Question

3) The armature of a permanent-magnet dc motor has a resistance of 0.72 and draws a current of 50 A when the speed is 1500 r/min when the armature is connected to a source of 48 V, calculate the following: a) The induced emf, EA, and the torque, T, produced at this speed. b) The motor speed, n (in rpm), when the torque is 2 Nm. c) The armature current at 1600 r/min. d) The torque at 1600 r/min. e) The power in, the power out, and the efficiency at 1600 r/min. What is the maximum torque that the motor can produce while maintaining an efficiency of at least 80 % ? g) If the motor operates in the range of loads from 0 to 1 Nm what is the speed regulation of the motor?

Answer 1

Te armature mature of or permanent magnet de motor esistance of 0.72 and draws an SOA on Steed 1500 men when w Connected to Source or calculate the following Hearma cato 0 48V here Armature Resistance (Ra)=0.722 Armature current To = 50A. Source voltage (v) = 48 volt @ Induced em.f (EA) = V - IaRa = 48 -0.72x50 = 12 volt speed is (N) = 1500 rpm Torque (T) = CAT 60 EA TX60 2AN 12 x 50x60 T= ZTX 1500 T= 3.8197 N.m Torque = 3.8197 N.m So

(6) The mo- The motor speed (in 8.p.m) when torque en 2N.m we know that, T= EA la 20/0 T = 12x50x60 20xN 12x50x60 2= 204N N= - 12x50x60 45 4 = 2864. 78 R.P.M. (e) The Armature curent at 1600 r/min N=1600 rp. m. Ea = 12 volt T = 3.8197 N.m Ea Ta 2AN/co we know that, T = = 3.8197 = 12% fax60 20X1600 -> fa - 27X1600x3.8197 12 x 60 - Ta = 53.33 Amp

(d) The torque at 1600 r/min Here, N=1600 rpm T= 200/60 7 T - 12 x 50x60 25 x 1600 9 T = 3.580 Nm Power in (Pin) = fala = 12x50 a 600 watt Pout = 2AIN Watt Here, N = 1600 RP.M. 20x3.580x1600 watt Pout - - 60 - 599.8 efficiency (n) = 599.8 x 100% 600 = 99.9%

What is the maximum torque thout the motor com produce while maintaining efficiency at least 88% - Pout x 100% 20 Timex N Po 805 60 Ta 20 Tmax x 1500 80 = 12x50 Tmax - 80x12850 2X 1500 Tmax - 5.09 N.m is Nnoload - Newload speed Regulation = Nfull load o-1 = -1