Assymetric Current - PEAK & RMS

[mrelectric]

Can anyone pls tell me how the following formula for assymetric current (rms) and (peak)from a power system manual has been obtained.

I asym (RMS) = I"k * sqroot (1+2e^[-4(3.14)(R/X)C]
I asym (peak) = I"k * sqroot(2) * (1+e^[-2(3.14)(R/X)C]

where C = cycle
I"k - initial symmetrical fault current
R/X - Complex or separate R/X ratio (ANSI)

I understand that the transient characteristic of a circuit at fault contains a forcing function(steady state) and a natural response function(dc component). I don't know how to obtain above formula using this concept.

Also, I have been doing waveform simulation using a SPICE software and have been observing that the instantaneous fault current is at MAX when the phase angle is at 90 degrees. Was the phase angle also taken into account when using the above formulas?

Many thanks.

 

[jbartos]

Suggestion: Please, would you elaborate on the brackets:
I asym (RMS) = I"k * sqroot (1+2e^[-4(3.14)(R/X)C]
I asym (peak) = I"k * sqroot(2) * (1+e^[-2(3.14)(R/X)C]
Compare to ANSI standards, e.g. ANSI/IEEE C37.13-1990 multiplying factors MP:
1. For unfused circuit breaker based on highest peak current:
MF=sqroot(2) * (1+e^[-(3.14)(R/X)])/2.29
2. For fused circuit breaker based on total rms current (asymmetrical):
MF=sqroot (1+2e^[-2(3.14)(R/X)])/1.25
Also, perform the above Advanced Search in this Forum

 

[mrelectric]

JBartos,

I obtained above equation from SKM Power Tools (DAPPER Comprehensive Short Circuit Reference and A_Fault Reference). It is a general equation which lets you compute for the RMS and PEAK current at any cycle or time frame.

For the 1st 1/2 cycle fault (where highest peak occurs), c = 1/2 and above formula becomes:

I asym (RMS@1/2) = I"k * sqroot (1+2e^[-2(3.14)(R/X)]
I asym (peak@1/2)= I"k * sqroot(2)*(1+e^[-(3.14)(R/X)]

The results are used to compare to the Asymmetric rating of the interrupting device (which can be calculated based symmetric rating and on the test X/R ratio of the fuse or breaker).

Compared to your formula,

1. For unfused circuit breaker based on highest peak current:
MF=sqroot(2) * (1+e^[-(3.14)(R/X)])/2.29
2. For fused circuit breaker based on total rms current (asymmetrical):
MF=sqroot (1+2e^[-2(3.14)(R/X)])/1.25

the MF are almost the same except for the divisors which I could not account for(2.29 & 1.25). My guess is that those numbers are connected to the TEST X/R ratio of the circuit breakers or fuse used in the standard. Just guessing....

Anyway could you please clarify if the formula you gave are used to compute for the asymetric current? Thanks....

 

[jbartos]

Comments: I noticed that you used the equations from SKM. There are two sets of them.
The first set pertains to the medium voltage systems, where circuit breakers have certain number of cycles, c, e.g. 2 cycles, 3 cycles, 5 cycles, 8 cycles, and more cycles, to open the faulty circuit.
The second set pertains to low voltage circuit breakers, e.g. molded case circuit breakers that are typically using c=1/2.
Generally,
Calculated Symmetrical Current x MF <= Circuit Breaker Interrupting Ratings
1.
1.25 denominator constant is derived for the circuit breaker tested for value of X/R=4.6
sqrt(1+2xe^(-2xpix(1/4.6)))~1.25
2.
2.29 denominator constant is derived for the circuit breaker tested for value of X/R=6.6
sqrt2 x [1+e^(pix(1/6.6))]~2.29