{ PRESENTATION ON TRANSFORMER. 1. INTRODUCTION: - TRANSFORMER : A static device- Utilized for raising or lowering the voltages. Transfers electrical energy.

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Presentation transcript:

{ PRESENTATION ON TRANSFORMER

1. INTRODUCTION: - TRANSFORMER : A static device- Utilized for raising or lowering the voltages. Transfers electrical energy from one circuit to other through inductive coupling. No change in frequency of the supply. Very efficient device.

2. PRINCIPLE USED: Inductive coupling(or) mutual inductance. Includes:- Faraday’s law of electromagnetism.

3. TRANSFORMER CONSTRUCTION: CONSISTS OF :- The a.c. supply, load. Laminated magnetic core with insulation. Two windings- i.e., the primary coil and the secondary coil windings.

4. TYPES CLASSIFIED, ACCORDING TO :- No. of windings or turns: STEP-UP TRANSFORMER i.e., (V2>V1). STEP DOWN TRANSFORMER i.e.,(V2<V1).

Placing of the coils: CORE-TYPE TRANSFORMER:- the windings surround a considerable part of the core and. SHELL-TYPE TRANSFORMER:- here, the core surrounds a considerable portion of the windings.

5. WORKING OPERATION: - A varying current in the first or primary winding creates a varying magnetic flux in the transfor- -mer's core and thus a varying magnetic field through the secondary winding. This varying magnetic field induces a varying Electromot- -ive force (EMF), or "voltage", in the secondary winding. This effect is called inductive coupling. Points:  Magnetizing current produced  megnetomotive force  magnetic flux.  Both self + mutual inductance takes place.  So, voltage energized or de-energized from one coil to other.  Purpose achieved.  Instead, some losses occur.  Efficiency decrease a bit but can be minimized.

6. IDEALLY: - The resistance of the coils are zero, perfect efficiency. Power in=Power out. The relative permeability of the core is infinite. No loss. No leakage flux considered. 7. EMF EQUATION OF A TRANSFORMER :- [ E1=4.44*f*N1*Bm*A.] And [E2=4.44*f*N2*Bm*A.]

8. VOLTAGE TRANSFORMATION RATIO:- [ Vs/Vp=Ns/Np=k ] Where, k is voltage transformation ratio. 9. TRANSFORMER FORMULAE: - Vs/Vp=Ns/Np=Ip/Is=sq.rt(Ls)/sq.rt(Lp) Then, magnetic flux in the core (Φ) lags 90o behind the source voltage waveform. The current drawn by the primary coil from the source to produce this flux is called the magnetizing current, and it also lags the supply voltage by 90o.

10. LOSSES:- IRON LOSS  includes: 1) HYSTERESIS LOSS. 2) EDDY CURRENT LOSS. COPPER LOSS  I^2 *R LOSS; (resistive heating occurs). MECHANICAL LOSSES. MEGNETOSTRICTION LOSS. MEGNETIC LEAKAGE INDUCTANCE  It merely affects the loss, but gives inductive impedance in series with the wire resistance.  LOSSES CAN BE MINIMIZED.

MEASURES FOR LOSS MINIMIZATION:- 1) Hysteresis loss  by using low hysteresis coefficient having materials like silicon steel. 2) Eddy current loss  circulating currents are reduced by laminating the core into thin sheets. 3) By using pure iron material. 4) Magnetic leakage  by winding the insulated coils on top of each other. 5) Copper loss  i.e., heat loss by using good quality windings.

11. TRANSFORMER EFFICIENCY :- It is important in terms of energy lost. Efficiency, η=(Power out/ Power in) x 100. Power in = Power out + total losses. total losses = Cu loss + Fe loss. Why is it Important to Consider Efficiency? What Causes Inefficiency and Power Loss in the Transformer?  All the losses due to which electrical energy is converted into heat energy.

12.APPLICATIONS: - Before the development of the transformer, electric power distribution prima- -rily used direct current. It was difficult for a DC Utility-power generation station to be more than a few kilometres from the user. It was understood that high voltages allowed long distance transmission with low amperage (250 volts at 5000 amps = volts at 50 amps) so the transmission wires can be smaller and less expensive, but it still needed to be stepped back down to utility voltage at the customer's location. So, transformers are used for efficient transmission.

AC Coil Example:- 120 volts at 60 Hz is applied to the coil. Since it has an iron core, a large alternating magnetic field is produced. The magnetic field alternates 60times per second, being produced by an AC, iron core coil. The changing magnetic field induces a voltage in the coil which is sufficient to light the bulb if it is close enough. When you bring the bulb close enough for it to produce light, we are demonstrating the action of a transformer, which uses the changing magnetic field produced by the current in one coil to induce a voltage into a second coil. This is an example of Faraday's law.

Tesla Coil:- High voltage discharges having an appearance similar to lightning can be obtained from a properly designed Tesla coil. The Tesla coil employs the principles of the transformer with a small number of turns on the primary and a large number of turns on the secondary. The high frequency input to the primary coil is transformed up to about a million volts, and therefore the current is very small. The high voltage is enough to produce spectacular electric discharges like those shown above.

THANK YOU THANK YOU