Download Chapter 1 - Transformer

EKT 451
CHAPTER 1
Transformer
The Power Grid
Diagram courtesy howstuffworks.com
Power Grid
3-phase
industrial
Intertie
Power Plants
Generator
0.5 – 5kV
Step-up
Transformer
Generator
0.5 – 5kV
Step-up
Transformer
Step-down
Transformer
Substation
Step-down
Transformer
100-750 kV
Step-down
Transformer
35-70 kV
4-15 kV
Step-down
Transformer
120/240V
residential
Transmission Tower
Substation
Utility Pole Carrying Three Phase
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.
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).
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.
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.
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.
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.
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.
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.
• 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.
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.
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.
• Inserting a magnet into a coil
also produces an induced
voltage or current.
• The faster speed of insertion/
retraction, the higher the induced
voltage.
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
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)
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
a > 1  Step down transformer
a < 1  Step up transformer
a = 1  Isolation Transformer
(1.5)
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)
fm
ax
 Substitute eqn(1.6) into eqn(1.2)
df
d
eN
 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)
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.
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.
Figure 1.6: An Ideal Transformer and the Schematic Symbols.
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.
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
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.
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 
4.11 * f * f max
220
 166turns.
3
(4.11)(60)(5 X 10 )
110
N2 
 83turns.
3
(4.11)(60)(5 X 10 )

EKT 451
CHAPTER 1
Transformer
17 jan 2007
1.5 The Ideal Transformer.
 An Ideal transformer is a lossless device with an input
winding and an output winding.
v p (t )
v s (t )

Np
Ns
a
Ip
1

Is a
where a is defined to be the turns ratio of the transformer.
1.5 The Ideal Transformer.
Ip
Vp
Is
a
Ep
Es = Vs
Equivalent circuit of an ideal transformer
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.
Cont’d…

The output power 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.
Cont’d…

The reactive power, Q, and the apparent power, S;
Qin  V p I p sin q
S in  V p I p
Qout  Vs I s sin q
S out  Vs I s
Qin  Qout
S in  S out
*
*
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.
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:
Cont’d…Example 2
.
1.7 The Equivalent Circuit of a Real
Transformer.
m
s_leakage
p_leakage
Leakage flux in the real transformer
 Copper losses are resistive loses in the 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.
1.7 The Equivalent Circuit of a Real
Transformer.
 The leakage flux in the windings is,
eLP (t )  L p
di p
dt
di
eLS (t )  Ls s
dt
 Core excitation effect can be model as
 i) magnetization reactance, XM
 ii) core resistance, Rc
1.7 The Equivalent Circuit of a Real
Transformer.
 Figure below is an exact model of a transformer.
Transferring impedances through a
transformer
IP
Vac
T
VP
Z load
IS
VS
Thévenin equivalents of transformer
circuit
Zload
ZP 
VS
 Zs 
IS
V
VP aVS

 a2 S
I P  IS 
IS
 
a
Z P  a 2 Z load
Transferring impedances through a
transformer
IP
Vac
VP
a2Zload
Equivalent circuit when secondary impedance is
transferred to primary side and ideal transformer
eliminated
IS
Vac/a
VS
Zload
Equivalent circuit when primary source is
transferred to secondary side and ideal transformer
eliminated
Cont’d…

To analyze practical circuits containing transformer, it is necessary
to convert the entire circuit to an equivalent circuit at a single
voltage level.

Therefore, the equivalent circuit must be referred either to its
primary side or to its secondary side in problem solving.
The Transformer Model Referred to its Primary Windings
Cont’d…
The Transformer Model Referred to its Secondary Voltage Level.
Cont’d…
The Transformer Model Referred to its Primary Windings
Cont’d…

Based on the above equations and assuming a zero degree
reference angle for V2, the phasor diagram is shown as
1.8 The Approximate Equivalent Circuit of
a Transformer.
 In the Approximate model the voltage drop in Rp and Xp is
negligible because the current is very small.
Approximate Transformer Model Referred to the Primary Side.
Cont’d…

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 referred to the primary side is;
Cont’d…
 Figure 1.11 shows the approximate equivalent circuit of
transformer referred to the secondary side.
Approximate Circuit Model of a Transformer Referred to the
Secondary.

The equivalent impedance is;
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%
 For ideal transformer, VR = 0. It is a good practice to have as small
voltage regulator as possible.
Cont’d…
 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;

P1  P1\losses
P2
P2


P1
P2  Plosses P2  Pcopper  Pcore
Example 3: Transformer Voltage Regulation.
Cont’d…Example 3
1.1
1
1.10 Open Circuit and Short Circuit.
- 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.
Cont’d…
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.
Cont’d…
Open circuit test evaluation
Cont’d…
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, 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.
Cont’d…
Equivalent Circuit of the Short-Circuit Test.
Psc  I Req _ HV
2
sc
Req _ HV
Psc
 2
I sc
Z eq _ HV
Vsc

I sc
X eq _ HV  Z eq _ HV  Req _ HV
2
2
Equivalent circuit obtained by measurement
Xe_s
Xm_p
Re_s
Rc_p
Equivalent circuit for a real transformer resulting from the
open and short circuit tests.
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.
Three-Phase Transformer Connections.
 A three-phase transformer consists of three transformers either
separate or combined on one core.
combined on one core.
3 separate core
Three-Phase Transformer Connections.
 There are four possible connections between the secondary and
primary of a three-phase transformer.
(1) Wye-Wye (Y-Y). - don't use causes harmonics problems
(2) Wye-Delta (Y-). - use : high voltage transmissions
(3) Delta-Wye (Y) - use : most common; commercial and industrial.
(4) Delta-Delta ( ). - use: industrial applications
Cont’d…
Figure 1.15: Three-Phase Transformer Connections and Wiring Diagram.
(1) Wye-Wye Connection.
VLP
a
VLS
Y Y
(2) Wye-Delta Connection.
VLP
 3a
VLS
Y 
(3) Delta-Wye Connection.
VLP
3

VLS
a
 Y
(4) Delta-Delta Connection.
VLP VfP

a
VLS VS

Cont’d…
Example 2.6: Three-Phase
Transformer.
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:
a) wye-delta, b) delta-wye, and
c) delta-delta?
Solution
For both delta and wye connections,
the line currents can be obtained as:
Example 7: Voltage
Regulation at Full Load.
A 7200V/208V, 50kVA, threephase 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.
Solution
Voltage Regulation.
.
Example 8: 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.
Solution
.
Tutorial 1
• Group Computer
– Monday 22/01/07
– Makmal Komputer 1, Kg Wai
• Group Communication
– Tuesday 23/01/07
– Makmal Komputer 2, Kg Wai