 # EEEB443 Control & Drives Dynamic Model of Induction Machine By

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EEEB443 Control & Drives Dynamic Model of Induction Machine By
Dr. Ungku Anisa Ungku Amirulddin Department of Electrical Power Engineering College of Engineering Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives Dr. Ungku Anisa, July 2008

Outline Introduction Three-phase Dynamic Model
Space Phasors of Motor Variables Three-phase to Two-phase Transformation (Stationary) Two-phase (Stationary) Dynamic Model (in dsqs frame) Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame) Voltage equations Torque equation Commonly-used Induction Motor Models Stationary (Stator) Reference Frame Model Rotor Reference Frame Model Synchronously Rotating Reference Frame Model Equations in Flux Linkages References Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Introduction Per phase equivalent circuit model only useful for analysing IM performance in steady-state all transients neglected during load and frequency variations used in scalar control drives which do not require good transient response example: drive systems for fans, blowers, compressors Dynamic model used to observe dynamic (steady-state and transient) behaviour of IM since: Considers instantaneous effects of varying: Voltages and currents Stator frequency Torque disturbances machine is part of the feedback loop elements to control the dynamics of the drive system High performance drive control schemes are based on dynamic model of IM Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Introduction Dynamic model – complex due to magnetic coupling between stator phases and rotor phases Coupling coefficients vary with rotor position and rotor position vary with time Dynamic behavior of IM can be described by differential equations with time varying coefficients Complexity of dynamic model can be reduced by employing space vector equations Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
Magnetic axis of phase B Magnetic axis of phase C IM consists of three-phase windings spaced at 120 apart Model windings using simplified equivalent stator winding located on the magnetic axis of each phase. ics ibs a b b’ c’ c a’ Magnetic axis of phase A Simplified equivalent stator winding ias Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
Similar model is applied to represent the rotor ‘windings’ Rotor rotates at speed r Rotor phase ‘a’ winding displaced from stator phase ‘a’ winding by angle r stator, b stator, c stator, a rotor, b rotor, a rotor, c r Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
Voltage equation for each stator phase: Similarly, voltage equation for each rotor phase: These equations can be written in a compact form. Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
Stator voltage equation (compact form): Rotor voltage equation (compact form): where:  abcs = stator flux linkage (flux linking stator windings ) abcr = rotor flux linkage (flux that links rotor windings) (1) (2) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
The displacements between 3 phase stator (rotor) windings are non-quadrature (i.e. not 90) Magnetic coupling exists between the 3 stator (or rotor) phases, i.e. the flux linkage each stator (or rotor) phase is sum of: fluxes produced by the winding itself fluxes produced from the other two stator (or rotor) windings fluxes produced by all three rotor (or stator) windings Example: Flux linkage for stator phase ‘a’ is sum of: Fluxes produced by stator phase ‘a’ winding itself Fluxes produced by stator phase ‘b’ and stator phase ‘c’ Fluxes produced by rotor phase ‘a’, ‘b’ and ‘c’ Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
In general, the stator flux linkage vector: (3) Flux linking stator winding due to stator currents Flux linking stator winding due to rotor currents Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
In general, the rotor flux linkage vector: (4) Flux linking rotor winding due to rotor currents Flux linking rotor winding due to stator currents Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
Self inductances in (3) and (4) consists of magnetising inductance and leakage inductance: For stator: For rotor: Due to symmetry in windings, mutual inductances between stator phases in (3)(and rotor phases in (4)) can be written in terms of magnetising inductances: Stator leakage inductances Rotor leakage inductances Note: Subsrcipt ‘s’ is replaced with ‘r’ for rotor phase leakage inductances, currents and flux linkage Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
Mutual inductances between the stator and rotor windings depends on rotor position r: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase Dynamic Model
Equations (1) – (4): completely describe dynamic characteristics of 3-phase IM consists of 6 equations (3 for stator and 3 for rotor), i.e. large number of equations all equations are coupled to one another Magnetic coupling complicates dynamic model in 3-phase! Better to develop model based on space phasors: reduces number of equations eliminates magnetic coupling between phases Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Space Phasors of Motor Variables
If xa, xb, and xc are the 3-phase IM quantities, whereby: The space phasor in the 3-phase system is obtained from the vectorial sum of the 3-phase quantities, i.e.: (5) x is called the space phasor or complex space vector , where a = ej2/3 Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase to Two-phase Transformation (Stationary)
Any three-phase machine can be represented by an equivalent two-phase machine using Park’s transformation Two-phase equivalent Three-phase easier way to obtain dynamics of IM There is magnetic coupling between phases There is NO magnetic coupling between phases (due to 90 angle between phases) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase to Two-phase Transformation (Stationary)
Dynamic model of IM usually obtained using the two-phase equivalent machine r stator, qs rotor,  stator, ds rotating r Two-phase equivalent rotor,  stator, b rotating r rotor, b rotor, a r stator, a Three-phase rotor, c stator, c Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase to Two-phase Transformation (Stationary)
All 3-phase system quantities have to be transformed to 2-phase system quantities Equivalence between the two systems is based on the equality of mmf produced and current magnitudes, i.e.: MMF produced by 2-phase system = MMF of 3-phase system current magnitude of 2-phase system = current of 3-phase system The use of space phasors enables the transformations from 3-phase to 2-phase system. Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase to Two-phase Transformation (Stationary)
In the stationary 2-phase system, the space phasor is defined as : (6) The space phasor in the 2-phase system must equal that in the 3-phase system. Hence, by comparing (5) and (6): Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase to Two-phase Transformation (Stationary)
The abc  dsqs transformation is given by: (7) Under balanced conditions, the zero-sequence component adds to zero, i.e.: Zero-sequence components, which may or may not be present. Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase to Two-phase Transformation (Stationary)
Assuming balanced conditions, the abc  dsqs transformation: (8) The inverse transform (dsqs  abc transformation) is given by: (9) where: (10) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase to Two-phase Transformation (Stationary)
Transformation equations (8) and (9) apply to all 3-phase quantities of the IM (i.e. voltages, current and flux linkages) Transformation matrices (Tabc and Tabc-1) given by (10) causes the space phasor magnitude to be equal to peak value of the phase quantities, i.e.: This is one of many abc  dsqs and dsqs  abc transformation matrices in literature, eg.: space phasor magnitude to be equal to 1.5 times peak value of the phase quantities ( ) space phasor magnitude to be equal to rms value of the phase quantities ( ) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

SPACE VECTORS Space vector representation of the mmf distribution in an AC machine created by balanced positive-sequence three-phase sinusoidal currents. Each of the ABC (RGB) space vectors pulsates along its respective axis. The resultant vector (in black), of 1.5 magnitude, rotates at the excitation frequency. Source: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

SPACE VECTORS This animation shows the motion of space vectors for the case of a balanced three- phase sinusoidal signal: fA = cos(ωt), fB = cos(ωt-α), fC = cos(ωt+α) where α = 2π/3. The corresponding space vector is obtained from fR = (fA + γ fB + γ2 fC) = ejωt where γ = ejα. Source: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Example 1 - Space Phasor & 3-2 phase transformation
An induction motor has the following parameters: Parameter Symbol Value Rated power Prat 30 hp (22.4 kW) Stator connection Delta () No. of poles P 6 Rated stator phase voltage (rms) Vs,rat 230 V Rated stator phase current (rms) Is,rat 39.5 A Rated frequency frat 60 Hz Rated speed nrat 1168 rpm Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Example 1 - Space Phasor & 3-2 phase transformation (contd.)
Assuming that the motor is operating under rated conditions, with stator and rotor current phasors of: Calculate the values the following stator current values at time t = 0: ias, ibs and ics (3-phase stator phase currents) isds and isqs (2-phase stator phase currents) iar, ibr and icr (3-phase rotor phase currents) irds and irqs (2-phase rotor phase currents) Show that the magnitude of isds and isqs is equal to the peak stator phase current Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase (Stationary) Dynamic Model (in dsqs frame)
From the three-phase dynamic model (eq. 1 and 2): Applying the transformation given by (8): Note: dsqs – stator equivalent two-phase winding  - rotor equivalent two-phase winding r stator, qs rotor,  stator, ds rotating r rotor,  Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase (Stationary) Dynamic Model (in dsqs frame)
The  rotor winding rotates at a speed r Hence, need to transform the rotor quantities from the  to the stationary dsqs frame. r stator, qs rotor,  stator, ds rotating r rotor,  r r xr qs ds Bring all rotor and stator quantities to be on the same axis! Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase (Stationary) Dynamic Model -   dsqs frame transform
xr qs xr xr On the  frame: On the dsqs frame: The angle between the two frames is r xrds r r xr qs ds xrqs Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase (Stationary) Dynamic Model -   dsqs frame transform
xr qs xr xr ds Therefore: More elegantly : (full derivation   dsqs frame transform) (11) r r xr qs ds xrqs (12) xrds Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase (Stationary) Dynamic Model (in dsqs frame)
Two-phase dynamic model: To transform rotor quantities from the   dsqs frame, from equation (12): Substituting these into the rotor voltage equation above.. Expressed in rotating frame Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase (Stationary) Dynamic Model (in dsqs frame)
Hence: Therefore, the two-phase dynamic model in the stationary dsqs frame: Expressed in rotating frame Expressed in stationary frame (13) (14) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase (Stationary) Dynamic Model (in dsqs frame)
The flux linkages in (13) and (14) are given by: where and Note that equations (13)-(16) each consists of two equations. One from equating real quantities One from equating imaginary quantities Final dynamic equations in the stationary dsqs frame is given by substituting (15) and (16) into (13) and (14) and separating the real and imaginary equations. (15) (16) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase (Stationary) Dynamic Model (in dsqs frame)
Final dynamic equations in the stationary dsqs frame: Lm = mutual inductance Lr’ = rotor self inductances referred to stator Rr’ = rotor resistance referred to stator Ls = stator self inductance vrd, vrq, ird, irq are the rotor voltages and currents referred to stator S = derivative operator (17) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Example 2 – Dynamic Model of Induction Motor in dsqs frame
The induction motor from the Example 1 has the following additional parameters: Parameter Symbol Value Rated torque Te,rat 183 Nm Stator resistance Rs 0.294  Stator self inductance Ls H Referred rotor resistance Rr’ 0.156  Referred rotor self inductance Lr’ H Mutual inductance Lm 0.041 H Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Example 2 - Dynamic Model of Induction Motor in dsqs frame (contd.)
Using the values of stator and rotor currents obtained in the Example 1 , calculate the stator flux s and rotor flux r vectors at time t = 0. Given that Calculate the torque produced by the motor using: Stator flux s and stator current Is vector Rotor flux r and stator current Is vector Stator current  Is and rotor current Ir vector Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

SPACE VECTOR DECOMPOSITION
Space vectors under balanced sinusoidal conditions, appears as constant amplitude vectors rotating at the excitation frequency (2f). In the stationary dsqs (αβ in the diagram) frame: dsqs components are time varying sinusoidal signals at stator frequency (2f) Source: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

SPACE VECTOR DECOMPOSITION
If we want the Induction Motor to behave like a DC motor, the two-phase components must be constant values. This can be achieved by having a two-phase frame that rotates together with the space vector. A rotating dq frame! Source: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
The dq referance frame is rotated at an arbitrary speed g On the dsqs frame: On the dq frame: The angle between the two frames is g where: g g d q x qs ds xqs xds Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives Dr. Ungku Anisa, July 2008

Two-phase Dynamic Model in Arbitrary Rotating Frame (dsqs  dq frame transform)
Therefore: More elegantly : (full derivation dsqs  dq frame transform) g g d q x qs ds xqs g (18) xd xq xds (19) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Hence, space vectors in the stationary dsqs frame will have to be transformed into the rotating dq frame using: Equation (20) will have to be employed onto equations (13) –(16) to obtain the IM dynamic model in the rotating dq frame. g g d q x qs ds xqs (20) xds Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Subst. (20) into (13), stator voltage equation: Subst. (20) into (14), rotor voltage equation: Expressed in stationary frame Expressed in arbitrary rotating frame (21) Expressed in stationary frame Expressed in arbitrary rotating frame (22) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Subst. (20) into the flux linkages equations of (15) and (16): Stator flux linkage: Rotor flux linkage: Airgap flux linkage: where and (23) (24) (25) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Note that equations (21)-(25) each consists of two equations. One from equating real quantities One from equating imaginary quantities Final dynamic equations in the arbitrary rotating dq frame is given by substituting (23)-(24) into the (21)-(22) and separating the real and imaginary equations. Final dynamic equations in the arbitrary dq frame: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives (26)

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Torque equation in the arbitrary dq frame: Product of voltage and current conjugate space vectors: It can be shown that for ias + ibs + ics = 0, Input power to the IM: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Torque equation in the arbitrary dq frame: Input power to the IM: If and then: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Torque equation in the arbitrary dq frame: The IM equation given by (26) can be written as: Note that the matrices: [R] = consists of resistive elements [L] = consists of coefficients of the derivative operator S [G] = consists of coefficients of the electrical rotor speed r [F] = consists of coefficients of the reference frame speed g Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Torque equation in the arbitrary dq frame: Hence, the input power is given by: Power associated with g – upon expansion gives zero Power Losses in winding resistance Rate of change of stored magnetic energy Mechanical power Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Torque equation in the arbitrary dq frame: The mechanical power is most important: By observing equation (26), [G] consists of terms associated with r : Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Torque equation in the arbitrary dq frame: Therefore, mechanical power: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame)
Torque equation in the arbitrary dq frame: Since m = r / (P/2), hence the electromagnetic torque: (27) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Dynamic Model in Arbitrary Rotating Frame (in dq frame) - Summary
Final dynamic equations in the arbitrary dq frame: Torque equation in the arbitrary dq frame: (26) (27) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Commonly-used Induction Motor Models - Stationary (Stator) Reference Frame
Speed of reference frame: Dynamic model in the stationary reference frame: Torque equation in the stationary reference frame: (28) (29) (30) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Three-phase to Two-phase Transformation (Stationary)
Assuming balanced conditions, vsds, vsqs, isds, isqs can be obtained from: (31) The inverse transform is given by: (32) where: and (33) vrds, vrqs, irds, irqs obtained from transforming vabcr  vr  vrdqs using equations (31), (32), (33) and (11) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Commonly-used Induction Motor Models - Stationary (Stator) Reference Frame
This model is used when: stator variables are required to be actual (i.e. same as in the actual machine stator) rotor variables can be fictitious Allows elegant simulation of stator-controlled induction motor drives phase-controlled and inverter-controlled IM drives (i.e. this IM model is used for variable voltage control at constant frequency) Input variables are well defined and can be used to find vsd and vsq easily Reduce computations leading to real-time control applications Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Commonly-used Induction Motor Models – Rotor Reference Frame
Speed of reference frame: Dynamic model in the rotor reference frame: Torque equation in the rotor reference frame: (34) (35) (36) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Transformation (from stationary stator dsqs frame to rotor drqr frame)
Assuming balanced conditions, vsdr, vsqr, isdr, isqr can be obtained from vsds, vsqs, isds, isqs using: (37) The inverse transform is given by: (38) Note: vsds, vsqs, isds, isqs can be obtained from vabcs and iabcs using (31), (32) and (33) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Commonly-used Induction Motor Models - Rotor Reference Frame
vrdr, vrqr, irdr, irqr obtained from transforming vabcr  vrdqr using equations (31), (32) and (33) This model is used when: Switching elements and power are controlled on the rotor side Example: for simulations of slip-energy recovery scheme Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Commonly-used Induction Motor Models – Synchronously Rotating Reference Frame
Speed of reference frame: Dynamic model in the synchronously rotating frame: Torque equation in the synchronously rotating frame: (39) (40) (41) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Two-phase Transformation (from stationary stator dsqs frame to synchronously rotating dq frame)
Assuming balanced conditions, vsd, vsq, isd, isq can be obtained from vsds, vsqs, isds, isqs using: (42) The inverse transform is given by: (43) Note: vsds, vsqs, isds, isqs can be obtained from vabcs and iabcs using (31), (32) and (33) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

Commonly-used Induction Motor Models - Synchronously Rotating Reference Frame
vrd, vrq, ird, irq obtained from transforming vabcr  vr  vrdqs  vrdq using equations (31), (32), (33), (11) and (42) Synchronous reference frame: transforms sinusoidal inputs into dc signals provides decoupled torque and flux channels Hence, IM control similar to separately excited DC motor achieved by employing vector control schemes Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

SPACE VECTOR DECOMPOSITION
Space vectors under balanced sinusoidal conditions, appears as constant amplitude vectors rotating at the excitation frequency (2f). In the stationary αβ (dsqs) frame: αβ (dsqs) components are time varying sinusoidal signals at stator frequency (2f) In the rotating synchronous dq frame: dq components are constant values depend on the orientation of the space vectors with respect to the dq axes. Source: Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

All dynamic equations presented are consists of 4 variables, i.e. vsdq, vrdq, isdq and irdq Note that in squirrel-cage IM: vrdq = 0 at all times If the equations are required to contain flux linkages (i.e. either sdq, rdq or odq), the dynamic model can be obtained by substituting irdq using the following equations respectively: (23) (24) (25) Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

References Krishnan, R., Electric Motor Drives: Modeling, Analysis and Control, Prentice-Hall, New Jersey, 2001. Bose, B. K., Modern Power Electronics and AC drives, Prentice-Hall, New Jersey, 2002. Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

  dsqs frame transform
xr qs xr xr ds xrqs xrds Back Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

dsqs  dq frame transform
g g d q x qs ds xqs g xd xq xds Back Dr. Ungku Anisa, July 2008 EEEB443 - Control & Drives

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