Lecture 5
Single and Three Phase Transformers
Single Phase and three phase Transformers
In the class the following topics will be covered:
1) A comprehensive overview of transformer, essentials, and how does it operate
2) Explain ideal and real transformers construction, equivalent circuit, phasor
diagrams
3) Transformer Design Equations, Performance of transformers,
4) Solving examples on these topics
By the end of this class you should be able to
1) Understand electromagnetism, induction and how these apply to transformer
operation
2) Understand how to analyze, and find the performance of Transformer
3) Design, calculate performance, and identify different types of Single phase
transformers
Introduction
•
For transmission and distribution networks to transfer large amounts of alternating
current electricity over long distances with minimum losses and least cost, different
voltage levels are required in the various parts of the networks.
•
For example, the transfer of electricity efficiently over a long transmission line requires
the use of high voltages. At the receiving end where the electricity is used, the high
voltage has to be reduced to the levels required by the consumer.
•
Transformers enable these changes in voltage to be carried out easily, cheaply and
efficiently.
•
A transformer used to increase the voltage is called a "step up" transformer, while that
used to decrease the voltage is called a "step down" transformer.
Introduction
•
What is a transformer?
– A device for increasing or decreasing an AC voltage
– Power Transformers, TV sets to provide High Voltage to picture tubes,
portable electronic device converters, transformers on the pole, etc. are few
examples
•
A transformer consists of two coils of wires known as primary and secondary
windings
– The two coils can be interwoven or linked by a laminated soft iron core to reduce
eddy current losses
Basic Components of Transformers
A transformer consists of two coils electrically separate but linked by a common magnetic circuit of
low reluctance formed by a laminated soft iron core. If one coil (the primary coil) is connected to an
AC supply, an alternating magnetic flux is set up in the iron core.
This alternating magnetic flux passes through the secondary coil and induces and alternating voltage
in the secondary coil.
The magnitude of the secondary voltage is directly proportional to the ratio of the number of turns in
the secondary and primary windings and to the primary voltage.
Figure 1
How do transformers work?
– A changing current through a coil of wire can create a changing magnetic
field.
– Currents can be induced in other wires by these changing magnetic field.
– Therefore, the primary coil current must have AC.
– The iron core of the transformer is not required but it does increases the
efficiency a great deal.
Theory of the Transformer
The operation of a transformer is based on two principles:
•
1. A voltage is induced in a conductor when the conductor passes through a magnetic
field. The same effect is produced if the conductor is stationary but the magnetic field
in which it is located varies; and
•
2. A current passing through a conductor will develop a magnetic field around the
conductor.
•
Note: In this discussion on transformers, the term magnetic "flux" will usually be used
instead of magnetic "field".
A magnetic field is the space or region surrounding a magnet or a current carrying
conductor, in which magnetic effects can be detected.
The strength of the magnetic field is generally expressed in terms of magnetic flux
density (magnetic flux per square meter). Magnetic flux refers to the magnetic lines of
force.
•
•
Transformer Equation
•
•
•
The transformer equation does not work for DC current since there is no
change of magnetic flux
If NS>NP, the output voltage is greater than the input so it is called a stepup transformer while NS<NP is called step-down transformer
Now, it looks like energy conservation is violated since we can get more
emf from smaller ones, right?
– Wrong! Wrong! Wrong! Energy is always conserved!
– A well designed transformer can be more than 99% efficient
– The power output is the same as the input:
–
V p I p = (V s I S )
(
– Transformer equation
Vp
Is
=
Vs
I
=
p
Np
Ns
Transformer Equations
• When an AC voltage is applied to the primary, the
changing B it produces will induce voltage of the same
frequency in the secondary wire
• So how would we make the voltage different?
– By varying the number of loops in each coil
– From Faraday’s law, the induced emf in the secondary is
Vs = N s
dθ B
dt
– The input primary voltage is
dθ B
Vp = N p
dt
– Since dΦB/dt is the same, we obtain
Vs N s
=
Vp N p
Transformer Equations
• Transformer Ratio, this is the most important relationship by which most
of the transformer variables are governed and it is denoted (a)
a=
Number of Turns in the Primary Coil
Number of Turns in the Secondary Coil
Np
a=
Ns
Induction and the Transformer
1000amp
2amp
120V AC
60,000V AC
Figure 2
Remember that the Relative Number of Turns Dictates the Output Current and Voltage
as seen above and by Transformer equations
Transformers Types
Transformers are manufactured in many types the most widely used in power systems are classified
with their core types as seen below. Core type where each of the windings are wound on one leg of
the core, while the shell core type, in which both windings are wound on the same leg. Each type
has its own advantages and disadvantages. Core type is very reliable and easy to maintain, but take
larger space, however, shell type is smaller but not reliable.
A) Core type
B) Shell type
Figure 3 Single phase transformer construction
Ideal Transformer: Transformer Equations
• Transformer Ratio, this is the most important
relationship by which most of the transformer variables are
governed and it is denoted (a)
Number of Turns in the Primary Coil
a=
Number of Turns in the Secondary Coil
Np
a=
Ns
Ideal Transformer
This is the most known relationships that relates the induction between the two coils
(primary/Secondary) of an ideal transformer to turns ratio and their currents
a =
Number
of Turns
N
p
N
s
=
E
p
E
s
Coil
Voltages
Is
=
Ip
Coil
Currents
Ideal Unloaded Transformers
•
•
Winding resistances are zero, no leakage inductance and iron loss
Magnetization current generates a flux that induces voltage in both windings
Ep =
Es =
N p Φm ω
2
Ns Φm ω
2
Figure 4 Current, voltages and flux in an unloaded ideal transformer
Ideal loaded transformer
•
Loaded transformer , means current will be also feeding the load and that
secondary side voltage, current (load Side) will be calculated using the
previous relations
Figure 5 Currents and fluxes in a loaded ideal transformer
Transformers
Remember that the following are mentioned before we bring them here to
emphasize their importance
–
–
Turn ratio is:
a = T =
N
N
p
s
=
Ep
Es
The input and output complex powers are equal
E p I = S p = Ss = Es I
*
p
–
*
s
Notice the direct relations between number of turns and voltages and reciprocal relations
with currents.
a =T =
Ep
Es
Is
=
Ip
Ideal Transformers Equivalent Circuit
Usually we find that ideal transformer symbol and its variables as depicted in figure 7
Ip = Is /T
T
Vp
Ep
Is
Es = V s
V p = Ep = T Es
Figure 7 Equivalent circuit of an ideal transformer
Transformers
Ideal transformer
•
Transferring impedances through a transformer
VP T VS
2 VS
=
=T
ZP =
IP ⎛ IS ⎞
IS
⎜ ⎟
⎝T ⎠
Z P = T 2 Z load
IP
Vac
T
VP
IS
VS
Zload
a) Original circuit with ideal
Figure 8-a Thévenin equivalents of transformer circuit
Transferring impedances through a transformer
We have seen how turns ratio (a) or (T) relates to voltages and currents, the presence of such
ideal transformer relations facilitates the development of real transformers. This is to say its
equivalent circuit and using such relations to transfer from primary to secondary and vice
versa. To transfer impedances of either side then the square of the turns ratio is used.
IP
Vac
VP
T2Zload
b) Equivalent circuit when secondary
impedance is transferred to primary side
and ideal transformer eliminated
(c) Equivalent circuit when primary source is
transferred to secondary side and ideal
transformer eliminated
Figure 8 b-c Thévenin equivalents of transformer circuit
Example 1
A 200 kV A, 6600 V/400 V, 50 Hz single-phase transformer
has 80 turns on the secondary. Calculate:
(i) the appropriate values of the primary and secondary currents;
(ii) the approximate number of primary turns
Solution
V1/V2=N1/N2=a
N1=V1N2/V2= (a) N2
6600/400 (80)= 1220 turns
Transformer rating=Primary Voltage x Primary current
= Secondary Voltage x Secondary Current
For Primary 200 KVA = 6600 v (I1)
I1 = 200KVA/6600v = 30.3 A
For Secondary 200 KVA= 400 I2
I2 = 200 KVA/400v = 500 A
Quiz
A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The
no-load current is 3 A at a power factor of 0.2 lagging.
Calculate the primary current and power factor when the secondary current is 280 A at a power
factor of 0.8 lagging. Assume the voltage drop in the windings to be negligible.
Answer: 58.3 A, 0.78 pf lagging
•
This lecture covers the first essentials that one should know on
transformer operation and ideal transformers. Next lecture we will go
over the practical transformer operation, equivalent circuit and
modeling