Diseño de instalaciones en media tensión

Design rules

0

Busbar calculation

Busbar calculation example
Here is a busbar calculation to check.

Exercise data
# Consider a switchboard comprised of at least 5 MV cubicles. Each cubicle has 3 insulators(1 per phase). Busbars comprising 2 bars per phase, inter-connect the cubicles electrically.

Busbar characteristics to check:
S d
l : : : : : : : busbar cross-section (10 •1)phase to phase distance distance between insulators on the same phase ambient température permissible temperature rise
(90-40=50)

10 18 70 40 50

cm2 cm cm °C °C

θn
(θ - θn) profile material
Cellule 1 Cellule 2 Cellule 3 Cellule 4 Cellule 5

flat busbars in copper 1/4 hard, with a permissible strain η = 1 200 daN/cm2 edge-mounted

arrangement:

number of busbar(s) per phase:

2d d

# The busbars must be able to withstand a rated current Ir = 2,500 A on a permanent basis and a short-time withstand current Ith = 31,500 A rms. for a time of tk = 3 seconds. # Rated frequency fr = 50 Hz # Other characteristics: 5 parts in contact with the busbars can withstand a maximum temperature of θmax = 100°C 5 the supports used have a bending resistance of F’ = 1 000 daN

1 cm1 cm

10 cm 5 cm

12 cm

d

d

TED300014EN_030_037.

Gamme

Schneider Electric

Design rules

Busbar calculation
0

Let’s check the thermal withstand of the busbars!

For the rated current (Ir)
The MELSON & BOTH equation allows us to define the permissible current in the conductor: 0.61 0 .5 0 .39

· 24.9 ( θ – θ n ) S p I = K------------------------------------------------------------------ρ20 [ 1 + α ( θ – 20 )]

with:
l : : : : : : permissible current expressed in amperes (A) ambient temperature permissible temperature rise* busbar cross-section busbar perimeter resistivity of the conductor at 20°C copper: 1.83 µΩ cm

θn
(θ - θn) S

40 50 10 22

°C °C cm2 cm

p
e

p20

a

α

:

temperature coefficient for the resistivity:

0.004e

K

:

condition coefficient product of 6 coefficients (k1, k2, k3, k4, k5, k6), described below

*(see table V in standard CEI 60 694 pages 22 and 23)

Definition of coefficients k1, 2, 3, 4, 5, 6:
# Coefficient k1 is a function of the number of bar strips per phase for: 5 1 bar (k1 = 1) 5 2 or 3 bars, see table below:

e /a
0.05 0.06 0.08 0.10 number of bars per phase 2 1.63 1.731.76 1.80 3 2.40 2.45 2.50 2.55 0.12 0.14 k1 1.83 1.85 2.60 2.63 0.16 1.87 2.65 0.18 1.89 2.68 0.20 1.91 2.70

In our case: e/a = number of bars per phase = giving k1 =
0.1 2 1.80

Schneider Electric

Gamme

AMTED300014EN_030_03

Design rules

0

Busbar calculation

# Coefficient k2 is a function of the surface condition of the bars: k2 = 1 5 bare: k2 = 1.15 5 painted: #Coefficient k3 is a function of the busbar position: k3 = 1 5 edge-mounted busbars: 5 1 bar flat-mounted: k3 = 0.95 k3 = 0.75 5 several flat-mounted bars: # Coefficient k4 is a function of where the bars are installed: k4 = 1 5 calm indoor atmosphere: 5 calm outdoor atmosphere: k4 = 1.2 5 bars in non-ventilated ducting: k4 = 0.80 # Coefficient k5 is a function of the artificial ventilation: k5 = 1 5without artificial ventilation: 5 cases with ventilation must be treated on a case by case basis and then validated by testing. # Coefficient k6 is a function of the type of current: 5 for alternatif current at a frequency of 60 Hz, k6 is a function of the number of busbars n per phase and of their spacing. The value of k6 for a spacing equal to the thickness of the busbars:
n k6 1 1 2 1 3 0.98

Inour case: n= 2

giving k6 = 1

In fact, we have: k = 1.80 • 1 •

1

•

0.8 •

1

•

1

= 1.44

0 . 61 0.5 0 . 39 • 10 • 22 24 . 9 ( 90 – 40 ) I = 1. 44 • ---------------------------------------------------------------------------------------------------------------------------------------1 . 83 [ 1 + 0 . 004 ( 90 – 20 ) ]

0 . 61 0.5 0 . 39 24 . 9 ( θ – θn ) •S •p I = K •...