Question

(d) The equivalent circuit parameters of the motor in (a), at 50Hz, are: R; = 1.50; X;= X, = 40; R = 20; and Xx = 1000 If the motor is to develop its maximum torque at standstill, calculate the output frequency and voltage (RMS) from the inverter, if the air-gap flux remains constant. [6 marks) Question 5 A 4-pole 3-phase, 50 Hz, Y-connected, non-salient pole synchronous motor has the following parameters: R=0.7/phase; X.. = 10 phase @ 50 Hz The motor induced emf phase, Ee, at base speed is 280 V. Vreed = 380 V/phase; Trated = 40 A/phase. The motor is driven from a current source inverter which supplies three-phase sinusoidal currents of variable frequency to the motor. The switching of the inverter is synchronized with rotor position in such a way that the stator current phasor of each phase can be set at any arbitrary commutation angle %, with respect to the back emf phasor E, of the corresponding phase. (a) Assuming that the motor is to be driven such that it develops its highest torque per ampere MTPA) characteristic, calculate the amplitude of the stator current when the motor develops 100 Nm of torque while operating at 1000 rev/min. [6 marks) 6) Calculate the power developed by the motor for this condition of operation. (4 marks] ©) Calculate the operating power factor of the motor. [6 marks] (d) Discuss, briefly, the effect of bringing the commutation angle y forward (i.e., 1, phasor leading the EA phasor) on the operation of the drive. (4 marks)

Can you please answer both (d) and Question - 5.

Answer 1

(d) RI P 11.52 Equivalent circuit of induction Motor ; jx, jX2 To j 40 ✓ 3jXm ၂၀၀ jur nim vlno At stond still S = 1 Rotor Terminal Thevenin Resistance a lyoss Rotor terminals- (Ritjx,) \ljxm . (Ritjx1) j xm Ritj CX) + Xm) (1.5 t; 4) * ;100 1.5-jloy is-jloy 1.5 +j104 ( 1.5+; 104) - 600.0 +15 600 225+j41600 2.25 +10816 15000 tj 41825 j 150 - 400 tj 10818.25 =(1.3865 +; 3.86615) Thevenin Voltage: With a Vi jxm Ritj(16*Xm) Nix j loo 1.5 tjloy Vilj 150 floyool 10818:25 Vth 1.3815 mm jy j 3.86 ണ jXth + j x2 Rth th condition for man tosque will be Variable Resistance - | constant Impedance ? max. power Transfer I pth + j(*th + x2)) 1(1.3865)*+(7.8072 2 2 s : 7.98 V Smt = 0.25 Aix gap flux Remain constant- && f V2 fi fz Vi V2 0.25x50 V2 = Vixo.25 tost VI 4 12:55 1 x50 = 12.5 Hz fa Y Nor given. xu, V, is V2 luonn : ( وله ) a) Torque developed (Ta) = speed 1000 rpon Motor e.mif. as- T 3 E a la Cosy with Tosque is Related ولا 28ou giver : ملا looore 60 ws Where ha is emf per phase Y is commutation angle. so rew/min .1000 x 2k rad./sec 104.72 Mad. / see Eg = 280V at base speed Base speed 120f Р 120x50 4 = 15oor pm X 1009 50 Hz 1503 frequency of operation & 1ooonpm At 2 33.33 Hz as Ems a 280 x 33.33 so ang @ 1000rpm $ 50 1860648V / phase 3x186.6 48 x la 100 = 104.72 ta 18.702 A cbi Equivalent ciyan't gf motor j6.67 I 0.7.12 + Xse Ra Ut=380v Lo &L-S [ xse af lor x 33-33 5o Xse at 33.33H2 = 6.672 = 1 86.648 v I efl Ital Ef - 18-702A Vt - Ia Ra - j Ia Xse Vt Lo - ErL-s la = Rot ja se I a I a 380 Loo - 186.648L-6 (0.7 tj 6. 16.67) 380 - 186.648 Cosb fj 186.64d sing 6.7 L84° 56.712-84° - 27.85 26-8-84°) 5-93 27-85cos (8 +084") + 27.8 5.93's (8 +840) 115.93 -27.05605C6 +849; 2 Ia = tý +(27.85 sin18+84°)? I a 18.702 - l I als solving for of as (18.702)2 = 810.76-330.3 cosis +84 6.-17.30 25.578 IQ = 5.93 - 27.05x0.395 ; -5.00 -j 25.578 -101.21 Ia - L1a - Power factor = cos (Ulut - 1.)) = cos 101.21° 0.1944 ly . Power developed will be = 3 Ea Ja - 3 x 18 6:648 X 18.702 - 10.472 KVA cd) of I phasor leads Et phas or -I Ra - Ia 14 Ia -jhaxs as Power factor will become lagging IE/<IVEI Here y - Internal power factor ongle. Drive will operate at less power so torque output will be less