0
\$\begingroup\$

I'm studying Fault Analysis Using Symmetrical Components and I have doubt over the single Phase to Ground Fault. Single Phase to Ground Fault

On the book is written that generally Is and It are equal to 0 during a single-phase fault ground, and I really don't get the reason behind this statement.

The line prior to the fault event is supposed to be symmetric and connected to a load, therefore after the fault I don't get why Is and It should be equal to 0. I would say that Is and It are affected by the fault but not necessary are equal to 0 and their value depends on the characteristic of the load (i.e if the load is grounded or not).

Furthermore, if we suppose that the system is not grounded (So R is very large), how does the fault evolve?

Hope that my questions are clear. Have a good day!

\$\endgroup\$
4
  • 1
    \$\begingroup\$ As per your diagram Is and It must be zero. \$\endgroup\$
    – Andy aka
    Jan 12, 2018 at 15:55
  • \$\begingroup\$ Thank you for replying, the problem is the my diagram is modelling an overhead line with a fault. So I don't know why in the diagram seems that no current flowing in It and Is. \$\endgroup\$
    – Tommaso Bendinelli
    Jan 12, 2018 at 16:01
  • 2
    \$\begingroup\$ There is nothing in the diagram that shows a load hence those currents are zero. \$\endgroup\$
    – Andy aka
    Jan 12, 2018 at 16:02
  • \$\begingroup\$ Maybe I dind't explain myself well, let's say there is a substation that feeds the overhead line and at end of it there is a load. if in normal condition the three phase are symmetric, then why when there is a fault we can just neglect the current through the unfaulted phase (Thévenin equivalent of unfaulted phase current I equal to 0) ?? \$\endgroup\$
    – Tommaso Bendinelli
    Jan 13, 2018 at 10:52

2 Answers 2

Reset to default
1
\$\begingroup\$

When earth fault is analyzed, you are looking on currents that are flowing to ground. So, you are analyzing fault current of phase R to ground, which is nonzero, because of fault. On other two phases, current to GROUND is zero (if we ignore capacitive currents) but current to customer is nonzero.

\$\endgroup\$
0
\$\begingroup\$

If the system is not grounded (isolated neutral) there is actually no short circuit current flowing during an earth fault. Because there is no return conductor to the starpoint available. However, a small capacitive earth fault current will flow through the earth capacitances.

One advantage of this kind of neutral point treatment is, that grids with isolated neutral point can be kept in operation even though a single-phase earth fault occurred.

Furthermore for small currents the fault will be self-exstingushing.

During an earth fault the voltages on the healthy phases will be increased by a factor of √3 which can lead to further earth faults (risk of double earth fault).

Usually this kind of neutral point treatment is used for small networks. Because the earth capacitance will increase with the increase of the power system. As a result, also the earth fault current will increase.

Network with isolated neutralpoint and an earth fault

Network with isolated neutralpoint and an earth fault

Source of the picture: http://www.yourelectrichome.com/2011/09/underground-or-isolated-eutral-system.html

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.