different currents on the parallel wires of a phase 2

[gaux]

hi, all i got a question that maybe is very easy for electric guys, we have a generator, 400 V 1500 A, and we use 6 wires to get out the current, and i got diferents currents on each wire, i tight them again but it´s not sucesful, and now i don´t know why...we cheek the terminals and all seems to be ok, does somebody know why...
the diference is about 10 % of the current....?

i hope somebody could help me cause i´m worried about some wire will burn.... thanks in advance

 

[electricpete]

I wasn't thinking ahead too far. This is probably 6 leads to carry 3 phases. Modelling 3 phase is more of a challenge than I am up to.

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[ScottyUK]

epete,

120mm[sup]2[/sup] cross section area, not 120mm diameter. Not far off 12mm diameter as it turns out.


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

 

[gaux]

120mm^2 that´s 6 cables for each phase, and three phases... total cables 18
thanks for your attention,

 

[stevenal]

How are they connected? If stacked, the outer most terminals see more contact resistance.

When dealing with impedances close to zero, it doesn't take much of a difference to get to 10%.

By the way, the individual strand currents are not equal either.

 

[electricpete]

Good point.

My general impression - for smaller leads, the impedance is primarily resisive and would be sensitive to contact resistance and temperature-coefficient of resistance.

For larger leads, the impedance becomes primarily inductive and no longer very sensitive to contact resistance (assuming the cable is reasonably long) or temperature. Now more sensitive to position, proximity effects, shielding effects from raceway etc.

I will try some more this weekend to do an analysis using FEMM. I have never done a 3-phase component, but would be valuable as a learning activity for me.

Scotty - how did you know that "mm" in this case referred to area rather than diameter?


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[PeterKsiazek]

I'm guessing Scotty could tell because there aren't too many 120mm (12cm, almost 5") diameter wires running around.

 

[rbulsara]

It's (mm) a common (mis)nomer among trade persons in IEC world. The correct term should be sq. mm. Slightly more intuitive than AWGs.

I have heard many people in the USA referring sq.ft. as just ft.

 

[electricpete]

Thanks. Apparently one of those IEC peculiarities. We have a few peculiar terminologies of our own of course...

I have done a 3-phase model of 6 solid (*) 10mm diameter copper onductors per phase spaced at 15mm C/C distance. Each phase carries 1000A 60hz (balanced).

(* Yes I know stranded will be different. Solid is easier to model/understand and a logical starting point to check the model... will try getting to stranded later).

I started with the absolute simplest geometry I can think of as shown in slide 1. A 6pack of phase C on the left, then 6-pack of phase B in the middle, the 6-pack of phase A on the right... all equally spaced (hmm I'm getting thirsty).

The results are shown on slides 2 and 3. It shows a dramatic (factor of 2) difference within the B phase between the left-side currents of B phase and the right side currents of B phase. It sounds very non-intuitive at first... after all this is a symmnetrical circuit... the effect on B phase from C on the left should be the same as the effect on B from A phase on the right, right?

I believe that intuition is wrong and the results are correct. Draw a vector diagram. The voltage drop in B phase associated with it's own resistance and inductance leads the current by somewhere less than 90 degrees (it would lead by 90 degrees if no resistance... but we have resistance). The voltage drop in B phase due to A and C phase currents leads those currents by exactly 90 degrees (there is no resistive effect here, only mutual inductance). So when you draw the vector diagram, the original voltage drop from phase B alone is closer to either A or C (not sure which one). I tested the theory by increasing the conductivity and the pattern became increasingly more symmetrical (and more around outside of conductors due to skin effect) and perfectly symmetrical when I made the conductors perfectly conducting. So the symmetric effect occurs only in absence of B phase resistance.

Agree/disagree...any other comments?

There are of course a lot of other possible effects especially if we add a grounded raceway. But I think for solid conductor (*) we have already demonstrated 10% difference is very very easily achievable.

Meanwhile will start working on stranded.

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[ScottyUK]

ePete,

It was an educated guess based on generator size and voltage, and I hear the colloquial "120 mil" as a description for 120mm[sup]2[/sup] often enough to recognise it. And I know you guys use "mil" where we use "thou" to describe 0.001" so that's another opportunity for confusion when we contract the word millimetres into 'mil'.

I have a suspicion that the six conductors in the OP might be per phase if the 120mm[sup]2[/sup] size is correct because two conductors of that size are inadequate to carry 1500A.


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[tonyjony]

Pete, can you set the phase sequence in that software? I'm sure that if you change the sequence to c-b-a, the results will be mirrored

With the stranded, you will get the same result, because they are all at the aprox. same voltage/sequence/electromagnetic field in each phase than the solid conductor.

 

[7anoter4]

I don't understand how 6*120 sqr.mm [2 conductors per phase] could withstand 1500 A.
If the cables run on free open-air at 30 oC the maximum permitted for a XLPE insulated copper conductor 600 V [or 0.6/1 Kv] will be about 350 A .One needs at least 4 such cables per phase.
If the cables are laying on an open cable tray-no sunshine exposure- at 40 oC air at a distance of 90 mm one with respect to another, then 2 cables of 1000 MCM [500 sqr.mm] will be fair enough.
If the cables are in the same order –group 1 phase A,B,C and group 2 A,B,C then the current in the phase conductor will be as follows:
1-phase A conductor 1-650 A
2-phase B conductor 1-750 A
3-phase C conductor 1-850 A
4-phase A conductor 2-850 A
5-phase B conductor 2-750 A
6-phase C conductor 2-650 A
If the order will be A,B,C,C,B,A each cable current will the same 750 A.

 

[tonyjony]

there are 6 conductors per phase

 

[electricpete]

fjkaller - Yes you are right that the general pattern flips if I change the phase sequence, as expected.

By physical reasoning, I agree with your comments about effect of stranding. I think it has a pretty dramatic effect on the internal current distribution of large conductors (if these conductors were stranded, you wouldn't see the large current density difference within a single conductor that shows in my powerpoint above), but it has a much smaller effect on sharing among different conductors. So the sharing among conductors predicted using solid conductors is probably not too far away from what we'd see for stranded.


I don't have an easy method for modelling twisted stranding unless I use the approach outlined in "Electromagnetic Modeling By Finite Element Methods" by Bastos and Sadowski. Specifically section 6.4 of that book is "The Skew Effect in Electrical Machines Using 2D Simulation". It provides a means to combine multiple 2-d simulation to create a 3-d simulation of geometry where the position of the conductors gradually changes as we move axially. In the context of the book it is used for modelling skewed squirrel cage rotor bars but I am pretty sure it could be applied to skewed (twisted) stranded wire. But it looks like way more work than I want to try right now. So I am satisfied to think current sharing for stranded is similar by physical reasoning as discussed above.

I have been thinking about whether there would be a way to estimate current sharing analytically without finite element modeling if we made some simplifying assumptions such as uniform current density within a given conductor.

Let's look at a simpler system 2 conductors per phase (although I realize the one in question is 6 conductors per phase). We have a total of 6 conductors c=1..6..call them IA1, IA2, IB1, IB2, IC1, IC2. The previous 6 currents represent 6 complex unknowns. I think we could also get 6 complex equations to solve the unknowns.

The first three equations simply state that the currents within a phase pair add up to the known phase current:
IA1 + IA2 = IAtot
IB1 + IB2 = IBtot
IC1 + IC2 = ICtot

The second three equations simply state that the voltages accross the two conductors of a phase pair is equal.
VA1 =VA2
VB1 = VB2
VC1 = VC2

Now how would we compute the voltage for example VA1:
VA1 = R * IA1 + j * L * IA1 + j * Sum (M1j * Ij) for j=2..6
The Sum represents the voltage induced by mutual inductance to each of the other 5 conductors.

6 complex equations in 6 complex unknowns... should be able to solve it.

L and Mij should be dependent upon the geometry. But finding them is not so straightfroward. L and Mij would be well defined if each of the current variables represented a loop, but in our case the circuits IA1, IA2 etc do not represent loop but instead represent current floiwng in one direction and so the selection of expressions for L and Mij is not so obvious. I think maybe a way around this would be to add a 7th conductor – call it an "imaginary return conductor" which we visualize to carry the return current for each of the conductors... which of course sums to 0 = -IA1 - IA2 - IB1 - IB2 - IC1 - IC2. Now we can view each current as a loop which goes out over one path (for example IA1) and returns over the imaginary return path, and this gives us the advantage of reformulating the problem in terms of loops so we can come up with expressions for L and M from the geometry of the rpoblem. When we add all the loops up of course we have the correct current distribution because the currents all sum to 0 so that current in that imaginary return conductor is 0.

I'll give it a try if I get a chance. Unless someone else has a suggetsion for an easier way to calculate current sharing from geoemtry under simplifying assumptions such as 1 – uniform current density within conductor; (and of course we also have thus far made simplifying assumption that there are no other current paths such as raceway or shields grounded at both ends).

Meanwhile, I am curious how 7anoter4 came up with his numbers for the particular current distributions discussed 27 Feb 09 10:13. Can you explain how you developed those numbers?


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[electricpete]

j was a bad choice for index of summation above.
"VA1 = R * IA1 + j * L * IA1 + j * Sum (M1c * Ic) for j=2..6"
should have been:
"VA1 = R * IA1 + j * L * IA1 + j * Sum (M1c * Ic) for c=2..6"

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[7anoter4]

I followed the EPRI calculation way [as per EPRI-Power Plant Electrical ref. series VOL.4-WIRE AND CABLES EQ.A-1 FROM APPENDIX A] :
Ea1=Ia1*(Ra1+jXa1)+Ia2(jXa1-a2)+Ib1(jXa1-b1)+...+Ia(Rl+jXl) where:
Ea1 = 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
Da1-a2=the distance between A1 and A2[typical]
Xa1-a2 the mutual reactance between conductor A1 and A2 .
Xa1-a2=LA/1000*.023*ln(1/Da1-a2) LA= minimum of La1 and La2
Xa1-b1 the mutual reactance between conductor A1 and B1 .
Xa1-b1=LA/1000*.023*ln(1/Da1-b1) LA= minimum of La1 and Lb1
Xa2-b1 the mutual reactance between conductor A2 and B1 .
Xa1-b1=LA/1000*.023*ln(1/Da2-b1) LA= minimum of La2 and Lb1
Xa1-b1- Xa1-b1=k*ln(Da2-b1/Da1-b1) where k=LA/1000*.023
Ia is phase A total current and 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)
For conductor A1 will be:
Ea1=Ia1*(Ra+jXa)+Ia2*(jXA1-A2)+Ib1(jXa1-b1)+Ic1(jXa1-C1)+Ia2(jXa1-a2)+Ib2(jXa1-b2)+Ic2(jXa1-c2)+Ia(Rl+jXl)
For conductor A2 will be:
Ea2=Ia2*(Ra+jXa)+Ia1*(jXa1-a2)+Ib1(jXa2-b1)+Ic1(jXa2-c1)+Ib2(jXa2-b2)+Ic2(jXa2-c2)+Ia(Rl+jXl)
Substracting Ea1-Ea2=(Ia1-Ia2)*[(Ra+jXa)-jXa1-a2]+Ib1*(j*Xa1-b1-jXa2-b1).....Ic2*(j*Xa1-c2-jXa2-c2)=0
Ia1={Ia2*[(Ra+jXa)-jXa1-a2]-Ib1*(jXa2-b1-jXa1-b1)-...Ic2*(jXa2-c2-jXa1-c2)}/ [(Ra+jXa)-jXa1-a2]
.................................................................................................................................
Ic2={Ic1*[(Rc+jXc)-jXc1-c2]-Ib1*(jXc2-b1-jXc1-b1)-...Ia2*(jXc2-a2-jXc1-a2)}/ [(Rc+jXc)-jXc1-c2]
Ia1=Ia2-[Sum(Ii*ln(Da1-ij/Da2-ij)|i=b,c j=1,2|]/[(Ra+jXa)-jXa1-a2]
.............................................................................
Ic2=Ic1+[Sum(Ii*ln(Da1-ij/Da2-ij)|i=a,b j=1,2|]/[(Rc+jXc)-jXc1-c2]
As you could see Xa-Xa1-a2 will be k*ln(Da1-a2/cond.radius).
As it is difficult to solve it by matrix [or determinant] I prefer iteration way .So I recalculate all this about 10-1000 times. At the beginning I take all current as equal [in module] and according to lag angle [angle 0 is phase A -120 degrees phase B and - 240 phase C] I extract Iactive and Ireactive. Of course I do it better using QB or VB instead of Excell .
I have a very old program translated from GWBasic in QBasic and I think with minor modifications may run again on GWBasic. [The last program in VB6 it is difficult to be inserted here].If it would not boring you I could post it soon.
Best Regards

 

[electricpete]

Very cool. I will have to study that.
Thanks!

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[mccla5214]

must have missed something early in the post. He has 6 wires per phase/ 3 phases / ? conduits. Each conduit has an equal number of phase A, B and C conductors. Yes measuring different conductors would result in different currents because of different loading on each phase. Also he never mentioned the phase to ground voltage of each phase. What if phase A was 500, phase B was 477 and Phase C was 460. According to OG&E our local public utility, this spread would be normal and they would not adjust the voltage but it would result in a large difference in phase currents. If he has access to a large clamp on CT he can put a 0-5 amp meter on it and clamp it around the entire bundle of wires and he should see 0 current if the sum of all currents balance out. If they don't then he could have a real problem because this will result in heating inside the conduit.

Let's hope to God that they did not run all 6 Phase A wires in the same conduit. I have seen that melt rigid conduits.

Mike

 

[electricpete]

Mike - look at the post dated 24 Feb 09 9:07 where he talks about conductors going to the same bus bar and the post 24 Feb 09 10:18 where he talks about current being "shared".
From that I concluded he is comparing currents within the same phase.

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[mccla5214]

I was assuming (I know) that he had 6-500 MCM conductors per phase. If that is the case then he has a signifacant difference in lead length since the conduits have to be spread apart. We dont know how many bends, junction boxes that his feeder runs through so we don't have realistic idea of the total length. I am still trying to wrap my mind around the idea of having 6 conduits so it makes me wonder if he has cable tray. Another question is are the conduits PVC which would allow the currents in all phases to interact. My bet is that the cables are at least 2 ft longer from conduit 1 to 6.

 

[electricpete]

Good point.

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