Handbook of Electrical Engineering

CABLES, WIRES AND CABLE INSTALLATION PRACTICES 227

The equivalent full-load impedanceZfltis,

Zflt=

Vos

√

3 Iflt

=

460

√

3 × 3137. 8

= 0 .08464 ohms/phase

This impedance also represents the 100% impedance of the transformer, hence by simple
proportion the ohmic resistanceRtand reactanceXtare,

Rt=Rpu×Zflt=

1. 08 × 0. 08464

10

= 0 .000914 ohms/phase

And

Xt=Xpu×Zflt=

6. 40 × 0. 08464

100

= 0 .005417 ohms/phase

The total impedanceZfupstream of the fuses when both transformers are operating is,

Zf=Zbs+

Zt

2

=Rbs+

Rt

2

+j

[

Xbs+

Xt

2

]

(9.8)

= 0. 000597 +j 0 .004113 ohms/phase at 460 V.

The magnitude of which is 0.004156 ohms/phase.

This impedance has anX/Rratio of 6.8894 which will give rise to a current ‘doubling factor’
Dof,

D=

√

2

[

1. 0 +e

−πR

x

]

=

√

2

[

1. 0 +e

−π

6. 8894

]

= 2. 3106 ,see sub-section 11.6.1 for an explanation ofD.

The prospective rms fault current at or near to the fuses isIf,

If=

Vos

√

3 Zf

=

460

√

3 × 0. 004156

= 63903 .1amps

Hence the peak value of the asymmetrical fault current Ifpka= 2. 3106 × 63903. 1 =
147 ,654 amps.

To ensure that a good cut-off occurs in the fuses, choose the cut-off currentIcoto be say 20%
of the peak fault currentIfpka,

Ico 0. 2 ×Ifpka= 0. 2 × 147 , 654

= 29 ,531 amps