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5 Copper Busbar Jointing Methods

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5 Copper Busbar Jointing Methods Posted  Jun 17 2015 by Edvard i n Energy and Power with 5 Comments 5 Copper Busbar Jointing Methods (photo redit! dona"mann#om$ Efcient joints in copper busbar conductors Efficient joints in copper busbar conductors can be made very simply by: 1# Bolting 
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2# Clamping 

%# Riveting 

&# Soldering 

5# Welding

Bolting and clamping are used extensively on-site. Shaped busbars may be prefabricated by using friction stir welding.

1. Bolted joints (most common)

Bolted joints are formed by overlapping the bars and bolting through the overlap area. They are compact, reliable and versatile but have the

disadvantage that holes must be drilled or punched through the conductors, causing some distortion of the current flow in the bar.

Bo"ted 'oints a"so tend to have a less uniorm contact pressure than those made by "amping but despite these issues bo"ted

 'oints are very ommon"y used and have proven to be reliable#

They can be assembled on-site without difficulty.

)igure 1 * + typia" bo"ted 'oint

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)igure % * + riveted 'oint

Go back to Methods ↑

$. Soldered or bra%ed joints (rare)

Soldered or bra$ed joints are rarely used for busbars unless they are reinforced with bolts or clamps since heating under short-circuit

conditions can ma!e them both mechanically and electrically unsound.

)igure & * + so"dered 'oint

Go back to Methods ↑

&. Welded joints (not ver' sae#)

elded joints are made by butting the ends of the bars and welding. They are compact and have the advantage that the current-carrying capacity is

unimpaired, as the joint is effectively a continuous copper conductor. %owever, it may not be safe or desirable to ma!e welded joints in situ.

elding of copper is discussed in opper !evelopment "ssociation #ublication $%, Cost-&ffective 'anufacturing( )oining of Copper and

Copper *lloys +!ownload here

.

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)igure 5 * + we"ded 'oint

Go back to Methods ↑

 oint Resistance Calculation

#n principle, a clamped or bolted joint is made by bringing together two flat surfaces under controlled +and maintained pressure, as shown in igure

.

)igure , * +n over"apped 'oint

The resistance of a joint is mainly dependent on two factors(

1#  -he stream"ine e.et or spreading resistance" Rs due to the diversion o/ the urrent ow through the 'oint

2#  -he ontat resistane or interace resistance o te joint" Ri#

*e total joint resistance" Rj" is given b'+ 

Rj = Rs+ Ri

This applies specifically to direct current applications. here alternating currents are flowing, the changes in resistance due to skin and

proximity effects in the joint &one must also be ta!en into account.

Go back to Methods ↑

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,ntroduction

hen a conductor carries a current it creates a magnetic field which interacts with any other magnetic field present to produce a force. hen the

currents flowing in two adjacent conductors are in the same direction the force is one of attraction, and when the currents are in opposite directions a

repulsive force is produced.

 In most busbar systems the current-carrying conductors are usually straight and parallel to one another.

*e orce produed by the two ondutors is proportiona" to the produts o/ their urrents#

 /ormally in most busbar systems the forces are very small and can be neglected, but under short-circuit conditions, they become large and must be

ta!en into account together with the conductor material fibre stresses when designing the conductor insulator and its associated supports to ensure

ade0uate safety factors.

The factors to be taken into account may be summarised as follows:

1# tresses due to diret "atera" attrative and repu"sive /ores#

2# 4ibrationa" stresses#

%# ongitudina" stresses resu"ting /rom "atera" deetion#

&# -wisting moments due to "atera" deetion#

#n most cases the forces due to short-circuits are applied very suddenly. 1irect currents give rise to unidirectional forces while alternating currents

 produce vibrational forces.

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-aimum stresses

hen a busbar system is running normally the interphase forces are normally very small with the static weight of the busbars being the dominant

component.

Under short-circuit conditions this is ery often not the case as the current rises to a peak of some thirty times its normal alue! falling after a few

cycles to ten times its initial alue.

These high transitory currents create large mechanical forces not only in the busbars themselves but also in their supporting system.

 -his means that the support insu"ators and their assoiated stee"wor3 must be designed to withstand these high "oads as we"" as

their norma" strutura" re6uirements suh as wind ie seismi and stati "oads#

The pea! or fully asymmetrical short circuit current is dependent on the power factor 'cos () of the busbar system and its associated connected

electrical plant. The value is obtained by multiplying the r.m.s. symmetrical current by the appropriate factor given in Balanced three-phase short-

circuit stresses.

#f the power factor of the system is not !nown then a factor of 2.33 will normally be close to the actual system value especially where generation is

concerned. /ote that the theoretical maximum for this factor is *+* or *,%*% where cos ( .. These pea! values reduce exponentially and after

approximately 45 cycles the factor falls to 4.5, i.e., the symmetrical r.m.s. short circuit current.

The pea! forces therefore normally occur in the first two cycles +5.56 s as shown in igure 4 below.

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)igure 1 * hortiruit urrent wave/orm

#n the case of a completely asymmetrical current wave, the forces will be applied with a fre0uency e0ual to that of the supply fre0uency and with a

double fre0uency as the wave becomes symmetrical. Therefore in the case of a 35 %$ supply these forces have fre0uencies of 35 or 455 %$.

The maximum stresses to which a bus structure is li!ely to be subjected would occur during a short-circuit on a single-phase busbar system in which

the line short-circuit currents are displaced by 4758.

8n a tree/pase s'stem a shortiruit between two phases is a"most identia" to the sing"e phase ase and a"though the

phase urrents are norma""y disp"aed by 1209 under shortiruit onditions the phase urrents o/ the two phases are a"most

1:09 out o/ phase# -he e.et o/ the third phase an be neg"eted#

 In a balanced three-phase short-circuit! the resultant forces on any one of the three phases is less than in the single-phase case and is dependent

on the relatie physical positions of the three phases.

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#n the case of a single-phase short-circuit, the forces produced are unidirectional and are therefore more severe than those due to a three-phase short-

circuit, which alternate in direction. The short-circuit forces have to be absorbed first by the conductor. The conductor therefore must have an

ade0uate proof strength to carry these forces without permanent distortion.

Copper satisfies this re0uirement as it has high strength compared with other conductor materials +Table 2 below.

/able * 0 /ypical relative properties of copper and aluminium

0roperties o Cu and l Copper(CW22$) luminium (1!&2) 3nits

E"etria" ondutivity (annea"ed$ 101 ,1 ; 8+C

E"etria" resistivity (annea"ed$ 1#72 2#:% <= m

 -emperature oe>ient o/

resistane(annea"ed$

0#00%? 0#00& @9 C

 -herma" ondutivity at 209C %?7 2%0 A@m 

Coe>ient o/ epansion 17 10*, 2% 10*, @9 C

 -ensi"e strength (annea"ed$ 200 * 250 50 * ,0 D@mm2

 -ensi"e strength (ha"/*hard$ 2,0 * %00 :5 * 100 D@mm2

0#2; proo/ stress (annea"ed$ 50 * 55 20 * %0 D@mm2

0#2; proo/ stress (ha"/*hard$ 170 * 200 ,0 * ,5 D@mm2

E"asti modu"u 11, * 1%0 70 3D@mm2

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pei heat %:5 ?00 J@3g  

Fensity :#?1 2#70 g@m%

Me"ting point 10:% ,,0 9C

Because of the high strength of copper, the insulators can be more widely spaced than is possible with lower-strength materials.

-etods o reducing conductor stresses

#n cases where there is a li!elihood of vibration at normal currents or when subjected to short- circuit forces causing damage to the conductor, the

following can he used to reduce or eliminate the effect(

a) Reduce the span between insulator supports.

This method can be used to reduce the effects of both continuous vibration and that due to short-circuit forces.

b) Increase the span between insulator supports.

This method can only be used to reduce the effects of vibration resulting from a continuous current. #t will increase the stresses due to a short-circuit

current.

c) Increase or decrease the fexibility o the conductor supports.This method will reduce the effects of vibration due to continuous current but has very little effect on that due to short-circuit forces.

d) Increase the conductor fexibility.

This can only be used to reduce the effects of vibration due to a continuous current. The short-circuit effect is increased.

e) Decrease the conductor fexibility.

This method will reduce the effects of vibration due to either a continuous current or a short-circuit.

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It will be noted that in carrying out the various suggestions above, changes can only be made within the overall

design requirements o the busbar system.

 Reference: Fundamentals of Power Sstem Protection


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