0195136047.pdf

11.4 TRANSFORMER PERFORMANCE 489

Note that the current at one-half rated load is half of the full-load current, and that the

copper loss is one-quarter of that at rated current value.

Core loss =Poc=173 W

which is considered to be unaffected by the load, as long as the secondary terminal voltage

is at its rated value. Then

Total losses at one-half rated load= 162. 5 + 173 = 335 .5W

Input= 15 , 000 + 335. 5 = 15 , 335 .5W

The efficiency at one-half rated load and 0.6 power factor is then given by

η=

15 , 000

15 , 335. 5

× 100 = 97 .8%

EXAMPLE 11.4.3

The distribution transformer of Example 11.3.1 is supplying a load at 240 V and 0.8 power factor
lagging. The open-circuit and short-circuit test data are given in Example 11.4.2.

(a) Determine the fraction of full load at which the maximum efficiency of the transformer

occurs, and compute the efficiency at that load.

(b) The load cycle of the transformer operating at a constant 0.8 lagging power factor is 90%

full load for 8 hours, 50% (half) full load for 12 hours, and no load for 4 hours. Compute

the all-day energy efficiency of the transformer.

Solution

(a) For maximum efficiency to occur at a certain load, the copper loss at that load should be

equal to the core loss. So,

k^2 Psc=Poc

wherekis the fraction of the full-load rating at which the maximum efficiency occurs.

Therefore,

k=

√

Poc

Psc

=

√

173

650

= 0. 516

The output power corresponding to this condition is

50 , 000 × 0. 516 × 0. 8 = 20 , 640 W

where 0.8 is the power factor given in the problem statement. Also, core loss=copper

loss=173 W. The maximum efficiency is then given by

ηmax=

output

output+losses

=

20 , 640

20 , 640 + 173 + 173

=

20 , 640

20 , 986

= 0. 9835 , or 98.35%