Question 3 of 7: Transformer Losses versus Frequency, and the Cantilever Model

Problem 3: Transformer Losses versus Frequency, and the Cantilever Model (20 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.

Given.

Find. How each class of transformer loss responds to frequency; then the ammeter and wattmeter readings for the short-circuit and open-circuit tests as they are actually connected in this paper, and the primary voltage at the stated load — all using the cantilever equivalent circuit.

Part (a) — frequency and the three loss families. A transformer loses power in three distinguishable ways, and they respond to frequency quite differently. The copper (I²R) loss in the windings is, to first order, independent of frequency: it depends on the load current and the dc winding resistance. It is not exactly independent, because skin effect and proximity effect crowd the current towards the conductor surface as frequency rises, so the effective ac resistance grows roughly as $\sqrt{f}$ at high frequency; in power transformers at 50–60 Hz this is a small correction, handled by transposing and by using stranded (Litz or continuously transposed) conductor.

The two core losses are strongly frequency-dependent. Hysteresis loss follows the Steinmetz law $P_h=k_h f B_{\max}^{\,n}$ with $n\approx1.6$–$2.0$, one loop of the B–H characteristic being traversed per cycle. Eddy-current loss follows $P_e=k_e f^{2}B_{\max}^{2}t^{2}$, the square arising because the induced eddy emf is itself proportional to frequency and the resulting loss is that emf squared over the core resistance; the lamination thickness $t$ appears squared for the same reason, which is why cores are laminated at all and why grain-oriented silicon steel is used.

The crucial point — and the one the question is really testing — is that $B_{\max}$ is not an independent variable. From Faraday’s law $V\approx4.44 f N A B_{\max}$, so the flux density is fixed by the volts-per-hertz ratio. Two cases must therefore be distinguished. If $V/f$ is held constant (the normal way to re-rate a transformer for a different system frequency), $B_{\max}$ is constant, so hysteresis loss rises linearly with $f$ and eddy loss rises with $f^{2}$; total core loss grows and the unit must be de-rated thermally. If instead the applied voltage is held constant and the frequency raised, $B_{\max}\propto1/f$, so hysteresis loss varies as $f^{\,1-n}$ — it actually falls for $n>1$ — while eddy loss becomes independent of frequency altogether. This is why a 50 Hz transformer operates happily and cooler on a 60 Hz supply at the same voltage, but a 60 Hz transformer placed on a 50 Hz supply at rated voltage saturates: its flux density rises by 20 per cent and the magnetising current can grow several-fold.

Approach. Lump the two series arms into $R_{eq}=R_1+R_2^{\prime}$ and $X_{eq}=X_{l1}+X_{l2}^{\prime}$. In the cantilever model a short on the primary shorts out the excitation branch, and an open secondary removes the series arm, so each test isolates exactly one part of the circuit; the load case is then a single phasor addition with no iteration.

1. Combine the series arms and fix the turns ratio. Referred to the high-voltage side, $$R_{eq}=R_1+R_2^{\prime}=2.45+2.45=4.90\ \Omega,\qquad X_{eq}=X_{l1}+X_{l2}^{\prime}=9.85+9.85=19.70\ \Omega$$ so $Z_{eq}=4.90+j19.70=20.3002\angle 76.03^{\circ}\ \Omega$, and $a=2200/220=10$. Referred to the low-voltage side the same impedance is $Z_{eq}/a^{2}=0.0490+j0.1970\ \Omega$.
2. Part (b) — recognise that shorting the PRIMARY removes the excitation branch. The cantilever model puts $R_c$ and $jX_m$ directly across the primary terminals. Short those terminals and the shunt branch is short-circuited: it carries no voltage and therefore no current, and it cannot influence the measurement. Looking in from the secondary, the only impedance left is the series arm. This is exactly why the question drives the test from the secondary — in the conventional T model a primary short would leave the magnetising branch in parallel with the referred series arm and the test would not be exact.
3. Refer the applied test voltage to the high-voltage side and get the current. Twenty-two volts on the 220 V winding is $22\times10=220\ \text{V}$ referred, so $$\left|I^{\prime}\right|=\frac{220}{20.3002}=10.8373\ \text{A (referred to HV)}\;\Longrightarrow\;\left|I_s\right|=a\left|I^{\prime}\right|\;\Longrightarrow\;\boxed{\ \text{ammeter}=108.4\ \text{A}\ }$$ The ammeter is on the secondary, so it reads the actual secondary current, not the referred one. As a sanity check the rated secondary current is $25\,000/220=113.6$ A, so 22 V is very nearly the true short-circuit (impedance) voltage of the unit — 10 per cent of rated, which is entirely typical.
4. Take the wattmeter reading, which is pure copper loss. All the power in this test goes into $R_{eq}$, and it may be evaluated on either side provided the current and resistance are referred consistently: $$P_{sc}=\left|I^{\prime}\right|^{2}R_{eq}=(10.8373)^{2}(4.90)=\left|I_s\right|^{2}\frac{R_{eq}}{a^{2}}=(108.373)^{2}(0.0490)\;\Longrightarrow\;\boxed{\ \text{wattmeter}=575.5\ \text{W}\ }$$ The test power factor is $\cos(76.03^{\circ})=0.2414$ lagging, which is why short-circuit wattmeters must be low-power-factor instruments.
5. Part (c) — an open secondary removes the series arm instead. With the secondary open no current flows in $R_{eq}+jX_{eq}$, so the 2200 V appears undivided across the parallel combination of $R_c$ and $jX_m$. The two branch currents are $$\begin{aligned}I_c&=\frac{2200}{38\,750}=0.056774\ \text{A}\\I_m&=\frac{2200}{24\,800}=0.088710\ \text{A}\end{aligned}$$ and they are in quadrature, the core-loss current in phase with the applied voltage and the magnetising current lagging it by $90^{\circ}$.
6. Combine the two branch currents and take the power. $$\left|I_{oc}\right|=\sqrt{I_c^{2}+I_m^{2}}=\sqrt{(0.056774)^{2}+(0.088710)^{2}}\;\Longrightarrow\;\boxed{\ \text{ammeter}=0.1053\ \text{A}\ }$$ $$P_{oc}=\frac{V^{2}}{R_c}=\frac{(2200)^{2}}{38\,750}\;\Longrightarrow\;\boxed{\ \text{wattmeter}=124.9\ \text{W}\ }$$ The no-load current is 0.93 per cent of the rated primary current of 11.36 A and the test power factor is $124.9/(2200\times0.1053)=0.539$, both squarely in the normal range for a distribution transformer.
7. Part (d) — the load current is the only current in the series arm. This is the payoff of the cantilever model: the excitation current is drawn from the source before the series impedance, so it produces no additional voltage drop. The secondary current is $15\,000/220=68.18\ \text{A}$, or $6.8182\ \text{A}$ referred to the high-voltage side, at an angle of $-\arccos(0.85)=-31.79^{\circ}$ against the secondary voltage taken as reference: $$\mathbf{I}^{\prime}=6.8182\angle{-31.79^{\circ}}=5.7955-j3.5917\ \text{A}$$
8. Add the series drop to the referred secondary voltage. With $\mathbf{V}_2^{\prime}=a V_2=10(220)=2200\angle 0^{\circ}\ \text{V}$, $$\mathbf{V}_1=\mathbf{V}_2^{\prime}+\mathbf{I}^{\prime}\left(R_{eq}+jX_{eq}\right)=2200+\left(5.7955-j3.5917\right)\left(4.90+j19.70\right)$$ $$=2200+\left(99.153+j96.571\right)=2299.15+j96.57\ \text{V}$$ $$\boxed{\ \left|V_1\right|=2301.2\ \text{V}\ \left(\angle 2.405^{\circ}\right)\ }$$ The voltage regulation is $(2301.2-2200)/2200=4.60$ per cent, consistent with a unit whose impedance voltage is about 10 per cent at a 0.85 lagging power factor and 60 per cent of rated load.