Ac machines notes

PrasadSolasa 2,560 views 101 slides Feb 20, 2022
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About This Presentation

Ac machines


Slide Content

LECTURE NOTES ON

ELECTRICAL
MACHINE-II

MODULE-IIt

Three Phase Induction Motors: Types, Construction and
principle of operation, 3 phase Induction Motor, general
phasor diagram, equivalent circuit, power and torque

relations, condition for maximum torque, cirle diagram,
Performance characteristics, effect of rotor resistance on
speed torque characteristics, stable & unstable region of
‘operation, Operation with unbalanced supply voltage
Starting: Stating of phase induction motors, high startin
torque motors, speed contro, rheosaic method, pol
changing method, cascade contol of speed, Double cage
induction motor, C ing of Induction motor,
Induction generator

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MODULE-III

THREE PHASE INDCTION MOTOR

Tree Phase Induction Motors: Types,
Construction and principle of operation, 3 phase
Induetion Motor, general phasor diagram,
equivalent circuit, power and torque relations,
condition for maximum torque, circle diagram,
Performance characteristics, effect of rotor
resistance on speed torque characteristics, table
& unstable region of operation, Operation with
unbalanced supply voltage. Stating: Stating of
3 phase induction motors, high starting torque
‘motors, speed control, cosaic method, pole
changing method cascade control of speed,
Double cage induction motor, Cogging and
Crawling of Induction motor, induction

Topics are arranged as per above

Winding

Rotor
Construction

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coc eer Surendra Sal Univers o Techo, Bra

coc eer Surendra Sal Univers o Techo, Bra

Ths flux density distribution in the air gap can bo expres

BÍ0)=B,

The air gap flux per pole

the axial Tel othe satin

y isthe radius of Y the ae op

de at ox =0") (Fig.3.52) and zero at of =90", The Max

¢ À (001) will vary as the cosine ofthe angle?

fore. the voltage induced in phase coil aa is obtained

The voltages induced in the other phase coils are al
sinusoidal, but phaseshifted from cach other by 120 electrical
degrees Ths,

sinfex -120)

infor +120).

the rms vals of the ind

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AL instant 1, où Therefore, the three
Axes are given b

sum of dy and 8 the resultant ox 4

The phasor sum of du. 6, and Gis the resultat Dus 6

Phasor sum of and 6, 4

Phasor sum of and

ote that resulant us is displaced 30° elockwise from position 1

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stant 3. co ~ 60°, Therefore, the tse
fuses are given b

The resultan fax the ph

Note that resultant fl i

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At instant 4, ot Therefore, the thee
fluxes are given

The phasor sum of the resultant Max a

Phasor sum of ~ gad = ats

Note that the resulant Max is donnant ie. i is displaced 90° «
From position 1

P
Cycles of current = À evolutions of it

‘or Cycles ofcument per second = © x revolutions of field per second

Sines revolutions por second is equal to the revolutions pr mints (N) divida]
0 andthe number of eveles per secon so freuen L

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) Ihe quantity N,N is sometimes called sip speed

Ina th rotor i stationary Lo N = 0). sip. s = 1 or 100
i) In an induction motor, the change in slip

im
ro N- Relative speed between magnetie Held and the wind
P= Number of poles
sed N, the relative speed between the rotating lus and the rotor is
quently, the rotor current frequency Ps given by
(S, NP
pan

120

ie, Rotor euren frequency = Fractional
When the rotor à at stands or stationary (ce ie frequen
rotor cent the same as that of supply frequency (P= SE be f= D,

ES

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(0) Tie fist por Ry is the rotor resistance’phase, and represents the rotor Cu

G The second part RA 1-1) is a variale-esistance load, The power

delivered to this load represents the tal mechanical power developed in
the rotor. Thus mechanical load on the induction motor can be replaced by

e-resistance load of value R( 1-1]. Tisis

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Toul electrical load = R’

Power input

There will be str core loss und stator Cu loss. The remaining power will
be the power mansferrd sctoss the airgap ic, put to the rotor

in Rotor input
Rotor Cut toss = A: PR
Toul mechanical power developed by the rotor is
Pa ~ Rotor input Rotor Culos

ar PR

AR 2 PR

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que develops bythe roto, then,

Note that shat torque Ta ill que required to mest

indage and fiction o

The mechanical power P availa electri motor can be

N vor inp
T = torque developed in N-
0 P io gsP
r= Fao ssh Nm
output of the rotor of an induction motor 5 Pa and its

Note. Since windage and tition loss is small Ta ~ Ta. This assumplion ba

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TFT, newton-meire isthe gross torque developed and N rpm. is the

Rotor input

Rotor Culos
Rotor input

Rotor Cu loss = à x Rotor inp
mor up, Pa = Rotor input = Rotor Cu k
Rotor input - s Rotor input
Rotor put (1-3)
Ror input
Rotor Culo:

the stator losses as well as fiction nd windage losses are neglected, then
pat = Useful output
Rotor inp Stator put
Use oput |

od = 1-6 = Eficieney

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Rotor input

60% Rotor input

ein = Rotor Close

Note that inthe the values Fa, Fs. Re and X; represent
the phase vas.

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‘eloped by the rotor is directly proportional 1

The torque T

(ii) rotores
(ii) power factor o

T«E
=KEL

1: = rotor eurent at standstill
ES = rotor ef. at standstill
pf at standstill

Note. The values of rotor emf, rotor curent and rotor power fi

for the given conditions

Rotor impedancs at standstill

Rotor eurent pha at standsül

at standstill

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inde of stating torque would depend upon the relative
ie. rotor. tesistancephase and standull rot

Diflereniain

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olor Resistance,

KEIR
REX

I

Since Ez «Supply vola

hore Ka is another cost

ev

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coc eer Surendra Sal Univers o Techo, Bra

coc eer Surendra Sal Univers o Techo, Bra

coc eer Surendra Sal Univers o Techo, Bra

or

E

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coc eer Surendra Sal Univers o Techo, Bra

Relation between starling and F.L. torques

orque (Ta) is

Ler us strate the above relation with a mumericat e Suppose Lu = $y
ull-load slip s,-0.04. Then,

34 SUPPLY

SERIES RESISTANCE
‘STARTING

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THAEE PHASE:
AUTO-TRANSFORMER

Relation between starting And FL. torques a star-comected

quirrelcage induetion motor. IF V is the line voltage, then voltage across motor
Iöhase on direct switching is V/Y3 and starting eurent is ly = La. In case of
autotransformer, fa tapping of transformation ratio K (a faction) is used, then

‘etovarstormer

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ing currentphase, I= VZ where V = lin

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coc eer Surendra Sal Univers o Techo, Bra

coc eer Surendra Sal Univers o Techo, Bra

coc eer Surendra Sal Univers o Techo, Bra

coc eer Surendra Sal Univers o Techo, Bra

LT mess

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MODULE-IV

SINGLE PHASE MOTORS

SYLLABUS/ TOPICS COVERED

Three Phase Induction Motors: Types
Construction and principle of operation, 3 pha
Induction Motor, general phasor diagram,
equivalent circuit, power and torque relations,
condition for maximum torque, ciel diagram,
Performance characteristics, effect of rotor
resistance on speed torque characteristics, stable
& unstable region of operation, Operation with
unbalanced supply voltage, Starting: Starting of
3 phase induction motors, high starting torque
motor, speed control, costat method, pole
method cascade control of speed,
and
‘of Induction motor, induction

generator

[Topics are arranged as per above sequence

One magenetic

10 be obtained fom this est ar
phase power P,

ase current

and, can be obtained using the following equation

ean be dome as fll

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o be obtained from this tes are
phase power P
©) Phase volle Y

D Phase eurent Za,

|
Backward

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Then, R, and À, can be obtained as allows

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\_Torquedue to\\

main winding

winding

Start & Run Capacitor

Centrifugal switch
Operation

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‘ompensating
indi

1phase
supply

Main field
vine Compensation
winding

1-phase
supply

DIRECT
CURRENT
OPERATION

ALTERNATINe|
CURRENT
OPERATION

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