Unit II - TRANSFORMERS

1.State the principle of working of a transformer?Transformer consists of two inductive coils which are electrically separated but magnetically linked through a path of low reluctance. The two coils possess high mutual inductance. If one coil connected to a source of alternating voltage, an alternating flux is set up in the laminated core, most of which linked with the other coil in which it produces mutually induced emf. If the second coil is closed current flows in it and so electrical energy is transferred from the first coil to the second coil.

2.List the types of transformers based on construction: Core type transformer Shell type transformer

3.State why the transformer core is made of magnetic materials?To provide a continuous magnetic path with the minimum air gap.

4..Define transformation ratio ?Transformation ratio is defined as the ratio of number of turns in the secondary winding to number of turns in primary winding . K= N2/N1

= E2/E1

Types :Step up transformerStep down transformer

5..Define voltage regulation of a transformer:The change is secondary terminal voltage from no load condition to full load condition is known as voltage regulation

6..State the methods of reducing leakage flux:To minimize the leakage flux, transformer core is sectionalized and by interleaving the primary and secondary windings.

7..List the various losses in a transformer and state the condition for maximum efficiency:

Core lossCopper lossCondition for maximum efficiency: Iron loss = copper loss

8..Define all-day efficiency of a transformer?All day efficiency is defined as the ratio of output in Kwh to input in Kwh

calculated for 24 hours

9. Define the purpose of breather used in a transformer?Complete air tight chambers provided to permit the oil inside the tank expand and contract as temperature increases or decreases10..State the test to be conducted to determine the efficiency of a transformer

OC test –iron lossesSC test – copper losses

11..State the various types of a transformers depending on its voltage transformation ratio:

Step-up transformerStep- down transformer

12.Write the emf equation of a transformer and explain the terms used:E=4.44 m f N (volts)

13 .why Core loss of a transformer practically constant ?Since the core flux is practically constant for all loads the core loss is practically same for all loads.14. What is the main difference between core type and shell type

transformers ?Core type:The coils wound are form woundand are cylindrical type.Wound

in helical layers insulated from each other by paper,cloth,mica.Used for low voltage transformers.Windings surround a considerable part of the core.

Shell type:coils are form wound but are multi layer disc type,insulated from each other by paper.Transformer core consists of laminations arranged in groups which radiate out from the centre.Used for high voltage transformers.Core surrounds a considerable portion of the winding.

15. Why the rating of the transformers is in KVA ?Opper loss of transformer depends on current and iron loss on voltage.Total loss depends on volt-ampere and not on phase angle.

16. In what way the iron losses can be minimised.Laminating the core,laminations being insulated from each other by varnish.

17. What are the two components of no load input current of a transformer

(i) Active or working or iron loss plus small quantity of primary cu loss

(ii) Quadrature with v1 known as magnetising component because of function to sustain the alternating flux in the core.

18. What are the different kinds of transformersPower transformer ,distribution transformer ,auto transformer

19. What is the need for conducting O.C and S.C tests in a transformer.?For the calculation of Iron loss and copper losses so as to determine the efficiency of the transformer.

20. Define the equivalent circuit of a transformer?Pictorial representation of the transformer,simplification can be done by transferring resistances,reactances,load,no load branch for easy calculation.

21. Define an Ideal transformer ?A perfect transformer considered to have no loss and having max efficiency which is practically not possible.

22. Explain how the flux in the core is practically maintained constant?The secondary current sets up its own flux which is in opposition to main flux called demagnetising amp turns.For a moment v1 gains E1 and hence more current to flow in primary.Additional current called i2’ known as load component of primary current.Magnetic effects of secondary current i2 are immediately neutralized by additional primary current i2’ hence the net flux is constant.

23. Define mutual inductionConsists of two inductive coils electrically separated but magnetically linked.If one coil is connected to the source alternating flux is set up which links the other coil also which produces mutually induced emf.If the second coil circuit id closed a current flows in it and so electrical energy is transferred from the first coil to second coil.

24. What is approximate equivalent circuit.?All the impedances combined neglecting the no load branch

25. Explain the term demagnetizing ampere turns.?The secondary current sets up its own flux which is in opposition to main flux

PART -- B

TRANSFORMERS

GROUP---A

1) (i)Derive the EMF equation of a Transformer? (6)

N1=No of turns in PrimaryN2=No of turns in secondaryΦm=Max flux in the core =Bm *AAverage rate of change of flux=Φm/1/4f=4fΦmwb/sAverage enf/turn=4fΦmvoltForm factor=R.M.S value/average value=1.11Rms value of emf/turn=1.11*4fΦm=4.44fΦmvRms value of the induced emf in the whole of primary winding=4.44fN1Φm=4.44fN1BmAE2=4.44fN2Φm=4.44fN2BmAE1/N1=E2/N2=4.44fΦm

(ii) A 5 KVA ,2300/230 V ,50 Hz transformer was tested for the iron losses with normal excitation and copper losses at full load and they were found to be 40 W and 112 W respectively.Calculate the efficiencies of the transformer at 0.8 p.f for the following KVA outputs 1.25,2.5,3.75,5.0,6.25 and 7.5 (10)

Solution: F.L CU loss:12 Iron loss:40W

Formulas: CU loss for given KVA =F.L Cu loss *[given KVA/Total KVA]2

Total loss=F.L Cu loss+Cu loss at given KVA

Output =Given KVA * given power factor

ή = output/output + Losses * 100

i) Cu loss at 1.5 KVA=7 W Total loss=47 W Output = 1000w Effeciency = 95.51%ii) ii) Cu loss at 2.5 KVA=28 W Total loss=68 W Output = 2Kw Effeciency = 96.71%iii) iii) Cu loss at 3.75 KVA=63 W Total loss=103 W Output = 3Kw Effeciency = 96.34%

iv) Cu loss at 5 KVA= 112W

Total loss=0.152KW Output = 4Kw Effeciency =96.34 %

v) Cu loss at 6.25 KVA= 175W Total loss= 125W Output =5Kw Effeciency =95.88%

vi) Cu loss at 7.5 KVA=252 W Total loss= 292W Output = 6Kw Effeciency =95.36 %

2) (i) Draw and explain the phasor diagram of a transformer with no load and on load.How it affects the power factor of the loaded transformer? (12)

Secondary current sets its own mmf,own flux Φ2Secondary amp-turns N2I2 known as demagnetising amp turns additional primary current by I2’Additional flux is produced which is in opposition,Hence the two cancel each other out.Net flux passing through the core is approximately the same as at no loadCore loss is practically same under all load conditionsn2i2=n1i2’i2’=n2/n1*i2=ki2Total primary current i1 is the vector sum of i0 and i2’ and lags behind v1 by an angle Φ1It is observed that Φ1 is slightly greater than Φ2If we neglect i0 Φ1=Φ2I2’/i2=i1/i2=n2/n1=kUnder full load conditions the ratio of primary and secondary currents are constantFor inductive load I2 lags E2 by Φ2

(ii) A single phase transformer has 400 primary and 1000 secondary turns .The net cross sectional area of the core is 60 sq cm.If the primary winding be connected to a 50 Hz supply at 520 V ,Calculate the peak value of flux density in the core and secondary voltage ? (4)

E1=4.44 f N1Φm520=88,800*ΦmΦm=5.85*10exp -3 wbBm= Φm/(60*10^-4)=.975wb/m2(ii) Secondary voltage v2/v1=n2/n1V2=n2/n1*v1=1300v

3) (i) Obtain the equivalent circuit of a single phase transformer (8)

X0=e1/i0 R0=e1/iwE2/E1=n2/n1=kPrimary eqvt of secondary induced vg E2’=E2/K V2’=V2/k i2’=ki2For transferring secondary impedance to primary multiply by k^2R2’=R2/K^2 X2’=X2/k^2 z2’=Z2/k^2Total eqvt circuit of transformer is obtained by adding in the primary impedance simplification can be made by transferring the exciting circuit across the terminals

Further simplification may be achieved by ommiting i0 altogether

(ii) A 230/230 ,3 KVA transformer gave the following results

O.C Test : 230V , 2 A,100 W S.C Test : 15V , 13 A,120 WDetermine the regulation and efficiency at full load 0 .8 P.F lagging (8)

Rated current=power/voltage=13ATotal loss=cu loss+core loss=220wPower output=Rated power*P.F=2400wRequired ή=2400/2400+220*100=91.6%From S.C test Z=V/I =1.154ΩP=I^2 RR=P/I^2=o.53ΩX=SQ Root Z^2-R^2=1.0251ΩApproximate Voltage RegulationIRCOSΦ+IXSinΦ=13.51v% Vg Regulation =App Vg Regulation/Rated Vg * 100=5.874%

4) Explain in detail about the conduction of open circuit and short circuit test on single phase transformers to predetermine the efficiency and regulation (16)To determine the no load loss or core loss ,high voltage winding is left open other connected to supply voltage.A wattmeter w1,voltmeter V1 and ammeter A are connected to L.V windingWith normal voltage applied to primary normal flux will be set up in the core normal iron loss will occur which are recorded by the wattmeterWattmeter reading represents practically core loss under no load conditionw=v1io cosΦ0 cos Φo=W/v1i0iµ=io sin Φ0 iw=i0 cos Φo xo=v1/iµRo=v1/iwExciting admittance yo of the transformer is given by io=v1yo yo=io/v1Exciting conductance go is given by w=v1^2 go or go=w/v1^2Exciting susceptance Bo=Sq root (40^2 – Go^2)

5) (i) Discuss about the different types of losses occurring in the transformer ? (8)Core loss: Due to alternating flux set up in the magnetic core of the transformer it undergoes a cycle of magnetisation and demagnetisation.Due to hysterisis effect there is loss of energy which is called as hysterisis loss.Hysterisis loss = Kh Bm ^ 1.67 f v

(ii) With neat diagram explain the working of a single phase transformer (8)

GROUP---B1) (i) Derive the condition for obtaining Maximum efficiency of a single phase transformer ? (6)

(ii) Explain the operation of single phase transformer on load condition ? (10)

3)(i) Explain the constructional details of a single phase transformers ? (10)

(ii) A 25 KVA single phase transformer has 250 turns on the primary and 40 turns on the secondary winding.The primary is connected to 1500 V ,50 Hz mains.Calculate (i) Primary and secondary currents on full load (ii) Secondary emf (iii) Maximum flux in the core (6)

4)Consider a 4 KVA ,200/400 single phase transformer supplying full load current at 0.8 lagging P.F .The O.C and S.C test results are as follows

O.C Test : 200 V , 0.8 A , 70 W (L.V Side)S.C Test : 20 V , 10 A ,60 W (H.V Side)

Calculate the efficiency ,Secondary voltage and current into the primary at the above load Calculate the load at unity P.F corresponding to Maximum effeciency

5) (i) Discuss about the regulation of a Transformer? (6)

(ii) A 100 KVA Transformer has 400 turns on the primary and 80 turns on the secondary .The primary and secondary resistances are 0.3Ω and 0.01Ω respectively and the corresponding leakage reactancec are 1.1 and 0.035 Ω respectively.The supply voltage is 2200V Calculate (i) Equivalent impedance referred to primary (ii)Voltage regulation and secondary terminal voltage for full load having a power factor of 0.8 leading. (12)