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Crossbonding

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Loor

Loor

Electrical
Jul 27, 2012
5
A circuit in trefoil and I would like to know how to do crossbonding on it. Should I apply this on a distance of 500m?
Any ideas suggestions on what to focus on?
 
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Is the total length of the trefoil system 500m? Usually cross bonding is only for long cable length (>1km) with high capacity and cross-sections greater than 630sqmm.
 
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no the length varie from 3km up to 10km.
Using cable of 800sqmm.
 
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Using above mentioned pdf article of Mitton Consulting Ltd. single point bonded system curve:
If cables are laid without any clearance-close one to another- and
if the rated current will be 800 A the build-up voltage for 1 km length will be 64 V[average].
For 500 m 900 A but one overall cable diameter clearance, the build-up induced voltage will be 12*9/2=54 V.
Since the maximum permissible induced voltage would be 65 V
I think you have to limit the minor section length to 1000 m- double as you proposed.

See : IEEE-575 Appendix D Calculation of Induced Voltages for induced voltage calculation.
See :
Raychem[Tyco Electronics Energy Division]Link boxes and sheath voltage limiters
 
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Hi 7anoter4 (Electrical)

Can you send me again the above mentioned pdf article of Mitton Consulting Ltd. single point bonded system curve.

I don't get the option to open it.

Do you also have the IEEE-575 Appendix D Calculation of Induced Voltages for induced voltage calculation?
I would be greatfull.

 
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Hi 7anoter4,

Thanks for the documents.
But I do have 1 question regarding the calculation
according to Prysmian.

That one is based on a flat formation.
As I want to do mine in trefoil formation, I see that
for the induced sheath voltage = Phase A, B & C = IXm

Now for the Metallic Shield Loss Formula for Solidly Grounded Shields: Phase A, B & C:
Are two calculations and Total loss:

Where P = Rs/Y and Q = Rs/Z
But my question is what is Y and what is Z.
What is the unit of P, Q, Y and Z.?
Thus why I can't continue the calculation.

Please do advice me on this.
 
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In case of equilateral cable geometry configuration nor P,Q neither Y you need in order to calculate induced voltage:
I*Xm -as you already said and for losses [in the case of cable shield both ends are short-circuited and grounded]
Ploss=I^2*Rs*[Xm^2/(Rs^2+Xm^2)]
Where I =rated cable current[ the current flowing through the cable conductor will heat it up to maximum permitted temperature]
 
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If you have to run 3*1*800 sqr.mm copper cables[10 kV] in trefoil arrangement underground 0.8 m depth RHO=90 Earth temperature 20oC
then calculated according to IEC 60287-1-1
ch.1.4.2 Buried cables where partial drying-out of the soil occurs
From different authors usually Δθx and v was employed.
Δθx is the critical temperature rise of the soil. This is the temperature rise of the boundary
between the dry and moist zones above the ambient temperature of the soil (θx- θa) (K);
Where θx=57 oC
v is the ratio of the thermal resistivities of the dry and moist soil zones (v = ρd/ρw)
v=2.26
See-for instance:
Improving the Under-Ground Cables Ampacity by using Artificial Backfill Materials
If only one end of the cable shield will be grounded maximum permitted current will be: 720 A per each conductor.
The induced voltage will be 65 V at 250-260 m cable length.
Both ends grounded- taken into consideration shield losses due to induced circulating current- the maximum permitted current
will be 608 A.
 
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Hi 7anoter4,

Is there any other software I can simulate with except vision. Can this be done as well in matlab?
I got 33.8V at 500m cable length.

Can you check if I'm doing it right?

Thanks.
 
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You are right. The xls file I sent it on August 4 was correct but for flat arrangement.
only voltAB is suitable for equilateral triangle.
Now I mixed inch and mm calculating Xm and I got an awful Xm.
I have a "software" handmade by my self in Visual Basic 6 but only for underground cable temperature calculation.
The shield induced voltage and losses I have not[it was a program in QBasic but it does not run in Win7].
I'll take your recommendation and I'll revise my "software".
If your cables run close one to another no clearance will remain then the distance will be only overall diameter.
Reducing Xm the shield losses will drop and the permissible current will raise. The new calculation reveals a non-significant
difference: 930 A with shield losses and 940 A without.
The conductor resistance increase due to proximity effect and Xm decrease then the shield losses also.
Using the above xls file for 1000 m and 900 A the induced voltage will be 44.8 volts.
 
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