Why is all-day efficiency preferred rather than commercial efficiency?

The ordinary or commercial efficiency of a transformer is given by the ratio

Output in watts

Input in watts

Again, for all-day efficiency,

¶=Output in Kwh/input in Kwh (For 24 hours)

Whereas core loss occurs throughout the day, the Cu loss occurs only when the transformers are loaded. Hence it is considered a good pracice to design such transformers so that Core loss is very low. The Cu losses are relatively less important, because they depend on the load. The performance of such a transformer should be judged by all day efficiency.

Math:

The energy output is 5*08*6=24 Kw=hr (operates at rated KVA)

½*5*.5*12=15 Kw-hr (one-half rated KVA)

Total Output = 39 KW-hr

The energy loss due to core loss is 35*24=840 W-hr=0.84 Kw-hr

The copper loss is 40*(1)2=40*6=0240 Kw-hr

¼*40*42=120 Kw-hr =0.120 Kw-hr

Total Copper energy loss = (0240+.120) Kw-hr

=0.360 Kw-hr

Total energy loss= (.840+.360) Kw-hr

=1.2 Kw-hr

Total energy input =39+1.2=40.2 Kw-hr

All-day efficiency= (total energy output)/ (total energy output)*100%

=39/40.2*

=97.01%

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