Ch 2 – Transformers

Chapter 2 – Transformers

BEE 4123Electrical Machines & Drives

Module Outlines

1. Why Transformers are Important to Modern Life?1. Why Transformers are Important to Modern Life?

2. Types and Construction of Transformers2. Types and Construction of Transformers

3. The Ideal Transformer 3. The Ideal Transformer

4. Theory of Operation of Real 1- Transformers 4. Theory of Operation of Real 1- Transformers

5. The Equivalent Circuit of a Transformer5. The Equivalent Circuit of a Transformer

6. Transformer Voltage Regulation and Efficiency 6. Transformer Voltage Regulation and Efficiency

7. Three-Phase Transformers 7. Three-Phase Transformers

Why Transformers are Important to Modern Life?

raise or lower the voltage or current in an ac circuit,

isolate circuits from one another, and

change the impedance of the load as seen by the source, thus enable load matching which results in maximum power transfer.

The transformer is one of the most useful electrical devices. It can:

Why Transformers are Important to Modern Life?

transmit electrical energy over great distances and to distribute it to the end-users.

The transformer gives to ac a feature lacking in dc power systems, i.e. efficiently change ac voltage from small amplitudes to large amplitudes and vice versa.

In general, the higher the voltage, the more insulation and switching costs, but less current used, so lower I2R loss and greater efficiency results.

Types & Construction of Transformers

Power Transformers, classified into 2 types of cores: Core Form Shell Form

Core Form Shell Form

Potential Transformer Sampling a high voltage and produce a low

secondary voltage proportionally. To handle very small current.

Current Transformer Provide much smaller secondary current than but

directly proportional to its primary current.

Types & Construction of Transformers

The Ideal Transformer

Assumptions made for an ideal transformer:

It is infinitely easy to set up a magnetic flux in the core. (Alternatively we could say that the core permeability is infinite; the reluctance of the core is zero; the magnetizing current required is zero).

No magnetic flux leaks out between the two windings (that is, all the flux links all the turns)

The winding resistance is zero, and There is no hysteresis or eddy-current power loss in the

core

iP(t)

vP(t) Ep Es

miS(t)

vS(t)NP NS2

mpp

NE

2

mpp

NE

2

mss

NE

2

mss

NE

Current, voltages and flux in an unloaded ideal transformer



Turn ratio, a

Power, P

p

s

s

p

s

p

I

I

V

V

N

Na

p

s

s

p

s

p

I

I

V

V

N

Na

pppin IVP cos pppin IVP cos

sssout IVP cos sssout IVP cos

pppin IVQ sin pppin IVQ sin

sssout IVQ sin sssout IVQ sin

outssppin SIVIVS outssppin SIVIVS


Transferring impedances through a transformer (load matching)

Vac Zload

a

VP VS

IP IS

Vac Zload

a

VP VS

IP IS

S

S

S

S

P

PP I

VIV

I

VZ 2a

a

a

S

S

S

S

P

PP I

VIV

I

VZ 2a

a

a

loadP ZZ 2a loadP ZZ 2a


Thévenin equivalents of transformer circuit

Vac a2ZloadVP

IP

Vac a2ZloadVP

IP

Vac/a ZloadVS

IS

Vac/a ZloadVS

IS

Equivalent circuit when secondary impedance is transferred to primary side and ideal transformer eliminated

Equivalent circuit when primary source is transferred to secondary side and ideal transformer eliminated

Example

A single-phase power system consists of a 480-V 60-Hz generator supplying a load Zload = 4 + j3 through a transmission line of impedance Zline = 0.18 + j0.24 . Answer the following questions:

a) If the power system is exactly as described above, what will the voltage at the load be? What will the transmission line losses be?

b) Suppose a 1:10 step up transformer is placed at the generator end of the transmission line and a 10:1 step down transformer is placed at the load end of the line. What will the load voltage be now? What will the transmission line losses be now?

Theory of Operation of Real Single-Phase Transformers Practical transformers differ from ideal

transformers in several ways: Resistance of the windings Flux leakage between the windings, leading to

inductive reactance effects The magnetizing current is not zero (but it may be as

little as 3% of the load currents). There is power loss in the iron core, due to hysteresis

and eddy current effects. Capacitive effects are present in high voltage

transformers, affecting their ability to withstand strikes of lightning.

Faraday’s law

Theory of Operation of Real Single-Phase Transformers

M

LP

LS

dt

deind

dtd

eind

N

N

dt

dNeind

dt

dNeind

Since,

Mutual flux Leakage flux

LPMP LPMP LSMS LSMS

Vp Vs

+

-

+

-

From the diagram, re-expressed Faraday’s law:


dt

dN

dt

dN

dt

dNtV

LPp

Mp

ppp

)(

dt

dN

dt

dN

dt

dNtV

LPp

Mp

ppp

)(

eP(t) eLP(t)

dt

dN

dt

dN

dt

dNtV

LSS

MS

SSS

)(

dt

dN

dt

dN

dt

dNtV

LSS

MS

SSS

)(

eS(t) eLS(t)

Due to mutual fluxDue to mutual flux

Current flows in primary circuit consists of: Magnetization current, iM to produce flux in the

transformer core Core-loss current, ih+e, to make up for hysteresis

and eddy current losses The average flux in the core,

If vP(t) = VM cos t,


dttvN PP

)(1 dttv

N PP

)(1

Wbsin cos1

tN

VdttV

N P

MM

P

Wbsin cos1

tN

VdttV

N P

MM

P

ehmex iii ehmex iii

excitation current

Current flows into dotted end


Ip

Vp Np NsLoad Vs

Is

Current flows out of dotted end

0 SSPPnet iNiN 0 SSPPnet iNiN

Net MMF in the core,

Nearly zero!

The Equivalent Circuit of A Transformer

The losses to be considered in any accurate model of transformer: Copper (I2R) losses – resistive heating losses in the

primary and secondary windings. Eddy current losses – resistive heating losses in the

core. Hysteresis losses – associated with the rearrangement

of the magnetic domains in core. Leakage flux – fluxes LP and LS that escape the core.


Equivalent circuit of a real transformer

IsIp

Vp Np

a

VsNs RcjXm

Im

jXp Rp jXs Rs Ist

IsIp

Vp Np

a

VsNs RcjXm

Im

jXp Rp jXs Rs Ist

Copper losses

Leakage fluxes

Core-loss current, ih+e


Transformer model referred to primary voltage level

Ip

Vp

jXp Rp

aVs

ja2Xs a2Rs

RcjXm

Im

Is/a

Ip

Vp

jXp Rp

aVs

ja2Xs a2Rs

RcjXm

Im

Is/a

Ip

Vp aVs

jXeqP ReqP

RcjXm

Is/a

ReqP = RP+a2RS

XeqP = XP+a2XS

Ip

Vp aVs

jXeqP ReqP

RcjXm

Is/a

ReqP = RP+a2RS

XeqP = XP+a2XS

Ip

Vp aVs

jXeqP ReqP Is/a

ReqP = RP+a2RS

XeqP = XP+a2XS

Ip

Vp aVs

jXeqP ReqP Is/a

ReqP = RP+a2RS

XeqP = XP+a2XS


Transformer model referred to secondary voltage level

aIp

Vp/a

jXp/a2 Rp/a

2

Vs

jXs Rs

Rc/a2jXm/a2

Im

Is

aIp

Vp/a

jXp/a2 Rp/a

2

Vs

jXs Rs

Rc/a2jXm/a2

Im

Is

aIp

Vp/a Vs

jXeqS ReqS

Rc/a2jXm/a2

Is

ReqS = RP/a2+RS

XeqS = XP/a2+XS

aIp

Vp/a Vs

jXeqS ReqS

Rc/a2jXm/a2

Is

ReqS = RP/a2+RS

XeqS = XP/a2+XS

aIp

Vp/a Vs

jXeqS ReqS Is

ReqS = RP/a2+RS

XeqS = XP/a2+XS

aIp

Vp/a Vs

jXeqS ReqS Is

ReqS = RP/a2+RS

XeqS = XP/a2+XS


Determining the values of components in the transformer model using, Open-circuit test – secondary winding is open-

circuited and primary winding connected to full-rated voltage.

Short-circuit test – secondary winding is short-circuited and primary winding connected to low voltage source. The voltage is adjusted until the current in the short-circuited windings is equal to its rated value. (Beware)


Total excitation admittance,

vP(t)

A

V

iP(t)Wattmeter

+

-

v(t)vP(t)

A

V

iP(t)Wattmeter

+

-

v(t)

Open-circuit test

MC

MCE

Xj

R

jBGY

11

MC

MCE

Xj

R

jBGY

11


The magnitude of the excitation admittance referred to primary circuit,

The open-circuit power factor (PF),

The admittance, YE (with its angle),

OC

OCE V

IY

OC

OCE V

IY

OCOC

OC

IV

PPF cos

OCOC

OC

IV

PPF cos

PFV

I

V

IY

OC

OC

OC

OCE

1cos PFV

I

V

IY

OC

OC

OC

OCE

1cos lagging


Since v(t) is low voltage, current flowing into excitation branch is negligible. Series impedances referred to primary,

vP(t)

A

V

iP(t)Wattmeter

+

-

v(t)

iS(t)

vP(t)

A

V

iP(t)Wattmeter

+

-

v(t)

iS(t)

Short-circuit test

SC

SCSE I

VZ

SC

SCSE I

VZ


Power factor of the current,

Overall impedance,

The series impedance ZSE is,

SCSC

SC

IV

PPF cos

SCSC

SC

IV

PPF cos

SC

SC

SC

SCSE I

V

I

VZ

0

SC

SC

SC

SCSE I

V

I

VZ

0

)()( 22SPSPSE

eqeqSE

XaXjRaRZ

jXRZ

)()( 22SPSPSE

eqeqSE

XaXjRaRZ

jXRZ

Transformer Voltage Regulation (VR) and Efficiency

Full-load voltage regulation is a quantity that compares VS, no load and VS, full load.

At no load, VS=VP/a

%100,

,, XV

VVVR

flS

flSnlS %100

,

,, XV

VVVR

flS

flSnlS

%100/

,

, XV

VaVVR

flS

flSP %100

/

,

, XV

VaVVR

flS

flSP

Transformer Voltage Regulation and Efficiency

Phasor diagram is use to determine VR.

aIp

Vp/a Vs

jXeqS ReqS Is

ReqS = RP/a2+RS

XeqS = XP/a2+XS

aIp

Vp/a Vs

jXeqS ReqS Is

ReqS = RP/a2+RS

XeqS = XP/a2+XS

SeqSeqSP IjXIRVa

V SeqSeqS

P IjXIRVa

V


VSReqIS

jXeqIS

VP/a

IS

VS ReqIS

jXeqIS

VP/a

IS

VSReqIS

jXeqISVP/a

IS

Lagging pf

VR=+ve

Unity pf

VR=+ve

Leading pf

VR=-ve


Transformer efficiency,

Since Pout = VSIScosS,

%100%100 XPP

PX

P

P

lossout

out

in

out

%100%100 X

PP

PX

P

P

lossout

out

in

out

%100cos

cosX

IVPP

IV

SScoreCu

SS

%100

cos

cosX

IVPP

IV

SScoreCu

SS

Example

A 15-kVA 2300/230-V transformer is to be tested to determine its excitation branch components, its series impedances and its voltage regulation. The following test data have been taken from the primary side of the transformer.

Open circuit test

Short-circuit test

Voc = 2300 V Vsc = 47 V

Ioc = 0.21 A Isc = 6.0 A

Poc = 50 W Psc = 160 W

a) Find the equivalent circuit of this transformer referred to the high voltage side

b) Find the equivalent circuit of this transformer referred to the low voltage side

c) Calculate the full-load voltage regulation at 0.8 power factor.

Documents

Ch 2 – Transformers