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ron los windage and friction loss mechanical power shaft active power to stator P. active power supplied to rotor

Question:

Answer 1

6-pole 3-phase f = 60HZ. 12of No = 120 X 60 = 120x 10 Ns = 1200Rpm Copper losses in stator (Pers) = 1.2kw Iron losses in sitator (Press) = 0.5kw copper losses in Rotor (peur ) = 1kw. Friction and windage losses = 1.5kw. * Irm lowes in poor is very heas Nalue because in indention Rotor operates at the Slip freguenty (Sf). Roter iven losses are divertly propertimal to stip Fregineinty

Active power to the Stator = bokw. Rotor (hg) 1) Active power supplied to the Pag = (Active power supply / Stator losses) to the stator Pas Psi - lones (skater copper losses + Stator Ixon) Lostater input power Ar gap power Pag = 60k - (1.2k + 0.5k) Pag = 58.3 kw) Rotov copper loves (Peux ) = 5 Pay Ik = sx 58.3K s = 58.3 Is = 0.0172.

speed of motor Ny ) = (1-5) Ns N = (1-5) Ne Ny = (1-0.0172) X1200 Ny .= 1179.36 N = 1180 rpm 10) mechanical power developed ( Panel ) = Pag - (Rotor copper lones) Prech - Pag - (Rotor copper corses) = (1-5) Pag Pimech - 58.3K - lk Pineth = 57.3 kw a Torque applied on the locou 60

t = shat - Pohy Pincel - shaft power ( Pears) = Petriction devindy loves P Meihanical l power , / Friction clindaje ( losses. Psh = 57.3k - 1.5k Poh = 55.8kW 54 x torque applied on load (T) = 2T NY wy, talento, los UVO TE 55.84103 X 60 277 x 1180 T = 451.57N-m

e) Efficiency of motor (rn) 4. N = shaft power 1. X100 active power to stator 1 55.8.K - X100 An = 60k Y n = 93%