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Crculating currents in a neutral

Crculating currents in a neutral

Crculating currents in a neutral

(OP)
Hi guys,

I have a case of circulating currents in a Neutral circuit.
The neutral circuit is made up of 4x500 single core cable on a 3 phase 4 wire TNC supply from a generator to a switchboard (Phase are 4x400sqmm per phase).

The 'combined' neutral curent on the N circuit was measured at 20A or so. The current on the 4 individual conductors of the neutral were (approx) 130A, 30A, 50A, 40A.
Oscilloscope reading show that the 130A is in phase opposition (aprox.) to the other 3, indicating that there is a circulating current on the cables in addition to the actual 'total' neutral current of 20A.

No harmonicw were measured (THD curent  = 1% or so)

Have you guys ever witnessed this ?

The cable run is aprox 100m long, cables are on cable trays in a cable trench run out of the building to the external generator sets.

Is this common ? Unusual ?

We suspect an inducted current caused by a "back-EMF" induced by a magnetic field over the surface of the cables (closed loop).  One of Maxwell equations shows this.

Would love to hear from you !
 

RE: Crculating currents in a neutral

Are the cables have gland earthing at only one end or both ends? It should be at only one end.

Have measured the current passing through gland earthing?

NC

RE: Crculating currents in a neutral

How exactly are the phases separated and laid in relation to the neutrals?

Keith Cress
kcress - http://www.flaminsystems.com

RE: Crculating currents in a neutral

What is the frequency of the measured current?

RE: Crculating currents in a neutral

I think this is not a problem of inequal impedance as what is described is one cable with opposite phase / direction of all other.

NC

RE: Crculating currents in a neutral

If the neutrals are grounded at both ends, then there will be currents induced in the neutrals from the phase conductors.  The induced current will depend on the spacing between each neutral and each phase.  

RE: Crculating currents in a neutral

(OP)
Additional information :

"Gland earthing" : I am assuming reference is made to the what we call the "gland plate" here in Australia, which is the plate through which the cable passes as it enters the switchboard. Both gland plates are earthed, and the slots are cut in the gland plate between each circular hole to prevent the formation of induced eddy currents around each hole (which heats up the sheet metal of the gland pate)
We had measured the earth current from the frame of the generator : 25A. Generator manaufacturer (Cummins) has advised this is normal and acceptable (1850 kVA 415V gen set)

Cable installtion: Cables (4x400sqmm per phase .. so a total of 12x400 for phase conductors... and 4x500sqmm neutral and 4x120sqmm for the earth) are installed on a 600mm cable tray in a pseudo trefoil arrangement : 4 x (Red-White-Blue-Neutal+Earth) in a square format each square spaced by about 50 mm, and installed so that the phases in adjoining squares are in proximity to one another (industry recommended installation).

Frequency : not 100% sure, but suspect 50Hz (50Hz system)

Impedance : Yes this does not appear a question of unequal impedance due to the phase current opposition.

Neutral grounding : the neutal is earhted (grounded) only at the swithboard end . It is not (i repeat not) grounded at the generator end.

Thanks for all your help fellow engineers, it is much appreciated !

RE: Crculating currents in a neutral

Gland plates should be earthed at both ends. It is correct.  No issue.

Are the cables armoured?  Does cbles have shielding?  Is the cable shielding grounded at both ends?  Is cable armour / Shieth connected to ground at both ends?

It should be at only one end.  Other wise you will have circuilating currents like this.

NC

RE: Crculating currents in a neutral

(OP)
The cables are all single-core double-insulated XLPE/PVC outer-sheath. No steel wire armour, and no metal outer shield, so nothing that can be earthed.

 

RE: Crculating currents in a neutral

Have you measured the phase currents in each of the four groups?  I could see that if there were a large phase imbalance in one group, a current could be induced in that group neutral through mutual coupling that may circulate in the other neutrals.  A large phase imbalance in one group may indicate a bad connection of one phase conductor.

 

RE: Crculating currents in a neutral

It is still difficult to understand the cable lay-out.
I guess it may be as per attached sketch.
The phase, neutral and grounding cable dimensions are as follows:
   sqr.mm    Cond.dia        ins.core    jacket     thick          o/dia
400    26.1    30.7    2.1    35.3
500    29.2    34.24    2.2    39.1
120    14.5    17.34    1.6    20.9
ZNeutral=0.003571+j0.003389 ohm per one neutral cable[100m].
As the distance between phase A[Red] to Neutral=distance between phase C[Blue] to Neutral, but the distance from B[White] to neutral is bigger, then a voltage is induced in the neutral.
For group1,2,4 current =2573.7/4=643.43 A, A[Red]angle=0 B angle=240o and C[Blue] angle=120o then Vneutral1=Vneutral2=Vneutral4=0.3418 V
For group3 current =2573.7/4=643.43 A, A[Red]angle=0 B angle=120o and C[Blue] angle=240o then Vneutral3=-.4591 V
Then the voltage all around neutral cables[connected both ends] will be 0.8088 V and the current through neutral 3 will be:
08088/(ZN+ZN/3)=122 A and the current through neutral 1,2,4 will be 122/3=41A [supposing they are identical]
Regards

RE: Crculating currents in a neutral

(OP)
7anoter4,

I am interested in your calculations.. can you detail the method ?

But to confirm the cable installation :
(a) there are 4 groups of cables
(b) each group contains 3 phase (400sqmm), 1 neutral (500sqmm) and 1 earth (120sqmm)
(c) by starting top left and going clockwise, the groups are :
G1 : A - B - C - N
G2 : B - A - N - C
G3 : as G1
G4 : as G2

The groups are spaced by one cable size (+/- 50mm)

Can you detail :
(a) how you calculate the induced voltage in each Neutral. What effect are you taking into account ?
(b) How do you calculate the total Voltage (you seem to do V1 - V3 = .34 + .45 = .8.. why ignore V2 and V4 ? Or am I interpreting your calcs incorrectly ?

What you are describing 'could' explain what we are seeing, but I would really like more detail !

Thanks !

Boris

RE: Crculating currents in a neutral

I would look for a loose connection in one of the phase cables. Check the current balance on the phase cables if you can.
The cable topography will result in the magnetic fields of the cables canceling each other quite well, but there will be a little unbalance inducing a small voltage in each neutral. However, if a phase cable in one group is not carrying current there will be an large unbalance in the magnetic field of that group. This may induce much larger voltages in the neutral. As the neutrals are connected together at each end, a larger voltage induced in one cable will drive a circulating current. This current will be in one direction in the cable causing it and in the opposite direction in the other cables, as you have observed. I believe that this is in accord with your measurements and observations.
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Crculating currents in a neutral

(OP)
Further investigation was conducted on site and this phenomena was discovered on all 3 generators !!!

I agree (in principle) that a loose phase cnnection could induce a similareffect, but the fact that 3 generators have the similar issue seems to imply that it is not linked to the cable termination.

I am inclined to believe it is linked to the way the cables are installed (ie physically laid), so I am interested in 7anoter4's ideas. I hope he can give us more information.

Thanks,

Boris

RE: Crculating currents in a neutral

I notice that the neutral is always adjacent to "A" phase. This is probably the cause.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Crculating currents in a neutral

(OP)
Waross,

Can you detail more ?
The 4 cables are in a "square" configuration, so the N conductor touches A  and C, and is fairly close to B.

I would suspect the problem is due to the fact that N is not as close to B (as suggests 7anoter4) as it is to A and C..

Boris

RE: Crculating currents in a neutral

Typical UK practice would install the three phase conductors in a trefoil group, the cores being held held equi-distant by the cleats, with the neutral and earth being carried on supplementary cleats. This method gives the closest grouping and thus the best cancellation of the external magnetic field assuming the load is balanced. Multiple groups in parallel should transpose the phase rotation. I have not seen the 'square' method with a central earthing conductor before but it appears to cause definite assymetry in the magnetic field.

I think 7anoter4 is likely on the right track with his analysis.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: Crculating currents in a neutral

In two groups the neutral is between "A" and "C", in two groups the neutral is between "A" and "B". The neutral is always adjacent to "A".

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Crculating currents in a neutral

Hi bfuchs
As Slava indicated in his the above post, I followed the EPRI calculation mode [as per
EPRI-Power Plant Electrical ref. series VOL.4-WIRE AND CABLES  EQ.A-1 FROM APPENDIX A] :
Ean=Ia1*(Ra1+jXa1)+Ia2(jXa1-a2)+Ib1(jXa1-b1)+...+Ia(Rl+jXl) where:
Ean = 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=La1/1000*.023*[k+ln(1/rc)]  and La1= cond. A1 length[ft] ; frq= 60 HZ
rc= cond. radius [inch]
k=.25 for concentric stranding
Xa1-a2 the mutual reactance between conductor A1 and A2[ let us say N1 to N2 in your case]
Xa1-a2=LA/1000*.023*ln(1/Da1-a2) LA= minimum of La1 and La2
Rl,Xl the load res. and react.
One can translate these formulas into SI if he put:
Xa1=La[m]*kx* [k+ln(1/rc)]   rc[mm]; kx= 4*frq*pi()/10^7
Xa1-a2= La[m]*kx*ln(1/Da1-a2)
As one could see, if I follow the EPRI calculation mode I cannot take more than 2 groups into consideration.
For conductor N1 will be:
Ean1=In1*(Rn+jXn)+In2*(jXn1-n2)+Ia1(jXn1-a2)+Ib1(jXn1-b1)+Ic1(jXn1-b1)+Ia2(jXn1-a2)+Ib2(jXn1-b1)+Ic2(jXn1-b1)+Ia(Rl+jXl)
Since the circuitry is limited to 4 neutral conductors connected together both ends Ean=0 and also Ia=0
So the "In1*(Rn1+jXn1)" remains now against the induced voltage produced by phase currents A1,B1,C1,A2,B2,C2 and N2.
Vn1=In2*(jXn1-n2)+Ia1(jXn1-a2)+Ib1(jXn1-b1)+Ic1(jXn1-b1)+Ia2(jXn1-a2)+Ib2(jXn1-b1)+Ic2(jXn1-b1).
I'll take phase A as origin of angles.
From my cable lay-out [see the above sketch] the unbalance was as I said.
But for the new lay-out the problem is different.
If I put now IN2= 32A [105 degrees] I get IN1=32.5 A [240 degrees].
So, if there is not any fault in this cable lay-out, through 2 neutral [1, 3] will flow 32.5 A [240 degrees] and through other 2 neutral 32 A [105 degrees].
Actually, I have to develop a software in order to manage four parallel neutral conductors at once, since, as you could see, the no.2 neutral-for instance- is more induced than no.1 and the unbalance may be different.
Regards
 

RE: Crculating currents in a neutral

(OP)
Thanks for that !

I tried searching for the document mentionned above (EPRI-Power Plant Electrical ref. series VOL.4-WIRE AND CABLES), and have not been successful in locating it. Is downloadable from their website ?
Where can I access it (free ? what price ?)

Cheers

 

RE: Crculating currents in a neutral

1) My EPRI handbook is dated 1982 .I'll try to find it using the link:
https://login.epri.com/login.asp
without any success. I think you have to contact EPRI at this address:
EPRI 3420 Hillview Avenue, Palo Alto, California 94304 USA
2) I think I exaggerated when I said "I'll develop a software". I am not a programmer. I am only an engineer using VisualBasic6 [or .NET] to write a modest program for my personal use [at my work place they are truly programmers [working with assembler or UNIX or else] and I never employ the Society software or other document outside.
3) I think that IEC 60287 vol.1- 3 is treating the parallel cable imbalance load case:
http://www.standardsdirect.org/standards/standards3/StandardsCatalogue24_view_20879.html
and also : "Calculation of current division in parallel single-conductor powercables for generating station applications"
IEEE Transactions on Power Delivery,
Volume 6, Issue 2, Apr 1991 Page(s):479 – 487
See: http://ieeexplore.ieee.org/Xplore/login.jsp?url=/iel3/61/3636/00131101.pdf?arnumber=131101
Regards
 

RE: Crculating currents in a neutral

(OP)
Thanks for the detail.
Reviewing it I have one question (at this stage).

The formula refer to "ln(1/Da1-a2)"

I do not understand this...

I assume we are talking about the N log function, and Da1-a2 is the distance between conductors..

But in what unit ?
millimeters ? Meters ? Inches ? feet ?

If Da1-a2 = 1 (m? mm?) .. then ln (1/1) = 0...
If Da1-a2 < 1 (m?, mm), then ln (1/Da1-a2) < 0...

I checked the IEEE document, and the formula you have written is there as well...Da1-a2 is given in inches.. so if the conductors are 1 inch apart (25.4 mm for us SI engineers), then there is ZERO mutual capacitve reactance... that does not sound right...

There is something 'obvious'I am missing.

I would never take the logarithmic of a distance... as it does not have a unit anymore... I don't mind taking the log of a ratio (for example r / D ).

In fact "Elements of power system analysis" by William D. Stevenson, Jr (the favourite book of our senior resident elec engineer) states that Xc = 2.862/f * 10e9 * ln (D/r) (and that gives a linear reactance value .. ie per meter)

I am sure you have the answer to this question,

Boris

RE: Crculating currents in a neutral

I hate xls but I have not any other possibility. So, if I did some mistake tell me pls. and I will fix it as soon as possible.
The EPRI is not the first specification dealing with parallel single phase cables imbalance. The first was F.Buller in 1946 but is impossible to find this book. His premise was:
IA1+IB1+IC1+IN1+IA2+IB2+IC2+IN2=0
That means: IN1=-(IA1+IB1+IC1+IA2+IB2+IC2+IN2)
He stated the formula for British Units, of course, but now I use SI as more convenient.
Magnetic Flux between A1 and N1 is: F[A1-N1]=miuo*IA/2/pi()*ln(DA1-1/kcd/rc)*La
 where miuo=4*pi()/10^7 H/m air permeability [in SI or as we say MKSA-meter, kilogram[mass],second and ampere].
kcd=.778 [instead to add 0.25 I prefer to multiply by kcd]
XA1-N1=-w*FA1-N1=-2*PI()*frq*4*pi()/10^7/2/pi()*ln(DA1-N1/kcd/rc) that means:
XA1-N1=-4*frq*pi()/10^7*ln(DA1-N1/kcd/rc)
XB1-N1=-4*frq*pi()/10^7*ln(DB1-N1/kcd/rc) and so on.
EN1=IN1*RN1+j*(IA1*XA1-N1)+...+j*IN2*(XN2-N1)+INconsumer*(Rcons+j*Xcons)
Where IN1,IA1...IN2 are vectors[I should say "phasors" instead "vectors" as they are not actually vectors as E [Electric Field Intensity] for instance].
IA1*XA1-N1=4*frq*pi()/10^7*La*(IA1*ln(DA1-N1/kcd/rc)= 4*frq*pi()/10^7*La*(IA1*LN(DA1-N1)-IA1*LN(kcd*rc))
IB1*XB1-N1=4*frq*pi()/10^7*IB1*La*ln(DB1-N1/kcd/rc)= 4*frq*pi()/10^7*IB1*La*LN(DB1-N1)-IB1*LN(kcd*rc)
IC1*XC1-N1=4*frq*pi()/10^7*IC1*La*ln(DC1-N1/kcd/rc)=4*frq*pi()/10^7*La*( IC1*LN(DC1-N1)-IC1*LN(kcd*rc))
IN2*XN2-N1=IN2*La*ln(DN2-N1/kcd/rc)= IN2*4*frq*pi()/10^7*La*LN(DN2-N1)-IN2*LN(kcd*rc)
IN1*RN1+j*(IA1*XA1-N1)+...+j*IN2*(XN2-N1)=*4*frq*pi()/10^7*La*(IA1*LN(DA1-N1)+IB1*LN(DB1-N1)+....+IN2*LN(DN2-N1)-(IA1+IB1+IC1+IA2+IB2+IC2+IN2)*LN(kcd*rc))
but IN1=-(IA1+IB1+IC1+IA2+IB2+IC2+IN2) then:
IN1*RN1+j*(IA1*XA1-N1)+...+j*IN2*(XN2-N1)=*4*frq*pi()/10^7*La*(IA1*LN(DA1-N1)+IB1*LN(DB1-N1)+....+IN2*LN(DN2-N1)+IN1*LN(kcd*rc))
EN1=IN1*(RN1+4*frq*pi()/10^7*LN(kcd*rc)+4*frq*pi()/10^7*(IA1*LN(DA1-N1)+IB1*LN(DB1-N1)+....+IN2*LN(DN2-N1))+INconsumer*(Rcons+j*Xcons)
Notes:
1)    Self-reactance does not exist without reference to the other conductors. This formula presented by you in the above post does not represent the self-reactance but the reactance from conductor to its shield[if it does exist] or to earth if the shield is grounded or a partial reactance used to be combined in order to find the reactance between phases or to earth.
2)    As you may see, the measuring units are not important, but they have to be the same for all conductor.
Regards
 

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