Transformers 02

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TRANSFORMERS Non-Ideal

Lecture By:

Dr. Walid A. M. Ghoneim,Associate Professor in Electrical Engineering

Electrical Engineering and Control Department

College of Engineering and Technology

Arab Academy for Science, Technology and Maritime Transport


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6 A Non-ideal Transformer:

To develop an equivalent circuit for a non-ideal transformer.

Every Non-Ideal (Practical) device contains an Ideal one.

Winding Resistances:

Each winding has some resistance.

Represented as a lumped R in series with each winding.

As shown in Figure, R1 and R2 are the winding resistances ofthe primary and the secondary, respectively.


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6 A Nonideal Transformer:

The inclusion of the winding resistances dictates that:

(a) The power input must be greater than the power output:

S1 > SL and P1 > PL

(b) The terminal voltage V is not equal to the induced emf E:

V1 > E1 and E2 > V2

(c) The efficiency of a non-ideal transformer is less than 100%:

Eff. = PL / P1 %


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Leakage Fluxes:

Not all of the flux created by a winding confines itself to themagnetic core on which the winding is wound.

Part of the flux, known as the leakage flux, does complete itspath through air.

Therefore, when both windings in a transformer carry currents,each creates its own leakage flux as illustrated in Figure.


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Leakage Fluxes:

The primary leakage flux set up by the primary does notlink the secondary: l1

Likewise, the secondary leakage flux restricts itself to thesecondary and does not link the primary: l2

The common flux that circulates in the core and links bothwindings is termed the mutual flux: m

1 = l1 + m also 2 = l2 + m

Although a leakage flux is a small fraction of the total flux, itdoes affect the performance of a transformer.

1 >> l1 also 2 >> l2


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We can model a winding as if it consists of two windings:

One winding is responsible to create the leakage flux through air.

The other encircles the core.

Such hypothetical winding arrangements are shown in thefigure for a two-winding transformer.

The two windings enveloping the core now satisfy theconditions of an idealized transformer.


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The leakage flux associated with either winding causes a voltagedrop across it, therefore, we can represent the voltage drop dueto the leakage flux by a leakage reactance.

If X1 and X2, are the leakage reactances of the primary andsecondary windings, a real transformer can then be represented

in terms of an idealized transformerwith windingresistances and leakage reactances as shown in Figure.

This is an EQUIVALENT CIRCUIT .


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The EQUIVALENT CIRCUIT Solution:

Usually starts from the Load

Then the secondary Circuit.

Then the Ideal Transformer

Finally the Primary Circuit


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Finite Permeability:

The core of a non-ideal transformer has finite permeability andcore loss ( Inf. AndIR 0 )

Therefore, even when the secondary is left open (no-loadcondition) the primary winding draws some current, known asthe excitation current I, from the source:

I2 = 0but I1 0, I1 = I

The excitation current, I, is the sum of two currents: the core-loss current Ic and the magnetizing current Im. I = Ic + Im


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How to represent Finite Permeability:

The core-loss component of the excitation current accounts forthe magnetic loss (the hysteresis loss and the eddy-currentloss) in the core of a transformer.

If E1 is the induced emf on the primary side and Rc is the

equivalent core-loss resistance, then the core-loss current, Ic is: Ic = E1 / Rc

The magnetizing component of the excitation current Im is

responsible to set up the mutual flux in the core. Since a current-carrying coil forms an inductor, then it can be

represented by a magnetizing reactance Xm. Thus:

Im = E1 / jXm


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The EXACT EQUIVALENT CIRCUIT:

We can now modify the equivalent circuit to include the core

loss resistance and the magnetizing reactance.

Such a circuit is shown in Figure:


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The EXACT EQUIVALENT CIRCUIT Referred to Primary:

Voltage Referring:

E2 = a . E2 = E1

Current Referring:

I2 = I2 / a = Ip

Impedance Referring:

Let ZLbe the load impedance on the secondary side, then,

ZL = V2 / I2 = (V1/a) / (a.I1)

= (V1/I1) / a2

= ZL / a2 (9)

Thus, ZL = a2 . ZL (10)

where ZL = V1 / I1 is the load impedance as referred to the

primary side.


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The EXACT EQUIVALENT CIRCUIT Referred to Primary:

Similarly, the values of the secondary circuit elements referredto primary circuit are:

R2 = a2 . R2 and X2 = a2 . X2


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The EXACT EQUIVALENT CIRCUIT Referred to Sec.:


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The Phasor Diagram

The load voltage is takenas a reference because it isknown.


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The APPROXIMATE EQUIVALENT CIRCUIT Referred toPrimary:

In a well-designed transformer, the core losses are low, whichimplies a high core loss resistance.

The high permeability of the core ensures high magnetizing

reactance.

So at Full-Load, Ip is very large compared to I, so we canmove the parallel branch representing the magnetic circuit tothe source side.


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From the Load Circuit:

I2 = SL / V2 or I2 = V2 / ZL

V2 = a.V2

I2 = I2/a

The Primary and Secondary winding impedances are now

SERIES, thus: Ze = Re + jXe

Re = R1 + R2 = R1 + a2 . R2

Xe = X1 + X2 = X1 + a2 . X2

V1 = V2 + Ip . Ze The Magnetic Circuit:

I1 = Ip + I I = Ic + Im

Ic = V1 / Rc Im = V1 / jXm

The overall circuit: V1 = V2 + Ip . Ze


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7 Voltage Regulation:

The primary voltage (V1) is adjusted so that it delivers the ratedload (I2) at the rated secondary terminal voltage (V2).

If the load is removed (I2 = 0), V2 changes because of thechange in the voltage drops across the winding resistances andleakage reactances.

The Voltage Regulation (VR) is the percentage change inthe secondary winding voltage from no load to full load for thesame primary winding voltage.

VR% = (V2N.L V2F.L) / V2F.L

The voltage regulation is the figure-of-merit of a transformer. The smaller the voltage regulation, the better the operation of

the transformer. In ideal transformer, VR = 0.

For approximate equivalent circuit referred to primary:

VR% = (V1 a.V2) / a.V2


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8 Determination of Trans. Parameters:

The equivalent circuit parameters (Rc, Xm, Re, Xe) of atransformer can be determined by performing two tests:

The open-circuit test.

The short-circuit test.


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The Open-Circuit Test:

The secondary winding of the transformer is left open while theprimary is connected to the source voltage.

The figure shows the circuit in this case:

We need to measure the open-circuit voltage (Vo.c), the open-circuit current (Io.c) and the active power (Po.c).

Thus, we have to use a voltmeter, an ammeter and a wattmeter.

The next figure shows the connection diagram.


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It is evident that the source supplies the excitation currentunder no load. I1 = I only , since Ip = 0.

One component of the excitation current is responsible for thecore loss (ACTIVE), whereas the other is responsible toestablish the required flux in the magnetic core (REACTIVE).

The Wattmeter measures Active Power (Po.c), which the coreloss component only.


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Calculations: The Open Circuit Apparent Power is: So.c = Vo.c * Io.c =V . I

The Open Circuit Active Power: Po.c =W= Vo.c *Io.c*cos(o.c)

Hence, the open circuit power factor can be calculated from:

cos (o.c) = Po.c / So.c =W / (V . I) The Real Component of I is Ic = I . cos(o.c)

The Imaginary Component of I is Im = I . sin(o.c)

The Core Loss Resistance Rc = Vo.c / Ic

The magnetization inductance Xm = Vo.c / Im


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The Short-Circuit Test:

The secondary winding of the transformer is shorted (ZL = 0)while the primary is connected to a small voltage, enough tooperate the transformer at full load (I1 = IF.L). WARNING !!!

Since the applied voltage is a small fraction of the rated voltage,

both the core-loss and the magnetizing currents are so smallthat they can be neglected.

The figure shows the approximated equivalent circuit of thetransformer in this case:


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We need to measure the primary short-circuit voltage (Vs.c),the short-circuit current (Is.c) and the active power (Ps.c).

Thus, we have to use a voltmeter, an ammeter and a wattmeter.

In this case, the wattmeter records the copper loss atfull load.

The next figure shows the connection diagram.


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Calculations: The Short Circuit Active Power:

Ps.c =W= I2s.c * Re = I2 * Re

Hence, the equivalent resistance can be calculated from:

Re = Ps.c / I2

s.c =W / I2

The total equivalent impedance is:

|Ze| = Vs.c / Is.c =V / I

But, |Ze|2 = Re2 + Xe2

Thus Xe = SQRT(|Ze|2

Re2

)

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