Unsymmetrical faults are more common in power systems than symmetrical faults. Understanding how to solve numerical problems related to these faults is crucial for analyzing power system behavior under abnormal conditions and ensuring system stability. This module focuses on the application of symmetrical components to solve these problems.

Unsymmetrical faults involve unequal currents and voltages in the three phases. The most common types are single line-to-ground (SLG) faults, line-to-line (LL) faults, and double line-to-ground (DLG) faults. These are analyzed using the method of symmetrical components, which decomposes unbalanced three-phase systems into three balanced sets of positive, negative, and zero sequence components.

The core principle is to represent unbalanced phase quantities (currents $I_a, I_b, I_c$ and voltages $V_a, V_b, V_c$) in terms of their symmetrical components: positive sequence ($V_a1, V_b1, V_c1$), negative sequence ($V_a2, V_b2, V_c2$), and zero sequence ($V_a0, V_b0, V_c0$). The transformation matrix is key:

$V_a = V_{a0} + V_{a1} + V_{a2}$ $V_b = V_{a0} + aV_{a1} + a^2V_{a2}$ $V_c = V_{a0} + a^2V_{a1} + aV_{a2}$

where $a = e^{j120^\circ}$ and $a^2 = e^{j240^\circ}$.

An SLG fault occurs when one phase conductor is connected to ground. The conditions at the fault point are: $I_b = I_c = 0$, $V_a = 0$. To solve this, the positive, negative, and zero sequence networks are connected in series. The fault current $I_f$ is equal to the positive sequence current at the fault point, $I_{f} = I_{a1} = I_{a2} = I_{a0}$. The fault current is given by $I_f = \frac{V_{a1}}{Z_1 + Z_2 + Z_0}$.

An LL fault occurs between two phase conductors, say phase 'b' and phase 'c'. The conditions at the fault point are: $I_a = 0$, $V_b = V_c$, $I_b = -I_c$. For this fault, the positive and negative sequence networks are connected in parallel, and the zero sequence network is not involved. The fault current is $I_f = I_{bc} = \frac{V_{a1}}{Z_1 + Z_2}$.

A DLG fault occurs when two phase conductors are connected to ground, say phases 'b' and 'c'. The conditions at the fault point are: $I_a = 0$, $V_b = 0$, $V_c = 0$. For this fault, the positive, negative, and zero sequence networks are connected in parallel. The fault current is $I_f = I_{b} = -I_{c} = \frac{V_{a1}}{Z_1 + \frac{Z_2 Z_0}{Z_2 + Z_0}}$.

To solve numerical problems:

1. Identify the fault type: SLG, LL, or DLG.
2. Draw the sequence networks: Positive, negative, and zero sequence networks for the given power system.
3. Determine sequence impedances: Calculate or obtain $Z_1, Z_2, Z_0$ for all components (generators, transformers, lines).
4. Connect the networks: Connect the sequence networks at the fault location according to the fault type.
5. Calculate fault current: Use the appropriate formula based on the network connection.
6. Calculate phase currents and voltages: Use the symmetrical component transformation to find the actual phase values.

Consider a simple power system with a generator connected to a transformer and a transmission line. If a double line-to-ground fault occurs at the end of the line, we would first calculate the positive, negative, and zero sequence impedances of each component. Then, we would sum these impedances appropriately for the parallel connection of the sequence networks to find the fault current. Finally, we would use the symmetrical component transformation to find the phase currents $I_a, I_b, I_c$.