A single-phase, 10-kVA, 2300:230-V, 60-Hz, two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2300 V to 2530 V.

(a) Draw a schematic diagram of the arrangement showing all the voltages and currents while delivering full load.

(b) Find the permissible kVA rating of the autotransformer if the winding currents are not to exceed those for full-load operation as a two winding transformer. How much of that is transformed by electromagnetic induction?

(c) Based on the data given for the two-winding transformer in Problem 11.4.1, compute the efficiency of the autotransformer corresponding to full load and 0.8 lagging power factor. Comment on why the efficiency of the autotransformer is higher than that of the two-winding transformer.
Problem 11.4.1

These data were obtained from tests carried out on a 10-kVA, 2300:230-V, 60-Hz distribution transformer:

• Open-circuit test, with low-voltage winding excited: applied voltage 230 V, current 0.45 A, input power 70 W

• Short-circuit test, with high-voltage winding excited: applied voltage 120 V, current 4.35 A, input power 224 W

(a) Compute the efficiency of the transformer when it is delivering full load at 230 V and 0.85 power factor lagging.

(b) Find the efficiency of the transformer when it is delivering 7.5 kVA at 230 V and 0.85 power factor lagging.

(c) Determine the fraction of rating at which the transformer efficiency is a maximum, and calculate the efficiency corresponding to that load if the transformer is delivering the load at 230 V and a power factor of 0.85.

(d) The transformer is operating at a constant load power factor of 0.85 on this load cycle: 0.85 full load for 8 hours, 0.60 full load for 12 hours, and no load for 4 hours. Compute the transformer's all-day (or energy) efficiency.

(e) If the transformer is supplying full load at 230 V and 0.8 lagging power factor, determine the voltage regulation of the transformer. Also, find the power factor at the high-voltage terminals.