Practical Transformer

 

 

24.1 Goals of the lesson

 

In practice no transformer is ideal. In this lesson we shall add realities into an ideal transformer for correct representation of a practical transformer. In a practical transformer, core material will have (i) finite value of μ_r , (ii) winding resistances, (iii) leakage fluxes and (iv) core loss. One of the major goals of this lesson is to explain how the effects of these can be taken into account to represent a practical transformer. It will be shown that a practical transformer can be considered to be an ideal transformer plus some appropriate resistances and reactances connected to it to take into account the effects of items (i) to (iv) listed above.

Next goal of course will be to obtain exact and approximate equivalent circuit along with phasor diagram.

Key words : leakage reactances, magnetizing reactance, no load current.

After going through this section students will be able to answer the following questions.

 

• How does the effect of magnetizing current is taken into account?

 

 

• How does the effect of core loss is taken into account?

 

 

• How does the effect of leakage fluxes are taken into account?

 

 

• How does the effect of winding resistances are taken into account?

 

 

• Comment the variation of core loss from no load to full load condition.

 

 

• Draw the exact and approximate equivalent circuits referred to primary side.

 

 

• Draw the exact and approximate equivalent circuits referred to secondary side.

 

 

• Draw the complete phasor diagram of the transformer showing flux, primary & secondary induced voltages, primary & secondary terminal voltages and primary & secondary currents.

 

 

24.2 Practical transformer

 

A practical transformer will differ from an ideal transformer in many ways. For example the core material will have finite permeability, there will be eddy current and hysteresis losses taking place in the core, there will be leakage fluxes, and finite winding resistances. We shall gradually bring the realities one by one and modify the ideal transformer to represent those factors.

Consider a transformer which requires a finite magnetizing current for establishing flux in the core. In that case, the transformer will draw this current I_m even under no load condition. The level of flux in the core is decided by the voltage, frequency and number of turns of the primary and does not depend upon the nature of the core material used which is apparent from the following equation:

maxφ=112VfNπ

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Hence maximum value of flux density Bmax _{is known from Bmax} = max,iAφwhere A_i is the net cross sectional area of the core. Now H_max is obtained from the B - H curve of the material. But we know H_max = 1max,miNIlwhere I_mmax is the maximum value of the magnetizing current. So rms value of the magnetizing current will be I_m = max.2mIThus we find that the amount of magnetizing current drawn will be different for different core material although applied voltage, frequency and number of turns are same. Under no load condition the required amount of flux will be produced by the mmf N₁I_m. In fact this amount of mmf must exist in the core of the transformer all the time, independent of the degree of loading.

Whenever secondary delivers a current I₂, The primary has to reacts by drawing extra current I'₂ (called reflected current) such that I'₂N₁ = I₂N₂ and is to be satisfied at every instant. Which means that if at any instant i₂ is leaving the dot terminal of secondary, 2i′ will be drawn from the dot terminal of the primary. It can be easily shown that under this condition, these two mmfs (i.e, N₂i₂ and 21iN′) will act in opposition as shown in figure 24.1. If these two mmfs also happen to be numerically equal, there can not be any flux produced in the core, due to the effect of actual secondary current I₂ and the corresponding reflected current 2I′

The net mmf therefore, acting in the magnetic circuit is once again I_mN₁ as mmfs 21IN′and I₂N₂ cancel each other. All these happens, because KVL is to be satisfied in the primary demanding φ_max to remain same, no matter what is the status (i., open circuited or loaded) of the secondary. To create φ_max, mmf necessary is N₁I_m. Thus, net mmf provided by the two coils together must always be N₁I_m - under no load as under load condition. Better core material is used to make I_m smaller in a well designed transformer.

Keeping the above facts in mind, we are now in a position to draw phasor diagram of the transformer and also to suggest modification necessary to an ideal transformer to take magnetizing current mI into account. Consider first, the no load operation. We first draw the maxφ phasor. Since the core is not ideal, a finite magnetizing current mI will be drawn from supply and it will be in phase with the flux phasor as shown in figure 24.2(a). The induced voltages in primary 1Eand secondary2E are drawn 90º ahead (as explained earlier following convention 2). Since winding resistances and the leakage flux are still neglected, terminal voltages 1V and 2V will be same as 1Eand 2E respectively. Figure 24.1: MMf directions by I2 and ′2I N1 ′12NiN2 N2 i2 i2 ′2i

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(a) No load

If you compare this no load phasor diagram with the no load phasor diagram of the ideal transformer, the only difference is the absence of mI in the ideal transformer. Noting that mI lags 1V by 90º and the magnetizing current has to supplied for all loading conditions, common sense prompts us to connect a reactance X_m, called the magnetizing reactance across the primary of an ideal transformer as shown in figure 24.3(a). Thus the transformer having a finite magnetizing current can be modeled or represented by an ideal transformer with a fixed magnetizing reactance X_m connected across the primary. With S opened in figure 24.3(a), the current drawn from the supply is 1I= mI since there is no reflected current in the primary of the ideal transformer. However, with S closed there will be 2I, hence reflected current 22/II′= will appear in the primary of the ideal transformer. So current drawn from the supply will be 12mIII′=+.

This model figure 24.3 correctly represents the phasor diagram of figure 24.2. As can be seen from the phasor diagram, the input power factor angle 1θ will differ from the load power factor 2θ in fact power factor will be slightly poorer (since θ₂ > θ₁).

Since we already know how to draw the equivalent circuit of an ideal transformer, so same rules of transferring impedances, voltages and currents from one side to the other side can be revoked here because a portion of the model has an ideal transformer. The equivalent circuits of the transformer having finite magnetizing current referring to primary and secondary side are shown respectively in figures 24.4(a) and (b). O 11V=E 22E=V φmΙ V1 Z2 S (a) with secondary opened Ideal Transformer mI Xm 1mI = I I2= 0 ′2I=0 V1 Z2 S (a) with load in secondary Ideal Transformer mIXm 1I′2I2IFigure 24.3: Magnetising reactance to take Im into account.

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24.2.1 Core loss

A transformer core is subjected to an alternating time varying field causing eddy current and hysteresis losses to occur inside the core of the transformer. The sum of these two losses is known as core loss of the transformer. A detail discussion on these two losses has been given in Lesson 22.

Eddy current loss is essentially2IR loss occurring inside the core. The current is caused by the induced voltage in any conceivable closed path due to time varying field. Obviously to reduce eddy current loss in a material we have to use very thin plates instead of using solid block of material which will ensure very less number of available eddy paths. Eddy current loss per unit volume of the material directly depends upon the square of the frequency, flux density and thickness of the plate. Also it is inversely proportional to the resistivity of the material. The core of the material is constructed using thin plates called lamination. Each plate is given a varnish coating for providing necessary insulation between the plates. Cold Rolled Grain Oriented, in short CRGO sheets are used to make transformer core.

After experimenting with several magnetic materials, Steinmetz proposed the following empirical formula for quick and reasonable estimation of the hysteresis loss of a given material.

maxnhhPkBf=

The value of n will generally lie between 1.5 to 2.5. Also we know the area enclosed by the hysteresis loop involving B-H characteristic of the core material is a measure of hysteresis loss per cycle.

24.3 Taking core loss into account

The transformer core being subjected to an alternating field at supply frequency will have hysteresis and eddy losses and should be appropriately taken into account in the equivalent circuit. The effect of core loss is manifested by heating of the core and is a real power (or energy) loss. Naturally in the model an external resistance should be shown to take the core loss into account. We recall that in a well designed transformer, the flux density level Bmax _{practically remains same from no load to full load condition. Hence magnitude of the core loss will be practically independent of the degree of loading and this loss must be drawn from the supply. To take this into account, a fixed resistance R}cl _{is shown connected in parallel with the magnetizing reactance as shown in the figure 24.5.} '2I V1 a2 Z2 (a) Equivalent circuit referred to primaryIm I1 2I1Va(b) Equivalent circuit referred to secondary I'm Figure 24.4: Equivalent circuits with Xm Xm '2V V2 Z2 m2Xaa = N1/N2

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It is to be noted that R_cl represents the core loss (i.e., sum of hysteresis and eddy losses) and is in parallel with the magnetizing reactance X_m. Thus the no load current drawn from the supply I_o, is not magnetizing current I_m alone, but the sum of I_cl and I_m with clI in phase with supply voltage 1V and mI lagging by 90º from 1V. The phasor diagrams for no load and load operations are shown in figures 24.6 (a) and (b).

It may be noted, that no load current I_o is about 2 to 5% of the rated current of a well designed transformer. The reflected current 2I′is obviously now to be added vectorially with oI to get the total primary current 1Ias shown in figure 24.6 (b).

24.4 Taking winding resistances and leakage flux into account

The assumption that all the flux produced by the primary links the secondary is far from true. In fact a small portion of the flux only links primary and completes its path mostly through air as shown in the figure 24.7. The total flux produced by the primary is the sum of the mutual and the leakage fluxes. While the mutual flux alone takes part in the energy transfer from the primary to the secondary, the effect of the leakage flux causes additional voltage drop. This drop can be represented by a small reactance drop called the leakage reactance drop. The effect of winding resistances are taken into account by way of small lumped resistances as shown in the figure 24.8. (a) Rcl for taking core loss into account'2IV1 a2 Z2 (b)Equivalent circuit referred to primary I1 2I 1Va Z2 (c) Equivalent circuit referred to secondaryI'1 V1 Z2 S Ideal Transformer Im Rc1 I1 Xm I0 Ic1 Rc1Xm X'm R'c1 Figure 24.5: Equivalent circuit showing core loss and magnetizing current. Im (a) phasor diagram: no load O11E=V22E=Vmaxφ mΙ(b) phasor diagram: with load power factor angle θ2 θ1 θ2 1Ι2Ι'2IFigure 24.6: Phasor Diagram of a transformer having core loss and magnetising current. E1 = V1E2 = V2I0 Icl φmaxθ0 0Ι0ΙclΙ

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The exact equivalent circuit can now be drawn with respective to various sides taking all the realities into account. Resistance and leakage reactance drops will be present on both the sides and represented as shown in the figures 24.8 and 24.9. The drops in the leakage impedances will make the terminal voltages different from the induced voltages.

It should be noted that the parallel impedance representing core loss and the magnetizing current is much higher than the series leakage impedance of both the sides. Also the no load current I₀ is only about 3 to 5% of the rated current. While the use of exact equivalent circuit will give us exact modeling of the practical transformer, but it suffers from computational burden. The basic voltage equations in the primary and in the secondary based on the exact equivalent circuit looks like:

1V=1111 EIrjIx++

2V=2222 EIrjIx−−

It is seen that if the parallel branch of R_cl and X_m are brought forward just right across the supply, computationally it becomes much more easier to predict the performance of the transformer sacrificing of course a little bit of accuracy which hardly matters to an engineer. It is this approximate equivalent circuit which is widely used to analyse a practical power/distribution transformer and such an equivalent circuit is shown in the figure 24.11. V1 Z'2 Im Rc1r1 Xm I0 Ic1 Equivalent circuit referred to primary x1 r'2 x'2 I'2 V'1 Z2 I'mR'c1r'1 X'm I'0 I'c1 Equivalent circuit referred to secondary x'1r2 x2 I2 Figure 24.9: Exact Equivalent circuit referred to primary and secondary sides. Leakage FluxMutual Flux Figure 24.7: Leakage flux and their paths. r1 Equivalent circuit showing both sides x1 V1 Im Rc1 Xm Ic1 I1 Z2 r2 x2 I2 I'2 E1 E2 Figure 24.8: Equivalent circuit of a practical transformer. Ideal Transformer

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I

The exact phasor diagram of the transformer can now be drawn by drawing the flux phasor first and then applying the KVL equations in the primary and in the secondary.

The phasor diagram of the transformer when it supplies a lagging power factor load is shown in the figure 24.10. It is clearly seen that the difference in the terminal and the induced voltage of both the sides are nothing but the leakage impedance drops of the respective sides. Also note that in the approximate equivalent circuit, the leakage impedance of a particular side is in series with the reflected leakage impedance of the other side. The sum of these leakage impedances are called equivalent leakage impedance referred to a particular side.

Equivalent leakage impedance referred to primary 11eerjx+=()(22121rarjxax+++

Equivalent leakage impedance referred to secondary 22eerjx+=112222rxrjxaa⎛⎞⎛+++⎜⎟⎜⎝⎠⎝

Where, a = 12NN the turns ratio.

24.5 A few words about equivalent circuit

Approximate equivalent circuit is widely used to predict the performance of a transformer such as estimating regulation and efficiency. Instead of using the equivalent circuit showing both the φ I0 θ0 I1 I2 θ1 θ2 E1 E2 -E1 I2 r2

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sides, it is always advantageous to use the equivalent circuit referred to a particular side and analyse it. Actual quantities of current and voltage of the other side then can be calculated by multiplying or dividing as the case may be with appropriate factors involving turns ratio a.

Transfer of parameters (impedances) and quantities (voltages and currents) from one side to the other should be done carefully. Suppose a transformer has turns ratio a, a = N₁/N₂ = V₁/V₂ where, N₁ and N₂ are respectively the primary and secondary turns and V₁ and V₂ are respectively the primary and secondary rated voltages. The rules for transferring parameters and quantities are summarized below.

 

1. Transferring impedances from secondary to the primary:

If actual impedance on the secondary side be 2Z, referred to primary side it will become 222ZaZ′=.

 

 

2. Transferring impedances from primary to the secondary:

If actual impedance on the primary side be 1Z, referred to secondary side it will become 211/ZZa′=.

 

 

3. Transferring voltage from secondary to the primary:

If actual voltage on the secondary side be 2V, referred to primary side it will become 22VaV′=.

 

 

4. Transferring voltage from primary to the secondary:

If actual voltage on the primary side be 1V, referred to secondary side it will become 11/VVa′=.

 

 

5. Transferring current from secondary to the primary:

If actual current on the secondary side be 2I, referred to primary side it will become 22/IIa′=.

 

 

6. Transferring current from primary to the secondary:

If actual current on the primary side be 1I, referred to secondary side it will become 11IaI′=.

 

In spite of all these, gross mistakes in calculating the transferred values can be identified by remembering the following facts.

 

1. A current referred to LV side, will be higher compared to its value referred to HV side.

 

2. A voltage referred to LV side, will be lower compared to its value referred to HV side.

 

3. An impedance referred to LV side, will be lower compared to its value referred to HV side.

 

24.6 Tick the correct answer

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1. If the applied voltage of a transformer is increased by 50%, while the frequency is reduced to 50%, the core flux density will become

 

[A] 3 times [B] 34 times [C] 13 [D] same

 

2. For a 10 kVA, 220 V / 110 V, 50 Hz single phase transformer, a good guess value of no load current from LV side is:

 

(A) about 1 A (B) about 8 A (C) about 10 A (D) about 4.5 A

 

3. For a 10 kVA, 220 V / 110 V, 50 Hz single phase transformer, a good guess value of no load current from HV side is:

 

(A) about 2.25 A (B) about 4 A (C) about 0.5 A (D) about 5 A

 

4. The consistent values of equivalent leakage impedance of a transformer, referred to HV and referred to LV side are respectively:

 

(A) 0.4 + j0.6Ω and 0.016 + j0.024Ω

(B) 0.2 + j0.3Ω and 0.008 + j0.03Ω

(C) 0.016 + j0.024Ω and 0.4 + j0.6Ω

(D) 0.008 + j0.3Ω and 0.016 + j0.024Ω

 

5. A 200 V / 100 V, 50 Hz transformer has magnetizing reactance X_m = 400Ω and resistance representing core loss R_cl = 250Ω both values referring to HV side. The value of the no load current and no load power factor referred to HV side are respectively:

 

(A) 2.41 A and 0,8 lag (B) 1.79 A and 0.45 lag

(C) 3.26 A and 0.2 lag (D) 4.50 A and 0.01 lag

 

6. The no load current drawn by a single phase transformer is found to be i₀ = 2 cos ωt A when supplied from 440 cos ωt Volts. The magnetizing reactance and the resistance representing core loss are respectively:

 

(A) 216.65 Ω and 38.2 Ω (B) 223.57 Ω and 1264 Ω

(C) 112 Ω and 647 Ω (D) 417.3 Ω and 76.4 Ω

24.7 Solve the problems

 

1. A 5 kVA, 200 V / 100 V, 50Hz single phase transformer has the following parameters:

 

HV winding Resistance = 0.025 Ω

HV winding leakage reactance = 0.25 Ω

LV winding Resistance = 0.005 Ω

LV winding leakage reactance = 0.05 Ω

Resistance representing core loss in HV side = 400 Ω

Magnetizing reactance in HV side = 190 Ω

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Draw the equivalent circuit referred to [i] LV side and [ii] HV side and insert all the parameter values.

 

2. A 10 kVA, 1000 V / 200 V, 50 Hz, single phase transformer has HV and LV side winding resistances as 1.1 Ω and 0.05 Ω respectively. The leakage reactances of HV and LV sides are respectively 5.2 Ω and 0.15 Ω respectively. Calculate [i] the voltage to be applied to the HV side in order to circulate rated current with LV side shorted, [ii] Also calculate the power factor under this condition.

 

 

3. Draw the complete phasor diagrams of a single phase transformer when [i] the load in the secondary is purely resistive and [ii] secondary load power factor is leading.

 

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