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Daniel Lang, P.Eng.
GP Technologies Ltd.
Induction Motor Circle Diagrams
The Equivalent Circuit Graphical Solution as a Function of Slip
Daniel Lang, P.Eng.
GP Technologies Ltd.
200 9016-51 Ave Edmonton, AB
dlang@gptechnologies.com
Abstract
The motor Circle Diagram (also known as a
Heyland, Behrend, or Ossanna diagram) is
simply a graphical representation of the locus of
induction motor stator current from zero speed
to synchronous speed. The concept of the Circle
Diagram was conceived in the late 19th
century,
but is today no longer commonly utilized by
engineers. With the advent of modern
computational techniques utilizing the motor
exact equivalent circuit, stator currents and
torques can be calculated faster and with more
accuracy than by graphical means. However, the
Circle Diagram is invaluable in providing an
intuitive understanding of motor characteristics
throughout the entire range of operation.
Introduction
The Circle Diagram is a graphical solution of the
motor equivalent circuit which traces the locus
of stator current, 𝐼̅1, at all values of slip,
0 ≤ 𝑠 ≤ 1, in the complex plane. The Circle
Diagram however is not a phasor diagram, as a
phasor diagram is the solution of an equivalent
circuit for a single operating condition. As a
motor will continuously change its operating
condition as it accelerates from zero speed to
full load speed, the equivalent circuit changes as
well.
Figure 1
From this single figure (Figure 1), the Circle
Diagram allows the engineer to graphically
visualize all motor parameters, such as real
power, reactive power, torque, current,
efficiency, power factor, and losses, for all
operating points.
Circle Diagram Required Information
As shown in Figure 2, the Circle Diagram is a
semi-circle drawn in the complex plane, located
in the first quadrant with the real and imaginary
components plotted on the vertical and
horizontal axes, respectively. As the diagram is
semicircular, two points are necessary to trace
the graph. The information required to construct
the diagram is acquired from the no load test and
the locked rotor test. The full load test can be
used in place of the locked rotor, although
considering the full load current is
approximately six times less in magnitude, it
may not yield as accurate a result.
Figure 2
Drafting of the Circle Diagram can be done
manually or by a computer, although manual
construction can provide a more intuitive
understanding.
Daniel Lang, P.Eng.
GP Technologies Ltd.
No-Load Test
At no load and rated voltage, motor losses are
comprised of stator copper losses, stator core
losses, and rotational losses. For the purposes of
the Circle Diagram, the test yields the
magnetizing branch current vector, 𝐼̅ 𝑚, as shown
in Figure 3, necessary to plot the first point, A.
Figure 3
As the motor has no load, the rotor will rotate
nearly at synchronous speed, with the non-zero
slip only being required to compensate for
rotational losses.
Figure 4
As shown in Figure 4, point A is actually the
point of minimum practical slip, where A’ is the
extrapolated current vector representing zero slip
(which can be achieved by driving the motor by
an auxiliary synchronous machine).
A’A’’ = Stator Core and Copper Losses
Figure 5
Locked-Rotor Test
During the locked rotor condition, current flow
is limited predominantly by the machine leakage
inductance, which is much smaller than the
magnetizing reactance. Due to this, locked rotor
current can typically reach 6 or more times the
normal operating current. For the purposes of
the circle diagram, the locked rotor test yields
the leakage branch current vector, 𝐼̅2, as shown
in Figure 6, necessary to plot the second point,
B.
Figure 6
Power input during the locked rotor condition is
comprised mainly of (above 99%) stator and
rotor 𝐼2
𝑅 losses.
Figure 7
Daniel Lang, P.Eng.
GP Technologies Ltd.
Constructing the Circle Diagram
From these two points, A and B, the circle can
now be drawn. As the stator core and copper
losses are constant throughout the range of
motor slip, the bottom of the circle is a
horizontal line parallel to the imaginary axis.
From point A, a circle is drawn through π/2
radians, which satisfies the equation of the rotor
referred stator current (1):
𝐼2 =
𝑉1
√(𝑅1+𝑅′
2 𝑠⁄ )2+(𝑋1+𝑋′
2)2
(1)
The impedance angle of the equivalent circuit
can be described by (3):
𝜃 = sin−1 (𝑋1+𝑋′
2)
√(𝑅1+𝑅′
2 𝑠⁄ )2+(𝑋1+𝑋′
2)2
(2)
Combining equations (1) and (2) yields the
equation of the circle diagram (3), which when
added vectorially to the no load current, traces
the locus of stator current as a function of slip.
𝐼2 =
𝑉1
(𝑋1+𝑋′
2)
sin 𝜃 (3)
The diameter of the circle is a function of the
total leakage reactance.
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 ≈
𝑉1
(𝑋1+𝑋2
′)
(4)
Note that the images in this paper have vastly
accentuated angles and magnitudes in order to
properly illustrate the components of the
diagram. In practice, many of the losses shown
would be very difficult to see and draw without
the aid of a computer.
Geometric Observations from the Circle
Diagram
The circle diagram is located in the first
quadrant only for convenience, but could be
projected in any direction. As the real axis is
vectorially in the same direction as the applied
voltage, any line segment drawn vertically
represents total real power (kW), while any line
segment drawn horizontally represents total
reactive power (kvar).
Line Segments at Locked Rotor (B)
Consider an induction motor at rated voltage
with a stationary rotor. The locked rotor current
vector is drawn to point B, as shown in Figure 7,
represents the mechanical power developed.
This line segment is thus referred to as the
mechanical power developed line. Since the
rotor is stationary, there is no mechanical power
output, and thus any input power must go
towards real losses. The line segment BB”’ must
then be the total stator and rotor losses.
BB’’’ = Stator and Rotor Losses
Figure 8
The line segment BB” can be split between the
rotor and stator losses by the following
equations:
𝑩𝑩′ ≅
𝐼 𝐿𝑅
2
𝑅1
𝑉1
(5)
BB’ = Rotor Copper Losses
𝑩′𝑩′′ ≅
𝐼 𝐿𝑅
2
𝑅′
2
𝑉1
(6)
B’B’’ = Stator Copper Losses
Equations 5 and 6 draw an important conclusion,
that being the ratio of the line segments BB’ and
B’B” is the ratio of rotor to stator resistances.
𝑩𝑩′
𝑩′ 𝑩′′ ≅
𝑅′
2
𝑅1
(7)
Daniel Lang, P.Eng.
GP Technologies Ltd.
As the amount of starting torque is directly a
function of the amount of rotor copper losses at
locked rotor, the length of the BB’ segment may
also represent the amount of starting torque at
rated voltage. Since the amount of torque
developed is dictated by the amount of real
power transferred across the air gap, a line
drawn from A’ to B’ is referred to as the torque
line or air gap line.
Maximum Developed Torque
As torque is a function of real power, and real
power is represented by vertical line segments,
the maximum developed torque can be found at
the point along the circle arc with the largest
real line segment to the torque line. This can be
found graphically by drawing a line from the
center of the circle base to the circle arc that is
normal to the torque line, as shown in Figure 9.
Figure 9
Note that the maximum torque obtained by
graphical means is not readily obtained
experimentally.
Maximum Developed Power
Referring back to the No-Load test, the
difference in current from point A to point A’
represents the motor rotational losses. Adding
the rotational losses to the maximum developed
power line yields total output. Thus, the output
line is a line segment drawn from point A
parallel to the maximum developed power line.
The maximum output power is found at the
point along the circle arc with the largest real
line segment to the output line. This can be
found graphically by drawing a line from the
center of the circle base to the circle arc that is
normal to the output line, as shown in Figure 10.
Figure 10
Similarly to maximum torque, the maximum
output power obtained graphically is also not
readily obtained experimentally.
Line Segments at the Operating Point (C)
The operating point is the steady state condition
where motor output torque is equal to the rated
load required torque. Once all the lines have
been plotted, a graphical solution of the motor
operating parameters can be found, as shown in
Figure 11.
Figure 11
CC’ = Total Input Power
CCO = Total Output Power
𝑪𝑪 𝑶
𝑪𝑪′ = Efficiency
𝑪𝑪′
𝑪𝑶
= Power Factor
𝑪 𝑷 𝑪 𝑻
𝑪𝑪 𝑻
= Slip (per-unit)
Note that slip is equivalent to the rotor copper
losses divided by air gap power.
Daniel Lang, P.Eng.
GP Technologies Ltd.
Assumptions and Error
The motor circle diagram constructed and
analyzed in this paper uses a perfect circle. In
practice however, this circle does not have a
constant radius or base.
Motor locked rotor tests are done similarly to
transformer percent impedance tests. That is, the
equivalent leakage reactance is found by
applying a voltage level such that rated stator
current flows, and then is extrapolated to the
rated voltage. When locked rotor current
actually flows in the windings, iron core
saturation will occur, reducing the leakage
reactances. This will distort the circumference of
the circle in the regions of high current flow.
Constructing the circle diagram makes use of the
induction motor approximate equivalent circuit
rather than the exact equivalent circuit. As the
magnetizing branch is moved ahead of the stator
impedance branch, this brings about error in the
stator core and copper losses from the no-load
test. As a result, the circle base is not actually
parallel to the imaginary axis.
References
1. Ketnere et. al., Simulation of Induction
Motor Characteristics Using a Circle
Diagram, 51st
Annual International
Scientific Conference of Riga Technical
University, October 2010.
2. Willem Van Den Burg, F., An Electrical
Transmission Shaft and Its Stability
Founded on a General Theory of Electrical
Machines, PhD Thesis, June 1961.
3. Cochran, P., Polyphase Induction Motors:
Analysis, Design, and Application, CRC
Press, May 1989.
4. Del Toro, V., Electric Machines and Power
Systems, Prentice-Hall, Inc., 1985.
About the Author
Daniel Lang holds a
Bachelor of Science in
Electrical Engineering
from the University of
Alberta (2010). He has
worked as an engineering
consultant in all areas of
power systems design and
analysis, from small
commercial buildings to
large scale heavy
industrial installations.
His work has included power system analyses and
design with an emphasis on protective relaying,
balanced and unbalanced fault analysis, steady-state /
loadflow and transient analysis, arc flash analysis, and
protection logic.
Daniel is a Professional Engineer registered in the
Province of Alberta, and is a member of the IEEE.

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Induction Motor Circle Diagrams_GP Technologies White Paper_Rev0_19 October 2016

  • 1. Daniel Lang, P.Eng. GP Technologies Ltd. Induction Motor Circle Diagrams The Equivalent Circuit Graphical Solution as a Function of Slip Daniel Lang, P.Eng. GP Technologies Ltd. 200 9016-51 Ave Edmonton, AB dlang@gptechnologies.com Abstract The motor Circle Diagram (also known as a Heyland, Behrend, or Ossanna diagram) is simply a graphical representation of the locus of induction motor stator current from zero speed to synchronous speed. The concept of the Circle Diagram was conceived in the late 19th century, but is today no longer commonly utilized by engineers. With the advent of modern computational techniques utilizing the motor exact equivalent circuit, stator currents and torques can be calculated faster and with more accuracy than by graphical means. However, the Circle Diagram is invaluable in providing an intuitive understanding of motor characteristics throughout the entire range of operation. Introduction The Circle Diagram is a graphical solution of the motor equivalent circuit which traces the locus of stator current, 𝐼̅1, at all values of slip, 0 ≤ 𝑠 ≤ 1, in the complex plane. The Circle Diagram however is not a phasor diagram, as a phasor diagram is the solution of an equivalent circuit for a single operating condition. As a motor will continuously change its operating condition as it accelerates from zero speed to full load speed, the equivalent circuit changes as well. Figure 1 From this single figure (Figure 1), the Circle Diagram allows the engineer to graphically visualize all motor parameters, such as real power, reactive power, torque, current, efficiency, power factor, and losses, for all operating points. Circle Diagram Required Information As shown in Figure 2, the Circle Diagram is a semi-circle drawn in the complex plane, located in the first quadrant with the real and imaginary components plotted on the vertical and horizontal axes, respectively. As the diagram is semicircular, two points are necessary to trace the graph. The information required to construct the diagram is acquired from the no load test and the locked rotor test. The full load test can be used in place of the locked rotor, although considering the full load current is approximately six times less in magnitude, it may not yield as accurate a result. Figure 2 Drafting of the Circle Diagram can be done manually or by a computer, although manual construction can provide a more intuitive understanding.
  • 2. Daniel Lang, P.Eng. GP Technologies Ltd. No-Load Test At no load and rated voltage, motor losses are comprised of stator copper losses, stator core losses, and rotational losses. For the purposes of the Circle Diagram, the test yields the magnetizing branch current vector, 𝐼̅ 𝑚, as shown in Figure 3, necessary to plot the first point, A. Figure 3 As the motor has no load, the rotor will rotate nearly at synchronous speed, with the non-zero slip only being required to compensate for rotational losses. Figure 4 As shown in Figure 4, point A is actually the point of minimum practical slip, where A’ is the extrapolated current vector representing zero slip (which can be achieved by driving the motor by an auxiliary synchronous machine). A’A’’ = Stator Core and Copper Losses Figure 5 Locked-Rotor Test During the locked rotor condition, current flow is limited predominantly by the machine leakage inductance, which is much smaller than the magnetizing reactance. Due to this, locked rotor current can typically reach 6 or more times the normal operating current. For the purposes of the circle diagram, the locked rotor test yields the leakage branch current vector, 𝐼̅2, as shown in Figure 6, necessary to plot the second point, B. Figure 6 Power input during the locked rotor condition is comprised mainly of (above 99%) stator and rotor 𝐼2 𝑅 losses. Figure 7
  • 3. Daniel Lang, P.Eng. GP Technologies Ltd. Constructing the Circle Diagram From these two points, A and B, the circle can now be drawn. As the stator core and copper losses are constant throughout the range of motor slip, the bottom of the circle is a horizontal line parallel to the imaginary axis. From point A, a circle is drawn through π/2 radians, which satisfies the equation of the rotor referred stator current (1): 𝐼2 = 𝑉1 √(𝑅1+𝑅′ 2 𝑠⁄ )2+(𝑋1+𝑋′ 2)2 (1) The impedance angle of the equivalent circuit can be described by (3): 𝜃 = sin−1 (𝑋1+𝑋′ 2) √(𝑅1+𝑅′ 2 𝑠⁄ )2+(𝑋1+𝑋′ 2)2 (2) Combining equations (1) and (2) yields the equation of the circle diagram (3), which when added vectorially to the no load current, traces the locus of stator current as a function of slip. 𝐼2 = 𝑉1 (𝑋1+𝑋′ 2) sin 𝜃 (3) The diameter of the circle is a function of the total leakage reactance. 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 ≈ 𝑉1 (𝑋1+𝑋2 ′) (4) Note that the images in this paper have vastly accentuated angles and magnitudes in order to properly illustrate the components of the diagram. In practice, many of the losses shown would be very difficult to see and draw without the aid of a computer. Geometric Observations from the Circle Diagram The circle diagram is located in the first quadrant only for convenience, but could be projected in any direction. As the real axis is vectorially in the same direction as the applied voltage, any line segment drawn vertically represents total real power (kW), while any line segment drawn horizontally represents total reactive power (kvar). Line Segments at Locked Rotor (B) Consider an induction motor at rated voltage with a stationary rotor. The locked rotor current vector is drawn to point B, as shown in Figure 7, represents the mechanical power developed. This line segment is thus referred to as the mechanical power developed line. Since the rotor is stationary, there is no mechanical power output, and thus any input power must go towards real losses. The line segment BB”’ must then be the total stator and rotor losses. BB’’’ = Stator and Rotor Losses Figure 8 The line segment BB” can be split between the rotor and stator losses by the following equations: 𝑩𝑩′ ≅ 𝐼 𝐿𝑅 2 𝑅1 𝑉1 (5) BB’ = Rotor Copper Losses 𝑩′𝑩′′ ≅ 𝐼 𝐿𝑅 2 𝑅′ 2 𝑉1 (6) B’B’’ = Stator Copper Losses Equations 5 and 6 draw an important conclusion, that being the ratio of the line segments BB’ and B’B” is the ratio of rotor to stator resistances. 𝑩𝑩′ 𝑩′ 𝑩′′ ≅ 𝑅′ 2 𝑅1 (7)
  • 4. Daniel Lang, P.Eng. GP Technologies Ltd. As the amount of starting torque is directly a function of the amount of rotor copper losses at locked rotor, the length of the BB’ segment may also represent the amount of starting torque at rated voltage. Since the amount of torque developed is dictated by the amount of real power transferred across the air gap, a line drawn from A’ to B’ is referred to as the torque line or air gap line. Maximum Developed Torque As torque is a function of real power, and real power is represented by vertical line segments, the maximum developed torque can be found at the point along the circle arc with the largest real line segment to the torque line. This can be found graphically by drawing a line from the center of the circle base to the circle arc that is normal to the torque line, as shown in Figure 9. Figure 9 Note that the maximum torque obtained by graphical means is not readily obtained experimentally. Maximum Developed Power Referring back to the No-Load test, the difference in current from point A to point A’ represents the motor rotational losses. Adding the rotational losses to the maximum developed power line yields total output. Thus, the output line is a line segment drawn from point A parallel to the maximum developed power line. The maximum output power is found at the point along the circle arc with the largest real line segment to the output line. This can be found graphically by drawing a line from the center of the circle base to the circle arc that is normal to the output line, as shown in Figure 10. Figure 10 Similarly to maximum torque, the maximum output power obtained graphically is also not readily obtained experimentally. Line Segments at the Operating Point (C) The operating point is the steady state condition where motor output torque is equal to the rated load required torque. Once all the lines have been plotted, a graphical solution of the motor operating parameters can be found, as shown in Figure 11. Figure 11 CC’ = Total Input Power CCO = Total Output Power 𝑪𝑪 𝑶 𝑪𝑪′ = Efficiency 𝑪𝑪′ 𝑪𝑶 = Power Factor 𝑪 𝑷 𝑪 𝑻 𝑪𝑪 𝑻 = Slip (per-unit) Note that slip is equivalent to the rotor copper losses divided by air gap power.
  • 5. Daniel Lang, P.Eng. GP Technologies Ltd. Assumptions and Error The motor circle diagram constructed and analyzed in this paper uses a perfect circle. In practice however, this circle does not have a constant radius or base. Motor locked rotor tests are done similarly to transformer percent impedance tests. That is, the equivalent leakage reactance is found by applying a voltage level such that rated stator current flows, and then is extrapolated to the rated voltage. When locked rotor current actually flows in the windings, iron core saturation will occur, reducing the leakage reactances. This will distort the circumference of the circle in the regions of high current flow. Constructing the circle diagram makes use of the induction motor approximate equivalent circuit rather than the exact equivalent circuit. As the magnetizing branch is moved ahead of the stator impedance branch, this brings about error in the stator core and copper losses from the no-load test. As a result, the circle base is not actually parallel to the imaginary axis. References 1. Ketnere et. al., Simulation of Induction Motor Characteristics Using a Circle Diagram, 51st Annual International Scientific Conference of Riga Technical University, October 2010. 2. Willem Van Den Burg, F., An Electrical Transmission Shaft and Its Stability Founded on a General Theory of Electrical Machines, PhD Thesis, June 1961. 3. Cochran, P., Polyphase Induction Motors: Analysis, Design, and Application, CRC Press, May 1989. 4. Del Toro, V., Electric Machines and Power Systems, Prentice-Hall, Inc., 1985. About the Author Daniel Lang holds a Bachelor of Science in Electrical Engineering from the University of Alberta (2010). He has worked as an engineering consultant in all areas of power systems design and analysis, from small commercial buildings to large scale heavy industrial installations. His work has included power system analyses and design with an emphasis on protective relaying, balanced and unbalanced fault analysis, steady-state / loadflow and transient analysis, arc flash analysis, and protection logic. Daniel is a Professional Engineer registered in the Province of Alberta, and is a member of the IEEE.