22-Elec-A6 Power Systems and Machines · December 2018
Question 2 of 5: Transformer Short-Circuit Test and Efficiency
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Closed-book, 3 hours. Five questions constitute a complete paper and all are of equal value (20 marks each); answer all five. All AC voltages and currents are rms, three-phase voltages are line-to-line, and power is total real power unless noted otherwise.
Reference texts (22-Elec-A6 Power Systems and Machines): S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); B. S. Guru & H. R. Hiziroglu, Electric Machinery and Transformers, 3rd ed. (Oxford); P. C. Sen, Principles of Electric Machines and Power Electronics, 2nd ed. (Wiley).
Question 2: Transformer Short-Circuit Test and Efficiency (20 marks)
Given. A single-phase transformer whose winding resistances are known, with a short-circuit test that fixes the series impedance.
Given data
Ratings
2300 / 240 V, 48 kVA, 60 Hz
Primary (HV) resistance
$R_1 = 0.6\ \Omega$
Secondary (LV) resistance
$R_2 = 0.025\ \Omega$
SC applied voltage (HV) for rated current
$V_{sc} = 238\ \text{V}$
Load condition (c)
rated kVA, 0.8 p.f. lagging
Find. (a) the equivalent series reactance $X_{eq}$; (b) the power drawn in the short-circuit condition (= full-load copper loss); (c) the efficiency at rated load, 0.8 p.f. lagging.
Approach. Refer both winding resistances to the HV side to get $R_{eq}$, use the short-circuit voltage and rated current to get $Z_{eq}$, then $X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2}$. The short-circuit input power is the copper loss, and — with core loss neglected — the efficiency follows directly.
Equivalent resistance referred to HV. Reflect $R_2$ by $a^2$ and add to $R_1$:
$$R_{eq} = R_1 + a^2 R_2 = 0.6 + (9.583)^2(0.025) = 0.6 + 2.296 = 2.896\ \Omega.$$
Equivalent impedance from the SC test. The 238 V drives rated HV current through the series branch:
$$Z_{eq} = \frac{V_{sc}}{I_1} = \frac{238}{20.87} = 11.40\ \Omega.$$
Power supplied in the SC test (b). All of the short-circuit input is dissipated as copper loss (the core is essentially unexcited at 238 V):
$$P_{sc} = I_1^2 R_{eq} = (20.87)^2(2.896) = 1261\ \text{W} \approx 1.26\ \text{kW}.$$
This equals the full-load copper loss $P_{cu}$.
Efficiency at rated load (c). Output power at rated kVA and 0.8 p.f.:
$$P_{out} = S\cos\theta = (48\,000)(0.8) = 38\,400\ \text{W}.$$
With core loss neglected, only copper loss subtracts:
$$\eta = \frac{P_{out}}{P_{out}+P_{cu}} = \frac{38\,400}{38\,400+1261} = 0.9682 = 96.8\%.$$
Check: Part (b) is read as the input power under the short-circuit test, which — because the applied 238 V is only ~10% of rated and produces negligible core flux — equals the full-load copper loss (1.26 kW). Part (c)'s instruction "losses are negligible" is taken as core loss negligible; copper loss is retained, otherwise the efficiency would be a trivial 100%.
