0195136047.pdf

488 MAGNETIC CIRCUITS AND TRANSFORMERS

(b) The corresponding phasor diagram is shown in Figure E11.4.1(b).

(c) The equivalent circuit of the transformer, along with the feeder impedance, referred to the

high-voltage side, neglecting the exciting current of the transformer, is shown in Figure

E11.4.1(c). The total series impedance is( 0. 5 +j 2. 0 )+( 1. 5 +j 2. 0 )= 2 +j 4 . Using

KVL,

V ̄S= 2400 0°+( 20. 8  − 36 .9°)( 2 +j 4 )= 2483. 5  0 .96° V

The voltage at the sending end is then 2483.5 V. The power factor at the sending end is

given by cos( 36. 9 + 0. 96 )°= 0 .79 lagging.

EXAMPLE 11.4.2

Compute the efficiency of the transformer of Example 11.3.1 corresponding to (a) full load,

0.8 power factor lagging, and (b) one-half load, 0.6 power factor lagging, given that the input

powerPocin the open-circuit conducted at rated voltage is 173 W and the input powerPscin the

short-circuit test conducted at rated current is 650 W.

Solution

(a) Corresponding to full load, 0.8 power factor lagging,

Output= 50 , 000 × 0. 8 = 40 , 000 W

TheI^2 Rloss (or copper loss) at rated (full) load equals the real power measured in the

short-circuit test atrated current,

Copper loss =IHV^2 Req HV=Psc=650 W

where the subscript HV refers to the high-voltage side. The core loss, measured at rated

voltage, is

Core loss=Poc=173 W

Then

Total losses at full load= 650 + 173 =823 W

Input= 40 , 000 + 823 = 40 ,823 W

The full-load efficiency ofηat 0.8 power factor is then given by

η=

output

input

× 100 =

40 , 000

40 , 823

× 100 =98%

(b) Corresponding to one-half rated load, 0.6 power factor lagging,

Output=

1

2

× 50 , 000 × 0. 6 = 15 ,000 W

Copper loss=

1

4

× 650 = 162 .5W