Single –phase transformer

Q.1 what is the transformer? What is the differentiate between step-Up and step-down transformer?

A transformer is an ac static device which transfers electric powerFrom one circuit to anther without changing it frequency.

When the transformer raises the voltage i.e. when the output voltageOf a transformer is higher than its input voltage, it is called the step-Up transformer.

When the transformer output voltage is lower than input voltage ,It is called the step-down transformer.

1

N2>N1

N1>N2

Step-down transformer

(Q.2) Describe the operation of a single –phase transformer, explaining Clearly the functions of the different parts. Why are the cores Laminated?

Working :-

When an alternating voltage V1 is applied to the primary, analternating flux ø is set up in the core .

This alternating flux links both the windings and induces ,e.m.f sE1 and E2 in them according to faraday’s law of electro -magnetic induction.The e.m.f E1 is termed as primary e.m.f and e.m.f E2 is termed as

Secondary e.m.f.

Clearly, E1= -N1xdø/dt

E2= -N2x dø/dt

And , E2/E1=N2/N1

Note that magnitude of E2 and E1 depend upon the numberOf turns of the secondary and primary respectively.

If N2>N1, then E2>E1 (or V2>V1) and we get a step-up transformer.On the other end if N2<N1,then E2<E1 (or V2<V1) and we get a step-down transformer.

If load is connected across the secondary winding , the secondarye.m.f E2 will cause a current I2 to flow through the load.

E.M.F EQUATION OF TRANSFORMER

(Q.3) WHAT IS AN IDEAL TRANSFORMER?WHAT ARE THE PROPERTIES OF IDEAL TRANSFORMER?

A transformer is said to be ideal if it satisfies following properties

(i) It has no losses.(ii) Its winding have zero resistance(iii)Leakage flux is zero (iv)Permeability of core is so high that negligible current is required to establish the flux in it.

(Q.4) A SINGLE –PHASE , 50HZ TRANSFORMER HAS 80 TURNS ON THE PRIMARY WINDING AND 400 TURNS ON THE SECONDARY WINDING .THE NET CROSS-SECTIONAL AREA OF THE CORE IS 200 Cm^2.

IF THE PRIMARY WINDING IS CONNECTED TO A 240 V, 50HZ SUPPLY,DETERMINE.

(i) The e.m.f induced in the secondary winding.

(ii) The maximum value of the flux density in the core.

Q.5 For s single phase transformer having primary and secondaryTurns of 440 and 880 respectively, determine the transformer KVA rating if half load secondary current is 7.5 A and maximum value ofCore flux is 2.25 mwb.

IDEAL TRANSFORMER ON NO LOAD

Phasor diagram for ideal transformer on no -load

IDEAL TRANSFORMER ON NO -LOAD

The no-load equivalent circuit for the transformer

PRACTICAL TRANSFORMER ON NO LOAD

PHASOR DIAGRAM FOR PRACTICAL TRANSFORMER ON NO LOAD

From the phasor diagram

THEORY EXPLANATION

PHASOR DIAGRAM OF TRANSFORMER AT NO LOAD

Q.4 Explain phasor of transformer on no-load?

A transformer is said to be on no-load when secondary windingIs open –circuited. The secondary current is thus zero. When analternating voltage is applied to the primary, a small current flows in the primary.The current is called the no-load current of the transformer.It is made up of two components (i) A purely reactive component called magnetizing component

0I

0I

mI

of no load current which required to produce the flux .this is Called watt less component.(ii) An active component which supplies total losses under no-load condition called power component of no load current.

wI

This is also called watt ful component or core loss component of no-Load current.

LOSSES IN TRANSFORMER

Core loss Copper loss

1.Hysteresis loss 2.eddy current loss

The flux Ø is taken as the reference phasor. since E1 and E2 are induced by the same flux , they will be in phaseWith each other.

(Q.3) A transformer takes 0.8 A when its primary is connected to 200V,50 HZ supply .The secondary is open-circuited. The power absorbed from the supply is 60 watts.

Determine the iron loss current and magnetising current.

(Q.4) A 2200/200v transformer takes no load current of 0.6 A and absorbs 400 watts . Find (i) magnetising current (iii) Iron-lossCurrent.

(Q.3) The no-load current of a transformer is 5A at 0.3 power factorlagging when supplied at 230V, 50HZ. The number of turns on the primary winding is 200.Calculate (i) the maximum value of flux in the core. (ii) the core loss (iii) the magnetising current.

Leakage flux in transformer

All the flux in transformer will not be able to link with boththe primary and secondary windings. A Small portion of flux will link either winding but not both. This portion of flux is called Leakage flux.

Both primary and secondary currents produce flux. The flux whichWhich links both the winding is the useful flux and is called mutualFlux.

(Q) WHAT ARE THE EFFECT OF LEAKAGE FLUX IN TRANSFORMER

Ans:- Due to this leakage flux in transformer there will be a Self –reactance in the concern winding.

This self-reactance of transformer is alternatively known as leakage reactance of transformer.

Resistance of transformer

Generally both primary and secondary windings of transformer are made of copper. These copper windings have own resistance.

IMPEDANCE OF TRANSFORMER

As we said ,both primary and secondary windings will have resistance and leakage reactance. These resistance and reactance will be in combination is nothing but impedanceOf transformer.

If R1 and R2 and X1 and X2 are primary & Secondary resistance and reactance of transformer respectively.So Z1 and Z2 impedance of primary and secondaryWinding are respectively.

Z1=(R+JX1) Z2=(R+JX2)

(Q.4)

Find the active and reactive components of no-load current, and the no-load current of a 440/220 v single-phase transformer if the Power input to the hv winding is 80 w. The low-voltage is kept open.The power factor of the no-load current is 0.3 lagging.Input power = iron loss of the transformer

0010 COSIVP

Practical transformer on load

Assume that having no winding resistance and leakage reactance

Theory of transformer on load, with resistance as well as leakage reactance in transformer.

For inductive load

Equivalent circuit of transformer

TRANSFORMER WITH RESISTANCE AND LEAKAGE REACTANCE

EQUIVALENT RESISTANCE

When secondary resistance is transferred

to the primary, it is divided by

2k

Referred to primary

Equivalent resistance of transformer referred to primary is

)/( 221

/2101 KRRRRR

Referred to secondary

When primary resistance is transferred to the secondary, it

Is multiplied by 2k

So equivalent resistance of transformer referred to the secondary is

)()( 212

/1202 KRRRRR

Similarly ,when primary reactance is transferred to the secondary,It is multiplied by .2KIt is then called equivalent primary reactance

and is denoted by /1X

SO EQUIVALENT REACTANCE OF TRANSFORMNER REFERRED TO SECONADRY IS

12

2/1202 XKXXXX

EQUIVALENT IMPEDANCE OF TRANSFORMER REFFERED TO SECONDARYIS

202

20202 XRZ

REFERRED TO PRIMARY

When secondary resistance or reactance is transferred to the primary,It is divided by

2K

Equivalent resistance of transformer referred to primary is

)/( 221

/2101 KRRRRR

EQUIVALENT RESISTANCE OF TRANSFORMER REFERRED TO PRIMATY IS

221

/2101 /KXXXXX

Equivalent impedance of transformer referred to primary is

201

20101 XRZ

1. A 6600/400 V, single phase transformer has primary resistance of 2.5Ω and secondary resistance of 0.01Ω.

Calculate total equivalent resistance referred to primary and Secondary.

K=voltage transformation ratio.K=N2/N1=V2/V1=E2/E1

GIVEN Data are R1=2.5Ω, R2=0.01Ω

2. A 100 KVA transformer has 400 turns on the primary and and 80 turns on the secondary.

The primary and secondary resistance are 0.3Ω and 0.01Ω respectivelyand the corresponding leakage reactance are 1.1Ω and 0.035Ω respectively.

Calculate the equivalent impedance referred to secondary circuit.

Sol:-

ASSIGNMENT

1. A 10 KVA, 2000/400 V single-phase transformer has the followingResistance and leakage reactances.

Primary winding : resistance 6Ω : leakage reactance 10 ΩSecondary winding: resistance 0. 8Ω : leakage reactance 0.4Ω

Calculate (i) equivalent resistance referred to secondary. (ii) equivalent leakage reactance referred to secondary.

2. A 15 kVA, 2200/110V transformer has R1= 1.75Ω, R2=0.0045Ω,theLeakage reactances are X1=2.6Ω and X2=0.0075Ω.calculate (a)Equivalent resistance referred to primary.(b) Equivalent resistance referred to secondary.(c) Equivalent reactance referred to primary.(d)Equivalent reactance referred to secondary.(e)Equivalent impedance referred to primary.(f) Equivalent impedance referred to secondary.(g)total copper loss

Open circuit test of transformer

A voltmeter , wattmeter and anammeter are connected in LV sideOf the transformer as shown

The HV side of the transformer is keptOpen. Now with the help of varic appliedIs slowly increase until the voltmeter givesreading equal to the rated voltage of theLV side .After reaching at rated LV side Voltage , all three instruments reading are recorded. The applied primary voltageV1 is measured by the voltmeter, the No-load current by ammeter and no-loadInput power by wattmeter as shown in fig.

Hence watt-meter will record the iron losses and small copper losses In the primary.

Since no-load current is vary small (usually 2-10 % of rated current)

Copper losses in the primary under no-load condition are negligibleas compared with iron losses.

Hence , wattmeter reading practically gives the core losses in the Transformer.

Short –circuit test of transformer

The connection diagram for short –circuit test on transformer is shown in figure.

In this test ,the secondary is short circuitedby a thick conductor and variable lowVoltage applied to the primary.

There is no output from the transformer under short –circuit Conditions. therefore, input power is all losses and this loss is Almost entirely copper loss.it is because iron loss in the coreIs negligibly small since the applied voltage is small.

Q-12 ADVANTAGESE OF TRANSFORMER TEST.

Q.13 EXPLAIN AUTO TRANSFORMER

Auto transformer is kind of electrical transformer where primary and secondary shares same common single winding.