X/R Calculation or %X

[gilbertomejiac]

Greetings

For a Short-Circuit calculation, the Utility provide me with the next data:

Voltage: 11 kV
Sk'': 248,20 MVAsc
Ik'': 13,027 < -86,24°


----

Phi-Ik'': -86,24°

---

With this data I want to calculate X/R value and %X. I don't have a Power Factor PF. I have short circuit magnutude and angle, and base voltage and short circuit power.

thanks.

 

[davidbeach]

It's just a tad bit of basic trig. If you have the angle and want the ratio of rise/run, what function would you use?

 

[Fuse4]

Define: Vk" = 11 kV < 0°
Sk" = SQRT(3)*Vk"*Ik" = 1.73*(11kV<0°)*(13,027kA<-86,24°)= 248,20MVA<-86,24°
Zk" = Vk"*Vk"/Sk" = (Rk"+jXk")Ω= 0.844Ω<86.24°=(0.055+j0.843)Ω
Rk" = 0.055 Ω
Xk" = 0.843 Ω
X/R = 15.327
%X = (0.843 Ω / 0.844Ω)*100% = 99.882%

 

[Jackreacher]

I think Zk" should be .48<-86.25

 

[Fuse4]

Jackreacher, was a mistake, thanks. You are with true. Here the values:

Define: Vk" = 11 kV < 0°
Sk" = SQRT(3)*Vk"*Ik" = 1.73*(11kV<0°)*(13,027kA<-86,24°)= 248,20MVA<-86,24°
Zk" = Vk"*Vk"/Sk" = (Rk"+jXk")Ω= 0.488Ω<86.24°=(0.032+j0.486)Ω
Rk" = 0.032 Ω
Xk" = 0.886 Ω
X/R = 15.216
%X = (0.486Ω/0.488Ω)*100% = 99.785%

 

[7anoter4]

In my opinion, per-unit[p.u.] calculation it is always possible since the base short-circuit apparent power is arbitrarily chosen as 100 MVA [for instance].
However, the percent impedance value it has to be referred to a 100% value.
Usually it could be a rated impedance [ of a transformer for instance].If in our case the short-circuit current is considered at the terminals of a 30 MVA rated transformer -considered high-voltage short-circuit power as infinite and 11.09% short-circuit voltage- then the rated impedance will be Zn=11^2/30=4.03 ohm and Xsc%=0.486/4.03*100=12.06%.

 

[7anoter4]

Sorry, I forgot : it was not an infinite power on high-voltage side considered
but 3000 MVA as IEC 60076-5 recommends for Europe up to 72.5 kV system.
For infinite power it has to be not more than 11.09%, of course.