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:
D2L = Q/ (Co ns ) m3
Co = 11 Bav q Kw η cosΦ x 10-3
= 11x 0.48
x 26000 x 0.955 x 0.88 x 0.86 x 10-3
= 99.2
and ns = 16.67
rps
D2L = 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) Bav = (π x 0.33 x 0.17/ 6) x 0.48 = 0.141
wb
Assuming
Eph =Vph = 440 volts
Tph
= Eph / 4.44fΦkw = 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 ns= 120
x 50 /(4 x 60) = 25 rps;
we have
output coefficient Co = out put / D2Lns = 15 /( 0.252 x 0.16 x 25) = 60
we have Co
= 11 Bav q Kw η cosΦ x 10-3
= 11 x Bav x 23000 x 0.955x 0.84 x 0.82 x
10-3
= 166.42
Bav
Hence Bav =
60/166.42 = 0.36 Tesla
Pole
pitch τp= π D/p = π x 0.25/4 = 0.196 m; L/ τp = 0.815
For 11kW
motor: the design data from 15 kW
machine has to be taken
So Bav = 0.36
Tesla; q = 23000 ac/m ; L/ τp =
0.815; and C0 = 60
Synchronous
speed = 120 x 50 / (6 x 60) = 16.667 rps;
D2L = Q/ (Co ns ) m3
= 11 /
(60 x 16.667) = 0.01099 m3
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 D2L product and solving for D and L
0.427 D3
= 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/ τs for slot pitch 1.5 cm, Ss = π x 30 /
1.5 = 63
For slot
pitch 2.5 cm Ss = π x 30 /
2.5 = 37 Hence number of slots must be between 37 & 63
Assuming
no. of stator slots /pole/phase = 3, Ss = 6 x 3 x 3 = 54
Flux per
pole = Bav
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 Tph= Eph/ (4.44 f Kw) = 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 Bav
= 0.5wb/m2, q = 30000 ac/m, efficiency = 90 % and power factor =
0.9, winding factor = 0.955, current density =3.5 a/mm2, 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 Tph = 32 x 12/2 = 192,
Assuming
full pitched coils, kw = 0.955, Eph = Vph and star connected stator winding,
Eph
= 3600/√3 = 2078 volts,
We have Eph = 4.44fΦTphkw, ie
Φ= Eph
/( 4.44fTphkw) =
2078 /( 4.44 x 50 x 192 x
0.955) = 0.051wb
(ii) Gap
flux density Ag = πDL/p = π x 0.67 x 0.38 / 8 = 0.1 m2
Bg
= Φ/ Ag
= 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 103 / 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/mm2, area of cross section of the
conductor = 53.47 /4.1 = 13.04 mm2 as the conductor section is >
5 mm2 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 mm2
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, lmt = 2L +
2.3 τp + 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 lmt x Tph
) / as = 0.021 x 1.6 x 192 / 13.2 = 0.49 ohm.
Total
copper losses = 3Is2rs = 3 x 53.472
x 0.49 =4203 watts
(vi) Flux
density in stator tooth
Diameter
at 1/3rd height, D' = D + 1/3 x hts x 2 = 0.67
+ 1/3 x 0.035 x 2 = 0.693 m
Slot
pitch at 1/3rd height = τ's = π x D' /Ss =
π x 0.693 /96 = 0.02268 m
Tooth
width at this section = b't = τ's – bs
= 0.02268 – 0.012 = 0.0168 m
assuming
3 ventilating ducts with 1cm width and iron space factor of 0.95 Iron length li = ( 0.38
-3 x 0.01) 0.95 = 0.3325 m
Area of
the stator tooth per pole A't
= b't x li x number
of teeth per pole
= b't
x li x Ss
/p = 0.01068 x 0.3325 x 96/8
= 0.04261 m2
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 dcs = ½ ( Do- D – 2 hss) = ½ (
0.86 - 0.67 – 2 x 0.035) = 0.06 m
Area of
stator core Ac = Li
x dcs = 0.3325 x 0.06 = 0.01995 m2
Flux in
stator core = ½ x Φ = ½ x
0.051 = 0.0255 wb
Flux
density in stator core, Bc = Φc/ Ac
= 0.0255/ 0.01995 = 1.28 Tesla
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