Question 3.24: The equivalent circuit parameters of a 300 kVA, 2200/200 V, ...
The equivalent circuit parameters of a 300 kVA, 2200/200 V, 50 Hz single-phase transformer are: primary winding resistance, R_{1} = 0.1 Ω; secondary winding resistance, R_{2} = 0.01 Ω; primary leakage reactance, X_{1} = 0.4 Ω; secondary leakage reactance, X_{2} = 0.03 Ω; resistance representing core loss, R_{c} = 6 × 10³ Ω, magnetizing reactance X_{m} = 2 × 10³ Ω. Calculate the voltage regulation and efficiency of the transformer at full load at 0.8 power factor lagging.
We will consider E_{1} = 2200 V and E_{2} = 220 V
Transformation ratio, K = \frac{N_{2}}{N_{1}}= \frac{E_{2}}{E_{1}}= \frac{220}{2200}= \frac{1}{10} = 0.1
By transferring the secondary quantities to the primary side we will calculate the equivalent resistance and equivalent reactance of the transformer as
R_{e}^{′} =R_{1} + \frac{R_{2}}{K^{2}} = 0.1 + \frac{0.01}{(0.1)^{2}} = 1.1 Ω
X_{e}^{′} =X_{1} + \frac{X_{2}}{K^{2}} = 0.4 + \frac{0.04}{(0.1)^{2}} = 1.4 Ω
kVA rating = 300
VA rating = 300 × 1000
I_{1} = \frac{VA}{E_{1}}=\frac{300 × 1000}{2200} = 136 A.
p.f = \cos Φ= 0.8; \sin Φ = 0.6
The equivalent circuit with the secondary quantities referred to the primary side and the phasor diagram have been shown in Fig. 3.65.
V_{1}^{2} = AC^{2} + CB^{2}
= (E_{1} \cos Φ + I_{1} R^{′}_{e})^{2} + (E_{1} \sin Φ + I_{1} X^{′}_{e})^{2}
=(2200 × 0.8 + 136 × 1.1)^{2} + (2200 × 0.6 + 136 × 1.4)^{2}
= (1909)^{2} + (1510.4)^{2}
V_{1} = 2400 V.
Voltage regulation =\frac{V_{1} – E_{1}}{V_{1}} × 100 = \frac{(2400 – 2200)}{2400} × 100
= 8.3 per cent.
To calculate efficiency, we need to calculate the copper loss and core loss.
Full- load copper loss, W_{cu} = I_{1}^{2} R_{e}^{′} = (136)² × 1.1 = 20.345 kW.
Core loss = V_{1} I_{c} = V_{1} \frac{ V_{1}}{R_{c}} = \frac{ V_{1}^{2}}{R_{c} }= \frac{(2400)²}{6 × 10³} = 960 W
= 0.96 kW.
Efficiency, \eta = \frac{Output × 100}{Output + W_{cu} + W_{c}}
= \frac{300 × 0.8 × 100}{300 × 0.8 + 20.345 + 0.96}
= \frac{240 × 100}{261.3} = 91.84 per cent.
