INDIAN INSTITUTE OF TECHNOLOGY ROORKEE EEN-112 Electrical Science Unit-VI: Single Phase Transformer Dr. Dheeraj K Khatod Department of Electrical Engineering Contents • Concept of magnetic circuit • Transformer – – – – – Basic constructional features Operating principle Equivalent circuit and phasor diagram Tests: Open circuit and short circuit tests Losses: Iron losses (Eddy current and Hysteresis) and Cu losses – Efficiency and voltage regulation 2 Concept of Magnetic Circuit 3 Magnetic Circuit • The path of a magnetic flux is known as magnetic circuit. • The flow of magnetic flux in a magnetic circuit is almost analogues to the flow of electric current in an electric circuit. • To carry electric current in an electric circuit, usually aluminum or copper wires are used because the resistance of these materials is comparatively much lower than other materials. Similarly, to carry magnetic flux, iron or soft steel circuits are used as “opposition” of these materials to flux is low in comparison with other materials. 4 Magnetic Circuit (cont.) • Consider a toroidal ring of ferromagnetic material of mean radius, π π and circular cross-section of diameter, ππ as shown in Fig. • The ring termed as core is excited by a coil wound round it with ππ turns carrying a current, ππ. • By virtue of symmetry, flux established in the magnetic core is circular in shape having the length of the mean flux path, ππ. • The cross sectional area of the magnetic core is π΄π΄. 5 Magnetic Circuit Definitions • Magnetic Flux – Represents the magnetic field – Magnetic flux lines do not have origins or terminating points but exist in continuous loops – Symbol: ππ – Unit: Weber, (Wb) • Magnetomotive Force (MMF) – – – – “Driving force” that causes a magnetic field Symbol: β± Unit: Ampere-Turns, (AT) Definition: β± = ππ × πΌπΌ 6 Magnetic Circuit Definitions (cont.) • Magnetic Field Intensity or Magnetizing Force – – – – MMF gradient, or MMF per unit length Symbol: β Unit: Ampere-Turns per meter (AT/m) Definition: β = β±οΏ½ππ = ππ × πΌπΌοΏ½ππ • Magnetic Field Density – – – – Concentration of the lines of force in a magnetic circuit Symbol: π΅π΅ Unit: Weber per square meter (Wb/m2) or T (Tesla) Definition: ππ π΅π΅ = οΏ½π΄π΄ 7 Magnetic Circuit Definitions (cont.) • Permeability – – – – Relates flux density and field intensity Symbol: ππ Unit: Weber per Ampere-Turns-meter (Wb/AT-m) Definition: ππ = π΅π΅οΏ½ β • Permeability of free space (air) – Symbol: ππ0 = 4ππ × 10−7 Wb/AT-m • Relative Permeability – Compares permeability of material with the permeability of free space (air) – Symbol: ππππ – Dimensionless – Definition: ππππ = πποΏ½ππ 0 8 Magnetic Circuit Definitions (cont.) • Reluctance – Measure of “opposition” that the magnetic circuit offers to the flux – Analogous to resistance in an electrical circuit – Symbol: β – Unit: Ampere-Turns per Weber (AT/Wb) – Definition: β = β±οΏ½ππ = ππ × πΌπΌοΏ½ππ = πποΏ½πππ΄π΄ • Permeance – Reciprocal of reluctance – Symbol: ℘ – Unit: Weber per Ampere-Turns (Wb/ AT) – Definition: ℘ = 1οΏ½β 9 Magnetic Circuit Definitions (cont.) • Leakage Flux – Flux that leaks through the core – Flux that lies in air • Fringing – The flux passing from the core to the air-gap cannot remain confined to the air-gap but would somewhat spread out. This effect is called fringing. – Average flux density in the air-gap is slightly less than the flux density in the core 10 DC Circuit Analog of Magnetic System • The concept of reluctance lumps the magnetic system into a circuit analogically expressed as a dc electric circuit as shown in Fig. • In this analog – β± ~ dc voltage (potential) – β ~ resistance – ππ ~ current 11 Magnetization Curve and Hysteresis Loop 12 Magnetic Hysteresis Loss • Hysteresis produces heat due to re-alignment of magnetic domains • Hysteresis loss varies directly with – Frequency of the flux density – πππ‘π‘π‘ power of the flux density ππ • Hysteresis loss, ππβ = ππβ οΏ½ ππ οΏ½ π΅π΅ππππππ οΏ½ ππ where ππβ is the characteristic constant of core material ππ is the frequency of the flux (Hz) π΅π΅ππππππ is the maximum value of the flux density ππ is Steinmetz exponent; ranges between 1.5–2.0; typical value is 1.6 ππ is the volume of core material. 13 Eddy-Current Loss • The voltage induced in core (made of conducting material) by alternating flux produces circulating currents in it. These are called eddy-currents. The flow of eddy current also causes loss (πΌπΌ 2 π π ) known as eddy-current loss. 2 • Eddy-current loss, ππππ = ππππ οΏ½ ππ 2 οΏ½ π΅π΅ππππππ οΏ½ ππ where ππππ is the characteristic constant of core material ππ is the frequency of the flux (Hz) π΅π΅ππππππ is the maximum value of the flux density ππ is the volume of core material. 14 Core Loss • Core loss, also referred to as iron loss, consists of hysteresis loss and eddy-current loss. • Ways to reduce hysteresis loss – By using core material (Cold Rolled Grain Oriented (CRGO) Silicon Steel) having less area of hysteresis loop • Ways to reduce eddy-current loss – By using high resistive core material by adding Si – By using a laminated core 15 Electromagnetic Induction • Flux Linkage – If flux ππ passes through all the N turns of a coil, the flux is said to be linked with the coil – The flux linkage of the coil, ππ = ππ × ππ Weber-Turns (Wb-T) 16 Electromagnetic Induction (cont.) • Faraday’s first law states that whenever magnetic flux linked with a close coil changes, an induced emf is set up in the coil and the induced emf lasts as long as the change in magnetic flux continues. • Faraday’s second law states that the magnitude of the induced emf is directly proportional to the time rate of change of the magnetic flux linked with a close coil. • Lenz’s law states that the induced emf would tend to cause a current flow in the coil which would oppose the change in flux (the original cause of emf induction). 17 Electromagnetic Induction (cont.) ππππ ππππ • Induced emf, ππ = − ππππ = −ππ ππππ • A current flows through the loop when a magnet is moved near it, without any batteries! Moving the magnet inwards the coil No movement of the magnet Moving the magnet outwards the coil 18 Inductance • Self-Inductance – The property of a current carrying coil by which it opposes the change in flux linkage through it due to the production of self induced emf is called self-inductance. – The role of self-inductance in an electrical circuit is the same as that of the inertia in mechanical motion. Thus the selfinductance of a coil is a measure of its ability to oppose the change in current through it and hence is also called electrical inertia. – Symbol, πΏπΏ – Unit, Henry (H) ππππ ππππ ππππ ππππ ππππ Induced emf, ππ = − ππππ = −ππ ππππ = −ππ ππππ οΏ½ ππππ = −πΏπΏ ππππ Self-inductance, πΏπΏ = ππ ππππ ππππ 19 Inductance (cont.) • Self-Inductance – For a linear B-H curve (material operated in the region of constant permeability or when the magnetic circuit has a dominant air-gap), Self-inductance, πΏπΏ = ππ ππ ππ = ππ ππ Flux linkage of the coil, ππ = πΏπΏ × ππ As already derived, Reluctance, β = β±οΏ½ππ = ππ×πποΏ½ππ = πποΏ½πππ΄π΄ Again self-inductance, πΏπΏ ππ = ππ ππ = ππ2 β = ππ 2 ππ π΄π΄ ππ – Self-inductance of a coil is independent of excitation current and depends upon the core geometry, permeability of the core’s magnetic material and number of coil turns. 20 Inductance (cont.) • Mutual-Inductance – If two coils are wound on a common core or placed close to each other, a part of the flux produced by one coil also links the other coil. Whenever a change in current occurs in a coil, an induced emf is set up in the other coil. This process is called mutual induction. – If a current πΌπΌ1 flows in Coil 1 (having ππ1 turns), the magnetic flux linked with Coil 2 (having ππ2 turns) is ππ2 = ππ2 × ππ21 = πΏπΏ21 × πΌπΌ1 where, ππ21 is the flux of Coil 1 linking with Coil 2 πΏπΏ21 is the mutual inductance of Coil 2 with respect to Coil 1. ππ2 πΏπΏ21 = πΌπΌ1 21 Inductance (cont.) • Mutual-Inductance – When both coils are carrying current, the total flux linkages are given by ππ1 = πΏπΏ11 × πΌπΌ1 ± πΏπΏ12 × πΌπΌ2 ππ2 = ±πΏπΏ21 × πΌπΌ1 + πΏπΏ22 × πΌπΌ2 where, πΏπΏ11 and πΏπΏ22 are self-inductance of the coils and πΏπΏ12 and πΏπΏ21 are mutual inductance of the coils (equal in a bilateral circuit) – The induced emf in each coil is given by πππΌπΌ1 πππΌπΌ2 ππ1 = πΏπΏ11 × ± πΏπΏ12 × ππππ ππππ πππΌπΌ1 πππΌπΌ2 ππ2 = ±πΏπΏ21 × + πΏπΏ22 × ππππ ππππ 22 Inductance (cont.) • Mutual-Inductance – The voltage induced by mutual flux may be negative or positive, therefore, the dot convention is used to indicate the polarity of voltage induced by mutual flux. – Dot convention • A dot is placed in the circuit at one end of each of the two magnetically coupled coils to indicate the direction of the magnetic flux if current enters that dotted terminal of the coil. • If a current enters the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is positive at the dotted terminal of the second coil. • Alternatively, if a current leaves the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is negative at the dotted terminal of the second coil. 23 Inductance (cont.) 24 Numerical Problem on Magnetic Circuit • A rectangular iron core has an air-gap. Find the exciting current needed to establish a flux density of 1.2 T in the air-gap. Given ππ = 400 turns and ππππ (iron) = 4000. 25 Numerical Problem on Magnetic Circuit (cont.) • Solution: – Given Flux density, π΅π΅ = 1.2 T No. of turns, ππ = 400 Relative permeability of iron, ππππ = 4000 Core length, ππππ = 2 [(20 – 4) + (16 – 4)] – 0.2 = 55.8 cm Cross-sectional area of core, π΄π΄ππ = 16 cm2 Core reluctance, π π ππ = 55.8 × 10–2 4ππ×10−7 ×4000×16×10–4 = 0.694 × 105 AT/Wb Air-gap length, ππππ = 0.2 cm Cross-sectional area of air-gap, π΄π΄ππ = π΄π΄ππ = 16 cm2 Air-gap reluctance, π π ππ = 0.2 × 10–2 4ππ×10−7 ×16×10–4 = 9.95 × 105 AT/Wb Total reluctance, π π ππ = π π ππ + π π ππ = 10.64 × 105 AT/Wb Flux in the magnetic circuit, ππ = π΅π΅ × π΄π΄ππ = 1.2 × 16 × 10–4 = 1.92 mWb Now, ππ × π π ππ = β± = ππ × ππ ⇒ ππ = ππ × π π ππ ππ = 1.92× 10–3 ×10.64×105 4000 = 5.11 A 26 Question on Magnetic Circuit • How does the inductor used in Lab Class of EEN-112 (study of RLC circuit) behave as a variable inductor? 27 Single Phase Transformer 28 Definition and Applications • A static electromagnetic devices (or electrical machine), which transfers electrical power from one circuit to another circuit using the principle of mutual induction without change in the frequency. • Transformer are used for a wide range of purposes, such as – Transmission and distribution of electricity for changing the voltage levels – Telecommunication circuits – Instrumentation circuits – Measurement of voltage, current, power, energy – Control circuits – DC supply such as chargers and adaptors 29 Application of Power Transformers in Power System 30 Operating Principle • The primary winding is connected to the single–phase ac supply, an ac current starts flowing through it, which produces an alternating flux in the core. • Most of this changing flux gets linked with the secondary winding through the core. • The varying flux will induce voltage into the secondary winding according to the faraday’s laws of electromagnetic induction. • Voltage level changes but frequency remains same. • There is no electrical contact between the two winding, an electrical energy gets transferred from primary to the secondary 31 Constructional Details • Windings – The primary and secondary windings are made of copper/aluminum conductors. • Core – The transformer core is made of highly permeable iron so that excitation current required to establish core flux is a small percentage (2 - 4) of the primary current, the rest being the useful component which corresponds to the load current. – High permeable core reduces the leakage flux. – To minimize the eddy current loss, the core is constructed with silicon steel laminations in form of rectangular strips (0.35 mm thickness for 50 Hz) insulated from one another by thin layers of varnish. 32 Constructional Details (cont.) • Core – Two types of geometrical core shapes and winding arrangements are used practically • Core Type – Two limbs or legs with half-LV and half-HV placed on each limb – Has a longer mean flux path but a shorter mean length of coil turn. – LV coils are placed inside (adjoining the core) and HV coils are placed outside. • Shell type – Three limbs with both windings placed on the central limb which forms a shell around the windings. – Half of the flux of the central limb is returned though each outer limb. – LV and HV coil packets are sandwiched. 33 Constructional Details (cont.) Core-type Shell-type 34 Constructional Details (cont.) 35 Constructional Details (cont.) • Some other parts such as suitable tank, conservator, bushings, breather, explosion vent etc. are also used along with the core and windings. 36 Ideal Transformer • A transformer is said to be ideal, if it has the following properties: – The transformer core material has infinite permeability so that it requires zero MMF to create flux in the core. – The transformer core losses are negligible. – Both coils are lossless i.e. ohmic power losses and resistance voltage drops in the actual transformer are neglected – The leakage flux is negligible, i.e. no reactive voltage drops in windings 37 Voltages and EMFs in an Ideal Transformer • When a sinusoidal voltage is applied to the primary winding, flux, ππ is established in the core linking both primary and secondary windings ππ = ππππππππ sin ππππ where, ππππππππ is the maximum value of flux; ππ = 2ππππ is the angular frequency of supply in rad/s; ππ is the supply frequency; π‘π‘ denotes the instant of time; and ππ1 and ππ2 are the number of turns in primary and secondary windings, respectively. 38 Voltages and EMFs in an Ideal Transformer (cont.) • Induced emf in primary winding balances the applied voltage as per KVL ππππ = ππππ1 ππππππππ = ππππ1 ππππππππ sin ππππ + 90° π£π£1 = ππ1 = ππ1 ππππ • Similarly, the induced emf in secondary winding becomes π£π£2 = ππ2 = ππππ2 ππππππππ cos ππππ = ππππ2 ππππππππ sin ππππ + 90° • In terms of rms values, ππ1 = πΈπΈ1 = ππππ1 ππππππππ = 2ππππππ1 ππππππππ 2 ππ2 = πΈπΈ2 = 2ππππππ2 ππππππππ • The core flux is dependent on the supply voltage and maximum flux is proportional to rms value of supply voltage. 39 Transformation Ratio and Phasor Diagram for an Ideal Transformer • The transformation ratio of rms values of voltages is πποΏ½1 πΈπΈοΏ½1 ππ1 πΈπΈ1 ππ1 = = = = = ππ πΈπΈ2 ππ2 πποΏ½2 πΈπΈοΏ½2 ππ2 where, ππ is the turn ratio. • Also, the voltage per turn for both the windings are same ππ1 πΈπΈ1 ππ2 πΈπΈ2 = = = ππ1 ππ1 ππ2 ππ2 • In an ideal transformer, the voltages are in direct ratio of turns with no change in phase angle 40 Current in an Ideal Transformer • The current drawn by the secondary load is ππ2 = πΌπΌ2 2 cos ππππ − ππ2 where, ππ2 = phase angle of load assumed lagging • By Lenz’s law, this current causes MMF to oppose the core flux ππ. In phasor terms, secondary MMF is πΉπΉοΏ½2 = ππ2 πΌπΌ2Μ where, πΌπΌ2Μ = πΌπΌ2 ∠ππ2 • As the core flux cannot change being governed by primary applied voltage and frequency, a current is drawn from the source to cause MMF, πΉπΉοΏ½1 equal and opposite to MMF, πΉπΉοΏ½2 πΉπΉοΏ½1 = πΉπΉοΏ½2 ⇒ ππ1 πΌπΌ1Μ = ππ2 πΌπΌ2Μ 41 Current in an Ideal Transformer (cont.) • The transformation ratio of rms values of currents is πΌπΌ1Μ πΌπΌ1 ππ2 1 = = = πΌπΌ2 ππ1 ππ πΌπΌ2Μ • An ideal transformer transforms the current in the inverse ratio of turns and the phase is preserved 42 Volt-Amperes in an Ideal Transformer • As derived • Therefore, πποΏ½1 ππ1 = οΏ½ ππ2 ππ2 πΌπΌ1Μ ππ2 and = Μ πΌπΌ2 ππ1 πποΏ½1 οΏ½ πΌπΌ1∗Μ = πποΏ½2 οΏ½ πΌπΌ2∗Μ ⇒ ππ1 + ππππ1 = ππ2 + ππππ2 • Input VA equals output VA, that is, there is change in VA in the ideal transformer. • Input active and reactive power equal output active and reactive power, that is, there is no loss in the ideal transformer. 43 Equivalent Circuit of an Ideal Transformer • Secondary voltage and current referred to the primary side ππ ππ πποΏ½2′ = 1 πποΏ½2 = πποΏ½1 and πΌπΌ2′Μ = 2 πΌπΌ2Μ = πΌπΌ1Μ ππ2 ππ1 ππ1 ππ2 • Primary voltage and current referred to the secondary side ππ ππ πποΏ½1′ = 2 πποΏ½1 = πποΏ½2 and πΌπΌ1′Μ = 1 πΌπΌ1Μ = πΌπΌ2Μ 44 Impedance Transformation in an Ideal Transformer • An impedance ππ2Μ connected across the secondary side of the ideal transformer as seen from the primary side becomes ππ2 οΏ½ 2 2 ππ2 πποΏ½1 πποΏ½1 ππ1 πποΏ½2 ππ1 ππ1 ππ2Μ = = = ⇒ = ππ2Μ = ππ2Μ ′ Μ Μ ππ1 Μ ππ ππ2 πΌπΌ2Μ πΌπΌ πΌπΌ 1 1 1 πΌπΌ1 ππ2 • ππ2Μ ′ is called ‘the secondary impedance referred to the primary side”. 45 Accounting for Finite Permeability and Core Loss • In a real transformer, the core has finite permeability and to establish flux in the core, the primary winding would draw a current component called magnetizing current from the source over and above the load current. • Assuming the core to be linear, the magnetizing current is then given by ππππ β β β = ππ = ππ sin ππππ = 2 ππ1 ππ1 ππππππ ππ ππ1 ππ sin ππππ 2 1 where, β is the core reluctance and πΌπΌππ is the rms value of magnetizing current. • Recalling ππ = ππππππππ sin ππππ π£π£1 = ππππ1 ππππππππ cos ππππ = ππππ1 ππππππππ sin ππππ + 90° ππππ1 ππππππππ 2ππ1 ⇒ ππππππππ = ππ1 = ππππ1 2 46 Accounting for Finite Permeability and Core Loss (cont.) • Magnetizing current is in phase with the core flux and lags the induced emf by 90°. It also depends on the supply voltage. Therefore, the effect of magnetizing current is modelled by a magnetizing reactance ππππ across the supply. • A real core will also have power losses (hysteresis and eddy-current losses) because it carries alternating flux. Since both hysteresis and eddy-current losses depend on maximum flux density (maximum flux per unit area) and maximum flux is proportional to rms value of supply voltage, the core losses can be modelled as a resistance π π ππ across the supply. 47 Accounting for Finite Permeability and Core Loss (cont.) • The rms values of net exciting current drawn by the primary to create core flux is Μ + πΌπΌππΜ πΌπΌ0Μ = πΌπΌππ Μ = where, magnetizing current, πΌπΌππ οΏ½1 ππ ππππππ οΏ½ ππ = and core (iron) loss current, πΌπΌππΜ = 1 π π ππ • The magnetizing current in a transformer is in the range 2–5% of the rated current. • On no-load (πΌπΌ2Μ = 0), the transformer primary would draw only the exciting current from the source. Therefore, the exciting current is also termed as no-load current (hence, the symbol πΌπΌ0Μ ). 1 π π ππ 1 Susceptance, π΅π΅ππ = ππππ Conductance, πΊπΊππ = 48 Accounting for Winding Resistance and Leakage Flux • Both primary and secondary windings of a transformer have resistances. • Apart from this the two windings have leakage flux (flux linking only one winding). These leakage fluxes do not contribute in the process of energy transfer, but these cause the primary and secondary windings to possess leakage inductances and, therefore, leakage reactances at steady sinusoidal operation. • The winding resistances and leakage reactances can be lumped in series with the windings along with the core in a circuit model. Let windings resistances be ππ1 , ππ2 and winding reactances (inductive) be π₯π₯1 and π₯π₯2 . 49 Equivalent Circuit of a Real Transformer πΌπΌ2′Μ ππ2 = πΌπΌ2Μ ππ1 πΌπΌ1Μ = πΌπΌ0Μ + πΌπΌ2′Μ πΈπΈοΏ½1 ππ1 = πΈπΈοΏ½2 ππ2 πποΏ½1 = πΈπΈοΏ½1 + πΌπΌ2′Μ × (ππ1 + πππ₯π₯1 ) πποΏ½2 = πΈπΈοΏ½2 − πΌπΌ2Μ × (ππ2 + πππ₯π₯2 ) 50 Equivalent Circuit of a Real Transformer (cont.) • By the technique of impedance transformation, the resistance and leakage reactance of one side can be transferred to other side of the transformer. • Then equivalent series resistance and reactance of the transformer referred to the primary side are Equivalent resistance, π π = ππ1 + ππ2′ = ππ1 + Equivalent resistance, ππ = π₯π₯1 + π₯π₯2′ = π₯π₯1 + ππ1 2 ππ2 ππ2 ππ 2 1 ππ2 π₯π₯2 ⇒ 51 Equivalent Circuit of a Real Transformer (cont.) ππ1 πποΏ½1 = πποΏ½2′ + πΌπΌ2′Μ × π π + ππππ = πποΏ½2 + πΌπΌ2′Μ × ππ2 = = = = ππ1 πποΏ½ ππ2 2 + πΌπΌ2′Μ × ππ1 + πππ₯π₯1 + πΌπΌ2′Μ × ππ1 ππ1 + ππ2 ππ1 2 ππ2 ππ2 + πππ₯π₯2 ππ1 ππ2 ππ1 2 ′Μ οΏ½ Μ ππ + πΌπΌ2 × ππ1 + πππ₯π₯1 + πΌπΌ2 × ππ2 ππ2 2 ππ1 ππ2 ππ1 πποΏ½2 + πΌπΌ2Μ × ππ2 + πππ₯π₯2 + πΌπΌ2′Μ × ππ1 + πππ₯π₯1 ππ2 πΈπΈοΏ½1 + πΌπΌ2′Μ × ππ1 + πππ₯π₯1 = πποΏ½1 2 ππ2 + πππ₯π₯2 = ππ1 πΈπΈοΏ½ ππ2 2 ππ1 + ππ π₯π₯1 + ππ2 2 π₯π₯2 + πΌπΌ2′Μ × ππ1 + πππ₯π₯1 52 Approximate Equivalent Circuit of a Real Transformer • The magnetizing shunt branches in the circuit model do not affect voltage computation and may therefore be ignored. 53 Tests on Transformer • Transformer parameter determination necessitates two following two tests – Open-Circuit (OC) or No-load test • Performed to determine core loss and parameters of the shunt branch of the transformer circuit model. – Short-Circuit (SC) test • Performed to determine Cu-loss and series parameters of the transformer circuit model. 54 Open-Circuit (OC) or No-Load Test • The transformer is excited at rated voltage (and frequency) at LV side while HV side is kept open-circuited. • The magnetizing current in a transformer is in the range 2–5% of the rated current. 55 Open-Circuit (OC) or No-Load Test (cont.) • Let the meter readings be – Voltage (V) = ππ1 – Current (A) = πΌπΌ0 – Power (W) = ππ0 = core loss (ππππ ) • The shunt branch parameters become ππ0 = πΊπΊππ = πΌπΌ0 ππ1 π΅π΅ππ = = ππ0 ππ1 2 ππ0 πΊπΊππ 2 2 + π΅π΅ππ − πΊπΊππ 2 2 56 Short-Circuit Test • The transformer is shorted on LV side and is excited from a reduced voltage (rated frequency) source from HV side with full-load current on HV side. • The voltage needed to circulate full-load current is only of the order of 5–8% of the rated voltage. 57 Short-Circuit Test (cont.) • At the reduced primary voltage, the exciting πΌπΌ0 gets reduced to 0.1% to 0.4% of the rated current. The magnetizing shunt branch of the circuit model can therefore be conveniently dropped. • Let the meter readings be – Voltage (V) = ππππππ – Current (A) = πΌπΌππππ – Power (W) = ππππππ = full load copper loss (ππππ ) 58 Short-Circuit Test (cont.) • The winding parameters become ππππππ ππ = = π π 2 + ππ πΌπΌππππ ππππππ π π = πΌπΌππππ 2 ππ = ππ 2 − π π 2 2 59 Losses in Transformer • The transformer has two major losses: – Core (iron) loss, ππππ is a constant loss (for constant voltage and frequency operation) – Copper (πΌπΌ 2 π π ) loss of both windings, ππππ is a variable loss. • Other transformer losses, which are insignificant for efficiency computation, are – Load (stray) loss which results from leakage fields inducing eddy currents in tank walls and conductors – Dielectric loss caused by leakage current in the insulating materials 60 Efficiency of Transformer • The efficiency of transformer, ππ Output power Output power = ππ = Input power Output power + Losses Output power, ππ2 = ππ2 οΏ½ πΌπΌ2 οΏ½ cos ππ2 where, ππ2 is the phase angle of load ππ2 οΏ½ πΌπΌ2 οΏ½ cos ππ2 ππ2 οΏ½ πΌπΌ2 οΏ½ cos ππ2 = ππ = ππ2 οΏ½ πΌπΌ2 οΏ½ cos ππ2 + ππππ + ππππ ππ2 οΏ½ πΌπΌ2 οΏ½ cos ππ2 + ππππ + πΌπΌ2 2 π π 2 where, π π 2 is the equivalent transformer resistance referred to secondary side. ππ = ππ2 οΏ½ cos ππ2 ππ ππ2 οΏ½ cos ππ2 + ππ + πΌπΌ2 οΏ½ π π 2 πΌπΌ2 61 Efficiency of Transformer (cont.) • For maximum efficiency of transformer, ππππππππ ππππ = 0 ⇒ ππππ = πΌπΌ2 2 π π 2 πππΌπΌ2 • Thus, for maximum efficiency of transformer, copper loss should be equal to iron loss or variable loss should be equal to constant loss. • The load corresponding to maximum efficiency Load current, πΌπΌ2 = ππππ π π 2 = πΌπΌ2,πΉπΉπΉπΉ ππππ ππππ,πΉπΉπΉπΉ Output power, ππ2 = ππ2 οΏ½ πΌπΌ2 οΏ½ cos ππ2 where, πΌπΌ2,πΉπΉπΉπΉ is the full load secondary current, and ππππ,πΉπΉπΉπΉ is the Cu loss at full load. 62 Voltage Regulation • Various loads require a nearly constant voltage supply. It is, therefore, essential that the output voltage of a transformer stays within narrow limits as load and its power factor vary. • The leakage reactance is the major cause of voltage drop in a transformer and must be kept as low as possible by design and manufacturing techniques. • Voltage regulation is an indication of change in output voltage of transformer with change in load. It is expressed in percentage of rated output voltage. 63 Voltage Regulation (cont.) • The voltage regulation of a transformer at a given power factor is defined as VR (in %) = ππ2,0 − ππ2,πΉπΉπΉπΉ × 100% ππ2,πΉπΉπΉπΉ where, ππ2,0 is the no-load secondary voltage, and ππ2,πΉπΉπΉπΉ is the full load secondary voltage. • As in a transformer, πΌπΌπΌπΌ and πΌπΌπΌπΌ voltage drops are much smaller in magnitude compared to ππ1 and ππ2 , the angle between πποΏ½1 and πποΏ½2 is only a few degrees such that ππ1 ≈ ππππ 64 Voltage Regulation (cont.) ππ1 − ππ2 ≈ π΅π΅π·π· = I π π cos ππ + ππ sin ππ , for lagging power factor = I π π cos ππ − ππ sin ππ , for leading power factor • When the load is thrown off, ππ2,0 = ππ1 , therefore ππ2,0 − ππ2 = ππ1 − ππ2 = I π π cos ππ ± ππ sin ππ πΌπΌ π π cos ππ ± ππ sin ππ VR (in %) = × 100% ππ2 65 Voltage Regulation (cont.) • For minimum voltage regulation in transformer π π cos ππ − ππ sin ππ = 0 (leading power factor) π π tan ππ = ππ ππ cos ππ = π π 2 + ππ 2 • For maximum voltage regulation in transformer ππππππ = 0 ⇒ −π π sin ππ + ππ cos ππ = 0 (lagging power factor) ππππ ππ tan ππ = π π π π cos ππ = π π 2 + ππ 2 66 Numerical Problems on Transformer 1. An ideal transformer has a turn-ratio of 100/300. The LV winding is connected to a source of 3.3 kV, 50 Hz. An impedance of (100 + j 35) Ω is connected across the secondary terminals. Calculate a) Value of maximum core flux, b) Primary and secondary currents, c) Real and reactive powers supplied by the source to the transformer primary, and d) Value of impedance which connected directly across the source would draw the same real and reactive power as in (c). 67 Numerical Problems on Transformer (cont.) 2. A 150 kVA, 2400/240 V single-phase transformer has the following parameters of the equivalent circuit (supplied from 2400 V side) ππ1 = 0.2 Ω; ππ2 = 2 × 10−3 Ω; π₯π₯1 = 0.6 Ω; π₯π₯2 = 6 × 10−3 Ω; π π ππ = 10 × 103 Ω; and ππππ = 1.6 × 103 Ω a) Calculate the equivalent resistance and leakage reactance as seen of the HV side. b) Convert the resistance and reactance values referred to LV side. c) At rated current, calculate the impedance voltage drop on the HV side. Also calculate the voltage drop as percentage of the rated voltage. d) With the secondary open (no load), what current will be drawn from HV side (2400 V source). What is its pf? 68 Numerical Problems on Transformer (cont.) 3. The resistances and leakage reactances of a 10 kVA, 50 Hz, 2300/230 V distribution transformer are (subscript 1 refers to HV and 2 to LV winding) ππ1 = 3.96 Ω; ππ2 = 3.96 × 10−2 Ω; π₯π₯1 = 15.8 Ω; and π₯π₯2 = 1.58 × 10−3 Ω; a) The transformer delivers rated kVA at 0.8 pf lagging to a load on the LV side. Find the HV-side voltage necessary to maintain 230 V across load terminals. Also find the percentage voltage regulation. b) If a capacitor bank is connected across the load, what should be the kVA capacity of the bank to reduce the voltage regulation to zero? What should be the HV-side voltage under these circumstances? 69 Numerical Problems on Transformer (cont.) 4. The test data obtained on a 15 kVA, 3000/250 V, 50 Hz distribution transformer are: OC test (LV side) 250 V 0.62 A 105 W SC test (HV side) 157 V 5.2 A 360 W a) Determine the equivalent circuit parameters referred to the HV side. b) If the transformer is carrying full-load at 250 V, 0.8 pf leading, calculate the transformer (i) voltage regulation, and (ii) efficiency. c) Find the pf of full-load for zero voltage regulation. d) Find the percentage load at 0.8 pf for maximum efficiency of transformer and its value. 70 Suggested Readings • Kothari D.P. , Nagrath I.J., “Basic Electrical Engineering”, Tata McGraw Hill Education (India) Private Limited, 4th Edition, 2022. • Vincent Del Toro, “Electrical Engineering Fundamentals”, Prentice Hall of India, 2002. • Alexander C.K., Sadiku M.N.O., “Fundamentals of Electric Circuits”, McGraw Hill, 5th Edition, 2012. • Chapman, Stephen, J., “Electric Machinery Fundamentals”, McGraw Hill Book Company, 1985. 71 Thanks 72
