Unit 1 Transformers

Unit-1 Transformers Objectives: After completing this unit, you will be able to: • State the applications of a transformer in electrical and electronic circuits. • State the principle of operation of a two-winding transformer. • Draw the circuit symbol of a transformer. • Derive the basic emf equation of a transformer. • State the four conditions for transformer to be ideal. • State the relations for transformation of voltage, current and impedance by a transformer in terms of its turns-ratio. • State why a transformer should have no-load current. • Define the two components of the no-load current. • Explain why the hysteresis and eddy-current losses occur in the core, and how these can be reduced. • Explain the construction of core-type and shell-type transformers. • State what is meant by load component of primary current. • State why in the equivalent circuit of a transformer we include a resistance and a leakage reactance in both the primary and secondary side. • State how we obtain a simplified equivalent circuit of a transformer as referred to the primary or to the secondary. • State the meaning of ‘regulation down’ and ‘regulation up’ of a transformer. • Derive the condition for zero regulation and condition for maximum regulation of a transformer. • Define ‘commercial efficiency’ and ‘all-day efficiency’ of a transformer. • Derive the condition of maximum efficiency of a transformer. • Explain how to convert a two-winding transformer into an autotransformer, and state the advantages and disadvantages of doing it. • Explain how to get the ‘equivalent circuit parameters’ of a transformer by conducting ‘open-circuit test’ and ‘shortcircuit test’.

1.1

INTRODUCTION A transformer is a highly efficient device for changing ac voltage from one value to another, without any change in its frequency. There exists no simple device that can accomplish such changes in dc voltages. Thus, the transformer has provided a feature to ac power system that lacks in dc power system. The general practice is to generate ac voltage at about 11 kV, then step up by means of a transformer to higher voltages of 132 kV, 220 kV and 400 kV for the transmission lines. This conversion aids the transmission of huge electrical power at low cost. High-voltage lines carry low currents, and hence the cost of lines and the power loss are tremendously reduced. At distribution points, other transformers are used to step the voltage down to 400 V or 220 V for use in industries, offices and homes. Since there are no moving parts in a transformer, it practically needs almost no maintenance and supervision. Apart from the above, the transformers are also used in communication circuits, radio and TV circuits, telephone circuits, instrumentation and control systems.

1.2 PRINCIPLE OF OPERATION A transformer operates on the principle of mutual induction between two coils. Figure 1.1a shows the general construction of a transformer. The vertical portions of the steel-core are termed limbs, and the top and bottom portions are called yokes. The two coils P and S, having N1 and N2 turns, are wound on the limbs. These two windings are electrically unconnected but are linked with one another through a magnetic flux in the core. The coil P is connected to the supply and is therefore called primary; coil S is connected to the load and is termed the secondary.

(a) Construction.

(b) Circuit symbol.

Fig. 1.1 A transformer. 1

Figure 1.1b shows the circuit symbol of a transformer. The thick line denotes the iron core. By having different ratios N1/N2 of the two windings, power at lower or at higher voltage can be obtained. When N2 > N1, the transformer is called a step up transformer; and when N2 < N1, the transformer is called a step down transformer. EMF Equation Consider a sinusoidally varying voltage V1 applied to the primary of the transformer shown in Fig. 1.1a. Due to this voltage, a sinusoidally varying magnetic flux is set up in the core, which can be represented as

Φ = Φm sin ωt = Φm sin 2π ft ...(1.1) where, Φm is the peak value of the flux and f is the frequency of sinusoidal variation of flux. As per the law of electromagnetic induction, the induced emf in a winding of N turns is given as

e = −N

dΦ d = − N (Φm sin ωt ) = − N ωΦm cos ωt = ω NΦm sin (ωt − π / 2) dt dt ...(1.2)

Thus, the peak value of the induced emf is Em = ωNΦm. Therefore, the rms value of the induced emf E is given as

E=

Em ω NΦm 2π fNΦm = = = 4.44 fNΦm 2 2 2 E = 4.44 fNΦm

or

...(1.3) This equation, known as emf equation of transformer, can be used to find the emf induced in any winding (primary or secondary) linking with flux Φ. Example 1.1 The primary of a 50-Hz step-down transformer has 480 turns and is fed from 6400 V supply. Find (a) the peak value of the flux produced in the core, and (b) the voltage across the secondary winding if it has 20 turns. Solution: (a) Using Eq. 1.3, we get

Φm =

E 6400 = = 0.06 Wb = 60 mWb 4.44 fN1 4.44 × 50 × 480

(b) The voltage induced in the secondary winding is given as

E = 4.44 fN 2Φm = 4.44 × 50 × 20 × 0.06 = 266.4 V 1.3

IDEAL TRANSFORMER We shall describe the physical construction and equivalent circuit of an actual transformer a little later. Here, we define the ideal transformer as a circuit element. We shall then explore its properties in voltage, current, and impedance transformation. Primary and secondary voltage and current variables are defined in Fig. 1.1b, which shows the circuit model of an ideal transformer. The complete behaviour of a practical transformer can be better understood by initially assuming the transformer to be ideal, and then allowing the imperfections of the actual transformer by suitably introducing some impedance. Conditions for Ideal Transformer: (i) The permeability (µ) of the magnetic circuit (the core) is infinite, i.e., the magnetic circuit has zero reluctance so that no magneto-motive force (mmf) is needed to set up the flux in the core. (ii) The core of the transformer has no losses. (iii) The resistance of its windings is zero, hence no I2R losses in the windings. (iv) Entire flux in the core links both the windings, i.e., there is no leakage flux. Thus, an ideal transformer has no losses and stores no energy. However, an ideal transformer has no physical existence. But, the concept of ideal transformer is very helpful in understanding the working of an actual transformer. Consider an ideal transformer whose secondary is connected to a load ZL and primary is supplied from an ac source V1 (Fig. 1.2a). The voltage across the load is V2. The primary and secondary windings of the ideal transformer have zero impedance. Hence, the induced emf E1 in the primary exactly counter balances the applied voltage V1, that is, V1 = -E1. Also, the induced emf E2 is the same as voltage V2, that is, E2 = V2. Here, E1 is called counter emf or back emf induced in the primary, and E2 called mutually induced emf in the secondary.

2

(a) The circuit.

(b) The phasor diagram.

Fig. 1.2 Ideal transformer. Figure 1.2b shows the phasor diagram of the ideal transformer. We have taken flux Φ as reference phasor, as it is common to both the primary and secondary. As per Eq. 1.2, the induced emfs E1 and E2 lag flux Φ by 90°. The voltage V1 is equal and opposite to emf E1. Thus, the applied voltage V1 leads the flux Φ by 90°. According to the first condition of ideality, the reluctance of the magnetic circuit is zero and hence the required magnetizing current to produce flux Φ is also zero. Transformation Ratio The ratio of secondary voltage to the primary voltage is known as transformation ratio or turns-ratio. It is denoted by letter K. Let N1 and N2 be the number of turns in primary and secondary windings, and E1 and E2 be the rms values of the primary and secondary induced emfs. Using Eq. 1.3, we can write E1 = 4.44 fN1Φ m ...(1.4)

and

E2 = 4.44 fN 2Φ m ...(1.5)

Then, the transformation ratio or turns-ratio can be expressed as

K=

V2 E2 N 2 = = V1 E1 N1

...(1.6) Thus, the side of the transformer with the larger number of turns has the larger voltage. Indeed, the voltage per turn is constant for a given transformer. By selecting K properly, the transformation of voltage can be done from any value to any other convenient value. There can be two1 cases: (i) When K > 1 (i.e., N2 > N1); V2 > V1: the device is known as step-up transformer. (ii) When K < 1 (i.e., N2 < N1); V2 < V1: the device is known as step-down transformer. In general, a transformer can have more than 2 windings. The windings of a three-winding transformer are called primary, secondary and tertiary. The primary is connected to an ac supply. Different loads may be connected across the secondary and tertiary2. The induced emf in a winding is still proportional to its number of turns,

E1 : E2 : E3 :: N1 : N 2 : N3 Volt-Amperes Consider again the two-winding transformer of Fig. 1.2a. For an ideal transformer, the current I1 in the primary is just sufficient to provide mmf I1N1 to overcome the demagnetizing effect of the secondary mmf I2N2. Hence,

∴

I1 N1 = I 2 N 2

or

I 2 N1 1 = = I1 N 2 K

...(1.7) Thus, we find that the current is transformed in the reverse ratio of the voltage. That is, the side of the transformer with the larger number of turns has the smaller current. For example, a step-up transformer would have a primary with few turns of thick wire (small voltage, large current) and the secondary would have many turns of thin wire (large voltage, small current). Combining Eqs. 1.5 and 1.7, we have

E1I1 = E2 I 2 Hence, in an ideal transformer the input VA and output VA are identical.

1

The third case, when K = 1 (i.e., N1 = N2) is not important. We hardly ever use a transformer with unity turns ratio. Such a transformer is used only when you need electrical isolation between two electrical circuits.

2

Sometimes, the tertiary winding has a centre-tap; the two halves having same number of turns, N3. The voltage of such a winding is then specified as E3/0/ E3, or E3 - 0 - E3.

3

Impedance Transformation Equations 1.6 and 1.7 reveal a very useful property of transformers, called impedance transformation. Figure 1.3 shows an ideal transformer. It has N1 and N2 turns in its primary and secondary windings respectively. A load impedance ZL is connected across its secondary, and an equivalent impedance Zeq is defined at its primary.

Fig. 1.3 The transformer changes the impedance ZL to equivalent impedance Zeq. The equivalent impedance Zeq as faced by a source V1 is given as Z eq = or

V1 V1 × (V2 I 2 ) ⎛ V1 ⎞ ⎛ I 2 ⎞ ⎛ V2 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ = = ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟ × ⎜ ⎟ × ZL I1 I1 × (V2 I 2 ) ⎝ V2 ⎠ ⎝ I1 ⎠ ⎝ I 2 ⎠ ⎝ K ⎠ ⎝ K ⎠ Z eq = Z L / K 2

...(1.8) Therefore, the impedance is transformed in inverse proportion to the square of the turns-ratio. The concept of impedance transformation is used for impedance matching. As per maximum power transfer theorem, the load impedance has to be properly matched with the source impedance, as illustrated in Example 1.3 given below. Example 1.2 A single-phase, 50-Hz transformer has 30 primary turns and 350 secondary turns. The net cross-sectional area of the core is 250 cm2. If the primary winding is connected to a 230-V, 50-Hz supply, calculate (a) the peak value of flux density in the core, (b) the voltage induced in the secondary winding, and (b) the primary current when the secondary current is 100 A. (Neglect losses.) Solution: (a) The peak value of the flux in the core is given as

Φm =

E1 230 = = 0.034534 Wb 4.44 fN1 4.44 × 50 × 30

Therefore, the peak value of the flux density in the core is

Bm =

Φm A

=

0.034534 = 1.3814 T 250 × 10 −4

(b) The voltage induced in the secondary winding is

E2 = E1 ×

N2 350 = 230 × = 2683.33 V = 2.683 kV N1 30

(c) The primary current is

⎛N ⎞ ⎛ 350 ⎞ I1 = I 2 ⎜ 2 ⎟ = 100 × ⎜ ⎟ = 1166.67 A 1.167 kA ⎝ 30 ⎠ ⎝ N1 ⎠ Example 1.3 A source with an output resistance of 50 Ω is required to deliver power to a load of 800 Ω. Find the turns-ratio of the transformer to be used for maximizing the load power. Solution: For delivering maximum power to the load, the equivalent resistance must be equal to the source resistance. This requires a resistance of 50 Ω looking into the primary of the transformer. That is,

Req = RL / K 2

or

50 = 800 / K 2

⇒

K = 800 / 50 = 16 = 4

Thus,

K=

N2 =4 N1

Example 1.4 Determine the load current IL in the ac circuit shown in Fig. 1.4a.

4

Ip 20 Ω j20 Ω

2:1

+ −

Ip 20 Ω j20 Ω

IL

2Ω 30 V

⇒

-j10 Ω

( 2)

+ −

2

×2 Ω

30 V

(a)

( 2 ) × ( − j10 ) Ω 2

(b)

Fig. 1.4 Solution: We first transform the load impedance into the primary to simplify the circuit, as shown in Fig. 1.4b. The primary current is then calculated as

Ip =

30∠0° = 0.872∠35.53° A 20 + j 20 + 22 (2 − j10)

The load current, which is the same as the secondary current, is given by Eq. 1.7 as

I L = 2 × I p = 2 × 0.872∠35.53° = 1.74∠35.53° A Example 1.5 A single-phase transformer has a core with cross-sectional area of 150 cm2. It operates at a maximum flux density of 1.1 Wb/m2 from a 50-Hz supply. If the secondary winding has 66 turns, determine the output in kVA when connected to a load of 4-Ω impedance. Neglect any voltage drop in the transformer. Solution:

Φm = Bm A = 1.1× 0.015 = 0.0165 Wb .

Since the voltage drop in the transformer is negligible, we have

V2 = E2 = 4.44 fN 2Φm = 4.44 × 50 × 66 × 0.0165 = 241.76 V The output current, I 2 =

∴

V2 241.76 = = 60.44 V ZL 4

Output volt-amperes = 241.76 × 60.44 = 14 612 VA = 14.612 kVA

Example 1.6 A single-phase, 50-Hz transformer has a square core having a net cross-sectional area of 9 cm2, and three windings designed for the following voltages: (i) Primary: 230 V; (ii) Secondary: 110 V; and (iii) Tertiary: 6/0/6 V. Find the number of turns in each winding if the flux density is not to exceed 1 T. Solution:

Φm = Bm A = 1× 9 × 10 −4 = 9 × 10−4 Wb . The tertiary winding is divided into two halves; each half having a

voltage E3 = 6 V. Thus, the number of turns in each half of the tertiary is

N3 = ∴

E3 6 = = 30 turns 4.44 f Φ m 4.44 × 50 × 9 × 10−4

Total number of turns on the tertiary winding = 2 × 30 = 60 turns .

We have seen that across 30 turns of the tertiary winding, the induced emf is 6 V. Therefore, the number of turns on the primary and secondary can be calculated as follows :

and

N1 E1 = N 3 E3

or

N1 =

N 3 E1 30 × 230 = = 1150 turns E3 6

N 2 E2 = N 3 E3

or

N2 =

N 3 E2 30 ×110 = = 550 turns E3 6

1.4 PRACTICAL TRANSFORMER AT NO LOAD In actual practice, a transformer can never satisfy any of the conditions specified above for the ideal transformer. Nevertheless, the concept of ideal transformer is helpful to understand the working of an actual transformer. We shall consider these conditions one by one, and see in what way a practical transformer deviates from the ideal transformer. In this Section, we shall consider only the first two ideality conditions. The remaining two conditions shall be considered in Section 1.7. Consider a transformer with its primary connected to an alternating voltage source V1, and no load connected across its secondary (Fig. 1.5a). With open circuit, the current in the secondary winding is zero. If the transformer were truly ideal, the primary current too would be zero, as per 713.5. But, in practice there does flow a little no-load current I0 in the primary. This current I0 is also called the exciting current of the transformer. Following are the two reasons why the no-load current I0 flows in the primary.

5

(1) Effect of Magnetization Consider the first ideality condition. No magnetic material can have infinite permeability so as to offer zero reluctance to the magnetic circuit. Hence, in a practical transformer a finite mmf is needed to establish magnetic flux in the core. As a result, an in-phase magnetizing current Im in the primary is needed to set up flux Φ in the core. This current Im is purely reactive and lags the voltage V1 by 90°. The flux Φ induces emfs E1 and E2 in the primary and the secondary windings. As per Eq. 1.2, both these emfs lag flux Φ by 90°, as shown in Fig. 1.5b.

(a) The circuit.

(b) The phasor diagram.

(c) The equivalent circuit.

Fig. 1.5 Transformer on no load. As the current I2 in the secondary is zero (no load connected), the voltage drop in the secondary winding is zero. Hence, V2 = E2. The induced emf E1 counter balances the applied voltage V1 and establishes an electrical equilibrium. If the third and fourth ideality conditions (i.e., the effect of the resistance of the winding and the leakage of flux) are ignored, the magnitude of V1 will be the same as that of emf E1. Thus, V1 = -E1. (2) Effect of Core Losses Let us now consider the second ideality condition. There exist two reasons (hysteresis and eddy current) for the energy loss in the core of the transformer. The source must supply enough power to the primary to meet the core losses. Therefore, a core-loss current Iw (in phase with V1) flows through the primary, as shown in the phasor diagram of Fig. 1.5b. Thus, we find that the no-load current I0 has two components, Im and Iw. The magnetizing current Im lags voltage V1 by 90° and the loss component Iw is in phase with voltage V1. The angle φ0 is the no-load phase angle. Thus, from the phasor diagram of Fig. 1.5b, we have`

I 0 = I w2 + I m2 ;

φ0 = tan −1 ( I m / I w );

and Input power = V1 I w = V1 I 0 cos φ0

In the equivalent circuit shown in Fig. 1.5c, the no-load current I0 is divided into two parallel branches. The component Iw accounts for the core loss, and hence is shown to flow through a resistance R0. The component Im represents magnetizing current. Hence, it is shown to flow through a pure inductive reactance X0. The R0-X0 parallel circuit is called exciting circuit of the transformer. Example 1.7 A single-phase, 230-V/110-V, 50-Hz transformer takes an input of 350 volt amperes at no load while working at rated voltage. The core loss is 110 W. Find the loss component of no-load current, the magnetizing component of no-load current and the no-load power factor. Solution: Given: V1 I 0

∴

= 350 VA . I0 =

VA 350 = = 1.52 A V1 230

The core loss = Input power at no load, Pi

= V1 I 0 cos φ0

Therefore, the power factor at no load is given as

pf = cos φ0 =

Pi 110 W = = 0.314 V1 I 0 350 VA

The loss component of no-load current is given as

I w = I 0 cos φ0 = 1.52 × 0.314 = 0.478 A The magnetizing component of no-load current is given as

I m = I 02 − I w2 = (1.52) 2 − (0.478) 2 = 1.44 A Since the core losses occur in the iron core, these are also called iron losses. These losses have two components: (i) Hysteresis Loss, and (ii) Eddy-Current Loss

6

(i) Hysteresis Loss: When alternating current flows through the windings, the core material undergoes cyclic process of magnetization and demagnetization. It is found that there is a tendency of the flux density B to lag behind the field strength H. This tendency is called hysteresis3. The effect of this phenomenon on the core material can be best understood from the B-H plot shown in Fig. 1.6.

Fig. 1.6 Hysteresis loop and energy relationship per half-cycle. During positive half-cycle, when H increases from zero to its positive maximum value, the energy is stored in the core. This energy is given by the area abda. However, when H decreases from its positive maximum value to zero, the energy is released which is given by the area bdcb. The difference between these two energies is the net loss and is dissipated as heat in the core. Thus, as H varies over one complete cycle, the total energy loss (per cubic metre) is represented by the area abcea of the hysteresis loop. The hysteresis loss (usually expressed in watts) is given as Ph = K h Bmn f V

...(1.9) where

Kh = hysteresis coefficient whose value depends upon the material (Kh = 0.025 for cast steel, Kh = 0.001 for silicon steel) Bm = maximum flux density (in tesla)

n = a constant, 1.5 ≤ n ≤ 2.5 depending upon the material f = frequency (in hertz) V = volume of the core material (in m3) This loss can be minimized by selecting suitable ferromagnetic material for the core. (ii) Eddy-Current Loss: The eddy currents are the circulating currents set up in the core due to alternating magnetic flux (shown with dotted lines in Fig. 1.7a). These currents (shown in Fig. 1.7b) may be quite high since the resistance of the iron is quite low. This results in unnecessary heating of the core and loss of power. The eddy-current loss (in watts) is given by

Pe = K e Bm2 f 2t 2V

...(1.10) where

Ke = a constant dependent upon the material t = thickness of laminations (in metre) The eddy-current loss can be minimized by dividing the solid iron core into thin sheets or laminations insulated from one another (Fig. 1.7c). The path of the induced eddy currents in the core is broken by the insulating material between the sheets. The eddy currents and hence the eddy-current loss is thus substantially reduced.

(a)

(b)

(c)

Fig. 1.7 Laminated core helps in reducing the eddy currents. 3

In Greek, hysterein means ‘to lag’.

7

Note that the eddy-current loss varies as the square of the frequency, whereas the hysteresis loss varies directly with the frequency. The total iron loss is given as

Pi = Ph + Pe ...(1.11) Example 1.8 A single-phase, 230-V/110-V, transformer has iron loss of 100 W at 60 Hz, and 60 W at 40 Hz. Determine the hysteresis and eddy-current losses at 50 Hz. Solution: We know that the hysteresis loss, Ph

∝ f

⇒

the eddy-current loss, Pe ∝ f

and

Ph = Af 2

⇒

Pe = Bf 2

Then, the total iron loss, Pi = Ph + Pe = Af + Bf . Therefore, at the given two frequencies, we have 2

At 60 Hz :

100 = 60 A + 3600 B 60 = 40 A + 1600 B

At 40 Hz : Solving the above two equations, we get A = 1.167 and B = 0.00834. We can now calculate the two losses at 50 Hz, Hysteresis loss at 50 Hz,

Ph = Af = 1.167 × 50 = 58.35 W

Eddy-current loss at 50 Hz, Pe = Bf

2

= 0.00834 × (50) 2 = 20.85 W

1.5 CONSTRUCTION OF TRANSFORMER The main parts of an actual transformer used in power circuits are as follows : (i) An iron core that provides a magnetic circuit. (ii) Two inductive coils wound on the core. These are suitably insulated from each other and also from the core. The individual turns of a coil are also insulated from each other. (iii) A suitable container for the assembled core and windings. (iv) A suitable medium (called transformer oil) for insulating the core and windings from the container. This medium also cools the windings and core of the transformer. (v) Suitable porcelain bushings for insulating and bringing out winding terminals from the tank. Core of Transformer The core is made of steel laminations so as to minimize eddy-current loss. The laminations are about 0.35 mm thick and are insulated from each other by a light coat of varnish on the surface. The laminations are pressed together so as to form a continuous magnetic path, with minimum air gap. Depending upon the construction of the core, there are two types of transformers: (1) Core Type Transformer: In this type, the windings surround a considerable part of the core. Both the windings are divided into two parts and half of each winding is placed on each limb, side by side (shown schematically in Fig. 1.8a). This is done to reduce the leakage of the magnetic flux. Furthermore, the low voltage (LV) winding is placed adjacent to the core and high voltage (HV) winding is placed around the LV winding, as shown in Fig. 1.8b. This is done to minimize the cost of insulation.

(a)

(b)

Fig. 1.8 Core type transformer. (2) Shell Type Transformer: It has three limbs. Both the windings are placed on the central limb (shown schematically in Fig. 1.9a). The LV and HV windings are sandwiched as shown in Fig. 1.9b. Here, the core surrounds a considerable part of the windings.

8

(a)

(b)

Fig. 1.9 Shell type transformer. In core type transformer, the flux has single path. But in shell type transformer, the flux divides equally in the central limb and returns through the outer two legs. Since there is more space for insulation in the core type transformer, it is preferred for high voltages. On the other hand, the shell type construction is more economical for low voltages.

1.6 TRANSFORMER ON LOAD Let us examine what happens when a load is connected to the secondary of the transformer. Note that for simplicity we are still considering a partially ideal transformer (i.e., a transformer satisfying only the ideality conditions (iii) and (iv) stated on page 000). Before connecting the load, there exists a flux Φ in the core due to the no-load current I0 flowing in the primary. On connecting the load, a current I2 flows through the secondary, as shown in Fig. 1.10. The magnitude and phase of I2 with respect to the secondary voltage V2 depends upon the nature of the load.

Fig. 1.10

Transformer on load. The current I2 sets up a flux Φ in the core, which opposes the main flux Φ. This momentarily weakens the main flux, and the primary back emf E1 gets reduced. As a result, the difference V1 - E1 increases and more current is drawn from the supply. This again increases the back emf E1, so as to balance the applied voltage V1. In this process, the primary current increases by I1’. This current is known as primary balancing current, or load component of primary current. Under such a condition, the secondary ’

'

ampere-turns must be counterbalanced by the primary ampere-turns. That is, N1 I1

= N 2 I 2 . Hence, we have

⎛N ⎞ I1' = ⎜ 2 ⎟ I 2 = KI 2 ⎝ N1 ⎠ ...(1.12) The total primary current I1 is the phasor sum of the no-load current I0 and the primary balancing current I1’. That is,

I1 = I 0 + I1'

...(1.13) Phasor Diagram The phasor diagrams for the (partial) ideal transformer with resistive, inductive and capacitive loads are shown in Fig. 1.11. We take flux Ф as reference phasor, as it is common to both the primary and secondary. The induced emfs E1 and E2 lag flux Ф by 90°. Assuming a step-up transformer, E2 > E1. The voltage V2 is same as E2. However, the voltage V1 has the same magnitude as E1 but opposite in phase. Note that the primary balancing current I1’ is always in phase opposition to the secondary current I2. For a step-up transformer, I1’ > I2. Since the no-load current I0 is negligibly small (compared to I1’), the primary current I1 is almost opposite in phase to the secondary current I2. (Note that, in the phasor diagrams of Fig. 1.11, for the sake of clarity the noload current I0 is shown much larger than the practical values.)

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(a) Resistive load.

Fig. 1.11

(b) Inductive load.

(c) Capacitive load.

Phasor diagrams for a transformer on load.

Example 1.9 A single-phase, 440-V/110-V, 50-Hz transformer takes a no-load current of 5A at 0.2 power factor lagging. If the secondary supplies a current of 120 A at a power factor of 0.8 lagging to a load, determine the primary current and the primary power factor. Solution:

φ0 = cos −1 0.2 = 78.46° and φ2 = cos −1 0.8 = 36.87°

The transformation ratio, K =

∴

V2 110 1 = = V1 440 4

I1' = K × I 2 = (1/ 4) ×120 = 30 A

The angle between I 0 and I1' ,θ = φ0 − φ2 = 78.46° − 36.87° = 41.59° ∴

I1 = I 02 + I1'2 + 2 I 0 I1' cos θ = 52 + 302 + 2 × 5 × 30 × cos 41.59° = 33.9 A

The phasor diagram is shown in Fig. 1.12. Note that for the sake of clarity, the phasors are not drawn to the scale. The angle between I0 and I1 is given as

α = tan −1 ∴

I1' sin θ 30sin 41.59° = tan −1 = 35.97° ' I 0 + I1 cos θ 5 + 30 cos 41.59°

angle φ1 = φ0 − α = 78.46° − 35.97° = 42.49°

Thus, Primary power factor =

cos φ1 = cos 42.49° = 0.737

Fig. 1.12

Phasor diagram.

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1.7 PRACTICAL TRANSFORMER ON LOAD We now consider the ideality conditions (iii) and (iv) stated on page 3. The effects of deviations from these conditions become more prominent when a practical transformer is put on load. Effect of Winding Resistance The windings are made of copper wire, since copper has good conductivity. In actual practice, each winding has some resistance. Current flow through the windings causes not only a voltage drop but also a power loss called I2R loss or copper loss. This effect is accounted for by including a resistance R1 in the primary and resistance R2 in the secondary, as shown in Fig. 1.13.

Fig. 1.13 Effect of winding resistance. Effect of Flux Leakage In a practical transformer, the entire magnetic flux does not remain confined to the magnetic core. Not all the flux produced by the primary winding links with the secondary. As shown in Fig. 1.14a, a part of it leaks through the air paths. The difference between the total flux linking with the primary and the useful mutual flux Φu linking with both the windings is called the primary leakage flux, ΦL1. Similarly, ΦL2 represents the secondary leakage flux.

(a) Its definition.

(b) Its effect accounted for.

Fig. 1.14

Leakage flux in a transformer. It is the useful mutual flux Φu that is responsible for the transformer action. The primary leakage flux ΦL1 induces an emf EL1 in the primary winding. Similarly, flux ΦL2 induces an emf EL2 in the secondary. The effect of flux leakage can then be accounted for by including reactances X1 and X2 in the primary and secondary windings, respectively (as shown in Fig. 1.14b), such that

EL1 = I1 X 1

and

EL 2 = I 2 X 2

The reactances X1 and X2 are called primary and secondary leakage reactances, respectively. Note that these reactances are fictitious quantities. These are introduced just as a convenience to represent the effect of the flux leakage. The reluctance of the paths of the leakage fluxes ΦL1 and ΦL2 is almost entirely due to the long air paths and is therefore practically constant. Consequently, the value of the leakage flux is proportional to the current. On the other hand, the value of the useful flux Φu remains almost independent of the load. Furthermore, the reluctance of the paths of the leakage flux is very high. Hence, the value of this flux is relatively small even on full load.

1.8 EQUIVALENT CIRCUIT OF A TRANSFORMER The function of an ideal transformer is to transform electric power from one voltage level to another without incurring any loss and without needing any magnetizing current. For such a transformer, the volt-amperes in the primary are exactly balanced by the volt-amperes in the secondary. An ideal transformer is supposed to operate at 100 percent efficiency. We stated the four conditions that must be satisfied by a transformer to be ideal. We then examined the effects of each of these conditions and explored how to account for the deviations in a practical transformer. Based on this, we can now draw the equivalent circuit of a practical transformer (Fig. 1.15). This circuit is merely a representation of the following KVL equations for the primary and secondary sides of the transformer.

V1 = I1 R1 + jI1 X 1 − E1 = I1 ( R1 + jX 1 ) − E1 ...(1.14) and

E2 = I 2 R2 + jI 2 X 2 + V2 = I 2 ( R2 + jX 2 ) + V2 ...(1.15)

11

Equation 1.14 states that the applied voltage V1 is the phasor sum of the negative of induced emf E1 and the voltage drops in primary resistance, R1, and leakage reactance, X1, due to the flow of current I1. The induced emf E2 forces a current I2 in the secondary circuit. Hence, Eq. 1.15 states that the induced emf E2 is phasor sum of the load voltage V2 and the voltage drops in secondary resistance, R2, and leakage reactance, X2, due to the flow of current I2.

Fig. 1.15 Equivalent circuit of a transformer. Furthermore, it can be seen from Fig. 1.15 that the primary current I1 is composed of two components, the no-load current I0 and the load component of primary current I1’. Moreover, the current I0 consists of two components Iw and Im. The current Iw flows through resistance R0 and accounts for the iron loss of the transformer. The current Im, called magnetizing current, is required to establish working magnetic flux in the core. The ac voltage source connected to the primary winding does not have to supply any power in making this current flow. Hence, in the equivalent circuit, it is shown to flow through a pure reactance X0. Phasor Diagram We can draw the phasor diagram of a transformer circuit with a given load, provided all its parameters (as used in the equivalent circuit of Fig. 1.15) are known. While drawing the phasor diagram, we should keep the following points in mind :

1.

It is most convenient to commence the phasor diagram with the phasor representing the quantity that is common to the two windings, namely, the flux Φ.

2. 3. 4. 5. 6. 7.

The induced emfs E1 and E2 lag behind flux Φ by 90°.

8.

The primary balancing current I1’ and the secondary current I2 are in inverse proportion to the number of turns on the primary and secondary windings.

The values of emfs E1 and E2 are proportional to the number of turns on the primary and secondary windings. The magnitude and phase of the current I2 is decided by the load. The resistive voltage drops are always in phase with the respective current phasor. The inductive voltage drops lead the respective current phasor by 90°. The secondary induced emf E2 is obtained by vector sum of the terminal voltage V2 and the impedance drop I2Z2. Hence, V2 must be drawn such that the phasor sum of V2 and I2Z2 is E2.

9. The primary current I1 is the vector sum of the no-load current I0 and the primary balancing current I1’. 10. The primary voltage V1 is obtained by adding vectorially the impedance drop I1Z1 to the negative of E1. 11. The phase angle φ1 between V1 and I1 is the power factor-angle of the transformer. The phasor diagrams for different types of loads (resistive, inductive and capacitive) are shown in Fig. 1.16.

(a) Resistive load.

(b) Inductive load.

Fig. 1.16

(c) Capacitive load.

Phasor diagrams for different types of loads.

Simplified Equivalent Circuit

12

Since the no-load current I0 of a transformer is only about 3-5 percent of the full-load primary current, not much error will be introduced if the exciting circuit R0-X0 in Fig. 1.15 is shifted to the left of impedance R1-X1. This results in a circuit shown in Fig. 1.17a.

(a) The exciting circuit shifted to the left.

(b) The impedances transferred from secondary side to the primary side.

(c) Equivalent resistance and reactance referred to the primary side.

Fig. 1.17

Simplified equivalent circuit of a transformer as referred to the primary side.

Using the impedance transformation, we can now transform the impedances from the secondary side to the primary side and remove the ideal transformer from the circuit, as shown in Fig. 1.17b. This can further be simplified by combining the two resistances together and the two leakage reactances together, as shown in Fig. 1.17c. Here, the total resistance and total leakage reactance as referred to primary are given as

Re1 = R1 + ( R2 / K 2 )

and

X e1 = X 1 + ( X 2 / K 2 )

...(1.16) Approximate Equivalent Circuit Compared to the full-load primary current, the no-load current of a transformer is very small (only 3-5 percent). Therefore, while considering the behaviour of a transformer on full-load, we can omit the exciting circuit R0-X0 without introducing much error. The resulting approximate equivalent circuit is shown in Fig. 1.18a.

(a) As referred to primary side.

(b) As referred to secondary side.

Fig. 1.18 Approximate equivalent circuit of a transformer. Alternatively, we could transfer the impedances from primary side to the secondary side, so as to get the approximate equivalent circuit as shown in Fig. 1.18b. Here, the total resistance and total leakage reactance as referred to secondary are given as Re 2 = K 2 R1 + R2

and

X e2 = K 2 X1 + X 2

...(1.17)

13

Example 1.10 A single-phase, 50-kVA, 4400-V/220-V, 50-Hz transformer has primary and secondary resistances R1 = 3.45 Ω and R2 = 0.009 Ω, respectively. The values of the leakage reactances are X1 = 5.2 Ω and X2 = 0.015 Ω. Calculate for this transformer (a) the equivalent resistance as referred to the primary, (b) the equivalent resistance as referred to the secondary, (c) the equivalent reactance as referred to the primary, (d) the equivalent reactance as referred to the secondary, (e) the equivalent impedance as referred to the primary, (f) the equivalent impedance as referred to the secondary, (g) the total copper loss first by using the individual resistances of the two windings and then by using the equivalent resistances as referred to each side. Solution: Full-load primary current, I1 =

kVA 50 000 = = 11.36 A V1 4400

Full-load secondary current, I 2 = Transformation ratio, K =

kVA 50 000 = = 227.27 A V2 220

V2 220 1 = = = 0.05 V1 4400 20

(a) Re1 = R1 + ( R2 / K ) = 3.45 + [0.009 /(0.05) ] = 7.05 Ω 2

2

(b) Re 2 = K R1 + R2 = (0.05) × 3.45 + 0.009 = 0.0176 Ω 2

2

(c) X e1 = X 1 + ( X 2 / K ) = 5.2 + [0.015 /(0.05) ] = 11.2 Ω 2

2

(d) X e 2 = K X 1 + X 2 = (0.05) × 5.2 + 0.015 = 0.028 Ω 2

2

(e)

Z e1 = Re21 + X e21 = (7.05) 2 + (11.2) 2 = 13.23 Ω

(f)

Z e 2 = Re22 + X e22 = (0.0176)2 + (0.028) 2 = 0.0331 Ω

(g) Total copper loss = I1 R1 + I 2 R2 = (11.36) × 3.45 + (227) × 0.009 = 909 W 2

2

2

2

By considering equivalent resistances, Total copper loss = I1 Re1 = (11.36) × 7.05 = 909.8 W 2

2

Total copper loss = I 2 Re 2 = (227.27) × 0.0176 = 909 W 2

2

1.9

VOLTAGE REGULATION OF A TRANSFORMER With the increase in the load on a transformer, there is a change in its secondary terminal voltage. The voltage falls if the load power-factor is lagging. It increases if the power factor is leading. The voltage regulation of a transformer is defined as the change in its secondary terminal voltage from no load to full load, the primary voltage being assumed constant. Let V2(0) = secondary terminal voltage at no load, and V2 = secondary terminal voltage at full load. Then, the voltage drop V2(0) - V2 is called the inherent regulation. We can compare this change in voltage with respect to either the no-load voltage or the full-load voltage, and it can be expressed as either per unit basis or percentage basis. Thus, we have (i)

Per unit regulation down = % regulation down =

(ii) Per unit regulation up =

% regulation up =

V2(0) − V2 V2(0) V2(0) − V2 V2(0)

× 100

V2(0) − V2 V2 V2(0) − V2 V2

×100

Normally, when nothing is specified, ‘regulation’ means ‘regulation down’. Approximate Voltage Drop At no load, V1

E1 and E2 = V2(0) .

Also, if K is the transformation ratio, the secondary terminal voltage at no

load, V2(0) = E2 = KE1 = KV1 . Thus, using the equivalent circuit referred to the secondary (shown in Fig. 1.18b), we can draw the phasor diagram of the transformer, as given in Fig. 1.19. In this diagram, OA represents the full-load terminal voltage, V2 AB represents the resistive voltage drop, I2Re2

14

BC represents the reactive voltage drop, I2Xe2 AC represents the total impedance voltage drop, I2Ze2 OC represents the no-load terminal voltage, V2(0) (same as the input voltage referred to the secondary, KV1)

Fig. 1.19 Phasor diagram of a transformer referred to the secondary. With point O as centre and OC as radius, draw an arc which intersects the extended OA at point G, so that OC = OG. From point C drop a perpendicular on OG, which intersects it at F. Thus, we have Exact voltage drop = V2(0) − V2 = OC − OA = OG − OA = AG = AF + FG Here, AF is the approximate voltage drop, which we intend to determine. The error in this approximation is FG. Since the impedance voltage drop I2Ze2 (= AC) is very small compared to the full-load voltage V2 (= OA), the error committed in the determination of the approximate voltage drop would be negligibly small. From point B, drop a perpendicular BE on OG, and draw BD parallel to AF. We can now write

Approximate voltage drop, AF = AE + EF = AE + BD = I 2 Re 2 cos φ + I 2 X e 2 sin φ

...(1.18) In case of leading power factor of the load (instead of lagging), we can determine the approximate voltage drop either by putting φ in place of φ in Eq. 1.18 or by redrawing the phasor diagram for leading power factor, to get

Approximate voltage drop, AF = AE − EF = AE − BD = I 2 Re 2 cos φ − I 2 X e 2 sin φ

...(1.19) Thus, in general, we can write

Approximate voltage drop = I 2 Re 2 cos φ ± I 2 X e 2 sin φ ...(1.20) in which + sign is to be used for lagging power factor and – sign for leading power factor. The voltage drop expressed in percentage is the percent regulation, and is given as

% Regulation =

I 2 Re 2 cos φ ± I 2 X e 2 sin φ × 100 V2(0)

= Vr cos φ ± Vx sin φ ...(1.21)

I 2 Re 2 ×100 V2(0)

where

Vr = % resistive drop =

and

I X Vx = % reactive drop = 2 e 2 × 100 V2(0)

...(1.22)

...(1.23) Exact Voltage Drop Referring to Fig. 1.19, the exact voltage drop is AG. We have already determined the approximate value of voltage drop given by AF. To determine FG, we proceed as follows. For the right angle triangle OFC, we can write

15

OC 2 = OF 2 + FC 2 or

OC 2 − OF 2 = FC 2

or

(OC − OF ) (OC + OF ) = FC 2

or

(OG − OF ) (OC + OF ) = FC 2 FG (2OC ) = FC 2 [Taking OF

or ∴

FG =

OC ]

FC ( DC − DF ) ( DC − BE ) ( I X cos φ − I 2 Re 2 sin φ ) 2 = = = 2 e2 2OC 2OC 2OC 2V2(0) 2

2

2

Thus, for lagging power factor,

Exact voltage drop = AF + FG ( I 2 X e 2 cos φ − I 2 Re 2 sin φ ) 2 = ( I 2 Re 2 cos φ + I 2 X e 2 sin φ ) + 2V2(0) For leading power factor,

Exact voltage drop = ( I 2 Re 2 cos φ − I 2 X e 2 sin φ ) +

( I 2 X e 2 cos φ + I 2 Re 2 sin φ ) 2 2V2(0)

Thus, in general, we have

Exact voltage drop = ( I 2 Re 2 cos φ ± I 2 X e 2 sin φ ) +

or

( I 2 X e 2 cos φ m I 2 Re 2 sin φ ) 2 2V2(0)

...(1.24) ( I R cos φ ± I 2 X e 2 sin φ ) ( I X cos φ m I 2 Re 2 sin φ ) 2 % Exact voltage drop = 2 e 2 × 100 + 2 e 2 × 100 V2(0) 2V2(0) × V2(0) = (Vr cos φ ± Vx sin φ ) +

1 (Vx cos φ m Vr sin φ ) 2 2V2(0)

...(1.25) Keep in mind that upper signs are to be used for lagging power factor, and the lower signs for leading power factor. Condition for Zero Regulation It is possible to obtain zero regulation for a transformer. For this, the voltage drop (as given by Eq. 1.20) from no-load to fullload should be zero. It is possible only if the sign in Eq. 1.20 is negative, i.e., only if the load has leading power factor. Thus, the condition of zero regulation is given as

I 2 Re 2 cos φ − I 2 X e 2 sin φ = 0

tanφ =

or

Re 2 X e2

...(1.26) Also, note that for leading power factor, if the magnitude of the phase angle φ is high, the magnitude of I 2 X e 2 sin φ may become greater than that of I 2 Re 2 cos φ . The regulation then becomes negative. It means that on increasing the load the terminal voltage increases. Condition for Maximum Regulation We can derive the condition for maximum regulation (the worst case) using Eq. 1.20. The maximum value of regulation occurs when the voltage drop is maximum (for which we use + sign). Therefore, the condition of maximum regulation can be obtained by differentiating Eq. 1.20 with respect to the phase angle φ and equating it to zero,

or

d ( I 2 Re 2 cos φ + I 2 X e 2 sin φ ) = 0 dφ X tan φ = e 2 Re 2

⇒

(− I 2 Re 2 sin φ + I 2 X e 2 cos φ ) = 0

...(1.27)

16

Example 1.11 A single-phase, 40-kVA, 6600-V/250-V, transformer has primary and secondary resistances R1 = 10 Ω and R2 = 0.02 Ω, respectively. The equivalent leakage reactance as referred to the primary is 35 Ω. Find the full-load regulation for the load power factor of (a) unity, (b) 0.8 lagging, and (c) 0.8 leading. Solution: Given : R1 = 10 Ω; R2 = 0.02 Ω; Xe1 = 35 Ω

∴

250 = 0.0379 6600 40 000 = 160 A the full-load current, I 2 = 250 Re 2 = K 2 R1 + R2 = (0.0379) 2 × 10 + 0.02 = 0.0343 Ω

and

X e 2 = K 2 X e1 = (0.0379) 2 × 35 = 0.0502 Ω

the turns-ratio, K =

(a) For power factor, cos φ = 1; sin φ = 0. Hence,

∴

% Regulation = =

I 2 Re 2 cos φ + I 2 X e 2 sin φ × 100 V2(0)

160 × 0.0343 ×1 + 0 ×100 = 2.195 % 250

(b) For power factor, cos φ = 0.8 (lagging, φ positive); sin φ = 1 − cos

∴

% Regulation =

2

φ = 0.6 .

Hence,

I 2 Re 2 cos φ + I 2 X e 2 sin φ × 100 V2(0)

160 × 0.0343 × 0.8 + 160 × 0.0502 × 0.6 × 100 = 3.68 % 250 (c) For power factor, cos φ = 0.8 (leading, φ negative); sin φ = −0.6 . Hence, =

∴

% Regulation = =

I 2 Re 2 cos φ − I 2 X e 2 sin φ × 100 V2(0)

160 × 0.0343 × 0.8 − 160 × 0.0502 × 0.6 ×100 = −0.172 % 250

1.10 EFFICIENCY OF A TRANSFORMER Like any other machine, the efficiency of a transformer is defined as

η=

Power output Power output Po = = Power input Power output +Power loss Po + Pl

1.10 There are two types of losses in a transformer: (i) Copper losses or I2R losses in the primary and secondary windings, given as

Pc = I12 R1 + I 22 R2 = I12 Re1 = I 22 Re 2

1.11 The copper losses are variable with current. Let us see by what factor the copper losses get reduced when the load on the transformer decreases. We know that the output power is given as

Po = VI cos φ = VI × pf

⇒

VI =

Po pf 1.12

Thus, for a given load we can find the volt-ampere (VA) of the transformer. Assuming the voltage to remain constant, the current is proportional to the VA of the transformer. Since the copper losses are proportional to the square of current, the value of the copper losses for a given load (and hence for given VA) can be calculated from 2

⎛ VA ⎞ Pc = ⎜ ⎟ Pc (FL) ⎝ VA FL ⎠

17

1.13 (ii) Iron losses or core losses, due to hysteresis and eddy-currents, given by Eqs. 1.7 and 1.8, respectively. That is, Pi = Ph + Pe. Since the maximum value of the flux Φm in a normal transformer does not vary more than about 2 % between no load and full load, it is usual to assume the core losses constant at all loads. The efficiency of a transformer can thus be written as

η=

Po Po V2 I 2 cos φ2 = = Po + Pl Po + Pc + Pi V2 I 2 cos φ2 + I 22 Re 2 + Pi

1.14 Condition for Maximum Efficiency Assuming that the transformer is operating at a constant terminal voltage and a constant power factor, we are interested to know for what load (i.e., what value of I2) the efficiency becomes maximum. To determine this, we first divide the numerator and denominator of Eq. 1.30 by I2, to get

η=

V2 cos φ2 V2 cos φ2 + I 2 Re 2 + Pi / I 2

Obviously, the efficiency will be maximum when the denominator of the above equation is minimum, for which we must have

d (V2 cos φ2 + I 2 Re 2 + Pi / I 2 ) = 0 dI 2 I 22 Re 2 = Pi

or

or

or

Re 2 −

Pi =0 I 22

Pc = Pi 1.15

Thus, the efficiency at a given terminal voltage and load power factor is maximum for such a load current I2 which makes the variable losses (copper losses) equal to the constant losses (iron losses). All-day Efficiency The efficiency defined in Eq. 1.26 is called commercial efficiency. This efficiency is not of much use in case of a distribution transformer. The primary of a distribution transformer remains energized all the time. But the load on the secondary is intermittent and variable during the day. It means that the core losses occur throughout the day, but the copper losses occur only when the transformer is loaded. Such transformers, therefore, are designed to have minimum core losses. This gives them better all-day efficiency, defined below.

ηall-day =

Output energy (in kW h) in a cycle of 24 hours Total input energy (in kW h) 1.16

Example 1.12 For a single-phase, 150-kVA transformer, the required no-load voltage ratio is 5000-V/250-V. Find (a) the number of turns in each winding for a maximum core flux of 0.06 Wb, (b) the efficiency at half rated kVA, and unity power factor, (c) the efficiency at full load, and 0.8 power factor lagging, and (d) the kVA load for maximum efficiency, if the full-load copper losses are 1800 W and core losses are 1500 W. Solution: (a) Using the emf equation, we have

E2 = 4.44 fN 2 Φ m ⇒ and

N2 =

E2 250 = = 18.8(say, 19 turns) 4.44 f Φ m 4.44 × 50 × 0.06

N1 =

5000 E1 N2 = × 19 = 380 turns 250 E2

(b) At half rated-kVA, the current is half the full-load current, and hence the output power too reduces by 0.5. Thus, Output power, Po = 0.5 × (kVA) × (power factor) = 0.5 × 150 × 1 = 75 kW Since copper losses is proportional to the square of current, we have Copper losses, Pc =

(0.5) 2 × (full-load copper loss) = (0.5) 2 × 1800 W = 0.45 kW

Iron losses being fixed, we have Iron losses, Pi = 1500 W = 1.5 kW

∴

η=

Po 75 × 100 = × 100 = 97.47 % Po + Pc + Pi 75 + 0.45 + 1.5

(c) At full load and 0.8 power factor, Output power, Po = (kVA) × (power factor) = 150 × 0.8 = 120 kW Copper losses, Pc = 1800 W = 1.8 kW

18

Iron losses, Pi = 1500 W = 1.5 kW

η=

∴

Po 120 × 100 = × 100 = 97.3 % Po + Pc + Pi 120 + 1.8 + 1.5

(d) Let x be the fraction of full-load kVA at which the efficiency becomes maximum (that is, when the variable copper losses are equal to the fixed iron losses). Then

Pc = Pi

x 2 × 1800 = 1500

or

x = 1500 /1800 = 0.913

Therefore, the load kVA under the condition of maximum efficiency is Load kVA = (Full-load kVA) × x = 150 × 0.913 = 137 kVA Example 1.13 For a single-phase, 200-kVA, distribution transformer has full-load copper losses of 3.02 kW and iron losses of 1.6 kW. It has following load distribution over a 24-hour day: (i) 80 kW at unity power factor, for 6 hours. (ii) 160 kW at 0.8 power factor (lagging), for 8 hours. (iii) No load, for the remaining 10 hours. Determine its all-day efficiency. Solution: (a) For 80 kW load at unity power factor (for 6 hours): Output energy = 80 × 6 = 480 kW h kVA =

Po 80 = = 80 kVA 1 pf

Using Eq. 1.29, we have 2

2

⎛ kVA ⎞ ⎛ 80 ⎞ ⎟ Pc (FL) = ⎜ ⎟ × (3.02) = 0.4832 kW ⎝ 200 ⎠ ⎝ kVA FL ⎠

Copper losses, Pc = ⎜

Iron losses, Pi = 1.6 kW Total losses, Pl = Pc + Pi = 0.4832 kW + 1.6 kW = 2.0832 kW

∴

Total energy losses in 6 hours = 2.0832 × 6 = 12.50 kW h

(b) For 160-kW load at 0.8 power factor (for 8 hours): Output energy = 160 × 8 = 1280 kW h kVA =

∴

Po 160 = = 200 kVA = kVA FL pf 0.8

Copper losses, Pc = Pc (FL) = 3.02 kW Iron losses, Pi = 1.6 kW Total losses, Pl = Pc + Pi = 3.02 kW + 1.6 kW = 4.62 kW

∴

Total energy losses in 8 hours = 4.62 × 8 = 36.96 kW h

(c) For the no-load period of 10 hours: Output energy = 0 Copper losses, Pc = 0 Iron losses, Pi = 1.6 kW Total losses, Pl = Pc + Pi = 0 + 1.6 = 1.6 kW

∴

Total energy losses in 10 hours = 1.6 × 10 = 16 kW h

For 24-hour period: Total output energy, Wo = 480 + 1280 = 1760 kW h Total energy losses, Wl = 12.50 + 36.96 +16 = 65.46 kW h

∴

All-day efficiency, ηall-day =

Wo 1760 × 100 = × 100 = 96.41% Wo + Wl 1760 + 65.46

1.11 AUTOTRANSFORMERS An autotransformer is a special transformer-connection that is useful in power systems, motor starters, variable ac sources, and other applications. Figure 1.20a shows the special connection with the primary and secondary drawn in the usual position. Figure 1.20b shows the autotransformer (in the step-down mode) drawn in a manner that clarifies its function.

19

(a) Special connection of a transformer

(b) Redrawn in the standard way

Fig. 1.20

A two-winding transformer converted into an autotransformer. Note that the primary and secondary windings are connected in series for the new primary; the secondary is the new secondary. Also, the primary and secondary are not electrically isolated from each other. Obviously, the voltage V2 = Vo. From Fig. 1.20b, it is obvious that

Vi = V1 + V2 = Vo =

or

N1 N + N2 V2 + V2 = 1 Vo N2 N2 N2 Vi N1 + N 2

1.17 Hence, the new turns-ratio becomes N 2 : ( N1 + N 2 ) . Thus, we find that an autotransformer works like a potential divider circuit, except that numbers of turns are to be used instead of resistances. The apparent power rating (kVA rating) of the transformer is increased by the special connection, as is illustrated in Example 1.14, given below.

Example 1.14 A single phase, 12-kVA, 120-V/120-V transformer is connected as an autotransformer to make a 240-V/120-V transformer. What is the apparent power rating of the autotransformer? Solution: Figure 1.21 shows the transformer connection with rated voltage and current. The current rating on both primary and secondary windings is

I1 = I 2 =

12 kVA =100 A 120 V

In the autotransformer mode, the input apparent power is 240 × 100 = 24 kVA , and the output apparent power is 120 × 200 = 24 kVA . Thus, the apparent power capacity of the 12-kVA transformer is doubled by the autotransformer connection. In effect, half the apparent power is transformed and half is conducted directly to the secondary side.

Fig. 1.21 Practical Autotransformers In practice, an autotransformer is made by winding a single coil XZ of N1 turns on a magnetic core, as shown in Fig. 1.22a. This winding is excited by a voltage V1, so that a flux is set up in the core and an emf E1 is induced in the winding. The winding is tapped at point Y, such that there are N2 turns between Y and Z. An emf E2 exists between the terminals Y and Z such that the ratio E2/E1 = N2/N1 = K becomes the turns-ratio of the autotransformer. For ideal conditions, the turns-ratio is given as

K=

N 2 V2 = N1 V1

20

If a load is connected across terminals Y and Z, a load-current I2 flows due to the emf E2. The mmf due to current I2 is counterbalanced by the mmf due to current I1. The portion YZ of the winding is common to both the primary and secondary sides. Hence, it is called common winding. The portion XY is called series winding. In variacs (variable autotransformers), point Y is made a sliding contact so as to give a variable output voltage.

(a) A step-down autotransformer.

Fig. 1.22

(b) A step-up autotransformer.

In practice, a single winding is used in making an autotransformer.

(a) Step-down Autotransformer: In Fig. 1.22a, N 2 < N1 , hence V2 < V1. Hence, this arrangement is a step-down autotransformer. The output current I2 is greater than the input current I1. The distribution of currents in the winding is shown in the figure. The apparent power (volt-amperes) on the two sides must be the same delivered to the load as

V1 I1 = V2 I 2 . We can write the volt-amperes

V2 I 2 = V2 I1 + V2 ( I 2 − I1 ) The part V2I1 represents the volt-amperes conductively transferred from ac source to the load through the winding portion XY. Only the remaining part V2 ( I 2 − I1 ) of the total volt-amperes is inductively transferred from ac source to the load through the winding portion YZ. (b) Step-up Autotransformer: In Fig. 1.22b, N 2 > N1 , hence V2 > V1. Hence, this arrangement is a step-up autotransformer. The output current I2 is less than the input current I1. The volt-amperes drawn from the ac source at the input of the autotransformer can be written as

V1 I1 = V1 I 2 + V1 ( I1 − I 2 ) The part V1I2 represents the volt-amperes conductively transferred to the load through the winding portion XY. The remaining part V1 ( I1 − I 2 ) is inductively transferred to the load through the winding portion YZ. Saving in Copper For the same voltage ratio and capacity (volt-ampere rating), an autotransformer needs much less copper (or aluminium) material compared to a two-winding transformer. The cross-sectional area of a conductor is proportional to the current carried by it, and its length is proportional to the number of turns. Therefore, Weight of copper in a winding For a two-winding transformer:

∝ NI = kNI

Weight of copper in primary = kN1 I1 Weight of copper in secondary = kN 2 I 2 Total weight of copper = k ( N1 I1 + N 2 I 2 ) For an autotransformer (see Fig. 1.22a): The portion XY of the winding has carries current I2 - I1. Therefore,

N1 − N 2 turns and carries current I1. The portion YZ of the winding has N 2 turns and

Weight of copper in portion XY = k ( N1 − N 2 ) I1 Weight of copper in portion YZ = kN 2 ( I 2 − I1 ) Total weight of copper = k ( N1 − N 2 ) I1 + kN 2 ( I 2 − I1 ) = k[( N1 − 2 N 2 ) I1 + N 2 I 2 ] Therefore, the ratio of copper- weights for the two cases is

21

⎡ ⎛ N 2 ⎞ ⎤ ⎛ I1 ⎞ ⎛ N 2 ⎞ ⎢1 − 2 ⎜ ⎟⎥ ⎜ ⎟ + ⎜ ⎟ k[( N1 − 2 N 2 ) I1 + N 2 I 2 ] ⎣ ⎝ N1 ⎠ ⎦ ⎝ I 2 ⎠ ⎝ N1 ⎠ [1 − 2 K ]K + K = = = 1− K k ( N1 I1 + N 2 I 2 ) K+K ⎛ I1 ⎞ ⎛ N 2 ⎞ ⎜ ⎟+⎜ ⎟ ⎝ I 2 ⎠ ⎝ N1 ⎠ Evidently, the saving is large if K is close to unity. A unity transformation ratio means that no copper is needed at all for the autotransformer. The winding can be removed all together. The volt-amperes are conductively transformed directly to the load ! Disadvantages: The use of autotransformer has following disadvantages: 1. No electrical isolation between the two sides. 2. Should an open-circuit occur between points Y and Z, full primary high voltage appears across the load. 3. The short-circuit current is larger than that in two-winding transformer. Applications: The autotransformers find applications in following areas: 1. Boosting or buckling of supply voltage by a small amount. 2. Starting of ac machines, where the voltage is raised in two or more steps. 3. Continuously varying ac supply as in variacs.

1.12 THREE-PHASE TRANSFORMERS Modern large transformers are usually of the three-phase core type, schematically shown in Fig. 1.23. Three similar limbs are connected by top and bottom yokes. Each limb has primary and secondary windings arranged concentrically. In Fig. 1.23, the primary is shown star-connected and the secondary delta-connected. In actual practice, the windings may be connected star/delta, delta/star, star/star or delta/delta, depending upon the conditions under which the transformer is to be used.

Fig. 1.23 Three-phase core-type star/delta connected transformer. Example 1.15 A three-phase, 50-Hz transformer has 840 turns on the primary and 72 turns on the secondary winding. The supply voltage is 3300 V. Determine the secondary line voltage on no load when the windings are connected (a) star/delta, (b) delta/star. Solution: (a) For star/delta connection:

VL1 3300 = = 1905.3 V 3 3 72 Secondary phase voltage, Vph2 = 1905.3 × = 163.3 V 840 Secondary line voltage, VL 2 = V ph 2 = 163.3 V Primary phase voltage, Vph1

∴

=

(b) For delta/star connection: Primary phase voltage, V ph1 = VL1 = 3300 V Secondary phase voltage, Vph2 = 3300 ×

∴

72 = 283 V 840

Secondary line voltage,VL 2 = V ph 2 × 3 = 283 × 3 = 490 V

1.13 TRANSFORMER TESTING There are two simple tests that may be conducted on a transformer to determine its efficiency and regulation. These are called open-circuit test and short-circuit test. The power required to carry out these tests is very small compared with the full-load output of the transformer.

22

(1) Open-Circuit Test This test determines the no-load current and the parameters of the exciting circuit of the transformer. The transformer is connected as shown in Fig. 1.24. Generally, the low voltage (LV) side is supplied rated voltage and frequency through an autotransformer (also called a variac). The high voltage (HV) side is left open. The ratio of the voltmeter readings, V2/ V1, gives the transformation ratio of the transformer. The reading of ammeter A, Io, gives the no-load current I0, and its reading is a check on the magnetic quality of the ferromagnetic core and joints. The primary current on no load is usually less than 5 per cent of the full-load current. Hence, the I2R loss on no load is less than 1/400 of the primary I2R loss of full load and is therefore negligible compared with the core loss. Hence the wattmeter reading, Wo, can be assumed to give the core loss of the transformer.

Fig. 1.24 Open-circuit test on a transformer. Various parameters of the transformer can be calculated as under. Pi = Wo ;

I0 = Io ;

Iw =

Wo ; V1

I m = I 02 − I w2 ;

R0 =

V1 ; Iw

X0 =

V1 Im

(2) Short-Circuit Test This test determines the equivalent resistance and leakage reactance of the transformer. The connections are made as shown in Fig. 1.25. Generally, the LV side of the transformer is short-circuited through a suitable ammeter A2. A low voltage is applied to the primary (HV) side. This voltage is adjusted with the help of a variac so as to circulate full-load currents in the primary and secondary circuits. The reading of ammeter A1, Isc, gives the full-load current in the primary winding. On the other hand, the core loss is negligibly small, since the applied voltage (and hence the flux) is less than about one-twentieth of the rated voltage. Hence, the wattmeter reading, Wsc, gives the copper loss (Pc).

Fig. 1.25 Short-circuit test on a transformer. Equivalent resistance, reactance and impedance as referred to the primary side can be calculated as under. Re1 =

Wsc ; I sc2

Z e1 =

Vsc ; I sc

X e1 = Z e21 − Re21

Example 1.16 A single-phase, 50-Hz, 12-kVA, 200-V/400-V transformer gives the following test results: (i) Open-circuit test (with HV winding open) : 200 V, 1.3 A, 120 W (ii) Short-circuit test (with LV winding short-circuited) : 22 V, 30 A, 200 W Calculate (a) the magnetizing current and the core-loss current, and (b) the parameters of equivalent circuit as referred to the low voltage winding. Solution: (a) The wattmeter reading, 120 W, in the open-circuit test gives the core losses. Therefore, the core-loss current is given as

Iw =

Wo 120 W = = 0.6 A V1 200 V

Hence the magnetizing current is given as

I m = I 02 − I w2 = (1.3) 2 − (0.6) 2 = 1.15 A (b) The parameters of the exciting circuit are given by the open-circuit test, as

R0 =

V1 200 V = = 333 Ω I w 0.6 A

and

X0 =

V1 200 V = = 174 Ω I m 1.15 A

23

The short-circuit test gives the equivalent resistance and reactance as referred to the primary side (high voltage winding). From the given specification of the transformer, The transformation ratio, K =

V2 200 V 1 = = V1 400 V 2

The rated full-load current in the high voltage side, I FL =

12 kVA = 30 A 400 V

This confirms that the short-circuit test has been done at the rated full-load. Thus,

Re1 =

∴

Wsc 200 W = = 0.222 Ω I sc2 (30 A) 2

and

Z e1 =

Vsc 22 V = = 0.733 Ω I sc 30 A

X e1 = Z e21 − Re21 = (0.733) 2 − (0.222) 2 = 0.699 Ω

We can now determine the equivalent resistance and reactance as referred to the secondary side (low voltage winding), as 2

⎛1⎞ Re 2 = K 2 Re1 = ⎜ ⎟ × 0.222 = 0.055 Ω ⎝2⎠

2

⎛1⎞ X e 2 = K 2 X e1 = ⎜ ⎟ × 0.699 = 0.175 Ω ⎝2⎠

and

ADDITIONAL SOLVED EXAMPLES Example 1.17 A 25-kVA transformer has 500 turns on the primary and 40 turns on the secondary winding. The primary winding is connected to a 3-kV, 50-Hz ac source. Calculate (a) the secondary emf, (b) the primary and secondary currents on full load, and (c) the maximum flux in the core. Solution: (a) The transformation ratio is given as

K=

∴

N2 40 = = 0.08 N1 500

Secondary emf, E2 = KE1 ≈ KV1 = 0.08 × 3000 = 240 V

(b) The primary and secondary full-load currents are given as

I1 =

kVA 25 kVA = = 8.33 A V1 3 kV

and

I2 =

I1 8.33 A = = 104.125 A K 0.08

(c) The maximum flux in the core is given by emf equation,

Φm =

E1 3000 = = 0.027 Wb 4.44 fN1 4.44 × 50 × 500

Example 1.18 A 230-V, 50-Hz, single-phase transformer has 50 turns on its primary. It is required to operate with a maximum flux density of 1 T. Calculate the active cross-sectional area of the core. Find suitable dimensions for a square core. Solution: From the emf equation, we have

Φm = ∴

E1 230 = = 0.02072 Wb 4.44 fN1 4.44 × 50 × 50

Active core area, A =

Φ m 0.02072 = = 0.02072 m 2 = 207.2 cm 2 Bm 1

Due to the insulation of laminations from each other, the gross area is about 10 % greater than the active area. Thus,

Gross area = 207.2 ×1.1 = 227.92 cm 2 If the core has square cross-section, the side of the square is

a = 227.92 = 15.09 ≈ 15 cm Example 1.19 A single-phase transformer has a core whose cross-sectional area is 150 cm2, operates at a maximum flux density of 1.1 Wb/m2 from a 50-Hz supply. If the secondary winding has 66 turns, determine the output in kVA when connected to a load of 4-Ω impedance. Neglect any voltage drop in the transformer. Solution:

Φ m = Bm A = 1.1× 0.015 = 0.0165 Wb .

24

E2 = 4.44Φm fN 2 = 4.44 × 0.0165 × 50 × 66 = 242 V = V2 (Neglecting the voltage drops) Secondary current, I 2 =

∴ output in kVA =

V2 242 = = 60.5 A ZL 4

V2 I 2 242 × 60.5 = = 14.6 kVA 1000 1000

Example 1.20 A 11-kV/400-V distribution transformer takes a no-load primary current of 1 A at a power factor of 0.24 lagging. Find (a) the core-loss current, (b) the magnetizing current, and (c) the iron loss. Solution: (a) The core-loss current, (b) The magnetizing current, (c) The iron loss,

I w = I 0 cos φ0 = 1.0 × 0.24 = 0.24 A

I m = I 02 − I w2 = (1)2 − (0.24)2 = 0.971 A

Pi = V1 I 0 cos φ0 = 11000 ×1.0 × 0.24 = 2640 W

Example 1.21 A two-winding, step-down transformer has a turns-ratio (N2/ N1) of 0.5. The primary winding resistance and reactance are 2.5 Ω and 6 Ω, whereas the secondary winding resistance and reactance are 0.25 Ω and 1 Ω, respectively. Its magnetizing current and core-loss current are 51.5 mA and 20.6 mA, respectively. While in operation, the output voltage for a load of 25∠30° Ω is found to be 50 V. Determine the supply voltage, the current drawn from the supply and the power factor. Solution: Given:

Z1 = (2.5 + j 6) Ω = 6.5∠67.38° Ω ; Z 2 = (0.25 + j1) Ω = 1.03∠75.96° Ω

Let us take V2 as the reference phasor, i.e.,

V2 = 50∠0° V .

V2 50∠0° V = = 2∠ − 30° A Z L 25∠30° Ω E 2 = V2 + I 2 Z 2 = (50 + j 0) + (2∠ − 30°)(1.03∠75.96°) = 51.45∠1.65° V

∴ I2 =

⎛N ⎞ E1 = E2 ⎜ 1 ⎟ = (51.45∠1.65°) × 2 = 102.9∠1.65° V; − E1 = 102.9∠181.65° V ⎝ N2 ⎠ ⎛N ⎞ I1' = −I 2 ⎜ 2 ⎟ = −(2∠ − 30°) × 0.5 = 1∠150° A ⎝ N1 ⎠ As seen from the phasor diagram of Fig. 1.5b, Im lags –E1 by 90° and Iw is in phase with –E1.

∴ I m = 0.0515∠(181.65° − 90°) = 0.0515∠91.65° A and I w = 0.0206∠181.65° A ∴ I1 = I1' + I m + I w = (1∠150° + 0.0515∠91.65° + 0.0206∠181.65°) = 1.044556∠148.17° A

V1 = −E1 + I1Z1 = 102.9∠181.65° + (1.044556∠148.17°)(6.5∠67.38°) = 108.601∠183.647° V The angle between V1 and I1 = 183.647° − 148.17° = 35.477°

∴ Primary power factor, pf = cos (35.477°) = 0.814 lagging Example 1.22 A single-phase, step-down transformer has turns-ratio of 4. The resistance and reactance of the primary winding are 1.4 Ω and 5.5 Ω, respectively, and those of the secondary winding are 0.06 Ω and 0.04 Ω, respectively. If the LV winding is short-circuited and the HV winding is connected to a 24-V, 50-Hz source, determine (a) the current in the LV winding, (b) the copper loss in the transformer, and (c) the power factor. Ignore the no-load current I0. Solution: Given: K = (1/4)=0.25. The equivalent resistance and reactance as referred to HV side are given as

Re1 = R1 + R2 / K 2 = 1.4 + (0.06) /(0.25) 2 = 2.36 Ω and

X e1 = X 1 + X 2 / K 2 = 5.5 + (0.04) /(0.25) 2 = 6.14 Ω

∴

Z e1 = Re21 + X e21 = (2.36) 2 + (6.14) 2 = 6.578 Ω

(a) The current in the HV winding,

I SC =

VSC 24 = = 3.648 A Z e1 6.578

25

I1 = I1' = 3.648 A

Therefore, ignoring I0, we have

Thus, the current in the LV winding, I 2

= I1' / K = 3.648 /(0.25) = 14.592 A

(b) The copper loss in transformer = I1 Re1 = (3.648) × 2.36 = 31.4 W 2

(c) The power factor = cos φ1 =

2

P 31.4 = = 0.3586 V1 I1 24 × 3.648

Example 1.23 The results of tests conducted on a single-phase, 20-kVA, 2200-V/220-V, 50-Hz transformer are given as under: OC test (HV winding open) : 220 V, 4.2 A, 148 W. SC test (LV winding short-circuited) : 86 V, 10.5 A, 360 W. Determine (a) the regulation and efficiency at 0.8 pf lagging at full load, and (b) the power factor on short-circuit. Solution: (a) From the short-circuit test, we have VSC = 86 V, ISC = 10.5 A, PSC = 360 W. Then,

Z e1 =

∴

VSC 86 V = = 8.19 Ω; I SC 10.5 A

Re1 =

PSC 360 W = = 3.265 Ω 2 I SC (10.5 A) 2

X e1 = Z e21 − Re21 = (8.19) 2 − (3.265)2 = 7.51 Ω VA 20 000 = = 9.09 A V1 2200

The full-load primary current, I1 =

cos φ = 0.8, and sin φ = sin[cos −1 0.8] = 0.6

Since power factor is 0.8, we have

Using Eq. 1.18, we can determine regulation in terms of quantities referred to the primary side, % Regulation =

I1 ( Re1 cos φ + X e1 sin φ ) × 100 V1

=

9.09(3.265 × 0.8 + 7.51× 0.6) × 100 = 2.94 % 2200

The short-circuit test has been conducted for a short-circuit primary current of 10.5 A, whereas full-load primary current is only 9.09 A. Therefore, the full-load copper loss is given as 2

⎛ 9.09 ⎞ ⎟ × 360 = 269.58 W ⎝ 10.5 ⎠

Full-load copper loss = ⎜

The open-circuit test gives the core loss. Hence, Pi = 148 W. The full-load output power, Po = VA × pf = 20 000 × 0.8 = 16 000 W

∴ % Efficiency, η =

Po 16 000 × 100 = × 100 = 97.45 % 16 000 + 269.8 + 148 Po + Pc + Pi

(b) The power factor on short-circuit is given as

pf = cos φSC =

Re1 3.265 = = 0.399 (lagging) Z e1 8.19

Example 1.24 A single-phase, 200-kVA transformer has an efficiency of 98 % at full-load. If the maximum efficiency occurs at three-quarter of full-load, calculate the efficiency at half of full-load current, assuming the power factor to be 0.8 lagging. Solution: At a power factor of 0.8, the full-load output power,

Po = (kVA) × (pf ) = (200 kVA)×0.8 = 160 kW Since the efficiency is only 98 %, the input power, Pin =

Po

η

=

160 kW = 163.26 kW 0.98

Therefore, total losses (copper loss and iron loss) on full load is

Pc + Pi = Pin − Po = 163.26 − 160 = 3.26 kW

…(i)

We know that maximum efficiency occurs when the variable loss (copper loss) equals the fixed loss (iron loss). Hence, we must have 2

⎛3⎞ ⎜ ⎟ Pc = Pi ⎝4⎠

…(ii)

26

Pc = 2.09 kW

Solving Eqs. (i) and (ii), we get

and

Pi = 1.17 kW

Now, at half load, the total loss is given as 2

2

⎛1⎞ ⎛1⎞ ⎟ Pc + Pi = ⎜ ⎟ × 2.09 + 1.17 = 1.69 kW ⎝2⎠ ⎝2⎠ Po (160 / 2) × 100 = = 97.93 % % Efficiency,η half-load = Po + Pl (160 / 2) + 1.69

Total losses, Pl = ⎜

∴

Example 1.25 A single-phase, 150-kVA, 5000-V/250-V, 50-Hz transformer has the full-load copper losses of 1.8 kW and core losses of 1.5 kW. Find (a) the number of turns in each winding for a maximum core flux of 60 mWb, (b) the efficiency at full rated kVA, with power factor of 0.8 lagging, (c) the efficiency at half the rated kVA, with unity power factor, and (d) the kVA load for maximum efficiency. Solution: (a) Since E = 4.44 fN Φ m , the number of turns on the secondary winding,

N2 = ∴

E2 250 = = 19 turns 4.44 f Φ m 4.44 × 50 × 0.06

N1 = N 2

5000 E1 = 19 × = 380 turns 250 E2

(b) At full rated kVA, the current is also full-load. Therefore, Power output, Po = V2 I 2 × cos φ = (150 kVA) × 0.8 = 120 kW Copper losses, Pc = 1.8 kW

∴ % Efficiency,η =

the iron losses, Pi = 1.5 kW

and

Po 120 × 100 = × 100 = 97.32 % 120 + 1.8 + 1.5 Po + Pc + Pi

(c) At half the rated kVA, the current is half the full-load current. Thus, Power output, Po = 0.5 × V2 I 2 × cos φ = 0.5 × (150 kVA) × 1 = 75 kW Copper losses, Pc = (0.5 I 2 ) Re 2 = 0.25 I 2 Re 2 = 0.25 × (1.8 kW)=0.45 kW 2

2

Iron losses, Pi = 1.5 kW

∴ % Efficiency,η =

(Iron losses do not change with load)

Po 75 × 100 = × 100 = 97.47 % 75 + 0.45 + 1.5 Po + Pc + Pi

(d) Let x be the fraction of the full-load kVA at which maximum efficiency occurs. Then, according to the condition for maximum efficiency, we should have

x 2 (copper losses at full load) = Pi

1.5 = 0.913 1.8 Hence, the required kVA load for maximum efficiency = (150 kVA) × 0.913 = 137 kVA or

x 2 ×1.8 = 1.5

⇒

x=

Example 1.26 A single-phase, 50-kVA, 2400-V/240-V, 50-Hz transformer is used to step down the voltage of a distribution system. The low tension voltage is required to be kept constant at 240 V. (a) What load impedance connected to the LV side will be loading the transformer fully at 0.8 power factor lagging? (b) What is the value of this impedance referred to the high voltage side? (c) What is the value of the current referred to the high voltage side? Solution: (a) Since the secondary voltage is required to be constant, the secondary current remains the same whatever be the value of the power factor. The full-load secondary current I2 can be calculated from the kVA ratings,

I2 =

50 kVA = 208.33 A 240 V

Therefore, the load impedance, Z L = (b) Transformation ratio, K =

V2 240 V = = 1.152 Ω I 2 208.33 A

V2 240 = = 0.1 V1 2400

27

The load impedance referred to the primary side,

Z eq = Z L / K 2 = 1.152 /(0.1)2 = 115.2 Ω (c) The current referred to the high voltage side,

I1' = KI 2 = 0.1× 208.33 = 20.833 A Example 1.27 A single-phase, 10-kVA, 2300-V/230-V, 50-Hz transformer is connected as an autotransformer with LT winding in series with the HT winding as shown in Fig. 1.26. The autotransformer is excited from a 2530-V source, and it is fully loaded such that the rated currents of the windings are not exceeded. Determine (a) the current distribution in the windings, (b) the kVA output, (c) the volt-amperes transferred conductively and inductively from the input to the output, and (d) the saving in copper as compared to the two-winding transformer for the same output.

Fig. 1.26 A two-winding transformer connected as autotransformer. Solution:

10 000 = 4.35 A 2300 10 000 Current rating of LT winding = = 43.48 A 230 Current rating of HT winding =

(a) The autotransformer can supply a load current I2 = 43.48 + 4.35 = 47.83 A at 2300 V, as shown in Fig. 1.26. The input current I1 = 43.48 A. The current distribution is shown in the figure.

2300 × 47.83 = 110kVA 1000 (c) The volt-amperes transferred conductively = V2 I1 = 2300 × 43.48 VA = 100 kVA (b) The kVA output =

The volt-amperes transferred inductively = V2 ( I 2 (d) Saving in copper = K =

− I1 ) = 2300 × 4.35 VA = 10 kVA

N 2 V2 2300 = = = 0.909 or 90.9 % N1 V1 2530

28

SUMMARY

1. 2.

A transformer operates on the principle of mutual induction between two coils called primary and secondary.

3.

A transformer is said to be ideal, if it has (i) infinite permeability, (ii) no core losses, (iii) no copper loss, and (iv) no leakage flux.

4.

Transformation ratio, K =

5. 6.

The exciting current or no-load current, I0 has two components: (i) magnetizing current, Im, and (ii) core-loss current, Iw.

7.

The hysteresis loss varies directly with the frequency, and is given as

The emf equation of a transformer: E = 4.44Φ m fN

V2 E2 N 2 I 2 N1 1 2 ; and Z eq1 = Z L 2 / K = = = = V1 E1 N1 I1 N 2 K

The exciting circuit in the equivalent circuit of a transformer has a resistance R0 (which accounts for the core losses) and reactance X0 (which accounts for the required magnetic flux in the core) in parallel.

Ph = K h Bmn f V

8.

The eddy-current loss varies directly with the square of the frequency, and is given as

Pe = K e Bm2 f 2t 2V

9.

In core-type transformer, the windings surround a considerable part of the core, whereas in the shell-type transformer, the core surrounds a considerable part of the windings.

10. On loading a transformer, the flux remains the same but the load current I2 gets reflected in the primary side as primary balancing current or load component of primary current, I1’.

11. To account for the copper loss, resistances R1 and R2 are included in the primary and secondary of the equivalent circuit. 12. To account for the leakage flux, reactances X1 and X2 are included in the primary and secondary of the equivalent circuit. 13. Total resistance and leakage reactance referred to primary are Re1 = R1 + ( R2 / K 2 ) and X e1 = X 1 + ( X 2 / K 2 ) 14. Total resistance and leakage reactance referred to secondary are Re 2 = K 2 R1 + R2 and X e 2 = K 2 X 1 + X 2 15. Per unit regulation down = 16. Per unit regulation up =

V2(0) − V2 V2(0)

V2(0) − V2 V2

17. Approximate voltage drop = I 2 Re 2 cos φ ± I 2 X e 2 sin φ (+ for lagging, −for leading pf). 18. Condition for zero regulation: tanφ =

Re 2 (occurs only for leading pf). X e2

19. Condition for maximum regulation: tan φ = 20. Efficiency,η =

X e2 Re 2

Power output Po = Power output +Power loss Po + Pl

21. Condition for maximum efficiency: Variable losses = Fixed losses, or Pc = Pi 22. An autotransformer has only one winding, a part of which is common to primary and secondary. 23. The OC test determines the following: W V V Pi = Wo ; I 0 = I o ; I w = o ; I m = I 02 − I w2 ; R0 = 1 ; X 0 = 1 V1 Iw Im 24. The SC test determines the following: W V Re1 = 2sc ; Z e1 = sc ; X e1 = Z e21 − Re21 I sc I sc

29

CHECK YOUR UNDERSTANDING Before you proceed to the next Chapter, take this Test. Give yourself two marks for each correct answer and minus one for each wrong answer. If your score is 12 or more, go to the next Chapter; otherwise study this Chapter again. No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Statement When the primary winding of a transformer is connected to an ac source, an emf is induced only in the secondary winding. In a step-up transformer, the current is stepped down. In a core-type transformer, the two windings are put on two separate limbs. The magnetizing current Im leads the mutual flux Φm by 90°. The no-load current in a transformer is about 2 – 5 % of the full-load current. If the power factor of the load on a transformer is lagging, it is possible to make voltage regulation zero. A transformer is designed such that it has maximum efficiency at full load. If it has total losses P at full load, the total losses at half load will be 0.625P. The primary and secondary leakage fluxes are simulated by primary and secondary leakage reactances in the equivalent circuit of a transformer. There is electrical isolation between primary and secondary windings of an autotransformer. In a short-circuit test of a transformer, usually the high voltage side is shorted.

True

False

Marks

Your Score Answers 1. False 6. False

2. True 7. True

3. True 8. True

4. False 9. False

5. True 10. False

REVIEW QUESTIONS

1.

Explain the principle of working of a transformer. What is meant by step-up and step-down transformers? Which quantity is being stepped up or stepped down?

2. 3.

Deduce the emf equation of a transformer.

4. 5.

How do you justify that the voltage per turn is constant for a given transformer.

6. 7.

What is meant by ‘resistance referred to the primary’ and ‘resistance referred to the secondary’?

8. 9.

Explain why the hysteresis loss and the eddy-current loss occur in the core of a transformer.

State the conditions to be satisfied by a transformer to be ‘ideal’. How is the concept of an ‘ideal transformer’ useful in understanding the behaviour of an actual transformer? What do you understand by ‘transformation ratio’? A resistive load is connected across the secondary. What will be its equivalent resistance as referred to the primary? Draw the no-load phasor diagram of a transformer. Express the magnetizing current and the core-loss current in terms of the no-load current and the power factor. Explain how the hysteresis loss and the eddy-current loss are related to the frequency of the ac supply. Also state how these losses can be reduced.

10. Give the constructional differences between a core-type and a shell-type transformer. 11. Assuming that the windings of a practical transformer have no resistance and there is no leakage of flux, draw complete phasor diagram for (a) a resistive load, (b) an inductive load, and (c) a capacitive load.

12. Explain what you understand by leakage flux and leakage reactance in reference to a transformer. 13. Explain how the effects of winding resistance and flux leakage represented in the equivalent circuit of a transformer. 14. Draw the complete equivalent circuit of a practical transformer. Show how this equivalent circuit can further be simplified without introducing much error.

30

15. Draw the equivalent circuit of a transformer with (a) the primary quantities referred to the secondary side, and (b) the secondary quantities referred to the primary side.

16. Define voltage regulation of a transformer. Explain what is meant by (a) inherent regulation, (b) percentage regulation up, and (c) percentage regulation down.

17. Assuming that the impedance voltage drop is quite small compared to the full-load voltage, derive an expression for the

voltage drop from the no-load condition to the full-load condition in a transformer for both the lagging and leading power factors.

18. Show that one can achieve zero voltage-regulation in a transformer only if the load has a leading power factor. Also, derive the condition for zero regulation.

19. Derive the condition for maximum regulation of a transformer. 20. Explain what you understand by the efficiency of a transformer. Deduce the condition for the maximum efficiency. 21. Explain how the all-day efficiency differs from the commercial efficiency in case of a transformer. For what application of a transformer, the all-day efficiency assumes more importance?

22. 23. 24. 25.

Explain how a two-winding transformer can be converted into an autotransformer. Deduce an expression for the amount of copper saved in an autotransformer. State the advantages, disadvantages and applications of an autotransformer. Explain how you can determine the efficiency and regulation of a transformer by conducting ‘open-circuit test’ and ‘shortcircuit test’ on it.

MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correct:

1.

The core of a transformer is assembled with laminated sheets so as to (a) reduce hysteresis loss (b) reduce eddy-current loss (c) reduce both the hysteresis and eddy-current losses (d) ensure good magnetic coupling between primary and secondary windings

2.

A close magnetic coupling between primary and secondary windings results in (a) high commercial efficiency (b) high all-day efficiency (c) good voltage regulation (d) none of the above

3.

The number of turns in the primary winding of a transformer depends on (a) the input voltage (b) the input current (c) both the input voltage and current (d) the kVA

4.

Cooling of transformers is required so as to (a) increase the efficiency (b) reduce the losses (c) reduces the humming (d) dissipate the heat generated in the windings

5.

The emf induced in the windings of a transformer (a) is in phase with the core flux (b) is out of phase with the core flux (c) lags the core flux by π/2 (d) leads the core flux by π/2

6.

A single-phase, 150-kVA, 1100-V/400-V transformer has 100 turns on the secondary winding. The number of turns on its primary winding will be (a) 5500 (b) 2200 (c) 550 (d) 275

7.

A single-phase, 5-kVA, 200-V/100-V transformer delivers 50 A at the rated voltage. The input current will be (a) 25 A (b) more than 25 A (c) less than 25 A (d) 100 A

8.

In a single-phase, 10-kVA, 230-V/1000-V transformer, the no-load current will be about (a) 0.5 A (b) 2 A (c) 10 A (d) 15 A

9.

A single-phase transformer has a turns-ratio of 4:1. If the secondary winding has a resistance of 1 ohm, this resistance as referred to the primary will be (a) 16 Ω (b) 4 Ω (c) 0.25 Ω (d) 0.0625 Ω

10. A transformer is supplying a unity power-factor load. The power factor at the primary terminals will be 31

(a) about 0.95 lagging (c) about 0.8 lagging

(b) about 0.95 leading (d) unity

11. When the secondary winding of a transformer is shorts-circuited, the power factor of the input is (a) unity

(b) about 0.8 leading

(c) about 0.8 lagging

(d) about 0.2 lagging

12. Under no-load condition, the power factor of a transformer is (a) unity

(b) zero

(c) about 0.4 lagging

(d) about 0.4 leading

13. If the full-load copper loss of a transformer is 100 W, its copper loss at half load will be (a) 200 W

(b) 100 W

(c) 50 W

(d) 25 W

14. If the full-load core loss of a transformer is 100 W, its core loss at half load will be (a) 200 W

(b) 100 W

(c) 50 W

(d) 25 W

15. A single-phase transformer is supplying power to a load at a terminal voltage of 11 kV. When the load is disconnected, the terminal voltage becomes 11.5 kV. The voltage regulation of this transformer for this load is (a) 55 % (b) 11.55 % (c) 5 % (d) 2.5 %

16. A transformer operates at maximum efficiency, when (a) its hysteresis loss and eddy-current loss are minimum (b) the sum of its hysteresis loss and eddy-current loss is equal to its copper loss (c) the power factor of the load is leading (d) its hysteresis loss is equal to its eddy-current loss

17. A distribution transformer should be selected on the basis of its (a) all-day efficiency (c) commercial efficiency

(b) regulation (d) all the above

18. In a transformer, the iron losses do not vary with load current because (a) the core area is constant (b) the core flux remains constant (c) the iron losses are equal to the copper losses (d) the iron losses are very small

19. It is economical to use an autotransformer when the turns-ratio is (a) low

(b) high

(c) more than 10

(d) none of the above

20. An auto-transformer steps down voltage V1 to V2. If an open-circuit develops in the common winding, the voltage across the load may become (a) V1

1. b 11. d

2. c 12. c

3. a 13. d

(b) V1 – V2 4. d 14. b

5. c 15. c

(c) V1 +V2 Answers 6. d 7. b 16. b 17. a

(d) V2 8. b 18. a

9. a 19. a

10. a 20. a

PROBLEMS (A) Simple Problems

1.

A single-phase, 250-kVA, 11-kV/415-V, 50-Hz transformer has 80 turns on the secondary. Calculate (a) the approximate values of the primary and secondary currents, (b) the approximate number of primary turns, and (c) the maximum value of the flux. [Ans.: (a) 22.7 A, 602 A; (b) 2121; (c) 23.4 mWb]

2. The primary winding of a 50-Hz transformer has 480 turns and is fed from a 6400-V supply. Determine (a) the peak value of the flux in the core, and (b) the secondary voltage if the secondary winding has 20 turns. [Ans.: (a) 0.06 Wb; (b) 266.4 V]

3. A single phase, 50-Hz 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 cm2. If the primary winding is connected to 240-V, 50-Hz supply, determine (a) the emf induced in the secondary winding, and (b) the maximum flux density in the core. [Ans.: (a) 1200 V; (b) 0.675 T]

4. A 10-kVA, single-phase transformer has its primary connected to a 2000-V supply. It has 60 turns on the secondary winding

and the voltage across it is found to be 240 V. Assuming the transformer to be ideal, calculate (a) the number of turns on its primary winding; (b) the full-load primary and secondary currents. [Ans.: (a) 500; (b) 5 A, 41.67 A]

5. A 200-kVA, 3300-V/240-V, 50-Hz, single-phase transformer has 80 turns on the secondary winding. Assuming an ideal transformer, calculate (a) primary and secondary currents on full load, (b) the maximum value of flux, and (c) the number of primary turns. [Ans.: (a) 60.6 A, 833.33 A; (b) 0.0135 Wb; (c) 1100]

6. A single-phase transformer with 10:1 turns-ratio and rated at 50 kVA, 2400-V/240-V, 50 Hz is used to step down the voltage of a distribution system. The low tension (LT) voltage is to be kept constant at 240 V. Find the value of the load impedance of the LT side so that the transformer is loaded fully. Find also the value of the maximum flux inside the core if the LT side has 23 turns. [Ans.: 1.152 Ω, 0.047 Wb]

32

7. The primary of a single-phase transformer takes 1 A at power factor of 0.4 when connected to a 240-V, 50-Hz supply and the secondary is on open circuit. The number of turns on the primary is twice that on the secondary. A load taking 50 A at a lagging power factor of 0.8 is now connected across the secondary. What is now the value of the primary current? (Neglect the voltage drops in the transformer.) [Ans.: 25.9 A]

8.

A single-phase, 50-Hz transformer has 100 turns on the primary winding and 400 turns on the secondary winding. The net cross-sectional area of the core is 250 cm2. If the primary winding is connected to a 50-Hz, 230-V supply, calculate (a) the emf induced in the secondary winding, and (b) the maximum value of the flux density in the core. [Ans.: (a) 920 V; (b) 0.414 T]

9.

The no-load current of a single-phase transformer is 5.0 A at 0.3 power factor when supplied from a 240-V, 50-Hz source. The number of turns on the primary is 200. Calculate (a) the maximum value of the flux in the core, (b) the core losses, and (c) the magnetizing current. [Ans.: (a) 5.4 mWb; (b) 360 W; (c) 4.77 A]

10. A transformer on no-load takes 1.5 A at a power factor of 0.2 lagging when its primary is connected to a 50-Hz, 230-V supply. Its transformation ratio (N2/N1) is 1/3. Determine the primary current when the secondary is supplying a current of 300 A at a power factor of 0.8 lagging. Neglect the voltage drops in the windings. [Ans.: 14.5 A]

11. The no-load current of a 50-Hz, 230-V transformer is 4.5 A at a power factor of 0.25 lagging. The number of turns on the primary winding is 250. Calculate (a) the magnetizing current, (b) the core loss, and (c) the maximum value of the flux in the core. [Ans.: (a) 4.35 A; (b) 258.75 W; (c) 4.14 mWb]

12. A 100-kVA transformer has 400 turns on the primary and 80 turns on the secondary winding. The primary and secondary resistances are 0.3 Ω and 0.1 Ω, respectively. The primary and secondary leakage reactances are 1.1 Ω and 0.035 Ω, respectively. Calculate the equivalent impedance referred to the primary side. [Ans.: 3.426 Ω]

13. A single-phase, 100-kVA, 1100-V/220-V transformer has following parameters: R1 = 0.1 Ω, X1 = 0.3 Ω, R2 = 0.004 Ω and

X2 = 0.012 Ω. Determine (a) the equivalent resistance and leakage reactance as referred to the high voltage winding, and (b) the equivalent resistance and leakage reactance as referred to the low voltage winding. [Ans.: (a) 0.2 Ω, 0.6 Ω; (b) 0.008 Ω, 0.024 Ω]

14. In a 50-kVA, 11-kV/400-V, single-phase transformer, the iron and copper losses are 500 W and 600 W, respectively under rated conditions. Calculate (a) the efficiency at unity power factor at full load, (b) the load for maximum efficiency, and (c) the iron and copper losses for this load. [Ans.: (a) 97.85 %; (b) 45.64 kVA; (c) 500 W, 500 W]

15. A 10-kVA, 200-V/400-V, 50-Hz, single-phase transformer gave the following test-results: OC test (HV winding open) : 200 V, 1.3 A, 120 W. SC test (LV winding shorted) : 22 V, 30 A, 200 W. Calculate (a) the magnetizing current, and (b) the equivalent resistance and leakage reactance as referred to the low voltage side. [Ans.: (a) 1.15 A; (b) 0.0555 Ω, 0.1745 Ω] (B) Tricky Problems

16. A single-phase, 50-Hz transformer has 30 turns on primary and 350 turns on secondary. The net cross-sectional area of the core is 250 cm2. If the primary winding is connected to a 230-V, 50-Hz supply, calculate (a) peak value of the flux density in the core, (b) the voltage induced in the secondary winding, and (c) the primary current when the secondary current is 100 A. Neglect the losses. [Ans.: (a) 1.3814 T; (b) 2683.33 V; (c) 1166.67 A]

17. A single-phase, 50-Hz, 100-kVA, 2400-V/240-V transformer has no-load current of 0.64 A and core loss of 700 W, when its

high voltage (HV) side is energized at rated voltage and frequency. Calculate the two components of the no-load current. If this transformer supplies a load current of 40 A at 0.8 lagging pf on its low voltage side, determine the primary current and its power factor. Ignore the resistance and leakage reactance drops. [Ans.: 0.5697 A, 0.2917 A; 4.584 A, 0.762 lagging]

18. An ideal 50-Hz, core-type transformer has 100 primary-turns and 200 secondary-turns. The primary rated voltage is 220 V.

If the maximum permissible flux density is 1.2 T, what should be the cross-sectional area? Also, find (a) the secondary voltage, and (b) the primary current in complex form with reference to the secondary voltage vector when secondary delivers −3

a current of 8 A at a lagging power factor of 0.8. [Ans.: 8.26 × 10 m ;(a) 440 V; (b) 16∠143.1° A ] 2

19. A 500-kVA, 11000-V/400-V, 50-Hz, single-phase transformer has 100 turns on the secondary winding. Calculate (a) the approximate number of turns on the primary winding, (b) the approximate values of the primary and secondary current, and (c) the maximum value of the flux in the core. [Ans.: (a) 2750; (b) 45.45 A, 1250 A; (c) 0.018 Wb]

20. A single-phase transformer has a turns-ratio of 144/432 and operates at a maximum flux of 7.5 mWb at 50 Hz. When working on no load, it takes 0.24 kVA at a power factor of 0.26 lagging from the supply. If it supplies a load of 1.2 kVA at a power factor of 0.8 lagging, determine (a) the magnetizing current, (b) the primary current, and (c) the primary power factor. [Ans.: (a) 0.97 A; (b) 5.8 A; (c) 0.731 (lagging)]

21. The primary and secondary windings of a 30-kVA, 6000-V/230-V, single-phase transformer have resistances of 10 Ω and

0.0016 Ω, respectively. The total reactance of the transformer as referred to the primary side is 23 Ω. Calculate the percentage regulation of the transformer when supplying full-load current at a power factor of 0.8 lagging. [Ans.: 2.54 %]

22. A transformer with turns-ratio 8: 1, has the resistances of the primary and secondary windings as 0.85 Ω and 0.012 Ω,

respectively, and the leakage reactances as 4.8 Ω and 0.07 Ω, respectively. Determine the voltage to be applied to the

33

primary to obtain a current of 150 A in the secondary when the secondary terminals are short-circuited. Ignore the magnetizing current. [Ans.: 176.6 V]

23. A single-phase, 50-Hz transformer has a turns ratio 6:1. The primary and secondary winding resistances are 0.90 Ω and

0.03 Ω, respectively, and leakage reactances are 5 Ω and 0.13 Ω, respectively. Find (a) the voltage to be applied to the high voltage side to get a current of 100 A in the low voltage winding on short-circuit, and (b) the primary power factor on shortcircuit. Neglect the no-load current. [Ans.: (a) 164.67 V; (b) 0.2 lagging] (C) Challenging Prolems

24. A single-phase transformer has Z1 = (1.4 + j 5.2) Ω and Z 2 = (0.0117 + j 0.0465) Ω . The input voltage is 6600 V and turns-ratio is 10.6:1. The secondary feeds a load which draws 300 A at 0.8 power factor lagging. Neglecting no-load [Ans.: 600 V, 144 kW] current I0, determine the secondary voltage and the output in kW.

25. A single-phase, 10-kVA, 4000-V/400-V transformer has following parameters: R1 = 13 Ω, X1 = 20 Ω, R2 = 0.15 Ω, X2 = 0.25

Ω, R0 = 12000 Ω and X0 = 6000 Ω. Determine (a) the equivalent resistance and leakage reactance as referred to the primary winding, (b) the input current with secondary terminals open-circuited, and (c) the input current when the secondary supplies a load current of 25 A at a power of 0.8 lagging. [Ans.: (a) 28 Ω, 45 Ω; (b) 0.745∠ − 63.5° A ; (c) 3.18∠ − 42.9° A ]

26. A single-phase, 200-V/2000-V transformer is fed from a 200-V supply. The equivalent winding resistance and leakage reactance as referred to the low voltage side are 0.16 Ω and 0.7 Ω, respectively. The resistance representing core losses is 400 Ω and the magnetizing reactance is 231 Ω. A load impedance of ZL = (596 + j444) Ω is connected across the secondary terminals. Calculate (a) the input current, (b) the secondary terminal voltage, and (c) the primary power factor. [Ans.: (a) 25.96∠ − 40.78° A ; (b) 1859 V; (c) 0.757 lagging]

27. A single-phase, 100-kVA, 2000-V/200-V, 50-Hz transformer has impedance drop of 10 % and resistance drop of 5 %. (a) What is the regulation at full-load 0.8 power factor lagging? (b) At what power factor is the regulation zero? [Ans.: (a) 9.196 %; (b) 0.866 leading]

28. The primary and secondary windings of a 500-kVA, 11-kV/415-V, single-phase transformer have resistances of 0.42 Ω and 0.0019 Ω, respectively. Its core losses are 2.9 kW. Assuming the power factor to be 0.8, calculate its efficiency on (a) full load, and (b) half load. [Ans.: (a) 98.39 %; (b) 98.13 %]

29. For the transformer in Prob. 28, assuming the power factor to be 0.8, find the output at which the efficiency is maximum and calculate its value.

[Ans.: 447 kVA, 98.4 %]

30. A single phase transformer has percentage regulations of 4 and 4.4 for lagging power factors of 0.8 and 0.6, respectively. The full-load copper loss is equal to the iron loss. Calculate (a) the lagging power factor at which the full-load regulation is maximum, and (b) the full-load efficiency at unity power factor. [Ans.: (a) 0.4472 lagging; (b) 96.15 %]

31. A 250-kVA, single-phase transformer has an efficiency of 96 % on full load at 0.8 power factor lagging and on half load 0.8 power factor lagging. Find iron loss and full-load copper loss.

[Ans.: 2.78 kW, 5.56 kW]

EXPERIMENTAL EXERCISE 1.1 Load Test on a Single-Phase Transformer OBJECTIVES:

1. 2. 3. 4.

To determine the polarity of the primary and secondary windings. To find the turns-ratio. To determine the efficiency at different loads.

To determine the voltage regulation. APPARATUS: Single-phase ac power supply 230 V; One single-phase transformer 2 kVA, 220-V/110-V; One Variac 0-270 V, 15 A; One wattmeter 10 A, 230 V; Two voltmeters (MI type) 0-230 V; Two ammeters (MI type) 0-10 A and 0-20 A; One resistive load. CIRCUIT DIAGRAMS: The circuit diagrams are shown in Fig. 1.27.

(a) Circuit for polarity test and voltage ratio test.

34

(b) Circuit for determining the efficiency and the voltage regulation. Fig. 1.27 Circuit diagrams for load test on a single-phase transformer. BRIEF THEORY: (i) When a transformer is used, a terminal of primary (as well as of secondary) winding is alternately positive and negative with respect to other terminal. It becomes important to know the relative polarities of the primary and secondary terminals in situations such as follows: (1) When two single-phase transformers are connected in parallel so as to share the total load on the system. (2) When three single-phase transformers are connected to make a three-phase transformer. The relative polarities can be determined by shorting one of the terminals of the primary and secondary windings (say, P2 and S2 in Fig. 1.27a), and then measuring the voltage across the other terminals of the primary and secondary windings (say, P1 and S1 in Fig. 1.27a). If P1 and S1 have same polarity (say, positive at an instant), the situation is as shown in Fig. 1.28a. Obviously, in such a case, the voltage V3 = V1 – V2. On the other hand, if P1 and S1 have opposite polarity (say, positive and negative, respectively, at an instant), the situation becomes as shown in Fig. 1.28b. Obviously, in such a case, the voltage V3 = V1 + V2.

(a) P1 and S1 of same polarity. (b) P1 and S1 of opposite polarity. Fig. 1.28 Reading of V3 depends on the relative polarity of primary and secondary. (ii) The induced emf in winding is proportional to the number of turns on it. Therefore, the turns-ratio is given as

K=

N 2 E2 = N1 E1

V2 V1

(iii) If Pin is the input power and Po is the output power of a transformer, its efficiency is given as

η=

Po Po = Pin Po + Pl

where, Pl represents copper loss and iron loss in the transformer. The copper loss in the windings is proportional to the square of the current. However, the iron loss (hysteresis and eddy-current loss in the core) remains constant. As the load is increased, the efficiency also increases. It becomes maximum when the variable copper-loss becomes same as the fixed iron-loss. On further increasing the load, the efficiency starts decreasing (Fig. 1.29a).

as

(a) Efficiency versus load-current. (b) Output voltage versus load-current. Fig. 1.29 Both efficiency and output voltage of the transformer vary with load current. (iv) If V2(0) is the output no-load voltage and V2(FL) is the output full-load voltage, the percentage voltage-regulation is defined

35

% Regulation =

V2(0) − V2(FL) V2(FL)

×100 %

PROCEDURE: (i) For Polarity and Turns-Ratio Test: 1. Make connections as given in Fig. 1.27a. 2. Switch on the ac supply and adjust the variac so that voltmeter V1 reads about 100 V. 3. Record the reading of voltmeter V3. 4. If V3 < V1, the terminals P1 and S1 have same polarity. 5. If V3 > V1, the terminals P1 and S1 have opposite polarity. 6. Using variac, vary the input voltage. 7. For various values of the input voltage V1, note down the readings of the output voltage V2. 8. Calculate the ratio V2/V1. 9. Switch off the supply. (ii) For Efficiency and Regulation Test: 1. Make connections as given in Fig. 1.27b. 2. Disconnect the load. 3. Switch on the ac supply and adjust the variac so that voltmeter V1 reads 220 V, the rated value for the given transformer. 4. Note down the reading of the ammeter A1. This gives no-load primary current. 5. Note down the reading of the voltmeter V2. This gives no-load output voltage V(0). 6. Note down the reading of the wattmeter W. This gives the iron loss Pi in the transformer. 7.

Switch on the load in parts till the full-load current [ I 2(FL) = (2 kVA/110 V)=18.18 A ] is reached. For each load,

note down the readings of wattmeter W, voltmeter V2 and ammeter A2. 8. Calculate the efficiency for each value of load and then plot the curve of efficiency versus the load current. 9. Calculate the percentage regulation. 10. Switch off the supply. OBSERVATIONS AND CALCULATIONS: (i) Polarity Test: Reading of the voltmeter V1 = Reading of the voltmeter V2 = Reading of the voltmeter V3 = If V3 < V1, the terminals P1 and S1 have same polarity, and if V3 > V1, the terminals P1 and S1 have opposite polarity. (ii) Table for Turns-Ratio Test: Sr. No.

V1

V2

Turns-Ratio = V 2/V 1

1 2 3 (iii) Table for Efficiency Test:

Sr. No.

W

V1

V2

Output = V 2 I 2

Efficiency = V 2I 1/W

Plot the graph of efficiency versus load current (as in Fig. 1.29a). (iv) Regulation Test: Secondary voltage on no-load, V2(0) = Secondary voltage on full-load, V2(FL) =

∴ % Regulation =

V2(0) − V2( FL ) V2( FL )

×100 % =

RESULTS:

36

1. 2. 3.

The polarities of the primary and secondary windings have been marked. The turns-ratio has been found almost same as the specified value. The efficiency versus load-current curve has been plotted. The maximum efficiency occurs at a load current of _____ amperes. PRECAUTIONS: 1. Before switching on the supply, the zero reading of the wattmeter, voltmeters and ammeters should be checked. 2. The meters of proper range should be selected. 3. While determining efficiency, use an ammeter in series with the wattmeter so as to keep a check that the current through the wattmeter does not exceed its rating. VIVA-VOCE: 1. Can you name some applications of transformers? Ans.: To step-up generated voltage before transmission and to step-down at consumer end. Current and voltage transformers are used in instrumentation. Small size transformers are used in electronics and communication circuits, e.g., radio, TV, etc. 2. Can you name the material normally used for making the core of transformers? Ans.: Cold rolled grain oriented (CRGO) silicon steel. 3. What is the purpose of adding silicon to steel? Ans.: By adding silicon (about 4 %) to the steel, its resistance increase. Hence the eddy-current loss is reduced. 4. Any harm if you add more than 4 % silicon? Ans.: By adding more silicon, the resistance increases further. But, the steel become very brittle. 5. Why do we use laminations for making the core of a transformer? Ans.: To decrease the eddy-current loss. 6. Usually, how much is the thickness of laminations used. Ans.: It is about 0.35 mm. 7. A transformer is rated as 0.5 kVA, 220 V/110 V, 50 Hz. Which winding, primary or secondary, is made of thinner wire? Ans.: Since the primary carries less current, it is made of thinner wire. 8. If I apply 220 V at 40 Hz, how much voltage will you get across the secondary? Ans.: Since voltage transformation ratio does not depend on the frequency of the supply, we will still get 110 V across the secondary. 9. If I apply 220 V at 0 Hz, will I still get 110 V across the secondary? Ans.: No, we will get 0 V, as the flux is not varying at all for zero frequency (it is a dc source). Moreover, the transformer will get burnt due to flow of large current, as there will be no back emf induced in the primary. 10. There are 780 turns on the primary and 240 turns on the secondary winding of a transformer. The primary winding is connected to an ac supply and secondary is connected to a load such that the primary current is 10 A. Now, suppose 100 turns of the secondary winding suddenly get short-circuited. Will the primary current increase or decrease? Ans.: The short-circuited 100 turns of the secondary will draw a large current. Hence the primary current will increase by large amount. EXPERIMENTAL EXERCISE 1.2 Open-Circuit and Short-Circuit Tests on a Transformer OBJECTIVES: 1. To determine the parameters Iw, Im, R0 and X0 of the transformer by conducting open-circuit test on it. 2. To determine the equivalent resistance Req, the equivalent reactance Xeq and the equivalent impedance Zeq of the transformer by conducting short-circuit test on it. APPARATUS: Single-phase ac power supply 220 V; One single-phase transformer 2 kVA, 220-V/110-V; One variac 0-270 V, 20 A; One wattmeter 20 A, 250 V; One voltmeter (MI type) 0-50 V; One voltmeter (MI type) 0-250 V; One ammeter (MI type) 0-2 A; One ammeter (MI type) 0-20 A. CIRCUIT DIAGRAMS: The circuit diagrams are shown in Fig. 1.30.

(a) Circuit for open-circuit test.

37

(b) Circuit for short-circuit test. Fig. 1.30 Circuit diagrams for testing a transformer. BRIEF THEORY: (i) In open-circuit test, one winding is left open and the other winding is applied rated voltage (see Fig. 1.30a). The readings of the wattmeter (Wo), the voltmeter (Vo) and the ammeter (Io) are noted. Since the no-load current is very small, the copper loss can be ignored. Therefore, the wattmeter indicates the iron losses, i.e., Iron losses, Pi = Wo The parameters Iw, Im, R0 and X0 can be calculated as follows:

Iw =

Wo ; Vo

I m = I 02 − I w2 ;

R0 =

Vo ; Iw

and

X0 =

Vo Im

(ii) In short-circuit test, one winding (preferably, low voltage) is short-circuited and the other winding is applied a low voltage such that full-load current flows in the windings. The readings of the wattmeter (Wsc), the voltmeter (Vsc) and the ammeter (Isc) are noted. Since the applied voltage is low, the iron losses are negligibly small compared to the copper loss. Therefore, the wattmeter indicates the full-load copper loss, i.e., Copper loss, Pc = Wsc The equivalent resistance Req, the equivalent reactance Xeq and the equivalent impedance Zeq, as referred to the winding on which the measurements are made, can be calculated as follows:

Re1 =

Wsc ; I sc2

Z e1 =

Vsc ; I sc

and

X e1 = Z e21 − Re21

PROCEDURE: (i) For Open-Circuit Test: 1. Connect the circuit as shown in Fig. 1.30a. 2. Put the variac at low voltage output. 3. Switch on the ac supply. 4. Adjust the variac to the rated voltage of the transformer. 5. Record the ammeter, wattmeter and voltmeter readings. 6. Switch off the supply. (ii) For Short-Circuit Test: 1. Connect the circuit as shown in Fig. 1.30b. 2. Put the variac at lowest voltage output. 3. Slowly increase the applied voltage till the ammeter reading equals the rated value. 4. Record the ammeter, wattmeter and voltmeter readings. 5. Switch off the supply. OBSERVATIONS: (i) Open-Circuit Test: Reading of the ammeter, Io = Reading of the wattmeter, Wo = Reading of the voltmeter, Vo = (ii) Short-Circuit Test: Reading of the ammeter, Isc = Reading of the wattmeter, Wsc = Reading of the voltmeter, Vsc = CALCULATIONS: (i)

Iw =

Wo = Vo

=

A;

I m = I 02 − I w2 =

( ) −( ) 2

2

=

A;

38

(ii)

R0 =

Vo = Iw

Re1 =

Wsc = I sc2

=

Ω; =

X0 =

Ω;

X e1 = Z e21 − Re21 =

Vo = Im

Z e1 =

( ) −( ) 2

2

=

=

Vsc = I sc

Ω. =

Ω;

Ω.

RESULTS: 1. The values of the equivalent circuit parameters are as follows:

R0 =

Ω;

X0 =

Ω;

Re1 =

Ω;

X e1 =

Ω;

Z e1 =

Ω.

The iron-loss current, Iw = A. The magnetizing current, Im = A; The no-load current, I0 = A. PRECAUTIONS: 1. Before switching on the supply, the zero reading of the wattmeter, voltmeters and ammeter should be checked. 2. The meters of proper range should be selected. 3. While conducting the short-circuit test, the voltage applied should be initially set at zero. It should then be increased slowly. If, by mistake, higher voltage than needed is applied there is every possibility that the transformer may be damaged. VIVA-VOCE: 1. What is the importance of open-circuit and short-circuit tests on a transformer? Ans.: The purpose of these tests is to find the iron loss, copper loss and hence the efficiency of the transformer at different loads. This is an indirect method of testing, since the transformer is not actually loaded during the testing. Especially, for large size transformers this type of indirect testing is important as the arrangement for actual loading just cannot be done. 2. How do you justify that the reading of the wattmeter in open-circuit test indicates the iron losses? Ans.: The iron losses (hysteresis and eddy-current) depend upon the magnetization level of the core, which in turn depends upon the voltage applied. In open-circuit test, full rated voltage is applied to the transformer. Hence, the iron losses occur at full rated value. Furthermore, as the secondary winding is open, the secondary current is zero. Hence there is no copper loss in the secondary winding. However, in the primary winding, there is small no-load current flowing. The copper loss in the primary due to this small current is negligible. Hence, the wattmeter reading gives the iron losses. 3. What does the reading of wattmeter indicate in the short-circuit test? Justify your answer. Ans.: It indicates the full-load copper losses of the transformer. In short-circuit test, full rated current is made to flow through both the secondary and primary windings, by applying very low voltage to the primary. Because of low voltage the magnetization level of the core is very low. Hence the iron losses are negligibly small. Thus, the wattmeter indicates only the full-rated copper losses of the transformer. 4. Suppose that the full rated voltage is applied in the short-circuit test. What do you expect to happen? Ans.: Since the secondary winding is shorted, a heavy current will be drawn by the transformer. As a result, the circuit breaker should trip off. If it does not, the transformer will get excessively overheated resulting in burning of the insulation. 5. How do the iron losses vary with the load on the transformer? Ans.: Iron losses do not depend on the load. These losses depend only on the magnetization level of the core, which in turn depend upon the voltage applied. 6. How do the copper losses vary with the load on the transformer? Ans.: Copper losses or I2R losses depend on the square of the currents flowing in the windings. 7. What is the phasor relationship between Iw, Im and I0? 2. 3.

Ans.: The current I0 is the phasor sum of Iw and Im. I 0 = 8. 9.

I w2 + I m2 . The iron-loss component Iw is in phase with the

applied voltage. The magnetizing component Im lags behind the applied voltage by 90°. What is the magnitude of no-load current as compared to the full-load current? Ans.: The no-load current is about 3 – 5 % of the full-load current. Of what order is the power factor of a transformer under no-load condition? Ans.: It is about 0.2 lagging. **********

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