Determination of Equivalent Circuit parameters and performance characteristics Circle Diagram

EEE 2003
ELECTROMECHANIC
AL ENERGY
CONVERSION
Dr. P. Vijayapriya
Associate Professor, SELECT
Module: 6 Testing of Induction Machine
Determination of Equivalent Circuit parameters – performance characteristics Circle Diagram –Speed
Control –Induction Generator Applications

Equivalent Circuit
The induction motor is similar to the transformer with the
exception that its secondary windings are free to rotate
When the rotor is locked (or blocked), i.e. s =1, the largest
voltage is induced in the rotor at highest frequency
(supply frequency f)
On the other side, if the rotor rotates at synchronous
speed, i.e. s = 0, the induced voltage and frequency in the
rotor will be equal to zero

Equivalent Circuit
So the voltage in rotor under running condition is given by
E2r = sE2
Where E2 is the rotor’s induced voltage obtained at s = 1(locked rotor)
The same is true for the frequency, i.e.
f` = sf
The same is true for the rotor reactance as frequency changes
X2r = sX2

Equivalent Circuit
Then, we can draw the rotor equivalent circuit as follows
Where E2r is the induced voltage in the rotor and R2 is the rotor resistance
jX2r = jsX2
E2r = sE2
R2
I2

Equivalent Circuit
Now we can calculate the rotor current as
Dividing both the numerator and denominator by s so nothing changes
we get
Where E2 is the induced voltage and X2 is the rotor reactance at blocked rotor
condition (s = 1)
2
2
2
2
2
2
2
2
)
(sX
R
sE
Z
E
I
r
r
r



2
2
2
2
2
2
)
(X
s
R
E
I r



Equivalent Circuit
• Now we can have the rotor equivalent circuit
E2
jX2
R2
S
I2

Equivalent Circuit
• Splitting the resistances in to two we get the modified equivalent
circuit as
E2
jX2
R2
S
I2
Actual rotor
resistance
Resistance
equivalent to
mechanical load

Equivalent Circuit
• We can rearrange the equivalent circuit as follows
Actual rotor
resistance
Resistance
equivalent to
mechanical load
R0 X0
I0
I`2

9
Problem-1
Dr. J. Belwin Edward, Asso. Prof., SELECT, VIT, Vellore
The maximum torque of a 3-phase induction motor occurs at a slip of 12%. The motor has an equivalent secondary
resistance of 0.08 /phase. Calculate the equivalent load resistance RL, the equivalent load voltage VLand the current at
this slip if the gross power output is 9,000 watts.

10
Problem - 2
A 220-V, 3-, 4-pole, 50-Hz, Y-connected induction motor is rated 3.73 kW. The equivalent circuit parameters are:
R1 = 0.45 , X1 = 0.8 ; R2’ = 0.4 , X2’ = 0.8 , BO = – 1/30 mho
The stator core loss is 50 W and rotational loss is 150 W. For a slip of 0.04,
find : input current
i. p.f.
ii. air-gap power
iii. mechanical power
iv. electro-magnetic torque
v. output power and
vi. efficiency.

11
Problem - 2

12
Problem - 3
A 440-V, 3-φ 50-Hz, 37.3 kW, Y-connected induction motor has the following parameters:
R1 = 0.1 , X1 = 0.4 , R2’ = 0.15 Ω, X2′ = 0.44 Ω
Motor has stator core loss of 1250 W and rotational loss of 1000 W. It draws a no-load line current of 20 A at a p.f. of
0.09 (lag). When motor operates at a slip of 3%,
calculate :
(i) input line current and p.f.
(ii) electromagnetic torque developed in N-m
(iii) output and
(iv) efficiency of the motor.

Determination of motor parameters
Due to the similarity between the induction motor equivalent circuit
and the transformer equivalent circuit, same tests are used to
determine the values of the motor parameters.
DC test: determine the stator resistance R1
No-load test: determine the rotational losses and magnetization current
(similar to no-load test in Transformers).
Locked-rotor test: determine the rotor and stator impedances (similar to
short-circuit test in Transformers).

DC test
• The purpose of the DC test is to determine R1. A variable
DC voltage source is connected between two stator
terminals.
• The DC source is adjusted to provide approximately
rated stator current, and the resistance between the
two stator leads is determined from the voltmeter and
ammeter readings.

DC test
• then
• If the stator is Y-connected, the per phase stator resistance is
• If the stator is delta-connected, the per phase stator resistance is
DC
DC
DC
V
R
I

1
2
DC
R
R 
1
3
2
DC
R R


No-load test
1. The motor is allowed to spin freely
2. The only load on the motor is the friction and windage
losses, so all Pconv is consumed by mechanical losses
3. The slip is very small

No-load test
4. At this small slip
The equivalent circuit reduces to…
2 2
2 2
(1 ) R (1 )
&
R s s
R X
s s
 
>> >>

No-load test
5. Combining Rc & RF+W we get……

No-load test
6. At the no-load conditions, the input power measured by meters must equal
the losses in the motor.
7. The PRCL is negligible because I2 is extremely small because R2(1-s)/s is very
large.
8. The input power equals
Where
&
2
1 1
3
in SCL core F W
rot
P P P P
I R P
  
 
&
rot core F W
P P P
 

Blocked-rotor test
• In this test, the rotor is locked or blocked so that it
cannot move, a voltage is applied to the motor, and
the resulting voltage, current and power are
measured.

Blocked-rotor test
• The AC voltage applied to the stator is adjusted so that the current
flow is approximately full-load value.
• The locked-rotor power factor can be found as
• The magnitude of the total impedance
cos
3
in
l l
P
PF
V I

 
LR
V
Z
I



• The core is very small as only small voltage is required to circulate the
rated current under blocked rotor condition and also there is no load
delivered by the motor
• Hence the entire power consumed by the motor is only to meet the
total copper loss (stator +rotor)

The Heyland diagram is an approximate representation
of circle diagram applied to induction motors, which
assumes that stator input voltage, rotor resistance and rotor
reactance are constant.
First conceived by A. Heyland in 1894 and B.A. Behrend
in 1895, the circle diagram is the graphical representation
of the performance of the electrical machine drawn in
terms of the locus of the machine's input voltage and
current.
CIRCLE DIAGRAM

1. Draw the voltage axis (Y axis) and X axis (Input line), mark intersection point as O
2. From the no load test and blocked rotor test calculate the following
3. Fix the current scale such that Isn does not exceed 25 cm
4. At an angle φo from Y axis, draw Io according to scale and mark O`
5. Draw O`B parallel to X axis which is the constant loss line
6. At an angle φsc from Y axis, draw Isn according to scale and mark the point A
7. Join O`A (which is the output line ) and draw perpendicular bisectors to O`A. Extend it to meet the constant loss line at C.
8. With C as centre and CO` as radius draw a semicircle. The circle will pass through A.
Step wise Procedure






 
0
0
0
1
0
3
cos
I
V
P
 





 
sc
sc
sc
sc
I
V
P
3
cos 1

sc
sc
rated
SN I
V
V
I 






sc
sc
rated
SN P
V
V
P 





 2
2

10. From A drop a perpendicular to X axis to intersect constant loss line at F. AF represents
the total cu loss ie power input at blocked rotor condition at rated voltage (ie Psn)
11. Fix the power scale as follows
12. Location of point E:
Check for data of R1 (stator resistance) If given calculate stator cu loss = 3 I2snR1.
Calculate
13. Or calculate EF = stator cu loss / power scale and mark E from the point F
14. If no data given, assume stator cu loss equals rotor copper loss and divide as AE = EF
15. Join EO` which is the torque line
cm
watts
AF
P
Powerscale
watts
P
AFcm
SN
SN
/


AF
y
EF
y
loss
cu
Total
loss
cu
stator
AF
EF
*





15. Join EO` which is the torque line.
16. For full load condition (if the power rating of the machine is given in horse power convert
to watts ( * 746 w (mks unit)). Find how many cm represents output power and extend it
from A to A`
17. From A` draw a line parallel to output line. It will intersect circle at maximum two
different places. Take the point near to Y axis as operating point H.
18. From H drop a perpendicular to X axis to meet at N. Mark K, L, M as intersection with
output line, torque line and constant loss line
19. Calculate the required parameters
20. Out put = HK x power scale
Input = HN x power scale
rotor input = HL x power scale
rotor cu loss = KL x power scale
rotor efficiency = HK / HL
slip = KL / HL
total efficiency = HK / HN
power factor = HN / OH
21. To find Maximum quantities:
draw a line parallel to the corresponding line and tangent to the circle. Drop a
perpendicular from the circle intersection point to the corresponding line. When
multiplied by power scale we get the maximum value of that particular quantity

O
V Draw the X Axis and Y Axis

ɸ0 O’
O
V Draw the Vector OO’ = I0 at ɸ0

ɸ0 O’
O
V Draw the Horizontal line from O’ parallel to X axis

ɸ0 O’
O
ɸsc
V
A
Draw the Vector OA = ISN at ɸSC

ɸ0 O’
O
ɸsc
V
A
Join O’A this is output line

ɸ0 O’
O
ɸsc
V
A
Draw the Perpendicular Bi-Sector of AO’, to
meet the horizontal line from O’ at C.
This is the centre of the circle
A
Radius
Centre

O’
Draw a Semi Circle to meet the horizontal line from O’ at B
with C as a centre and CO’ as radius

O’
D
Draw perpendicular from A on X-axis, meeting at point D

O’
D
E
F
Draw perpendicular from A on X-axis, meeting at point D
AD = AE + EF + FD

O’
D
F
E
AF = AE + EF
AF = WSN = Y1 cm
Power Scale = WSN /AF
Calculate stator cu loss = 3 I2
SNR1
Where R1 = stator resistance / phase
ISN = stator current / phase under short
circuit with normal voltage
EF = Stator cu loss : Measure EF from F
Power scale
Join O`E which is the torque line
AD = Total Losses
AE = Rotor Cu Loss
EF = Stator Cu Loss
FD = Fixed Loss

O’
P
Q
S
T
R
AF = WSN = Y1 cm
D
F
E
Base Line
X-Axis
AF = AE + EF
A’
Draw the line from A’
parallel to the output line,
it intersects the circle at P
AA’ = Output Power / Power Scale
For Rated output :

O’
P
Q
S
T
R
AF = WSN = Y1 cm
D
F
E
Total motor input = PT x Power scale
Fixed loss = ST x Power scale
Stator copper loss = SR x Power scale
Rotor copper loss = QR x Power scale
Total loss = QT x Power scale
Rotor output = PQ x Power scale
Rotor input = PQ + QR = PR x Power scale
Draw the Vertical Line from
P to intersect output line at Q
intersect torque line at R
intersect base line at S
intersect X-axis at T
Base Line
X-Axis
A’

O’
P
Q
S
T
R
AF = WSN = Y1 cm
D
F
E
Slip s = QR/PR = rotor cu loss / rotor input
Rotor output = PQ x Power scale (PQ = AA`)
Power factor cos = PT/OP
Motor efficiency = Output / Input = PQ/PT
ɸ
A’
Rated Current = OP x Current scale

O’
P
Q
S
T
R
AF = WSN = Y1 cm
D
F
E
Slip s = QR/PR
Rotor output = PQ x Power scale
Power factor cos = PT/OP
Motor efficiency = Output / Input = PQ/PT
ɸ
J
K
Max Torque = JK * power scale
A’

Construction of circle diagram
Radius
Centre
C
O
O’
Io
I2’
ISN
A
F
E
B
V
φo
φsc
Input Line
I1
Stator
Cu loss
Rotor
Cu loss
Core loss
H
K
L
M
N
P’
P
S’
S
T
T’






 
0
0
0
1
0
3
cos
I
V
P







 
sc
sc
sc
sc
I
V
P
3
cos 1

sc
sc
rated
SN I
V
V
I 






sc
sc
rated
SN P
V
V
P 





 2
2
cm
watts
AF
P
Powerscale
watts
P
AFcm
SN
SN
/


)
25
(
1
cm
than
more
not
is
I
suchthat
xA
cm
le
Currentsca
SN


Out put = HK x power scale
Input = HN x power scale
rotor input = HL x power scale
rotor cu loss = KL x power scale
slip = KL / HL
total efficiency = HK / HN
power factor = HN / OH

10. From A drop a perpendicular to X axis to intersect constant loss line at F. AF represents
the total cu loss ie power input at blocked rotor condition at rated voltage (ie Psn)
11. Fix the power scale as follows
12. Location of point E:
Check for data of R1 (stator resistance) If given calculate stator cu loss as 3 I2snR1.
Calculate
EF = 3I2
SNR1
Power scale
13. If not assume stator cu loss equals rotor copper loss and divide as AE = EF
14. Join EO` which is the torque line
cm
watts
AF
P
Powerscale
watts
P
AFcm
SN
SN
/



Problem 1
• A 3 phase, 400V, star connected induction motor gave the following
test reading.
• No load : 400V 9A 1250W
• Blocked : 150V 38A 4kW
• Draw the circle diagram. IF the normal rating is 20.27 hp find from
the circle diagram, the full load values of current, power factor slip,
efficiency and rotor efficiency. Also find all the maximum values.
Assume rotor copper loss equals stator copper loss.

ISN = Short circuit @ Normal Voltage
WSN = Short circuit power / blocked rotor input with normal voltage

𝑁𝑠 =
120𝑓
𝑃
• NS – Synchronous Speed
• f – Supply frequency in Hz
• P – No. of Stator Poles
Speed Control of 3-Phase IM
• D.C. shunt motors can be made to run at any speed within wide limits, with good
efficiency and speed regulation, merely by manipulating a simple field rheostat,
the same is not possible with induction motors.
• In their case, speed reduction is accompanied by a corresponding loss of
efficiency and good speed regulation.
• That is why it is much easier to build a good adjustable-speed d.c. shunt motor
than an adjustable speed induction motor.

Speed control of
3-Phase Induction
Motor
Stator Side
Control
by changing the
applied voltage
by changing the
applied frequency
by changing the
number of stator
poles
Rotor Side Control
rotor rheostat
control
by operating two
motors in
concatenation or
cascade
by injecting an
e.m.f. in the rotor
circuit.
Kramers Scherbius
Speed Control Methods

Speed Control
• There are 2 types of speed control of 3 phase induction
machines
• From stator side
i. Variation in supply voltage
ii. Variation in supply frequency
iii. Variation in number of poles
• From Rotor side
i rotor rheostat control
Ii cascaded operation
Iii by injecting emf in rotor circuit

• Maximum torque changes
• The speed which at max torque occurs is
constant (at max torque, XR=RR/s
• Relatively simple method – uses power
electronics circuit for voltage controller
• Suitable for fan type load
• Disadvantages :
• Large speed regulation since ~ ns
T
ns~nNL
T
nr1
nr2
nr3
n
nr1> nr2 > nr3
V1
V2
V3
V1> V2 > V3
V decreasing
Varying supply Voltage

• The best method since supply voltage
and supply frequency is varied to keep
V/f constant
• Maintain speed regulation
• uses power electronics circuit for
frequency and voltage controller
• Constant maximum torque
T
nNL1
T
nr1
nr2
nr3
n
f decreasing
nNL2
nNL3
Varying supply voltage and supply frequency

Rotor Rheostatic Control
• One serious disadvantage of this method is that with increase in rotor resistance, I2R losses also increase which
decrease the operating efficiency of the motor. In fact, the loss is directly proportional to the reduction in the
speed.
• The second disadvantage is the double dependence of speed, not only on R2 but on load as well. Because of the
wastefulness of this method, it is used where speed changes are needed for short periods only.
T
ns~nNL
T
nr1
nr2
nr3
n
nr1< nr2< nr3
R1
R2
R3
R1< R2< R3

Cascade or Concatenation or Tandem Operation
In this method, two motors are used and are ordinarily mounted on the same shaft, so that both run at the same speed
(or else they may be geared together).

Injecting an e.m.f. in the Rotor Circuit - Kramer
In this method, the speed of an induction motor is controlled by injecting a voltage in the rotor circuit, it being of course,
necessary for the injected voltage to have the same frequency as the slip frequency. There is, however, no restriction as to the
phase of the injected e.m.f.
One big advantage of this method is that any
speed, within the working range, can be
obtained instead of only two or three, as with
other methods of speed control.
Yet another advantage is that if the rotary
converter is over-excited, it will take a leading
current which compensates for the lagging
current drawn by main motor M and hence
improves the power factor of the system.

Injecting an e.m.f. in the Rotor Circuit – Scherbius System
The slip energy is not converted into d.c.
and then fed to a d.c. motor, rather it is fed
directly to a special 3-phase (or 6-phase)
a.c. commutator motor-called a, Scherbius
machine.
The polyphase winding of machine C is
supplied with the low-frequency output of
machine M through a regulating
transformer RT.
The commutator motor C is a variable-
speed motor and its speed (and hence that
of M) is controlled by either varying the
tappings on RT or by adjusting the position
of brushes on C.

63
• Same basic construction as squirrel-cage induction motors
• Drive at a speed greater than the synchronous speed
• Not started as a motor
• Operated by wind turbines, steam turbines, etc.
Induction Generator

64
Motor – to – Generator Transition

65
1 - Motor shaft coupled to a steam turbine
Initially, the turbine
valve is closed
2 - Motor started at full
voltage by closing the
breaker
3 –Motor drives the turbine at less than synchronous speed
Typical setup for induction-generator operation

66
Operation as an Induction-Generator continued
Gradually open the turbine valve, causing
a buildup of turbine torque, adding to the
motor torque, resulting in an increase in
rotor speed.

67
When the speed approaches synchronous speed, the slip = 0, Rs/s becomes infinite, rotor current Ir =
0, and no motor torque is developed. (The motor is neither a motor or a generator – it is “floating” on
the bus. The only stator current is the exciting current to supply the rotating magnetic field and the iron
losses.

68
The speed of the rotating flux is independent of the rotor speed – only a function of the
number of poles and the frequency of the applied voltage. Increasing the rotor speed above
the synchronous speed causes the slip [(ns – nr)/ns] to become negative! The gap power, Pgap
= Prcl/s becomes negative, now supplying power to the system!

Applications of Induction Generator
• The use of induction generator started in the early twentieth
century. In 1960’s and 1970’s its usage has become very less but
later it usage started again.
• They are used with the alternative energy sources, such as
windmills (WEGS), Hydro Electric Power Plants, Diesel Generators
(DGs).
• They are also used to supply additional power to a load in a remote
area that is being supplied by a weak transmission line.
• Energy recovery systems in the industrial processes. Externally
excited generators are widely used for regenerative breaking of
hoists driven by the three phase induction motors.

70

71

72

73

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