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Equivalent circuit and half-load efficiency of 50 kVA transformer

Short answer

From open-circuit test on HV: R0=20833 Ω, X0=9092 Ω, referred to secondary become 208.33 Ω and 90.92 Ω. Short-circuit test gives R01=1.25 Ω, X01=4.841 Ω, referred to secondary: R02=0.0125 Ω, X02=0.0484 Ω. At half load, copper loss=125 W, iron loss=300 W, efficiency (upf)=98.33%.

A 50 kVA, 2500V/250V, 50 Hz single phase transformer draws a current of 0.3 Amp at no load and consumes 300 Watts. When the primary winding is supplied by 100 V with secondary winding short circuited, the primary draws a current of 20 Amp and consumes 500 watts. Calculate the equivalent circuit parameters of the transformer referred to secondary side. Also calculate the efficiency of the transformer at half load.

Interpretation: the no-load test (0.3 A, 300 W) is at rated primary voltage, 2500 V, and the SC test is on the primary (HV) with the secondary shorted. Rated primary current =50000/2500=20= 50000/2500 = 20 A, so the SC test is at full load. Factor HV to LV: (2502500)2=0.01\left(\frac{250}{2500}\right)^2 = 0.01.

Shunt branch (no-load test on HV: 2500 V, 0.3 A, 300 W)

cos⁡ϕ0=3002500×0.3=0.4,sin⁡ϕ0=0.9165Iw=0.3×0.4=0.12 A,Iμ=0.3×0.9165=0.275 AR0=25000.12=20833 Ω,X0=25000.275=9092 Ω (HV side)\begin{aligned} \cos\phi_0 &= \frac{300}{2500 \times 0.3} = 0.4, \quad \sin\phi_0 = 0.9165 \\ I_w &= 0.3 \times 0.4 = 0.12\ \text{A}, \quad I_\mu = 0.3 \times 0.9165 = 0.275\ \text{A} \\ R_0 &= \frac{2500}{0.12} = 20833\ \Omega, \quad X_0 = \frac{2500}{0.275} = 9092\ \Omega \ (\text{HV side}) \end{aligned}

Referred to secondary:

R0′=20833×0.01=208.33 Ω,X0′=9092×0.01=90.92 ΩR_0' = 20833 \times 0.01 = 208.33\ \Omega, \qquad X_0' = 9092 \times 0.01 = 90.92\ \Omega

Series branch (SC test on HV: 100 V, 20 A, 500 W)

Z01=10020=5 Ω,R01=500202=1.25 ΩX01=52−1.252=4.841 Ω\begin{aligned} Z_{01} &= \frac{100}{20} = 5\ \Omega, \quad R_{01} = \frac{500}{20^2} = 1.25\ \Omega \\ X_{01} &= \sqrt{5^2 - 1.25^2} = 4.841\ \Omega \end{aligned}

Referred to secondary:

R02=1.25×0.01=0.0125 ΩX02=4.841×0.01=0.0484 ΩZ02=5×0.01=0.05 Ω\begin{aligned} R_{02} &= 1.25 \times 0.01 = 0.0125\ \Omega \\ X_{02} &= 4.841 \times 0.01 = 0.0484\ \Omega \\ Z_{02} &= 5 \times 0.01 = 0.05\ \Omega \end{aligned}

Equivalent circuit referred to secondary

 I1 ->        I0                 I2
 o-----+--------+---[ R02 ]---[ jX02 ]---o
 +     |        |                        +
     [ R0 ]  [ jX0 ]                   load
 V1'   |        |                       V2
 -     |        |                        -
 o-----+--------+------------------------o
 All values referred to secondary (250 V):
 R0 = 208.33 ohm, X0 = 90.92 ohm
 R02 = 0.0125 ohm, X02 = 0.0484 ohm

Efficiency at half load

Iron loss Pi=300P_i = 300 W; full-load copper loss =500= 500 W (test at rated current). Assuming unity pf:

Output=0.5×50000×1=25000 WPcu=0.52×500=125 Wη=2500025000+300+125×100=98.33%\begin{aligned} \text{Output} &= 0.5 \times 50000 \times 1 = 25000\ \text{W} \\ P_{cu} &= 0.5^2 \times 500 = 125\ \text{W} \\ \eta &= \frac{25000}{25000 + 300 + 125}\times 100 = 98.33\% \end{aligned}

Answer: Referred to secondary: R0=208.33 ΩR_0 = 208.33\ \Omega, X0=90.92 ΩX_0 = 90.92\ \Omega, R02=0.0125 ΩR_{02} = 0.0125\ \Omega, X02=0.0484 ΩX_{02} = 0.0484\ \Omega; half-load efficiency (upf) = 98.33%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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