Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Professional Engineers Ontario — 16-Elec-A6 Power Systems and Machines, December 2017. Closed-book, formula sheet supplied; three hours. FIVE questions constitute a complete paper and all are of equal value (20 marks). All AC voltages and currents are rms; three-phase voltages are line-to-line and power is total real power unless stated. All five questions are solved below.
Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill) — DC machines ch. 8–9, transformers ch. 2, synchronous machines ch. 4–5, induction machines ch. 6. P. C. Sen, Principles of Electric Machines and Power Electronics, 3rd ed. (Wiley). T. Wildi, Electrical Machines, Drives, and Power Systems, 6th ed. (Pearson). Formula sheet as printed on pages 5–6 of the exam.
Given. Open- and short-circuit test data for a 100 kVA, 7200/240 V single-phase transformer.
Given data — transformer tests
Test
Voltage
Current
Power
Open-circuit
$V_{OC}=7200$ V
$I_{OC}=0.45$ A
$P_{OC}=355$ W
Short-circuit
$V_{SC}=250$ V
$I_{SC}=13.88$ A
$P_{SC}=1275$ W
Find. The side on which each test was run; the approximate equivalent-circuit parameters referred to the HV side; and the full-load voltage regulation and efficiency at 0.8 leading pf.
Figure 3.1 — Approximate equivalent circuit referred to the HV (7200 V) side: shunt magnetizing branch $R_c\parallel jX_m$ at the input, series equivalent impedance $R_{eq}+jX_{eq}$ to the load.
Approach. Identify the test side from the applied voltage/current versus the ratings, get the shunt branch from the OC test and the series branch from the SC test (both on the HV side), then apply $V_1 = V_2' + I\,Z_{eq}$ for regulation and the loss totals for efficiency.
(a) Which side. The rated HV current is $I_{HV}=100{,}000/7200 = 13.89\ \text{A}$; the SC test injects $I_{SC}=13.88\ \text{A}$ at only 250 V — that is rated HV current, so the short-circuit test was on the HV side. The OC test applies the full 7200 V (rated HV voltage), so the open-circuit test was also on the HV side. Both tests were taken on the high-voltage side.
(b) Shunt branch from OC (HV side). Core-loss conductance current $I_c=P_{OC}/V_{OC}=355/7200 = 0.0493\ \text{A}$, so $R_c=V_{OC}/I_c=\boxed{146.0\ \text{k}\Omega}$. Magnetizing current $I_m=\sqrt{I_{OC}^2-I_c^2}=\sqrt{0.45^2-0.0493^2}=0.4473\ \text{A}$, so $X_m=V_{OC}/I_m=\boxed{16.10\ \text{k}\Omega}$.
(b) Series branch from SC (HV side). $R_{eq}=P_{SC}/I_{SC}^2=1275/13.88^2=\boxed{6.618\ \Omega}$; $Z_{eq}=V_{SC}/I_{SC}=250/13.88 = 18.01\ \Omega$, so $$X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}=\sqrt{18.01^2-6.618^2}=\boxed{16.75\ \Omega}.$$ These are the equivalent series parameters referred to the HV side (Figure 3.1).
(c) Voltage regulation, 0.8 leading. Full-load HV current $I=13.89\ \text{A}$ at 0.8 leading pf, i.e. $I=13.89\,\angle{+}36.87^\circ$. Taking the referred load voltage $V_2'=7200\ \text{V}\angle 0$, $$V_1=V_2'+I\,Z_{eq}=7200+(13.89\,\angle 36.87^\circ)(6.618+j16.75)=7138\,\angle 1.94^\circ\ \text{V}.$$ Hence $$\mathrm{VR}=\frac{|V_1|-V_2'}{V_2'}=\frac{7138-7200}{7200}=\boxed{-0.86\%}.$$ The regulation is negative — the terminal voltage rises on a leading-pf load.
(c) Efficiency, 0.8 leading, full load. Output $P_{out}=S\cdot\text{pf}=100{,}000(0.8)=80{,}000\ \text{W}$. The losses are the core loss ($P_{OC}=355\ \text{W}$) plus full-load copper loss ($P_{SC}=1275\ \text{W}$, since $I_{SC}\approx$ rated). Therefore $$\eta=\frac{P_{out}}{P_{out}+P_{core}+P_{cu}}=\frac{80{,}000}{80{,}000+355+1275}=\boxed{98.00\%}.$$
