Field System

8/13/2019 Field System

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Magnetic circuit

The path of magnetic flux is

called magnetic circuit

Magnetic circuit of dc machine

comprises of yoke , poles,

airgap, armature teeth and

armature core

Flux produced by field coils

emerges from N pole and cross

the air gap to enter thearmature tooth. Then it flows

through armature core and

again cross the air gap to enter

the S pole

YokeFlux Path

Pole BodyArmature Core

N

SS

N

Magnetic Circuit of 4-Pole DC Machine

hpl

hplly

lc


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Let BgMax. flux density in the coreKg- Gap contraction factor

lcLength of magnetic path in the core

l yLength of magnetic path in the yoke

ds

- Depth of the slot

dc - Depth of core

hpl - Height of field pole

DmMean diameter of armature

When the leakage flux is neglected magnetic circuit of a DC machine consists

of following:i. Yoke

ii. Pole and pole shoe

iii. Air gap

iv. Armature teeth

v. Armature core

Magnetic circuit


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Total MMF to be developed by each pole is given by the sum of

MMF required for the above five sections.

MMF for air gapATg=800000 BgKglg

MMF for teethATt=att Xds

MMF for coreATc=atcXlc/2

MMF for poleATp

= atpXh

plMMF for yokeATy= atyXly/2

att , atc , atp , aty - are determined B-H curves

lc= Dm/P = (D 2dsdc)/P

ly= Dmy/P = (D+ 2lg+ 2hpl+dy)/PAT total =ATg+ ATt + ATc+ ATp+ATy


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Design of field system

Consists of poles, pole shoe and field winding.

Types:

Shunt field

Series field

Shunt field windinghave large no of turns made of thinconductors ,because current carried by them is very low

Series field winding is designed to carry heavy current

and so it is made of thick conductors/strips

Field coils are formed, insulated and fixed over the fieldpoles


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Factors to be considered in design:

MMF/pole &flux density

Losses dissipated from the surface of field coil

Resistance of the field coil

Current density in the field conductors



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Let ,

ATfl -MMF developed by field winding at full load

Qf - Copper loss in each field coil(W)

qf - Permissible loss per unit winding surface for normal temperature rise(W/m2 )

Sf - Copper space factor

- Resistivity (m)

hf - Height of winding(m)

df - Depth of winding(m)

S - Cooling surface of field coil(m2)

Lmt -Length of mean turn of field winding(m)Rf - Resistance of each field coil (ohms)

Tf - Number of turns in each field coil

Af - Area of each conductor of field winding(m2)

If - Current in the field winding (A)

f - Current density in the field winding(A/mm2 )

Design of field systemTentative design of field winding


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To have temperature rise within the limit, the copper loss should be equal to the

permissible loss.

Using Eqns. (2) & (6),

2Lmthf qf =f2Lmt (Sfhf df) =>

MMF per metre height of field winding


)(dS

q

ff

f

f 72

(8)--10heightmeterperMMF

1022

2

4

8

fff

fff

ff

ff

f

fff

f

ffff

f

fff

f

ff

f

fl

dSq

]m.[dSq

dSdS

qdS

h

hdS

h

Ta

h

TI

h

AT


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Normal values:

Permissible loss, qf -700W/m2

Copper Space factor, Sf:

Small wires: 0.4

Large round wires: 0.65 Large rectangular conductors: 0.75

Depth of the field winding, df :


Armature Dia (m) Winding Depth (mm)

0.2 30

0.35 35

0.5 40

0.65 45

1.00 50

1.00 and above 55


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Height of field,

Total height of the pole,

hpl=hf+hs+ height for insulation and curvature of yoke

where,

hs- Height of the pole shoe (0.1 to 0.2 of the pole height)


fff

fl

f

fl

f

dSq

ATh

ATh

410

(8),EqnUsing

heightmeterperTurnsAmpere


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Design of shunt field winding

Involves the determination of the following information

regarding the pole and shunt field winding

Dimensions of the main field pole ,

Dimensions of the field coil ,

Current in shunt field winding,

Resistance of coil,

Dimensions of field conductor,

Number of turns in the field coil ,

Losses in field coil.

Dimensions of the main field pole For rectangular field poles

o Cross sectional area, length, width , height of the body

For cylindrical pole

o Cross sectional area, diameter, height of the body


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Area of the pole body can be estimated from the knowledgeof flux per pole , leakage coefficient and flux density in the

pole

Leakage coefficient (Cl) depends on power output of the

DC machine Bpin the pole 1.2 to 1.7 wb/m2

p = Cl.

Ap= p/Bp

When circular poles are employed, C.S.A will be a circle Ap = dp

2 /4

/Ap4dp



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When rectangular poles employed, length of pole is chosen

as 10 to15 mm less than the length of armature

Lp=L (0.001 to 0.015)

Net iron length Lpi= 0.9 Lp

Width of pole, bp= Ap/Lpi

Height of pole body hp= hf + thickness of insulation and

clearance

Total height of the pole hpl = hp+ hs



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Field coils are former wound and placed on the poles

They may be of rectangular or circular cross section

depends on the type of poles

Dimensions Lmt, depth, height, diameter

Depth(df) depends on armature

Height (hf) - depends on surface required for cooling the

coil and no. of turns(Tf)

hf, Tfcannot be independently designed



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Lmt

- Calculated using the dimensions of pole and depth of

the coil

For rectangular coils

Lmt =2(Lp+ bp+ 2df) or (Lo+Li)/2

Where Lolength of outer most turn & Lilength of inner most turn

For cylindrical coils

Lmt= (dp+df)

No of turns in field coil: When the ampere turns to be

developed by the field coil is known, the turns can be

estimated

Field ampere turns on load, ATfl= If. Tf

Turns in field coil, Tf= ATfl/If



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Power Loss in the field coil:

Power loss in the field coil is copper loss, depends onResistance and current

Heat is developed in the field coil due to this loss and it is

dissipated through the surface of the coil

In field coil design , loss dissipated per unit surface area isspecified and from which the required surface area can be

estimated.

Surface area of field coildepends on Lmt, depth and height of

the coil



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Lmtestimated from dimensions of pole

Depthassumed (depends on diameter of armature)

Heightestimated in order to provide required surface area

Heat can be dissipated from all the four sides of a coil. i.e,

inner , outer, top and bottom surface of the coil

Inner surface area= Lmt(hfdf)

Outer surface area = Lmt(hf+ df)

Top and bottom surface area = Lmtdf

Total surface area of field coil, S= Lmt(hf

df)+ = Lmt(hf+ df)+Lmtdf + Lmtdf

S= 2Lmthf+Lmtdf = 2Lmt(hf+df)

Permissible copper loss, Qf=S.qf [qf -Loss dissipated/ unit area]



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Substitute S in Qf,

Qf= 2Lmt(hf+df).qf

Actual Cu loss in field coil=If2Rf=Ef

2/Rf

Substituting Rf=(LmtTf)/ af ,Actual Cu loss in field coil=Ef

2.af /(LmtTf)


fmt

f

2

ffffmt

TL

aE)d(hq2L

fff

ff

dhS

coilfield

ofsection-XofAreaXfactorspaceCopper

coilField

inareaConductor

aT

conductorfield

ofsection-XofAreaXturnsNo.of

coilfield

inareaConductor


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Procedure for shunt field design

Step1 : determine the dimensions of the pole. Assume a

suitable value of leakage coefficient and B = 1.2 to 1.7 T

p= Cl.

Ap=

p/BpWhen circular poles are employed, C.S.A will be a circle

Ap= dp2 /4 : dp=(4Ap/) When rectangular poles

employed, length of pole is chosen as 10 to15 mm less than

the length of armatureLp=L (0.001 to 0.015)

Net iron length Lpi = 0.9 Lp

Width of pole = Ap/Lpi


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Step 2 : Determine Lmt of field coil

Assume suitable depth of field windingFor rectangular coils

Lmt=2(Lp+ bp+ 2df) or (Lo+Li)/2

For cylindrical coils Lmt= (dp+df)

Step 3: Calculate the voltage across each shunt field coilEf = (0.8 to 0.85) V/P

Step 4 : Calculate C.S.A of filed conductor

Af = LmtATfl/Ef

Step 5:Calcualate diameter of field conductordfc=(4af/)

Diameter including thickness dfci = dfc+ insulation thickness

Copper space factor Sf= 0.75(dfc/dfci)2



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Step 9 : Check for desired value of ATATactual= If.Tf

ATdesired- 1.1 to 1.25 times armature MMF at full load

When ATactual exceeds the desired value then increase the

depth of field winding by 5% and proceed again.



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Check for temp rise:

Actual copper loss = If2Rf

Surface area = S = 2Lmt(hf+ df)

Cooling coefficient C = (0.14 to 0.16)/(1 + 0.1 Va)

m = Actual copper loss X (C/S)

If temperature rise exceeds the limit , then increase the depth

of field winding by 5% and proceed again.


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Design of Series Field Winding

Step 1: Estimate the AT to be developed by series field coil,

AT /pole = (Iz. (Z/2))/PFor compound m/c, ATse= (0.15 to .25) (Iz. Z)/2P

For series m/c, ATse = (1.15 to 1.25) (Iz. Z)/2P

Step 2: Calculate the no. of turns in the series field coil,

Tse= ATse/Ise (Corrected to an integer)

Step 3: Determine cross sectional area of series field conductor,

ase= Ise/se

Normally, se - 2 to 2.3 A /mm2


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Step 4 : Estimate the dimension of the field coil

Conductor area of field coil = Tse.ase

Also Conductor area of field coil = Sfse.hse.dse

When circular conductors are used

Sfse= 0.6 to 0.7For rectangular conductors, Sfsedepends on thickness and

type of insulation

On equating above two expressions,

Tse.ase= Sfse.hse.dse

hse= (Tse.ase)/(Sfse.dse)

Design of Series Field Winding


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Design of commutator and brushes Commutator and brush arrangement are used to convert the

bidirectional current to unidirectional current

Brushes are located at the magnetic neutral axis ( mid way

between two adjacent poles)

The phenomenon of commutation is affected by resistance of

the brush , reactance emf induced by leakage flux, emf inducedby armature flux.


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Classification of commutation process

1. Resistance commutation

2. Retarded commutation

3. Accelerated commutation

4. Sinusoidal commutation

Commutator is of cylindrical in shape and placed at one end of thearmature

Consists of number of copper bars or segments separated from oneanother by a suitable insulating material of thickness of 0.5 to 1mm

Number of commutator segments = no. of coils in the armature

Materials used : Commutator segments: Hard Drawn Copper or Aluminum Copper

Insulation :Mica, Resin Bonded Asbestos

Brushes:Natural Graphite, Hard Carbon , Electro Graphite, Metal Graphite

Design of Commutator and brushes


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Design formulae

1. No. of commutator segments, C = u.Sawhere, ucoils sides/slot

Sano. of armature slots

2. Minimum no. of segments = Ep/15

3. Commutator segment pitch = c= Dc/Cwhere,

Commutator Diameter Dc60% to 80% of diameter of armature

c 4mm

4. Current carried by each brush Ib= 2Ia/P for lap winding

Ib= Ia for wave winding5. Total brush contact area/spindle Ab= Ib/b6. Number of brush locations are decided by the type of winding

Lap winding: No of brush location = no. of poles

Wave winding : No of brush location =2



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7. Area of each individual brush should be chosen such that , it does not carry

more than 70ALet ,

abContact area of each brush

nbNumber of brushes / spindle

Contact area of brushes in a spindle, Ab= nb. ab

also ab= wb.tbAb = nb. wb.tb

Usually, tb= (1 to 3) cwb= Ab/ nb. Tb= ab/tb

8. Lcdepends on space required for mounting the brushes and to dissipate the

heat generated by commutator lossesLc = nb(wb+ Cb) + C1 + C2

where, Cb- Clearnace between brushes (5mm)

C1- Clearance allowed for staggering of brushes (10mm, 30mm)

C2Clearance for allowing end play (10 to 25 mm)



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9. Losses :

Brush contact losses: depends on material, condition, quality of

commutation

Brush friction losses

Brush friction loss Pbf = pbAB.Vc

Coefficient of frictionpb-Brush contact pressure on commutator (N/m

2)

AB - Total contact area of all brushes (m2)

AB =P Ab(for lap winding)

= 2 Ab(for wave winding)

VcPeripheral speed of commutator (m/s)



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Design of Interpoles

Interpoles: Small poles placed between main poles

Materials Used: Cast steel (or) Punched from sheet steelwithout pole shoes

Purposes: To neutralize cross magnetizing armature MMF

To produce flux density required to generate rotational voltage in thecoil undergoing commutation to cancel the reactance voltage.

Since both effects related to armature current, interpolewinding should be connected in series with armaturewinding

Average reactance voltage of coil by PitchelmayersEquationis, Erav= 2Tc

ac Va.L .

Inductance of a coil in armature =2Tc2 .L .


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Normally, Length of interpole = length of main pole

Flux density under interpole, Bgi= ac. .(L/Lip)

where, Lip- length of interpole

In general,

Bgi= 2 Iz. Zs. (L/Lip). (1/Va.Tc).

MMF required to establish Bgi= 800000Bgi.Kgi.lgi


reactionarmature

comeovertorequiredmmf

Bestablish

torequiredmmfATi

gi

winding)ngcompensatiwith(2P

.ZI)-(1

winding)ngcompensatiwithout(2P

.ZI

reactionarmature

overcometorequiredMMF

z

z


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Losses and efficiency :

1. Iron Loss - i)Eddy current loss ii) Hysteresis loss

2. Rotational losses - Windage and friction losses

3. Variable or copper loss

Condition for maximum efficiency :

Constant Loss= Variable Loss

IA

conductor,interpoleofsection-XofArea

A/mm4to2.5,windinginterpole

indensityCurrent

I

ATturnsNo.of

i

aip

2

i

a

i


Field System

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