Question 2 of 5: Single-phase transformer performance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Professional Engineers Ontario — 16-Elec-A6 Power Systems and Machines, May 2017. Closed-book; formula sheet supplied. Five questions of equal value; all five constitute a complete paper, and all are solved here. All ac quantities are rms; three-phase quantities are line-to-line and total real power unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers ch. 2, DC machines ch. 8, synchronous machines ch. 4–5, induction machines ch. 6); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis, ch. 2).
Given. A 100 kVA, 7200/480 V single-phase transformer with the series and shunt parameters below, supplying half rated current at 0.75 pf lagging with 480 V at the load.
Given data
Quantity
Value
Rating / voltages
100 kVA, 7200/480 V, 60 Hz
HV series $R_1,X_1$
3.06 Ω, 6.05 Ω
LV series $R_2,X_2$
0.014 Ω, 0.027 Ω
Shunt (HV) $R_c,X_m$
71,400 Ω, 17,809 Ω
Load
480 V, half rated I, 0.75 pf lagging
Find. Voltage regulation, efficiency, the no-load secondary voltage after the load is removed, and the secondary current giving maximum efficiency.
Approximate equivalent circuit (shunt branch at the input, series R_eq+X_eq), referred to HV.
Approach. Refer the series branch to the HV side, evaluate the load current phasor, then apply the phasor drop for regulation, a power balance for efficiency, and the $P_{cu}=P_{core}$ condition for maximum efficiency.
(a) Turns ratio and approximate circuit on both sides. $a=7200/480=15$. HV side: referring the LV series values by $a^2=225$, $R_{eq,HV}=R_1+a^2R_2=3.06+225(0.014)=6.21\,\Omega$ and $X_{eq,HV}=X_1+a^2X_2=6.05+225(0.027)=12.125\,\Omega$, with the shunt branch $R_{c,HV}=71{,}400\,\Omega$ ∥ $X_{m,HV}=17{,}809\,\Omega$ across the input. LV side: the same circuit with every impedance divided by $a^2$: $R_{eq,LV}=6.21/225=0.0276\,\Omega$, $X_{eq,LV}=12.125/225=0.0539\,\Omega$, $R_{c,LV}=71{,}400/225=317.3\,\Omega$, $X_{m,LV}=17{,}809/225=79.15\,\Omega$, source voltage $V_s/a$ and load 480 V.
Load current (referred to HV). Rated HV current $=100{,}000/7200=13.89$ A, so half rated is $I=6.944\,\text{A}$. At 0.75 pf lagging, $\theta=-41.41^\circ$, giving $I=6.944\angle{-41.41^\circ}$ A. The load voltage referred to HV is $7200\angle 0^\circ$ V.
(b)(i) Source voltage and voltage regulation. $V_s=7200+I\,(R_{eq}+jX_{eq})=7288\angle 0.27^\circ$ V, so $$\boxed{\,VR=\dfrac{|V_s|-7200}{7200}\times100\%=1.22\%\,}$$
(b)(ii) Efficiency. Output $P_{out}=480\times104.17\times0.75=37{,}500$ W. Copper loss $P_{cu}=I^2R_{eq}=6.944^2(6.21)=299.5$ W; core loss $P_{core}=V^2/R_c=7200^2/71{,}400=726.1$ W. Hence $\eta=37{,}500/(37{,}500+299.5+726.1)=97.34\%$. (Core loss is taken at rated voltage; evaluating it at the input-terminal voltage of the approximate circuit, $7288^2/71{,}400=743.9$ W, gives 97.29% and moves the part-(iv) current to 164.2 A — a negligible difference.)
(b)(iii) Secondary voltage, load removed. With no load current there is no series drop, so the secondary rises to the source value referred back: $V_{2,nl}=|V_s|/a=7288/15=485.9$ V.
(b)(iv) Current for maximum efficiency. Maximum efficiency occurs where copper loss equals the (constant) core loss: $I_{HV}^2R_{eq}=P_{core}\Rightarrow I_{HV}=\sqrt{726.1/6.21}=10.81$ A, i.e. $I_{LV}=a\,I_{HV}=162.2$ A (about 0.78 of rated secondary current).
