Unit-3
Electrical Machines
*Transformer
*DC Machine
*Induction Motor
*Synchronous Machine
Single-phase Transformer
Transformer
• A transformer is a static device/machine which converts ac
electric power from one voltage level to ac electric power to
another level through the action of a magnetic field at same
frequency.
•
•
•
•
•
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The main tasks perform by a transformer are:
Changing the voltage and current levels.
Maintaining constant frequency, power and flux.
It can be either step up or step down.
Input kVA= output KVA always true in transformer
Transformer is a two-port network used in transmission line.
Transformer
used in
transmission
line:
• This is a basic connection of today’s distribution system.
• At first, step-up transformer is used to convert generated voltage at high voltage for
proper transmission of the electrical energy.
• After that, we use the step-down transformer so that we can distribute the obtained
electricity in houses and industries.
Working of Transformer
• Transformer works on the principle of Faraday’s law
of electromagnetic induction.
• Faraday’s law of electromagnetic induction:
“Whenever a carrying conductor is placed in a
varying magnetic field, an electromotive force is
induced. Likewise, if the conductor circuit is closed, a
current is induced, which is called induced current.”
• Transformer consists of two inductive coils
electrically separated but magnetically linked through
a path of low reluctance.
• AC voltage source is given at one coil an alternating
flux is setup in the laminated core most of which is
linked with other coil in which it produces mutually
induced emf (according to the faraday’s law).
• If the second coil is closed a current flows in it and so,
electric energy is transferred.
Core type:
• Here two limbs or legs are there, and
the windings are present in both the
limbs.
• Amount of copper used is more
• Series magnetic circuit
• Smaller cross section of area.
Types of
transformer:
Shell type:
• Core surrounds the windings.
• Three limbs or legs placement
of windings are in the middle
so that amount of copper used
is less here.
• Parallel magnetic circuit
• Larger cross section area.
Construction
Construction of Single-Phase Transformer
• Core:
• core is a magnetic link between the two systems connected to the transformer. It
contains a lot of energy, so it is an active energy
• Requirement of core material
• Maximum flux
• Minimum magnetizing current
• Minimum core or iron loss
• Windings:
• Windings are made up of solid and stranded copper or aluminum strip conductors.
• Several small parallel strips are preferred to one large strip
• Insulation:
• Major insulation is provided between windings and earthed parts. Minor insulation
includes the insulation provided between the elements of a given windings such as
conductor insulation, insulation of turns etc.
• Transformer oil /insulation oil:
• It acts as a coolant and an insulant.
• Natural cooling: by air convection and natural radiation
• Oil natural cooling: by conduction.
Tank:
Tank is container for assembled core and windings. It
must be capable of withstand the stresses developed
by jacking.
Conservator/expansion tank:
Conservator is an additional tank provided
with transformer which stores oil when it gets
expanded due to temperature rise. It also serves
another important purpose that is, as a reservoir of
transformer oil.
Emergency vent:
It relieve the pressure inside the tank in case the
pressure inside the transformer rises to a danger
point.
Breather:
When transformer becomes warm the oil and gas
expands the gas at the top of oil expelled out when
the transformer cools air is drawn into the
transformer. The moisture is drawn during this
process called breathing.
Buchholz relay:
Used to protect the transformer against internal fault.
It causes tripping for major fault and alarm for minor
fault. This type of relay is only possible when we have
conservators .
Emf Equation of Transformer:
•
When the primary winding is excited by an ac voltage V1, it circulates an alternating flux (fi), the primary has N1 or T1 turns.
•
The alternating flux (fi) linking with the primary winding itself induces an emf E1.
•
The flux linking with N2 or T2 produces mutually induced emf E2 in the secondary winding.
Where ,
 E1 , E2 are emf of primary and secondary
windings
 f is the frequency
Volt-ampere rating of a transformer:
• The rating of transformer is specified as the product of voltage and current called
VA rating. On both sides, primary and secondary VA remains same.
• The rating is generally is expressed as KVA.
KVA input = KVA output
V1*I1 = V2*I2
The full load primary and secondary currents indicates
the safe maximum values of a transformer of specified
KVA rating.
Leakage flux is defined as the magnetic flux which
does not follow the particularly intended path in a
magnetic circuit.
Most of the flux is set up in the core of the solenoid and
passes through the particular path that is through the air
gap and is utilized in the magnetic circuit. This flux is
known as Useful flux φu.
Fringing in
transformer:
• As practically it is not possible that all the flux in the circuit follows a particularly
intended path and sets up in the magnetic core and thus some of the flux also sets
up around the coil or surrounds the core of the coil, and is not utilized for any
work in the magnetic circuit.
• This type of flux which is not used for any work is called Leakage Flux and is
denoted by φ
• Therefore, the total flux Φ produced by the solenoid in the magnetic circuit is the
sum of the leakage flux and the useful flux and is given by the equation shown
below:
l.
ASSUMPTIONS
Winding resistance are neglected
The copper losses are negligible
Assumptions of Ideal
transformer on Noload:
Core/ iron losses are negligible.
The magnetization curve for the core is linear.
All the flux setup by the primary links the secondary windings i.e all
the flux is confined to the magnetic core and there is no leakage flux.
Ideal transformer with finite permeability on no load:
The supply voltage is V1 and as it is on no load the
secondary current I2 =0 the primary draws a current
I1 which is just necessary to produce flux in the core.
As it is magnetizing the core, it is called as
magnetizing current and denoted as Im.
This magnetizing current is very small and lags V1 by
90 degrees( pure inductive windings) this Im will be in
phase with flux.
The Fig-1 shows a phasor diagram of transformer at
no load.
Fig-1:phasor diagram of transformer at no load
Practical transformer at noload:
• In the practical transformer
at no-load io breaks in two
components working current
iw and magnetizing current
im.
• Iw or ic is the active
component which produces
core loss during no load.
Practical transformer at load:
• For any load condition, no load to full load flux in the
core is always constant.
• I1 mentioned in the figure is the total current in the
primary windings.
• The second fig. shows the loaded condition phasor
diagram for resistive inductive and capacitive load.
Equivalent circuit diagram of transformer:
• The given fig shows the equivalent circuit diagram of a onload transformer.
• R1 & R2 are the winding resistances.
• X1, X2 are the leakage reactance's, used to compensate
leakage emf’s from leakage flux.
• Ro & Xo are the no-load components.
• I2’ is the current opposite to secondary current I2.
• E1 is the self emf and E2 is the mutual emf.
• V1 is the source voltage.
To simplify the equivalent circuit diagram of transformer we
refer the circuit to either primary side or secondary side.
Referred Circuits
Per unit quantities
• Definition: The per-unit value of any quantity is defined as the ratio of actual value in any unit to
the base or reference value in the same unit.
Voltage Regulation
Voltage Regulation in Transformer
• It is defined as the change in
magnitude of secondary terminal
voltage, expressed as percentage
(or per unit) of the secondary
rated voltage, when a load at a
given power factor is reduced to
zero, with primary applied
voltage held constant.
Derivation of
Voltage Regulation
Fig: Impedance triangle
 Only possible at lagging power factor
Maximum
Voltage
Regulation
Voltage
Regulation
Curve
Losses in Transformer
 IRON/CORE LOSS:
• Hysteresis and eddy current losses are there, both depend upon the magnetic properties of materials used to construct the
core of the transformer.
• They are considered as CONSTANT LOSS
Iron Loss (Pi) = Hysteresis loss (Ph) + Eddy current loss (Pe)
 COPPER LOSS:
• Copper loss is the term often given to heat produced by electrical currents in the form of energy of transformer winding,
or other electrical devices.
• Copper loss are undesirable transfer of energy.
• It depends upon the load current hence it is called as VARIABLE LOSS
Hysteresis & Eddy Current Loss:
Hysteresis loss: this loss is related with the structure of B-H
curve. the formula for the hysteresis is as given where the
symbols have their usual meanings,
Ƞ is the hysteresis constant. Sometimes we also define it as kh
Eddy current loss: eddy current loss arise because of
changing flux in core. to reduce this loss we use thin
laminations in the core. Laminations means core is sliced and
each piece is laminated from the next and hence it cuts the path
of eddy current
The formula to define eddy current loss is as follows
The term efficiency is defined as the ratio of output power to the input power
Efficiency of
Transformer
Condition for Maximum Efficiency
𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑉2 𝐼2 𝑐𝑜𝑠 𝛷2
If R02 is the total resistance of the transformer referred to secondary, then,
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑝𝑝𝑒𝑟 𝑙𝑜𝑠𝑠, 𝑃𝐶 = 𝐼22 . 𝑅02
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 𝑃𝑖 + 𝑃𝐶
𝑉2 𝐼2 𝑐𝑜𝑠 𝛷2
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 =
𝑉2 𝐼2 𝑐𝑜𝑠 𝛷2 + 𝑃𝑖 + 𝐼22 . 𝑅02
𝑉2 𝑐𝑜𝑠 𝛷2
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 =
𝑃
𝑉2 𝑐𝑜𝑠 𝛷2 + 𝐼 𝑖 + 𝐼2 . 𝑅02
2
Condition for Maximum Efficiency
𝑑
𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 = 0
𝑑𝐼2
𝑑
Pi
V2 cos Φ2 + + I2 . R 02 = 0
𝑑𝐼2
I2
𝑃𝑖
0 − 2 + R 02 = 0
𝐼2
𝑃𝑖 = 𝐼22 R 02
• i.e, Iron loss = Copper loss
Condition for Maximum Efficiency
• Hence efficiency of a transformer will be maximum when copper
losses are equal to iron losses.
• From equation 𝑃𝑖 = 𝐼22 R 02 , the load current I2 corresponding to
maximum efficiency is given by,
𝐼2 =
𝑃𝑖
R 02
Output kVA Corresponding to Maximum Efficiency
• PC = Copper losses at full-load kVA
• Pi= Iron losses
• X = Fraction of full-load kVA at which efficiency is maximum
• Total Cu losses = X2PC
• For maximum efficiency, Pi = X2 PC
∴ X=
Pi
PC
Output kVA Corresponding to Maximum Efficiency
Output kVA corresponding to max. efficiency = 𝑋 × Full load kVA
Pi
Output kVA corresponding to max. efficiency = Full load kVA ×
PC
• It may be noted that the value of kVA, at which the efficiency is
maximum, is independent of p.f. of the load.
Tests on Transformer
• Tests on transformer are used to define the circuit constants, efficiency and
regulation without loading the transformer.
Open Circuit Test
Calculation for open circuit test
• The open circuit test is carried out at rated voltage and rated
frequency it is performed at low voltage(LV) side with high
voltage(HV) side open.
Short Circuit Test
• Short circuit test is carried out at rated current to
determine the full load copper loss
• It also determines the equivalent resistance and
leakage reactance
Obtain the equivalent circuit of a 200 / 400 V, 50 Hz, 1 phase
transformer from the following test data:
O.C. test: 200 V, 0.7 A, 70 W – on L.V side.
S.C. test: 15 V, 10 A, 85 W – on H.V side.
Calculate the secondary voltage when delivering 5 kW at 0.8 p.f
lagging, the primary voltage being 200 V.
From OC Test
From SC Test
P0 = V0 I0 cos ϕ0
P0
70
cos ϕ0 =
=
V0 I0 200 × 0.7
cos ϕ0 = 0.5
sin ϕ0 = 0.866
Iw = I0 cos ϕ0 = 0.7 × 0.5 = 0.35𝐴
2
Psc = Isc
R 02
R 02
Psc
85
= 2 = 2 = 0.85 Ω
Isc 10
Z02
Vsc 15
=
=
= 1.5 Ω
Isc 10
Im = I0 sin ϕ0 = 0.7 × 0.866 = 0.606𝐴
X02 =
Z02 2 − R 02 2
V0
70
R0 =
=
= 200 Ω
Iw 0.35
X02 =
1.52 − 0.852
V0
70
X0 =
=
= 115.5 Ω
Im 0.606
X02 = 1.235 Ω
Equivalent Circuit Referred to Primary Side
400
𝐾=
=2
200
R 01
0.85
= 2 = 0.212 Ω
2
X01
1.235
=
= 0.308 Ω
2
2
Equivalent Circuit Referred to Secondary Side
400
𝐾=
=2
200
𝑅0′ = 200 × 22 = 800 Ω
𝑋0′ = 115.5 × 22 = 462 Ω
• Load kVA corresponding to 5 kW is,
5000
=
= 6250 𝑉𝐴
0.8
• Load current I2 while delivering 6250 VA is,
6250
=
= 15.625 𝐴
400
• Total voltage drop in secondary when it carries 15.625 A is,
= 𝐼2 𝑅02 𝑐𝑜𝑠 𝜙2 + 𝑋02 𝑠𝑖𝑛 𝜙2
= 15.625 0.85 × 0.8 + 1.235 × 0.6
= 22.20 𝑉
• Hence the secondary voltage is,
𝑉2 = 400 − 22.2 = 377.8 𝑉
All Day Efficiency
• The ordinary or commercial efficiency of a transformer is defined as
the ratio of output power to the input power i.e.,
Output power
Commercial efficiency =
Input power
Primaries of distribution transformers are energized all the 24 hours in
a day but the secondary windings supply little or no load during the
major portion of the day.
All Day Efficiency
• Constant loss occurs during the whole day but copper loss occurs only
when the transformer is loaded.
• The performance of such transformers is judged on the basis of
energy consumption during the whole day (i.e., 24 hours).
• This is known as all-day or energy efficiency.
All Day Efficiency
• The ratio of output in kWh to the input in kWh of a transformer over
a 24-hour period is known as all-day efficiency i.e.,
𝜂𝑎𝑙𝑙−𝑑𝑎𝑦
𝑘𝑊ℎ 𝑜𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 24 ℎ𝑜𝑢𝑟𝑠
=
𝑘𝑊ℎ 𝑖𝑛𝑝𝑢𝑡 𝑖𝑛 24 ℎ𝑜𝑢𝑟𝑠
• In the design of such transformers, efforts should be made to reduce
the iron losses which continuously occur during the whole day.
All Day Efficiency
• A 40kVA distribution transformer has iron loss of 500 W and full load
copper loss of 500 W. the transformer is supplying a lighting load. The
load cycle is as under: Full load for 4 hours, half load for 8 hours and
no load for 12 hours. Calculate the all day efficiency.
All Day Efficiency
• A transformer has its maximum efficiency of 0.98 at 15 kVA at UPF.
During the day it is loaded as follows:
Duration
Load
Power Factor
12 hours
2 kW
at 0.5 p.f
6 hours
12 kW
at 0.8 p.f
4 hours
18 kW
at 0.9 p.f
2 hours
No load
• Find the “All Day Efficiency”.
Introduction to
Auto-transformer
Auto Transformer
• An autotransformer has a single winding on an iron core and a part of
winding is common to both the primary and secondary circuits.
Auto Transformer
• Primary and secondary windings are connected electrically as well as
magnetically.
• Therefore, power from the primary is transferred to the secondary
conductively as well as inductively (transformer action).
• The voltage transformation ratio K of an ideal autotransformer is,
𝐸1
𝑁1
𝑉1
𝐼2
=
=
=
=𝐾
𝐸2
𝑁2
𝑉2
𝐼1
Theory of Autotransformer
• Winding 1-3 - N1 turns - pri winding
• winding 2-3 - N2 turns - sec winding
• Input current is I1
• Output current is I2
• Portion 1-2 of the wdg has N1 - N2 turns and voltage across this portion
of the winding is V1 - V2.
• The current through the common portion of the winding is I2 - I1.
Theory of Autotransformer
• The equivalent circuit of the autotransformer.
• From this equivalent circuit, we have,
V1 − V2 N1 − N2
=
V2
N2
N1 − N2 V2 = V1 − V2 N2
V2 N1 − V2 N2 = V1 N2 − V2 N2
V2 N1 = V1 N2
N2
V2
=
=K
N1
V1
Output of Autotransformer
• Output apparent power = V2 I2
Apparant power transferred inductively
= V2 I2 − I1 = K. V1
= K. V1 I1
I1
− I1
𝐾
1
−1
𝐾
1−𝐾
= K. V1 I1
= V1 I1 (1 − 𝐾)
𝐾
= 𝐼𝑛𝑝𝑢𝑡 × (1 − 𝐾)
Apparant power transferred conductively = Input − Input × 1 − 𝐾
= Input 1 − (1 − 𝐾) = 𝐼𝑛𝑝𝑢𝑡 × 𝐾
• Suppose the input power to an ideal autotransformer is 1000 W and its voltage
transformation ratio K = 0.25. Then,
Apparant power transferred inductively = Input × 1 − 𝐾 = 1000 × 1 − 0.25
= 750 𝑊
Apparant power transferred conductively = Input × 𝐾 = 1000 × 0.25 = 250 W
Saving of Copper in Auto Transformer
• For the same output and voltage transformation ratio, an autotransformer
requires less copper than an ordinary 2-winding transformer.
• Weight of Cu required in a winding is α current X turns
Saving of Copper in Auto Transformer
Two winding transformer
• Weight of Cu required α (I1N1 + I2N2)
Autotransformer
• Weight of Cu required in section 1-2 α I1 (N1 – N2)
• Weight of Cu required in section 2-3 α (I2 – I1) N2
• Total weight of Cu required α I1 (N1 – N2) + (I2 – I1) N2
Saving of Copper in Auto Transformer
Saving of Copper in Auto Transformer
 Thus if K = 0.1, the saving of Cu is only 10% but if K = 0.9, saving
of Cu is 90%.
 Therefore, saving of Cu is more when K is nearer to 1.
Advantages of Autotransformers
• An autotransformer requires less Cu than a two -winding transformer
of similar rating.
• Autotransformer operates at a higher efficiency than a two-winding
transformer of similar rating.
• An autotransformer has better voltage regulation than a two-winding
transformer of the same rating.
• An autotransformer has smaller size than a two-winding transformer
of the same rating.
Advantages of Autotransformers
• An autotransformer requires smaller exciting current than a twowinding transformer of the same rating.
• These advantages decrease as the ratio of transformation increases.
So an autotransformer has advantages only for low values of
transformation ratio.
Disadvantages of Autotransformers
• There is a direct connection between the primary and secondary.
Therefore, the output is no longer isolated from the input.
• It is not safe for stepping down a high voltage to a low voltage.
• The short - circuit current is much larger than for the two-winding
transformer of the same rating.
• This reduces the effective resistance and reactance.
Applications of Autotransformers
• 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 continuous variable supply.
KVA of Two-winding
Transformer and
Auto Transformer
For safely starting the machines like induction motors, synchronous motors i.e
as a starter.
To give small boost to a distribution cable to compensate for a voltage drop i.e
as a booster.
Application of
Auto-transformer
As a furnace transformer to supply power to the furnace at the required supply
voltage.
For interconnecting the systems which are operating to the furnace at the
required supply voltage.
It can be used to vary the voltage to the load, smoothly at the same voltage
level. Such a device giving smooth and continuous supply using an autotransformer is called ”variac”
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