20 kVA 250/2500 V tests, equivalent circuit

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

Open-circuit test (on the LV side)

The no-load power is almost entirely core loss, so the shunt branch is found from it.

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{105}{250\times1.4} = 0.3 \\ I_w &= I_0\cos\phi_0 = 0.42\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.336\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 595.2\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 187.2\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

Short-circuit test (on the HV side)

At the reduced test voltage the core loss is negligible, so the wattmeter reading is the copper loss.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{120}{8} = 15\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{320}{8^2} = 5\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 14.14\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the LV side

The turns ratio is $K = 2500/250 = 10$, so $K^2 = 100$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$\begin{aligned} R_{eq} &= 5 \div 100 = 0.05\ \Omega \\ X_{eq} &= 14.14 \div 100 = 0.1414\ \Omega \end{aligned}$$

Equivalent circuit referred to the LV side

   Req=0.05 ohm   Xeq=0.141 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=595.2] [X0=187.2]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 0.05\ \Omega$, $X_{eq} = 0.1414\ \Omega$ ($Z_{eq} = 0.15\ \Omega$), shunt branch $R_0 = 595.2\ \Omega$ in parallel with $X_0 = 187.2\ \Omega$.

Answer (referred to primary): $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$.