Question 2 of 5: Single-Phase Transformer Tests and Performance
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 2015. Closed-book; formula sheets supplied. FIVE questions constitute a complete paper and all are of equal value (20 marks each). All ac quantities are rms; three-phase voltages are line-to-line and power is total unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill) — Ch. 2 transformers, Ch. 6–7 induction machines, Ch. 5 synchronous machines, Ch. 8–9 DC machines; A. E. Fitzgerald, C. Kingsley & S. Umans, Electric Machinery, 6th ed.; P. C. Sen, Principles of Electric Machines and Power Electronics, 3rd ed.
Question 2: Single-Phase Transformer Tests and Performance (20 marks)
Given. A 50 kVA, 2400/240 V, 60 Hz single-phase transformer, with the standard open-circuit (no-load) and short-circuit tests below. Both tests are taken on the high-voltage side (the OC voltage is rated 2400 V and the SC current 20.83 A equals rated HV current $50{,}000/2400$).
Given data
Quantity
Value
Rated apparent power $S_r$
50 kVA
Voltage ratio
2400 / 240 V ($a=10$)
Open-circuit test (HV)
2400 V, 0.9 A, 395 W
Short-circuit test (HV)
195 V, 20.83 A, 950 W
Find. (a) the series and shunt equivalent-circuit parameters referred to the 2400 V side; (b) the efficiency and voltage regulation at rated kVA, 0.8 pf lagging; (c) the load apparent power that maximizes efficiency.
Figure Q2. Approximate equivalent circuit referred to the 2400 V side. Series branch from the short-circuit test; shunt branch from the open-circuit test.
Approach. The short-circuit test fixes the series branch $R_{eq}+jX_{eq}$; the open-circuit test fixes the shunt (magnetizing) branch $R_c\parallel jX_m$. Efficiency uses fixed core loss plus load-dependent copper loss; voltage regulation uses the series drop at rated current; maximum efficiency occurs where copper loss equals core loss.
Series branch from the SC test (referred to HV). The short-circuit test drives rated current through the series impedance with the shunt branch negligible:$$Z_{eq}=\frac{V_{sc}}{I_{sc}}=\frac{195}{20.83}=9.36\ \Omega,\quad R_{eq}=\frac{P_{sc}}{I_{sc}^2}=\frac{950}{20.83^2}=2.19\ \Omega,$$$$X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}=\sqrt{9.36^2-2.19^2}=9.10\ \Omega.$$
Shunt branch from the OC test (referred to HV). The open-circuit test applies rated voltage with the series drop negligible. With $\cos\theta_{oc}=P_{oc}/(V_{oc}I_{oc})=395/2160=0.183$,$$R_c=\frac{V_{oc}^2}{P_{oc}}=\frac{2400^2}{395}=14.58\ \text{k}\Omega,\quad X_m=\frac{V_{oc}}{I_{oc}\sin\theta_{oc}}=\frac{2400}{0.9(0.983)}=2712\ \Omega.$$These pin the equivalent circuit: $\boxed{R_{eq}=2.19\ \Omega,\ X_{eq}=9.10\ \Omega,\ R_c=14.58\ \text{k}\Omega,\ X_m=2712\ \Omega}$ (all referred to the 2400 V side).
Efficiency at rated load, 0.8 pf lagging (part b-i). At rated kVA the load draws rated current, so the copper loss equals the SC-test loss and the core loss equals the OC-test loss:$$P_{out}=S_r\cos\theta=50{,}000(0.8)=40{,}000\ \text{W},\quad P_{core}=395,\ P_{cu}=950\ \text{W},$$$$\eta=\frac{P_{out}}{P_{out}+P_{core}+P_{cu}}=\frac{40{,}000}{41{,}345}=\boxed{96.75\%}.$$
Voltage regulation (part b-ii). Referred to the HV side the load current at rated is $I=20.83\angle{-36.87^\circ}$ A (0.8 lag) against a terminal voltage $V_2'=2400\angle 0^\circ$. The required primary voltage is$$V_1=V_2'+I\,(R_{eq}+jX_{eq})=2400+20.83\angle{-36.87^\circ}\times 9.36\angle 76.47^\circ=2553\ \text{V},$$$$\text{VR}=\frac{|V_1|-|V_2'|}{|V_2'|}=\frac{2553-2400}{2400}=\boxed{6.39\%}.$$
Load for maximum efficiency (part c). Copper loss varies as the square of load, so maximum efficiency occurs when the (variable) copper loss equals the (fixed) core loss:$$\left(\tfrac{S}{S_r}\right)^2 P_{sc}=P_{oc}\;\Rightarrow\; S=S_r\sqrt{\frac{P_{oc}}{P_{sc}}}=50\sqrt{\frac{395}{950}}=\boxed{32.2\ \text{kVA}}.$$
