Lecture 6 EE743 - PowerPoint PPT Presentation

1 / 33
About This Presentation
Title:

Lecture 6 EE743

Description:

... from a north pole of a magnet into the air toward the south pole (stator winding ... enters the air gap from the north pole of the magnetic system of the rotor. ... – PowerPoint PPT presentation

Number of Views:30
Avg rating:3.0/5.0
Slides: 34
Provided by: ValuedGate741
Category:
Tags: ee743 | lecture | north | pole

less

Transcript and Presenter's Notes

Title: Lecture 6 EE743


1
Lecture 6 - EE743
Single-Phase Reluctance Machine
Professor Ali Keyhani
2
Single-Phase Reluctance Machine
  • Consider a single-phase 1- ? reluctance machine
    shown below,

.
Te
?r
as
?r
as_axis
x
as
3
Single-Phase Reluctance Machine
  • The equations that describe a single-phase
    reluctance machine are,

4
Single-Phase Reluctance Machine
  • Note that when ? r 0, the reluctance is maximum
    along the magnetic as_axis.
  • The torque expression - Te - is valid for the
    transient and steady-state conditions.

5
Single-Phase Reluctance Machine
  • Consider the case when iasconstant,

.
Reluctance Te ?r
rotor position
as
Maximum 0 0
as_axis (magnetic axis)
x
as
6
Single-Phase Reluctance Machine
.
Reluctance Te ?r
rotor position
as
k
as_axis (magnetic axis)
x
as
7
Single-Phase Reluctance Machine
.
Reluctance Te ?r
rotor position
as
0
as_axis (magnetic axis)
x
as
8
Single-Phase Reluctance Machine
.
Reluctance Te ?r
rotor position
as
-k
as_axis (magnetic axis)
x
as
9
Single-Phase Reluctance Machine
.
Reluctance Te ?r
rotor position
as
Maximum 0
as_axis (magnetic axis)
x
as
10
Single-Phase Reluctance Machine
  • Assume that there is no external torque on the
    shaft.
  • There is a force created to minimize the
    reluctance of the magnetic system.
  • For at the time when ias is
    increased from zero to a constant value, the
    torque Te would become .

11
Single-Phase Reluctance Machine
  • Thus, a positive Te would cause the rotor to
    rotate to . (Te gt 0 when ias ? 0)

12
Single-Phase Reluctance Machine
  • From the previous graph, we can deduce that
    is stable point of operation for the
    single-phase reluctance machine, that is, the Te
    0 at this position.
  • Analysis
  • If (decreases very slighty).
    Then, Te gt 0 . The Te increase back
    to .

13
Single-Phase Reluctance Machine
  • Analysis
  • If (increases very slightly).
    Then, Te lt 0 . The Te increase back
    to .
  • Therefore, are stable operating
    points.

14
Single-Phase Reluctance Machine
  • The previous analysis, will led us to ask what
    would happen if the rotor is initially positioned
    at when we increase ias from zero to a
    constant value.
  • No external torque is applied to the rotor.
  • Ias goes from zero to a constant value
  • If

15
Single-Phase Reluctance Machine
  • Stability Test
  • If ?r is increased slightly from zero, Te becomes
    positive, which would cause ?r to increase to a
    stable point ?/2.
  • If ?r is decreased slightly from zero, Te lt 0
    (negative torque), which would cause ?r even
    further and the rotor will come to rest at a
    stable point -?/2.

16
Single-Phase Reluctance Machine
  • Therefore,
  • If we set ?r ? ? and increase ias from 0 to a
    constant value.
  • The rotor would theoretically remain at rest
    ?r ? ? .
  • However, any disturbance, even a slightly one,
    will cause the rotor to rotate away from ?r ? ? .

17
Single-Phase Reluctance Machine
  • There is a 50-50 chance for the rotor to move
    counter-clockwise, or clockwise.

18
Single-Phase Reluctance Machine
  • Assume
  • and

19
Single-Phase Reluctance Machine
  • For steady-state operation (?e constant),
  • Recall that

20
Single-Phase Reluctance Machine
  • After some work, we will have,

21
Single-Phase Reluctance Machine
  • Term A is an average steady-state torque
    produced when ?r0.
  • Term A or Term B will produce an average torque
    when ?r ?e or when ?r - ?e .
  • ?r rotor or shaft speed
  • ?e synchronous speed

22
Single-Phase Reluctance Machine
  • Let ?r ?e ,

23
Single-Phase Reluctance Machine
  • The steady-state torque will pulsate at 4 ?e and
    2 ?e , and the average torque is a double-angle
    sinusoidal function of

24
Windings in relative motion
  • Consider the following machine,

.
Te
?r
as_axis 2
1
?r
as_axis 1
x
1
25
Windings in relative motion
  • The equivalent magnetic circuit is shown below

26
Windings in relative motion
27
Windings in relative motion
  • If we assume

28
Windings in relative motion
  • Note
  • For the stator, the flux issues from a north pole
    of a magnet into the air toward the south pole
    (stator winding for a positive current).
  • For the rotor, the flux produced by a positive
    current flowing in the rotor winding, enters the
    air gap from the north pole of the magnetic
    system of the rotor.

29
Windings in relative motion
.
.
Sr
Nr
Ns Ss
Te
Ns Ss
Nr
Sr
s
x
s
x
.
.
Sr Nr
Ns Ss
Nr Sr
Sr
Nr
s
x
30
Windings in relative motion
  • An electromagnetic torque is produced in attempt
    to align the magnetic system established by the
    currents flowing in the stator and rotor windings
    to align the magnetic axes.
  • The stable operation is reached when ?r 0
  • The unstable operation is reached when ?r ?

31
Windings in relative motion
  • If we assumed that
  • The, for the steady-state operation,

32
Windings in relative motion
33
Windings in relative motion
  • An average torque is produced if ?e ?r or ?e
    - ?r
  • If ?e ?r
Write a Comment
User Comments (0)
About PowerShow.com