• High starting current: on DOL starting the motor draws a starting (locked rotor) current of 5 to 8 times full-load current. For a large motor this can be several thousand amperes.
• Voltage dip: this large inrush current causes a heavy voltage drop in the generator and cable impedance. The busbar voltage may fall below the acceptable limit (typically not below 85 to 90% of rated), which can cause contactors to drop out, other motors to stall, and sensitive electronic equipment to malfunction.
• Generator overload: the sudden load may exceed the capacity of the running generators, causing the prime mover to slow, frequency to fall, and possibly the overload protection to trip the generator.
• Mechanical shock: DOL starting applies full torque suddenly, causing high mechanical stress on the motor shaft, coupling, gearbox and the thruster unit, and can cause excessive wear.
• Thermal stress: the high starting current produces high I squared R heating in the windings; frequent DOL starts can overheat the motor.
• Starting torque may be too high for the driven load, causing damage to the propeller or thrust unit.
• The supply system (cables, switchgear, fuses) must be rated for the starting current, which is uneconomic for a large motor.

• A main contactor (line contactor) connecting the supply to the motor.
• A star contactor which connects the three motor phase windings in star during starting.
• A delta contactor which connects the windings in delta for running.
• A timer (time delay relay) which changes over from star to delta after the motor has accelerated.
• Overload relay, fuses or MCCB, start and stop push buttons, and an indicating lamp.
• The motor has six terminals (U1 V1 W1 and U2 V2 W2) brought out to the starter.
• During starting the star contactor closes so each winding receives line voltage divided by root 3 (phase voltage = line/1.732), reducing starting current to about one third of the DOL value. After a set time the star contactor opens and the delta contactor closes, reconnecting the windings in delta so full line voltage is applied for running.

• On pressing start, the line and star contactors close. The motor windings are connected in star, so each phase winding is subjected to phase voltage (line voltage / root 3). The starting current is therefore reduced to approximately one third of the DOL starting current, and the starting torque is reduced to about one third of the DOL torque.
• The motor accelerates under reduced voltage. After a preset time (set by the timer, typically a few seconds, allowing the motor to reach near rated speed), the star contactor opens and the delta contactor closes.
• The windings are now connected in delta, so each winding receives full line voltage and the motor runs at its normal rated condition.
• The changeover is timed so that the motor has accelerated sufficiently to avoid a large current surge when delta is applied. The overload relay protects the motor against sustained overload, and the timer prevents too early or too late a changeover.

• A single-speed three-phase squirrel-cage induction motor (the drive motor) is connected to the three-phase supply and mechanically coupled to a d.c. generator.
• The d.c. generator armature is connected directly to the armature of the d.c. winch motor (the two armatures are connected in series, forming a closed loop).
• The field winding of the d.c. generator is supplied from a d.c. exciter (or from a controlled rectifier) through a field rheostat / reversing switch.
• The field winding of the d.c. winch motor is separately excited from the same d.c. source (constant field).
• The induction motor runs continuously at constant speed, driving the d.c. generator. By controlling the generator field current, the generator e.m.f. and hence the voltage applied to the winch motor armature is varied, giving smooth speed control of the winch motor from zero to full speed in either direction.
• The circuit: 3-phase supply -> squirrel cage motor -> d.c. generator (armature) -> d.c. winch motor (armature) -> back to generator. Generator field circuit with reversing switch; motor field circuit separately excited.

• The direction of rotation of a d.c. motor depends on the relative direction of the armature current and the field flux. In the Ward Leonard system the winch motor field is kept constant, so reversal is achieved by reversing the direction of the armature current.
• This is done by reversing the polarity of the generator field current using a reversing switch (or by reversing the generator field connections). Reversing the generator field reverses the polarity of the generated e.m.f., which reverses the direction of current in the armature loop, and hence reverses the direction of rotation of the winch motor.
• Because the generator field is a low-power circuit, reversal is easy and can be done smoothly. The motor can also be brought to rest and reversed by reducing the generator field to zero and then building it up in the opposite direction, giving smooth, controlled reversal without large current surges.

• Very smooth, stepless speed control from zero to full speed in both directions.
• Excellent speed regulation and high torque at low speed, ideal for winches and windlasses.
• Easy reversal with low-power control circuits.
• Regenerative braking is possible (the motor can act as a generator and return power to the system).
• High starting torque with controlled current.

Disadvantages:

• Low overall efficiency because power is converted three times (electrical to mechanical in the drive motor, mechanical to electrical in the generator, electrical to mechanical in the winch motor).
• High initial cost and large physical size (three machines plus exciter).
• Requires more maintenance (commutators and brushes on two d.c. machines).
• The drive motor runs continuously even when the winch is idle, wasting power.
• Slow response compared to modern thyristor/static drives.

• A single NPN transistor (or PNP) with the emitter common to input and output.
• Bias resistors: R1 and R2 form a potential divider across the supply (Vcc) to set the base voltage and hence the quiescent operating point. Rc is the collector load resistor, Re is the emitter resistor (with a decoupling capacitor Ce across it) for d.c. stabilisation.
• Coupling capacitors: C1 couples the input signal to the base, and C2 couples the amplified output from the collector to the next stage or load. These block d.c. and pass only the a.c. signal.
• Decoupling capacitor: Ce across Re bypasses the a.c. signal to earth so that the a.c. gain is not reduced by negative feedback from Re, while still providing d.c. bias stabilisation.
• Supply: Vcc connected through Rc to the collector; emitter through Re to earth; base biased by R1 (to Vcc) and R2 (to earth).
• Output taken from the collector.

• R1 and R2: potential divider sets the base potential so the transistor operates in the active region (base-emitter junction forward biased, base-collector reverse biased), giving a stable quiescent point.
• Rc (collector load): converts the collector current variation into a voltage variation at the output. The amplified output voltage is developed across Rc.
• Re (emitter resistor): provides d.c. negative feedback for bias stabilisation - if collector current rises, the voltage across Re rises, reducing base-emitter voltage and opposing the rise, stabilising the operating point against temperature and transistor variations.
• Ce (emitter decoupling capacitor): presents a low impedance to the a.c. signal, so the a.c. signal is not attenuated by Re; it restores the full a.c. voltage gain while keeping d.c. stabilisation.
• C1 (input coupling capacitor): blocks the d.c. bias from the signal source and passes only the a.c. input signal to the base.
• C2 (output coupling capacitor): blocks the d.c. collector potential from the load and passes only the amplified a.c. signal.
• The transistor: amplifies the small base current/voltage signal; a small change in base current produces a much larger change in collector current (current gain beta), giving voltage and power amplification. The output at the collector is 180 degrees out of phase with the input.

A brushless alternator is a type of electrical generator that produces alternating current (AC) without the need for brushes and commutators.

• Rotor: Instead of using brushes and a commutator, a brushless alternator has a rotor with permanent magnets or electromagnets. These magnets generate a rotating magnetic field when electricity is supplied to them.
• Stator: The stator consists of coils of wire arranged around the rotor. As the magnetic field of the rotor rotates, it induces an alternating current in the stator windings through electromagnetic induction.
• Rectifier: The alternating current produced in the exiter rotor windings is then converted into direct current (DC) by a rectifier assembly and fed to the main rotor. This DC is necessary for the excitation of the rotor's magnets.
• Excitation: The DC is fed to the rotor's electromagnets or permanent magnets, creating a steady magnetic field. This field interacts with the rotating magnetic field of the rotor, inducing a three-phase AC current in the stator windings.
• Output: The three-phase AC output from the stator windings is then available for use in powering electrical devices or for distribution in an electrical grid.

• A rectifier is a device (using diodes or thyristors) that converts alternating current (a.c.) into direct current (d.c.). It allows current to flow in one direction only, so the output is a unidirectional (pulsating) d.c. voltage. Rectifiers are used to supply d.c. for battery charging, excitation of generators, d.c. motors, and electronic equipment.
• Effect of overcurrent: excessive current heats the semiconductor junction beyond its rated temperature, causing thermal breakdown and permanent damage to the diode/thyristor. It can also blow fuses, damage the transformer and cause the device to fail short-circuit. Overcurrent must be limited by fuses, circuit breakers or current-limiting circuits.
• Effect of overvoltage: if the reverse voltage across a diode exceeds its peak inverse voltage (PIV) rating, the junction breaks down (avalanche) and conducts in the reverse direction, causing overheating and destruction. Overvoltage can also be caused by switching surges, lightning or inductive loads; surge suppressors (varistors, RC snubbers) are fitted to protect the rectifier. Both overcurrent and overvoltage must be limited to protect the semiconductor devices.

Let each resistor have resistance R and the line voltage be V.

(i) Star connection:

• With three resistors in star, each phase voltage = V / root 3. Power per resistor = (V/root3)^2 / R = V^2 / (3R). Total power P3 = 3 x V^2/(3R) = V^2 / R.
• With one resistor removed, two resistors remain in star. Power P2 = 2 x V^2/(3R) = 2V^2/(3R).
• Reduction in load = P3 - P2 = V^2/R - 2V^2/(3R) = V^2/(3R).
• Fractional reduction = (V^2/(3R)) / (V^2/R) = 1/3 = 33.3%.

(ii) Delta connection:

• With three resistors in delta, each phase voltage = line voltage V. Power per resistor = V^2/R. Total P3 = 3V^2/R.
• With one resistor removed, two resistors remain in delta. Power P2 = 2V^2/R.
• Reduction = 3V^2/R - 2V^2/R = V^2/R.
• Fractional reduction = (V^2/R)/(3V^2/R) = 1/3 = 33.3%.

So in both cases the load is reduced by one third (33.3%) when one of three equal resistors is removed.

Forward bias characteristics:

• The diode conducts when the p-side is connected to the positive terminal and the n-side to the negative terminal.
• A noticeable current flow begins once the forward bias voltage reaches approximately 0.5V to 0.7V (depending on the material, silicon, or germanium).
• After this threshold, a small increase in voltage results in a large increase in current.

Reverse bias characteristics:

• When the p-side is connected to the negative terminal and the n-side to the positive terminal, the diode does not conduct (except for a small leakage current).
• For Zener diodes, conduction occurs in reverse bias after the breakdown voltage is reached.

Breakdown characteristics:

• In reverse bias, if the reverse voltage exceeds a specific value (breakdown voltage), the diode may allow a large current to flow, potentially damaging the diode (unless it is a Zener diode designed for this purpose).

Dynamic Resistance (Rd):

• This represents the diode's resistance to AC voltage. It's calculated as the change in AC voltage divided by the change in AC current. It varies depending on the operating point on the diode's I-V curve.

Static Resistance (Rs):

• This is the resistance to DC voltage. It's calculated as the DC voltage across the diode divided by the DC current through it. Similar to dynamic resistance, it also depends on the operating point on the diode's I-V curve.

• The saturation (magnetisation) curve is a graph of generated e.m.f. E against field current If, with the armature on open circuit and driven at constant rated speed.
• Starting from zero field current, a small residual e.m.f. is generated due to residual magnetism in the pole cores.
• As field current increases, the e.m.f. rises almost linearly at first (the iron is unsaturated, so flux is proportional to field current).
• As the field current is increased further, the iron begins to saturate and the e.m.f. rises more slowly, the curve bending over (knee of the curve).
• At high field currents the curve flattens off as the iron is fully saturated and further field current produces little increase in flux or e.m.f.
• The curve is used to determine the operating point, the field current required for a given voltage, and the effect of armature reaction and demagnetisation. The knee represents the most economical operating point (maximum e.m.f. per ampere of field current).

• Synchronous impedance Zs = open-circuit e.m.f. / short-circuit current (same field current) = 50 / 200 = 0.25 ohm.
• Synchronous reactance Xs = sqrt(Zs^2 - Ra^2) = sqrt(0.25^2 - 0.1^2) = sqrt(0.0625 - 0.01) = sqrt(0.0525) = 0.229 ohm.
• Induced voltage to deliver 100 A at 0.8 p.f. lagging with terminal voltage 200 V:

Assume a three-phase star-connected alternator. Phase voltage Vph = 200 / root 3 = 115.5 V. Load current per phase I = 100 A (line = phase for star). cos phi = 0.8, sin phi = 0.6.

• E = sqrt[(Vph cos phi + I Ra)^2 + (Vph sin phi + I Xs)^2]
• = sqrt[(115.5 x 0.8 + 100 x 0.1)^2 + (115.5 x 0.6 + 100 x 0.229)^2]
• = sqrt[(92.4 + 10)^2 + (69.3 + 22.9)^2]
• = sqrt[102.4^2 + 92.2^2] = sqrt[10486 + 8501] = sqrt[18987] = 137.8 V per phase.
• Line value of induced e.m.f. = 137.8 x root 3 = 238.6 V.

So the alternator must be excited to give an induced e.m.f. of about 137.8 V per phase (238.6 V line).

Rectifying Action of a Commutator:

The armature windings of a DC generator produce an AC voltage. To obtain a unidirectional (DC) voltage at the output terminals, a commutator is used. The commutator consists of multiple copper segments insulated from each other and mounted on the shaft. The ends of the armature coils are connected to these segments. Carbon brushes rest on the commutator, making contact with different segments as the commutator rotates.

As the armature rotates, the voltage induced in each coil alternates. However, the commutator segments are arranged such that when the voltage in a coil reverses, the brushes switch to contact a different set of commutator segments, connected to the coil's opposite ends. This switching action effectively reverses the coil's connections to the output terminals, thereby rectifying the alternating voltage into a pulsating direct current.

In a simple DC generator with a single coil, the output voltage would be highly pulsating. To achieve a smoother, more uniform DC output, multiple coils and commutator segments are used. The coils are arranged around the armature such that their voltages add up to produce a relatively constant output voltage, even with a pulsating waveform. The more coils and segments, the smoother the DC output becomes. This smoother output is a result of the commutator's continuous switching action between different coil windings as they pass through their peak AC voltages. The brushes are strategically positioned at the neutral points on the commutator, minimising sparking and ensuring smooth current flow.

• The stator input power P1 is the electrical power drawn from the supply.
• Stator losses (stator copper loss and iron/core loss) are subtracted to give the air-gap power Pg (also called the rotor input power), which is the power transferred across the air gap to the rotor by electromagnetic induction.
• Pg = P1 - stator losses.
• The air-gap power is divided into two parts: the rotor copper loss (I2^2 R2) and the mechanical power developed (gross mechanical power Pm).
• Pg = rotor copper loss + gross mechanical power.
• Rotor copper loss = s x Pg (where s is the slip), and gross mechanical power = (1 - s) x Pg.
• The useful (shaft) power is the gross mechanical power minus the rotational losses (friction, windage and iron losses in the rotor). So the useful power = Pg(1 - s) - rotational losses.
• The useful power is the portion that appears as mechanical output at the shaft.

• Total load: kW = 400 kW. Total kVA = 400/0.8 = 500 kVA. Total kVAr (lagging) = 500 x 0.6 = 300 kVAr (since sin phi = 0.6).
• Given: kW on A = 150 kW; kVAr on B = 150 kVAr (lagging).
• kW on B = 400 - 150 = 250 kW.
• kVAr on A = 300 - 150 = 150 kVAr (lagging).
• Machine A: kVA = sqrt(150^2 + 150^2) = sqrt(45000) = 212.1 kVA. p.f. = 150/212.1 = 0.707 lagging.
• Machine B: kVA = sqrt(250^2 + 150^2) = sqrt(85000) = 291.5 kVA. p.f. = 250/291.5 = 0.858 lagging.
• Current loading: I = S / (root 3 x V).
• I_A = 212100 / (1.732 x 440) = 212100 / 762.1 = 278.3 A.
• I_B = 291500 / 762.1 = 382.5 A.

So A supplies 150 kW at 0.707 p.f. lagging, current 278 A; B supplies 250 kW at 0.858 p.f. lagging, current 382 A.

at the start of the motor, the average torque on the rotor is zero. This is because when a DC supply is applied to the stationary rotor, the unlike poles try to attract each other, causing the rotor to be subjected to an instantaneous torque in one direction. However, the rotor's inertia prevents it from rotating, and as the stator poles continue to rotate, the direction of the torque on the rotor changes. This cycle continues, resulting in an average torque on the rotor of zero, so an external force is required to bring the motor up to the synchronous speed.

Ways to start a synchronous motor:

Pony Motor

• A smaller auxiliary motor (the "pony motor"), either AC or DC, is mechanically coupled to the synchronous motor. The pony motor accelerates the synchronous motor to a speed slightly above synchronous speed. Once this speed is reached, the pony motor is disconnected, and the synchronous motor's field is energized, allowing it to lock into synchronism with the AC supply.

Induction Motor Starting (Damper Windings)

• The rotor of the synchronous motor can be equipped with a "cage winding," essentially an embedded squirrel cage. This cage winding enables the motor to operate as an induction motor during the starting phase. The induction motor action accelerates the rotor up to near synchronous speed. Once close to synchronous speed, the DC field is applied, pulling the rotor into synchronism and allowing it to operate as a synchronous motor

Variable Frequency Drive (VFD)

• A VFD gradually increases the supply frequency from zero, enabling the synchronous motor to accelerate smoothly without additional starting mechanisms. This method provides precise control over the motor's acceleration and is commonly used in modern applications.

This ensures that any sparks or hot surfaces produced by the equipment, even under fault conditions (e.g., short circuits, open circuits, component failures), cannot ignite the surrounding atmosphere.

Key characteristics of intrinsic safety:

• Energy Limitation: Voltage and current are strictly limited using components such as resistors, Zener diodes, and fuses.
• No Ignition Source: The design ensures that no spark or hot surface can release enough energy to ignite the hazardous atmosphere.
• Fault Tolerance: The system remains safe even with multiple independent faults (e.g., “ia” protection level withstands two faults).
• Use of Barriers: Intrinsically safe circuits are typically connected to non-IS circuits in a safe area through IS barriers that limit energy transfer.

Although intrinsically safe circuits are designed to remain non-igniting even under fault conditions—making them theoretically safe for live work in hazardous areas—live maintenance is generally NOT permitted or recommended on ships or in most industrial settings.

Reasons include:

• Regulatory Restrictions: Marine regulations, company safety systems, and permit-to-work procedures typically prohibit live work in hazardous areas regardless of equipment protection type. The standard requirement is to de-energize and obtain proper permits.
• Risk of Misidentification: Personnel may accidentally work on a non-IS circuit or introduce non-IS test equipment/tools into the hazardous zone.
• Unpredictable Faults: Incorrect installation, hidden damage, or unexpected circuit faults may compromise intrinsic safety.
• Best Practice: Always isolate, de-energize, and verify zero energy before maintenance in hazardous areas.

Thus, while IS circuits are designed to make live work safe, practical safety protocols and regulations almost universally prohibit live maintenance.

(c) Intrinsically safe equipment used onboard ship.

Intrinsically safe (IS) equipment is essential for safe operation in hazardous areas on ships—particularly tankers, gas carriers, and vessels transporting dangerous goods. Hazardous zones include cargo tanks, pump rooms, cofferdams, gas-dangerous deck zones, paint lockers, and battery rooms.

Common types of intrinsically safe equipment used onboard:

• Portable Gas Detectors: Used to check for flammable gases, toxic gases, or oxygen deficiency before entering enclosed or hazardous spaces. Designed so that batteries, sensors, and circuits cannot ignite vapours.
• Portable Two-Way Radios: Specially designed walkie-talkies preventing sparks or RF energy from igniting the atmosphere.
• Portable Lighting (Torches/Hand Lamps): Battery-operated lights engineered to prevent sparks at switches, filaments, or terminals.
• Tank Gauging Systems: Intrinsically safe sensors and transmitters installed in cargo tanks or pump rooms for measuring level, temperature, and pressure.
• Fixed Gas Detection Systems: Permanently installed sensors in hazardous spaces (e.g., pump rooms) that continuously monitor for gas leaks, with IS-certified local wiring.
• Process Control Instrumentation: Pressure transmitters, temperature sensors, valve indicators, and similar devices used in cargo handling systems within hazardous zones.
• Personal Electronic Devices: IS-certified phones, tablets, and cameras used during inspections or data collection.

All intrinsically safe equipment carries an ‘Ex’ marking, specifying the explosion protection type (e.g., Ex i, Ex ia, Ex ib) and the applicable gas group and temperature class.

• In this system, the generator star point is connected directly to the ship’s hull.
• Any earth leakage current will therefore complete a circuit back to the generator.
• A Neutral Earthing Resistor (NER) is fitted to limit the earth fault current to about 5 Amps.
• This limited current is detected using a Current Transformer (CT), as illustrated in the sketch.
• The CT output is then connected to protection and alarm systems to indicate the fault.

(ii) Insulated Neutral System:

• An instrument is used which injects a DC voltage into the busbars through a resistor (R1) and a diode.
• No earth leakage condition:
  • No return path exists, hence no current flows through the circuit.
  • Voltage on both sides of R1 remains equal.
  • The Operational Amplifier (Op-Amp) detects no potential difference (PD), so the output remains zero.
• Earth leakage condition (resistance Re):
  • A return path is created through the ship’s hull.
  • Current now flows through R1, causing a voltage drop across it.
  • The Op-Amp detects a PD: one terminal sees full voltage while the other sees reduced voltage.
  • This imbalance causes the Op-Amp to send a signal to the meter/alarm system.
  • The magnitude of earth leakage determines the current flow and the PD across R1, allowing fault severity to be measured.

(i) Voltage Stabilization:

This refers to the process of maintaining a constant output voltage despite variations in the input voltage or load current. A stable voltage is required for the proper operation of electronic devices, as many are sensitive to voltage fluctuations. Methods for voltage stabilization include using Zener diodes, which maintain a constant voltage across them once a certain reverse bias voltage (breakdown voltage) is exceeded. Other methods involve using integrated circuits and feedback control loops to dynamically adjust the output voltage.

(ii) Filter Choke:

A filter choke is an inductor used in power supplies to smooth out the pulsating direct current (DC) produced by rectification. Inductors resist changes in current, so the choke helps to reduce the ripple voltage, resulting in a more stable DC output. The effectiveness of the filtering depends on the inductance of the choke and the frequency of the ripple. Often, filter chokes are used in conjunction with capacitors for optimal filtering.

(iii) Impedance:

Impedance is the measure of opposition that a circuit presents to the flow of alternating current (AC). It's a complex quantity that includes both resistance (which converts electrical energy into heat) and reactance (which stores energy in electric or magnetic fields and returns it to the circuit). Reactance, in turn, has two components: capacitive reactance (opposition due to a capacitor) and inductive reactance (opposition due to an inductor). Impedance in AC circuit analysis affects the current flow and power distribution in the circuit. Matching impedance between different parts of a circuit (e.g., a transmitter and an antenna) is essential for efficient power transfer.

(iv) Rectification:

Rectification is the process of converting alternating current (AC), which periodically reverses direction, into direct current (DC), which flows in one direction only. This is essential because many electronic devices require DC power. Rectification is usually achieved using diodes, semiconductor devices that allow current to flow easily in one direction but block it in the opposite direction. Different rectifier configurations (half-wave, full-wave, bridge) exist, each with its own characteristics regarding efficiency and ripple voltage (unwanted AC component in the DC output). Following rectification, filtering is often used to smooth the DC output.

(v) Grid Bias Voltage:

Grid bias voltage is the voltage applied to the grid of a vacuum tube (triode or other multi-element tube) relative to its cathode. This voltage controls the flow of electrons between the cathode and the anode (plate), acting as a gate to regulate the output current. A negative grid bias voltage reduces the flow of current, while a less negative or positive bias increases the current. Grid bias is essential for establishing the operating point of the vacuum tube, determining its amplification characteristics and preventing distortion in the output signal. The concept is analogous to the base-emitter voltage in transistors, controlling the collector current.

P-Type and N-Type Semiconductor Materials and the Mechanism of Current Conduction

A semiconductor is a material whose electrical conductivity lies between that of a conductor and an insulator. Pure semiconductor materials such as silicon (Si) and germanium (Ge) have limited conductivity. Their conductivity can be greatly increased and controlled by adding a small amount of a suitable impurity. This process is known as doping.

Depending on the type of impurity added, a semiconductor becomes either an N-type or a P-type semiconductor.

1. N-Type Semiconductor

An N-type semiconductor is formed by adding a small quantity of a pentavalent impurity to a pure semiconductor such as silicon or germanium. Pentavalent impurities have five valence electrons. Examples include phosphorus, arsenic and antimony.

The pure semiconductor atom forms four covalent bonds with neighbouring atoms. When a pentavalent impurity atom replaces one of these atoms:

• Four of its five valence electrons form covalent bonds with neighbouring atoms.
• The fifth electron is weakly bound and becomes a free electron.
• This free electron can move through the crystal structure when an electric field is applied.

Therefore, an N-type semiconductor has an excess of free electrons.

• Majority charge carriers: Free electrons.
• Minority charge carriers: Holes.

The letter "N" denotes that the majority charge carriers are negative electrons. However, the semiconductor material as a whole remains electrically neutral.

Current Conduction in an N-Type Semiconductor

When a voltage is applied across an N-type semiconductor, an electric field is established.

The free electrons gain energy from the electric field and drift through the crystal towards the positive terminal (anode). Their movement through the conduction band constitutes the main mechanism of current conduction.

Thus:

Applied voltage → Electric field → Movement of free electrons → Current flow

Although conventional current is considered to flow from positive to negative, the actual electrons move in the opposite direction, from the negative terminal towards the positive terminal.

2. P-Type Semiconductor

A P-type semiconductor is formed by adding a small quantity of a trivalent impurity to a pure semiconductor. Trivalent impurities have three valence electrons. Examples include boron, gallium and indium.

When a trivalent impurity atom is introduced into the semiconductor crystal:

• Its three valence electrons form covalent bonds with neighbouring atoms.
• One bond remains incomplete because there is a shortage of one electron.
• This missing electron position is called a hole.

A hole behaves as a positive charge carrier because it represents a deficiency of an electron.

Therefore, a P-type semiconductor has an excess of holes.

• Majority charge carriers: Holes.
• Minority charge carriers: Free electrons.

The letter "P" denotes that the majority charge carriers are effectively positive holes. However, the semiconductor material as a whole remains electrically neutral.

Current Conduction in a P-Type Semiconductor

When a voltage is applied across a P-type semiconductor, an electric field is established.

The holes act as the main charge carriers. However, the actual physical movement is still carried out by electrons. An electron from a neighbouring covalent bond moves to fill a nearby hole. This movement leaves a new hole at the electron's original position.

The process continues as follows:

Electron fills a hole → A new hole is created → Another electron fills the new hole → Progressive movement of holes

As a result of this continuous electron movement, the holes appear to move through the material towards the negative terminal (cathode). The resulting progressive movement of holes in the valence band constitutes the main current flow in a P-type semiconductor.

Summary of P-Type and N-Type Semiconductors

Temperature Sensor:

• Measures the temperature of the system. Several types exist, including Resistance Temperature Detectors (RTDs), Thermistors, and filled-tube thermometers. The sensor provides a signal representing the measured value (MV) of the temperature.

Transmitter:

• The transmitter receives the signal from the temperature sensor. Its primary function is to amplify and condition this signal, making it suitable for comparison and processing by the controller. It converts the sensor's output into a standardized signal (e.g., 4-20 mA).

Comparator:

• The comparator compares the measured value (MV) signal from the transmitter with the set value (SV) or desired temperature. The difference between the MV and SV is the error signal. This signal reflects how far the actual temperature deviates from the desired temperature.

Temperature Controller:

• This is the brain of the system. It receives the error signal from the comparator and uses a control algorithm (often a PID – Proportional, Integral, Derivative – controller) to determine the appropriate corrective action. The PID algorithm adjusts the output signal to minimize the error.

Signal Converter:

• This component takes the output signal from the controller and converts it into a form suitable to operate the actuator. For example, it might convert an electrical signal into a pneumatic signal (compressed air pressure) or a hydraulic signal.

Actuator (or 3-way Valve):

• The actuator is the final control element. It receives the converted signal and makes adjustments to the system to correct the temperature. Examples include pneumatic diaphragm control valves, which control the flow of a heating or cooling medium. A higher signal might open the valve to allow more heat, while a lower signal would reduce the flow.

Star connection:

• The three phase windings have one end of each connected to a common neutral point; the other ends form the three line terminals.
• Line current = phase current: IL = Iph (the same current flows through the winding and the line).
• Line voltage = root 3 x phase voltage: VL = root 3 x Vph. The line voltage is the phasor difference of two phase voltages, which are 120 degrees apart, giving a magnitude of root 3 times the phase voltage.
• Vector diagram: three phase voltages VRN, VYN, VBN at 120 degrees; the line voltage VRY is the phasor sum of VRN and (-VYN), equal to root 3 Vph and leading the phase voltage by 30 degrees.

Delta connection:

• The three windings are connected end to end to form a closed loop, the junctions forming the three line terminals.
• Line voltage = phase voltage: VL = Vph (each winding is directly across two lines).
• Line current = root 3 x phase current: IL = root 3 x Iph. The line current is the phasor difference of two phase currents, giving root 3 times the phase current.
• Vector diagram: three phase currents at 120 degrees; the line current is root 3 times the phase current and lags the phase current by 30 degrees.
• Power in both cases: P = root 3 x VL x IL x cos phi.

• |Z| = sqrt(8^2 + 12^2) = sqrt(64 + 144) = sqrt(208) = 14.42 ohm.
• cos phi = R/|Z| = 8/14.42 = 0.555.

(i) Phase current: in delta, phase voltage = line voltage = 415 V. Iph = Vph/|Z| = 415/14.42 = 28.78 A.

(ii) Line current: IL = root 3 x Iph = 1.732 x 28.78 = 49.85 A.

(iii) Power consumed by each phase: Pph = Iph^2 x R = 28.78^2 x 8 = 828.3 x 8 = 6626 W (6.63 kW).

• Total power = 3 x 6626 = 19.88 kW (also = root 3 x 415 x 49.85 x 0.555 = 19.88 kW).

1. Field Diverter Method:

A variable resistor, known as a diverter, is connected in parallel with the series field winding. By adjusting the diverter's resistance, a portion of the current is shunted away from the field winding, reducing the field current and, consequently, the magnetic flux. This reduction in flux leads to an increase in motor speed, as speed is inversely proportional to flux.

2. Tapped Field Control:

In this method, the field winding is divided into sections with taps at various points. A selector switch allows the operator to choose different numbers of turns in the field winding, effectively varying the total ampere-turns. Selecting fewer turns reduces the magnetic flux, resulting in higher motor speed. This method provides discrete speed settings based on the available taps.

3. Armature Diverter Method:

Here, a variable resistor is connected in parallel with the armature winding. Adjusting this resistor changes the current distribution between the armature and the diverter. For a constant load torque, reducing the armature current increases the field current, enhancing the magnetic flux and decreasing the motor speed. Conversely, increasing the armature current reduces the field current, decreasing the flux and increasing the speed.

$$R_a \space = \space 0.025Ω$$

$$R_{sh} \space = \space 230Ω$$

$$I_1 \space = \space 5A$$

$$V \space = \space 230V$$

$$N_1 \space = \space 1000rpm$$

$$I_{sh} \space = \space {{230} \over 230} \space = \space 1A$$

$$I_{a1} \space = 5 - 1 \space = \space 4A$$

$$E_{b1} \space = \space V - I_{a1}R_a$$

$$= \space 230 - 4 \times 0.025$$

$$= \space 229.9V$$

On load,

$$I_2 \space = \space 41A$$

$$I_{sh} \space = \space 1A$$

$$I_{a2} \space = \space 41 - 1 \space = \space 40A$$

$$N_2 \space = \space ?$$

$$E_{b2} \space = \space V - I_{a2}R_a$$

$$= \space 230 - 40 \times 0.025$$

$$= \space 229V$$

Since Ф remains the same,

$$E_b \space ∝ \space N$$

$${{N_2} \over N_1} \space = \space {{E_{b2}} \over E_{b1}} \space$$

$$N_2 \space = \space {{229} \over 229.9} \times 1000$$

$$N_2 \space = \space 996.08rpm$$

Drop in speed:

$$= \space 1000 - 996.08 \space = \space 3.92rpm$$

• Supply Frequency (f)
• Number of Poles (P)

The relationship between these factors and the synchronous speed (Ns) is given by the formula:

$$N_{s}=\frac{120f}{P}$$

Where Ns is the synchronous speed in revolutions per minute (RPM), f is the supply frequency in hertz, and P is the number of poles.

For example,

• A 2-pole motor (P=2) running on a 50 Hz supply (f=50) will have a synchronous speed of 3000 RPM.
• A 4-pole motor (P=4) on the same supply will have a synchronous speed of 1500 RPM.

In practical applications, especially for induction motors, the actual operating speed is slightly less than the synchronous speed due to a phenomenon known as slip, which is necessary for torque production.

For synchronous motors, the rotor locks in step with the stator's rotating magnetic field, and thus operates precisely at the synchronous speed.

Given:

$$Output=20000W$$

$$Iron\:loss=780W$$

$$Friction\:loss\:\left(mech\operatorname{loss}\right)=590W$$

$$R_{a}=0.15\Omega$$

$$R_{se}=0.025\Omega$$

$$R_{int}=0.028\Omega$$

$$R_{sh}=115\Omega$$

$$I_{sh}=\frac{V}{R_{sh}}=\frac{200}{115}$$

$$I_{sh}=1.74A$$

$$Copper\:loss\:in\:stator=I^2R$$

$$C_{S}=1.74^2\times115$$

$$C_{S}=348W$$

$$Gen\:output\:=\:V\times I_{L}=20000W\:\left(given\right)$$

$$200\times I_{L}=20000$$

$$I_{L}=100A$$

$$I_{a}=I_{sh}+I_{L}$$

$$=1.74+100$$

$$I_{a}=101.74A$$

$$Copper\:loss\:in\:stator=I^2R=I_{a}^2\left(R_{se}+R_{a}+R_{int}\right)$$

$$C_{R}=101.74^2\times\left(0.025+0.15+0.028\right)$$

$$C_{R}=2101W$$

$$Total\:Copper\:loss=C_{S}+C_{R}$$

$$=348+2101$$

$$=2449W$$

$$\eta=\frac{Output}{Input}=\frac{Output}{Output+losses}$$

$$=\frac{20000}{20000+780+590+2449}$$

$$\eta=83.96\%$$

is an alternator that is directly coupled (without any intermediate gearing mechanisms like belts or chains) to its driving source, such as a diesel engine. This integration allows the alternator to be mounted directly on the extension shaft of the engine. Such alternators are typically used in portable engine-driven applications and are generally small in size, with power ratings ranging from 1 to 1.75 kW. These systems are often air-cooled and compact, making them suitable for mobile and low-power setups.

(ii) In a direct-connected alternator, the stator (or armature winding) may be either single-phase or three-phase with distributed winding. The rotor (field winding) is typically a silent pole design, often using permanent magnets for field excitation.

The arrangement ensures stable performance with:

• Voltage variations within ±5% from no-load to full-load conditions.
• Frequency variation limited to ±1% of its rated value.

Temperature Sensor:

• Measures the temperature of the system. Several types exist, including Resistance Temperature Detectors (RTDs), Thermistors, and filled-tube thermometers. The sensor provides a signal representing the measured value (MV) of the temperature.

Transmitter:

• The transmitter receives the signal from the temperature sensor. Its primary function is to amplify and condition this signal, making it suitable for comparison and processing by the controller. It converts the sensor's output into a standardized signal (e.g., 4-20 mA).

Comparator:

• The comparator compares the measured value (MV) signal from the transmitter with the set value (SV) or desired temperature. The difference between the MV and SV is the error signal. This signal reflects how far the actual temperature deviates from the desired temperature.

Temperature Controller:

• This is the brain of the system. It receives the error signal from the comparator and uses a control algorithm (often a PID – Proportional, Integral, Derivative – controller) to determine the appropriate corrective action. The PID algorithm adjusts the output signal to minimize the error.

Signal Converter:

• This component takes the output signal from the controller and converts it into a form suitable to operate the actuator. For example, it might convert an electrical signal into a pneumatic signal (compressed air pressure) or a hydraulic signal.

Actuator (or 3-way Valve):

• The actuator is the final control element. It receives the converted signal and makes adjustments to the system to correct the temperature. Examples include pneumatic diaphragm control valves, which control the flow of a heating or cooling medium. A higher signal might open the valve to allow more heat, while a lower signal would reduce the flow.

An Automatic Voltage Regulator (AVR) regulates the generator terminal voltage by controlling the amount of current supplied to the generator field winding by the exciter.

The AVR controls the alternator output voltage by automatic adjustment of the exciter stator field strength. The AVR provides closed-loop control by sensing the alternator output voltage at the main stator windings and adjusting the exciter stator field strength

• Design a separate HV side panel for the bow thrusters, isolated from adjacent LV panels.
• Ensure independent earthing system for the HV side, connecting to the ship's hull.
• Incorporate harmonic filters on both ends of the HV busbar to avoid waveform distortion.
• MSB should be robust to withstand maritime conditions such as vibration, humidity, high temperatures, and exposure to sea water.
• Implement a neutral earth resistor (NER) for neutral point earthing.
• Choose circuit breakers suitable for HV, such as vacuum or SF6 types, with manual or motor-driven closing mechanisms.
• Integrate standard interlocks and trips for under voltage, overcurrent, and short circuit protection.
• Include monitoring instruments for incoming voltage, frequency, current, in/out power lamps, and load status lamps.
• Additional features like anti-condensation heaters and interlocks for main-emergency power and door closure are recommended.

• Utilize a combined motor protection relay offering overcurrent (inverse time relay), differential current monitoring, and earth fault monitoring.
• Connect the relay to a 'Lock-out' relay to trip the motor power in the event of overload.
• Implement thermistor trip/protection for single phasing and high-temperature conditions.

• Ensure proper cable rating based on the electrical system specifications (1900/3300V for earthed neutral system; 3300/3300V for insulated neutral system).
• Choose conductors rated for short circuit and maximum fault current.
• Select high-quality insulation materials such as EPR (ethylene propylene rubber) or XLPE (cross-linked polyethylene).
• Use sheet material with ratings for short circuit, heat, oil, chemical resistance, and flame retardancy.
• Plan cable tray routing and ensure proper junctions with terminals.
• Consider the complexity of insulation design for HV cables, emphasizing clearance and creepage distance.
• Recognize that less copper area is required for HV conductors, allowing for space and weight savings during installation.

• In the event of a single earth fault, no earth fault current flows through the ship's hull due to the insulated neutral, minimizing fire hazards.
• The hull does not carry current, ensuring safety from electrical currents passing through the structure.
• A single earth fault does not cause generator breaker tripping, avoiding sudden blackouts or operational disruptions.
• Harmonic currents caused by third harmonics in the generated voltage are prevented from flowing through the neutral, protecting the generator windings from overloading.

1. Only one system voltage (line-to-line) is possible, unlike earthed neutral systems which also provide line-to-neutral voltages.
2. While an earth fault alarm and phase indicator are triggered, locating the exact fault location requires a time-consuming trial-and-error process.
3. In cases of inductive or capacitive faults to earth, surge voltage can rise 3.5 to 4 times the system voltage, risking insulation failure and system collapse.

A metallic resistor is inserted between the neutral point and the ship’s hull to limit earth fault current.

The resistor’s value is determined by:

$$R=\frac{V}{\sqrt3I}\:$$

Where,

• V = Line voltage,
• I = Full load current.

Metallic resistors are used for their stability, low maintenance, and ability to prevent arcing grounds.

The risk of electric shock is considered equally dangerous in both earthed and insulated neutral systems. In an insulated system, normal leakage currents from capacitance and surface leakage, along with the possibility of earth faults, mean that touching live parts still carries a considerable shock risk. Similarly, in an earthed system, line-to-neutral voltages (even those as low as 110 or 250 volts) can be lethal under certain shipboard conditions, making neither system inherently safer regarding electric shock than the other. Appropriate safety precautions are essential for both systems.

• The resistance R is independent of frequency; it is a horizontal straight line on the graph of reactance/resistance against frequency.
• Inductive reactance XL = 2 pi f L increases linearly with frequency (a straight line through the origin, rising with f).
• Capacitive reactance XC = 1/(2 pi f C) decreases with frequency (a hyperbola, falling as f increases).
• The net reactance X = XL - XC. At low frequency XC dominates (capacitive); at high frequency XL dominates (inductive).
• At the resonant frequency f0, XL = XC, the net reactance is zero, and the circuit behaves as purely resistive. The impedance is a minimum (equal to R) and the current is a maximum.
• Below resonance the circuit is capacitive (current leads voltage); above resonance it is inductive (current lags voltage).
• The graph shows R as a horizontal line, XL rising linearly, XC falling hyperbolically, and the two curves crossing at resonance.

• Capacitive reactance XC = 1/(2 pi f C) = 1/(2 x 3.1416 x 50 x 30 x 10^-6) = 1/0.009425 = 106.1 ohm.
• Current in resistor: IR = V/R = 230/130 = 1.769 A (in phase with voltage).
• Current in capacitor: IC = V/XC = 230/106.1 = 2.168 A (leads voltage by 90 degrees).
• Total current: I = sqrt(IR^2 + IC^2) = sqrt(1.769^2 + 2.168^2) = sqrt(3.129 + 4.700) = sqrt(7.829) = 2.798 A.
• Phase angle: cos phi = IR/I = 1.769/2.798 = 0.632, so phi = 50.8 degrees (current leading voltage, since capacitive).
• Power consumed: P = V x IR = 230 x 1.769 = 406.9 W (only the resistor consumes power; also P = V I cos phi = 230 x 2.798 x 0.632 = 406.9 W).

So IR = 1.77 A, IC = 2.17 A, total current = 2.80 A, phase angle = 50.8 degrees leading, power = 407 W.

Types of Rotors in Three-Phase Induction Motors:

Squirrel-Cage Rotor:

• Consists of laminated steel sheets with parallel slots carrying heavy copper or aluminum bars. The ends of these bars are short-circuited by end rings, forming a closed loop resembling a squirrel cage.
• The simplicity and ruggedness of the squirrel-cage rotor make it the most commonly used type in induction motors.

Wound Rotor (Slip-Ring Rotor):

• Features a laminated iron core with three-phase windings placed in the slots. These windings are connected to external slip rings via brushes, allowing for external connections.
• The wound rotor design allows for external resistances to be added to the rotor circuit, enabling control over the motor's starting torque and speed. This feature is particularly useful in applications requiring precise speed control and higher starting torque.

In a practical transformer, iron (core) losses remain nearly constant from no-load to full-load operation. This means that the power loss occurring in the transformer core does not change significantly with changes in the load current.

Iron losses consist of two main components:

1. Hysteresis loss
2. Eddy current loss

Both of these losses are produced by the alternating magnetic flux in the transformer core and depend mainly on the supply voltage, frequency, and core material properties, rather than on the load current.

1. Hysteresis Loss

Hysteresis loss is caused by the continuous reversal of magnetization in the iron core as the alternating current produces an alternating magnetic field.

During every AC cycle, the core undergoes repeated magnetization and demagnetization. This repeated reversal requires energy because of the inherent magnetic properties of the core material, and the energy is dissipated as heat.

The magnitude of hysteresis loss depends on:

• The area of the hysteresis loop of the core material.
• The frequency of the alternating supply.
• The magnetic properties of the core, such as coercivity and magnetic permeability.

Since these factors remain practically constant for a transformer operating at constant supply voltage and frequency, hysteresis loss is essentially independent of the load current.

2. Eddy Current Loss

Eddy current loss is caused by circulating currents induced within the transformer core due to the alternating magnetic flux.

These induced currents flow through the resistance of the core material, producing heat and resulting in power loss.

The magnitude of eddy current loss depends on:

• The electrical resistivity of the core material.
• The frequency of the alternating supply.
• The core geometry and thickness of the laminations.

Since these factors are determined by the transformer design and operating frequency, eddy current loss also remains practically independent of the load current.

• Starting current: at standstill (speed = 0) the starting current is high (5 to 8 times full-load current). As the motor accelerates the current falls, and at synchronous speed it would be zero (in practice small no-load current). The current curve falls from a high value at zero speed to a low value near synchronous speed.
• Torque: at standstill the starting torque is moderate (about 1.5 to 2 times full-load torque). As speed increases the torque rises to a maximum (pull-out torque) at a speed corresponding to the slip for maximum torque, then falls to zero at synchronous speed. The torque-speed curve rises from the starting value, peaks, then drops to zero at synchronous speed.
• The two curves are plotted against speed from 0 to synchronous speed.

• As a motor: input power = 10/0.85 = 11.765 kW. Line current = 11765/230 = 51.15 A.
• Shunt field current (motor) = 230/220 = 1.045 A. Armature current (motor) = 51.15 - 1.045 = 50.1 A.
• Armature copper loss (motor) = Ia^2 Ra = 50.1^2 x 0.15 = 2510 x 0.15 = 376.5 W.
• Back e.m.f. (motor) E = V - Ia Ra = 230 - 50.1 x 0.15 = 230 - 7.5 = 222.5 V.
• As a generator driven at 1080 rev/min with field current 1.1 A:
• E.m.f. is proportional to speed and flux. Flux is proportional to field current (assumed linear). E_g = E_m x (1080/1000) x (1.1/1.045) = 222.5 x 1.08 x 1.0526 = 252.9 V.
• Armature copper loss kept the same as when running as a motor (376.5 W): Ia^2 x 0.15 = 376.5, so Ia = sqrt(376.5/0.15) = sqrt(2510) = 50.1 A.
• Terminal voltage of generator V = E_g - Ia Ra = 252.9 - 50.1 x 0.15 = 252.9 - 7.5 = 245.4 V.
• Load current = Ia - field current = 50.1 - 1.1 = 49.0 A.
• Output power = V x I_load = 245.4 x 49.0 = 12.02 kW.

So the expected generator output is about 12 kW.

Differences between squirrel cage and wound rotor motors

(a) Rotor Construction

Squirrel Cage Motor:

• Rotor consists of aluminum or copper bars embedded in a laminated steel core.
• These bars are short-circuited at both ends using end rings, forming a cage-like structure.
• Construction is simple, robust, and cost-effective.

Wound Rotor Motor:

• Rotor contains a three-phase winding similar to the stator winding.
• The winding is connected to slip rings and brushes.
• Slip rings enable connection of external resistors to the rotor, allowing for control of rotor current.
• Construction is more complex and expensive compared to squirrel cage motors.

Squirrel Cage Motor:

• Provides low to moderate starting torque.
• Exhibits low slip at full load, resulting in nearly constant speed operation.
• Suitable for applications where high starting torque is not required and simple, reliable operation is preferred.

Wound Rotor Motor:

• Has higher starting torque compared to squirrel cage motors due to the addition of external resistance in the rotor circuit.
• Allows torque control by varying external rotor resistance.
• Suitable for high-inertia loads and applications requiring smooth starting and controlled acceleration.

(c) Speed Variations

Squirrel Cage Motor:

• Speed is almost constant at a fixed frequency due to low slip.
• Speed regulation is poor, with minor variations under load.
• Limited speed control, typically achieved through supply voltage variation or by using variable frequency drives (VFDs).

Wound Rotor Motor:

• Allows wide speed variation by adjusting external resistance connected to the rotor winding.
• Offers good speed regulation with proper control methods.
• Capable of adjustable speed, making it suitable for applications requiring speed control.

Type and Location:

• Usually a diesel-driven internal combustion engine connected to an alternator.
• Installed in a separate compartment, typically on the upper deck or superstructure, and outside the main machinery space to ensure isolation from hazards such as fire or flooding in the engine room.

Automatic and Manual Operation:

• Equipped with an automatic starting system, which activates within 45 seconds of main power failure.
• A manual start option must also be available in case of automatic system failure.

Fuel Supply:

• Supplied by a dedicated day tank, ensuring at least 18 hours (for Cargo ships) and 36 hours (for passenger ships) of continuous operation under full emergency load conditions.

Emergency Switchboard:

• The generator supplies power to an emergency switchboard, from which electrical power is distributed to essential services such as
  • Emergency lighting
  • Fire detection and alarm systems
  • Emergency communication systems
  • Steering gear and navigation equipment
  • Fire pumps and bilge pumps

  Battery Backup:

  • In addition to the generator, emergency batteries are provided to supply immediate power to lighting, communication, and control systems during the delay in generator starting.

  Capacity:

  • The emergency generator is sized adequately to supply simultaneous power to all essential systems required for the safety of the ship and personnel during emergencies.
  Part (b)

  List of Essential Services That Must Be Supplied Simultaneously1. Emergency Lighting

  • Navigational bridge
  • Engine control room and Engine room
  • Escape routes and stairways
  • Emergency generator room
  • Emergency lights throughout vital areas

  2. Navigation and Control Equipment

  • At least one steering gear motor
  • Navigation lights and signal lights
  • Bridge control and monitoring instruments
  • Navigation equipment

  3. Communication Systems

  • Internal communication systems (PA system, intercom)
  • External communication systems (GMDSS)
  • Emergency alarms (general, fire, CO₂ warning)

  4. Fire Detection and Firefighting Systems

  • Emergency fire pump or fire pumps
  • Sprinkler / Hi-fog / water spray systems
  • Fire detection panels and fire detectors

  5. Emergency Machinery and Systems

  • Emergency air compressor
  • BA (Breathing Apparatus) compressor
  • CO₂ room exhaust fan
  • One engine room vent fan
  • Emergency generator fuel oil pumps and ventilation
  • Engine room pumps and systems required for first start from dead ship condition
  • Essential ventilation and fuel pumps for emergency equipment

  6. Lifesaving Equipment

  • Lifeboat davits
  • Watertight door control systems
  • Bilge alarm systems

  7. Electrical and Monitoring Systems

  • Emergency battery charging circuits
  • Battery chargers
  • UPS (Uninterruptible Power Supply) system
  • Engine room alarm system
  Part (c)

  Periodic Testing of Emergency Installation

  1. Weekly Testing (No Load / Manual Start)

  • The emergency generator is started manually and run without load.
  • Primary and secondary starting systems are tested (if available).
  • Parameters such as voltage, frequency, oil level, and fuel level are checked.
  • Exhaust temperature and sump oil level are monitored.
  • The automatic starting system is tested by simulating a power failure to verify functionality.

  2. Monthly Testing (Simulated Automatic Start)

  • A simulated power failure is carried out by opening the interconnector breaker between the main and emergency switchboards.
  • The emergency generator should start automatically and connect to the emergency switchboard.
  • Battery voltage and electrolyte levels are also checked.

  3. Quarterly Testing (On Load)

  • The emergency generator is operated on load for at least 30 minutes.
  • As much of the emergency load as safely possible is connected.
  • This test confirms the generator's ability to supply essential services and reach normal operating temperatures and pressures.
  • Transfer switches are also tested to ensure seamless transition.

  4. Annual Testing (Class Survey / Blackout Test)

  • A controlled blackout test may be performed by shutting down the main power (if safe and permitted).
  • The test confirms that the generator automatically starts, connects to the emergency switchboard, and restores all essential services.
  • A full inspection or overhaul of the emergency generator is carried out.
  • Auto-start and auto-transfer logic are verified.
  • The emergency switchboard, wiring, and circuits are thoroughly inspected.

  5. Battery Testing

  • Accumulator or emergency batteries are tested for:
    • Charge levels
    • Discharge capability
    • Terminal condition and connections
  • Controlled discharge tests (typically done in port or during drydock) may be used to assess actual capacity.

  6. Testing of Automatic Transfer Switches (ATS)

  • Automatic transfer switches are checked to ensure they:
    • Detect power failure
    • Initiate generator start
    • Transfer load smoothly

    7. Fuel Supply Verification

    • Regular checks to confirm:
      • Adequate fuel quantity
      • Correct fuel quality
      • Cleanliness of fuel tank and supply lines

      8. Log Book Entry

      • All tests must be logged with:
        • Date and time
        • Load details
        • Duration
        • Any observed faults or anomalies

(a) Overcoming High Starting Current:

(i) Star-Delta Starting:

• The stator windings have end connections brought out to a starter box with six terminals.
• These are first connected in a star configuration for starting, which reduces the voltage and hence the current.
• Once the rotor comes up to speed, the windings are automatically reconfigured to delta using a timer circuit.
• Interlocks are provided to prevent both star and delta contacts from closing together, ensuring safe switching.

(ii) Auto-Transformer Starting:

• An autotransformer with tapping points is used to provide reduced voltage during starting.
• Initially, reduced voltage is supplied to the motor through the autotransformer, which lowers the starting current.
• As the rotor picks up speed, the voltage is gradually increased using higher tapping points.
• Once full voltage is reached, the mains supply is directly connected to the motor, and the autotransformer is isolated.

(b) Overcoming Low Starting Torque:

(i) Wound Rotor Motor:

• The rotor has three windings connected at one end and brought out through slip rings.
• External variable resistances are connected through brushes and slip rings.
• At starting, current passes through these resistances, producing high starting torque.
• As speed increases, the resistance is reduced and eventually short-circuited by a common connection.

(ii) Double Cage Rotor:

• The rotor is designed with two sets of bars:
  • Outer cage: small cross-section, high resistance.
  • Inner cage: large cross-section, low resistance.
• At startup, most current flows in the high-resistance outer cage, developing high starting torque.
• As the speed increases, the slip decreases, the inner cage reactance reduces, and it takes over torque production efficiently.

(i) Hazards and problems associated with electric cable insulation in the event of fire:

The insulation of electric cables is typically made from rubber or plastic. The type and quantity of smoke produced during the combustion of plastic materials depend on various factors, such as:

• The nature of the plastic
• The presence of additives
• Whether the fire is flaming or smouldering
• The availability of ventilation

Most plastics decompose when heated, producing dense to very dense smoke. Ventilation may help in dispersing the smoke, but usually not enough to maintain clear visibility. Plastics that burn cleanly emit less dense smoke when subjected to heat and flame.

Urethane foam, when exposed to both flaming and non-flaming heat, generally produces dense smoke, and visibility can be lost within seconds.

Hydrogen chloride, a deadly gas with a pungent and irritating odour, is released during the combustion of chlorine-containing plastics such as PVC—commonly used in electrical wiring insulation.

Burning rubber produces dense, black, oily smoke, which has toxic properties. Two of the harmful gases released in the combustion of rubber are hydrogen sulphide and sulphur dioxide, both of which are dangerous and potentially lethal.

(ii) Remedies for these problems:

• Use cables with Fire-Resistant (FFR) insulation combined with flame-retardant sheathing, such as FEP or XLPE, and stainless steel (SS) armouring.
• The SS armouring must be properly earthed.
• The combustibility of insulation material is assessed by its oxygen index number, which represents the minimum percentage of oxygen required to sustain combustion:
  • Materials with an oxygen index below 21 will continue to burn.
  • Materials with an oxygen index of 27 or above are self-extinguishing.
• Therefore, insulation materials should have an oxygen index greater than 27 to ensure fire resistance.

• All electric cables installed externally to equipment must be of flame-retardant type and installed in a way that preserves their flame-retarding properties.
• Cables and wiring serving essential or emergency power, lighting, internal communications, or signals should, wherever possible, be routed away from high-risk areas such as galleys, laundries, refrigerated cargo (r/c) spaces of category 'A', their casings, and other hazardous zones.
• In hazardous areas where cables could cause fire or explosions during an electrical fault, special precautions must be taken.
• Cables should be installed and supported in a manner that avoids chafing or other physical damage.
• Terminations and joints must maintain the fire-resistant properties of the original cable.
• Every individual circuit should be protected against short-circuiting and overloading.
• When a cable passes through a bulkhead or exits a gland box, a fireproof compression gland must be fitted to prevent the spread of fire.

Electrical Causes:

• Overcurrent due to overvoltage, defective insulation, or overloading.
• Unbalanced supply voltage.
• Single phasing (loss of one phase in a three-phase system).

Mechanical Causes:

• Overloading (mechanical or electrical).
• Misalignment of the motor.
• Bearing troubles.
• Vibrations.

Environmental Causes:

• High ambient temperature.
• Improper ventilation.

Other Causes:

• Damaged insulation of windings.

Overload protection:

• Thermal Overload Relays: These devices monitor the motor's current and disconnect the power supply if the current exceeds a preset limit for a specified duration, preventing overheating.
• Magnetic Overload Relays: They respond to excessive currents by utilizing magnetic fields to trip the circuit, offering rapid protection against short circuits.

Overcurrent protection:

• Fuses and Circuit Breakers: Installed in the motor's power supply line, they interrupt the circuit during overcurrent situations, safeguarding the motor and associated wiring.

Environmental Protection:

• Proper Enclosures: Selecting appropriate motor enclosures shields the motor from dust, moisture, and other environmental factors that could cause damage.
• Regular Maintenance: Routine inspections and maintenance, such as checking for condensation and ensuring proper ventilation, help maintain motor health.

Temperature Monitoring:

• Thermistors and Temperature Sensors: Embedded in the motor windings, these devices monitor temperature and can trigger alarms or shutdowns if overheating is detected.

Proper Installation and Alignment:

• Alignment Checks: Ensuring the motor is correctly aligned with the driven equipment reduces mechanical stress and prevents premature wear.
• Vibration Monitoring: vibration analysis can detect misalignment or imbalance issues early, allowing for corrective action before significant damage occurs.

Fuse backup protection serves as a secondary line of defence against severe faults. If a short circuit occurs in the motor starter or supply cable, it can generate a massive fault current. This current poses a significant risk of damaging the motor windings and cables. The fuses, placed upstream of the contactor, act as a fast-acting protective device. They instantly trip, disconnecting the power supply and thus preventing extensive damage. These fuses are specifically designed with a time/current characteristic that allows them to tolerate the brief high current surge during direct-on-line (DOL) motor starting without blowing, while rapidly responding to sustained short circuit currents. The coordination between the overcurrent relays (OCR) and the fuses is crucial. The contactor should trip based on thermal overload detected by the OCR, while the fuses handle short circuit fault currents.

A three-phase AC supply energises the three stator windings, creating a rotating magnetic field. This field rotates at a synchronous speed determined by the supply frequency and the number of motor poles. As this rotating magnetic field sweeps across the rotor conductors (in a squirrel cage rotor), it induces an alternating electromotive force (EMF). Because the rotor conductors are shorted, these induced EMFs create rotor currents. These rotor currents, in turn, generate a magnetic field that interacts with the rotating stator field, producing a torque. This torque forces the rotor to rotate in the same direction as the rotating magnetic field. The direction of rotation can be determined using Fleming's left-hand rule.

The condition for maximum running torque in an induction motor is achieved when the rotor's resistance equals the rotor's reactance (R_r = X_r). This situation creates the maximum interaction between the rotor and stator fields, leading to the highest possible torque output.

However, it's important to note that maximum torque occurs at a specific slip (difference between synchronous speed and actual rotor speed) and not necessarily at the motor's rated speed.

The form factor of an AC waveform is the ratio of its RMS value to its average value. For a perfect sine wave, the form factor is approximately 1.11 (RMS value/Average Value = 1.11). This factor indicates how closely a waveform resembles a pure sine wave; a form factor closer to 1.11 suggests a waveform that is more sinusoidal.

• A peak rectifier consists of a diode in series with a capacitor and a load resistor. The diode rectifies the a.c. input and charges the capacitor to the peak value of the input voltage.
• Circuit: a.c. source -> diode -> node (capacitor to earth, load resistor to earth). The output is taken across the capacitor/load.
• Operation: during the positive half cycle the diode conducts and charges the capacitor to the peak value Vm. When the input falls below the capacitor voltage, the diode becomes reverse biased and stops conducting; the capacitor discharges slowly through the load resistor. If the time constant (R x C) is large compared with the period, the capacitor holds the voltage near Vm, so the output is approximately the peak value.
• If the supply voltage is v(t) = Vm sin(wt), the voltage across the load resistor is approximately the peak value Vm (for an ideal diode with negligible forward drop and a large time constant). The output is a d.c. voltage close to Vm, with a small ripple.

• The circuit is a half-wave (or full-wave) rectifier feeding a battery through a variable resistor R. The diode forward resistance is negligible and reverse resistance infinite; battery internal resistance negligible.
• The charging current is to be 1.0 A. The battery has a fixed e.m.f. (say E_b). The rectified supply provides a peak voltage Vm. The charging current flows only when the instantaneous rectified voltage exceeds the battery e.m.f.
• For a half-wave rectifier, the mean charging current is given by the average of (v - E_b)/R over the conducting period.
• The necessary value of R is found from: R = (V_mean - E_b) / I_charge, where V_mean is the mean rectified voltage available. For example, if the supply peak is Vm and the battery e.m.f. is E_b, then R = (Vm - E_b)/1.0 ohm (for a simple d.c. equivalent), or using the mean value of the rectified waveform.
• The variable resistor is adjusted so that the charging current is exactly 1.0 A. (The exact numerical value depends on the supply voltage and battery e.m.f. given in the figure; the method is to set R so that the mean charging current equals 1.0 A.)

minimize energy losses (copper losses due to winding resistance and iron losses due to core magnetization). This translates directly to lower operating costs and reduced environmental impact due to less wasted energy. High-efficiency transformers typically achieve 95.5% efficiency for 5 kVA units and up to 97.5% for units up to 1 MVA.

Lower energy losses mean less heat is generated within the transformer. This reduces the risk of overheating, extending the lifespan of the equipment and preventing potential fire hazards. Overheating can damage the insulation and reduce the lifespan of the transformer. Lower operating temperatures contribute to enhanced reliability and a longer operational life for the transformer.

Low voltage regulation ensures that the output voltage remains relatively constant even under varying load conditions. This stability is essential for the stable operation of equipment connected to the transformer. Voltage fluctuations can lead to wear and tear on connected equipment. Maintaining a constant voltage extends the lifespan of the equipment.

• Starting torque is proportional to the square of the supply voltage. A reduced supply voltage will significantly reduce the starting torque.
• Higher rotor resistance leads to higher starting torque but also lower efficiency and higher rotor heating at running speeds. This is because higher resistance allows for greater current flow in the rotor at standstill.
• Increasing rotor resistance enhances starting torque by improving the power factor, but it also leads to higher losses and reduced efficiency during normal operation.
• Changes in frequency affect the motor's magnetic field and slip, impacting starting torque. Generally, lower frequency results in higher torque.
• The design of the stator windings, including the number of turns, winding configuration (star or delta), and the shape and placement of the stator poles, significantly affects the strength of the magnetic field produced, thus affecting torque.
• A smaller air gap between the stator and rotor generally leads to higher starting torque due to increased magnetic coupling.

Comparison with Rated Torque:

Typically, the starting torque of a standard three-phase induction motor is designed to be equal to or slightly higher than the rated (full-load) torque. This design ensures that the motor can overcome initial inertia and start the load effectively. However, the exact ratio varies based on motor design and application requirements. For instance, some motors may have a starting torque that is 1.25 times the full-load torque, while others, especially those designed for high starting torque applications, may have even higher ratios.

While higher starting torque is beneficial for initiating motion, it often comes with an increased starting current, which can impact the electrical supply system. Therefore, motor designs aim to balance starting torque and current to suit specific application needs.

$$Resistance \space of \space solenoid \space = \space 35Ω$$

$$Supply \space voltage \space = 220V, \space 50Hz$$

$$Initial \space current \space = \space 2A$$

$$Current \space in \space "Full-in" \space position \space = \space 0.7A$$

When plunger is out (initial stage),

$$Impedance \space (Z) \space = \space {{220} \over 2} \space = \space 110Ω$$

$$Inductive \space reactance \space (X) \space = \space \sqrt{Z^2 - R^2}$$

$$X \space = \space \sqrt{110^2 - 35^2} \space = \space 104.28Ω$$

$$Also, \space X \space = \space 2 \pi fl$$

$$l \space = \space {{X} \over 2 \pi f} \space = \space {{104.28} \over 2 \times 3.14 \times 50} $$

$$l \space = \space 0.33H$$

Inductance of solenoid when plunger is out = 0.33H

When "Full-in",

$$Impedance \space = \space {{V} \over I} \space = \space {{220}\over 0.7} \space = \space 314.286Ω$$

$$∴ \space X \space = \space \sqrt{314.286^2 - 35^2} \space = \space 312.33Ω$$

$$l \space = \space {{312.33} \over 2 \times 3.14 \times 50 } \space = \space 0.994H $$

$$Also, L \space = \space {{N \phi} \over I_{peak}}$$

$$∴ \space N \phi \space = \space L \space I_{peak}$$

$$∴ \space N \phi \space = \space 0.994 \times \sqrt{2} \times 0.7$$

$$∴ \space N \phi \space = \space 0.984 \space weber-turn$$

For a star-connected system:

$$V_{line}=\sqrt3\:V_{phase}$$

$$\frac{V_{line}}{V_{phase}}=\sqrt3\:=\:1.732$$

In a star connection, the line current is equal to the phase current:

$$I_{line}=I_{phase}$$

• In the event of a single earth fault, no earth fault current flows through the ship's hull due to the insulated neutral, minimizing fire hazards.
• The hull does not carry current, ensuring safety from electrical currents passing through the structure.
• A single earth fault does not cause generator breaker tripping, avoiding sudden blackouts or operational disruptions.
• Harmonic currents caused by third harmonics in the generated voltage are prevented from flowing through the neutral, protecting the generator windings from overloading.

1. Only one system voltage (line-to-line) is possible, unlike earthed neutral systems which also provide line-to-neutral voltages.
2. While an earth fault alarm and phase indicator are triggered, locating the exact fault location requires a time-consuming trial-and-error process.
3. In cases of inductive or capacitive faults to earth, surge voltage can rise 3.5 to 4 times the system voltage, risking insulation failure and system collapse.

A metallic resistor is inserted between the neutral point and the ship’s hull to limit earth fault current.

The resistor’s value is determined by:

$$R=\frac{V}{\sqrt3I}\:$$

Where,

• V = Line voltage,
• I = Full load current.

Metallic resistors are used for their stability, low maintenance, and ability to prevent arcing grounds.

The risk of electric shock is considered equally dangerous in both earthed and insulated neutral systems. In an insulated system, normal leakage currents from capacitance and surface leakage, along with the possibility of earth faults, mean that touching live parts still carries a considerable shock risk. Similarly, in an earthed system, line-to-neutral voltages (even those as low as 110 or 250 volts) can be lethal under certain shipboard conditions, making neither system inherently safer regarding electric shock than the other. Appropriate safety precautions are essential for both systems.

Power factor correction refers to the process of improving the power factor of an electrical system to bring it closer to unity (1 or 100%). It involves reducing the phase difference between voltage and current, which is caused by inductive loads like motors, transformers, and fluorescent lighting. These loads consume reactive power, leading to a lagging power factor. By adding components like capacitors, synchronous condensers, or phase advancers, reactive power is compensated, and the power factor is improved.

In practical terms, power factor correction aims to minimize the inefficiencies in the electrical system, reduce energy losses, and ensure optimal utilization of the power supplied by the generator or grid.

• By improving power factor, the current flow in the system is reduced, leading to lower I²R losses in cables, transformers, and other distribution components.
• With a higher power factor, the electrical system operates more efficiently, ensuring better utilization of the generated power.
• Improved power factor reduces the apparent power (kVA) requirement, allowing for smaller-sized generators, transformers, and cables, thus reducing capital costs.
• Higher power factor ensures better voltage stability across the system, preventing voltage drops and protecting sensitive equipment from under-voltage issues.
• By reducing reactive power, the system can handle more active power (real load) within the same capacity of the equipment, maximising output.
• With reduced current and heat generation, the wear and tear on electrical components are minimized, extending their lifespan.
• Higher efficiency in power usage reduces the overall energy demand, lowering fuel consumption and greenhouse gas emissions in power generation.

A three-phase system typically has an inductive load (e.g., motors), causing a lagging power factor. Capacitors can provide leading reactive power to compensate for this. The capacitors are connected in parallel with the inductive load.

• The size (capacitance) of each capacitor is calculated based on the size of the inductive load and the desired power factor improvement. Specialised software or calculation methods are often used for accurate determination.
• The capacitors are connected in a star or delta configuration, matching the load's connection. They should be appropriately rated for the voltage and current of the system.
• The leading reactive power supplied by the capacitors cancels out some of the lagging reactive power from the inductive load, effectively reducing the overall reactive power and improving the power factor.

Besides capacitors, synchronous motors can also be used for power factor correction. Synchronous motors can be operated at leading power factor, effectively counteracting the lagging power factor of inductive loads. These motors can contribute both real power and leading reactive power to the system. However, synchronous motors are more complex and expensive than capacitors. They are often used in larger industrial installations where the power factor correction requirements are significant.

The frequency response of a transistor amplifier refers to how its voltage gain changes with signal frequency. Ideally, the gain should remain constant over the desired frequency range. However, in practice, the gain deteriorates at both low and high frequencies due to various circuit components and effects.

The two main factors that cause deterioration are:

1. Coupling and Bypass Capacitors (Affecting Low-Frequency Response):

At low frequencies (below f₁, the lower cut-off frequency), the following capacitive components contribute to the drop in gain:

• Input Coupling Capacitor (Cin) and Output Coupling Capacitor (Cout):
• These capacitors have high reactance at low frequencies, which restricts the AC signal from passing through effectively. As a result, only a small part of the signal reaches the amplifier, reducing the voltage gain.
• Emitter Bypass Capacitor (Cb):
• At low frequencies, the reactance of this capacitor is also high, and it cannot effectively bypass the emitter resistor (Re). This causes more of the input signal to drop across Re, reducing the output.

Effect on Performance:

• Causes reduced signal transfer through the amplifier.
• Leads to drop in amplifier gain at frequencies below the lower cut-off frequency (f₁).

2. Diffusion and Stray Capacitances (Affecting High-Frequency Response):

At high frequencies (above f₂, the upper cut-off frequency), internal capacitances cause deterioration:

• Base-Emitter Diffusion Capacitance:
• At high frequencies, this capacitance presents low reactance, increasing base current. This results in reduced current amplification (β) and thus lowers the voltage gain.
• Collector-to-Earth Stray Capacitance:
• Also decreases in reactance with frequency, introducing a loading effect and further lowering the amplifier’s output gain.
• Cout at High Frequencies:
• Acts as an additional load to the next stage, further reducing the voltage gain.

Effect on Performance:

• Results in loss of gain due to internal capacitance effects.
• Leads to drop in amplifier gain at frequencies above the upper cut-off frequency (f₂).

Conclusion:

• At low frequencies (< f₁): Gain drops due to high reactance of coupling and bypass capacitors (Cin, Cout, Cb).
• At high frequencies (> f₂): Gain drops due to low reactance of base-emitter diffusion capacitance and stray capacitances.

The maximum, stable gain is maintained in the mid-frequency range, and the cut-off frequencies f₁ and f₂ are defined where the gain falls to 70.7% of its maximum value. For optimal performance, the amplifier should be designed to operate within this mid-frequency range.

Single phasing, also known as phase failure or phase loss, is a condition in a three-phase electrical system where one of the three supply phases becomes disconnected or open-circuited while the remaining two phases continue to supply power.

This condition may occur due to:

• A blown fuse in one phase
• Broken conductors or cables
• Faulty contactors
• Loose electrical connections
• Open circuit in the supply line

When single phasing occurs, a three-phase motor connected to the supply may continue operating on the remaining two phases. However, the supply becomes unbalanced, resulting in abnormal current distribution and unsafe operating conditions.

Single phasing is particularly dangerous for induction motors because it can cause overheating and serious damage if not detected quickly.

1. Motor Continues Running

Unlike some other electrical faults, the motor may not stop immediately.

• The motor continues running because the remaining two phases still produce an unbalanced rotating magnetic field.
• However, motor performance becomes poor and unstable.

2. Excessive Current in the Remaining Phases

To maintain the required load torque, the current in the remaining energized phases increases considerably.

In a delta-connected motor:

• One winding effectively loses its proper supply.
• The other two windings attempt to carry the entire load.

As a result:

• Current in the energized windings may rise to approximately 1.73 to 2 times the normal full-load current, depending on the load condition.

3. Reduction in Torque and Speed

The motor is unable to develop normal torque under single phasing conditions.

This causes:

• Reduction in torque output
• Slight drop in motor speed
• Difficulty in carrying the connected load

The motor may struggle to continue operating efficiently.

4. Severe Overheating

The most serious effect of single phasing is rapid overheating of the motor windings.

Due to excessive current:

• Copper losses ((I^2R) losses) increase sharply.
• Winding temperature rises rapidly.

This overheating can damage insulation very quickly.

5. Winding Burnout and Motor Failure

If the condition continues:

• Insulation breakdown occurs,
• Windings may short circuit,
• The motor may eventually burn out completely.

Although thermal overload relays may trip the motor, they may not act quickly enough to prevent serious damage, especially when the motor is heavily loaded.

The most effective method of protecting a motor against single phasing is by using a Phase Failure Relay (also called a Phase Loss Relay or Single Phasing Relay).

Working Principle of Phase Failure Relay

1. Continuous Monitoring

The relay continuously monitors:

• Voltage in all three phases, or
• Current in all three phases

to ensure balanced supply conditions.

2. Detection of Phase Failure

If:

• One phase is lost,
• Voltage becomes severely unbalanced, or
• Current imbalance occurs,

the relay detects the abnormal condition immediately.

3. Tripping Action

Once single phasing is detected:

• The relay operates its contacts,
• The control circuit of the motor contactor is interrupted,
• The contactor coil becomes de-energized.

4. Disconnection of the Motor

When the contactor drops out:

• The main power contacts open,
• The motor is disconnected from the supply.

This prevents overheating and protects the motor windings from damage.

Modern Motor Protection Relays

Modern electronic motor protection relays often include additional protective features such as:

• Overcurrent protection
• Under-voltage protection
• Phase sequence protection
• Earth fault protection
• Thermal protection

These provide complete protection for the motor.

Limitations of Thermal Overload Relays

Thermal overload relays provide only indirect protection against single phasing.

• They operate based on overheating,
• But overheating may already have caused insulation damage before the relay trips.

Therefore, a dedicated phase failure relay is preferred because it detects the fault condition itself and acts much faster.

The circuit breaker used for an A.C. generator is typically an Air Break Circuit Breaker. It is frame-mounted and of the draw-out type, allowing it to be isolated from the busbar and alternator input contacts by moving it horizontally. An interlock ensures the breaker is turned off before being drawn out.

Main Components:

• Contacts: High-conductivity, silver-coated copper contacts ensure efficient current flow.
• Arcing Contacts: Separate arcing contacts protect the main contacts by taking the arc during the operation. These arcing contacts open slightly after the main contacts and are replaceable.
• Arc Chutes and Splitter Plates: Electromagnetic forces guide the arc to the arcing horn, where the arc is elongated and quenched. The arc is divided into sections by splitter plates, effectively extinguishing it.
• Anti-Bouncing Devices: These prevent rebound of the contacts, ensuring a clean break.

• The circuit breaker's protection system (e.g., overcurrent relay) detects an excessive current flow indicating a short circuit.
• The breaker opens, interrupting the flow of current and preventing further damage. If only one generator is operating, a complete blackout occurs. If generators are in parallel, the load is transferred to the other generator(s), potentially causing an overload trip.
• In a parallel system, the remaining generator(s) assume the load, which may overload them and trigger a trip.
• This can lead to a complete system shutdown (blackout).

Checks following the event:

• Open the backside of the switchboard associated with the tripped generator and inspect for short circuits.
• Check all outgoing feeders individually to locate and clear the fault.
• Inspect the affected generator’s armature, field circuit, AVR, and connections for insulation resistance, physical damage, and overheating.

• Immediately operate the emergency manual trip mechanism to isolate the affected generator.
• Completely isolate the generator from the system by switching off its supply and load. This is the most important step to prevent further damage.
• Open the generator's field circuit supply to cease voltage generation.
• Once the generator is fully isolated, proceed to locate and clear the fault.
• After the fault is resolved, perform insulation resistance tests on the armature, field, and AVR circuits.
• Thoroughly inspect all components of the generator for any damage or overheating.
• Once the fault is rectified and all checks are satisfactory, the alternator can be safely brought back into service.

1. Waveform

A waveform is the shape or pattern obtained when an alternating voltage or current is plotted against time.

For a sinusoidal alternating current, the waveform is a sine wave, in which the magnitude and direction of current vary continuously with time.

2. Frequency

Frequency is the number of complete cycles of an alternating quantity occurring in one second.

The unit of frequency is hertz (Hz).

$$f=\frac{1}{T}$$

where:

• f = frequency in Hz
• T = time period of one complete cycle in seconds

3. Average Value

The average value of an alternating quantity is the arithmetic mean of its instantaneous values over a specified period.

For a symmetrical sinusoidal AC waveform, the average value over a complete cycle is zero, because the positive and negative half-cycles cancel each other.

For a rectified waveform, the average value is obtained by considering the rectified current over the complete cycle.

Given:

• AC supply voltage = 110 V RMS
• Resistance = 50 Ω
• Resistance to current in the reverse direction = infinite
• Therefore, current flows through the circuit in only one direction.

Hence, the current is a half-wave rectified sine wave.

Step 1: Calculate the Peak Voltage

The given 110 V is the RMS value of the sinusoidal AC supply.

For a sinusoidal waveform:

$$V_m=\sqrt{2}\times V_{rms}$$

Therefore:

$$V_m=\sqrt{2}\times110$$

$$V_m=155.56\ V$$

Step 2: Calculate the Peak Current

Using Ohm's law:

$$I_m=\frac{V_m}{R}$$

$$I_m=\frac{155.56}{50}$$

$$I_m=3.11\ A$$

Therefore:

$$\boxed{I_m=3.11\ A}$$

This current flows only during one half-cycle because the rectifier blocks current in the opposite direction.

The current waveform is therefore a half-wave rectified sine wave.

(i) Ammeter Readings

Moving Coil Ammeter

A moving coil ammeter responds to the average value of current.

For a half-wave rectified sine wave:

$$I_{avg}=\frac{I_m}{\pi}$$

$$Substituting\:I_{m}=3.11\ A$$

$$I_{avg}=\frac{3.11}{\pi}$$

$$I_{avg}=0.99\ A$$

Therefore, the moving coil ammeter reads:

$$\boxed{I_{MC}=0.99\ A}$$

Thermal Ammeter

A thermal ammeter operates on the heating effect of current and therefore indicates the RMS value of current.

For a half-wave rectified sine wave:

$$I_{rms}=\frac{I_m}{2}$$

Therefore:

$$I_{rms}=\frac{3.11}{2}$$

$$I_{rms}=1.555\ A$$

Hence, the thermal ammeter reads:

$$\boxed{I_{thermal}=1.56\ A}$$

(ii) Form Factor and Peak Factor

Form Factor

The form factor is defined as:

$$Form\ Factor=\frac{RMS\ value}{Average\ value}$$

For a half-wave rectified sine wave:

$$Form\ Factor=\frac{I_m/2}{I_m/\pi}$$

Therefore:

$$Form\ Factor=\frac{\pi}{2}$$

$$\boxed{Form\ Factor=1.57}$$

Peak Factor

The peak factor is defined as:

$$Peak\ Factor=\frac{Maximum\ value}{RMS\ value}$$

For the half-wave rectified sine wave:

$$Peak\ Factor=\frac{I_m}{I_m/2}$$

Therefore:

$$\boxed{Peak\ Factor=2.0}$$

Final Answers

• Supply voltage: 110 V RMS
• Peak voltage: 155.56 V
• Peak current: 3.11 A
• Moving coil ammeter reading: 0.99 A
• Thermal ammeter reading: 1.56 A
• Current waveform: Half-wave rectified sine wave
• Form factor: 1.57
• Peak factor: 2.0

Therefore:

$$\boxed{I_{MC}=0.99\ A}$$

$$\boxed{I_{thermal}=1.56\ A}$$

$$\boxed{Form\ Factor=1.57}$$

$$\boxed{Peak\ Factor=2.0}$$

Note: The values 1.1 A, 1.11 and 1.414 are not applicable to the stated half-wave rectified circuit. For the given circuit, the correct values are 1.56 A, 1.57 and 2.0, respectively.

In an iron-core transformer, when the primary winding is connected to a supply voltage, a current flows through the winding depending on the load and the winding resistance. This current generates a magnetic field, forming a magnetic north and south pole, thereby creating magnetic flux between them.

The magnetic flux consists of lines that ideally pass through the iron core and link both the primary and secondary windings. This mutual flux is responsible for the energy transfer from the primary to the secondary winding and is essential for transformer operation.

However, not all magnetic flux produced by the primary winding links to the secondary winding. Some of the magnetic lines spread into the surrounding air space and do not pass through the secondary coil. This portion of the magnetic flux is known as leakage flux.

Leakage Flux:

• Leakage flux is the portion of the magnetic flux generated by the primary winding that does not couple with the secondary winding.
• It occurs due to the physical separation between the windings and non-ideal magnetic coupling.
• The leakage flux induces an electromotive force (emf) in the primary winding itself, which opposes the current flow in the primary. This results in an additional voltage drop.

How It Is Accounted for in Analysis:

• The opposition caused by the leakage flux is modeled as an inductive reactance (i.e., a leakage inductance) in series with the primary winding resistance.
• This series inductance creates a voltage drop equal to the emf generated by the leakage flux.
• In transformer equivalent circuits, the leakage inductances of both primary and secondary windings are included to reflect this non-ideal behavior.

Minimising Leakage Flux:

Interleaving Technique:

• Leakage flux can be minimized by arranging the primary and secondary windings closely together on the same leg of the transformer core, a method known as interleaving.
• This improves magnetic coupling and enhances energy transfer efficiency.

Silicon Controlled Rectifier (SCR)

A Silicon Controlled Rectifier (SCR) is a four-layer, three-terminal semiconductor switching device belonging to the thyristor family. Its four semiconductor layers are arranged in the PNPN configuration. The three terminals are Anode (A), Cathode (K), and Gate (G).

An SCR is a unidirectional device and is mainly used for switching, rectification, and controlling electrical power. It is normally triggered by applying a small current to the gate terminal.

symbol:

Modes of Operation

An SCR has three main modes of operation:

1. Forward Blocking Mode (OFF state): The SCR is forward biased but remains in the non-conducting state.
2. Forward Conducting Mode (ON state): When a suitable gate current is applied, the SCR is triggered and conducts current from anode to cathode. Once turned ON, it remains conducting until the current falls below the holding current.
3. Reverse Blocking Mode (OFF state): When reverse biased, the SCR blocks the flow of current, apart from a small leakage current.

Breakover Voltage of SCR

The breakover voltage (VBOV_{BO}) is defined as the minimum anode-to-cathode voltage at which an SCR changes from the OFF state (high impedance) to the ON state (low impedance) without any gate current being applied.

When the anode-to-cathode voltage exceeds the breakover voltage, the SCR turns ON automatically due to avalanche breakdown, even though the gate has not been triggered.

• Applications: SCRs are widely used in motor speed control, light dimming, and controlled rectifier circuits.

Given:

Armature voltage, $$V = 480\ V$$

Armature current, $$I_a = 110\ A$$

Armature circuit resistance, $$R_a = 0.2\ \Omega$$

Number of poles, $$P = 6$$

Flux per pole, $$\Phi = 0.05\ Wb$$

Number of armature conductors, $$Z = 864$$

Lap-connected armature winding, therefore number of parallel paths, $$A = P = 6$$

(i) Speed of the Motor

For a DC motor:

$$V = E_b + I_aR_a$$

Therefore, the back EMF is:

$$E_b = V-I_aR_a$$

$$E_b=480-(110\times0.2)$$

$$E_b=480-22=458\ V$$

The EMF equation of a DC machine is:

$$E_b=\frac{P\Phi ZN}{60A}$$

For a lap winding, $$A=P$$. Therefore:

$$458=\frac{6\times0.05\times864\times N}{60\times6}$$

Since 6 cancels:

$$458=\frac{0.05\times864\times N}{60}$$

Therefore:

$$N=\frac{458\times60}{0.05\times864}$$

$$N=636.02\ rpm$$

Therefore:

$$N\approx636\ rpm$$

(ii) Gross Torque Developed by the Armature

The gross mechanical power developed by the armature is:

$$P_g=E_bI_a$$

The torque is given by:

$$T_g=\frac{60P_g}{2\pi N}$$

Therefore:

$$T_g=\frac{60E_bI_a}{2\pi N}$$

Substituting the values:

$$T_g=\frac{60\times458\times110}{2\pi\times636.02}$$

$$T_g\approx754.3\ N\cdot m$$

The same result can be obtained directly from the torque equation:

$$T_g=\frac{PZ\Phi I_a}{2\pi A}$$

Substituting:

$$T_g=\frac{6\times864\times0.05\times110}{2\pi\times6}$$

$$T_g\approx754.3\ N\cdot m$$

Therefore:

$$T_g\approx754\ N\cdot m$$

Final Answers

Speed of the motor = $$636\ rpm$$

Gross torque developed by the armature = $$754\ N\cdot m$$

1. Stator Supply Voltage (V1): The maximum torque is directly proportional to the square of the stator supply voltage. A higher supply voltage leads to a significantly higher maximum torque.
2. Stator Resistance (R1): Higher stator resistance tends to reduce the maximum torque. It contributes to the overall impedance that limits current flow.
3. Stator Leakage Reactance (X1): Higher stator leakage reactance reduces the maximum torque. Leakage reactance represents the magnetic flux that does not link both stator and rotor windings.
4. Rotor Leakage Reactance (X2'): Higher rotor leakage reactance (referred to the stator) reduces the maximum torque. Similar to stator leakage reactance, it impedes the flow of rotor current and thus the torque production.
5. Supply Frequency (f): The maximum torque is inversely proportional to the supply frequency. This is because reactances (X = 2πfL) are directly proportional to frequency, and synchronous speed (ωs = 2πf/P) is also directly proportional to frequency.
6. Number of Poles (P): The maximum torque is directly proportional to the number of poles of the motor. This is due to its inverse relationship with the synchronous speed (ωs = 2πf/P).

Given

• Three-phase balanced star-connected load
• Resistance per phase, R = 50 Ω
• Inductance per phase, L = 0.2 H
• Supply voltage, VL = 415 V
• Frequency, f = 50 Hz

Step 1: Calculate inductive reactance

$$X_L = 2\pi fL$$

$$X_L = 2 \times \pi \times 50 \times 0.2$$

$$X_L = 62.83\,\Omega$$

Step 2: Calculate phase impedance

$$Z = \sqrt{R^2 + X_L^2}$$

$$Z = \sqrt{50^2 + 62.83^2}$$

$$Z = \sqrt{2500 + 3947.6}$$

$$Z = 80.3\,\Omega$$

Step 3: Calculate phase voltage

For a star-connected load,

$$V_{ph} = \frac{V_L}{\sqrt{3}}$$

$$V_{ph} = \frac{415}{1.732}$$

$$V_{ph} = 239.6\,V$$

Step 4: Calculate phase current

$$I_{ph} = \frac{V_{ph}}{Z}$$

$$I_{ph} = \frac{239.6}{80.3}$$

$$I_{ph} = 2.98\,A$$

For a star-connected load,

$$I_L = I_{ph}$$

$$\boxed{I_L = 2.98\,A}$$

$$\cos\phi = \frac{R}{Z}$$

$$\cos\phi = \frac{50}{80.3}$$

$$\boxed{\cos\phi = 0.623 \text{ lagging}}$$

Step 1: Calculate active power

$$P = \sqrt{3}V_LI_L\cos\phi$$

$$P = 1.732 \times 415 \times 2.98 \times 0.623$$

$$P = 1335\,W$$

Step 2: Initial reactive power

$$\phi_1 = \cos^{-1}(0.623)$$

$$\phi_1 = 51.46^\circ$$

$$Q_1 = P\tan\phi_1$$

$$Q_1 = 1335 \times \tan(51.46^\circ)$$

$$Q_1 = 1673\,VAR$$

Step 3: Reactive power at desired power factor

$$\phi_2 = \cos^{-1}(0.9)$$

$$\phi_2 = 25.84^\circ$$

$$Q_2 = P\tan\phi_2$$

$$Q_2 = 1335 \times \tan(25.84^\circ)$$

$$Q_2 = 647\,VAR$$

Step 4: Capacitor VAR required

$$Q_C = Q_1 - Q_2$$

$$Q_C = 1673 - 647$$

$$Q_C = 1026\,VAR$$

Step 5: Calculate capacitance per delta-connected capacitor

For a delta-connected capacitor bank,

$$Q_C = 3V_L^2\omega C$$

where

$$\omega = 2\pi f$$

$$\omega = 2\pi \times 50$$

$$\omega = 314.16\,rad/s$$

Therefore,

$$C = \frac{Q_C}{3V_L^2\omega}$$

$$C = \frac{1026}{3 \times 415^2 \times 314.16}$$

$$C = 6.33 \times 10^{-6}\,F$$

$$\boxed{C = 6.33\,\mu F}\:per\:capacitor$$

At the improved power factor of 0.9,

$$P = \sqrt{3}V_LI_{new}\cos\phi_2$$

$$1335 = 1.732 \times 415 \times I_{new} \times 0.9$$

$$I_{new} = 2.06\,A$$

$$\boxed{I_{new} = 2.06\,A}$$

The circuit breaker used for an A.C. generator is typically an Air Break Circuit Breaker. It is frame-mounted and of the draw-out type, allowing it to be isolated from the busbar and alternator input contacts by moving it horizontally. An interlock ensures the breaker is turned off before being drawn out.

Main Components:

• Contacts: High-conductivity, silver-coated copper contacts ensure efficient current flow.
• Arcing Contacts: Separate arcing contacts protect the main contacts by taking the arc during the operation. These arcing contacts open slightly after the main contacts and are replaceable.
• Arc Chutes and Splitter Plates: Electromagnetic forces guide the arc to the arcing horn, where the arc is elongated and quenched. The arc is divided into sections by splitter plates, effectively extinguishing it.
• Anti-Bouncing Devices: These prevent rebound of the contacts, ensuring a clean break.

• The circuit breaker's protection system (e.g., overcurrent relay) detects an excessive current flow indicating a short circuit.
• The breaker opens, interrupting the flow of current and preventing further damage. If only one generator is operating, a complete blackout occurs. If generators are in parallel, the load is transferred to the other generator(s), potentially causing an overload trip.
• In a parallel system, the remaining generator(s) assume the load, which may overload them and trigger a trip.
• This can lead to a complete system shutdown (blackout).

Checks following the event:

• Open the backside of the switchboard associated with the tripped generator and inspect for short circuits.
• Check all outgoing feeders individually to locate and clear the fault.
• Inspect the affected generator’s armature, field circuit, AVR, and connections for insulation resistance, physical damage, and overheating.

• Immediately operate the emergency manual trip mechanism to isolate the affected generator.
• Completely isolate the generator from the system by switching off its supply and load. This is the most important step to prevent further damage.
• Open the generator's field circuit supply to cease voltage generation.
• Once the generator is fully isolated, proceed to locate and clear the fault.
• After the fault is resolved, perform insulation resistance tests on the armature, field, and AVR circuits.
• Thoroughly inspect all components of the generator for any damage or overheating.
• Once the fault is rectified and all checks are satisfactory, the alternator can be safely brought back into service.

(a) Analogue vs Digital Measuring Instruments and Their Working Principles

Analogue Instruments

Definition:

• An analogue measuring instrument displays the measured value as a continuous movement of a pointer over a graduated scale.

Working Principle:

• The instrument converts the measurand (input quantity) into a proportional mechanical deflection.
  • In a typical analogue electrical meter:
    • The current flowing through a coil generates a magnetic torque.
    • This torque causes the pointer to move across the scale.
    • A spring provides a balancing torque.
    • The steady deflection of the pointer is proportional to the input signal (e.g., current or voltage).

  Digital Instruments

  Definition:

  • A digital measuring instrument displays the measured value as numerical digits on an electronic display (such as an LCD).

  Working Principle:

  • The instrument works by sampling the input signal, converting it into digital form, and processing it electronically to produce a precise reading.
    • In a typical digital meter:
      • The input signal passes through protection and signal conditioning circuits.
      • An Analogue-to-Digital Converter (ADC) converts the input voltage into a stream of digital bits.
      • A microcontroller or processor computes the final value.
      • The processed measurement is displayed on the LCD screen.

    (b) Examples of Analogue and Digital Instruments Used Onboard

    1. Analogue Instrument: Bourdon Tube Pressure Gauge

    Working Principle:

    • The Bourdon gauge measures fluid pressure using the elastic deformation of a C-shaped metal tube.
      • When internal pressure increases, the curved tube tends to straighten.
      • This motion is transmitted through a link and sector gear mechanism to a pointer, which moves proportionally across a calibrated dial.
    • Applications Onboard:
      • Commonly used in lube oil, fuel oil, and cooling water lines.
      • Advantages:
        • Rugged construction and no power requirement.
        • Provides an instant visual indication and helps monitor trends easily.

      2. Digital Instrument: Digital Multimeter

      Working Principle:

      • A digital multimeter measures voltage, current, and resistance electronically.
        • The input passes through protection and range selection networks.
        • The signal is digitised by an ADC.
        • The internal microprocessor computes the corresponding electrical value.
        • The result is shown numerically on the LCD display.
        • For AC measurements, a true-RMS converter or sampling algorithm ensures accurate readings even for non-sinusoidal waveforms.
      • Applications Onboard:
        • Checking 24V DC control circuits.
        • Verifying generator phase voltages.
        • Measuring sensor loop currents such as 4–20 mA signals in control systems.

With respect to power transformers, the following protections:

• Protects the transformer against sustained overcurrent above its rated capacity, which would cause overheating of windings and insulation and shorten life.
• Usually provided by thermal overload relays or by an oil/winding temperature indicator (WTI/OTI) which operates on the temperature of the oil or windings. The relay has an inverse time characteristic - the higher the overload, the faster it operates.
• The overload relay gives an alarm first and trips the circuit breaker if the overload persists. It allows short-term overloads (e.g. during motor starting) but trips on sustained overload.

• Protects against short circuits between phases (phase-to-phase faults), which produce very high fault currents.
• Provided by overcurrent relays (IDMT - inverse definite minimum time) connected to current transformers on the primary and/or secondary. These have an inverse time-current characteristic so that larger fault currents trip faster.
• The relay operates the circuit breaker to isolate the transformer quickly, limiting damage. Instantaneous elements may be added for very high fault currents.

• Protects against faults between a winding and earth (ground), which may not be detected by phase overcurrent relays if the fault current is small.
• Provided by earth fault relays connected in the residual circuit of the current transformers (the residual current is the vector sum of the three phase currents, which is zero under balanced conditions but non-zero on an earth fault).
• Alternatively, a core-balance (zero-sequence) current transformer surrounds all three conductors and detects any imbalance due to earth leakage. The relay trips the breaker on an earth fault.

• Compares the current entering the transformer primary with the current leaving the secondary, using current transformers on both sides.
• Under normal and through-fault conditions the currents are balanced (allowing for the turns ratio and vector group) and the relay does not operate. On an internal fault (winding-to-winding or winding-to-earth inside the transformer), the currents are unbalanced and the relay operates to trip the breaker.
• This gives fast, sensitive protection for internal faults and is the main protection for large power transformers. It is biased to prevent operation on through-faults and magnetising inrush current.

• Detects the direction of power flow and operates only when the fault current flows in a particular direction.
• Used where a fault could be fed from more than one source, e.g. in parallel feeders or ring systems. The directional relay compares the phase of the current with the voltage (or with a reference) to determine the direction of the fault.
• It ensures that only the circuit breaker on the faulted side operates, isolating the fault while leaving healthy sections in service. It is used for busbar and feeder protection where discrimination by time alone is not sufficient.

The regulation state that the voltage must recover within 1.5 seconds. However an acceptable recovery time would be 0.5 seconds for a brushless and 0.2 seconds or less for a compounded machine.

(ii) Effect on small power installation:

When a large load is suddenly applied, the electrical system experiences a significant voltage dip. Initially, there is a sharp drop in voltage due to the high inrush current drawn by the load. This is followed by a slower decrease as the alternator's reactance and power factor affect the voltage. During this period, the alternator’s excitation system, AVR (Automatic Voltage Regulator), and prime mover governor work to restore the voltage.

The sudden load causes a drop in power factor, increasing the reactive power demand on the system. If the voltage dip is significant and prolonged, sensitive equipment may malfunction, and other connected loads might experience disruptions. The system's ability to recover depends on the alternator's capacity, excitation response, and governor speed control.

In this method, the output voltage of the bus bar is continuously measured and compared to the normal rated voltage. Any deviation from the desired voltage generates an error signal, which is sent to the excitation system. This error signal adjusts the excitation to regulate the output voltage. For instance, if the voltage drops, the excitation current is increased, and if the voltage rises, the excitation is reduced.

Examples of error-operated voltage control include brushless alternators with an Automatic Voltage Regulator (AVR) and alternators using a carbon pile AVR and DC exciter.

(ii) Functional Voltage Control:

This type of voltage control is directly based on the instantaneous value of the voltage. If the voltage falls, the excitation is increased proportionally to the amount of voltage drop, and vice versa. Since the excitation is a direct function of the voltage, it is referred to as functional voltage control.

Static excitation systems are an example of functional voltage control. These systems offer faster response times compared to error-operated methods, making them suitable for applications requiring precise and rapid voltage regulation.

• Electrical currents induced in the shafting of rotating machinery flow through the bearings, journals and the machine frame. As the current passes through the bearing oil film it can cause sparking (electric discharge machining), which pits and scores the bearing surfaces and the journal.
• This leads to rapid bearing wear, overheating, and eventual bearing failure. The pitting (frosting) of the bearing and shaft surfaces is characteristic of shaft currents.
• In d.c. machines, shaft currents can also cause sparking at the commutator and damage to the brushes.
• The currents are caused by magnetic asymmetry in the machine (e.g. unbalanced magnetic pull, eccentric rotor, segmented stator laminations, or a circulating flux linking the shaft) which induces an e.m.f. along the shaft.

(i) d.c. machines:

• The shaft is insulated from the frame at one end by fitting an insulated bearing (a bearing with an insulating layer between the bearing housing and the frame, or an insulated bearing liner). This breaks the circulating current path through the shaft and frame.
• The other bearing is left earthed (metallic) so that any residual current has a defined path and does not pass through the insulated bearing.
• A brush (earthing brush) may be fitted to the shaft to collect and earth any residual shaft current, preventing it from passing through the bearings.
• Ensuring the magnetic circuit is symmetrical and the air gap is uniform reduces the unbalanced magnetic pull that induces shaft currents.

(ii) Main shafting fitted with a bronze propeller:

• The bronze propeller and the steel shaft form a galvanic couple in seawater, and the shaft can carry current due to the propeller earthing effect and any stray currents.
• The shaft is insulated from the propeller (insulating coupling or insulating sleeve between the propeller and the shaft) to break the electrical path.
• An earthing brush (shaft earthing brush) is fitted to the shaft to provide a low-resistance path to earth, so that any current is conducted to earth through the brush rather than through the bearings and stern gland.
• The shaft earthing brush also prevents electrolytic corrosion of the propeller and shaft and reduces the risk of bearing damage from shaft currents.

The form factor of an AC waveform is the ratio of its RMS value to its average value. For a perfect sine wave, the form factor is approximately 1.11 (RMS value/Average Value = 1.11). This factor indicates how closely a waveform resembles a pure sine wave; a form factor closer to 1.11 suggests a waveform that is more sinusoidal.

Supply voltage, $$V_{rms} = 110\ V$$

Diode internal resistance, $$R_D = 20\ \Omega$$

Load resistance, $$R_L = 1000\ \Omega$$

The total resistance in the conducting circuit is:

$$R_T = R_D + R_L$$

$$R_T = 20 + 1000 = 1020\ \Omega$$

(i) Peak Load Current

The peak value of the supply voltage is:

$$V_{peak} = \sqrt{2}\,V_{rms}$$

$$V_{peak} = 1.414 \times 110 = 155.56\ V$$

Therefore, the peak load current is:

$$I_{peak} = \frac{V_{peak}}{R_T}$$

$$I_{peak} = \frac{155.56}{1020} = 0.1525\ A$$

Peak load current:

$$I_{peak}\approx0.153\ A=153\ mA$$

(ii) DC Load Current

For a half-wave rectifier, the average or DC value of current is:

$$I_{DC} = \frac{I_{peak}}{\pi}$$

$$I_{DC} = \frac{0.1525}{3.142} = 0.0485\ A$$

DC load current:

$$I_{DC}\approx0.0485\ A=48.5\ mA$$

(iii) AC Load Current

For a half-wave rectified current, the RMS load current is:

$$I_{RMS} = \frac{I_{peak}}{2}$$

$$I_{RMS} = \frac{0.1525}{2} = 0.07625\ A$$

The AC component of the load current is:

$$I_{AC} = \sqrt{I_{RMS}^{2}-I_{DC}^{2}}$$

$$I_{AC} = \sqrt{(0.07625)^2-(0.0485)^2}$$

$$I_{AC} \approx 0.0588\ A$$

• A peak rectifier (peak detector) consists of a diode in series with a capacitor and a load resistor. The diode rectifies the a.c. input and charges the capacitor to the peak value of the input voltage.
• Circuit: a.c. source -> diode -> node (capacitor to earth, load resistor to earth). Output taken across the capacitor/load.
• During the positive half cycle the diode conducts and charges the capacitor to the peak value Vm. When the input falls below the capacitor voltage the diode becomes reverse biased and stops conducting; the capacitor discharges slowly through the load. If the time constant R x C is large compared with the period, the output is held near Vm.
• If the supply voltage is v(t) = Vm sin(wt), the voltage across the load resistor is approximately the peak value Vm (for an ideal diode and large time constant), i.e. a d.c. voltage close to Vm with small ripple.

• Back e.m.f. E = V - Ia Ra = 480 - 110 x 0.2 = 480 - 22 = 458 V.
• For a lap-connected armature, number of parallel paths A = number of poles P = 6.
• E.m.f. equation: E = (P x Z x phi x N) / (60 x A). Since A = P, E = (Z x phi x N)/60.
• (i) Speed: N = (E x 60)/(Z x phi) = (458 x 60)/(864 x 0.05) = 27480/43.2 = 636.1 rev/min.
• (ii) Gross torque developed: T = (P x Z x phi)/(2 pi A) x Ia = (6 x 864 x 0.05)/(2 x 3.1416 x 6) x 110 = (259.2/37.70) x 110 = 6.876 x 110 = 756.4 N m.
• (Check: armature power = E x Ia = 458 x 110 = 50,380 W; angular speed = 2 pi x 636.1/60 = 66.6 rad/s; T = 50380/66.6 = 756.5 N m.)

So speed = 636 rev/min and gross torque = 756 N m.

• The applied voltage (starting torque is proportional to the square of the applied voltage, T proportional to V^2).
• The rotor resistance (increasing rotor resistance increases the starting torque up to a maximum, and shifts the maximum-torque point towards standstill).
• The rotor and stator reactances (leakage reactance) - higher leakage reactance reduces the starting torque.
• The number of poles and the synchronous speed (torque depends on the air-gap power and synchronous speed).
• The rotor and stator winding resistances and the turns ratio.
• The supply frequency.
• The air-gap flux.
• Comparison with rated torque: the starting torque of a standard squirrel-cage motor is typically about 1.5 to 2 times the full-load (rated) torque. It is generally greater than the rated torque so that the motor can start the load, but not excessively high. For a slip-ring motor the starting torque can be increased up to the maximum torque by adding rotor resistance.

• (i) Synchronous speed Ns = 120 f / P = 120 x 50 / 4 = 1500 rev/min.
• (ii) Speed at 4% slip: N = Ns (1 - s) = 1500 x (1 - 0.04) = 1500 x 0.96 = 1440 rev/min.
• (iii) Rotor frequency when rotor speed is 600 rev/min:
• Slip s = (Ns - N)/Ns = (1500 - 600)/1500 = 900/1500 = 0.6.
• Rotor frequency fr = s x f = 0.6 x 50 = 30 Hz.

So synchronous speed = 1500 rev/min, rotor speed at 4% slip = 1440 rev/min, rotor frequency at 600 rev/min = 30 Hz.

The magnetic flux consists of lines that ideally pass through the iron core and link both the primary and secondary windings. This mutual flux is responsible for the energy transfer from the primary to the secondary winding and is essential for transformer operation.

However, not all magnetic flux produced by the primary winding links to the secondary winding. Some of the magnetic lines spread into the surrounding air space and do not pass through the secondary coil. This portion of the magnetic flux is known as leakage flux.

Leakage Flux:

• Leakage flux is the portion of the magnetic flux generated by the primary winding that does not couple with the secondary winding.
• It occurs due to the physical separation between the windings and non-ideal magnetic coupling.
• The leakage flux induces an electromotive force (emf) in the primary winding itself, which opposes the current flow in the primary. This results in an additional voltage drop.

How It Is Accounted for in Analysis:

• The opposition caused by the leakage flux is modeled as an inductive reactance (i.e., a leakage inductance) in series with the primary winding resistance.
• This series inductance creates a voltage drop equal to the emf generated by the leakage flux.
• In transformer equivalent circuits, the leakage inductances of both primary and secondary windings are included to reflect this non-ideal behavior.

Minimising Leakage Flux:

Interleaving Technique:

• Leakage flux can be minimized by arranging the primary and secondary windings closely together on the same leg of the transformer core, a method known as interleaving.
• This improves magnetic coupling and enhances energy transfer efficiency.

• Supply Frequency (f)
• Number of Poles (P)

The relationship between these factors and the synchronous speed (Ns) is given by the formula:

$$N_{s}=\frac{120f}{P}$$

Where Ns is the synchronous speed in revolutions per minute (RPM), f is the supply frequency in hertz, and P is the number of poles.

For example,

• A 2-pole motor (P=2) running on a 50 Hz supply (f=50) will have a synchronous speed of 3000 RPM.
• A 4-pole motor (P=4) on the same supply will have a synchronous speed of 1500 RPM.

In practical applications, especially for induction motors, the actual operating speed is slightly less than the synchronous speed due to a phenomenon known as slip, which is necessary for torque production.

For synchronous motors, the rotor locks in step with the stator's rotating magnetic field, and thus operates precisely at the synchronous speed.

(b) Given:

• Number of poles, P = 12
• Speed, N = 600 rev/min
• Line voltage, VL = 440 V
• Resistance per phase, R = 35 Ω
• Inductive reactance per phase, XL = 25 Ω
• The alternator is delta connected and supplies a balanced star-connected load.

(i) Frequency of Supply

The frequency of an alternator is given by:

$$f = \frac{P \times N}{120}$$

Substituting the given values:

$$f = \frac{12 \times 600}{120} = 60\ Hz$$

Answer: Frequency = 60 Hz

(ii) Current in Each Coil

Since the load is star connected, the phase voltage is:

$$V_{ph} = \frac{V_L}{\sqrt{3}}$$

$$V_{ph} = \frac{440}{1.732} \approx 254.03\ V$$

The impedance of each coil is:

$$Z = \sqrt{R^2 + X_L^2}$$

$$Z = \sqrt{35^2 + 25^2}$$

$$Z = \sqrt{1850} \approx 43.01\ \Omega$$

The current through each coil is:

$$I_{coil} = \frac{V_{ph}}{Z}$$

$$I_{coil} = \frac{254.03}{43.01} \approx 5.91\ A$$

Answer: Current in each coil = 5.91 A

(iii) Current in Each Phase of the Alternator

For a star-connected load:

$$I_L = I_{coil} = 5.91\ A$$

Since the alternator is delta connected, the phase current is:

$$I_{phase} = \frac{I_L}{\sqrt{3}}$$

$$I_{phase} = \frac{5.91}{1.732} \approx 3.41\ A$$

Answer: Current in each phase of the alternator = 3.41 A

(iv) Total Power Supplied to the Load

First, calculate the power factor:

$$cos\phi = \frac{R}{Z}$$

$$cos\phi = \frac{35}{43.01} \approx 0.814$$

Total three-phase power is given by:

$$P = \sqrt{3} \times V_L \times I_L \times cos\phi$$

$$P = 1.732 \times 440 \times 5.91 \times 0.814$$

$$P \approx 3662.4\ W$$

$$P \approx 3.66\ kW$$

Answer: Total power supplied = 3662.4 W (approximately 3.66 kW)

Losses in a Squirrel-Cage Induction Motor

When a squirrel-cage induction motor operates on load, part of the electrical input power is lost in the motor due to the following:

1. Stator Copper Loss (I²R Loss): Loss due to the resistance of the stator windings. It is proportional to the square of the stator current.
2. Rotor Copper Loss (I²R Loss): Loss occurring in the rotor bars and end rings due to their electrical resistance. It is proportional to the square of the rotor current.
3. Core (Iron) Losses: These occur mainly in the stator core and consist of:
   • Hysteresis Loss – caused by repeated magnetisation and demagnetisation of the iron core.
   • Eddy Current Loss – caused by circulating currents induced in the laminated iron core.
4. Rotor core losses also occur, but under normal operating conditions they are very small and are generally neglected because the rotor/slip frequency is low.
5. Mechanical Losses: These consist of:
   • Friction Loss – caused mainly by friction in the bearings and other rotating contact points.
   • Windage Loss – caused by air resistance acting on the rotating rotor and cooling fan.
6. Stray Load Losses: These are additional small losses caused by leakage flux, harmonic magnetic fields, and non-uniform current distribution in the stator and rotor.

(i) Losses independent of load current and speed

• Stator Core (Iron) Losses – Hysteresis and Eddy Current Losses

These losses depend mainly on the supply voltage and frequency. Therefore, when the supply voltage and frequency remain constant, they are approximately independent of load current and motor speed.

(ii) Losses dependent on load current

• Stator Copper Loss
• Rotor Copper Loss
• Stray Load Loss

The stator copper loss varies as I₁²R₁, while the rotor copper loss varies as I₂²R₂. Therefore, these losses increase as the motor load and current increase. Stray load losses also increase approximately with load current.

(iii) Losses dependent on speed

• Friction Loss
• Windage Loss

These mechanical losses vary with the speed of rotation. Friction loss generally increases with speed, while windage loss increases more rapidly with speed and is approximately proportional to the cube of speed.

Note: Rotor core loss also varies with rotor/slip frequency (fᵣ = s × f) and is therefore related to slip and rotor speed. However, under normal operating conditions it is very small and is generally neglected.

In the event of a failure of the main electrical power supply on a ship, certain circuits must be fed from the emergency source of power to ensure the safety and operational capability of critical systems. These circuits are essential for maintaining the functionality of key equipment and systems that are essential for the safety of the vessel and its crew. The selection of these circuits is based on several factors aimed at prioritising important functions and ensuring the vessel's ability to respond to emergencies effectively.

1. Steering Gear Motor: The steering gear motor is essential for controlling the direction of the vessel, ensuring manoeuvrability and avoiding collisions, especially in emergency situations.
2. Emergency Fire Pump: The emergency fire pump is vital for supplying water to firefighting systems in case of fire onboard, helping to contain and extinguish fires to prevent them from spreading.
3. Emergency Air Compressor: The emergency air compressor provides compressed air for starting from a dead ship or if all the air from the main air bottle is lost.
4. Breathing air compressor: for filling SCBA bottles that can be used during fire fighting and entry into enclosed spaces.
5. Sprinkler/Hi-fog Pump: These pumps are responsible for spraying water for fire suppression to control and extinguish fires in different areas of the vessel.
6. Fire Detectors: Fire detection systems continuously monitor various areas of the ship for signs of fire or smoke, providing early warning to enable prompt response and evacuation if necessary.
7. Navigation Equipment: Navigation equipment, including radar, GPS, and gyrocompass systems, is essential for maintaining situational awareness, determining the vessel's position, and navigating safely, especially in adverse weather conditions or restricted visibility.
8. Communication Equipment: Communication systems, such as radios, satellite communication terminals, and distress alert systems, enable the crew to communicate with shore authorities, other vessels, and emergency responders in case of distress or emergencies.
9. Watertight Doors: Watertight doors are important in maintaining the vessel's watertight integrity and preventing the ingress of water in case of flooding or damage to the hull.
10. Lifeboat Davits: Lifeboat davits are used for launching lifeboats and rescue boat, providing a means of evacuation for the crew and passengers in emergencies such as abandon ship.
11. CO2 Room Exhaust Fan: The CO2 room exhaust fan is essential for ventilating the spaces where carbon dioxide (CO2) fire suppression systems are installed, ensuring that the room is safe to enter.
12. Engine Room Vent Fan: One of the Engine room blower power is supplied from the Emergency generator, which helps in air supply to E/R and provides safe entry to the Engine room.
13. E/R Pumps and Systems for First Start from 'Dead Ship': These pumps and systems are necessary for restarting the required pumps and systems in the engine room, enabling the vessel to restore power and propulsion from a state of complete power loss ('Dead Ship').
14. Emergency Lights: Emergency lights are provided in the Engine room, accommodation, upper deck, escape routes, stairwells, and muster stations, ensuring visibility during power outages or emergencies.
15. Battery Chargers: Battery chargers maintain the charge of essential batteries, including those for emergency lighting, communication equipment, and control systems, ensuring their readiness for use in emergencies.
16. E/R Alarm System: The engine room alarm system monitors various parameters and conditions in the engine room, providing early warning of abnormalities, malfunctions, or hazards that could jeopardise the safety and operation of the vessel.

B)It is phase-angle firing that thyristor controllers commonly adopt to precisely regulate the power of electric heaters.

Phase-angle firing, also known as phase cutting, phase-angle control, or phase-fired control (PFC), is a power-limiting method used for AC voltages. It regulates power by triggering a thyristor, SCR, triac, thyratron, or similar gated device into conduction at a specific phase angle of the AC waveform.

In the heating power controller, the thyristor acts like a switch that controls the on/off state of the heater in the main circuit. As shown in the figure of the APR3H series three-phase power controller, each phase is equipped with one thyristor. During each half AC cycle, the thyristor is triggered into conduction at a certain phase angle between 0 and 180 degrees, and then it naturally turns off at 180 degrees. The earlier the thyristor is triggered within the phase angle, the greater the power the controller delivers to the heater. Zero-crossing means triggering at the zero phase angle, where the thyristor conducts for the entire half cycle.

Waveform with Phase-Angle Control

The output voltage waveform of a thyristor power controller is clipped depending on the firing angle. At higher firing angles, the waveform is clipped more, resulting in a smaller area under the curve and, therefore, less average power delivered to the heater.

Summary

SCR (SCR Power Controller): Ideal for applications requiring precise power and temperature control, such as industrial heating systems. SCR allows for smooth, continuous modulation of power, making it suitable for systems needing fine adjustments.

Speed Change:

The synchronous speed of an induction motor is governed by the formula:

$$N_{s}=\frac{120f}{P}$$

Where,

Ns = Synchronous speed.

f = Frequency of power supply.

P = number of poles.

Methods to Achieve Speed Change:

Multiple Stator Windings:

• Two sets of windings are installed on the stator, each designed for different pole numbers. Only one winding is energized at a time, allowing a change in speed.

Consequent Pole Method:

• A single stator winding is divided into coil groups. By altering the connections (series or parallel), the number of poles is changed, resulting in different speeds.

Pole Amplitude Modulation (PAM):

• Used when a speed ratio other than 2: 1 is required. The winding is split into parts that can be connected in series or parallel. The current direction in specific parts of the winding determines the pole configuration, allowing finer speed adjustments.

Braking Mechanism:

Braking is used to reduce the torque and stop the motor.

Plugging:

• Plugging is a braking method where the power supply to the motor is switched over in a way that two phases are interchanged. This creates a reverse torque that quickly reduces the motor speed. Once the speed becomes negligible, the power is switched off to prevent the motor from running in the opposite direction. An electromagnetic brake is then applied to stop the motor.

Rheostatic Braking:

• In this method, the motor is switched off, and all three phases are shorted through rheostats. The rheostats act as resistors, dissipating the kinetic energy of the motor in the form of heat through copper losses. The resistance provided by the rheostats slows down the motor and brings it to a stop.

Regenerative Braking:

• For regenerative braking, the motor is switched off from the A.C. power supply, and the stator winding is provided with a D.C. supply from batteries. The fixed magnetic flux of the D.C. tries to create a magnetic locking with the rotating rotor poles, generating a retarding torque that reduces the motor speed. When the speed drops to zero, the D.C. supply is switched off, and an electromagnetic brake is applied to stop the motor.

Condition Monitoring of Electrical Machinery

Two important parameters recorded for condition monitoring of electrical machinery on board a ship are:

1. Temperature

Monitoring the temperature of electrical machinery is essential because it provides valuable information about the health and operating condition of the equipment.

Electrical machines such as motors, generators, and transformers generate heat during normal operation. By recording and analysing temperature trends, abnormal heating patterns or excessive temperature rise can be detected.

Excessive temperature may indicate:

• Inadequate cooling
• Insulation degradation
• Bearing problems
• Overloading

If not corrected in time, overheating can lead to serious damage and eventual failure of the machinery.

2. Vibration

Vibration analysis is another important parameter used for monitoring the condition of rotating electrical machinery.

Vibration sensors measure:

• Magnitude
• Frequency
• Vibration patterns

Excessive vibration often indicates mechanical or electrical faults such as:

• Misalignment
• Imbalance
• Bearing wear
• Mechanical looseness

By continuously monitoring vibration levels, deviations from normal operating conditions can be detected early, allowing corrective maintenance before major damage occurs.

1. Vibration Measurement and Defects Revealed

Vibration is measured using transducers such as:

• Accelerometers
• Velocity pick-ups
• Seismic transducers

These sensors are mounted on:

• Machine casing
• Bearing housing
• Rotor assembly

They detect vibration signals at various frequencies, which are analysed to identify specific faults.

Defects Identified Through Vibration Analysis

• Imbalance: A vibration peak at shaft speed frequency (1X) indicates rotor imbalance.
• Misalignment: Vibrations at 1X, 2X, and 3X shaft speed frequencies usually indicate misalignment.
• Bearing Damage: High-frequency peaks between 2 kHz and 5 kHz (depending on shaft speed and transducer resonance) indicate bearing defects.
• Electrical Problems: Synchronous frequency components and sidebands in the vibration signal suggest electrical faults.
• Gear Damage: Gear mesh frequency and its harmonics (depending on shaft speed and number of gear teeth) indicate gear defects.
• Cracked or Bent Shaft: Vibrations at 2X and 3X shaft speed frequencies may indicate a cracked or bent shaft.

2. Temperature Measurement and Defects Revealed

Temperature is measured using:

• Thermocouples
• Infrared cameras

Sensors are installed near:

• Bearings
• Windings
• Electrical components

Temperature monitoring helps detect abnormal heating and potential failure.

Defects Revealed Through Temperature Monitoring

• Bearing Failure: Rising bearing temperature indicates increased friction and possible bearing damage.
• Insulation Deterioration: Temperature rise on the surface of insulating materials may indicate insulation breakdown.
• Overload or Loose Connections: Hot spots detected on electrical panels using infrared cameras may indicate overload conditions or loose electrical connections.

The form factor of an AC waveform is the ratio of its RMS value to its average value. For a perfect sine wave, the form factor is approximately 1.11 (RMS value/Average Value = 1.11). This factor indicates how closely a waveform resembles a pure sine wave; a form factor closer to 1.11 suggests a waveform that is more sinusoidal.

Operation and Output Voltage Across Load Resistor

Let the input supply be: v(t)=Vm​sin(ωt)

• During the positive half-cycle, the diode conducts whenever v(t) exceeds the voltage across the capacitor. The capacitor charges instantly (ideal case) to the peak value Vm​.
• After reaching the peak, the diode becomes reverse biased (since v(t)<Vm​), and the capacitor holds its charge.
• The load resistor (R) is connected in parallel with the capacitor; thus, the voltage across Ris the same as across C.
• If we ignore the small discharge of C(assuming R is large and/or C is large so discharge is negligible between cycles):

Voltage Across Load Resistor

The voltage across R after initial peak is: vR​(t)≈Vm​That is, the output is a DC voltage nearly equal to the peak value of the AC input.

Mathematical Expression

If capacitor discharge is considered negligible: vout​(t)=Vm​If the discharge is not negligible (realistic case), the output will have slight ripple and can be given by: vout​(t)≈Vm​−ΔVwhere ΔV is the ripple voltage (depends on R, C, and frequency f).

minimize energy losses (copper losses due to winding resistance and iron losses due to core magnetization). This translates directly to lower operating costs and reduced environmental impact due to less wasted energy. High-efficiency transformers typically achieve 95.5% efficiency for 5 kVA units and up to 97.5% for units up to 1 MVA.

Lower energy losses mean less heat is generated within the transformer. This reduces the risk of overheating, extending the lifespan of the equipment and preventing potential fire hazards. Overheating can damage the insulation and reduce the lifespan of the transformer. Lower operating temperatures contribute to enhanced reliability and a longer operational life for the transformer.

Low voltage regulation ensures that the output voltage remains relatively constant even under varying load conditions. This stability is essential for the stable operation of equipment connected to the transformer. Voltage fluctuations can lead to wear and tear on connected equipment. Maintaining a constant voltage extends the lifespan of the equipment.

• The square of the applied voltage (T proportional to V^2).
• The rotor resistance (increasing rotor resistance increases starting torque up to a maximum).
• The rotor and stator leakage reactances.
• The number of poles / synchronous speed.
• The supply frequency.
• The air-gap flux and the winding turns ratio.
• Comparison with rated torque: the starting torque of a standard squirrel-cage motor is typically about 1.5 to 2 times the full-load torque, so that the motor can start its load. It is generally greater than rated torque but not excessive.

• (i) Synchronous speed: Ns = 120 f / P = 120 x 60 / 8 = 900 rev/min = 15 rev/s.
• (ii) Rotor copper loss:
• Slip s = (Ns - N)/Ns = (15 - 14.4)/15 = 0.6/15 = 0.04.
• Shaft output = 8 kW. Rotor gross mechanical power = shaft output + rotational losses = 8 + 0.8 = 8.8 kW.
• Air-gap power Pg = rotor gross power / (1 - s) = 8.8 / 0.96 = 9.167 kW.
• Rotor copper loss = s x Pg = 0.04 x 9.167 = 0.367 kW (367 W).
• (iii) Input power to the motor = air-gap power + stator loss = 9.167 + 1 = 10.167 kW.
• (iv) Motor current: I = P_in / (root 3 x V x cos phi) = 10167 / (1.732 x 440 x 0.85) = 10167 / 648.1 = 15.69 A.

So synchronous speed = 900 rev/min, rotor copper loss = 0.367 kW, input power = 10.17 kW, motor current = 15.7 A.

• The basic equation is that the resistance between any two terminals must be the same in both the star and delta networks.
• For a delta network with resistances R12 (between terminals 1-2), R23 (2-3), R31 (3-1), and a star network with resistances R1, R2, R3 (each connected to a terminal and a common centre point):
• Resistance between terminals 1 and 2: in delta it is R12 in parallel with (R23 + R31); in star it is R1 + R2. Equating:

R1 + R2 = R12 (R23 + R31) / (R12 + R23 + R31)

R2 + R3 = R23 (R31 + R12) / (R12 + R23 + R31)

R3 + R1 = R31 (R12 + R23) / (R12 + R23 + R31)

• Solving these gives the delta-to-star conversion:

R1 = R12 R31 / (R12 + R23 + R31)

R2 = R12 R23 / (R12 + R23 + R31)

R3 = R23 R31 / (R12 + R23 + R31)

• And the star-to-delta conversion:

R12 = (R1 R2 + R2 R3 + R3 R1) / R3

R23 = (R1 R2 + R2 R3 + R3 R1) / R1

R31 = (R1 R2 + R2 R3 + R3 R1) / R2

• For equal resistances: delta R = 3 x star r (R = 3r), and star r = R/3.

• Battery A: 105 V, internal 0.25 ohm. Battery B: 100 V, internal 0.2 ohm. Battery C: 95 V, internal 0.25 ohm.
• Let the node voltages at the positive terminals be Va, Vb, Vc (common negative = 0).
• Current from A into the 0.5 ohm resistor: (105 - Va)/0.25 = (Va - Vb)/0.5.

105 - Va = 0.5(Va - Vb) -> 1.5 Va - 0.5 Vb = 105. (1)

• Current from C into the 0.3 ohm resistor: (95 - Vc)/0.25 = (Vb - Vc)/0.3.

95 - Vc = 0.8333(Vb - Vc) -> 5 Vb + Vc = 570. (2)

• At node B: (Va - Vb)/0.5 + (100 - Vb)/0.2 = (Vb - Vc)/0.3.

2(Va - Vb) + 5(100 - Vb) = 3.333(Vb - Vc)

2 Va - 10.333 Vb + 3.333 Vc = -500. (3)

• From (1): Va = 70 + 0.3333 Vb. From (2): Vc = 570 - 5 Vb.
• Substitute into (3): 2(70 + 0.3333 Vb) - 10.333 Vb + 3.333(570 - 5 Vb) = -500

140 + 0.6667 Vb - 10.333 Vb + 1900 - 16.667 Vb = -500

2040 - 26.333 Vb = -500 -> Vb = 2540/26.333 = 96.46 V.

• Va = 70 + 0.3333 x 96.46 = 70 + 32.15 = 102.15 V.
• Vc = 570 - 5 x 96.46 = 570 - 482.3 = 87.7 V.
• Current in 0.5 ohm resistor: I_AB = (Va - Vb)/0.5 = (102.15 - 96.46)/0.5 = 5.69/0.5 = 11.38 A (from A to B).
• Current in 0.3 ohm resistor: I_BC = (Vb - Vc)/0.3 = (96.46 - 87.7)/0.3 = 8.76/0.3 = 29.2 A (from B to C).
• Power in 0.5 ohm: P = I^2 R = 11.38^2 x 0.5 = 129.5 x 0.5 = 64.8 W.
• Power in 0.3 ohm: P = 29.2^2 x 0.3 = 852.6 x 0.3 = 255.8 W.

So the 0.5 ohm resistor carries 11.4 A (64.8 W) and the 0.3 ohm resistor carries 29.2 A (255.8 W).

The difference between half wave and full wave rectification:

Half-wave rectification:

• Uses a single diode to allow only one-half of the AC waveform (either the positive or negative half-cycle) to pass through. The other half is blocked.
• Results in a pulsating DC output with a significant amount of ripple (variation in voltage). The average DC voltage is lower compared to the input AC voltage.
• Simpler to implement, requiring only one diode, but less efficient because it wastes half of the input power.
• The average output current is 0.318 times the peak input current. The RMS value is 0.5 times the peak input current.

Half-wave rectification is not well adopted because:

• Less Average current
• Less average RMS
• High pulsation output
• Lower voltage developed
• More ripple as compared to others
• Efficiency is less as compared to others.

Full-wave rectification:

• Uses either two diodes in a centre-tapped transformer configuration or four diodes in a bridge rectifier configuration to utilize both halves of the AC waveform.
• Produces a pulsating DC output with less ripple than half-wave rectification, resulting in a smoother DC output and a higher average DC voltage.
• More efficient as it uses both halves of the input AC waveform.
• Requires more components (two or four diodes), but provides a more efficient and improved DC output. The average output current is 0.6365 times the peak input current, and the RMS value is 0.707 times the peak input current.

Sketch of bridge connection for full wave rectification:

To form a star connection, one end of each phase winding is joined together to create a neutral point. The other ends of the windings are connected to outgoing conductors leading to the bus bar. This neutral point can either be insulated or connected to a neutral line, depending on the system design.

On ships, the neutral point is usually insulated to prevent the system from tripping in the event of a single earth fault. This is critical for maintaining power continuity to essential equipment like the steering gear, navigation systems, and emergency lighting.

By insulating the neutral, the system can tolerate one earth fault without immediate interruption, allowing time to locate and rectify the fault while ensuring continuous power supply. Only if a second earth fault occurs, creating a short circuit, will the protection system trip. This arrangement allows the ship to maintain essential operations.

In systems where the neutral point is earthed, any earth fault in the system will immediately create a fault current, causing the circuit protection (e.g., breakers or fuses) to trip. This configuration is common in high-voltage systems to ensure that faults are quickly isolated, preventing damage to equipment and reducing the risk of electric shock or fire.

Earthed neutral systems also simplify fault detection and protection mechanisms, making them suitable for vessels with high-voltage installations where rapid fault isolation is a priority.

• Voltage: Voltage should be equal to or slightly higher than the busbar voltage. This is checked using a voltmeter.
• Frequency: The frequency should be equal to or slightly higher than the busbar frequency. In practice, the frequency of the incoming alternator is kept slightly higher so that when load is applied, it matches the busbar frequency. The synchroscope should move clockwise at a slow speed.
• Phase Angle: There should be no phase angle between the incoming and running generator. The synchroscope pointer should be at 12 o'clock, indicating a zero or acceptable phase angle difference between the incoming alternator and the busbar.

• Mechanical Surge Torque: A significant surge of torque is exerted on the rotor. This can cause damage to the rotor shaft (twisting, keyway damage), coupling (breakage), and stator windings (deformation). The stator core might also shift relative to its frame.
• Electrical Surge: A surge of current and power circulates through the system. This greatly strains the entire system, potentially leading to overheating and component failure. The sudden inrush of current could lead to circuit breakers tripping to protect the system.

• Automatic Synchronization: Systems with automatic synchronization pre-program the correct voltage, frequency, and phase angle, greatly reducing the chances of errors.
• Manual Synchronization with Synchroscope: With manual synchronisation, a synchroscope carefully compares the incoming alternator's frequency and phase angle to the busbar's. Adjust the incoming alternator’s voltage to match the busbar. When the synchroscope pointer moves slowly clockwise and approaches the 12 o'clock position, close the alternator breaker to ensure proper synchronisation.

(i) Reduced torque from the prime mover of one machine:

If one alternator's prime mover (the engine driving the alternator) experiences reduced torque, that alternator will begin to reduce its load contribution to the busbar. The other alternator will compensate for the reduced output, taking on the additional load. If the torque continues to decrease on the first alternator, it will eventually draw power from the busbar, acting as a motor rather than a generator. This will trip a reverse power relay, shutting down the affected alternator for protection.

(ii) Reduced excitation on one machine:

If the excitation of one alternator is reduced, its generated voltage decreases. This creates a circulating current between the alternators, almost 90 degrees out of phase, due to the inductive nature of alternator windings. The other alternator carries both its original load current and the circulating current, leading to an increased current and a more lagging power factor. The affected alternator will have a reduced current and a less lagging power factor. Both will continue to share the load (kW) despite operating at different currents and power factors. The reduced excitation can lead to instability and, in some cases, result in the alternator becoming overloaded and tripping offline.

(a) Rectifier Circuit

• Converts AC (Alternating Current) into DC (Direct Current).
• Input: AC signal.
• Output: DC signal.
• Operation: Conducts during the positive half cycle of the input signal (half-wave rectifier) or during both half cycles (full-wave rectifier).
• Types: Half-wave, full-wave, bridge rectifier.
• Feedback: No feedback involved.
• Use Case: Used continuously for powering DC loads.

The sketch shows a simplified representation of an AC input waveform being converted into a pulsating DC waveform by a rectifier. A smoothing capacitor is added to reduce the pulsations and produce a more constant DC output.

(b) Amplifier Circuit

• Amplifies the amplitude of a weak signal without altering its waveform.
• Input: Weak signal to be amplified.
• Output: Amplified version of the input signal.
• Operation: Amplifies signals during both positive and negative cycles.
• Types: Categorized by frequency (audio, RF), or by physical placement (voltage, current amplifiers).
• Feedback: Uses negative feedback to stabilize gain.
• Use Case: Repeatedly used in circuits to maintain signal strength.

This sketch illustrates a generic amplifier. The input signal is smaller than the output signal.

(c) Oscillator Circuit

• Generates periodic, oscillating electronic signals such as sine waves or square waves.
• Input: DC supply.
• Output: AC signal.
• Operation: Converts DC into AC using positive feedback.
• Types: Linear (sine wave oscillators) and non-linear (square wave, sawtooth oscillators).
• Feedback: Uses positive feedback to sustain oscillations.
• Use Case: Used initially in circuits to provide a signal source.

Requirements for Unattended Machinery Space (UMS) Ship:

1. Fire Precaution

• Arrangements should be provided on the UMS ship to detect and give an alarm in case of fire.
• In the boiler, air supply casing and uptake.
• In scavenge space of propulsion machinery.
• In engines of power, 2250 kW and above or cylinders having bore more than 300mm should be provided with an oil mist detector for the crankcase or bearing temperature monitor or either of two.

2. Centralized control & instruments are required in Machinery Space

• UMS ships must have a centralised control room that is easily accessible and equipped with adequate instrumentation and equipment to monitor and operate all main and auxiliary machinery. A system must be provided to call the engineers to the machinery space in case of emergency

3. Protection against flooding:

• UMS ships must have bilge wells that are located and designed to detect the accumulation of liquid at a normal angle of heel and trim and to accommodate the drainage of liquid during an unattended period. If the bilge pump starts automatically, an alarm must indicate that the flow of liquid pumped is more than the capacity of the pump.

4. Automatic Fire Detection

• Alarms and detection should operate very rapidly and effectively. It should be placed at numerous well-sited places for quick response of the detectors.

5. Fire Extinguishing System

• There should be arrangements for a fire extinguishing system other than the conventional hand extinguishers, which can be operated remotely from machinery space. The station must give control of emergency fire pumps, generators, valves, extinguishing media, etc.

6. Alarm System

• A comprehensive alarm system must be provided for control & accommodation areas.

7. Automatic Start of Emergency Generator

• Arrangements for the starting of an emergency generator and automatic connection to the bus bar must be provided in case of a blackout condition, apart from that, the following points are also to be noted.
• 8. Local hand control of essential machinery like steering, emergency generator starting, emergency start for main engine, etc. 8. Adequate settling tank storage capacity. 9. Regular testing & maintenance of machinery alarms & instruments.

• Immediately Inform the Chief Engineer and Master about the situation.
• Man the Engine Room, Increased vigilance is necessary, and a dedicated person should be positioned at the ECR station.
• Conduct significantly more frequent fire rounds, examining all areas for any signs of fire or overheating. This involves visual inspection, checking temperatures using infrared thermometers, and detecting unusual smells or sounds.
• Monitor all machinery parameters closely for any signs of abnormality, such as temperature increases, unusual vibrations, or unusual pressure changes. This includes checking oil and gas leakages.
• Ensure all containers of lubricating oil, diesel oil, fuel oil, and chemicals are properly secured to prevent spillage or movement.
• Check that all other machinery alarms are functioning correctly, and report any additional issues to the Chief Engineer and Master.
• Given the failure of the main system, consider the use of alternative detection methods. This could include enhanced visual inspections and the use of portable smoke detectors.
• If the situation remains unresolved, or other safety concerns arise, it might be necessary to consider diverting to the nearest port for repairs and improved safety.

(b) given:

$$V=230V$$

$$N_1=900rpm$$

$$I_{a1}=30A$$

$$R_{a1}=0.4\Omega$$

$$N_2=500rpm$$

$$I_{a2}=25A$$

$$R_{a2}=?$$

$$E_{b1}=V-I_{a}R_{a1}$$

$$=230-30\times0.4$$

$$E_{b1}=218V$$

$$\frac{E_{b1}}{E_{b2}}=\frac{N_1}{N_2}$$

$$E_{b2}=\frac{E_{b1}\times N_2}{N_1}$$

$$E_{b2}=\frac{218\times500}{900}$$

$$E_{b2}=121.11V$$

$$E_{b2}=V-I_{a2}R_{a2}$$

$$R_{a2}=\:\frac{V-E_{b2}}{I_{a2}}$$

$$R_{a2}=\frac{230-121.11}{25}$$

$$R_{a2}=4.35\Omega$$

$$R\:to\:be\:added=4.35-0.4=3.95\Omega$$

$$R_{a2}=3.95\Omega$$

• Consider a coil of N turns rotating in a uniform magnetic field of flux density B. The flux linking the coil is phi = B A cos(wt), where A is the coil area and w the angular velocity.
• By Faraday's law, the induced e.m.f. is e = -N d(phi)/dt = -N d(B A cos wt)/dt = N B A w sin(wt) = Em sin(wt).
• The maximum e.m.f. Em = N B A w = N phi_m w, where phi_m = B A is the maximum flux linking the coil.
• For a machine with Z conductors, P poles, flux per pole phi, speed N rev/min, the average e.m.f. per conductor is E_av = 2 phi N P / 60, and the generated e.m.f. is:

E = (P x phi x Z x N) / (60 x A) volts,

where A is the number of parallel paths (A = 2 for wave winding, A = P for lap winding).

• The r.m.s. value of the generated e.m.f. per phase for a distributed winding is E = 4.44 x f x phi x T x k_w, where T is the number of turns per phase, f the frequency, and k_w the winding factor.

• Synchronous speed Ns = 1000 rev/min (6-pole at 50 Hz). Angular speed (mechanical) w = 2 pi x 1000/60 = 104.72 rad/s.
• Per-unit synchronous reactance Xs = 0.25 p.u. (based on 3000 kVA).
• Synchronizing power: for a small angular displacement, the synchronizing power per phase = (Ef V / Xs) x (electrical angle in rad). At no load Ef = V.
• In per unit, Ef V / Xs = 1 x 1 / 0.25 = 4 p.u. of the per-phase base power. Per-phase base power = 3000/3 = 1000 kW, so Ef V/Xs = 4 x 1000 = 4000 kW per phase; total for 3 phases = 12000 kW.
• One mechanical degree = (P/2) electrical degrees = 3 electrical degrees = 3 x pi/180 = 0.05236 rad.
• Synchronizing power Ps = 12000 x 0.05236 = 628.3 kW.
• Synchronizing torque Ts = Ps / w = 628300 / 104.72 = 6000 N m.

So the synchronizing power for one mechanical degree of displacement is about 628 kW and the corresponding synchronizing torque is about 6000 N m.

Given:

$$Output=20000W$$

$$Iron\:loss=780W$$

$$Friction\:loss\:\left(mech\operatorname{loss}\right)=590W$$

$$R_{a}=0.15\Omega$$

$$R_{se}=0.025\Omega$$

$$R_{int}=0.028\Omega$$

$$R_{sh}=115\Omega$$

$$I_{sh}=\frac{V}{R_{sh}}=\frac{200}{115}$$

$$I_{sh}=1.74A$$

$$Copper\:loss\:in\:stator=I^2R$$

$$C_{S}=1.74^2\times115$$

$$C_{S}=348W$$

$$Gen\:output\:=\:V\times I_{L}=20000W\:\left(given\right)$$

$$200\times I_{L}=20000$$

$$I_{L}=100A$$

$$I_{a}=I_{sh}+I_{L}$$

$$=1.74+100$$

$$I_{a}=101.74A$$

$$Copper\:loss\:in\:stator=I^2R=I_{a}^2\left(R_{se}+R_{a}+R_{int}\right)$$

$$C_{R}=101.74^2\times\left(0.025+0.15+0.028\right)$$

$$C_{R}=2101W$$

$$Total\:Copper\:loss=C_{S}+C_{R}$$

$$=348+2101$$

$$=2449W$$

$$\eta=\frac{Output}{Input}=\frac{Output}{Output+losses}$$

$$=\frac{20000}{20000+780+590+2449}$$

$$\eta=83.96\%$$

• Supply Frequency (f)
• Number of Poles (P)

The relationship between these factors and the synchronous speed (Ns) is given by the formula:

$$N_{s}=\frac{120f}{P}$$

Where Ns is the synchronous speed in revolutions per minute (RPM), f is the supply frequency in hertz, and P is the number of poles.

For example,

• A 2-pole motor (P=2) running on a 50 Hz supply (f=50) will have a synchronous speed of 3000 RPM.
• A 4-pole motor (P=4) on the same supply will have a synchronous speed of 1500 RPM.

In practical applications, especially for induction motors, the actual operating speed is slightly less than the synchronous speed due to a phenomenon known as slip, which is necessary for torque production.

For synchronous motors, the rotor locks in step with the stator's rotating magnetic field, and thus operates precisely at the synchronous speed.

Power factor correction refers to the process of improving the power factor of an electrical system to bring it closer to unity (1 or 100%). It involves reducing the phase difference between voltage and current, which is caused by inductive loads like motors, transformers, and fluorescent lighting. These loads consume reactive power, leading to a lagging power factor. By adding components like capacitors, synchronous condensers, or phase advancers, reactive power is compensated, and the power factor is improved.

In practical terms, power factor correction aims to minimize the inefficiencies in the electrical system, reduce energy losses, and ensure optimal utilization of the power supplied by the generator or grid.

• By improving power factor, the current flow in the system is reduced, leading to lower I²R losses in cables, transformers, and other distribution components.
• With a higher power factor, the electrical system operates more efficiently, ensuring better utilization of the generated power.
• Improved power factor reduces the apparent power (kVA) requirement, allowing for smaller-sized generators, transformers, and cables, thus reducing capital costs.
• Higher power factor ensures better voltage stability across the system, preventing voltage drops and protecting sensitive equipment from under-voltage issues.
• By reducing reactive power, the system can handle more active power (real load) within the same capacity of the equipment, maximising output.
• With reduced current and heat generation, the wear and tear on electrical components are minimized, extending their lifespan.
• Higher efficiency in power usage reduces the overall energy demand, lowering fuel consumption and greenhouse gas emissions in power generation.

A three-phase system typically has an inductive load (e.g., motors), causing a lagging power factor. Capacitors can provide leading reactive power to compensate for this. The capacitors are connected in parallel with the inductive load.

• The size (capacitance) of each capacitor is calculated based on the size of the inductive load and the desired power factor improvement. Specialised software or calculation methods are often used for accurate determination.
• The capacitors are connected in a star or delta configuration, matching the load's connection. They should be appropriately rated for the voltage and current of the system.
• The leading reactive power supplied by the capacitors cancels out some of the lagging reactive power from the inductive load, effectively reducing the overall reactive power and improving the power factor.

Besides capacitors, synchronous motors can also be used for power factor correction. Synchronous motors can be operated at leading power factor, effectively counteracting the lagging power factor of inductive loads. These motors can contribute both real power and leading reactive power to the system. However, synchronous motors are more complex and expensive than capacitors. They are often used in larger industrial installations where the power factor correction requirements are significant.

To form a star connection, one end of each phase winding is joined together to create a neutral point. The other ends of the windings are connected to outgoing conductors leading to the bus bar. This neutral point can either be insulated or connected to a neutral line, depending on the system design.

On ships, the neutral point is usually insulated to prevent the system from tripping in the event of a single earth fault. This is critical for maintaining power continuity to essential equipment like the steering gear, navigation systems, and emergency lighting.

By insulating the neutral, the system can tolerate one earth fault without immediate interruption, allowing time to locate and rectify the fault while ensuring continuous power supply. Only if a second earth fault occurs, creating a short circuit, will the protection system trip. This arrangement allows the ship to maintain essential operations.

In systems where the neutral point is earthed, any earth fault in the system will immediately create a fault current, causing the circuit protection (e.g., breakers or fuses) to trip. This configuration is common in high-voltage systems to ensure that faults are quickly isolated, preventing damage to equipment and reducing the risk of electric shock or fire.

Earthed neutral systems also simplify fault detection and protection mechanisms, making them suitable for vessels with high-voltage installations where rapid fault isolation is a priority.

• Voltage: Voltage should be equal to or slightly higher than the busbar voltage. This is checked using a voltmeter.
• Frequency: The frequency should be equal to or slightly higher than the busbar frequency. In practice, the frequency of the incoming alternator is kept slightly higher so that when load is applied, it matches the busbar frequency. The synchroscope should move clockwise at a slow speed.
• Phase Angle: There should be no phase angle between the incoming and running generator. The synchroscope pointer should be at 12 o'clock, indicating a zero or acceptable phase angle difference between the incoming alternator and the busbar.

• Mechanical Surge Torque: A significant surge of torque is exerted on the rotor. This can cause damage to the rotor shaft (twisting, keyway damage), coupling (breakage), and stator windings (deformation). The stator core might also shift relative to its frame.
• Electrical Surge: A surge of current and power circulates through the system. This greatly strains the entire system, potentially leading to overheating and component failure. The sudden inrush of current could lead to circuit breakers tripping to protect the system.

• Automatic Synchronization: Systems with automatic synchronization pre-program the correct voltage, frequency, and phase angle, greatly reducing the chances of errors.
• Manual Synchronization with Synchroscope: With manual synchronisation, a synchroscope carefully compares the incoming alternator's frequency and phase angle to the busbar's. Adjust the incoming alternator’s voltage to match the busbar. When the synchroscope pointer moves slowly clockwise and approaches the 12 o'clock position, close the alternator breaker to ensure proper synchronisation.

(i) Reduced torque from the prime mover of one machine:

If one alternator's prime mover (the engine driving the alternator) experiences reduced torque, that alternator will begin to reduce its load contribution to the busbar. The other alternator will compensate for the reduced output, taking on the additional load. If the torque continues to decrease on the first alternator, it will eventually draw power from the busbar, acting as a motor rather than a generator. This will trip a reverse power relay, shutting down the affected alternator for protection.

(ii) Reduced excitation on one machine:

If the excitation of one alternator is reduced, its generated voltage decreases. This creates a circulating current between the alternators, almost 90 degrees out of phase, due to the inductive nature of alternator windings. The other alternator carries both its original load current and the circulating current, leading to an increased current and a more lagging power factor. The affected alternator will have a reduced current and a less lagging power factor. Both will continue to share the load (kW) despite operating at different currents and power factors. The reduced excitation can lead to instability and, in some cases, result in the alternator becoming overloaded and tripping offline.