Transformer1. What is transformer ? explain working principle of transformer.

Electrical power transformer is a static device which transforms electrical energy from one circuit to another without any direct electrical connection and with the help of mutual induction between two windings. It transforms power from one circuit to another without changing its frequency but may be in different voltage level.

Working Principle of Transformer: The working principle of transformer is very simple. It depends upon Faraday's law of electromagnetic induction. Actually, mutual induction between two or more winding is responsible for transformation action in an electrical transformer. According to these Faraday's laws,"Rate of change of flux linkage with respect to time is directly proportional to the induced EMF in a conductor or coil". The alternating current through the winding produces a continually changing flux or alternating flux that surrounds the winding. If any other winding is brought nearer to the previous one, obviously some portion of this flux will link with the second. As this flux is continually changing in its amplitude and direction, there must be a change in flux linkage in the second winding or coil.

According to Faraday's law of electromagnetic induction, there must be an EMF induced in the second. If the circuit of the later winding is closed, there must be an current flowing through it. This is the simplest form of electrical power transformer and this is the most basic of working principle of transformer.

2.Derive the e. m. f. equation for 1 ph transformer.

Let's say, T is number of turns in a winding,

Φm is the maximum flux in the core in Wb.

As per Faraday's law of electromagnetic induction,

Where φ is the instantaneous alternating flux and represented as,1 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

As the maximum value of cos2πft is 1, the maximum value of induced emf e is,

To obtain the rms value of induced counter emf, divide this maximum value of e by √2.

This is EMF equation of transformer.

If E1 & E2 are primary and secondary emfs and T1 & T2 are primary and secondary turns then, voltage ratio or turns ratio of transformer is,

Transformation Ratio of TransformerThis constant is called transformation ratio of transformer , if T2>T1, K > 1, then the transformer is step up transformer. If T2 < T1, K < 1, then the transformer is step down transformer.

Voltage Ratio of TransformerThis above stated ratio is also known as voltage ratio of transformer if it is expressed as ratio of the primary and secondary voltages of transformer.

Turns Ratio of TransformerAs the voltage in primary and secondary of transformer is directly proportional to the number of turns in the respective winding, the transformation ratio of transformer is sometime expressed in ratio of turns and referred as turns ratio of transformer .2 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

Objective Questions on Transformer (MCQs)

3.Explain Transformer Construction in detail.

The construction of a simple two-winding transformer consists of each winding being wound on a separate limb or core of the soft iron form which provides the necessary magnetic circuit. This magnetic circuit, know more commonly as the “transformer core” is designed to provide a path for the magnetic field to flow around, which is necessary for induction of the voltage between the two windings.

To provide a low reluctance path for the magnetic field, the core is designed to prevent circulating electric currents within the iron core itself. Circulating currents, called “eddy currents”, cause heating and energy losses within the core decreasing the transformers efficiency.

These losses are due mainly to voltages induced in the iron circuit, which is constantly being subjected to the alternating magnetic fields setup by the external sinusoidal supply voltage. One way to reduce these unwanted power losses is to construct the transformer core from thin steel laminations. In the “closed-core” type (core Type) transformer, the primary and secondary windings are wound outside and surround the core ring. In the “shell type” transformer, the primary and secondary windings pass inside the steel magnetic circuit (core) which forms a shell around the windings as shown below.

Transformer Core Construction

The coils are not arranged with the primary winding on one leg and the secondary on the other but instead half of the primary winding and half of the secondary winding are placed one over the other concentrically on each leg in order to increase magnetic coupling.

3 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

Shell type transformer cores overcome this leakage flux as both the primary and secondary windings are wound on the same centre leg or limb which has twice the cross-sectional area of the two outer limbs.

Transformer Laminations : In both the shell and core type transformer constructions, in order to mount the coil windings, the individual laminations are stamped or punched out from larger steel sheets and formed into strips of thin steel resembling the letters “E’s”, “L’s”, “U’s” and “I’s” as shown below.

Transformer Core Types

These lamination stampings when connected together form the required core shape. These individual laminations are tightly butted together during the transformers construction to reduce the reluctance of the air gap at the joints producing a highly saturated magnetic flux density.

Transformer core laminations are usually stacked alternately to each other to produce an overlapping joint with more lamination pairs being added to make up the correct core thickness. This alternate stacking of the laminations also gives the transformer the advantage of reduced flux leakage and iron losses.

The insulation used to prevent the conductors shorting together in a transformer is usually a thin layer of varnish or enamel in air cooled transformers. This thin varnish or enamel paint is painted onto the wire before it is wound around the core.

4.Explain Transformer Losses.The ability of iron or steel to carry magnetic flux is much greater than it is in air, and this ability to allow magnetic flux to flow is called permeability. Most transformer cores are constructed from low carbon steels which can have permeabilities in the order of 1500 compared with just 1.0 for air.

There are two types of losses 1) core losses ( constant losses) a) hysteresis b) eddy current losses

2) copper losses( variable) a) primary b) secondary

Core losses : a) Hysteresis Losses

4 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

Transformer Hysteresis Losses are caused because of the friction of the molecules against the flow of the magnetic lines of force required to magnetize the core, which are constantly changing in value and direction first in one direction and then the other due to the influence of the sinusoidal supply voltage. ( Magnetic reversals)

Wh = Kn Bm1.6 f v Watts

Where f-frequency v- volume of magnetic material Bm – maximum flux density

Kh – hysteresis coefficient

b) Eddy Current Losses

Transformer Eddy Current Losses on the other hand are caused by the flow of circulating currents induced into the steel caused by the flow of the magnetic flux around the core. These circulating currents are generated because to the magnetic flux the core is acting like a single loop of wire. Since the iron core is a good conductor, the eddy currents induced by a solid iron core will be large.

We = KeBm2 f2 t2 V Watts Where ke – const t- thickness of laminations

Laminating the Iron Core

Eddy current losses within a transformer core can not be eliminated completely, but they can be greatly reduced and controlled by reducing the thickness of the steel core. Instead of having one big solid iron core as the magnetic core material of the transformer or coil, the magnetic path is split up into many thin pressed steel shapes called “laminations”.

The losses of energy, which appears as heat due both to hysteresis and to eddy currents in the magnetic path, is known commonly as “transformer core losses”. Since these losses occur in all magnetic materials as a result of alternating magnetic fields. Transformer core losses are always present in a transformer whenever the primary is energized, even if no load is connected to the secondary winding. Also these hysteresis and the eddy current losses are sometimes referred to as “transformer iron losses”, as the magnetic flux causing these losses is constant at all loads.

2) Copper Losses

But there is also another type of energy loss associated with transformers called “copper losses”. Transformer Copper Losses are mainly due to the electrical resistance of the primary and

5 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

secondary windings. Most transformer coils are made from copper wire which has resistance in Ohms, ( Ω ). This resistance opposes the magnetizing currents flowing through them.

When a load is connected to the transformers secondary winding, large electrical currents flow in both the primary and the secondary windings, electrical energy and power ( or the I2 R ) losses occur as heat. Generally copper losses vary with the load current, being almost zero at no-load, and at a maximum at full-load when current flow is at maximum.

Total cu losses = I12R1 + I2

2R2 ( Watts)

Increasing the rate of heat dissipation (better cooling) by forced air or oil, or by improving the transformers insulation so that it will withstand higher temperatures can also increase a transformers VA rating.

Then we can define an ideal transformer as having:

No Hysteresis loops or Hysteresis losses → 0 Infinite Resistivity of core material giving zero Eddy current losses → 0 Zero winding resistance giving zero I2R copper losses → 0

5. Explain efficiency of transformer.

It is defined as ratio of input power to input power both in watt. Normally it is expressed as percentage.Efficiency ( η ) = output powerinput power

Efficiency ( η ) = output poweroutput power+Losses

Efficiency ( η ) = output poweroutput power+Wi+Wcu

Efficiency ( η ) = rating∈VA∗p . f .rating∈VA∗p . f .+Wi+Wcu

Efficiency changes with load

Condition for maximum efficiency

6 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

Load current

η

Load current

Capacitive load

Resistive load

Inductive load

Efficiency depends on load condition on no load efficiency is zero. At light load core ( constant ) losses being large as compared to input. So efficiency is low. As load is increased efficiency rises until it reaches max value. It then fall as load increases.Efficiency is max for that load which makes variable cu losses equal to constant losses.

Condition for max efficiency---- variable losses = constant losses

Load at max efficiency = full load KVA x √ iron lossfull load cu losses

6.Explain Regulation of transformer. The voltage regulation is the percentage of voltage difference between no load and full load voltages of a transformer with respect to its full load voltage. Change in secondary voltage from no load to full load is called regulation.

As secondary voltage depends on magnitude of load current but also depends on p.f. of load i. e. load may be resistive, inductive, capacitive.

As shown in fig. for inductive load (motors) and resistive load sec. terminal voltage decreases with increase in load on transformer.

7 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

Regulation always positive for inductive load with lagging p.f. and resitive load unity p.f. and may be negative for capacitive load with leading p.f.

For capacitive load it may increases. Transformer having less regulation is always considered good transformer.

7.Explain Transformer on NO Load.

There will always be losses associated with the transformers loading as the transformer is put “on-load”. But what do we mean by: Transformer Loading.

when it is in “no-load” condition, that is with no electrical load connected to its secondary winding and therefore no secondary current flowing.

A transformer is said to be on “no-load” when its secondary side winding is open circuited, in other words, nothing is attached and the transformer loading is zero. When an AC Sinusoidal Supply is connected to the primary winding of a transformer, a small current, IO will flow through the primary coil winding due to the presence of the primary supply voltage.

With the secondary circuit open, nothing connected, a back EMF along with the primary winding resistance acts to limit the flow of this primary current. Obviously, this no-load primary current ( Io ) must be sufficient to maintain enough magnetic field to produce the required back emf. Consider the circuit below.

Transformer “No-load” Condition

The ammeter above will indicate a small current flowing through the primary winding even though the secondary circuit is open circuited. This no-load primary current is made up of the following two components:

1). An in-phase current, IE = Iw which supplies the core losses (eddy current and hysteresis).

2). A current, IM at 90o to the voltage which sets up the magnetic flux.

8 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

Note that this no-load primary current, Io is very small compared to the transformers normal full-load current. Also due to the iron losses present in the core as well as a small amount of copper losses in the primary winding,

8.Explain open circuit test and short circuit test . . 1. O.C. and S.C. Tests on Single Phase Transformer The efficiency and regulation of a transformer on any load condition and at any power factor condition can be predetermined by indirect loading method. In this method, the actual load is not used on transformer. But the equivalent circuit parameters of a transformer are determined by conducting two tests on a transformer which are,1. Open circuit test (O.C Test)2. Short circuit test (S.C.Test) The parameters calculated from these test results are effective in determining the regulation and efficiency of a transformer at any load and power factor condition, without actually loading the transformer. The advantage of this method is that without much power loss the tests can be performed and results can be obtained. Let us discuss in detail how to perform these tests and how to use the results to calculate equivalent circuit parameters.9.Explain Open Circuit Test (O.C. Test) .

this test gives core losses (constant losses) The experimental circuit to conduct O.C test is shown in the Fig. 1.

The transformer primary is connected to a.c. supply through ammeter, wattmeter and variac. The secondary of transformer is kept open. Usually low voltage side is used as primary and high voltage side as secondary to conduct O.C test.

9 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

. The wattmeter measures input power. The ammeter measures input current. The voltmeter gives the value of rated primary voltage applied at rated frequency. Sometimes a voltmeter may be connected across secondary to measure secondary voltage which is V2 = E2 .

The observation table is as follows

Vo = Rated voltage Wo = Input power Io = Input current = no load current As transformer secondary is open, it is on no load. So current drawn by the primary is no load current Io. where cos Φo = No load power factorAnd hence power input can be written as, Wo = Vo Io cos Φo

The phasor diagram is shown in the Fig. 2.

Fig. 2

As output power is zero and copper losses are very low, the total input power is used to supply iron losses. This power is measured by the wattmeter i.e. Wo. Hence the wattmeter in O.C. test gives iron losses which remain constant for all the loads.... Wo = Pi = Iron losses Calculations : We know that, Wo = Vo Io cos Φ cos Φo = Wo /(Vo Io ) = no load power factorOnce cos Φo is known we can obtain, Iw = Io cos Φo and Im = Io sin Φo

Once Iw and Im are known we can determine exciting circuit parameters as, Ro = Vo /Ic Ω and Xo = Vo /Im Ω

Key Point : The no load power factor cos Φo is very low hence wattmeter used must be low power factor type otherwise there might be error in the results.

7.Explain Short Circuit Test (S.C. Test).

10 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

This test gives copper losses (variable losses) In this test, primary is connected to a.c. supply through variac, ammeter and voltmeter as shown in the Fig. 3.

The secondary is short circuited with the help of thick copper wire or solid link. As high voltage side is always low current side, it is convenient to connect high voltage side to supply and shorting the low voltage side. As secondary is shorted, its resistance is very very small and on rated voltage it may draw very large current. Such large current can cause overheating and burning of the transformer. To limit this short circuit current, primary is supplied with low voltage which is just enough to cause rated current to flow through primary which can be observed on an ammeter. The low voltage can be adjusted with the help of variac. Hence this test is also called low voltage test or reduced voltage test. The wattmeter reading as well as voltmeter, ammeter readings are recorded. The observation table is as follows,

Hence the wattmeter reading is the power loss which is equal to full load copper losses as iron losses are very low.... Wsc = (Pcu) F.L. = Full load copper loss Calculations : From S.C. test readings we can write, Wsc = Vsc Isc cos Φsc

... cos Φsc = Vsc Isc /Wsc = short circuit power factor Wsc = Isc

2 R1e = copper loss... R1e =Wsc /Isc

2 while Z1e =Vsc /Isc = √(R1e

2 + X1e

2)... X1e

= √(Z1e2 - R1e

2) Calculation of Efficiency from O.C. and S.C. Tests We know that, From O.C. test, Wo = Pi

From S.C. test, Wsc = (Pcu) F.L.

11 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

Thus for any p.f. cos Φ2 the efficiency can be predetermined. Similarly at any load which is fraction of full load then also efficiency can be predetermined as,

where n = fraction of full load

where I2= n (I2) F.L.

Calculation of Regulation From S.C. test we VFL and from open circuit test we get VNL.

Voltage regulation(%) = V NL−V FVF L

∗100

10.Explain direct loading test . Being a static device it has a very high efficiency as compared to rotating machine of same rating as the losses are less. Make experimental arrangement as shown in fig. we can calculate efficiency and regulation by using this method.

Wattmeter’s are connected in primary giver Input power = W1 We can also connect wattmeter for secondary which will give secondary power instead that we can also calculate secondary power by measuring secondary voltage and current. poweroutput ¿ the transformer=W 2=V 2∗I 2 ( cosØ being unity for lamp bank load)

12 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur

Now percentage efficiency:

s.no. V1 I1 W1 V2 I2

Percentageefficiency=W 2W 1

∗100

FOR regulation measure secondary voltage at no load Vnl and at full load Vfl

Thus, percentage voltage regulation=Vnl−VflVfl

∗100

The voltage regulation should be as small as possible. This method is not accurate for measuring efficiency as compare to oc and sc test.

13 | Mrs. V.S.Patki(M.E.Electrical)Electronics Department, WIT, Solapur