Induction Motor Modeling _steady State

ECE 8830 - Electric Drives

Topic 3: Induction Motor Modeling Steady State

Spring 2004

Introduction Induction machines are the most widely used of all electric motors. They offer the following attractive features: 

     

Generally easy to build and cheaper than corresponding dc or synchronous motors Rugged and require little maintenance Offer reasonable asynchronous performance A manageable torque-speed curve Stable operation under load Generally satisfactory efficiency Range in size from few Watts to several MW

Introduction (cont’d) Some disadvantages of induction motors are:  



Speeds not as easily controlled as dc motors Draw large starting currents, typically 6-8 x their full load values Operate with a poor lagging power factor when lightly loaded

Introduction (cont’d) New designs for high performance induction machines, such as in high speed motors for gas compressors, will be required to have new characteristics from existing machines, it is important to have a good fundamental understanding of these types of machines. Goal: To develop a “simple” model for the induction machine that is useful for control and simulation.

Structure of an Induction Machine Two types of induction machine: Wound rotor or squirrel cage rotor

Rotating Magnetic Field and Slip We previously showed that a balanced set of three-phase currents flowing in a set of symmetrically placed, three-phase stator windings produces a rotating mmf given by:

34N F (θ , t ) = I m cos(θ ae − ω et ) 2π P e a

[eq. (6.1) Ong, eq. (2.9) Bose]

where θae is the electrical angle measured from the a-phase axis and ωe is the angular speed of the stator mmf in electrical

Rotating Magnetic Field and Slip (cont’d)

Rotating Magnetic Field and Slip (cont’d) In mechanical radians/sec. the synchronous speed is related to the electrical speed by:

ωsm

2 = ωe P

If the rotor is rotating at an angular speed ωrm the slip speed is simply equal to ωsm - ωrm. The “slip”,s, is the normalized slip speed and is given by: ω sm − ω rm ω e − ω r s= = ω sm ωe

Torque Production The torque produced by an induction motor may be derived and expressed by the following equation: (see ref. [1] in Bose)

P Te = π   lrB p Fp sin δ 2

where P= # of poles l = axial length of motor r = radius of motor Bp= peak air-gap flux density Fp=π peak value of rotor mmf and δ = 2 + θ r

Per-Phase Equivalent Circuit Model A per-phase transformer-like equivalent circuit is shown below:

Per-Phase Equivalent Circuit Model (cont’d) Synchronously rotating air gap flux wave generates a counter emf Vm. This in turn is converted to a slip voltage in the rotor phase, Vr’ = nsVm, where n=rotor:stator turns ratio, and s=normalized slip. Stator terminal voltage, Vs = Vm + VRs +VLls where VRs=voltage drop across stator resistance (Rs) and VLls=voltage drop across stator leakage inductance (Lls).

Per-Phase Equivalent Circuit Model (cont’d) Excitation current, I0 = Ic + Im where Ic is core loss current (=Vm/Rm)

and Im is magnetizing current (=Vm/ ω e Lm)

Rotor-induced voltage, Vr’ = VRr’ + VLl’ where VRr’ = voltage drop across rotor resistance and VLl’ = voltage drop across rotor leakage inductance The induced voltage in the rotor leads to a rotor

current Ir’ at slip frequency ωsl.

Per-Phase Equivalent Circuit Model (cont’d) The stator current, IS = I0 + Ir where Ir is the rotor-reflected current induced in the stator. I0

Per-Phase Equivalent Circuit Model (cont’d) 2 n sVm Vm ' I r = nI r = ' = ' Rr + jω sl Llr  Rr    + jω e Llr  s 

Rr' Rr = 2 n L'lr Llr = 2 n

Per-Phase Equivalent Circuit Model (cont’d)

Per-Phase Equivalent Circuit Model (cont’d) Torque expression can be written as:

3 P ˆ Te =  ψˆ m I r sin δ 2 2  where ψˆ m = peak value of air gap flux linkage/pole and Iˆr = peak value of rotor current

Per-Phase Equivalent Circuit Model: Power Expressions Input Power: Pin = 3Vs I s cos φ Stator copper loss:

where cosφ is input PF

Pls = 3I s2 Rs Plr = 3I Rr 2 r

Rotor copper loss: Plc = 3Vm2 / Rm Core loss: Pg = 3I r2 Rr / s Power across air gap: Po = Pg − Plr = 3I r2 Rr (1 − s / s ) Output power: Psh = Po − PFw Shaft Power: where P is friction and

Per-Phase Equivalent Circuit Model: Torque Expression The torque can be expressed as:

Po 3 2 1− s  P  2 Rr Te = = I r Rr = 3  Ir ωm ωm s  2  sω e 2 2 where ω m =  P  ω r =  P  (1 − s)ω e is the rotor    

mechanical speed (radians/sec.)

Per-Phase Equivalent Circuit Model: Torque Expression (cont’d) Using a little algebra (see Bose) it can be shown that the torque may be further expressed as:

P Te = 3   Lm I m I a 2

where I a = I r sin δ . This torque expression is similar to that for a dc motor, where Im = magnetizing component of stator current and Ia = armature component of stator current.

Simplified Per-Phase Equivalent Circuit A simplified circuit dropping Rm and shifting Lm to the input is applicable to integral horsepower machines.

Performance of this equivalent circuit is typically within 5%.

Simplified Per-Phase Equivalent Circuit Model (cont’d) The current Ir in this circuit is given by: Ir =

Vs ( Rs + Rr / s ) 2 + ω e2 ( Lls + Llr ) 2

The torque of the motor using this circuit is given by: Vs2  P  Rr Te = 3   2 2 2 2 s ω ( R + R / s ) + ω ( L + L )   e s r e ls lr

Example of Calculating Efficiency of an Induction Motor

Example 5.1 Krishnan

Flowchart for Computing Steady State Performance of Induction Motor

Ref: R. Krishnan, “Electric Motor Drives”

Torque-Speed Curve of Induction Motor The torque-speed curve as a function of slip can be calculated from the equation two slides back.

Torque-Speed Curve of Induction Motor (cont’d) Three regions in torque-speed curve: 1) Plugging (braking) region (1<s<2) Rotor rotates opposite to direction of air gap flux. Can happen, for example, if stator supply phase sequence reversed while rotor is moving. 2) Motoring region (0<s<1) Te=0 at s=0. As s increases (speed decreases), Te increases until max. torque (breakdown torque) is reached. Beyond this point, Te decreases with

Torque-Speed Curve of Induction Motor (cont’d) 3) Regenerating Region (s<0) Here the induction machine acts as a generator. Rotor moves faster than air gap flux resulting in negative slip.

Torque-Speed Curve of Induction Motor (cont’d)

Ref: R. Krishnan, “Electric Motor Drives”

Performance Characteristics of Induction Motor

Ref: R. Krishnan, “Electric Motor Drives”

Starting Torque of Induction Motor The starting torque of an induction motor is given by substituting for s=1 and is given by:

Vs2  P  Rr Te = 3   2 2 2 2 ω ( R + R ) + ω ( L + L )   e s r e ls lr

Starting Torque of Induction Motor (cont’d) This torque can be enhanced for line start motors (ones started directly with full line voltage) by increasing the rotor resistance. This can be achieved by connecting external resistors in the case of slip ring rotors. However, with squirrel cage rotors where the rotor is shorted, deep bar or double-cage rotors can be used to increase starting torque.

Characterizing Induction Motors One way to characterize an induction motor is with the No-load/blocked rotor tests which yield the per-phase equivalent circuit model shown earlier (see figure below). ias

vas

iar

=M

Characterizing Induction Motors (cont’d) We can characterize an induction motor with the variables Rs, Lls, M, Llr, Rr determined through lab tests using balanced 3Φ excitation. This circuit described the impedance perceived per phase on a line-neutral connected machine. Everything in the dashed box is a rotor quantity that has been “referred” to the stator by the ideal transformer in the machine model. From now on, assume that Llr, Rr and iar are referred to the stator.

Characterizing Induction Motors (cont’d) 

No-Load Test (s=0) Equivalent circuit:

Ref: R. Krishnan, “Electric Motor Drives”

Characterizing Induction Motors (cont’d) 

No-Load Test (s=0) yields: In sinusoidal steady state, ignoring resistances: vas = ias X s + ibs X ab + ics X ab Las

But -ias = (ibs+ics)

3   ∴ vas = ias [ X s − X ab ] = iasω  Lls + Lsr  2   From transformer model: vas = iasω [ Lls + M ]

=>

3 M = Lsr 2

Characterizing Induction Motors (cont’d) 

Blocked rotor test (s=1) yields estimates of Lls and Llr. Equivalent circuit at standstill is shown below:

Ref: R. Krishnan, “Electric Motor Drives”

Characterizing Induction Motors (cont’d) 

Ohmmeter/Power loss tests give Rs and Rr.

So, with Llr, Rr and all ir’s understood as referred rotor quantities, the “stator-side” tests identify all the model parameters for the induction motor.

Example of Determining Induction Motor Model Parameters

Example 5.2 Krishnan

NEMA Classification of Induction Motors The National Electrical Manufacturers Association (NEMA) has classified induction motors based on their torque-slip characteristics. (see text for details)

Circuit Model of a Three-Phase Induction Machine (State-Space Approach)

Voltage Equations Stator Voltage Equations: d λas vas = ias rs + dt d λbs vbs = ibs rs + dt d λcs vcs = ics rs + dt

Voltage Equations (cont’d) Rotor Voltage Equations: d λar var = iar rr + dt d λbr vbr = ibr rr + dt d λcr vcr = icr rr + dt

Flux Linkage Equations

Model of Induction Motor To build up our simulation equation, we could just differentiate each expression for λ, e.g.

d λas d [first row of matrix] vas = = dt dt

But since Lsr depends on position, which will generally be a function of time, the trig. terms will lead to a mess! Park’s transform to the rescue!