Problems on Transformer

Lesson-45

Problems on Transformer

7.Ina25KVA2000/200Vsinglephasetransformer,theironandfullloadcopperlossesare350Wand400Wrespectively.Calculatetheefficiencyatunitypowerfactoronfullloadandhalfload. [Dec2014/Jan2015]

KVA=25KVA E1=2000V E2=200V Wi=350W

Wcu=400W

I2= (25×103)/200=125A

Ƞ=

Ƞ=97.08%

Ƞ=

Ƞ=96.52%

8.An8polealternatorrunsat750rpmandsuppliespowertoa6poleinductionmotor

whichrunsat970rpm.Whatistheslipoftheinductionmotor?[Dec2014/Jan2015] Alternator data: poles=8;N=750rpm

So, f =Hz

==50Hz

Induction motor: poles = 6;N =970rpm

Ns ===1000rpm

S =* 100 =* 100=0.03 =3%

9. A600KVAtransformerhasanefficiencyof92%atfullload,unitypowerfactorandhalffullload,0.9pf.Determineitsefficiencyat75%offullload,0.9pf. [Dec2014/Jan2015]

S = 600 KVA, %

On fullload,%

= 92% onfull loadand halfloadboth

=

0.92=

= 52173.913

On half load,n == 0.5

%=

0.92 =

0.25= 23478.26…………..(2)

Subtracting(2) from(1),

0.75= 28695.64

= 38260.86 watts

and= 13913.04 watts

Nown=0.75i.e., 75%of full load and cos= 0.9

()new=n2()F.L. = (0.75)2×.

%=

=

= 91.95%

10.Findthenumberofturnsontheprimaryandsecondarysideofa440/230V,50Hzsingle

phasetransformer,ifthenetareaofcrosssectionofthecoreis30cm2andthemaximumfluxdensityis1Wb/m2. [June/July2014]

V1= 440V

V2= 230V

f =50 Hz

A =30cm2= 30×10-2×10-2m2

Bm=1 Wb/m2

E1= 4.44fN1V

= BmA

E1= 4.44fBmA N1V

400 = 4.44 *50*N1* 30×10-2×10-2

N1 =660

= 5.63×10-3Wb

=

N2=345

11.Asinglephasetransformerworkingat0.8pfhasefficiency94%atboththreefourthfullloadandfullloadof600kW.Determinetheefficiencyathalffull–load,unitypowerfactor. [June/July2014]

On full load, %=

0.94=

= 30638.3…………(1)

On half load,n == 0.75

%=

92=

0.5625 = 22978.72…………..(2)

Subtracting(2) from(1),

= 17.51K watts

and= 13.13K watts

Nown=0.5i.e., 75%of full load and cosΦ= 1

% =

=

= 90.73%

12. Asinglephase,20KVAtransformerhas1000primaryturnsand2500secondaryturns.Thenetcrosssectionalareaofthecoreis100cm2.whentheprimarywindingisconnected

to500V,50Hzsupply,calculate)themaximumvalueofthefluxdensityinthecore,thevoltageinducedinthesecondarywindingandtheprimaryandthesecondaryfullloadcurrents. [June/July2013]

KVA =20KVA

N1= 1000V

N2= 2500V

f =50 Hz

A =100cm2= 100×10-2×10-2m2

V1 =500V

E1= 4.44fN1V

500 = 4.44*50**1000

= 2.25mWb

=

E2=1250V

I1=== 40A