IEEE 605 Fault Current for force calculations

[rockman7892]

When performing basic force equation calculations per IEEE605 is it typically the RMS symmetrical fault current only that is used for force equation calculations on rigid bus?

In looking at the standard it appears there is a somewhat conservative force equation that uses the RMS symmetrical value only but then there are some "less conservative" equations that make use of a decrement factor and components that are associated with asymmetrical portion of fault current.

I was wondering what the standard approach was for use of symmetrical vs asymmetrical fault current and corresponding equations in the standard. Is there one that is more typically applied that the other?

Is there any "rules of thumb" where below a certain fault current threshold the fault current will be too low to have any impact on bus forces and thus can be considered negligible for such evaluations?

 

[waross]

Force is proportional to current.
Using RMS current will result in RMS forces, which may not be useful.
Using peak asymmetric current will result in peak forces.
It may be a peak force that breaks things.
A fully offset asymmetric current and the resulting forces may approach 2 times root 2 of the RMS value or 2.828 times the RMS value.
The difference is not trivial nor may conservative be a correct description of RMS versus peak.
Conservative to me means safe.
Using fully offset values in place of actual offset values is, to me, conservative, or safe.
Using RMS values in place of fully offset or actual offset values to calculate forces may be dangerously unsafe.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[7anoter4]

See chapter 11.3.3 Simplified calculations for short circuit loads on rigid buses 11.3.3.1 Basic force equation between infinitely long parallel conductors:
The equation for the force between parallel, infinitely long conductors in a flat configuration due to a fully asymmetrical short circuit current is as follows: [eq.14 or eq.15]
However, in these equations Isc is the symmetrical RMS fault current, A ,but using the factor Γ the fully asymmetrical short circuit factor is included.

 

[cuky2000]

 

[iop95]

Maybe need to add an additional correction factor if operate at (very) low temperature where resistances are lower (higher Isc).

 

[cuky2000]

IOP95, In high-voltage substation application, the bus resistance is not a significant factor in determining the available short circuit current. The SC current tends to be the dominant factor in determining the bus load forces.

 

[iop95]

Yes, but in some special conditions (SC at short time after power-up a cold transformer) may gain some significance.

 

[cuky2000]

No sure about if special low temperature conditions will impact significantly the SC current level. Even in cryogenic application, with superconductor at very low temperature, the bus resistance is not a significant factor. In HV systems, the impedance is dominated by the reactance witch is much larger than the resistance in all cases including the one for power sources such as transformer and generators.