Phan Anh Bien AP Transformers

7/31/2019 Phan Anh Bien AP Transformers

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TRANSFORMERS

Transformers are key elements in power systems. In order to

effectively transmit power over long distances without prohibitive line

losses, the voltage from the generator (a maximum of output voltage ofapproximately 25-30 kV) must be increased to a significantly higher level

(from approximately 150 kV up to 750 kV). Transformers must also be

utilized on the distribution end of the line to step the voltage down (in

stages) to the voltage levels required by the consumer.

Transformers also have a very wide range of applications outside the

power area. Transformers are essential components in the design of DC

power supplies. They can provide DC isolation between two parts of a

circuit. Transformers can be used for impedance matching between sourcesand loads or sources and transmission lines. They can also be used to

physically insulate one circuit from another for safety.

Fundamentally, the transformer consists of two or more windings that

are magnetically coupled using a ferromagnetic core. For a two-winding

transformer, the winding connected to the AC supply is typically referred

to as theprimary while the winding connected to the load is referred to as

thesecondary. A time-varying current passing through the primary coil

produces a time-varying magnetic flux density within the core. According

to Faradays law, the time-changing flux passing through the secondary

induces a voltage in the secondary terminals.


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IDEAL TRANSFORMER

The basic two-winding (single-phase) transformer is shown below.

To simplify the initial analysis, the transformer will be assumed to be ideal.

The following assumptions are made in the analysis of an ideal transformer:

(1) The transformer windings are perfect conductors (zero winding

resistance).

(2) The core permeability is infinite (the reluctance of the core is

zero).

(3) All magnetic flux is confined to the transformer core (no

leakage flux).

(4) Core losses are assumed to be zero.

The figure above shows the common convention for the primary and

secondary voltage polarities and current directions. The voltage polaritiesand current directions shown above yield positive input power on the

primary and positive output power on the secondary. The actual polarity

relationship between the primary voltage and the secondary voltage is

dictated by the orientation of the primary and the secondary coils. The

transformer voltage relationship is obtained by applying Faradays law.


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Applying Faradays law to both the primary and the secondary (noting the

possibility of sources applied to either winding), yields

where the line integrals of the electric fields are along the primary and

secondary windings from the ! terminal to the + terminal, and thecorresponding surface integrals of the magnetic flux densities are over the

cross-sections of the primary and secondary coils. The directions of the

differential lengths and differential surfaces are related by the right-handrule. The total magnetic flux passing through the primary coil also passes

through the secondary coil, assuming an ideal transformer (all of the flux

is confined to the transformer core).

Note that the orientations of the given coils yield differential surface

vectors that point in the same direction around the magnetic circuit, such

that


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Dividing the equations forv1 and v2 gives

where the ratio of the primary turns to the secondary turns (defined as theturns ratio a) is equal to the ratio of the primary and secondary voltages.

According to the turns ratio equation, a transformer with more secondary

turns than primary turns yields a secondary voltage that is larger than the

primary voltage (step-up transformer) while a transformer with fewer

secondary turns than primary turns yields a secondary voltage that is

smaller than the primary voltage (step-down transformer).

If the orientation of one of the transformer coils is reversed, then the

differential surface vectors for the primary and secondary would be in

opposite directions yielding

If we apply Amperes law around the transformer core (clockwise, on

the centerline of the core), we find

Note that withL clockwise, the normal to the surface Sis inward, so that

currents inward are positive and currents outward are negative. The total

enclosed current is then


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Assuming an ideal transformer core (:r= 4), the magnetic field inside thecore is zero (similar to the fact that the electric field is zero inside a perfectconductor with F = 4, but carries a current on the surface of the conductor).Thus,

The conservation of power relationship for the ideal transformer (power in

equals power out, given no losses) can be stated by multiplying the voltage

ratio by the current ratio:

Assuming sinusoidal excitation, the ideal transformer can be

analyzed using phasor techniques. The ratios of the phasor voltages and

phasor currents satisfy the same turns ratio relationships as the time-domain

values.


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Note that the complex power relationship is also valid for the ideal

transformer.


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Polarity of Transformer Windings

The operation of the transformer depends on the relative orientation

of the primary and secondary coils. We mark one of the terminals on the

primary and secondary coils with a dot to denote that currents enteringthese two terminals produce magnetic flux in the same direction within the

transformer core.

The equivalent circuit diagram and phasor equations for this idealtransformer are

If either coil orientation is reversed, the dot positions are reversed and the

current and voltage equations must include a minus sign.


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Input Impedance (Ideal Transformer)

Consider an arbitrary load (Z2) connected to the secondary terminals

of the ideal transformer as shown below.

The input

impedance seen looking into the primary winding is given by

Thus, the input impedance seen looking into the primary of the ideal

transformer is the load impedance times a2. Using this property, the

secondary impedance of the ideal transformer can be reflected to the

primary.


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In a similar fashion, a load on the primary side of the ideal transformer can

be reflected to the secondary.


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Example (Ideal transformer)

Determine the primary and secondary currents for the ideal

transformer below ifZs = (18!j4) S and Z2 = (2+j1) S .

The load impedance reflected to the primary of the transformer is

The primary current is then

The primary voltage is

The secondary voltage is

The secondary current is


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TRANSFORMERRATING

Transformers carry ratings related to the primary and secondary

windings. The ratings refer to the power in kVA and primary/secondary

voltages. A rating of 10 kVA, 1100/110 V means that the primary is rated

for 1100 V while the secondary is rated for 110 V (a =10). The kVA ratinggives the power information. With a kVA rating of 10 kVA and a voltage

rating of 1100 V, the rated current for the primary is 10,000/1100 = 9.09 A

while the secondary rated current is 10,000/110 = 90.9 A.

NON-IDEAL TRANSFORMEREQUIVALENT CIRCUITS

The non-ideal transformer equivalent circuit below accounts for all

of the loss terms that are neglected in the ideal transformer model. The

individual loss terms in the equivalent circuit are:

Rw1,Rw2 - primary and secondary winding resistances

(losses in the windings due to the resistance of the wires)

Xl1,Xl2 - primary and secondary leakage reactances

(losses due to flux leakage out of the transformer core)

Rc1 - core resistance(core losses due to hysteresis loss and eddy current loss)

Xm1 - magnetizing reactance

(magnetizing current necessary to establish magnetic flux

in the transformer core)


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Using the impedance reflection technique, all the quantities on the

secondary side of the transformer can be reflected back to the primary side

of the circuit. The resulting equivalent circuit is shown below. The primed

quantities represent those values that equal the original secondary quantity

multiplied by a (voltages), divided by a (currents) or multiplied by a2

(impedance components).

Approximate Transformer Equivalent Circuits

Given that the voltage drops across the primary winding resistance

and the primary leakage reactance are typically quite small, the shuntbranch of the core loss resistance and the magnetizing reactance (excitation

branch) can be shifted to the primary input terminal. The primary voltage

is then applied directly across the this shunt impedance and allows for the

winding resistances and leakage reactances to be combined.


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A further approximation to the transformer equivalent circuit can be

made by eliminating the excitation branch. This approximation removes

the core losses and the magnetizing current from the transformer model.

The resulting equivalent circuit is shown below.

Note that this equivalent circuit

is referred to the primary side of

the transformer (V1 and VN2).This circuit can easily be

modified so that it is referred to

the secondary side of the

transformer (VN1 and V2).


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DETERMINATION OF EQUIVALENT CIRCUIT PARAMETERS

In order to utilize the complete transformer equivalent circuit, the

values ofRw1,Rw2,Xl1,Xl2,Rc1,Xm1 and a must be known. These values canbe computed given the complete design data for the transformer including

dimensions and material properties. The equivalent circuit parameters can

also be determined by performing two simple test measurements. These

measurements are the no-load(or open-circuit) testand theshort-circuit

test.

No-Load Test - The rated voltage at rated frequency is applied to

the high-voltage (HV) or low-voltage (LV) windingwith the opposite winding open-circuited.

Measurements of current, voltage and real power

are made on the input winding (most often the LV

winding, for convenience).

Short-Circuit Test - Either the LV or HV winding is short-circuited and

a voltage at rated frequency is applied to the

opposite winding such that the rated current results.

Measurements of current, voltage and real power

are made on the input winding (most often the HV

winding, for convenience, since a relatively low

voltage is necessary to obtain rated current under

short-circuit conditions).


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Example (Equivalent circuit parameters / no-load / short-circuit tests)

The approximate equivalent circuit parameters for a single-phase

10kVA, 2200/220, 60 Hz transformer are required.

The rated currents and voltages for the transformer windings are:

VH,rated= 2200 V IH,rated= 10000/2200 = 4.55 A

VL,rated= 220 V IL,rated= 10000/220 = 45.5 A

No-load and short-circuit tests are performed on the transformer with the

following results:

No-load test (HV winding open, VL = VL,rated= 220 V)

IL = 2.5 A,PL = 100 W

Short-circuit test (LV winding shorted,IH=IH,rated= 4.55 A)

VH= 150 V,PH= 215 W

(a.) Determine the approximate equivalent circuit parameters from the test

data (use the approximate equivalent circuit that includes core losses).

Draw the equivalent circuit for this transformer referred to the LV side.

(b.) Draw the equivalent circuit for this transformer referred to the HV

side. (c.) From the no-load test results, express the excitation current as a

percentage of the rated current in the LV winding. (d.) Determine the

power factor for the no-load and short-circuit tests.

Referred to LV side Referred to HV side


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Equivalent circuit for no-load test (determineRcL andXmL)


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Equivalent circuit for short-circuit test (determineReqHandXeqH)

The values measured on the HV winding (primary) in the short-circuit

test need to be referred to the LV side. Note that our turns ratio is given by


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(a.)

Referred to LV side

(b.) To obtain the same equivalent circuit referred to the HV side(primary), we simply multiply all impedances by a2 (100). The

resulting equivalent circuit is

Referred to HV side

(c.) From the no-load test results, the total excitation current is 2.5 A

while the rated current in the LV winding is 45.5 A. Thus the

excitation current is (2.5/45.5) or 5.5% of the rated current.


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TRANSFORMERVOLTAGE REGULATION

For a given input (primary) voltage, the output (secondary) voltage

of an ideal transformer is independent of the load attached to the secondary.

As seen in the transformer equivalent circuit, the output voltage of arealistic transformer depends on the load current. Assuming that the

current through the excitation branch of the transformer equivalent circuit

is small in comparison to the current that flows through the winding loss

and leakage reactance components, the transformer approximate equivalent

circuit referred to the primary is shown below. Note that the load on the

secondary (Z2) and the resulting load current (I2) have been reflected to the

primary (ZN2, IN2).

The percentage voltage regulation (VR) is defined as the percentage change

in the magnitude of the secondary voltage as the load current changes from

the no-load to the loaded condition.

The transformer equivalent circuit above gives only the reflected secondary

voltage. The actual loaded and no-load secondary voltages are equal to the

loaded and no-loaded refelcted secondary values divided by the turns ratio.


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Thus, the percentage voltage regulation may be written in terms of the

reflected secondary voltages.

According to the approximate transformer equivalent circuit, the reflected

secondary voltage under no-load conditions is equal to the primary voltage,

so that

The secondary voltage for the loaded condition is taken as the rated

voltage.

Inserting the previous two equations into the percentage voltage regulation

equation gives

Note that this equation is defined in terms of the voltages given in the

transformer approximate equivalent circuit. Also note that the rated

secondary voltage reflected to the primary is the rated primary voltage.

To determine the percentage voltage regulation, we may use the reflected

secondary voltage as the voltage reference,

and determine the corresponding value of *V1* from the approximateequivalent circuit.


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The voltagesV1 andVN2 in the approximate equivalent circuit are related by

where

The reflected secondary current can be written as

The expression forV1 becomes

We can draw the phasor diagram relating the voltages V1 and VN2 to

determine how the phase angles of the load and the transformer impedanceaffect the percentage voltage regulation. Note that the percentage voltage

regulation can be positive or negative and the sign ofVR is affected by the

phase angle in the expression above. Thus, the power factor of the load

will affect the voltage regulation of the transformer.


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The percentage voltage regulation is positive if*V1* > V1,ratedand negativeif*V1* < V1,rated. Note that with the limits on the angles of

the worst case scenario for the percentage voltage regulation occurs when

or when the load has a lagging power factor with the power factor angle

equal to the transformer impedance angle ofZeq1.


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Example (Transformer voltage regulation)

Using the transformer for which the approximate equivalent circuits

were found based on no-load and short-circuit test results, determine the

percentage voltage regulation for (a.) a load drawing 75% of rated currentat a power factor of 0.6 lagging (b.) a load drawing 75% of rated current at

a power factor of 0.6 leading.

The approximate equivalent circuit for the transformer (10kVA,

2200/220, 60 Hz) referred to the high voltage winding was found to

be (neglecting the excitation branch of the model)

Assume:Primary = HV winding

Secondary = LV winding

a =N1/N2 = 10

VH,rated= 2200 V

IH,rated= 10000/2200 = 4.55 A

VL,rated= 220 V

IL,rated= 10000/220 = 45.5 A

(a.) The reflected voltage VLN on the HV side is given by

so that the circuit to be analyzed becomes


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Based on the information provided in the problem statement, the

magnitude of the load current (and thus the current ILN) is 0.75 timesthat of the rated value.

The phase angle ofILN is given by the load power factor.

so that the phasor current ILN is

The voltage VHand VLN are related by

The percentage voltage regulation is thus

(b.) For the leading power of 0.6, 2i = +53.13o so that the phasor currentILN is

and VH is given by


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The voltage regulation is

The percentage voltage regulation results for these two cases shows

that if this transformer is providing 75% of rated current (3.41 A-rms) to

a load with a power factor of 0.6 lagging, and the load is suddenly

removed, the load voltage magnitude rises from 220 V to 230.70V. For the

load with a power factor of 0.6 leading, the load voltage magnitude drops

from 220 V to 213.79 V.


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TRANSFORMEREFFICIENCY

The efficiency (0) of a transformer is defined as the ratio of the outputpower (Pout) to the input power (Pin). The output power is equal to the input

power minus the losses (Ploss) in the transformer. The transformer losspower has two components: core loss (Pcore) and so-called copper loss (Pcu)

associated with the winding resistances. The transformer efficiency in

percent is given by

Assuming a relatively constant voltage source on the primary of the

transformer, the core loss can be assumed to be constant and equal to power

dissipated in the core loss resistance (Rc1

) of the equivalent circuit for the

no-load test. The copper loss in a transformer may be written in terms of

both the primary and secondary currents, or in terms of only one of these

currents based on the relationshipI2 = aI1.

The output power of the transformer can be written in terms of the

secondary voltage and current (real part of output complex power).


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The transformer efficiency, written in terms of secondary values, is

It can be shown that the maximum transformer efficiency occurs when the

core losses equal the copper losses and the power factor is unity.

Example (Transformer efficiency)

Using the transformer for which the approximate equivalent circuits

were found based on no-load and short-circuit test results, determine (a.)

the transformer efficiency at 75% of rated output power with a power factor

of 0.6 lagging (b.) the output power at maximum efficiency, the value of

maximum efficiency, and at the percentage of full load power where

maximum efficiency occurs.

(a.) The rated power for this transformer is 10 kW at a power factor of

unity. IfPout is 75% of the rated value at a power factor of 0.6, then

From the no-load test, the core losses werePcore = 100 W. The copper

losses for this transformer are

The transformer efficiency is


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(b.) At maximum efficiency YPCu=IL2ReqL =Pcore = 100 W,PF= 1

The maximum efficiency is

The maximum efficiency occurs at

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Phan Anh Bien AP Transformers