Question

A de shunt motor drives a centrifugal pump at a speed of 1000 rpm when the terminal voltage and line currents are 200 V and 50 A, respectively. The arma- ture and field resistances are 0.1 and 100 2, respectively. a. Design a starting resistance for a maximum starting current of 120 A in the armature circuit. b. What resistance should be added to the armature circuit to reduce the speed to 800 rpm? C. If the terminal voltage is reduced by 25%, what is the speed of the motor?

Answer 1

Speed N = 1000 rpm V = 200V, I_L = 50A R_a = 0.1 ohm, R_sh = 100 ohm I_sh = V/R_sh = 200/100 = 2 ohm I_G = I_L - I_sh = 50 - 2 = 48A Therefore E_b = V - I_a R_a = 200 - 48 times 0.1 = 200 - 4.8 = 195.2 volts (a) for a dc shunk motor, at starting E_b = 0 E_b = V - I_a R_a implies 0 = V - I_a R_a V = I_a R_a Here I_a = I_st at starting V = I_st middot (R_st + R_a) R_at R_st = V/I_st = 200/120 = 1.66 ohm R_st = 1.66 - 0.1 = 1.56 ohm (b) We have E_b = phi PNZ/60 pi implies E_b proportionalto N Therefore E_b_1/E_b_2 = N_1/N_2 E_b_1 = 195.2 volts & N_1 = 1000 rpm (given in data) \r\n Let R be the required additional resistance E_b_2 = V - I_a (R_a + R)