Appendices

Summary of Methods of Busbar Rating

The following examples summarise the rating methods details in section 3 and section 4 for typical cases. UnlessOtherwise stated, a temperature rise of 50°C above an ambient of 4°C and a frequency of 50 Hz have been assumed. The Ratings may be increased by blackening the busbar surfaces. (See Radiation on Page 18)

Case I d.c., single rectangular-section bar on edge in still air.

Apply formula 4 or read direct from Table 12, for standard sizes.

Example:Copper bar 100mm x 6.3 mm (A =630 mm², p-212.6mm)I = 7.73 (630)0.5 (212.6) 0.39 = 1570 A(or read direct form Table 12).

Case II d.c., single circular -section bar (solid or hollow) in still air.

Apply formula 6 or read direct from Table 16, for standard sizes.

Example:50mm diameter copper rodI = 8.63 (1964)0.5 (157) 0.39 = 2360 A(or read direct form Table 16).

Case III d.c., laminated bars in still air.

a) Apply formula 4 or read direct from Table 12, for one bar.b) Multiply by appropriate factor from section 3

Example:4 copper bars 100mm x 6.3 mm with 6.3 mm spacing.I = 1570 A per bar.Multiply factor for 4 bars = 3.20.Hence I = 3.2x 1570 = 5020 A

Case IV a.c., single rectangular-section bar in still air.

Divide d.c. rating by appropriate value of √Rf/Ro as obtained from Figure 7

Example:Copper bar 100 mm x 6.3 mm (a/b = 100/6.3 = 16)d.c. rating 1579 A (Case i).Rf/Ro = 1.12 from figure 7√1.12 = 1.058Hence I = 1570/1.058 = 1480 A

Case V a.c., single rectangular-section bar in still air.

Divide d.c. rating by appropriate value of √Rf/Ro as obtained from Figure 4 (solid rods or tubes).

Example:50mm diameter copper rod.

d.c Rating = 2360 A (Case II)

x = 1.207 x 10-2 √Af

= 1.207 x 10-2 √1964x50

= 3.782

Hence Rf/Ro = 1.61, from Figure 4

2360Hence= I √1.61

I = 1860 A

Case VI a.c., laminated bars, in still air.

a) Determine rating of one bar as for Case IV .b) Multiply by appropriate factor, Table 8

Example:4 Copper bar 100 mm x 6.3 mm with 6.3 mm spacing d.c. rating per bar = 1570 A (as Case I).a.c. rating per bar = 1480 A (as Case IV).Multiplying factor for 4 bars = 2.3Hence I = 2.36 x 1480 =3404 A

Case VII Enclosed busbars

a) Multiply still air appropriate consultant (see Enclosed copper conductors) i.e.. by0.6 to 0.65 for conductor configurations largely dependent on air circulation (e.g.,Modified hollow square arrangement, Figure 9 (c)), or by 0.7 for tubular conductors or Closely grouped flat laminations.

b) Multiply by further 0.85 if enclosure of thick magnetic material.

Example:4 Copper bar 100 mm x 6.3 mm with arranged as in figure 9 (c), to carry a.c.d.c. rating , single bar = 1570 A (as in Case I).a.c. rating , single bar = 1480 A (as in Case IV).Multiplying factor for 4 laminations (Table 8) = 2.3Multiplying factor for configuration of figure 9 (c), (see Figure 11) = 1.28 Hence still air rating for this configuration = 1480 x 2.3 x 1.28 = 4360 AMultiplying factor for non-magnetic enclosure (Enclosed copper conductors) = 0.60Hence enclosed rating = 4360 x0.6 = 2610 AMultiplying factor for magnetic enclosure = 0.85Hence rating in magnetic enclosure = 2610 x 0.85 = 2220 A

Case VIII Economical use of busbar configurations

Example:

Two channels, each 100 mm high x 45 mm flange width x 8.6 mm thick (A = 1430 mm2 per channel). a.c. 60 Hz, 30o C rise on 40o C ambient in still air. From table 15, rating based on50o C rise on 40oC ambient. = 5550A

Use re-rating formula (equation 8) to obtain rating for 70oC working temperature and 40oC ambient.

Hence rating under conditions specified = 5550 x 0.756 = 4195A

Equivalent 4-bar laminated configuration for same cross-sectional area= 118mm x 6.3mmper bar (A= 743 mm2, P= 249 mm)

Hence d.c., rating per bar for 50oC rise on 40oC ambient.= 1300A (from equation 4, and application of appropriate conversion constant as above).

a/b = 118/6.3 = 18.7(see figure 7)

=1.08 (from figure 7 for 60Hz).

Hence a.c rating per bar = 1300/1.08 = 1190 A

Multiplying factor for 4 laminations = 2.3 (Table 8)

Hence a.c. rating for 4 laminations = 1190 x 2.3 = 2760 A

Thus the double channel arrangement is able to carry more current than laminated bars, in The ratio 1.52:1 for this cross-sectional area. This corresponds to the factor given in figure

11. For larger cross-sectional areas this factor would be still greater, for smaller sections the

Increase would be rather less than this, the exact value depends on the ratio of web to flange Lengths of the channel used, and on the thickness of web and channel; a rather wide spacingBetween “go” and “return” conductors is also assumed in Table 15, in order to approximate toThe “equi-inductance line” condition (see condition for minimum loss).

Current –carrying capacity of Busbars

(a) Flat bars on edge:

A05 p039θ0.61

I = 1.02 [(1+αθ)p]0.5

I = Current, AA = Cross-sectional area, mm2

P = Perimeter of conductor, mmΘ = Temperature difference between conductor and the ambient air, o Cα = Resistance temperature coefficient of copper at the ambient temperature,

per o C.Ρ = Resistivity of copper at the ambient temperature, µΩcm

If the temperature rise of the conductor is 50oC above ambient of 40oC and the resistivity of the copper at 20oC Is 1.724 µΩcm, then the above formula become:

(i) Flat bars:I = 7.73A05p039