Question 3 of 7: Transformer losses versus frequency and a 50-to-60 Hz transfer

Question 3: Transformer losses versus frequency and a 50-to-60 Hz transfer

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) — how frequency acts on each loss mechanism

Transformer losses divide into core (no-load) losses and winding (load) losses, and frequency touches them through quite different mechanisms.

Hysteresis loss is the energy consumed in reversing the domain structure of the core once per cycle, so it is proportional to frequency and to the area of the hysteresis loop: $P_h = k_h f B_{\max}^{n}$, with the Steinmetz exponent $n$ between 1.6 and 2.0 for modern grain-oriented steel. Eddy-current loss arises from circulating currents induced in the laminations and depends on the square of the induced voltage per lamination, giving $P_e = k_e f^{2} B_{\max}^{2} t^{2}$ for lamination thickness $t$. Raising frequency at constant flux density therefore raises hysteresis loss linearly and eddy loss quadratically, and eddy loss comes to dominate at high frequency — which is why laminations are thinned, or ferrite replaces steel, as design frequency rises.

The important qualification is that flux density is not independent of frequency. From Faraday's law the peak core flux is

so at constant applied voltage flux density falls as $1/f$. Substituting, hysteresis loss varies as $f^{1-n}$, that is it decreases with frequency, while eddy loss becomes independent of frequency altogether. This is the practical case for the present problem: moving a 50 Hz transformer to 60 Hz at the same voltage reduces the flux by one sixth, moves the core further from saturation and reduces total core loss. The reverse move, 60 Hz equipment onto a 50 Hz supply at rated voltage, raises flux by 20 per cent, drives the core towards saturation and can multiply the magnetising current several times over — the reason nameplates specify a volts-per-hertz limit rather than a voltage alone.

Winding loss is nominally $I^{2}R$ and frequency-independent, but the a.c. resistance rises with frequency through skin and proximity effects, and the leakage flux that produces stray loss in tanks, clamps and windings scales with frequency as well. Eddy losses in the winding conductors themselves rise roughly as $f^{2}$. Reactances scale directly with frequency, $X = 2\pi fL$, so both the leakage reactance that governs regulation and the magnetising reactance that governs no-load current increase by the frequency ratio — the fact Part (b) rests on. The net effect of a 50 Hz unit run at 60 Hz is lower core loss, slightly higher stray and eddy winding loss, lower magnetising current and poorer voltage regulation because of the larger leakage reactance.

Parts (b) and (c) — scaling to 60 Hz and the open-circuit test

Given. A single-phase transformer whose reactances were measured at 50 Hz and which is to be operated at 60 Hz.

Find. The magnetising reactance referred to the low-voltage winding and the leakage reactance of each winding at 60 Hz, then the primary current and the open-circuit secondary voltage with 240 V applied to the low-voltage side.

Approach. Every reactance is $2\pi fL$ with $L$ fixed by geometry and turns, so each scales by the frequency ratio 60/50. The open-circuit condition then removes the secondary entirely, leaving a single series loop, and the secondary terminal voltage is the magnetising-branch voltage multiplied by the turns ratio.

1. Establish the scaling rule. Inductance is a function of permeability, turns and geometry only, so it does not change with supply frequency. Therefore $$\frac{X_{60}}{X_{50}}=\frac{2\pi(60)L}{2\pi(50)L}=\frac{60}{50}=1.20$$ Every reactance in the model is multiplied by 1.20, and no resistance changes.
2. Scale the three reactances. Applying the factor to each measured value, $$X_m=46.2(1.20)=\boxed{55.44\ \Omega\ \text{referred to the 230 V winding}}$$ $$\begin{aligned} X_{\ell 1} &= 27.8\times10^{-3}(1.20)=33.36\ \text{m}\Omega \\ X_{\ell 2} &= 25.3(1.20)=30.36\ \Omega \end{aligned}$$ The second of these belongs to the 6.6 kV winding and is quoted on that side. Referred to the low-voltage side it would be $30.36/a^{2}$ with $a = 6600/230 = 28.696$, that is 36.87 mΩ — comparable with the primary's 33.36 mΩ, which is the expected near-equal split of leakage between the two windings of a concentrically wound transformer.
3. Reduce the open-circuit network. With the secondary open no current can flow in $X_{\ell 2}$, so the ideal transformer and the secondary leakage branch drop out entirely. The applied 240 V appears across $X_{\ell 1}$ in series with $X_m$: $$I_1=\frac{V_1}{X_{\ell 1}+X_m}=\frac{240}{0.03336+55.44}=\frac{240}{55.4734}$$ $$\boxed{I_1 = 4.326\ \text{A}}$$ That is 2.21 per cent of the rated low-voltage current $45\,000/230 = 195.65$ A — a normal exciting current for a unit of this size.
4. Find the voltage across the magnetising branch. The leakage reactance takes a very small share of the applied voltage: $$E_1=I_1X_m=(4.3264)(55.44)=239.86\ \text{V}$$ so only 0.14 V of the 240 V is lost in the primary leakage reactance.
5. Refer to the open secondary. The ideal transformer scales $E_1$ by the turns ratio, and with no secondary current there is no secondary leakage drop: $$V_2=aE_1=\frac{6600}{230}(239.86)=28.696(239.86)$$ $$\boxed{V_2 = 6883\ \text{V} = 6.88\ \text{kV}}$$
6. Check the result against the volts-per-hertz rule. Rated operation is $230/50 = 4.60$ V/Hz; the present condition is $240/60 = 4.00$ V/Hz, that is 87 per cent of rated flux. The core is well below saturation, so treating $X_m$ as a constant is legitimate here — the assumption would have been unsafe in the opposite direction, where the same transformer on 50 Hz at 240 V would run at 113 per cent of rated flux and $X_m$ would collapse with saturation.