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Transformers Mechanical and Electrical Systems SKAA 2032

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Presentation on theme: "Transformers Mechanical and Electrical Systems SKAA 2032"— Presentation transcript:

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

2 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: The ideal transformers Practical transformers Special transformers Three phase transformers

3 Transformers

4 Transformers

5 Transformers

6 Applications of the transformer
A typical power system consists of generation, transmission and distribution Power from plant/station is generated around kV (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.

7 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

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

9 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

10 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.

11 Configuration of Single phase

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

13 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.

14 Construction of Transformer
Core and Shell Type Transformer Shell Type Core Type

15 Construction of a Small Transformer

16 Transformer with Cooling System

17 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

18 The Ideal Transformer Primary and secondary posses N1 and N2 turns respectively Primary is connected to a sinusoidal source Eg Magnetizing current Im creates a flux of Φm

19 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

20 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

21 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 N1 is the number of winding turns in primary winding, Фm the maximum (peak) flux and f is the frequency of the supply voltage

22 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:

23 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 = Number of turns on the primary

24 Example problem A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240V, determine the secondary voltage, assuming an ideal transformer N1=500, N2=3000, V1=240V For ideal transformer, V1=E1 and V2=E2

25 Ideal Transformer Under Load: Current Ratio
Let us connect a load Z across the secondary of the ideal transformer. A secondary current I2 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 E1 remain fixed. Consequently, Фm also remains fixed. It follows that E2 also remain fixed We conclude that E2 remains fixed whether a load is connected or not

26 Ideal Transformer Under Load: Current Ratio
Let us now examine the mmf created by the primary and secondary windings. First, current I2 produces a secondary mmf N2I2. 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 N2I2 at every instant. Thus, a primary current I1 must flow so that

27 Ideal Transformer Under Load: Current Ratio
To obtain the required instant-to-instant bucking effect, current I1 and I2 must increase and decrease at the same time (a)Ideal transformer under load (b) Phasor relationships under load

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

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

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

31 Example problem

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

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

34 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, rather there is loss of flux due to leakage. These effects are modeled as leakage reactance in the equivalent circuit representation.

35 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

36 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.

37 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: The effective voltage across the secondary terminals The peak voltage across the secondary terminals. The instantaneous voltage across the secondary when the instantaneous voltage across the primary is 37 V. Ans: 3000V, 4242 V, 925 V.

38 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 mA. The peak flux linked by the secondary winding. Ans: 50 A, 2.5 A, 10 mWb

39 Example Problem A 125 kVA transformer has 500 turns on the primary and 80 turns on the secondary. The primary and secondary resistances are 0.5 Ω and Ω respectively, and the corresponding leakage reactances are 2.5 Ω and Ω respectively. The supply voltage is 2.2 kV. Calculate the voltage regulation and the secondary terminal voltage for full load having a power factor of 0.85 lagging

40 Example Problem The primary and secondary windings of a 400 kVA transformer have resistances of Ω and Ω respectively. The primary and secondary voltages are 15 kV and 0.4 kV respectively. If the core loss is 2.5 kW and the power factor of the load is 0.80, calculate the efficiency of the transformer on full load.

41 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

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