Single Phase Transformer

Prepared By:

RITU K.R.

Assistant Professor (EE)

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Coil 1

i1(t) i2(t)

Coil 2

M

e1(t) e2(t)

S1 S2i1(t) i2(t)

The Transformer

(Primary has N1 turns) (Secondary has N2 turns)

V2

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Transformer Construction

Left: Windings shown only on one legRight: Note the thin laminations

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Transformer Construction

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• The source side is called Primary • The load side is called Secondary

• Ideally

1. The resistance of the coils are zero.

2. The relative permeability of the core in infinite.

3. Zero core or iron loss.

4. Zero leakage flux

The Transformer

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The Transformer

i) Switch ‘S1’ is closed and ‘S2’ is open at t=0

The core does not have a flux at t=0

The voltage induced across each coil is proportional to its number of turns.

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The Transformer

ii) Switch ‘S2’ is now closed

A current now starts to flow in resistance R. This currentis i2(t) (flows out of the dotted terminal).

R

)t(2V

R

)t(e)t(i 2

2

Thus a MMF N2i2(t) is applied to the magnetic circuit. This will immediately make a current i1(t) flow into the dot of the primaryside, so that N1i1(t) opposes N2i2(t) and the original flux in the core remains constant. Otherwise, N2i2(t) would make the core flux change drastically and the balance between V1 and e1(t) will be disturbed.

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The Transformer

Observation: It was shown that the flux in the core ism Sin(t). Since the permeability of the core is infinite ideallyzero current can produce this flux! In actuality, a current Im, knownas magnetizing current is required to setup the flux in the transformer. This current is within 5% of the full load current ina well designed transformer.

21

11

;1 N

LL

rmsVIm

L1 is the primary side self inductance.

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Polarity (dot) convention

Terminals of different windings are of same polarity if currentsentering (or leaving) them produce flux in the same direction in the core.

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How to check polarity?

1) Measure e12 and e34

2) Connect 2 and 4 and measure e13

3) If e13= e12+e34, 1 and 4 have same polarity

4) If e13= e12-e34, 1 and 4 have different polarity

The transformer’s equivalent circuit

To model a real transformer accurately, we need to account for the following losses:

1.Copper losses – resistive heating in the windings: I2R.2.Eddy current losses – resistive heating in the core: proportional to the square of voltage applied to the transformer.3.Hysteresis losses – energy needed to rearrange magnetic domains in the core: nonlinear function of the voltage applied to the transformer.4.Leakage flux – flux that escapes from the core and flux that passes through one winding only.

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Transformer Equivalent Circuit

The Exact Equivalent Circuit Of A Real Transformer

The transformer’s equivalent circuit

However, the exact circuit is not very practical.

Therefore, the equivalent circuit is usually referred to the primary side or the secondary side of the transformer.

Equivalent circuit of the transformer referred to its primary side.

Equivalent circuit of the transformer referred to its secondary side.

Approximate Equivalent Circuit Of A Transformer

For many practical applications, approximate models of transformers are used.

Referred to the primary side.

Without an excitation branch referred to the primary side.

Referred to the secondary side.

Without an excitation branch referred to the secondary side.

The values of components of the transformer model can be determined

experimentally by an open-circuit test or by a short-circuit test.

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Open circuit Test

•It is used to determine Lm1 (Xm1)and Rc1

•Usually performed on the low voltage side

•The test is performed at rated voltage and frequency underno load

Open Circuit TestFinds Iron Losses

(Fe)

Full Supply Voltage

SecondaryOpen Circuit

Wattmeter indicates Iron Losses (Fe)

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Short circuit Test

•It is used to determine Llp (Xeq) and Rp(Req)

•Usually performed on the high voltage side

•This test is performed at reduced voltage and rated frequency with the output of the low voltage winding short circuited such that rated current flows on the high voltage side.

Short Circuit TestCopper Losses

(Cu)

SecondaryShort

Circuited

Limited

Supply Voltage ≈ 5-10 %

Wattmeter indicates Copper Losses (Cu)

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Transformer Regulation

•Loading changes the output voltage of a transformer.Transformer regulation is the measure of such a deviation.

Definition of % Regulation

100*|V|

|V||V|

load

loadloadno

Vno-load =RMS voltage across the load terminals without load

V load = RMS voltage across the load terminals with a specifiedload

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Maximum Transformer Regulation

1212

1

011

02

'2

0'21

;0

max

.0

eqeq

eqeq

or

whenimumisVClearly

ZIVV

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Transformer Losses and Efficiency•Transformer Losses

•Core/Iron Loss =V12 / Rc1

•Copper Loss = I12 R1+ I2

2 R2

Definition of % efficiency

100*222

222

CosIVLosses

CosIV

100*/ 2222

221

211

21

222

CosIVRIRIRV

CosIV

c

2Cos = load power factor

100*/ 2222

221

21

222

CosIVRIRV

CosIV

eqc

Transformer Losses

Copper Losses (Cu)

•Varies with load current

•Produces HEAT •Created by resistance of windings

•Short circuit test supplies copper losses

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Transformer LossesIron Losses

(Fe)•Fixed

•Always present

•Related to transformers construction

Eddy Currents

Reduced by laminations

Produces HEAT

Hysteresis

Reduced by using special steels in laminations

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Efficiency

PowerInputPowerOutput

η

LossesOutputInput

LossesPowerOutput

PowerOutputη

Ratio between Input power and Output Power

PowerInputLossesPowerInput

η

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Transformer Efficiency

PowerIn

PowerOut

OvercomeIron

Losses

Overcome Copper Losses

Some Poweris used to:

η = 100%

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Maximum Transformer Efficiency

The efficiency varies as with respect to 2 independent quantitiesnamely, current and power factor•Thus at any particular power factor, the efficiency is maximum if

core loss = copper loss .This can be obtained by differentiating theexpression of efficiency with respect to I2 assuming power factor, andall the voltages constant.

•At any particular I2 maximum efficiency happens at unity power factor.This can be obtained by differentiating the expression of efficiency with respect to power factor, and assuming I2 and all the voltages constant.

•Maximum efficiency happens when both these conditions are satisfied.

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100

0

% full load current

pf=1

pf= 0.8

pf= 0.6

At this load currentcore loss = copper loss

Maximum efficiency point

All Day Efficiency

• Most Transformers are connected permanently

• The time that the transformer has to be calculated when determining efficiency

• Able to determine the best transformer for the

application by its efficiency

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All Day EfficiencyTransformer A

Sout = 300 kVA Fe = 1.25 kVA Cu = 3.75 kVAHours Load kW kWh % Load Cu Loss Cu kWh Fe kWh Losses kWh Input kWh

1.00 6.00 5 100 500.0 33.33 0.42 2.08 6.25 8.33 508.336.00 7.00 1 200 200.0 66.67 1.67 1.67 1.25 2.92 202.927.00 8.00 1 300 300.0 100.00 3.75 3.75 1.25 5.00 305.008.00 9.00 1 360 360.0 120.00 5.40 5.40 1.25 6.65 366.659.00 12.00 3 300 900.0 100.00 3.75 11.25 3.75 15.00 915.0012.00 14.00 2 280 560.0 93.33 3.27 6.53 2.50 9.03 569.0314.00 18.00 4 300 1200.0 100.00 3.75 15.00 5.00 20.00 1220.0018.00 20.00 2 360 720.0 120.00 5.40 10.80 2.50 13.30 733.3020.00 22.00 2 280 560.0 93.33 3.27 6.53 2.50 9.03 569.0322.00 1.00 3 200 600.0 66.67 1.67 5.00 3.75 8.75 608.75

5900.0 5998.02

98.37

Time Period

P out kWh = P in kWh =

% Eff =29SISTec-e

Thank You

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