Transformer and Machine

EE2901s Basic Electricity and Electronics

Subject lecturer: Dr. Zhao XU

Department of Electrical Engineering

Hong Kong Polytechnic University

Email: [email protected]

Room: CF632

Tel: 27666160

1 EE2901s Basic Electricity &

Electronics

EE2901s Basic Electricity & Electronics

2

Transformer Outline

(a) Mutual inductance and Coupling coefficient

(b) Dot convention

(c) Ideal transformer

(d) Practical transformer

(e) Use of Transformer


3

Magnetic Field of a Straight Wire

• The magnetic field lines around a

long wire which carries an electric

current form concentric circles

around the wire.

• If a wire is grasped with the thumb

of your right hand pointing in the

current direction, the fingers encircle

the wire in the direction of the

magnetic field.


4

Magnetic Field of Current Loop

Electric current in a circular

loop creates a magnetic field

which is more concentrated in

the center of the loop than

outside the loop.


5

Magnetic Field in a Solenoid

Stacking multiple loops

concentrates the field even

more into what is called a

solenoid.


6

Magnetic Field in a Bar Magnet

• The lines of magnetic field from a

bar magnet form closed lines.

• By convention, the field direction

is taken to be outward from the

North pole and in to the South pole

of the magnet.


7

Magnetic Field of The Earth

The earth's magnetic field is

similar to that of a bar

magnet tilted 11 degrees

from the spin axis of the

earth.


8

Electromagnets

• Electromagnets are usually in the form of

iron core solenoids.

• The ferromagnetic property of the iron

core causes the internal magnetic domains

of the iron to line up with the smaller

driving magnetic field produced by the

current in the solenoids.

• The effect is the multiplication of the

magnetic field by factors of tens to even

thousands.


9

Coupled Inductors and Ideal Transformer

When we place two inductors in close proximity or wound

them around a single ferromagnetic core experimental

evidence shows that a change in i1 will generate a voltage v2

across the open circuit.

Each inductor is said to be magnetically coupled.


10

Dot Convention

Current entering the (un)dotted terminal of one coil produces a voltage that is

sensed positively at the (un)dotted terminal of the second coil.


11

Physical Basis of the Dot Convention

From the consideration of direction

of magnetic flux produced by each

coil, dots may be placed either on

the upper terminal of each coil or

the lower terminal of each coil.


12

Coupled Inductors (1)

Faraday’s law:

v1 N1

d

dt1

N1

d

dt11 N1

d

dt12 (1)

where 1 11 12 is the total flux linking Coil 1


13


For linear magnetic materials, flux is proportional to current.

We can write flux Φ11 in terms of i1 and L1.

Similarly, we can write flux Φ12 in terms of i2 and M12 .

11 L1i1

N1

(2)

12 M12i2

N1

(3)


14


Substitute (2) and (3) into (1), we have:

Similarly, we can obtain v2 as:

In general, M12 = M21 = M Mutual inductance of

the coupled inductors

v1 L1

d

dti1 M12

d

dti2

v2 L2

d

dti2 M21

d

dti1


15

Circuit Symbol for Coupled Inductor

Phasor equations:

Each parameter depends

on the magnetic properties

and geometry of the core. Coupling coefficient

V1 jL1I1 jMI2

V2 jL2I2 jMI1

L1L2 M 2

k2

k M

L1L2


16

Example 1 (1)

The coupled inductors have parameters: L1 = 8 H, L2 = 6 H, and

M = 4 H. Find i1(t) and i2(t) if i2(0-) = 0 and S is closed at t = 0.

In Mesh 1, applying KVL, we have:

i1R1 v1 E

i1R1 L1

di1

dt M

di2

dt E

4i1 8di1

dt 4di2

dt 36


17

Example 1 (2)

0462

0

0

:have weKVL, applying 2,Mesh In

122

12222

222

dt

di

dt

dii

dt

diM

dt

diLRi

vRi


18

Example 2 (1)

Find v2(t).

212

211

128

816

IjIjV

IjIjV


19

Example 2 (3)

Solving for I2 and V2 gives:

I2 0.138 1410 A

V2 1.656 1410 V

or v2 1.656cos(4t 1410 ) V


20

Ideal Transformer (1)

If the coupled inductors have unity coupling, i.e., M2 = L1L2,

or the coupling coefficient k = 1, then

If the coupling coils of the inductors have infinite mutual and

self-inductances.

Turn ratio v1(t)

v2 (t)L1

MM

L2

a

i1(t)

i2 (t)

1

a


21

Ideal Transformer (2)

Two coupled coils are said to be ideal transformer if they

satisfy:

Circuit symbol

aN

N

ti

ti

aN

N

tv

tv

1

)(

)(

)(

)(

1

2

2

1

2

1

2

1


22

Properties of a Transformer

1. A transformer consists of two or more coils wound on

the same core.

2. The basic property of a transformer is to change AC

voltage.

3. A transformer cannot change direct current voltage.

4. A step down transformer has a lower AC output voltage

at its secondary winding than the AC input voltage to its

primary winding.

5. Conversely, a step up transformer has a higher secondary

than primary voltage.


23

A Transformer Circuit

A current in the primary

coil produces a magnetic

field, like a solenoid.

The magnetic field

couples around through

the secondary coil.

A voltage is induced

in the secondary coil.


24

Impedance Transformation

Zin v1

i1

Zin av2

i2

a

a2 v2

i2

a2 Z


25

Example 1

Find v1 and v2.

The equivalent resistance as

seen from the primary: Rpr

200

52 8

v1 Rpr

Rpr 2 vs

8

8 2 50cos1000t

40cos1000t V

v2 5 v1

200cos1000t V


26

Example 2

k 10

2

2

1pr

N

NR

Find the turn ratio (N1:N2) to achieve maximum transfer to

the 10 kΩ.

The equivalent resistance as

seen from the primary:

For max. power transfer, we

want Rpr = 50 Ω.

14

1000050

1

2

2

2

1

N

N

N

N


27

Example 3 (1)

Find Vc.

The equivalent resistance as seen from the primary:

The voltage across

the primary:

Zpr (5 j8) 32

45 j72

v1 Zpr

Zpr (30 j20) 80 500

45 j72

75 j52 80 500

84.9 580

91.26 34.730 80 500

74.42 73.270


28

Example 3 (2)

The voltage across the secondary:

The capacitor voltage Vc: VC j8

5 j8 v2

8900

9.43 580 24.80 73.270

21.0 74.730

v2 v1

3

74.42 73.270

3 24.80 73.270


29

Small transformer construction

a) Lamination, b) Iron core with winding

Iron core

Terminals

Secondary

winding

Insulation

Physical Structure

The transformer laminations (or coating of shellac, enamel or

varnish) is to insulate adjacent turns from shorts between

winding.

Core type-Reduce heat loss


30

Summary of Objectives

You should be able to:

1. Understand the dot convention and mutual inductance.

2. Analyze circuits with coupled inductors and transformer.

3. Understand the properties of ideal and linear

transformers.

4. Use transformer for impedance matching.

5. Use transformer for voltage level adjustment.


31

Machine Outline

(a) DC machine fundamental

(b) DC machine torque relation (dynamics)

(c) AC machine fundamental

(d) AC machine torque relation (dynamics)

DC Motor Experiment

F=BLi

L

F

Mercury - kind of inductor

F=BLI

Homemade DC Motor

Rotate

Two-Pole DC Motor

StatorRotor

Mech Power Elect Power Generator

Mode

Field Current to create

flux

Mech Power Elect Power Motor

Mode

Field Current to create

flux

DC Machines

DC Motor Types

Separately Excited (field excitation by a separate electric circuit)

Self-excited

Shunt

Series

Compound

Parallel+series

Connect in series

ic

L m

One turn on the armature

(As the armature rotates the wire ab is moved in the stator

field)

a

b

et

a’

b’

One Turn on the Armature

Torque on one Turn

c c cT f r BLi r Torque on ab

Lorentz Force on ab

c cf BLi

Magnetic Flux Density, B BA

Area under each pole, A 2 rLA

p

E

cc

p iT

p - no. of pole in the stator field

Per pole

New Tc = 2Tc (old)

Armature Torque

E

cc

p iT

Torque on one Turn

Torque on N turns cc

Np iT

Coil current ac

Ii

a

Number of parallel paths, a = 2 for Wave winding

a = p for Lap winding

)(a

pNK

IKT

a

aa

Armature Back emf

Voltage induced on one turn 2 2c me BLV BL r

Total induced voltage

a a mE K

Varying magnetic field-＞voltae induced

DC Motor Equations

a aT K IArmature Torque, Nm =

Armature back e.m.f. =

Armature constant =

a

pNK

pa

a a mE K

Mechanical Power =

m a a mW T K I

We use Pm more frequently than W

Armature Circuit

t a a aV E I R

Example 1-DC motor fundamental The speed at no load for a separately-excited DC motor is 1200 RPM

when the armature is connected to 115 Volts.

What armature voltage should we apply to change the speed

• to 1800 RPM?

• to 100 RPM?

Vt=Ea=Ka(fine)Wm

Example 1-DC motor fundamental

N

S

N

S

A 4-pole dc machine has a wave winding of 300 turns.

The flux per pole is 0.025 Wb.

The dc machine rotates at 1000 RPM.

a. Determine the generated voltage.

b. Determine the mechanical power (kW) generated if the current

through each turn is 25 A.

c. Solve (a) and (b) for lap-wound armature.

Current - Speed Relation

Ea

Ra

If

+

-

Ia

Field Winding

Vt

+

-

at a a a

dIV I R E L

dt

0

t aa

a

V EI

R

t a ma

a

V KI

R

Torque - Speed Relation

t a ma

a

V KI

R

Ea

Ra

If

+

-

Ia

Field

Winding

Vt

+

-

m

a

a

a

at

R

K

R

KVT

2)(

Torque-Speed Curve

5aK 4aK

3aK 2aK

m

a

a

a

at

R

K

R

KVT

2)(

Speed-Torque Curve

2aK 3aK 4aK 5aK

The torque-speed relation above can be rearranged as

Motor Acceleration

static

dT T J C

dt

where J is the combined inertia of the rotor and the load; C is the

viscous damping coefficient; and Tstatic

is the static load

AC machine fundamental Ladder under a magnet

Ladder tries to but can never catch the magnet.

The ladder will be slipping behind

1. Metal ladder moving left at v;

2. Voltage & current (Faraday’s law);

3. Lorentz force on Ladder as shown

Operation principle-Ac motor

Squirrel Cage

Squirrel Cage Motor

Squirrel-cage motors- most common type

Wound rotor motor- frequent start &stop

Synchronous motor –large size, low speed (<300rpm)

Single-Phase Motors- small applications

Stator field in 3-Phase induction motor

Two-pole three-phase stator

A

A

B

B

C

C

2s

fn

p

synchronous speed-speed at which the stator field rotates

Example : The synchronous speed for a three-phase two-pole AC motor in

Australia is 50 Hz.

f = frequency of the source Hz p=Number of stator

poles per phase

Slip ratio

s

s

n ns

n

n -rotor speed, ns-the synchronous speed

SNs=Ns-N=frequence of rotor current

Example- ns

Example : Calculate the synchronous speed of a four-pole induction motor excited by a

3f 50-Hz source.

What would be the synchronous speed for the same motor in USA where the line

frequency is 60 Hz?

Winding Connection Windings can be connected to 3-phase AC sources. Which can be connected in different ways

120)120cos(2

120)120cos(2

)cos(2

rmscrmsc

rmsbrmsb

rmsarmsa

VVortVv

VVortVv

VVortVv

c

a

bwye or star

c

a

b

Delta (D)

Y- Connection

3

Linephase

phase Line

VV

I I

c

a

bwye or star

Delta (D) Connection

c

a

b

3

Linephase

phase Line

II

V V

Similarly, the load can be connected in Y/Delta

Example- slip ratio

A 3- induction motor is running at almost 1500 RPM under no load

conditions. The speed is reduced to 1400 under full load.

Calculate

a. The number of poles?

b. The slip ratio at full load?

A. 2f/p=1500rpm=1500/60 p=4

B. S=Ns-N/Ns=0.05

Example- slip ratio

The rotor of this motor can be represented by

the following LR component

What is the rotor impedance under full load

conditions?

What is it under no-load conditions?

Full loadNs-N/Ns=sS*1400/60=FR2piFR=WRNo load S*1500/60=FRWR*LR=impedance

Mechanical Model

s s sP TStator Power

Under steady-state conditions

s mT T

Imagine, we turn the stator mechanically

Mechanical Power

m m mP T

R s m m s mP P P T

Rotor Losses

or

s m

R s m s sm

s

P T s T sP

s

*****a missing slide before this slidePls refer to the updated notesTalking about source ofbdifferent power loss

Power Relations

Rs

PP

s

1m R

sP P

s

The motor power depends on the slip. The power varies because the power factor varies.

Power Factor

0.2

0.4

0.6

0.8

ns

Rotar speed

Stator Equivalent Circuit

RsLS

RC LmE1

V1

I1

IC Im

Leakage inductance per phase

Stator magnetizing component

Induced voltage in stator phase

Core loss(small)

Statir loss(small)

Simplified Stator Circuit

LmE1

V1

I1

LS

Blocked Rotor

EBR

IR

LR

RR

First, consider the rotor being blocked

21

1

BR

NE E

N

This is at frequency s

Normally, the rotor would rotate at R

2s R

BR BR

s

E E sE

Rotor Equivalent Circuit

E =sE2 BR

IR

LR

RR

BRR

RS R

EI

Rj L

s

BRR

R s R

sEI

R js L

Divide by s

Rotor Equivalent Circuit

EBR

IR

X =j LR S R

R /sR

(1 )R RR

R RR s

s s

Power to the rotor

2 2 2 (1 )RR R R R R

R sI I R I R

s s

Air gap Power Rotor Loss MechanicalPower

Internal resistance

Combined Circuit

Ac machine

Simple Equivalent Circuit

1R

RE

VI

RjX

s

Total impedance

Example- AC machine model A 3-phase 450-V 50-Hz six-pole 10-hp squirrel cage induction motor has been tested to

determine the following circuit parameters:

R'R

=0.50 Ohms

XE

= XS

+ X'R

= 1.3 Ohms

Xm

= 10 Ohms

Assuming a D-connected stator, calculate the stator per-phase line current and the power

factor at a slip of 10%.

What should be XE

to have a power factor of at least 0.9 ?

Very easy

Equation of Motion

m load

dT T J C

dt

Motor Torque = Load Torque + Inertial Torque + Friction

Motor Torque (per phase)

2

1 R R

m

S

I RT

s

Air gap power

Motor Torque Motor Torque (per phase)

2

1 R R

m

S

I RT

s

Substitute the current

2

1

22

2

Rm

RSS R

R VT

RsX X

s

Wrong equationShd be square of the whole equation

Total Motor Torque

For three phases

2

1

22

2

3 Rm

RSS R

R VT

RsX X

s

Simple equivalent

circuit

Torque-Speed Curve

s=1 s=0

Max

Torque

smax

Find maximum torque

2

1

22

2

3 Rm

RSS R

R VT

RsX X

s

T-s equation

To find maximum, differentiate wrt s

2

1

22

2

10m R

s RE

dT R V d

ds ds Rs X

s

Maximum Torque

Rmax

E

R

Xs s

2

1max

2 s E

VT

X

Maximum Torque

The slip when the torque is maximum

This is what can be put onto the motor shaft by the motor. The motor will stall if the load torque exceeds

the motor torque.

Another T-s relation

2

1

22

2

3 Rm

RSS R

R VT

RsX X

s

Rmax

E

R

Xs s

2

1max

2 s E

VT

X

max

2 2

max max

2s sT

T s s

Example- AC motor dynamics


79

The stall torque (maximum torque) for a 3f four-pole 50-Hz motor is 40 N-m. The slip

ratio at that torque is 5%.

a. Calculate the speed at stall conditions.

b. If the slip at full load is 2%, determine the ratio between the start-up current and the

full-load current.

c. determine the operating speed if the full-load torque is slowly increased to 20 N-m

AC Motor Classes

Class A

Quick Start

Class B

Common Duty

Class C

Common Duty

Class D

Rugged

Low Efficiency

Documents

Transformer and Machine