transformers and power flow.pptx

7/30/2019 Transformers and Power Flow.pptx

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A transformer is an electrical device that isused to raise or lower the voltage (or current)level.

A single-phase transformer has two electricallyisolated windings.

The winding that is connected to the electricalpower source is called the primary winding.

The winding that is used to draw electricalpower is called the secondary winding.

Syed A Rizvi ENS 441/ELT 437 2


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Figure below shows a schematic diagram of anideal single-phase transformer.


EP

= induced voltage on the primary

side (RMS)

ES = induced voltage on the secondary

side (RMS)

NP = Number of turns on the primary

windingsNS = Number of turns on the secondary

windings

m = magnetic flux created by the

source voltage E

m


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When the primary winding is connected to an ACpower source it produces an instantaneousmagnetic flux m(t) linking the primary as well

as secondary windings of the transformer. The flux linkage induces an instantaneous voltage

Ep(t) in the primary winding in accordance withthe Faradays law of electromagnetic induction.

That is,

() = ()



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Since the voltage applied at the primary issinusoidal, therefore, the magnetic flux m(t) isalso sinusoidal and mathematically it can be

expressed as = sin .

Therefore,

() = ()

= (sin )

or() = cos .



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Substituting cos = sin( + 90) and = 2 we get

= 2 sin( + 90) .

Above Eq. shows that the induced voltage Ep(t) leads the

magnetic flux m(t) by 90o. The RMS value of Ep(t),denoted by Ep is given by

= =

2=

2

2

or

= 4.44 . (1)



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Since the magnetic flux m

(t) also links the secondarywinding that has Ns turns, the RMS value of the inducedvoltage at the secondary side of the transformer, denotedby Es is given by

= 4.44 . (2)

Comparing Eqs. (1) and (2) yields

4.44 =

4.44

or

=



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or

=

=

or =

where, a is called the turn ratio of thetransformer.



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Figure below shows an ideal transformer with aload on the secondary side.



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Under ideal conditions no current flows through theprimary winding and the primary current IP is zero whensecondary side is open.

When a load is connected to the secondary side, a current,

Is,flows out of the secondary windings.

The secondary current, Is,creates a magnetic flux s in thetransformers core.

Since the source voltage, E, is unchanged, which mean the

flux, m, must remain the same as it was before the load wasconnected.

Therefore, the new flux (s) due to the secondary thecurrent, Is, must be countered by an equal and opposite flux.



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The counter flux p is produced by causing acurrent, Ip, toflow into the primary windings.

Under ideal conditions, the fluxes s and p canceleach other and the total flux through the core

remains m. Result: Primary and secondary induced voltages

(Ep and Es) remain unchanged. The flow of the primary current, Ip, represents the

power transfer from primary side to the secondaryside of the transformer to support the load Zs.

Under ideal conditions there is no loss of powerduring the power transfer from the primary side tothe secondary side of the transformer.



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Therefore, the total power, Sp, on the primary side of thetransformer must be equal to the total power, Ss, on thesecondary side of the transformer, where

= and =

but Sp = Ss, therefore,

= .

But = , therefore,

= or

=1

.



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The load impedance as seen by the primary side, denoted byZp, can be given by

=

=

=

but

=

.

Therefore,

=

.

Above Eq. shows an important property of the transformer;that is, it can be used to change the effective impedance of anyload and thus can also act as an impedance transformer.



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Using impedance transformation, the impedances on thesecondary side can be expressed as the equivalentimpedances on the primary side to greatly simplify thecircuit analysis involving transformers.



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The ideal transformer is lossless.

In a real transformer, however, the primarycurrent is not zero under no-load condition.

Reason: Iron losses and magnetizing reactance

Causes of iron loss:

Eddy currents

Hysteresis losses



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Eddy Current Losses:

When time-varying magnetic flux passes through asolid metallic core, the solid metallic core acts likea bundle of concentric conductors and a voltage isinduced in each of these sections of the metalliccore.

In a solid metal plate each of these sections wouldbehave like a short-circuited conductor and,

therefore, the induce voltages would cause astrong currents which swirl back and forth in thecore.

These currents are called eddy currents.



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Eddy currents are a major source of generating heat intransformers. In a solid metal core these currents can be sostrong that they render the core red hot.

Solution: Design the core withthin laminated strips to insulatesthe strips from each other.

Result: The length of the

conducting path is greatlyreduced.

Magnitudes of the inducedvoltages in the core therebydiminishing eddy current losses.


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Hysteresis LossesThe magnetic flux is caused by magnetic field strength, H. Theresulting flux density, B, in any material is related to themagnetic field strength (H) by the following equation:

=

where,B = magnetic flux density (weber/m2 or tesla)H = magnetic field strength (A-turns/m)o = permeability of free space = 4 10

r

= relative permeability of the material

The value of r changes with flux and, therefore, therelationship between B and H is expressed in terms of a B-Hcurve.



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Figure below shows the closed B-H curve calledhysteresis loop.


The area under this curve

represents the energy thatchanges into heat duringeach cycle (joules per cubicmeters).

Iron losses are representedas resistance (Rm) inparallel to the primaryterminals of thetransformer.


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Magnetizing Inductance: In order to generate themagnetic flux m, a current must flow through the primarywindings.

This current is called magnetizing current and depends on

the permeability of the material used to build the core. Higher permeability means lower magnetic current and vice

versa.

Permeability can be represented by a parallel inductance

(Xm) to the primary terminal because magnetizing currentdoes not affect the amount of induced voltage.

This inductance is called magnetizing inductance.



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Winding Resistance:

In a real transformer, the winding of thetransformer have finite resistance that must be

taken into account while analyzing ormodeling a practical transformer.

Since these winding resistances result in apotential drop when current passes through

the windings, they are represented as the seriesresistors (Rp and RS), both at the primary aswell as at the secondary side of thetransformer.



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Leakage Flux: In a real transformer some of s and p does leak

in the air. This flux leakage induces voltages both in the

primary winding and secondary windings, whichis counter to the voltages induced by the flux m. It results in a voltage drop both at the primary as

well as the secondary side. Accordingly, the effect of the leakage flux is

represented as the series impedances (Xlp and Xls)at the primary as well as the secondary side of thetransformer in the model of a practicaltransformer.



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Figure below shows the complete model of practicaltransformer that incorporate the iron losses, magnetizingcurrent, primary and secondary winding resistances, andthe effect of the leakage flux under load conditions.



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The model of the practical transformer is easier toanalyze if the elements on the secondary side areshifted to the primary side as shown in Fig. below.



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No-load Condition:

Under no-load condition the secondary is open and,therefore, no current flows through the secondary

terminals.



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Full-load Condition:

Under the full-load condition and, therefore,

we can ignore Rm and Xm and the model for the

practical transformer reduces to the one shown below.



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The model of the practical transformer can furtherbe simplified by adding the resistances andinductances as follows:

=

+

and = +

.

Furthermore, Rlp and Xlp can be represented by

the complex impedance Zp as follow:

= +



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Figure below shows the simplified circuit for thepractical transformer with full load.



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Example: Specifications of a power transformer are asfollows:

Compute the full-load Vs, Is, as well as heat dissipation inthe transformer using the simplified model. Also,

compute Iron losses using the model under no-load

condition .


Sn (KVA) Enp (V) Ens (V) Inp (A) Ins (A) Rp ()1000 69000 6900 14.5 145 27.2Rs() Xlp () Xls () Xm() Rm() Io (A)0.25 151 1.5 505000 432000 0.210


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Solution: Lets assume its a pure resistive load and full-loadsecondary voltage is the same as Vns and full-load secondarycurrent is the same as Ins. Now we can calculate the loadresistance, RL, as follows:

=

=

. = 47.6

The turn ratio is given by

=

=

69000

6900= 10

Now we can compute Rep and Xep

= + = 27.2 + 10

0.25 = 52.2

= + = 151 + 10

1.5 = 301



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The resistance shifted to primary, RLp, is given by

= = 10

47.6 = 4760 .

Now we can compute the total impedance of the model,Z

Tp, as follows

= + + = 52.2 + 4760 + 301

= 4812.2 + 301 = 4821.6 0.0186

Now we can compute primary current, Ip, primary

voltage, Ep, secondary current, Is, and secondary voltage,Es as follows

=

=

69000

4821.6= 14.31



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= = 14.31 4821.6 = 68118.4

=

=

68118.4

10= 6811.84

=

=

6811.84

47.6 = 143.1

Now we can compute the heat loss due to the load as follows

= = 14.31

52.2 = 10689.3 = 10.69

Finally, well compute iron losses at no load and at full load asfollows

=

=

69000

432000= 11020.83 = 11.02



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=

=

68118.4

432000= 10741 = 10.74

Total power loss at full load = 10.69 + 10.74 =21.43 KW.

This power loss produces significant amount of

heat that must be removed from the transformerto avoid any damage to transformer.



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There are a number of techniques that are usedin the transformers to remove heat.

Through natural circulation of air (type AA),

Through forced air circulation (type AFA), Oil-immersed self-cooled (OA),

Oil immersed self-cooled/forced air cooled

(type OA/FA) to name a few.



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Figure below show the front view of an OA/FAtype power transformer.



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Figure next shows the rear view ofthe same transformer wherecooling fans can be seen whichremove heat from the transformerthrough forced air circulation.

A transformer is called a step-down transformer if thesecondary voltage, ES, is lower

than the primary voltage, EP. Onthe hand, if the secondary voltageis higher than the primary voltage,it is called a step-uptransformer.


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Terminal Marking on Power Transformers:

In power transformers, the high voltage (HV)terminals are marked by symbols H1 and H2.

The low voltage (LV) terminal are marked bysymbols X1 and X2.

These terminals are mounted on the

transformer tank in a way that they either haveadditive polarity or they have subtractivepolarity.



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Figure below shows the terminal marking andpolarity of the power transformers.

It is imperative to know the correct polarity of

the transformers if you are connecting them in


parallel orconfiguring a 3-phase

transformerbank usingsingle phasetransformers.


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Voltage Regulation:

Voltage regulation is measure of variation inthe secondary voltage from no-load to full-load

condition when the primary voltage is keptconstant.

Voltage regulation is expressed in percent (%)

and is defined by the following equation

= () ()

() 100



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In AC systems the power is divided into threedifferent categories:

Apparent Power:

It is the product of the voltage and complexconjugate of the current and is expressed in volt-amperes or VA.

If the current I = IR + j IX then the conjugate of I,denoted by I*, is given by I*= IR j IX. The apparentpower, PA, is given by

=



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Real Power:

The real power (PR) is the power utilized by auser and is expressed in watts (W).

When the apparent power is expressed inrectangular form, the real part of the apparentpower is the real power.

The real power always flows from the source tothe user.



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Reactive Power:

The reactive power (PX)is needed to establishmagnetic field in inductors.

Electric field in a capacitor acts as a source ofreactive power.

Reactive power can be supplied as well asabsorbed by an electric power source. It meansthat the reactive power can flow from thesource to the circuit or from the circuit to thesource.



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When the reactive power flows from the sourceto the circuit it is considered positive reactivepower.

However, it is considered negative reactivepower when it flows from the circuit to thesource.

Mathematically, the apparent power(PA), the

real power (PR), and the reactive power (PX) arerelated as follow:

=

transformers and power flow.pptx

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