8 kVA 400/120 V tests, parameters on HV side

Given: 8 kVA, 400/120 V. HV = 400 V, LV = 120 V, $K = 400/120 = 3.333$, $K^2 = 11.11$. Rated HV current $= 8000/400 = 20$ A, so the SC test is at 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.

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{75}{120\times4} = 0.1562 \\ I_w &= I_0\cos\phi_0 = 0.625\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 3.951\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 192\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 30.37\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

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.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{9.5}{20} = 0.475\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{110}{20^2} = 0.275\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 0.3873\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the HV side

The turns ratio is $K = 400/120 = 3.333$, so $K^2 = 11.11$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$\begin{aligned} R_0' &= 192 \times 11.11 = 2133\ \Omega \\ X_0' &= 30.37 \times 11.11 = 337.5\ \Omega \end{aligned}$$

Equivalent circuit referred to the HV side

   Req=0.275 ohm   Xeq=0.387 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=2133] [X0=337.5]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 0.275\ \Omega$, $X_{eq} = 0.3873\ \Omega$ ($Z_{eq} = 0.475\ \Omega$), shunt branch $R_0 = 2133\ \Omega$ in parallel with $X_0 = 337.5\ \Omega$.

Efficiency at half load, 0.8 pf lagging

$$\begin{aligned} P_{out} &= 0.5\times8000\times0.8 = 3200\ \text{W} \\ P_{cu} &= 0.5^2\times110 = 27.5\ \text{W},\qquad P_i = 75\ \text{W} \\ \eta &= \frac{3200}{3200 + 27.5 + 75}\times100 = 96.90\% \end{aligned}$$

Answer (referred to HV): $R_0 = 2133\ \Omega$, $X_0 = 337.5\ \Omega$, $R_{eq} = 0.275\ \Omega$, $X_{eq} = 0.3873\ \Omega$; efficiency at half load $= 96.90\%$