20.1 Power System Protection & Protective Relaying

20.1 Power System Protection & Protective Relaying

Power system protection, protective relaying coordination, instrument transformer application, and circuit breaker rating constitute major core subjects on the PRC Registered Electrical Engineer (REE) Licensure Examination. Protective relaying systems act as the automated intelligence of electric power networks, continuously detecting abnormal voltage, current, or frequency conditions and isolating faulted apparatus to preserve equipment integrity and grid stability.

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1. Fundamentals of Protection & Instrument Transformers

Core Protective Relaying Criteria

Every protective relaying scheme must satisfy five essential operational requirements:

1. Selectivity (Coordination): Isolates only the minimum faulted portion of the system, keeping healthy feeders energized.
2. Sensitivity: Operates reliably under minimum fault current conditions (e.g., high-resistance ground faults at light system loading).
3. Speed: Disconnects short circuits as rapidly as possible to limit thermal/mechanical equipment damage and maintain power system transient stability.
4. Reliability: Consists of Dependability (certainty of operating when a fault occurs) and Security (certainty of NOT operating when no fault exists).
5. Simplicity: Minimizes secondary wiring complexity and relay components to reduce failure modes.

Instrument Transformers: CTs and PTs

Protective relays operate on reduced secondary current and voltage signals supplied by Current Transformers (CTs) and Potential Transformers (PTs / VTs).

Current Transformers (CTs)

• Standard secondary current rating is $5\ \text{A}$ (or $1\ \text{A}$ in modern numerical European/IEC standards).
• CT Ratio ($N_{\text{ct}}$): $N_{\text{ct}} = \frac{I_{\text{primary}}}{I_{\text{secondary}}}$. For an $800/5\ \text{A}$ CT, $N_{\text{ct}} = 160$.
• ANSI Accuracy Classes (C-Class): Designates transformer secondary terminal voltage at 20 times nominal secondary current without exceeding $10%$ ratio error. For example, a C400 CT maintains ratio accuracy up to $V_{\text{sec}} = 400\ \text{V}$ at $20 \times 5\ \text{A} = 100\ \text{A}$ secondary current ($Z_{\text{burden, max}} = \frac{400\ \text{V}}{100\ \text{A}} = 4.0\ \Omega$).
• CT Burden ($Z_B$): Total external impedance connected across secondary terminals, including relay coil impedance, lead wire resistance, and terminal contacts: $Z_B = R_{\text{ct,sec}} + 2 R_{\text{lead}} + Z_{\text{relay}}$

CRITICAL REE SAFETY RULE: Never open-circuit the secondary terminals of an energized Current Transformer! Open-circuiting a CT causes infinite secondary impedance, inducing extreme high-voltage spikes ($V = N \frac{d\Phi}{dt} \ge 10\ \text{kV}$) across secondary terminals that cause lethal electric shock and core insulation breakdown.

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2. Overcurrent Protection (ANSI 50/51) & Relaying Curves

Overcurrent relays detect phase short circuits and ground faults when current exceeds a predetermined threshold.

• ANSI 50: Instantaneous Overcurrent Relay (operates with no intentional time delay, $t < 50\ \text{ms}$).
• ANSI 51: AC Inverse Time Overcurrent Relay (operating time decreases non-linearly as fault current magnitude increases).

Relay Pickup & Plug Setting

• Plug Setting (PS): Percentage tap setting on the relay ($50% \text{ to } 200%$ in steps of $25%$ for phase relays; $20% \text{ to } 80%$ for ground relays).
• Pickup Current ($I_{\text{pu}}$): The minimum secondary current required to initiate relay operation: $I_{\text{pu}} = I_{\text{ct,sec}} \times \text{PS} = 5\ \text{A} \times \left( \frac{\text{PS}\%}{100} \right)$
• Primary Pickup Current ($I_{\text{pu,pri}}$): $I_{\text{pu,pri}} = I_{\text{pu}} \times N_{\text{ct}}$

Multiple of Pickup Current ($M$)

$M = \frac{I_{\text{fault,sec}}}{I_{\text{pu}}} = \frac{I_{\text{fault,pri}}}{I_{\text{pu,pri}}}$

Time Dial Setting (TDS / TMS) & Standard Inverse Curves

Relay operating time $t$ is calculated using standardized mathematical curves (IEC 60255 / IEEE C37.112):

Coordination Time Interval (CTI): When coordinating series relays along a distribution feeder, a selective time margin $\text{CTI} \approx 0.30 \text{ to } 0.40\ \text{seconds}$ is maintained between downstream feeder relays and upstream main circuit breaker relays to account for breaker clearing time and relay overshoot.

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3. Differential Protection (ANSI 87) & Transformer Relaying

Differential relays operate on the unit protection principle: current entering a protected zone must equal current exiting under normal or external fault conditions.

Percentage Differential Relay Principle

                  PERCENTAGE DIFFERENTIAL RELAY SCHEMATIC (ANSI 87)

          Line In I1                                             Line Out I2
         -------------[ CT1 ]-------------------[ CT2 ]-------------
                         |                         |
                        i1                        i2
                         |                         |
                         +----[ Restraint Coil ]---+
                                     |
                             [ Operating Coil ] (Id = |i1 - i2|)
                                     |
                                    ==== Ground

• Differential Operating Current ($I_d$): $I_d = |i_1 - i_2|$
• Restraint Current ($I_{\text{restraint}}$): $I_{\text{restraint}} = \frac{|i_1| + |i_2|}{2}$
• Trip Condition: Relay trips when operating current exceeds the percentage restraint slope $k$ ($15% \text{ to } 40%$): $I_d > k \cdot I_{\text{restraint}} + I_0$

Transformer Differential Protection Challenges

1. Ratio Mismatch: Primary and secondary transformer full-load currents differ due to transformation ratio $N_1 / N_2$. CT ratios $N_{\text{ct1}}$ and $N_{\text{ct2}}$ are chosen so secondary currents $i_1$ and $i_2$ match closely under rated load.
2. Phase Shift in $\Delta\text{--}Y$ Transformers: A $\Delta\text{--}Y$ transformer introduces a $30^\circ$ phase shift between primary and secondary line currents. In electromechanical protection, CTs on the $\Delta$ side are connected in $\text{Y}$, and CTs on the $\text{Y}$ side are connected in $\Delta$ to cancel the $30^\circ$ shift and supply $\sqrt{3}$ current magnitude compensation. Numerical relays perform phase shift compensation via software matrix multiplication.
3. Magnetizing Inrush Current: Energizing a transformer creates high transient inrush current (up to $8\text{--}12 \times I_{\text{rated}}$) that flows only through the primary winding, producing false differential current. Inrush current contains a high 2nd harmonic component ($15%\text{--}20%$ of fundamental). Numerical relays use 2nd harmonic blocking to prevent false tripping during energization.

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4. Distance Protection (ANSI 21) & Transmission Line Protection

Distance (impedance) relays protect high-voltage transmission lines by measuring the ratio of secondary line voltage to line current:

$Z_{\text{seen}} = \frac{V_{\text{relay}}}{I_{\text{relay}}} = z_{\text{line}} \times d$

where $z_{\text{line}}$ is positive-sequence impedance per km, and $d$ is distance to the fault.

Relay Characteristics on R-X Complex Plane

• Plain Impedance Relay: Circular characteristic centered at the origin ($|Z_{\text{seen}}| < Z_{\text{set}}$). Non-directional; requires additional directional element.
• Mho Relay: Circular characteristic passing through the origin. Inherently directional; ideal for long high-voltage transmission lines.
• Reactance Relay: Horizontal line characteristic on R-X plane ($X_{\text{seen}} < X_{\text{set}}$). Resists fault arc resistance ($R_{\text{arc}}$); used for short lines and ground fault protection.

Three-Zone Distance Protection Scheme

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5. Circuit Breakers & Short-Circuit Interrupting Capacity

Circuit breakers interrupt heavy fault currents under extreme arc plasma conditions ($> 5000^\circ\text{C}$).

Arc Quenching Media

• Vacuum Circuit Breakers (VCB): Used in medium-voltage distribution ($4.16\ \text{kV} \text{ to } 34.5\ \text{kV}$). High dielectric recovery rate, zero fire hazard.
• Sulfur Hexafluoride (SF6) Breakers: Used in high-voltage transmission ($69\ \text{kV} \text{ to } 500\ \text{kV}$). Superior electronegative gas that rapidly absorbs free electrons during current zero crossing.

Asymmetrical Fault Current & Making Capacity

Total short-circuit current contains an AC symmetrical component $I_{\text{sym, RMS}}$ and a decaying DC offset $i_{\text{dc}}(t)$:

$i_{\text{fault}}(t) = \sqrt{2} I_{\text{sym, RMS}} \left[ \cos(\omega t + \alpha - \theta) - e^{-t/\tau} \cos(\alpha - \theta) \right]$

where $\tau = \frac{L}{R} = \frac{X}{2 \pi f R}$ is the DC attenuation time constant.

Peak Making Current ($I_{\text{making}}$):

The maximum peak current occurring at the first half-cycle ($t \approx 10\ \text{ms}$ at $60\ \text{Hz}$) under maximum DC offset:

$I_{\text{making}} = 2.55 \times I_{\text{sym, RMS}} \quad (\text{Peak Amperes})$

Short-Circuit Symmetrical Breaking Duty ($I_{\text{sc}}$):

$I_{\text{sc}} = \frac{S_{\text{fault, 3-phase}}}{\sqrt{3} \times V_{\text{line, nominal}}} \quad (\text{kA RMS})$

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Solved Board Exam Examples

Example 1: Overcurrent Relay Pickup & Operating Time Calculation

Problem: A $13.8\ \text{kV}$ distribution feeder is protected by an inverse time overcurrent relay (ANSI 51) connected through an $800/5\ \text{A}$ CT. The relay plug setting is $\text{PS} = 125%$ and the Time Dial Setting is $\text{TDS} = 4$. If a 3-phase fault produces a primary short-circuit current $I_{\text{fault}} = 6,000\ \text{A}$, calculate using the Standard Inverse curve ($t = \text{TDS} \times \frac{0.14}{M^{0.02} - 1}$): (a) Relay pickup current $I_{\text{pu}}$, (b) Primary pickup current, (c) Multiple of pickup current $M$, and (d) Relay operating time $t$.

Solution:

1. Calculate secondary pickup current ($I_{\text{pu}}$): $I_{\text{pu}} = 5\ \text{A} \times 1.25 = 6.25\ \text{A}$
2. Calculate primary pickup current ($I_{\text{pu,pri}}$): $N_{\text{ct}} = \frac{800}{5} = 160 \implies I_{\text{pu,pri}} = 6.25\ \text{A} \times 160 = 1,000\ \text{A}$
3. Calculate multiple of pickup current ($M$): $M = \frac{I_{\text{fault,pri}}}{I_{\text{pu,pri}}} = \frac{6,000\ \text{A}}{1,000\ \text{A}} = 6.0$
4. Calculate relay operating time $t$ for $\text{TDS} = 4$: $M^{0.02} = (6.0)^{0.02} = 1.03643$ $t = 4 \times \left( \frac{0.14}{1.03643 - 1} \right) = 4 \times \left( \frac{0.14}{0.03643} \right) = 4 \times 3.843 = 15.37\ \text{seconds}$

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Example 2: Transformer Differential Relay CT Rating & Mismatch

Problem: A $3$-phase, $30\ \text{MVA}$, $115\ \text{kV} / 13.8\ \text{kV}$, $\Delta\text{--}Y$ power transformer is protected by percentage differential relays. CTs on the $115\ \text{kV}$ $\Delta$ side are connected in $\text{Y}$, and CTs on the $13.8\ \text{kV}$ $\text{Y}$ side are connected in $\Delta$. Standard CT ratios available are $200/5\ \text{A}$ on the high-voltage side and $1500/5\ \text{A}$ on the low-voltage side. Calculate CT secondary currents under rated full load and determine any current mismatch.

Solution:

1. Calculate full-load primary line currents on high-voltage (HV) and low-voltage (LV) sides: $I_{\text{HV, rated}} = \frac{30 \times 10^6}{\sqrt{3} \times 115,000} = 150.61\ \text{A}$ $I_{\text{LV, rated}} = \frac{30 \times 10^6}{\sqrt{3} \times 13,800} = 1,255.11\ \text{A}$
2. Calculate CT secondary currents entering relay restraint coils:
   • HV Side ($,\Delta$ transformer side, $\text{Y}$-connected CTs): $i_{\text{sec, HV}} = 150.61 \times \left( \frac{5}{200} \right) = 3.765\ \text{A}$
   • LV Side ($,Y$ transformer side, $\Delta$-connected CTs): Line current exiting $\Delta$-connected CT secondary contains a $\sqrt{3}$ factor: $i_{\text{sec, LV}} = 1,255.11 \times \left( \frac{5}{1500} \right) \times \sqrt{3} = 4.184 \times 1.73205 = 7.247\ \text{A}$
3. Evaluate differential mismatch current: $\text{Mismatch Current } I_d = |7.247 - 3.765| = 3.482\ \text{A}$

   Conclusion: Numerical relays apply internal digital scaling factors ($a_{\text{HV}} = 1.0, a_{\text{LV}} = 0.5195$) to balance secondary currents perfectly to $3.765\ \text{A}$, eliminating spill current under normal full-load operation.