Set 4 — Hysteresis and Eddy-Current Losses
An alternating flux is not free. Taking the iron round its hysteresis loop costs the loop's area in joules per cubic metre every cycle, and the changing flux drives currents in the iron itself which cost \(I^2R\) on top. This set puts numbers to both: a loop area measured off a recorder trace, the Steinmetz law that scales it with flux density, the derivation that produces the \(t^2\) of laminating, and the two-frequency test that pulls the two losses apart.
The one idea that ties the set together is that flux density is not a free variable. It is fixed by the terminal voltage and the frequency through \(B_{max} \propto V/f\), so every question about changing the supply is answered by tracking volts per hertz first and applying the loss laws second. Problem 6 shows what falls out of that: the eddy-current loss of a voltage-fed core does not depend on frequency at all.
The hysteresis loop is an energy diagram. Taking the iron once round the loop costs an energy equal to the enclosed area, in joules per cubic metre per cycle:
\[ w_h = \oint H\,dB \quad \left[\text{J/m}^3\ \text{per cycle}\right] \]Multiply by volume for joules per cycle and by frequency for watts. Everything else in this set is that statement with a scaling law attached.
Steinmetz scales the loop with flux density. Over the working range of electrical steel the area grows as \(B_{max}^{n}\) with \(n \approx 1.6\):
\[ P_h = \eta\,B_{max}^{1.6}\,f\,V \]Eddy loss comes from induced circulating currents, so it carries a square on both the density and the frequency, and a square on the lamination thickness:
\[ P_e = k_e\,B_{max}^{2}\,f^{2}\,t^{2}\,V, \qquad \frac{P_e}{V} = \frac{\pi^2 B_{max}^2 f^2 t^2}{6\rho} \]The two losses are separated by their frequency dependence. At constant \(B_{max}\), divide the measured total by \(f\) and plot against \(f\): the result is a straight line whose intercept is the hysteresis term and whose slope is the eddy term:
\[ P_i = Af + Bf^2 \;\Longrightarrow\; \frac{P_i}{f} = A + Bf \]Holding \(B_{max}\) constant means holding \(V/f\) constant. From \(E = 4.44fN\Phi_{max}\),
\[ B_{max} = \frac{V}{4.44\,f\,N\,A_c} \;\propto\; \frac{V}{f} \]So a two-frequency loss test must be done at proportionally reduced voltage, or the two readings are not comparable.
Written in terms of the terminal quantities, the two losses behave quite differently — and the eddy loss stops depending on frequency at all:
\[ P_h \propto \frac{V^{1.6}}{f^{0.6}}, \qquad P_e \propto V^{2} \]Core loss is not a copper loss. It depends on voltage and frequency, not on load current, so it is constant from no load to full load. That is why it is measured by an open-circuit test and treated as a fixed loss in every efficiency calculation later in the book.
The hysteresis loop of a sample of transformer steel is traced on a recorder at a peak density of 1.2 T. The plot has the following scales, and the loop encloses 12 cm2 of area.
| Quantity | Value |
|---|---|
| Horizontal scale (\(H\)) | 1 cm = 250 A/m |
| Vertical scale (\(B\)) | 1 cm = 0.1 T |
| Enclosed loop area | 12 cm2 |
| Core cross-section | 60 cm2 |
| Core mean length | 2.0 m |
| Density of the steel | 7650 kg/m3 |
A core of the dimensions given is built from this material and operated at 1.2 T. Find:
- the hysteresis energy per cubic metre per cycle;
- the energy lost per cycle in the whole core;
- the hysteresis loss at 50 Hz, in watts and in watts per kilogram;
- the loss if the same core is operated at 60 Hz and the same peak density.
What one square centimetre of the plot is worth. The loop area is \(\oint H\,dB\), so the two axis scales multiply:
The units work because \(\text{A/m} \times \text{T} = \text{A/m} \times \text{Wb/m}^2 = \text{J/m}^3\). Checking that is the safest way to be sure the two scales have been multiplied and not divided.
Energy per cycle per unit volume:
The volume of iron, and the energy it costs per cycle:
Watts is joules per cycle times cycles per second:
Between 1 and 2 W/kg at 1.2 T and 50 Hz is what a good grade of transformer steel gives, so the figure is a sanity check on the whole calculation as much as an answer.
At 60 Hz with the same peak density, the loop is the same loop — only traversed more often:
Strictly proportional to frequency, which is exactly what separates hysteresis from the eddy loss of Problem 3.
The core of Problem 1 — volume 0.012 m3, 91.8 kg — loses 300 J/m3 per cycle to hysteresis at a peak density of 1.2 T. The material obeys the Steinmetz law with an exponent of 1.6. Find:
- the Steinmetz coefficient \(\eta\);
- the hysteresis loss at 50 Hz when the peak density is raised to 1.5 T;
- the peak density at which the 50 Hz hysteresis loss would be double its 1.2 T value;
- why an exponent of 2 is not used, given that the loop is roughly a parallelogram.
Fit the coefficient to the one measurement available. The Steinmetz law written per cycle per unit volume is \(w_h = \eta B_{max}^{1.6}\):
\(\eta\) is a property of the material and of nothing else. Its awkward unit is the price of an empirical exponent, and it is why the coefficient is almost always eliminated by taking ratios instead, as in part (c).
The loss at 1.5 T, keeping 50 Hz and the same core:
The same answer without ever finding \(\eta\). Ratios are quicker and carry no unit trouble:
A 25% rise in flux density brings a 43% rise in hysteresis loss. This is the reason transformer designers do not chase the last tesla of core utilisation.
The density that doubles the loss:
On paper. In practice 1.85 T is deep into saturation, where the exponent itself climbs towards 2 and beyond, so the true loss at that density would be higher still. The Steinmetz law is a fit over a working range, not a law of nature.
Why 1.6 and not 2. If the loop simply scaled up in both directions — every dimension proportional to \(B_{max}\) — the enclosed area would go as \(B_{max}^2\). It does not, because the loop widens in \(H\) much more slowly than it grows in \(B\) until saturation is approached. The measured exponent for good silicon steel over 0.2–1.5 T therefore lands between 1.5 and 1.7, and 1.6 is the conventional value. Above the knee it rises towards 2.
The core loss of a single-phase transformer is measured by an open-circuit test at two frequencies. To hold the peak flux density constant the applied voltage is reduced in proportion to the frequency.
| Test | Applied voltage | Frequency | Measured core loss |
|---|---|---|---|
| 1 | 250 V | 50 Hz | 65 W |
| 2 | 125 V | 25 Hz | 25 W |
Separate the hysteresis and eddy-current components at 50 Hz, predict the total loss at 60 Hz and 300 V, and find the frequency at which the two components are equal.
Check that the test is valid before using it. The separation works only if \(B_{max}\) is the same in both readings, and
Equal volts per hertz, so equal flux density. Had both tests been done at 250 V the 25 Hz reading would correspond to twice the flux density and the two equations would describe different materials' worth of loss.
Write the model and linearise it. With \(B_{max}\) fixed, the two constants absorb it:
Dividing by \(f\) is the whole trick: it turns a quadratic into a straight line whose intercept is the hysteresis constant and whose slope is the eddy constant.
Form the two data points:
Split the 50 Hz loss:
| Frequency | \(P_h = 0.7f\) | \(P_e = 0.012f^2\) | Total | Measured |
|---|---|---|---|---|
| 25 Hz | 17.5 W | 7.5 W | 25.0 W | 25 W |
| 50 Hz | 35.0 W | 30.0 W | 65.0 W | 65 W |
| 60 Hz | 42.0 W | 43.2 W | 85.2 W | — |
Both measured points are reproduced exactly, as they must be — two constants fitted to two readings. The value of the model is the third row.
60 Hz at 300 V. Check the volts per hertz first: \(300/60 = 5\) V/Hz, the same as before, so \(B_{max}\) is unchanged and the fitted constants still apply:
Had the question said 250 V at 60 Hz, the density would have fallen and the constants would have needed rescaling — that is Problem 6.
Where the two components cross:
Below 58 Hz hysteresis dominates; above it, eddy currents do. This is why the laminations of a 400 Hz aircraft transformer are far thinner than those of a 50 Hz power transformer, while the grade of steel matters comparatively less.
A core built from 0.50 mm laminations has a total loss of 90 W at 50 Hz and 1.1 T, made up of 60 W of hysteresis and 30 W of eddy-current loss. The core is to be rebuilt from the same grade of steel and to the same overall dimensions, using thinner sheet. Find:
- the total loss if 0.35 mm laminations are used;
- the thickness needed to bring the eddy loss down to 10 W;
- the eddy loss the core would have if it were a solid block 50 mm thick instead of laminated;
- why the answer to (c) is an overestimate.
Only one of the two losses moves. Hysteresis is a property of the material and the flux density; slicing the same iron into thinner sheets does not change how much iron there is or what density it works at. Eddy loss, by contrast, carries \(t^2\):
At 0.35 mm:
A 30% reduction in thickness halves the eddy loss but cuts the total by only 17%, because two-thirds of the loss was never eddy loss to begin with. Thinner laminations have diminishing returns for exactly this reason.
The thickness for 10 W of eddy loss:
About 0.29 mm, near the thinnest sheet normally rolled for 50 Hz power work. Below this the stacking factor falls, the burr and the interlaminar insulation take a larger fraction of the stack, and the effective iron area — and therefore the flux density — changes, which the simple \(t^2\) scaling does not account for.
The solid block, by the same square law:
Against 60 W of hysteresis. The core would destroy itself in seconds, and this single ratio is the entire justification for laminating.
Why 300 kW is too large a figure to believe. The \(t^2\) law is derived on the assumption that the flux density is uniform across the thickness. In a solid block it is not: the eddy currents themselves oppose the flux and confine it to a surface layer of the order of the skin depth,
A 50 mm block is some forty-five skin depths thick, so most of the iron carries no flux at all. The real loss is far below 300 kW — but so is the useful flux, which is the more serious objection. The \(t^2\) law is trustworthy only while \(t \ll \delta\), which is exactly the regime laminations are chosen to sit in.
Derive the eddy-current loss per unit volume of a lamination of thickness \(t\) carrying a sinusoidal flux density of peak value \(B_{max}\) at frequency \(f\), and hence evaluate it for the following core.
| Quantity | Value |
|---|---|
| Lamination thickness \(t\) | 0.35 mm |
| Resistivity \(\rho\) | \(0.50\ \mu\Omega\cdot\text{m}\) |
| Peak flux density | 1.2 T |
| Frequency | 50 Hz |
| Core volume | 0.012 m3 |
| Density | 7650 kg/m3 |
The lamination is wide and long compared with its thickness, so the induced currents may be taken to flow in flat rectangular loops in planes parallel to the flux. Give the loss in W/m3, W/kg and watts, and state what the answer becomes at 0.50 mm.
Set up the elementary loop. Take the lamination of thickness \(t\), width \(w\) and length \(l\), with \(B\) along its length. Consider the thin loop lying at \(\pm x\) from the mid-plane, of thickness \(dx\). It encloses an area \(2x\,l\), so the emf round it is
The resistance of that loop. The current runs the length \(l\) on each side, through a cross-section \(w\,dx\); the short returns across the width are neglected because \(w \gg t\):
Integrate across the half-thickness and divide by the volume \(lwt\):
There is the \(t^2\) of Problem 4, and there is the reason resistivity appears in the denominator — which is why silicon is alloyed into the steel: 3% silicon roughly quadruples \(\rho\) and quarters this loss.
Evaluate it. With \(t = 0.35\times10^{-3}\) m and \(\rho = 0.50\times10^{-6}\ \Omega\cdot\text{m}\):
Around 0.2 W/kg is what a datasheet quotes for 0.35 mm grain-oriented sheet at 1.2 T and 50 Hz, so the derivation lands where it should.
At 0.50 mm, by the square law just derived:
What the derivation quietly assumed. Three things: that \(B\) is uniform across the thickness, which requires \(t\) to be small against the skin depth; that the induced currents do not themselves alter the flux, which is the same condition stated differently; and that the laminations are electrically separate, which fails if burrs from the punch bridge the insulation. A stack with shorted laminations behaves partly as a thicker sheet and can lose several times the calculated figure.
A transformer designed for 240 V, 50 Hz has a total core loss of 200 W, of which 120 W is hysteresis and 80 W eddy-current loss. It is proposed to operate it on a 60 Hz supply. Take the Steinmetz exponent as 1.6.
- Find the core loss at 240 V, 60 Hz (constant voltage).
- Find the core loss at 288 V, 60 Hz (constant volts per hertz).
- Show that the eddy-current loss depends on the applied voltage alone, and explain why.
- Comment on which of the two operating conditions is safer for the core.
Everything turns on the flux density, and the flux density is set by the terminal conditions, not chosen:
Since \(N\) and \(A_c\) are fixed by the iron already wound, the ratio \(V/f\) is the only handle on \(B_{max}\).
Case (a): 240 V, 60 Hz. The voltage is held and the frequency raised, so the flux density falls:
A 6% fall in total loss, entirely from the hysteresis term.
Case (b): 288 V, 60 Hz. Here \(V/f = 288/60 = 4.8 = 240/50\), so \(B_{max}\) is exactly what it was designed for and only the frequency ratio acts:
Collect the two cases against the original:
| Condition | \(V/f\) | \(B_{max}\) (relative) | \(P_h\) | \(P_e\) | Total |
|---|---|---|---|---|---|
| 240 V, 50 Hz | 4.80 | 1.000 | 120.0 W | 80.0 W | 200.0 W |
| 240 V, 60 Hz | 4.00 | 0.833 | 107.6 W | 80.0 W | 187.6 W |
| 288 V, 60 Hz | 4.80 | 1.000 | 144.0 W | 115.2 W | 259.2 W |
Why the eddy loss did not move in case (a). Substitute \(B_{max} \propto V/f\) into each law and watch the frequency cancel:
The eddy loss of an iron core fed from a voltage source is a function of voltage alone. Case (a) held 240 V, so \(P_e\) stayed at exactly 80 W — not a coincidence but an identity, and a useful check on any answer of this type.
Which condition is safer. Case (a): lower flux density, lower total loss, lower magnetising current, and the core further from the knee — entirely benign, and the reason a 50 Hz transformer runs happily on 60 Hz at rated voltage. Case (b) restores the design flux density but raises the loss by 30%, so the core runs hotter and the rating must be reviewed for temperature rise even though the iron itself is no more heavily worked.
The dangerous case is the reverse one, not asked here: a 60 Hz transformer on a 50 Hz supply at rated voltage, where \(V/f\) rises by 20%, the core saturates, and the magnetising current can multiply several times over.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Loop area as energy | \(w_h = \oint H\,dB\) | J/m3 per cycle — Problem 1 |
| Graphical scale factor | \(1\ \text{cm}^2 \equiv (\text{A/m per cm})(\text{T per cm})\) | Multiply the two axis scales — Problem 1 |
| Hysteresis power | \(P_h = \eta B_{max}^{1.6} f V\) | Linear in \(f\) — Problems 1, 2 |
| Steinmetz ratio form | \(\dfrac{P_{h2}}{P_{h1}} = \left(\dfrac{B_2}{B_1}\right)^{1.6}\dfrac{f_2}{f_1}\) | No coefficient needed — Problems 2, 6 |
| Eddy-current power | \(P_e = k_e B_{max}^2 f^2 t^2 V\) | Square in \(f\) and \(t\) — Problems 3, 4 |
| Eddy loss density | \(\dfrac{P_e}{V} = \dfrac{\pi^2 B_{max}^2 f^2 t^2}{6\rho}\) | Derived, not fitted — Problem 5 |
| Loss separation | \(P_i = Af + Bf^2\) | Constant \(B_{max}\) assumed — Problem 3 |
| Linearised form | \(\dfrac{P_i}{f} = A + Bf\) | Intercept \(A\), slope \(B\) — Problem 3 |
| Equal-component frequency | \(f = A/B\) | Above it eddy loss dominates — Problem 3 |
| Flux density from terminals | \(B_{max} = \dfrac{V}{4.44 f N A_c}\) | \(B_{max} \propto V/f\) — Problems 3, 6 |
| Voltage form of the losses | \(P_h \propto \dfrac{V^{1.6}}{f^{0.6}}, \quad P_e \propto V^2\) | Eddy loss is frequency-free — Problem 6 |
| Lamination scaling | \(\dfrac{P_{e2}}{P_{e1}} = \left(\dfrac{t_2}{t_1}\right)^{2}\) | Hysteresis unaffected — Problem 4 |
| Skin depth | \(\delta = \sqrt{\dfrac{\rho}{\pi f\mu}}\) | Validity limit of the \(t^2\) law — Problems 4, 5 |
| Specific loss | \(P/m = P/(\rho_m V)\) | W/kg, the datasheet quantity — Problems 1, 5 |
Common Mistakes
Reading the loop area as a power. It is an energy density, in J/m3 per cycle. Watts appear only after multiplying by volume and by frequency — Problem 1.
Dividing the two axis scales instead of multiplying them. \((\text{A/m})\times(\text{T}) = \text{J/m}^3\); check the units on the scale factor before touching the area — Problem 1.
Comparing two loss readings taken at the same voltage. Halving the frequency at constant voltage doubles \(B_{max}\), so the two readings belong to different flux densities and the \(A + Bf\) fit is meaningless. The voltage must be scaled with the frequency — Problem 3.
Fitting \(P_i\) against \(f\) instead of \(P_i/f\) against \(f\). The first is a parabola through the origin and cannot be read off a ruler; the second is a straight line whose intercept is the answer — Problem 3.
Scaling hysteresis loss with the square of flux density. The exponent is about 1.6 over the normal working range, so a 25% rise in \(B\) gives 43%, not 56% — Problem 2.
Changing the hysteresis loss when the laminations are made thinner. Thickness enters the eddy term only. The same mass of iron at the same density loses the same hysteresis energy per cycle — Problem 4.
Trusting the \(t^2\) law outside its range. It assumes uniform flux across the sheet, which fails once the thickness approaches the skin depth — the 300 kW of Problem 4 is nominal, not real.
Forgetting that eddy loss is fixed by voltage alone. Raising the frequency at constant voltage leaves \(P_e\) untouched, because \(B^2f^2 \propto V^2\). An answer that shows \(P_e\) rising with \(f\) at constant \(V\) has ignored the flux-density change — Problem 6.
Treating core loss as load-dependent. It follows voltage and frequency, both of which are held constant in service, so it is the same at no load and at full load — the premise of every open-circuit test in Part 3.
Mixing \(\rho\) the resistivity with \(\rho\) the mass density. One is \(0.5\ \mu\Omega\cdot\text{m}\) and divides; the other is 7650 kg/m3 and converts W/m3 to W/kg — Problem 5.
Sets 1 to 3 asked what ampere-turns a flux costs. This set asked what an alternating flux costs, and the answer came in two parts with two different signatures: an area traced once per cycle, proportional to \(f\), and a circulating current driven by \(dB/dt\), proportional to \(f^2\). Two loss readings at two frequencies separate them, and every later efficiency calculation in this book — transformer, dc machine, induction motor — uses that split as its fixed loss.
Notice what has quietly entered the argument. Faraday's law was used to find the emf round an eddy loop, and again to turn a terminal voltage into a flux density. The same law, applied to a winding rather than to the iron, defines inductance; applied to a second winding on the same core, it defines mutual inductance; and applied to a conductor moving through a field, it produces the \(Blv\) of every generator in the book.
Next: Set 5 — Self, Mutual Inductance and Induced EMF, where \(L = N^2/\mathcal{R}\) is put to work on real cores, the coupling coefficient is measured by a series-aiding and series-opposing test, and dynamically induced emf is set beside statically induced emf as two readings of one law.