Magnetic circuit The path of magnetic flux is called magnetic circuit Magnetic circuit of dc machine comprises of yoke, poles, airgap, armature teeth

D.C MACHINES

Design of Field Poles & Field Coils Design of Commutator & Brushes

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 the armature tooth. Then it flows through armature core and again cross the air gap to enter the S pole

YokeFlux Path

Pole Body

Armature Core

N

SS

N

Magnetic Circuit of 4-Pole DC Machine

hpl

hpl

ly

lc

Let Bg – Max. flux density in the core

Kg- Gap contraction factor

lc – Length of magnetic path in the core

l y – Length of magnetic path in the yoke

ds - Depth of the slot

dc - Depth of core

hpl - Height of field pole

Dm – Mean 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

Total MMF to be developed by each pole is given by the sum of MMF required for the above five sections.

MMF for air gap ATg=800000 Bg Kg lg

MMF for teeth ATt=att X ds

MMF for core ATc=atc X lc/2

MMF for pole ATp = atp X hpl

MMF for yoke ATy= aty X ly/2

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

lc = πDm/P = π(D – 2ds – dc)/P

ly = πDmy/P = π(D+ 2lg + 2hpl +dy)/P

AT total =ATg + ATt + ATc + ATp +ATy

Design of field system Consists of poles, pole shoe and field winding. Types:

Shunt field Series field

Shunt field winding – have large no of turns made of thin conductors ,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 field poles

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

Design of field system

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

ρ - 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)

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

Design of field systemTentative design of field winding

Cooling surface of the field winding, S=2Lmthf -- (1)

Permissible copper loss in each field coil, Sqf=2Lmthfqf -- (2)

Area of X-section of field coil=hfdf -- (3)

Area of copper in each section=Sfhfdf -- (4)

i.e, Tfaf=Sfhfdf -- (5)

Copper loss in each field coil, Qf=If2 Rf=If

2 (TfLmt)/af

i.e., Copper loss f2 (Square of the current density)


Copper of Volume

62

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22

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mtffff

f

mtfff

f

mtfff

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a

LTa

a

LTIQ

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

Using Eqns. (2) & (6),

2Lmt hf qf =f2

Lmt (Sfhf df ) =>

MMF per metre height of field winding


)(dS

q

ff

ff 7

2

(8) -- 10heightmeter per MMF

1022

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8

fff

fff

ffff

ffff

f

ffff

f

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ff

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fl

dSq

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qdS

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hdS

h

Ta

h

TI

h

AT

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

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

flf

flf

dSq

ATh

ATh

410

(8),Eqn Using

heightmeter per Turns Ampere

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 poleo Cross sectional area, diameter, height of the body

Area of the pole body can be estimated from the knowledge of flux per pole , leakage coefficient and flux density in the pole Leakage coefficient (Cl) depends on power output of the DC machine Bp in the pole 1.2 to 1.7 wb/m2 Фp = Cl. Ф Ap = Фp/Bp When circular poles are employed, C.S.A will be a circle

Ap = πdp2 /4 /Ap4dp

Design of shunt field winding

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


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, Tf – cannot be independently designed


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 Lo – length of outer most turn & Li – length of inner most turn

For cylindrical coils Lmt = π(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


Power Loss in the field coil:• Power loss in the field coil is copper loss, depends on

Resistance 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 is

specified and from which the required surface area can be estimated.

• Surface area of field coil – depends on Lmt, depth and height of the coil


• Lmt – estimated from dimensions of pole• Depth – assumed (depends on diameter of armature)• Height – estimated 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 (hf – df)Outer surface area = Lmt (hf + df)Top and bottom surface area = Lmt df

Total surface area of field coil, S= Lmt (hf – df)+ = Lmt (hf + df)+ Lmt df + Lmt df

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

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


Substitute S in Qf,

Qf= 2Lmt (hf +df).qf

Actual Cu loss in field coil=If2Rf=Ef

2/Rf

Substituting Rf=(Lmt Tf)/ af ,

Actual Cu loss in field coil=Ef2 .af /(Lmt Tf)


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f2f

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fff

ff

dhS

coil fieldof section-Xof Area

X factor space Coppercoil Field

in area Conductor

aT

conductor fieldof section-Xof Area

X turnsNo.of coil field

in area Conductor

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 TФp= Cl. Ф

Ap = Фp/Bp

When circular poles are employed, C.S.A will be a circleAp = πdp

2 /4 : dp =Ѵ(4Ap/π) 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

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 = π(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 conductorAf = ρLmt ATfl/Ef

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

Diameter including thickness dfci = dfc + insulation thickness

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


Step 6 : Determine no. of turns (Tf) and height of coil (hf)They can be determined by solving the following two equations 2Lmt(hf + df) = Ef

2 af/ρLmt Tf

Tf.af = Sf.hf.df

Step 7 : Calculate Rf and If : Rf = Tf. ρLmt /af

If = Ef/Rf

Step 8 : Check for δf

δf = If / af

δf – not to exceed 3.5A/mm2 .

If it exceeds then increase af by 5% and then proceed again


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

ATdesired- 1.1 to 1.25 times armature MMF at full loadWhen ATactual exceeds the desired value then increase the depth of field winding by 5% and proceed again.


Check for temp rise:Actual copper loss = If

2 Rf

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.

Design of Series Field Winding Step 1: Estimate the AT to be developed by series field coil,

AT /pole = (Iz . (Z/2))/P

For 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 /δse

Normally, δse - 2 to 2.3 A /mm2

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 0.7

For rectangular conductors, Sfse – depends 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

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 induced by armature flux.

Classification of commutation process1. Resistance commutation

2. Retarded commutation

3. Accelerated commutation

4. Sinusoidal commutation

Commutator is of cylindrical in shape and placed at one end of the armature

Consists of number of copper bars or segments separated from one another 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

Design formulae

1. No. of commutator segments, C = ½ u.Sa

where, u – coils sides/slot

Sa – no. of armature slots

2. Minimum no. of segments = Ep/15

3. Commutator segment pitch = βc = πDc/C

where,

Commutator Diameter Dc – 60% to 80% of diameter of armature

βc ≥ 4mm

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

Ib= Ia for wave winding

5. Total brush contact area/spindle Ab= Ib/δb

6. 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


7. Area of each individual brush should be chosen such that , it does not carry more than 70A

Let ,

ab – Contact area of each brush

nb – Number of brushes / spindle

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

also ab = wb.tb

Ab = nb. wb.tb

Usually, tb = (1 to 3) βc

wb = Ab/ nb. Tb = ab/tb

8. Lc – depends on space required for mounting the brushes and to dissipate the heat generated by commutator losses

Lc = nb(wb + Cb) + C1 + C2

where, Cb - Clearnace between brushes (5mm)

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

C2 – Clearance for allowing end play (10 to 25 mm)


9. Losses : Brush contact losses: depends on material, condition, quality of

commutation Brush friction losses

Brush friction loss Pbf = μ pb AB.Vc

μ – Coefficient of friction

pb-Brush contact pressure on commutator (N/m2)

AB - Total contact area of all brushes (m2)

AB =P Ab (for lap winding)

= 2 Ab (for wave winding)

Vc – Peripheral speed of commutator (m/s)


Design of Interpoles Interpoles: Small poles placed between main poles Materials Used: Cast steel (or) Punched from sheet steel

without pole shoes Purposes:

To neutralize cross magnetizing armature MMF To produce flux density required to generate rotational voltage in

the coil undergoing commutation to cancel the reactance voltage. Since both effects related to armature current, interpole

winding should be connected in series with armature winding

Average reactance voltage of coil by Pitchelmayer’s Equation is, Erav = 2Tc

ac Va.L .λ

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

Normally, Length of interpole = length of main poleFlux density under interpole, Bgi = ac. λ .(L/Lip)

where, Lip- length of interpole

In general, Bgi = 2 Iz. Zs. (L/Lip). (1/Va.Tc).λ

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

Design of Interpoles

reaction armature

come over to required mmf

B establish

to required mmf ATi

gi

winding) ngcompensati with ( 2P

.ZI )-(1

winding) ngcompensati without ( 2P

.ZI

reaction armature

overcome

to required MMF

z

z

Losses and efficiency : 1. Iron Loss - i)Eddy current loss ii) Hysteresis loss 2. Rotational losses - Windage and friction losses3. Variable or copper loss

Condition for maximum efficiency : Constant Loss= Variable Loss

δI

Aconductor, interpole

of section-X of Area

A/mm4 to 2.5 δ,winding interpole

indensity Current

IAT

turns No.of

i

aip

2i

a

i

Design of Interpoles

Documents

Magnetic circuit The path of magnetic flux is called magnetic circuit Magnetic circuit of dc machine comprises of yoke, poles, airgap, armature teeth