Question

A 460-V 60-Hz four-pole Y-connected induction motoris rated at 25 hp(note: this is the nominal output power at the nominal speed (or nominal slip).The equivalent circuit parameters areR1= 0.15 R2= 0.154 XM= 20 X1= 0.852 X2= 1.066 PF&W= 400 WPmisc= 150 WPcore= 400 WFor a slip of 0.02, find(Importantnote: this 2% slip mayor may notbe the nominal slip of the motoras it is not explicitly stated. Hence,the output power associated with this slip maybe different than the given nominal output power (i.e. 25 hp)).(a)The line current IL(b)The stator power factor (c)The rotor power factor (d)The rotor frequency (e)The stator copper losses SCLP(f)The air-gap power PAG(g)The power converted from electrical to mechanical form Pconv(h)The induced torque ind(i)The load torque load(j)The overall machine efficiency (k)The motor speed in revolutions per minute and radians per second(l)What is the starting line current of this motor?(Hint: starting slip is one, now redraw your circuit and solve it for the starting current)(m) What is the slip at the pullout torque? What is the pullout torque of this motor?

Answer 1

Step-by-step explanation:

Please find the solution to the problem in the file attached below