20 kVA 250/2500 V tests, regulation, efficiency

Given: 20 kVA, 250 V/2500 V. Primary = LV (250 V), $K = 2500/250 = 10$, $K^2 = 100$.

Equivalent circuit parameters (referred to primary)

OC test (LV side): 250 V, 1.4 A, 105 W

$$\begin{aligned} \cos\phi_0 &= \frac{105}{250\times1.4} = 0.3,\quad I_w = 0.42\ \text{A},\quad I_\mu = \sqrt{1.4^2 - 0.42^2} = 1.3355\ \text{A} \\ R_0 &= \frac{250}{0.42} = 595.2\ \Omega,\qquad X_0 = \frac{250}{1.3355} = 187.2\ \Omega \end{aligned}$$

SC test (HV side): 120 V, 8 A, 320 W. (The rated HV current is $20\,000/2500 = 8$ A, so this is a full-load test.)

$$\begin{aligned} Z_{eq,H} &= \frac{120}{8} = 15\ \Omega,\quad R_{eq,H} = \frac{320}{8^2} = 5\ \Omega,\quad X_{eq,H} = \sqrt{15^2 - 5^2} = 14.14\ \Omega \end{aligned}$$

Referred to primary (LV) by dividing by $K^2 = 100$:

$$R_{01} = 0.05\ \Omega,\qquad X_{01} = 0.1414\ \Omega,\qquad Z_{01} = 0.15\ \Omega$$

   R01=0.05 ohm   X01=0.1414 ohm
 o--+-----/\/\/\-----UUUU------o
    |
   [R0=595.2] [X0=187.2]   Load
    |
 o--+---------------------------o

Voltage regulation at half load, 0.8 pf lagging

Rated primary current $I_1 = 20\,000/250 = 80$ A; half load gives $I_1 = 40$ A.

$$\begin{aligned} I_1(R_{01}\cos\phi + X_{01}\sin\phi) &= 40(0.05\times0.8 + 0.1414\times0.6) = 4.994\ \text{V} \\ \text{VR} &= \frac{4.994}{250}\times100 = 2.0\% \end{aligned}$$

Efficiency at half load, 0.8 pf lagging

$$\begin{aligned} P_{out} &= 0.5\times20\,000\times0.8 = 8000\ \text{W} \\ P_{cu} &= 0.5^2\times320 = 80\ \text{W},\qquad P_i = 105\ \text{W} \\ \eta &= \frac{8000}{8000 + 80 + 105}\times100 = 97.74\% \end{aligned}$$

Answer: $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$, $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$; regulation $\approx 2.0\%$; efficiency $= 97.74\%$