Transcript Lecture 7 - UniMAP Portal

DKT 123
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: Block Diagrams of Transformer.
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1.2 Applications of Transformer.
Why do we need transformer?
The Power Grid
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1.2 Applications of Transformer.
Why do we need transformer?
The Transmission Tower
(a) Step Up.
The Power Grid
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.
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1.2 Applications of Transformer.
Why do we need transformer?
The Substation
The Power Grid
The Utility poles
(b) Step Down.
The transformer will stepped down the voltage to the 12kV
to 34.5kV range for local Distribution.
In homes, offices and factories stepped down to 240V.
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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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Cont’d…
A) Core type
B) Shell type
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.
Shell form.
 The shell form construction consists of a three-legged
laminated core with the winding wrapped around the
center leg.
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Cont’d…
Figure 1.3: A Simple Transformer.
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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Cont’d…
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.
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Cont’d…
Operation.
 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.
FARADAY’S LAW
If current produces a
magnetic field, why can't a
magnetic field produce a
current ?
Michael Faraday
In 1831 two people, Michael Faraday in
the UK and Joseph Henry in the US
performed experiments that clearly
demonstrated that a changing magnetic
field produces an induced EMF (voltage)
that would produce a current if the circuit
was complete.
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• When the switch was closed, a momentary deflection was
noticed in the galvanometer after which the current returned
to zero.
• When the switch was opened, the galvanometer deflected
again momentarily, in the other direction. Current was not
detected in the secondary circuit when the switch was left
closed.
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An e.m.f. is made to happen (or induced) in a
conductor (like a piece of metal) whenever it
'cuts' magnetic field lines by moving across
them. This does not work when it is stationary. If
the conductor is part of a complete circuit a
current is also produced.
• Faraday found that the induced e.m.f. increases if
(i) the speed of motion of the magnet or coil increases.
(ii) the number of turns on the coil is made larger.
(iii) the strength of the magnet is increased.
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Faraday’s Law
f
EN
Δt
•
•
•
•
E = Electromotive force (emf)
Φ = Flux
N = Number of turn
t = time
• Any change in the magnetic environment of a coil of wire
will cause a voltage (emf) to be "induced" in the coil. No
matter how the change is produced, the voltage will be
generated.
• The change could be produced by changing the magnetic
field strength, moving a magnet toward or away from the
coil, moving the coil into or out of the magnetic field, rotating
the coil relative to the magnet, etc.
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• Inserting a magnet into a coil
also produces an induced
voltage or current.
• The faster speed of insertion/
retraction, the higher the induced
voltage.
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Figure 1.4: Basic Transformer
Components.
 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,
d
d
e
 N
dt
dt
df
e1  N 1 ;
dt
df
e2  N 2
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
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Cont’d…
 Lenz’s Law states that the direction of e1 is such to produce a
current that opposes the flux changes.
 If the winding resistance is neglected, then equation (1.1) become;
v1  e1  N
df
1( );
dt
df
v 2  e 2  N 2( )
dt
(1.2)
 Taking the voltage ratio in equation (1.2) results in,
N 1 e1

N 2 e2
(1.3)
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Cont’d…
 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
(1.5)
a > 1  Step down transformer
a < 1  Step up transformer
a = 1  Isolation Transformer
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Cont’d…
 According to Lenz’a Law, the direction of e is oppose
the flux changes, and the flux varies sinusoidally such
that
f = fmax sin t
(1.6)
fmax
 Substitute eqn(1.6) into eqn(1.2)
eN
df
d
 N (f max sin 2ft )
dt
dt
(1.7)
 The rms value of the induce voltage is;
E
Nf max
2
 4.44 fNf max
(1.8)
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Cont’d…
 Losses are composed of two parts;
(a) The Eddy-Current lost.
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.
(b) The Hysteresis loss.
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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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 below shows the relationship of input voltage
and output voltage of the ideal transformer.
An Ideal Transformer and the Schematic Symbols.
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Cont’d…
 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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Cont’d…
 The relationship between current into the primary side,
Ip(t), of transformer versus the secondary side, Is(t), of
the transformer;
N p I p (t )  N s I s (t )
I p (t )
1

I s (t ) a
 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
a
Ip
1

Is a
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Cont’d…

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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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
E1
N1 
4.44 * f * fmax
220
 166turns.
3
(4.44)(60)(5 X 10 )
110
N2 
 83turns.
3
(4.44)(60)(5 X 10 )

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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 q 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 q s
where, qs is the angle between the secondary voltage and the
secondary current.
 Voltage and current angles are unaffected by an ideal transformer ,
qp – qs  q. The primary and secondary windings of an ideal
transformer have the same power factor.
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Cont’d…

The power out of a transformer;
Pout  Vs I s cos q
- apply Vs= Vp/a and Is= aIp into the above equation gives,
Pout 
Pout
Vp
(aI p ) cos q
a
 V p I p cos q  Pin
- The output power of an ideal transformer is equal to the input
power.
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Cont’d…

The reactive power, Q, and the apparent power, S;
Qin  V p I p sin q  Vs I s sin q  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.
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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,
Solution:
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Cont’d…Example 2
.
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1.7 The Exact Equivalent Circuit of a Real
Transformer.
 Figure below is an exact model of a transformer.
Model of a Real 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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1.7 The Exact Equivalent Circuit of a Real
Transformer.
 To analyze the transformer it is necessary to convert the
entire circuit to an equivalent circuit at a single voltage level.
(a) The Transformer Model Referred to its Primary Windings.
(b) The Transformer Model Referred to its Secondary Windings.
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Cont’d…
 Symbols used for the Exact Equivalent Circuit above;
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Refer to primary side
V  aV2
'
2
1
I  I2
a
'
2
Z 2'  a 2 Z 2
Refer to Secondary side
I1'  aI1
1
V  V1
a
'
1
1
Z  2 Z1
a
'
1
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The Approximate Equivalent Circuit of a Transformer
Approximate Transformer Model Referred to the Primary Side.

The equivalent impedance for the circuit is;
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Approximate Circuit Model of a Transformer Referred to the Secondary.
 The equivalent impedance for the circuit 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
VR  a
 Vs , fl
Vs , fl
100%
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1.9 Transformer Voltage Regulation and
Efficiency.
 In the per-unit system;
VR 
V p , pu  Vs , fl , pu
Vs , f , pul
 100%
 For ideal transformer VR=0. It is a good practice to have as
small voltage regulator as possible.
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Cont’d…
Example of Transformer Voltage Regulation.
46
Cont’d…
 Transformer Efficiency, efficiency of a transformer is defined
as follows;
Output Power P2


input 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).
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Cont’d…
 Thus, in terms of the total losses, Plosses, the above equation
may be expressed as;
P1  Plosses
P2
P2



P1
P2  Plosses P2  Pcopper  Pcore
 The winding or copper loss is load dependent, whereas the
core loss is constant and almost independent of the load on
the transformer.
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Example 3: Transformer Voltage Regulation.
49
Cont’d…Example 3
1.1
1
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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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Cont’d…
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, high-current 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 high-voltage side of the transformer.
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