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.
• 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.
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 .
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.
DELTA-DELTA CONNECTION
• 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.
DELTA-DELTA CONNECTION
• Example 1
DELTA-DELTA CONNECTION
DELTA-DELTA CONNECTION
DELTA-DELTA CONNECTION
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.
DELTA-WYE CONNECTION
• 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
DELTA-WYE CONNECTION
• 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.
DELTA-WYE CONNECTION
• Example 2
DELTA-WYE CONNECTION
Fig.5
DELTA-WYE CONNECTION
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.
OPEN-DELTA CONNECTION
• Example 3
OPEN-DELTA CONNECTION
• 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.
OPEN-DELTA CONNECTION
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.
THREE-PHASE TO 2-PHASE TRANSFORMATION
• 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.
THREE-PHASE TO 2-PHASE TRANSFORMATION
(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
THREE-PHASE TO 2-PHASE TRANSFORMATION
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
THREE-PHASE TO 2-PHASE TRANSFORMATION
THREE-PHASE TO 2-PHASE TRANSFORMATION
Fig.16c
THREE-PHASE TO 2-PHASE TRANSFORMATION
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
3-PHASE TO 2-PHASE CONVERSION
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
3-PHASE TO 2-PHASE CONVERSION
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
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
TORQUE, POWER AND THEVENIN’S
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
TORQUE, POWER AND THEVENIN’S
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
TORQUE, POWER AND THEVENIN’S
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
TORQUE, POWER AND THEVENIN’S
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
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
INDUCTION-MOTOR PARAMETERS
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
254
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
INDUCTION-MOTOR PARAMETERS
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
INDUCTION-MOTOR PARAMETERS
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
273
274
275
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