Transcript Lecture 04 Transformer

mohd hafiz
ismail
hafizism, january 2007
• 1.0 Transformer
 Introduction to Transformer.
 Applications of Transformer.
 Types and Constructions of Transformers.
 General Theory of Transformer Operation.
 The Ideal Transformer.
 Real Single-Phase Transformer.
 The Exact Equivalent Circuit of a Real Transformer.
 The Approximate Equivalent Circuit of a Transformer.
 Transformer Voltage Regulation and Efficiency.
 Open Circuit and Short Circuit.
 Three Phase Transformer.
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• 1.1 Introduction to Transformer.
 Transformer is a device that
changes ac electrical power at one
voltage level to ac electric power at
another voltage level through the
action of magnetic field.
 Figure 1.1 is the block diagrams of;
(a) transformer,
(b) electric motor
(c) generator.
Figure 1.1: Block Diagrams of
Transformer, an Electric Motor and a
Generator..
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Figure 1.1a: A 500 MVA Power Transformer. (left)
A Pole-Mount 15 kVA Distribution Transformer. (right) (Courtesy of ABB)
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• 1.2 Applications of Transformer.

Why do we need transformer ?
(a) Step Up.
 Step up transformer, it will decrease the current to keep the power
into the device equal to the power out of it.
 In modern power system, electrical power is generated at voltage of
12kV to 25kV. Transformer will step up the voltage to between
110kV to 1000kV for transmission over long distance at very low
lost.
(b) Step Down.
 The transformer will stepped down the voltage to the 12kV to 34.5kV
range for local distribution in the homes, offices and factories as low
as 120V (America) and 240V (Malaysia).
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• 1.3 Types and Constructions of Transformer.
Power transformers are constructed on two types of
cores;
(i) Core form.
(ii) Shell form.
A) Core type
B) Shell type
Figure 1.2: Core Form and Shell Form.
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Core form.
 The core form construction consists of a simple rectangular
laminated piece of steel with the transform winding wrapped
around the two sides of the rectangle.
A) Core type
B) Shell type
Shell form.
 The shell form construction consists of a three-legged laminated
core with the winding wrapped around the center leg.
 In both cases the core is constructed of thin laminations electrically
isolated from each other in order to minimize eddy current.
 The coils are usually not directly connected.
 The common magnetic flux present within the coils connects the
coils.
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Construction.
 Transformer consists of two or more coils of wire wrapped around a
common ferromagnetic core. The coils are usually not directly
connected.
 The common magnetic flux present within the coils connects the
coils.
 There are two windings;
(i) Primary winding (input winding); the winding that is connected
to the power source.
(ii) Secondary winding (output winding); the winding connected to
the loads.
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Operation.
• When AC voltage is applied to the primary winding of the
transformer, an AC current will result iL or i2 (current at load).
• The AC primary current i2 set up time varying magnetic flux f in the
core. The flux links the secondary winding of the transformer.
• From the Faraday law, the emf will be induced in the secondary
winding. This is known as transformer action.
• The current i2 will flow in the secondary winding and electric
power will be transfer to the load.
• The direction of the current in the secondary winding is determined
by Len’z law. The secondary current’s direction is such that the flux
produced by this current opposes the change in the original flux
with respect to time.
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1.4 General Theory of Transformer Operation.

According to the Faraday’s law of electromagnetic induction, electromagnetic force
(emf’s) are induced in N1 and N2 due to a time rate of change of fM,
e
d
d
 N
dt
dt
e1  N 1
d
;
dt
e2  N 2
d
dt
Where,
(1.1)
e = instantaneous voltage induced by magnetic field (emf),
 = number of flux linkages between the magnetic field and the electric circuit.
f = effective flux
 Lenz’s Law states that the direction of e1 is such to produce a current that opposes
the flux changes.
Figure 1.4: Basic Transformer
Components.
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 If the winding resistance is neglected, then equation (1.1) become;
v1  e1  N 1( ddt );
v 2  e 2  N 2(
d
)
dt
(1.2)
 Taking the voltage ratio in equation (1.2) results in,
N1
e1

N 2 e2
(1.3)
 Neglecting losses means that the instantaneous power is the same on both sides of
the transformer;
e1i1  e 2i 2
(1.4)
 Combining all the above equation we get the equation (1.5) where a is the turn ratio
of the transformer.
N 1 v1 i 2
a
 
N 2 v 2 i1
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(1.5)
 The flux varies sinusoidally such that;
 = max sin t
m
ax
 The rms value of the induce voltage is;
eN
E
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d
d
 N ( max sin 2ft )
dt
dt
N max
2
 4.44 fN max
 Losses are composed of two parts;
(a) The Eddy-Current lost.
(b) The Hysteresis loss.
 Eddy current lost is basically loss due
to the induced current in the
magnetic material. To reduce this lost,
the magnetic circuit is usually made
of a stack of thin laminations.
 Hysteresis lost is caused by the energy
used in orienting the magnetic
domains of the material along the
field. The lost depends on the
material used.
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Figure 1.5: A Magnetic Hysteresis or B-H
Curve of Core Steel.
Example 1: Transformer.
How many turns must the primary and the secondary windings of a
220 V-110 V, 60 Hz ideal transformer have if the core flux is not
allowed to exceed 5mWb?
Solution:
For an ideal transformer with no losses,
E1  V1  220V
E 2  V2  110V
From the emf equation, we have
N1 
E1
4.11 * f *  max
220
 166turns.
( 4.11)( 60)(5 X 10 3 )
110
N2 
 83turns.
( 4.11)( 60)(5 X 10 3 )

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1.5 The Ideal Transformer.
 An Ideal transformer is a lossless device with an input winding
and an output winding.
 Zero resistance result in zero voltage drops between the terminal
voltages and induced voltages
 Figure 1.6 shows the relationship of input voltage and output
voltage of the ideal transformer.
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Figure 1.6: An Ideal Transformer and the Schematic Symbols.
• The relationship between voltage and the number of turns.
•
Np , number of turns of wire on its primary side.
•
Ns , number of turns of wire on its secondary side.
•
Vp(t), voltage applied to the primary side.
•
Vs(t), voltage applied to the secondary side.
v p (t )
v s (t )
•

Np
Ns
a
where a is defined to be the turns ratio of the transformer.
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• The relationship between current into the primary side, Ip(t), of
transformer versus the secondary side, Is(t), of the transformer;
I p (t )
1

I s (t ) a
N p I p (t )  N s I s (t )
• In term of phasor quantities;
•
-Note that Vp and Vs are in the same phase angle. Ip and Is are in
the same phase angle too.
•
- the turn ratio, a, of the ideal transformer affects the magnitude
only but not the their angle.
Vp
Vs
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a
Ip
1

Is a
•
•
•
The dot convention appearing at one end of each winding tell the
polarity of the voltage and current on the secondary side of the
transformer.
If the primary voltage is positive at the dotted end of the winding
with respect to the undotted end, then the secondary voltage will
be positive at the dotted end also. Voltage polarities are the same
with respect to the doted on each side of the core.
If the primary current of the transformer flow into the dotted end
of the primary winding, the secondary current will flow out of the
dotted end of the secondary winding.
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1.5.1 Power in an Ideal Transformer.
 Power supplied to the transformer by the primary circuit is given by
;
Pin  V p I p cos  p
where, qp is the angle between the primary voltage and the
primary current.
 The power supplied by the transformer secondary circuit to its loads
is given by the equation;
Pout  Vs I s cos  s
where, qs is the angle between the secondary voltage and the
secondary current.
 Voltage and current angles are unaffected by an ideal transformer ,
qs – qs = q. The primary and secondary windings of an ideal
transformer have the same power factor.
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•
The power out of a transformer;
Pout  Vs I s cos 
•
Apply Vs= Vp/a and Is= aIp into the above equation gives,
( aI p ) cos 
a
 V p I p cos   Pin
Pout 
Pout
•
•
Vp
The output power of an ideal transformer is equal to the input power.
The reactive power, Q, and the apparent power, S;
Qin  V p I p sin   Vs I s sin   Qout
S in  V p I p  Vs I s  S out
•
•
•
In term of phasor quantities;
Note that Vp and Vs are in the same phase angle. Ip and Is are in the same phase
angle too.
The turn ratio, a, of the ideal transformer affects the magnitude only but not the
their angle.
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Example 2: Ideal Transformer.
Consider an ideal, single-phase 2400V-240V transformer. The primary is
connected to a 2200V source and the secondary is connected to an impedance
of 2 W < 36.9o, find,
(a) The secondary output current and voltage.
(b) The primary input current.
(c) The load impedance as seen from the primary side.
(d) The input and output apparent power.
(e) The output power factor.
Solution:
(a) The ratio of rated terminal
voltage equal to the actual
turns-ratio as;
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1.6 Real Single-Phase Transformer.
 The ideal transformer in Section 1.5 can never been made. The real
transformer has many imperfections.
 The real transformers consists of two or more coils of wire
physically wrapped around the ferromagnetic core. The real
transformer approximate the characteristic of the ideal transformer.
 Operation if the real transformer;
(i) It consists of two coils of wire wrapped around a transformer
core.
(ii) The primary of the transformer is connected to an ac power
source, and the secondary winding is an open-circuited.
(iii) Figure 1.5 is the hysteresis of the transformer.
(iv) Basic operation from the faraday law,
eind
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d

dt
•
•
l is the flux linkage in the coil across which the voltage is being
induced.
The sum of the flux passing through each turn in the coil added
over all the turns of the coil is;
N
   i
i 1
•
The average flux per turns is given by ;
 
•

N
And Faraday’s law can be written as ,
eind
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d
N
dt
1.7 The Exact Equivalent Circuit of a Real
Transformer.
 Copper losses are resistive loses in the primary and the secondary
windings of the transformer core.
 Copper losses 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 in the primary windings is,
e LP (t )  L p
di p
dt
di s
e LS (t )  Ls
dt
 Figure 1.7 is an exact model of a transformer. To analyze the
transformer it is necessary to convert the entire circuit to an
equivalent circuit at a single voltage level as in Figure 1.8.
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Figure 1.7: Model of a Real Transformer.
Figure 1.8: (a) The Transformer Model Referred to its Primary Windings (top).
hafizism, january 2007 (b) The Transformer Model Referred to its Secondary Voltage Level (bottom).
•
Symbols used for the Exact Equivalent Circuit above;
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•
•
•
The major use of the exact equivalent circuit of a transformer is to
determine the characteristics such as voltage regulation and efficiency.
A phasor diagram for the circuit of Figure 1.8, for lagging power factor
can be obtained by using the following equations:
Based on the above equations and assuming a zero degree reference
angle for V2, the phasor diagram is shown in Figure 1.9 for the exact
equivalent circuit model of a transformer.
Figure 1.9: RMS Phasor Diagram
for the Exact Equivalent Circuit
Model of a Transformer.
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1.8 The Approximate Equivalent Circuit of a
Transformer.
 The Exact transformer in Section 1.6 is complex for a practical
engineering applications.
 In the Approximate model the voltage drop in Rp and Xp is negligible
because the current is very small. Figure 1.10 is the Approximate
equivalent circuit referred to the primary side.
Figure 1.10: Approximate Transformer Model Referred to the Primary Side.
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•
•
•
•
The voltage in the primary series impedance (r1 + jx1) is small,
even at full load. Also, the no load current (I0) is so small that its
effect on the voltage drop in the primary series impedance is negligible.
Therefore, it matters little if the shunt branch of Rc in parallel
with Xm is connected before the primary series impedance or
after it. The core loss and magnetizing currents are not greatly
affected by the move.
Connecting the shunt components right at the input terminals
has the great advantage of permitting the two series impedance
to be combined into one complex impedance.
The equivalent impedance for the circuit in Figure 1.11 is;
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•
•
The value of this equivalent impedance of a particular transformer
depends, of course, on whether the model used is referred to the primary
or secondary.
Figure 1.11 shows the approximate equivalent circuit of transformer
referred to the secondary side.
Figure 1.11: Approximate Circuit Model of a Transformer Referred to the Secondary.
•
The equivalent impedance for the circuit in Figure 1.12 is;
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1.9 Transformer Voltage Regulation and Efficiency.
 Voltage regulation is a measure of the change in the terminal voltage
of the transformer with respect to loading. Therefore the voltage
regulation is defined as:
VR 
Vs ,nl  Vs , fl
Vs , fl
 100%
 At no load, Vs = Vp/a and the voltage regulation can also be express
as;
Vp
 In the per-unit system;
VR  a
 Vs , fl
Vs , fl
 100%
 For ideal transformer VR=0. It is a good practice to have as small
voltage regulator as possible.
VR 
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V p , pu  Vs , fl , pu
Vs , f , pul
 100%
Figure 1.12: Example of Transformer Voltage Regulation.
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 Transformer Efficiency, efficiency of a transformer is defined as
follows;

Output
Output
Power P2

Power P1
 For Non-Ideal transformer, the output power is less than the input
power because of losses. These losses are the winding or I2R loss
(copper losses) and the core loss (hysteresis and eddy-current losses).
 Thus, in terms of the total losses, Plosses, the above equation may be
expressed as;
 The winding or copper loss is load dependent, whereas the core
loss is constant and almost independent of the load on the
transformer.
P1  P1\losses
P2
P2



P1
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P2  Plosses
P2  Pcopper  Pcore
 The efficiency can also be obtained by using the per-unit system.
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1.10 Open Circuit and Short Circuit.
Open Circuit Test.
 The open circuit test is conducted by applying rated voltage at rated
frequency to one of the windings, with the other windings open
circuited. The input power and current are measured.
 For reasons of safety and convenience, the measurements are
made on the low-voltage (LV) side of the transformer.
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Figure 1.13: Equivalent Circuit of the Open-Circuit Test.
• Figure 1.13 is the equivalent circuit for the open-circuit test.
• The high voltage (HV) side is open, the input current is equal to
the no load current or exciting current (I0), and is quite small.
• The voltage drops in the primary leakage reactance and winding
resistance may be neglected and so may the primary loss (I12r1).
• The input power is almost equal to the core loss at rated voltage
and frequency.
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•
qoc is the angle by which Io_LV lags Voc. The core loss current, Ic is in phase with
Voc while Im lags Voc by 90°. Then;
•
The core-loss current, Ic, may be found from above equation, then Rc_LV may be
calculated by the equation below,
•
The magnetization current Im is given by the above equation or may be found
from Ioc and Ic using
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Short Circuit Test.
• The short-circuit test is used to determine the equivalent series
resistance and reactance.
• One winding is shorted at its terminals, and the other winding is
connected through proper meters to a variable, low-voltage, highcurrent source of rated frequency.
• The source voltage is increased until the current into the
transformer reaches rated value. To avoid unnecessary high
currents, the short-circuit measurements are made on the highvoltage side of the transformer.
• The test circuit with the effective equivalent circuit is shown in
Figure 1.14.
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Figure 1.14: Equivalent Circuit of the Short-Circuit Test.
• Neglecting I0, the input power during this test is consumed in the
equivalent resistance referred to the primary or high-voltage side,
Req_HV. Then
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1.11 Three Phase Transformer.
 Almost all the major power generation and distribution systems in
the world today are three-phase ac system.
 Two ways of constructing transformer of three-phase circuit;
(i) Three single phase transformers are connected in three-phase bank.
(ii) Make a three-phased transformer consisting of three sets of
windings wrapped on a common core.
 The three-phased transformer on a common core (ii) is preferred
because it is lighter, smaller, cheaper and slightly more efficient.
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1.11.1 Three-Phase Transformer Connections.
 A three-phase transformer consists of three transformers either
separate or combined on one core.
 There are four possible connections between the secondary and
primary of a three-phase transformer.
(1) Wye-Wye (Y-Y).
(2) Wye-Delta (Y-D).
(3) Delta-Wye (D-Y).
(4) Delta-Delta (D -D).
 Figure 1.15 are these four possible transformer connections.
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Figure 1.15: Three-Phase Transformer Connections and Wiring Diagram.
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(1) Wye-Wye Connection.
VLP
a
VLS
Y Y
(2) Wye-Delta Connection.
 A three-phase transformer consists of three transformers either
separate or combined on one core.
VLP
 3a
Y 
VLS
(3) Delta-Wye Connection.
VLP
3

VLS
a
(4) Delta-Delta Connection.
VLP VP

a
VLS VS
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 Y

Example 5: Three-Phase Transformer.
•
a)
b)
c)
What should be the ratings (voltages and currents) and turns
ratio of a three-phase transformer to transform 10 MVA from 230
kV to 4160 V, if the transformer is to be connected:
wye-delta,
delta-wye, and
delta-delta?
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Example 6: Voltage Regulation at Full Load.
•A 7200V/208V, 50kVA, three-phase distribution
transformer is connected delta-wye. The transformer
has 1.2% resistance and 5% reactance.
Find the voltage regulation at full load, 0.8 power factor
lagging.
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Example 7: Transformer Efficiency.
•If the core loss of the transformer in Example 7 is 1kW, find the
efficiency of this transformer at full load and 0.8 power factor.
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