By the end of this chapter you should be able to:
Explain why magnetising inrush occurs and estimate its magnitude.
Describe its unipolar waveform and its two very different decay time constants.
Explain second-harmonic restraint and why it distinguishes inrush from a fault.
List the methods of reducing inrush.
Explain why tap changing is needed and where the taps are placed.
Distinguish off-circuit from on-load tap changers.
Compute the effect of tap position on secondary voltage and on core flux density.
Estimate the temperature rise and interpret the four-letter cooling code.
Inrush: The Mechanism
Chapter 42 established the cause. Since the flux is the integral of the applied voltage,
the instant of switching decides what flux the core is asked to carry.
Switch at a voltage peak and the flux starts from zero exactly as the steady state requires — no transient at all.
Switch at a voltage zero and the flux must integrate over a full half cycle from zero, reaching \(2\Phi_m\).
Add residual flux left from the previous switch-off and the demand rises further, to as much as \(2.5\Phi_m\).
The core saturates heavily, its permeability collapses towards that of air, and the current becomes enormous.
This chapter takes the story forward: what the current actually looks like, how long it lasts, why protective relays must be taught to ignore it, and what can be done to reduce it.
Inrush is not a fault. It is the normal, healthy behaviour of a perfectly sound transformer being energised. Nothing is wrong, nothing is damaged, and the current subsides on its own.
Yet it is the largest current a healthy transformer ever draws — comparable with a short-circuit current — which is precisely why it causes so much trouble for protection.
Shape and Decay of the Inrush
The waveform has three features that together identify it.
- It is unipolar. The flux offset is in one direction, so the core saturates on one half cycle only. The current appears as a series of one-sided pulses that never cross zero.
- It is peaky. Current flows only while the core is saturated, which is a small part of each cycle, so each pulse is narrow and tall with long gaps between.
- It decays slowly, and at two different rates. The winding resistance dissipates the offset, but the effective inductance changes by orders of magnitude as the core comes out of saturation.
Deep in saturation, \(L\) is little more than the leakage inductance — the flux path is effectively air — so \(\tau\) is a fraction of a cycle and the first few peaks fall fast.
As the current decays the core emerges from saturation, \(L\) climbs towards the magnetising inductance, and \(\tau\) grows with it. The tail therefore lingers for hundreds of cycles, even though the first peaks collapsed within a few.
Problem. For the 50 kVA, 2200/220 V transformer used throughout Chapters 41 to 46 — \(I_0 = 1.2\) A, \(R_{01} = 1.36~\Omega\), \(X_{01} = 4.60~\Omega\), \(X_0 = 1867~\Omega\) referred to HV — estimate the inrush peak if it reaches ten times the full-load peak, compare it with the normal magnetising current, and find the two decay time constants.
Full-load and magnetising peaks.
Inrush peak.
Saturated time constant. With the flux path effectively air, take \(L\) from the leakage reactance:
Unsaturated time constant. With the core restored, \(L\) comes from the magnetising reactance:
Comment. The current is 189 times the transformer's normal no-load current, and comparable with a short-circuit current — from a machine in perfect condition.
The two time constants differ by a factor of
The ratio of the time constants is just the ratio of the shunt to series reactance — the same 400-to-1 separation that made the approximate equivalent circuit of Chapter 44 so accurate. Here it explains why the inrush has a fast head and a very slow tail: the transformer traverses the whole range between the two as the core desaturates.
In practice neither figure applies alone. The decay begins near \(\tau_{\text{sat}}\) and lengthens continuously, so a distribution transformer settles in a few tenths of a second while a large power transformer can take several seconds.
Consequences and Mitigation
Spurious tripping of differential and overcurrent protection, the most common consequence.
Mechanical stress on the windings, since force varies as the square of current.
Voltage dip on the supply, visible as a brief flicker of lights.
Harmonic distortion, briefly, from the peaky unipolar waveform.
Sympathetic inrush in transformers already energised on the same bus.
Point-on-wave switching: close the breaker near a voltage peak, where the steady-state flux is zero.
Pre-insertion resistors in the breaker, shorted out a few cycles later.
Controlled residual flux, by managing the instant of de-energisation.
Series reactors or a deliberately soft supply.
Energising from the HV side, where the current is smaller in absolute terms.
A differential relay compares the current entering with the current leaving. During energisation all the current enters and none leaves — which looks exactly like an internal fault.
The distinguishing feature is the waveform. Because the inrush is unipolar, it is rich in even harmonics, and the second harmonic is typically 15 to 60 % of the fundamental. A genuine fault current is symmetrical and contains almost none.
The relay blocks itself whenever the second-harmonic content is high, and trips normally otherwise.
The connection to Chapter 42 is worth noting. There the magnetising current's third harmonic mattered, because the steady-state waveform is symmetrical and therefore contains odd harmonics only. Inrush is asymmetrical, so it generates even harmonics — and that difference is exactly what the relay exploits.
Problem. A differential relay with a 16 % second-harmonic restraint setting measures the following. Decide in each case whether it restrains or trips.
| Case | Fundamental \(I_1\) | Second harmonic \(I_2\) | \(I_2/I_1\) | Decision |
|---|---|---|---|---|
| (a) Energisation | 400 A | 88 A | 22.0 % | Restrain |
| (b) Internal fault | 400 A | 20 A | 5.0 % | Trip |
| (c) Marginal | 400 A | 64 A | 16.0 % | Restrain (at the setting) |
Working, case (a).
Comment. Both (a) and (b) show 400 A of differential current, and on magnitude alone the relay could not tell them apart. Only the harmonic content separates them.
Case (c) shows the difficulty of the setting. Too low and genuine faults are blocked, leaving the transformer unprotected while it burns; too high and every energisation trips the breaker. Modern relays therefore combine harmonic restraint with waveform recognition, looking for the flat gaps between the inrush pulses, which no fault current has.
Note also that a real internal fault often shows a few percent of second harmonic from current-transformer saturation — which is why (b) is given 5 % rather than zero.
Why Tap Changing Is Needed
Chapter 45 showed that the secondary voltage of a fixed-ratio transformer varies with both the magnitude and the power factor of the load. For the 50 kVA machine the regulation ran from \(+3.975\,\%\) at 0.8 lagging to \(-1.727\,\%\) at 0.8 leading — a swing of over 5 % from the load alone.
The supply itself adds more. A distribution feeder's sending-end voltage varies through the day as the system load rises and falls, and the consumer must still receive voltage within a statutory band, commonly \(\pm6\,\%\).
A transformer cannot change its regulation, which is fixed by its impedance. What it can do is change its no-load ratio, shifting the whole voltage band up or down until it sits inside the statutory limits.
Taps alter \(N_1\), and so the first term. The second term is untouched.
Taps are almost always brought out of the high-voltage winding, for three reasons:
The HV winding carries the smaller current, so the tap leads, contacts and switch can be lighter and cheaper.
The HV winding has more turns, so a step of 2.5 % is a whole number of turns rather than a fraction.
It is usually the outer winding, so tappings are physically accessible.
Typical provision is \(\pm5\,\%\) in steps of 2.5 % for a distribution transformer, and a much wider range — often \(\pm15\,\%\) in 1.25 % steps, seventeen positions or more — for a power transformer with an on-load changer.
Off-Circuit and On-Load Tap Changers
| Off-circuit (off-load) | On-load (OLTC) | |
|---|---|---|
| Operated | With the transformer disconnected | With the transformer carrying load |
| Frequency of use | Seasonally, or once at commissioning | Many times a day, automatically |
| Mechanism | A simple rotary or linear selector | Selector plus diverter switch with transition impedance |
| Range | Typically \(\pm5\,\%\) in 2.5 % steps | Typically \(\pm15\,\%\) in 1.25 % steps |
| Cost | Low | High — often a fifth of the transformer |
| Typical use | Distribution transformers | Transmission and grid transformers |
Moving from one tap to the next on a loaded winding poses two contradictory requirements:
The load current must not be interrupted — breaking it would draw an arc and cause a supply interruption.
The two taps must not be solidly connected together — that would short-circuit the turns between them, and the resulting current would be limited only by the leakage impedance of those few turns.
The resolution is a transition impedance — a centre-tapped reactor or, more usually, a pair of resistors — which is briefly inserted so that both taps are connected but through enough impedance to limit the circulating current. The changeover takes a few tens of milliseconds.
Because the diverter switch makes and breaks current, its contacts erode and its oil carbonises. It is therefore housed in a separate compartment with its own oil, so that arcing products cannot contaminate the main tank — and the count of operations is logged, since maintenance is scheduled by operation count rather than by elapsed time.
Tap Position, Turns and Flux
Two consequences follow from changing \(N_1\), and the second is easy to overlook.
More primary turns give less secondary voltage.
Fewer primary turns give more flux, and risk saturation.
A tap labelled "+5 %" means 5 % more turns, which gives 5 % less output voltage. It is the tap to select when the supply voltage is high.
This catches students and engineers alike, because the label describes the winding, not the output. The safe way to think about it is always in turns: more turns, less volts, less flux.
Problem. A 2200/220 V transformer has \(N_1 = 1000\) and \(N_2 = 100\) turns, with HV taps at \(\pm2.5\,\%\) and \(\pm5\,\%\). The nominal flux density is 1.1 T. Tabulate the secondary voltage and flux density at each tap for a 2200 V supply, then find the flux density on the lowest tap when the supply is 10 % high.
| Tap | \(N_1\) | \(V_2\) | \(B_m\) |
|---|---|---|---|
| +5.0 % | 1050 | 209.52 V | 1.0476 T |
| +2.5 % | 1025 | 214.63 V | 1.0732 T |
| Nominal | 1000 | 220.00 V | 1.1000 T |
| −2.5 % | 975 | 225.64 V | 1.1282 T |
| −5.0 % | 950 | 231.58 V | 1.1579 T |
Sample working, the −5 % tap.
Lowest tap with a 10 % high supply.
Comment. The full tap range moves the output from 209.5 V to 231.6 V, a span of 22.1 V or just over 10 % — which is what \(\pm5\,\%\) of turns buys.
The flux column is the one to watch. The \(-5\,\%\) tap alone raises \(B_m\) by 5.26 %, and a 10 % overvoltage on top brings it to 1.274 T, nearly 16 % above nominal. Still short of saturation at 2.0 T, but the margin is being spent — and the exciting current, which depends on \(B\) through the sharply bending B-H curve of Chapter 5, will have risen far more than 16 %.
This is why transformers are specified for a maximum volts-per-turn rather than simply a maximum voltage, and why over-fluxing (volts/hertz) protection is fitted to large units — the same relay also catches the other route to saturation, a fall in frequency at constant voltage.
The Thermal Problem
Every watt of loss computed in Chapter 44 becomes heat inside the tank, and must find its way out. The rating of a transformer is set almost entirely by this.
The core and windings could carry more flux and more current than they are asked to. What cannot be exceeded is the temperature of the insulation, because insulation ages chemically, and the rate roughly doubles for every 6 to 8 °C.
A transformer run 12 °C above its rated hot-spot temperature lasts a quarter as long. This is the Montsinger rule, and it is why temperature rise, not saturation or current density, is the quantity guaranteed on the nameplate.
Heat leaves the tank by two mechanisms, in roughly equal measure for a plain surface:
A useful distinction: fins and radiators add convecting area but little radiating area, because adjacent fins face each other and radiate to one another rather than to the surroundings. Corrugated tanks are therefore rated on their convection contribution alone.
Problem. The 50 kVA transformer of Chapter 44 has an iron loss of 500 W and a full-load copper loss of 702 W. Its plain tank has a dissipating surface of 1.4 m². Find the temperature rise, and the area needed to limit it to 50 °C.
Total loss.
Temperature rise of the plain tank.
Area required for a 50 °C rise.
Comment. The plain tank runs 68.7 °C above ambient, comfortably beyond the 50 to 55 °C oil rise usually permitted. The transformer is not electrically overloaded — it is thermally overloaded, and the cure is sheet metal, not copper.
Adding 0.523 m² of fins or a set of radiator tubes solves it, which is why distribution transformers are corrugated rather than smooth. Note how modest the underlying quantity is: 1202 W, about the rating of a domestic kettle, dictates the entire external form of a 50 kVA machine.
Note also the coupling to the load. Copper loss varies as the square of loading, so at 125 % load the total becomes \(500 + (702)(1.25)^2 = 1597\) W and the required area rises to 2.56 m². An overload rating is really a cooling rating.
Cooling Methods and the Four-Letter Code
Beyond a few hundred kVA a plain or finned tank is not enough, and oil is circulated through radiators. The method is described by a standard four-letter code.
| Position | Meaning | Letters |
|---|---|---|
| 1st | Internal cooling medium | O mineral oil, K ester, A air |
| 2nd | Internal circulation | N natural, F forced, D directed |
| 3rd | External medium | A air, W water |
| 4th | External circulation | N natural, F forced |
So ONAN is oil-natural, air-natural; ONAF adds fans; OFAF adds an oil pump as well; and ODAF directs the pumped oil into the winding ducts instead of merely stirring the tank.
| Code | Description | Typical range |
|---|---|---|
| AN | Dry type, air natural | Up to a few hundred kVA, indoors |
| ONAN | Oil natural, air natural — radiators only | Up to about 30 MVA |
| ONAF | Fans blow air over the radiators | Adds roughly 33 % |
| OFAF | Oil pumped through radiators, fans on air side | Adds roughly 67 % |
| ODAF | Oil directed into the winding ducts | Largest units |
| OFWF | Oil pumped to a water heat exchanger | Generator transformers, underground |
A large transformer commonly carries a dual or triple rating such as ONAN/ONAF/OFAF 20/26.7/33.3 MVA — one machine with three nameplate outputs, selected by how much of the cooling plant is running.
Problem. A transformer is rated ONAN/ONAF/OFAF 20/26.7/33.3 MVA. Its ONAN copper loss at 20 MVA is 90 kW and its iron loss is 20 kW. Find the copper loss at each rating, and the life expectancy if the OFAF rating is exceeded so that the hot spot runs 12 °C high.
Loading in per unit of the ONAN rating.
Copper loss, which varies as the square.
| Class | MVA | Per unit | Copper loss | Total loss |
|---|---|---|---|---|
| ONAN | 20.0 | 1.000 | 90.0 kW | 110.0 kW |
| ONAF | 26.7 | 1.335 | 160.4 kW | 180.4 kW |
| OFAF | 33.3 | 1.665 | 249.5 kW | 269.5 kW |
Life at a 12 °C excess.
Comment. Going from ONAN to OFAF raises the output by 66.5 % but the copper loss by 177 % — the total loss rises from 110 to 269.5 kW. The cooling plant must remove nearly two and a half times as much heat to deliver two-thirds again as much power, which is why the pumps and fans are worth their cost only when the extra capacity is actually needed.
The life figure explains the seriousness of the rating. Twelve degrees is a small excursion — an unnoticed fan failure on a hot afternoon would do it — yet it cuts the expected life to a quarter. A transformer meant for 30 years would then be used up in about 7.5.
This is also why cooling plant is duplicated and alarmed, and why a fan or pump failure raises an alarm and an automatic reduction in permitted load rather than being left to the next inspection.
Oil, Conservator, Breather, Buchholz
The oil does two jobs at once, and its condition governs both.
Oil has roughly three times the dielectric strength of air and fills every void in the winding, where air would ionise and start a discharge.
Oil carries heat from the windings to the tank wall by convection, a far better path than conduction through solid insulation alone.
Dry oil withstands perhaps 30 kV across a 2.5 mm gap. A few parts per million of dissolved water can halve that, and water also accelerates the ageing of the paper insulation, which is the part that cannot be replaced.
Since oil expands and contracts as the load and ambient change, the tank must breathe — and every breath draws in humid air. The conservator and breather exist to manage exactly this.
| Fitting | Purpose |
|---|---|
| Conservator | A small tank above the main tank, part full of oil. Accommodates expansion and contraction, and keeps the main tank completely full so the windings never meet air. |
| Breather | Admits air to the conservator through silica gel, which absorbs moisture. The gel is blue when dry and turns pink when saturated — a visual maintenance check. |
| Buchholz relay | Fitted in the pipe between tank and conservator. Slow gas accumulation raises an alarm; a surge of oil trips the transformer. It detects incipient faults nothing else can see. |
| Explosion vent | A diaphragm that ruptures to relieve pressure from a severe internal fault, directing it safely away. |
| Temperature indicators | Oil temperature directly; winding temperature by a heater coil fed from a current transformer, imitating the winding's rise above the oil. |
Summary and Key Formulas
Inrush arises because flux is the integral of voltage. Switching at a voltage zero demands \(2\Phi_m\), and residual flux takes it to \(2.5\Phi_m\); the core saturates and the current becomes enormous.
The waveform is unipolar, peaky, and decays with two very different time constants — saturated \(L\) gives a fast head, unsaturated \(L\) a very slow tail.
Being asymmetrical, the inrush is rich in even harmonics; a fault current is not. Relays use second-harmonic restraint at 15 to 20 %.
Inrush is reduced by point-on-wave switching, pre-insertion resistors, and control of residual flux.
Taps change the no-load ratio to re-centre the output band; they cannot change the regulation. They are placed on the HV winding.
Off-circuit changers are cheap and moved rarely; on-load changers use a diverter switch with a transition impedance, so the circuit never opens and the taps never short.
More turns give less voltage and less flux. Tapping down raises \(B_m\) and risks saturation, especially with a high supply.
The rating is set by insulation temperature, not magnetics. Life halves for every 6 °C of excess.
A plain tank dissipates about 12.5 W/m²/°C; fins add convection but little radiation.
The four-letter code gives internal medium, internal circulation, external medium, external circulation — ONAN, ONAF, OFAF, ODAF.
Oil insulates and cools; the conservator allows expansion, the breather keeps moisture out, and the Buchholz relay detects gas from incipient faults.
| Quantity | Formula | Notes |
|---|---|---|
| Flux from voltage | \(\Phi = \dfrac{1}{N_1}\displaystyle\int v\,\mathrm{d}t\) | why the instant matters |
| Worst-case flux | \(\Phi_{\text{peak}} = 2\Phi_m + \Phi_r\) | zero-crossing plus residual |
| Decay time constant | \(\tau = L/R\) | two values, ratio \(X_0/X_{01}\) |
| Restraint criterion | \(I_2/I_1 \gt 0.15\ \text{to}\ 0.20\) | restrain if exceeded |
| Tapped ratio | \(V_2 = V_1N_2/N_1\) | more turns, less volts |
| Flux at a tap | \(\Phi_m = \dfrac{V_1}{4.44fN_1}\) | fewer turns, more flux |
| Temperature rise | \(\Delta\theta = \dfrac{P_i + P_c}{kA}\) | \(k \approx 12.5\) W/m²/°C |
| Insulation life | \(\text{Life} \propto 2^{-\Delta\theta/6}\) | Montsinger rule |
| Loss at a loading | \(P = P_i + x^{2}P_c\) | cooling must remove it |
Common Mistakes
Treating inrush as a fault. It is normal behaviour of a sound transformer.
Thinking the worst instant is the voltage peak. It is the voltage zero — flux lags voltage by 90°.
Expecting inrush to contain a third harmonic. It is asymmetrical, so even harmonics dominate; the third belongs to the symmetrical steady-state magnetising current.
Using one time constant for the whole decay. The inductance changes by orders of magnitude as the core desaturates.
Reading a "+5 %" tap as raising the output. It adds turns and lowers the output.
Forgetting that tapping down raises the flux. This, not the winding, is what limits the tap range.
Believing an on-load changer simply switches between taps. Without a transition impedance it would either open the circuit or short-circuit the tap turns.
Putting taps on the LV winding. The current is larger and the turns fewer, so the switch is bigger and the steps coarser.
Assuming fins add radiating area. They face each other, so they add convection only.
Treating a small temperature excess as harmless. Twelve degrees quarters the life.
Reading the cooling code in the wrong order. First pair is internal, second pair external.
Chapter Review
For tap problems, work in turns rather than percentages — it removes the sign confusion. For thermal problems, remember that copper loss scales as the square of loading.
P50.1 A transformer is switched on at a voltage zero with a residual flux of \(0.6\Phi_m\). If the design flux density is 1.3 T, what density would be demanded, and is it attainable?
Show answer
\[\Phi_{\text{peak}} = \Phi_m + \Phi_m + 0.6\Phi_m = 2.6\Phi_m\]Not attainable — saturation is near 2.0 T. The core saturates, the permeability collapses towards that of air, and the current rises out of all proportion.\[B = (2.6)(1.3) = 3.38~\mathrm{T}\]P50.2 A transformer has \(R_{01} = 2.0~\Omega\), \(X_{01} = 6.0~\Omega\) and \(X_0 = 2400~\Omega\) at 50 Hz. Find both decay time constants and their ratio.
Show answer
\[L_{\text{sat}} = \frac{6.0}{314.16} = 0.01910~\mathrm{H} \quad\Longrightarrow\quad \tau_{\text{sat}} = \frac{0.01910}{2.0} = 9.55~\mathrm{ms}\]\[L_{\text{unsat}} = \frac{2400}{314.16} = 7.639~\mathrm{H} \quad\Longrightarrow\quad \tau_{\text{unsat}} = \frac{7.639}{2.0} = 3.820~\mathrm{s}\]\[\frac{3.820}{0.00955} = 400 = \frac{2400}{6.0} \quad\checkmark\]P50.3 A relay set at 18 % restraint measures 250 A fundamental with 52 A of second harmonic. Restrain or trip?
Show answer
The waveform is consistent with energisation, not an internal fault.\[\frac{52}{250} = 0.208 = 20.8\,\% \gt 18\,\% \quad\Longrightarrow\quad \text{restrain}\]P50.4 A 6600/440 V transformer has \(N_1 = 1500\) turns with taps at \(\pm2.5\,\%\). Find \(N_2\), and the secondary voltage on each tap at nominal supply.
Show answer
\[N_2 = (1500)\frac{440}{6600} = 100~\text{turns}\]\[+2.5\,\%:\ N_1 = 1537.5, \quad V_2 = (6600)\frac{100}{1537.5} = 429.27~\mathrm{V}\]\[\text{Nominal}:\ V_2 = 440.00~\mathrm{V}\]\[-2.5\,\%:\ N_1 = 1462.5, \quad V_2 = (6600)\frac{100}{1462.5} = 451.28~\mathrm{V}\]P50.5 For P50.4, if the nominal flux density is 1.4 T, find it on the \(-2.5\,\%\) tap with the supply 8 % high.
Show answer
A rise of 10.77 % over nominal. Still below saturation, but the exciting current will have risen far more than 10.77 % because the B-H curve is bending sharply by this point.\[B_m = (1.4)\frac{1500}{1462.5}(1.08) = (1.4)(1.02564)(1.08) = 1.5508~\mathrm{T}\]P50.6 A transformer dissipates 3.5 kW of total loss. Its tank has 3.0 m² of surface. Find the temperature rise, and the area needed for 45 °C.
Show answer
\[\Delta\theta = \frac{3500}{(12.5)(3.0)} = \frac{3500}{37.5} = 93.3\,^\circ\mathrm{C}\]More than double the plain-tank area — radiators are essential.\[A = \frac{3500}{(12.5)(45)} = \frac{3500}{562.5} = 6.222~\mathrm{m^2}\]P50.7 A transformer runs 18 °C above its rated hot spot. By what factor is its life reduced, and how long would a 30-year design last?
Show answer
\[2^{-18/6} = 2^{-3} = 0.125\]An eighth of the intended life — which is why hot-spot temperature is monitored continuously on large units.\[(30)(0.125) = 3.75~\text{years}\]P50.8 Explain why the inrush current is unipolar and why that matters for protection.
Show answer
The flux offset produced at switching is in one direction, so the core is driven deep into saturation on one half cycle and not at all on the other. Current flows only while saturated, so the pulses appear on alternate half cycles only and never cross zero.Why it matters. A waveform with unequal half cycles is asymmetrical, and asymmetry means even harmonics — principally the second, typically 15 to 60 % of the fundamental.
A genuine fault current is symmetrical and contains almost no second harmonic. This gives protection a criterion that magnitude alone cannot supply, since energisation and an internal fault can produce identical differential current magnitudes.
Hence second-harmonic restraint: the relay blocks itself when \(I_2/I_1\) exceeds about 15 to 20 %.
P50.9 Explain the difficulty of changing taps on load and how it is resolved.
Show answer
Two requirements conflict:The load current must not be interrupted, or an arc is drawn and supply is lost.
The two taps must not be solidly bridged, since that short-circuits the turns between them; the current would be limited only by the leakage impedance of those few turns and would be very large.
The resolution is a transition impedance — a centre-tapped reactor, or more usually a pair of resistors — inserted for the few tens of milliseconds during which both taps are connected. Current flows continuously, but the circulating current between taps is limited to an acceptable value.
The switching is done by a diverter switch operating at speed, with a separate selector that pre-positions the next tap while it is still off-circuit.
Because the diverter makes and breaks current, it arcs. It is therefore housed in its own oil compartment, so carbon and gas cannot contaminate the main tank, and its maintenance is scheduled by operation count rather than elapsed time.
P50.10 Why is the rating of a transformer a thermal quantity rather than a magnetic or electrical one?
Show answer
Because nothing electrical or magnetic fails first. The core could carry more flux before saturating, and the conductors could carry more current before melting — by a wide margin in both cases.What limits the machine is the chemical ageing of the insulation, whose rate roughly doubles for every 6 to 8 °C, by the Montsinger rule \(\text{Life} \propto 2^{-\Delta\theta/6}\). The paper insulation cannot be replaced, so its life is the transformer's life.
Consequences. The nameplate guarantees a temperature rise, not a flux density. A machine can be overloaded briefly with little harm if it starts cold, since the thermal time constant is hours. And the same transformer has several ratings depending on how hard the cooling plant is run — ONAN/ONAF/OFAF — which would be meaningless if the limit were magnetic.
It also explains why a transformer's external form is dominated by radiators and fins that have nothing to do with its electromagnetic design.
MCQ 1. The worst instant to energise a transformer is at a supply voltage:
(a) peak (b) zero (c) 45° point (d) any instant is the sameShow answer
(b) zero, since the flux must then integrate over a full half cycle to \(2\Phi_m\).MCQ 2. Inrush current is:
(a) a fault condition (b) normal behaviour of a healthy transformer (c) caused by the load (d) caused by poor insulationShow answer
(b) normal behaviour of a healthy transformer.MCQ 3. The inrush waveform is rich in:
(a) third harmonic (b) fifth harmonic (c) second harmonic (d) no harmonicsShow answer
(c) second harmonic, because the waveform is unipolar and therefore asymmetrical.MCQ 4. A differential relay restrains when the second-harmonic ratio is:
(a) low (b) high (c) zero (d) negativeShow answer
(b) high — high second harmonic indicates energisation, not a fault.MCQ 5. Taps are normally provided on the:
(a) LV winding (b) HV winding (c) core (d) tankShow answer
(b) HV winding — smaller current, more turns, and accessible.MCQ 6. Selecting a tap with more primary turns gives a secondary voltage that is:
(a) higher (b) lower (c) unchanged (d) zeroShow answer
(b) lower, and the core flux is lower too.MCQ 7. An on-load tap changer needs a transition impedance to:
(a) reduce losses (b) avoid shorting the tap turns while keeping the circuit closed (c) cool the oil (d) filter harmonicsShow answer
(b) — it satisfies both conflicting requirements at once.MCQ 8. In the code ONAF, the letter F refers to:
(a) forced oil circulation (b) forced air circulation (c) oil filtering (d) frequencyShow answer
(b) forced air circulation — it is the fourth letter, describing the external medium.MCQ 9. The Buchholz relay responds to:
(a) overcurrent (b) gas generated inside the tank (c) overvoltage (d) low oil temperatureShow answer
(b) gas generated inside the tank, detecting incipient faults before any electrical protection could.MCQ 10. A transformer running 6 °C above its rated hot spot has its life reduced to about:
(a) 90 % (b) 75 % (c) 50 % (d) 25 %Show answer
(c) 50 % — the Montsinger rule, \(2^{-6/6} = 0.5\).
Explain the cause of magnetising inrush and why the switching instant matters.
Describe the inrush waveform and account for its two decay time constants.
Explain second-harmonic restraint and why magnitude alone cannot distinguish inrush from a fault.
List the methods of reducing inrush and explain the principle of each.
Explain why taps are needed and why they are placed on the HV winding.
Compare off-circuit and on-load tap changers, and explain the transition impedance.
Explain how tap position affects the core flux, and why this limits the tap range.
Explain the four-letter cooling code and why one transformer can carry three ratings.
Chapter 51 develops the per-unit system, which this book has already been using informally. Chapter 43 quoted the percentage impedance; Chapter 46 measured it; Chapter 49 converted percentage impedances to a common base to decide how two transformers share load. The per-unit system makes that a general method.
Parameters are referred to chosen base values — \(S_{base}\), \(V_{base}\), and from them \(I_{base}\) and \(Z_{base}\) — so that the scaling factors vanish from the calculation and the units disappear with them. The result worth waiting for is this: the per-unit impedance of a transformer is the same whichever side it is referred to, so the turns ratio disappears from the equivalent circuit altogether, and a network of many transformers at many voltages becomes a single circuit with no ratios in it at all.
Part 3 then turns to three-phase transformers: their construction, the star, delta and zigzag winding connections, the vector groups that Chapter 49's fourth condition referred to, and the open-delta connection that lets two single-phase units do the work of three.