20 kVA 2200/220 V tests, equivalent circuit LV side

Given: 20 kVA, 2200/220 V, so HV = 2200 V, LV = 220 V, $K = 10$, $K^2 = 100$. (Rated HV current is $20\,000/2200 = 9.09$ A; the SC test was at 8.4 A, near full load.)

Open-circuit test (on the LV side)

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

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{125}{220\times1.1} = 0.5165 \\ I_w &= I_0\cos\phi_0 = 0.5682\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 0.9419\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 387.2\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 233.6\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

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.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{52.7}{8.4} = 6.274\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{287}{8.4^2} = 4.067\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 4.777\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the LV side

The turns ratio is $K = 2200/220 = 10$, 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.

$$\begin{aligned} R_{eq} &= 4.067 \div 100 = 0.04067\ \Omega \\ X_{eq} &= 4.777 \div 100 = 0.04777\ \Omega \end{aligned}$$

Equivalent circuit referred to the LV side

   Req=0.0407 ohm   Xeq=0.0478 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=387.2] [X0=233.6]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 0.04067\ \Omega$, $X_{eq} = 0.04777\ \Omega$ ($Z_{eq} = 0.06274\ \Omega$), shunt branch $R_0 = 387.2\ \Omega$ in parallel with $X_0 = 233.6\ \Omega$.

Answer (referred to LV side): $R_0 = 387.2\ \Omega$, $X_0 = 233.6\ \Omega$, $R_{eq} = 0.04067\ \Omega$, $X_{eq} = 0.04777\ \Omega$ ($Z_{eq} = 0.0627\ \Omega$)