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Topic Title: Fault Currents
Topic Summary: Are the traditional equations correct?
Created On: 13 December 2010 12:14 PM
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 13 December 2010 12:14 PM
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williamjohn

Posts: 178
Joined: 22 November 2010

Please may I have the views of other engineers on fault calculations.

Traditional equations for the fault current are;
Id = Id" exp(-t/Td") + Id' exp(-t/Td') + Isync
where Td" = Tds" Xd'(Xd" + X)/[Xd"(Xd' + X)]
Td' = Tds' Xd(Xd' + X)/[Xd'(Xd + X)]
Id" = E/(Xd" + X) - E/(Xd' + X)
Id' = E/(Xd' + X) - E/(Xd + X)

First problem
Td" and Td' by these equations do not always come in the range Tds" - Tdo" and Tds' - Tdo'.
Putting X very large gives relations between Tdo", Tds", Tdo', Tds' and the reactances. All these parameters can be measured and in my experience manufacturer's figures do not agree with these relationships.

The fault current links with the flux and in effect has a positive feedback that speeds up the decay. It therefore seems logical that Td" and Td' are between their respective Tdo and Tds in proportion to the (armature current resolved along the pole axis)/(armature current with X zero).

Second problem
(Id"/Id' when X large) = Xd"/Xd times (Id"/Id when X is zero)

Initial values of the current induced in the damper winding and the current induced in the main field have the same ratio for all values of X. This ratio depends on their turns linkage and impedances which do not depend on X. Id"/Id' depend on the interaction of these currents. It seems logical that Id"/Id' should not depend on X.

Third problem
Xd" = leakage reactance and is a real reactance. Xd' on the other hand is merely a convenient way to define Id' in the particulr case where X = 0. I cannot see any reason why Xd' should be the same for any other value of X.

Fourth problem
Some faults have resistance in the impedance. This must affect the time constants.

If my logic is correct, then a bit of maths comes up with;
Td" = Tdo" Tds" /[Tds" + (Tdo" - Tds") Xd" cos(a) /(Z + Xd")]
Td' = Tdo' Tds' /[Tds' + (Tdo' - Tds') Xd' cos(a) /(Z + Xd')]
Tq" = Tqo" Tqs" /[Tqs" + (Tqo" - Tqs")Xq" sin(a) /(Z + Xq")]
where Z is the fault impedance including resistance and a is the phase angle between the fault current and the pole axis.

Id" and Id' can be evaluated as;
mmf due to Id" + mmf due to Id' = change in direct axis mmf due to the armature from the on load value to (on load + fault current) value.
and Id" /Id' = (Xd' - Xd") Xd /[(Xd - Xd') Xd"]
Similarly mmf due to Iq" = change in quadrature axis mmf from the on load value to the (on load + fault) value.

Is my logic correct?
Are the tradional equations really wrong or where is the error in my logic.

Does no one have any comment on this posting. It is a serious issue. If the traditional equations are indeed wrong, then the actual fault current may decay more rapidly than the value calculated by traditional methods. In this case there is the possibility that the fault may fall below the overload setting and not be cleared.

Best regards
John

Edited: 12 January 2011 at 11:14 PM by williamjohn
IET » Energy » Fault Currents

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