18EE402-Circle diagram.pdf

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

2
➢ No Load Test
Tests on Induction Motor

3
➢ No Load Test
➢ As the motor is running at no load, the total input power is equal to the constant iron loss,
friction and windage losses of the motor.
➢ Vinl is the input line voltage
➢ Pinl is the total three-phase input power at the no load
➢ I0 is the input line current
➢ Vip is the input phase voltage
➢ 𝑃𝑖𝑛𝑙 = 3𝑉𝑖𝑛𝑙𝐼0 cos 𝜑0
➢ 𝐼𝜇 = 𝐼0 sin 𝜑0 , 𝐼𝑤 = 𝐼0 cos 𝜑0
➢ 𝑅0 =
𝑉𝑖𝑝
𝐼𝑤
, 𝑋0 =
𝑉𝑖𝑝
𝐼𝜇

4
➢ No Load Test: Separation of Losses
➢ Friction and windage loss can be separated from the no-load loss P0. At no load various
readings of the No load loss are taken at the different stator applied voltages. The readings
are taken from rated to the breakdown value at rated frequency.

5
➢ Blocked Rotor (Locked Rotor) Test

6
➢ Blocked Rotor (Locked Rotor) Test:
➢ Vscl is the short circuit line voltage, Iscl is the short circuit line current
➢ 𝑃𝑠𝑐 = 3𝑉𝑠𝑐𝑙𝐼𝑠𝑐𝑙 cos 𝜑𝑠𝑐
➢ 𝑅𝑒1
=
𝑃𝑠𝑐𝑝
𝐼𝑠𝑐𝑝
2 , 𝑍𝑒1
=
𝑉𝑠𝑐𝑝
𝐼𝑠𝑐𝑝
, 𝑋𝑒1
= 𝑍𝑒1
2
− 𝑅𝑒1
2
➢ The slip of the induction motor varies between 2 to 4 percent, and the resulting rotor
frequency is in the range of 1 to 2 hertz for the stator frequency of 50 hertz at the normal
conditions.
➢ In order to obtain the accurate results, the Blocked Rotor Test is performed at a
frequency 25 percent or less than the rated frequency. The leakage reactances at the rated
frequency are obtained by considering that the reactance is proportional to the frequency.
➢ However, for the motor less than the 20-kilowatt rating, the effects of the frequency are
negligible, and the blocked rotor test can be performed directly at the rated frequency.

7
➢ Circle diagram of an induction motor can be drawn by using the data obtained from
(1)No-load test
(2)Short-circuit test or Blocked rotor test
(3)Stator Resistance Test
Construction of the Circle Diagram

8
Step No. 1
• From no-load test, I0 and φ0 can be calculated. Hence, as shown in fig, vector for I0
can be laid off lagging φ0 behind the applied voltage V.
V
X
I0
φ0
O
O’

9
V
X
I0
φ0
Step No. 2
• From blocked rotor test or short-circuit test, short circuit current ISN
corresponding to normal voltage and φS are found.
• The vector OA represents ISN
• 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

10
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

11
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.

12
V
X
φ0
O
φS
A
O’ C D
P
• The line O′A is known as output line.
• It should be noted that as the voltage vector is drawn vertically, all vertical distances
represent the active or power or energy components of the currents.
• the vertical component O′P of no-load current OO′ represents the no-load input, which
supplies core loss, friction and windage loss and a negligibly small amount of stator copper
loss

13
V
X
φ0
O
φS
A
O’ C D
P G

14
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
• When the rotor is locked, then all the power supplied to the motor goes to meet
core losses and Cu losses in the stator and rotor windings.
• 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.

15
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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(ii) Wound Rotor. In this case, rotor and stator resistances per phase r2 and r1 can be easily computed.
For any values of stator and rotor currents I1 and I2 respectively, 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 Ws is 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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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 cosφ1
• 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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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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Maximum Quantities
(i) Maximum Output
• It occurs at point where the tangent is parallel to output line O’A.
.
(ii) Maximum Torque or Rotor Input
• It occurs at point where the tangent is parallel to torque line O′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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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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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) Maximum Torque or Rotor Input
• It occurs at point T where the tangent is parallel to torque line O′E.
or
• point T may be found by drawing CT perpendicular to the torque line.
• Its value is represented by TT’ . (From point T, draw a line vertically downwards till it
intersects torque line i.e. point T’)
T
T’
• Maximum torque is also known as stalling or pull-out torque.
Max. Torque

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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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Reference:
A Textbook of Electrical Technology: Volume 2 - AC and DC Machines by B.L. Theraja and A.K. Theraja

24

25

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➢ The following test results refer to a 3-phase, 20hp(metric), 440V, delta connected, 50Hz, 4 pole
induction motor.
• Running light test: 440V, 10A (line), 1.5kW (input)
• Locked rotor tests: 120V 30A (line), 2.25kW (input)
Draw the circle diagram of this induction motor and determine therefrom
a. Full load current and power factor
b. Maximum possible power output
c. The best possible operating power factor
➢ Solution
No load power factor 𝑐𝑜𝑠∅0 =
𝑃𝑜
3𝑉𝐼𝑜
=
1500
3×440×10
= 0.1968
No load phase angle ∅0 = cos−1 0.1968 = 78.649

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On short circuit
Short circuit current at normal voltage
𝐼𝑆𝑁 = 𝐼𝑆
𝑉
𝑉𝑆
= 30 ×
440
120
= 110𝐴
short circuit input power at normal voltage 𝑃𝑆𝑁 = 𝑃𝑆
𝐼𝑆𝑁
𝐼𝑆
2
= 2250 ×
110
30
2
= 30.25𝑘𝑊
Short Circuit power factor 𝑐𝑜𝑠∅𝑆 =
𝑃𝑆𝑁
3𝑉𝐼𝑆𝑁
=
30250
3×440×110
= 0.3608
Short Circuit phase angle ∅𝑆 = cos−1
0.3608 = 68.85
Let the current scale be 5A/cm
OO’=10A=2cm at 78.650
OA=110A=22cm at 68.850
O’A=I2’=20cm = 100A

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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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➢ Solution
To determine the power scale a perpendicular AF is drawn. AF represents total input on short
circuit with normal voltage applied i.e 30250 Watts
Since AF=7.9 cm
Power Scale, 1cm = 1𝑐𝑚 =
30250
7.9
= 3829 𝑤𝑎𝑡𝑡𝑠
Power Output = 20hp = 20X735.5=14710 Watts which will be represented by
14710
3829
= 3.84 𝑐𝑚
FA line is extended to point T, so that AT=3.84cm. From point T line TH is drawn parallel to
output line O’A intersecting the circle at H.
Point H is joined to the origin O and perpendicular HN is drawn.

30
➢ Solution
a) Full load current = OH =5.35 cm = 5.35 X 5=26.75A
Full load power factor, cos ∅𝑓𝑙 =
𝑁𝐻
𝑂𝐻
=
4.5
5.35
= 0.86
Full load phase angle ∅𝑓𝑙 = 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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➢ 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 ∅𝑚 = 300
Best possible operating power factor: cos ∅𝑚 = cos 30 = 0.866

32
➢ Draw the circle diagram from no-load and short-circuit test of a 3-phase. 14.92 kW, 400-V, 6-
pole induction motor from the following test results (line values).
No-load : 400-V, 11 A, p.f. = 0.2
Short-circuit : 100-V, 25 A, p.f. = 0.4
➢ Rotor Cu loss at standstill is half the total Cu loss. From the diagram, find (a) line current,
slip, efficiency and p.f. at full-load (b) the maximum torque.
➢ Solution
No load power factor =0.2 ; ∅0 = cos−1
0.2 = 78.5
Short Circuit power factor =0.4 ; ∅𝑠 = cos−1
0.4 = 66.4
Short circuit current at normal voltage
𝐼𝑆𝑁 = 𝐼𝑆
𝑉
𝑉𝑆
= 25 ×
400
100
= 100𝐴
short circuit power input with this current 𝑃𝑆𝑁 = 3𝑉𝐿𝐼𝐿 cos ∅𝑠 = 27.71 𝑘𝑊

33
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 78.50 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 :

34
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

35
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 32.90 = 0.84
ϕ

36
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%

37
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

38
➢ Draw the circle diagram for a 5.6 kW, 400-V, 3-φ, 4-pole, 50-Hz, slip-ring induction motor from the
following data :
No-load readings : 400 V, 6 A, cos φ0 = 0.087 : Short-circuit test : 100 V, 12 A, 720 W.
➢ The ratio of primary to secondary turns = 2.62, stator resistance per phase is 0.67 Ω and of the rotor
is 0.185 Ω. Calculate (i) full-load current (ii) full-load slip (iii) full-load power factor (iv) maximum
torque full - load torque (v) maximum power..
➢ Solution
No load power factor cos ∅0 = 0.087, ∅0 = 85
Short Circuit power factor; ∅𝑠 = cos−1 720
3×100×12
= 69.4
Short-circuit current with normal voltage 𝐼𝑆𝑁 = 𝐼𝑆
𝑉
𝑉𝑆
= 12 ×
400
100
= 48𝐴
short circuit power input with this current 𝑃𝑆𝑁 = 3𝑉𝐿𝐼𝐿 cos ∅𝑠 = 27.71 𝑘𝑊

39
G
O O’
C
B Output Line
V
A
85
69.4
Y
X
With C as the centre and 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

40
G
O O’
C
B Output Line
V
A
85
69.4
Y
X
D
Now AD = 8.6 cm and represents 11.52 kW power scale = 11.52/8.6 = 1.34 kW/cm. 1 cm = 1.34 kW
Given K = 2.62 R1 = 0.67 Ω R2 = 0.185 Ω
𝑟𝑜𝑡𝑜𝑟 𝑐𝑢 𝑙𝑜𝑠𝑠
𝑠𝑡𝑎𝑡𝑜𝑟 𝑐𝑢 𝑙𝑜𝑠𝑠
= 2.622
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

41
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 =
𝐹𝐻
𝑃𝐻
=
0.2
4.25
= 0.047 or 4.7%
S
(iii) Power factor =
𝑃𝑆
𝑂𝑃
=
4.6
5.75
= 0.8

42
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
full load torque
=
𝑀𝐾
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

43
➢ A 3 phase 400V, 50Hz, 6 pole Y connected IM has the following test data:
No-load readings : 400 V, 9 A, 1250W : Blocked rotor test : 200 V, 50 A, 6930 W.
➢ Construct the circle diagram. For an operating line current of 30A, obtain from circle diagram (i)
operating power factor (ii) slip and (iii) efficiency. Assume stator cu loss equal to rotor cu loss

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