9.4 Double Line-to-Ground (DLG) Faults & Bus Impedance Matrix (Zbus) Methods

9.4 Double Line-to-Ground (DLG) Faults & Bus Impedance Matrix (Zbus) Methods

In complex, interconnected power systems containing dozens or hundreds of substations, manual network reduction (series-parallel and delta-wye transformations) is completely impractical. Modern power engineering and automated relay setting software rely on the Bus Impedance Matrix ($Z_{bus}$) to perform rapid, exact short-circuit calculations across entire power grids.

On the NCEES PE Electrical and Computer: Power examination, fault analysis questions in this domain test your capability to analyze Double Line-to-Ground (DLG) sequence network interconnections, compute ground return fault currents, interpret diagonal ($Z_{kk}$) and off-diagonal ($Z_{ik}$) elements of $Z_{bus}$, and calculate post-fault bus voltages and line currents without recalculating network equivalents.

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1. Double Line-to-Ground (DLG / 2LG) Fault Analysis

A Double Line-to-Ground fault occurs when two phase conductors (conventionally Phase B and Phase C) short-circuit to each other and simultaneously contact earth ground through a fault impedance $Z_f$ and neutral/ground impedance $Z_g$.

               DOUBLE LINE-TO-GROUND (DLG) FAULT TOPOLOGY

          Phase A  o----------------------------------o  I_a = 0 (Open)

          Phase B  o----------------+-----------------o  Fault Current: I_b
                                    |
                                   +---+ 
                                   |   | Z_f
                                   +---+ 
                                    |
          Phase C  o----------------+-----------------o  Fault Current: I_c
                                    |
                                   +---+ 
                                   |   | Z_f
                                   +---+ 
                                    |
                                   +---+ 
                                   |   | Z_g (Ground Return Path)
                                   +---+ 
                                    |
                                   ===

Boundary Conditions (Bolted Fault on Phases B and C to Ground: $Z_f = 0, Z_g = 0$)

1. $I_a = 0$
2. $V_b = 0$
3. $V_c = 0$

Sequence Voltage and Current Relationships

Applying the inverse symmetrical component transformation to phase voltages ($V_a \ne 0, V_b = 0, V_c = 0$):

$$\begin{bmatrix} V_{a0} \\ V_{a1} \\ V_{a2} \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix} \begin{bmatrix} V_a \\ 0 \\ 0 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} V_a \\ V_a \\ V_a \end{bmatrix}$$ $$V_{a0} = V_{a1} = V_{a2} = \frac{V_a}{3}$$

From the current boundary condition $I_a = 0$:

$$I_a = I_{a0} + I_{a1} + I_{a2} = 0 \implies I_{a0} + I_{a1} + I_{a2} = 0$$

Sequence Network Interconnection: PARALLEL Connection of All Three Networks

Because the sequence voltages at the fault point are identical ($V_{a0} = V_{a1} = V_{a2}$) and the three sequence currents sum to zero, all three sequence networks (Positive, Negative, and Zero) must be connected in PARALLEL at the fault point.

               DLG SEQUENCE NETWORK INTERCONNECTION (PARALLEL)

                   +---[ Z_1 ]---(~) V_f (Positive Sequence)
                   |               |
             +-----+---------------+-----+  I_a1 (Inflow)
             |                           |
             |     +---[ Z_2 ]-----+     |  I_a2 (Outflow)
             |     | (Negative Seq)|     |
             +-----+---------------+-----+ 
             |                           |
             |     +---[ Z_0 + 3Z_g ]--+ |  I_a0 (Outflow)
             |     | (Zero Sequence)   | |
             +-----+-------------------+-----+ 
             |                               |
             +-------------------------------+ Reference Bus

Master DLG Sequence Current Formulas

From the parallel combination of the negative and zero sequence branches:

$$Z_{parallel} = Z_2 \parallel (Z_0 + 3 Z_f + 3 Z_g) = \frac{Z_2 (Z_0 + 3 Z_f + 3 Z_g)}{Z_2 + Z_0 + 3 Z_f + 3 Z_g}$$ $$I_{a1} = \frac{V_f}{Z_1 + Z_{parallel}} = \frac{V_f}{Z_1 + \frac{Z_2 (Z_0 + 3 Z_f + 3 Z_g)}{Z_2 + Z_0 + 3 Z_f + 3 Z_g}} \quad [\text{pu}]$$

Applying current division to find the negative and zero sequence currents:

$$I_{a2} = -I_{a1} \left( \frac{Z_0 + 3 Z_f + 3 Z_g}{Z_2 + Z_0 + 3 Z_f + 3 Z_g} \right) \quad [\text{pu}]$$ $$I_{a0} = -I_{a1} \left( \frac{Z_2}{Z_2 + Z_0 + 3 Z_f + 3 Z_g} \right) \quad [\text{pu}]$$

Total Ground Return Current ($I_{ground}$)

$$I_{ground} = I_b + I_c = 3 I_{a0} = -3 I_{a1} \left( \frac{Z_2}{Z_2 + Z_0 + 3 Z_f + 3 Z_g} \right) \quad [\text{pu}]$$

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2. Master Comparative Summary of All Four Fault Types

+---------------------------------------------------------------------------------------------------+
|                         POWER SYSTEM FAULT COMPARISON MASTER TABLE                                |
+---------------------------------------------------------------------------------------------------+
| Feature           | 3-Phase (3Φ)      | Single Line-Ground (SLG)| Line-to-Line (L-L) | Double Line-Ground (DLG)|
| :---              | :---              | :---                   | :---               | :---                    |
| **Symbolic Code** | $3\text{P} / 3\text{LG}$ | $1\text{LG}$           | $2\text{L}$        | $2\text{LG}$            |
| **Statistical %** | $\sim 5\%$        | $\sim 70\%$            | $\sim 15\%$        | $\sim 10\%$            |
| **Boundary**      | $V_a = V_b = V_c$ | $I_b = 0, I_c = 0$     | $I_a = 0, I_b=-I_c$| $I_a = 0$               |
| **Conditions**    | $I_a+I_b+I_c=0$   | $V_a = 0$              | $V_b = V_c$        | $V_b = 0, V_c = 0$      |
| **Sequence**      | $I_{a2} = 0$      | $I_{a0}=I_{a1}=I_{a2}$ | $I_{a0} = 0$       | $V_{a0}=V_{a1}=V_{a2}$  |
| **Relationships** | $I_{a0} = 0$      | $= I_a / 3$            | $I_{a1} = -I_{a2}$ | $I_{a0}+I_{a1}+I_{a2}=0$|
| **Network**       | Positive Sequence | **SERIES** of Pos,     | **PARALLEL** of    | **PARALLEL** of Pos,    |
| **Topology**      | Only              | Neg, and Zero Networks | Pos & Neg (Zero Open)| Neg, and Zero Networks|
| **Positive Seq.** | $I_{a1} = \frac{V_f}{Z_1}$| $I_{a1} = \frac{V_f}{Z_1+Z_2+Z_0}$| $I_{a1} = \frac{V_f}{Z_1+Z_2}$| $I_{a1} = \frac{V_f}{Z_1 + (Z_2 \parallel Z_0)}$ |
| **Current (I_a1)**|                   |                        |                    |                         |
| **Phase Fault**   | $I_{f} = \frac{V_f}{Z_1}$ | $I_a = 3 I_{a0}$       | $|I_b| = \sqrt{3} |I_{a1}|$ | $|I_b|, |I_c|$ from     |
| **Current**       |                   | $= \frac{3V_f}{Z_1+Z_2+Z_0}$ | $= \frac{\sqrt{3}V_f}{Z_1+Z_2}$ | transformation matrix |
| **Ground Current**| $0$               | $I_{ground} = I_a$     | $0$                | $I_{ground} = 3 I_{a0}$  |
| **(I_ground)**    |                   | $= 3 I_{a0}$           |                    |                         |
+---------------------------------------------------------------------------------------------------+

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3. The Bus Impedance Matrix ($Z_{bus}$) Method

In nodal network analysis, the power system is represented by the Bus Admittance Matrix ($Y_{bus}$), which is sparse and computationally efficient for load flow studies. However, for short-circuit calculations, the Bus Impedance Matrix ($Z_{bus}$) is the primary analytical tool because it directly contains all system Thevenin and transfer impedances.

$$Z_{bus} = Y_{bus}^{-1}$$ $$\mathbf{V}_{bus} = Z_{bus} \mathbf{I}_{bus}$$

                    STRUCTURE OF THE BUS IMPEDANCE MATRIX (Z_bus)

                        Bus 1     Bus 2     ...     Bus k     ...     Bus N
               Bus 1  [ Z_11      Z_12      ...     Z_1k      ...     Z_1N ]
               Bus 2  [ Z_21      Z_22      ...     Z_2k      ...     Z_2N ]
                 :    [  :         :        ...      :        ...      :   ]
       Z_bus = Bus k  [ Z_k1      Z_k2      ...     Z_kk      ...     Z_kN ]
                 :    [  :         :        ...      :        ...      :   ]
               Bus N  [ Z_N1      Z_N2      ...     Z_Nk      ...     Z_NN ]

Physical Significance of Matrix Elements

1. Diagonal Elements ($Z_{kk}$ - Driving-Point Impedance): $Z_{kk}$ is the exact Thevenin equivalent impedance seen looking into Bus $k$ with all system voltage sources shorted to ground ($Z_{th,k} = Z_{kk}$).
2. Off-Diagonal Elements ($Z_{ik}$ - Transfer Impedance): $Z_{ik}$ represents the open-circuit voltage response produced at Bus $i$ due to a unit current injected at Bus $k$ ($Z_{ik} = \frac{V_i}{I_k}\Big|_{I_m = 0, m \ne k}$). By reciprocity in bilateral networks, $Z_{ik} = Z_{ki}$.

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4. Calculating Fault Currents, Bus Voltages, and Line Flows Using $Z_{bus}$

Consider an $N$-bus power system operating at prefault bus voltages $\mathbf{V}^{(0)} = \begin{bmatrix} V_1^{(0)} & V_2^{(0)} & \dots & V_N^{(0)} \end{bmatrix}^T$ (typically $1.0\angle 0^\circ\text{ pu}$).

+---------------------------------------------------------------------------------------------------+
|                      Z_bus FAULT CALCULATION ALGORITHM (BUS k FAULT)                              |
+---------------------------------------------------------------------------------------------------+
| Step 1: Three-Phase Symmetrical Fault Current at Bus k:                                           |
|                                                                                                   |
|             I_f,k = V_k^(0) / (Z_kk + Z_f)                                                        |
|                                                                                                   |
| Step 2: Post-Fault Voltages at EVERY Bus i (i = 1, 2, ..., N):                                     |
|         Applying superposition with fault current injection (-I_f,k) at Bus k:                   |
|                                                                                                   |
|             V_i^(f) = V_i^(0) - Z_ik * I_f,k = V_i^(0) - (Z_ik / (Z_kk + Z_f)) * V_k^(0)          |
|                                                                                                   |
|         * Note: At the faulted bus itself (i = k) for a bolted fault (Z_f = 0):                   |
|                 V_k^(f) = V_k^(0) - Z_kk * (V_k^(0) / Z_kk) = 0.0 pu                             |
|                                                                                                   |
| Step 3: Post-Fault Branch Currents in Line connecting Bus i to Bus j:                             |
|                                                                                                   |
|             I_ij^(f) = (V_i^(f) - V_j^(f)) / z_ij,line                                            |
|             where z_ij,line is the actual series branch impedance of line i-j.                    |
+---------------------------------------------------------------------------------------------------+

Sequence $Z_{bus}$ Matrices for Unsymmetrical Faults

To perform unsymmetrical fault calculations at Bus $k$, we construct the three sequence bus impedance matrices:

• Positive-Sequence Matrix: $Z_{bus}^{(1)} = [Y_{bus}^{(1)}]^{-1}$
• Negative-Sequence Matrix: $Z_{bus}^{(2)} = [Y_{bus}^{(2)}]^{-1}$
• Zero-Sequence Matrix: $Z_{bus}^{(0)} = [Y_{bus}^{(0)}]^{-1}$

$$\text{Single Line-to-Ground (SLG) Fault at Bus } k: \quad I_{f,k,SLG} = \frac{3 V_k^{(0)}}{Z_{kk}^{(1)} + Z_{kk}^{(2)} + Z_{kk}^{(0)} + 3 Z_f}$$ $$\text{Line-to-Line (L-L) Fault at Bus } k: \quad |I_{f,k,LL}| = \frac{\sqrt{3} |V_k^{(0)}|}{|Z_{kk}^{(1)} + Z_{kk}^{(2)} + Z_f|}$$ $$\text{Double Line-to-Ground (DLG) Fault at Bus } k: \quad I_{a1,k} = \frac{V_k^{(0)}}{Z_{kk}^{(1)} + \frac{Z_{kk}^{(2)} (Z_{kk}^{(0)} + 3 Z_f)}{Z_{kk}^{(2)} + Z_{kk}^{(0)} + 3 Z_f}}$$

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5. Step-by-Step Worked Mathematical Example

Problem Statement

A 3-bus, $60\text{ Hz}$ power transmission system has a base of $100\text{ MVA}$ and nominal line-to-line voltage of $138\text{ kV}$ ($I_{base} = 418.37\text{ A}$). The positive-sequence and zero-sequence bus impedance matrices ($Z_{bus}^{(1)}$ and $Z_{bus}^{(0)}$) in per-unit are:

$$Z_{bus}^{(1)} = Z_{bus}^{(2)} = \begin{bmatrix} j0.12 & j0.08 & j0.04 \\ j0.08 & j0.20 & j0.10 \\ j0.04 & j0.10 & j0.16 \end{bmatrix} \quad [\text{pu}], \qquad Z_{bus}^{(0)} = \begin{bmatrix} j0.24 & j0.10 & j0.05 \\ j0.10 & j0.30 & j0.12 \\ j0.05 & j0.12 & j0.25 \end{bmatrix} \quad [\text{pu}]$$

Assume prefault voltages at all buses are $V_1^{(0)} = V_2^{(0)} = V_3^{(0)} = 1.0\angle 0^\circ\text{ pu}$.

A bolted short circuit occurs on Bus 2 ($Z_f = 0$).

Calculate:

1. The bolted Three-Phase fault current ($I_{f,2,3\phi}$) at Bus 2 in per-unit and physical Amperes.
2. The post-fault bus voltages ($V_1^{(f)}, V_2^{(f)}, V_3^{(f)}$) at all three buses during the 3-phase fault at Bus 2.
3. The fault current flowing in Transmission Line 1-2 during the Bus 2 fault, given the actual series branch line impedance is $z_{12,line} = j0.125\text{ pu}$.
4. The sequence currents ($I_{a1}, I_{a2}, I_{a0}$), total ground return current ($I_{ground}$), and phase currents ($I_b, I_c$) if the fault at Bus 2 is a bolted Double Line-to-Ground (DLG) fault.

=========================================================================================
CALCULATION WORKFLOW & DETAILED STEP-BY-STEP SOLUTION:
=========================================================================================

Step 1: Three-Phase Symmetrical Fault Current at Bus 2
  The Thevenin driving-point impedance at Bus 2 is the diagonal element Z_22^(1):
    Z_th,2 = Z_22^(1) = j0.20 pu

  Fault Current:
    I_f,2 = V_2^(0) / Z_22^(1) = (1.0 /_ 0°) / (j0.20) = -j5.0000 pu = 5.0000 /_ -90° pu

  Actual Physical Current:
    I_base = (100 * 10^6 VA) / (sqrt(3) * 138,000 V) = 418.37 A
    I_f,2,actual = 5.0000 * 418.37 A = 2,091.85 A ≈ 2.09 kA rms

Step 2: Post-Fault Bus Voltages During Bus 2 Three-Phase Fault
  Using transfer impedances from column 2 of Z_bus^(1) (Z_12 = j0.08, Z_22 = j0.20, Z_32 = j0.10):

  Bus 1 Post-Fault Voltage:
    V_1^(f) = V_1^(0) - Z_12^(1) * I_f,2
            = 1.0 - (j0.08) * (-j5.0000)
            = 1.0 - (0.4000) = 0.6000 /_ 0° pu

  Bus 2 Post-Fault Voltage (Faulted Bus):
    V_2^(f) = V_2^(0) - Z_22^(1) * I_f,2
            = 1.0 - (j0.20) * (-j5.0000)
            = 1.0 - (1.0000) = 0.0000 pu (Zero voltage at bolted fault)

  Bus 3 Post-Fault Voltage:
    V_3^(f) = V_3^(0) - Z_32^(1) * I_f,2
            = 1.0 - (j0.10) * (-j5.0000)
            = 1.0 - (0.5000) = 0.5000 /_ 0° pu

Step 3: Line 1-2 Current Flow During Fault at Bus 2
    I_12^(f) = (V_1^(f) - V_2^(f)) / z_12,line
             = (0.6000 - 0.0000) / (j0.125)
             = 0.6000 / (j0.125) = -j4.8000 pu = 4.8000 /_ -90° pu

  Actual Current in Line 1-2:
    I_12,actual = 4.8000 * 418.37 A = 2,008.18 A ≈ 2.01 kA rms

Step 4: Double Line-to-Ground (DLG) Bolted Fault at Bus 2
  Driving point impedances at Bus 2:
    Z_22^(1) = j0.20 pu
    Z_22^(2) = j0.20 pu
    Z_22^(0) = j0.30 pu

  Parallel combination of Negative and Zero sequence branches:
    Z_parallel = (Z_22^(2) * Z_22^(0)) / (Z_22^(2) + Z_22^(0))
               = (j0.20 * j0.30) / (j0.20 + j0.30)
               = (-0.0600) / (j0.5000) = j0.1200 pu

  Positive-Sequence Current:
    I_a1 = V_2^(0) / (Z_22^(1) + Z_parallel)
         = (1.0 /_ 0°) / (j0.20 + j0.120)
         = 1.0 / (j0.3200) = -j3.1250 pu = 3.1250 /_ -90° pu

  Negative-Sequence Current (Current Division):
    I_a2 = -I_a1 * [ Z_22^(0) / (Z_22^(2) + Z_22^(0)) ]
         = -(-j3.1250) * [ j0.30 / j0.50 ]
         = +j3.1250 * 0.6000 = +j1.8750 pu = 1.8750 /_ 90° pu

  Zero-Sequence Current (Current Division):
    I_a0 = -I_a1 * [ Z_22^(2) / (Z_22^(2) + Z_22^(0)) ]
         = -(-j3.1250) * [ j0.20 / j0.50 ]
         = +j3.1250 * 0.4000 = +j1.2500 pu = 1.2500 /_ 90° pu

  Check Sum of Sequence Currents: I_a = I_a0 + I_a1 + I_a2 = j1.250 - j3.125 + j1.875 = 0.00 pu (CONFIRMED)

  Total Ground Return Current:
    I_ground = 3 * I_a0 = 3 * (+j1.2500 pu) = +j3.7500 pu = 3.7500 /_ 90° pu
    I_ground,actual = 3.7500 * 418.37 A = 1,568.89 A ≈ 1.57 kA rms

  Phase Fault Currents (I_b, I_c):
    I_b = I_a0 + a^2 * I_a1 + a * I_a2
        = j1.2500 + (-j3.1250 /_ 240°) + (j1.8750 /_ 120°)
        = j1.2500 + (3.1250 /_ 150°) + (1.8750 /_ 210°)
        = j1.2500 + (-2.70633 + j1.56250) + (-1.62380 - j0.93750)
        = -4.33013 + j1.87500 pu
    |I_b| = sqrt( (-4.33013)^2 + (1.87500)^2 ) = sqrt( 18.7500 + 3.5156 ) = 4.7186 pu
    I_b,actual = 4.7186 * 418.37 A = 1,974.13 A ≈ 1.97 kA rms

    I_c = I_a0 + a * I_a1 + a^2 * I_a2 = +4.33013 + j1.87500 pu
    |I_c| = 4.7186 pu (1,974.13 A ≈ 1.97 kA rms)

  Check Ground Current: I_b + I_c = (-4.33013 + j1.87500) + (4.33013 + j1.87500) = +j3.7500 pu = I_ground (CONFIRMED)
=========================================================================================

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6. Common PE Exam Traps & Tactical Pitfalls

• Confusing Diagonal ($Z_{kk}$) and Off-Diagonal ($Z_{ik}$) Roles: Using $Z_{ik}$ to compute the fault current at Bus $k$. The fault current at Bus $k$ depends strictly on the driving-point impedance $Z_{kk}$. The off-diagonal element $Z_{ik}$ is used exclusively to compute the voltage dip at remote Bus $i$ caused by the fault at Bus $k$.
• Using Line Impedance Instead of $Z_{kk}$ for Bus Faults: Calculating fault current at a bus using the series impedance of the transmission line connected to it. The bus fault current is governed by the entire system Thevenin equivalent $Z_{kk}$, which accounts for all parallel paths and generation sources across the entire network.
• Sign Error in Post-Fault Bus Voltage Calculation: Writing $V_i^{(f)} = V_i^{(0)} + Z_{ik} I_{f,k}$. Because fault current flows out of the system into the short circuit, the voltage response is a subtraction: $V_i^{(f)} = V_i^{(0)} - Z_{ik} I_{f,k}$.
• DLG Sequence Current Inversion: Forgetting the negative sign in the current division formulas for negative and zero sequence currents ($I_{a2} = -I_{a1} \frac{Z_0}{Z_2 + Z_0}$). Because $I_{a1}$ flows into the parallel network, $I_{a2}$ and $I_{a0}$ flow out, requiring an explicit negative sign to satisfy $I_{a0} + I_{a1} + I_{a2} = 0$.