Topic 1 : Magnetic Concept and Transformer

Introduction

Two winding transformersConstruction and principlesEquivalent circuitDetermination of equivalent circuit parametersVoltage regulation EfficiencyAuto transformer3 phase transformer

Introduction

Different variety of transformers

Introduction

Introduction

The word “Transformer” means an electromagnetic device which transforms electrical power from one end to another at different voltages and different currents keeping frequency constant.

Unlike motor and generator it is static machine with different turns ratio of primary and secondary windings through which voltage/current is changed.

The transfer of energy takes place through the magnetic field and all currents and voltages are AC.

The rating of transformer is either in kVA or MVA because load to be connected is unknown

Introduction

Transformers are adapted to numerous engineering applications and may be classified in many ways:Power level (from fraction of a volt-ampere (VA) to over a

thousand MVA), Application (power supply, impedance matching, circuit

isolation), Frequency range (power, audio, radio frequency (RF)) Voltage class (a few volts to about 750 kilovolts) Cooling type (air cooled, oil filled, fan cooled, water cooled,

etc.)-ONAN, ONAF Purpose (distribution, rectifier, arc furnace, amplifier output,

etc.).

Introduction

Examples of transformer classifications:Power Three phase transformers (Step up) used for transmission

of power (3-phase) at a distanceDistribution transformer (Step down) used for utilization of power

3-pahse/1-phaseInstrument Transformer (VT & CT) used for

measurement/practicalAuto Transformer (Single limb, electrically connected) used for

measurement, practical, supply/utilizationIsolation Transformer (having winding ratio of 1:1) used for safety

of human and equipment for sensitive appliances or practical purpose

Introduction

The invention of transformer caused transmission of heavy AC electrical power possible thus plays important role in electrical power technology

Functions of transformer:Raise or lower voltage or current in AC circuitIsolate circuit from each otherEnable to transmit electrical power energy over large

distances at about >1200kVProvides electrical power according to the utilization

needs

Power transmission

Transformer- Introduction

Introduction

A typical power system consists of generation, transmission and distribution.

Power 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 losses.

Its distribution is made through step down transformer according to the consumer demand.

Here again at this stage, a transformer play an important role to reduce the voltage to suit the consumer need.

Introduction

Power Transmission

Introduction

Transformer is a device that makes use of the magnetically coupled coils to transfer energy.

It is typically consists of one primary winding coil and one or more secondary windings.

The primary winding and its circuit is called the Primary Side of the transformer.

The secondary winding and its circuit is called the Secondary Side of the transformer.

A magnetic circuit provides the link between primary and secondary.

Introduction

When an AC voltage Vp is applied to the primary winding of the transformer, an AC current Ip will result.

Ip sets up a time-varying magnetic flux Ф in the core.

A voltage Vs is induced in the secondary circuit according to the Faraday’s law.

Construction

The magnetic (iron) core is made of thin laminated steel sheet. to minimize the eddy current loss by

reducing thickness.

There are two common cross section of coresquare or (rectangular) for small

transformerscircular (stepped) for the large and 3

phase transformers.

Construction

Core (U/I) Type: Constructed from a stack of U and I shaped laminations.The primary and secondary windings are wound on two

different legs of the core.

Shell Type: Constructed from a stack of E and I shaped laminations.The primary and secondary windings are wound on the

same leg of the core, as concentric windings, one on top of the other.

Construction

Construction

Construction

Construction

Ideal Transformer

Winding resistances are zero, no leakage inductance and iron loss

Magnetization current generates a flux that induces voltage in both windings

Current, voltages and flux in an unloaded ideal transformer

Ideal Transformer

Currents and fluxes in a loaded ideal transformer

Ideal Transformer

Turn ratioIf the primary winding has N1 turns and secondary

winding has N2 turns, then:

The input and output complex powers are equal

Ideal Transformer

Functional description of a transformer:a = 1

Isolation Transformer

| a | < 1Step-Up TransformerVoltage is increased from Primary side to secondary side

| a | > 1Step-Down TransformerVoltage is decreased from Primary side to secondary side

Ideal Transformer

Transformer RatingPractical transformers are usually rated

based on:Voltage Ratio (V1/V2) which gives us

the turns-ratioPower Rating, small transformers are

given in Watts (real power) and Larger ones (Power Transformers) are given in kVA (apparent power)

Ideal Transformer

Example 1Determine the turns-ratio of a 5 kVA 2400V/120V Power Transformer

Turns-Ratio = a = V1/V2 = 2400/120 = 20/1 = 20This means it is a Step-Down transformer

Ideal Transformer

Example 2A 480/2400 V (r.m.s) step-up ideal transformer delivers 50 kW to a resistive load. Calculate:

the turns ratio,

(0.2)the primary current,

(104.17A)the secondary current.

(20.83A)

Ideal Transformer

Exercise 1A 250kVA, 1100V/400v, 50Hz single-phase transformer has 80 turns on the secondary. Calculate:

the approximate values of the primary and secondary currents

(227A, 625A)

the approximate number of primary turns(220)

the maximum value of the flux(22.5mWb)

Ideal Transformer

Nameplate of a transformer

Ideal Transformer

Equivalent circuit of an ideal transformer

Ideal Transformer

Equivalent circuit of an ideal transformerTransferring impedances through a transformer

Ideal Transformer

Equivalent circuit when secondary impedance is transferred to primary side and ideal transformer eliminated.

Equivalent circuit when primary source is transferred to secondary side and ideal transformer eliminated.

Practical Transformer

Equivalent Circuit

In a practical magnetic core having finite permeability, a magnetizing current Im is required to establish a flux in the core. This effect can be represented by a magnetizing

inductance Lm.

The core loss can be represented by a resistance Rc.

Equivalent Circuit

Rc :core loss component

Xm : magnetization component

Equivalent Circuit

Phasor diagram of an unloaded transformer

Equivalent Circuit

Winding resistance and leakage flux

The effects of winding resistance and leakage flux are respectively accounted for by resistance R and leakage reactance X (2πfL).

Equivalent Circuit

Rc :core loss component

Xm : magnetization component

R1 and R2 are resistance of the primary and secondary winding

X1 and X2 are reactance of the primary and secondary winding

Equivalent Circuit

Phasor diagram of a loaded transformer (secondary)

Equivalent Circuit

Phasor diagram of a loaded transformer (primary)

Approximate Equivalent Circuit

Since no load current is very small(3-5% of full load), the parallel circuit of Rc and Xm can be moved close to the supply without significant error in calculation.Calculations becomes easier

Approximate Equivalent Circuit

Calculations will be much more easy if the primary and secondary circuit are combined.Transfer the secondary circuit to the primary circuit

Approximate Equivalent Circuit

Phasor diagram of a loaded transformer (primary)

Approximate Equivalent Circuit

For convenience, the turns is usually not shown

The resistance and reactance can be lumped together

We can also transfer the primary circuit to the secondary circuit

Approximate Equivalent Circuit

Example 3 A 100kVA transformer has 400 turns on the primary and 80

turns on the secondary. The primary and secondary resistance are 0.3 ohm and 0.01 ohm respectively and the corresponding leakage reactances are 1.1 ohm and 0.035 ohm respectively. The supply voltage is 2200V. Calculate:

the equivalent impedance referred to the primary circuit

(2.05 ohm) the equivalent impedance referred to the secondary circuit

(0.082)

Transformer Test

The equivalent circuit model for the actual transformer can be used to predict the behavior of the transformer.

The parameters Rc, Xm, R1, X1, R2, X2 and N1/N2 must be known so that the equivalent circuit model can be used.

These parameters can be directly and more easily determined by performing tests:No load test (or open circuit test)Short circuit test

Transformer Test

No load/Open circuit testProvides magnetizing reactance (Xm) and core loss

resistance (Rc)

Obtain components are connected in parallel

Short circuit testProvides combined leakage reactance and winding

resistanceObtain components are connected in series

Transformer Test – Open Circuit

Equivalent circuit for open circuit test, measurement at the primary side

Simplified equivalent circuit

Transformer Test – Open Circuit

Open circuit test evaluation

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Transformer Test – Short Circuit

Short circuit testSecondary (normally the LV winding) is shorted, that

means there is no voltage across secondary terminals; but a large current flows in the secondary.

Test is done at reduced voltage (about 5% of rated voltage, with full-load current in the secondary. Hence the induced flux are also 5%) The core losses is

negligible since it is approximately proportional to the square of the flux.

So, the ammeter reads the full-load current; the wattmeter reads the winding losses, and the voltmeter reads the applied primary voltage.

Transformer Test – Short Circuit

Equivalent circuit for short circuit test, measurement at the primary side

Simplified equivalent circuit

Transformer Test – Short Circuit

Simplified circuit for calculation of series impedance

Primary and secondary impedances are combined

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Transformer Test – Short Circuit

Short circuit test evaluation

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Transformer Test

Equivalent circuit for a real transformer resulting from the open and short circuit tests.

Transformer Test

Example 4Obtain the equivalent circuit of a 200/400V, 50Hz 1-phase transformer from the following test data:-

O/C test : 200V, 0.7A, 70W - on L.V. side(LV data)S/C test : 15V, 10A, 85W - on H.V. side(HV data)

(Rc =571.4 ohm, Xm=330 ohm, Re=0.21ohm, Xe=0.31 ohm)

Voltage Regulation

Most loads connected to the secondary of a transformer are designed to operate at essentially constant voltage. However, as the current is drawn through the transformer, the load terminal voltage changes because of voltage drop in the internal impedance.

To reduce the magnitude of the voltage change, the transformer should be designed for a low value of the internal impedance Zeq

The voltage regulation is defined as the change in magnitude of the secondary voltage as the load current changes from the no-load to the loaded condition.

Voltage Regulation

The voltage regulation is expressed as follows:

V2NL= secondary voltage (no-load condition)

V2L = secondary voltage (full-load condition)

Voltage Regulation

For the equivalent circuit referred to the primary:

V1 = no-load voltage

V2’ = secondary voltage referred to the primary (full-load condition)

Voltage Regulation

Consider the equivalent circuit referred to the secondary,

(-) : leading power factor (+) : lagging power factor

Voltage Regulation

Consider the equivalent circuit referred to the primary,

(-) : leading power factor (+) : lagging power factor

Voltage Regulation

Example 5Based on Example 3 calculate the voltage regulation and the secondary terminal voltage for full load having a power factor of

0.8 lagging

(0.0336pu,425V)0.8 leading

(-0.0154pu,447V)

Approximate Equivalent Circuit

Example 3 A 100kVA transformer has 400 turns on the primary and 80

turns on the secondary. The primary and secondary resistance are 0.3 ohm and 0.01 ohm respectively and the corresponding leakage reactances are 1.1 ohm and 0.035 ohm respectively. The supply voltage is 2200V. Calculate:

the equivalent impedance referred to the primary circuit

(2.05 ohm) the equivalent impedance referred to the secondary circuit

(0.082)

Voltage Regulation

Efficiency

Losses in a transformerCopper losses in primary and secondary

windingsCore losses due to hysteresis and eddy

current. It depends on maximum value of flux density, supply frequency and core dimension. It is assumed to be constant for all loads

Efficiency

Equipment is desired to operate at a high efficiency.Efficiency is defined as

Since it is a static device, losses in transformers are small

The losses in the transformer are the core loss (Pc) and copper loss (Pcu)

Efficiency

The copper loss can be determined if the winding currents and their resistances are known:

The copper loss is a function of the load current.

The core loss depends on the peak flux density in the core, which in turn depends on the voltage applied to the transformerSince a transformer remains connected to an essentially constant

voltage, the core loss is almost constant

Efficiency

If the parameters of the equivalent circuit of a transformer are known, the efficiency of the transformer under any operating condition may be determined

.

Normally, load voltage remains fixedTherefore efficiency depends on load current and load

power factor

Efficiency

Efficiency on full load

where S is the apparent power (in volt amperes)

Efficiency for any load equal to n x full load

where corresponding total loss =

Efficiency

Example 6The following results were obtained on a 50 kVA

transformer: open circuit test – primary voltage, 3300 V; secondary

voltage, 400 V; primary power, 430W.Short circuit test – primary voltage, 124V;primary current,

15.3 A; primary power, 525W; secondary current, full load value.

Calculate the efficiency at full load and half load for 0.7 power factor.

(97.3%, 96.9%)

Efficiency

Exercise 2The primary and secondary windings of a 500kVA transformer have resistance of 0.42 ohm and 0.0019 ohm respectively. The primary and secondary voltages are 11000V and 400V respectively and the core loss is 2.9kW, assuming the power factor of the load to be 0.8. Calculate the efficiency on

Full load half load

(98.3%, 98.1%)

Efficiency

For constant values of the terminal voltage V2 and load power factor angle θ2 , the maximum efficiency occurs when

If this condition is applied, the condition for maximum efficiency is

that is, core loss = copper loss.

Efficiency

Exercise 3Assuming the power factor of the load to be 0.8, find the output power at which the efficiency of the transformer of Exercise 2 is a maximum and calculate its value

(346.4kW, 98.4%)