Utility Fault Current Calculations

[rancam]

I've searched through the forum and found some close examples but for some reason my values are not coming out. Here is what is provided by the utility:

Primary Voltage = 24kV P to P
Fault Current LLL = 4863
Fault Current LG = 4364
X1/R1 = 9.62
X0/R0 = 4.56

My end goal is to develop the inputs for SKM which requires:
Three Phase Current and X/R
Line to Ground Current and X/R

The three phase current and X/R are a given since X/R in a three phase only has the positive sequence impedance X1/R1.

Line to Ground current is given so all I have to do is calculate SLG X/R ratio, this is giving me fits.

This is what I've done:

I3sc=4863<-84deg

Z1=Z2=Vll/(sqrt(3)*I3sc)
Z1=Z2=24kV/(1.73*4863<-84deg)
Z1= 0.2946+2.834j

Islg=4364<-77.63deg
Zg=3*Vln/Islg=3*13856<0/4364<-77.63deg
Zg=2.04042+9.304j

Z0=Zg-2*Z1
Z0= 1.45122+ 3.6362j

and finally, I thought at least
SLG X/R=(x0+2*x1)/(r0+2*r1)
SLG X/R=4.42

and as you can see I just calculated in a circle my original X0/R0 value!

Any direction in what I am messing up here?

Thanks in advance for any help here.

 

[davidbeach]

I'm pretty sure that what SKM wants is X0/R0.

 

[submonkey]

Hi rancam,

You have correctly used the arctan of X1/R1 to predict the current angle for a three-phase fault. What you have done is valid because X1 and R1 are the total of the impedance in the equivalent circuit.

You can't do exactly the same thing to predict the current angle for a phase-earth fault. The impedance in the equivalent circuit is Z1 + Z2 + Z0, but you have assigned an angle based only on X0/R0.

The positive, negative and zero sequence networks are connected in series for an earth fault, so:

(4364/3) = (24kV/sqrt(3)) / abs(Z0 + Z1 + Z2)
abs(Z0 + Z1 + Z2) = 9.525

In summary:

1. magnitude of (Z0 + Z1 + Z2) = 9.525 (above)
2. Z1 = Z2 = 2.849 < 84.07 deg (you have calculated this already)
3. Angle of Z0 is atan(4.56) = 77.63 deg (from arctan of X0/R0)

If I sketch that out and solve it with trigonometry, I get something like:

Z0 = 3.840 < 77.63

Which means that for a SLG fault, the current
will be 4365 < -81.48 deg which suggests a thevenin X/R of 6.68.

Does that help?

Thanks,
Alan

 

[odlanor]

submonkey,
in expression: (4364/3) = (24kV/sqrt(3)) / abs(Z0 + Z1 + Z2
you are assuming IFG=3I0. This is only true with single grounding system
In multiple grounding system IFG= I1 + I2 + I0 = 2I1 + I0.

 

[rancam]

Okay, I understand what I did wrong. I didn't think the vector addition though for Zt=Z0+Z1+Z2. From the information I had I could calculate Z1 with angle, Z0 angle only, and magnitude of Zt.

I drew it out like you did and through wiki found the formula I needed to solve the triangle with (Z1+Z2), Z0, and Zt. The Law of Cosines worked to solve the Z0 magnitude:

a=b * cos(angle) + sqr(c^2 - b^2*(sin(angle))^2)
where a=|Z0|, b=(|Z1|+|Z2|), c=Zt and angle= angle between Z0 and Z1

Using this I calculated |Z0| = 3.84075 and with the X0/R0 ratio the angle is given at 77.63deg.

And finally what I was looking for:

SLG X/R = ((2*X1)+X0))/((2*R1)+R0)
SLG X/R = 6.64.

Thanks for the help submonky, you cleared up my conceptual error. I can't believe there isn't an easier method to calculate it. I will pass the formulas around to the other engineers to educate them also.

rancam

 

[dpc]

On the plus side, this value does not really have much impact on anything we are normally concerned about when it comes to the actual short circuit currents. The positive sequence X/R is much more important.

 

[rancam]

Interestingly the local utility says to calculate the SLG X/R from the following equation:

SLG X/R = {(2*x1/r1)+x0/r0}/3

When I use this I get 7.93 vice 6.67 from utilizing the vectors. It must use assumptions to come up with the formula.

 

[jghrist]

The local utility is just using the equivalent impedance of 2·Z1 + Z0 and assuming that the X/R ratios follow the same equation. As Alan's calculations show, the local utility is wrong.

 

[rancam]

That is what I was thinking. Once again a thanks to Alan and the rest of you for helping on this.