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How to apply F. Buller’s formula to get current imbalance for multiple conductors per phase? 1

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Krustabas

Electrical
Nov 5, 2008
22
There are different ways to estimate current imbalance. The most recent and easiest to manage is given in [1, 2, and 3]. Earlier approach from [4, 5] can also be mentioned.

All these methodologies employ simple reactance formulas (ohms per 1000 feet at 60Hz):
Xii = 0.023*(K’+ln(1/rc)) (1)
Xij = 0.023*ln(1/Dij) (2)

It is often mentioned that the earliest technique is given in Buller’s paper [6] and reprinted in [7]. Unfortunately I don’t have neither [6] (can’t find anywhere) nor [7] (too expensive).

In [8] and [9] I have found the following F. Buller’s reactance formula (see Formula.png):
Xi=0.023*{K’+A+Sigma(from k=2 to n)[ln(Dik/Di(k-1))*Sigma(from j=1 to k-1)(Ij/Ii)] (3)

I am trying to understand how to apply (3) to get the current sharing among cables. The problem is that (3) contains both the unknown reactance and the unknown currents. In [9] it is said that: “…To derive the actual current sharing in a group of cables you must first calculate the current sharing based on cable impedances derived from an initial balanced current basis, then repeat the same process with the currents thus obtained…”. I guess I can calculate all the necessary reactances using IA = 1 pu, IB = a^2, IC = a but I can’t figure out how that can help me to get further (to obtain the actual unbalanced currents).

[1] IEC 60287-1-3. Electric cables - Calculation of the current rating - Part 1-3: Current rating equations (100 % load factor) and calculation of losses - Current sharing between parallel single-core cables and calculation of circulating current losses, 2002.
[2] G. Anders, “Rating of Electric Power Cables: Ampacity Computations for Transmission, Distribution, and Industrial Applications,” 1st Edition, 1997. ISBN-13: 978-0070017917.
[3] PES-ICC C35D reports ( 2011-2012.
[4] EPRI Power Plant Electrical Reference Series, Volume 4, Wire and Cable, 1987. ISBN-13: 978-0803350038.
[5] K.A. Petty, “Calculation of current division in parallel single-conductor power cables for generating station applications”, Power Delivery, IEEE Transactions on Vol. 6, Issue 2, pp. 479 - 487, Apr 1991
[6] Buller, F. H., “A Technique for Calculating Inductance, Reactance, Impedance, and Sheath Voltages of Single-Conductor Cable in Duct Banks,” General Electric Review, Vol. 52, pp. 40 - 44, March 1949.
[7] Underground Systems reference Book. Washington, D.C.: Edison Electric Institute, 1957.
[8] Adamson, Colin, “Loci of reactances of multiple-conductor circuits in the complex plane,” Proceedings of the Institution of Electrical Engineers, Vol. 114, Issue 9, September 1967.
[9] Alex Y. Wu, “Single-Conductor Cables in Parallel,” IEEE Transactions on Industry Applications, Vol. IA-20, Issue 2, March 1984.
 
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"I guess I can calculate all the necessary reactances using IA = 1 pu, IB = a^2, IC = a" - You really can't do that because it says you don't have any phase imbalance, just phase shift.

I think a better way would be to use, 1, a and a^2 for the per unit source voltages. Then solve for the voltage drops. You'll have the conductor impedances and the load impedances. The voltage drop for each sub conductor per phase will be the same from source to load. This will be a matrix of impedances for the system. If you have 4 sets of conductors per phase, you'll have a total of 12 conductors and their mutual terms to deal with. The geometric layout will determine the mutual terms.

For one phase, you can write the voltage drop for each subconductor and set them equal to each other.
 
To solve for the voltage drops (as in [4] and [5]) I need self and mutual reactances already calclated. It can be easily done using (1) and (2) and then proceed exactly as you suggested. Generally I prefer [1,2,3] where reactances are calculated with (1) and (2) and balanced current sources are used instead of voltage sources.

But with (3) the problem is that I can't calculate reactances without knowing or assuming some currents (and vice versa). The methodology involves several steps and I can't figure them out. I'm not going to use it in practice but I would like to know it.
 
I checked the aforementioned thread before I posted my question. I am familiar with [1-5] and it is possible that equations (1) and (2) are also given in Buller’s paper [6] and his methodology is similar to the one in [4 and 5]. The problem is that the cornerstone of the Buller’s approach is equation (3) developed by him and I don’t know how to apply it.

In [9] it is said about (3) that: “…To derive the actual current sharing in a group of cables you must first calculate the current sharing based on cable impedances derived from an initial balanced current basis, then repeat the same process with the currents thus obtained…”. In [4] some constants (B and C) as well as iterations are mentioned.
 
Sorry, I tried to get this thread but I failed. It was in October 26,2007 and I followed here EPRI method. First of all it is not on this forum but [237] in Electric motors, generators & controls engineering Forum.
"IMBALANCE CALCULATION ACCORDING TO EPRI-Power Plant Electrical ref. series" VOL.4-WIRE AND CABLES EQ.A-1 FROM APPENDIX A :
If there are only 2 parallel single core cables per phase then:
Ea=Ia1*(Ra1+jXa1)+Ia2(jXa1-a2)+Ib1(jXa1-b1)+...+Ia(Rl+jXl)
Ea = source phase A to neutral voltage
Ia1 = line current in conductor A1
Ra1 = conductor A1 ac res. at operating temp.
Xa1 =the self-reactance of conductor A1
Xa1-a2 the mutual reactance between conductor A1 and A2
Rl,Xl the load res. and react
Xa1=La1/1000*.023*[k+ln(1/rc)] La1= cond. A1 length[ft]
rc= cond. radius [inch]
k=.25 for concentric stranding
Xa1-a2=LA/1000*.023*ln(1/Da1-a2) LA= minimum of La1 and La2[ft] or:
Xa1-a2=LA/10^7*4*pi*F*Ktrans*ln(1/Da1-a2) LA= minimum of La1 and La2 [Ktrans=measure unit/meter]
Da1-a2= the distance betw. A1 and A2 conductors
n=no. of parallel cables per feeder
for n=2
Ia1={Ea-[Ia2(jXa1-a2)+Ib1(jXa1-b1)+...+Ia(Rl+jXl)]}/(Ra1+jXa1)
If there n cables/phase then:
Ia1={Ea-[Ia2(jXa1-a2)+... Ian(jXa1-an)+Ib1(jXa1-b1)+...+Ia(Rl+jXl)]}/(Ra1+jXa1)
Ia2={Ea-[Ia1(jXa1-a2)+... Ian(jXa21-an)+Ib1(jXa2-b1)+...+Ia(Rl+jXl)]}/(Ra2+jXa2)
--------------------------------------------------------------------------------------------------------
Ian={Ea-[Ia1(jXan-a1)+...Ian-1(jXan-an-1)+Ib1(jXan-b1)+...+Ia(Rl+jXl)]}/(Ran+jXan)
If Ra1=Ra2=...Ran and Xa1=Xa2=...Xan then Za1=Za2=....Zan.
Let's define Ea1:
Ea1=SumIki(jXa1-ki)| for i<>1 to n if k=a;i=1 to n if k<>a then:
Ia1=(Ea-Ea1-Ia*Za)/Za1 the same for Ia2.... Ian
Ia2=(Ea-Ea2-Ia*Za)/Za2
Ian=(Ea-Ean-Ia*Za)/Zan
Ia1-Ia2=(Ea2-Ea1)/Za1
----------------------------
Ia1-Ian=(Ean-Ea1)/Zan
Total on left side= Sum[(Ia1-Ia2)...(Ia1-Ian)]=(n-1)*Ia1-(Ia-Ia1)=n*Ia1-Ia
SUM(Iai)=Ia
n*Ia1-Ia= Sum[Ea1+Ea2+...Ean]-(n-2)*Ea1
Ia1={Ia+Sum[Ea1+Ea2+...Ean]-(n-2)*Ea1}/n or generally:
Iai={Ia+Sum[Ea1+Ea2+...Ean]-(n-2)*Eai}/n i=1 to n.
Ibi={Ib+Sum[Eb1+Eb2+...Ebn]-(n-2)*Ebi}/n
Ici={Ic+Sum[Ec1+Ec2+...Ecn]-(n-2)*Eci}/n
Usually |Ia|=|Ib|=|Ic|.However, this way of calculation is available even if the total currents per phase are not symmetric.

 
Thank you. I think I understand it know.
 
Sorry. Instead of:
“n*Ia1-Ia= Sum[Ea1+Ea2+...Ean]-(n-2)*Ea1
Ia1={Ia+Sum[Ea1+Ea2+...Ean]-(n-2)*Ea1}/n or generally:
Iai={Ia+Sum[Ea1+Ea2+...Ean]-(n-2)*Eai}/n i=1 to n.
Ibi={Ib+Sum[Eb1+Eb2+...Ebn]-(n-2)*Ebi}/n
Ici={Ic+Sum[Ec1+Ec2+...Ecn]-(n-2)*Eci}/n”
It has to be:
n*Ia1-Ia= Sum[Ea1/Za1+Ea2/Za2+...Ean/Zan]-Ea1[1/Za2+...1/Zan]
Since –per phase-all impedances are the same Za1=Za2=...Zan=Za then:
Ia1=Ia/n+{Sum[Ea1+Ea2+...Ean]-(n-2)*Ea1}/Za/n
Iai=Ia/n+{Sum[Ea1+Ea2+...Ean]-(n-2)*Eai}/Za/n
Ibi=Ib/n+Sum[Eb1+Eb2+...Ebn]-(n-2)*Ebi}/Zb/n
Ici=Ic/n+{Sum[Ec1+Ec2+...Ecn]-(n-2)*Eci}/Zc/n
I apologize for my mistake!

 
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