Question

A 5 KVA, 500/250 V, 50 Hz, single phase transformer gave the following reading,
O.C Test : 500 V, 1 A, 50 W (L.V side open)
S.C Test : 25 V, 10 A, 60 W (L.V side shorted)
Determine the following:
a) The efficiency of full load, 0.8 lagging p.f.
b) The voltage regulation on full load, 0.8 leading p.f.
c) The efficiency on 60% of full load, 0.8 leading p.f.
d) Draw the equivalent circuit referred to primary and insert all the values in it.​

Answer 1

Given:

A single-phase transformer with a rating

Power =$5KVA$,

Voltage= $500/250 V$

frequency=$50 Hz$

O.C Test:

$V_{o}$$=500 V$

$I_{o}=1A$

$P_{o}=50 W$

S.C Test:

$V_{SC}=25 V$

$I_{SC}=10 A$

$P_{SC}=60 W$

To Find:

a) The efficiency of full load, at 0.8 lagging p.f.

b) The voltage regulation on full load, at 0.8 leading p.f.

c) The efficiency on 60% of full load, at 0.8 leading p.f.

d) Draw the equivalent circuit that is referred to primary and put all the values in it.​

Solution:

It is observed that in both the tests, meters are on the H.V. side that is primary of the transformer so the parameters obtained from these results will be on the primary

For the O.C. Test

$cos$Ф$_{o}$$=\frac{P_{o} }{V_{o}\times I_{o} }$

$=\frac{50}{500\times1}$

$=0.1$

The currents

$I_{c}=I_{o}cos\Phi_{o}$

$=0.1 A$

$I_{m}=I_{o}sin\Phi_ {o}$

$=0.9949$ $A$

Resistance

$R_{0} =\dfrac{V_{0} }{I_{c} }\\=\dfrac{500}{0.1}\\=5000 \Omega$

Reactance

$X_{0}=\dfrac{V_{0} }{I_{m} }\\=\dfrac{500}{0.9949}\\=502.52 \Omega$

iron losses $P_{i}=P_{o}=50$ $W$

For the S.C Test,

Resistance

$R_{1e} =\dfrac{P_{sc} }{I_{sc}^{2} }\\=\dfrac{60}{(10)^{2} }\\=0.6 \Omega$

Impedance

$Z_{1e}=\dfrac{V_{sc}}{I_{sc} }\\=\dfrac{25}{10}\\=2.5 \Omega$

$X_{1e}=\sqrt{2.5^{2} -(0.6)^{2} }\\=2.4269 \Omega$

Full load current

$I_{1fl}=\frac{VA rating}{V_{1}}\\=\dfrac{5\times10^{3} }{500}\\=10 A$

$P_{sc} =(P_{cu} )F.L\\.=60 W$

a) The efficiency of full load at 0.8 lagging power factor

$\eta=\dfrac{(VA rating)cos\Phi_{2} }{(VA rating)cos\Phi_{2}+P_{i}+(P_{cu})F.L. }\times100\\=\dfrac{5\times10^{3}\times0.8 }{5\times10^{3}\times0.8+50+60 }\times100$

$=97.32$ %.

(b) voltage Regulation on full load,at cos$\Phi$= 0.8 leading p.f.

$\%R=\dfrac{I_{1fl} R_{1e}cos\Phi-I_{1fl}X_{1e}sin \Phi}{V_{1} }\times100$

$\%R=\dfrac{I_{1fl} R_{1e}cos\Phi-I_{1fl}X_{1e}sin \Phi}{V_{1} }\times100\\=\dfrac{10\times0.6\times0.8-10\times2.4269\times0.6}{500}\times100$

$=-1.95\%$

(c) The efficiency on 60% of full load at 0.8 leading p.f.

Copper loss in the new load

$P_{cu}=$ $\eta^{2}\timesP(cu)F.L$

$=(0.6)^{2}\times60\\=21.6 W$

$\%\eta=\dfrac{\eta(VA rating)cos\Phi_{2} }{\eta (VA rating)cos\Phi_{2}+P_{i}+\eta^{2}(P_{cu} F.L. ) }\times100\\=97.103 \%$.

(d) The equivalent circuit can be drawn as in the figure in the image