Fault level calculation 2

[cc22cc]

Can someone help me to solve my problem?

I have the 3ph faultlevel of our system (incoming) in MVA and in kV. (1830 MVA on 66kV) I know I have to translate it to a common base with the pu-method. (100MVA and 11kV) My problem is I want the faultlevel on a point 10km away, I have the ACSR-characteristics of the line, but do not know how to use it:

158mm² conductor
R1 - 0.188 ohm/km
X1 - 0.418 ohm/km

R1 - 0.00432 pu on 100MVA base, 66kV
X1 - 0.009598 pu on 100MVA base, 66kV

Thanks
cc22cc

 

[DanDel]

This requires a certain amount of research on your part, or consultation with an experienced engineer.
Refer to Stevenson's 'Elements of Power System Analysis', or IEEE Buff Book(Std 242) for a description of the procedures of performing a Short Circuit Study.
Many other books are available, as well as other posts on this site, which are available from the search function.

 

[jlazucena]

You have short circuit available at 66kV equal to 1830 MVA, and you are using power base equal to 100 MVA and voltage base 11 kV (Is this because you have a transformer? Or should be 66kV?), then
You may do it without pu analysis
Short circuit available (I = S / (1.732 V)
System impedance will be
Z line-to-ground = (system voltage^2) / (System capacity) Or
Z line-to-ground = ((kV line-to-line ^2) / (System capacity in MVA)
Z line-to-ground = (66kV)^2 / (1830MVA)
Z line-to-ground = 2.38 ohms
Note: you should have X/R ratio to know the angle
Line is 10 km with R1 - 0.188 ohm/km and X1 - 0.418 ohm/km, then
Z line = 1.88 + j 4.18 ohms = 4.58 ohms
If you neglect the angles
Z total line-to-ground = 2.38 + 4.58 = 6.96 ohms
Then you can find short circuit current
Isc = Vline-to-ground / Z line to ground

 

[jghrist]

If your fault level is given in MVA at 66 kV, it is probably at the 66 kV side of the 66/11 kV transformer. You will need the transformer impedance. Also, the conductor reactance you gave appears to be the reactance at one meter spacing. This needs to be adjusted for the equivalent spacing of the 11 kV conductors.

You will need to calculate the 66 kV system impedance. This is:

ZS (in pu) = MVAbase/MVAfault = 100/1830 = 0.0546

assuming that you were given the fault MVA at a 100 MVA base.

You will have to add the transformer impedance, changed to a 100 MVA base:

ZT (in pu at MVAbase) = ZT1 * (MVAbase/MVArated)

where ZT1 is the nameplate pu impedance (%/100) MVArated is the base transformer rating.

Both ZS and ZT are magnitudes, but will be approximately all reactive.

Total pu impedance to the fault = Ztotal = sqrt(Rline^2 + (Xline + XS + XT)^2)

Three-phase fault in pu = IFpu = 1/Ztotal

IF (amps at 11 kV)= IFpu * (100000 kVA)/(sqrt(3) * 11 kV)

Single phase faults are more complicated.

 

[jlazucena]

jghrist
I neglected the evaluation of the transformer because the impedance of the line was referred to 66kV voltage base.
If the system has a transformer then your analysis applies.

Now, the system short circuit capacity is independent of the per unit bases selected (The same value of short circuit capacity will be if you choose 10 or 40 or 100 MVA base), and the equation “Z system = MVA base / MVA scc “ is an approximation that neglects the pre-fault voltage

Assuming pre-fault voltage equal to 1.0 in pu the following applies for 10, 40, and 100 MVA base
Z system (10 MVA base) = 10 MVA / 1830 MVA sc = 0.0055 in pu
Z system (40 MVA base) = 40 MVA / 1830 MVA sc = 0.0219 in pu
Z system (100 MVAbase) = 100 MVA/1830 MVA sc = 0.0546 in pu

If the pre-fault voltage is 1.05 pu and the system base is 100 MVA then
Z system = (100 MVA / 1830 MVA sc) x 1.05 = 0.0574 in pu

 

[jghrist]

jlazucena,

I missed the fact that the line pu impedance was given at 66 kV. The impedances on the 11 kV side should be on an 11 kV base.

Because you missed the fact that the line was 11 kV, you also neglected to convert the 66 kV system impedance to the 11 kV side. The 66 kV system impedance of 2.38 ohms has to be converted to the 11 kV side (11/66)^2 if it is to be added to the 11 kV line impedance in ohms. This is why the pu system is easier; you don't have to convert pu impedances from one voltage to the other if the base voltages on each side match the transformer voltages.

 

[DanDel]

cc22cc, perhaps now, from looking at the other posts, you can see why I made my original suggestions.

 

[cc22cc]

Thanks all

DanDel, I must admit, I do have a copy of "A practical guide to short-circuit calculations" by Conrad St Pierre, but it seems greek to me. Most of the calculations are on bigger systems, and without anyone to show you what to do, a nightmare starts. I think the best would be - as you pointed out - to visit a experienced engineer. The only problem is that I am far away from such engineers, (Africa) and consultation costs a lot of money. Also, I am more in computer-stuff, but I will try my best to find a copy of Stevenson's book.

Thanks again

cc22cc

 

[peebee]

66kV equipment costs a lot of money, too. Hire an engineer, they're cheap compared to the equipment costs, not to mention the possible risks to life safety. Medium & high voltage installations is not something to be attempted by do-it-yourselfers.

Your location and proximity to engineers is really not an issue. As you have already obviously realized, you can probably handle all communications with your consulting engineer on this issue via phone, fax, & email, and hire an engineer anywhere in the world you please.