Ex.1.
During
the stator design of a 3 phase, 30 kW, 400volts, 6 pole, 50Hz,squirrel cage
induction motor following data has been obtained. Gross length of the stator =
0.17 m, Internal diameter of stator = 0.33 m, Number of stator slots = 45,
Number of conductors per slot = 12. Based on the above design data design a
suitable rotor.
Soln:
(i)
Diameter of the rotor
Length of
the air gap lg = 0.2 + 2 √DL mm
= 0.2 + 2 √0.33 x 0.17 mm
= 0.67 mm
Outer
diameter of rotor Dr = D - 2 lg
= 0.33 – 2 x 0.67
x 10-3
= 0.328 m
(ii)
Number of rotor slots
(a) Ss
> Sr
(b) To avoid
cogging and crawling: Sr ≠ Ss, Ss - Sr ≠ ±3P Sr ≠ 45, Ss - Sr ≠ ± 3P → 45 – 18 ≠ 27,
(c) To avoid
synchronous hooks and cusps in torque speed characteristics Ss - Sr ≠ ±P,
±2P, ±5P
Ss - Sr ≠ (45 – 6), (45 – 12), (45 – 03) ≠ 39, 33, 15
To avoid noisy operation Ss - Sr ≠ ±1, ±2, (±P ±1), (±P ±2) Ss - Sr ≠ (45 – 1) , (45 – 2), (45 – 7),
(45 – 8)
Considering
all the combination above Sr = 42
Rotor
slot pitch = πDr / Sr =
π x 32.8 / 42 = 2.45 cm (quite
satisfactory) (iii) Rotor bar current
Assuming
star – delta connection for stator winding
Vph = 400
volts
Assuming η = 88 %
and p.f = 0.86
Motor
input = 30/0.88 = 30.1 kW
Full load stator current = input / 3 vph cosΦ
= 30.1 x
103/ 3 x 440 x 0.86
= 33 amps
I'r
= 0.85 Is
= 0.85 x 33
= 28 amps
Assuming
Kws = 0.955 & No. of rotor cond/slot = 1
Ib
= ( Kws x Ss x Z's
) x I'r
/ ( Kwr x Sr x Z'r
)
= (0.955 x
45 x 12) x 28 /( 1 x 42 x 1)
343.8amps
(iv) Size
of rotor bar and slot
Assuming
the current density in rotor bars = 6.0
amps/mm2
Ar = Ir / δr mm2
Ar
= 343.8/
6.0
= 57. 3
mm2
Selecting
rectangular standard conductor available
Area of
conductor = 57.6 mm2
Hence
standard conductor size = 13 mm x 4.5 mm
Size of
rotor slot to fit the above cond = 13.5 mm x 5 mm
(v)
Resistance of rotor bar
Length of
rotor bar lb = L + allowance for skewing + allowance between
end rings and rotor core
lb
= 0.17
+0.05 =0.22 m
Rotor bar
resistance = 0.021 x lb / Ab
= 0.021 x
0.22 / 57.6
= 8.02 x 10-5
ohm
Copper
loss in rotor bars = Ib2 x rb x number
of rotor bars
= 343.82 x 8.02 x 10-5 x 42
= 398 watts
(vii) End
ring current Ie =
1/π x Sr/P x Ib
= 1/π x 343.8 x 7
= 765.8
amps
(viii) Area of
cross section of end ring Assuming a current density of 6.5 Amp/mm2
Area of
each end ring Ae = Ie / δe mm2,
= 765.7 /
6.5
= 117.8 mm2
(ix) Rotor dia
Dr = 32.8 cm,
Assuming
Dme 4.8 cms less than that of the rotor Dme = 28 cms
Mean
length of the current path in end ring lme
= πDme = 0.88 m
Resistance
of each end ring re = 0.021 x lme / Ae
= 0.021 x 0.88
/117.8
= 1.57 x 10-4 ohms
Total
copper loss in end rings = 2 x Ie2 x re
= 2 x
765.72 x 1.57 x 10-4
= 184 watts
(x) Equivalent
rotor resistance
Total
copper loss = copper loss in bars + copper loss in end rings
= 398 +
184 = 582 watts
Equivalent
rotor resistance r' = Total
rotor copper loss / (3 x Ir'2)
= 582 / ( 3
x 282)
= 0.247
ohms
Ex.2. A 3 phase
200 kW, 3.3 kV, 50 Hz, 4 pole induction motor has the following dimensions. Internal
diameter of the stator = 56.2 cm, outside diameter of the stator = 83cm, length
of the stator = 30.5 cm, Number of stator slots = 60, width of stator slot =
1.47 cm, depth of stator slot = 4.3 cm, radial gap = 0.16 cm, number of rotor
slots = 72, depth of rotor slot 3.55 cm, width of rotor slots = 0.95 cm.
Assuming air gap flux density to be 0.5 Tesla, calculate the flux density in
(i) Stator teeth (ii) Rotor teeth (iii) stator core.
Soln: (i) Flux
density in Stator teeth
Internal
diameter of stator = 56.2 cm, Depth of stator slot = 4.3 cm,
Diameter
at 1/3rd height from narrow end of the stator teeth D' = D + 1/3 x hts x 2
= 56.2 +
1/3 x 4.3 x2
= 59.1 cm
Slot
pitch at 1/3rd height τ's = π x D' /Ss
= π x 59.1/ 60 = 3.1 cm
Tooth
width at this section b't
= τ's – bs
= 3.1
-1.47
=
1.63 cm
Area of
one stator tooth a't
= b't x li
li = ki(L – nd x wd ) = 0.93(30.5 – 3 x 1) = 25.6 cm
Area of
stator tooth A't = b't x li
= 25.6 x
1.63
= 0.00418
m2
Number of
stator teeth per pole = 60 /4 =15
Air gap
area = π DL = π x 0.562 x 0.305 = 0.535 m2
Total
flux = Bav x π DL = 0.5 x 0.535 = 0.2675 wb
Hence
flux per pole 0.2675/4 = 0.06679 wb
Mean flux
density in stator teeth B't =
Φ / (A't x
no of teeth per pole)
= 0.0669
/(0.00418 x 15)
= 1.065
Tesla
Max flux density in stator teeth = 1.5 x 1.065 =
1.6 Tesla. (ii) Flux density in rotor teeth
Diameter
of the rotor = D – 2lg = 56.2 -2 x 0.16 = 55.88 cm Depth of rotor slot = 3.55
cm
Diameter
at 1/3rd height Dr' = D - 2/3 x htr x 2 =
55.88 - 2/3 x 3.55 x 2 =51.14 cm
Slot
pitch at 1/3rd height = τ'r = π x Dr' /Sr = π x 51.14
/72 = 2.23 cm
Width of
the rotor slot = 0.95 cm
Tooth
width at this section = b'tr = τ'sr – bsr = 2.23 –
0.95 = 1.28 cm
Iron
length li
= 25.6 cm
Area of
one rotor tooth = a'tr = b'tr x li
= 1.28 x 25.6 = 32.8 cm2 = 0.00328 m2
Number of
rotor tooth per pole = 72/4 = 18
Area of
all the rotor tooth / pole A'tr = b't
x li x Sr /P =
0.00328 x 18 = 0.05904 m2
Mean flux
density in rotor teeth B'tr = Φ / A'tr
= 0.0669 / 0.05904 = 1.13 Tesla
Maximum
flux density in the rotor teeth = 1.5 x 1.13 = 1.69 Tesla (iii) Flux density in
Stator core
Depth of
the stator core dc = ½ ( D0 – D – 2ht ) = ½ (
83 -56.2 – 2 x 4.3) = 9.1 cm
Area of
stator core Ac = li x dc = 25.6 x 9.1 =
233 cm2 = 0.0233 m2
Flux in
stator core = ½ c =0.5 x 0.0669 = 0.03345 wb
Flux
density in stator core = c /
Ac = 0.03345 / 0.0233 = 1.435 Tesla
Ex.3. A 3 phase
3000 volts 260 kW, 50 Hz, 10 pole squirrel cage induction motor gave the following
results during preliminary design.
Internal
diameter of the stator = 75 cm, Gross length of the stator = 35 cm, Number of
stator slots = 120, Number of conductor per slot =10. Based on the above data
calculate the following for the squirrel cage rotor. (i) Total losses in rotor
bars, (ii) Losses in end rings, (iii) Equivalent resistance of the rotor.
Soln. (i) Total
losses in rotor bars Number of stator slots = 120,
To
confirm to the requirements the rotor slots can be selected in the following
way
Number of rotor slots a) Ss > Sr
(b ) To avoid cogging and crawling: Sr ≠ Ss, Ss - Sr ≠ ±3P Sr ≠ 120, Ss - Sr ≠ ± 3P → 120 – 30 ≠ 90,
(c) To avoid
synchronous hooks and cusps in torque speed characteristics Ss - Sr ≠ ±P,
±2P, ±5P
Ss - Sr ≠ (120 – 10), (120 – 20), (120 –
50) ≠ 110, 100, 70
(d) To avoid
noisy operation Ss - Sr ≠ ±1, ±2,
(±P ±1), (±P ±2)
Ss - Sr ≠ (120 –
1) , (120 – 2), (120 – 11), (120 – 12) ≠ 119,
118, 109, 108
Considering
all the combination above Sr = 115
Rotor
slot pitch = πD / Sr = π x 75 / 115 = 2.048 cm (quite
satisfactory)
Rotor bar
current
Assuming η = 90 %
and p.f = 0.9
Motor
input = 260/0.9 = 288.88 kW
Assuming
star connection
Full load
stator current = input / (√3 VL cosΦ )
=
288.88 x 103/ (√3 x 3000 x 0.9)
= 61.5 amps
I'r =
0.85 Is = 0.85 x 61.5 = 52.275 amps
Assuming
Kws = 0.955 & No. of rotor cond/slot = 1
Ib
= ( Kws x Ss x Z's
) x I'r
/ ( Kwr x Sr x Z'r
)
= (0.955 x
120 x 10) x 52.275 /( 1 x 115 x 1)
= 521 amps
Area of
rotor bar
Assuming
the current density in rotor bars = 6.5
amps/mm2
Ab = Ib / δb mm2
Ab
= 521/ 6.5
= 80.2 mm2
Length of
rotor bar lb = L + allowance for skewing + allowance between
end rings and rotor core
lb
= 0.35
+0.05 =0.4 m
Rotor bar
resistance = 0.021 x lb / Ab
= 0.021 x
0.4 / 80.2
= 1.05 x 10-4
ohm
Copper
loss in rotor bars = Ib2 x rb x number
of rotor bars
= 5212 x 1.05 x 10-4 x 115
= 3278 watts
(ii)
Losses in end rings
End ring
current Ie = 1/π x Sr/P x Ib
= 1/π x (115/10) x 521
= 1906 amps
Area of
cross section of end ring
Assuming
a current density of 6.5 Amp/mm2
Area of
each end ring Ae = Ie / δe mm2,
= 1906/6.5
= 293.2 mm2
Air gap length lg
= 0.2 + 2√DL
= 0.2 +2√0.75 x 0.35
= 1.22 mm
Rotor diameter Dr
= D -2 lg
= 75 – 0.122
= 74.878 cm
Rotor
dia Dr = 74.878 cm,
Assuming
Dme 6.878 cms
less than that of the rotor Dme = 68 cms
Mean
length of the current path in end ring lme
= πDme = 2.136 m
Resistance of each end ring re = 0.021 x lme / Ae
= 0.021 x 2.136
/293.2
= 1.529 x 10-4 ohms
Total
copper loss in end rings = 2 x Ie2 x re
= 2 x 19062
x 1.529 x 10-4
= 1111.55
watts
(iii) Equivalent rotor resistance
Total
copper losses in the rotor = Copper loss in bars + copper loss in end rings
= 3278
+1111.55
= 4389.55
watts
Equivalent
Rotor resistance = Rotor cu loss / ( 3 I’
r2 )
= 4389.55/(3
x 52.2752)
= 0.535 ohm
Ex.4. Following
design data have been obtained during the preliminary design of a 3 phase, 850 kW,
6.6 kV, 50 Hz, 12 pole slip ring induction motor. Gross length of stator core =
45 cm, internal diameter of the stator core = 122 cm, number of stator slots =
144, Number of conductors per slot = 10. For the above stator data design a
wound rotor for the motor.
Soln : (i) Diameter of the rotor
Length of
the air gap lg = 0.2 + 2 √DL mm
= 0.2 + 2 √1.22 x 0.45 mm
= 1.68 mm
Outer
diameter of rotor Dr = D - 2 lg
= 1.22 – 2
x 1.68 x 10-3
= 1.217 m
(ii) Number of
rotor slots : Considering all the factors for selection of number of rotor
slots, and selecting fractional slot winding, assuming number of rotor slots
per pole per phase as 3½
Total
number of rotor slots = 3.5 x 12 x 3 = 126
Rotor
slot pitch = π Dr
/ Sr
= π x 1.217 / 126
= 0.0303 m
(quite satisfactory)
(iii) Number of
rotor turns: For this motor the voltage
between slip rings must be less than
1000
volts. Assume the voltage between slip rings as 600 volts.
Assuming
star connection for stator winding Es = 6600/√3 = 3810 volts, Assuming Kws
= Kwr =1
Rotor
winding will always be star connected
Total number
of stator conductors = 144 x 10
Total
number of stator turns per phase = 144 x 10 / (3 x 2) = 240
Rotor
turns per phase Tr = (Er/Es) x (Kws/Kwr)
Ts
= 600/√3 x 1 x 240 / 3810
= 22 turns
Rotor
conductors per phase = 44,
Number of
slots per phase = 126/3 = 42,
Therefore
number of conductors per slot = 1.
Final
rotor turns/phase = number of conductors
per phase / 2 = 42/ 2 = 21
(iv)
Rotor current
As the
motor is of 850 kW, efficiency will be high, assuming an efficiency of 92% and
cos = 0.91
Input to
the motor = 850/0.92 = 924 kW,
Full load
stator current per phase Is = 924 x 103 / (3 x 3180 x 0.91)
= 88.8 amps
Equivalent
rotor current Ir'
= 0.85 Is = 0.85 x 88.8 =75.5
amps
Ir = (Kws x Ss x Z's ) x I'r / ( Kwr x Sr x Z'r ) ;
= (144 x 10
x 75.5) / 126 x 1
= 863 amps
(v) Size of
rotor conductors
Assuming
a current density of 5 Amp/ mm2
for the rotor conductors ,
Cross
sectional area of the rotor conductor = 863/5 = 172.6 mm2
Size of
the rotor conductors is too large and this conductor can not be used as it is
and hence has to be stranded. Stranding the conductors into 4 rectangular
strips of each area 43.1 mm2, in parallel,
Standard
size of the rectangular strip selected = 11 mm x 4 mm,
Thus
sectional area of the rectangular conductor
43.1 x 4 = 172.4 mm2
Size of
the rectangular conductor with insulation = 11.5 mm x 4.5 mm
(vi) Size
of the rotor slot
Four
strips of the rectangular conductor are arranged as 2 strips widthwise and 2
strips depthwise, with this arrangement the size of the slot can be estimated
as follows
(a) width
of the slot
Space
occupied by the conductor
2 x4.5 9.0 mm
Slot
liner 2 x 1.5 3.0 mm
Clearance 1.0 mm
Total
width of the slot 13.0 mm
(b) Depth
of the slot
Space
occupied by the conductor 2 x11.5 23.0 mm
Slot
liner 3 x 1.5 4.5 mm
Wedge 3.5 mm
Lip 1.0 mm
Clearance 1.0 mm
Total
depth of the slot 34.0 mm
Thus size
of the rotor slot = 13 mm x 34 mm
(vi)
Resistance and copper losses
Length of
the mean Turn lmt = 2L + 2.3 τp + 0.08 m
lmt = 2x 0.45
+ 2.3 ( π x 1.22 / 12 ) + 0.08 m = 1.72 m
Resistance
of rotor winding is given by Rr = (0.021 x lmt x Tr
) / Ar
= (0.021
x 1.72 x 21) / 172.4 = 0.0044 ohm
Total
copper loss = 3 Ir2 Rr
Watts
= 3 x 8632
x 0.0044
= 9831
watts
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