General ObjectiveTo understand the basic principles of electromagnetism.
1.0 INTRODUCTION Arah uratdaya magnetThe pattern of magnetic field of bar a magnet.
TarikanTolakan Attraction and Repulsion ( tarikan dan tolakan )
1.1 CURRENT-CARRYING CONDUCTOR AND ELECTROMAGNETISM ( keelektromagnetan ke atas pengalir yang membawa arus )• A flow of current through a wire produces a magnetic field in a circular path around the wire.The direction of magnetic line of flux around the wire is best remembered by the screw rule or the grip rule.
Arus masukArus keluar The field pattern of current flowing in the wire
tarikan tolakan( a) flow in the same direction ( b) opposite direction
If two closed current-carrying conductors flow in the same direction, magnetic flux around that conductor will combine tocreate attraction between them. If closed current-carrying conductors flow in opposite direction, these two conductors will repulse each other
1.2. MAGNETIC FIELD STRENGTH, H (MAGNETISING FORCE)• Magnetic field strength is defined as magnetomotive force, Fm Fm NI H= = ampere turn / metre l l N = bilangan lilitan pengalir I = arus yang mengalir l = panjang bahan magnet
1.3. MAGNETIC QUANTITY AND THEIR RELEVANT FORMULAE• 1.3.1 Magnetic Flux and Flux density -Magnetic flux,Φ is the amount of magnetic filed produced by a magnetic source. - Flux density,B is the amount of flux passing through a defined area unit for flux is the weber, wb
• Example 1.1• A current of 500mA is passed through a 600 turn coil wound of a iron of mean diameter 10cm. Calculate the magnetic field strength.
Example 1.2• An iron ring has a cross-sectional area of 400 mm2 and a mean diameter of 25 cm. it is wound with 500 turns. If the value of relative permeability is 250, find the total flux set up in the ring. The coil resistance is 474 Ω and the supply voltage is 20 V.
● 1.3.2 Permeability ( ketelapan ) ( Kebolehan sesuatu bahan magnet untuk menghasilkan uratdaya magnet ) the ratio of magnetic flux density to magnetic field strength is constant B = a constant H
For air, free space and any othernon-magnetic medium, the ratio B = μ 0 = 4π x 10-7 H/m HFor all media other than free pace, B = μ0μr H
Cast iron μr = 100 – 250Mild steel μr = 200 – 800Cast steel μr = 300 – 900 μr for a vacuum is 1 μ - absolute permeability μr - relative permeability μo - air permeability where μ = μoμr
• 1.3.3 Reluctance ( Engganan ) Reluctance, S is the magnetic resistance of a magnetic circuit Fm Hl H l 1 l S = = = = Φ BA B A μ ομ r A unit for reluctance is H-1
Perbadingan di antara Litar Elekrik Dengan Litar magnet Litar Elektrik Litar Magnet1. Arus 1. Uratdaya ( Fluks )2. Dge 2. Dgm3. Rintangan 3. Engganan
• Example 1.3 A magnetic pole face has rectangular section having dimensions 200mm by 100mm. If the total flux emerging from the the pole is 150μWb, calculate the flux density. Example 1.4 A flux density of 1.2 T is produced in a piece of cast steel by a magnetizing force of 1250 A/m. Find the relative permeability of the steel under these conditions.
• Example 1.5 Determine the reluctance of a piece of metal of length 150mm, and cross sectional is 100mm2 when the relative permeability is 4000. Find also the absolute permeability of the metal. Exersice 1 The maximum working flux density of a lifting electromagnet is 1.8 T and the effective area of a pole face is circular in cross-section. If the total magnetic flux produced is 353 mWb, determine the radius of the pole.
1.4 ELECTROMAGNETIC INDUCTIONWhen a conductor is moved across a magneticfield so as to cut through the flux,an electromagnetic force (e.m.f.) is producedin the conductor. This effect is known aselectromagnetic induction. The effect ofelectromagnetic induction will causeinduced current.
Two laws of electromagnetic induction i. Faraday’s law Conductor cuts flux
This induced electromagnetic field is given by θ° Where , B = flux density, T = length of the conductor in the magnetic field, m v = conductor velocity, m/s
ii. Lenz’z Law Bar magnet move in and move out from a solenoid
Example 1.6A conductor 300mm long moves at a uniformspeed of 4m/s at right-angles to a uniformmagnetic field of flux density 1.25T. Determinethe current flowing in the conductor when i. its ends are open-circuited ii. its ends are connected to a load of 20 Ωresistance.
Exersice 2 A conductor of length 0.5 m situated in and at right angles to a uniform magnetic field of flux density 1 wb/m2 moves with a velocity of 40 m/s. Calculate the e.m.f induced in the conductor. What will be the e.m.f induced if the conductor moves at an angle 60º to the field.
Solution to Example 1.3Magnetic flux, Φ = 150 μWb = 150 x 10-6 WbCross sectional area, A = 200mm x 100mm = 20 000 x 10-6m2 Φ 150 × 10 − 6Flux density, B = = A 20000 × 10 − 6 = 7.5 mT
Solution to Example 1.4 B = μ0μr H B 1.2 μr = = μ 0 H (4π × 10−2 )(1250) = 764
Solution to Example 1.5 l Reluctance, S = μ 0 μ r A 150 × 10 − 3 = ( 4π × 10 − 7 )( 4000 )(100 × 10 − 6 ) = H-1 Absolute permeability, μ = μ0μr ( 4π × 10 − 7 )( 4000 ) = 5.027 x 10-3 H/m
Solution To Example 1.6 i. If the ends of the conductor are open circuited, no current will flow even though 1.5 V has been induced. ii. From Ohm’s law E 1 .5 I = = 75 mA R 20
• Whenever a conductor is moved within a magnetic field in such a way that the conductor cuts across magnetic lines of flux, voltage is generated in the conductor. - Faraday’s Law
• The POLARITY of the voltage depends on the direction of the magnetic lines of flux and the direction of movement of the conductor. To determine the direction of current in a given situation, the RIHGT-HAND RULE FOR GENERATORS is used.
• The AMOUNT of voltage generated depends on : i. the strength of the magnetic field, ii. the angle at which the conductor cuts the magnetic field, iii. the speed at which the conductor is moved, and iv. the length of the conductor within the magnetic field.
Bahagian Angker( Gelung Angker ) Bahagian Stator (Gelung Medan)
Types of DC Generator• Separately-excited generators• Self-excited generators i. Shunt-wound generator ii. Series-wound generator iii. Compound-wound generator a. Short compound generator b. Long compound generator
1. Penjana Ujaan Berasingan Angker Medan DC Power Supply Penjana Ujaan Berasingan
2. Penjana Ujaan Diri 1. Series-wound generator
Example 2.1 A shunt generator supplies a 20 kW load at 200 V. If the field winding resistance, Rf = 50Ω and the armature resistance Ra = 40 mΩ, determine (a) the terminal voltage (b) the e.m.f. generated in the armature
2.3. E.m.f generated ( Voltan janaan, dge ) 2p Φ Zn generated e.m.f, Eg = c Where ; Z = number of armature conductors, Φ = useful flux per pole in Webers Ρ = number of pairs of poles n = armature speed in rev/s ( c=2 for a wave winding and c= 2p for a lap winding )
Example 2.2. An 8-pole generator, wave winding connected armature has 600 conductor and is driven 625 rev/min. If the flux per pole is 20mWb, determine the generated e.m.f.
SolutionZ = 600, c = 2 for a wave windingP = 4 pairs, n = 625/60 rev/min, Φ = 30 × 10-3 Wb 2p Φ Zn Dge, Eg = c 625 2(4)(20 × 10 )( -3 ) 60 2
Example 2.3. A 4-pole generator has a lap winding armature, with 50 slots and 16 conductors per slot. The useful flux per pole is 30mWb. Determine the speed at which the machine must be driven to generate an e.m.f. of 240 volts.
E = 240 V, Z = 50 x 16 = 800c = 2p (for a lap winding), Φ = 30 × 10-3 Wb Ans : ( 10 rev/s )
2.4 Power Losses and Efficiency For any type of machine, output power is different from input power. The difference is caused by power losses that had happened whenever one type of energy is converted or delivered to the other type.
The principal losses of machine are:• Copper loss ( I2R )• Iron losses, due to hysteresis and eddy current• Friction and windage losses• Brush and contact losses ( vB )
2.4.1. Efficiency of DC generator The efficiency of an electrical machine is the ratio of the output power and input. The greek letter ‘η’ (eta) is used to signify efficiency, the efficiency has no units.
output powerefficiency, η = ( ) × 100 % input power Vo η =( ) × 100% Vo + VD VL I L η= ( ) × 100 % VL I L + I a R a + I f V f + C 2
Example 2.4 A shunt generator supplies 96 A at a terminal voltage of 200 volts. The armature and shunt field resistances are 0.1Ω and 50Ω respectively. The iron and frictional losses are 2500 W. Find : (i) e.m.f generated. (ii) copper losses (iii) efficiency
Example 2.5 A 75 kW shunt generator is operated at 230 V. The stray losses are 1810 W and shunt field circuit draws 5.35 A. The armature circuit has a resistance of 0.035 Ω and brush drop is 2.2 V. Calculate : 1. total losses 2. input of prime mover 3. efficiency at rated load.
DC motors are very useful in many applications of our everyday life, forexample controlling, such as crane, tape driver, lift system and others.
OBJECTIVES• General Objective To apply the basic principles of DC motor operation, types of DC motor and their application
• Specific Objectives• Explain the principle operation of DC motor• List the types of DC motor• State the left-handed rule for motors• List the advantages and disadvantages of the different types of DC motors.• State typical applications of DC motors
3.1 INTRODUCTIONDC Motor is a machine that converts electrical energy into mechanical energy.
Electrical MechanicalLoad Energy Energy Blok Diagram
TEST YOUR UNDERSTANDING 1 . What is a DC motor? 2. State the uses of DC motors. 3. What are the parts of DC motors?
3. 3 PRINCIPLE OF OPERATIONFleming′s Left Hand Rule
When a wire carrying current sits into a magnetic field, a force is created on the wire causing it to move perpendicular ( tegak lurus ) to the magnetic field. Thegreater the current in the wire, or the greater the magnetic field, the faster the wire moves because of the greater force created. If the wire sits parallel with the magnetic field, there will be no force on the wire.
conductor( field ) Left-hand rule for DC motors
Example A 350 V shunt motor runs at its normal speed of 12rev/s when the armature current is 90 A. The resistance of the armature is 0.3 Ω. Find the speed when the armature current is 45 A and a resistance of 0.4 Ω is connected in series with the armature, the shunt field remaining constant.
3.6 FACTORS THAT INFLUENCE SPEED CONTROL OF DC MOTOR The speed of a dc motor is changed by changing the current in the field or by changing the current in the armature.
3.7 REVERSE DIRECTION METHOD• The direction of rotation of a dc motor depends on the direction of the magnetic field and the direction of current flow in the armature.
3.8 EFFICIENCY AND POWER LOSSESKecekapan output power efficiency, η = × 100% input power VI − I R a − I f V − C 2 η =( ) × 100 % a VI
Example 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.
OBJECTIVESTo analyze the basic principles of operation of an AC generator and the differences between DC generator and AC generator by using commutator and slip ring.
4.1 INTRODUCTIONAn electric generator is a device used toconvert mechanical energy into electricalenergy.The generator is based on the principle of"electromagnetic induction" discovered byMichael Faraday Law’s.
The amount of voltage generated depends on the following:• The strength of the magnetic field.• The angle at which the conductor cuts the magnetic field.• The speed at which the conductor is moved• The length of the conductor within the magnetic field.
4.2 THE DIFFERENCES BETWEEN AC GENERATOR AND DC GENERATOR The difference between AC and DC generator is that the DC generator results when you replace the slip rings of an elementary generator with commutator, changing the output from AC to pulsating DC. AC generator is also called Alternator.
4.3 E.M.F. Equation of an Alternator E rms / phase = 2.22 KdKp f Φ Z volts where, Z = No. of conductors or coil Φ = Flux per pole in webers P = Number of rotor poles N = Rotor speed in r.p.m Kd = Distribution factor Kp = Pitch factor ⎛ 120 f ⎞ ⎜N = ⎜ ⎟ ⎟ ⎝ p ⎠
Example 4.1 A 3-phase, 50 Hz star-connected alternator has 180 conductors per phase and flux per pole is 0.0543 wb. Find:- a) e.m.f. generated per phase b) e.m.f. between line terminals. Assume the winding to be full pitched and distribution factor to be 0.96.
Exersice 1 Find the number of armature conductors in series per phase required for the armature of a 3-phase, 50Hz, 10-pole alternator. The winding is star-connected to give a line voltage of 11000. The flux per pole is 0.16 wb. Assume Kp = 1 and Kd = 0.96.
Differences between AC Motor and DC motorIn general, AC motors cost less than DCmotors. Some types of AC motors do notuse brush carbon and commutators. What is the advantage of AC motor over DC motor ?
OBJECTIVESTo understand the basic principles of a transformer, construction principle, transformer ratio, current and core, type of transformer and uses.
At the end of the unit you will be able to :• explain the operating principles of a transformer• describe transformer construction• explain transformer ratio for voltage, current and winding coil.• calculating of the efficiency.• describe auto transformer
Primary Secondarywinding winding Core Coil /Winding transformer construction
5.1 Introduction The basic transformer is an electrical device that transfers alternating-current energy from one circuit to another circuit by magnetic flux of the primary and secondary windings of the transformer.
• Ep = 4.44 Np f Φm volts• Es = 4.44 Ns f Φm volts• If K < 1 i.e. Ns < Np : step-down• If K > 1 i.e. Ns > Np : step-up• If K = 1 i.e. Ns = Np : coupling
equations of ideal transformer V p Np Is = = Vs Νs I pTransformer rating : The rating of the input power of the transformer. example : 25 kVA ( kV x Arus )
Example 5.1 A 2000/200V, 20kVA transformer has 66 turns in the secondary. Calculate (i) primary turns (ii) primary and secondaryExample 5.2 A 250 kVA, 1100 V / 400 V, 50 Hz single-phase transformer has 80 turns on a secondary. Calculate : a) the values of the primary and secondary currents. b) the number of primary turns. c) the maximum values of flux.
Example 5.3 An ideal 25 kVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 V, 50 Hz supply. Calculate (i) primary and secondary currents (ii) secondary e.m.f. and (iii) the maximum core flux
Advantages and disadvantages of autotransformers- a saving in a cost- less volume, hence less weight.- higher efficiency- a continuously variable output- a smaller percentage voltage regulation.
5.4 Efficiency and losses of a transformer The losses which occur in the transformer :- i. copper losses, I2R ii. Core losses , Pc - hysterises - eddy current
output powerEfficiency = input power output power = output power + losses I sVs × p.f. = I sVs × p.f. + Pc + I p R p + I s Rs 2 2
Example 5.4 The primary and secondary windings of a 500 kVA transformer have resistances of 0.42 Ω and 0.0019 Ω respectively. The primary and secondary voltages are 11 000 V and 400 V respectively and the core loss is 2.9 kW, assuming the power factor of the load to be 0.8. Calculate the efficiency on : i. full load ii. half load
‘Manusia tidak pernah merancang untuk gagal, tetapi gagal untuk merancang’ “Selamat maju jaya”