# High-Efficiency transformers For Energy Savings

In the example thus far, we have used a 150°C-rise, aluminum-wound, 1,500 kVA dry-type transformer. Its efficiency of 98.47% would appear to be quite high. However, let’s consider a premium-efficiency, 80°C-rise, copper-wound unit, with an efficiency of 99.02%, and carrying a price premium of \$5,900. Let’s further assume we have already gone through a savings and payback analysis of standard and high-efficiency motors, have chosen the efficient one, and have selected the larger wire size based on energy efficiency. The additional savings for selecting the high-efficiency transformer can be approximated as follows:

 transformer Loss = (load of motor + loss in wiring) x (transformer inefficiency) Loss in high-efficiency transformer based on efficient motor and wire size: = (14.94 kW + 0.159 kW) x (1 - 0.9902) = 0.148 kW, or 888 kWh/yr Loss in standard transformer based on efficient motor and wire size: = (14.94 kW + 0.159 kW) x (1 - 0.9847) = 0.231 kW, or 1,386 kWh/yr Additional savings due to more efficient transformer: 1,386 - 888 = 498 kWh/yr
at \$.09 per kWhat \$.07 per kWh
Value of Annual Savings in transformer: \$44.82 \$34.86

The high-efficiency transformer’s cost premium of \$5,900 must be spread over our example motor plus all other loads on the transformer, since this is only a small part of its total load. For simplicity, we’ll estimate what fraction of the transformer’s output is required by our motor, and reduce the transformer’s cost premium proportionately. The differential cost associated with the increment of the 25 hp high-efficiency motor is determined from:

Motor kVA = 30 FLA x 0.460 kV x √3 = 23.9 kVA(2)

 \$5,900 cost premium x 23.9 kVa 1500kVa = \$94 cost premium for one motor
at \$.09 per kWhat \$.07 per kWh
Value of Annual Savings \$44.82 \$34.86
Simple Payback 2.1 yr 2.7 yr

(2) 30 full load amps (FLA) is taken from the motor catalogue.