Question

T u resistance in a DC is decreased while the motor is driving a ii. A 100-hp, 250-V DC shunt motor with compensating winding has an am 0.04 12 and a variable shunt field resistance R; 92. The magn ompensating winding has an armature resistance of shunt field resistance R, 12. The magnetization curve of this motor is shown in Fig. 1. At no load, the field resistor has been a . At no load, the field resistor has been adjusted to make the motor run at 1200 rpm. Mechanical and core losses may be assumed to be negligible. (a) What is the shunt field current, Isand the shunt field resistance, R, in this machine at no load? (b) The motor is assumed to be initially driving a load with an input line current (IL) of 125 A and a speed of 1103 r/min at the field resistance found in part (a). Assume that the amount of armature current (L.) drawn remains constant. If the field resistance is raised to 6012, find its speed, load torque and efficiency. 300 n =1200 rpm Open-circuit terminal voltage (V) FA 9.0 10.0 8.0 7.0 3.0 2.0 4.0 5.0 6.0 Field current (A) 1.0 0.0

Answer 1

il gineos Vic 250: V. Raco.oun naagneti carve. I we have to find @ 86 and Re at no load (&co). LU A from the crecenta to &q Rq & Eco ut Eat & Ral o sine I = Ratio at no load incolaco from egun at no load 1 e Estrada I c . Eanefend 2.250V.. from magnetizing Cince Cassume asolinear gat o no toad erage : from 'encant que com write to Also So, 25ofa SAX RA Roc :2500 Critical resultance at No 1200 PM

D PL = 125 = Ratlf 2 125 = Ig AS Ra = 1204 qa, cat speedHe 110 3 opon. v = Eqt la X Ro 2 950 = E +120xboy taerren. 18c252 I le (= 2 45:2 V fn, 2482 60 when af → inereate Roomi (son toson) thien 2 = 250 = 4.166A Jazz 120 A (consty here given for 24, then the 245.2 we know that a Lond Rp N approx z = Bp X N 2454 a 5 x 1103 Ez le X H2 2454 - 4:06 x Na N22 1103x5-1323 . 4.166 теr Torque e ta la 245.2 x 120 en x 1323.6 60 2 12 28 Nom I torguen