Transformer (Mumbai University)

8/13/2019 Transformer (Mumbai University)

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8/14/2013 unit 4-Transformer

Transformer

The transformer is a static device, (the

one which does not contain anyrotating or moving parts) which is usedto transfer electrical energy from oneac circuit to another ac circuit, with

increase or decrease in voltage/currentbut without any change in frequency.


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Ac input Ac output

I1, V1 I2, v2

Transformer


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Application of a Transformer


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Transformer Symbol

(Iron core)

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Schematic symbol for transformer consists of twoinductor symbols, separated by lines indicating aferromagnetic core.

The two inductor coils are easily distinguished in the symbol. Thepair of vertical lines represent an iron core common to bothinductors. While many transformers have ferromagnetic corematerials, there are some that do not, their constituent inductorsbeing magnetically linked together through the air.


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Laboratory transformer

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Generator transformer

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Substation transformer

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3 Phase 3 wire transmission / 3 phase 4 wire utilization


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Transformer can be step up or stepdown transformer. The principle ofoperation is same.

In step up the secondary coil consistsof large no of turns of insulated copper

wires Low voltage high current ac

high voltage low current


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AUTO-TRANSFORMER

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ISOLATION TRANSFORMER

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P i i ti


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Princip e o operation examEnergy is being transferred from the primary coil tothe secondary coil by means of a mutual magneticfields.


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Can D.C. Supply be used forTransformer?

No The transformer works on the principle of mutual

induction, for which current in one coil must changeuniformly. If d.c. supply is given, the current will notchange due to constant supply and transformer will

not work. Practically winding resistance is very small. Ford.c., the inductive reactance XL is zero as d.c. has nofrequency. So total impedance of winding is very lowfor d.c. Thus winding will draw very high current ifd.c. supply is given to it. This may cause the burningof windings due to extra heat generated and may

cause permanent damage to the transformer. There can be saturation of the core due to which

transformer draws very large current from the supplywhen connected to d.c.


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Construction of transformer


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Construction contd.

1) Magnetic circuit consisting of limbs(core), yokes and clamping structures(providing the flux path)

2) Electric circuit consisting of primaryand secondary windings

3) Dielectric circuit consisting ofinsulation in different forms and used

at different places in the transformeri.e core to primary winding, primarywinding to secondary winding, etc)


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Types of Transformers

Depending on construction:

Core type ,Shell type

Depending on transformation ratio: Step-up ,Step-down,Isolation

&Auto-Xmer

Depending on supply:

Single phase & three phase

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Core type and shell typetransformer
http://en.wikipedia.org/wiki/File:Transformer_winding_formats.jpg


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Core Type Transformer

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Shell Type Transformer

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Core type transformer


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Core type transformer

A single phase core type transformer consists ofa magnetic frame with two cores, upper yokeand bottom yoke. The primary and secondarycoils are split into two half turns of the primaryand half the secondary turns are placed on each

core (limb) A 3-phase core type transformer consists of

minimum 3 cores, each provided with theprimary and secondary coils of one phase.

Circular coils (windings) are used in the coretype transformer, which indicate theoretically

that a circular core should be used. It is very complicated to manufacture circular

core as a result, the stepped core is generallyused.


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Shell type Transformer


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Shell type transformer

It consists of magnetic frame with a central core(limb) and two side cores, completing the pathof magnetic flux.

Primary and secondary coils (windings) are

placed on the central core in a particular fashionas shown in figure. Such an arrangement formsa shell of iron around copper.

The central leg flux is divided at the yokesection, half i.e. /2 towards each side leg.

As the flux in the section of yoke and side coresis only half, the cross section of the yokes andside cores is approximately half the section ofthe central limb


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Comparison

Core Type

1. The windingsenclose the wholecore.

2. Average length ofthe core is more.

3. Magnetic flux hasonly one continuouspath.

4. More suitable forHV transformers.

5. Easy to repair.

Shell Type

1. The core enclosesthe windings.

2. Average length ofthe core is less.

3. Magnetic flux isdistributed into twoparts.

4. More suitable for LVtransformers.

5. Difficult to repair.

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Ideal Transformer (exam)

An Ideal transformer is an imaginary transformerstudied to understand the concept easily.

Characteristics

1) An ideal transformers core is highly permeable

so that it requires very small magneto motiveforce (mmf) to set up the flux in the core.

2) Its leakage flux is zero, that is the entire flux isconfined to the core and links with both thewindings

3) The resistance of the primary and secondarywinding is negligible (zero).

Therefore E1=V1 and E2= V2


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Characteristics of idealtransformer

4) There are no losses due toresistance, hysteresis and eddycurrents.

5) Efficiency of an ideal transformeris 100% as no losses takes place.

6) Voltage regulation is 0 %. Thatmeans secondary voltage willremain constant irrespective ofthe load current.


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Practical Transformer

A practical transformer possesses the

following characteristics:

1. There are Cu & Iron losses.2. There is leakage flux.

3. Its windings contain ohmic

resistance.

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Comparison

Practical Transformer

1. There are Cu & ironlosses.

2. There is leakage flux.

3. Its windings containohmic resistance.

4. Voltage regulation isnever 0%.

5. Efficiency is 93-97 %.

6. All constructedtransformers arepracticaltransformers.

Ideal Transformer1. There is no loss.

2. There is no leakageflux.

3. Its windings consist of

purely inductive coils,wound on a losslesscore.

4. Voltage regulation is0%.

5. Efficiency is 100%.6. It is impossible toconstruct an idealtransformer.

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EMF equation of a transformer (e)


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EMF equation of a transformer

The various quantities which affect themagnitude of the induced e.m.f. are :

= Fluxm = Maximum value of flux

N1 = Number of primary winding turnsN2 = Number of secondary winding turns

f = Frequency of the supply voltageE1 = R.M.S. value of the primary induced

e.m.f.E2 = R.M.S. value of the secondary induced

e.m.f.


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refer notes


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Losses in transformer

An ideal transformer has no loss.

But in practical transformercopper losses (Pcu ) and ironlosses (Pi) takes place. Iron lossis further classified into two typesnamely hysteresis loss and eddy

current loss.


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Copper Loss

Copper losses occur in all those parts of atransformer that carry electric current.

It is the total power loss taking place in the

winding resistances of a transformer Copper Loss = Power loss in the primary

resistance + Power loss in the secondaryresistance.

Cu loss can be calculated by conducting

SC test on the transformer.

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Copper Loss

Should be kept as low as possible toincrease efficiency.

To reduce this reduce R1 & R2.

Also called as variable loss-dependenton square of load current.

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Iron Losses (or Core Losses)

Hysteresis loss occurs in the magneticframe of the transformer.

Due to time varying flux, there isinduced emf in the core which causeseddy currents to flow through the corebody.

Iron loss can be calculated byconducting OC test on transformer.

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Losses in a Transformer

Hysteresis loss depends on area of hysteresis loopof magnetic material used for frame, volume ofcore, frequency of magnetic flux reversal. Thisloss can be minimized by selecting a material forthe core that has a low hysteresis coefficient(silicon steel)

Eddy current loss depends on thickness of lamination of core, frequency of flux reversal,maximum value of flux density in core, volume ofcore, quality of magnetic material used for themagnetic frame. This loss can be minimized byusing laminated core.

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Ideal Transformer(no load)

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Conditions for an ideal or loss freetransformer

The primary and secondary windingsdo not have any resistance i.e.winding resistance of primary and

secondary should be zero.

The losses taking place in the corei.e. hysteresis loss and eddy current

loss should be zero. There should not be any leakage flux


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Phasor Diagram

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Ideal transformer on load

Refer notes

Practical transformer on no load

Refer notes


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When some load is connected between thesecondary terminals of the transformer, thetransformer is said to be loaded or on load.

Due to load on the secondary , a finitesecondary current starts flowing.

If the load is R+L type the I2 will lagbehind V2 by an angle 2.

As per the lenzs law, the secondary currentI2 will oppose the cause producing it.Hence it opposes the magnetic flux. This iscalled demagnetizing effect of I2


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I2 = I2 * N2/N1 =kI2 I2 = kI2 is 180 degree out of phase

with I2.

The net primary current is 1 =2 + mag Thus due to load on secondary side,

the primary current of thetransformer increases to supply theadditional power to the load


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Practical transformer on NO load

When the practical transformer is onNo load, the secondary current will bezero. Hence the copper loss in the

secondary winding is zero. However a small primary current does

flow under the no load condition. Dueto small primary resistance, a small

primary copper loss takes place evenat no load


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Practical transformer on NO load

The primary current under no loadcondition has to supply the ironlosses and a small primary copper

loss. The primary current under the no

load condition is denoted by o. o =m +c m is the magnetizing component c is the active component


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Practical transformer no No load

Im is the purely reactive component.

It magnetizes the core and producesflux in the core.

This component is at 90 degree w.r.tto E1.

Im is also called as wattlesscomponent.


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Practical transformer no No load

Ic is the active component of no loadcurrent Io.

Its job is to supply the total loss

under no load conditions. It is therefore called as the power

component or core less component ofIo.

It is at 90 degree w.r.t themagnetizing current Im.


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Practical Transformer(no load)

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o =m +c Im and Ic are 90degree phase shifted

w.r.t each other and Io is the resultantof the two.

No laod power factor = coso

Im = Io sino

Ic= Io coso

Io=Im2 + Ic2 o = tan-1(Im/Ic)


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Total input power on No load

Wo = V1 Io coso

Io coso =Ic

Wo= V1 Ic primary current Io is very small therefore

primary copper loss is very very small. Icis called as core loss or iron losscomponent.

Wo represents the core loss or iron loss ascopper is small.

Wo- iron loss= V1 Ic


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Practical Transformer(on load)-w/o resistance and leakage

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Practical transformer on load

When some load is connected between thesecondary terminals of the transformer, itis said to be on load.

Due to the load on secondary, a finitesecondary current starts flowing.Depending on the type of load (resistive,inductive , capacitive) the secondary

current I2 will be in phase, or lag or leadthe load voltage V2.


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Due to loading of the transformer, the primary currentincreases above its no load value due to the followingfactors:

1) when the transformer is loaded, the load current I2 willstart flowing. Due to increase in load current I2 thesecondary ampere turns N2I2 will also increase.

2)This increased secondary (mmf) N2I2 will increase theflux 2 set up by the secondary current.

3) This flux opposes the main flux 1 set up in the core bythe current flowing through the primary winding. Hence thesecondary mmf N2 I2 is called as the demagnetizingampere turns.

Due to reduction in the main flux 1, the induced emf in theprimary winding E1 will also reduce. Hence the differencebetween Vi and E1 will increase and the primary current willincrease.


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Effect of additional primary current

The additional current drawn by the primary windingdue to the loading is called as the load component I2

I2 is 180 degree out of phase with load current I2.The current I2 develop its own magnetic flux 2.

2 is in opposite direction to that of 2. Hence it helps the main flux 1.

1 and 2 are in the same direction Thus the reduction in the main flux due to 2 is

compensated by 2 and the core flux 1 will almost

remain constant. Therefore for any load between no load to full load,

the core flux will always remain constant.


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Phasor Diagram

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Effect of leakage reactance
http://en.wikipedia.org/wiki/File:Transformer_flux.gif


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notes

P ti l T f ( l d)


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Practical Transformer(on load)-with resistance and leakage

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Ph di f it


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Phasor diagram for unitypower factor load

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Ph di f l i


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Phasor diagram for laggingpower factor

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Ph di f l di


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Phasor diagram for leadingpower factor

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T f P t


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Transformer Parameters1. Effect OF Winding Resistances

High voltage side Low current side High resistancesideLow voltage side High current side Low resistanceside

Sejal Chopra,DBIT,Kurla 64


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Equivalent Impedance

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i l d


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Equivalent Impedance

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Rules for Transferring Parameters and Quantities



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Rules for Transferring Parameters andQuantities



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Rules for Transferring Parameters andQuantities

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Equivalent circuit of


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qTransformer

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No load equivalent circuit

Equivalent circuit of


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qTransformer

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Exact equivalent circuit


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Exact equivalent circuitreferred to primary

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R2'= R2/K2 , X2' = X2/K

2' , Z2' = Z2/K2

While E2' = E2/K' I2' = K I2Where K = N2 /N1

Exact equivalent circuit


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Exact equivalent circuitreferred to secondary

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R1' = K2 R1 , X1' = K

2 X1, Z1' = K2 Z1

E1'= K E1, Io' = I1 /K' Io' = Io /K


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Calculation of Regulation

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Efficiency of a Transformer Power output = Power input Total

losses

Power input = Power output + Totallosses

Power input = Power output + Pi + Pcu = Power output/power input

= Power output/(power output + Pi +Pcu )

Now power output = V2 I2 cos

where cos = Load power factor75


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Efficiency of a Transformer


This is full load percentage efficiency with,

I2 = Full load secondary current


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Let n =Fraction by which load is lessthan full load = Actual load/Full load

when load changes, the load current

changes by same proportion. new I2= n (I2) F.L.Similarly the output V2 I2 cos2 also

reduces by the same fraction. new Pcu = n

2 (Pcu ) F.L.

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Condition for Maximum


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Condition for MaximumEfficiency

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for maximum efficiency,

d /d I2 = 0

Now = (V2I2cos

2)/(V

2I2cos

2+ Pi + I22 R2e)

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(V2 I2 cos 2 + Pi + I22 R2e)(V2 cos 2)- (V2 I2 cos 2)(V2 cos 2 + 2I2 R2e) =0

Cancelling (V2 cos 2) from both theterms we get,

V2 I2 cos 2 + Pi +I22 R2e - V2 I2 2 - 2I22

R2e = 0... Pi - I2

2 R2e= 0

... Pi = I22 R2e = Pcu

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Load Current I2m at Maximum Efficiency


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For max, I22 R2e = Pi but I2 = I2mI2m

2 R2e = PiI2m = (Pi/ R2e)

This is the load current at max,

Let (I2)F.L. = Full load current

... I2m/(I2) F.L.= (1/(I2) F.L.)(Pi/ R2e)

... I2m/(I2) F.L.= (Pi )/({(I2) F.L.}2

R2e)

= (Pi )/((Pcu) F.L.)

... I2m = (I2 )F.L.(Pi )/((Pcu) F.L.)

\ This is the load current at max intermsof full load current.

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KVA supplied at maximum Efficiency

For constant V2 the KVA supplied isthe function of.

KVA at max = I2m V2= V2 (I2) F.L. x

(Pi)/((Pcu)F.L.)KVA at max = (KVA rating) x (Pi)

/((Pcu)F.L.)

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KVA supplied at maximum


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KVA supplied at maximumEfficiency

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Open Circuit Test (O.C. Test)

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Short Circuit Test (S.C. Test)

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Transformer (Mumbai University)