Transformer Losses versus Frequency and a 50 Hz Unit Run at 60 Hz: 22-Elec-B7 Power Systems Engineering, December 2015, Q3

Problem 3: Transformer Losses versus Frequency and a 50 Hz Unit Run at 60 Hz (25 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) — how frequency acts on each loss mechanism

Transformer losses divide into the core (no-load) losses and the copper (load) losses, and frequency touches them in quite different ways. It is essential to say at what flux density the comparison is being made, because that is where most of the confusion arises: at a fixed applied voltage the peak flux is inversely proportional to frequency, since $V \approx 4.44 f N \Phi_m$, so raising the frequency at constant voltage reduces the flux.

Hysteresis loss follows the Steinmetz relation $P_h = k_h f B_m^{n}$ with the exponent $n$ between about 1.6 and 2.0. At constant flux density it is directly proportional to frequency, because each cycle traverses the same hysteresis loop and there are more cycles per second. At constant applied voltage, however, $B_m \propto V/f$, so $P_h \propto f (V/f)^{n} = V^{n} f^{1-n}$, which falls as frequency rises. Eddy-current loss follows $P_e = k_e f^{2}B_m^{2}t^{2}$ and so rises with the square of frequency at constant flux density, but at constant voltage it becomes $P_e \propto f^{2}(V/f)^{2}=V^{2}$, that is, essentially independent of frequency. Together these give the practical rule that a transformer moved from 50 Hz to 60 Hz at its rated voltage runs with about 17 per cent less peak flux and correspondingly lower core loss, and further from saturation.

Copper loss $I^{2}R$ is nominally frequency-independent, but only nominally. The effective winding resistance rises with frequency through skin effect and proximity effect, and stray loss in the tank, clamps and core-frame — produced by leakage flux linking structural steel — rises roughly with the square of frequency. For a distribution transformer at 50/60 Hz these effects are small, of the order of a few per cent of the copper loss, but they become dominant for units serving rectifier or drive loads where harmonic currents at several hundred hertz are present; that is the entire motivation for the K-factor and factor-K derating standards. The other frequency-linked consequence is reactance: every reactance in the equivalent circuit is proportional to frequency, so a 50 Hz transformer used at 60 Hz has 20 per cent higher leakage reactance and therefore 20 per cent higher percentage impedance, giving worse regulation but proportionally lower through-fault current. The dangerous direction is the reverse one — running a 60 Hz unit on 50 Hz at rated voltage drives the flux up by 20 per cent and can push the core into saturation, with an exciting current several times normal.

Parts (b), (c) and (d) — the 60 Hz equivalent circuit and two tests

Given.

Quantity	Value at 50 Hz
Rating	45 kVA, 230 V : 6.6 kV
Magnetising reactance, from the 230 V terminals	46.00 Ω
Leakage reactance, 230 V winding	28.00 mΩ
Leakage reactance, 6.6 kV winding	25.00 Ω
New supply frequency	60 Hz
Test voltages applied to the 230 V winding	240 V (open circuit), 24 V (short circuit)

Find. All three reactances at 60 Hz, then the primary current and secondary voltage on open circuit at 240 V, and the primary current on short circuit at 24 V.

Figure 3.1 — the T equivalent circuit referred to the 230 V winding, with every reactance already scaled to 60 Hz. Resistances are neglected, as the question's data implies.

Approach. Every reactance scales directly with frequency, so multiply each by 60/50; refer the high-voltage leakage reactance to the low- voltage side through the square of the turns ratio; then solve the open-circuit case as a simple series divider and the short-circuit case as a series-parallel combination.

Part (b) — rescale every reactance to 60 Hz. Since $X = 2\pi f L$ and the inductances are fixed by the geometry and the winding turns, each reactance is multiplied by the frequency ratio $k = 60/50 = 1.20$: $$\begin{aligned} X_m &= 46.00 \times 1.20 = 55.20\ \Omega\\ x_1 &= 28.00\ \text{m}\Omega \times 1.20 = 33.60\ \text{m}\Omega\\ x_2 &= 25.00 \times 1.20 = 30.00\ \Omega \end{aligned}$$ $$\boxed{X_m=55.20\ \Omega,\quad x_1=33.60\ \text{m}\Omega,\quad x_2=30.00\ \Omega}$$ The first two are already referred to the low-voltage winding; $x_2$ is the actual reactance of the 6.6 kV winding.
Refer the high-voltage leakage reactance to the low-voltage side. The turns ratio is $$a=\frac{6600}{230}=28.696$$ and impedance transfers as the square of that ratio: $$x_2^{\prime}=\frac{x_2}{a^{2}}=\frac{30.00}{823.44}=36.43\ \text{m}\Omega$$ It is worth noticing that the two leakage reactances are now almost equal (33.60 and 36.43 milliohms), which is the usual design outcome and a good sign the referral has been done the right way round.
Part (c) — open-circuit test at 240 V. With the secondary open there is no current in the referred secondary leakage branch, so that branch is left out of the loop entirely and the circuit collapses to $x_1$ in series with $X_m$: $$I_1=\frac{V_1}{x_1+X_m}=\frac{240}{0.0336+55.20}=\frac{240}{55.2336}$$ $$\boxed{I_1=4.345\ \text{A}}$$ That is 2.22 per cent of the rated low-voltage current $45\,000/230 = 195.7$ A, which is a plausible exciting current for a unit of this size.
Secondary open-circuit voltage. The induced voltage is the voltage across the magnetising branch, not the whole 240 V, because a little is lost in the primary leakage reactance: $$V_m=I_1X_m=(4.345)(55.20)=239.85\ \text{V}$$ The 0.146 V shortfall is exactly $I_1x_1$. Stepping up through the ideal transformer, $$V_2=aV_m=(28.696)(239.85)$$ $$\boxed{V_2=6883\ \text{V}=6.883\ \text{kV}}$$ The secondary sits about 4.3 per cent above its 6.6 kV nameplate, which simply mirrors the 240 V applied to a 230 V winding.
Part (d) — short-circuit test at 24 V. Shorting the secondary places the referred secondary leakage branch in parallel with the magnetising branch: $$X_{\text{par}}=\frac{X_mx_2^{\prime}}{X_m+x_2^{\prime}} =\frac{(55.20)(0.036433)}{55.20+0.036433}=36.41\ \text{m}\Omega$$ so the total seen by the source is $$X_{\text{tot}}=x_1+X_{\text{par}}=0.03360+0.036409=70.01\ \text{m}\Omega$$ and $$I_1=\frac{24}{0.070009}$$ $$\boxed{I_1=342.8\ \text{A}}$$ Because $X_m$ is more than fifteen hundred times $x_2^{\prime}$, dropping the magnetising branch altogether would have given 342.7 A — a difference of less than 0.1 per cent, which is the standard justification for the approximate short-circuit equivalent circuit. The 342.8 A is 1.75 times rated current at only 10.4 per cent of rated voltage, consistent with a percentage impedance of about 6.0 per cent at 60 Hz (0.0700 ohm on a 1.176 ohm low-voltage base).

Check: the magnetising reactance is treated as constant between the two tests. That is legitimate here because both tests reduce the core flux relative to the nameplate. The volts-per-hertz figure applied in part (c) is 240 / 60 = 4.00 against a rated 230 / 50 = 4.60, so the core runs at 87 per cent of rated flux and is further from saturation than it was designed to be. Had the machine been moved the other way — a 60 Hz unit put on 50 Hz at rated voltage — the flux would have risen 20 per cent, X_m would have collapsed with saturation, and no linear calculation of this kind would be defensible.

Final Results.

Quantity	Value at 60 Hz
Magnetising reactance X_m (LV side)	55.20 Ω
Leakage reactance, 230 V winding	33.60 mΩ
Leakage reactance, 6.6 kV winding	30.00 Ω (36.43 mΩ referred to LV)
Open-circuit primary current at 240 V	4.345 A
Open-circuit secondary voltage	6883 V (6.883 kV)
Short-circuit primary current at 24 V	342.8 A
Corresponding secondary short-circuit current	11.94 A

Topic. Frequency scaling of transformer parameters, and the open- and short-circuit tests.

Key relations. $V\approx 4.44fN\Phi_m$ so $B_m\propto V/f$; $P_h=k_hfB_m^{n}$ ($n\approx 1.6$–2.0); $P_e=k_ef^{2}B_m^{2}t^{2}$; $X=2\pi fL$ so $X\propto f$; impedance referral $x^{\prime}=x/a^{2}$ with $a=N_2/N_1$; open circuit isolates $x_1+X_m$; short circuit gives $x_1+(X_m\parallel x_2^{\prime})\approx x_1+x_2^{\prime}$.

Why this works. The two standard tests are designed so that each one makes a different branch of the T circuit disappear. On open circuit the secondary current is zero, so the referred secondary leakage reactance carries nothing and the source sees only the primary leakage in series with the large magnetising reactance; the current is small and the flux is full, which is why this test measures core loss. On short circuit the terminal voltage collapses to a few per cent of rated, so the flux and hence the magnetising current are negligible, the magnetising branch is effectively open, and the source sees the two leakage reactances in series; the current is near rated, which is why this test measures copper loss. Frequency enters only through X = 2 pi f L for the reactances and through B = V / (4.44 f N) for the flux, and keeping those two statements separate is what makes the whole family of frequency-change problems straightforward.

Common pitfall. Leaving the referred secondary leakage branch in the loop during the open-circuit calculation. It carries no current, so including it inflates the series reactance and understates the exciting current. The mirror-image error is to scale the magnetising reactance by the turns ratio as well as by frequency; the question already states that the 46 ohms was measured from the 230 V terminals, so it is referred to the low-voltage side already.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, rev. ed. (IEEE Press), ch. 5 (transformers); Chapman, Electric Machinery Fundamentals, 5th ed., ch. 2 and 3 (transformer equivalent circuit and testing); Glover, Sarma & Overbye, Power System Analysis and Design, 6th ed., ch. 3.

Question 3 of 7: Transformer Losses versus Frequency and a 50 Hz Unit Run at 60 Hz

Problem 3: Transformer Losses versus Frequency and a 50 Hz Unit Run at 60 Hz (25 points)

Part (a) — how frequency acts on each loss mechanism

Parts (b), (c) and (d) — the 60 Hz equivalent circuit and two tests