Transformers

TransformersWorkshop on Basic Electrical Engineering held at

VVCE, Mysuru, on 30-April-2016

R S Ananda Murthy

Associate ProfessorDepartment of Electrical & Electronics Engineering,

Sri Jayachamarajendra College of Engineering,Mysore 570 006

R S Ananda Murthy Transformers

Learning Outcomes

After completing this lecture the student should be able to –State why transformers are necessary in power systems.Describe the principle of operation of transformers.Describe the construction of core and shell types ofsingle-phase transformers.Calculate the induced E.M.F. in a transformer.Define voltage regulation of a transformer.Describe various power losses in a transformer.Find the efficiency of a transformer.


Photograph of a Power Transformer

Power transformers are used at transmission level.


Photograph of a Distribution Transformer


Necessity of Transformers in Power Systems

Transformers are necessary to step up from generationvoltage which is typically 11 kV to transmission voltagelevels which are typically 220 kV or 400 kV.They are also required to step down voltage fromtransmission levels which are typically 220 kV or 400 kV todistribution levels which are typically 11 kV, 415 V(three-phase) or 230 V (single-phase).


Core and Shell Types

Core TypeCross-section of core type transformer

L.V. Winding H.V. Winding

L.V. Winding H.V. Winding

Shell Type Cross-section of shell type transformer.

Cor

e

Yoke

Yoke

Cor

e

Laminations

Yoke

Yoke

Cor

e

Cor

e

Cor

e

Laminations


Core Type Transformer


Shell Type Transformer


Principle of Operation of Transformer

Load

When supply is given to the primary winding having T1 turns,an alternating flux is established in the transformer core. Due tothis, there will be change of flux linkage in the primary and alsoin the secondary winding. Then, according to Faraday’s law,voltage is induced in both the primary and secondary windings.


Induced E.M.F. in Transformer

When the primary applied voltage is sinusoidal, the mutual fluxφ also varies sinusoidally. So, let

φ = Φmax sinωt (1)

where ω = 2πf and f = supply frequency. Let the flux linkage ofthe primary and secondary be

λ1 = φT1 and λ2 = φT2 (2)

According to Faraday’s law, when φ changes, voltage isinduced in both primary and secondary. The voltage induced inthe primary is given by

e1 =−dλ1

dt=−T1

ddt

(Φmax sinωt) = ωT1Φmax sin(ωt−90◦) (3)


Induced E.M.F. in Transformer

Similarly the voltage induced in the secondary is given by

e2 =−dλ2

dt=−T2

ddt

(Φmax sinωt) = ωT2Φmax sin(ωt−90◦) (4)

This shows that both e1 and e2 lag behind the mutual flux φ by90◦. The R.M.S. values of e1 and e2 are given by

|E1| =2πf ΦmaxT1√

2= 4.44f ΦmaxT1 (5)

|E2| =2πf ΦmaxT2√

2= 4.44f ΦmaxT2 (6)


Voltage Transformation Ratio

In a practical transformer, the voltage drop in primary andsecondary winding impedances is very small — typically lessthan 5% of the rated voltage. So, by applying K.V.L. to theprimary and secondary we get

V1 ≈ E1 and V2 ≈ E2 (7)

Hence we can write

|V1||V2|

≈ |E1||E2|

=T1

T2= a12 (8)

This is known as voltage transformation ratio.


Current Transformation Ratio

Due to good magnetic material used for the core, it can beshown that the reluctance of the core is almost zero. Thismeans that

F1 = F2 =⇒ |F1|= T1|I1|= |F2|= T2|I2|

where F1 is the primary M.M.F. and F2 is the secondary M.M.F.Therefore,

|I1|T1 ≈ |I2|T2 =⇒ |I1||I2|

=T2

T1= a21 =

1a12

(9)

This shows that current transformation ratio is the reciprocal ofvoltage transformation ratio.


Voltage Regulation

Voltage regulation of a transformer is defined as

Regulation =|V2NL|− |V2FL||V2FL|

×100%

where V2NL = secondary terminal voltage when there is noload, V2FL = secondary terminal voltage when there is fullload.Ideally the regulation should be zero which means that thesecondary terminal voltage should not vary from no-load tofull-load.Typically the regulation of practical transformers would beless than 5%.


Power Losses in Transformer

Copper Loss (Pcu) in primary and secondary windings.Iron Loss (Pi) which has two components

Hysteresis loss Ph = khtfB1.6max

Eddy current loss Pe = kef 2t2B2max

where Pe = eddy current loss, Ph = hysteresis loss, t =thickness of laminations used in the core. kh = hysteresis lossconstant, and ke =eddy current loss constant which areconstants for a particular material used for the core.Bmax = Φmax/Ai is the maximum flux density of in the core.

This shows that by using very thin laminations, iron loss can bereduced and by using good conducting material like copper oraluminium for the windings, copper loss can be reduced.


Copper Loss (Pcu)

This is nothing but I2R loss occurring in the primary andsecondary winding resistances. Since I2R power loss isproportional to square of current in the windings, it is obviousthat this power loss varies with load.

If Pcu(r) is the copper loss in the transformer at full-load current,then, at any multiple x of full load, the copper loss will bePcu = x2Pcu(r).


Iron Loss (Pi)

Maximum value of flux in the transformer core is given by

Φmax ≈|V1|

4.44fT1(10)

Since the primary supply voltage |V1| and frequency f areconstant, Φmax will also be constant.The iron loss in a transformer is given by

Pi = Pe + Ph = kef 2t2B2max + khtfB1.6

max (11)

This shows that, at constant frequency and supply voltagethe iron loss Pi will remain practically constant even if theload varies.


Efficiency of Transformer

The efficiency of transformer at any multiple x of full-load isgiven by

η =Pout

Pin=

x |V2r | · |I2r |cosθ

x |V2r | · |I2r |cosθ + Pi + x2Pcu(r)(12)

which is same as

η =x |Sr |cosθ

x |Sr |cosθ + Pi + x2Pcu(r)

where x = any multiple of full load, |Sr |= rated volt-amperes ofthe transformer, cosθ = power factor of the load, Pi = iron lossof the transformer which is constant, and Pcu(r) = copper loss inthe transformer at rated load.


Condition for Maximum Efficiency

At maximum efficiency, dη/dx = 0. Applying this condition weget the condition for maximum efficiency as

Pi = x2Pcu(r) =⇒ x =

√Pi

Pcu(r)

This gives the multiple x of full load at which efficiency will bemaximum. In this equation, for full load we have to take x = 1,for half-load x = 0.5, for quarter load x = 0.25, and so on.


License

This work is licensed under aCreative Commons Attribution 4.0 International License.


Transformers

Engineering

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