power point slides to chapter 13 transformers

Friday, April 7, 2023 Ch. 13 Transformers 1

Topics to be Discussed Introduction. Principle of Operation. Step-Up and Step-Down

Transformer. EMF Equation. Effect of Frequency. Conditions for Ideal

Transformer. Drawing the Phasor Diagram. Volt-Amperes (in Ideal

Transformer). Impedance Transformation, Practical Transformer at no

Load.1. Effect of Magnetization.2. Effect of Core Losses.

Iron Losses. Hysterisis Loss. Eddy current Loss.

Construction of a Transformer.

Laminations. Core Type Transformer. Shell Type Transformer.

How I0 changes on Loading ? Practical Transformer on

Load.1. Effect of Winding

Resistance.2. Effect of Flux Leakage.

Leakage flux in a transformer

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Equivalent Circuit. Simplified Equivalent

Circuit. Approximate Equivalent

Circuit. Voltage Regulation.

Condition for Zero Regulation.

Condition for Maximum Regulation.

Efficiency of a Transformer. Power Losses in

Transformers. Condition for Maximum

Efficiency. All-day Efficiency.

Autotransformers. Applications. Saving of copper. Advantages. Disadvantages.

Transformer Testing.(1) Open-Circuit Test.(2) Short-Circuit Test.

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Introduction A transformer is a highly efficient (about 99.5 %) static

(non-moving) device. It transfers electrical energy form one circuit to another

(usually from one ac voltage level to another), without any change in its frequency.

There exists no simple device that can accomplish such changes in dc voltages.

Transformation of voltage is necessary at different stages of the electrical network consisting of generation, transmission and distribution.

Small-size transformers are used in communication circuits, radio and TV circuits, telephone circuits, instrumentation and control systems.

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Principle of Operation It operates on the principle of mutual induction between two coils.

When two coils are inductively coupled and if current in one coil is changed uniformly, then an EMF gets induced in the other coil. This EMF can drive a current, when a closed path is provided to it.

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It consists of two inductive coils electrically separated but magnetically linked through a common magnetic circuit.

Coil in which electrical energy is fed is Primary Winding.

Coil in which other load is connected is called as Secondary Winding.

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Friday, April 7, 2023 Ch. 13 Transformers 6 Next

http://en.wikipedia.org/wiki/Image:Transformer3d_col3.svg


(a) Construction.

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N1 : Number of turns in the Primary

N2 : Number of turns in the Secondary

E1 : EMF Induced in the Primary

E2 : EMF Induced in the Secondary

(b) Symbol.

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Step-Up and Step-Down Transformer

If N1 > N2 E1 > E2 Step down

E1 < E2 Step up If N1 < N2

The transformation ratio,

2 2

1 1

N EK

N E

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EMF EquationDue to the sinusoidally varying voltage V1 applied to the primary voltage, the flux set up in the core,

m msin sin 2t ft The resulting induced emf in a winding of N turns,

m

m m

( sin )

cos sin ( / 2)

d de N N t

dt dtN t N t

Thus, the peak value of the induced emf, Em = ωNΦm.

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m m mm

24.44

2 2 2

E N fNE fN

mor 4.44E fN

Therefore, the rms value of the induced emf E,

This equation, known as emf equation of transformer.

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Effect of Frequency At a given flux, emf of a transformer increases with

frequency. By operating at higher frequencies, transformers

can be made physically more compact. Because a given core is able to transfer more power

without reaching saturation. Aircraft and military equipments employ 400-Hz

power supplies which reduces size and weight. Disadvantage : The core loss and conductor

resistance increases due to skin effect.

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Conditions for Ideal Transformer :1. The permeability (μ) of the core is infinite, (i.e., the

magnetic circuit has zero reluctance so that no mmf is needed to set up the flux in the core).

2. The core of the transformer has no losses.

3. The resistance of its windings is zero, hence no I2R losses in the windings.

4. Entire flux in the core links both the windings, i.e., there is no leakage flux.

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Ideal transformer

(a) The circuit.

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• We take flux Φ as reference phasor, as it is common to both the primary and secondary.

• EMF E1 and E2 lag flux Φ by 90°.

• The emf E1 in the primary exactly counter balances the applied voltage V1. Hence, E1 is called counter emf or back emf .

(b) The phasor diagram.

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Flux,

V1 = -E1

E1

E2

O90

Drawing the Phasor Diagram

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Volt-Amperes (in Ideal Transformer) The current I1 in the primary is just sufficient to provide mmf

I1N1 to overcome the demagnetizing effect of the secondary mmf I2N2. Hence,

2 11 1 2 2

1 2

1or

I NI N I N

I N K

Note that the current is transformed in the reverse ratio of the voltage. If V2 > V1, then I2 < I1. Also, we have

1 1 2 2E I E I

Hence, in an ideal transformer the input VA and output VA are identical.

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Impedance Transformation

1 1 2 2 1 2 2eq L

1 1 2 2 2 1 2

( ) 1 1

( )

V V V I V I VZ Z

I I V I V I I K K

2

eq Lor /Z Z K

The concept of impedance transformation is used for impedance matching.

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Example 1 A single-phase, 50-Hz transformer has 30

primary turns and 350 secondary turns. The net cross-sectional area of the core is 250 cm2. If the primary winding is connected to a 230-V, 50-Hz supply, calculate

(a) the peak value of flux density in the core,

(b) the voltage induced in the secondary winding, and

(c) the primary current when the secondary current is 100 A. (Neglect losses.)

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Solution :

(a) The peak value of the flux,

1m

1

2300.034534 Wb

4.44 4.44 50 30

E

fN

mm 4

0.034534

250 10B

A

1.3814 T

(b) The voltage induced in the secondary,

22 1

1

350230 2683.33 V .

30

NE E

N 2 683 kV

(c) The primary current,

21 2

1

350100 1166.67 A .

30

NI I

N

1 167 kA

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Example 2Determine the load current IL in the ac circuit shown.

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Transforming the load impedance into the primary

p 2

30 00.872 35.53° A

20 20 2 (2 10)j j

I

L p2 2 0.872 35.53 I I 1.74 35.53° A

Solution :

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Practical Transformer at no Load There are following two reasons why the no-load

current (also called exciting current) I0 is not zero :

1. Effect of Magnetization : No magnetic material can have infinite permeability. A finite mmf is needed to establish magnetic flux in

the core. An in-phase magnetizing current Im in the

primary is needed. Im is purely reactive (current Im lags voltage V1 by

90°). This effect is modelled by putting X0 in parallel with

the ideal transformer.

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2. Effect of Core Losses : There exist hysteresis and eddy current losses for

the energy loss in the core. The source must supply enough power to the

primary to meet the core losses. These iron losses can be represented by putting a

resistance R0 in parallel.

The core-loss current Iw flowing through R0 is in phase with the applied voltage V1,

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(a) The circuit.

(b) The equivalent circuit.

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• The R0-X0 circuit is called exciting circuit.

2 2 10 w m 0 m w

1 w 1 0 0

; tan ( / );

and Input power Iron loss

cos

I I I I I

V I V I

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Modified phasor diagram

O Flux,

E2 = V2

E1

V1 = -E1

Iw

Im

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Iron Losses

The core losses occur in iron core, hence these are also called iron losses.

There are two reasons for these losses:

1. Hysteresis Loss.

2. Eddy current loss.

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Hysterisis Loss

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When alternating current flows through the windings, the core material undergoes cyclic process of magnetization and demagnetization.

h h mnP K B f V

Kh = hysteresis coefficient whose value depends upon the material(Kh = 0.025 for cast steel, Kh = 0.001 for silicon steel)Bm = maximum flux density (in tesla)n = a constant, depending upon the material f = frequency (in hertz)V = volume of the core material (in m3)

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Eddy current Loss2 2 2

e e mP K B f t V

where Ke = a constant dependent upon the material t = thickness of laminations (in metre)

i h eP P P Next


Construction of a Transformer There are two basic parts of a transformer :

Magnetic Core Winding or Coils

Limb

Yoke

Magnetic core Winding or coil

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Laminations The core of a

transformer is usually laminated to reduce the eddy currents.

These laminations may be different sections of E,I,T,F.

They are stacked finally to get the complete core of the transformer.

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Two Types of Transformers (1) Core Type Transformer :

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The windings surround a considerable part of the core.

Both the windings are divided into two parts and half of each winding is placed on each limb, side by side.

This is done to reduce the leakage of the magnetic flux.

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Practically, the windings are placed as follows.

To minimize the cost of insulation, the low voltage (LV) winding is placed adjacent to the core and high voltage (HV) winding is placed around the LV winding

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(2) Shell Type Transformer The core surrounds a considerable part of the windings. It has three limbs. Both the windings are placed on the central limb. The flux divides equally in the central limb and returns

through the outer two legs.

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How I0 changes on Loading ?

Φ Φ’

V1 E2E1 V2

N1 N2

I0 + I1’ I2

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Before connecting the load, there exists a flux Φ requiring current I0 in the primary.

On connecting the load, a current I2 flows in the secondary.

The magnitude and phase of I2 with respect to V2 depends upon the nature of the load.

The current I2 sets up a flux Φ’, which opposes the main

flux Φ. Hence, it is called demagnetizing flux. This momentarily weakens Φ, and back emf E1 gets

reduced. As a result, V1 - E1 increases and more current is drawn

from the supply. This again increases E1 to balance the applied voltage

V1. Next


In this process, the primary current increases by I1’. This current is known as primary balancing

current, or load component of primary current. Under such a condition, the secondary ampere-turns

must be counterbalanced by the primary ampere-turns.

'1 1 2 2N I N I

21 2 2

1

' NI I KI

N

and

'1 0 1 I I I

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Volt Ampere Rating of a Transformer Output power depends on cos2 ( power factor of

secondary). As pf can change depending on the load, the rating is

not specified in watts or kilowatts. But is indicated as a product of voltage and current

called VA RATING.

2211 IVIV

100022 IV

1000

IV rtransforme a of rating kVA 11

1load) (full1

1000 ratingkVA

VI

2load) (full2

1000 ratingkVA

VI

Fore ideal transformer :

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Transformers rating in kVA ?

• Transformers are rated in VA, because the manufacturer does not know the power factor of the load which you are going to connect.

• So the customer should not exceed the VA rating of the transformer.

• In case of motors, the manufacturer knows exactly the power factor at full load.

• That is why motors are rated in kW.

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Example 3 A single-phase, 230-V/110-V, 50-Hz transformer takes an

input of 350 volt amperes at no load while working at rated voltage. The core loss is 110 W. Find

(a) the no-load power factor,

(b) the loss component of no-load current, and

(c) the magnetizing component of no-load current.

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Solution : (a) Given : 1 0 350 VAV I

01

3501.52 A

230

VAI

V

The core loss = Input power at no load, 1 0 0cosiP V I

01 0

110 Wcos

350VAiP

pfV I

0.314

(b) The loss component of no-load current,

w 0 0cos 1.52 0.314I I 0.478 A

(c) The magnetizing component of no-load current,

2 2 2 2m 0 w (1.52) (0.478)I I I 1.44 A

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Example 4 A 100-kVA, 4000-V/200-V, 50-Hz, single-phase

transformer has 100 secondary turns. Determine

(a) the primary and secondary currents,

(b) the number of primary turns, and

(c) the maximum value of the flux.

Solution : The kVA rating = V1I1 = V2I2 = 100 kVA.

11

22

kVA rating 100000

4000

kVA rating 100000

200

IV

IV

25 A

500 A

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1 1

2 2

11 2

2

Since

4000100

200

N V

N V

VN N

V

2000

(b)

(c)2 2

2

2

4.44

200

4.44 4.44 50 100

m

m

E f N

E

fN

9.01 mWb

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Flux,

E1

E2= V2

OI0

0

I1I1

I2

V1 = -E1

1

Phasor Diagram for Resistive Load

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Flux,

E1

E2 = V2

OI0

0

I1I1

I2

V1 = -E1

1

Phasor Diagram for Inductive Load

2

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Flux,

E1

E2 = V2

OI0

0

I1

I1

I2

V1 = -E1

1

Phasor Diagram for Capacitive Load

2

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(a) Resistive. (b) Inductive. (c) Capacitive.

Phasor Diagrams for Different Types of Loads

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Is it ever possible that the load connected to the secondary is capacitive but the overall power factor is inductive ?

Ans. : Yes. See the phasor diagram for capacitive load.

Is it ever possible that the load connected to the secondary is inductive but the overall power factor is capacitive?

Ans. : No. Not possible.

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Example 5 A single-phase, 440-V/110-V, 50-Hz

transformer takes a no-load current of 5 A at 0.2 power factor lagging. If the secondary supplies a current of 120 A at a power factor of 0.8 lagging to a load, determine the primary current and the primary power factor. Also, draw the phasor diagram.

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Solution :1 1

0 2cos 0.2 78.46 and cos 0.8 36.87

2

1

110 1

440 4

VK

V

' '1 2 1(1/ 4) 120 30 A; 30 36.87 AI K I I

'1 1 0 30 36.87 5 78.46 I I I 33.9 42.49° A

Primary power factor,

1cos cos 42.49pf 0.737(lagging)

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Practical Transformer on Load We now consider the deviations from the last two

ideality conditions :

1. The resistance of its windings is zero.

2. There is no leakage flux.

The effects of these deviations become more prominent when a practical transformer is put on load.

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(1) Effect of Winding Resistance Current flow through the windings causes a power

loss called I2R loss or copper loss. This effect is accounted for by including a resistance

R1 in the primary and resistance R2 in the secondary

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(2) Effect of Flux Leakage The difference between the total flux linking with the

primary and the useful mutual flux Φu linking with both the windings is called the primary leakage flux, ΦL1.

Similarly, ΦL2 represents the secondary leakage flux.

Flux leakage results in energy being alternately stored in and discharged from the magnetic fields with each cycle of the power supply.

It is not directly a power loss, but causes the secondary voltage to fail to be directly proportional to the primary voltage, particularly under heavy loads.

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Leakage flux in a transformer

(a) Its definition. (b) Its effect accounted for.

• The useful mutual flux Φu is responsible for the transformer action.

• The leakage flux ΦL1 induces an emf EL1 in the primary winding.

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Similarly, flux ΦL2 induces an emf EL2 in the secondary.

Hence, we include reactances X1 and X2 in the primary and secondary windings, in the equivalent circuit.

The paths of leakage fluxes ΦL1 and ΦL2 are almost entirely due to the long air paths and are therefore practically constant.

The reluctance of the paths being very high, X1 and X2 are relatively small even on full load.

However, the useful flux Φu remains almost independent of the load.

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Equivalent Circuit of a Transformer It is merely a representation of the following KVL equations :

1 1 1 1 1 1 1 1 1 1( )I R jI X I R jX V E E

2 2 2 2 2 2 2 2 2 2( )I R jI X I R jX E V V

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m

E2

E1

-E1

I1R1

I1X1V1

I1

I0

I1'

I2

V2

I2R2

I2X2

0

1

Pha

sor

Dia

gram

for

Pra

ctic

al T

rans

form

er o

n R

esis

tive

Lo

ad

1 1 1 1 1

2 2 2 2

( )

( )

I R jX

V I R jX

2

V E

E

O

I1Z1

I2Z2

Next


Pra

ctic

al T

rans

form

er o

n In

du

ctiv

e L

oad

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Pra

ctic

al T

rans

form

er o

n C

apac

itiv

e L

oad

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Simplified Equivalent Circuit The no-load current I0 is only about 3-5 % percent of

the full-load current. The exciting circuit R0-X0 in is shifted to the left of

impedance R1-X1.

Transforming the impedances from the secondary to the primary side.

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Equivalent resistance and reactance referred to the primary side

2 2e1 1 2 e1 1 2( / ) and ( / )R R R K X X X K

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Approximate Equivalent Circuit

As referred to primary side.

As referred to secondary side.

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Example 5 A single-phase, 50-kVA, 4400-V/220-V, 50-Hz

transformer has R1 = 3.45 Ω, R2 = 0.009 Ω, X1 = 5.2 Ω and X2 = 0.015 Ω. Calculate

(a) the Re as referred to the primary,

(b) the Re as referred to the secondary,

(c) the Xe as referred to the primary,

(d) the Xe as referred to the secondary,

(e) the Ze as referred to the primary,

(f) the Ze as referred to the secondary, and

(g) the total copper loss.

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Solution : Full-load primary current,

11

kVA 5000011.36 A

4400I

V

Full-load secondary current, 22

kVA 50000227.27 A

220I

V

2

1

220 10.05

4400 20

VK

V

(a) 2 2e1 1 2( / ) 3.45 [0.009 /(0.05) ]R R R K 7.05 Ω

(b)

(c)

2 2e2 1 2 (0.05) 3.45 0.009R K R R 0.0176 Ω

2 2e1 1 2( / ) 5.2 [0.015 /(0.05) ]X X X K 11.2 Ω

(d) 2 2e2 1 2 (0.05) 5.2 0.015X K X X 0.028 Ω

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(e) 2 2 2 2e1 e1 e1 (7.05) (11.2)Z R X 13.23 Ω

(f)

(g) Total copper loss

2 2 2 2e2 e2 e2 (0.0176) (0.028)Z R X 0.0331 Ω

2 2 2 21 1 2 2 (11.36) 3.45 (227) 0.009I R I R 909 W

Alternatively, by considering equivalent resistances, total copper loss

2 21 e1 (11.36) 7.05I R 909.8 W

2 22 e2 (227.27) 0.0176I R 909 W

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Voltage Regulation

V2(0) = secondary terminal voltage at no load,

and V2 = secondary terminal voltage at full load.

The voltage regulation of a transformer is defined as the change in its secondary terminal voltage from no load to full load, the primary voltage being assumed constant.

The voltage drop V2(0) - V2 is called the inherent regulation.

2(0) 2

2(0)

( ) Per unit V V

iV

regulation down

2(0) 2

2(0)

% 100V V

V

regulation down

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2(0) 2

2

( ) Per unit V V

iiV

regulation up

2(0) 2

2

% 100V V

V

regulation up

Normally, when nothing is specified, ‘regulation’ means ‘regulation down’.

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Exact voltage drop =

2(0) 2 OC OA OG OA AG AF+FGV V

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In case of leading power factor,

2 e2 2 e2

Approximate voltage drop, AF AE EF AE BD

cos sinI R I X

2 e2 2 e2

Approximate voltage drop, AF AE EF AE BD

cos sinI R I X

In general,

2 e2 2 e2Approximate voltage drop cos sinI R I X

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2 2 2 2

2(0)

cos sin% Regulation 100

cos sin

e e

r x

I R I X

V

V V

Condition for Zero Regulation :

Possible only if the load has leading power factor.

2 2 2 2cos sin 0e eI R I X 2

2

tan e

e

R

X

Use + sign for lagging power factor and – sign for leading power factor.

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2 e2 sinI X 2 e2 cosI R

Note that for leading power factor, if the magnitude of the phase angle is high, we may have

• The regulation then becomes negative.

• It means that on increasing the load, the terminal voltage increases.

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Condition for Maximum Regulation

2 2 2 2( cos sin ) 0e e

dI R I X

d

2 2 2 2( sin cos ) 0e eI R I X

2

2

tan e

e

X

R

Maximum regulation can occur only for inductive load. The voltage drop is maximum when

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Example 6

Solution :

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Example 7

Solution :

2the load voltage, 240 6V 234V

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Example 8 A single-phase, 40-kVA, 6600-V/250-V,

transformer has primary and secondary resistances R1 = 10 Ω and R2 = 0.02 Ω, respectively. The equivalent leakage reactance as referred to the primary is 35 Ω. Find the full-load regulation for the load power factor of

(a) unity,

(b) 0.8 lagging, and

(c) 0.8 leading.

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2 2e2 1 2 (0.0379) 10 0.02 0.0343R K R R

2 2e2 e1and (0.0379) 35 0.0502X K X

(a) For power factor, cos = 1; sin = 0. Hence,

2 2 2 2

2(0)


160 0.0343 1 0100

250

e eI R I X

V

2.195 %

Solution : Given : R1 = 10 Ω; R2 = 0.02 Ω; Xe1 = 35 Ω

2

250 the turns-ratio, 0.0379

660040000

the full-load current, 160 A250

K

I

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(b) For power factor, cos = 0.8 (lagging, positive);

(c) For power factor, cos = 0.8 (leading, negative);

sin 0.6 2 e2 2 e2

2(0)


160 0.0343 0.8 160 0.0502 0.6100

250

I R I X

V

0.172 %

2sin 1 cos 0.6

2 e2 2 e2

2(0)


160 0.0343 0.8 160 0.0502 0.6100

250

I R I X

V

3.68 %

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Efficiency of a Transformer Like any other machine, the efficiency of a transformer is defined as

o

o l

Power output Power output

Power input Power output + Power losses

P

P P

• Large-size transformers are designed to be more efficient ( > 98 %)

• But, the efficiency of small transformers (used in power adapters for charging mobile phones) is not more than 85 %.

Power lost

Outputpower

Inputpower

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Power Losses in Transformers(i) Copper losses or I2R losses :

The copper losses are variable with current. Assuming the voltage to remain constant, the current is proportional to the VA. Therefore, the copper losses for a given load (and hence for given VA) is given as

2

c c(FL)FL

VA

VAP P

2 2 2 2c 1 1 2 2 1 e1 2 e2P I R I R I R I R

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(ii) Iron losses or core losses :

Due to hysteresis and eddy-currents. Pi = Ph + Pe.

Since the flux Φm does not vary more than about 2 % between no load and full load, it is usual to assume the core losses constant at all loads.

In general, the efficiency,

o o 2 2 22

o l o c 2 2 2 2 2 i

cos

P cosi e

P P V I

P P P P V I I R P

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Condition for Maximum EfficiencyAssuming the operation at a constant voltage and a constant power

factor, for what load (i.e., what value of I2) the efficiency becomes maximum ?

Let us first divide the numerator and denominator by I2, to get

The efficiency will be maximum when the denominator of the above equation is minimum,

i2 2 2 e2 2 e2 2

2 2

22 e2 c i

( cos / ) 0 or 0

or or

i

i

d PV I R P I R

dI I

I R P P P

2 2

2 2 2 e2 i 2

cos

cos /

V

V I R P I

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Copper loss = Iron loss

Condition :

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All-day Efficiency The efficiency defined above is called commercial

efficiency. In a distribution transformer, the primary remains

energized all the time. But the load on the secondary is intermittent and variable during the day.

The core losses occur throughout the day, but the copper losses occur only when the transformer is loaded.

Such transformers, therefore, are designed to have minimum core losses. This gives them better all-day efficiency, defined below.

all-day

Output energy (in kW h) in a cycle of 24 hours

Total input energy (in kW h)

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Example 9 For a single-phase, 50-Hz, 150-kVA transformer,

the required no-load voltage ratio is 5000-V/250-V and the full-load copper losses are 1800 W and core losses are 1500 W. Find (a) the number of turns in each winding for a maximum core flux of 0.06 Wb, (b) the efficiency at half rated kVA, and unity power factor,(c) the efficiency at full load, and 0.8 power factor lagging, and (d) the kVA load for maximum efficiency.

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Solution :

2 2 m

22

m

4.44

25018.8(say, )

4.44 4.44 50 0.06

E fN

EN

f

19 turns

11 2

2

5000and 19

250

EN N

E 380 turns

0.5 (kVA) (power factor) 0.5 150 1 75 kWoP 2 2(0.5) (full-load copper loss) (0.5) 1800 W 0.45 kWcP

Iron losses (fixed), Pi = 1500 W = 1.5 kW

(a) Using the emf equation, we have

(b) At half rated-kVA, the current is half the full-load current, and hence the output power too reduces by 0.5. Thus,

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o

o c i

75100 100

75 0.45 1.5

P

P P P

97.47 %

(c) At full load and 0.8 power factor,

(kVA) (power factor) 150 0.8 120 kWoP 1800 W 1.8 kW; and 1500 W 1.5kWc iP P

o

o c i

120100 100

120 1.8 1.5

P

P P P

97.3 %

(d) Let x be the fraction of full-load kVA at which the efficiency becomes maximum

2c i or 1800 1500 1500 /1800 0.913P P x x

(Full-load kVA) 150 0.913x 137 kVA

Therefore, the load kVA under the condition of maximum efficiency,

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Example 10 For a single-phase, 200-kVA, distribution transformer

has full-load copper losses of 3.02 kW and iron losses of 1.6 kW. It has following load distribution over a 24-hour day :

(i) 80 kW at unity power factor, for 6 hours.

(ii) 160 kW at 0.8 power factor (lagging), for 8 hours.

(iii) No load, for the remaining 10 hours.

Determine its all-day efficiency.

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Solution :

(i) For 80 kW load at unity power factor (for 6 hours) :

Output energy 80 6 480 kW h

o 80kVA 80kVA

1

P

pf

2 2

c(FL)FL

kVA 80(3.02) 0.4832 kW

kVA 200cP P

Iron losses, Pi = 1.6 kW

Total losses, Pl = Pc + Pi = 0.4832 kW + 1.6 kW = 2.0832 kW

Total energy losses in 6 hours 2.0832 6 12.50 kW h Next

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Output energy 160 8 1280 kW h

FL

160200kVA kVA

0.8oP

kVApf

c c(FL)Copper losses, 3.02 kWP P

Total energy losses in 8 hours 4.62 8 36.96 kW h

(ii) For 160-kW load at 0.8 power factor (for 8 hours) :

Iron losses, Pi = 1.6 kWTotal losses, Pl = Pc + Pi = 3.02 kW + 1.6 kW = 4.62 kW

(iii) For the no-load period of 10 hours :

Output energy Po = 0

Copper losses, Pc = 0

Iron losses, Pi = 1.6 kW

Total losses, Pl = Pc + Pi = 0 + 1.6 = 1.6 kW Next

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Total energy losses in 10 hours 1.6 10 16 kW h

Thus, for 24-hour period :

Total output energy, Wo = 480 + 1280 = 1760 kW h

Total energy losses, Wl = 12.50 + 36.96 +16 = 65.46 kW h

oall-day

o l

All-day efficiency,

1760100 100

1760 65.46

W

W W

96.41%

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Autotransformer It is a special transformer that is useful in power

systems, motor starters, variable ac sources, etc. An autotransformer is a transformer which has a part

of its winding common to the primary and secondary circuits.

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Applications Boosting or buckling of supply voltage by a

small amount. Starting of ac machines, where the voltage is

raised in two or more steps. Continuously varying ac supply as in variacs.

(a) Step-down (b) Step-up

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The turns-ratio is given as

2 2

1 1

N VK

N V

• The portion YZ of the winding is called common winding.

• The portion XY is called series winding. • In variacs (variable autotransformers), point Y is made

a sliding contact so as to give a variable output voltage.

Consider the Step-down autotransformer :

The volt-amperes on the two sides must be the same,

1 1 2 2V I V I 2 2 2 1 2 2 1( )V I V I V I I

• The part V2I1 is conductively transferred through the winding XY.

• The remaining part is inductively transferred through the winding YZ.

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Saving of copper in an autotransformer

• For the same voltage ratio and capacity (volt-ampere rating), an autotransformer needs much less copper compared to a two-winding transformer.

• The cross-sectional area of a conductor is proportional to the current carried by it, and its length is proportional to the number of turns. Therefore,

Weight of copper NI kNI

For a two-winding transformer :

1 1

2 2

1 1 2 2

Weight of copper in primary

Weight of copper in secondary

Total weight of copper ( )

kN I

kN I

k N I N I

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For a step-up autotransformer :

1 2 1

2 2 1

1 2 1 2 2 1

1 2 1 2 2

Weight of copper in portion XY ( )

Weight of copper in portion YZ ( )

Total weight of copper ( ) ( )

[( 2 ) ]

k N N I

kN I I

k N N I kN I I

k N N I N I

Therefore, the ratio of copper- weights for the two cases is

2 1 2

1 2 11 2 1 2 2

1 1 2 2 1 2

2 1

1 2[( 2 ) ] [1 2 ]

1( )

N I NN I Nk N N I N I K K K

Kk N I N I K KI N

I N

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• Evidently, the saving is large if K is close to unity. • A unity transformation ratio means that no copper is

needed at all for the autotransformer. • The winding can be removed all together. • The volt-amperes are conductively transformed

directly to the load !

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Advantages of autotransformers

A saving in cost since less copper is needed. Less volume, hence less weight. A higher efficiency, resulting from lower I2R

losses. A continuously variable output voltage is

achievable if a sliding contact is used. A smaller percentage voltage regulation. Higher VA Rating.

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Disadvantages of autotransformersThe primary and secondary windings are not electrically separate, hence if an open-circuit occurs in the secondary winding the full primary voltage appears across the secondary.

Low impedance hence high short circuit currents for short circuits on secondary side.

No electrical separation between primary and secondary which is risky in case of high voltage levels.

Economical only when the voltage ratio is less than 2.

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Transformer Testing There are two simple tests to determine the equivalent-

circuit parameters and its efficiency and regulation: Open-circuit test (OC Test) Short-circuit test (SC Test)

Advantage of these tests is without actually loading the transformers, we can determine the Losses and Regulation, for full-load.

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(1) Open-Circuit Test This test determines the no-load current and the

parameters of the exciting circuit of the transformer. Generally, the low voltage (LV) side is supplied rated

voltage through a variac. The high voltage (HV) side is left open.

Open Circuit

Low voltage

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• The I2R loss on no load is negligibly small compared with the core loss.

• Hence the wattmeter reading, Wo, can be assumed to give the core loss of the transformer.

Calculations :

2i o 0 o

1

2 2ow m 0 w

1

1 10 0

w m

; ;

; ;

;

VP W I I K

V

WI I I I

V

V VR X

I I

Next


(2) Short-Circuit Test This test determines the equivalent resistance and leakage

reactance. Generally, the LV side of the transformer is short-circuited

through a suitable ammeter A2. A low voltage is applied to the primary (HV) side. This voltage is adjusted with the help of a variac so as to

circulate full-load currents in the primary and secondary circuits.

Short Circuit

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• The reading of ammeter A1, Isc, gives the full-load current in the primary winding.

• Since the applied voltage (and hence the flux) is small, the core loss is negligibly small.

• Hence, the wattmeter reading, Wsc, gives the copper loss (Pc).

Calculations :

2 2sc sce1 e1 e1 e1 e12

sc sc

; ;W V

R Z X Z RI I

Next


Example 11 A single-phase, 50-Hz, 12-kVA, 200-V/400-V transformer

gives the following test results :

(i) Open-circuit test (with HV winding open) : 200 V, 1.3 A, 120 W

(ii) Short-circuit test (with LV winding short-circuited) : 22 V, 30 A, 200 W

Calculate :

(a) the magnetizing current and the core-loss current, and

(b) the parameters of equivalent circuit as referred to the low voltage winding.

Next


Solution :(a) The wattmeter reading, 120 W, in the open-circuit test gives the core losses. Therefore, the core-loss current is given as

(b) The parameters of the exciting circuit are given by the open-circuit test, as

FL

12 kVAand 30 A

400 VI

This confirms that the short-circuit test has been done at the rated full-load

ow

1

120 W

200 V

WI

V 0.6 A

2 2 2 2m 0 w (1.3) (0.6)I I I 1.15 A

1 10 0

w m

200 V 200 V and

0.6 A 1.15 A

V VR X

I I 333 Ω 174 Ω

2

1

200 V 1Now,

400 V 2

VK

V

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sc sce1 e12 2

sc sc

200 W 22 V0.222 and 0.733

(30 A) 30 A

W VR Z

I I

The equivalent resistance and reactance as referred to the secondary side (low voltage winding),

22

e2 e1

10.222

2R K R

0.055 Ω

22

e2 e1

1and 0.699

2X K X

0.175 Ω

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power point slides to chapter 13 transformers

Documents

shell type

hz transformer

leading power

applied voltage

unity power

eddy current

load power

low voltage