Summary of Methods of Busbar Rating

The following examples summarise the rating methods detailed in section 3 and section 4 for typical cases. Unless otherwise stated, a temperature rise of 50°C above an ambient of 40°C and a frequency of 50 Hz have been assumed. The ratings may be increased by blackening the busbar surfaces. (see Radiation)
Apply formula 4 or read direct from Table 12, for standard sizes.
Example:
Copper bar l00 mm x 6.3 mm (A = 630 mm2, p= 212.6 mm)
I = 7.73 (630)0.5 (212.6)0.39 = 1570 A
(or read direct from Table 12).

Apply formula 6 or read direct from Table 16 for standard sizes.
Example:
50 mm diameter copper rod
I = 8.63 (1964)0.5 (157)0.36 = 2360 A
(or read direct from Table 16).

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

Divide d.c. rating by appropriate value of
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
Ö1.12= 1.058
Hence I = 1570/1.058 = 1480 A

a) Divide d.c. rating by appropriate value of
as obtained from Figure 4 (solid rods or tubes).
Example:
50 mm diameter copper rod.
d.c. rating = 2360 A (Case II)
Hence Rf/Ro = 1.61, from Figure 4
Hence

a) Determine rating of one bar as for Case IV.
b) Multiply by appropriate factor, Table 8
Example:
4 copper bars 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 = 2.3
Hence I = 2.3 x 1480 = 3404A

a) Multiply still air rating by appropriate constant (see Enclosed copper conductors) i.e.. by 0.6 to 0.65 for conductor configurations largely dependent on air circulation (e.g., modified hollow square arrangement, Figure 9c), 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 bars 100 mm x 6.3 mm arranged as in Figure 9c, 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) = 2.3
Multiplying factor for configuration of Figure 9c, (see Figure 11) = 1.28
Hence still air rating for this configuration = 1480 x 2.3 x 1.28 = 4360 A
Multiplying factor for non-magnetic enclosure (Enclosed copper conductors) = 0.60
Hence enclosed rating = 4360 x 0.6 = 2610 A
Multiplying factor for magnetic enclosure = 0.85
Hence rating in magnetic enclosure =2610 x0.85 = 2220 A
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, 30°C rise on 40°C ambient in still air. From Table 15, rating based on 50°C rise on 40°C ambient. = 5550 A
Use re-rating formula (equation 8) to obtain rating for 70°C working temperature and 40°C ambient.
Hence rating under conditions specified = 5550 x 0.756 = 4195 A
Equivalent 4-bar laminated configuration for same cross-sectional area = 118 mm x 6.3 mm per bar (A = 743 mm2, p = 249 mm).
Hence d.c., rating per bar for 50°C rise on 40°C ambient. = 1300 A (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 60 Hz).
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 spacing between "go" and "return" conductors is also assumed in Table 15, in order to approximate to the "equi-inductance line" condition (see Condition for minimum loss).

Geometric Mean Distance Formulae

Rectangular Bars


Ref. Dwight 'Electrical Coils and Conduits' 1946, p. 143

Three Phase conductors


Ref. Dwight ' Geometric Mean Distance for Rectangular Conductors' 1946

Figure 18 - Geometric Mean Distance - two rectangular bars (Apologies for quality of diagrams. Please contact CDA UK if you need better quality)

Capacitance Formulae of Busbars

The capacitance of an a.c. system can be of great importance when designing the protection equipment for the busbars and associated electrical plant. Capacitances for several configurations of busbars are as follows, where
permittivity E = E0Er
where
Er = relative permittivity of the material.
Isolated twin line

Line above a conducting earth

Twin line above a conducting earth

Isolated three-phase line with transposition

Concentric cylinders

Inductance Formulae of Busbars

The development of inductance formulae is mathematically complex and is the subject of many electrical engineering papers and books. To enable many of the normal busbar configuration inductances to be calculated for self and mutual inductances, the following formulae have been included.
It should be noted that self inductance LS is the inductance due to a single conductor, assuming that it is effectively outside the flux range of all other conductors. Mutual inductance M is the inductance resulting from the flux from other conductors.

Single Conductor
where
Ls = self inductance, mH
Ds = 0.2235 (a + b)
= g.m.d. (geometric mean distance from itself), cm
l = length of busbar, cm
Two Parallel Conductors
where
M = Mutual inductance, mH
Dm = g.m.d between bars
Obtain Dm from figure 18 or formulae at end of this Section)
l = length of busbar, cm
Go-and-Return Conductors
The inductance L per conductor includes both self and mutual components and therefore becomes equal to LS - M, i.e.,
where the conductor spacing is small compared with the conductor length, or
where b is small compared with d.
Single Conductor

Two Parallel Conductors

Go-and-return conductors
As before,
L = Ls - M
if the conductor spacing is small in comparison with its length and in comparison with d.

Busbar Impedance

The busbar reactance is not normally sufficiently large to affect the total reactance of a power system and hence is not included in the calculations when establishing the short-circuit currents and reactive volt drops within a power system. The exception to this is when considering certain heavy current industrial applications such as furnaces, welding sets, or roll heating equipment for steel mills. In these cases the reactance may be required to be known for control purposes, or to obtain busbar arrangements to give minimum or balanced reactance. This may be important because of its effect on both volt drop and power factor, and hence on the generating plant kVA requirement per kW of load, or on the tariffs payable where the power is purchased from outside.
The busbar impedance is made up of three components: resistance, inductance and capacitance. The values of these components are given an ohmic value which in the case of inductance and capacitance is dependent on the frequency of the system. They are defined as follows:
Resistance:
where Rf = resistance at frequency f (Hz), W
Ro = d.c. resistance
S = skin effect ratio
K = proximity ratio
Inductance:
where XL = inductive reactance, W
f = frequency, Hz
L = inductance, H
Capacitance:
where Xc = capacitive reactance, W
f = frequency, Hz
C = capacitance, F
Impedance:
where X = XL - XC
The value of XC is usually very much smaller than XL, and XL is usually much larger than Rf. The value of X is taken to be positive with the sign of XL - XC to indicate whether the system has a positive or negative power factor.
The volt drop in a busbar system is estimated as follows from the usual formula:
VB = I ZB
where VB = volt drop, V
I = current flowing in the conductor, A
ZB = busbar impedance, W
However, to find the magnitude of the load voltage VL available, the busbar volt drop VB must be subtracted vectorially from the supply voltage VS:
VS = supply voltage, V qL = angle of load, °
VB = busbar volt drop, V fB = angle of busbar, °
VL = load voltage, V RB = busbar resistance, W
I = current, A XB = busbar reactance, W
The apparent volt drop in the busbar trunking, phase to neutral, is given by:
Multiply by Ö3 for phase to phase volt drop.
The above formula gives a very close approximation as long as the busbar system volt drop remains small in comparison to the system voltage.

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