22-Elec-A6 Power Systems and Machines · December 2013
Question 3 of 6: Single-Phase Transformer — Tests, Efficiency, 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, Fall 2013 (closed book, formula sheet supplied). Six questions of equal value; any five constitute a complete paper. All ac voltages/currents are rms; three-phase voltages are line-to-line unless noted. All six questions are solved below as a complete study resource.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); T. Wildi, Electrical Machines, Drives, and Power Systems, 6th ed. (Pearson); Fitzgerald, Kingsley & Umans, Electric Machinery, 6th ed.
Given. Single-phase transformer, 60 VA, 208 V (HV) / 120 V (LV), 60 Hz. Turns ratio $a=208/120=1.733$.
Given test data
Test
Side
V
I
P
Open-circuit
LV (120 V)
120 V
25.07 mA
2 W
Short-circuit
HV (208 V)
16.85 V
300 mA
4.7 W
Find. (a) series and shunt parameters referred to the HV side; (b) full-load efficiency at 0.8 leading pf; (c) percentage voltage regulation.
Approximate ("cantilever") equivalent circuit referred to the 208 V HV side: shunt excitation branch at the input, series equivalent impedance to the load.
Approach. The short-circuit test (on the HV side) yields the series equivalent impedance directly. The open-circuit test (on the LV side) yields the shunt magnetizing branch, which is then referred to HV by $a^2$. Efficiency uses core loss (from OC) plus full-load copper loss (from $I_{rated}^2R_{eq}$); regulation uses the series drop at rated leading-pf current.
Series branch from the SC test (already on HV).
$$Z_{eq}=\frac{V_{sc}}{I_{sc}}=\frac{16.85}{0.300}=56.17\ \Omega,\qquad R_{eq}=\frac{P_{sc}}{I_{sc}^2}=\frac{4.7}{0.300^2}=52.22\ \Omega$$
$$\boxed{X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}=\sqrt{56.17^2-52.22^2}=20.68\ \Omega\ \text{(HV)}}$$
Shunt branch from the OC test (on LV). $R_{c,LV}=\dfrac{V_{oc}^2}{P_{oc}}=\dfrac{120^2}{2}=7200\ \Omega$. The admittance is $Y_{oc}=I_{oc}/V_{oc}=2.089\times10^{-4}\ \text{S}$, $G_c=P_{oc}/V_{oc}^2=1.389\times10^{-4}\ \text{S}$, so $B_m=\sqrt{Y_{oc}^2-G_c^2}=1.561\times10^{-4}\ \text{S}$ and $X_{m,LV}=1/B_m=6408\ \Omega$.
Refer the shunt branch to HV (multiply by $a^2=1.733^2=3.004$):
$$\boxed{R_{c}=7200(3.004)=21.63\ \text{k}\Omega,\qquad X_{m}=6408(3.004)=19.25\ \text{k}\Omega\ \text{(HV)}}$$
This completes the equivalent circuit: series $52.22+j20.68\ \Omega$ with a shunt $21.63\,\text{k}\Omega\,\Vert\,j19.25\,\text{k}\Omega$.
Full-load losses (part b). Rated HV current $I_{HV}=\dfrac{60}{208}=0.2885\ \text{A}$. Full-load copper loss $P_{cu}=I_{HV}^2R_{eq}=0.2885^2(52.22)=4.35\ \text{W}$; core loss $P_{core}=P_{oc}=2.0\ \text{W}$. Output at 0.8 pf: $P_{out}=S\,\text{pf}=60(0.8)=48\ \text{W}$.
Voltage regulation (part c), 0.8 leading pf, referred to HV. Rated load current leads the load voltage: $I=0.2885\angle{+}36.87^\circ$ A, $V_{load}=208\angle0^\circ$ V. Primary voltage:
$$V_{p}=V_{load}+I\,Z_{eq}=208+ (0.2885\angle36.87^\circ)(56.17\angle21.6^\circ)=216.5+j13.8=216.9\ \text{V}$$
$$\boxed{\text{VR}=\frac{|V_p|-V_{load}}{V_{load}}=\frac{216.9-208}{208}=+4.29\%}$$
The regulation is positive (not negative) because the large equivalent resistance ($R_{eq}\gt X_{eq}$) dominates the drop even at leading pf.
