AS3851 Unable to calculate X/R ratio

[silkwormfrog]

Hi

I have trying to calculate the fault level (ground fault) by using AS3851 method.

In section 8.5.2.1, the factor k for the R/X and X/R ratios can be obtained using
k = 1.02 + 0.98 * e ^(-3R/X).

I have been given the following:
the source 3 phase max. fault level 3800A
the source 3 phase max. fault level 3600A

the source 1 phase max. fault level 930A.
the source 1 phase max. fault level 910A

So for 3 phase fault, k = 0.746.
The problem is I cannot obtain R/X ratio using k = 1.02 + 0.98 * e ^(-3R/X) due to obvious reasons.

Is there anyway to get around this problem. Please note I need to follow AS3851 standard. Many thanks

 

[ijl]

Could it be that you have got it the wrong way around? k is needed for the peak current of 3-phase fault. k is calculated using the equation you refer to. You need R/X for k. In IEC-world the corresponding standard gives three ways to calculate R/X. I guess the same applies to AS-world. In addition k = 0.746 does not look correct.

 

[healyx]

Reading between the lines - are you saying the peak (Ip) current is 3800A and rms current (Ik") is 3600A? If so, that peak current is way too low. I think it is probably out by a factor of sqrt(2). If so, the real peak current would be 5374A and X/R would be 4.11.

This is close to what would be expected as an X/R ratio at a final distribution transformer (0.4kV), although those currents look a bit low for that.

 

[silkwormfrog]

Thanks for the comments.

The fault level is given by the supply authority. And this is a 22kV distribution network.

 

[sibeen]

What are those fault levels supposed to mean?

Is the first one the peak?

I really don't understand what you're asking. Section 8.5 of AS3851 is written so you can work out the peak current (i_p).

 

[EddyWirbelstrom]

Utility max and min source positive, negative, zero sequence impedances are required to determine the Utility contribution to a short-circuit at the Utility connection.
If the Utility only provides short-circuit current contribution, then the following is required:
- Max and Min PPP and PE short-circuit contributions from the Utility ( preferably in amps ),
- X/R for PPP = X1/R1,
- X/R for PE = X1+X2+Xo / R1+R2+Ro.
- Was the Utility contribution calculated using the voltage 'c' Factor ?