Question 3 of 7: Transformer Losses, Tests and Regulation

Problem 3: Transformer Losses, Tests and Regulation (5 + 5 + 5 + 5 points)

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.

Part (a) — The effect of frequency on transformer losses

Transformer losses divide into core (no-load) losses and winding (load) losses, and frequency enters each differently. The core losses are hysteresis and eddy current. Hysteresis loss obeys Steinmetz's law, $P_h = k_h f B_{\max}^{\,n}$, where the exponent $n$ lies between 1.5 and 2.0 for modern grain-oriented silicon steel. Eddy-current loss in a lamination of thickness $t$ obeys $P_e = k_e f^{2}B_{\max}^{2}t^{2}$. At first sight both rise with frequency, but that reading is incomplete, because in a transformer supplied from a constant voltage the flux density is not an independent variable: Faraday's law gives $V = 4.44 f N A B_{\max}$, so $B_{\max}\propto V/f$.

Substituting that constraint is what makes the answer physically useful. At constant applied voltage the hysteresis loss varies as $$P_h\propto f\left(\frac{V}{f}\right)^{n}=V^{n}f^{-(n-1)}$$ so with $n \approx 1.6$ it falls roughly as $f^{-0.6}$ when the frequency is raised. The eddy-current loss varies as $$P_e\propto f^{2}\left(\frac{V}{f}\right)^{2}=V^{2}$$ so it is essentially independent of frequency at constant voltage. Total core loss therefore decreases modestly as frequency rises and increases sharply as it falls. The practical corollary is the one that matters on an exam and in the field: operating a 60-Hz transformer at 50 Hz at rated voltage raises the flux density by 20 per cent, drives the core toward saturation, and can multiply the magnetising current several-fold — which is why nameplates specify volts per hertz and why over-excitation (volts/hertz) protection exists on generator step-up units. The converse also holds: a 50-Hz design is safe on 60 Hz at rated voltage.

The winding losses behave in the opposite sense. The direct-current copper loss $I^{2}R$ is frequency-independent, but skin and proximity effects raise the effective conductor resistance approximately with $\sqrt{f}$ at power frequencies, and the stray losses produced by leakage flux in the tank, clamps and core edges scale roughly as $f^{2}$. In a 60-Hz power transformer these effects are small; in a transformer feeding a harmonic-rich load such as a variable-frequency drive they dominate, which is the reason for the K-factor and factor-K derating conventions. Finally, the leakage reactances themselves are proportional to frequency, so the per-unit impedance and hence the voltage regulation of the unit change with frequency even when the losses do not.

Parts (b), (c) and (d) — Cantilever model calculations

Given.

Find. The ammeter and wattmeter readings for the short-circuit test conducted from the secondary, the same two readings for the open-circuit test conducted from the primary, and the primary voltage when the transformer supplies 15 kVA at 220 V and 0.8 power factor lagging.

Approach. In the cantilever model the whole excitation branch sits across the primary terminals and the combined series impedance sits between that node and the referred secondary. Each test therefore isolates exactly one branch: shorting the primary kills the excitation branch, and opening the secondary kills the series branch.

1. Combine the series elements once, for use in parts (b) and (d). Because the two halves of the winding impedance are in series in the cantilever arrangement, $$\begin{aligned}R_{eq}&=R_1+R_2'=2.7+2.7=5.4\ \Omega \\ X_{eq}&=X_{l1}+X_{l2}'=10.5+10.5=21.0\ \Omega\end{aligned}$$ $$Z_{eq}=5.4+j21.0=21.6832\angle 75.579^{\circ}\ \Omega\ \text{(referred to the 2200-V side)}$$
2. Part (b) — Recognise what the short-circuit test measures. With the primary terminals shorted, the excitation branch — which in the cantilever model is connected directly across those terminals — has zero voltage across it and therefore draws no current. The only impedance the secondary source sees is $Z_{eq}$ referred down through the square of the turns ratio: $$\begin{aligned}Z_{eq,\text{sec}}&=\frac{Z_{eq}}{a^{2}}=\frac{5.4+j21.0}{100} =0.05400+j0.21000\ \Omega \\ |Z_{eq,\text{sec}}|&=0.216832\ \Omega\end{aligned}$$
3. Evaluate the two secondary-side instrument readings. The ammeter reads the magnitude of the current and the wattmeter reads the real power, which in the absence of any excitation current is entirely $I^{2}R$ dissipated in the equivalent resistance: $$\begin{aligned} I_{sc}&=\frac{V_{sc}}{|Z_{eq,\text{sec}}|}=\frac{20}{0.216832}=92.237\ \text{A}\\ P_{sc}&=I_{sc}^{2}R_{eq,\text{sec}}=(92.237)^{2}(0.0540)=459.42\ \text{W} \end{aligned}$$ $$\boxed{\begin{aligned}\text{ammeter}&=92.237\ \text{A} \\ \text{wattmeter}&=459.42\ \text{W}\end{aligned}}$$ The reading is credible: rated secondary current is $25\,000/220=113.64$ A, so 20 V drives the test to 81.2 per cent of rated current, which is where a short-circuit test is normally run.
4. Part (c) — Recognise what the open-circuit test measures. With the secondary open no current can flow in the series arm, so the applied 2250 V appears entirely across the parallel combination of $R_c$ and $jX_m$. The two branch currents are in quadrature, the core-loss current being in phase with the applied voltage and the magnetising current lagging it by $90^{\circ}$: $$\begin{aligned} I_{h+e}&=\frac{V_{oc}}{R_c}=\frac{2250}{37\,500}=0.0600\ \text{A}\\ I_m&=\frac{V_{oc}}{X_m}=\frac{2250}{20\,000}=0.1125\ \text{A} \end{aligned}$$
5. Evaluate the two primary-side instrument readings. The ammeter reads the phasor sum of two quadrature components, and the wattmeter reads only the component in phase with the voltage: $$I_{oc}=\sqrt{I_{h+e}^{2}+I_m^{2}}=\sqrt{(0.0600)^{2}+(0.1125)^{2}}=0.12750\ \text{A}$$ $$P_{oc}=\frac{V_{oc}^{2}}{R_c}=\frac{(2250)^{2}}{37\,500}=135.0\ \text{W}$$ $$\boxed{\begin{aligned}\text{ammeter}&=0.12750\ \text{A} \\ \text{wattmeter}&=135.0\ \text{W}\end{aligned}}$$ The no-load power factor is $0.0600/0.12750=0.4706$ lagging, and the exciting current is only 1.12 per cent of the rated primary current $25\,000/2200=11.36$ A — both are the expected orders of magnitude for a distribution transformer of this size, which is the check that the branch assignment is right.
6. Part (d) — Refer the load current to the primary side. At 15 kVA and 220 V the secondary current is $15\,000/220=68.182$ A, lagging the secondary voltage by $\cos^{-1}0.8=36.870^{\circ}$. Taking the referred secondary voltage $aV_s=10(220)=2{,}200$ V as the reference phasor, the referred current is $$I_s/a=\frac{68.182}{10}\angle-36.870^{\circ}=6.8182\angle -36.870^{\circ}\ \text{A}$$
7. Add the series drop to obtain the primary voltage. In the cantilever model the primary terminal voltage is the referred secondary voltage plus the drop across the series arm; the excitation current does not enter, because it is drawn from the primary node itself and produces no drop in $Z_{eq}$: $$V_p=aV_s+\left(\frac{I_s}{a}\right)Z_{eq} =2{,}200+(6.8182\angle -36.870^{\circ})(21.6832\angle 75.579^{\circ})$$ $$V_p=2{,}200+115.364+j92.455=2{,}315.364+j92.455\ \text{V}$$ $$\boxed{V_p=2{,}317.21\angle 2.287^{\circ}\ \text{V}}$$ The magnitude is 2,317.2 V, so the transformer needs 5.33 per cent above the referred secondary voltage to deliver this load — a normal regulation for a 25-kVA unit at 0.8 lagging power factor, and noticeably worse than it would be at unity power factor because the reactive component of the current works against the dominant reactance of $Z_{eq}$.

Final results.