electrical machine design solved problems on dc machine magnetic circuit continuation of chapter.pdf

8/17/2019 Electrical Machine Design SOLVED PROBLEMS ON DC MACHINE MAGNETIC CIRCUIT Continuation of chapter.pdf

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SOLVED PRO

Example.1

Calculate the ampere-turns f

Gross core length = 40cm,= 1.0cm, slot pitch = 6.5cm,=0.63T. Field form factor =Carter’s coefficient=0.82 for

Note: If the Carter’s coefficiCarter’s coefficient given iTherefore and are t

less like 1, 2 or 3 and the Caclose to zero, then it may be

LEMS ON DC MACHINE MAGNETIC CI

r the air gap of a dc machine given the followi

air gap length = 0.5cm, number of ducts =slot opening = 0.5cm, average value of flux de0.7, Carter’s coefficient = 0.72 for opening/ga

opening/gap-length = 1.0.

ent given is greater than 1.0, then it may be s less than 1.0, then it may be or

be found out to find out .When the ratio

ter’s coefficient given is close to 1.0, then it ma , or .

CUIT

g data.

5, width of ductnsity in the air gapp-length = 2.0 and

. If the, or . o r is

y be or . If it is


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7

7

3

7 9

9 379

Example.2

Calculate the ampere turns required for the air gap of a DC machine given the following data.

Gross core length = 40cm, air gap length = 0.5 cm, number of ducts = 5, width of each duct= 1.0cm, slot pitch = 6.5cm, average value of flux density in the air gap = 0.63T. Field formfactor = 0.7, Carter’s coefficient = 0.82 for opening/gap length = 1.0 and Carter’s coefficient= 0.82 for opening/gap length = 1.0, and Carter’s coefficient = 0.72 for opening/gap length =2.0.

= Carter 's gap expansion coefficient=

= Carter 's gap expansion coefficient for the slots =λ s

λ s- b os (1- δ s)

Since = 1.0/0.5 = 2.0, corresponds to ducts, opening/gap-length = 1.0, must correspond to

slots.Therefore, opening of the slot = x 1.0 = 0.5x1.0=0.5cm.

At 0 50 5 1 0,δ s 0

=6.5

6.5 -0.5 (1- 0.82 ) 1 014

= Carter 's gap expansion coefficient for the ducts = − (1 − )

At 1 00 5 0, 0 7


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= 4040 − 5 1(1− 0 7 ) 1 04

= 1 014 1 04 1 054

Maximum value of flux density in the air gap 0 630 7 0 9

00000 0 5 10 1 054 0 9 3794 4 Example.3

Find the ampere-turns/pole required for a dc machine from the following data. Radical lengthof the air gap = 6.4mm, tooth width = 18.5 mm, slot width = 13.5mm, width of core packets= 50.8mm, width of ventilating ducts = 9.5mm, Carter’s coefficient for slots and ducts = 0.27and 0.21, maximum gap density = 0.8T. Neglect the ampere turns for the iron parts.

AT pole AT + AT + AT + AT + AT

+

as the ampere turns for the iron parts is to be negle

00000 =

=λ s

λ s- bos (1- δ s)

Slot pitch + 1 5+ 13 5 3 mm Opening of the slot 13 5mm if an open slot is assumed

At 13 56 4 1,(1 δ s) 0 7


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= 33 − 13 5 0 7 1 13

= − (1 − )

width of the core packets + width of ventilating ducts i e 50 + 9 5 60 3 mm

= 60 360 3 − 9 5 0 1 1 03 = 1 13 1 03 1 17

00000 6 4 10 1 17 0 479 3 Example.4

Find the ampere turns required for the air gap of a 6pole, lap connected dc machine with thefollowing data. No load voltage = 250V, air gap length = 0.8cm, pole pitch = 50cm, pole arc= 33cm, Carter’s coefficient for slots and ducts = 1.2, armature conductors = 2000, speed= 300RPM, armature core length = 30cm.

00000 =

ψ

φ

φ 60 60 50 6000 300 6 0 0 5Wb

Since , 50 6 300cm

6 0 0 5 0 3 0 17T

ψ ℎ 3350 0 66


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0 170 66 0 5 OR

φ ψ

φ ψ

φ ψ

0 0 50 33 0 3

0 5T

00000 0 10 1 0 5 19 0 Example.5

Calculate the ampere turns for the air gap of a machine using the following data. Core length= 32cm, number of ventilating ducts = 4, width of duct = 1.0cm, pole arc of ventilating ducts =4, width of duct = 1.0cm, pole arc = 19cm. Slot pitch = 5.64 cm, semi-closed slots with slotopening = 0.5cm, air gap length = 0.5cm, flux/pole = 0.05Wb.

00000 =

=λ s

λ s- bos (1- δ s)

At 0 50 4 1 0,(1 δ s) obtained from Carter e e e 0 1

=5 64

5 64 − 0 5 0 1 1 016

= − (1 − )


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At1 00 5

0,(1 − ) e m C e e e e

e e l 0 C e e e l e e

e m l e e l

=3

3 − 4 1 0 1 04

= 1 016 1 04 1 06

ψ

φ

ψ

φ

ψ

φ ψ

0 05

0 1 0 3 0

00000 0 5 10 1 06 0 34 6

Example.6

A DC machine has an armature diameter of 25cm, core length of 12cm, 31 parallel slots 1.0cmwide and 3.0cm deep. Insulation on the lamination is 8.0%. The air gap is 0.4cm long and thereis one radial duct 1cm wide in the core. Carter’s coefficient for the slots and the duct is 0.68.Determine the ampere turns required for the gap and teeth if the flux density in the gap is 0.7T.The magnetization curve for the iron is:

Flux density in tesla 1.4 1.6 1.8 2.0 2.1 2.2 2.3ampere- turns/cm 18 30 65 194 344 630 1200

00000

=

=λ s

λ s- bos (1- δ s)

Slot pitch 5

31 53cm

ℎ

ℎ 1 0

= 5353− 1 0(1− 0 6 ) 1 15


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= − (1 − )

1

1 − 1 1(1 − 0 6 ) 1 03

= 1 15 1 03 1 1

ψ 00 1 0 , with the assumption thatψ 0 7

00000 0 4 10 1 1 1 0 3776 ℎ

Flux density in the tooth at 1 3 height from the root of the tooth,

φ

φ Wb

Width of the tooth at 1 3 height from the root of the tooth,

( − ℎ ) − 5 −

3

31 − 1 0 1 13cm

( − ) 0 9(1 − 1 1) 10 1

0 066

0 0113 0 101 1 6

At 1 6T, att, the ampere turns obtained from the magnetization cdrawn to scale is 140/cm

Therefore, 140 3 4 0

140

1.86

Fluxdensity

aampereturns/cm


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Example.7

A shunt generator with lap cfollowing data relating to its

Yoke (CaPole (StallGap (EffeTeeth (MeArmature

Leakage coefficient = 1.2. Carmature has 232 conductors

nnected armature rated 100kW, 125V, 450rpmmagnetic circuit.

Part Area in cm Length in cmt iron) 350 40oy) 650 22tive) 900 0.86an) 380 2.9core (Stalloy) 300 15.2

alculate the ampere turns/pole required for 125.

, and 4pole has the

V at no load. The


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AT pole AT + AT + AT + AT + AT

) :

Flux density in the yoke φ /

φ 60 60 1 5 43 450 4 ≈ 0 07 Wb

0 07 1 /350 10 1 3

The ampere turns per metre at,obtained from the magnetization cu

to cast steel at B 1 3 1350. mean length of the lux path in the yoke

1350 0 4 540 ) :

Flux density in the pole φ 0 07 1650 10 1 33

The ampere turns per metre at,obtained from the magnetization cu to stalloy at B 1 33 00.

mean length of the lux path in the poleℎ 00 0 176

) : 00000 Effective air gap length 0 6cm

ψ φ ψ

φ

effective gap area /pole 0 07900 10 0 T

00000 0 6 10 0 5504 ) :


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Flux density in the tooth at 1 3 height from the root of the tooth,

φ

φ

mean area of the teeth / pole 0 073 0 10 1 9

The ampere turns per metre at,obtained from the magnetization cu to stalloy at 1 9 30000.

mean length of the lux path in the toothℎ 30000 0 0 9 70

) :

Flux density in the armature core φ / 0 07 /300 10 1

The ampere turns per metre at,obtained from the magnetization cu to stalloy at B 1 400.

mean length of the lux path in the armature co 400 0 153 60

ThereforeAT pole 540 + 176 + 5504 + 70 + 60 7Example.8

Find the ampere turns/pole required to drive the flux through the teeth using Simpson’s rulewith the following data: flux/pole = 0.07Wb, core-length = 35cm, number of ducts = 4, widthof each duct = 1.0cm, slot pitch at the gap surface = 2.5cm, slot pitch at the root of the tooth =2.3cm, dimensions of the slot = 1.2cm x 5cm, slots/pole-pitch = 12

ℎ

16( + 4 + )

1, φ

( − ) 0 9(35 − 4 1) 7 9


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0 070 013 0 79 1 1 6

The ampere turns per metre H,obtained from the magnetization cu

to stalloy at B 1 6 6000.

lux density at section , φ

Over a slot depth of 5cm,the tooth width changes by(1 3 1 0) 0 cm Therefore

for a slot depth of 5 cm, the tooth width changes by0 55 0 1 0cm Thus the

tooth width at section is Bt 1 3 0 1 1 cm

0 070 01 0 79 1 1 74

The ampere turns per metre H,obtained from the magnetization cu to stalloy at B 1 74 15000.

3, φ 0 070 011 0 79 1 1 9

The ampere turns per metre H,obtained from the magnetization cu to stalloy at B 1 9 30000.

ℎ 16(6000 + 4 15000 + 30000) 16000

16000 0 05 00 Example.9

Find the ampere turns required to drive the flux through the teeth with the following data usinggraphical method. Minimum tooth width = 1.1cm, maximum tooth width = 1.5cm, slot depth= 4.0cm, maximum value of flux density at the minimum tooth section = 2.0T. Material usedfor the armature is Stalloy.


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Since the height of the tooth is 4cm, let the tooth be divided into 4 parts. Therefore, accordingto graphical method,

+ ℎ4 + + ℎ4 +

+ ℎ4 + + ℎ4

5, 0 The ampere turns per metre H,obtained from the magnetization curve c to stalloy at B 0 65000.

φ φ

,

lux density at section 4,

Over a tooth height of 4cm,the tooth width changes by(1 5 1 1) 0 4cm Therefore

for every 1 0cm the tooth width changes by0 4 1 04 0 0 1cmThus the tooth width at section 3 is bt3 1 +0 1 1 3cm section is bt 1 3+0 1 1 4cm

0 1 11 1 3

0 1 11 3 1 7 , 0 1 1

1 4 1 57 0 1 1

1 5 1 46

The ampere turns per metre H,H,H and H obtained from the magnetizatio

corresponding to stalloy at respective lux densities are 00

+ ℎ4 + + ℎ4 +

+ ℎ4 + + ℎ4

[ + + + + + + + ] ℎ4


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[500 + 5000 + 5000 + 10 00 + 10 00 + 0000 + 0000 + 65000 ] 0 044

695 5 Example.10

Calculate the apparent flux density at a section of the tooth of the armature of a DC machinewith the following data at that section. Slot pitch = 2.4cm, slot width = 1.2 cm, armature corelength including 5 ducts each 1.0cm wide = 38cm, stacking factor = 0.92, true flux density inthe teeth at the section is 2.2T for which the ampere turns/m is 70000.

+ ( − 1)

( − ) ( − 4 3

( 4− 1 ) 0 9 (3 − 5 5

+ 4 10 70000( 5 − 1) 33 Example.11

Calculate the apparent flux-density at a particular section of a tooth from the following data.Tooth width = 12mm, slot width = 10mm, gross core length = 0.32mm, number of ventilatingducts = 4, width of the duct each = 10mm, real flux density = 2.2T, permeability of teethcorresponding to real flux density = 31.4x10-6H/m. Stacking factor = 0.9.

+ ( − 1)

, 31 4 10 70063 7

(1 + 1 0) 31 0 9(3 − 4 1) 33

+ 4 10 70063 7( 33− 1) 3 Example.12

The armature core of a DC machine has a gross length of 33cm including 3 ducts each 10mmwide, and the iron space factor is 0.9.If the slot pitch at a particular section is 25 mm and theslot width 14mm, estimate the true flux density and the mmf/m for the teeth at this sectioncorresponding to an apparent flux/density of 23T. The magnetization curve data for thearmature stamping is,

B in tesla 1.6 1.8 1.9 2.0 2.1 2.2 2.3At/m 3700 10000 17000 27000 41000 70000 109000


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Since, + ( − 1) − ( − 1)

Slot factor 5 33( 5− 1 4) 0 9(33− 3 1)

3 − 4 10 ( − 1) 3 − 6 10 and is an equation of a straight line

The intersection ofthis line and the magnetization curve leadand

When 0, 3 and when 0,

6 10 1017 7 10

Since 1017 7 10is too a large value to locate on the x axis, le de inite value of H be considered to locate another point and line With that, when 70000, 3 − 6 10 70000 14T

It is clear at the intersection point, 1 59000 *********

electrical machine design solved problems on dc machine magnetic circuit continuation of chapter.pdf

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