25 kVA 250/2500 V tests, equivalent circuit

Given: 25 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): $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$.

Note: the 25 kVA rating is not needed for the parameters. The SC test current of 8 A is the rated HV current only for a 20 kVA unit (a 25 kVA unit has 10 A on the HV side), so the test is below full load. The equivalent circuit parameters are the same, because they do not depend on the test current if the transformer is linear. The copper loss at the true full load would be $320\times(10/8)^2 = 500$ W.