Introduction to Transformer
A transformer is a device that changes ac electric energy
at one voltage level to ac electric energy at another
voltage level through the action of a magnetic field.
BASIC ELECTRICAL MACHINE
DMT 232/3
Chapter 4:
Single Phase Transformer
1
Why do we need transformer?
2
Use of power transformers
(a) Step Up.
3
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
4
Distribution …
Converts electricity from high voltage and low current to
low voltage and high current … or … vice versa
Electricity is better generated and used at low voltage,
and better transported at high voltage…
Electricity and Magnetism work together to produce
“transformer action”…
(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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6
1
Types and Construction of Transformer
Power transformers are constructed on one of two types of
cores.
1. Core form
Consists of a simple rectangular
laminated piece of steel with
the
transformer
winding
wrapped around two sides of
the rectangle.
2. Shell Form
Consists of a three-legged laminated
core with the windings wrapped around
the center leg.
7
A ) C o re ty pe
9
B ) S h ell ty p e
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
10
to the loads.
General Theory of Transformer Operation.
Operation.
-When AC voltage is applied to the primary winding of the transformer, an
FARADAY’S LAW
AC current will result iL or i2 (current at load).
-The AC primary current i1 set up time varying magnetic flux ф in the core.
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.
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
Michael Faraday
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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12
2
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.
• 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.
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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14
Faraday’s Law
E=N
∆φ
∆t
•
•
•
•
E = Electromotive force (emf)
Φ = Flux
N = Number of turn
t = time
• Inserting a magnet into a coil also
produces an induced voltage or
current.
• 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.
• The faster speed of insertion/
retraction, the higher the induced
voltage.
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16
Cont’d…
Lenz’s Law states that the direction of e1 is such to produce a
current that opposes the flux changes.
Basic Transformer
Components.
If the winding resistance is neglected, then equation (1.1)
become;
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 фM,
e=±
dλ
dΦ
= ±N
dt
dt
e1 = N 1
dφ
;
dt
e2 = N 2
dφ
dt
v1 ≅ e1 = N1( ddtφ );
v2 ≅ e2 = N 2(
dφ
)
dt
(1.2)
Taking the voltage ratio in equation (1.2) results in,
(1.1)
Where,
e = instantaneous voltage induced by magnetic field (emf),
N 1 e1
=
N 2 e2
(1.3)
λ = number of flux linkages between the magnetic field and the
electric circuit.
ф = effective flux
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18
3
Cont’d…
Cont’d…
- According to Lenz’a Law, the direction of e is oppose the flux
changes, and the flux varies sinusoidally such that
Neglecting losses means that the instantaneous power is the
same on both sides of the transformer;
e1i1 = e2i 2
φ = φmax sin ωt
(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
a>1
Step down transformer
a<1
Step up transformer
a=1
Isolation Transformer
(1.6)
φmax
- Substitute eqn(1.6) into eqn(1.2)
(1.5)
e=N
dφ
d
= N (φ max sin 2πft )
dt
dt
(1.7)
- The rms value of the induce voltage is;
E=
ωNφ max
2
=
2πf
Nφ max = 4.44 fNφ max
2
(1.8)
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20
-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 produced to the secondary side.
Ideal Transformer
An ideal transformer is a lossless device with an input winding
and output winding.
v p (t )
vs (t )
=
Np
Ns
=a
where a is defined to be the turns ratio of the transformer.
Relationship between the current in the transformer,
Sketch of an ideal transformer
N p i p (t ) = N s is (t )
i p (t )
i s (t )
Symbols of a transformer
=
1
a
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22
Power in an Ideal Transformer.
Dot convention
-
Power supplied to the transformer by the primary circuit is given by ;
Pin = V p I p cos θ p
1. 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
dots on each side of the core.
where, Ѳp 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 θ s
2. If the primary current of the transformer flows
into the dotted end of the primary winding, the
secondary current will flow out of the dotted
end of the secondary winding.
where, Ѳ s is the angle between the secondary voltage and the secondary
current.
-
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Voltage and current angles are unaffected by an ideal transformer , Ѳ p –
Ѳs = Ѳ. The primary and secondary windings of an ideal transformer
have the same power factor.
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4
Cont’d…
-
Cont’d…
The power out of a transformer;
-
Pout = Vs I s cosθ
The reactive power, Q, and the apparent power, S;
Qin = V p I p sin θ = Vs I s sin θ = Qout
- apply Vs= Vp/a and Is= aIp into the above equation gives,
Pout =
Vp
Sin = V p I p = Vs I s = S out
(aI p ) cos θ
a
Pout = V p I p cos θ = Pin
-
- The output power of an ideal transformer is equal to the
input power.
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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26
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?
Power in ideal transformer
Solution:
Pout = Pin = V p I p cos θ
For an ideal transformer with no losses,
losses
1 : .
E1 ≈ V1 = 220V
E x a mp l e
Qout = Qin = V p I p sin θ
S out = Sin = V p I p
E 2 ≈ V2 = 110V
From the EMF equation, we have
N1 =
E1
4.44 * f * φmax
220
= 166turns.
( 4.44)(60)(5 X 10 −3 )
110
N2 =
= 83turns.
(4.44)(60)(5 X 10 −3 )
=
Where θ is the angle between voltage and
current
(a) The secondary output current and voltage.
E x a m p l
:
2
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 < 36.9o, find,
e
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(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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5
Impedance transformation through the
transformer
E
x
a
m
p
l
e
The impedance of a device – the ratio of the phasor
voltage across it in the phasor current flowing through
it:
ZL =
2
:
VL
IL
Z L = a2Z L
'
ZL’-apparent
impedance of the
primary
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31
E
x
a
m
p
l
e
A single phase power system consists of a 480V 60Hz generator
supplying a load Zload=4+j3 through a transmission line
ZLine=0.18+j0.24 . Answer the following question about the
system.
a) If the power system is exactly as described below (figure
1(a)), what will be the voltage at the load be? What will the
transmission line losses be?
E
x
a
m
p
l
e
ILine
IG
ZLoad=0.18+j0.24Ω
ILoad
+
VLoad
V=480∠00V
ZLoad=4+j3Ω
-
Figure 1 (a)
3
:
b) Suppose a 1:10 step-up transformer is placed at the
generator end of the transmission line and a 10:1 step down
transformer is placed at the load end of the line (figure 1(b)).
What will the load voltage be now? What will the
transmission line losses be now?
3
:
T1
IG
1:10
ILine
T2
ILoad
ZLine=0.18+j0.24Ω
+
10:1
V=480∠00V
VLoad
Figure 1 (b)
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-
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Solution Example
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(a) From figure 1 (a) shows the power system without
transformers. Hence IG = ILINE = ILoad. The line current in this
system is given by:
I line =
V
Z line + Z load
480 ∠ 0 °V
( 0 .18 Ω + j 0 .24 Ω ) + ( 4 Ω + j 3Ω )
480 ∠ 0 °
480 ∠ 0 °
=
=
4 .18 + j 3 . 24 Ω 5 .29 ∠ 37 .8 °
= 90 .8 ∠ − 37 .8 °
=
3
:
3
:
Therefore the load voltage is:
Vload = I line Z load
= (90 .8 ∠ − 37 . 8 ° A )( 4 Ω + j 3Ω )
= (90 .8 ∠ − 37 . 8 ° A )( 5∠ 36 .9 °Ω )
= 454 ∠ − 0 .9 °
and the line losses are
Ploss = ( I line ) 2 Rline
= (90 .8 A ) 2 ( 0 .18 Ω )
= 1484 W
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Solution Example
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:
Solution Example
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(b) From figure 1 (b) shows the power system with the
transformers. To analyze the 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’s line voltage level.
ii) Eliminate transformer T1 by referring the transmission line’s
elements and the equivalent load at the transmission line’s
voltage over to the source side.
3
:
The value of the load’s impedance when reflected to the
transmission system’s voltage is
Z 'load = a 2 Z load
The total impedance at the transmission line level is now:
Z eq = Z line + Z 'load
= 400 . 18 Ω + j 300 .24 Ω
= 500 . 3∠ 36 .88 °Ω
The total impedance at the transmission line level (Zline+Z’load) is now
reflected across T1 to the source’s voltage level:
Z 'eq = a 2 Z eq
= a 2 ( Z line + Z 'load )
1 2
) ( 0 .18 Ω + j 0 .24 Ω + 400 Ω + j 300 Ω )
10
= ( 0 .0018 Ω + j 0 .0024 Ω + 4 Ω + j 3Ω )
=(
10
= ( ) 2 ( 4 Ω + j 3Ω )
1
= 400 Ω + j 300 Ω
= 5 .003 ∠ 36 .88 ° Ω
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Solution Example
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Solution Example
Notice that Z’’load = 4+j3 Ω and Z’line =0.0018+j0.0024 Ω. The
resulting equivalent circuit is shown below. The generator’s
current is:
IG =
480 ∠ 0 °V
= 95 .94 ∠ − 36 .88 ° A
5 .003 ∠ 36 .88 ° Ω
a) System with the load referred
to the transmission system
voltage level
3
:
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:
b) System with the load and
transmission line referred to the
generator’s voltage level
Knowing the current IG, we can now work back and find
Iline and ILoad. Working back through T1, we get:.
N p 1 I G = N s1 I line
I line =
N p1
N S1
IG
1
(95 .94 ∠ − 36 .88 ° A )
10
= 9 .594 ∠ − 36 .88 ° A
=
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Solution Example
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:
Solution Example
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Working back through T2 gives:
N p 2 I line = N s 2 I load
I load =
N p2
N s2
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I line
10
)( 9 . 594 ∠ − 36 .88 ° A )
1
= 95 .94 ∠ − 36 . 88 ° A
=(
3
:
It is now possible to answer the questions. The load
voltage is given by:
Vload = I load Z load
and the line losses are given by:
Ploss = ( I line ) 2 Rline
= ( 9 .594 A ) 2 ( 0 .18 Ω )
= 16 .7W
Notice that raising the transmission voltage of the power
system reduced transmission losses by a factor of nearly 90.
Also, the voltage at the load dropped much less in the
system with transformers compared to the system without
transformers.
= (9 .594 ∠ − 36 .88 ° A )( 5∠ 36 .87 °Ω )
= 479 .7 ∠ − 0 .01 °V
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7
The equivalent circuit of a transformer
Real Transformer
The major items to be considered in the construction of such a
model are:
43
Nonideal or actual transformer
•
Copper (I2R) losses: Copper losses are the resistive heating in the
primary and secondary windings of the transformer. They are
proportional to the square of the current in the windings.
•
Eddy current losses: Eddy current losses are resistive heating losses
in the core of the transformer. They are proportional to the square of
the voltage applied to the transformer.
•
Hysteresis losses: Hysteresis losses are associated with the
arrangement of the magnetic domain in the core during each half
cycle. They are complex, nonlinear function of the voltage applied to
the transformer.
•
Leakage flux: The fluxes ΦLP and ΦLS which escape the core and pass
through only one of the transformer windings are leakage fluxes.
These escaped fluxes produce a self inductance in the primary and
secondary coils, and the effects of this inductance must be accounted
for.
44
Winding resistance & magnetic leakage
Mutual flux
R1
V1
X1
I1
X2
E1
E2
R2 I2
V2
R1 & R2 : resistances of primary & secondary windings
respectively.
X1 & X2 : leakage reactances of primary & secondary windings
respectively.
45
Nonideal or actual transformer
46
The model of a real transformer
Transformer equivalent circuit, with secondary impedances
referred to the primary side
Ep = primary induced voltage
Vp = primary terminal voltage
Ip = primary current
Ie = excitation current
XM = magnetizing reactance
RC = core resistance
Rs = resistance of the secondary winding
Xs = secondary leakage reactance
Es = secondary induced voltage
Vs = secondary terminal voltage
Is = secondary current
IM = magnetizing current
IC = core current
Rp = resistance of primary winding
Xp = primary leakage reactance
47
48
8
Exact equivalent circuit the actual transformer
a) The transformer model referred to primary side
b) The transformer model referred to secondary side
49
Exact equivalent circuit of a transformer
Approximate equivalent circuit the actual transformer
a) The transformer model referred to primary side
secondary side
b)The transformer model referred to
c) With no excitation branch referred to primary side
secondary side
d) with no excitation branch referred to
Primary side
Secondary side
I p = Ie + Is / a
E S = I s ( Rs + jX s ) + Vs
Ie = IC + I M
Vs = I s Z L
V p = I p ( R p + jX p ) + E p
a=
E p = I C RC
Ep = primary induced voltage
Vp = primary terminal voltage
Ip = primary current
Ie = excitation current
XM = magnetizing reactance
RC = core resistance
Rs = resistance of the
secondary winding
Xs = secondary leakage reactance
Es = secondary induced voltage
Vs = secondary terminal voltage
Is = secondary current
IM = magnetizing current
IC = core current
Rp = resistance of primary winding
Xp = primary leakage reactance
a2Xs
Xp
Is/a
Ep
=
Es
=
Is N p
=
I p Ns
E p = I e ( RC // jX M )
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52
Approximate equivalent circuit of a transformer
referred to primary side
Ip
Ip
Vs
E p = I M ( jX M )
Exact equivalent circuit of a transformer referred to
primary side
Rp
Vp
a2Rs
Reqp
jXeqp
Is/a
+
+
Reqp=Rp+a2Rs
Xeqp=Xp+a2Xs
Ie
Vp
Vp
Ep
Rc
aVs
jXM
aVs
-
Rp/a2
aIp
Xp/a2
Rs
Is
-
Approximate equivalent circuit of a transformer
referred to secondary side
Exact equivalent circuit of a transformer referred to
secondary side
aIp
Xs
Reqs
+
jXeqs
Is
+
aIe
Vp/a
aIc
Rc/a2
Reqs=Rp/a2+Rs
Xeqs=Xp/a2+Xs
aIm
Ep/a = Es
Vs
Vp/a
XM/a2
Rc
53
/a2
jXM/a2
Vs
54
9
Open Circuit and Short Circuit.
Cont’d…
Determination of transformer parameter
by measurement
Open Circuit Test.
Provides magnetizing reactance and core loss resistance
Obtain components are connected in parallel
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.
Equivalent Circuit of the Open-Circuit Test.
The secondary / 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 input power is almost equal to the core loss at rated voltage
and frequency.
55
Cont’d…
Open circuit test evaluation
Short Circuit Test.
The short-circuit test is used to determine the equivalent series
resistance and reactance.
Provides combined leakage reactance and winding resistance
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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58
Cont’d…
Equivalent Circuit of the Short-Circuit Test.
Psc = I sc2 Req _ HV
Req _ HV =
Psc
I sc2
Z eq _ HV =
Vsc
I sc
X eq _ HV = Z eq _ HV − Req _ HV
2
2
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10