Fundamentals of Electrical Drivess - Chapter 7 pps

The IRTF generic module as given infigure 7.5a is shown with a stator to rotor coordinate flux conversion moduleand rotor to stator current conversion module.. In the example given here

INTRODUCTION TO ELECTRICAL MACHINES

This chapter considers the basic working principles of the so-called sical’ set of machines This set of machines represents the asynchronous (in-duction), synchronous, DC machines, and variable reluctance machines Thelatter will be discussed in the book ‘Advanced Electrical Drives’ currently un-der development by the authors of this book Of these classical machines, theasynchronous machine is most widely used in a large range of applications.Note that the term ‘machine’ is used here, which means that the unit is able tooperate as a motor (converting electrical power into mechanical power) or as agenerator (converting mechanical power into electrical power) The machinecan be fed via a power electronic converter or connected directly to an AC or

‘clas-DC supply

Central to this chapter is the development of an ‘ideal rotating transformer’,which is in fact a logical extension of the two-phase ITF module discussed inchapter 6 We will then look to the conditions required for producing constanttorque in an electrical machine This in turn will allow us to derive the principle

of operation for the three classical machine types A general model conceptwill be introduced at the end of this chapter which forms the backbone of themachine models discussed in this book

The fundamental building block for rotating machines used in this book is theIRTF module which is directly based on the work by A Veltman [Veltman, 1994]which is directly derived from the two-phase space vector ITF concept given

in figure 6.1 The new IRTF module, shown in figure 7.1, differs in two points.Firstly, the inner (secondary) part of the transformer is assumed to be able to

rotate freely with respect to the outer (primary) side The airgap between thetwo components of this model remains infinitely small Secondly, the number

of ‘effective’ turns on the primary and secondary winding are assumed to be

equal, i.e n1 = n2 = n Furthermore, the windings are (like the ITF) taken

to be sinusoidally distributed (see appendix A) This implies that the windingrepresentation as shown in figure 7.1 is only symbolical as it shows where themajority of conductors for each phase are located In the future the primaryand secondary will be referred to as the stator and rotor respectively A second

Figure 7.1 Two-phase IRTF

model

complex plane (in addition to the stator based complex plane) with axis
xy,

 xy is introduced in figure 7.1 which is tied to the rotor Note the use of the

superscript xy which indicates that a vector is represented in rotor coordinates.

For a stationary coordinate system, as used on the stator side, we sometimes

use the superscript αβ However, in most cases this superscript is omitted

to simplify the mathematical expressions Hence, no superscript implies astationary coordinate based vector

The angle between the stationary and rotating complex plane is given as θ

and this is in fact the machine shaft angle of rotation (relative to the stationary

part of the motor) If the angle of rotation θ is set to zero then the IRTF module

is reduced to the two-phase ITF concept (with n1 = n2) as given by figure 6.1.The symbolic representation of the IRTF module as given in figure 7.2 showssimilarity with the ITF module (see figure 6.2(a))

The IRTF is a three-port unit (stator circuit, rotor circuit and machine shaft)

In figure 7.2 a symbolic shaft (shown in ‘red’) is introduced which is physically

Figure 7.2 Symbolic IRTF

An illustration of the flux-linkage seen by the rotor and stator winding is given

in figure 7.3(a) The relationship between the stator and rotor oriented

Figure 7.3. Flux-linkage and current space vector diagrams

linkage space vectors is given as

be projected onto a rotating or stationary complex reference frame The

rela-tionship between for example the variables i α , i β and i x , i y , θ of equation (7.3)

can also be written as

) (

i x iy

)

(7.4a)

The symbolic diagram of figure 7.2 can also be shown in terms of its spacevector components as was done for the ITF case (see figure 6.3) The result,

given in figure 7.4, shows that the rotor side of the IRTF now rotates with the

rotor shaft given that it is physically attached to it The energy balance for the

Figure 7.4

Three-dimen-sional IRTF representation

IRTF module is found by making use of the power expressions (5.30), (5.33)

which are directly linked with the incremental energy dW = pdt The input (stator) power p infor the IRTF module is of the form

p in =
 u



Note that  u andi are taken to be time dependent complex numbers as indicated

by equation (5.8) on page 123 Equation (7.5) can with the aid of  u = ψ m

torque T e (Nm) and shaft speed ω m(rad/sec) If this product is positive themachine is said to operate as a motor The total (mechanical plus electrical)output power is now of the form

pout=
 u xy



i xy ∗

The incremental mechanical and electrical output energy linked with the rotor

side of the IRTF can with the aid of  u xy = ψ m xy

dt be written as

dW out=
d  ψ m xy

i xy ∗

The overall IRTF incremental energy balance can with the aid of equations (7.6)and (7.8) and use of the energy conservation law be written as

d  ψ m

i ∗−
d  ψ m xy

i xy ∗

The left hand side of equation (7.9) shows the electrical incremental energy

IRTF components In equation (7.9), the term d  ψ m xy may be developed furtherusing equation (7.2) and the differential ‘chain rule’, which leads to

equa-d  ψ m xy

i xy ∗

= d  ψ m i ∗ − j ψm i ∗ dθ (7.11)Substitution of equation (7.11) into equation (7.9) leads to the following ex-pression for the electromechanical torque on the rotor

Equation (7.12) can with the aid of expression
jab ∗

= a ∗ bbe ten as

rewrit-Te= ψ  ∗

Hence, the torque acting on the rotor is at its maximum value in case the two

vectors  ψ m ,i as shown in figure 7.3, are perpendicular with respect to each other.

Under these circumstances the torque is directly related to the product of therotor radius and the Lorentz force The latter is proportional to the magnitudes

of the flux and current vectors The generic diagram of the IRTF module thatcorresponds to the symbolic representation shown in figure 7.2 is based on theuse of equations (7.2), (7.3) and (7.13) The IRTF generic module as given infigure 7.5(a) is shown with a stator to rotor coordinate flux conversion moduleand rotor to stator current conversion module The two coordinate conversionmodules can also be reversed as shown in figure 7.5(b) The IRTF version used

is application dependent as will become apparent at a later stage The torquecomputation is not affected by the version used Nor for that matter is the torque

affected by the choice of coordinate system The rotor angle θ required for the

IRTF module must be derived from the mechanical equation-set of the machinewhich is of the form

T e − Tl = J dωm

(a) IRTF-flux (b) IRTF-current

Figure 7.5. Generic representations of IRTF module

with T l and J representing the load torque and inertia of the rotor/load

combi-nation respectively (as discussed in section 1.4.2) Finally, it is noted that theIRTF has a unity winding ratio, which implies that an inductance component(unlike a resistive component) may be moved from one side to the other withouthaving to change its value

The IRTF module forms the backbone to the electrical machine conceptspresented in this book Consequently, it is particularly important to fully un-derstand this concept In the example given here we will discuss how statorcurrents and torque can be produced in the event that stator windings are con-nected to a voltage source and the rotor windings to a current source as shown

in figure 7.6) In the discussion to come we will make use of figure 7.1 in astylized form for didactic reasons Furthermore, we will assume that we canhold the motor shaft at any desired position The voltage source shown in fig-

Figure 7.6 IRTF module

current source

Figure 7.7 Voltage

excita-tion and flux-linkage for the α

winding

ure 7.6 delivers a pulse to the α winding at t = t oas shown in figure 7.7 The

flux-linkage ψ mα versus time waveform which corresponds with the applied

pulse is also shown in figure 7.7 The β stator winding is short circuited with condition ψ mβ = 0 We will consider events after t = t o + T in which case

a flux distribution will be present in the IRTF where the majority of the flux isconcentrated along the
α,β axis as shown in figure 7.8(a) Furthermore, the

flux-linkage value will be equal to ψ mα = ˆψ m The corresponding space vector

representation will be of the form  ψ m= ˆψ m

We would like to realize a rotor excitation of the formi xy = jˆi, which implies

that the y rotor winding must carry a current i y = ˆi We have omitted for didactic

reasons the x winding from figure 7.8 because this winding is not in use (open

circuited) We will now examine the IRTF model and corresponding spacevector diagrams for the excitation conditions indicated above and three rotorpositions For each case (rotor postion) we will assume that the rotor is initiallyset to the required rotor position after which the stator and rotor excitation asdiscussed above is applied The aim is to provide some understanding withrespect to the currents which will occur on the stator side and the nature oftorque production based on first principles To assist us with this discussion two

‘contours’ namely x and y are introduced in figure 7.8 which are linked to the

rotating complex plane These contours are helpful in determining the currentswhich must appear on the stator side If we assume that such a contour represents

a flux tube then there would need to be a corresponding resultant MMF within thecontour in case the latter would contain some form of magnetic reluctance Themagnetic reluctance of the IRTF model is zero (infinite permeability materialand infinitely small airgap), hence the MMF ‘seen’ inside either contour mustalways be zero

(a) θ = 0, i α = 0, i β = ˆi, T e= ˆψ m ˆi

Rotor position θ = 0: if we consider the x contour in figure 7.8(a), then it

coincides with the flux distribution that exists in the model The MMF seen

by this contour is equal to ni α The excited rotor winding cannot contribute

(given both halves are in the contour) to this contour Hence, the current i α

must be zero By observing the y contour and the MMF ‘seen’, we note the presence of the imposed rotor current i y = ˆi The number of winding turns

on rotor and stator are equal, hence a stator current i β = ˆi (with the direction

shown) must appear in the short-circuited β coil to ensure that the zero MMF

condition with this contour is satisfied The space vector representation ofthe current and flux as given in figure 7.8(a), shows that they are π2 radian

apart Note (again) that there is only one set of space vectors and their

components may be projected onto either the rotor or stator complex plane.The torque according to equation (7.13) will under these circumstances (with

the current vector leading the flux vector) equal to T e = ˆψm ˆi From first

principles (see figure 1.9) we note that forces will be exerted on the rotorwinding in case the latter carries a current and is exposed to a magnetic field

In this case the flux and flux density distributions are at their highest level

along the α axis (see appendix A) The y winding carries a current in the

direction shown and this will cause a force on the individual conductors of

the y rotor winding and thus a corresponding torque in the anti-clockwise

(positive) direction

Rotor position θ = π4: if we consider the x contour in figure 7.8(b), then we note that a MMF due to the α winding would be less than ni α The reason

for this is that the contour also encloses part of α winding which gives a

negative MMF contribution to the total MMF The resultant MMF is found

by integrating equation (A.9) over the angle range π/4 → π, −π → −3π/4

and comparing the outcome of this integral with the integral over the range

0→ π This analysis will show that the MMF is reduced by a factor 1/ √2

(when compared to the previous case) hence the resultant MMF of the α winding is equal to ni α/ √

2 The same x contour also encloses part of

the β winding and its MMF contribution is equal to ni β / √

2 There are

no other contributions, hence the sum of these two MMF’s is of the form

niα/ √

2 + ni β / √

2 = 0 because the MMF in the contour must be zero

From this analysis it follows that, for the given angle, the currents must be

in opposition, i.e i α=−iβ If we now consider the MMF enclosed by the

y contour we note that the α will contribute a MMF component −niα / √

2

(now negative because the other side of this winding), while the β will contribute a component ni β / √

2 Furthermore, the y winding will add a

component nˆi The resultant MMF (the sum of these three contributions), gives together with the condition i α=−iβ , the required stator currents i α =

−ˆi/ √ 2, i β = ˆi/ √

2 respectively The space vector representation shown in

figure 7.8(b) confirms the presence of the two current components Notealso that the angle between the current and flux vectors is equal to3π4 radian.The torque according to equation (7.13) will under these circumstances be

equal to T e = ˆψ m ˆi/ √

2 The torque must be less (than the previous case)

because a number of the conductors of the y winding now ‘see’ a flux density

value which is in opposition to that seen by the majority of the conductors

Rotor position θ = π2: if we consider the x contour in figure 7.8(c), then

we can conclude that the current i βmust be zero (the other winding cannotcontribute, given both halves are in the contour) The reason for this is that

the MMF seen by this contour must be zero By observing the y contour and

the MMF ‘seen’ by this contour we note the presence of the rotor current

iy = ˆi The number of winding turns on rotor and stator are equal, hence a

stator current i α =−ˆi must appear to ensure that the zero MMF condition

with this contour is satisfied The space vector representation of the current

and flux as given in figure 7.8(c) shows that they are π radian apart The

torque according to equation (7.13) will under these circumstances be equal

to T e = 0 From first principles we note that half the conductors of the y

winding will experience a force in opposition to the other half, hence thenet torque will be zero

7.3 Conditions required to realize constant torque

The ability of a machine to produce a torque which has a non-zero averagecomponent is of fundamental importance In this section we will consider theconditions under which electrical machines are able to produce a constant (timeindependent) torque For this analysis it is sufficient to re-consider the modelaccording to figure 7.6 The rotor is again connected to a current source which

in this case is assumed to be of the form ˆie j(ω r t+ρ r), where ω rrepresents the

angular velocity of this vector relative to the rotor of the IRTF The ITF current

i xy is therefore of the form

The stator is connected to a three-phase sinusoidal voltage source such that

the voltage vector is taken to rotate at a constant angular velocity ω s The

corresponding flux vector  ψm can then be found using  u = ψ m

dt which will ingeneral terms result in a rotating flux vector of the form

The importance of the angle variable ρ mwill be discussed at a later stage The

Figure 7.9 Simplified

two-phase, machine model

induced voltage on the rotor side in figure 7.9 is in this case  e xy

m = ψ m xy

dt , whichwith the aid of equations (7.16), (7.2) and (7.17) may also be written as

 m xy = d  ψ

xy m

dt = e

j((ω s −ω m )t −ρ m)d ˆ ψm

dt + j (ω s − ωm )  ψ m xy (7.18)

in which the term d ˆ ψ m

dt is taken to be zero given that a quasi-steady-state tion is assumed, i.e we assume that the flux amplitude ˆψ mis constant On thebasis of this assumption equation (7.18) reduces to

The torque produced by this machine is found using equation (7.13) Furthermathematical handling of this torque equation and use of equations (7.16) (inrotor coordinates), (7.2) and (7.17) leads to

Te= ψmˆ ˆie j((ω r +ω m −ω s )t+ρ r +ρ m)

(7.20)

where it is emphasized that the angle variables ρ r , ρ mshown in equation (7.20)are not a function of time Equation (7.20) is of prime importance as it showsthat a time independent torque value is obtained in case the following speedcondition is met

it is instructive to look at the vector diagram given in figure 7.10 which sponds to machine configurations discussed in this section The diagram shows

corre-Figure 7.10. Space vector diagram for machine model

the stationary complex plane (
αβ, αβ) and the rotating (with constant rotor

speed ω m ) rotor coordinate frame which is displaced by an angle θ = ω mt+ρm

with respect to the former Also shown in the diagram is the rotating (with

angu-lar speed ω s ) flux vector  ψ m, which is caused by the presence of the three-phase

voltage supply connected to the stator Finally, the current vectori xyis shown

which rotates at an angular speed ω r relative to the rotating reference frame.

This means that this vector rotates at a speed ω r + ω m relative to the statorbased (stationary) reference frame

In the example shown, the current vector leads (we assume positive tion as anti-clockwise) the flux vector This combination of the two vectorscorresponds to a positive torque condition as will be shown below A timeindependent torque will be realized if the angle between the current and fluxvectors remains constant This is the case when condition (7.21) is met Fur-thermore, we also assume that the magnitude of the two vectors is constant Ifcondition (7.21) is met, the angle between the two vectors is equal to the sum

direc-of ρ m and ρ r In the example shown, the sum of these two angles is taken to

be positive, which according to equation (7.22) gives a positive torque, i.e the

machine acts as a motor The induced voltage vector  e m(written here in statorcoordinates) is orthogonal to the flux vector The projection of the current vec-

tor on this voltage vector shown as i torque = ˆisin (ρ r + ρ m) (see figure 7.10), is

according to equation (7.22) proportional to the torque, provided the condition

as given by equation (7.21) is met The largest motor torque value, equal to

ˆ

T e = ˆψ m ˆi, which can be delivered is realized when the rotor current vector

leads the flux vector by π2rad, in phase with the voltage dψ m

dt We will use thisdiagram again in the next three chapters in different forms consistent with thethree main types of machines in use today A tutorial at the end of this chapter

is given to reinforce the concepts discussed in this section