22-Elec-A6 Power Systems and Machines · December 2016
Question 2 of 6: Single-Phase Transformer Parameters, Efficiency and Regulation
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, December 2016. Closed-book; formula sheet supplied. Six questions of equal value; candidates answer any five. All six are solved here as a study resource. All ac voltages and currents are rms; three-phase voltages are line-to-line unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers, induction/synchronous machines, dc motors); J. D. Glover et al., Power System Analysis and Design, 6th ed. (three-phase power, two-wattmeter method, power-factor correction).
Question 2: Single-Phase Transformer Parameters, Efficiency and Regulation (20 marks)
Given. A 100 kVA, 7200/240 V, 60 Hz single-phase transformer.
Given data
Test
Voltage
Current
Power
Open-circuit (primary)
7200 V
0.65 A
425 W
Short-circuit (primary)
250 V
13.89 A
1420 W
Find. The equivalent-circuit parameters referred to the 7200 V side, then the efficiency, voltage regulation and primary-side phasor diagram at full load, 0.9 pf lagging.
Approximate equivalent circuit referred to the primary (shunt excitation branch, series leakage impedance).
Approach. The open-circuit test (rated voltage, negligible current) yields the shunt magnetizing branch; the short-circuit test (rated current, small voltage) yields the series resistance and leakage reactance. Because both tests are taken at 7200 V / 13.89 A (the primary ratings), the parameters are already referred to the primary.
Shunt branch from the OC test. The core-loss and magnetizing components are
$$R_c=\frac{V_{oc}^2}{P_{oc}}=\frac{7200^2}{425}=1.22\times10^{5}\ \Omega,\qquad
I_c=\frac{P_{oc}}{V_{oc}}=0.059\ \text{A}.$$
The magnetizing current is $I_m=\sqrt{I_{oc}^2-I_c^2}=\sqrt{0.65^2-0.059^2}=0.647\ \text{A}$, so
$$\boxed{R_c=122\ \text{k}\Omega,\qquad X_m=\frac{V_{oc}}{I_m}=\frac{7200}{0.647}=11.1\ \text{k}\Omega.}$$
Series branch from the SC test.
$$R_{eq}=\frac{P_{sc}}{I_{sc}^{2}}=\frac{1420}{13.89^{2}}=7.36\ \Omega,\qquad
Z_{eq}=\frac{V_{sc}}{I_{sc}}=\frac{250}{13.89}=18.0\ \Omega,$$
$$\boxed{X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}=\sqrt{18.0^2-7.36^2}=16.4\ \Omega.}$$
Full-load losses (part b-i). "100% of rated power" is full load, so the current equals rated ($I=13.89$ A) and the copper loss equals the SC input: $P_{cu}=1420$ W. The core loss is the OC input, $P_{core}=425$ W (constant at rated voltage). Output power $P_{out}=S_{rated}\cos\theta=100\,(0.9)=90$ kW. Hence
$$\boxed{\eta=\frac{P_{out}}{P_{out}+P_{cu}+P_{core}}=\frac{90\,000}{90\,000+1420+425}=97.99\%.}$$
Voltage regulation (part b-ii). With the load current $I=13.89\angle{-25.84^\circ}$ A (0.9 lag) referred to the primary and the rated primary voltage taken as the load-referred output, the no-load primary emf is
$$E_1=V+IZ_{eq}=7200+ (13.89\angle{-25.84^\circ})(7.36+j16.4)=7391.5+j160.8=7393\angle 1.25^\circ\ \text{V}.$$
$$\boxed{VR=\frac{|E_1|-V}{V}\times100=\frac{7393-7200}{7200}\times100 = 2.68\%.}$$
Phasor diagram (part b-iii). Drawn below: $V$ along the reference, the load current $I$ lagging it by $25.8^\circ$, and $E_1$ obtained by adding the series drops $IR_{eq}$ (in phase with $I$) and $jIX_{eq}$ (leading $I$ by $90^\circ$). $E_1$ leads $V$ slightly and exceeds it in magnitude, consistent with the small positive regulation.
Primary-side phasor diagram at full load, 0.9 pf lagging (series drop exaggerated for clarity).
