Question

Problem 7 (Chapter 5: Three-Phase Synchronous Motors) A 440 V, 124 HP, 4 poles, 50 Hz, 0.9 leading Power Factor, A-connected synchronous motor has an armature resistance of 0.05 12, and a synchronous reactance of 1.5 12. This motor is operating at its rated conditions with an efficiency of 92 %. 1. Calculate the electrical input power. 2. Find the RMS phase and line currents. 3. What is the reactive power consumed by or supplied by this motor? 4. Calculate the copper losses. 5. How much power is converted from electrical to mechanical by this machine? 6. How much torque is induced by this machine? 7. The field current is changed until this motor operates at unity power factor under the same rated load. Find the new armature current, and internal generated voltage, and the torque angle.

Answer 1

Greetings of the day

2 ] Synchronous motor - = 92.4668 KW 92.4668 KW & Cinen in A40V, 124 HP 20 Hz 2009 Leading (A connected) Ra = 0.05€ Xs -1.5e Efficiency 92% 1. Electrical Input Power : 12 HP 124 10341 We knowo 1 kW = 10341 HP * Pina 2. RMs Phase and here Current Bella connected Ž IL = V3 Iphase Ve Vpnose VE V IL COO = Pin * V3 X 440 * I_ x0,9 * ILS 92.4668 X103 134.8124 A I. Phase = 77.834 A IL 13 * Iphase - 77.834 A

* Q. - 44.783 on supplied. 3. Reactive power consumed by the Synchronous Moles : - 8 = r2 VI, sino V2 x 440 X 134.812 (VT – 0.92) V3 x 440 X 134.812 X 0.4359 KVAR Supplies Beactive power (KVAR). &ynchronous moto workes at leading pf, it 1. Calculate the lopper loss. n of @yudironous moter - Pin - Losses in Pin no 92%. when X100 92.4668 - losses X150 92.4668 85.0694 - 92.4668 - Losses. Losses La 7.3973 kw 908.72 Watt - *Cu toss a 908.72 Watt copper losses 3 Iph ? Ra ( for 3 phase) 5. Power converted from elechical to Mechanical by the machine -Losses Qutfue Penver = Mechanical Power = Porvel) Sospent 85.0694 kW 85.0694 KW *Pnech 6. Torque induced by the machine Pinech = ' induced & Wm.

Nm = speed of mota - Wm 2aNm 60 120x5 тохо 21560 P 4 Spm = 27x is end - Son. - 157.0796 x 60 rad. 85069.4 = 157.0796 x Zinduced Pinech * Tinduced = 541.57 N-m field current is changed pf = 1. hated load. Iainew) = ? , & internal voltage) = ? S (torque ange) = ? Now, Pin 92.4668 KW Ii z v I caso 92.4668 x103. 92.4668 X103 V3 x 440 X 1 * Tanow - IL CRMS) - 121.331 A Ia (Phase) = 70.05A Ey - V - I (Ra+jxs) - 44040 - Ja (Phases less ci) (0.05 +5115 = 44060 - 70.05.0 (0.05+J1.5) * Ef (Phase voltage = Lure vollage] 448.966 1-13.5350 Volt * Torque angle (8) - 13.5350

Answer 2

Greetings of the day

2 ] Synchronous motor - = 92.4668 KW 92.4668 KW & Cinen in A40V, 124 HP 20 Hz 2009 Leading (A connected) Ra = 0.05€ Xs -1.5e Efficiency 92% 1. Electrical Input Power : 12 HP 124 10341 We knowo 1 kW = 10341 HP * Pina 2. RMs Phase and here Current Bella connected Ž IL = V3 Iphase Ve Vpnose VE V IL COO = Pin * V3 X 440 * I_ x0,9 * ILS 92.4668 X103 134.8124 A I. Phase = 77.834 A IL 13 * Iphase - 77.834 A

* Q. - 44.783 on supplied. 3. Reactive power consumed by the Synchronous Moles : - 8 = r2 VI, sino V2 x 440 X 134.812 (VT – 0.92) V3 x 440 X 134.812 X 0.4359 KVAR Supplies Beactive power (KVAR). &ynchronous moto workes at leading pf, it 1. Calculate the lopper loss. n of @yudironous moter - Pin - Losses in Pin no 92%. when X100 92.4668 - losses X150 92.4668 85.0694 - 92.4668 - Losses. Losses La 7.3973 kw 908.72 Watt - *Cu toss a 908.72 Watt copper losses 3 Iph ? Ra ( for 3 phase) 5. Power converted from elechical to Mechanical by the machine -Losses Qutfue Penver = Mechanical Power = Porvel) Sospent 85.0694 kW 85.0694 KW *Pnech 6. Torque induced by the machine Pinech = ' induced & Wm.

Nm = speed of mota - Wm 2aNm 60 120x5 тохо 21560 P 4 Spm = 27x is end - Son. - 157.0796 x 60 rad. 85069.4 = 157.0796 x Zinduced Pinech * Tinduced = 541.57 N-m field current is changed pf = 1. hated load. Iainew) = ? , & internal voltage) = ? S (torque ange) = ? Now, Pin 92.4668 KW Ii z v I caso 92.4668 x103. 92.4668 X103 V3 x 440 X 1 * Tanow - IL CRMS) - 121.331 A Ia (Phase) = 70.05A Ey - V - I (Ra+jxs) - 44040 - Ja (Phases less ci) (0.05 +5115 = 44060 - 70.05.0 (0.05+J1.5) * Ef (Phase voltage = Lure vollage] 448.966 1-13.5350 Volt * Torque angle (8) - 13.5350