14.2 Protective Relaying Schemes (Differential 87, Distance 21, Undervoltage 27)

14.2 Protective Relaying Schemes (Differential 87, Distance 21, Undervoltage 27)

Executive Overview: Modern power systems rely on specialized relaying schemes to protect apparatus and maintain grid stability. While simple overcurrent relays provide non-unit backup, critical infrastructure—transformers, generators, transmission lines, and major switchgear buses—demands unit protection (ANSI 87 Differential), impedance-based protection (ANSI 21 Distance), and voltage/frequency health monitoring (ANSI 27/59, ANSI 81). On the PE Power examination, candidates must understand ANSI device numbering, dual-slope percentage differential calculations, harmonic restraint for inrush, stepped distance zone parameterization, and teleprotection pilot schemes.

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1. ANSI Standard Device Numbers & Function Taxonomy

The standard IEEE/ANSI C37.2 device numbers categorize protective relay functions across electrical substations:

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2. Differential Protection (ANSI 87)

Differential protection is based on Kirchhoff's Current Law: under normal or external through-fault conditions, the sum of currents entering a bounded zone equals the sum of currents leaving the zone. If a fault occurs inside the zone, current flows into the fault from one or both ends, creating an imbalance.

Percentage Differential Relay Schematic (External Through-Fault vs. Internal Fault):

External Through-Fault (NO TRIP):
  Zone In -------------------[ Protected Equipment ]-------------------> Zone Out
     I_1 = 1000 A --->                                   I_2 = 1000 A --->
    (CT 1) 1000:5                                       (CT 2) 1000:5
       |                                                   |
   i_1 = 5 A                                           i_2 = 5 A
       +-----[ Restraint 1 ]----o----[ Restraint 2 ]-------+
                                |
                          i_op = i_1 - i_2 = 0 A
                           [ Operating Coil ]
                                |
                           (No Trip: i_op = 0)

Internal Fault (TRIP):
  Zone In -------------------[ Fault: 2000 A ]<------------------------- Zone Out
     I_1 = 1000 A --->               |                   I_2 = 1000 A <---
    (CT 1) 1000:5                    v                  (CT 2) 1000:5
       |                                                   |
   i_1 = 5 A                                           i_2 = -5 A
       +-----[ Restraint 1 ]----o----[ Restraint 2 ]-------+
                                |
                          i_op = i_1 - (-i_2) = 10 A
                           [ Operating Coil ]
                                |
                           (TRIP: i_op > Pickup)

Dual-Slope Percentage Differential Characteristic

CT errors, CT saturation during asymmetric through-faults, and transformer LTC tap changes create fictitious differential currents during external faults. To prevent misoperation, differential relays utilize a restraint current ($I_{rest}$) and a dual-slope operating curve:

$$\mathbf{I}_{diff} = |\mathbf{I}_1 - \mathbf{I}_2| \quad [\text{Operating Current}]$$ $$I_{rest} = \frac{|\mathbf{I}_1| + |\mathbf{I}_2|}{2} \quad \text{or} \quad I_{rest} = \max(|\mathbf{I}_1|, |\mathbf{I}_2|) \quad [\text{Restraint Current}]$$

Dual-Slope Operating Characteristic:

I_diff (Operating Current) ^
                           |                   / Slope 2 (50% - 80%)
                           |                  /  (Severe CT Saturation Region)
                           |                 /
                           |    Slope 1     / 
                           |   (20% - 40%) /
             I_pickup ---- |----.--------o (Break Point I_BP)
                           |   /         |   
                           |  / TRIP     |     BLOCK (Restraint Region)
                           | /           |
                           +-------------+---------------------> I_rest (Restraint)
                                       I_BP

1. Slope 1 Region ($I_{rest} < I_{BP}$): Set low ($20\% - 35\%$) to provide high sensitivity for high-impedance internal faults while accommodating minor CT ratio mismatch and transformer LTC tap excursions (typically $\pm 10\%$).
2. Slope 2 Region ($I_{rest} \ge I_{BP}$): Set steep ($50\% - 80\%$) to prevent false tripping during heavy through-faults where severe DC offset causes CT core saturation and large false differential currents.

Transformer Differential Protection (87T) Challenges

Transformer differential protection requires four critical compensations:

1. Voltage & Current Ratio Matching: CT ratios must be balanced so that secondary currents equal each other at rated MVA: $I_{sec1} \approx I_{sec2}$.
2. Phase Angle Displacement: A Delta-Wye transformer introduces a $30^\circ$ phase shift ($\{DAB, Y\} = 30^\circ$ lag). Legacy electromechanical relays used Delta-connected CTs on the Wye winding and Wye-connected CTs on the Delta winding to cancel the shift. Modern microprocessor relays use Wye-Wye CTs everywhere and execute numerical matrix phase-angle compensation internally.
3. Zero-Sequence Filtering: External ground faults on the Wye-grounded side create zero-sequence current $I_0$ that does not pass through the Delta winding, producing a false differential current. Numerical algorithms filter out zero-sequence currents ($I_{a}' = I_a - I_0$).
4. Magnetizing Inrush & Overexcitation Restraint:
   • Magnetizing Inrush (2nd Harmonic): Energizing a transformer creates high, unipolar inrush currents ($8-12 \times FLA$) rich in second harmonic content. Relays block tripping if $\frac{I_{2nd}}{I_{fundamental}} \ge 15\%$.
   • Overexcitation (5th Harmonic): System overvoltage or low frequency ($V/Hz$) drives core saturation, generating fifth harmonic currents. Relays block or alarm if $\frac{I_{5th}}{I_{fundamental}} \ge 30\%$.

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3. Distance Protection (ANSI 21)

Distance relays measure the ratio of local voltage to local current to calculate the apparent positive-sequence impedance from the relay terminal to the fault point:

$$\mathbf{Z}_{app} = \frac{\mathbf{V}_{relay}}{\mathbf{I}_{relay}} = z_{line} \cdot d_{fault} \quad [\Omega]$$

Since transmission line impedance is strictly proportional to physical length ($z_{line} = r + jx \; \Omega/\text{mile}$), the apparent impedance directly indicates the distance to the fault.

Complex R-X Impedance Plane and Mho Operating Characteristic:

          +jX (Reactance) ^
                          |        Zone 3 Reach (100% Line 1 + 120% Line 2, t3 = 0.8s)
                          |      .-----------------------------.
                          |    /     Zone 2 Reach (120% Line 1, t2 = 0.35s)
                          |   /    .---------------------.
                          |  /   /     Zone 1 Reach       \
                          | /   /    (80-85% Line 1, t1=0) \
                          | |  |    .-------------.
       Line Angle Theta ->| |  |   /               \
               . - - - - -|-+--|--o-----------------o
              /           | |  |   \               /
             /            | |  |    `-------------'
            /             |  \  \
           /              |   \   `---------------------'
          /               |    \                               /
  -------o----------------+------------------------------------> +R (Resistance)
    Relay Location (0,0)  |      `-----------------------------'
                          |

Stepped Distance Zones of Protection

Pilot Teleprotection Communication Schemes

To achieve $100\%$ instantaneous clearing across the entire line length without waiting for Zone 2 time delays, relays at both line terminals exchange logic signals over fiber optic, power line carrier (PLC), or microwave channels:

1. Permissive Overreaching Transfer Trip (POTT): Both ends set Zone 2 (overreaching, $>100\%$). If terminal A sees an in-zone fault, it sends a permissive key to terminal B. Terminal B trips instantaneously only if it simultaneously senses a forward Zone 2 fault and receives the permission signal.
2. Directional Comparison Blocking (DCB): Relays look reverse into their own system. If terminal B sees a reverse external fault, it immediately transmits a high-speed BLOCK signal to terminal A to prevent terminal A's overreaching Zone 2 element from tripping.
3. Line Current Differential (87L): Direct digital exchange of synchronized current phasors via fiber-optic communication. Enforces strict unit differential protection without impedance reach limitations.

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4. Voltage, Frequency & Auxiliary Relaying

Undervoltage Protection (ANSI 27)

• Motor Stall Protection: Severe voltage depressions ($V < 70-80\%$) cause induction motors to decelerate toward high-current stall conditions. ANSI 27 trips feeder breakers or unlatches contactors.
• Under-Voltage Load Shedding (UVLS): Deployed across transmission load pockets experiencing reactive power deficits to prevent catastrophic system voltage collapse.

Frequency Protection (ANSI 81U / 81O)

• Operating Dynamic: System frequency reflects active power balance: $\frac{df}{dt} = \frac{f_0}{2H} (P_{gen} - P_{load})$. Sudden loss of generation causes system frequency to plunge.
• NERC Under-Frequency Load Shedding (UFLS): Regional entities mandate stepped automatic customer load shedding to arrest frequency decline:
  • Stage 1: $59.5\text{ Hz}$, drop $10\%$ total system load.
  • Stage 2: $59.1\text{ Hz}$, drop additional $10\%$ load.
  • Stage 3: $58.7\text{ Hz}$, drop additional $10\%$ load.

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5. Comprehensive Worked Calculations

Part A: Dual-Slope Transformer Differential (87T) Evaluation

A $30\text{ MVA}$, $115\text{ kV} / 13.8\text{ kV}$ Delta-Wye transformer has:

• High-Voltage CT Ratio: $200:5\text{ A}$ ($CTR_1 = 40$)
• Low-Voltage CT Ratio: $1,500:5\text{ A}$ ($CTR_2 = 300$)
• Differential Relay Settings: $I_{pickup} = 0.30\text{ A sec}$, $\text{Slope 1} = 30\%$, $I_{BP} = 2.0\text{ A sec}$, $\text{Slope 2} = 60\%$.
• Restraint current formulation: $I_{rest} = \frac{|I_{1,sec}| + |I_{2,sec}|}{2}$.

Calculate:

1. Full-load primary and secondary currents on HV and LV sides.
2. Relay tap compensation multipliers ($CTC$) required to balance secondary currents.
3. Differential current $I_{diff}$ and restraint current $I_{rest}$ during an external through-fault producing $I_{HV,pri} = 1,800\text{ A}$ and a $10\%$ CT saturation error on the LV side ($I_{LV,sec} = 0.90 \times I_{balanced}$). Determine if the relay restrains properly.

============================== STEP-BY-STEP SOLUTION ==============================

Step 1: Compute Rated Full-Load Currents
  HV Full-Load Primary Current:
    I_FLA,HV = S / (sqrt(3) * V_HV) = 30,000,000 / (sqrt(3) * 115,000) = 150.61 A pri
  LV Full-Load Primary Current:
    I_FLA,LV = S / (sqrt(3) * V_LV) = 30,000,000 / (sqrt(3) * 13,800) = 1,255.1 A pri
  
  Secondary Currents at Rated Load:
    I_sec,HV = 150.61 A / (200 / 5) = 150.61 / 40 = 3.765 A sec
    I_sec,LV = 1,255.1 A / (1500 / 5) = 1,255.1 / 300 = 4.184 A sec

Step 2: Establish Digital Relay Tap Balancing (CTC Factors)
  To equalize secondary currents, define matching factors:
    Tap_HV = 3.765 A, Tap_LV = 4.184 A
    Normalized per-unit secondary current = 1.0 pu at rated load for both sides.

Step 3: Evaluate External Through-Fault with 10% CT Saturation Error
  HV Through-Fault Current: I_fault,HV = 1,800 A pri (11.95x rated FLA)
  Ideal LV Through-Fault Current: I_fault,LV = 1,800 * (115 / 13.8) = 15,000 A pri
  
  Secondary Currents entering relay:
    I_1 = 1,800 A / 40 = 45.0 A sec
    Normalized I_1,pu = 45.0 / 3.765 = 11.952 pu
    
    With 10% saturation error on LV side:
    Normalized I_2,pu = 0.90 * 11.952 pu = 10.757 pu
  
  Compute Differential and Restraint Currents in per-unit:
    I_diff = |I_1,pu - I_2,pu| = |11.952 - 10.757| = 1.195 pu
    I_rest = (I_1,pu + I_2,pu) / 2 = (11.952 + 10.757) / 2 = 11.355 pu

Step 4: Verify Relay Restraint Threshold
  Since I_rest = 11.355 pu > I_BP (where I_BP = 2.0 / 3.765 = 0.531 pu):
  Operating Threshold I_diff,trip = I_pickup + Slope 1 * I_BP + Slope 2 * (I_rest - I_BP)
  Let base threshold = Slope 2 * I_rest = 0.60 * 11.355 = 6.813 pu
  
  Compare Operating vs Trip Threshold:
    Actual I_diff = 1.195 pu
    Required Trip I_diff = 6.813 pu
    Since 1.195 pu << 6.813 pu, the relay RESTRAINS (Does NOT false trip!).
===================================================================================

Part B: Transmission Line Distance (ANSI 21) Mho Reach Sizing

A $230\text{ kV}$, $60\text{ Hz}$, $40\text{ mile}$ overhead transmission line connects Substation A to Substation B. An adjacent line of $30\text{ miles}$ extends from Substation B to Substation C.

• Line positive-sequence impedance: $\mathbf{z}_1 = 0.12 + j0.68\; \Omega/\text{mile}$
• $CTR = 1200:5 = 240$, $PTR = 230,000:115 = 2000$

Calculate:

1. Total primary impedance $\mathbf{Z}_{AB,pri}$.
2. Secondary impedance scale factor $k_Z = \frac{CTR}{PTR}$.
3. Zone 1 reach in primary and secondary ohms ($85\%$ reach).
4. Zone 2 reach in primary and secondary ohms ($125\%$ reach).

============================== STEP-BY-STEP SOLUTION ==============================

Step 1: Compute Total Primary Line Impedance (Z_AB)
  Length l_AB = 40 miles
  Z_AB,pri = (0.12 + j0.68) * 40 = 4.80 + j27.20 ohms = 27.620 /_ 79.99 deg ohms pri

Step 2: Determine Impedance Conversion Factor (k_Z)
  k_Z = Z_sec / Z_pri = CTR / PTR = 240 / 2,000 = 0.120
  
  Z_AB,sec = Z_AB,pri * 0.120 = (4.80 + j27.20) * 0.120 
           = 0.576 + j3.264 ohms = 3.314 /_ 79.99 deg ohms sec

Step 3: Calculate Zone 1 Mho Reach Setting (85% Reach, t1 = 0 s)
  Primary Reach: 
    Z_Z1,pri = 0.85 * Z_AB,pri = 0.85 * 27.620 ohms = 23.477 ohms pri
  Secondary Reach Setting:
    Z_Z1,sec = 0.85 * Z_AB,sec = 0.85 * 3.314 ohms = 2.817 ohms sec

Step 4: Calculate Zone 2 Mho Reach Setting (125% Reach, t2 = 0.35 s)
  Primary Reach:
    Z_Z2,pri = 1.25 * Z_AB,pri = 1.25 * 27.620 ohms = 34.525 ohms pri
  Secondary Reach Setting:
    Z_Z2,sec = 1.25 * Z_AB,sec = 1.25 * 3.314 ohms = 4.143 ohms sec
===================================================================================

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6. Common Exam Traps & Strategic Pitfalls

• Setting Distance Zone 1 to 100%: Attempting to protect $100\%$ of the transmission line with Zone 1 instantaneous tripping. Transient overreach caused by DC offset, CVT transient response, and line data errors will cause the relay to trip instantaneously on out-of-zone faults past the remote bus.
• Overlooking Phase Angle Compensation in Delta-Wye 87T Relays: Failing to recognize that a Delta-Wye transformer shifts currents by $30^\circ$. If not compensated numerically or via CT connections, full load current creates a $52\%$ false differential current ($|1 - 1\angle 30^\circ| = 0.518$), causing instant trip upon energization.
• Inverting the Impedance Ratio ($k_Z$): Multiplying primary ohms by $PTR/CTR$ instead of $CTR/PTR$. Because secondary voltage is small ($1/2000$) and secondary current is moderate ($1/240$), secondary impedance is always much smaller than primary impedance: $Z_{sec} = Z_{pri} \times (CTR/PTR)$.
• Confusing 2nd Harmonic Inrush with 5th Harmonic Overexcitation: 2nd harmonic ($120\text{ Hz}$) indicates magnetizing inrush during transformer energization; 5th harmonic ($300\text{ Hz}$) indicates severe core overfluxing ($V/Hz$) from overvoltage.