Reactive Power Generation of 34.5 kV Underground Distribution Lines 1

[Wloria]

Hi All,

I am building a power plant reactive power calculator to determine what the reactive power at the point of interconnection would be. I have read other threads on this topic and I wanted to confirm my answers.

Coming off a MV Xfmer we have 34.5 kV. The data that I gathered from several underground xple cable manufactures says that a typical capacitance and inductance for the size cable (~700MCM) would be in the ball park of 0.300 uF/km and 0.3mH/km. For a short run of 5-16 miles, and transmitting between 10-80 MVA, my quick math tells me that the line would be very capacitive, almost to the point where I could neglect the inductance. The two port pi model calculations also show that the receiving end voltage and current would be higher which indicate the same thing.

For the quick math I used: Qc=(V^2/1.732)*2*pi*60*C*length
and Ql=I^2*2*pi*60*L*length

So, my question: Does this seem correct for underground cables in a trefoil configuration buried in a duct? Does the capacitance increase so much that the overall characteristic would be capacitive? I know it depends on the architecture of the cabling but even using the worse case scenario of 1 long line that carries all the amperage I find capcitive qualities. I am so used to lines being inductive that, I just wanted to run this by people who have hands-on experience and/or deal with these types of calculations every day.

Any feedback is greatly appreciated!

 

[davidbeach]

At high enough voltages the system will be seen as capacitive, but I doubt that 34.5kV is high enough.

 

[Wloria]

Thanks for the reply. I guess I will recheck my math, and cable datasheets again.

 

[mgtrp]

A couple of notes:
1/ Try (V/1.732)^2
2/ Xc = 1/(2*pi*f*C)

 

[7anoter4]

If we shall follow UL 1072 recommendations the 34.5 kV cable 750 MCM XLPE insulated the conductor diameter is 25 mm and the insulated core shield diameter is 44 mm.
If frequency is 60 Hz and the cable rated current Irat= 1000 A then one km inductive reactance will be:
XL=2*pi*60*2*ln(2*44/25+1/4)/10^4=0.100059098 ohm/km
The inductive reactive power of the cable in series with the load will be:
QL=3*XL*I^2=3*0.1*1000^2/1000= 300 kVAR.
The capacity of one km three-core cable will be:
Cs=eps.r/18/ln(D/d) microF/km where: eps.r=3.5; D=44; d=25 mm.
Cs=3.5/18/ln(44/25)/10^6= 3.44E-07 F/km
XC=1/2/PI/frq/Cs=7711.9 ohm/km
The capacitive reactive power parallel with the load will be:
QC=kV^2/XC=34.5^2/7711.9*1000=154.3 kVAR
The load reactive power will be Qload=sqrt(3)*kV*Irat*sin(fi)
Let's say load cos(fi)=0.9 then sin(fi)=sqrt(1-0.9^2)=0.43589
Qload=sqrt(3)*34.5*1000*0.43589=26047 kVAR
QC/Qload*100=154.3/26047*100=0.5925%
The inductive load will not turn to capacitive in any case.
However, in no-load case the cable-as always it wills be-it is a capacitance, of course.