chapter 5 - transformer (jun 2011)

7/31/2019 CHAPTER 5 - Transformer (Jun 2011)

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A device which uses the phenomenon ofmutual induction to change the values

of alternating voltages and currentsAdvantages:

› low losses› high efficiency› long life› very stable



Types:

Three-phase core

Helix and disc coils on single-phaseCore and windings of three-phase

Single-phase shell



Consists of two windings connected by a

magnetic coreOne winding is connected to a power supply (AC supply) and the other to a load

N1 N2

AC supplyLoad

Flux

Secondary windingPrimary winding

Core



When the secondary is an open-circuit andan alternating voltage V 1 is applied to aprimary winding, a small current flows whichset up a magnetic flux in the core.This alternating flux links with both primary

and secondary coils and induces in theme.m.f.’s of E 1 and E 2. The induced e.m.f., E in a coil of N turns isgiven by

d E N dt

where= rate of change of flux

d

dt

Unit: Volt



In an ideal transformer, losses areneglected

Considered to be 100% efficient.In ideal transformer, the rate of changeof flux is the same for both primary andsecondary thus

The induced e.m.f. per turn is constant.

1 2

1 2

E E

N N



Assuming no losses, E 1 = V 1 and E 2 = V 2

Hence, Input power = Output power V1 I1 = V 2 I2

Primary and secondary volt-amperes areequal,

1 1

2 2

V N

V N or 1 1

2 2

V N V N

1 1

2 2

V I

V I



Hence combining equations,

1 1 1

2 2 2

V N I V N I

where;V₁ = primary voltage (volt, V)

V₂ = secondary voltage (volt, V)N₁ = primary windings (turns)N₂ = secondary windings (turns)I₁ = primary current (ampere, A)I₂ = secondary current (ampere, A)



Example1. A transformer is to be used to provide a 60

V output from a 240 V A.C supply.Calculate

a. the turns of ratio requiredb. the number of primary turns, if the secondary is

wound with 500 turns.

2. A 2000/200V, 20kVA transformer has 66 turnsin the secondary. Calculate:

a. primary turns

b. primary and secondary full-load currents

Neglect the losses.



Two types of single-phase double-woundtransformer construction

› the core type

› the shell type



Transformer ratio is the comparisonbetween primary and secondary

elements

Np NsVp Vs

Secondary windingPrimary winding

Flux produced by primary windinginduces secondary winding via core

s s s

p p pTransformer ratio,

N E V K N E V

If K < 1 i.e. Ns < Np : this transformer is called step-down transformer If K > 1 i.e. Ns > Np : this transformer is called step-up transformer

If K = 1 i.e. Ns = Np : this transformer is called coupling transformer



Step-up transformer

Step-down transformer

Coupling transformer

Np

Np

Np

Ns

Ns

Ns



Example

1. The number of windings for the threetransformers are

a. Np = 100, Ns = 2000b. Np = 3000, Ns = 2000c. Np = 100, Ns =100

Calculate the value of K for eachtransformer then determine the type of

transformer and draw the symbol oftransformer to differentiate the number of windings



R.M.S value of e.m.f induced in primarywinding

E1 = 4.44 N 1 f m volts

R.M.S value of e.m.f induced in

secondary windingE2 = 4.44 N 2 f m volts



Example1. A 250 kVA, 1100 V / 400 V, 50 Hz single-phase

transformer has 80 turns on a secondary.Calculate :a. the approximate values of the primary and

secondary currents.b. the approximate number of primary turns.c. the maximum values of flux.

2. An ideal 25 kVA transformer has 500 turns on theprimary winding and 40 turns on the secondarywinding. The primary is connected to 3000 V, 50Hz supply. Calculate

a. primary and secondary currents on full-loadb. secondary e.m.f. andc. the maximum core flux



losses which occur in a transformer onload can be divided into two groups:

›

Copper losses› Core losses (hysteresis and eddy currents)



Variableresult in a heating of the conductors,

due to the fact that they possessresistanceIf R1 and R 2 are the primary and

secondary winding resistances then thetotal copper loss isPcu = I1

2R1 + I22R2



Hysteresis loss› the heating of the core as a result of the internal

molecular structure reversals which occur as themagnetic flux alternates.

› proportional to the area of the hysteresis loop and

thus low loss nickel iron alloys are used for the coresince their hysteresis loop have small areas.Eddy current loss

› the heating of the core due to e.m.f . ‘s beinginduced not only in the transformer windings but alsoin the core.

› These induced e.m.f.’s set up circulating currents calleddy currents. Owing to the low resistance of thecore, eddy currents can be quite considerable andcan cause a large power loss and excessive heatingof the core.



Assume the core loss constant at all loads› (the maximum value of the flux in a normal

transformer does not vary by more than about 2per cent between no load and full load)

If P c = total core loss, total losses intransformer;

Total losses = P c + P cu

Total losses = P c + I21 R1 + I2

2 R2



lossesPP

in

out

lossespowerinputpoweroutput,Efficiency

22

212

122

22

p.f.

p.f.

, R I R I PV I

V I

Efficiency c

where;I₁ - primary current (A)

I₂ - secondary current (A)V₁ - primary voltage (V)V₂ - secondary voltage (V)Pc – core lossesR₁ - primary load ( Ω)R₂ - secondary load ( Ω)

p.f – power factor



powerinput

losses-powerinput

powerinput

poweroutput,Efficiency

powerinput

losses

1,Efficiency



Example1. The primary and secondary windings of a 500

kVA transformer have resistances of 0.42 and 0.0019 respectively. The primary andsecondary voltages are 11 000 V and 400 Vrespectively and the core loss is 2.9 kW,assuming the power factor of the load to be

0.8. Calculate the efficiency on :a. full loadb. half load

2. In a 50 kVA transformer, the iron loss is 500 Wand full-load copper loss is 800W. Find theefficiency at full-load and half-load at 0.8 p.f.lagging.



Advantages:› enable efficiency and voltage regulation to

be a calculated without actually loading thetransformer › Higher accuracy is possible by direct

measurement of input and output powers

and voltages› Power required to carry out these tests is very

small compared with the full-load output oftransformer.



The transformer is connected to a supply at the ratedvoltage and frequency, namely the voltage and thefrequency given on the name plate.The ratio of the voltmeter readings, V 1 / V 2, gives theratio of the number of turnsAmmeter A gives the no-load current, and its readingis a check on the magnetic quality of theferromagnetic core and joints.The primary current on no load is usually less than 5per cent of the full-load currentI 2 R loss on no load is less than 1/400 of the primary I 2 R

loss on full load and is therefore negligible comparedwith the core loss.Wattmeter reading can be taken as the core loss ofthe transformer.



Secondary winding is short-circuited through asuitable ammeter A 2Low voltage is applied to the primary circuit.This voltage should, if possible, be adjusted tocirculate full-load current in the primary andsecondary circuits.Assuming this to be the case, the I 2 R loss in thewindings is the same as that on full load.On the other hand, the core loss is negligibly small,since the applied voltage is lowFlux are only about one-twentieth to one-thirtieth ofthe rated voltage and fluxThe core loss is approximately proportional to thesquare of the flux.Hence the power registered on wattmeter W can betaken as the I 2 R loss in the windings.



A transformer having a part of its windingcommon to the primary and secondary

circuits



The nearer the ratio of transformation is tounity, the greater is the economy of

conductor materialMainly used for interconnecting systemsthat are operating at roughly the samevoltage and starting cage-type inductionmotors.Should an auto transformer be used tosupply a low voltage system from a highvoltage system, it is essential to earth thecommon connection

(to avoid serious shock)An auto transformer should not be used for interconnecting high voltage and lowvoltage systems.



Advantages Disadvantages1. Save cost (less copper core

is needed)1. A smaller percentagevoltage regulation

2. Less volume 2. The primary and secondarywindings are not electricallyseparate

3. Less weight

4. High efficiency

5. Low copper losses

6. A continuously variableoutput voltage is achievable ifa sliding contact is used.

chapter 5 - transformer (jun 2011)

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