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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 = (0