Open and short-circuit tests performed on a 500 kVA, 6600/2300 V, 50 Hz transformer yielded the following data:
No-load loss = 3 kW
Full-load short circuit loss = 4 kW
(a) Calculate the load (kVA) at which the transformer efficiency would be maximum for a given power factor. Calculate this efficiency for a pf of 0.85.
(b) The transformer supplies the following load cycle.
12 hours, full load 0.8 pf.
12 hours, half full load 0.9 pf.
alculate the energy efficiency of the transformer.