“ELECTROHEAT describes any heating process where electricity is the primary energy source”. I.e. an electromagnetic field (or (or electromagnetic radiation) interacts interacts with the part to be heated (otherwise known as “the work-piece”), and causes it to be heated. Generally, the frequency of the electromagnetic field interacting with the work piece characterises the heating process. The frequency of the electromagnetic radiation can stretch from DC to beyond daylight. DC
Through
Resistance (DC) g n i t a e H R 2 I
through
Mains-frequency Induction (50Hz) Medium-frequency Induction (1-20kHz) y c n e u q e r F
Radio-frequency Induction (50kHz-10MHz)
Deep shallow
surface
Dielectric (27-48MHz) (27-48MHz)
through
Microwave (GHz)
shallow surface
Infrared heating Daylight
n o i t a r t e n e P t a e H
Surface
This figure shows most of the different categories of electro-heat, arraigned in an ascending order of frequency.
At DC (or low frequency AC), direct resistance heating heats
conductive work pieces by attaching an electrode to each end of of the material to be heated. A large current is then passed through through the work piece. This method features high power densities and relatively high efficiencies.
A typical use for this heating method is for
resistance welding of car bodies, etc. Direct resistance welding is limited to relatively few materials and heating geometries, however, as the current will flow through the shortest path between the electrodes, and the highest resistance part of the circuit will w ill get hottest, so if the work piece is a lower resistivity than the electrodes, the electrodes will be heated more! Billet heating by direct resistance heating is rather wasteful for this reason, as the part of the material under and beyond the electrodes is not heated, and would be scrapped in a forging application.
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A further limitation on the application of direct resistance heating is the contact pressure required between the electrode and the work piece. This can be be considerable, and can damage some more fragile materials. Direct resistance heating is a DC process. As the electro-heat frequency increases the heating is done via a time-changing magnetic field inducing a current in a conducting work 2
piece. This current produces I R heating in the work piece, and the process is known as induction heating. It can be seen that induction heating is split into three categories: Mains frequency, medium frequency and radio radio frequency. This is for two reasons. The first is that the equipment for generating the excitation current for the coil (or work head) that generates the time-varying magnetic field has historically been very different. MAINS FREQUENCY induction heating systems are excited by direct connection to the utility supply. The current penetrates deeply into the the work-piece and they tend to be used in very high power systems. MEDUIM FREQUENCY induction heating systems (500Hz-10kHz) have suffered a long evolution. They are used for smaller work-pieces work-pieces and the current penetrates about 1cm into the work-piece. In their inception, in the 1920’s, the excitation was provided provided by the use of a motor-generator set. As suggested by by their name, these work by mechanically coupling a high-frequency generator to the shaft of a mains-powered mains-powered motor. They are primitive, primitive, noisy and inefficient. They are also simple, and easy to service, service, and are hence still found in some less technologically advanced countries. The advent of the THYRISTOR in the 1960’s saw most MG sets being replaced by various different topology inverters. inverters. Thyristor-based inverters are still still in production, and modern thyristors can carry several thousand amps and can block several thousand volts, whilst offering a very low forward conducting conducting voltage drop drop (1.4V). This capability means that medium-frequency induction heaters with powers in excess of 500kW are often best manufactured with IGBT-based inverters. Across the power range 50kW upwards, thyristor thyristor based inverters are still being manufactured, and are characterised by their reliability and simplicity, and their manufacturing expense and their infle xibility. Recently (within the last 5 or so years), design emphasis has moved from thyristor bases inverters to IGBT based inverters. IGBTs offer three principal principal advantages over thyristors: The first advantage is that, unlike thyristors, IGBTs turn off when their gate signal is removed. Once turned on, on, a thyristor thyristor will remain remain conducting until it is reverse biased. This Page 2 of 2
means that the power-electronic topology of an IGBT-based inverter can be considerably simpler.
The second major advantage of an IGBT over a thyristor is the packaging packaging
technology is considerably easier to use. These two improvements improvements make the manufacturing costs of the inverter inverter considerably lower. The final improvement that IGBTs offer offer is that they can have considerably faster switching speeds, so the operating frequency range of an IGBT based system will be extended. The operation of these inverters will be discussed in more detail at the end of this course section. Finally, we have RADIO-FREQUENCY induction heating, which covers the range 50kHz – 10MHz +, although it gets difficult to make work-head coils that work at the highest frequencies. From the 1920’s, 1920’s, the excitation has been provided by thermionic valve-based oscillators, originally originally made from modified radio radio transmitters.
Themionic valves valves are
characterised by very fast switching times (ns) and very large forward conduction voltage drops (100V). To make a workable induction heating system, a very high voltage voltage DC power supply is used (up (up to 12kV). Even with this high supply supply voltage, the efficiencies of these “generators” is only in the region of 60%, so, with powers of 2½kW to 2MW, it can be seen that a considerable amount of power is i s wasted. This inefficiency, coupled with the safety and reliability concerns associated with a high voltage DC link, compounded with the high unit cost for a power triode, have led, since the 1980’s to the increasing use of MOSFET MOSFET based inverters. Currently, commercially available inverters can deliver 1MW at 800kHz, and the frequency range is expanding. So, to summarise, for medium frequency systems, the THYRISTOR is being replaced by the IGBT, and for radio frequency systems, the TRIODE is almost completely superseded by the MOSFET. Having briefly explored how the different frequencies are generated, we should now look at why the different frequencies are generated. This is all related to current current penetration. At low frequencies, the magnetic field penetrates deep into the work-piece, which means that the induced current will flow deep into the work-piece. If the field can penetrate further than half way through the material, then the field coming from the opposite direction will cancel it out, as it will be acting in the opposite opposite direction. Therefore, for a given piece of material, there is a minimum frequency that can be used to heat it. It is this effect that allows the use of a laminated steel magnetic core on a transformer.
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Conversely, if the frequency is too high, only the surface of the material is heated, which may mean that the applied power has to be limited such that the outside is not overheated (melted or burnt) before the inside is heated. It may be, of course, that you only want to heat the outside! Therefore, the frequency controls the depth of heating. Moving up through the electromagnetic spectrum, we come to DIELECTRIC HEATING. Dielectric heating uses the application of a high frequency electric field to align and move the molecules of a non-conductive material. This movement causes friction within the material, which causes it to heat. This process is not affected by penetration depth, but as the intensity of the electric field is inversely proportional to the distance between the electrodes, the thickness of the heated part is restricted. The most common use for dielectric heating is therefore plastic welding, which is use to weld thin plastic sheet in many applications, such as pencil cases, plastic sacks, etc. Moving further up the electromagnetic spectrum, we get to possibly the best-known use of electro-heat: the microwave. Microwaves are used in industrial heating as well as domestic, for example drying paper as it is produced. As a far-field (i.e. wave) electromagnetic source is used, care has to be taken when using it, as if any leakage occurs, operators can become injured. Microwave heating systems work by causing certain molecules to vibrate at a resonant frequency (in the case of a domestic microwave oven, water molecules at 2.4GHz). Once again, as in dielectric heating, this vibration then causes heating essentially by friction. Microwaves are a shallow heating mechanism, with most of the incident energy being absorbed within the first ~4cm of the target thickness (dependent on the material and the frequency). Moving further up the electromagnetic spectrum, we get to infrared heating.
Infrared
heating describes the use of using the infrared radiation from a hot body to heat a colder body. As radiation is a surface phenomenon, the surface of the emitter is cooled by the radiation, and the surface of the target is heated, and the thermal conductivity of the heated material carries the heat into the work-piece. There are three sub-categories of IR heating long wave, medium wave and short wave, which correspond to increasing surface temperatures of the emitting body. As the wavelengths increase, the heat tends to be able to pass further through various materials. The heat rate is governed by the equation
Page 4 of 4
P
= ε A(θ s4 − θ T 4 )
where epsilon is a material constant called the emissivity. Having defined electro heat in terms of operation, its use must be justified. The most common criticism of electro heat in most forms is it’s cost, both operational and in terms of capital plant cost. An example of this can be seen in the small company of R.S. Hall Engineering Ltd, the small company that I worked in between finishing my degree and starting my PhD. Seven years ago, they invested in an induction heater to replace a gas furnace, for heat-treating steel. The induction heater heats steel bar from ambient to 1200C. As the bar that they were heating could be a minimum of 9.5mm in diameter, they had to use a 30kHz induction heater (we’ll derive this frequency later), and, to get the throughput they needed, they chose a 30kW power unit. This cost them £22000. It was estimated that, if they had rebuilt their gas-powered furnace from scratch, it would have only cost them £8000. Clearly, for them, the initial outlay was offset by other factors. One major factor is the energy cost. This will be shown (on the next slide) to be reduced. From a process point of view, one of the most attractive features of electro heat in general is the quality of the heat control. Most electro heat processes allow an instantaneous power control with continual variability. Compare this with a coal-powered furnace that can take several hours to warm up.
Further, the power put into the process can be accurately
controlled with an electro heat system, whereas most combustion-based methods are subject to fuel quality variations. A further advantage of electro heat is that the power density within the work piece can be very high. This is because the heat is developed within the work piece and therefore no conduction process is required. The fact that the heat is developed in the work piece means that it, rather than a surrounding furnace, is the hottest part in the system. These factors combine to reduce the heating time, which reduces the material losses due to oxidisation and chemical (metallurgical) changes such as decarburisation, which traditionally account for a high proportion of the lost material in combustion-based heating processes. As well as being able to modulate the power in the time-domain, the special control of the heat available with induction, dielectric, and, to a lesser extent direct conduction heating, is considerably better than any other means (such as an oxy-acetylene flame). This has its advantages: the profiles of heated parts can be accurately controlled, such as the seam of a plastic weld, or the part of a pop-rivet that bends.
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A further tangible improvement that electro heat offers over traditional heating processes it the elimination of soot and smoke from the workplace. Most electro heat methods are also much quieter than combustion processes, and therefore the replacement of combustion heating processes with electro-heat processes results in a workplace that is considerably more pleasant. Again, the main criticism of electro heat is the cost. To make a true comparison of each energy source, it is important to consider it’s calorific value and to compare the cost on an equal plane. Oil=20p/40MJ=0.5p/MJ Gas=34p/106MJ=0.32p/MJ Coal=4800p/27500MJ=0.17p/MJ Electricity=4p/3.6MJ=1.1p/MJ It is therefore obvious that coal is the cheapest form of energy calorifically, and electricity is the most expensive, by a factor of about 6. However, when we compare process efficiencies, a more realistic picture emerges. Although gas has a higher calorific value per unit cost than electricity, the burner efficiency is limited as gas is mixed with air rather than oxygen, so some of the energy goes into heating the nitrogen. This is translated into hot exhaust gasses rather than hot material. Further, getting the perfect mix between air and fuel is impossible, leaving some of the gas partially un-burnt (carbon monoxide or soot). An efficiency of 60% for combustion is therefore rather optimistic. All we have done so far is to heat a flame – the flame then has to heat up a billet, generally an inefficient process, with half of the heat going up the chimney. Finally, the equipment utilisation has to be taken account of. With a gas furnace, a warm-up time of over an hour is unexceptional, which means that the furnace will be left on over coffee breaks etc, wasting energy. The overall efficiency can be worked out 60/100*50/100*60/100=18% …and the cost of the gas per useful MJ ca n then be found 0.32/0.18=1.77p/MJ
Page 6 of 6
A similar analysis can be carried out on the same process if it were to be done using direct resistance heating. It can be seen that the system is not 100% efficient either. The “burner” 2
(i.e. energy conversion) efficiency is not 100%. This is due to I R heating of the contact electrodes etc, and is a function of the relative resistivity of the work piece and the electrodes. The heat transfer, though, is 100% efficient, as the heat is generated in the work piece. The heat utilisation is not 100% because of the contact area. The metal under the contact area is not heated uniformly, and may have to be scrapped, so the energy that has gone into that part of the work-piece is wasted. This gives an overall efficiency of 72%, and a cost per useful MJ of 1.52p, considerably cheaper than the cost of gas! The electro-heat process suddenly seem more economically viable! Unfortunately, with direct resistance heating, the ends of the work-piece don’t get hot. In a forging application, this can mean that 10% of the m aterial can be wasted. Alternatively, in a zone-heating application, this is a good thing. For the forging case, the unheated part of the billet is generally scrap metal, which introduces further costs into the equation. However, metal loss as scale (oxides), which can be significant in a gas-fired system, is almost eliminat ed. With induction heating, the heated length can be as long as the work-piece. The inverter can be up to 98% efficient, and the coil can be over 90% efficient. The near-instantaneous power means that the heat utilisation can be 100%, but there is loss due to radiation etc. from the work piece. It is therefore possible to get a total energy efficiency of 80% from an induction heating system with only minimal material wastage. Both combustion furnaces and direct resistance heaters can give rise to a significant degree of material wastage: 3% due to oxidation and metallurgical change in the case of combustion furnaces, and up to 10% due to poor current distribution in the case of resistance heaters. This increased material wastage can mean that the direct resistance heater is not an economical choice for a process heating for billet forging applications. With its lack of contact and increased power density, coupled with it’s ability to heat 100% of the billet length with good heat distribution, induction heating can be a m ore economical choice.
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This picture shows a work head connected to a 7½ kW 150kHz (i.e. RF) induction heater, o
heating an M10 bolt to well over 1000 C. As we shall see over the next few weeks, the efficiency is a function of the material being heated, along with the heating profile, and, as such, is variable. With billet heating, a heating efficiency of 70% is reasonable. With induction heating, as with direct resistance heating, the heat is directly generated within the material, so the heat transfer efficiency is 100%. The heat utilisation, like the efficiency, is a function of the required heating pattern. It is easy to heat 100% of the length of a billet; you simply put it in a long work-head coil. It is more difficult to heat a narrow band, as focussing the magnetic flux can only be done to a certain degree, with fringing over the (required) air-gap between the work-head and workpiece spreading the heating somewhat. This means that although it is possible to make a hot spot, it is difficult to avoid a warm area around it. As induction heaters tend to give instantaneous heating control and can draw no quiescent power, the time utilisation of energy can be 100% (i.e. when it’s not heating, it’s not drawing power) Therefore, for billet heating, we can get a 70% total energy efficiency, coupled with a sub1% total material wastage, if scale/decarburisation wastage is accounted for. So, how does induction heating actually work? As briefly discussed in the last lecture, induction heating works by inducing a current into a conducting object. It does it like this: First, a copper coil (often a solenoid, but not exclusively), has a large, time varying current set up in it by the imposition of a time-varying voltage across it (generally in the form of a sine wave).
Page 8 of 8
This current then creates a time-varying magnetic field (for solenoid H = cause a time varying flux ( B
NI l
), which will
= µ H ).
I
E
If a conducting object is placed in the field, then a voltage will be induced around it ( E =
d Φ dt
,Φ
= BA ).
If the conducting object is a closed ring, then, like a shorted turn in a transformer, then the voltage will cause a current to flow around the outside of the object. V = I ( R + jX ) …. I =
V R + jX
The allowance for an impedance has to be made as this is an AC system:- If it were Dc, then the rate of change of flux with time (
d Φ dt
) would be zero, so no current would be induced.
2
Finally, this induced current causes I R losses in the work piece, which makes this method of heating effectively a resistance heating method, albeit with the current flowing at right angles to that of direct resistance heating (i.e. around the billet rather than along it). Having shown in some more detail how the basic principle of induction heating works by basically considering the current flow in a thin tubular foil, we shall now look at what happens to the induced current when induction-heating a solid work piece. The answer to this question is fairly mathematically heavy, and to go too deeply into it would be a bit of a waste of time. Therefore, I will simply give you a “hand-waving” account of how the field, and therefore the current, is distributed throughout the materials, followed by the analytical solution. This way we avoid vector integration, Bessel functions etc. Page 9 of 9
H o
H 1
z
x
H 2 H 3
y
To avoid having to discuss flux return paths and end effects, we consider a semi-infinite slab of material being heated by an infinite 2-diamentional sheet of current just above it. This diagram shows a finite part of a cross-section of the infinite set-up. The sheet of current that represents the work head stretches infinitely to the right and left (in the x direction), and infinitely forward and backward, or into and out of the page (the z direction). It occupies no space a all in the y direction. The semi-infinite slab that represents the work-piece also extends infinitely into and out of the page, and infinitely to the left and right, but it goes from y=0 to y=-∞. To see where the current goes, we ca n turn the homogeneous slab into a series of thin slices. ˆ cos(ω t ) acting on it. In the Consider the top slice. It has a time varying magnetic field, H 0 same way as the last slide, this will have a current density induced in it, J 0 cos(ω t + θ ) . The phase shift (lag) is due to the inductance of the slab causing a lag between the EMF induced in the slice and the current flowing through it. This current density in the slab creates an opposing magnetic field in the slab, marked as H 1. The resistivity and the inductance of the slab reduce the magnitude of the current, and hence the field, so H1 is smaller than H0. Now consider the strip below. It sees a field that is equal to the vector sum of H0+H1 so, this strip sees an attenuated field, as H 0 opposes H1. This strip will therefore have an attenuated current density induced in it J 1 cos(ω t + θ 1 ) . This attenuated current density creates a magnetic field H 2. The third strip down will see a field made up of the vector sum of H 0,H1 and H2, i.e. a further attenuated field, which will induce a smaller still current density, making the resultant total field smaller and smaller as you go down through the y axis. This effect, known as the ‘skin effect’, means that the field, and hence the loss, or heating effect, is concentrated on the surface of the work-piece.
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It can therefore be shown, by letting the thickness of the slices tend to zero, and solving the resultant differential equation, that the field in the x direction, the current in the z direction, and the flux in the x direction all follow this form: H x ( y ) = H x (0 )
− y δ y cos J z ( y ) = J z (0 ) e ω t − δ Φ x ( y ) = Φ x (0) i.e. they are all of the form e
−
y
cos ω t −
δ
y
δ
which is an exponential decay multiplied by an oscillating term with a variable pha se shift. 1 − y
0.8 |
0
J / J
|
Magnitude attenuation = e
δ
0.6 0.4
1/ e
0.2 0
0.3679
δ 0
1
2
3
4
5
y/ δ
cos ω t −
y →
δ
The oscillating term (note shifting phase)
This assumes that the current flowing through the coil is sinusoidal. Generally, this is the case, with only a small distortion. The reason for this will be made more clear when we have discussed how the coil is connected to the excitation circuit. These terms are only true for a semi-infinite slab, so they have no direct use. However, they are simple, and most of induction heating theory is based on the m. The most important part of the equations is the term
δ. This is the skin depth, or the
penetration depth, and is the depth into the slab at which the current etc. has fallen to 1/e (i.e. 1/natral number) of it’s surface value.
δ =
2 ρ
ωµ
On inspection of the formula for the penetration depth, it can be seen that the heating depth is a function of the resistivity, the permeability and the frequency. As the resistivity and the Page 11 of 11
permeability of the work-piece are fixed by its material, the only way to actively control the penetration depth of the current into the material is to alter the frequency. This is why induction heating systems are divided into the three different f requency bands: Mains – used for through heating large pieces of metal (gas cylinders etc) MF – used for through heating smaller billets and strips – down to 15mm R.F – used for surface heating or for heating very small pieces. Although the current (and therefore the heat) is induced into the very surface of the material under an RF magnetic field, RF induction heaters can be used for through heating by allowing the heat to be conducted in through the material. This limits the rate at which the material can be heated, as too high a power will result in the material melting on the surface before the inside is even warm! One other thing to note about the equations is the phase shift in the oscillating term. As the y position decreases, the current, H-field and flux become more retarded. The total current (per unit length) in the slab can be found by integrating the current density from the surface to –infinity, with respect to the semi-infinite axis, y. The current density has been defined as: J z ( y ) = J z (0)e
− y
δ
cos ω t −
y
δ
JZ(0) is the surface current density. The phase shift on the current with depth has an effect on the integration, which evaluates to: I =
J z (0)
2
cosω t −
π
4
This total current I can be considered to be flowing in 1 skin depth δ. Therefore, as far as our semi-infinite slab goes, we have made the y-direction into a sheet of thickness
δ that
carries all of the current in a uniform manner. This is only a definition, remember, but it is extremely useful. The most important bit is that it allows the surface power density to be defined, and from that, an equivalent circuit. So far, we have got as far as finding the total current in the work piece, and stating (although proof can be found in the “induction heat ing handbook” by John Davies and Peter Simpson), that the total current Page 12 of 12
J z (0)
∠45 0
2
can be considered to be uniformly concentrated in t he outer skin depth of the material 2 ρ
δ =
ωµ
If this is true, then the surface power density can be found, by the use of R
l
= ρ A
So, for a surface area of 1m ×1m,
1m
δ 1m A=δ×1m, l=1m 2
So, using P=I R, 2
J (0) l = 1m P= z × × ρ = resistivit y 2 A = δ × 1m
↑=I2 -2
…in Wm
Now, it’s all very well knowing the power density in the work piece as a function of the surface current in the work piece, but this has no real connection with induction heating: the power is provided by a magnetic field, not by direct connection. To relate the power to a coil, consider a long solenoid. The total current density in the coil is equal to the coil current multiplied by the number of turns divided by the length. J 0
=
I c N l
-1
Am
We can substitute this value for J 0 into the surface power density to get Page 13 of 13
2
2
I N / l ρ I c N ρ P= c = l δ 2δ 2
-2
Wm
From this, we can find the power density due to the coil current in a long solenoid. If we now consider the field due to a solenoid H 0
=
NI l
-1
Am
this is equal to the surface current density. Therefore substituting H 0 into the surface power density equation, 2 P = H RMS
ρ 2δ
The RMS symbol is inserted as a reminder that the rms value of the field must be used = ˆ H 2 To show the effect of the work piece properties on the surface power density, three -1
examples are worked through, each with the same field magnitude (100kAm ) and frequency (50Hz), but with varying permeabilities and resistivities.
Material
Mild Steel
Temperature θ θw Resistivity, ρ ρ ρ Relative
Mild Steel o
20 C
o
800 C
20 C
o
200nΩm
1.1 µΩm
17nΩm
50 (
1 (>Curie)
1 (non magnetic)
permeability µ µ µ r Applied Field H 0
Copper
-1
-1
-1
100kAm
100kAm
100kAm
Frequency f
50Hz
50Hz
50Hz
Skin depth δ δ
0.0045 m
0.0747
0.0093
Surface densityPs
Page 14 of 14
Power
-2
222kWm
-2
73kWm
-2
9.16kWm
The first case is for mild steel at room temperature. The steel is below the Curie point o
(~720 C), at which the ferromagnetic properties of the steel stop, and therefore a magnet will not stick to it. This means that the relative permeability of the steel is relatively high. Electrical engineers amongst you will notice that the permeability of the steel is considerably lower than you would expect, with
µr being well over 1000 normally.
This discrepancy can be explained by considering the given
µr as a ‘large signal’ quantity,
µr being a small signal quantity. If the expected µr of 1000 is substituted
and the expected into B
= µ H
then you get -7
B=1000×4×π×10
×100×103=124T
At the surface of the steel, but steel saturates at about 2T. The effect of the saturation is to reduce the incremental permeability. The effective relative permeability is therefore reduced to take this saturation into account – i.e. the large signal incremental permeability is used. Effective values of 20 to 50 are often used for steel under lower frequency induction heating fields. Anyway, to find the surface power density, we first need to find the current penetration depth
δ =
2 ρ
µω
=
2 × 200 × 10 −9 50 × 4π × 10
−7
× 50 × 2π
=0.0045m=4.5mm
Substituting the skin depth into the surface power density equation
ρ (100 × 10 3 ) × 200 × 10 −9 -2 = =222kWm 2δ 2 × 0.0045 2
P = H
2 rms
Can you work out the other two situations?
Note, then, that the hot steel has a much deeper current penetration than the cold, due both the (relatively slow) increase in the resistivity with te mperature, and to the rapid reduction in the permeability at curie. Although the resistivity of the hot steel is over 5 times greater than the resistivity of the cold steel, the losses in the steel are lower, as the current flows in an area that is over 16 times Page 15 of 15
greater. Remember that the total current in the material is only a function of the H-field, so the current is the same. The resistance is equal to the resistivity multiplied by the length divided by the area, so if the penetration depth is smaller, the resistance, and hence the loss, is higher. From this, we can see that the steel is heated much faster below the Curie point than above it for a fixed field strength. This is a general conclusion, and is something to consider when dimensioning an induction heating system. Moving on now to the copper load, we can see that the penetration depth is just over twice that for the cold steel load.
Although the conducting area is similar, the per-unit surface
area effective resistance of the copper is much smaller that that of the steel. It can be seen that the losses in copper, under the same field strength as the steel, are considerably smaller. This can cause problems when induction heating copper! On the other hand, if the H-field source (i.e. the work head) is also considered to be a semiinfinite slab, starting where the work piece semi-infinite slab ends, then the analysis is valid for the work head as well as the work piece. This allows the work head losses to be estimated. As the H-field at the very surface of the load slab is equal to the H-field at the very surface of the source slab, all of the previous calculations can be used. H0(workhead)=H0(workpiece) Almost invariably, the work-head is made out of copper, although above 10MHz, the copper is silver plaited to increase the surface conductivity. As all of the losses worked out on the previous slide were for the same field intensity, the relative efficiencies can be found for heating steel above and below Curie, and also for heating copper. Eff=Heat into load÷total heat Total heat= Heat into load + Heat into work head The theoretical efficiencies can be found First for cool steel E =
222 222 + 9.16
Page 16 of 16
× 100% =96%
similarly, Hot steel=89% Cold copper=50% This calculation makes the assumption that the length and the width of the work piece and the work head are equal. Whilst the width of a practical work piece may be equal to the width of a practical work head, the length of the work head has to be longer, as it is wrapped around the work piece. The efficiency is therefore always worse than this calculated value. By substituting the skin depth equation into the surface power density equation, and substituting this into the efficiency equation, both in terms of the work-head and the work piece, then the best-case efficiency can be shown to be 1
Eff =
1+
ρ c ρ w µ rw
It is easy to draw the conclusion that for efficient heating, you need a highly conductive coil and a highly resistive work piece, and if the work piece has a high permeability, then the efficiency is improved. Of course, many assumptions have been made here, not least the semi-infinite slab model: this result suggests that you could induction heat air (resistivity infinite) with 100% efficiency. This is clearly not the case. Also, the analysis assumes that the work head is an homogeneous sheet of current, rather that a series of discrete turns separated by layers of insulation. This means that the area that the current flows in the work head is reduced, meaning that the local current density is higher, increasing the losses. This is compounded by the fact that the length of the current path around a practical work head is longer than the length of the current path around most practical work pieces. To make a better model of the work head / work piece pairing; which includes the reactive as well as the resistive losses, we can create an equivalent circuit by considering the magnetic flux. We are first going to derive the EMF induced in the work piece by considering the magnetic flux in the work piece. If we start with the flux density. This follows the general form:
Page 17 of 17
B = B0 e
−
y
δ
cosω t −
y
δ
where we have the surface value modified in amplitude and phase as you pass through the material. If we assume that the material is linear, and that B
= µ H
holds, then we can substitute µH0 for B0 and relate the flux density to the applied H-field. We can then find the flux in the work piece by noting that the flux density is the rate of change of flux with respect to space, so flux is the integral of flux density, with respect to space. Therefore, the total flux in the work piece can be found by integrating the flux density over whole depth of the semi-infinite slab.
Φ x = H 0
δµ 0 µ r 2
cos ω t −
π
4
(note the similarity with the total current) From the basic electromagnetic definitions, the E MF in a coil is equal to the number of turns multiplied by the rate of change of the flux contained in the coil with respect to time. E = N
d Φ dt
Of course, with the work piece, N=1, but we can refer this voltage back to the coil by substituting the number of turns into N. So, differentiating the total flux with respect to time, and multiplying by the number of coil turns, we get the coil EMF due to the flux in the work piece. E = − N ω
δµ H 0 2
sin ω t +
π
-1 Vm 4
or, in cosine form E = N ω
δµ H 0 2
cos ω t +
π
-1 Vm 4 o
i.e. The voltage leads the H-field by 45 . If we consider that in the case of a long solenoid
Page 18 of 18
H 0
=
NI l
-1
Am
we can substitute for H to get 2
E =
I 0δµ N ω
2l
cos ω t +
π
-1 Vm . 4 o
I.e. the voltage leads the current by 45 , or, more conventionally, the current lags the voltage o
by 45 . If we use the form P=VI cos(φ) where cos(φ) is the power factor, it can be seen that the power factor of the work piece is o
cos(45 ), or 1/ √2 Alternatively, if we re-write the EMF equati on so that E and I are complex AC Signals rather than the time-domain signals used so far, we get E = I
δµ N 2ω 1
1 + j Vm-1 2 l 2 2
If we consider this to be in the form of Ohm’s law, then we can see that the impedance of the work piece, referred to the coil terminals, is Z =
δµ N 2ω 1
1 + j Ωm-1 2 l 2 2
allowing equivalent values for the resistance and the reactance to be found: R =
Ωm-1, X =
δµ N 2ω 2l
Ωm-1
Further, if we consider the reactance of an inductor XL=ωL The reactance can be expressed as an equivalent inductor L =
δµ N 2 2l
-1
Hm
Page 19 of 19
δµ N 2ω 2l
This per-unit equivalent circuit can be used to analyse a very big (d>20 δ) cylinder with a solenoid wrapped tightly around it.
=
Rload
δµ N 2ω 2l
π D Ω
where πD is the length of the electrical path around the work-piece. Similarly X load
=
δµ N 2ω 2l
π D Ω
Using the concept that the coil can be approximated to a semi-infinite slab, providing that the conductor is over 10 times thicker than the skin depth of the current, it is possible to define the equivalent circuit components in a similar manner. Rcoil
= k c
δµ N 2ω 2l
π D , X coil = k c
δµ N 2ω 2l
π D
Inspection of the equation shows that a coil factor, k c, has been included. This is to account for the fact that the coil is made up of discrete turns rather than a homogeneous mass. k c is normally of the range 1.1 to 2, depending on the shape and spacing of the coil conductors.
The lowest end of the range, being the most efficient, is for rectangular
conductors with minimal turn-on-turn insulation.
If round tube is used, then the efficiency is compromised by the reduction in the conducting area, where only the bottom of the tube caries the full skin-depth of current, so a fair bit of the area facing the work piece does not carry any current.
Page 20 of 20
Once again, this analysis assumes that the coil is touching the work piece, and that the diameter is much larger than the penetration depth. If the coil is not touching the work piece, then there will be some ‘stray’ flux in the air gap between the coil and the work piece. This stray flux will cause an EMF to be induced in the coil, just like the flux in the work piece and the flux in the coil. We can consider the field H0 to be constant across the air-gap by considering the penetration depth. 2 ρ
δ =
ωµ
the resistivity of air is infinite when it is not ionised (i.e. a spark or plasma). Therefore the penetration depth of magnetic field into air is infinite, which means the rate of change of the H-field with space is zero across the airgap. This means that the identity H0(coil)=H0(workpiece) Remains true in the presence of an air gap. Further, the equations for Rc, Xc, Rw and Xw also remain correct. However, we now have to consider the flux in the air gap, which will contribute to the coil terminal voltage. The flux in the air-gap ( Φ gap) can be found from B= µH
Φ gap = µ 0 H 0 Ag Where Ag, the area of the air-gap, is equal to the area enclosed by the coil minus the area of the work piece. Ag
=
π 4
(d − d 2 coil
2 wokrpiece
)
The coil voltage due to the air-gap can be found by finding the rate of change of the flux in the air-gap with respect to time. V g
V g
= N =
d Φ gap dt
N 2 I l
d NI = N cos(ω t )µ 0 Ag dt l
sin (ω t )ωµ 0 Ag
Alternatively, in orthogonal complex AC form
Page 21 of 21
V g
= j
N 2 I ωµ 0 Ag l
i.e. the voltage in the coil due to the air-gap is totally reactive, which makes sense as the air does not get hot! The equivalent reactance of the air-gap can be found by dividing the voltage by the current. 2
X g
=
N ωµ 0 Ag l
Although the equivalent component for the air-gap is totally reactive, it does indirectly increase the losses in the coil, as it increases the length of the current path in the work head. Therefore, induction heating systems with a large air-gap are less efficient than those with a small air-gap. As the total flux within a coil gives rise to the EMF induced in the coil, it appears as if all of the EMFs, and therefore all of the equivalent components, are in series. Therefore we have derived a full equivalent circuit for a coil heating an electrically large load.
Rc
Xc
Xg
Lw
Rw
Example This example is for a long thick bar of steel (300mm*2m) being heated from room o
temperature to 1200 C under a field being excited at 50Hz. The coil is 500mm in diameter and has 100 turns. It is the same length as the steel bar. Similar bars of steel are heated at chesterfield cylinders at a power of 1MW prior to being forged to make gas cylinders. Initially, we will find the equivalent circuit of the system with the steel at room temperature. If we compare the skin depth (4.5mm at room temp), with the diameter (300mm), then it c an be seen that the condition given for the work pie ce to be electrically large is true (20δ
Page 22 of 22
Following the formula, Rcoil can be found:
=
Rc
=
Rc
δ c µ 0 N 2ω 2l
π d c k c
9.3 × 10 −3 × 4π × 10 −7 × 10000 × 2π × 50 × π × 500 × 10 −3 × 1.5 2× 2
=0.0216Ω Xc=0.0216Ω also Lc=Xc / ω=69µH. We have now found the equivalent component values for the coil. Moving on to the work piece: Rw
=
=
δ w µ w N 2ω 2l
π d w
4.5 × 10 −3 × 50 × 4π × 10 −7 × 10000 × 2π × 50 × π × 300 × 10 −3 2×2
=0.209Ω Xw=0.209Ω also Lw=Xw / ω=666µH. Finally, the inductance due to the air gap. This is due to the area inside the coil that is not accounted for by the work piece. X g
= N 2 µ 0ω Ag
Ag
= Ac − Aw =
Ag
=
X g
= 10000 × 4π × 10 −7 × 2π × 50 × 0.126 =0.25Ω
π 4
π 4
(d − d ) 2 c
2 w
(0.5 − 0.3 ) =0.126m 2
2
Lg=Xg / ω=790µH
So, if we were to apply 1000V to the termina ls of the coil, what power would we get?
Page 23 of 23
First find the coil current I =
V Z
Z = R 2
+ X 2 = ( Rw + Rc )2 + ( X w + X c + X g )2
Z =
(0.209 + 0.022)2 + (0.209 + 0.022 + 0.25)2 =0.53Ω
∴ I =
1000V 0.53Ω
=1874A 2
then find the power by using P=I R 2
Power in work piece=I Rw=734kW 2
Power in work head=I Rc=77kW The efficiency can be found by finding the useful power divided by the total power. E =
Pw Pw
+ Pc
× 100% =734/(734+77)×100%=90.5%
The semi-infinite slab example, which does not account for the air gap, gives an efficiency of 96%. It can be seen therefore that the air gap does indeed increase the losses in the coil, despite dissipating no power itself. It is all well and good being able to calculate the equivalent circuit values for an electrically large work-piece in a long solenoid: this covers about 35% of all induction heating cases. A further 35% of induction heating cases are for systems that are electrically small, and the rest are for cases where the work head is short – it can be down to 1 turn. One of the best conditions for rapid through heating is to have the skin-depth similar in magnitude to the diameter, such that the majority of the load is being directly induction heated, rather than being indirectly heated by thermal conduction from the heated surface. Having an increased heating penetration gives a smaller temperature gradient through the material for a given surface power density, allowing the middle to be heated at a much higher rate than would be the case with an electrically large work piece, without the surface melting or becoming degraded. As well as increasing the throughput, this is useful, as a rapidly heated product will spend less time at an elevated temperature, so less material will be lost through oxidation or metallurgical change.
Page 24 of 24
To enable analysis of magnetically smaller parts, the equations for the resistance and reactance due to the load are modified to include a correction variable. R w
=
ωµ N 2 pAw lc
, X w
=
ωµ N 2 qAw lc
,
Where Aw is the area of the workpiece Aw
=
π 4
d w2 .
Substituting Aw
=
π 4
d w2 into the new equations for R w and Xw, and then comparing these
equations with those for electrically large cylinders, it can be seen that p
as the effective width of the current flow in terms of the losses, and q
d w
2
d w
has replaced δ
2
has replaced δ as
the effective width of the current flow in terms of the magnetic stored energy. These modifiers (p & q) are due to the work-piece not being adequately approximated by a semi-infinite slab. Their values have been derived from the solution of the Poncare integrals used to analyse the magnetic field distribution in a finite geometric shape. As such, they are a function of component shape and relative electrical size. For cylinders with δ d w being below 8, a graphical lookup of the p and q values is often used. For values of d/ δ >8, the solution approximates to q
=
2 d δ
, p
=
2 1.23 + d δ
, and for
values of d/ δ >20, the solution approximates to the semi-infinite slab assumption, so p
=q=
2 d δ
.
Page 25 of 25
1 0.9
q 0.8 0.7
s e u 0.6 l a v q 0.5
d n a 0.4
p
p
0.3 0.2 0.1 0 0
1
2
3
4
5
6
7
8
d/ δ
The induction heating handbook, amongst other books, has the curves in. These curves can also b found in some of the more advanced electromagnetic field textbooks, although in some cases, p and q are reversed in meaning. Example o
This example is the same set up as before, except that the steel is at 800 C. The steel will be past the Curie point, so the current penetration w ill be much deeper. First, the relative current penetration depth must be found. The current penetration depth, given by
δ =
2 ρ
ωµ
,
has already been found to be 0.0747m or 74.7mm. The diameter of the steel bar is 300mm, so the steel is not electrically large in diameter. The ratio of the skin depth to the diameter has to be found: 300/75=4. This is lower than the simple approximation, so the graphical lookup approach is used. First, we shall find the equivalent resistance, Rw.
Page 26 of 26
1 0.9
q 0.8 0.7
s e u 0.6 l a v q 0.5
d n a 0.4
p
p
0.3 0.2 0.1 0 0
1
2
3
4
5
6
7
8
d/ δ
Using the graphical lookup curve, we can find a value for p=0.38. This can be substituted into the equivalent resistance formula R w
=
ωµ N 2 pAw lc
The area of the work piece is also required Aw
=
π 4
2
d w2 =0.0707m .
Remember that N is the number of turns in the work head coil, as the voltage is referred to the work head coil terminals Rw
=
2π × 50 × 4π × 10 −7 × 10000 × 0.38 × 0.0707 2
=0.053Ω
Moving on to Lw, the equivalent inductance of the work piece. Both d/ δ and Aw are common, but q is different from p. Therefore, for an electrically small work piece, the reactance is different from the resistance. Inspection of the p and q curves reveals that the q curve is always higher than the p curve, with the two asymptotically converging.
This means that the work-piece is always more reactive than resistive,
especially with small electrical diameter parts. First, q needs to be found. Using the lookup curve, q is 0.54.
Page 27 of 27
Substituting all of the relevant data into the X w formula X w
=
ωµ N 2 qAw lc
=
2π × 50 × 4π × 10 −7 × 10000 × 0.54 × 0.0707 2
=0.075Ω
Using L=X/ ω, the equivalent inductor can be found =0.075/(2×π×50)=240µH As the frequency and the coil and work-piece dimensions are the same as in the previous example, then the equivalent circuit values for the coil and the air-gap will be the same. Therefore the power and the efficiency can be found, again with a 1000Vrms excitation voltage. First finding the current I=V/Z Z = R 2
+ X 2 = ( Rw + Rc )2 + ( X w + X c + X g )2
Z =
(0.053 + 0.022)2 + (0.075 + 0.022 + 0.25)2 =0.354Ω
∴ I =
1000V 0.354Ω
=2820A
2
Using P=I R, the losses in the work-piece and in the coil can be found: Pw=421kW Pc=172kW The efficiency can be found by dividing the work-head power by the total power E =
Pw Pw
+ Pc
=421/(421+172)=71% - compare with 89% for the semi-infinite sl ab
This reduction in efficiency is due to the airgap, although the efficiency actually peaks when the work piece is not electrically large, by concentrating the current. The peak efficiency and therefore the best heating occur around d/ δ=3.5, at the peak in the p-curve. This is very important – it does not occur at d/ δ=2, which is predicted by assuming that all of the current flows in the first skin depth.
Page 28 of 28
The reason for the lack of power for skin-depths of less than 1/3 of the diameter is a phenomenon known as ‘flux cancellation’, where flux from one side cancels out flux from the other side, causing the current flow around the inner circuits of the work-piece to be reduced. This makes the work-piece a net reactive element in the equivalent circuit. If the equations for the coil and the air-gap are compared to the new equations for the workpiece, bearing in mind that p
d w
2
can replace
δ as the effective width of the current flow in
terms of the losses it can be seen that they all share a number of common terms. They are all in the same coil, so the number of turns and the coil length are common, and they are excited at the same frequency. Further, the permeability of free space can be taken out. This can massively simplify the calculation by pre-calculatin g a common constant: 2
K =
2π f µ 0 N c lc
R w
= K µ r pAw , X w = K µ r qAw
Rc
= X c =
X g
= KAg
1 2
Kk r π d cδ c
So far, although we have been calculating values for the equivalent reactance of the workhead system, we have not really considered its impact, which is to increase the apparent power (Vrms×Irms), without increasing the actual power (I
×R), for a given excitation
2 rms
current. This means that the power factor is reduced by the reactance (p.f.=P/(VI)). 2
Given that the real power=I R 2
And the apparent power=I Z, Power factor=
Rtot Z tot
Example Find the power factor of the example coil equivalent circuit before and after curie. Before: Z tot
=
2 2 Rtot + X tot
Rtot=0.209+0.0216=0.231Ω Page 29 of 29
Xtot=0.209+0.0216+0.25=0.481Ω Ztot=0.533Ω (already found to get the coil current) p.f.=Rtot /Ztot=0.231/0.533=0.433 After: Rtot=0.053+0.022=0.075Ω Ztot=0.354Ω(already found to get the coil current) p.f.=0.21 So, after curie, ≈1/5 of the terminal VA of the coil is used in the heating, so the coil would need five times more VA to heat the work-piece than direct resistance heating would, despite comparable efficiencies. As the air-gap adds only to the reactance, the power factor of a real coil is always less than 1/ √2, and decreases as the air-gap grows. If we examine the resistance and reactance due to a large cylindrical work piece (or coil), substituting the skin depth into the equations R, X =
2 ρ µ N 2ωπ d π N 2 d ρµω = . l ωµ 2l 2
we can see that the resistance and reactance due to the work piece and the coil are proportional to the square-root of the frequency, whereas examination of the equation for the reactance due to the air-gap shows that Xg is proportional to the frequency. This suggests that as the frequency increases, the power factor will fall. It is not uncommon for RF induction heating coils to have power factors of le ss than 1/15. At the time of writing these notes, I was designing a 100kW induction heater to heat chain o
o
from 720 C to over 1200 C in a coil ~1m long with 20 turns. This machine was to harden chain for use in anchoring for ships. Due to the non-uniform shape of the coil, and its high temperature, the area of the air-gap between the steel and the coil was considerably larger than the cross-sectional area of the chain segments. I measured the power factor to be 0.063. If this coil was to be directly connected to the 100kVA inverter, a power of only 6.3kW would be developed. This would be an incredibly expensive heating method, in terms of plant cost (£4000/kW). This poor power factor was corrected by the standard method: A capacitor was inserted into the circuit to supply negative kVARs to cancel out the positive reactive power provided by the coil. The capacitor and inductor form a resonant circuit, and if this circuit is excited at Page 30 of 30
its resonant frequency, the power factor at its terminals will be 1. The actual form of the circuit is dictated by the nature of the inverter used to supply it. There are two main types of inverter: voltage-source and current-source.
Voltage-source inverters produce a voltage
square-wave across their output terminals and the nature of the coil circuit determines the current. Current-source inverters try to output a square-wave of current through their output terminals, and the output circuit determines their output voltage.
If you were to connect a capacitor across the output terminals of a voltage-source inverter, then the rapid rate of change of voltage with respect to time at the edge of each square-wave half cycle would cause a large current spike through the capacitor ( I = C
dv dt
). This spike
would create large losses in the switching devices within the inverter, and the transistors would probably rapidly fail. Therefore, for a voltage source inverter, the power factor correction capacitor is often connected in series with the induction-heating coil. The workhead circuit and the capacitor form a resonant circuit, and tend to filter out all but the fundamental of the excitation voltage. This means that although the output voltage of the inverter is a square wave, the output current is a fairly pure sinusoid. The disadvantage of this connection is that the inverter has to provide all of the coil current, which can be very high if few turns are used. In this case, a transformer is used to scale the output voltage and current appropriately.
Q1
SW1
D1
SW 4
Q2
100
D4
Cfilt
) A ( t n e r r u C
Q1
SW2
D2
SW 3
Q2
D3
50 0 -50
-100 0
Tank current Link current 60
120 180 240 normalised time (deg.)
300
360
60
120 180 240 normalised time (deg.)
300
360
400 ) V ( e g a t l o V
200 0
-200 -400 0
TANK CAPACITOR CT
WORK HEAD L2+RL
Conversely, the current-source inverter provides an essentially square-wave of current to the load circuit. If a series-compensated circuit were used, the high di/dt at the edges of the Page 31 of 31
square-wave would cause large voltage spikes across the inductance of the work-head circuit, which would be carried through the capacitor and the resistance to the inverter terminals, and would probably rapidly destroy the switching elements in the inverter. A parallel compensation circuit is therefore used for current-source induction heating inverters. Once again, this circuit forms a resonant circuit that filters out all but the fundamental of the excitation waveform, making the voltage across the inverter output terminals a fairly pure sine-wave. LSupply Q
D1
D4
SW 1
SW 4
Q
1000 ) V ( e g a t l o V
500 0 -500
-1000
D2 Q
SW 2
0
D3 Q
Tank voltage Link voltage 60
120 180 240 Normalised time (deg.)
300
360
60
120 180 240 Normalised time (deg.)
300
360
100
SW 3 ) A ( t n e r r u C
TANK WORK CAPACITOR HEAD
50 0 -50
-100 0
In both cases, for MF and RF systems, the reactive elements in the circuit dominate the resistive. This makes the resultant behaviour of both circuits highly resonant. As the phase difference between the fundamental of the square-wave quantity (i.e. voltage for a voltage source) and the sine wave (i.e. current for a voltage source) is zero when the switching frequency of the inverter is at the resonant frequency of the network, the power factor is at a maximum. As the waveforms of the current and the voltage are different, the power factor cannot be 1. In fact, for a sinusoidal current and a square wave voltage, the in-phase power factor is 4/(π√2) = 0.9. The square-wave component of the output is often scaled by this value to give an equivalent sinusoidal voltage or current to make the maximum power factor be equal to 1 when the current is in phase with the voltage. Often, the system is run at a slightly inductive power factor to allow the inverter to commutate correctly, but this is beyond the scope of this course. We will analyse the circuit at it’s resonant frequency, as it is much simpler and yields sufficiently accurate results. Consider the series resonant (voltage source) case. At resonance, the voltage across the o
capacitor is equal in magnitude, but 180 out of phase, to the voltage across the total inductive component of the equivalent work-head circuit. Page 32 of 32
This means that the voltage
across Ccomp and Ltot cancel out, which means that the voltage across the resonant circuit terminals is equal to the voltage across the equivalent series resistance. Therefore, the resonant impedance of the circuit is equal to Rtot. At resonance: V C= -V L
∴V R=V in I in = I R
=
V R Rtot
=
V in Rtot
∴ Z T=R tot The resonant frequency can be found by equating the reactance of the capacitor with the negative reactance of the inductance and rearranging. It can be seen that this frequency is the undamped resonant frequency no matter what the l oading is: from V C= -V L Solve for ω
ω 0 =
1 LC
The parallel case is a little more complex. Due to the circuit topology, the voltage across the capacitor is equal to the voltage across the work-head. o
At resonance, the current through the capacitor is equal in magnitude and 180 out pf phase with the reactive current flowing through the work-head equivalent circuit – this means that the reactive currents will cancel and the phase shift will be zero. Solving this equality for ω yields the resonant frequency.
I work =
V in ( R − j L ) R 2 + ω 2 L2
At resonance: IC=V in jω C=-Im( I work )
V in jω C =
V in jω L R 2 + ω 2 L2
Solve for ω
Page 33 of 33
ω 0 =
L − CR
2
2
L C -1
If the resonant frequency (in radians s ) is substituted into the equation for I work , then the terminal current will be equal to the real component of Iwork , as the imaginary component will be cancelled out by the parallel capacitor’s current at resonance. Z T
=
CR L
In the case of the parallel-tuned circuit, the tuning capacitor and the inductance of the workhead circuit affect the impedance.
The definition of resonant impedance brings us on to the last part of this course: Matching. Although an inverter will have a nominal power specification, it will also have a maximum voltage and a maximum current rating, and will only be able to deliver the rated power at the maximum voltage and current. The maximum inverter voltage divided by the maximum inverter current gives the inverter’s nominal output impedance.
If the load circuit’s
impedance is above the inverter’s nominal impedance then the output current will not be high enough to give the full power even when the voltage is turned up to the maximum level. Alternatively, if the load circuit’s impedance is lower than the inverter’s nominal impedance, then the output current will equal the maximum inverter output current before the voltage is turned up to full, again limit ing the power. If the output circuit’s impedance is wildly wrong, then the inverter may be damaged, although most inverters have limiting circuits to prevent over currents. Although it is sometimes possible to choose the number of coil turns etc to make the coil impedance match the inverter, geometrical, mechanical or metallurgical constrains may force the coil dimensions to be such that the required impedance is not directly attainable. In this case, a transformer is often used.
There are 2 circuit positions to insert the
transformer, the first being between the work-head and the capacitor (i.e. inside the resonant circuit):
Page 34 of 34
Placing a transformer inside the resonant circuit is useful if the required work-head current or voltage can’t be matched to a standard capacitor. These are often used for single-turn coils, where currents in excess of 20,000A are required at only 10-100V. MF induction heating capacitors are generally limited by their terminal current, with 200A/stud being a common limit, and 8 studs being used to terminate the capacitor. These capacitors often cost in excess of £1000, and are rated in voltage at 600 to 1200V, depending on capacitance. Generating the required compensation reactive current by the direct use of such capacitors would clearly be expensive, as well as technically difficult, so a transformer is often used, with a ratio of as much as 30:1, to scale the current down and the voltage up, such that the capacitors and the inverter are better utilised. A transformer inserted between the capacitor and the work head coil does have to carry the full kVA of the coil, and with power factors being as low as 0.05, this can be up to 20 times the nominal heating power. This circuit topology is not very efficient due to the high stress on the transformer, so the transformers tend to be large and an efficiency of 60% or less can be expected from the resonant circuit alone.
N
1
Where practical, therefore, a transformer is placed between the inverter terminals and the resonant circuit. This topology is much more efficient, as the transformer only carries the heating power, with the reactive power being contained within the resonant circuit. The ratio of the transformer can be simply dimensioned. If ZT is the terminal impedance of the resonant circuit, and ZI is the nominal impedance of the inverter, then the ratio can be found: N =
Z I Z T
This can be easily proven: VI and II represent the voltage and the current at the inverter terminals Page 35 of 35
VT and IT represent the voltage and current at the resonant circuit terminals First take the voltage across the resonant circuit V T
=
V I N
Then find the current into the resonant circuit I T
=
V T Z T
=
V I Z T N
The current into the transformer is the current into the resonant circuit divided by the ratio I I
=
I T N
=
V I Z T N 2
Re-arraigning Z T N 2
=
V I I I
= Z I
Solving for N N =
Z I Z T
So, we have found the equivalent circuit for a work-piece in a multi-turn coil, found out how to chose a value of compensation capacitor to tune the coil to a specific frequency, and chosen a transformer ratio to match the coil to the inverter. This is the end of the theoretical part of the course. Now for some examples: Example: Chain Heating
Page 36 of 36
Freq Chain Bar ∅ Link length Link width Chain Material
θ Chain (start) θ Chain (end)
30kHz 10mm 40mm 20mm Mild Steel o 750 C o
950 C
Power to chain
40kW
ρ Chain µ Chain
1.1×10 1
d coil l coil
70mm 1.2m
ρ coil
0.017×10 12
-6
N coil
-6
What power inverter is required to heat the w ork-piece? What rating capacitor is required? What transformer ratio is required to match to a 600Vrms Voltage-source output?
Solution: First find the common constant: 2
K =
2π f µ 0 N c lc
=28.42
Assume chain can be represented by k long round bars of same diameter as the wire used to make the links (k=2 –i.e. 2 bars!) find skin depth in bar 2 ρ
δ =
ωµ
=
2 × 1.1e − 6 2π × 30e3 × 4π × 1e − 7
= 3.0476mm.
find the ratio diameter: skin-depth d
δ
= 3.3
Therefore the graphical look-up table should be used p=0.38, q=0.67 Resistance of chain = k ×resistance of bar Aw
=
π D 2 4
2
=7.9e-5m
Page 37 of 37
= K µ r pAw =28.42×1×0.38×7.9e-5=0.848mΩ
Rw
Rchain=1.69mΩ Reactance of chain = k ×reactance of bar
= K µ r qAw =28.42×1×0.67×7.9e-5=1.50mΩ
X w
Xchain=2.99mΩ Airgap area= Acoil-k(Abar)
=
Aw
π D 2 4
π D 2
=
Acoil
2
=7.9e-5m
4
2
=3.845e-3m 2
Agap=3.69845e-3m
= KAg =104.9mΩ
X g
Resistance and reactance of coil standard (assume k c=1.2) 2 ρ
δ c = Rc
ωµ 0
= X c =
=0.379mm 1 2
Kk r π d c δ c =0.5×28.42×1.2×π×70e-3×0.379e-3=1.42mΩ
It can be seen that the air-gap dominates the behaviour, and that the coil is not very efficient at all. Power2
To find the current to achieve the specified power, use P load =I Rload. I coil
Pload
=
Rload
=
40e3 1.69e − 3
=4.87kA
2
Use Ptot=I Rtot to find the inverter power Rtot=1.69e-3+1.42e-3=3.11mΩ 2
Ptot= I Rtot=73.6kW
Capacitorre-arrange ω 0 C =
=
1 LC
to find C
1 Lω 02
L must be found Ltot
=
X tot
ω
=(2.99e-3+104.9e-3+1.42e-3)/(2π×30e3)= 0.58µH
Page 38 of 38
