22-Elec-A6 Power Systems and Machines · May 2016

Question 4 of 7: Single-Phase Transformer — OC/SC Tests

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, Spring 2016. Closed book (formula sheets supplied); do any five of seven questions, all of equal value (20 marks each). Three hours. All ac quantities are rms; three-phase voltages are line-to-line and power is total real power unless noted. All seven questions are solved here as a complete study resource.

Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (DC machines ch. 8, transformers ch. 2, induction motors ch. 7, synchronous machines ch. 4–5, parallel operation ch. 5); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis ch. 2); T. Wildi, Electrical Machines, Drives, and Power Systems. Canadian frame: EGBC/PEO national exam; SI units.

Question 4: Single-Phase Transformer — OC/SC Tests (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Given data — Question 4
Rating	15 kVA, 2300/230 V
Open-circuit test	2300 V, 0.21 A, 50 W
Short-circuit test	47 V, 6.52 A, 160 W
Rated HV current	15000/2300 = 6.52 A
Load condition (c)	full load, 0.8 pf leading

Find. (a) side of each test; (b) parameters and sketch of the approximate circuit referred to HV; (c) full-load voltage regulation and efficiency at 0.8 leading pf.

Approach. Identify the test side from the applied voltage and current relative to the ratings. The OC test gives the shunt (core) branch and the SC test gives the series (leakage) branch; then apply the series drop for regulation and add core plus copper loss for efficiency.

Test sides (a). The OC voltage 2300 V equals the high-voltage rating, and the SC current 6.52 A equals the rated HV current (15 kVA / 2300 V). Both tests were therefore performed on the high-voltage (2300 V) side, so the parameters are already referred to HV.
Core (shunt) branch from OC (b). $R_c=\dfrac{V_{OC}^2}{P_{OC}}=\dfrac{2300^2}{50}=105.8\ \text{k}\Omega$. The exciting angle is $\theta=\cos^{-1}\!\dfrac{50}{2300(0.21)}=84.06^\circ$, so the magnetizing reactance is $X_m=\dfrac{V_{OC}}{I_{OC}\sin\theta}=\dfrac{2300}{0.21(0.9946)}=11.0\ \text{k}\Omega.$
Series branch from SC (b). $Z_{eq}=\dfrac{47}{6.52}=7.21\ \Omega$, $R_{eq}=\dfrac{P_{SC}}{I_{SC}^2}=\dfrac{160}{6.52^2}=3.76\ \Omega$, and $X_{eq}=\sqrt{7.21^2-3.76^2}=6.15\ \Omega.$ The approximate equivalent circuit (HV side) is drawn below.
Voltage regulation at 0.8 leading (c). The rated HV current is $6.52\angle{+36.87^\circ}$ A (leading). Referring the load to $2300\angle 0^\circ$, the primary voltage is $$V_p=2300+I(R_{eq}+jX_{eq})=2300+(6.52\angle 36.87^\circ)(7.21\angle 58.53^\circ)=2296\angle 1.17^\circ\text{ V},$$ giving $$\boxed{\text{VR}=\frac{2296-2300}{2300}=-0.17\%.}$$ The negative regulation is the signature of a leading load, where the capacitive current raises the output voltage above the input.
Efficiency at full load, 0.8 pf (c). Output $P_{\text{out}}=15\,000(0.8)=12\,000$ W; copper loss at rated current $=P_{SC}=160$ W; core loss $=P_{OC}=50$ W. Thus $$\boxed{\eta=\frac{12\,000}{12\,000+160+50}=98.28\%.}$$

Figure 4 — approximate equivalent circuit referred to the 2300 V side: series R_eq+jX_eq with the shunt core branch R_c ∥ jX_m.

Final results — Question 4
Quantity	Value
(a) Test sides	both on the HV (2300 V) side
(b) Series branch (HV)	R_eq = 3.76 Ω, X_eq = 6.15 Ω
(b) Core branch (HV)	R_c = 105.8 kΩ, X_m = 11.0 kΩ
(c) Voltage regulation (0.8 leading)	−0.17%
(c) Efficiency (full load, 0.8 pf)	98.28%

Topic: Transformer parameter identification from OC/SC tests, and regulation at a leading power factor.

Key relations: OC → $R_c=V^2/P$, $X_m=V/(I\sin\theta)$; SC → $Z_{eq}=V/I$, $R_{eq}=P/I^2$, $X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}$; $\text{VR}=(V_p-V_s)/V_s$; $\eta=P_{\text{out}}/(P_{\text{out}}+P_{\text{cu}}+P_{\text{core}})$.

Why this works: The OC test applies rated voltage with no load, so almost all current is the small exciting current through the high-impedance shunt branch; the SC test applies a small voltage to drive rated current through the low-impedance series branch with negligible excitation. Each test therefore isolates one branch. At a leading pf the reactive drop can be negative, so the output voltage exceeds the input — a negative regulation.

Common pitfall: Assuming the OC test is always on the LV side; here the OC voltage of 2300 V unambiguously identifies the HV side. Also, forgetting that copper loss at full load equals the SC test power (taken at rated current).

Source: Chapman, Electric Machinery Fundamentals §2.6–2.7 (transformer testing, regulation, efficiency).

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