D.C. Machine Design Problem (EE Electrical Machine Design I) By Pratik Mochi CSPIT, CHARUSAT

Transcription

1 D.C. Machine Design Problem (EE Electrical Machine Design I) By Pratik Mochi CSPIT, CHARUSAT 1

2 2 Cross Section View of 4 pole DC Machine

3 Design Problem Design a 250kW, 400V, 625A, 600 rpm, lap wound dc compound generator, provided with suitable commutating winding to ensure sparkless commutation and permissible limit of following quantities should not exceed the value given below: 1. Peripheral speed of armature, 25m/sec 2. Voltage between commutator segments, 15V 3. Iron losses, 3kW 4. Total losses at full load, 18kW 5. Temperature rise of any part, 50 C. 3

4 Solution Steps DC Machine Design Process (Rotating part) Design of Armature Design of Armature Winding Design of Armature core Performance of Armature Basic Equations No of poles Check for P, D, L No of Conductors No of slots Size of Conductor Design of slots & check Φ, β, d, D of core Armature losses Cu loss Iron loss 4

5 Solution Steps DC Machine Design Process (Stationary part) Design of field system Design of field Winding Design of Commutator & Brushes Design of Interpole Overall Performance Main Pole Yoke AT for iron part Airgap Shunt Field Series Field Commutator & Brush Dimensions Losses Temp. rise Dimensions Airgap Density Interpole Winding Losses Efficiency 5

6 Design of Armature Important Equations Assumptions made Bav = 0.63 Tesla, ac = amp conductors/m Field Armature Cu loss = 5% of output Ratio of pole arc to pole pitch = 0.7 Gross length of armature, L = pole arc (square pole face) Power developed by armature, Pa = 1.05 x 250 = kw Output coefficient C0 = D*D*L = m2 6

7 Design of Armature No of Poles Assuming frequency of flux reversal to be 30Hz. P = 120F/N = (120*30)/600 = 6 Poles Gross length of armature L = pole arc = 0.7(πD/6) π = 0.367D From D*D*L= m2 And L = 0.367D D = 0.72m L = 0.25m 7

8 Design of Armature Check on the selection of P, D and L 1. Peripheral speed = πdn 2. Voltage between commutator segment = 2 Bg*L*v 3. Pole pitch = πd/p Current per Brush arm = load current/pair of poles Armature ampere turns per pole = πd*ac/2p Net length of armature Li = (0.25-4*0.01)*0.9 = 0.19m Which means 4 ventilating ducts of 0.01m length are provided with iron factor =

9 Design of Armature Winding Type of winding assumed- Simple lap, double layer No of conductors Z = E*60*A/φ*p*Nφ Voltage drop= 3% of terminal voltage (assumed) E = 400+(400*0.03) = 412 V For lap winding A = P = 6 Flux per pole φ = Bav*(πDL/p) π = wb (equation of specific magnetic loading) Z = 704 9

10 Design of Armature Winding No of slots (slot pitch assumed to be 2.6cm) No of slots = πd/slot pitch = 88 No of conductors per slot = 704/88 = 8 (2x4) Using single turn coil, no of coils = 704/2 = 352 Current loading in slot = (625/6)*8 = 832A Size of conductor Full load current of generator = 625A Armature current = 625+5= 630A (Assuming Ish = 5 A) Current per conductor = 630/6 = 105A Area of conductor= 105/5= 21mm2 (15mm x 1.4mm without insulation) (Assuming current density=5a/mm2) (15.4mm x 1.8mm with insulation) 10

11 Design of Armature Winding Dimensions of slots Thickness of insulation in slot = 0.4mm (Assumed) Insulation between two layers = 0.5mm (Assumed) Wedge space = 3mm (Assumed) Slot Width: Space occupied by insulated conductor, 4x1.8= 7.2mm Insulation in slot, 2x0.4 = 0.8mm 11 Allowance for slackness = 1.0mm Total slot width = 9.0 mm Slot Depth: Space occupied by insulated conductor,2x15.4= 30.8mm Insulation I slot, 3x0.4 = 1.2mm Insulation between layers = 0.5mm Wedge spaces = 3.0mm Allowance for slackness = 2.5mm Total slot depth = 38.0 mm

12 Design of Armature Winding Check on design of slot Average flux density in teeth Btav = φ/ati = 1.41Tesla Maximum flux density in teeth = 1.41/0.7 = 2.02 Tesla 12

13 Design of Armature core Flux in the core section φc = /2 = wb Flux density in the core Bc = 1.3 Tesla (Assumed) Sectional area of the core Ac = /1.3 = m2 Depth of core dc = Ac/L = /0.19 = 0.12m Mean diameter of core Dmc = D-2hs-dc = 0.72-(2*0.038)-0.12=0.524m Internal diameter of armature stamping di = D-2(hs+dc) = ( ) = 0.4m 13

14 Performance of Armature Armature copper loss Armature Cu loss = Ia*Ia*Ra = 630*630*0.0153(calculated) = 6.1kW Eddy current losses in conductor may be assumed 20% of armature cu loss. Hence, Total cu loss= = 7.32kW Iron Losses Weight of core = 289kg (calculated) Hysteresis loss per kg at core flux density of 1.3 Tesla and freq 30Hz can be determined from graph to be = 2.8 W. Total Hysteresis loss = 2.8*289 = 809 W 14

15 15 Armature of 500HP DC Machine

16 Design of Field System Design of Main pole To avoid magnetic centering of armature, axial length of pole is kept 1.5cm lesser than length of armature. Thus axial length of main pole = = 23.5 cm 1.Width of pole Leakage factor = 1.2 (Assumed) Pole flux on full load = 1.2* = 0.07wb Flux density in pole body = 1.6 Tesla (Assumed) Net area of pole = 0.07/1.6 = m2 Gross area of pole = /0.95 = m2 (Assume iron factor=0.95) Width of pole = 0.046/0.235 = 0.196m 16

17 Design of Field System 2.Height of pole Height of pole = Height of field winding + Height of pole shoe + space wasted due to insulation and curvature of yoke (0.1 times pole pitch) Field AT per pole on full load (Calculated) /AT per meter height of coil (Calculated) = 6160/44243 = 0.14m = 14 cm = height of field coil Height of pole shoe = 4 cm Total height of pole = 14+4+(0.1*37.7) = 22 cm (Approx) 17

18 Design of Field System Dimension of yoke Material for yoke = cast steel Flux density in yoke section = 1.2 Tesla (Assumed) Flux in the yoke = 0.07/2 = 0.035wb Area of yoke Ay = 0.035/1.2 = wb/m2 Axial length of yoke is assumed to be 1.5times the length of armature. Ly = 1.5*0.25 = 0.375m Depth of yoke Dy = Ay/Ly = /0.375 = 0.08m 18

19 Design of Field System Calculation of AT for iron parts 1. AT for pole (flux density = 1.6 Tesla) AT per meter length = 4800 (From BH curve) Height of pole = 0.22m ATp = 4800*0.22 = AT for armature teeth (flux density = 2.02 Tesla) AT per meter length = (From BH curve) Flux path in teeth = 3.8cm ATt = 21500*0.038 = AT for armature core (flux density = 1.3 Tesla) AT per meter length = 400 (From BH curve) Flux path in core = 0.137m ATc = 400*0.137 = 55 19

20 Design of Field System Calculation of AT for iron parts 4. AT for yoke (flux density = 1.2 Tesla) AT per meter length = 1230 (From BH curve) Flux path in yoke = 1.24m ATy = 1230*1.24 = 400 Total field AT for iron parts = ATp + ATt + ATc + Aty = = 2328 Total field AT of pole (calculated before) =

21 Design of Field System Length of Airgap Out of total AT, 15% is used to overcome armature reaction. So, AT needed on load = 0.85*6160 = 5236 Out of 5236, 2328 needed for iron parts. So ATg = = 2908 Max. flux density Bg= Bav / Kf = 0.63/0.7 = 0.9 Tesla. Atg = Bg*Kg*Lg*0.796*10^6 Lg = 0.35cm 21

22 Design of Field Winding The field winding will be designed to develop 6160 AT on full load. Out of 6160, shunt field winding will be designed for 5236 AT which is sufficient to generate an emf of 400V on no load. The series field winding is then designed for remained 924 AT. Design of shunt field winding Height of field coil = 14cm Assuming 80% of above for shunt field winding, the space available = 0.8*14 = 11.2cm 1. Cross section area of conductor Assuming 15% of voltage drop in field,vsh = 0.85*400 = 340V Voltage across each field coilvc= 340/6 = 56.7 V. AT for each field coil =

23 Design of Field Winding Depth of field winding df = 4.5cm (Assumed) Thickness of insulation on pole body ti= 1.0cm (Assumed) Length of mean turn of shunt field winding Lmt = 2(bp+lp)+ π(df +2ti) = 1.07 m Cross section area of conductor Ash = Þ*Lmt*AT/Vc = 1.98mm2 2. Permissible loss Cooling surface of shunt field coil = 2*lmt*height of shunt coil = 2*1.07*0.112 = 0.24m2 Permissible losses per sq. meter= 725W/m2 (Assumed) Permissible losses per exciting coil = 725*0.24 = 174W Total permissible loss of shunt field winding = 174*6 = 1044W 23

24 Design of Field Winding 3. Exciting current Ish = 1044/340 = 3.1A 4. Number of turns per exciting coil = 5236/3.1 = Arrangement of winding Bare dia of round conductor = 1.6mm (From standard table) Area of standard conductor = mm2 Insulated dia of conductor with insulation = 1.71mm (From standard table) No of turns in available height of 11.2cm = 11.2/0.171= 65 No of turns per layer = 65 No of layers = 1690/65 = 26 Actual depth of winding = 26*0.171 = 4.45cm 24

25 Design of Field Winding 6. Resistance of winding Total turns of shunt field winding = 1690*6 = Resistance of winding, Rsh = Þ*Lmt*Tsh/Ash = 0.02*1.07*10140/ = 108ΩΩ 7. Losses in shunt field winding = (3.1*3.1)*108 = 1038W Design of series field winding AT to be developed by winding = 924, Ise = Ia = 630A. No of turns per pole = 924/630 = 2 (Approx) Modified AT for this winding = 2*630 = 1260 Assumed current density = 2.1 A/mm2 Cross section area of conductor = 630/2.1 = 300mm2 25

26 Design of Field Winding Size of copper strap = 15mm X 20mm Rse = Ω (calculated) Cu loss in series field winding = 630*630* = 341W 26

27 Design of Commutator and Brushes 27 Cut Section of DC Machine

28 Design of Commutator and Brushes Voltage drop per brush set = 2V (Assumed) 1. No of commutator segments = No of armature coils = Dia of commutator = 70% of dia of armature = 0.7*0.72 = 0.5m 3. Dimension of brush Number of segments covered by brush = 3 Approximate thickness of brush = 3*0.446 = 1.3cm Standard size of brush selected = 12.5mm thick, 20mm width, 32 mm height 4. No of brushes per spindle Area of one brush = 12.5*20 = 250mm2 = 2.5cm2 28

29 Design of Commutator and Brushes Total current per brush set = 625/3 = 208A Working current density of brush = 6 A/cm2 (Assumed) Total area per spindle = 208/6 = 34.7 cm2 No of brushes per spindle = 34.7/2.5 = 14 5.Length of commutator Thickness of brush box = 0.5cm (Assumed) Allowance of staggering of brushes = 4cm (Assumed) Clearance at the end of commutator = 2cm (Assumed) Allowance of riser = 3cm (Assumed) Space occupied by one brush box = = 2.5cm Total space required by 14 brush boxes = 14*2.5 = 35cm Length of commutator = = 44cm 29

30 Design of Commutator and Brushes 30 6.Commutator losses Brush Friction loss = µ*ab *p *Vc where, coefficient of friction, µ = 0.2 (Assumed) Brush pressure on commutator, p = 12000N/m2 (Assumed) Total contact area of all brushes = 2.5*14*6 = 210cm2 = 0.021m2 Thus, brush friction loss = 0.2*0.021*12000*15.7 = 791W And, Brush contact drop=2*625 = 1250W Total commutator loss = = 2041W 7. Temperature rise of commutator Temp rise ϴ = 120*losses in w/m2 / 1+(0.1*Vc) cooling surface of commutator = πdc*lc = π*50*43 = 6758cm2 Losses in w/m2 = 2041/6758 = Temp rise ϴ = 120*0.305/(1+0.1*15.7) = 14.1ºC

31 Design of Interpole 31 Dimensions 1.Axial length of interpole Lcp = 0.64 times gross armature length = 0.64*25 = 16cm 2. Width of interpole shoe Bcp = [(x-1)τc + tb - tm] Da/Dc x = no of coil sides per layer in armature slot = 4 τc = commutator segment pitch = 0.446cm Thickness of brush tb = 12.5mm = 1.25cm Thickness of mica insulation between segments = 0.8mm =0.08cm Da = Dia of armature, Dc = Dia of core, Bcp = 3.6cm 3.Airgap under interpole Lgc= 1.5*Lg = 1.5*0.35 = 0.52cm

32 Design of Interpole 4. Height of interpole hcp= [(Dyi-D)/2] Lgc Internal dia of yoke Dyi = D+2Lg+2hp = 116.7cm hcp = 21.8cm Flux density in the airgap under interpole Flux density Bcp = (L/Lcp)*(2Ia Zsλ/βxa) Ia = current in armature conductor = 105A Zs = no of conductors per slot = 8 32 βxa = distance moved by armature during commutation of x coil sides in slot = 3.6cm λ = [0.4π(hc/3bs + ht/bs + bcp/6lgc) + (lfr/l)*{0.23log(lfr/b) }]*10^-6 wb/m where, hc = copper height in armature slot = 3.08cm

33 Design of Interpole 33 ht = height above copper in armature slot = 0.55 cm bs = width of slot = 0.9 cm b = periphery of four coil sides in layer of slot = 45.2mm lfr = free length of turn = 57.4 cm λ = 4.39*10^-6 wb/m Bcp = 0.32 Tesla Interpole Winding 1. No of turns AT for interpole = ATa *Bcp*Kgcp*Lgcp*10^6 = 7580 Current in interpole winding = Ia = 630A No of turns in each interpole = 7580/630 = 12 (Approx) Modified AT = 12*630 = Conductor cross section area Acp = 630/2.3 = 274 mm2 (current density assumed to be 2.3A/mm2)

34 Design of Interpole 3. Arrangement of winding Height of interpole = 21.8cm Space for insulation = 3.8 cm, available space for 12 turns = 18cm Space available for one turn = 18/12 = 1.5cm Size of conductor selected = 14.5mm x 9.5mm Size of conductor with insulation = 15mm x 10mm Depth of winding = 2x10 = 20mm So, Lmtc = 47.4cm (calculated from above data) 4. Resistance of winding Rcp = (calculated) 5. Losses in interpole winding = 630*630* = 1064 W 34

35 Overall Performance 1. Losses Constant losses = = kw Variable losses = = kw Total losses at full load = = 17.25kW 2. Efficiency At full load η = 250/( ) = 93.5% For maximum efficiency, a*a*9.975=7.274 a = 0.85 time full load So, load at maximum efficiency = 0.85*250 = 212.5kW ηmax = 212.5/( ) = 93.6% 35

36 36 M&V Patel Department of Electrical Engineering