Unbalanced fault analysis is the ultimate filter in power system engineering exams and real-world relay coordination studies. While balanced three-phase faults are straightforward, Single Line-to-Ground (SLG) faults require Fortescue’s Theorem to decouple the system into positive, negative, and zero sequence networks. The most common point of failure for students and junior engineers isn't the basic algebra—it's misinterpreting transformer winding connections in the zero-sequence network.

This guide walks through a classic, high-yield symmetrical components short circuit calculation involving a step-up transformer. We will build the sequence networks, expose the most common exam trap, and verify the results using boundary conditions.

Problem Statement & System Parameters

Exam Problem: A 100 MVA, 13.8 kV generator is connected to a 13.8/69 kV step-up transformer. The transformer has a Yg-Delta connection (solidly grounded Wye on the 13.8 kV side, Delta on the 69 kV side). A solid Single Line-to-Ground (SLG) fault occurs on Phase A at the 13.8 kV generator bus. Assume the pre-fault voltage is 1.0 pu. Calculate the subtransient fault current in Amperes. Identify the zero-sequence network topology.

To solve this, we first establish our per-unit base values and gather the equipment sequence impedances. We will use the generator ratings as the system base.

Component Rating / Base X1 (Positive) X2 (Negative) X0 (Zero)
System Base 100 MVA, 13.8 kV - - -
Generator (Bus 1) 100 MVA, 13.8 kV, Yg j0.15 pu j0.15 pu j0.05 pu
Transformer (Bus 1-2) 100 MVA, 13.8/69 kV, Yg-Δ j0.10 pu j0.10 pu j0.10 pu
69kV Transmission Line 69 kV (Base adjusted) j0.20 pu j0.20 pu j0.60 pu

Step-by-Step Algebraic Solution & The Delta Trap

Which method applies and why? We use Fortescue’s Theorem (Symmetrical Components) because an SLG fault is an unbalanced condition. Symmetrical components allow us to decouple the unbalanced three-phase system into three independent, balanced sequence networks (Positive, Negative, Zero) that are connected in series at the fault point for an SLG fault.

⚠️ The Trap in This Problem: Most students look at the zero-sequence network and simply add the Generator X0, Transformer X0, and Line X0 in series. This is incorrect. The fault is on the 13.8 kV bus. The transformer is connected Yg on the 13.8 kV side and Delta on the 69 kV side. Zero-sequence currents require a ground return path. While current can flow from the fault into the grounded Yg winding of the transformer, the Delta winding on the 69 kV side acts as an open circuit to zero-sequence line currents. The zero-sequence current circulates inside the Delta but cannot pass through to the 69 kV transmission line. Therefore, from the perspective of the 13.8 kV fault bus, the transformer's zero-sequence impedance (X0_xfmr) is in parallel with the generator's zero-sequence impedance (X0_gen), and the line impedance is completely excluded from the zero-sequence network.

Let's calculate the Thevenin equivalent sequence impedances at the fault bus (Bus 1).

Step 1: Positive and Negative Sequence Networks
For the positive and negative sequences, the transformer does not block current, but since there is no source on the 69 kV side (assuming a radial system feeding into an infinite bus or simply looking at the generator contribution for this specific subtransient snapshot), the sequence impedances are just the generator's subtransient reactances.
Z1 = X1_gen = j0.15 pu
Z2 = X2_gen = j0.15 pu

Step 2: Zero Sequence Network (Applying the Trap Fix)
The zero-sequence current from the fault splits: some flows through the generator neutral, and some flows into the transformer's grounded Yg neutral. Because the Delta winding blocks the current from flowing further, the transformer X0 acts as a shunt path to ground.
Z0 = X0_gen || X0_xfmr
Z0 = (j0.05 * j0.10) / (j0.05 + j0.10)
Z0 = -0.005 / j0.15 = j0.03333 pu

Step 3: Calculate Sequence Fault Currents
For a solid SLG fault on Phase A, the sequence networks are connected in series. The sequence currents are equal:
I_a1 = I_a2 = I_a0 = V_f / (Z1 + Z2 + Z0)
Assuming pre-fault voltage V_f = 1.0 ∠0° pu:
I_a1 = 1.0 / (j0.15 + j0.15 + j0.03333)
I_a1 = 1.0 / j0.33333 = -j3.0 pu

Step 4: Calculate Total Phase Fault Current in Per-Unit
I_a = I_a1 + I_a2 + I_a0
I_a = 3 * I_a0 = 3 * (-j3.0) = -j9.0 pu

Step 5: Convert to Actual Amperes
First, find the base current at the 13.8 kV fault bus:
I_base = S_base / (√3 * V_base)
I_base = 100,000,000 VA / (1.73205 * 13,800 V) = 4,183.7 A
Now, multiply by the per-unit fault current:
I_fault_actual = 9.0 pu * 4,183.7 A = 37,653 A (or 37.65 kA)

Sanity Checks, Verification & Exam FAQ

Never hand in an exam or submit a relay coordination report without running a sanity check. Here is how to verify your answer independently.

1. Order of Magnitude & Units Check
A 100 MVA generator at 13.8 kV typically has a short circuit capacity yielding fault currents in the 30 kA to 50 kA range. Our answer of 37.65 kA fits perfectly within standard 13.8 kV switchgear ratings (usually 40 kA or 50 kA). If your calculation yielded 376 kA, you likely forgot to divide by √3 when calculating the base current. If you got 12.5 kA, you incorrectly added Z0 in series (j0.05 + j0.10 = j0.15) instead of parallel.

2. Independent Verification via Boundary Conditions
The fundamental boundary condition for a solid Phase-A-to-Ground fault is that the voltage at the fault point must be zero (V_a = 0). Let's prove our math holds up by calculating the sequence voltages at the fault bus and summing them.

  • Positive Sequence Voltage: V_a1 = V_f - (I_a1 * Z1) = 1.0 - (-j3.0 * j0.15) = 1.0 - 0.45 = 0.55 pu
  • Negative Sequence Voltage: V_a2 = 0 - (I_a2 * Z2) = 0 - (-j3.0 * j0.15) = -0.45 pu
  • Zero Sequence Voltage: V_a0 = 0 - (I_a0 * Z0) = 0 - (-j3.0 * j0.03333) = -0.10 pu

Summing the sequence voltages:
V_a = V_a1 + V_a2 + V_a0 = 0.55 - 0.45 - 0.10 = 0.0 pu.
The boundary condition holds perfectly. Your network topology and algebra are verified.

Frequently Asked Questions: Symmetrical Component Exam Traps

Q: What if the transformer was Yg-Yg instead of Yg-Delta?
A: If both neutrals are solidly grounded, the zero-sequence current can pass completely through the transformer. In that case, the 69 kV line's X0 and any connected grid sources would be included in the zero-sequence network, typically in series with the transformer X0, and then placed in parallel with the generator X0 depending on the exact fault location.

Q: Does the Delta winding contribute to the fault at all?
A: Yes, but internally. The zero-sequence flux induced by the Yg winding currents creates a circulating zero-sequence current inside the closed Delta loop. This provides the necessary ampere-turn balance for the transformer core, which is why the transformer X0 appears as a finite shunt impedance rather than an open circuit in the zero-sequence network.

Q: How do I verify this in industry practice?
A: In professional settings, engineers use software like ETAP, SKM PTI, or ASPEN OneLiner. However, understanding the manual sequence network topology is critical because software will happily simulate an incorrectly grounded transformer if the user inputs the wrong winding connection code. For deeper study on protective relaying applications using these components, refer to technical resources from Schweitzer Engineering Laboratories and the IEEE Power and Energy Society.