Question

A 3300 V, 3-phase, 50 Hz, star-connected synchronous motor has a synchronous impedance of 2+j 15 ohms per phase. Operating with a line emf of 2500 V, it just falls out of step at full load.

a)  Draw the equivalent circuit of a synchronous motor.                   

b) To what open-circuit emf will it have to be excited so that it will just remain in synchronism at 50% above rated torque?                                

c)   With this emf , what will then be: the input power; current; and               Power factor at full load. 

Answer 1

(b) answer:

Given, Voltage of 33001 Cline voltage) Synchronous impedance =(75)=0(2+3" phare =(75)=0(2+;15) =(15.132/82.409 - Iphone (25) Emf induce d(E) 25001 Cline voltage) out of falls is given that , motor just step on Step at hull load ut ap This implies at (Maximum torque) (It is will happens maximum Bwer ) (Pman) (Maximum torque we But (Pmax) occurs will at get at (8=0) impedance angle) (load Angle) = cos (0-5) a coso (Pm) en E | Zs=15-132/82-40) 10=82-40°) - COSO = COST=1 (Por max = (2500) cos(82.40° (2500) (3300) (2500) 15-132 15.132 (Pm) max; 490 576. 101 Watte [Maximum Torque] = Paman (N = Speed) not given J to 50 %. What is Emf (E) Tor que increased TP the Power in created by 50%. incre abed by 50% - A Torque | (Pm) = 1:5 (Pm many 100%. + 50% - 150% = 15 (times)

» (15) Porar, coso 222 (BB009 E, Cos (829 2 Cos (82-40°) =) (1.5) (490576.101) = Ez (3300) 15.132 15-132 -> (15)(490576.101)(15132) - E, (3300) - Ey 6.13225639 0.13225639 + 4.5 490576. loY15.)32) EO - (3300) E => 0.13225639 6,2 - (3300) Es + 11135096.34 = 0 By solving above equation - 20928066 V, E = 4022.86 Volta greater Be | E= 20928.66 is | So, È' cannot be taken than terminal voltage), this value 40 22.86 Volts J Ez a 4KV) (This is is but to be excited to stand 50% exces torque)