13.2 Unsymmetrical Fault Analysis (SLG, L-L, DLG Fault Calculations)

13.2 Unsymmetrical Fault Analysis (SLG, L-L, DLG Fault Calculations)

Executive Overview: While three-phase faults represent the most severe balanced condition, the vast majority of real-world power system faults are unsymmetrical: Single Line-to-Ground (SLG, $\sim 70%$ of all faults), Line-to-Line (L-L, $\sim 15%$), and Double Line-to-Ground (DLG, $\sim 10%$). Analyzing unsymmetrical faults requires Fortescue symmetrical component transformations to decouple mutually coupled three-phase networks into independent Positive ($1$), Negative ($2$), and Zero ($0$) sequence networks. On the NCEES PE Power exam, candidates must master the boundary conditions, sequence network interconnections, ground potential rise ($GPR$), and unfaulted phase voltage shifts across all fault types.

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1. Symmetrical Components Matrix Foundations

In 1918, Charles LeGeyt Fortescue proved that any unbalanced set of three-phase phasors ($\mathbf{I}_a, \mathbf{I}_b, \mathbf{I}_c$) can be decomposed into three symmetrical sequence components:

$\begin{bmatrix} \mathbf{I}_a \\ \mathbf{I}_b \\ \mathbf{I}_c \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a^2 & a \\ 1 & a & a^2 \end{bmatrix} \begin{bmatrix} \mathbf{I}_{a0} \\ \mathbf{I}_{a1} \\ \mathbf{I}_{a2} \end{bmatrix} = \mathbf{A} \mathbf{I}_{012}$

where the complex rotation operator $a = 1\angle 120^\circ = -0.5 + j\frac{\sqrt{3}}{2} = -0.5 + j0.8660$ has properties:

• $a^2 = 1\angle 240^\circ = 1\angle -120^\circ = -0.5 - j0.8660$
• $a^3 = 1\angle 360^\circ = 1.0\angle 0^\circ$
• $1 + a + a^2 = 0$

Inverting the transformation matrix $\mathbf{A}$ gives sequence currents from phase currents:

$\mathbf{I}_{012} = \mathbf{A}^{-1} \mathbf{I}_{abc} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix} \begin{bmatrix} \mathbf{I}_a \\ \mathbf{I}_b \\ \mathbf{I}_c \end{bmatrix}$

$\begin{aligned} \mathbf{I}_{a0} &= \frac{1}{3} (\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) \quad &[\text{Zero Sequence: In-phase, returns via ground}] \\ \mathbf{I}_{a1} &= \frac{1}{3} (\mathbf{I}_a + a\mathbf{I}_b + a^2\mathbf{I}_c) \quad &[\text{Positive Sequence: Normal } a-b-c \text{ phase rotation}] \\ \mathbf{I}_{a2} &= \frac{1}{3} (\mathbf{I}_a + a^2\mathbf{I}_b + a\mathbf{I}_c) \quad &[\text{Negative Sequence: Reversed } a-c-b \text{ phase rotation}] \end{aligned}$

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2. Single Line-to-Ground (SLG) Fault Analysis

Consider an SLG fault on phase $a$ through fault impedance $Z_f$ to ground:

Boundary Conditions in Physical Domain

1. $\mathbf{I}_b = 0$
2. $\mathbf{I}_c = 0$
3. $\mathbf{V}_a = \mathbf{I}_a Z_f$

Sequence Domain Transformations

Substituting boundary conditions into $\mathbf{A}^{-1}$:

$\mathbf{I}_{a0} = \frac{1}{3}(\mathbf{I}_a + 0 + 0) = \frac{1}{3}\mathbf{I}_a, \qquad \mathbf{I}_{a1} = \frac{1}{3}\mathbf{I}_a, \qquad \mathbf{I}_{a2} = \frac{1}{3}\mathbf{I}_a$

$\therefore \mathbf{I}_{a0} = \mathbf{I}_{a1} = \mathbf{I}_{a2} = \frac{1}{3}\mathbf{I}_a$

Because the sequence currents are strictly identical, the positive, negative, and zero sequence networks are connected in SERIES.

Sequence Network Connection for Single Line-to-Ground (SLG) Fault:

      (+) o---(~) V_F ---[ Z_1 ]---o
                                   |
      (+) o--------------[ Z_2 ]---o  (SERIES CONNECTION)
                                   |
      (+) o--------------[ Z_0 ]---o
                                   |
                                  [ 3*Z_f ]
                                   |
      (-) o------------------------o Neutral Reference Bus

SLG Fault Current Formulas

$\mathbf{I}_{a0} = \mathbf{I}_{a1} = \mathbf{I}_{a2} = \frac{\mathbf{V}_F}{\mathbf{Z}_1 + \mathbf{Z}_2 + \mathbf{Z}_0 + 3 Z_f}$

$\mathbf{I}_f = \mathbf{I}_a = \mathbf{I}_{a0} + \mathbf{I}_{a1} + \mathbf{I}_{a2} = 3\mathbf{I}_{a0} = \frac{3\mathbf{V}_F}{\mathbf{Z}_1 + \mathbf{Z}_2 + \mathbf{Z}_0 + 3 Z_f}$

Ground Potential Rise & Unfaulted Phase Voltage Swell

When an SLG fault occurs, current returning through grounding resistance $R_g$ elevates substation ground grid voltage:

$GPR = |\mathbf{I}_g| \cdot R_g = 3|\mathbf{I}_{a0}| \cdot R_g$

On ungrounded systems ($Z_0 \to \infty$), the fault current is limited only by distributed line-to-ground capacitance ($I_f \approx 3 I_{C0}$), but the voltages of unfaulted phases $b$ and $c$ shift from line-to-neutral to full line-to-line magnitude:

$V_{b,fault} = \sqrt{3} V_{LN} \angle -150^\circ, \qquad V_{c,fault} = \sqrt{3} V_{LN} \angle 150^\circ \quad \Longrightarrow \quad |V_{unfaulted}| = \sqrt{3} |V_{LN}| = |V_{LL}|$

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3. Line-to-Line (L-L) Fault Analysis

Consider an L-L fault between phases $b$ and $c$ through fault impedance $Z_f$:

Boundary Conditions in Physical Domain

1. $\mathbf{I}_a = 0$
2. $\mathbf{I}_b = -\mathbf{I}_c$
3. $\mathbf{V}_b - \mathbf{V}_c = \mathbf{I}_b Z_f$

Sequence Domain Transformations

$\mathbf{I}_{a0} = \frac{1}{3}(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = \frac{1}{3}(0 + \mathbf{I}_b - \mathbf{I}_b) = 0$

$\mathbf{I}_{a1} = \frac{1}{3}(0 + a\mathbf{I}_b - a^2\mathbf{I}_b) = \frac{a - a^2}{3}\mathbf{I}_b = \frac{j\sqrt{3}}{3}\mathbf{I}_b = \frac{j}{\sqrt{3}}\mathbf{I}_b$

$\mathbf{I}_{a2} = \frac{1}{3}(0 + a^2\mathbf{I}_b - a\mathbf{I}_b) = -\mathbf{I}_{a1}$

Because $\mathbf{I}{a0} = 0$ and $\mathbf{I}{a1} = -\mathbf{I}_{a2}$, the zero-sequence network is completely isolated (open-circuit), and the positive and negative sequence networks are connected in PARALLEL.

Sequence Network Connection for Line-to-Line (L-L) Fault:

      (+) o---(~) V_F ---[ Z_1 ]---+---o
                                   |   |
                                  [Z_f]|
                                   |   |
      (+) o--------------[ Z_2 ]---+---o  (PARALLEL: Pos || Neg)

      (+) o--------------[ Z_0 ]   (OPEN CIRCUIT: I_a0 = 0)

L-L Fault Current Formulas

$\mathbf{I}_{a1} = -\mathbf{I}_{a2} = \frac{\mathbf{V}_F}{\mathbf{Z}_1 + \mathbf{Z}_2 + Z_f}, \qquad \mathbf{I}_{a0} = 0$

$\mathbf{I}_b = -\mathbf{I}_c = -j\sqrt{3}\mathbf{I}_{a1} = \frac{-j\sqrt{3}\mathbf{V}_F}{\mathbf{Z}_1 + \mathbf{Z}_2 + Z_f}$

$|\mathbf{I}_{f,LL}| = \sqrt{3}|\mathbf{I}_{a1}| = \frac{\sqrt{3}|\mathbf{V}_F|}{|\mathbf{Z}_1 + \mathbf{Z}_2 + Z_f|}$

[!NOTE] L-L vs. 3-Phase Ratio: If $Z_1 = Z_2$ and $Z_f = 0$, the bolted line-to-line fault current is: $|\mathbf{I}_{f,LL}| = \frac{\sqrt{3} V_F}{2 Z_1} = \frac{\sqrt{3}}{2} \left( \frac{V_F}{Z_1} \right) = \frac{\sqrt{3}}{2} I_{f,3\phi} \approx 0.8660 \cdot I_{f,3\phi}$

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4. Double Line-to-Ground (DLG) Fault Analysis

Consider a DLG fault where phases $b$ and $c$ contact each other and ground through fault impedance $Z_f$:

Boundary Conditions in Physical Domain

1. $\mathbf{I}_a = 0$
2. $\mathbf{V}_b = \mathbf{V}_c = (\mathbf{I}_b + \mathbf{I}_c) Z_f = \mathbf{I}_g Z_f$

Sequence Domain Transformations

Connecting positive, negative, and zero sequence networks in PARALLEL:

Sequence Network Connection for Double Line-to-Ground (DLG) Fault:

               +---[ Z_1 ]---(~) V_F
               |
      (+) o----+---[ Z_2 ]-----------o (-)  (PARALLEL COMBINATION)
               |
               +---[ Z_0 + 3*Z_f ]---o

DLG Fault Current Formulas

$\mathbf{I}_{a1} = \frac{\mathbf{V}_F}{\mathbf{Z}_1 + \left( \frac{\mathbf{Z}_2 (\mathbf{Z}_0 + 3Z_f)}{\mathbf{Z}_2 + \mathbf{Z}_0 + 3Z_f} \right)}$

Using current division to find negative and zero sequence currents:

$\mathbf{I}_{a2} = -\mathbf{I}_{a1} \left( \frac{\mathbf{Z}_0 + 3Z_f}{\mathbf{Z}_2 + \mathbf{Z}_0 + 3Z_f} \right), \qquad \mathbf{I}_{a0} = -\mathbf{I}_{a1} \left( \frac{\mathbf{Z}_2}{\mathbf{Z}_2 + \mathbf{Z}_0 + 3Z_f} \right)$

Total ground return current:

$\mathbf{I}_g = \mathbf{I}_b + \mathbf{I}_c = 3\mathbf{I}_{a0} = -3\mathbf{I}_{a1} \left( \frac{\mathbf{Z}_2}{\mathbf{Z}_2 + \mathbf{Z}_0 + 3Z_f} \right)$

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5. Fault Severity Comparison ($Z_0$ vs. $Z_1$ Dynamics)

Many engineers intuitively assume that three-phase faults always produce the highest current. However, in solidly grounded transmission substations and near generating stations, the zero-sequence impedance is frequently lower than positive-sequence impedance ($Z_0 < Z_1$).

When is SLG Fault Current Greater than 3-Phase Fault Current?

$I_{f,SLG} = \frac{3 V_F}{Z_1 + Z_2 + Z_0} > \frac{V_F}{Z_1} = I_{f,3\phi}$

Assuming $Z_1 = Z_2$:

$\frac{3}{2 Z_1 + Z_0} > \frac{1}{Z_1} \implies 3 Z_1 > 2 Z_1 + Z_0 \implies \mathbf{Z}_0 < \mathbf{Z}_1$

$\begin{array}{|l|c|c|c|} \hline \textbf{Fault Type} & \textbf{Sequence Interconnection} & \textbf{Fault Current Formula } (Z_f = 0) & \textbf{Ratio to } I_{3\phi} \text{ (when } Z_1 = Z_2 = Z_0) \\ \hline \textbf{3-Phase Bolted } (3\phi) & \text{Positive Sequence Only} & I_f = \frac{V_F}{Z_1} & 1.000 \\ \hline \textbf{Single Line-to-Ground } (SLG) & \text{Series: } Z_1 + Z_2 + Z_0 & I_f = \frac{3 V_F}{Z_1 + Z_2 + Z_0} & 1.000 \\ \hline \textbf{Line-to-Line } (L-L) & \text{Parallel: } Z_1 \parallel Z_2 & I_f = \frac{\sqrt{3} V_F}{Z_1 + Z_2} & 0.866 \\ \hline \textbf{Double Line-to-Ground } (DLG) & \text{Parallel: } Z_1 \parallel Z_2 \parallel Z_0 & I_f = \text{Phase Current: } I_b, I_c & 0.866 - 1.732 \\ \hline \end{array}$

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6. Comprehensive Worked Calculation: Multi-Fault Analysis

Problem Statement

A $13.8\text{ kV}$ solidly grounded substation bus has the following Thevenin sequence impedances:

• $\mathbf{Z}_1 = j0.10\text{ pu}$
• $\mathbf{Z}_2 = j0.10\text{ pu}$
• $\mathbf{Z}_0 = j0.05\text{ pu}$ ($Z_0 < Z_1$ due to proximity of a large solidly grounded $Y_g-\Delta$ generator step-up transformer)
• System Base: $S_{base} = 100\text{ MVA}$, $V_{base} = 13.8\text{ kV}$
• Prefault Voltage: $\mathbf{V}_F = 1.0\angle 0^\circ\text{ pu}$
• Fault Impedance: $Z_f = 0$ (bolted faults)

Calculate for all four fault types:

1. Symmetrical three-phase ($3\phi$) bolted fault current.
2. Single line-to-ground (SLG) fault current and ground current.
3. Line-to-line (L-L) fault current.
4. Double line-to-ground (DLG) sequence currents, phase currents, and total ground return current.

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

Step 1: Compute System Base Current at 13.8 kV
  I_base = S_base / (sqrt(3) * V_base) = 100,000 kVA / (sqrt(3) * 13.8 kV)
         = 4,183.70 A = 4.1837 kA

Step 2: Symmetrical Three-Phase Fault (3ph)
  I_a1 = V_F / Z_1 = 1.0 / (j0.10) = -j10.00 pu = 10.00 /_ -90 deg pu
  I_f,3ph = 10.00 pu
  Physical Current: I_f,3ph = 10.00 * 4.1837 kA = 41.84 kA

Step 3: Single Line-to-Ground Fault (SLG on Phase a)
  Z_total = Z_1 + Z_2 + Z_0 = j0.10 + j0.10 + j0.05 = j0.25 pu
  I_a0 = I_a1 = I_a2 = V_F / Z_total = 1.0 / (j0.25) = -j4.00 pu = 4.00 /_ -90 deg pu
  
  Total Fault Current in Phase a:
  I_f,SLG = I_a = 3 * I_a0 = 3 * (-j4.00 pu) = -j12.00 pu = 12.00 /_ -90 deg pu
  Physical Current: I_f,SLG = 12.00 * 4.1837 kA = 50.20 kA
  
  >> CRITICAL OBSERVATION: I_SLG (50.20 kA) is 20% LARGER than I_3ph (41.84 kA) 
     because Z_0 (0.05 pu) < Z_1 (0.10 pu)!

Step 4: Line-to-Line Fault (L-L between Phases b and c)
  I_a0 = 0
  I_a1 = -I_a2 = V_F / (Z_1 + Z_2) = 1.0 / (j0.10 + j0.10) = 1.0 / (j0.20)
       = -j5.00 pu = 5.00 /_ -90 deg pu
  
  Fault Current Magnitude:
  |I_f,LL| = sqrt(3) * |I_a1| = sqrt(3) * 5.00 pu = 8.660 pu
  Physical Current: I_f,LL = 8.660 * 4.1837 kA = 36.23 kA
  Check Ratio: I_LL / I_3ph = 36.23 / 41.84 = 0.8660 = sqrt(3)/2  (Exact)

Step 5: Double Line-to-Ground Fault (DLG on Phases b and c)
  Parallel combination of Z_2 and Z_0:
  Z_parallel = (Z_2 * Z_0) / (Z_2 + Z_0) = (j0.10 * j0.05) / (j0.10 + j0.05)
             = -0.005 / (j0.15) = j0.03333 pu
  
  Positive-Sequence Current:
  I_a1 = V_F / (Z_1 + Z_parallel) = 1.0 / (j0.10 + j0.03333) = 1.0 / (j0.13333)
       = -j7.500 pu = 7.500 /_ -90 deg pu
  
  Negative and Zero Sequence Currents (Current Divider):
  I_a2 = -I_a1 * [ Z_0 / (Z_2 + Z_0) ] = -(-j7.500) * [ 0.05 / 0.15 ]
       = j2.500 pu = 2.500 /_ 90 deg pu
  I_a0 = -I_a1 * [ Z_2 / (Z_2 + Z_0) ] = -(-j7.500) * [ 0.10 / 0.15 ]
       = j5.000 pu = 5.000 /_ 90 deg pu
  
  Verify Phase a Current: I_a = I_a0 + I_a1 + I_a2 = j5.0 - j7.5 + j2.5 = 0 pu (Verified)
  
  Phase b and Phase c Fault Currents:
  I_b = I_a0 + (a^2)*I_a1 + a*I_a2
      = j5.0 + (-0.5 - j0.8660)*(-j7.5) + (-0.5 + j0.8660)*(j2.5)
      = j5.0 + (-6.495 + j3.750) + (-2.165 - j1.250)
      = -8.660 + j7.500 pu = 11.456 /_ 140.89 deg pu
  
  |I_b| = |I_c| = 11.456 pu
  Physical Phase Fault Current: I_b = 11.456 * 4.1837 kA = 47.93 kA
  
  Total Ground Return Current:
  I_g = 3 * I_a0 = 3 * (j5.000 pu) = j15.000 pu = 15.000 /_ 90 deg pu
  Physical Ground Current: I_g = 15.000 * 4.1837 kA = 62.76 kA

============================== SUMMARY OF RESULTS ==============================
  - 3-Phase Bolted Fault Current (I_3ph)  = 10.00 pu = 41.84 kA
  - Single Line-to-Ground Current (I_SLG) = 12.00 pu = 50.20 kA  (+20.0% vs 3ph)
  - Line-to-Line Fault Current (I_LL)     =  8.66 pu = 36.23 kA  (-13.4% vs 3ph)
  - DLG Phase Fault Current (|I_b|, |I_c|)= 11.46 pu = 47.93 kA  (+14.6% vs 3ph)
  - DLG Ground Return Current (I_g)       = 15.00 pu = 62.76 kA  (+50.0% vs 3ph)
===================================================================================

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7. Common Exam Traps & Strategic Pitfalls

• The "Factor of 3" Ground Current Omission: Calculating $I_{a0}$ in an SLG fault and forgetting to multiply by $3$ to obtain the physical phase/ground fault current ($I_f = 3 I_{a0}$). Relay engineers size ground overcurrent elements (50N/51N) based on $3 I_{a0}$.
• Assuming Three-Phase Fault Current is Always the Highest: Sizing circuit breakers exclusively for $I_{3\phi}$ without checking $I_{SLG}$. If $Z_0 < Z_1$, the SLG fault produces the maximum interrupting duty.
• Applying Zero-Sequence Networks to Line-to-Line Faults: Incorrectly including $Z_0$ in L-L fault calculations. Because the fault does not touch ground and $I_b = -I_c$, no zero-sequence current can flow ($I_{a0} = 0$).
• Confusing Fault Impedance Factor $3Z_f$ with $Z_f$: In SLG and DLG sequence networks, the physical fault impedance to ground $Z_f$ must be multiplied by $3$ in the zero-sequence branch ($3Z_f$) because $3I_{a0}$ flows through the neutral ground path.