Design of rotating electrical machines

This course includes a detailed step by step approach to design rotating electrical machines like induction motor and DC machines.

1. 1




Design of Rotating
Electrical Machines
(EET3186)
By
Dr. Binod Kumar Sahu
Associate Professor, Department of Electrical
Engineering, ITER, Siksha ‘O’ Anusandhan University,
Bhubaneswar, Odisha, India

2. 2

Chapter-1
Principal Laws and Methods in Electrical Machine Design
1.1. Electromagnetic Principles: -
Basic quantities involved in the electromagnetic phenomena are:
Quantity Unit
i. Electric field strength (E) V/m
ii. Magnetic field strength (H) H/m
iii. Electric flux density (D) C/m
2

iv. Magnetic flux density (B) Wb/m
2

v. Current density (J) A/m
2

vi. Electric charge density  C/m
3

Presence of an electric and magnetic field can be analyzed from the force exerted by the
field on a charged object or a current-carrying conductor. Force experienced by an infinitesimal
charge of ‘dQ ’ Coulomb moving at a speedsec/''mv is given by Lorentz’s force equation i.e.   BIdldQEB
dt
dl
dQdQEBdQvdQEBvEdQdF 
Newton (1)
Above equation is basically used for computing torque of various electrical
machines. F
I
B
B
I
F
dl
dl

Figure 1 Force experienced by a differential length ‘dl’ of a conductor carrying an electric current ‘i’ in
magnetic field ‘B’.

3. 3

Example 1: Calculate the force exerted on a conductor 0.1 m long carrying a current of
10 A at an angle of 80
0
with respect to a field density of 1 T.
Solution: -
Force experienced by the conductor,
NBliBildF 98.080sin11.010sin..
0
 
Other fundamental laws related to electrical engineering theory are as follows:
t
J




. Continuity equation (2)
t
B
E


 (3)
t
D
JH


 (4)
D (5)
0B (6)
Equations 3-6 are Maxwell’s equations in differential form. Equation (3) is
Faraday’s induction law that describes how a changing magnetic flux creates an electric
field around it. Equation (4) describes the situation where a changing electric flux and
current produce magnetic field strength around them. This is Ampere’s law. Equation (5)
is known as Gauss’s law of electric field which indicates that there can be a source (positive
charge) and a sink (-ve charge) in an electric field. Equation (6) indicates that magnetic
flux is always a circulating with no starting or end point (i.e. source or sink).
Equation (3) in integral form can be written as
 

l S
dt
d
dSB
dt
d
dlE

. (7)
Equation (7) states that the change of a magnetic flux ‘Φ’ penetrating an open
surface ‘S’ is equal to negative of line integral of the electric field strength along the line ‘l’
around the surface.

4. 4


Figure 2 Illustration of Faraday’s induction law.
The arrows in the circles point the direction of the electric field strength E in the
case where the flux density B inside the observed area is increasing. If we place a short-
circuited metal wire around the flux, we will obtain an integrated voltage 

l
dlE in the
wire, and consequently also an electric current. This current creates its own flux that will
oppose the change of flux penetrating through the coil.
If there are several turns ‘N’ of winding, the flux does not link all these turns ideally,
but with a ratio of less than unity. Hence we may denote the effective turns of winding by
‘Nk
w ’ 1
w
k . Where ‘w
k ’ is known as the winding factor. For ‘N’ number of turns, induced
EMF equation (7) becomes 

S
ww
dt
d
dt
d
NkdsB
dt
d
Nke

(8)
Where ‘ ’ is the flux linkage and is given as LINk
w
 . It may be said that the
inductance ‘L’ describes the ability of a coil to produce flux linkage ‘ ’.
Example 2: There are 100 turns in a coil having a cross-sectional of 0.0001 m
2
. There is
an alternating magnetic field of peak flux density ‘1 T’ linking the turns of the coil with a
winding factor kw = 0.9. Calculate the induced EMF in the coil when the flux density
variation has a frequency of 100 Hz.
Solution:
Induced EMF,

5. 5


)2cos(2))2sin(1()( ftfNSkft
dt
d
NSkSB
dt
d
NkdsB
dt
d
Nke
ww
S
ww   
)cos(655.5)2cos(10020001.01009.0 wtft 
So the effective value of the voltage isV4
2
655.5
 .
Ampere’s law in integral from can be written as


S
e
Sl
dt
d
tidSD
dt
d
dSJdlH

)( (9)

Figure 3 Application of Ampere’s law in the surroundings of a current-carrying conductor. The line ‘l’
defines a surface ‘S’, the vector ‘dS’ being perpendicular to it.
Equation (9) indicates that a current ‘i(t)’ penetrating a surface ‘S’ and including
the charge of electric flux is equal to line integral of the magnetic flux ‘H’ along the line ‘l’
around the surface ‘S’. Equation (9) is often represented in static or quasi-static form and
becomes


encl
Sl
ItidSJdlH )( (10)
Term quasi-static indicates that the frequency of the phenomenon is low enough
to neglect displacement current. The phenomena occurring in electrical machines meet
the quasi-static requirement. In integral form Gauss’s law can be written as


S V
v
dVdSD . (11)

6. 6

Equation (11) indicates that a charge inside a closed surface ‘S’ that surrounds a
volume ‘V’ creates an electric flux density ‘D’ through the surface. Gauss’s Law for
magnetic field in integral form can be written as 

S
dSB 0.
(12)
Equation (12) indicated that there is no source for a magnetic flux. In electrical
machines, this means for instance that the main flux encircles the magnetic circuit of the
machine without a starting or end point.
The permittivity, permeability and conductivity ε, μ and σ of the medium
determine the dependence of the electric and magnetic flux densities and current density
on the field strength. ED
(13) HB
(14) EJ
(15)
Example 1.3: Calculate the electric field density ‘D’ over an insulation layer 0.3 mm
thick when the potential of the winding is 400 V and the magnetic circuit of the system is
at earth potential. The relative permittivity of the insulation material is εr = 3.
Solution: -
Electric field strength,
mkV
d
V
E /33.133
003.0
400
 .
Electric flux density,
212
0 /542.333.133310854.8 mCED
r  
 .

Example 1.4: Calculate the displacement current over the slot insulation of the previous
example at 50 Hz when the insulation surface is 0.01 m
2
.
Solution: -
Electric flux over the insulation,
nCSD 42.3501.010542.3
6


 .
Time dependent electric flux over the insulation is
tt 314sin21042.35)(
9


 .

7. 7

nCtt 314sin09.50)( .
Displacement current Id is
AtnAt
dt
td
i
d


314cos73.15314cos31409.50
)(
 .
Effective current over the insulation is Ai
d
12.11
2
73.15


From the above examples it is seen that the displacement current is insignificant
from the viewpoint of the machine’s basic functionality. However, when a motor is
supplied by a frequency converter and the transistors create high frequencies, significant
displacement currents may run across the insulation.
1.2 Numerical Solution: -
Accurate performance of the machine is usually evaluated using different
numerical methods. With these numerical methods, the effect of a single parameter on
the dynamical performance of the machine can be effectively studied. The most widely
used numerical method is the finite element method (FEM), which can be used in the
analysis of two- or three-dimensional electromagnetic field problems.
There are five common methods to calculate the torque from the FEM field
solution. The solutions are (1) the Maxwell stress tensor method, (2) Arkkio’s method, (3)
the method of magnetic coenergy differentiation, (4) Coulomb’s virtual work and (5) the
magnetizing current method.
The magnetic fields of electrical machines can often be treated as a two-
dimensional case, and therefore it is quite simple to employ the magnetic vector potential
in the numerical solution of the field. In many cases, however, the fields of the machine
are clearly three dimensional, and therefore a two-dimensional solution is always an
approximation. Relationship between vector magnetic potential ‘A’ and magnetic flux
density ‘B’ is given by:
AB (16)
The magnetic flux penetrating a surface is easy to calculate with the vector
potential using the relation,


lS
dlAdSB . (17)

8. 8

1.3 The Most Common Principles Applied to Analytic Calculation
Design of an electrical machine involves the quantitative determination of the
magnetic flux of the machine. In the design of a magnetic circuit, the precise dimensions
for individual parts are determined, the required current linkage for the magnetic circuit
and also the required magnetizing current are calculated, and the magnitude of losses
occurring in the magnetic circuit are estimated. Design of a magnetic circuit of an
electrical machine is based on Ampere’s law. Total magneto-motive force (MMF) required
for a magnetic circuit is 

l
T
linkageCurrentNIdlHMMF .

Example 1.5: Consider a C-core inductor with a 1 mm air gap. In the air gap, the flux
density is 1 T. The ferromagnetic circuit length is 0.2 m and the relative permeability of
the core material at 1 T is μr = 3500. Calculate the field strengths in the air gap and the
core, and also the magnetizing current needed. How many turns N of wire carrying a 10
A direct current are needed to magnetize the choke to 1 T? Fringing in the air gap is
neglected and the winding factor is assumed to be kw = 1.
Solution: -
Magnetic field strength in the air gap,
mkA
B
H
g
/77.795
104
1
7
0




 .
Magnetic field strength in the core,
mA
B
H
r
i
/36.227
3500104
1
7
0




 .
Total MMF required i.e. current linkage,
ATlHlHMMF
iiggT
24.8412.036.2271011077.795
33


For a coil carrying a current of 10 A, number of turns ‘N’ required are
turns
I
MMF
N
T
85124.84
10
24.841


9. 9

1.3 Determination of Force and Torque
Voltage equation of a winding can be written as
dt
d
NRiLi
dt
d
Ri
dt
d
Riv

 (18)
So the required power in the winding dt
d
NiRiviP


2
(19)
Change in energy required in the winding
NiddtRividtPdtdW 
2 (20)

10. 10

Windings of Electrical Machines
The operating principle of electrical machines is based on the interaction between the
magnetic fields and the currents flowing in the windings of the machine. The winding
constructions and connections together with the currents and voltages fed into the windings
determine the operating modes and the type of the electrical machine. According to their different
functions in an electrical machine, the windings are grouped for instance as follows:
i. armature windings;
ii. other rotating-field windings (e.g. stator or rotor windings of induction motors);
iii. field (magnetizing) windings;
iv. damper windings;
v. commutating windings; and
vi. compensating windings.
Armature winding: -
Armature winding is a winding in a synchronous, DC or single-phase commutator machine,
which, receives active power from or delivers active power from/to the external electrical system.
This definition also applies to a synchronous compensator if the term ‘active power’ is replaced
by ‘reactive power’. Configuration of a symmetrical poly-phase winding depends on
i. Even distribution of air-gap periphery for determination of pole pitch which covers
180
0
electrical.
ii. Division of the periphery of machine into phase zones of positive and negative
values.
Pole pitch in terms of number of slots is calculated as
P
S
p
 , where ‘S’ is total number of slots and ‘P’ number of poles.
Phase zone distribution is expressed as
m
p
v

 , where ‘m’ is the number of phases.
Number of zones aremP and number of slots for each such zone (number of slots per
pole per phase) is expressed as
Pm
S
s



11. 11

If ‘s’ is integer, the winding is called as integral slot winding and if ‘s’ is a fraction, then
the winding is known as fractional slot winding.

Figure 2.1 Division of the periphery of a three-phase, four-pole machine into phase zones of positive and negative
values.
The phase zones are distributed symmetrically to different phase windings so that the phase
zones of the phases U, V, W, . . . are positioned on the periphery of the machine at equal distances
in electrical degrees. In a three-phase system, the angle between the phases is 120 electrical
degrees. This is illustrated by the periphery of Figure 2.1, where we have 2 × 360 electrical degrees
because of four poles. Now, it is possible to label every phase zone. We start for instance with the
positive zone of the phase U. The first positive zone of the phase V will be 120 electrical degrees
from the first positive zone of the phase U. Correspondingly, the first positive zone of the phase
W will be 120 electrical degrees from the positive zone of the phase V and so on. In Figure 2.1,
there are two pole pairs, and hence we need two positive zones for each phase U, V and W. In the
slots of each, now labelled phase zones, there are only the coil sides of the labelled phase coil, in
all of which the current flows in the same direction. Now, if their direction of current is selected
positive in the diagram, the unlabelled zones become negative. Negative zones are labelled by
starting from the distance of a pole pitch from the position of the positive zones. Now U and −U,
V and −V, W and −W are at distances of 180 electrical degrees from each other.

12. 12

Main Dimensions of a Rotating Machine
The design of a rotating electrical machine can be commenced by defining certain basic
characteristics, the most important of which are:
 Machine type (synchronous, asynchronous, DC, reluctance machine, etc.).
 Type of construction (external pole, internal pole, axial flux, radial flux machine, etc.).
 Rated power:
 For electric motors, the shaft output power P0 in W is given.
 For synchronous motors, also the power factor (cosϕ) is given.
 For induction and DC generators, the electric output power P0 in W is given.
Induction generators take reactive power from the network according to their
power factor. The reactive power must usually be compensated by capacitor banks.
 For synchronous generators, the apparent power ‘S’ is given in VA. The power factor
(typically cosϕ = 0.8 overexcited) is also given.
 Rated rotational speed ‘N’ of the machine or rated angular speed ‘w’.
 Number of poles ‘P’ of the machine (with frequency converter drives, this is also a subject
of optimization).
 Rated frequency ‘f’ of the machine (with frequency converter drives this is also a subject
of optimization).
 Rated voltage ‘V’ of the machine.
 Number of phases ‘m’ of the machine.
 Intended duty cycle.
 Additional information, such as efficiency, required locked rotor torque, pull-up torque,
peak torque, locked rotor current, speed-controlled drive etc.
In machine design, there are a considerable number of free parameters. When aiming for
an optimal solution, the task becomes extremely complicated unless the number of these free
parameters is somehow limited. Many free parameters vary only slightly, and therefore, to simplify
the task, these parameters can be assumed constant. The following 10 parameters can be selected
as free parameters:

13. 13

 outer diameter of the stator stack (with the standard IEC frames, this parameter is often
fixed to certain values);
 length of the stator stack;
 width of the stator slot;
 height of the stator slot;
 diameter of the air gap;
 air-gap length;
 peak value of the air-gap flux density;
 width of the rotor slot;
 height of the rotor slot;
 pole number and frequency.
Actual machine design starts with the selection of the main dimensions of the machine.
The term ‘main dimensions’ refers to the air-gap diameter Ds measured at the stator bore and the
equivalent core length l’. The equivalent length of the core takes into account the influence of the
flux fringing at possible cooling ducts of the machine and also at the ends of the machine. In
electrical machine design, there are certain empirically defined variation ranges of current and flux
densities, which can be applied in the preliminary phase of the design. Tables 1 and 2 introduce
some values of electromagnetic loadings for well-designed standard electrical machines.
Table 1

14. 14

Table 2

Table 3

15. 15

The local value for the tangential stress depends on the local linear current density A(x)
and the local flux density B(x), σFtan(x) = A(x)B(x). Average tangential stress 2
cos
2
cos
tan


mmm
F
ABBA

(1)
Where Am and Bm are the peak values, A is the rms value of linear current density. This
tangential stress produces torque of the machine. Torque developed in the rotor is rFrFrrFrrF VlrrlrrST
tan
2
tantantan 2)'(2)'2()(  
(2)
Where rr is the radius of the rotor, 'l equivalent length of rotor and r
V is the effective
volume of the rotor.
Internal apparent power,
ImES
phi
 , where ‘m’ is the number of phases, Eph is the induced EMF per phase and ‘I’
is the stator current per phase. 1
111
'2
2
1
2
2
1
2
2
44.4
wph
pm
wph
m
wphavwphavph
KT
lB
w
KTwKTfKTfE










(3)
Where P
D
p

 is the pole pitch. pmmmm
lB
P
D
lBAB 

 ''
RMS value of linear current density (A) may be defined from the stator slot pitch
S
 and
RMS value of slot current (Is).
Slot pitch, S
D
S

 , where ‘S’ is the total number of slots.
Slot current, slotperconductorsofNumberphasepercurrentStatorI
S

Number of conductors/slot,
S
mT
phasepoleslotsofNo
Z
Z
phph
S


2
//. (4)
Linear current density,
D
mTI
S
D
S
mT
I
ZII
A
ph
ph
s
S
s
s







2
2 (5)

16. 16

Stator current per phase,
mT
DA
I
ph


2
 (6)
Internal apparent power,
ImES
phi
 (7)
Putting the values of Eph and ‘I’ from equations (3) and (6) in equation (7) we get,
Smw
wmS
ph
wph
m
s
ph
wph
pm
phi
nlDABK
AKDlBn
mT
DA
KT
P
D
lB
Pn
m
mT
DA
KT
lB
wmImES
'
2
'
2
1
2
'2
2
2
2
1
2
'2
2
1
2
1
2
1
22
1
1


















 (8)
Where, ‘nS’ is the synchronous speed in rps.
Equation (8) can be expressed as
S
i
Si
Cn
S
lD
nlCDS


'
'
2
2 (9)
Where, mmwmw BAKABKC
1
2
1
2
22

 (10)
In the above expression ‘C’ is known as the machine constant, Am is the peak value of
linear current density in Amp/m and Bm is the peak value of air-gap flux density in Wb/m
2
.
Rotor volume required for certain apparent power can be expressed as,
S
i
r
Cn
S
lDV 
4
'
4
2 (Putting the value of '
2
lD from equation (9)) (11)
For a rotating field motor, mechanical power
smech
ph
i
m nlDC
E
S
VVIpowerinputP 'coscos
2
  (12)

17. 17

Example # 1: -
The diameter of the rotor of a 4 kW, 50 Hz, two-pole induction motor is 98 mm and the
length is 82 mm. Assume 1% rated slip calculate the machine constant and the average tangential
stress.
Solution: -
Mechanical constant,
./58.101
60
3000
082.0098.0
104
3
2
3
2
mkWs
nlD
P
C
s
m
mech






Torque producing area of the rotor is,
.0252.010821098
233
mDlS
r



Speed of the machine, .29703000)01.01()1( rpmNsN
S

Rated torque of the motor,
.86.12
29702
10460
2
60
60
2
3
Nm
N
P
N
P
w
P
T 













Tangential force on the rotor,
.45.262
10
2
98
86.12
3
tan
N
r
T
F
r









Average tangential stress,
./415.10
0252.0
45.262
2tan
tan
mN
S
F
r
F


Example # 2
Repeat Example 1 for a 30 kW, 50 Hz, four-pole, induction motor running at 1450 rpm.
The rotor diameter is 200 mm and the rotor length 206 mm.

18. 18

Mechanical loadability: -
In addition to temperature rise, the output power and maximum speed of a machine are
restricted by the highest permissible mechanical stresses caused by the centrifugal force, highest
permissible electrical and magnetic loadings. The highest stress caused by the centrifugal force in
the rotor is proportional to the square of the angular speed and is expressed as: 22
max'
mrwrC
(13)
Where,
8
3
'
v
C

 for a smooth cylinder.
4
3
'
v
C

 for cylinder with a small bore.
1'C for thin cylinder.
rr is the rotor radius.
m
w is the mechanical angular speed.
 is the density of the material. v
is the Poisson’s ratio (ratio of lateral contraction to longitudinal extension in the
direction of the stretching force)
When the maximal mechanical stress of the rotor material is known, the equation can be
used to determine the maximum allowable radius r of the rotor. Typically, a safety factor has to be
used so as not to exceed the integrity of the rotor material. Poisson’s ratios vary slightly for
different materials. Table 4 lists some ratios for pure materials.
Table 4

19. 19

Example # 3
Calculate the maximum diameter for a smooth steel cylinder having a small bore. The
cylinder is rotating at 15000 rpm. The yield strength for the material is 300 N/mm
2
= 300 MPa.
The density of the steel is ρ = 8760 kg/m
3
.
Solution: -
Poisson’s ratio for steel is 0.29 and for a cylinder with steel bore, 8225.0
4
29.03
4
3
' 




v
C
.13.0
60
150002
87608225.0
10300
'
'
2
6
2
max
max
22
max
m
wC
r
wrC
m
r
mr
















Electrical Loadability: -
Resistive losses Pcu in a winding are proportional to the square of the current density J and
to the mass of the conductors. If the rotor dimensions (diameter, length, etc.) of the machine are
assumed to be variable over the scale  (lengths and diameters are proportional to , areas to 2

and volumes to3
 ), resistive loss,
322
 JmJP
cucu (13)
Thermal resistance between conductors and teeth is
ii
i
th
S
d
R

 (14)
where di is the thickness of the slot insulation, i
 is the thermal conductivity of the
insulation and Si the area of the slot wall. The thickness of the insulation di is constant independent
of the machine size (it depends on the rated voltage) and hence 2
1


th
R
(14)
Temperature difference between conductors and the teeth is



2
2
32 1
JJRPT
thcu
 (15)

20. 20

For a given temperature difference, 
2
J is constant, therefore


1
J (16)
The linear current density A in this case may be calculated as the total RMS current in a
slot divided by the slot pitch.






2
,
1



S
ScuS
S
S
SJI
A (17)
The linear current density A and the current density J are dimensions of the electrical
loading of a machine. The product of A and J is 


1
AJ
Constant. (18)
Thus AJ is independent of the size of the machine; it depends only on the effectiveness of
the cooling of the machine. For totally enclosed machines, AJ is smaller than for open circuit
cooling. In air-cooled machines, the product AJ ranges from 10×10
10
A
2
/m
3
to 35×10
10
A
2
/m
3
,
but in different machines with the same kind of cooling, the product is approximately the same
independent of the machine size. In the case of direct water cooling, AJ attains essentially higher
values. Table 5 depicts the value of ‘AJ’ for different machines:
Table 5: Value of product AJ for different machines.

21. 21

Magnetic Loadability:
The air-gap flux density and the supply frequency of a machine determine the magnetic loading of
the machine.
Iron loss, femfe
VfBP
22

Where Bm is the maximum flux density, ‘f’ is the supply frequency and Vfe is the volume of the
iron core.
More is the dimension of the machine less is the speed and vice-versa, therefore maximum speed
of the machine is inversely proportional to the linear dimension, i.e.


1
max
n (19)
max
2
3
max
2
max
2 11
n
B
n
nBP
mmfe
 (20)
Temperature rise of a machine depends on power loss per cooling area. Assuming constant
temperature rise,
max
2
max
max
2 11
n
BConstn
n
BConst
S
P
mm
fe
 (21)
So maximum power available by increasing the speed of the machine is
3
max
max
max
max
2
maxmaxmax
max
2
max
1
1111
'
n
P
n
nnnn
nlDABP
m



 (22)
The maximum length of the rotor lmax that guarantees operation below the first critical speed is
empirically defined as S
EI
kw
nl
m

2
22
max
(23)
Where
S is the area of the cross-section of the cylinder (m
2
),
E is the modulus of elasticity (Young’s modulus) of the rotor material, typically 190–210
GPa for steel,
I is the second moment of inertia of area (m
4
),
n is the order of the critical rotation speed,
k is the safety factor (the ratio of the n
th
critical angular speed to rated angular speed),

22. 22

is the density of material.
The ratio of length to diameter of various machines are depicted in table 6.
Table 6:

Air gap: -
An air gap δ of a 50 Hz asynchronous machine can be calculated in metres as a function of
power P with the equations:
m
P
1000
01.02.0
4.0

 for machines with P = 2.
(23) m
P
1000
006.018.0
4.0


for machines with P > 2.
(24)
In drives for extremely heavy duty, the air gap is increased by 60%.
Example: - What is a suitable air gap for a 110 kW, six-pole, heavy-duty induction motor?
Solution: - Air gap, mm
P
285.1
1000
)10110(006.018.0
6.1
1000
006.018.0
6.1
4.034.0






23. 23

Design Processes
The design process of a rotating electrical machine can be carried out for instance in the
following order. This procedure can be directly applied to asynchronous motors, but it is also
applicable to the design of other type of machines.
1. The initial design parameters are checked.
2. The tangential stresstan
 or the machine constant C (from table 3 of previous chapter)
are determined according to the power, speed and cooling method of the machine.
3. The rotor size is determined according to the required torque using equation (2) of
previous chapter i.e.
r
VT
tan
2 (1)
Then a suitable ratio of length to air-gap diameter ( ) is selected (using table 6 of
previous chapter). This ratio, then, determines the air-gap diameterD and the
equivalent length 'l of the machine.
From equation (1), volume of the rotor
tan
2
tan
2
'
42 


T
lD
T
V
r
 (2)
D
l'
 (3)
Solving equations (2) and (3), diameter and effective length of the machine is found
out.
4. A suitable air-gap is defined for the machine (using equations (23) & (24) of previous
chapter).
5. A suitable stator winding is selected for the machine. This is a decisive phase with
respect to the final characteristics of the machine. A guiding principle is that a poly-
phase winding produces the more sinusoidal current linkage; the more slots there are
in the stator. The winding factors for a slot winding usually become lower for the

24. 24

harmonics as the number of slots increases. Recommended slot pitches for different
types of machines are depicted in table 1
Table # 1

6. Since the tangential stress or the machine constant has already been selected, the air-
gap flux density Bm has to correlate with the selected machine constant. The initial
value employed in the calculation can be selected according to Table 1 of previous
chapter.
7. As the main dimensions, the winding method and the air-gap density have been
selected, the required number of coil turns per phase ph
T can be defined with the desired
emf. The emf per phase Em=ωΨm induced by the air-gap flux linkage (Ψm = ImLm) can
first be estimated from the RMS value of the fundamental terminal voltage (V):
For induction motors, Eph ≈ 0.96V − 0.98V
For generators, Eph ≈ 1.03V − 1.06V
The main flux penetrating a winding varies almost sinusoidally as a function of time
and can be expressed as:
wt
m
sin (4)
Instantaneous value of induced emf/phase,
wphpmiwphmiwphavph
kTlBwkTfKTfE 

 '
2
1
2
2
44.4  (5)


2

i for sinusoidal flux density distribution.
If the machine is dimensioned in such a way that its iron parts saturate at the peak value of
the flux density, the flux density distribution is flattened. In ordinary network-supplied induction
motors, both the stator and rotor teeth are saturated at the peak value of the flux density. This leads
to a higher reluctance of these teeth when compared with other teeth, and thusi
 takes notably

25. 25

higher values than the value corresponding to a sinusoidal distribution. The factori
 has to be
iterated gradually to the correct value during the design process. The value 64.0
i
 of an
unsaturated machine can be employed as an initial value. The theoretical maximum value of i
 is
1. In practice, i
 should not exceed 0.85. In that case, the MMF difference in the stator (s
MMF )
and rotor (rMMF ) teeth is already higher than the MMF of the air gap  
a
MMF . To simplify the
machine design, the factor i
 can be determined beforehand for different phases of saturation.
Figure below illustrates i


as a function of saturation factorsat
K .

Figure 7.1 Effect of the saturation factor on the factor. a
rs
sat
MMF
MMFMMF
K


(6)
From equation (5) number of turns per phase can be determined as:
wpmi
ph
ph
klBw
E
T
'
2
 (7)
Note that the peak fluxΦm is calculated per single pole. In a symmetrical machine, the
flux of each pole is of equal magnitude. Tph is the number of coil turns in series
needed for the assembly. This number can also, if required, be divided into several
pole pairs. If there are two pole pairs in the machine, for instance, they can be connected

26. 26

in parallel if desired, the number of turns now being Tph in both pole pairs. The number
of parallel paths in such a case is a = 2. If the pole pairs are connected in series
instead, the number of turns becomes Tph /2 for both pole pairs, the total number
being Tph. If possible, it is advisable to connect the coils in series, since in that case
the possible asymmetries between the pole pairs do not cause circulating currents in the
machine. For example for a 4-pole machine if Tph = 400, if the windings of each pole
pair are to be connected in parallel then number of turns for each pole pair is 400. If
windings are to be connected in series, then number of turns for each pole pair should
be Tph/2 i.e. 200.
8. Next, we have to find a suitable integer closest to the previously calculated number of
turns Tph. In a phase winding, there are Tph turns in series. A single coil turn is composed
of two conductors in slots, connected by the coil ends. In a single-phase winding, there
are thus 2Tph conductors in series. With m phases in a machine, the number of
conductors becomes 2mTph. There may be ‘a’ number of parallel paths in a winding, in
which case the number of conductors is 2amTph. The number of conductors per slot
becomes: S
amT
Z
ph
S
2

(8)
Where S is the number of slots and ZS should be an integer.
9. The selected number of turns for the phase winding has an effect on the value of
maximum flux density Bm. For the newly calculated turns per phase, new value of Bm
can be calculated using equation (7).
Example # 1: - Find a suitable number of stator turns for the motor with m = 3, P = 4, Bm = 1 T,
l’ = 0.2 m, τp = 0.2 m, f = 50 Hz, phase voltage Vph = 400V, and q = 4 with no short pitching.
Takei
 = 0.72.
Solution: - Distribution factor,)2/sin(
)2/sin(


m
m
k
d
 .
Slot angle, 0
000
15
2
4
434
360
2
360
2
360





















P
Pmq
P
S
 .

27. 27

So, 958.0
)2/15sin(4
)30sin(
0
0

dk .
Number of tu rns in series, .643.63
958.02.02.0172.0502
40097.02
'
2





wpmi
ph
ph
klBw
E
T

If all the coils are connected in series there will be 32 turns per pole pairs with turns. If
they are arranged in two parallel paths there will be 64 turns per pole pairs.
So for series connection of all turns for one phase, number of conductors in one slot
.8
44
264






Pq
Z
Z
ph
S
If the windings are arranged in two parallel paths for one phase, number of conductors in
one slot
.16
44
2264






Pq
Z
Z
ph
S
If a series connection is possible, it is always safest to select it, because with parallel paths,
there exists a possibility of circulating current, if any non-symmetry is present in the construction.

10. The air-gap flux density Bm being determined, the stator and rotor teeth are dimensioned
next. The flux densities of the stator and rotor teeth are chosen for normal machines
according to the permitted values presented in Table 6.1. In high-frequency machines,
it may be necessary to select values notably lower than the values presented in Table
6.1 to avoid excessive iron losses. When the apparent reference flux densities are
selected for the stator and rotor teeth, the widths bds and bdr are calculated for the stator
and rotor teeth using the selected flux densities at the teeth.
11. In order to determine the dimensions of the stator and rotor slots, we have to estimate
the stator and rotor currents. In synchronous and asynchronous motors, the stator
current Is is obtained with the shaft power P0, the stator phase voltage Vph, the efficiency
η and the power factor cos ϕ.

28. 28

For electric motors, stator current,
cos
0
ph
S
Vm
P
I
For electric generators,
cos
0
ph
S
mV
P
I
In an induction motor, the rotor current referred to the stator is approximately of the
same magnitude as the real component of the stator current (since the magnetizing current
flows only in the stator). cos
'
sr
II

The real rotor current is defined by the transformation ratio between the rotor and the stator.
The current of the bar of the cage winding of an induction motor is written as:
cos
s
r
sS
r
I
S
S
A
Z
I

When the stator and rotor currents have been resolved, the areas of the conductors are to
be determined next. Cross-sectional areas of stator and rotor can be calculated as follows: SS
S
S
JA
I
a

rr
r
r
JA
I
a
Then stator slot area and rotor slot area can be determined by taking stator and rotor space
factors into account.
Stator solt area,
Scs
SS
Sta
K
aZ
A
,

Rotor slot area,
rcs
rr
Rot
K
aZ
A
,


29. 29

The space factor KCu of the slot depends principally on the winding material, the voltage
level and the winding type of the machine. The windings of small electrical machines are
usually made of round wire. In that case, the space factor of an insulated wire in a free slot
(with the area reserved for the slot insulation subtracted) varies, depending on the quality of
winding assembly, from 60 to 66%. If aluminium bars are die cast in the induction motor, the
space factor becomes kCu,r = 1.
12. When the air-gap diameter, the air gap, the peak value of the air-gap flux density and
the dimensions of the stator and rotor slots of the machine are known, we may start to
calculate the magnetomotive force over the air gap and the teeth.

30. 30

Design Problem
Design a 2-phase, 90 W, 230 V, 4-pole, 50 Hz, 1400 rpm Induction Motor.
Solution: -
1. 2-phase, 230 V, 0.43 A, 90 W, 4-pole, 50 Hz, 1400 rpm Induction Motor.
2. Assume that the tangential stress, 2
tan /12000mN .
3. Torque experienced by the rotor, N
N
P
N
P
w
P
T 6139.0
1400
90
55.955.9
60
2
000

 .
4. We know that, r
VT
tan
2 .
Let 2
tan /12000mN (Assuming mkAAmWbB
m /30&/7.0
2
 from table 6.1, 6.2
and 6.3)
So, 3232
335
tan
57.32'58.25'
4
)(58.2510558.2
120002
6139.0
2
cmlDcmlD
Calculatedcmm
T
V
r






 (1)
(But diameter of the rotor is 6.27 cm and length is 2.7 cm)
So actual volume of the rotor is, 32
3322
14.106'
37.837.227.6
4
'
4
cmlD
cmcmlDV
r



Ratio of length to diameter of the machine is 43.0
'

D
l

(2)
Since actual volume of the machine is bigger than the calculated one, the tangential stress
is much less than the assumed value 12000 N/m
2
.
Actual stress experienced by the rotor is, 2
6
,
tan /78.3681
1037.832
6139.0
2
mN
V
T
actualr





(3)
So the assumed values of Bm and A are not same as the values taken in step 4. To determine
these values we need to cross check with the number of turns per phase of the machine.

31. 31

There are 6 coils for each phase and each coil consists of nearly 300 turns. So actual
number of turns/phase of the machine is 1800 (approx.).
Assume that VoltVE
ph
8.22023096.096.0  .
But wphpmiwphmiwphavph
kTlBwkTfKTfE 

 '
2
1
2
2
44.4  (4) 

2

i
for sinusoidal flux density distribution.
So number of turns/phase,
1800
'
2

wpmi
ph
ph
klBw
E
T
 (5)
In equation (5) cml3'
cm
P
D
p
92.4
4
27.6





Winding factor ddpw
kkkk  (assuming full pitch winding)
)2/sin(
)2/sin(


q
q
k
d

3
24
24
// 

 phasepoleslotsofNumberq
Slot angle electrical
P
S
0
00
30
2
4
24
360
2
360

So distribution factor, 911.0
)2/30sin(
)45sin(
)2/sin(
)2/sin(
0
0

qq
q
k
d


From equation (5) number of turns/phase,
TB
B
T
m
m
ph
645.0
911.00492.003.01800
2
502
8.2202
1800
911.00492.003.0
2
502
8.2202












Using this value of peak flux density and tangential stress, linear current density (A) can
be found.

32. 32

Tangential stress,
mAA
AB
m
/74.10090
8.0645.0
78.36812
cos
2
tan 


 
5. Number of conductors per slot, 300
24
18002122



S
amT
Z
ph
S

6. In order to determine the dimensions of the stator slot, we have to first estimate the stator
currents. In synchronous and asynchronous motors, the stator current I is obtained.
Stator current/phase, A
V
P
I 256.0
85.02308.02
90
cos2
0





 .
Area of bare copper conductor, 22
0569.0000569.0
4501
256.0
mmcm
Ja
I
a
cus





So the copper wire selected for the machine is 32 SWG.
Standard Wire
Gauge (SWG)
Diameter
in mm
Cross-sectional
area in mm
2

Resistance per length
in Ω/Km in µΩ/cm
31 0.295 0.0682 14.8 148
32 0.274 0.0591 21.3 213
33 0.254 0.0507 26.3 263

Taking space factor as 0.3, slot area required to accommodate )300(
S
Z conductors,
2
591.0
3.0
000591.0300
3.0
cm
aZ
a
cusS
S





7. Air gap required for the machine, mmmm
P
19.0
1000
4006.018.0
1000
006.018.0
4.04.0






8. Inner diameter of the stator, cmd
is
31.6019.0227.6 

Slot pitch, cm
S
d
is
s
826.0
24
31.6





Width of the slot, cmb5.0
4
Depth of the slot, cmh
s
18.1
5.0
591.0


33. 33

Width of teeth, cmbb
sds
326.05.0826.0
4



Figure 1 Slot width, slot height and overall dimension of the stator of a rotating machine.
9. Outer diameter of the stator, cmhhdD
yssisos
47.114.1218.1231.6 

10. Final dimensions: -
Rotor inner diameter = 6.27 cm (rotor)
Rotor length = 2.7 cm
Slot dimension
Width of slot = 0.5 cm
Depth of slot = 1.18 cm
Teeth dimension
Width of teeth = 0. 326 cm
Depth of teeth =1.18 cm
Outer diameter of stator = 11.47 cm