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Unit II Design of Electrical Machines
- 1. DC MACHINES Power developed by the armature=CoD²Ln kW Output coefficient = π²Bav ac *10^-2 Maximum flux density Bg = Bav/Kf+Bav/ψ Kf = field form factor Ψ = Ratio of pole arc to pole pitch Pa = P/η P= Rated power output η = Efficiency of the machine
- 2. CHOICE OF Bаv Flux density in the tooth Frequency Size of the machine Bav = 0.4 to 0.8 Wb / m² Bg = .55 to 1.15 Wb/m²
- 4. CHOICE OF AMPERE CONDUCTOR PER METER Temperature rise Speed of the machine Voltage Size of the machine Armature reaction Commutation ac = 5000 to 50,000
- 15. The Carter’s gap co-efficient ( Kcs) depends on the ratio of slot width to gap length.
- 29. SELECTION OF NUMBER OF POLES Frequency Weight of iron parts Weight of copper Length of commutator Labour charges Distortion of field form
- 34. GUIDANCE FACTOR FOR CHOICE OF NUMBER OF POLES Frequency of flux reversals f = pn/2 frequency should not exceed 30 Hz Current per brush arm Ib = 2Ia/p Ib should not be equal to or greater than 400 ampere Armature mmf per pole Ata = acζ / 2 ζ = ΠD / p ac should not be greater than 1000 amperes Air gap mmf ATg = 50% to 60% of Ata =8,00,000 Kg Bg Lg
- 36. Number of Armature Slots 3611/12/2019
- 37. Guiding Factors for No. of Slots 3711/12/2019
- 40. Depth of Armature Core 4011/12/2019
- 45. Lap and Wave Winding - Comparison 4511/12/2019
- 52. Winding Pitches for lap winding 5211/12/2019
- 53. Designing of Lap Winding- Procedure 5311/12/2019
- 55. Simplex Lap Winding Diagram- Procedure 5511/12/2019
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- 73. Wave Winding Design - Procedure 11/12/2019 73
- 75. Guidelines- Drawing Wave Winding 11/12/2019 75
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- 88. Design of Commutator & Brushes 11/12/2019 98
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- 93. ARMATURE DESIGN Find Z and conductor per slot Zs E = PΦzn/a Bw = pΦ/πDL Choose current density δ =1.25 to 2.5 A / mm² Find conductor area area of the conductor = Iz/ δ
- 94. LIMITING VALUES OF L AND B Factor of consideration when selecting L ventilation cost L = ez / (Bav*Va*Ta*Na) Bav = .7 Wb/m² ez=conductor emf at no load 7.5 Va = 30 m /sec Tc = turns per coil Nc = Number of coil Tc=Nc=1 L=7.5/(.7*30*1*1) = .36 m
- 95. CORE DIAMETER Factor of consideration Peripheral speed Pole pitch Limiting values Output P = EIa *10^-3 kW P = (ez Z/a)Ia *10^-3 =ez*π*D*ac*10^-3 D = P*10^-3/ π*ac*ez
- 96. LENGTH OF AIR GAP(lg) Cooling Noise Pulsation loss •Cooling α lg •Noise, pulsation loss α (1/lg) •When air gap is less Pulsation of flux produces noise.So to reduce noise air gap is increased, circulating current increases and so loss is more
- 97. SELECTION OF SLOTS Slot pitch (Yss) = πD/S S = Number of slots Iz Zs should not be greater than or equal to 1500 ampere Conductor per slot (Zs) = Z/S Iz = current through each conductor Slots per pole = 8 to 16 Slots per pole pair = odd pair
- 98. SLOT DIMENSION Slot area = Ws* Ds Ds/Ws = 2 to 4 Insulation area = 10% to 20% of slot area Area of core = Dc*Li = Φc/Bc Bc=1.2 to 1.5 tesla
- 99. POLE DESIGN height of pole= hf + hs + height of insulation height of insulation is 15% to 20% hf hs = 20% to 40% of hf hp = hf + insulation height DESIGN OF FIELD SYSTEM Area of the pole Ap = Lpi *Bp Lpi = lp*0.9 =0.9*(L-(.001 to .015)) Ap = Φp/Bp Bp = 1.2 to 1.7 tesla Φp = Φ*Cl Cl = leakage coefficient =1.12 to 1.7 telsa
- 100. NOTATION S → cooling surface available for dissipation of heat S=2Lmt*hf(excluding top and bottom) S=2Lmt(hf+df)(including top and bottom) Tf → Number of turns af → conductor area Lmt → mean length Rf → resistance of coil If → current across each coil Ef → voltage across each coil
- 101. Qf → Total copper loss qf → specific copper loss df → depth of field coil hf →height of field coil Sf → space factor Sf = 0.4 to 0.75 ρ → resistivity of material used Copper = 2*10^8 Ωm
- 102. Area of copper in field coil =Tf*af=hf*df*Sf Qf = qf*S=If ²* Rf Qf = δf ²*af ² *ρ* Lmt *Tf / af qf *S = δf ²*af *ρ* Lmt *Tf qf(2Lmt*hf) = δf ²*af *ρ* Lmt *Tf δf = √((qf*2*hf)/(af* ρ*Tf)) Ρ = 2.8*10^8 Ωm δf =10^4 √((qf*hf)/(af*Tf)) δf =10^4 √((qf*hf)/(df*hf *Sf)) δf =10^4 √((qf)/(df *Sf))
- 103. ATfl/hf=If*Tf/hf = δf (af*Tf)/hf = δf (Sf*hf*df)/hf = δf (Sf*df) =10^4√((qf(Sf*df)/(Sf*df)) =10^4√(qf*Sf*df) hf = ATfl*10^-4/√(qf*Sf*df) qf should not be greater than 700 Wb/m² df = 40 to 50 mm Sf=0.8(d/d1) ²
- 104. DESIGN OF FIELD COIL SHUNT FIELD Assume df Find Ef = (0.8 to 0.85)rated voltage/P Rf = Ef/If =ρ Lmt Tf/af af = ρ Lmt Tf If/Ef =ATfl ρ Lmt /Ef Tf = Sf hf df /af
- 105. Qf = If ²*Rf If = Ef / Rf Cooling coefficient c= .14 to .16/(1+.1Va) Temperature rise θ =Qf * c/S θ should be within temperature limit If θ exceeds the limit then change the value of df and proceed SERIES FIELD ATse = (.1 to .2)Ata=Ise*Tse Ise = Ia δseries = δarmature = Ise/ase
- 106. DESIGN OF FIELD SYSTEM Total mmf required at no load ATf = ATg+ATt+ATo+ATy+ATp Mmf required for air gap ATg = 800000 Kg Bg Lg = Φg*Lg/µoAg Mmf required for teeth ATt = att *ds = Φt*lt/(µoµr*Ws*L*Ki) Φt =Φ /((S/P)*ψ) ds = depth of slot Mmf required for core ATc = atc*lc = Φc*lc/(µoµrAc) Φc = Φ/2 lc = π(D-2S)/P
- 107. contd.… Mmf required for yoke ATy = aty *ly = Φy*ly/(µoµrAy) Φy = Φ/2 ly = π(D+2lg+2hp)/2P P number of poles µo = 4 π*10^-7 µr = permeability of the material
- 108. ARMATURE WINDING Lap winding Yb = back pitch = 2c/p + 1 Yf = front pitch = 2c/p -1 Y = winding pitch = Yb – Yf Yc = +1 progressive winding Yc = -1 retrocressive winding coil pitch =slots/poles
- 109. Wave winding Yb , Yf must be an odd Y = odd integer Y = Yb +Yf = winding pitch • Progressive Yc = Y/2 = (c+1) /(P/2) Yb = (2c/P)+k Y = (2c+2) /(P/2) • Retrogressive Yc = Y/2 = (c-1)/(P/2) Yb = (2c/P)-k Y = (2c -2)/(P/2)