Electrical Machine

EPO540 ELECTRICAL MACHINES TRANSFORMERS (I)

prepared by Kanendra

1. INTRODUCTION  

 

Static machine Not an energy conversion device - necessary & essential in many energy conversion Consist of two or more windings coupled by a mutual magnetic field Ferromagnetic cores usually used to provide tight magnetic coupling and high flux densities – IRON CORE



IRON CORE used in high-power applications



AIR-CORE used in low power electronic circuits

Schematic representation of a two winding transformer prepared by Kanendra

1. INTRODUCTION

Core type

  

Shell Type

Primary function is to change voltage level – step up/step down Low power electronic circuit – isolation or impedance matching Measure voltage and current – instrument transformers prepared by Kanendra

2. IDEAL TRANSFORMER 

Ideal transformer has the following properties:

1.

Winding resistances are negligible

2.

All fluxes confined to the core and link both windings. No leakage fluxes are present. Core loss = 0.

3.

Permeability of core is infinite ( μ = 0) & net mmf required to establish flux is zero (∑ MMF = 0)

Ideal transformer prepared by Kanendra

3. PRACTICAL TRANSFORMER 

Practical transformers – winding resistances, not all windings link the same flux, permeability of the core material is not infinite, core losses occur

prepared by Kanendra

3. PRACTICAL TRANSFORMER E1 = E2’ = aE1 V2’ = aV2 I2’ = I2/a Xl2’ = a2Xl2 R2’ = a2R2 Transformer equivalent circuit

Approximate equivalent circuit prepared by Kanendra

4. DETERMINATION OF EQUIVALENT CIRCUIT PARAMETERS

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In order to predict the behavior of the transformer, circuit parameters (R1, X1, R2’, Xl2’, Rc, Xm) has to be known These parameters can be calculated from the dimensions and properties of the materials used – design data of transformer Difficulty in obtaining data and some required parameters A simple and direct method is used to obtain these parameters– performing two different tests: “No Load (Open Circuit) Test” & “Short Circuit Test”

prepared by Kanendra

5. No-Load Test 

Open circuit HV side and apply low voltage on LV side



Low voltage power supply is readily available  =  =

  

=>  =

  

 

 =  =

 2 −  2

Wiring diagram

  Equivalent circuit prepared by Kanendra

6. Short-Circuit Test 

Short circuit LV side and apply low voltage on HV side



Low voltage power supply is readily available 2



 =   =>  =    =

 =

 

 2 −  2

Wiring diagram

Equivalent circuit prepared by Kanendra

7. VOLTAGE REGULATION 

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Loads connected to secondary are usually designed to operate at constant voltage However, current drawn from the transformer results in a voltage drop in the internal impedance of the transformer (Zeq)



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When switch is open (no load condition) => 2 | =

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When switch is closed (with load) => 2 | = 2 | ± ∆2

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



Voltage change depends on the nature of the load – due to IZ (internal impedance of transformer) Large voltage change is undesirable prepared by Kanendra

7. VOLTAGE REGULATION 

Figure of merit used to identify characteristics of voltage change in transformer is called voltage regulation

   =

2



− 2

2







VR can be positive or negative, depending on the nature of the load

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Voltage regulation should be as small as possible (< 10%)

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Equivalent circuit referred to primary:

 =

2 ′

− 2 ′ 2 ′ 





Load voltage is normally taken at rated voltage => 2 ′ 

If load is removed (no load condition), 2 ′

 =





= 2 ′



= 

− 2 ′ 2 ′ 







prepared by Kanendra

7. VOLTAGE REGULATION

Phasor diagram

 = 2  2 ′ + j2 ′ θ = angle of load impedance 2

θeq1 = angle of transformer equivalent impedance, Zeq1

prepared by Kanendra

8. EFFICIENCY 

Losses in transformers are small – static device, no rotational losses

=



  =    

Losses – core losses (Pc) & copper loss (Pcu)

=





    

Core loss (Pc) depends on peak flux density in the core. Transformer connected to constant voltage supply, therefore core loss is constant– obtained from No Load Test

prepared by Kanendra

8. EFFICIENCY 

Copper loss (Pcu) function of load current - determined through winding currents and resistances Pcu = I12 R1 + I22 R2 = I12 Req1 = I22 Req2

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Therefore, given any load condition (with power factor) Pout = V2 I2 cos ϴ2 η

=

V2 I2 cos 2 V2 I2 cos 2 + PC + I22 Req2

prepared by Kanendra

9. AUTOTRANSFORMER 



Common winding mounted on a core, secondary is taken from tap on the winding Advantages:

1.

Reduced weight and size

2.

Lower losses

3.

Lower exciting current

4.

Lower cost – less copper used

5.

Variable output voltage



1.

Disadvantages:

Autotransformer

Direct connection between primary and



secondary windings / no isolation

2 2  1 = 2 

=



=

prepared by Kanendra