Mechanical and Electrical Systems SKAA 2032 Transformers Dr

Mechanical and Electrical Systems SKAA 2032 Transformers Dr. Asrul Izam Azmi Faculty of Electrical Engineering Universiti Teknologi Malaysia

Transformers • A transformer is a static machine which change voltage: step up or down. • There is no electromechanical energy conversion. • The transfer of energy takes place through the magnetic field and all currents and voltages are AC • Transformer can be categorized as: ü The ideal transformers ü Practical transformers ü Special transformers ü Three phase transformers

Transformers

Transformers

Transformers

Applications of the transformer • A typical power system consists of generation, transmission and distribution • Power from plant/station is generated around 11 -1320 -30 k. V (depending upon manufacturer and demand) • This voltage is carried out at a distance to reach for utilization through transmission line system by step up transformer at different voltage levels depending upon distance and losses • The distribution is made through step down transformer according to the consumer demand.

Functions of transformer • Raise or lower voltage or current in AC circuit • Isolate circuit from each other • Increase or decrease the apparent value of a capacitor, inductor or resistor • Enable transmission of electrical energy over great distances • Distribute safely in homes and factories Transformer is one of the most useful electrical devices ever invented

Principles of Transformer • A transformer consists of two electric circuits called primary and secondary • A magnetic circuit provides the link between primary and secondary

Principles of Transformer The basic idea: • AC voltage is applied to the primary circuit, AC current Ip flows into the primary circuit. • Ip sets up a time-varying magnetic flux Ф in the core • An AC voltage is induced to the secondary circuit, Vs according to the Faraday’s law

Core Types of Transformer • The magnetic (iron) core is made of thin laminated steel sheet. The reason of using laminated steel is to minimize the eddy current loss by reducing thickness • There are two common cross section of core which include square or (rectangular) for small transformers and circular (stepped) for the large and 3 phase transformers.

Configuration of Single phase

Three Phase Transformer The three phase transformer iron core has three legs A phase winding is placed in each leg

Construction of Transformer • Core (U/I) Type: Is constructed from a stack of U and I shaped laminations. In a core-type transformer, the primary and secondary windings are wound on two different legs of the core • Shell Type: Is constructed from a stack E and I shaped laminations. In a shell-type transformer, the primary and secondary windings are wound on the same leg of the core, as concentric windings, one on top of the other.

Construction of Transformer • Core and Shell Type Transformer Shell Type Core Type

Construction of a Small Transformer

Transformer with Cooling System

The Ideal Transformer For an ideal transformer, we assume: • No losses • Flux produced by the primary is completely linked by the secondary and vice versa • Core is infinitely permeable • No leakage flux of any kind

The Ideal Transformer • Primary and secondary posses N 1 and N 2 turns respectively • Primary is connected to a sinusoidal source Eg • Magnetizing current Im creates a flux of Φm

The Ideal Transformer • The flux is completed linked by the primary and secondary windings – mutual flux • Flux varies sinusoidally and reaches a peak value of Φm

The Ideal Transformer • A transformer with more turns in its primary than its secondary coil will reduce voltage and is called a step-down transformer • One with more turns in the secondary than the primary is called a step-up transformer

The Ideal Transformer • The sinusoidal current Im produces a sinusoidal mmf (magnetomotive force )= NIm which in turn creates a sinusoidal flux. The flux induces an effective voltage E across the terminals of the coil Induces Voltages: • The effective induced emf in primary winding is • Where N 1 is the number of winding turns in primary winding, Фm the maximum (peak) flux and f is the frequency of the supply voltage

The Ideal Transformer • This equation shows that for a given frequency and a given number of turns, Фm varies in proportion to the applied voltage, Eg • This means that if Eg is kept constant, the peak flux must remain constant • Similarly, the effective induced emf in secondary winding:

The Ideal Transformer - at No Load a = Turn ratio = Voltage induced in the primary [V] = Voltage induced in the secondary [V] = Number of turns on the primary

Example problem A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240 V, determine the secondary voltage, assuming an ideal transformer N 1=500, N 2=3000, V 1=240 V For ideal transformer, V 1=E 1 and V 2=E 2

Ideal Transformer Under Load: Current Ratio Let us connect a load Z across the secondary of the ideal transformer. A secondary current I 2 will immediately flow. 1) In an ideal transformer the primary and secondary windings are linked by the mutual flux, Фm, consequently voltage ratio will be the same as at no load 2) If the supply voltage Eg is kept fixed, then the primary induced voltage E 1 remain fixed. Consequently, Фm also remains fixed. It follows that E 2 also remain fixed We conclude that E 2 remains fixed whether a load is connected or not

Ideal Transformer Under Load: Current Ratio • Let us now examine the mmf created by the primary and secondary windings. First, current I 2 produces a secondary mmf N 2 I 2. If it acted alone, this mmf would produce a profound change in the Фm. But we just saw that Фm does not change under load. • We conclude that flux Фm can only remain fixed if the primary develops a mmf which exactly counterbalances N 2 I 2 at every instant. Thus, a primary current I 1 must flow so that

Ideal Transformer Under Load: Current Ratio To obtain the required instant-to-instant bucking effect, current I 1 and I 2 must increase and decrease at the same time (a)Ideal transformer under load (b) Phasor relationships under load

Ideal Transformer Under Load: Current Ratio I 1 I 2 Secondary Primary = Primary current [A] Single phase transformer = Secondary current [A] = Number of turns on the primary = Number of turns on the secondary INSPIRING CREATIVE AND INNOVATIVE MINDS

Transformer Impedance Primary Impedance: Primary impedance in terms of secondary impedance : Primary Voltage : Primary Current : INSPIRING CREATIVE AND INNOVATIVE MINDS

Example problem An ideal transformer has a turns ratio of 8: 1 and the primary current is 3 A when it is supplied at 240 V. Calculate the secondary voltage and current.

Example problem

Example problem A transformer coil possesses 4000 turns and links an ac flux having a peak value of 2 m. Wb. If the frequency is 60 Hz, calculate the effective value of the induced voltage E.

Example problem A coil having 90 turns is connected to a 120 V, 60 Hz source. If the effective value of the magnetizing current is 4 A, calculate the following: a. The peak value of flux b. The peak value of the mmf c. The inductive reactance of the coil d. The inductance of the coil.

Real Transformer Leakage Flux: Not all of the flux produced by the primary current links the winding, but there is leakage of some flux into air surrounding the primary. Similarly, not all of the flux produced by the secondary current (load current) links the secondary, rathere is loss of flux due to leakage. These effects are modeled as leakage reactance in the equivalent circuit representation.

Voltage Regulation • Voltage Regulation is defined as the change in the magnitude of the secondary voltage as the load current changes from the no-load to full load • The primary side voltage is always adjusted to meet the load changes; hence V’s and Vs are kept constant

Efficiency of Transformer • As always, efficiency is defined as power output to power input ratio • Pcopper represents the copper losses in primary and secondary windings. There are no rotational losses.

Example Problem A not-quite-ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 120 V, 60 Hz source. The coupling between the primary and the secondary is perfect but the magnetizing current is 4 A. calculate: a. The effective voltage across the secondary terminals b. The peak voltage across the secondary terminals. c. The instantaneous voltage across the secondary when the instantaneous voltage across the primary is 37 V. Ans: 3000 V, 4242 V, 925 V.

Example Problem An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 200 V, 50 Hz source. The load across the secondary draws a current of 2 A at a power factor of 80 per cent lagging. Calculate : • The effective value of the primary current • The instantaneous current in the primary when the instantaneous current in the secondary is 100 m. A. • The peak flux linked by the secondary winding. Ans: 50 A, 2. 5 A, 10 m. Wb

Example Problem A 125 k. VA transformer has 500 turns on the primary and 80 turns on the secondary. The primary and secondary resistances are 0. 5 Ω and 0. 025 Ω respectively, and the corresponding leakage reactances are 2. 5 Ω and 0. 025 Ω respectively. The supply voltage is 2. 2 k. V. Calculate the voltage regulation and the secondary terminal voltage for full load having a power factor of 0. 85 lagging

Example Problem The primary and secondary windings of a 400 k. VA transformer have resistances of 0. 3 Ω and 0. 0015 Ω respectively. The primary and secondary voltages are 15 k. V and 0. 4 k. V respectively. If the core loss is 2. 5 k. W and the power factor of the load is 0. 80, calculate the efficiency of the transformer on full load.

Summary At the end of this topic, you should know: • The functions of a transformer in electrical distribution system • The working principles of a transformer • The voltage and current ratios • The difference between the ideal and real transformer