Short-Circuit Test — Equivalent Reactance and Efficiency of a Single-Phase Transformer: 16-Mechatronics-B10, December 2019, Q2

Question 2: Short-Circuit Test — Equivalent Reactance and Efficiency of a Single-Phase Transformer (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.

Quantity	Value
Rated voltages, V₁/V₂	2300 V / 240 V
Rated apparent power, S	48 kVA
Frequency, f	60 Hz
Primary (HV) winding resistance, R₁	0.6 Ω
Secondary (LV) winding resistance, R₂	0.025 Ω
Short-circuit test voltage (for rated current), V_SC	238 V

Find. (a) equivalent reactance X_eq referred to the primary; (b) power drawn during the short-circuit test at rated current; (c) efficiency at rated load, 0.8 p.f. lagging, core loss neglected.

Approach. Refer the secondary resistance to the primary (HV) side by the turns-ratio squared to form R_eq1, use the short-circuit test (applied voltage at rated current) to get the total series impedance Z_eq1, then separate X_eq1; the short-circuit wattmeter reading at rated current is exactly the rated copper loss used for the efficiency in (c).

Turns ratio and equivalent resistance referred to the primary. $$a=\frac{V_1}{V_2}=\frac{2300}{240}=9.583$$ $$R_{eq1}=R_1+a^2R_2=0.6+(9.583)^2(0.025)=\boxed{2.896\ \Omega}$$
Rated primary current. $$I_{1,rated}=\frac{S}{V_1}=\frac{48{,}000}{2300}=20.87\ \text{A}$$
Total series impedance from the SC test. $$Z_{eq1}=\frac{V_{SC}}{I_{1,rated}}=\frac{238}{20.87}=11.404\ \Omega$$
(a) Equivalent reactance. $$X_{eq1}=\sqrt{Z_{eq1}^2-R_{eq1}^2}=\sqrt{11.404^2-2.896^2}=\boxed{11.03\ \Omega}$$
(b) Power for rated current/voltage (SC-test copper loss). $$P_{SC}=I_{1,rated}^2R_{eq1}=(20.87)^2(2.896)=\boxed{1261\ \text{W}}$$
(c) Efficiency at rated load, 0.8 p.f. lagging (core loss neglected). $$P_{out}=S\cos\theta=48{,}000(0.8)=38{,}400\ \text{W}$$ $$\eta=\frac{P_{out}}{P_{out}+P_{cu}}=\frac{38{,}400}{38{,}400+1261}=\boxed{96.82\%}$$

Quantity	Value
(a) X_eq (referred to primary)	11.03 Ω
(a) R_eq (referred to primary)	2.896 Ω
(b) Power at rated current (SC test)	1261 W
(c) Efficiency, rated load 0.8 pf lagging	96.82%

Concept & Theory

Topic: Transformer short-circuit test and the series (leakage) branch of the approximate equivalent circuit.

Key relations: R_eq1=R₁+a²R₂; Z_eq1=V_SC/I_rated; X_eq1=√(Z_eq1²−R_eq1²); η=P_out/(P_out+P_cu+P_core).

Why this works: With the secondary shorted, the small applied SC voltage drives negligible current through the (comparatively huge) magnetizing branch, so essentially all the applied voltage and all the wattmeter reading belong to the series resistance/reactance alone — the SC test isolates R_eq/X_eq and, run at rated current, hands over the full-load copper loss directly.

Common pitfall: referring R₂ by a instead of a² before adding it to R₁; and treating the SC-test wattmeter reading as the transformer's total loss (it excludes core loss, correctly neglected here only because the question says so — normally supplied by a separate open-circuit test as in Question 3).

Source: Chapman, Electric Machinery Fundamentals, Ch. 2 §2.7 (transformer testing, efficiency).

Question 2 of 5: Short-Circuit Test — Equivalent Reactance and Efficiency of a Single-Phase Transformer

Question 2: Short-Circuit Test — Equivalent Reactance and Efficiency of a Single-Phase Transformer (20 marks)