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SAB 2032 ELECTRICAL TECHNOLOGY. Weeks 13-14 Transformer. INSPIRING CREATIVE AND INNOVATIVE MINDS. Transformers. INSPIRING CREATIVE AND INNOVATIVE MINDS. Transformers. A transformer is a static machine which step voltage/current up or down - PowerPoint PPT Presentation
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SAB 2032ELECTRICAL TECHNOLOGY
Weeks 13-14
Transformer
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
Transformers
INSPIRING CREATIVE AND INNOVATIVE MINDS
Transformers
INSPIRING CREATIVE AND INNOVATIVE MINDS
A transformer is a static machine which step voltage/current up or downUnlike in rotating machines, there is no electromechanical energy conversion
The transfer of energy takes place through the magnetic field and all currents and voltages are AC
Transformer can be categorized as: The ideal transformers Practical transformers Special transformers Three phase transformers
INSPIRING CREATIVE AND INNOVATIVE MINDS
Transformers
INSPIRING CREATIVE AND INNOVATIVE MINDS
Transformers
INSPIRING CREATIVE AND INNOVATIVE MINDS
Transformers
INSPIRING CREATIVE AND INNOVATIVE MINDS
Applications of the transformer
A typical power system consists of generation, transmission and distributionPower from plant/station is generated around 11-13-20-30kV (depending upon manufacturer and demand)
This voltage is carried out at a distance to reach for utilization through transmission line system by step up transformer at different voltage levels depending upon distance and lossesIts distribution is made through step down transformer according to the consumer demand
Transformers
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
Transformer is one of the most useful electrical devices ever inventedFunctions of transformer:Raise or lower voltage or current in AC circuitIsolate circuit from each otherIncrease or decrease the apparent value of a capacitor, inductor or resistorEnable to transmit electrical energy over great distancesDistribute safely in homes and factories
Transformers
Principles of Transformer
INSPIRING CREATIVE AND INNOVATIVE MINDS
A transformer consists of two electric circuits called primary and secondaryA magnetic circuit provides the link between primary and secondaryWhen an AC voltage is applied to the primary winding Vp of the transformer,
an AC current Ip will result
Ip sets up a time-varying magnetic flux Ф
in the coreA voltage is induced to the secondary circuit Vs according to
the Faraday’s law
INSPIRING CREATIVE AND INNOVATIVE MINDS
Core Types of Transformer
The magnetic (iron) core is made of thin laminated steel sheet. The reason of using laminated steel is to minimize the eddy current loss by reducing thicknessThere are two common cross section of core which include square or (rectangular) for small transformers and circular (stepped) for the large and 3 phase transformers.
INSPIRING CREATIVE AND INNOVATIVE MINDS
PrimaryWinding
SecondaryWinding
Multi-layerLaminatedIron Core
X1X2H1 H2
WindingTerminals
Configuration of Single phase transformer
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
3 Phase Transformer
The three phase transformer iron core has three legsA phase winding is placed in each leg
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
Construction of Transformer
This can be divided into 2 types: Core (U/I) Type: Is constructed from a stack of U and I shaped
laminations. In a core-type transformer, the primary and secondary windings are wound on two different legs of the core
Shell Type: Is constructed from a stack E and I shaped laminations. In a shell-type transformer, the primary and secondary windings are wound on the same leg of the core, as concentric windings, one on top of the other.
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
Construction of a Small Transformer
Iron core
Terminals
Secondarywinding
Insulation
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
Transformer with Cooling System
Oil tank
Coolingradiators
High voltagebushing
Low voltagebushing
Winding
Iron corebehind the steel
bar
Radiator
Steeltank
Insulation
Bushing
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
The Ideal Transformer
For an ideal transformer, we assume No losses Core is infinitely permeable Flux produced by the primary is completely linked by the
secondary and vice versa No leakage flux of any kind
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
The Ideal Transformer
Primary and secondary posses N1 and N2 turns respectively
Primary is connected to a sinusoidal source Eg Magnetizing current Im creates a flux of Φm
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
The Ideal Transformer
The flux is completed linked by the primary and secondary windings – mutual flux
Flux varies sinusoidally and reaches a peak value of Φm
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
The Ideal Transformer
A transformer with more turns in its primary than its secondary coil will reduce voltage and is called a step-down transformer
One with more turns in the secondary than the primary is called a step-up transformer
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
The Ideal Transformer
The sinusoidal current Im produces a sinusoidal mmf NIm which in turn creates a sinusoidal flux. The flux induces an effective voltage E across the terminals of the coil
Induces Voltages: The effective induced emf in primary winding is
Where N1 is the number of winding turns in primary winding, Фm the maximum (peak) flux and f the frequency of the supply voltage
mfN.E 11 444
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
The Ideal Transformer
This equation shows that for a given frequency and a given number of turns, Фm varies in proportion to the applied voltage Eg
This means that if Eg is kept constant, the peak flux must remain constant
Similarly, the effective induced emf in secondary winding:
mfN.E 22 444
INSPIRING CREATIVE AND INNOVATIVE MINDS
The Ideal Transformer - at No Load
aNN
EE
2
1
2
1
= Voltage induced in the secondary [V] 1E
2E
1N
2N
a = Turn ratio= Voltage induced in the primary [V]
= Number of turns on the primary= Number of turns on the primary
INSPIRING CREATIVE AND INNOVATIVE MINDS
Ideal Transformer under load: Current ratio
Let us connect a load Z across the secondary of the ideal transformer. A secondary current I2 will immediately flow.Does E2 change when we connect the load?
1) In an ideal transformer the primary and secondary windings are linked by the mutual flux, Фm, consequently voltage ratio will be the same as at no load
2) If the supply voltage Eg is kept fixed, then the primary induced voltage E1 remain fixed. Consequently, Фm also remains fixed. It follows that E2 also remain fixed
We conclude that E2 remains fixed whether a load is connected or not
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
Let us now examine the mmf created by the primary and secondary windings. First, current I2 produces a secondary mmf N2I2. If it acted alone, this mmf would produce a profound change in the Фm. But we just saw that Фm does not change under load.
We conclude that flux Фm can only remain fixed if the primary develops a mmf which exactly counterbalances N2I2 at every instant. Thus, a primary current I1 must flow so that
2211 ININ
Ideal Transformer under load: Current ratio
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
To obtain the required instant-to-instant bucking effect, current I1 and I2 must increase and decrease at the same timeIn other words, the current must be in phase
(a)Ideal transformer under load (b)Phasor relationships under load
Ideal Transformer under load: Current ratio
INSPIRING CREATIVE AND INNOVATIVE MINDS06/07/2009
aNN
II 1
2
1
1
2
1I = Primary current [A]
2I = Secondary current [A] 1N = Number of turns on the primary2N = Number of turns on the secondary
Ideal Transformer under load: Current ratio
INSPIRING CREATIVE AND INNOVATIVE MINDS
Transformer Impedance
Primary Impedance: Primary impedance in terms of secondary impedance : Primary Voltage : Primary Current :
P
PL I
V'Z
LL Za'Z 2
SP aVV
aII S
P
INSPIRING CREATIVE AND INNOVATIVE MINDS
Example problem
A transformer coil possesses 4000 turns and links an ac flux having a peak value of 2 mWb. If the frequency is 60 Hz, calculate the effective value of the induced voltage E.
Ans: 2131V
INSPIRING CREATIVE AND INNOVATIVE MINDS
A coil having 90 turns is connected to a 120V, 60 Hz source. If the effective value of the magnetizing current is 4 A, calculate the following:a. The peak value of fluxb. The peak value of the mmfc. The inductive reactance of the coild. The inductance of the coil.
Example problem
INSPIRING CREATIVE AND INNOVATIVE MINDS
Real Transformer
Leakage Flux: Not all of the flux produced by the primary current links the winding, but there is leakage of some flux into air surrounding the primary. Similarly, not all of the flux produced by the secondary current (load current) links the secondary, rather there is loss of flux due to leakage. These effects are modelled as leakage reactance in the equivalent circuit representation.
INSPIRING CREATIVE AND INNOVATIVE MINDS
Magnetization Current in a Real Transformer
Although the output of the transformer is open circuit, there will still be current flow in the primary windings
• Magnetization current, iM • Core-loss current, ic
The relation between current and flux is proportional since
NSi
SNIF
INSPIRING CREATIVE AND INNOVATIVE MINDS
Therefore, in theory, if the flux produce in core is sinusoidal, than the current should also be a perfect sinusoidal. Unfortunately, this is not true.
Current in a transformer has the following characteristics:• It is not sinusoidal but a combination of high
frequency oscillation
• The current lags the voltage at 900
• At saturation, the high frequency components will be
extreme
Magnetization Current in a Real Transformer
INSPIRING CREATIVE AND INNOVATIVE MINDS
Core-loss current:• Eddy current loss- is dependent upon the rate of change of
flux• Hysteresis loss - a non linear lossAt no-load, the primary windings is known as the excitation current
cm iii
Magnetization Current in a Real Transformer
INSPIRING CREATIVE AND INNOVATIVE MINDS
The equivalent circuit of a transformer
Taking into account real transformer, there are several losses that has to be taken into account in order to accurately model the transformer, namely:
1- Copper (I2R) Losses 2- Eddy current Losses 3- Hysteresis Losses 4- Leakage flux
INSPIRING CREATIVE AND INNOVATIVE MINDS
Equivalent circuit of a Real Transformer
Equivalent Circuit of a Two-winding, 1-phase:
Rc : Core loss component, this resistance models the active loss of the coreXm : magnetization component, this resistance models the reactive loss of the coreRp and Xp : are resistance and reactance of the primary windingRs and Xs : are resistance and reactance of the secondary winding
INSPIRING CREATIVE AND INNOVATIVE MINDS
Equivalent circuit of a Real Transformer
Impedance Transfer:
To model a transformer, it is important to understand how impedance are transferred from one side to another, that is primary to secondary or secondary to primary
Impedance transfer is used to calculate the current/voltage easier and also to get the voltage and current ratio for the rest of the calculation
In general any impedance transferred from secondary side to primary side must be multiplied by the square of the turn ratio, a2
INSPIRING CREATIVE AND INNOVATIVE MINDS
Equivalent circuit-seen from primary side
The terminal voltage (Vp,Vs) is not constant and changes depends on the load current.
Secondary parameters transferred to primary
INSPIRING CREATIVE AND INNOVATIVE MINDS
Equivalent circuit of a Real Transformer
Equivalent Circuit
a) referred to primary side b) referred to secondary side
INSPIRING CREATIVE AND INNOVATIVE MINDS
Approximate equivalent circuit
Approximate equivalent circuit
c) Referred to primary side (no exicitation) d) Referred to secondary side (no excitation)
INSPIRING CREATIVE AND INNOVATIVE MINDS
Approximate equivalent circuit
In this model, the parameters from secondary are transferred to primary side.
where, Rep and Xep are equivalent resistance and reactance from primary winding side
INSPIRING CREATIVE AND INNOVATIVE MINDS
Voltage Regulation
Is defined as the change in the magnitude of the secondary voltage as the load current changes from the no-load to full load
The primary side voltage is always adjusted to meet the load changes; hence V’s and Vs are kept constant.
100VVV
100aVaVV V
VVVR%
's
'sp
ssp
NL
FLNL
100
INSPIRING CREATIVE AND INNOVATIVE MINDS
Efficiency of Transformer
Pcopper represents the copper losses in primary and secondary windings. There are no rotational losses.
As always, efficiency is defined as power output to power input ratio
coppercoreoutin
inout
PPPP%PP
100
43
Example ProblemA not-quite-ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 120 V, 60 hz source. The coupling between the primary and the secondary is perfect but the magnetizing current is 4 A. calculate:a. The effective voltage across the secondary
terminalsb. The peak voltage across the secondary terminals.c. The instantaneous voltage across the secondary
when the instantaneous voltage across the primary is 37 V.
Ans: 3000V, 4242 V, 925 V.
44
Example Problem
An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 200 V, 50 Hz source. The load across the secondary draws a current of 2 A at a power factor of 80 per cent lagging. Calculate :a. The effective value of the primary currentb. The instantaneous current in the primary when the
instantaneous current in the secondary is 100 mA.c. The peak flux linked by the secondary winding.
Ans: 50 A, 2.5 A, 10 mWb
INSPIRING CREATIVE AND INNOVATIVE MINDS
QUESTION
06/07/2009
46
Example Problem• A 125 kVA transformer has 500 turns on the primary
and 80 turns on the secondary. The primary and secondary resistances are 0.5 Ω and 0.025 Ω respectively, and the corresponding leakage reactances are 2.5 Ω and 0.025 Ω respectively. The supply voltage is 2.2 kV. Calculate:
The voltage regulation and the secondary terminal voltage for full load having a power factor of 0.85 lagging
47
Example Problem• The primary and secondary windings of a 400 kVA
transformer have resistances of 0.3 Ω and 0.0015 Ω respectively. The primary and secondary voltages are 15 kV and 0.4 kV respectively. If the core loss is 2.5 kW and the power factor of the load is 0.80, calculate the efficiency of the transformer on full load.
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