Question 3 of 6: Single-Phase Transformer — Tests, Efficiency, Regulation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, Spring 2013. Closed-book; five of the six questions constitute a complete paper (all of equal value, 20 marks each). All voltages and currents are rms values; three-phase voltages are line-to-line unless noted otherwise. All six questions are solved below as a complete study resource.
Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill): DC machines (Ch. 9), transformers (Ch. 2), induction machines (Ch. 7), synchronous machines (Ch. 4–5), magnetic circuits (Ch. 1). J. D. Glover, M. S. Sarma & T. J. Overbye, Power System Analysis and Design (Cengage): per-phase and balanced three-phase network analysis (Ch. 2–3). T. Wildi, Electrical Machines, Drives, and Power Systems (Pearson).
Find. The approximate HV-referred equivalent circuit, the full-load efficiency, and the percentage voltage regulation.
Figure 3.1 — approximate (cantilever) equivalent circuit referred to the HV side.
Approach. The short-circuit test at rated current yields the series (winding) impedance; the open-circuit test yields the shunt (core) branch. Efficiency and regulation then follow at the stated full-load, 0.90-lagging condition.
Rated HV current. $I_{rated} = \dfrac{S}{V_{HV}} = \dfrac{1000}{240} = 4.17\ \text{A}$ — the short-circuit test was run at rated current, so its copper loss is the full-load copper loss.
Series impedance from the SC test (a).
$$R_{eq} = \frac{P_{sc}}{I_{sc}^2} = \frac{11.75}{4.17^2} = 0.676\ \Omega,\qquad Z_{eq} = \frac{V_{sc}}{I_{sc}} = \frac{10.8}{4.17} = 2.59\ \Omega,$$
$$X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{2.59^2 - 0.676^2}.$$
$$\boxed{R_{eq} = 0.676\ \Omega,\quad X_{eq} = 2.50\ \Omega}$$
Shunt branch from the OC test (a). The core-loss resistance and magnetizing reactance referred to HV are
$$R_c = \frac{V_{oc}^2}{P_{oc}} = \frac{240^2}{5.2} = 11.1\ \text{k}\Omega,\qquad
B_m = \sqrt{\Big(\tfrac{I_{oc}}{V_{oc}}\Big)^2 - \Big(\tfrac{P_{oc}}{V_{oc}^2}\Big)^2},\quad X_m = \frac{1}{B_m} = 2.46\ \text{k}\Omega.$$
$$\boxed{R_c = 11.1\ \text{k}\Omega,\quad X_m = 2.46\ \text{k}\Omega}$$
Full-load efficiency (b). Output $P_{out} = S\cdot\mathrm{pf} = 1000(0.90) = 900\ \text{W}$; copper loss at full load $= 11.75\ \text{W}$ (the SC power at rated current); core loss $= 5.2\ \text{W}$. Thus
$$\eta = \frac{P_{out}}{P_{out} + P_{cu} + P_{core}} = \frac{900}{900 + 11.75 + 5.2}.$$
$$\boxed{\eta = 98.2\%}$$
Voltage regulation (c). Referred to HV, the full-load current is $I = 4.17\,\angle{-25.84^\circ}\ \text{A}$ (0.90 lagging). The primary emf is
$$E = V + I\,Z_{eq} = 240 + (4.17\,\angle{-25.84^\circ})(0.676 + j2.50) = 247.2\,\angle 1.9^\circ\ \text{V},$$
so
$$\mathrm{VR} = \frac{|E| - V}{V}\times 100 = \frac{247.2 - 240}{240}\times 100.$$
$$\boxed{\mathrm{VR} = 3.0\%}$$
(The approximate formula $\mathrm{VR}\approx \tfrac{I(R_{eq}\cos\theta + X_{eq}\sin\theta)}{V} = 2.95\%$ agrees.)
Both figures are exactly what one expects of a small, well-designed distribution unit: an efficiency above 98 % at unity-ish loading and a regulation of only a few percent, dominated here by the reactive voltage drop across $X_{eq}$ because the load is lagging.
