Single-Phase Transformer Construction

i Learning Objectives

By the end of this chapter you should be able to:

Explain the limitations of the DC system that made the transformer necessary.
State what a transformer is and why it is a transformer and not a converter.
State the magnetic properties required of the core material and why CRGO steel is used.
Explain the trade-off in lamination thickness and compute the stacking factor.
Distinguish core-type from shell-type construction.
Derive the cruciform core dimensions and its utilisation of the circular window.
Describe concentric and sandwich winding arrangements.
Explain the cooling methods and the function of the conservator, breather and Buchholz relay.

Section 40-1

A Look Back in History

William Stanley, who built the first reliable commercial transformer in 1886 — William Stanley.

Who invented the transformer?

Otto Bláthy, Miksa Déri and Károly Zipernowsky first designed it. Later Lucien Gaulard, Sebastian Ferranti and William Stanley perfected the design.

When and where?

In 1886 William Stanley built the first reliable commercial transformer, used at Great Barrington, Massachusetts. In 1891 Mikhail Dolivo-Dobrovolsky demonstrated three-phase transformers at Frankfurt, Germany.

Michael Faraday, who discovered electromagnetic induction in 1831 — Michael Faraday.

Michael Faraday established the principle of electromagnetic induction in 1831 — the law of Chapter 10, on which everything here rests.
The finding forms the basis for many magneto-electric machines.
The earliest use of the phenomenon was in the development of induction coils, used to generate high-voltage pulses to ignite explosive charges in mines.

Section 40-2

The Limitations of the DC System

Comparison of AC and DC power systems — The AC and DC systems.

A DC power system was in use at the time. It suffered two limitations that no refinement could remove.

For economic transmission of power, the generating station and the load centre had to be close to each other.
DC generators cannot be scaled up, because of the limitations of the commutator — Chapter 31's reactance voltage sets a ceiling on the current a commutator can handle, and Chapter 23's segment voltage sets one on the voltage.

⚡

Why Voltage Level Decides Everything

Loss falls as the square of the transmission voltage

For a given power \(P\) transmitted at voltage \(V\), the current is \(P/V\), so the line loss is

\[P_{\text{line}} = I^{2}R = \left(\frac{P}{V}\right)^{2}R \propto \frac{1}{V^{2}}\]

Raising the transmission voltage tenfold cuts the line loss to one hundredth. That is the whole economic case for high-voltage transmission — and until the transformer, there was no way to raise it.

The world therefore looked for other efficient methods of bulk power generation and transmission. During the second half of the nineteenth century the alternator, the transformer and the induction motor were invented — all working on alternating supply.

With the transformer it became possible to separate three choices that DC had forced together. A moderate voltage for generating AC power, a high voltage for transmitting it over long distances, and a small, safe voltage at the user's end. Each can now be chosen on its own merits, which is why the transformer is called the heart of the AC system.

Video · Transformer Construction

Section 40-3

What a Transformer Is

A large power transformer — A power transformer.

A transformer is a static device — nothing rotates.
It changes the voltage and current levels keeping the power invariant.
It consists of two electrical circuits linked by a common magnetic circuit.
One coil generates a time-varying magnetic field; the second links that field and has a voltage induced in it.
The magnitude of the induced EMF is decided by the number of turns in each coil, so the voltage level is changed by changing the turns.
The excitation winding is called the primary and the output winding the secondary.
There is no conductive connection between the two electrical circuits, so the transformer provides electrical isolation.
The frequency on the two sides is the same.
The efficiency of the conversion is extremely high — commonly 98 to 99.5 %.

? Why "Transformer" and Not "Converter"

Because there is no change in the nature of the power. Electrical power goes in and electrical power comes out, at the same frequency; only the ratio of voltage to current is altered.

Contrast the DC machines of Part 2, which convert between electrical and mechanical form, and a rectifier, which converts AC to DC. The transformer transforms; it does not convert.

This also explains the absence of a whole category of loss. With nothing rotating there is no friction and no windage, which is why transformer efficiencies exceed those of any rotating machine.

Section 40-4

Scale and Ubiquity

A power system showing generation, transmission and distribution — The place of transformers in a power system.

Transformers are not limited to power systems. Special transformers serve electronic supplies, rectification, furnaces and traction.
Operating frequencies range from a few hertz to several megahertz.
Power ratings range from a few milliwatts to several hundred megawatts — a span of eleven orders of magnitude.
Electric power generation demand doubles roughly every decade in a developing country.
For every MVA of generation, the installed capacity of transformers grows by about 7 MVA.

That last figure repays a moment's thought. A unit of energy is transformed several times on its way from the generator to the lamp — stepped up at the station, down at the grid substation, down again at the distribution substation, and once more at the pole. There is seven times as much transformer capacity installed as generating capacity, which is why the efficiency of a transformer matters so much: the losses are incurred repeatedly on every unit delivered.

Section 40-5

Core Materials

The construction of a transformer differs with its end use, but the principle of operation is the same. Focusing on the power transformer, the constructional aspects fall under three categories — core construction, winding arrangements and cooling — which are the subjects of this and the following sections.

🧲

What the Magnetic Material Must Provide

Four requirements, each with a reason

High permeability \(\mu_r \gt 1000\) — giving low reluctance, so the flux confines itself to the iron core rather than leaking through air.
High saturation flux density — allowing a small core cross-section for a given flux.
Small area under the B-H loop — permitting a high working flux density with low magnetising current and low hysteresis loss (Chapter 6).
High resistivity — reducing eddy-current loss (Chapter 7).

Silicon steel in the form of thin laminations meets all four. Silicon raises the resistivity and narrows the hysteresis loop; the lamination attacks the eddy currents geometrically.

Table 40.1 — The development of core steel.
Material	Character	Note
Hot-rolled non-oriented	Same properties in all directions	Earliest
Hot-rolled grain-oriented	Grains aligned with the rolling direction	Better along the grain
Cold-rolled grain-oriented (CRGO)	Strong preferred direction, low loss	Present standard

The silicon content is about 3.5 %. Above this the steel becomes very brittle and very hard to cut. The saturation flux density of present-day laminations is about 2 T, and cores are worked at roughly 1.5 to 1.7 T — close enough to gain the small core, far enough to keep the magnetising current reasonable.

! Why Grain Orientation Must Be Respected

CRGO steel is anisotropic: its loss and permeability are far better along the rolling direction than across it. The core must therefore be built so that the flux everywhere follows the grain.

This is why modern cores use mitred joints, cut at 45°, rather than square butt joints. At a square joint the flux must turn a right angle across the grain, raising both the loss and the magnetising current locally.

Section 40-6

Laminations and Stacking Factor

Transformer core laminations — Core laminations.

Laminations are coated with a thin layer of insulating varnish, oxide or phosphate, so that eddy currents cannot pass from one sheet to the next.
Their thickness has progressively fallen from over 0.5 mm to about 0.25 mm per lamination.
Eddy loss varies as the square of the thickness, so halving it quarters that loss.
But if the lamination is made too thin, the production cost of the steel rises, and more sheets are needed for a given stack.
For very small transformers, from a few VA to a few kVA, hot-rolled silicon steel laminations in \(E\;\&\;I\), \(C\;\&\;I\) or \(O\) shapes are used, and the core cross-section is square or rectangular.

📐

Stacking Factor

The insulation occupies space but carries no flux

\[k_s = \frac{\text{net iron area}}{\text{gross core area}} = \frac{t}{t + t_i}\]

where \(t\) is the steel thickness and \(t_i\) the coating. Thinner laminations mean a worse stacking factor, because the coating thickness does not shrink with the steel. Typical values are 0.90 to 0.97.

1 Worked Example 40.1 — The Lamination Trade-Off

Problem. Laminations are reduced from 0.5 mm to 0.25 mm, the insulating coating being 0.01 mm in each case. Find the change in eddy-current loss and in stacking factor, and hence the net change in eddy loss if the total flux is unaltered.

Eddy loss on thickness alone. From Chapter 7, \(P_e \propto t^{2}\):

\[\frac{P_{e2}}{P_{e1}} = \left(\frac{0.25}{0.50}\right)^{2} = 0.250 \quad\Longrightarrow\quad \text{a 75 % reduction}\]

Stacking factor.

\[k_1 = \frac{0.50}{0.51} = 0.9804, \qquad k_2 = \frac{0.25}{0.26} = 0.9615\]

\[\text{net iron area falls by } \left(1 - \frac{0.9615}{0.9804}\right) = 1.92\,\%\]

Consequence for flux density. The same flux in a smaller iron area means a higher \(B\):

\[\frac{B_2}{B_1} = \frac{k_1}{k_2} = 1.0196 \quad\text{(a 1.96 % rise)}\]

Net effect. Since \(P_e \propto t^{2}B^{2}\):

\[\frac{P_{e2}}{P_{e1}} = (0.250)(1.0196)^{2} = 0.2599 \quad\Longrightarrow\quad \text{a 74.0 % reduction}\]

Comment. The stacking penalty eats only a percentage point of the benefit — 74.0 % instead of 75 % — so thinning the laminations is overwhelmingly worthwhile. What stops the process is cost, not physics.

Note the hysteresis loss moves the other way. With \(P_h \propto B^{1.6}\) (Chapter 6), the 1.96 % rise in \(B\) increases hysteresis loss by 3.16 %. Since hysteresis usually dominates eddy loss in a well-laminated core, this partly offsets the gain — which is exactly why the switch to CRGO steel, which attacks the hysteresis loop directly, mattered as much as thinner sheets.

Section 40-7

Core Type and Shell Type

Single-phase transformers are built in two arrangements, distinguished by whether the winding surrounds the core or the core surrounds the winding.

Core-type and shell-type single-phase construction.

Table 40.2 — Core type against shell type.
	Core type	Shell type
Arrangement	Winding surrounds core	Core surrounds winding
Limbs	Two, each carrying half of each winding	Three; winding on the central limb
Flux path	Single path	Divides, Φ/2 in each outer limb
Central limb area	—	Twice each outer limb
Mean length of flux path	Longer	Shorter
Mean length of turn	Shorter	Longer
Leakage reactance	Higher	Lower
Mechanical bracing	Less natural support	Winding well braced by core
Repair access	Easier	Harder
Suits	High voltage, lower current	Low voltage, high current

Note the reciprocal geometry, which explains most of the table. In the core type the iron path is long and the copper path short; in the shell type the reverse. Whichever material wraps around the other has the longer mean path — so the core type uses less copper and more iron, and the shell type the opposite.

In both, the low-voltage winding is placed nearest the core. It requires the least insulation to the earthed core, so putting it innermost minimises the total insulation and hence the window space.

Section 40-8

Stepped and Cruciform Cores

Windings are wound as cylinders, so the space available to the core limb is a circle. A square limb wastes much of it. Larger transformers therefore use a stepped or cruciform section, built from laminations of two or more widths.

Utilisation of the circular winding space by square and stepped cores.

🔲

Utilisation Factor

How much of the circle the iron occupies

\[\text{utilisation} = \frac{\text{net core area}}{\pi d^{2}/4}\]

More steps fill the circle better but need more lamination widths, so the gain must be set against manufacturing cost. Beyond about four steps the return is not worth the complication.

2 Worked Example 40.2 — Deriving the Cruciform Core

Problem. A cruciform core is inscribed in a circle of diameter \(d\). Find the two dimensions that maximise the core area, and compare the utilisation with that of a square core.

Setting up. The cruciform is the union of two rectangles, \(a \times b\) and \(b \times a\), both centred and both inscribed, so \(a^{2} + b^{2} = d^{2}\). Their union has area

\[A = 2ab - b^{2}\]

Maximising. Substituting \(a = \sqrt{d^{2} - b^{2}}\) and setting \(\mathrm{d}A/\mathrm{d}b = 0\):

\[d^{2} - 2b^{2} = b\sqrt{d^{2} - b^{2}}\]

Squaring and collecting terms gives a quadratic in \(b^{2}\):

\[5b^{4} - 5d^{2}b^{2} + d^{4} = 0 \quad\Longrightarrow\quad b^{2} = \frac{5 - \sqrt{5}}{10}d^{2} = 0.27639d^{2}\]

Dimensions.

\[b = 0.5257d, \qquad a = \sqrt{1 - 0.27639}\,d = 0.8507d\]

Area and utilisation.

\[A = 2(0.8507)(0.5257)d^{2} - 0.27639d^{2} = 0.61803d^{2}\]

\[\text{utilisation} = \frac{0.61803}{\pi/4} = \frac{0.61803}{0.78540} = 78.69\,\%\]

Compare a square. Its side is \(d/\sqrt{2}\):

\[A_{\square} = 0.5000d^{2}, \qquad \text{utilisation} = \frac{0.5000}{0.78540} = 63.66\,\%\]

Comment. The cruciform gives \(0.618/0.500 = 1.236\), that is 23.6 % more iron inside the same winding. For the same flux the core can therefore be smaller, or for the same core the flux density lower — and with it the iron loss.

Note that \(0.61803\) is \((\sqrt{5}-1)/2\), the golden ratio, which appears here purely as an accident of the quadratic. Adding a third step raises the utilisation to 85.1 % and a fourth to 88.6 %, so the returns diminish steadily.

3 Worked Example 40.3 — From Core Dimensions to Turns

Problem. A cruciform core is inscribed in a 200 mm circle, with a stacking factor of 0.90 and a working flux density of 1.6 T at 50 Hz. Find the net iron area, the flux, the volts per turn, and the turns needed for an 11 000 V / 415 V transformer.

Gross and net area.

\[A_{\text{gross}} = 0.61803(0.200)^{2} = 0.02472~\mathrm{m^{2}}\]

\[A_{i} = (0.90)(0.02472) = 0.02225~\mathrm{m^{2}}\]

Flux.

\[\Phi_m = BA_i = (1.6)(0.02225) = 0.03560~\mathrm{Wb}\]

Volts per turn. From the EMF equation of Chapter 8, which Chapter 41 will derive again for the transformer:

\[E_t = 4.44f\Phi_m = (4.44)(50)(0.03560) = 7.903~\mathrm{V/turn}\]

Turns.

\[N_{HV} = \frac{11\,000}{7.903} = 1392~\text{turns}, \qquad N_{LV} = \frac{415}{7.903} = 52.5 \to 53~\text{turns}\]

Comment. The whole design chain runs one way — circle diameter to core area to flux to volts per turn to turns — and every transformer design begins exactly here.

Note the awkwardness at the LV end. 52.5 turns is not available, so rounding to 53 shifts the ratio slightly: the actual no-load secondary voltage becomes \((53)(7.903) = 418.9\) V rather than 415 V, about 0.9 % high. That is deliberate — a small excess compensates the voltage drop the transformer will suffer on load, which is Chapter 45's subject.

Section 40-9

Winding Arrangements

Cutaway of an electrical transformer showing core and windings — Core and windings of a transformer.

The second of the three constructional categories. Windings are made of copper — occasionally aluminium — either as round wire for small ratings or as rectangular strip for large currents.

Concentric (cylindrical)

The LV winding is wound directly over the limb, and the HV winding over it, separated by an insulating cylinder.

Standard for core-type transformers. Simple, and the LV sits where the insulation to earth is least demanding.

Sandwich (interleaved)

LV and HV coils are stacked alternately along the limb, with an LV half-coil at each end.

Standard for shell-type transformers. Interleaving reduces the leakage flux, and hence the leakage reactance, because each HV coil is closely flanked by LV.

! Why Interleaving Reduces Leakage Reactance

Leakage flux is the flux that links one winding but not the other, and it is driven by the MMF between them. The more finely the two windings are interleaved, the smaller the region over which that MMF can build up, and the less leakage flux results.

This is a design lever, not merely a detail. Leakage reactance sets the voltage regulation (Chapter 45) and limits the short-circuit current — so a low value is good for regulation and bad for fault levels. The designer chooses the interleaving to strike a balance.

Other winding types met in practice include helical windings for high-current LV, disc windings for high-voltage HV, and cross-over windings for small distribution units. All are variants on the two basic arrangements above.

Section 40-10

Cooling and Accessories

The third category. Even at 99 % efficiency a 10 MVA transformer dissipates 100 kW, and that heat must leave without the winding insulation exceeding its temperature limit.

Table 40.3 — Cooling methods. The four letters give the internal medium and its circulation, then the external medium and its circulation.
Code	Meaning	Typical rating
AN	Air natural — dry type, no oil	Up to a few hundred kVA
AF	Air blast, fan assisted	Dry type, larger
ONAN	Oil natural, air natural	Distribution, up to ~10 MVA
ONAF	Oil natural, air forced by fans	Up to ~30 MVA
OFAF	Oil forced by pump, air forced	Large power transformers
OFWF	Oil forced, water forced	Very large, generator transformers

🌡

The Oil Does Two Jobs

Insulation and heat transport

Transformer oil has a far higher dielectric strength than air, allowing closer spacing of the windings, and it carries heat from the windings to the tank walls by natural convection.

Its enemy is moisture: a trace of water collapses the dielectric strength. Hence the breather and the conservator described below.

Tank Accessories

Conservator

A small drum above the tank, partly filled, that accommodates the expansion and contraction of the oil with temperature while keeping the main tank completely full.

Breather

As the oil contracts, air is drawn into the conservator. The breather passes it through silica gel, which removes moisture. The gel is blue when dry and pink when saturated.

Buchholz relay

Fitted in the pipe between tank and conservator. Internal faults decompose the oil into gas; slow gas accumulation raises an alarm, and a sudden surge of oil trips the transformer.

Bushings and tap changer

Bushings bring the connections through the tank without flashover. The tap changer alters the turns ratio in small steps to hold the secondary voltage as the load varies.

Larger units add radiators or cooling tubes to multiply the tank's dissipating surface, an explosion vent, and a temperature indicator for oil and winding.

4 Worked Example 40.4 — Sizing the Cooling Surface

Problem. A 1000 kVA transformer is 98.5 % efficient at full load. Its plain tank has a dissipating surface of 20 m², and a plain surface dissipates about 12.5 W/m² per °C of temperature rise. Find the temperature rise, and the surface needed to limit it to 40 °C.

Loss to be dissipated.

\[P_{\text{loss}} = (1000\,000)(1 - 0.985) = 15\,000~\mathrm{W} = 15~\mathrm{kW}\]

Temperature rise with the plain tank.

\[\theta = \frac{P_{\text{loss}}}{Ac} = \frac{15\,000}{(20)(12.5)} = \frac{15\,000}{250} = 60~^{\circ}\mathrm{C}\]

Surface required for 40 °C.

\[A = \frac{15\,000}{(40)(12.5)} = \frac{15\,000}{500} = 30~\mathrm{m^{2}}\]

\[\text{additional surface} = 30 - 20 = 10~\mathrm{m^{2}}\]

Comment. The plain tank gives 60 °C, well above the usual 40 to 50 °C limit, so cooling tubes or radiators are essential — they must provide half as much surface again as the tank itself.

Note how the requirement scales. Loss grows roughly with the volume of the machine, that is with the cube of a linear dimension, while tank surface grows only with the square. Larger transformers therefore need disproportionately more cooling surface, which is why a small distribution unit has a plain tank, a medium one has tubes, and a large one needs pumps and fans.

5 Worked Example 40.5 — Choosing a Construction

Problem. Two single-phase transformers are to be built: (a) 25 kVA, 11 000/240 V for pole mounting; (b) 500 kVA, 415/33 000 V feeding a furnace, subject to frequent heavy overloads. Recommend core or shell type, winding arrangement and cooling for each, with reasons.

(a) 25 kVA, 11 000/240 V distribution unit.

Core type. The high-voltage side dominates the design, and the core type's shorter mean turn saves copper on the many HV turns. Its better access also suits a unit that will be repaired in the field.
Concentric windings, LV innermost, standard for core type.
ONAN cooling with a plain or lightly tubed tank. At 25 kVA and, say, 98 % efficiency the loss is only 500 W.
Square or lightly stepped core — at this size the cost of extra lamination widths outweighs the iron saved.

(b) 500 kVA furnace transformer with heavy overloads.

Shell type. The LV side carries a very large current, and the shell type braces the windings mechanically within the core — which matters because the forces between windings rise as the square of the current, and this unit is overloaded often.
Sandwich windings, giving low leakage reactance and good mechanical support.
ONAF cooling, since the duty is heavy and intermittent; fans can be switched in on the overload peaks.
Stepped core, worthwhile at this rating.

Comment. The two go opposite ways on every count, and one principle decides most of it: core type where the voltage is high and the current modest; shell type where the current is high and mechanical robustness matters.

The overload point deserves emphasis. A short circuit on a 500 kVA transformer can produce twenty times rated current, and the resulting forces are proportional to \(I^{2}\) — four hundred times normal. Mechanical design, not thermal, is what limits a transformer's fault withstand.

Section 40-11

Summary and Key Formulas

DC systems were limited by the need for generation near the load and by the commutator, which caps a DC machine's voltage and current.
Line loss varies as \(1/V^{2}\), so high-voltage transmission is essential — and only the transformer makes the voltage choosable.
A transformer is a static device: two electrical circuits linked by one magnetic circuit, changing voltage and current at constant power and constant frequency, with electrical isolation.
Core material needs high permeability, high saturation density, a narrow B-H loop and high resistivity. CRGO silicon steel with about 3.5 % silicon meets these; saturation is around 2 T.
Laminations have fallen from over 0.5 mm to about 0.25 mm. Eddy loss \(\propto t^{2}\), but the stacking factor \(k_s = t/(t + t_i)\) worsens as sheets get thinner.
Core type: winding surrounds core, two limbs, longer iron path, shorter mean turn. Shell type: core surrounds winding, three limbs with \(\Phi/2\) in each outer, lower leakage reactance, better bracing.
A cruciform core fills 78.7 % of the circular winding space against a square's 63.7 % — 23.6 % more iron for the same coil.
Concentric windings suit core type, sandwich windings suit shell type and give lower leakage reactance.
Cooling runs from AN through ONAN, ONAF, OFAF to OFWF. The oil both insulates and carries heat, and must be kept dry by the breather.

Table 40.4 — Formulas of this chapter.
Quantity	Formula	Notes
Transmission line loss	\(P_{\text{line}} = \left(P/V\right)^{2}R\)	∝ 1/V²
Stacking factor	\(k_s = \dfrac{t}{t + t_i}\)	typically 0.90–0.97
Net iron area	\(A_i = k_sA_{\text{gross}}\)	—
Eddy loss	\(P_e \propto t^{2}B^{2}f^{2}\)	Chapter 7
Hysteresis loss	\(P_h \propto B^{1.6}f\)	Chapter 6
Cruciform dimensions	\(a = 0.8507d,\ b = 0.5257d\)	inscribed in circle \(d\)
Cruciform area	\(A = 0.61803d^{2}\)	78.69 % of circle
Square core area	\(A = 0.5000d^{2}\)	63.66 % of circle
Volts per turn	\(E_t = 4.44f\Phi_m\)	\(\Phi_m = BA_i\)
Tank temperature rise	\(\theta = \dfrac{P_{\text{loss}}}{Ac}\)	\(c \approx 12.5\) W/m²/°C

Section 40-12

Common Mistakes

Calling a transformer a converter. Electrical power in, electrical power out, same frequency — nothing is converted.
Expecting the frequency to change. It is the same on both sides, always.
Using gross core area for the flux. Use the net iron area \(k_sA_{\text{gross}}\); the insulation carries no flux.
Forgetting that thinner laminations worsen the stacking factor. The coating thickness does not shrink with the steel.
Confusing core type with shell type. In the core type the winding surrounds the core.
Giving the shell-type outer limbs the same area as the centre. They each carry \(\Phi/2\), so each has half the area.
Placing the HV winding next to the core. The LV goes innermost — it needs the least insulation to earth.
Assuming more core steps are always better. Each extra step needs another lamination width; beyond about four the gain does not pay.
Reading saturation flux density as a percentage. It is about 2 tesla.
Thinking the conservator cools the oil. It accommodates expansion; the tank, tubes and radiators do the cooling.

Section 40-13

Chapter Review

Practice Problems

For core problems, work from the circle diameter through the utilisation factor and the stacking factor to the net iron area.

P40.1 Laminations of 0.35 mm with 0.01 mm coating are used. Find the stacking factor and the net iron area if the gross area is 0.0180 m².

Show answer

\[k_s = \frac{0.35}{0.36} = 0.9722\]

\[A_i = (0.9722)(0.0180) = 0.01750~\mathrm{m^{2}}\]
P40.2 A square core is inscribed in a 150 mm circle. Find its side, its area, and the utilisation factor.

Show answer

\[a = \frac{150}{\sqrt 2} = 106.1~\mathrm{mm}, \qquad A = 11\,250~\mathrm{mm^{2}}\]

\[\frac{11\,250}{\pi(150)^{2}/4} = \frac{11\,250}{17\,671} = 63.66\,\%\]
P40.3 Replace the square of P40.2 by a cruciform. Find its dimensions and the extra iron gained.

Show answer

\[a = (0.8507)(150) = 127.6~\mathrm{mm}, \qquad b = (0.5257)(150) = 78.9~\mathrm{mm}\]

\[A = (0.61803)(150)^{2} = 13\,906~\mathrm{mm^{2}}\]

\[\frac{13\,906}{11\,250} = 1.236 \quad\Longrightarrow\quad 23.6\,\%\ \text{more iron}\]
P40.4 For the cruciform of P40.3 with \(k_s = 0.92\) and \(B = 1.5\) T at 50 Hz, find the volts per turn.

Show answer

\[A_i = (0.92)(0.013906) = 0.012793~\mathrm{m^{2}}\]

\[\Phi_m = (1.5)(0.012793) = 0.019189~\mathrm{Wb}\]

\[E_t = (4.44)(50)(0.019189) = 4.260~\mathrm{V/turn}\]
P40.5 For P40.4, find the turns for a 6600 V / 230 V transformer.

Show answer

\[N_{HV} = \frac{6600}{4.260} = 1549~\text{turns}, \qquad N_{LV} = \frac{230}{4.260} = 54.0 \to 54~\text{turns}\]
The LV rounds almost exactly, which is fortunate; more often a small adjustment of \(B\) or of the core area is made to land on whole turns.
P40.6 A 250 kVA transformer is 98 % efficient. Its tank has 9 m² of surface. Find the temperature rise at 12.5 W/m²/°C, and the surface needed for 45 °C.

Show answer

\[P_{\text{loss}} = (250\,000)(0.02) = 5000~\mathrm{W}\]

\[\theta = \frac{5000}{(9)(12.5)} = \frac{5000}{112.5} = 44.4~^{\circ}\mathrm{C}\]

\[A = \frac{5000}{(45)(12.5)} = 8.89~\mathrm{m^{2}}\]
The plain tank is just adequate — 44.4 °C against a 45 °C limit — with no margin for a hot day.
P40.7 Why are transformer cores laminated, and why is the lamination not made indefinitely thin?

Show answer
Why laminated: the alternating flux induces EMFs in the core itself, and in a solid core these drive large circulating eddy currents. Dividing the core into thin sheets, each insulated from the next, breaks the paths available to those currents. Since \(P_e \propto t^{2}\), halving the thickness quarters the loss.
Why not indefinitely thin: three reasons. The cost of rolling and handling very thin steel rises steeply; the number of sheets to be stacked and insulated multiplies; and the stacking factor falls, because the coating thickness stays the same while the steel shrinks, so an ever-larger fraction of the core carries no flux.

Around 0.25 to 0.35 mm the marginal gain no longer justifies the cost, which is where the industry has settled.
P40.8 Compare core-type and shell-type construction, and state where each is preferred.

Show answer
Core type: the winding surrounds the core. Two limbs joined by yokes, with half of each winding on each limb. The iron path is long and the mean length of turn short, so it uses less copper and more iron. Leakage reactance is higher, mechanical bracing weaker, but access for repair is better.
Shell type: the core surrounds the winding. Three limbs, the whole winding on the central one, the flux dividing so each outer limb carries \(\Phi/2\) and needs half the area. Short iron path, long mean turn — more copper, less iron. Lower leakage reactance and much better mechanical support.

Preference: core type for high voltage and moderate current, where copper on the many HV turns is the expense; shell type for low voltage and high current, and wherever heavy fault or overload forces demand mechanical robustness.
P40.9 Explain the function of the conservator, the breather and the Buchholz relay.

Show answer
Conservator: a partly filled drum above the tank that takes up the expansion and contraction of the oil with temperature, so the main tank stays completely full and no air contacts the windings.
Breather: as the oil cools and contracts, air must enter the conservator. The breather passes it through silica gel to remove moisture, since even a trace of water sharply reduces the oil's dielectric strength. The gel is blue when dry, pink when saturated.

Buchholz relay: mounted in the pipe between tank and conservator. An internal fault decomposes the oil into gas; slow accumulation raises an alarm for incipient faults, while a sudden surge of oil operates a second contact that trips the transformer.

All three exist because of the oil — which is present for insulation and cooling, and brings its own requirements with it.
P40.10 Why does a large transformer need proportionally more cooling surface than a small one?

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Because loss scales with volume and cooling with area. If every linear dimension is multiplied by \(x\), the volume of iron and copper — and therefore roughly the loss — grows as \(x^{3}\), while the tank surface grows only as \(x^{2}\).
The loss per unit of dissipating surface therefore rises as \(x\), and the temperature rise with it.

This is why the cooling method escalates with rating: a plain tank for a small distribution unit, tubes or radiators at a few hundred kVA, forced air at some megavolt-amperes, and forced oil with water cooling for the largest generator transformers. It is the same square-cube argument that governs the cooling of every machine in this book.

Multiple-Choice Questions

MCQ 1. A transformer changes:
(a) voltage and frequency (b) voltage and current at constant power (c) power (d) frequency only

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(b) voltage and current at constant power, and at constant frequency.
MCQ 2. Transmission line loss varies as:
(a) \(V\) (b) \(V^{2}\) (c) \(1/V\) (d) \(1/V^{2}\)

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(d) \(1/V^{2}\) — the whole economic case for high-voltage transmission.
MCQ 3. The silicon content of core steel is about:
(a) 0.35 % (b) 3.5 % (c) 35 % (d) 50 %

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(b) 3.5 %. Above this the steel becomes brittle and hard to cut.
MCQ 4. The saturation flux density of modern laminations is about:
(a) 0.2 T (b) 2 T (c) 20 T (d) 2 %

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(b) 2 T. Cores work at roughly 1.5–1.7 T.
MCQ 5. Eddy-current loss varies as the lamination thickness to the power:
(a) 1 (b) 1.6 (c) 2 (d) 3

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(c) 2. Halving the thickness quarters the loss.
MCQ 6. In a core-type transformer:
(a) the core surrounds the winding (b) the winding surrounds the core (c) there are three limbs (d) there is no yoke

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(b) the winding surrounds the core. The shell type is the other way round.
MCQ 7. In a shell-type transformer each outer limb carries:
(a) \(\Phi\) (b) \(\Phi/2\) (c) \(2\Phi\) (d) zero

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(b) \(\Phi/2\), so each outer limb has half the area of the central one.
MCQ 8. A cruciform core fills about what fraction of the circular winding space?
(a) 63.7 % (b) 78.7 % (c) 85.1 % (d) 100 %

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(b) 78.7 %, against 63.7 % for a square and 85.1 % for a three-stepped core.
MCQ 9. The low-voltage winding is placed nearest the core because:
(a) it is heavier (b) it needs least insulation to earth (c) it carries less current (d) it runs cooler

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(b) it needs least insulation to earth, so the total insulation and window space are minimised.
MCQ 10. The breather contains silica gel in order to:
(a) cool the oil (b) remove moisture from incoming air (c) detect faults (d) absorb expansion

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(b) remove moisture from incoming air, since water destroys the oil's dielectric strength.

Conceptual Questions

Explain the limitations of the DC system and how the transformer removed them.
Define a transformer and justify the name in preference to "converter".
State the four magnetic requirements of the core material and the reason for each.
Explain the trade-off in lamination thickness, including the stacking factor.
Compare core-type and shell-type construction under at least six headings.
Derive the dimensions of a cruciform core inscribed in a circle of diameter \(d\).
Describe concentric and sandwich windings and explain why interleaving lowers leakage reactance.
Describe the cooling methods and the function of the conservator, breather and Buchholz relay.

Looking Ahead

The transformer has now been described as an object. Chapter 41 turns it into a circuit, starting from the ideal transformer — one with no losses, no leakage and infinite core permeability — and deriving the EMF equation that Example 40.3 already used:

\[E = 4.44fN\Phi_m\]

From it follow the turns ratio, the transformation of voltage, current and impedance, and the notion of referring a quantity from one side to the other. That last idea does more work than any other in Part 3: it lets a two-winding machine be replaced by a single equivalent circuit.

Chapters 42 and 43 then relax the idealisations one at a time — the magnetising current and core loss on no load, then the leakage reactance and winding resistance on load — arriving at the full equivalent circuit of Chapter 44. Every departure from the ideal traces back to something in this chapter: the core steel sets the magnetising current and iron loss, the winding arrangement sets the leakage reactance, and the conductor size sets the resistance.