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# Solved problems: Design of Stator - Induction Motors

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

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 = 0.135PL

= 0.135 x 6L

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, Ssπ x 30 / 1.5 = 63

For slot pitch 2.5 cm     Ssπ 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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Design of Electrical Machines : Induction Motors : Solved problems: Design of Stator - Induction Motors |