TRANSFORMER

C

vp

ip

P

Np

= Npip/S …….(1)where S is the reluctance

A transformer consists of a core made of laminated iron separated by insulators and a coil of Np turns wound around the core. This coil is supplied with an a.c voltage supply vp which then produces a current ip. Due to this current , a flux is produced which is given by an equation

dt

dN pp

v

)(dt

dNp

2p

p iv

Since current varied with time , also varied with time. A back electromagnetic force (e.m.f) will be produced which is given by the equation.

Substitute = Npip/S into the above equation , then

……(2)

……(3)

dt

2sindNpp

ftv m

= m sin 2ft

If ip is sinusoidal, the flux produced also sinusoidal, i.e

therefore

vp = NP2fmcos 2ft = NP2fmsin (2ft + /2)

The peak value = Vpm = NP2fm

and vp is leading the flux by /2.

mmpm

p f44.4πf2707.02

VV PP NNThe rms value

……(4)

……(5)

……(6)

……(7)

dt

dNss

v ……(8)

C

vp

ip

vs

is

LoadP S

N

s

N

p

primary secondary

When another coil is wound on the other side of the core with no of turns Ns , then the fux will induce the e.m.f vS as given by

From (2) and (8) we get

p

s

p

s

N

N

V

V

…….(9)

p

s

p

s

N

N

v

v

Or in rms value

With load , is will flow in the load, mmf at load will equal to the mmf at input, then (in rms value)

NpIp = NsIs

rearrange

p

s

s

p

N

N

I

I

…….(10)

…….(12)

…….(11)

For ideal transformer, the energy transferred will be the same as input. Thus power at primary is same power at secondary.

Pp = Ps

or IpVp = IsVs

NP : NS

VP VS

Primary Secondary

Symbol for ideal transformer

A 250 kVA,11000V/400V, 50Hz single –phase transformer has 80 turns on the secondary. Calculate(a) The appropriate values of the primary and secondary currents;(b) The approximate number of primary turns;(c) the maximum value of the flux.

(a) Full-load primary current

AI s 625400

10250

V

P -3

s

AV

PI

pp 7.22

11000

10250 3

Full-load secondary current

220011000400

80V

V

NN P

s

sP

p

s

p

s

N

N

V

V

mf N44.4E

mWbfm 5.22

508044.4

400

N4.44

E

s

s

(b) Number of primary turns

recall

(c) Maximum flux

recall

EP VP VSNSNP

IO

Ideal transformer with no load

Io is the no load current when the secondary is open

circuit. This current consists of Iom that is required to

produce the flux in the core (it is in phase) and Io1 is to

compensate the hysteresis and eddy current losses.

VP= emf of supply to the primary coil

and 90o leads the flux.

EP=emf induced in the primary coil and

same phase as VP.

VS=emf induced in the secondary coil

and 90o lags the flux.

Iom=magnetizing current to produce flux

and it is in phase with flux.

Io1=current to compensate the losses due

to hysteresis and eddy current.

Io=the no load current and given by

21

2I oomo II

o

oo I

I 1cos Power factor

Phasor diagram for no load transfomer

IOm

IOI

EP

VP

IO

VS

Transformer 1 Transformer 2

Low voltage generator

Low voltage load

High voltage line

Transformer converts the energy to high electrical voltage and transmits in the high voltage line. At the load, the high voltage energy is converted to low voltage. In this way, it will compensate the losses during transferring of the voltage energy.

A single-phase transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in core is 1.8m and the joints are equivalent to the airgap of 0.1mm. The value of the magnetic field strength for 1.1 T in the core is 400A/m, the corresponding core loss is 1.7W/kg at 50Hz and the density of the core is 7800kg/m3.If the maximum value of the flux is to be 1.1T when a p.d of 2200V at 50Hz is applied to the primary, calculate:(a) the cross-sectional area of the core;(b) the secondary voltage on no load;(c) the primary current and power factor on no load

AB

recall mf N44.4E

Wbfm 0206.0

5048044.4

2200

4.44N

E

p

p

20187.01.1

026.0A m

Bm

recall

(a)

(b)p

s

p

s

N

N

V

Vrecal

l

V4132200480

90V

V

NN P

s

sP

Practically 10% more allow for insulator

AC 7208.1400HC

AB

ao

aa 5.870001.0104

1.1H

7

A5.8075.87720

AN

HIom 682.1

480

5.807

INH

AIomom 19.1682.1707.0707.0I

(c) magnetomotive force (mmf) for the core is

mmf for the airgap is

Total mmf is

Maximum magnetizing current

recall

Rms value

30337.00187.08.1 mA

kgdensityVol 26378000337.0.

W4477.1263

AV

PI

po 203.0

2200

4471

laggingI

I

o

168.021.1

203.0cos 01

AIII oomo 21.1203.019.1 2221

2

Volume of core

mass of core

Core loss= loss rate x mass

Core-loss component of current

No load current

Power factor

E1 V1 V2

I1 I2

L(2)

Loaded transformer

L(2)= load with power factor of cos 2

V1 = emf at supply

E1=induced voltage at primary

V2=emf at load

E2=induced voltage at secondary

I1= primary current

I2=secondary current

Phasor diagram

Io

V 1 , E 1

- I 2 I 1

2

1

O

V 2 , E 2

I2

‘

A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no load current is 3A at a power factor 0.2 lagging when secondary current is 280A at a power factor of 0.8 lagging. Calculate the primary current and the power factor. Assume the voltage drop in the windings to be negligible.

A562801000

200I

N

NI S

P

SP

Recall Equation 12p

s

s

p

N

N

I

I

therefore

ooIII sinsinsin '2'211

ooIII coscoscos '2'211

6.0sin '2

98.0sin o2.0c oos

8.0c '2 os

A4.452.038.056

'50381o

AI 3.5854.364.45 221

laggingos o 78.050'38cosc 1

805.04.45

54.36tan 1

A54.3698.036.056

Io

V 1 , E 1

- I 2 I 1

2

1

O

V 2 , E 2

I2

‘

Solve for horizontal and vertical components

Power factor

Path of leakage

V2’

R1 L1 L2 R2

RCLm V2E1

I1’I1

Equivalent circuit of transformer

Flux leakage is due to secondary current that produce flux which encounter the primary flux. Some of the flux will link to its own windings and produce induction. This is represented by inductance L1. Similarly with the flux in secondary and represented by L2.

There are four main losses•Dissipated power by wire resistance of the windings (I2R)•Power due to hysteresis•Power due to eddy current•Power via flux leakages.

R1= wire resistance of primary windingsL1=inductance due to leakage flux in primary windingsRC=resistance represent power loss due to in hysteresis and eddy currentLm= inductance due to magnetizing current Iom

L2=inductance due to leakage flux in secondary windingsR2=wire resistance of secondary windings

E2

R1 L1 L2 R2

RCLm V2V1

I1’I1

Equivalent circuit of transformer

E1

Phasor diagram for a transformer on load

V1

E1

I2

-I2’

I1X1

V2

E2

I1

I0

I1R1

I1Z1

I2X2

I2Z2

I2R2

1

2

2

2

12

2

2

12'2

V

VX

N

NXX

2

2

12

2

1

22'2

V

VR

I

IRR

222'2

21 RIRI

We can replace R2 by inserting R2’ in the primary thus the equivalent resistance is

Giving us

Similarly

2

2

121'21

V

VRRRRRe

22eee XRZ

e

e

R

Xtan e

eee ZX sin

eee ZR cos

then

and

where

2

2

121'21

V

VXXXXX e

(b)

Transformer simplified circuit

Ze

V1 E1=V2’

I1 I2

To loadE2=V2

Phasor diagram of simplified equivalent circuit of transformer

I1R e

I1Z e

I1X e e

Magnified Ze portion

I1X e

I1

E 1=V 2'

V 1

I1Z e

I1R e

E 2 ,V 2

I2

e

Complete

o-2

voltageload-no

voltageload-fullvoltageload-noregulationVoltage

1

212 N

NVV

p

s

p

s

N

N

V

Vrecall

Secondary voltage on no-load

V2 is a secondary terminal voltage on full load

1

21

21

21

regulationVoltage

NN

V

VNN

VSubstitute we have

1

221 sincos

V

XRIregulationvoltageunitPer ee

centperV

NN

VV

1001

2

121

unitperV

NN

VV

1

2

121

regulationVoltage

Or

From phasor diagram can be proved that

1

21 cos

V

ZIregulationvoltageunitPer ee

Or

A 100kVA transformer has 400 turns on the primary and 800 turns on the secondary. The primary and secondary resistances are 0.3 and 0.01 respectively, and the corresponding leakage reactances are 1.1and 0.035 respectively. The supply voltage is 2200V. Calculate:

(a)The equivalent impedance referred to the primary circuit;(b)The voltage regulation and the secondary terminal voltage for

full load having a power factor of (i) 0.8 lagging and (ii) 0.8 leading.

(c)The percentage resistance and leakage reactance drops of the transformer

AV

PcurrentprimaryloadFull 45.45

2200

10100 3

unitper0336.0

2200

6.0975.18.055.045.45

975.1

80

400035.01.1

22

2

121 V

VXXX e

05.2975.155.0 2222eee XRZ

55.0

80

40001.03.0

22

2

121 V

VRRRe

(a)

(b) (i)

1

221 sincos

V

XRIregulationvoltageunitPer ee

centper36.3

V78.44678.6440

VN

NV

P

SP 440

400

802200

centper54.1

(b) (ii) power factor 0.8 leading

unitperregulationVoltage 0154.0

2200

6.0975.18.055.045.45

V8.140336.0440

V78.60154.0440

Sec. terminal voltage on no-load

The decreasing of full-load voltage is

V2.4258.14440 Therefore the secondary full-load voltage

The increasing of full-load voltage is

Therefore the secondary full-load voltage

1

e1

V

RI

2200

55.045.45 0.0114

0.0114%

Per unit

ltageprimary vo

primary toreferred

resistance equivalent

currentprimary

load-full

unitper drop Resistance

load-noon voltagesecondary

secondary toreferred

resistance equivalent

currentsecondary

load-full

unitper drop ResistanceOr

drop Resistance

Alternative

Full load secondary current

Recall Equation 12p

s

s

p

N

N

I

I

Equivalent resistance referred to secondary

AS 2.22745.4580

400I

N

NI P

S

P

022.0

400

803.001.0

22

1

212 N

NRRRe

VN

NV

P

SP 440

400

802200 Secondary voltage on no-load

440

022.02.227

load-noon voltagesecondary

secondary toreferred

resistance equivalent

currentsecondary

load-full

unitper drop Resistance

1.14% unit per 0.0114

2200

975.145.45

ltageprimary vo

primary toreferred

resistance leakage equivalent

currentprimary

load-full

unitper drop ractance Leakage

975.1

80

400035.01.1

22

2

121 V

VXXX e

%08.4unitper 0408.0

primary toreferred resistance leakage equivalent

AV

PcurrentprimaryloadFull 45.45

2200

10100 3