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Tuesday, 24 March 2015

Electric Field and Magnetic Field

When charges are separated, a space is created where forces are exerted on the charges. An electric field is such a space. Depending upon the polarity of the charges, the force is either attractive or repulsive. Therefore, we can say that static charges generate an electric field. An electric field influences the space surrounding it. Electric field strength is determined in terms of the force exerted on charges. A capacitor is a reservoir of charge. The two parallel plates of a capacitor, when connected to a voltage source, establishes an electric field between the plates. The positive terminal, or pole of the voltage source will draw electrons from plate 1 whereas the negative pole will push extra electrons on to plate 2. Voltage across the capacitor will rise. The capacitor gets charged equal to the voltage of the source. The capacitance of a capacitor is a measure of its ability to store charge. The capacitance of a capacitor is increased by the presence of a dielectric material between the two plates of the capacitor.

A current-carrying conductor or a coil produces magnetic field around it. The strength of the magnetic field produced depends on the magnitude of the current flowing through the conductor or the coil. There is presence of magnetic field around permanent magnets as well.

A magnet is a body which attracts iron, nickel, and cobalt. Permanent magnets retain their magnetic properties. Electromagnets are made from coils through which current is allowed to flow. Their magnetic properties will be present as long as current flows through the coil.

The space within which forces are exerted by a magnet is called a magnetic field. It is the area of influence of the magnet.

Atomic Structure and Electric Charge

Several theories have been developed to explain the nature of electricity. The modern electron theory of matter, propounded by scientists Sir Earnest Rutherford and Niel Bohr considers every matter as electrical in nature. According to this atomic theory, every element is made up of atoms which are neutral in nature. The atom contains particles of electricity called electrons and protons. The number of electrons in an atom is equal to the number of protons.

The nucleus of an atom contains protons and neutrons. The neutrons carry no charge. The protons carry positive charge. The electrons revolve round the nucleus in elliptical orbits like the planets around the sun. The electrons carry negative charge. Since there are equal number of protons and electrons in an atom, an atom is basically neutral in nature.

If from a body consisting of neutral atoms, some electrons are removed, there will be a deficit of electrons in the body, and the body will attain positive charge. If neutral atoms of a body are supplied some extra electrons, the body will attain negative charge. Thus, we can say that the deficit or excess of electrons in a body is called charge.

Charge of an electron is very small. Coulomb is the unit of charge. The charge of an electron is only 1.602 × 10^–¹⁹ Coulomb (C). Thus, we can say that the number of electrons per Coulomb is the reciprocal of 1.602 × 10^–19 which equals approx. 6.28 × 10¹⁸ electrons. Therefore, charge of 6.28 × 10¹⁸ electrons is equal to 1C. When we say that a body has a positive charge of 1C, it is understood that the body has a deficit of 6.28 × 10¹⁸ electrons.

Any charge is an example of static electricity because the electrons or protons are not in motion. You must have seen the effect of charged particles when you comb your hair with a plastic comb, the comb attracts some of your hair. The work of combing causes friction, producing charge of extra electrons and excess protons causing attraction.

Charge in motion is called electric current. Any charge has the potential of doing work, i.e., of moving another charge either by attraction or by repulsion. A charge is the result of separating electrons and protons. The charge of electrons or protons has potential because it likes to return back the work that was done to produce it.

SUMPNER’S TEST

To determine the rise of maximum temperature of a transformer, its load test is of utmost importance. Using suitable load impedance, small transformers can be put on full load. The full-load test of large transformers is not possible because considerable wastage of energy occurs and it is difficult to get a suitable load for absorbing full-load power. Sumpner’s test is used to put large transformer on full load. This test can also be used to determine the efficiency of a transformer. Figure 1.45 shows the schematic diagram of Sumpner’s test. This test is also known as back to back test or load test.

This test requires two identical transformers. The two primaries are connected in parallel and are energized at rated voltage and rated frequency. The wattmeter W₁ records the reading of core loss of both the transformers. Next the two secondaries are connected in series in such a way that their polarities are in phase opposition and the reading of the voltmeter V₂ becomes zero. With the help of voltage regulator fed from source, a voltage is injected to the secondary of the transformers, which is adjusted until the rated secondary current flows. The voltmeter reads a voltage, which is the leakage impedance drop of the two transformers. The reading of W₁ remains unaltered. The wattmeter W₂ reads the total copper (Cu) loss of the two transformers. Although the transformers are not supplying any load current, this test measures the full iron loss as well as copper loss of the transformers. The net input during this test is W₁ + W₂. To measure the temperature rise, the two transformers are kept under rated loss conditions for several hours.

Figure 1.45 Sumpner’s Test

POLARITY TEST OF A SINGLE-PHASE TRANSFORMER

Polarity testing of transformers is vital before connecting them in parallel. Otherwise, with incorrect polarity, it is not possible to connect them in parallel. The rated voltage is applied to the primary and its two terminals are marked as A₁ and A₂, respectively, as shown in Figures 1.44(a) and 1.45(b), respectively. The secondary winding terminals are also marked as a₁ and a₂, shown in Figures 1.44(a) and 1.45(b), respectively. Now a voltmeter is connected across A₂ and a₂. if it measures the difference of E₁ and E₂, A₂ and a₂ are of the same polarity. If it measures the addition of E₁ and E₂, A₂ and a₂ are of opposite polarity.

Figure 1.44 Polarity Test of a single-phase Two Winding Transformer

ALL-DAY EFFICIENCY

The ratio of output in watts to input in watts is called commercial efficiency of a transformer. Distribution transformers are used for supplying lighting and general networks. Distribution transformers are energized throughout the day. Their secondaries are at no load most of the time in a day except during the hours of lighting period. Core loss occurs throughout the day. Copper loss occurs only when they are loaded and hence is less important. To judge their performance, all-day efficiency or operational efficiency is calculated. The all-day efficiency is defined by

The all-day efficiency is less than the commercial efficiency of a transformer.

Example 1.16 A 200 kVA single-phase transformer is in circuit throughout 24 hours. For 8 hours in a day, the load is 150 kW at 0.8 power factor lagging and for 7 hours, the load is 90 kW at 0.9 power factor. Remaining time or the rest period, it is at no-load condition. Full-load Cu loss is 4 kW and the iron loss is 1.8 kW. Calculate the all-day efficiency of the transformer.

Solution

Full-load output = 200 kVA, Full-load Cu loss = 4 kW, Iron loss = 1.8 kW.

EFFICIENCY OF A TRANSFORMER

Due to the losses in a transformer, its output power is less than the input power.

∴ Power output = Power input – Total losses

∴ Power input = Power output + Total losses = Power output + P_i + P_Cu

The ratio of power output to power input of any device is called its efficiency (η).

Output power of a transformer at full-load = V₂I_2ftcosθ, where cosθ is the power factor of the load, I_2ft is the secondary current at full load and V₂ is the rated secondary voltage of the transformer.

Full-load copper loss of the transformer = I_2ftR₀₂.

∴ Efficiency of the transformer at full load is given by

Now V₂I_2ft = VA rating of the transformer.

∴

i.e.,

Efficiency of the transformer at any load m is given by

where m=

and P_Cuft is the Cu loss of the transformer at full load.

1.34 CONDITION FOR MAXIMUM EFFICIENCY

During working of a transformer at constant voltage and frequency, its efficiency varies with the load. Its efficiency increases as the load increases. At a certain load, its efficiency becomes maximum. If the transformer is further loaded, its efficiency starts decreasing. Figure 1.43 shows the plot of efficiency versus load current.

Figure 1.43 Comparison Efficiency and Current

To determine the condition of maximum efficiency, let us assume that the power factor of the load remains constant and the secondary terminal voltage (V₂) is constant. Therefore, efficiency becomes only a function of load current (I₂).

For maximum efficiency

Now,

∴

i.e.,

i.e., V₂I₂cosθ+P_i+I₂²R₀₂–V₂I₂cosθ–2I₂²R₀₂=0

i.e., P_i=I₂²R₀₂ (1.66)

To achieve maximum efficiency, Iron loss = Cu loss

i.e., Constant loss = Variable loss

1.34.1 Load Current at Maximum Efficiency

Let I_2M be the load current at maximum efficiency.

∴ I_2M²R₀₂=P_i

i.e.,

Let I_2ft be the full-load current.

∴

i.e.,

Equation (1.67) shows the load current in terms of full-load current at maximum efficiency.

1.34.2 kVA Supplied at Maximum Efficiency

For constant V₂ the kVA supplied is the function of load current only.

∴

In general,

where

LOSSES IN A TRANSFORMER

Two types of losses occur in a transformer:

Core loss or iron loss occurs in a transformer because it is subjected to an alternating flux.
The windings carry current due to loading and hence copper losses occur.

1.32.1 Core or Iron Loss

The separation of core losses has already been introduced. The alternating flux gets set up in the core and it undergoes a cycle of magnetization and demagnetization. Therefore, loss of energy occurs in this process due to hysteresis. This loss is called hysteresis loss (P_h), which is expressed by

P_h=K_hB_m^1.6fV W (1.59)

where K_h is the hysteresis constant depending on the material, B_m is the maximum flux density, f is the frequency and V is the volume of the core.

The induced emf in the core sets up eddy current in the core, and hence eddy current loss (P_e) occurs, which is given by

P_e=K_eB_m²f²t² W per model W (1.60)

where K_e is the eddy current constant and t is the thickness of the core.

Since the supply voltage V₁ at rated frequency f is always constant, the flux in the core is almost constant. Therefore, flux density in the core remains constant. Hence, hysteresis and eddy current losses are constant at all loads. Thus, the core loss or iron loss is also known as constant loss. The iron loss is denoted by P_i.

Iron loss is reduced using high-grade core material such as silicon steel having very low hysteresis loop for reducing hysteresis loss and laminated core for reducing the eddy current loss.

1.32.2 Copper Loss

The loss of power in the form I²R due to the resistances of the primary and secondary windings is known as copper losses. The copper loss also depends on the magnitude of currents flowing through the windings. The total Cu loss is given by

Copper losses are determined on the basis of R₀₁ or R₀₂ which is determined from short circuit test. Since the standard operating temperature of electrical machine is taken as 75 °C, it is then corrected to 75 °C.

The copper loss due to full-load current is known as full-load Cu loss. If the load on the transformer is half, the Cu loss is known as half-load Cu loss, which is less than the full-load Cu loss. The Cu loss is also known as variable loss.

There are two other losses known as stray loss and dielectric loss. Since leakage field is present in a transformer, eddy currents are induced in the conductors, tanks walls and bolts etc. Stray losses occur due to this eddy currents. Dielectric loss occurs in insulating materials coil and solid insulation. These two losses are small and hence neglected.

Therefore, the total loss of the transformer = Iron loss + Cu loss = P_i + P_Cu

Kapp’s Regulation

Kapp had designed a diagram shown in Figure 1.42 to determine the regulation at any power factor. The description of the construction of the diagram is shown below. images

Figure 1.42 Kapp’s Diagram

Load current (I₂) is taken as a reference phasor. OA representing V₂ is drawn at angle θ₂ with I₂. AB represents I₂R₀₂ drawn parallel to I₂, whereas BC represents I₂X₀₂ drawn perpendicular to AB, i.e., I₂. Here OC represents secondary emf (₀V₂ = E₂) at no-load. The circle 1 known as often circuit EMF circle is drawn with O as centre and OC as radius. The line OO^′ is drawn parallel to AC representing I₂Z₀₂. With O^′ as centre and OA as radius, the circle 2 known as terminal voltage circle is drawn, which intersects with circle 1 at the points D and E. The region above and below the reference line represents the lagging and leading power factors region, respectively. The point D is the point corresponding to zero regulation. The intercept FG gives the maximum regulation, which is drawn through O and drawn parallel to AC. The regulation at any power factor angle θ is obtained by extending OA to meet the outer circle at H. The regulation at the required power factor cosθ is represented by AH, which is the intercept between the two circles.

CALCULATION FOR VOLTAGE REGULATION

The voltage regulation up is expressed mathematically by

Positive sign is for lagging power factor and negative sign is for leading power factor.

1.31.1 Zero Voltage Regulation

For lagging power factor and unity power factor, ₀V₂ > V₂. Therefore, we get positive voltage regulation. For leading power factor, V₂ starts increasing. At a certain leading power factor, ₀V₂ = V₂ and hence regulation becomes zero. If the load power factor is further increased, ₀V₂ becomes less than V₂ and hence regulation becomes negative. For zero voltage regulation, we have

₀V₂ – V₂ = 0

i.e., I₂(R₀₂cosθ–X₀₂sinθ)=0

Equation (1.56) shows the leading power factor at which voltage regulation becomes zero.

1.31.2 Condition for Maximum Voltage Regulation

Maximum voltage regulation can be obtained for lagging power factor. For maximum voltage regula-tion, we have

i.e., –R₀₂sinθ+X₀₂cosθ=0

i.e.,

This is satisfied only when the power factor of the load is lagging. The regulation is maximum when the load power factor angle is equal to the impedance angle of the transformer.

i.e.,

Equation (1.58) shows the power factor of the load at which voltage regulation is maximum.

Figure 1.41 shows the variation of regulation with the power factor of the load. Figure 1.41 also shows that regulation becomes maximum when the power factor of the load is lagging and it becomes zero when the power factor is leading.

VOLTAGE REGULATION OF TRANSFORMER

With constant voltage applied in primary, the secondary terminal voltage will decrease due to voltage drop across its internal resistance and leakage reactance.

Let ₀V₂ and V₂ be the secondary terminal voltages at no load and on load respectively. There are three kinds of voltage regulation, which are discussed below.

1.30.1 Inherent Voltage Regulation

The difference ₀V₂ – V₂ is known as inherent voltage regulation of the transformer.

1.30.2 Voltage Regulation Down

If inherent voltage drop is divided by ₀V₂, it is known as voltage regulation down. Mathematically, we can write

and also

1.30.3 Voltage Regulation Up

If inherent voltage drop is divided by V₂, it is known as voltage regulation up. Mathematically, we can write it as

and also

The secondary terminal voltage not only depends on load current but also on the power factor of the load. The regulation is said to be at full load provided V₂ is determined for full load and at specified power factor condition. V₂ drops more and more with increasing load current. For lagging power factor load, V₂ < E₂, the voltage regulation is positive. For leading power factor load, V₂ > E₂, the voltage regulation is negative.

To maintain constant secondary terminal voltage on load, the primary terminal voltage is adjusted. It is expected that voltage drop would be as small as possible. Therefore, the lesser the value of regula-tion, the better is the performance of a transformer.

PER UNIT RESISTANCE, LEAKAGE REACTANCE AND IMPEDANCE VOLTAGE DROP

Full-load voltage of a transformer can be expressed as a fraction of the full-load terminal voltage.

Let I_1fl be the full-load primary current, I_2fl be the full-load secondary current, V₁ be the rated pri-mary voltage and V₂ be the rated secondary voltage.

Per unit resistance drop of a transformer

∴

Per unit reactance drop of a transformer

Per unit reactance drop of a transformer is called per unit reactance and it is given by

Per unit impedance drop of a transformer is called per unit impedance and it is given by

EXACT VOLTAGE DROP

In Figure 1.40, the exact voltage drop is AH instead of AG. During calculation of approximate voltage drop, AG has already been calculated. If GH is being added to AG, the exact voltage drop can be obtained.

Consider the right-angled triangle OCG. We have

OC² = OG² + GC²

i.e. OC² – OG² = GC²

i.e. (OC – OG)(OC + OG) = GC²

i.e. (OH –OG)(OC + OG) = GC²

i.e. GH.2.OC= GC² [Taking OC ≍ OG]

i.e.

For lagging power factor, the exact voltage drop is

For leading power factor, the exact voltage drop is given by

In general, the voltage drop is

and percentage voltage drop is

It may be noted that the upper sign is to be used for lagging power factor and the lower sign for leading power factor.

Example 1.11 The OC and SC tests on a 300/600 V, 50 Hz, single-phase transformer gave the fol-lowing results:

OC test (LV side): 300 V, 0.8 A, 70 W

SC test (HV side): 20 V, 12 A, 90 W

where LV and HV is low voltage and high voltage, respectively.

Find the equivalent circuit of the transformer referred to as LV side and also calculate the secondary voltage when delivering 6 kW at 0.8 p.f. (power factor) lagging.

Solution

OC test (LVside):

Instruments are placed on LV side and HV side is kept open.

V₁ = 300 V, I₀ = 0.8 A, W₀ = 70 W

Now,