Answer / kashifhabib

Efficiency = (Output/Input)*100
&
Output (KW)= KVA * Power Factor
Input = Output + Iron losses + Copper losses
Iron losses are due to Primary voltage so remains constant
can be calculated by open circuit test
Copper losses are due to secondary current I*I*R
can be calculated by short circuit test
Exp: Full load copper losses 800 W
1/2 load copper losses will be 200 watts

Answer / vinod parasa

Efficiency = (Output/Input)*100
output=(x)(KVA * Power Factor)
input=(x)(KVA * Power Factor)+(X)^2 copperlosses+core losses

the volue of x depend on load
x=1 for full load
x=0.75 for (3/4)th load
x=0.5 for half load
x=0.25 for quarter load

copper losses are variable losses and iron losses are
constant losses.

at the time of full load copperlosses=iron losse
Full load copper losses 800 W
half load copper losses=(X)^2 copperlosses
x=0.5 for half load
=(0.5)^2 *400
=0.25*400
half load copper losses=200

Answer / n.makali

POWER Transformer is 55MVA, 27/11.5KV ,efficiency =99.49

Find the Losses of Transformer

Answer / s.ravi

Efficiency =
(Output/Output+Iron losses+Copper losses)*100
Iron losses = No load test,
Copper losses = Short circuit test
Regulation = (Impedance voltage/Rated input voltage)*100
Impedance voltage = Short circuit test(Voltage drop)

Answer / prof dr lehaz kakakhel

go for two tests
open and short circuits tests
From the open, u compute the core power loss and from the
short, u get copper losses. then the formula is
efficiency = output/ (output + the above two losses)
SIMPLE

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