The Transformer Trainer Tt179

THE TRANSFORMER TRAINER TT179

CONTENTS Introductory Survey Specification Operating Notes Student Assignments 1. Familiarization and Ration test 2. Transformer Polarity 3. Series and Parallel connections 4. The Magnetic Circuit 5. Phasor Diagram on No-load 6. No-load Losses 7. Open-circuit test 8. Short-circuit test – transformer efficiency 9. The Transformer efficiency 10. Complete Equivalent Circuit – Voltage Regulation 11. Current Transformer Installation and Maintenance Appendix 1 – The Magnetic Circuit

The Health and Safety at Work Act 1974 We are required under the Health and Safety at Work Act 1974, to make available to users of this equipment certain information regarding its safe use. The equipment, when used in normal or prescribed applications within the parameters set for its mechanical and electrical performance, should not cause any danger or hazard to health and safety if normal engineering practices are observed and they are used in accordance with the instructions supplied. If, in specific cases, circumstances exist in which a potential hazard may be brought about by careless or improper use, these will be pointed out and the necessary precautions emphasized. While we attempt to give the fullest possible user inform- ation in our handbooks, if there is any doubt whatsoever about any aspect relating to the proper use of this equipment the user should contact the Product Safety Officer at Feedback Instruments Limited, Crowborough. Component replacement Although this Feedback manual was believed to be correct at the time of printing, components supplied may differ slightly from those described. We endeavour to improve our equipment continually by incorporating the latest developments and components, even up to the time of despatch. If it is practicable we include such new or revised information in the manual. Whenever possible, replacement components should be similar to those originally supplied. These may be ordered direct from Feedback Instruments Limited or its agents by quoting the following information: 1. 2. 3. 4.

Equipment type Equipment serial number Component reference Component value

Standard components can often be replaced by alternatives available locally. COPYRIGHT NOTICE © Feedback Instruments Limited 1982 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of Feedback Instruments Limited.

Introductory Survey The transformer effect was first observed by Michael Faraday in 1831 when experimenting with two coils wound on an iron core. He found that on applying a voltage to one of the coil an voltage was induced in the second coil although there was no electrical connection between them.

Fig. 1 It was not until the 1880’s that transformers were used for the distribution of electrical power. They are now among the most widely used of all types of electrical machine. Transformers range in size from the very large 3-phase types used in power stations to step up the generator voltage to the level required for transmission over long distances, down to the small single-phase transformers used in audio frequency work, which may weigh only a few grammes.

Fig. 2a. 240 megavolt-ampere power transformer The transformer as an Electrical Machine A machine may be defined as an apparatus which accepts energy in some form and transfers that energy, possibly in an alternative form, to a load. Mechanical machines include pulleys, gear drive and various types of engine. Some typical electrical machines are described below:

Fig 2b. 2.4 volt-ampere power supply transformer A synchronous generator, driven from a mechanical source of power at a fixed rotation speed will supply electrical power at a fixed frequency. A direct-current motor connected to an electrical power supply will produce torque over a range of shaft speeds.

A frequency converter accepts power at a given frequency an supplies power at another set frequency. A power transformer accepts electrical power at a given a.c voltage and supplies electrical power at a higher or lower a.c voltage. It may also be used electrically to isolate one electrical circuit from another, possibly using the same input and output voltage. A transformer is analogous to a gear drive, which changes shaft speed, usually by a fixed ratio, without changing the form of energy transferred from the input to the output shaft. The transformer provides an excellent introduction to the general study of electrical machines, since its construction is relatively simple and having no rotating parts its assembly and testing is not complicated. Despite its relative simplicity it can be used as a means of understanding the significance of the magnetic circuit, losses, efficiency, temperature rise, phasor diagrams and equivalent circuits. All of these concepts are relevant to the study of motors, generators and other electromagnetic machines. Transformer Principles A transformer in its basic form consists of two coils fitted over a magnetic steel core, as shown in fig 3. If one of these coils, which we will call the primary winding, is connected to an a.c power supply it will be found that an alternating voltage will also be present across the terminals of the other, or secondary winding. If we connect the secondary winding to an electrical load it will supply current to it and a corresponding current will be drawn into the primary from the supply. The transformer therefore transfers a.c voltage and current from the primary circuit to the secondary with no electrical connection between them. This is a useful facility but, more important, by having a different number of turns on the secondary to those on the primary we can change, or transform the secondary voltage to whatever level we require. In other words, we can step up or step down the mains supply voltage, but the product of voltage and current, or volt-amperes, in the primary is the same as the volt-amperes in the secondary, apart from some losses within the transformer itself. Although there is no electrical connection between the two windings, there is a magnetic flux which links them through their common core and which alternates at the same frequency as the a.c mains supply. It is this changing flux which causes a voltage to be induced in the secondary winding.

The voltage per coil turn will be the same for the primary and secondary windings. To take an example, consider a transformer which has 500 primary turns and 250 secondary turns. If we apply 100 volts to the primary winding the volts per turn will be 100 ÷ 500 = 0.2. From this we can say that the secondary voltage will be 0.2 x 250 = 50 volts. This shows that the voltage ratio of the transformer is the same as the turns ratio so that where

V2/V1 = T2/T1 V1 = primary voltage V2 = secondary voltage T1 = primary turns T2 = secondary turns

If now the secondary winding is connected to an electrical load, .current will flow from the secondary to the load and this will cause current to be drawn from the mains supply into the primary. Ignoring transformer losses at this stage we can equate the primary and secondary voltamperes V 1 I1 = V 2 I2 Where I1 = primary current I2 = secondary current

Fig. 3 Therefore from this equation and that relating to the voltage and turns ratio we can write

V2 T2 I = = 1 V1 T1 I2

for an ideal transformer

Transformer Construction There are three principal forms of construction which cover all except very specialised transformers. These are the core, shell and toroidal types. Single-phase Core Type In this construction U-and I-shaped steel laminations are built up to form a rectangular core with the windings supported on two of its limbs as shown in fig 4.

Fig. 4 It is usual for each winding to be split so that each limb supports half of the primary turns and half of the secondary turns, which are wound concentrically. The low voltage windings are placed nearer to the core to reduce the voltage gradient across the insulation (see notes under Transformer Windings' later in this review). The U- and I-laminations are fitted alternately to distribute the small air gaps and the core structure is held rigid by bolts or rivets. Large three-phase power and distribution transformers are generally of the core type, whereas small single-phase transformers use shell-type construction. Single-phase Shell Type The laminated magnetic core in this case has a central limb which supports all the windings and two outer limbs to complete the flux path, as shown in fig 5.

Fig. 5 This construction is most widely used for small power transformers as it is compact and economical to produce. The magnetic core is built up from either U and T-sections or E and 1-sections, fitted alternately. Single-phase Toroidal Type This has a ring-shaped core, often formed from helically wound steel strip, resin bonded to form a rigid structure. The primary-and secondary are wound over the whole of the ring surface with the low-voltage winding nearer to the core, as shown in fig 6.

Fig. 6 Because of the setting up time and special coil winding equipment which is required, this type of transformer is only economical when made in fairly large production quantities. However it has two features which may sometimes cause it to be preferred to conventional types: there is no flux leakage outside the magnetic path and it has an inherently low profile, or pancake shape, which can be advanTaQeOus where space is limited. Transformer Windings Although a designer's first concern is with the electrical requirements of the windings he must also 'ensure that they have adequate mechanical strength and that the heat developed

internally is carried away to the outside air or other cooling medium. Fig 7 shows some of the winding arrangements used in large and small transformers.

Fig.7 In large power system transformer the copper conductors will usually be of rectangular cross section, covered with impregnated paper or fabric as an insulant. The coils are wound on the resin bonded mica, glass fibre or pressboard cylinders. Disc and cylindrically wound coils, fig 7a and b) have their turns distributed so that cooling air, gas or oil can circulate within the winding. Rigid separators prevent voltage breakdown between adjacent layers and give the mechanical support needed to withstand the large forces which can be applied under fault conditions. In small power transformer, fig 7c enamel-covered round copper wire is used for the windings. This may be wound on a bobbin made from a resin bonded fabric or a thermally stable plastic material such as Nylon. Alternatively the coils may be wound directly on to a sleeve consisting of several layers of impregnated paper. To prevent voltage breakdown within the winding a paper or fabric sheet is inserted between layers of turns as the winding proceeds. Additional insulation is placed between the primary and secondary windings. To isolate further the two sets of windings, sheet copper can be fitted between the primary and copper can be fitted between the primary and secondary to act as screen which is then connected to an external earth. This must not be closed upon itself as it would then act as a shorted turn of low resistance which would cause the transformer to over heat. Fig 8 gives an extract from a wire manufacturer’s table, shown the relationship between wire diameter, current rating and electrical resistance.

Conductor diameter

Sectional Weight area per km 2 mm kg Equivalent inch

Nom mm

Max mm

Min mm

1.600

1.616

1.584

0.0630

2.011

17.87

1.500

1.515

1.485

0.0591

1.767

15.71

1.400 1.320 1.250 1.180 1.120 1.060 1.000 0.950 0.900 0.850 0.800 0.750 0.710 0.670 0.630 0.600 0.560 0.530

1.414 1.333 1.263 1.192 1.131 1.070 1.010 0.960 0.909 0.859 0.808 0.758 0.717 0.677 0.636 0.606 0.566 0.536

1.386 1.307 1.237 1.168 1.109 1.049 0.990 0.940 0.891 0.841 0.792 0.742 0.703 0.663 0.624 0.594 0.554 0.527

0.0551 0.0520 0.0492 0.0465 0.0441 0.0417 0.0394 0.0374 0.0354 0.0335 0.0315 0.0295 0.0280 0.0264 0.0248 0.0236 0.0220 0.0209

1.539 1.368 1.227 1.094 0.9852 0.88255 0.7854 0.7088 0.6362 0.5675 0.5027 0.4418 0.3959 0.3526 0.3117 0.2827 0.2463 0.2206

13.69 12.17 10.91 9.722 8.758 7.845 6.928 6.301 5.656 5.045 4.469 3.928 3.520 3.134 2.771 2.514 2.190 1.961

Nominal Current resistance at 20oC rating at 4.65 a Per Per kg mm-2 metre ohms amperes ohms 0.00857 0.4799 9.649 5 0.00975 0.6211 8.217 7 0.01120 0.8181 7.158 0.01260 1.035 6.364 0.01405 1.288 5.706 0.01577 1.622 5.085 0.01750 1.998 4.581 0.01954 2.491 4.103 0.02195 3.144 3.652 0.02432 3.860 3.296 0.02710 4.791 2.958 0.03038 6.022 2.639 0.03430 7.675 2.337 0.03903 9.936 2.054 0.04355 12.37 1.841 0.04890 15.60 1.639 0.05531 19.96 1.449 0.06098 24.26 1.315 0.07000 31.96 1.145 0.07814 39.85 1.026

Fig. 8 t4.65 A mm-2 is equivalent to 3000 amperes per square inch. Local conditions may necessitate some variation in this figure. The conductor sizes in italic print are preferred sizes and should be used wherever possible. Classification

Class A Class B

Max. continuous operating temperature 105oC 130oC

Typical Materials

Typical Dielectric Strength

Impregnated cotton, silk, paper Bonded mica, glass fibre, asbestos

4 to 18kV/mm 8 t o 30kV/mm

Class H

180oC

Silicone and epoxy resisns, fluorochemicals, e.g polytetrafluoroethylene bonded mica.

12 t o 30kV/mm

Fig. 9 The temperature at which a winding can operate is largely determined by the properties of the insulation materials used. Manufacturers work to a classification system from which the list given in fig 9 is extracted. In this table the dielectric strength is the maximum voltage gradient that can be applied to an insulation material without breakdown, expressed as volts per mm thickness of material. Measured values of dielectric strength are dependent on the thickness of the sample material and the test conditions. The figures given are for guidance only. Transformer Core The magnetic core in large and most small transformers is built up from steel laminations normally using two of the shapes shown in fig 10. Steels used for transformer cores are special alloys of Iron and Carbon with the addition of a small percentage of Silicon, or in some special cases, of Nickel - typically a coldrolled Carbon steel containing up to 3% Silicon would be used. The subject of power loss in the transformer core is treated in greater detail in Assignment 6 of this manual but briefly we can say that Silicon steels are used to reduce the hysteresis component of total core loss and a laminated structure is employed to reduce the eddy- current component. The lamination thickness is dependent on the power supply frequency - for 50 or 60Hz applications this would be typically 0.3mm (O.014in). Each lamination has an insulating coating applied to one surface. Lamination stacks for large transformers are held rigid by insulated bolts which pass through the core and an insulated clamping frame. I n the smaller transformers the core is riveted or simply held together by an insulated clamp. Impregnation To assist the transfer of heat from the inner parts of the winding and to improve the overall insulation, small transformers are taken through an impregnation process after assembly of the core and windings. The impregnant is a synthetic resin which may contain a solvent to thin the resin and improve penetration. The impregnant must have good dielectric strength, good mechanical strength when set and be capable of acting as a barrier to

moisture and dirt. It should also be a relatively good conductor of heat and of course be capable of operating well above the transformer running temperature.

Fig. 10 In one typical impregnation process the transformer is first dried in an oven, then cooled under vacuum in a tank. The transformer is then completely immersed in the impregnant for up to ten minutes, after which it is removed from the tank and allowed to drain and dry off. A baking period follows, terminating at the cure temperature of the resin. In addition to the impregnation process, transformers which are to operate in tropical conditions are given an envelopment dip. This provides a relatively thick coating over the windings and core to make doubly sure that there is no moisture penetration and to resist fungus growth. Winding Connections Conventional transformers are double wound the primary and one or more secondaries are separate with no electrical connection between them. To enable the primary to be connected to a number of different mains supply voltages, or to allow some variation in the secondary terminal voltage, it is usual to provide either tapped connections to the main windings or to have separate tapping coils which may be connected so that the voltage across a particular winding can be increased or decreased. Typical tapped windings for small power transformers are shown in fig 11. In dealing with tapped windings we should keep in mind that the volts per turn on all windings is constant. In fig 11 a the primary winding is tapped so that it can accept mains supply voltages of 240, 200, 120 and 100 volts. Note that if the volt-ampere rating of the transformer is to be the same for all connections, the primary current on the 100 or 120-volt tappings will be approximately twice that at 200 or 240 volts and therefore the wire size must be increased proportionately. In fig 11 b the secondary has tapping points, so that when operating from a fixed mains supply the transformer can provide an output of 10, 15 or 20 volts.

In fig 11c the primary consists of two separate coils which may be connected in series for a 240-volt supply or in parallel for a 120-volt supply. This arrangement is more economical than the tapped primary of fig 11 a, since irrespective of the connection used each primary winding need only be designed to carry half the rated volt-amperes of the transformer. I n fig 11 d both the primary and secondary have two separate coils, giving a choice of mains supply voltage and of secondary output voltage. I n this example the transformer can accept a mains supply of 240 or 120 volts and can provide an output of 20 volts at 1 A or 10 volts at 2A. Other Transformer types Auto-transformers In some transformer applications it is not necessary to provide separate primary and secondary windings. In this case a single tapped winding is used which will act as the equivalent of a double wound transformer, as shown in fig 12.

Fig. 11

Fig. 12 In this figure we have taken as an example a step-down transformer whose secondary voltage is 1/3 that of the primary. For the double wound transformer, fig 12a. Vab = 3 Vcd and Iab = 1/3 Icd also

Tab = 3 Tcd

where Vab = primary volts Vcd = secondary volts Iab = primary current Icd = secondary current Tab = turns on winding ab Tcd = turns on winding cd I n the auto transformer we fix the tapping point to give the same turns ratio as in the double wound version, fig 12b. Then Vab = 3 Veb Iab = 1/3 Ieb and

Tab = 3 Teb

where Veb = secondary volts Ieb = secondary current

Tab = turns on winding ab Teb = turns on winding eb The auto wound transformer has one important advantage over the double wound version. The currents lab and leb flow in opposite directions through the common winding eb. This part of the winding therefore carries 3A - 1 A = 2A, whereas in the double wound version the whole primary carries 1 A and the secondary carries 3A. There is, therefore, a substantial saving in copper, leading to a reduction in cost and weight. Despite this, the auto-transformer is not very V widely used since it provides no isolation between the supply and the output. Three-phase Transformers If three identical single-phase transformers have their primaries connected to a three phase supply, as shown in fig 13a, each transformer secondary will produce the same output voltage amplitude but the three will be displaced in phase by 1200 from one another in the same way as the supply line voltages. Instead of having three separate transformer cores we can mount their windings on a single three-phase core, as shown in fig 13b. This is a three-phase version of the coretype transformer and is widely used for power system applications. The connections to the primary and secondary windings can be made in a variety of ways. Three of these are shown in fig 14. I n this diagram the high-voltage windings are marked A, B, C and the low-voltage windings a, b, c. If a 3-phase transformer has the outer terminals of its high- voltage or low-voltage windings connected to the line, and the inner terminals connected together, that winding is said to be star-connected and the common connection point is called the neutral or star point, fig 14a. If the winding terminals are connected in sequence across three-phase line pairs they are said to be mesh or delta- connected and there is no neutral or star point, fig 14b. For the star/star and delta/delta connections shown in fig 14 there is no phase displacement between the terminals A2 to a2, B2 to b2 and C2 to C2. In one of the many alternative methods of Connection the high voltage winding is deltaconnected and the low voltage winding is star-connected as shown in fig 14c. This produces a 300 phase shift between terminals A2 to a2, B2 to b2, C2 to C2. It is important in power system work to establish the correct phase relationship between transformers which are to be connected in parallel.

Fig. 13

Fig. 14 (continued overleaf)

Fig. 14 (continued) Current Transformers In power system protection and metering circuits it is necessary to measure the currents in high voltage lines and convey this information to a control or switching station. It would be obviously impracticable to connect the measuring equipment directly to the high voltage line, and for this reason a transformer is interposed between the line and the measuring circuit, as in fig 15a. Apart from safety considerations this has the advantage that by suitable choice of turns ratio the current in the low-voltage winding may be made many times lower than the high voltage line current. A good current transformer will have adequate insulation to isolate the high-voltage and low- voltage circuits, it will have a linear high voltage to low voltage current relationship (low current ratio error), and the phase angle of the measured current should be close to that of the line current (low phase angle error). In power system work the current transformer may consist of a single heavily insulated bar conductor which is connected in series with the line. This bar will pass through the centre of a toroidally wound secondary winding, as shown in fig 15b. When the secondary terminals are not connected to the measuring circuit they should be shorted out, otherwise dangerously high voltages can be produced across the terminals due to the high turns ratio involved. Audio-frequency Transformers Applications involving audio-frequency transformers include microphone amplifiers, power amplifiers and line matching in telephone systems. Transformers used for good quality significant attenuation of any signals within tat band. For telephone operation a frequency range of 100 to 3000Hz may be adequate.

The transformer core is usually of shell or toroidal form and may be constructed from a high permeability nickel-iron alloy in the form of laminations 0.1 mm (0.04inl thick or from foil of thickness down to 0.003mm (0000125in). Audio frequency transformers are physically small - the dimensions of a typical power amplifier output transformer being 75 x 65 x 90 mm (3.0 x 2.6 x 3.6inl and those of a micro- phone transformer being 21 x 20 x 18mm (0.84 x 0.80 x 072inl.

Fig. 15

SPECIFICATION The Transformer Trainer consists of a free-standing Measurements Console and a separate Dissectable Transformer. The front panel of the Measurements Console contains AC voltage and current meters and has a large mimic diagram from which connections are made to the panel meters, Dissectable Transformer, and an external Load Unit. All front panel connections are made to 4mm sockets using the leads and links provided. An external Power Supply Unit is connected to the rear of the Measurements Console via a 3-pin plug and flying lead. The equipment is designed to provide maximum safety without detracting from its inherent convenience in use. Measurements Console This is the central unit to which the Dissectable Transformer, Power Supply and Load Unit are connected. Safety Features Transparent hinged cover: raising the cover operates a microswitch to disconnect the power supply to the test circuit and dissectable transformer, giving protection against contact with live terminals. Prominent main 'power off' pushbutton which disconnects power to the test circuit and dissectable transformer. Overload cut-out which disconnects the power supply if the primary current in the dissectable transformer exceeds its full load rating. After the fault is cleared the reset button restores the power. Control circuit which requires positive operation of the main 'power on' pushbutton to restore the supply after any disconnection. Front Panel Meters and Switches Primary and secondary switched ammeter ranges: 0·1 A, 0-5A. Primary switched voltmeter ranges: 0-35 volts, 0-140 volts, 0-280 volts. Secondary switched voltmeter ranges: 0-70 volts, 0-140 volts. External wattmeter 3-position switch : connects wattmeter to primary circuit, secondary circuit or 'off' (see under 'Ex- ternal Units'). Power-On illuminated pushbutton switch: makes connection between external power supply and test circuit provided the safety cover is down and the overload cut-out is not in the tripped condition. Power-Off pushbutton switch: disconnects external power supply from test circuit. Mains supply to Measurements Console 100 to 125 volts, 50/60Hz, 160mA

or

190 to 250 volts, 50/60Hz, 160mA

For mains supply to external variable AC power supply unit see under ‘External Units’ Dissectable Transformer This is a free-standing unit consisting of a two-part laminated and bonded core, a primary coil and two secondary coils, all wound on separate bobbins. The transformer can be taken apart by operating quick-release fasteners attached to the clamp and base. When the clamp is released the core and coils can be removed. Component Parts The Dissectable Transformer contains the following components: Bonded E-section lamination package, overall dimensions 30 x 22 x 133mm (1.20 x 0.88 x 5.32in) Bonded I-section lamination package, overall dimensions 30 x 89 x 133mm (1.20 x 0.36 x 5.32in) Primary winding : 270 turns of 0.90mm wire, centre tapped, with flying leads j terminated by 4mm plugs. Two Secondary windings: each 148 turns of 0.56mm wire, with flying leads terminated by 4mm plugs. Two non-magnetic spacers Spare unwound bobbin Clamping frame Mounting base Electrical Specification Continuous full-load rating : 240 volt-amperes, 50/60Hz Primary voltage : 120 volts, centre tapped for 60 volts. Secondary 1 voltage and current: 60 volts, 2A. Secondary 2 voltage and current: 60 volts, 2A. Regulation, (percent of open-circuit secondary voltage) : 15% Efficiency at full load: 83% Maximum ambient temperature: 70°C

Overall dimensions: 140 x 140 x 145mm (5.6 x·5.6 x 5.8in) Additional Items Supplied with TT179 The items listed below are supplied in a plastic storage box. Mercury/glass thermometer - range 0 to 120°C Connecting leads and links : For interconnection to sockets on front panel of Measurements Console, (all terminated in 4mm plugs): QTY COLOUR 4 Yellow 4 Yellow 2 Yellow 2 Yellow 1 Blue 1 Orange 2 Black 1 Red 1 Pink 6 White 1 Green

LENGTH mm 65 115 155 190 915 915 915 915 1700 Connecting bar Earthing lead

One three-core lead for connecting the external power supply to the rear panel of the Measurements Console. Terminated at the power supply end by 4mm plugs and at the Console end by a Belling Lee 3-pin plug, type 1722/P L. Spare fuse links - Two 160mA anti-surge 20 x 5mm, Beswick type TDC123 External Units These units are not provided as part of the TT179 Transformer Trainer but, with the exception of the oscilloscope, are manufactured and supplied by Feedback Instruments Ltd. Load Unit LU 178 A free-standing unit with a ventilated rear section containing two separate banks of resistors, individually switched, each bank ranging from 30 ohms at 2.5A to 300 ohms at 0.17 A. The banks are separately fused and have flying lead connections terminated in 4mm plugs. Power Supply PS189 Mains operated unit which provides a continuously variable AC output of 0 to 135 volts at 5A. The unit can also be switched to provide a rectified DC output of 0 to 120 volts at 5A.

The unit contains an isolating transformer, over-load cut-out and an indicating meter in addition to a variable output transformer. Mains supply required 100 to 125 volts, 50/60Hz, l0A. or 190 to 250 volts, 50/60Hz, .5A. Electronic Wattmeter EW604 Mains operated unit which has eight voltage settings, ranging from 5 to 1000 volts and eight current settings, ranging from 0.05 to l0A. Measures DC power and AC power at frequencies up to 20kHz. Mains supply required 100 to 125 volts, 50/60Hz or 190 to 250 volts, 50/60Hz Oscilloscope A twin-beam oscilloscope is required having both time base and X-Y display, preferably with a line trigger facility to lock the time base to the 50 or 60Hz supply. A suitable oscilloscope can be supplied by Feedback Instruments Ltd. as part of a laboratory package.

Operating Notes There are two parts to the Transformer Trainer TT179: the measurements console and the dissectable transformer, The following notes explain the purpose of the Trainer - how it is operated and the procedure for assembling the transformer. Power for the transformer is provided by the Feedback Power Supply PS189 while the Feedback Load Unit LU 178 dissipates the output power from the transformer. Included in the notes are instructions on how to operate these two units with the TT179. However if these units are not available, the notes set out the types of power supply and load resistors required. Transformer Trainer TT179 1. The Measurements Console All the units are connected to the measurements console as shown in Fig 16. It contains the power-on/power-off switches, voltage and current meters, mimic diagram as shown in Fig 17. On the mimic diagram are set the patching sockets, overload protection and safety circuits, When the power-on button is pressed it will be illuminated provided the safety cover is down and the overload cut-out is in its normal closed condition. The main supply is disconnected when either a) The power-off button is pressed b) The overload cut-out operates c) The safety cover on the TT179 is lifted d) The main supply is interrupted Power is only restored by removing the cause of disconnection and pressing the power-on button. The four a.c meters used for measuring the primary and secondary voltages and currents in the dissectable transformer are connected into circuit by the patching sockets on the mimic diagram. An external wattmeter is used in some assignments and is connected via sockets on the front panel. The dissectable transformer is connected into circuit by the bank of 4mm sockets marked A to G on the lower part .of the front panel, corresponding to similarly marked sockets on he mimic diagram, Connection to the load unit is made via a load on/off switch and the 4mm sockets marked K, L, Nand P on the right of the front panel. Connection to the variable a.c power supply output is made via the plug and socket connectors provided on the rear panel of the measurements console.

Earthed sockets connected to the instrument case are provided in the bottom corners of the front panel. These should be connected to the core of the Dissectable Transformer and to the similar socket on the Load Unit LU178 to complete the earth bonding, for safety. (Even where no earth connection is available the bonding connections between units reduce risk, provided that the main protection, a properly isolated power supply unit, is in use).

Fig. 16

Fig. 17 Measurement console front panel mimic diagram

Fig. 18 Rear panel layout and connections 2. The Dissectable Transformer This transformer (see fig 19) is designed so that it can be assembled and taken apart easily, enabling the individual component parts to be examined.

On a transformer of this size the primary and secondary windings would normally be wound concentrically on a single former. However, as this transformer is made to be dissectable, separate coils are used. Each coil lead is identified by a letter and as an aid to establishing coil polarities the connecting leads are colour coded. The core of the transformer is made up from E· and I-shaped electrical steel laminations, stacked and bonded together. Normally the laminations· would be interleaved but since the transformer is dissectable they are formed as two separate stacks. Quick .release fasteners are used to attach the clamp to the base.

Fig. 19 Dissectable Transformer component parts Flying leads, terminated in 4mm plugs are used to connect the transformer windings to the measurements console at the, sockets marked A to G. Taking the transformer apart Remove the dips from the coil connection leads. Release the two fasteners from the mounting base and remove the clamp from the lamination stack.

Fig. 20 Separate the E and I stacks and remove the coils from the E stack. Assembling the transformer -: Fit the primary and secondary coils on the centre pole of dle E-Iamination stack, as shown in fig 19. See that the white dot on each coil faces the closed end of the stack. . Place the l-larni nation stack centrally on the mounting base, smooth side uppermost. Place the E stack and coils on top of the I stack, centralise, and clamp in position using the quick release fasteners. Bunch the coil connection leads together and fit the plastic clips over them.

3. Load Unit LU178 This is used to apply a load to the dissectable transformer. It contains two banks of resistors as shown in fig 21, with separate fuses. Each resistor can be switched into circuit individually to give a range of load values. Flying leads terminating in 4mm plugs are used to connect the load unit to the measurements console at the sockets marked K, L, N and P. Other forms of load may be used, provided that the range of resistance values corresponds to that of the LU178, and that the load is suitable for the applied voltage. See, for example, fig 1.8 in assignment 1 for details of LU178 ratings.

Fig. 21 4. Power Supply PS189 This unit, shown in fig 22, provides the isolated a.c power supply to the dissectable transformer. It can be switched to give either an a.c or unsmoothed d.c output, and in each case the voltage load can be set by the voltage control knob on the front panel. For the assignments in this manual only the a.c output is required. When power is switched off at the TT179 measurements console, either by pressing the ‘power off’ switch or by the action of the safety devices, the output of the power supply unit is disconnected.

Fig. 22 This unit is connected to the measurements console by the lead provided. This has two 4mm plugs for connection to the sockets on the front panel of the power supply unit and a three-way plug for connection to the socket at the rear of the measurements console. The output of the PS189 is isolated from the main supply. Other forms of variable transformer may only be used if an isolating transformer or equivalent is also used. Direct use of an autotransformer (including a variable transformer) will by-pass the safety features of the TT179, and in the case of Assignment 6 especially will cause breakdown and/or danger to the student. 5. Wattmeter Connections a. Dynamometer Wattmeter. A typical terminal arrangement is shown in fig 23. In this the voltage and current coils have separate terminals with an external link joining one side of the voltage coil to either the supply or load side of the current coil. With this type of instrument it is conventional to have the voltage coil on the supply side and the leads from the TT179 Measurements Console will be connected as shown in fig 23.

Fig. 23 b. Electronic Wattmeter EW604 In this instrument the high impedance voltage input is always connected across the load side of the current sensor and the leads from the TT179 Measurements Console will be connected as shown in fig 24.

Fig. 24

Familiarization Assignment 1 Object To familiarise the student with the measurements console and dissectable transformer which make up the Transformer Trainer. To carry out a simple test which will demonstrate the principles of transformer operation. To measure the ratio of the dissectable transformer windings when connected in different ways. Equipment Required Qty Apparatus 1 TT179 Transformer Trainer comprising a) Measurements console with inter-connecting leads b) Dissectable Transformer with inter-connecting leads 1 PS189 Power Supply Unit * 1 LU178 Load Unit * Time Required Two hours Pre-requisites Some knowledge of transformer construction, as covered in the Introductory Survey. *For alternative units see Operating Notes Th e Tra n s f o rm e r Tra in e r TT1 7 9 M e a s u re m e n t C o n s o le

P o w e r S u p p ly P S 1 8 9

L o a d U n it L U 1 7 8

D is s e c ta b le Tra n s f o rm e r

Fig. 1.1 Preliminary Procedure Arrange the units on your bench as shown in Fig 1.1.

IMPORTANT Before starting the assignment read through the Operating Notes and make sure that you under-stand the purpose of, and know how to use, the Measurements console and Dissectable transformer. Discussion and Experimental Procedure Before we start our study of the transformer in detail it will be useful to look at what a transformer is and how it works. The simplest practical transformer consists of two coils placed together and linked by a closed magnetic core which passes through the centre of the coils, fig 1.2.

Fig. 1.2 Each coil consists of insulated copper wire wound round a bobbin and protected by an insulating outer cover. The coils may have many turns of fine wire for low current work or fewer turns of large diameter wire or copper strip capable of passing heavy currents. In power transformers one coil or winding will be designated the primary and this will be connected to the a.c mains supply. There will be one or more secondary windings, each supplying power to its load at a chosen voltage. The core of a power transformer is constructed from insulated steel laminations bonded and clamped together. This laminated structure is used to reduce internal heating in the core. When current flows through the turns of a coil, lines of flux are set up which pass through the centre of the coil and form closed loops outside the coil. Constant current will produce a steady flux, alternating current will produce an alternating flux. If the coil is wound round a closed steel core the flux level will be increased many times and the lines of flux will be confined to the core path. Iron and many of its alloys are ferromagnetic

materials and have the valuable property of raising the value of flux produced by a current-carrying coil, and in reducing flux leakage. As in the simple transformer of fig 1.2 a second coil may be fitted over the core so that the flux links both coils. A steady flux will produce no voltage in the seconding coil but any change of flux will cause a voltage to be induced in it. When an alternating voltage is connected to the primary winding, a flux is set up in the core which will alternate in direction at the same frequency as the supply voltage. The flux will link with the turns of the secondary winding and cause an a.c voltage to be induced in it whose value will depend on the number of turns in the winding and the rate of change of flux in the core. The effect of the magnetic steel core is to increase the flux produced by current in the primary and keep it within the core path which links the two windings. In fig 1.3(a) two similiar coils of the dissectable transformer are shown fitted to their core. When one coil is energised from an a.c supply the second coil will produce a voltage equal to the supply voltage. If the core is removed as in fig 1.3(b) the flux linking. The two windings will be much reduced and the secondary voltage will fall to a low value. Separating the coils as in fig 1.3(c) will further reduce the flux linkage and cause the secondary voltage to become negligibly small. We will now use the TT179 to illustrate these effects. It is not necessary in these first tests to record meter readings - we need only observe the change in readings which occur when the construction is altered. Practical 1.1 Familiarisation Test TT179 Assembly Take the dissectable transformer apart as instructed in the Operating Notes. Remove the large coil from the core and reassemble with the two small coils in position. Using fig 1.4, set the meter ranges and make the mimic diagram connections shown. Plug the dissectcble transformer coil leads into the sockets marked 0, E, F, G on the console. PS189 Connections Turn the output control to zero. Set the AC/DC switch to AC. Connect the output terminals to the rear socket of the TT179 using the lead provided. Switch off the TT179, remove the coils from the core but keep them together. Switch on the TT179, reset the primary voltage to 5V and read the secondary voltage V 2; it will have fallen to a very low value, hardly readable.

Place the T laminations through the centre of the two coils and observe a small increase in secondary voltage. The drop in secondary voltage resulting from the removal of the transformer core is caused by the reduction in flux passing through the secondary coil. The core has two effects - it produces a much greater flux for a given current in the primary and it confines the flux to its own path.

Fig 1.3

Fig. 1.4

Without the core the flux is reduced in value and there is a considerable flux leakage between the coils. By placing laminations through the coil centres we improved the flux linkage or coupling between the coils. Separate the two coils and observe that the secondary voltage becomes negligibly small. Switch off the power 1:0 the TT179 by pressing the 'power off' button when you have finished these tests. Q1.1

If the two coils used in the previous test were separated but still fitted to a closed steel core, would you expect the voltage on the secondary coil to be similar to the primary voltage or much lower?

Voltage ratio of the Dissectable Transformer In the previous test we saw how a simple transformer produces a secondary voltage which depends on the ratio of the primary to secondary turns and on the magnetic flux which links the two windings. As previously discussed the current drawn by the primary winding from the a.c supply produces an alternating flux which links all the coils which embrace the core. The voltage induced in the secondary can be measured directly but a voltage will also be induced in the primary winding since its turns are linked by the same flux. The induced voltage in the primary is referred to as the back emf. It is almost equal in value to the a.c supply voltage and will always act in opposition to it, so tending to reduce the current, taken from the supply. The voltage in each turn of both the primary and secondary windings will be the same, consequently the value of the voltage induced in the secondary will depend on the ratio of the number of turns in the secondary to those in the primary. When there are more secondary than primary turns the transformer is said to be ‘step-up’ and the secondary voltage is higher than that of the primary. With fewer secondary than primary turns the transformer is ‘step-down’ and output voltage is lower than the supply. For the same number of turns in each winding, the transformer is ‘one-to-one’. I n the tests which follow we shall first apply a low a.c supply voltage to one winding of the transformer and measure the voltages which are produced across the other windings when the transformer is not supplying a load. From these measurements we will work out the no-load voltage and current ratios of the transformer. A further test will then be made in which the voltage and current ratios are measured with the transformer supplying a load.

Practical 1.2 Voltage ratios on no-load TT179 Raise the safety cover on the measurements console. Make the connections to the mimic diagram and set the meter range switches as shown in fig 1.5. Assemble the transformer according to instructions in the Operating Notes. Plug the dissectable transformer leads into the sockets marked A to G on the front panel, ensuring that the identifying letters on the leads and sockets correspond. PS189 Turn the output control of the PS189 to zero. Lower the safety cover and switch on the TT179 using the ‘power on’ button. The voltage induced by the alternating flux in any coil wound round the core may be expressed as a given value of volts per turn. Let number of turns in primary = T1 number of turns in secondary = T2 volts per turn = k Then primary volts, V1 = k T1 Secondary volts, V2 = k T2 Taking k as constant, V1/V2 = T1/T2 = k……… (1) Load current in the secondary will produce a flux which tends to reduce the main flux. This leads to an increase in primary current sufficient to maintain the flux in the core at its original value. To do this, the (additional current) x (number of turns in the primary) will become equal to the (load current) x (number of turns in the secondary). This may be expressed as primary ampere-turns = secondary ampere-turns I1 T 1 = I 2 T 2 and

I1/I2 = T2/I1

………………….. (2)

combining equations (1) and (2) V1/V2 = I2/I1

Fig. 1.5 Voltmeter 1 Connection Reading AC 20 AC 20 AC 20 BC 20

Voltmeter 2 Connection Reading DE FG DG (EF linked) DG (EF linked)

Voltage Ratio on noload DE/AC = FG/AC = DG/AC = DC/BC =

So for an ideal transformer V 1 I1 = V 2 I2 In the next test we shall measure the voltage and current in the primary and secondary windings and work out the voltage and current ratios. Turn the output control on the PS 189 until a reading of 20 volts is obtained on the primary voltmeter V1. Measure the voltage on the secondary voltmeter V2 and record this in the first row of a table like fig 1.6. Repeat this procedure for all the mimic diagram connections given in fig 1.6. It is not necessary to switch off the mains supply when changing connections as the supply will be disconnected when the safety cove is raised. To restore the mains supply lower the safety cover and press the ‘power on’ button. At the end of the test switch off the TT179. Exercise 1.1 The test results recorded in fig 1.6 give us all the information needed to /work out the voltage ratios of the transformer windings on no-load. Using this information draw a transformer diagram similar to fig 1.7 and mark in the voltage appearing across each winding. Q1.2

In the transformer formed by the windings AG and FG we are told that there are 270 turns in winding AG. What are the approximate number of turns in winding FG?

Fig. 1.7

Now calculate the ratios of secondary to primary volts to complete each row of your table (fig. 1.6.) Q1.3

Winding BC is used as a primary winding. Terminals E and F are linked, and windings DG are the secondary. Would you describe this as a step-up or a stepdown transformer:

Practical 1.3 Voltage and Current ratios on load Connect the LU178 load unit to sockets K, L, N and P on the TT179 measurements console. Switch all the load resistors to ON and the load switch on the TT179 to ON. On the TT179 measurements console, make the connections to the mimic diagram and set the meter range switches as shown in fig 1.8. Lower the safety cover and press the ‘power on’ button. Turn the output control on the PS189 to obtain a reading of 100 volts on the primary voltmeter V1. Measure the voltmeter and ammeter readings as given in the table of fig 1.9. Voltmeter 1 100V

Ammeter 1

Voltmeter 2

Ammeter 2

Load Resistance ohms 60

Fig. 1.8

Exercise 1.2 From the results recorded in fig 1.9 calculate a)

The voltage ratio V2/V1 and the current ratio I1/I2

b)

The primary and secondary volt-amperes V1 I1 and V2 I2.

In an ideal transformer, V1 I1 would equal V2 I2, but in a practical transformer this is not the case, partly due to losses within the transformer, as will be discussed in later assignments. Exercise 1.3 With a transformer energised and supplying the load called for in the table of fig 1.9 disconnect the load, using the switch on the front panel of the measurements console. Look at the primary and secondary voltmeters while switching off the load - you will see that both rise to a higher voltage. Again, this is partly due to losses within the transformer, but also to a change in supply voltage when the transformer is on load. Repeat the test but this time adjust the output' of the PS 189, after operating the load switch, to keep a constant reading of 100 volts on the primary voltmeter V1. Any change in secondary voltage due to switching in the load is now due to the transformer itself. This effect is well known in transformer design. It is called ‘regulation’ and is dealt with in more detail in a later assignment. Q 1.4 If a manufacturer describes a power supply transformer as providing a secondary voltage of 24 volts when connected to a 240-volt mains supply, do you think he is referring to the full load or no-load secondary voltage? Practical Applications Most power transformers are used to provide either a step-up or step-down of secondary voltage. Large electricity supply transformers are used to change the relatively low voltage produced by the power station generators to a much higher level, for transmission of power over long distances, e.q power may be generated at 11,000 volts and transmitted at 275,000volts. At a sub-station, situated near to a town or city where the electricity will be used, several step-down transformers will be connected between the transmission lines and the distribution system, followed by further step-down transformers which bring the mains

supply to the nominal value specified by the supply authorities, for example 240 volts in domestic and light industrial applications. In many types of electrical equipment such as industrial electronics instruments, stepdown transformers are used to provide the low-voltage power supplies required for transistors and integrated circuits. Stepped ratio transformers enable a small motor or possibly a heating element to be powered from a range of different supply voltages. I n these cases a mains voltage selector switch may be connected to tapping points on the primary winding for 100V, 120V, 200V, 240V or other supply voltage levels. Answers Q1.1

The voltage on the secondary coil would remain similar to that of the primary coil because the majority of the flux produced by the primary passes round the closed core and is thus linked with the secondary.

Q1.2

Since the secondary voltage is approximately half the primary Voltage, the secondary turns must be approximately half the primary turns, that is 270/2 = 135 turns.

Q1.3

Since the secondary voltage is higher than the primary voltage it is a step-up transformer.

Q1.4

The secondary voltage is usually specified at full load, since this is the condition which is normally of interest when the transformer is in use. Likewise the VA rating is the product of secondary voltage and current at full load.

Typical results Exercise 1 .2 Voltage ratio V2/V1 = 0.98 Current ratio I1/I2 = 1.25 Practical 1.2 Voltmeter 1 Connection Reading AC 20 AC 20 AC 20 BC 20

Voltmeter 2 Connection Reading DE 10.5 FG 10.5 DG (EF linked) 21.8 DG (EF linked) 44.2

Voltage Ratio on noload DE/AC = 0.52 FG/AC = 0.52 DG/AC = 1.1 DC/BC = 2.21

Practical 1.3 Voltmeter 1

Ammeter 1

Voltmeter 2

Ammeter 2

100V

2A

98V

1.6A

Load Resistance ohms 60

Object To clarify the meaning of the term ‘polarity’ when applied to transformers and to show how the polarity of windings can be established. Equipment required 1 - TT179 Transformer Trainer 1 - PS189 Power Supply * Time Required One hour Prerequisites Reading of Introductory Summary and Assignment 1. *For alternative unit, see Operating Notes.

Fig. 21 Discussion and Experimental Procedure A simple transformer with two separate windings is shown in fig 2.1 a. To illustrate the principles involved the same transformer is re-drawn in fig 2.1 b with the primary and secondary coils wound in the same direction on a common core, and with the ‘start’ and ‘finish’ wires of each coil labelled.

If an alternating voltage is now connected to coil A-C the resulting current flow will set up an alternating flux in the magnetic core. This causes an induced voltage to be produced in coil AC which acts to reduce the flow of current through that coil. This induced voltage is called the ‘back emf’ or counter emf. The relationship between the supply voltage and the back emf in the primary can be likened to that of two cells connected positive to positive, negative to negative, as shown in fig 2.1 c. They have the same polarity, but it should be noted that their terminal voltages are in opposition to one another and as a result no current flows between them. In the transformer we can say that the polarity of the back emf in the primary winding is the same as that of the supply voltage, again the two voltages oppose one another. The effect of the back emf is to reduce the flow of current into the transformer. The secondary winding is wound around the same core as the primary and consequently an induced voltage will also be set up within its windings. This is the output voltage of the transformer - its frequency is the same as the supply voltage frequency and its magnitude will depend on the number of turns in its winding. Referring to fig 2.1 b, the polarity of the induced secondary voltage across terminals D to E will be the same as the polarity of the primary voltage across terminals A to C. The polarity of the secondary coil can be reversed by reversing the connections to its terminals. During manufacture it is probable that all coils in the transformer will be wound in the, same direction of rotation, clockwise or anticlockwise. If this is so then the start of winding leads will always have the same polarity. Establishing the polarity of a transformer winding If the transformer winding details are not given we can carry out a test to establish their polarity. In this test one of the primary coil terminals is made common with one of the secondary coil terminals. A low a.c voltage is applied to the primary and the voltage difference between the two non-commoned terminals is measured. We can then work out the relative polarity of the two windings. The connection that gives minimum voltage across the windings is that which gives equivalent polarity. If the voltage ratio of the two windings is one to one, the voltage across the windings is zero when their polarities are the same. For other ratios, the voltage across the windings is the difference between the individual winding voltages (same polarity) or the sum of the individual voltages (opposite polarity). Where a transformer has more than one secondary the same test can be extended to identify the polarity of all the windings. The circuits used for a.c polarity tests are given in fig 2.2.

Fig. 2.2 Practical 2.1 Winding Polarity Switch off the mains supply to the TT179 and PS189. On the TT179 measurements console, set the meter ranges, make the mimic diagram connections and connect the dissectable transformer as shown in fig 2.3. Lower the safety cover.

Fig. 2.3 On the PS189, set the output control to zero and the AC/DC switch to AC. Use the lead provided to connect the a.c output to the measurements console. Switch on the TT179 and PS189. Set the PS189 output to 20 volts, as indicated on voltmeter V1 of the measurements console. Read the voltage between the ends of the transformer windings which have not been commoned as measured on V 2 and record this in a copy of the table of fig 2.4. Repeat the test with all the connections listed. We now have all the information needed to establish the polarity of the primary and two secondary windings of the dissectable transformer. In the first two tests the minimum voltage across the winding occurs when the connections C and E are commoned. Therefore the secondary connections D to E have the same polarity as the primary connections A to C. The remaining tests establish the polarity of the secondary winding FG with reference to the primary winding and the polarity of the two secondaries to one another.

Exercise 2.1 Complete the columns headed ‘polarity’ in your copy of fig 2.4. Draw the dissectable transformer circuit diagram and indicate by arrows the polarity of all three windings. Q2. 1 Would you expect the polarity of winding 8 to C to be the same as A to C? Explain how you would check this. Voltmeter 1 Connection Reading AC AC AC AC AC AC

20 20 20 20 20 20

Link between primary and secondary C–E C–D C–G C–F E–G E–F

Voltmeter 2 Connection Reading

Voltage Ratio on no-load

AD AE AF AG DF DG

Fig. 2.4 Practical considerations and applications in most transformer applications it is not necessary to know the polarity of the windings, but where two transformers are to be connected in parallel it is essential to establish correct polarity, otherwise destructively large currents would circulate in the two secondaries. This is of particular importance in power distribution systems. I n a transformer supplying industrial loads, e.g motors, heaters etc. two or more secondaries may be connected in series or in parallel to provide a range of voltage and current levels. Again, the polarity of the secondary winding must be known. Answer Q2.1

Yes, one would expect winding BC to have the same polarity as winding AC because the winding ABC is continuous with all the turns in the same direction and linking a commond flux path. This may be checked by applying a voltage between two of the terminals and then measuring the three voltages between them, obtaining the result VBC + VAB + VAC.

Fig. 2.5 Results Practical 2.1 Voltmeter 1 Connection Reading AC AC AC AC AC AC

20 20 20 20 20 20

Link between primary and secondary C–E C–D C–G C–F E–G E–F

Voltmeter 2 Connection Reading AD AE AF AG DF DG

8.5V 31.2 8.5V 31.2 0 22

Voltage Ratio on no-load same opposite same opposite same opposite

Object To show the effect on output voltage and of connecting the secondary windings of a transformer in series and in parallel. Equipment Required 1 – TT179 Transformer Trainer 1 – PS 189 Power Supply * 1 – LU178 Load Unit * Time Required One hour Prerequisites Assignment 1–Familiarization and Assignment 2 ‘Transformer Polarity’. *For alternative units, see Operating Notes.

Discussion and Experimental Procedure In Assignments 1 and 2 we carried out tests to find the ratio and polarity of the dissectable transformer. These tests and the principles involved are, of course, applicable to power transformers which have two or more secondaries of the same voltage ratio. In this assignment we will make use of the information obtained and extend it to show that by connecting the secondaries in series or in parallel we can change the output voltages and currents.

The information gained from series or parallel connection of the secondaries of a single transformer also applies where two or more transformers of appropriate voltage ratio are connected to a common power supply.

Fig. 3.1 Fig 3.1 shows the principle. used in the tests which follow. If the two secondary windings are to be connected in parallel we must first ensure: a)

that they have the same voltage ratio - if this is not so a circulating current will flow through the windings, even if their polarities are correct.

b)

that they are connected to have the same polarity - this is essential to avoid the flow of a destructively heavy current through the windings.

Both these requirements are met by an initial check on the voltage across the open ends of the secondary windings when their other terminals are commoned. If this voltage is zero the two windings have the same voltage ratio and the same polarity. Given that each secondary can supply up to 2 amperes at 60 volts, two secondaries in parallel will give 4 amperes at 60 volts (240 VA) and in series will give 2 amperes at 120 volts (240 V A) without exceeding the transformer power rating. Practical 3.1 Check on voltage ratio and polarity

Before switching on the mains supply, refer to fig 3.2. Set the meter ranges and make the mimic diagram connections shown. Connect the dissectable transformer, PS 189 variable a.c power supply, and LU 178 load unit to the measurements console. Switch on the PS 189 and set its output to zero. Lower the safety cover and switch on the mains supply to the TT179. Raise the output from the PS189 to 120 volts as read on the voltmeter V 1 and check that the voltage reading on V2 is approximately 60 volts. Change the connections of voltmeter V2 from O-E to F-G, switch on, and check that V2 reads approximately 60 volts. Change the connections of voltmeter V 2 from F-G to O-F, switch on, and check that V 2 reads zero. This shows that the secondaries have the same output voltage and the correct polarity. Practical 3.2 Parallel Connection Connect both secondaries in parallel, as shown in fig 3.3. Make the connections shown from the LU 178 to the TT 179. Set the meter ranges and put the load switch to OFF. Press the POWE R ON button and raise the supply voltage from the PS189 to 120 volts, as read on V1. Secondary Connection Parallel Series

V1

I1

Fig. 3.5

V2

I2

Load 15Ω 60Ω

Fig. 3.2

Fig. 3.3

Fig. 3.4

Switch all the load resistors on the LU178 to ON. Set the load switch on the TT179 to ON - re-adjust the supply voltage 120V. Read V1, I1, V2 and I2 and record your results as indicated in fig 3.5. Practical 3.3 Series Connection Connect both secondaries in series, as shown in fig 3.4. Make the connections shown from the LU 178 to TT179. Set the meter ranges and put the load switch to OFF. Press the POWE R ON button and adjust the supply voltage from the PS 189 to 120 volts, as read on V1. Set the load switch on the TT179 to ON and re-adjust the voltage supply to 120 volts. Read VI' II' V2 and 12 and record your results as in fig 3.5. Switch off the TT179 and the PS189. Exercise 3.1 From the information recorded in fig 3.5 calculate the power output of the transformer when the secondaries are connected in parallel in series. Q3.1

Referring to the parallel connection test, we could disconnect one of the secondary windings and still supply power to the two banks of load resistors. The voltage and current would be only a little less than those obtained with both windings connected. What would be wrong in doing this?

Practical considerations and applications In a transformer supplying industrial loads, e.q motors, heaters, etc. two or more secondaries may be connected in series or in parallel to provide a range of voltage and current levels. A test procedure similar to that given in this assignment is necessary to ensure correct polarity and to check the final voltage and current on load so that the transformer output rating is not exceeded. Typical results Secondary connection Pract 3.2, parallel Pract 3.3, series

V1 120V 120V

Exercise 3.1 Power output, parallel secondaries series secondaries

I1 2.4A 2.42A 217W 225W

V2 57V 118.3V

I2 3.8A 1.9A

Load 15 60

Answer Q3.1 The current in the single winding would be doubled, causing the heat generated in it to be quadrupled, almost certainly damaging the winding insulation Object To introduce and illustrate the principal terms used in the study of the magnetic circuit of a transformer. To show the relationship between flux density, field strength and magnetizing current for a typical transformer core. Equipment required Qty 1 1 1 1

Apparatus TT179 Transformer Trainer PS189 Power Supply* LU178 Load Unit* Oscilloscope

Time Required Two hours Prerequisites Some knowledge of transformer construction as covered by Introductory Survey. *For alternative units, see Operating Notes

Discussion and Experimental Procedure In earlier assignments we have looked mainly at the voltage relationship between the primary and secondary windings of the transformer and only briefly at the magnetic path taken by the, flux which links those windings. In this assignment we shall study the magnetic circuit in more detail. A simple transformer with two coils wound on a common magnetic core is shown in fig 4.1. If we connect the primary coil to an electrical supply the flow of current through its windings will set up a magnetic field which produces a flux in the core.

Fig. 4.1 There is an important law of electro-magnetism which states ‘whenever the flux through a circuit changes, an e.m.f is induced in that circuit’. If the primary winding of our simple transformer is connected to an alternating current source, the flux through the magnetic core will alternate at the same frequency as the source. This will cause voltages to be induced in both the primary and secondary windings. In the primary winding this voltage is called the back e.m. f. It opposes the supply voltage and reduces the current flowing in the primary. The voltage induced in the secondary has the, same frequency as the supply voltage and its value will depend on the turns ratio of the primary and secondary windings, assuming that all the flux which links the primary will also link the secondary. In practice there is always some flux leakage, and this will be considered more fully in a later assignment. The action of the transformer is therefore seen to depend on the magnetic flux which links the two windings. The strength of this flux will itself depend on four main factors which are taken into account when deriving the magnetic circuit:

the value of current flowing in the primary 1 the number of turns in the primary the dimensions of the magnetic core the materials which form the magnetic core. The Magnetic Circuit The path taken by the flux in an electrical machine is called the magnetic circuit. Transformers, motors and generators, solenoids and relays all obey similar rules and are made, up from similar magnetic materials. By under- standing the principles of the magnetic circuit of a transformer we can extend our knowledge of other electrical machines. Definitions of the terms used in this section are set out in Appendix 1. However, before studying these in detail we can obtain an understanding of a typical magnetic circuit by comparing it with an electrical circuit as in fig 4.2.

Fig. 4.2 In the electrical circuit a voltage V is connected: to a resistor R, causing a current I to flow round the circuit. In this case if R is fixed the current is directly proportional to applied voltage. In the magnetic circuit a coil of T turns is wound around one section of a magnetic core. When there is current / flowing in the coil it will produce a magnetic field and this will cause a flux to be set up in the magnetic material which forms the core. We say that current flowing through the turns of the coil produces a magnetomotive force F (expressed in ampere-turns) and that the field strength due to this is H (ampere-turns per metre of the flux path length). The flux set up in the magnetic core due to this field is Φ (expressed in weber). Looking again at the magnetic and electrical circuits of fig 4.2, we can see that the relationship between flux Φ and magnetomotive force F is similar to that between current I and voltage V in an electrical circuit with resistive elements. However, there is one basic difference between the two circuits. I n the electrical circuit

the graph of current against voltage is a straight line through the origin, whereas in a magnetic circuit the graph of flux against magnetomotive force is not linear, fig 4.3.

Fig. 4.3 Electrical Circuit I= V I R where the resistance R does not change appreciably with current Magnetic Circuit Φ = F/Rm where Rm is the reluctance of the magnetic circuit expressed as ampere-turns per weber Here we have used the reluctance, Rm as an element in the magnetic circuit. Although this corresponds approximately to resistance in the electrical circuit we can see that the ratio Rm = F/Φ is not constant, the reluctance increases rapidly as the flux in the core is progressively increased. Q4.1

We are told that the primary coil of the transformer shown in fig 4.1 has 270 turns and requires a magnetizing current of 0.2A to produce a flux in the core of 1.3 milli-weber. What is the reluctance of the magnetic path at that flux level?

Transformer with no secondary load When an a.c voltage is applied to the primary of a transformer and there is no load on the secondary, the current which flows is limited by the impedance of the primary. There is a small voltage drop due to the resistance of the winding and a much larger drop due to the induced voltage in the primary. The induced voltage is caused by the changing flux in the core and always opposes the applied voltage. This is a direct application of Lenz’s Law ‘induced voltages act in such a direction as to oppose the action by which they are due’. Here the applied alternating voltage causes current to flow in the primary coil which produces an alternating flux in the core. The voltage induced in the primary will be proportional to the rate of change of flux in the core, therefore sufficient current must flow in the primary to produce the required value of flux. Phase Relationship between Core Flux and Induced Voltage We can show that the rate of change of any sine wave can be represented by another sinusoidal curve, but displaced by 90°.

Fig. 4.4 In the full-line sine curve of fig 4.4a the rate of change, or slope, is a maximum where the curve crosses the zero axis and is a minimum at the crests of the sine wave. The broken line curve, fig 4.4b, represents the rate of change of the sine curve and is itself sinusoidal (it is a cosine curve) and leads by π/2 radians or 90°. If curve (a) represents flux Φ in the core and curve (b) the induced voltage E in either the primary or secondary due to the rate of change of flux, we can see that the flux will lag the \ voltage by 90°. The magnetizing current is in) phase with the flux which it produces and also lags the induced voltage by 90°.

Magnetizing Current Waveform We have seen in fig 4.3 that the curve of flux Φ against magnetomotive force F is not linear. In this curve, if the flux is progressively increased the magnetomotive force required to produce that flux will increase in greater proportion. From this we can see that as the voltage applied to the primary is increased the core flux will have to increase by the same amount to maintain equality between the induced voltage and the applied voltage. This will require a proportionately greater increase in magneto motive force and hence in the current flowing in the primary, particularly as the core approaches saturation. The applied alternating voltage has a sinusoidal waveform and the flux in the core will also be sinusoidal, but the current producing that flux will have a waveform which will depend on how near the core is to saturation, fig 4.5.

Fig. 4.5 I n the test which follows we shall measure the current taken by the primary, with the secondary on no-load, for increasing values of applied voltage. By connecting an oscilloscope in series with the primary we can also look at the shape of the current waveform and note how it develops a peak in each half cycle as the applied voltage is increased. Fig 4.5 shows the circuit diagram used in this test.

Fig. 4.6

Practical 4.1 Primary Current and Voltage - Transformer on No-load Switch off the Mains supply to the TT179 and PS189. Refer to fig 4.7, and on the measurements console: Set the meter ranges Make the mimic diagram connections Connect the dissectable transformer, oscilloscope, and PS189 to the TT179. As both oscilloscope input terminals are live it is important to ensure that the oscilloscope mains earth is not connected. Lower the safety cover. Set the PS 189 output to zero and the AC/DC switch to AC. Switch on the PS189 and TT179 but not the oscilloscope. Starting from zero, raise the voltage applied to the primary winding in 20V steps as measured by V1, until you reach 140V or the nearest attainable value. For each step record the voltage V1 and the corresponding primary current I 1 as indicated in fig 4.8. For the last reading the range of the ammeter may need to be changed from 1 A to 5A. V1 I1

0

20

40

60

80

100

120

140

Fig. 4.8 Reduce the PS189 output to zero and switch on the oscilloscope: Set the time base to 5ms per division Set the Y amplifier to-Zv per division Raise the output from the PS189 continuously from zero to 140 volts (or the nearest attainable value) and observe the change in current wave-form as the voltage is increased. Q4.2

Why does the current waveform develop peaks as the voltage is increased?

Exercise 4.1 Plot the results from the table of fig 4.8. The resulting graph should have a similar shape to the Φ/F curve in fig 4.3. Practical considerations and applications Working flux level

One of the most important points in the design of a transformer is to fix the working flux level, since on this depends the current taken from the supply oil load and off load. If the working flux is too high and the core becomes saturated the input current will be high and the transformer will over-heat. If the working flux is low the transformer will be larger and more costly than it need be.

Fig. 4.7

The designer makes a compromise, usually fixing the working flux at a point just beyond the linear part of the flux - mmf curve. When a transformer has to cover a wide range of applied Voltages, tappings are provided on the primary winding so that saturation of the core is avoided. Typical results Practical 4.1 V1 I1

0 0

20 0.115

40 0.2

60 0.29

80 0.405

100 0.545

120 0.79

140 1.55A

Answers Q4.1

Rm =

Reluctance is given by Rm = F/Φ where. F = IN In this case F = 0.2 x 270 = 54A turns 54 1.3 ×10 −3

= 4.15 x 104 ampere-turns per weber Note that 51 does not recognise ‘turns’ as a unit, so that any of the values 4.15x 104 AWb-1 41.5 kA Wb-1 41.5 A/mWb may be considered correct. Q4.2

As the applied voltage is increased the current adjusts itself so that a balance is maintained between the applied voltage and the emf induced by flux changes in the primary coil. This implies a larger amplitude of sinusoidal flux change, extending into the region of saturation, which requires a large magnetising current at the peak of the flux wave, as shown in fig 4.9.

Fig. 4.9 Object To show the phase relationship between terminal voltage, current and flux in a transformer with no-load applied to its secondary winding, assuming that the winding resistance and core losses are negligible. Equipment Required Qty Apparatus 1 TT179 Transformer Trainer 1 PS189 Power Supply* 1 Oscilloscope with mains earth not connected Time Required One hour Pre-requisites Introductory Survey and the Magnetic Circuit -. Assignment 4. *For alternative unit see Operating Notes

Discussion and Experimental Procedure In all alternating current electrical machines, knowledge of the phase relationship between the voltage, current and magnetic flux waveforms is essential to the designer and of considerable assistance to the user in furthering his under- standing of the machine. Fortunately, in power transformers we are mainly concerned with sinusoidal waveforms these are more easily analysed and measured than the complex wave-forms encountered in, for example, a pulse transformer. It is often useful to draw the required waveforms in full, showing amplitude and phase for one or more cycles, but where a number of sine waves is involved it is more convenient to use a phasor diagram. The phasor diagram is one method by which we can show the relationship between alternating currents or voltages in an electrical circuit. We need to show their amplitudes and how they lead or lag one another in time. By constructing a phasor diagram we can often either check the accuracy of a calculation or more easily understand the circuit to which it applies. We shall first use an oscilloscope to examine the aveform and phase relationship between the primary terminal voltage V1, the secondary terminal voltage V2, the primary current I1. and from our observations draw the phasor diagram. For this test we will use a double-beam oscilloscope to check the phase relationship between the primary and secondary terminal voltages, with the start and finish of each winding connected to give the same polarity. We will then check the phase relationship between the secondary voltage and primary current and at the same time observe the waveform of the primary current. Fig 5.1 shows the circuit diagram used for this test. We use the secondary voltage as a reference in both tests to avoid connection difficulties where the Y1 and Y2 inputs on the oscilloscope have the same common terminal.

Fig. 5.1

It is important to disconnect the mains earth lead from the oscilloscope as in the second test both terminals of input 1 are live. Practical 5.1 Phase relationship of Primary and Secondary voltages Switch off the mains supply to the TT179 and PS189. On the Measurements Console, referring to fig 5.2 Set the meter ranges as shown Make the mimic diagram connections Connect the Dissectable Transformer, Oscilloscope and PS 189 to the TT179 Lower the safety cover Set the PS189 output to zero and the AC/DC switch to AC. Switch on the PS189, TT179 and Oscilloscope On the Oscilloscope set input 1 arid input 2 to 50V /division. Raise the voltage output of the PS 189 to 120 volts as read on V1. Observe that the primary voltage V1 and the secondary voltage V2 are sine waves and that they are in phase with one another, as shown in fig 5.6a. Draw a diagram of the two waveforms from the oscilloscope traces, then switch off the TT179. Practical 5.2 Phase Relationship of Primary current and Secondary voltage Q5.1

Do you remember and can you explain the phase relationship between the voltage applied to the primary winding and the current in it?

Q5.2

Can you predict the phase of the secondary voltage relative to the primary current?

Change the oscilloscope connections to monitor the secondary voltage and primary current, as shown in fig 5.3. Set the oscilloscope input 1 to 2V/division and. input 2 to 50V/division. Switch on the TT179 and if necessary readjust voltage V 1 to 120V. Observe that the primary current waveform is not sinusoidal and that it lags the secondary voltage by approximately 90 electrical degrees. The word ‘phase’ has exact meaning only in relation to a true sine wave, but it will suffice for the moment to talk of the phase of the current while ignoring its peaky waveshape.

Fig. 5.2

Fig. 5.3

Fig. 5.4 Phasor Diagrams A rod, pivoted at one end and rotating at uniform velocity will have a vertical projection which increases to a maximum twice in each revolution, fig 5.4. If we measure the length of these projections at intervals throughout one revolution and plot the measured values against angular position we will obtain a sine curve, as in fig -5.4b, whose . amplitude corresponds to the length of the rod. We can therefore represent any sine wave by a line which is assumed to rotate at a constant angular velocity determined by frequency. Two or more sine waves, which may correspond to voltage, current or flux can be represented by. radial lines, or phasors, whose lengths are proportional amplitude and whose angular positions correspond to phase angle. Fig 5.5a shows a phasor diagram for the voltage across a pure inductance and the current flowing through it. In this case the current lags the voltage by 90° in phase.

Fig. 5.5

The phasor diagram provides useful information in a concise form which can often be an aid to understanding the behaviour of an electrical circuit. In some circuits we may need to obtain the resultant of two sinusoidal quantities - this can be readily carried out by means of a phasor diagram. Fig S.Sb shows the phasor addition of two a.c voltages. Phasor Diagram for Transformer on No-Load We now have most of the information needed to construct the phasor diagram. We have not measured the flux in the core, since this would require special instruments, but we can say that it has a sinusoidal waveform and is in phase with the primary current. Using this information we can draw the phasor diagram as shown in fig 5.6b. The current wave-form is not sinusoidal due to the onset of core saturation at the applied voltage peaks. It is therefore not strictly correct to include this in a simple phasor diagram but for the present purpose the primary current can be treated as a sine-wave.

Exercise 5.1 Using graph paper draw the waveforms which were displayed on the oscilloscope, similar to those shown in fig 5.6a. Construct the phasor diagram, firstly as shown in fig 5.6b and then re-draw, this time assuming that the polarity of the secondary winding has been reversed.

Fig. 5.6 Practical Applications We have seen that the phasor diagram enables us to show in a concise way several important points relating to the dissectable transformer: a)

the secondary voltage on no-load is in phase with the primary voltage,

b)

the amplitude of each secondary voltage is approximately half that of the primary voltage, and

c)

the primary current on no-load lags the primary voltage by about 90° (this shows that on no-load the transformer behaves as an inductor with low series resistance)

At this stage the phasor diagram is not complicated and can easily be extended to show, for example, the relationship between the components of primary current and primary voltage when the transformer is supplying power to an external load. This is dealt with in Assignment 7. Answers Q5.1

The primary current lags 90 degrees on the applied voltage. This is because the primary current is in phase with the flux, whose rate of change is in phase with the voltage.

Q5.2

Since the flux changes in the same way in both windings the phase of the secondary voltage must be the same as the phase of the primary voltage (assuming like polarities of connection). The primary voltage and therefore also the secondary voltage leads the primary current by 90 degrees.

Fig. 5.7 Object To introduce the two principal causes of power loss in a transformer connected to an a.c supply but with no-load applied to its secondary windings. To display the total loss loop on an oscilloscope. Equipment Required Qty Apparatus 1 TT179 Transformer Trainer* 1 PS189 Power Supply*

1

Oscilloscope for XY operation 1V/cm sensitivity on X and Y channels.

Time Requjred One and a half hours Pre-requisites Reading of Introductory Survey, Appendix 1 and Assignment 5. *For alternative units, see Operating Notes

Discussion and Experimental Procedure In the work carried out so far we have seen how a transformer produces a voltage at its secondary terminals which is dependent on the turns ratio and how the current taken from the supply is affected by the magnetic properties of the core. We have not previously taken into account the power dissipated in the transformer itself. All machines, electrical or mechanical, are less than 100% efficient; they take more power from the source than they give out to their load. The difference is a power loss which is dissipated as heat or noise by the machine. There are two ways in which power is lost in the transformer - copper loss due to current flow in the windings and iron or core loss due to the alternating flux in the core. These losses cause the temperature of the windings and core to rise so that power is dissipated by thermal radiation and convection. Copper loss varies with load current and core loss with applied voltage. As the electrical supply voltage is usually fixed the core loss is almost independent of load. In later assignments we shall measure the iron and copper losses separately but in this assignment we will make a preliminary study of the cause of power loss in the transformer core and display the total no-load loss curve of a transformer on an oscilloscope. No-load Current When a transformer is connected to an a.c supply with no-load applied to its secondary the current in the primary will have two components. a)

The magnetizing current - this produces the magnetomotive force F which sets up the working flux in the transformer core. This current, as we have seen in Assignment 5 lags the primary voltage by almost 90°. When an alternating voltage and current in the same circuit are 90° out of phase (in quadrature) there is no power consumed.

b)

The core loss current - this is in phase with the applied voltage and causes power to be lost, which is dissipated as heat in the core. This power loss is due to the behaviour of a ferromagnetic material when carrying an alternating flux. I n this assignment we will look at the two principal effects hysteresis and eddy current losses.

Hysteresis Loss If a coil linking a ferromagnetic core is supplied from a d.c source and the current through it is progressively increased to a given maximum, the field strength H will increase proportionately. The curve relating flux density B and field strength H will be nearly linear until the core approaches saturation, as shown by the dotted line in fig 6.1.

Fig. 6.1 If the current through the coil is now reduced through zero to the given maximum in the. reverse direction, the resulting B/H curve will not follow the original curve but that shown as a full line in fig 6.1. Again, taking the current through zero to its maximum in the forward direction causes B to complete its cycle. Repeating the cycle of current reversals will cause the same B/H loop to be retraced. This is known as the hysteresis loop; different types of magnetic material will have different forms of hysteresis loop. The hysteresis loop provides useful information on one of the principal sources of power loss in the core. When the core is carrying an alternating flux, as is the case in most transformers and many other electrical machines, there will be a power loss proportional to the area enclosed by the hysteresis loop and the number of times per second that the loop is traced out. I n a trans- former the loop will be completed once per cycle of the a.c supply voltage. The area enclosed by the hysteresis loop can be expressed in terms of a power of Bm, leading to the empirical formula Ph = Kh f Bm1.6 where Ph is the hysteresis loss per unit volume of core Kh is a constant for a given core material f = frequency of supply Bm = maximum flux density The area enclosed by the hysteresis loop is dependent on how far the core is taken into saturation in each cycle and on the properties of the ferromagnetic material which forms the core. A power transformer will be designed to have a working flux which will approach but not reach saturation when the flux density is at its maximum value so reducing the width of the hysteresis loop. Also, the laminations which form the core will be chosen from a range of steels which have an inherently low hysteresis loss. Steel alloys with up to 3%

Silicon content are frequently used since they have a narrow hysteresis loop as compared, for instance, with mild steel. The hysteresis curves for silicon steel and mild steel are shown in fig 6.2.

Fig. 6.2 Eddy Current Loss We have previously seen that a coil linked by a changing flux will have a voltage induced in it. The steel which makes up the core of the transformer is itself a conductor and when carrying an alternating flux will act as a single turn coil closed upon itself. As a result circulating currents will flow in the core material, as shown in fig 6.3(a). These are known as eddy currents and if the magnetic core were to be made from solid steel there would be a substantial power loss and temperature rise due to this effect. To reduce eddy current loss the core of a power transformer is built up from: laminated steel punchings insulated from one another and held together as a stack by rivets, bolts or by resin impregnation. Eddy current flow in a laminated core will be reduced by the substantial increase in resistance of each current path, fig 6.3(b). For an operating frequency of 50 or 60Hz a lamination thickness of 0.3mm (0.012in) will normally be used, and for a frequency of 400Hz the lamination thick- ness will be around 0.1 mm (0.004in). Eddy Current loss per unit volume is given by the formula Pe = Ke, f2. 8m2 where Ke is a constant, proportional to the square of lamination thickness and inversely proportional to resistivity. f = supply frequency Bm = maximum flux density A further advantage in the use of Silicon steel is that the resistively of this material is greater than that of mild steel. As a result the magnitude of the eddy currents is reduced and losses are lower.

Q6.1

Why are thinner steel laminations used at high supply frequencies?

Core Loss Loop An oscilloscope can be used to display the total core-loss loop of the transformer when operating on no-load but with a rated supply voltage applied to its primary windings. A low value resistor is connected in series with the primary winding and the voltage drop across it, which is directly proportional to primary current, is applied to the X amplifier of the CRO. The secondary winding is connected to the Y-amplifier of the CRO via an RC integrating circuit. The voltage across the secondary winding is proportional to the rate of change of flux in the transformer core and the integral of this voltage therefore represents the flux directly. By this means we can display the relationship between flux in the core (integral of secondary voltage - Y axis) and the primary current (mainly magnetizing current - X axis), and from these the loss in the core can be calculated. If the dimensions of the core are known it is possible to scale the resulting loop in terms of flux density B and field strength H, from which one can calculate the core loss per unit volume, which can then be used as design data for other cores made of the same material. However loss measurements based on the loss loop are not very accurate, and it is better to measure the losses directly using a wattmeter (as described in Assignment 8).

Fig. 6.3 The shape of the core-loss loop is very dependent Ion the level of voltage applied to the primary from the a.c source. At the higher values of applied voltage the peak flux density in the core approaches saturation in each half cycle. In the practicals which follow, the

supply voltage is raised from below to above the rated value and the effect on the coreloss loop can be observed.

This is carried out first with the dissectable transformer assembled normally and then with a strip of non-magnetic material placed between the E- and l-sections of the core to give the same effect as an air gap in the magnetic circuit. Fig 6.4 shows the circuit diagram used in these practicals. Q6.2

Give one source of error in measuring the no-load losses by the oscilloscope method.

Practical 6.1 Core-loss Loop - No Added Gap Switch off the TT179 and PS189. On the Measurements Console, refer to fig 6.5: Make the mimic diagram connections shown Set the meter: ranges Connect the Dissectable Transformer, PS189, and Oscilloscope, to the Measurements Console Lower the safety cover Set the X-channel sensitivity to 1.0 volt per division. Set the Y-channel sensitivity to 1.0 volt per division. Set the PS189 output to zero and the AC/DC switch to AC. Switch on the PS189, TT179, and Oscilloscope. Raise the PS 189 output to 70 volts as read on meter V1 and observe the core-loss loop on the Oscilloscope. Centre the loop about zero and, using tracing paper or plotting point by point, record the. core-loss loop. Repeat with the PS 189 output set to 100 and 120 volts. The curves obtained should be similar to those of fig 6.6a. Practical 6.2 Core-loss Loop - Air Gap Added Switch off the TT179 and PS189. Release the fasteners on the Dissectable Transformer frame, remove the E-section of the core with its attached coils and place the insulation strip over the I-section, Replace the E-section with its coils in position and clamp the transformer frame, using the quick-release fasteners. Repeat the procedure given in Practical 6.1 and make records of the core-loss loop for applied voltages of 70, 100 and 120 volts. The curves obtained should be similar to those of fig 6.6b. Exercise 6.1 Scale the records taken in Practicals 6.1 and 6.2, using the following approximations

X axis: 0.5 mill i-weber per input volt Y axis: 250 ampere-turns per input volt. If we compare the two sets of records, taken with and without the non-magnetic gap in the transformer core, it can be seen that the ampere-turns required to produce a given flux level differ considerably. With no gap the magnetomotive force or ampere-turns applied to the core is used to set up flux in a closed magnetic path of relatively low reluctance. When there is even a small gap - in this case 0.25mm (0.010in) the reluctance of the flux path is increased substantially. We can think of the magnetic circuit as several series elements, each with their own reluctance, across which a certain number of ampere-turns are dropped, rather like voltage drops across resistors in series in an electrical circuit (see appendix).

Fig. 6.4

Fig. 6.5

Q6.3

Why is it important to keep the air gaps in the core of a power transformer as low as possible?

Practical Considerations Core loss in a power transformer is usually constant, since most of these operate with a fixed voltage applied to the primary. It is important to reduce the core loss to as low a value as practicable, in order to improve efficiency and reduce the temperature rise associated with any power loss. This is achieved by using a laminated core to reduce eddy-current loss and choosing a transformer steel which has a low hysteresis loss. It is also an aim of the designer to reduce the current taken by the transformer when it is supplying no external load, i.e the current required to set up the working flux in the core. This is partly brought about by minimizing the air gaps between laminations. Although the dissectable transformer is made up of two separate E- and I-sections for ease of assembly, in practice these would be assembled alternately to distribute the gaps and the whole core clamped firmly before riveting and bonding the structure as a whole. The core-loss loop, as displayed on an oscilloscope, is not suited to the accurate measurement of losses, but does provide a useful means of comparing different core structures and gives some insight into the factors which determine the no-load losses, for example the effect of introducing an air gap into the magnetic circuit.

Fig. 6.6

Answers Q6.1

Eddy current loss is proportional to the square of frequency and the square of lamination thickness. To avoid excessive loss at high frequency the thickness must be reduced to compensate. (However there is a practical limit to this, since the core must be sufficiently rigid to support itself and the windings, and the magnetic properties are impaired by rolling steel very thin).

Q6.2

The area of the core-loss loop display is seriously affected by any unwanted phase shift in the oscilloscope or the integrating network, in the same manner as any Lissajous diagram changes shape. Other sources of error are tolerances in the oscilloscope scaling and the components, in the measuring circuit. In calculating core loss per unit volume further errors arise because of uncertainties in dimensions, especially in relation to the effective cross-sectional area of the core, for which the space occupied by interlamination insulation has to be taken into account.

Q6.3

It is important to keep the air gaps in the core of a power transformer small in order to keep the magnetising current small.

Object To derive an electrical circuit which will ac as a model of the transformer on no-load an which can then be extended to represent the complete transformer. Equipment Required Qty Apparatus 1 TT179 Transformer Trainer 1 PS189 Power Supply* 1 EW604 Electronic Wattmeter* Time Required One hour Pre-requisites Introductory Survey, Assignment 5 and 6. * For alternative units, see Operating Notes

Discussion and Experimental Procedure Many electrical machines, particularly trans- formers, a.c motors and generators can be more easily analysed and their performance better understood with the aid of an equivalent circuit in which the effects of core loss, copper loss, flux leakage, etc, are represented by electrical resistances and inductances. An equivalent circuit will produce the same phasor diagram as the machine itself and the values used to construct either diagram are derived from tests on the actual machine. In Assignment 5 we drew a phasor diagram for a transformer on no-load, showing the relationship between primary voltage, primary current and flux in the core where it was assumed that the power loss in the core was negligible. In this assignment measured values of core loss, primary voltage and current will be used to construct an equivalent circuit and a more detailed phasor diagram. The Equivalent Circuit A real transformer with no-load on its secondaries may be represented as an ideal transformer with no core loss and which requires zero magnetizing current plus two parallel elements Rc and Xm, as in fig 7.1.

Fit. 7.1 The resistance Rc is the core loss element. The current through this will be in-phase with the applied voltage and will dissipate' power equivalent to that of the core at a specified voltage and frequency. The reactance Xm is the magnetizing element. Its current will lag the applied voltage by 90°, i.e it is in quadrature with the applied voltage, and no power is dissipated. The current taken by Xm produces the magnetomotive force which sets up the flux in the core. By measuring the current and power taken from the supply, as shown in fig 7.2, we can derive values for the elements of the equivalent circuit and construct the phasor diagram.

Fig. 7.2 Practical 7.1 Open circuit test Switch off the mains supply to the TT179 and PS189. On the Measurements Console, referring to fig 7.3; Set the meter ranges as shown Make the mimic diagram connections Connect the Dissectable Transformer, Watt-meter and PS 189 to the TT179 Replace the safety cover Set the PS 189 output to zero and the AC/DC switch to AC. Set the Wattmeter ranges to 1 ampere and 200 volts. Switch on the PS189 and TT179. Raise the PS 189 output to 120 volts and read on V1. Measure the voltage, current and power in the primary and the voltage of the secondary winding. Record your results as in fig 7.4. Primary volts V1

Primary current Io

Power P1

Secondary volts V2

120V Fig 7.4 These test results can be used to construct a phasor diagram similar to that shown in fig 7.5. We will first calculate the phase angle between the current I1 and the primary voltage V1, then derive values for the core loss and magnetizing currents.

Fig 7.3

Fig. 7.5 Let

P1 = power input (wattmeter W1) V1= voltage applied to primary (meter V1) V2= voltage across secondary (meter V2) IO = total primary current on no load (meter I1) Ic = in-phase component of current 10 (core loss component) Im= quadrature component of current 10 (magnetizing component) Φ = phase angle between V1 and IO

We can now calculate the currents through the core-loss resistance Re and the magnetizing reactance Xm, also the phase relationship between these currents and the primary voltage. A word of explanation is needed as to why we refer to the total primary current on noload as 10 instead of I1. IO represents the phasor sum of the core-loss current I c and the magnetizing current 1m as shown in fig 7.5. When the trans- former is supplying no external load this is the total current taken by the primary, therefore for this condition IO = I1. When the transformer is supplying load there is a large additional current flowing in the primary and in this case II is not equal to 10 but to the phasor sum of 10 and primary load current component. The volt-amperes taken by the primary on no- load is VA1 = V1 IO and the power input is P1 = V 1 X 10 cos Φ ∴cos φ =

and

P1 V1VO

Ic = IO cos Φ (core-loss component in phase with applied voltage)

Im = IO sin Φ (magnetizing component in quadrature with applied voltage) Exercise 7.1 Calculate cos Φ, the angle Φ, I c and 1m from the test results recorded in fig 7.4 and construct the phasor diagram. Values of Rc and Xm Now that the currents Ic and Xm have been evaluated we can find the values of the equivalent core-loss resistance Rc and magnetizing reactance Xm. Rc =

V1 Ic

:

Xm =

V1 Im

Exercise 7.2 Calculate values for Rc and Xm. Draw the equivalent circuit for the transformer on noload and insert these values into it. Referred values When a measurement has been made on one winding of a transformer, it is often convenient to think in terms of the equivalent effect as it would apply to another winding. For instance if the transformer you have been using were to be used in reverse, with the secondary winding now used as the primary, the current required to support the same flux in the core would be multiplied by the factor T 1/T2. Also the voltage supported by that flux would be multiplied by T2/T1. Exercise 7.3 Deduce the turns ratio from your results of fig 7.4. Use the result to calculate the values of Rc and Xm which would have been found if you had performed the open-circuit test on the secondary winding instead of the primary. The equivalent circuit which you have constructed in this exercise provides a model of the transformer on no-load which is compatible with the phasor diagram. At this stage it does not provide a complete model, since the effect of load current in the winding impedances has not been considered. In the next assignment we will add the circuit elements necessary to complete the equivalent circuit of the transformer.

Object To measure the power taken by the transformer with secondary short-circuited and passing load current. To predict the efficiency of the transformer when operating over a range of output loads. Equipment Required Qty Apparatus 1 Transformer Trainer TT179 1 Power Supply PS189* 1 Electronic Wattmeter EW604* Time Required One and half hour s Pre-requisites Assignments 5, 6 and 7 * For alternative units, see Operating Notes

Typical results Practical 7.1 Primary volts V1

Primary current IO

Power P1

Secondary volts V2

120V

0.78A

24W

63.5V

Specimen calculations Exercise 7.1 cos φ =

=

P1 V1 I O 24 120 × 0.78

= 0.256 Φ = arc cos 0.256 = 75o Ic

= IO cos Φ = 0.78 x 0.256 = 0.20A

Im

= IO sin Φ = 0.78 sin 75o = 0.75A

Exercise 7.2 Rc = V1/Ic = 120/0.2 = 600Ω Xm = V1/Im = 120/0.75 = 160Ω

Exercise 7.3 The turn The turns ratio from Practical 7.1 is T2/T1 = 63.5/120 = 0.53

The secondary voltage would therefore be 0.53 times as great for the same core flux, and the magnetizing current would have been 1/0.53 times greater. Thus R c and Xm would each have been (0.53)2 times as great. The referred values would have been: for Rc, (0.53)3 x 600 = 169 ohms for Xm, (0.53)2 x 160 = 45 ohms Object To measure the power taken by the transformer with secondary short-circuited and passing load current. To predict the efficiency of the transformer when operating over a range of output loads. Equipment Required Qty Apparatus 1 Transformer Trainer TT179 1 Power Supply PS189* 1 Electrdnic Wattmeter EW604* Time Required One and a half hours Prerequisites Assignments 5, 6 and 7 * For alternative units, see Operating Notes

Discussion The last assignment examined the behaviour of the transformer on no load. Now suppose that a load current 12 flows in the secondary winding. It will tend to reduce the flux in the core and hence the emf’s including the back-emf in the primary.1he result is an increase in primary current which can be calculated from the changes in mmf (ampere-turns). On no load the primary mmf was IOT1. Additional mmf required is I2 T2 to cancel the secondary's demagnetising mmf. Thus the primary must supply a total mmf of IOT1 +I2T2. The current in the T1 turns of the primary thus becomes I1 = IO + I2 T2/T1. From this we can see that the load on the secondary has- in effect been transferred to the primary and the additional primary current is taken from the power supply. Winding resistance Each winding of a transformer has resistance. When a current flows this gives rise to a voltage drop and to a power loss, usually referred to as the ‘copper loss’. Copper loss = I12 R1 + I22 R2 Leakage Reactance The flux produced by current flow in the primary turns does not all link with the secondary turns, as shown in fig 8.1 a. Similarly, the counter-flux produced by load current in the secondary turns does not all link with primary turns. The effect of leakage flux in a real transformer is similar to having an ideal transformer in which all the flux links both windings, plus separate inductors connected in series with the primary and secondary as in fig 8.1 b. In the equivalent circuit reactances X 1 and X2 will represent the leakage components in series with the primary and secondary windings. Completing the Equivalent Circuit By adding the winding resistances and leakage reactances to the equivalent circuit of the transformer on no-load we can extend it to form a complete model of the transformer. The degree of complexity to which the equivalent circuit is taken will depend on the particular application for which it is to be used.

Short-circuit test The magnitude of the total effective winding resistance and leakage reactance can be found by placing a short-circuit across the secondary, and finding out how much voltage at the primary terminals is needed to drive rated current through the transformer. This voltage will be quite small, so that the core losses (the effect of R c and Xm) will be negligible. The required circuit is shown in fig 8.3.

Fig. 8.2

Fig. 8.3 There is however a preliminary test which we shall undertake in order to overcome some limitations in our laboratory situation. In industry the testing of a transformer would be carried out with high-accuracy instruments, regularly recalibrated. Also it happens that in large transformers (for which open- and short-circuit tests are most valuable) the results are not usually nearly so dependent on measurement errors as in our small one. The

reasons will become apparent later. But to ensure that you get good results it will be advisable to check that the voltmeter ammeter and wattmeter are consistent with one another. Practical 8.1 Meter check Switch off the TT179 and set the PS189 output to zero. Connect the wattmeter and primary meters to the load unit, as indicated in fig 8.4. Note that the Dissectable Transformer is not in circuit. Set the meter ranges as shown, and set the wattmeter ranges to 20V and 2A. Switch ‘on’ all the load resistors on the LU188 load unit. Lower the safety cover and switch on the TT179. Set the TT179 load switch to ‘on’ and raise the output of the PS189 gradually until the indicated current is 1.5A. Then read the voltmeter and the wattmeter, recording all three readings. Calculate the power from the product of the voltage and current, and compare this result with the power shown by the wattmeter. Unless they are within 1% of each other, calculate the factor which must be applied to the wattmeter reading to make the result agree with the volt- ampere product. This factor will be used later. Practical 8.2 Short-circuit Test Switch off the TT179 and set the PS189 output to zero. On the Measurements Console, referring to fig 8.5: Set the meter ranges as shown Make the mimic diagram connections Set the Wattmeter ranges to 2 amperes, 20 volts. Switch on the TT179 and raise the PS189 output voltage gradually until the current through the ammeter 12 reads 2A. Measure the Voltage, current and power in the primary winding and record your results as in fig 8.6. Increase the secondary current in 0.5A steps up to 2.5A recording the primary voltage current and power at each step. The power reading should be corrected by applying the factor (if any) which you found in Practical 8.1. Primary Voltage V1

Primary Current I1

Power W1 (PSC)

Fig 8.6 Short-circuit test

Secondary Current I2 2A

That completes the short-circuit test as It is normally performed, but in order to discover a little more about it we shall do a further related test. Practical 8.3 Core loss during short-circuit test Switch off the TT179. Without disturbing anything else, raise the protective cover, remove a link to break the secondary circuit and replace the cover. Switch on once more and read the power indicated by the wattmeter. You will find it appropriate to change the current range. Make a note of the power as ‘core loss with the short-circuit removed’. The value of core loss during the short-circuit test would have been much less than the figure you have just measured, since the core flux will have been roughly halved by the reduction in voltage due to current in the primary resistance and leakage inductance. Since eddy current loss is proportional to the square of the flux density, and hysteresis loss nearly so, a fair estimate of the core loss in the short-circuit test is about one quarter of the loss you measured with the short-circuit removed.

Fig. 8.5

Q8.1

What percentage error arises if copper loss during the short-circuit test is as sumed equal to the total loss during the test?

Calculations The power W1 which you have measured in the short-circuit test is unlikely to be the same as V1 I1 calculated from your measurements. Let us suppose for the moment that the impedance at the primary terminals is a series combination of resistance R 1' and reactance X1'. Then, since no power is dissipated in a reactance, so that

W1 = I12 R1’ R1’ = W1/I12

also the total impedance is V1 = I1

R1 ' 2 + X 1 ' 2

so that X1’ can be calculated as X 1 ' = ( R1 / R2 ) 2 − R1 ' 2

Notice that only a single resistance and a single reactance value are obtained from these calculations. We must find out how they are related to the individual primary and secondary resistances and reactances. Referred values In just the same way as the core losses in Assignment 7 were referred to the secondary, having been established for the primary, so with the winding resistances and reactances the secondary contribution can be referred to the primary (or vice versa). The same factor is applied, the square of the turns ratio. Thus, if the core magnetising and loss components are neglected, the equivalent circuits of the transformer shown in figs 8.7a, band c are equally valid. Fig 8.7b is the circuit relevant to your practical results, since if in this circuit the ideal transformer is shortcircuited at the secondary terminals the primary of it becomes effectively a short-circuit also (since no voltage can exist across it).

Fig. 8.7 Equivalent circuit (neglecting core loss) The winding resistance and reactance referred to the primary are T  R1 ' = R1 +  1  T2 

and

2

T  X1'= X1 +  1  T2 

R2 2

X2

The complete equivalent circuit of course requires the components representing the core magnetisation and losses, Rc and Xm, to be added.

An exact analysis would show that slightly different values would be appropriate in each different equivalent circuit, but in a typical transformer the values of Rc and Xm are so much higher than the series impedances referred to the primary that the same value very nearly is appropriate when added to any of the circuits infig8.7. Exercise 8.1 Calculate values of R1' and X1' from fig 8.6. Draw the equivalent circuit of the transformer in the form having all impedances on the primary side of the ideal transformer. Mark in the values you have found, including values of Rc and Xm from Assignment 7 (your own results if you have them, or if not take the results printed at the end of Assignment 7). Practical Considerations It may seem that it would be far simpler to measure the winding resistance with an ohmmeter, and the leakage reactance may seem rather uncertain from your measurements. But consider the situation with a very large transformer. Large transformers have thick conductors in order to pass large currents without excessive heating. The resistance is often too low to measure with an ordinary ohmmeter, and in any case may differ between a.c and d.c values, so R 1' is better determined by the ammeter and wattmeter method using a.c. You may have found the, reactance uncertaion because its calculation involved a small difference between two large quantities. But in a large transformer the reactance is usually much larger than the resistance, and in this case the errors of measurement matter far less. Practical 8.1 Meter check Ideally, with 1.5A in 15n., the remaining readings would be 22.5V and 33.75W. In practice errors of ±2% full scale are likely in the voltmeter and ammeter, and ±5% full scale is possible in the wattmeter. Taking the worst case the errors must be calculated as full scale x the stated percentage reading when applied to the reading. Thus the worst case would be 35 5 40 × 2% + × 2% + × 5% = 15% 22.5 1.5 33.75

Such a bad result is in practice unlikely.

Practical 8.2 Short-circuit test Primary Voltage V1 17.7V

Primary Current I1

Power W1 (PSC)

2.13A

34.4W

Secondary Current I2 2A

Practical 8.3 Core loss during short-circuit test 0.36W loss with the short-circuit removed. Q8.1

Assuming that during the short-circuit test the core loss is one-quarter of the value measured in Practical 8.3, i.e 0.09W, it will be 0.3% of the power measured, and will therefore contribute a negligible error.

Exercise 8.1 R1’ = W1/I12 = 34.4/(2.13)2 = 7.6Ω X1’ = =

(V1 / I 1 ) 2 − R1 ' 2 (17.7 / 2.13) 2 −(7.6) 2

= 3.4Ω

Fig 8.8

Object To measure the temperature rise of a transformer on load and to calculate its efficiency and voltage regulation when hot. Equipment Required Qty Apparatus 1 Transformer Trainer TT179 1 Power Supply PS189* 1 Load Unit LU 178* 1 Electronic Wattmeter EW604* 1 Mercury Thermometer (0 to 100°C) 1 Clock (to measure ambient temperature) Clock (to measure the duration of the test in minutes.) Time required Two and a half hours Prerequisites Reading of Introductory Survey, Assignments 7 and 8. *For alternative units, see Operating Notes

Discussion and Experimental Procedure The principal aim of this assignment is to measure the temperature rise of a transformer from first switching on the load until a steady temperature has been reached. We shall also calculate efficiency and regulation at the end of the heat run. In the practical which follows, the primary voltage will be set to its nominal value of 120volts at the start of the test. As the test proceeds, the primary voltage will be maintained at its nominal value and readings will be taken of core temperature, primary current, primary power together with secondary voltage, current and power until the transformer has reached its steady temperature. From these readings the efficiency and temperature rise above ambient can be calculated. At the conclusion of this practical the load is switched off and a reading taken of the secondary voltage on open circuit with the primary voltage set to its nominal value. This enables the voltage regulation to be calculated. The circuit diagram for the load test is given in fig 9.1. This shows two wattmeters, but when using the TT:179 only one wattmeter is required, as this can be switched into either the primary or secondary circuit. Practical 9.1 Place the thermometer to be used for measuring the ambient temperature in a sheltered position about one metre from the dissectable transformer. Before switching on the mains supply refer to the connection diagram, fig 9.2. Set the meter ranges and make the mimic diagram connections shown. Connect the dissectable transformer, wattmeter, variable A.C power supply and load unit to the measurements console. Place the me rcury /glass thermometer supplied into the hole at the top of the dissectable transformer. Set the output of the PS189 Variable AC Power Supply to zero and switch on. Switch on the Electronic Wattmeter EW6Q4 and set the ranges to 1A and 200V. Switch all the load resistors on the LU178 Load Unit to ‘on’ and set the load switch on the front panel of the measurements console to ‘off’. Set the wattmeter switch initially to WI. Take initial readings of the ambient and transformer core temperatures - these should be the same. Record them in the first line of a table like fig 9.3.

Lower the safety cover on the measurements console and switch on the mains supply. Increase the output of the PS189 to obtain 120 volts across the primary of the dissectable transformer as read by Voltmeter V1. Quickly read the secondary voltage V2 and record it and the time in your table. Then turn the load switch ‘on’, readjust the primary voltage to 120V, and take readings for Tcore, Tamb, V1 , I1, W1, V2 , I2, W2. Repeat these readings at each time interval given in fig 9.3, maintaining the primary voltage V1 at the nominal value of 120 volts. The test can be continued beyond the limit of two hours shown in the table, but the additional temperature rise should only be three or four degrees. At the conclusion of the load test, switch the secondary to open circuit using the load switch on the front panel of the measurements console. Quickly readjust the primary voltage V1 to its nominal value of 120 volts and measure the secondary voltage V2.

Fig. 9.1 Circuit diagram

Fig. 9.2 Connection diagram

Time Interval min min 0 5 10 15 25 35 45 60 75 90 105 120

Tcore o C

Tamb o C

Trise o C

V1 I1 volts amp 120 120 120 120 120 120 120 120 120 120 120 120

W1 rdg

W1 W2 I2 W1 W1 watts volts amp rdg watts

0

Fig. 9.3 Temperature Rise of Core Record time, V1, I1, W1 and V2 for this last set of readings. Complete the table of fig 9.3 by filling in values for temperature rise, throughout the load test and if necessary convert the wattmeter readings to watts (where a scale factor is involved). Exercise 9.1 Calculate and record the core temperature rise above ambient from readings taken during the load test: Trise = Tcore - Tambient (oC). Plot core temperature rise and ambient temperature as shown in fig 9.4. Calculate and record the efficiency from readings taken at the beginning and at the end of the load test: Efficiency =

output power ×100(%) input power

Calculate and record the voltage regulation from readings of secondary voltage taken at the beginning and at the end of the test: Regulation =

open circuit voltage − ful load voltage ×100(%) open circuit voltage

Q9.1

Referring back to the equivalent circuit of the transformer, as given in Assignment 8, give one reason for the drop in secondary voltage which occurs as the temperature of the transformer increases.

Heat Transfer from the Transformer When a transformer is connected to its supply, heat will be generated by iron losses within the core (these are constant for a fixed primary voltage) and by copper losses due to current flowing in the primary and secondary windings (these are proportional to the square of the load current). This internal power loss will cause the transformer temperature to rise until the rate at which heat is generated internally is equalled by the rate at which heat is removed by convection, conduction and radiation. Heat produced by iron losses is transferred by conduction to the core surface, while that due to copper loss by conduction through the copper conductors and surrounding insulation is transferred to the winding surface. Most of the heat is then removed by natural convection to the surrounding air in the case of small transformers and by natural or forced convection to the cooling oil in the larger power transformers. As the heat flow is outward from the region where heat is generated the hottest parts will be the least accessible and the designer must take account of the difference between the temperature rise at the point of measurement and the temperature rise at the ‘hot spots’ of the transformer. The rate at which heat is removed is proportional to the temperature difference between the transformer and the cooling medium. On first switching on to a steady load the transformer will be at the same temperature as its surroundings and initially most of the heat produced by the iron and copper losses will be stored in the body of the transformer, causing its temperature to increase. As its temperature continues to rise the temperature difference between the transformer and the cooling medium also increases (assuming that the cooling air or fluid is at a steady temperature) causing a proportionate increase in the rate at which heat is removed. As the transformer approaches its steady temperature more heat is transferred to the cooling medium and less into raising the body temperature. Eventually the temperature rise will be sufficient to cause all the internal heat to be transferred to the coolant and the body of the transformer will then remain at a steady temperature. It can be seen that improving the rate at which heat is removed from a transformer leads to a reduction in its operating temperature. For most small power transformers which are mounted in an enclosure it is important to provide adequate ventilation holes to allow cooling air to pass freely over the core and windings (some small transformers can be cooled by conduction of heat to a metal case which itself is adequately ventilated). We can summarise this by saying that in a transformer (or other machine) the final steady temperature will depend on the rate at which heat is generated- this is proportional to the internal power losses which are related to the loading and efficiency of the transformer, and on the rate at which heat is transferred to the surroundings - this is dependent on the

temperature difference between the transformer and its coolant, on the surface area exposed to the coolant and on the rate at which fresh coolant can be brought into contact with the transformer. Q9.2

A small power trensformer, supplying an external load, is operating at a temperature higher than that specified by the manufacturer. Assuming the transformer is not faulty, give two possible causes and steps which could be taken to reduce the operating temperature.

Practical Considerations & Applications The user of a power transformer will be mainly concerned with its efficiency, voltage regulation and temperature rise, so that he can work out how much power will be drawn from the supply, what voltage will be obtained from the secondary and what the operating temperature will be for a given supply voltage, applied load and ambient temperature. The dissectable transformer is deliberately designed to have higher copper and iron losses than would normally be specified for a 120VA transformer. This is to enable their effect on voltage regulation and efficiency to be clearly shown and more easily measured. It is worth mentioning at this point some of the tests which are carried out by the manufacturer to meet a standard specification (such as British Standards Institute BS2214 - Performance of Power Transformers not Exceeding 2kVA). These will include a fullload heat run, which requires the transformer to be operated at full- load and at rated primary voltage until the rate of change of surface temperature does not exceed 1°C per hour. The temperature rise of the windings is calculated from measurements of the winding resistance made at ambient temperature and at the final temperature.  R − R1   (T1 + 234.5) Temperature rise ( o C) =  2  R1 

Where T1 is the ambient temperature R1 is the resistance at ambient temperature R2 is the resistance at final temperature The mean temperature of the windings for a given ambient temperature can be derived from this calculation and should be within a specified value (in BS2214 this is 110°C). The secondary voltage, again at the rated primary voltage, is measured at the start and finish of the full-load heat run and should be within ±5% of its nominal value for windings giving an output voltage of more than 100 volts. When specifying voltage regulation it is usual to express this as the change in secondary voltage between no-load and full-load as a percentage of the no-load voltage:

Voltage regulation =

no − load voltage − ful − load voltage ×100(%) no − load voltage

An alternative definition which may be encountered in some transformer specifications, expresses voltage regulation as the change in v secondary voltage between no-load and full-load as a percentage of the full-load voltage. I n the heat run carried out in this assignment we have simplified the experimental procedure by using a fixed value of load resistance throughout the test and by measuring the core temperature only, omitting measurement of the hot and cold winding resistance. The Application Note-which follows the assignments in this manual gives the procedure for measuring winding resistance and for calculating the mean temperature rise of the windings.

Fig. 9.4

Results Practical 9.1 Time min

Interval min 0 5 10 15 25 35 45 60 75 90 105 120

Tcore o C 18 25 30 35 45 54 61 69 76 79 81 83 83

Tamb o C 18 18 18 18 18 18 18 18 19 19 19 19 19

Trise o C 0 7 12 17 27 36 43 51 57 60 62 64 64

V1 volts 120 120 120 120 120 120 120 120 120 120 120 120 120 120

I1 amp 2.50 2.50 2.50 2.45 2.45 2.40 2.40 2.40 2.40 2.40 2.40 2.40 0.80

W1 rdg 0.71 0.70 0.69 0.69 0.68 0.68 0.68 0.68 0.68 0.68 0.68 0.68

W1 watts 284 280 276 276 272 272 272 272 272 272 272 272 24

Exercise 9.1 Efficiency at the start was

Regulation at the start was

232 × 100% = 81.7% 284

128 − 115 × 100% = 10.2% 128

Efficiency at the end was

Regulation at the end was

216 × 100% = 79.4% 272

127 − 111 × 100% = 12.6% 127

W2 volts 125 115 114 114 113 113 112 112 112 112 112 111 111 127

I2 amp 0 1.90 1.85 1.85 1.85 1.85 1.80 1.80 1.80 1.80 1.80 1.80 1.80 0

W1 rdg 0.5 8 0.5 6 0.5 6 0.5 6 0.5 5 0.5 5 0.5 4 0.5 4 0.5 4 0.5 4 0.5 4 0.5 4

W1 watts 232 224 224 224 220 220 216 216 216 216 216 216

Object To calculate the voltage regulation and efficiency of a transformer at different loads from measurements made in open-circuit and short-circuit tests. Equipment Required Qty Apparatus 1 TT179 Transformer Trainer 1 PS189 Power Supply* 1 EW604 Electronic Wattmeter* Time Required Two hours Pre-requisites Assignments 8 and 9 * For alternative units see Operating Notes

Discussion In small power transformers it is not difficult to measure the regulation (voltage drop with load) and the efficiency directly. With large transformers there are difficulties. It is not always easy to provide either the power or the load; and efficiency measurements make excessive demands for measurement accuracy. A large transformer might be typically 98% efficient. If wattmeter measurements of input and output power were each subject to as little as ±1 % possible error, the losses might be assessed as anything from zero to twice the true value. In practice therefore the equivalent circuit is extensively used as a means of predicting the regulation and t he efficiency of large trans- formers. As Assignments 7 and 8 have pointed out, the equivalent circuit is ascertained by means of the-open- and short-Circuit tests. By extending the short-circuit test useful results of direct relevance to the efficiency calculations can be obtained. Also by referring the series elements of the equivalent circuit to the secondary, as indicated in fig 10.1, the calculations on regulation are simplified.

Fig. 10.1 In this assignment we shall first measure the secondary voltage on open-circuit with rated supply voltage applied to the primary, fig 10.2(a). We shall then measure the power and current in the primary circuit with the secondary passing various currents on short-circuit, fig 0.2(b). Practical 10.1 Open-circuit Test For the open-circuit test we apply 120 volts to the primary and measure the secondary voltage on no load and the core loss. Set up the equipment as shown in fig 10.3. Set the meter ranges. Wattmeter settings-are 200V, 2A. Make the mimic diagram connections Connect the Dissectable Transformer, Watt-meter and PS189 to the TT179. Set the PS189 output initially to zero and the AC/DC switch to AC.

Lower the safety cover, switch on the PS189 and TT179. Raise the PS189 output to 120 volts as read on meter V1. Measure the secondary voltage on meter V2 and the power shown on wattmeter W 1 corrected if necessary using the factor found in Practical 8.1. Record the results as in fig 10.4. This completes the open-circuit test. Switch off the TT179 and reduce the PS189 output to zero.

Fig. 10.2 Test OC SC

Primary voltage V1 120V

Primary current I1 -

Primary power W1 -

Secondary voltage V2 0 0 0 0 0

Fig. 10.4

Secondary current I2 0 0.5A 1.0A 1.5A 2.0A 2.5A

Fig. 10.3

Practical 10.2 - Short-circuit test For the short-circuit test make the following changes to the connections and settings of fig 10.3. Disconnect voltmeter V2 Link terminals Hand R Set the load switch to ‘on’ Set ammeters A1 and A2 to the 5A range Set voltmeter V1 to 35V range Set the wattmeter switch to W1' Switch on the TT179 and carefully raise the output of the PS189 until ammeter A2 reads 0.5A. Measure the primary voltage, current and power and record them in the table. Repeat these measurements for secondary currents at 0.5A intervals up to 2.5A. Transformer Efficiency Efficiency η = output power/input power

(1)

From this equation we can see that efficiency could be measured directly by taking wattmeter readings of input and output power with the transformer on full load. However, this method can lead to inaccuracies, as the wattmeter error on full load will be significant, particularly for transformers whose efficiency is over 90%. A more accurate method is given below. Frome quation (1), η =

output power input power

but output = input – losses therefore η =

input losses losses =1 − input input

this can be exp ressed as η = 1 −

losses output + losses

  ( POC + PSC ) therefore η % = 100 × 1 −   output + ( POC + PSC ) 

(2)

where POC and PSC are the losses found in the open- and short-circuit tests respectively. . .

Exercise 10.1 Full load is 2A at 120V. Complete a table as in fig 10.5, using the watt- meter readings of fig 10.4 for POC and PSC. Calculate the efficiency using equation 2. Exercise 10.2 Plot a graph of efficiency against load similar to that shown in fig 10.6.

Fig. 10.6 Percent Full Load 25% 50% 75% 100% 125%

Output Watts

Core Loss POC

Copper Loss PSC

POC + PSC (A)

Output + (POC + PSC) (B)

60 120 180 240 300 Fig. 10.5 Efficiency/Load Table

Evaluating R2’ and X2’

A/B

Efficiency 100 (1 – A/B)

In our evaluation of R2’ and X2’ we need to know the turns ratio of the secondary to the primary windings. In practice it is not always possible to obtain the number of turns on the primary and secondary whereas we can easily measure the voltage ratio directly from the open-circuit test. The two ratios should correspond well to one another. T T2 ≈ 20 where V10 and V20 are the primary and secondary terminal voltage on openT1 T10 circuit.

The open-circuit test also gives the no-load secondary voltage with rated voltage applied to the primary. This is required in our calculation of percentage regulation. From the measurements taken in the short-circuit test we can first evaluate the impedance circuit test we can first evaluate the impedance Z 1’ and resistance R1’ as seen at the primary side Z1 ' =

V ISC I ISC

R1 ' =

W ISC ( I ISC ) 2

and then refer these to the secondary by multiplying by the square of the turns ratio (or voltage ratio). Knowing Z2’ and R2’ we can then derive X2 V Z 2 ' = ISC I ISC

 T2  T  1

WISC R2 ' = ( I ISC ) 2

   

2

 T2   T1

  

2

X 2 ' = ( Z 2 ' ) 2 − ( R2 ' ) 2

where Z2’ = impedance referred to secondary R2’ = winding resistance referred to secondary X2’ = leakage reactance referred to secondary T2/T1 = turns ratio (for this we shall use the voltage ratio on open-circuit, V20/V10) VISC, IISC and WISC are values obtained in the short-circuit test. From these expressions and the measurements recorded in fig 10.4 we can complete the table given in fig 10.7. Voltage V20/V10 = n

Ratio Z2’ = VISC/IISC x n2

R2’ = WISC/(IISC)2 x n2

X 2 ' = ( Z 2 ' ) 2 − ( R2 ' ) 2

Fig. 10.7 Exercise 10.3 Using your results for I2 = 2A in the short-circuit test, complete a table as indicated in fig 10.7, and draw the equivalent circuit as in fig 10.1 showing your values of R2’ and X2’. (These will be used for regulation calculations, in which the core loss elements can be neglected). Voltage Regulations We have now reached a point where we can calculate the voltage drop across the impedance Z2’ for a given value of load current in the secondary winding. We can use this to find the regulation as a percentage of the no-load secondary voltage. The load applied to the transformer secondary may be purely resistive or have an inductive or capacitive component, e.g an induction motor or a long unloaded transmission line. Our calculation of voltage regulation must therefore take account of the nature of the load. In fig 10.8(a) the transformer on load is shown as an ideal transformer with zero winding impedance plus components representing leakage reactance and winding resistance referred to the secondary side.

Fig. 10.8 The voltage V2 applied to the load at the secondary terminals will be the voltage E 2 produced across the secondary of the ideal transformer but reduced by the voltage drop across the impedance formed by R2' and X2'. From the equivalent circuit we can derive a phasor diagram such as that of fig 10.8(b) showing these voltage components and their phase relationship. In this diagram the load is inductive and the secondary current 1 2 will lag the secondary terminal voltage V2 by angle Φ. We can see that if the load is disconnected there will be no voltage drop across R2' or X2' and the no-load or open-circuit voltage V20 across the secondary terminals will be the voltage E2. On load the internal voltage E2 is equal to the phasor sum of the terminal voltage V2, the voltage drop across the winding resistance which is in phase with the load current, and that due to the leakage reactance which is in quadrature with it. As the inherent regulation is the difference between the terminal voltage at no-load V20 and the terminal voltage on load V2 we can use the phasor diagram to obtain an equation for the percentage regulation in terms of load current I2, phase angle Φ, and the impedance formed by , R2' and X2'. However this will produce a fairly complex expression and for most purposes we can derive a simplified equation. From the phasor diagram of fig 10.8(b) we can see that the expression (V 2 + I2 R2' cos γ + I2 X2' sin γ) is nearly equal in magnitude to V20 Hence V20 – V2 ≈ I2R2’ cos γ + I2X2’ sin γ

V20 − V2 ×100 V20

% Re gulaton =

≈

I 2 R2 cos γ + I 2 X 2 sin γ V20

As an example of the use of this equation let us first calculate the percentage regulation when the transformer is supplying a purely resistive load (phase angle γ is zero). We shall assume values for V20, R2' and X2' which should be similar to those obtained in your practical results. Let

V20 = 128 volts (no-load terminal voltage) I2 = 2.0 A (secondary load current) R2’ = 8.6 ohms (referred winding resistance) X2’ = 3.9 ohms (referred leakage reactance). γ = 0o (phase angle)

% Re gulaton ≈

I 2 R2 cos γ + I 2 X 2 sin γ × 100 V20

Resistive Load Cos γ = 1, sin γ = 0 Re gulation =

2.0 × 8.6 ×1 + 0 ×100% 128

= 13.5% For inductive or capacitive loads the secondary current will lag or lead the terminal voltage by phase angle γ. This is sometimes expressed in terms of ‘power factor’ where power factor = cos γ. Thus a power factor of 0.5 lagging represents an inductive load whose phase angle is 60°. Again using the values given above let us calculate the voltage regulation for a power factor of 0.6 lagging and 0.5 leading. Inductive Load Cos γ = 0.6, sin γ = 0.8 Re gulation =

2.0 × 8.6 × 0.6 + 2.0 × 3.9 × 0.8 ×100% 128

= 12.9% Capacitive Load

Cos γ = 0.5, sin γ = 0.87 In this case the term involving sind γ will be negative as shown by the phasor diagram in fig 10.9 Re gulation =

2.0 × 8.6 × 0.5 − 2.0 × 3.9 × 0.87 ×100% 128

= 1.3%

Fig. 10.9 Phasor diagram for capacitive load Exercise 10.4 Use the results obtained in Practical 10.1 to calculate the voltage regulation of the Dissectable Transformer for a load current of 2.0A at 0.6 power factor, lagging. Q10.2 Do you think it is possible for a transformer to have a full-load terminal voltage which is greater than its no-load terminal voltage? What load condition might produce this result?

Practical Considerations In many transformer applications it is important to limit the variation in secondary voltage which will occur as the load is changed. In power system work the electricity supply authority will normally specify the maximum and minimum voltage which will be provided at the consumer terminals (in the UK this is 240 volts ± 6%). In power system transformers variation in secondary voltage may be limited by automatic tap-changing equipment which senses the output voltage and effectively alters the turns ratio to suit the load. However the attainment of low inherent regulation can only be brought about by reducing the winding resistance and leakage reactance of the transformer. This in turn requires an increase in the cross-sectional area of copper in the winding and of iron in the core. Both these steps would lead to an increase in size and cost, so a compromise is reached and a value of regulation is specified which is acceptable to the user. British Standard 2214 ‘The Performance of Power Transformers not Exceeding 2kV A’ calls for a regulation not greater than 10% unless otherwise specified by the purchaser. In assessing regulation, direct measurement from load tests is to be preferred over calculated values derived from open- and short-circuit tests. In: small transformers this is often possible but for large transformers it may not be practicable to provide the load required; indirect methods must then be used. The method described in this assignment is in common use though a lengthier and more precise expression for percentage regulation is sometimes employed, particularly where this figure is less than 5%. The measuring instruments will also be calibrated against recognized standards at regular intervals. It will be noted that the efficiency of the dissectable transformer was greatest at its fullload rating. It can be shown that efficiency is a maximum when the variable copper loss is equal to the fixed core loss. In power transformers which operate over a range of load conditions it is usual to design the transformer so that the copper loss is equal to the core loss at around 50% of full load, giving an efficiency/load curve similar to fig 10.10. For this reason, when specifying a transformer it is best to provide information on the load range over which it will operate.

Fig. 10.10 Results Practicals 10.1 and 10.2 Test OC

Primary voltage V1 120V

Primary current I1 -

Primary power W1 -

Secondary voltage V2 127V

Secondary current I2 0

SC

3

0.58

1.85

0

0.5

7

1.13

7.2

0

1.0

10.8

1.68

16.3

0

1.5

14.75

2.22

32

0

2.0

19.5

2.78

49

0

2.5

Exercise 10.1 Percent Full Load 25%

Output Watts

Core Loss POC

Copper Loss PSC

POC + PSC (A) 25.85W

Output + (POC + PSC) (B) 85.85W

60

24@

1.85W

50%

120

24

7.2

A/B 0.30

Efficiency 100 (1 – A/B) 70%

31.2

151.2

0.206

79.4%

75%

180

24

16.3

40.3

220.3

0.183

81.7%

100%

240

24

32

56

296

0.189

81.1%

125%

300

24

49

73

373

0.196

80.4%

Exercise 10.3 Voltage V20/V10 = n 1.058

Ratio Z2’ = VISC/IISC x n2

R2’ = WISC/(IISC)2 x n2

X 2 ' = ( Z 2 ' ) 2 − ( R2 ' ) 2

7.27Ω

1.56Ω

7.44Ω Fig. 10.11

Object To investigate the use of a transformer for establishing a desired ratio between a primary input current and a secondary load current. Equipment required Qty Apparatus 1 TT179 Transformer Trainer 1 LU178 Load Unit* 1 PS189 Power Supply* 1 AC ammeter 0 to 100mA. Time Required One hour Prerequisites Operating notes, Assignments 1 and 6.

Discussion and Experimental Procedure Current transformers are simply transformers designed and/or applied with an emphasis on the ratio of the currents in the primary and secondary windings, rather than on the voltage ratio. As mentioned in the Introductory Survey, applications are usually in the fields of measurement and protection, and therefore require the secondary current to be accurately related to the primary current in a known ratio. The ratio will usually be chosen to bring the secondary current to a convenient value for use in instruments and protective equipment. Current transformers may be constructed in any of the forms, core, shell and toroid which are used for other transformers and described in the Introductory Survey, differences in design being largely a matter of choosing the proportions to give a different distribution of losses between iron and copper, to suit the different application. The current transformer with a bar primary, fig 11.1, is a common form, and an extreme example-of this different distribution of losses.

Fig. 11.1 The one-turn primary with its great length of conductor may have large voltage-drop and associated power losses in the conductors, but this is not relevant to its function as a current transformer. Ratio of cu rrent transformer

As you have already discovered, the primary and secondary ampere-turns for an ideal transformer are equal, i.e

I1 T 1 = I 2 T 2 I1/I2 = T2/T1

To see physically why this is so, consider fig 11.2, in which a current I 1 flowing in the primary of the current transformer is to be measured by the meter in the secondary circuit. Suppose that when I1 is 50A the meter is to show its full scale deflection, requiring a secondary current of 5A, and suppose also that a primary winding of two turns is chosen.

Fig 11.2 Some (usually small) voltage is required to drive 5A through the ammeter. If we assume that the secondary has 20 turns this determines the amplitude of flux change required. The primary current has to produce ampere-turns: to magnetise the core, say 10 ampere-turns in this case to cancel the demagnetising ampere-turns produced by the secondary winding, i.e 100 ampere-turns. If the corresponding primary currents were in phase, the discrepancy in this assumed case would be only 10% between turns ratio and current ratio. If the core-magnetising current were (as it ideally should be) in quadrature, the error would be only %%. An exact expression for the current ratio in the presence of losses becomes very complicated. However, for the majority of cases the current transformation ratio R defined as I1/I2 is very nearly approximated by R = N2/N1 + Io/Is sin (α + δ)

Fig. 11.3

where Io is the exciting current, i.e the resultant of the primary current necessary to magnetise the core and supply its losses, δ is the angle between the secondary current and secondary induced voltage, i.e the phase angle of the secondary circuit load impedance including the secondary leakage reactance of the transformer, and a is the angle between 10 and the working flux. The turns ratio and the current ratio can be seen to differ by an amount dependent on the exciting current. For the particular case of δ = 0, a resistive burden, R = N2/N1 + Io/Is sin α = N2/N1 + Ic/Is where Ic is the component of the exciting current which supplies the core losses. It is clearly important to keep these small. Practical 11.1 - Ratio of a current transformer Suppose that you wish to use a 100mA a.c ammeter for measuring currents up to 5A, using the dissectable transformer. The required current ratio R is 50:1, and none of the existing combinations of windings can provide this. If one of the secondary windings available is used, having 148 turns, and losses in the transformer are neglected, the primary turns required are T1 = T2/R = 148/50 = 2.96 = 3 approximately. Use the unwound bobbin and the pink lead from the accessories box to make the primary coil. Wind three turns of the lead on to the bobbin, adjust the ends to approximately equal length and secure them by knotting the last turn once. Select one of the 148-turn secondary windings, and assemble it with the three-turn winding on to the centre limb of the transformer core so that the smooth sides of the bobbins lie together for minimum separation. Assemble the transformer fully Connect the TT179, transformer, LU 178 and milliammeter as shown in fig 11.3. Make all the settings shown there, and if the milliammeter has more than one range ensure that it is correctly set to measure 100mA a.c. Connect the PS189 to the rear of the TT179Console using the lead provided. Set the PS189 for an a.c output and set its voltage to zero. Check that all switches on the LU178 are on. Switch on the equipment and adjust the voltage of the PS189 output to give a reading of 0.5A on ammeter A1. Measure the current in the secondary circuit of the current

transformer, indicated by the 100mA meter. Record the measurement in a table like fig 11.4. T1 = 3 turns I1 0.5A 1.0A 1.5A 2.0A 2.5A . . . .

I2

Fig. 11.4 Increase the primary circuit current in steps of 0.5A, taking readings of the secondary current and recording them at each step, until you reach full scale on one of the two meters you are using. Q11.1 If the primary winding is altered to one turn, what effect will this have on the current in the 100mA meter? Disconnect and dismantle the Dissectable Transformer. Rewind the primary winding with only one turn, then reassemble and reconnect as before. Switch on the equipment, and following the same steps as before increase the primary current and record readings of the secondary current. On compilation set the PS 189 output to zero and switch off the equipment. Practical considerations and applications Applications of current transformers include all kinds of situations in which a measurement or implied measurement of an alternating current is required, where the current in question is of an inconvenient value, or is at an inconvenient potential, or must be added to or subtracted from some other alternating current. The latter case may proivde occasion for the use of more than one primary winding on a single current transformer, or alternatively separate current transformers may have their secondaries interconnected to combine the secondary currents in the required way. The measurement may be direct and give rise to an indication, such as the use of an ammeter in the way you have seen in the practical, or it may take various indirect forms. Control systems for motors and the like often require signals proportional to current, for torque control, or for stabilising feedback. Protective equipment may, through the current transformer, monitor

the current in the main circuit and open the circuit if that current becomes excessive. Wattmeters, phase meters, watt-hour meters, reactive-VA meters and others may take the measure of primary current represented by the secondary current and combine it with other signals to form composite outputs which are recorded or indicated or otherwise processed as required. For a resistive burden it is mostly the core losses which must be kept low to maintain a current ratio close to the turns ratio. However it may be shown that the phase angle is (for a resistive burden) most affected by the magnetising component of the exciting current. Since the phase of the current is important in power, measurements, it follows that a current transformer is widely useful only if both components of the exciting ampere-turns are small. Current, transformers are often required to possess robustness far exceeding that required for a power-transformer of similar rating. If the current is being measured in a highvoltage line, the current transformer must be insulated to withstand the highest voltage on that line, notwithstanding that neither primary nor secondary windings may develop more than a few millivolts. Also current transformers in power systems and elsewhere must withstand any fault currents which could occur in the system: the requirement may be merely to survive, or it may be to continue working (perhaps with reduced accuracy) while passing currents perhaps fifty times the normal maximum for brief periods. Another situation in which a fault can be potentially dangerous is a break in the secondary circuit of a current transformer. In this case the primary current will not change significantly (unlike the situation in a power transformer). Instead the secondary winding develops a very large Voltage, which may break it down, or present a hazard to personnel working in the neighbourhood of the (normally lowvoltage) instrument wiring. A spark gap or other voltage- limiting device is sometimes placed directly across the secondary terminals of a current transformer to limit the excess voltage in these cases. An interesting application of the current-transformer principle is the clip-on current transformer. This device, shown in principle in fig 11.5 is a current transformer secondary with a core which can be opened on a hinge, and then closed around a conductor in which it is required to measure the current. This conductor forms the primary winding. The secondary is connected to a suitable meter. The advantage is that the instrument can be used for taking measurements at several different places in the circuit in turn without the necessity for breaking the circuit or making any electrical connections.

Fig. 11.5 Exercise 11.1 Using the results obtained in Practical 11.1 for each primary winding, plot graphs of the secondary current against the primary current. Label each graph with the number of primary turns used. From each graph calculate the current ratio R = I 1/I2, Compare this with the turns ratio T2/T1.

Fig M2

and to the main electricity supply. Set the PS189 a.c/d.c switch to a.c and set its variable control to zero, then plug it in to the electricity supply and switch it on. Press the 'on' button on the Measurement Console, which should light up. If it fails to remain alight when released, check that the transparent safety cover over the front panel is fully closed, and that the circuit breaker has not been operated. Switch on the PS189 and slowly raise its output voltage, indicated on Vl, to 120V. Watch the ammeters while you are doing this. Ammeter A1 should rise slowly until at 120V it indicates a current not exceeding 800mA. With 120V indicated by Vl, also check the reading on V2, which should be 126V. Reduce the output voltage of the PS189 to zero. Switch on the ‘load’ switch on the Measurement Console, then very gradually raise the output of the PS189. This time the voltmeter readings should be very low (zero for V2). Raise the current shown on A 1 to 5 amperes. The current shown on A2 should then be a little over 4 amperes. Switch off after completing the tests. Although they do not check every possible fault they do check everything which is likely to have been upset by transport or by normal use. On successful completion of the tests the equipment should therefore be ready for use in the assignments described in this handbook. Circuit Description The TT179 Measurement Console in use is connected to external equipment as indicated in fig M3. The circuit diagram of the Console itself is given in fig M4. The diagram may be considered in three parts, starting at the top, which is a control and protective circuit. Its essential function is to switch, via relay contacts RL 1/2 and R L 1 /3, the input from the PS189 to the Dissectable Transformer circuit, which extends along the centre of the diagram. The lower part of the diagram is concerned with metering. Power on-off circuit To switch the equipment on, SW1 ‘ON’ pushbutton is operated. This applies voltage from the mains supply via fuse F1 to the series combination of a selected resistor and a 24V pilot lamp which illuminates SW1, and to a series circuit including the coil of relay RL1. This series circuit includes resistance when the selected supply voltage is between 200 and 250V, but none when it is 100 to 120V, since the relay operates on this voltage directly. The other items in series with the relay coil are listed below, together with the corresponding functions. SW4,5

Microswitches closed by fully closing the safety cover

Fig. M3

SW12/2

Normally closed auxiliary contact on the circuit breaker. Opens when the Dissectable Transformer circuit is overloaded. Manually reset.

SW2

Normally closed 'OFF' button.

Thus pushing SW1 operates RL1 if the supply is present, the fuse is intact, the cover is closed, there is no overload and the ‘OF F’ button is released. Contact RL 1/1 in parallel with SW1 will then hold the relay so long as those conditions persist. If any of them is altered RL1 will be released and will remain so until SW1 is again operated, with all the conditions again met. Fuse F 1 is always rated at 160mA. The resistances selected by the voltage selector when set at the various voltages are as follows: Voltage range 100/120 200/220 220/250

In series with RL1 Zero R25, 5.6k R21, 6.8k

Lamp R22, 2.2k R20, 4.3k R20, 4.3k

Dissectable Transformer circuit The supply from the PS 189 enters the un it at SK1. The working conductors are marked Land N to correspond with the socket markings, but the Dissectable Transformer circuit is isolated from earth and (provided that a PS 189 or equivalent supply is used) from the main electricity supply. The 5A circuit-breaker SW12 has a thermal element which carries all the supply current in the circuit. If this current exceeds 5A for any significant time the current is directly interrupted by contact SW12/1, and contact SW12/2 releases the control relay RL1. Therefore following an overload it is necessary to reset the circuit breaker and to operate the ‘ON’ button to restore the supply. The Dissectable Transformer is, in use, connected to the row of sockets on the front panel below the mimic diagram. These sockets are internally linked to the sockets shown on the mimic diagram as the' terminations of the Dissectable Transformer. R23, used for monitoring the primary current, is mounted on the base of the console. SW11, a three-position five-pole switch, enables one wattmeter connected to the wattmeter sockets in the bottom left corner of the front panel to be switched into circuit in either of the positions indicated as W1, W2 on the mimic diagram. With SW11 at ‘OFF’ the wattmeter is disconnected from the circuit (although its current and voltage terminals remain linked). The load switch SW6 is a manually operated double-pole switch at the right end of the panel above the mimic diagram.

Metering circuits The range of each voltmeter is determined by the resistor selected by a single-pole rotary switch. When a voltage is applied the resulting current is rectified by a diode bridge, so that d.c flows in the meter which is a moving-coil type of 1mA sensitivity. In each of two identical ammeter circuits the input to the ammeter is the primary winding of a current transformer having a nominal ratio 10A/45mA. The secondary current is rectified by silicon rectifiers, and the resulting d.c current flows in a burden comprising principally a 750-ohm resistor. Across this resistor a moving- coil meter is connected in series with a resistance selected according to the desired current range by a rotary switch. Each resistance is made up of a fixed and an adjustable resistor, the latter being used to set the full-scale range of the ammeter accurately. Integrating Circuit R24 and C1, C2 provide an integrating circuit of time constant 300ms, which is used in the assignments for monitoring the flux in the transformer core. Maintenance Very little maintenance to the TT179 is required, apart from occasional lubrication of the stays of the protective cover and their pivots with a drop or two of light machine oil. To remove finger marks from the paintwork and protective cover, use a soft cloth lightly damped with a soap and water solution. For any other maintenance it is necessary to remove the case. Before removing the case, see that the equipment is entirely disconnected from all electricity supplies. Remember that when the case is off, many of the conductors exposed will carry dangerous voltage in the event of the supply being reconnected. Removal of the case is accomplished by removing first three Posidriv screws in each side panel, then seven screws from the rear of the cover. After this the cover may be slid straight back, away from the front panel.