Factors | DC Machine | Transformers | Induction Machines | Synchronous Machines |
Output Equation | Pa=CoD2Ln, where Pa=P/h for generators, Pa=P for motors | For Single Phase Q=2.22 f Bm Ai Kw Aw d 10-3 For Three Phase Q=3.33 f Bm Ai Kw Aw d 10-3 | Q=CoD2 L ns KVA Input Q= HP * 0.746 / Cos f * h | Q=CoD2 L ns KVA Input Q= HP * 0.746 / Cos f * h For Turbo alternators Q=1.11Bavac KwsVa2 L 10-3/ns |
Output Coefficient | Co=Bav ac* 10-3 where Bav-magnetic loading and ac - electric loading | DNA | Co=11 Kws Bav ac 10-3 | Co=11 Kws Bav ac 10-3 |
Choice of Magnetic Loading | Flux Density in Teeth Frequency of Flux Reversals Size of machine | DNA | Magnetizing current, Flux Density, Iron loss | Iron loss, Stability, Voltage Rating, Parallel Operation, Transient ShortCircuit current |
Choice of Electric Loading | Temperature rise, speed of machine, Voltage, Armature reaction, Commutation | DNA | Overload Capacity, Copper losses, Temperature rise, Leakage Reactance | Copper loss, Synchronous reactance, Temperature rise, Stray Load losses, Voltage rating |
Flux Density | Yoke – 1.3 to 1.6 Wb/m2 Pole – 1.2 to 1.7 Air Gap – 0.4 to 0.6 Armature teeth – 1.5 – 2.2 Armature core – 1.0 to 1.5 | 1. 0 to 1.4 wb/m2 for Distribution Transformer 1.2 to 1.5 for power Transformers | Stator tooth1.3 to 1.7 Wb/m2Rotor tooth1.3 to 1.7 Wb/m2 | |
Current Density | Large Machine with strap wound conds – 4.5A/mm2 Small M/c with wire wound conds – 5A/mm2 High speed - 6 to 7A/mm2 General - 4 to 7A//mm2 | 1.1 to 2.2 A/mm2- small Tr.. 2.2 to 3.2 – large power Tr.. 5.4 to 6.2 – Large power Tr with Forced circulation | In rotor bar 4 to 7A/mm2 | Current density in armature conductor d = 3 to 5A/mm2 |
Main Dimension | D- Diameter of the armature, L- Armature Length | Hw (Height of window) and Ww (width of Window) | D –diameter of stator bore L- Length of Stator core | D –diameter of stator bore L- Length of Stator core |
L/Ï„ Ratio | 0.45 to 1.1 b/Ï„ =0.64 to 0.72 (for Square pole face and Square Pole Section) | DNA | minimum cost L/t =1.5- 2 good pf L/t = 1 – 1.25 good h L/t =1.5 overall design L/t =1 best pf t =Ö0.18L | L/t = 0.6 to 0.7 L/t = 1 to 5 |
Choice of Number of Poles | Frequency Between 25 to 50 HZ Current per parallel path is limited to 200A. The Armature MMF Should not be large. | DNA | DNA | For Bolted poleVa = 50 m/sDovetail & T Head Va=80 m/s |
Length of Air Gap | lg=(0.5 to 0.7)* ac* Ï„ * 1.6*106 Bg Kg | DNA | Lg= 0.2 + 2ÖDL Lg=0.2 + D Lg=0.125+0.35D+L+0.015Va Lg=(1.6ÖD) – 0.25 | M/c with open type slots L g/t = 0.01 to 0.015 M/c with maximum o/p L g /t = 0.02 Turbo Alternator Lg=0.5SCR act Kf10-6/Kg Bav |
Slot Information | Slot Area = Conductor area/ slot space Factor | DNA | Stator Slot Pitch Yss=p D/Ss | Values of Stator slot pitch Yss < 25mm – Low Volt M/c Yss < 40mm – for 6KV & less Yss < 60mm - m/c upto 15kV |
Tooth Information | DNA | DNA | width of stator tooth Wts min = fm / 1.7(Ss/p) Li Width of rotor toothWtr min = fm / 1.7(Sr/p) Li | DNA |
Core | D=Di+2dc+2ds Di Inner Dia.of armature dc Depth of core ds Depth of slot | Square Core Kc = 0.45 Two stepped core Kc = 0.56 Three stepped core Kc=0.6 Kc – core area factor = Ai/d2 | Depth of stator coredcs = fm / 2 Bcs LiDepth of rotor coredcr = fm / 2 Bcr Li | DNA |
Armature MMF/Pole | Up to 100Kw- 5000 or less 100 to 200Kw- 5000 –7000 500 to1000Kw 7500-10000 over1500Kw –Upto 12,500 | DNA | DNA | Armature MMF/poleATa = 2.7 Ip Tph Kws / pField MMFATf =SCR * Ata |
Dispersion Coefficient | DNA | DNA | s = Im/Isci Im-Magnetizing current Isci – ideal Short ckt current | DNA |
Short Circuit Ratio | DNA | DNA | DNA | The ratio of field current required to produce rated voltage on OC to field current required to circulate rated current at SC. SCR = 1/Xd |
Slot Loading | DNA | DNA | Slot Loading = Zss Is Conds/slot Zss=6Ts/Ss Stator Conds = Ss Zss | DNA |
Additional Information | Current/Parallel path = Ia/p For Wave Winding Ia/2 | number of tubes = [(Pi +Pc/q)-12.5 St] /8.8p dt lt | Rotor Bar currentIb = 0.85 6Is Ts/SrEnd ring currentIe=Sr Ib/p p(Ss-Sr) should not be equal to 0,±p, ±2p, ±3p, ±5p, ±1, ±2, ±(p±1) ±(p±2) | Current thru the conductor Iz = Iph/a Peripheral Speed Va = p D ns |
Reference: Electrical Machine design by A. K Sawhney
sir,
ReplyDeleteyour work is excellent, but if we consider in design aspect, you should mention nomenclature also, other wise different text books will follow different nomenclatures.
Hmm Yes. But I prepared this 13 years back, I think it still works. Now I am in Computer Science and Engg. So Lost the touch in this subject.
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