Single Phase Transformer© - Weebly

1 Prepared By: Nafees Ahmed www.eedofdit.weebly.com

Single Phase Transformer©

By

Nafees Ahmed Asstt. Prof. Department of Electrical Engineering

DIT, University, Dehradun, Uttarakhand

References:

1. V. Del Toro “Principles of electrical engineering” Prentice Hall International

2. B. L Theraja “ Electrical Technology” S. Chand

3. www.google.com

4. www.wikipedia.org

http://www.goole.com/


It is a static device which transfers energy form one electrical circuit to another electrical

circuit without change in frequency. Very high efficiency (up to 96% or up to 98%, Because there is no rotating part)

Types of transformer:

Depending on types of phase 1. 1-ϕ transformer 2. 3- ϕ transformer

Depending on use a. Step up transformer: Output voltage is higher then applied voltage b. Step down transformer: Output voltage is less then applied voltage

Depending of construction 1. Shell type transformer 2. Corel type transformer

Construction and working principle: Construction:

Laminated core of Si-Steel (Soft Steel) Soft Steel is used to reduce hysteresis losses

L V

L V

H V

H V

L V

L V

H V

H V

Figure 1

Core type Shell type

HV

LV

HV

LV

LV

HV

LV

HV

LV

LV

Laminated Core

Primary (LV/HV)

Secondary (HV/LV)

Secondary (HV/LV)

Flux

Figure 2


Laminated core is to reduced eddy current losses Principle: Based on low of electromagnetic induction. Ideal Transformer: Conditions for ideal transformer are

1. Copper losses are zero: => Winding resistance = 0 2. Iron (Core) losses are zero: => Hysteresis & eddy current losses =0 3. Leakage flux is zero: => Permeability of iron is infinity

=>

4. Magnetizing current is zero EMF Equation: (Prove that emf per turn for a transformer is constant)

Let an alternating voltage (v1) is applied to the primary side So an alternating flux will produce in the core

Let

Because of this an emf will induce in primary as well as in secondary windings Primary induced emf

Or )2(2/11 tSinEe m Where

01

A

lS

dt

dNe

11

)1( tSinm

dt

tSindNe m

11

tCosNe m 11

2/11 tSinNe m

Figure 3: Core type 1-Ph transformer

N1 N2 e1 e2 v1 v2

ϕ


Constant44.42

2

1

1 mfN

E

N

E

Ksay Constant 1

2

1

2

1

2 N

N

E

E

V

V

K2

1

1

2

1

2

1

2 I

I

N

N

E

E

V

V

So its RMS value

Similarly RMS values of secondary induced emf

For ideal transformer Induced emf in primary (E1) = Applied Voltage (V1) Induced emf in secondary (E2) = Terminal Voltage (V2) From equations (3) & (4) Hence emf/turn = constant Voltage and current transformation ratio: Again from equations (3) & (4) K= Voltage (Current) transformation ratio For ideal transformer Input VA = Output VA V1I1 = V2I2

Hence

Transformer on NO load: Consider an ideal transformer with magnetizing current: Ideal transformer without iron loss:

Im & ϕ will be in same phase V1 & E1 will be equal and opposite

emfinducedofValueMaxNE mm 11

)3(44.4 11 NfE m

1111

1 44.42

2

22Nf

fNNEE m

mmm

)4(44.4 22 NfE m

2

1

1

2

I

I

V

V


Ideal transformer with iron loss: + a little copper loss No load current (I0) will be very less (2% - 5% of full load current) No load power factor angle (ϕ0) will be very high ( 780 to 870) Im= Magnetizing or wattles component of no load current (I0) Ie= Active or wattfull component of no load current (I0)

No load power Neglecting copper loss Note: No load power factor is very less because ϕ0 is very high. (ϕ0 is high because Ie<<Im)

Im

E1

E2

V1= -E1

ϕ

Figure 4b: Phasor diagram of ideal transformer

Without Iron loss

RI 2

0

V1 E1 R0

X0

I0

Ie Im

I0

Figure 4e: Equivalent Circuit

lossiron Total =010 CosIVP o

00 CosIIe

00 SinIIm

eI

VR 1

0

mI

VX 1

0

Figure 4a

N1 N2 E1 E2 V1 V2

ϕ

Im

Figure 4c

N1 N2 E1 E2 V1 V2

ϕ

I0

Im=I0Sinϕ0

E1

E2

V1= -E1

ϕ

Figure 4d: Phasor diagram of ideal transformer

with Iron loss

Φ0 Ie=I0Cosϕ0

I0


22

'

11 ININ

Transformer on load:

MMF induced in primary winding =MMF induced in secondary winding

Note: I1

’ and I2 will be in phase opposition Load may be pure resistive, inductive or capacitive resulting in unity, lagging and leading

power factor respectively. So the phasor diagram may be on unity, lagging and leading power factor.

Resistance and leakage reactance: In transformer windings do have some resistance Let R1 = Resistance of primary winding

R2 = Resistance of secondary winding The flux which is linked with primary as well as secondary windings is known as common flux.

Figure 5a

N1 N2 E1 E2 V1 V2

ϕ

I1 =I0 + I1’

L

O

A

D

I2

22

1

2'

1 KIIN

NI

E1

E2=V2

V1= -E1

ϕ

Figure 5c: Phasor diagram at lagging PF

Φ0 I0

I1

Φ1

I2

I1’ =KI2

Φ2

E1

E2=V2

V1= -E1

ϕ

Figure 5b: Phasor diagram at Unity PF

Φ0 I0

I1 I1

’ =KI2

Φ1

I2 Φ2=0 E1

E2=V2

V1= -E1

ϕ

Figure 5d: Phasor diagram at leading PF

Φ0 I0

I1

Φ1

I2

I1’ =KI2

Φ2


There is some flux which does not link with primary or secondary winding know as leakage flux as shown in figure 6a. Let ϕL1 = leakage flux of primary winding (Flux which does not link with secondary winding)

ϕL2 = leakage flux of secondary winding (Flux which does not link with primary winding)

So at no load I1 & I2 both are negligible hence leakage flux is negligible. Leakage flux produces a self induced back emf in their respective windings. They are

therefore equivalent to small reactance in series with the respective windings. This reactance is known as leakage reactance.

Let X1 = Leakage reactance of primary winding X2 = Leakage reactance of secondary winding

Note: V1≠ E1 & V2 ≠ E2 as we are taking previously.

2211 IandI LL

Figure 6a

V1 V2

Common Flux

I1

L

O

A

D

I2

Leakage Flux

R1 R2

V1 V2

I1

L

O

A

D

I2 R1 R2

E1 R0 X0

I1’

Ie Im

I0

Figure 6c: Equivalent Circuit

E2

X2 X1

Figure 6b

V1 V2

Common Flux

I1

L

O

A

D

I2 R1 R2

E1 N1 E2 N2

X1 X2


Phasor diagram of actual transformer ON Load: Load may be pure resistive load( Unity power factor), inductive load (Lagging power factor) and capacitive load (Leading power factor) so as the phasor diagram.

1. Phasor diagram for pure resistive load (at unity power factor):

Note:

a. Resistive drop I2R2 is parallel to I2 and inductive drop I2X2 is perpendicular (leading) to I2.

b. Similarly resistive drop I1R1 is parallel to I1 and inductive drop I1X1 is perpendicular (leading) to I1.

Figure 7a

V1 V2

Common Flux

I1

L

O

A

D

I2 R1 R2

E1 N1 E2 N2

X1 X2

E1

E2=V2

-E1

ϕ

Figure 7b: Phasor diagram at Unity PF

Φ0 I0

I1 I1

’ =KI2

Φ1

I2

Φ2=0

V2 I2R2

I2X2

V1

I1R1

I1X1


c. This is also true for rest of following two diagrams (lagging and leading pf). If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then

Φ1 = Φ2 = Φ1’= 0 or say Φ

Neglecting 21111 SinRICosXI

2. Phasor diagram for inductive load (at lagging power factor):

222

2

2222 XIRIVE

2222 RIVE

K

EE 2

1

2

1111

2

111111 SinRICosXISinRICosRIEV

SinRICosRIEV 111111

01111 RIEV

E1

E2=V2

-E1

ϕ

Figure 7c: Phasor diagram at lagging PF

Φ0 I0

I1’ =KI2

Φ1

I2

Φ2

V2 I2R2

I2X2

V1

I1R1

I1X1

I1


If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then

Φ1 = Φ2 = Φ1’= Φ (say)


3. Phasor diagram for capacitive load (at leading power factor):

If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then Φ1 = Φ2 = Φ1

’= Φ (say)

2222222

2

22222222 SinRICosXISinXICosRIVE

22222222222222 neglecting SinRICosXISinXICosRIVE

K

EE 2

1

2

1111

2


SinRICosRIEV 111111

E1

E2=V2

-E1

ϕ

Figure 7d: Phasor diagram at leading PF

Φ0 I0

I1’ =KI2

Φ1

I2

Φ2

V2

I2R2

I2X2

V1 I1R1

I1X1 I1

2222222

2

22222222 SinRICosXISinXICosRIVE

22222222222222 neglecting SinRICosXISinXICosRIVE


KI

I

N

N

E

E

V

V

2

1

1

2

1

2

1

2


Note: The above results can be obtained directly if we replace Φ with – Φ in the case of lagging power factor.

Equivalent circuit of transformer:

Let R1 & X1 are resistance and leakage reactance of primary windings and R2 & X2 are resistance and leakage reactance of secondary windings respectively.

We know

K

EE 2

1

21111

2


SinRICosRIEV 111111

V1 V2

I1

L

O

A

D

I2 R1 R2

E1 R0 X0

I1’

Ie Im

I0

Figure 8a: Transformer Equivalent Circuit

E2

X2 X1

2

2

121

1V

N

NV

KV

V1

I1

L

O

A

D

R1

R0 X0

I1’

Ie Im

I0

Figure 8b: Transformer Equivalent Circuit Referred to primary side

X1

2

2

1

1E

N

NE

2

2

2

1 XN

N

2

2

2

1 RN

N

2

2

1 VN

N

2

1

2 IN

N


Equivalent resistance and reactance of transformer: Let R1 = Resistance of primary winding

R2 = Resistance of secondary winding X1 = Leakage reactance of primary winding X2 = Leakage reactance of secondary winding I1R1 = Resistive drop in primary winding I2R2 = Resistive drop in secondary winding I1X1 = Reactive drop in primary winding I2X2 = Reactive drop in secondary winding

1. Equivalent resistance and reactance referred to secondary side:

V1

I1

L

O

A

D

R1

R0 X0

I1’

Ie Im

I0

Figure 8c: Approximately Equivalent Circuit Referred to primary side after shifting parallel branch

X1

2

2

1

1E

N

NE

2

2

2

1 XN

N

2

2

2

1 RN

N

2

2

1 VN

N

2

1

2 IN

N

V1

I1

L

O

A

D

R1

Figure 8d: Approximately Equivalent Circuit Referred to primary side after neglecting parallel branch

X1

2

2

2

1 XN

N

2

2

2

1 RN

N

2

2

1 VN

N

2

1

2 IN

N

KV1

I1/K L

O

A

D

K2R1

Figure 9a: Approximately Equivalent Circuit Referred to secondary side

K2X1 R2 X2 I2

V2


2

2

21

22

2 RIRKI

21

22

2 RRKI

2

2

21

22

2 XIXKI

21

22

2 XXKI

The above circuit is obtained after considering the following relations Total resistive drop

Where R02 = Equivalent resistance referred to secondary side Total reactive drop

Where X02 = Equivalent reactance referred to secondary side

If ϕ = 0

KI

I

N

N

V

V

2

1

1

2

1

2

K

IIKVV 1

212 &

02

2

2 RI

02

2

2 XI

KV1

L

O

A

D

Figure 9b: Approximately & simplified Equivalent Circuit Referred to secondary side

X02 R02 I2

V2

2022022

2

02202221 SinRICosXISinXICosRIVKV

SinRICosXINeglectingSinXICosRIVKV 02202202202221

SinXICosRIKVV 02202212

02212 RIKVV

ϕ I2 I2R02

I2X02

KV1

V2

Figure 9c: Phasor Diagram (Assuming lagging load)


2

2

2

11

2

1 / KRIRI

2

21

2

1 / KRRI

2

2

2

21

2

1 / KXIXI

2

21

2

1 / KXXI

If ϕ = 900

If ϕ is leading Replace ϕ with – ϕ in above equation, so

2. Equivalent resistance and reactance referred to primary side:

The above circuit is obtained after considering the following relations

Total resistive drop

Where R01 = Equivalent resistance referred to primary side Total reactive drop

Where X01 = Equivalent reactance referred to primary side

02212 XIKVV



KI

I

N

N

V

V

2

1

1

2

1

2

2121 &1

KIIVK

V

01

2

1 RI

01

2

1 XI

V1

I1

L

O

A

D

R1

Figure 9d: Approximately Equivalent Circuit Referred to primary side

X1

2

2 / KX

2

2 / KR

KV /2

2KI


If ϕ = 0 If ϕ = 900

If ϕ is leading Replace ϕ with – ϕ in above equation, so

Voltage regulation: It is change in the voltage of secondary side from no load to full load. Let V1 = Full load secondary voltage E2 = No load secondary voltage So % voltage regulation

2011011

2

01101121 / SinRICosXISinXICosRIKVV

SinRICosXINeglectingSinXICosRIKVV 01101101101121 /

0112

1 RIK

VV

0111

1 XIK

VV

SinXICosRIKVV 01101121 /

SinXICosRIKVV 01101121 /

ϕ I1 I1R01

I1X01

V1

V2/K

Figure 9f: Phasor Diagram (Assuming lagging load)

V1

L

O

A

D

Figure 9e: Approximately & simplified Equivalent Circuit Referred to primary side

X01 R01 KI2

V2/K


1002

22

E

VE

NfBAArea

Em44.4

)(

Where Cos ϕ = Power factor R02 = Equivalent resistance referred to secondary side

X02 = Equivalent reactance referred to secondary side +Ve sign for lagging power factor and –Ve sign for leading power factor.

Condition for zero voltage regulation:

From equation (1) it is clear that voltage regulation will be zero for leading power factor only. So condition of zero voltage regulation

Losses in transformer: The losses are of two type core loss and copper loss

(1) Core or iron losses: It is also known as fixed losses, again of two type (a) Eddy current losses: Given by

Where Ke = Constant Bm = Maxi value of flux density f = frequency t = thickness of laminations v = Volume of core We know NfE m44.4

Put the value of Bm in equation (1)

100

voltageloadNo

voltageloadFullvoltageloadNo

100sec

2

E

windingindropVoltage

)1(1002

022022

E

SinRICosRI

02

022022

E

SinRICosRI

02

02tanX

R

)1(222 WattvtfBKP mee

NAArea

fAArea

E m

)(44.4

)(

)2(44.4

/

fN

AEBm


For a given transformer N, A, t and V are constants so

So eddy current losses are independent of frequency.

(b) Hysteresis losses: Given by

Where η = Constant Again put the value of Bm from equation (2) to equation (3)

For a given transformer N, A and V are constants so

So eddy current losses dependent of voltage and frequency both.

(2) Copper Losses: Also know as variable losses because they depend on load current.

R1 = Resistance of primary winding R2 = Resistance of secondary winding I1 = Full load current in primary winding I2 = Full load current in secondary winding

Full load copper losses

Copper losses at x time of full load

vtffN

AEKP ee

22

2

44.4

/

vtEAN

KP ee

22

2

44.4

1

2EPe 2VPe )(VoltageVE

)3( WattfvBP x

mh

fvfN

AEP

x

h

44.4

/

vfEAN

P xx

x

h

1

44.4

1

xx

h fEP 1

x

h fVP 12)(VoltageVE

etcxloadhalfforcurrentloadFull

currentloadAnyxWhere

2

1;

2

2

21

2

1 RIRIPC

CCx PxP 2


Tests on transformer 1. Open Circuit (OC) test or No load test:

By OC test we can find out Iron losses (Pi) No load current (I0) Cosϕ0, Ie, Im, R0 & X0

Iron losses Pi = Reading of wattmeter (P0) No load current I0 = Reading of Ammeter Let V = Reading of voltmeter P0 = Pi = VI0Cosϕ0

Ie = I0Cosϕ0 Im = I0Sinϕ0

Note:

(i) Rated voltage is applied at LV side. (ii) This test is generally done on LV side (Why?)

2. Short Circuit (SC) test:

By OC test we can find out Copper losses (PC) Equivalent resistance or leakage reactance (R01 & X01 OR R02 & X02)

referred to metering side.

Full load Cu losses PC = Reading of wattmeter (Wsc) Let Short circuit current Isc = Reading of Ammeter

0

0VI

PCos i

eI

VR 0

mI

VX 0&

0201

2 RorRRRIW eqeqscsc

0201 ZorZZZIW eqeqscsc

Figure: 10

A

V

LV HV

1-P

hase A

uto

Auto

Tran

sform

er

Rated Voltage

Wattmeter

M L

COM V

PC

CC


Note: (i) Rated Current is applied at HV side.

(ii) This test is generally done on HV side (Why?) (iii) Why the position of ammeter and voltmeter is changed as compared to OC

test? Efficiency of transformer:

Efficiency at full load

Efficiency at x time of full load

0201

22 XorXXRZX eqeqeqeq

Figure: 11

A

V

HV LV

1-P

hase A

uto

Auto

Tran

sform

er

Rated Current

Wattmeter

M L

COM V

PC

CC

Shorted

d

100/

/

powerPI

powerPO

100/

/

lossespowerPO

powerPO

100/

/

lossCulossIronpowerPO

powerPO

100222

222

Ci PPCosIV

CosIV

22222

2

2 100 CosVARatedCosIVPWherePPP

P

Ci

PFLoadCosHerePxPxP

xP

Ci

22

2

2 )1(100


Condition for maximum efficiency

Differentiating equation (1) w. r. t “x” and putting dη/dx=0

Example 1: A 50 KVA, 4400/220 V transformer has R1 = 3.45 Ω , R2 = 0.009 Ω. The values of reactance are X1 = 5.2 Ω, X2 = 0.015 Ω. Calculate

a. Equivalent resistance as referred to primary b. Equivalent resistance as referred to secondary c. Equivalent reactance as referred to primary d. Equivalent reactance as referred to secondary e. Equivalent impedance referred to both side f. Total copper loss first using individual resistance of two windings and

secondly as referred to each sides. Solution:

a. Equivalent resistance as referred to primary

b. Equivalent resistance as referred to secondary

0100

222

2

2

2

2

Ci

CCi

PxPxP

xPPxPPPxPxP

dx

d

022

2

2 CCi xPPxPPPxPxP

iC PPx 2

C

i

P

Px

lossConstantlossVariableorlossIronlossCu

05.020

1

4400

220

1

2

1

2 V

V

N

NK

2

2101 / KRRR

201 20/1/009.045.3 R

AnsΩ7.05R01

21

2

02 RRKR

009.045.320/12

02 R

AnsΩ0.0176R02


c. Equivalent reactance as referred to primary

d. Equivalent reactance as referred to secondary

e. Equivalent impedance referred to both side

f. Total copper loss first using individual resistance of two windings and

secondly as referred to each sides.

Taking referred values

2

2101 / KXXX

201 20/1/015.02.5 X

AnsΩ11.2X 01

21

2

02 XXKX

015.02.520/12

02 X

AnsΩ0.028X 02

Ansj11.2Ω7.05Z01

AnsΩ j0.0280.0176Z02

2

2

21

2

1 RIRIlossCopper

AV

RatingVAI 3636.11

4400

1050 3

1

1

AV

RatingVAI 2727.227

220

1050 3

2

2

W009.02727.22745.33636.1122

AnsW910.38

01

2

1 RIlossCopper

W05.73636.112

AnsW910.38

02

2

2 RIlossCopper

W0176.02727.2272

ionapproximattodue ErrorAnsW909.09


AI 09.911000

10001001

Example 1: Following results were obtained on a 100 KVA, 11000/220 V, single phase transformer

(i) OC Test (LV Side) 220 V, 45 A, 2 KW (ii) SC Test (HV Side) 500 V, 9.09 A, 3 KW

Determine equivalent circuit parameter of transformer referred to low voltage side and efficiency at full load unity power factor. Solution: Equivalent circuit parameters:

From OC test: Pi = P0 = 2 KW, I0 = 45 A, V2 = 220 V

P0 = Pi = V2I0Cosϕ0

Ie = I0Cosϕ0=9.09 A Im = I0Sinϕ0=44.07 A

From SC test: PC = Wsc = 3 KW, Isc = 9.09 A, Vsc = 500 V

Efficiency at full load and unity power factor:

Note:

Full load HV side current Here SC Test is done on full load, so 3 KW is the full load cu loss.

202.002

0 IV

PCos i

AnsΩ24.20I

VR

e

20 AnsΩ5

I

VX&

m

20

01

2 RIW scsc

01ZIV scsc

AnsΩ36.31R01

AnsΩ55Z 01

2

01

2

0101 RZX

AnsΩ41.31X 01

1002

2

2

Ci PxPCosRatedVAx

CosRatedVAx

10031211001

110012

Ans95.24%η

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Single Phase Transformer© - Weebly