Transformers (3.15) PDF

TRANSFORMERS SUBMODULE: 3.15 LEVEL: 2 PRESENTER:SMRUTI GHADI WHERE ARE TRANSFORMERS USED IN AIRCRAFT?  What are current transformers (CTs) used for?  Electrocube current sensing transformers, also known as instrument transformers, are used primarily for power system wir...

TRANSFORMERS SUBMODULE: 3.15 LEVEL: 2 PRESENTER:SMRUTI GHADI WHERE ARE TRANSFORMERS USED IN AIRCRAFT?  What are current transformers (CTs) used for?  Electrocube current sensing transformers, also known as instrument transformers, are used primarily for power system wiring protection and power supply control in commercial jetliners, military aircraft and land-based defense facilities.  A Transformer Rectifier Unit (TRU) combines the functions of a Transformer and a Rectifier into one unit.  In aircraft applications, the TRU converts the 115V AC power generated by the engine or APU generators or provided by a Ground Power Unit (GPU) to 28V DC power for use by various electrical components.  Each AC bus feeds a TRU which feeds a DC bus.  A transformer is a static (or stationary) piece of apparatus by means of which electric power in one circuit is transformed into electric power of the same frequency in another circuit.  It can raise or lower the voltage in a circuit but with a corresponding decrease or increase in current.  The physical basis of a transformer is mutual induction between two circuits linked by a common magnetic flux.  A transformer is a device that:  Transfers electric power from one circuit to another  It does so without a change in frequency  It accomplishes this by electromagnetic induction &  Where the two electric circuits are in mutual inductive influence of each other.  PRINCIPLE OPERATION OF A TRANSFORMER  A transformer changes electrical energy of a given voltage into electrical energy at a different voltage level.  It consists of two coils that are not electrically connected, but are arranged so that the magnetic field surrounding one coil cuts through the other coil.  When an alternating voltage is applied to (across) one coil, the varying magnetic field set up around that coil creates an alternating voltage in the other coil by mutual induction.  A transformer can also be used with pulsating DC, but a pure DC voltage cannot be used, since only a varying voltage creates the varying magnetic field that is the basis of the mutual induction process.  A transformer consists of three basic parts:  An iron core which provides a circuit of low reluctance for magnetic lines of force  A primary winding which receives the electrical energy from the source of applied voltage  A secondary winding which  When an AC voltage is connected across the primary terminals of a transformer, an alternating current will flow and self induce a voltage in the primary coil that is opposite and nearly equal to the applied voltage.  The difference between these two voltages allows just enough current in the primary to magnetize its core.  This is called the exciting, or magnetizing, current.  The magnetic field caused by this exciting current cuts across the secondary coil and induces a voltage by mutual induction.  If a load is connected across the secondary coil, the load current flowing through the secondary coil will produce a magnetic field which will tend to neutralize the magnetic field produced by the primary current.  This will reduce the self-induced (opposition) voltage in the primary coil and allow more primary current to flow.  The primary current increases as the secondary load current increases, and decreases as the secondary load current decreases.  When the secondary load is removed, the primary current is again reduced to the small exciting current sufficient only to magnetize the iron core of the transformer.  BASIC CONSTRUCTION OF A TRANSFORMER  Basically a transformer consists of two inductive windings and a laminated steel core.  The coils are insulated from each other as well as from the steel core.  A transformer may also consist of a container for winding and core assembly (called as tank), suitable bushings to take our the terminals, oil conservator to provide oil in the transformer tank for cooling purposes etc.  The sheets are cut in the shape as E,I and L. To avoid high reluctance at joints, laminations are stacked by alternating the sides of joint. That is, if joints of first sheet assembly are at front face, the joints of following assembly are kept at back face.  TYPES OF TRANSFORMERS  (A) On the basis of construction, transformers can be classified into two types as;  (i) Core type transformer and  (ii)Shell type transformers  (I) Core Type Transformer  Windings surrounds a considerable part of core.  In core type transformer, windings are cylindrical former wound, mounted on the core limbs. The cylindrical coils have different layers and each layer is insulated from each other. Materials like paper, cloth or mica can be used for insulation. Low  TYPES OF TRANSFORMERS  (II) Shell Type Transformer  The core surrounds a considerable portion of windings  The coils are former wound and mounted in layers stacked with insulation between them. A shell type transformer may have simple rectangular form , or it may have a distributed form.  TYPES OF TRANSFORMERS  (B) On the basis of their purpose  TURNS RATIO  The factor that determines whether a transformer is a step-up, or step-down type is the “turns” ratio.  The turns ratio is the ratio of the number of turns in the primary winding to the number of turns in the secondary winding.  TURNS RATIO: N1/N2  For example, the turns ratio of the step- down transformer is 5 to 1, since there are five times as many turns in the primary as in the secondary.  VOLTAGE TRANSFORMATION RATIO (K)  This constant K is known as voltage transformation  ratio.  (i) If N2 > N1 i.e. K > 1, then transformer is called step- up  transformer.  (ii) If N2 < N1 i.e. K < 1, then transformer is known as  step-down transformer.  Again, for an ideal transformer, input VA = output VA.  Hence, currents are in the inverse ratio of the (voltage) transformation ratio. E.M.F. EQUATION OF A TRANSFORMER  Let N1 = No. of turns in primary  N2 = No. of turns in secondary  Φm = Maximum flux in core in webers = Bm × A  f = Frequency of a.c. input in Hz  Flux increases from its zero value to maximum value Φm in one quarter of the cycle i.e. in 1/4 f second.  ∴ Average rate of change of flux = Φm 1/4f  = 4 f Φm Wb/s or volt If flux Φ varies sinusoidally, then r.m.s. value of induced e.m.f. is obtained by multiplying the average value with form factor. Form factor = r.m.s. value= 1.11 average value ∴ r.m.s. value of e.m.f./turn = 1.11 × 4 f Φm = 4.44 f Φm volt Now, r.m.s. value of the induced e.m.f. in the whole of primary winding = (induced e.m.f/turn) × No. of primary turns  COEFFICIENT OF COUPLING  No transformer can be constructed that is 100 percent efficient, although iron core transformers can approach this figure.  This is because all the magnetic lines of force set up in the primary do not cut across the turns of the secondary coil.  A certain amount of the magnetic flux, called leakage flux, leaks out of the magnetic circuit. The measure of how well the flux of the primary is coupled into the secondary is called the “coefficient of coupling.”  For example, if it is assumed that the primary of a transformer develops 10,000 lines of force and only 9,000 cut across the secondary, the coefficient of coupling would be 0.9 or, stated another way, the transformer would be 90 percent efficient.  (C) On the basis of type of supply SINGLE PHASE THREE PHASET TRANSFORMER RANSFORMER  (D) On the basis of their use  Power transformer: Used in transmission network, high rating  Distribution transformer: Used in distribution network, comparatively lower rating than that of power transformers.  Instrument transformer: Used for measuring large alternating voltages and currents. They are of two types  Current transformer (CT) Potential transformer (PT)  INSTRUMENT TRANSFORMERS  In d.c. circuit when large currents are to be measured, it is usual to use low-range ammeters with suitable shunts. For measuring high voltages, low-range voltmeters are used with a high resistance connected in series with them.  But it is not convenient to use this methods with alternating current and voltage instruments. For this purpose, specially constructed accurate ratio instrument transformers are employed in conjunction with standard low-range a.c. instruments. These instrument transformers are of two kinds :  (i) current transformers for measuring large alternating currents and  (ii) potential transformers for measuring high alternating voltages.  CURRENT TRANSFORMERS  These transformers are used with low-range ammeters to measure currents in high-voltage alternating-current circuits where it is not practicable to connect instruments and meters directly to the lines.  In addition to insulating the instrument from the high voltage line, they step down the current in a known ratio.  The current (or series) transformer has a primary coil of one or more turns of thick wire connected in series with the line whose current is to be measured.  The secondary consists of a large number of turns of fine wire and is connected across the ammeter terminals.  As regards voltage, the transformers is of step-up variety but it is obvious that current will be stepped down.  Thus, if the current transformer has primary to secondary current ratio of 100 : 5, then it steps up the voltage 20 times whereas it steps down the current to 1/20th of its actual value.  Hence, if we know current ratio (I1/I2) of the transformer and the reading of the a.c. ammeter, the line current can be calculated. PROBLEM: A 100 : 5 transformer is used in conjunction with a 5-amp ammeter. If the latter reads 3.5 A, find the line current. Solution. Here, the ratio 100 : 5 stands for the ratio of primary-to-secondary currents i.e.  It should be noted that, since the ammeter resistance is very low, the current transformer normally works short circuited.  If for any reason, the ammeter is taken out of the secondary winding, then this winding must be short-circuited with the help of short-circulating switch S.  If this is not done, then due to the absence of counter amp-turns of the secondary, the unopposed primary m.m.f. will set up an abnormally high flux in the core which will produce excessive core loss with subsequent heating and a high voltage across the secondary terminals.  This is not the case with ordinary constant- potential transformers, because their primary current is determined by the load in their secondary whereas in a current transformer,  POTENTIAL TRANSFORMERS  These transformers are extremely accurate- ratio step-down transformers.  In voltage transformers, the primary coils are connected in parallel across the supply.  They are used in conjunction with standard low-range voltmeters (usually 150-V) whose deflection when divided by voltage transformation ratio,gives the true voltage on the high voltage side.  For safety, the secondary should be completely insulated from the high-voltage primary and should be, in addition, grounded for affording protection to the operator. POTENTIAL TRANSFORMER AS A VOLTMETER CURRENT POTENTIAL TRANSFORMERS TRANSFORMERS  (E) On the basis of cooling employed  Oil-filled self cooled type used for small to medium sized transformers.  Oil-filled water cooled type used for large transformers such as high voltage transmission lines.  Air blast type (air cooled) for transformers operating at voltages below 25,000 V. THEORY OF AN IDEAL TRANSFORMER  An ideal transformer is one which has no losses i.e. its windings have no ohmic resistance, there is no magnetic leakage and hence which has no I2R and core losses.  In other words, an ideal transformer consists of two purely inductive coils wound on a loss-free core.  An ideal transformer i.e. one in which there were no core losses and copper losses.  It may, however, be noted that it is impossible to realize such a transformer in practice, yet for convenience, we will start with such a transformer and step by step approach an actual transformer. TRANSFORMER WITH LOSSES BUT NO MAGNETIC LEAKAGE  Consider two cases (i) when such a transformer is on no load and (ii) when it is loaded.  1)Transformer on No-load  2) Transformer on Load  TRANSFORMER ON NO- LOAD  When an actual transformer is put on load, there is iron loss in the core and copper loss in the windings (both primary and secondary) and these losses are not entirely negligible.  Even when the transformer is on no-load, the primary input current is not wholly reactive.  The primary input current under no-load conditions has to supply (i) iron losses in the core i.e. hysteresis loss and eddy current loss and (ii) a very small amount of copper loss in primary (there being no Cu loss in secondary as it is open).  Hence, the no-load primary input current I 0 is not at 90° behind V1 but lags it by an angle φ0 < 90°.  No-load input power  Wo = V1Iocosφo  where cosφo is primary power factor under no-load conditions.  Primary current Io has two components :  (i) One in phase with V1.This is known as active or working or iron loss component Iw because it mainly supplies the iron loss plus small quantity of primary Cu loss.  Iw = Io cosφo  (ii) The other component is in quadrature with V1 and is known as magnetizing component Iμ because its function is to sustain the alternating flux in the core. It is wattless.  Iμ = Io sin φo  Obviously, I0 is the vector sum of Iw and Iμ, hence  Io = (Iμ2 + Iω2).  The following points should be noted carefully :  1. The no-load primary current Io is very small as compared to the full-load primary current.  It is about 1 per cent of the full-load current.  2. As Io is very small, the no-load primary Cu loss is negligibly small which means that no-load primary input is practically equal to the iron loss in the transformer.  3. As it is principally the core-loss which is responsible for shift in the current vector, angle φ0 is known as hysteresis angle of advance.  TRANSFORMER ON LOAD  When the secondary is loaded, the secondary current I2 is set up. The magnitude and phase of I2 with respect to V2 is determined by the characteristics of the load. Current I2 is in phase with V2 if load is non-inductive, it lags if load is inductive and it leads if load is capacitive.  The secondary current sets up its own m.m.f. (=N2I2) and hence its own flux Φ2 which is in opposition to the main primary flux Φ which is due to Io.  The secondary ampere-turns N2 I2 are known as demagnetising amp-turns. The opposing secondary flux Φ2 weakens the primary flux Φ momentarily, hence primary back e.m.f. E1 tends to be reduced.  For a moment V1 gains the upper hand over E1 and hence causes more current to flow in primary.  the additional primary current be I2′. It is known as load component of primary current. This current is antiphase with I2′.  The additional primary m.m.f. N1I2 sets up its own flux Φ2′ which is in opposition to Φ2 (but is in the same direction as Φ) and is equal to it in magnitude. Hence, the two cancel each other out.  So, we find that the magnetic effects of secondary current I2 are immediately neutralized by the additional primary current I2′ which is brought into existence exactly at the same instant as I2.  Hence, whatever the load conditions, the net flux passing through the core is approximately the same as at no-load.  An important deduction is that due to the constancy of core flux at all loads, the core loss is also practically the same under all load conditions  Full load copper losses are determined by this method.  TRANSFORMER TESTS  The four important parameters are equivalent resistance, the equivalent leakage reactance, the core-loss conductance and the magnetizing susceptance.  These constants or parameters can be easily determined by two tests (i) open-circuit test and (ii) short circuit test.  OPEN-CIRCUIT OR NO-LOAD TEST:The purpose of this test is to determine no-load loss or core loss.  SHORT-CIRCUIT OR IMPEDANCE TEST: This method is used for determining the following :  (i) Equivalent impedance, leakage reactance and total resistance of the transformer as referred to the winding in which the measuring instruments are placed.  (ii) Cu loss at full load (and at any desired load). This loss is used in calculating the efficiency of the transformer.  (iii) Knowing the equivalent resistance ,the total voltage drop in the transformer as referred to primary or secondary can be calculated and hence regulation of the transformer determined. WHY TRANSFORMER RATING IN KVA ?  As seen, Cu loss of a transformer depends on current and iron loss on voltage. Hence, total transformer loss depends on volt-ampere (VA) and not on phase angle between voltage and current i.e. it is independent of load power factor.  That is why rating of transformers is in kVA and not in kW.  LOSSES IN A TRANSFORMER  In a static transformer, there are no friction or windage losses.  Hence, the only losses occuring are :  (i) Core or Iron Loss : It includes both hysteresis loss and eddy current loss. Because the core flux in a transformer remains practically constant for all loads.  These losses are minimized by using steel of high silicon content for the core and by using very thin laminations respectively.  Iron or core loss is found from the O.C. test. The input of the transformer when on noload measures the core loss.  (ii) Copper loss. This loss is due to the ohmic resistance of the transformer windings.  Total Cu loss= I12R1 + I22R2.  It is clear that Cu loss is proportional to (current)2 or kVA2.  In other words, Cu loss at half the full- load is one-fourth of that at full-load.  The value of Cu loss is found from the short-circuit test.  ‘OHMIC’ OR ‘COPPER LOSS’  Copper loss is I2R loss, in primary side it is I12R1 and in secondary side it is I22R2 loss, where I1& I2 are primary and secondary current of transformer and R 1 and R2 are resistances of primary & secondary winding.  As the both primary & secondary currents depend upon load of transformer, copper loss in transformer vary with load.  ‘CORE’ LOSS OR ‘IRON’ LOSS  Hysteresis loss and eddy current loss, both depend upon magnetic properties of the materials used to construct the core of transformer and its design.  So these losses in transformer are fixed and do not depend upon the load current.   Where, Kh = Hysteresis constant. Ke = Eddy current constant. Kf = form constant.  HYSTERESIS LOSS  The magnetic core of transformer is made of ′Cold Rolled Grain Oriented Silicon Steel′.  Steel is very good ferromagnetic material. That means, whenever magnetic flux would pass through, it will behave like magnet. Ferromagnetic substances have numbers of domains in their structure. In other words, the domains are like small permanent magnets situated randomly in the structure of substance.  These domains are arranged inside the material structure in such a random manner, that net resultant magnetic field of the said material is zero. Whenever external magnetic field or mmf is applied to that substance, these randomly directed domains get arranged themselves in parallel to the axis of applied mmf.  After removing this external mmf, maximum numbers of domains again come to random positions, but some of them still remain in their changed position. Because of these unchanged domains, the substance becomes slightly magnetized permanently.  This magnetism is called " Spontaneous Magnetism". To neutralize this magnetism, some opposite mmf is required to be applied. The magneto motive force or mmf applied in the transformer core is alternating.  For every cycle due to this domain reversal, there will be extra work done. For this reason, there will be a consumption of electrical energy which is known as Hysteresis loss of transformer.  EDDY CURRENT LOSS  In transformer, we supply alternating current in the primary, this alternating current produces alternating magnetizing flux in the core and as this flux links with secondary winding, there will be induced voltage in secondary, resulting current to flow through the load connected with it.  Some of the alternating fluxes of transformer; may also link with other conducting parts like steel core or iron body of transformer etc.  As alternating flux links with these parts of transformer, there would be a locally induced emf.  Due to these emfs, there would be currents which will circulate locally at that parts of the transformer. These circulating current will not contribute in output of the transformer and dissipated as heat.  This type of energy loss is called eddy current loss of transformer. METHODS TO OVERCOME CORE LOSSES  The steel used is having high silicon content and sometimes heat treated, to provide high permeability and low hysteresis loss.  Laminated sheets of steel are used to reduce eddy current loss. EFFICIENCY OF A TRANSFORMER  Efficiency = output / input.  Transformers are the most highly efficient electrical devices. Most of the transformers have full load efficiency between 95% to 98.5%.   CONDITION FOR MAXIMUM EFFICIENCY  Cu loss = I12Ro1 or I22Ro2 = Wcu  Iron loss = Hysteresis loss + Eddy current loss  = Wh + We = Wi  Condition for Maximum Efficiency is  Cu loss = Iron loss  If we are given iron loss and full load Cu loss, then the load at which two losses would be equal (i.e. corresponding to maximum efficiency) is given by  = Full load × ( Iron loss ) √ F.L. Cu loss  AUTO TRANSFORMERS  It is a transformer with one winding only, part of this being common to both primary and secondary.  Obviously, in this transformer the primary and secondary are not electrically isolated from each other as is the case with a 2- winding transformer.  But its theory and operation are similar to those of a two-winding transformer. Because of one winding, it uses less copper and hence is cheaper.  It is used where transformation ratio differs little from unity. (near to unity)  AUTO TRANSFORMERS It can be proved that power transformed inductively is input (1 − K). The rest of the power = (K × input) is conducted directly from the source to the  ADVANTAGES OF AUTO TRANSFORMER  For transformation ratio = 2, the size of the auto transformer would be approximately 50% of the corresponding size of two winding transformer. For transformation ratio say 20 however the size would be 95 %. The saving in cost of the material is of course not in the same proportion. The saving of cost is appreciable when the ratio of transformer is low, that is lower than 2. Thus auto transformer is smaller in size and cheaper.  An auto transformer has higher efficiency than two winding transformer. This is because of less ohmic loss and core loss due to reduction of transformer material.  Auto transformer has better voltage regulation as voltage drop in resistance and reactance of the single winding is less.  ADVANTAGES OF AUTO TRANSFORMER  The kVA rating of an ordinary 2- winding transformer is increased when connected as an autotransformer because of the establishment of conductive link between primary and secondary.  Hence the energy is transferred both inductively and conductivity  The saving in Cu achieved by converting a 2-winding transformer into an autotransformer is determined by voltage transformation ratio DISADVANTAGES OF AUTO TRANSFORMER  Because of electrical conductivity of the primary and secondary windings the lower voltage circuit is liable to be impressed upon by higher voltage. To avoid breakdown in the lower voltage circuit, it becomes necessary to design the low voltage circuit to withstand higher voltage.  The leakage flux between the primary and secondary windings is small and hence the impedance is low. This results into severe short circuit currents under fault conditions.  The connections on primary and secondary sides have necessarily needs to be same, except when using interconnected starring connections. This introduces complications due to changing primary and secondary phase angle particularly in the case of delta / delta connection.  ISOLATION TRANSFORMERS  An Isolation transformer is a transformer used to transfer electrical power from a source of alternating current (AC) power to some equipment or device while isolating the powered device from the power source, usually for safety reasons.  Isolation transformers provide galvanic isolation; no conductive path is present between source and load.  This isolation is used to protect against electric shock, to suppress electrical noise in sensitive devices, or to transfer power between two circuits which must not be connected.  A transformer sold for isolation is often built with special insulation between primary and secondary, and is specified to withstand a high voltage between windings.  Isolation transformers block transmission of the DC component in signals from one circuit to the other, but allow AC components in signals to pass.  Transformers that have a ratio of 1 to 1( turns ratio N1/N2=1) between the primary and secondary windings are often used to protect secondary circuits and individuals from electrical shocks between energized conductors and earth ground. THANK YOU!!!