(i) 30% from Practical # 20%: Lab Reports # 10%: Lab Test(ii) 20% from Written # 15%: Test 1 & Test 2 # 5%: Tutorials, Attendance & Quizzes

TransformerDC MachinesAC Machines

DC and AC MetersDC and AC BridgesSensor and Transducers

Lab 0 – Lab Introduction

Lab 1 – Single Phase Transformer; Voltage and Current Ratio

Lab 2 – DC Series Motor

Lab 3 – Three Phase AC Induction Motor

Lab 4 – d’Arsonval Galvanometer

Lab 5 – The Basic Voltmeter Design

Lab 6 – The Wheatstone Bridge

DC SystemThomas A. EdisonFor incandescent

bulb

Low Voltage (120-V)

High Current

High Voltage

(12 to 25-kV)

Very Low Power Loss in Transmission Line

Power Distribution System

Huge Power Loss in Transmission Line

Step Up Voltage

(for Transmission over Long Distance)

110 to 1000-kV

Step Down Voltage

(for Final Use)

12 to 34.5-kV

AC System(Transformer)

Decrease Current

A transformer is a device that changes ac electric power at one voltage level to another voltage level through the action of a magnetic field.

Basic design of a transformer consists of one core and two coils.The core can be air, soft iron or steel.

The core types of a transformer:

1. Core form

2. Shell form

Core form is a construction of two-legged laminated core with the coils wound on two legs.

Shell form is a construction of three-legged laminated core with the coils wound on center legs.

Windings of shell form are arranged concentrically to minimize the flux leakages.

Fig.1-1 The transformer construction of (a) Core-form and (b) shell-form

(a)

(b)

Figure 1-2 Core construction of (a) hollow type and (b) shell-type.

(b)(a)

Air core Iron core

Figure 1-3 Schematic symbols of transformer

~ RL

Primary windings Secondary windings

Figure 1-4 Simple circuit of transformer

Part FunctionCore Provides a path for the magnetic lines of flux

Primary winding Receives energy from the ac source

Secondary winding Receives the energy from primary winding and delivers it to the load

Enclosure Protects the above components from dirt, moisture, and mechanical damage

Table 1-1: Principle Parts of a Transformer

Core

Primary Windings

Secondary Windings

~AC Source

RL

►

►

Core

Fig.1-5 Paths of mutual and leakage flux

The flux produced by a transformer is divided into two components:

Mutual flux: the portion of flux that goes from the primary winding to secondary winding through the core. It remains in the core and links both windings.

Leakage flux: the portion of flux that goes through one of the windings but not the other one. It returns through the air.

► ► ► ►

►►

►►

ΦLP

ΦLS

~

►►

N1 N2RL

►ΦM►

The total average primary flux:

where ФM = flux component linking both primary and secondary coils

ФLP = primary leakage flux

LPMP

The total average secondary flux:

LSMS

where ФLP = secondary leakage flux

(1-1)

(1-2)

Total power losses in a transformer are a combination of four types of losses:

1. Leakage flux: fluxes which escape the core and pass through only one of the transformer windings.

2. Copper losses: resistive heating losses in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings.

3. Eddy current losses: resistive heating loss in the core of transformer. They are proportional the square of the voltage applied to the transformer.

4. Hysteresis losses: losses caused by the rearrangement of the magnetic domains in the core.

By definition, an ideal transformer is a device that has no core loss and no leakage flux due to its core is infinitely permeable. Any flux produced by the primary is completely linked by the secondary, and vice versa.

No current is(t) flowing out of the secondary side.

Relationship between voltage applied to the primary side of the transformer, VP(t) and the voltage produced on the secondary side, VS(t) is equal to the ratio of the numbers of turns on the primary and on the secondary.

aN

N

tv

tv

S

P

S

P )(

)(

where a is defined to be the turns ratio of the transformer.

(1-3)Fig.1-6 Schematic symbol of an ideal transformer at no-load.

+

-

►iP(t) NP

+

-

►iS(t) = 0NS

►

►

VPvS(t)~vP(t) VS

There is current iS(t) flowing out of the secondary side. The relationship between the current

iP(t) flowing into the primary side of the transformer and the current iS(t) flowing out of the secondary side is

aN

N

ti

ti

P

S

S

P 1

)(

)(

Fig.1-7 Schematic symbol of an ideal transformer under load.

where, iP(t) = primary current (A)

iS(t) = secondary current (A)

(1-4) +

-

►iP(t) NP +

-

►iS(t)NS

►

►

VP VS~ RL

In terms of phasor quantities, the equation 1-2 and 1-3 can be written as follow:

aN

N

I

I

V

V

S

P

P

S

S

P

Phase angle of VP is the same as the angle of VS and the phase angle of IP is the same as the phase angle of IS.

(1-5)

Fig.1-7 Schematic symbol of an ideal transformer under load.

+

-

►iP(t) NP +

-

►iS(t)NS

►

►

VP VS~ RRL

The dots appearing at one end of each winding in Fig. 1-3 tell/”state” the polarity of the voltage and current on the secondary side of the transformer. This utilize the dot convention.

Dot convention:

1. If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are same with respect to the dots on each side of the core.

2. If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.

The power supplied by the ac generator to the transformer is given by the equation:

PPPin IVP cos

where θp is the angle between the primary voltage and the primary current.

The power supplied by the secondary transformer to its load is given by the equation:

SSSout IVP cos

where θs is the angle between the secondary voltage and the secondary current.

Cos θp = power factor of the primary windings and cos θs = power factor of the primary windings.

(1-6)

(1-7)

Fig.1-7 Schematic symbol of an ideal transformer under load.

+

-

►iP(t) NP +

-

►iS(t)NS

►

►

VP VS~ RRL

For an ideal transformer, θP = θS. Therefore, applying the turns-ratio equations gives VS = VP/a and IS = aIP, so the output power of an ideal transformer is equal to its input power:

(1-8)Pout = Pin

Fig.1-7 Schematic symbol of an ideal transformer under load.

+

-

►iP(t) NP +

-

►iS(t)NS

►

►

VP VS~ RRL

The impedance of a device or an element is defined as the ratio of the voltage across it to the current flowing through it.

Thus, the impedance of the load connected to the secondary winding of a transformer can be expressed as:

s

sL I

VZ

Fig.1-8 (a) Impedance scaling through a transformer. (b) Definition of impedance.

+–

IP

VAC

••ZL

IS

VSVP

IP

VAC

VP+– LL ZaZ 2'

(1-9)

The impedance of the primary circuit of the transformer is given by:

LP

PL Za

I

VZ 2' (1-10)

Fig.1-9

a. The actual circuit showing the actual voltages and currents on the secondary side.

b. Impedance Z2 is shifted to the primary side. Note the corresponding change in V2

and I2.

+–

I1

VAC

••Z1

V2I2 V4

V3

Z2

Z3

Z4I4

I3

+–

I1

VAC

Z1

aV2

I2/a

V4

V3

Z3

Z4 I4

I3••a2Z2

(a)

(b)

Objectives:

•To solve for all the voltages across each element.

•To calculate the current flowing each element.

aV3

+–

I1

VAC

Z1

aV2

I2/a

V4 Z4 I4

a2Z3

I3/a••

a2Z2

+–

I1

VAC

Z1

aV2

I2/a

aV4

I4/a

aV3

a2Z3

I3/aa2Z2

••a2Z4

I = 0 I = 0

Fig.1-9

c. Impedance Z3 is shifted to the primary side. Note the corresponding change in V3

and I3.

d. Impedance Z4 is shifted to the primary side. Note the corresponding change in V4 and I4. The currents in T are now zero.

(c)

(d)

+

–

I1

VAC

Z1

aV2

I2/a

aV4

I4/a

aV3

a2Z3

I3/a

a2Z2 a2Z4

Fi.1-9

• All the impedances are now transferred to the primary side and the transformer is no longer needed.

With shifting impedances from secondary to primary, we will find that:

The impedance values is multiplied by a2.

The real voltage across the transferred impedance increases by a factor a.

The real current flowing the transferred impedance decreases by a factor a.

+–

I1

VAC

••Z1

V2I2 V4

V3

Z2

Z3

Z4I4

I3

+–

VAC

•• aI1

Z1/a2

V2I2 V4

V3

Z2

Z3

Z4I4

I3

V1/a

Fig.1-10

a. The actual circuit showing the actual voltages and currents on the primary side.

b. Impedance Z1 is transferred to the secondary side. Note the corresponding change in V1

and I1.

(a)

(b)

V1

VAC

aI1

Z1/a2

V2I2 V4

V3

Z2

Z3

Z4I4

I3

V1/a

+–

••

VAC/a

aI1

Z1/a2

V2I2 V4

V3

Z2

Z3

Z4I4

I3

V1/a

+–

Fig.1-10

c. The source is transferred to the secondary side. Note the corresponding change in VAC. Note also that the currents in T are zero.

d. All the impedances and even the source are now on the secondary side. The transformer is no longer needed because the currents are zero.

(d)

(c)

With shifting impedances from primary to secondary, we will find that:

The impedance values is divided by a2.

The voltage across the transferred impedance is lower than the real voltage.

The real current flowing the transferred impedance is higher than the real current.

Fig.1-11 AC power system (a) without and (b) with transformer

+–

Iline jXR

+

–

Zload

IloadT1 T2

••••IG

VAC

Zline

1 2

3

Vload

Based on the use in AC power system, a transformer is differed in a various names:

1. Unit transformer:

a transformer connected to the output of a generator and used to step up the voltage to

transmission line (110+ kV).

2. Substation transformer:

a transformer at the end of the transmission line, which steps down the voltage from

transmission levels to distribution levels (from 2.3 to 34.5 kV).

3. Distribution transformer:

a transformer that takes the distribution voltage and steps down it to the final voltage at

which the power is actually used (110, 208, 220 V, etc.).

All these transformers are essentially same—the only difference among them is their intended

use.

For your first exercise, open the text book of “Electrical Machinery Fundamentals-Fourth Edition by Stephen J. Chapman, Pg. 73-76 (Example 2-1). You try to understand the example first and after that you just change the values of V, Zload and Zline and then you answer all questions like in the example. OK…

Remember….I’ll test you

Fig.1-12 The power system (a) without and (b) with transformer

+–

Iline jXR

+

–

Zload

Iload

Vload

T1 T2

••••IG

VAC

+

–

VloadVAC

Iload

Zload

Iline jXR

+–

IGZline

Zline

(a)

(b)

VAC

+–

Iline jXRT1

••IGZline

'loadZ

Equivalent circuit

VAC

jXR

+–

IG 'lineZ

Equivalent circuit

''loadZ

•

•

•

•

Fig.1-13 (a) System referred to the transmission system voltage level. (b) System and transmission line referred to the generator’s voltage level

ο

ο

VS

I0

VP

IP

VAC

+–

IS = 0RP RS

• •►

►m

NP NS

Fig. 1-14 Complete circuits of a real transformer for (a) open and (b) short circuits

VS

I0

VP

IP

VAC

+–

IS ≠ 0RP RS

• •►

►m

NP NS

(a)

(b)

jXP

jXP

jXS

jXS

To construct a complete circuit of a real transformer:

each imperfection (core losses) and permeability are taken in to

account;

their effects are included in the transformer model.

Imperfection and permeability can be represented by the values of resistance and reactance, respectively, by doing two tests:

open-circuit test

short-circuit test

I0

VP

IP = 0

ο

IS

VAC

+–

I0

Excitation branch

RC jXMImIf

ο

VS

• •►

►m

RP jXP RS jXS

Fig.1-15 An imperfect core represented by a reactance Xm and a resistance Rm

The resistance, RC shows the core losses.

Magnetizing reactance, XM represents a measure of the permeability of

the transformer core.

The current, Im represents the magnetizing current needed to create the

magnetic flux Φm.

IOC

VP

IP = 0

ο

IS

VAC

+–

IOC

Excitation branch

RC jXMImIf

ο

VS

• •►

►m

RP jXP RS jXS

Fig.1-15 An imperfect core represented by a resistance RC and a reactance XM

Because of RP and XP are too small in comparison to RC and XM to cause a significant voltage drop, so all the input voltage is dropped across the excitation branch.

Therefore, RC and XM can be obtained with measuring the magnitude of input voltage (VOC), input current (IOC) and input power (POC).

Wattmeter

ip (t)

v (t) V

A+

-

vp (t)

is (t)

~+-

Wattmeter

ip (t)

v (t)

A+

-

vp (t)

is (t)

V~+-

Fig. 1-16 (a) open-circuit test and (b) short-circuit test

(a)

(b)

Using the input voltage, input current and input power of the open-circuit test , power factor (PF) can be determined as follows:

OCOC

OC

IV

PPF cos

The power factor angle (θ) is given by

OCOC

OC

IV

P1cos

The power factor is always lagging for a real transformer, so the angle of the current always lags the angle of the voltage by θ degrees. Therefore, the excitation admittance YE is

)sin()cos( OC

OC

OC

OC

OC

OCE V

Ij

V

I

V

IY

(1-11)

(1-12)

(1-13)

Beside that, the excitation admittance can be also obtained using the formula:

MCMCE X

jR

jBGY11

where, GC = conductance of the core-loss resistor

BM = susceptance of the magnetizing inductor

RC = resistance in the transformer

XM = reactance in the transformer

Thus, we find

)cos(1

OC

OC

C V

I

R

)sin(1

OC

OC

M V

I

X

(1-14)

(1-15)

(1-16)

From the values of resistance (RC) and reactance (XM), we can determine the core losses and reactive power following equations:

C

pM R

VP

2

M

pM X

VQ

2

where, RM = resistance representing the core losses (Ω)

Xm = magnetizing reactance of the primary winding (Ω)

Vp = primary voltage (V)

PM = core losses (W)

QM = reactive power needed to set up the mutual flux Φm (W)

(1-17)

(1-18)

Fig. 1-17 Complete circuit of a real transformer for short-circuit

VS

I0

VP

IP

VAC

+–

IS ≠ 0RP RS

• •►

►m

NP NS

jXP jXS

Apart from the magnitude of input voltage (VSC), input current (ISC) and input power (PSC), resistance and inductance values can be obtained with constructing the equivalent circuit.

Fig. 1-18 Equivalent circuit of a real transformer for short-circuit

I0

VAC

+–

IS/aRP a2RSjXP ja2XS

Equivalent resistance and reactance are

speqp RaRR 2

speqp XaXX 2

(1-19)

(1-20)

The series impedance ZSE is equal to

eqeqSE jXRZ

)()( 22SPSpSE XaXjRaRZ

The power factor of the current is given by:

SCSC

SC

IV

PPF cos

and is lagging. The current angle is thus negative, and the overall impedance angle θ is positive:

SCSC

SC

IV

P1cos

(1-21)

(1-22)

(1-23)

Series impedance can be calculated as follows:

sincos0

SC

SC

SC

SCo

SC

SCo

SC

oSC

SE I

Vj

I

V

I

V

I

VZ

Therefore, we get

cosSC

SCeq I

VR

sinSC

SCeq I

VX

(1-24)

(1-25)

(1-26)

Per-unit (pu) system of measurements is an another method to solve the circuits containing transformers. In this method, the impedance transformations can be avoid. Thus, circuits containing many transformers can be solved easily with less chance of error.

In pu system, the voltages, currents, powers, impedances, and other electrical quantities are not measured in SI units (volts, amperes, watts, ohms, etc.) but measured in decimal fraction.

Any quantity can be expressed on a per-unit basis by the equation:

quantityofvaluebase

valueActualunitperQuantity

where actual value is a value in volts, amperes, ohms, etc.

(1-27)

To define a pu system, two quantities need to select. The ones usually selected are voltage and real or apparent power.

In a single-phase system, these relationships are

basebasebasebasebase IVSorQP ,,

base

base

base

basebase S

V

I

VZ

2

base

basebase V

IY

(1-28)

(1-29)

(1-30)

Voltage regulation is a quantity that compares the secondary voltage of a transformer at no load with the secondary voltage at full load. It is defined by the equation:

%100,

,, xV

VVVR

flS

flSnlS

where, VS,nl = secondary voltage at no-load (V)

VS,fl = secondary voltage at full-load (V)

(1-31)

Since at no load, VS = VP/a, the voltage regulation can also be expressed as

%100,

,/ xV

VVVR

flS

flSap (1-32)

If the transformer equivalent circuit is in the per-unit system, then voltage regulation can be expressed as

%100,,

,,, xV

VVVR

puflS

puflSpuP (1-33)

For an ideal transformer, VR = 0.

The efficiency of a devices (motors, generators as well as transformers) is defined by the equation:

%100xP

P

in

out

%100xPP

P

lossout

out

At full-load, a transformer has the total loss (Ploss) in Watts:

Cucoreloss PPP

(1-34)

(1-35)

(1-36)

where,

Pcore = input power in Watts on the open-circuit test = core loss

PCu = input power in Watts on the short-circuit test with full-load currents

= I2R loss on full load

and the output power (Pout) in Watts is given by:

sssout IVP cos

So, the efficiency of the transformer can be written as

%100cos

cosx

PPIV

IV

Cucoress

ss

(1-37)

(1-38)