pyrhonen squirrel cage motor calculation mathcad13

8/12/2019 Pyrhonen Squirrel Cage Motor Calculation Mathcad13

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Design of Rotating Electrical Machines

Juha Pyrhnen, Tapani Jokinen, Valria Hrabovcov

Three-phase squirrel cage motor with a two-layer integral slot winding and totally enclosed fancooling

This calculation sheet is designed for analytical evaluation of a three-phase squirrelcage motor with a two-layer integral slot winding and totally enclosed fan cooling.

While every effort has been made to ensure the accuracy of the information and technical data in thiscalculation sheet, the authors cannot accept any liability for damage, injury, breakdown, or poor

performance arising from the application of this machine design or data contained in it. It is alsoemphasized that the calculation example is simplified and does not take all phenomena into account. Thereader is responsible to check and supplement the calculations and the technical data and conduct suchtests as may be necessary to ensure that the machine design suggested by this publication is suitable for the use to which it is put.

The design process follows the outline presented in Chapter 7. The table and equation numbers refer tothe numbering in the textbook.

1. Initial data of the motor

Power, W

Synchronous speed, 1/min

Line-to-line voltage star connected, V

Phase voltage, V

Number of phases

Number of pole pairs

Frequency, Hz

Stator angular frequency, rad/s

Rated power factor, estimated

Rated efficiency, estimated

Permeability of vacuum, VsA -1m-1

Temperature rise in the windings, K

Conductivity of copper at 20 degrees C, S/m

P 30 10 3:=

n syn 1500:=

U 690:=

U sphU

3:= U sph 398.3717= V

m 3:=

p 2:=

f n syn

60 p:= f 50= Hz

2 f := 314.1593= 1 / s

cos n 0.92:=

0.927:=

0 4 107

:=

80:=

Cu20C 57 106

:=


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Temperature coefficient of copper,

Density of copper, kg/m 3

Conductivity of aluminium at 20 degrees C,S/m

Temperature coefficient of aluminium,

Density of aluminium, kg/m 3

Space factor of stator and rotor core

Density of iron, kg/m 3

S1 continuous drive

Enclosure IP 55, dust proof, water tolerant

Thermal class 130

Cooling method, surface-cooled IC 41 with anexternal fan attached on the motor shaft

Cu 3.81 103

:=

Cu 8960:=

Al20C 37 106

:=

Al 3.7 103

:=

Al 2700:=

k Fe 0.97:=

Fe 7600:=

Subscripts:

stator s

rotor r

We introduce the BH curve of the M800-50A laminationmaterial in T and A/m

B

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.70.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

2.1

:= H

0

84.5

107

121

133

145

156

168180

194

209

228

254

304

402

660

1480

3710

7300

15000

30000

100000

:=

The specific iron loss per mass at 1.5 T and 50 Hz with M800-50 A

P 15 6.6:= W /kg


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0 5 .104 1 .1050

1

2

B

H

T

BH curve of M800-50

A / m

Additional variables for drawing thefigures

j 0 10..:= k 0 19..:=

k sat

0

0.05

0.1

0.16

0.25

0.37

0.5

0.670.875

1.2

1.7

:= i

2 1

0.66

0.68

0.70

0.72

0.74

0.76

0.780.80

0.82

0.84

:=

B y

0

0.1

0.2

0.3

0.4

0.5

0.6

0.70.8

0.9

1.0

1.1

1.2

1.3

1.41.5

1.6

1.7

1.8

1.9

:= c

0.72

0.72

0.72

0.72

0.72

0.72

0.71

0.700.67

0.63

0.57

0.48

0.40

0.33

0.260.20

0.17

0.16

0.15

0.14

:=

0 0.5 1 1.5 20.6

0.8 i j

k sat j


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Influence of the maximum flux density of the stator or rotor yoke on thedefinition of the coefficient c, applied in the determination of magneticvoltage (see Fig. 3.17 in the textbook)

0 0.5 1 1.5 20

0.5

1

ck

B yk 2. Tangential stress

The determination of the main dimensions starts with choosing the appropriatetangential stress according to the machine type (Table 6.3).

Ftan 15000cos n

0.8:= Ftan 1.725 10

4= Pa

3. Rotor size

The rotor volume V r can be solved from Eq. (6.2)

2r

tan tan r ' 2 ,

2 F F D

T l V = =

where D r is the outer rotor diameter and l' the equivalent length of the rotor.Before calculating the rotor volume we need an estimation of the rotor torque. Weassume that the rated speed of the motor is

nn 1474:= min -1

The rated torque estimate is

T P

2 nn

60

:= T 194.3547= Nm

tan

2r

r 2'

4

F

T l

DV

==The volume V r is

V r T

2 Ftan:= V r 5.6335 10

3= m3


5/35

The ratio of equivalent core length and air-gap diameter is according to Table 6.5:

3

2

' p

p Dl =

3 p

2 p

:= 0.9895=

From the equations above we can solve the rotor diameter and the equivalent ironcore length:

D r

3 4 V r

:= D r 0.1935= m

We select a rounded number

D r 0.200:= m

l D r := l 0.1979= m

4. Air gap and core length

The air-gap length is, depending on the number of pole pairs, calculatedfrom Eqs. (6.41) and (6.42):

m100001.02.0 4.0 P +

= for p = 1

m1000

006.018.0 4.0 P += for p > 1

In this case, p = 2 and the air gap is calculated from

0.18 0.006 P 0.4+

1000:= 5.5068 104

= m

In heavy-duty applications, the air gap is increased by 60 %. We use in thiscase an increased air gap

1.60.18 0.006 P 0.4+

1000:= 8.8109 10 4= m

We select a rounded number for the air gap


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0.0008:= m

The inner diameter of the stator is

D s D r 2 +:= D s 0.2016= m D s 1000 201.6= mm

As there are no cooling channels, we have nv 0:= bv 0:= bve 0:=

Core length in a machine with no cooling channels is

l l n v b ve 2 +:= l 0.1963= m

5. Stator and rotor windings

The number of stator slots per pole and phase q is chosen

q 4:=

and the winding pitch W = 5/6 p.

W p5

6:=

The number of stator slots:

Q s 2 p m q:= Q s 48=

The stator slot pitch:

us D s

Q s:= us 0.0132= m

The stator pole pitch:

p D s

2 p

:= p 0.1583= m

We select a skewed rotor. According to Tables 7.3 and 7.5, we select the number of rotor slots to be

Q r 44:=

The rotor slot pitch ur will be

ur D r

Q r := ur 0.0143= m 1000 ur 14.28= mm


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According to Table 7.1, the slot pitch should be between 7 and 45 mm. Hence weaccept the value found above.

6. Air-gap flux density and linear current density

Since the tangential stress has already been selected, the amplitude of the

fundamental of the air-gap flux density has to correlate with the selected stressvalue, which was the average value. According to Table 6.1, the peak value of the fundamental air-gap flux density varies normally from 0.7 T to 0.9 T. Let uschoose for the fundamental air-gap flux peak density

B1peak 0.80:= T

According to Table 6.2, the linear current density RMS value for air-cooled inductionmachines varies between 30 and 65kA/m.

Tangential loading assuming sinusoidal flux density waveform to estimate themechanical torque is according to the textbook

2cos

2cos

tan

B A B A

F ==

We may solve for the RMS linear current density in Mathcad format

A2 Ftan

B1peak cos n:= A 3.3146 10 4= A /m

This value corresponds well to the values given in Table 6.3.

7. Number of coil turns in a phase winding

The number of coil turns in series in a phase winding can be calculated accordingto Eq. (7.7) in the textbook

'2

2

pi1w

m

m1w

m

l Bk E

k E

N

==

where E m is the air-gap-induced voltage that is here assumed to be

E m 0.94 U

3:= E m 374.4694= V

and k w1 the winding factor for the fundamental (Eq. (2.61) = 1)


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=

Q p

mpQ

mW

sin

2

sin

2

sin2

p

d pw k k k =

k w1

2 sin W p 2 sin m 2

Q s sin p

Q s

m p

:= k w1 0.925=

1i2

:= Initial value of the saturation factor i

N s 2 E m k w1 1i B1peak p l

:= N s 114.1857=

8. Number of conductors in a slot

The number of parallel branches is selected as

a 1:=

The number of conductors in a slot is

z Qs 2 a m N s

Q s:= z Qs 14.2732=

z Q has to be an even number in a two-layer winding. (In a one-layer winding, z Qhas to be an even or odd number, i.e. an integer number.) As we have a four-polemachine, we can select two parallel paths

a 2:=

z Qs 2 a m N s

Q s:=

z Qs 28.5464=

we select z Qs 28:=

and the number of coil turns in the phase winding will be

N sQ s z Qs

2 a m:= N s 112=


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9. New B1peak

The rounding of z Q influences the air-gap flux density. The saturation factor i in Eq. (7.7) has

to be iterated (We return from Item 13 to this point) . The original unsaturated value was i =

2/. We insert the value found in Item 13 (e.g. after the third iteration round)

2i 0.681:=

The new air-gap flux density is

B1peak 2 E m

N s k w1 2i p l := B1peak 0.7625= T

10. Width of the stator slot

According to Table 6.1, the flux density of a stator tooth varies normally from1.4 T to 2.1 T and the rotor tooth from 1.5 to 2.2 T. Let us choose for theapparent flux densities

Bdapps 1.6:= T

Bdappr 1.6:= T

( )' ud

Fe v v d

' l B Bk l n b b

=

The apparent flux density according to the textbook is

The tooth width is

bdsl us

k Fe l n v b v( ) B1peak

Bdapps:=

bds 6.5351 103= m 1000 bds 6.5351= mm

The rotor tooth width is correspondingly

bdr l ur

k Fe l n v b v( ) B1peak

Bdappr := bdr 7.0726 10

3= m 1000 bdr 7.0726= mm

11. Slot dimensions

To determine the stator slot dimensions, we have first to estimate the stator current, Eq. (7.9a). The initial guess for the stator current is


10/35


11/35

h4

b4

b5

b1

h5

h2

h6

h3

h '

b5c

b4c

h1We can choose the following dimensions

b1s 0.003:= m

h1s 0.001:= m

h2s 0.002:= m

h3s 0.005:= m

h6s 0.0005:= m

h s 0.0005:= m

Other dimensions are determined as follows

b4s D s 2 h1s h2s+( )+

Q sbds:= b4s 7.0523 10

3= m

b4cs b4s2 h3s

Q s+ 2 h6s:= b4cs 6.7068 10

3= m

The height h5 is determined so that the tooth width bd is constant and the woundarea of the slot S C us has the value calculated above. The following two equations

are used to solve h5

55c 4

2c

hb b

Q= +

24c 5ccus 5 5

2 8 cb b

S h b+= +

The slot separator h' is assumed zero. Its influence has been taken into account in the spacefactor k Cus . The user must give an iteration value for h5s to match S Cus

h5s 0.02:= m S Cus 1000000 186.8803= mm 2

b5cs b4cs2 h5s

Q s+:= b5cs 9.3248 10

3= m

S Cusb4cs b5cs+

2

h5s

8b5cs

2+:= 1000000 S Cus 194.4626= mm 2


12/35

The value h5s given above satisfies the equations with sufficient accuracy. b5s gets the value

b5s b5cs 2h6s+:= b5s 0.0103= m

h4s h5sb5cs

2+:= h4s 2.47 10

2= m

The total area of the slot is needed later in the calculations

S slot b1s h 1s h2sb4s

2

b1s

2+

+ h3s b4s h3s

Q s+

+ b4s b5s+

2

h 5s+

8b5s

2+:=

S slot 2.6558 104= m2

S slot 1000000 265.5843= mm 2

The rotor slot dimensions are calculated next.

The rotor current referred to the stator is approximately

I' r I s cos n:= I' r 27.0789= A

The transformation ratio between the rotor and stator currents is (Eq. 7.47)

K rs12 m k w1 N s

Q r := K rs1 14.1277=

The squirrel cage bar current is

I bar K rs1 I' r := I bar 382.5644= A

The rotor aluminium current density is selected according to Table 6.2

J bar 3.5 106

:= A / m 2 (rotor winding) 3 ... 8 A / mm 2 for copper 3 ... 6.5 A / mm 2 for aluminium

J ring 4 106

:= A / m 2 (short-circuit rings)

As the die-cast-rotor slot space factor is k Alr = 1, the rotor slot area is equal to the bar cross-sectional area


13/35

S bar I bar

J bar := S bar 1.093 10

4= m2 S bar 1000000 109.3041= mm 2

S Alr S bar :=

The rotor short-circuit ring current is calculated as (Eq. 7.38)

r u

u

bar ring

2;

2sin2 Q

p I I

==

u 2

Q r := u 0.1428= rad

I ring I bar

2 sin

u

2

:= I ring 2.6813 103= A

The end ring cross-sectional surface is

S ring I ring

J ring := S ring 6.7033 10

4= m2 S ring 1000000 670.3265= mm 2

The rotor bar is described in the figure below

h4

b4

b5

b1

h5

h2

h1

We can choose the following dimensions

b1r 0.003:= m


14/35

h1r 0.001:= m

h2r 0.002:= m

Other dimensions are determined similarly as for the stator.

bdr 7.0726 103= m

b4r D r 2 h1r h2r +( )

Q r bdr := b4r 6.779 10

3= m

The height h5r is determined so that the tooth width bdr remains constant and thewound area of the slot S Alr has the value calculated above. The followingequations are used to solve h5r . We start with an initial guess for h5r

h5r 0.0165:= m S Alr 1000000 109.3041= mm2

b5r b4r 2 h5r

Q r := b5r 4.4228 10

3= m

h4r h5r b5r

2+:= h4r 0.019= m

S Alr b4r b5r +

2

h5r

8

b5r 2+

b1r b4r +

2

h2r +:= 1000000 S Alr 109.8755= mm 2

We see that h5r guessed above gives the right solution for S Alr with sufficientaccuracy.

12. Magnetic voltages over the stator and rotor teeth

The flux density in the tooth is obtained by solving the intersection of the BH curveof the electric sheet in question and the line

d0d

u'dd H S

S B B =

where

( )1

'

dvvFe

u

d

u

=bbnl k

l S S

In our example, the number of ventilation channels nv = 0 and


15/35

u u

d Fe d

'1

S l S k lb

=

Using the BH curve of M800-50A we get for the field strengths in the teeth

vs lspline H B,( ):= lspline returns a vector, which interp uses to create a cubic, piecewise polynomial that passes through all the (x,y) data points

Initial guess

Bes Bdapps:= H es interp vs B, H , Bes,( ):= H es 1.48 10 3= A/m

Ber Bdappr := H er interp vs B, H , Ber ,( ):= H er 1.48 10 3= A/m

Given

Bdappsl us

k Fe

l bds

1

0 interp vs B, H , Bes,( ) Bes=

Bds Find B es( ):= Bds 1.598= T

H ds interp vs B, H , Bds,( ):= H ds 1.4634 10 3= A/m

Given

Bdappr l ur

k Fe l bdr 1

0 interp vs H , B, Ber ,( ) Ber =

Bdr Find B er ( ):= Bdr 1.6= T

H dr interp vs B, H , Bdr ,( ):= H dr 1.48 10 3= A/m

The solution for the tooth magnetic voltage is

=

d

0 ddm, d

hU l H

We simplify and calculate as

U mds H ds h3s h5s+( ):= U mds 36.5859= A

U mdr H ds h5r ( ):= U mdr 24.1467= A


16/35

13. Magnetic voltage of the air gap and saturation factor

The magnetic voltage of the air gap is calculated according to Eqs. (3.7b) and (3.8)

2

atan

b1s

2 ( )

2

b1s ln 1

b1s

2

2

+

:= 0.4322=

k C1s us

us b1s:= k C1s 1.109=

The equivalent air gap

e k C1s := e 8.8717 104= m

r 2

atan

b1r

2 e( )

2 e

b1r ln 1

b1r

2 e

2

+

:= r 0.4058=

k C1r ur

ur b1r := k C1r 1.0999=

The equivalent air gap

e k C1s k C1r := e 9.7576 104= m

The magnetic voltage of the air gap is according to Eq. (3.35)

U m e B1peak

0 e:= U m e 592.0373= A

According to Eq. (7.6), the saturaton factor is calculated as

k sat1U mds U mdr +

U m e:= k sat1 0.1026=

vs lspline k sat i,( ):=

With Fig 7.2 we get

i2 interp vs k sat , i, k sat1,( ):= i2 0.681=


17/35

If this value differs from the value of 2i used in Item 9, we must return to Item 9 and

give 2i the value of i2 and repeat the iteration until 2i = i2

14. Stator and rotor yokes

The maximum flux densities of the stator and rotor yokes are selected according toTable 6.1.

B ys 1.3:= B yr 1.3:=

The height of the stator yoke hys and the height of the rotor yoke hyr are solvedfrom Eqs. (3.48) and (3.49):

( ) ysvvFem

ys

mys 2

2

hbnl k S B

==

( ) yr vvFem

yr

m

yr 2

2

hbnl k S B

==

where the air-gap flux

m 2i B1peak p l := m 0.0163= Vs

h ys m

2k Fe l n v b v( ) B ys:= h ys 0.0329= m

h yr

m

2k Fe l n v b v( ) B yr := h yr 0.0329= m

The magnetic voltages are according to Eqs. (3.51) and (3.52)

m,ys s ys ys U c H =

m,yr r yr yr U c H =

Stator diameters

D ys D s 2 h1s h2s+ h3s+ h4s+ h6s+( )+ h ys+:= D se D ys h ys+:=

where

ys D ys

2 p:= ys 0.2362= D ys 0.3008= m D se 0.3337= m

D yr D r h yr 2 h1r h2r + h4r +( ):= D yr 0.1237= m


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yr D yr

2 p:= yr 0.097= m

The coefficient c can be taken from the figure below

0

0.2

0.4

0.6

0.8

c

0 0.5 1.0 1.5 2.0T B B /, ysyr

From the BH curve of M800-50A, we get for the maximum field strengths in the stator and rotor yokes

vs lspline B H ,( ):=

B ys 1.3= H ymaxs interp vs B, H , B ys,( ):= H ymaxs 304= A /m B yr 1.3= H ymaxr interp vs B, H , B yr ,( ):= H ymaxr 304= A /m

Using Fig. 3.17, we get the correction factor c:

vs lspline B y c,( ):=c s interp vs B y, c, B ys,( ):= c s 0.33=

cr interp vs B y, c, B yr ,( ):= cr 0.33=

The stator and rotor magnetic voltages are calculated as

U mys c s H ymaxs ys:= U mys 23.6995= A

U myr cr H ymaxr yr := U myr 9.7475= A

15. Total magnetic voltage and magnetizing current

The magnetic voltages are


19/35

U m e 592.0373= A

U mds 36.5859= A

U mdr 24.1467= A

U myr 9.7475= A

U mys 23.6995= A

The total magnetic voltage is

U mtot U m e U mds+ U mdr + U mys

2+

U myr

2+:= U mtot 669.4933= A

According to Eq. (2.13), the current linkage of a rotating field winding is calculatedaccording to the textbook as

I p

Nk I

p Nk m

2

32

24

2 w1w1

1 ==

from which the effective value of magnetizing current can be calculated in Mathcad format

I mU mtot p

3 N s k w1 2:= I m 9.5701= A

16. Stator resistance

The average length l av of a coil turn (Eq. (5.2))

l av 2l 2.4 W p p+ 0.1+:= l av 0.8093= m

The conductivity of copper wire at 100 degrees C (the temperature rise =80 K) is

Cu Cu20C

1 Cu+:= Cu 4.3685 10

7= S /m

The DC resistance of a phase winding

R s N s l av

Cu a S cs:= R s 0.2467=

17. Rotor resistance referred to stator

The conductivity of aluminium at 100 degrees C (the temperature rise = 80


20/35

K) is

Al Al20C

1 Al +:=

The DC resistance of a rotor bar is according to Eq. (5.1)

Rbar l

S Alr Al := Rbar 6.258 10

5=

We assume that the end ring has the same height as the rotor bar h4 + h2and is located at depth h1. The average end-ring diameter is

D ring D r 2 h1r h2r h4r +

2+

:= D ring 0.1773= m

l ring D ring

Q r := l ring 0.0127= m

The cross-sectional area of the ring is

S ring 1000000 670.3265= mm 2 S ring 6.7033 104= m2

Rring l ring

S ring Al := Rring 6.6145 10 7=

Rr Rbar Rring

2 sin p

Q r

2

+:= Rr 7.891 105=

The resistance of the rotor may be referred to the stator by multiplying the resistance with Eq.(7.53)

2

r sq

sws

r

s4

=

k

k N Qm

According to Eq. (4.71), the skewing factor is calculated as


21/35

2

2

sin

p

sq

p

sq

sq

s

s

k

=

The skewing is carried out in the rotor so that it corresponds to one stator slot pitch (Fig. 4.16)

s sq D s

Q s:= k sq

sin s sq

p

2

s sq

p

2

:= k sq 0.9971=

4m

Q r

N s k w1

k sq

2

:= 2.9441 103=

R r Rr := R r 0.2323=

18. Magnetizing inductance and reactance

The effective air gap

ef

U m e

U mds

+ U mdr

+ U mys

2+

U myr

2+

U m e e:= ef 1.1034 10

3= m

Calculation of magnetizing inductance (Eq. 3.110)

Lmm

2

2

0 l

1

2 p

4

p

ef k w1 N s( )2:= Lm 1.1644 10 1= H

The mangnetizing reactance is

X m Lm := X m 36.5791= N s 112=

19. Air-gap leakage inductances and reactances

The stator leakage factor (Eq. 4.18)


22/35

2

11w

w

+=

=

=

k k

Pitch factor (Eq. 2.32) Distribution factor (Eq. 2.24)

=

2

sin

p

p

W

k ( )

( )2/sin

2/sin

u

ud

q

qk =

us p 2

Q s:= us 0.2618= (Eq. 2.2)

The air-gap leakage is calculated in two parts for 600 harmonics

k 1

1

300

k

sin 1 2 k m+( ) W p

2

sin 1 2 k m+( ) q

us2

q sin 1 2 k m+( ) us

2

1 2 k m+( ) k w1

2

=

:=k 1 2.8543 10

3=

k 12

1

300

k

sin 1 2 k m+( ) W p

2

sin 1 2 k m+( ) q us

2

q sin 1 2 k m+( ) us

2

1 2 k m+( ) k w1

2

=

:=k 12 3.3376 10

3=

Thus, the air-gap leakage factor is

s k 1 k 12+:= s 6.1918 103=

In asynchronous machines with a cage winding, the cage damps harmonics, and consequently,the air-gap inductance becomes less significant. This can be estimated empirically by multiplyingthe inductance obtained from Eq. (4.16) with a damping factor, usually of the magnitude 0.8.Hence, the air-gap leakage inductance is

L s 0.8 s Lm:= L s 5.7676 104= H

and the stator air-gap leakage reactance


23/35

X s L s 2 f := X s 0.1812=

The rotor air-gap leakage is calculated with Eq. (4.21)

2

r

2

2

2

r

2

r

r 3

1

1

1

=

+

= Q p

cQ p

pcQ cc

r

2

3

p

Q r

2:= r 6.7972 10

3=

L r r Lm:= L r 7.9144 104= H

and the rotor air-gap leakage reactance referred to the stator

X r L r 2 f := X r 0.2486=

20. Slot leakage inductances and reactances

Eq. (4.51)

1 W p:= 0.1667=

Eq. (4.53)

k 1 19

16 := k 1 0.9063=

k 2 13

4 := k 2 0.875=

Eq. (4.49)

us k 1h4s h s

3 b4s k 2

h3s

b4s

h1s

b1s+

h2s

b4s b1s ln

b4s

b1s

+

+ h s

4 b4s+:= us 2.3339=

Eq. (4.30)

Lus4m

Q s 0 l N s

2 us:= Lus 1.8202 10

3= H


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X us 2 f Lus:= X us 0.5718=

The leakage of a single slot

ur 2Q0u1 ' z l L =

In the rotor slot, there is only one bar, z Q = 1.

According to (4.32) we get for the rotor slot leakage

ur h4r

3b4r

h1r

b1r + 0.66+:=

Lur 0 l 12 ur := Lur 4.7586 10

7= H

X ur 2 f Lur := X ur 1.495 10 4=

21. Tooth tip leakage inductances and reactances

Eq. (4.62)

ds k 2

5

b1s

5 4

b1s+

:= ds 0.1923= where k 2 is given by Eq. (4.48)and ' g ' in it by Eq. (4.41)

Eq. (4.30)

L ds4m

Q s 0 l ds N s

2:= L ds 1.4998 10

4= H

X ds 2 f L ds:= X ds 0.0471=

dr k 2

5

b1r

5 4 b1r +

:= dr 0.1923=

L dr 0 l dr := L dr 4.7827 108= H

X dr 2 f L dr := X dr 1.5025 105=

22. End winding leakage inductances and reactances

The average length l w of the stator end winding


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2=qQ z

l eW

W eW

Q z

l ew 0.025:= m l wl av

2l := l w 0.2083= m

W ew l w 2l ew:= W ew 0.1583= m

The permeance factors according to Table 4.1

lew 0.5:=

W 0.2:=

ws2 l ew lew W ew W +

l w:= ws 0.272=

The end winding leakage inductance (Eq. 4.64)

Lws4m

Q sq N s

2 0 l w ws:= Lws 8.9326 10

4= H

X ws 2 f Lws:= X ws 0.2806=

The rotor short-circuit ring leakage is calculated according to (4.67) as

( )

+= p

Dl l

pmQ

L2

''

31 r

s bar 2s

r 0ring

is 0.36 for p = 1 is 0.18 for p > 1

Lring 0Q r

m p2

1

3 l l ( ) 0.18

D ring 2 p

+

:= Lring 3.6037 10

8= H

23. Skew leakage inductance

The skew leakage inductance is found as (Eq. 4.81)


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sq 1 k sq2:=

Eq. (4.80)

L sq sq Lm:= L sq 6.6351 104= H

24. Inductances and reactances referred to the stator

The stator leakage inductance (Eq. 4.7)

L s L s Lus+ L ds+ Lws+ L sq+:= L s 4.1037 103= H

and the leakage reactance

X s L s := X s 1.2892=

The magnetizing inductance Lm was calculated in Item 18

Lm 0.1164= H

L r 7.9144 104= The rotor bar leakage inductance (see Items 20 and 21)

Lur 4.7586 107=

Lbar Lur L dr +:= Lbar 5.2369 107=

L dr 4.7827 108= The rotor end winding leakage is (see Item 22)

Lring 3.6037 108= H

The rotor total leakage inductance (Eq. 7.92)

Lr Lbar Lring

2 sin p

Q r

2

+:= Lr 1.4133 106=

The inductance may be referred to the stator by multiplying with Eq. (7.53)

2

r sq

sws

r

s4

=

k

k N Qm

L r Lr L r +:= L r 4.9525 103= H

25. Core and mechanical losses


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To calculate the core losses, we need the masses of different iron parts.

The total volume of the stator is

V s

4 D se

2 D s2( ) l := V s 0.0109= m3

The volume of the stator yoke

V ys D se

2

2 D se

2h ys

2

l :=

V ys 6.0964 103= m3

The mass of the stator yoke

m ys V ys k Fe Fe:= m ys 44.9427= kg

The total volume of the stator slots

V slots Q s S slot l :=

V slots 2.5025 103= m3

The volume of the teeth

V ds V s V ys V slots:=

V ds 2.2988 103= m3

The total mass of the teeth

mds V ds k Fe Fe:= mds 16.9467= kg

The stator teeth (only the height h5s)

mdsloss

k Fe

Fe Q

s b

ds h

5s l := m

dsloss9.0791= kg

Iron loss correction coefficients (Table 3.2)

k Fed 1.8:=

k Fey 1.5:=

The core loss in the stator yoke is


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P Feys k Fey P 15 B ys

1.5

2

m ys f

50

3

2:= P Feys 334.1942= W

The core loss of the tooth area is calculated using the mass mdsloss defined above

P Feds k Fed P 15

Bds

1.5

2

mdsloss

f

50

3

2

:= P Feds 122.411= W

The total iron losses are

P Fe P Feys P Feds+:= P Fe 456.6052= W

Because of the low rotor fundamental frequency, the rotor iron losses are taken into account in theadditional losses.

The mechanical losses consisting of windage and ventilator losses are calculatedfrom an experimental equation (Eq. (9.19), Table 9.2)

vr n syn

60 D r := vr 15.708= m/s n syn 1.5 10

3= min -1

k 15:= Ws2/m4

P k D r l 0.6 p+( ) vr 2:= P 215.6334= W

26. Equivalent circuit parameters

j 1:= The imaginary unit is used in the impedance calculations

The rotational speeds used in the following calculations vary from zero tosynchronous speed (from 0 min -1 to 1500 min -1)

x 1 60 f

p..:= This x is seen as a subscript in the following calculations

n x60 f

p

x

1500:= n1500 1.5 10

3= min -1

slip(The latter small number isinserted to avoid numerical

problems) s x

60 f n x p60 f

10 10+:=

The rotor skin effect is evaluated assuming the rotor bar to be arectangle. The resistance factor is (Eqs. 5.24 and 5.26)


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bb

h cc0c 21

= ( )

2cos2cosh2sin2sinh

R +==k

We use the total conductor height in the calculation

hc h2r h4r +:= x hc1

2 s x 0 Al 1:=

k R x x

sinh 2 x( ) sin 2 x( )+cosh 2 x( ) cos 2 x( ):=

Rotor slip frequency: fr x s x f :=

The corresponding rotor leakage inductance skin effect factor is (Eqs. (4.57) and (4.58))

=

2cos2cosh

2sin2sinh23 Lk

k L x3

2 x

sinh 2 x( ) sin 2 x( )cosh 2 x( ) cos 2 x( ):= k L1500 1=

The skin effect factors for the whole range of slip are depicted in the graph below and used for the nextcalculations .

0 0.2 0.4 0.6 0.80.8

1

1.2

1.4

Motor slip

s k i n e f

f e c t

f a c t o r s

k R

k L

The rotor resistance referred to the stator is


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R r x Rbar k R x

Rring

2 sin p

Q r

2

+

:= R r 1486

0.2323=

and the rotor leakage inductance referred to the stator is

L r x Lbar k L x

Lring

2 sin p

Q r

2

+

:= X r x L r x :=

R r x R r x

:=

X' r x X r x

:=

Rotor circuit impedance:

Z r x

R r x s x

j X' r x+:= Z r 1474 13.4048 1.3072j+=

Core loss resistance:

R Fe3 E m

2

P Fe:= R Fe 921.3254=

Magnetizing circuit impedance:

Z m R Fe X m j

R Fe X m j+:= Z m 1.45 36.5216j+=

Stator circuit impedance:

Z s R s X s j+:=

Z s 0.2467 1.2892j+=

Total impedance of the equivalent circuit:

Z x Z s Z m Z r x

Z m Z r x+

+:= Z 1486 16.3203 12.7902j+=

27. Rated load, stator and rotor current


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The motor currents and the output power are calculated at all relevant speeds nx

Stator current phasor: The index, e.g. 1474 ( the value must begiven ), corresponds to a speed in min -1. Thesentence below returns the correspondingcurrent absolute value

I s x

U sph

Z x:= I s1 155.8465= I s1474 30.8324= A

Air gap voltage:

E m xU sph I s x

Z s:= E m1474 373.4674= V E m1474

U sph0.9375=

If the air-gap voltage differs by more than 1 % of the value used in Item 7, we should return toItem 7 and repeat the calculations

Rotor current phasor:

I r x I s x

Z m

Z m Z r x+:= I r 1474 27.7291= A

No-load current:

A I 0 x I s x

I r x:= I 01500 10.5253=

Mechanical losses:

P mech x P

n x

nn

3

:=

P mech 1475216.0725= W

Output power: P mech 1474

215.6334= W

P x m I r x( )2

1 s x

s x R r x

P mech x:= P 1474 3.0169 10

4= W

P 1475 2.9132 104= W

P 1475 P 1474+2

2.9651 10 4= W

We see that the rated output speed is between 1474 and 1475 min -1

28. Losses, efficiency, power factor and torque

P 1474 3.0169 104= W

Resistive losses of the stator

P Cus xm I s x( )

2 R s:= P Cus1474 703.6409= W


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Resistive losses of the rotor

P Cur xm I r x( )

2 R r x

:= P Cur 1474 535.965= W

Power factor:

cos x Re I s x( )

I s x

:= cos 1474 0.8706=

cos 1475 0.8679=

Additional losses:

P ad x3 I s x U cos x 0.5 10

2:= P ad 1474 277.8111= W

P ad 1475 268.1306= WCore losses:

P Fe xm

E m x( )2

R Fe:= P Fe 1474 454.165= W

Total losses:

P losstot x P Cus x P Cur x+ P Fe x+ P ad x+:= P losstot 1474 1.9716 10 3= W

Efficiency:

x P x

P x P losstot x+

:= 1474 0.9387=

1475 0.9394=

Torque:

T x

m I r x( )2

1

s x R r x P mech x

2 f

p

:= T 1474 195.4768= Nm

T 1475 188.63= Nm


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Maximum torque per rated torque:

max T ( ) 504.5188= Nm. The value is found at T 1364 504.517= Nm

max T ( )

T 14742.581=

Starting current:

I s1155.8465= A

NOTE

Please note that this calculation example is simplified and doesnot take all phenomena into account. For example, the effectsof harmonics on the torque and the effects of leakagesaturation at start are neglected.

If the per unit starting current is high and the torque low:

-The rotor could be equipped with a double cage.

-The rotor resistance could be increased by using less aluminium. This, however, results in a smaller efficiency.

-The rotor inductance could also be increased. This results

in a smaller starting torque.

I s1 I s1474

5.0546=

Starting torque:

T 1 134.7773= Nm

Starting torque per rated torque:

T 1T 1474

0.6895=

Efficiency:Torque:

0 7500 1.5 .104 2.25 .104 3 .1040.7

0.77

0.85

0.93

1

Output power / W

E f f i c i e n c y

0 500 1000 15000

500

rotational speed 1/s

T o r q u e

/ N m


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Stator current Power factor:

0 500 1000 15000

200

rotational speed 1/s

S t a t o r c u r r e n

t / A

1 .104 2 .104 3 .1040.4

0.6

0.8

Output power / W

P o w e r

f a c t o r

The motor rated parameters differ from the initial data to some degree. It is up to the designers to decidewhether the results are valid to them. For example the calculated efficiency is better than the target, whereasthe power factor is lower. Both values are strongly affected by the air-gap length, which was selected greater than recommended.

The figure below shows a FEM*-calculated flux plot of the machine at the ratedoperation point.*FCSMEK software


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pyrhonen squirrel cage motor calculation mathcad13

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