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Ejemplo 02: 1HP 4polos 50 Hz

Dimensionamiento del estator  Pn= 1HP U=220/380 V conexión: Y/D f=50 Hz F.P. = 0.79 eff = 71.7 % n= 1380 rpm Cálculo provisorio del estator U=380 V Conexión: Estrella 1)Potencia aparente: Q=(1 HP * 0.746)/(0.79*0.717)= 1.317 KVA 2)# polos  n_s=1500 ---> p = 4 3)Velocidad sincrónica Ns=2*f/p=25rps 4)Adopto: Lambda = 1.25 ; Bav=0.5 T ; ac=20000 A.c/m ; kws=0.955 5)Constante de salida Co=1.11*(Pi^2)*Bav*ac*kws*10^-3=104.6227 KVA.seg.m^-3 6)Dimensiones principales Como: Lambda = L/Tau ---> L = Lambda * Tau = 1.25*Tau Pero: Tau=Pi*D/4 Luego: L=1.25*Pi*D/4 Entonces: Co=Q/(D^2*L*Ns) ---> Q/Co = D^2*(1.25*Pi*D/4)*Ns = (5/16)*Pi*D^3*Ns Despejando D: D^3=Q/[Co*(5/16)*Pi*Ns] ---> D = 0.0799 m ~ 8 cm = 80mm Tau=Pi*D/4=6.283 cm ~ 63 mm L=1.25*Tau=7.854 cm ~ 78.5 mm 7)Adopto: q=2 ranuras/polo*fase # Ranuras: Ss=q*polos*fases=2*4*3=24 Paso de ranura:  y_ss=(Pi*D)/Ss=10.47 cm = 10.5 mm 8)Flujo/polo:  F_p=Bav*Tau*L=2.47275*10^-3 Wb 9)# vueltas por fase Uph=380/sqrt(3)=219...
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Example 01 : 22KW 6 poles 415 V 50 Hz

Youtube link: 22 KW 30HP 72 Slot 6 Pole 975 RPM full Motor Winding with Connection Diagram DoubleDelta Connection 1) Nameplate data: U = 415 V I = 42.6 A f = 50 Hz P.F. = 0.79 eff = 90.9% conn: Delta TClass/Rise: F amb: 50°C weight: 230 Kg n = 975 rpm IP 55  Duty S1 2) Data from video:  Di = 200 mm;  Dout = 300 mm;  L = 250 mm;  72 slots;  18 sets; 2 coils/set (pitch: 1-10 & 1-12); 26 turns/coil (3 wires in parallel: SWG #19 #20 #21) Copper Weight: 800 grams/set  Double Delta connection 3) Approximate calculations Assume: kws = 0.955 KVA  Rating Q=22/(0.79*0.909)=30.63597 KVA Sync speed n=975 rpm ---> n_s=1000 rpm ---> Ns=1000/60=16.666 rps # poles p = 2*50/16.666 = 6 poles Polar pitch Tau = (Pi*Di)/p = 0.104719 m Form factor  Lambda = L / Tau = 0.25 / 0.104719 = 2.387 ~ 2.4 Phase current Iph=42.6/sqrt(3)=24.59512 A 2 Delta connection ----> Paths = 2 Conductor current Iz = Iph/2 = 12.2975 A Conductor area as = #19 + #20 + #21 = 0...
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