ashishsoniwifistudy

UPPSC -Assistant Engineer

Transformers• A transformer is a static device that transforms electric energy from one

ac voltage level to another. It is this device that has made the electric system almost universally AC. The electric power is generated at relatively low voltages (up to a maximum of 33 kV) which then is raised to very high voltages (e.g. 756 kV) by means of a transformer and then transmitted.

• High voltages are associated with low currents and reduced transmission losses.

• The core which supports the transformers mechanically and conducts their mutual flux, is normally made of highly permeable iron or steel alloy (cold-rolled, grain oriented sheet steel). Such a transformer is generally called an iron-core transformer.

• However in special cases, the magnetic circuit linking the windings may be made of non-magnetic material, in which case transformer is referred to as an air-core transformer.

• Air core transformer used in radio devices and in certain types of measuring and testing instruments.

• The magnetic core of the transformer is made up of stacks of thin lamination of 0.35 mm thickness of CRGO lightly insulated with varnish, this material allow the use of high flux density [ 1-1.5 T] and its low loss property together with laminated core reduces the core loss to fairly low values.

Core Type transformer: -

• In core type transformer windings are around two legs or three legs (depend on their phases) of the rectangular magnetic core.

• Rectangular magnetic core is made by using the E and L shape of the sheets.

• Though most of the is confined to a high permeability core, some flux always leaks through the core and lies in air called leakage flux.

• Leakage is reduced by bringing the two coils closer. In core type we achieve this by using L.V. and H.V. on each limb of the core.

Shell type transformers: -

• In this type of the transformer windings are wound on the central leg of a three legged core.

• Leakage in shell type transformer is reduced by sub-dividing each winding into sub-section and interleaved in L.V. and H.V winding.

Definition of Core type Transformer: -

• The magnetic core of the transformer is made up of laminations to form a rectangular frame. The laminations are cut in the form of L-shape strips shown in the figure below. For avoiding the high reluctance at the joints where laminations are butted against each other, the alternate layer is stacked differently to eliminate continues joints.

• The primary and secondary windings are interleaved to reduce the leakage flux. Half of each winding is placed side by side or concentrically on the leg of the core as shown in the figure below. For simplicity, the primary and secondary winding is located on the separate limbs of the core.

• The insulation layer is provided between the core and lower winding and between the primary and the secondary winding. For reducing the insulation, the low winding is always placed near to the core. The winding is cylindrical, and the lamination is inserted later on it.

Definition of Shell Type Transformer: -

• The laminations are cut in the form of a long strip of E’s, and I’s as shown in the figure below. To reduce the high reluctance at the joints where the lamination are butted against each other, the alternate layers are stacked differently to eliminate continuous joint.

The shell type transformer has three limbs or legs. The central limb carries the whole of the flux, and the side limb carries the half of the flux. Hence the width of the central limb is about to double to that of the outer limbs.

• The primary and secondary both the windings are placed on the central limbs. The low voltage winding is placed near the core, and the high voltage winding is placed outside the low voltage winding to reducing the cost of insulation placed between the core and the low voltage winding. The windings are cylindrical, and the core laminations are inserted on it.

Key Differences Between Core Type and Shell Type Transformer: -

• In core type transformer the core surrounds the windings whereas in shell type transformer the winding surrounds the core of the transformer.

• In core type transformer the lamination is cut in the form of L-shape whereas, in shell type transformer, the laminations are cut in the E and L shapes.

• The cross-section area of the core type transformer is rectangular, whereas the cross-section area of the shell type transformer is square, cruciform two slipped, or three stepped in shapes.

• The core type transformer requires more copper conductor as compared to shell type transformer because in core type transformer the winding is placed on the separate limbs or legs.

• The core type transformer is also called cylindrical or core winding transformer because their windings are arranged as the concentric coil. In shell type transformer, the low voltage winding and the high voltage winding are put in the form of the sandwich, and hence it is called the sandwich or disc winding transformer.

• The core type transformer has two limbs, whereas the shell type transformer has three limbs.

• The mechanical strength of the core type transformer is low as compared to shell type transformer because the shell type transformer has bracings.

• The core type transformer required less insulation as compared to shell type transformer because shell type transformer has three limbs.

• In core type transformer the flux is equally distributed to the side limb of the transformer whereas, in shell type transformer, the central limb carries the whole of the flux and the side limbs carry the half of the flux.

• In core type transformer both the primary and the secondary windings are placed on the side limbs whereas, in shell type transformer, the windings are placed on the central limbs of the transformer.

• The core type transformer has two magnetic circuits whereas the shell type transformer has one magnetic circuit.

• The losses in a core type transformer are more as compared to shell type transformer because the core type transformer consists two magnetic circuits.

• In core type transformer few windings are removed for maintenance. In shell type transformer numbers of the winding are required to remove for the maintenance.

• The output of the core type transformer is less because it has more losses as compared to the shell-type transformer.

• The winding of the shell type transformer is distributed type, and hence heat is dissipated naturally, whereas, in core type transformer, the natural cooling is not possible.

The laminations of both the transformer are made up of high-grade silicon steel. The lamination reduces the eddy current loss. and silicon steel reduces the hysteresis loss. The laminations are insulated from each other by using the enamel insulation coating.

Construction: -

Conservator: -

Power transformer is provided with a conservative through which transformer breaths into air. It is a small size tank placed on the top of main tank. It prevents fast oxidization and consequent deterioration of insulating properties of oil.

Explosion vent tube: -

Purpose of this is to prevent damage of transformer tank be releasing any excessive pressure generated inside the transformer.

Radiator: -

• When an electrical transformer is loaded, the current starts flowing through its windings. Due to this flowing of electric current, heat is produced in the windings, this heat ultimately rises the temperature of transformer oil. Hence, if the temperature rise of the transformer insulating oil is controlled, the capacity or rating of transformer can be extended up to significant range. The radiator of transformer accelerates the cooling rate of transformer. Thus, it plays a vital role in increasing loading capacity of an electrical transformer. This is basic function of radiator of an electrical power transformer.

• The working principle of radiator is very simple. It just increases the surface area for dissipating heat of the oil. It is used in large size transformers.

Breather: -

• When the temperature change occurs in transformer oil, the oil expands or contracts. There is an exchange of air also occur when transformer is fully loaded.

• When the transformer gets cooled then oil level gets down and when it goes down it absorbs air this process is called breathing.

• Silica gel breather controls the level of moisture. Silica gel is used to absorb moisture content from air.

• When silica gel absorbs moisture it becomes pink. Generally its color is blue.

Transformer core: -

The magnetic core of transformer is made up of stacks of thin laminations [0.35 mm] thickness of CRGO lightly insulated with varnish, this material allow the use of high flux density [1-1.5 tesla] and its low loss property, together with lamination reduce the core loss to fairly low values.

Important points about transformers: -

• Changing voltage and current levels in electric power systems.

• Matching source and load impedances for maximum power transfer.

• Provides electrical isolation [ isolating one circuit from another or isolating dc while maintaining ac continuity between two circuits].

• Transformer transfer power and transforms both voltage and current.

• A transformer in its simplest form consists essentially of two insulating winding interlinked by a common or mutual magnetic field established in a core of magnetic material.

• One of the winding termed as primary is connected to an alternating voltage source, an alternating flux sets-up in the core with an amplitude depending on the primary voltage, frequency and number of turns.

• This mutual flux links with the other winding called secondary. A voltage is induced in the secondary of the same frequency as the primary voltage but its magnitude depends on the number of secondary turns.

• When the no. of primary and secondary turns are in proper proportion for almost any desired voltage ratio then ratio of transformation can be achieved.

• The transformer being an electromagnetic device, its analysis greatly aids in understanding the operation of electromechanical energy conversion devices which also use magnetic field but the interchange of energy is between electrical and mechanical parts.

• If the secondary voltage is greater than primary value the transformer is called step up transformer, if it is less it is known as step down transformer.

• If primary and secondary voltage are equal the transformer is said to have a one to one ratio (1:1).

• One to one transformer are used to electrically isolate two parts of a circuit. Booster transformer used in electric traction have one to one ratio.

• The core which supports them mechanically and conducts their mutual flux is normally made of high permeable iron or steel alloy called CRGO [ cold rolled grain oriented sheet steel].

• If the core is made up of non magnetic material then the core is referred as air core transformer. It is mainly used in radio devices (high frequency device).

Transformer at no-load (ideal transformer): -

• The primary winding of this transformer draws a small amount of current of instantaneous value Io, called the exciting current from the voltage source with positive direction as indicated.

• When the transformer is operating at no load, the secondary winding is open circuited, which means there is no load on the secondary side of the transformer and, therefore, current in the secondary will be zero, while primary winding carries a small current I0 called no load current which is 2 to 10% of the rated current. This current is responsible for supplying the iron losses (hysteresis and eddy current losses) in the core and a very small amount of copper losses in the primary winding. The angle of lag depends upon the losses in the transformer. The power factor is very low and varies from 0.1 to 0.15.

The no load current consists of two components: -

• Reactive or magnetizing component Im

(It is in quadrature with the applied voltage V1. It produces flux in the core and does not consume any power)

• Active or power component Iw , also know as working component

(It is in phase with the applied voltage V1. It supplies the iron losses and a small amount of primary copper loss)

The following steps are given below to draw the phasor diagram: -

• The function of the magnetizing component is to produce the magnetizing flux, and thus, it will be in phase with the flux.

• Induced emf in the primary and the secondary winding lags the flux ϕ by 90 degrees.

• The primary copper loss is neglected, and secondary current losses are zero as I2 = 0. Therefore, the current I0 lags behind the voltage vector V1 by an angle ϕ0 called no-load power factor angle shown in the phasor diagram above.

• The applied voltage V1 is drawn equal and opposite to the induced emf E1 because the difference between the two, at no load, is negligible.

• Active component Iw is drawn in phase with the applied voltage V1.

• The phasor sum of magnetizing current Im and the working current Iw gives the no load current I0.

E.M.F. equation: -

When a sinusoidal voltage is applied to the primary winding of a transformer, alternating flux ϕm sets up in the iron core of the transformer. This sinusoidal flux links with both primary and secondary winding. The function of flux is a sine function. The rate of change of flux with respect to time is derived mathematically.

• The derivation of EMF Equation of the transformer is shown below. Let ϕm

be the maximum value of flux in Weber. f be the supply frequency in Hz. N1is the number of turns in the primary winding. N2 is the number of turns in the secondary winding.

• As shown in the above figure that the flux changes from + ϕm to – ϕm in half a cycle of 1/2f seconds.

By Faraday’s Law,

Let E1 is the e.m.f induced in the primary winding

E1 = −𝒅𝝀

𝒅𝒕

Where 𝛌= N1ɸ

Therefore, E1 = − N1𝐝ɸ

𝒅𝒕

Since ϕ is due to AC supply ϕ = ϕm Sinωt

E1 = − N1𝒅

𝒅𝒕(ϕm Sinωt)

E1 = − N1 ϕm ω Cosωt

E1 = N1 ϕm ω Sin (ωt− 90°)

So the induced e.m.f lags flux by 90 degrees.

Maximum valve of e.m.f,

E1 = N1 ϕm ω

But ω = 2πf

(E1)max = 2πf N1 ϕm

Root mean square RMS value is

E1= (𝐄𝟏)𝐦𝐚𝐱

√𝟐…………….(1)

Putting the value of (E1)max in equation (1) we get

E1 = 4.44 f N1ϕm……………..(2)

Similarly,

E2 = 4.44 f N2ϕm…………….(3)

Now, equating the equation (2) and (3) we get

• The above equation is called the turn ratio where K is known as transformation ratio.

• Turns ratio is defined as the ratio of high voltage turns to the low voltage turns.

Transformer at full load on lagging power factor: -

φo

φ2

φ1

V1= E1

IW

Im

I2’=I2/a

I1I1

Io

Hysteretic angle

φ= constant

V2= E2

• Figure shows a practical transformer on load. Both primary and secondary have finite resistance R1 and R2 gives rise to associated copper loss.

• Leakage flux φL1 cause by primary mmf (N1I1) links with primary winding itself.

• Leakage flux φL2 cause by N2I2 links the secondary winding thereby causing self linkages of two windings.

Equivalent circuit of transformer: -

• The equivalent circuit diagram of any device can be quite helpful in predetermination of the behavior of the device under the various condition of operation. It is simply the circuit representation of the equation describing the performance of the device.

• The simplified equivalent circuit of a transformer is drawn by representing all the parameters of the transformer either on the secondary side or on the primary side. The equivalent circuit diagram of the transformer is shown below: -

• Let the equivalent circuit of a transformer having the transformation ratio K = E2/E1

• The induced e.m.f E1 is equal to the primary applied voltage V1 less primary voltage drop. This voltage causes current I0 no load current in the primary winding of the transformer. The value of no-load current is very small, and thus, it is neglected. Hence, I1 = I1’. The no load current is further divided into two components called magnetizing current (Im) and working current (Iw).

• The secondary current I2 is: -

• The terminal voltage V2 across the load is equal to the induced e.m.f E2 in the secondary winding less voltage drop in the secondary winding.

Equivalent circuit when all the quantities are referred to primary side: -

• In this case to draw the equivalent circuit of the transformer all the quantities are to be referred to the primary as shown in the figure below

The following are the values of resistance and reactance given below: -

• Secondary resistance referred to primary side is given as: -

The equivalent resistance referred to primary side is given as

Req = R1 + R2’

Secondary reactance referred to primary side is given as: -

The equivalent reactance referred to primary side is given as

Xeq = X1 + X2’

E2= V2+ I2R2+ jI2X2

Multiply both side by a,

aE2 = a ( V2+ I2R2+ jI2X2)

𝐍𝟏

𝐍𝟐=𝐄𝟏𝐄𝟐

=𝑽𝟏𝑽𝟐

=𝐈𝟐𝐈𝟏

= a

aE2 = aV2+ aI2R2+ jaI2X2

E1= E2’= V2

’+ 𝐈𝟐𝐚× 𝒂𝟐𝑹𝟐+ ja2X2

𝐈𝟐

𝐚

E1= E2’ = V2

’ + I2‘ R2’ + j I2

‘X2’

Ro representing core loss component of resistance.

Xo representing the core loss component of reactance.

• Any leakage flux has no role in power transformers its simply creates the voltage drops on the systems.

• I1= I2’+ Io

• Further simplification of the equivalent circuit of the transformer can be done by neglecting the parallel branch consisting R0 and X0.Because the no load current is 2% to 5% of full load current. The simplified circuit diagram of the transformer is shown below: -

O.C. test and S.C. test: -• The aim of carrying out O.C. test and S.C. test on a transformer is to predict its

performance without actually loading it.O.C. Test: -• O.C. test is carried out at rated frequency and rated voltage to determine the

core loss. The iron loss is thus is treated as constant, in spite of minor voltage variation in voltage and frequency during actual operation.

• This test is carried out with the instruments placed on low voltage side while the high voltage side is left open circuited.

• This is done because it is easier to manage rated voltage supply at low voltage level rather than at high voltage level. Also the instruments used are economic in cost and it is easier to work on low voltage side.

• Therefore no load current is limited to 5% of full load current, a primary winding copper loss is ignored, also the primary impedance drop at such low current is neglected.

• Because no load power factor is very low, it is recommended that a low power factor wattmeter to be used.

Then the iron loss of the transformer Pi = W0 and

The no-load power factor is

S.C. Test: -

• S. C. test is carried out at rated current to determine the full load copper loss.

• This test is carried out with the instrument placed on high voltage side while the low voltage side is short circuited by a thin a conductor (so that wire will not burn). This is because a rated current is lower on low voltage side as compared to high voltage side.

• Consequently the instruments are economic in cost since the voltage required to circulate full load current at circuited would be about 10% of rated voltage.

• The core loss under this low voltage condition is ignored. Also the exciting current at such low value of voltage will be completely neglected.

• Short circuit test need not to be carried out strictly at rated frequency because the copper loss that depends upon the winding resistance is independent of the frequency of the supply as the skin effect in transformer at power frequency is negligible.

• Wattmeter used in this test is of high power factor.

Voltage regulation: -

Voltage regulation is defined as the change in magnitude of the secondary(terminal) voltage, when full load at a specified power factor supplied at rated voltage is thrown off, i.e. reduced to no-load with primary voltage (and frequency) held constant, as percentage of the rated load terminal voltage.

% voltage regulation = 𝐕𝐍𝐋−𝐕𝐅𝐋

𝐕𝐅𝐋× 100

VNL = No load voltage

VFL = Full load voltage

• Regulation is only defined at full load: -

Zeq

Approximate voltage regulation = Zp.u. Cos (θeq - ɸ)

= Zp.u. [Cos θeq Cos ɸ + Sin θeq Sin ɸ]

= Zp.u. Cos θeq Cos ɸ + Zp.u. Sin θeq Sin

Vreg = Rp.u. Cos ɸ ± Xp.u. Sin ɸ

Condition for maximum voltage regulation: -

θeq - ɸ = 0

θeq = ɸ

Vreg = Zp.u.

Power factor for maximum voltage regulation: -

Cos θeq = Cos ɸ = Req/Zeq lagging

Condition for Zero voltage regulation: -

θeq - ɸ = 90°

Power factor for zero voltage regulation:

Sin θeq = Xeq/Zeq Leading

Condition for minimum voltage regulation: -

Minimum voltage regulation can be achieved at ɸ = 90° leading

Corresponding regulation = - Xeq at zero p.f. leading

Important points regarding voltage regulation: -

Regulation is a figure of demerit of a transformer and its low value is desirable.

Voltage regulation can be reduce by reducing resistance R or reactance X or both.

• R is already kept at optimum value due to efficiency consideration. This means a leakage reactance is to be reduced for reducing voltage regulation.

• Leakage reactance can be reduced by reducing leakage flux (leakage flux can be reduced by keeping the low voltage winding and high voltage winding physically as close as possible (air gap should be minimum).

• This physical proximity is obtained in core type transformer by using concentric cylindrical coil.

• In shell type transformer this physical proximity is obtained by using sandwich winding, also pan cake winding or interleave winding.

• In fact it is possible to grade leakage reactance in a shell type transformer by varying thickness of H.V. and L.V. windings.

• In a core type transformer leakage reactance can also be reduced by increasing the core height to width ratio.

• A power transformer operates on full load or is switch off therefore voltage regulation as a performance index has no significance of power transformer.

• However the load on distribution transformer depends upon the consumer and may therefore vary between full load or no load, thus voltage regulation has very significant factor in distribution transformer.

• Accordingly the per unit impedance of distribution transformer may be as low as 0.015p.u.

• Whereas the per unit impedance of a power transformer may be as high as 0.15 p.u.

• A high value of per unit impedance of a power transformer has the advantage that it induces the fault MVA level of the power system.

• As compared to a transformer of low voltage rating a high voltage transformer has high leakage reactance, as the thicker insulation makes the two winding further apart. Therefore voltage regulation high.

Efficiency of the transformer: -In open circuit test, output is zero because current is zero. And in short circuit test, output is zero because voltage is zero; hence efficiency is zero in both the cases.From the O.C. test we find the iron losses or core losses which divided in two parts:(1) Hysteresis loss (2) Eddy current lossHysteresis loss ∝ f Bm

n

Here n is called Steinmetz constant and the value of n is 1.54 to 2.5.Hysteresis loss (Ph) = kh f Bm

n VHere V= Volume of the coreEddy current loss ∝ f2 Bm

2

Eddy current loss (Pe) = f2 Bm2 t2 V

V = volume of the coret = thickness of the lamination

Therefore iron loss (Pi) = Ph + Pe

• From the S.C. test we find the copper losses which depend on load current value and are proportional to square of the load current.

• Pcu ∝ I2 ∝ (KVA)2

• η = 𝐕𝐈 𝐂𝐨𝐬ɸ

𝐕𝐈 𝐂𝐨𝐬ɸ+ 𝐏𝐢+𝐏𝐜𝐮

• If the load power factor is variable and the load current is constant then the maximum efficiency is obtained when the load power factor is unity.

However if the load power factor is constant but the load current is variable then the condition for maximum efficiency is;

Copper losses = Iron losses

Condition for current at maximum efficiency: -

I2Req= Pi

Iηmax= 𝐏𝐢

𝐑𝐞𝐪

So the maximum efficiency: -

KVA at maximum efficiency: -

Sηmax = Sj𝐏𝐢

𝐏𝐜𝐮𝐣

Here j is fraction of full load.

η = 𝐕𝐈 𝐂𝐨𝐬ɸ

𝐕𝐈 𝐂𝐨𝐬ɸ+ 𝟐𝐏𝐢=

Sηmax𝐂𝐨𝐬ɸ

Sηmax𝐂𝐨𝐬ɸ+𝟐𝐏𝐢

Alternate method for ηmax

Pcu∝ S2

Pcu(ηmax)

𝐏𝐜𝐮𝐣=

𝑺ηmax𝟐

𝑺η𝐣𝟐

Efficiency considerations in power and distribution transformers: -

(1) Maximum efficiency of a transformer is obtained when iron loss becomes equal to copper loss.

(2) A power transformer either operates at full load or not at all. Hence the maximum efficiency of a power transformer should be design for full load of operation, and therefore iron loss of power transformer is designed to be full load for full load copper loss therefore is high.

𝐒η(max)= 𝐒η𝐣𝐏𝐜𝐮(ηmax)

𝐏𝐜𝐮𝐣

(3) A distribution transformer has a load that depends on consumers demand and is found to be 70 to 75% of its full load rating.

(4) Accordingly the maximum efficiency of the distribution transformer should be designed for its average loading.

(5) Consequently a distribution transformer should be designed with iron loss = copper loss at its average loading.

(6) Obviously therefore the iron loss in distribution transformer is designed to be low.

(7) The iron loss in distribution transformer is reduced by increasing the cross sectional area of the core.

(8) This results into a higher iron to copper ratio in distribution transformer as comparable rating.

(9) Consequently the physical size of a distribution transformer is larger than of power transformer of same rating and capacity.

Q: - The maximum efficiency η of a 500 kVA, 3300/500 V, 50 Hz single phase transformer is 97% and occurs at 75% of full load at unity power factor. Find the iron losses.

(A) 0.116 p.u. (B) 0.0116 p.u.

(C) 1.16 p.u. (D) 1.06 p.u.

Q: - A 500 kVA transformer has an η of 95% at full load and also at 60% of full load both at unity p.f.

(1) Separate out the losses and determine the η of the transformer at ¾th of full load.

All day efficiency of the transformer: -

The all day efficiency of the transformer is the ratio of total energy output in a 24 hour day to the total energy input in the same time.

(η) all day = 𝐨𝐮𝐭𝐩𝐮𝐭 𝐤𝐰𝐡 𝐢𝐧 𝟐𝟒 𝐡𝐨𝐮𝐫𝐬

𝐢𝐧𝐩𝐮𝐭 𝐤𝐰𝐡 𝐢𝐧 𝟐𝟒 𝐡𝐨𝐮𝐫𝐬

(η) all day =𝐨𝐮𝐭𝐩𝐮𝐭 𝐤𝐰𝐡 𝐢𝐧 𝟐𝟒 𝐡𝐨𝐮𝐫𝐬

𝐨𝐮𝐭𝐩𝐮𝐭 𝐤𝐰𝐡 𝐢𝐧 𝟐𝟒 𝐡𝐨𝐮𝐫𝐬+ 𝐏𝐢 𝟐𝟒 𝐡𝐨𝐮𝐫𝐬 +𝐏𝐜𝐮( 𝟐𝟒 𝐡𝐨𝐮𝐫𝐬)

• Cu losses are varying with load but core losses are independent from load.

Q: - A 500 kVA transformer has a maximum efficiency of 98.6% at 350 kVA at unity p.f. during the day it is as follows: -

6 hours: 300 kVA, 0.8 p.f. lag

4 hours: 240 kW, 0.6 p.f. lag

5 hours: No load

9 hours: 225 kVA, unity p.f.

• Autotransformer: -

• An auto transformer is a transformer in which a part of the winding is common between the both primary and secondary. Unlike a two winding transformer where the power transfer is only inductive.

• An auto transformer provides inductive transfer as well as conductive transfer.

• This results into a higher output in an autotransformer as compared to a two winding transformer that uses a same material.

• In other words an auto transformer that uses the same material as a two winding transformer has higher output higher efficiency and lower per unit impedance and therefore lower voltage regulation. However a lower value of per unit impedance results into higher short circuit current.

• If a two winding transformer is to be reconnected as autotransformer then its low voltage winding insulation should be withstand with the high voltage expected during operation as autotransformer.

• The advantage in an auto transformer is most visible when the voltage ratio as autotransformer is closed to 1.

• If an autotransformer is compared with a two winding transformer for the same duty then the auto transformer is bound to use less copper and less iron also the exciting current of an autotransformer is lower, the efficiency is higher and the voltage regulation is lower.

• Infect the copper saving in the auto transformer is directly related to the conductive transfer and the ratio for copper saving in auto transformer and in two winding transformer is the same as the ratio of conductive transfer.

VH

VL(IL-IH)

C

A

B

IH

IL= (IL- IH+ IH)

(NH- NL) (VH- VL)

(NH)

(NL)

AC= Complete windingBC = Common windingAB = Auto transformer winding

𝐕𝐇𝐕𝐋

=𝐍𝐇

𝐍𝐋=𝐀𝐚𝐮𝐭𝐨

(1) At mmf balance

(NH- NL) IH= (IL- IH) NL

IHNH- NLIH = NLIL- IHNL

IHNH = ILNL

𝐍𝐇

𝐍𝐋=𝐈𝐋

𝐈𝐇Similarly power at both sides are equal: -

SHV = SLV

• Copper weight = Copper volume × Copper density

• Copper weight ∝ conductor cross sectional area ∝ conductor length

• Copper weight ∝ I ∝ N

• Copper weight ∝ N.I

• Copper weight ∝ mmf

Copper comparison: -𝐂𝐮𝐚𝐮𝐭𝐨

𝐂𝐮𝟐𝐰𝐝𝐠=

𝐍𝐇−𝐍𝐋 𝐈𝐇+𝐍𝐋(𝐈𝐋−𝐈𝐇)

𝐍𝐇𝐈𝐇+𝐍𝐋𝐈𝐋

= 𝐍𝐇 𝐈𝐇−𝐍𝐋𝐈𝐇+𝐍𝐋𝐈𝐋−𝐍𝐋𝐈𝐇

𝐍𝐇𝐈𝐇+𝐍𝐋𝐈𝐋

NHIH = NLIL ( in auto and 2 winding transformer)

=𝟐(𝐍𝐇𝐈𝐇−𝐍𝐋𝐈𝐇)

𝟐𝐍𝐇𝐈𝐇

𝐂𝐮𝐚𝐮𝐭𝐨

𝐂𝐮𝟐𝐰𝐝𝐠=

𝐍𝐇−𝐍𝐋

𝐍𝐇= (1-

𝟏

𝒂𝒂𝒖𝒕𝒐)

Copper saving: -

Copper saving = 𝐂𝐮𝟐𝐰𝐝𝐠−𝐂𝐮𝐚𝐮𝐭𝐨

𝐂𝐮𝟐𝐰𝐝𝐠

Copper saving = 𝟏 −𝐂𝐮𝐚𝐮𝐭𝐨

𝐂𝐮𝟐𝐰𝐝𝐠

= 1 −[1-𝟏

𝒂𝒂𝒖𝒕𝒐]

Copper saving = 𝟏

𝒂𝒂𝒖𝒕𝒐

Component of power transfer: -

SLV= VL IL = VL [(IL- IH) + IH]

SLV= VL (IL- IH) + VLIH

inductive transfer Conductive transfer

SAB= VL (IL- IH) = VLIL – VHIH

If VLIL = VHIH

SAB = (VH – VL) IH

The ratio = 𝐒𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐢𝐯𝐞

𝐒𝐭𝐨𝐭𝐚𝐥=

𝐕𝐋𝐈𝐇

𝐕𝐇𝐈𝐇=

𝟏

𝒂𝒂𝒖𝒕𝒐= copper saving

𝐒𝐢𝐧𝐝𝐮𝐜𝐭𝐢𝐯𝐞

𝐒𝐭𝐨𝐭𝐚𝐥= 1 -

𝟏

𝒂𝒂𝒖𝒕𝒐

2 winding transformer using the material of autotransformer: -

(NH- NL) : NL

VL

VH- VL

IH

IL- IH

IL- IH

Sauto= [𝐚𝐚𝐮𝐭𝐨

𝐚𝐚𝐮𝐭𝐨−𝟏] S2wdg

Q: - A 25 kVA, 2500/250 V, 2 winding transformer is to be reconnected as auto transformer determine the kVA output and voltage ratio as autotransformer for all possible connections.

Applications of auto transformer:-

• To connect two power systems at different voltage level where the voltage ratio is limited to 3:1.

• Example: 765/400 kV, 400/220 kV, 220/132 kVA

• As booster for line drop compensation in electric traction supply system.

• To start a 3 phase induction motor usually of the squirrel cage type having large rating.

• In manual and automatic and servo voltage stabilizer for domestic, commercial and industrial use.

• As continuously variable transformer for laboratory applications.

Tertiary windings: -

This is the 3rd winding that is additional winding besides the usual primary and secondary windings.

It is used to provide a 3rd voltage level for a special requirements such as: -

(A) To connect 3 power systems at different levels. Example: - 765 kV /400kV/220 kV, 400 kV/220 kV/ 132 kV.

(B) To provide an additional auxiliary voltage in unit auxiliary transformers of power stations.

(C) To connect reactive power compensating equipment in substations.

(D) To provide low voltage supply or lightning and water requirements in high voltage substations.

(E) Tertiary winding is also used by a star-connected transformers to overcome problems related to harmonics related communication interference, oscillating neutral and unbalanced loading.

Such a tertiary delta winding if unloaded is called stabilizing winding.

3-Phase transformer: -

• In generation, transformation, transmission and utilization of electric energy it is economical to use 3-phase system rather than 1-phase.

• For 3 phase transformation, three single-phase transformers are needed.

• Two arrangements are possible a bank of three, single-phase transformer or a single, 3-phase transformer on three legs of a common core.

• The three single-phase transformer unit costs about 15% less than that of a bank and further more the single unit occupy less space.

• The bank also offers the advantage of de-rated open delta operation when one single unit becomes inoperative.

• In a 3 phase bank phases are electrically connected but the three magnetic circuit are independent.

3 phase transformer connections: -

Group 1: - Yy0, Dd0

Group 2: - Yy6, Dd6

Group 3: - Yd1, Dy1 (30 degree lag connection)

Group 4: - Yd11, Dy11 (30 degree lead connection)

• Application of different 3 phase connections: -

• According to general recommendation star connection is used for high voltage and low capacity application.

• Since the kVA capacity of the transformer of the transformer is the same on both sides, it would be natural to connect the H.V. side in star and L.V. side in delta.

• Accordingly Δ-Υ transformers are used in step up application while Υ-Δconnection is used for step down application.

• However there is an exception in secondary distribution system a neutral is required to feed single phase loads.

• Because he secondary distribution system is used for mix loading. Since a neutral can be provided only by star the secondary distribution transformer is connected in Δ-Υ or step down applications.

• A Δ- Δ connection is used for low voltage high capacity applications, it has the advantage that if 3 single phase transformer are used in a 3 phase bank and one transformer is need to be removed then the remaining transformer may be used in open Δ also called V-V connection to continue to feed at a reduced capacity of 57.7%.

• A Δ- Δ should only be used for 3-phase loads, although a Υ-Υ connection appears to be quite attractive for extra high voltage applications. They are rarely used without a tertiary Δ winding because of problems associated with magnetizing current phenomena related harmonics and unbalanced loading.

Open delta or V-V connection: -

• The open-delta, also known as the V-V connection, is a 3-phase arrangement that makes use of only two, instead of three, single-phase transformers.

• An open delta connection system is also called a V-V system. Open delta connection systems are usually only used in emergency conditions, as their efficiency is low when compared to delta-delta (closed delta) systems (which are used during standard operations).

• The transformer output power (in VA) is for a balanced transformer system for the closed delta connection (using phase current), this give: -

SΔ = 𝟑 VLIL

SV-V = 𝟑 VL(IL/ 𝟑)SV−V

SΔ=

𝟏

𝟑= 0.577

i.e., 57.7%.

SV-V = 57.7% SΔ

Scott connection: -

The Scott-T Connection is the method of connecting two single phase transformer to perform the 3-phase to 2-phase conversion and vice-versa. The two transformers are connected electrically but not magnetically.One of the transformers is called the main transformer, and the other is called the auxiliary or teaser transformer.

• The figure below shows the Scott-T transformer connection. The main transformer is center tapped at D and is connected to the line B and C of the 3-phase side. It has primary BC and secondary a1a2. The teaser transformer is connected to the line terminal A and the centre tapping D. It has primary AD and the secondary b1b2

• The teaser transformer has the primary voltage rating that is √3/2 or 0.866 of the voltage ratings of the main transformer. Voltage VAD is applied to the primary of the teaser transformer and therefore the secondary of the voltage V2t of the teaser transformer will lead the secondary terminal voltage V2m of the main transformer by 90º as shown in the figure below.

• Turns ratio of main transformer am = N1/N2

• Turns ratio of teaser transformer aT = 𝟑

𝟐

N1

N2=

𝟑

𝟐am

• Rating of the main transformer = V1I1

• Rating of the teaser transformer = 𝟑

𝟐V1I1

𝐒𝐌

𝐒𝐓=

𝟐

𝟑= 1.15

Parallel operation of transformer: -• When the load outgrows the capacity of an existing transformer, it may be

economical to install another one rather than replacing it with a single larger unit.

• Also sometimes in a new installation, two units in parallel, though more expensive may be preferred over a single unit for reasons of reliability.

• Half of the load can be supplied with one unit output. Further the cost of maintaining a spare is less with two units in parallel.

The satisfactory and successful operation of transformer connected in parallel on both sides requires that they full fill the following conditions: -(A) Equal voltage ratio (Must)Small difference in voltage ratio may be permitted if unavoidable.(B) Same polarity (Must)(C) Same per unit impedance (Desirable)(D) Same X/R ratio, same impedance angle for same power factor operation (Desirable).

Applicable condition for 3-phase transformer only: -

(1) Same phase sequence (Must)

(2) Zero phase difference (Must)

This means that transformer belonging to the same phasor group may alone be parallel.(i.e. Yd1 , Dy1).

IC = circulating current

Two transformers with equal voltage ratio: -

VZ= IA ZA= IB ZB= IL ×𝐙𝐀 𝐙𝐁

𝐙𝐀+𝐙𝐁

Where ZA and ZB are in ohms.

However if ZA and ZB are to be expressed in per unit then their per unit value should be, on a common base impedance, so that the ohmic ratio remains unchanged.

SA = V IA

SA = P- jQA= V × 𝐈𝐀

IA = 𝐙𝐁

𝐙𝐀+𝐙𝐁IL

IB = 𝐙𝐀

𝐙𝐀+𝐙𝐁IL

SA = V 𝐙𝐁

𝐙𝐀+𝐙𝐁IL

𝐒𝐀 =𝐒𝐋𝒁𝑩

𝒁𝑨+𝒁𝑩

Or

𝐒𝐁 =𝐒𝐋𝒁𝑨

𝒁𝑨+𝒁𝑩

Per unit load on transformer: -(Equal voltage ratio)

𝐈𝐣𝐙𝐣 = constant

𝐈𝐣 ∝𝟏

𝐙𝐣

Multiply by 𝑽𝟏*

𝑽𝟏* 𝐈𝐣 ∝

𝟏

𝒁𝒋

𝐒𝐣* =

𝟏

𝐙𝐣

Since ohmic value = per unit value × Base value

𝐒𝐣* ∝

1

𝐙𝐣(𝐩.𝐮.)×Zj(base)

𝐒𝐣* ∝

1

𝐙𝐣(𝐩.𝐮.)×𝐕𝐫𝐚𝐭𝐞𝐝𝟐

𝐒𝐣 𝐫𝐚𝐭𝐞𝐝

∵ Zj (base) =𝐕𝐫𝐚𝐭𝐞𝐝𝟐

𝐒𝐣(𝐫𝐚𝐭𝐞𝐝)

𝐒𝐣* ∝

Sj(rated)

𝐙𝐣(𝐩.𝐮.)×𝐕𝐫𝐚𝐭𝐞𝐝𝟐

This means that the transfer with the lowest per unit impedance on its own base would reach full load first.

• For proportional load sharing: -

𝐒𝐣(p.u.) ∝ 𝐒𝐣(𝐫𝐚𝐭𝐞𝐝)

Actual kVA must be proportional to its capacity.

𝐒𝐣(p.u.)

𝐒𝐣(𝐫𝐚𝐭𝐞𝐝)

= constant

𝐒𝐣∗

𝐒𝐣(𝐫𝐚𝐭𝐞𝐝)∝

𝟏

Zp.u.

𝐒𝐣∗

(p.u.) ∝𝟏

Zp.u.

• This means that for proportional load sharing p.u. impedance loading on each transformer should be the same and this would be possible only when their p.u. impedances are the same.

Magnetization current phenomenon: -

• If the applied voltage to a transformer is sinusoidal then the core flux should also be sinusoidal.

• If the magnetization curve of the core material would have been linear then the magnetizing current should have been sinusoidal.

• Due to economic reasons, modern transformers are operated with high flux density, which drives the transformer core in deep saturation; obviously the magnetization curve becomes highly non-linear.

• With such a non-linear magnetization curve, a sinusoidal flux may only be obtain with a peaky containing dominant peaking 3rd harmonic component.

Note: - Any symmetrical component can never contain even harmonics.

• The third harmonic component of the magnetizing current can only flow if the electric circuit permits. Hence it can easily flow in single phase transformers employed in 1- phase circuits.

• However in 3-phase transformers the 3rd harmonic component can flow only in a star connection if the star neutral is connected to source neutral or it can flow only in a closed delta.

• If the electric circuit does not permit the flow of 3rd harmonic component of magnetizing current then the magnetizing current would remain sinusoidal if higher harmonics are ignored.

• A sinusoidal magnetizing current results into a flat topped flux containing prominent depressing 3rd harmonic flux component.

• The third harmonic flux can get established in the core only if the magnetic circuit permits.

Hence it can easily get established in those transformers that have magnetically independent circuits, such as in 3 single phase transformers in 3-phase bank.

• The presence of 3rd harmonic flux in magnetically independently transformers results into peaky induced e.m.f in both windings that stresses the insulation and results into a highly objectionable phenomenon known as oscillating neutral.

• The above problem of oscillating neutral and highly insulating stress in transformers having independent magnetic circuits drop-up due to the fact that the electric circuits did not give permission for the flow of 3rd

harmonic current.

• Obviously the solution lies in providing in the path for 3rd harmonic currents in the 3-phase transformer connections.

• One of the solution in a star-star connection is to provide a return path for 3rd harmonic currents by joining the star neutral to the source neutral.

• However this option is not exercised because the 3rd harmonic generated voltage of the generator phase would now appear across the primary winding of the transformer.

• Obviously this 3rd harmonic voltage could be transferred to the secondary side resulting into flow of 3rd harmonic current in the transmission line causing objectionable telephonic communication interference.

• Therefore the only option left is to provide a delta at least on one side of the transformer. Thus star/delta, delta/star or delta/delta connections should be used.

However if a star/star connection has to be used then it should be provided with a tertiary delta.• The presence of tertiary delta is star/star connection helps in minimizing shifting

neutral problem, prevents current choking and facilitates single phase loading (unbalance loading).

• The presence of a delta winding at least on any one side ensures a closed path for 3rd harmonic current that replace the missing 3rd harmonic current in the primary lines. This 3rd harmonic current in a closed delta, therefore follows Lenz law and opposes the third harmonic flux resulting into restoration of harmonic variation in the core flux.

• It is interesting to know that only 1 to 2% 3rd harmonic flux is sufficient to generate the 3rd harmonic voltage in the closed delta to circulate this 3rd

harmonic magnetizing current.

• Obviously the presence of only 1% to 2% of 3rd harmonic flux is good as good as the core flux being sinusoidal.

• The above problems are encountered in these 3-phase transformers that have magnetically independent circuits.

• Specifically the Star-star connection in such transformers required a tertiary delta winding.

• A 3-limb core type transformer has magnetically interlink circuit, hence the 3rd harmonic flux components finds a very high reluctance path through the large air gap and tank walls.

• Consequently its strength remains negligible and therefore the core flux remains also sinusoidal.

• However even this slight present in 3rd harmonic current causes a tank wall heating. Large transformers therefore are provided with a copper ring around the core to prevent 3rd harmonic flux from reaching the tank wall and hence tank wall heating minimized.

• Thus it can be concluded that a star-star connection without tertiary delta can be used in 3 limb core type transformers. However it is recommended that if the unbalanced in load is expected to exceed 10% then a tertiary delta winding may be provided.

• In a 3 limb core type transformer the absence of low reluctance path for 3rd

harmonic flux causes the magnetization current to contain considerable 5th

and 7th harmonics.

• Since 5th and 7th harmonic currents cannot be suppressed by any electrical connection, if these harmonics are in themselves objectionable. Then a 4 limb or 5 limb core construction may be provided.

Magnetizing inrush current: -

• If the applied voltage to transformer is sinusoidal then under steady state the core flux should rise from -ɸm to +ɸm in the positive half cycle of the voltage waveform (flux is lagging by 90 degree).

• If the transformer is switched on at an instant when the instantaneous voltage is say at its positive peak, then the core flux would rise naturally from 0 to +ɸm in a quarter cycle. This natural rise of core would ensure a trouble free and normal switching (on/off) of the transformer.

• However if the instantaneous value of the applied voltage at the switching instant is zero and say going towards positive, then the core flux would have to +2ɸm in half cycle to balance the applied voltage.

• It may be remembered any residual flux in the core is being ignored. This is famously called doubling effect and it is accompanied by a huge magnetizing inrush current that may reach 5 times the full load current, resulting into massive winding forces and a possible voltage dip in the power system.

• The magnetizing inrush is a highly unsymmetrical current that decays according to the time constant of the system and may stay for a few cycles.

• It may stay longer and decay slowly if the transformer is switched ON, on no-load or with inductive load. On the other hand it is expected to decay quickly if switched on with resistive load or capacitive load.

• Since magnetizing inrush current is a primary phenomenon, the differential protection scheme of a transformer may cause unintended tripping when the transformer is switched ON.

• Hence large transformers are provided with a second harmonic restrained feature that prevents tripping due to switching OFF transformer without any fault on the system.

Extra points about transformers: -

• If the linear dimension of the transformer is increased by k times then kVA rating of the transformer is increased by k4 times, and losses are increased by k3 times, surface area increases by K2 times and moment of inertia is increased by k5 times.

• Disc winding is used for HV coils (cooled effectively).

• Helical winding is used for LV coils.

• Spiral windings are used in small transformers for HV coils.

Sumpner’s test: -

The back to back test determines the maximum temperature rise in a transformer and hence the load is chosen accordingly to the capability of the transformer.

It requires two identical transformer primary of the two transformer are connected in parallel and supplied at rated voltage and rated frequency.

Secondary are connected in series with their polarities in phase opposition which can be checked by voltmeter. The range of this voltmeter should be double the rated voltage of either transformer secondary.

If voltmeter at the secondary side reads zero means they are in phase opposition and terminals A and B are used for the test.

• If the primary circuit is closed then the total voltage of the secondary which is connected in series will be zero. There will be no current in the secondary winding of the transformer 1 and 2 and they behave as an open circuit.

• The reading of W1 gives the iron losses of both the transformer.

• A small voltage is injected into the secondary circuit by a regulating transformer excited by the main supply.

• Magnitude of injected voltage adjusted till A2 reads full load current.

• The current will follow a circulating path through the main bus bar as shown by a dotted line.

• The reading of wattmeter in the secondary side will not be affected by this current.

• Wattmeter in the secondary side will give the full load copper loss of the two transformer.

• A2 gives the no load current of the two transformer.

• The temperature rise of the transformer can be determine by operating these transformer back to back for a long time say 48 hours. And measuring the oil temperature at periodic interval say 1 hour.

Determination of Iron Loss: -

• The wattmeter W1 measures the power loss which is equal to the iron loss of the transformer. For determining the iron loss, the primary circuit of the transformer is kept closed. Because of the primary closed circuit, no current flows through the secondary windings of the transformer. The secondary winding behaves like an open circuit. The wattmeter is connected to their secondary terminal for the measurement of iron loss.

Determination of Copper Loss: -

• The copper loss of the transformer is determined when the full load current flows through their primary and secondary windings. The additional regulating transformer is used for exciting the secondary windings. The full load current flows from the secondary to the primary winding. The wattmeter W2 measures the full load copper loss of the two transformers.

Q: - A transformer has a maximum efficiency of 98% at 3/4th of its full load at unity p.f. The iron losses equal 314 watts. Compute the efficiency of the transformer at 50% and 100% rated full load at the same power factor.

SSC JE-2019

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Electrical Engineering by ASHISH SIR

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Electrical Engineering by ASHISH SIR

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Electrical Engineering by ASHISH SIR

Electrical Engineering by ASHISH SIR

Electrical Engineering by ASHISH SIR

(28) In a steam power plant…………heats the feed water on its way to the boiler by delivering heat from the flue gases: -

(A) Superheater (B) Economizer

(B) Preheater (D) Turbine

(29) Power generation of thermal power plants is based on: -

(A) Rankine cycle

(B) Otto cycle

(C) Diesel cycle

(D) Carnot cycle