Page 243 - Handbook of Electrical Engineering

CABLES, WIRES AND CABLE INSTALLATION PRACTICES 227

The equivalent full-load impedance Z flt is,

460
V os
Z flt = √ = √ = 0.08464 ohms/phase
3I flt 3 × 3137.8
This impedance also represents the 100% impedance of the transformer, hence by simple
proportion the ohmic resistance R t and reactance X t are,

1.08 × 0.08464
R t = R pu × Z flt = = 0.000914 ohms/phase
10

And
6.40 × 0.08464
X t = X pu × Z flt = = 0.005417 ohms/phase
100

The total impedance Z f upstream of the fuses when both transformers are operating is,

Z t R t X t
Z f = Z bs + = R bs + + j X bs + (9.8)
2 2 2
= 0.000597 + j0.004113 ohms/phase at 460 V.

The magnitude of which is 0.004156 ohms/phase.

This impedance has an X/R ratio of 6.8894 which will give rise to a current ‘doubling factor’
D of,

√ −πR √ −π
D = 2 1.0 + e x = 2 1.0 + e 6.8894
= 2.3106, see sub-section 11.6.1 for an explanation of D.

The prospective rms fault current at or near to the fuses is I f ,

460
V os
I f = √ = √ = 63903.1amps
3Z f 3 × 0.004156

Hence the peak value of the asymmetrical fault current I fpka = 2.3106 × 63903.1 =
147,654 amps.
To ensure that a good cut-off occurs in the fuses, choose the cut-off current I co to be say 20%
of the peak fault current I fpka ,

I co 0.2 × I fpka = 0.2 × 147,654

= 29,531 amps