18EE402-Circle diagram.pdf

Circle diagram

1. 1
5/24/2021
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
by
18EE402
Transformers and Induction Machines
Pradeep Kumar
Assistant Professor
Department of
Electrical and Electronics Engineering
NMAM Institute of Technology

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢NoLoadTest
Tests on Induction Motor

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢NoLoadTest
➢Asthemotorisrunningatnoload,thetotalinputpowerisequaltotheconstantironloss,
frictionandwindagelossesofthemotor.
➢V
inlistheinputlinevoltage
➢P
inlisthetotalthree-phaseinputpoweratthenoload
➢I
0istheinputlinecurrent
➢V
ipistheinputphasevoltage
➢??????
??????��=3�
??????��??????
0cos??????
0
➢??????
??????=??????
0sin??????
0,??????
??????=??????
0cos??????
0
➢??????
0=
??????
??????�
�??????
,�
0=
??????
??????�
�??????
Tests on Induction Motor

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢NoLoadTest:SeparationofLosses
➢Frictionandwindagelosscanbeseparatedfromtheno-loadlossP
0.Atnoloadvarious
readingsoftheNoloadlossaretakenatthedifferentstatorappliedvoltages.Thereadings
aretakenfromratedtothebreakdownvalueatratedfrequency.
Tests on Induction Motor

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢BlockedRotor(LockedRotor)Test
Tests on Induction Motor

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢BlockedRotor(LockedRotor)Test:
➢V
sclistheshortcircuitlinevoltage,I
sclistheshortcircuitlinecurrent
➢??????
????????????=3�
????????????�??????
????????????�cos??????
????????????
➢??????
�1
=
�????????????�
�
????????????�
2,�
�1
=
??????????????????�
�
????????????�
,�
�1
=�
�1
2
−??????
�1
2
➢Theslipoftheinductionmotorvariesbetween2to4percent,andtheresultingrotor
frequencyisintherangeof1to2hertzforthestatorfrequencyof50hertzatthenormal
conditions.
➢Inordertoobtaintheaccurateresults,theBlockedRotorTestisperformedata
frequency25percentorlessthantheratedfrequency.Theleakagereactancesattherated
frequencyareobtainedbyconsideringthatthereactanceisproportionaltothefrequency.
➢However,forthemotorlessthanthe20-kilowattrating,theeffectsofthefrequencyare
negligible,andtheblockedrotortestcanbeperformeddirectlyattheratedfrequency.
Tests on Induction Motor

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢Circlediagramofaninductionmotorcanbedrawnbyusingthedataobtainedfrom
(1)No-loadtest
(2)Short-circuittestorBlockedrotortest
(3)StatorResistanceTest
Construction of the Circle Diagram

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
StepNo.1
•Fromno-loadtest,I0andφ0canbecalculated.Hence,asshowninfig,vectorforI0
canbelaidofflaggingφ0behindtheappliedvoltageV.
V
X
I0
φ0
O
O’

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
I0
φ0
Step No. 2
•Fromblockedrotortestorshort-circuittest,shortcircuitcurrentISN
correspondingtonormalvoltageandφSarefound.
•ThevectorOArepresentsISN
•ISN= Is×V/Vs
•where ISN= short-circuit current obtainable with normal voltage V
•Is= short-circuit current with voltage VS
•Vector O′A represents rotor current I2′ as referred to stator.
O
φS
ISN
A
O’
I2′
•Power factor on short-circuit is found from

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’
•For finding the centre C of this circle,
•Draw a line perpendicular to voltage vector from O’
•chord O′A is bisected at right angles–its bisector giving point C.
C

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
•The diameter O′D is drawn perpendicular to the voltage vector
•Scale of current vectors should be so chosen such that the diameter is more than 20 cm
•With centre C and radius = CO′, the circle can be drawn.

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P
•ThelineO′Aisknownasoutputline.
•Itshouldbenotedthatasthevoltagevectorisdrawnvertically,allverticaldistances
representtheactiveorpowerorenergycomponentsofthecurrents.
•theverticalcomponentO′Pofno-loadcurrentOO′representstheno-loadinput,which
suppliescoreloss,frictionandwindagelossandanegligiblysmallamountofstatorcopper
loss

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P G

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P G
F
Step No. 3
Torque line: This is the line which separates the stator and the rotor copper losses
fixed losses
•Whentherotorislocked,thenallthepowersuppliedtothemotorgoestomeet
corelossesandCulossesinthestatorandrotorwindings.
•The vertical component AG of short-circuit current OA is proportional to the motor
input on short circuit.
•Out of this, FG (= O′P) represents fixed losses i.e. stator core loss and friction and
windage losses.

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P G
F
•AF is proportional to the sum of the stator and rotor Cu losses.
•The point E is such that
•line O′E is known as torque line.
•This is the line which separates the stator and the rotor copper losses.
E
fixed losses
Stator Cu losses
Rotor Cu losses

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
(ii) Wound Rotor. In this case, rotor and stator resistances per phase r
2and r
1can be easily computed.
For any values of stator and rotor currents I
1and I
2respectively, we can write
How to locate point E ?
(i) Squirrel-cage Rotor. Stator resistance/phase i.e. R1 is found from stator-resistance test.
Now, the short-circuit motor input Wsis approximately equal to motor Cu losses (neglecting iron losses)
Value of K may be found from short-circuit test itself by using two ammeters, both in stator and rotor circuits.

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P G
F
E
fixed losses
Stator Cu losses
Rotor Cu losses
•Let us assume that the motor is running under certain load and taking a current OL at pf of cosφ
1
•Then,
•the perpendicular JK represents fixed losses,
•JN is stator Cu loss,
•NL is the rotor input,
•NM is rotor Cu loss,
•ML is rotor output
•LK is the total motor input.
L
φ1
M
N
J
K
•Draw a line from point L vertically downwards till it intersects horizontal axis.

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5/24/2021
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P G
F
E
fixed losses
Stator Cu losses
Rotor Cu losses
L
φ1
M
N
J
K

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
Maximum Quantities
(i) Maximum Output
•It occurs at point where the tangent is parallel to output line O’A.
.
(ii)MaximumTorqueorRotorInput
•ItoccursatpointwherethetangentisparalleltotorquelineO′E.
(iii) Maximum Input Power
•It occurs at the highest point of the circle i.e. at a point where the tangent to the circle
is horizontal.

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5/24/2021
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P G
F
E
fixed losses
Stator Cu losses
Rotor Cu losses
L
φ1
M
N
J
K
Maximum Quantities
(i) Maximum Output
•It occurs at point S where the tangent is parallel to output line O’A.
S
S’
Max. Output
•Or Point S may be located by drawing a line CS from point C such that it is perpendicular to
the output line O′A.
•Maximum output is represented by the vertical SS’. (From point S, draw a line vertically
downwards till it intersects output line i.e. point S’)

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P G
F
E
fixed losses
Stator Cu losses
Rotor Cu losses
L
φ1
M
N
J
K
Maximum Quantities
S
S’
Max. Output
(ii)MaximumTorqueorRotorInput
•ItoccursatpointTwherethetangentisparalleltotorquelineO′E.
or
•pointTmaybefoundbydrawingCTperpendiculartothetorqueline.
•ItsvalueisrepresentedbyTT’.(FrompointT,drawalineverticallydownwardstillit
intersectstorquelinei.e.pointT’)
T
T’
•Maximumtorqueisalsoknownasstallingorpull-outtorque.
Max. Torque

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V
X
φ0
O
φS
A
O’ C
D
P G
F
E
fixed losses
Stator Cu losses
Rotor Cu losses
L
φ1
M
N
J
K
Maximum Quantities
S
S’
Max. Output
T
T’
Max. Torque
(iii) Maximum Input Power
•It occurs at the highest point of the circle i.e. at point R where the tangent to the circle is horizontal.
•It is proportional to RR’.(From point R, draw a line vertically downwards till it intersects horizontal axis
i.e. point R’)
•As the point R is beyond the point of maximum torque, the induction motor will be unstable here.
However, the maximum input is a measure of the size of the circle and is an indication of the ability of
the motor to carry short time over-loads. Generally, RR’ is twice or thrice the motor input at rated
load.
R
R’
Max. Input (RR’)

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5/24/2021
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
Reference:
A Textbook of Electrical Technology: Volume 2 -AC and DC Machines by B.L. Therajaand A.K. Theraja

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢Thefollowingtestresultsrefertoa3-phase,20hp(metric),440V,deltaconnected,50Hz,4pole
inductionmotor.
•Runninglighttest:440V,10A(line),1.5kW(input)
•Lockedrotortests:120V30A(line),2.25kW(input)
Drawthecirclediagramofthisinductionmotoranddeterminetherefrom
a.Fullloadcurrentandpowerfactor
b.Maximumpossiblepoweroutput
c.Thebestpossibleoperatingpowerfactor
➢Solution
Noloadpowerfactor????????????�∅
0=
�
�
3??????�
�
=
1500
3×440×10
=0.1968
Noloadphaseangle∅
0=cos
−1
0.1968=78.649

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
Onshortcircuit
Shortcircuitcurrentatnormalvoltage
??????
??????�=??????
??????
??????
????????????
=30×
440
120
=110??????
shortcircuitinputpoweratnormalvoltage??????
??????�=??????
??????
�????????????
�??????
2
=2250×
110
30
2
=30.25��
ShortCircuitpowerfactor????????????�∅
??????=
�????????????
3??????�????????????
=
30250
3×440×110
=0.3608
ShortCircuitphaseangle∅
??????=cos
−1
0.3608=68.85
Letthecurrentscalebe5A/cm
OO’=10A=2cmat78.65
0
OA=110A=22cmat68.85
0
O’A=I
2’=20cm=100A

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
E
A
O O’
H
R
S’
N
C
B
S
Output Line
Max Output
T
V
F
78.65
68.85

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢Solution
TodeterminethepowerscaleaperpendicularAFisdrawn.AFrepresentstotalinputonshort
circuitwithnormalvoltageappliedi.e30250Watts
SinceAF=7.9cm
PowerScale,1cm=1??????�=
30250
7.9
=3829??????����
PowerOutput=20hp=20X735.5=14710Wattswhichwillberepresentedby
14710
3829
=3.84??????�
FAlineisextendedtopointT,sothatAT=3.84cm.FrompointTlineTHisdrawnparallelto
outputlineO’AintersectingthecircleatH.
PointHisjoinedtotheoriginOandperpendicularHNisdrawn.

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢Solution
a)Full load current = OH =5.35 cm = 5.35 X 5=26.75A
Fullloadpowerfactor,cos∅
��=
��
��
=
4.5
5.35
=0.86
Fullloadphaseangle∅
��=cos
−1
0.86=30.68
b)To determine the maximum possible power output, draw the CS’ line perpendicular to
output line OO’. Drop perpendicular from S’ on the line OF meeting output line at S.
Now SS’ represents the maximum power output
Maximum power output = SS’ X power scale = 7.3 X 3829 = 27.95kW

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢Solution
c)The best possible operating power factor is obtained by drawing a line from point O and
tangential to the semicircle. From circle diagram ∅
�=30
0
Best possible operating power factor: cos∅
�=cos30=0.866

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢Drawthecirclediagramfromno-loadandshort-circuittestofa3-phase.14.92kW,400-V,6-
poleinductionmotorfromthefollowingtestresults(linevalues).
No-load :400-V, 11A,p.f.=0.2
Short-circuit:100-V, 25A,p.f.=0.4
➢RotorCulossatstandstillishalfthetotalCuloss.Fromthediagram,find(a)linecurrent,
slip,efficiencyandp.f.atfull-load(b)themaximumtorque.
➢Solution
Noloadpowerfactor=0.2;∅
0=cos
−1
0.2=78.5
ShortCircuitpowerfactor=0.4;∅
??????=cos
−1
0.4=66.4
Shortcircuitcurrentatnormalvoltage
??????
??????�=??????
??????
??????
????????????
=25×
400
100
=100??????
shortcircuitpowerinputwiththiscurrent??????
??????�=3�
�??????
�cos∅
??????=27.71��

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B
Output Line
V
A
78.5
66.4
Y
X
With C as the centre and CO′ as radius, a semicircle is drawn as shown.
OO′ represents 11 A=11/5 = 2.2 cm is drawn at an angle of 78.5
0
with OY.
Vector OA represents 100 A and measures 100/5 = 20 cm. It is drawn at an angle of 66.4º with OY.
O′G is drawn parallel to OX. BC is the right angle bisector of O′A.
Assume a current scale of 1 cm = 5 A.* The circle diagram is constructed as follows :

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B
Output Line
V
A
78.5
66.4
Y
X
E
F
AF represents power input on short-circuit with normal voltage applied.
It measures 8.1 cm (as measured) and represents 27,710 W.
Hence, power scale becomes 1 cm = 27,710/8.1 = 3,421 W

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B
Output Line
V
A
78.5
66.4
Y
X
E
F
Full Load motor output is given as = 14,920 W. The power scale = 3,421 W,
Hence Full Load Output represents = 14,920/3,421=4.36 cm Extend A to S such that AS=4.36 cm.
Draw a parallel line to output line at S such that the line cuts the semicircle at P.
Point P represents the full-load operating point.
S
P
Line current = OP = 6.5 cm
=6.5 ×5 = 32.5 A.
which means that full-load line current
ϕ= 32.9º (by measurement)
cos 32.9
0
= 0.84
ϕ

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B
Output Line
V
A
78.5
66.4
Y
X
E
F
Given rotor cu loss = ½ total cu loss. Locate H such that AH= ½AE=7.7/2=3.85 cm. Join O’H (torque line)
S
P
ϕ
H
Torque Line
Draw line PL perpendicular to X axis. K –Output line, M –Torque Line.
PM=rotor input, PL=Motor input, KM=rotor cu loss, PK=Motor output
K
M
L
slip = KM/ PM= 0.3 / 4.65 = 0.064 or 6.4% ;
η = PK / PL = 4.36 / 5.48 = 0.795 ~ 80%

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B
Output Line
V
A
78.5
66.4
Y
X
E
F
(b) For finding maximum torque, line CN is drawn ⊥to torque line O′H.
NT is the vertical intercept between the semicircle and the torque line and represents the maximum torque
of the motor in synchronous watts
Now, NT = 7.9 cm (by measurement) ∴Tmax= 7.9 ×3,421 = 27,026 synch. watt
S
P
ϕ
H
Torque Line
K
M
L
N
T
90

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢Drawthecirclediagramfora5.6kW,400-V,3-φ,4-pole,50-Hz,slip-ringinductionmotorfromthe
followingdata:
No-loadreadings:400V,6A,cosφ
0=0.087:Short-circuittest:100V,12A,720W.
➢Theratioofprimarytosecondaryturns=2.62,statorresistanceperphaseis0.67Ωandoftherotor
is0.185Ω.Calculate(i)full-loadcurrent(ii)full-loadslip(iii)full-loadpowerfactor(iv)maximum
torquefull-loadtorque(v)maximumpower..
➢Solution
Noloadpowerfactorcos∅
0=0.087, ∅
0=85
ShortCircuitpowerfactor;∅
??????=cos
−1
720
3×100×12
=69.4
Short-circuitcurrentwithnormalvoltage??????
??????�=??????
??????
??????
??????
??????
=12×
400
100
=48??????
shortcircuitpowerinputwiththiscurrent??????
??????�=3�
�??????
�cos∅
??????=27.71��

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B Output Line
V
A
85
69.4
Y
X
With C as the centreand CO′ as radius,
a semicircle is drawn as shown.
In the circle diagram, OO’ = 3 cm and inclined at 85º with OV.
Line OA represents short-circuit current with normal voltage. It measures 48/2 = 24 cm and represent 48 A.
It is drawn at an angle of 69.4º with OY.
Assume a current scale of 1 cm = 2 A.* The circle diagram is constructed as follows :
O’A is the output line AD is perpendicular to OX.
D

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B Output Line
V
A
85
69.4
Y
X
D
Now AD = 8.6 cm and represents 11.52 kWpower scale = 11.52/8.6 = 1.34 kW/cm. 1 cm = 1.34 kW
Given K = 2.62 R
1= 0.67 Ω R
2= 0.185 Ω
�??????�??????�??????��??????��
����??????�??????��??????��
=2.62
2
0.185
0.67
=1.9 AE=8.25 cm represents total Cu loss and is divided at point T in
the ratio (rotor cu loss, AT: stator cu loss, TE = 1.9 : 1
E
T
Torque Line

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B Output Line
V
A
85
69.4
Y
X
D
Full-load output = 5.6 kW It is represented by a line = 5.6/1.4 =4 cm
AD is extended to R such that AR = 4 cm. Line RP is parallel to output line and cuts the circle at P.
OP represents full-load current. PS is drawn vertically.
E
T
Torque Line
P
R
F
H
(i) F.L. current = OP = 5.75 cm = 5.75 ×2 = 11.5 A
(ii) F.L. slip =
??????�
��
=
0.2
4.25
= 0.047 or 4.7%
S
(iii) Power factor =
�??????
��
=
4.6
5.75
= 0.8

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
G
O O’
C
B Output Line
V
A
85
69.4
Y
X
D
Point W represents Maximum Output CW ⊥O’P (output line) and WL ⊥OX
E
T
Torque Line
P
R
F
H
(v) Maximum output is represented by WL = 7.6 cm.∴Max. output = 7.6 ×1.4 = 10.64 kW
(iv)
max.torque
fullloadtorque
=
��
PH
=
10
4.25
= 2.35
S
W
90
L
Point M represents Maximum torque CM ⊥O’T (torque line) and MK ⊥OX
90
M
K

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
➢A3phase400V,50Hz,6poleYconnectedIMhasthefollowingtestdata:
No-loadreadings:400V,9A,1250W:Blockedrotortest:200V,50A,6930W.
➢Constructthecirclediagram.Foranoperatinglinecurrentof30A,obtainfromcirclediagram(i)
operatingpowerfactor(ii)slipand(iii)efficiency.Assumestatorculossequaltorotorculoss