Transformer Series Impedance and Efficiency: 22-Elec-A6 Power Systems and Machines, May 2018, Q2

Question 2: Transformer Series Impedance and Efficiency (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. A single-phase transformer with winding resistances referred to their own sides, and a short-circuit (impedance) test taken from the high-voltage side. The printed assumption "$P_{cu}$ -$P_c$" has lost its relational symbol; it is read here as $P_{cu}=P_c$, which sets the core loss equal to the full-load copper loss for the efficiency calculation.

Check: The assumption line in Question 2 is printed as "Assume that P_cu -P_c" with the symbol between the two terms missing. The reading used here is $P_{cu}=P_c$ (core loss equal to full-load copper loss), giving $\eta=93.8\%$. On that reading the core loss is zero and only the copper loss remains, so $\eta=38{,}400/(38{,}400+1261)=96.8\%$. Parts (a) and (b) are the same under both readings.

Given data
Quantity	Symbol	Value
Rated voltages (HV/LV)	$V_1/V_2$	2300 / 240 V
Rating	$S$	48 kVA
Primary (HV) resistance	$R_1$	0.6 Ω
Secondary (LV) resistance	$R_2$	0.025 Ω
SC voltage for rated current	$V_{sc}$	238 V (HV side)
Load power factor	$\cos\theta$	0.8 lagging

Find. (a) the equivalent leakage reactance $X_{eq}$; (b) the input power $P$ drawn in the short-circuit test at rated current; (c) the full-load efficiency at 0.8 p.f. lagging with $P_c=P_{cu}$.

Series (approximate) equivalent circuit referred to the HV side: the two winding resistances combine to $R_{eq}=R_1+a^2R_2$; the leakage reactance $X_{eq}$ follows from the short-circuit test.

Approach. Refer $R_2$ to the HV side through the turns ratio squared to get $R_{eq}$; obtain the total series impedance from the SC test $Z_{eq}=V_{sc}/I_{1,\text{rated}}$; resolve $X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}$. The SC input power is the copper loss; combine with the assumed equal core loss for efficiency.

Turns ratio and equivalent resistance (HV side). With $a = V_1/V_2 = 2300/240 = 9.583$, $$R_{eq}=R_1+a^2R_2 = 0.6 + (9.583)^2(0.025) = 0.6+2.296 = 2.90\ \Omega.$$
Rated primary current and series impedance. Rated HV current $$I_{1}=\frac{S}{V_1}=\frac{48000}{2300}=20.87\ \text{A},\qquad Z_{eq}=\frac{V_{sc}}{I_1}=\frac{238}{20.87}=11.41\ \Omega.$$
(a) Equivalent reactance. The reactance is the quadrature part of the series impedance: $$X_{eq}=\sqrt{Z_{eq}^{2}-R_{eq}^{2}}=\sqrt{11.41^{2}-2.90^{2}}=\boxed{11.03\ \Omega.}$$
(b) Power required for rated current (SC test input). At the reduced SC voltage the core is essentially unexcited, so the wattmeter reads the full-load copper loss: $$P = I_1^{2}R_{eq}=(20.87)^2(2.90)=\boxed{1261\ \text{W}\approx1.26\ \text{kW}.}$$ (Equivalently $P=V_{sc}I_1\cos\theta_{sc}=238\times20.87\times0.254=1261\ \text{W}$.)
(c) Full-load efficiency at 0.8 p.f. lagging. Output power at rated load $$P_{out}=S\cos\theta = 48000\times0.8 = 38{,}400\ \text{W}.$$ With copper loss $P_{cu}=1261\ \text{W}$ and the stated $P_c=P_{cu}=1261\ \text{W}$, total loss $=2P_{cu}=2522\ \text{W}$: $$\eta=\frac{P_{out}}{P_{out}+P_{cu}+P_c}=\frac{38400}{38400+2522}=\boxed{93.8\%.}$$

Final Results — Question 2
Quantity	Result
Equivalent resistance $R_{eq}$ (HV)	2.90 Ω
Equivalent impedance $Z_{eq}$ (HV)	11.41 Ω
Equivalent reactance $X_{eq}$ (HV)	11.03 Ω
SC input power / copper loss	1261 W
Full-load efficiency (0.8 pf lag)	93.8 %

Topic: Transformer series model and the short-circuit (impedance) test.

Key relations: $a=V_1/V_2$; $R_{eq}=R_1+a^2R_2$; $Z_{eq}=V_{sc}/I_{rated}$; $X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}$; $P_{sc}=I^2R_{eq}$; $\eta=P_{out}/(P_{out}+P_{cu}+P_c)$.

Why this works: In the SC test the secondary is shorted and only a few percent of rated voltage is needed to circulate rated current, so the magnetizing branch draws negligible current and the meters see the series leakage impedance alone. The input power is therefore the copper loss at that current. Referring quantities across the ideal ratio uses $a^2$ for impedances because voltage scales by $a$ and current by $1/a$. Maximum efficiency occurs when copper loss equals core loss, which is precisely the condition the problem imposes.

Common pitfall: forgetting the $a^2$ factor when combining $R_1$ and $R_2$ (they are on different sides), or adding the core loss to the input rather than to the losses in the efficiency denominator.

Source: Chapman, Electric Machinery Fundamentals, 5th ed., §2.6–2.7 (per-unit and transformer efficiency).

Quantity	Symbol	Value
Rated voltages (HV/LV)	\(V_1/V_2\)	2300 / 240 V
Rating	\(S\)	48 kVA
Primary (HV) resistance	\(R_1\)	0.6 Ω
Secondary (LV) resistance	\(R_2\)	0.025 Ω
SC voltage for rated current	\(V_{sc}\)	238 V (HV side)
Load power factor	\(\cos\theta\)	0.8 lagging

Quantity	Result
Equivalent resistance \(R_{eq}\) (HV)	2.90 Ω
Equivalent impedance \(Z_{eq}\) (HV)	11.41 Ω
Equivalent reactance \(X_{eq}\) (HV)	11.03 Ω
SC input power / copper loss	1261 W
Full-load efficiency (0.8 pf lag)	93.8 %

Question 2 of 5: Transformer Series Impedance and Efficiency

Question 2: Transformer Series Impedance and Efficiency (20 marks)