- #1

Curiouschap101

- 4

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## Homework Statement

A 50 KVA,6360 V/230 V transformer is tested on open and short-circuit to obtain its efficiency,the results being as follows:

O.C.:

Primary Voltage-6360 V,

Primary Current-1 A,

Power Input - 2 kW

S.C:

Voltage across primary winding-180 V

Current in secondary winding-175 A

Power Input-2 kW.

Find the efficiency of the transformer when supplying full load at a power input of 0.8 lagging.

Also make the phasor diagram (ignoring impedance drops)for this condition.

## Homework Equations

Efficiency(η)=P

_{out}/(P

_{out}+P

_{OC}+n

^{2}P

_{SC})

n=applied load/full load

## The Attempt at a Solution

n=1

P

_{out}=50*0.8=40 kW

P

_{OC}=2 kW

P

_{SC}=2 kW*(230/180)

^{2}=3.265432099..kW

η=40/(40+2+3.265432099)=0.88367...p.u.

The solution in the book is given to be 0.887 p.u.(Maybe they have approximated P

_{OC}that's why.Please do check.)

But I am confused by the way the tests are conducted e.g. Why apply the open circuit test on L.V. side(kept as secondary in the first test if I am interpreting the problem correctly)?(My instructor told that O.C. test needs to be done with the H.V. side kept open(I forget the reason).On second thought power loss in core is said to be nearly invariant of the setup if I am correct right? ).As far as I know O.C Test can be used to determine the core loss and core impedance and the short circuit test is used to determine the copper loss and series impedance in circuit.Can you find the full load constraints(say full load current etc.) from any of these?