Answer / sathish n p

Take an example let tr rating is 2500KVA, 11KV/415v

TO calculate maximum loading on primary side
2500 X 10^3=1.732 X 11 X 10^3 X I

I=131.01Amps
Similarly Secondary side
2500 X 10^3=1.732 X 415 X I

I=3478 Amps

Answer / debashis

For determining the optimum loading of a transformer we need
to determine the load for which my copper loss would be equal to my iron loss.

At this point the loss would be at its minimum level and hence
the efficiency would be maximum.

Remember Iron loss is independent of load and copper loss depends on the load.

Answer / elminshawy

Sorry Vikadeny
But how it is optimum if I bought a 100MVA transformer and
I only loaded it to 50MVA or 70MVA ?

for optimum loading of transformers please refer to IEC
standard (60354 Loading guide for oil-immersed power
transformers )

Answer / rajendran

I think question refers to the optimum loading at which the
efficiency of the transformer is more/maximum...


if it is like that then we can refer the loading curves
provided by the manufacturer of the transformer....

generally as a thumbrule the optimum loading point is
around 85 to 95 percent of the capacity..

Answer / vikadent

As a rule of thumb, the optimum loading is between 50% to 70
% and my reference is Electrical installation guide by
Schneider and this is the range where the losses are minimum
inside the transformer.

Answer / khalid hussain shaikh

POWER=1.732*VOLTAGE*CURRENT*POWER FACTOR
HERE POWER=2500KVA
VOLTAGE=11 KV
POWER FACOR AS STANDARD=0.85
I=?
I=2500KVA/1.732*11KV*0.85
KV CANCELLED EACH DENOMINATOR AND NUMARATOR,
I=2500/1.732*11*0.85
I=154 AMP (11 KV SIDE)
I=2500 KVA/1.732*415 V*0.85
V CANCELLED EACH DENOMINATOR AND NUMARATOR
I=2500 K/610.963
I=4.091 KA
I=4091 AMP

hi frnds can any one help me ? plz send me placement paper of IOCL of Electrical streem .... plz send ....its urgent....

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