DC Machines (a)�DC Generator

Module 3

Contents

• Introduction
• Principle Of Operation
• Construction Of DC Generators
• Expression For induced e.m.f
• Types of DC Generators
• Relation between induced e.m.f and terminal voltage

Introduction

• An electrical machine, deals with the energy transfer either from mechanical to electrical form or from electrical to mechanical form .
• This process is called Electromechanical Energy Conversion.

• Electrical Machine – 2 Types
• Electric Generator
• Electric Motor
• An electrical machine which converts mechanical energy into an electrical energy is called an Electric Generator.
• An electrical machine which converts an electrical energy into mechanical energy is called an Electric Motor.

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DC Machines

• DC Generator: These machines converts mechanical input power into dc electrical power.
• DC Motor :These machines converts dc electrical power into mechanical power.
• The construction of both the types of dc machines remains same.

DC Generator

Principle Of Operation

• All generators work on the principle of dynamically induced emf.
• This principle is nothing but the Faraday’s Law Of Electromagnetic Induction.
• It states that whenever the number of magnetic lines of force i.e flux linking with the a conductor or a coil changes , an electromotive force is set up in that conductor or coil.

• A generating action requires following basic components to exist,
• The conductor or a coil
• The flux
• The relative motion between conductor and flux

• To have large voltage as the output, the number of conductors are connected together in a specific manner, to form a winding known as armature winding.
• The part on which winding is kept is called armature of DC Machine.
• Armature is rotated with the help of external device called a Prime Mover.

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• The necessary magnetic flux is produced by current carrying winding which is called field winding.

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• The induced emf is given by

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Construction

Yoke

• The outermost cover of the Dc machine.
• It provides mechanical support to the poles.
• It forms a part of the magnetic circuit.
• It is prepared by using cast iron.
• For large machines rolled steel, cast steel, silicon steel is used.

Poles

• Each pole is divided into two parts:
• Pole core
• Pole shoe
• Pole core basically carries a field winding which is necessary to produce the flux.
• Pole shoes enlarges the area of armature core to come across the flux.
• It is made up of cast iron or cast steel.

Field Winding

• The field winding is wound on the pole core with a definite direction.
• To carry current due to which the pole core, on which the field winding is wound behaves as an electromagnet.
• It is made up of aluminium or copper.
• Field windings is divided into various coils called field coils.

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Armature

• Armature is further divided into two parts
• Armature core
• Armature winding
• Armature core provides house for armature winding.

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Commutator

• The basic nature of emf induced in the armature conductors is alternating.
• This needs rectification in case of dc generator, which is possible by a device called Commutator.

Brushes

• Brushes are stationary and resting on the surface of the commutator.

Bearings

Types Of Armature Windings

1. Lap
1. In lap type the connections overlap each other as the winding proceeds.
2. Number of parallel paths ⇒ A=P

2. Wave

• In wave type the winding travels ahead avoiding the overlapping in a progressive fashion.
• Number of parallel paths ⇒ A=2

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Comparison Between Lap And Wave Type Windings

�E.M.F Equation Of DC Generator

Numerical 1

• A 4 pole, 1500 rpm dc generator has a lap wound armature having 24 slots with 10 conductors per slots. If the flux per pole is 0.04wb, calculate the emf generated in the armature. What would be the generated emf if the winding is wave connected?

Numerical 2

• An 8 pole, lap connected armature has 40 slots with 12 conductors per slot, generates a voltage of 500V. Determine the speed at which it is running if the flux per pole is 50mWb.

Symbolic Representation Of DC Generator

Types Of DC Generator

Separately Excited Generator

Self Excited Generator

• Based on how field winding is connected to the armature to derive its excitation , this type is further divided into,
• Shunt Generator
• Series Generator
• Compound Generator

Shunt Generator

Series Generator

Compound Generator

• In this type, the part of the field winding is connected in parallel with armature and part in series with the armature.
• Both series and shunt field windings are mounted on the same poles.
• There are two types:
• Long Shunt Compound Generator
• Short Shunt Compound Generator

Long Shunt Compound Generator�

Short Shunt Compound Generator�

Applications Of DC Generators

• Separately Excited Generators:
• As a seperate supply is required to excite field, the use is restricted to some special applications like electro plating , electro refining of materials etc

• Shunt Generators:
• Commonly used in battery charging and ordinary lighting purposes.
• Series Generators:
• Commonly used as boosters on dc feeders, as a constant current generators for welding generator and arc lamps.

• Compound Generators:
• These are used for domestic lighting purposes and to transmit energy over long distance.
• Also used in electric arc welding.

Efficiency Of A Dc Machine

1. Mechanical Efficiency
2. Electrical Efficiency
3. Commercial Efficiency

Mechanical Efficiency�

• It is defined as the ratio of total watts generated (E_g I_a ) in the armature to the total mechanical power supplied by the prime mover.

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Electrical Efficiency�

• It is defined as the ratio of total watts available in the load circuit (V_t I_L ) to the total watts generated in the armature (E_g I_a ).

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Commercial Efficiency�

• It is defined as the ratio of total watts available in the load circuit (V_t I_L ) to the total mechanical power supplied by the prime mover.

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• The Overall Efficiency is

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• Practically efficiency of a generator is nothing but commercial or overall efficiency and can also be expressed as

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Example 1

• A 30kW, 300V DC shunt generator has armature and field resistances of 0.05Ω and 100 Ω respectively. Calculate the total power developed by the armature when it delivers full output power.

Example 2

• A dc series generator has armature resistance of 0.5Ω and series field resistance of 0.03Ω. It drives a load of 50A. If it has 6 turns/coil and total 540 coils on the armature and is driven at 1500 rpm, calculate the terminal voltage at the load. Assume 4 pole, lap type winding, flux per pole as 2mWb and total brush drop as 2V.

Example 3

• A short shunt compound dc generator supplies a current of 75A at a voltage of 225V. Calculate the generated voltage if the resistance of armature, shunt field and series field windings are 0.04Ω, 90Ω, 0.02Ω respectively.

MODULE 3

DC Machines (b)

DC Motor

DC Motor

• Principle Of Operation
• Back EMF
• Torque Equation
• Types Of DC Motors
• Characteristics Of DC Motors
• Applications

Principle Of Operation

• The principle of operation of dc motor can be stated in a single statement as ‘ when a current carrying conductor is placed in a magnetic field , it experiences mechanical force’.

• There are two fluxes present:
• The flux produced by the permanent magnet called main flux.
• The flux produced by the current carrying conductor.

Back E.M.F In A DC MOTOR

• In Motor there is an induced emf in the rotating armature conductors according to Faraday’s Law of Electromagnetic Induction.
• This induced emf always acts in the opposite direction to the supply voltage.
• This is according to the Lenz’s Law.

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Voltage Equation Of DC Motor

• The voltage equation for a DC Motor can be written as,

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• If Brush drop is neglected , the armature current can be expressed as,

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Significance Of Back EMF

• DC Motor behaves like a regulating Machine.

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• Back EMF regulates the flow of armature current to meet the load requirement.
• At start speed of the motor is zero hence the back emf is also zero.

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Power Equation Of A DC Motor

• The voltage equation of a DC motor can be given by,

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• This equation is called Power Equation of a DC Motor.
• VI_a = Net electrical power input to the armature in watts
• = Power loss due to the resistance of the armature called armature copper loss.
• is called electrical equivalent of gross mechanical power developed by the armature, denoted by P_m .

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Torque Equation Of A DC Motor

• The turning and twisting force about an axis is called torque.

Types Of DC Motor

DC Shunt Motor

DC Series Motor

DC Compound Motor

• Long Shunt DC Compound Motor
• Short Shunt DC Compound Motor

Long Shunt DC Compound Motor�

Short Shunt DC Compound Motor�

DC Motor Characteristics

• The performance of a DC Motor under various conditions can be judged by the following characteristics,
• Torque - Armature Current Characteristics (T Vs I_a )
• Speed - Armature Current Characteristics (N Vs I_a )
• Speed – Torque Characteristics (N Vs T)

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Characteristics Of DC Shunt Motor

1. Torque - Armature Current Characteristics (T Vs I_a )

Characteristics Of DC Shunt Motor

2. Speed - Armature Current Characteristics (N Vs I_a )

Characteristics Of DC Shunt Motor

3. Speed – Torque Characteristics (N Vs T)

1. Speed – Torque Characteristics (N Vs T)

Characteristics Of DC Series Motor

1. Torque - Armature Current Characteristics (T Vs I_a )

Characteristics Of DC Series Motor

2. Speed - Armature Current Characteristics (N Vs I_a )

Characteristics Of DC Series Motor

3. Speed – Torque Characteristics (N Vs T)

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Why series motor is never started on no load?

Applications Of DC Motors

Numerical 1

• Find the useful flux per pole on no load of a 250V, 6 pole shunt motor having a two circuit connected armature winding with 220 conductors. At normal working temperature, the ovarall armature resistance including brushes is 0.2Ω. The armature current is 13.3 A at the no load speed of 908 rpm.

Numerical 2

• A 500V shunt motor has 4 poles and wave connected winding with 492 conductors. The flux per pole is 0.05 Wb. The full load current is 20A.The armature and the shunt field resistances are 0.1Ω and 250Ω respectively. Calculate the speed and the developed torque.

Numerical 3

• A 4 pole, 220V, lap connected , DC shunt motor has 36 slots , each slot containing 16 conductors. It draws a current of 40A from the supply. The field resistance and the armature resistance are 110Ω, 0.1Ω respectively. The motor develops an output power of 6kW. The flux per pole is 40 mWb. Calculate a)the speed b)The torque developed by the armature c) the shaft torque.

Numerical 4

• A dc series motor is running with a speed of 1000rpm while taking a current of 22A from the supply. If the load is changed such that the current drawn by the motor is increased to 55A, Calculate the speed of the motor on new load. The armature and series winding resistances are 0.3Ω and 0.4Ω respectively. Assume supply voltage as 250V.

MODULE 3

Single Phase Transformer

Single Phase Transformers

• Introduction
• Necessity Of Transformer
• Principle Of Operation
• Types & Construction Of Transformers
• Emf Equation
• Losses
• Variation Of Losses With Respect To Load
• Efficiency
• Condition For Maximum Efficiency

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Introduction

Necessity Of Transformer

• What is transformer?
• What are its functions?
• What are its application in a.c. Transmission?

Definition:

“ The transformer is a static device by means of which an electrical power is transferred from one alternating current circuit to another with the desired change in voltage and current, without any change in the frequency.”

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Principle Of Operation

• The transformer works on the principle of mutual induction.

Mutual Induction:

“When two coils are inductively coupled and if current in one coil is changed uniformly then an e.m.f gets induced in the other coil”.

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Can D.C Supply Be Used For Transformers?

Construction

• There are two basic parts of a transformer
• Magnetic Core
• Winding or Coils

Types Of Transformers

• Based on construction of core , single phase transformers are classified as
• Core type transformer
• Shell type Transformer

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Core Type Transformer

Shell Type Transformer

Comparison Of Core Type & Shell Type

E.M.F Equation Of A Transformer

Concept Of Ideal Transformer

What is ideal Transformer?

A transformer is said to be ideal if it satisfies following properties:

• It has no losses.
• Its windings have zero resistance.
• Leakage flux is zero.
• Permeability of core is so high that negligible current is required to establish the flux in it.

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Ratios Of a Transformer

1. Voltage Ratio
2. Current Ratio

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Voltage Ratio

• The ratio of secondary induced emf to primary induced emf is known as voltage transformation ratio denoted as K.

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Thus,

• If K>1, Then the transformer is called step up transformer.
• If K< 1, Then the transformer is called step down transformer.
• If K=1, The transformer is called 1:1 transformer.

Current Ratio

• For an ideal transformer there are no losses. Hence the product of primary voltage V₁ and primary current I₁ , is same as the product of secondary voltage V₂ and the secondary current I₂ .
• If V₁ I₁ is Input VA and V₂ I₂ is Output VA ,
• For an ideal transformer,

V₁ I₁ = V₂ I₂

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• Hence the currents are in the inverse ratio of the voltage transformation ratio.

Volt-Ampere Rating

• As losses depend on V and I only , the rating of the transformer is specified as a product of these two parameters VxI.
• Hence the transformer rating is specified as the product of voltage and current is called VA rating.
• On both sides primary and secondary VA rating remains same. This rating is generally expressed in KVA (kilo volt ampere rating)

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Full Load Currents

• This is the safe maximum current limit transformer may carry , keeping temperature rise below its limiting value.

Example 1

• Find the number of turns on the primary and secondary side of a 440/230 V , 50Hz single phase transformer, if the net area of cross section of the core is 30cm² and the flux density is 1 Wb/m².

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Example 2

• A single phase 20kVA transformer has 1000 primary turns and 2500 secondary turns. The net cross sectional area of the core is 100 cm² . When the primary winding is connected to 500 V, 50 Hz supply, calculate i) The maximum value of the flux density in the core ii) The voltage induced in the secondary winding and iii) The primary and secondary full load currents.

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Example 3

• A 200 kVA, 10000/400 V, 50 Hz single phase transformer has 100 turns on the secondary. Calculate: i)The primary and secondary currents ii) The number of primary turns iii) The maximum value of flux.

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Losses

Losses In A Transformer

• In a transformer there exists two types of losses .
• The core gets subjected to an alternating flux , causing core losses.
• The windings carry currents when transformer is loaded causing copper losses.
• Core Loss is also known as iron loss.
• Hysteresis Loss
• Eddy Current Loss
• Copper Loss is also known as I² R Loss.

������������� Efficiency�Variation Of Losses With Respect To Load

Efficiency Of A Transformer

• Due to the losses in a transformer, the output power of a transformer is less than the input power supplied.
• The efficiency of transformer is defined as the ratio of the power output to power input.

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Efficiency Expressions

• Full load percentage efficiency,

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• For fractional load the efficiency is given by,

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Condition For Maximum Efficiency

Condition For Maximum Efficiency

• The maximum efficiency is denoted by η_max.
• The load current at which the efficiency attains maximum value is denoted as I_2m .
• Condition to achieve maximum efficiency is ,

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Other Expressions

• The Load current I_2m at Maximum Efficiency,

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• The kVA supplied at maximum efficiency,

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Example 1

• A transformer is rated at 100 kVA. At full load its copper loss is 1200 W and its iron loss is 960 W. Calculate

i) The efficiency at full load, unity power factor

ii) The efficiency at half load, 0.8 p.f.

iii) The load kVA at which maximum efficiency will occur

iv) Maximum efficiency at 0.85 p.f.

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Example 2

• The maximum efficiency at full load and unity power factor of a single phase 25 kVA, 500/1000 V, 50Hz, transformer is 98%. Determine its efficiency at,

i) 75% load, 0.9 p.f.

ii) 50% load, 0.8 p.f.

iii) 25% load, 0.6 p.f.

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Example 3

• A 600 kVA transformer has an efficiency of 92% at full load, unity p.f and at half load, 0.9 p.f. Determine its efficiency at 75% of full load and 0.9 p.f.

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Voltage Regulation Of A Transformer

• It is defined as change in the magnitude of the secondary terminal voltage when full load, with primary voltage maintained constant expressed as the percentage of the rated terminal voltage.

Example 4

A 50kVA, 400/200V single phase transformer has an efficiency of 98% at full load and 0.8pf, while its efficiency is 96.9% at 25% of full load and unity pf. Determine the iron and full load Cu-losses and voltage regulation, if the terminal voltage on full load is 195V.