Design of Commutator amp 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 ID: 1030896
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1. D.C MachinesDesign of Field Poles & Field Coils Design of Commutator & Brushes
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4. Magnetic circuit The path of magnetic flux is called magnetic circuitMagnetic circuit of dc machine comprises of yoke , poles, airgap, armature teeth and armature coreFlux 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 PathPole BodyArmature CoreNSSNMagnetic Circuit of 4-Pole DC Machinehplhpllylc
5. 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: YokePole and pole shoe Air gap Armature teeth Armature core Magnetic circuit
6. 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 lgMMF for teeth ATt=att X dsMMF for core ATc=atc X lc/2MMF for pole ATp = atp X hplMMF for yoke ATy= aty X ly/2 att , atc , atp , aty - are determined B-H curveslc = πDm/P = π(D – 2ds – dc)/P ly = πDmy/P = π(D+ 2lg + 2hpl +dy)/P AT total =ATg + ATt + ATc + ATp +ATy
7. Design of field systemConsists of poles, pole shoe and field winding.Types:Shunt field Series fieldShunt field winding – have large no of turns made of thin conductors ,because current carried by them is very lowSeries field winding is designed to carry heavy current and so it is made of thick conductors/stripsField coils are formed, insulated and fixed over the field poles
8. Factors to be considered in design:MMF/pole &flux density Losses dissipated from the surface of field coilResistance of the field coil Current density in the field conductorsDesign of field system
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10. Let ,ATfl -MMF developed by field winding at full loadQf - Copper loss in each field coil(W)qf - Permissible loss per unit winding surface for normal temperature rise(W/m2 )Sf - Copper space factorρ - 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 coilAf - 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
11. 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=If2 (TfLmt)/af i.e., Copper loss f2 (Square of the current density)Design of field system
12. 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 windingDesign of field system
13. Normal values:Permissible loss, qf -700W/m2Copper Space factor, Sf :Small wires: 0.4Large round wires: 0.65Large rectangular conductors: 0.75Depth of the field winding, df :Design of field systemArmature Dia (m)Winding Depth (mm)0.2300.35350.5400.65451.00501.00 and above55
14. Height of field, Total height of the pole, hpl=hf+hs+ height for insulation and curvature of yokewhere, hs - Height of the pole shoe (≈0.1 to 0.2 of the pole height)Design of field system
15. Design of shunt field winding Involves the determination of the following information regarding the pole and shunt field windingDimensions 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 polesCross sectional area, length, width , height of the bodyFor cylindrical poleCross sectional area, diameter, height of the body
16. Area of the pole body can be estimated from the knowledge of flux per pole , leakage coefficient and flux density in the poleLeakage coefficient (Cl) depends on power output of the DC machine Bp in the pole 1.2 to 1.7 wb/m2 Фp = Cl. ФAp = Фp/BpWhen circular poles are employed, C.S.A will be a circleAp = πdp2 /4Design of shunt field winding
17. 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 LpWidth of pole, bp = Ap/LpiHeight of pole body hp = hf + thickness of insulation and clearance Total height of the pole hpl = hp + hsDesign of shunt field winding
18. Field coils are former wound and placed on the polesThey may be of rectangular or circular cross section depends on the type of polesDimensions – Lmt, depth, height, diameter Depth(df) – depends on armatureHeight (hf) - depends on surface required for cooling the coil and no. of turns(Tf)hf, Tf – cannot be independently designedDesign of shunt field winding
19. Lmt - Calculated using the dimensions of pole and depth of the coilFor rectangular coilsLmt =2(Lp + bp + 2df) or (Lo +Li)/2Where Lo – length of outer most turn & Li – length of inner most turnFor 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 estimatedField ampere turns on load, ATfl= If. Tf Turns in field coil, Tf = ATfl/If Design of shunt field winding
20. Power Loss in the field coil:Power loss in the field coil is copper loss, depends on Resistance and currentHeat is developed in the field coil due to this loss and it is dissipated through the surface of the coilIn 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 coilDesign of shunt field winding
21. Lmt – estimated from dimensions of poleDepth – 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 coilInner surface area= Lmt (hf – df)Outer surface area = Lmt (hf + df)Top and bottom surface area = Lmt dfTotal 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]Design of shunt field winding
22. Substitute S in Qf, Qf= 2Lmt (hf +df).qfActual Cu loss in field coil=If2Rf=Ef2/Rf Substituting Rf=(Lmt Tf)/ af ,Actual Cu loss in field coil=Ef2 .af /(Lmt Tf)Design of shunt field winding
23. Procedure for shunt field designStep1 : determine the dimensions of the pole. Assume a suitable value of leakage coefficient and B = 1.2 to 1.7 T Фp= Cl. Ф Ap = Фp/Bp When circular poles are employed, C.S.A will be a circle Ap = πdp2 /4 : dp =Ѵ(4Ap/π) 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 = Ap/Lpi
24. Step 2 : Determine Lmt of field coil Assume suitable depth of field winding For rectangular coils Lmt =2(Lp + bp + 2df) or (Lo +Li)/2For cylindrical coils Lmt = π(dp +df)Step 3: Calculate the voltage across each shunt field coil Ef = (0.8 to 0.85) V/PStep 4 : Calculate C.S.A of filed conductor Af = ρLmt ATfl/EfStep 5:Calcualate diameter of field conductor dfc =Ѵ(4af/π) Diameter including thickness dfci = dfc + insulation thickness Copper space factor Sf = 0.75(dfc/dfci)2 Procedure for shunt field design
25. 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) = Ef2 af/ρLmt Tf Tf.af = Sf.hf.dfStep 7 : Calculate Rf and If : Rf = Tf. ρLmt /af If = Ef/RfStep 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 againProcedure for shunt field design
26. Step 9 : Check for desired value of AT ATactual= 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.Procedure for shunt field design
27. Check for temp rise: Actual copper loss = If2 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.
28. 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)/2PStep 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
29. 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, 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
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32. 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.
33. Classification of commutation processResistance commutation Retarded commutation Accelerated commutationSinusoidal 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 1mmNumber of commutator segments = no. of coils in the armatureMaterials used : Commutator segments: Hard Drawn Copper or Aluminum Copper Insulation :Mica, Resin Bonded AsbestosBrushes :Natural Graphite, Hard Carbon , Electro Graphite, Metal GraphiteDesign of Commutator and brushes
34. Design formulae No. of commutator segments, C = ½ u.Sa where, u – coils sides/slot Sa – no. of armature slotsMinimum no. of segments = Ep/15Commutator segment pitch = βc = πDc/C where, Commutator Diameter Dc – 60% to 80% of diameter of armature βc ≥ 4mmCurrent carried by each brush Ib= 2Ia/P for lap winding Ib= Ia for wave windingTotal brush contact area/spindle Ab= Ib/δbNumber 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 =2Design of Commutator and brushes
35. 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/tbLc – 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)Design of Commutator and brushes
36. Losses : Brush contact losses: depends on material, condition, quality of commutationBrush 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 Commutator and brushes
37. Design of Interpoles Interpoles: Small poles placed between main polesMaterials Used: Cast steel (or) Punched from sheet steel without pole shoesPurposes: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
38. Normally, Length of interpole = length of main pole Flux 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
39. Losses and efficiency : Iron Loss - i)Eddy current loss ii) Hysteresis loss Rotational losses - Windage and friction lossesVariable or copper lossCondition for maximum efficiency : Constant Loss= Variable Loss Design of Interpoles