Question

A 100 KVA ,50 HZ, 440/11,000V, SINGLE PHASE TRANSFORMER HAS AN EFFICIENCY OF 98.5% when supplying full load current at 0.8 p.f logging and an efficiency of 99% when suppling half full load current at unity power factor .find the core losses and the copper losses corresponding to full load current?​

Answer 1

Answer:

Explanation

power =100KVA

v1/v2=440/110000

efficiency1=98.5%

full load=n=1

powerfactor =0.8

efficiency2=99%

half load=n=0.5

powerfactor=1

pc=copper loss

pi=iron loss

Answer 2

Answer: Pi=$0.26kw$ and Pc=$0.950Kw$

Explanation:

Here, Pc=copper loss

Pi=iron loss

P=$100KvA$

$\frac{V_{1} }{V_{2} } = \frac{440}{1100}$

n=$98.5$% - Full load,$n=1,Pf=0.8$

n=$99.1$%= half load,$n=0.5,Pf=1$

$n=\frac{nV_{2}I_{2}Cos(theta) }{nV_{2}I_{2}Cos(theta)+Pi+Pc(n^{2} )}$

$= \frac{1*100*0.8}{1*100*0.8+Pi+Pc(1)} = 0.985$

Upon solving,

Pi+Pc=$1.218$

Similarly,the other n upon solving is $0.99$

The deduced equation is $50.5050-50=Pi+0.35Pc\\0.5050=Pi+0.25Pc\\Pi=0.267\\Pc=0,950$

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