Design & Construction Distribution Transformer
Of
100kva
Aluminium
Wound
Basic Concept of a Transformer Introduction This project paper starts with the basic concept of the transformer and transformer related theory before undertaking a commercial working design. Chapter 1 contains an introductory discussion relevant to the principles and practices of transformer design and construction. Some of the articles of this chapter have essentially dealt with relevant aspects of magnetism, while the rest dealt with the practices of transformer construction including types, regulation and efficiency. In a transformer design, the selection of number of turns is a key factor, which has to be estimated first. The selection of low voltage and high voltage conductors and their sizes are chosen on the basis of assumed current density. Then, estimating the coil configuration till arriving at the window height of the core frame. Based on the calculated window height, the design of the secondary coil is done. Further, the core diameter, step width, core stack, core area, flux density, etc. are calculated with the design output available. The next step of the design is the formation of the coil diameter and limb center of core. Based on the window height and limb center of the core frame, a detailed design of the core up to the weight of the :complete set are estimated. Working details of low voltage and high voltage windings, tapping, placements, internal clearances, weight of, or conductor etc. are also done. Next comes the calculation of performance figures. The method of calculation of resistance, losses, no-load current, percentage impedance, efficiency, regulation etc. have been dealt with. After completion of the internal active part design, the chapter has covered the procedure of designing the tank with different types of radiators. Apart from the conventional elliptical tube radiators has been discussed. Also, added to the procedure of designing the core frame, conservator, tank etc. Calculations of the oil volume and overall weight and dimensions have also been discussed.
1.1 Definition of a Transformer A transformer is a static (a stationary) piece of apparatus (machine) by means of which electric power in one circuit is transformed into electric power of the same frequency in another circuit. It can raise or lower the voltage in a circuit but with corresponding decrease or increase in current; for which the total volt-ampere (VA or KVA) remains the same.
Fig. l. l Two circuit linked by a common magnetic flux ( Ď• ). The physical basis of a transformer is mutual induction between two circuits linked by a common magnetic flux ( Ď• ). In brief, a transformer is a device that transfers electrical power from one electrically isolated circuit to another, under the following conditions: a) It does so without any change of frequency. accomplishes this by electromagnetic induction. b) The two electrical circuits are mutually coupled. The first coil, in which the electrical energy is fed from the A.C. supply mains, is called primary winding and the other from which energy is drawn out, is called secondary winding. 1.2 Magnetism The name magnetism is derived from magnetite, which is an iron-oxide mineral whose magnetic properties were discovered before man grasped the basic principles of electricity. It was found that a magnet had the property of attracting iron and similar materials; the word ferromagnetism is frequently used in association with the magnetic properties of iron. To understand the nature of a magnet we must take a look at the composition of iron. In an atom, the spinning motion of the electrons around the atomic nucleus is equivalent to an electric current in a loop of wire. In turn, this electric current produces a small magnetic effect. There are many electrons associated with each atom, and the magnetic effect of some electron spins, giving rise to an overall magnetic effect. Groups of atoms appear to club
together to produce a small permanent magnet, describe by scientists and engineers as a magnetic domain. Although a magnetic domain contains billions of atoms, it is smaller in size that the point of a needle. Each domain in a piece of iron has a N-pole but, in a demagnetized piece of iron, the domains point in random directions so that the net magnetic field is zero, as shown in Figure 1.2(a). Each arrow in the figure represents a magnetic domain. When an external magnetic field is applied to the iron in the direction shown in Figure 1.2(b), some of the magnetic domains turn in the direction of the applied magnetic field and remain in that direction after the field is removed. That is the external magnetic field causes the piece of iron to be partially magnetized so that it has a N-pole and S-pole. Repeated application of the external magnetic field results in the iron becoming progressively more magnetized as more of the domains align with the field. A bar of iron can be magnetized in this way simply by stroking it along its length in one direction with one pole of a permanent magnet. If the magnetic field intensity is increased, all the domains ultimately align with the applied magnetic field as shown in Figure 1.2Š. When this occurs, the iron produces its maximum field strength and the bar is said to be magnetically saturated.
Fig. l.2 Magnetizing a piece of iron
Since no further domains remain to be aligned, any further increase in the external magnetic field does not produce any significant increase in the magnetism of the iron. 1.3 Magnetic coupling If a portion of the magnetic flux established by one circuit interlines with a second circuit, the two circuits are coupled magnetically and energy may be transferred from one circuit to the other by way of the magnetic field, which is common to the two circuits. The practical operation of many devices depends upon this type of coupling.
Fig. l.3 Illustrating the four component fluxes ϕ11, ϕ12, ϕ22 and ϕ21
Separation of Magnetic Flux into Hypothetical Components: Magnetic coupling between two individual circuits is shown in Fig. 1.3. For the purpose of analysis, the total flux, which is established by i 1, namely, is ϕ 1, is divided into two components. One component of ϕ 1, is that part which links with circuit 1 but not with circuit 2, namely, ϕ 11. The second component of ϕ 1 is ϕ 12, that part which links with both circuit 2 and circuit 1. In a similar manner, the flux established by i 2 is separated in to two components for the sake of detailed analysis. By definition: ϕ 1= ϕ 11+ ϕ 12 ----------------(1) ϕ 2= ϕ 22+ ϕ 21 ----------------(2) and The four component fluxes are shown in Fig. 1.3 and a recapitulation of their definitions is given below: ϕ 1= The total flux which is established by i 1. ϕ 1= The total flux which is established by i 2, ϕ 1=The fractional part of ϕ 1, which links only with the turns of circuit-1. This is the leakage flux of circuit -1 with respect to circuit-2. ϕ 12 = The fractional part of ϕ 1, which links with the turns of circuit-2. This is the mutual flux produced by circuit -1. ϕ 22 = The fractional part of ϕ 2, which links with the turns of circuit-2. This is the leakage flux of circuit-2 with respect to circuit-1. ϕ 21 = The fractional part of ϕ 2, which links with the turns of circuit-1. This is the mutual flux produced by circuit-2
1.4 Mutual Inductance In order to describe the magnetic interaction between circuits or between portions of the same circuit, the circuit parameter M is introduced. It is called the co-efficient of mutual inductance, or simply mutual inductance, and is dimensionally equivalent to the coefficient of self-inductance, L. The similarity between the concept in mutual inductance of (or between) two circuits and the concept of self-inductance may be shown in the following manner refer to Fig.l.3. For the purpose at hand we shall define the selfinductance of circuit 1 as: N1ϕ1 L1 = ----------- (3) I1 (Flux linkages of circuit-1 per unit current in circrit-1) On the same basis of reckoning, the mutual inductance of circuit-1 with respect to circuit-2 is: M
21
=
N1ϕ21 ----------- (4) I2
(Flux linkages of circuit-1 per unit current in circuit-2) N1ϕ12 M 12 = ----------- (5) I1 (Flux linkages of circuit-2 per unit current in circuit-1) If the characteristics in equations (3), (4) and (5) are not straight lines then L 1, M21 and M12 are variable circuit parameters and for certain types of analysis can best be written in the forms: d1ϕ1 L1 = N 1 ----------- (3a) di1 d1ϕ21 M 21 = N 1 ----------- (4a) di2 (M
12
=N2
d1ϕ12 ----------- (5a) di1
If however, the flux is proportional to the current (i.e. permeability constant), both selfinductance and mutual inductance in equations (3), (4) and (5) are constant and as such are very useful circuit parameters in classical circuit theory. Under the condition of constant permeability, the reluctance of the mutual flux path (T21 or 912) is a fixed quantity and T21 = cp12. N1ϕ21 KN1 N 2 M 21 = = ----------- (6) i2 ϕ21 N 2ϕ12 KN 2 N1 M 21 = = ----------- (7) i1 ϕ12 Where K is a constant which depends for its value upon the units employed in evaluating ϕ = KN1 . R Therefore if the permeability of the mutual flux path is constant, M 2, and M 12 are constant and M-1 I = M, 2= M.
This fact may also be proved in terms of the energies stored in the magnetic field when both circuits are energized. If the permeability of the mutual flux path is not constant, neither M Z, not M12 will be constant and the following method of representing mutually induced voltages in terms of M losses much of its effectiveness. Unless otherwise stated, absence of ferromagnetic material will be assumed, in which case M21 = M,Z= M. The units in which mutual inductance is expressed are identical with the units in which selfinductance is expressed, usually the henrys or millinery. If the flux linkages in equations (4) or (5) are expressed in Weber-turns (10 8 Maxwell-turns) and the current is expressed in amperes, M is given in henrys. Mutual Reactance XM It is evident that any change in i2 of Figure.2 will cause a corresponding change in cp2,. in accordance with Lenz’s law, any time rate of change of cp 21 will manifest itself in circuit-1 in the form of a generated or induced voltage the value of which is : dϕ21 dϕ21 e12 = N1 or v12 = N1 ----------- (8) dt dt Where e12 is considered as a voltage rise or generated voltages and v 12 is considered as a voltage drop. Similarly, any change in i, will manifest itself in circuit-2 as : dϕ12 dϕ12 e21 = N2 or v21 = N2 ----------- (9) dt dt It is through the agency of these mutually induced voltages that the phenomenon known as mutual inductance can be taken into account in circuit analysis. 1.5 Magnetization curve for iron and other ferromagnetic material If a coil has an air core Fig.1.4, and if the current in the coil (and therefore the magnetic field intensity) is gradually increased from zero, the magnetic flux density in the air core increases in a linear manner as shown by the straight line graph at the bottom of Figurel.6. The curve relating the flux density B to the magnetic field intensity H is known as the magnetization curve of the material. Since in an air core B = µ o H, the magnetization curve is a straight line having a slope of µ o ( µ o is called the permeability of free space, µ o = 4 π x 10-7 H/m).
Fig. - 1.4 Coil an air core.
Fig.- 1.5 Coil an iron core
If the air is replaced by a ferromagnetic material such as iron, we find that the flux density in the core increases very rapidly between 0 and A (see Figure 1.6), after which the slope of the
curve reduces. Finally, at a very high value or H (corresponding to large value of excitation current), the magnetization curve for the iron becomes parallel to that for the air core. As the excitation current in the coil surrounding the iron increases, the magnetic domains in the iron begin to align with the magnetic field produced by the current in the coil. That is, the magnetism of the domains adds to the magnetic field of the excitation current; as the current increases, more and more of the domains in the iron align with the external magnetic field, resulting in a rapid increase in the flux density in the iron.
Fig. 1.6 Magnetization curves for air and iron However, when point A in Figure 1.6 is reached, most of the domains have aligned with the filed produced by the magnetizing current, and beyond this point the rate of increase of the flux density reduces. This denotes the The remaining few domains coming into alignment with the applied magnetic field causes any further increase in flux density. Finally, the magnetic flux density increases very slowly, and is due only to the increase in excitation current (since all the domains are aligned with the field, no more magnetism can be produced by this means); at this point the magnetization curve for iron remains parallel to the curve for air.
Fig. 1.7 The B/H curves of some typical magnetic materials. Note: One of the most important points in the design of a transformer is to fix the working flux level. If the working flux is too high and the core becomes saturate the input current will be high and distorted. If the working flux is low the transformer will larger and more costly than it need be. The designer makes a compromise, usually fixing the working flux at a point just beyond the linear part of the flux B-H curve. 1.6 Magnetic Hysteresis or B-H Loop (B =Flux density, H= Field strength) An unmagnified bar of iron magnetized by placing it within the field of a solenoid (Fig 1.8). The field strength produced by the solenoid, is given byH =
NI I
The value of H can be increased or decreased increasing or decreasing current through the coil. Let H be increased in steps from zero up to a certain maximum value and the corresponding values of flux density (B) be noted. If we plot the relation between H and B, a
curve as shown in Fig. 1.9. is obtained. The material becomes magnetically saturated for H = OM and has at that time a maximum flux density of B max established through it.
Fig 1.8 Magnetic Field of a solenoid Fig. 1.9 B/H curve If H is now decreased gradually (by decreasing the solenoid current), flux density B still not decrease along AO, as might be expected, bur will decrease less rapidly along AC. When H is zero, B is zero not but has a definite value B r = OC(T). It means that on removing the magnetizing force H, the iron bar is not completely demagnetized. This value of B measures the retentively of the material and is called residual flux density Br. To demagnetize the iron bar, we have to apply the magnetizing force in the reverse direction. When H is reversed (by reversing current through the solenoid), then B is reduced to zero at point D where H=OD. This value of H required to wipe off residual magnetism is known as coercive force (HO). If, after the magnetization has been reduced to zero, value of H is further increased in the negative i.e. reversed direction, the iron bar again reaches a, state of magnetic saturation represented by point L. By taking H back form its value corresponding to negative saturation (OL) to its value for positive saturation (OM), a similar curve EFGA is obtained. If we again start from G, the same curve GACDEFG is obtained once again. It is seen that B always lags behind H. The two never attain zero value simultaneously. This lagging of B behind H is given the name hysteresis which literally means to lag behind. The closed loop ACDEFGA which is obtained when iron bar is taken through one complete cycle or magnetization is known as hysteretic loop. Area of Hysteretic Loop: Just as the area of an indicator diagram measures the energy made available in a machine when taken through one cycle of operation, so also the area of the hysteresis loop represents the net energy spent in taking the iron bar through one cycle or magnetization.
According to Weber's Molecular Theory of magnetism, when a magnetic material is magnetized, its, molecules are forced along a straight line. So, energy is spent in this process. Now, if iron had no retentivity, then energy spent in straightening the molecules could be recovered by reducing H to zero in the same way as the energy stored up in a spring can be recovered by allowing the spring o release its energy by driving some kind of load. Hence, in the case of magnetization of a material of high retentivity, all the energy put into it originally for straightening the molecules is not recovered when H is reduced to zero. We will now proceed to find this loss of energy per cycle of magnetization. Let I = mean length of the iron bar A = its area of cross-section N = No. of turns of wire of the solenoid. If B is the flux density at any instant, then cp = BA. When current through the solenoid changes, then flux also changes and so produces an induced e.m.f, whose value is dϕ e = volt dt dϕ dB e =N (BA)= NA dt dt
volt
Now, H =
NI ×V I
I =
HI dϕ e = N dt
The power or rate of expenditure of energy in maintaining the current I against induced e.m.f. is P = ei =
HI × NA N
dB dB =AIH watt. dt dt
Energy spent in time dt AIH
dB × dt = AIH .dB joule dt
Total network done for one cycle of magnetization is W = AI ∫ HdB joule. Where
∫
stands for integration over the whole cycle.
Now, H dB represents the shaded area Hence,
∫
HdB
= area of the loop
i.e. the area between the B/H curve and the B-axis. Therefore, Work done/Cycle Now, A1 3 done/cycle/m = (Loop area) joule Wh = (area of B/H loop) Joule/m3/cycle.
= Al X(area of the loop) joule = volume of the material Therefore, Net work
1.7 Fundamental principle of operation. When an electric circuit is supplied with a voltage, that circuit must take sufficient current from the source to build up a counter electromotive force equal and opposite to the voltage applied. The primary of a transformer is a circuit wound on an iron core. The resistance of the
winding is purposely made low. Thus the counter electromotive force is made up essentially of voltage produced by the action of the flux on the primary turns. The instantaneous value of this counter electromotive force is dϕ e1 = - N l volt -------------------- (1) dt
Where N, is the number of turns on the primary winding and 9 is the instantaneous value of the flux in the core. Assuming the primary resistance to be zero, the applied voltage v, (instantaneous value) is equal and opposite to e1 , Or dϕ v1 = - N l volt -------------------- (1) dt
If the applied voltage is sinusoidal, that is, v1 = v1max sin 2 π ft Then, ϕ = ϕ max sin 2 π ft and el = - Nl ϕ max cos 2 π ft x 2 π ft volt ----------(2) Thus if vl and el are sinusoidal, the flux wave must be a cosine wave. These quantities man be represented as vectors as is shown in Fig 1.10.
Fig 1.10 Vector relations of voltage, winding current, and flux. Where Vl and El are root mean square values of vl and el . To obtain these root mean square values, divide the maximum value of eq. 2 by 2 then 2π El = ∫ N ϕ max Volt ---------------- (3) 2
The cosine term has no significance except to derive instantaneous values. It is not needed to obtain the root mean square valre, but it shows that since the flux wave was assuced to be a sine function and the voltage wave appeared as a cosine function, there exists a 90 ° relationship between ϕ and El. This will appear on the vector diagrams. El = 4.44fNl ϕ max Volt ------------------------- (4) Or, El = 4 (form factor) fNl ϕ max Volt -------------------(5) The flux is kept at its maximum value in this equation. Since this is the most convenient way to designate it. Since the flux in the core is produced by a current, it must be in time phase with it. The current Iex (exciting current) is there fore shown in Fig. 1.10 in time phase with the flux.
It is now evident from the above equations that a given value of the rms voltage V, will require the same rms value El. The flux vector must 90° ahead of the voltage vector E1 and the flux cp must be fixed for a given value of Vl, and f . Removal of the iron core would not decrease the flux. This operation would only affect the magnitude of the exciting current, and the voltage vectors of Fig. 1.10 (b) would remain fixed in magnitude and direction. If a secondary winding having N2 turns is now placed on the core as shown in Fig. l .11 (a), a voltage will produced in it by the flux ϕ .
Fig.l.ll(a) Two-winding transformer showing currents and voltages existing at one instant. (b) Vector diagram showing time-phase relationships of voltage and current with resistances and leakage reactance neglected. The magnitude of this voltage E2 is directly proportional to the number of turns NI EI/E2 = NI/N2 -------------------- (6) And since both voltages EI and E2 are produced by the same flux, they must be in time phase as shown in Fig. 1.11(b). For the present, let it be assumed that all the flux Ď• links both windings NI and N 2. Then if a resistance is connected to the secondary, a current I I, will flow from it in time phase with E 2This current will tend to produce a flux Ď• 2 However, the primary must keep the total flux
constant at a value ϕ in the direction shown. This is accomplished by the primary only through its production of another flux, - ϕ 2. To produce - ϕ 2, the primary must carry another current in the direction of - ϕ 2. The magneto motive force of the primary N I 1L, necessary to counteract the magneto motive force N 2I2 requires a load component of current IL, such that IL = 1 2.N 2/NI Thus the ratio of primary load current to secondary current can be expressed as IL/1 2 =N 2/NI -------------------- (7) The total current 1 1 in the primary, when N 2 is supplying the above load, is the vector sum of IL and 1ex and the resultant flux under condition of load is ϕ . The primary winding is therefore self-regulating in that it draws a load component of current only when it is needed to neutralize magneto motive forces of secondary circuit or circuits. Further analysis will show that the power and power factor associated with V 1 and IL are equal to those associated wit E2 and I2. The above analysis is based on the ideal transformer with no winding resistances or core losses and no leakage fluxes. When these are taken into consideration the above analysis is slightly modified. 1.8 Classification of transformers. Transformers can be classified from different points of view. According to construction transformers are of three general types, distinguished from each other merely by the manner in which the primary and secondary coils are placed around the laminated steel core. General types of transformer are: 1) Core-type 2) Shell-type 3) spiral-core type Another means of classifying of transformers, According to the type of cooling employed in transformer. The following types are in common use: 1) Oil-filled self-cooled 2) Oil-filled water-cooled 3 ) Air-blast cooled. According to the manner of using the transformer: 1) Power transformer 2) Distribution transformer 3 ) Auto transformer. According to facilitate the instrumentation, they are: 1) Current transformer (C.T) 2) Potential transformer ( P.T). 1.9 Construction of transformer : There are three distinct type of construction as stated above: i)Core-type ii) Shell-type iii) spiral-core type. Core type: Half of the primary Fig 1.12 (a) winding and half of the secondary winding is placed round each limb. This reduces the effect of flux leakage, which would seriously affect the operation if the primary and secondary were each wound separately on different limbs. The limbs are joined together by an iron Yoke.
Fig. 1.12 Construction of single-phase transformer Core-type, (b) Shell-type and (c) Berrytype (Plane view). Shell type: In shell type Fig. 1.12(b) both windings are placed round a central limb, the two outer limps acting simply as a low reluctance flux path Berry type: In berry type Fig. 1.12Š of construction, the core cay be considered to be placed round the windings. It is essentially a shell-type of construction with the magnetic paths distributed evenly round the windings Owing to construction difficulties it is not so common as the first two. According to the manner of using the transformer: (1) Power transformer (2) Distribution transformer (1) Power transformer: Power transformer from 3 MVA to 25 MVA of 33 KV application are either with off load tap changer or on load tap changer. Transformer may have automatic mixed cooling system like ONAN/ONAF by the cooling fan, which runs automatically by sensing the winding temperature. 25 MVA and above 132/33 kV transformers are star/ star connected with on load tap changer. (2) Distribution transformer : Distribution transformer ranges from 25 KVA to 3000 KVA, having system voltage either 11 kV or 33 kV. Pole type transformers are mainly designed for rural electrification board. These transformer comply with ANSI, BSS or IEC specification. Construction of distribution transformer : The core each built of grain oriented cold rolled silicon steel to give to low no-load losses and to reduce noise levels as oil. Yoke laminations are clamped between frames with core clamping bolt, and the frames also support the winding through end blocks and rings. The construction ensures low iron losses and magnetizing currents. Windings: High voltage coil up to 11 kV is continuous disc wound or cross over type using paper insulated Copper/Aluminum conductor. Low voltage coil are normally larger layer wounds using paper insulating Copper/Aluminum conductor. Core-coil Assemble:
Being the only active part of the transformer, it has to withstand all the ranges of short circuit stresses apart from the impulse voltage and thermal stresses. The design of the clamping structure takes due care of the axial short circuit forces. Careful design of the winding supports and spacer enable the windings to withstand radial as well as loop stress in the outer winding and compressive stress in the inner winding. Tapping: Tapping are normally provided in the high voltage windings to cater either for high voltage or low voltage variation. Tapping are adjusted by off circuit tap changer in case of distribution transformer of 11 kV class assembled from top cover.] Processing & Oil filling : To improve the dielectric characteristic of the insulation between the coil and core, coil assemble of transformer are dehydrated in the vacuum autoclaves or air-drying oven. The transformer oil is processes by high vacuum oil centrifuging plant. Before pouring the oil under vacuum, when the transformer is perfectly dried up, the oil is tested for dielectric breakdown at the test laboratory. Tank and Cover: Mild steel sheet made tanks and covers are fabricated by welding. Each tank is tested under pressure for detecting any leakage. Tanks employed for housing the core-oil assemble are mechanically strong. Tanks for distribution transformers are suitable to withstanding full vacuum. The steel parts before painted are shot or sand blasted to remove weld splatters, mill scales etc. Radiators: Pressed steel radiators are specially made from "D" grade CRCA sheets with profile formed and the elements are resistance welded with seam welding machine. Each individual element and radiator under assembled condition is tested with air pressure of approximately 3.0 kg/cm2. Conservator Tank: Transformer is provided with an overhead conservator tank. It is detachable by providing a flange at its point of connection to the tank cover. The connection to the main tank cover will be at the highest point to prevent trapping of air in conservator tank, which shall be detachable to facilitate cleaning. Insulating oil specification: Transformers are delivered with initially oil filling. Transformer oil is so that no further drying will be required before putting into operation. The characteristics of the insulating oil are as follows: Specific gravity at 15째C : 0.87 Viscosity at 75째C : 5.5 Centistokes (Max.) 째 Viscosity at 30 C : 19.0 Centistokes (Max.) Flash point : 145째C (Min.) Reaction : Neutrality. Dielectric strength : 50 kV (Min.) 12.5 mm diameter spare electrode gap length 2.5 mm. Finishing: The paint of transformer is not only enhancing its appearance but also the more important things is to protect the exposed metal surfaces from corrosion. Transformer tank, top cover, conservator etc. are provided with one coat of anti-rust primer and two coats of synthetic enamel paint. Tests:
Tests are carried out rigorously at every stage of assembly. Transformers are qualified for dispatch after successful completion of routine test as per BS 171/197 or IEC 76. Routine Test : (i) Insulation resistance measurement. (ii) Measurement of voltage ratio and checking voltage vector relationship. (iii) Measurement of winding resistance. (iv) No-load current and No-load loss measurement at 90%, 100%,150% of rated voltage. (v) Measurement of impedance voltage and measurement of loadloss. (vi) Separate source voltage withstand test. (vii) Induced voltage withstand test. (viii) Di-electric strength of oil. (ix) Oil leakage test for transformer tank. Name plate: Transformer have a nameplate of durable non- corrosive metal with all letters and makings are permanently readable and can not be removed by cleaning or polishing. The name plate will show the complete information of the transformer, in English language as follows: (a) Rating of transformer (Power of transformer). (b) Manufacture's name. (c) Manufacture's serial no. (d) Year of manufacture. (e) Nos. of phase. (f) Rated power (g) Rated frequency (h) Rated voltage at tap position. (i) Rated current. (j) Vector group with symbol. (k) Percentage impedance. (1) Total mass (m) Volume of oil. Terminal Marking: HV terminals are designated as A B,C from the right hand side of the case when facing the highest voltage side. LV terminals are designated as a,b,c,n being from the left hand side of the case when the facing the low voltage side. Terminals are marked as per IEC and VDE standard. Accessories: The following accessories are furnished with transformers. (i) Drain valve. (ii) Filter valve. (iii) Thermometer pocket. (iv) Earth terminal. (v) Lifting lugs. (vi) Radiator valve. (vii) Four jacking lugs. (viii) Pressure relief device. (ix) Winding temperature indicators.
(x) Buchholz relay. (xi) HV and LV bushing. (xii) Oil temperature indicator. (xiii) Oil level gauge.
Fig. The distribution transformer 1.10 Regulation and efficiency of a transformer: Regulation of a transformer: When a transformer is loaded with a constant primary voltage, the secondary voltage decreases (assuming lagging power factor. It will increases if power factor is leading) because of its internal resistance and leakage reactance. Let V 2' = Secondary terminal voltage at no load. V 2 = secondary terminal voltage on full load. The change in secondary terminal voltage from no-load to full load is = V 2' - V 2 This change divided by V Z" is known as regulation `down'. If the change is divided by V 2 i.e. full load secondary terminal voltage, then it is called regulation ‘up’. V2 '−V2 X 100 V2 ' V2 '−V2 %Reg. Up = X 100 V2 In further treatment, unless stated otherwise, regulation is to be taken as regulation down.
%Reg. Down =
Efficiency of a Transformer: The efficiency of a transformer at a particular load and power factor is defined as the output divided by the input - the two being measured in the same units (either watts or kilowatts). Efficiency = Output/Input. But a transformer being a highly efficient piece of equipment, has very small loss, hence it is impractical to try to measure efficiency by measuring input & output. These quantities are
nearly of the same size. A better method is to determine the losses and then to calculate the efficiency form: Efficiency = Output/ Output+ Losses. But total losses = Copper losses + iron or core losses. 1.11 Application: Transformer is used in different fields. In generation station, the generated voltage rating is normally 11 KV. The transmission line voltage is 132 KV. So at the transmitting substation, transformer is used to raise the generated voltage 132 KV. At the receiving substation transformer is used to step down the incoming voltage to a desired value. The usable voltage rating is 220 Volts line to ground and 440 Volts line to line. Distribution transformer is used to meet the demand of 220/440 Volts. In electronic field, it has also wide use. But its power is very small compare with the transformer used in substations. a) In rectifier circuit, it is used as a step down transformer. b. For circuit isolation, it is also used is variable power supply unit. c. For impedance matching between two circuits or elements, transformer is used. d. It used in amplifier and oscillator circuits. e. In radio receiver, it is used as intermediate frequency (LF) transformer and audio frequency (A.F) transformer. f. Audio transformers are used to transfer complex signals containing energy at a large number of frequencies from on circuit to the other. g. A.F (output) transformer are used to change the impedance level of the output signal to the impedance level of load and/or to provide d.c isolation from the amplifier. 1.12 Objectives: 1) Design of a good cost-effective transformer. 2) To lower the production cost of aluminium wound transformer compare to copper wound transformer. 3) To compute the other components of the transformer. 4) To construct the above designed transformer. 5) To test the above-mentioned transformer. 1.13 Methodology: Collection of data from related transformer manufacturing company in Bangladesh, like Energy pack Engineering for copper wound 100KVA 11/.415KV distribution transformer. And the help of text books. Comparison of collected data with product of Lams Electronics. Collection of data for aluminium wound 100KVA distribution transformer and compared with testing panel in Lams Electronics. Calculation of losses and Efficiency of the transformer. Finally design and construct the transformer. Chapter 2 Design & Construction of Transformer
2.1 Design and construction of the 100 KVA, 11I0.415 kV Distribution transformer: We would like to select the 100 KVA, 1110.415 kV transformer to design and construction. 2.2 Description of a 100 KVA Transformer: Before a starting with a working design, a brief specification of a 100 KVA, 11 /.415 KV aluminum-wound transformer should be available. The first step of the design is to select the number of turns of coils and proceed further towards estimating the coil configuration till arriving at the window height of the core frame. Based on the calculated window height, the design of the secondary coil is done. Further, the core diameter, step width, core stack, core area, flux density etc. are calculated with the design output available. The next step of the design is the formation of the coil diameter and limb center of core. Based on the window height and limp center of the core frame, a detailed design of the core up to the weight of the complete set are estimated. Working details of low voltage and high voltage windings, tapping, placements, internal clearances, weight of conductor etc. are also done. Next comes the calculation of performance figures. The method of calculation of resistance, losses, no-load current, percentage impedance, ratio error, efficiency, regulation etc. have been dealt with. After completion of the internal active part design, it has covered the procedure of designing the tank with different types of radiators. Apart from the conventional elliptical tube and pressed steel radiators. The corrugated wall panel has been discussed. Corrugated wall radiators have some advantage over the other two types of radiators in respect of overall dimension, weight and cost. Reduced weight and dimensions have inspired the manufacturers to use a corrugated wall panel in export consignments for achieving better container load ability. A few paragraphs have also been added to designing the core frame, core stud, tie rod, footplate conservator, etc. Calculations of the oil volume and overall weight and dimensions have also been discussed. Proposal Specification: Here, we shall establish a working design of a 100 KVA transformer, the brief specifications of which are as follows: Table 2.1 Specification of a 100 KVA transformer Rating Rating No load voltage ratio 11000/415 Volts. No. of phase/frequency 3 Ph/50Hz Connection Delta/Star- Dyn11 Winding material Aluminum Tapping on HV At + 2.5%, +5% for RV variation No-load and load loss (maximum) 260/1760 W (maximum) Impedance 4.5% Maximum flux density 1.6 tesla Maximum current density 1.5 A / Sq. mm. Temperature rise 40/500C 2.3 Primary Coil: Table 2.2 Primary coil (connected in Delta) Voltage per phase
Vp = 1 1000 Volts
Current per phase
Ip =
Current density (assumed)(Cd) Conductor area
100 =3.03A 3Χ11
1.5 A/sq mm (maximum) 3.03 =2.02 sq mm (minimum) 1.5 πd 2
Equivalent conductor diameter
4
=2.02 , d =1.6037 mm
Let us propose to select the next round figure: d = 1.7 mm. Area of the proposed conductor
πd 2 4
Therefore, the working current density is
=
π ×1.7
2
4
=227 sq mm.
3.03 =1.335 A/sq mm 2.27
Number of Turns: Voltage per turn, Et = K Q, where, Q is the rated KVA. The value of ‘K’ in between 0.32 to 0.35 for aluminum-wound transformer may be taken. For our calculation, let us assume, K = 0.33 Therefore, Et = 0.33 100 = 3.3 No. of secondary turns =
Secondary phase voltage 415/ 3 = =72.06 turns E, 3.3
Let us round it off to 73T. Primary phase voltage
No. o f secondary turns at normal= Secondary phase voltage Χ Secondary turns =
11000 Χ73T 415/ 3
= 46x73T = 3358T Additional turns towards 5% tapping voltage =
3358 T ×5 =168 T 100
Total primary turns = 3358 + 168 = 3526 T No. of coil per phase (assumed) = 4 Nos. Turns per coils = 3526/4 = 882 T The HV winding design may proceed as follows: Table:2.3 Vp Ip Wirer size Area Cd (Current density) Turns (N)=
11000 V 3.03 A 1.7 mm diameter 2.27 sq mm 1.335 A/sq mm
11000 3 ×73T =3358 T 415/
5% turns towards tapping =
3358T × 5 =168 T 100
Total HV turns per phase = (3358T + 168T) = 3526 T. Coil per phase = 4 Nos.
Turns per coil = 882 T Table: 2.4 Description (1) Bare conductor (2) Covering thickness towards DPC Covered conductor 3) Gap between two consecutive conductors (assumed)
Insulated size of conductor with tolerance
Length 1.7 mm (+)0.2 1.9 mm
Radial 1.7 mm (+)0.2 1.9 mm
(+)0.05
(+)0.05
working 1.95 mm x 53
(4) Turns per layer: 882117+1) = 53 X 2.05 mm & 17 layer
1.95 mm (+)0.1 (4 mil inter- layer insulation)
53 2.05 mm (x) 17 (No. of layer – 17 which must be odd)
Axial length of HV coil Rounded-off to
103.35 =104 mm
34.85 mm =35 (Radial build of HV coil)
As of now we have concluded that there are 4 HV coils per phase, each having 882 turns, the winding length of each coil being 104 mm, the radial build of coil being 35 mm, and the inter-layer insulation 4 mil Kraft paper. Estimating the core window height: (i) Total axial length of four coils: 4x 104 mm = 416 mm (ii) Gap between yoke to top and bottom coil: 2x25 mm = 50 mm (iii) Gap between two tap coils at the center: 1 X 10 mm = 10 mm (iv) Gap between plain coils 2x7 mm = 14 mm Therefore, core window height = 490 mm Secondary Coil: Table 2.5 Secondary coil (connected in Star) The LV winding design may proceed as follows: 415 Voltage per phase Vs = =239.6V Current per phase Current density (assumed) Conductor area
Is =
3 100 =139.12A 3 Ă—0.415
Cd = 1.5 A/sq mm (maximum) 139.12 1.5
No. of strips proposed to be used in parallel
2 Nos.
= 93 sq mm (minimum)
Approx. area of each strip Disposition of strips No. of turns per phase (as calculated earlier) Space required to accommodate 38 T Since the strips are placed one above the other, we need to keep some space for transposition, which creates the need for space for one moue additional turns, as such, space required to accommodate transposition. Gap between LV coil to yoke at the top and the bottom. Window height of core (W/H) as calculated before Axial length available to accommodate 40T Approximate width of LV strip, including DPC covering and with working tolerance Air gap between two consecutive conductors (assumed) Therefore, width of the bare conductor taking in to consideration of covering thickness 0.4 mm Let us conceive a strip having bare width of 11 mm (which is a round figure) Area of each strip (as calculated before) Therefore, the approximate depth of the bare conductor
93/2 = 46.5 sq mm (minimum) 1 width × 2depth 73 T 38+1=39T 39+1=30T
10 mm 490mm 490 –(2x10) =470mm 470/40 =11.75mm .01mm 11.75 –(0.4+0.1) =11.25mm 11mm 44.5 sq mm 44.5/11 = 4.05 mm
We now propose to use two strips in parallel having a bare size 11 × 4.5 mm and placed as (1 W × 2D) Area of proposed strip We need to reduce the area by its roundingoff factor
11 × 4.5 mm × 2 nos. placed l WX2D
11x 4.5 = 49.5 sq mm 0.21 sq mm up to a depth of 1.6 mm 0.36 sq mm up to a depth of 2.24 mm 0.55 sq mm up to a depth of 3.25 mm 0.86 s mm above 3.25 mm Since, in this case, the depth is 4.5 mm, the 49.5-0.86=48.64 sq mm rounding-off factor is 0.86 sq mm. Therefore, the net area of one strip, after taking into consideration the rounding-off factor, is Total area of two strips 48.64 × 2=97.28 sq mm Working current density
13.34 =1.37 A/sq 97.28
The LV winding design may proceed as follows: Table:2.6 Voltage per phase Current per phase Strip size Area of strip Current density (Cd) Turns per phase Nos. of coil per phase Turns per coil No. of layers Turns per layer Transposition
Vs = 239.6V 1s = 139.12V 11 × 4.5 mm × 2 nos. in parallel 97.28 sq mm 1.37 A/sq mm 73 T 1 ns. 73 T 2 layers 38 turns To be provided
Table: 2.7
Description (1) Bare strip size (2) Covering thickness towards DPC Covered strip size (3) Gap between two consecutive strips (assumed) (4) Insulated size of strip with working tolerance (5) Placement of strip (1 W × 2 D) (6) Effective dimension of each turn (7) Space required to accommodate 38 turns i.e. (38+2=40 turns)
Length 11 mm (+) 0.4 11.4 mm (+) 0.1
Radial 4.5 mm +0.4 4.9 mm (+) 0.1
11.5mm
5.0 mm
×1 W 11.5 mm
×2 D 10 mm (build for 1st layer) + 0.25 mm (10 mil interlayer insulation) + 10 mm (build for 2nd layer) 20.25 21.00 (rounded off)
× 40
(8) Length of LV coil 460 mm (9) Use end packing of 5 mm on (+)10 mm either side of the coil i.e.( 5 × 2 = 10mm) (10) length of LV coil with packing 470 mm
21.0 (Radial build of LV coil)
2.5 Core diameter The approximate core diameter may be calculated as follows: voltage per turn, E1 =
415 / 3 = 3.282 73
flux density (available from specification, Bm = 1.6 tesla stacking factor (assumed) = 0.97 Gross core area (to be calculated) = A g in sq cm The gross core area may be calculated by using the formula:
Et = Phase voltage/turns = 4.44 × f × Bm × A g × 0.97 × 10 -4 Where, f = 50 Hz. The above equation may be rewritten after putting the value of frequency as: Et= 4.44 × 50 × Bm × Ag × 0.97 × 10 -4 Et = 2.22 × 102 × Bm × Ag × 0.97 × 10 -4 Et = 2.22 × B m × A g × 0.97 × 10 -2 Et ×102 2.22 ×Bm ×0.97
sq cm =
3.282 ×10 2 95.5 sq mm (minimum) 2.22 ×1.6 ×0.97
Let us conceive a core stack of 9 steps having rounding-off factor 0.935 (assumed) πd 2 4
d=
× 0.935 = 95.5 (where d = core diameter)
95.5 × 4 =11.403 cm π × 0.935
let us conceive a round figure of core diameter = 114 mm. 2.6 Step Width The core limb has a diameter of 114 mm and has 9 steps. The first and 9` h steps may be taken as 110 mm and 40 mm respectively. The balance 7 steps may be chosen as 105 mm, 100 mm, 90 mm, 80 mm, 70 mm, 60 mm and 50 mm. While selecting the step width, we must bear in mind that there should be at least a difference of 5 to 10 mm between the two consecutive steps and they should be in descending order. Let us re-write the step width as under: 1st step 2nd step 3rd step 4th step 5th step 6th step 7th step 8th step
110 mm (L1) 105 mm (L2) 100 mm (L3) 90 mm (L4) 80 mm (L5) 70 mm (L6) 60 mm (L7) 50 mm (L8)
9th step
40 mm (L9)
Core stack calculation of core stack (K) k = d 2 - L2 where d is the core diameter and L is the step width. 1st step (L1=110 mm), k1 = 114 2 - 110 2 = 29.93 mm =29.93 mm 2nd step (L2=105 mm), k2 = 114 2 - 105 2 (-)29.93 = (44.39-29.93) mm=14.46 mm 3rd step (L3=100 mm),
k3 = 114 2 - 100 2 (-)44.39 = (54.37-44.39) mm=10.34 mm 4th step (L4=90 mm), k4 = 114 2 - 90 2 (-)54.73 = (69.97-54.73) mm=15.24 mm 5th step (L5=80 mm), k5 = 114 2 - 80 2 (-)69.97 = (81.21-69.79) mm=11.24 mm 6th step (L6=70 mm), k6 = 114 2 - 80 2 (-)81.21 = (89.97-81.21) mm=18.76 mm 7th step (L7=60 mm), k7 = 114 2 - 60 2 (-)89.97 = (96.93-89.97) mm=6.96 mm 8th step (L8=50 mm), k8 = 114 2 - 50 2 (-)96.93 = (102.45-96.93) mm=5.52 mm 9th step (L9=40 mm), k9 = 114 2 - 40 2 (-)102.45 = (106.75-102.45) mm=4.30 mm K = K1+ K2+ K3+ K4+ K5+ K6+ K7+ K8+ K9 K =29.93+14.46+10.34+15.24+11.24+8.76+6.96+5.52+4.30 K =106.75 2.8 Core area: The gross core area can be calculated from core steps and core stacks as follows: Step No.
Step width (mm)
Core stack (mm)
Gross core area (sq mm)
Total gross core area (sq mm) 9594.2
1 110 (L1) 29.93(K1) 3292.3 (L1 × K1) 2 105 (L1) 14.46(K2) 1518.3 (L1 × K1) 3 100 (L1) 10.34(K3) 1034.0 (L1 × K1) 4 90 (L1) 15.24(K4) 1371.6 (L1 × K1) 5 80 (L1) 11.24(K5) 899.2 (L1 × K1) 6 70 (L1) 8.76(K6) 613.2 (L1 × K1) 7 60 (L1) 6.96(K7) 417.6 (L1 × K1) 8 50 (L1) 5.52(K8) 276.0 (L1 × K1) 9 40 (L1) 4.30(K9) 172.0 (L1 × K1) The net core area can be calculated as: Net core area = Gross core area × Stacking factor The stacking factor may be assumed as 0.97 for all practical purposes Therefore, the net core area = 9s94.2X0.97 = 9306.374 sq mm. 2.9 Flux density: Flux density can be calculated as follows:
Et = Phase voltage/turns =4.44 × f × Bm × Ag × 0.97 × 10-4 Bm=
Et ×10 4 tesla 4.44 × ∫ ×Ag ×0.97
where, Et = 239.6/73 = 3.282 f = 50 Hz Ag = 9594.2 sq mm = 95.942 sq cm 3.282 ×10 4
Therefore, Bm= 4.44 ×50 ×95.942 ×0.97 tesla = 1.592 tesla. 2.10 Core diameter and Core limb center Coil diameter and core limb center can be determined as under: Core diameter: 114 mm (previously calculated) Radial build of LV coil: 21 mm (previously calculated) Radial build of HV coil : 35 mm (previously calculated) Radial clearance between core to LV coil : 3 mm (assumed) Radial clearance between LV and HV For 11 KV transformer) : 3 mm (assumed) Based on the above parameters, the coil diameter and limb center can be calculated as follows: Table 2.9 Core diameter R × 2 Radial gap between core to LV coil
Radius 57 × 2 (+)3
LV coil inside diameter Radial build of LV coil
60 × 2 (+)21
LV coil outside diameter Radial gap between LV and HV coils
81 × 2 (+)10
HV coil inside diameter Radial build of HV coil
91 × 2 (+)35
Diameter 114 mm (+)3 117 (+)3 120 mm (+)21 141 mm (+)21 162 mm (+)10 172 (+)3 182 mm (+)35 252 mm (+)13
HV coil outside diameter Gap between HV limb between the phases Core limb centre 2.11 Core Details Since the mitered cut core has enormous advantages over the conventional rectangular cut core, we have conceived a core frame, details of which have been shown as follows: Core frame size and other core details:
Core diameter== 114 mm Window height (W/H) = 490 mm Limb center (C/L) = 265 mm Grade of core = 27 - M4 Gross core area = 95.942 sq cm Nos. of core steps = 9 Core step width: 110/105/100/90/80/70/60/50/40 mm Core stack = 29.93/14.46/10.34/15.24/11.24/8.76/6.96/5.52/4.30 mm Total core stack = 106.75 mm
2.12 Approximate weight of core Without going for detailed calculations the approximate core weight may be calculated with the following formula: Weight of the complete set of core in kg =[(3 × W/H+4 XC/L) + (2 × Width of 1st step × 0.86)] × Gross core area × Density of core material × 0.97 × 10-3 Where, W/H, C/L and width of the 1 st step are in cm, gross area in sq cm and density of material 7.65 g/cc. Therefore, approximate weight of the complete set of core: =[(3 × 49+4 × 26.5) + (2 × 11 × 0.86)] × 95.942 × 7.6 × 0.97 × 10-3 = 193.6 kg. 2.13 Step-wise Weight Calculation of core: Details of Step (A) i.e. two side limps:
Fig. 5.3 Details of step A The shape of lamination - A has been shown in Fig. 2.3. While calculating the number of pieces, the thickness of individual lamination will be taken as 0.27 mm The length of the lamination is represented by `L' and the width by ‘W’ The length can be calculated as: (W/H +2W) mm Where, W/H and W are in mm. The weight of each step can be calculated as: (L-W)XWXKX7.65X0.97X10-3kg Where, L, W and K are in cm. Details of Step `A' Table 2.10 Note: The number of pieces should be adjusted in such a way that it is divisible by `4', as Step No. 1. 2. 3. 4. 5. 6. 7. 8. 9.
Step width (W (mm) 110 105 100 90 80 70 60 50 40
Step length (L) Step stack(K) Nos. of pieces Weight (mm) (mm) (no.) (kg) × 710 2 29.93 220 29.32 700 2 × 14.46 108 13.41 690 2 × 10.34 76 9.05 670 2 × 15.24 112 11.81 650 2 × 11.24 84 7.61 630 2 × 8.76 64 5.10 610 2 × 6.96 52 3.41 × 590 2 5.52 40 2.21 570 2 × 4.30 32 1.35 Total 2 × 106.75 778 83.27 such, the calculated figures are rounded-off to the nearest numbers, which are divisible by `4'. Details of Step ‘B’ i.e. center limb: The shape of lamination - B has been shown in Fig. 2.4. The length can be calculated as: (W/H + W) mm Where, W/H and W are in mm. The weight of each step can be calculated as:
(L-1/2W)XWXKX7.65X0.97X10 -3kg Where, L, W and K are in cm.
Fig. 5.4
Details of step B
Details of Step `B' Table 2.11 Note: The number of pieces should be exactly half of A-step. Step No. 1. 2. 3. 4. 5. 6. 7. 8. 9.
Step width (W (mm) 110 105 100 90 80 70 60 50 40
Step length (L) Step stack(K) Nos. of pieces Weight (mm) (mm) (no.) (kg) 600 29.93 110 13.31 595 14.46 54 6.11 590 10.34 38 4.14 580 15.24 56 5.44 570 11.24 42 3.54 560 8.76 32 2.39 550 6.96 26 1.61 540 5.52 20 1.05 530 4.30 16 0.65 Total 106.75 Total 38.24 Details of step ‘C’ i.e. top and bottom yoke core: The shape of lamination - C has been shown in Fig. 2.5. The length can be calculated as: (2 × C/L + W) mm The weight of each step can be calculated as: [(L - W) × W - 1 /2 W2] × K × 7.65 × 0.97 × 10-3 kg Where, L, W and K are in cm.
Fig. 5.5 Details of step C
Details of Step `B' Table 2.11 Step No. 1. 2. 3. 4. 5. 6. 7. 8. 9.
Step width (W (mm) 110 105 100 90 80 70 60 50 40
Step length (L) Step stack(K) Nos. of pieces Weight (mm) (mm) (no.) (kg) 640 2 × 29.93 220 24.55 635 2 × 14.46 108 11.35 630 2 × 10.34 76 7.75 620 2 × 15.24 112 10.33 610 2 × 11.24 84 6.81 × 600 2 8.76 64 4.66 590 2 × 6.96 52 3.19 580 2 × 5.52 40 2.12 570 2 × 4.30 32 1.33 Total 2 × 106.75 Total 72.09 Note: The number of pieces should be exactly same as that of A-step Total weight of core : Weight of step A = 83.27 kg Weight of step B = 38.24 kg Weight of step C = 72.09 kg Total weight core = 193.60 kg Core chart specification : Material : CRGO, Grade - M4 Thickness :0.27 mm Core diameter : 114 mm Window height :490 mm Limb centre :265 mm Core stack : 106.75 mm Weight/ set : 193.6 kg. 2.14 Winding details
Table 2.13 (a) Low voltage windings No Description 1. Conductor material 2. Type of coil 3. Connection 4. Size of bare conductor 5. Covering 6. Size of covered conductor 7. Conductor disposition 8. Transposition (if provided) 9. Turns per phase 10. Number of coils per phase 11. Turns per coil 12. Number of layers 13. Turns per layer 14 Inter-layer- insulation 15. Tapping details 16. Inside diameter of coil 17. Outside diameter of coil 18. Winding length of coil 19. End packing details 20. Overall length of coil 21. Approximate bare weight of conductor per transformer 22. Approximate covered weight of conductor per transformer including leads
Design parameters Aluminum, electrolytic grade Spiral Star 11 X 4.5 mm X 2 in parallel DPC - 0.4 mm 11.4 X4.9 mm 1 width X2 depth (1 WX2D) Yes, at the centre of each layer 73 T 1 No. 73 T 2 layers 38 T 10 mil Nil 120 mm 162 mm (maximum) 460 mm 5 mm on either side 470 mm 27.1kg 27.1 Ă— 1.07 = 29.0 kg.
High voltage windings:
Picture.5.2 High voltage Winding Table 2.13 (b) High voltage windings No 1.
Description Conductor material
2. 3. 4. 5. 6. 7. 8.
Type of coil Connection Size of bare conductor Covering Size of covered conductor Conductor disposition Transposition (if provided)
Design parameters Aluminum, electrolytic grade Cross-over or sectional Delta 1.7 mm diameter DPC - 0.2 mm 1.9 mm diameter nil nil
9.
Turns per phase
10. 11. 12. 13. 14 15. 16.
Number of coils per phase Turns per coil Number of layers Turns per layer Inter-layer- insulation Tapping details Inside diameter of coil
17. 18. 19. 20. 21.
Outside diameter of coil Winding length of coil End packing details Overall length of coil Approximate bare weight of conductor transformer Approximate covered weight of conductor per 45 transformer including leads
22.
3358+168 for tapping = 3526 T (total) 4 Nos. 882 T 17 layers 52/51 T (average) 4 mil (0.1 mm) 0- (710)-(794)-(878)(F) (3) (4) (5) (8) (7) (6) 182 mm 252 mm 104 mm nil per 45 kg
× 1.14 = 21.5 kg.
2.15 Coil Assembly Table 2.14 Coil assembly specification 100 KVA ,11000/433 KV No. of LV coil/limb. No. of LV coil/limb. (plain) No. of LV coil/limb.(tap) HV coil LD. × O.D. × Length LV coil LD. × O.D. × Length Gap between plain coil Gap between tap coil Gap between HV to yoke Gap between LV to yoke Size of plain block Between L V to yoke Size of dovetailed block between HV coil
1 No 2 Nos. 2 Nos. 182(DX252(DX104 mm 120(DX 162(DX470 mm 7 mm 10 mm 25 mm 10 mm N/A 40 × 65 mm 40 × 40 mm × 8 nos./cycle.
Delta wire size Size of thimble to be used Size of LT busbers Tapping switch, if any
2.3 Φ (Cu) 95 sq mm N/A 5 position, 30A/11 KV
2.16 Weight of LV and HV Aluminum: Table 2.15 Weight calculation of LV and HV aluminum Particulars Inside diameter of coil (DI Outside diameter of coil (D2) Mean diameter of coil (Dm) [Dm = (D1+D2)/2] Mean length of turns (L 1)
LV coil 120 mm 162 mm 141 mm
HV coil) 182 mm 252 mm 217 mm
443 mm
682 mm
[L I = π X Dm] No. of turns (T) Total length of conductor (L) (L = L1 × T) Specific gravity of aluminum (s) Cross-sectional area of the conductor (A) Bare weight of conductor being used in one phase (L × A × S × 10-6) Bare weight of conductor being used for 3 phases Weight of the covered conductor (considering a coverage of 7% for LV strip and 14 % for HV wire, including leads).
73 T 33668 mm
3358+168 3526 T 2395184 mm
2.76 g/cc 97.28 sq mm 9.0 kg
2.76 g/cc 2.27 sq mm 15.0 kg
27 kg
45 kg
27 × 1.07 29 kg
45 × 1.14 51.5 kg
2.17 Winding Resistance and load loss Table 2.16 Calculation of winding resistances and load loss Particulars Inside diameter of coil (DI Outside diameter of coil (D2) Mean diameter of coil (Dm) [Dm = (D1+D2)/2] Mean length of turns (L 1) [L I = π X Dm] No. of turns at normal (T) Total length of conductor (L) (L = L1 × T) Resistively of electrolytic aluminum (k) Cross-sectional area of the conductor (A) Winding resistance per phase at 75 oC R= (L × K × 10-3)/A Current per phase (I) 12R × 3 phase at 750c Total 12 R (LV + HV) Approximate stray loss for 100 KVA Load loss at rated Load and at 75 0C Load loss (guaranteed)
LV coil 120 mm 162 mm 141 mm
HV coil) 182 mm 252 mm 217 mm
443 mm
682 mm
73 T 33668 mm
3358 T 2280608 mm
0.0345 Ohm-cm at 750C 97.28 sq mm 0.012 Ohms
0.0345 Ohm-cm at 750C 2.27 sq mm 34.7 Ohms
139.12 A 214 × 3=642 W 642+957 1599 W 100 W 1599+100= 1699 W 1760 Watts
3.03 A 319 × 3 = 957 W
Maximum
2.18 No-Load Loss Table 2.17 Calculation of no-load loss at rated volta2e and frequency Core diameter Core window height Core limb centre
114 mm 490 mm 265 mm
Gross core area Net core area Working flux density Grade of core being used Specific loss (watts/kg) at 1.6 teals for 27 - M4 grade of core. (Value taken from standard core characteristics curve available from Nippon Steel Corporation book, a specimen curve has been shown in Fig. 2.8) Handing factor (assumed) (on account of slitting, shearing, knotching, air gap and human error during assembly, core loss tends to increase by 25% than that of the specified value). Effective core loss per kg, including the handling factor Total weight of core being used Calculated no-load loss No-load loss (guaranteed)
95.942 sq cm 93.06 sq cm 1.592 tesla 27 – M4 1.0 W/kg
25%
1 X 1.25 = 1.25 W/kg 193.6 kg 193.6 Ă— 1.25 = 242 W 260 W (maximum)
Fig: 2.8 2.19 Percentage reactance, Resistance and Impedance Calculation of percentage reactance, percentage resistance and percentage impedance a) Percentage reactance: The formula commonly being used for calculating reactance is as follows: Percentage reactance = x (%) 7.91× ∫ ×1s × T2 ×π × D b +b × a + 1 2 × 10-6 X (%) = Vs × A1 3 Where, f = Rated frequency = 50 Hz.
IS = Rated secondary current = 139.12 A No. of secondary turns per phase = 73 T D = Mean diameter of LV and HV coil (141+217)/2 = 179 mm Vs= Rated secondary phase voltage = 415 3 = 239.6 Volts. A1 =Average stack length of LV and HV coil =(460+440)/2 = 450 mm a = Radial gap between LV and HV coil in cm = 1.0 cm b1 = Radial build of LV coil in cm = 2.1 cm b2= Radial build of HV coil in cm = 3.5 cm 7.91 ×50 ×139.12 × 732 ×π ×179 1.0 + 2.1 + 3.5 × × 10-6 Therefore, X (%) =
239.6 ×450
3
= 4.36 % (Multiply the above by 0.95 towards leakage flux.) Therefore, the calculated value of x (%) = 4.36x0.95 = 4.14% b) Percentage reactance: The formula commonly being used for calculating reactance is as follows: R (%) =
Calculated loss in kW × 100 Rated KVA1
Where, Calculated load loss = 1.696 kW and Rated KVA = 100 KVA Therefore, R (%) =
1.696 × 100 100
= 1.696 % C) Percentage Impedance: Percentage Impedance is the vectorial sum of percentage reactance and Percentage resistance, i.e. Z (%) = x(%)2 + R(%)2 = 4.14 2 +1.6962 = 4.47% The guaranteed percentage impedance is 4.5 % , with a tolerance +10% in accordance with IS-2026. 2.20
Efficiency
Calculation of efficiency at different percentage of load and at different power factors Output
Efficiency is defined as Input × 100 and is expressed as a percentage. Output
The above expressed may be re-written as Output + Losses × 100 Where, Output is the rated KVA and losses are equal to the sum total of no-load loss and load loss in kW Efficiency is the rated load and at unity power factor =
100 × 100 = 98.1% (100 + 0.242 +1.696)
Efficiency at 75% load and at unity power factor
=
100 × 0.75 × 100 = 98.43% (100 × 0.75) + 0.242 + (0.75 2 ×1.696)
Efficiency at 50% load and at unity power factor =
100 × 0.50 × 100 = 98.69% (100 × 0.50) + 0.242 + (0.50 2 ×1.696)
Efficiency at 25% load and at unity power factor =
100 × 0.25 × 100 = 98.63% (100 × 0.25) + 0.242 + (0.25 2 ×1.696)
Efficiency at any percentage of loading (n) and at unity power Factor =
100 × n × 100 (100 × n) + 0.242 + (n 2 ×1.696)
Efficiency at rated and at 0.8 power factor =
100 × 0.8 × 100 = 97.63% (100 × 0.8) + 0.242 +1.696
Efficiency at 75% load and at 0.8 power factor =
100 × 0.75 × 0.8 × 100 = 98.05% (100 × 0.75 × 0.8) + 0.242 + (0.75 2 ×1.696
Efficiency at 50% load and at 0.8 power factor =
100 × 0.50 × 0.8 × 100 = 98.36% (100 × 0.50 × 0.8) + 0.242 + (0.50 2 ×1.696
Efficiency at 25% load and at 0.8 power factor =
100 × 0.25 × 0.8 × 100 = 98.29% (100 × 0.25 × 0.8) + 0.242 + (0.25 2 ×1.696
Efficiency at any % age of loading (n%) and at any power factor =
100 × n × pf × 100 (100 × n × pf ) + 0.242 + ( n 2 ×1.696
2.21 Maximum Efficiency No-load loss of the transformer is constant and does not change with the variation of load. However, load loss changes in respect of the lording pattern. The transformer will yield maximum efficiency at a load when the no-load loss and load loss are equal and may be represented as: K = Load at which maximum efficiency will occur : And max. efficiency =
No-load loss load loss
KVA × K × 100 ( KVA × K ) + No-load loss + ( K 2 × load loss )
In this case, no-load loss = 0.242 kW And load loss = 1.696 kW There the load at which maximum efficiency will occur (k) =
0.242 1.696
= 0.378 i.e. at 37.8% load, maximum efficiency will occur. And maximum efficiency = 2.22
Regulation
100 × 0.378 × 100 =98.74% (100 × 0.378) + 0.242 + (0.378 2 ×1.696)
The voltage ratio defined for any transformer is at no-load. During lording the load voltage drops down, based on its percentage reactance and resistance. For any assumed load other than the rated load and any power factory, the percentage regulation is approximately equal to : (n.E r %.cosθ + n.E x . sin θ +
(n.E x .% cosθ − n.E x % sin θ ) 2 × 100 = 98.74% 200
Where, n = percentage of loading Er % = percentage resistance Ex % = percentage reactance Cos θ = power factor Sin θ = ‘sine’ component of power factor angle (a) Regulation at rated load and at unity power factor : Where, n = 1.0 Er % = 1.696 Ex % = 4.14 Cos θ =0.8 ( θ = 36.860 ) Cos θ = 0.6 Therefore, regulation at rated load and at 0.8 pf : (1.696 × 0.8 + 4.14 × 0.6) +
(4.14 × 0.8 − 1.696 × 0.6) 2 = 3.87% 200
2.23 The Design of Tank: Though the tend acts as an external enclosure to the active part (core coil assembly), still, the performance of a transformer mostly depends on the right kind of tank design. Various internal gaps and clearances, easy and safe termination arrangement, locking, fixing etc. are some of the areas where much attention is needed. The following assumptions are made while calculating the tank dimensions: (i) Gap between HV coil to the inside of tank on length side - 25 mm (ii) Gap between HV coil to the inside of tank on width side - 40 mm (iii) Gap between core yoke to tank bottom - 40 mm (iv) Gap between core yoke to ratio switch base - 20 mm (v) Height of ratio switch - 90 mm (vi) Gap between ratio switch top to the inside of tank cover - 40 mm Based on the above internal clearances, let us form the tank dimensions as follows: Length of tank (inside): 2 × C/L of core + HV coil O.D.+ 2 × 25 mm :2 × 265+253+50 : 833 mm 835 mm (after being rounded-off (b) Breadth of tank (inside) : HV coil O.D. + 2 × 40 mm : 253 + 80 :333 mm : 335 mm (after being rounded off) © Height of tank (up to tank flange) : 40 + botton yoke insulation + W/H of core + 2 × widts of 1st core step +
20 + R/S height + 40 mm = 40+5+490+2 Ă— 110+20+90+40 =905 mm Therefore, the final tank dimensions are : Length = 835 mm Breadth = 335 mm Height = 905 mm 2.24 Radiators Use of radiators During service, the tank body can dissipate a total loss equivalent to 500 W/sq m (maximum) of the tank surface area. In case the total loss, i.e. (no-load + load loss) is more than the loss dissipated by the tank surface, the loss in excess is required to be dissipated with the help of additional cooling surface which is commonly called as radiator. Classification of radiators Radiators commonly being used are of three different types: (i) Conventional round / elliptical tube radiator. (ii) Pressed steel radiator. Corrugated wall panel. Round tube radiators are almost obsolete these days since its efficiency with respect to an elliptical tube is poor. Moreover, it contains more of oil. As such, we shall discuss here the construction of elliptical tube radiators only. Elliptical tube radiators and pressed steel radiators operate on the convection process of cooling, whereas a corrugated wall panel performs cooling by radiation only. Corrugated wall panel radiators are commonly being used for sealed type transformers and also on such places where there are restrictions on overall dimensions. These radiators are widely used for transformers built for export, as corrugated wall panel transformers occupy less space and can accommodate more transformers in one container during transport. The design approach of each of the above radiators are different. Conventional elliptical tube radiators The standard length of an elliptical tube available in the market is 6.1 meters. It is cut into a number of pieces and welded together to form a radiator bank. Each radiator bank has an inlet and an outlet for free flow of oil. When a transformer is in service, it emits losses, which transform into heat energy. As we know, liquid, when heated up, becomes lighter since it losses its density and causes increase in volume. The same phenomena is applicable for transformer oil also. The heated oil become lighter and tries to come-up by displacing the heavier oil on the top. The heavier oil, in the process of displacement, has no other alternative but to push through the inlet of the header pipe. When the heated oil gets into the inlet header pipe, it losses its heat and becomes heavier again. Due to gravity, this oil will fall down through the pipe and reach the outlet header pipe. Thus, the sequence of oil flow by natural convection is completed. The circulation of flow of oil has been shown in Fig. 2.14. In a transformer, we are required to provide more tubes, having certain fixed length according to the tank height as indicated in Fig. 2.15.
Let us select the length of each tube as 590 mm, which will yield 10 tubes from a 6.1 m length without much wastage. The number of elliptical tubes required can be calculated from the following formula: No of tubes =
1 K −12.5 × A 8.8 × X ×Y L
Where, A = Tank surface area in sq metre (only side wall to be considered). K = Total loss in watts (i.e. guaranteed no-load plus load loss). L = Average oil temp. rise in degrees C (maximum guaranteed oil temperature rise multiplied by 0.8). Y = Unit length of each tube in metre. X = Surface length of elliptical tube in metre. Tank dimensions are : Length = 835 mm = 0.835 m Breadth = 335 mm = 0.335 m Height = 905 mm = 0.905 m Therefore, tank surfaces area for side walls only : A A = 2x(0.835+0.335) X0.905 = 2.1177 sq.m (ii) Guaranteed no-load+ load loss: K K =260 + 1760 =2020 W (iii) Average oil temperature rise : L L = 40x0.8 = 32°C (iv) Unit length of each tube in meter: Y Y = 0.590 m (v) Surface length of elliptical tube : X X = 2x(75+15) x0.9 =162mm=0.162m (vi) Therefore, (vii)
No of tubes =
1 2020 −12.5 × 2.1177 8.8 ×0.162 ×0.590 32
= 43.58 nos. We propose to use two radiators, each having 22 nos. tubes as shown in Fig. 2.17. 2.25 Weight of Tank Weight of tank may be calculated with + 5% accuracy. The weight of individual parts shall be calculated and added together to form the total weight. A standard table may also be consulted while calculating the weight of channels, flats, pipe, tubes etc. For general information, the approximate weight of each meter of elliptical tube having section 57 is approximately 1.5 kg. 2.26 Volume of Oil (a) Calculation of oil in the tank only: Volume of oil in the tank = Tank volume - approximate volume u: coil assembly. Tank volume = Length × Breadth × Height × 10-3 litres, Where the dimensions are in centimeters. Therefore, Tank volume = 83.5 × 33.5 × 90.5 × 10-3 litres, = 253 litres. And approximate volume of core coil assembly
= Volume of core + volume of aluminium + volume of core titti m. ,N - volume of insulating materials. 193.6 ( 29 + 51.5) 30 + + + 10 = 68.3 rounded-off to 68 litres = 7.65 2.76 7 .8 Therefore, volume of oil in lthe tank = 253- 68 =185 litres. (a) Volume of oil in the elliptical tube radiator: The volume of each metre of elliptical tube is 0.9 litre. We have provided 26 meters tube; which means that oil in the tube = 264.9 = 23.4 litres, Therefore, oil in the tube = 24 litres (Approx.) Volume of oil in the radiator header is approx. = 6 litres. Conservator: Conservator is an oil chamber placed on top of the transformer to increase the oil height. Since the bushing being provided are oil communicating type, with the addition of conservator, oil will go up to the bushing top, thereby reducing the risk of voltage failure between the lead and the bushing turret. Volume of the conservator is generally taken as 10% of the volume of the oil in the tank and radiators. In case of a tank with elliptical tube, radiators, the volume of conservator will be 10% of 215 liters. A suitable conservator diameter and length has to be estimated to yield a volume of 21.5 liters. Let us estimate D = 235 mm and L = 500 mm. Volume of conservator = π /4 × 23.52 × 50 × 10-3 = 21.7 liters. Oil in the conservator should be 1/3rd the volume of the conservator i.e. 21.7/3 = 7.2 liters (rounded-off to 8 liters. Table 2.26 Total volume of oil in the transformer 1. Volume of oil in tank 2. Volume of oil in (radiator +header) 3. 4. 5.
185 liters 24+6 = 30 liters (for elliptical tube radiator) 8 liters 3 liters (Approx.)
Volume of oil in conservator Volume of oil in LV Box Others misc. like HV pocket , conservator 4litres(approx.) connection pipe explosion vent etc. Total 230 liters.
2.27 Overall weights and dimensions Overall weight is calculated by adding the weight of individual component and making approximation in some cases. Table 2.27 Overall weight Overall dimensions (with elliptical tube radiators) 1. 2. 3. 4. 5.
Weight of core Weight of aluminum: LV HV Insulation (approx.) Core fittings (Approx.) Other misc. like R/S . leads and cost analysis support etc.
193.6 kg 29.0 kg 51.5 kg 15.0 kg 30.0 kg 3.0 kg
(a) weight of core and winding
322.1 kg = 322 kg
(b)(i) weight of tank
187 kg
(ii) weight of fittings like bushing with fittings, breather, hard wares etc. (iii) WT of tank fittings Š Oil (litters/kg) weight of complete transformer.
25 kg 212 kg 230 litres/196 kg 730 kg
Calculation of overall dimensions Overall length = (200 + 70 + 835 + 7 + 70 + 200) mm = 1382 mm Overall breadth = (140 + 70 + 335 + 7+ 40) mm = 592 mm Overall height = (75 + 5 +905 +40 +280 +200 +60) mm = 1565 mm Cost analysis: The comparative study of copper & aluminium wound transformer is shown bellow. Type of Transformer No load loss Full load loss Efficiency atPrice at localRemarks unity powermarket factor Copper wound210 Watts (Max) 1550 Watts98.8% TK. 2,50,000 100KVA Distribution (Max) Transformer Aluminium wound260 Watts (Max) 1696 100KVA Distribution (Max) Transformer
Watts98.1%
TK. 1,90,000
The efficiency of the aluminium wound transformer & the copper wound transformer is same, the price of aluminium wound transformer is cheaper than the copper wound transformer. So it is a cost effective transformer. Discussions:
From the analyzed of the losses, it was found that the efficiency although the aluminium wound transformer and the copper wound transformer is same but the price of aluminium wound transformer is cheaper than the copper wound transformer. The 100KVA aluminium wound distribution transformer is already be used in Digilab Medical Service Ltd. at mirpur-10, Dhaka. Conclusions: This project paper has covered the design and construction of a 100 KVA, 11/0.415 KV Aluminium wound distribution transformer. These types of transformers are widely used in Electrical power supply and distribution system. We were thinking that, it would be appropriate to discuss briefly a few basic concept of the transformer related theory before undertaking a commercial working design. After that, we discuss in details with figures and tables of the selected 100 KVA, 11/0.415 kV Aluminium wound Distribution Transformer, which is designed and constructed by us. To make a good transformer, we definitely need a good cost-effective design. But along with a good design, we need good machines, good materials and reasonably skilled workmen. We must ensure the availability of all these four things before undertaking manufacturing activities. In fact, once all these four things are available, the percentage of rework/rejection will be less, which in turn reduces the ultimate cost of the product. This project paper has prepared to think about the future growth of power generation with the increasing demand of transformers in Bangladesh. We may hope that, when our designing Aluminium wound transformer will go through the manufacturing, then the costing of this transformer will be cheaper than the copper wound transformer, which is imported or already manufactured by the company in Bangladesh. Recommendations: Normally transformer are wound by copper wire but from our project it was found that it can also be wound with aluminium wire. The efficiency is approximately same but it is cost-effective. So other type of transformer can be wound with aluminium wire. Bibliography: 1) Alfred Still & Charles S. Siskind, Elements of Electrical Machine Design, Year of pubication-1958. McGraw Hill Publishing Co. Ltd. Edition- 1969. 2) B. L Thzraia & A. h Thzraja. A text book of Electrical technology, Year of pubication-195 9, Nirja Construction and Development Co (Pvt) Ltd. RamNagar, New Delhe, India, Edition- 1989. 3 ) Indrajit Dasgupta, Design of transformer, Year of pubication 1990, McGraw Hill Publishing Co. Ltd, Edition- 2002. 4 ) Mohammad Azizul Islam, Ph.D, Electromagnetic Theory, Year of pubication-1969, EP University of Engineering & Technology, Dacca, East Pakistan, Edition-1969. 5) M.V. Deshpande, Design and testing of electrical machines, Year of pubication-1988, McGraw Hill Publishing Co. Ltd, Edition2001.