electrical drives.ppt
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PresenstationTRANSCRIPT
EE-283
Electrical Drives and InstrumentationBy
Dr. Mohammed Moshiul Hoque
DC Generator
Generator Principle
04/08/23 CSE, CUET 2
MachineMechanicalenergy
Electricalenergy
Faraday’s Law’s of Electromagnetic
Induction
Two Essential Conditions
• A magnetic filed (flux)• A Conductor must be move such a way so that
it cut the flux.
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Simple Loop GeneratorConstruction
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Copper coil: ABCD
Slip-ring
Slip-ring
Brushes (carbon/copper)
Slip-ring: The function is To collect the currentInduced in the coil andTo convey it to the externalLoad resistance (R)Armature: The rotating coil.Field Magnets: Magnets
Simple Loop Generator
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Slip-ring
Slip-ring
Brushes (carbon/copper)
Working Principle
• Assumption: coil is rotating in clockwise direction• When the coil is moving in the magnetic filed, the flux
linkage with it changes due to the successive position changes.
• Induced emf
dt
dNe
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Working Principle
The amount of voltage generated depends on
• (1) the strength of the magnetic field • (2) the angle at which the conductor cuts the
magnetic field • (3) the speed at which the conductor is moved• (4) the length of the conductor within the
magnetic field.
Direction of Induced emf• The Polarity of the voltage
depends on(i) Direction of the magnetic
lines of flux(ii) Direction of movement of
the conductor. To determine the direction of
current in a given situation, -LEFT-HAND RULE FOR
GENERATORS is used.
Working Principle
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At pos 1, 00: the flux linkage is Maximum butthe rate of change Of flux is minimum.Emf=0
At pos 3, 900: the rate ofChange of flux (emf)Increases, till position 3 is Reached. Flux linkage is Minimum but rate of changeOf flux is maximum.Emf=maximum
Working Principle
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At pos 3 TO 5, 900~1800:the flux linkage with theCoil gradually increasesBut the the rate of change Of flux is decreases.Emf= decreases graduallyAt pos 5, emf=0.
Pos1: emf=0/minimumPos3: emf=maximumPos5: emf=0/minimum
The direction of this induced emf can be found by apply the Fleming’s RHR (A-B & C-D)
Direction: ABMLCD
Working Principle
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At pos 5 TO 9, 1800~3600:the variations in the magnitude of emf are similar to those in the first half cycle.Pos7: emf= maximum Pos1: emf=0/minimum.
Direction: reverseD-C and B-ADCLMBA
** The current reverses its direction after every Half cycle. This type of periodic reversals is Known as the alternating current (AC)
How Can converted these AC to DC?
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•Slip-rings are replaced by Split-rings
Split-ringSplit-ring
How Can converted these AC to DC?
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•The split-rings are made out of a conducting cylinder Which is cut into two halves/Segments insulated from eachOther by a thin sheet of Mica/some insulating materials.•The coil ends are joined to these segments on which rest the carbon/copper Brushes.
How Can converted these AC to DC?
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•1st cycle, the current flows along ABMLCD brush#1 in contact with segment a• +ve end of the supply
•2nd half cycle, direction is reversed •At the same time, the positionsOf the segments (a & b) have also reversed.
How Can converted these AC to DC?
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•Brush#1 comes in touch with that Segment which is +ve (Segment b)Current in the R again flows from M to L
The current is unidirectional but not continuous like pure DC (Rectified AC)•It is due to the rectifying action of the split-rings (commutator) that it becomesUnidirectional in the external circuit.
Pulsating DC
Parts of the Generator
1. Magnetic Frame/Yoke2. Pole-cores and Pole Shoes3. Pole Coils or Field Coils4. Armature Core5. Armature Windings/Conductors6. Commutator7. Brushes and Bearings
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Yoke
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02 purposes:1. It provides
mechanical support for the poles and acts as a protecting
cover for the whole machine
2. It carries the magnetic flux
produced by the poles.
Pole Cores/Poles Shoes
• 02 Purposes:1. They spread out the flux
in the air gap and also, being of larger cross-section, reduce the reluctance of the magnetic path
2. They support the exciting coils/field coils.
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Pole Coils
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•Consists of copper wire/strip, are former- wound for the correctDimension.
•When the current is passed through these coils, they electro-magnetize the poles which produce the necessary flux that are cut by the revolving armature conductors.
Armature Core
• It houses the armature conductors or coils and causes them to rotate and hence cut the magnetic flux of the field magnets.
• To provide a path of very low reluctance to the flux through the armature from a N-pole to S-pole.
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Armature Windings• Usually former-wound.• These are first wound in the form of flat rectangular coils and are
then pulled into their proper shape in a coil puller.• Various conductors of the coils are insulated from each other.• The conductor are placed in the armature slots which are lined
with tough insulating material.• This slot insulation is folded over above the armature conductors
placed in the slot and is secured in place by special hard wooden/fiber wedges.
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Commutator
• To facilitate collection of current from the armature conductors.
• It rectifies/converts the AC induced in the armature conductors into unidirectional current in the external load circuit.
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Commutator
• It is of cylindrical structure and is built up of wedge-shaped segments of high conductivity hard-drawn/drop forged copper.
Brushes and Bearings• To collect current from
commutator• Carbon/graphite made• Rectangular block.• Brushes are housed in
brush-holders usually of the box-type variety.
• Because of their reliability, ball bearings are frequently employed though for heavy duties, roller bearings are preferable.
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Field Excitation
• When a dc voltage is applied to the field windings of a dc generator, current flows through the windings and sets up a steady magnetic field. This is called FIELD EXCITATION.
• This excitation voltage can be produced by the generator itself or it can be supplied by an outside source, such as a battery.
Types of Generators
• According to the way of field excitation:1. Separately-excited-filed magnets are energized from an
independent external source (dc current)
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ADC
Types of Generators
2. Self-excited-field magnets are energized by the current
produced by the generators themselves.-due to the residual magnetism, there is always
present some flux in the poles.
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Phenomenon of Self Excitation• Residual Magnetism: Self-excitation is possible only if the
field pole pieces have retained a slight amount of permanent magnetism
• When the generator starts rotating, the weak residual magnetism causes a small voltage to be generated in the armature.
• This small voltage applied to the field coils causes a small field current.
• Although small, this field current strengthens the magnetic field and allows the armature to generate a higher voltage.
• The higher voltage increases the field strength, and so on. This process continues until the output voltage reaches the rated output of the generator.
Types of Generators
Types of self-excited generators
1. Shunt wound-filed windings are
connected across/in parallel with the armature conductors
-have the full voltage of the generator applied across them
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A
Ia = I + Ish
I = Ia-Ish
Types of Generators2. Series wound-field windings are connected
in series with the armature conductors.
-They carry the full load current, they consist of relatively few turns of thick wire/strips.
-Such generators are rarely used except for special purpose (as boosters)
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Ia = I
A
Types of Generators3. Compound wound:-combination of a few series and a few shunt windings
and can be either short-shunt or long-shunt.-the shunt field is stronger than the series field.
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A A
Short-shunt
Long-shunt
Types of Generators
Commutatively Compounded: series aids the shunt field
Differentially Compounded: series field opposes the shunt field
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Generated E.M.F. or E.M.F. Equation of a Generator
• Let, =flux/pole (wb)• Z= total no. Of armature conductors=No.of
slots X No.of conductors/slot• P=No. of generator poles• A = No. of parallel paths in armature• N= Armature rotation in rev/min (rpm)• E= emf induced in any parallel path in armature
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Generated EMF
• Generated emf, Eg = emf generated in any one parallel paths
• Average emf generated/conductor,
• Flux cut/conductor in one rev,
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)1(
Ndt
d
dt
dNE
)/( polefluxPd
Generated EMF
• N=rev/min• Rev/sec = • Time for one rev, second• Generated emf/conductor= (1)
• For a simplex wave-wound generator-No. of parallel paths = 2-No. of conductors (in series) in one path=Z/2
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60
N
Ndt
60
voltPN
N
P
dt
d
60/60
Generated EMF
voltPN
N
P
dt
d
60/60
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Generated emf/conductor= (1)
Generated emf/Path= (2)voltZPNZPN
120260
For a Simplex lap-wound generatorNo.of parallel paths = PNo. of conductors in one path= Z/P
(1) Implies, (3)voltZPN
P
ZPN
6060
Generated EMF
voltA
PZNE
g
60
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A = 2 (for simplex wave-winding) = P (for simplex lap-winding)
NKEag
A
ZPK
a
-For a given dc m/c, Z, P, and A are constant-N (rps)
Loss in Generator• Occur in different parts of the machine.• All appear as heat, i.e. they represent conversion to unless thermal
energy. Losses has two major effects:(i) Losses raise the temperature inside the m/c, and thus affect the
performance/life of the materials of the m/c, particularly insulation. Therefore losses determine the upper limits on machine rating.
(ii) Losses are a waste of energy, and energy costs money; therefore losses result in a waste of money( in the operating cost of the machine).
$$ Losses cannot be eliminated, but they can be reduced by proper design; -design must also provide for ventilation to disperse the heat generated.
Thus, losses have a significant effect on the initial cost of the machine.
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Copper Losses• The power lost in the form of heat in the armature winding of
a generator is known as COPPER LOSS. • Heat is generated any time current flows in a conductor. • Copper loss = I2R , which increases as current increases. • The amount of heat generated is also proportional to the
resistance of the conductor. • The resistance of the conductor varies directly with its length
and inversely with its crosssectional area. • Copper loss is minimized in armature windings by using large
diameter wire.• R L/A, A, R
Copper Losses1. Armature copper loss:- Wa = Ia
2Ra
- 30 to 40% of full-load losses
2. Field copper loss, -Wf = Ish
2Rsh/Ise2Rse
- 20 to 30% of full-load losses
3. Brush contact loss- Due to the brush contact resistance
Magnetic/Iron/Core Losses
1. Hysteresis Loss-Wh Bmax
1-6f
2. Eddy current loss-We B2
maxf2
- Both losses total up to 20 to 30% of full-load losses
Hysteresis Loss• Hysteresis loss is a heat loss caused by the magnetic
properties of the armature. • When an armature core is in a magnetic field, the magnetic
particles of the core tend to line up with the magnetic field. • When the armature core is rotating, its magnetic field keeps
changing direction. • The continuous movement of the magnetic particles, as they
try to align themselves with the magnetic field, produces molecular friction.
• This, in turn, produces heat. This heat is transmitted to the armature windings.
• The heat causes armature resistances to increase.
How to reduce Hysteresis Loss?
• To compensate for hysteresis losses, heat-treated silicon steel laminations are used in most dc generator armatures.
• After the steel has been formed to the proper shape, the laminations are heated and allowed to cool.
• This annealing process reduces the hysteresis loss to a low value.
Eddy Current Loss
• Eddy Current• The core of a generator armature is made from soft
iron, which is a conducting material with desirable magnetic characteristics.
• Any conductor will have currents induced in it when it is rotated in a magnetic field.
• These currents that are induced in the generator armature core are called EDDY CURRENTS.
• The power dissipated in the form of heat, as a result of the eddy currents, is considered a loss.
Eddy Current Loss
• Affected by the resistance of the material in which the currents flow.
• The resistance of any material is inversely proportional to its cross-sectional area (R L/A, A, R)
Eddy Current• Fig. B, shows a soft iron core of the
same size, but made up of several small pieces insulated from each other. This process is called lamination.
• The currents in each piece of the laminated core are considerably less than in the solid core.
• The currents in the individual pieces of the laminated core are so small
• Sum of the individual currents is much less than the total of eddy currents in the solid iron core.
R L/A, A , R I
How can eddy current be reduced?• As you can see, eddy current losses are kept low when the core
material is made up of many thin sheets of metal. • Laminations in a small generator armature may be as thin as
1/64 inch. • The laminations are insulated from each other by a thin coat of
lacquer or, in some instances, simply by the oxidation of the surfaces.
• Oxidation is caused by contact with the air while the laminations are being annealed.
• The insulation value need not be high because the voltages induced are very small.
• Most generators use armatures with laminated cores to reduce eddy current losses.
Mechanical Losses1. Friction loss at bearings and commutator2. Air-friction/Windage loss of rotating armature-10 to 20% of full load lossesStray Losses: Magnetic + Mechanical losses-Rotational LossesConstant/Standing Losses: (Shunt Cu + Stray)
Losses
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Total Loss = Variable Loss + Constant Loss
Ia2Ra
Shunt/Compound
Total Losses: Summary
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Classification of LossesLoss type rotati
onalWith
load
dependence
Armature circuit copper loss E No variable IA2
Series field copper loss E No variable IA2
Shunt field copper loss E No constant Vt2
Brush contact loss E No variable Ia
Hysteresis loss M Yes constant fBXmax
Eddy current loss M Yes constant f2B2max
Friction loss ME Yes constant Power of n
Windage loss ME Yes constant Power of n
Stray load loss E+ME
Yes variable indeterminate
Power Stages
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A B C
Generator Efficiency
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1.Mechanical Efficiency
2.Electrical Efficiency
3. Overall/Commercial Efficiency
engine driving ofoutput
IE
suppliedpower mechanical
armaturein generated wattstotal agm
aIgEVI
e generated wattstotal
circuit loadin available wattsBC
suppliedpower mechanicalcircuit loadin available watts
AC c
emc B
C
A
B
A
C
~95% for good generator
Condition for Max’m Efficiency
• Generator Output Wo = VI
• Generator input, Wi= output + losses
= VI + (Ia2Ra+Wc)=VI + (I + Ish)2Ra + Wc [Ia = I + Ish]
-Ish is negligible as compared to load current (I)
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Ia = I + Ish= I
Condition for Max’m Efficiency
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caa WRIVI
VI
2in
o
Winput,
Woutput, ,Efficiency
IIWRIVI
VIa
ca
2
-(1)---1
12
VI
W
V
IR
VI
W
VI
RI
VIVI
VIVI
caca
I
1
1
Condition for Max’m Efficiency• Efficiency is maximum when denominator is
minimum
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0I or,
0VI
RI or,
0V
R or,
0
2
2a
2
2a
ca
c
c
ca
WR
W
VI
W
VI
W
V
IR
dI
dcWaR 2I
Thus, Generator Efficiency is maximum whenVariable loss (I2Ra) = constant Loss (Wc)
aRcW
aRcW
I or,
2I
Example
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In a long-shunt compound generator, the terminal voltage is 230 V when generator delivers 150 A. Determine (i) induced emf (ii) total power generated and (iii) Distribution of this power. Given that shunt field, series field, divertor and armature resistances are 92, 0.015 , 0.03 , and 0.032 respectively.
Given, V=230 VI = 150 ARsh = 92 Rse = 0.015 R = 0.03 Ra = 0.032
Eg = Ptotal= Pdist=
Solution
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A
Rsh
=92
Rd=
0.03
Rse
=0.
015
V=230 V
Ra=
0.03
2
I = 150 AIsh = 2.5 A
V 4.2364.6230VVE voltage,Generatd
V4.6042.05.152RIVRat drop Voltage
042.001.0032.0RRR
01.003.0015.0
03.0015.003.0015.0RRR
A 5.1525.2150III
A 5.292
230
R
VI
RI V (i)
atotalg
atotalaatotal atotal,
caatotal
dsec
sha
shsh
shsh
Solution
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Watt5.1524.236aIgEaP
Armaturein GeneratedPower Total (ii)
36,051
3605134500575232744
PPPPP power, ofon distributi Total
34500150230P load, todeliveredPower
Watt5755.2230VIP ing,shunt windin lossPower
Watt23201.05.152RIP Divertor, and seriesin lossPower
Watt744032.05.152RI P armature,in lossPower (iii)
Loadshcadist
Load
shsh
2
c
2
ac
2
a
2
aa
VI
Self Study
• Solve the Following Problems:• Example 24.3, 24.4, 24.5 24.7, 24.21, 24.22,
24.23, 24.24, 24.25• Exercise: 2, 9
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CT#01
• 3/E, Monday• Syllabus: Up to 36
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