Power System Protection (EX-701) - Unit 4 Short Notes

UNIT 4: POWER SYSTEM PROTECTION - EXAM-FOCUSED SHORT NOTES

I. FAULT ANALYSIS & SYMMETRICAL COMPONENTS

Types of Power System Faults

• Symmetrical Fault: Three-phase fault (LLLG). All phases affected equally. System remains balanced. Highest fault current.

• Unsymmetrical Faults:

  • Single Line-to-Ground (SLG/LG): Most common (~70%).

  • Line-to-Line (LL): Second most common.

  • Double Line-to-Ground (DLG/LLG):

• Fault Level / Fault MVA: The maximum power (in MVA) that can flow into a fault at a given point. It determines the required breaking capacity of circuit breakers.

$$ \text{Fault MVA} = \sqrt{3} \times V_{\text{base}} \times I_{\text{fault}} $$

Symmetrical Components (Krenz's Theorem)

• Significance: Any set of three unbalanced phasors (voltages/currents) can be resolved into three sets of balanced phasors:

  1. Positive Sequence ($$\displaystyle V_1, I_1 $$): Balanced, phase sequence A-B-C.

  2. Negative Sequence ($$\displaystyle V_2, I_2 $$): Balanced, phase sequence A-C-B.

  3. Zero Sequence ($$\displaystyle V_0, I_0 $$): All three phasors in phase.

• Simplification: Allows analysis of complex unsymmetrical faults by solving three simple, independent, balanced systems.

Sequence Networks

• Positive Sequence Network: Represents the normal balanced system. Contains internal EMFs ($$\displaystyle E_f $$). Reactance = $$\displaystyle X_d'' $$ (sub-transient) for generators during faults.

• Negative Sequence Network: No internal EMFs ($$\displaystyle E_2 = 0 $$). Reactance = $$\displaystyle X_d'' $$ (for synchronous machines, often same as positive sequence for faults).

• Zero Sequence Network: Path depends on grounding. For a generator with solidly grounded neutral, $$\displaystyle Z_0 = Z_{g} + 3Z_n $$ (where $$\displaystyle Z_g $$ is generator zero seq. reactance, $$\displaystyle Z_n $$ is neutral grounding impedance). No zero sequence current flows if neutral is isolated.

Fault Current Derivation (at Generator Terminals, Unloaded, Solid Ground)

General Method: Interconnect sequence networks based on fault type. Solve for $$\displaystyle I_0, I_1, I_2 $$. Fault currents are derived from these.

1. Single Line-to-Ground (SLG) Fault (Phase A to Ground):

   • Connection: Series: $$\displaystyle Z_1 + Z_2 + Z_0 $$.

   • Sequence Current:

$$ I_1 = I_2 = I_0 = \frac{E_f}{Z_1 + Z_2 + Z_0} $$

*   **Fault Current (Phase A):** 

$$ I_f = I_a = 3I_0 = \frac{3E_f}{Z_1 + Z_2 + Z_0} \boxed{} $$

2. Line-to-Line (LL) Fault (Phase B & C):

   • Connection: Parallel: $$\displaystyle I_1 = -I_2 $$, $$\displaystyle I_0 = 0 $$. Networks in parallel: $$\displaystyle Z_1 + Z_2 $$.

   • Sequence Current:

$$ I_1 = -I_2 = \frac{E_f}{Z_1 + Z_2} $$

*   **Fault Current (e.g., $$\displaystyle I_b $$):** 

$$ I_f = I_b = \sqrt{3} I_1 \angle 30^\circ \boxed{} $$

3. Double Line-to-Ground (DLG) Fault (Phases B & C to Ground):

   • Connection: $$\displaystyle I_1 + I_2 + I_0 = 0 $$. Networks in parallel: $$\displaystyle Z_1 \parallel (Z_2 + Z_0) $$.

   • Sequence Currents:

$$ I_1 = \frac{E_f}{Z_1 + \frac{Z_2 Z_0}{Z_2 + Z_0}} $$

$$ I_2 = -I_1 \frac{Z_0}{Z_2 + Z_0}, \quad I_0 = -I_1 \frac{Z_2}{Z_2 + Z_0} $$

*   **Fault Current (Phase A):** 

$$ I_a = 0, \quad I_b = I_1 (\sqrt{3} \angle 30^\circ - \frac{Z_2}{Z_2+Z_0}), \quad I_c = I_1 (\sqrt{3} \angle -30^\circ - \frac{Z_0}{Z_2+Z_0}) $$

4. Three-Phase Fault (LLLG):

   • Connection: Only positive sequence network. $$\displaystyle I_1 = E_f / Z_1 $$, $$\displaystyle I_2 = I_0 = 0 $$.

   • Fault Current:

$$ I_f = I_a = I_1 = \frac{E_f}{Z_1} \boxed{} $$

Effect of Fault Impedance ($$\displaystyle Z_f $$)

• Adds to the impedance in the fault path.

• SLG Fault: $$\displaystyle I_f = \frac{3E_f}{Z_1 + Z_2 + Z_0 + 3Z_f} $$

• LL Fault: $$\displaystyle I_f = \frac{\sqrt{3} E_f}{Z_1 + Z_2 + Z_f} $$

• DLG Fault: $$\displaystyle Z_f $$ appears in parallel with $$\displaystyle Z_0 $$ in the negative-zero sequence loop.

• Result: Fault current decreases as $$\displaystyle Z_f $$ increases.

System Representation for Fault Studies

• Per-Unit System: Simplifies calculations by eliminating voltage/transformer ratio conversions.

• Base Quantity Selection: Choose common $$\displaystyle S_{base} $$ and $$\displaystyle V_{base} $$ for the entire system.

• Conversion of Reactances:

$$ X_{\text{p.u., new}} = X_{\text{p.u., old}} \times \frac{(S_{base,\ \text{new}} / S_{base,\ \text{old}})}{(V_{base,\ \text{new}} / V_{base,\ \text{old}})^2} $$

• Example: For multiple generators on a common bus, convert all reactances to a common system base before drawing the reactance diagram.

II. PROTECTIVE RELAYS: OPERATING PRINCIPLES & CHARACTERISTICS

Fundamental Relay Concepts

• Operating Principle:

  • Electromagnetic Attraction: plunger or armature attracted by magnetic force (DC/AC).

  • Electromagnetic Induction: Rotating magnetic field induces torque in a disc or cup (AC only).

• Characteristics:

  • Definite Time (DT): Fixed time delay after pickup.

  • Inverse Definite Minimum Time (IDMT): Operating time inversely proportional to fault current magnitude. Has a minimum time (dial setting).

• Key Terms:

  • Pick-up Value: Minimum current/voltage to operate relay.

  • Time Dial Setting (TMS): Multiplier for the basic time curve.

  • Plug Setting Multiplier (PSM): $$\displaystyle \text{PSM} = \frac{\text{Fault Current}}{\text{Relay Pick-up Current}} $$

IDMT Relay Operation Time Calculation

Standard IEC/BS curves: Standard Inverse (SI), Very Inverse (VI), Extremely Inverse (EI).

• Formula (for Standard Inverse Curve):

$$ t = \frac{0.14 \times \text{TMS}}{(\text{PSM})^{0.02} - 1} \ \text{seconds} \boxed{} $$

(Note: Exponent varies with curve type).

• Steps:

  1. Find relay primary pick-up current from CT ratio & relay setting.

  2. Calculate PSM = Fault current / Relay pick-up current.

  3. Apply formula with given TMS.

Electromechanical Relays

• Induction Disc Relay:

  • Construction: Electromagnet, aluminum disc, spring, coil, gear.

  • Principle: Two out-of-phase fluxes (from split poles) induce eddy currents in disc, producing torque.

$$ T \propto \phi_1 \phi_2 \sin \delta $$

*   **Application:** Overcurrent, directional, distance, reverse power.

*   **Minimize Overrun (disc inertia):** Use **braking magnet** (permanent magnet) to provide braking torque near operation point.

• Induction Cup Relay:

  • Construction: Stationary electromagnets, cylindrical cup (conducting), rotor with control spring.

  • Principle: Similar to disc, but cup has lower inertia → faster operation, higher sensitivity.

  • Reset to Pick-up Ratio: Typically > 0.9 (very close to 1), meaning it resets at a current very near the pickup value. Better than disc relay.

Static Relays (Electronic Relays)

• Advantages: No moving parts → faster, more accurate, adjustable characteristics, multi-function, self-test.

• Functional Building Blocks:

  • Level Detector: Outputs 1 if input > set value, else 0.

  • Comparator: Compares two inputs (amplitude or phase).

    • Amplitude Comparator: $$\displaystyle |A| > |B| $$?

    • Phase Comparator: $$\displaystyle \angle(A) - \angle(B) > \theta $$?

  • Logic Circuits: AND, OR, NOT for combining comparator outputs.

• Conversion:

  • Amplitude to Phase: Use a time delay on one input. $A \sin \omega t$ vs $A \sin (\omega t + \phi)$ after delay becomes phase comparison.

  • Phase to Amplitude: Use a phase-shifting network (RC) to convert phase difference into amplitude difference.

• Specific Relays:

  • Static Directional Relay: Uses phase comparator. Compares fault current ($I$) and reference voltage ($V$). Operates if $\angle(V) - \angle(I)$ is within set directional window (e.g., $$\displaystyle -90^\circ $$ to $$\displaystyle +90^\circ $$).

  • Static Distance Relay: Measures impedance $$\displaystyle Z = V/I $$. Uses amplitude comparator (circle characteristic) or phase comparator (quadrilateral).

Microprocessor/Numerical Relays

• Block Diagram:

  CT/VT → Analog Input → Anti-aliasing Filter → **ADC** → **CPU/Microprocessor** (runs protection algorithm) → Memory (settings, data) → Output Logic → Trip Signal

• Software Development: Implements complex algorithms (Fourier for phasors, distance impedance calculation, differential restraint). Enables self-monitoring, communication (IEC 61850), and multiple functions in one unit.

• Logic Circuits: Provide security (blocking signals, timers) and dependability (multiple criteria must be met).

Relay Characteristics on R-X Diagram

• Impedance Relay (Plain):

  • Characteristic: Circle with center at origin. $$\displaystyle |Z| < Z_{set} $$ → Trip.

  • Operation: Measures impedance. Trips for faults within the circle.

  • Application: Simple distance protection, but non-directional.

• MHO Relay (Admittance/Offset Impedance):

  • Characteristic: Circle passing through origin. Diameter along the line $$\displaystyle Z = Z_{set} e^{j(\theta - 90^\circ)} $$.

  • Directional: Only operates for faults in the forward direction (inside the circle and in direction of line).

  • Application: Primary protection for long transmission lines (Zone 1). Resists power swings (swing impedance locus is a line through origin, may not enter MHO circle).

• OFF-SET MHO Relay:

  • Characteristic: MHO circle offset from origin towards the resistive axis.

  • Need: To cover high-resistance faults (large R component) that a plain MHO might miss. Used for Zone 2/3 settings.

III. CIRCUIT BREAKERS: THEORY, TYPES & SELECTION

Arc Phenomenon & Interruption Theory

• Arc Formation: When contacts separate, ionization of air/gas creates a low-resistance path → high current arc.

• Theory of Current Interruption (Energy Balance):

  • At current zero (AC), arc attempts to extinguish.

  • Restriking Voltage ($$\displaystyle v_r $$): Transient voltage across contacts immediately after current zero. High $di/dt$ → high $dv/dt$ → may restrike.

  • Recovery Voltage ($$\displaystyle v_{rec} $$): Sustained power frequency voltage across contacts after arc extinction. Must withstand this.

  • Condition for Successful Interruption: Dielectric strength of arc space must build up faster than restriking voltage rise.

• Methods to Increase Dielectric Strength: Rapid cooling (SF6, Vacuum), lengthening arc (oil, air blast), deionization.

Arc Quenching Media & Methods

• Methods:

  1. High Resistance Method: Increase arc resistance (air blast, SF6 puffer).

  2. Low Resistance (Zero-Current) Method: Use medium where arc extinguishes naturally at current zero (vacuum, SF6 thermal blast).

  3. Oil Decomposition: Gas bubbles from oil decomposition provide cooling and insulation (oil CB).

Types of Circuit Breakers

Circuit Breaker Ratings & Selection

• Breaking Capacity: Maximum RMS current (kA) it can interrupt at specified recovery voltage. Most important rating. Based on symmetrical RMS current.

• Making Capacity: Maximum peak current (kA) it can close onto a fault. Making Capacity = √2 × Breaking Capacity × (1 + X/R factor). Usually higher than breaking capacity.

• Selection Factors:

  1. Voltage Rating (system voltage).

  2. Breaking Capacity (≥ maximum 3-phase fault current at point).

  3. Making Capacity.

  4. Type of System (grounded/ungrounded).

  5. Duty Cycle (number of operations, CO, OCO).

  6. Installation Conditions (indoor/outdoor, altitude).

• Importance of Testing: To verify ratings (short-circuit, mechanical, dielectric). Types: Type Test (design), Routine Test (each unit), Commissioning Test.

IV. APPLICATION-SPECIFIC PROTECTION SCHEMES

Generator/Alternator Protection

• Common Faults:

  • Stator: Phase-to-phase, phase-to-ground (winding faults).

  • Rotor: Field winding faults (ground, turn-to-turn), loss of excitation.

  • External: Back-up protection for system faults.

• Stator Protection (Percentage Differential - Merz-Price):

  • Principle: Compare currents entering/leaving stator winding via CTs. Under normal/through fault, currents balance. Internal fault → unbalance → relay operates.

  • Slope: Characteristic has a slope (e.g., 15-30%) to allow for CT mismatch & magnetizing current during external faults. Stability during through faults is key.

  • CT Ratio Calculation: Ensure $$\displaystyle I_{CT,\ HV} \times \text{CT ratio}_{HV} = I_{CT,\ LV} \times \text{CT ratio}_{LV} $$ for balanced 3-phase.

• Rotor Protection:

  • Loss of Excitation: Detects transition from inductive to capacitive power factor (impedance relay in $R-X$ plane).

  • Field Failure: Uses reverse power relay or offset MHO relay.

• Generator-Transformer Unit Protection: Treats generator + transformer as one unit. Uses biased differential (high set for unit faults, low set for transformer inrush) or percentage differential with harmonic restraint.

• Buchholz Relay (for Transformer/Unit): Gas-actuated relay in oil-filled tank. Operates for:

  • Minor Faults: Gas accumulation → alarm.

  • Major Faults: Oil surge → trip.

Transformer Protection

• Internal Faults: Winding shorts, inter-turn faults.

• External Faults: Back-up overcurrent.

• Differential Protection (Merz-Price):

  • Challenge: Star-Delta connection introduces 30° phase shift between HV & LV currents.

  • CT Ratio Calculation: To compensate, CT ratios must be chosen such that:

$$ \frac{I_{HV}}{I_{LV}} = \frac{\text{CT Ratio}_{LV}}{\text{CT Ratio}_{HV}} = \frac{V_{LV}}{V_{HV}} \times \frac{\sqrt{3}}{1} \ \text{(for Star-Delta)} \boxed{} $$

    (For Star-Star, ratio = $$\displaystyle V_{LV}/V_{HV} $$).

*   **Procedure:** Calculate secondary currents for both sides under balanced load. Set CT ratios to make them equal.

• Other Protections:

  • Overcurrent: Back-up.

  • Restricted Earth Fault (REF): Sensitive protection for earth faults near neutral (only senses faults within transformer zone).

  • Buchholz Relay: For oil-filled transformers.

  • Temperature Monitoring: Winding temperature, oil temperature.

Busbar Protection

• Importance: Busbar fault → massive outage, loss of multiple circuits.

• Differential Protection Scheme:

  • CTs on all incoming/outgoing circuits.

  • Sum of CT currents should be zero under normal condition.

  • Internal fault → sum ≠ 0 → relay operates.

  • Challenge: CT saturation during external faults can cause false operation.

• Frame Leakage Protection Scheme:

  • CTs on busbar frame/ground only.

  • Under normal, no current in frame CT.

  • Internal fault → fault current flows through frame → frame CT operates.

  • Simple, but not fail-safe (if frame insulation fails).

• Other Methods:

  • High-Impedance Differential: Uses high series impedance to prevent maloperation during CT saturation.

  • Biased Differential: Similar to transformer differential, with slope to handle through faults.

Transmission Line Protection

• Pilot Relaying Schemes: Use communication channel (pilot) between line ends for fast, selective tripping.

  • Types of Pilots:

    • Pilot Wires: Dedicated copper pairs. Reliable, fast, but expensive for long distances.

    • Power Line Carrier (PLCP): Uses transmission line itself as channel. Frequency: 30-500 kHz. Merits: No extra wires. Demerits: Attenuation, noise, requires coupling capacitors & line traps. Used for 33 kV to 220 kV.

    • Microwave: Point-to-point radio. Fast, reliable, independent of line. Used for EHV (400 kV+).

  • Phase Comparison Scheme (Carrier):

    • Compares phase of currents at both ends.

    • Operation: For internal fault, currents at both ends are in phase (both flow towards fault). Carrier signal sent → tripping. For external fault, currents out of phase → blocking.

    • Sketch: Shows CTs, phase comparator at each end, carrier transmitter/receiver, blocking/tripping logic.

• Distance Protection:

  • Measures impedance $$\displaystyle Z = V/I $$ to fault.

  • MHO Relay is primary for Zone 1 (80-90% line length). OFF-SET MHO for Zones 2 & 3 (reach beyond line, with offset for high-resistance faults).

  • Protection Zones:

    • Zone 1: 80-90% of line. Instantaneous.

    • Zone 2: Covers remaining line + backup for next line (120-150% of line). Time delay (e.g., 0.5 s).

    • Zone 3: Backup for remote lines, and as last resort. Long reach, long delay (e.g., 1.0 s).

Current Limiting Reactors

• Types: Air-core (dry, no saturation) and Iron-core (saturates at high current).

• Construction: Series reactor installed in each feeder or at busbars.

• Application: Limit fault current magnitude during short circuits. Protect equipment (breakers, cables) from high thermal/mechanical stress. Used in generator neutral, bus-section, and feeder applications.

• Role: Increases system impedance, thereby reducing $$\displaystyle I_{fault} = E / (Z_{sys} + Z_{reactor}) $$.

V. AUXILIARY & OTHER PROTECTION DEVICES

High Rupturing Capacity (HRC Fuses)

• Function: Fast, high-current interrupting device for LT switchgear.

• Principle: Fusible element melts under overcurrent. Sand/quenching material surrounds element to cool arc and prevent restrike.

• Application: Transformer protection (LT side), motor protection, capacitor banks.

Current Transformers (CTs) & Voltage Transformers (VTs)

• CT Ratio Selection for Differential Protection:

  • Goal: Make CT secondary currents equal under normal load.

  • For Star-Delta Transformer:

    • HV side (Star): $$\displaystyle I_{HV, ph} = \frac{S}{\sqrt{3} V_{HV}} $$

    • LV side (Delta): $$\displaystyle I_{LV, line} = \frac{S}{\sqrt{3} V_{LV}} $$, $$\displaystyle I_{LV, ph} = I_{LV, line} / \sqrt{3} $$

    • Therefore, $$\displaystyle \frac{I_{HV, ph}}{I_{LV, ph}} = \frac{V_{LV}}{V_{HV}} \times \sqrt{3} $$

    • CT ratios must satisfy: $$\displaystyle \frac{\text{CT Ratio}_{HV}}{\text{CT Ratio}_{LV}} = \frac{V_{LV}}{V_{HV}} \times \sqrt{3} $$

• CT Saturation: Occurs when primary current is so high that core flux exceeds saturation limit. Effect: CT secondary current is distorted (flattened) and inaccurate. Can cause failure of differential protection (through fault) or false operation (internal fault). Use Class X/TP CTs for protection.

Buchholz Relay (Detailed)

• Location: Installed in the pipe between transformer tank and conservator.

• Working Principle:

  1. Gas Accumulation (Alarm): Minor internal faults generate gas. Gas collects in relay chamber → float drops → closes alarm contact.

  2. Oil Surge (Trip): Major fault generates large volume of gas rapidly → oil surges towards conservator → deflects baffle plate → closes trip contact.

• Protects: Against internal faults with oil decomposition (winding shorts, core faults). Does not protect against external faults or minor insulation deterioration without gas.

VI. SYSTEM & MODERN PROTECTION CONCEPTS

Security and Reliability

• Reliability (Dependability): Relay must operate when required (for faults in its zone). False trips are unacceptable.

• Security: Relay must not operate when not required (for faults outside zone or no fault). Failure to trip is unacceptable.

• How Ensured:

  • Redundancy: Dual protection schemes (e.g., two independent differential relays).

  • Diversity: Different operating principles (e.g., differential + overcurrent).

  • Proper Setting & Coordination: Ensure selectivity.

  • Testing & Maintenance: Regular functional tests.

  • Logic in Numerical Relays: Blocking signals, timers, harmonic restraint (for transformer inrush).

Microprocessor-Based Protection Schemes

• Advantages over Conventional:

  • Multiple functions in one unit (OC, distance, differential, autoreclose).

  • Adaptive settings (change with system conditions).

  • Self-monitoring & diagnostics.

  • Communication (SCADA, IEC 61850) for data, control, and event recording.

  • Advanced algorithms (Fourier, traveling wave).

  • Lower burden on CTs/VTs.

  • No drift (digital).

• Implementation: Software executes protection logic. Hardware: ADC, DSP/CPU, memory, I/O, communication port.

Software Development for Protection

• Role in Security & Reliability:

  • Secure Code: Prevents bugs causing false trips.

  • Verification & Validation (V&V): Extensive testing (simulation, hardware-in-loop).

  • Setting Management: Secure storage, easy retrieval, version control.

  • Fail-Safe Design: Watchdog timers, default safe states.

  • Cyber Security: Authentication, encryption for communication.

VII. CALCULATION & DESIGN PROBLEMS (INTEGRATED)

1. CT Ratio Calculation for Transformer Differential (Star-Delta)

Problem: 3-phase transformer 0.4 kV (Star) / 11 kV (Delta). CT ratio on LV side = 500/5. Find CT ratio on HV side.

• Step 1: $$\displaystyle V_{LV} = 0.4 \ \text{kV} $$ (phase), $$\displaystyle V_{HV} = 11 \ \text{kV} $$ (line). For Star-Delta, $$\displaystyle V_{HV, ph} = 11 \ \text{kV} $$.

• Step 2: Current ratio (line) $$\displaystyle I_{LV} / I_{HV} = 11 / 0.4 = 27.5 $$.

• Step 3: CT secondary current should be equal. So, $$\displaystyle I_{HV, ph} \times \text{CT Ratio}_{HV} = I_{LV, line} \times \text{CT Ratio}_{LV} $$.

• Step 4: But $$\displaystyle I_{HV, ph} = I_{HV, line} / \sqrt{3} $$. So:

$$ \frac{I_{HV, line}}{\sqrt{3}} \times \text{CT Ratio}_{HV} = I_{LV, line} \times 500/5 $$

$$ \Rightarrow \text{CT Ratio}_{HV} = \sqrt{3} \times \frac{I_{LV, line}}{I_{HV, line}} \times \frac{500}{5} = \sqrt{3} \times 27.5 \times 100 = 4762.5 $$

• Answer: $$\displaystyle \boxed{4762.5/5 \ \text{A} \ \text{or standard} \ 5000/5 \ \text{A}} $$

2. Fault Current Calculation for Unloaded Alternator

Problem: 25 MVA, 13.2 kV, solidly grounded. $$\displaystyle X_d'' = 0.25 $$ p.u., $$\displaystyle X_2 = 0.35 $$ p.u., $$\displaystyle X_0 = 0.1 $$ p.u. SLG fault at terminals. Find $$\displaystyle I_f $$ and line-to-line voltage.

• Step 1: $$\displaystyle E_f = 1.0 $$ p.u. (reference). $$\displaystyle Z_1 = j0.25 $$, $$\displaystyle Z_2 = j0.35 $$, $$\displaystyle Z_0 = j0.1 $$.

• Step 2: SLG fault: $$\displaystyle I_1 = I_2 = I_0 = \frac{1.0}{j(0.25+0.35+0.1)} = \frac{1}{j0.7} = -j1.4286 $$ p.u.

• Step 3: Fault current: $$\displaystyle I_f = 3I_0 = 3 \times 1.4286 = 4.2857 $$ p.u.

• Step 4: Convert to actual: Base current $$\displaystyle I_{base} = \frac{25 \times 10^6}{\sqrt{3} \times 13.2 \times 10^3} = 1092.4 \ \text{A} $$. $$\displaystyle I_f = 4.2857 \times 1092.4 = 4682 \ \text{A} $$.

• Step 5: Line-to-line voltage (e.g., $$\displaystyle V_{bc} $$) during SLG fault (A-G):

$$ V_{bc} = V_{b0} + V_{b1} + V_{b2} $$

$$ V_1 = E_f - I_1 Z_1 = 1 - (-j1.4286 \times j0.25) = 1 - 0.3571 = 0.6429 \ \text{p.u.} $$

$$ V_2 = -I_2 Z_2 = -(-j1.4286 \times j0.35) = -0.5 \ \text{p.u.} $$

$$ V_0 = -I_0 Z_0 = -(-j1.4286 \times j0.1) = -0.1429 \ \text{p.u.} $$

For phase B: $$\displaystyle V_{b1} = a^2 V_1 $$, $$\displaystyle V_{b2} = a V_2 $$, $$\displaystyle V_{b0} = V_0 $$.

$$ V_{bc} = (V_{b1} - V_{c1}) + (V_{b2} - V_{c2}) + (V_{b0} - V_{c0}) $$

Since $$\displaystyle V_0 $$ is zero-sequence, $$\displaystyle V_{b0} - V_{c0} = 0 $$.

$$ V_{bc} = V_1(a^2 - a) + V_2(a - a^2) = (V_1 - V_2) j\sqrt{3} $$

$$ V_{bc} = (0.6429 - (-0.5)) j\sqrt{3} = 1.1429 \times j1.732 = j1.98 \ \text{p.u.} $$

Magnitude = **1.98 p.u.** (Line-to-line voltage rises during SLG fault!).

3. IDMT Relay Operation Time

Problem: 5 A, 2.2 s IDMT relay, plug setting 125%, TMS=0.6. CT 400/5. Fault current = 4000 A.

• Step 1: Relay primary pick-up = $$\displaystyle 5 \ \text{A} \times 1.25 = 6.25 \ \text{A} $$.

• Step 2: Fault current seen by relay = CT secondary current = $$\displaystyle 4000 \times (5/400) = 50 \ \text{A} $$.

• Step 3: PSM = Fault current / Relay pick-up = $$\displaystyle 50 / 6.25 = 8 $$.

• Step 4: For Standard Inverse (SI) curve: $$\displaystyle t = \frac{0.14 \times \text{TMS}}{(\text{PSM})^{0.02} - 1} = \frac{0.14 \times 0.6}{8^{0.02} - 1} $$.

  $$\displaystyle 8^{0.02} = e^{0.02 \ln 8} = e^{0.02 \times 2.079} = e^{0.04158} = 1.0424 $$.

  $$\displaystyle t = \frac{0.084}{0.0424} = 1.98 \ \text{s} $$.

• Answer: $\boxed{t \approx 1.98 \ \text{seconds}}$

4. Percentage Differential Relay Stability Check

Problem: Stator protection, 15% slope. Fault: $$\displaystyle I_{CT1} = 400 \ \text{A} $$, $$\displaystyle I_{CT2} = 320 \ \text{A} $$. CT ratio = 500/5. Will relay trip?

• Step 1: Bias/Through Current $$\displaystyle I_{through} = \frac{I_{CT1} + I_{CT2}}{2} = \frac{400+320}{2} = 360 \ \text{A} $$.

• Step 2: Operating Current $$\displaystyle I_{op} = |I_{CT1} - I_{CT2}| = |400 - 320| = 80 \ \text{A} $$.

• Step 3: Slope Setting: 15% means relay operates if $$\displaystyle I_{op} > 0.15 \times I_{through} $$.

  $$\displaystyle 0.15 \times 360 = 54 \ \text{A} $$.

• Step 4: Since $$\displaystyle I_{op} (80 \ \text{A}) > 54 \ \text{A} $$, relay will trip.

• Note: CT ratio is same on both sides, so actual primary currents are proportional. Check is valid on secondary currents.

5. Per-Unit Reactance Conversion

Problem: Three generators on common bus. Convert to base: 200 MVA, 35 kV.

*   G1: 100 MVA, 33 kV, 10%

*   G2: 150 MVA, 32 kV, 8%

*   G3: 110 MVA, 30 kV, 12%

• Step 1: For each generator, calculate its own base current: $$\displaystyle I_{base,G} = S_{base,G} / (\sqrt{3} V_{base,G}) $$.

• Step 2: Convert each reactance to new base using formula:

$$ X_{\text{new}} = X_{\text{old}} \times \frac{S_{base,\ \text{new}}}{S_{base,\ \text{old}}} \times \left( \frac{V_{base,\ \text{old}}}{V_{base,\ \text{new}}} \right)^2 $$

• Calculations:

  • G1: $$\displaystyle X_1 = 0.10 \times \frac{200}{100} \times \left( \frac{33}{35} \right)^2 = 0.10 \times 2 \times 0.888 = 0.1776 $$ p.u.

  • G2: $$\displaystyle X_2 = 0.08 \times \frac{200}{150} \times \left( \frac{32}{35} \right)^2 = 0.08 \times 1.333 \times 0.836 = 0.0892 $$ p.u.

  • G3: $$\displaystyle X_3 = 0.12 \times \frac{200}{110} \times \left( \frac{30}{35} \right)^2 = 0.12 \times 1.818 \times 0.7347 = 0.160 $$ p.u.

• Result: Reactance diagram: Three generators in parallel at common bus (35 kV), with $$\displaystyle X_1=0.1776 $$, $$\displaystyle X_2=0.0892 $$, $$\displaystyle X_3=0.160 $$ p.u. on 200 MVA, 35 kV base.

[!TIP]

Exam Focus: Be prepared to derive fault current expressions (especially DLG), calculate CT ratios for transformer differentials (Star-Delta is a favorite), compute IDMT times (memorize SI curve formula), and draw sequence networks for all fault types. For circuit breakers, compare SF6, VCB, and OCB. Always box final formulas and numerical answers.