sab 2032 electrical technology

SAB 2032ELECTRICAL TECHNOLOGY

Weeks 13-14

Transformer

INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009

Transformers

INSPIRING CREATIVE AND INNOVATIVE MINDS

Transformers


A transformer is a static machine which step voltage/current up or downUnlike in rotating machines, 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


Transformers


Applications of the transformer

A typical power system consists of generation, transmission and distributionPower from plant/station is generated around 11-13-20-30kV (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 lossesIts distribution is made through step down transformer according to the consumer demand

Transformers


Transformer is one of the most useful electrical devices

ever invented

Functions of transformer:Raise or lower voltage or current in AC circuitIsolate circuit from each otherIncrease or decrease the apparent value of a capacitor, inductor or resistorEnable to transmit electrical energy over great distancesDistribute safely in homes and factories

Transformers

Principles of Transformer


A transformer consists of two electric circuits called primary and secondary

A magnetic circuit provides the link between primary

and secondary

When an AC voltage is applied to the

primary winding Vp of the transformer,

an AC current Ip will result

Ip sets up a time-varying magnetic flux Ф

in the core

A 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.


PrimaryWinding

SecondaryWinding

Multi-layerLaminatedIron Core

X1X

2H1 H2

WindingTerminals

Configuration of Single phase transformer


3 Phase Transformer

The three phase transformer iron core has three legs

A phase winding is placed in each leg


Construction of Transformer

This can be divided into 2 types: 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 a Small Transformer

Iron core

Terminals

Secondarywinding

Insulation


Transformer with Cooling System

Oil tank

Coolingradiators

High voltagebushing

Low voltagebushing

Winding

Iron corebehind the steel

bar

Radiator

Steeltank

Insulation

Bushing


The Ideal Transformer

For an ideal transformer, we assume No losses Core is infinitely permeable Flux produced by the primary is completely linked by the

secondary and vice versa No leakage flux of any kind



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



The flux is completed linked by the primary and secondary windings – mutual flux

Flux varies sinusoidally and reaches a peak value of Φm



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 sinusoidal current Im produces a sinusoidal mmf 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 the frequency of the supply voltage

mfN.E 11 444



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:

mfN.E 22 444


The Ideal Transformer - at No Load

aN

N

E

E

2

1

2

1

= Voltage induced in the secondary [V] 1E

2E

1N

2N

a = Turn ratio= Voltage induced in the primary [V]

= Number of turns on the primary= Number of turns on the primary


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.Does E2 change when we connect the load?

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


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

2211 ININ



To obtain the required instant-to-instant bucking effect, current I1 and I2 must increase and decrease at the same timeIn other words, the current must be in phase

(a)Ideal transformer under load (b)Phasor relationships under load



aN

N

I

I 1

2

1

1

2

1I = Primary current [A]

2I = Secondary current [A]

1N = Number of turns on the primary

2N = Number of turns on the secondary



Transformer Impedance

Primary Impedance: Primary impedance in terms of secondary impedance : Primary Voltage : Primary Current :

P

PL I

V'Z

LL Za'Z 2

SP aVV

a

II S

P


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.

Ans: 2131V


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:a. The peak value of fluxb. The peak value of the mmfc. The inductive reactance of the coild. The inductance of the coil.

Example problem


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 modelled as leakage reactance in the equivalent circuit representation.


Magnetization Current in a Real Transformer

Although the output of the transformer is open circuit, there will still be current flow in the primary windings

• Magnetization current, iM • Core-loss current, ic

The relation between current and flux is proportional since

N

Si

SNIF


Therefore, in theory, if the flux produce in core is sinusoidal, than the current should also be a perfect sinusoidal. Unfortunately, this is not true.

Current in a transformer has the following characteristics:• It is not sinusoidal but a combination of high

frequency oscillation

• The current lags the voltage at 900

• At saturation, the high frequency components will be

extreme



Core-loss current:• Eddy current loss- is dependent upon the rate of change of

flux• Hysteresis loss - a non linear lossAt no-load, the primary windings is known as the excitation current

cm iii



The equivalent circuit of a transformer

Taking into account real transformer, there are several losses that has to be taken into account in order to accurately model the transformer, namely:

1- Copper (I2R) Losses 2- Eddy current Losses 3- Hysteresis Losses 4- Leakage flux


Equivalent circuit of a Real Transformer

Equivalent Circuit of a Two-winding, 1-phase:

Rc : Core loss component, this resistance models the active loss of the coreXm : magnetization component, this resistance models the reactive loss of the coreRp and Xp : are resistance and reactance of the primary windingRs and Xs : are resistance and reactance of the secondary winding



Impedance Transfer:

To model a transformer, it is important to understand how impedance are transferred from one side to another, that is primary to secondary or secondary to primary

Impedance transfer is used to calculate the current/voltage easier and also to get the voltage and current ratio for the rest of the calculation

In general any impedance transferred from secondary side to primary side must be multiplied by the square of the turn ratio, a2


Equivalent circuit-seen from primary side

The terminal voltage (Vp,Vs) is not constant and changes depends on the load current.

Secondary parameters transferred to primary



Equivalent Circuit

a) referred to primary side b) referred to secondary side


Approximate equivalent circuit


c) Referred to primary side (no exicitation) d) Referred to secondary side (no excitation)



In this model, the parameters from secondary are transferred to primary side.

where, Rep and Xep are equivalent resistance and reactance from primary winding side


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.

100VVV

100aVaVV

V

VVVR%

's

'sp

ssp

NL

FLNL

100


Efficiency of Transformer

Pcopper represents the copper losses in primary and secondary windings. There are no rotational losses.

As always, efficiency is defined as power output to power input ratio

coppercoreoutin

inout

PPPP

%PP

100

43

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

terminalsb. 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: 3000V, 4242 V, 925 V.

44

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 :a. The effective value of the primary currentb. The instantaneous current in the primary when the

instantaneous current in the secondary is 100 mA.c. The peak flux linked by the secondary winding.

Ans: 50 A, 2.5 A, 10 mWb


QUESTION

06/07/2009

46

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 0.025 Ω respectively, and the corresponding leakage reactances are 2.5 Ω and 0.025 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

47

Example Problem

• The primary and secondary windings of a 400 kVA transformer have resistances of 0.3 Ω and 0.0015 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.

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