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DESIGN DATA

SYMBOL UNITS

-

n -

A mm2

d mm

Rdc ohm/m

c / C

To C

Tc C

Ts C

r -

V m/s

e -

S' W/m2

a -

CALCULATION

Continous Current Rating

The ampcity of the conductor is calculated based on thermal rating as follows:

I2Rac + Ha = Hc + Hr

where

PARAMETER VALUE DESCRIPTION

Hard drawn Aluminium Material of Conductor

2 No. of Conductor per Phase (feeder)

300 Conductor cross Area

22.61 Conductor Diameter

9.48E-05 Conductor DC Resistance @ 20 C

0.00403 Temperature Coefficient of Resistance

35 Ambient Temperature

85 Max. Conductor Operating Temperature

200 Final Conductor Short - Circuit Temperature

1.023 Skin Effect Constant

1.2 Velocity of Wind

0.6 Enissivity coefficient of Conductor

1200 Solar radiation intensity

0.8 Solar Absorption Coefficent

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I = Current Rating of ConductorRac = a.c resistance at operating temperature

Ha = Heat gained by solar radiation

Hc = Heat loss by convection

Hr = Heat loss by radiation

Resistance value Rac at conductor operating temperature

At 20 degC

Rdc = 9.48E-05 W/m

At 85 degC

Rac = r * Rdc * 1 + c * (Tc-To)

= 1.023 * 9.48E-05 * 1 + 0.00403 * 50= 1.023 * 9.48E-05 * 1 + 0.2015

= 1.023 * 9.48E-05 * 1.2015

= 1.17E-04 ohm/m

Heat loss by convection (W/m)

Hc = 0.387 * ( V* d )0.448

* (Tc - To )

= 0.387 * 27.1320.448

* (Tc - To )

= 0.387 * 55.560.448

* 50

= 0.387 * 6.04858074 * 50

= 117.04 W/m

Where,

v = Wind Velocity (m/sec)

d = Conductor Diameter (mm)

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Heat loss by radiation ( W/m )

Hr = 5.70E-11 * e * ( Tc + 273)4

- (To + 273 )4

* pi * d

= 5.70E-11 * 0.6 * 16426010896 - 8999178496 * 3.1415 * 22.61

= 5.70E-11 * 0.6 * 7426832400 * 3.1415 * 22.61

= 18.04 W/m

Heat loss by absorption (W/m )

Ha = 1.00E-03 * a * S' * d= 1.00E-03 * 0.8 * 1200 * 22.61

= 2.17E+01 W/m

Conductor Current Rating

From equation

I = Hc + Hr - Ha

Rac=

=

= 986 Amp per conductor

Short - Circuit current Rating

According to Annex A-9 of IEC 865-1

Rated short - time withstand current density S thr is given by:

S thr = K

Tkr

1.13E+02

1.17E-04

9.73E+05

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Where K = K20CP ln 1+a20 (Tc - 20 C)

a20 1+a20 (Ts - 20 C)

and Tkr = Rated short time

For Aluminium Conductor

C = 910 J/kg C

P = 2700 kg/m

3

K20 = 9.48E-05 /Wmm

a20 = 0.004 / C

Substituting the above values into equiation

Where K = K20CP ln 1+a20 (Ts - 20 C)

a20 1+a20 (Tc - 20 C)

= 2.33E+02 ln 1.72

0.004 1.26

= 5.82E+04 ln 1.36508

= 5.82E+04 * 0.31121

= 1.81E+04

= 134.62 As/mm2

From equation

S thr = K

Tkr

= 134.62

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3= 77.72 A/mm

2

Minimum conductor cross- area for 40kA, 3 sec. Rating

Amin = 31500

77.72

= 405.29 mm2

Result

For bus coupler,4 x 800mm2

Aluminium Conductor is used,

Continuous Current rating = 4 x 1577.8 X = 1,972.81 A > 2000A

Short time current rating = 4 x 800 x 77.72 = 47 kA > 31.5kA

Hence, the current rating is sufficient