Handbook of Electrical Engineering

226 HANDBOOK OF ELECTRICAL ENGINEERING

The base currentIbaseat 11,000 V is,

Ibase=

Sf× 106

√

3 ×Vbase

=

150 × 106

√

3 × 11000

= 7872 .9amps

The base impedanceZbaseis,

Zbase=

Vbase

√

3 ×Ibase

=

11000

√

3 × 7872. 9

= 0 .8067 ohms/phase

This impedance has anX/Rratio of 10, its resistanceRbaseand reactanceXbaseare,

Zbase=

√

R^2 base+Xbase^2 =

√(

Xbase

10

) 2

+X^2 base

=Xbase

√

0. 12 + 1. 02 = 1. 00499 Xbase

Xbase=

0. 8067

1. 00499

= 0 .8027 ohms/phase

And

Rbase=

Xbase

10

= 0 .08027 ohms/phase

Transfer these components to the secondary circuit at 460 volts, and call themZbs,RbsandXbs.

The transformation ratioupsof the impedance is,

ups=

4602

110002

= 0. 001749

Zbs=Zbase×ups= 0. 8067 × 0. 001749 = 0 .001411 ohms/phase

Similarly Rbs= 0 .000140 ohms/phase
And Xbs= 0 .001404 ohms/phase

The ohmic impedance of the load on the transformerZfltseen at its secondary winding is
found as follows:-

Full-load currentIfltof the transformer secondary winding is,

Iflt=

St× 106

√

3 ×Vos

Where,Stis the MVA rating of the transformer, andVosis the open-circuit line voltage of the
secondary winding.

Iflt=

2. 5 × 106

√

3 × 460

= 3137 .8 amps/phase