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Losses and Efficiency

Author: E.E. Kimberly

A transformer energized from a power line and supplying no load, but in constant readiness to do so, draws an exciting current which is necessary to maintain the core flux, and this flux in turn maintains the secondary voltage. The exciting current has a power component which supplies the hysteresis and eddy-current losses in the iron and also a very small amount of resistive power loss in the primary winding. These losses are almost constant and are independent of the load, and constitute a cost to the power system which must be passed on to the consumer as a part of the readiness-to-serve charge in power contracts, as discussed in Chapter 24.

When a transformer is loaded, the secondary current and the added primary current cause an additional resistive loss which is proportional to the square of the load current. The losses may be summarized as follows:

1. Core loss (hysteresis and eddy-current)

2. Copper loss



The power efficiency, in per cent, is:


The power efficiency is maximum when the load is great enough to make the fixed losses equal to those which are variable functions of the load. The energy efficiency is of great importance in the case of any device, such as a distribution transformer, which must be kept energized and ready for use without notice but is actually used only a small part of the time. The energy efficiency is:


Example 17-3. - A 100-kv-a, 2300/230-volt, single-phase transformer, when excited at normal rated voltage with no load, has a power input of 1000 watts. The resistance of the primary coil is 0.243 ohm and that of the secondary coil is 0.00243 ohm. Calculate the power efficiency when the secondary current is 200 amp and the load power factor is 100%.

Solution. - Core loss = 1000 watts; secondary copper loss = Is2Rs = 2002 x 0.00243 = 97.2 watts; primary current = 1/10 x 200 = 20 amp; primary copper loss = Ip2Rp = 202 x 0.243 = 97.2 watts. Hence,


Last Update: 2010-10-06