Induction Motor Design (3-Phase)

[Pages:20]3-Phase Induction Motor Design

(? Dr. R. C. Goel & Nafees Ahmed )

By

Nafees Ahmed

Asstt. Prof. Department of Electrical Engineering DIT, University, Dehradun, Uttarakhand

References: 1. Notes by Dr. R. C. Goel 2. Electrical Machine Design by A.K. Sawhney 3. Principles of Electrical Machine Design by R.K Agarwal 4. VTU e-Learning 5. 6.

1 Prepared by: Nafees Ahmed

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OUTPUT EQUATION: - It gives the relationship between electrical rating and physical dimensions (Quantities)

Output of a 3 phase IM is Q 3VPh1 IPh1 Cos 103

Where VPh1= Stator phase voltage IPh1= Stator Phase current

Cos Stator power factor

KW (1)

Efficiency of motor

Or equation (1) can be written as Q 3(4.44 K pd1 f 1 NPh1) IPh1 Cos 103 KW (2)

( VPh1 4.44 K pd1 f 1 NPh1)

Where

f = frequency of supply =PN/120

P =No of Poles

N =Speed in RPM

Kpd1= Winding factor =0.955

1

B

P

L

B

D P

L

=Average

value

of

fundamental

flux

B = Average value of fundamental flux density

P =Pole pitch = D P

D = Inner diameter of stator

L = Length of the IM

Total No of Conductors on Stator 3 2NPh1 6NPh1

Total Ampere Conductors on Stator 6NPh1 I Ph1

Total Ampere conductors is known as total electric loading

Specific electric loading

It is defined as electric loading per meter of periphery, denoted by ac .

ac

6N Ph1I Ph1

D

Or

N Ph1I Ph1

ac

D 6

Putting the values of f, 1 & NPh1IPh1 in equation 2 we get

Q

3

4.44

0.955

(

NP

)

(

B

D

L)

(

ac

D ) Cos 103

KW

120

P

6

Q (17.4 105 B ac Cos )D2LN

Or Q CD2LN KW

Where

C Output Co efficient 17.4105 B ac Cos

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CHOICE OF MAGNETIC LOADING ( B ):

( B is average value of fundamental flux density in the air gap)

1. Magnetizing current :

Lower B

2. P.F

:

Lower B

3. Iron Loss

:

Lower B

4. Heating & Temp rise :

5. Overload Capacity

:

Lower B

Higher B

We know

VPh1 4.44 K pd1 f 1 NPh1

If voltage is constant so for Higher B , Nph1 will be less.

And we know

Leakage reactance

N

2 Ph1

Leakage Reactance

Isc is more Dia of circle diagram Overload Capacity

6. Noise & Vibration

:

Lower B

7. Size

:

Higher B

8. Cost

:

Higher B

Range of B = 0.3 to 0.6 Tesla

CHOICE OF SPECIFIC ELECTRIC LOADING:

1. Copper Losses

:

Lower ac

2. Heating & Temp Rise :

Lower ac

3. Overload Capacity

:

Lower ac

If ac NPh1

And we know

Leakage reactance

N

2 Ph1

Leakage

reac tan ce

Isc is more Dia of circle diagram Overload Capacity

4. Size 5. Cost

:

Higher ac

:

Higher ac

Suitable values of ac are

ac =10,000 to 17,500

=20,000 to 30,000 =30,000 to 45,000

Amp Cond/meter

Amp Cond/meter Amp Cond/meter

up to 10 KW

up to 100 KW > 100 KW

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MINI AND MAXI VALUE OF C: We know

C 17.4 105 B ac Cos Cmin 17.4 105 0.30 10000 0.80 0.85

Cmin 0.35 Cmax 17.4 105 0.60 45000 0.85 0.88

Cmin 3.5

( let Cosmin 0.80 & min 85%) ( let Cosmax 0.85 & max 88%)

EFFECT OF SPEED ON COST AND SIZE OF IM:

D2L Q Represents the volume of Machine CN

So for higher speed IM, volume is inversely proportional to speed. Hence High speed means less volume that is low cost

ESTIMATION OF MAIN DIMENSIONS (D, L):

We know

D2L Q (1) CN

L P

1 1 1.25

: Good Overall Design : for Good PF

(2)

1.5

: for higher

1.5 2.0 : Overall Economical Design

Solving equation (1) & (2) we can find out D & L. Alternate method: Fitting the design into the "Standard frame size".

LENGTH OF AIR GAP:

0.2 2 DL

mm

min 0.25 mm For medium rating machines

2 3 mm

Note: D & L are in Meters

Our effort is to keep the length of the air gap as small as possible. If air gap length is higher, then magnetizing current will be more it will result in poor power factor. EFFECTIVE LENGTH OF MACHINE:

Generally l1= l2= l3=.......... = ln

l1

l2

l3

ln

Let

nov =No of ventilating ducts

bv

bv = Width of one ventilating duct

L

(Generally for every 10 cm of core length there used to be 1 cm ventilating duct)

Gross Iron length

l = l1+ l2+ l3+.......... +ln

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Actual Iron length

Where

li =Ki*l Ki =Stacking factor

=0.90 to 0.92

Overall length

L = l + nov*bv

Effective length

Le L nov bv'

Where

bv'

bv

5 5 bv

=Effective width of ventilating duct (< bv due to fringing)

DESIGN OF STATOR:

Stator and rotor with semi-closed slots

(1) Shapes of stator slots:

May be

(i) Open Slot: Used for Synchronous M/Cs

(ii) Partially Closed Slot: Used for Induction M/Cs

(2) No of Stator Slots S1: Two approaches

a.

Slot pitch

sg1

15

20

mm

D S1

Let

q1=

16 3

5

1 3

y=3

If poles are 4 then pole pairs=2

Select

q1=

5

1 2

y=2

So

S1

D sg1

For 3-Phase IM having P-poles

S1 3q1P

Where

q1

S1 3P

No of

slots

per

pole

per

phase

Winding may be integral (q1 is integer) or fractional (q1 is fractional) slot winding.

If q1 is fractional, say

q1

x y

m

n y

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Then for windings to be symmetrical it is essential that the denominator `y' should be such that the no of pole pair is divisible by `y'. If double layer winding is to be use then `y' should be divisible by 2. Hence S1 is estimated.

b. Select q1=3 to 10 and then find S1.

(3) Estimation of No of turns per Phase (Nph1), Total no of conductors (Z1) & No of conductors per slots (Nc1):

We know

VPh1 4.44 K pd1 f 1 NPh1 (1)

So

N Ph1

VPh1 4.44 K pd1

f

1

(2)

Where

1

B P

le

B

D P

le

(3)

Z1 3 2NPh1 6NPh1

(4)

Nc1

Z1 S1

(5)

Nc1 Must be an integer and divisible by 2 for double layer windings. If not an integer make it integer

and hence find the corrected value of Nc1 that is Nc1,corrected . Also find out the corrected values of

Followings

Z1,corrected

Using equation (5)

N Ph1,corrected Using equation (4)

1,corrected

Using equation (1)

B corrected

Using equation (3)

(4) Sectional area of stator conductor (Fc1):

Per phase stator current

I Ph1

Q 103 3VPh1 Cos

So

FC1

I Ph1 1

mm2

Where

1 Current density 3 4 A / mm2

From ICC (Indian Cable Company) table, find dc corresponding to Fc1

SWG |

Fc1 (mm)

|

dc (mm)

|

doverall (mm)

50 |

|

0.025

|

25 |

|

0.5

|

1|

|

7.62

|

(5) Stator slot design: Let nv = No of conductors vertically nh = No of conductors horizontally So Nc1= nv*nh ------------------------- (1)

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nv 3 5 (2) nh Solving equation (1) & (2) find out nv & nh.

Height of slot hs1 = nv*dc + 3*0.5 + 3.5 + 1.5 + 2 mm

Width of the slot

bs1 = nh*dc + 2*0.5 + 2

mm

Slot opening

b01

2 5 bs1

Ratio

hs1 3 5 bs1

hs1

(0.5 mm is insulation thickness and 2 mm for slack & tolerance) (0.5 mm is insulation thickness and 2 mm for slack & tolerance)

b01 1.5 mm Tooth lip 3.5 mm Wedge

0.5 mm

bs1

Partially closed slot for 400Volts IM Thickness of insulation With mica or leatheroid insulation for small rating machines KV 0.4 1.1 3.3 6.6 11 15 mm 0.5 0.75 1.5 2.5 4 5.5

With improved insulation (Semica Therm) KV 2 3 6 10 16 25 mm 1.1 1.4 1.8 2.8 4.0 6.0

Thickness = 0.215 KV +0.7 mm Advantages of Semica Therm: (a) Much better heat is dissipated for higher rating machines due to less thickness of wall

insulations. (b) Insulation occupies little less space in the slot.

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(6) Length of mean turns (Lmt1)

Where

Lmt1 2 L 2.3 P 0.24

P

Pole

pitch

D P

L Coil Span

< Pole pitch

(7) Resistance of stator winding per phase (RPh1)

RPh1

0.021106

Lmt1 Fc1

N Ph1

(8) Total copper loss in the stator winding

3

I2 Ph1

RPh1

(9) Flux density in stator tooth

One Turns

hy Tooth

Slot Yoke

Maximum flux density in stator tooth should not exceed 1.8T; otherwise iron losses and magnetizing current will be abnormally high. (So if flux density >1.8T, change slot dimensions) Mean flux density in the stator tooth is calculated at 1rd of tooth height from the narrow end of the

3 stator tooth.

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