single phase transformer construction & operation

UNIT- I

SINGLE PHASE TRANSFORMER

CONSTRUCTION & OPERATION

INTRODUCTION

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

There are two or more stationary electric circuits that are coupled magnetically.

It involves interchange of electric energy between two or more electric systems

Transformers provide much needed capability of changing the voltage and current levels easily.

They are used to step-up generator voltage to an appropriate voltage level for power transfer.

Stepping down the transmission voltage at various levels for distribution and power utilization.

WHAT IS TRANSFORMER

A transformer is a static piece of apparatus by means of which an electrical power is transferred from one alternating current circuit to another electrical circuit

There is no electrical contact between them

The desire change in voltage or current without any change in frequency

Symbolically the transformer denoted as

NOTE :

It works on the principle of mutual induction

STRUCTURE OF TRANSFORMER

The transformer two inductive coils ,these are electrical separated but linked through a common magnetic current circuit

These two coils have a high mutual induction

One of the two coils is connected of alternating voltage .this coil in which electrical energy is fed with the help of source called primary winding (P) shown in fig.

The other winding is connected to a load the electrical energy is transformed to this winding drawn out to the load .this winding is called secondary winding(S) shown in fig.

The primary and secondary coil wound on a ferromagnetic metal core

The function of the core is to transfer the changing magnetic flux from the primary coil to the secondary coil

The primary has N1 no of turns and the secondary has N2 no of turns the of turns plays major important role in the function of transformer

WORKING PRINCIPLE

The transformer works in the principle of mutual induction

When the alternating current flows in the primary coils, a changing magnetic flux is generated around the primary coil.

The changing magnetic flux is transferred to the secondary coil through the iron core

The changing magnetic flux is cut by the secondary coil, hence induces an e.m.f in the secondary coil

“The principle of mutual induction states that when the two coils are inductively coupled and if the current in coil change uniformly then the e.m.f. induced in the other coils. This e.m.f can drive a current when a closed path is provide to it.”

Now if load is connected to a secondary winding, this e.m.f drives a current through it

The magnitude of the output voltage can be controlled by the ratio of the no. of primary coil and secondary coil

The frequency of mutually induced e.m.f as same that of the alternating source which supplying to the primary winding b

CONSTRUCTION OF TRANSFORMER

These are two basic of transformer construction

Magnetic core

Windings or coils

Magnetic core

The core of transformer either square or rectangular type in size

It is further divided into two parts vertical and horizontal

The vertical portion on which coils are wounds called limb while horizontal portion is called yoke. these parts are

Core is made of laminated core type constructions, eddy current losses get minimize.

Generally high grade silicon steel laminations (0.3 to 0.5mm) are used

Transformer Classification

In terms of number of windings

Conventional transformer: two windings

Autotransformer: one winding

Others: more than two windings

In terms of number of phases

Single-phase transformer

Three-phase transformer

Depending on the voltage level at which the winding is operated

Step-up transformer: primary winding is a low voltage (LV) winding

Step-down transformer : primary winding is a high voltage (HV) winding

WINDING

Conducting material is used in the winding of the transformer

The coils are used are wound on the limbs and insulated from each other

The two different windings are wounds on two different limbs

The leakage flux increases which affects the performance and efficiency of transformer

To reduce the leakage flux it is necessary that the windings should be very close to each other to have high mutual induction

CORE TYPE CONSTRUCTION

In this one magnetic circuit and cylindrical coils are used

Normally L and T shaped laminations are used

Commonly primary winding would on one limb while secondary on the other but performance will be reduce

To get high performance it is necessary that other the two winding should be very close to each other

SHELL TYPE CONSTRUCTION

In this type two magnetic circuit are used

The winding is wound on central limbs

For the cell type each high voltage winding lie between two voltage portion sandwiching the high voltage winding

Sub division of windings reduces the leakage flux

Greater the number of sub division lesser the reactance

This type of construction is used for high voltage

EMF EQUATION

EMF EQUATION

E2/E1=K TRANSFORMATION RATIO

IDEAL TRANSFORMERS

No iron and copper losses No leakage fluxes A core of infinite magnetic permeability and of infinite

electrical resistivity Flux is confined to the core and winding resistances are

negligible

An ideal transformer is a lossless device with an input winding and an output winding. It has the following properties:

IDEAL TRANSFORMERS

An ideal transformer is a lossless device with an input winding and an output winding.

The relationships between the input voltage and the output voltage, and between the input current and the output current, are given by the following equations.

atiti

tv

tv

p

s

s

p In instantaneous quantities

fM

IDEAL TRANSFORMERS

Np: Number of turns on the primary winding Ns: Number of turns on the secondary winding vp(t): voltage applied to the primary side vs(t): voltage at the secondary side a: turns ratio ip(t): current flowing into the primary side is(t): current flowing into the secondary side

a

N

N

titi

tv

tv

s

p

p

s

s

p

aII

V

V

p

s

s

p In rms quantities

DERIVATION OF THE RELATIONSHIP

a

N

N

titi

tv

tv

aN

N

titi

tiNtiN

aN

N

tv

tvdt

tdN

dttd

tv

dttd

Ndt

tdtv

s

p

p

s

s

p

s

p

p

s

sspp

s

p

s

p

Ms

ss

Mp

pp

f

f

From Ampere’s law

Dividing (1) by (2)

…………….. (1)

…………….. (2)

………………......……….. (3)

…………………..……….. (4)

………………….. (5) Equating (3) and (4)

POWER IN AN IDEAL TRANSFORMER

outssppin

outsss

sppin

inpppp

sssout

sp

pppin

SIVIVS

QIVa

IaVIVQ

PIVaIa

VIVP

IVP

sincossin

coscoscos

cos

Real power P supplied to the transformer by the primary circuit

Real power coming out of the secondary circuit

Thus, the output power of an ideal transformer is equal to its input power.

The same relationship applies to reactive Q and apparent power S:

PHASOR DIAGRAM ON NO LOAD

THEORY OF OPERATION OF SINGLE-PHASE REAL TRANSFORMERS

Leakage flux: flux that goes through one of the transformer windings but not the other one Mutual flux: flux that remains in the core and links both windings

LSMS

LPMP

fff

fff

fp: total average primary flux fM : flux linking both primary and secondary windings fLP: primary leakage flux fS: total average secondary flux fLS: secondary leakage flux

THEORY OF OPERATION OF SINGLE-PHASE REAL TRANSFORMERS

MAGNETIZATION CURRENT

When an ac power source is connected to a transformer, a current flows in its primary circuit, even when the secondary circuit is open circuited. This current is the current required to produce flux in the ferromagnetic core and is called excitation current. It consists of two components:

1. The magnetization current Im, which is the current required to produce the flux in the transformer core

2. The core-loss current Ih+e, which is the current required to make up for hysteresis and eddy current losses

E1

THE MAGNETIZATION CURRENT IN A REAL TRANSFORMER

Io

Ic

IM

E1

f

o

When an ac power source is connected to the primary of a transformer, a current flows in its primary circuit, even when there is no current in the secondary. The transformer is said to be on no-load. If the secondary current is zero, the primary current should be zero too. However, when the transformer is on no-load, excitation current flows in the primary because of the core losses and the finite permeability of the core.

Magnetization current IM (current required to produce flux in the core)

Core-loss current Ih+e (current required to make up for hysteresis and eddy current losses)

Excitation current, Io

IM is proportional to the flux f Ic = Ih+e = Core loss/E1

IDEAL V/S PRACTICAL TRANSFORMER

A transformer is said to be ideal if it satisfies the following properties, but no transformer is ideal in practice.

It has no losses

Windings resistance are zero

There is no flux leakage

Small current is required to produce the magnetic field

While the practical transformer has windings resistance , some leakage flux and has lit bit losses

PHASOR DIAGRAM ON LOAD

RESISTIVE LOAD

INDUCTIVE LOAD

LOSSES IN TRANSFORMER

Copper losses : It is due to power wasted in the form of I2Rdue to

resistance of primary and secondary. The magnitude of copper losses depend upon the current flowing through these coils.

The iron losses depend on the supply voltage while the copper depend on the current .the losses are not dependent on the phase angle between current and voltage .hence the rating of the transformer is expressed as a product o f voltage and current called VA rating of transformer. It is not expressed in watts or kilowatts. Most of the timer, is rating is expressed in KVA.

Hysteresis loss :

During magnetization and demagnetization ,due to hysteresis effect some energy losses in the core called hysteresis loss

Eddy current loss :

The leakage magnetic flux generates the E.M.F in the core produces current is called of eddy current loss.

THE EXACT EQUIVALENT CIRCUIT OF A TRANSFORMER

Modeling the copper losses: resistive losses in the primary and secondary windings of the core, represented in the equivalent circuit by RP and RS.

Modeling the leakage fluxes: primary leakage flux is proportional to the primary current IP and secondary leakage flux is proportional to the secondary current IS, represented in the equivalent circuit by XP (=fLP/IP) and XS (=fLS/IS).

Modeling the core excitation: Im is proportional to the voltage applied to the core and lags the applied voltage by 90o. It is modeled by XM.

Modeling the core loss current: Ih+e is proportional to the voltage applied to the core and in phase with the applied voltage. It is modeled by RC.

THE EXACT EQUIVALENT CIRCUIT OF A TRANSFORMER

Although the previous equivalent circuit is an accurate model of a transformer, it is not a very useful one. To analyze practical circuits containing transformers, it is normally necessary to convert the entire circuit to an equivalent circuit at a single voltage level. Therefore, the equivalent circuit must be referred either to its primary side or to its secondary side in problem solutions.

Figure (a) is the equivalent circuit of the transformer referred to its primary side. Figure (b) is the equivalent circuit referred to its secondary side.

APPROXIMATE EQUIVALENT CIRCUITS OF A TRANSFORMER

APPLICATION AND USES

The transformer used in television and photocopy machines

The transmission and distribution of alternating power is possible by transformer

Simple camera flash uses fly back transformer

Signal and audio transformer are used couple in amplifier

Todays transformer is become an essential part of electrical engineering

ALL-DAY EFFICIENCY

-> is defined as the ratio of the energy (kilowatt-hours)

delivered by the transformer in a 24-hour period to the

energy input in the same period of time.

-> to determine the all-day efficiency, it is necessary to know

how the load varies from hour to hour during the day.

Example:

The transformer of example 18 operates with the following

loads during a 24-hr period: 1 ½ times rated kva, power

factor = 0.8, 1hr; 1 ¼ times rated kva, power factor = 0.8,

2hr; rated kva, power factor = 0.9, 3hr; ½ rated kva, power

factor = 1.0, 6hr; ¼ rated kva, power factor = 0.8; no-load,

4hr. Calculate the all-day efficiency.

Solution:

Energy output, kw-hr Energy losses, kw-hr

W1 = 1.5 x 5 x 0.8 x = 6.0 (1 ½)2 x 0.112 x 1 =

0.252

W2 = 1.25 x 0.8 x 2 = 10.0 (1 ½)2 x 0.112 x 2 =

0.350

W3 = 1 x 5 x 0.9 x 3 = 13.5 1 x 0.112 x 3 = 0.336

W6 = 0.5 x 5 x 1.0 x 6 = 15.0 (1/2)2 x 0.112 x 6 =

0.168

W8 = 0.25 x 5 x 1.0 x 8 = 10.0 (1/4)2 x 0.112 x 8 =

0.056

____

Total. . . . . . . . 54.5 Iron = 0.04 x 24 =

0.960

_____

Total. . . . . . . . .. . . . 2.122

All-day Efficiency = (1 – 2.122/54.5 + 2.122) x 100 = 96.25%

UNIT - II

TESTING OF

SINGLE PHASE TRANSFORMER

&

AUTOTRANSFORMER

Short Circuit

Open Circuit

Transformer

Short Circuit Test

Open Circuit Test

Conclusion

Source

A short circuit is an electrical circuit that allows

a current to travel along an unintended path,

often where essentially no (or a very low)

electrical impedance is encountered.

In circuit analysis a short circuit is a connection

between two nodes that forces them to be at the

same voltage.

In an ideal short circuit, this means there is no

resistance and no voltage drop across the short.

In real circuits, the result is a connection with

almost no resistance. In such a case, the current

that flows is limited by the rest of the circuit.

An electrical circuit is an "open circuit" if it lacks a

complete path between the terminals of its power

source; in other words, if no true "circuit" currently

exists, because for instance a power switch is turned

off.

The electrical opposite of a short circuit is an "open

circuit", which is an infinite resistance between two

nodes.

The open circuit test, or "no-load test", is one of the

methods used in electrical engineering to determine

the no load impedance in the excitation branch of a

transformer.

.

A transformer is a static electrical device that transfers energy by inductive coupling between its winding circuits.

A varying current in the primary winding creates a varying magnetic flux in the transformer's core and thus a varying magnetic flux through the secondary winding. This varying magnetic flux induces a varying electromotive force (emf) or voltage in the secondary winding.

In electrical engineering, two conductors are referred to as mutual-inductively coupled or magnetically coupled when they are configured such that change in current flow through one wire induces a voltage across the ends of the other wire through electromagnetic induction. The amount of inductive coupling between two conductors is measured by their mutual inductance.

These two tests are performed on a transformer to

determine:-(i) equivalent circuit of transformer

(ii) voltage regulation of transformer

(iii) efficiency of transformer.

The power required for these Open Circuit test and

Short Circuit test on transformer is equal to the

power loss occurring in the transformer.

A voltmeter, wattmeter, and an ammeter are

connected in LV side of the transformer as shown in

the figure below.

The voltage at rated frequency is applied to that LV

side with the help of a variac of variable ratio auto

transformer.

The HV side of the transformer is kept open. Now with

help of variac applied voltage is slowly increase until

the voltmeter gives reading equal to the rated voltage

of the LV side.

After reaching at rated LV side voltage, all three

instruments reading (Voltmeter, Ammeter and

Wattmeter readings) are recorded.

The ammeter reading gives the no load current Ie.

As no load current Ie is quite small compared to rated current of the transformer, the voltage drops due to this electric current then can be taken as negligible.

Since, voltmeter reading V can be considered equal to secondary induced voltage of the transformer. The input power during test is indicated by watt-meter reading.

As the transformer is open circuited, there is no output hence the input power here consists of core losses in transformer and copper loss in transformer during no load condition.

The no load current in the transformer is quite small compared to full load current so copper loss due to the small no load current can be neglected.

Hence the wattmeter reading can be taken as equal to core losses in transformer.

LET US CONSIDER WATTMETER READING IS PO.

These values are referred to the LV side of

transformer as because the test is conduced on LV

side of transformer. These values could easily be

referred to HV side by multiplying these values

with square of transformation ratio.

Therefore it is seen that the open circuit test on

transformer is used to determine core losses in

transformer and parameters of shunt branch of the

equivalent circuit of transformer.

OPEN CIRCUIT POWER FACTOR

ococ

oc

IV

PcosPF

Open circuit Power Factor Angle

ococ

oc

IV

Pcos 1

A voltmeter, wattmeter, and an ammeter are connected in

HV side of the transformer as shown in figure.

The voltage at rated frequency is applied to that HV side

with the help of a variac of variable ratio auto

transformer.

The LV side of the transformer is short circuited . Now with help of variac applied voltage is slowly increase until the ammeter gives reading equal to the rated current of the HV side

After reaching at rated current of HV side, all three instruments reading (Voltmeter, Ammeter and Watt-meter readings) are recorded

The ammeter reading gives the primary equivalent

of full load current IL.

As the voltage, applied for full load current in short circuit test on transformer, is quite small compared to rated primary voltage of the transformer, the core losses in transformer can be taken as negligible here.

Let’s, voltmeter reading is VSC . The input power

during test is indicated by watt-meter reading.

As the transformer is short circuited, there is no

output hence the input power here consists of copper

losses in transformer

Since, the applied voltage Vsc is short circuit voltage

in the transformer and hence it is quite small

compared to rated voltage so core loss due to the small

applied voltage can be neglected.

Hence the wattmeter reading can be taken as equal to

copper losses in transformer.

LET US CONSIDER WATTMETER READING IS PSC .

These values are referred to the HV side of

transformer as because the test is conduced on HV

side of transformer.

These values could easily be referred to LV side by

dividing these values with square of transformation

ratio.

Therefore it is seen that the Short Circuit test on

transformer is used to determine copper loss in

transformer at full load and parameters of

approximate equivalent circuit of transformer.

POWER FACTOR OF THE CURRENT

scsc

sc

IV

PcosPF

Angle Power Factor

scsc

sc

IV

Pcos 1

the open circuit test on transformer is used to

determine core losses in transformer and

parameters of shunt branch of the equivalent

circuit of transformer.

the Short Circuit test on transformer is used to

determine copper loss in transformer at full load

and parameters of approximate equivalent circuit

of transformer.

’

TRANSFORMERS

INTRODUCTION -

The Sumpner's test (back to back test) is the very

practical, convenient, efficient and minimum power

consumption test which is done without actual

loading to find regulation and efficiency of large

power transformer.

Two

Basic

Tests

Open

Circuit

Test

Short

Circuit

Test

Test to determine the iron loss/core losses. In this test the secondary of the transformer is kept open circuited.

Test to determine the copper losses. In this test the secondary of the transformer is kept short-circuited.

ALREADY TWO TESTS..!

WHY DO WE NEED SUMPNER’S TEST

TO DETERMINE THE SAME THINGS …..??

In O.C. test, there is no load on the transformer while in S.C.

circuit test only fractional load gets applied. In all O.C. and S.C.

tests, the loading conditions are absent. Hence the results are

inaccurate.

In open and short circuit test iron losses and copper losses are

determined separately but in actual use both losses occurs

simultaneously.

The temperature rise in the transformer is due to total loss that

occurs simultaneously during actual use, it cant be determined by

O.C and S.C tests.

SUMPNER’S TEST

Its a improved method of testing transformer efficiency & other parameters. This test gives the value of total loss accurately as it occurs when it is in actual use. Sumpner's test or back to back test requires two transformers. Both transformers are connected to supply in such a way that one transformer is loaded on another.

Circuit diagram of

Sumpner’s test :

OPERATIONS OF SUMPNER’S

TEST :

1st the primaries of the

two identical

transformers are

connected in parallel

across the supply V1 and

secondary's are

connected in series

opposition.

The switch S2 is opened

and switch S1 is closed.

Now the no load current

I0 flows in primaries and

I2 is zero.

Then switch S2 also

ADVANTAGES

The power required to carry out the test is small.

The transformers are tested at full-load

conditions.

As the test results gives the value of core and

copper losses occurring simultaneously so heat

run test can be conducted on two transformers.

The secondary current(i.e I2) can be varied to

any value using regulating transformer. Hence

we can determine the copper losses at full load

condition or at any load.

Drawbacks Only limitation is that two identical transformers

are required. In practice exact identical

transformers cannot be obtained and as two

transformers are required, the test is not

economical.

CONCLUSION:

In many electrical machines its seen that sumpner's test or back to back test is done in one or other way. As it is important to test every electrical machines at its rated capacity and its inconvenient for machines of large rating to actually fully load the equipment's and test. So for all electrical machines some form of back to back test becomes important.

PARALLEL OPERATIONS OF

TRANSFORMER

For supplying a load in excess of the rating of an existing

transformer, two or more transformers may be connected in

parallel with the existing transformer. The transformers are

connected in parallel when load on one of the transformers is

more than its capacity. The reliability is increased with parallel

operation than to have single larger unit. The cost associated

with maintaining the spares is less when two transformers are

connected in parallel.

• It is usually economical to install another

transformer in parallel instead of replacing the

existing transformer by a single larger unit. The

cost of a spare unit in the case of two parallel

transformers (of equal rating) is also lower than

that of a single large transformer. In addition, it

is preferable to have a parallel transformer for

the reason of reliability. With this at least half

the load can be supplied with one transformer

out of service.

CONDITION FOR PARALLEL OPERATION

OF TRANSFORMER

• For parallel connection of transformers, primary windings of the Transformers are connected to source bus-bars and secondary windings are connected to the load bus-bars.

• Various conditions that must be fulfilled for the successful parallel operation of transformers:

• Same voltage Ratio & Turns Ratio (both primary and secondary Voltage Rating is same).

• Same Percentage Impedance and X/R ratio.

• Same KVA ratings.

• Same Frequency rating.

• Same Polarity.

1.SAME VOLTAGE RATIO & TURNS RATIO

• If the transformers connected in parallel have slightly different voltage ratios, then due to the inequality of induced emfs in the secondary windings, a circulating current will flow in the loop formed by the secondarywindings under the no-load condition, which may be much greater than the normal no-load current.

• The current will be quite high as the leakage impedance is low. When the secondary windings are loaded, this circulating current will tend to produce unequal loading on the two transformers, and it may not be possible to take the full load from this group of two parallel transformers (one of the transformers may get overloaded).

• A small voltage difference may cause sufficiently high circulating current causing unnecessary extra I2R loss.

• The ratings of both primaries and secondary’s should be identical.

2. SAME PERCENTAGE IMPEDANCE AND

X/R RATIO

• If two transformers connected in parallel with similar per-unit impedances they will mostly share the load in the ration of their KVA ratings. Here Load is mostly equal because it is possible to have two transformers with equal per-unit impedances but different X/R ratios. In this case the line current will be less than the sum of the transformer currents and the combined capacity will be reduced accordingly.

• A difference in the ratio of the reactance value to resistance value of the per unit impedance results in a different phase angle of the currents carried by the two paralleled transformers; one transformer will be working with a higher power factor and the other with a lower power factor than that of the combined output. Hence, the real power will not be proportionally shared by the transformers.

• The current shared by two transformers running in parallel should be proportional to their MVA ratings.

• The current carried by these transformers are inversely proportional to their internal impedance.

• From the above two statements it can be said that impedance of transformers running in parallel are inversely proportional to their MVA ratings. In other words percentage impedance or per unit values of impedance should be identical for all the transformers run in parallel.

SAME POLARITY

• Polarity of transformer means the instantaneous direction of induced emf in secondary. If the instantaneous directions of induced secondary emf in two transformers are opposite to each other when same input power is fed to the both of the transformers, the transformers are said to be in opposite polarity.

• The transformers should be properly connected with regard to their polarity. If they are connected with incorrect polarities then the two emfs, induced in the secondary windings which are in parallel, will act together in the local secondary circuit and produce a short circuit.

• Polarity of all transformers run in parallel should be same otherwise huge circulating current flows in the transformer but no load will be fed from these transformers.

• If the instantaneous directions of induced secondary emf in two transformers are same when same input power is fed to the both of the transformers, the transformers are said to be in same polarity.

SAME PHASE SEQUENCE

The phase sequence of line voltages of both the

transformers must be identical for parallel

operation of three-phase transformers. If the

phase sequence is an incorrect, in every cycle

each pair of phases will get short-circuited.

This condition must be strictly followed for

parallel operation of transformers.

SAME KVA RATINGS

• If two or more transformer is connected in parallel, then load sharing % between them is according to their rating. If all are of same rating, they will share equal loads

• Transformers of unequal kVA ratings will share a load practically (but not exactly) in proportion to their ratings, providing that the voltage ratios are identical and the percentage impedances (at their own kVA rating) are identical, or very nearly so in these cases a total of than 90% of the sum of the two ratings is normally available.

• It is recommended that transformers, the kVA ratings of which differ by more than 2:1, should not be operated permanently in parallel.

• Transformers having different kva ratings may operate in parallel, with load division such that each transformer carries its proportionate share of the total load To achieve accurate load division, it is necessary that the transformers be wound with

• the same turns ratio, and that the percent impedance of all transformers be equal, when each percentage is expressed on the kva base of its respective transformer. It is also necessary that the ratio of resistance to reactance in all transformers be equal. For satisfactory operation the circulating current for any combinations of ratios and impedances probably should not exceed ten percent of the full-load rated current of the smaller unit.

OTHER NECESSARY CONDITION FOR

PARALLEL OPERATION

• All parallel units must be supplied from the same network.

• Secondary cabling from the transformers to the point of paralling has approximately equal length and characteristics.

• Voltage difference between corresponding phase must not exceed 0.4%

• When the transformers are operated in parallel, the fault current would be very high on the secondary side. Supposing percentage impedance of one transformer is say 6.25 %, the short circuit MVA would be 25.6 MVA and short circuit current would be 35 kA.

• If the transformers are of same rating and same percentage impedance, then the downstream short circuit current would be 3 times (since 3 transformers are in Parallel) approximately 105 kA. This means all the devices like ACBs, MCCBs, switch boards should withstand the short-circuit current of 105 kA. This is the maximum current. This current will get reduced depending on the location of the switch boards, cables and cable length etc. However this aspect has to be taken into consideration.

• There should be Directional relays on the secondary side of the transformers.

• The percent impedance of one transformer must be between 92.5% and 107.5% of the other. Otherwise, circulating currents between the two transformers would be excessive.

ADVANTAGES OF TRANSFORMER

PARALLEL OPERATION

1) Maximize electrical system efficiency: • Generally electrical power transformer gives the maximum

efficiency at full load. If we run numbers of transformers in parallel, we can switch on only those transformers which will give the total demand by running nearer to its full load rating for that time.

• When load increases we can switch no one by one other transformer connected in parallel to fulfil the total demand. In this way we can run the system with maximum efficiency.

2) Maximize electrical system availability: • If numbers of transformers run in parallel we can take

shutdown any one of them for maintenance purpose. Other parallel transformers in system will serve the load without total interruption of power.

3) Maximize power system reliability:

• If nay one of the transformers run in parallel, is tripped due to fault other parallel transformers is the system will share the load hence power supply may not be interrupted if the shared loads do not make other transformers over loaded.

4) Maximize electrical system flexibility:

• There is a chance of increasing or decreasing future demand of power system. If it is predicted that power demand will be increased in future, there must be a provision of connecting transformers in system in parallel to fulfil the extra demand because it is not economical from business point of view to install a bigger rated single transformer by forecasting the increased future demand as it is unnecessary investment of money.

• Again if future demand is decreased, transformers running in parallel can be removed from system to balance the capital investment and its return.

DISADVANTAGES OF TRANSFORMER

PARALLEL OPERATION

Increasing short-circuit currents that increase necessary

breaker capacity.

The risk of circulating currents running from one transformer to another Transformer. Circulating currents that diminish load capability and increased losses.

The bus ratings could be too high.

Paralleling transformers reduces the transformer impedance significantly, i.e. the parallel transformers may have very low impedance, which creates the high short circuit currents. Therefore, some current limiters are needed, e.g. reactors, fuses, high impedance buses, etc

CONCLUSION

Loading considerations for paralleling transformers are simple unless kVA, percent impedances, or ratios are different. When paralleled transformer turn ratios and percent impedances are the same, equal load division will exist on each transformer. When paralleled transformer kVA ratings are the same, but the percent impedances are different, then unequal load division will occur.

The same is true for unequal percent impedances and unequal kVA. Circulating currents only exist if the turn ratios do not match on each transformer. The magnitude of the circulating currents will also depend on the X/R ratios of the transformers. Delta-delta to delta-wye transformer paralleling should not be attempted.

AUTO- TRANSFORMER

WHAT IS TRANSFORMER ??

A transformer is a

static device which is

use to convert high

alternatic voltage to a

low alternatic voltage

and vice versa,

keeping the frequency

same.

PRINCIPLE OF OPERATION

Transformer works on the

principle of mutual

induction of two coils.

When current in the primary

coil is changed the flux

linked to the secondary coil

also changes. Consequently

an EMF is induced in the

secondary coil.

WHAT IS INDUCTION LAW ??

Faraday’s law states that:

Vs=Ns.dΦ/dt where VS is the instantaneous

secondry winding voltage.

NS is the number of turns in the

secondary coil.

CONSTRUCTION OF TRANSFORMER

Mainly Transformers have two types of construction….

• CORE type construction

• SHELL type construction

A wide variety of transformer designs are used

for different applications. Auto-transformer

Poly-phase transformer

Instrument transformers

AUTO-TRANSFORMERS

• An autotransformer (sometimes called auto step down transformer)is an electrical transformer with only one winding. The "auto" (Greek for "self") prefix refers to the single coil acting on itself and not to any kind of automatic mechanism.

http://en.wikipedia.org/wiki/Transformer

http://en.wikipedia.org/wiki/Coil

http://en.wiktionary.org/wiki/auto-

http://en.wikipedia.org/wiki/Automation

• N1=primary turn(1-3) • N2=secondary turn(2-3) • I1=primary current • I2=secondary current • V1=primary voltage • V2=secondary votage

THEORY OF AUTOTRANSFORMER

From the above fig. We get

(I/P=O/P)

OUT PUT The primary and secondary windings of an autotransformer are Connected magnetically as well as electrically. So the power transferred primary to secondary inductively as well as conductively.

COPPER SAVING IN AUTO TRANSFORMER • The same output and voltage transformation

ratio an autotransformer requires less copper than the 2-winding transformer

TYPES OF AUTOTRANSFORMER

Step UP Transformer :

A transformer in which Ns>Np is called a step up transformer. A step up transformer is a transformer which converts low alternatic voltage to high alternatic voltage.

Step DOWNTransformer :

A transformer in which Np>Ns is called a step down transformer. A step down transformer is a transformer which converts high alternating voltage to low alternating voltage.

CONVERSION OF 2-WINDING TRANSFORMER INTO

AUTOTRANS FORMER

ADDITIVE POLARITY (STEP-UP)

SUBSTRACTIVE POLARITY (STEP DOWN)

In this case common current flow towards the common terminal

ADDITIVE POLARITY

SUBSTRACTIVE POLARITY

In this case common current flow away from common terminal

ADVANTAGES • An autotransformer requires less Cu than a two-winding

transformer of similar rating.

• An autotransformer operates at a higher efficiency than a two-winding transformer of similar rating.

• An autotransformer has better voltage regulation than a two-windingtransformer of the same rating.

• An autotransformer has smaller size than a two-winding transformer of the same rating.

• An autotransformer requires smaller exciting current than a two-windingtransformer of the same rating.

DISADVANTAGES

• There is a direct connection between the primary and secondary. Therefore,

the output is no longer d.c. isolated from the input.

• An autotransformer is not safe for stepping down a high voltage to a low voltage. As an illustration.

If an open circuit develops in the common portion of the winding, then full-primary voltage will appear across the load. In such a case, any one coming in contact with the secondary is subjected to high voltage. This could be dangerous to both the persons and equipment. For this reason, autotransformers are prohibited for general use.

• The short-circuit current is much larger than for the two-winding transformer of the same rating. So that a short-circuited secondary causes part of the primary also to be short-circuited. This reduces the effective resistance and reactance.

APPLICATION

• Autotransformers are used to compensate for voltage drops in transmission and distribution lines. When used for this purpose, they are known as booster transformers.

• Autotransformers are used for reducing the voltage supplied to a.c.motors during the starting period.

• Autotransformers are used for continuously variable supply.

• On long rural power distribution lines, special autotransformers with automatic tap-changing equipment are inserted as voltage regulators, so that customers at the far end of the line receive the same average voltage as those closer to the source. The variable ratio of the autotransformer compensates for the voltage drop along the line.

• In control equipment for 1-phase and 3-phase electrical locomotives.

http://en.wikipedia.org/wiki/Voltage_drop

LIMITATION Because it requires both fewer windings and a smaller core,

an autotransformer for power applications is typically lighter and less costly than a two-winding transformer, up to a voltage ratio of about 3:1; beyond that range, a two-winding transformer is usually more economical.

Like multiple-winding transformers, autotransformers operate on time-varying magnetic fields and so will not function with DC.

A failure of the insulation of the windings of an autotransformer can result in full input voltage applied to the output. Also, a break in the part of the winding that is used as both primary and secondary will result in the transformer acting as an inductor in series with the load .

http://en.wikipedia.org/wiki/Magnetic_field

http://en.wikipedia.org/wiki/Voltage

CONCLUSION

TO ABOVE STUDY WE CONCLUDE THAT

AUTOTRANSFORMER HAVE LESS AMOUNT OF CU.

LOSS REQUIRED.HIGH EFFICIENCY,POSSIBLE TO

GET SMOOTH AND CONTINUOES VARIATION

VOLTAGE.

UNIT- III

POLYPHASE TRANSFORMERS

INTRODUCTION

• The transformers may be inherently 3-

phase,

having three primary windings and three

secondary windings mounted on a 3-legged

core.

• The same result can be achieved by using

three single-phase transformers connected

together to form a 3-phase transformer bank.

1. BASIC PROPERTIES OF 3-PHASE TRANSFORMER BANK

• When three single-phase transformers are used to transform a 3-

phase voltage, the windings can be connected in several ways. the

ratio of the 3-phase input voltage to the 3-phase output voltage

depends not only upon the turns ratio of the transformers, but also

upon how they are connected.

• A 3-phase transformer bank can also produce a phase shift between

the 3-phase input voltage and the 3-phase output voltage. The amount

of phase shift depends upon

- the turns ratio of the transformers

- how the primaries and secondaries are interconnected

BASIC PROPERTIES OF 3-PHASE

TRANSFORMER BANK

• The phase shift feature enables us to change the number of

phases a 3-phase system can be converted into a 2-phase, a 5-

phase, a 6-phase, or a 12-phase system by an appropriate choice of

single-phase transformers and interconnections.

BASIC PROPERTIES OF 3-PHASE

TRANSFORMER BANK

• The basic behavior of balanced 3-phase transformer banks is

basedon the following simplifying assumptions:

(1) The exciting currents are negligible.

(2) The transformer impedances, due to the resistance

and leakage reactance of the windings, are negligible.

(3) The total apparent input power to the transformer

bank is equal to the total apparent output

power.

2. DELTA-DELTA CONNECTION

Fig.1 Delta-delta connection of three

single-phase transformers. The incoming

lines (source) are A, B, C and the outgoing

lines (load) are 1, 2, 3.

DELTA-DELTA CONNECTION

Fig.2 Schematic diagram of a delta-delta connection

and associated phasor diagram.


• In such a delta-delta connection, the voltages between the respective

incoming and outgoing transmission lines are in phase.

• If a balanced load is connected to lines 1-2-3, the resulting line

currents are equal in magnitude. This produces balanced line currents

in the incoming lines A-B-C.

• The power rating of the transformer bank is three times the rating

of a single transformer.


• Example 1

3. DELTA-WYE CONNECTION

Fig.3 Delta-wye connection of three single-phase

transformers.

DELTA-WYE CONNECTION

Fig.4 Schematic diagram of a delta- wye connection and associated phasor diagram.


• The voltage across each primary winding is equal to the incoming

line voltage.

• However, the outgoing line voltage is 3 times the secondary voltage

across each transformer.

• The line currents in phases A, B and C are 3 times the currents in

the primary windings.

•A delta-wye connection produces a 30° phase shift between the line

voltages of the incoming and outgoing transmission lines


• If the outgoing line feeds an isolated group of loads, the

phase shift creates no problem. But, if the outgoing line has to be

connected in parallel with a line coming from another source, the 30°

shift may make such a parallel connection impossible, even if the line

voltages are otherwise identical.

• One of the important advantages of the wye connection is that it

reduces the amount of insulation needed inside the transformer. The

HV winding has to be insulated for only 1/3, or 58 percent of the line

voltage.


• Example 2


Fig.5

4. WYE-DELTA CONNECTION

• The currents and voltages in a wye-delta connection are identical to

those in the delta-wye connection. The primary and secondary

connections are simply interchanged.

• There results a 30° phase shift between the voltages of the incoming

and outgoing lines.

5. WYE-WYE CONNECTION

• When transformers are connected in wye-wye, special

precautions have to be taken to prevent severe distortion of the line-

to-neutral voltages.

(1) connect the neutral of the primary to the neutral of the source,

usually by way of the ground

Fig.6 Wye-wye connection with neutral of the primary connected to the neutral of the source.

WYE-WYE CONNECTION

(2) provide each transformer with a third winding,

called tertiary winding.

Fig.7 Wye-wye connection using a tertiary winding.

WYE-WYE CONNECTION

• Note that there is no phase shift between the incoming

and outgoing transmission line voltages of

a wye-wye connected transformer.

6. OPEN-DELTA CONNECTION

• It is possible to transform the voltage of a 3-phase system by

using only 2 transformers, connected in open-delta.

• The open-delta arrangement is identical to a delta-delta

connection, except that one transformer is absent.

• The open-delta connection is seldom used because the load

capacity of the transformer bank is only 86.6 percent

of the installed transformer capacity.

OPEN-DELTA CONNECTION

• The open-delta connection is mainly used in emergency

situations. Thus, if three transformers are connected in delta-delta

and one of them becomes defective and has to be removed, it is

possible to feed the load on a temporary basis with the two

remaining transformers.

Fig.8a Open-delta connection.


• Example 3


• The current Is in lines 1, 2, 3 cannot, therefore, exceed 250

A (Fig.8b). Consequently, the maximum load that the

transformers can carry is

Fig.8b Open-delta connection Associated schematic and phasor diagram.

THREE-PHASE TRANSFORMERS

• A transformer bank composed of three single-phase transformers

may be replaced by one 3-phase transformer.

• For a given total capacity, a 3-phase transformer is always

smaller and cheaper than three single-phase transformers.

• Nevertheless, single-phase transformers are sometimes

preferred, particularly when a replacement unit is essential.

THREE-PHASE TO 2-PHASE TRANSFORMATION

• The voltages in a 2-phase system are equal but displaced

from each other by 90°.

• There are several ways to create a 2-phase system from a 3-

phase source.

(1) Use a single-phase autotransformer having taps at 50

percent and 86.6 percent.

(2) Scott connection.

THREE-PHASE TO 2-PHASE TRANSFORMATION (ตอ)

(1) Use a single-phase autotransformer having taps at 50 percent

and 86.6 percent.

Fig.15

THREE-PHASE TO 2-PHASE TRANSFORMATION (ตอ)

• The ratio of transformation (3-phase voltage to

2-phase voltage) is fixed and given by EAB/EAT =

100/86.6 = 1.15.


• The Scott connection has the advantage of isolating the 3-phase

and 2-phase systems and providing any desired voltage ratio

between them.

• Except for servomotor applications, 2-phase systems are seldom

encountered today.


(2) Scott

connection : It

consists of two

identical single-

phase

transformers, the

one having a 50

percent tap and

the other an 86.6

percent tap on the

primary winding.

Fig.16


Example 5: A 2-phase, 7.5 kW (10 hp), 240 V, 60 Hz motor has an

efficiency of 0.83 and a power factor of 0.80. It is to be fed from a

600 V, 3-phase line using a Scott-connected transformer bank

(Figure 16c).

Calculate

(a) The apparent power drawn by the motor

(b) The current in each 2-phase line

(c) The current in each 3-phase line


Fig.16c

Scott Connection

OBJECTIVES

On The Completion Of This Period

You Will Be Able To Know

The Scott Connection

Advantage And Disadvantages

Application Of Scott Connection

SCOTT CONNECTIONS

Scott Connections:-a Scott- T-transformer Calso

Called A Scott Connections Is A Type Of Circuit

Used To Derive Three-phase(3- ) Current From A

Two Phase Source Or Vice-verse

Can Be Used For 3 Phase To Two Phase

Connections Also

3-P TO 3-P COVERSION

Scott-t-t Connection Consists Of A Center – Tapped 1:1 Ration Main Transformer And An86.6%(0.5 *1.73)ratio Teaser Trensformer

Cont……

One End Of The Teaser Transformer Is Joined To The

Center Tap Of The Main Transformer

Current In Teaser Transformer Is In Phase With

Voltage

Current Leads The Voltage By 30in Half Portion Of

Main Transformer And Lays The Voltage By 30 In The

Rest Of The Winding

3-PHASE TO 2-PHASE CONVERSION

This Conversion Is Required To Supply Two Phase Furnaces

To Link Two –Phase Circait With 3-phase System And Also

To Supply A 3-phase Appartus From A 2-phase Supply

Source

The Connection reqires 2 transformer Whice May Be

Identical But Having Suitable Tapings


CONNECTION DIAGRAMS

If The Secondary Of Both The Transformes Have

The Some Unmbe Of Turns ,Then Secondary Voltage

Will Be Equal In Magnitade Thus Resuiting In A

Symmetrical 2-phase ,3-wire Sysem

conn…..

The No. Of Turns Between A And D Should Be Also

(1.73/2) N, For Making Voltage Turn The Same In

Both Primaries

Then For Secondary Having Equal Turns The

Secondary Terminal Voltages Will Be Equal In

Magnitude Although In Phase Quadrature

ADVANTAGES OF SCOTT CONNECTIVE

The Scott Connection Evenly Distributes a

Balanced Load Between The Phase Of The

Source

Trans Formers Can Deliver 92.8% Of Their

Capacity

There Is A Cost Saving Due to the 2-coil T

Connection to the traditional three –Coil Primary

to 3 Coil Secondary Transformer

With the help Of two transformer 3-phase Supply

Can Be Maintained

DISADVANTAGES OF SCOTT CONNECTION

The Full Rating Of The Transformers Is Not

Utilized

The Teaser Trans Former Operates At Only

0.866 Of Its Rated Voltage

UNIT - IV

POLYPHASE INDUCTION MOTORS

CONTENTS Introduction Construction Parts of induction motor Rotor construction Rotating magnetic field(RMF) Principle of operation Equivalent circuit Power losses Power flow in induction motor Torque speed characteristics Speed control Advantage Application.

INTRODUCTION Induction Motors transform electrical energy

into mechanical energy. simple design, Cheap,robust, low-price, easy

maintenance. wide range of power ratings: fraction

horsepower to 10 MW run essentially as constant speed from no-

load to full load. Its speed depends on the frequency of the

power source.

CONSTRUCTION The three basic parts of an Induction motor are the

rotor, stator, and enclosure. The stator and the rotor are electrical circuits that

perform as electromagnets.

PARTS OF INDUCTION MOTOR

Stator Stamping

Tooth Slots

Stator Stamping

Tooth Slots

Stator has three main parts: Outer Frame – It is the outer body of the of the

motor.

It protects the inner part of the machine.

Stator Core – Built up of high grade silicon steel.

Carries the alternating magnetic field.

Stator winding – Has a three phase winding.

ROTOR CONSTRUCTION

The rotor is the rotating part of the electromagnetic circuit.

It can be found in two types: Squirrel cage

Wound rotor

However, the most common type of rotor is the “squirrel cage” rotor.

Fig.1.Squirrel cage

rotor Fig.2.Wound rotor

SQUIRREL CAGE

ROTOR

It consists of a laminated cylindrical core having semi closed circular slots at the outer periphery.

Copper or aluminum bar conductors are placed in these slots and short circuited at each end by copper or aluminum rings called short circuiting rings.

The rotor winding is permanently short circuited and it is not possible to add any external resistance.

The rotor slots are not parallel to the shaft but skewed to –

Reduce humming .

Provide smoother torque for different positions of rotor.

Reduce magnetic locking of stator and rotor.

PHASE WOUND ROTOR

It is also called SLIP RING ROTOR

Consists of a laminated core having semi closed slots at the outer periphery and carries a 3-phase insulated winding.

The rotor is wound for the same number of poles as that of stator.

The three finish terminals are connected together

forming a star point and the three star terminals are connected to three slip rings fixed on the shaft.

ROTATING MAGNETIC FIELD

When a 3 phase stator winding is connected to a 3 phase voltage supply, 3 phase current will flow in the windings, which also will induced 3 phase flux in the stator.

These flux will rotate at a speed called a synchronous speed, ns. The flux is called as rotating magnetic field

Synchronous speed: speed of rotating flux

p

fns

120

ROTATING MAGNETIC FIELD • Balanced three phase windings,

i.e. mechanically displaced 120 degrees form each other, fed by balanced three phase source

• A rotating magnetic field with constant magnitude is produced, rotating with a speed

Where fe is the supply frequency and

P is the no. of poles and nsync is called the synchronous speed in rpm (revolutions per minute)

120 esync

fn rpm

P

ROTATING MAGNETIC FIELD

PRINCIPLE OF OPERATION • When a 3 phase stator winding is connected to a 3 phase voltage

supply, 3 phase current will flow in the windings, hence the stator is energized.

• A rotating flux φ is produced in the air gap. The flux Φ induces a voltage ea in the rotor winding (like a transformer).

• The induced voltage produces rotor current, if rotor circuit is closed. • The rotor current interacts with the flux φ, producing torque. The rotor

rotates in the direction of the rotating flux.

INDUCTION MOTOR SPEED

• At what speed will the IM run? – Can the IM run at the synchronous speed, why? – If rotor runs at the synchronous speed, which is the

same speed of the rotating magnetic field, then the rotor will appear stationary to the rotating magnetic field and the rotating magnetic field will not cut the rotor. So, no induced current will flow in the rotor and no rotor magnetic flux will be produced so no torque is generated and the rotor speed will fall below the synchronous speed

– When the speed falls, the rotating magnetic field will cut the rotor windings and a torque is produced

INDUCTION MOTOR SPEED

• So, the IM will always run at a speed lower than the synchronous speed

• The difference between the motor speed and the synchronous speed is called the Slip

Where nslip= slip speed

nsync= speed of the magnetic field

nm = mechanical shaft speed of the motor

slip sync mn n n

THE SLIP

sync m

sync

n ns

n

Where s is the slip

Notice that : if the rotor runs at synchronous speed

s = 0

if the rotor is stationary

s = 1

Slip may be expressed as a percentage by multiplying

the above eq. by 100, notice that the slip is a ratio and

doesn’t have units

INDUCTION MOTORS AND TRANSFORMERS

• Both IM and transformer works on the principle of induced voltage – Transformer: voltage applied to the primary

windings produce an induced voltage in the secondary windings

– Induction motor: voltage applied to the stator windings produce an induced voltage in the rotor windings

– The difference is that, in the case of the induction motor, the secondary windings can move

– Due to the rotation of the rotor (the secondary winding of the IM), the induced voltage in it does not have the same frequency of the stator (the primary) voltage

FREQUENCY

• The frequency of the voltage induced in

the rotor is given by

Where fr = the rotor frequency (Hz)

P = number of stator poles

n = slip speed (rpm)

120r

P nf

( )

120

120

s mr

se

P n nf

P snsf

FREQUENCY

• What would be the frequency of the rotor’s induced voltage at any speed nm?

• When the rotor is blocked (s=1) , the frequency of the induced voltage is equal to the supply frequency

• On the other hand, if the rotor runs at synchronous speed (s = 0), the frequency will be zero

r ef s f

EQUIVALENT CIRCUIT The induction motor is similar to the transformer with the exception that its secondary

windings are free to rotate

As we noticed in the transformer, it is easier if we can combine these two circuits in one circuit but there are some difficulties

EQUIVALENT CIRCUIT When the rotor is locked (or blocked), i.e. s =1, the

largest voltage and rotor frequency are induced in the

rotor, Why?

On the other side, if the rotor rotates at synchronous

speed, i.e. s = 0, the induced voltage and frequency in

the rotor will be equal to zero, Why?

Where ER0 is the largest value of the rotor’s induced voltage obtained at s = 1(loacked rotor)

0R RE sE

EQUIVALENT CIRCUIT The same is true for the frequency, i.e.

It is known that

So, as the frequency of the induced voltage in the rotor

changes, the reactance of the rotor circuit also changes

Where Xr0 is the rotor reactance

at the supply frequency

(at blocked rotor)

r ef s f

2X L f L

0

2

2r r r r r

e r

r

X L f L

sf L

sX

EQUIVALENT CIRCUIT

Then, we can draw the rotor equivalent circuit as

follows

Where ER is the induced voltage in the rotor and RR is

the rotor resistance

EQUIVALENT CIRCUIT

Now we can calculate the rotor current as

Dividing both the numerator and denominator by

s so nothing changes we get

Where ER0 is the induced voltage and XR0 is the rotor

reactance at blocked rotor condition (s = 1)

0

0

( )

( )

RR

R R

R

R R

EI

R jX

sE

R jsX

0

0( )

RR

RR

EI

RjX

s

EQUIVALENT CIRCUIT Now we can have the rotor equivalent circuit

EQUIVALENT CIRCUIT

Now as we managed to solve the induced voltage

and different frequency problems, we can

combine the stator and rotor circuits in one

equivalent circuit

Where

22 0

22

2

1 0

eff R

eff R

R

eff

eff R

Seff

R

X a X

R a R

II

a

E a E

Na

N

POWER LOSSES IN INDUCTION MACHINES

Copper losses

Copper loss in the stator (PSCL) = I12R1

Copper loss in the rotor (PRCL) = I22R2

Core loss (Pcore)

Mechanical power loss due to friction and

windage

How this power flow in the motor?

POWER FLOW IN INDUCTION MOTOR

POWER RELATIONS

3 cos 3 cosin L L ph phP V I V I 21 13SCLP I R

( )AG in SCL coreP P P P

22 23RCLP I R

conv AG RCLP P P

( )out conv f w strayP P P P convind

m

P

EQUIVALENT CIRCUIT

We can rearrange the equivalent circuit as

follows

Actual

rotor

resistance

Resistance

equivalent to

mechanical load

POWER RELATIONS

3 cos 3 cosin L L ph phP V I V I 21 13SCLP I R

( )AG in SCL coreP P P P

22 23RCLP I R

conv AG RCLP P P

( )out conv f w strayP P P P

conv RCLP P 2 223

RI

s

2 22

(1 )3

R sI

s

RCLP

s

(1 )RCLP s

s

(1 )conv AGP s P conv

indm

P

(1 )

(1 )AG

s

s P

s

POWER RELATIONS

AGP

RCLP

convP

1

s

1-

s

: :

1 : : 1-AG RCL convP P P

s s

TORQUE, POWER AND THEVENIN’S

THEOREM

Thevenin’s theorem can be used to transform the network to the left of points ‘a’ and ‘b’ into an equivalent voltage source VTH in series with

equivalent impedance RTH+jXTH


THEOREM

1 1( )M

THM

jXV V

R j X Xf

1 1( ) //TH TH MR jX R jX jX

2 21 1

| | | |( )

MTH

M

XV V

R X Xf


THEOREM

Since XM>>X1 and XM>>R1

Because XM>>X1 and XM+X1>>R1

1

MTH

M

XV V

X Xf

2

11

1

MTH

M

TH

XR R

X X

X X


THEOREM

Then the power converted to mechanical (Pconv)

2 222

2( )

TH TH

T

TH TH

V VI

Z RR X X

s

2 22

(1 )3conv

R sP I

s

And the internal mechanical torque (Tconv)

convind

m

P

(1 )

conv

s

P

s

2 223

AG

s s

RI Ps


THEOREM 2

2

222

2

3

( )

THind

s

TH TH

V R

sRR X X

s

2 2

222

2

31

( )

TH

inds

TH TH

RV

s

RR X X

s

MAXIMUM TORQUE

Maximum torque occurs when the power

transferred to R2/s is maximum.

This condition occurs when R2/s equals the

magnitude of the impedance RTH + j (XTH + X2)

max

2 222( )TH TH

T

RR X X

s

max

2

2 22( )

T

TH TH

Rs

R X X

MAXIMUM TORQUE The corresponding maximum torque of an induction

motor equals

The slip at maximum torque is directly proportional to

the rotor resistance R2

The maximum torque is independent of R2

2

max 2 22

31

2 ( )TH

s TH TH TH

V

R R X X

MAXIMUM TORQUE Rotor resistance can be increased by inserting

external resistance in the rotor of a wound-rotor

induction motor.

The

value of the maximum torque remains unaffected

but

the speed at which it occurs can be controlled.

MAXIMUM TORQUE

Effect of rotor resistance on torque-speed

characteristic

TORQUE-SPEED CHARACTERISTICS

COMMENTS 1. The induced torque is zero at synchronous speed.

Discussed earlier.

2. The curve is nearly linear between no-load and full

load. In this range, the rotor resistance is much

greater than the reactance, so the rotor current,

torque increase linearly with the slip.

3. There is a maximum possible torque that can’t be

exceeded. This torque is called pullout torque and is

2 to 3 times the rated full-load torque.

COMMENTS

4. The starting torque of the motor is slightly

higher than its full-load torque, so the motor

will start carrying any load it can supply at full

load.

5. The torque of the motor for a given slip varies

as the square of the applied voltage.

6. If the rotor is driven faster than synchronous

speed it will run as a generator, converting

mechanical power to electric power.

UNIT- V

CHARACTERISTICS

OF

INDUCTION MOTORS

NEED OF DEEP BAR DOUBLE CAGE ROTOR

DEEP BAR ROTOR

SPEED- TORQUE CHARACTERISTICS

DOUBLE CAGE ROTOR

SLIP- TORQUE CHARACTERISTICS

3 PHASE INDUCTION MOTOR STARTER

STARTER

NEED OF STARTER

• If a rated stator voltage is applied to the motor at the time of starting, then the motor will draw heavy starting current.

• This will lead to excess i2R losses in the winding which will overheat the motor.

• Secondly due to a heavy current drawn from the AC supply voltage will reduce.

• The heavy starting current may damage the motor windings.

• In order to avoid these problems, we can use some kind of a starter to start the induction motor safely.

TYPES OF STARTER

• Stator resistance starter

• Auto transformer starting

• Star-delta starter

• Rotor resistance starter

• Direct on line (DOL) starter

TYPES OF STARTER FOR 3-PH INDUCTION MOTORS

For slip-ring induction motors:

Rotor rheostat starter

For squirrel cage induction motors:

D.O.L starter

Stator resistance starter

Auto transformer starter

Star delta starter

STATOR RESISTANCE STARTER

• A starter resistance is connected in each line in series with each phase winding of the stator.

• Initially all the starter resistance are kept in “Start” position so that they offer their maximum resistance .

• The switch is turned ON to connect the three phase AC supply to the stator winding.

• Due to starter resistance in series, each phase winding will receive a reduced voltage. Due to reduction in the value of V1

, the starting current is limited to a safe value.

• As the motor accelerates, the starter resistance is reduced by moving the variable contact of the resistance towards the “Run” position.

• In the “Run” position, the starter resistance is shorted out and full stator voltage is applied across the stator winding.

AUTO-TRANSFORMER STARTER • An autotransformer is used to apply a low voltage to the stator

winding at the time of starting. When the motor speed reaches the desired level, autotransformer is disconnected and motor is connected directly across the supply.

• The stator of the motor is connected through a 6-way double throw switch.

• While starting, the switch is thrown to ‘Start’ side so that a reduced voltage is applied to stator. This keeps the starting current safe limits.

• Once motor take up the speed, the switch is throw to ‘Run’ side so that full supply voltage is applied to stator.

• A specific advantage of this starter is that reduction in voltage during starting, can be done to any desired level by selecting proper tapping of the autotransformer.

AUTO-TRANSFORMER STARTER

STAR DELTA STARTER

• Most induction motors are started directly on line, but when very large motors are started that way, they cause a disturbance of voltage on the supply lines due to large starting current surges.

• To limit the starting current surge, large induction motors are started at reduced voltage and then have full supply voltage reconnected when they run up to near rotated speed.

STAR-DELTA STARTER

This is very commonly used starter, compared to the other types of the starters.

Star-delta starter can be used, provided the stator of the 3-Ø induction motor is designed for delta connection during its normal operation.

At starting, the stator winding is connected in star, therefore the applied voltage to each phase of winding is 1/√3 of the rated voltage of the motor.

When the motor has picked-up the speed(say 70 to 80% of its normal speed ) the phases of the stator winding are connected in delta.

Now full supply voltage is applied across the stator windings.

This method is cheap but limited to applications where high

starting torque is not necessary e.g., machine tools, pumps, motor-generator sets etc.

The method is unsuitable for motors for voltage exceeding

3000 V because of the excessive number of stator turns needed for delta connection.

Such starters are employed for starting 3-phase squirrel cage

induction motors of rating between 4 and 20 k W.

ADVANTAGES OF STAR-DELTA STARTER:

The operation of the star-delta method is simple and rugged

It is relatively cheap compared to other reduced voltage methods.

Good Torque/Current Performance.

It draws 2 times starting current of the full load ampere of the motor connected

MOTOR STARTING CHARACTERISTICS OF STAR-DELTA STARTER:

• Available starting current: 33% Full Load Current.

• Peak starting current: 1.3 to 2.6 Full Load Current.

• Peak starting torque: 33% Full Load Torque.

DISADVANTAGES OF STAR-DELTA STARTER:

• Low Starting Torque, only 33% starting torque • Break In Supply – Possible Transients • Six Terminal Motor Required (Delta Connected). • It requires 2 set of cables from starter to motor. • The delta of motor is formed in starter and not on motor

terminals. • Applications with a load torque higher than 50 % of the motor

rated torque will not be able to start using the start-delta starter.

• Low Starting Torque: reduction of the line voltage by a factor of 1/√3 (57.7%) to the motor and the current is reduced to 1/3 of the current at full voltage, but the starting torque is also reduced 1/3 to 1/5 of the DOL starting torque .

MOTOR STARTING CHARACTERISTICS ON DOL STARTER:

• Available starting current: 100%.

• Peak starting current: 6 to 8 Full Load Current.

• Peak starting torque: 100%

ADVANTAGES OF DOL STARTER:

• Most Economical and Cheapest Starter

• Simple to establish, operate and maintain

• Simple Control Circuitry

• Easy to understand and trouble‐shoot.

• It provides 100% torque at the time of starting.

• Only one set of cable is required from starter to motor.

• Motor is connected in delta at motor terminals.

DISADVANTAGES OF DOL STARTER:

• It does not reduce the starting current of the motor.

• High Starting Current: Very High Starting Current (Typically 6 to 8 times the FLC of the motor).

• Mechanically Harsh: Thermal Stress on the motor, thereby reducing its life.

• Voltage Dip: There is a big voltage dip in the electrical installation

• High starting Torque: Unnecessary high starting torque, even when not required by the load.

SUITABILITY

• DOL is Suitable for:

• Small water pumps, compressors, fans and conveyor belts.

• Motor rating up to 5.5KW

• DOL is not suitable for:

• The peak starting current would result in a serious voltage drop on the supply system

• Motor rating above 5.5KW

ROTOR RESISTANCE STARTER

•This starter is used with a wound rotor induction motor. It uses an external resistance/phase in the rotor circuit so that rotor will develop a high value of torque. •High torque is produced at low speeds, when the external resistance is at its higher value. •At start, supply power is connected to stator through a three pole contactor and, at a same time, an external rotor resistance is added.

DIFFERENCE BETWEEN DOL/STAR DELTA /AUTOTRANSFORMER

Sr

.

DOL Starter Star delta starter Auto transformer starter

1 Used up to 5 HP Used 5 HP to 20HP

Used above 20 HP

2 Does not decrease the starting current

Decreases the starting current by 1/3 times

Decreases the starting current as required

3 It is cheap It is costly It is more costly

4 It connects directly the motor with supply for starting as well as for running

It connects the motor first in star at the time of starting in delta for running

It connects the motor according to the taping taken out from the auto transformer

249

DETERMINATION OF

INDUCTION-MOTOR PARAMETERS

DC Test

Determines R1

Connect any two stator leads to a variable-voltage DC

power supply

Adjust the power supply to provide rated stator

current

Determine the resistance from the voltmeter and

ammeter readings

250

DCDC

DC

VR

I

25

1

FOR A Y-CONNECTED STATOR

1,

1,

2

2

DC wye

DCwye

R R

RR

25

2

FOR A DELTA-CONNECTED STATOR

1 11

1 1

1

2 2

2 3

1.5

DC

DC

R RR R

R R

R R

253

DETERMINATION OF


Blocked-Rotor Test

Determine X1 and X2

Determines R2 when combined with data from the

DC Test

Block the rotor so that it will not turn

Connect to a variable-voltage AC supply and adjust

until the blocked-rotor current is equal to the rated

current

25

5

SIMPLIFIED EQUIVALENT CIRCUIT

Neglect the exciting current under blocked-rotor

conditions – remove the parallel branch

256

IEEE test code recommends that the blocked-rotor

test be made using 25% rated frequency with the

test voltage adjusted to obtain approximately rated

current.

A 60-Hz motor would use a 15-Hz test voltage.

The calculated reactance is corrected to 60-Hz by

multiplying by 60/15.

Calculated resistance is correct.

257

1 2 ,15

,15,15

,15

,15,15 2

,15

2 ,15 1

BR

BRBR

BR

BRBR

BR

BR

R R R

VZ

I

PR

I

R R R

258

2 2,15 ,15 ,15

2 2,15 ,15 ,15

,60 ,15

,60 1 2

60

15

BR BR BR

BR BR BR

BR BR

BR

Z R X

X Z R

X X

X X X

25

9

HOW IS THE BLOCKED-ROTOR

IMPEDANCE DIVIDED?

,60 1 2BRX X X

If the NEMA-design letter of the motor is known,

use Table 5.10 to divide the impedances. Otherwise,

divide the impedances equally.

260

DETERMINATION OF


No-Load Test

Determine the magnetizing reactance, XM and

combined core, friction, and windage losses.

Connect as for blocked-rotor test (next slide).

The rotor is unblocked and allowed to run unloaded

at rated voltage and rated frequency.

261

Electrical connection for the No-Load Test is the

same as for the Blocked-Rotor Test

262

DETERMINATION OF


At no-load, the speed is very close to synchronous

speed – the slip is =0, causing the current in R2/s

to be very small, and will be ignored i the

calculations.

IM>>Ife, so I0 = IM.

263

The equivalent circuit for the no-load test is shown.

Ignore

264

2 2

2 2

2

2

1

NL NL NL

NL NL NL

NL NL NL

NL NL NL

NLNL

NL

NL M

S V I

S P Q

Q S P

Q I X

QX

I

X X X

Substitute X1 from the blocked-rotor test to

determine the value of XM.

265

EXAMPLE 5.16

The following data were obtained from no-load,

blocked-rotor, and DC tests of a three-phase,

wye-connected, 40-hp, 60-Hz, 460-V, design B

induction motor whose rated current is 57.8A.

The blocked-rotor test was made at 15 Hz.

266

Blocked-Rotor No-Load DC

Vline = 36.2V Vline = 460.0V VDC = 12.0V

Iline = 58.0A Iline = 32.7A IDC = 59.0A

P3phase = 2573.4W P3phase = 4664.4W

a) Determine R1, X1, R2, X2, XM, and the combined

core, friction, and windage loss.

b) Express the no-load current as a percent of rated

current.

267

,15

,15

,15

2573.4857.80

336.2

20.903

58.0

4664.41554.80

3460

265.5813

32.7

BR

BR

BR

NL

NL

NL

WP W

VV V

I A

WP W

VV V

I A

Convert the AC test data to corresponding phase

values for a wye-connected motor.

268

Determine R1

1,

12.00.2034

59.0

0.102 /2

DCDC

DC

DCwye

V VR

I A

RR phase

Determine R2

,15,15

,15

,15,15 2 2

,15

2 ,15 1,

20.900.3603 /

58.0

857.80.2550 /

(58 )

0.2550 0.102 0.153 /

BRBR

BR

BRBR

BR

BR wye

V VZ phase

I A

P WR phase

I A

R R R phase

269

Determination of X1 and X2

2 2 2 2,15 ,15 ,15

,60 ,15

(0.3603) (0.255) 0.2545

60 60(0.2545) 1.0182

15 15

BR BR BR

BR BR

X Z R

X X

From Table 5.10, for a design B machine,

X1 = 0.4XBR,60 = 0.4(1.0182) = 0.4073Ω/phase

X2 = 0.6XBR,60 = 0.6(1.0182) = 0.6109/phase

270

Determination of XM

2 2 2 2

2 2

1

1

(265.581 )(32.7 ) 8684.50

(8684.50) (1554.8) 8544.19

8544.197.99

(32.7)

7.99 0.4073 7.58 /

NL

NL NL NL

NL NL

NLNL

NL

NL M

M NL

S V I V A VA

Q S P VARS

QX

I

X X X

X X X phase

271

Determination of combined friction, windage, and

core loss:

21, ,

2,

,

1554.8 (32.7) (0.102)

1446 /

NL NL wye core f w

core f w

core f w

P I R P P

P P

P P W phase

b) Express the no-load current as a percent of rated

current.

32.7% 100% 100% 56.6%

57.8NL

NLrated

II

I

CIRCLE DIAGRAM

Tests required

No load test

Blocked Rotor test

272

UNIT-VI

Speed control METHODS

27

7

Speed control of three

phase induction motor

Introduction

Requirement of Speed control

Types of Methods to control the speed of Induction motor

Advantages & disadvantages

Industrial applications of AC drives

Conclusion

Research

Agenda

A three phase induction motor is basically a

constant speed motor .

It is widely used in industry due to low cost and

rugged construction .

The speed control of induction motor is done at

the cost of decrease in efficiency and low

electrical power factor.

INTRODUCTION

Speed control means change the drive speed as

desired by the process to maintain different

process parameter at different load .

Energy Saving.

Speed control is a different concept from speed

regulation where there is natural change in

speed due change in load on the shaft.

Speed control is either done manually by the

operator or by means of some automatic control

device.

Low speed starting requirement.

REQUIREMENT OF SPEED

CONTROL

Stator voltage Control

Stator Frequency Control

Stator Current Control

V/F Control

Static rotor resistance control

METHODS OF SPEED CONTROL OF

INDUCTION MOTORS

Synchronous speed Ns = 120 f

P

Slip = Ns-N

Ns

Torque =

Where E2 is the rotor emf

Ns is the synchronous speed

R2 is the rotor resistance

X2 is the rotor inductive reactance

STATOR VOLTAGE CONTROL

Rotor resistance R2 is constant and if slip s is small then sX2 is so small that it can be neglected. Therefore, T ∝ sE2

2 where E2 is rotor induced emf and E2 ∝ V And hence T ∝ V2, thus if supplied voltage is decreased, torque decreases and hence the speed decreases.

This method is the easiest and cheapest, still rarely used because-

A large change in supply voltage is required for relatively small change in speed.

Large change in supply voltage will result in large change in flux density, hence disturbing the magnetic conditions of the motor.

CONTINUE…..

Variable Terminal Voltage Control

ms

mLT

T

V decreasing

Synchronous speed of induction motor Ns = 120 f

P

where, f = frequency of the supply and P = number of stator

poles.

Thus, synchronous speed changes with change in supply

frequency, and thus running speed also changes.

This method is not widely used. This method is used where,

only the induction motor is supplied by a generator (so that

frequency can be easily change by changing the speed of

prime mover).

FREQUENCY CONTROL

By changing the frequency we can control the speed above

and below the rated speed.

It offers high range of speed control.

CONTINUE…..

STATOR CURRENT CONTROL

Starting torque of IM Ts is proportional to square of stator

current.

It is independent of supply frequency.

It is independent of rotor resistance.

A constant current for 3 phase IM can be obtained from 3

phase CSI .

Inductor convert the dc voltage as constant current source.

CSI regulate the output frequency and hence torque of

induction motor.

CONTINUES…..

HOW SPEED IS CONTROLLED

USING VFD Rectifier: The rectifier in a VFD is used to convert incoming ac

power into direct current (dc) power.

DC bus:

DC output of rectifier flows through the dc link to inverter input.

Inverter:

The “Insulated Gate Bipolar Transistor” (IGBT) is a common choice in modern VFDs.

The IGBT can switch on and off several thousand times per second and precisely control the power delivered to the motor.

The IGBT uses a method named “pulse width modulation” (PWM) to simulate a current sine wave at the desired frequency to the motor.

A slip ring motor or a phase wound motor is an

induction motor which can be started with full

line voltage, applied across its stator terminals.

The value of starting current is adjusted by

adding up external resistance to its rotor circuit.

290

STATIC ROTOR RESISTENCE

CONTROL

CONTINUE…..

Advantages of ac drives For the same rating, ac drives are lighter in weight as

compared to dc drive.

AC drive require low maintenance.

AC drives are less expensive.

Provides the most efficient means of motor speed control.

Reduces the thermal and mechanical stresses on the motor.

Provides low speed motor starting facility.

Saves more energy

ADVANTAGES AND DISADVANTAGES OF

AC DRIVES

Disadvantage of ac drives Power converters for the control of ac motors are more

complex.

Power converter for ac drives are more expensive.

Power converters for ac drives generate harmonics in the supply system and load circuit.

CONTINUE…..

Induction motors with squirrel cage rotors are the workhorse of industry .

When Squirrel cage induction machine is operated directly from the line voltages an Induction motor is operated at constant speed. However in the industry we required to vary the speed of an Induction motor. This can be done by Induction motor drive.

Fans, Compressor, Pumps, blowers, machine tools like lathe, drilling machine, lifts, conveyer belts etc.

INDUSTRIAL APPLICATIONS

In Case of Squirrel cage induction motor the slip cannot be increase above certain limit, the operating speed range is very less. By applying the V/F control we can get the large operating range by keeping V/F ratio constant.

CONCLUSION

single phase transformer construction & operation

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