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Maskinmestereksamen practice questions

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Question 1Diesel 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 2Diesel 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 3Diesel 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 4Pumps, separation, freshwater, refrigeration and heat exchange

What is the physical root cause and symptom of cavitation in a marine centrifugal seawater cooling pump?

  1. Local liquid pressure at the impeller eye falls below the seawater saturation vapor pressure, causing vapor cavities to form and violently implode against the impeller vanes, resulting in pitting erosion and loud gravel-rattling noises Correct answer.
  2. Air is sucked through the packing gland, creating excessive discharge pressure and high motor amperage
  3. High seawater salinity causes galvanic electrolytic etching on the pump casing wear rings
  4. The discharge check valve chatters because the pump is operating at excessively high suction head

Explanation

Cavitation occurs when the absolute fluid pressure at the pump suction eye drops below the vapor pressure of the pumped liquid at that temperature. Vapor bubbles form in the low-pressure region and are swept into higher-pressure zones along the impeller vanes, where they collapse violently within microseconds. The resulting micro-jets generate localized shock pressures exceeding 1,000 bar, pitting and eroding the bronze or stainless impeller metal while producing characteristic loud cracking or gravel-rattling sounds.

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

A marine centrifugal seawater cooling pump operates at 1,800 rpm, delivering 200 m³/h against a 30 m head while drawing 30.0 kW of electrical power. If a variable frequency drive reduces the pump rotational speed to 1,440 rpm (80% of rated speed), what is the new power absorbed by the pump (assuming constant efficiency)?

  1. 15.36 kW Correct answer.
  2. 24.00 kW
  3. 19.20 kW
  4. 12.28 kW

Explanation

According to the pump affinity laws, power absorbed by a centrifugal pump varies with the cube of the rotational speed ratio: P2 = P1 × (N2 / N1)³. The speed ratio is 1,440 / 1,800 = 0.80. Substituting the values: P2 = 30.0 kW × (0.80)³ = 30.0 kW × 0.512 = 15.36 kW. This cubic relationship explains the substantial energy savings achieved by installing VFDs on marine cooling pumps.

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

What is the key functional and structural difference between operating a centrifugal disc-stack separator as a 'purifier' versus as a 'clarifier'?

  1. A purifier separates two immiscible liquids (oil and water) plus solids using a water seal and gravity disc, whereas a clarifier separates solids and trace water from a single liquid without a water seal using a blind discharge ring Correct answer.
  2. A purifier rotates at twice the rotational speed of a clarifier
  3. A purifier can only process lubricating oil, whereas a clarifier can only process heavy fuel oil
  4. A clarifier uses chemical solvent injection to dissolve asphaltenes during rotation

Explanation

When arranged as a purifier, the separator separates three phases: heavy liquid (water), light liquid (oil), and solids. It requires an initial water seal (priming water) and a properly sized gravity disc to position the oil-water interface line outside the disc stack. When configured as a clarifier, the water outlet is completely blocked with a blind ring, and no water seal is used; it is designed to extract remaining fine solid particulates and microscopic water droplets from already purified oil.

Question 7Hull form, free surfaces and bounded stability calculations

What does the block coefficient (Cb) of a ship's hull describe, and what typical values distinguish a full-form ship from a fine-form ship?

  1. The ratio of the underwater volume of the hull to the volume of a rectangular block of the same length, breadth and draught; loaded tankers and bulk carriers run around 0.80 or higher, while fine-form container ships and ferries are considerably lower. Correct answer.
  2. The ratio of the area of the immersed midship section to the area of the waterplane; values above 0.90 indicate a full-form ship.
  3. The ratio of the ship's deadweight to her lightship displacement; values above 3.0 indicate a full-form ship.
  4. The proportion of enclosed watertight hull volume above the waterline that provides reserve buoyancy; values above 0.80 indicate a full-form ship.

Explanation

Block coefficient compares immersed hull volume with length times breadth times draft under the stated definitions. A fuller immersed form has a larger coefficient in that comparison. Typical ship-type ranges do not establish every vessel’s squat direction, resistance or safety; use actual hull and operating information for those questions.

Question 8Hull form, free surfaces and bounded stability calculations

In naval architecture, what longitudinal structural bending condition is referred to as 'hogging', and what stresses are developed in the hull girder?

  1. The ship is heavily loaded at the forward and aft ends with excess buoyancy in the midbody; upper deck plating experiences tensile stress while the bottom shell plating experiences compressive stress. Correct answer.
  2. The ship is heavily loaded amidships; deck plating is in compression while bottom plating is in tension.
  3. The ship's bow twists relative to the stern due to quartering seas.
  4. Local hydrostatic water pressure pushes the forward hull plating inward between frames.

Explanation

Hogging occurs when weights at the fore and aft ends exceed local buoyancy while buoyancy amidships exceeds midship weight, bending the vessel like an arch (convex upward). The upper deck plating is stretched in tension, while the keel and bottom shell plating are squeezed in compression. Sagging is the opposite condition.

Question 9Hull form, free surfaces and bounded stability calculations

A ship has a total displacement of 12,000 tonnes in seawater. A rectangular double-bottom fuel oil tank is slack, with length L = 15.0 m, breadth B = 12.0 m, and contains fuel oil of density 0.95 t/m^3. What is the reduction in metacentric height (virtual loss of GM) caused by the free surface effect of this tank?

  1. Approximately 0.171 m. Correct answer.
  2. Approximately 0.342 m.
  3. Approximately 0.085 m.
  4. Approximately 0.512 m.

Explanation

The second moment of area of the rectangular liquid surface is i = (L * B^3) / 12 = (15.0 * 12.0^3) / 12 = (15.0 * 1,728) / 12 = 2,160 m^4. The virtual loss of GM (GGv) = (i * rho_liquid) / Delta_ship = (2,160 * 0.95) / 12,000 = 2,052 / 12,000 = 0.171 m.

Question 10Programme scope, watchkeeping and accountable ship systems

Under IMO enclosed-space entry guidance, what is the vital rule governing the designated attendant stationed at the entrance during an entry?

  1. The attendant must remain continuously stationed outside the entrance, maintain constant communication with entrants, monitor environmental conditions, and NEVER enter the space to attempt a rescue alone if an entrant collapses Correct answer.
  2. The attendant must enter the space every 15 minutes to inspect the work area personally
  3. The attendant may leave the entrance unattended if called to assist with mooring stations
  4. The attendant is permitted to smoke near the entrance provided a fire extinguisher is within reach

Explanation

The attendant remains at the entrance, maintains agreed communication, monitors the entry, and summons the trained rescue response if needed. An unprotected impulsive rescue often creates additional casualties; the attendant must not enter alone. The entry permit and vessel procedures specify the communication, monitoring, and rescue equipment for the particular space.

Question 11Electrical circuits, machines and digital logic

A marine diesel generator is designed to supply a standard 60 Hz electrical grid at an operational speed of 900 rpm. How many magnetic poles are wound on the alternator rotor?

  1. 8 poles Correct answer.
  2. 6 poles
  3. 4 poles
  4. 10 poles

Explanation

Electrical frequency f, rotational speed n (in rpm), and number of pole pairs p are related by f = (p × n) / 60, or in terms of total poles P: f = (P × n) / 120. Rearranging for total poles gives P = (120 × f) / n. Substituting the given values: P = (120 × 60) / 900 = 7,200 / 900 = 8 poles.

Question 12Electrical circuits, machines and digital logic

A 440 V, three-phase marine synchronous generator supplies a balanced line current of 800 A to the shipboard distribution network at a lagging power factor of 0.80. What is the active electrical power (kW) delivered by the generator?

  1. 487.6 kW Correct answer.
  2. 609.6 kW
  3. 352.0 kW
  4. 281.6 kW

Explanation

Active three-phase electrical power is given by P = √3 × V_line × I_line × cos(φ). Substituting the given values: P = 1.73205 × 440 V × 800 A × 0.80 = 487,645 W, or 487.6 kW. The apparent power is S = √3 × 440 × 800 = 609.6 kVA.

Question 13Electrical circuits, machines and digital logic

In a supplied illustrative generator model, the reverse-power trip threshold is 8 percent of a 1000 kW rating with the stated delay. What threshold power corresponds to that percentage, rather than an exact prediction of the transient motor power or overshoot?

  1. 80 kW Correct answer.
  2. 8 kW
  3. 125 kW
  4. 800 kW

Explanation

Eight percent of 1000 kW is 80 kW. That is the supplied relay threshold, not proof the actual power will stay exactly there throughout the delay or transient. Reverse real-power flow indicates the alternator can motor its prime mover; actual protection settings and behavior require the approved system basis.

Question 14Measurement, feedback, PLC and automation

In marine instrumentation, why does standard analog process transmission utilize a 4–20 mA current loop with 4 mA representing the 'live zero' rather than a 0–20 mA loop?

  1. Because a live zero allows the control system to instantly distinguish between a valid zero process reading (4 mA) and a broken cable or transmitter power failure (0 mA) Correct answer.
  2. Because 4 mA is the minimum current required to heat the thermocouple wires to their operating temperature
  3. Because maritime classification rules prohibit currents below 4 mA due to galvanic hull corrosion
  4. Because an electrical current below 4 mA causes capacitive charging of the PLC analog input card

Explanation

A live-zero 4 to 20 mA range makes a valid zero-scale value distinguishable from an out-of-range zero-current condition. A zero current can indicate several wiring, supply or instrument faults, so it is not a unique diagnosis of a broken cable. The actual system and diagnostic conventions still matter.

Question 15Measurement, feedback, PLC and automation

A 4–20 mA pressure transmitter is calibrated to measure main engine starting air receiver pressure over a range of 0 to 16.0 bar. If the transmitter outputs a steady current of 12.0 mA to the automation system, what is the measured air pressure?

  1. 8.0 bar Correct answer.
  2. 12.0 bar
  3. 9.6 bar
  4. 6.4 bar

Explanation

The transmitter current span is 20 mA - 4 mA = 16 mA representing a pressure span of 16.0 bar. At an output of 12.0 mA, the current above live zero is 12.0 mA - 4.0 mA = 8.0 mA. The fraction of full scale is 8.0 mA / 16.0 mA = 0.50 (50% of span). Multiplying by the 16.0 bar measurement range yields 0.50 × 16.0 bar = 8.0 bar.

Question 16Measurement, feedback, PLC and automation

In the supplied stable, unsaturated loop with suitable integral action and a constant disturbance, which listed description identifies the integral term’s intended steady-state function?

  1. To continuously accumulate the process error over time and adjust the output until the steady-state error (offset) is reduced to exactly zero Correct answer.
  2. To provide an instantaneous output spike proportional to the acceleration of the setpoint knob
  3. To act as a low-pass filter to dampen mechanical vibration from the control valve stem
  4. To predict future temperature trends based on the mathematical second derivative of the error

Explanation

Integral action accumulates error and can remove the relevant steady-state offset in a suitably stable, unsaturated arrangement. It does not guarantee every plant becomes stable or reaches exactly zero error under actuator limits and measurement faults. Tuning and anti-windup require the actual engineering model and safe assessment.

Question 17Thermal Physics

In a marine vapor-compression provision refrigeration system, why does the thermostatic expansion valve (TXV) maintain a designated degree of superheat (typically 4°C to 8°C) at the evaporator outlet?

  1. To ensure that all liquid refrigerant has completely vaporized before entering the compressor suction, preventing catastrophic liquid slugging and mechanical valve breakage Correct answer.
  2. To maximize the condensation rate of refrigerant inside the sea-water cooled condenser
  3. To prevent lubricating oil from circulating through the expansion valve orifice
  4. To heat the cold room air to prevent frost from accumulating on shipboard vegetables

Explanation

The TXV’s intended outlet-superheat control helps limit liquid carryover in the assessed refrigeration arrangement. Actual target values depend on the refrigerant, equipment, operating condition and manufacturer data. A superheat reading alone is not a guarantee against every compressor fault, and the question does not authorize valve adjustment.

Question 18Thermal Physics

A shipboard provision refrigeration compressor draws 15.0 kW of electrical power while the evaporator successfully extracts 45.0 kW of thermal heat from the meat and fish cold storage rooms. What is the operational Coefficient of Performance (COP) of the refrigeration plant?

  1. 3.0 Correct answer.
  2. 0.33
  3. 4.0
  4. 2.5

Explanation

The Coefficient of Performance (COP) of a refrigeration cycle is defined as the useful cooling effect (heat extracted by the evaporator) divided by the net mechanical/electrical work input to the compressor: COP = Q_cooling / W_input. Substituting the given values: COP = 45.0 kW / 15.0 kW = 3.0. This indicates that 3.0 kW of heat is removed from the cold spaces for every 1.0 kW of electrical power consumed.

Question 19Thermal Physics

For the supplied ideal refrigeration diagram, enthalpy is on the horizontal axis and pressure on the vertical axis. How is the modeled constant-enthalpy throttling process shown?

  1. As a vertical line moving downward at constant enthalpy (isenthalpic throttling, h3 = h4) Correct answer.
  2. As a horizontal line moving left at constant pressure with heat rejection
  3. As a curved line moving upward at constant entropy (isentropic compression)
  4. As a parabolic path reflecting reversible adiabatic work production

Explanation

Throttling through an expansion valve or capillary tube is an irreversible adiabatic process with no shaft work produced (W = 0) and negligible external heat transfer (Q = 0). Applying the first law of thermodynamics to steady flow yields h_in = h_out (h3 = h4). On a standard P-h diagram, where enthalpy is plotted on the horizontal axis and pressure on the vertical axis, this isenthalpic pressure drop appears as a straight vertical line descending from condensing pressure to evaporating pressure.

Question 20Maintenance 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 21Maintenance 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 22Maintenance 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 23Diesel 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 24Diesel 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%.

Question 25Diesel propulsion mechanisms and supplied performance models

An out-of-phase (draw) indicator diagram taken from a medium-speed diesel engine reveals that the peak cylinder pressure (P_max) occurs prematurely at 2° before Top Dead Center (TDC), accompanied by severe mechanical knocking. What is the most likely cause?

  1. Fuel injection timing is excessively advanced Correct answer.
  2. Fuel injection timing is excessively retarded
  3. The fuel injector nozzle holes are partially blocked
  4. The scavenge air cooler is heavily fouled on the air side

Explanation

The listed injection, combustion or sealing mechanism is a plausible explanation for the specified evidence, not a unique diagnosis from one observed symptom. Compare relevant measurements, sensor condition, operating context and manufacturer information before assigning an actual fault. This written inference does not authorize adjustment or an emergency operating action.

Question 26Diesel propulsion mechanisms and supplied performance models

What thermodynamic mechanism causes turbocharger compressor surging during heavy sea conditions or severe air cooler fouling?

  1. The mass airflow drops below the aerodynamic stability limit for the given pressure ratio, causing flow separation on the impeller blades and cyclic airflow reversal Correct answer.
  2. The turbine nozzle ring becomes starved of exhaust gas, causing the turbine wheel to rotate in the reverse direction
  3. Excessive scavenge air receiver pressure forces high-pressure gas back through the exhaust gas economizer
  4. The charge air cooler water mist catcher freezes, causing a complete vacuum at the compressor inlet

Explanation

Compressor surging occurs when the operating point on the compressor performance map crosses to the left of the surge line. When downstream air resistance rises (e.g. from fouled cooler fins or engine load reduction) or airflow fluctuates rapidly in heavy seas, the air velocity through the impeller blades becomes insufficient to overcome the high backpressure. Aerodynamic stalling occurs, causing high-pressure air to surge violently backward through the compressor with loud banging and severe mechanical vibrations.

Question 27Diesel propulsion mechanisms and supplied performance models

Why are large marine two-stroke diesel engines designed with a crosshead and guide slippers rather than a trunk piston construction?

  1. To transfer transverse connecting-rod side thrust entirely to the engine frame guide shoes, allowing the piston to move with pure axial reciprocating motion and zero side load on the cylinder liner Correct answer.
  2. To allow the piston to rotate freely inside the cylinder liner during each stroke to equalize wear
  3. To eliminate the need for high-pressure lubricating oil in the bottom-end crankpin bearings
  4. To reduce the overall vertical engine height and optimize cargo hold volume above the machinery space

Explanation

The crosshead guides carry the relevant connecting-rod side thrust in this arrangement, reducing the piston’s direct side-thrust role compared with a trunk-piston design. This is a mechanical design distinction, not a claim of literally zero lateral force or perfect alignment in every real engine. Actual wear and condition still matter.

Question 28Diesel propulsion mechanisms and supplied performance models

A marine propulsion engine indicates a total indicated power of 10,000 kW on indicator cards. The engine mechanical efficiency is rated at 90%. What is the brake power delivered to the shaft, and how much power is dissipated as internal friction and pumping losses?

  1. Brake power is 9,000 kW; internal losses are 1,000 kW Correct answer.
  2. Brake power is 11,111 kW; internal losses are 1,111 kW
  3. Brake power is 9,500 kW; internal losses are 500 kW
  4. Brake power is 8,100 kW; internal losses are 1,900 kW

Explanation

Mechanical efficiency is defined as the ratio of brake power to indicated power: η_m = P_b / P_i. Therefore, brake power delivered to the shaft coupling is P_b = P_i × η_m = 10,000 kW × 0.90 = 9,000 kW. Friction and pumping power losses represent the difference between indicated power and brake power: P_loss = P_i - P_b = 10,000 kW - 9,000 kW = 1,000 kW.

Question 29Diesel propulsion mechanisms and supplied performance models

Compared with traditional camshaft-driven injection, what primary operational advantage is shared by electronically controlled injection systems such as MAN ME-C engines and common-rail systems such as WinGD RT-flex?

  1. Independent and flexible control of fuel injection timing, injection pressure, and rate-shaping profiles across all engine load ranges Correct answer.
  2. Complete elimination of high-pressure fuel piping inside the engine machinery space
  3. Direct combustion of unheated high-viscosity heavy fuel oil without preheating or viscosity control
  4. Replacement of all exhaust valves with two-stroke reed valves to simplify head construction

Explanation

Electronic control decouples key injection and exhaust-valve settings from a fixed mechanical cam profile, allowing timing and injection pattern to be optimized across the load range. RT-flex uses a common rail, while MAN ME-C uses electronically controlled hydraulic actuation and should not itself be labelled a fuel common-rail design. The precise controllable parameters depend on the engine model.

Question 30Diesel propulsion mechanisms and supplied performance models

Why is it critical to inspect and drain the scavenge air cooler condensate drains and water mist catchers regularly, especially in tropical climates?

  1. To prevent entrained water droplets from entering the scavenge receiver and cylinders, which causes hydraulic cylinder washing, lube oil film breakdown, and severe liner corrosion Correct answer.
  2. To recover freshwater for direct feeding into the high-pressure propulsion boiler feed system
  3. To prevent water from backing up into the turbocharger exhaust gas turbine casing
  4. To maintain high relative humidity inside the scavenge receiver to suppress NOx formation

Explanation

When warm, humid tropical ambient air is compressed by the turbocharger and cooled in the charge air cooler, its temperature frequently falls below the dew point, condensing substantial volumes of water. If water mist catchers or drains become clogged, water droplets are carried into the scavenge receiver and swept into the cylinders. This moisture washes away the alkaline cylinder lubricant film, causing rapid corrosive wear, scuffing, and abrasive ring failure.

Question 31Diesel propulsion mechanisms and supplied performance models

A marine propulsion engine produces a brake power of 15,000 kW with an overall brake thermal efficiency of 48%. If the exhaust gas stream carries away exactly 28% of the total heat energy supplied by the fuel, what is the thermal power carried by the exhaust gas?

  1. 8,750 kW Correct answer.
  2. 4,200 kW
  3. 7,200 kW
  4. 10,500 kW

Explanation

Total heat input from fuel is calculated from brake power and thermal efficiency: Q_in = P_b / η_th = 15,000 kW / 0.48 = 31,250 kW. The heat energy carried away in the exhaust gas stream is 28% of the total energy input: Q_exhaust = 31,250 kW × 0.28 = 8,750 kW. This substantial waste heat is typically utilized in an exhaust gas economizer to generate shipboard auxiliary steam.

Question 32Diesel propulsion mechanisms and supplied performance models

During an in-service inspection of a large two-stroke crosshead engine, an engineer observes heavy black carbon buildup in scavenge space #3, accompanied by hissing sounds and elevated scavenge air temperatures. What defect does this indicate?

  1. Severe piston ring blow-by caused by worn, broken, or stuck piston rings Correct answer.
  2. A crack in the cylinder cover exhaust valve cooling water jacket
  3. Excessive cylinder oil feed rate delivering unburned lubricant into the receiver
  4. Fatigue failure of the bottom-end connecting rod bearing shells

Explanation

The listed injection, combustion or sealing mechanism is a plausible explanation for the specified evidence, not a unique diagnosis from one observed symptom. Compare relevant measurements, sensor condition, operating context and manufacturer information before assigning an actual fault. This written inference does not authorize adjustment or an emergency operating action.

Question 33Diesel propulsion mechanisms and supplied performance models

What is the primary function of the cylinder relief (safety) valve fitted on each cylinder head of a marine diesel engine?

  1. To protect the cylinder head, liner, and running gear against dangerous mechanical overpressure caused by hydraulic lock or severe detonation Correct answer.
  2. To relieve excess cylinder pressure during routine engine starting with compressed air
  3. To act as the primary exhaust port for expelling burnt gases during the scavenge period
  4. To vent unburned fuel gas safely into the engine room bilges during emergency shutdown

Explanation

Cylinder relief valves are spring-loaded safety valves set to lift automatically at approximately 20% to 30% above the normal maximum cylinder firing pressure (typically 150 to 180 bar depending on engine rating). They protect structural engine components from catastrophic damage if excessive pressures develop due to hydraulic lock (water or fuel accumulated in the combustion space) or extreme combustion anomalies.

Question 34Diesel propulsion mechanisms and supplied performance models

When calibrating a cylinder liner of a large two-stroke marine engine with an inside micrometer, where is the maximum radial wear and ovality typically found?

  1. At the upper region of the liner near the Top Dead Center position of the top piston ring Correct answer.
  2. At the bottom of the liner near the scavenge port cutouts
  3. Uniformly distributed along the exact middle of the piston stroke
  4. Around the lower scraper ring band below the stuffing box gland

Explanation

Maximum liner wear occurs at the very top of the piston stroke (TDC of the top piston ring). At this point, gas pressure behind the top ring is at its maximum (forcing the ring hard against the liner wall), gas temperatures are highest (thinning the lubricating oil film), and piston sliding velocity slows to zero, causing hydrodynamic lubrication to break down into boundary lubrication.

Question 35Diesel propulsion mechanisms and supplied performance models

In a medium-speed four-stroke marine diesel engine burning residual fuel oil (HFO), why does the crankcase lubricating oil require continuous centrifugal purification and regular Base Number (BN) monitoring?

  1. Because the trunk piston design allows combustion blow-by and unburned acidic fuel residues to enter the crankcase directly, contaminating the system oil Correct answer.
  2. Because four-stroke engines do not utilize oil filters in the main lubrication gallery
  3. To prevent the system oil from cooling below the sea water injection temperature
  4. To extract dissolved refrigerant gases leaking from the charge air cooler circuit

Explanation

Unlike crosshead engines which physically separate the cylinders from the crankcase via a diaphragm and stuffing box, four-stroke engines use trunk pistons where the piston skirt slides directly against the liner inside the crankcase space. Combustion blow-by containing soot, unburned fuel residues, and sulfuric acid inevitably contaminates the crankcase oil. Continuous centrifugal separation and regular BN testing are essential to remove insolubles, maintain acid neutralization capacity, and prevent corrosion.

Question 36Diesel propulsion mechanisms and supplied performance models

A container ship's main engine operates continuously at 18,000 kW with an average SFOC of 170 g/kWh. How many metric tonnes of fuel oil are consumed by the main engine during a standard 24-hour sea passage?

  1. 73.44 tonnes Correct answer.
  2. 61.20 tonnes
  3. 85.60 tonnes
  4. 51.84 tonnes

Explanation

Total fuel consumption over 24 hours is calculated as: Fuel (tonnes) = (Power in kW × SFOC in g/kWh × Operating hours) / 1,000,000 g/tonne. Substituting the values: (18,000 kW × 170 g/kWh × 24 h) / 1,000,000 = 73,440,000 g / 1,000,000 = 73.44 metric tonnes per day.

Question 37Diesel propulsion mechanisms and supplied performance models

What is the physical sequence of events that leads to a primary and secondary crankcase explosion in a marine diesel engine?

  1. An overheated bearing vaporizes lubricating oil; the vapor condenses into a flammable fine white mist which ignites; the primary explosion creates a partial vacuum that sucks in fresh air, fueling a far more destructive secondary explosion Correct answer.
  2. High scavenge pressure forces exhaust flames past the stuffing box, igniting the oil sump in a single detonation
  3. Excessive fuel pump delivery pressure causes fuel line rupture, spraying diesel fuel onto the turbocharger turbine casing
  4. Water leakage into the crankcase causes rapid steam generation that overpressurizes the crankcase relief valves

Explanation

When a mechanical component (such as a crosshead, crankpin, or main bearing) overheats due to friction or lubrication failure, it reaches temperatures exceeding 400°C. Sump oil splashing onto this hotspot vaporizes and, upon moving into cooler crankcase regions, condenses into a dense white oil mist. When mist concentration reaches the lower flammable limit (~50 mg/L), the hotspot ignites it. The primary explosion forces gas out through crankcase relief valves, creating a subsequent partial vacuum that draws fresh air inside, fueling an catastrophic secondary explosion.

Question 38Pumps, separation, freshwater, refrigeration and heat exchange

In an electro-hydraulic marine steering gear system, what is the mechanical function of the floating lever (hunting gear) mechanism?

  1. It acts as a mechanical feedback linkage that gradually returns the variable-delivery hydraulic pump swashplate to neutral (zero delivery) as the rudder approaches the commanded angle, stopping rudder motion smoothly Correct answer.
  2. It locks the rudder mechanically to the hull frame when steering power fails
  3. It balances hydraulic pressure between port and starboard steering cylinders during heavy rudder yawing
  4. It automatically purges entrained air from the hydraulic ram cylinders during rudder movement

Explanation

The floating lever connects three points: the telemotor/control actuator (command input), the variable-displacement pump control spindle (output), and the rudder tiller linkage (mechanical feedback). When the helm is moved, the lever tilts, shifting the pump swashplate off-center to deliver oil to the rams. As the rudder moves toward the requested angle, the feedback link moves the floating lever back in the opposite direction, restoring the pump swashplate precisely to zero stroke (neutral) when the rudder reaches the commanded helm angle.

Question 39Pumps, separation, freshwater, refrigeration and heat exchange

A marine centrifugal bilge pump experiences persistent loss of suction and severe cavitation when attempting to pump warm engine room bilge water through a dirty strum box. From an NPSH perspective, what is occurring?

  1. Frictional losses across the fouled strainer and the higher vapor pressure of warm water reduce Net Positive Suction Head Available (NPSHa) below Net Positive Suction Head Required (NPSHr) Correct answer.
  2. High suction head increases NPSHa to a level that exceeds the pump motor torque capacity
  3. The pump impeller diameter expands due to heat, rubbing against the wear rings
  4. Atmospheric air dissolved in the water undergoes catalytic breakdown into pure oxygen gas

Explanation

Net Positive Suction Head Available is given by NPSHa = (P_atm - P_vapor) / (ρ × g) - h_suct - h_friction. A dirty suction strum box drastically increases suction friction losses (h_friction), while warmer water exhibits significantly higher vapor pressure (P_vapor). Both factors cause NPSHa to plummet below the pump manufacturer's minimum required head (NPSHr). When NPSHa < NPSHr, fluid vaporizes at the impeller entrance, triggering severe cavitation and loss of suction prime.

Question 40Pumps, separation, freshwater, refrigeration and heat exchange

Why must all marine positive-displacement pumps (such as gear, screw, or reciprocating piston pumps) be fitted with an integral or external pressure relief valve?

  1. Because positive-displacement pumps deliver a fixed liquid volume per revolution regardless of resistance, meaning an obstructed or closed discharge valve causes near-instantaneous, dangerous overpressurization and mechanical rupture Correct answer.
  2. To prevent seawater from siphoning backward when the electric drive motor is stopped
  3. To aerate lubricating oil before it enters the engine oil coolers
  4. To allow the pump to reverse its rotational direction without stopping the motor

Explanation

Unlike centrifugal pumps (where fluid simply churns against a closed valve and pressure rises only to the shut-off head), positive-displacement pumps force a constant volume of incompressible fluid through the discharge port with every turn. If the discharge line is closed or blocked, fluid has no escape path, causing discharge pressure to escalate instantly until the drive motor stalls, shafts snap, or pump casings and piping burst catastrophically. A spring-loaded relief valve bypassing fluid back to suction is mandatory.

Question 41Electrical circuits, machines and digital logic

What is the primary operational function of an Automatic Voltage Regulator (AVR) fitted to a marine synchronous alternator?

  1. To automatically modulate rotor field excitation current to keep terminal voltage constant under varying electrical loads and ensure balanced reactive power (kVAR) sharing Correct answer.
  2. To adjust the fuel rack of the diesel engine to maintain constant rotational speed
  3. To open the main generator breaker when short-circuit current exceeds 50 kA
  4. To invert DC battery power into 440 V AC during emergency blackout conditions

Explanation

An AVR changes excitation to regulate the intended voltage and related reactive behavior. Appropriate sharing among parallel machines also needs the actual coordination, droop or compensation arrangement; an AVR alone does not guarantee equal reactive sharing. Do not confuse excitation/reactive control with the governor’s real-power function.

Question 42Electrical circuits, machines and digital logic

What is the operating principle and safety purpose of a marine switchboard preferential tripping system?

  1. It automatically sheds non-essential electrical loads in predetermined timed stages when an alternator is sustainedly overloaded, protecting essential machinery from total blackout Correct answer.
  2. It trips the main steering gear pump first when bus frequency drops below 58 Hz
  3. It disconnects all navigation and emergency radios immediately upon starting a large deck crane
  4. It isolates the neutral earthing resistor when an insulation fault is detected on deck lighting

Explanation

When an online generator exceeds approximately 110% of its rated current, the preferential trip relay initiates timed stage tripping of non-essential consumers (e.g., Stage 1: galley equipment, air conditioning; Stage 2: cargo ventilation, engine room exhaust fans). By shedding secondary loads before the generator circuit breaker can trip on thermal overload, essential ship services (steering gear, propulsion auxiliaries, lube oil pumps) remain continuously powered.

Question 43Electrical circuits, machines and digital logic

A four-pole, 60 Hz three-phase induction motor driving a seawater circulating pump rotates at a full-load shaft speed of 1,746 rpm. What is the operational rotor slip percentage of the motor?

  1. 3.0% Correct answer.
  2. 4.5%
  3. 1.5%
  4. 6.0%

Explanation

The synchronous speed of the stator rotating magnetic field is n_s = (120 × f) / P = (120 × 60) / 4 = 1,800 rpm. The slip s is defined as the difference between synchronous speed and rotor speed divided by synchronous speed: s = (n_s - n) / n_s = (1,800 - 1,746) / 1,800 = 54 / 1,800 = 0.03, or 3.0%.

Question 44Electrical circuits, machines and digital logic

Under the supplied ideal motor comparison, use the same line voltage and winding impedance with delta direct-on-line versus star starting. Which listed ratio describes star line current and starting torque relative to delta?

  1. Both the starting line current and the starting torque are reduced to exactly 1/3 (33.3%) of their direct-on-line values Correct answer.
  2. The starting line current is halved (50%), while starting torque remains 100%
  3. Starting current is reduced to 1/3 (33.3%), but starting torque increases by √3 (173%)
  4. Both starting current and torque are reduced to 1/√3 (57.7%) of their direct-on-line values

Explanation

In star connection, phase voltage applied across each winding is V_line / √3. Because winding current is proportional to phase voltage, phase current drops to 1/√3 of its delta value. Because line current in star equals phase current (whereas line current in delta is √3 × phase current), the resulting line current in star is (1/√3) × (1/√3) = 1/3 of the DOL line current. Since motor torque is proportional to the square of the applied winding voltage: T ∝ V_phase² = (V_line / √3)² = 1/3 T_DOL.

Question 45Electrical circuits, machines and digital logic

What are the three core functional stages of a modern marine Pulse Width Modulation (PWM) Variable Frequency Drive (VFD)?

  1. A diode or thyristor rectifier stage, a filtered DC intermediate link, and an insulated-gate bipolar transistor (IGBT) inverter stage Correct answer.
  2. A step-up transformer, a rotary slip-ring converter, and a magnetic amplifier
  3. A silicon controlled rectifier (SCR), a high-pressure oil cooling pump, and a synchronous spark gap
  4. An induction voltage regulator, an autotransformer, and a mechanical centrifugal switch

Explanation

A modern VFD consists of: (1) Rectifier (converter) that rectifies three-phase AC into pulsating DC; (2) DC intermediate bus (link) featuring capacitors and choke inductors to smooth voltage ripple and store energy; and (3) Inverter utilizing high-speed IGBT transistors modulated by PWM algorithms to synthesize variable-voltage, variable-frequency AC to control motor speed and torque efficiently.

Question 46Electrical circuits, machines and digital logic

In a standard 440 V insulated neutral (IT) marine distribution system fitted with a three-lamp earth indication panel, what visual symptom indicates a solid earth fault on Phase B (L2)?

  1. The lamp connected to Phase B goes completely dark, while the lamps for Phase A and Phase C glow noticeably brighter than normal Correct answer.
  2. All three lamps extinguish simultaneously and the bus-tie breaker trips
  3. The Phase B lamp flickers rapidly while Phase A and C remain at half brilliance
  4. All three lamps glow with blinding maximum brilliance and the preferential trip buzzer sounds

Explanation

In an insulated system, three identical lamps are connected in star between each line and the ship's hull. Normally, each lamp experiences line-to-earth voltage (440 / √3 ≈ 254 V) and glows with equal half-brilliance. When Phase B is solidly shorted to hull earth, the voltage across Lamp B drops to 0 V, extinguishing it completely. Concurrently, the neutral-to-hull potential shifts, subjecting Lamps A and C to full line-to-line voltage (440 V), causing them to glow at full brilliance.

Question 47Electrical circuits, machines and digital logic

In marine electrical distribution protection, what is the principle of selective discrimination (grading)?

  1. Coordinating circuit breaker trip thresholds and time delays so that only the protective device immediately upstream of a fault opens, leaving the rest of the electrical system undisturbed Correct answer.
  2. Ensuring that the largest main generator breaker always trips first to protect the engine crankshaft
  3. Distributing active loads equally between all online generators regardless of prime mover rating
  4. Selecting different color-coded cables for high-voltage and low-voltage cable trays

Explanation

Selective discrimination ensures that protective devices (fuses, MCCBs, and ACBs) are arranged with graded time and current settings. When a short circuit occurs on a branch feeder (e.g. a fan motor), only that feeder's local breaker trips instantaneously, while upstream sub-switchboard and main generator breakers remain closed. This isolates the fault locally and prevents widespread machinery blackout.

Question 48Electrical circuits, machines and digital logic

How do magnetic blowout coils and de-ion arc chutes collaborate to interrupt high short-circuit currents inside a low-voltage Air Circuit Breaker (ACB)?

  1. The magnetic field generated by the fault current forces the electric arc upward into split metal plates that lengthen, cool, and de-ionize the arc plasma until it extinguishes at current zero Correct answer.
  2. High-pressure SF6 gas is injected across the opening contacts to liquefy the ionized copper vapors
  3. A permanent magnet clamps the moving contacts together until the generator governor reduces engine speed
  4. A dielectric vacuum bottle encloses the contacts to prevent any electrical spark from forming

Explanation

When an ACB interrupts a fault, an electric arc forms between the parting contacts. The fault current flowing through blowout coils produces a strong magnetic field that exerts a Lorentz force on the charged arc plasma, propelling it into the arc chute. Inside the chute, a series of insulated metal splitter plates divide the single arc into numerous series sub-arcs, rapidly extracting heat, lengthening the path, and de-ionizing the plasma so it cannot re-ignite after alternating current passes through zero.

Question 49Electrical circuits, machines and digital logic

After one supply phase is lost, the supplied heavily loaded three-phase motor is observed still rotating. Which listed consequence is the relevant overheating risk unless appropriate protection acts?

  1. The motor continues to rotate but draws severe overcurrent in the remaining two healthy phases, causing rapid stator winding overheating unless tripped by thermal overload or negative-phase-sequence protection Correct answer.
  2. The motor immediately reverses its rotational direction and generates regenerative power back to the switchboard
  3. The motor stops instantaneously and the mechanical shaft locks via dynamic braking
  4. The motor continues running with zero change in current, torque, or temperature

Explanation

When a three-phase motor loses one phase while running, the revolving magnetic field collapses into a pulsating single-phase field containing strong backward-rotating negative sequence components. To deliver the required mechanical torque, current in the two remaining phases surges to approximately 1.73 to 2.4 times normal full load current. This produces intense localized resistive heating in the stator windings and rotor bars, leading to insulation burn-out within minutes unless tripped by motor protection.

Question 50Electrical circuits, machines and digital logic

What is the operational function of the interconnecting tie-breaker between the main switchboard (MSB) and the emergency switchboard (ESB)?

  1. To supply power from the MSB to the ESB during normal sea conditions, and to trip automatically upon loss of MSB voltage to prevent the emergency generator from backfeeding the dead main switchboard Correct answer.
  2. To allow the emergency generator to supply all cargo cranes and bow thrusters during port operations
  3. To continuously step up 440 V emergency power to 6.6 kV for main propulsion auxiliary drives
  4. To synchronize the emergency generator with the shore power connection without an operator present

Explanation

Under normal operating conditions, the emergency switchboard is fed directly from the main switchboard via the interconnecting tie-breaker. Upon failure of the main power supply, interlocks immediately trip this tie-breaker before the emergency generator breaker can close. This essential interlock prevents the emergency generator from attempting to energize the entire un-shed main switchboard, which would cause an immediate overload trip.