Here, Diameter of stator Length of Stator No. of stator turns per phase No. of the stator slots No. of rotor slots Area of Cross-section of Stator conductor Area of Cross-section of Rotor Bars(as squirrel cage) Area of the cross-section of End-Ring Length of the Air-gap are calculated step by step .

1. Project On : Induction motor design
Course No : EE-3220
Group No : 16
Md. Ataur rahman 1703002
Shetu Mohanto 1703018
Md. Nafis Jawad 1703026
Md. Sabbir Mahmud 1703072
Submitted By :

2. DESIGN THE MAIN DIMENSION AND ROTOR OF
A 22KW, 415V, 3‐PHASE, 50HZ, 970 RPM,
SQUIRREL CAGE TYPE INDUCTION MOTOR
HAVING A FULL‐LOAD EFFICIENCY OF 90%
AND POWER FACTOR IS 0.9.
THE MACHINE IS TO BE STARTED BY A
STAR‐DELTA STARTER.

3. Determination of magnetic & Electrical loading:
As stated the motor must be started by star delta starter, so the
input side ( stator) must be delta connected.
The INPUT KVA Q
i = Output real power / ( full load efficiency *
full load P.F.)
= (22* 103)/(0.9*0.9) = 27.1605 KVA
Considering it is a 6 pole machine,
Synchronous speed of field in armature = (120* supply
freq.)/no. of pole = 1000 rpm = 16.67 rps

4. The winding factor depends on the distribution factor & pitch
factor.
Assuming, winding factor Kw = 0.95
The flux density depends on the material used for the stator core
as the maximum flux density must lie on the linear region.
Here, the flux density is chosen as,
Bav =0.45 Wb ;

5. Ampere conductor is defined as the average value of flux density
over the whole surface of air gap in the machine. It’s value varies
from 5000 to 450000 .
So, Ampere conductor is chosen as ,ac = 26000

6. Determination of output co-efficient:
The output co efficient Co ,
= 11 * Bav * Kw* ac * 10-3
=11*0.45*0.955*26000* 10-3
= 122.91 KVA / m3 - rps

7. Determination of D & L :
KVA input Qo = Co * D2 * L * Ns
=> D2 *L = 27.1605 / ( 122.91*16.67) = 0.0132 m3
For overall good design assumed , L/τ = 1
so, L = τ = (pi*D)/P
Then D2 * (pi*D/P) = 0.0132 => D3 = (0.0132*6)/ 3.1416
=> D = 0.293 m
Now , L = (pi*D)/P = 0.154 m

8. Determination of No. of turns per phase :
Since the stator is delta connected, the line voltage is equal to
phase voltage.
Maximum flux per pole φ
m = Bav*pi*D*L/P = 0.0106 Wb
Stator turns per phase Tp = Es/( 4.44*f* Kw* φ
m )
= 415/(4.44*50*0.95*0.0106)
= 185.4 ≈ 186

9. Determination of No. of slots per pole per phase :
No of stator slot Ss = no. of phase * pole * no slot per pole per
phase
= 3*6*X [where x is the unknown]
Stator slot pitch Ys =( pi*D )/ Ss
where Ss = no. of stator slots
stator slot pitch should lie between,
= 10 to 15mm ( for single layer winding )
= 15 to 25mm ( for double layer winding )

10. considering double layer winding ,
(pi*D)/(3*6*X) = 15 to 25 mm
=> X = 2.04 to 3.4.
So the no of slot range can vary from 3.4 to 2.04 which is
fractional value.
Taking integer value X=3,
the no. of total stator slots = 3*6*3 = 54
So , total no. of conductors = no of conductor in one turn * no of
phase * stator turns per phase
= 2*3*186 = 1116

11. Correction in the value of total no. of turns per phase :
Conductors per slot Zss = 1116 / 54 = 20.67
Considering Zss =21,
New value of total turn per phase = no. of slot * conductor per
slot / (conductor per turn * no of input phase in stator)
= 54* 21 / (2*3)
= 189

12. Total Power = P = 3 x Vph x Iph
Phase current in stator Ip = KVA / (3* Vph) = 27160.5/(3*415) =
21.816 A [ for delta , Vph = VL ]
Check for slot loading :
Slot loading = Iz* Zss = 21.816*21 = 458.136 amp conductor (ac)
Considering conductor current density 3.5 A /mm2,
Stator conductor cross section area =( phase current / density)
= 21.816/3.5 = 6.23 mm2

13. Air Gap calculation :
Length of Air gap Lg = 0.2+2*sqrt (DL) [For small size motor ]
= 0.2 + 2*sqrt(0.293*0.154)
= 0.625 mm

14. Calculation No. of Rotor slots :
Let Sr = Number of Rotor slots , Ss = No. of stator slots
We know that , (Ss – Sr) cannot be 0 , ± p , ± 2p , ± 3p , ± 5p , ±
1, ±2, ±(p ±1), ±(p+2)
Here P = 6
So, (Ss - Sr) cannot be 0, ±6, ±12, ±18, ±30, ±1, ±2, ±5, ±7, ±4, ±
8
Here, Ss - Sr can be ± 3, ±9, ±10, ±11, ±13 , ……….

15. Let Ss - Sr = ±10
So Sr can be = 44 or 64
Let the no of rotor bar is 44.
Rotor bar cross section area calculation :
Rotor bar current Ib = 0.85*(6* Tp* Ip) / Sr = 477.92 A
Let, current density in rotor bar δb
= 5.5 A / mm2 [rotor current density can be taken from
4 to 7 A/mm2]

16. Area of cross section of rotor bar ab = Ib / δb = 477.92/ 5.5
= 86.89 mm2 .
End ring cross section area calculation :
End ring current Ie = Sr * Ib / pi*pole = 44*477.92/(3.1416*6) =
1115.60 amp
Let δe = 5.5 A/ mm2
Area of cross section of end ring , ae = Ie / δe = 202.84 mm2
Let, ae = 205 mm2

17. Result
❖ Diameter of Stator =0.293 m
❖ Length of Stator=0.154 m
❖ Turns per phase=189
❖ No. of stator slots=54
❖ 5.No. of rotor slots=44
❖ Area of cross section of stator conductor= 6.23 mm2
❖ Area of cross section of Rotor Bar=86.89 mm2
❖ Area of cross section of End ring= 205 mm2
❖ Length of Airgap=0.625 mm