: CHAPTER 5 Design of Transformers PART A : CONSTRUCTIONAL DETAILS 5.1 INTRODUCTION A transformer is essentially a static electromagnetic device consisting of two or more windings which link with a common magnetic field. One of these windings, the primary, is connected to an alternating voltage source, an alternating flux is produced whose amplitude depends on the primary voltage and number of turns. The primary induced voltage is Z, = 44.4 f4,, T, where f = frequency, 4,, = mutual flux and 7, = number of turns in the primary winding. This flux linking with the secondary winding induces in it a voltage whose value depends on the amplitude of flux and the number of secondary winding turns. The induced voltage is the secondary winding is E, = 4.44 f®,, T, where T, = number of turns of secondary winding. Ratio of voltages is E,/E, = T,/T,. Therefore, any desired value of secondary voltage can be attained by using a suitable number of turns, It must be understood that a transformer is not an energy conversion device but a device that transforms electrical energy from one or more primary a.c. circuits to one or more secondary ac. circuits with changed values of voltage and current. The main reason for extensive use of a.c. power systems is on account of transformers. This is because the transformers allow the power to be transferred from the most economical generator voltage the most economical transmission voltage and power utilization at the most suitable voltage required for different applications. Presently, most of the electric power for industrial and utility purposes, is generated by large hydro-electric plants and steam power stations in the form of three phase a.c. at a frequency of 50 Hz. The voltages of the generators installed at the power plants usually varies from 6.6 kV to 22 kV. To transmit the power over long distances the voltage of the generators has to be increased depending upon quantum of power and the distance in order to reduce transmission losses and to effect economy. On the other hand, the voltage is decreased by the distribution substations to 3.3 KV, 6.6 kV and 11 kV in rural, urban and industrial areas. Ultimately, the voltage used in industrial and domestic premises has to be dropped to 433 V(3ph) or 230 V(1 ph). ‘The raising and lowering of a.c. supply voltages is accompolished by transformers. Therefore, as regards this application, the transformers may be classified as (i) Step-up transformers—transformers which raise the voltage, and (ii) Step-down transformers—transformers which lower the voltage. However, basically each transformer may be used as both a step-up and a step-down transformer because it is a reversible device. ‘The power transformers have a remarkably high efficiency ranging from 95 to 99.5 percent, depending upon the power rating. The greater the power rating of the transformer the higher the efficiency. (5.1)5.2 ELECTRICAL MACHINE DESIGN 5.2 CLASSIFICATION OF TRANSFORMERS The transformer is basically a very simple device. It consists of windings wound on a laminated magnetic core and insulated from it and from each other. The core is actually a magnetic circuit (eg. iron) which serves as a path for the mutual flux. Therefore, the windings encircle the core and the core encircles the windings. There are two general types of constructions employed to achieve this in transformers. Consequently, depending upon the type of construction used, the transformers are classified into two categories as : (i) core type, and (ji) shell type. 5.2.1 Core Type Transformers The magnetic core is built of laminations to form a rectangular frame and the windings are arranged concentrically with each other around the legs or limbs of the core as shown in Fig. 5.1(a). The top and bottom members, called the yokes, connect the two limbs and have a cross- sectional area equal to or greater than that of the limbs. HV. Winding Je wei HV. Winding if Lv. Winding Core Core Yoke (a) Core type (6) Shell type Fig. 5.1. Single phase transformer connection. A single phase transformer may be designed with primary winding wound on one limb and secondary winding on the other limb. This arrangement results in a large separation between the primary and the secondary windings and hence a large leakage reactance. In actual practice, each limb carries one half of the primary winding and one half of the secondary winding so that the two windings can be closely coupled together to keep the leakage reactance low. ‘The low voltage (I.v.) winding is wound on the inside nearer to the core while the high voltage (h.v) winding is wound over the l.v. winding away from core in order to reduce the amount of insulating materials required. 5.2.2 Shell Type Transformers In shell type transformers the windings are put around the central limb and the flux path is completed through two side limbs as shown in Fig. 5.1 (6). The central limb carries total mutual flux while the side limbs forming a part of a parallel magnetic circuit carry half the total flux. Consequently, the cross-sectional area (and hence width) of the central limb is twice that of cach of the side limbs, Both high voltage (h.v) and low voltage (I.v.) windings are divided into a number of coils. The h.v. and Lv. coils are shaped like pancakes and are arranged longitudinally along the coreDESIGN OF TRANSFORMERS 5.3 alternately. This gives rise to a sandwich winding with hv. coils sandwiched between Lv. coils. : In the core type the impression is created that the windings surroundsthe core, whereas with the shell type that the cores surround the windings. 5.2.3 Comparison of Core and Shell Types of Transformers 1, Construction. Core type transformers are much simpler in design and permit easier assembly and insulation of windings. Also, the core type of transformers are easier to dismantle for repair work. 2. The force produced between current carrying windings is proportional to the product of the currents carried by them. These currents tend to be very large under fault conditions. Consequently very large electromagnetic forces are produced when the’ secondary winding -is- short circuited with the primary winding energized. Since, the windings carry currents in opposite direction, there exists a force of repulsion between them. Hence, the inner winding experiences a compressive force crushing it on to the core; the outer winding experiences a tensile force pulling it away as shown in Fig. 5.2. inner winding Tensile force Tensile core force a Compressive force Za Outer winding Fig. 6.2. Electromagnetic forces on transformer windings. In modern power networks, the reliability of transformer operation is very important and therefore the design of the transformers should be such that the windings suffer no damage when short circuited. It is amply clear from Fig. 5.1 that windings in a shell type transformer have greater capability of withstanding forces produced under short circuit conditions as these windings are surrounded and thus braced (or supported) by the core over a large portion of length. On the other hand, the windings in core type construction have a poorer mechanical strength, because they (the windings) are not braced or supported. Therefore, the windings in core type transformers are more susceptible to damage under short circuit conditions, than the windings of a shell type transformer. 3. Leakage Reactance. Due to large space required between the high and low voltage windings, it is not easily possible to subdivide the windings to a great extent in the case of core type transformers, while, in the shell type, the windings can be easily subdivided by using sandwich coils. Thus it is possible to reduce the leakage reactance of shell type transformers to any desired value. 4, Repairs. The windings of a core type transformer are completely accessible except for a small portion in the window. This is of a great advantage in repair work because the coils can be easily inspected. Also, the core type transformer is easy to dismantle for repairs.54 ELECTRICAL MACHINE DESIGN In case of shell type transformers, the coils are surrounded by core for a large length and therefore there is great difficulty in inspection and repair of coils. 5. Cooling. In the case of core type transformer, the windings surround the core. The windings are exposed and therefore the cooling is better in windings than in core. In the case of shell type transformers, the core is exposed and therefore cooling is better in core than in windings. The ‘most vulnerable part of a transformer is the insulation of windings. Therefore, core type of construction is universally followed because it affords better heat dissipation facilities from a part which is most prone to damage on account of heat developed. 5.3 SINGLE AND THREE PHASE TRANSFORMERS Single phase transformers. The cross-sectional view of the windings and core of a I-phase core type transformer is shown in Fig. 5.3(a). Single phase core type transformers use two legged iron frame with one half of the primary winding and one half of the secondary winding wound on each leg. The low voltage (Lv.) and high voltage (h.v.) windings are concentric with each other with lv. winding placed on the inner side nearer to the core. HY. LV. Leg limb) we (a) Core type Window a Lo Lv. Lv. HY. HY. Lv. Lv. Side a firms HY. HY. Lv. la Lv. i Central limb (b) Shell type Fig. 6.3. Single phase core and shell type transformers. Fig. 5.3(b) shows the cross-sectional view of a single phase shell type transformer. The low voltage (I.v.) and the high voltage (h.v.) coils are sandwiched between each other.DESIGN OF TRANSFORMERS: ant ‘Three phase transformers. The generation, transmission and power utilization of a.c. electric energy almost invariably involves the use of three phase networks. The voltage transformation in a 3 phase network can be obtained by using (@ three single phase transformers connected to form a 3-phase bank, (i) a 8-phase transformer. A transformer bank used on 3 phase systems consists of three independent 1-phase transformers with their primary and secondary windings connected either in star or in delta. The 3-phase transformation is also possible through the medium of one 3-phase transformer having a magnetic circuit common to all the three phases. A three phase magnetic circuit for transformation of three phase voltages may be obtained by combining three individual single phase core or shell type magnetic circuits into a common magnetic circuit with suitable changes in the configuration. Fig. 5.4(a) shows a 3 phase core type of transformer. The core consists of three legs with the magnetic circuit completed through two yorks, one at the top and the other at the bottom. ‘A primary and a secondary windings of one phase are wound on one leg. Flux flows up each leg in turn and down the other two legs in general, so that the magnetic circuits of different phases are Windows Top yoke Windings Leos (a) Three phiase coretype transformer in series and therefore independent. Fig. 5.4(b) shows the instant where the flux in the leg carrying the winding of phase R is positive (upwards) maximum while the flux in the other two legs carrying windings phases Y and B is half of the negative (downwards) maximum. It should be noted that the transformer has only two windows. Each of two windows contains . two primary and two secondary windings. 120° 120 Fig. 5.5 shows the less commonly used 3-phase shell type transformer whose construction appears like three 1-phase shell-type cores built on top of 120 one another. The windings of the middle core, phase Y, are * reversed so that the parts of the core carry flux O{2 + Oy? or Oy/2 + Dy/2 instead of flux @p/2 — o,/2 or out ofan arabe 0 affect fconere ia core (6) Phasor diagram cross-section since ®,/2 + ®,/2 is smaller than _ Fig. 5.4, Three phase coretype transformer ©,/2-O,/2 and also O/2 + Dp/2 < OY2 + Oy/2. and its phasor diagram.5.6 ELECTRICAL MACHINE DESIGN TN UR UZZZZA N N N \Y (@) Construction (6) Phasor diagram Fig. 5.5, Three phase shell type transformers. The reversal in the flux ©, in the central limb is brought about/by reversing the windings wound on the central limb so that the flux carried corresponds to the phasor sum of ®,/2 and ,/2 (or ®y and p/2) instead of their difference. This is illustrated by the phasor diagram of Fig. 5.5. It should be noted that (,/2 + @,/2) < (Og/2 - @,/2). In the case of 8 phase shell type of transformers the phase magnetic circuits are in parallel and therefore, independent, if saturation effects in the common paths are neglected. The advantages of a 3-phase transformer over a bank of three I-phase transformers are : (iA 8-phase transformer is lighter, occupies lesser space, cheaper and more efficient than a bank of 1-phase transformers. The reasons for lower cost of 3-phase transformers are : savings in cost of the iron core of the tank and oil (since there is only one tank whose volume is much smaller) of the bushings (since the connections can be made internally and therefore the number of bushings is smaller) and of the auxiliary apparatus.” The higher efficiency of a 3-pliase transformer is on account of the fact that it has a shorter magnetic path and consequently the volume of core and hence the core (iron) loss is smaller. i) In case of 3-phase transformers, there is only one unit to install and operate. Hence, the. installation and operational costs are smaller for 3-phase units. The disadvantages of 3-phase transformers are : ‘The 3-phase transformers are built as a single unit whereas an equivalent bank of single phase transformers consists of three separate units. The weight and hence the cost per unit is thus higher in the case of 8 phase transformers. The three phase unit costs about 15% lesser than the bank. : @ It is more difficult to transport a 3 phase transformer as the weight per unit is more, (ii) In the event of a fault in any phase of a 3 phase transformer, the fault is transferred to the other two phases. Therefore, the whole unit needs replacement. Hence, in 3-phase transformer installations, a more expensive unit must be available for speedy replacement in the event of a brealedown. However, in tlie case of banks of single phase transformers, only a much smaller and less costlier single phase unit has to be kept in spare to replace the faulty unit. In the case of delta connected banks, the faulty unit is simply disconnetted with continuity of supplyDESIGN OF TRANSFORMERS 57 maintained by operating the bank in open delta at 58% of the original power capacity till the faulty single phase unit is replaced. (The practice of operating two I-phase transformers in open delta at 58% of rated load is quite common in some western countries) The choice of transformer arrangement for 3-phase systems is mainly governed by the relative importance of the factors listed above. For example, in mines three 1-phase units may be preferred to a 3-phase transformer on account of ease of transport. 5.4 THREE PHASE TRANSFORMER CONNECTIONS A 3-phase transformer or a bank of three 1-phase transformers may be connected in star or delta. When star connection is used, the phase voltage is 1/V3 times the line voltage while in delta connection the phase voltage is equal to the line voltage. Therefore, the number of turns per phase in a star connected winding are 1/V3 times the number of turns per phase in a delta connected winding. On the other hand, the current in each phase (and hence in each conductor) of a star connected winding is V3 times the current in each phase (and hence in each conductor) of a delta connected winding. Thus we conclude that a star connected winding is characterised by a small number of turns with conductors having a large area of cross-section while a delta connected winding has a large number of turns with a small area of cross-section. The larger number of turns of lower cross-sectional area used in delta connected transformers obviously require greater amount of insulation. The advantages of star connected transformers are : (@ the phase voltage is smaller, and hence the major insulation required between windings and earth is smaller. Another advantage of star connected winding with earthed neutral is that the maximum voltage to the core (which is at the earth’ potential) is limited to 58% of line voltage, whereas with a delta connected winding, in case of a line to earth fault the maximum voltage between windings and core increases to full line voltage. (@@ the number of turns is smaller with the result the amount of insulation used is smaller. The use of star connection at high voltages results in reduction in cost especially at higher voltages since at higher voltages the cost of insulation becomes significant. At very high voltages, the savings on account of star connection in the cost of insulation may amount to as much as 10%. The savings in the cost of insulation on account of connections may not be significant below a voltage of 11 kV. (ii) In the case of star connection, both phase and line voltages are available for four wire supply. This is useful for distribution purposes at lower. voltages where both 3-phase and 1-phase loads have to be supplied. ‘The delta connection behaves better under conditions of unbalance and in fact a delta connected primary is essential where the low voltage secondary is star connected four wire supply to mixed 3-phase and 1-phase loads. 5.5 THREE WINDING TRANSFORMERS The use of multi-circuit transformers in large power networks, the power generated at power stations is transmitted to consumers located at widely different distances. The most economic transmission voltage depends upon the distance of the load from the power station. Therefore, in situations where the load centres are located at varying distances from the power generation centre, it becomes imperative (due to economic considerations) to transmit power at different voltages. . There are certain important consumers and systems which require to be fed from two or more independent distribution systems. Also, it is often desirable to interconnect several systems having different voltages into a common grid network in order to economically distribute electric energy.58. ELECTRICAL MACHINE DESIGN In all such applications two-winding transformers having two independent circuits i.e. primary and secondary, may be used. However, it is more convenient apd economical to use multi-circuit transformers. These transformers have three or more independent circuits with different transformation ratios, so that the windings operate at three or more different voltage levels. A three winding transformer consists of three sets of windings primary, secondary and tertiary. Three winding transformers may be either 3-phase units or 1-phase units connected in a three phase bank. These transformers have a large kVA rating. Since the three windings are operated at three different voltage levels, they may be called as high voltage (h.v.), medium voltage (m.v.) and low voltage (I.v.) windings. Two alternative arrangements of windings of a 3-phase transformer are shown in Fig. 5.6(a) and (6). The high voltage (h.v.) winding in both cases is the outermost winding away from the core because of insulation considerations. The Lv. and mv. windings are Core kept adjacent to each other with either of two nearer to the core. The Lv. and m.v. windings are kept near to each other in order to reduce the leakage reactance between this pair of windings. Thus the two possible winding arrangements for windings starting from core (a) Lv., mwv., hv. @)mv.,Lv, hv, outwards are : Fig. 5.6. Winding arrangement of 3-winding ‘transformers, (@) Lv. my. and h.v. and (ii) my., Lv. and hv. Ty. Ty. h. ma Ty. Core In two winding transformers, both the primary and the secondary windings have the same rated kVA rating. However, in a three winding transformer, the kVA ratings of three windings may be unequal. The rated kVA of the transformer is considered to be equal to the largest rated KVA of any of its windings. ‘There are many other reasons why three winding transformers are built with an additional winding, the tertiary winding, apart, of course, from the traditional primary and secondary windings of a two winding transformer. These reasons are : () To supply small additional load at a different voltage. (ii) To supply phase compensating devices such as capacitors required for power factor improvement which can be operated at different voltage. (ii) In star/star or star/zigzag transformers, a delta connected teritary winding reduces the zero-phase-sequence impedance and allows adequate earth fault current to flow for the operation of protective device in order to limit the voltage imbalance which may be produced when the load is unbalanced. (iv) To indicate voltage in a h.v. testing transformer. The teritary winding is called an auxiliary winding when it is used for supplying an additional small load at a different voltage. Qn the other hand, it is called a stabilizing winding, when it is used for limiting the short circuit. current as in (ii). Tertiary windinges are normally connected in delta so that when line to earth or line to line faults occur on the primary or the secondary windings, the considerable unbalance in phase voltages may be compensated by the circulating currents flowing in the closed delta. The reactance of the windings should be large enough to limit the circulating current in order that there is no overheating of the windings.DESIGN OF TRANSFORMERS 5.9 5.6 DISTRIBUTION AND POWER TRANSFORMERS, ‘The transformers used in power systems may be di upon the type of service. These are : (@ Distribution transformers and (ii) Power transformers. Distribution Transformers. Transformers upto a size of about 500 kVA, used to step down the distribution voltage to a standard service voltage or from transmission voltage to distribution voltage are usually known as distribution transformers. They are kept in operation all the 24 hours a day whether they are carrying any load or not. Energy is lost in iron losses throughout the day while the copper losses account for loss in energy when the transformer is loaded. Therefore, distribution transformers should have their iron losses small as compared with full load copper losses. In other words, they should be designed to have maximum efficiency at a load much lower than fall load (about 50 per cent). Owing to low iron loss, the distribution transformers have a good all day efficiency. Distribution transformers should have a good voltage regulation and therefore they should be designed for a small value of leakage reactance. Power Transformers. They have a rating above 500 kVA and are used in generating stations and substations at each end of a power transmission line for stepping up or stepping down the voltage. Actually the, power rating of the power transformer are in the order of MVA. ‘They may be either single phase or three phase units. They are put in operation during load periods and are disconnected during light load periods. Therefore power transformers should be designed to have maximum efficiency at or near full load. Power transformers are designed to have considerably greater leakage reactance than is permissible in distribution transformers as in the case of power transformers inherent voltage regulation is less important than the current limiting effect of the higher leakage reactance. 5.7 CORE The transformer core is a closed magnetic circuit through the mutual flux i.e. the flux which links with both the windings passes. The core material and construction should be such that both the magnetizing current and the core losses are minimum. The cores of transformers are laminated in order to reduce the eddy current losses. The eddy current loss is proportional to the square of thickness of laminations. This apparently implies, that the thickness of the laminations should be extremely small in order to reduce the eddy current losses to a minimum. However, there is a practical limit beyond which the thickness of the laminations cannot be decreased further on account of mechanical considerations. This practical limit of thickness is 0.3 mm. The laminations are made 0.33—0.5 mm thick. The thickness should not be reduced below 0.3 mm because in that case, the laminations become mechanically weak and tend to buckle, These laminations are made of the so called transformer grade steel containing 3—5% silicon. The higher content of silicon increases the resistivity of the core, thereby reducing the eddy current core loss. High content silicon steel is a soft iron material having a narrow hysteresis losses are also small. This material has a high permeability and hence the magnetizing current is also small. The steel used for transformer cores may be hot rolled or cold rolled. The hot rolled steel which permitted a maximum flux density of 1.45 Wb/m? was in use for a considerable length of time. In recent years this type of steel has completely been superseded by 0.33 mm (or 0.35 mm) thick cold rolled steel allowing much higher flux densities upto 1.8 Wb/m? to be used. Although, cold rolled steel is 25—35% more expensive than the hot rolled steel, the increase in value of maximum flux density makes it possible to reduce the amount of core material. Cold Rolled Grain Oriented (CRGO) steel sheet with an approximate silicon content of 8% is typically used for magnetic circuits of transformer. CRGO steel offers following advantages : (@ Magnetic induction is maximum and the loop of BH curve is large (i) Core loss during no load operation of the transformer is low. (iii) Reactive power input at no load operation of the transformer is low. ided into two categories depending5.10, ELECTRICAL-MACHINE DESIGN (iv) Magnetostriction is low. (v) Good mechanical properties. + The ferrous base of CRGO steel offers maximum magnetizability and the rolling direction becomes the direction of maximum magnetic properties and thus CRGO steel approaches the ideal properties of the individual crystallite. Based on grain orientation CRGO steel is also categorized as conventional type and HI-B type. HI-B type has low watts loss/kg. Because of grain orientation, hysteresis loss is lower for HI-B type CRGO steel. Eddy current loss is usually reduced by applying a surface coating of insulation. Reduction in core-loss in CRGO steel has been achieved with reduction in thickness of lamina. Magnetostriction is a phenomenon that is produced due to the change in configuration of a magnetizable body in a magnetic field leading to periodic changes in the length of the body in an alternating field. The frequency of magnetostriction is twice as large as that of existing alternating field. Magnetostriction gives rise to the noise in the core of the transformer. Magnetostriction is minimum in the rolling direction and is maximum in the 90° direction. To keep magnetostriction well within limits, high induction amplitudes are prevented in the rolling direction. Magnetostriction of HI-B steel is lower. However, the use of cold rolled steel involves a more complicated core construction and requires new methods for machining the laminations. The hot rolled steel is sheared to size by power guillotines and then punched in multiple presses. With cold rolled steel, rolls of mass upto 2 tons are slit into widths by gang operated slitters. This working of cold rolled steel impairs its property and therefore it is annealed to relieve stresses. The annealing process involves heating sheets or complete small cores at 800°C in an inert atmosphere (to avoid oxidation and carbon contamination). The insulation on the surface of laminations is kaolin or varnish in the case of hot rolled steel but for cold rolled oriented steel, phosphate-base coating is used. This coating is done by the makers of the steel and can withstand the annealing process. Cores of small transformers need no further insulation if made of c.r.0.8. However, the transformers of capacity 10 MVA and above kaolin or varnish must be applied to the lamination (in addition to phosphate-base coating which already exists). 5.8 CORE CROSS-SECTION Small core type transformers have rectangular section limbs with rectangular coils as shown in Fig. 5.7. However, in large capacity transformers, the economic use of core material requires that the cross-section of the core should ideally be a circle since a circle has the minimum periphery for a particular area and hence, the windings which are put around the core have a minimum length of mean turn resulting in reduced amount of conductor material thereby reducing the costs. A circular core, however, involves the use of an unmanageably large number of laminations of different sizes. The use of laminations of different sizes is possible but is highly time-consuming and uneconomical on account of the obvious difficulties in core assembly and increased labour costs. A compromise is achieved by arranging the core section in steps in such a way that the net sectional area is maximum for the number of steps employed and the corners of the steps are so arranged that they lie on a circle known as a circumscribing circle of predetermined diameter. Fig. 5.8 shows a two-stepped core which is also known as cruciform core. A two stepped core requires two sizes of laminations. However as the number of steps increases, the number of different sizes of laminations also increases. With large number of steps, there is a reduction in the length of mean turn of windings and consequently there is a reduction in the cost of conductor material and the /°R losses but there is extra cost involved in shearing and assembling of different sizes of laminations. Therefore, while designing a core section, balance should be struck between the cost of conductors and core and also the labour charges. Cores for shell type transformers are usually of simple rectangular cross-section.DESIGN OF, TRANSFORMERS : hw. 5.11 duct hw ; iw ears N duct : ol care Circumscribing circle Fig. 5.7. Rectangular core. Fig. 5.8. Two-stepped (cruciform) core. 5.9 CORE CONSTRUCTION WITH HOT ROLLED LAMINATIONS In small transformers, the complete magnetic circuit can be punched as a whole but this process would involve too much wastage of sheet if followed for large transformers. Moreover, there would be great difficulty in putting the winding as each turn will have to be separately threaded through the window which is impossible for large transformers. Therefore, the coils are made separately and are then placed on assembled cores. The magnetic circuit is made of different types of laminations to give it a proper shape. Fig. 5.9 (a) shows a core built from I shaped laminations. Joints at the junctions of different strips introduce air gaps in the magnetic circuit resulting in increased magnetizing current as air has a much lower permeability than iron. Therefore the joints should be made as tight as possible and the joints in adjacent laminations (a) I type laminations. (b) E type laminations. Fig. 5.9. Core construction with different types of laminations. should be overlapped. This gives a smaller effective air gap for the flux to jump. In small transformers the path of flux through iron is small, and the air gap at the joints is of more relative importance than in large transformers. Therefore in order to decrease, the magnetising current in small transformers, we have to make the laminations of special shape. In the case of shell type of transformers, the laminations can take the form of an E as shown in Fig. 5.9(b). L type laminations can be used for building up cores of both core and shell of transformers as shown in Figs. 5.10(a) and (6). (a) Core type (®) Shell type Fig. 5.10. Core built with L type laminations.5.12, ELECTRICAL MACHINE DESIGN In the case of large core type transformers, I type laminations are used and the joints between limbs and yokes are interleaved. If the magnetic properties of the circuit are the only consideration, the best arrangement is obtained by interleaving one plate at a time. Fig. 5.11 shows the alternate layers of laminations of interleaved core of a single phase transformer while Fig. 5.12 shows the same for a three phase transformer. It is usually desirable for mechanical reasons to interleave the plates two, three or four at a time. As many as 20 widths of steel strip may be interleaved at a time for large power transformers as this accelerates the magnetic (a) Istposition(odd (6) 2nd position (even (a) 1st position (odd (b) 2nd position (even numbered lami- numbered lamina- numbered lami- numbered lamina- nations) tions) nations) tions) Fig. 5.11, Interleaving core of 1-phase Fig. 5.12. Interleaving core of 3-phase core type transformers. core type transformers. circuit assembly and minimizes the risk of imperfect interleaving on account of buckling of laminations. Fig. 5.13 shows the interleaving of core plates using a number of laminations at a time. In this case the interleaving patterns remain unchanged although the thickness of each Air gop Fig. 5.13. Interleaving of core Fig. 5.14. Effect of air gap on flux laminations. patterns of ore joints. layer grows accordingly two or three times. It should be pointed out here that the increased thickness of the layer increases the no load current and also the iron loss of the transformer because of increased reluctance offered to the magnetic lines of force bypassing the lamination joints as shown in Fig. 5.14. It is clear that larger the number of laminations in a layer, the greater is the number of magnetic lines of force which go through the air gaps between leg and yoke laminations. Since the flux has to pass through the air gaps in the joints of the laminations, the gaps left during the interleaving of the core result in large magnetizing current of the transformer, Therefore, in order to reduce magnetizing current, the interleaving at the lamination joints should be done with utmost care. The gaps between laminations must not be greater than 1—2 mm.DESIGN QF TRANSFORMERS 5.13 5.10 YOKE CROSS-SECTION : ‘The cross-section of the yoke is made about 15 per cent greater than that of the core in transformers using hot rolled steel. This reduces the flux density in the yoke which reduces the magnetizing current and iron losses. The core is made multi-stepped in order to reduce the length of mean turn of the windings. However, since there are no windings around the yoke it can be made of a larger cross-section in order to reduce no load current and also the iron losses. Also in order to simplify the core construction and to cut down labour costs, the yoke need not to be of multi-stepped cross- section. It can be made rectangular or two-stepped cross-section in order to simplify the construction. This construction leads to cross-fluxing between core and yoke giving rise to additional iron losses. The phenomenon of cross-fluxing can be explained with the help of a analogy in hydraulic systems. Let us consider three channels of equal area with equal water as shown in Fig. 5.15. Let them be joined together at a water head and then be trifurcated into three channels again. It is observed that water in a incoming channel goes into a corresponding outgoing channel straight across without cutting across and without causing turbulence in the common water head. Ideally there are no hydraulic losses since there is no turbulence of water. Incomit i ye Z = E SS 1 —— Outgoing a Turbulonce. — Outgoing channels channels Fig. 5.15, Hydraulic analogy for cross-fluxing. Let us consider another case wherein there are three incoming channels pouring water into a common water head with only two outgoing channels. In this case, since the water of three incoming channels has to go to two outgoing channels, water has to cut across the common water head and therefore increased turbulence is caused with resultant loss. Similarly, when the core is three stepped and the yoke is two stepped, the flux will cut across the laminations at the joints of core and yoke resulting in cross-fluxing and consequent increase in iron loss. Common water head The iron losses in a magnetic core, where the flux crosses from one part to another, are minimum if the flux density in all the packets comprising the magnetic circuit is the same. In a transformer, this is possible only if the core and the yoke are of the same shape so that the additional iron losses on account of cross-fluxing are avoided. When the core and the yoke cross-sections are of different forms, cross fluxing occurs resulting in non-uniform flux density distribution which gives rise to additional iron losses and local overheating of the core. It is not always possible to use the same cross-section for both yoke and core owing to certain technological reasons. The yoke is made of much simpler cross- section which may be either rectangular or cruciform or with a step upwards or downwards as shown in Fig. 5.16. Yoke stepping, to some extent, equalises the flux density distribution in the core packets and reduces the additional iron loss as compared with the loss in rectangular section yokes.5.14. o ELECTRICAL MACHINE DESIGN Yokes with rectangular cross-section as shown in Fig. 5.16(a) are used, for small capacity transformers, while the yoke cross-sections shown in Figs. 5.16(b), (c) and (d) are used for large capacity transformers. PLM (a) Rectangular (6) Cruciform (c) Withastepupwards (d) Withastep downwards Fig. 5.16. Yoke cross-section. The yoke cross-section with a step downwards as shown in Fig. 5.16(d) is commonly used in large power transformers and has the advantage of providing additional space along the height for location of tapings in the winding. 5.11 CLAMPING OF CORE In small types of transformers, the laminations are held together by either string or by a strong cotton webbing, But this method does not give much strength and rigidity. Cores can also Channel section for clamping yoke Flitch plate Core clamping bolts Silicon steel laminations Bracket for supporting winding end insulation Fig. 5.17. Clamping arrangement for transformer cores.DESIGN QF TRANSFORMERS 5.15 be clamped between iron frames. In bigger transformers, cores are kept in position by side plates bolted together at intervals along the limbs and the yoke. Holes are punched out in the laminations in order to accommodate the bolts. These bolts, which necessarily pass through the cores must be insulated both from the side plates and the laminations, while the side plates are insulated from the laminations. The isolation is necessary as otherwise the bolts would short circuit the laminations and would provide paths for the eddy currents. In order to provide more rigidity to the core and to prevent bulging of core between bolts, flitch plates are used. Fig. 5.17 shows the clamping arrangements for a core. 5.12 CORE CONSTRUCTION OF MODERN CORE TYPE POWER TRANSFORMERS Cold rolled grain oriented steel laminations are used for cores of all modern power transformers. This is because it permits the use of flux densities between 1.6 to 1.8 Wb/m? as compared with 1.3 Wb/m? for hot rolled steel and consequently the weight of both core and winding is reduced. Prior to introduction of grain oriented silicon steel, transformer core design was characterised by rectangular interleaved corners, larger yoke than core (limb) cross-sections, and clamping by bolts passing through the active core area. With the advent of cold grain oriented steel the three limbed construction for 3 phase transformers (Fig. 5.4) use identical yoke and limb cross-sections. With very large 3 phase transformers, the yoke sections become very heavy resulting in increased height of transformer. A limitation on the height of transformer may be imposed by the loading gauge of the railways route or the road along which the transformer is to be hauled. ‘Thus it may become necessary to reduce the overall height of transformer. The height of yoke is reduced by using a 5 limbed construction in place of the conventional three limbed construction. In five limbed construction (Fig. 5.18) the top and the bottom yokes can be made of smaller of cross section (than the principal three limbs) by providing two extra limbs which provide extra return paths for the flux. Top and the bottom yoke- sections are 58 percent and the Fig. 5.18. Five limbed three phase transformer. vertical return limbs are 45 percent of principal limb cross-section. | | | | | | | | | | | | | | | | | | i i T T The type of construction to be used whether say “core type” or “shell type” is usually an invariable function of the design policy, factory layout and construction. In India the core type of construction is used almost exclusively.5.16, ELECTRICAL MACHINE DESIGN When very large transformers are manufactured, their overall dimensions must be taken into consideration from the point of view of transportation. The height ofthe transformers has to be reduced on account of transportation difficulties like height of the bridges on the way to installation sites. The reduction of overall height of transformers requires the height of yoke to be reduced. It is a common practice today to manufacture, 1-phase generator transformers upto 800 MVA with cores of star form or cruciform plan; three or four “two limb” cores are assembled with one central combined limb for the windings. Fig. 5.19 shows a transformer core arrangement with a central leg around which both primary and secondary windings are wound and the return path for the flux is through four yokes. This arrangement reduces the height of the yokes by four times. The reduction of height of yoke facilitates the use of either a circular or a square tank. Central core Fig. 6.19. Four yoke 1-phase core. In cold rolled grain oriented steel the grains are along the direction of rolling. The cold rolled grain oriented steel has the maximum permeability in the direction of the grain orientation. Also:the specific iron loss is minimum if the flux lines are along the direction of rolling. Therefore, if the cores and the yokes are stacked of interleaved right angled laminations the magnetic lines of force at the corners will turn at an angle to the direction of grain orientation as shown in Fig. 5.20(a). This gives rise to increased iron loss and increased no load current. Gi) Flux fines Mitre (bevelled) core Grains Grains (a) Conventional core. (b) Mitred core. Fig. 5.20. Stacking of core using cold rolled oriented steel.DESIGN OF TRANSFORMERS 5.17 ‘The iron loss and the no load current can be decreased if the flux lines are made to flow along the direction of the grain orientation. This is possible if mitred (bevelled) joints are used for cores and yoke as shown in Fig. 5.20(6). The commonly used forms of mitred joints are (i) 45° mitre and (ii) 35°/55° mitre. These joints are shown in Figs. 5.21(a) and (6) respectively. The full lines show the joints in one layer of laminations and the dotted lines show the joints in the subsequent layer, The 35°/55° mitre improves the flow of the flux around the corner while the 45° mitre reduces waste of core material during the cutting process. (a) 45°Mitre (®) 35°/55* Mitre Fig. 6.21, Mitred joints for cores using cold rolled oriented steel (c.0.r-s.). When the core (circumscribing) circle diameter exceeds about 0.8 m the commercially available laminations are too narrow to span the full diameter and therefore splitting of the core becomes essential. Therefore, bridging sections as shown in Fig. 5.22 are needed between the two halves of the core at the central limb. Incidently, an advantage of splitting the core in two halves is improvement in cooling of the core because the area of the laminations exposed to the coolant gets doubled. In order to make the best use of the grain oriented silicon steel, it is necessary to make the flux run parallel to the direction of rolling (.e. the direction of grain orientation) for as much of the magnetic path as possible. This is made by designing yoke and limb sections of identical cross-section and shape in order to avoid cross fluxing and additional iron loss. The laminations are assembled with mitred (bevelled) F Fig. 5.22. Splitted core with bridge of 35% joints. 55° mitre corners. The use of mitred joints assures that the flux flows along the direction of grain orientation thereby minimising both the core loss and the magnetising current. However, the use of mitred joints complicates the core construction. The core may be clamped with bolts. The bolts are insulated from the core by core of s.r.b.p. or with larger cores with higher operating temperatures, these are often of epoxy resin bonded fibre glass. However, in order to retain the advantages of cold rolled oriented steel, a boltless construction is used as the clamping bolts distort the flux path and increase the eddy current loss. The core is bound under pressure by synthetic-resin-bonded glass or Terelene which is then heated to set the resin. The outer packets of laminations are then bounded by a suitable adhesive such as Araldite in order to spread the pressure more evenly.5.18 ELECTRICAL MACHINE DESIGN 5.13 COOLING OF CORES In transformers of medium and high capacity with diameter of circumscribing circle D > 0.35 m the cores have the relatively small surface/volume ratio so that the SI temperature gradient in the core is excessive. In such cases the cooling must be augmented by dividing the core into different stacks with longitudinal oil ducts (usually 6 mm (a) Longitudinal ducts. (b) Longitudinal and transverse ducts. wide) running parallel to the Fig. 6.23. Core ducts. laminations as shown in Fig. 5.23(a). In transformers of very high capacity (D > 0.8 m) longitudinal duets may not be sufficient and as heat flows more readily along the laminations, than between insulated laminations, it is necessary to increase the area of lamination edges by using transverse duets which may be 10-12 mm wide as shown in Fig. 5.23(6). ‘The magnetic ¢ircuit'is therefore, divided into packets insulated from each other and to ensure good electrical continuity between packets, tinned copper strip bridging pieces are used. 5.14, CORE EARTHING . With the exception of individual laminations and core bolts, all internal metal parts of the transformer require earthing. Due care must be taken in the design of the earthing system to avoid multiple paths which may initiate partial discharges because of the circulating currents inducing relatively high voltages across high impedance sections of an earth path. 5, 15 TRANSFORMER WINDINGS ‘The windings used in transformers are of different types and employ different arrangements for coils. . Shell type transformers use sandwich type of winding with coils shaped as pancakes. In this type of winding both low and high voltage windings are split up into a number of coils. Each high voltage coil lies (or is sandwiched) between two low voltage coils as shown in Fig. 5.3(b). The two low voltage coils at the ends have half the turns of a normal low voltage coil and therefore these coils are called half coils. The subdivision of low and high voltage windings into a number of coils gives a better coupling between the two windings and therefore results in lower leakage flux thereby reducing the leakage reactance. The leakage flux and leakage reactance of the windings depends upon the number of sections in which the windings are divided; the larger the number of coils (and hence sections), the lower is the leakage reactance. Therefore, the advantage of sandwich coil is that with their use the leakage reactance of the transformer can be controlled to any desired value with a suitable division of windings. Core type of transformers use concentric type of windings. Each limb is wound with a group of coils consisting of both-primary and secondary windings which are concentric to each other as shown in Fig. 5.3(a). The low. voltage winding is placed next to the core (which is at the earth potential) and the high voltage winding is placed on the outside. However, the low voltage and the high voltage windings can be alternately interlaced so as to reduce the leakage reactance.DESIGN OF TRANSFORMERS 5.19 ‘The type and arrangement used for windings used for core type of transformers depend upon many factors. Some of these factors are : . @ current rating, i) short circuit strength, temperature rise, (iv) impedance, (») surge voltage and (vi) transport facilities. ‘The windings used for core type of transformers are of the following types 1. Cylindrical windings 2, Helical windings 3. Double helical windings 4, Multi-layer helical windings. 5. Cross-over windings 6. Dise and continuous dise windings 7. Aluminium foil windings. Following types of conductors are used for making the different types of windings as listed above : 1, rectangular copper conductor insulated by paper coverings, 2. rectangular copper conductors bunched together and having twin paper coverings, 3. CTC conductor having paper covering, 4. twin transposed copper conductor bunched together, twin rectangular copper conductors bunched together and having a common paper covering with a epoxy coated glued paper strip between the two conduction. 6. epoxy coated CTC. Comparison reveals that conductors mentioned at 2 to 4 above can improve the winding space factor. Configuration given in 5 and 6 improve the mechanical strength of the winding and improve its space factor. The copper strips are made of electrolytic grade copper wire bars with high conductivity and are annelid. Sharp edges are normally avoided and normal shop is given at the corners. Since the transformer windings require to withstand different high and power frequency voltage hence it is required that the surface of these conductors are smooth. High air permeability paper is used for covering. All the layers except the outermost are butt wound. The outermost layer is overlap wound 1, Cylindrical windings. These windings are layered type and use either rectangular or round conduetors. A cylindrical winding using rectangular conductors is shown in Fig. 5.24. The conductors are wound on the flat side with their longer sides parallel to the core axis as shown in Fig. 6.25(a). However, sometimes they are wound on the rib side ie. their longer sides are perpendicular to the core axis as shown in Fig, 5.25(6). The winding using rectangular conductors may be simultaneously wound from one or more parallel conductors placed flatwise or edgewise.5.20 | ELECTRICAL MACHINE DESIGN The layered winding may have to conductors wound in one, two or more layers and is, therefore, accordingly called the one, two or multilayer winding. The windings using rectangular conductors are usually two layered type because in this case it is easier to secure the lead-out ends. The two layers are separated by an oil duct. The windings designed for heavy currents are 4 Conductor 1 2 3 4 5 6 Insulation Core FRPTTL b i @ @) (a)ontheflatside _(b) on therrib side , Fig. 5.24, Cylindrical windings Fig. 5.25. Positions of rectangular conductors, wound with a number of conductors connected in parallel located side by side in one layer. The parallel conductors have the same length and are located in the magnetic field of leakage flux of almost the same flux density, and hence it is not necessary to make any transposition of the conductors, A wedge-shaped packing is used at each of the two extreme ends of winding in order to level it. The packing is made either pressboard strips or rings cut from a bakelite cylinder. Cylindrical windings employing rectangular conductors are used mainly as low voltage windings upto 6.6 kV for kVA ratings upto 600—750. However, their main use is for voltages upto 400 V. Cylindrical windings using circular conductors are multilayered as shown in Fig. 5.26. ‘They are wound on a solid paper bakelite cylinder. QQ Q@QOQ ZIWQao QZWQP SKY Fig. 6.26. Insulation between layers and on the butt-end of the cylindrical windings of circular conductors.DESIGN OF TRANSFORMERS 5.21 In order to improve the cooling conditions of the inner layer, the cylindrical windings using circular conductors are often wound on vertical strips forming an oil duct between winding and the insulating cylinder as shown in Fig. 5.27(b). Sometimes the winding is divided into two parts by an additional oil duct. This oil duct is usually located nearer to the inner winding as shown in Fig. 5.27(c). om Oil ducts a 2 MOOOOO O000 So OO @ ) (o) Fig. 5.27. Arrangements of the cylindrical windings with circular conductors, The cylindrical windings employing circular conductors are mainly used for high voltage windings with voltages 6.6, 11 and 33 kV for ratings upto 600—100 kVA 2, Helical windings. The helical windings are of two types (i) single helical winding. (ii) double helical winding. Single helical windings. The single helical or simply a helical winding has its turns wound in an axial direction along a screw line with an inclination corresponding to the height of a conductor and an oil duct between turns, There is only one turn in each winding layer as shown in Fig. 5.28(a). The winding consists of a single section conductor or a number of strands in parallel wound in the form of a continuous helix. The conductor is rectangular in cross- section and is paper covered. The oil is mounted on a thick pressboard or s.r.b.p. cylinder. Helical windings are used for low voltage winding medium and high capacity transformers where the number of winding turns is small but the current is high. fii Therefore, low voltage windings of medium and co high capacity transformers require the use of a com conductor made of a large number of strips connected in parallel. The parallel connected (a) Helix (6) Disc-helix strips are placed side by side in a radial direction Fig. 5.28. Helical windings. so that cach conductor occupies the total radial depth of the winding to form a Dise-helical winding as shown in Fig. 5.28(b). The individual strips can be assembled in a radial pack, either as a single column or as two columns in parallel. Helical windings are usually wound on the battens around the bakelite cylinder with insulating cylinders placed between the turns. The continuous helical winding exhibits high5.22 ELECTRICAL MACHINE DESIGN axial mechanical strength and therefore finds wide application in low voltage windings of large size power transformers. A distinguishing feature of the helical winding is the use of transposed conductors by changing the relative position of individual conductors or groups of conductors. The transposition is essential for equalizing the resistance and leakage reactance of each of parallel conductors, In the absence of transposition these conductors will be of different length and being situated in the leakage field having unequal flux densities, will have different resistance and leakage reactance. This would lead to non-uniform distribution of current in parallel conductors thereby overloading of portions of conductors and causing additional eddy current losses in conductors. The transpositions in this winding are made in the following manner. The whole of the winding turns along axial length are subdivided into four equal sections, The transpositions are made between these sections, at three different points of the winding. The three transpositions include one standard transposition and two partial transpositions as shown in Fig, 5.29. The 1/4 Height Partial Transposition sis[s[s[2[3 dow ‘ght 1/2 Partial 10 Transposition 1/4 Height Fig. 5.29, Transpositions in Helical windings.DESIGN GF TRANSFORMERS 5.23 standard transposition is done in the middle of winding with each conductor varying its position symmetrically relative to the middle point i.e. the conductor on the extreme right being transposed to the extreme left, the second conductor from right being transposed to second position from left and so on. The two partial transpositions are done at a distance of 1/4 height from top and bottom ends of the windings. In partial transpositions two halves of parallel conductors are interchanged in positions with the right half the conductors being transposed to left half positions and vice-versa. ‘The disadvantage of the simplex helical winding is that ampere turns are as if thinned out at the points of transpositions, which leads to unequal distribution of mmf throughout the height, of winding. 3. Double-helical windings. The double-helical winding is used in low voltage windings of high power ratings where the number of winding turns is small and a single helical winding with normal width of oil ducts does not fill up properly a window height, while the current and the number of parallel conductors required are very large. Turn Jeross— section Turn cross— a section Spacer cylinder Cylinder Batten Batten (@ ) Fig. 5.80. Cross-section of helical windings, Fig. 5.30(a) shows a single helical winding. All the conductors forming one turn are situated side by side in one layer, the adjacent turns separated by spacers along the axial length. In contrast, the parallel conductors of a double helical winding are divided into two parallel circuits and are situated in two layers shifted in axial direction as shown in Fig. 5.30(b). The advantage of double helical winding is the reduced eddy current loss in conductors. This is on account of the reduced number of parallel conductors situated in radial direction. This can be illustrated by simple example. Suppose there are 8 parallel conductors which form one turn. If a single helical winding is used, there will be 8 conductors placed radially, while there will be only 4 conductors in the radial direction if double-hexial winding is employed. The magnetic field is non-uniform in the radial direction and, therefore, there is greater magnetic assemetry between the conductors when the single helix winding is employed, resulting in greater /R loss and increased leakage reactance. In double-helix winding the transposition is obtained without using an axial space on interchanging the conductors, because the turn in its cross-section consists of two groups of conductors. ‘The transposition in this winding is termed uniformly distributed and is made in the following way. The entire winding (turns) is sub divided into equal sections. The number of these sections should be equal to the total number of parallel conductors. The conductors are then changed5.24 y ee { iT2T3T+ & TL 1/16 (ler ha az Tt 1/8 (7Ts{sT4} 3 e 7isti{2 / TT 1/8 GENER Be) ¢ D) 1/8 S[4]3]2 / i S[s(718 TL 1/8 GbE Bs r Ist) 1), L Got) 2 ° SPST TT ote 78 3 1/8 1/8 Fig. 5.31. Transpositions in double layer helical windings. ELECTRICAL MACHINE DESIGN Fig. 5.32. Transposed conductors according to equally distributed transposition diagram. Fig. 5.33, Multi-layer helical winding. Core Fig. 5.34. Cross-over winding,DESIGN OF TRANSFORMERS 5.25 between these sections; the top conductor of one group is placed on top of the second group while at the same time and in the same portion of the winding the bottom conductor of the second group is inserted underneath the first group. + The procedure for making a transposition in the double-helical winding is shown in Fig 5.31. The conductors are transposed in the gap between the spacers as shown in Fig. 5.32. A double-helical winding is used for the same range of voltages as a single-helical winding. However, the current rating for the double helical winding is twice as that of single helical winding. . Helical windings are used in power transformers with outputs ranging from 150 kVA to 30 MVA at voltages from 400 V to 11 kV and sometimes upto 33 kV. 4. Multilayer helical windings. Multilayer helical windings are commonly used as high voltage windings for 110 kV and above. This type of winding consists of several cylindrical layers concentrically wound and connected in series as shown in Fig. 6.33. The number of layers depends upon the voltage, higher the voltage, the larger the number‘ of layers. All the layers are wound on paper cylindets and are separated from each other by vertical strips forming vertical oil ducts. One line terminal is connected to the outermost layer while the innermost layer adjoining the low voltage winding is grounded. ; ‘The outer layers are made shorter than the inner ones thereby distributing the capacitance uniformly. This winding is primarily used for improving the surge behaviour of transformers, ‘The improvement in surge behaviour requires large capacitance, which means decrease in radial depth of winding. The decrease in radial dimensions means poor mechanical strength and also low leakage reactance. Therefore, multilayer helix windings are subject to large short cireuit forces and since they are inherently weak (mechanically), their use is restricted. 5. Cross-over windings. These windings are used for high voltage windings of small transformers. It has been mentioned earlier that cylindrical windings (especially cylindrical windings using circular conductors), are used for high voltage windings of low rating transformers. However, the use of cylindrical multi-layer winding at high voltages, the voltages between adjacent layer become too high and it becomes difficult to select proper thickness for the interlayer insulation. Thus, it becomes imperative to reduce interlayer insulation and therefore the winding is axially separated into several multilayer coils. The winding is divided into a number of coils separated from each other by insulating washers or oil duets formed by spacers. In cross-over windings, the conductors are paper covered round wires or strips. The coils are wound over formers with side cheeks (U-pieces) and each coil consists of a number of layers with a number of turns per layer. The complete winding consists of a number of coils connected in series. Two ends of each coil are brought out, one from inside and one from outside. The inside end of a coil is connected to the outside end of the adjacent coil. The actual axial length of each coil is about 50 mm while the spacing between the adjacent coils is about 6 mm to accommodate blocks of insulating material and to allow free circulation of oil. The width of coils is 25 to 50 mm. Fig. 5.34 shows cross-over coils. Cross-over windings are used in the same range of ratings as the cylindrical windings. The ver winding has higher strength than the cylindrical winding under normal operating ns. However, this winding has a lower impulse strength than the cylindrical winding and also is more labour-consuming. 6. Disc and continuous disc windings. Disc windings are primarily used in high capacity transformers. The winding consists of a number of flat coils or discs connected in series or parallel. The coils are formed with rectangular strips wound spirally from centre5.26 ELECTRICAL MACHINE DESIGN outwards in the radial direction as shown in Fig. 5.35. The conductor used is in such lengths as are sufficient for complete winding or section of winding between tappings. The conductor can be a single strip or a number of strips in parallel, wound on the flat side* This gives a robust construction for each of the discs. The dises are wound on a insulating cylinder spaced from it by strips along the length of cylinder. The discs are separated from each other with press-board sectors attached to vertical strips. The vertical and horizontal spacers provide radial and axial duets for free circulation of oil which comes in contact with every turn. Core Coil Fig. 5.35. Continuous disc windings. ‘The disc coils are usually assembled into double coils because this arrangement leads to a more convenient connection of the inner ends. The double disc windings are wound from an entire length of conductor. Each coil is wound starting from the middle, ie., from the connection point of coils, until a double coil is formed. This disposition of coils formed in pairs is achieved through pairs of coils with requisite turns for one disc being loosely wound so that the conductor finishes in a position to provide the start of the inside turn of the adjacent dise, which is then wound from inside outwards. The first disc is then rearranged in such a manner that the start is located as an outside turn. In case, winding consists of a’ number of discs connected in series, it can be wound continuously without breaking the conductor between the separate disc coil. Thus, following the formation of one dise, the procedure is repeated without cutting the conductor, thereby saving jointing and joint space. This is an advantage, especially for the winding which is placed on the inside. A distinguishing feature of the continuous disc windings is the transposition of the coils. The purpose of these coils is clear from Fig. 5.36. These coils are initially wound in the ordinary manner, beginning from the cylinder and outward and then these coils are transposed in the reverse order. The conductors are slackened somewhat in order to make the reversing, easier and the conductor running from the drum is again tensioned. This facilitates the continuous inter-connection of coils without any soldered jointsDESIGN OF TRANSFORMERS 5.27 The advantage of disc and continuous windings is their greater mechanical axial strength and cheapness, . ©.) (ct 5 5 2 4 4 3 3 3 + 2 2 5 1 7 3 6 { Fig. 5.36. Transposition of continuous disc winding. 7. Aluminium foil windings. Aluminium, in place of copper can be used as a material for conductors in any of the above mentioned windings. Aluminium, when used as a single section has many disadvantages as compared with copper such as poor mechanical strength, poor workability and increased cross-secting, However, it is uniquely employed in foil windings because it can be rolled to thinner and more flexible sheets than copper. The aluminium sheet is used in bobbin type coils of one or more turns per layer. ‘The types of windings used in core type of transformers are listed in Table.6.1. Table 5.1, Windings used in core type transformers Type of Rating ELV. Winding LV. Winding Service MVA Voltage Type Voltage Type kV kV Distribution upto 1133 Foil, 048 Helix Cross-over or Multilayer System 1-30 3336 Dise n Dise or Helix Transmission | 30and 132500 Dise or 11, 33,66 Disc or Dise- upwards Multilayer Helix Generator 80 and 132500 Disc or 122 Disc-Helix upwards Multilayer 5.16 CONTINUOUSLY TRANSPOSED CONDUCTOR (CTC) WINDINGS ‘There has been a steady increase in the demand for electrical power. This situation, calls for a commensurate increase in the power ratings and operating voltages of electrical power systems. The transformers used in power systems are now required to withstand higher electrical, mechanical and thermal ratings.” The development of continuously transposed conductor (CTC) has provided the transformer industry a winding material that can be used for ever rising system voltages and ampacity. The windings constructed with CTC have been found beneficial for high voltage high power transformers. Continuously transposed conductor (CTC) used for transformer windings has a high thermo-mechanical strength, excellent insulation characteristics and low stray load5.28, loss. The stray losses in large transformers using conv tend to be high. Therefore, a need arises to decrease windings of transformer. ELECTRICAL MACHINE DESIGN ventional configuration for the conductors these losses by guitably designing the Stray load loss has two components, (i) eddy current loss, and (ii) circulating current loss. Eddy current loss within each strand can be reduced by using thinner insulated conductors while the circulating current loss between strands can be reduced through frequent transposition of conductors. The advantage of using continuously transposed conductors is that the stray load loss is reduced since CTC uses a large number of conductor strips which are continuously transposed. The transposed conductor consists of odd number of copper strips connected in parallel. The number of strips ranges from 6 to 31. The cross-section of a CTC is shown in Fig. 5.37 (a). Each strip is insulated with polyrinyl-acetal based enamel. One layer of interleave paper is inserted in parallel between two parallel stacks of strips to avoid damage to insulation during transposition. ‘The transposed conductor is butt-lapped with paper tape. The transposition of individual strips help in equalizing their lengths in the stack thereby Interleave reducing the circulating currents and the consequent loss. The transposition is done after 16 times the width of conductor or 60 mm whichever is greater. Paper Fig. 5.87. Continuously transposed conductor (cTO). 7 2 1 5 7 6 3 3 2 1 1 5 4 4 3 3 Fig. 5.87 (a) Transposition of strips in a CTC. Fig. 5.37 (a) shows the transposition of strips in a CTC. At least one full transposition should be made ii in a full turn of winding. The minimum winding diameter (D,,) is governed by the following relationship. p, = Ditch of transposition x number of conductors 7 where pitch of transposition is the distance between adjacent cross overs measured on one side of the conductors and is normally the 15 times the strip width. In Fig. 5.37 (a), H = radial height of CTC W = axial width of CTCDESIGN OF TRANSFORMERS 5.29 thickness of individual strip w = width of individual strip t, = thickness of enamel 1, = thickness of paper added. The advantages of CTC over conventional paper covered conductors are : (@ The use of CTC reduces the overall size of transformer on account of the compactness of strands and consequent reduction in the weight of core, windings, tank and oil (ii) The winding time is reduced due to the use of assembled transposed conductors in place of parallel strips. (iii) The winding has an improved mechanical strength due to composite construction of the transposed conductor. (iv) The individual strips use thin enamel insulation. Therefore, cooling of the conductors is improved on account of increased heat dissipation. (v) The conductors have an increased mechanical strength. 5.17 COOLING OF TRANSFORMERS The transformer is a static device which converts energy at one voltage level to another voltage level. During this process of energy transfer, losses occur in the windings and core of the transformer. These losses appear as heat. The heat developed in the transformers is dissipated to the surroundings. The coolants used in transformers are : @ air, and (ii) oil. ‘The transformers using air as the coolant are called dry type transformers while transformers which use oil as the coolant are called oil immersed transformers. In dry type transformers, the heat generated is conducted across the core and windings to be dissipated from the outer surfaces of windings to the surrounding air through convection. In the case of oil immersed transformers, the heat produced inside the core and the windings is conducted across them to their surfaces. This heat is transferred by the oil to the walls of the tank through convection. Finally, the heat is transferred from the tank walls to the surrounding air by radiation and convection. It must be understood that cooling of transformers differs from that of rotating machines and presents greater problems since there are no moving parts in a transformer, that are responsible for inbuilt cooling of rotating machines. 5.18 METHODS OF COOLING OF TRANSFORMERS ‘There are a number of methods used for cooling of transformers. The choice of method depends upon the size, type of application and the type of conditions obtaining at the site where the transformer is installed. The large number of methods used for dissipation of heat generated in transformers make it necessary to use a concise standard designation for them. The letter symbols used for designating these methods depend upon (i) medium of cooling used, and (ii) the type of circulation employed. 1. Medium. The cooling mediums (coolants) used for transformers along with symbols used for designating them are : (i) Air—A, (ii) Gas—G, (iii) Synthetic oil—L, (iv) Mineral oil—O, (v) Solid insulation—S, and (vi) Water—W. 2, Circulation. The circulation of the cooling medium (coolant) may be through natural means or there may be a forced circulation of the coolant. Accordingly the symbols used are : () Natural—N, and (ii) Foreed—F. ‘There are two ways of cooling a transformer : (@ The coolant circulating inside the transformer comes in contact with the windings and cores and transfers all the heat entirely to the tank walls from where it is dissipated to the surrounding medium.5.30 | ELECTRICAL MACHINE DESIGN Gi) The coolant circulating inside the transformer comes in contact with windings and cores. The coolant partly transfers the heat generated to the transformer tank walls with the major portion of the heat generated inside the transformer being taken up by the coolant circulating inside the transformer, to be dissipated away later in an external heat exchanger. The coolant circulating inside the transformer gets heated and is cooled in the heat exchanger. The heat exchanger may employ air or water in order to dissipate the heat of the coolant circulated inside the transformer. ‘The cooling methods are designated by symbols. Each of these letters is significant of some characteristic of the method of cooling. The cooling methods which do not employ an external heat exchanger are designated by two letters. The order in which letters are used to designate methods of cooling without external heat exchangers is @ the medium in contact with the windings, and (i the circulation of the coolant in contact with the windings. These methods, therefore, are designated by two letters. The order in which letters are used designate methods of cooling using external heat exchangers is : @ the medium in contgct with the windings, Gi) the cieslation of dhe colent in contact with the windings, iii) the medium used in the external heat exchanger, and (iv) the circulation of the coolant in the external heat exchanger. The cooling methods used for dry type transformers are : 1. Air Natural (AN). This method uses the ambient air as the cooling medium. The natural circulation of surrounding air is utilized to carry away the heat generated by natural convection. A sheet metal enclosure is used to protect the windings against mechanical damage. ‘This method is used for small low voltage transformers. However, the development of insulating materials like glass and silicone resins class C materials which can withstand higher temperature (150° C) makes the method suitable for transformers of ratings up to 1.5 MVA. The high rating transformers are used in special applications like in mines where fire is a great hazard. 2. Air Blast (AB). Cooling by natural circulation of air becomes inadequate to dissipate heat from large transformers and hence for circulation of air (air blast) is employed in order to keep the temperature rise within limits. The forced air circulation improves the heated dissipation. In this method, the transformer is cooled by a continuous blast of cool air forced through the cores and the windings. The air blast is produced by external fans. ‘The improvement in heat dissipation caused by air blast allows higher specific loadings to be used in dry type transformers. The use of higher specific loading results in lower size for the transformers. The air supply must be filtered to prevent accumulation of dust particles in the ventilating ducts. The cooling methods used for oil immersed transformers are 1, Oil Natural (ON). The cooling by air is not so effective and proves insufficient for transformers of medium sizes. Oil as a coolant has two distinct advantages : @ It is a better conductor of heat than air, and Gi) it has a high co-efficient of volume expansion with temperature. Therefore, substantial circulation is easily obtained on account of the natural “thermal head” produced due to convection so long as the cooling ducts in the cores and windings are not undually restricted. Hence, almost all transformers (except for the transformers used for special applications like mines where there is a fire hazard) are oil immersed. The assembly of an oil immersed transformer is shown in Fig. 5.38.DESIGN OF TRANSFORMERS: 5.31 The transformer is immersed in oil and the heat generated in cores and windings is passed on to oil by conduction. Oil in contact with the héated parts rises and its place is taken by cool oil from the bottom. The heated oil transfers its heat to the tank walls from where it (heat) is taken away to the ambient air. The heated oil thereby gets cooler and falls to the bottom. Therefore, a natural thermal head is created which transfers heat from the heated parts to the tank walls from where it is dissipated to the surrounding air. The 1. ik surface is the best dissipator of heat but in the case of large rating transformers the transformer tank will have to be excessively large, if used without any auxiliary means of heat dissipation. The reason for this is explained below Consider a transformer A with its linear dimensions & times the linear dimensions of another smaller but a similar unit B employing the same type of cooling technique. The rating of transformer A ig &* times that of transformer B. The losses in a transformer are proportional to volume and hence losses in transformer A are k? times those in transformer B. The heat dissipating area of transformer A is k? times that of transformer B. ‘The temperature rise 8 = Qc/S where Q = losses, S = heat dissipating area, and c = cooling co-efficient. Therefore, the temperature rise of transformer A is k*/A? = h times that of B because the value of cooling co-efficient c remains the same if similar cooling techniques are used. Thus we conclude that temperature rise increases linearly with increase in dimensions. ‘The above can be explained with a simple example. Suppose, transformer B is designed for a rating of 10 kVA with temperature rise say, 40°C which is the maximum permissible. Let transformer A be designed with every linear dimension being twice of corresponding linear dimension of transformer B, Therefore, the rating of transformer A is 24 = 16 times that of B, which is 160 kVA. The losses in transformer A are 2° = 8 times that in B, and the surface area of Ais 2? = 4 times that of B. Consequently, the temperature rise of transformer A is 8/4 = 2 times that of B, ie., the temperature rise of transformer A is therefore, 80°C. This temperature rise is twice than the maximum permissible. Hence, we conclude, that with increase in rating of transformers, the temperature rises and means must be adopted in order to keep the temperature rise within permissible limits. The temperature rise is given by the relationship 0 = Qc/S. It is evident, that the temperature rise can be decreased and brought within limits by two means. These are : (i increasing S, the area of heat dissipation, and (ii) decreasing the cooling co-efficient, c. The temperature rise of a transformer is inversely proportional to S, its heat dissipating area. Thus if the size of the tank is increased, the dissipating area increases and hence the temperature rise decreases. Thus, in the example above if the size of the tank is increased such that its heat dissipating area is twice that which is otherwise demanded, the temperature rise of the transformer is brought down to 40°C. Therefore, this calls for an excessively large tank. Corrugations Ribs or fins Tubes Lr Tank wall Fig 5.38, Oil-circulation in a transformer Fig. 5.39. Transformer tanks with using cooling tubes. fins and corrugations.5.32 ELECTRICAL MACHINE DESIGN Increasing the size of tank out of proportion with the increase in transformer rating and consequent increase of dimensions, is obviously no solution for the temperature rise. The plain walled tank cannot be used beyond a particular rating. This is becauge if the large rating transformers used plain walled tank, the size of the tank would become enormously large. A large sized tank needs large volume of oil and hence it will result higher cost and weight of transformer. Also, beyond a particular size it would become impossible to transport the transformer (due to its large size) from the place of manufacture to the site of installation due to the limitations imposed by the gauge of the rail line or the road. ‘The solution to the problem of decreasing the temperature rise of large transformers lies in decreasing c, cooling co-efficient. The value of cooling co-efficient can be decreased by augmenting the cooling by using auxiliary means, It should be understood that as the rating of the transformer increases, the value of cooling co-efficient has to be decreased. The reduction in the value of cooling co-efficient can be brought out by the use of sophisticated methods of cooling. Therefore, as the rating of transformers increase we have to use improved methods of cooling in order to keep the temperature rise within limits. The higher rating transformers require increasingly improved methods of cooling. Oil immersed transformers of ratings upto 30 kVA use plain walled tanks. The heat dissipating capability of transformers of ratings higher than 30 kVA is increased by providing corrugations, fins, tubes and radiator tanks. Fig. 5.39 shows fins and corrugations provided on four walls. Fig. 5.40 shows a transformer provided with cooling tubes. These tubes are welded to the tank walls at the top and bottom, The use of cooling tubes provides additional cooling surface but also improves the circulation of oil due to increase in thermal head. For larger sizes of transformers, radiator tanks with fins or corrugations are employed. Fig. 5.41 shows a transformer provided with external radiators. It is clear, that in the method described above the oil in the transformer is circulated on account of natural thermal head. The oil takes away the heat from inside the transformer to outside. The oil is cooled in tubes or external radiators by natural circulation of air. Therefore, the methods described above can be termed as ONAN (oil natural and air natural). SOSOSSSSOOOO gon Radiator tonk Tubes 75 mm Or dia. 1 O 0mm dia. | 75 mm Fig. 5.40, Tank with tubes. Fig. 5.41. Tank with external radiators. ‘Transformers upto a capacity of about 5 MVA or a loss of upto 50 kW use tanks with tubes, The tubes are usually round and are 55 mm in diameter and are arranged in one to three rows. Elliptical tubes are also used.DESIGN OF TRANSFORMERS 5.33 2. Oil Natural Air Forced (ONAF). In this method the oil circulating under natural head transfers heat to the tank walls. The transformer tank is made hollow and air is blown through the hollow space to cool the transformer. The heat removed from the inner tank walls can be increased to five or six times that dissipated by natural means and therefore very large transformers can be cooled by this method. However, the normal way of cooling the transformers by air blast is to use radiator banks of corrugated or elliptical tubes separated from the transformer tank and cooled by air blast produced by fans. 3, Oil Natural Water Forced (ONWF). In this method, copper cooling coils are mounted above the transformer core but below the surface of oil. Water is circulated through the cooling coils to cool the transformer. ‘This method proves to be cheap where a natural water head is already available. The method has, however, the serious disadvantage that it employs a cooling system which carries water inside the oil tank. Since the water is at higher head than oil, therefore, in case of leakage water in the cooling tubes will enter the transformer tank contaminating oil and redueing its dielectric strength. - Since heat passes three times as rapidly from copper cooling tubes to water as from oil to copper tubes, the tubes are provided with fans to increase conduction of heat from oil to tubes. The water inlet and outlet pipes are lagged in order to prevent the moisture in the ambient air from condensing on the pipes and getting into the oil 4, Forced Circulation of Oil (OF). In large transformers the natural circulation of oil is insufficient for cooling the transformer and forced circulation is employed. Oil is circulated by a motor driven pump from the top of a transformer tank to an external cooling plant (heat exchanger or refrigerator) where the oil is cooled. The cold oil enters the transformer at the bottom of the tank. The method of cooling oil in the heat exchanger depends upon the condition obtained at the site. The methods of cooling transformers by forced circulation of oil are classified accordingly as: (@ Oil Forced Air Natural (OFAN). In this method oil is circulated through the transformer with the help of a pump and cooled in a heat exchanger by natural circulation of air. This method is not commonly used. However, this method proves very useful where the coolers have to be well removed from the transformer. (ii) Oil Forced Air Forced (OFAF). The method is depicted in Fig. 5.42. The oil is cooled in external heat exchangers using air blast produced by fans. It is interesting to note that the oil Bushings Conservator eS Nema Oil/heat |r IL I Ht Se Fon oil pump Fig. 5.42. Oil forced air forced cooling (OPAP).5.34° ELECTRICAL MACHINE DESIGN pump and fans may not be used all the time. At low loads, the losses are small and therefore natural circulation of oil with an ONAN condition may be sufficient to cool the transformer. At higher loads, the pump and the fans may be switched on by temperature sensing elements. Therefore mixed cooling conditions are used, the transformer working with ONAN conditions upto 50% of rating and OFAF conditions at higher loads. This arrangement results in higher efficiency for the system. (iii) Oil Forced Water Forced (OFWF). In this method the heated oil is cooled in a water heat exchanger. The pressure of oil is kept higher than that of water and therefore any leakage that occurs is from oil to water. Also there are no condensation problems. At sites, where the cooling water has a considerable head, it is usual to employ cascaded heat exchangers i.e. oil/water and water/water with the intermediate water circuit being at a low pressure. This cooling method is suitable for banks of transformers, but from the system reliability considerations not more than, say, three tanks should be connected in one cooling pump circuit. The advantages of OFWF method over ONWF are that the transformer is smaller and the transformer tank does not have to contain cooling coils carrying water. ‘The use of water as a coolant is common at generating stations, particularly hydro-electric stations, where large supply of water is available. ‘Transformers with a capacity of upto 10 MVA have a cooling radiator system with natural cooling. ‘The forced oil and air circulation (OFAF) method is the usual one for transformers of capacities 30 MVA upwards. As stated earlier, the forced oil and water (OFWF) is used for transformers designed for hydro-electric plants 5.19 TRANSFORMER TANK Tank bodies for most of the transformers are made from rolled steel plates which are fabricated to form the container. Small tanks are welded from steel plates while larger ones are assembled from boiler plates. The tanks are provided with lifting lugs. Small transformers have cooling tubes let into the vertical sides, but large transformers require separate banks of cooling tubes. Such transformers have plain tanks with provision for pipe and valves to direct and control the oil flow. While designing tanks for transformers, a large number of factors have to be considered. These factors include keeping the weight, stray load losses and cost a minimum, and it is obvious that these are requirements contradictory. ‘The tanks should be strong enough to withstand stresses produced by jacking and lifting. The size of the tank must be large enough to accommodate cores, windings, internal connections and also must give the requisite clearance between the windings and the walls. Aluminium is increasingly being used for transformer tanks as a mean’ of reducing weight. ‘The use of aluminium in piace of steel reduces the stray magnetic fields (since it is a non- magnetic material) and consequently the stray load loss, However, aluminium tanks are costlier. Also the use of aluminium necessitates special lifting arrangements in order to prevent stressing of tank. However, usually aluminium tanks are made of cast aluminium parts mounted on a shallow mild steel tray. The mild steel tray is arranged to carry the main lifting and jacking members. Where mild steel tanks are used for units with high leakage flux, electromagnetic screens or shunts are used to reduce eddy current losses.DESIGN, OF TRANSFORMERS 5.35 5.20 COOLING DUCTS In large transformers, the cooling surface of the cores must be augmented otherwise temperature rise will be excessive owing to smail surface/volume ratio of the cores. The additional surface is provided by cooling ducts. The cooling may be (i) horizontal or (ii) vertical. The vertical cooling ducts are along the direction of laminations and hence can be easily provided. The horizontal ducts are across the laminations and therefore require special punching of core. The oil flowing through these ducts takes away the heat. In the case of vertical ducts, the heat is conducted across the laminations and since the thermal resistivity across the laminations is high, there is a large temperature gradient between the hot spot in the core and the duct. Hence provision of vertical ducts does not improve the cooling of core. However, when horizontal ducts are provided, the heat is conducted along the laminations and since the thermal resistivity along the laminations is only 1/20 of that across the core, the temperature gradients are extremely small and therefore cooling is significantly improved. | O ——— wom — (ff J UL ve } ODDO H { L- Hortzontal SS = I Coil Sno Gylinder —_—————— a snr oll mam (Goouo _—— SS OOOOUo Fig. 6.43, Directed flow of oil in transformers,5.36 - ELECTRICAL MACHINE DESIGN ‘The oil must have a free access to all:parts of winding. The multilayer helical windings have an advantage in this respect since they are made coils of relatively small radial depth and, therefore, the majority of the coil surface is exposed to oil in the vertical cooling ducts. ‘The large disc type of windings have a large radial depth and they have a disadvantage that the majority of the coil is exposed to oil in the horizontal ducts. Where the vertical ducts on each side of the winding are of equal width, the oil flows up these duets under the influence of pumps or natural convection and will not tend to enter the horizontal ducts particularly when the radial depth of the winding is large. Therefore, baffles are used to direct the rising stream of oil to flow in the horizontal duets. This arrangement is shown in Fig. 5.43. 5.21 TRANSFORMER INSULATION 5.21.1 Oil One of the most important factors which determines the life and satisfactory operation of a transformer is the oil in which it is immersed. The transformer oil has two prime functions (® To create an acceptable level of insulation in conjunction with insulated conductors and coils. (ii) To provide a cooling medium capable of extracting quantities of heat without deterioration as an insulating medium. ‘Transformer oil consists of mainly four generic class of organic compounds namely : aromatics, paraffins, napthenes and olefines. This makes the transformer will to be a pure hydrocarbon mineral oil. For a fresh sample of new transformer oil, it is desirable to have more of saturated paraffins, less of aromatic and napthenes and olefines. This increases the insulation property. On the other hand, for better stability property, it is necessary to have more aromatics and napthenes. By careful refining of crude petroleum, it is possible to have optimum mix of the four organic components such that the transformer oil does have both insulating property and stability. Vegetable and animal oils are not used in transformer as they form fatty acids and attacks the fibrous insulating materials. Some of the insulating properties/characteristics of the transformer oil (as per IS 335) are discussed below : PHYSICAL PROPERTIES (i) Oil density—The maximum value of density is fixed at 29.5C as 0.89 gm/ce as p at this temperature the water in the forms of ice present in the oil (if any) settles at the bottom. and does not float on the oil upto—temperature of - 10°C. (ii) Viscosity—It is the measure of oil resistance to the continuous flow of oil without the effect of external forces. Viscosity increases with decrease in temperature and it is necessary that viscosity remains low at low temperature. The permitted value of viscosity is 27 cSt at 27°C (max) (iii) Interfacial tension—It is the intermolecular attractive forces between the oil and water molecules at their interface. With ageing, soluble contaminants and different chemical products develop in the oil (called sludging) and the intermolecular for: between oil and water molecular deteriorate. Interfacial tension gives an indication of degree of sludging in transformer oil. The permitted value of this tension is 0.04 N/wt at 27°C (min).DESIGN OF TRANSFORMERS 5.37 (iv) (v) (vi) Flash point—It represents a temperature at which oil gives an optimum quantity of oil vapour, which when mixed with air, forms an ignitable mixture and may cause a momentary flash. A minimum flash point has been specified in order to prevent the risk of fire resulting from accidental ignition. The flash point as specified is 140°C (min) Moisture Constant—It represents the amount of force and dissolved water present in transformer oil and expressed as ppm by weight (mg/kg). Presence of moisture degrades the insulation property of the oil and winding insulation. The specified moisture constant is 50 ppm by weight (max). Pour point—I is that temperature at which oil just commences its flow under prescribed conditions. The specified pour point is - 6°C (max). CHEMICAL PROPERTIES (wii) (viii) (ix) @) Neutralisation value—It is a measure of acidity of oil. It represents the measure of force organic and inorganic acids present in the oil and is expressed in terms of mg of base (KOH) required to neutralise the total force acids in one gram of oil. The recommended value is 0.03 mg KOH/gm (max). Higher value of neutralisation means higher acidity of oil and it may lead to higher sludge formation. Corrosive Sulphur—The transformer oil being derived from petroleum, if contains traces of sulphur compounds and sulphur. Presence of these chemicals in oil leads to formation of a black deposit on the surface of base copper parts in the transformer. This deposit adversely affects the dissipation of heat and consequently the performance of the transformer is affected. Hence corrosive sulphur is not recommended to be present in oil. Oxidation Stability—This parameter is a measure of neutralisation value and sludge after ageing of the oil. This test is very important for new oil. For this parameter, the neutralisation value after oxidation is 0.4 mg (KOH)/gm (max) and total sludge after oxidation (max) is 0.1% by weight. Sediment and precipitable sludge—These are basically oil deterioration products which are insoluble after dilution of oil under prescribed condition. Oil is considered not useable once sediment or precipitable sludge is detucted. Sludge is poor conductor of heat and produces sharp temperature gradients across oil and winding insulation, Moreover it does not allow the oil to flow in ducts and impairs cooling. ELECTRICAL PROPERTIES (i) (wii) (xiii) Breakdown voltage (BDV)—It is the voltage at which electric breakdown occurs between two electrodes under prescribed conditions. It represents the electric strength of oil. The electric strength of oil deteriorates with presence of contaminating agents, moisture, sludge and sediment. The minimum breakdown voltage of used transformer oil is 30 kV (rms) while for the new sample or a sample after filtration is 60 kVirms). Resistivity—It is one of the most sensitive property of transformer oil. It not only characteristics the insulating property of the oil, the insulating property of the winding insulation is also dependent upon the resistivity of oil. The recommended min. resistivity at 90°C is 30 x 101? QO cm and at 27°C is 1500 x 10!? Q em. electric dissipation factor (DDF). It is numerically the sine of the loss angle and is approximately equal to tangent of loss angle for dielectrics. A nigh value of DDF indicates an inferior quality of oi] marked by presence of contaminants such as oxidation products, moisture ete. The DDF (or turn delta) factor of vil at 90°C is recommended to be 0.002 (5.38 ELECTRICAL MACHINE DESIGN ‘The physical constants, recommended for the transformer oil is shown in the following table : Physical Constants of Transformer Oil Property Recommended Value (a) Permittivity 2.2 () Thermal conductivity 0.12 Wim C* (c) Specific heat 2.06 kJ/kg C° (d) Coefficient of expansion 0.00078/deg C (e) Mean density correction factor 0.00065/deg C Transformer oil tends to deteriorate in service, but this tendency can be greatly reduced by paying attention to transformer operating conditions and to oil itself when this is shown to be necessary as the result of regular tests. The most important factors are : (@ operating temperature, (ii) atmospheric conditions particularly inside stations, (ii) electrie strength, (iv) moisture and other contamination (v) sludge formation. If the oil is found to contain moisture or suspended contaminants it should be filtered or if this treatment is considered to be inadequate, the oil should be replaced by fresh charge. Transformer oil is normally tested once every year and, if found below standard, may be treated by a centrifuge or filter unit. 5.21.2 Other Insulating Materials A Insulating Paper Vegetable fibres are fitted to form a sheet of paper. The fibres mainly consist of cellulose, molecular formulae being (C,H,,0,),. The paper for insulation is prepared by a complex chemical process and some of the chief characteristics of the insulating paper are highlighted below : (@) Grammage. It is the ratio of mass to the area and influences most of the electrical and mechanical properties. It's recommended value is 100 + 5 gm/mt? for 125 ym thick paper. (ii) Density. It represents the mass per unit volume of paper. Paper of density range 0.6 to 1.3 gm/c. is recommended in transformer use. (iii) Moisture content. Paper being hygroscopic, moisture ingression takes place with usage and it reduces the dielectric property of the paper. Maximum moisture content of 8% is allowed. (iv) Oil and water absorption. Paper attains a high value of electric strength when impregnated in oil under vacuum. However, water absorption is not recommended through 6 to 10% of water absorption is permitted at maximum. (v) Air permeability. It indicates the rate at which paper allows air to penetrate through it, Electrical strength of the paper is inversely proportional to its air permeability. Recommended value is 0.2-0.5 umt/Pascal. (vi) Tensile Strength and elongation. Paper tape is wound over the conductor and the paper must be able to withstand tension exerted during its wrapping. Recommended value is 78- 85 N-mt/gm (mm) in wrapping direction and 25-30 N-mt/gm (min.) in cross direction. (vii) Tearing resistance. It is the load under which the paper would just tear off. Its value ranges from 40-120 gmv/mt? (viii) Heat stability. It is the capability of the paper to withstand thermal stress. The maximum recommended value is 20%.DESIGN OF TRANSFORMERS 5.39 ‘The electrical properties are indicated below : (a) Breakdown voltage ~ 7-7.5 kV/mm (min) in air at 90°C. * (®) Dissipation Factor ~ 0.003 (max) for oil impregnated paper. (c) Conductivity - 10 ns/mt (max). B. Pressboard Like paper, pressboard is also made of vegetable fibres and contains cellulose. Solid pressboard upto 6 mm thick or precompressed pressboard upto 8 mm thick is ordinarily made. Since the most difficult insulation problem in HT transformer occur at the ends of the windings and at the lead outs from the windings hence moulded pressboards are widely used in these parts for insulation. Synthetic resin bonded paper based laminates are used in voltage stressed zones. Properties like density, tensile strength, elongation, conductivity, oil absorption, moisture content ete. are also specified for pressboards. In addition, compressibility, shrinkage and edgewise electric strength are the important parameters. If the compressibility of the pressboard is more than the value taken in design, the coil height will be less than the design value. This reduction can head to serious problems of short circuit forces if not corrected. Shrinkage is a measure of reduction in length, width and thickness under specified conditions. Edgewise electric strength measures the creep strength across the glue line of laminated pressboard. 5.22 TERMINALS AND LEADS The connections to the windings are of insulated copper rods or bars. The shape and size of leads is important in high voltage transformers owing to dielectric stress and corona which are caused at bends and corners. Sharp edges and corners should be avoided. 5.23 BUSHINGS ‘Transformers are connected to high voltage lines, and, therefore, care must be taken to prevent flashover from the high voltage connection to the earthed bank. Connections for cables are made in cable boxes, but overhead connections must be brought through bushings specifically designed for various voltage classes. The bushing consists of a current carrying part in the form of a conducting rod, bus or cable, a porcelain cylinder installed in a hole in the transformer cover and used for isolating the current carrying part. The simplest bushing is a moulded, high quality glazed porcelain insulator with a conductor through its centre, This bushing is used for voltages upto 33 kV. The porcelain bushings used for indoor use have a smooth surface or slightly finned surface. The outside (upper part) of the bushing used for transformers working outdoors is made with petticoats to protect the lower fins against water in rainy weather. The bushings used for transformers having voltages above 36 kV are either oil filled or capacitor type. The oil filled bushing (Fig. 5.44) consists of a hollow two part porcelain cylinder with a conductor, usually a cylinder, passing through its axis. The space between the conductor and the inner surface of the porcelain is filled with oil. The oil is contained separately from the oil in the transformer tank. The top of the bushing is connected to a small expansion chamber required to accommodate variations in the volume of the oil due to change in operating temperature. There is a provision for current transformer at the lower end of the bushing. The arrangement is such that the bushing can be removed without disturbing the current transformer. The capacitor bushing is made up of layers of synthetic resin bonded paper (s.r.b.p.) interleaved with thin layers of metal foil for paper impregnated with conducting material. The result is a5.40 + ELECTRICAL MACHINE DESIGN series of capacitors with a capacitor formed-by two layers of metal foil with s.r-b.p. cylinder in between. The variation in length of metal foils and the thickness of sz.b.p. cylinders is so arranged that there is a uniform distribution of dielectric stress throughout the radial depth i.e. along the radius of the bushing. Ventilating breather Oil level indicator il expansion chamber Top arcing One piece horn insulator Tubular conductor Spacers anh. sirb.p. Paper Bottom arcing horn Current transformer and housing Lower ingulator Entry for flexible cable Fig. 5.44, High-voltage oil-filled bushing,DESIGN OF TRANSFORMERS 5.41 5.24 TAPPINGS AND TAP CHANGING Windings ‘The voltage of power networks supplied by transformers can be controlled by changing the ratio of transformation of the transformers. The change in ratio of transformation can be affected by providing tappings on the transformer windings. The tappings are connections provided at different places in the windings and therefore, the number of turns included in the circuit at one tap is different from the number of turns at another tap. Hence, the turns ratio is different at different tapings and as different, voltages are obtained at different tapings. Fig. 5.45. Transformer tappings. The tappings used in a transformer are shown in Fig. 5.45. Consider a 8 phase, 11 kV/0.4 KV, distribution transformer. There is always @ tapping on the h.v. winding which when connected to rated voltage (11 kV) gives rated voltage (400 volt in this case) on the Lv. side. This tapping is called the principal tapping. The principal tapping is that to which the rating of the winding is related, The number of turns included at other taps may be either greater or lesser than the number of turns used at principal tapping. The tapping at which the turns included are more than that of the principal tapping is called positive tapping, while a tapping at which the number of turns included is less than the number of turns included at principal tapping is called negative tapping. ‘The tappings are provided on the high voltage (h.v.) winding because of the following reasons : Toppings @ the number of the turns in h.v. winding is large and therefore a fine voltage regulation is obtained. Gi) it may not be possible to provide correct véltage regulation by using tapings on the Lv. side because of the smaller number of turns. Suppose, it is desired to obtain + 24% voltage regulation in a 3 phase, 6.6 kV/400 V, delta/star connected transformer designed for 11.55 V/turn. The voltage per phase on the l.v. side is 400/V3 = 231 V. The number of turns are, 231/11.55 = 20. It is possible to tap a whole number of turns. The minimum number of turns that is possible to tap is 1. Therefore, the minimum voltage regulation possible in this transformer by providing tappings on the Ly. side is + 11.55 V or + 5% and hence it is not possible to obtain a voltage regulation of + 214% with taps on the secondary side. (iii) The current on the Lv. side of high capacity transmission and generation transformers is very high. Therefore, provision of tappings on the l.v. windings for these transformers is impracticable on account of the difficulties encountered in interruption of high currents. (iv) The 1.v. winding is placed on the inside nearer to the cores while the h.v. winding is placed on the outside. Therefore, on account of the practical considerations, it is simpler to provide tappings on the high voltage winding. (v) There is an additional advantage of providing tappings on high voltage winding of step down transformers. The voltage on low voltage side of these transformers increases on light loads. Therefore, in order to decrease this voltage, the tapping on the high voltage side is adjusted to such a position where the number of turns is large. The large number of turns decreases the flux and the flux density. This reduces the core loss which in turn increases the efficiency of the transformers at low loads.5.42 ELECTRICAL MACHINE DESIGN The voltage control in electric supply networks is required on account of many reasons. These include : . @ adjustment of voltage at consumers’ premises within statutory limits. (i) control of active and reactive power, and (ii) adjustments of short period (1—2%), daily (85%) and seasonal (5—10%) voltage variations in accordance with variations of load. Location of tapings is partly a constructional question. The tapings can be provided at the phase ends, at the neutral point, or in the middle of the windings. The advantage of providing tappings at phase ends is that the number of bushing insulators is reduced. This is important, where the transformer cover space is limited. Some transformers have reinforced insulation at the phase ends, it is essential that in such cases either the tappings should not be provided or the re-inforcement should be carried beyond the lowest tap. When the tappings are made at the neutral point the insulation between various parts is small. This arrangement is economical especially in the case of high voltage transformers. When a large voltage variation is required, tappings should be near the centres of phase windings to reduce magnetic asymmetry. However, this arrangement cannot be used on l.v. windings placed next to the core. The different methods of providing tapings on a transformer are shown in Fig. 5.46. ‘The tappings are potential source of axial magnetic asymmetry in a transformer. This is on account of the fact that the number of turns in the winding that is provided with tapings is altered while the number of turns in the other winding remain the same. Therefore, it is clear that cutting in or cutting out a part of the transformer winding would cause a mmf unbalance, thereby producing a magnetic asymmetry in the axial direction. This magnetic asymmetry produces large mechanical forces and consequent displacement of windings in the axial direction in case of faults. ‘The tappings are on one end of the windings as shown in Fig. 5.46(a) in the case of small transformers while they are arranged at the centre of the windings of larger transformers as shown in (6) and (c). The axial mmf unbalance can be significantly reduced by thinning a part of the Lv. winding which is radically opposite to the part of the h.v. where the tapings are provided. HV Lv LT HIE ELE : (0) () © @ ©) © @) Fig. 5.46. Position of tapings. This is shown in (@). The axial mmf unbalance is also reduced by adopting winding arrangements where parts of the winding are balanced more symmetrically as shown in (d) and (e). The axial mmf balance may also be achieved by splitting the untapped winding into a number of parts and then connecting these parts in parallel. This arrangement is shown in (g). The tappings may be changed when the transformer is disconnected from the supply. This is called off-circuit tap changing. The off-circuit tap changing is used for occasional adjustments, as in distribution transformers which are provided with # 5% and + 24% taps. The tappings may also be changed while the transformer is energized or on load. This is known as on load tap changing. Daily and short time voltage adjustments are done with the help of on-load tap- changing gear.DESIGN OF TRANSFORMERS 5.43 Selector The of-circuit and on load tap changing arrange- switch ments are described below. 43 Off circuit tap changing. As the name suggests, 5 the tappings are changed by disconnecting the transformer from the supply. This adjustment is carried 96 out by tapping the respective windings as required and bringing the connections of tappings to some position near the top of the transformer, The change of tappings is done manually through hand holes provided in the cover. Another arrangement employs reconnection that can be made by carrying the tapping leads through the cover for changing either by hand or by manually operated. ? switches. The commonly used switches are : (i) vertical 1 O2 03 O4 O56 tapping switches and (ii) faceplate switches. One form of selector switch used for off-load tap changers is shown in Fig. 5.47. This arrangement is commonly used for providing + 5% tappings in steps of + 244%. Six brass or copper terminals are mounted on an insulating base and a contactor is mounted on an arm attached to the shaft. Taps are brought out __Fig. 6.47. Off-load tap changer. of middle part of the winding and connected to the terminals of tap changer. The shaft is tuned from one position to the next, the contactor connects adjacent pair of the stationary terminals. With the contactor in the position shown, the selector switch connects taps 3 and 4 and therefore all the turns in both parts of the winding are in use. If the contactor is moved one point to the right, it makes a connection between taps 2 and 4, thus cutting out part of the winding between taps 2 and 3. The next step connects taps 2 and 6 and cuts part of the winding between taps 4 and 5. These parts of the winding are cut out in steps, until the final position which connects taps 1 and 6. This position leaves minimum number of turns in the winding. There are five positions of the switch. These positions correspond to five different voltages. Supposing it is desired to obtain + 5% regulation in steps of + 244%. The switch which corresponds to normal voltage is on taps 2 and 5. Assuming that the winding which is tapped is the primary winding, the regulation at other switch position is Taps 4,2 [34 as! 1,6 Regulation —2%% | -1% | +2%% + 5% 2, On-load tap changing. When a transformer is connected to a system it is necessary that arrangements be provided to vary the voltage on the secondary side in order to maintain normal voltage under load conditions. In the case of on-load tap changing, this variation is made when the load is on and hence the tap changing gear must be capable of changing the turns ratio without interruption of supply. It is invariable practice to connect the tapings at the neutral end of high voltage windings of a generator transformer. The tapings on the windings are brought out through a terminal board to separate oil filled compartment in which the on-load tap-changer switch is housed. The tap changer is in the form of a selector switch. The tap changer is operated by a motor operated driving mechanism by local or remote control and a handle is fitted for manual operation in case of an emergency. ‘The essential feature of an on-load tap changing gear is the maintenance of circuit continuity throughout the tap changing operation, The circuit must not be broken otherwise there will be discontinuity of supply to load. Therefore, as the sector switch must not break current, a additional separate oil filled compartment is used to house the diverter switch which breaks the load5.44, ELECTRICAL MACHINE DESIGN current by an interrupted are. This causes formation of carbon and therefore the oil in the diverter switch compartment must be prevented from mixing with the oil,in the main tank. The oil in the selector switch tank may be connected directly to main transformer oil through the conservator. * As mentioned earlier, the tap changing takes place when the load is on and hence in order to maintain continuity of supply to the load, before one tapping is left. opened, contact must of made to the next tapping. Therefore, the selectors witch is on-load tap changers is a make before a break switch and during the period of transition from one tap to another, momentary connection must be made between the adjacent taps. This results in short selector circuiting of turns between the adjacent tappings. gwitch Therefore, the short circuit current must be limited Sz by including resistors or reactors. Reactors have been used fort his purpose earlier, but in modern equipment it is usual to use two resistors for this purpose. Fig. 6.48 shows a typical winding connection for a high speed resistor type on-load tap changer provided at the neutral end of each phase of windings of a star connected 3 phase transformer. terminal Selector ‘switch Sy One selector switch, S, is on tap 1 and the other S, DWerter on tap 2, The diverter switch, S,, is shown red connecting tap 1 to the Neutral neutral point of the transformer winding and terminal the switching sequence for change over to tap 2 is as follows : Fig. 6.48. On-load neutral terminal tap changer. 1. Contacts a and 6 are closed (resistance R, shorted) as shown. The load current flows from tap through contact 6. This is the running position at tap 1. 2, An external mechanism moves the diverter switch S,, contact 6 opens. The load current from tap 1 now flows through resistance R, and contact a. 3. As the moving contact of S, continues it travels further to the left, contact closes and resistance R, is open circuited. Both the resistances R, and R, are connected across taps 1 and 2 and the load current flows through these resistances to their mid-point (junction of a and d). 4, When S, moves still further to the left, contact a is opened. The load, current flows from tap 2 through resistance R, and contact d. 5, Finally, as the diverter switch S, reaches the extreme left position, contact ¢ closes and resistance R, is short circuited. The load current from tap 2 flows through contact a. This is the running position for tap 2. It is observed that tap change from tap 1 to tap 2 as described above does not involve any movement of selector switches S, ‘and S,. However, a further tap change in the same direction, i.e. from tap 2 to tap 3, is required, the selector switch S, is moved to tap 3 before the diverter switch, S, moves. The diverter switch then follows the sequence described above but in the reverse order. In order to limit the loss of energy, it is essential that the resistors be kept in circuit for as minimum of time as possible. The resistors are designed for short time rating for economical considerations and therefore, it is desirable to minimize their time of duty. Therefore, some form of energy storage must be incorporated in the driving mechanism to ensure that tapDESIGN QF TRANSFORMERS 5.45 change, once initiated, shall be completed even in case of failure of auxiliary control supply. All modern on load tap changers use springs energy as storage elements. They reduce the time that a resistor is in the circuit to a few periods. Such a tap changer is compact in size, and high speed breaking reduces the contact wear. ‘The current breaking is eased by the fact that the short circuit resistor current has unity power factor. 5.24A CONSERVATOR AND BREATHER: ‘The satisfactory operation of transformers depends so largely on the condition of oil and therefore devices and methods for keeping the oil clean and dry are of prime importance. ‘The oil level of a transformer changes with changes the temperature rise of oil which in turn depends upon the load on the transformer. The oil expands if the load increases and contracts when it decreases. Therefore, provision must be made to take up this expansion and contraction of oil. Smaller transformers are not totally filled with oil and some space is left between oil level and the tank cover. This space is taken by air. The tank is connected to the atmosphere through a vent pipe. When the oil expands, air is expelled out while if it contracts air is drawn in from the atmosphere. This is called breathing of transformer. The air entering the transformer is passed through an apparatus called breather for the purposes of extracting moisture from it. A breather consists of a small container connected to the vent pipe and contains a dehydrating material like silica gel crystals impregnated with cobalt chloride. The material is blue when dry and a whitish pink when damp. The colour can be observed through a glass window provided in front of the container. Breathers alone are not sufficient for protection of large and important power transformers because : (@ these transformers are liable to overloads which may overheat the oi] and consequently there is sludge formation if air is present. (ii) Oceasionally such transformers also suffer short circuits and temperature rise becomes very high. This causes vaporization of a part of the oil. The oil vapours form explosive mixture with air which ignites and can cause considerable damage. For these reasons oil is prevented from having contact with air as well as moisture. Conservators are used for this purpose. The function of the conservator is to take up expansion and contraction of the oil without allowing it to come in contact with ambient air, from which it might absorb moisture. ‘The conservator is an airtight cylindrical drum mounted on or near the cover of the transformer and connected to it through a small pipe as shown in Fig. 5.49. The oil is set so that the transformer tank is entirely full with oil and the conservator is about half full. The interior of the conservator above the oil level is connected to the atmosphere through a breather having dehydrating material. With the use of conservators, interchange of oil between conservator and main tank as a result of temperature changes is slow. Also dry is in contact with much smaller surface of relatively cool oil. Hence the sludge formation is considerably reduced and whatever sludge is formed remains in the conservator there being no sludge formation in the main tank. This is a great improvement over the ordinary tank with air space above the oil. 5.25 EXPLOSION VENT In order to guard against the possibility of a sudden high pressure caused by a breakdown or a short circuit in the transformer winding, a diaphragm relief device is used. This device consists of a large opening to the atmosphere covered by a thin non-metallic diaphragm. The diaphragm bursts if the pressure inside the tank becomes excessive. The relief device must be above the level of oil in the conservator in order to prevent an overflow of oil in case the device operates.5.46 ELECTRICAL MACHINE DESIGN 5.26 TEMPERATURE INDICATORS ‘The most obvious indicator of transformer temperature is the tempesature of the hot oil. The oil temperature is measured by a dial type thermometer. The bulb of the thermometer is mounted in the oil and the dial is mounted outside the tank. However, oil temperature is not a reliable measure of the winding temperature especially under sudden over-loads which cause the winding temperature to rise more rapidly than the oil temperature. Therefore, it is desirable to use ari indicator which will show the actual temperature of hot spot in the windings. Winding temperature indicator is a thermometer with a bulb. The thermometer is immersed in oil and the bulb is heated by heaters which carry a current proportional to the winding current. Therefore, the reading of the thermometer is an analogue indication of winding temperature. 5,27 BUCHHOLZ RELAY It is a gas and oil actuated protective device and is used practically in all oil immersed transformers with the exception of smaller distribution transformers. Buchholz relay is used for protection of transformer against faults developed in side the transformer. The device relies on the fact that an electrical fault inside the transformer tank is accompanied by generation of gas, and if the fault current is high enough by a surge of oil from the tank to the conservator. Bushings veri Conservator Bushings: ey Buchholz relay Transformer tank (a) Fig.5.49. Buchholz relay.DESIGN OF TRANSFORMERS 5.47 ‘The Buchholz relay is particularly useful in that it is eapable of detecting fault conditions of very low magnitude such as interturn faults, incipient winding faults, and core faults due, for example, to core bolt insulation failure which gives rise to short circuit and subsequent arcing and gas. ‘The use of a Buchholz relay is possible only with transformers having conservators and the relay is placed between transformer tank and the conservator as shown in Fig. 5.49(a). A Buchholz relay consists of an oil filled chamber as shown in Fig. 5.49 (6). It contains two floats, the top float F, and the bottom float F,. Both the floats are hinged so as to be pressed by their buoyancy against two stops. When.these floats sink, each of them short circuits two contacts thereby closing a circuit which gives a warning about the fault. ‘The operation of the Buchholz relay can be explained as under : In case a fault occurs, gas bubbles are generated in the transformer tank on account of increased heating produced by faults currents. These bubbles rise and go towards the conservator through the pipe line, The gas bubbles are trapped in the upper part of the relay chamber and since the chamber is full of oil, the oil is displaced which lowers the top float F,. The float sinks ultimately thereby causing the closure of contacts of the “alarm circuit”. A small window in the wall of the relay chamber shows the amount of gas trapped and its colour. A sample of the gas may be withdrawn and analysed. The amount of gas is indicative of the severity of the fault while the colour of the gas indicates the nature of the fault since the faults occurring in cores and windings produce gases of different colour. In case the fault is not severe, the generation of gas is not sufficient so as to lower the level of oil to the extent to affect the bottom float F,, Therefore, in case of mild faults inside the transformer, the bottom float F, remains unaffected. However, in the case of severe faults, such as short circuits between phases or to earth and faults in the tap changing gear, the gas production becomes violent and cause a surge which cause the bottom float F, to fall and close the contacts of a “trip circuit”. This trip circuit energizes the relay of a circuit breaker which open circuits the transformer thereby clearing the fault. 5.28 SEALING MATERIALS Sealing materials are used to avoid oil leakage at joints, gasket and moulded components and are made from nitrile rubber or nitrile rubber bonded cork polyamide washers, tape etc are used for plugs and valves. The sealing material is used in form of a gasket with a metal limiter to avoid excessive compression. The corked gasket is suitable to even a rough un-machined surface, The sealing material used must be clear, uniformly granulated and soft grade type. it offers a tight seal even under a soft and light load. The sealing gasket even allows considerable distributor of gasket without crack and damage. The pressure of cork allows compression without excessive spreading.5.48, ELECTRICAL MACHINE DESIGN For easy understanding and to get a pictorial view, a complete assembly of a transformer with details of core, windings, tank, connections and major accessories is,sHown in Fig. 5.50 g _ T= Cover ng lugs a Thermometer alk AA Air vent pocket (Breather) ; Cold oil level Olt gauge Cooling tube: Tie rods Tapping = switches Fabricated steel plate tonk Radial coi! ‘spacers . Ineulating tube Core leg clamp Insulated core plates bolts ° | ° Earthing socket on LY. side Ol drain valve ° ° Rollers @ Cable socket Tapping switches (Lv.) on Tap indicator plate Porcelain insulated terminal with cable Core & coils socket (HV.) fitting lug Compound filed cable box Top core ‘clamp Tapping leads to switches LV. Insulating Insulation cylinder HY. Insulating cylinder Bottom core clamps &) Fig. 6.60. Transformer AssemblyDESIGN OF TRANSFORMERS 5.49 PART B : DESIGN DETAILS 5.29 OUTPUT OF TRANSFORMER * Let ©,, = main flux, Wb; B,, = maximum flux density, Wb/m” current density A/m? ; A,, = gross core area, m? ; net core area, m? = stacking factor x gross core area ; A, = area of copper in the window, m?; A,, = window area, m?; D = distance between core centres, m ; d = diameter of circumscribing circle, m ; K,, = window space factor ; f = frequency, Hz, E, = emf per turn, V ; T,, T, = number of turns in primary and secondary windings respectively ; 1, 1, = current in primary and secondary windings respectively, A ; V,, V, = terminal voltage of primary and secondary windings respectively, V ; 4, @, = area of conductors of primary and secondary windings respectively, m? 1, = mean length of flux path in iron, m ; L, = length of mean turn of transformer windings,\m ; G, = weight of active iron, kg ; G, = weight of copper, kg ; &; = Weight per m® of iron, kg ; g, = weight per m? of copper, kg ; P; = loss in iron per kg, W ; p, = loss in copper per kg, W. (@ Single Phase Transformers. The voltage induced in a transformer winding with T turns and excited by a source having a irequency f Hz is given by : E Voltage per turn —E, 444 f.©,, (6.1) ‘The window in a single phase transformer contains one primary and one secondary winding. -. Total copper area in window A, = copper area of primary winding + copper area of secondary winding, = primary turns x area of primary conductor + secondary turns x area of secondary conductor =T,a,+T, 4, "py + Ty Oy Taking the current density 5 to be the same in both primary and secondary windings. a, = 1/8 and a= 1/5 +. Total conductor area in window A, =, .L,/3 +7, . 1/8 = (Tl, + Til,Vo 2AaT 5 as TI, = TJ, = AT if we neglect magnetising mmf. The window space factor K,, is defined as the ratio of copper area in window to total area of window, conductor area in window A, a of window Ay Conductor area in window A, = KA, (5.8 From Eqns. 5.2 and 5.3, vr © = total are or (5.4)5.50 ELECTRICAL MACHINE DESIGN Rating of a single phase transformer in kVA Q=V,I,.10%=E,I,.10- (as V, is approximately equal to E,) =E,T,J, 10° = E, AT 10° <8, Kede® 99-440, Kede® 102 = 2.22/0,, KA, 6. 10% ABB) But ®,, = maximum flux density x net area of core = B,,A; Q=2.22/B,, 8K,A,, A; x 10° kVA (5.6) Gi) Three Phase Transformers. In the case of three phase transformers, each window contains two primary and two secondary windings. Proceeding as in the case of single phase transformers. Total conductor area in each window A, = 2(a,7, + @,7,) 2U,T, + 1.7.8 Total conductor area is also equal to K,A,. -. 4AT/8 = KA, or AT = Bes (5.8) Rating of a three phase transformer in kVA Q=3V,J, 10%=3EJ, 10%=3 ET, =3 E,AT ‘ly =3x 4440, x Skt x 10-= 3.380, KA, 6x 10% — ...(5.9) = 3.33 /B,, K,A,A; x 10% (5.10) Equations (5.6) and (5.10) hold good for both core and shell type transformers. 5.30 OUTPUT EQUATION—VOLT PER TURN Considering the output of one phase. kVA rating of one phase Q=1LY,.10%=1,x444/0, 7, 10%=4.44f0, AT.10% (6.11) The ratio ©,/AT is a constant for a transformer of a given type, service and method of construction. Let ,/AT = r where r is a constant. From Eqn. 5.11 Voltage per turn E, 44 f 0, = 4.44 (eam 2 = y44afr.10 . (Q =K/@ (5.12) a where Kz 444 fr. = (444 f ue 108) (6.13) As the ratio ©,/AT depends upon type of transformer and therefore K is also a constant whose value depends upon type, service conditions and method of construction.DESIGN OF TRANSFORMERS 5.51 Table 5.2 gives values of constant K for different types of transformers. Table 5.2 * Type K Single phase shell type 10t012 Single phase core type 0.75 100.85 ‘Three phase shell type a ‘Three phase core type (distribution) 045, ‘Three phase core type (power) 0.6t00.7 5.31 RATIO OF IRON LOSS TO COPPER LOSS Copper loss per m° = pé? on(5.14) Taking resistivity of copper as 2.1 x 10 Om at 75° and its density as 8.9 x 10° kg/m’. Copper loss per kg at 75°C i.e. specific copper loss = 21x10" oo 32 32 P= $9 x10 = 2.36 x 10-1 6? Whkg (5.15) where 6 = current density, A/m?. In addition to above we must take into consideration the stray load loss which may be 5 to 25 per cent of copper loss. Total copper loss W. = p,G,. Total iron loss per kg i.e. specific iron loss p; can be found from the loss curves. Total iron loss W, = pj, Ratio of iron loss to copper loss PB PG P. = 2G (5.16) ‘When the densities in the iron and copper are fixed, the loss per kg for iron and copper can readily be determined. The ratio of weights for a given ratio of losses can easily be calculated from Eqn. 5.16. The ratio of weight of iron to weight of copper generally lies between 1.5 to 3.0 for distribution transformers. For small capacity, single phase core type transformers, the ratio of weights is often less than 1.5. For high voltage power transformers, it may be twice the values given above. 5.32 RELATION BETWEEN CORE AREA AND WEIGHT OF IRON AND COPPER From Eqn. 5.6, kVA output of a single phase transformer is given by Q =2.22/B,, K,A,A; x 10% Now weight of iron G, = A; /,g, and weight of copper G, = 24, T,, L,,&, if the weights of the primary and the secondary windings are taken to be equal. G Abg | _ Ads . G, * Bay Ry LB Ke Ay lm Be Ae If the ratio of mean length of the magnetic circuit to the mean length of turn of the winding is assumed to be constant, which is approximately true for a given type of transformer, then Ratio o,f, ie G, 4 Rea and therefore K,A,, = C. con where C, = rs Substituting the value of K,A,, in Eqn. 5.6, @ 22 B,C, & AZ x 103552, ELECTRICAL MACHINE DESIGN 51D) Aa 1 where = (fm fy 1) (5.18) s (222% gi Taking g; = 7.8 x 10? kg/m? and g, = 8.9 x 10° kg/m? C= 22.67 Jin, (5.19) For three phase transformers 1_ Ent Be 193 = {Poe &e 5103 (6.20) ae % Bi ) 620) =13.1 (En /h (5.21) ‘Typical values of ratio L,,/l; for different types of transformers are : Single phase core type—0.8 to 0.55 Three phase core type—0.17 to 0.5 Single phase shell type—1.2 to 2. 5.33 OPTIMUM DESIGNS ‘Transformers may be designed to make one of the following quantities as minimum. (i) Total volume (ii) total weight, (ii) total cost, (iv) total losses. In general, these requirements are contradictory and it is normally possible to satisfy only one of them. All these quantities vary with ratio r = ©,/AT. If we choose a high value of r, the flux becomes larger and consequently a large core cross-section is needed which results in higher volume, weight and cost of iron and also gives a higher iron loss. On the other hand, owing to decrease in the value of AT the volume, weight and cost of copper required decrease and also the [?R losses decrease. Thus we conclude that the value of r is a controlling factor for the above mentioned quantities. 5.33.1. Design for Minimum Cost Let us consider a single phase transformer. In kVA output is Q = 2.22 B,,5K,A,A; x 10° = 2.22 f B,,dA,A; x 10°. Assuming that the flux and current densities are constant, we see that for a transformer of given rating the product 4,A, is constant. Let this product A.A, = M? Ai) The optimum design problem is, therefore, that of determining the minimum value of total cost: Now, r JAT and, =D,A, and AT = 5K,A,/2 = 8AJ2 2BA A, oA aa r=B (id) where fi is a function of r only as B,, and 5 are constant. Thus from (7) and (ii) we have Mi and A, = M/ Ji Let total cost of transformer active materials,DESIGN OF TRANSFORMERS 5.53 C, = total cost of iron, and C, = total cost of conductor C,=C,+C.= ¢G,+¢,G, = CBA, + Cbg where ¢; and c, are the specific costs of iron and copper respectively. Now, C,= 8), M YB + 68, Ly MIB Differentiating C, with respect to B, a 1 =i yv2_ 1 32 ap = BK MOY — = 68 Lge MB Por minima cont £2 «0 og or 68; 1;A; = Cc Lge Ac OF 6/6; = 0,6, Be Lime BY OF ,8;1; = C8. Lime a or C=C, Hence, for minimum total cost, the cost of iron must equal the cost of conductor. Now G,/G, = cJe; for minimum cost. Knowing the value of specific costs of iron and conductor the ratio of weight of iron to conductor can be determined. This can be substituted in Eqn. 5.17 to determine the core area which gives minimum cost for the transformer. Similar conditions apply to other quantities eg., For minimum volume of transformer, Volume of iron = volume of conductor Glg,=G,le, ie, GIG, = 8/8, ore weight of iron = weight of conductor ie, G; = G,. For minimum losses in transformer i.e., for maximum. efficiency, iron loss = PR loss in conductor or P, = x°P, 5.33.2 Design for Minimum Loss or Maximum Efficiency Total losses at full load = P, + P, At any fraction x of full load, the total losses are P, +x? P, If Q is the output at full load, the output at fraction x of full load is x@. .. Efficiency at output xQ, 1, = +27 = Differentiating n, we have te = G@+R +x RQ OQ Osh) dx (2Q+ P+ x?P,)? For maximum efficiency, (x@ + P, + x? P,) Q- xQ(Q + 2xP.) = or P,=2x2P, So that the maximum efficiency is obtained when the variable losses are equal to the constant losses.5.54 ELECTRICAL MACHINE DESIGN From Eqn. 5.16, we have : AL BG . PPG, x2 PG gp Gi «32 Pe foe maximum efficiency. Pe Ge G. Pi Now knowing the values of densities in iron and copper the specific loses p, and p, can be determined and the value of x ic., the fraction of full load where the maximum efficiency occurs depends upon the service conditions of the transformer and is, therefore, known. Thus ratio GJG, is known and its value is put in Eqn. 5.17 to get the core area for maximum efficiency. 5.34 VARIATION OF OUTPUT AND LOSSES IN TRANSFORMERS WITH LINEAR DIMENSIONS Consider two transformers of same type with all their linear dimensions in the ratio x : 1 and having the same flux density, current density, frequency and window space factor. Let the transformer with dimensions x times be called A and the other transformer B. Output. From Eqns. 5.6 and 5.10, output of single and three phase transformers is Output « fB,, 5K A,A,. Now f,B,, 8 and K,, are constants, window area A,, « x? and net iron area A, « x2. Output cx? x x? « a4, Hence output of transformer A is x‘ times that of B. Losses, Total R loss = PR loss in primary + PR loss in secondary 22k 2, pop Phmts Pay oe ay CLruapr Enus = length of mean turn of primary and secondary windings respectively). Now I, = 8a, and I, =6a, :. Total PR loss = 8% (4,T,Lgyy + @,T Lp) = 8% x volume of copper. Now 8 and p are constants and volume of copper « x2. Hence total FR loss = x Thus PR loss of transformer A is x° times that of transformer B. The specific iron loss i.e. loss per unit volume is constant if flux density and frequency are constant. Total Iron loss = loss per unit volume x volume « volume « x3. Both FR and iron losses vary as the third power of linear dimensions. :. Total losses « 2°. This means that losses of transformer A are x° times that of transformer B. 5.35 DESIGN OF CORE ‘The net cross sectional area is obtained from the dimensions of various packets and an allowance is made for the space lost between laminations. This allowance is known as stacking factor and for sheet stee! of 0.28 mm thickness with a coating of insulation it becomes 0.96. Area is also deducted for oil ducts. The ratio of net cross sectional area to the gross area of the core circle is known as utilisation factor (UF). UF increases if the number of core steps are increased (however, this increases the manufacturing cost). Usually optimum number of steps is 6 for smaller transformer and 16 for large transformers. Improvement in UF increase the core area and hence increases volts/turns for any particular core diameter and specified flux density. This results in the reduction in winding forms and reduction of copper. Thus core area optimization results in better economy.DESIGN OF TRANSFORMERS 5.55 The core section for core type of transformers may be rectangular, square or stepped. Shell type transformers use cores with rectangular cross-section, 5.35.1 Rectangular Core For core type distribution transformers and small power transformers for moderate and low voltage, the rectangular shaped core section may be used. The ratio of depth to width of core varies between 14 to 2. Rectangular shaped coils are used for rectangular cores. For a shell type transformer width of central limb is 2 to 3 times the depth of core. 5.35.2 Square and Stepped Cores When circular coils are required for high voltage distribution and power transformers, square and stepped cores are used. Circular coils are preferred because of their superior mechanical characteristics. A transformer coil, under mechanical stresses produced by excessive leakage flux due to short circuits, tends to assume a circular form. On circular coils, these forces are radial and there is no tendency for the coil to change its shape; on rectangular coils the forces are perpendicular to the conductors and tend to give the coils a circular form, thus deforming it. With core type transformers of small sizes, simple rectangular core can be used with either circular or rectangular coils. As the size of the transformer increases, it becomes wasteful to use rectangular cores. For this purpose the cores are square shaped as shown in Fig. 5.51. ‘The circle represents the inner surface of the tubular form carrying the windings. This circle is known as the circumscribing circle. Clearly a lot of useful space is wasted, the length of circumference of circumscribing circle being large in comparison with its cross-section. This means that the length of mean turn of winding is increased giving rise to higher R losses and conductor costs. With larger transformers, cruciform cores, which utilize the space better, are used as shown in Fig. 5.51. As the space utilization is better with cruciform cores, the diameter of circumscribing circle is smaller than with square cores of the same area, Thus the length of mean turn of copper is reduced with consequent reduction in cost of copper. It should be kept in mind that two different sizes of laminations are used in cruciform cores. With large transformers, further steps are introduced to utilize the core space which reduces the length of mean turn with consequent reduction in both cost of copper and copper loss. bed ° 51. Square core section. Fig. 5.52. Cruciform core. It would seem that we can go on introducing steps with resultant reduction in cost of winding. However, with larger number of steps a large number of sizes of laminations have to be used. This results in higher labour charges for shearing and assembling different types of laminations. Thus the reduction in winding costs with a certain number of steps has to be balanced with the extra labour costs. The number of steps to be used for a particular transformer have to be decided by the above considerations.5.56 - ELECTRICAL MACHINE DESIGN 1, Square Cores. Referring to Fig. 5.51, Gross area of core Ay: (0.71d? = 0.54? : where a= side of the square and d = diameter of circumseribing circle. Net iron area A, = stacking factor x gross iron area = 0.9 x 0.5d? = 0.45 d? (taking stacking factor as 0.9) net core area 045d? Ratio V/(800-1000). cross-over winding. When a continuous type of winding is used, the number of coils should be chosen in such a way as to obtain the necessary winding height utilizing the standard size of conductors. Windings designed for voltages between 33—110 kV normally use 60 to 80 coils. It should be noted that the number of coils of a continuous type winding with tapping at the middle should be a multiple of four.5.66. ELECTRICAL MACHINE DESIGN 4, Helical winding. A helical winding is used for l.v. windings of the power transformers with outputs ranging from 160 kVA to some ten thousand kVA at voltages from 230 V to 15 kV and sometimes upto and including 33 kV. In order to secure adequate mechanical strength the cross-sectional area of a strip not made less than 75-100 mm?. The maximum number of strips used in parallel to make up a conductor is 16 Helical windings are used for current range 300 A to 2400 A, the maximum limit is based on the assumption that the maximum cross-sectional area of a strip normally does not exceed 50 mm’, A double helical winding is used for the same range of voltages as a single helical one, but the upper limit of currents for this winding is about two times higher. In windings using rectangular conductors the radial width of a strip should not exceed 3.5 mm in order to limit the eddy current losses. 5.40 SURGE PHENOMENON Transformers used in power systems are subject to surge overvoltage. These surge overvoltages, initiated on overhead transmission lines are caused by switching, faults or lighting discharges. The surges may have a steep wavefront with a rate of rise 1000 times as great as the peak rate of normal operating frequency thereby imposing intense and rapidly changing electric stresses within the transformer winding. The surge phenomenon is particularly important in the case of high voltage transformers. Not so long ago, design of high voltage windings of transformers exceeding 110 kV was governed wholly by power frequency voltages. The surge and impulse voltages were regarded as destructive and the designer was not expected to anticipate them. However, this attitude has undergone a radical change and the surge behaviour of the transformer has not only gained recognition as an essential part of designer's responsibility, but to assume such significance that it has become the foremost consideration in the design of high voltage windings. The effect of a surge arriving at a transformer has an exceedingly complex effect on the transformer winding. The steeper the front and flatter the tail of the wave, the more severe is its effect and therefore we will consider the effect of the application of a rectangular wave i.e. a step function voltage. Theoretically, the incidence of such a wave means that the line terminal is raised in potential at an infinite rate to the crest value and then held there. However, the transformer insulation may be considered as a network of capacitances with interturn insulation acting in series and each turn having a shunt capacitance to earth as shown in Fig. 5.58. Tank’ fea Neutral —_ Cool Fig. 5.58. Transformer model for analysis Fig. 5.59. Voltage and charge distribution of surge phenomenon, ina winding element.DESIGN OF TRANSFORMERS 5.67 Let C, = total series capacitance of the winding, C, = total shunt capacitance (i.e. capacitance of winding with respect to ground), otal length (axial height) of winding. Let us consider a winding element of length dr located at a distance x froin the neutral end having a voltage v and a charge q as shown in Fig. 5.59. The shunt capacitance of a winding element relative to ground is proportional to its length and the series capacitance is inversely proportional to the length. Therefore, the shunt capacitance of the winding element shown in Fig. 5.59 is (C,/1) dx and the series capacitance is (C,)/dx. If the charge of the adjacent winding element is (q — dq), the charge of the shunt capacitance (Cy? de i, therefor, dg. The relationship between voltage and charge for any capacitance is v =giC. The relationships for shunt and series capacitances of the element considered, are : and ve (C,/Ddx ~ C," do 1 dg bt Ce 1G, du a Gy de’ Cl’ dk Ci’ lt “4 °C,” de FC, Let a = /C,/C, . Therefore, the above differential equation can be written as 2g? wee co aT ‘The solution of differential equation is v= Act! + Beel!, The remote end (neutral end) of the winding is earthed v=0 at x =0 O=A+B or At the line end a step voltage of V is applied Ate= e VeAet-e) or A= Vee) Hence, voltage at the element vee! ee) Vsin hx/D 5.23) (ee ~ sinha Initial Distribution. Eqn. 5.23 shows that the voltage distribution along a winding depends upon the value of a = JC, /C, . If the capacitance relative to ground is zero i.e. when a = 0, the inh (x /D, ha from which we have 0 . expression is of the form 7. The equation can be solved by using L’ Hospitals’ rule, « cosh @ v= Vie/l) (5.24) From above 2 Y constant dx i ‘Thus, it is evident from Eqn. 5.24 that if the shunt capacitance is zero, the voltage is uniformly distributed over the winding. However, in modern power transformers a varies from 5 to 15 and hence the initial voltage distribution is considerably different from the linear distribution indicated by Eqn. 5.24.5.68 . ELECTRICAL MACHINE DESIGN ‘The voltage gradient at the line-end of the winding is 7 Vater REL Va coth * (5.25) G,_, V0! “Gaba |... = Ve cot AB. As a> 8, cotha = J, do) = Ve nlB.2 ael,_, (a (5.26) ie, the voltage gradient at the line end is « times the value of gradient corresponding to the uniform voltage distribution (the voltage gradient corresponding to uniform distribution is V volt/metre). Fig. 5.60 shows the initial distribution of a surge voltage over a uniform winding with neutral. The voltage distribution is uniform for a = 0 but in the presence of earth capacitances a may typically assume value of 10, and in that case, the voltage distribution becomes nonuniform with most of the voltage dropped across a fraction of the line end of the winding, in which the voltage between adjacent turns becomes excessive and may require additional insulation to withstand the high voltage stresses produced. 1.25 0.75) 0.50 2 g 3 3 Voltage ratio, UV ° 1.0 0.75 0.50 0.25 0 Ratio x/t Ratio x/t Fig. 5.60. Initial voltage distribution Fig. 5.61, Oscillations in winding subjected with earthed neutral. to voltage surge. Final Voltage Distribution. The tail of the step voltage is equivalent to a sustained d.c. voltage of magnitude V after the all transients have died down and the system has settled to steady state conditions. The voltage distribution therefore becomes v= Veil ‘This distribution is obtained with time ¢ = «. It is clear from above, that the final voltage distribution is a straight line giving a uniform voltage gradient in the winding. This distribution is identical to the one obtained with a = 0 (See Fig. 5.61). Intermediate Distribution. The transition from initial to final voltage distribution is always accompanied by oscillations on. account of inductance and capacitance of transformer winding. There is a constant interchange of stored energy between capacitors and winding inductance resulting in complex oscillations at a variety of natural frequencies. As a result of damped transference of electrostatic and electromagnetic energy during complex oscillations, all parts of winding may be severely stressed (i.e. have large voltage gradients) at differertt - instants of time as shown in Fig. 5.61. Initially concentrations of voltage may appear at the line end of the winding; during transitional period concentrations may appear at the neutral endDESIGN OF TRANSFORMERS 5.69 whilst voltages to earth considerably in excess of the incident surge may develop in the main body of winding, Therefore, reinforcement of enid turns of the winding are of little assistance as far as the protection of windings against surge voltages is concerned. 5.40.1 Surge Protection In a general case, under steady state conditions, equal voltages are induced between turns and consequently, ideally, equal amounts of insulation are required between turns. To utilize this uniformly disposed insulation to best advantage, the voltage appearing between turns throughout the winding under surge conditions should also be able to approach this ideal, in which case oscillation voltages are completely eliminated, the initial distribution, like the final must be uniform. Before the effect of surge voltages was cléarly understood, it was a universal practice to reinforce the insulation of a few turns (say 5%) on the line end of the transformer to withstand the impulsive voltage gradients. However, this results in decrease in the series capacitance resulting in increase of « and hence unequal voltage gradients. The use of reinforced turns at winding ends, instead of mitigating the trouble, intensifies it. External surge absorbers may be connected between the transmission line and the transformer terminals to reduce the steepness of wavefront and to dissipate some of the energy of the surge wave. However, the real solution to the problem of surge voltage distribution lies in designing the winding in such a way that the voltage distribution is more or less uniform and no part of the winding insulation is unduly stressed. The initial distribution of voltage is determined wholly by the capacitance network, The voltage distribution depends upon a which is equal to {, /C, . Consequently two circuit elements are available for controlling and improving the initial response. It has been stated earlier that when a = 0 (i.e, when the shunt capacitance C, = 0 or when series capacitance C, = »), the initial distribution is uniform and is coincident with final distribution. Thus, it can be safely concluded that the distribution can be improved by decreasing shunt capacitance and/or by increasing the series capacitance. The methods adopted for securing uniform voltage distribution are : 1. Shielded Windings. The basic principle of protection of transformers rated at 110 kV and more involves the arrangement of an additional capacitance network connected to the line input terminals in order to neutralize the winding to earth capacitance. An electrostatic shield is provided along the axial length of h.v. winding as shown in Fig. 5.62. The provision of the electrostatic shield introduces capacitances between the line terminal and the winding in a graded diminishing proportion as in a transmission line insulating string, thereby helping to neutralize the winding to ground capacitances. pcesin = Line SS (a) Unshielded (6) Shielded Fig. 5.62. Winding with static shield. Fig. 5.63. Surge voltave distribution.— 5.70 ELECTRICAL MACHINE DESIGN ‘The use of the shield permits the shunt currents to flow directly from the line, thereby enabling the series currents to have the same value in all the sections. Hence a uniform voltage distribution is obtained. The surge voltage distribution with and without eldetrostatic shields is shown in Fig. 5.63, the figures indicating percentage equipotentials. The surge protection of transformers by an electrostatic shield involves the following procedure: @ The shield is provided on the six terminal coils (for 110 to 220 kV) at increasing distances. (i) A capacitive ring is placed over the winding to equalize the voltages across the turns of the input coil. The ring (or collar) is off pressboard with copper or aluminium foil and is securely insulated. The capacitive ring is connected to the line terminal of the winding. 2. Centre point dise winding. In this method, the high voltage disc winding is divided in two halves which are connected in parallel. Each half starts from opposite end and finishes at the centre point. The centrepoint becomes the h.v. terminal of the winding (Fig. 5.64). The h.v. winding is provided with radial and complete axial shields and the arrangement results in uniform initial and final distributions, 8. Interleaved windings. Two normal disc coils are shown in Fig. 5.65(a). If the order of the turn interconnection is changed to have interleaved coils as in Fig. 5.65(6), the interturn or series capacitance C, increases thereby reducing a. Therefore, voltage distribution becomes more uniform. Windings of interleaved double disc type can withstand impulse voltages better than the non- interleaved type. Partially interleaved windings are also used with interleaved coils only at the ends of h.v. windings. Fig. 5.64. Centre point disc winding. 4, Layer type windings. Recent developments in the design of high voltage transformers emphasise the difficulties of obtaining adequate impulse strength with continuous disc type winding. Modern trend is to use layer type of winding as shown in Fig. 5.66. Shields oon Neutral Neutrot i " Line s{4{3f2]a 3]7f2le[s I shield Core ie s|7]}8]}9 fio afs]o]s fio = @ @) Fig. 5.65. Normal and Interleaved coils. Fig. 5.66. Layer type winding.DESIGN OF TRANSFORMERS 571 The high voltage winding is divided into concentric layers, separated by oil ducts, each diminishing in length. Two shields are used, one at the line end and the other at the neutral end. On account of the provision of shields on the two ends, it is possible to neglect the capacitance - between winding and earth (C,) in comparison with capacitance between adjacent turns (C,). Therefore, « is small for these windings and hence the voltage distribution is almost uniform. Fig. 5.67 shows the voltage distribution in a layer type winding. It is clear that in a layer type winding the voltage distribution initially and under transient conditions is almost uniform. The layer type of winding is thus lightning proof. The same conditions exist in the windings of shell type power transformers because the capacitances between large flat coils considerably exceed the capacitances relative to earth and therefore, the initial and the final voltage distributions practically caincide, i TTT TTT IPeak of potentials | | Zoe 0 gos a i ib Wd 304 +-+ 7 . 4 wat tieeiem (Ti 3 02-7 Gradient R | | o_ LIL 0.25 0:50 0.75 1.0 a Ratio x/t Fig. 5.67. Initial voltage distribution and peaks Fig. 5.68. Transformer frame. of potentials in layer type windings. 5.41 WINDOW SPACE FACTOR The window space factor is defined as the ratio of copper area in the window of the total window area. It depends upon the relative amounts of insulation and copper provided, which in turn depends upon the voltage rating and output of transformers. The following empirical formulae may be used for estimating the value of window space factor : K,, = 1080 + kV) 5.27) where kV is the voltage of h.v. winding in kilo-volt. The above formula is for transformers of ratings between 50 to 200 kVA. Space factor is larger for large outputs and smaller for small outputs. For a transformer of about 1000 kVA rating K,, = 12/30 + kV) and for transformers of about 20 kVA rating K,, = 8(30 + kV). The values of space factor for intermediate ratings can be interpolated. 5.42 WINDOW DIMENSIONS ‘The leakage reactance is affected by the distance between adjacent limbs. If this distance is relatively small, the width of the winding is limited and this must be counter balanced by increasing the height of the winding. Thus the windings are long and thin. This arrangement leads to a low value of leakage reactance. If the height of the window is limited, the width of the window has to be increased in order to accommodate the coils. This results in short and wide coils giving a large value of leakage reactance. ‘The height and width of the window can be adjusted to give a suitable arrangement of windings and also to give a desired value of leakage reactance. ‘The area of the window depends upon total conductor area and the window space factor.5.72. ELECTRICAL MACHINE DESIGN of window A, = {2telL conductor area ‘window space factor . 2a, T, = —2-? for single phase transformers wus(5.28) = 447, for three phase transformers (5.29) Area of window A, = height of window x width of window = H,, x W,, The ratio of height to width of window, H,,/W,, is between 2 to 4 Assuming a suitable value for ratio H,/W,,, the height and width of window can be calculated. 5.43 WIDTH OF WINDOW FOR OPTIMUM OUTPUT Let D be the distance between adjacent limbs as shown in Fig. 5.68. Now, D = width of iron + width of bare conductors + width of insulation and clearance. Let m be the space occupied by insulation and clearance ete. along the width . Width occupied by copper plus iron D’ = D ~ m. Width of bare conductors in window = width occupied by iron plus conductors — width occupied by iron = D‘ - d. Let S be the output in VA of transformer per unit height of window. 5 S =E,T,, I where T,, = turns per unit height. Now E,=4.44f0,,=444/B,, A, = 4.44/B,, kd? =. For a fixed value of frequency and flux density, E, is proportional to d?. Now IT, = mmf per unit height = 8 a T,, where a = area of each conductor and 5 = current density. Now aT), = height of conductors x width of conductors = width of conduetors (as height is unity). :. IT), is proportional to (6 x width of copper in window) or IT,, is proportional to (D' — d) for a constant value of current density. S = Ad? (D' - d) = Ad? D’ - Ad? where A is a constant. In order to determine the maximum output for a given value of D’, S is differentiated with respect to d. or 28. 2 24nd 3 Ad? For maximum output, & = 0 dd dd or 2AD'd = 3Ad? 1. Ds 18d. Now D = D’ +m. The value of m can be taken as 0.2d with normal designs. or D=lid :. The width of window which gives the maximum output is W, = D-d=0.7d 5.44 DESIGN OF YOKE The area of the yoke is taken as 15 to'25 percent larger than that of core for transformers using hot rolled silicon steel. This reduces the value of flux density obtaining in the yoke and therefore there is reduction in the iron losses and the manetising current. For transformers using cold rolled grain oriented steel in the area of yoke is taken equal to that of the core. ‘The section of the yoke can either be taken as rectangular or it may be stepped. The yoke sections are shown in Fig.-5.16. In the case of rectangular section yokes, the depth of the yoke is equal to the depth of the core. This depth of the core is equal to the width of the largest stamping when square or stepped cores are used. For example, in the case of cruciform core (Fig. 5.52), the depth of yoke is equal to a.DESIGN OF TRANSFORMERS For rectangular section yokes, Area of yoke Ay = depth of yoke x height of yoke =Dyx Hy where Dy = width of largest core stamping a. (1.15 to 1.25) A,, for transformers using hot rolled’ steel =A,; for transformers using grain oriented steel. 5.45 OVERALL DIMENSIONS When dealing with overall dimensions in transformer problems, refer to the following details and diagrams : ‘dth of largest stamping, iameter of circumscribing circle, istance between, centres of adjacent limbs, width of window, Height of window, length of limb, H,, = height of yoke, H = Overall height of transformer over yokes or overall height of frame, a da D Wy Hy Width over two_limbs Fig. 6.69. Single phase core type transformer. a tht a : Kf ke Width over three limbs Fig. 5.70. Three phase core type transformer. Depth over winding Fig. 6.71. Single phase shell type transformer.5.74 ELECTRICAL MACHINE DESIGN W = length of yoke = overall length of frame. We have the following relations for single phase core type transfdrmers (Fig. 5.69). D=d+W,,Dy=a, H=H,,+2Hy, =D+a, Width over two limbs = D + outer diameter of h.v. winding, Width over one limb = outer diameter of h.v. winding. We have, for a 3 phase core type transformers (Fig. 5.70). D=d+W,;Dy=4;H=H,,+2H,;W=2D +a, Width over 3 limbs = 2D + outer diameter of h.v. winding. Width over one limb = outer diameter of h.v. winding For single phase shell type referring to Fig. 5.71 Dy =, Hy =a, W = 2W,, + 4a, H=H,, +20. 5.45.1 Simplified Steps for Transformer Design (a) Steps of Determination of Main Dimensions for Core, Window and Yoke Step-1. Calculate voltage per turn using the formulae E,=K JQ volts. Step-2. Find net cross sectional area of the core A, from equation E, = 4.44 f,,= 4.44 (B,, A; where 4,= —HL_— mt? 444 7B, Step-8 Determine the diameter of the circumscribing circle using the formula A,=Kd? where ae fe mt K Step-4. Find the width of the window from the equation Wy = D-d 1D representing the distance between core centres) Step-5. Obtain window area from the following equations : (@ for single phase transformer Q = 2.22/B, K, 5A, A; x 10° kVA 4-2 w* 222 7B, K, 8A; (ii) for 3 phase transformer Q = 3.33 /B,, 6K, A, A, x 10° kVA es 333 /B,, 5K, A; Step-6. Find height of the window as x 10° mt?, giving A, = x 103 mt?, x, = fe mtDESIGN OF TRANSFORMERS 5.75 lif d is not given, first A, (area of window) is calculated using step 5 instead of step 4. Then we use the concepts that usually the ratio and + A, =H, x W,, Hy _ Ww, =2to4, ‘Thus height of the window (H,,) is obtained. Since W,, = D - d hence with known values of W, and D, we can find d] Step-7. Obtain the depth and height of the yoke for the stepped core using the formula D, (depth of yoke) = a H (height of yoke) = a where a is the width of largest stamping. Step-8. Obtain the overall height and length of the frame from the following relations (@ for simple phase core type transformer H=(H, +2H,) mt; W=(D+a) mt (ii) for three phase core type transformer H=(H,+2H,) mt; W=(2D +a) mt (iii) for single phase shéll type transformer H=(H,, +20) mt; W = (2W, + 4a) mt [In case a is not given, we assume 0.9 d instead of a in the above equations for stepped core, 0.85 d for cruciform core and 0.71 d for square core) (b) Design Steps for LV Winding Design Step-1. Calculate number of turns/phase using the equation T, = Ye; where V, = secondary voltage A E, = voltage per turn. [&, can be obtained using the formula E, = k {@] It may be noted here the number of turns calculated above may be a fraction which should be rounded of suitably. This rounding off does not affect the voltage ratio as the turna/phase of the h.v. side will be calculated on the final value of turns per phase of |.v. winding and voltage ratio. Step-2. Find secondary phase current J, x10 using aa (for 3-phase transformer) kVA 108 and HVA Gor 1 phase transformer). I, is obtained in Amps. Step-3. Obtain the cross-sectional area (in mm®) of the secondary conductor given by zB , where 8, is the assumed current density in Amp/mm?. Please refer to text about suitable values of 8 for assumption. Step-4. Use Table 23.1 (Is 1897-1962) to find the suitable dimension of the conductor. If the cross-sectional area of the secondary conductor exceeds 50 mm, it is better to take stranded5.76 ELECTRICAL MACHINE DESIGN parallel conductors (multiple layers). Also calculate the secondary current density 5, again with the selected conductor area (since the conductor dimension is selected from the table as per availability hence the area selected may be slightly more than the theoretically calculated area; thus we need to obtain the current density again. Step-5. The conductors are paper insulated. This increases the thickness of the conductor. Allow 0.5 mm for increase in thickness of the conductor for single paper covering. Conductor dimension becomes [Conductor width x (Conductor thickness + 0.5 mm)]. Step-6. Consulting Table 5.4, the type of winding is selected. If it is a helical winding keep 1 turn more along the axial depth Step-7. Determine axial depth (L.,) of L.v. winding; L,, = number of secondary turns x axial depth of conductor (width of conductor). Next obtain window clearance as below Window clearance = (hight of the window ~ L,y2 {Minimum clearance is 6 mm for windings having voltage below 500 V.] Step-8. Next find radial depth of 1.v. winding as follows : b, = number of layers x radial depth of conductor + insulation between layers Step-9. Next the inside and outside diameters of the L.v. winding can be found out as follows : inside dia, = dia. of circumscribing circle + 2 x pressboard thickness for insulation between lv. winding and core. Outside dia = inside dia +b, (c) Design Steps for HV Winding Design Step-1. Number of turns/phase (7,) in HV side can be obtained from the following relation _ HT side phase voltage °” Voltage per turn Voltage rating of HV side ‘Voltage rating of LV side If the result is a fractional number, it must be rounded up. Step-2. If # 5% tappings are to be provided, the number of turns is to be increased to Tyogy,= 1.05 x T, (number of turns are rounded off to an integer.) Step-3. Next HT side current is calculated from kVA x 10° 2” 3x HT side voltage rating A current density of 6, is assumed [since the cooling conditions are better for hy winding than that of lv winding, hence higher value of current density can be used. This causes the saving in copper material]. For distribution transformers, at HT side 6 can be around 2 to 2.5 mm? while for power transformers it can be taken between 2.4 to 3.5 Amp/mm?. Step-4. Next the conductor is selected. In HT winding, cross over type, cylindrical type with circular conductor and continuous dise type are generally used. The type of winding is first selected depending upon the rating of the transformer and design limitations. The conductor section is then selected depending upon the type of winding. The conductor section may be circular or rectangular. For small and medium transformers with comparatively lesser conductor section, circular conductors are selected. For large transformers, because of higher current ratings, rectangular conductors are preferred. = Number of LT side turns x amp.DESIGN OF TRANSFORMERS 5.77 When circular conductor is selected, the dimension can be obtained from the nearest value of the cross-section from Table 23.4. The actual current density is recalculated on the basis of new or modified cross-sectional area. : HT side phase current Modified area of the conductor * In case of selection of rectangular conductor, the winding height may be assumed to be 70% of the window height. Actual current density = h,=0.7H,. Thus, the axial length for one coil is given by (h,/number of coils). ‘Number of turns/coil (axially) axial length for one coil ~ diameter of the insulated conductor and axial length of each coil = number of turns per coil x diameter of insulated conductor. Say axially 16 coils are given. Then axial length of 16 coils = 16 x axial length of each coil calculated above. For spacers (each of 6 mm thick) = (15 x 0.6) mm. Assuming 30 mm is end insulation, we are to add 60 mm axial end insulations for two ends. We can add a little slackness factor and add these axial lengths to find net axial length. To get the net radial length, we are to add with outside dia of lv winding the depth of oil ducts between lv and hv winding and between hv winding and hv insulating cylinder. Also outside diameter of hv winding to be added. To get net radial length, we are to add with outside dia of lv winding the oil duct between lv and hv winding, outside dia of hv winding. Example 5.1. Estimate the reduction in volume, expressed as a percentage of original volume of (i) iron core (ii) conductors, in a transformer when hot rolled silicon steel laminations worked at a flux density of 1.2 Wb/m? are replaced by cold rolled grain oriented silicon steel laminations worked at 1.8 Wo /m?. Circular coils are used and the thickness of winding is negligible as compared with diameter of circumscribing circle. The length of flux path, area of conductor and the number of turns are the same for both cases. Solution. Let A,, and A,, be the net iron areas with hot rolled and cold rolled laminations respectively. ‘The net iron area is inversely proportional to flux density. A,, = (1.2/1.8) Aj, = 0.667 A,, Hence, reduction-in net, iron area = (1 - 0.667) x 100 = 33.3% Let d, and d, be the diameters of circumscribing circle with hot rolled and cold rolled laminations respectively. Diameter of circumscribing circle is proportional to JA, . d.= J0667 4, = 0.817 d,. ‘The length mean turn is nd. Therefore, length of mean turn of conductors wound on cold rolled steel is 0.817 times that of conductors wound on hot rolled steel. Since the number of turns and area of each conductor are the same in both cases, the reduction in volume of conductors with the use of cold rolled steel is (1 - 0.817) x 100 = 18.3%. Example 5.2. Show that the output of a 3 phase core type transformer is : Q = 5.23 f B,, Hd? H, x 102 RV A where f = frequency, Hz ; B,, = maximum flux density, Wo/m? ; d = effective diameter of core, m; H = magnetic potential gradient in limb, Alm ; H,, = height of limb (window), m.5.78, ELECTRICAL MACHINE DESIGN Solution. kVA output of a three phase transformer Q = 3EI x 10 =3x4.44f0, TI x 10%=3 x 4.44/B,A,TIx 103 + In a three phase core type transformer each limb has one primary and one secondary winding wound on it and therefore total mmf over one limb = 271. mmf 2M i ao lagnetic potential gradient H Tight of Tb yoo or T= Bw Also A, = (a4 dd? Hy 2 Substituting the value of A, and 77 in the expression for Q, we have H, x 10% Q=3x 4.44 /B, x 2 dx H 2 = 5.23 f B,, Hd? H,, x 10 kVA. Example 6.3. Calculate the kVA output of a single phase transformer from the following data : core height 2g, diameter of circumscribing circle _ 4 5 distance between core centres 7 distance between core centres net iron area aa area of circumscribing circle" current density = 2.3 A/mm?, window space factor = 0.27, frequency = 50 Hz, flux density of core = 1.2 Wb/m?, distance between core centres = 0.4 m. Solution. Distance between core centres D = 0.4 m. » Core height (window height) H,, = 2.8 x 0.4 = 1.12 m. Diameter of circumscribing circle d = 0.56 x 0.4 = 0.224 m. Width of window W,, = Dd = 0.4 ~ 0.224 = 0.176 m Area of window A, = H,, x W,, = 0.112 x 0.176 = 0.197 m? Area of circumscribing circle = (n/4) d? = 0.0894 m? °. Net iron area A; = 0.7 x 0.0394 = 0.0276 m? From Eqn. 5.6, for a single phase transformer Q =222/B,, K, 84, A; x 107 KVA = 2.22 x 50 x 1.2 x 0.27 x 2.3 x 108 x 0.197 x 0.0276 x 10 = 450 kVA. Example 5.4. Determine the dimensions of core and yoke for a 200 kVA, 50 Hz single phase core type transformer. A cruciform core is used with distance between adjacent limbs equal to 16 times the width of core laminations. Assume voltage per turn 14 V, maximum flux density 1.1 Wb/m?, window space factor 0.32, current density 3 Almm?, and stacking factor = 0.9. The net iron area is 0.56 d? in a cruciform core where d is the diameter of circumscribing circle. Also the width of largest stamping is 0.85 d. Solution. Voltage per turn E, 14 444 x 50x12 -, Diameter of circumscribing circle d = A,/056 = /005737056 = 0.32 m. 44 ©, = 4.44 £B,, A; <. Net iron area A, = 0.0573 m?DESIGN OF TRANSFORMERS 5.79 Width of largest stamping a = 0.85 d = 0.85 x 0.32 = 0.272 m. Distance between core centres D = 1.6 a (given) = 1.6 x 0.272 = 0.485,m. Width of window W,, = D -d = 0.435 - 0.32 = 0.115 m. From Eqn. 5.6, for a single phase transformer, Q =2.22/B,, K,5A,A; x 10% 200 = 2.22 x 50 x 1.1 x 0.32 x 3 x 108 x A,, x 0.0573 x 10-8 :. Window area A,, = 0.0298 m? :. Height of window H,, = 0.0298/0.115 = 0.26 m. Using the same stepped section for the yoke as for core Depth of yoke D, 0.272 m and height of yoke H, Referring to Fig. 5.70, Overall height of frame H = H,, + 2H, = 26 + 2 x 0.272 = 0.804 m. Overall length of frame W = D +a = 43.5 + 0.272 = 0.737 m Example 5.5. Calculate approximate overall dimensions for a 200 kV A, 600/440 V, 50 Hz, 3 phase core type transformer. The following data may be assumed : emf per turn = 10 V; maximum flux density = 1.3 Wb/m?; current density = 2.5 A/mm? ; window space factor = 0.3 overalll height = overall width ; stacking factor = 0.9. Use a 3 stepped core. For a three stepped core : Width of largest stamping = 0.9 d, and Net iron area = 0.6 d? where d is the diameter of circumscribing circle. 10 444fB, 444% 50x13 0.272 m. Solution. Net iron area A; = 0.0347 m?. Diameter of circumscribing circle d = J00347/06 = 0.24 m, and width of largest stamping a = 0.9 x 0.24 = 0.216 m. Using @ 3 stepped section for the yoke Height of yoke H, = a = 0.216 m, depth of yoke D, = a = 0.216 m. From Eqn. 5.10, for a 3 phase transformer, Q = 3.33 fB,, K, 5A,A; x 10° or 200 = 3.33 x 50 x 1.3 x 0.3 x 2.5 x 10° x A, x 0.0847 x 10° 2 Window area A, = 0.0355 m? or H,, x W,, = 0.0355 m? The given condition is, overall height = overall width or H= W Now, referring to Fig. 5.71 =H, + 2H, =H, + 2x 0.216 =H, + 0.432 W =2D+a=2W,, +d) +a =2W,, + 0.48 + 0.216 = 2W,, + 0.696 As H = W, we have : H,, + 0.432 = 2W,, + 0.696 or H,, = 2W,, + 0.264 (2W,, + 0.264) W,, = 0.0355 or —2W,,? + 0.264 W,, — 0.0355 = 0 or width of window W,, = 0.083 m and height of window H,, = ee = 0.428 m. Thus the dimensions of core are : Distance between adjacent core centres D = W,, + d = 0.323 m. Overall height H =H, + 2H, = 0.86 m. Overall width W = 2D +a = 0.862 m.5.80 ; ELECTRICAL MACHINE DESIGN Example 5.6. The ratio of flux to full load mmf in a 400 kVA, 50 Hz, single phase core type power transformer is 2.4 x 10-6, Calculate the net iron area and the window arga of the transformer. Maximum flux density in the core is 1.8 Wb/m?, current density 2.7 Almm? and window space factor 0.26. Also calculate the full load mmf. Solution. The Eqn. 5.13, K = 444 f (®,/ AT) 10° = 4.44 x 50 x 24 x 10° x 10° = 0.732 Voltage per turn E, =K @ = 0.732 400 = 14.64 V. E, 14.64 -. Fi ©, = apts => = = 0.066 Wh. we Gad F daa xc60 ~ 0088 r ®, _ 0.066 A Net iron area A,= By "73 7 ).0507 m?. Window area of single phase transformer —__ 2 2.22 f By Ky 6A; x10 400 —— 2 = 222% 60 x13 x 0.26 x 27x10" «00507 x10 = 20777 ®, 0.066 =o y= A Full load mmf AT = 8 gato 7 27500 Example 5.7. Determine the main dimensions of the core, the number of turns and the cross-section of the conductors for a5 kV A, 11000/400 V, 50 Hz, single phase core type distribution transformer. The net conductor area in the window is 0.6 times the net cross-section of iron in the core. Assume a square cross-section for the core, a flux density 1 Wb/m2, a current density 1.4 Almm?, and a window space factor 0.2. The height of window is 3 times its width. Solution. Given that Net conductor area = 0.6 x net iron area or <. Window area From Eqn. 5.6, for a single phase transformer, Q=2.22/B,, K, 54, A; x 10° or 5 = 2.22 x 50 x 1.0 x 0.2 x 14x 10° x 3A; x A; x 109 0.00732 or net iron area A; = 0.00732 m*. Gross iron area A,; = 0.00814 m2, <. Width of core a = /000814 = 0.09 m. Gross iron area provided = 0.0081 m?. Net iron area provided = 0.00729 m?, Window area A, = 3 x 0.00729 = 0.02187 m?. Height of window H,, = 3W, ». 3W,? = 0.02187, But HH, x W, =A, or width of window W,, = 0.085 m, and height of window H,, = 3 x 0.085 = 0.255 m. The yoke has the same gross area as the core. Gross area of yoke A, = 0.081 m?. 0.0081 009 Depth of yoke D, = a= 0.09 m, :. Height of yoke H, = A; = 1.0 x 72.9 x 10° = 7.29 x 10-9 Wb. = 0.09 m. Flux 9,DESIGN OF TRANSFORMERS 5.81 Voltage per tum E, = 4.44 f ©, = 4.44 x 50 x 7.29 x 10 = 1.625 V. _ Primary voltage 11000 Primary turns = Ce dary voltage * Secondary turns = [oo x 246 = 6765 (The turns of the low voltage should be calculated first and that of high voltage afterwards by using the voltage ratio). secondary voltage 400 Secondary turns 7, = 246, eeondary turns 7, =~ oitage per turn 1628 ~ “° . 5000 Primary winding current J, = 20 = 0.455 A. ny i current “= 71000 455 Area of primary winding conductor a, = 14 Using circular conductors, diameter of primary conductor = (oa24x 47x = 0.642 mm Secondary winding current = 5000 105A. 400 Area of secondary winding conductor a, = 4 Be 8.93 mm? Using a square conductor 3 x 3 mm?. Area of secondary-conductor a, = 9 mm. Overall dimensions of core—Referring to Fig. 5.69, Distance between core centres D = a + W,, = 0.09 + 0.085 = 0.175 m. Length of frame W =D +a = 0.175 + 0.09 = 0.265 m. Height of frame H =H,, + 2H, = 0.255 + 2 x 0.09 = 0.435 m. Example 5.8. Calculate the main dimensions and winding details of a 100 kV A 2000/400 volt, 50 Hz, single phase shell type, oil immersed, self cooled transformer. Assume : Voltage per turn, 10 V flux density in core, 1.1 Wo/m? ; current density, Almm? window space factor, 0.33. The ratio of window height to window width and ratio of core depth to width of central limb = 2.5. The stacking factor is 0.9. 10 Solution. Net iron area A= Tog = ae eae = Ol mit Gross iron area Ay = 29% = 0.0555 m2, 09 Referring to Fig. 5.71, - = 2.5 (given), We have, gross iron area A,, = 2a x b 2.5 (2a)? = 0.04555 _or width of central limb, 2a = 0135 m. Core depth b = 2.5 x 0.135 = 0.3475 m. ‘The yoke carries half of the flux in the central limb. Assuming the same flux density in the core as in the limb, the area of yoke is equal to half the area of the central limb. Gross area of yoke A, = 204858 = 227.75 x 10° m?, 22.275 x 10 0.3375 Depth of yoke D, = 0.3875 m, Height of yoke H, = 0.0675 m.582, . ELECTRICAL MACHINE DESIGN ‘The side limbs carry haif of the flux in the central limb. Therefore, the width of side limb is half of the width of central limb. Width of side limb a = 0.0675 m. From Eqn. 5.6 for a single phase transformer, Q=2.22/B,,K,5A,A, x 10% 100 = 2.22 x 60 x 1.1 x 0.33 x 2 x 10% x 0.041 x A, x 10% We have, H,, x W,, = 0.0303 and He Window area A,, = 0.0303 m?. . or 3W,,? = 303 x 10+, Thus width of window W,, = 0.1 m. Height of window =H, =0.3.m Referring to Fig. 5.71 Overall height of frame H =H, + 2H, = 0.3 + 2 x 0.0675 = 0.435 m. Overall length of frame W = 2W,, + 4a = 2x 0.1 +4 x 0.0675 = 0.47 m, Overall depth of frame = 0.3375 m. Windings HLV. winding turns mee = 200, L.V, winding turns 2 = 400. 100 x 1000 FLY. wind t 200%1000 = 50.4. winding curren hie HLV. winding conductor area a, = 2 = 25 mm’. LV. winding current 100 x 1000 _ 960 4. 400 LV. winding area 250 = 105 mm?. 2 Example 5.9. Calculate the core and window areas of a 400 kVA, 50 Hz, single phase, core type power transformer. The following data may be assumed : Ratio of weight of iron to weight of copper = 4, Ratio of length of mean turn of copper to length of mean flux path = 0.5 ; maximum flux density = 1.5 Wb/m? ; current density = 2.2 Almm! ; density of copper = 8.9 x 10° kg/m ; density of iron = 7.8 x 103 hg/m?, copper space factor = 0.12. Solution. From Eqn. 5.18, wa 1 x109 og 89x10" | 493) al =16 (222 7.65 x 10° J eeas0neeee (50 x 15 x 22 x 10° v2 ‘) = 0.0478 m? From Eqn. 5.6, window area jae Pere 2.22 f B,, Ky 8A, x10" 400 = yee = 5 = Fod% 50x15 x 012% 22% 10° ~ 00498 x10 = 0-783 m- Ay =DESIGN OF TRANSFORMERS 5.83 Example 5.10. Calculate the ratio of weight of iron to weight of copper, net iron area, voltage per turn and the constant K (where voltage per turn E, = K JRVA) for a,500 kVA, 50 Hz, single phase care type power transformer for (i) maximum efficiency to occur at 90 percent of full load (ii) minimum cost (iii) minimum weight (iv) minimum volume. Assume : maximum flux density 1.5 Wb/m? ; current density = 2.75 A/mm? ; resistivity of copper at 75°C = 2.1 x 10° Qm; density of iron = 7.65 x 10° kglm* ; density of copper = 8.9 x 10° kg/m: ; ratio of specific cost of copper to specific cost of iron for built up cores = 4 ; ratio of length of mean length of turn of windings to length of flux path = 0.5 ; stray load loss = 10 percent of full load copper loss ; iron loss per kg for 1.5 Wo/m? = 1.23 W. Assume an extra loss for joints = 20 percent of total iron loss. Solution. Specific iron loss p; =1.2 x 1.23 = 1.475 Wrkg. Bp _ (275 x 10%)? x 21x10 cape = 17.8 W 89x 10° 89x 10° eg Specific copper loss = Specific copper loss including stray load loss p, = 1.1 x 17.8 = 19.6 Wkg. Ratio of weight of iron to weight of copper to give maximum efficiency at 90% full load. G,_ xp, 196 Sia FPe 297 x 22 = 1057. Gin GE + 09? x Figg ~ 1087 2 Now ce en Bex 10) 2221 g e108 12 = [= x05 x 82210 10%) = 36 222 7.65 x 10° we we Net iron area A, =e 2 4) = 06 epg 1073) 7 78,8 G, 50 x 15 x 2.75 x 10° = 00819 m?, Voltage per turn E, = 4.44) B,, A; = 4.44 x 60 x 1.5 x 0.0819 = 27.27 V. E, _ 2727 :. Constant Ke7e= = 1.22, Ye ~ Y500 (ii) Ratio weight of iron to weight of copper to give minimum cost is G,/G, 500x4 oo | = (0.05 m?. Es x15 x 275 x | = = 444 x 50 x 1.5 x 0.05 = 16.65 V :, Net iron area A 16.65 and K = 0.745 ‘500 P G Gd) For minimum weight 6° = A; = 0.025 m2, E, = 8.34 V and K = 0.375 (iv) For minimum volume GB. 785 Gi-B _ 185 _ 9 56 G. Be 89 0.023 m? ; E, .7 Vand K = 0.344,5.84 . . ELECTRICAL MACHINE DESIGN Example 5.11. A h.v. disc winding has 9 coils each separated by an oil duct as shown in Fig. 5.62. Each coil has 10 turns as shown. Compare the series capacitance of the interleaved dise coil pair as shown in Fig. 5.65 (b) with that of normal pair as shown in-Fig. 5.65(a). Solution. (i) Normal Coil. Let E be the voltage per coil pair and C, the turn to turn capacitance. <. Voltage between adjacent turns E, = E/10 as there are 10 turns in each coil. +» Electric field energy between successive and adjacent turns 1 Q (zy nosed ro 2 “0, There are 8 complete pair of ducts (as there are 9 coils) and therefore the total electrie field energy W=8x i C,B/10) = 0.04 CE wi) Let C, be the total effective series capacitance between turns 1 and 10. .. Total electric field energy W = i ,, B ii) Equating (i) and (ii), we have : C,, = 0.08 C, (ii) Interleaved coil. The geometry of the turns is unchanged and therefore C, remains the same. However, the interleaving effect increases the voltage between adjacent (but no longer successive) turns to 5 E, (eg. turn 1 to 6, 2 to 7 etc.) or to 4 Eeg. 6 to 2 and 7 to 8 etc.) :. Total electrostatic field energy We 3 C,[A(5E/10)? + 4(4B/109] = 0.82 C,E? alii) Let C,, be the effective series capacitance in this case. Therefore, total field energy = Fly Ee wiv) Equating (iii) and (iv), the effective series capacitance C,, = 1.64 C, Hence, ratio of effective series capacitance for interleaved and normal coils is C., _ 1640, C,, 008¢, ‘Therefore, a for the interleaved coils is 1/J205 = 0.22 times that for normal coils. Hence the voltage distribution in interleaved coils with surge voltages is much more uniform than that in normal coils. = 20.5 PART C : OPERATING CHARACTERISTICS 5.46 RESISTANCE OF WINDING Let Lage L, ‘mip? mts ty, length of primary and secondary windings respectively, m; resistance of primary and secondary windings respectively, m. Ti, Lin TL, To bmp and r, = p Ems ap as Total /°R loss in windings P.= 1%, +127, Total resistance (per phase) of transformer referred to primary side 2 raryt (2) , (6.30) Per unit resistance 6.31)DESIGN OF TRANSFORMERS 5.85 5.47 LEAKAGE REACTANCE OF WINDING The estimation of leakage reactance is primarily the estimation of the distribution of leakage - flux and the resulting flux leakages of the primary and the secondary windings. The distribution of the leakage flux depends upon the geometrical configuration of the coils and the neighbouring iron masses and also on the permeability of the latter. ‘The most common arrangements of the core and the windings are of those of core and shell types as shown in Fig. 5.72. Fig. 5.72(a) shows the case of cylindrical core type of windings of equal axial length. ‘The leakage field is mainly packed into the space between the windings (i.e. into the duct) and runs parallel with the core for nearly the full length of the coils. The distribution of leakage flux in the case of shell type of transformers having sandwich coils is shown in Fig, 5.72(6). The field is mainly packed into the ducts between the windings, and runs parallel along the width of the coils in this case. In both of the above two cases, the field is sufficiently symmetrical and geometrical and therefore considerable simplifications can be used to evolve (q)Concentrie Winding _(b) Sandwich coils usable approximate mathematical relationships. Fig. §.72. Leakage flux in transformer windings. 5.47.1 Leakage Reactance of Core Type Transformers The arrangement of windings of a core type 4 fo0(4, transformer is shown in Fig. 5.73. The problem of calculation of leakage flux and consequent leakage reactance is greatly simplified by making the following Core assumption Wind 1, The primary and the secondary windings have ened an equal axial length. Le 2. The flux paths are parallel to the windings along the axial height. 3. The permeance of the leakage flux path external to the winding L, is taken to be s0 x large as to require the expenditure of a 7777. 7 negligible mmf i.e. whole of the winding mmf —| 1 is expended on the length of flux path of length L,; this path being entirely through air. be [a | bp Therefore, in fact, the assumption is that the Tt | mmf required for iron parts is negligible. 4, The primary winding mmf I,1, is equal to AT=Ip tp =I Ts secondary winding (mmf I,7,) - Therefore, the magnetizing mmf and hence magnetizing | current is equal to zero. Fig. 5.73. Leakage flux and mmf distribution, The total mmf AT = 1,7, = 1,7, in core type transformers. 5. Half of the leakage flux in the duct links with each winding. 6. The lengths of mean turn of the windings are equal. 7. The reluctance of flux path through yokes is negligible. Therefore, the reluctance of yoke does not affect the flux distribution. 8. The windings are uniformly distributed and therefore the winding mmf varies linearly from zero from one end to AT at the other end.5.86. ELECTRICAL MACHINE DESIGN Three leakage flux paths ©,, 0, and.®, are shown, ©,, ©, are the leakage fluxes in the primary and the secondary windings respectively while ©, is the flux through the duct. Let : L, = mean circumference of duct, I, = axial height of windings, 8, ,b, = radial width of primary and secondary windings respectively, a = width of radial duct. ‘The flux leakages of the windings can be found as under. Conductor portion. Consider an infinitesimal strip of width dx at a distance x from the edge of primary winding along its width. Mnf acting across the strip= ;= .I,T,= ,T,) 5~ by by Permeance of strip = Hy fe dx Flux in the strip = U,7,) ee X Hy fue =p 22 1,7, fs de. ‘This flux links with @/b,)T, aria” L <. Flux linkages of the strip = dy, = 4 —” 1,7, - dx <1, ” Hence, flux linkages of primary ae due to flux in the strip L “= favs oho PP Ty! Duct portion. Mmf acting across duct = [,7,. Permeance of duct = Hp ia a. ' Lo ZL, Flux in duct ©, = 1,7, * 72 @= 91,7, 7° @ Half of the duct flux links with each of the two windings or duct flux linking with primary 1 L, =z Moht Te ‘This flux links with the entire primary winding. winding 1 L, -. Flux linkages of primary winding due to duct fux vg = > 4 J,T, 7* @%T, 1 ole gk Hod, IT,’ le a, Henee, total flux linkages of primary winding y, pee (m2 2 +bL, 2) ‘The above expression is simplified by a Liga = Lap = L, LyDESIGN OF TRANSFORMERS 5.87 Leakage inductance of primary winding Leakage reactance of primary winding Metso X, = On fugl,? tee( 2 2) --(6.73) The leakage reactance of the secondary winding can be similarly calculated by assuming eae ws Leakage reactance of secondary winding Lm (>, @ = anfugt 2 Lae { be. 3) (6.74 = 2nf gD in(*e 8 (6.74) Leakage reactance of secondary winding referred to primary side 1,¥ Lm (%p , @ x, «(2 =n fuel? fa (2g -+ Total reactance of transformer (per phase) referred to primary side Lue Kenta tefusty Sat (6.18) Per unit reactance = AT Ln (| = 2fto gs Beery (5.78) as T,/V, = E, and 1,7, = AT where AT is the total mmf per limb of either coil. In some cases, one of the windings, usually the h.v. is split up into two parts as shown in Fig. 5.74. Papo ; [ tas Group 1] Group 2 far sar Fig. 5.74. Mmf distribution in transformers Fig. 5.75, Mmf distribution in core type transformers with h.v. winding split in two parts. with h.v. winding sandwiched between two sections of .v. winding.5.88" ELECTRICAL MACHINE DESIGN ‘The leakage reactance of transformer (per phase) referred to primary side in this case can L, bp’ +b," +d, be written as : X,= 2x fuyT,? fau{o. toe ots) 6.17) bp'+b, +b, a! as the p.u. reactance [« + aan saan a where a close regulation of voltage is essential as in distribution transformers used for purely domestic purposes, it is necessary to make the leakage reactance small so that the voltage remains within 5% of the rated value. The obvious way to achieve this is to sandwich the h.v. winding between two sections of l.v. winding as shown in Fig. 5.75. For the purposes of analysis the actual arrangement of the windings on each leg can be considered as consisting of two groups of coils connected in series, with each group consisting of half the total number of turns per phase for each winding. This is a modified arrangement. From Eqn. 5.76, total leakage reactance of transformer (per phase) referred to primary side Pec by +b, saemree() “eilosts") and with the modified arrangement the total leakage reactance of transformer (per phase) referred to primary side becomes Ine ( , opt bs X, = 2X1 =f gl? tt a (5.79) Comparing Eqn. 5.79 with Eqn. 5.75, it can be safely concluded that the leakage reactance is reduced to nearly one fourth by subdivision and interlacing. 5.47.2 Leakage Reactance of Sandwich Coils ‘The idealized flux distribution in shell type transformers using sandwich coils is shown in Fig. 5.76. Let the h.v winding have n coils. Each of the n coils jandwiched between two coils of I.v. winding. This requires the l.v. winding to have two half coils, one at each end of the winding and n - 1 full coils spread in between. Each half coil of Iv. winding contains half the number of turns of that of a full Lv. coil. ‘Thus, if there are n coils in the h.v. winding, there will be n — 1 full coils and 2 half coils in the l.v. winding. The half coils have half the turns of a full coil. The winding can be considered as consisting of 2n units connected in series with each unit consisting of a half Lv. coil and a half h.v. coil. Each of these Fig. 6.76. Sandwich winding. units can be treated on the same basis as that of cylindrical concentric winding. The width of the coil, W, can be taken analogous to axial length L,, of the cylindrical windings. Leakage reactance of each unit (per phase) referred to primary side with analogy to Eqn. 5.75. 1T,) L, ( ‘X= 21 P|} Smt 2 fg Te) Lalas as T,/2n = number of turns in each half coil of primary winding. +b.) JDESIGN OF TRANSFORMERS 5.89 . Total reactance of transformer (per phase) referred on the primary side 7) L, 2 = fg Has Te (a, 24) (6.80) Wn 6 byt Per unit reactance fig AT tala +B (5.81) ne, W é Example 5.12. The current densities in the primary and secondary windings of a transformer are 2.2 and 2.1 A/mm? respectively. The ratio of transformation is 10 : 1 and the length of mean turn of the primary is 10 per cent greater than that of the secondary. Calculate the resistance of the secondary winding given that primary winding resistance is 8 Q. Solution, Resistance of primary winding r = pZelne = pln = pre bmisbp ° a, 1,785 Ty Similarly, resistance of secondary winding r= penuh as 2 2 2) (8) (ex) la 8p) Lip) ? 10. = 0.0694 Q. Example 5.13. A 300 kV A, 6600/400 V, 50 Hz, delta/star 3-phase core type transformer has the following data : width of h.v. winding = 25 mm; width of Lv. winding = 16 mm ; height of coils = 0.5 m ; length of mean turn = 0.9 m ; hv winding turns = 830 ; width of duet between h.v. and Lv. windings = 15 mm. : (a) Calculate the leakage reactance of the transformer referred to the h.v. side. (b) If the Lv. coil is split into two parts with one part on each side of the h.v. coil, calculate the leakage reactance referred to the h.v. side. Assume that there is a duct 15 mm wide between hv. winding and each part of l.v. winding: Solution. Leakage reactance referred to the primary side by +b, 3 x,=20fugty Mat (a+5.90 ELECTRICAL MACHINE DESIGN 015+ 2149. = Br 5 x An 107 x (90 x O8( 9025 +0016) (6) The 1.v. winding divided into two parts, one on each side of h.v. winding and therefore Eqn. 5.79 is used. 3") = mx 50 x 4m x 10-7 x (830)* x 22 (aos + 05 Kye ntngt 2 | > 0025 + 0016) 5.360. Example 5.14. A 100 kV A, 2000/400 V, 50 Hz, single phase shell type transformer has sandwich coils. There are two full h.v. coils, one full L.v. coil and two half 1.v. coils. Calculate the value of leakage reactance referred to h.v. side. Also calculate p.u. leakage reactance. The data given is : depth of h.v. coil = 40 mm, depth of L.v. coil = 36 mm, depth of duct between hv. and Lv. = 16 mm, width of winding = 0.12 m, length of mean turn = 1.5 m. The number of turns in h.v. winding are 200. Solution. From Eqn. 5.80, leakage reactance referred to h.v. side, Ty? (by tby X, =f ty Lat Fe fa) Won 6 2 =x 50x 4x x 107 x 3a x Ome ( 016 + 204 +0088) = 1410. 012 2 HLV. winding current at full load= 200% 1000 _ 59 4. 2000 :. Per unit leakage reactance = 50141 _ 9.9353, 2000 5.48 REGULATION Fig. 5.77 (a) shows an approximate equivalent circuit of transformer with parameters referred to primary side. Rp X Y '» a, % Vp ve js ip TpRe (a) () Fig. 5.77. Simplified equivalent circuit and phasor diagram with lagging load. On no load the secondary terminal voltage V,’= V,. The drop in secondary terminal voltage from no load to full load can be calculated by drawing the phasor diagram. Fig. 5.77(b) shows the phasor diagram at a lagging power factor cos ¢. From the phasor diagram :DESIGN OF TRANSFORMERS 5.1 Assuming that the angle @ between V, and Y,’ is very small, we have : V, =V,'+1,R, cos 6+1X, sin’ or V,— ~V,) = LR, cos 6 + 1X, sin 6. The p.x. regulation, for full load rated output @ and full load eurrent J, is Vp-Vp _ IpRp cos +1,X, sind ; e 7. 2 v. =e, cos +e, sin > -6.81) 'p > If the regulation is large and the phase shift between V, and V,’ is not justified than for this case the following relationship should be used : 008 6 +6, sing + z , cos 6—€, sin 6), «(6.82) Example 5.15. Estimate the per unit regulation, at full load and 0.8 power factor aging, for a 300 kVA, 50 Hz, 6600/400 V, 3 phase, delta/star, core type transformer. The data given is : LV. winding— Outside diameter = 0.36 in, inside diameter = 0.29 m, area of conductor = 5.4 mm, L.V. winding— Outside diameter = 0.26'm, inside diameter = 0.22 m, area of conductor = 170 mm?. Length of coils = 0.5 m, voltage per turn = 8 v, resistivity = 0.21 Q/m/mm?. Solution. L.V. voltage per phase V, ae = 231 V. LV. turns.per phase _ T, se = = 29, HLV, voltage per phase V, = 6600 V. i 6600 HY. ti e = = jurns per phase 7, = Se x 29 = 926 ‘Mean diameter of lv. winding = ee = 0.24 m. Length of mean turn of L.v. winding = x x 0.24 = 0.752 m. 0.021 x 29 x 0.752 Resistance of l.v. winding 70 = 0,00269 2. Mean diameter of h.v. winding = (0.36 + 0.29V/2 = 0.825 m. Length of mean turn of h.v. winding = x x 0.325 = 1.02 m. Resistance of h.v. winding Resistance of transformer referred to primary R,, 300 x 1000 9.28 + 0.00269 (826/29)? = 5.47 Q. HLY. winding current per phase J, = 020° = 15.1.4. :. P.U, resistance = 11x57 _ o.o106, 6800 Mean diameter 0.36 + 0,222 = 0.29 m, Length of mean turn Liq, =x x 0.29 = 0.91 m Width of Lv. winding 8, = (0.26 - 0,222 Width of h.v winding 6, = (0.36 - 0.29/2 = 0.035 m, Width of duct a = (0.29 - 0.262 = 0.015 m.5.92 ELECTRICAL MACHINE DESIGN ‘Leakage reactance of transformer referred to primary side X, = 2x x 50 x dn x 10-7 x (826) x 92 (0015+ 0.035 +0: 05 151x173 and p.u, leal ct = 181x178 _ 0.0395 Pu, leakage reactance ¢, = W2AX From Eqn. 5.81, per unit regulation € =e, cos $ +, sin ¢ = 0.0126 x 0.8 + 0.0895 x 0.6 = 0.0338. 5.49 MECHANICAL FORCES It is well known that parallel conductors repel or attract each other according to whether the currents they carry are in the opposite or in the same direction. Mechanical forces exist in transformers (even though they are stationary) on account of the interaction of the current in the windings with the leakage field surrounding them (windings). It should be noted that the two windings carry currents in the opposite direction. ni it (a) Leakage field (b) Axial leakage field (ec) Radial leakage field Fig. 5.78. Leakage fields and mechanical forces. The leakage field in a core type transformer is shown in Fig. 5.78(a). The leakage field can be resolved into its axial and radial components as shown in Fig. 5.78(6) and (c). The mechanical forces produced on account of interaction of two components of leakage flux with current are consequently in two different ‘perpendicular directions. The forces in two different directions are: th Fr Fr Z (@ radial forces and (ii) axial forces. 1. Radial Forces. Fig. 5.78(b) shows the interaction of axial component of leakage flux with the current carrying windings. Applying Flaming’s left hand rule it is found that force acts in the radial direction. The nature of radial force is such that it tries to burst the outer winding (subjecting it to tensile force) and to crush the inner winding (subjecting it to compressive force). ‘This is obvious, because the primary and the secondary windings of a transformer carry currents in the opposite direction and therefore produce a force of repulsion tending to pull the outer winding and to compress the inner winding. 2. Axial Forces. Fig. 6.78(c) shows the interaction of radial component of leakage flux with the current carrying windings. Applying Fleming’s left hand rule, it is found that forces produced due to this interaction act in the axial direction. The forces acting on the transformer windings in the axial direction are compressive in nature. However, if there is an axial magnetic asymmetry between the two windings, a tensile axial force is produced. ‘The magnitude of force acting on a conductor is proportional to the product of current in it and the intensity of magnetic field due to currents in the neighbouring conductors, and since the latter is itself proportional to the current, the forces are proportional to square of the current.DESIGN OF TRANSFORMERS 5.93 ‘The mechanical forces with normal load currents are low compared to the strength of windings and hence, are not noticeable. However, when large transformers are subjected to short circuit conditions at full voltage, the short circuit currents may reach 10 to 25 times full load value with forces increasing to 100 to 625 times under normal conditions. Under fault conditions (such as short circuits or other severe faults) the mechanical forces may become so great as to damage the transformer permanently unless the windings are solidly braced ie. mechanically reinforced. 5.50 CALCULATION OF MECHANICAL FORCES ‘The mechanical forces produced in a transformer may be classified as : (@ radial forces, (ii) axial forces and (iii) unbalanced axial forces due to magnetic asymmetry. 5.50.1 Radial Force Duct Consider a core type transformer with cylindrical windings as shown in Fig. 5.79. The winding has T turns and carries an instantaneous current i. The instantaneous mmf acting across the duct is i7. The flux density in the duct is B, = pH = py iTIL,. The flux density in the windings varies from Secondory 0 to B,. Therefore, the average value of flux density nding in the duct is B/2 or B,, = Uy iT/2L,. -, Radial force acting on a strip situated at a mean radius R within an angle d9 of a coil is dF, = turns x flux density x current x length of strip = Tx B,, xix Rdd ‘e Fig. 5.79. Radial forces on windings. = foam & 2 iT? L. dg Hence, total instantaneous radial force acting on the coil a Rt ar F, aF,= 40 are [ d= 42 ar? J cao eeasels J eee But 2nR = mean periphery of coil = Ly ae F, = © @r? =e (6.83) =o are a Now, short circuit fault current is I/e, where I is the rms value of full load current and ¢, is the per unit leakage reactance. (The resistance of the winding is neglected). On worst fault conditions the instantaneous value of current may become equal to twice the maximum value ot current under short circuit conditions.5.94 ELECTRICAL MACHINE DESIGN :. Under worst fault conditions i= Qvale, Hence, maximum radial force under worst fault conditions is ‘ I L, Fears ta (aL) t= ee = by (=) fm (6.84) by ‘e The ratio of maximum radial force under worst possible fault conditions to maximum, radial force under normal conditions is : Frimasip _ 2o(V2T fey)? Lm (Le _ 2 Frmaxin He(V2I)" Loge / Le ee It is clear from above, that the radial force is inversely proportional to square of the p.u. reactance. Let consider a transformer with ¢, = 0.05. The maximum radial force acting in this transformer is 2(0.05)? = 800 times that maximum radial force under normal operation. The average radial force under normal operating conditions is 2 Pegg) = 22 Ene, 1 fing? nan) Do f 'T? dwt) En Ho Lm 1 in? aaiad JPton ot dot) = Ho ar Eu (6.85 are (5.85) The windings of a transformer must be designed to withstand radial forces under worst fault conditions. It is clear that circular coils should be preferred as they are the strongest to withstand radial forces. This is because the tendency of the radial forces is to make the coils circular and as they are already circular there is no deformation. However, if rectangular or square coils are used, they would be deformed by the radial forees. In shell type transformers, the coils are rectangular but these coils are mechanically braced for a considerable part of their length and hence are mechanically strong. 5.50.2 Axial Forces ‘The winding currents interact with the radial component of leakage flux to produce axial compressive forces tending to squeeze the windings together at the middle. With symmetrical windings these forces are not noticeable even at short-circuits. In shell type transformers having sandwich coils symmetrically arranged as in Fig. 5.80, there is a force of repulsion between the coils of each of pairs 1-2, 3-4, __ | | fi] als} (als (s}7] [e} 5-6 and 7-8, The force on coil 2 is opposite in direction to that acting on 3, likewise for coils 4-5 and 6-7. Thus there is force acting on coils 1 and 8 only. The magnitude of this force is computed as follows fh] als) [els] [el7 le} - Average value of flux densitv iu. the winding, 1 Fig. 5.80. Axial forces on windings.DESIGN OF TRANSFORMERS 5.95 Instantaneous axial force due to interaction of magnetic field and the current i in T., turns of end coil is f= QV yy Ty i 2 But F= {i oa? 5.50.3 Forces Due to Assymmetry The analysis carried out till now has been for symmetrical arrangements of windings. The axial length of windings is assumed equal with windings placed symmetrically with respect to each other. The mmf distribution in a symmetrical arrangement is uniform and therefore the magnetic centres of the windings are coincident. Fig. 5.81 shows a winding arrangement where the two windings have an axial asymmetry as a result of which the magnetic centres are not coincident, The force, F, between the two windings can be resolved into its radial and axial components. The radial component of force is F cos a and the axial component is F sin a. The latter may attain large 49 ir. Lng GT) - Ent Ho at Tis Bo T= = as 2n is the total number of half coils. a) (ie) = som) Average value of this force is: Fijq,y a Gr)3 =e fn (5.86) (5.87) F sing. F sina F cosa F cova, Fig. 5.81. Forces due to magnetic assymmetry. values under fault conditions causing movement of winding in the axial direction which would increase the axial force still further. Therefore, in order to prevent axial movement of the windings, adequate bracing of the winding is necessary. The causes of axial asymmetry between windings of a transformer are provision of tapings for voltage regulation, manufacturing process of certain types of windings and nonuniform turn insulation. These are explained as under : 1. One of the windings of the transformer, usually the h.v. winding, is provided with tappings. These tapings include and exclude a part of the winding during service. This causes mmf unbalance between the windings in the axial direction. 2. When calculating the axial length of helical windings, one turn is added to the number of turns due to helical wind of the turns. However, at a greater width h, of the turn, an empty space, free of a conductor material, is formed at the ends of helical (or cylindrical) windings as shown by hatched area. This causes shortening of winding height by approximately h/2. 3._ The end turns of the high voltage are reinforced to withstand the effect of surge voltages. Therefore, the end coils of the windings have lesser hy/2 a = Lf ; turns and hence the mmf distribution of the h.v. windings is not uniform along axial direction. Fig. 5.82, Helical winding. ‘The case of axial asymmetry caused by provision of tappings in the windings is shown in Fig. 5.83. Due to cutting out of a section of one of the windings to provide voltage regulation, the5.96 ELECTRICAL MACHINE DESIGN axial length of the winding provided with tappings becomes smaller as compared with the axial length of the other winding. The axial forces produced are on account of radial leakage field resulting from axial asymmetry of windings. Fig. 5.83(a) represents worst éonditions where the tappings are provided at one end of winding. The coils get further displaced due to assymmetry resulting in further intensification of forces. If the tapings are provided at both the ends, we have an arrangement shown in Fig. 5.83(b). There is thus ideally a lesser tendency to aggravate the effect. Fig. 5.83(c) shows the effect of providing tappings at the centre of a winding. 7 HI I? HU} ® ) © Fig. 5.83, Axial forces of asymmetry. The force acting between coils of unequal length can be calculated by resolving the leakage flux into its axial and radial components. Fig. 5.84(a) shows the case of two windings having unequal axial length, with each winding having an mmf AT. The arrangement is resolved as in (6) into two superposed mmf distribution (é) and (ii) in accordance with the length wise unbalance &. The first case (i) is dealt with as if it comprised balanced windings of equal axial lengths and latter (ii) in a corresponding way but with different orientation and appropriate physical dimensions. ‘The windings are assumed to be enclosed between two iron surfaces separated by a distance 2/ as shown in Fig. 5.84(c). i ee (1-WAT |e @ e 2] wat J 21} lat 7 @ Gi) @ () © Fig. 6.84. Resolution of mmf of axially assymmetrical windings into two superposed mmfs. where lza+b,+b, Average value of radial flux density B,.,) = dno ar Total instantaneous axial force acting on the coil Bo KET) ana ee 2) Ln oS (5.88) a+b, +b,) Lyy (iT) = $M wT?DESIGN OF TRANSFORMERS 5.97 L, —_—— a+b, +) S Fig. 5.85 shows the different cases of assymmetry in windings along axial length on account of provision of tappings. Both radial and axial forces are produced. The values of p.u. reactance and the values of radial and axial forces produced compared with those produced with symmetrical windings are given. (| Oh il Ot LN) oe Not Ct | (ot tooF Average axial force Fy...) = imo RUT? (5.89) p.u. reactance or 014 oun 0.105 100 50 83 a1 0 18 9 5 Fig. 6.85. Axial assymmetry in cylindrical coils. 5.51 BRACING OF WINDINGS Inward axial forces are taken by windings themselves. Inward radial forces are passed on to the formers, packing pieces and cores. Outward axial forces must be withstood by end insulation. A well constructed transformer should have suitable choice of conductor dimensions and interturn insulation and the coils well supported and braced with compressive stresses kept in view. The end supports are not easily arranged to give good insulation and at the same time good mechanical strength. Therefore, the need for second quality (good mechanical strength) should be avoided as far as possible by having good magnetic symmetry and the unpreventive asymmetry caused by provision of tapings should be suitably distributed. Example 5.16. A 1000 kVA 6600/400 V, 50 Hz, 3 phase delta/ star, core type, oil immersed transformer has 500 turns on the h.v. winding. The height of winding is 0.6 m and the length of ‘mean turn 1.3 m. Caleulate the instantaneous radial force on the h.v. winding if a short cireuit occurs at the terminals of the l.v. winding with h.v. energised. The leakage impedance is 5 percent. Take the doubling effect multiplier as 1.8. Also calculate the force at full load. Solution. Full load current per phase on hav. side = 22°*2000 _ 595 4. 8 x 6600 -. Instantaneous peak value of short circuit current = 18x JQ x (1/0.05) x 50.5 = 2580 A. 52, the total instantaneous radial force on the h.v. c From Eqn. Ho gy? Lm _ 4x 107 13 iT)? St es x 500)? x “= = 2.33 x 108 2 ar)? (2580 x 5007? x GE = 2.99 x 10 N F, e107 Force at full load, F, = wa (50.5 « 500)? x 2 = 866 N. This shows that the forces under short circuit conditions are considerably large as compared with forces at full load.5.98 . ELECTRICAL MACHINE DESIGN Example 6.17. A 575 BVA, 750/495 V; 60 Hz, single phase, core type transformer has the following data : Width of hv. winding = 27 mm, width of lv. winding = 23 mm, width of duct = 15 mm, height of coils = 0.35 m, length of mean turn = 1.25 m, number of turns in h.v. winding = 190, per unit impedance = 0.036; doubling effect multiplier = 1.8. (a) Find the instantaneous radial force on h.v. winding under short circuit conditions if the height of R.v. and Lv. winding is equal. (b) Find the instantaneous axial force on h.v. winding under short circuit conditions if the h.v. winding is 5 percent shorter than the l.v. winding at one end. Solution. Rms value of full load current = 575 x 1000/7500 = 6.7 A. Peak current under short circuit conditions, .B x JB x 1/0.036 x 76.7 = 5420 A. (a) From Eqn. 5.52, radial force F, Sur fe _ 4nxi07 2 125 . = J * (6420 x 1907 x oe = 2.38 x 10°N, (6) From Eqn. 5.58, axial force F, L, > ag mt 2 MET? oo4b, +b) Anxlo7 f 1.25 * 0.05 x (6420 x 190 x Toss o0n7 + 0029) = 32x 10°N, This shows that there is a very large axial force, even though one of the windings is only 5 percent shorter than the other at one end. 5.52 NO LOAD CURRENT The no load current J, consists of two components : (i) magnet component J, and its value is given by Ip = (T,2 + 12)¥2, Thus the estimation of no load current J, requires the calculation of its two components ,, and I, ing current [,, (ii) loss 5.52.1 No load Current of Single Phase Transformers Let 1.. L, = length of flux path through core and yoke respectively (I, = H,, 1, = w), at,, at, = mmf/metre for flux densities in core and yoke respectively. Total magnetizing mmf AT, = 2 at, 1, + 2 at, 1, + mmf required for joints (5.90) The values of at, and at, are taken from the B -H curves for transformer steel (given in chapter 3), The joints in a magnetic circuit may be taken as short air gaps in parallel with iron paths. The calculation of total mmf is based upon the maximum value of flux density. Rms value of magnetising current J,, = AT, //3T,) (5.91) But the magnetising is not sinusoidal and therefore the peak factor K,, should be used in place of /3 Ip = AT [Ky Ty) (5.92) Let the iron losses be P,, the loss component (6.93)DESIGN OF TRANSFORMERS 5.99 The iron losses are calculated by finding the weight of cores and yokes. The loss per kg is taken from relevant curves given in chapter 3. 5 5.52.2 No load. Current of Three Phase Transformers Total magnetising mmf required for the transformer =3at, 1, +2 at, l, + mmf required for joints. Total magnetising mmf required per phase 3at,l, + 2at,l, + mmf required for joints AT, 5.94) 3 + Magnetising current per phase ,, = ATy( 2 T,) or I, = AT y/(Ky4 T,) (5.95) ‘The values of at, and at, are taken from relevant graphs given in chapter 3. Let P; be the total iron loss for the three phases. 2 1=P/V, It is not usual to calculate the no load current in transformers as in a normally designed transformer, the no load current would be below 5 percent and-a considerable variation can be made in its value without effecting the transformer performance. 5.52.3 Magnetizing Volt-ampere We have E, 44 fT, B, A; AT Magnetizing volt. ampere (VA), = B,J, = (4.44 FT, By A) | Fm 44 f BnA(AT)) B Now, AT, = magnetizing mmf per metre x length of path in iron = at,/, 4. A ee 44 Ba Yat I Weight of iron = A, J, x 7.8 x 108, if the density of iron is assumed as 7.8 x 10° kg/m? Magnetizing volt-ampere/kg 444 f By ty Beth = 04 FBy tty X10. Now a curve can be plotted between B,, and magnetizing volt ampere/kg from B-H curve of the material (value of at,, for different values of B,, will be known from this B-H curve). Manufactures supply the magnetizing VA/kg-B,, characteristics and from it the magnetizing current can be known. magnetizing VA / kg x weight of core Magnetizing current Jt m ~ ~Tamber of phases x voltage/ phase (5.96) Example 5.18. A single phase, 400 V, 50 Hz, transformer is built from stampings having a relative permeability of 1000. The length of the flux path is 2.5 m, the area of cross-section of the core is 2.5 x 10 m? and the primary winding has 800 turns. Estimate the maximum flux and no load current of the transformer. The iron loss at the working flux density is 2.6 W/kg. Iron weighs 5.8 x 10° kg/m’. Stacking factor is 0.9. Solution. Net iron area A; = 0.9 x 2.5 x 10-9 = 2.25 x 108 m? We have, E, = 4.440, 7, = 4.44/B,, A, T,, B, = 1.0 Wb/m?, © 4.44 x 50 x 2.25 x 10° x 8005.100 . . ELECTRICAL MACHINE DESIGN ., Flux in the core = B,, A; = 1 x 2.25 x 10° = 2.25 x 10° Wb. Magnetizing mmf AT, = reluctance x flux = 25 x1 * 1000 x 4x x 10 . AT) __ 1980 Magnetising current In = 59" 5 600 7 Volume of core = 2.25 x 10° x 2.5 = 5.625 x 10% 2. Weight of core = 5.625 x 10-S x 7.8 x 10° = 43.8 kg and iron loss = 2.6 x 43.8 = 144 W. Bus —— =0.25A. vo” 400 “9? -. No load current Ty = 1.75" +0285? = 1.77 A. Example 5.19. Calculate the active and reactive components of no load current of a 100 V, 50 Hz, single phase transformer having the following particulars : Core of transformer steel; Stacking factor = 0.9; density = 7.8 x 10° kg/m; length of mean flux path 2.2 m; gross iron section 10 x 10 m? ; primary turns 200; joints equivalent to 0.2 mm air gap. Use the following data : 75 A. ‘Loss component of no load current I, = oe 09 10 “12 13 14 Weim? Mf 130 20 420 660 3300 Alm eres 08 13 19 24 29 Wrke Solution. Gross iron area = 10 x 10-8 m®. Net iron area = 9 x 10-9 m?. 400 som x 10-3 Wb. Flux in core ©, = Taprsoxgoo = 902 Flux density B, 02 x 10-99 x 10% = 1 Wb/m?. m= = 1.0 Wh/m?, mmf/metre = 210 A, loss per kg = 1.3 W. 210 x 2.2 = 462 A. Corresponding to Mnf for iron path Maf for joints = 800,000 BI, = 800,000 x 1 x =S. = 160 A. Total magnetising mmf, AT, = 462 + 160 = 622 A. 622 iv ising) current, I, = Jo——— = 2.2 A. Reactive (magnetising) current, I,, = T5559 Volume of core = 2.2 = 0,009 = 0.0198 m°. Weight of core = 7.8 x 10° x 0.0198 = 155 kg. Total iron loss = 155 x 1.3 =201.5 W. 2015 400 Loss (active) component of no load current J, = 0.5 A No load current = 7 y22? 405? = 2.26 A.DESIGN OF TRANSFORMERS: . 5.101 Example 5.20. A 6600 V, 60 Hz single phase transformer has a core of sheet steel. The net iron cross-sectional area is 22.6 x 10~ m®, the mean length is 2.23 m, and there are four lap + Joints. Each lap joint takes 1/4 times as much reactive mmf as is required pér metre of core. If B,, = 1.1 Wb/m?, determine (a) the number of turns on the 6600 V winding and (b) the no load current. Assume an amplitude factor of 1.52 and that for given flux density, mmf per metre = 232 Alm; specific loss = 1.76 W/kg. Specific gravity of plates = 7.5. EB. 6600 . Solution. (a) Number of tums T= Ta = Tea penny” 100. (b) Mf required for iron parts = 232 x 2.23 = 517 A. Mmf required for joints 4x i x 232 = 232 A, 17 + 232 = 749 A. ATy 149 = 0.318 A. Kyly ~ 152 J2x1100 (as peak factor K,, = amplitude factor x JZ). ‘Weight of core = 2.23 x 22.6 x 10° x 7.5 x 10 = 378 kg. Total magnetizing mmf AT, ‘Magnetizing current Total iron loss P, 1.76 x 378 = 665 W. Loss component J, = =0.1A, 665 6600 (0.318)? + (0.1)? = 0.833 A. Example 5.21. A 15000 kV A, 33/6.6 kV, 3 phase star/delta, core type transformer has the following data : Net iron area of each limb = 1.5 x 10% m®, net area of yoke = 1.8 x 10% m?, mean length of flux path in each limb = 2.3, mean length of flux path in each yoke = 1.6, number of turns in h.v. winding = 450. Calculate the no load current. Use the data given in example 5.19 for mmf per metre and loss per kg. Solution, H.V. winding voltage per phase = 33300/ 19100 m © 4.44 x 50 x 450 0.191 Flux density in the limbs, B,, (limb) = “j= = 1.27 Wb/m?. Corresponding to this flux density, at, = 560 A/m ; p, = 2.25 Wrkg. 0191 018 = 2.25 Wig. No load current I, = 19100 V. Flux ©, = 0.191 Wb. Flux density in the yokes, B,, (yoke) 1.6 Wb/m?. For this flux density, at, = 260 A/m ; Total mmf for three limbs = 8 x 560 x 2.3 = 3860 A. ‘Total mmf for two yokes 2 x 260 x 1.6 = 832 A. Total mmf for limbs and yokes = 3860 + 832 = 4692 A. Magnetizing mmf per phase AT, = 4692/3 = 1564 A. Magnetizing current per phase J, = 1564/V2 x 450) = 2.46 A. Volume of 3 limbs = 3 x 2.3 x 0.15 = 1.035 m’. Weight of 8 limbs = 1.036 x 7.8 x 10 = 8.08 x 10 kg.5.102 ELECTRICAL MACHINE DESIGN Volume of two yokes = 2 x 1.6 x 0.18 = 0.576 m’, Weight of two yokes = 0.576 x 7.8 x 10° = 4.49 x 10% kg. Loss in limbs = 8.08 x 108 x 2.25 = 18.2 x 10° W. Loss in yokes = 4.49 x 108 x 14 = 6.3 x 109 W. ‘Total iron loss = 18.2 x 10 + 6.3 x 10° = 24.5 x 10° W. ‘Loss per phase = 24.5 x 10/3 = 8.16 x 10° W. :. Loss component per phase J, = 8.16 x 10/1910 = 0.427 A. No load current I, = 27 = 25 A. 5.53 CHANGE OF PARAMETERS WITH CHANGE OF FREQUENCY 1. Effect on Losses. The specific iron loss is given by p; = K,/B,,? + K, f° B,,? Wikg. From this relationship it is evident that change of either for B,, or both, will, in general, result in change of iron losses. As the transformer is designed on the basis of a definite heat dissipation, the /°R loss should be re-adjusted if the total losses are to remain the same i.e. if the core loss under changed conditions is greater than that in the original design the /R loss will have to be made smaller and vice versa. With the new /°R loss and effective resistance of winding, the new current rating is easily obtained. Let us examine the effect of change of frequency on iron losses if the voltage remains the same. We have voltage Z = 4.44 f B,,A,7 and eddy current loss P, = K, f* B,,?. ‘Therefore as long as E remains constant product (f B,,) remains constant and therefore, eddy current losses remain constant even though the frequency is changed. We have hysteresis loss P, = K, f B,,2. Now if applied voltage is kept constant product f B,, remains constant. Let f B,, = K. K _ K,K? WKB, = KK. = : Hysteresis loss = K,(fB,,) By, f From above it is clear that the hysteresis losses will decrease with increase in frequency if the voltage is kept constant. Therefore, the total iron losses decrease if the frequency is increased and the app! voltage is kept constant. Hence with increased frequency, we can afford to have more [°R loss and thus for the same loss (i.e., the same temperature rise) the rating of the transformer can be increased. This fact is illustrated by example 5.23. 2. Effect on voltage. If the flux density is also changed, the new voltage rating is determined from the emf equation. For the new voltage total iron loss and the magnetising current and the no load current can be calculated. 3. Effect on leakage reactance and resistance of windings. The change in frequency will not have much effect on leakage inductance and therefore leakage reactance will increase linearly with frequency. Due to skin effect, the effective resistance increases with increase in frequency. This effect is not normally important with small changes in frequency. Example 5.22. A 11 kV, 25 Hz transformer has IR, hysteresis and eddy current losses 1.6, 0.6 and 0.4 per cent of the output. What will be the percentage losses if the transformer is connected to 22 kV, 50 Hz supply assuming the full load current to remain the same ? Solution. Subscript 1 refers to case with 11 kV and 25 Hz. Subscript 2 refers to case with 22 kV and 50 Hz (a) PR loss. As the full load current is same in both the cases, the copper loss remains the same.DESIGN OF TRANSFORMERS 5.103 - “ PR losses = P, = P, = 1.6 per cent of output with 11 kV. ‘The transformer is assumed to be single phase. 44 f,®,, Land By = 4.44 fq oT Ez Omfo 22 _ On, , 50 E,* Onh "Te, * 25 or On, = Bm, 8 Bay = Bay Py | KaBn? fe _ fy (0) Hysteresis loss. P, = K,B,,?f or = mu Py EyBm ff 50 or B,, = 0.6 x 5 = 1.2% of output with 11 KV. (©) Bad; 1 loss. P, = KB, 2 2 Py KeBng he _ pf © current loss. or = . see Per KBy he AF 50 or P,, = 04x Der = 1.6% of output with 11 kV. The output with 22 kV is double that with 11 kV as the current is the same in both the cases. Therefore, PR, hysteresis and eddy current losses are respectively 0.8%, 0.6% and 0.8% of output with 22kV. Example 5.23. A 40 Hz transformer is to be used on a 50 Hz system. Assuming the Steinmetz’s co-efficient as 1.6 and losses at lower frequency 1.2%, 0.7% and 0.5% for PR, hysteresis and eddy currents respectively. Find (a) losses on 50 Hz for the same supply voltage and current (®) output at 50 Hz for the same total losses as on 40 He. Solution. (a) The voltage and current at both 40 Hz and 50 Hz are the same and therefore the output in both the cases is the same. Subscript 1 refers to 40 Hz and subscript 2 refers to 50 Hz. ‘The current at both 40 Hz and 50 Hz is the same and therefore the R loss is same in both the cases, B= P., = 1.2%. Ep _ 444 fy %p,.T We have, = Om das E, = Ey. Ey” 444f, 0,7 and as a= Ay Gq, = Dp, 0.8 Op, oF Bp, = 0-8 By, « K,By, 1%) (08 B,, 2° Now, hysteresis loss P, = K,B,1°f or Pa KiB he, O8Bn) he Py EnBm A Bm hh or Ph, = Ph, x 0.81 x a = 0.875 F,, = 0.875 x 0.7 = 0.61%, i : Py _KeButht 50)? Eddy current loss P, = K, By, 9f2 on Bae? = (0.8% i) 215.104 . ELECTRICAL MACHINE DESIGN (6) If the rating of the transformer is changed, eddy current and hysteres the same while there is a change in °R loss. Total loss in Ist case = 1.2+0.7+0.5 = 2.4% ‘Total loss in 2nd case = 1.2 + 0.61 + 0.5 = 2.31%. In order that the loss should be equal in both the cases, the /?R loss has to be increased by 2.4 - 2.81-= 0.09% in the 2nd case, -. PR loss in 2nd case= 1.2 + 0.09 = 1.29%. losses remain P, PR loss varies as the square of the output [2 = ( output in 2nd suse) output in Ist ease or output in second case = output in Ist case x JP, /P., = output in Ist case x J729/12 = 1.038 x output in Ist cas ‘Thus an increase in frequency results in increase in rating. 5.54 TEMPERATURE RISE OF TRANSFORMERS The problem of temperature rise and cooling of transformers is essentially the same as that of rotating machinery. Similar to the latter, the losses developed in the transformer cores and windings during conversion are converted into thermal energy and cause heating of corresponding transformer parts. From its source the heat is directed, due to thermal gradients, to the places where it may be transferred to a cooling medium i.e. to air, or water, depending upon the method of transformer cooling. Heat dissipation occurs in the same way as in electrical machines, i.e. by way of radiation and convection. The path of heat flow is : @ From the internal most heated spots of a given part (of core or winding) to their outer surfaces in contact with the oil. (i) From the outer surface of a given transformer part to the oil that cools it. (ii) From the oil to the walls of a cooler, for instance, of the tank. (iv) From the walls of the cooler to the cooling medium—air or water. In section (i) the heat is transferred by conduction. In sections (ii) and (iii), the heat is transferred by convection of the oil. In section (iv), the heat is dissipated by both convection and radiation. 8.54.1 Transformer Oil Cooling Medium ‘Tests have shown that an average working temperature of the oil 0, = 50 to 60°C and oil viscosity corresponding to this temperature, the specific heat dissipation due to convection of oil is eony = 40.3(0/H)™4 Wim? - °C where 0 = temperature difference of the surface relative to oil, H = height of dissipating surface, m. If we assume an average @ = 20°C and H = 0.5 to 1m, hryq, = 80 to 100 Wim? — °C. ‘The corresponding figure for convection due to air is 8 W/m? - °C. Thus the convection due to oil is 10 times and above that with air. This constitutes a major valuable property of oil as a cooling medium.DESIGN OF TRANSFORMERS 5.105 5.54.2 Temperature Rise in Plain Walled Tanks, The walls of tank dissipate heat by both radiation and convection. Jt has been found experimentally that a plain tank surface dissipates 6 and 6.5 W/m? —°C by radiation and convection respectively (for a temperature rise of nearly 40°C above an ambient temperature of 20°C. Thus the total loss dissipation is 12.5 W/m? - °C. total loss _ +P ; specific heat dissipation x surface ~ 125 S, where S, = heat dissipating surface of tank. The surface, to be considered in applying the above formula, is the total area of the vertical sides plus one half area of the cover, unless the oil is in contact with the cover in which case whole area of the lid should be taken. The area of bottom of the tank should be neglected as it has very little cooling effect. For transformers of low output, the plain walled tanks large are enough to accommodate the transformer and oil have sufficient surface to keep the temperature rise within limits. But for transformers of large output, the plain walled tanks are not sufficient to dissipate losses. This is because volume and hence losses increase as cube of linear dimensions While the dissipating surface increases as the square of linear dimensions. Thus an increase in rating results in an increase in loss to be dissipated per unit area giving a higher temperature rise. ‘Modern oil immersed power transformers with natural oil cooling and a plain tank may be produced for outputs not exceeding 20—30 kVA. Transformers rated for larger outputs must be provided with means to improve the conditions of heat dissipation. This may be done by providing corrugations, tubing or radiators where feasible. 5.55 DESIGN OF TANK WITH TUBES If the temperature rise as calculated with plain tank exceeds the specified limits, it can be brought down by provision of tubes. The provision of tubes increases the dissipating area but the increase in dissipation of heat is not proportional to area because tube screen some of the tank surface preventing radiation from there. So there is no change in surface as far as dissipation of heat due to radiation is concerned. But the increase i dissipation of heat is more than what is justified by the increase in surface area. The circulation of oil is improved due to more effective heads of pressure produced by columns of oil in tubes. An addition of about 35 per cent should be made to tube area in order to take into account this improvement in dissipation of loss by convection. Let the dissipating surface of the tank be S, It will dissipate (6 + 6.5) S, = 12.5 S, W/C’. Let the area of tubes = xS,. Loss dissipated by tubes by convection = 1.35 x 6.5 xS, = 88 x S, WAC :. Total loss dissipated by tank wall and tubes 12.5 S, + 88 xS, (12.5 + 8.8 x) WAC “Temperature rise 0 = (5.97) ‘Total area of tank walls and tubes =S,+28,=S(1 +2). -. Loss dissipated = = 025 +882) Win _ 06 (5.98) x+1 B+P. 1(R+R T wit = it == £-125) (6.99) ‘emperature rise with tubes 6 Bassey * 4 on) ) (5.99) 1(B+R Total area of tubes aA -125 s) Let J, and d, be the length and diameter of each tube respectively. ‘Area of each tube = ndJ,. Hence, number of tubes n, (Ae -125 8) + (6.100)5.106 , ELECTRICAL MACHINE DESIGN ‘The area of the tubes can be found out by using the above expression. The diameter of tubes, normally used, is 50 mm and they are spaced at 75 mm, Elliptical tubes with pressed radiators are increasingly being used as they give a greater dissipating surface for smaller volume of oil. The inner dimensions of the transformer tank are fixed by the active dimensions of the transformer and cleatances between windings and grounded parts of transformer. Width of tank = W, = 20 + D, + 2 (for three phase) =D+D,+2% (for single phase) where _D = distance between adjacent limbs, D, = external diameter of h.v. winding and b = clearance between h.v, winding and tank. Length of tank —_L, = D, + 2 where I = clearance on each side between the winding and tank along the width, [gee Be ee | He H Sa { y+] L | Tog] Fig. 5.86. Tank dimensions. Height of transformer tank H, = H +h where H = height of transformer frame and h = clearance (height) between the assembled transformer and the tank. This includes the clearance at the base, oil height above the assembled transformer space for terminals and tap changing gear. Typical values of clearances 6, / and h are given below Voltage Rating Clearance Ww VA mim b 1 h 11 kV or less less than 1000 40 50 450 about 11 kV 1000-5000 0 ” 420 and upto 33 kV less than 1000 5 100 550 1000-5000 85 125 550DESIGN OF TRANSFORMERS 5.107 5.56 AIR BLAST COOLING At present time the radiators are forced cooled by small fans mounted on each radiator. Compared with natural cooling, air blast cooling of a tank increases the heat dissipation by 50 to 60 per cent or 2. = loss dissipated W/m? ~ °C by both convection and radiation with air bla = (1.5 to 1.6) x 12.5 = 20 W/m? ~ °C. ‘Transformers upto a capacity of 10 MVA have a cooling radiator system with natural cooling. For 10 MVA upwards air blast cooling of radiator is used. 5.57 FORCED OIL CIRCULATION ‘The velocity with which natural oil circulation takes place is very low, of the order of a few mm/s. Investigations of the problem have shown that with an increase in velocity of oil circulation by m times, the transformer output, for the same winding temperature rise, increases (m)“4 times. If for example m = 3, the transformer output rises by about 30 per cent. An excessive increase of oil circulating rate is unsuitable because this involves large energy loses in the pumping unit. To cool the oil, it is circulated through a special oil cooler. In an oil cooler with natural air cooling the flow rate of the circulating oil is of the order of 12 litre per minute per kW of losses. When the cooler is air blast cooled, the transformer output increases roughly to the same extent as when the tank is air blast cooled. In transformers with water cooling of the circulating oil, the pipe coils and tube coolers are employed working on the counterflow principle. Cooler surface per 1 kW of iosses ranges from 0.18 to 0.25 m?, The flow rate of the circulating oil per kW of losses, equals approximately 6 to 8 litre per minute. The water flow rate is about 1.6 litre per minute. The temperature difference between the incoming and outgoing water is usually taken to be nearly 10°C. Example 5.24. A 250 kVA, 6600/400 V, 3 phase core type transformer has a total loss of 4800 W at full load. The transformer tanks is 1.25m in height and 1 m x 0.5 m in plan. Design a suitable scheme for tubes if the average temperature rise is to be limited to 35° C. The diameter of tubes is 50 mm and are spaced 75 mm from each other. The average height of tubes is 1.05 im. Specific heat dissipation due to radiation and convection is respectively 6 and 6.5 W/m? -*C. Assume that convection is improved by 35 per cent due to provision of tubes. Solution. Area of plane tank S, = 2(1 + 0.5) x 1.25 = 3.75 m? Let the tube area be xSt. - Total dissipating surface = (1 + x) S, = 3.75 (1 +x) 4800 365 S760 +x) x35 14x From Eqn. 5.98 loss dissipated = En 2. ‘Wim? ~°C Specific loss dissipation = Wim? — 125+88x _ 365 = or x = 2.78 I+x Ixx Area of tubes = 2.73 x 3.75 = 10.23 m®, Wall arca of each tube = nd, J, = x x 0.05 x 1.05 = 0.165 m?, . Total number of tubes to be provided = 10.23/0.165 = 62. ‘The tubes are spaced 75 mm apart. Therefore in 1 m along the width of tank, we can accommodate 12 tubes leaving 90 mm on each side. In 0.5 m along the depth of tank we can accommodate 5 tubes with 100 mm space on each side, The total tubes provided in the first row along width and depth are 2 x 12 + 2 x 5 = 34, The balance 62—34 = 28 tubes can be provided in second row at the back.5.108 . ELECTRICAL MACHINE DESIGN 11 tubes can be provided in a staggered fashion in each of the two long sides and 4 in each of the two short sides. + . Total tubes provided =2x 12+2x11+2x5+2x4=64, Fig. 5.87 shows the arrangement for tubes. 0 000000000 oo00000000 1000 ooo0o°o oo00°0 oo00o00 ooo0°0o o00000000000 oo0000000000 Fig. 5.87. Arrangement of cooling tubes. Example 5.25. A 1000 kV A, 6600/440 V, 50 Hz, 3 phase, delta/ star, core type, oil immersed natural cooled (ON) transformer. The design data of the transformer is : Distance between centres of adjacent limbs = 0.47 m, outer diameter of high voltage winding = 0.44 m, height of frame = 1.24 m. Core loss = 3.7 kW and PR loss = 10.5 kW. Design a suitable tank for the transformer. The average temperature rise of oil should not exceed 35°C. The specific heat dissipation from the tank walls is 6W/m? - °C and 6.5 W/m? - °C due to radiation and convection respectively. Assume that the convection is improved by 35% due to convection, Solution. The clearance between windings and tank is assumed as 70 mm and 90 mm on each side along width and length respectively. Width of tank W, = 2D + D, + 2b = 2x 0.47 + 0.44 + 2 x 0.07 = 1.52 m. Length of tank L, = D, + 2l = 0.44 + 2 x 0.09= 0.62 m. Allowing 50 mm for the base and 300 mm for oil above the frame, Height of oil level from the bottom of tank = 1.24 + 0.05 + 0.3 = L6 m. A further height of 300 mm is necessary for the leads ete. Height of tank H, = 1.6 + 0.3 = 1.9 m. Dissipating surface of plain tank S, = 2(1.52 + 0.62) x 1.9 = 8.132 m®, Let the area of tubes be x S,. °. Total dissipating area = a +x) S, = 8.32 a +x) m*, 6+65+(1+035)xS, _ 125+ 88x lex I+z Specific loss dissipation 2 = ‘Wm? - °C.DESIGN OF TRANSFORMERS 5.109 Total loss = 3.7 + 10.5 = 14.2 kW. The temperature rise is to be kept below 35°C. 7 14.2+10° 125 + 88x Specific loss dissipati Mas, 125+ Be peaate Boss GSSPANON = B32 (Lea) x35 Lee Hence, x = 4.25. . Area of tubes = x S, = 4.25 x 8.132 = 34.56 m?, The average length of tubes is approximately 1.4 m and the diameter of each tube is 50 mm. :, Dissipating area of each tube = x d, J, = x 0.05 x 1.4 = 0.22 m?. 34.56 022 Hence, number of tubes required N, 153 The tubes are spaced 75 mm apart. The arrangement of tubes is : Along length : 4 rows on each side with 18, 17, 16 and 15 tubes Along width : 4 rows on each side with 7, 6, 5 and 4 tubes ‘Total number of tubes used = 2(18 + 17 + 16 + 15) + U7 +6 +5 + 4) = 156. 5.58 THERMAL RATING The rating of a transformer is based exclusively on thermal basis, the limitation being imposed by the maximum working temperature of the winding which will afford a reasonable life to the insulation. The critical component of most transformer insulation is cellulose, which deteriorates physically or mechanically at a rate determined by moisture content, electric stress and oil purity. The most important factor is temperaturé. Insulating materials immersed in oil maintained continuously at a temperature 75°C may have a life of 50 years, but low limit of temperature would be uneconomical from the rating view point. If the insulation is kept in oil maintained at a temperature of 110°C continuously, the life may be less than 6 years. Thus this would require transformer replacement after a very short time. Therefore, the transformer should not be operated at a high temperature (for the sake of getting higher output) otherwise its life will be very short. The reference ambient temperatures assumed are : (a) maximum ambient air temperature—45°C (6) maximum daily average air temperature—35°C . (c) maximum yearly average air temperature—30°C. Operation of a transformer at its rated kVA provides normal life expectancy if the temperature of the cooling air does not, at any time, exceed the reference ambient temperature. This results in copper in the winding attaining a temperature of 90° to 100°C (i.e. 55° C above maximum reference ambient temperature of 35°C daily average and 45°C maximum) and an assumed hot spot temperature in the windings of about 105°C. A rough and ready rule concerning the deterioration of insulating materials at high temperatures which has received wide acceptance is that the rate of deterioration of class. A insulation in oil highly doubles for every 8°C rise in operating temperature. From this it is apparent that operation of a transformer for any given period in a cooling air temperature in excess of reference ambient temperatures such as to produce a rise in the temperature of insulation of 8°C above normal is equivalent to its operation under standard ambient conditions for roughly double the period. This would, of course, mean a corresponding reduction in the life expectancy of transformer. On the other hand, it is possible to operate a transformer in cooling air temperatures exceeding reference ambient temperatures whilst still maintaining its normal life expectancy by suitably reducing its load from its rated KVA. As a very rough guide, it may be stated that for a transformer covered by IS : specification 2026—1962, a reduction in load from its rated kVA of5.110 . ELECTRICAL MACHINE DESIGN approximately 2 per cent is necessary for this purpose for each 1°C by which the actual cooling temperature exceeds the reference ambient temperature. This rule does not hold good if the cooling air temperature exceeds the reference ambient temperature by more than 10°C. If the cooling air temperature is less than the reference ambient temperature it is possible to operate the transformer at a load above its rated kVA by about 1 per cent per degree centigrade by which cooling air temperature is less than the reference air ambient temperature. This holds good if the cooling air temperature does not differ from the reference ambient temperature by more than 10°C. Whenever the transformer has a temperature rise less than its specified limit, it may be subjected to overloads on thermal considerations. The time for which such an overload can be sustained depends obviously on the initial temperature and on the time constant. The greater the weight of the transformer, the greater will be its thermal capacity and time constant. Thus a transformer with large weight has a smaller temperature rise for short time overloads. It is possible to construct a transformer of small weight to confirm with the standard rating in respect of temperature rise of windings and oil by providing ample tank cooling surface and adequate ducts in the windings. But the performance of such a low weight transformer will be unsatisfactory on short time overloads owing to its small time constant. ‘However, it is incorrect to compare two transformers in this respect solely on the basis of weights without a knowledge of their cooling. The ability of a transformer to withstand overloads depends upon the efficiency of the cooling system as well as on weight. The overload that can be imposed upon a transformer depends upon the ratio full load copper loss P. to iron loss P;. Let us compare two transformers, one with P/P, = 1 and other with P/P, = 2. At 200 per cent overload, the first transformer has a loss ratio of total loss at over load _ (2)*x1+1 _» e total loss at full load T+t and the second transformer has a loss ratio of total loss at overload _ (2)? x1+1 total loss at full load 241 Therefore overload losses in two cases are 2.5 : 3; so that a transformer with a greater PJP, ratio is less capable of sustaining overloads. 5.59 MOMENTARY OVERLOADS Transformers in service must be capable of withstanding short-circuits at normal line voltage without injury. The duration of short circuits according to IS : 2026—1962 is : | Percent impedance 4or less | 5 6 7and above Duration of short circuit s 2 3 | 4 5 The calculated winding temperature must not exceed 25 °C starting from an initial value of 90°C for water cooling and 105°C for air cooling. It is assumed that the heat generated is stored in the copper and there is no dissipation as the interval is very small. at( 24. +at Seok.) c Temperature rise 0 = at| [oer "2h sat (5.101)DESIGN OF TRANSFORMERS 5.111 where t ‘ime, s; T=, + 235°; initial temperature, °C + K, = eddy current loss ratio at 75°C .0025 x loss in Wkg at 8, = 1.9 6°7, x 10° and 6= current density, A/mm?, 5.60 HEATING TIME CONSTANT OF TRANSFORMERS ‘The heating time constant is T), = Gh/Sh (see Chapter 4). This means that the heating time constant 7, is directly proportional to weight G and specific heat A while it is inversely proportional to specific heat dissipation 4. The value of heating time constant lies between one to five hours. Natural cooling and high voltage insulation, which try to give a reduced value of specific heat dissipation 2, produce longer time constants. A transformer with large weight has a longer time constant and is thus capable of withstanding momentary overloads better than a light weight transformer (as in a transformer with a longer time constant, it takes longer time for the temperature to rise). A transformer is not a homogeneous body and therefore the above relationship is not strictly applicable to it. As a transformer consists essentially of a core, a winding and oil, all having different weights, specific heats etc. we have to deal with the heating time constants of core T,, the winding T,, and the oil T,, Example 5.26 shows how greatly the heating time constants differ for the different parts of a transformer. Example 5.26. A 3 phase 5000 kVA, 33 RV, 50 Hz, oil immersed transformer has the following data : Weight : core = 5200 kg, copper = 1200 kg, oil = 5500 kg Core losses = 18 kW, copper losses = 57 RW. The specific heats of core steel, copper and oil are respectively 480, 390 and 1670 J/kg~ Assuming that the temperature rise of core above oil 8, = 20°C, of the winding above oil 6,= 20°C, and the temperature rise of oil 8, = 40°C. Find the heating time constants for core, winding and oil. Comment upon the results Solution. Heating time constant 7, = Gh/Si but Q= S18 or Si= Qf 7, = GhoiQ Q stands for loss in the above relationship. Heating time constant for : Core : 5200x480 x20 _ 9719 5 = 45.3 minutes. 16000 Winding : 1200 39020 _ 164 5 = 2.44 minutes. 57000 ; 5500 x 1670 x 40 . Oil: 1. = “Tg5007 57000” 7 1900 8 = 81.7 minutes. ‘Thus the heating time constant for oil is 30 times that of windings. This makes it legitimate to assume that during rapid changes in load, we may without any significant error, neglect the oil temperature variation as compared with that in the windings. Therefore, the oil temperature is no indicative of the winding temperature during rapid changes of in load. Hence, in addition to a temperature indicator immersed in oil another temperature indicator must be used to monitor the temperature of the windings.5.112 . ELECTRICAL MACHINE DESIGN DESIGN PROBLEMS Problem I. Design a 25 kVA, 11000/433 V, 50 Hz, 3 phase, delta/star, core type, oil immersed. natural cooled distribution transformer. The transformer is provided with tappings + 2% + 5% on the h.v. winding. Maximum temperature rise not to exceed 45°C with mean temperature rise of oil 35°C. Solution. Core Design. The value of K is taken from Table 5.2. K = 0.45 for three phase core type distribution transformers. Voltage per turn E,=K JQ = 0.45 /25 = 2.25 V. . Flux in the core ®,, = 0.010135 Wb. Hot rolled silicon steel grade 92 is used. The value of flux density B,, is assumed as 1.0 Wo/m?. Net iron area A, = ee = 0.01035 m? = 10.135 x 10° mm? Using a cruciform core, A; = 0.56 d? Diameter of circumscribing circle d = 10135 x 105/056 = 134.5 mm. Reference Figs. 5.52 and 5.53(a), widths of laminations a = 85, d = 0.85 x 135.8 = 114.8 mm, b = 0.53 d = 0.53 x 135.8 = 71.6 mm. ‘The laminations are punched from 750 mm wide plates and the nearest standard dimensions are a = 114 mm and 6 = 73 mm. Window Dimensions. The window space factor for a small rating transformer is 0.195. The value assumed is K, = 0.18 ‘The current density in the windings is taken as 2.3 A/mm?. Output of transformer Q=3.33/B,,K, 54, 4, x 10% 83 x BO x 1 x 0.18 x 2.3 x 108 x A, x 0.010135 x 10-8 or window area A,, = 0.0358 m? = 35.8 x 10° mm?. Taking the ratio height to width of window as 2.5, H,, x W,, = 35.8 x 10° or 2.5 W,? = 3.8 x 108, . Width of window W,, = 120 mm and height of window = 300 mm. Area of window provided A,, = 300 x 120 = 36 x 10° mm? = 0.036 m?. Distance between adjacent core centres D = W, +d = 120 + 185 = 255 mm. Yoke Design. The area of yoke is taken as 1.2 times that of limb. :. Flux density in yoke = 1/1.2 = 0.0833 Wb/m?. Net area of yoke 1.2 x 10.135 x 10° = 12.16 x 10° mm?. Gross area of yoke 12.16 x 107/0.9 = 13.5 x 10° mm’. ‘Taking the section of the yoke as rectangular, Depth of yoke D, =a = 114 mm 2. Height of yoke H, = 13.5 x 104114 = 114mm Overall Dimensions of Frame. Reference Fig. 5.70. Height of frame Width of frame 2D +a = 2x 255 + 114 = 624 mm. Depth of frame D, =a = 114 mm.DESIGN OF TRANSFORMERS 5.113 L.V. Winding : Secondary line voltage 433 V ; Connection = star. + Secondary phase voltage V, = 433/\3 = 250 V. Number of turns per phase T, = V/E, = 250/2.26 = 111. 25 x 1000 3x 250 A current density of 2.3 A/mm? is used. ‘Area of secondary conductor a, = 33.9/2.3 = 14.48 mm?. From Table 23.1 (IS : 1897-1962), using a bare conductor of 7.7 x 2.2 mm. Area of bare conductor a, = 14.9 mm? Current density in secondary windings 6, = 33.4/14.9 = 2.23 A/m? The conductors are paper covered. The increase in dimensions on account of paper covering is 0.5 mm. .. Dimensions of insulated conductor = 7.5 x 2.7 mm?, ‘Using three layers for the winding. Helical winding is used. Therefore space has to be provided for (37 + 1) = 38 turns along the axial depth. Axial depth of L.v. winding L,, = 38 x axial depth of conductor = 38 x 7.5 = 285 mm. ‘The height of window is 300 mm. This leaves a clearance of (300 ~ 285/2 = 7.6 on each side of the winding. (The minimum clearance should be 6 mm for windings having voltages below 500 V.) Using 0.5 mm pressboard cylinders between layers, Radial depth of lv. winding number of layers x radial depth of conductor + insulation between layers 2.7 42x 0.6 = 9.1 mm. 5.88 shows a cross-section through Lv. coil, Secondary phase current 1, = 33,3 A, O.5mm Press boord 0.25mm Paper tope Fig. 5.88. L.V, winding (All dimensions in mm),5.114, . ELECTRICAL MACHINE DESIGN Diameter of circumscribing circle d = 185 mm. Using pressboard wraps 1.5 mm thick as insulation between l.v. winding and core. Inside diameter of L.v. winding = 135 + 2 x 1.5 = 138 mm. Outside diameter of Lv. winding = 138 + 2 x 9.1 = 156.2 mm. HLV. Winding : Primary line voltage = 11000 V. Connection = Delta Primary phase voltage V, = 11000 V. Number of turns per phase T,, = 11000 x 111/250 = 4884. As £ 5% tappings are to be provided, therefore, the number of turns is increased to T, = 1.05 x 4884 = 5128. The voltage per coil is about 1500 V. . Using 8 coils, Voltage per coil = 11000/8 = 1375 V. Turns per coil = 5158/8 = 641. Using 7 normal coils of 672 turns and one reinforced coil of 424 turns. Total h.v. turns provided 7,, = 7 x 672 + 424 = 5128. Taking 24 layers per coil. Turns per layer = 674/24 = 28. Maximum voltage between layers = 2 x 28 x 2.25 = 126 V, which is below the allowable limit. 25 x 1000 HLV. winding phase current I, = 5°" 955 = 0.787 A. As the current is below 20 A, cross-over coils are used for h.v. winding. Taking a current density of 2.4 A/mm’, . Area of h.v. conductor a, = 0.757/2.4 = 0.316 mm’. Diameter of bare conductors (4/x x 0.316)? = 0.635 mm. Using paper covered conductors. From Table 23.4 (BIS : 3454 — 1966) the nearest standard conductor size has bare diameter = 0.63 mm, insulated diameter = 0.805 mm with fine covering. Modified area of conductor a, = wW4 x (0.63)? 312 mm?. Actual value of current density used oan 0.757/0.312 = 2.43 A/mm?. Axial depth of one coil = 28 x 8.05 = 22.6 mm. The spacers used between adjacent coils are § mm in height. Axial length of h.v. winding jumber of coils x axial depth of each coil + depth of spacers 8 x 226 + 8 x 5 = 221 mm The height of window is 300 mm and therefore, the space left between winding and window is 300-221 = 79 mm. This is occupied by insulation and axial bracing of the coil. The clearance left on each side is 39.5 mm, which is sufficient for 11 kV transformers. The insulation used between layers is 0.3 mm thick paper. Radial depth of h.v. coil b, = 24 x 0.805 + 23 x 0.3 = 26.22 mm. From Eqn. 5.22 the thickness of insulation between h.v. and lv. winding 5 + 0.9 kV = 5 +0.9 x 11 = 15 mm, This includes the width of oil duct also. ‘The insulation between h.v. and Lv. winding is a 5 mm thick bakelized paper cylinder. The h.v. winding is wound on a former 5 mm thick and the duct is 5 mm wide, making the total insulation between h.v. and l.v. windings 15 mm,DESIGN OF TRANSFORMERS 5.115, :, Inside diameter of h.v. winding yutside diameter L.v. winding + 2 x thickness of insulation ci 156.2 + 2 x 15 = 186.2 mm. - Outside diameter of h.v. winding = 186.2 + 2 x 26.22 = 238.64 mm = 239 mm. Clearance between windings of two adjacent limbs = 255 ~ 239 = 16 mm. Resistance : 1862 + 239 Mean diameter of primary winding = 212 mm. Length of mean turn of primary winding Ly, = x 212 x 10 = 0.666 m. Ty Lin Resistance of primary winding at 75°C r, Zeb Emp. oy = 4884 x 0021 0666 _ p199 0 0312 Mean diameter of secondary winding = 3841062 = 149 mm. Length of mean turn of secondary winding L,,,, = x 149 x 10% = 0.468 m. 111 x 0.021 x 0.468 149 Resistance of secondary winding at 75°C, r, = 0.0732 2 2 x 0.0732 = 364 ©. . Total resistance referred to primary side R, = 219.2 + (4884 P.U, resistance of transformer ¢, = Leakage Reactance : Mean diameter of windings = (138 + 239)/2 = 188.5 mm. Length of mean turn L,,, = x x 188.5 x 10° = 0.592 m. Height of winding L, = (L,, + L,,V2 = (221 + 2852 = 253 mm. Leakage reactance of transformer referred to primary side Im (, . ves 2nf by T,? ve + ) ( (a5. 2! = 2m x 5O x 4m x 10-7 x (4884)? x (2) x( 15 2822498) x 10- = 590 ©. x PU. leakage reactance ¢, = 0.757 x 590/11000 = 0.0406. P.U. impedance «, = 40.025)? + (0.0406)? = 0.0477. Regulation. P.U. regulation = ¢, cos $ + €, sin 6. Per unit regulation at unity power factor € lagging © = 0.025 x 0.8 x 0.0406 x 0.6 = 0.0444. 0.025, at zero p.f. lagging ¢ = 2 0.406, at 0.8 pf. Losses : PR Loss ER loss at 75°C, = 31,? R, = 3 x 0.757? x 364 =5.116. ELECTRICAL MACHINE DESIGN ‘Taking stray load loss 15% of above. . Total PR loss including stray load loss P, = 1.15 x 626 = 720 W. Core Loss Taking the density of laminations as 7.6 x 10° kg/m’, weight of 3 limbs = 3 x 0.3 x 0.010185 x 7.6 x 10% = 69.3 kg. ‘The flux density in the limbs is 1 Wb/m? and corresponding to this density, specific core loss is 1.2 W/kg (See Fig. 4.29). Core loss in limbs = 69.3 x 1.2 = 83.2 W. Weight of two yokes = 2 x 0.624 x 0.01216 x 7.6 x 10% = 115.3 kg Corresponding to a flux density of 0.833 Wb/m? in the yoke, the specific core loss = 0.85 W. Core loss in yoke = 115.3 x 0.85 = 98 W. Total core losses P, = 83.2 + 98.0 = 181 W. Efficiency. Total losses at full load = 181 + 720 = 901 W. 25000 Efficiency at full load and unity pli = 07 apy x 100 = 96.5 per cent For maximum efficiency x? P. = P, «a= (PTR, = Jigis720 = 0.0501 Thus maximum efficiency occurs at 50.1 percent of full load. This is a good figure for distribution transformers. No load current. Corresponding to flux densities of 1 Wb/m? and 0.833 Wb/m? in core and yoke respectively at, = 120 A/m and at, = 80 A/m. (See Fig. 3.3) --Total magnetizing mmf = 3 x 120 x 0.3 + 2 x 80 x 0.624 = 207 A. ‘Magnetizing mmf per phase AT, = 207/3 = 69 A. Magnetizing current I,, = ATy(V2 T,) = 62/(V2 x 4884) = 5.5 x 10° A, Loss component of no load current I; = 181/3 x 11000 = 5.5 x 10° A. No load current I = y(10 x 10)? + (55 x 10 No load current as a percentage of full load current = 14x 103A, 114 x10 0757 Allowing for joints etc. the no load current will be about 2.5% of full load current. Tank. Height over yoke H = 328 mm. Allowing 50 mm at the base and about 150 mm for oil. Height of oil level = 528 + 50 + 150 = 728 mm. Allowing another 200 mm height for leads etc., height of tank H, = 728 + 200 = 928 mm. The height of tank is taken as 0.95 m or H, = 0.95 m. Assuming a clearance of 40 mm along the width on each side Width of tank W, = 2D + D, + 2U = 2 x 255 + 239 + 2 x 40 = 829 mm, ‘The width of tank W, is taken as 0.84 m. ‘The clearance along the length of the transformer is greater than that along the width. This is because additional space is needed along the length to accommodate tappings ete. The clearance used is approximately 50 mm on each side. Length of tank L, = D, + 2b = 239 + 2 x 50 = 339 mm. The length of tank L, is taken as 0.35 m. Total loss dissipating surfae of tank S, = 2(0.84 + 0.35) x 0.95 = 2.26 m?, Total specific loss dissipation due to radiation and convection is 12.5 W/m? — ‘Temperature rise = 901/(2.26 x 12.5) = 31.9°C This is below 35°C and therefore plain tank is sufficient for cooling and no tubes are required. Fig. 5.89 shows core and winding details of the transformer. x 100 = 1.5%.DESIGN (OF TRANSFORMERS DESIGN SHEET 5.117 kVA 25 Phase 3 Frequency—50 Hz * Delta/Star hv. 11000. hv. 11000. Line voltage Ie aa Phase voltage [i Ea [hw 181A, hv. 0.757 A, Line current liv. 36a. Phase current f aera ‘Type—core ‘Type of cooling—ON Core 1. Material 0.35 mm thick 92 Grade 2. Output constant K 0.25, 3, Voltage per turn E, 2.25V 4, Circumscribing circle diameter d 135mm 5. Number ofsteps 2 6. Dimensions a 4mm b 73mm 7. Netiron area A, 10.135 x 10? mm? 8. Fluxdensity By 1.0 Wom? 9. Flux bn 10.135 m Wb 10. Weight 69.3kg 11. Specific iron loss 12Whke 12. Iron loss 83,.2W Yoke 1. Depth of yoke D, 114mm 2. Height of yoke i, 114mm 3. Net yoke area 12.16 « 10? mm? 4. Fluxdensity 0.833 Wom? ‘5. Flux 10.135 m Wb 6. Weight 115.3kg 7. Specific iron loss 0.8 Wikg 8, Iron loss 98 W Windows 1. Number 2 2. Window space factor K, 0.18 3. Height of window A, 300mm 4. Width of window Ww, 120mm 5, Window area Ay 36 x 10? mm? Frame 1, Distance between adjacent limbs D 255mm 2. Height of frame H 536mm 3. Width of frame w 624mm 4, Depth of frame D, 14mm5.118 . ELECTRICAL MACHINE DESIGN Fig. 5.89. 25 kVA, 3 Phase, 100/433 Volt, 50 Hz, Core Type Distribution TransformerDESIGN OF TRANSFORMERS 5.119 Windings Ly. EV 1. Type of winding helical Cross-over 2 Connections Star Delta 3. Conductor Dimensions-bare 7.0x2.2mm* Diameter = 0,68 mm insulated 7.6% 2.7 mm? Diameter =0.814mm Area 149mm? 0.912mm* Number in parallel None None 4. Currentdensity 2.23 A/mm? 2.43 A/mm? 5. Turns per phase uu 4884 (5128 at -5% tap) 6. Coils total number 3 3x8 per core leg 1 8 7. ‘Turns per coil m of 672 turns, 1 of 424 turns perlayer wu 8B 8 Number oflayers 3 m 9. Height of winding 285mm 221mm 10. Depth of winding 9.1mm 26mm 11, Insulation Between layers 0.5 mm press board 0.3 mm paper Between coils 5.0mm spacers, 12. Coil Diameters Inside 138 mm 186.2 mm Outside 156.2mm_ 239mm 13. Length of mean turn 0.468 m 0.666 m 14, Resistance at 75°C 0.07329 219.20 Insulation 1. Between Lv. winding and core ‘= press board wraps 1.5 mm 2. Between Lv. winding and h.v. winding = bakelized paper 5.0mm. 3. Widthofduct between lv.andhy. = 5mm Tank 1. Dimensions: height H, 0.95m length L, 035m width W, 0.84m 2 OilLevel 0.728 m 3, Tubes Nil 4, ‘Temperature rise 3190 Impedance 1. PU. Resistance 0.025 2 P.U. Reactance 0.0406 3. P.U. Impedance 0444 Losses 1, Total core loss 181W 2. Total copper loss 720 8, Total losses at full load 901W 4, Efficiency at full load and u pf. 96.5%5.120 ELECTRICAL MACHINE DESIGN Problem II. Design a 500 kVA, 50 Hz, 600/400 V, single phase core type, oil immersed, natural cooled power transformer. The mean temperature rise of oil not ta exceed 35°C. Solution. Core Design For this transformer we use 56 Grade (G.K.W.) cold rolled oriented steel laminations. The following assumptions are made : B,, = 1.5 Whim? ; 6 = 2.75 A/mm? ; g, = 8.9 x 10% kg/m? ; g, = 7.65 x 10° kg/m? ; ratio L_d/l; 0.5. The values of constant K for various conditions have been worked out in Example 5.10. The value of K is respectively 1.22, 0.745, 0.375 and 0.344 for maximum efficiency at 90% full load, minimum cost, minimum weight and minimum volume. It is clear from above that if the transformer is designed for maximum efficiency to occur at 90% full load, the cost would be exorbitant. Therefore, keeping the cost of the transformer the main consideration and keeping in view to some extent, the efficiency, volume and weight of transformer the value of K is assumed as 0.8 Voltage per turnE, = {@ =0.8 500 = 17.9 V. Flux 7.94.44 x 50) = 0.0806 Wb Net iron area 0.0806/1.5 = 0.0537 m? = 53.7 x 10° mm?, A 3-stepped core is used. From Table 5.3, for a 3-stepped core A, = 0.6 d? or diameter of circumscribing circle d = 3. x10°/ The dimensions of the laminations are : @ = 0.9 x 300 = 270 mm, 6 = 0.7 x 300 = 210 mm, c = .42 x 300 = 126 mm. (See Fig. 5.53). The standard sizes of laminations used are 270 mm, 215 mm and 135 mm. Window Dimensions. As the transformer is rated at 500 kVA, we take window space factor x -—10 10 « " 30+kV 30+66 From Eqn. 5.6, for a single transformer, Q = 2.22 /B,, K,, 5A, A; x 10% =0.27 or 500 = 2.22 x 50 x 1.5 x 0.27 x 2.75 x 10° x A, x 0.0537 x 10° , Window area A, = 75.3 x 10? mm? Taking the ratio height to width of window = 2.5 22.5 W,? = 75.3 x 10° :. Width of window W,, = 175 mm and height of window H,, = 437.5 mm Distance between adjacent limbs D = d + W,, = 300 + 175 = 475 mm. Yoke Design The area of yoke is taken to be the same as that of core. Assuming rectangular cross- section for the yoke, height of yoke H, = a = 270 mm and depth of yoke D, = a = 270 mm. Frame Height of frame = H,, + 2H, = 487.5.+ 2 x 270 = 977.5 mm. Width of frame = D +a = 475 + 270 = 745 mm. Depth of frame a = 270 mm. Windings L.V. Winding LV. winding turns _T, = 400/17.9 = 22.3 say 22. Modified value of core flux density B,, = 1.5 x 22.8/22 = 1.52 Wb/m’,DESIGN OF, TRANSFORMERS 5.121 LY. winding current I, sense = 1250 A. 3250 455 mm’, 75, Helical winding is used. The l.v. conductor consists of 20 strips connected in parallel. Each strip is 6 x 4 mm in cross-section with an area of 23.4 mm? (see Table 23.1). Total atea provided a, = 20 x 23.4 = 468 mm, Actual current density used 8, = 1250/468 = 2.67 A/mm?, The arrangement of twenty conductors is shown in Fig. 5.90 The insulated conductors have dimensions 6.5 x 4.5 mm. ‘Axial depth of one turn = 5 x 6.5 + 1 mm slack = 33.5 mm. Radial width of one turn = 5 x 4.5 + 1 mm slack = 19 mm. The winding is applied on both the limbs with 11 turns per limb. Therefore spac has to be provided for 11 + 1 ie. 12 turns, Axial height of Lv. coil L,, = 12 x $8.5 = 402 mm, 48) ‘The height of window is 437.5 mm and the balance height of 35.5 mm is occupied by end insulation and bracings etc. 65 60) Radial depth or width of Lv. coil b, = 19 mm, The insulation between core and lv. winding is bakelite former 5 mm thick. Inside diameter of former d = 300 mm. 33.5 Outside diameter of former = 310 mm. Inside diameter of Lv. winding = 310 mm. Inside diameter of l.v. winding = 310 mm. Outside diameter of l.v. winding = 310 + 2 x 19 = 348 mm Area of Lv. conductor a, b—13 41 Mean diameter of 1.v. winding (810 + 348)/2 = 329 mm. Fig. 6.90, L.V. conductor. Length of mean turn of Lv. winding L,,,, = * x 0.829 = 1.03 m. 22 x 0.021 x 468 PR losses in l.v. winding= (1250)? x 0.001 = 1562 W. LV. Winding HLV. winding turns _ T,, = (6600/40) x 22 = 363. Using 182 turns on one limb and 181 on the other. 500 x 1000 6600 Area of h.v. conductor a, =75.75/2.75 = 27.5 mm?. The available height of window is not-sufficient for a single coil. Therefore a double helical winding is used with 91 turns in helix. (On one limb there are 91 turns per helix but on the other there are 91 in one and 90 in the other helix). ‘The cross-section of bare conductor is 3.5 x 8 mm. It is wound on edge with 0.5 mm thick press board between turns. Resistance of Lv. winding r, = = 0.001 2. HLV. winding current J, = = 75.75 A.5.122, . ELECTRICAL MACHINE DESIGN Area of h.v. conductor a, = 27.4 mm?. (see Table 23.1) Radial width of each coil = 8 mm, Axial height of each turn =.3.5 + 0.5 = 4 mm. “Axial depth of 91 turns = 92 4 Insulation and bracing etc. Axial height of hv. coil, L,, = 368 + 9.5 = 377.5 mm. Insulation between h.v. and Lv. windings =5 +0.9kV=5 40.9 x 66=11 mm. Using a duct 6 mm wide and a former 5 mm thick for first coil (helix) of h.v. winding. Inside diameter of first h.v. former = 348 + 2 x 6 = 360 mm. Inside diameter of 1st h.v. coil = outside diameter of first h.v. former 360 + 2x 5 = 370 mm. Outside diameter of first h.v. coil = 870 + 2 x 8 = 386 mm. Allow a duct 8 mm wide including the former for second h.v. coil. Inside diameter of second h.v. coil = 386 + 2 x 8 = 402 mm. Outside diameter of second h.v, coil = 402 + 2 x 8 = 418 mm. Distance between h.v. windings on adjacent limbs = 475 - 418 = 567 mm. ‘Mean diameter of h.v. winding = (360 + 418)/2 = 889 mm. Length of mean turn of h.v. winding Ljy, = % x 389 mm = 1.22 m. Resistance of h.v. winding at 76°C, r, = sence 20.342 PR loss in h.v. winding = (75.75)? x 0.34 = 1951 W. Resistance Total PR loss = 1562 + 1951 = 3513 W. ‘This is increased by about 10 percent to account for stray load loss. Total F°R loss including additional loss P, = 1.1 x 9518 = 3865 W. z+ Resistance of transformer referred to primary (h.v.) side 3865 R= ES, = 0.6742. P (18.75) 78.118 x 0674 P.U. resistance of transformer ¢, = = 0.0077 Leakage Reactance : Length of mean tun L,,, = % Grox8) x10 = 1.14 m. Mean height of coil L, ‘The windings are divided equally over two limbs and therefore the leakage reactance of the transformer is equal to twice the leakage reactance of windings arranged on one limb. Further the h.v, coil is divided into two parts and therefore Eqn. 5.78 is used.DESIGN OF TRANSFORMERS 5.123 Leakage reactance of winding on each limb bp thy +b a! * oe wernt tat {os tects she) 114 (90313-0008 09082 0019, 0008) =n 80 4 107 52t HE 3 + 4 ‘ 20.9822 .. Total leakage reactance of transformer referred to primary xX, = 2 x 0,982 = 1964.9. 18.75 x 1964 P.U. leal cts CE a kage reactance ¢, P.U. Impedance e, = Jo.0082? +0026? = 0.0236. Losses : PR loss. Total copper loss P, = 3865 W. Core loss. As mentioned earlier 0.33 mm thick 56 grade (G.K.W.) laminations are used. Length of mean flux path J, = 2(437.5 + 475 + 270) x 10-8 = 2.37 m Weight of iron = 2.37 x 0.0587 x 7.65 x 10° = 974 kg. The specific loss for B,, = 1.52 Wh/m?, from Fig. 3.26, is 1.25 Wkg. Core loss = 1.25 x 974 = 1217.4 W. Making an allowance of 20% for joints, total core loss P, = 1.2 x 1217.5 = 1460 W. Total loss at full load = 3865 + 1460 = 6325 W. = 0.025, Efficiency : Efficiency at full load and unity p.f 500 x 108 = — 00x — x 100 = 98: 500 « 108+ 5806 * 10° = 98.95% Regulation : Regulation at full load and 0.8 pf. lagging « = ¢, cos 4 +e, sin ¢. = 0.0077 x 0.8 + 0.0226 + 0.6 = 0.0197 p.v. or 19.7%, Tan! Height of frame H = 977.5 mm. Allowing 60 mm for the base and 260 mm for oil +. Oil level = 977.6 + 50 + 250 = 1300 mm. Allowing 300 mm further for leads ete. Height of tank H, = 1800 + 800 = 1600 mm = 1.6 m. Allowing a clearance of 80 mm at each along the length. Length of tank L, = 475 + 418 + 2 x 80 mm = 1,05 m. : A clearance of about 200 mm is allowed along the width to accommodate taps ete. ». Width of tank W, = 418 + 200 mm = 0.62 m Loss dissipating surface of tank S, = 2(1.05 + 0.62) x 1.6 = 5.34 m? The mean temperature rise of oil is to be limited to 35°C. Let the area of tubes be xS,. 5325 285 Badd +a) x58 ive Wm? -"C :. Specific loss dissipation =5.124 ELECTRICAL MACHINE DESIGN OOO00N000000 OQOOCOOOOOOOONO _- | BOOOOOOOOOOO OOOOOCO000O0 1050 Fig. 5.91. Core and winding details of 500 kVA, Single Phase 6600/400 volt, ‘50 Hz core type power transformer (dimensions in mm) or 285 _ 125+ 88% (geo Eqn. 5.8) 1. x = 1.82 l+x l+x .. Area of tubes needed = 1.82 x 6.34 = 9.72 m? Using 50 mm diameter tubes spaced 7 mm apart. The average length of tubes as 1.35 m. Dissipating area of each tube = x x 0.05 x 1.85 = 0.212 m? Number of tubes to be provided = 9.72/0.212 = 48. Arrangement of tubes : Along length—2 rows—12 and 11 tubes. The details of core windings and tank are given in Fig. 5.91. Problem IIL Design a 150 kV A, 600/400 V, 50 Hz, single phase shell type, oil immersed, self cooled power transformer. Average temperature rise of oil not to exceed 35°C. Solution. Core Design. Front Table 5.1, the value of output coefficient X is taken as 1. Voltage per turn E, = K/@ =1x Ji50 = 12.25 V. E, 1225 Flux &, = 797 = Gapegy 7 920552 Wh. ‘Taking the value of maximum flux density B,, = 1.25 Wb/m?, Net iron area A, = 0.0552/1.25 = 0.0442 m? = 44.2 x 109 mm?, ‘Taking the iron stacking factor = 0.9. Gross iron area A,; = 44.2/0.9 = 49 x 10° mm?*, The ratio of b/2a is taken as 2.5 (See Fig. 6,71). 2.5(2a)? = 49 x 10° or width of central limb 2a = 140 mm. Depth of frame b = 2.5 x 140 = 350 mm. The side limbs carry half of the flux and therefore their width is equal to half of the width of central limb or width of side limbs a = 70 mm.DESIGN OF TRANSFORMERS 5.125 Window Dimensions : 10 10 * ‘Window space factor K, = 0.27. = 20+KV 30466 For the sake of calculation of window area, a current density of 2.3 A/mm? is assumed. Now, Q=222/B,,K, 54,4, x 10% or 150 = 2.22 x 50 x 1.25 x 0.27 x 2.8 x 108 x 0.0442 x A, x 10°, :. Window area A,, = 0.0394 m? = 39.4 x 10° mm? ‘The height of the window is taken as twice the width of window, or 2 W,? ‘Width of window = 140 mm, and height of window H,, = 280 mm, Yoke Design. The flux in the yoke is half of the flux in the central limb. The flux density in the yoke is taken same as that for limbs. -. Height of yoke H, 0 mm. Frame. Reference Fig. 5.92. Height of frame H =H, + 2H, = 280 +2 x 70 = 420 mm. 19.4 x 108 bh 1 420 Lv. 140 140 560 100 Fig. 5.92. Frame of shell type transformer Fig. 6.93, Arrangement of coils. (All dimensions in mm) (All dimensions in mm) Width of frame W = 4a + 2W,, = 4x 70 +2 x 140 = 560 mm. Depth of frame D, = 6 = 50 mm.5.126 . . ELECTRICAL MACHINE DESIGN Windings : Number of turns in l.v. winding T, = 440/12.26 = 36. This number is divisible by four to allow the subdivision of l.v. winding into 4 sections of 9 turns each. ‘Number of turns in the h.v. winding T, = 36 x 6600/440 = 540. This is divided into two coils of 150 turns each and two of 129 turns each with reinforced end turns. ‘The arrangement of the coils is shown in Fig. 5.93. The main insulation, height wise in the window is given below : Insulation between end Lv. coils and core = 2 x 10 = 20 mm. Spacers between h.v. and Lv. coils = 4 x 20.= 80 mm. Spacers between Lv. coils = 1 x 8 = mm. Spacers between h.v. coils = 2 x 10 = 20 mm. Total height of insulation = 20 + 80 + 8 + 20 = 128 mm. -. Height available for conductors = 280 - 128 = 152 mm. Total number of hy. and Ly. coils = 8. _. Height of each coil = 152/8 = 19 mm. LV. Winding 150 x 1000 6600 Area of h.v, winding conductor a, = 22.7/2.3 = 9.9 mm?. ‘The h.v. conductor comprises of two paper covered strips 8 x 0.6 mm bare in parallel. Area of h.v. conductor used a, = 2 x 8 x 0.6 = 9.6 mm?, ‘Actual value of current density used in hiv. winding = 22.796 = 2.97 A/mm?. Space taken by h.v. conductor in a coil height wise = 2 x 8 = 16 mm. The balance height (19 ~ 16) = 3 mm is taken up by paper. Space taken by h.v. bare conductors width wise = 150 x 0.6 = 90 mm. ‘The space taken up by insulation between h.v. and core width wise = 2 x 20 = 40 mm. Space for winding along width of window = 140 - 40 = 100 mm. orwidth of hv. coil = 100 mm. The balance space = 100 - 90 = 10 mm is taken by insulation between h.v. conductors. LV. Winding HLY. winding current I, = = 227A. L.V. winding current I, = soni = 341A. 1 Area of Lv. winding conductor a, = a = 148 mm. Using 4 strips of 7.5 x 6 mm each in parallel. Area of L.v. winding conductor used a, = 4 x 7.5 x 6 = 150 mm?, Actual value of current density used in l.v. winding 5, = 941/150 = 2.28 A/mm?, Space occupied by bare conductors in a coil height wise = 2 x 7.5 = 15 mm. ‘The balance space (19 - 15) = 4 mm is taken by insulation etc. Space occupied by bare conductors, in a section, width wise =9x2x5=90mm.DESIGN OF, TRANSFORMERS 5.127 Using 1 mm thick press board width wise between each conductor, ©. Width of Lv. coil = 9+ 9x1 = 100 mm, * The balance space 140 ~ 100 = 40 mm along the width is taken up by insulation between core and l.v. winding with 20 mm on each side. Resistance. The approximate length of mean turn = Lyyy = Lyy, = 1.58 m. Resistance of hv. winding r, = S02 x 560 158 = 1879. Resistance of lv. winding r, = ORI HE = 15S = 0,008 2. Resistance of I.v. winding referred to hv. side r,’ = 0,008 x (540/36)? = 1.8 9. Total resistance of transformer referred to primary side R, = 1.87 + 1.8 = 3.679 = 8.85 Q (taking loss in connections ete.) _ 227% 385 6600 Leakage Reactance. For sandwich coils, leakage reactance of transformer referred to primary side : P.U. resistance = 0.0133. Lime To? by +b, X,=afh bo aan (o-2eth (See Eqn. 5.80) 0.0236 x 2 158 , 540" ( #1739. = mx 50 x 4x x 10-7 x OL PLU, reactance c, = 222%173 _ 9.9596. P.U, impedance ¢, = 0.012" + 0.0595 Regulation. The regulation of transformer at full load and 0.8 p-f. lagging. € =e, cos } +6, sin } = 0.0125 x 0.8 + 0.0595 x 0.6 = 0.0457 = 4.57% Losses : FR Lose PR loss in hv. winding = 22.7? x 1.87 = 964 W. PR loss in lv. winding = 341? x 0,008 = 228 W, Total JPR loss= 964 + 928 = 1892 W. The total /2R loss including loss in connections ete, may be taken as P, = 2000 W. Core Loss. Referring to Fig. 5.98. Volume of iron = 14 x 35 x 28 + 2x 7x 35 x 66 +2 x 7x 35 x 28 = 0.05488 m* Weight of iron = 0.05488 x 7.6 x 10° = 417 kg. ‘The specific iron loss for a flux density of 1.25 Wh/m? is 1.8 W. Total core loss = 417 x 1.8 = 750 W. The total core loss including additional losses may be taken as 840 W. 0.046 + 9042) .061. Efficiency : Total loss at full load = 2000 + 840 = 2840 W 150 x 10° 150 x 10° + 2840 PU, load for maximum efficiency = (84072000 = 0.65 Load for maximum efficiency = 0.65 x 150 = 97.5 kVA. Efficiency at full load unity power factor = x 100 = 98.3%5.128, . ELECTRICAL MACHINE DESIGN 5.61 DESIGN OF SMALL SINGLE PHASE TRANSFORMERS 5.61.1 Core Design The starting point in the design of small transformers is the choice of turns per volt. The values of turns per volt are given in Table 5.7. Table 6.7. Turns per volt T, VA Turns per volt VA Turns per volt 10 23.3 200 3.5 15 175 28 20 14.0 300 28 25 wt 400 2.3 50 7.0 500 20 6 56 17 100 46 16 150 4.0 1 Now E=4.44/0, 7 «Turns per volt 7, = TIE = 14.44 f©,,. Flux in the core ©,, = 1/4.44 fT. The frequency of the transformer is specified and the value of turns per volt T, is taken from Table 5.7. Therefore flux ®,, in the core is known. Net area of core A; = ©,/B,. The value of maximum flux density B,, is assumed to be 1 Whim?. Gross area of core A,; = A; = A/0.9. (Assuming a stacking factor of 0.9) A shell type of construction is normally used for small transformers. The core is made up of any of the following combination of stampings : @ Eand I Fig. 5.94(a)), (i) Tand U (Fig. 5.946) (ii) E used in pairs (Fig. 5.94(c)) D A D ra | Lo | [Js | (a) E-Itype. () T-U type. (0) Etype. Fig. 6.94. Laminations for small transformers Tables 5.8, 5.9 and 5.10 give information about standard stampings manufactured by the manufacturers for small transformers and choks.DESIGN OF TRANSFORMERS Table 5.8. E-I Stampings [(Fig. 5.94(a)] 5.129 Dimensions * ‘Remarks No. A B c D E Ww ve | oii | ova wa" var 2a | se | 17 | 19n6 ss" we" 2 58" 2 208" sas" se" 10 ser | 296" | 218 we" ae" 10A se | 298 | 21e" ae 3B" 1 suez | ai7sx | zy" we" ss" " aie" | 2une | 1avse | 182" 182" 3 sar | aver | 178" ae" ae" u aa" s ous" a8" a8 nA s 2.58" se" ae 2 x s se" ae 30 6omm | 50mm | 10mm | 10mm holes 5/82" dia. a1 2ee" | 2a6" me" me" 45 256" | zane" mer ms" 4holes 5/82" dia. 15 x ue hg wz 4tholes 7/32" dia. “ r 3 ue ve ver Aholes 7/32" dia. “4 y | aene | 258" qae" ve holes 7/32" dia. 4 y | sasne | sasney | i72" ver 4holes 7/32" dia. 33 28mm | 64mm | 70mm | 4mm | 14mm holes 11/64" dia. 3 aie | aaar | sue 58" se 4 holes 7/32" dia. 13 ua | 4 sae" ver er holes 7/32" dia 4A asne | aus" | 3.76 ause" ays" 4 holes 3/16" dia 16 Lue | ave | saa aia" ar ‘holes 7/32" dia. 5 uve | ase | saan aan aa ‘4tholes 17/64” dia. 6 Lue | st ue aan war ‘4tholes 9/52" dia, 7 z e | 4.5ne ry Yr 4tholes 17/64" dia. 43 r e s y r ‘holes 17/64" dia. 8 ez | 7a | eae r r holes 9/8" dia. Table 5.9. E-U Stampings [(Fig. 5.94(6)] Dimensions Remarks No. A B c D E 34 se] ae" | ae ver we holes 5/32" dia. 9 ws" | 3ua" | 2uer mer ms" 4 holes 7/32" dia 9a ms | 3us | our mer me 4 holes 5/32" dia. 4ax | isne" | sane | sane me me 4 holes 5/82" dia % r 3.318" ve ve" 4 Uholes 1/4" dia, 35a | ae | ous Bua" aa aa" i5.130 . ELECTRICAL MACHINE DESIGN - Table 5,10. E-E Stampings [(Fig. 5.94(c)] Dimensions Remarks No. A a [| c¢ pd [| £ a ve ver ae ve } used in pair var r va vs" vs" J 2B | se | ize | 2ane wns" ane" 12¢ se | 178° 2 sis" sie" 20 se | ave | 156 38" 38" a 79 | 309" | 1313" 0475" 0475" 718/32" dia holes a7 | 309 | 0.585" 0.485" 0.475" 48/82" diaholes 1-816" dia hole A square section is normally used for the central limb, i.e. the depth of the core is made equal to the width of the central limb or width of central limb A = Aj, A standard stamping giving a width A nearly equal to the value caleulated above may be used. 5.61.2 Winding Design Current in the primary winding J, = VA/n. ‘The efficiency of small transformers varies from 80 to 96 per cent. Area of primary winding conductors a, = 1/8, mm®, where 6, is the current density in the primary winding conductors in A/mm?. A value of 2.3 A/mm? may be used. Enamelled round conductors are used for the windings of small transformers. A standard size of conductor is selected by referring to relevant tables given in chapter 23. Turns in primary winding 7, = V,7,. Current in secondary winding I, = VA/V Area of secondary winding conductor a, = 1/8, mm*. When calculating the number of secondary winding turns, an allowance of 5 percent extra turns is made to compensate for the voltage drop in the windings. Secondary winding turns T, = 1.05 V,T.. 5.61.3 Window Area Space is required in the window for : (@ primary winding (éi) secondary winding (iii) insulation and the former (bobbin) on which the windings are supported. Tpty Space factor Teas pace Space required for primary winding = Space required for secondary winding = ctor Space factor S = 0.8 (d/d,) where d = diameter of bare conductor and d, = diameter of insulated conductor. The space required for insulation and former is estimated as 20% of that required for the windings.DESIGN OF TRANSFORMERS 5.131 + Window area required A,, = 1.2 (window area required for primary and secondary windings). + It should be checked that the stamping used, gives a higher value of window area than the value calculated above. Design Problem IV. Design a single phase transformer to be connected to a 230 V, 50 Hz supply. The transformer is to deliver 3 A at 50 V. Solution, Core. Volt ampere rating of transformer = 50 x 3 = 150 VA. From Table 5.7, turns per volt T, = 40. 1 1 = Gaa fT, 444x504 Taking a flux density of 1.0 Wb/m?, -3 Net iron area of core, A, = Ga, HB AIO Gross core area A, = 1.125 x 10°/0,9 = 1.255 x 10° mm’. Taking a square section for the central limb. ‘Width of central limb A:= 12.5 x 10° = 35.4 mm?*, Primary winding. The efficiency of this transformer is assumed as 92%. VA 150 ©, ‘Wh = 1.125 m Wb. m? = 1.125 x 10° mm?, Py — = =0/ vimary winding current J, = 507°" Gap, o95 = 0-7 Taking a current density of 2.3 A/mm? Area of primary winding conductor a, = ae = 0.309 mm?, Diameter of bar conductor = 0.626 mm. Using enamelled conductors. From Table 23.7, the nearest standard conductor has bare diameter = 0.63; mm. The diameter of insulated conductor is 0.707 mm. Space factor for primary winding = 0.8 (0.63/0.707)? = 0.635. Area of primary conductor used a, re (0.63)? = 0.312 mm?. Number of primary winding turns 7, = V,T, = 230 x 4 = 920. Tray 920x012 S 0635 Secondary winding. Secondary winding current J, = 3 A. Window space required by primary winding = = 452 mm?, ‘ 3 Area of secondary winding conductor a, = 55 = 13 mm, Diameter of bare conductor = 1.285 mm. Using enamelled conductor for secondary winding. ‘The nearest standard conductor has bare diameter = 1.32 mm. diameter of insulated conductor = 1.42 mm. Space factor for secondary winding = 0.8 (1.32/1.42)? = 0.69. Area of secondary winding conductor a, = i (1.32)? = 1.37 mm?, Number of secondary winding turns T, 05 V,T, = 1.05 x 50 x 3 = 210,5.132. : ELECTRICAL MACHINE DESIGN Tra, _ 210x137 Wind ired b; dary winding = indow space required by secondary winding = “5° oy = 417 mm? Stamping Size. Total window space required A,, = 1.2 (space required for primary and secondary winding) 1.2 (452 + 417 = 1045 mm?) ‘We have now to select a stamping which gives : width of central limb A = 35.4 mm = 1.39" and area of window A, = 1045 mm? = 1.62 sq. inch. Using a combination of E and J stampings. From Table 5.8, selecting No. 16 A= Fad c= 3/4", D = 3/4", B= 3/4" From Fig. 5.94(a), width of window W,, Bod 2e = 9/4", and height of window -H,, = C- 2B = at : 1 .. Area of window provided A, = W,, x H, = 3/4” x 25 = 1.69 sq. inch. This is more than the required window space. z+ Width of central limba = uz = 38.1 mm. EXERCISES What are the classifications of transformers ? Distinguish between them. Draw a schematic of a three phase core type transformer as well as a shell type transformer alongwith respective phasor diagram. Compare between three phase star and delta connections for transformer windings. ‘What is a three winding transformer ? What are the categories of transformers used in power systems ? Why do you require to use special shaped core laminations in power transformers ? Why interleaving is done in magnetic cores ? Explain the term “Cross-fluxing” in transformers. & Which type of material is preferred for core laminations ? What is the advantage of using mitred joints in core construction ? Pree pe Which types of windings are commonly used in shell as well as in core type transformers ? 10. Why transposition is done in helical windings ? 11. Write short notes on (i) Cross-over windings (ii) Dise windings. 12, What is CTC winding ? Why it is preferred ? 13, What are the different methods of cooling of transformers ? 14, What physical and electrical properties need attention for insulating oil of transformer ? 15, Write short notes on (i) transformer bushings (ii) transformer tappings. 16, What are the functions of conservator and breather in transformer ? 17, How do you give incipient fault protection for transformers ? 18 Develop the output equation for a single phase as well as a three phase transformer.DESIGN OF TRANSFORMERS 5.133 19. 20, BERBER BEE s 3. ). Optimise the transformer design from the point of view of (i) minimum cost (ii) minimum loss. Write a comparative note on “choice of flux density” in power transformers. Why should be the current densities of two windings in a transformer should be equal ? ‘What are the methods by which you could provide protection from surges for a transformer ? ‘What is window space factor ? Find the width of window for optimum output of transformer. Derive the expression for total leakage reactance of a transformer referred to primary side. Derive an expression of Regulation for a single phase transformer. What type of mechanical forces are developed in transformer windings ? Calculate the no-load current and magnetising voltamp for a single phase transformer. Write a short note on “change of parameters of a transformer with change of frequency.” Why do transformer with a greater copper to iron loss ratio is less capable of switching overloads Calculate the core and window areas required for a 1000 kVA, 6600/400 V, 50 Hz, single phase core type transformer. Assume a maximum flux density of 1.25 Wb/m? and a current density of 2.5 A/mm?. Voltage per turn = 30 V, and window space factor = 0.32. (Ans. 108 x 10? mm? ; 83.4 x 10° mm] A8 phase, 50 Hz, oil cooled core type transformer has the following dimensions : distance between core centres = 0.2m; height of window = 0.24 m; diameter of cireumscribing circle = 0.14 m. The flux density in the core is 1.25 Whim? and the current density in the conductors is 2.5 A/mm®, Estimate the kVA rating. Assume a window space factor of 0.2 and a core area factor = 0.56. The core is 2 stepped. (Ans. 16.5KVA] Determine the dimensions for core and yoke for a 5 kVA, 50 Hz, single phase core type transformer. A rectangular core is used with long side twice as long as short side. The window height is 3 times the width. ‘Voltage per turn is 1.8 V; space factor 0.2; current density 1.8 A/mm?; flux density 1 Wb/m?. Ans. A, = 8100 mm? ; core 135 x 67.5 mm? ; A, = 15.6 x 10° mm? ; window 72.5 x 217.5 mm] Estimate the main dimensions including winding condictor areas of a 3 phase delta/star core type transformer rated at 300 kVA, 600/440 V, 50 Hz. A suitable core with three steps having a circumscribing circle of 0.25 m diameter and a leg spacing of 0.4 m is available. The emf per turn is 8.5 V. Assume a current density of 2.5 A/mm?, a window space factor of 0.28 and a stacking factor of 0.9. {Ans. T, = 776, 4, =6 mm? ; T, = 30, a, = 157.5 mm? ; window 0.15 x 0.45 m? ; core = 37.5 x 10° mm] Determine the dimensions of the core, the number of turns and the cross-sectional area of conductors in. the primary and secondary windings of a 100 kVA, 2200/480 V single phase core type transformer to operate at a frequency of 50 Hz, assuming the following data; approximate voltage per turn, 7.5 V; maximum flux density, 1.2 Wb/m?; ratio of effective cross-sectional area of core to square of diameter of circumscribing circle, 0.6 ; ratio of height to width of window, 2 ; window space factor, 0.28 ; current density, 2.5 A/mm?, (Ans, 28.2 10° mm? ; 294, 64 : 18.2 mm? ; 83,2 mm?] Calculate the dimensions of the core the number of turns and the cross-sections of the conductors for a 100 kVA, 230/400 V, 50 He single phase shell type transformer, assuming : ratio of magnetic and electric loadings (ie. lux and secondary rimf at full load), 480 x 108 ; maximum flux density 1.1 Wb/m? ; current density, 2.2 x 10° A/m? ; window space factor, 0.3 ; ratio of depth of stacked core to width of central limb, 2.6 ; ratio of height to width of window, 2.5 ; stacking factor 0.9. (Ans. 185 x 340 mm? ; 230, 40 ; 19.8 mm?, 114 mm?) ‘Two single phase transformers with AN cooling having linear dimensions in the ratio: 1, are designed to work at the same current density, flux density and frequency. Calculate relative ratings, losses and total weights per kilovolt ampere of the two transformers. Comment briefly on the temperature rise of the first transformer for increasing values of x, assuming that the second transformer operates with permissible temperature rise on full load. WAns.xt: 15x35 12-4; 1)5.134 ELECTRICAL MACHINE DESIGN ‘A750 kVA, 6600 V, 50 Hs, three phase delta/star core type transformer has the following data ; width of 1. winding, 30 mm ; width ofh.v, winding, 25 mm ; width of dust between h.v, and Lv. windings, 16 mm ; height of windings, 0.4 m ; length of mean turn 1.6 m, h.v. winding turns, 217, Estimate the leakage reactance of the transformer referred to h.v. side. Estimate the per unit regulation of the transformer at full load and 0.8 power factor lagging ifthe resistance per phase referred to the h. 0.82. tAns. 2.920, 0.03471 Derive an expression for the leakage inductance per limb of the windings ofa single phase transformer referred to the primary side in terms of the relevant quantities, and calculate this inductance in a given ‘case where the windings have a mean length of turn 1.5 m and each winding has a radial depth of 30 mam. ‘The radial clearance between windings on the same limb is 85 mm and the axial height of each winding is 0.6m, The primary winding consists of 240 turns per limb. (Ans. 9.98 mH] Inatransformer having a 10: 1 turns ratio, the copper loss in the primary winding is 15 per cent less than that in the secondary winding. The resistance per turn of the primary is 0.00394 Q and there are 1485 ‘more turns in the primary than in the secondary. Caleulate the resistance of secondary winding. (Ans. 0.0675 0] ‘A.100KVA, 2000/40 V 50 He, single phase shell type transformer has two full h.. coils, one full. coil and two halflv. coils. Calculate the value of average axial force acting on the end coils under short circuit at full voltage ifits leakage reactance is 0.035 p.u. Length of mean turn 1.5m, width of coils, 0.12 m ; hv. winding turns, 200 ; pu. resistance of transformer, 0.01. (Ans. 35.2 x 10° N] ‘A.220/110 V, 1 kVA, 50 Hz single phase transformer has a core with a uniform cross-sectional ares of 2500 mm, an effective magnetic core length of 0.4 m and a core weight of 8 kg. Ifthe core is worked at ‘a maximum flux density, B, of 1.2 Wb/m?, the corresponding magnetizing force, H, is 200 A/m and the specific core loss is 1.0 W/kg, determine (a) the transformer no load current when the h.v. is fed at 220 V. and (b) the corresponding magnetizing reactance and the equivalent shunt resistance to represent the core loss. fAns. 0.175 A ; 1287 0 ; 6600.2) ‘A300 VA, 6600/40 V, three phase delta/star core type transformer has a maximum flux density of 1.35 Wh/m? and the total weight of core is 650 kg. The magnetizing VA/kg and the iron loss/kg corresponding to 1.35 Wh/m? are 30 and 2.5 W respectively. Calculate the no load current ifthe mmf required for joints is 2.5 percent of that for iron. (Ans. 0.695 A per phase] ‘The tank of a 1250 kVA natural oil cooled transformer has the dimensions length, width and height as 1.55 m x 0.65 m x 1.85 m respectively. The full load loss is 13.1 kW. Find the number of tubes for this transformer assuming ; W/m? °C due to radiation = 6 ; W/m?—*C due to convection = 6.5 ;improvement in convection due to provision of tubes = 40 per cent ; temperature rise = 40°C ; length of each tube = 1m; diameter of tubes = 50 mm. Neglect the top and bottom surfaces of the tank as regards cooling. {Ans. 190 tubes} With 100 A taken from the terminals ofa certain transformer connected to a 25 Hz primary supply circuit, the resulting total reactive drop is 5 per cent. Calculate the percentage reactive drop ifthe same transformer is rewound with 15 per cent more turns in both windings and connected to a 60 Hz circuit with a load drawing 110 A from secondary winding. Assume the terminal voltage and the dimensions of all leakage paths to remain unaltered. (Ans. 17.5 per cent}