

Question 23Diesel propulsion mechanisms and supplied performance models

Why is lubrication of the crosshead (top end) bearing in a large 2-stroke engine more demanding than lubrication of the crankpin and main journal bearings?

  1. The crosshead pin runs at three times crankshaft speed and generates large centrifugal loads.
  2. The bearing is exposed directly to combustion gas passing the piston skirt.
  3. The crosshead pin is made of cast lead, which melts at normal running temperatures.
  4. The pin only oscillates through a small arc rather than rotating, and in a single-acting engine the load never fully reverses, so a hydrodynamic wedge cannot form and re-form as it does in a fully rotating bearing. Correct answer.

Explanation

A journal bearing builds its oil film because the shaft rotates continuously and drags oil into a converging wedge, and because the load direction reverses each cycle so the film can be replenished. The crosshead pin does neither: it swings through only a small angle, and gas and inertia loads in a single-acting engine keep it pressed against the lower shell for almost the whole cycle. Manufacturers therefore supply the crosshead through a dedicated booster pump at high pressure, feeding oil into grooves and pockets that lift the pin and re-establish the film during the brief period of reduced load near the top of the stroke.

Question 24Diesel propulsion mechanisms and supplied performance models

A 12,000 kW two-stroke propulsion engine runs at its full continuous rating with a cylinder oil feed rate of 0.80 g/kWh. If the cylinder oil has a density of 0.92 kg/litre, what is the daily consumption of cylinder oil in litres over 24 hours?

  1. 250.4 litres per day Correct answer.
  2. 230.4 litres per day
  3. 272.2 litres per day
  4. 184.3 litres per day

Explanation

Daily work output = 12,000 kW × 24 h = 288,000 kWh. Mass of cylinder oil = 288,000 × 0.80 = 230,400 g = 230.4 kg. Converting mass to volume at the stated density gives 230.4 / 0.92 = 250.4 litres per day. Keeping this figure under review against the measured feed and the drain oil analysis is how the engineer confirms that the feed rate is right for the fuel sulphur and the engine load.

Question 25Diesel propulsion mechanisms and supplied performance models

In an electronically controlled 2-stroke marine diesel engine of the camshaft-less type, how is the exhaust valve opened and closed?

  1. Opened mechanically by a camshaft and pushrod, closed by a heavy helical steel spring.
  2. Opened and closed electromagnetically by solenoid coils on the cylinder head.
  3. Opened by exhaust gas backpressure and closed by gravity counterweights.
  4. Opened hydraulically by high-pressure servo oil metered through an electronically controlled valve, and closed by a pneumatic air spring. Correct answer.

Explanation

In camshaft-less engines the mechanical camshaft is replaced by a hydraulic cylinder unit under electronic control. High-pressure servo oil is admitted through a proportional or fast-acting control valve to an actuator that pushes the exhaust valve open at whatever crank angle the control system selects. Closing is by an air spring - a piston fed with control air - which gives a progressive closing force that cushions the valve onto its seat and avoids the fatigue and breakage risk of a large steel spring, while allowing the valve timing to be varied freely with load.

Question 26Diesel propulsion mechanisms and supplied performance models

In a propulsion shafting installation fitted with a tilting-pad thrust bearing, how is propeller thrust transmitted into the ship's structure?

  1. Through magnetic repulsion coils that float the shaft in a vacuum housing.
  2. Directly through the crankshaft webs into the engine bedplate holding-down bolts.
  3. Through helical springs housed in the stern tube after seal.
  4. Through a thrust collar forged on the shaft that bears against stationary white-metalled pads free to tilt on their supports, so that a converging oil wedge carries the load into the thrust block and thence into the double bottom seating. Correct answer.

Explanation

The thrust collar rotates with the shaft against a ring of sector-shaped pads. Because each pad is free to tilt slightly on its support, it takes up an angle that creates a converging wedge of oil in the direction of rotation, and the hydrodynamic pressure in that film carries the axial load without metal-to-metal contact. The load passes from the pads into the thrust block housing and from there through heavy chocks and holding-down bolts into the double bottom structure, which is why that seating is of substantial construction and its bolts are checked for tightness.

Question 27Diesel propulsion mechanisms and supplied performance models

What is the principal hazard in a marine diesel engine's starting air system, and what protective devices are required on the starting air line?

  1. Explosion in the starting air pipework, when a leaking or stuck cylinder starting air valve admits hot combustion gas into a line whose interior carries a film of compressor lubricating oil; guarded against by non-return valves, flame arresters, relief valves, and bursting discs. Correct answer.
  2. Freezing of the starting air lines through adiabatic expansion; guarded against by steam tracing.
  3. Hydraulic lock of the air receiver from condensed sea water; guarded against by vacuum breakers.
  4. Overspeed of the starting air compressor; guarded against by centrifugal flyweights.

Explanation

Compressed air carries over a fine mist of compressor lubricating oil that gradually coats the inside of the starting air pipework. If a cylinder starting air valve leaks or sticks open while the engine is firing, hot combustion gas blows back into that branch and can ignite the oil film in an atmosphere of compressed air, propagating a detonation along the manifold. The safeguards are a non-return valve at the branch to each cylinder, a flame arrester, and relief valves or bursting discs to vent the pressure. The practical defences are equally important: draining the receivers regularly, keeping compressor oil carryover low, and feeling the branch pipes for abnormal heat after every manoeuvre.

Question 28Diesel propulsion mechanisms and supplied performance models

On a 2-stroke propulsion engine, when do the electrically driven auxiliary scavenge blowers cut in and out, and why are they needed?

  1. They run continuously at all loads in order to cool the piston crowns.
  2. They start automatically when engine load or scavenge pressure falls below the manufacturer's threshold - commonly around 30% to 40% of the maximum continuous rating - because at low load the exhaust energy is insufficient to drive the turbocharger, and they stop again once the turbocharger can sustain the scavenge pressure on its own. Correct answer.
  3. They operate only during a crash stop to reverse the direction of propeller rotation.
  4. They run only in port with the main engine stopped, to circulate dry air through the crankcase.

Explanation

Auxiliary scavenge blowers support the specified engine where exhaust-driven air delivery is insufficient at lower loads. Actual cut-in and cut-out conditions depend on the engine and approved control arrangement; the illustrative load range is not a universal setpoint. This function does not authorize manual blower changes during a fire.

Question 29Diesel propulsion mechanisms and supplied performance models

What is the risk of bunkering two chemically incompatible residual fuel oils into the same tank, and which laboratory parameters measure the stability reserve?

  1. Asphaltenes drop out of solution as heavy sludge that overloads the purifiers, blinds the filters, and can starve the engine of fuel; the reserve is assessed by the total sediment potential and by the spot test for compatibility. Correct answer.
  2. The mixture self-ignites in the double bottom tanks through exothermic oxidation.
  3. The flash point rises above 300°C so that the fuel will not burn in the engine.
  4. The sulphur content increases logarithmically beyond the MARPOL limit.

Explanation

A residual fuel holds high molecular weight asphaltenes in colloidal suspension, stabilised by the aromatic fraction around them. Blending a paraffinic fuel into an asphaltenic one destroys that solvency balance and the asphaltenes flocculate, producing a heavy sludge that the purifiers cannot handle, blocks filters within hours, and in the worst case stops the engine. The condition is assessed by the total sediment potential, which measures the sediment after ageing, and by the spot test that rates the appearance of a drop of blended fuel on filter paper. The operational defence is to segregate new bunkers from the fuel already on board wherever the tank arrangement allows.

Question 30Diesel propulsion mechanisms and supplied performance models

Using full-load frequency as the specified reference, a generator governor has 61.8 Hz at no load and 60.0 Hz at full load. Which listed droop and conceptual sharing explanation fits that supplied convention?

  1. 1.8%, needed to hold the power factor at unity.
  2. 3.0%; the downward-sloping frequency-against-load characteristic makes each machine take a definite share of any load change, so that generators on the same bus divide load stably in proportion to their ratings instead of hunting. Correct answer.
  3. 5.5%, needed to prevent reverse current into the battery charger.
  4. 0.0% isochronous, because any droop would trip the generator on overload.

Explanation

Speed droop is the fall in speed or frequency from no load to full load, expressed as a percentage of the rated value: (61.8 − 60.0) / 60.0 × 100 = 3.0%. The sloping characteristic is what makes stable parallel operation possible. If bus frequency dips, every machine on the bus moves along its own droop line and picks up a share of the extra load determined by its slope, and the frequency settles at a new steady value. Two governors with no droop would each try to hold frequency exactly and would fight one another, swinging load back and forth between them.

Question 31Pumps, separation, freshwater, refrigeration and heat exchange

Why do marine starting air compressors delivering to 30 bar receivers use two or three stages of compression with intercoolers rather than a single stage?

  1. So that the compressor can be run from a single-phase emergency lighting circuit.
  2. Intercooling brings the process closer to isothermal compression, reducing the work required, keeps discharge temperatures well below the temperature at which the lubricating oil would ignite, and condenses moisture out between stages so it can be drained. Correct answer.
  3. So that liquid oxygen can be produced for engine starting.
  4. So that mechanical suction and discharge valves become unnecessary.

Explanation

Compressing air to 30 bar in one stage means a very high pressure ratio and, with it, a discharge temperature high enough to carbonise and even ignite the oil film on the valves and in the delivery pipe. Splitting the compression into stages and cooling between them holds each stage's temperature rise to a manageable value, which is a safety requirement as much as an efficiency one. It also cuts the total work, because cooling between stages moves the process towards the isothermal ideal, and it drops the air below its dew point so the separators can drain out water that would otherwise corrode the receivers and the starting air lines.

Question 32Pumps, separation, freshwater, refrigeration and heat exchange

What is the effect of an excessive clearance volume in the cylinder of a reciprocating air compressor?

  1. It raises the delivery pressure above the setting of the relief valves.
  2. It causes the driving motor to run at twice its synchronous speed.
  3. It forces lubricating oil from the crankcase up into the suction filter.
  4. It reduces volumetric efficiency, because the high-pressure air trapped in the clearance space must first re-expand below suction pressure before the suction valve can open, so less fresh air is drawn in on each stroke. Correct answer.

Explanation

Clearance volume is the space left between the piston crown at top dead centre and the cylinder head and valve pockets. At the end of delivery it is full of air at discharge pressure. As the piston descends, that air must expand until its pressure falls slightly below suction pressure before the suction valve can lift, and only the remainder of the stroke draws in fresh air. The larger the clearance volume - or the higher the pressure ratio - the greater the fraction of the stroke consumed by re-expansion, and the lower the mass of air delivered per revolution.

Question 33Pumps, separation, freshwater, refrigeration and heat exchange

What does the thermostatic expansion valve do in a marine refrigeration plant, and what happens if its sensing bulb loses its charge?

  1. It regulates cooling water flow through the condenser; loss of charge trips the compressor on high oil pressure.
  2. It cycles the compressor on and off; loss of charge makes the compressor run in reverse.
  3. It meters liquid refrigerant into the evaporator so as to hold a constant degree of superheat at the evaporator outlet; if the bulb loses its charge the opening force disappears and the spring closes the valve, starving the evaporator. Correct answer.
  4. It purges non-condensable gases from the receiver; loss of charge causes the evaporator to burst.

Explanation

The valve balances three forces: bulb pressure acting to open it, and evaporator pressure plus spring force acting to close it. By admitting only as much liquid as the evaporator can boil off, it maintains a stable superheat at the outlet, typically a handful of degrees, which both makes full use of the evaporator surface and guarantees that only dry vapour reaches the compressor suction. If the bulb or its capillary is damaged and the charge escapes, the opening force is lost, the spring shuts the valve, the evaporator is starved, and suction pressure falls away until the low-pressure cut-out operates.

Question 34Pumps, separation, freshwater, refrigeration and heat exchange

In a plate-type vacuum fresh water generator heated by main engine jacket water, how is low-temperature boiling achieved and what does the salinity monitor do?

  1. Sea water is compressed to 100 bar so that it boils at room temperature; the salinity monitor injects chlorine into the fresh water tanks.
  2. An ejector maintains a high vacuum in the shell so that sea water boils at around 45°C, which allows the engine's jacket water to act as the heating medium; the salinity monitor measures the conductivity of the distillate and operates a dump valve if the salt content exceeds the set limit. Correct answer.
  3. The plates are electrically heated; the salinity monitor controls the speed of the brine ejector pump.
  4. The vacuum raises the boiling point to 150°C to sterilise the water; the salinity monitor measures pH only.

Explanation

Jacket cooling water leaves the engine at around 80°C to 85°C, far too cool to boil sea water at atmospheric pressure. A sea water driven ejector draws a deep vacuum on the shell, lowering the saturation temperature to roughly 45°C, at which the jacket water becomes an ample heat source. The low temperature has a second benefit: it keeps the brine below the point at which hard calcium sulphate scale forms on the plates. The distillate passes a conductivity cell, and if the salt content rises above the set value - because of a leaking plate or carryover from over-production - a three-way valve dumps the distillate rather than sending it to the domestic tanks.

Question 35Pumps, separation, freshwater, refrigeration and heat exchange

Why is a coordinated phosphate treatment maintained in the drum of an auxiliary marine boiler?

  1. To lower the latent heat of vaporisation and raise the evaporation rate.
  2. It converts calcium hardness into a soft, non-adherent sludge that can be removed by blowdown instead of forming hard scale on heating surfaces, while buffering the boiler water alkalinity so that free caustic soda cannot concentrate and cause caustic attack. Correct answer.
  3. To prevent soot accumulating on the fire side of the tubes.
  4. To convert sulphur dioxide into sulphuric acid for cleaning purposes.

Explanation

Any calcium entering with the feed will otherwise deposit on the hottest surfaces as adherent calcium sulphate or carbonate scale, which insulates the tube and leads to overheating and failure. Dosing orthophosphate precipitates the calcium instead as a soft hydroxyapatite sludge that settles in the drum and headers and is removed by regular bottom blowdown. The coordinated part of the treatment matters just as much: by holding the phosphate and alkalinity in the right relationship, free sodium hydroxide is prevented from concentrating in crevices and under deposits, where it would cause caustic gouging or stress corrosion cracking.

Question 36Pumps, separation, freshwater, refrigeration and heat exchange

What is the function of a counterbalance (overcentre) valve in the hydraulic circuit of an anchor windlass or a deck crane winch motor?

  1. It doubles the oil flow during high-speed hoisting.
  2. It converts alternating current to direct current for the solenoid valves.
  3. It prevents the motor running away under an overhauling load when lowering, by holding back-pressure on the return side that only releases in proportion to the pilot pressure on the supply side, and it locks the load in place if supply pressure is lost. Correct answer.
  4. It drains contaminated hydraulic oil to the deck scuppers.

Explanation

The specified counterbalance arrangement resists an overhauling load under its actual pilot and return-pressure conditions. It is not an unconditional guarantee that every load remains locked despite leakage, failure or an unsuitable circuit. Real lifting systems require their complete assessed protective arrangements.

Question 37Pumps, separation, freshwater, refrigeration and heat exchange

During the automatic desludging cycle of a self-cleaning disc-stack separator, what happens inside the operating water mechanism?

  1. The bowl stops rotating, a hinged door opens, and the bowl tips into the sludge tank.
  2. A scraper blade rotates inside the bowl and pushes the solids into the clean oil line.
  3. The closing water holding the sliding bowl bottom up against its seal is released, so hydrostatic pressure inside the bowl drives the bowl bottom down for a fraction of a second, uncovering the peripheral ports through which the sludge is ejected, after which closing water is immediately restored. Correct answer.
  4. Steam at 20 bar is injected into the disc stack to melt the solids.

Explanation

The sliding bowl bottom is a hydraulically operated piston. In normal separation, closing water in the space beneath it produces an upward force that holds it firmly against the seal ring in the bowl hood. To discharge, operating water is directed to a valve arrangement that drains the closing water away; the force disappears and the pressure of the liquid inside the bowl drives the bowl bottom down, opening the peripheral ports so that the accumulated sludge is thrown out by centrifugal force. The whole event takes a fraction of a second, and closing water is immediately readmitted to re-seat the bowl. Correct operation depends on clean operating water and sound seal rings, which is why those are routine overhaul items.

Question 38Hull form, free surfaces and bounded stability calculations

In damage stability calculations under SOLAS Chapter II-1, what is compartment permeability and why is a machinery space assigned a relatively high value?

  1. It measures how quickly paint dissolves in sea water.
  2. It is the proportion of a compartment's volume that floodwater can occupy, and machinery spaces are assigned a high value because the engines, boilers, and tanks in them occupy a smaller fraction of the gross volume than their crowded appearance suggests, leaving most of the space open to water. Correct answer.
  3. It is the ratio of engine room air pressure to atmospheric pressure.
  4. It is the proportion of the crew certificated in survival craft handling.

Explanation

Permeability describes the fraction of the defined compartment volume available to floodwater in the actual model. Machinery does not necessarily occupy as much solid volume as its visual complexity suggests. Applicable damage-stability methods supply their own values and assumptions; this concept does not prescribe one universal set of current regulatory compartment percentages.

Question 39Electrical circuits, machines and digital logic

On tankers, what is the engineering distinction between flameproof (Ex d) and intrinsically safe (Ex ia) protection for electrical equipment in hazardous areas?

  1. Ex d is waterproof for diving work while Ex ia is intended for accommodation spaces.
  2. Ex d uses helium purging and Ex ia uses vacuum insulation.
  3. Ex d permits open sparking while Ex ia operates only on 440 V power circuits.
  4. Ex d contains an internal explosion within a robust enclosure whose flame paths cool the escaping gases below the ignition temperature of the surrounding atmosphere; Ex ia limits the electrical and thermal energy in the circuit so that no spark or hot surface can ignite the atmosphere even with two independent faults present. Correct answer.

Explanation

The two techniques accept opposite premises. Flameproof protection assumes the flammable atmosphere will enter the enclosure and may be ignited there, and therefore builds the enclosure strong enough to contain the explosion, with machined flanged gaps long and narrow enough to quench the flame and cool the gases as they escape. That suits motors, light fittings, and junction boxes. Intrinsic safety takes the opposite approach and limits the energy available in the circuit, using a barrier or galvanic isolator so that neither a spark nor a hot surface can release enough energy to ignite the gas, with the 'ia' level maintaining that guarantee under two independent faults. It suits instrumentation, current loops, and portable equipment.

Question 40Electrical circuits, machines and digital logic

Two identical generators on the same balanced sinusoidal voltage bus share 400 kW each, but draw different currents. With measurement and harmonic issues excluded, which listed sharing explanation fits unequal power factors?

  1. The governor droop on A is set too high; correct it by increasing the fuel rack on B.
  2. A has broken rotor bars; correct it by renewing the crankshaft.
  3. The machines are sharing real power equally but not reactive power, because their excitation levels differ; correct it by adjusting the automatic voltage regulator settings or the quadrature droop and cross-current compensation so that the power factors match. Correct answer.
  4. The busbar frequency has fallen to 50 Hz on one side of the board.

Explanation

Real power is set by the prime mover governors and reactive power by the excitation. Since both machines carry 400 kW, the governors are correctly matched and the fault must lie in the excitation. The over-excited machine supplies more lagging reactive power than its share, and a circulating reactive current flows between the two stators, showing up as a large difference in ammeter readings with no corresponding difference in kilowatts. The cure is to trim the voltage setting of the automatic voltage regulators, or to check the quadrature droop or cross-current compensation arrangement that is supposed to equalise reactive sharing automatically.

Question 41Electrical circuits, machines and digital logic

For the stated steerable direct-drive podded electric arrangement with suitable motor control, which listed set describes its possible layout and maneuvering advantages?

  1. The propulsion motor sits in a submerged steerable pod driving the propeller directly, which removes the long shaft line, stern tube, and reduction gearing, dispenses with a rudder by vectoring thrust through 360°, frees hull volume, and gives full torque at zero shaft speed. Correct answer.
  2. It reduces the installed electrical generating capacity by 80%.
  3. It propels the ship without consuming fuel or electrical power.
  4. It connects the diesel engine directly to the propeller by a mechanical chain drive.

Explanation

In a podded unit the electric motor is inside a streamlined housing suspended below the hull, with the propeller mounted directly on the motor shaft, and the whole assembly rotates about a vertical axis. That gives full thrust in any direction, which transforms low-speed manoeuvring and removes the need for a rudder and, often, for stern thrusters. It also removes the shaft line, intermediate bearings, stern tube, and gearbox from the hull, releasing space that on a cruise ship is used for cabins and on other types for cargo. In ice, the ability to direct thrust astern and mill ice with the propeller is a further advantage. The cost is the conversion losses of the electrical chain and the need to lift the vessel to overhaul the pod seals and bearings.

Question 42Electrical circuits, machines and digital logic

What do the arc chutes and magnetic blow-out arrangements in a low-voltage air circuit breaker do?

  1. They lubricate the toggle linkage with vaporised copper.
  2. They extinguish the arc drawn as the contacts part under fault current, by driving it upward into a stack of steel splitter plates that divide it into many short series arcs and cool and de-ionise it so that it cannot restrike after the current passes through zero. Correct answer.
  3. They store compressed air for closing the breaker.
  4. They convert 440 V alternating current to 24 V direct current for the trip coil.

Explanation

Interrupting a fault current does not happen simply by separating the contacts: an arc forms and will continue to conduct until it is deprived of the conditions it needs. The magnetic field produced by the fault current itself, sometimes reinforced by dedicated blow-out coils, exerts a force that drives the arc off the arcing contacts and up into the chute. There the stack of steel plates splits it into many short arcs in series, each of which needs its own arc voltage, so the total voltage required to sustain the arc exceeds what the system can supply. At the same time the plates absorb heat and de-ionise the gas, so that when the alternating current passes through its natural zero the arc cannot re-establish itself.

Question 43Electrical circuits, machines and digital logic

In a ship fitted with a shaft generator arranged for power take-off and power take-in, how is power transferred between the propeller shaft and the 60 Hz ship's network as engine speed varies?

  1. Through a continuously variable chain drive to the bridge gyrocompass.
  2. By keeping the main engine at exactly 3,600 rev/min at all times.
  3. A bidirectional static frequency converter decouples shaft speed from network frequency: in take-off mode the variable-frequency output of the shaft machine is rectified and inverted to constant 60 Hz for the switchboard, and in take-in mode power flows the other way so that the shaft machine acts as a motor boosting the propeller. Correct answer.
  4. Through a vacuum hose using static electricity.

Explanation

A shaft generator driven directly by a slow-speed main engine produces a frequency that varies with propeller revolutions, which is useless to a fixed-frequency network. Interposing a static frequency converter - a rectifier feeding a direct current link, feeding an inverter - breaks that dependence, so the machine can supply constant-voltage, constant-frequency power to the board over a wide range of engine speeds. Because the converter can pass power in either direction, the same machine can be run as a motor, taking power from the auxiliary generators to add torque to the shaft, which is used for take-home capability and for boosting in heavy weather. The commercial attraction is that the ship's electrical load is then carried at the main engine's low specific fuel consumption.

Question 44Measurement, feedback, PLC and automation

A supplied assessed safety loop uses normally closed contacts and de-energizes to trip. Which listed explanation describes its open-wire response, rather than claiming every actual alarm and overspeed sensor has this architecture?

  1. Normally closed contacts consume no energy while normally open contacts overheat the input card.
  2. Classification rules require them to prevent radio interference.
  3. They allow the controller to operate without any software.
  4. So that the system fails safe: current flows continuously in the healthy state, and a broken wire, a loose terminal, or a loss of loop supply opens the circuit and produces the trip, instead of leaving the protection silently inoperative. Correct answer.

Explanation

A protective input is only worth having if its failure modes are known. With a normally closed contact the healthy state is a continuous current, so anything that interrupts that current - the operator pressing the button, a wire chafing through, a terminal working loose with vibration, or the loop supply failing - is interpreted by the controller as a demand to trip. With normally open contacts the healthy state is no current, which is indistinguishable from a broken wire, so the first anyone would know of the fault is when the protection failed to act. Testing these circuits, including by simulating open circuits, is part of the routine safety system checks.

Question 45Measurement, feedback, PLC and automation

A supplied UV-flame-detection burner system defines a failed required flame signal as a fuel-trip and lockout demand under its approved logic. Which listed description matches that specified function, rather than a universal light-up time?

  1. It measures the boiler casing temperature to trim the fuel heater steam valve.
  2. It detects the ultraviolet radiation characteristic of a hydrocarbon flame while ignoring the infrared glow of hot refractory; if no flame is detected within the few seconds allowed at light-up, or if the flame is lost in service, the burner management system locks out and shuts the fuel quick-closing solenoid valves so that unburnt oil cannot accumulate in the furnace. Correct answer.
  3. It photographs the flame and uploads the images by satellite.
  4. It injects liquid nitrogen into the burner to extinguish soot fires.

Explanation

The configured detector and burner-management logic prevent continued fuel admission without the required flame evidence under the specified design. Actual detection, timing, reset and ventilation requirements depend on the approved system. This question gives no permission or executable procedure to restart, purge or bypass a burner.

Question 46Diesel propulsion mechanisms and supplied performance models

In a modern low-speed two-stroke crosshead diesel engine, what is the primary function of the piston rod stuffing box (scraper gland)?

  1. To prevent scavenge air and combustion sludge from entering the crankcase while scraping lubricating oil back down into the crankcase Correct answer.
  2. To lubricate the crosshead pin by directing high-pressure system oil into the crosshead bearings
  3. To provide lateral alignment for the piston rod to prevent side thrust against the cylinder liner
  4. To act as a pressure relief barrier in the event of a scavenge air trunk explosion

Explanation

The piston rod stuffing box is mounted in the diaphragm plate separating the scavenge air receiver from the crankcase. Its upper scraper and sealing rings prevent contaminated cylinder oil and pressurized scavenge air from entering the crankcase, while the lower scraper rings prevent crankcase system oil from being carried up into the scavenge space. This keeps the crankcase system oil clean and minimizes cylinder oil contamination.

Question 47Diesel propulsion mechanisms and supplied performance models

A mechanical indicator diagram taken from a cylinder of a two-stroke marine engine has an area of 480 mm² and an overall length of 80 mm. The indicator spring scale constant is 1.5 bar per mm. What is the mean indicated pressure (p_mi) of the cylinder?

  1. 9.0 bar Correct answer.
  2. 6.0 bar
  3. 12.0 bar
  4. 7.5 bar

Explanation

The mean height of the indicator diagram is calculated by dividing the area by the length: 480 mm² / 80 mm = 6.0 mm. Multiplying this average diagram height by the spring constant yields the mean indicated pressure: 6.0 mm × 1.5 bar/mm = 9.0 bar. This value represents the average constant pressure acting on the piston throughout the stroke.

Question 48Diesel propulsion mechanisms and supplied performance models

A six-cylinder, two-stroke marine diesel engine has a cylinder bore of 0.60 m and a piston stroke of 2.0 m. During sea trials at 100 rpm, the measured mean indicated pressure is 15.0 bar (1500 kPa). What is the total indicated power developed by the engine?

  1. 8,482 kW Correct answer.
  2. 4,241 kW
  3. 16,965 kW
  4. 1,414 kW

Explanation

Cylinder cross-sectional area is A = (π / 4) × (0.60 m)² = 0.28274 m². For a two-stroke engine running at 100 rpm, the firing frequency per cylinder is n = 100 / 60 = 1.6667 strokes per second. The indicated power per cylinder is P_i = p_mi × L × A × n = 1500 kPa × 2.0 m × 0.28274 m² × 1.6667 s⁻¹ = 1,413.7 kW. For 6 cylinders, total indicated power is 1,413.7 kW × 6 = 8,482 kW (approximately 8.48 MW).

Question 49Diesel propulsion mechanisms and supplied performance models

A main propulsion engine delivers a steady brake power of 12,000 kW while consuming 2,160 kg of heavy fuel oil per hour. What is the specific fuel oil consumption (SFOC) of the engine under these operating conditions?

  1. 180 g/kWh Correct answer.
  2. 165 g/kWh
  3. 216 g/kWh
  4. 195 g/kWh

Explanation

Specific fuel oil consumption is defined as fuel mass flow rate divided by brake power: SFOC = m_fuel / P_b. Converting the hourly consumption into grams gives 2,160 kg/h × 1,000 g/kg = 2,160,000 g/h. Dividing by 12,000 kW yields 2,160,000 / 12,000 = 180 g/kWh, which is a standard baseline value for modern large marine engines.

Question 50Diesel propulsion mechanisms and supplied performance models

A marine diesel engine operates with a specific fuel oil consumption (SFOC) of 175 g/kWh using fuel oil with a lower calorific value (LCV) of 42.7 MJ/kg. What is the brake thermal efficiency of the engine?

  1. 48.2% Correct answer.
  2. 42.7%
  3. 52.5%
  4. 38.5%

Explanation

One kilowatt-hour equals 3,600 kJ. The heat energy supplied to generate one kWh of brake work is given by SFOC × LCV = 0.175 kg/kWh × 42,700 kJ/kg = 7,472.5 kJ/kWh. Brake thermal efficiency is the work output divided by heat input: η_th = 3,600 kJ / 7,472.5 kJ = 0.4817, or 48.2%.