This module focuses on solving numerical problems related to symmetrical faults in power systems, a crucial topic for the GATE Electrical Engineering exam. We will cover the fundamental concepts and apply them to practical scenarios.

Symmetrical faults, also known as balanced three-phase faults, are the most severe type of fault and occur when all three phases are short-circuited simultaneously. They are relatively rare but important for system design and stability analysis. The analysis assumes balanced conditions, meaning positive, negative, and zero sequence impedances are equal.

To solve numerical problems, we need to understand several key concepts:

• Per Unit System: Simplifies calculations by normalizing system parameters.
• Impedance Diagram: A simplified representation of the power system network using impedances.
• Sequence Impedances: Positive sequence impedance ($Z_1$), negative sequence impedance ($Z_2$), and zero sequence impedance ($Z_0$). For symmetrical faults, $Z_1 = Z_2 = Z_0$ is often assumed for simplicity, though this is not strictly true for all components.
• Bus Impedance Matrix (Zbus): A matrix representing the driving point and transfer impedances between buses in a network.

The magnitude of the fault current ($I_f$) for a symmetrical three-phase fault at a bus 'k' is given by:

$I_f = \frac{V_{ph}}{Z_{eq}}$

where $V_{ph}$ is the pre-fault phase voltage at the fault location, and $Z_{eq}$ is the equivalent positive sequence impedance of the system looking from the fault point towards the source.

Let's consider a scenario: A generator with $X_g = 0.2$ pu is connected through a transformer with $X_t = 0.1$ pu to a transmission line with $X_l = 0.3$ pu. A symmetrical fault occurs at the end of the line. The system base is 100 MVA and 13.8 kV for the generator. The transformer is rated 100 MVA, 13.8 kV/69 kV. The transmission line is rated for 69 kV.

<b>Step 1: Convert all impedances to a common base.</b> Base MVA = 100 MVA Generator base kV = 13.8 kV Transformer base kV = 13.8 kV / 69 kV Line base kV = 69 kV

All impedances are already in per unit on their respective bases, and the bases align for the fault location. Thus, the total equivalent impedance is: $Z_{eq} = X_g + X_t + X_l = 0.2 + 0.1 + 0.3 = 0.6$ pu

<b>Step 2: Calculate the fault current in per unit.</b> Assuming the pre-fault voltage at the fault location is 1.0 pu (infinite bus assumption): $I_f ( ext{pu}) = \frac{1.0 ext{ pu}}{Z_{eq} ( ext{pu})} = \frac{1.0}{0.6} = 1.667$ pu

<b>Step 3: Convert the fault current to amperes.</b> The base current for the high-voltage side (69 kV) is: $I_{base} = \frac{S_{base}}{\sqrt{3} imes V_{base}} = \frac{100 imes 10^6}{\sqrt{3} imes 69 imes 10^3} = 836.9$ A

Therefore, the fault current in amperes is: $I_f ( ext{A}) = I_f ( ext{pu}) imes I_{base} = 1.667 imes 836.9 \approx 1395$ A

For more complex networks, the Bus Impedance Matrix (Zbus) method is often employed. This method involves constructing the Zbus matrix for the positive sequence network and then directly calculating the fault current using the voltage at the faulted bus. The Zbus method is particularly useful when dealing with multiple fault locations or when the system configuration changes.