UPPSC -Assistant Engineer

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.ke

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.

Test Series of SSC JE -2019

Started

22-02-2020

Benefits

Electrical Engineering by ASHISH SIR

Personalized Learning: -

Electrical Engineering by ASHISH SIR

Regular PDF update

Electrical Engineering by ASHISH SIR

Electrical Engineering by ASHISH SIR

Electrical Engineering by ASHISH SIR

Electrical Engineering by ASHISH SIR

Electrical Engineering by ASHISH SIR

Electrical Engineering by ASHISH SIR

Thank-You.If you have any Query then call844-844-7780

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