Question 3.25: A 10 kVA 440/220 V, 50 Hz single-phase transformer gave the ...
A 10 kVA 440/220 V, 50 Hz single-phase transformer gave the following test results when both the following tests were conducted on the high-voltage side:
Open circuit test : 440 V, 1.0 A, 100 W
Short circuit test : 20 V, 22.7 A, 130 W.
Using the test data, calculate the efficiency and voltage regulation at 0.8 power factor lagging.
Full-load current on the high-voltage side = \frac{10 × 1000}{440} = 22.7 A
This shows that the short-circuit test has been conducted on full load. The wattmeter reading, therefore, represents the full-load copper loss.
W_{cu} = 130 W and from oc test data,
W_{c} = 100 W
Full-load efficiency is calculated as
\eta = \frac{Output}{Output + W_{cu} + W_{c}} × 100
= \frac{10 × 0.8}{10 × 0.8 + 0.13 + 0.1} × 100
= 97.2 per cent
Calculation of voltage regulation
Current, I_{1}= \frac{kVA}{V_{1}} = \frac{10 × 1000}{440} = 22.7 A.
From the short-circuit test data, wattmeter reading can be taken as equal to copper losses in the windings.
W_{SC} = 130 = I_{1}^{2} R_{e}^{′} = ( 22.7)² R_{e} ^{′}
Therefore, R_{e}^{′} =\frac{130}{( 22.7)²} = 0.25 Ω.
Z_{e} ^{′} = \frac{V_{SC}}{I_{SC}} = \frac{ 20}{22.7} = 0.88 Ω.
Z_{e} ^{2} = R_{e} ^{2}\ ^{′} + X_{e} ^{2}\ ^{′}
X_{e}^{′} = \sqrt{(0.88)² – (0.25)²} = 0.84 Ω.
\cosΦ = 0.8, \sinΦ = 0.6
Regulation = \frac{(I_{1} R_{e}^{′} \cosΦ + I_{1} X_{e}^{′} \sinΦ) }{V_{1}} × 100
= \frac{(22.7 × 0.25 × 0.8 + 22.7 × 0.84 × 0.6)}{440} × 100
= \frac{(4.54 + 11.44)}{440} × 100 = 3.63 per cent