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UNIT – 4 1. Explain the construction of Single phase transformer with neat diagrams. ANS: TRANSFORMER CONSTRUCTION: There are two general types of transformers 1. Core type transformer 2. Shell type transformer These two differ by the manner in which the windings are wound around the magnetic core. The magnetic core is a stack of thin silicon-steel laminations about 0.35 mm thick for 50 Hz transformer. In order to reduce the eddy current losses, these laminations are insulated from one another by thin layers of varnish. In order to reduce the core losses, transformers have their magnetic core made from cold-rolled grainoriented sheet steel (C.R.G.O). This material, when magnetized in the rolling direction, has low core loss and high permeability. CORE TYPE TRANSFORMER: In the core-type, the windings surround a considerable part of steel core as shown in fig (a). The core type transformers require more conductor material and less iron when compared to shell-type. The vertical portions of the core are usually called limbs or legs and the top and bottom portions are called the yoke. For single phase transformers, core-type has two legged core. In order to reduce leakage flux, half of the L.V. winding is placed over one leg and other half over other leg. For H.V. winding also, half of the winding is placed over one leg and the other half over the other leg. L.V. winding is placed adjacent to the steel core and H.V. winding outside, in order to minimize the amount of insulation required. Shell Type Transformer: In the core-type, the steel core surrounds a considerable part of the windings as shown in fig (b). Shell-type transformer has three legged core. The L.V. and H.V. windings are wound on the central limb. In order to reduce leakage flux, the windings are interleaved or sandwiched. The shell type transformers require more iron and less conductor material when compared to coretype. There are two types of windings employed for transformers. 2. Explain the principle of operation of a Transformer in detail with neat diagrams. ANS: WORKING PRINCIPLE OF A TRANSFORMER A transformer is a static device which transfers electric energy from one circuit to another circuit without changing the frequency of the system. It works on electromagnetic induction principle. According to this principle, an e.m.f. is induced in a coil if it links a changing flux. Consider two coils 1 and 2 wound on a simple magnetic circuit as show in fig.1.1. These two coils are insulated from each other and there is no electrical connection, but magnetically coupled. The two coils posses high mutual inductance. If one coil is connected to a source of alternating voltage, an alternating flux is set up in the laminated core, most of which is linked with the other coil, in which it produces mutually induced e.m.f. according to Faraday’s law of electromagnetic induction. If the secondary coil circuit is closed a current flows in it and so an electrical energy transfers from the first coil to the second coil. Coil 1 which receives energy from the source of ac supply is called the primary coil or primary winding and coil 2, which is connection to load and delivers energy to the load is called the secondary coil or secondary winding. The symbolic representation of a two winding transform is as shown in fig.1.1. The two vertical lines are used to represent magnetic core, which signify the tight magnetic coupling between the windings. 3. Write the classifications of Transformer. ANS: TRANSFORMER CLASSIFICATION 1. According to number of phases (a)Single phase transformer (b)Three phase transformer 2. According to construction (a) Core type transformer (b) Shell type transformer (c) Berry type transformer 3. According to Function (a) Power transformer (i)Step-up transformer (ii)Step-down transformer (b) Distribution transformer (c) Instrument transformer (i) Current transformer (CT) (ii) Potential transformer (PT) 4. According to type of cooling (a) Air cooled transformer (b) Oil cooled transformer 5. According to frequency (a) Low frequency (50 Hz) (b) High frequency (HFT’s pulse transformer) 6. Special purpose transformer (a) Scott connection transformer (b)V-connection transformer (c) Pulse transformer (d)Auto transformer (e)Welding transformer 7. According to location (a) Indoor transformer (b) Outdoor transformer 8. According to rating (a) Low rating transformer (b) Medium rating transformer (c) High rating transformer 4. Derive the EMF equation of a Transformer. ANS: E.M.F. EQUATION OF A TRANSFORMER When the primary winding is excited by an alternating voltage, it circulates alternating current and hence alternating flux is produced in the core. Let Ф = Фm Sin ωt. Where Фm = Maximum value of flux f= Frequency of supply voltage N1= Number of primary winding turns N2= Number of secondary winding turns E1= r.m.s value of the primary induced e.m.f. E2= r.m.s. value of the secondary induced e.m.f. According to Faraday’s Law of electromagnetic induction, E.M.F induced is given by eN d d N m Sin t N Cost N m Sin (t 90 0 ) dt dt It is clear from the above equation that maximum value of induced e.m.f. is Emax N m The r.m.s. value induced e.m.f. is E rms E max 2 N m 2 N m 2f 2 4.44 N f m RMS value of e.m.f. induced in the primary winding Similarly RMS value of e.m.f. induced in the secondary winding E1= 4.44 N1 f Фm Volts E2= 4.44 N2 f Фm Volts 5. Explain the Transformer action when at NO-Load and When Loaded. Also draw the phasor diagram for different loaded conditions. ANS: PRACTICAL TRANSFORMER ON NO-LOAD In the case of ideal transformer, losses are neglected, but practically it is not possible. When the primary of a transformer is connected to ac supply source, and the secondary is open, the transformer is said to be on no-load. When the transformer is on no-load, the primary current is not completely reactive. But it has to supply iron losses in the core and a very small amount of copper losses in primary (there being no copper loss in secondary as it is open). Hence, the no load primary input current I0 is not at 900 behind V1 but lags it by an angle Ф0 <900. No load input power of the transformer is W0 =V1I0 CosФ0 The phasor diagram under noload condition is drawn as shown in fig(b). From the pharos diagram Io has two components i) Working or active component, Iw = I0CosФ0 , which is in phase with V1 ii) Magnetizing component, Im = I0CosФ0 , which is in quadrature with V1 I 0 I m2 I w2 PRACTICAL TRANSFORMER ON LOAD When the transformer is loaded, the secondary current, I 2 is set up. I2 will be in phase with V2 if the load is resistive, it lags V2 if the load is inductive, it leads V2 if the load is capacitive. The secondary current I2 sets up its own m.m.f.(=N2I2) and hence it produces flux Ф2, which is in opposition to the main primary flux Ф , which is due to I0. Secondary flux Ф2 weakens the main flux Ф momentarily and hence primary back e m f E1 tends to be reduced. For a moment V1 gains the upper hand over E1 and hence causes additional current I2’ to flow in primary and hence flux Ф2’ (due to m.m.f. N1 I2’ ) which counter balance the secondary flux Ф2. 2' 2 N1 I 2' N 2 I 2 I 2' where K = Transformation Ratio N2 I2 K I2 N1 I 2' is known as load component of primary current. This current is in anti-phase with I1. I 1 I 0 I 2' Here The phasor diagram of a transformer under load condition can be drawn as shown below. 6. Draw the Equivalent circuit of the Transformer when referred to primary and secondary. ANS: EQUIVALENT CIRCUIT OF A TRANSFORMER In transformers, the problems concerning voltages and currents can be solved by use of phasor diagrams. However, it is more convenient to represent the transformer by an equivalent circuit. The term equivalent circuit of a transformer means the combination of fixed and variable resistances and reactances, which exactly simulates performance and working of the machine. If an equivalent circuit is available, the computations can be done by the direct application of circuit theory. The transformer shown diagrammatically in fig. can be resolved into an equivalent circuit in which the resistance and leakage reactance of the transformer are imagined to be external to the windings whose function is only to transform the voltage. For a transformer, no-load primary current I0 has two components I m I 0 Sin 0 Magnetizin g Component I w I 0 Cos 0 Working Component Im produces the flux and is assumed to flow through reactance X 0 called no-load reactance while Iw is active component representing core losses hence is assumed to flow through resistance R0. R0 and X0 are connected in parallel across the primary circuit as shown in fig. Equivalent Circuit of a Transformer From the equivalent circuit we can write R0 E1 Iw and X0 E1 Im To make the calculations simpler, it is preferable to transfer voltage, current and impedance to the one side either primary or secondary of transformer. (i) Equivalent circuit when referred to Primary Let us see how to transfer secondary values to primary side: E 2' The primary equivalent of secondary E.M.F. = E2 E1 K V2' The primary equivalent of secondary terminal voltage i.e. I1 V2' I 2 V2 V2' V2 K I2 V V2 2 I1 K I 2' The primary equivalent of the sec ondary current K I 2 i.e. V1 I 2' V2 I 2 I 2' V2 I2 K I2 V1 R2 K2 X X 2' The primary equivalent of secondary reactance 22 K Z Z 2' The primary equivalent of secondary Impedance 22 K R2' The primary equivalent of secondary resistance R2 X ; X 01 X 1 22 ; 2 K K R01 j X 01 R01 R1 Z 01 where K V2 E N I 2 2 1 Transformation Ratio V1 E1 N1 I2 Thus the equivalent circuit of the transformer when referred to primary is as shown in the below fig. Equivalent circuit when referred to primary Approximate equivalent circuit of a transformer: The no-load current I0 usually less than 5% of the full load primary current. The voltage drop produced by I0 in (R1+jX1) is negligible for practical purposes. Therefore, it is immaterial that the shunt branch R0//X0 is connected before the primary series impedance (R1+jX1) or after it. The currents Im and Iw are not much affected. Therefore, the equivalent circuit can be further simplified by shifting the no load branch containing R 0 and X0 to the supply terminals as shown in fig. By doing this, we are creating an error that the drop across R1 and X1 due to I0 is neglected. Hence equivalent circuit is called approximate equivalent circuit. Approximate equivalent circuit when referred to primary The equivalent circuit can be further simplified by combining parameters R1 and and combining parameters X1 and R21 as R01 X 2' as X01 Where R2 X ; X 01 X 1 22 ; 2 K K R01 j X 01 R01 R1 Z 01 The circuit is shown in the fig. Approximate equivalent circuit when referred to primary In the similar fashion the approximate equivalent circuit referred to secondary can also be obtained (ii) EQUIVALENT CIRCUIT WHEN REFERRED TO SECONDARY Similarly if all the primary quantities are transferred to secondary, we get the equivalent circuit of transformer referred to secondary as shown in fig. R1' R1 K 2 X 1' X 1 K 2 Z1' Z1 K 2 I I I 1' 1 I 0' 0 E1' E1 K K K 2 2 R02 R2 K R1 X 02 X 2 K X 1 Z 02 R02 j X 02 1 1 Similarly the exciting circuit parameters also gets transformed to secondary as R0 and X 0 The equivalent circuit referred to secondary is as shown in fig. Equivalent circuit when referred to secondary (a) (b) Equivalent circuit when referred to secondary (c) 7. Write the efficiency equation of a transformer. ANS: EFFICIENCY OF A TRANSFORMER Efficiency of a transformer at a given load and power factor is defined as the ratio of output power to the input power, both the quantities expressed in the same units. Efficiency, Output Power Output power Input Power Output power Losses V2 I 2 * power factor V2 I 2 * power factor I 12 R1 I 22 R2 Wi It may appear that efficiency can be determined by directly loading the transformer and measuring the input power and output power. However, this method has the following drawbacks: Since the efficiency of a transformer is very high, even 1% error in each wattmeter (output and input) may give ridiculous results. This test, for instance, may give efficiency higher than 100%. Since the test is performed with transformer on load, considerable amount of power is wasted. For large transformers, the cost of power alone would be considerable. The test gives no information about the proportion of various losses. Due to these drawbacks, direct loading method is seldom used to determine the efficiency of a transformer. In practice, the efficiency can be calculated by determining iron loss from open circuit test and copper loss from short circuit test. 8. Explain briefly about the regulation of an Alternator. ANS: REGULATION OF A TRANSFORMER When the transformer is loaded its terminal voltage falls from no load to full load. The change in secondary terminal voltage from no load to full load and expressed as secondary no load voltage is known as regulation down. V V2 * 100 Voltage Re gulation down 0 2 V 0 2 If the change in voltage is expressed as secondary full-load voltage then it is known as regulation up V V2 * 100 Voltage Re gulation up 0 2 V2 Where 0V2 = No-load secondary voltage=KV1 V2= Secondary voltage on load Unless stated otherwise, regulation is to be taken as regulation down. Lesser the regulation better the transformer performance. It may be noted that percentage voltage regulation of the transformer will be the same whether primary or secondary side. From the phasor diagram considering OE ≈OF (by neglecting EF) ∴OC = OE =OA + AD + DE 0V2 = V2+ AD+ DE AD From Δle ADB, AB AD AB *Cos2 I 2 R02 Cos 2 Cos 2 From Δle BCG, Cos (90 2 ) BG BC Sin 2 DE BC DE I 2 X 02 Sin 2 ∴ From above equations, 0 V2 V2 AD DE V2 I 2 R02 Cos 2 I 2 X 02 Sin 2 0V2 V2 I 2 R02 Cos 2 I 2 X 02 Sin 2 In general the approximate voltage drop is given by 0 V2 V2 I 2 R02 Cos 2 I 2 X 02 Sin 2 Note: + ve sign for lagging p.f & - ve sign for leading p.f V V2 I R Cos 2 I 2 X 02 Sin 2 * 100 2 02 Voltage Re gulation 0 2 * 100 0 V2 0 V2 9. What is ALL DAY EFFICIENCY? ANS: ALL-DAY OR ENERGY EFFICIENCY The ordinary or commercial efficiency of a transformer is defined as the ratio of output power to the input power i.e., Commercial Efficiency, Output Power Output power Input Power Output power Losses There are certain types of transformers whose performance cannot be judged by this efficiency. For instance, distribution transformers used for supplying lighting loads have their primaries energized all the 24 hours in a day but the secondaries supply little or no load during the major portion of the day. It means that a constant loss (i.e., iron loss) occurs during the whole day but copper loss occurs only when the transformer is loaded and would depend upon the magnitude of load. Consequently, the copper loss varies considerably during the day and the commercial efficiency of such transformers will vary from a low value (or even zero) to a high value when the load is high. The performance of such transformers is judged on the basis of energy consumption during the whole day (i.e., 24 hours). This is known as all-day or energy efficiency. All-day efficiency is defined as the ratio of output in kWh (i.e output energy) to the input in kWh (i.e input energy) of a transformer over a 24-hour period. i.e. all day kWh Output in 24 hrs kWh Input in 24 hrs This efficiency is always less than the commercial efficiency of a transformer. To find the allday efficiency, we have to know the load cycle on the transformer i.e., how much and how long the transformer is loaded during 24 hours. All-day efficiency is of special importance for those transformers whose primaries are never open-circuited but the secondaries carry little or no load much of the time during the day. In the design of such transformers, efforts should be made to reduce the iron losses which continuously occur during the whole day. Note: Efficiency of a transformer means commercial efficiency unless stated otherwise. 10. What are the different tests to be performed on the transformer to find out IRON and COPPER losses in a transformer. ANS: TRANSFORMER TESTS The circuit constants, efficiency and voltage regulation of a transformer at any load and p.f. condition can be predetermined without actually loading the transformer. But the equivalent circuit parameters of a transformer are determined by conducting two simple tests (i) opencircuit test and (ii) short-circuit lest. These tests give more accurate results than those obtained by taking measurements on fully loaded transformers. Further, the power required to carry out these tests is very small as compared with full-load output of the transformer. These tests consist of measuring the input voltage, current and power to the primary first with secondary open-circuited (open-circuit test) and then with the secondary short-circuited (short circuit test). OPEN CIRCUIT TEST This test is conducted to determine iron loss (or core loss) and no load current, I0 which is helpful in finding R0 and X0. In this test, the rated voltage is applied to the primary (usually low-voltage winding) while the secondary is left open circuited. A wattmeter ‘W’ a voltmeter ‘V’ and an ammeter ‘A’ are connected in the low voltage side i.e primary winding in the present case. The applied primary voltage V1 is measured by the voltmeter, the no load current I0 by ammeter and no-load input power W0 by wattmeter as shown in fig. With normal voltage applied to primary, normal flux will set up in the core and hence normal iron losses will occur, which are recorded by the wattmeter ‘W’. As the primary no load current I0 (2 to 10% of rated current) is small, copper loss is negligibly small in the primary and nill in the secondary (being opened). Hence the O.C test gives core losses alone practically (i.e wattmeter reading) and is same for all loads. Iron loss, Wi= Wattmeter reading= W0 No-load current= Ammeter reading=I0 Applied voltage=Voltmeter reading=V1 Input power, W0 V1 I 0 Cos 0 no load p. f Cos 0 I m I 0 Sin 0 , R0 V1 , Iw W0 V1 I 0 I w I 0 Cos 0 X0 V1 Im Thus open-circuit test enables us to determine iron losses and parameters R 0 and X0 of the transformer. Point to be remembered: The OC test is always conducted on LV side of a transformer. In OC test, the equipments are connected in LV side, since we need to get rated voltage. If it is connected in HV side the value of voltage will be more and current will be less. Hence the ordinary instruments cannot measure this value of current. The man aim of connecting meters on LV side is to use low range meters and low voltage supply, which are easily available and safe. SHORT CIRCUIT TEST This test is conducted to determine R 01 (or R02), X01 (or X02) and full-load copper losses of the transformer. In this test, the secondary (usually low-voltage winding) is short circuited by a thick conductor and variable low voltage is applied to the primary as shown in fig. A wattmeter ‘W’ a voltmeter ‘V’ and an ammeter ‘A’ are connected in the high voltage winding i.e primary winding in the present case. A low voltage (usually 5 to 10% of normal primary voltage) is applied through a variac to the primary and is gradually increased till the ammeter ‘A’ indicates full load current I1 in the primary. Since the applied voltage is very low, so flux produced is very small. Hence, the iron losses are so small that these can be neglected, with the result, the wattmeter ‘W’ reads total full load copper loss of the transformer. Full load Cu loss, Wcu= Wattmeter reading= Wsc Full load primary current= Ammeter reading=Isc Applied voltage=Voltmeter reading=Vsc 2 2 2 2 WSC I SC R1 I SC R2 I SC ( R1 R2 ) I SC R01 Total impedance when referred to primary, Z 01 R01 WSC 2 I SC VSC I SC Total leakage reactance when referred to primary, X 01 2 2 Z 01 R01 Thus short-circuit lest gives full-load Cu loss, R01 and X01 Note: The short-circuit test will give full-load Cu loss only if the applied voltage Vsc is such so as to circulate full-load currents in the windings. If, in a short circuit test, current value is other than full-load value, the Cu loss will be corresponding to that current value. Point to be remembered: The SC test is always conducted on HV side of a transformer. If the measurements were made on LV side, the voltage needed would be inconveniently low and the current would be inconveniently high. 11. Pre-determination of efficiency and Voltage Regulation of a transformer. ANS: Predetermination of efficiency and voltage regulation Knowing the equivalent resistance and reactance referred to primary (or secondary) from Short Circuit test, the voltage regulation of the transformer at any p.f. can be determined by Voltage Re gulation I 1 R01 Cos I 1 X 01 Sin * 100 V1 I 2 R02 Cos 2 I 2 X 02 Sin 2 * 100 0 V2 By performing open circuit and short circuit test we can find transformer total losses If W0 =Input power in watts from OC test =Iron losses, Wi Wsc =Input power in watts from SC test with full load current =Full load copper losses, Wcu Then the total losses on full load=Wi + Wcu Full load efficiency is given by full load Full load kVA * Cos Full load kVA * Cos Wi Wcu Efficiency at any load is given by at any load x S * Cos x S * Cos Wi x 2 Wcu Where x=Fraction of F.L at which the transformer is working S=Full load kVA of the transformer