Dear friends,
Electrical Power System is wide subject in which systems are included from generation, distribution and transmission.
Transformer is the major equipment which is used in every phase of power systems ie. in +generation+ distribution+ transmission. The different type of transformers used depends on its voltage level and MVA/KVA ratings, and type of constructions. Apart from this instrument transformer used for metering and protection in system .
For example- Current Transformer- CT
Potential Transformer- PT
The subject is quite wide, so here I am providing the details of +power+ transformer +substation+design+busbarschemes+layouts
Inside Transformers
How a transformer works, and how to design them. Dissection of a transformer.
Ideal Transformer
A transformer is a device in which two circuits are coupled by a magnetic field that is linked to both. There is no conductive connection between the circuits, which may be at arbitrary constant potentials. Only changes in one circuit affect the other. The circuits often carry at least approximately sinusoidal currents, and the effect of the transformer is to change the voltages, while transferring power with little loss. Sinusoidal excitation is not necessary, and transformers may handle arbitrary signals, in which the action can be considered as a transformation of impedances. The magnetic field coupling the circuits can be in air, but is usually in a ferromagnetic material, the core, in which the field can be thousands of times greater than it would be in air, making the transformer efficient and small. The transformer is an honorary electrical "machine" in which the flux changes occur by variation in currents with time, instead of by motion.
The diagram shows the usual schematic way to represent a transformer. In an actual transformer, the windings are wound on top of each other, not on separate legs, to reduce leakage flux. In the usual shell-type transformer, both primary and secondary are on one leg, and are surrounded by the core. A core-type transformer has windings covering the core legs.
In order to design a transformer, or to examine in more detail how it departs from ideality, it is necessary to understand how a transformer works, not just how to express its terminal relations in an approximate way. It is also important to know how the properties of the iron core affect the performance of the transformer. A real transformer becomes hot because of losses, and the ouput voltage may vary with load even when the primary voltage is held constant.
The mutual flux φ is the means of transfer of energy from primary to secondary, and links both windings. In an ideal transformer, this flux requires negligibly small ampere-turns to produce it, so the net ampere-turns, primary plus secondary, is about zero. When a current is drawn from the secondary in the positive direction, ampere-turns decrease substantially. This must be matched by an equal increase in primary ampere-turns, which is caused by an increase in the current entering the primary in the positive direction. In this way, the back-emf of the primary (the voltage induced in it by the flux φ) equals the voltage applied to the primary, as it must. This fundamental explanation of the operation of a transformer must be clearly understood before proceeding further.
Magnetic Fields
Since magnetic fields play an essential role in transformers, we must review them first. They exert forces on magnetic poles, but here the important thing is that they induce electric fields when they move. A magnetic field is produced surrounding any flow of electric charge, or current. The relation between the magnetic field and the electric current can be most compactly expressed by Ampère's Law, which states that the line integral of the component of the magnetic field along a closed curve is proportional to the current linked by the closed curve. The value of this line integral is called the magnetomotive force, or mmf, around the curve. There are equivalent formulas that give the magnetic field at a point directly in terms of the currents. These also require integration, and are much more difficult to use in the simple cases we require.
The question of units always comes up, and the question is not a pretty one. We'll treat the matter in some detail here, because it is a frequent source of confusion and anguish. If we wish simple formulas, without a lot of constants, we have to massage the units so that it all comes out straight. If we make the choice shown in the box, and measure current in units called absolute amperes, or abamperes, then the magnetic field H comes out in units called oersteds if distance is in centimetres. The value of the mmf can be called so many gilberts if you like, and then oersteds are gilberts per centimeter. The basic concept is current linked per unit length of path, however expressed. We presume that there is some way to define the absolute ampere independently of Ampère's Law.
The unit for current you know is probably ordinary amperes. Then, if we use an even simpler expression for the mmf, omitting the 4π, and use metres for distance, we get a magnetic field in amperes per metre. Here, amperes means the total current. If we have N turns of wire carrying I amperes, we have NI amperes total linking the magnetic field. Hence, people usually speak of ampere-turns in this case. It is the omission of the 4π that makes things confusing, though you can easily work out the relation between ampere-turns/metre and oersted by calculating the field for the same current in the two cases, and comparing the results.
Less weird, perhaps, was the practice in American electrical engineering of using the inch as the length unit. Now we have gilberts per inch for the magnetic field, which is not an oersted, but 1/2.54 of an oersted. One can also speak of ampere-turns per inch in the same way.
Now there is the matter of magnetic flux. Flux is another manifestation of the magnetic field, defined by its ability to induce an electric field when it changes. The basic relation is Faraday's Law. Note the analogy with Ampère's Law. The negative sign (another fertile source of confusion) is there so energy will be conserved, and depends on the positive directions shown in the figure. This requires that if φ is increasing, any current that flows due to the induced voltage must act to decrease φ (Lenz's Law). Check that this is so in the diagram.
The integral of the electric field around a closed curve is called the electromotive force, emf, and is probably a more familiar quantity than the mmf. As in the case of Ampère's Law, we presume that there is some independent way to define an emf, and the unit here is the abvolt. This unit is, in fact, defined by Coulomb's Law, at least in principle. Now, we choose the simplest relation for Faraday's Law, emf = -dφ/dt. The unit of flux is such that a change of one unit per second induces an emf of one abvolt. The unit is called a line or a maxwell. Let's use line here, and remember that a magnetic field is not made of physical lines of anything. "Line" is just a conventional name, recalling the representation of a field by lines of force. The flux density B is in lines per square centimetre, units called gauss with G as the symbol.
We now have two quantities, the abampere of current or mmf, and the abvolt of potential or emf, defined in terms of the forces between currents, and between charges, respectively. We also have two definitions of magnetic field, one in terms of mmf per unit length, and the other in terms of emf times seconds. There is really only one magnetic field with two manifestations or properties, though early mechanical theories of electromagnetism considered that one caused the other, like stress caused strain. This interpretation has been fossilized in the Giorgi system of units that is now standard in engineering, and which has caused a great deal of confusion and unnecessary anxiety, especially in magnetism. Things are actually simpler than they have been made out to be.
The big advantage of the definitions of the magnetic field in terms of the abampere and abvolt is that the magnetic field H in oersted comes out to be exactly the magnetic flux density B in gauss. Naturally: they are the same thing, after all. In all the other systems we have skirted around, this is not true. You already know that B = μ0H in Giorgi units. The μ0 is simply a number, not a physical quantity, that adjusts for arbitrary choices of units. It has no physical interpretation whatsoever.
We are rightly very attached to our volt and ampere, the practical units arbitrarly defined by the potential of a Daniell cell and the resistance of a column of mercury early in the history of electromagnetism. Giorgi found that if we took 10 ampere = 1 abampere, and 1 volt = 108 abvolt, approximately, that everything came out wonderfully consistently, so long as the metre was the length unit, not the centimetre. There were some small adjustments in the previous practical units, but this was a small price to pay for having them join an absolute, very scientific, system of units. The opportunity was also taken to remove the factors of 4π from some equations, and add them to other equations, which has complicated things. For any overall view, it is best to consider Maxwell's Equations and the force equations, but what we have done here will do us.
The older way was to use oersted and gauss, with volts and amperes, and inches as the length unit, at least in English-speaking countries. The gauss and oersted have not disappeared, even with Giorgi units everywhere else. In fact, 104 G = 1 Weber, and 1 Tesla = 108 lines. Obviously, 1 T/s = 1 V. If it weren't for the damned 4π, all the new units would differ only by powers of 10 from the old! The Giorgi system is just the old absolute electromagnetic system, tarted up, really. All the fuss over units is actually a blessing; when you understand the units, you will understand electromagnetism.
We all are comfortable with an electric circuit. The current stays in the wires, and the applied emf is equal to the sum of the voltage drops, and the sum of the currents entering a junction is zero. A resistive element has a voltage drop that is proportional to the current. This analogy to friction is actually rather complicated, but at least it looks simple, and all we need is the resistance, R = l / σ A, where σ is the conductivity of the material of uniform cross section A and length l. What happens is that the electric field resulting from the emf causes free electrons to move along the wire. There is an electric field outside the wire as well, but it finds no movable electrons.
An mmf also causes a magnetic field to occur, and it usually just spreads out in space. There are no little free magnetic poles for it to act on, but in iron there are electrons that can be called to attention, so that they all point in roughly the same direction, and hold themselves there by atomic forces. The electrons are spinning charges, and the net effect is that of a lot of small current loops causing a very strong magnetic field. That is, the H field associated with the mmf calls up a total magnetic field in the same direction that we must call B, because it will certainly induce an electric field according to its magnitude. H is only a partial field, produced by whatever makes the mmf, while B is the total field. This is the usual interpretation of H and B, which is quite useful. H and B are not really different kinds of field, except in the imagination. In the Giorgi system, they are expressed in different units, however, which has no physical significance.
So, an mmf is produced by ampere-turns, and in a uniform iron core creates a uniform H field, which snaps the electrons into order making a flux density B μ times larger, and a total flux φ = BA. Just as in the electrical case, there is a magnetic field outside the iron core, but no electrons are there to join the army, so not much flux results. The small flux that does go this way is called leakage flux. Leakage flux is important in transformers. The whole thing is remarkably like a circuit, with mmf in place of emf, and flux in place of current. The analog of resistance is called reluctance, given by the same formula as resistance, but with permeability replacing conductance. The ratio of the permeability in the iron to that in air is about 1000. For electric currents, the ratio of conductivities is a million times greater.
The ideal is to have all the magnetic flux produced by one winding link the other. This is not attained in practical transformers. Flux that does not link both windings is called leakage flux and has the effect of adding an inductance that produces a voltage drop when current is present. Leakage is affected by the shape of the core and by the arrangement of the windings. To reduce leakage, the core must be compact and the windings close together.
Iron
The conductivity of a conductor is practically constant, independent of the current. The permeability of iron is by no means constant. The flux density B of a sample is a nonlinear function of H, given by the magnetization curve. Worse than this, the flux density depends on what has happened in the past, as well as to the current value of H. If H is varied sinusoidally, B follows an S-shaped curve called the hysteresis loop, which is different for every different maximum value of H. B lags behind H in its variation, and energy is lost in a kind of internal friction. The energy lost in each cycle is proportional to the area inside the hysteresis loop. The slope of the approximately linear portions of this loop is the value to be used in estimating the reluctance, which we will discuss later. It is not, incidentally, the slope of the magnetization curve. Refer to books on magnetism for a full discussion of magnetization and hysteresis. There is a lot of lore here.
An ideal core material would have a hysteresis loop with no area inside it. In this case, B would be a single-valued function of H, as shown in the diagram on the left. At high values of H, when all the electrons were aligned, the curve would bend over and saturate. Although the iron saturates, the flux continues to increase as in air, so the B curve is never horizontal. It is most economical to operate magnetic materials at as high a flux density as possible, usually near the knee of the curve. Note that the curve in the figure is a hysteresis curve (alternating flux), not a magnetization curve.
In actual magnetic materials, the flux does not drop to zero when the mmf returns to zero, but there is a remanent flux. To reduce the flux to zero, an mmf in the reverse direction called the coercive force must be applied. Now the B vs H curves are moved left and right, without changing their shapes much. The horizontal distance between the curves is twice the coercive force. The ends of the loop depend on the maximum flux density, so there are different hysteresis curves for each maximum flux density. Since the flux lags the current, there is now component of the voltage in phase with the applied current, which means an energy loss, the hysteresis loss. The amount of loss, in ergs per cycle per cubic centimeter, is the area within the hysteresis loop divided by 4π. (Axes are gauss and Oersted.)
If a sinusoidal alternating voltage is applied to the winding creating the emf, the flux is sinusoidal, but the current is not. The current must vary in just such a way as to produce the sinusoidal flux, which is 90° out of phase with the applied voltage. The figure shows a sketch of the current waveform that results. This current, called the magnetization current, will be almost 90° out of phase with the applied voltage, as shown, with a small in-phase component representing hysteresis loss. The magnetizing current shows a strong third harmonic component. The prominent peaking at maximum B is the result of the bending over of the magnetization curve. Note that the flux is a maximum when the voltage is zero, and that the coercive force can be observed in the current waveform, showing that there is a component of the current in phase with the voltage, representing an energy loss. A transformer is designed so that the magnetizing current is a small fraction of the normal load current, usually only a few per cent. Since the flux is accurately sinusoidal, the secondary voltage (and current) will also be sinusoidal. The peculiarities of the magnetizing current will not be reflected in the transformer output, and will not cause distortion.
The magnetizing current can be observed with an oscilloscope using the circuit at the right. The voltage across the 10Ω current-sensing resistor will be negligible. Channel 1 will be set to 50 V/division, Channel 2 to perhaps 0.2 V/division. If you feed the circuit with a Variac, make QUITE sure that the connection shown grounded is indeed grounded, so it will not conflict with the oscilloscope ground. To be safe, use an isolation transformer. If you know the constructional details of the transformer tested (see the paragraph below on dissecting a transformer), you can find B from the voltage, number of turns and core area, while H can be found from the current, number of turns, and magnetic circuit length. This makes it possible to sketch a hysteresis curve for the transformer core.
As the flux φ varies with time, there will certainly be an induced electric field in the core, and if it is conducting, currents will flow. These are called eddy currents, and are precisely analogous to a short-circuited secondary winding. They will cause heating, and will represent an energy loss. They are minimized by decreasing the conductivity of the iron (silicon iron has a smaller conductivity), and by making the core from a pile of thin laminations insulated from each other. They cannot be reduced to zero, but can be made quite small. Like hysteresis, eddy currents will produce a current in phase with the applied voltage, that is included in the magnetization current. Iron losses, as they are called, are reflected in the in-phase component of the magnetization current.
Ferromagnetic materials saturate at a certain flux, as has been mentioned, and are usually used at fluxes not far below saturation. If the primary of a transformer carries direct current, which is not resisted by inductance but only by the resistance of the winding, which is always low, it is easy to drive the core into saturation for at least part of the cycle. The result is excessive current, first appearing as current spikes, that causes the transformer to heat, and, of course, the transformer action to fail. Transformers must not be subjected to direct currents. This includes secondary windings as well as primary windings, of course.
Some iron-core inductors are designed to carry direct currents. These are chokes that block alternating currents. To prevent saturation, the core is provided with an air gap. Even a small air gap dominates the reluctance and prevents saturation at reasonable currents (the flux is proportional to the ampere-turns). The flux density is kept lower than the saturation value for the iron, of course, but now the inductance is independent of the direct current. In audio transformers for push-pull amplifiers, direct currents flow in the divided primary in opposite directions, so that their mmf's cancel, while the alternating audio signal flows in the whole primary.
Losses and Iron
Early transformers used pure iron for their cores. Pure iron has a hysteresis loss of about 600 ergs per cycle per cubic centimeter, a relatively small value, and a coercive force of 0.2 Oe from 10 kG, so it is magnetically quite "soft." Commercial sheet iron has a larger, but still small, hysteresis loss. However, it is a fairly good electrical conductor, with a resistivity of 7.64 μΩ-cm. Therefore, it acts like a short-circuited turn, and currents are induced that circle round the magnetic field as it changes. These eddy currents dissipate large amounts of energy, making a solid core very inefficient. This can be overcome by laminating the core, and interrupting the possible current paths. Bundles of wire were used as cores for this purpose, but thin sheet laminations are now the universal practice. The eddy-current loss is proportional to the square of the thickness of the laminations. It's proportional to the squares of the frequency and the maximum flux density as well.
Around 1903, Sir Robert Hadfield discovered that adding silicon to steel greatly increased its resistivity, while affecting the magnetic properties very little. A silicon steel with 4.25% Si has a resistivity of 60 μΩ-cm. The hysteresis loss is about 1800 ergs per cycle per cubic centimeter, and the coercive force is about 0.4 Oe. The only drawback of silicon is that it makes the steel brittle in such concentrations, making punching and other mechanical operations more difficult. The laminations must be insulated from each other, of course. It seems that originally rust was the insulation, but since rust contributes greatly to hysteresis loss, laminations are now pickled and insulated by a compound. The density of silicon steel is 7.5 g/cc for Si < 2%, and 7.7 g/cc for Si > 2%.
Serendipitously, it was found that steel containing more than 1% Si was not subject to the phenomenon of "ageing" in which the coercive force (and so the hysteresis loss) increased as much as 100% with time, when the iron operated at temperatures above 150 °F (as is often the case) or was subject to mechanical actions. All transformer cores for power frequencies, and parts of rotating machinery subject to alternating fields, are now made of silicon steel, for this and the other reasons mentioned.
With the eddy-current loss problem solved by silicon steel and laminations, the larger part of core loss is hysteresis loss, which is proportional to the frequency (since a fixed amount of energy is lost in each hysteresis cycle). Steinmetz found that the hysteresis loss was W = A B1.6, where B is the maximum flux density. The constant A is about 0.001 for silicon steel, 0.013 for cast iron, and 0.004 for sheet iron. The large value for cast iron should be noted. Magnetic cores for early rotating machines were typically cast iron, which has a relatively high electrical resistance (reducing eddy-current losses) but large hysteresis loss. The eddy-current and hysteresis losses can be separated experimentally by tests at different frequencies, because of their distinct frequency dependences.
Core losses generally determine the power ratings of large transformers, which often must be artificially cooled by forced ventilation or circulating fluids. It is more efficient to design transformers for high maximum flux densities, but this also increases the core losses.
Copper losses are the I2R losses in the windings, and are relatively more important in small transformers than in large. It was the practice to design windings at 700 c.m./A (c.m. = circular mils, the square of the diameter in thousandths of an inch), or alternatively at 500 c.m./A. Many transformers are now designed with as little copper as 340 c.m./A.
Transformer Operation
In a real transformer, we must include the magnetizing current, as well as the effects of leakage flux. Leakage flux links only one winding, and is completely independent of the other. Its effect is to act like an inductance in series with a winding. In addition, there is the IR drop in the copper of a winding when it carries current. These are the things that make a real transformer behave differently from an ideal transformer. If we just talk about these things, we soon are mired in a swamp of phase angles and relative directions that make understanding very difficult. The best way to understand things is to draw a phasor diagram of the currents, voltages and fluxes. In doing this, we draw the diagram in stages, showing how to proceed step by step so that it would be possible to make accurate numerical estimates of each quantity. The magnetizing current is not well-represented by a phasor, since it is nonsinusoidal, but this causes no serious inaccuracy, since most of the quantities in a transformer under load are sinusoidal.
Before a phasor diagram can be drawn, a circuit diagram showing the connections and the positive directions of each circuit quantity is necessary. Such a diagram is shown at the right. It is not really complete, since it does not show the connection through the flux φ that reflects equal induced voltages and currents from secondary to primary. This is hinted at by the generators marked E. When the phasor diagram is drawn, we will reverse the positive directions in the primary to separate the two parts of the diagram so that they will not fall on top of each other. The diagram shows the winding resistances R, and the reactances due to the leakage fluxes X, as well as a current generator for the magnetizing current I0. We will assume that the turns ratio is unity for convenience, so the vectors turn out to be about the same lengths on the same scale. If the turns ratio were not unity, the voltages and currents would be reflected into the primary multiplied by the proper ratios.
The phasor diagram for the transformer at full load is shown at the left. This complicated diagram can be understood by following through how it was constructed. We start by assuming that the secondary is supplying a current I2 at a terminal voltage V2 with phase angle θ. These are the first two phasors drawn. Now, to V2 we add I2R2 in phase with I2, and I2X2 in quadrature, to find the induced voltage E in the secondary. The voltage -E is induced in the primary. The flux φ is at right angles to E, as shown. The current I0 is necessary to create the flux, and is drawn with its proper relation to φ. The current I2 is reflected to the primary as -I2, and added to I0 to find the total primary current I1. Now that we know I1, we can add I1R1 and I1X1 to -E to find the primary terminal voltage, V1. In this diagram, the magnetizing current, and the voltages due to leakage flux and winding resistance, are greatly magnified so their effects can easily be seen. As an exercise, draw a phasor diagram for an open secondary, when I2 is zero, and find V1 for this case. Also, draw diagrams for various phase angles θ.
Try to visualize what happens when the secondary current decreases. It may help to draw additional phasor diagrams. The difference between the terminal voltage V2 and the induced voltage E becomes less, and they approach one another. The magnetizing current becomes a larger part of the smaller primary current, which changes in phase accordingly. The flux, of course, remains the same. The input voltage decreases, and becomes equal to V2. One of the important characteristics of a transformer is its regulation, the change in output voltage between no load and full load, divided by the full-load output voltage (or some such definition), expressed as a percentage. This is practically the same as the difference between the full-load and no load input voltages divided by the no-load input voltage, holding the output voltage constant, and this latter quantity can be read off the full-load vector diagram. The no-load input voltage is V2, and the full-load input voltage is V1. Typical power transformer regulation is around 1% or even less at unity power factor, falling off to a few percent at 0.8 pf. Such transformers are called constant-potential transformers. Regulation is improved by decreasing leakage flux and winding resistance. If poor regulation is desirable, it can be obtained by increasing the leakage flux without added resistive heating.
The efficiency of a transformer is quite high, usually over 98% at full load. It increases at partial load, sometimes to over 99%. A real transformer is not very far from an ideal transformer, in fact. These figures for efficiency and regulation apply to large power and distribution transformers, say above about 5 kVA capacity (product of rated voltage and current) with silicon steel cores. Smaller transformers used in electronics and consumer products are slightly less ideal, largely because of economy in use of copper and the proportionately greater leakage flux. It is difficult to measure the efficiency of a transformer directly (by comparing output with input), so the copper and core losses are usually measured or estimated instead. Core losses are usually only about 1/3 of the total.
Transformer Design
Transformer design is a well-developed subject, and the reader is referred to the many textbooks that treat it. It is important not simply to design a transformer that will do, but one that is economical, efficient and makes the best use of available materials. However, it is not difficult to design a serviceable transformer, and by understanding how this is done we will know better why transformers are as they are. As in all design processes, there are numerous trade-offs between competing requirements, and some sort of optimum is sought. On the other hand, the properties of available materials rigidly limit the possibilities.
The size, shape and material of the core must be chosen, and the number of turns of the primary and secondary windings. The size of the wire and the insulation determines if the windings will fit in the space available. The windings must be arranged for minimum leakage flux. The insulation determines the permissible temperature rise, and means of cooling must balance the loss of energy under load. We will treat only the core and the number of turns, the fundamental parameters, in our analysis. A complete design must, of course, include all these features.
Two basic equations are used in transformer design. The first is essentially Faraday's law, E = dΦ/dt x 108 V, where Φ is maxwells (gauss times square centimeters). For transformers, this is written Bmax = √2 E 108 / 2πfNAK, where E = rms voltage in a winding, N = number of turns, f = frequency (Hz), A = area of core (cm2), K = stacking factor (the proportion of A occupied by iron). A sinusoidal variation in flux is assumed, which is a reasonable assumption when the core does not saturate, but by no means exact. Bmax is generally assumed as a design parameter, as well as a value for E/N in volts per turn, and the necessary A results. A typical value for a small transformer is E/N = 0.1 V/turn.
The second is Ampere's law, H = 0.4πNI/l, where l is the length of the magnetic circuit. From the value of Bmax, the value of Hmax can be found from the magnetization curve of the core material, and this equation used to determine N, when a reasonable value of I is assumed (say, 5% of the full-load current).
Let's try to proportion a transformer for 120 V, 60 Hz supply, with a full-load current of 10 A. All the AC values we use will be effective values. The core material is to be silicon-steel laminations with a maximum operating flux density Bmax = 12,000 gauss. This is comfortably less than the saturation flux density, Bsat. The first requirement is to ensure that we have sufficient ampere-turns to magnetize the core to this level with a permissible magnetizing current I0. Let's choose the magnetizing current to be 1% of the full-load current, or 0.1 A. The exact value is not sacred; this might be thought of as an upper limit. From past experience, we should have some idea of the size of core that will be required. Here, we will assume a simple, uniform magnetic circuit for simplicity. In an actual case, a more complicated magnetic circuit would have to be considered. If l is the length of the magnetic circuit, H is 0.4πN(√2I0)/l, and the magnetization curve for the core iron gives the H required for the chosen Bmax. From this, we can find the number of turns, N, required for the primary.
We could also estimate the ampere-turns required by using an assumed permeability μ. Experience will furnish a satisfactory value. It is not taken from the magnetization curve, but from the hysteresis loop. Let's take μ = 1000. Then, N = Bmaxl / 0.4π√2 μI0. If we estimate l = 20 cm, the number of primary turns required is N = 1350. The rms voltage induced per turn is determined from Faraday's Law: √2 e = (2πf)BmaxA x 10-8. Now, e must be 120 / 1350 = 0.126 V/turn, f is 60, and Bmax = 12,000 gauss. We know everything but A, the cross-sectional area of the core. We find A = 2.8 cm2.
This may, or may not, be an acceptable result. If not, we simply change our assumptions and try again. Design, after all, is an iterative process. By considering the above calculations, we can appreciate the changes that a different frequency, voltage and current rating, permeability or maximum flux density would produce. The number of secondary turns is now easily found from the volts/turn and the desired ratio. A 24 V secondary would have 24 / 0.126 = 190 turns.
The number of primary turns is determined so that the magnetizing current is limited to an acceptable value, and depends on the length of the core. The area of the core is determined by the required volts per turn, now that the total number of turns is known. These are the things that determine the size and weight of a transformer.
Toroidal ferrite or powdered iron cores are now easily available. Ferrite is a high-permeability, high-resistance material that has acceptable losses at high frequency. Powdered iron has granules insulated from each other for high resistivity (low eddy-current losses) and is good at moderate frequencies. Such cores can be used to experiment with transformer design, using a signal generator as a power source. The length, area and permeability of the core is now known at the beginning, making the above calculations somewhat easier. An oscilloscope can be used for measurements.
Powdered iron and ferrite cores have low Bsat and permeability values. A type 43 ferrite has Bsat = 2750 gauss, but a maximum permeability of 3000, and is recommended for frequencies from 10 kHz to 1 MHz. Silicon iron is much better magnetically, but cannot be used at these frequencies. The approximate dimensions of an FT-114 ferrite core (of any desired material) are OD 28 mm, ID 19 mm, thickness 7.5 mm. The magnetic dimensions are l = 74.17 mm, A = 37.49 mm2, and volume 2778 mm3. Similar information is available for a wide range of cores. There are tables showing how much wire can be wound on them, and even the inductance as a function of the number of turns. Consult the ARRL Radio Amateur's Handbook, or specifications from Amidon Associates.
Transformers with air cores are used in radio work. They have no core losses, and are not limited in frequency (which is why they are used). A little calculation will show how hopeless an air-core transformer is at power frequencies. Air core transformers have large leakage fluxes, which cannot be avoided, and therefore poor regulation. Windings must be carefully designed to give the largest mutual flux possible.
Dissecting a Transformer
There is a silver lining if you happen to burn out a small transformer--you can take the transformer apart to see what is inside, and analyze its design. I disassembled a transformer with a 24 VCT, 1 A secondary and 117 V primary. The primary had a resistance of 48.8Ω, the secondary 2.7Ω. The first step is to remove the mounting U bracket by bending back the tabs holding it in place. This leaves you with the core and windings, with the core tightly held by the plastic bobbin holding the windings. The windings on this transformer are wound one over the other, the secondary on top of the primary. Some transformers now use separate bobbins for primary and secondary, which are stacked on the core. This arrangement offers advantages in manufacture, but the leakage flux is probably greater than with windings one over the other.
The next step is to remove the core, which is done lamination by lamination. The core consists of E and I pieces, alternating so that successive I pieces do not fall one over the other. A suitable tool for the disassembly, which is not easy, is a small, sharp cold chisel driven by a light hammer. The first E lamination can be driven out by using the chisel at the edge of the bobbin. The laminations are glued together by the insulating substance, and have to be pried apart before they can be removed. After the first E lamination on each side is removed, things become a little easier. The idea is to chisel off the I piece, then pry the E piece away so that it comes out. I discovered 37 laminations in the core (37 E, 35 I). The area of the magnetic circuit was A = 2.88 cm2, and its length l = 9.6 cm, found by measuring the laminations. The stacking factor K could not be determined accurately, so we shall assume that it is 1.00 without much error. The laminations were 0.50 mm thick, close to USS gauge #26. Use the DMM to verify that the surfaces of the laminations are not conducting. The total weight of the core was 224 g, about half a pound.
The two general configurations of iron and copper in a transformer are shown at the right. Most small power transformers are shell-type, in which the iron surrounds the copper. The ideal would be for the iron to completely surround the windings, but this is impractical. The compromise is to divide the magnetic circuit into two return paths on opposite sides of the core, as is done with the E and I laminations, as shown. Another configuration, the core type, in which the copper surrounds the iron, is common with larger high-voltage transformers, as well as with small ferrite-core toroidal transformers. Note that both primary and secondary windings are placed on each leg. The laminations are L-shaped to make a core of uniform cross-section with a central rectangular window. Core transformers are easier to insulate and to cool. If the primary winding were placed on one leg and the secondary on the other, as in the diagrams of transformers in texts, leakage flux would be excessive, and the transformer would have poor regulation (variation of voltage with load). There are other special types of construction, but all can be classified as shell or core.
Note how connections are made, and the provision for insulation. When the outside insulating material has been removed, the secondary winding can be unwound. Note the number of turns, which came out at 240 for my transformer, so that there are 10 turns per volt rms. It is now possible to estimate Bmax from Faraday's law, with the result 13 000 G, a quite reasonable result. The primary should have 1170 turns (I did not unwind the primary to find out--this would result in far too much fine wire). The magnetization curve shows that about 9.5 Oe is required to produce this flux. From Ampere's law, the primary magnetizing current should be 44 mA (rms), a reasonable figure. The primary inductance is about 7 H, from this result. The actual transformer had a measured magnetizing current of 63 mA rms. This is a relatively large value, but of little consequence in a transformer of this size.
The primary winding consisted of 0.50 mm enameled wire, about AWG #25, while the primary was 0.21 mm, about AWG #31. From the current rating of the transformer, these sizes correspond to 340 c.m./A, a somewhat adventurous figure. Note that the size of the wire has nothing to do with the magnetic design of the transformer, only with the copper loss and heating, and determines what will fit in the space provided by the core. The matter of insulation between the windings is of importance. In this small transformer, the critical point appeared to be where the primary wires were led out past the secondary winding. An extra piece of insulating cloth was used to protect the leads.
Using Transformers; Phasing
Three rules should always be observed in using transformers: first, a voltage greatly in excess of the rating should not be applied to any winding; second, a significant direct current should not be allowed to flow through any winding not designed for it; third, the frequency should not be significantly less than the design frequency. If you make the primary of a small 120 V transformer carry, say, 50 mA direct current, and then connect it to the 120 V line, you will have smoke and combustion! Applying 120 V to a 12 V secondary in hopes of getting 1200 V at the primary will have a similar result, accompanied by insulation failure which will add further fireworks. Merely overloading a transformer is not so serious--it will simply get hot and eventually burn up, or, more likely, a wire will fuse. There is normally little chance of running across a lower power frequency, such as 25 Hz. A 60 Hz transformer will draw too much magnetizing current on 25 Hz, and will run quite hot.
However, you can always apply less than the rated voltage to any winding, and use any winding as primary or secondary, so long as the current rating is not exceeded. A few milliamperes of direct current is no problem, either. Any frequency higher than the rated frequency is also safe.
Many transformers are made with duplicate windings, to permit flexibility in operation. Two 110 V primaries can be connected in parallel for 110 V, or in series for 220 V. Even more common are dual secondaries, which can be connected in series or parallel. In parallel, there will be a center tap as well. The windings cannot arbitrarily be connected to each other, because of the problem of phasing. The voltages across the windings will either be in phase, which is the desired state, or in antiphase. If connected in antiphase, the result is a short circuit.
Two windings on a core are shown at the right. The black dot at one terminal or the other of each winding is called a phase or polarity mark. Currents entering the marked terminals cause magnetic flux in the same direction in the core. An increasing current entering the marked terminal as shown will cause a positive voltage at the marked terminal of the other winding. If the unmarked terminal of the winding on the left is connected to the marked terminal of the winding on the right, the two windings will be in phase and their ampere-turns will add. If they are connected in the opposite sense, their ampere-turns will cancel, and there will be no inductance. Unfortunately, transformers are not provided with phase marks, and you have to figure them out for yourself. This process is called phasing.
Of course, most transformers come with numbered terminals and wiring diagrams so that phasing is not required. Still, it is a good idea to check the phasing in any case, and to know how to do it in case you have a strange transformer without instructions. An obvious method is to use an oscilloscope and a function generator, which is easy and completely safe.
However, it is possible to phase a transformer with a voltmeter alone. I use a Variac to apply a low voltage, but a series resistance can be used to limit the current in case of an unexpected short. Apply a voltage to one of the windings, here winding 1-2. We can arbitrarily assign a phase mark to one end of this winding, say terminal 1. Connect the other end of this winding to one end of the other winding. There are only two possibilities, that we have connected to the marked or the unmarked terminal, as shown. Now measure the voltage between terminal 1 and the free terminal of the other winding (this cannot result in a short circuit because of the high resistance of the voltmeter). If you find twice the applied voltage, then conditions are as at the left. If you find zero volts, then conditions are as at the right. Now the second winding can be marked, and the proper connections determined.
To phase the secondaries, measurements can be made in exactly the same way. One winding is marked arbitrarily, the other connected to its other end, and the voltage measured to the open terminal. The secondaries can be phased relative to the primary just as if they were more primary windings, provided low voltages are used and the turns ratio taken into consideration. In all these measurements, we make use of the fact that the phase relations can be only in phase or in antiphase, which give distinctly different resultant voltages.
Autotransformers
In the diagram at the right, a step-down transformer is shown. The secondary voltage is half the primary, so the currents are doubled, in this example. The ampere-turns of the secondary, carrying 20 A, are opposed by the same number of ampere-turns of the whole primary. In the middle diagram, the secondary load is shown attached to the primary winding at its mid-point. Now, the ampere-turns below the connection are opposed by the ampere-turns above the connection, each excited by 10 A, but in opposite directions. The currents add at the node to give 20 A in the secondary load, as required. Conditions are just as in the case of separate windings, but without the secondary winding carrying 20 A. This is called an autotransformer, and the savings involved in its use are obvious. The disadvantage, of course, is that the primary and secondary are no longer isolated and share a metallic connection. In many applications, this is no matter, and an autotransformer can be used.
At the right is shown a "Variac" (a trade name belonging to General Radio Corporation, but now used like Kleenex and Vaseline) or variable transformer, a useful device that provides an adjustable AC voltage. For safety, the common connection must be connected to the grounded or white wire of the AC lines. If, somehow, the connection is reversed, then everything connected to the Variac becomes "hot" and hazardous. The common 120/240V converters are autotransformers, and present the same hazard.
Some early AC/DC radios (the name for those without power transformers) had only one wire in the line cord, the hot wire. The other connection was to be made to a handy pipe, which was the service ground. If the plug were inserted incorrectly, no power would be supplied, and everything would be safe. With today's polarized plugs, this is not necessary--unless an amateur electrician wired the receptacle
3 Phase Electrical Power Transformer
A 3 phase transformer, there is a three-legged iron core as shown below. Each leg has a respective primary and secondary winding.
Most power is distributed in the form of three-phase AC. Therefore, before proceeding any further you should understand what is meant by 3 phase power. Basically, the power company generators produce electricity by rotating (3) coils or windings through a magnetic field within the generator . These coils or windings are spaced 120 degrees apart. As they rotate through the magnetic field they generate power which is then sent out on three (3) lines as in three-phase power. 3 phase transformers must have (3) coils or windings connected in the proper sequence in order to match the incoming power and therefore transform the power company voltage to the level of voltage we need and maintain the proper phasing or polarity.
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3 Phase Power Is More Efficient Than Single Phase
Three phase electricity powers large industrial loads more efficiently than single-phase electricity. When single-phase electricity is needed, It is available between any two phases of a three-phase system, or in some systems , between one of the phases and ground. By the use of three conductors a 3 phase system can provide 173% more power than the two conductors of a single-phase system. Three-phase power allows heavy duty industrial equipment to operate more smoothly and efficiently. 3 phase power can be transmitted over long distances with smaller conductor size.
Also read about 3 phase isolation transformers here. For an excellent source for these all transformer types check out TEMCo 3 phase transformers. Or check with Isolation Transformer Sales for 3 phase isolation transformers. These two companies manufacture some of the most recognized high quality 3 phase transformers available today.
In a three-phase transformer, there is a three-legged iron core as shown below. Each leg has a respective primary and secondary winding.
The three primary windings (P1, P2, P3) will be connected at the factory to provide the proper sequence (or correct polarity) required and will be in a configuration known as Delta. The three secondary windings (S1, S2, S3) will also be connected at the factory to provide the proper sequence (or correct polarity) required. However, the secondary windings, depending on our voltage requirements, will be in either ?Delta? or a ?Wye? configuration.
3 Phase Transformer Delta and Wye Connections
In a 3 phase transformer, there is a three-legged iron core as shown below. Each leg has a respective primary and secondary winding.
3 Phase Transformer Winding Combination
As can be seen, the three-phase transformer actually has 6 windings (or coils) 3 primary and 3 secondary. These 6 windings will be pre-connected at the factory in one of two configurations:
Configuration 1. Three primary Windings in Delta and Three Secondary Windings in Wye
Note: These are the designations which are marked on the leads or terminal boards provided for customer connections and they will be located in the transformer wiring compartment.
In both single and 3 phase transformers, the high voltage terminals are designated with an “h” and the low voltage with an “X” Configuration 2. Three Primary Windings in Delta and Three Secondary Windings in Delta
Note: These are the designations which are marked on the leads or terminal boards provided for the customer connections and they will be located in the transforming wiring compartment.
In both single and three-phase transformers, the high voltage terminals are designated with an “H” and the low voltage with an “X”.
3 Phase Transformer Voltage in Delta and Wye Connections
Different brands of 3 phase transformers handle the windings in different manners. All Federal Pacific 3 phase transformers have their primary windings pre-connected in a Delta configuration. Therefore, when connected to a three-phase source, each primary winding will have the same voltage across it.
For Example: 480V 3 Phase Source If the secondary windings are also connected Delta then they have equal voltages across each winding. Of course, this voltage will be either higher or lower than the primary depending upon the “turns ratio”.
480V Primary Source with 240V Secondary Output @ 2/1 Turns Ratio (Delta-Delta)
Note: it is important to note that three-phase transformers with Delta-connected primaries when connected to a 30, 4-wire supply system do not utilize the 4th wire or neutral.
Wye: If the secondary is not connected in Delta it will be pre-connected at the factory as a Wye secondary. All Wye connections provide two voltages due to the common point or neutral connection. A typical rating would be 208/120V. The 208Y indicates the voltage between phases of the secondary windings.
For Example:
The 120 volt portion indicates the voltage from each phase to the common point or neutral
For Example:
This Phase-to-Neutral voltage in a Wye is always equal to the Phase-to-Phase voltage divided by
For Example:
Therefore a 3 phase transformer with its secondary connected in a Wye configuration for 208Y/120 volts will have the available: Common Three-Phased Transformer Voltage Combinations
Special Three Phase Delta Connected Transformers
There are certain situations where only a very small portion of a building loads require 120V single-phase . A special transformer is available and you should be familiar with it.
The 240 Volt 30 Delta Connected Secondary With 120 Volt 10 Lighting Tap
As you can see there is no point in a Delta at which an equal potential to all three lines and the grounded neutral can be made. This is a disadvantage of a Delta compared to a Wye secondary connection
This Delta secondary connection has only one winding (S3) with a neutral conductor. The mid-point of winding S3 is tapped which gives the XI and X3 to neutral a voltage reading of 120 volts. In a 3-phase system, winding S3 is the workhorse; it has to carry all the 120V lighting and appliance loads plus one-third of all the 3 phase loads. (The 120V loads must not exceed 5% of the nameplate KVA, and the total of the nameplate KVA must be derated by 30%). Winding S1 and S2 cannot carry any 120 volt loads as there is no neutral connection to these windings. Windings S1 and S2 can only carry one-third of the three-phase loads each, and the 240 volt single-phase loads. *Caution: A240 volt Delta connected transformer with a 120 volt neutral tap creates a condition called “high leg” As indicated in the above diagram, the voltage between Phase B (X2) and the neutral tap will be 208 volts; therefore, no 120 volt single-phase loads can be connected between X2 and the neutral tap.
Single Phase Transformers Connected to Form Three Phase Bank
Normally , when 3 phase is required, a single enclosure with three primary and three secondary windings wound on a common core is all that is required. However three single-phase transformers with the same rating can be connected to form a three-phase bank. Since each single-phase transformer has a primary and a secondary winding, then 3 single-phase transformers will have the required 3 primary and 3 secondary windings and can be connected in the field either Delta-Delta or Delta-Wye to achieve the required 3 phased transformer bank, as shown below.
3 Phase Transformer: Delta-Delta
Utilizing 3 single-phase transformers is normally not done because it is more expensive than utilizing 1 three-phase transformer. However, there is an advantage which is called the open Delta or V-Connection and it functions as follows: A defective single-phase transformer in a Delta-Delta 3 phase bank can be disconnected and removed for repair. Partial service can be restored using the remaining single-phase transformer open-Delta until a replacement transformer is obtained. With two transformers three-phase is still obtained, but at reduced power. 57.7 of original power. This makes it a very practical transformer application for temporary emergency conditions
Open Delta 57.7%
3 Phase Loads and Single Phase Loads
If the load is 3 phase, then both the supply and the transformer must be in three-phase. If the load is single-phase the supply can either be single or 3 phase but the transformer need only be single-phase with the primary being connected to two lines on the three phase circuit. With single-phase loads, an attempt to use a transformer with three-phase input and only one phase connected at the output to convert the loading on the line to 3 phase is not practical.
3 Phase Transformer Sizing with 3 Phase Loads
1) Determine electrical load
A. Voltage required by load. B. Amperes or KVA required by load. C. Frequency in Hz (cycles per second). D. Verify load is designed to operate on three phase.
All the above information is standard data normally obtained from equipment nameplates or instruction manuals.
2) Determine supply voltage
A. Voltage of supply (source). B. Frequency in Hz (cycles per second).
The frequency of the line supply and electrical load must be the same. A 3 phase transformer is selected which is designed to operate at this frequency having a primary (input) equal to the supply voltage and a secondary (output) equal to the voltage required by the load.
3) If the load nameplate expresses a rating in KVA, a transformer can be directly selected from the charts in the catalog. Choose from the group of transformers with primary and secondary voltages matching that which you have just determined.
A. Select a 3 phase transformer with a standard KVA capacity equal to or greater than that needed to operate the load.
B. Primary taps are available on most models to compensate for line voltage variations. (Refer to question #2 in the Transformer Question and Answer Section of Acme's marketplace.
C. When load ratings are given only in amperes, the following formulas below may be used to determine proper KVA size for the required transformer.
(1) To determine three phase KVA when volts and amps are known:
Three Phase KVA =Volts x Amps x 1.73 /1000
(2) To determine Amperes when KVA and volts are known: Amps = 3 Phase KVA x 1000 /Volts x 1.73
Three Phase Example
Question:
Select a transformer to fulfill the following conditions. Load is a three phase induction motor, 25 horsepower @ 240 volts, 60 Hz and a heater load of 4 kilowatts @ 240 volts single phase. The supply voltage is 480Y/277, three phase, 4 wire.
Answer: Compute the KVA required.
28.2 KVA =240 volts x 68 amps x 1.73 /1000 Heater - 4 KVA
A three phase transformer must be selected so that any one phase is not overloaded. Each phase should have the additional 4 KVA rating required by the heater even though the heater will operate on one phase only. So, the transformer should have a minimum KVA rating of 28.2 - 4 + 4 + 4 or 40.2 KVA.
A 480 delta primary - 240 delta secondary transformer may be used on a 4 wire, 480Y/277 volt supply. The fourth wire (neutral) is not Connected to the transformer. To not overload the transformer, a 45 KVA transformer should be selected.
NOTE: Any two wires of the 240 volts, 3 phase developed by the secondary of the transformer may be used to supply the heater. Any 2 wires of a 3 phase system is single phase.
Three Phase Transformers Overview
Three phase transformers are used throughout industry to change values of three phase voltage and current. Since three phase power is the most common way in which power is produced, transmitted, an used, an understanding of how three phase transformer connections are made is essential. In this section it will discuss different types of three phase transformers connections, and present examples of how values of voltage and current for these connections are computed.
3 Phase Transformer Construction:
A three phase transformer is constructed by winding three single phase transformers on a single core. These transformers are put into an enclosure which is then filled with dielectric oil. The dielectric oil performs several functions. Since it is a dielectric, a nonconductor of electricity, it provides electrical insulation between the windings and the case. It is also used to help provide cooling and to prevent the formation of moisture, which can deteriorate the winding insulation.
3-Phase Transformer Connections:
There are only 4 possible transformer combinations:
Delta to Delta - use: industrial applications
Delta to Wye - use : most common; commercial and industrial
Wye to Delta - use : high voltage transmissions
Wye to Wye - use : rare, don't use causes harmonics and balancing problems.
3 phase transformers are connected in delta or wye configurations. A wye-delta transformer has its primary winding connected in a wye and its secondary winding connected in a delta (see figure 1-1). A delta-wye transformer has its primary winding connected in delta and its secondary winding connected in a wye (see figure 1-2).
Delta Connections:
A delta system is a good short-distance distribution system. It is used for neighborhood and small commercial loads close to the supplying substation. Only one voltage is available between any two wires in a delta system. The delta system can be illustrated by a simple triangle. A wire from each point of the triangle would represent a three-phase, three-wire delta system. The voltage would be the same between any two wires (see figure 1-3).
Wye Connections:
In a wye system the voltage between any two wires will always give the same amount of voltage on a three phase system. However, the voltage between any one of the phase conductors (X1, X2, X3) and the neutral (X0) will be less than the power conductors. For example, if the voltage between the power conductors of any two phases of a three wire system is 208v, then the voltage from any phase conductor to ground will be 120v. This is due to the square root of three phase power. In a wye system, the voltage between any two power conductors will always be 1.732 (which is the square root of 3) times the voltage between the neutral and any one of the power phase conductors. The phase-to-ground voltage can be found by dividing the phase-to-phase voltage by 1.732 (see figure 1-4).
Connecting Single-Phase Transformers into a 3 phase Bank:
If three phase transformation is need and a three phase transformer of the proper size and turns ratio is not available, three single phase transformers can be connected to form a three phase bank. When three single phase transformers are used to make a three phase transformer bank, their primary and secondary windings are connected in a wye or delta connection. The three transformer windings in figure 1-5 are labeled H1 and the other end is labeled H2. One end of each secondary lead is labeled X1 and the other end is labeled X2.
Figure 1-6 shows three single phase transformers labeled A, B, and C. The primary leads of each transformer are labeled H1 and H2 and the secondary leads are labeled X1 and X2. The schematic diagram of figure 1-5 will be used to connect the three single phase transformers into a three phase wye-delta connection as shown in figure 1-7.
The primary winding will be tied into a wye connection first. The schematic in figure 1-5 shows, that the H2 leads of the three primary windings are connected together, and the H1 lead of each winding is open for connection to the incoming power line. Notice in figure 1-7 that the H2 leads of the primary windings are connected together, and the H1 lead of each winding has been connected to the incoming primary power line.
Figure 1-5 shows that the X1 lead of the transformer A is connected to the X2 lead of transformer c. Notice that this same connection has been made in figure 1-7. The X1 lead of transformer B is connected to X1, lead of transformer A, and the X1 lead of transformer B is connected to X2 lead of transformer A, and the X1 lead of transformer C is connected to X2 lead of transformer B. The load is connected to the points of the delta connection.
Open Delta Connection:
The open delta transformer connection can be made with only two transformers instead of three (figure 1-8). This connection is often used when the amount of three phase power needed is not excessive, such as a small business. It should be noted that the output power of an open delta connection is only 87% of the rated power of the two transformers. For example, assume two transformers, each having a capacity of 25 kVA, are connected in an open delta connection. The total output power of this connection is 43.5 kVA (50 kVA x 0.87 = 43.5 kVA).
Another figure given for this calculation is 58%. This percentage assumes a closed delta bank containing 3 transformers. If three 25 kVA transformers were connected to form a closed delta connection, the total output would be 75 kVA (3 x 25 = 75 kVA). If one of these transformers were removed and the transformer bank operated as an open delta connection, the output power would be reduced to 58% of its original capacity of 75 kVA. The output capacity of the open delta bank is 43.5 kVA (75 kVA x .58% = 43.5 kVA).
The voltage and current values of an open delta connection are computed in the same manner as a standard delta-delta connection when three transformers are employed. The voltage and current rules for a delta connection must be used when determining line and phase values of voltage current.
Closing a Delta:
When closing a delta system, connections should be checked for proper polarity before making the final connection and applying power. If the phase winding of one transformer is reversed, an extremely high current will flow when power is applied. Proper phasing can be checked with a voltmeter at delta opening. If power is applied to the transformer bank before the delta connection is closed, the voltmeter should indicate 0 volts. If one phase winding has been reversed, however, the voltmeter will indicate double the amount of voltage.
It should be noted that a voltmeter is a high impedance device. It is not unusual for a voltmeter to indicate some amount of voltage before the delta is closed, especially if the primary has been connected as a wye and the secondary as a delta. When this is the case, the voltmeter will generally indicate close to the normal output voltage if the connection is correct and double the output voltage if the connection is incorrect.
Overcurrent Protection for the Primary:
Electrical Code Article 450-3(b) states that each transformer 600 volts, nominal or less, shall be protected by an individual overcurrent device on the primary side, rated or set at not more than 125% of the rated primary current of the transformer. Where the primary current of a transformer is 9 amps or more and 125% of this current does not correspond to a standard rating of a fuse or nonadjustable circuit breaker, the next higher standard rating shall be permitted. Where the primary current is less than 9 amps, an overcurrent device rated or set at not more than 167% of the primary current shall be permitted. Where the primary current is less than 2 amps, an overcurrent device rated or set at not more than 300% shall be permitted.
Example #1:
What size fuses is needed on the primary side to protect a 3 phase 480v to 208v 112.5 kVA transformer?
* Important when dealing with 3 phase applications always use 1.732 (square root of 3).
To solve: P / I x E
112.5 kVA X 1000 = 112500 VA
112500 VA divided by 831 (480 x 1.732) = 135.4 amps
Since the transformer is more than 9 amps you have to use 125 %.
135.4 X 1.25 = 169 amps.
Answer: 175 amp fuses (the next higher standard, Electrical Code 240-6).
Example #2:
What size breaker is needed on the primary side to protect a 3 phase 208v to 480v 3kVA transformer?
To solve: P / I x E
3kVA X 1000 = 3000 VA
3000 VA divided by 360 (208 x 1.732) = 8.3 amps
Since the transformer is 9 amps or less you have to use 167%.
8.3 X 1.67 = 13.8 amps
Answer: 15 amp breaker (preferably a 20 amp breaker)
Electrical Code Article 450-3(b)(2) states if a transformer 600 v, nominal, or less, having a an overcurrent device on the secondary side rated or set at not more than 125% of the rated secondary current of the transformer shall not be required to have an individual overcurrent device on the primary side if the primary feeder overcurrent device is rated or set at a current value not more than 250% of the rated primary current of the transformer.
Overcurrent Protection for the Secondary:
Electrical Code Article 450-3(b)(2) states that a transformer 600 v, nominal, or less, shall be protected by an individual overcurrent device on the secondary side, rated or set at not more than 125% of the rated secondary current of the transformer. Where the secondary current of a transformer is 9 amps or more and 125% of this current does not correspond to a standard rating of a fuse or nonadjustable circuit breaker, the next higher standard rating shall be permitted. Where the secondary current is less than 9 amps, an overcurrent device rated or set at not more than 167% of the secondary current shall be permitted.
Example:
What size breaker is needed on the secondary side to protect a 3 phase 480v/208v 112.5 kVA transformer?
To solve : P / I x E
112.5 kVA x 1000 = 112500 VA
112500 divided by 360 (208 x 1.732) = 312.5 amps
312.5 X 1.25 = 390.6 amps
Answer: 400 amp breaker
Transformer Protection- Existing practices and new trends
1.0 General: Transformers are the most important main equipment in any Power Transmission & Distribution network. The performance of the transformers depends upon how well they are maintained & protected against all possible fault conditions that can arise in the installation, in the network and the ambient environment.
The following sections describe the role of Protective relays in assuring the satisfactory performance of transformers both from fault clearance and maintenance point of views.
2.0 Types of Transformers : Transformers in a power system can be divided into three major categories:
Apart from the above one may come across special types of transformers like rectifier transformers, reactors etc.
Management of sub-stations with large transformers, from a remote location is becoming a major activity. This function is now being integrated into the protection system of transformers.
3.0 Protection philosophy : The type and extent of protection for for transformers depends upon :
a) the size and importance of duty performed
b) the location of the transformer in the power system
Transformers are two winding machines- hence they will need two sets of protections – one on the primary side and the other on the secondary side.
Transformers have to be protected both for external faults ( faults occurring outside the terminals of the transformer) as well as the Internal faults ( faults occurring within the transformer).
Normal over current + Earth fault relays are adequate for protection against external faults. Special relays like differential and REF relays are required for protection against internal faults. Sections 4,5 &6 explain the types of faults, method of protections for each fault, recommended types of relays etc.
Apart from fault conditions, which are severe abnormalities in electrical parameters, there are three major killers of transformers in the present day transmission system. These are :
a) Over load conditions : These produce excessive heat which causes rise in operation temperature. Every 10 degree rise in temperature ( beyond the withstand limit specified) results in 50% reduction in life of transformer insulation.
b) Single phasing conditions : There are increasing incidences of single phasing in transformers . The main reasons are poor maintenance of transmission lines and circuit breakers.
c) Unbalanced loads : Any unbalance in the three phase currents of a transformer will cause over heating , even if the currents are within rated values. Certain level of unbalance can be tolerated by transformer design – however we have to worry about unbalances more than 20%. Large unbalances can cause neutral shift , which may be harmful to end users. If excessive neutral shift takes place, there can be flashover in sub-station.
These three conditions are on the rise in many substations – including some of the industrial plants. Necessary care has to be incorporated in the protection systems to handle these situations.
4.0 Protection of small transformers (Less than 1MVA) :
The following figures show the SLD and the list of protections. Bare
minimum protections are envisaged – since economy of protections is the
major factor in deciding the extent of protection.
a) Low set Over Current Protection (51) : Used to protect the transformer from over currents in Py and Sy side. Pick levels are normally around 140% to 150%. Normal Inverse IDMT characteristics are followed for trip time.
b) Highset Over current Protection (50) : Used to protect from high level fault currents of the order of 300% and above. Always instantaneous trip.
c) Under Voltage protection (27) : This is a bus level protection – pick up levels are normally 85% and below.
d) Over voltage protection (59) : This is a bus level protection – pick up levels are around 110%.
e) Restricted Earth fault protection (64) : Normally provide on the star connected side – for protection transformer from internal faults.
5.0 Protection of medium size transformers (1 to 10 MVA):
Please refer the SLD and the list of protections, shown below:
Since the transformer is handling a higher power and it is in a key location
like the incomer of a substation or an industry, following additional
protections are advised.
a) Thermal Overload protection (49) : Let us consider a case where a normal over current relay with pick up level of 140% is used. It should be noted that the transformer is in the over load region between 105% to 140%. If the load is around 135%, the O/C relay will not protect – but the transformer will get hot and loose its life. Thermal overload protection will help in this case.
It is also beneficial to monitor the overload conditions in the winding and the core separately. The copper portion will get hot faster – for a given overload current, trip time will have to be faster than that for iron core.
b) Current Unbalance protection (46) : This will protect transformers against heavy unbalances. In case of unbalance currents, the negative sequence component will increase – resulting in over heating of transformers. It is advisable to have two levels of unbalance protection – one for alarm and other for trip.
d) I2T Protection : This protection is very useful for rectifier transformers – where the currents will be fluctuating . In this case the energy dissipated for given over current condition is set as trip limit. If this energy level is exceeded, the transformer is tripped earlier than the IDMT over current trip for the same value.
6.0 Protection of Large size transformers (above 10 MVA) : What we are talking about here are very large bulk power handling transformers where the criticalities are very high. Consequently more protections , than those listed in section 5 above, are envisaged.
The extra protections are in the form of differential and over fluxing protections which are mainly internal faults.
a) Differential Protection (87) : This is one of the major protections for large transformers. This protects the transformers whenever there is an internal fault . As shown in the SLD, this protection needs two additional sets of CTs, which are perfectly matched and have adequate knee point voltage to drive a relay measuring circuit. It should be noted that:
- a differential relay should trip only for an internal fault
- a differential relay should never trip for an external/through fault.
For this reason a percentage biased relay, with dual slope facility will be the best choice. This will have a very good through fault stability.
It may so happen that a differential relay can trip whenever the transformer is switched on. This is due to the magnetizing inrush current flowing only in primary side of the transformer. To avoid this , the relay should have a second harmonic restraint facility. Similarly a 5th harmonic restraint facility in the relay will help avoiding a differential trip during temporary over fluxing conditions.
b) Voltage unbalance protection (47V) : Voltage unbalance in large transformers are good indication of a grid disturbance. Can be used as an alarm .
c) Over fluxing protection (24) : This is to monitor the flux levels inside the large transformer. If the per unit ratio of V/Hz goes beyond a value 1.05, the transformer will go into an over fluxing condition – this will cause over heating even when the currents are within limits. Hence the need to monitor separately.
7.0 Monitoring of E/F in ungrounded transformers : In case of transformers, predominantly medium size, there can be installations where the neutral is grounded through an impedance or high resistance. In this when an earth fault occurs, normal E/F relays will not work – since the required relay operating current will not flow in the ground path. Consequently, a different method has to be adopted - monitoring the zero sequence voltage . The zero sequence voltage is a good indication of a neutral shift, which happens when there is an earth fault.
There are two schemes for monitoring the neutral shift –
a) use an open delta transformer + a low cost voltage relay. In this case the open delta transformer may become expensive.
b) use normal star connected bus PT – but with a relay which calculates zero sequence voltage by numerical methods.
8.0 Power Management concept : One of the main concerns of power transmission is the poor power factor conditions at the HT level. Many substations are resorting to adding HT capacitor banks for improving the pF – particularly at 33kV and 11 kV levels. Special relays like Capacitor bank protection relays, Reactive power measuring relays, Voltage & PF monitoring relays will be required here.
9.0 Grid Islanding & Load shedding : This requirement is very important to keep power transmission stable, within a specified area where there is a reasonable power generation available, when there is a large scale grid disturbance. In this case the entire grid , under disturbed conditions, is islanded into small networks so that the smaller networks can continue with power availability with their own generation capacities. This way total collapse is avoided.
The key parameters for detecting grid disturbances are:
- rate of change of frequency (dF/dT)
- Over / Under voltage
- Over / Under Frequency
- Heavy fault current which flows from the substation to grid
- reverse power flow from substation to grid
- large unbalance in grid voltage
- Vector shift in grid voltage
It is advisable to have a protection scheme to monitor all the above parameters – particularly for a transformer close to a generating station. It will help in islanding the power station from grid disturbances.
10.0 Advantages of Numerical relays in Transformer Protection : It has been a practice to use electro-mechanical / solid state relays relays for all above protections. The present trend is to use Numerical relays which offer many advantages as shown in the following table, over the earlier technology.
The usual worry that Numerical relays are very expensive is now removed- continuous production improvement techniques have made numerical relay affordable – some times cheaper from the over all protection perspective. Above all, with features listed as above, Numerical relays are more user friendly and are gaining popularity every where.
11.0 Conclusion : Transformer protection plays a major role in ensuring consistent power transmission and distribution. This paper is a brief attempt to bring out the various protections required for transformers. The protections are based on size and location. Numerical relays offer better solutions for transformer protection.
TESTING OF TRANSFORMERS
1.0 AIM: The purpose of this document is to explain the brief procedure for testing the power
transformer at site.
2.0 TESTING EQUIPMENTS REQUIRED :
2.1 Insulation tester
2.2 Micro ohm meter 1ph, 230v, 1A.
2.3 Multi-meter – 2 no’s
2.4 Mini-clamp meter – 0-50A
2.5 Clamp meter – 0-1000A
2.6 Transformer Test board
3.0 TESTING OF TRANSFORMER :
3.1 CHECK FOR COMPLETENESS OF INSTALLATION AND EARTHING.
3.2 INSULATION REISISTANCE TEST:
The insulation resistance of each winding in turn to all other windings, core, frame and tank connected together and to earth shall be measured by standard megger.
A 2.5 KV mains (or rotor) operated megger may be used. The line lead of the megger to be connected to the bushing of the winding under test and the earth leads are connected to the terminals of the transformer tank. The one minute insulation value is to be taken and recorded. A minimum IR value of 2 mega ohm per KV is acceptable.
For the transformer the following insulation resistance value are to be taken.
3.2.1 HV to earth 1000 V 2.5 / 5 KV
3.2.2
3.3.3 HV to
The Polarization Index (P.I) is the ratio of the IR at the tenth minute to the IR value at first minute and mass be taken and recorded for all winding. A P.I value of around 1.5 is good for oil immersed winding like the transformer. Ten minutes reading is taken for large transformer.
3.3 TRANSFORMER OIL TEST:
The dielectric strength- Break Down Voltage (B.D.V) of the transformer oil sample has to be tested before and after circulation.
3.3.1 Transformer top sample
3.3.2 Transformer bottom sample
3.3.3 Radiator sample
3.3.4 Conservator sample
3.4 RATIO TEST:
With 415V applied on high voltage side, measure the voltage between all phases on low voltage side for every tap position, Result should tally with name plate of the transformer. The permissible tolerance is + 0.5 % of the declared ratio. This test should be conducted for position of all tap.(Refer fig 1)
3.5 WINDING RESISTANCE TEST:
Micro-ohm meter should be used for measurement of winding resistance, tapped winding resistance should be measured at all tap position on HV side. This also helps to know whether OLTC tap changer contact are making properly. The micro-ohm meter gives the winding resistance value directly on
Resistance per winding = 1.5 x measured resistance.
The average of 3 sets of measurement shall be reported as the tested values. Before recording cold resistance, the transformer must be in oil without excitation or load for sufficient time to ensure that windings are at this same temperature as the surrounding oil. The resistance of each winding at principle maximum and minimum tap shall be measured by micro-ohm meter; any stable stabilized DC supply can be used for resistance measurement. To reduce the high inductive effect it is advisable to use high current to saturate the core. This will reduce the time required to get a stabilized reading.
3.6 VECTOR GROUP VERIFICATION:
This test is necessary to check whether the windings are connected internally as declared in the name plate.
The connections made for ratio test by voltage measurement can be used for this test. The primary and secondary windings are connected together at one point as indicated in figure and 3 phase LT supply is applied to the HV terminals. Voltage measurements are than taken between various pairs of terminals as indicated in the diagrams and the readings obtained should be the vector sum of the separate voltages of each winding under consideration.
3.6.1 Inference: (For Dyn11) Refer fig 2
Rr = 0 ; Yy = Yb ; By > Bb ; RY = Rn + Yn ;
3.7 MAGNETISING CURRENT TEST (MEASUREMENT OF NO LOAD CURRENT):
This test is to measure the iron – loss, dielectric loss and I2R loss No-load current shall also be recorded for 415V, 3 ph, 50Hz. This can be used the waveform of the applied voltage shall be approximately sinusoidal. No-load current shall also be recorded for 415V, 3ph, 50Hz supply. This can be used as guide at the time of commissioning. 3ph, 415V supply is given to the HV winding of 3ph transformers and simultaneous current readings of the three phases are taken. As the middle limb has a lower net reluctance (as the fluxes traverse a shorter distance), the magnetizing current drawn by the middle limb winding will be lower. Value ranges are in mA. (Refer Fig. 1)
3.8 MAGNETIC BALANCE TEST:
The purpose of this test is to check whether the flux path in the core is uniform and properly distributed in the core limbs. This test is done preferably on
The No-load current shall be measured at rated voltage and frequency.
The alternative type of testing procedure also can be done in a transformer, by applying the phase to phase voltage of 440V AC to primary winding of Y & B phases, and measure the voltage of primary winding ‘RY’ and ‘YB’ keep the ‘R’ phase switch in ‘OFF’ position. Similarly by sequence the other two phases (i.e. ‘Y’ & ‘B’ in ‘OFF’ condition) test should be repeated and voltage distribution of adjacent phases are recorded for proper distribution.(Refer fig 3)
3.9 SHORT CIRCUIT TEST:
Before commencing the test, the short circuit current is to be calculated for the available mains voltage from the percentage impedance of the transformer. Based on the calculated short circuit currents for the available mains voltage, the instruments to read the HV and
The transformer board used in the voltage ratio test can be used for the short circuit test. The test can be commenced by keeping the transformer tap at one extreme position. Usually, the HV short circuit current can be read directly through Ammeter / multi meter and the
4.0 REMARKS : Verify all test results with the factory test reports, wherever possible .
 | |||||||
POWER TRANSFORMER
LIST OF RELATED INDIAN STANSARDS
IS 1885 Part XXXVIII : Electrotechnical vocabulary: Part XXXVIII Transformers
IS 1886 : Code of practice for installation and maintenance of transformers ( Not active & substituted by 10028/III )
IS 2026 Part-I : Specification for power transformers: General
IS 2026 Part-II : Specification for power transformers: Temperature-rise
IS 2026 Part-III : Specification for power transformers: Insulation levels and dielectric tests
IS 2026 Part-IV : Specification for power transformers: Terminal markings, tapings and connections
IS 2026 Part-V : Transformer/Reactor bushings minimum external clearance in air-specification
IS 6600 : Guide for loading of oil immersed transformers
IS 10028 Part 1 : Code of Practice for Selection, Installation and Mtc of Transformers : Selection
IS 10028 Part 2 : Code of practice for selection, installation and maintenance of transformers: Installation
IS 10028 Part 3 : Code of practice for selection, installation and mtc of transformers: Maintenance (superseding IS:1886)
IS 10561 : Application guide for power transformers
IS 12676 : Oil impregnated paper insulated condenser bushings - Dimensions and requirements
IS 11171 : Specification for Dry-Type Power Transformers
IS 3347 Part I to V : Dimensions for porcelain transformer bushings
IS 2099 : Bushings for alternating voltages above 1 000 Volts (For bushings which are supplied separately).
IS 8468 : On-load tap changers
IS 8478 : Application guide for on-load tap-changers
IS 3639 : Specification for Fittings and Accessories for Power Transformers
IS 335 : Specification for new insulating oils for transformers and switchgears
IS 1866 : Code of pratice for maintenance and supervision of insulating oil in service
What is Substation ?
Substation
is an Electrical installation where power is controlled for transmission &
distribution purpose.
Substations can be categorized as:
qPower evacuation substation
qSubstation part of Transmission system
qSubstation part of Distribution system
Power Evacuation Substation
qLocated adjacent to the Power Plant
qPreferred voltage level
–420 kV
–245 kV
–123,145 kV
–72.5, 66 kV
qEvacuation voltage level depends on
–Quantum of power to be evacuated
(size of power plant)
–Distance of Transmission
–Gird network voltage of
surrounding transmission system
qNormally built by
–Utilities (e.g. NTPC, NHPC,
Captive Power Plant by industries)
Independent Power Producer (IPP)
Transmission Substation
qLocation is decided based on Transmission
Grid Network
qPreferred voltage level
–420kV
–245kV
–145kV
qTransmission voltage depends on
–Quantum of power to be received /
transmitted
–Length of transmission line
qNormally built by
Utilities (e.g. Power Grid, State
Electricity Boards
Distribution Substation
qLocation is near the load
qPreferred voltage level
–245kV
–145kV
–72.5/36 kV
qStation voltage level depends on
–Demand of power
–Utility norms of distribution
qNormally built by
–Utilities (e.g. State Electricity
Broads)
–Industries
What is Busbar Configuration ?
Busbar
configuration or Bus switching scheme is the circuit adopted for substation
based on following:
–System reliability
–Operational flexibility
–Ease of maintenance
–Limitation of fault level
–Simplicity of Protection system
–Ease of extension
–Availability of Land
–Cost
Various Busbar Configuration
Single Bus Scheme
qApplication
–Industrial stations with voltage
level generally up to 145kV
–Also used for 245kV station with
2/3 bays
qFeatures
–Simple system
•Ease of Operation &
maintenance
•Single level bus layout
–Very simple Control &
Protection philosophy
–Large saving in space
No redundancy
Double Bus Scheme
qApplication
–Normally this is used for most
industrial stations and some time small power evacuation system
qFeatures
–Simpler system
–Better availability as additional
bus is provided
–Breaker can be taken out for
maintenance where transfer bus feature is provided
–
Three Bus Scheme
qApplication
–Power evacuation station
–Interconnecting substation for
transmission lines
–Heavy Industrial stations viz.
Steel, Aluminium, Petro-chemicals
qFeatures
–Independent two buses sharing the
feeder
–Better load management
–Better reliability
–One transfer bay is in-built for
redundancy of one bay
–Complex Control & Protection
system
ADVANTAGE AND DISADVANTAGE OF ONE AND HALF BREAKER SCHEME
One & half breaker scheme
qApplication
–Mostly this configuration is
adopted where high reliability is required
–Power evacuation station for big
power plant
–Interconnecting transmission
substations with 420/245kV level
qFeatures
–Very high reliability
–Costly because of increase no. of
Circuit Breaker
–Complex Control & Protection
philosophy
ADVANTAGES:
1. THREE
CIRCUIT BREAKERS IN A FULL DIA ARE ALWAYS INSERVICE.
2. IF
MAIN CB PROBLEM / FOR PERIODICAL MAINTANENCE, DURING THAT PERIOD THE FEEDER IS FEEDING FROM TIE CB. NEED NOT
REQUIRE TRANSFER THE FEEDER.
3. IF
TIE CB PROBLEM / FOR PERIODICAL MAINTANENCE, DURING THAT PERIOD THE FEEDER IS
FEEDING FROM MAIN CB.
4. EVEN
IF BOTH MAIN BREAKERS UNDER TROUBLE, DURING THAT PERIOD ONE FEEDER IS WORKS AS
INCOMING AND OTHER FEEDER IS WORKS AS OUTGOING VIA TIE CB. (WITHOUT INTERUPTION
TO FEEDERS)
5. IF
BUSBAR-1 OPERATED / FOR PERIODICAL MAINTANENCE, ONLY THE MAIN BREAKERS
CONNECTED TO BUSBAR-1 WILL TRIP. DURING THAT PERIOD THE FEEDERS CONNECTED TO
BUS-1 ARE FEEDING VIA TIE CB FROM BUS-2.
6. IF
BUSBAR-2 OPERATED / FOR PERIODICAL MAINTANENCE, ONLY THE MAIN BREAKERS
CONNECTED TO BUSBAR-2 WILL TRIP. DURING THAT PERIOD THE FEEDERS CONNECTED TO
BUS-2 ARE FEEDING VIA TIE CB FROM BUS-1.
7. IN
THIS BUSBAR PROTECTION IS SIMPLE AND NEED NOT REQUIRE SELECTION OF ISOLATOR.
8. IT IS
MORE RELIABLE SYSTEM FOR OPERATION & MAINTANENCE POINT OF VIEW.
B. DISADVANTAGES
1. CT CONNECTIONS &
PROTECTION TRIP LOGICS ARE SOME
WHAT COMPLICATED.
2. COST TOWARDS CTs,
CIRCUIT BREAKERS & PANELS ARE TO BE INCREASED DUE TO EXTRA BAY. THE COST
COMPARISION TABLE SHOWN IN NEXT PRESENTATION.
3. THE OPERATIONS FOR
OPERATOR POINT OF VIEW IS LITTLE BIT COMPLICATED WHEN COMPARED TO CONVENTIONAL
SYSTEM.
4. SPACE OCUPATION FOR
BAY IS MORE i.e COST OF LAND IS INCREASE.
Busbar configuration adopted by..
qDistribution Substations
–Single Bus with / without sectionaliser
–Double bus scheme
–Three bus scheme
•Two Main & one Transfer
•One Main & Main cum Transfer
Main Equipment
qCircuit Breaker
qInstrument Transformer (CT/CVT)
qSurge Arrester
qIsolator
qTransformer & Reactor
qBus Support Insulator
qLine Traps
qControl & Protection system
qPower Line Carrier Communication system
Engineered items
qBusbar
Materials
–Strung bus conductor (ACSR)
–Rigid bus conductor (AL Tube)
–Clamps & Connectors
–String Insulator & Hardware
qEarthing
Materials
–Main earthing materials (Rod or Flat)
–Earth Pits
qCables & accessories
–LT Power & Control cable
–HT Power Cable
–Cable terminations, gland, lugs,
ties, marker etc.
–Cable supporting materials
Auxiliary Equipment
qLTAC
Switchgear panel
qAC Distribution board
qBattery & Battery Charger
qDC Distribution board
qAuxiliary LT transformer
qLighting Transformer
qMain Lighting Distribution board
qEmergency Lighting Distribution board
qDiesel Generator
qUPS
Auxiliary System
qIllumination system
–Indoor (Control Room Building,
DG, FFPH Building.)
–Outdoor switchyard
qFire fighting and fire detection system
qAir conditioning and ventilation system
Substation Structures
qMain Column and Beam structures
qEquipment supporting structures
Substation Civil
qFoundation for gantry column & equipment
qFoundation for Transformer / Reactor
qControl room building / DG, FFPH building
qOutdoor switchyard cable trench
qSurface drainage
qRoads
qFencing and gate
COST OF EQUIPMENT IN SUBSTATION
A > Transformer /
Reactor----------------------------------------- 27%
B > EHV Circuit Breaker, CT, CVT, LA, Isolator,
BPI------- 22%
C > Control, Protection &
PLCC-----------------------------------
7%
D > Medium Voltage
Switchgear----------------------------------
5%
E > Cables, AC/DC Aux., Lighting, Busbar
etc.--------------- 16%
F >
Structures---------------------------------------------------------- 6%
G >
Civil------------------------------------------------------------------ 13%
H >
Erection------------------------------------------------------------- 4%
What is Substation Layout ?
Substation
Layout is the conception which transforms to reality through engineering,
execution & commissioning.
Layout Engineering . . . .
qis the basic document to be sent to Customer
/ Consultants for approval
qforms the basis of entire project scope
qforms basis for engineering estimates
qform input for civil & structure design
qbasic document for statutory approval
Layout Engineering
Following
details form the input for layout engineering:
qBay & Busbar configuration
qBay width & depth, Height of busbar,
Distance to fence, Building and total dimensions.
qStatutory clearances (Phase to Phase, Phase
to Earth, Sectional clearance)
qLightning Mast, Gantries (Column & Beam),
String Insulators, Busbar, Equipment etc.
qPosition of various drive box, junction box,
Marshalling box.
Bay Width it depends on...
qPhase to Phase & Phase to Earth
clearances
qSectional Clearances
qTower width
qSwing of conductor
qBird clearances
qCable trench routing
qMaintainability of equipment
qTandem Isolator arrangement
Equipment Distances it
depends on...
qPhase to Phase & Phase to Earth
clearances
qSectional Clearances
qPosition
of switch cubicle, junction box, drive box etc.
Maintainability of equip
DESIGN ENGINEERING TOOLS
–STANDARD SOFTWARE
–AutoCAD 2000
–STAAD-III
–ETAP
–MS OFFICE ’97
–CUSTOMISED SOFTWARE
–AUTO SWITCHYARD: FOR ENGINEERING CALCULATION
–SAG-TENSION CALCULATION
–CABLE SIZING
–SOFTWARE FOR CABLE ENGINEERING
RELAY SETTING