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# Solved problems: Design of Rotor - Induction Motors

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

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 + 2DL

=  0.2 +20.75 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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Design of Electrical Machines : Induction Motors : Solved problems: Design of Rotor - Induction Motors |