Question

A single phase transfermer is rated at100kva at Full load it's copper loss is 1200 watts and it's iron loss is 960wts calculate the efficiency at 0.75 of full load 0.7 power factor

Answer 1

Answer:

Given

KVA=2KV

Iron loss(Pi) =100W

Copper loss(Pcu) =200W

AT full load x =1

Unit power factor(Pf)=1

Let x*full load the efficiency is maximum then this value x can be

x=square root of Pi/Pcu =0.707

KVA at maximum efficiency =2*10^3*0.707=1414.21w

then maximum efficiency = (KVA at max. effi.*power factor)/(KVA at max. effi.*power factor+Pi+x^2Pcu) times 100%

hence (1414.21*1)/(1414.21*1+100+(0.707)^2*200) =0.876*100 =87.6%

therefore Maximum efficiency =87.6%