Section 15.4: Performance Calculations cont.
Pd = 25 KW = (1 - s) ´ 27 KW = (1 - s)Pg
s = 1 - 25/27 » 0.074 (7.4%)
The synchronous speed is:
ns = 120f/p = 120 ´ 60 Hz. / 2 poles = 3600 RPM (377 rad./sec.)
The shaft speed is n = (1 - s)ns = 3334 RPM (349 rad./sec.)
Te = 3Pd /wm = 25 KW / 349 rad./sec. = 71.63 N×m
Likewise, Tw&f = Pw&f /wm = 400 W / 349 rad./sec. = 1.15 N×m
\ Tout = Te - Tw&f = (71.63 - 1.15)N×m = 70.48 N×m
A two-pole three-phase 60 Hz. induction motor develops 25 KW (about 33.5 hp) of electromagnetic power at a certain speed. The rotational mechanical loss at this speed is 400 W. (0.54 hp). If the power crossing the air gap is 27 KW (36 hp), calculate the slip and output torque.