الفلسطينية الوطنية السلطة

العالي والتعليم التربية وزارة

والتكنولوجيا العلوم كلية

التطبيقية والفنون الهندسية العلوم قسم

المباني هندسة بكالوريوس

التطبيقية الكهرباء لمساق المنهجي الكتاب

جبر أحمد : م. سمير إعداد

الثاني الفصل2007/2008

TABLE OF CONTENTS:

UNIT 1 : AC, DC and Electrical Signals pp 3UNIT 2 : DC Laws and Theorems pp 16UNIT 3 : Single Phase AC pp 28UNIT 4 : Three Phase AC System pp 50UNIT 5 : Electrical Transformers pp 58 UNIT 6 : Electrical Motors pp 72 UNIT 7 : Switching and Controlling of AC Motors pp 101UNIT 8 : Introduction To Protection pp 113

2

UNIT 1 AC, DC and Electrical Signals

AC means Alternating Current and DC means Direct Current. AC and DC are also used when referring to voltages and electrical signals which are not currents! For example: a 12V AC power supply has an alternating voltage (which will make an alternating current flow). An electrical signal is a voltage or current which conveys information, usually it means a voltage. The term can be used for any voltage or current in a circuit.

1.1 Direct Current (DC)

Direct Current (DC) always flows in the same direction,

but it may increase and decrease.

A DC voltage is always positive (or always negative),

but it may increase and decrease. Electronic circuits

normally require a steady DC supply which is constant

at one value or a smooth DC supply which has a small

variation called ripple.

Cells, batteries and regulated power supplies provide

steady DC which is ideal for electronic circuits.

Power supplies contain a transformer which converts the

mains AC supply to a safe low voltage AC. Then the AC

is converted to DC by a bridge rectifier but the output is

varying DC which is unsuitable for electronic circuits.

Some power supplies include a capacitor to provide

smooth DC which is suitable for less-sensitive electronic

circuits, including most of the projects on this website.

Lamps, heaters and motors will work with any DC supply.

Applications

Direct-current installations usually have different types of sockets, switches, and fixtures, mostly due to the low voltages used, from those suitable for alternating current. It is usually important with a direct-current appliance not to reverse polarity unless the device has a diode bridge to correct for this (most battery-powered devices do not).

3

Steady DCfrom a battery or regulated power supply,

this is ideal for electronic circuits.

Smooth DCfrom a smoothed power supply,

this is suitable for some electronics.

Varying DC

from a power supply without smoothing,this is not suitable for electronics.

This symbol is found on many electronic devices that either require or produce direct current

DC is commonly found in many low-voltage applications, especially where these are powered by batteries, which can produce only DC, or solar power systems, since solar cells can produce only DC. Most automotive applications use DC, although the alternator is an AC device which uses a rectifier to produce DC. Most electronic circuits require a DC power supply. Applications using fuel cells (mixing hydrogen and oxygen together with a catalyst to produce electricity and water as byproducts) also produce only DC.

Many telephones connect to a twisted pair of wires, and internally separate the AC component of the voltage between the two wires (the audio signal) from the DC component of the voltage between the two wires (used to power the phone).

Telephone exchange communication equipment, such as DSLAM, uses standard -48V DC power supply. The negative polarity is achieved by grounding the positive terminal of power supply system and the battery bank. This is done to prevent electrolysis depositions.

An electrified third rail can be used to power both underground (subway) and overground trains.

1.2 Alternating Current (AC)

Alternating Current (AC) flows one way, then the other way, continually reversing direction.

An AC voltage is continually changing between positive (+) and negative (-).

The rate of changing direction is called the frequency of the AC and it is measured in hertz (Hz) which is the number of forwards-backwards cycles per second.

Mains electricity in the USA and Canada have a frequency 60Hz while other counties have a frequency of 50Hz.

AC from a power supplyThis shape is called a sine wave.

This triangular signal is AC because it changesbetween positive (+) and negative (-).

4

An AC supply is suitable for powering some devices such as lamps and heaters but almost all electronic circuits require a steady DC supply (see below).

1.3 Properties of electrical signals

An electrical signal is a voltage or current which conveys information, usually it means a voltage. The term can be used for any voltage or current in a circuit.

The voltage-time graph on the right shows various properties of an electrical signal. In addition to the properties labelled on the graph, there is frequency which is the number of cycles per second.

The diagram shows a sine wave but these properties apply to any signal

Amplitude is the maximum voltage reached by the signal. It is measured in volts, V.

Peak voltage is another name for amplitude. Peak-peak voltage is twice the peak voltage (amplitude). When reading an oscilloscope trace it is

usual to measure peak-peak voltage. Time period is the time taken for the signal to complete one cycle.

It is measured in seconds (s), but time periods tend to be short so milliseconds (ms) and microseconds (µs) are often used. 1ms = 0.001s and 1µs = 0.000001s.

Frequency is the number of cycles per second. It is measured in hertz (Hz), but frequencies tend to be high so kilohertz (kHz) and megahertz (MHz) are often used. 1kHz = 1000Hz and 1MHz = 1000000Hz.

frequency = 1 and time period = 1 time period frequency

Mains electricity in the Palestine has a frequency of 50Hz, so it has a time period of 1/50 = 0.02s = 20ms

1.4 Root Mean Square (RMS) Values

The value of an AC voltage is continually changing from zero up to the positive peak, through zero to the negative peak and back to zero again. Clearly for most of the time it is less than the peak voltage, so this is not a good measure of its real effect.

Instead we use the root mean square voltage (VRMS) which is 0.7 of the peak voltage (Vpeak):

VRMS = 0.7 × Vpeak and Vpeak = 1.4 × VRMS (1.1)

These equations also apply to current. They are only true for sine waves (the most common type of AC) because the 0.7 and 1.4 are different values for other shapes. The RMS value is the effective value of a varying voltage or current. It is the equivalent steady DC (constant) value which gives the same effect.

5

For example a lamp connected to a 6V RMS AC supply will light with the same brightness when connected to a steady 6V DC supply. However, the lamp will be dimmer if connected to a 6V peak AC supply because the RMS value of this is only 4.2V (it is equivalent to a steady 4.2V DC).

You may find it helps to think of the RMS value as a sort of average, but please remember that it is NOT really the average! In fact the average voltage (or current) of an AC signal is zero because the positive and negative parts exactly cancel out!

What do AC meters show, is it the RMS or peak voltage?AC voltmeters and ammeters show the RMS value of the voltage or current. DC meters also show the RMS value when connected to varying DC providing the DC is varying quickly, if the frequency is less than about 10Hz you will see the meter reading fluctuating instead.

What does '6V AC' really mean, is it the RMS or peak voltage?If the peak value is meant it should be clearly stated, otherwise assume it is the RMS value. In everyday use AC voltages (and currents) are always given as RMS values because this allows a sensible comparison to be made with steady DC voltages (and currents), such as from a battery.

For example a '6V AC supply' means 6V RMS, the peak voltage is 8.6V. The Palestinian mains supply is 220V AC, this means 220V RMS so the peak voltage of the mains is about 311V.

1.5 Insulators and Conductors

An Insulator is a material that resists the flow of electric current. It is an object intended to support or separate electrical conductors without passing current through itself. An insulation material has atoms with tightly bonded valence electrons. The term electrical insulation has the same meaning as the term dielectric.

Some materials such as silicon dioxide or teflon are very good electrical insulators. A much larger class of materials, for example rubber-like polymers and most plastics are still "good enough" to insulate electrical wiring and cables even though they may have lower bulk resistivity. These materials can serve as practical and safe insulators for low to moderate voltages (hundreds, or even thousands, of volts

Electrical conductor: In science and engineering, conductors, such as copper or aluminum, are materials wAll conductors contain movable electric charges which will move when an electric potential difference (measured in volts) is applied across separate points on a wire (etc) made from the material. This flow of charge (measured in amperes) is what is meant by electric current. In most materials, the amount of current is proportional to the voltage (Ohm's law) provided the temperature remains constant and the material remains in the same shape and state. The ratio between the voltage and the current is called the resistance (measured in ohms) of the object between the points where the voltage was applied. The resistance across a standard mass (and shape) of a material at a given temperature is called the resistivity of the material. The inverse of resistance and resistivity is conductance and conductivityith atoms having loosely held valence electrons. See electrical conduction.

Since all conductors have some resistance, and all insulators will carry some current, there is no theoretical dividing line between conductors and insulators. However, there is a large gap between the conductance of materials that will carry a useful current at working voltages and those that will carry a negligible current for the purpose in hand, so the categories of insulator and conductor do have practical utility.

Thermal and electrical conductivity often go together (for instance, most metals are both electrical and thermal conductors). However, some materials are practical electrical conductors without being a good thermal conductor.

6

Conductor size

In many countries, conductors are measured by their cross section in square millimeters.

However, in the United States, conductors are measured by American wire gauge for smaller ones, and circular mils for larger ones.

Conductor materials

Of the metals commonly used for conductors, copper, has a high conductivity. Silver is more conductive, but due to cost it is not practical in most cases. However, it is used in specialized equipment, such as satellites, and as a thin plating to mitigate skin effect losses at high frequencies. Because of its ease of connection by soldering or clamping, copper is still the most common choice for most light-gauge wires.

Conductor voltage

The voltage on a conductor is determined by the connected circuitry and has nothing to do with the conductor itself. Conductors are usually surrounded by and/or supported by insulators and the insulation determines the maximum voltage that can be applied to any given conductor

1.6 Electrical Resistance

Electrical resistance is a measure of the degree to which an object opposes an electric current through it. Its reciprocal quantity is electrical conductance (provided the electrical impedence is real) measured in siemens. Assuming a uniform current density, an object's electrical resistance is a function of both its physical geometry and the resistivity of the material it is made from:

(1.2)

where

" " is the length "A" is the cross sectional area, and "ρ" is the resistivity of the material

Electrical resistance shares some conceptual parallels with the mechanical notion of friction. The SI unit of electrical resistance is the ohm, symbol Ω.

The electrical resistivity ρ (rho) of a material is given by

(1.3)

Where : ρ is the static resistivity (measured in ohm metres, Ω-m); R is the electrical resistance of a uniform specimen of the material (measured in ohms, Ω); is the length of the piece of material (measured in metres, m);

A is the cross-sectional area of the specimen (measured in square metres, m²).

Electrical resistivity can also be defined as

(1.4)

Where : E is the magnitude of the electric field (measured in volts per metre, V/m); J is the magnitude of the current density (measured in amperes per square metre, A/m²).

7

Electrical resistivity is also defined as the inverse of the conductivity σ (sigma), of the material, or

(1.5)

Table (1): Resistivity of Famous Conducting and Insulating Materials

Material Resistivity (Ω-m) at 20 °C CoefficientSilver 1.59×10−8 0.0038

Copper 1.72×10−8 0.0039Gold 2.44×10−8 0.0034

Aluminum 2.82×10−8 0.0039Tungsten 5.60×10−8 0.0045

Nickel 6.99×10−8 ?Brass 0.8×10−7 0.0015Iron 1.0×10−7 0.005Tin 1.09×10−7 0.0045

Platinum 1.1×10−7 0.00392Lead 2.2×10−7 0.0039

Manganin 4.82×10−7 0.000002Constantan 4.9×10−7 0.00001

Mercury 9.8×10−7 0.0009Nichrome 1.10×10−6 0.0004

Carbon 3.5×10−5 -0.0005Germanium 4.6×10−1 -0.048

Silicon 6.40×102 -0.075Glass 1010 to 1014 ?

Hard rubber approx. 1013 ?Sulfur 1015 ?

Quartz (fused) 7.5×1017 ?Teflon 1022 to 1024 ?

Example 1.1 :What is the value of the resistance of a copper wire with length 250m and cross section area equals 2.5 mm²Solution :Plug into the formula and using the value of the resistivity of the copper from the table :

8

1.7 Ohm's law

Ohm's law states that in an electrical circuit, the current passing through a conductor between two points is directly proportional to the potential difference (i.e. voltage drop or voltage) across the two points, and inversely proportional to the resistance between them, and for the same temperature.

The mathematical equation that describes this relationship is:

(1.6)

where I is the current in amperes, V is the potential difference between two points of interest in volts, and R is a circuit parameter, measured in ohms (which is equivalent to volts per ampere), and is called the resistance. The potential difference is also known as the voltage drop, and is sometimes denoted by U, E or emf (electromotive force) instead of V.

Ohm’s law defines the relationship between current, voltage, and resistance. There are three ways to express Ohm’s law mathematically.1. The current in a circuit is equal to the voltage applied to the circuit divided by the resistance of the circuit:

(1.7)

2. The resistance of a circuit is equal to the voltage applied to the circuit divided by the current in the circuit:

(1.8)

3. The applied voltage to a circuit is equal to the product of the current and the resistance of the circuit:(1.9)

where I = current, A R = resistance, V = voltage, VExample 1.2 :What is the value of the resistance if the current is 18 mA and the voltage is 229 mV?Solution :First, convert these values to amperes and volts. This gives I = 0.018 A and E = 0.229 V. Then plug into the equation R= E/I = 0.229/0.018 = 13 . You’re justified ingiving your answer to two significant figures, because the current is only given to that many digits.

Example 1.3 :Suppose the ammeter reads 52 and the voltmeter indicates 2.33 kV. What is the resistance?Solution :Convert to amperes and volts, getting I = 0.000052 A and E = 2330 V. Then plug into the formula: R = 2330/0.000052 = 45,000,000 = 45 M

Example 1.4 :Find I when V = 120V and R = 30Solution :Using Ohm's Law

Example 1.5 :Find V when I = 120nA and R = 25Solution :

9

Convert to amperes and ohms, getting and R = . Then plug into the formula:

Example 1.6 :Find V when I = 85A and R = 6.5Solution :Plug into the formula:

Example 1.7 :Find I when V = 22kV and R = 440Solution :Using Ohm's Law Eq. (1.7)

1.8 Electrical PowerThe electric power P used in any part of a circuit is equal to the current I in that part multiplied by the voltage V across that part of the circuit. Its formula is:

P = VI (1.10)

where P = power, W V = voltage, V I = current, AOther forms for P = VI are I = P/V and V = P/I.If we know the current I and the resistance R but not the voltage V, we can find the power P by using Ohm’s law for voltage, so that substituting V = IR into (1.10) we have

(1.11)

In the same manner, if we know the voltage V and the resistance R but not the current I, we can find the

power P by using Ohm’s law for current, so that substituting into (1.10) we have

(1.12)

Example 1.8 : The current through a 100Ω resistor to be used in a circuit is 0.20 A. Find the power rating of the resistor.Solution :Since I and R are known, use Eq. (1.11) to find P.

Example 1.9 If the voltage across a 25000 Ω resistor is 500 V, what is the power dissipated in the resistor?Solution :Since R and V are known, use Eq. (1.12) to find P.

1.9 Electrical energyElectrical energy in any part of the circuit is that power of that part multiplied of the time of consumption,

and it can be found from the formula:

10

Electrical energy U = Power x time (1.13)

U = VIt Joules (1.14)

Although the unit of energy is the joule, when dealing with large amounts of energy, the unit used is the

kilowatt hour (kWh) where

1 kWh = 1000 watt hour

= 1000 x 3600 watt seconds or joules

= 3 600 000 J

Example 1.10 A source e.m.f. of 5 V supplies a current of 3 A for 10 minutes. How much energy is

provided in this time?

Solution :

Energy = power x time, and power = voltage x current. Hence

U = VIt = 5 x 3 x (10 x 60) = 9000 Ws or J

= 9 kJ

Example 1.11: An electric heater consumes 1.8 MJ when connected to a 250 V supply for 30 minutes.

Find the power rating of the heater and the current taken from the supply.

Solution :

Power rating of heater =

Thus

Example 1.12: An electric bulb has a power 100 W. If it works 10 hours per a day. Find the energy

consumed in 1 month.

Solution :

Since energy = power x time

Then, Energy U = P x t = 100 x 10 x 30 = 30000 Wh

= 30 kWh

11

1.10 Series and Parallel Networks

Series circuits: The Figure shows three resistors R1, R2 and R3 connected end to end, i.e., in series, with a battery source of V volts. Since the circuitis closed a current I will flow and the p.d. across each resistor may be determined from the voltmeter readings V1, V2 and V3

In a series circuit(a) the current I is the same in all parts of the circuit and hence the same reading is found on each of the two ammeters shown, and(b) the sum of the voltages V1, V2 and V3 is equal to the total applied voltage, V, i.e.

From Ohm’s law:

where R is the total circuit resistance.Since V = V1 + V2 + V3then IR = IR1 + IR2 + IR3Dividing throughout by I gives

(1.15)Thus for a series circuit, the total resistance is obtained by adding together the values of the separate resistances.

Example 1.13: For the circuit shown in the illustrated figure, determine(a) the battery voltage V, (b) the total resistance of the circuit, and (c) the values of resistance of resistors R1, R2 and R3, given that the p.d.’s across R1, R2 and R3 are 5 V, 2 V and 6 V respectively.Solution :

Voltage divider The voltage distribution for the circuit shown in the Figure is given by:

(1.16)

(1.17)

12

Example 1.14: Two resistors are connected in seriesacross a 24 V supply and a current of 3 A flows in thecircuit. If one of the resistors has a resistance of 2 Ω determine: (a) the value of the other resistor, and (b) the volt. difference. across the 2 Ω resistor.Solution :(a) Total circuit resistance

Value of unknown resistance,

(b) Volt. Difference across 2 Ω resistor, V1 = IR1 = 3 x 2 = 6 VAlternatively, from above,

Parallel networks The Figure shows three resistors,R1, R2 and R3 connected across each other, i.e.,in parallel, across a battery source of V volts.In a parallel circuit:(a) the sum of the currents I1, I2 and I3 is equal to the total circuitcurrent, I, i.e. I = I1 + I2 + I3, and(b) the source p.d., V volts, is the same across each of the resistors.From Ohm’s law:

where R is the total circuit resistance.Since I = I1 + I2 + I3 then,

Dividing throughout by V gives:

(1.18)This equation must be used when finding the total resistance R of a parallel circuit. For the special case of two resistors in parallel

(1.19)Example 1.15: For the circuit shown in the Figure, find(a) the value of the supply voltage V (b) the value of current I.Solution :(a) V.d. across 20 Ω resistor = I2R2 = 3 x 20 = 60 V, hencesupply voltage V = 60 V since the circuit is connected in parallel.

13

Current I = I1 + I2 + I3 and hence I = 6 + 3 + 1 = 10 AAnother solution for part (b):

Current division: For the circuit shown in the Figure, the total circuit resistance, RT is given by:

Summarizing, with reference to the above Figure

(1.20)

(1.21)

Example 1.16: For the circuit shown in the Figure, find the current Ix

Solution :Commencing at the right-hand side of the arrangement shown in the Figure below, the circuit is gradually reduced in stages as shown in Figure (a)–(d).

14

From Figure (d)

From Figure (b)

From the above Figure

15

UNIT 2 DC Laws and Theorems

Introduction The laws which determine the currents and voltage drops in d.c. networks are:(a) Ohm’s law, (b) the laws for resistors in series and in parallel , (c) Kirchhoff’s laws (d) the superposition theorem (e) Th´evenin’s theorem

)f (Norton’s theorem

2.1 Kirchhoff’s lawsKirchhoff’s laws state:

(a) Current Law. At any junction in an electric circuit the total current flowing towards that junction is equal to the total current flowing away from the junction, i.e. ∑I = 0Thus, referring to Figure 2.1:I1 + I2 = I3 + I4 + I5 or I1 + I2 - I3 - I4 - I5 = 0

)b (Voltage Law. In any closed loop in a network, the algebraic sum of the voltage drops (i.e. products of current and resistance) taken around the loop is equal to the resultant e.m.f. acting in that loop.Thus, referring to Figure 2.2:

E1 - E2 = IR1 + IR2 + IR3(Note that if current flows away from the positive terminal of a source, that source is considered by convention to be positive. Thus moving anticlockwise around the loop of Figure 2.2, E1 is positive and E2 is negative.)

Example 2. 1. (a) Find the unknown currents marked in Figure 2.3(a). (b) Determine the value of e.m.f. E in Figure 2.3(b).

Figure 2.3Solution:(a) Applying Kirchhoff’s current law:For junction B: 50 = 20 + I1. Hence I1 = 30 AFor junction C: 20 + 15 = I2. Hence I2 = 35 AFor junction D: I1 = I3 + 120i.e. 30 = I3 + 120. Hence I3 = −90 A(i.e. in the opposite direction to that shown in Figure 2.3(a))For junction E: I4 + I3 = 15i.e. I4 = 15 – (-90). Hence I4 = 105 AFor junction F: 120 = I5 + 40. Hence I5 = 80 A

16

Figure 2.1

Figure 2.2

(b) Applying Kirchhoff’s voltage law and moving clockwise around the loop of Figure 2.3(b) starting at point A:3 + 6 + E - 4 = (I)(2) + (I)(2.5) + (I)(1.5) + (I)(1) = I(2 + 2.5 + 1.5 + 1)i.e. 5 + E = 2(7), since I = 2 AHence E = 14 - 5 = 9 V

Example 2. 2. Use Kirchhoff’s laws to determine the currents flowing in each branch of the network shown in Figure 2.4.

Solution:Procedure1. Use Kirchhoff’s current law and label current directions on the original circuit diagram. The directions chosen are arbitrary, but it is usual, as a starting point, to assume that current flows from the positive terminals of the batteries. This is shown in Figure 2.5 where the three branch currents are expressed in terms of I1 and I2 only, since the current through R is I1 + I2.2. Divide the circuit into two loops and apply Kirchhoff’s voltage law to each. From loop 1 of Figure 2.5, and moving in a clockwise direction as indicated (the direction chosen does not matter), gives

(1)From loop 2 of Figure 2.5, and moving in an anticlockwise direction as indicated (once again, the choice of direction does not matter; it does not have to be in the same direction as that chosen for the first loop), gives:

(2)3. Solve equations (1) and (2) for I1 and I2.

(3)

(4)

(i.e. I2 is flowing in the opposite direction to that shown in Figure 2.5.)From (1)

Hence

Current flowing through resistance R is

17

Figure 2.5Figure 2.4

Note that a third loop is possible, as shown in Figure 2.6,giving a third equation which can be used as a check:

Example 2. 3. Determine, using Kirchhoff’s laws, each branch current for the network shown in Figure 2.7.

Solution:1. Currents, and their directions are shown labelled inFigure 2.8 following Kirchhoff’s current law. It is usual,although not essential, to follow conventional current flowwith current flowing from the positive terminal of the source.

2. The network is divided into two loops as shown in Figure 2.8.Applying Kirchhoff’s voltage law gives:For loop 1:

(1)For loop 2:

Note that since loop 2 is in the opposite direction to current

the volt drop across is by convention negative).

Thus (2)

3. Solving equations (1) and (2) to find I1 and I2:

(3)(2) + (3) gives :

From (1):

Current flowing in

18

Figure 2.6

Figure 2.7

Figure 2.8

Example 2. 4. For the bridge network shown in Figure 2.9 determine the currents in each of the resistors.

Solution:Let the current in the 2Ω resistor be I1, then by Kirchhoff’s current law, the current in the 14 Ω resistor is (I - I1). Let thecurrent in the 32 Ω resistor be I2 as shown in Figure 2.10. Then the current in the 11 resistor is (I1 - I2) and that in the3 Ω resistor is (I - I1 + I2). Applying Kirchhoff’s voltage law to loop 1 and moving in a clockwise direction as shown in Figure 2.10 gives:

(1)

Applying Kirchhoff’s voltage law to loop 2 and moving in an anticlockwise direction as shown in Figure 2.10 gives:

However

Hence

(2)Equations (1) and (2) are simultaneous equations with two unknowns, I1 and I2.

(3)

(4)(4) – (3) gives:

Substituting for I2 in (1) gives:

Hence,

the current flowing in the 2 Ω resistor = I1 = 5 A

the current flowing in the 14 Ω resistor = I - I1 = 8 - 5 = 3 A

the current flowing in the 32 Ω resistor = I2 = 1 A

the current flowing in the 11 Ω resistor = I1 - I2 = 5 - 1 = 4 A and

the current flowing in the 3 Ω resistor = I - I1 + I2 = 8 - 5 + 1 = 4 A

19

Figure 2.9

Figure 2.10

2.2 The Superposition TheoremThe superposition theorem states:‘In any network made up of linear resistances and containing more than one source of e.m.f., the resultant current flowing in any branch is the algebraic sum of the currents that would flow in that branch if each source was considered separately, all other sources being replaced at that time by their respective internal resistances.’

Example 2. 5. Figure 2.11 shows a circuit containing two sourcesof e.m.f., each with their internal resistance. Determine the currentin each branch of the network by using the superposition theorem.

Solution Procedure:1. Redraw the original circuit with source E2 removed, being replaced by r2 only, as shown in Figure 2.12(a).2. Label the currents in each branch and their directions as shown inFigure 2.12(a) and determine their values. (Note that the choice ofcurrent directions depends on the battery polarity, which, by conventionis taken as flowing from the positive battery terminal as shown.)

R in parallel with r2 gives an equivalent resistance of:

From the equivalent circuit of Figure 2.12(b)

From Figure 2.12(a)

and

3. Redraw the original circuit with source E1 removed, being replacedby r1 only, as shown in Figure 2.13(a).4. Label the currents in each branch and their directions as shown inFigure 2.13(a) and determine their values.r1 in parallel with R gives an equivalent resistance of:

From the equivalent circuit of Figure 2.13(b)

From Figure 2.13(a)

20

Figure 2.11

Figure 2.12

Figure 2.13

5. Superimpose Figure 2.13(a) on to Figure 2.12(a) as shown in Figure 2.14.

6. Determine the algebraic sum of the currents flowing in each branch.Resultant current flowing through source 1, i.e.

Resultant current flowing through source 2, i.e.

Resultant current flowing through resistor R, i.e.

The resultant currents with their directions areshown in Figure 13.15.

2.3 Th´evenin’s TheoremTh´evenin’s theorem states:‘The current in any branch of a network is that which would result if an e.m.f. equal to the p.d. across a break made in the branch, were introduced into the branch, all other e.m.f.’s being removed and represented by the internal resistances of the sources.’

The procedure adopted when using Th´evenin’s theorem is summarized below. To determine the current in any branch of an active network (i.e. one containing a source of e.m.f.):(i) remove the resistance R from that branch,(ii) determine the open-circuit voltage, E, across the break,(iii) remove each source of e.m.f. and replace them by their internal resistances and then determine the resistance, r, ‘looking-in’ at the break,(iv) determine the value of the current from the equivalent circuit shown

in Figure 2.16, i.e.

Example 2. 6. Use Th´evenin’s theorem to find the current flowing in the 10 Ω resistor for the circuit shown in Figure 2.17

21

Figure 2.14

Figure 2.15

Figure 2.16

Figure 2.17

Solution:Following the above procedure:(i) The 10Ω resistance is removed from the circuit as shown inFigure 2.18

(ii) There is no current flowing in the 5 Ω resistor and current I1 is given by:

P.d. across R2 = I1R2 = 1 x 8 = 8 VHence p.d. across AB, i.e. the open-circuit voltage across the break,E = 8 V.

(iii) Removing the source of e.m.f. gives the circuit of Figure 2.19.Resistance

(iv) The equivalent Th´evenin’s circuit is shown in Figure 2.20Current

Hence the current flowing in the 10 resistor of Figure 2.17 is 0.482 A

Example 2.7. A Wheatstone Bridge network is shown inFigure 2.21(a). Calculate the current flowing in the 32Ω resistor,and its direction, using Th´evenin’s theorem. Assume the source ofe.m.f. to have negligible resistance.

22

Figure 2.19

Figure 2.20

Figure 2.18

Figure 2.21

Solution:Following the procedure:(i) The 32Ω resistor is removed from the circuit as shown inFigure 2.21(b)

(ii) The p.d. between A and C,

The p.d. between B and C,

Hence the p.d. between A and B = 44.47 - 8.31 = 36.16 VPoint C is at a potential of +54 V. Between C and A is a voltage drop of 8.31 V.Hence the voltage at point A is 54 - 8.31 = 45.69 V. Between C and B is a voltage drop of 44.47 V. Hence the voltage at point B is 54 - 44.47 = 9.53 V.Since the voltage at A is greater than at B, current must flow in the direction A to B.

(iii) Replacing the source of e.m.f. with a short-circuit (i.e. zero internal resistance) gives the circuit shown in Figure 2.21(c). The circuit is redrawn and simplified as shown in Figure 2.21(d) and (e), fromwhich the resistance between terminals A and B,

(iv) The equivalent Th´evenin’s circuit is shown in Figure 13.32(f), from which,Current

Hence the current in the 32 Ω resistor of Figure 2.21(a) is 1 A, flowing from A to B

Example 2.8. Use Th´evenin’s theorem to determine the currentflowing in the 3 Ω resistance of the network shown inFigure 2.22. The voltage source has negligible internalresistance.

Solution:Following the procedure:(i) The 3 Ω resistance is removed from the circuit as shown inFigure 2.23(ii) The Ω resistance now carries no current.

= 16V

Hence p.d. across AB, E = 16 V

23

Figure 2.22

Figure 2.23

(iii) Removing the source of e.m.f. and replacing it by its internal resistance means that the 20 Ω resistance is short-circuited as shown in Figure 2.24 since its internal resistance is zero. The 20 Ω resistance may thus be removed as shown in Figure 2.25.

From Figure 2.25 Th´evenin’s resistance,

(iv) The equivalent Th´evenin’s circuit is shown in Figure 2.26From which

2.4 Norton’s TheoremNorton’s theorem states:‘The current that flows in any branch of a network is the same as that which would flow in the branch if it were connected across a source of electrical energy, the short-circuit current of which is equal to the current that would flow in a short-circuit across the branch, and the internal resistance of which is equal to the resistance which appears across the open-circuited branch terminals.’The procedure adopted when using Norton’s theorem is summarized below.To determine the current flowing in a resistance R of a branch AB of an active network:(i) short-circuit branch AB(ii) determine the short-circuit current ISC flowing in the branch(iii) remove all sources of e.m.f. and replace them by their internalresistance (or, if a current source exists, replace with an open circuit),then determine the resistance r,‘looking-in’ at a break madebetween A and B(iv) determine the current I flowing in resistance R from the Nortonequivalent network shown in Figure 13.33, i.e.

Example 2.9. Use Norton’s theorem to determine the current flowing in the 10 Ω resistance for the circuit shown in Figure 2.27

24

Figure 2.24

Figure 2.25

Figure 2.26

Figure 2.26

Figure 2.27

Solution:Following the procedure:

(i) The branch containing the 10 Ω resistance is short-circuited asshown in Figure 2.28

(ii) Figure 2.29 is equivalent to Figure 2.28. Hence

(iii) If the 10 V source of e.m.f. is removed from Figure 2.29 theresistance ‘looking-in’ at a break made between A and B is given by:

(iv) From the Norton equivalent network shown in Figure 2.30 thecurrent in the 10 Ω resistance, by current division, is given by:

as obtained previously in Example 2.6 using Th´evenin’s theorem.

Example 2.10. Use Norton’s theorem to determine the current flowing in the 3 resistance of the network shown in Figure 2.31(a). The voltage source has negligible internal resistance.

25

Figure 2.28

Figure 2.29

Figure 2.30

Figure 2.31

Solution:Following the procedure:(i) The branch containing the 3 resistance is short-circuited as shown in Figure 2.31(b).(ii) From the equivalent circuit shown in Figure 2.31(c),

(iii) If the 24 V source of e.m.f. is removed the resistance ‘looking-in’ at a break made between A and B is obtained from Figure 2.31(d) and its equivalent circuit shown in Figure 2.31(e) and is given by:

(iv) From the Norton equivalent network shown in Figure 2.31(f) the current in the 3 Ω resistance is given by:

as obtained previously in Example 2.8 using Th´evenin’s theorem.

Extra Example:Example 2.11. Use Kirchhoff's Laws to find the current in each resistor for the circuit in Figure 2.32

Solution:1. Currents, and their directions are shown labelled in Figure 2.33 following Kirchhoff’s current law.

Applying Kirchhoff’s current law for node A gives

and for node B gives

2. The network is divided into three loops as shown in Figure 2.33.Applying Kirchhoff’s voltage law for loop 1 gives: ,then simplify : divide by 5 :

26

Figure 2.32

Figure 2.33

Applying Kirchhoff’s voltage law for loop 2 gives: ,then simplify : divide by 10 : Applying Kirchhoff’s voltage law for loop 3 gives: ,then simplify : divide by 5 : Arrange the three equations in matrix form :

A

A

A

A, AUNIT 3

Single Phase AC

Introduction: In this unit inductors' and capacitor's reactances, AC circuit reactance, electrical power, power factor, and power triangle will be explained.

3.1 InductorsInductance: The ability of a conductor to induce voltage in itself when the current changes is its self-inductance, or simply inductance. The symbol for inductance is L, and its unit is the henry (H). One henry is the amount of inductance that permits one volt to be induced when the currentchanges at the rate of one ampere per second. The formula for inductance is:

(3.1)

where L = inductance, H = induced voltage across the coil, V

=rate of change of current, A/s The self-induced voltage from Eq . (3.1)

(3.2)

Physical Characteristics:A coil's inductance depends on how it is wound, the core material on which it is wound, and the number of turns of wire with which it is wound.

27

1. Inductance L increases as the number of turns of wire N around the core increases. Inductance increases as the square of the turns increases. For example, if the number of turns is doubled ( 2 x ), inductance increases or ( 4 x ), assuming the area and length of the coil remain constant .2. Inductance increases as the relative permeability , of the core material increases.3. As the area A enclosed by each turn increases, the inductance increases. Since the area is a function of the square of the diameter of the coil, inductance increases as the square of the diameter.4. Inductance decreases as the length of the coil increases (assuming the number of turns remainsconstant) .

Inductive Reactance:Inductive reactance XL is the opposition to ac current due to the inductance in the circuit. The unit of inductive reactance is the ohm. The formula for inductive reactance is:

(3.3)

where, XL = inductive reactance, Ω f = frequency, Hz L = inductance, HExample 3.1: A choke coil of negligible resistance is to limit the current through it to 50 mA when 25 V is applied across it at 400 kHz. Find its inductance. Solution : Find XL by Ohm's law and then find L.

Inductors in Series and ParallelIf inductors are spaced sufficiently far apart so that they do not interact electromagnetically with each other, their values can be combined just like resistors when connected together. If a number of inductors are connected in series (Fig. 3.1), the total inductance LT is the sum of the individual inductances, orSeries: (3.4)

Series: (3.4)If a number of inductors are connected in parallel (Fig. 3.2), their total inductance LT is

Parallel: (3.5)

3.2 Capacitors28

Figure 3.2

Figure 3.1

A capacitor is an electrical device which consists of two conducting plates of metal separated by an insulating material called a dielectric (Fig. 3.3a). Schematic symbols shown (Fig. 3.3b and c) apply to all capacitors.

Electrically, capacitance is the ability to store an electric charge. Capacitance is equal to the amount of charge that can be stored in a capacitor divided by the voltage applied across the plates

(3.6)

where C = capacitance, F Q = amount of charge, C V = voltage, VThe unit of capacitance is the farad (F). The farad is that capacitance that will store one coulomb of charge in the dielectric when the voltage applied across the capacitor terminals is one volt.

Types of CapacitorsCommercial capacitors are named according to their dielectric. Most common are air, mica, paper, and ceramic capacitors, plus the electrolytic type. These types are compared in Table 3-1. Most types of capacitors can be connected to an electric circuit without regard to polarity. But electrolytic capacitors and certain ceramic capacitors are marked to show which side must be connected to the more positive side of a circuit.

Table 3.1Dielectric Construction Capacitance RangeAir Meshed plates 10-400 pFMica Stacked sheets 10-5000 pFPaper Rolled foil 0.001-1 µFCeramic Tubular 0.5-1600 pF

Disk 0.002-0.1 µ FElectrolytic Aluminum 5-1000 µ F

Tantalum 0.01-300 µ F

Capacitors in Series and Parallel

When capacitors are connected in series (Fig. 3-4), the total capacitance CT is

Series: (3.7)

When capacitors are connected in parallel (Fig. 3-5), the total capacitance CT is the sum of the individual capacitances.

29

Figure 3.3

Figure 3.4

Parallel: (3.8)

Capacitive Reactance:Capacitive reactance Xc is the opposition to the flow of ac current due to the capacitance in the circuit. The unit of capacitive reactance is the ohm. Capacitive reactance can be found by using the equation:

(3.9)

where Xc = capacitive reactance, Ω f = frequency, Hz C = capacitance, FExample 3.2 A 120 Hz, 25 mA ac current flows in a circuit containing a 10 µF capacitor. What is the voltage drop across the capacitor?. Solution : Find Xc and then Vc by Ohm's law.

3.3 Series a.c. circuits(a) Pure resistanceIn an a.c. circuit containing resistance R only (Figure 3.6(a)), the current IR is in phase with the applied voltage VR as shown in the phasor diagram of Figure 3.6(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 3.6(c). The impedance Z of the circuit is given by:

(3.10)

(b) Pure inductanceIn an a.c. circuit containing pure inductance L only (Figure 3.7(b)the current IL lags the applied voltage VL by 90° as shown in the phasor diagram of Figure 3.7(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 3.7(c). The impedance Z of the circuit is given by :

30

Figure 3.5

Figure 3.6

(3.11)

(c) Pure capacitanceIn an a.c. circuit containing pure capacitance only (Figure 3.8(a)), the current IC leads the applied voltage VC by 90° as shown in the phasor diagram of Figure 3.8(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 3.8(c). The impedance Z of the circuit is given by:

(3.12)

(d) R–L series circuitIn an a.c. circuit containing resistance R and inductance L in series (Figure 3.9(a)), the applied voltage V is the phasor sum of VR and VL as shown in the phasor diagram of Figure 3.9(b). The current I lags the applied voltage V by an angle lying between 0° and 90°—the actual value depending on the values of VR and VL, which depend on the values of R and L. The circuit phase angle, i.e., the angle between the current and the applied voltage, is shown as angle in the phasor diagram. In any series circuit the current is common to all components and is thus taken as the reference phasor in Figure 3.9(b). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 3.9(c), where it may be seen that in complex form the supply voltage V is given by:

(3.13)

31

Figure 3.7

Figure 3.8

Figure 3.9

Figure 3.8

Figure 3.10(a) shows the voltage triangle that is derived from the phasor diagram of Figure 3.10(b) (i.e. triangle Oab). If each side of the voltage triangle is divided by current I then the impedance triangle of Figure 3.10(b) is derived. The impedance triangle may be superimposed on the Argand diagram, as shown in Figure 3.10(c), where it may be seen that in complex form the impedance Z is given by:

(3.14)

Thus, for example, an impedance expressed as (3 + j4) Ω means that the resistance is 3 Ω and the inductive reactance XL is 4 Ω.

In polar form, where, from the impedance triangle, the modulus of impedance

and the circuit phase angle lagging.

(e) R–C series circuit

In an a.c. circuit containing resistance R and capacitance C in series (Figure 3.11(a)), the applied voltage V is the phasor sum of VR and VC as shown in the phasor diagram of Figure 3.11 (b). The current I leads the applied voltage V by an angle lying between 0° and 90°—the actual value depending on the values of VR and VC, which depend on the values of R and C. The circuit phase angle is shown as angle in the phasor diagram. The phasor diagram may be superimposed on the Argand diagram as shown in Figure 3.11 (c), where it may be seen that in complex form the supply voltage V is given by:

(3.15)

32

Figure 3.10

Figure 3.11

Figure 3.12a) shows the voltage triangle that is derived from the phasor diagram of Figure 3.12(b). If each side of the voltage triangle is divided by current I, the impedance triangle is derived as shown in Figure 3.12(b). The impedance triangle may be superimposed on the Argand diagram as shown in Figure 3.12(c), where it may be seen that in complex form the impedance Z is given by:

(3.16)Thus, for example, an impedance expressed as (9 - j14) Ω means that the resistance is 9 Ω and the capacitive reactance XC is 14 Ω .In polar form, where, from the impedance triangle,

and the circuit phase angle leading.

(f) R–L–C series circuitIn an a.c. circuit containing resistance R, inductance L and capacitance C in series (Figure 3.13(a)), the applied voltage V is the phasor sum of VR, VL and VC as shown in the phasor diagram of Figure 3.13(b) (where the condition VL > VC is shown). The phasor diagram may be superimposed on the Argand diagram as shown in Figure 3.13(c), where it may be seen that in complex form the supply voltage V is given by:

(3.17)From the voltage triangle the impedance triangle is derived and superimposing this on the Argand diagram gives, in complex form,impedance (3.18)

where, and the circuit phase angle

When VL = VC, XL = XC and the applied voltage V and the current I are in phase. This effect is called series resonance.

(g) General series circuit33

Figure 3.12

Figure 3.13

In an a.c. circuit containing several impedances connected in series, say, Z1, Z2, Z3, . . ., Zn, then the total equivalent impedance ZT is given by:

(3.19)

Example 3.3: A 200 V, 50 Hz supply is connected across a coil of negligible resistance and inductance 0.15 H connected in series with a 32 Ω resistor (Figure (3.14)). Determine (a) the impedance of the circuit,(b) the current and circuit phase angle, (c) the p.d. across the 32 Ω resistor, and (d) the p.d. across the coil.

Solution :(a) Inductive reactance Impedance

(b) Current

i.e., the current is 3.51 A lagging the voltage by 55.81°(c) P.d. across the 32Ω resistor, i.e. (d) P.d. across the coil, i.e.The phasor sum of VR and VL is the supply voltage V as shown in the phasor diagram of Figure 3.15.

Hence

Example 3.4: A The impedance of an electrical circuit is (30 - j50) ohms. Determine(a) the resistance, (b) the capacitance, (c) the modulus of the impedance, and (d) the current flowing and its phase angle, when the circuit is connected to a 240 V, 50 Hz supply.

Solution :(a) Since impedance Z = (30 - j50), the resistance is 30 Ω and the capacitive reactance is 50 Ω

(b) Since , then

(c) The modulus of impedance

(d) Impedance

34

Figure 3.14Figure 3.15

Hence current

Example 3.5: For the circuit shown in Figure 3.16, determine the value of impedance Z2

Solution :

Total circuit impedance

Total impedance from which, impedance

Example 3.6: A coil of resistance R ohms and inductanceL henrys is connected in series with a 50 μF capacitor (Fig 3.17).If the supply voltage is 225 V at 50 Hz and the current flowing in the circuit is 1.5 30° A, determine the values of R and L. Determinealso the voltage across the coil and the voltage across the capacitor.Solution :Circuit impedance

Capacitive reactance

Circuit impedance Z = R + j(XL - XC)

i.e. 129.9 + j75.0 = R + j(XL – 63.66)

Equating the real parts gives: resistance R = 129.9 Ω

Equating the imaginary parts gives: 75.0 = XL - 63.66,

from which, XL = 75.0 + 63.66 = 138.66 Ω

Since , inductance

The circuit diagram is shown in Figure 3.17.

Voltage across coil, VCOIL = I ZCOIL

ZCOIL = R + jXL = (129.9 + j138.66) or 190 46.87° Ω

Hence VCOIL = (1.5 -30°)(190 46.87°)

= 285∟ 16.87° V or (272.74 + j 82.71)V

Voltage across capacitor, VC = I XC = (1.56 -30°)(63.66 -90°)

= 95.49 −120° V or (− 47.75 − j 82.70)V

35

Figure 3.16

Figure 3.17

Example 3.7: For the circuit shown in Figure 3.18, determineThe values of voltages V1 and V2 if the supply frequency is 4 kHz. Determine also the value of the supply voltage V and the circuit phase angle. Draw the phasor diagram.

Solution : For impedance Z1

Hence Z1 =(8 - j15) or 17 -61.93° Ωand voltage V1 = I Z1 = (6 0°)(17 -61.93°) = 102 −61.93° V or (48 − j 90)VFor impedance Z2,

Hence Z2 = (5 + j12) Ω or 13 67.38° Ω

and voltage V2 =I Z2 = (6 0°)(13 67.38°)

= 78 67.38° V or (30 + j 72)V

Supply voltage, V = V1 + V2 = (48 - j90) + (30 + j72)

= (78 − j 18)V or 80 −13° V

Circuit phase angle, = 13° leading.

The phasor diagram is shown in Figure 3.19.

3.4 Parallel a.c. circuitsAs with series circuits, parallel networks may be analyzed by using phasor diagrams. However, with parallel networks containing more than two branches this can become very complicated. It is with parallel a.c. network analysis in particular that the full benefit of using complex numbers may be appreciated. Before analyzing such networks admittance, conductance and susceptance are defined.

Admittance, conductance and susceptanceAdmittance is defined as the current I flowing in an a.c. circuit divided by the supply voltage V (i.e. it is the reciprocal of impedance Z). The symbol for admittance is Y. Thus

(3.20)

The unit of admittance is the Siemen, S.An impedance may be resolved into a real part R and an imaginary part X, giving Z = R ± jX. Similarly, an admittance may be resolved into two parts—the real part being called the conductance G, and the imaginary part being called the susceptance B—and expressed in complex form.

(3.21) (a) R–L parallel circuitFor parallel circuits with R and XL (Fig. 3.20a), the same applied voltage VT is across R and XL since both are in parallel with VT. There is no phase difference between these voltages. Therefore, VT will be used as the reference phasor. The resistive branch current IR = VT/R is in phase with VT The inductive branch current IL = VT/XL lags VT by 90° (Fig. 3.20b) because the current in an inductance lags the voltage across it by 90°. The phasor sum of IR and IL equals the total line current IT (Fig. 3.20c), or

(3.22)

36

Figure 3.18

Figure 3.19

(3.23)

(3.24)

(b) R–C parallel circuitFor parallel circuits with R and XC (Fig. 3.21a), the same applied voltage VT is across R and XC since both are in parallel with VT. There is no phase difference between these voltages. Therefore, VT will be used as the reference phasor. The resistive branch current IR = VT/R is in phase with VT The capacitive branch current IC = VT/XC leads VT by 90° (Fig. 3.21b) because the current in an capacitance leads the voltage across it by 90°. The phasor sum of IR and IC equals the total line current IT (Fig. 3.21c), or

(3.25)

(3.26)

(3.27)

37

Figure 3.20

Figure 3.21

(c) R–L–C parallel circuitA three-branch parallel ac circuit (Fig. 3.22a) has resistance in one branch, inductance in the second branch, and capacitance in the third branch. The voltage is the same across each parallel branch, so VT = VR = VL = VC. The applied voltage VT is used as the reference line to measure phase angle θ. The total current IT is the phasor sum of IR, IL, and IC. The current in the resistance IR is in phase with the applied voltage VT (Fig. 3.22b). The current in the inductance IL lags the voltage VT by 90". The current in the capacitor Ic leads the voltage VT by 90°. IL and IC are exactly 180° out of phase and thus acting in opposite directions (Fig. 3.22b). When IL > IC, IT lags VT (Fig. 3.22c) so the parallel RLC circuit is considered inductive.

If IC > IL, the current relationships and phasor triangle (Fig. 3.23) show that IT now leads VT so this type of parallel RLC circuit is considered capacitive.

When IL > IC, or XL < XC , circuit is inductive and(3.28)

38

Figure 3.22

Figure 3.23

(3.29)

(3.30)

When IC > IL, or XC < XL , the circuit is capacitive and(3.31)

(3.32)

(3.33)

Example 3.8: A 500 Ω R is in parallel with 300 Ω XL. Find IT, θ, and ZT . Assume VT = 500VSolution:

Example 3.9: A 15Ω resistor and a capacitor of 20Ω capacitive reactance are placed in parallel across a 120-V ac line. Calculate IR, IC, IT, θ, and Z. Draw the phasor diagramSolution:

39

Figure 3.24

Figure 3.24 shows the phasor diagram.

Example 3.10: A 400 Ω resistor, a 50 Ω inductive reactance, and a 40 Ω capacitive reactance are placed in parallel across a 120-V ac line (Fig. 3.25a). Find the phasor branch currents, total current, phase angle, and impedance. Draw the phasor diagram.Solution:

Find IR, IL, IC

Find IT, θ. Since XL > XC (50Ω > 40Ω) or IC > lL (3.0A > 2.4A), the circuit is capacitive

Figure 3.25(b) shows the phasor diagram.

40

(a) (b)Figure 3.25

Example 3.11: For the circuit shown in Figure 3.26, determine(a) The current I(b) the voltage across the coil, and (c) the voltage across the capacitor

Solution : The coil is in series with the 20Ω resistor, and their impedance Z1 is:

Z1 = (20 + j40) Ω

The capacitor is in series with the 5Ω resistor, and their impedance Z2 is:

Z2 = (5 – j15) Ω

Z1 is in parallel with Z2 and their equivalent Z3 can be found by:

ZT =14 + 12 – j16 + 4 = (30 – j16) Ω = 34 -28.07° Ω

Therefore the total current I can be found by:

It can be directly found from:

This current is divided into the branch of the coil and the branch of the capacitor, and the current in each

branch can be found using current division law:

41

Figure 3.26

VCOIL = ICOIL x (j40)

= (0.046 – j1.788)(j40) = (71.52 + j1.84) V or = 71.54 1.47° V

VCAPACITOR = ICAPACITOR x(–j15)

= (–j15) = (55.02 – j52.26) V or = 75.88 –43.52° V

Example 3.12: For the circuit shown in Figure 3.27, determine the current I1, the current I2, the current IT, and the equivalent impedance ZT.

Solution : In the first branch the coil 87mH is in series with the 42Ω resistor, and their impedance is Z1

In the second branch the capacitor 10µF is in series with the 65Ω resistor and the coil 72mH, and their

impedance is Z2

Z1 is in parallel with Z2 and their equivalent is ZT

Z1 = (42 + j27.33) Ω = 50.11 33.05° Ω

Z2 = (65 + j22.62 – j318.31) Ω = (65 – j295.69) Ω = 302.75 –77.6° Ω

42

Figure 3.27

Example 3.12: For the circuit shown in Figure 3.28, determine (a) the value of the impedance Z1, (b) If the supply frequency is 5 kHz, determine the value of the components comprising impedance Z1

Solution : (a) Total circuit admittance,

YT = Y1 + Y2 + Y3

(b) Since Z1 = (1.6 – j1.2) resistance = 1.6 Ω and capacitive reactance, XC = 1.2 Ω

Since , then

43

Figure 3.28

3.5 Power in a.c. circuitsAlternating currents and voltages change their polarity during each cycle. It is not surprising therefore to find that power also pulsates with time. The product of voltage v and current i at any instant of time is called instantaneous power p, and is given by:

p = vi (3.34)

For purely resistive circuits the current and the voltage are in the same phase, and considering RMS values of current and voltage then the average power is given by:

(3.35)

For purely inductive circuits the current lags the voltage by angle 90°, so the average power is zero. Also, for purely capacitive circuits the current leads the voltage by angle 90°, so the average power is zero.For circuits containing resistors and inductors, the current lags the voltage by angle Φ which is between 0° and 90°, and also for circuits containing resistors and capacitors, the current leads the voltage by angle Φ which is between 0° and 90°, so the average power is given by:

(3.36)

A phasor diagram in which the current I lags the applied voltage V by angle (i.e., an inductive circuit) is shown in Figure 3.29(a). The horizontal component of V is (V cos ), and the vertical component of V is (V sin ). If each of the voltage phasors of triangle Oab is multiplied by I, Figure 3.29(b) is produced and is known as the ‘power triangle’. Each side of the triangle represents a particular type of power:

True or active power P = VI cos watts (W) Apparent power S = VI voltamperes (VA) Reactive power Q = VI sin (VAR)

The power triangle is not a phasor diagram since quantities P, Q and S are mean values and not rms values of sinusoidally varying quantities.Superimposing the power triangle on an Argand diagram produces a relationship between P, S and Q in complex form, i.e.,

S = P + jQ (3.37)

Apparent power, S, is an important quantity since a.c. apparatus, such as generators, transformers and cables, is usually rated in voltamperes rather than in watts. The allowable output of such apparatus is usually limited not by mechanical stress but by temperature rise, and hence by the losses in the device. The losses are determined by the voltage and current and are almost independent of the power factor. Thus the amount of electrical equipment installed to supply a certain load is essentially determined by the voltamperes of the load rather than by the power alone. The rating of a machine is defined as the maximum apparent power that it is designed to carry continuously without overheating.The reactive power, Q, contributes nothing to the net energy transfer and yet it causes just as much loading of the equipment as if it did so. Reactive power is a term much used in power generation, distribution and utilization of electrical energy. Inductive reactive power, by convention, is defined as positive reactive power; capacitive reactive power, by convention, is defined as negative reactive power. The above relationships derived from the phasor diagram of an inductive circuit may be shown to be true for a capacitive circuit, the power triangle being as shown in Figure 3.30.

44

Figure 3.29

Power factor is defined as:

(3.38)

(3.39)

A circuit in which current lags voltage (i.e., an inductive circuit) is said to have a lagging power factor, and indicates a lagging reactive power Q.A circuit in which current leads voltage (i.e., a capacitive circuit) is said to have a leading power factor, and indicates a leading reactive power Q.

Example 3.13: For the circuit shown in Figure 3.31, determine the active power developed between points (a) A and B, (b) C and D, (c) E and F.

Solution : Circuit impedance,

(a) Active power developed between points A and B = I²R = (8.24)² (25) = 339.5 W(b) Active power developed between points C and D is zero, since no power is developed in a pure capacitor.(c) The current I1 is calculated first to calculate the power between E and F.

45

Figure 3.30

Figure 3.31

Hence the active power developed between points E and F

Example 3.14: The circuit shown in Figure 3.32 dissipatesan active power of 400 W, and has a power factor of 0.766 lagging. Determine (a) the apparent power, (b) the reactive power, (c) the value and phase of current I, and (d) the value of impedance Z.

Solution : Since power factor = 0.766 lagging, the circuit phase angle , i.e., = 40° lagging which

means that the current I lags voltage V by 40°.

(a) Since power, P = VI cos , the magnitude of apparent power,

(b) Reactive power Q = S sin = 522.2 sin 40° = 335.66 VAR

Since the power factor is lagging the reactive power is inductive

i.e. positive value. Figure 3.33 shows the power triangle.

(c) Since VI = 522.2 VA,

magnitude of current

Since the voltage is at a phase angle of 30° (Figure 3.32) and current lags voltage by 40°, the phase

angle of current is 30° – 40° = –10°.

Hence current I = 5.222 −10° A

(d) Total circuit impedance

or

Hence impedance Z = ZT – 4 = (14.67 + j12.31) – 4

= (10.67 + j 12.31) Ω or = 16.29 49.08° Ω

Example 3.15: For the circuit of example 3.12 find the active power, reactive power, and apparent power that dissipated by the circuit, then draw the power triangle.

Solution:In the solution of the example the total current of the circuit is found that:

Then the value of active power P can be found:

P = VI cos Φ = (220)(4.19) cos(23.61°) = 844.64 W

Also, the value of reactive power Q can be found:

Q = VI sin Φ = (220)(4.19) sin(23.61°) = 369.19 VAR

The value of apparent power S can be found:

46

Figure 3.32

Figure 3.33

Figure 3.34

S = P + jQ = 844.64 + j369.19 = 921.8 23.61° VA

Or it can be found by

S = V I = (220) (4.19) = 921.8 VA

The power triangle is illustrated in figure 3.34.

Power Factor Improvement

For a particular active power supplied, a high power factor reduces the current flowing in a supply system and therefore reduces the cost of cables, transformers, switchgear and generators. Supply authorities use tariffs which encourage consumers to operate at a reasonably high power factor. One method of improving the power factor of an inductive load is to connect a bank of capacitors in parallel with the load. Capacitors are rated in reactive voltamperes and the effect of the capacitors is to reduce the reactive power of the system without changing the active power. Most residential and industrial loads on a power system are inductive, i.e. they operate at a lagging power factor.A simplified circuit diagram is shown in Figure 3.35(a) where a capacitor C is connected across an inductive load. Before the capacitor is connected the circuit current is ILR and is shown lagging voltage V by angle in the phasor diagram of Figure 3.35 (b). When the capacitor C is connected it takes a current IC which is shown in the phasor diagram leading voltage V by 90°. The supply current I in Figure 3.35(a) is now the phasor sum of currents ILR and IC as shown in Figure 3.35(b). The circuit phase angle, i.e., the angle between V and I, has been reduced from to and the power factor has been improved from cos

to cos .Figure 3.35(a) shows the power triangle for an inductive circuit with a lagging power factor of cos . In Figure 3.35(b), the angle has been reduced to , i.e., the power factor has been improved from cos to cos by introducing leading reactive voltamperes (shown as length ab) which is achieved by connecting capacitance in parallel with the inductive load. The power factor has been improved by reducing the reactive voltamperes; the active power P has remained unaffected. Power factor correction results in the apparent power S decreasing (from 0a to 0b in Figure 3.35(b)) and thus the current decreasing, so that the power distribution system is used more efficiently.Another method of power factor improvement, besides the use of static capacitors, is by using synchronous motors; such machines can be made to operate at leading power factors.

47

Figure 3.35

Example 3.16: A circuit has an impedance Z = (3 + j4) Ω and a source p.d. of 50 30° V at a frequency of 1.5 kHz. Determine(a) the supply current, (b) the active, apparent and reactive power,(c) the rating of a capacitor to be connected in parallel with impedance Z to improve the power factor of the circuit to 0.966 lagging, and (d) the value of capacitance needed to improve the power factor to 0.966 lagging.

Solution:

(a) Supply current,

(b) Apparent power, S = V I = (50 30°) (10 23.13°) = 500 53.13° VA = (300 + j400) VA = P + jQ

Hence active power, P = 300 W apparent power, S = 500 VA and reactive power, Q = 400 VAR lagging.

The power triangle is shown in Figure 3.36.

(c) A power factor of 0.966 means that cos = 0.966.Hence angle = arccos 0.966 = 15° .To improve the power factor from cos 53.13°, i.e. 0.60, to 0.966, the power triangle will need to change from 0cb (Figure 3.37) to 0ab, the length (ca) representing the rating of a capacitor connected in parallel with the circuit. From Figure 3.37, tan 15° = ab/300, from which, ab = 300 tan 15° = 80.38 VAR.Hence the rating of the capacitor, ca = cb – ab = 400 – 80.38

= 319.6 VAR leading.

(d) Current in capacitor

Capacitive reactance,

Thus , from which,

required capacitance

48

Figure 3.36

Figure 3.37

UNIT 4 Three Phase AC System

Introduction

This unit introduces the three phase AC connection, the relation between line and phase voltage or current, and the star and delta connections and their relations.

4.1 Characteristics of Three Phase Systems

A three-phase (3-Φ) system is a combination of three single-phase (1-Φ) systems. In a 3-Φ balanced system, the power comes from an ac generator that produces three separate but equal voltages, each of which is out of phase with the other voltages by 120° (Figure. 4.1). Although 1-Φ circuits are widely used in electrical systems, most generation and distribution of alternating current is 3-Φ. Three-phase circuits require less weight of conductors than 1-Φ circuits of the same power rating; they permit flexibility in the choice of voltages; and they can be used for single-phase loads. Also, 3-Φ equipment is smaller in size, lighter in weight, and more efficient than 1-Φ machinery of the same rated capacity. A three-phase a.c. supply is carried by three conductors, called ‘lines’ which are colored red, yellow and blue. The currents in these conductors are known as line currents (IL) and the v.d.’s between them are known as line voltages (VL). A fourth conductor, called the neutral (colored black, and connected through protective devices to earth) is often used with a three-phase supply. The phase-sequence is given by the sequence in which the conductors pass the point initially taken by the red conductor. The national standard phase sequence is R, Y, B. To reduce the number of wires it is usual to interconnect the three phases. The three phases of a 3-Φ system may be connected in two ways. If the three common ends of each phase are connected together at a common terminal marked N for neutral, and the other three ends are connected to the 3-Φ line, the system is wye- or Y-connected (Figure. 4.2a). If the three phases are connected in series to form a closed loop, the system is delta- or Δ-connected (Figure. 4.2b).

49

(a) (b)Figure 4.1

(a) (b)

Figure 4.2

4.2 Star Connection

(i) A star-connected load is shown in Figure 4.3 where the three line conductors are each connected to a load and the outlets from the loads are joined together at N to form what is termed the neutral point or the star point.(ii) The voltages, , , and are called phase voltages or line to neutral voltages. Phase voltages are generally denoted by (iii) The voltages, , and are called line voltages and generally denoted by .(iv) From Figure 4.3 it can be seen that the phase currents (generally denoted by ) are equal to their respective line currents , and , i.e. for a star connection:

(4.1)For a balanced system: = = = =

= = = = and the current in the neutral conductor, When a star connected system is balanced, then the neutral conductor is unnecessary and is often omitted.(vi) The line voltage, , shown in Figure 4.4(a) is given by = − ( is negative since it is in the opposite direction to ). In the phasor diagram of Figure 4.4(b), phasor is reversed (shown by the broken line) and then added phasorially to (i.e. = + (− )). By trigonometry, or by measurement, = , i.e. for a balanced star connection:

(4.2)

(vii) The star connection of the three phases of a supply, together with a neutral conductor, allows the use of two voltages—the phase voltage and the line voltage. A 4-wire system is also used when the load is not balanced. The standard electricity supply to consumers in Palestine is 380/220 V, 50 Hz, 3-phase, 4-wire alternating current, and a diagram of connections is shown in Figure 4.5.

50

Figure 4.3

Figure 4.4

Example 4.1:A star-connected load consists of three identical coils each of resistance 30Ω and inductance 127.3 mH. If the line current is 5.08 A, calculate the line voltage if the supply frequency is 50 Hz.Solution:Inductive reactance Impedance of each phase

For a star connection

Hence phase voltage Line voltage

Example 4.2:A balanced, three-wire, star-connected, 3-phase load has a phase voltage of 240 V, a line current of 5 A and a lagging power factor of 0.966. Draw the complete phasor diagram.Solution:The phasor diagram is shown in Figure 4.6Procedure to construct the phasor diagram:(i) Draw = = = 240 V and spaced 120° apart. (Note that is shown vertically upwards—this however is immaterial for it may be drawn in any direction.)(ii) Power factor = cos Φ = 0.966 lagging. Hence the load phase angle is given by arccos 0.966, i.e. 15° lagging. Hence = = = 5 A, lagging , and respectively by 15°.(iii) = - (phasorially). Hence is reversed and added phasorially to . By measurement,

= 415 V (i.e. and leads by 30°. Similarly, = - and = - .

51

Figure 4.5

Example 4.3:A 415 V, 3-phase, 4 wire, star-connected system supplies three resistive loads as shown in Figure 4.7. Determine (a) the current in each line and (b) the current in the neutral conductor.Solution:For a star-connected system

Hence

Since current

Then

and

(b) The three line currents are shown in the phasor diagram of Figure 4.8. Since each load is resistive the currents are in phase with the phase voltages and are hence mutually displaced by 120°.The current in the neutral conductor is given by:

= + + phasorially

52

Figure 4.6

Figure 4.8 Figure 4.9

Figure 4.7

Figure 4.9 shows the three line currents added phasorially. oa represents in magnitude and direction. From the nose of oa, ab is drawn representing in magnitude and direction. From the nose of ab, bc is drawn representing in magnitude and direction. oc represents the resultant, .By measurement, Alternatively, by calculation, considering at 90°, at 210° and at 330°:

Total horizontal component = 100 cos 90° + 75 cos 330° + 50 cos 210° = 21.65

Total vertical component = 100 sin 90° + 75 sin 330° + 50 sin 210° = 37.50

Hence magnitude of

4.3 Delta connection

(i) A delta (or mesh) connected load is shown in Figure 4.10. Where the end of one load is connected to the start of the next load.(ii) From Figure 4.10, it can be seen that the line voltages , and are the respective phase voltages, i.e. for a delta connection

(4.3)(iii) Using Kirchhoff’s current law in Figure 4.10, = - = + .From the phasor diagram shown in Figure 4.11, by trigonometry or by measurement, = , i.e. for a delta connection:

(4.4)

Example 4.4:Three identical capacitors are connected in delta to a 415 V, 50 Hz, 3-phase supply. If the line current is 15 A, determine the capacitance of each of the capacitors.Solution:For a delta connection

Hence phase current

Capacitive reactance per phase, (since in delta connection )

Hence

from which capacitance,

Example 4.5:Three coils each having resistance 3Ω and inductive reactance 4Ω are connected (i) in star and (ii) in delta to a 381 V, 3-phase supply. Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.Solution:

53

Figure 4.10 Figure 4.11

(i)For a star connection: and (a) A 381 V, 3-phase supply means that the line voltage,

Phase voltage,

(b) Impedance per phase,

Phase current,

Line current, (ii) For a delta connection: and (a) Line voltage, Phase voltage

(b) Phase current,

Line current,

4.4 Power in Three-Phase Systems

The power dissipated in a three-phase load is given by the sum of the power dissipated in each phase. If a load is balanced then the total power P is given by: P = 3 x power consumed by one phase.The power consumed in one phase = (where is the phase angle between and ).

For a star connection, and hence

(4.5)

For a delta connection, and hence

(4.6)

Hence for either a star or a delta balanced connection the total power P is given by: (4.7)

or (4.8)

Total volt-amperes, (4.9)

Total reactive power, (4.10)

Example 4.6:The input power to a 3-phase a.c. motor is measured as 5 kW. If the voltage and current to the motor are 400 V and 8.6 A respectively, determine the power factor of the system.Solution:Power, P = 5000 W; Line voltage = 400 V; Line current, = 8.6 APower,

Hence power factor =

Example 4.7:Three identical coils, each of resistance 10Ω and inductance 42 mH are connected (a) in star and (b) in delta to a 415 V, 50 Hz, 3-phase supply. Determine the total power dissipated in each case.Solution:

54

(a) Star connectionInductive reactance Phase impedance

Line voltage = 415 V and phase voltage,

Phase current,

Line current,

Power factor =

Power dissipated, = (Alternatively,

(b) Delta connection, , and (from above)

Phase current,

Line current, Power dissipated, = (Alternatively,

Hence loads connected in delta dissipate three times the power than when connected in star, and also take a line current three times greater.

55

Example 4.8:A 381 V, 3-phase a.c. motor has a power output of 12.75 kW and operates at a power factor of 0.77 lagging and with an efficiency of 85%. If the motor is delta-connected, determine (a) the power input, (b) the line current and (c) the phase current.Solution:

(a) Efficiency = hence

from which, power input = = 15000 = 15 kW

(b) Power,

line current,

(c) For a delta connection,

hence phase current,

56

UNIT 5 Electrical Transformers

Introduction

This unit introduces the construction and principle of operation for single phase and three phases transformers, star and delta connections for three phases transformers.

5.1 Transformer's principle of operation

A transformer is a device which uses the phenomenon of mutual induction to change the values of alternating voltages and currents. In fact, one of the main advantages of a.c. transmission and distribution is the ease with which an alternating voltage can be increased or decreased by transformers.Losses in transformers are generally low and thus efficiency is high. Being static they have a long life and are very stable.Transformers range in size from the miniature units used in electronic applications to the large power transformers used in power stations. The principle of operation is the same for each. A transformer is represented in Figure 5.1(a) as consisting of two electrical circuits linked by a common ferromagnetic core. Magnetic coupling is used to transfer electric energy from one coil to another. The coil which receives energy from an ac source is called the primary. The coil which delivers energy to an ac load is called the secondary. The core of transformers used at low frequencies is generally made of magnetic material, usually sheet steel. Cores of transformers used at higher frequencies are made of powdered iron and ceramics, or nonmagnetic materials. Some coils are simply wound on nonmagnetic hollow forms such as cardboard or plastic so that the core material is actually air. A circuit diagram symbol for a transformer is shown in Figure 5.1(b).

When the secondary is an open-circuit and an alternating voltage V1 is applied to the primary winding, a small current—called the no-load current —flows, which sets up a magnetic flux in the core. This alternating flux links with both primary and secondary coils and induces in them e.m.f.’s of E1 and E2 respectively by mutual induction. The induced e.m.f. E in a coil of N turns is given by:

(5.1)

where is the rate of change of flux. In an ideal transformer, the rate of change of flux is the same for

both primary and secondary and thus , i.e. the induced e.m.f. per turn is constant.

Assuming no losses, and , Hence

(5.2)

or (5.3)

57

(a) (b)

Figure 5.1

is called the voltage ratio, and is called the turns ratio, or the ‘transformation ratio’ of the

transformer. If is less than then is less than and the device is termed a step-down transformer. If is greater then then is greater than and the device is termed a step-up transformer. When a load is connected across the secondary winding, a current flows. In an ideal transformer losses are neglected and a transformer is considered to be 100% efficient.

Hence input power = output power, or = , i.e., in an ideal transformer, the primary and secondary volt-amperes are equal.

Thus (5.4)

Combining equations (5.3) and (5.4) gives:

(5.5)

The rating of a transformer is stated in terms of the volt-amperes that it can transform without overheating. With reference to Figure 5.1(a), the transformer rating is either or , where is the full-load secondary current.

5.2 Transformers construction

(i) There are broadly two types of single-phase double-wound transformer constructions—the core type and the shell type, as shown in Figure 5.2. The low and high voltage windings are wound as shown to reduce leakage flux.

(ii) For power transformers, rated possibly at several MVA and operating at a frequency of 50 Hz in Palestine, the core material used is usually laminated silicon steel or stalloy, the laminations reducing eddy currents and the silicon steel keeping hysteresis loss to a minimum.Large power transformers are used in the main distribution system and in industrial supply circuits. Small power transformers have many applications, examples including welding and rectifier supplies, domestic bell circuits, imported washing machines, and so on.

(iii) For audio frequency (a.f.) transformers, rated from a few mVA to no more than 20 VA, and operating at frequencies up to about 15 kHz, the small core is also made of laminated silicon steel. A typical application of a.f. transformers is in an audio amplifier system.

(iv) Radio frequency (r.f.) transformers, operating in the MHz frequency region have either an air core, a ferrite core or a dust core. Ferrite is a ceramic material having magnetic properties similar to silicon steel, but having a high resistivity. Dust cores consist of fine particles of carbonyl iron or permalloy (i.e.

58

(a) Core Type (b) Shell Type

Figure 5.2

nickel and iron), each particle of which is insulated from its neighbors. Applications of r.f. transformers are found in radio and television receivers.

(v) Transformer windings are usually of enamel-insulated copper or aluminum.

(vi) Cooling is achieved by air in small transformers and oil in large transformers.

5.3 E.m.f. equation of a transformer

The magnetic flux set up in the core of a transformer when an alternating voltage is applied to its primary winding is also alternating and is sinusoidal.Let be the maximum value of the flux and f be the frequency of the supply. The time for 1 cycle of the

alternating flux is the periodic time T, where seconds.

The flux rises sinusoidally from zero to its maximum value in ¼ cycle, and the time for ¼ cycle is

seconds.

Hence the average rate of change of flux = = Wb/s

and since 1 Wb/s = 1 volt, the average e.m.f. induced in each turn = volts.As the flux varies sinusoidally, then a sinusoidal e.m.f. will be induced in each turn of both primary and secondary windings.

For a sine wave, form factor = = 1.11

Hence, RMS value = form factor x average value = 1.11 x average valueThus RMS e.m.f. induced in each turn = 1.11 x volts = 4.44 voltsTherefore, RMS value of e.m.f. induced in primary,

volts (5.6)Therefore, RMS value of e.m.f. induced in secondary,

volts (5.7)

5.4 Equivalent circuit of a transformer

Figure 5.3 shows an equivalent circuit of a transformer. and represent the resistances of the primary and secondary windings and and represent the reactances of the primary and secondary windings, due to leakage flux.

59

Figure 5.3

The core losses due to hysteresis and eddy currents are allowed for by resistance R which takes a current , the core loss component of the primary current. Reactance X takes the magnetizing component .

In a simplified equivalent circuit shown in Figure 5.4, R and X are omitted since the no-load current Io is normally only about 3–5% of the full load primary current.It is often convenient to assume that all of the resistance and reactance as being on one side of the transformer.

Resistance in Figure 5.4 can be replaced by inserting an additional resistance in the primary circuit such that the power absorbed in when carrying the primary current is equal to that in due to the secondary current, i.e.,

from which

Then the total equivalent resistance in the primary circuit is equal to the primary and secondary resistances of the actual transformer.Hence

(5.8)

By similar reasoning, the equivalent reactance in the primary circuit is given by:

(5.9)

The equivalent impedance of the primary and secondary windings referred to the primary is given by: (5.10)

If is the phase angle between and the volt drop then

(5.11)

The simplified equivalent circuit of a transformer is shown in Figure 5.5.

60

Figure 5.4

Figure 5.5

5.5 Regulation of a transformer

When the secondary of a transformer is loaded, the secondary terminal voltage, , falls. As the power factor decreases, this voltage drop increases. This is called the regulation of the transformer and it isusually expressed as a percentage of the secondary no-load voltage, . For full-load conditions:

(5.12)

In words regulation is found from:

The fall in voltage, , is caused by the resistance and reactance of the windings.Typical values of voltage regulation are about 3% in small transformers and about 1% in large transformers.

5.6 Transformer losses and efficiency

There are broadly two sources of losses in transformers on load, these being copper losses and iron losses.(a) Copper losses are variable and result in a heating of the conductors, due to the fact that they possess resistance. If and are the primary and secondary winding resistances then the total copper loss is

(b) Iron losses are constant for a given value of frequency and flux density and are of two types—hysteresis loss and eddy current loss.

(i) Hysteresis loss is the heating of the core as a result of the internal molecular structure reversals which occur as the magnetic flux alternates. The loss is proportional to the area of the hysteresis loop and thus low loss nickel iron alloys are used for the core since their hysteresis loops have small areas.(ii) Eddy current loss is the heating of the core due to e.m.f.’s being induced not only in the transformer windings but also in the core. These induced e.m.f.’s set up circulating currents, called eddy currents. Owing to the low resistance of the core, eddy currents can be quite considerable and can cause a large power loss and excessive heating of the core. Eddy current losses can be reduced by increasing the resistivity of the core material or, more usually, by laminating the core (i.e., splitting it into layers or leaves) when very thin layers of insulating material can be inserted between each pair of laminations. This increases the resistance of the eddy current path, and reduces the value of the eddy current.

Transformer efficiency

(5.13)

and is usually expressed as a percentage. It is not uncommon for power transformers to have efficiencies of between 95% and 98%.Output power = ,total losses = copper loss + iron losses,and input power = output power + losses

5.7 Auto transformers

61

An auto transformer is a transformer which has part of its winding common to the primary and secondary circuits. Figure 5.6(a) shows the circuit for a double-wound transformer and Figure 5.6(b) that for an auto transformer. The latter shows that the secondary is actually part of the primary, the current in the secondary being ( - ). Since the current is less in this section,the cross-sectional area of the winding can be reduced, which reduces theamount of material necessary.

Saving of copper in an auto transformerFor the same output and voltage ratio, the auto transformer requires less copper than an ordinary double-wound transformer. This is explained below. The volume, and hence weight, of copper required in a winding is proportional to the number of turns and to the cross-sectional area of the wire. In turn this is proportional to the current to be carried, i.e., volume of copper is proportional to NI.Volume of copper in an auto transformer

(since )

Volume of copper in a double-wound transformer

(Also, since )

If = x then

(volume of copper in an auto transformer) = (1 − x) (volume of copper in a double-wound transformer) (5.14)

If, say, x = , then (volume of copper in auto transformer) =

(1- ) (volume of copper in a double-wound transformer)

= (volume in double-wound transformer) i.e., a saving of 80%

Similarly, if x = , the saving is 25%, and so on.

The closer is to , the greater the saving in copper.

62

Figure 5.6

Advantages of auto transformersThe advantages of auto transformers over double-wound transformers include:1. a saving in cost since less copper is needed (see above)2. less volume, hence less weight3. a higher efficiency, resulting from lower losses4. a continuously variable output voltage is achievable if a sliding contact is used5. a smaller percentage voltage regulation.

Disadvantages of auto transformersThe primary and secondary windings are not electrically separate, hence if an open-circuit occurs in the secondary winding the full primary voltage appears across the secondary.

Uses of auto transformersAuto transformers are used for reducing the voltage when starting induction motors and for interconnecting systems that are operating at approximately the same voltage.

5.8 Three-phase transformers

Three-phase power may be transformed by a bank of single-phase transformers or by a single three-phase transformer. A three-phase transformer is essentially three wound on a common core. They basically consist of three pairs of single-phase windings mounted on one core, as shown in Figure 5.7. The primary and secondary windings in Figure 5.7 are wound on top of each other in the form of concentric cylinders, similar to that shown in Figure 5.2(a). The geometry of the core is such that the fluxes of the phases share common paths. As a result, the volume of iron is less than that of three single-phase transformer of the same total rating.The three-phase transformer do not provide the flexibility of a set of single-phase transformer. For example, one single-phase transformer in a bank may have a higher kVA rating than the others, to serve an unbalanced load. In case of failure of the transformer serving one phase only that one transformer need to replace; whereas it is most likely that three-phase transformer damaged in one phase will have to be completely removed from service, at least temporarily. However, three-phase transformer, in addition to being lighter than the equivalent bank of the single-phase units, take less place, are less expensive, and involve much less external wiring. Their efficiency is slightly better. Improved construction and better means of protection from over-voltages and short circuits have made transformer failure a very rare occurrence. As a result, three phase banks of single-phase transformers are seldom used in new installations except in distribution circuits where a combination of single and three-phase loads to be served.

In three-phase transformer, the primary an secondary windings may be connected independently in either delta or why. The possible combinations are Δ-Δ, Δ-Y, Y-Y, Y-Δ. Figure 5.8 shows diagrams of these four connections.

63

Figure 5.7

The Y-Y connection is to avoided unless a very solid neutral connection is made between the primary and the power source. If a neutral is not provided, the phase voltages tend to become severely unbalanced when load is unbalanced. These problems do not exist when one of the sets of the windings is in delta. When it is necessary to have a Y-Y connection with a weak primary neutral, or none, each phase transformer is provided with a third winding in addition to the primary and secondary, called a "tertiary". The tertiaries are connected in delta. This is a relatively expensive arrangement. The connection is called "why-delta-why". The tertiary winding terminals are often brought out to supply auxiliary power (e.g., lights, fans, and pumps) for the station. When a Y-Δ or Δ-Y connection is used, the Y is preferably on the high-voltage side, and the neutral is grounded. The transformer insulation may be designed for times the line voltage, rather than for the full line voltage. Sometimes it is necessary to have the Y on the low-voltage side, if a neutral is required for the low-voltage circuit.Y-Δ and Δ-Y connections result in a 30° phase displacement between the primary and secondary voltages. It is standard practice to connect these transformers in such a way that the secondary voltage lags the primary voltage by 30°.

3.9 Current transformers

For measuring currents in excess of about 100 A a current transformer is normally used. With a d.c. moving-coil ammeter the current required to give full scale deflection is very small—typically a few milli-amperes. When larger currents are to be measured a shunt resistor is added to the circuit. However, even with shunt resistors added it is not possible to measure very large currents. When a.c. is being measured a shunt cannot be used since the proportion of the current which flows in the meter will depend on its impedance, which varies with frequency.

In a double-wound transformer:

from which, secondary current

In current transformers the primary usually consists of one or two turns whilst the secondary can have several hundred turns. A typical arrangement is shown in Figure 20.19. If, for example, the primary has 2 turns and the secondary 200 turns, then if the primary current is 500 A,

64

Figure 5.8

secondary current,

Current transformers isolate the ammeter from the main circuit and allow the use of a standard range of ammeters giving full-scale deflections of 1 A, 2 A or 5 A.For very large currents the transformer core can be mounted around the conductor or bus-bar. Thus the primary then has just one turn. It is very important to short-circuit the secondary winding before removing the ammeter. This is because if current is flowing in the primary, dangerously high voltages could be induced in the secondary should it be open circuited.

3.10 Voltage transformers

For measuring voltages in excess of about 500 V it is often safer to use a voltage transformer. These are normal double-wound transformers with a large number of turns on the primary, which is connected to a high voltage supply, and a small number of turns on the secondary. A typical arrangement is shown in Figure 5.10.

Since

the secondary voltage,

Thus if the arrangement in Figure 5.10 has 4000 primary turns and 20 secondary turns then for a voltage of 22 kV on the primary, the voltageon the secondary,

Solved Examples:Example 5.1 : A 100 kVA, 4000 V/200 V, 50 Hz single-phase transformer has 100 secondary turns. Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.Solution:

(a) Transformer rating =

Hence primary current,

and secondary current,

(b) primary turns,

(c) As , and assuming

from which, maximum flux

Example 5.2 : A single-phase, 50 Hz transformer has 25 primary turns and 300 secondary turns. The cross-sectional area of the core is 300 cm². When the primary winding is connected to a 250 V supply, determine (a) the maximum value of the flux density in the core, and (b) the voltage induced in the secondary winding.Solution:

65

Figure 5.9

Figure 5.10

(a) assuming

However,

Hence

(b)

(i.e. the induced voltage on the secondary windings = 3 kV)

Example 5.3 : A transformer has 600 primary turns and 150 secondary turns. The primary and secondary resistances are 0.25 Ω and 0.01 Ω respectively, and the corresponding leakage reactances are 1.0 Ω and 0.04 Ω respectively. Determine (a) the equivalent resistance referred to the primary winding, (b) the equivalent reactance referred to the primary winding, (c) the equivalent impedance referred to the primary winding, and (d) the phase angle of the impedance..Solution:

(a) From equation (5.8), equivalent resistance

i.e.

= 0.41 Ω

(b) From equation (5.9), equivalent reactance,

i.e.

(c) From equation (5.10), equivalent impedance,

(d) From equation (5.11),

then,

Example 5.4 : The open circuit voltage of a transformer is 240 V. A tap changing device is set to operate when the percentage regulation drops below 2.5%. Determine the load voltage at which the mechanism operates.Solution:

Hence

from which, load voltage,

Example 5.5 : A 400 kVA transformer has a primary winding resistance of 0.5 Ω and a secondary winding resistance of 0.001 Ω. The iron loss is 2.5 kW and the primary and secondary voltages are5 kV and 320 V respectively. If the power factor of the load is 0.85, determine the efficiency of the transformer (a) on full load, and (b) on half load

66

Solution:(a) Transformer rating =

Hence primary current,

and secondary current,

Total copper loss = = (80)² (0.5) + (1250)² (0.001) = 4762.5 wattsOn full load, total loss = copper loss + iron loss = 4762.5 + 2500 = 7262.5 watts = 7.2625 kWTotal output power on full load = Input power = output power + losses = 340 kW + 7.2625 kW = 347.2625 kW

Efficiency x 100%

x 100% = 97.91%

(b) Since the copper loss varies as the square of the current, then total

copper loss on half load =

Hence total loss on half load = 1190.625 + 2500 = 3690.625 W or 3.691 kW

Output power on half full load =

Input power on half full load = output power + losses

= 170 kW + 3.691 kW = 173.691 kW

Hence efficiency at half full load, x 100%

x 100% = 97.87%

Example 5.6 : A 3-phase, 11kV/660V, Y-Δ transformer is connected to the far end of a distribution line for the near-end voltage is maintained at 11kV. The effective leakage reactance and resistance per-phase of the transformer are respectively 0.25Ω and 0.05 Ω referred to the low-voltage side. The reactance and resistance of each line are respectively 2 Ω and 1 Ω. It is required to maintain the terminal voltage at 660V when a line current of 260A at 0.8 lagging power factor is drawn from the secondary winding. What percentage tapping must be provided on the H.V. side of the transformer to permit the necessary adjustment? The transformer magnetizing current may be neglected and approximate expression for the regulation may be used. Neglect also the changes to the impedance due to the alteration of turns ration.Solution:

67

Figure 5.11

The equivalent circuit per phase is shown on the figure (5.11). The line impedance can also be referred to the secondary side and included in the regulation expression.

Referred line impedance = (0.0108 + j0.0216)Ω per phase

The total impedance referred to the secondary = (0.0108 + j0.0216)+(0.05 + 0.25) = (0.0608 + j0.2716)Ω

From figure 5.12 the value of voltage regulation

Per phase can be calculated as:

= 31.76 V (as )

So the open circuit of the transformer must be =

660 + 31.76 = 691.76 V

and turns ratio must be instead of

(i.e. 9.18 instead of 9.623)

So high voltage tapping must be at = 95.4%

Example 5.7 : A 10 kVA,, 2400V/240V, single phase transformer is connected has the following resistances and leakage reactances. Find the primary voltage required to produce 240V at the secondary terminals at full load, when the power factor is : (a) 0.8 lagging , (b) 0.8 leading

,,

Solution:The equivalent impedance referred to the secondary side is:

where,

= (0.06 + j 0.30) Ω

Ω

(a) 0.8 power factor lagging

With as a references, the full load secondary current is

68

Figure 5.12

=

=

= 240 + (9.506 + j8.506) V

= 249.506 + j8.506 V

=

= 10 x 249.65 = 2496.5 V

(b) 0.8 power factor leading

=

=

=

= 240 + (-5.505 + j11.51) V

= 234.5 + j11.51 V

=

= 10 x 243.78 = 2437.8 V

Example 5.8 : A, 7200V/208V, 50 kVA, 3-phase distribution transformer is connected Δ-Y . The equivalent leakage reactance and resistance per-phase of the transformer are respectively 0.0104Ω and 0.0433 Ω referred to the low-voltage side. Find the voltage regulation at full load, 0.85 power factor, lagging.Solution:The primary side is connected Δ, so the primary voltage and current are:

The secondary side is connected Y, so the secondary voltage and current are:

Turns ratio

69

The equivalent impedance per phase referred to the secondary side is

At full load 0.85 power factor lagging,

Applying Kirchhoff's voltage low toe equivalent circuit:

=

=

=

=

=

No-load secondary phase voltage = 124.57 V

No-load secondary line voltage = 124.57 = 215.76 V

=

70

UNIT 6 Electrical Motors

Introduction

This unit introduces the construction and principle of operation for DC motors, and AC single phase and three phases motors, their equivalent circuits, and efficiency.

6.1 Basic Motor Principles

All motors can be classed into two categories, AC and DC. The basic motor principles are alike for both the AC and DC motor.Magnetism is the basis for all electric motor operation. It produces the forces necessary for the motor to run. There are two basic types of magnets, the permanent magnet and the electromagnet. The electromagnet has the advantage over the permanent magnet in that the magnetic field can be made stronger. Also the polarity of the electromagnet can easily be reversed. The construction of an electromagnet is simple. When a current is passed through a coil of wire, a magnetic field is produced. This magnetic field can be made stronger by winding the coil of wire on an iron core (Fig. 6.1). One end of the electromagnet is a north pole, while the other end is a south pole. These poles can be reversed by reversing the direction of current in the coil of wire.The basic principle of all motors can be easily be shown using two electromagnets and a permanent magnet. Current is passed through coil #1 and coil #2 in such a direction that north and south poles are generated next to the permanent magnet, as shown in Figure (6.2.a) A permanent magnet with a north and south pole is the moving part of this simple motor. In Figure (6.2.a) the north pole of the permanent magnet is adjacent to the north pole of the electromagnet. Similarly, the south poles are adjacent to each other. Like magnetic poles repel each other, causing the movable permanent magnet to begin to turn. After it turns part way around, the force of attraction between the unlike poles becomes strong enough to keep the permanent magnet rotating. The rotating magnet continues to turn until the unlike poles are lined up. At this point the rotor would normally stop because of the attraction between the unlike poles (Fig. 6.2 (b)). If, however, the direction of currents in the electromagnetic coils was suddenly reversed, thereby reversing the polarity of the two coils, then the poles would again be opposites and repel each other (Fig.6.2(c)). The movable permanent magnet would then continue to rotate. If the current direction in the electromagnetic coils was changed every time the magnet turned 180° or halfway around, then the magnet would continue to rotate. This device is a motor in its simplest form. An actual motor is more complex than the simple device shown above, but the principle is the same.

71

(a) (b) (c)

Figure 6.2

Figure 6.1

6.2 Direct Current (DC) Motor

6.2.1 DC General Construction

A typical DC generator or motor usually consists of: An armature core, an air gap, poles, and a yoke which form the magnetic circuit; an armature winding, a field winding, brushes and a commutator which form the electric circuit; and a frame, end bells, bearings, brush supports and a shaft which provide the mechanical support (Figure 6.3).

Armature Core or StackThe armature stack is made up thin magnetic steel laminations stamped from sheet steel with a blanking die. Slots are punched in the lamination with a slot die. Sometimes these two operations are done as one. The laminations are welded, riveted, bolted or bonded together.

Armature WindingThe armature winding is the winding, which fits in the armature slots and is eventually connected to the commutator. It either generates or receives the voltage depending on whether the unit is a generator or motor. The armature winding usually consists of copper wire, either round or rectangular and is insulated from the armature stack.

Field PolesThe pole cores can be made from solid steel castings or from laminations. At the air gap, the pole usually fans out into what is known as a pole head or pole shoe. This is done to reduce the reluctance of the air gap. Normally the field coils are formed and placed on the pole cores and then the whole assembly is mounted to the yoke.

Field CoilsThe field coils are those windings, which are located on the poles and set up the magnetic fields in the machine. They also usually consist of copper wire are insulated from the poles. The field coils may be either shunt windings (in parallel with the armature winding) or series windings (in series with the armature winding) or a combination of both.

YokeThe yoke is a circular steel ring, which supports the field, poles mechanically and provides the necessary magnetic path between the pole. The yoke can be solid or laminated. In many DC machines, the yoke also serves as the frame.

72

Figure 6.3

CommutatorThe commutator is the mechanical rectifier, which changes the AC voltage of the rotating conductors to DC voltage. It consists of a number of segments normally equal to the number of slots. The segments or commutator bars are made of silver bearing copper and are separated from each other by mica insulation.

Brushes and Brush HoldersBrushes conduct the current from the commutator to the external circuit. There are many types of brushes. A brush holder is usually a metal box that is rectangular in shape. The brush holder has a spring that holds the brush in contact with the commutator. Each brush usually has a flexible copper shunt or pigtail, which extends to the lead wires. Often, the entire brush assembly is insulated from the frame and is made movable as a unit about the commutator to allow for adjustment.

InterpolesInterpoles are similar to the main field poles and located on the yoke between the main field poles. They have windings in series with the armature winding. Interpoles have the function of reducing the armature reaction effect in the commutating zone. They eliminate the need to shift the brush assembly.

Frame, End Bells, Shaft, and BearingsThe frame and end bells are usually steel, aluminum or magnesium castings used to enclose and support the basic machine parts. The armature is mounted on a steel shaft, which is supported between two bearings. The bearings are either sleeve, ball or roller type. They are normally lubricated by grease or oil.

Back End, Front EndThe load end of the motor is the Back End. The opposite load end, most often the commutator end, is the Front End of the motor.

6.2.2 Shunt, series and compound windings

When the field winding of a d.c. machine is connected in parallel with the armature, as shown in Figure 6.4(a), the machine is said to be shunt wound. If the field winding is connected in series with the armature, as shown in Figure 6.4(b), then the machine is said to be series wound. A compound wound machine has a combination of series and shunt windings.Depending on whether the electrical machine is series wound, shunt wound or compound wound, it behaves differently when a load is applied. The behaviour of a d.c. machine under various conditions is shown by means of graphs, called characteristic curves or just characteristics. The characteristics shown in the following sections are theoretical, since they neglect the effects of armature reaction.Armature reaction is the effect that the magnetic field produced by the armature current has on the magnetic field produced by the field system. In a generator, armature reaction results in a reduced output voltage, and in a motor, armature reaction results in increased speed. A way of overcoming the effect of armature reaction is to fit compensating windings, located in slots in the pole face.

73

Figure 6.4

6.2.3 E.m.f. generated in an armature winding

Let Z = number of armature conductors, Φ = useful flux per pole, in webers p = number of pairs of polesand n = armature speed in rev/s

The e.m.f. generated by the armature is equal to the e.m.f. generated by one of the parallel paths. Each conductor passes 2p poles per revolution and thus cuts 2pΦ webers of magnetic flux per revolution. Hence flux cut by one conductor per second = 2pΦn Wb and so the average e.m.f. E generated per conductor is given by:E = 2pΦn volts (since 1 volt = 1 Weber per second)Let c = number of parallel paths through the winding between positive and negative brushesc= 2 for a wave windingc= 2p for a lap winding

The number of conductors in series in each path = The total e.m.f. between brushes = (average e.m.f./conductor)(number of conductors in series per path)

=

i.e. generated e.m.f., (6.1)

Since Z, p and c are constant for a given machine, then .However is the angular velocity in radians per second, hence the generated e.m.f. is proportional to and , i.e., generated e.m.f., .

6.2.4 Motor Principle

Although the mechanical construction of dc motors and generators is very similar, their functions are different. The function of a generator is to generate a voltage when conductors are moved through a field, while that of a motor is to develop a turning effort, or torque, to produce mechanical rotation.

Direction of Armature Rotation

The left-hand rule is used to determine the direction of rotation of the armature conductors. The left-hand rule for motors is as follows: With the forefinger, middle finger, and thumb of the left hand mutually perpendicular, point the forefinger in the direction of the field and the middle finger in the direction of the current in the conductor; the thumb will point in the direction in which the conductor tends to move (Fig. 6.5a). In a single-turn rectangular coil parallel to a magnetic field (Fig. 6.5b), the direction of current in

74

(a) single conductor (b) two conductors of a single loop coil

Figure 6.5

the left-hand conductor is out of the paper, while in the right-hand conductor it is into the paper. Therefore, the left-hand conductor tends to move upward with a force F1, and the right-hand conductor tends to move downward with an equal force F2. Both forces act to develop a torque which turns the coil in a clockwise direction. A single-coil motor (Fig. 6.5b) is impractical because it has dead centers and the torque developed is pulsating. Good results are obtained when a large number of coils is used as in a four-pole motor. As the armature rotates and the conductors move away from under a pole into the neutral plane, the current is reversed in them by the action of the commutator. Thus, the conductors under a given pole carry current in the same direction at all times.

6.2.5 DC Motor Equivalent Circuit

The equivalent circuit of the d.c. motor consists from a d.c. source E and connected with it in series a resistance . Voltage and current relationships of a dc motor equivalent circuit (Fig. 6.6) are as follows:

(6.2)

6.2.6 Torque of a d.c. Machine

From equation (6.2), for a d.c. motor, the supply voltage is given by:

Multiplying each term by current gives:

The term is the total electrical power supplied to the armature, the term is the loss due to armature resistance, and the term is the mechanical power developed by the armature.If T is the torque, in newton meters, then the mechanical power developed is given by wattsHence, , from which

Torque Newton.meters (6.3)

From equation (6.1), the e.m.f. E generated is given by:

Then, Torque Newton.meters

i.e. Torque Newton.meters (6.4)

For a given machine, Z, c and p are fixed valuesHence torque, (6.5)

75

Figure 6.6

6.2.7 Types of d.c. motor and their characteristics

(a) Shunt-wound motor

In the shunt wound motor the field winding is in Parallel with the armature across the supply as shown in Figure 6.7.For the circuit shown in Figure 6.7,Supply voltage, or generated e.m.f., Supply current,

CharacteristicsThe two principal characteristics are the torque/armature current and speed/armature current relationships. From these, the torque/speed relationship can be derived.(i) The theoretical torque/armature current characteristic can be derived from the expression . For a shunt-wound motor, the field winding is connected in parallel with the armature circuit and thus the applied voltage gives a constant field current, i.e. a shunt-wound motor is a constant flux machine. Since

is constant, it follows that , and the characteristic is as shown in Figure 6.8.(ii) The armature circuit of a d.c. motor has resistance due to the armature winding and brushes, ohms, and when armature current is flowing through it, there is a voltage drop of volts. In Figure 6.7 the armature resistance is shown as a separate resistor in the armature circuit to help understanding. Also, even though the machine is a motor, because conductors are rotating in a magnetic field, a voltage, , is generated by the armature conductors. From equation (6.2) , or However, as , hence, i.e.

speed of rotation, , (6.6)

For a shunt motor, V, and are constants, hence as armature current increases, increases and decreases, and the speed is proportional to a quantity which is decreasing and is as shown in Figure 6.9. As the load on the shaft of the motor increases, increases and the speed drops slightly. In practice, the speed falls by about 10% between no-load and full-load on many d.c. shunt-wound motors. Due to this relatively small drop in speed, the d.c. shunt-wound motor is taken as basically being a constant-speed machine and may be used for driving lathes, lines of shafts, fans, conveyor belts, pumps, compressors, drilling machines and so on.(iii) Since torque is proportional to armature current, (see (i) above), the theoretical speed/torque characteristic is as shown in Figure 6.10.

(b) Series-wound motorIn the series-wound motor the field winding is in series with the armature across the supply as shown in Figure 6.11.For the series motor shown in Figure 6.11,

76

Figure 6.7

Figure 6.8 Figure 6.9 Figure 6.10

Figure 6.11

Supply voltage or generated e.m.f.

CharacteristicsIn a series motor, the armature current flows in the field winding and is equal to the supply current, I.(i) The torque/current characteristicIt is shown in Section (5.2.6) that torque . Since the armature and field currents are the same current, I, in a series machine, then over a limited range, before magnetic saturation of the magnetic circuit of the motor is reached, (i.e., the linear portion of the B–H curve for the yoke, poles, air gap, brushes and armature in series). Thus and . After magnetic saturation, almost becomes a constant and . Thus the theoretical torque/current characteristic is as shown in Figure 6.12.

(ii) The speed/current characteristic

It is shown in equation (6.6) that . In a series motor, and below the magnetic

saturation level, . Thus where R is the combined resistance of the series

field and armature circuit. Since IR is small compared with , then an approximate relationship for the

speed is since is constant. Hence the theoretical speed/current characteristic is as

shown in Figure 6.13. The high speed at small values of current indicate that this type of motor must not be run on very light loads and invariably, such motors are permanently coupled to their loads.

(iii) The theoretical speed/torque characteristic may be derived from (i) and (ii) above by obtaining the torque and speed for various values of current and plotting the co-ordinates on the speed/torque characteristics. A typical speed/torque characteristic is shown in Figure 6.14.A d.c. series motor takes a large current on starting and the characteristic shown in Figure 6.12 shows that the series-wound motor has a large torque when the current is large. Hence these motors are used for traction (such as trains, milk delivery vehicles, etc.), driving fans and for cranes and hoists, where a large initial torque is required.

77

Figure 6.12 Figure 6.13 Figure 6.14

(c) Compound-wound motorThere are two types of compound wound motor:(i) Cumulative compound, in which the series winding is so connected that the field due to it assists that due to the shunt winding.(ii) Differential compound, in which the series winding is so connected that the field due to it opposes that due to the shunt winding. Figure 6.15(a) shows a long-shunt compound motor and Figure 6.15(b) a short-shunt compound motor.

CharacteristicsA compound-wound motor has both a series and a shunt field winding, (i.e. one winding in series and one in parallel with the armature), and is usually wound to have a characteristic similar in shape to a series wound motor (see Figures 6.12–6.14). A limited amount of shunt winding is present to restrict the no-load speed to a safe value. However, by varying the number of turns on the series and shunt windings and the directions of the magnetic fields produced by these windings (assisting or opposing), families of characteristics may be obtained to suit almost all applications. Generally, compound-wound motors are used for heavy duties, particularly in applications where sudden heavy load may occur such as for driving plunger pumps, presses, geared lifts, conveyors, hoists and so on.Typical compound motor torque and speed characteristics are shown in Figure 6.16.

6.2.8 Changing direction of rotation of a d.c. motor

The direction of rotation of a d.c. motor can be done simply by changing the direction of one of the magnetic fields either in stator or rotor. It is not convenient to change the direction of the magnetic field of the stator especially in the compound wound stators. Changing direction of rotation is achieved by changing the polarity of brushes which result in changing the direction of current in the armature making the direction of magnetic field to be changed. There are some mechanical or electronic circuits that can change the polarity of the brushes manually or automatically as requested. In every time of changing direction of rotation the motor must be braked before changing circuit begins.

6.2.9 The efficiency of a d.c. motorThe efficiency of a d.c. machine is given by:

Also, the total losses = (for a shunt motor) where C is the sum of the iron, friction and windage losses.For a motor, the input power =

78

Figure 6.15

Figure 6.16

and the output power = – losses =

Hence efficiency (6.7)

The efficiency of a motor is a maximum when the load is such that:

Solved Examples:

Example 6.1: A six-pole lap-wound motor is connected to a 250 V d.c. supply. The armature has 500 conductors and a resistance of 1Ω. The flux per pole is 20 mWb. Calculate (a) the speed and (b) the torque developed when the armature current is 40 A.Solution :

= 250 V, Z = 500, = 1 Ω, Wb, = 40 A, c = 2p for a lap winding(a) Back e.m.f. = 250 – (40)(1) = 210 V

E.m.f

i.e.

Hence speed rev/s

or = 21 x 60 = 1260 rev/min

(b) Torque N.m

Example 6.2: An 8-pole d.c. motor has a wave-wound armature with 900 conductors. The useful flux per pole is 25 mWb. Determine the torque exerted when a current of 30 A flows in each armature conductor.Solution :p = 4, Z = 900, Wb, = 30 A, c = 2 for a wave winding

From equation (6.4), torque

N.m

Example 6.3: A 220 V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The armature circuit resistance is 0.4 Ω. Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and (b) the steady state value of the armature current at the new value of flux, assuming the shaft torque of the motor remains constant.Solution :(a) For a d.c. shunt-wound motor, . Hence initial generated e.m.f., V. The generated e.m.f. is also such that , so at the instant the flux is reduced, the speed has not had time to change, and E = 208 x 90/100 = 187.2 V.Hence, the voltage drop due to the armature resistance is (220 – 187.2), i.e., 32.8 V. The instantaneous value of the current is 32.8/0.4, i.e., 82 A. This increase in current is about three times the initial value and causes an increase in torque, ( ). The motor accelerates because of the larger torque value until steady state conditions are reached.(b) , and since the torque is constant, then , The flux is reduced by 10%, hence Thus,

79

i.e. the steady state value of armature current, A

Example 6.4: A series motor has an armature resistance of 0.2 Ω and a series field resistance of 0.3 Ω. It is connected to a 240 V supply and at a particular load runs at 24 rev/s when drawing 15 A from the supply.(a) Determine the generated e.m.f. at this load.(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A. Assume that this causes a doubling of the flux. Solution :(a) With reference to Figure 6.11, generated e.m.f., E, at initial load, is given by:

= 240 – 15 (0.2 + 0.3) = 240 – 7.5 = 232.5 V(b) When the current is increased to 30 A, the generated e.m.f. is given by:

= 240 – 30 (0.2 + 0.3) = 240 – 15 = 225 VNow e.m.f.

thus

i.e., as

Hence speed of motor, rev/s

As the current has been increased from 15 A to 30 A, the speed has decreased from 24 rev/s to 11.6 rev/s. Its speed/current characteristic is similar to Figure 6.13.

Example 6.5: A 320 V shunt motor takes a total current of 80 A and runs at 1000 rev/min. If the iron, friction and windage losses amount to 1.5 kW, the shunt field resistance is 40 Ω and the armature resistance is 0.2 Ω, determine the overall efficiency of the motor.Solution :

Field current, A

Armature current = 80 – 8 = 72 AC = iron, friction and windage losses = 1500 W

Efficiency,

80

Example 6.6: A d.c. series motor drives a load at 30 rev/s and takes a current of 10 A when the supply voltage is 400 V. If the total resistance of the motor is 2 Ω and the iron, friction and windage losses amount to 300 W, determine the efficiency of the motor.Solution :

Efficiency,

6.3 Alternating Current (AC) Motor

The AC motors are mainly two types, synchronous motors and induction motors or asynchronous motors. Also, induction motors may be three phases or single phase. Only induction motors, three phases and single phase will be explained.In d.c. motors, introduced in section 5.2, conductors on a rotating armature pass through a stationary magnetic field. In a three-phase induction motor, the magnetic field rotates and this has the advantage that no external electrical connections to the rotor need be made. Its name is derived from the fact that the current in the rotor is induced by the magnetic field instead of being supplied through electrical connections to the supply. The result is a motor which: (i) is cheap and robust, (ii) is explosion proof, due to the absence of a commutator or slip-rings and brushes with their associated sparking, (iii) requires little or no skilled maintenance, and (iv) has self-starting properties when switched to a supply with no additional expenditure on auxiliary equipment. The principal disadvantage of a three-phase induction motor is that its speed cannot be readily adjusted.

6.3.1 Basic AC Motor Operation

An AC motor has two basic electrical parts: a "stator" and a "rotor" as shown in Figure 6.17. The stator is in the stationary electrical component. It consists of a group of individual electro-magnets arranged in such a way that they form a hollow cylinder, with one pole of each magnet facing toward the center of the group. The term, "stator" is derived from the word stationary. The stator then is the stationary part of the motor. The rotor is the rotating electrical component. It also consists of a group of electro-magnets arranged around a cylinder, with the poles facing toward the stator poles. The rotor, obviously, is located inside the stator and is mounted on the motor's shaft. The term "rotor" is derived from the word rotating. The rotor then is the rotating part of the motor. The objective of these motor components is to make the rotor rotate which in turn will rotate the motor shaft. This rotation will occur because of the previously discussed magnetic phenomenon that unlike magnetic poles attract each other and like poles repel. If we progressively change the polarity of the stator poles in such a way that their combined magnetic field rotates, then the rotor will follow and rotate with the magnetic field of the stator.

81

Figure 6.8

Figure 6.17

This "rotating magnetic fields of the stator can be better understood by examining Figure 6.18. As shown, the stator has six magnetic poles and the rotor has two poles. At time 1, stator poles A-1 and C-2 are north poles and the opposite poles, A-2 and C-1, are south poles. The S-pole of the rotor is attracted by the two N-poles of the stator and the N-pole of the rotor is attracted by the two south poles of the stator. At time 2, the polarity of the stator poles is changed so that now C-2 and B-1 and N-poles and C-1 and B-2 are S-poles. The rotor then is forced to rotate 60 degrees to line up with the stator poles as shown. At time 3, B-1 and A-2 are N. At time 4, A-2 and C-1 are N. As each change is made, the poles of the rotor are attracted by the opposite poles on the stator. Thus, as the magnetic field of the stator rotates, the rotor is forced to rotate with it.

6.3.2 Production of a rotating magnetic field

One way to produce a rotating magnetic field in the stator of an AC motor is to use a three-phase power supply for the stator coils. What, you may ask, is three-phase power? The answer to that question can be better understood if we first examine single-phase power. Figure 7 is the visualization of single-phase power. The associated AC generator is producing just one flow of electrical current whose direction and intensity varies as indicated by the single solid line on the graph. From time 0 to time 3, current is flowing in the conductor in the positive direction. From time 3 to time 6, current is flowing in the negative. At any one time, the current is only flowing in one direction. But some generators produce three separate current flows (phases) all superimposed on the same circuit. This is referred to as three-phase power. At any one instant, however, the direction and intensity of each separate current flow are not the same as the other phases. This is illustrated in Figure 6.19. The three separate phases (current flows) are labeled A, B and C. At time 1, phase A is at zero amps, phase B is near its maximum amperage and flowing in the positive direction, and phase C is near to its maximum amperage but flowing in the negative direction. At time 2, the amperage of phase A is increasing and flow is positive, the amperage of phase B is decreasing and its flow is still negative, and phase C has dropped to zero amps. A complete cycle (from zero to maximum in one direction, to zero and to maximum in the other direction, and back to zero) takes one complete revolution of the generator. Therefore, a complete cycle, is said to have 360 electrical degrees. In examining Figure 10, we see that each phase is displaced 120 degrees from the other two phases. Therefore, we say they are 120 degrees out of phase.

82

Figure 6.18

To produce a rotating magnetic field in the stator of a three-phase AC motor, all that needs to be done is wind the stator coils properly and connect the power supply leads correctly. The connection for a 6 pole stator is shown in Figure 6.20. Each phase of the three-phase power supply is connected to opposite poles and the associated coils are wound in the same direction. As you will recall from Figure 4, the polarity of the poles of an electro-magnet are determined by the direction of the current flow through the coil. Therefore, if two opposite stator electro-magnets are wound in the same direction, the polarity of the facing poles must be opposite. Therefore, when pole A1 is N, pole A2 is S. When pole B1 is N, B2 is S and so forth.Figure 6.21 shows how the rotating magnetic field is produced. At time1, the current flow in the phase "A" poles is positive and pole A-1 is N. The current flow in the phase "C" poles is negative, making C-2 a N-pole and C-1 is S. There is no current flow in phase "B", so these poles are not magnetized. At time 2, the phases have shifted 60 degrees, making poles C-2 and B-1 both N and C-1 and B-2 both S. Thus, as the phases shift their current flow, the resultant N and S poles move clockwise around the stator, producing a rotating magnetic field. The rotor acts like a bar magnet, being pulled along by the rotating magnetic field.

83

Figure 6.19

Figure 6.20

Figure 6.21

Up to this point not much has been said about the rotor. In the previous examples, it has been assumed the rotor poles were wound with coils, just as the stator poles, and supplied with DC to create fixed polarity poles. This, by the way, is exactly how a synchronous AC motor works. However, most AC motors being used today are not synchronous motors. Instead, so-called "induction" motors are the workhorses of industry.

6.3.2 Synchronous speed

The rotating magnetic field produced by three phase windings could have been produced by rotating a permanent magnet’s north and south pole at synchronous speed, (shown as N and S at the ends of the flux phasors in Figures 6.18). For this reason, it is called a 2-pole system and an induction motor using three phase windings only is called a 2-pole induction motor.If six windings displaced from one another by 60° are used, as shown in Figure 6.22 (a), by drawing the current and resultant magnetic field diagrams at various time values, it may be shown that one cycle of the supply current to the stator windings causes the magnetic field to move through half a revolution. The current distribution in the stator windings are shown in Figure 6.22(a), for the time t shown in Figure 6.22(b).It can be seen that for six windings on the stator, the magnetic flux produced is the same as that produced by rotating two permanent magnet north poles and two permanent magnet south poles at synchronous speed. This is called a 4-pole system and an induction motor using six phase windings is called a 4-pole induction motor. By increasing the number of phase windings the number of poles can be increased to any even number. In general, if f is the frequency of the currents in the stator windings and the stator is wound to be equivalent to p pairs of poles, the speed of revolution of the rotating magnetic field, i.e., the synchronous speed, is given by:

rev/s (6.8)

6.4 Three Phase Induction Motors

6.4.1 Basic Construction and Operating Principles

Like most motors, an AC induction motor has a fixed outer portion, called the stator and a rotor that spins inside with a carefully engineered air gap between the two. Virtually all electrical motors use magnetic field rotation to spin their rotors. A three-phase AC induction motor is the only type where the rotating magnetic field is created naturally in the stator because of the nature of the supply. DC motors depend either on mechanical or electronic commutation to create rotating magnetic fields. A single-phase AC induction motor depends on extra electrical components to produce this rotating magnetic field.

Stator

84

Figure 6.22

The stator is made up of several thin laminations of aluminum or cast iron. They are punched and clamped together to form a hollow cylinder (stator core) with slots as shown in Figure 6.23. Coils of insulated wires are inserted into these slots. Each grouping of coils, together with the core it surrounds, forms an electromagnet (a pair of poles) on the application of AC supply. The number of poles of an AC induction motor depends on the internal connection of the stator windings. The stator windings are connected directly to the power source. Internally they are connected in such a way, that on applying AC supply, a rotating magnetic field is created.The number of slots must be even number and so that if it is divided on three (no. of phases) then the result is divided on the number of poles must produce a result without fraction. The result of division is the number of slots per phase per pole (so it must be integer). For example if the number of slots is 36, and the motor has 4 poles, then the number of slots per phase per pole is 3.

Squirrel-Cage RotorThe rotor is made up of several thin steel laminations with evenly spaced bars, which are made up of aluminum or copper, along the periphery. In the most popular type of rotor (squirrel cage rotor), these bars are connected at ends mechanically and electrically by the use of rings. Almost 90% of induction motors have squirrel cage rotors. This is because the squirrel cage rotor has a simple and rugged construction. The rotor consists of a cylindrical laminated core with axially placed parallel slots for carrying the conductors. Each slot carries a copper, aluminum, or alloy bar. These rotor bars are permanently short-circuited at both ends by means of the end rings, as shown in Figure 6.24. This total assembly resembles the look of a squirrel cage, which gives the rotor its name. The rotor slots are not exactly parallel to the shaft. Instead, they are given a skew for two main reasons. The first reason is to make the motor run quietly by reducing magnetic hum and to decrease slot harmonics. The second reason is to help reduce the locking tendency of the rotor. The rotor teeth tend to remain locked under the stator teeth due to direct magnetic attraction between the two. This happens when the number of stator teeth are equal to the number of rotor teeth. The rotor is mounted on the shaft using bearings on each end; one end of the shaft is normally kept longer than the other for driving the load. Some motors may have an accessory shaft on the non-driving end for mounting speed or position sensing devices. Between the stator and the rotor, there exists an air gap, through which due to induction, the energy is transferred from the stator to the rotor. The generated torque forces the rotor and then the load to rotate. Regardless of the type of rotor used, the principle employed for rotation remains the same.

85

Figure 6.24

Figure 6.23

Wound RotorThe wound rotor or slip-ring rotor differs from the squirrel-cage rotor in the rotor winding. The rotor winding consists of insulated coils, grouped to form definite polar areas of magnetic force having the same number of poles as the stator. The ends of these coils are brought out to slip-rings. By means of brushes, a variable resistance is placed across the rotor winding (Figure 6.25). By varying this resistance, the speed and torque of the motor is varied. The wound rotor motor is an excellent motor for use on applications that require an adjustable-varying speed (an adjustable speed that varies with load) and high starting torque.

6.4.2 Slip

The force exerted by the rotor bars causes the rotor to turn in the direction of the rotating magnetic field. As the rotor speed increases, the rate at which the rotating magnetic field cuts the rotor bars is less and the frequency of the induced e.m.f.’s in the rotor bars is less. If the rotor runs at the same speed as the rotating magnetic field, no e.m.f.’s are induced in the rotor, hence there is no force on them and no torque on the rotor. Thus the rotor slows down. For this reason the rotor can never run at synchronous speed.When there is no load on the rotor, the resistive forces due to windage and bearing friction are small and the rotor runs very nearly at synchronous speed. As the rotor is loaded, the speed falls and this causes an increase in the frequency of the induced e.m.f.’s in the rotor bars and hence the rotor current, force and torque increase. The difference between the rotor speed, , and the synchronous speed, , is called theslip speed, i.e.

slip speed = - rev/s (6.9)The ratio ( - )/ is called the fractional slip or just the slip, s, and is usually expressed as a percentage. Thus

slip, s = (6.10)

86

Figure 6.25

6.4.3 Induction motor torque–speed characteristics

The torque-speed and torque-slip characteristics of an induction motor are as shown in Figure 6.26.

Starting Characteristic

Induction motors, at rest, appear just like a short circuited transformer and if connected to the full supply voltage, draw a very high current known as the “Locked Rotor Current.” They also produce torque which is known as the “Locked Rotor Torque”. The Locked Rotor Torque (LRT) and the Locked Rotor Current (LRC) are a function of the terminal voltage of the motor and the motor design. As the motor accelerates, both the torque and the current will tend to alter with rotor speed if the voltage is maintained constant.The starting current of a motor with a fixed voltage will drop very slowly as the motor accelerates and will only begin to fall significantly when the motor has reached at least 80% of the full speed. The actual curves for the induction motors can vary considerably between designs but the general trend is for a high current until the motor has almost reached full speed. The LRC of a motor can range from 500% of Full-Load Current (FLC) to as high as 1400% of FLC. Typically, good motors fall in the range of 550% to 750% of FLC.The starting torque of an induction motor starting with a fixed voltage will drop a little to the minimum torque, known as the pull-up torque, as the motor accelerates and then rises to a maximum torque, known as the breakdown or pull-out torque, at almost full speed and then drop to zero at the synchronous speed. The curve of the start torque against the rotor speed is dependant on the terminal voltage and the rotor design. The LRT of an induction motor can vary from as low as 60% of FLT to as high as 350% of FLT. The pull-up torque can be as low as 40% of FLT and the breakdown torque can be as high as 350% of FLT. Typically, LRTs for medium to large motors are in the order of 120% of FLT to 280% of FLT. The PF of the motor at start is typically 0.1-0.25, rising to a maximum as the motor accelerates and then falling again as the motor approaches full speed.

87

Figure 6.26

Running Characteristic

Once the motor is up to speed, it operates at a low slip, at a speed determined by the number of the stator poles. Typically, the full-load slip for the squirrel cage induction motor is less than 5%. The actual full-load slip of a particular motor is dependant on the motor design. The typical base speed of the four pole induction motor varies between 1420 and 1480 RPM at 50 Hz, while the synchronous speed is 1500 RPM at 50 Hz.The current drawn by the induction motor has two components: reactive component (magnetizing current) and active component (working current). The magnetizing current is independent of the load but is dependant on the design of the stator and the stator voltage. The actual magnetizing current of the induction motor can vary, from as low as 20% of FLC for the large two pole machine, to as high as 60% for the small eight pole machine. The working current of the motor is directly proportional to the load. The tendency for the large machines and high-speed machines is to exhibit a low magnetizing current, while for the low-speed machines and small machines the tendency is to exhibit a high magnetizing current. A typical medium sized four pole machine has a magnetizing current of about 33% of FLC.A low magnetizing current indicates a low iron loss, while a high magnetizing current indicates an increase in iron loss and a resultant reduction in the operating efficiency.Typically, the operating efficiency of the induction motor is highest at 3/4 load and varies from less than 60% for small low-speed motors to greater than 92% for large high-speed motors. The operating PF and efficiencies are generally quoted on the motor data sheets.

Load Characteristic

In real applications, various kinds of loads exist with different torque-speed curves. For example, Constant Torque, Variable Speed Load (screw compressors, conveyors, feeders), Variable Torque, Variable Speed Load (fan, pump), Constant Power Load (traction drives), Constant Power, Constant Torque Load (coiler drive) and High Starting/Breakaway Torque followed by Constant Torque Load (extruders, screw pumps).The motor load system is said to be stable when the developed motor torque is equal to the load torque requirement. The motor will operate in a steady state at a fixed speed. The response of the motor to any disturbance gives us an idea about the stability of the motor load system. This concept helps us in quickly evaluating the selection of a motor for driving a particular load.In most drives, the electrical time constant of the motor is negligible as compared to its mechanical time constant. Therefore, during transient operation, the motor can be assumed to be in an electrical equilibrium, implying that the steady state torque-speed curve is also applicable to the transient operation.As an example, Figure 6.27 shows torque-speed curves of the motor with two different loads. The system can be termed as stable, when the operation will be restored after a small departure from it, due to a disturbance in the motor or load. For example, disturbance causes a reduction of in speed. In the first case, at a new speed, the motor torque ( ) is greater than the load torque ( ). Consequently, the motor will accelerate and the operation will be restored to X. Similarly, an increase of in the speed, caused by a disturbance, will make the load torque ( ) greater than the motor torque ( ), resulting in a deceleration and restoration of the point of operation to X. Hence, at point X, the system is stable.In the second case, a decrease in the speed causes the load torque ( ) to become greater than the motor torque ( ), the drive decelerates and the operating point moves away from Y. Similarly, an increase in the speed will make the motor torque ( ) greater than the load torque ( ), which will move the operating point further away from Y. Thus, at point Y, the system is unstable.This shows that, while in the first case, the motor selection for driving the given load is the right one; in the second case, the selected motor is not the right choice and requires changing for driving the given load.

88

Figure 6.27

6.4.4 Starting methods for induction motors

Squirrel-cage rotor(i) Direct-on-line startingWith this method, starting current is high and may cause interference with supplies to other consumers.(ii) Auto transformer startingWith this method, an auto transformer is used to reduce the stator voltage, E1, and thus the starting current. However, the starting torque is seriously reduced, so the voltage is reduced only sufficiently to give the required reduction of the starting current. A typical arrangement is shown in Figure 6.28. A double-throw switch connects the auto transformer in circuit for starting, and when the motor is up to speed the switch is moved to the run position which connects the supply directly to the motor.(iii) Star-delta startingWith this method, for starting, the connections to the stator phase winding are star-connected, so that the voltage across each phase winding is (i.e. 0.577) of the line voltage and the line current which equals the phase current is one third of the line current of the delta connection . For running, the windings are switched to delta-connection. This method of starting is less expensive than by auto transformer.

Wound rotorWhen starting on load is necessary, a wound rotor induction motor must be used. This is because maximum torque at starting can be obtained by adding external resistance to the rotor circuit via slip rings. A face-plate type starter is used, and as the resistance is gradually reduced, the machine characteristics at each stage will be similar to Q, S, R and P of Figure 6.29. At each resistance step, the motor operation will transfer from one characteristic to the next so that the overall starting characteristic will be as shown by the bold line in Figure 6.29. For very large induction motors, very gradual and smooth starting is achieved by a liquid type resistance.

6.4.5 Changing direction of rotation of a

three phase induction motor

Changing the direction of rotation in three phase induction motors is simply done by changing the direction of the rotating field. It simply achieved by changing two feeding phases by each other keeping

89

Figure 6.28

Figure 6.29

the third feeding phase in its place. As an example changing phase R with phase Y keeping phase B in its place. There are some mechanical or electronic circuits that can change two phases manually or automatically as requested. In every time of changing direction of rotation the motor must be braked before changing circuit begins.

6.4.6 Induction motor losses and efficiency

Figure 6.30 summarizes losses in induction motors.

(6.11)

(6.12)

(6.13)

where, is the input power to the motor, is the input power to the rotor, or the air-gap power is the developed mechanical power of the rotor , and is the output power of the rotorRotor copper losses = s (Rotor input power ( )) (6.14)

= (1 - s) Rotor input power for 3-phase induction motor (6.15)

(6.16)where , T is the developed torque in N.m and is rotor angular velocity

Efficiency (6.17)

90

Figure 6.30

Solved Examples:

Example 6.7: The power supplied to a three-phase induction motor is 32 kW and the stator losses are 1200 W. If the slip is 5%, determine (a) the rotor copper loss, (b) the total mechanical power developed by the rotor, (c) the output power of the motor if friction and windage losses are 750 W, and (d) the efficiency of the motor, neglecting rotor iron loss.

Solution :(a) Input power to rotor = stator input power - stator losses

= 32 kW - 1.2 kW = 30.8 kW As, Rotor copper losses = s (Rotor input power ( )) then from which, rotor copper loss = (0.05)(30.8) = 1.54 kW(b) Total mechanical power developed by the rotor = rotor input power - rotor losses = 30.8 - 1.54 = 29.26 kW

Example 6.8: An induction motor draws 25A from a 460 V, three phase line at a power factor of 0.85, lagging. The stator copper loss is 1000 W, and the rotor copper loss is 500 W. The rotational losses are friction and windage = 250 W, core loss = 800 W, and stray load loss = 200 W. Calculate (a) the air-gap power , (b) the developed mechanical power , (c) the output horsepower and (d) the efficiency.Solution :(a) the air-gap power,

= (460)(25)(0.85) – 1000 = 16931 – 1000 = 15931 W(b) the developed mechanical power,

= 15931 – 500 = 15431 W(c) the output power,

= 15431 – ( 250 + 800 + 200) = 15431 – 1250 = 14181 W

the output horsepower = = 19 hp

(d) Efficiency = x 100 = 83.8 %

Example 6.9: If the frequency of the source in Example 6.8 is 50 Hz, and the motor has four poles, find (a) the slip, (b) the operating speed, and (c) the output torque.Solution:(a) Rotor copper losses = s (Rotor input power ( ))

Then the slip, s = = 0.0314

(b) the synchronous speed = = 25 rev/s = 1500 rev/min

As the slip, s =

Then the rotor speed, = 1500(1 – 0.0314) = 1452.9 rev/min(c) the output power,

Then, the output torque

= = 157.08 rad/s

= 152.15 rad/s91

the output torque, = = 93.2 N.m

Example 6.10: In a 3-phase, 380 V, Δ-stator connection, 6-pole, 50-Hz induction motor the useful torque on full-load is 135.66 N.m on a slip = 0.03. The total mechanical power loss is 1.2 hp and the stator copper loss is 750 W. Calculate (a) the rotor copper loss , (b) the motor input power (c) efficiency, and (d) the line input current if the motor operates on a 0.84 p.f .Solution :(a) the output power,

= = 104.72 rad/s

= 101.58 rad/s the output power, = 135.66 x 101.58 = 13780.3 W the total mechanical power loss = 1.2 x 746 = 895.2 W the developed mechanical power of the rotor = + total mechanical power loss = 13780.3 + 895.2 = 14675.5 W As, = (1 - s) Rotor input power , and Rotor copper losses = s (Rotor input power ( ))

then,

Rotor copper losses = rotor developed mechanical power

= x 14675.5 = 453.9 W

(b) Stator losses = 750 W The motor input power = the developed mechanical power + rotor copper losses + stator losses = 14675.5 + 453.9 + 750 = 15879.4 W

(c) The efficiency = x 100 = 86.78 %

(d) The input power

Then, the line current = 28.72 A

92

6.5 Single-Phase Induction Motor

There are probably more single-phase AC induction motors in use today than the total of all the other types put together. It is logical that the least expensive, lowest maintenance type motor should be used most often. The single-phase AC induction motor best fits this description.As the name suggests, this type of motor has only one stator winding (main winding) and operates with a single-phase power supply. In all single-phase induction motors, the rotor is the squirrel cage type.The single-phase induction motor is not self-starting. When the motor is connected to a single-phase power supply, the main winding carries an alternating current. This current produces a pulsating magnetic field. Due to induction, the rotor is energized. As the main magnetic field is pulsating, the torque necessary for the motor rotation is not generated. This will cause the rotor to vibrate, but not to rotate. Hence, the single There are probably more single-phase AC induction motors in use today than the total of all the other types put together. It is logical that the least expensive, lowest maintenance type motor should be used most often. The single-phase AC induction motor best fits this description. As the name suggests, this type of motor has only one stator winding (main winding) and operates with a single-phase power supply. In all single-phase induction motors, the rotor is the squirrel cage type.The single-phase induction motor is not self-starting. When the motor is connected to a single-phase power supply, the main winding carries an alternating current. This current produces a pulsating magnetic field. Due to induction, the rotor is energized. As the main magnetic field is pulsating, the torque necessary for the motor rotation is not generated. This will cause the rotor to vibrate, but not to rotate. Hence, the single phase induction motor is required to have a starting mechanism that can provide the starting kick for the motor to rotate.The starting mechanism of the single-phase induction motor is mainly an additional stator winding (start/ auxiliary winding) as shown in Figure 6.31. The start winding can have a series capacitor and/or a centrifugal switch. When the supply voltage is applied, current in the main winding lags the supply voltage due to the main winding impedance. At the same time, current in the start winding leads/lags the supply voltage depending on the starting mechanism impedance. Interaction between magnetic fields generated by the main winding and the starting mechanism generates a resultant magnetic field rotating in one direction. The motor starts rotating in the direction of the resultant magnetic field.Once the motor reaches about 75% of its rated speed, a centrifugal switch disconnects the start winding. From this point on, the single-phase motor can maintain sufficient torque to operate on its own.Except for special capacitor start/capacitor run types, all single-phase motors are generally used for applications up to 3/4 hp only.Depending on the various start techniques, single phase AC induction motors are further classified as described in the following sections.

93

Single-Phase AC Induction Motor Single-Phase AC Induction Motor

without Start Mechanism with Start Mechanism

Figure 6.31

6.5.1 Split-Phase AC Induction Motor

The split-phase motor is also known as an induction start/induction run motor. It has two windings: a start and a main winding. The start winding is made with smaller gauge wire and fewer turns, relative to the main winding to create more resistance, thus putting the start winding’s field at a different angle than that of the main winding which causes the motor to start rotating. The main winding, which is of a heavier wire, keeps the motor running the rest of the time. Figure 6.32 shows a typical AC split-phase induction motor.The starting torque is low, typically 100% to 175% of the rated torque. The motor draws high starting current, approximately 700% to 1,000% of the rated current. The maximum generated torque ranges from 250% to 350% of the rated torque (see Figure 6.37 for torque-speed curve). Good applications for split-phase motors include small grinders, small fans and blowers and other low starting torque applications with power needs from 1/20 to 1/3 hp. Avoid using this type of motor in any applications requiring high on/off cycle rates or high torque.

6.5.2 Capacitor Start AC Induction Motor

This is a modified split-phase motor with a capacitor in series with the start winding to provide a start “boost.” Like the split-phase motor, the capacitor start motor also has a centrifugal switch which disconnects the start winding and the capacitor when the motor reaches about 75% of the rated speed. Figure 6.33 shows a typical capacitor start AC induction motor. Since the capacitor is in series with the start circuit, it creates more starting torque, typically 200% to 400% of the rated torque. And the starting current, usually 450% to 575% of the rated current, is much lower than the split-phase due to the larger wire in the start circuit. Refer to Figure 6.37 for torque-speed curve.A modified version of the capacitor start motor is the resistance start motor. In this motor type, the starting capacitor is replaced by a resistor. The resistance start motor is used in applications where the starting torque requirement is less than that provided by the capacitor start motor. Apart from the cost, this motor does not offer any major advantage over the capacitor start motor.They are used in a wide range of belt-drive applications like small conveyors, large blowers and pumps, as well as many direct-drive or geared applications.

6.5.3 Permanent Split Capacitor (Capacitor Run) AC

Induction Motor

A permanent split capacitor (PSC) motor has a run type capacitor permanently connected in series with the start winding. This makes the start winding an auxiliary winding once the motor reaches the running speed. Since the run capacitor must be designed for continuous use, it cannot provide the starting boost of a starting capacitor. The

94

Figure 6.32

Figure 6.33

Figure 6.34

typical starting torque of the PSC motor is low, from 30% to 150% of the rated torque. PSC motors have low starting current, usually less than 200% of the rated current, making them excellent for applications with high on/off cycle rates. Figure 6.34 shows a typical PSC induction motor. Refer to Figure 6.37 for torque-speed curve.

Permanent split-capacitor motors have a wide variety of applications depending on the design. These include fans, blowers with low starting torque needs and intermittent cycling uses, such as adjusting mechanisms, gate operators and garage door openers.

6.5.4 Capacitor Start/Capacitor Run AC

Induction Motor

This motor has a start type capacitor in series with the auxiliary winding like the capacitor start motor for high starting torque. Like a PSC motor, it also has a run type capacitor that is in series with the auxiliary winding after the start capacitor is switched out of the circuit. This allows high overload torque. Figure 6.35 shows a typical capacitor start / capacitor run AC induction motor.This type of motor can be designed for lower full-load currents and higher efficiency (see Figure 6.37 for torque-speed curve). This motor is costly due to start and run capacitors and centrifugal switch.It is able to handle applications too demanding for any other kind of single-phase motor. These include woodworking machinery, air compressors, high-pressure water pumps, vacuum pumps and other high torque applications requiring 1 to 10 hp.

6.5.5 Shaded-Pole AC Induction Motor

Shaded-pole motors have only one main winding and no start winding. Starting is by means of a design that rings a continuous copper loop around a small portion of each of the motor poles. This “shades” that portion of the pole, causing the magnetic field in the shaded area to lag behind the field in the un-shaded area. The reaction of the two fields gets the shaft rotating. Figure 6.36 shows a typical shaded-pole AC induction motor. Because the shaded-pole motor lacks a start winding, starting switch or capacitor, it is electrically simple and inexpensive. Also, the speed can be controlled merely by varying voltage, or through a multi-tap winding. Mechanically, the shaded-pole motor construction allows high-volume production. In fact, these are usually considered as “disposable” motors, meaning they are much cheaper to replace than to repair.The shaded-pole motor has many positive features but it also has several disadvantages. It’s low starting torque is typically 25% to 75% of the rated torque. It is a high slip motor with a running speed 7% to 10% below the synchronous speed. Generally, efficiency of this motor type is very low (below 20%).The low initial cost suits the shaded-pole motors to low horsepower or light duty applications. Perhaps their largest use is in multi-speed fans for household use. But the low torque, low efficiency and less sturdy mechanical features make shaded-pole motors impractical for most industrial or commercial use, where higher cycle rates or continuous duty are the norm.Figure 6.37 shows the torque-speed curves of various kinds of single-phase AC induction motors, while Figure 6.38 shows the phase relationship of various kinds of single-phase AC induction motors.

95

Figure 6.35

Figure 6.36

6.5.5 Universal Motor

A variant of the wound field DC motor is the universal motor. The name derives from the fact that it may use AC or DC supply current, although in practice they are nearly always used with AC supplies. The principle is that in a series wound field DC motor the current in both the field and the armature (and hence the resultant magnetic fields) will alternate (reverse polarity) at the same time, and hence the mechanical force generated is always in the same direction. In practice, the motor must be specially designed to cope with the AC current (impedance must be taken into account, as must the pulsating force), and the resultant motor is generally less efficient than an equivalent pure DC motor. Operating at normal power line frequencies, the maximum output of universal motors is limited and motors exceeding one kilowatt are rare. But universal motors also form the basis of the traditional railway traction motor in electric railways. In this application, to keep their electrical efficiency high, they were operated from very low frequency AC supplies, with 25 Hz and 16 2/3 hertz operation being common. Because they are universal motors, locomotives using this design were also commonly capable of operating from a third rail powered by DC.The advantage of the universal motor is that AC supplies may be used on motors which have the typical characteristics of DC motors, specifically high starting torque and very compact design if high running speeds are used. The negative aspect is the maintenance and short life problems caused by the

96

Figure 6.37

Phase Relationship Phase Relationship Phase Relationship of Split Phase Motor of Capacitor Start Motor of PSC Motor

Figure 6.38

commutator. As a result such motors are usually used in AC devices such as food mixers and power tools which are used only intermittently. Continuous speed control of a universal motor running on AC is very easily accomplished using a thyristor circuit, while stepped speed control can be accomplished using multiple taps on the field coil. Household blenders that advertise many speeds frequently combine a field coil with several taps and a diode that can be inserted in series with the motor (causing the motor to run on half-wave rectified AC).Universal motors can rotate at relatively high revolutions per minute (rpm). This makes them useful for appliances such as blenders, vacuum cleaners, and hair dryers where high-speed operation is desired. Many vacuum cleaner and weed trimmer motors exceed 10,000 rpm, Dremel and other similar miniature grinders will often exceed 30,000 rpm. Motor damage may occur due to over-speed (rpm in excess of design specifications) if the unit is operated with no significant load. On larger motors, sudden loss of load is to be avoided, and the possibility of such an occurrence is incorporated into the motor's protection and control schemes. Often, a small fan blade attached to the armature acts as an artificial load to limit the motor speed to a safe value, as well as provide cooling airflow to the armature and field windings.

6.5.6 Changing direction of rotation of a single phase induction motor

Changing the direction of rotation in single phase induction motors is simply done by changing the direction of the rotating fields. It is achieved by changing the direction of one of the magnetic fields of two windings. It can be done by changing the direction of current in auxiliary winding by making the current to enter from the other side of the winding. There are some mechanical or electronic circuits that can change the current to enter from the other side of the auxiliary winding manually or automatically as requested. In every time of changing direction of rotation the motor must be braked before changing circuit begins.

6.5.7 Single-phase Induction motor losses and efficiency

The losses and efficiency of single phase motors is the same that of three phase motors, which can be understood from Figure 6.30.

(6.18)

(6.19)

(6.20)

where, is the input power to the motor, is the input power to the rotor, or the air-gap power is the developed mechanical power of the rotor , and is the output power of the rotorRotor copper losses = s (Rotor input power ( )) (6.21)

= (1 - s) Rotor input power (6.22) for single-phase induction motor (6.23)

(6.24)where , T is the developed torque in N.m and is rotor angular velocity

Efficiency (6.25)

97

Solved Examples:

Example 6.11: The name plate of a single phase, 4-pole induction motor gives the following data: Output ½ hp; 230 V; 50Hz; input current 2.9 A; Power factor 0.71; speed 1410 rpm. Calculate (a) the efficiency, (b) the slip of the motor when delivering the rated output. If a capacitor of 15.9 µF is connected across the motor while delivering the rated output , calculate (c) the new current taken by the motor, and (d) the new power factor. Solution :Rated output power = ½ x 746 = 373 WInput power = 230 x 2.9 x 0.71 = 473.6 W

(a) the efficiency

= = 78.8%

(b) slip, s =

= = = 25 rev/s = 1500 rpm

= 1410 rpm

s = = 0.06

(c) As shown in Figure 6.39 (a), the capacitor has been connected across the motor.

= = 200 Ω

= 1.15 A

This current leads the applied voltage by 90° as shown in Figure 6.39 (b). The motor current lags the voltage by the angle = 45° The motor current and capacitor current when combined vectorially give the line current The x-component = = 2.9 x 0.71 = 2.06 A The y-component = - = 1.15 - 2.06 = - 0.91 A = = 2.25 A

(d) the new power factor = = 0.916 lagging

98

(a) (b)Figure 6.39

Example 6.12: A PSC single phase, 4-pole induction motor has the following data: Output 2.5 hp; 220 V; 50Hz; It is found that it drives a current for the main winding = 10 A, and drives a current for the auxiliary winding = 4 A at full load and rotor speed = 1440 rpm. Calculate (a) the efficiency, (b) the slip of the motor when delivering the rated output. Solution:

The x-component of = + = 10 cos (-30) + 5 cos (75) = 9.954 AThe y-component of = + = 10 sin (-30) + 4 sin (75) = - 1.136 A

= = 10.02 A

= - 6.5°

Input power = 220 x 10.02 x cos (6.5) = 2190.23 WRated output power = 2.5 x 746 = 1865 W

(a) the efficiency

= = 85.2 %

(b) slip, s =

= = = 25 rev/s = 1500 rpm

= 1410 rpm

s = = 0.04

99

UNIT 7 Switching and Controlling of AC Motors

Introduction

This unit introduces the construction of relay, contactor and timer. It introduces also, switching on-off and controlling of three phases motors, switching three phases motors star-delta, switching three phases motors with timers and switching single phase motors

7.1 Relay

A relay is used to isolate one electrical circuit from another. It allows a low current control circuit to make or break an electrically isolated high current circuit path.

7.1.1 Construction

The basic relay consists of a coil and a set of contacts. The most common relay coil is a length of magnet wire wrapped around a metal core. When voltage is applied to the coil, current passes through the wire and creates a magnetic field. This magnetic field pulls the contacts together and holds them there until the current flow in the coil has stopped. In Figure 7.1, the relay's coil is energized by the low-voltage (12 VDC) source, while the single-pole, single-throw contact interrupts the high-voltage (220 VAC) circuit. It is quite likely that the current required to energize the relay coil will be hundreds of times less than the current rating of the contact. Typical relay coil currents are well below 1 amp, while typical contact ratings for industrial relays are at least 10 amps.

One relay coil/armature assembly may be used to actuate more than one set of contacts. Those contacts may be normally-open (NO), normally-closed (NC), or any combination of the two. As with switches, the "normal" state of a relay's contacts is that state when the coil is de-energized, just as you would find the relay sitting on a shelf, not connected to any circuit.There are two specifications that you must consider when selecting a relay, the coil voltage and the current carrying capability of contacts. The coil voltage for relays may be DC voltage with rating 6, 9, 12, and 24 V, or AC voltages with ratings 220 V. This means that if you apply 12 volts to the coil, it will pull in and stay there until the applied voltage is removed from the coil. The current rating on relay contacts tells how much current can be passed through the contacts without damage to the contacts. Some relays have different current ratings for the NC contacts (which are held together by spring tension) and the NO contacts (which are held together by the electromagnet). If you need to pass significant current through the NC contacts, you may want to check the manufacturers specifications for the relay.Pull in Voltage: The pull in voltage is the minimum voltage required for the relay coil to pull the contacts together.Drop out Voltage: The drop out voltage is the voltage at which the energized coil will release the movable contact.

100

Figure 7.1

Coil Resistance: In a DC relay coil, the coil resistance determines the current flow through the coil. In an AC relay coil, the resistance does not solely determine the current flow through the coil because the coil has inductance. The inductive reactance along with the DC resistance work together to limit the current flow through the coil.

7.1.2 Types of relay

(a) Latching relay:

A latching relay has two relaxed states (bistable). These are also called 'keep' or 'stay' relays. When the current is switched off, the relay remains in its last state. This is achieved with a solenoid operating a ratchet and cam mechanism, or by having two opposing coils with an over-center spring or permanent magnet to hold the armature and contacts in position while the coil is relaxed, or with a remnant core. In the ratchet and cam example, the first pulse to the coil turns the relay on and the second pulse turns it off. In the two coil example, a pulse to one coil turns the relay on and a pulse to the opposite coil turns the relay off. This type of relay has the advantage that it consumes power only for an instant, while it is being switched, and it retains its last setting across a power outage.

(b) Reed relay

A reed relay has a set of contacts inside a vacuum or inert gas filled glass tube, which protects the contacts against atmospheric corrosion. The contacts are closed by a magnetic field generated when current passes through a coil around the glass tube. Reed relays are capable of faster switching speeds than larger types of relays, but have low switch current and voltage ratings.

(c) Mercury-wetted relay

A mercury-wetted reed relay is a form of reed relay in which the contacts are wetted with mercury. Such relays are used to switch low-voltage signals (one volt or less) because of its low contact resistance, or for high-speed counting and timing applications where the mercury eliminates contact bounce. Mercury wetted relays are position-sensitive and must be mounted vertically to work properly. Because of the toxicity and expense of liquid mercury, these relays are rarely specified for new equipment.

(d) Machine tool relay

A machine tool relay is a type standardized for industrial control of machine tools, transfer machines, and other sequential control. They are characterized by a large number of contacts (sometimes extendable in the field) which are easily converted from normally-open to normally-closed status, easily replaceable coils, and a form factor that allows compactly installing many relays in a control panel. Although such relays once were the backbone of automation in such industries as automobile assembly, the programmable logic controller (PLC) mostly displaced the machine tool relay from sequential control applications.

(e) Contactor relay

A contactor is a very heavy-duty relay used for switching electric motors and lighting loads. With high current, the contacts are made with pure silver. The unavoidable arcing causes the contacts to oxidize and silver oxide is still a good conductor. Such devices are often used for motor starters. A motor starter is a contactor with overload protection devices attached. The overload sensing devices are a form of heat operated relay where a coil heats a bi-metal strip, or where a solder pot melts, releasing a spring to operate auxiliary contacts. These auxiliary contacts are in series with the coil. If the overload senses excess current in the load, the coil is de-energized. Contactor relays can be extremely loud to operate, making them unfit for use where noise is a chief concern.

(f) Buchholz relay

A Buchholz relay is a safety device sensing the accumulation of gas in large oil-filled transformers, which will alarm on slow accumulation of gas or shut down the transformer if gas is produced rapidly in the transformer oil.

101

(g) Forced-guided contacts relay

A forced-guided contacts relay has relay contacts that are mechanically linked together, so that when the relay coil is energized or de-energized, all of the linked contacts move together. If one set of contacts in the relay becomes immobilized, no other contact of the same relay will be able to move. The function of forced-guided contacts is to enable the safety circuit to check the status of the relay. Forced-guided contacts are also known as "positive-guided contacts", "captive contacts", "locked contacts", or "safety relays".

(i) Solid-state relay

A solid state relay (SSR) is a solid state electronic component that provides a similar function to an electromechanical relay but does not have any moving components, increasing long-term reliability. With early SSR's, the tradeoff came from the fact that every transistor has a small voltage drop across it. This collective voltage drop limited the amount of current a given SSR could handle. As transistors improved, higher current SSR's, able to handle 100 to 1,200 amps, have become commercially available. Compared to electromagnetic relays, they may be falsely triggered by transients.

(j) Overload protection relay

One type of electric motor overload protection relay is operated by a heating element in series with the electric motor . The heat generated by the motor current operates a bi-metal strip or melts solder, releasing a spring to operate contacts. Where the overload relay is exposed to the same environment as the motor, a useful though crude compensation for motor ambient temperature is provided.

7.2 Contactor

A contactor is an electrically controlled switch (relay) used for switching a power circuit. A contactor is activated by a control input which is a lower voltage / current than that which the contactor is switching. Contactors come in many forms with varying capacities and features. Unlike a circuit breaker a contactor is not intended to interrupt a short circuit current.( See Figure 7.2 )Contactors range from having a breaking current of several amps and 220 volts to thousands of amps and many kilovolts. The physical size of contactors ranges from a few inches to the size of a small car.Contactors are used to control electric motors, lighting, heating, capacitor banks, and other electrical loads.

7.2.1 Construction

A contactor is composed of three different systems. The contact system is the current carrying part of the contactor. This includes Power Contacts, Auxiliary Contacts, and Contact Springs. The electromagnet system provides the driving force to close the contacts. The enclosure system is a frame housing the contact and the electromagnet. Enclosures are made of insulating materials like Bakelite, Nylon 6, and thermosetting plastics to protect and insulate the contacts and to provide some measure of protection

102

Figure 7.2

against personnel touching the contacts. Open-frame contactors may have a further enclosure to protect against dust, oil, explosion hazards and weather.Contactors used for starting electric motors are commonly fitted with overload protection to prevent damage to their loads. When an overload is detected the contactor is tripped, removing power downstream from the contactor.Some contactors are motor driven rather than relay driven and high voltage contactors (greater than 1000 volts) often have arc suppression systems fitted (such as a vacuum or an inert gas surrounding the contacts).Magnetic blowouts are sometimes used to increase the amount of current a contactor can successfully break. The magnetic field produced by the blowout coils force the electric arc to lengthen and move away from the contacts. The magnetic blowouts in the pictured Albright contactor more than double the current it can break from 600 Amps to 1500 Amps.Sometimes an Economizer circuit is also installed to reduce the power required to keep a contactor closed. A somewhat greater amount of power is required to initially close a contactor than is required to keep it closed thereafter. Such a circuit can save a substantial amount of power and allow the energized coil to stay cooler. Economizer circuits are nearly always applied on direct-current contactor coils and on large alternating current contactor coils.Contactors are often used to provide central control of large lighting installations, such as an office building or retail building. To reduce power consumption in the contactor coils, two coil latching contactors are used. One coil, momentarily energized, closes the power circuit contacts; the second opens the contacts.A basic contactor will have a coil input (which may be driven by either an AC or DC supply depending on the contactor design) and generally a minimum of two poles which are controlled.The top three contacts switch the respective phases of the incoming 3-phase AC power, typically at least 380 Volts for motors 1 horsepower or greater. The lowest contact is an "auxiliary" contact which has a current rating much lower than that of the large motor power contacts, but is actuated by the same armature as the power contacts. The auxiliary contact is often used in a relay logic circuit, or for some other part of the motor control scheme, typically switching 220 Volt AC power instead of the motor voltage. One contactor may have several auxiliary contacts, either normally-open or normally-closed, if required.Three-phase, 380 volt AC power comes in to the three normally-open contacts at the top of the contactor via screw terminals labeled "L1," "L2," and "L3". Power to the motor exits the overload heater assembly at the bottom of this device via screw terminals labeled "T1," "T2," and "T3."

7.2.2 Operating Principle

Unlike general-purpose relays, contactors are designed to be directly connected to high-current load devices, not other control devices. Relays tend to be of much lower capacity and are usually designed for both Normally Closed and Normally Open applications. Devices switching more than 15 amperes or in circuits rated more than a few kilowatts are usually called contactors. Apart from optional auxiliary low current contacts, contactors are almost exclusively fitted with Normally Open contacts.When current passes through the electromagnet, a magnetic field is produced which attracts ferrous objects, in this case the moving core of the contactor is attracted to the stationary core. Since there is an air gap initially, the electromagnet coil draws more current initially until the cores meet and reduct the gap, increasing the inductive impedance of the circuit.For contactors energized with alternating current, a small part of the core is surrounded with a shading coil, which slightly delays the magnetic flux in the core. The effect is to average out the alternating pull of the magnetic field and so prevent the core from buzzing at twice line frequency.Most motor control contactors at low voltages (600 volts and less) are "air break" contactors, since ordinary air surrounds the contacts and extinguishes the arc when interrupting the circuit. Modern medium-voltage motor controllers use vacuum contactors.Motor control contactors can be fitted with short-circuit protection (fuses or circuit breakers), disconnecting means, overload relays and an enclosure to make a combination starter. In large industrial plants many contactors may be assembled in motor control centers.

103

7.2.3 Ratings

Contactors are rated by designed load current per contact (pole), maximum fault withstand current, duty cycle, voltage, and coil voltage. A general purpose motor control contactor may be suitable for heavy starting duty on large motors; so-called "definite purpose" contactors are carefully adapted to such applications as air-conditioning compressor motor starting. North American and European ratings for contactors follow different philosophies, with North American contactors generally emphasizing simplicity of application while European rating philosophy emphasizes design for the intended life cycle of the application. A contactor basically consists of two parts; signaling and actual.

7.3 Timer

A timer is a specialized type of clock. A timer can be used to control the sequence of an event or process.A simple digital timer. The internal components—including the circuit board with control chip and LED display, a battery, and a buzzer—are visible.Timers can be mechanical, electromechanical, digital, or even software, since most computers include digital timers of one kind or another.

7.3.1 Mechanical Timers

Early mechanical timers used typical clockwork mechanisms, such as an escapement and spring to regulate their speed. Inaccurate, cheap mechanisms use a flat beater that spins against air resistance. Mechanical egg-timers are usually of this type.More accurate mechanisms resemble small alarm clocks. The chief advantage is that they require little battery/electrical power, and can be stored for long periods of time. The most widely-known application is to control explosives.

7.3.2 Electromechanical timers

Electromechanical timers have two types. A thermal type has a metal finger made of two metals with different rates of thermal expansion (steel and bronze are common). An electric current flows through this finger, and heats it. One side expands less than the other, and an electrical contact on the end of the finger moves away from an electrical switch contact, or makes a contact (both types exist). The most common use of this type is now in the "flasher" units that flash turn signals in automobiles, or sometimes in Christmas lights.Another type of electromechanical timer (a cam timer) uses a small synchronous AC motor turning a cam against a comb of switch contacts. The AC motor is turned at an accurate rate by the alternating current, which power companies carefully regulate. Gears slow this motor down to the desired rate, and turn the cam. The most common application of this timer now is in washers, driers and dishwashers. This type of timer often has a friction clutch between the gear train and the cam, so that the cam can be turned to reset the time.Electromechanical timers survive in these applications because mechanical switch contacts are still less expensive than the semiconductor devices needed to control powerful lights, motors and heaters.In the past these electromechanical timers were often combined with electrical relays to create electro-mechanical controllers. Electromechanical timers reached a high state of development in the 1950s and 60s because of their extensive use in aerospace and weapons systems. Programmable electromechanical timers controlled launch sequence events in early rockets and ballistic missiles.

7.3.3 Digital Timers

Digital timers can achieve higher precision than mechanical timers because they are quartz clocks with special electronics. Integrated circuits have made digital logic so inexpensive that an electronic digital timer is now less expensive than many mechanical and electromechanical timers. Individual timers are implemented as a simple single-chip computer system, similar to a watch. Watch technology is used in these devices.However, most timers are now implemented in software. Modern controllers use a programmable logic controller rather than a box full of electromechanical parts. The logic is usually designed as if it were

104

relays, using a special computer language called ladder logic. In PLCs, timers are usually simulated by the software built into the controller. Each timer is just an entry in a table maintained by the software.Digital timers can also be used in safety device such as a Gas Timer.

7.3.4 Computer timers

Most computer systems have one to sixteen electronic timers. These are usually just digital counters that are set to a number by software, and then count down to zero. When they reach zero, they interrupt the computer.Another common form of timer is a number that is compared to a counter. This is somewhat harder to program, but can be used to measure events or control motors.Embedded systems often use a hardware timer to implement a list of software timers. Basically, the hardware timer is set to expire at the time of the next software timer of a list of software timers. The hardware timer's interrupt software handles the house-keeping of notifying the rest of the software, finding the next software timer to expire, and resetting the hardware timer to the next software timer's expiration.

7.4 Switching and Controlling of Three Phase Induction Motors

For every case there are two circuits, one is for control and the other is for power or working of motor. Also, switching on the motor is achieved by a bush-button switch, in which its contacts are normally opened, and switching off the motor is achieved by a bush-button switch, in which its contacts are normally closed.

7.4.1 Switching Three Phase Induction Motors ON-OFF

Actuation of pushbutton ON energizes the coil of contactor C1. The contactor switches on the motor and maintains itself after the button is enables via its own auxiliary contact C1/14-13 and pushbutton OFF (three-wire control contact). Contactor C1 is de-energized, in the normal course of events, by actuation of pushbutton OFF. In the event of an overload, it is de-energized via the normally closed contact 95-96 on the overload relay F2. The coil current is interrupted, and contactor C1 switches the motor off. Figure 7.3 shows the control circuit while Figure 7.4 shows the power circuit.

105

Figure 7.3

Figure 7.4

7.4.2 Switching Three Phase Induction Motors ON For A Fixed Period

Actuation of pushbutton ON energizes the bridging contactor C1 which then maintains itself after the button is enables via its own auxiliary contact C1/14-13, normally closed contacts of timing relay T1/16-15 and pushbutton OFF. At the same time, voltage is applied to the timing relay T1. When the set time has elapsed, the bridging contactor C1 is disconnected by T1/16-15 which is opened after time elapsed. T1 is likewise disconnected and, exactly as C1, can only be energized again after the motor has been switched off by pressing pushbutton OFF. In the event of an overload, normally closed contact 95-96 on the overload relay O.L effects de-energizing. Figure 7.5 shows the control circuit while Figure 7.6 shows the power circuit.

7.4.3 Switching Three Phase Induction Motors In Two Directions

Actuation of pushbutton ON Clockwise energizes the coil of contactor C1. It switches on the motor running clockwise and maintains itself after button ON Clockwise is enabled via its own auxiliary contact C1/14-13, normally closed contacts of C2/22-21, pushbutton OFF and normally closed contacts of O.L 96-95. The normally closed contact C1/22-21 (which is opened after energizing of C1) electrically inhibits switch on of contactor C2. When pushbutton ON Unti-Clockwise is pressed, contactor C2 closes (the motor runs in the anticlockwise direction).

106

Figure 7.5 Figure 7.6

Figure 7.7

It maintains itself after button ON Unti-Clockwise is enabled via its own auxiliary contact C2/14-13, normally closed contacts of C1/22-21 and pushbutton OFF. The normally closed contact C2/22-21 (which is opened after energizing of C2) electrically inhibits switch on of contactor C1. Depending on the circuit, direction can be changed from clockwise to anticlockwise either after pressing pushbutton OFF, then pressing the other ON button, or by directly pressing the pushbutton for the reverse direction. In the event of an overload, normally closed contact 95-96 on the overload relay O.L effects de-energizing. Figure 7.7 shows the control circuit while Figure 7.8 shows the power circuit. It is recommended to switch the motor off and bringing it to stop before switching the reverse ON pushbutton.

7.4.4 Switching Three Phase Induction Motors Star Delta Manually

Star delta power circuit consists from three contactor, the main contactor C2 , star only contactor C1, and delta only contactor C3. First, when the motor is switched on, C1 and C2 are energized and the motor runs and accelerates in star connection. After running to about 80% from rated speed, C1 is de-energized, and C3 is energized making the motor to run in delta connection via C2 and C3. Figure 7.9 shows the power circuit. Figure 7.10 shows the control circuit for running the motor in star then delta connections manually. Actuation of pushbutton ON energizes the bridging contactor C2 which then closes its own auxiliary contact C2/14-13. At the same time, voltage is applied to contactor C1 making it to be energized. C2 maintains itself after the button is enables via auxiliary contact C1/14-13, its own auxiliary contact C2/14-13, normally closed contacts of C1/22-21 pushbutton OFF 1 and normally closed contacts of O.L 96-95, while C1 maintains itself after the button is enables via its own auxiliary contact C1/14-13, pushbutton OFF 1 and normally closed contacts of O.L 96-95. The motor works in star via contactors C1 and C2. After reaching 80% from the motor speed, pushbutton OFF2 ( off star – on delta ) is actuated which de-energize C2 making its main and 14-13 contacts to released and 22-21 contacts to return to its normally close. At the same time, voltage is applied to contactor C3 making it to be energized. The motor then

107

Figure 7.8

works delta via contactors C3 and C1. The normally closed contact C3/22-21 (which is opened after energizing of C3) electrically inhibits switch on of contactor C2. The motor then can't work in star unless it is switched off by OFF pushbutton and restarted by ON pushbutton again. In the event of an overload, normally closed contact 95-96 on the overload relay O.L effects de-energizing.

108

Figure 7.10

Figure 7.9

7.4.5 Switching Three Phase Induction Motors Star Delta Automatically With Timer

The power circuit is the same in the previous section (Figure 7.9). Figure 7.11 shows the control circuit for running the motor in star then delta connections automatically with timer. Actuation of pushbutton ON energizes the bridging contactor C2 which then closes its own auxiliary contacts C2/14-13. At the same time, voltage is applied to the auxiliary contactor Ch, making it to be energized which then closes its own auxiliary contacts Ch/14-13 making contactor C1 to be energized which then closes its own auxiliary contacts C1/14-13. Voltage is also applied to the timing relay T1 at the same time making it to be energized. C1 maintains itself after the button is enables via auxiliary contact C2/14-13, its own auxiliary contact C1/14-13, normally closed contacts of T1/15-16, normally closed contacts of C3/22-21, pushbutton OFF 1 and normally closed contacts of O.L 96-95, while C2 maintains itself after the button is enables via its own auxiliary contact C2/14-13, auxiliary contact Ch/14-13, pushbutton OFF 1 and normally closed contacts of O.L 96-95. The motor works in star via contactors C1 and C2. When the set changeover time has elapsed, contactor C1 is disconnected by T1/16-15 which is opened after time elapsed making its own auxiliary normally closed contacts C1/22-21 to be closed. At the same time contacts T1/15-18 are closed making contactor C3 to be energized. The motor works then in delta via contactors C3 and C2. The motor cannot be started up again unless it has previously been disconnected by pushbutton OFF., or in the event of an overload, by the normally closed contact 95–96 of the overload relay O.L and restarted by ON pushbutton again.

109

Figure 7.11

7.5 Switching and Controlling of Single Phase Induction Motors

For motors more than 2 hp it is recommended to use contactors and pushbuttons to switch motors on or off, but for small motors it is not necessary to use contactors for switching.

7.5.1 Switching Single Phase Induction Motors ON-OFF or For A Fixed Period

When using contactors for switching, control circuits for switching the motor on-off or for a fixed period of time is the same as that for three phase motors (Figures 7.3, and 7.5), but the power circuit uses one line as shown in Figure 7.12

7.5.2 Switching Small Motors ON-OFF

For small motors it is not necessary to use contactors to switch the motors, but relays or switches are used. The coil of the relay is energized from 220 V AC, or from 24 or 12 or 5 V DC according to the voltage of the control circuit. One line of the motor is connected to the N.O. point of the relay and the other is to neutral of the mains as shown in Figure 7.13. When the switch is closed the voltage is applied to the coil of the relay making it to be energized and changing its contacts. As a result the AC voltage is applied to the motor making it to run. Neutral of the motor may be connected via another relay that switched at the same time for the main relay, or by using double circuit relay to switch phase and neutral leads.

110

Figure 7.12

Figure 7.13

7.5.3 Switching Single Phase Induction Motors in Two Directions

A double circuit relay is used in this circuit as shown in Figure 7.14. When the switch S1 is opened the current passes from the phase lead to the auxiliary winding then to the capacitor then to the neutral lead. After the switch S1 is closed the current passes from the phase lead to the capacitor then to the auxiliary winding from the other side then to the neutral lead. As a result switching-on the switch S1 makes the motor to run in the reverse direction.

111

Figure 7.14

UNIT 8 Introduction To Protection

Introduction

This unit introduces an introduction to protection of electrical transformers and motors. It introduces a brief presentation about protection devices like fuses, circuit breakers and over-load relays. Also, ratings of transformers and motors calculations for protection will be explained.

8.1 Protection Devices

The over-current devices are an essential part of a power distribution system to prevent fire or damage. When too much current flows through a wire, it may overheat and be damaged, or even start a fire. Wiring regulations give the maximum rating of a device for protection of a particular circuit. There are three protection devices that can be used to protect transformers and motors which are fuses, circuit breakers and over-load relays.

8.1.1 Fuses

An expendable protective device that eliminates overload on an electric circuit. The fuse is connected in series with the circuit being protected. The components of a typical low-voltage high-power fuse are a fuse element or wire, an insulating material support and housing, two metal end fittings, and a filler.The fuse element is a silver strip or wire that melts when the current is higher than the rated value. The melting of the wire generates an electric arc. The extinction of this arc interrupts the current and protects the circuit. The fuse element is connected to the metal end fittings which serve as terminals. The filler facilitates the arc extinction. The most commonly used filler is sand, which surrounds the fuse element. When the fuse element melts, the heat of the arc melts the sand near the element. This removes energy from the arc, creating a channel filled with the mixture of melted sand and metal. The metal particles from the melting fuse wire are absorbed by the melted sand. This increases the channel resistance, which leads to the gradual reduction of the current and the extinction of the arc. The insulating support and the tubular housing holds the fuse elements and the filler, which also serves as insulator after The interruption time is the sum of the melting and the arcing time. It is inversely proportional to the current, that is, a higher current melts the wire faster. The fuse operates in a time-current band between maximum interruption time and minimum meeting time. It protects the electric circuit if the fault current is interrupted before the circuit elements are overheated. The arc extinction often generates over-voltages, which produce flashovers and damage. A properly designed fuse operates without over-voltage, which is controlled by the shape of the fuse element and by the filler.Fuses come in a vast array of sizes & styles to cater for the immense number of applications in which they are used. While many are manufactured in standardized package layouts to make them easily interchangeable, a large number of new styles are released into the marketplace every year. Fuse bodies may be made of ceramic, glass, plastic, fiberglass, Molded Mica Laminates, or molded compressed fibre depending on application and voltage class.Cartridge (ferrule) fuses have a cylindrical body terminated with metal end caps. Some cartridge fuses are manufactured with end caps of different sizes to prevent accidental insertion of the wrong fuse rating in a holder. An example of such a fuse range is the 'bottle fuse', which in appearance resembles the shape of a bottle.Fuses used in circuits rated 200-600 volts and between about 10 and several thousand amperes, as used for industrial applications such as protection of electric motors, commonly have metal blades located on each end of the fuse. Fuses may be held by a spring loaded clip or the blades may be held by screws. Blade type fuses often require the use of a special purpose extractor tool to remove them from the fuse holder. While glass fuses have the advantage of a fuse element visible for inspection purposes, they have a low breaking capacity which generally restricts them to applications of 15 A or less at 250 VAC. Ceramic fuses have the advantage of a higher breaking capacity facilitating their use in higher voltage/ampere circuits. Filling a fuse body with sand provides additional protection against arcing in an over-current situation.

112

Cartridge fuses are generally measured as the overall length and diameter of the fuse. Due to the large variety of cartridge fuses available, fuse identification relies on accurate measurements as fuses can differ by only a few millimeters between types. 'Bottle style' cartridge fuses also require the measurement of the cap diameter as this varies between ampere ratings. Figure 8.1 shows the construction of a cartridge fuse.Other fuse packages can require a variety of measurements such as;

body (width x height x depth) blade or tag (width x height x depth) overall length of the fuse (when the fuse features blades

or tags) overall width of the fuse (when the fuse features 2

bodies) width of the mounting holes (when the fuse features

tags) distance between blades (when radially configured) fixing centre (when the fuse features tags - see below)

Fuses fitted with tags require the fixing centre measurement. This measurement is the distance between the tag mounting holes on either end of the fuse as measured from the centre of each mounting hole.Most fuses are marked on the body, or end caps to markings show their ratings. Surface mount technology "chip type" fuses feature little or no markings making identification very difficult.When replacing a fuse, it is important to interpret these markings correctly as fuses that may look the same, could be designed for very different applications. Fuse markings will generally convey the following information;

Ampere rating of the fuse Voltage rating of the fuse Time-current characteristic ie. element speed Approvals Manufacturer / Part Number / Series Breaking capacity

Interrupting ratingA fuse also has a rated interrupting capacity, also called breaking capacity, which is the maximum current the fuse can safely interrupt. Generally this should be higher than the maximum prospective short circuit current. Miniature fuses may have an interrupting rating only 10 times their rated current. Fuses for small low-voltage wiring systems are commonly rated to interrupt 10,000 amperes. Fuses for larger power systems must have higher interrupting ratings, with some low-voltage current-limiting "high rupturing capacity" (HRC) fuses rated for 300,000 amperes. Fuses for high-voltage equipment, up to 115,000 volts, are rated by the total apparent power (megavolt-amperes, MVA) of the fault level on the circuit.

Voltage ratingAs well as a current rating, fuses also carry a voltage rating indicating the maximum circuit voltage in which the fuse can be used. For example, glass tube fuses rated 32 volts should never be used in line-operated (mains-operated) equipment even if the fuse physically can fit the fuseholder. Fuses with ceramic cases have higher voltage ratings. Fuses carrying a 250 V rating may be safely used in a 125 V circuit, but the reverse is not true as the fuse may not be capable of safely interrupting the arc in a circuit of a higher voltage. Medium-voltage fuses rated for a few thousand volts are never used on low voltage circuits, due to their expense and because they cannot properly clear the circuit when operating at very low voltages.Figure 8.2(a) shows a marked fuseand (b) shows two fuses with dimensionsidentifications, while Figure 8.3 shows 4 shapes of fuses.

113

(a) (b)

Figure 8.2

Figure 8.1

8.1.2 Circuit Breakers

A device to open or close an electric power circuit either during normal power system operation or during abnormal conditions. A circuit breaker serves in the course of normal system operation to energize or de-energize loads. During abnormal conditions, when excessive current develops, a circuit breaker opens to protect equipment and surroundings from possible damage due to excess current. These abnormal currents are usually the result of short circuits created by lightning, accidents, deterioration of equipment, or sustained overloads.Unlike a fuse, which operates once and then has to be replaced, a circuit breaker can be reset (either manually or automatically) to resume normal operation. Circuit breakers are made in varying sizes, from small devices that protect an individual household appliance up to large switchgear designed to protect high voltage circuits feeding an entire city.Magnetic circuit breakers are implemented using a solenoid (electromagnet) whose pulling force increases with the current. The circuit breaker's contacts are held closed by a latch and, as the current in the solenoid increases beyond the rating of the circuit breaker, the solenoid's pull releases the latch which then allows the contacts to open by spring action. Some types of magnetic breakers incorporate a hydraulic time delay feature wherein the solenoid core is located in a tube containing a viscous fluid. The core is restrained by a spring until the current exceeds the breaker rating. During an overload, the solenoid pulls the core through the fluid to close the magnetic circuit, which then provides sufficient force to release the latch. The delay permits brief current surges beyond normal running current for motor starting, energizing equipment, etc. Short circuit currents provide sufficient solenoid force to release the latch regardless of core position thus bypassing the delay feature. Ambient temperature affects the time delay but does not affect the current rating of a magnetic breaker (see Figure 8.4 for construction).Thermal breakers use a bimetallic strip, which heats and bends with increased current, and is similarly arranged to release the latch. This type is commonly used with motor control circuits. Thermal breakers often have a compensation element to reduce the effect of ambient temperature on the device rating (see Figure 8.5 for construction).Thermo-magnetic circuit breakers, which are the type found in most distribution boards, incorporate both techniques with the electromagnet responding instantaneously to large surges in current (short circuits) and the bimetallic strip responding to less extreme but longer-term over-current conditions.Circuit breakers for larger currents are usually arranged with pilot devices to sense a fault current and to operate the trip opening mechanism (see Figure 8.6 for construction).

114

Figure 8.3

Under short-circuit conditions, a current many times greater than normal can flow. When electrical contacts open to interrupt a large current, there is a tendency for an arc to form between the opened contacts, which would allow the flow of current to continue. Therefore, circuit breakers must incorporate various features to divide and extinguish the arc. In air-insulated and miniature breakers an arc chute structure consisting (often) of metal plates or ceramic ridges cools the arc, and blowout coils deflect the arc into the arc chute. Larger circuit breakers such as those used in electrical power distribution may use vacuum, an inert gas such as sulfur hexafluoride or have contacts immersed in oil to suppress the arc.The maximum short-circuit current that a breaker can interrupt is determined by testing. Application of a breaker in a circuit with a prospective short-circuit current higher than the breaker's interrupting capacity rating may result in failure of the breaker to safely interrupt a fault. In a worst-case scenario the breaker may successfully interrupt the fault, only to explode when reset, injuring the technician.Small circuit breakers are either installed directly in equipment, or are arranged in a breaker panel. Power circuit breakers are built into switchgear cabinets. High-voltage breakers may be free-standing outdoor equipment or a component of a gas-insulated switchgear line-up. Components:

1. spring2. electrical contacts3. tie4. latch5. hinge6. iron piece 7. electromagnetic coil8. spring

115

Figure 8.4

)b (Up-normal Condition (a) Normal Condition Figure 8.5

) b (Up-normal Condition (a) Normal Condition Figure 8.6

Domestic circuit breakersThe 10 ampere DIN rail mounted thermal-magnetic miniature circuit breaker is the most common style in modern domestic consumer units and commercial electrical distribution boards throughout Europe. The design includes the following components:

1. Actuator lever - used to manually trip and reset the circuit breaker. Also indicates the status of the circuit breaker (On or Off/tripped). Most breakers are designed so they can still trip even if the lever is held or locked in the on position. This is sometimes referred to as "free trip" or "positive trip" operation.

2. Actuator mechanism - forces the contacts together or apart. 3. Contacts - Allow current to flow when touching and break the flow of current when moved apart. 4. Terminals 5. Bimetallic strip 6. Calibration screw - allows the manufacturer to precisely adjust the trip current of the device after

assembly. 7. Solenoid 8. Arc divider / extinguisher

Rated currentInternational Standard IEC 60898-1 and European Standard EN 60898-1 define the rated current In of a circuit breaker for household applications as the current that the breaker is designed to carry continuously (at an ambient air temperature of 30 °C). The commonly-available preferred values for the rated current are 6 A, 10 A, 13 A, 16 A, 20 A, 25 A, 32 A, 40 A, 50 A, 63 A, 80 A and 100 A (Renard series, slightly modified to include current limit of British BS 1363 sockets). The circuit breaker is labeled with the rated current in ampere, but without the unit symbol "A". Instead, the ampere figure is preceded by a letter "B", "C" or "D" that indicates the instantaneous tripping current, that is the minimum value of current that causes the circuit-breaker to trip without intentional time delay (i.e., in less than 100 ms):B indicates tripping above 3In up to and including 5In,C indicates tripping above 5In up to and including 10In, andD indicates tripping above 10In up to and including 20InTable 8.1 contains the standard ratings of fuses, Siememens 1, 2, and 3 poles miniature circuit breakers, Siememens one, two, and three poles circuit breakers.

8.1.3 Thermal Overload Relays

Thermal overload relays prevent an electric motor from drawing too much current and overheating. Thermal overload conditions are the most likely faults to be encountered in industrial motor applications. They result in a rise in the motor running current, which produces an increase in the motor's thermal dissipation and temperature. Overload protection prevents an electric motor from drawing too much current, overheating, and literally burning out. Thermal overload relays can be bimetallic relays, eutectic alloy relays, temperature control or probe relays, and solid-state relays. A bimetallic device is made up of two strips of different metals. The dissimilar metals are permanently joined. Heating the bimetallic strip causes it to bend because the dissimilar metals expand and contract at different rates. The bimetallic strip applies tension to a spring on a contact. If heat begins to rise, the strip bends and the spring pulls the contacts apart, breaking the circuit. A melting alloy (or eutectic) overload relay consists of a heater coil, a eutectic alloy, and a mechanical mechanism to activate a tripping device when an overload occurs. The relay measures the temperature of the motor by monitoring the amount of current being drawn. This is done indirectly through a heater coil. Temperature control relays are used to protect the motor by directly sensing the temperature of the windings using thermistor or RTD probes. The motor must have one or more positive temperature coefficient (PTC) thermistor probes embedded in its windings. When the nominal operating temperature of the probe is reached, its resistance increases rapidly. This increase is detected by a threshold circuit, which controls a set of relay contacts. Solid-state relays have no moving or mechanical parts. The relay calculates the average temperature within the motor by monitoring its starting and running currents. A solid-state relay is a type of over-current relay.

116

Important performance characteristics to consider when searching for thermal overload relays include full current load range, trip class, and temperature trip range. The adjustable current value allows the relay to be set to the full-load current shown on the motor rating plate. This is the current range of the thermal component that can be adjusted to the desired trip point. The trip class is the maximum time in seconds at which the overload relay will trip when the carrying current is at 600% of its current rating. Bimetallic overload relays can be rated as Class 10, meaning that they can be counted on to break the circuit no more than 10 seconds after a locked rotor condition begins. Melting alloy overload relays are generally Class 20. Although thermal overload relays are designed to protect motors against overload currents, they must be capable of handling large currents without tripping for short periods during motor starting (run-up). They should, however, trip quickly if the starting currents last too long. The temperature trip range is the trip point setting range for relays designed to monitor the temperature of the motor stator windings.Other important specifications to consider when searching for thermal overload relays include motor load phase, motor voltage, control circuit voltage, contact or output ratings, pole specifications, features, and environmental operating parameters. The motor load phase can be single-phase protection or three-phase protection. The maximum motor voltage is the applicable maximum motor voltage. The control voltage, if different from the motor voltage. This is called "separate control" and means that the control circuit gets its power form a separate source; usually lower in voltage form the motor's power source. Contact output ratings include the contact current rating and contact maximum rated voltage. Poles can be single pole, double pole, triple pole, four pole, or greater than four pole. Common features for thermal overload relays include ambient temperature compensation, automatic reset, built-in emergency override, built-in trip indicator, hermetically sealed, programmable or adjustable trip time, ground fault detection, phase loss detection, phase reversal detection, unbalance protection, and under-load protection. An important environmental parameter to consider is operating temperature.They are provided with 1NO + 1NC potentially free contacts which can be used for signaling. The relay can be used in either Auto or Manual reset mode (see Figure 8.7).

8.2 Determining Protection Ratings

Before inducing any protective device, the current rating of the device should be determined first, which depends mainly on the load. Current ratings of loads without starting current, like transformer can be determined easier than rating of loads with starting current like motors.

8.2.1 Current Rating for Transformers

The name plate of the transformer contains the full load power rating, primary and secondary line voltages, primary and secondary wires for connection, power factor, efficiency at full load, etc… The name plate of the three phase transformers also include the type of connection for primary and secondary, like Y-Δ .

(8.1)The power transformers ratings are usually in VA or KVA. For step down transformers is less than , so the primary winding wire is thinner than that of secondary. A short circuit in the secondary circuit could damage primary circuit. As a result protective device for transformer should be induced before the primary. Protection of secondary part is also preferred to be induced.

117

Figure 8.7

For three phase transformers the line current is the dominant in calculating current ratings. The power of three phase transformer is given for star or delta connections by:

(8.2)From transformers manufactures, it is known that transformer can handle extra 20% from the full load (i.e. it can work on 120% of full load). It means that the rated current should multiplied with 1.2 to find the maximum rating current. After determining primary and secondary rating currents including extra loads, the protection device rating is chosen to be the nearest higher value from the standard ratings.

8.2.2 Current Rating for Motors

The name plate of the motor contains the full load output power rating in hp , line voltages, primary and secondary wires for connection, power factor, efficiency at full load, type of the motor AC or DC, induction or synchronous, single or three phase etc… The name plate of the three phase transformers also include the type of connection for continuous , like Δ . From these numbers the full load line current can be calculated.For DC motors the input power is found by :

(8.3)For single phase induction motors the input power is found by :

(8.4)For three phase induction motors the input power is found by :

(8.5)

The input power can calculated from the output power and the efficiency by:

(8.6)

After that the full load current can be determined.Also like transformers, from motor manufactures, it is known that motors can handle extra 20% from the full load (i.e. it can work on 120% of full load). It means that the rated current should multiplied with 1.2 to find the maximum rating current. Also, starting current should be considered for three phase induction motors mainly. The calculated maximum rated current must be multiplied by 1.5 – 2 to consider the starting current.After determining rating currents including extra loads and starting, the protection device rating is chosen to be the nearest higher value from the standard ratings.

For transformers it is recommended to choose fuses or thermal circuit breakers and for motors it is recommended to choose fuses or thermal circuit breakers type C or D or thermal overload relays for protection.

118

Table 8.1: Fuses and Circuit Breakers Ratings for 240V

Fuse Ratings

(A)

Miniature Circuit Breaker Rating &

Types1, 2, 3 Poles (A)

One Pole Circuit Breaker Ratings &

Types (A)

Two Pole Circuit Breaker Ratings &

Types (A)

Three Pole Circuit Breaker Ratings&

Types (A)

2 1 B, C, D 15 B, C, D, H, Q 15 B, C, D, E, P, Q 15 B, C, E, P4 2 B, C, D 20 B, C, D, H, Q 20 B, C, D, E, P, Q 20 B, C, E, P6 4 B, C, D 30 B, C, D, H, Q 30 B, C, D E, P, Q 30 B, C, E, P8 6 B, C, D 35 B, C, D, H, Q 35 B, C, D, E, P, Q 35 B, C, E, P10 10 B, C, D 40 B, C, D, H, Q 40 B, C, D, E, P, Q 40 B, C, E, P12 16 B, C, D 50 B, C, D, H, Q 50 B, C, D, E, P, Q 50 B, C, E, P16 20 B, C, D 60 B, C, D, Q 60 B, C, D, E, P, Q 60 B, C, E, P, Q20 25 B, C, D 70 B, C, 70 B, C, E, P 70 B, C, E, P, Q25 32 B, C, D 80 B, C 80 B, C, E 80 B, C, E, Q32 40 B, C, D 90 B, C 90 B, C, E 90 B, C, E, Q40 50 B, C, D 100 B, C 100 B, C, E, J, P, Q 100 B, C, E, J, P, Q50 63 B, C, D ------ 125 E, J, Q 125 E, J, Q80 ------ ------ 150 E, F, J, K, Q 150 E, F, J, K, Q100 ------ ------ 175 E, F, J, K 175 E, F, J, K, Q125 ------ ------ 200 E, F, J, K, Q 200 E, F, J, K, L, Q160 ------ ------ 225 E, F, J 225 E, F, J, Q200 ------ ------ 250 F 250 F250 ------ ------ 300 C, E 275 E315 ------ ------ 350 C, E 300 C, E400 ------ ------ 400 C, E, H 350 C, E500 ------ ------ 450 E, H 400 C, E, H630 ------ ------ 500 E, H 450 C, E , H800 ------ ------ 550 E, 500 C, E, H1000 ------ ------ 600 E, H 550 C, E1250 ------ ------ 700 E, 600 C, E, H------ ------ ------ 800 E, 700 C, E, H------ ------ ------ ------ 800 C, E, H------ ------ ------ ------ 900 H------ ------ ------ ------ 1000 H

119