Transformers

Objectives
• Understand the single phase transformers
• Understand the equivalent circuit of a transformer
• Determine the equivalent circuit parameters
• Explain the losses in a transformer
 Transformer
• 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.
• Consists of two or more coils of wire wrapped around a common ferromagnetic
core.
• One of the transformer windings is connected to a source of ac electric power.
• Second transformer winding supplies electric power to loads
• There are many uses of transformers, transfer of power from one voltage level to
another – Power Transformers – Ex. Generator Transformers (Unit Transformers) ,
Grid Substation Transformers , Distribution Transformers, etc., - Voltage
Transformers – for Voltage sampling for measurements, - Current Transformers
for current sampling and special purpose transformers.
Transformer cont’d…
• Unlike in rotating machines, there is no energy conversion from one
form to another
• A transformer is a static device and all currents and voltages are AC
Two Types of Transformer Windings Core Type and Shell Type
Transformers are constructed in 'core form' or 'shell form'.
When windings surround the core, the transformer is core form;
when windings are surrounded by the core, the transformer is shell
form.
Shell form design may be more prevalent than core form design for
distribution transformer applications due to the relative ease in
stacking the core around winding coils.
 Core form design tends to, as a general rule, be more economical,
and therefore more prevalent, than shell form design for high voltage
power transformer applications at the lower end of their voltage and
power rating ranges (less than or equal to, nominally, 230 kV or
75 MVA). At higher voltage and power ratings, shell form
transformers tend to be more prevalent.
 Shell form design tends to be preferred for extra-high voltage and
higher MVA applications because, though more labour-intensive to
manufacture, shell form transformers are characterized as having
inherently better kVA-to-weight ratio, better short-circuit strength
characteristics and higher immunity to transit damage.
Some Pictures of Transformers
 Ideal Transformer
–no real power losses
–magnetic core has infinite permeability
–no leakage flux
• The Power Transformer are generally classified as step-
up transformers or step-down transformers : where in
step-up transformers the low voltage side is on the
primary and in step-down transformers the higher
voltage is the on primary side .

7
 Ideal Transformer Parameters Useful Relationships
Turns Ratio:

Since there is no power loss:

Hence :

And finally where Vp,Vs,Ip,Is are phasor quantities . Neither the Phase angles of
voltage nor current change due to transformer action.
Schematic Diagram of an Ideal Transformer
Dot Convention:
Impedance Transformation Explanation Using the
schematic diagram
The Load Impedance:
The Load Impedance Referred to
the Primary side or Apparent
Impedance of the Primary :
Primary Voltage:
Primary Current:
Hence the Primary Apparent
Impedance:
Ideal Transformer Relationships
Theoretical Considerations
Assume we have flux m in magnetic material.
Then flux linking coil 1 having N1 turns is:
1  N1m , and similarly 2  N 2m
d 1 d m d 2 d m
v1   N1 , v2   N2
dt dt dt dt
dm v1 v2 V1 N1
     a = turns ratio
dt N1 N2 V2 N2
Current Relationships
To get the current relationships use ampere's law
with path around core having total length L:
mmf    HdL  N1i1  N 2i2'

H  L  N1i1  N 2i2'
BL
 N1i1  N 2i2'

Assuming uniform flux density in the core
having area A, then   B  A
L
 N1i1  N 2i2'
A 12
Current/Voltage Relationships
If  is infinite then 0  N1i1  N 2i2' . Hence
i1 N2 i1 N2 1
'
  or   , where i2  i2'
i2 N1 i2 N1 a
I1 1
Then:  and:
I2 a
 a 0
V1    V2 
I   1 I 
 1  0   2
 a

13
Impedance Transformation Theoretical Consideration
An impedance load Z on the secondary can be referred to the primary voltage and
current in the following manner:

1 V2
V1  aV2 I1 
aZ
V1
 a 2 Z  primary referred value of
I1
secondary load impedance

14
Power in an Ideal Transformer
Power Supplied to the Primary winding:
Power Supplied by the secondary of the Transformer:
Since = and

Hence:

It also follows:
Ideal transformer

• No winding Resistance
• No leakage flux
• No iron losses
Equivalent Circuit of an Ideal Transformer

• Where Z1=a²ZL Z1 in this case is called the Equivalent Impedance of

secondary referred to Primary side.
Analysis of Circuits Containing Ideal Transformers
• Replace (one side) the Secondary side of the Transformer with its equivalent
referred to the (other side) the Primary side.
• Solve the equivalent circuit using one circuit to obtain the voltages and the
currents.

• Example: A single-phase power system consists of a 480-V 50-Hz generator

supplying a load ZLoad = 4 + j3 Ω through a transmission line of impedance
Z Line = O.I8 + jO.24 Ω. Answer the following questions about this system.
(i)If the power system is exactly as described above, what will the voltage at the load
be? What will the transmission line losses be.
(ii) Suppose a I: 10 step-up transformer is placed at the generator end of the
transmission line and a 10: I step-down transformer is placed at the load end of the
line . What will the load voltage be now? What will the transmission line losses be
now?
Example (i)
Figure shows the power system without transformers. Here IG = Iline = I load The line current in this
system is given by
Example (ii)
Figure shows the power system with the transformers. To analyze this system, it is necessary to convert it to a common voltage level. This is
done in two steps:
I. Eliminate transformer T2 by referring the load over to the transmission line's voltage level.
2. Eliminate transformer TI by referring the transmission line's elements and the equivalent load at the transmission line's voltage over to the
source side.
The value of the load's impedance when reflected to the transmission system's
voltage is:
System with the load and transmission line referred to the generator voltage level.

The total impedance at the transmission

Notice that
line level is now reflected across Tl
to the source's voltage level: and The resulting
equivalent circuit
is shown in Figure below The generator
current is
• Knowing the current IG we can now work back and find Iline and Iload working back
through T1 , we get

• working back through T2 , we get

• The load voltage is given by

• and the line losses are given by

• Notice that raising the transmission voltage of the power system reduced
transmission losses by a factor of nearly 90! i.e 1484W Vs. 16.7W
-Also, the voltage at the load increased in the system with transformers
compared to the system without transformers.
• This simple example dramatically illustrates the advantages of using high-voltage
transmission lines as well as the extreme importance of transformers in modern
power systems.
 Real Transformer
• Real transformers-two or more coils of wire
physically wrapped around a ferromagnetic
core.
• The characteristics of a real transformer
approximate the characteristics of an ideal
transformer, but only to a degree.
• Not all the flux produced by either coil pass
through the other coil some amount of flux
travel through the air this is called the
leakage flux.
• The smaller the leakage fluxes of the
transformer , the closer the transformer
voltage ratio approximates that of the ideal
transformer
Real Transformer
• 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 a real ferromagnetic core. Once the peak
flux reaches the saturation point in the core, a small increase in peak flux requires
a very large increase in the peak magnetization current.
• The other component of the no-load current in the transformer is the current
required to supply power to make up the hysteresis and eddy current losses in
the core. – core loss
Real Transformer
• Hence the primary current when the secondary is open circuit consist of
I. The magnetization current iM, which is the current required to produce the flux in
the transformer core and
2. The core-loss current ih+e which is the current required to make up for hysteresis
and eddy current losses

The total no- load current in the core is called the excitation current of the
transformer. It is just the sum of the magnetization current and the core- loss
current in the core:
THE EQUIVALENT CIRCUIT OF A TRANSFORMER
• When deriving an equivalent circuit of a real transformer the following
characteristics will have to be considered in addition to the magnetising
current :
THE EQUIVALENT CIRCUIT OF A TRANSFORMER
• Copper Loss : Copper losses are resistive losses in the primary and secondary
windings of the transformer core. They are modeled by placing a resistor Rp in
the primary circuit of the transformer and a resistor Rs in the secondary circuit.
• the leakage flux will be modelled by primary and secondary inductors
• The magnetization current im is a current proportional (in the unsaturated
region) to the voltage applied to the core and lagging the applied voltage by
90˚, so it can be modelled by a reactance XM connected across the primary
voltage source. The core- loss current ih+e is a current proportional to the voltage
applied to the core that is in phase with the applied voltage, so it can be
modelled by a resistance Re connected across the primary voltage source.
Practical Transformer
• Has : Core losses
Winding resistance
Magnetic leakage
Transformer Losses
• Core losses or Iron losses
- eddy current and hysteresis losses together known as core losses
- Depend on
- The supply frequency,
- Maximum flux density in the core
- Volume of the core
• Copper losses – losses due to the resistance in the windings.
• Flux leakage
Equivalent Circuit of the real transformer

• Ideal Transformer
Equivalent circuit refer to the primary
Equivalent circuit refer to the Secondary
Approximate equivalent circuit referred to the Primary and
Secondary
The excitation branch has a very small current compared to the load current of the transformers. In fact,
it is so small that under normal circumstances it causes a completely negligible voltage drop in Rp and
Xp. hence, a simplified equivalent circuit can be produced that works almost as well as the original
model. The excitation branch is simply moved to the front of the transformer, and the primary and
secondary impedances are left in series with each other. These impedances are just added, creating
the approximate equivalent circuits

Approximate equivalent circuit referred to the Primary and Secondary

Determine the equivalent circuit parameters Open Circuit
Test

• Applying Rated Line Voltage to the Primary and keep the secondary open.
• Measure the Power using a Watt meter get the Real Power component
• Get the power factor by measuring the voltage and the current in the primary
• you can get the Rc and XM
• Generally the OC test is performed in the LV side i.e. High Voltage side is
Opened as we do not need to have high Voltages to be applied.
Determine the equivalent circuit parameters Short Circuit Test

• Apply a very small voltage to the primary until the rated current flows in the transformer, If the
excitation current is ignored, then all the voltage drop in the transformer can be attributed to the
series elements in the circuit.
• Measure the power the real power dissipated.
• Get the power factor using the Voltage and Current.
• Get the impedance and the primary referred series impedance of the transformer can be derived.
• Generally the Short Circuit test is performed on the HV side i.e. low Voltage side short circuited
using an ammeter which is used to get rated current in the secondary side as the short circuit
rated current is very small in the high voltage side and sufficiently large in LV side.
Question
• A 15-kVA, 2300/230-V transformer is to be tested to determine its
excitation branch components, its series impedances. The following
test data have been taken from the primary side of the transformer:

• Open- Circuit Test Short-Circuit Test

= 2300 V Vsc = 47V
= 0.21 A Isc = 6.0A
= 50W Psc = I60W
The data have been taken by using the connections shown in Figures 1
and 2.

(a) Find the equivalent circuit of this transformer referred to the

high-voltage side.

(b) Find the equivalent circuit of this transformer referred to the low-
voltage side.

(c) What is the efficiency of the transformer at full load with a power
factor of 0.8 lagging?
Figure 1

Figure 2
The excitation branch values of the transformer equivalent circuit can be calculated
from the open-circuit test data, and the series elements can be calculated
from the short-circuit test data.

From the open-circuit test data,

the open-circuit Impedance angle

And 1
ZE

Hence
The Calculation of Series Elements:
• From the short-circuit test data,
the short-circuit impedance angle is
The Equivalent Circuit Referred to the Primary Side
The Equivalent Circuit Referred to the Secondary Side
The Secondary side quantities are obtained by using the turns ratio a=2300/230=10
to determine the current and the voltage referred to secondary parameters
and dividing the impedance elements by 100 as a=10
Efficiency of a transformer
• The ratio of output power to input power.
Calculation of the Efficiency of the Transformer
To find the Efficiency of the Transformer we have to calculate the losses. First we calculate
the full Load current We use the equivalent circuit referred to the Secondary side

Then we calculate the losses in the Transformer,

The copper losses are given by:

The Core Loss is Given by:

Calculation of the Efficiency of the Transformer
• The Power Factor is 0.8 and Cos Ѳ =0.8 and Ѳ = 36.9˚
- Hence
Three Phase Transformers.
• Three Phase Transformers are the most common Transformers in most applications
especially in the Power systems and in the Industry.
• Three Phase transformers are constructed in one of two ways :
- Use three single Phase Transformers connected in a three phase arrangement or
- wrap the three sets (primary and the secondary) windings in a common core

Advantages of three single Phase Transformers and their uses :

1. Each transformer is light weight hence is the preferred option when transporting long
distances in rugged terrain.
2. Each transformer can be replaced in case of a fault in any one of them.
Advantages of single three phase Transformer:
3. Efficient than the three single phase transformers.
4. Cheaper and
5. Total weight is lower than the total weight of the three single phase transformers.
The Two Types of Arrangements of Three Phase Transformers
A Three Phase Transformer Bank A Three Phase Transformer wound in a
composed of three single Phase common Core
transformers.
Three Phase Transformer Connections
• The three Phase Transformer, either a three single phase transformer bank or
a single three phase transformer can be connected in one of four
combinations of the primary and the secondary they are:
1. Wye – wye (Y-Y)
2. Wye – delta (Y-Δ )
3. Delta – wye (Δ-Y)
4. Delta- delta(Δ- Δ)
The Y – Y connection of the three Phase Transformer
In the Primary

In the Secondary

The Main disadvantage of Y – Y connection is if the

load of the transformer is unbalanced the voltages on
the phases of the transformer can be severely
unbalanced
The Y – Δ connection of the three Phase Transformer
In the Primary

In the Secondary

The advantage of Y – Δ connection is even if the

load of the transformer is unbalanced the voltages on
the phases of the transformer can be partially
balanced. However the secondary voltage is shifted by
30˚ (lagging) relative to the Primary
The Δ –Y connection of the three Phase Transformer
In the Primary V LP/ VLS = a/√3 Δ-Y

In the Secondary
a = [VLP/ VLS ] * √3
The advantage of Δ –Y connection is the same as in
the case of the Y – Δ connection the secondary
voltage is shifted by 30˚ (lagging) relative to the Primary
The Δ –Δ connection of the three Phase Transformer
In the Primary

In the Secondary

The advantage of Δ –Δ connection is that the secondary

voltage is in phase with the Primary and it has no
problems with the unbalanced loads either.