X, R, for utility transformers

[Peter John]

Hello, I'm trying to perform fault current calculations for a commercial building in Kirkland, WA. The Utility transformer is a 150KVA, 12.5KV primary, 208V 3 phase secondary system with a 26,900 fault current rating. PSE(the electrical utility) provides the R/X ratio .3 and %Z 1.55 but per my fault current calculation formula I need the individual X & R values.

Any help would be appreciated, thank you!

 

[jim dungar]

Hello, I'm trying to perform fault current calculations for a commercial building in Kirkland, WA. The Utility transformer is a 150KVA, 12.5KV primary, 208V 3 phase secondary system with a 26,900 fault current rating. PSE(the electrical utility) provides the R/X ratio .3 and %Z 1.55 but per my fault current calculation formula I need the individual X & R values.

Any help would be appreciated, thank you!

What software are you using? Most packages will accept X/R and %Z. I can't remember the last time I entered actual X and R values.

 

[Peter John]

Hi Jim, I'm trying to do the calculations manually as I run a small electrical shop, but if there is a good program you know of to do them I would love more info!

Thanks

 

[jim dungar]

Hi Jim, I'm trying to do the calculations manually as I run a small electrical shop, but if there is a good program you know of to do them I would love more info!

Thanks

Have you searched the internet? For example:

The Importance of the X/R Ratio in Short Circuit Calculations

The X/R ratio can affect how two impedances are added together. The X/R defines the impedance angle and should be considered for short circuit calculations

brainfiller.com

 

[dkidd]

Θ = Arctan X/R
X = Sin θ * Z
R = Cos θ * Z or X / (X/R)



Transformer R and X
%Z = 5.75 and X/R = 7
Inverse Tangent 7.00 = 81.8699°
X = Sin 81.8699* 5.75 % = 0.9899 * 5.75 % = 5.6919%
R = Cos 81.8699* 5.75 % = 0.1414 * 5.75 % = 0.8131%

The Importance of the X/R Ratio in Short Circuit Calculations

The X/R ratio can affect how two impedances are added together. The X/R defines the impedance angle and should be considered for short circuit calculations

brainfiller.com

 

[electrofelon]

Hello, I'm trying to perform fault current calculations for a commercial building in Kirkland, WA. The Utility transformer is a 150KVA, 12.5KV primary, 208V 3 phase secondary system with a 26,900 fault current rating. PSE(the electrical utility) provides the R/X ratio .3 and %Z 1.55 but per my fault current calculation formula I need the individual X & R values.

Any help would be appreciated, thank you!

What exactly are you trying to do? If you are given the fault current of 26,900, what more do you need? I have the old mike holt fault current calculator (I think its no longer on his "free stuff" page and hard to find) and it doesnt ask for XR but it gives 29,847 using 90% PF. Whether its 26,900 or 29,847 does that really make any difference?

 

[wbdvt]

First off, understand that the fault current you cite is the infinite bus fault current from the 150kVA transformer. This is adequate to use for sizing the SCCR and AIC for equipment but is totally wrong to use for any arc flash calculations.

Not sure what fault current calculations you need but seems like the first equipment downstream from the transformer would need to be sized greater than 27kA and then I would suspect the next level down you could go with 22kA rated equipment but your modeling will tell you for sure.