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Chapter.4

The different parts of the dc

teeth and armature core. The

magnetic circuit is the sumThat is,

AT / pole = ATy + ATp+ AT

Note:

1. Leakage factor or Le

All the flux produced by theof the flux produced by the

through the air gap and cut

leaks away from the desired

Thus

As leakage flux is generally

1. Yoke, 2. Pole, 3. Air gaab: Mean length of the flux p

AGNETIC CIRCUIT OF A D.C. MACHI

machine magnetic circuit / pole are yoke, pol

refore, the ampere-turns /pole to establish the

f the ampere-turns required for different parts

+ ATt + ATc

agnetic circuit of a 4 pole DC machine

kage coefficient LC.

pole p will not pass through the desired pathpole will be leaking away from the air gap. T

y the armature conductors is the useful flux

ath is the leakage flux1

.

round (15 to 25) % of,

, 4. Armature teeth, 5. Armature core, 6. Leakage

ath corresponding to one pole

1

E

, air gap, armature

equired flux in the

mentioned above.

i.e., air gap. Somehe flux that passes

and that flux that

lux

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p = + (0.15 to 0.25)

= LC x

where LC is the Leakage fac

2. Magnitude of flux in

a) Flux in the yo

b) Flux in the po

c) Flux in the air

d) Flux in the ar

e) Flux in the ar

3. Reluctance of the air

Reluctance of the air ga

where

lg = Length of air gap

= Width (pole arc) over

L = Axial length of the ar

L = Air gap area

Because of the chamfering o

center of the pole to

'

gl >lgcalculation of air gap relucta

.The length of air gap at the

of the pole.

Because of the fringing of fl

but it is more than that.

or or Leakage coefficient and lies between (1.1

different parts of the magnetic circuit

ke y = ( LC) /2

le p = LC

gap =

ature teeth =

ature core = / 2

gap

p S =ll g

=a ( L) r0 0

as r =1.0 for air

which the flux is passing in the air gap

ature core

pole over which the flux is passing in the air g

the pole, the length of air gap under the pole

at the pole tip. The length of air gap to bence is neither lg nor

'

gl , but has to be a value in

tips is generally 1.5 to 2 times the air gap leng

x, the width over which the flux passes throug

2

5 to 1.25).

gap

ap

aries from lg at the

considered for thebetween lg and

'

gl

th under the center

h the air gap is not

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The effect of variation in aappears in the numerator

reluctance.

While calculating the relucta

armature must also be consid

Effect of slots on the reluct

Consider a smooth surface a

reluctance of the air gap in t

lgS = ------- (1)SSA L

0S

Over the same slot pitch coinstead passing only over the

width over which the flux i

fringing coefficient for slots.

gap length and can be obtain

The reluctance of the

lgS =

AWS (b + b ) L s st 0

Dividing 2 by 1,

g t s sAWS

SSA g s 0

l (b b )S

S l / L

/ + =

ir gap length and fringing of flux can be ignand the latter in the denominator of the

nce of the air gap, effect of the presence of slo

ered.

nce of the air gap

rmature (SSA) i.e. having no slots and ducts.

e presence of smooth surface armature

nsider a slot and tooth. Because of the crowdtooth width bt, passes over some portion of the

s passing is equal to (bt + bs s ) where s is

It is less than 1.0 and depends on the ratio of

d from the Carters fringing coefficient curve.

air gap in the presence of arma

------- (2)

0

L

3

red as the formerxpression for the

ts and ducts on the

ver a slot pitchs

,

ng effect, the fluxslot also. Thus the

called the Carters

slot opening to air

ture with slots

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s SSAAWS

t s s

s SSA

t s s s

SS

(b b )

S

b b b

=

+

=

+ +

s SSAAWS

s s s

SS

- b (1 - )

= =

whereKgs is called the Carter

It is clear from the above ex

the air gap by a factor Kgs

smooth surface armature.

Effect of ventilating ducts

Consider a smooth surface aof the air gap, in the presenc

g

SSA

0

lS =

DL--------- (3)

Reluctance of the air gap in t

g

AWD

v v v

lS

D [L - n b ( l - )]=

wherev

is the carters frin

ratio opening of the duct to a

curve.

s

after adding and subtracting b in the denomisb

gs SSAK S

s gap expansion coefficient for slots and is gre

pression that the effect of the slots is to increas

s compared to the reluctance of the air gap i

n the reluctance of the air gap

mature (SSA) i.e. armature having no slots anof a smooth surface armature

he presence of the armature with ducts (AWD)

0

--------- (4)

ing coefficient for ducts. It is less than 1.0 a

ir gap length and is obtained from the Carters

4

nator

ter than 1.0.

e the reluctance of

the presence of a

ducts. Reluctance

nd depends on the

ringing coefficient

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Dividing 4 by 3,

g vAWD

SSA g

l D [ L - nS

S l /

/ =

SSAAWD

v v v

L SS K

L - n b (1- )

= =

whereKgv is called the CarThus the effect of ducts is to

to the reluctance of the air ga

Combined effect of slots an

The presence of slots and drespectively. Together theyexpansion coefficient (or ext

sg gs gv

s os

K K K

- b ( 1

= =

wherebos = opening of the slot

= width of the slot bs for

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Calculation of ampere-tur

The total ampere turns / pol

flux,

AT / pole = Sum of the ampepole, air gap, armature teeth

= ATy + ATp+ AT

a)ampere turns for the yoke

Flux density in the yoke By

Let atybe the ampere turns pthe yoke material, at By.

s per pole for the magnetic circuit of a DC m

e required for the magnetic circuit of a DC m

re turns required to over come the reluctance ond armature core+ ATt + ATc

/ pole ATy :

y

LC / 2

A

tesla

r metre, obtained from the magnetization curv

6

achine

achine to establish

the yoke,

e corresponding to

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NOTE:

Ly = Axial length of the yokedy = Depth of the yoke

Ay = Cross-sectional area of

bp = Width of the pole

Dy = Mean diameter of the y

fg = Pole pitch at mean diam

Mean length of the flux pathly = abc = abcde / 2

= (fg 2fb + 2ab) /2

y p y

y p

y

D 2 b 2 d- -

P 4 2

D b- - d / 2

P 2

=

=

Total ampere-turns for the y

b) ampere turns for the

Flux density in the pole Bp =

Let atpbe the ampere turns p

the pole material, at Bp.

Note:

Lp = Axial length of the pole

Lpi = Net iron length of the php = Height of the pole inclu

pole shoe height

Lpi = KiLp

D = Diameter of the armaturelg = Length of air gap

yoke = dyLy hp = Height of the pole

ke = (D + 2lg + 2hp + dy)

eter of the yoke = Dy / P

in the yoke

/ 2

ke / pole ATy = atyly

ole ATp :

p

LC

A

tesla

r metre, obtained from the magnetization curv

di = Diameter of the pole

ole Ap = Cross-sectional area of the poleding = bpLpi in case of square or

rectangular laminated poles

= d 2i/4 in case of circular poles

7

e corresponding to

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Mean length of the flux path

Total ampere turns for the po

c) ampere turns for the

Since flux = mmf or AT / rel

ATg = x reluctance.

Though the reluctance of th

Carters gap expansion coe

ducts. Therefore,

ATg = l K lg g g= L 4 0

whereBgis the maximum valpole.

That is,P

B = =g D L L D

P

a v e r a g e

f ie ld f o rm f ac to r K af

=

=

d) ampere turns for the

Flux density in the armaturheight from the root of the to

in the pole = pole height hp

le / pole ATp = atphp

ir gap / pole ATg :

ctance, ampere turns for the air gap per pole

air gap under a pole is0

g

L

l

, it is to b

ficient Kg = KgsKgv in order to account the

Bg g

- 7x 1 0

= 800,000lg KgBg (approximately)

ue of the flux density in the air gap along th

L

v a l u e o f t h e fl u x d e n s i t y B a v

n d i s a p p r o x i m a t e ly e q u a l t o p o l e e n c l o s u r e

a vB

=

rmature teeth / pole ATt:

tooth (in case of a parallel sided slot and taoth

8

multiplied by the

effect of slots and

center line of the

ered tooth) at 1/3

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Bt1/3 b L S /P

t 1/3 i

=

where bt 1/3 = width of the t

= ( D - 4 /

S

Li = Net iron length of the ar

Let attbe the ampere turns pthe armature core material, a

Mean length of the flux path

Total ampere turns for the ar

e) ampere turns for the

Flux density in the armature

Let atcbe the ampere turns pthe armature core material, a

Note:

dc = Depth of the armature c

Ac = Cross-sectional area ofMean length of the flux path

(D - 2h - dPQ R tl =c2 2P

=


Thus the total ampere-turns r

AT / pole = ATy + ATp+ AT

oth at 1/3 height from the root of the tooth3 h )t - b s

mature core = Ki (L nvbv)

r metre, obtained from the magnetization curvBt 1/3.

in the tooth = height of the tooth ht

ature teeth / pole ATt = att ht

rmature core / pole ATc :

ore Bc = / 2A c

tesla

r metre, obtained from the magnetization curvBc.

re

he armature core = dc Liin the armature core

)c

ature core / pole ATc = atc lc

equired for the magnetic circuit of the DC mac

g + ATt + ATc

9

e corresponding to

e corresponding to

ine

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Methods of calculating the

For a parallel sided slot, the

section of the tooth will be d

where the flux enters the too

minimum. Since the variatioiron, the calculation of ampe

Different methods available

1. Graphical method

2. Simpsons method and3. Bt 1/3 method

Graphical method

In this method the tooth is dsection is calculated. Corr

magnetization curve. Assumi

Note:ht = height of the tooth

bt1, bt2, bt3 etc., are the

Flux density at section l, Bt1

Let the ampere turns / metre,

Flux density at section 2, Bt2


Flux density at section 3, Bt3


Similarly let H4 be the ampe

Total ampere turns for the te

ampere turns for the armature teeth:

tooth is tapered and therefore the flux density

ifferent. The flux density is least at the air gap

h and maximum at the root of the tooth where

of flux density in the tooth is non-linear becae turns becomes difficult.

or the calculation of ATt are

ivided into a number of equal parts and flux dsponding to each flux density, At / m is

ng linearity between the sections considered, A

or depth of the slot

tooth width at different sections 1, 2, 3 etc.

=b L S / P

t1 i

obtained from the magnetization curve is H1 or

=

b L S / Pt2 i


=

b L S / Pt3 i


e turns / metre at Bt4 etc.

th / pole

10

at each and every

urface of the tooth

the tooth section is

se of saturation of

nsity at each toothbtained from the

Tt is calculated.

at1 at Bt1.

at2 at Bt2.

at3 at Bt3.

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ATt =2

HH 21

+ n

th

+

where n is the number of par

Simpsons method

In this method the tooth is

calculated and the correspo

curve.

Note:bt1, bt2, bt3 are the widt

Let H1 be the AT/m correspo



According to Simpsons rule

Hav =6

1 ( H1 + 4H2 + H3)


Bt 1/3 method

In this method, ATt is obtai

tooth.

Flux density in the tooth at 1

2

HH 32

+ n

th

+2

HH 43

+ n

th

etc.,

s by which the tooth is divided.

ivided into two equal parts. The flux density

ding ampere turns / metre are obtained from

of the tooth at section 1, 2 and 3

nding to the flux density Bt1 =t1 i

b L S /P

nding to the flux density Bt2=t2 i

b L S /P

a

nding to the flux density Bt3=

t3 i

b L S /P

, average ampere turns / m

ature teeth / pole ATt = Hav ht

ed considering the flux density at 1/3 height f

3 height from the root of the tootht 1/3

t 1 /

B =b

11

at each section is

the magnetization

at section 1.

t section 2.

t section 3.

om the root of the

i

L S /P

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Let attbe the ampere turns pe

the armature core material, a

Total ampere turns for the te

[Note: In all the above thre

words all the flux under a slo

Real and Apparent Flux de

When the iron is not saturate

pitch will be passing through

of the iron increases conside

and tooth paths.

Thus the flux density =iron

in the tooth, but it is an appa

be equal to f l u x i n t h e tn e t i r o n a r e

density Bapp.

r metre, obtained from the magnetization curve

Bt 1/3.

th / pole ATt = att ht .

methods, the effect of saturation of iron is

t pitch is assumed to be passing through the too

nsities

d, reluctance of the iron will be less and all the

the tooth only. However, when the iron gets s

rably and the flux over the slot pitch divides its

s

i

area of the tooth A

is not the real or actual

ent flux density. The real flux density Brealwill,

i

i

t o o t h o r i r o n p a t h

a o f t h e t o o t h A

and will be less than

12

corresponding to

eglected. In other

th only].

fluxs

over a slot

turated, reluctance

lf to take both slot

flux density

however,

the apparent flux

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Area of iron or tooth area ov

Total area over the slot pitch

s i n iapp

i i i

+B = = = +

A A A

wheren 0 r 0

B H = H=

equal to An / Ai . The magnet

Therefore Bapp = Breal + H0

If the slot factor Ks = 1 + K

Thus Bapp = Breal + H0 (Ks

[Note: Since the actual value

and therefore the AT / m i.e

Bapp = Breal + H0 (Ks 1) h

However the values of Bre

magnetization curve. The inprovides the values of Breal a

The co-ordinates, of the inte

values of Breal and H.

Therefore the total ampere-t

r which flux is passing Ai = bt Li

s L = area of iron Ai + area of non-magnetic p

n n nreal real n

i n i

A= B + x = B + B K

A A A

is the flux density in the non-magnetic path

izing force H is the ampere turns / metre to esta

K

(1 +i

n

A

A) =

it

s

i

ni

Lb

L

A

AA =

+then K = (Ks

1) and is an equation of straight line.

of flux passing through the slot or tooth is not

. H to establish Bn or Breal are also not known.

as two unknowns Breal and H. Thus the equatio

l and H can be found by plotting the abov

tersection point of the magnetization curve ad H.]

section point of magnetization curve and straig

rns for the armature teeth / pole ATt = H ht .

13

ath An

nd K is a constant

blish Breal or Bn.

1).

nown, Bn and Breal

ence the equation

cannot be solved.

e equation on the

d the straight line

t line provides the

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No-load, Magnetization or

Since the OCC is a plot of e

flux are found out by calc

increases as the number of v

Information that can be obtai

1) The value of critical2) shunt and series field3) Effect of armature re

Open circuit characteristic (OCC)

f induced and AT, ampere-turns for different a

lating ATy ,ATp , ATg, ATt and ATc. Acc

ltages considered increases.

ned from the open circuit characteristic is,

ield resistance.

ampere-turns

ction in conjunction with the internal character

******************

14

ssumed voltages or

racy of the curve

stic.

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