Design of Electrical Machines - Induction Motors - Solved problems: Design of Stator - Induction Motors

**Problems**

**Ex. 1. **Obtain
the following design information for the stator of a 30 kW, 440 V, 3Φ, 6 pole, 50** **Hz delta
connected, squirrel cage induction motor, (i) Main dimension of the stator,
(ii) No. of turns/phase (iii) No. of stator slots, (iv) No. of conductors per
slot. Assume suitable values for the missing design data.

**Soln: **Various
missing data are assumed from referring to Design data Hand Book or tables in** **Text Book
considering the size, economics and performance

Specific
Magnetic loading, Bav = 0.48 Tesla

Specific
Electric loading, q = 26000 ac/m

Full load
efficiency, η = 0.88

Full load
power factor cosΦ = 0.86

Winding
factor Kw = 0.955

(i) Main dimensions

We have
from output equation:

D** ^{2}**L = Q/ (C

C**o** = 11 B** _{av}** q K

= 11x 0.48
x 26000 x 0.955 x 0.88 x 0.86 x 10^{-3}

= 99.2

and n** _{s}** = 16.67
rps

D** ^{2}**L = 30/(99.2 x 16.67) = 0.0182 m3

Designing
the m/c for bets power factor

D =
0.135P√L

= 0.135 x
6√L

Solving
for D and L D = 0.33 m and L = 0.17 m

(ii) No.
of stator turns

Φ = (πDL/p) B_{av} = (π x 0.33 x 0.17/ 6) x 0.48 = 0.141
wb

Assuming
E_{ph} =V_{ph} = 440 volts

T_{ph}
= E_{ph} / 4.44fΦk_{w} = 440/(4.44 x 50 x
0.0141 x 0.955)

= 148

(iii) No. of
stator slots

Assuming
no of slot/pole/phase =3 Total no. of slots = 3 x 3 x 6 = 54 (iv) No of
conductors /slot

Total no
of conductors = 148 x 2 = 296

No. of
conductors /slot = 296/54 = 5.5

Assuming
76 conductors/ slot

Total no.
of conductors = 54 x 6 = 324

Revised
no. of turns/phase = 162

**Ex. 2 **A 15 kW
440m volts 4 pole, 50 Hz, 3 phase induction motor is built with a stator bore** **of 0.25 m
and a core length of 0.16 m. The specific electric loading is 23000 ac/m. Using
data of this machine determine the core dimensions, number of slots and number
of stator conductors for a 11kW, 460 volts,6 pole, 50 Hz motor. Assume full
load efficiency of 84 % and power factor of 0.82. The winding factor is 0.955.

**Soln:** For 15 kW motor:

Motor
Input = 15 /0.84 = 17.857
kW ; Synchronous speed n_{s}= 120
x 50 /(4 x 60) = 25 rps;

we have
output coefficient C_{o} = out put / D** ^{2}**Ln

we have C_{o}
= 11 B** _{av}** q K

= 166.42
B_{av}

Hence B** _{av}** =
60/166.42 = 0.36 Tesla

Pole
pitch τ_{p}= π D/p = π x 0.25/4 = 0.196 m; L/ τ_{p} = 0.815

For 11kW
motor: the design data from 15 kW
machine has to be taken

So B** _{av}** = 0.36
Tesla; q = 23000 ac/m ; L/ τ

Synchronous
speed = 120 x 50 / (6 x 60) = 16.667 rps;

D** ^{2}**L = Q/ (C

= 11 /
(60 x 16.667) = 0.01099 m^{3}

L/ (π D /p) = 0.815 , So L/D = 0.815 x π /6 = 0.427
or L = 0.427 D

Substituting
this value in D** ^{2}**L product and solving for D and L

0.427 D^{3}
= 0.01099 hence D = 0.30 m and L =
0.125 m

Number of
slots: Considering the slot pitch at the air gap between 1.5 cm and 2.5 cm

Number of
slots = π x D/ τ_{s} for slot pitch 1.5 cm, S_{s} = π x 30 /
1.5 = 63

For slot
pitch 2.5 cm S_{s} = π x 30 /
2.5 = 37 Hence number of slots must be between 37 & 63

Assuming
no. of stator slots /pole/phase = 3, S_{s} = 6 x 3 x 3 = 54

Flux per
pole = B_{av}
x D x L / p = 0.36 x π x 0.3 x 0.125/6 = 7.07 x 10^{-3}
wb

Assuming
star delta connection for the machine, under running condition using Delta
connection

Stator
turns per phase T_{ph}= E_{ph}/ (4.44 f K_{w}) = 460 /(4.44 x 50 x 7.07 x 10^{-3} x
0.955) =307

Number
conductors/phase = 307 x 2,

Total
number of stator conductors = 307 x 2 x 3 =1872

Number of
conductors per slot = 1872/54 = 34.1 ≈ 34

Hence
total number of conductor = 34 x 54 =1836.

**Ex. 3 **Determine
main dimensions, turns/phase, number of slots, conductor size and area of slot** **of 250
HP, 3 phase, 50 Hz, 400 volts, 1410 rpm, slip ring induction motor. Assume B_{av}
= 0.5wb/m^{2}, q = 30000 ac/m, efficiency = 90 % and power factor =
0.9, winding factor = 0.955, current density =3.5 a/mm^{2}, slot space
factor = 0.4 and the ratio of core length to pole pitch is 1.2. the machine is
delta connected.

**Ex. 4. **During
the preliminary design of a 270 kW, 3600 volts, 3 phase, 8 pole 50 Hz slip ring** **induction
motor the following design data have been obtained.

Gross
length of the stator core = 0.38 m, Internal diameter of the stator = 0.67 m,
outer diameter of the stator = 0.86 m, No. of stator slots = 96, No. of
conductors /slot = 12, Based on the above information determine the following
design data for the motor. (i) Flux per pole (ii) Gap density (iii) Conductor
size (iv) size of the slot (v) copper losses (vi) flux density in stator teeth
(vii) flux density in stator core.

**Soln. **(i) Flux
per pole

Number of
slots per phase 96/3 = 32

Number of
turns per phase T_{ph} = 32 x 12/2 = 192,

Assuming
full pitched coils, k_{w} = 0.955, E_{ph} = V_{ph} and star connected stator winding,

E_{ph}
= 3600/√3 = 2078 volts,

We have *E _{ph}*

Φ*= E _{ph}*

(ii) Gap
flux density A_{g} = πDL/p = π x 0.67 x 0.38 / 8 = 0.1 m^{2}

B_{g}
= Φ*/* A_{g}
= 0.051/ 0.1 =0.51 Tesla

(iii)
Conductor size

Assuming
an efficiency of 91% and a full load
power factor of 0.89

Input power
to the motor = 270 x 10^{3} / 0.91
= 296703 w

Full load
current per phase = 296703 / ( 3 x 2078 x 0.89) = 53.47 amps

Assuming
a current density of 4.1 amp/mm^{2}, area of cross section of the
conductor = 53.47 /4.1 = 13.04 mm^{2} as the conductor section is >
5 mm^{2} rectangular conductor is selected. Standard size of the
conductor selected satisfying the requirements is 2.5 mm x 5.5 mm.

Thus
sectional area of the conductor 13.2 mm^{2}

Size of
the conductor with insulation thickness of 0.2 mm is 2.9 mm x 5.9 mm

(iv) size
of the slot

12
conductors per slot are arranged in two layers with 6 conductors in each layer.
Six conductors in each layer are arranged as 2 conductors depth wise and 3
conductors width wise. With this arrangement the width and depth of the slot
can be estimated as follows.

(a) Width of the slot

Space
occupied by insulated conductor, 3 x 2.9 8.7
mm

Coil
insulation, 2 x 1.0 2.0 mm

Slot
liner, 2 x 0.2 0.4 mm

Clearance 0.9 mm

Total
width of the slot 12.0 mm

(b) Depth
of the slot

Space
occupied by insulated conductor, 4 x 5.9

23.6 mm

Coil
insulation, 4 x 1.0 4.0 mm

Slot
liner, 3 x 0.2 0.6 mm

Coil
separator, 1 x 1.0 0.5 mm

Top
liner, 1 x 1.0 0.5 mm

Wedge 3.0 mm

Lip 1.5 mm

Clearance 1.3 mm

Total
height of the slot 35.0 mm

Thus the
dimension of the slot 12.0 mm x 35.0 mm

(v)
Copper losses in stator winding

Length of
the mean turn, *l*_{mt} = 2L +
2.3 τ_{p} + 0.24 =
2 x 0.38 + 2.3 x π x 0.67/8
+ 0.24 = 1.6 m Resistance per phase = (0.021 x *l*_{mt} x T_{ph}
) / a_{s} = 0.021 x 1.6 x 192 / 13.2 = 0.49 ohm.

Total
copper losses = 3I_{s}^{2}r_{s} = 3 x 53.47^{2}
x 0.49 =4203 watts

(vi) Flux
density in stator tooth

Diameter
at 1/3^{rd} height, D^{'} = D + 1/3 x h_{ts} x 2 = 0.67
+ 1/3 x 0.035 x 2 = 0.693 m

Slot
pitch at 1/3^{rd} height = τ^{'}_{s} = π x D^{'} /S_{s} =
π x 0.693 /96 = 0.02268 m

Tooth
width at this section = b^{'}_{t} = τ^{'}_{s} – b_{s}
= 0.02268 – 0.012 = 0.0168 m

assuming
3 ventilating ducts with 1cm width and iron space factor of 0.95 Iron length *l*_{i} = ( 0.38
-3 x 0.01) 0.95 = 0.3325 m

Area of
the stator tooth per pole A^{'}_{t}
= b^{'}_{t} x *l*_{i} x number
of teeth per pole

= b^{'}_{t}
x *l*_{i} x S_{s}
/p = 0.01068 x 0.3325 x 96/8

= 0.04261 m^{2}

Mean flux
density in stator teeth B'_{t} =
Φ / A^{'}_{t} =
0.051/ 0.04261 = 1.10 Tesla

Maximum
flux density in stator tooth =1.5 x 1.10 = 1.65 Tesla

(vii) Flux density in stator core

Depth of
the stator core d_{cs} = ½ ( D_{o}- D – 2 h_{ss}) = ½ (
0.86 - 0.67 – 2 x 0.035) = 0.06 m

Area of
stator core A_{c} = L_{i}
x d_{cs} = 0.3325 x 0.06 = 0.01995 m^{2}

Flux in
stator core = ½ x Φ = ½ x
0.051 = 0.0255 wb

Flux
density in stator core, B_{c} = Φ_{c}/ A_{c}
= 0.0255/ 0.01995 = 1.28 Tesla

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