Transformers

TransformersTransformerAn A.C. device used to change high voltage low current A.C. into low voltage high current A.C. and vice-versa without changing the frequencyIn brief,1. Transfers electric power from one circuit to another2. It does so without a change of frequency3. It accomplishes this by electromagnetic induction4. Where the two electric circuits are in mutual inductive influence of each other.Principle of operation

It is based on principle of MUTUAL INDUCTION. According to which an e.m.f. is induced in a coil when current in the neighbouring coil changes.Constructional detail : Shell type Windings are wrapped around the center leg of a laminated core.

Core type

Windings are wrapped around two sides of a laminated square core.Sectional view of transformersNote: High voltage conductors are smaller cross section conductors than the low voltage coils

Shell typeThe HV and LV windings are split into no. of sectionsWhere HV winding lies between two LV windingsIn sandwich coils leakage can be controlled

Fig: Sandwich windingsWorking of a transformer1. When current in the primary coil changes being alternating in nature, a changing magnetic field is produced2. This changing magnetic field gets associated with the secondary through the soft iron core3. Hence magnetic flux linked with the secondary coil changes.4. Which induces e.m.f. in the secondary.

Ideal Transformers

Zero leakage flux:-Fluxes produced by the primary and secondary currents are confined within the coreThe windings have no resistance:- Induced voltages equal applied voltagesThe core has infinite permeability- Reluctance of the core is zero- Negligible current is required to establish magnetic fluxLoss-less magnetic core- No hysteresis or eddy currents

Ideal Transformers

For ideal transformer, the energy transferred will be the same as input. Thus power at primary is same power at secondary.Pp = PsorIpVp = IsVs Demagnetizing mmf of secondary is equal in magnitude but opposite in polarity to magnetizing mmf of primary.IpNp = IsNs

Emf Equation of TransformerPractical transformerThere is always leakage of flux.The windings have resistance:The core has finite permeability- Reluctance of the core is not zero- current is required to establish magnetic flux i.e. magnetizing component of currentMagnetic core is not Loss-less- Hysteresis or eddy currentsTransformer on No-loadSecondary current is zero.Small current (no load current) flows in primary winding consisting of two components:Magnetizing or reactive component of no load current Power or active component of no load currentPhasor diagram: Transformer on No-load

Leakage ReactanceActual flux set up transformer consists of two components:Useful flux linking with both winding which practically remains constant at all values of load.Leakage flux linking with one winding only. It is dependent on load.At no load leakage flux is negligible.As load increases leakage flux increases.Leakage flux produces a elf induced back emf in respective windings.They are equivalent to small choke in series with respective winding, the reactance of which is called leakage reactance. Transformer on load

Fig. a: Ideal transformer on loadFig. b: Main flux and leakage flux in a transformerEquivalent Circuit of Transformer

Transformer Equivalent Referred to Primary Side

Transformer Equivalent Referred to Secondary Side

Approximate Equivalent Circuit

Example:A 150 kVA, 2400/240 V single phase transformer has following parameters: r1=0.2 r2= 2X10-3 x1=0.6 x2= 6X10-3 and Rc=10 k Xm = 1.6k Calculate the equivalent resistance and reactance as seen on HV side.Calculate the equivalent resistance and reactance as seen on LV side.With secondary open what current will be drawn from HV side. Also find pf.

Transformer on load assuming no voltage drop in the windingFig shows the Phasor diagram of a transformer on load by assumingNo voltage drop in the windingEqual no. of primary and secondary turns

Transformer Phasor on Load

Transformer LossesGenerally, there are two types of losses;Iron losses :- occur in core parametersCopper losses :- occur in winding resistance

Iron Losses

ii Copper Losses

Transformer Efficiency

Where, if load, n = , load, n= , 90% of full load, n =0.9Transformer ratingTransformer rating is normally written in terms of Apparent Power.Apparent power is actually the product of its rated current and rated voltage.Example is A 20kVA, 220/2200V , 50 Hz single phase transformer.Transformer ratingThere are two type of losses in a transformer;1. Copper Losses2. Iron Losses or Core Losses or Insulation LossesCopper losses ( IR)depends on Current which passing through transformer winding while Iron Losses or Core Losses or Insulation Losses depends on Voltage.So total losses in transformer are dependent on voltage and current, not on phase difference between voltage and current.Thats why the rating of Transformer in kVA, not in kW.

Open Circuit Test

Open Circuit Test Performed on low voltage winding at rated voltage and frequency.High voltage winding is kept open.A small current flows i.e. no load current.As no load current is 1 to 3% of full load current, so copper loss is negligible. Short Circuit Test

Short Circuit TestPerformed on High voltage winding.Low voltage winding is short circuited through ammeter.Input voltage is increased until full load current flows short circuited winding.Normally applied voltage is 5 to 7% of rated voltage.So flux established in core is quite small and hence core losses are negligible.Open Circuit TestShort Circuit TestPerformed on LV side.At rated voltage.Copper losses are negligible.Gives shunt branch elements i.e. Rc and Xm.Performed on HV side.At rated current.Core losses are negligible.

Gives series branch resistance and reactance Req and Xeq.Example A 50 kVA, 2200/220 V, 50 Hz, single phase transformer gave the following results during no load and short circuit test. Open circuit test: 220 V, 5 A, 405 W (LV side) Short Circuit Test: 95 V, 20.2 A, 805 W (primary side)Calculate:The no load parameters R0 and Xm.Equivalent resistance and reactance referred to primary.Draw circuit model referred to HV side.Efficiency at full load and 0.8 pf lagging.