250/500 V test; equivalent circuit, max efficiency output

Given: 250/500 V, 50 Hz. Primary = 250 V (LV), $K = 2$. The kVA rating is not given, so the SC-test current (12 A on the HV side) is taken as the rated current, giving $S = 500\times12 = 6$ kVA. (The output at maximum efficiency found below does not actually depend on this assumption.)

Open-circuit test (on the LV side)

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

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{80}{250\times1} = 0.32 \\ I_w &= I_0\cos\phi_0 = 0.32\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 0.9474\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 781.25\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 263.9\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

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.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{20}{12} = 1.667\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{100}{12^2} = 0.6944\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 1.515\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the LV side

The turns ratio is $K = 500/250 = 2$, so $K^2 = 4$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$\begin{aligned} R_{eq} &= 0.6944 \div 4 = 0.1736\ \Omega \\ X_{eq} &= 1.515 \div 4 = 0.3788\ \Omega \end{aligned}$$

Equivalent circuit referred to the LV side

   Req=0.174 ohm   Xeq=0.379 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=781.25] [X0=263.9]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 0.1736\ \Omega$, $X_{eq} = 0.3788\ \Omega$ ($Z_{eq} = 0.4167\ \Omega$), shunt branch $R_0 = 781.25\ \Omega$ in parallel with $X_0 = 263.9\ \Omega$.

Output power for maximum efficiency at 0.9 lag

The core loss is the OC wattmeter reading, $P_i = 80$ W. The copper loss at the SC-test current is the SC wattmeter reading, 100 W, which is taken as the full-load copper loss $P_{cu,fl}$.

Maximum efficiency occurs when copper loss equals iron loss:

$$\begin{aligned} x^2P_{cu,fl} &= P_i \Rightarrow x = \sqrt{\frac{80}{100}} = 0.8944 \\ \text{kVA at } \eta_{max} &= 0.8944\times6 = 5.367\ \text{kVA} \\ P_{out} &= 5.367\times0.9 = 4.830\ \text{kW} \end{aligned}$$

Without assuming a rating: at maximum efficiency the copper loss must equal 80 W, so the HV current is $12\sqrt{0.8} = 10.73$ A and $P_{out} = 500\times10.73\times0.9 = 4.83$ kW, the same result.

Check of efficiency: $\eta_{max} = \dfrac{4830}{4830 + 2\times80}\times100 = 96.8\%$.

Answer: output power at maximum efficiency $\approx 4.83$ kW (5.37 kVA at 0.9 pf lagging); equivalent circuit parameters are as above