5 kVA 220/400 V efficiency and regulation

Given: 5 kVA, 220/400 V. The SC test is on the HV side: 40 V, 11.4 A, 200 W. The open circuit data is not printed, so the core loss $P_i$ cannot be calculated from the question. I derive the voltage regulation (which needs only the SC test) and give the efficiency with $P_i$ left as a symbol, followed by a worked value for an assumed $P_i$.

Equivalent impedance from the SC test (HV side)

$$\begin{aligned} Z_{eq} &= \frac{40}{11.4} = 3.509\ \Omega \\ R_{eq} &= \frac{200}{11.4^2} = 1.539\ \Omega \\ X_{eq} &= \sqrt{3.509^2 - 1.539^2} = 3.153\ \Omega \end{aligned}$$

Voltage regulation at full load, 0.9 pf lagging

Rated HV current: $I = \dfrac{5000}{400} = 12.5$ A, $\cos\phi = 0.9$, $\sin\phi = 0.4359$.

$$\begin{aligned} I(R_{eq}\cos\phi + X_{eq}\sin\phi) &= 12.5(1.385 + 1.374) = 34.49\ \text{V} \\ I(X_{eq}\cos\phi - R_{eq}\sin\phi) &= 12.5(2.838 - 0.671) = 27.09\ \text{V} \end{aligned}$$

Taking the no-load voltage as 400 V, the full-load terminal voltage $V_2$ follows from $400^2 = (V_2 + 34.49)^2 + 27.09^2$, which gives $V_2 = 364.6$ V, so

$$\text{VR} = \frac{400 - 364.6}{400}\times100 = 8.85\%$$

(The approximate formula gives $34.49/400 = 8.62\%$.)

Efficiency at full load, 0.9 pf lagging

The SC test current (11.4 A) is slightly below the rated 12.5 A, so the full-load copper loss is

$$P_{cu,fl} = 12.5^2\times1.539 = 240.5\ \text{W}$$ $$P_{out} = 5000\times0.9 = 4500\ \text{W},\qquad \eta = \frac{4500}{4500 + P_i + 240.5}$$

For example, if the OC test had given $P_i = 100$ W, then $\eta = \dfrac{4500}{4840.5}\times100 = 92.97\%$. For any other core loss, substitute it in the formula above.

Answer: voltage regulation $\approx 8.85\%$ (lagging); efficiency $= \dfrac{4500}{4740.5 + P_i}$, which is 92.97% for $P_i = 100$ W