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Từ trường đập mạch phân tích thành hai từ trường có biên độ nửa biên độ từ trường đập mạch, quay tốc độ theo hai chiều ngược Φ A1 B1 C1 Φ ΦA = Φ ΦB = ΦA n A2 B2 C2 n ΦB Như máy khơng có từ trường quay nên khơng có mơ men quay Nếu roto quay theo chiều đó(bằng tay chẳng hạn) hình bên, Trong dẫn Φroto xuất dẫn roto cắt qua d s.đ.đ dòng điện Dòng điện tạo từ thông Φ q Φ d gọi từ thông dọc trục Φ q – từ thông ngang trục Φd Φq Quan hệ pha Φ d, Φ d s.đ.đ: Φq E&2 I&2 &q Φ &d Φ Φd ωt 315 270 90 45 0oo 180 225 135 oo Như quay roto tạo từ trường ngang Từ trường kết hợp với từ trường dọc trục tạo từ trường quay theo hướng quay ban đầu Từ trường gia tốc roto lên đến tốc độ gần Động phân pha Để tạo từ trường quay cần có dây quấn stato thiết bị phân pha Xf Xc Rf PS Phase splitter Rc Dây quấn tạo Φ d cịn dây quấn phụ đặt lệch 90o điện so với dây quấn tạo Φ q Dây quấn phụ gọi dây quấn The phase splitter is connected in series with the auxiliary winding and causes the current in the auxiliary winding to be out of phase with the current in the main winding The splitter is normally resistance or Locked-rotor torque The locked-rotor torque of a split-phase motor is proportional to the product of the magnitude of the locked-rotor current in each winding times the sin of the angle of phase displacement between two currents M mm = kI cI f sin α mathematically: Ic – current in main winding If – current in auxiliary winding Expressed Example: The main and auxiliary windings of a 220V, 50Hz split-phase motor have the following locked-rotor parameters: Rc = 3.94Ω, Xc = 4.2Ω, Rf = 8.42Ω, Xf = 6.28Ω Determinate winding, two locked rotor current phase-displacement currents, locked-rotor in each angle between torque, external resistance required in series with the auxiliary winding in order to obtain a 30o phase The impedances of the two windings: Zc = 3.94 + j4.2 = 5.7588∠46.83o Ω Zf = 8.42 + j6.28 = 10.5∠36.71o Ω Currents in the two windings: o & U 120 ∠ o I&c = = = 20.84 ∠ − 46.83 A o Zc 5.7588∠46.83 o & U 120 ∠ o I&f = = = 11.42 ∠ − 36.71 A o Zf 10.5∠36.71 Phase displacement angle between the currents in the two windings: α = 46.83o – 36.71o = o 10.12 Locked-rotor torque: Mmm = k× 20.84× 11.42× sin10.12o = In41.82k order to obtain a 30o phase displacement between the currents in the two windings the phase angle of the current in the auxiliary winding If must be: o o θ = 46.83 - 30 = 16.83o Xf tgθ = Rf + Rx Zf θ Rf + Xf Xf 6.28 − Rf = − 8.42 = 12.34Ω Rx = o tgθ tg16.83 With the auxiliary resistance Rf, the current If is: o & U 120∠0 & If = = = 5.5327∠ − 16.83o A Zf (12.34 + 8.42) + j6.28 And the locked-rotor torque in this condition is: Mmm = k× 20.84× 5.5327× sin16.38o = 32.52k Resistance-start split-phase motor • The circuit diagram for resistance-start motor: Main winding • The auxiliary winding is wound with smaller diameter wire than the Aux winding main winding, causing the auxiliary winding to have a higher ratio of resitance to reactance than the main winding • The switch auxiliary in the circuit is magnetic relay, a solidstate switch centrifugally switch • A or a Main winding Aux winding operated & U representative phase diagram for the motor is shown in the I&f I&c &d Φ &q Φ • The torque-speed M characteristic is shown in the next figure Capacitor-start split-phase motor s • The circuit diagram of the motor is shown in the • fig.:The displacement the phasebetween currents is Main winding C Aux windin g • A phasor diagram and torque-speed characteristic for the motor is as follows: &q M I&f Φ & U I&c &d Φ s • With the capacitor, α = (77 ÷ 88)o then the locked-rotor torque greater than the one of the resistance-start motor Example: The main and auxiliary windings of a 220V, 50Hz split-phase motor have the following locked-rotor parameters: Rc = 3.94Ω, Xc = 4.2Ω, Rf = 8.42Ω, Xf = 6.28Ω Determinate the capacitance required in series with the auxiliary winding in order to obtain a 90o phase displacement between the current in the main winding and the current in the The impedances of the windings: Zc = 3.94 + j4.2 = 5.7588∠46.83o Ω Zf = 8.42 + j6.28 = 10.5∠36.71o Ω The current in the main winding: o & U 120 ∠ o I&c = = = 20.84 ∠ − 46.83 A o Zc 5.7588∠46.83 The current in the auxiliary winding: o & U 120 ∠ o I&f = = = 11.42 ∠ − 36.71 A o Zf 10.5∠36.71 For the capacitance in series with the auxiliary winding , to obtain the 90o phase displacement, the phase angle of the current in the auxiliary I&f winding must o θ = 90 – be: 46.83o = 43.17Xo − X c tgθ = f Rf θ 46 83 o X c = X f − R tgθ = 6.28 − 8.42tg(−43.17o ) = 14.1786Ω I&c & U 1 C= = = 224.5àF 2fX c 2ì ì 50ì 14.1786 The current in the auxiliary winding If: o & U 120∠0 & If = = = 19.06∠ − 43.17o A Zf 8.42 + j(6.28 − 14.1786) Locked-rotor torque is : Mmm = k× 20.84× 19.06 = 397.2k Shaded-pole motors • The shaded-pole motor utilizes a shortcircuited coil or copper ring, called shading coil • This coil wound around a part of the pole face • The current flow in the stator winding and produces magnetic flux Φ in poles: &= Φ &c + Φ &n Φ a Φc Φ n β Φ c - magnetic flux don’t pass through the short- circuited turn Φ n - magnetic flux pass through the shortcircuited turn Φ n induces in the shading &n Φ coil a e.m.f En In the &c Φ shading coil appears a current In This α & Φ E&n &s Φ I&n current creates a flux Φ s The resultant flux through the shading coil isΦΦ is shifted in time relative to Φ c an angle of α and in space an angle of β Therefore in the air gap there is a ellipse rotating magnetic field 6 Operation of three-phase induction motors from single-phase lines • Sometimes, three-phase induction motors can oprerate from single-phase lines • We can use the circuit diagram as follows: C1 ã Normally we use: C1 + C2 230àF/hp C1 ≈ 26.5µF/hp ≈ C2 C1 – running capacitance C2 – starting capacitance P1f = P3f Single phasing(a fault condition) • Single phasing is a fault condition in which a three- phase motor is operating with one line • If motor is running when single phasing open occurs, it will continue to run as long as the shaft load is less than 80 percent rated load and remaining single-phase is normal • Assuming that the A line is opened and the shaft load remains the same when single phasing as when operating three phase: 3UI 3f cosϕ3f = UI 1f cosϕ1f • The current for this fault condition: cosϕ3f I 1f = 3I 3f cosϕ1f • Thus, for the wye connection, the phase current and the line current are the same: IA = IB = • For delta connection: I A = I B = I 1f I C = I 1f • If single phasing occurs while operating at or near rated load, the increase in phase current will cause rapid heating of the winding • Therefore, protective devices must be provided to trip the machine from the supply lines or severe damage to stator and rotor ... Như máy khơng có từ trường quay nên khơng có mơ men quay Nếu roto quay theo chiều đó(bằng tay chẳng hạn) hình bên, Trong dẫn Φroto xuất dẫn roto cắt qua d s.đ.đ dòng điện Dịng điện tạo... stato thiết bị phân pha Xf Xc Rf PS Phase splitter Rc Dây quấn tạo Φ d dây quấn phụ đặt lệch 90o điện so với dây quấn tạo Φ q Dây quấn phụ gọi dây quấn The phase splitter is connected in series
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