Main dimension & rotor design of squirrel cage Induction Motor.pdf

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. Ataurrahman 1703002
ShetuMohanto 1703018
Md. NafisJawad 1703026
Md. SabbirMahmud 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.
THEMACHINE IS TO BE STARTED BY A
STAR‐DELTA STARTER.

3. Determination of magnetic & Electricalloading:
As stated the motor must be started by star delta starter, so the
inputside ( stator) must be delta connected.
The INPUT KVA
Q
i = Output real power / ( full load efficiency *
full load P.F.)
= (22*10
3
)/(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,
B
av=0.45 Wb ;

5. Ampere conductoris defined as the average value of flux density
over the whole surface of air gap in themachine. 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 C
o ,
= 11 * B
av* K
w* ac *10
-3
=11*0.45*0.955*26000*10
-3
= 122.91 KVA /m
3
-rps

7. Determination of D & L :
KVA input Q
o= C
o* D
2
* L * N
s
=> D
2
*L = 27.1605 / ( 122.91*16.67) = 0.0132 m
3
For overall good design assumed ,L/τ= 1
so, L = τ= (pi*D)/P
Then D
2
* (pi*D/P) = 0.0132 => D
3
= (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= B
av*pi*D*L/P = 0.0106 Wb
Stator turns per phase T
p= E
s/( 4.44*f*K
w*
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 S
s= 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
whereSs= 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 totalstatorslots = 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. Correctioninthevalueoftotalno.ofturns per phase :
Conductors per slot Z
ss= 1116 / 54 = 20.67
Considering Z
ss=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 V
phx I
ph
Phase current in stator I
p= KVA / (3*V
ph) = 27160.5/(3*415) =
21.816 A [ for delta , V
ph= V
L]
Check for slot loading :
Slot loading = I
z* Z
ss= 21.816*21 = 458.136 amp conductor (ac)
Consideringconductor current density 3.5 A /mm
2
,
Statorconductorcross section area =( phase current / density)
= 21.816/3.5 = 6.23mm
2

13. Air Gap calculation :
Length of Air gap L
g= 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 S
r= Number of Rotor slots , S
s= No. of stator slots
We know that ,(S
s –S
r) cannot be 0 , ±p , ±2p , ±3p , ±5p , ±
1, ±2, ±(p ±1), ±(p+2)
Here P = 6
So, (S
s-S
r) cannot be 0, ±6, ±12, ±18, ±30, ±1, ±2, ±5, ±7, ±4, ±
8
Here, S
s-S
rcan be ±3, ±9, ±10, ±11, ±13 , ……….

15. Let S
s-S
r= ±10
So S
rcan be = 44 or 64
Let the no of rotor bar is 44.
Rotor bar cross section area calculation :
Rotor bar current I
b= 0.85*(6*T
p*I
p) / S
r= 477.92 A
Let, current density in rotor bar δ
b
= 5.5 A /mm
2
[rotor current density can be taken from
4 to 7 A/mm
2
]

16. Area of cross section of rotor bar a
b= I
b/ δ
b= 477.92/ 5.5
= 86.89 mm
2 .
Endringcross section area calculation :
End ring current I
e= S
r*I
b/ pi*pole = 44*477.92/(3.1416*6) =
1115.60 amp
Let δ
e= 5.5 A/mm
2
Area of cross section of end ring , a
e= I
e/ δ
e= 202.84 mm
2
Let, a
e= 205 mm
2

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 mm
2
❖Area of cross section of Rotor Bar=86.89mm
2
❖Area of cross section of End ring= 205 mm
2
❖Length of Airgap=0.625 mm