Calculating R and X cable for LDC using actual measurement

[anggap]

Hi All,

I am trying to calculate LDC setting for a voltage regulator 34.5 kV that supplying a bus in radial, this voltage regulator connected to receiving bus using a 123,000 ft length subsea cable.
The problem that I can't find the cable specification since it is an old facility. Is it possible to calculate the R and X using power and voltage measurement at sending and receiving bus?

Thanks in advance
Angga

 

[HamburgerHelper]

I think though you don't and can't ignore the admittance and might want to use a medium or to be safe long line model. If those cables are close to each other there is going to be a significant amount charging current between the phases. I don't know if this helps you with what you need to put into the LDC.

http://www.nptel.ac.in/courses/117104071/chap2.pdf

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If you can't explain it to a six year old, you don't understand it yourself.

 

[anggap]

Thanks for your response, unfortunately I don't have the voltage angle data. All available data are in magnitude only. Below are the available data:
Bus 1 (sender):
Vs = 32.5 kV
Is = 212 A
Ss = 12.2 MVA
Ps = 10 MW
Qs = 7.1 MVAr

Bus 2 (receiver)
Vr = 30.1 kV
Ir = 250 A
Ss = 13.1 MVA
Ps = 9.3 MW
Qs = 9.4 MVAr

 

[HamburgerHelper]

If you know the cables spacing and conductor diameter and material, you can back into the impedance Z and admittance Y. From there, you can use the long line transmission model to see if it matches your above results.

ATP-EMTP has a free module for giving you transmission line parameters. It might be quicker just to call up a cable vendor and ask. They probably get asked this all the time.

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If you can't explain it to a six year old, you don't understand it yourself.

 

[anggap]

I found the typical datasheet of the cable (R= 0.1 ohm/km ; X = 0.135 ohm/km) and tried to calculate the voltage drop using this voltage drop formula:
Vs-Vr = IR Cos (phi) + IX Sin (phi)

but the calculated VD value didn't match the actual measurement value

 

[anggap]

one more thing, looking at the Reactive Power flow, this cable has negative loss, reactive power at receiving end is greater than the sender. Is it mean that this cable act as capacitor?

 

[HamburgerHelper]

Yeah, there is a lot of capacitance between the phases with cables. When I have seen 138kV and 345kV underground, they often had shunt reactors to bleed off vars to prevent the voltage from rising too much.

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If you can't explain it to a six year old, you don't understand it yourself.

 

[7anoter4]

At first you can calculate the cable resistance and capacitance. The cable resistance at 90oC [at 35 kV for a EPR insulated cable] will be Ploss/3/I^2. Ploss=10-9.3=0.7 MW.
The current will be the average of 212 and 250 A[231 A] R=4.37 ohm R/123=0.035551 ohm/kft
For 20oC R20oC=0.035551*(234.5+20)/(234.5+90)= 0.027882 ohm/kft=0.09147572 ohm/km
R=1/58*1000/SCU ohm/km scu==conductor cross section area [mm^2]
scu=1000/58/R=188 mm^2 [371 kcml]
You may take 3 cables of 350 MCM copper conductor EPR insulated 35 kV[single core].
90oC underground Table 310.60(C)(77) Det.1 -35 kV 90oC 390 A and Table 9 350 kcmls R=0.125 ohm/km at [60 Hz 75oC].
If 90-20=70oC for Tc=90oC and 390 A then DT=(231/390)^2*70=24.5 oC. Let’s say DT=30oC Tc=30+20=50 the resistance will be (234.5+50)/(234.5+75)*0.125=0.114903069 ohm/km [0.0350022 ohm/kft].
Ploss=3*0.0350022*231^2*123/10^6=6.895 MW
Conductor diameter 0.616 in.
The shield diameter has to be 0.62+2*0.42+0.01=1.47 in.
The overall diameter is 2.00” approx.
Cap=3.4/18/ln(1.47/0.616)=0.2172 μF/km [or 0.0662 μF/kft]
If the phase cables are in triangle configuration then
XL=2*pi*60 *(0.2*ln(2*2/0.616)+0.0528)/10^3 ohm/km
XL=0.049002261 ohm/kft
XC=10^6/(2*PI*60*0.0662)=40069 ohm/kft
The inductive reactive power will be QL=3*XL*I^2*123 [I=(212+250)/2=231A]
QL=0.956 MVAr
QC=32.5^2/40069*123=3.242 MVAr
Then Qb=7.1+3.242-0.956=9.38 MVAr
For recalculation of Vs and Is from Vr and Ir the IEEE 141 formula it not good enough.
You need the general formula of a transmission line :
Vs=Vr*ch(g)+Ir*Z*sh(g)
Is=Ir*ch(g)+Ur/Z*sh(g) where:
g=γ*length
γ=√[(R+jωL)x(G+jωCap)] and Z=√[(R+jωL)G+jωCap)]
ch=coshyperbolic
sh=sinhyperbolic
R=0.027882 ; ωL=XL=0.0490022; ωCap=1/XC=2.49568E-05 ;G=0
Vs=32.49;Is=214

 

[7anoter4]

I have misspelled the result. Actually Ploss=3*0.0350022*231^2*123/10^6=0.6895 MW,of course.

 

[7anoter4]

I still think that the 3 * 185 18/30[36] kV submarine cable is more suitable[for 50 Hz Systems]. See:

https://www.nexans.co.uk/Germany/2013/SubmPowCables_FINAL_10jun13_engl.pdf