Calculating Cable Reactance for unequal conductor sizes 1

[healyx]

Hi All,

Does anyone know how to calculate reactance for 2 cables in a closed loop (e.g. phase and earth cable) where one cable is larger than the other?

I use the equation X = 0.0628ln(GMD/GMR).

Cable 1 (Phase Cable is 240mm2 - GMR = 7.7)
Cable 2 (Earth Cable is 95mm2 - GMR = 4.8)

Distance between conductor centres (GMD) is 25mm (close to touching)

For Cable 1, X = 0.0628ln(25/7.7) = 0.074 Ohm/km
For Cable 2, X = 0.0628ln(25/4.8) = 0.104 Ohm/km

Can anyone confirm/deny that this is correct?

 

[7anoter4]

In a loop consisting of two conductors-the same current will flow through both only in opposite directions- the total magnetic flux will be the sum of two fields each one produced by the currents. So the reactance will be:
X12=2*pi()*frq*miuo/2/pi()*ln(D12^2/re1/re2) miuo=4*pi()/10^7 H/m D12=25 mm re1=7.7 re2=4.8 X12=4*pi()*50*ln(25^2/7.7/4.8)/10^4=0.1777 ohm/km.

 

[jghrist]

So the reactance will be:X12=2*pi()*frq*miuo/2/pi()*ln(D12^2/re1/re2) miuo=4*pi()/10^7 H/m D12=25 mm re1=7.7 re2=4.8 X12=4*pi()*50*ln(25^2/7.7/4.8)/10^4=0.1777 ohm/km.

Which you will note is the sum of your reactances for the two cables. This is because ln(D12^2/re1/re2) = ln(D12/re1) + ln(D12/re2)

 

[7anoter4]

You are right, of course, jghrist.Thank you.

 

[healyx]

OK,

So my example had Cable 1 Reactance + Cable 2 reactance equal to 0.178 Ohm/km, which is what 7anoter4 calculated as the sum of both.

My question is, have I correctly calculated the individual reactance of each cable in the pair? (e.g. their proportional share of the total reactance?

Thanks!

 

[jghrist]

You can break the reactance equation into two parts mathematically (see my 2 Nov post), but in physical reality, there is no such thing as the individual reactance of one conductor alone, there is only the reactance of the circuit.

 

[healyx]

Thanks jghrist.

The reason I am asking for separate because I want to know what impedance is for individual segments for fault current calculation. I guess it works out the same regardless because you need need to sum them anyway to determine the fault loop impedance.

 

[7anoter4]

In order to calculate Positive Sequence Impedance and Negative Sequence Impedance you don't need the Neutral cable. For the Zero Sequence Impedance if the neutral is not grounded and you don't expect a Ground Return Current you could take the neutral as the 4th phase and you could calculate an average Xo.
If FiR is the magnetic flux connected with cable R then:
FiR=KRS*(IR+IS)+KRT*(IR+IT)+KRN*IR+KNR*IN where KRS=2/10^4*LN(DRS/reR) and so on.
Where reR=(Conductor diameter of phase R)/2*0.779
We intend to use a Kav=(KRS+KRT+KRN+KSN+KTN)/6 in order to fit a linear equation with IR.
Kav=2/10^4*LN[(DRS*DRT*DTS*DRN*DSN*DTN)^(1/6)/(reR*reS*reT*SQRT(reR*reN)*SQRT(reS*reN)*SQRT(reT*reN))^1/6)) or:
Kav=2/10^4*LN(GMD/GMR) where:
GMD=(DRS*DRT*DTS*DRN*DSN*DTN)^(1/6)
GMR=(reR*reS*reT*SQRT(reR*reN)*SQRT(reS*reN)*SQRT(reT*reN))^(1/6) and since reR=reS=reT
GMR=(reR^3*SQRT(reR*reN)^3)^(1/6)
FiR=Kav*(IR+IS)+Kav*(IR+IT)+Kav*(IR+IN)
FiR=Kav*(3*IR+IS+IT+IN)=Kav*(2*IR) as IR+IS+IT+IN=0 –IR= IS+IT+IN
LRo=2*Kav
If the Neutral is grounded that means it is a ground return current the calculation is more complicated. If you are interested you could follow the indications from:
"OVERHEAD TRANSMISSION LINES AND UNDERGROUND CABLES" by R.Rivas

http://tkne.net/downloads/power/transmissionlines1/transmission lines/OVERHEAD.pdf

For short-circuit calculation Xo=2*X1 is a good approximation I think. This will be a little conservative of course.

 

[7anoter4]

There is no book or article treating underground cable without earth return path.
So we have to solve this problem by us ourselves.
In my above post I missed: the zero sequence currents are equal [IR=IS=IT=Io] in each phase except neutral where is 3*Io. Then the magnetic flux will be spread other way.
Since the currents are equals and in the same direction the magnetic flux between phase R and S,S and T,T and R will be null and have to be extracted from the R to N, S to N and T to N.
FiRo=FiRN-FiSR-FiTR+FiSN+FiTN+FiNR
FiSo=FiSN+FiTN+FiRN-FiRS-FiTS+FiNS
FiTo=FiTN+FiSN+FiRN-FiRT-FiST+FiNT
FiRo=Io*(LRN+LSN+LTN-LSR-LTR+3*LNR)
FiSo=Io*(LRN+LSN+LTN-LRS-LTS+3*LNS)

FiTo=Io*(LRN+LSN+LTN-LTR-LTS+3*LNT)
LRo=FiRo/Io
As typical LRN=2/10^4*LN(DRN/reR) LNR=2/10^4*LN(DRN/reN) and so on then:
LRo=2/10^4*LN((DRN*DSN*DTN*DNR^3)/(reR*reN^3*DSR*DTR))
LSo=2/10^4*LN(DRN*DSN*DTN*DNS^3/(reR*reN^3*DRS*DTS)
LTo=2/10^4*LN(DRN*DSN*DTN*DNT^3/(reR*reN^3*DTR*DTS))
In your case:
DRN= 71 mm DSN= 46 mm DTN= 23 mm DNR= 71 mm DNS= 46 mm
DNT= 23 mm DSR= 25 mm DTR= 50 mm DTS= 25 mm reR=7.7 reN=4.8
LRo=0.002027 H/km LSo=0.001906 LTo=0.001351
The average will be :Loav=(LRo+LSo+LTo)/3=0.001761 H/m and Xo=4*pi()*50*0.001761=1.1064 ohm/km.

 

[healyx]

Thanks 7anoter4.

I asked the original question because I wanted to find the reactance of the line and earth (or earth/Neutral) cable during an earth fault so I could calculate fault current. Everything I am doing is LV. It is not common to break things into symmetrical components in LV, assuming the feed transformer is Delta/Wye with earth neutral (Dyn) - e.g. no zero-sequence contribution above transformer. The fault loop for a phase to earth fault is the upstream network impedance (tiny with X/R = 10) + transformer impedance + sum of line cable impedances + sum of earth/neutral cable impedances (return path back to transformer). Don't usually use symmetrical components for this, as zero-sequence parameters are hard to obtain for cables(as far as I know). This simplified method works out very close to detailed symetrical component method anyway. So to find out Max and Min fault current for an Earth fault, I need to work out the impedance between Line Cable and Earth Cable - hence my original question.
To simplify this, I was going to say the maximum reactance for a cable set during an earth fault is if you assumed faulted phase was furthest from the Earth cable and you assumed the 2 non faulted phases (and neutral) had no current (effectively single phase problem). E.g Cable layout is flat ABCNE. If A has a fault to E - highest reactance is achieved if there are approximately 4 cable diameters separating them and you assume B,C and N don't have any current. In practice, B,C and N would likely have current but this would reduce the reactance obtained for A and E. Agree?
The minimum reactance is if you assume faulted phase and earth cable are touching e.g. EABCN, where A is the faulted phase.
So in this situation, the minimum total reactance of the A-E (used for maximum fault calc) is:

X = 0.0628[ln(25/7.7) + (25/4.8)] where 7.7 is GMRc for A and 4.8 is GMRc for E and 25 is centre/centre spacing.

The maximum total reactance of the A-E (used for maximum fault calc) is:

X = 0.0628[ln((25*4)/7.7) + ((25*4)/4.8)] where 7.7 is GMRc for A and 4.8 is GMRc for E and (25*4) is centre/centre spacing (4 x larger cable diameter).

 

[healyx]

Correction - i missed some ln's in last post, should be:

X = 0.0628[ln(25/7.7) + ln(25/4.8)] where 7.7 is GMRc for A and 4.8 is GMRc for E and 25 is centre/centre spacing.

The maximum total reactance of the A-E (used for maximum fault calc) is:

X = 0.0628[ln((25*4)/7.7) + ln((25*4)/4.8)] where 7.7 is GMRc for A and 4.8 is GMRc for E and (25*4) is centre/centre spacing (4 x larger cable diameter).

 

[7anoter4]

It is very difficult to calculate a single phase-to-Ground short circuit without symmetrical components. The maximum short-circuit current is employed in order to check
the EPR and touch and step voltage and also in order to check the breaker open and close rated short-circuit where usually 3 phase solid metallic short-circuit is good enough.
The minimum short-circuit you need in order to fit the protection. I think the protection device manufacturer could indicate how to calculate this minimum current.
With all these approximations I am not sure you cannot be somewhere at a lower value and the protection will not sense the short-circuit.
If you'll follow the IEC 60909 methodology you have to employ the symmetrical components.
Now, in order to calculate the cable Zo with Ground Return Path there are some simplified formula you can use [similar to those used for Overhead Transmission Line even the Ground position with respect the underground cable is not the same].

 

[zaidi289]

YOU CAN ALSO CALCULATE BY 0.002893.F.LN GMD/GMR*OHMS*PHASE*KM

 

[healyx]

Thanks for all your help 7anoter4. I owe you a beer.