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Single Line to Ground Faults

Introduction - Single line to ground faults are the most common type of fault in a power system. On overhead transmission and distribution lines, these are often caused by a tree branch coming into contact with a power line and the fault is typically intermittent. On industrial systems, these kinds of faults are commonly caused by a breakdown in insulation on one conductor leading to an unintended pathway to ground. Regardless of the type of power system, single line to ground faults can be a huge problem.


Figure 1: A Single Line to Ground Fault Circuit Diagram



Why study Single Line to Ground Faults? - The interrupting rating of overcurrent protection devices and the withstand duty of equipment is generally referenced to symmetrical faults (bolted three phase faults). However, proper analysis of a power system requires a detailed look at single line to ground faults as well. If you need a refresher on bolted three phase faults, check out my article here.


The first reason why single line to ground fault analysis is important is because of grounding design. During a three phase bolted fault, fault current is carried only in the phase conductors; none travels in the ground (equipment or earth ground). On the other hand single line to ground fault current all flows through the earth return path, meaning it is generally the largest shock hazard for people. Ground grid and equipment grounding conductors need to have sufficient short circuit withstand capabilities in accordance with the single line to ground faults. Additionally, ground grids need to be sized with sufficient material and appropriate geometry to ensure the safety of personnel per IEEE 80.



Symmetrical Components - The single line to ground fault current is typically calculated using symmetrical components, an alternative representation of power systems for unbalanced fault analysis. A detailed explanation of symmetrical components won't be provided here, but for a quick refresher:

  • Positive Sequence Impedance is the impedance of a power system in response to normal phasor rotation voltage sources. Positive sequence impedance is the "normal" impedance of a system that is used for things like voltage drop calculations and symmetrical fault analysis. When we say "impedance" without any modifiers, we're almost always talking about positive sequence impedance.

  • Negative Sequence Impedance is the impedance of a power system in response to reverse phasor rotation voltage sources. For common transformers and cables, the positive sequence impedance is the same as the negative sequence impedance.

  • Zero Sequence Impedance is the impedance of a power system in response to the application of a single phase voltage source. Zero sequence impedance is unique because there is no concept of phase rotation.


Figure 2: Symmetrical Components Visualized



Calculating the Single Line to Ground Fault Current - Performing detailed analysis of symmetrical components by hand can be cumbersome. Typically, manual analysis is only concerned with simplified systems, usually involving a transformer, like the case in Figure 3. For more information on transformer impedances in the sequence domain, see my article here.




Figure 3: The Typical Infinite Bus Condition for Single Line to Ground Faults


Figure 4 is a representation of a single line to ground fault as viewed in the sequence domain. For this type of fault, the positive sequence, negative sequence, and zero sequence networks are all connected in series. For typical power systems, the only source voltage will be the positive-sequence (normal operating) voltage. The current that flows in the sequence domain network is the zero-sequence current, I0. The actual single line to ground fault current will be three times this value.



Figure 4: A Single Line to Ground Fault in the Sequence Domain


Mathematically, we can use the network above and our relationship between the zero sequence current and the single line to ground fault current to get a relationship between our system parameters and our fault current magnitude:


Where:

  • Z1 is the positive sequence impedance in Ohms

  • Z2 is the negative sequence impedance in Ohms

  • Z0 is the zero sequence impedance in Ohms, accounting for transformer winding configurations

  • VLN is the line-to-neutral voltage in Volts

  • ISLG is the single line to ground fault current magnitude in Amps


In the case of a single line to ground fault through a transformer like in Figure 3 (assuming no source impedance), all of these impedances refer to transformer values that can be tested during a factory acceptance test. The line to neutral voltage refers to the value on the faulted winding.


 

Example: Compute the symmetrical fault current and the single line to ground fault current for the power system below. Assume an infinite grid (no source impedance).


Solution: This is an easy one! Start by computing the symmetrical fault current (the bolted three phase fault current). We do that by using the positive sequence impedance:

Where:

  • S is the transformer apparent power rating in VA

  • VLL is the line to line voltage on the faulted winding in Volts

  • Z1 is the positive sequence impedance of the transformer in %

  • Ibf is the bolted three-phase fault current magnitude in Amps


Plugging in our values:


Ibf = 100 / (√3 * 34.5 * .1) = 16.7 kA


The single line to ground fault current is computed by first recognizing that the faulted winding is connected in a solidly grounded wye configuration. This means that ground fault current can flow on the secondary bus (unlike in a delta or ungrounded wye winding). To begin solving this problem, we need to convert transformer percent impedances into Ohms. This starts with identifying the base value of the impedance at the 34.5 kV side of the transformer:


Zbase = 34.5 ^ 2 / 100 = 11.9 Ohms


Then, convert the impedances to Ohms from percent:


Z1 = Z2 = .1 * 11.9 = 1.19 Ohms

Z0 = .08 * 11.9 = .952 Ohms


Lastly, we can apply our equation for single line to ground fault current:


ISLG = 3 * (34.5 / √3) / (1.19 + 1.19 + .952) = 17.9 kA


These are the final results of this calculation. Notice that the single line to ground fault current is actually higher than the three phase bolted fault current. This is due to the lower zero sequence impedance of the transformer. When a system is being designed, we have to consider the potential impacts of this single line to ground fault current on our equipment ratings and fault clearing plans.

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