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WE generators – Coupling Fields
EE 559
2016
Iowa State University
James D. McCalley
Harpole Professor of Electrical &
Computer Engineering
Torque production
Let’s devise a way so that a similar rotating magnetic field is
produced by the windings on the rotor, such that it rotates at
the same speed as that from the stator. Thus, we now have
two rotating magnetic fields.
The two rotating magnetic fields, that from the rotor and
the composite field from the armature, are “locked in,”
and as long as they rotate in synchronism, a torque
(Torque=P/ωm=Force×radius, where Force is tangential
to the rotor surface), is developed.
This torque is identical to that which would be
developed if two magnetic bars were fixed on the same
pivot as shown on the next slide.
Torque production
In the case of generator operation, we can think of bar
A (the rotor field) as pushing bar B (the armature field),
as in Fig. a. In the case of motor operation, we can
think of bar B (the armature field) as pulling bar A (the
rotor field), as in Fig. 11b.
To compute the torque,
we must make use of
coupling fields and coenergy, according to:
S
S
S
Bar B
Bar A
N
N
Fig a: generator operation
N
Bar B
Bar A
S
N
Fig b: Motor operation
Wc
Te 

Torque production
One can make conceptual use of a version of the Lorentz
Force Law, below, in computing torque exerted on a current
-carrying conductor in the presence of a magnetic field.
  
F  il  B
However, most electromechanical energy-conversion
devices contain magnetic material; in these systems, forces
of the B-field act magnetically on the magnetic material, and
also on any current-carrying conductors. So the above
relation is not enough for us to compute force.
Instead, we turn to an energy method, which makes use of
W   Pdt
d
P  T  T
dt
The above two relations will allow us to express torque via energy relations.
Coupling fields
In electromechanical systems, we have an electrical
system, a mechanical system, and a coupling field. We
assume in our work that the coupling field is magnetic.
Most of the coupling
field energy is stored in
airgap because it has
very large “resistance”
(reluctance) to the flux.
Coupling
field
Electrical
system
Mechanical
system
Example of an elementary translational-motion electromechanical system with
one electrical input, one mechanical input, and one coupling field.
P. Krause, O. Wasynczuk, & S. Sudhoff, “Analysis
of Electric Machinery,” IEEE Press, 1995.
Coupling fields
Coupling
field
Mechanical system
Coupling
field
Electrical system
Example of an elementary rotational-motion electromechanical system with
one electrical input, one mechanical input, and one coupling field.
A. Fitzgerald, C. Kingsley, and A. Kusko,
“Electric Machinery, 3rd edition, 1971
Coupling fields
Mechanical system is the rotational motion of the rotor.
Coupling field is
in the flux path
of the iron
path/air gap
Electrical
systems are the
two coils.
Example of an elementary rotational-motion electromechanical system with
two electrical inputs, one mechanical input, and one coupling field.
P. Krause, O. Wasynczuk, & S. Sudhoff, “Analysis
of Electric Machinery,” IEEE Press, 1995.
Coupling fields: torque
The importance of the coupling field is that this is where the
energy conversion takes place. If losses in the coupling
field (eddy current, hysteresis) are neglected or assumed
to be modeled external to the coupling field, then
Energy stored in
the coupling field
OR
=
Energy transferred to
the coupling field by
the electrical system
+
Energy transferred to
the coupling field by
the mechanical system
W f  We  Wm
dW f  dWe  dWm
In differential form,
The last equation enables computation of torque.
We investigate dWm and dWe next. We consider a rotational
system with just 1 electrical input and 1 mechanical input.
Coupling fields: torque
Generalization of systems with 1 electrical input,
1 mechanical input, and 1 coupling field.
This is neglected, or
it is represented
within the model of
the electrical or the
mechanical system.
A. Fitzgerald, C. Kingsley, and A. Kusko,
“Electric Machinery, 3rd edition, 1971
Coupling fields: torque
dW f  dWe  dWm
From previous slide, we observe the electrical terminals at
the coupling field receive current i with voltage e. Thus power
is ei and the total energy supplied by the electrical source is:
We   eidt
dWe  eidt
In electromagnetic systems, the coupling element is a
machine winding, and e will be an induced voltage according
to Faraday’s Law:
d
e
Therefore:
dt
d
We  
idt   id 
dt
d
dWe 
idt  id
dt
For the following development, see (a) P. Krause, O. Wasynczuk, & S. Sudhoff, “Analysis of Electric
Machinery,” IEEE Press, 1995; (b) A. Fitzgerald, C. Kingsley, and A. Kusko, “Electric Machinery, 3 rd
edition, 1971; (c) J. Meisel, “Principles of electromechanical-energy conversion,” McGraw-Hill, 1966.
Coupling fields: torque
dW f  dWe  dWm
The mechanical torque Tm and the electromagnetic torque Te
are assumed positive in the same rotational direction. We
neglect torque due to shaft twist.
d 2
d
Tm  Te  J 2  D
dt
dt
2
d
d
Tm  J 2  D
 Te
dt
dt
J: moment of inertia
D: damping
Then the total energy supplied by the mechanical source is
This equation is the rotational
analogue to the one for linear motion,
which is WM  Fmdx

d 2
d
WM   Tm d   J 2 d   D
d   Te d
dt 
dt
 
Term 3
Term1
Term 2
Term 1 on RHS is stored inertial energy; Term 2 is losses.
Only term 3 is energy transferred to coupling field. So…
Wm    Te d
dWm  Te d
Coupling fields: torque
dW f  dWe  dWm (a)
dWe  id (b)
dWm  Ted (c)
Substituting expressions (b) and (c) into (a) results in
dW f  id  Te d
Solving for the torque term results in:
Te d  id   dW f
(*)
Assume independent variables of our system are i, θ. Then:
   i, 
W f  W f i, 
Taking the total
differential


d 
di 
d
i

W f
W f
dW f 
di 
d
i

Substitute these last equations into (*)…
Coupling fields: torque


di 
d
i

W f
W f
dW f 
di 
d
i

d 
W f
W f


Te d  id  dW f  i
di  i
d 
di 
d
i

i

Gather like terms in di and dθ:
  W f 
  W f 
di   i
d
Te d   i


i 
 
 i
 
Equate coefficients:
  W f 
  W f 
d
di
Te d   i

0   i

 
i 
 
 i
The first expression provides a way to compute torque:
 W f
Te  i



(T1)
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Coupling fields: torque
Define co-energy: Wc  i  W f
Recalling that independent variables of our system are i, θ:
Wc
 W f
i




Solve for the last term on the right:
W f
 Wc
i


 
Substitute into (T1):

 Wc
 W f
Te  i
i

Te  i



 (T1)

 
0
Wc
Te 

(T2)
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Coupling fields: torque
What is co-energy? Wc  i  W f
An expression for which torque computation is convenient
Wc
(T2)
Te 

To better understand co-energy, consider the relation for the
coupling field energy:
W f  We  Wm
Set the system so that initially, Wf=0 (no stored energy), and
fix θ so that no energy can be added to the coupling field via
mechanical means, i.e., Wm=0. Then increase the current to
a value ia, establishing a corresponding flux linkage of λa.
There is then energy in the coupling field, but only via
electrical means, i.e.
W f  We
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Coupling fields: torque
But recall that:
d
We   eidt  
idt   id
dt
And since Wm=0, this is also the
energy of the coupling field:
W f   id 
We observe the corresponding area
in the λ-i curve in the figure.
Recall the definition of co-energy: Wc  i  W f
Since iλ is the area of the (shaded) box, then Wc must be the
area below the curve. Wc is therefore given by
Wc   di
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Coupling fields: torque
If the medium is magnetically linear
(no saturation), then the λ-i curve is
just a diagonal through the iλ
shaded box, as shown. In this case,
the area above the λ-i is the same
as the area below it, and we have:
W f  Wc
Wc
Recall (T2), which is Te 

Therefore, under condition of magnetic linearity, we have:
W f
(T3)
Te 

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Coupling fields: energy
Recall from basic physics that the energy stored in a
winding of self-inductance Lpp carrying current i is given by:
1
W f  L ppi 2
2
with Lpp in henries and defined by Lpq=λp/Iq=NpNq/Rpq for
linear medium; Rpq is path reluctance (like resistance).
Generalization: For a linear electromagnetic system with J
electrical inputs (windings), the total field energy is given by:
1 J J
W f   Lpqi piq
2 p 1 q 1
where Lpq is the winding’s self inductance when p=q and
when p≠q, it is the mutual inductance between the two
Derivation: Given pp 22-24 of Krause, Wasynchzuk, and
windings.
Sudhoff, “Analysis of electric machinery,” 1995.
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Example
This device consists of
two conductors, #1 is on
the stator; #2 is on the
rotor. The magnetic
system is assumed linear.
So Wf is given by (with J=2):
1 J J
1
1
1
W f   Lpqi piq  L11i1i2  2 L12i1i2  L22i1i2   L11i1i2  L12i1i2  L22i1i2
2 p 1 q 1
2
2
2
The self-inductances, given by Lpp=λp/ip, are constant,
independent of θ, because the reluctance of the path seen by
the winding does not change as the rotor turns.
But the mutual inductances are not constant.
19
Example
The mutual inductances,
given by Lpq=λp/iq, i.e., the
amount of flux seen by
winding p due to a current
in winding q, are not
constant. The extreme
conditions of maximum & minimum linkages are given below.
θ=0° condition
(maximum
positive
linkage)
θ=90° condition
(zero linkages)
θ=0° condition
(maximum
negative
linkage)
The mutual inductance goes from max positive to 0 to max
negative to 0 and back to max positive. Thus, L12, L21 are:
L12  L21  M cos
20
Example
θ=0° condition
(maximum
positive
linkage)
θ=90° condition
(zero linkages)
θ=0° condition
(maximum
negative
linkage)
The mutual inductance goes from max positive to 0 to max
negative to 0 and back to max positive. Thus, L12, L21 are:
L12  L21  M cos
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Example
Summarizing, we have.
L11, L22 are constant.
L12  L21  M cos
1
1
W f  L11i1i2  L12i1i2  L22i1i2
2
2
1
1
Substituting, we obtain W f  L11i1i2  M cos i1i2  L22i1i2
2
2
Recalling Te 
W f

, we obtain Te   Mi1i2 sin 
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Summary
The previous procedure can be applied to a three-phase
induction machine to obtain an expression for its torque.
The effort requires a coordinate transformation which is
involved, and so we will not do it.
The resulting relation for torque (see Meisel, pp. 321322, and chapter 12) may be used to derive the torque
expression for steady-state conditions. We will use this
relation in the next set of material.
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