 Topic 1 : Magnetic Concept and Transformer

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Topic 1 : Magnetic Concept and Transformer

Introduction Two winding transformers Construction and principles
Equivalent circuit Determination of equivalent circuit parameters Voltage regulation Efficiency Auto transformer 3 phase transformer

Introduction Different variety of transformers

Introduction

Introduction The word “Transformer” means an electromagnetic device which transforms electrical power from one end to another at different voltages and different currents keeping frequency constant. Unlike motor and generator it is static machine with different turns ratio of primary and secondary windings through which voltage/current is changed. The transfer of energy takes place through the magnetic field and all currents and voltages are AC. The rating of transformer is either in kVA or MVA because load to be connected is unknown

Introduction Transformers are adapted to numerous engineering applications and may be classified in many ways: Power level (from fraction of a volt-ampere (VA) to over a thousand MVA), Application (power supply, impedance matching, circuit isolation), Frequency range (power, audio, radio frequency (RF)) Voltage class (a few volts to about 750 kilovolts) Cooling type (air cooled, oil filled, fan cooled, water cooled, etc.)-ONAN, ONAF Purpose (distribution, rectifier, arc furnace, amplifier output, etc.).

Introduction Examples of transformer classifications:
Power Three phase transformers (Step up) used for transmission of power (3-phase) at a distance Distribution transformer (Step down) used for utilization of power 3-pahse/1-phase Instrument Transformer (VT & CT) used for measurement/practical Auto Transformer (Single limb, electrically connected) used for measurement, practical, supply/utilization Isolation Transformer (having winding ratio of 1:1) used for safety of human and equipment for sensitive appliances or practical purpose

Introduction The invention of transformer caused transmission of heavy AC electrical power possible thus plays important role in electrical power technology Functions of transformer: Raise or lower voltage or current in AC circuit Isolate circuit from each other Enable to transmit electrical power energy over large distances at about >1200kV Provides electrical power according to the utilization needs

Transformer- Introduction
Power transmission

Introduction A typical power system consists of generation, transmission and distribution. Power from plant/station is generated around kV (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 losses. Its distribution is made through step down transformer according to the consumer demand. Here again at this stage, a transformer play an important role to reduce the voltage to suit the consumer need.

Introduction Power Transmission

Introduction Transformer is a device that makes use of the magnetically coupled coils to transfer energy. It is typically consists of one primary winding coil and one or more secondary windings. The primary winding and its circuit is called the Primary Side of the transformer. The secondary winding and its circuit is called the Secondary Side of the transformer. A magnetic circuit provides the link between primary and secondary.

Introduction When an AC voltage Vp is applied to the primary winding of the transformer, an AC current Ip will result. Ip sets up a time-varying magnetic flux Ф in the core. A voltage Vs is induced in the secondary circuit according to the Faraday’s law.

Construction The magnetic (iron) core is made of thin laminated steel sheet. to minimize the eddy current loss by reducing thickness. There are two common cross section of core square or (rectangular) for small transformers circular (stepped) for the large and 3 phase transformers.

Construction Core (U/I) Type: Shell Type:
Constructed from a stack of U and I shaped laminations. The primary and secondary windings are wound on two different legs of the core. Shell Type: Constructed from a stack of E and I shaped laminations. The primary and secondary windings are wound on the same leg of the core, as concentric windings, one on top of the other.

Construction

Construction

Construction

Construction

Ideal Transformer Winding resistances are zero, no leakage inductance and iron loss Magnetization current generates a flux that induces voltage in both windings Current, voltages and flux in an unloaded ideal transformer

Ideal Transformer Currents and fluxes in a loaded ideal transformer

Ideal Transformer Turn ratio
If the primary winding has N1 turns and secondary winding has N2 turns, then: The input and output complex powers are equal

Ideal Transformer Functional description of a transformer: a = 1
Isolation Transformer | a | < 1 Step-Up Transformer Voltage is increased from Primary side to secondary side | a | > 1 Step-Down Transformer Voltage is decreased from Primary side to secondary side

Ideal Transformer Transformer Rating
Practical transformers are usually rated based on: Voltage Ratio (V1/V2) which gives us the turns-ratio Power Rating, small transformers are given in Watts (real power) and Larger ones (Power Transformers) are given in kVA (apparent power)

Ideal Transformer Example 1
Determine the turns-ratio of a 5 kVA 2400V/120V Power Transformer Turns-Ratio = a = V1/V2 = 2400/120 = 20/1 = 20 This means it is a Step-Down transformer

Ideal Transformer Example 2
A 480/2400 V (r.m.s) step-up ideal transformer delivers 50 kW to a resistive load. Calculate: the turns ratio, (0.2) the primary current, (104.17A) the secondary current. (20.83A)

Ideal Transformer Exercise 1
A 250kVA, 1100V/400v, 50Hz single-phase transformer has 80 turns on the secondary. Calculate: the approximate values of the primary and secondary currents (227A, 625A) the approximate number of primary turns (220) the maximum value of the flux (22.5mWb)

Ideal Transformer Nameplate of a transformer

Ideal Transformer Equivalent circuit of an ideal transformer

Ideal Transformer Equivalent circuit of an ideal transformer
Transferring impedances through a transformer

Ideal Transformer Equivalent circuit when secondary impedance is transferred to primary side and ideal transformer eliminated. Equivalent circuit when primary source is transferred to secondary side and ideal transformer eliminated.

Practical Transformer

Equivalent Circuit In a practical magnetic core having finite permeability, a magnetizing current Im is required to establish a flux in the core. This effect can be represented by a magnetizing inductance Lm. The core loss can be represented by a resistance Rc.

Equivalent Circuit Rc :core loss component
Xm : magnetization component

Equivalent Circuit Phasor diagram of an unloaded transformer

Equivalent Circuit Winding resistance and leakage flux
The effects of winding resistance and leakage flux are respectively accounted for by resistance R and leakage reactance X (2πfL).

Equivalent Circuit Rc :core loss component
Xm : magnetization component R1 and R2 are resistance of the primary and secondary winding X1 and X2 are reactance of the primary and secondary winding

Equivalent Circuit Phasor diagram of a loaded transformer (secondary)

Equivalent Circuit Phasor diagram of a loaded transformer (primary)

Approximate Equivalent Circuit
Since no load current is very small(3-5% of full load), the parallel circuit of Rc and Xm can be moved close to the supply without significant error in calculation. Calculations becomes easier

Approximate Equivalent Circuit
Calculations will be much more easy if the primary and secondary circuit are combined. Transfer the secondary circuit to the primary circuit

Approximate Equivalent Circuit
Phasor diagram of a loaded transformer (primary)

Approximate Equivalent Circuit
For convenience, the turns is usually not shown The resistance and reactance can be lumped together We can also transfer the primary circuit to the secondary circuit

Approximate Equivalent Circuit
Example 3 A 100kVA transformer has 400 turns on the primary and 80 turns on the secondary. The primary and secondary resistance are 0.3 ohm and 0.01 ohm respectively and the corresponding leakage reactances are 1.1 ohm and ohm respectively. The supply voltage is 2200V. Calculate: the equivalent impedance referred to the primary circuit (2.05 ohm) the equivalent impedance referred to the secondary circuit (0.082)

Transformer Test The equivalent circuit model for the actual transformer can be used to predict the behavior of the transformer. The parameters Rc, Xm, R1, X1, R2, X2 and N1/N2 must be known so that the equivalent circuit model can be used. These parameters can be directly and more easily determined by performing tests: No load test (or open circuit test) Short circuit test

Transformer Test No load/Open circuit test Short circuit test
Provides magnetizing reactance (Xm) and core loss resistance (Rc) Obtain components are connected in parallel Short circuit test Provides combined leakage reactance and winding resistance Obtain components are connected in series

Transformer Test – Open Circuit
Equivalent circuit for open circuit test, measurement at the primary side Simplified equivalent circuit

Transformer Test – Open Circuit
Open circuit test evaluation .

Transformer Test – Short Circuit
Short circuit test Secondary (normally the LV winding) is shorted, that means there is no voltage across secondary terminals; but a large current flows in the secondary. Test is done at reduced voltage (about 5% of rated voltage, with full-load current in the secondary. Hence the induced flux are also 5%) The core losses is negligible since it is approximately proportional to the square of the flux. So, the ammeter reads the full-load current; the wattmeter reads the winding losses, and the voltmeter reads the applied primary voltage.

Transformer Test – Short Circuit
Equivalent circuit for short circuit test, measurement at the primary side Simplified equivalent circuit

Transformer Test – Short Circuit
Simplified circuit for calculation of series impedance Primary and secondary impedances are combined .

Transformer Test – Short Circuit
Short circuit test evaluation .

Transformer Test Equivalent circuit for a real transformer resulting from the open and short circuit tests.

Transformer Test Example 4
Obtain the equivalent circuit of a 200/400V, 50Hz 1-phase transformer from the following test data:- O/C test : 200V, 0.7A, 70W - on L.V. side(LV data) S/C test : 15V, 10A, 85W - on H.V. side(HV data) (Rc =571.4 ohm, Xm=330 ohm, Re=0.21ohm, Xe=0.31 ohm)

Voltage Regulation Most loads connected to the secondary of a transformer are designed to operate at essentially constant voltage. However, as the current is drawn through the transformer, the load terminal voltage changes because of voltage drop in the internal impedance. To reduce the magnitude of the voltage change, the transformer should be designed for a low value of the internal impedance Zeq The voltage regulation is defined as the change in magnitude of the secondary voltage as the load current changes from the no-load to the loaded condition.

Voltage Regulation The voltage regulation is expressed as follows:

Voltage Regulation For the equivalent circuit referred to the primary:
V1 = no-load voltage V2’ = secondary voltage referred to the primary (full-load condition)

Voltage Regulation Consider the equivalent circuit referred to the secondary, (-) : leading power factor (+) : lagging power factor

Voltage Regulation Consider the equivalent circuit referred to the primary, (-) : leading power factor (+) : lagging power factor

Voltage Regulation Example 5
Based on Example 3 calculate the voltage regulation and the secondary terminal voltage for full load having a power factor of 0.8 lagging (0.0336pu,425V) 0.8 leading ( pu,447V)

Approximate Equivalent Circuit
Example 3 A 100kVA transformer has 400 turns on the primary and 80 turns on the secondary. The primary and secondary resistance are 0.3 ohm and 0.01 ohm respectively and the corresponding leakage reactances are 1.1 ohm and ohm respectively. The supply voltage is 2200V. Calculate: the equivalent impedance referred to the primary circuit (2.05 ohm) the equivalent impedance referred to the secondary circuit (0.082)

Voltage Regulation

Efficiency Losses in a transformer
Copper losses in primary and secondary windings Core losses due to hysteresis and eddy current. It depends on maximum value of flux density, supply frequency and core dimension. It is assumed to be constant for all loads

Efficiency Equipment is desired to operate at a high efficiency.
Efficiency is defined as Since it is a static device, losses in transformers are small The losses in the transformer are the core loss (Pc) and copper loss (Pcu)

Efficiency The copper loss can be determined if the winding currents and their resistances are known: The copper loss is a function of the load current. The core loss depends on the peak flux density in the core, which in turn depends on the voltage applied to the transformer Since a transformer remains connected to an essentially constant voltage, the core loss is almost constant

Efficiency If the parameters of the equivalent circuit of a transformer are known, the efficiency of the transformer under any operating condition may be determined . Normally, load voltage remains fixed Therefore efficiency depends on load current and load power factor

where S is the apparent power (in volt amperes) Efficiency for any load equal to n x full load where corresponding total loss =

Efficiency Example 6 The following results were obtained on a 50 kVA transformer: open circuit test – primary voltage, 3300 V; secondary voltage, 400 V; primary power, 430W. Short circuit test – primary voltage, 124V;primary current, A; primary power, 525W; secondary current, full load value. Calculate the efficiency at full load and half load for 0.7 power factor. (97.3%, 96.9%)

Efficiency Exercise 2 The primary and secondary windings of a 500kVA transformer have resistance of 0.42 ohm and ohm respectively. The primary and secondary voltages are 11000V and 400V respectively and the core loss is 2.9kW, assuming the power factor of the load to be 0.8. Calculate the efficiency on Full load half load (98.3%, 98.1%)

Efficiency For constant values of the terminal voltage V2 and load power factor angle θ2 , the maximum efficiency occurs when If this condition is applied, the condition for maximum efficiency is that is, core loss = copper loss.

Efficiency Exercise 3 Assuming the power factor of the load to be 0.8, find the output power at which the efficiency of the transformer of Exercise 2 is a maximum and calculate its value (346.4kW, 98.4%)