2

Design of Magnetic
Circuits

2.1 Introduction
All electrical machines use magnetic materials for directing and shaping the magnetic field,
which acts as a medium for energy conversion. While most of the rotating machines use
ferromagnetic materials along with air, the transformer uses ferromagnetic materials only as
the medium. The magnetic circuit acts as a medium for conversion of electrical to electrical
energy in case of transformers, electrical to mechanical energy in case of motors and
mechanical to electrical energy in case of generators. Hence, analysis of magnetic circuits is
essential for understanding the machines.
Magnetic circuit is defined by the path travelled by magnetic flux. Magnetic flux traces
a closed path, returning back to its initial point, analogous to electric current in an electric
circuit. Magnetic flux is established and maintained by a Magneto Motive Force (MMF), in
any magnetic material.
The significant terms used in the magnetic circuit design are defined as follows.
Magneto motive force (MMF):  It is the force behind the production of magnetic flux in the
magnetic circuit. It is represented by Ampere turn (AT).
Magnetic flux:  It is the magnetic lines of force established by MMF. It is represented by φ.
The unit is Wb.
Magnetic field intensity:  At any point, it is specified by both direction and magnitude or
magnetic field strength. It is measured in amperes per metre (A/m). The magnetic field
intensity around a closed contour as defined by ampere’s law is equal to total current passing
through any surface linking that contour and is given by ∫ Hdl = ΣI.
2.2
2.2  Design
DesignofofMagnetic
MagneticCircuits
Circuits

Magnetic flux density

Magnetic flux (B): ItItisisdefined
density (B): definedasasthe
the force
force acting
acting per
per unit
unit current
current perper unit
unit length
length on
on a wire placed at right angles to the magnetic field. Like H, B is also a vector
a wire placed at right angles to the magnetic field. Like H, B is also a vector quantity and quantity andis
is measured
measured in in Tesla(T).
Tesla (T).Magnetic
Magneticflux fluxdensity
densityisisrelated
relatedtoto magnetic
magnetic field
field intensity
intensity by
by the
the
relation B =μ
relation B =µH where μ
H ,, where µ isis the
the permeability
permeability of of the
the magnetic
magnetic material.
material.
Reluctance: ItItis
Reluctance:  is the
the property
property of of aa magnetic
magnetic circuit
circuit that
that opposes
opposes the
the flow
flow ofof magnetic
magnetic flux.
flux. It
It
is the ratio of MMF to magnetic flux. It is represented by S. S .The
Theunit
unitisisAT/Wb.
AT/Wb.
Permanence: ItItisisthe
Permanence:  the property
property ofof aa magnetic
magnetic circuit,
circuit, which allows the flow of magnetic flux.
It is the ratio of magnetic flux to MMF. It is also given by the reciprocal of reluctance. It is
represented by Λ Λ.. The
The unit
unit is
is Wb/AT.
Wb/AT.
The
The magnetic
magnetic flux,
flux, MMF,
MMF, reluctance
reluctance and
and permeance
permeance areare related
related by
by the
the following
following relations:
relations:
= φ=
AT AT S φS

φ =φΛ=ATΛ AT
Further, magneticcircuits
Further, magnetic circuitscan
canbebe categorized
categorized into
into twotwo types,
types, namely
namely simple
simple and com-
and composite
posite
magnetic magnetic
circuits, circuits,
based on based on the
the number number materials
of magnetic of magneticused.materials
A simple used.
magneticA simple
circuit
­m agneticofcircuit
consists single consists
magneticofmaterial.
single magnetic material.
A composite A composite
magnetic magnetic
circuit consists circuit con-
of minimum of
sists of m
­ inimum
two different of two(either
materials different materials
magnetic (either magnetic
or non-magnetic) or non-magnetic)
of different magneticof different
properties.
magnetic
The magneticproperties.
circuit The magnetic
equivalent of circuit equivalent
transformer, of transformer,
induction induction motor,
motor, synchronous motor syn-
and
chronous
dc motor are motor and
dealt in dc motor
detail areforthcoming
in the dealt in detail in the forthcoming
chapters. The following chapters. The follow-
section analyses the
ing sectionmagnetic
composite analyses circuits
the composite magnetic
with series circuitsconnections.
and parallel with series and parallel connections.
It
It should
should also
also be
be noted
noted that
that the
the analysis
analysis based
based on on magnetic
magnetic circuit
circuit theory
theory isis the
the most
most
widely
widely used.
used. The
The analysis
analysis can
can also
also bebe done
done based
based on
on magneto-static
magneto-static finite
finite element
element
analysis.
analysis. The
The magnetic
magnetic model
model based
based onon finite
finite element
element analysis
analysis is
is generally
generally more
more accurate
accurate
than
than the
the model
model based
based on on magnetic
magnetic circuit
circuit theory,
theory, though
though at
at the
the expense
expense ofof complexity
complexity of of
programming.
programming.

2.1.1 Analysis
2.1.1 analysis of
of Series
series Composite
composite Magnetic
Magnetic Circuit
circuit
Considering
Considering aa series
series composite
composite magnetic
magnetic circuit
circuit as
as shown
shown in in Fig.
Fig. 2.1,
2.1, it
it is
is assumed
assumed that
that
there
there are three different magnetic materials of different relative permeabilities present
are three different magnetic materials of different relative permeabilities present in
in the
the
magnetic
magnetic circuit
circuit along
along with
with an
an air
air gap.
gap.

D φ E

I
F
lg

C B

fig. 2.1 
Fig. 2.1 | 
| Series
Series composite
composite magnetic
magnetic circuit
circuit
For the series composite magnetic circuit in Fig. 2.1, Table 2.1 gives the required parameters
For the series composite magnetic circuit in Fig. 2.1, Table 2.1 gives the required parameters
to establish a relation with the MMF, reluctance and magnetic flux.
to establish a relation with the MMF, reluctance and magnetic flux.
 Introduction  2.3

Table 2.1  |  Parameters of series composite magnetic circuit

Part of circuit Flux Length Cross-sectional Magnetizing Reluctance
area force
ABC φ l1 a1 = a H1 S1
CD φ l2 a2 = a H2 S2
DEF φ l3 a3 = a H3 S3
FA φ l4 ag = a Hg Sg

From the magnetic circuit represented in Fig. 2.1, it is observed that the flux is constant;
therefore, the total reluctance is equal to the sum of individual reluctances of different parts.

Hence, total reluctance, S = S1 + S2 + S3 + Sg .

We know that, generally

l
Reluctance, S = ,
μ0 μr a

where l is the length, μ0 is the permeability of free space, μr is the relative permeability
and a is the cross-sectional area.
Therefore, substituting the length, permeability and cross-sectional area for different
parts, we get

l1 l2 l3 lg
S= + + +
μ0 μr1 a1 μ0 μr 2 a2 μ0 μr 3 a3 μ0 μr g a g

Total MMF in the circuit,

AT = Flux × reluctance
= φ S1 + φ S2 + φ S3 + φ Sg (2.1)
 φl φ l2 φ l3 φ lg 
=  1
+ + + 
 μ0 μr1 a1 μ0 μr 2 a2 μ0 μr 3 a3 μ0 μr g a g 
B1 B2 B3 Bg
= × l1 + × l2 + × l3 + × lg
μ0 μr1 μ0 μr 2 μ0 μr 3 μ0

 
 ∵ B = φ and μrg = 1 for air
 a 
The MMF can also be represented in terms of magnetizing force and length as follows:
 
Total MMF = H1 l1 + H 2 l2 + H 3 l3 + H g lg  ∵ H = B 
 μ0 μr 
It can be observed from Eq. (2.1) that it is similar to that of emf equation in an equivalent
electrical circuit. Therefore, redrawing the series composite magnetic circuit analogous to
2.4
2.4  Design
Designofof
Design Magnetic
ofMagnetic
Magnetic Circuits
Circuits
Circuits

series
series electrical
electrical circuit
circuit with
with three
three resistances
resistances as
as shown
shown in
in Fig.
Fig. 2.2,
2.2, we
we define
define the
the electrical
electrical
analogy
analogy toto magnetic
magnetic circuit.
circuit.

RR33

RR22
RRgg
II
++
emf
emf −
−

RR11

fig.
fig. 2.2
Fig. 2.2 || 
2.2  | Equivalent
Equivalent electrical
electrical circuit
circuit

From
From Fig.
Fig. 2.2,
2.2, itit is
is observed
observed that
that the
the total
total resistance
resistance of
of the
the equivalent
equivalent electric
electric circuit
circuit is
is
equal
equal to
to the
the individual
individual sum sum of
of various
various resistance
resistance values,
values,
i.e.,
i.e.,
i.e., R=
R R111+
=R +RR222 +
+RR333 +
+RRggg
Therefore,
Therefore,
Therefore,
Total
Total emf,
Total emf, E=
emf, EE = IR11+
=IR
IR + IR22 +
+IR
IR + IR33 +
+IR
IR +IR
+ IRgg
IR
1 2 3 g
= Current ×
= Current × Total
Total resistance
resistance
Thus, by analogy with respect to electric circuit,
Thus, by analogy with respect to electric circuit,
Thus, by analogy with respect to electric circuit,
Total
Total MMF
Total MMF in
MMF in magnetic
in magnetic circuit =
magnetic circuit
circuit = AT for
= AT
AT for series
for series paths
series paths ++
paths ATfor
+ AT
AT forair
for airgap
air gap
gap

2.1.2
2.1.2 analysis
analysis of
2.1.2 Analysis of Parallel
Parallel composite
composite Magnetic
Composite Magnetic circuits
Circuits
circuits
Considering
Considering aa parallel
parallel composite
composite magnetic
magnetic circuit
circuit as
as shown
shown inin Fig.
Fig. 2.3,
2.3, itit is
is assumed
assumed that
that
there
there are
are three
three different
different magnetic
magnetic materials
materials of
of different
different relative
relative permeabilities
permeabilities present
present in
in the
the
magnetic
magnetic circuit
circuit along
along with
with an
an air
air gap.
gap.

VV φφ22 UU φφ11 ZZ

φφSS

II

NN

φφSS
φφ22 φφ11

W
W XX YY

fig.
fig. 2.3
Fig. 2.3 || 
2.3  | Parallel
Parallel composite
composite magnetic
magnetic circuit
circuit
 Introduction 2.5
Introduction  2.5
Table 2.2  | 
table 2.2 Parameters of
| Parameters of parallel
parallel composite
composite magnetic
magnetic circuit
circuit
Part of circuit Flux Length Cross-sectional Magnetizing Reluctance
Part of circuit flux Length cross-sectional Magnetizing reluctance
area force
area force
UZYX φ1 l1 a1 H1 S1
UZYX φ1 l1 a1 H1 S1
UVWX φ2 l2 a2 H2 S2
UVWX φ2 l2 a2 H2 S2
UX φs ls as Hs Ss
UX φs ls as Hs Ss
For the considered parallel composite magnetic circuit in Fig. 2.3, Table 2.2 gives the
For the
required considered
parameters to parallel
establishcomposite magnetic
relation with circuit
the MMF, in Fig. and
reluctance 2.3, magnetic
Table 2.2 flux.
gives the
required parameters
In Table 2.2, a1 = to
a2 establish
= as = a relation
[ ∵ Areawith the MMF, reluctance
of cross-section is the sameandinmagnetic flux.of the
all the parts
In Table
magnetic 2.2, a1 = a2 = as = a [∵ Area of cross-section is the same in all the parts of the
circuit].
magnetic circuit].
We know that the total flux in the circuit is given by
We know that the total flux in the circuit is given by
φs = φ 1+ φ 2 (2.2)
φs = φ 1+ φ 2 (2.2)
MMF = φ S + φ S = φ S + φ S (2.3)
MMF = φ11S11 + φssSss = φ12S22 + φssSss (2.3)
Also, as B = μ H and φ = Ba , we get
Also, as B = µ H and φ = BA, we get
l
φ = μ Ha and S = l
ϕ = µ Ha and S =μ a
µa
The MMF can be represented in terms of magnetizing force and length as
The MMF can be represented in terms of magnetizing force and length as
MMF = H l + H l = H l + H l
MMF = H11l11+ H sslss = H 22l22 + H sslss
From
From Eqs.
Eqs. (2.2)
(2.2) and
and (2.3),
(2.3), it
it can
can be
be observed
observed that
that the
the equation
equation is
is similar
similar to
to that
that of
of emf
emf
equation in an equivalent electrical circuit. Therefore, redrawing the parallel
equation in an equivalent electrical circuit. Therefore, redrawing the parallel composite composite
magnetic
magnetic circuit
circuit analogous
analogous to to parallel
parallel electrical
electrical circuit
circuit with
with three
three resistances
resistances as
as shown
shown in
in
Fig. 2.3, we define the electrical analogy to magnetic circuit.
Fig. 2.3, we define the electrical analogy to magnetic circuit.

V U Z

I2 IS I1

RS

R1 R2
+
− emf

IS
I2 I1

W X Y

fig. 2.4 
Fig. 2.4 | 
| Equivalent electrical circuit
2.6  Design of Magnetic Circuits

Table 2.3  |  Parameters of equivalent electrical circuit

Part of circuit Current Resistance
UZYX I1 R1
UVWX I2 R2
UX Is Rs

The parameters referring to the equivalent electric circuit shown in Fig. 2.4 are tabulated
in Table 2.3.
Total current in the circuit is given by I s = I1 + I 2

The emf in the circuit is = I1R1 + I s Rs = I 2 R2 + I s Rs

Thus, by analogy with electric circuit,
Total MMF required = AT for common section (UX)
+ AT for any one of the parallel paths (VW or ZY)
In a parallel magnetic circuit, the total flux exists in a common section of the magnetic
circuit, which contains the exciting coil. It divides into two parts, follows the different paths
and recombines at the other end of the common section.
From the aforementioned analysis of series and parallel composite magnetic circuits,
Krichoff’s laws for magnetic circuits can be stated as follows:
First Law:  The total flux towards a node is equal to the total flux away from the node in any
magnetic circuit, i.e., ∑ magnetic flux = 0.

Second Law:  In any magnetic circuit, the sum of the product of the magnetizing force in
each part of the magnetic circuit and the length of that part is equal to the resultant MMF, i.e.,
∑ MMF = ∑ (reluctance × magnectic flux).
2.1.3 Comparison Between Magnetic Circuit and Electric Circuit
The comparison between magnetic circuit and electric circuit is shown in Table 2.4.

Table 2.4  |  Comparison of magnetic and electric circuits

S. Magnetic circuit Electric circuit
No.
1. The closed path traced by magnetic flux is The closed path traced by electric current
known as magnetic circuit. is known as electric circuit.
2. The driving force is magneto motive force The driving force is electromotive force
(MMF)AT. (emf)V.
3. Magnetic flux is opposed by reluctance Electric current is opposed by resistance
(ℜ) At/Wb , (R)Ω .

(Continued)
 Determination of Reluctance and MMF of Air Gap  2.7

Table 2.4  | (Continued)

S. Magnetic circuit Electric circuit
No.
1 ρ
S= R= where ρ is the resistivity
μ0μr a a
1
where is the relativity
μ0μr
4. MMF emf
Magnetic flux = Wb Electric current = A
reluctance resistance
6. Ohm’s law is given by Ohm’s law is given by
MMF = magnetic flux × reluctance emf = electric current × resistance
7. Kirchoff’s laws are given by Kirchoff’s laws are given by
∑ magnetic flux = 0 and ∑ electric current = 0 and
∑ MMF = ∑(reluctance × magnectic flux) ∑ emf = ∑(resistance × electric current )
8. Permanence is given by Conductance is given by
1 1
P = Wb/At G= 
R R
9. Magnetic flux is established in a circuit Electric current flows in a circuit, i.e., due
and does not actually flow. to the movement of electrons.
10. Energy is required to establish the Energy is required to maintain the flow of
magnetic flux, not to maintain it. electric current.

2.2  Determination of Reluctance and MMF of Air Gap

The study of magnetic circuits is important in the study of electrical apparatus since their
operation can be characterized efficiently using magnetic circuits. So, the reluctance and MMF
determination form a crucial aspect in analysis of any electric machine. The determination
of reluctance and MMF of air gap requires prior knowledge of the geometrical dimensions,
in order to account for the presence of slotting, ventilating ducts for cooling and effect of
saliency. Apart from the above factors, an effect of ‘fringing’, which will be discussed in the
following sections, is to be considered.
Let us begin with a smooth armature surface corresponding to dc machine as shown in
Fig. 2.5(a) and (b).
For the smooth armature surface shown in Fig. 2.5(b), the reluctance of air gap with
respect to one slot pitch is given by

lg
Sg =  (2.4)
μ0 Ly s
2.8  Design of Magnetic Circuits

ys

A is gap lg

Slot Tooth

(b)
(a)

Fig. 2.5  |  (a) Simple dc machine. (b) Zoomed view of smooth armature surface

where lg − length of air gap, μ0 − permeability of free space, L − length of core and ys − slot
points.
A slotted armature surface corresponding to dc machine is shown in Fig. 2.6(a) and
(b).

yS

Air gap lg lg

Slot Tooth

wS wt

(b)
(a)

Fig. 2.6  |  (a) Simple dc machine. (b) Zoomed view of slotted armature surface
 Determination
Determination of
of Reluctance
Reluctance and
and MMF
MMF of
of Air Gap  2.9
Air Gap 2.9
Determination of Reluctance and MMF of Air Gap 2.9
For
For the
the slotted
slotted armature
armature shownshown in in Fig.
Fig. 2.6(b),
2.6(b), the the reluctance
reluctance of of air
air gap
gap with
with an
an assumption
assumption
that For theflux
slotted armature shown in pole
Fig. 2.6(b), the reluctance of airany
gapdeviation,
with an assumption
that all the flux from the teeth links the pole surface inparallel without any deviation, is
all the from the teeth links the surface inparallel without is given
given by by
that all the flux from the teeth links the pole surface inparallel without any deviation, is given by
llgg llgg
SSgg == lg =
= lg (2.5) (2.5)
Sg = µ μ00LL( yyss −−w wss) = µ μ00LwLwtt (2.5)
µ0 L ( y s − ws ) µ0 Lwt
where wss − − slot
slot width
width and
and w wtt −− tooth
tooth width.
width.
where
But w s − slot width
But practically,
practically, the flux
the flux wt − the
andfrom
from tooth
the teethwidth.
teeth is under
is under the the effect
effect of of ‘fringing’
‘fringing’ wherein
wherein the
the flux
flux atat
the But
the both practically,
both ends
ends of the
of teeth flux
teeth curves from
curves and the
and links teeth
theispole
links the under
pole the effect
surface
surface of ‘fringing’
as shown
as shown in Fig.
in wherein
Fig. 2.7.
2.7. Hence,
Hence,the
influx
in order
order at
the both
to account
to ends
account this of teeth
this fringing curves
fringing effect, and
effect, Fig. links
Fig. 2.7
2.7 is the pole surface as shown
is redrawn as Fig. 2.8, with yyss being in Fig. 2.7.
being the Hence,
the new in order
new contracted
contracted
to account
slot
slot pitch this fringing
pitch wherein
wherein the effect,
the total
total flux
flux Fig.is 2.7
is is redrawn
assumed
assumed beasconfined
to be
to Fig. 2.8, in
confined with ys being
in linking
linking thethe
the new
pole
pole contracted
surface
surface from
from
slot pitch
armature in
armature wherein
in parallel.
parallel.the total flux is assumed to be confined in linking the pole surface from
armature in parallel.
yS
yS

lg
Air gap
lg
Air gap

Slot Tooth
Slot Tooth

wS wt
wS wt

fig.
Fig. 2.7
2.7  ||  Fringing
Fringing of
of flux
flux in
in slotted
slotted armature
armature
fig. 2.7 | Fringing of flux in slotted armature
yS
yS
y´S
y´S
lg
Air gap
lg
Air gap

Tooth
Tooth
Slot
Slot

wS wt
wS wt

fig. 2.8 | Slotted armature with fringing of flux accounted for

fig. 2.8 
Fig. 2.8 | 
| Slotted
Slotted armature
armature with
with fringing
fringing of
of flux
flux accounted
accounted for
for
2.10  Design of Magnetic Circuits

Hence, the reluctance of air gap in this case is given by

lg
Sg = (2.6)
μ0 Ly s′
where y′s = wt + fraction of ws = wt + xws(2.7)

Adding and subtracting ‘ws’ to the RHS of Eq. (2.7), we get

y′s = wt + xws + ws − ws
On simplifying the above equation, we get
y′s = wt + ws + xws − ws
= ys + (x – 1)ws [∵ wt + ws = ys]
y′s = ys – (1 − x)ws = ys – Kcs ws(2.8)
where Kcs is the Carter’s gap coefficient. It is determined by an empirical formula,
1
Kcs =
5lg
1+
ws
Carter’s coefficient for parallel-sided open slots is given by
2  −1 1 
Kcs =  tan y − log 1 + y 2 
π y 
ws
where y = .
2l g
Carter’s coefficient for open and semi-closed slots with respect to slot opening/gap
length ratio is represented in Fig. 2.9.
0.9
0.8
ts ts
slo n slo
0.7 se
d Ope
clo
mi
0.6 Se
Carrier’s co-efficient

0.5
0.4
0.3
0.2
0.1

0 1 2 3 4 5 6 7 8 9 10 11 12
Slot opening
Gap length

Fig. 2.9  |  Carter’s coefficient for open and semi-closed slots

 Determination of Reluctance and MMF of Air Gap  2.11

Substituting Eq. (2.8) in Eq. (2.6), we get

lg
Sg = (2.9)
μ0 L ( y s − Kcs ws )
Equation (2.9) gives the reluctance of air gap of slotted armature with the effect of fringing
accounted for.
We proceed to define a term ‘Gap contraction factor’, ‘Kgs’, which is the ratio of reluctance
of air gap of slotted armature to reluctance of air gap of smooth armature. The gap contraction
factor is given by
Sg (slotted armature)
K gs = (2.10)
Sg (smooth armature)

lg
μ0 L ( y s − Kcs ws ) ys
K gs = = (2.11)
lg y s − Kcs ws
μ0 Ly s
From Eq. (2.10), it is observed that the reluctance of air gap of slotted armature is Kgs
times the reluctance of air gap of smooth armature. From Eq. (2.11), it is observed that Kgs is
always greater than 1.
Proceeding further, the effect of ventilating ducts on reluctance of air gap is analyzed.
From Chapter 1, the necessity of ventilating ducts is observed.
The ventilating ducts in radial direction causes flux contraction in axial direction. Similar
to that of fringing effect shown in Fig. 2.10, there is a reduction in effective axial length of
machine and leads to increase in reluctance of air gap.

L
Air gap
lg

Core stack

Ducts, wd : wide

Fig. 2.10  |  Ventilating ducts in a machine

Contracted on net axial length,

L′ = 3wt + ndxwd Or 3wl

Adding and subtracting ndwd on the RHS of the above equation, we get

L ′ = 3 wt + nd xwd + nd wd − nd wd
2.12  Design of Magnetic Circuits

= 3wt + nd wd + nd xwd − nd wd

= L + ( x − 1)nd wd [ ∵ 3wt + nd wd = L]
= L − (1 − x) nd wd

= L − Kcd nd wd (2.12)

where Kcd is the Carter’s coefficient for ventilating ducts, wd is the width of ducts and nd is
the number of ducts.
To calculate the effect of ventilating ducts on the air gap MMF, we define a term ‘Gap
contraction factor for ventilating ducts’, ‘Kgd’, which is the ratio of reluctance of air gap of
smooth armature with ducts to reluctance of air gap of smooth armature without ducts.

lg
μ0 Ly s
K gd =
lg

μ0 (L − Kcd nd wd ) y s
L (2.13)
=
L − Kcd nd wd
Reluctance of air gap with ventilating ducts and slotted armature is given by

lg
Sg =
μ0 L ′y s′

lg
=
μ0 (L − Kcd nd wd )( y s − Kcs ws )

Reluctance of air gap with ventilating ducts and smooth armature is given by

lg
Sg =
μ0 L ′y s

lg
=
μ0 (L − Kcd nd wd ) y s

We define another term ‘Total gap contraction factor for slots and ducts’, ‘Kg’, which is the
ratio of reluctance of air gap of slotted armature with ducts to reluctance of air gap of smooth
armature without ducts.
lg
μ0 (L − Kcd nd wd )( y s − Kcs ws )
Kg =
lg

μ0 Ly s
Ly s
=
(L − Kcd nd wd )( ys − Kcs ws )
 Determination of Reluctance and MMF of Air Gap  2.13

L ys
= × = K gs × K gd (2.14)
L ′ y s′

2.2.1 Contraction of Air Gap Area Per Pole (Effective Air Gap Area)
We know that the magnetic field intensity H and the magnetic flux density B are related by
permeability μ of the material:

B =μ H
For air, μ = 4π × 10−7, the air gap MMF per metre,
B
H= = 795774.71B
μ
Hence, the MMF per metre for air gap can be approximated as 800,000 B(2.15)
For smooth armature, with air gap length, lg,

MMF, (ATg)smooth = 800,000 Blg (2.16)

[∵ MMF = MMF/m × air gap length]

Similarly, for slotted armature, with air gap length, lg,

MMF, (ATg)slotted = 800,000 Kg Blg(2.17)

where Kg is the total gap contraction factor.

φ
Substituting B = in Eq. (2.17), we get
Ag

φ
(ATg)slotted = 800 , 000 K g lg  (2.18)
Ag

Rewriting Eq. (2.18), we get

φ
Ag
(ATg)slotted = 800 , 000 lg
Kg

φ  A 
= 800 , 000 lg    ∵ A ′ = g  (2.19)
Ag′  g
K g 
 
From Eq. (2.19), it can be observed that air gap has decreased (or contracted) to a value
′
Ag .
Proceeding to determine the contracted air gap area per pole, we get
Ag
Ag′ =
Kg
2.14  Design of Magnetic Circuits

Expressing Ag in terms of slot pitch, the number of slots per pole and length of core and
K g from Eq. (2.14), we get

S
× ys × L
S
Ag′ = P = y s′ L ′ (2.20)
Ly s P
L ′y s′

Equation (2.20) describes the effective air gap area per pole.
From Eqs. (2.16) and (2.17), it can be stated that the effective air gap length has increased
by the multiplication of total gap contraction factor, ‘Kg’ with lg in Eq. (2.17), compared to
Eq. (2.16).
Another perspective can be drawn in terms of change in air gap length, i.e., instead of
1
stating that the air gap for slotted armature area has decreased by times air gap area for
Kg
smooth armature, the air gap length of slotted armature has increased K g times the air gap
length of smooth armature. In the above case, K g can be termed as ‘air gap expansion
factor’.

2.2.2 Effect of Pole Saliency

The air gap length is not a fixed value over the pole pitch for a salient pole machine.
Therefore, to determine the reluctance of air gap, the magnetic flux distribution is required
to be known.
The typical magnetic flux distribution of a salient pole machine is represented in
Fig.  2.11(a) and (b). The flux distribution is approximated as a rectangle and is shown in
Fig. 2.11(c).
Therefore, for the flux distribution considered in Fig. 2.11 (c),

MMF of air gap, ATg = flux × reluctance

where flux = flux density × area = Bg × area

Effective air gap length
Reluctance =
μ0 × area
K g lg
=
4π ×10−7 × area
K g lg
⇒ ATg = Bg × area×
4π ×10−7 × area
= 800 , 000 Bg K g lg

In order to provide a measure for the average and maximum flux in salient pole machines,
a term ‘Field form factor’ is defined and given by the ratio of average flux density over pole
pitch to maximum flux density in air gap.
 DeterminationofofReluctance
Determination Reluctanceand
andMMF
MMFofofAir
AirGap 2.15
Gap  2.15

Pole pitch

Pole arc
Pole body

lg

(a) Armature surface

Flux distribution curve

Bg
Bg
Bav

(b)

Flux distribution curve

Bg
Bg
Bav

(c)

fig.
Fig.2.11
2.11 || (a)
(a)and
and(b)
(b)Magnetic
Magneticflux
fluxdistribution
distributionofofaasalient
salientpole
polemachine.
machine.(c)
(c)Flux
Fluxdistribution
distribution
approximation
approximation

BK = Bav
K f = av f
Bg
Bg
Theabove
The aboveexpression
expressioncan
canbe
beapproximated
approximatedas
as
pole arc
K f  K  pole =φ arc
, =φ ,
f pitch
pole pole pitch

ififthe
theeffect
effectof
offringing
fringingisisneglected.
neglected.
2.16  Design of Magnetic Circuits

Problems on Magnetic Circuits

Example 2.1: Determine the effective length of air gap of a machine having a stator
with smooth surface and rotor with open slots devoid of radial ducts,
with tooth width, wt = 15 mm, slot width, ws = 13 mm, air gap length,
1
lg = 3 mm and Carter’s coefficient = .
5l g
1+
ws

Solution:  Given
Tooth width, wt = 15 mm
Slot width, ws = 13 mm
Air gap length, lg= 3 mm
1
Carter’s coefficient, Kcs = (1)
5l g
1+
ws
Substituting ws and lg in the above equation, we get
1
Kcs = = 0.4642
5× 3
1+
13
To determine effective air gap length, it is required to find gap contraction factor for slots,
which is given by
ys
Gap contraction factor, K gs = (2)
y s − Kcs ws
In the above equation, slot pitch, ys is given by
y s = ws + wt

= 13 + 15 = 28 mm
Substituting ys, kcs and ws in Eq. (2),
28
K gs =
28 − (0.4642×13)
= 1.2747

As the radial ducts are not present, gap contraction factor for ducts, Kgd = 1
The total gap contraction factor (Kg) is given by

Kg = Kgs × Kgd = 1.2747 × 1

= 1.2747

Effective air gap length, lgs = Kglg

 Determination of Reluctance and MMF of Air Gap  2.17

Substituting Kg and lg in the above equation,

lgs = 1.2747 × 3 = 3.8241 mm

Example 2.2: Determine the MMF of air gap for an induction machine with the
length of air gap = 4 mm, slot pitch = 63 mm, slot opening = 4.5 mm,
pole arc  =  180 mm, flux per pole = 45 × 10-3 Wb, length of core =
300 mm, number of ducts = 4, width of ducts = 8 mm and Carter’s
coefficient being 0.2 for slot opening/air gap length = 1.125 and 0.24 for
slot opening/air gap length = 2.

Solution:  Given
length of air gap, lg = 4 mm,
slot pitch, ys = 63 mm,
slot opening, wo = 4.5 mm,
pole arc = 180 mm,
flux/pole = 45 mWb,
length of core, L = 300 mm,
number of ducts, nd = 4,
duct width, wd = 8 mm
0.2, for slot opening/air gap length = 1.125
Carter’s coefficient = 
0.24 , for slot opening/air gap length = 2

In order to determine the MMF of air gap, it is required to find total gap contraction factor,
Kg, length of air gap and flux density.
We know that
Gap contraction factor for slots,
ys
K gs = (1)
y s − Kcs wo
And,
slot opening 4.5
=
air gap length 4
= 1.125
Hence,

Carter’s coefficient, Kcs = 0.2

Substituting ys, Kcs and wo in Eq. (1), we get

63
K gs =
63 − 0.2× 4.5
= 1.0144
2.18  Design of Magnetic Circuits

Also,
Gap contraction factor for ducts,
L
K gd = (2)
L − Kcd nd wd
And,
duct width 8
= =2
air gap length 4
Hence, Carter’s coefficient for ducts, Kcd = 0.24
Substituting L, Kcd, wd and nd in Eq. (2), we get
L
K gd =
L − Kcd nd wd
300
=
300 − 0.24 × 6 × 8
= 1.0399

Total gap contraction factor,

K g = K gs × K gd

= 1.0144 ×1.0399 = 1.0548

Flux density at pole centre,

Flux/pole
Bg =
pole arc × length of core

45×10−3
=
180 ×10−3 × 300 ×10−3
= 0.8333 Wb/m2
MMF of air gap,
ATg = 800 , 000 Bg K g lg

⇒ ATg = 800 , 000 × 0.8333 ×1.0548 × 4 ×10−3

= 2812.687 A

Example 2.3: Determine an average air gap flux density of an alternator, with rating
150 MVA, having number of poles = 10, length of core = 1.6 m, diameter
of core = 6 m, total MMF per pole is 18,000 A, MMF required for air gap
is 0.8 times of total MMF per pole, field form factor = 0.65 m, slot width
= 20 mm, slot pitch = 60 mm and length of air gap at the centre of pole
= 25 mm. The type of stator slots used is parallel-sided open slots.
 Determination of Reluctance and MMF of Air Gap  2.19

Solution:  Given
Power rating = 150 MVA
Number of poles = 10
Length of core, L = 1.6 m = 1.6 × 103 mm
Diameter of core, D = 6 m = 6 × 103 mm
Total MMF per pole, AT = 18,000 A
MMF required for air gap, ATg = 0.8 AT
Field form factor, Kf = 0.65
Slot width, ws = 20 mm
Slot pitch, ys = 60 mm
Duct width, wd = 8 mm
Length of air gap, lg = 25 mm
Number of radial ventilating ducts, nd = 40
Since the type of slator slots used is parallel-sided open slots, Carter’s coefficient is given by

2  −1 1 
Kcs =  tan y − log 1 + y 2 
π  y 

 ws
 for slots
 2lg
where y =
 wd
 for ducts
 2lg
For slots,
ws 20
⇒ y= = = 0.4
2lg 2× 25
Hence, Carter’s coefficient,
2  −1 1 
Kcs =  tan 0.4 − log 1 + 0.4 2 
π 0.4 
2
=  0.3805
 − 0.0805
π  radians 
= 0.1909

In order to determine the average flux density, it is required to find MMF of air gap, gap
contraction for slots, ducts and total gap contraction factor.
For ducts,
w 8
y= d = = 0.16
2l g 2× 25

Hence, Carter’s coefficient for ducts,

2  −1 1 
kcd =  tan 0.16 − log 1 + 0.16 2 
π  0.16 
2.20  Design of Magnetic Circuits

2
=  0.1586
 − 0.0343
π  radians 
= 0.0791

Gap contraction factor for slots,

ys
K gs =
y s − Kcs ws
60
=
60 − 0.1969× 20
= 1.0679

Gap contraction factor for ducts,

L
Kgd =
L − Kcd nd wd
1600
=
1600 − 0.0791× 40 × 8
= 1.016

Total gap contraction factor,

Kg = Kgs Kgd = 1.0679 × 1.016

= 1.0849

MMF required for air gap,

ATg = 0.8 AT

Substituting the value of AT in the above equation,

ATg = 0.8 × 18,000

= 14,400 A
Also,
ATg = 800,000 KgBglg
ATg
⇒ Bg =
800000 K g lg
14400
=
800000 ×1.0849× 25

= 6.6365×10−4 Wb/mm 2

= 0.6636 Wb/m 2

 Determination of Reluctance and MMF of Air Gap  2.21

Average flux density,

Bav = k f × Bg

⇒ Bav = 0.65 × 0.6636

= 0.4313 Wb/m2

Example 2.4: Determine the length of air gap of dc machine having core length =
0.1 m, slot pitch = 22 mm, slot width = 8 mm, with 2 ducts each 8 mm
wide, flux density at centre of pole = 0.65 Wb/m2, MMF/pole = 3500 AT
and MMF requied for iron = 780 AT. Assume Carter’s coefficient for
slots and ducts to 0.28.

Solution:  Given,
Core length, L = 0.1 m
Slot pitch, ys = 22 mm
Slot width, ws = 8 mm
Number of ducts, nd = 2
Width of ducts, wd = 8 mm
Flux density, Bg = 0.65 Wb/m2
MMF/pole = 3500 AT
MMF for iron = 780 AT
Carter’s coefficient = 0.28 (for slots and ducts)
In order to determine the air gap length, it is required to find MMF of air gap and total gap
contraction factor, constituted by a gap contraction factor for slots and ducts.

MMF of air gap,

ATg = MMF/pole − MMF of iron
= 3500 − 780 = 2720 AT
Gap contraction factor for slots,
ys
K gs =
y s − Kcs ws
22
= = 1.1133
22 − (0.28 × 8)
Gap contraction factor for ducts,
L
K gd =
L − Kcd nd wd
0.1×10 3
=
0.1×10 3 − (0.28 × 2× 8)
= 1.0469
Total gap contraction factor,
K g = K gs × K gd
2.22  Design of Magnetic Circuits

= 1.1133 ×1.0469

= 1.1655

We know that MMF of air gap

ATg = 800,000 BgKgLg
⇒ Length of air gap,
ATg
lg =
800 , 000 × Bg K g

Substituting the values of ATg, Bg and Kg in the above equation, we get

2720
lg =
800 , 000 × 0.65×1.1655
= 4.488 ×10−3 m
= 4.488 mm

2.3  Determination of MMF of Teeth

The accurate determination of MMF in teeth is an arduous task due to the following reasons:
•• Uniform values of flux density cannot be obtained in tapered teeth with parallel-
sided slots, as the area of the flux path uniformly varies changing the flux density
values along the portion of the teeth.
•• The flux path can branch into slots in parallel to the teeth, creating two paths for
the flow of flux. This happens due to the operation of machine in saturation region,
making the permeability low in teeth causing some amount of flux to flow through
depths of the slots.

Graphical method
It is based on the construction of graph representing the variation of MMF and flux density
with respect to distance from one end to the other end of the teeth (i.e., length of the teeth) as
shown in Fig. 2.12(a)–(c). The mean value of MMF is calculated by integrating H over length
of the teeth. Hence, the total MMF of teeth is determined by multiplying the mean value of
MMF and length of the teeth (or depth of the slot).
Total MMF of the teeth,
Att = mean ordinate×length of the teeth (or depth of the slot)

= atmean × lt = atmean × ds

This method is applicable to all forms of teeth, with and without taper.
Bt1/3 method
It is based on the assumption that the mean MMF of whole tooth, is the MMF, ‘at’
corresponding to the flux density at 1/3 tooth height from the narrow end.
 Determination of
Determination of MMF
MMF of
of Teeth
Teeth  2.23

P1 = ds

Flux density

at

at mean

Distance from root

fig.
Fig. 2.12
2.12  ||  (a),
(a), (b)
(b) and
and (c)
(c) Graphical
Graphical method
method

Total MMF
Total of teeth,
MMFof teeth,
Att Att = at×
1/3 × lt = at1/3 × ds
= at1/3 lt = at1/3 × ds

where 1/3 is the value of MMF/m for Bt1/3

where at1/3 1/3
,, which
whichisisthe
the flux
flux density
density at
at 1/3
1/3 tooth
tooth height
height
from
from the
the narrow
narrow end.
end.
2.24
2.24  Design
Design ofofMagnetic
Magnetic Circuits
Circuits

It
It is
is aa simple
simple method,
method, applicable
applicable to to teeth
teeth with
with small
small taper.
taper. It
It is
is also
also suited
suited for
for operation
operation
under
under low low saturation.
saturation.
Simpson’s
Simpson’s rule rule (three
(three ordinate
ordinate method)
method)
This
This method is based on the
method is based on the formulation
formulation that
that the
the B-H
B-H curve
curve relating
relating thethe flux
flux density
density and
and
the
the MMF is a parabola. The total MMF of teeth is the mean value of MMF obtained at
MMF is a parabola. The total MMF of teeth is the mean value of MMF obtained at three
three
points
points of of the
the B-HB-H curve,
curve, which
which areare equidistant
equidistant as as shown
shown inin Fig.
Fig. 2.12,
2.12, with
with respect
respect to
to ends
ends ofof
the teeth and centre of the
the teeth and centre of the teeth. teeth.
From
From Fig.Fig. 2.13,
2.13, the
the mean
mean value
value of
of MMF
MMF is is given
given by
by
at + 4 at + at
atmean = =1at1 + 42at2 +3at3
atmean 6 6

where at
where at1,,atat2correspond
correspondtotoMMF
MMFatatends
ends
of of
thethe teeth
teeth andand
at3 at 3 corresponds
corresponds to the
to the mmfmmf at
at the
1 2
the centre of the
centre of the teeth. teeth.
It is
It is used
used for for teeth
teeth with
with small
small taper
taper and
and other
other primitive
primitive types.
types.

at

at3

at1
at2

Distance from root

dt = ds

fig.
Fig. 2.13
2.13  ||  Simpson’s
Simpson’s rule
rule

Example 2.5:
Example 2.5: 
Determine the
Determine the MMF
MMF of of tapered
tapered teeth
teeth of
of an
an electrical
electrical machine
machine using
using
Simpson’s rule following the data: length of teeth = 20 mm, maximum
Simpson’s rule following the data: length of teeth = 20 mm, maximum
width = 1.5
width = 1.5 times
times the
the minimum
minimum width,
width, mean
mean flux
flux density = 1.2
density = Wb/m22..
1.2 Wb/m
The B-at
The B-at curve
curve isis given
given by
by
 Determination of MMF of Teeth  2.25

B(wb/m2) 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7

‘at’ (A/m) 190 208 227 265 366 633 1490 3670

1.8

1.6
1.4
1.2
B (Wb/m2)

0.8

0.6

0.4

0.2

0 500 1000 1500 2000 2500 3000 3500 4000

H (AT/m)

Fig. 2.14  |  B-at curve

Solution:  Given
Length of teeth, lt = 30 mm
Maximum width = 1.5 minimum width
Mean following density = 1.2 Wb/m2
By Simpson’s rule,
at1 + 4 at2 + at3
atmean = (1)
6

We know that at2 = 227 A / m corresponding to Bt 2 = 1.2 Wb/m 2

In order to determine ‘at’ mean, it is required to find at1 and at3 , which in turn depends on
Bt 1 and Bt 3 , dependent on proportionality with tooth widths.
We know that,
wt1 = 1.5wt3 (2)

Let wt1 , wt2 and wt3 be the maximum, mean and minimum tooth widths.

wt1 + wt3
Also, wt2 = (3)
2
2.26  Design of Magnetic Circuits

Substituting Eq. (2) in Eq. (3), we get

1.54 wt3 + wt3
wt2 =
2
2.5wt3
= = 1.25wt3  (4)
2
And,

Flux density at nay region of teeth,

flux/teeth
Bt =
Sectional area of teeth

[where sectional area of teeth = net iron length × teeth width]

1
Hence, from the above equation, it is observed that Bt ∝ .
teeth width

1 1 1
Hence, Bt1 ∝ , Bt2 ∝ and Bt3 ∝
wt1 wt2 wt3
So, it can be stated that
Bt1 wt2 Bt2 wt3
= and = (5)
Bt2 wt1 Bt3 wt2

Substituting the value of Bt3 , Eqs. (2), (4) in Eq. (5), we get

Bt1 1.25 wt3 1.2 wt3

= and =
1.2 1.5 wt3 Bt3 1.25wt3

1.25
⇒ Bt1 = 1.2× = 1 Wb/m 2
1.5

and Bt3 = 1.2×1.25 = 1.5 Wb/m 2

From the B-at curve, the corresponding at1 and at3 to Bt1 and Bt3 are 190 AT/m and 633
AT/m.
Using at1 , at3 and at2 in Eq. (1), we get

190 + 4 (227 ) + 633

atmean =
6
= 288.5 AT / m
The total MMF required = atmean × length of teeth (lt)

= 288.5 × 0.03 = 8.655 A

 Determination of MMF of Teeth  2.27

Example 2.6: Determine theMMF of air gap and teeth using Simpson’s rule and Bt 1
3

method in a dc machine, having the following length of core = 170 mm,

diameter = 250 mm, number of slots = 26, type of slots-parallel, depth
of slot = 20 mm, width of slot = 5.5 mm, number of ducts = 1, duct
width = 12 mm, air gap length = 3.5 mm, maximum flux density =
1 Wb/m2, insulation of stampings = 0.12 times thickness of stampings
and Carter’s coefficient is 0.28 for slots and 0.35 for ducts.

The B-at curve is as follows:

B (Wb/m2) 0 0.8962 1.139 1.22 1.341 1.4957 1.503 1.625 1.6624 1.706 1.7194 1.8645 1.907
H (AT/m) 0 48.70 72.36 110.5 245.62 400 450 1809 2000 4000 4100 14000 15000

2
1.8
1.6
1.4
B (Wb/m2)

1.2
1
0.8
0.6
0.4
0.2

0 2000 4000 6000 8000 10000 12000 14000

H (AT/m)

Fig. 2.15  |  B-at curve

Solution:  Given
Length of core, L = 170 mm
Insulation of stampings = 0.12 thickness of stampings
Diameter of core, D = 250 mm
Number of slots, s = 26
Depth of slot, ds = 20 mm
Slot width, ws = 5.5 mm
Number of ducts, nd= 1
Duct width, wd = 12 mm
Air gap length, lg = 3.5 mm
Maximum flux density, Bg= 1 Wb/m2
Insulation of stampings = 0.12 (thickness of stampings)
2.28  Design of Magnetic Circuits

0.28 for slots

Carter’s coefficient = 
0.35 for ducts
In order to determine the MMF for air gap and teeth, it is required to find the total gap
contraction factor, flux density for different parts of tooth using tooth widths, slot pitch and
diameter in each case.
Gap contraction factor for slots,
ys
K gs =  (1)
y s − Kcs ws
In the above equation, slot pitch (gap surface)
πD
ys =
s
Substituting the values for D and s in the above equation, we get
π × 250
ys = = 30.2076 mm
26
Substituting the values of ys, Kcs and ws in Eq. (1), we get
30.2076
K gs =
30.2076 − (0.28 × 5.5)
= 1.0537
Similarly,
Gap contraction factor for ducts,
L
K gd =
L − Kcd nd wd
Substituting the values of L, Kcd, nd and wd in the above equation, we get
170
K gd =
170 − (0.35×1×12)
= 1.0253
We know that total gap contraction factor Kg=KgsKgd
Substituting the values of Kgs and Kgd in the above equation, we get

K g = 1.0537 ×1.00253 = 1.0803

We know that MMF of air gap,

ATg = 800 , 000 Bg K g lg

Substituting the values of Bg , K g and lg in the above equation, we get

ATg = 800 , 000 ×1×1.0803 × 3.5×10−3

= 3024.84 A
 Determination of MMF of Teeth  2.29

Proceeding to determine the MMF for teeth, we use Bt 1 method and Simpson’s rule.
3

Bt 1 method:
3

1
In this method, it is required to determine the flux density at of tooth height, for which
3
the calculations of wt1 , wt 1 , Dt 1 and Bt1 are to be done.
We know that 3 3

Bt 1
wt
3
= 1 (2)
Bt1 wt 1
3

where wt1 = y s − ws

Substituting the values of ys and ws in the above equation, we get

wt1 = 30.2076 − 5.5 = 24.776 mm

And
wt 1 = y s 1 − ws  (3)
3 2

where
πD 1
3
ys1 =  (4)
3
S
2
and D 1 = D − 2× ds
3
3

Substituting the values of D and ds in the above equation, we get

 2 
D 1 = 250 − 2× × 20
 3 
3
= 223.3333 mm
Substituting the values of D 1 and s in Eq. (4), we get
3
π × 223.3333
ys =
1 26
3

= 26.9854 mm

Substituting the values of y s 1 and ws in Eq. (3), we get

3
wt 1 = 26.9854 − 5.5 = 21.4854 mm
3
2.30  Design of Magnetic Circuits

Bt 1 is determined as follows:
φt
Bt 1 = (5)
Li wt1

where φt = Bg y s L

Li = Ki (L − nd wd )

wt1 = y s − ws

Therefore,

φt = 1× 30.2076 ×10−3 ×170 ×10−3

= 5.1352 mWb

And Li = Ki (170 ×10−3 − (1×12)×10−3 )

 insulation of stampings 
where Ki = 0.88 ∵  = 0.12
 thickness of stampings 
 
⇒ Li = 0.88(170 × 10-3 − 12 × 10-3)
= 0.1390
And
wt1 = y s − ws = 30.2076 − 5.5

= 24.7076 ×10−3 m = 0.02470 m

Substituting the values of φt , Li and wt1 in Eq. (5), we get

5.1362×10−3
Bt1 = = 1.4957 Wb/m 2  (6)
0.1390 ×.00247

Substituting the values of wt 1 , wt1 and Bt1 in Eq. (2), we get

3

Bt 1
0.02470
3
=
1.4957 21.4854 ×10−3

⇒ Bt 1 = 1.7194 Wb/m 2
3

From the following table, the corresponding MMF is

at 1 = 4100 A/m
3
 Determination of MMF of Teeth  2.31

MMF for teeth,

AT = 4100 × ds = 4100 × 20 × 10−3 = 824

Total MMF of air gap and teeth = 3024.84 + 82 = 3106.84 A

Simpson’s rule:

Top of teeth Centre portion of tooth Bottom of tooth

From Eq. (6), Dcentre or (Ds2 ) = D − ds Dbottom or (Ds3 ) = D − 2ds
Bt1 = 1.4957 Wb/m 2 = 250 − 20 = 250 − 2 (20)
corresponding = 230 mm = 210 mm
at1 = 400 A/m
πDcentre πDbottom
y s2 = y s2 =
S S
π × 230 π × 210
=   =  
26 26
= 2.791 mm = 25.3744 mm

wt2 = y s2 − ws wt3 = y s3 − ws
= 27.791 − 5.5 1 = 25.3744 − 5.5
= 22.291 mm = 19.8744 mm

wt1 wt1
Bt2 = Bt1 × Bt3 = Bt1 ×
wt2 wt3
24.776 24.776
= 1.4957 × = 1.4957 ×
22.291 19.8744
= 1.6624 Wb/m 2 = 1.8645 Wb/m 2

Corresponding Corresponding

at2 = 2000 AT/m at3 = 14000 AT/m

Therefore,
at1 + 4 at2 + at3
atmean =
6
400 + ( 4 × 20000) + 14000
=
6
= 3733.3333 A/m
MMF for teeth = 3733.3333 × ds
2.32 Design of Magnetic Circuits
2.32
2.32  Design
Design ofof
Magnetic Circuits
Magnetic Circuits
−3
= 3733.3333 × 20 × 10-3
= 3733.3333×
=3733.3333 ×2020×
×1010−3
= 74.6666 A
=
=74.6666
74.6666AA
Total MMF
Total MMF of
of air
air gap
gap and
and teeth = 3024.84
teeth = + 74.6666
3024.84 + 74.6666
Total MMF of air gap and teeth = 3024.84 + 74.6666
= 3099.5066 A
= 3099.5066 A

2.4 
2.4 Realreal Flux
flux Density
Density and and Apparent
apparent Flux
flux Density
Density
2.4 real flux Density and apparent flux Density
In case of high flux density in teeth, the MMF is quite large and acts across the slots with
In case of high flux density in teeth, the MMF is quite large and acts across the slots with
slots
slots positioned
In case
positioned parallel
of high flux
parallel to
to teeth.
density Hence,
in teeth,
teeth. Hence, thethisMMF
this slot
slotisflux under
quite
flux large
under saturation
and acts conditions
saturation across the cannot
conditions be
slots with
cannot be
neglected
slots or (omitted)
positioned in
parallel calculation
to teeth. of
Hence, flux density.
this slot This
flux leads
under to two different
saturation flux
conditions
neglected or (omitted) in calculation of flux density. This leads to two different flux densities, densities,
cannot be
namely
namely real and apparent flux densities represented in Fig. 2.16, where real flux density is
real
neglected orand apparent
(omitted) in flux densities
calculation of represented
flux density. in
This Fig.
leads2.16,
to where
two real
differentflux
fluxdensity
densities,
is
less
less than
namely apparent
thanreal flux
flux density
and apparent
apparent flux at
density all
all times.
densities
at times. represented in Fig. 2.16, where real flux density is
less than apparent flux density at all times.
Ideal condition Links and flows in
Ideal condition Links
ironand flows in
paths
iron paths
Flux from stator to
Flux through
rotor from stator to
airgap
rotor through airgap Takes an alternative
Takes an alternative
path through slot
Practical condition path through slot
Practical condition
fig. 2.16 
Fig. 2.16 | 
| Real
Real and
and apparent
apparent flux
flux densities
densities
fig. 2.16 | Real and apparent flux densities
The flux
The flux linking
linking the
the slot
slot and
and teeth
teeth under
under practical
practical conditions
conditions is is represented
represented in
in Fig.
Fig. 2.17.
2.17.
The flux linking the slot and teeth under practical conditions is represented in Fig. 2.17.
yS
yS

Air gap lg L
Air gap lg L

Tooth Li
Slot Tooth Li
Slot

wS wt
wS wt

fig. 2.17 | Flux distribution in teeth and slot under practical conditions
fig. 2.17 
Fig. 2.17 | | Flux
Flux distribution
distribution inin teeth
teeth and
and slot
slot under
under practical
practical conditions
conditions
The corresponding representation of real and apparent flux density is as follows:
Thecorresponding
The
The corresponding
real flux density representation
representation
is defined as the ofreal
of real
ratioand
and apparent
of apparent
actual flux
fluxflux density
density
in tooth to is isas
theasarea
follows:
follows:
of tooth.
Thereal
The realflux
fluxdensity
densityisisdefined
definedas asthe
theratio
ratioof ofactual
actualfluxfluxin intooth
toothto tothe
thearea
areaofoftooth.
tooth.
 Real Flux Density and Apparent Flux Density  2.33

Actual flux in tooth

Real flux density Breal =  (2.21)
Area of tooth

The apparent flux density is defined as the ratio of total flux in slot pitch to the area of tooth.

Total flux in slot pitch

Apparent flux density Bapp = (2.22)
Area of tooth

Proceeding to find the relation between real and apparent flux densities, the following
steps are followed.
We know that
Area of tooth (iron path),
Ai = tooth width × net iron length = Liwt (2.23)
Area of air path, Aa = Total area – Area of iron (or tooth)
= [Core length × slot pitch] − Liwt
Ag = Lys − Liwt (2.24)
Total flux in slot pitch, fs = fi + fa(2.25)
where fi – flux in iron path, fa – flux in air path.
From Eq. (2.25), we can state that fi – Actual flux in tooth
Using Eq. (2.25) in Eq. (2.22), we get
φs φi + φa φ φ
Bapp = = = i + a
Ai Ai Ai Ai
φa  φi 
= Breal + ∵ = Breal 
Ai  Ai 

Multiplying and dividing by Aa for the second term in the above equation, we get
φa Aa
= Breal + ×
Ai Aa
φa Aa
= Breal + ×
Aa Ai
Aa
= Breal + Ba ×
Ai

[where Ba – flux density in air = μ0H = μ0atreal]

Bapp = Breal + Ba K

Aa Ly s − Li wt
[where K = = ratio of air area to iron area]
Ai Li wt
2.34  Design of Magnetic Circuits

Also,

Bapp = Breal + Ba (K s − 1) (2.26)

where
Total area
Ks = K + 1 =
Iron area

Ly s
=
Li wt

Hence,
Breal = Bapp − Ba (K s − 1) (2.27)

Substituting Ba = μ0 atreal in the above equation, we get

Breal = Bapp − μ0 atreal (K s − 1) (2.28)

Example 2.7: Determine the permeability of teeth of dc machine with length of core

(gross) = 350 mm, slot pitch = 20 mm, width of the teeth = 10 mm, real flux
density = 2.2 T, apparent flux density = 2.5 T and stacking factor = 0.85.

Solution:  Given
Gross length of core, L= 350 mm
Slot pitch, ys= 20 mm
Width of the teeth, wt = 10 mm
Real flux density, Breal = 2.2 T
Apparent flux density, Bapp= 2.5 T
Stacking factor, Sf = 0.85
In order to determine the permeability, it is required to find MMF for which the calculation
of Ks is required.

Bapp = Breal + Ba (K s − 1) (1)

where
Ba = μ0 × at (2)

Ly s
Ks = (3)
Li wt

and Li = S f L (4)

Substituting Eq. (4) in Eq. (3), we get

Ly s y
⇒ Ks = = s
S f Lwt S f wt

20
= = 2.2222
0.9×10
 Real Flux Density and Apparent Flux Density  2.35

Substituting Ks and Ba in Eq.(1), we get

Bapp = Breal + μ0 at(2.2222 − 1)

Bapp − Breal
⇒ 1.2222 at =
μ0

Substituting the values for Bapp, Breal and μ0 in the above equation, we get
2.5 − 2.2
at =
1.2222× 4π ×10−7
= 195330.0725 AT/m
Permeability of teeth,
Breal 2.2
μ= =
at 195330.0725

= 11.2629×10−6 H/m

= 11.2629 μH/m.

Example 2.8: Determine the apparent flux density for a teeth of dc machine with
length of core = 400 mm, slot width = 11 mm, slot pitch = 20 mm,
number of ducts = 6, duct width = 8 mm, stacking factor = 0.85, real
flux density = 1.8 Wb/m2 and MMF = 80000 AT/m.

Solution:  Given
Length of core, L = 400 mm
Slot width, wt = 11 mm
Slot pitch, ys= 20 mm
Number of ducts, nd = 6
Duct width, wd = 8 mm
Stacking factor, Sf = 0.85
Real flux density, Breal = 1.8 Wb/m2
MMF, at= 80000 AT/m
In order to determine the apparent flux density, it is required to find Ks, for which the net
length of iron and tooth width are to be found.
We know that
Net length of iron,
Li = Ki(L – ndwd) = 0.85(400 − 6 × 8)
= 299.2 mm
Tooth width, wt = ys − ws = 20 − 11 = 9 mm
Ly s
Also, Ks =
Li wt
2.36  Design of Magnetic Circuits
2.36 Design of Magnetic Circuits
Substituting the values of L, ys, Li and wt in the above equation, we get
Substituting the values of L, ys, Li and wt 400
in the above equation, we get
× 20
Ks = = 2.9708
299 2×
400.× 209
Ks = = 2.9708
Apparent flux density, 299.2× 9
Apparent flux density,
Bapp = Breal + μoat (Ks− 1)
Bapp = Breal + µoat (Ks− 1)
Substituting the values of Breal, μo at and Ks in the above equation, we get
Substituting the values of Breal, µo at and Ks in the above equation, we get
B = 1.8 + (( 4π ×10− 7 
−7 ))× 80000 ((2.9708 − 1))
app = 1.8 +  4π ×10
Bapp × 80000 2.9708 − 1 
= 1.9981
= Wb/m22
1.9981 Wb/m

2.5 
2.5 Iron
iron Loss
Loss Calculation
calculation
When
When aa core
core of
of ferromagnetic
ferromagnetic material
material isis subjected
subjected toto aa changing
changing magnetic
magnetic field,
field, as
as in
in the
the
case
case of transformers, induction motors and alternators, some of the power transferred is lost
of transformers, induction motors and alternators, some of the power transferred is lost
in
in the
the core.
core. These
These losses
losses are
are called
called core
core losses
losses or
or iron
iron losses.
losses. The
The two
two components
components of of iron
iron or
or
core loss are as follows:
core loss are as follows:
(a) (a) Hysteresis
Hysteresis lossloss
(b)(b)
EddyEddy current
current lossloss

2.5.1 
2.5.1 Hysteresis
Hysteresis Loss
Loss
The
The repeated
repeated (or
(or cyclic)
cyclic) magnetization
magnetization ofof ferromagnetic
ferromagnetic material
material results
results in
in loss
loss termed
termed asas
hysteresis
hysteresis loss,
loss, which
which is
is proportional
proportional to
to area
area of
of the
the hysteresis
hysteresis loop
loop as
as represented
represented in in Fig.
Fig. 2.18
2.18
and
and the
the quality
quality ofof the
the material.
material.

B
a

c
H
H 0 f

e
d

fig. 2.18 
Fig. 2.18 | 
| Hysteresis
Hysteresis loop
loop of
of aa ferromagnetic
ferromagnetic material
material

Hysteresis loss,
Hysteresis loss,
h =
php= KAf (2.29)
KAf (2.29)
 Iron
Iron Loss Calculation 2.37
Loss Calculation  2.37

where
where K K –– constant
constantaccounting
accountingfor
forthe
thequantities,
quantities, AA ––area
areaof
ofloop abcdef, ff ––frequency.
loopabcdef, frequency.
For the Steinmetz relationship,
For the Steinmetz relationship,
Hysteresis
Hysteresis loss,
loss,
kk (2.30)
phh = Khh f Bm
m (2.30)

where B
where Bm –– maximum
maximum flux
flux density,
density, K
K h –– hysteresis
hysteresis coefficient,
coefficient, kk –– Steinmetz
Steinmetz coefficient,
coefficient,
m h
varies between 1.5 and 2.5 and f – frequency.
varies between 1.5 and 2.5 and f – frequency.
Hysteresis loss
Hysteresis loss is
is expressed
expressed in W/m33 or
in W/m or W/kg.
W/kg.
Hysteresis loss
Hysteresis loss can
can be
be minimized
minimized in
in the
the following
following ways:
ways:
•• Use
• Use of
of air
air core
core transformer
transformer reduces
reduces hysteresis
hysteresis loss,
loss, but
but increases
increases leakage
leakage flux.
flux.
•• Use of soft magnetic materials such as silicon steel, steel alloys, ferrite, etc., with low
coercivity and remanent magnetic flux density reduces hysteresis loss.

2.5.2 
2.5.2 Eddy
eddy Current
current Loss
According
According to to Faraday’s
Faraday’s law
law of
of electromagnetic
electromagnetic induction,
induction, when
when an
an alternating
alternating magnetic
magnetic field
field
is
is applied to any magnetic material, an emf is induced in the material. The emf circulates
applied to any magnetic material, an emf is induced in the material. The emf circulates
currents
currents inin the
the material.
material. These
These circulating
circulating currents
currents are
are called
called Eddy
Eddy Currents,
Currents, produces
produces aa loss
loss
(I22R loss) in the magnetic material known as an Eddy Current Loss.
(I R loss) in the magnetic material known as an Eddy Current Loss.
Eddy
Eddy current
current losses
losses can
can be
be minimized
minimized in in the
the following
following ways:
ways:
•
•• Use
Use of
of laminated
laminated cores
cores
•• Reduction of thickness of stampings
• Reduction of thickness of stampings
Consider a thin sheet as shown in Fig. 2.19, which when subjected to a magnetic flux,
Consider a thin sheet as shown in Fig. 2.19, which when subjected to a magnetic flux,
causes flow of eddy currents.
causes flow of eddy currents.

a b
h I
d c
l
y
w

fig. 2.19 
Fig. 2.19 | 
| Eddy
Eddy currents
currents in
in thin
thin plates
plates of
of laminations
laminations

The loss contributed by the eddy current is calculated as follows.

The loss contributed by the eddy current is calculated as follows.
Flux in the path abcda,
Flux in the path abcda,
φ = B × A (2.31)
φ = B × A(2.31)
where B = Bm sin ωt
where B = Bm sin ωt
2.38  Design of Magnetic Circuits

A = h ×( y + y ) = 2 hy

Substituting the values of B and A in Eq. (2.31), we get

φ = 2Bm hy sin ωt

Instantaneous emf induced in path abcda,

dφ
e=
dt
d
=
dt
(2Bm hy sin ωt)
= 2Bm hy cos ωt

= em cos ωt (2.32)

where em = 2Bmhy ω

RMS value of emf induced,

em
E=
2
= 2Bm hy ω  (2.33)

Resistance of eddy current path,

ρ × length
R=
area
ρ ×( h + h + y + y + y + y )
=
l × dy
ρ ×(2 h + 4 y )
=
l × dy
2ρh
 [ ∵ h  y ] (2.34)
ldy

Eddy current,
E
Ieddy =
R
2Bm h y ω
=
2ρ h
l × dy
 Iron Loss Calculation  2.39

Bm y dy l ω
=
2ρ
Substituting ω = 2π f in the above equation, we get
Bm y dy l ×2π f
I eddy =
2ρ
2Bm y dy lπ f
= (2.35)
ρ
Eddy current loss in the path abcda,

dPeddy = I 2 R
2
2Bm 2 y 2 (dy ) l 2 π 2 f 2 2ρ h
= ×
2 ldy
ρ

4Bm 2 l π 2 f 2 h y 2 dy
=
ρ
Total eddy current loss,

Peddy = ∫ dPeddy
w
y=
2
4Bm 2 lπ 2 f 2 h y 2 dy
= ∫ ρ
y =0

w
4B 2 lπ 2 f 2 h  y 3  2
= m  3 
ρ  0

4Bm 2 lπ 2 f 2 h w 3
= ×
ρ 3×8

Bm 2 π 2 f 2 l h w 3
Peddy = (2.36)
6ρ
We know that volume of plate,

V=l×h×w

Hence, eddy current loss per unit volume is given by

Bm 2 π 2 f 2 l h w 3 1
Peddy = ×
6ρ lhw
2.40  Design of Magnetic Circuits

π 2 Bm 2 f 2 w 2
=
6ρ

= ke f 2 Bm 2 (2.37)

π2w2
where ke =
6ρ

2.5.3  Total Iron or Core Loss

The total iron or core loss is given by the sum of hysteresis and eddy current loss and is
expressed as

k
pi = ph + pe = K h f Bm + K e f 2 Bm
2
(2.38)

2.5.4  Pulsation Loss

Apart from the hysteresis and eddy current loss, air gap flux pulsations in electric machine
cause losses termed as pulsation loss. These losses are mainly present due to slotted armature
causing changes in reluctance, leading to change in air gap flux changes. The losses are
pronounced in induction machines, where the air gap is small compared to slot openings,
leading to harmonic fields with high frequencies.

Example 2.9: Determine the specific iron loss of alloy steel with the following data:
frequency = 60 Hz, maximum flux density = 2.5 Wb/m2, thickness of
sheets = 0.3 mm, resistivity = 0.2 μΩm, density = 5 × 103 kg/m2 and
hysteresis loss/cycle = 300 J/m3.

Solution:  Given,
Frequency, f = 60 Hz
Maximum flux density, Bm = 2.5 Wb/m2
Thickness of sheets, t = 0.3 mm= 0.3 × 10−3 m
Resistivity, ρ = 0.2 × 10−6 Ωm
Density = 5 × 103 kg/m3
Hysteresis loss/cycle = 300 J/m3
We know that
Eddy current loss,
π 2t 2 f 2 Bm 2
Peddy =
ρ

Substituting the values of t, f, Bm and ρ in the above equation, we get

2
π 2 × 60 2 × 2.52 ×(0.3 ×10−3 )
Peddy =
6 × 0.2×10−6
 Iron Loss Calculation  2.41

= 16654.9574 W

Hysteresis loss,
Ph = 300 × 60 = 1800 W
Total iron loss,
Ptotal = Pi + Ph = 16654.9574 + 1800

= 18454.9574 W

Ptotal
Specific iron loss =
density
18454.9574
=
5×10 3
= 3.6909 W/kg

16654.9574
Peddy (in W/kg) =
5×10 3
= 3.3309 W/kg

3000
Ph (in W/kg ) = = 0.6 W/kg
5×10 3

Example 2.10: Determine the hysteresis coefficient for silicon steel with hysteresis
loss of 3 W/kg at a frequency of 60 Hz, maximum flux density of
2 Wb/m2 and specific gravity of 7.55. Also determine the hysteresis
loss kg at a frequency of 30 Hz and flux density of 1 Wb/m2. Use
Steinmetz coefficient as 1.6.

Solution:  Given
Hysteresis loss = 3 W/kg
Frequency, f = 60 Hz
Maximum flux density, Bm = 2 Wb/m2
Specific gravity = 7.55

⇒ Density = 7.55 × 103 kg/m3

We know that
Hysteresis loss,
k
Ph = k h f Bm
Hysteresis loss/kg,
k
Ph = k h f Bm ×7500
2.42  Design of Magnetic Circuits

Substituting the values of Ph, f, k and Bm in the above equation, we get

3 = k h × 60 × 21.6 × 7500

⇒ k h = 2.199×10−6

Hence, hysteresis loss/kg at 30 Hz, and flux density of Bm = 1 Wb/m 2 is

1.6
Ph = k h f Bm ×7500

= 2.199 × 10−6 × 30 × 11.6 × 7500

= 0.494775 W/kg

Example 2.11: The total iron loss for a synchronous is 300 W at 500 rpm and 200 W at
400 rpm. Determine the total iron loss if the thickness of laminations
is increased by 50%, the maximum flux density decreased by 10% and
the speed is at 250 rpm. Use Steinmetz coefficient as 2.

Solution:  Given
Total iron loss at 500 rpm, Ptotal = 300 W

Total iron loss at 400 rpm, Ptotal = 200 W

We know that
Total iron loss,

Ptotal = Ph + Peddy (1)

k 2
where Ph = k h f Bm = k h fBm [ ∵ k = 2] (2)

Peddy = ke f 2 Bm 2 (3)

Also,
We know that

Speed of machine,
120 f
N=
P
PN
⇒ f=
120
P
⇒ f = xN, where x =
120
Using the value of f in Eqs. (2) and (3), we get
2
Ph = k h xNBm = k h′ N  (4)
 Magnetic Leakage  2.43

Peddy = ke x 2 N 2 Bm
2
= ke′ N 2  (5)

where k h′ = k h xBm and ke′ = ke x 2 Bm 2 are constants with B and x are fixed.
Using Eqs. (4) and (5) in Eq. (1), we get

Ptotal = K h′ N + ke′ N 2 (5a)

Substituting Ptotal for different values of speed, we get

2
300 = k h′ (500) + ke′ (500)  (6)

2
200 = k h′ ( 400) + ke′ ( 400)  (7)

By solving Eqs. (6) and (7), we get

k h′ = 0.1

ke′ = 0.001
Proceeding to determine the total iron loss for 0.9 Bm, 1.5t and speed of 1500 rpm and using
k h′ and ke′ in Eq. (5a), we get
2 2 2 2
Ptotal = 0.1×(250)×(0.9) + 0.001×(250) ×(0.9) (1.5)

= 134.1562 W

2.6  Magnetic Leakage

Not all flux flowing in a magnetic circuit is useful. Some amount of flux does not link the
components of the magnetic circuit contributing towards the leakage flux, which affects the
performance of the machine. The leakage flux of alternating nature can induce a voltage termed
as leakage reactance voltage. Table 2.5 describes the effects of leakage flux in various machines.

Table 2.5  |  Effects of leakage flux in various machines

Type of machine Effect of leakage flux
DC machine Makes commutation difficult due to reactance voltage
Salient pole Modifies the field excitation demand
synchronous machine
Transformer •  Affects the voltage regulation
•  May cause circulating currents to flow in tank walls
• Causes forces to develop between windings in short circuit
conditions
Induction machine Causes force to develop between windings in short circuit conditions
Alternator Affects the voltage regulation
2.44   Design
Design ofofMagnetic
MagneticCircuits
Circuits

The leakage
The leakage flux
flux in
in armature
armature ofof the
the rotating
rotating machines
machines can
can be
be classified
classified into
into the
the following
following
categories as shown in Fig. 2.20.
categories as shown in Fig. 2.20.

Tooth top

Skew Zigzag

Types of
leakage
flux Differential
Overhang or
harmonic

Slot Peripheral

Fig. 2.20
fig. 2.20  ||  Types of armature leakage flux

•• Tooth top leakage flux

• Tooth top leakage flux
It passes from top of one teeth to another teeth’s top as shown in Fig. 2.21. It is pronounced
It passes from top of one teeth to another teeth’s top as shown in Fig. 2.21. It is pronounced
in dc and synchronous machines, since the machines have a large air gap.
in dc and synchronous machines, since the machines have a large air gap.
Flux
Flux
Airgap
Airgap

Teeth
Teeth

Fig. 2.21  |  Tooth top leakage flux

fig. 2.21 | Tooth top leakage flux
 Magnetic Leakage  2.45

•• Zig-zag leakage flux

It passes from one teeth to another teeth in zig-zag manner as shown in Fig. 2.22. Its
existence is based on relative to the positions of tooth tips of stator and rotor and length of
air gap.

Flux

Stator

Teeth

Slot

Airgap

Teeth
Slot

Rotor

Fig. 2.22  |  Zig-zag leakage flux

•• Differential or harmonic or belt leakage flux

It is present due to dissimilarity in harmonic contents of MMF of primary (stator)
and secondary (rotor), where spatial distributions between MMFs are not the same as
shown in Fig. 2.21. Though squirrel cage induction machines are devoid of this flux, due
to absence of harmonic leakage flux, its contribution is significant in other machines
(Fig. 2.23).

•• Peripheral leakage flux

It flows in the air gap throughout the circumference. It does not link any windings. It is
insignificant in almost all the machines.

•• Slot leakage flux

It takes a path travelling from one teeth to another teeth traversing the slot, linking the
slice of conductor beneath it coming back through iron as shown in Fig. 2.22. It is present in
all machines both ac and dc (Fig. 2.24).
2.46
2.46  Design
2.46 Design
Designofof
Magnetic
of Circuits
Magnetic
Magnetic Circuits
Circuits

Primary
Primary
winding
winding
II11

Air gap
Air gap
Secondary
Secondary
winding
winding

ll 2l
2l ll ll 2l
2l ll

8I1
8I 8I1
8I MMF
MMF
1 1
5I1
5I 6I1
6I Unbalanced
Unbalanced
1 1

5I2
5I 2I2
2I
2
5I2
2 6I2
6I
5I2
2

(a) Similar
(a) Similar mmf
mmf distribution
distribution (b) Dissimilar
(b) Dissimilar mmf
mmf distribution
distribution

fig. 2.23
fig.
Fig. 2.23 || 
2.23  | Differential
Differential leakage
leakage flux
flux for
for (a)
(a) similar
similar and
and (b)
(b) dissimilar
dissimilar MMF
MMF distribution
distribution

Slot
Slot
Teeth
Teeth

(a)
(a)

flux flux
flux
flux
Teeth
Teeth Teeth
Teeth Teeth
Teeth Teeth
Teeth

Conductor
Conductor
Conductor
Conductor

Slot
Slot Slot
Slot

(b)
(b) (c)
(c)

fig. 2.24
Fig.
fig. 2.24 || 
2.24  | (a)
(a) General
General representation
representation of
of slot
slot leakage
leakage flux,
flux, slot
slot leakage
leakage flux
flux in
in (b)
(b) stator
stator and
and (c)
(c) rotor
rotor
 Estimation
Estimation of
of Specific
Specific Permeance
Permeance and
and Leakage Reactance 2.47
Leakage Reactance  2.47
•
•• Overhang
Overhang leakage
leakage flux
flux
It
It is
is present
present due
due toto overhang
overhang region
region of
of armature
armature windings.
windings. It
It is
is aa unique
unique type
type of
of flux,
flux,
present due to grouping of overhang and vicinity of metal masses (including core
present due to grouping of overhang and vicinity of metal masses (including core stiffnessstiffness
and
and end
end covers)
covers) as
as shown
shown inin Fig.
Fig. 2.25.
2.25.

φ0

Overhang

Overhang

Overhang
leakage flux

Coil sides

fig. 2.25 
Fig. 2.25 | 
| Overhang
Overhang leakage
leakage flux
flux

•
•• Skew
Skew leakage
leakage flux
flux
It
It is
is present
present in
in skewed
skewed slots
slots of
of motors.
motors. It
It is
is unique
unique to
to induction
induction machines
machines in
in which
which rotors
rotors
are
are skewed.
skewed.

2.7 Estimation
2.7 estimation of Specific
specific Permeance and Leakage Reactance
reactance
The
The following
following assumptions
assumptions are
are made
made in
in estimation
estimation of
of specific
specific permeance
permeance and
and leakage
leakage
reactance
reactance of
of slots:
slots:
•
•• Uniform
Uniform distribution
distribution ofof current
current throughout
throughout thethe area
area of
of slot
slot conductors.
conductors.
•
•• The
The path of leakage flux through the slot and around the iron
path of leakage flux through the slot and around the iron at
at the
the bottom
bottom is
is straight.
straight.
•
•• Determination of permeance is performed for air
Determination of permeance is performed for air paths. paths.
•
•• The
The reluctance
reluctance of
of iron
iron is
is assumed
assumed to to be
be zero.
zero.
2.48
2.48  Design
DesignofofMagnetic
MagneticCircuits
Circuits

The
The cross-section
cross-section of
of non-conductor
non-conductor portion
portion of
of the
the slot
slot with
with the
the leakage
leakage flux
flux is
is shown
shown in
in
Fig. 2.26.
Fig. 2.26.

dx h
x

fig. 2.26 | Leakage flux representation in non-conductor portion of the slot

Fig. 2.26  |  Leakage flux representation in non-conductor portion of the slot
From Fig. 2.26,
From Fig. 2.26,
L – Depth of flux path, h – Height of flux path, y – Length of flux path.
L – Depth of flux path
On considering a small length dx in the region as shown in Fig. 2.26, the permeance of
h – Height of flux path
section is given by
y – Length of flux path
On considering a small length dx∧in = Ldx as shown in Fig. 2.26, the permeance of
theµ0region
section is given by
∫ sa
y
Ldxof slot is calculated by integrating the

above equation and is given by
∫
The total permeance of non-conductor
∧sa portion
= μ0
y

The total permeance of non-conductorhportion h

Ldx of slot isdxcalculated by integrating the
above equation and is given by ∧sa = ∫ µ0 = µ0 L ∫
y y
0 h 0 h
Ldx dx
∧sa
The specific permeance, defined as the = ∫ μ0 per
permeance μ0 L ∫length of slot (or armature) or
=unit
y y
depth of flux path is given by 0 0

The specific permeance, defined as the permeance h per unit length of slot (or armature) or
∧sa dx
depth of flux path is given by λsa = = µ0 ∫ (2.39)
L y
0 h
∧sa dx

where Zs – Total number of conductors per λsa = = μZ0 ∫– Number
slotLand  (2.39)
of conductors till height,
x y
x, from the bottom of slot. 0
The flux
where Zs(dfx) in this
– Total region
number ofisconductors
given by per slot and Zx – Number of conductors till height,
x, from the bottom of slot.
dφx = MMF × permeance
The flux (dfx) in this region is given by
dφ = MMF × permeance
Ldx
= IxZ Zx µo
y Ldx
= I Z Zx μo
Flux linkage linked with Zx conductors is given by y
Flux linkage linked with Zx conductors
dλ is= given
Z dφ by
Zx x x
dλZx = Zx dφx
 Estimation of Specific Permeance and Leakage Reactance  2.49

 Ldx 
= Zx I z Zx μ0 
 y 
dx
= μ0 Zx2 LI z
y

The total flux linkage in the conductor portion of slot is calculated by integrating the
above equation and is given by
h
dx
λZx = ∫ μ0 Zx2 LI z
y
0
h
dx
= μ0 LI z ∫ Zx2
y
0

Hence,
λzx  Total flux linkage 
effective flux =  
Zs  Total number of conductors per slot 

h
μ0 LI z 2 dx
=
Zs ∫ Zx y
0

The total permeance of conductor portion of slot is given by

Effective flux
λsc =
Total MMF
h
μ0 LI z 2 dx
Zs ∫ Zx y
0
=
I z Zs
h 2
 Z  dx
= μ0 L ∫  x 
 Zs  y
0

The specific permeance of conductor portion of a slot is given by

h 2
 Z  dx
λsc = μ0 ∫  x  (2.40)
 Zs  y
0

2.7.1  Parallel-sided Slot

A parallel-sided slot is represented in Fig. 2.27.
2.50
2.50  Design
DesignofofMagnetic
MagneticCircuits
Circuits

w0

h4

h3

w1
h2

h1 dx

wS

fig. 2.27 
Fig. 2.27 | 
| Parallel-sided
Parallel-sided slot
slot

The specific
The specific permeance
permeance is
is calculated
calculated as
as shown
shown in
in Table.
Table. 2.6.
2.6.

table 2.6 | Specific permeance calculation

Table 2.6  |  Specific permeance calculation
Parameters λ (or λ )
sa y h Zx Derivation
Parameters λsa (or λscsc) y hZx Derivation
Conductor λsc =λ1 ws h1 xxZs x 2
Conductor λsc = λ1 ws h1 h1 Zs h1  xZs  2 h1 2
portion h h Z  dx xh dx
portion h1 λ1 = µ0 ∫ 1 1h1  s  dx = µ0 ∫ 21 x 2 dx
 s
0 ∫ 
λ1 = μ0  Zs  w
Zs  ws
= μ0h1 ws
0 ∫ h12 ws
0h 0
3  h1
µ0  x 
1
µ
= 2 μ ∫ hx 1
2
dx =  3  h1
2 μ0 3  x 
2
=h1 ws 0 ∫ x dx h=1 ws   0
 
h12 ws 30 h12 ws  3  0
µ  h  h
= 2 0  1  3= µ0 1
h wμ0  3 h1  3 ws h1
=1 s   =μ 0
h12 ws  3  3 ws
Non- λsa =λ2 ws h2 - h2 h2
Non-
conductor λsa = λ2 ws h2 – dx µ0
λ2 = µ0 ∫ h2 dx= μ∫ dxh2
conductor w w 0
portion
portion
λ =μ2 s
0 =∫w
s dx
ws
0 ∫
µ 0h s 0
= 0 [ x ]02
μ
=ws 0 [ x ]02
h
whs
= µ0 2
wsh2
= μ0
ws
(Continued)

(Continued)
 Estimation of Specific Permeance and Leakage Reactance  2.51
Estimation of Specific Permeance and Leakage Reactance 2.51
Table 2.6  | (Continued)
table 2.6 | (Continued)
Parameters λsa (or λsc) y h Z Derivation
Parameters λsa (or λsc) y h Zx x Derivation
λsa = λ3 w +w h – h3 h3
wo o+ ws s h 3 - h3 dx 2μ h3
= 2µ0 0 ∫ dx
λ3 3= µ0 ∫0 ∫ wo + w=s
λsa =λ3 2 3 λ =μ dx
wo +ows ∫ s 0
2 w + w dx
0wo + ws
2 2
0 0

2µ20μ0 [hx3 ]h3 = μ 2h32 h3

== [ x ] 0= µ0 0
w
wo + o+wsws 0 +ow+
wo w w
s s

λ
λsa = λ4 wwo o h4h4 - – h h
sa =λ4 h4 4 dx h4 4
dx =µ0 μ0 dx
λ4 4= µ0 ∫0 ∫ w= w∫ dx
λ = μ ∫
w o w o
0 0 o o 0 0
µ μ0 h h4 = μh4 h4
== 0 [ x[]0x4]0= µ0 0 w
wwo o wo o

From Table 2.6, the total specific permeance is given by

λs = λ1 +λλs 2=+λλ13++λλ2 4+ λ3 + λ4
 h h h1 2h23 2hh  h 
= µ0  1=+ μ0 2 + + + + 43 + 4 
 3w  +sws wow+o ws wo 
 s w s3 ws wo w

2.7.2 
2.7.2 Parallel-sided Slot
slot with Double Layer Windings
A
A parallel-sided
parallel-sided slot
slot with
with double
double layer
layer windings
windings is
is represented
represented in
in Fig.
Fig. 2.28.
2.28.

w0

h3

h4

h3 Top coil
side

h1 A
dx

h2

h1 B
Bottom
coil side
wS

fig. 2.28 
Fig. 2.28 | 
| Parallel-sided slot with double layer windings

The specific permeance is calculated as shown in Table 2.7.

2.52  Design of Magnetic Circuits

Table 2.7  |  Specific permeance calculation

Parameters λsa (or λsc) y h Zx Derivation
x x 2
Conductor λsc = λ1 ws h1 Zs h1  Zs 
portion (A) h1 h  dx
λ1 = μ0 ∫  1 
 Zs  ws
0
h1 h1
x 2 dx μ 2
= μ0 ∫ = ∫x dx
0
h12 ws h12 ws 0

μ0  x 3  h1
=  
 3 
h12 ws 0

μ0  h13  h1
=  3  = μ0 3 w
2
h1 ws   s

x x 2
Conductor λsc =λ3 ws h3 Zs h3  Zs  h3 2
portion (B) h3 h  dx x dx
λ1 = μ0 ∫  3  = μ0∫ 2
 Zs  ws h w
0 0 3 s
h3
μ 2
=
h3 2 ws
∫x dx
0

μ0  x 3  h3
=  
 3 
h32 ws 0

μ0  h3 3  h3
= 2  3  = μ0 3 w
h3 ws   s

Non- λsa = λ2 ws h2 – h2 h2
conductor dx μ0
λ2 = μ0 ∫ = ∫ dx
portion ws ws
0 0
(above
μ h h
conductor A) = 0 [ x ]02 = μ0 2
ws ws

λsa = λ4 wo h4 – h4 h4
dx μ
λ4 = μ0 ∫ = 0 ∫ dx
wo wo
0 0
μ h h
= 0 [ x ]04 = μ0 4
wo ws

(Continued)
 Estimation of Specific Permeance and Leakage Reactance 2.53
Estimation of Specific Permeance and Leakage Reactance  2.53
Table 2.7  | (Continued)
table 2.7 | (Continued)
Parameters λsa (or λsc) y h Z Derivation
Parameters λsa (or λsc) y h x Zx Derivation
λsa =λ5 wo + ws h5 – h5 h5
λsa =λ5 wo + ws h5 -
h5 dx 2μ0 h5
2
2
λ = μ ∫
5 λ =0 µ
5 0∫wow++
dx
ws
= =
2 µ 0 ∫
wwo ++wws ∫ dx
dx
0 o ws o 0
s 0
0 2
2
2μ 22hh5
= = 20µ0 [ x[]x05]h=
h
0
5
=μµ
00
5
wow+o w
+sws wwoo++wwss

λsa λ=saλ=6 λ6 wo wo h6 h6 –- h6 h h6h

dxdx μ µ
6 6
µ h
λ6 λ=6 μ=0 µ∫0 ∫ == 0 0∫∫dx dx = 0 [ x ]06
w w ww w
0 0o o oo 0
0
o

μ hh6 h
= =0µ[0x ]06 = μ0 6
wo w o wo

From Table 2.7, the total specific permeance is given by

λsλ=+
λs = λ1 + λ1 + λ2 + λ3 + λ4 + λ5 + λ6
2 λ3 + λ4 + λ5 + λ6
 
 h=
 1 μ0
h 2h1 +h3h2 +h4h3 + 2hh4 5+ 2hh65  + h6 
= µ0 +  3 w+ w + 3 w+ w + +w
 3w  s s 3 ws s ws s wo +s wswo w  s wo 
 s w o

2.7.3  tapered
2.7.3 Tapered slot
Slot
A tapered
A tapered slot
slot is
is represented
represented in
in Fig.
Fig. 2.29.
2.29.

w0

h4

h3

w1
h2

w2

h1 dx

wS

fig.
Fig. 2.29
2.29  ||  Tapered slot

The
The specific
specific permeance
permeance is
is calculated
calculated as
as shown
shown in
in Table
Table 2.8.
2.8.
2.54  Design of Magnetic Circuits

Table 2.8  |  Specific permeance calculation

Parameters λsa (or λsc) y h Zx Derivation

w2 + ws x x 2 dx
Conductor λsc = λ1 h1 Zs h1  Zs 
portion 2 h1 h  w2 + ws
λ1 = μ0 ∫  1 
 Zs  2
0
h1
2 x2
= μ0
w2 + ws ∫ h12
dx
0
h1
2 2
= μ0
(w2 + ws ) h12
∫x dx
0

2  x 3  h1
= μ0  
 3 
(w2 + ws ) h12 0

2 h 3 
= μ0  1 
2  3 
(w2 + ws ) h1
2 h1
= μ0
3 ( w2 + ws )

w1 + w2 h2 h2
Non- λsa = λ2 h2 –  dx  2
2 λ2 = μ0 ∫   = μ0 ∫ dx
conductor  w1 + w2  w1 + w2
0   0
portion  2 
2 [ x ]h2
= μ0
(w1 + w2 ) 0
2 h2
= μ0
w1 + w2
λsa =λ3 w0 + w1 h3 – h h
3
 dx  2
3

λ3 = μ0 ∫ 
w0 + w1 ∫
2  = μ0 dx
 w0 + w1 
0   0
 2 
2 2 h3
= μ0 [ x ]h3 = μ0
0
(w0 + w1 ) w0 + w1
λsa = λ4 w0 h4 – h4 h4
dx 1
λ4 = μ0 ∫ = μ0 ∫ dx
wo w0
0 0
1 [ ]h4 h
= μ0 x = μ0 4
w0 0 w0
 Reactance 2.55
Estimation of Specific Permeance and Leakage Reactance 

From Table 2.8, the total specific permeance is given by

λsλ=
s=λ1λ+ 1λ+2λ+ +3λ+
2λ 3λ+4λ4
  2 h2 h1 2 h2 h2 2 h2 h3 h h4 
==μ0µ0  1 + + 2 + + 3 + +4  
 3 ( 3w(2w+
2+wsw) s ) w1w+1+w2w2 wow+o+w1w1 wow o 

2.7.4 
2.7.4 Circular
circular Slot
slot
A
A circular
circular slot
slot is
is represented
represented in
in Fig.
Fig. 2.30.
2.30.

w0

r
d

θ
dx

fig. 2.30 
Fig. 2.30 | 
| Circular slot

The specific permeance calculations are as follows.

From Fig. 2.29, it is observed that
y y
sin θ = 2= y 2 y (2.41)
r sin2θr = r = 2r  (2.41)

r−x r−x
cosθ = cosθ =  (2.42)
r r
r2 [ r 22[θ ]
Area of segment
Area ofat x =height
height at
segment 2θx−=sin 2θ − sin 2θ ]
2 2
Conductor portions
Conductor portions producing
producing flux
flux in
in the
the strip
strip

r2 [ r2
2θ − sin 2[θ2]θ − sin 2θ ]
1 [ 1
= 2 =2 2 Zs = 2Zθs−=sin 2[θ2]θZ−s sin 2θ ] Zs
2 2π 2 π
πr πr
1 sin 21θ  sin 2θ 
= θ − = π θ Z−s 2  Zs
π  2   
2.56  Design of Magnetic Circuits

Flux in the strip, dfx = MMF× permeance  (2.43)

1 sin 2θ 
MMF = θ −  Zs I z  (2.44)
π  2 
Ldx
Permeance = μ0  (2.45)
y
From Eq. (2.41),
y = 2r sin θ (2.46)
Differentiating Eq. (2.42) with respect to θ, we get
d[ d r−x
cos θ ] =  
dθ dθ  r 
d r d x
− sin θ =  −  
dθ  r  dθ  r 
1 dx
⇒ − sin θ = 0 −
r dθ
1 dx
⇒ sin θ =
r dθ

⇒ dx = r sin θdθ  (2.47)

Substituting Eqs. (2.47) and (2.46) in Eq. (2.45), we get

μ0 Lr sin θ dθ
Permeance =
2r sin θ
Ldθ
= μo (2.48)
2
Substituting Eqs. (2.44) and (2.48) in Eq. (2.43), we get
1 sin 2θ  Ldθ
dφx = θ −  Zs I z × μ0
π  2  2
μ0 LZs I z  sin 2θ 
= θ −  dθ
2π  2 
Flux linkage in strip,

dψx = dθx × conductor portions producing flux in the strip

μ0 LZs I z  sin 2θ  1 sin 2θ 

= θ −  dθ × θ −  Zs
2π  2  π  2 
 Estimation of Specific Permeance and Leakage Reactance  2.57

2
μ0 LZs2 I z  sin 2θ 
= θ −  dθ
2π 2  2 

Total flux linkage in the conductor portion of the slot,

ψ = ∫ dψx

π 2
μ0 LZs2 I z  sin 2θ 
=∫ θ −  dθ
0
2π 2  2 

π 2
μ0 LZs2 I z  sin 2θ 
=
2π 2
∫ θ − 2 
 dθ
0
π  
μ0 LZs2 I z 2
θ 2 + sin 2θ − 2θ sin 2θ  dθ
=
2π 2
∫  2 2 
0

= 0.623μ0 LZs2 I z

Effective permeance of conductor portion,

ψ
∧c = = 0.623 μ0 L
I z Zs2

Specific permeance of conductor portion,

∧c
λc = = 0.623 μ0
L
Specific permeance of slot opening,
h
dx h
λ1 = ∫ μ0 = μ0
y wc
0
Total specific slot permeance,

λs = λc + λ 1

h
= 0.623μ0 + μ0
wc
 h 
= μ0  0.623 +   (2.49)
 wc 
2.58  Design
Designofof
Magnetic Circuits
Magnetic Circuits

2.7.5 
2.7.5 Tt Bar Slot
slot (Induction
(induction Motor)
A
ATT bar
bar slot
slot is
is represented
represented in
in Fig.
Fig. 2.31.
2.31. The
The specific
specific permeance
permeance calculations
calculations are
are as
as follows:
follows:

w0

h3
Area b dy

h2
y

dx
h1
x
Area a

wS

fig. 2.31 
Fig. 2.31 | 
| TT bar
bar slot
slot

The
The specific
specific permeance
permeance calculation
calculation is
is shown
shown in
in Table
Table 2.9.
2.9.

table 2.9 
Table 2.9 | 
| Specific
Specific permeance
permeance calculation
calculation
Parameters λλ
Parameters (orλλ
(or
sasa sc)sc) y y hh ZZx x Derivation
Derivation

Conductor λsc1 = λ1 xx a a  x x a a 2 2

Conductor λsc1 = λ1 s h1 h
ww Zs Z
s b b s h1 h Z  
 1h1 ha +ab+ bs Zsdx
1
portion (a) h1ha1 +a +
portion    dx
1 0μ∫0 ∫
λ1λ==
µ   1
  Zs Z  ws w
0  s  s
0
h1 h12
1 x 2 a2
a2dx
µ0μ0 1∫ 2 x
==
∫
wsw h1 h( a + b)2 2
2
dx
0s 0 1 ( a + b )
h1 h
1 1 1
a2 a2
µ0μ0 2 ∫ x∫2 x 2
== 2
dx dx
h1s h012 ( a +( ab+
wsw ) b)2
0
h
h13 1
a2 a  x 3  x 
2
1 1
µ0μ0 2
== 2   
h1s h(12a +
wsw ( ab+) b )32  03  0
2  3
1 1 a 2 a  h13 h 1 
==
µ0μ0 2 2   
( ab+)2b )23   3 
h1s h(1a +
wsw
2
h h1 a 2 a
==
µ0μ0 1
( ab+)2b)2
3 w3sw(sa +

(Continued)
(Continued)
 Magnetic Pull  2.59
Magnetic Pull 2.59
Table 2.9  | (Continued)
table 2.9 | (Continued)
Parameters λsa (or λsc) y h Zx Derivation
Parameters λsa (or λsc) y h Zx Derivation
y b 2
Conductor h y b  y b a 2 
λ =λ
λsc2 =sc2λ2 2 wowo h2 2 h a + bZZ s+
+ h2y b Zs + a  Zs 
portion (b) h22a + b
s 
h2   Z
h a + sb + a +sb  dy
Z
λ2 = μ0∫h2 a2+ b a + b  dy
aa Zs λ2 = µ0 ∫   Z   wo
Zs  0 Zs s  wo
a
a+b+ b 0
 2
h 2 a + ab 2+ b / 3  2 
=µ= μh02  a2 + ab + b / 3  
0
wo w  o  ( a +( ab)+
2 b )2 
 


Non-
Non- λsa=λ3
λsa=λ w0w0 h3 h3 -– h3 h3
h3 h3
dx dx = μ1 1
3
conductor
conductor
portion
portion
λ3λ= ∫ wo wo w0 ∫w0dx∫ dx
µ0μ
3 = 0 ∫ = µ0 0
0 0 0 0
1 1h h
µ0μ0 [ x ]0[3x ]03
==
w 0w 0
h h
µ0μ03 3
==
w 0w 0

2.8  Magnetic
2.8 Magnetic Pull
Pull
Magnetic pull
Magnetic pull is
is the
the force
force exerted
exerted between
between two
two poles
poles of
of magnet,
magnet, when
when separated
separated by
by aa
distance as shown in Fig. 2.32.
distance as shown in Fig. 2.32.

Magnetic force between two poles

N
S

Ig dx
Airgap

fig.
Fig. 2.32
2.32  ||  Magnetic
Magnetic force
force exerted
exerted between
between two
two poles
poles

The force can be determined from change in magnetic field energy stored in terms of
The force can be determined from change in magnetic field energy stored in terms of
energy density and volume change and is given by
energy density and volume change and is given by
Change in magnetic field energy = Energy density × Volume change (2.50)
Change in magnetic field energy = Energy density × Volume change  (2.50)
2.60  Design of Magnetic Circuits

where

Change in magnetic field energy = Fdy = work done (when one pole is
moved by distance dy)
1 1
Energy density = BH = μ0 H 2
2 2

1 B2
=
2 μ0
[μ0 is used since it is a free space and relative permeability of air, μ0 = 1]
Volume change = Ady

and F – force between two poles, dy – distance by which one pole is moved, B – magnetic flux
density, H – magnetic field intensity and A – area of one pole.
Substituting the above values in Eq. (2.50), we get

1 B2
Fdy = × Ady
2 μ0

1 B2
⇒ F= A (2.51)
2 μ0

F 1 B2
⇒ =
A 2 μ0

1 B2
⇒ Pm = (2.52)
2 μ0

Thus from Eq. (2.52), it is observed that, force per unit area or magnetic pull depends on
magnetic flux density of air gap which in turn depends on MMF of source of excitation.
The MMF is given by

AT = ATi + ATg

where ATi – MMF of iron and ATg – MMF of air gap.

Generally, if there is no saturation in the portion of iron, MMF ATi is very small. Hence,
MMF can be approximated as AT  ATg.
Therefore, expressing magnetic flux density in terms of MMF, we get

μ0 ATg  AT 
B=  ∵ B = μ0 H , where H = 
lg  l 

where lg – length of air gap.

 Magnetic Pull  2.61

Substituting the value of B in Eq. (2.51), we get

2
1  μ0 ATg  A
F =   ×
2  lg  μ0

2
1  ATg 
= μ0  
 A
2  lg 

The above expression gives magnetic force of attraction existing between poles in terms
of MMF of source of excitation.

2.8.1  Radial Magnetic Forces

In a rotating electrical machine, the magnetic force of attraction exists between stator and
rotor poles, about the direction perpendicular to axis of shaft of rotor. This force of attraction
when considered for a two-pole machine as shown in Fig. 2.32 is given by
F = FT – FB (2.53)
where FT – force of attraction for stator and rotor poles at the top, given by
2
1  AT 
FT = μ0   A
2  lg 
FB – force of attraction for stator and rotor poles at the bottom, given by
2
1 A 
FB = μ0  T  A
2  lg 

Substituting the values of FT and FB in Eq. (2.53), it is observed that

2 2
1 A  1 A 
F = μ0  T  A − μ0  T  A
2  lg  2  lg 

F = 0 (2.54)
Equation (2.54) is valid for the following conditions:
(a) Rotor is symmetrical and concentric with respect to stator by being a uniform
cylinder.
(b) Length of air gap is uniform at all places.
(c) MMF of iron is negligible.
(d) Flux distribution is uniform.
From Eq. (2.54), it can be stated that resultant magnetic force of attraction is zero, resulting
in zero or nil radial magnetic pull on the rotor,
F
since F = 0  ⇒  Pm = = 0
A
2.62  Design of Magnetic Circuits

2.8.2 Radial Magnetic Forces and Unbalanced Magnetic Pull

In case of ac machine, magnetic flux density, B = Bm sin θ. Consider a small portion from
D
Fig. 2.33, with area, A = Ldθ
2

F1
y Stator

N
lg

S
Air gap
dθ
θ
X X´

N
lg
S Rotor

Fig. 2.33  |  Symmetrical machine with radial magnetic forces

The radial force, giving the force of attraction at top stator and rotor poles, for this small
element is given by
2
1 (Bm sinθ ) DL
FT = dθ
2 μ0 2
1
= Bm sin 2θDL dθ
4μ0
1
= Bm DL sin 2 θdθ
4μ0

The vertical component of the above equation is given by

1
FT(vertical) = Bm DL sin 2θ dθ × sin θ
4μ0
1
= Bm DL sin 3θ dθ  (2.55)
4μ0

The resultant magnetic pull acting vertically is determined by integrating the above
equation over area swept between 0 and π.
 Magnetic Pull  2.63

π
1
⇒ FT (vertical) = ∫ Bm DL sin 3θ dθ
4μ0
0

1 1 π
= Bm DL×  cos 3 θ − cos θ + C 
4μ0  3  0
 1 
  ∵ ∫ sin 3 xdx = cos 3 x − cosx + C 
 3 

1
Substituting C = 0 and cos 3θ = (cos θ + 3 cos θ ) in the above equation and substituting
the limits of integration 4

1 1 1 π
= Bm DL  × [ cos 3θ + 3cosθ ]− cos θ 
4μ0  3 4  0

1 1 1 
= Bm DL  (−4) − (−1) −  ( 4) − 1
4  
 12 
 12
1  4 1
= Bm DL   = Bm DL
4μ0  3  3μ0

Similarly, for bottom stator and rotor poles,

1
FB(vertical) = Bm 2 DL
3μ0

The net resultant radial magnetic pull is given by

F = FT – FB = 0 (2.56)

Equation (2.56) can be arrived at integrating Eq. (2.55) over limits 0 to 2π, covering whole
machine with two poles and is given by
2π
1
∫ Bm 2 DL sin 3θdθ = 0  (2.57)
4μ0
0

Equations (2.56) and (2.57) hold good for conditions defined in the previous section.
In practical cases, manufacturing tolerances are allowed in dimensions defining the
machine. The manufacturing defects can be present causing a non-uniform air gap to be
present between a stator and a rotor. In such cases, the resultant magnetic pull is not zero
and is given by difference between the magnetic forces of attraction existing between the
number of poles involved in stator and rotor of machine. The resultant unbalanced magnetic
pull is determined as follows.
Consider an unsymmetrical electric machine as shown in Fig. 2.34.
2.64  Design of Magnetic Circuits

F1
y

N
lg1 Stator
S Air gap

Rotor
X X´
Shaft
N
lg2
S

y´
U.M.P
F2

Fig. 2.34  |  Unsymmetrical machine with unbalanced magnetic pull

It is observed that the length of air gap at the top of the machine is larger than the length
of air gap at the bottom of the machine, i.e., ( lg 1 > lg 2 ).
The force of attraction at top stator and rotor poles is given by
2
1  AT 
FT = μ0   A
2  lg 1 

The force of attraction at bottom stator and rotor poles is given by

2
1  AT 
FB = μ0   A
2  lg 2 

The resultant magnetic pull or force is given by

F = FT – FB
2 2
1  AT  1  AT 
= μ0   − μ0    (2.58)
2  lg 1  2  lg 2 

1  1 1 
= μ0 AT 2  2 − 2  (2.59)
2 
 lg 1 lg 2 

From Eq. (2.59), it is observed that FT > FB since lg 2 > lg 1 . This resultant magnetic pull or
force is called unbalanced magnetic pull and in the above case it sets in the downward direction.
Apart from non-uniformity in air gap, asymmetrical magnetic circuit or placement of
windings can result in unbalanced magnetic pull.
 Magnetic Pull  2.65

2.8.3  Determination of Unbalanced Magnetic Pull

Consider a rotating machine with diameter of rotor (D), stack or core length (L) as shown in
Fig. 2.35(a) and (b).

Stator

lg + e = lg1 lg Normal rotor

lg2 = lg1 − e lg Displayed rotor

(a)

lg
A
e
dθ lg2

θ
e

lg1

A´

(b)

Fig. 2.35  |  (a) and (b) Unbalanced magnetic pull in electric machine with vertical rotor
displacement

From Fig 2.33(a) and (b), it is observed that the new values of
Modified length of air gap in the vertical axis at top = lg + e
2.66  Design of Magnetic Circuits

Modified length of air gap in the vertical axis at bottom = lg − e

where e is the displacement of rotor from its original position in downward direction (or
known as eccentricity).
For the machine to be rotating, the length of air gap is modified by angle q measured with
respect to horizontal axis.
Hence,
Modified length of air gap for top half of the rotor along X – X′,

lg1 = lg + e sin θ

Modified length of air gap for bottom half of the rotor along X – X′,

lg 2 = lg − e sin θ

Also,
The magnetic pull is represented by Pm.

lg2
The pull per unit area is given by Pm ×
lg2(modified)

⇒ Pull per unit area along X − X′

  lg2 
  
= Pm  Change in  
  lg(modified) 
2
 

 2  2 
 lg lg
= Pm   −   
 lg − e sin θ   lg + e sin θ  
 
 1 1 
= Pmlg 2  2
− 2

  l − e sin θ   lg + e sin θ  
  g    

 l2 + e 2 sin 2 θ + 2lg e sin θ −  lg2 + e 2 sin 2 θ − 2lg e sin θ  

2 g  
= Pmlg 

  lg − e sin θ  2  lg + e sin θ  2 

   

 4lg e sin θ 
 Pmlg2  4
 [∵ lg  e sin θ]

 l g 

Therefore, pull per unit area along X − X′ due to two poles of rotor

4Pm e sinθ
=
lg
 Magnetic Pull  2.67

Vertical component of pull per unit area

4Pm e 4P e
= sin θ × sin θ = m sin 2 θ
lg lg

1 B2
Substituting Pm = in the above equation, we get
2 μ0
Vertical component of pull per unit area
e e
= 4 × Pm × sin 2 θ = 4 Pm sin 2θ
lg lg

Therefore, the vertical component of pull

e
= 4 Pm sin 2 θ × area
lg

D
Substituting area = L dθ in the above equation, as we consider a small area defined by dq,
2
the vertical component of pull
e D
= 4 Pm sin 2θ × L dθ
lg 2
e
= 2Pm sin 2θ DL dθ
lg

As the unbalanced magnetic pull is acting in the downward direction,

Total pull,
π
e
P = ∫ 2 Pm sin 2 θ DL dθ
lg
0
π
e 1 − cos 2θ
= 2 Pm DL ∫ dθ
lg 2
0
π
e  sin 2θ 

= Pm DL θ −
lg  2  0
e
= Pm DL[ π − 0 − 0 + 0 ]
lg
πPm eDL
=
lg

1 B2
Substituting Pm = in the above equation, we get
2 μ0
2.68  Design of Magnetic Circuits

1 πB2 e DL
P=
2 μ0 lg

1 B2 e
= π DL (2.60)
2 μ0 lg

As we know that area per pole,

πDL
A= .
2
B2 e eA
Hence, P= A = 2 Pm
μ0 lg lg

e  1 B2 
= 2 A Pm    ∵ Pm =  (2.61)
lg  2 μ0 

Equation (2.61) is valid for a two-pole machine. For a machine with ‘P’ number of poles,
Eq. (2.61) becomes unbalanced magnetic pull
eA
= P × 2 Pm
lg

Bm
For sinusoidal flux distribution, substitute B = in Eq. (2.62).
2
Therefore, unbalanced magnetic pull
2
1B  e
=  m  πDL
2  2  μ0 lg

1 Bm 2 e
= π DL (2.62)
4 μ0 lg

2.8.4 Significance and Minimization of Unbalanced Magnetic Pull

Table 2.10 provides the significant effects and prevention methods of unbalanced magnetic
pull.

Table 2.10  |  Significant effects and prevention methods of unbalanced magnetic pull
Effect Prevention method
Unbalanced magnetic pull is pronounced in •  Decrease in stack length
induction motors due to small air gap •  Use of high-quality ball bearings
• Use of stator winding with equalizer
connections

(Continued)
 Review Questions  2.69

Table 2.10  | (Continued)

Effect Prevention method
Unbalanced magnetic pull is significant for Choice of slot numbers in stator and rotor
certain combination of stator and rotor slots, should properly made
causing vibration and noise
Presence of homopolar flux in two-pole Use of stator winding with parallel paths and
machines leads to asymmetry in air gap equalizer connections

Review Questions

Multiple-choice Questions
1. In DC machines, the total mmf produced by each pole is .
(a) the mmf for pole and pole shoe (b) the mmf for air gap and teeth
(c) the mmf for yoke and armature core (d) all the above
2. Carter’s coefficient for open slots is given by .
2 1  2 1 
(a) Kcs =  tan−1 y − log 1 + y 2  (b) Kcs =  tan−1 y − log y 2 
π  y  π  y 
π  −1 1  2 1 
(c) Kcs =  tan y − log 1 + y 2  (d) Kcs =  tan−1 y + log 1 + y 2 
2 y 
 
π y 

3. The mmf for air-gap in a salient pole machine is given by .
(a) 80,000 BgKglg (b) 800,000 BgKglg
(c) 800,000 BgKg (d) 800,000 Bglg
4. Simpson’s rule is also known as .
(a) Bt1/3 method (b) two ordinate method
(c) three ordinate method (d) graphical method
5. The real and apparent flux densities are related by Bapp = Breal + Ba (Ks −1), where Ks
is .
(a) ratio of air area to iron area (b) ratio of iron area to air area
(c) ratio of iron area to total area (d) ratio of total area to iron area
6. Hysteresis loss can be expressed in .
(a) Watts per cubic metre (W/m ) 3 (b) Watts per kilogram (W/kg)
(c) Both (a) and (b) (d) none of the above
7. Hysteresis loss can be minimized using .
(a) soft magnetic materials such as Si steel
(b) air core transformers
(c) both (a) and (b)
(d) none of the above