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Module 5
TRANSFORMERS
CONTENTS
1. Necessity of transformer
2. Principle of operation of transformer
3. Construction of transformers
- Core and Shell types.
4. EMF equation of transformer
5. Losses in Transformer,
6. Efficiency of transformer
7. Voltage regulation
Illustrative problems on e.m.f equation and efficiency only
COURSE OUTCOMES
At the end of the course the student will be able to:
1. State the necessity of transformer in electrical and electronic circuits.
2. State the principle of operation of transformer.
3. Explain the construction and derive the e.m.f equation of transformer.
4. Explain why the hysteresis and eddy current loss occur in the core, and
how these can be reduced.
5. Define efficiency and state the condition for maximum efficiency of transformer.
6. Define regulation of transformer.
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1. Necessity of transformers
For the power grid, transformers are essential for power transmission over long distances.
Without them, our power grid simply would not be able to scale up to meet growing demand.
This is because long lines enable power generation to be located at its energy or cooling source
(a hydroelectric dam or nuclear facility, say). The generated power can then be efficiently sent to
where it's needed (major population centers).
Here's how it works. High-current low voltage AC power at the generator is stepped up with
transformers into high voltage /low current - 220 kilovolts or more at 100A or so - then sent
down the high-tension lines.
The lower current is key: it allows practical wire sizes to be used in the overhead lines, and it
reduces wire losses (sometimes called I2R or voltage drop loss) and allows more of the generated
power to make it to the load and for lower cost.
But 220kV voltage isn't practical to use in the home. So transformers are used to step the voltage
back down to manageable levels. This is done in two stages:
* Receiving substations use transformers to step down the voltage to 13kV – 430 V or so
for local distribution on utility poles or underground.
* Buildings take their feeds from still another set of transformers that take the utility voltage
down to 430 to 220V. These transformers are the familiar canister-shaped devices mounted on
the power poles.
In electronics, small transformers perform a variety of duties: mixing, coupling. impedance
matching, isolation and power supply. These jobs aren't quite as dramatic as those done by the
large power grid transformers but they're no less essential for dealing with AC in these systems.
What is a Transformer?
•
Electrical power transformer is a static device which transforms electrical energy from
one circuit to another without any direct electrical connection and with the help of mutual
induction between two windings.
It transforms power from one circuit to another without changing its frequency but may be in
different voltage level
2. PRINCIPLE OF OPERATION
The working principle of transformer is: It depends upon Faraday's law of electromagnetic
induction. Actually, mutual induction between two or more winding is responsible for
transformation action in an electrical transformer.
The basic concept of the transformer: Whenever we apply alternating current to an electric coil,
there will be an alternating flux surrounding that coil. If we bring another coil near the first one,
there will be an alternating flux linkage with that second coil. As the flux is alternating, there
will be a rate of change in flux linkage with respect to time in the second
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coil. As per Faraday's law of electromagnetic induction an e.m.f
will be induced in it.
The winding which takes electrical power from the
source, is known as primary winding of transformer.
The winding which The winding which gives the desired
output voltage due to mutual induction in the transformer,
is known as secondary winding of transformer.
In open air very tiny portion of the flux of the first winding
will link with second coil.
The current that flows through the closed circuit of second
coil, will be so small in amount that it will be difficult to
measure. The rate of change of flux linkage depends upon
the amount of linked flux with the second winding.
We desire almost all flux of primary winding to link
with the secondary winding. This is effectively and
efficiently done by placing one low reluctance path common
to both of the winding.
This low reluctance path is core of transformer, through which maximum number of flux
produced by the primary is passed through and linked with the secondary winding.
3. CONSTRUCTION OF TRANSFORMER
The three main parts of a transformer are:
•
Primary winding of transformer - which produces magnetic flux when it is connected
to electrical source.
•
Magnetic Core of transformer - the magnetic flux produced by the primary winding,
that will pass through this low reluctance path linked with secondary winding and create
a closed magnetic circuit.
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•
Secondary Winding of transformer - the flux, produced by primary winding, passes
through the core, will link with the secondary winding. This winding also wounds on the
same core and gives the desired output of the transformer
Magnetic Core of transformer
Transformers core is constructed by assembling (stacking) laminated sheets of steel, with
minimum air-gap between them (to achieve continuous magnetic path). The steel used is
having high silicon content and sometimes heat treated, to provide high permeability and
low hysteresis loss. Laminated sheets of steel are used to reduce eddy current loss. The
sheets are cut in the shape as E,I and L. To avoid high reluctance at joints, laminations
are stacked by alternating the sides of joint.
TYPES OF TRANSFORMERS
Transformers can be classified on different basis, like types of construction, types of cooling etc.
On the basis of construction,
1. Core type transformer
2. Shell type transformer
On the basis of their purpose
1. Step up transformer:
2. Step down transformer
On the basis of type of supply
1. Single phase transformer
2. Three phase transformer
On the basis of their use
1. Power transformer
2. Distribution transformer
3. Instrument transformer
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On the basis of cooling
employed
1. Oil-filled self cooled type
2. Oil-filled water cooled type
3. Air blast type (air cooled)
Core type transformer
In core type transformer, windings are cylindrical former wound, mounted on the core limbs as
shown in the figure above. The cylindrical coils have different layers and each layer is insulated
from each other. Materials like paper, cloth or mica can be used for insulation. Low voltage
windings are placed nearer to the core, as they are easier to insulate.
Shell type transformer
The coils are former wound and mounted in layers stacked with insulation between them. A shell
type transformer may have simple rectangular form (as shown in above fig), or it may have a
distributed form
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Comparison between Core and Shell type transformers: Following is the comparison
between core and shell type transformer:
Core Type Transformer
Shell Type Transformer
The core has only one magnetic circuit. The core
cross section is uniform.
Two paths for the flux in the core.
Core has two limbs.
Core has three limbs.
Magnetic flux is uniform at all sections of the core.
Central limb carry double the flux compared to
outer limbs.
Coils are placed on two limbs
Coils are placed on central limb.
The winding surrounds considerable part of core.
Core surrounds the winding.
Better cooling since more coil surface is exposed to
atmosphere.
Cooling is less as less surface of the coil is exposed
to atmosphere.
Natural cooling is provided.
Difficult to provide natural cooling.
Easy access to the winding makes maintenance
easier
Less access to the winding makes maintenance
difficult.
Mechanical protection to coil is less
Mechanical protection to coil is more
The cross section of the central leg will be twice the
cross section of outer limbs.
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4. E.M.F EQUATION OF TRANSFORMER
When a sinusoidal voltage V1 is applied to the primary of the transformer, a sinusoidally varying
magnetic field is set up in the core. This flux is given by: ϕ = ϕm sin ωt = ϕm sin 2πft with ϕm ,
the peak value of the flux and f the frequency of sinusoidal variation of flux.
The induced e.m.f in the primary winding of T1 turns is:
e1 = - T1(dϕ/dt) = - T1 [d(ϕm sin ωt )/dt]
= T1ω ϕm cosωt = T1ω ϕm sin (ωt – π/2)
The peak value of the induced e.m.f, Em = T1ω ϕm
The rms value of the induced e.m.f E1 = Em /√2 =(2π f T1ϕm )/√2 = 4.44 f T1ϕm
The same flux which is linking with the primary winding passes through the core and links with
the secondary winding of T2 turns. The secondary winding will have change in the flux linkages
and therefore an e.m.f induced in it.
The rms value of this e.m.f is: E2 = 4.44 f T2ϕm
Volt per turn = E2/T2 = E1/T1 = 4.44 f ϕm
Volt per turn is constant for a given transformer.
Transformation ratio
The ratio of secondary voltage to the primary voltage is called the transformation ratio or turns
ratio, K.
The rms value of the induced e.m.f E1 = 4.44 f T1ϕm
The rms value of this e.m.f is: E2 = 4.44 f T2ϕm
Transformation ratio, K = E2/E1 = T2 / T1
•
If transformation ratio, K > 1 the transformer is step up transformer.
•
If transformation ratio, K < 1 the transformer is step up transformer.
5. LOSSES IN TRANSFORMER
In transformer, 'loss' can be defined as the difference between input power and output power.
An electrical transformer is an static device, hence mechanical losses (like windage or friction
losses) are absent in it. The losses in a transformer are: iron losses and copper losses.
Iron loss or Core Loss: The core loss has two components:
1.Hysteresis loss
2. Eddy current loss
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Eddy current loss and hysteresis loss depend upon the magnetic properties of the material used
for the construction of core. These losses are known as core losses or iron losses.
1. Hysteresis loss in transformer:
Hysteresis loss is due to reversal of magnetization in the transformer core. This loss depends
upon the volume and grade of the iron, frequency of magnetic reversals and value of flux
density.
It can be given by, Steinmetz formula: Wh= ηBmax1.6 f V (watts)
where, η = Steinmetz hysteresis constant,
V = volume of the core in m3
2. Eddy current loss in transformer:
In transformer, AC current supplied to the primary winding sets up alternating magnetizing flux.
This flux also gets linked with other conducting parts like steel core, which will result in induced
e.m.f in the core, causing small circulating current. This e.m.f does not contribute anything
towards the externally connected load or the output power and is dissipated in the form of heat
energy. The current induced in the core is called eddy current. Due to these eddy currents, some
energy will be dissipated in the form of heat. This heat is called eddy current loss.
It can be given by the formula: We= KeBmax2 f2 V t2 (watts)
where, Ke = eddy current loss constant
t = thickness of lamination.
Bmax = Maximum value of the flux density.V = volume of core
Eddy current losses within a transformer core cannot be eliminated completely, but they can be
greatly reduced and controlled by reducing the thickness of the steel core. Instead of having one
big solid iron core as the magnetic core material of the transformer or coil, the magnetic path is
split up into many thin pressed steel shapes called “laminations”. The laminations used in a
transformer construction are very thin strips of insulated metal joined together to produce a solid
but laminated core. These laminations are insulated from each other by a coat of varnish or paper
to increase the effective resistivity of the core thereby increasing the overall resistance to limit
the flow of the eddy currents.
The result of all this insulation is that the unwanted
induced eddy current power-loss in the core is greatly
reduced, and it is for this reason why the magnetic
iron circuit of every transformer and other
electro - magnetic machines are all laminated.
Using laminations in a transformer construction
reduces eddy current losses
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Copper Losses
Transformer Copper Losses are mainly due to the electrical resistance of the primary and
secondary windings. Most transformer coils are made from copper wire which has resistance.
This resistance opposes the magnetizing currents flowing through them. When a load is
connected to the transformers secondary winding, large electrical currents flow in both the
primary and the secondary windings, electrical energy and power ( or the I2 R ) losses occur as
heat.
Copper losses vary with the load current, being almost zero at no-load, and at a maximum at fullload when current flow is at maximum.
Core Losses for a given transformer depends on the supply voltage as the frequency and other
parameters are constant. As the voltage applied voltage to the primary winding is constant, these
loses are constant. Hysteresis loss can be minimized by selecting suitable core material.
Laminating the core reduces the eddy current loss.
For a given transformer, under all load conditions, the supply voltage and the frequency of
reversals of magnetic flux lines to which the core is subjected are constant the core losses are
constant. For the transformer the primary current and the secondary current are related through
the transformation ratio. When the secondary current changes the primary current also changes.
As load on the transformer vary, these currents also vary, therefore the losses also vary. Thus
copper loss in the transformer is variable losses, whereas core losses are constant losses.
6. EFFICIENCY OF TRANSFORMER
Efficiency of a transformer can be defined as the output power divided by the input power.
Efficiency = output / input.
Transformers are the most highly efficient electrical devices. (full load efficiency between 95%
to 98.5%). A better method to find efficiency of a transformer is using,
Efficiency = (input - losses) / input
= 1 - (losses / input).
As the core loss depends on the voltage and copper loss depends on current, the transformer
power rating will be in volt ampere. The output and input are measured in volt ampere. The
losses are measured in Watt.
For a load at a given power factor, the output of transformer is given by:
= Secondary Voltage x Secondary current x Cosine of the angle between voltage and current.
Full load output of the transformer = V2I2 Cos ϕ2 Watt
Core loss = Wi Watt
Full load Copper loss = Wcu
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Full load Efficiency =
At any load other than full load output of the transformer will vary, Copper losses will vary. The
iron loss will remain constant. For a load factor x (secondary current at any load/ full load
secondary current)
Output = x V2I2 Cos ϕ2 Watt
Core loss = Wi
Copper loss = x2 Wcu (copper loss will vary as square of the current)
Efficiency at any load x =
If the full load current is 25 Ampere and the transformer is supplying 15 ampere to a load, then
load factor x = (15 / 25 ) = 0.6
x is 0 for no load and 1 for full load.
Condition for Maximum Efficiency
Efficiency can be computed from the primary also.
Efficiency = (Input – Total losses)/ Input
Input = V1I1 Cos ϕ1 Watt
Core loss = Wi Copper loss = Wcu
Efficiency =
=
In the above equation the current I1 is variable. For maximum efficiency, the derivative of
efficiency with respect to I1 must be equal to zero.
d(efficiency)/dt = 0. On differentiating,
variable loss = constant loss
When Copper loss = core loss, the efficiency will be maximum.
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7. VOLTAGE REGULATION OF TRANSFORMER
An electrical power transformer is open circuited, means load is not connected with secondary
terminals. In this situation, the secondary terminal voltage of the transformer will be its
secondary induced e.m.f E2. Whenever full load is connected to the secondary terminals of the
transformer, rated current I2 flows through the secondary circuit and voltage is dropped.
The primary winding will also draw equivalent full load current from source.
The voltage drop in the secondary is I2Z2 where Z2 is the secondary impedance of transformer.
At this loading condition, the voltage between secondary terminals is V2 across load terminals
which will be less than no load secondary voltage E2 and this is because of I2Z2,voltage drop in
the transformer.
The voltage regulation is the percentage of voltage difference between no load and full load
voltages of a transformer with respect to its full load voltage.
Expression of Voltage Regulation of Transformer, represented in percentage, is
Voltage regulation = [(E2 – V2)/V2 ] x 100.
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PROBLEMS ON TRANSFORMERS
Problem 1
A single phase, 20 KVA transformer has 1000 primary turns and 2500 secondary turns.
The net cross sectional area of the core is 100 sq.cm. When the primary winding is
connected to 500 V, 50 Hz, find (a) the maximum flux density in the core (b) the voltage
induced in the secondary winding and (c) the primary and secondary full load currents.
Solution:
Rating of transformer = 20 KVA
Tp primary turns = 1000 and Ts secondary turns = 2500.
The net cross sectional area of the core Ai = 100 sq.cm. primary winding voltage = 500 V, 50
Hz,
(a) The maximum flux density in the core = 500/(4.44 x f x Ai x Tp) = 0.2252 Weber
(b) the voltage induced in the secondary winding = (v2/V1) = Ts/ Tp
= 500 X (2500/1000) = 1250 V
(c) the primary and secondary full load currents:
Ip = 20000/500 = 40 A and Is = 20000/1250 = 16 A
Problem:2
Find the number of turns on the primary and secondary side of a 440 / 230 V, 50 Hz single
phase transformer if the net area of cross section of the core is 30 cm2 and maximum flux
density is 1 Weber/m2.
Solution:
Voltage ratio = 440/230 Supply frequency = 50 Hz
net area of cross section of the core Ai =30 cm2 = 30 x 10 – 4 m2
maximum flux density Bm =1 Weber/m2
Induced voltage in the primary = 440 = 4.44 x f x Bm x Ai x Tp
Primary turns Tp = 440/(4.44 x 50 x 1 x 30 x 10 – 4) = 661 turns
Induced voltage in the secondary = 230= 4.44 x f x Bm x Ai x Ts
Secondary turns Ts = 230/(4.44 x 50 x 1 x 30 x 10 – 4) =346
For a transformer V2/V1 = Ts/ Tp
Ts = Tp(V2/V1) = 661 x(230/440) = 346
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Problem 3
The primary winding of a transformer is connected to a 240 V, 50Hz supply. The
secondary winding has 1500 turns. If the maximum value of the core flux is 0.00207 Weber,
determine (a) Secondary induced e.m.f (b) number of turns in the primary (c) cross
sectional area of core, if the flux has maximum value of 0.465 Tesla.
Solution:
The primary voltage, Vp = 240 V, 50Hz.
The secondary winding turns, Ts = 1500 turns. If the maximum value of the core flux, ϕm =
0.00207 Weber,
(a) Secondary induced e.m.f Vs = 4.44 x f x ϕm x Ts = 4.44 x 50 x 0.00207 x 1500 = 689.31 V
(b) number of turns in the primary Tp:
Ts/Tp = (V2/V1) Tp = Ts (V1/V2) = 1500 x (240/689.31) = 523.
(c) cross sectional area of core, if the flux density has maximum value of 0.465 Tesla.
Ai = ϕm /Bm = 0.00207/ 0.465 = 44.516 x 10 – 4 m2
Problem 4
A single phase transformer has 1000 turns on its primary and 400 turns on the secondary
side. An ac voltage of 1250 V, 50 Hz is applied to its primary side, with secondary open
circuited. Calculate: (a) Secondary e.m.f (b) maximum value of the flux density given that
the cross sectional area is 60 cm2
Solution:
Primary turns Tp = 1000.
Secondary turns Ts = 400.
Primary voltage = 1250 V, 50 Hz .with secondary open circuited.
(a) Secondary e.m.f V2: (V2/V1) = (Ts/Tp); V2 = V1(Ts/Tp)
= 1250 x (400/ 1000) = 500 Volts.
(b) maximum value of the flux density given that the cross sectional area is 60 cm2
Induced e.m.f = 4.44 x f x Bm x Ai x Ts
500 = 4.44 x 50 x Bm x 60 x 10 – 4 x 400
Bm = 0.9384 Tesla.
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Problem 5
A single phase transformer has 400 primary turns and 1000 secondary turns. The net cross
sectional area of the core is 60 cm2. If the primary winding be connected to a 50 Hz supply
at 500 V, calculate (a) the peak value of the flux density in the core and (b) the voltage
induced in the secondary.
Solution:
primary turns Tp = 400
secondary turns Ts = 1000
The net cross sectional area of the core = 60 cm2. If the Supply voltage to primary = 50 Hz
supply at 500 V,
(a) the peak value of the flux density in the core
primary voltage, Vp = 4.44 x f x Ai x Bm x Tp
500 = 4.44 x 50 x 60 x 10 – 4 x Bm x 400
Bm = 0.9383 Tesla
(b) the voltage induced in the secondary.
(V2/V1)= (Ts/Tp); V2 = V1(Ts/Tp) = 500 (1000/400) = 1250 V
Problem 6
The required no load ratio in a single phase, 50 Hz, core type transformer is 6000 / 250.
Find the number of turns in each winding if the flux is to be about 0.06 Weber.
Solution:
The required no load ratio = 6000/ 250. Supply frequency = 50 Hz,
flux ϕm = 0.06 Weber.
primary voltage, Vp = 4.44 x f x Ai x Bm x Tp
6000 = 4.44 x 50 x ϕm x Tp
Primary turns = 6000/(4.44 x 50 x 0.06) = 451
Secondary turns Ts = Tp(V2/V1) = 451 x(250/6000) = 19.
Problem 7
A single phase 50 Hz core type transformer has a square core of 20 cm side. The
permissible maximum density is 1 Weber/ m2. Calculate the number of turns per limb on
the high and low voltage sides for a 3000 / 220 V ratio.
Solution:
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Frequency = 50 Hz
core cross section = square core of 20 cm side.
Net core area = 0.9 x gross area = 0.9 x 0.2 x 0.2 = 0.036m2
The permissible maximum density = 1 Weber/ m2.
Primary voltage Vp = 3000; Secondary voltage Vs = 220.
No. of turns on high voltage side:
Vp = 4.44 x f x Ai x Bm x T p
Tp = Vp/(4.44 x f xBm x Ai)
= 3000/(4.44 x 50 x 1 x 0.036) = 376 turns
Turns of low voltage side: Vs/Vp = Ts/Tp ; Ts= Tp(Vs/Vp)
= 376( 220/3000) = 28
Problem 8
A 50 KVA, 400 / 200 V, single phase transformer has an efficiency of 98% at full load and
0.8 power factor while its efficiency is 96.9 % at 25 percent full load and unity power
factor. Determine the iron loss and full load copper loses and voltage regulation if the
voltage on full load is 195 V.
Solution:
Rating of the transformer = 50 KVA,
400 / 200 V, single phase transformer
full load efficiency at 0.8 power factor = 98%
0.98 = 50000 x 0.8/[50000 x 0.8 +Wi + Wcu]
Wi + Wcu = 816.32
At 25 % full load and unity power factor efficiency= 96.9 %.
0.969 = 0.25 x 50000/[0.25 x 50000 +Wi + 0.252 x Wcu]
Wi + 0.252 x Wcu = 399.896; solving
Wi = 372.135 W & Wcu = 444.18 W
voltage regulation = (no load voltage – full load voltage)/voltage on full load
= [ (200 – 195) / 195] x 100 = 2.564 %
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Problem 9
The maximum efficiency at full load and unity power factor of a single phase, 25 KVA, 500
/ 1000 V, 50 Hz transformer is 98%. Determine its efficiency at (a) 75% load, 0.9 power
factor, (b) 50% load, 0.8 power factor (c) 25% load, 0.6 power factor.
Solution:
Rating of transformer = 25kVA
The maximum efficiency at full load and unity power factor = 98%.
0.98 = 25000 / (25000 + Wi + Wcu)
For max. efficiency copper loss = core loss
Full load copper loss Wcu = 255.1 W
Iron loss Wi = 255.1
Load
Power factor
Efficiency
0.75
0.9
0.9769
0.5
0.8
0.9690
0.25
0.6
0.9326
Problem 10
A transformer is rated at 100KVA. At full load its copper loss is 1200 W and its iron loss is
960 W. Calculate (a) the efficiency at full load, unity power factor (b) efficiency at half full
load, 0.8 power factor (c) the load KVA at which maximum efficiency will occur and (d)
Maximum efficiency at 0.85 power factor.
Solution:
Transformer rating 100KVA.
Full load copper loss is 1200 W; Iron loss is 960 W.
(a) the efficiency at full load, unity power factor =
100 /(100 + 0.96 + 1.2) = 0.9788
(b) efficiency at half full load, 0.8 power factor =
(100 x 0.5 x 0.8)/[100 x 0.5x0.8)+0.96 + (0.5)2x1.2] = 0.9694
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(c) the load KVA at which maximum efficiency will occur
= 100 x √(960/1200) = 89.44 kVA
(d) Maximum efficiency at 0.85 power factor
= 89.44 x 0.85 /[89.44 x0.85 + 0.96 + 0.96]= 0.9753
Problem 11
A 600KVA, single phase transformer has an efficiency of 92% both at full load and half full
load, unity power factor. Determine the efficiency at 75% full load, 0.9 power factor.
Solution:
Rating = 600KVA,
Efficiency at full load unity power factor = 0.92 and
Efficiency at half full load, unity power factor = 0.92.
0.92 = 600 x 1/[600 x 1 +Wi + Wcu]
Wi + Wcu = 52.174kW
0.92 = 600 x 0.5/[600 x 0.5 +Wi +(0.5)2 Wcu]
Wi +(0.5)2 Wcu = 26.087kW
Solving: Wi = 17.391kW and Wcu = 34.782kW
The efficiency at 75% full load, 0.9 power factor =
600 x 0.75 x 0.9 /[600 x 0.75 x 0.9 + 17.391 + (0.75)2x 34.782]
= 91.638
Problem 12
A single phase transformer working at 0.8 power factor has an efficiency 94% at both three
fourth full load and full load of 600KW. Determine the efficiency at half full load, unity
power factor.
Solution:
Full load output = 600 kW
efficiency at three fourth full load 0.8 power factor = 0.94
efficiency at full load 0.8 power factor = 0.94
0.94 = 600 /[600 +Wi +Wcu]
Wi + Wcu = 38.297kW
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0.94 = 600x0.75/[600x0.75 +Wi +(0.75)2Wcu]
Wi +(0.75)2Wcu = 28.723kW.
Solving: Wi = 16.413kW & Wcu = 21.883kW
Efficiency at half full load, unity power factor =
600 x 0.5/[600 x 0.5 + 16.413 +(0.5)2 x 21.883] = 0.932
Problem 13
A 40 KVA single phase transformer has core loss of 450 W and full load copper loss of 850
W. If the power factor of the load is 0.8 calculate: (a) full load efficiency, (b) maximum
efficiency at unity power factor and (c) load for maximum efficiency.
Solution:
Full load output = 40kVA = 40000 VA
Iron loss = 450 W
Copper losses = 850 W respectively.
The efficiency on 0.8 power factor at:
(a) full load = [ ( 40000 x0.8/( 40000 x0.8 + 450 +850)] x 100 = 96.09%
(b) Maximum efficiency = [(40000/(40000+450+450)] x 100 = 97.799%
(c) The load for maximum efficiency = 40 x√(450/850) = 29.1kVA
Problem 14
A 1kW single phase transformer has core loss of 15 Watts and full load copper loss of 20
Watts, Calculate the efficiency at (a) full load 0.9 power factor lagging (b) Half full load
unity power factor and (c) ¾ full load 0.707 power factor leading.
Solution:
Output of transformer = 1kW = 1000 Watt
core loss of 15 Watts and
full load copper loss of 20 Watts,
The efficiency =
Load factor x full load output x power factor/[(load factor x output x power factor + iron loss+
(load factor)2x full load copper loss)]
The efficiency for the given load conditions are given in the following table:
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Load
Power factor
Efficiency
Full load
0.9
0.9625
Half full load
1
0.9615
¾ full load
0.707
0.9528
Problem 15
A single phase 25 KVA, 1000 / 2000 V, 50 Hz transformer has maximum efficiency of 98 %
at full load unity power factor. Determine its efficiency at (a) ¾ full load unity power factor
(b) ½ full load 0.8 power factor (c) 1.25 full load 0.9 power factor.
Solution:
Full load output = 25 KVA,
Maximum efficiency of 98 % occurs at full load unity power factor.
0.98 = 25000/(25000 + Wi +Wcu)
At maximum efficiency copper loss and core loss are equal.
Wi + Wcu = 2(Wcu) = 510.2.
Therefore Wcu = Wi = 255.1 W
The efficiency for the given load conditions are given in the following table:
load
Power factor
efficiency
¾ full load
unity
0.9791
½ full load
0.8
0.9690
1.25 full load
0.9
0.9773
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Problem 16
A 250 KVA single phase transformer has 98.135 % efficiency at full load and 0.8 power
factor lagging. The efficiency at half load and 0.8 power factor lagging is 97.751%.
Calculate the iron loss and full load copper loss.
Solution:
Full load output = 250 KVA
efficiency at full load and 0.8 power factor lagging = 0.98135. efficiency at half load and 0.8
power factor lagging= 97.751%
0.98135 = (250 x 0.8)/[250 x0.8 + Wi +Wcu]
Wi +Wcu = 3.8
0.97751 = (250 x0.8x0.5)/[(250 x 0.8 x 0.5)+ Wi +(1/2)2 Wcu]
Wi +0.25Wcu = 2.3
Solving : Wcu = 2 and Wi = 1.8
The iron loss = 1.8kW and full load copper loss = 2 kW.
Problem 17
A 600 KVA transformer has a maximum efficiency of 92% at full load and half load unity
power factor, the power factor is 0.9 at half load. Determine its efficiency at 75% of full
load and 0.9 power factor.
Solution:
Rated output = 600 KVA
maximum efficiency = 92% at full load unity power factor
Half load efficiency = 92% , at 0.9 power factor
0.92 = 600 x 1000 / (600x 1000 + Wi + Wcu)
Wi + Wcu = 600 x 1000[(1/0.92) - 1] = 52,174 W
0.92=( 600x1000x0.9x1/2)/[600x1000x0.9x1/2)+Wi+(1/4)Wcu]
= 23478.26 W. Solving Wi = 13913W & Wcu = 38261W. Efficiency at 75% of full load
and 0.9 power factor =
(600 x1000x0.75x0.9)/[(600 x1000x0.75x0.9)+13913+(0.75)2x38261] = 0.9195
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Problem 18
Find the efficiency of a 150 KVA transformer at 25%, 38% and 100% full load, (a) at unity
power factor, (b) at power factor 0.8 lagging, if the copper loss is 1600 W at full load and
iron loss is 1400 W.
Solution:
Output of transformer at full load = 150 kVA.
The full load copper loss =1600 W. The iron loss is 1400 W.
Efficiency =
{ (x x full load output in volt ampere x power factor) / [(x x full load output in volt ampere x
power factor) + iron loss + x2 x full load copper loss]} x 100
x = ½ , 0.38, 1.0 and power factors are 1 and 0.8
Load factor x
unity power factor
0.8 power factor
0.25
96.15
95.23
0.38
97.22
96.54
1
98
97.56
Problem 19
In a 25 KVA, 2000 / 200 V transformer the iron and copper losses are 400 and 850 W
respectively. Calculate the efficiency on unity power factor at (a) full load and (b) half load,
(c) Determine the load for maximum efficiency and the iron loss and the copper loss in this
case
Full load output = 25kVA = 25000 VA
Iron loss = 400 W
Copper losses = 850 W respectively.
The efficiency on unity power factor at:
(a) full load = [ ( 25000 /( 25000 + 400 +850)] x 100 = 95.23%
(b) half load=[(1/2x25000/(25000+400+(1/2)2x 850)] x 100 = 95.32%
(c) The load for maximum efficiency = 25 x√(400/850) = 17.15kVA
iron loss = 400 W and the copper loss = 400 W
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Problem 20
Calculate the efficiencies at half, full and 1.25 load of a 100 KVA transformer for power
factors (a) unity (b) 0.8. The copper loss is 1000 W at full load and the iron loss is 1000 W.
Output of transformer at full load = 100 kVA.
The full load copper loss =1000 W. The iron loss is 1000 W.
Efficiency =
{ (x x full load output in volt ampere x power factor) / [(x x full load output in volt ampere x
power factor) + iron loss + x2 x full load copper loss]} x 100
x = ½ , 1, 1.25 for half , full and 1.25 load respectively power factors are 1 and 0.8
Load factor x
unity power factor
0.8 power factor
½
97.56
96.96
1
98.039
97.56
1.25
97.99
97.5
Theory questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
In a transformer the iron loss is the constant loss and copper loss is the variable loss.
Justify.
Explain the operation of the transformer giving its no load vector diagram
Mention various power losses in a transformer and explain how they are minimized in
practice.
In what way does the core type transformer differ in construction from shell type
transformer? With figures compare the magnetic circuits.
Derive an expression for electromotive force induced in secondary winding of a
transformer.
What are the losses in a transformer? On what factors do they depend? How they are
minimized.
How is efficiency of transformer pre determined? Obtain the condition for maximum
efficiency.
Define voltage regulation of transformer. What is its importance?
Explain the construction and operation of transformer. Show that the voltage ratio of
primary and secondary winding is same as their turns ratio.
Explain the construction and principle of operation of a core type transformer
Derive the e.m.f equation of a single phase transformer.
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