Prysmian- wrong calculation of reactance 1

[odlanor]

I’m trying to calculate the correct X (reactance) for conductors formed by more than one cable per phase and I’m using a Prysmian e-file “General Calculations - Excerpt from Prysmian’s Wire and Cable Engineering Guide”. In this e-file, is given (page 3 – Revision 4 – 06/21/2006) the following formula to calculate geometric mean radius (GMR):

g = √(n&a.b.….z_n )

and are given the meanings of all symbols used verbatim:
a = distance from conductor a to b;
b = distance from conductor b to c;
zn = distance from conductor n-1 to n.

Please note that: a = distance between the first and second conductor (distance 1);
b = distance between the second and third conductor (distance 2);
zn = distance between the penultimate and the last conductor (distance n-1).

Kindly, confirm zn is the distance from conductor n-1 to n, because the number of the radicand terms of the geometric mean must be equal to the index of root and the radicand (a.b. ... . zn) is formed by (n-1) terms. Thus, to be correct, zn should be the distance from conductor n to a.
In the nomogram on page 20 of the referred publication, the spacing equivalent of three phase conductors (GMR) is stated as:
A = distance between the first and second conductor ( = distance from conductor a to n-1)
B = distance between the second and third conductor ( = distance from conductor n-1 to n)
C = distance between the third and first conductor (distance from n conductor to a).

I would be very grateful to hear from you on the above comments.

 

[odlanor]

Thank you guess. But I received an answer from Prysmian on the subject and thought I should transcribe for you.

Prysmian answer:
You are correct the number of radicand terms MUST equal to the index of the root.

Very simply, the radicand terms are comprised of the distances between EVERY conductor to EVERY other conductor. For example if you had two cables per phase, we'll call them A1, B1, C1, A2, B2, and C2, the equation would be the 30th root of the distances:
A1 to B1
A1 to C1
A1 to A2
A1 to B2
A1 to C2
B1 to A1*
B1 to C1
B1 to A2
B1 to B2
B1 to C2
C1 to A1*
C1 to B1*
C1 to A2
C1 to B2
C1 to C2
A2 to A1*
A2 to B1*
A2 to C1*
A2 to B2
A2 to C2
B2 to A1*
B2 to B1*
B2 to C1*
B2 to A2*
B2 to C2
C2 to A1*
C2 to B1*
C2 to C1*
C2 to A2*
C2 to B2*

Now, redundant distances (marked with an asterisk) can be removed, and index reduced accordingly - which would be the 15th root of the remaining 15 radicands.

As you can imagine the complexity increases significantly with the number of cables (per phase) involved.

I hope this will help you.

Carroll Lindler
Manager of Applications Engineering

carroll.lindler@prysmian.com

 

[7anoter4]

I have a great consideration to Prysmian and I think a mistake could happen to anyone. I saw myself this but I didn't think to fix it. I agree with you odlanor and I thank to Prysmian for the answer. From now I'll endeavor to fix the math formula everywhere.

 

[healyx]

I thought you only needed to consider the geometric relationship between each set of parallel phases to determine reactance. So if you had 2 parallel cable sets (6 conductors) and you laid them all flat in a line a1,b1,c1,a2,b2,c2. I thought you only needed to work out the reactance of a1,b1,c1 as a group, then for the parallel set, the actual reactance (and resistance) for the segment would be half this. I've seen it calculated this way alot.

I just did a calc with 240mm2 XLPE/PVC cables.

Laid flat touching one phase set is 0.097ohm/km. So the total reactance of the two combined sets in 0.0485ohm/km (e.g. half).

But if you have to consider all 6 together the reactance is 0.126ohm/km and I'm not even sure if you divide this by 2 anymore.

Which way is the correct way?

 

[7anoter4]

I am afraid your second reactance value is correct. The magnetic flux in a cable is produced by the current flowing through itself and the currents flowing in all other
parallel cables and the reactance will be direct proportional with the total magnetic flux. The average reactance will be calculated as the GMD on a double- circuit overhead transmission line.
So, for your low voltage 0.6/1 kV 6*240 sqr.mm cables [approx.overall 30 mm] the GMD=69120^(1/16)*30=60.2 mm and the reactance will be indeed 0.126639 ohm/km.

 

[healyx]

Thanks for the reply 7anoter4.

Wow, every software package I have ever used has never calculated the reactance of all cables sets by considering them as a whole! Schneider's Ecodial program doesn't work it out this way. The CENELEC (UTE C 15-500) standard for how computer programs should perform these calculations doesn't work it out this way. How can they all be wrong?

So in my example, 0.126ohm/km is the reactance per set - so is the total reactance for the 2 set segment this divided by 2 (0.063)?

By considering all conductors, you can obtain the same result with no regard to phase ordering. For example if 3 parallel sets are considered (9 conductors), you could arrange the sets in 3 spaced triangles (A1, A2, A3), (B1, B2, B3) and (C1, C2, C3) which I thought would produce a much higher reactance than (A1, B1, C1), (A2, B2, C2) and (A3, B3, C3)? And why wouldn't you also need to include the conductors of other unrelated circuits in the vicinity of the parallel sets?

This just doesn't seem intuitive to me????

Everything I do is low voltage, so this only really matters in larger cable sizes - but they are usually the ones you need to parallel. Is it just generally ignored/simplified in LV?

 

[7anoter4]

This is the average mutual reactance. You may use this for short-circuit calculations or line voltage drop. For an actual voltage per individual cable -in order to calculate the unbalance between the cables of the same phase- you need to consider the phase order, of course. See EPRI-Power Plant Electrical ref. series" Wire and Cables vol. 4 eq.A-1 from APPENDIX A or IEEE "Single-Conductor Cables in Parallel"

http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=4504423

or IEC 60287-1-3

http://webstore.iec.ch/preview/info_iec60287-1-3{ed1.0}en_d.pdf

For example, the total voltage [ Ean=VL_0 finally] of cable no.1 phase A will be:
Ean=Ia1*(Ra1+jXa1)+Ia2(jXa1-a2)+Ib1(jXa1-b1)+...+Ia(Rl+jXl)
where Ia1 is the current flowing in this cable Ia is the total current of phase A flowing through the load R and X , Xa1_a2= K*ln(1/distance a1_a2) and Xa1=K*ln(1/re) re =equivalent conductor radius [GMR].The currents are complex values, of course.

 

[7anoter4]

Two corrections [with respect the first post of 27 oct]:
1) There are only 15 single combination possible C2/6=(6*5)/(1*2)=15.
But we'll get the same result: 34560^(1/15)=2.007 and multiplying by 30=60.2 mm.
2) Verifying the calculation of double-circuit overhead line it is clear that only distance between different phases it is required to be in the multiplying chain. So they are only 12 factors.
In this case the result is 1280^(1/12)=1.81524 and the GMD=1.81524*30=54.4572 mm.
The reactance will be 0.12034 ohm/km [95% from the latter].

 

[healyx]

Thanks again 7anoter4!

More questions as I ponder...

(1) How far apart do parallel sets need to spaced to be able to be treated in isolation without contribution of other phase sets. With the equation as given, increasing distance between adjacent sets just keeps increasing the reactance - although the rate of increase reduces with increase group separation. Surely at some point this no longer holds? Using the equation I could separate the set groups by the width of a country and reactance just keeps increasing!
(2) By considering only distance between different phases, you can get different answers depending on phase sequence. For:
(i) ABCABC, spacing factor = 1.82 X=0.120 Ohm/km
(ii) ABCCBA, spacing factor = 1.91 X=0.123 Ohm/km
(iii) AABBCC, spacing factor = 2.39 X=0.137 Ohm/km
(iv) Any combination if distance between same phases is considered, spacing factor = 2.01 X=0.126 Ohm/km
Do you definitely disregard same phase spacings? There is alot of variation in those combinations.
(3) Going back to our ongoing example and assuming you only consider the distances between different phases. A single laid flat touching 240mm2 set gives 4^(1/3)= 1.26 => X = 0.097 ohm/km.
2 parallel sets laid flat touching ABCCBA gives spacing factor of 1.91 => X = 0.123 Ohm/km. This represents the "average" reactance of each parallel set right? So to work out the voltage drop and fault current over the whole circuit (segment) you divide this number by the number of parallel sets (2 in this case to give 0.123/2 = 0.0615). So using the "wrong" method of treating each phase set in isolation gives a reactance of 0.097/2 = 0.0485 Ohm/km for the combined sets. The "right" method gives 0.123/2 = 0.0615 Ohm/km. So in this case the mutual inductance of the parallel set increases reactance by 27%. Is this right?
(4) Do you disregard other cables in the vicinity of the parallel set?

 

[7anoter4]

In order to simplify the calculation an average value is employed for mutual reactance and so the phase voltage will depend only on the same phase current.
But the average could be done in different ways. Here is an arithmetic average:
(XRS+XST+XTR)/3=K*Ln((DRS*DST*DTR)^(1/3)).For 6 cable could be 6 ,9,12,15,or 30factors in an arithmetic average. ABB and Siemens use other type of average.
So, average mutual reactance could be different from a way to another.
The vicinity of other cables which are not in a closed loop will produce a magnetic flux in each closed loop but will decrease with the distance from this cable to the loop surface as the magnetic flux density B=miuo*I/2/PI()/dist. If this distance will be more than 10 times the distance between this cable and the other phase of the same system the influence will be negligible.
But for the cables of the same system-having a common source and a common load-the minimum reactance will be when they are in touch.

 

[healyx]

Thanks 7anoter4!

Any chance you can confirm item (3) from my previous post?

I have another question about calculating reactance in two unequal cable sizes but I might start a new thread for that - I have hijacked this thread enough

 

[7anoter4]

I think the formula shown in Prysmian Wire and Cable Engineering Guide it is not so bad. But it is referred only to 3 phase single line n=3. The last factor of g has to be "c= the distance between conductor c and a" instead of "zn= distance between conductor n and n-1".
I was wrong with the interpretation from the beginning.
The actual formula for 2 parallel lines it is [in SI units system]:
X=4*pi()/10^4*ln(GMD/GMR) ohm/km.
GMD=(Da1b1*Da1*c1*…)^(1/12)=0.04766 m.
GMR=((r*.779)^3*Da1a2*Db1b2*Dc1c2)(1/6)=21.79 mm=0.02179 m
See "OVERHEAD TRANSMISSION LINES AND UNDERGROUND CABLES" by R.Rivas

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

So for 6 cables1*240 sqr.mm of 19.8 mm conductor diameter r*.779=19.8/2*.779=7.7mm in a flat formation ABCCBA
X=4*pi()*50/10^4*ln(GMD/GMR)=4*pi()*50/10^4*ln(0.04766/0.02179)=0.04918 OHM/KM
This is the per-phase reactance [it is not" per-cable"].
For 3 cable [one cable per phase]
X=4*pi()*50/10^4*ln(S/r)+1/4) S=(25*50*25)^(1/3)= 31.5 mm.
X=0.08843 ohm/km per cable or 0.044215 per phase.

 

[healyx]

Hi 7anater4.

Thanks again for the post.

I looked at that reference and have followed through your calc. I agree with everything you said except the last line.

For the single circuit with 3 phase conductors laid flat touching X = 0.008843 ohm/km. This for both per cable and per phase, because there is only one cable per phase.

For the 2 circuit laid flat touching case ABCCBA we get 0.04918 ohm/km. This is per phase, not per circuit.

Getting back to my original issue with this, it that is it common to take the single circuit case and divide the reactance by the number of sets to get the per phase reactance for a n circuit set. Based on these calcs this would result in an under estimation of true "per phase" reactance of about 10% (for the 2 circuit case as described).
True reactance per phase if calculated correctly is 0.04918 ohm/km. If simplied by saying 0.08843/2 we get 0.04422 ohm/k. So using the simplified result means true reactance is 11.2% higher than the simplified estimate. Do you agree? This means it isn't as bad as first thought, but still out by enough to be concerned.

Thanks!

 

[7anoter4]

Of course, I agree with you. The above "exact" calculation is based on arithmetic average that means it is "an approximation". You are right, also, about the last sentence. I intended to compare the "exact" calculation of 2 cables per phase with the reactance of single cable per phase divided by 2 as you yourself said but I forgot to finish the sentence.

 

[healyx]

Hi 7anoter4.

Do you know how to adapt the above solution to case where there are 3 or 4 parallel sets.

I think I know how to calculate GMD. It it resolves to (X)^(1/27), where X is all distances multiplied (except a-a, b-b and c-c distances).

I can't work out how Rivas derived GMR for the general case.

For 2 parallel sets GMR is:

GMR = [(GMRc)^3(Da1a2)(Db1b2)(Dc1c2)]^(1/6)

What is this for 3 or 4 parallel sets?

My wild guess for 3 parallel sets would be:

GMR = [(GMRc)^9(Da1a2)(Da1a3)(Da2a3)(Db1b2)(Db1b3)(Db2b3)(Dc1c2)(Dc1c3)(Dc2c3)]^(1/18)

Any ideas?

 

[healyx]

I had another attempt at deriving this for the 3 parallel set case. Consider 3 phase sets the first set is A1, B1, C1, where A,B,C are the phases. Set 2 is A2, B2, C2, set 3 is A3, B3, C3.

I'll try and work out GMR for each group of conductors sharing the same phase. GMRd is the GMR of the conductor.

GMRa = [(GMRd^3)(Da1a2^2)(Da1a3^2)(Da2a3^2)]^(1/9)
GMRb = [(GMRd^3)(Db1b2^2)(Db1b3^2)(Db2b3^2)]^(1/9)
GMRc = [(GMRd^3)(Dc1c2^2)(Dc1c3^2)(Dc2c3^2)]^(1/9)

GMRtotal = [GMRa*GMRb*GMRc]^(1/3)

GMRtotal = [(GMRd^9)((Da1a2)(Da1a3)(Da2a3)(Db1b2)(Db1b3)(Db2b3)(Dc1c2)(Dc1c3)(Dc2c3))^2]^((1/27)

If this is correct, the estimate where the reactance of a parallel is the single phase set reactance divided by the number of sets seems to be OK. In fact (assuming all flat touching - which would not be done in practice) the reactance per circuit of a parallel set can be lower than the reactance of a single set. The increase in GMD is matched by the increase in GMR. I suspect my derivation is wrong.

 

[healyx]

I've looked into this a bit more.

With the 2 parallel set case shown by R.Rivas text:
(1) ABCCBA produces per circuit reactance 10% HIGHER than single set ABC circuit.
(2) ABCABC produces per circuit reactance 10% LOWER than single set ABC circuit.

So using the R.Rivas formula for 2 parallel sets you can conclude on average dividing the single set reactance by 2 will give an appropriate per phase reactance for the parallel set. The +/-10% is entirely dependent on phase ordering.

With my above equation (last post) for 3 parallel set case:
(1) ABCABCABC (not recommended in practice) produces per circuit reactance 13% LOWER than single set ABC circuit.
(2) A A A (Recommended Ordering)
B C C B B C with no spacing produces per circuit reactance 14% HIGHER than single trefoil ABC set.
(3) A A A
B C B C B C with no spacing produces per circuit reactance 10% LOWER than single trefoil ABC set.
(4) A A A (Recommended Ordering)
B C C B B C with 1D space between sets, phases touching, produces per circuit reactance 6% HIGHER than single trefoil ABC set.

So my derived equations for 3 parallel sets appears to produce similar results to the R.Rivas 2 parallel set equation. It looks like the number of sets has little bearing on the per circuit (per set) reactance (which is approx equal to the corresponding single set case). It looks like Xphase = Xcircuit/n where n is the number of parallel sets and Xcircuit is the single set reactance. The variation comes from phase ordering, with recommended phase orderings resulting in higher average reactance. This is without considering unequal phase currents as set out in IEC60287-1-3.

Going back to my original posts in this thread from Oct 26 and 27, the simplification used by many software programs where single set reactance is divided by n to obtain parallelled phase reactance appears to be a "reasonable" estimate/simplification. They should probaby include a +10% safety factor for voltage drop and minimum fault current calcs on paralleled sections. The increase in GMD caused by paralleling appears to be largely offset by the increase in GMR - even if you just look at the 2 set case outlined in the R.Rivas text (Assuming it is right).

Any opinions on this?

 

[7anoter4]

If the cables are laid with regular transposition then the "exact" calculation could be indeed exact calculation .For the cables without transposition the result is approximate. So I think your consideration is correct. By-the-way, the reactance shown in IEC 60287-1-3 [eq.9] is not "our" reactance but a kind of reactance used to
calculate the currents unbalance. It represents actually the difference between 2 reactances of 2 conductors of the same phase.