Section 15.7: Starting of Induction Motors cont.
Key assumption: |Is| » |Ir|, which is reasonable since the (rotating) transformer’s |jXm||Rc| >> |Rr/s + jXr|
The full-load (s = 0.05) torque is: TFL = Pg,FL/ws = (|Ir,FL|2Rr/sFL)/ws » (|Is,FL|2Rr/0.05)/ws (I.)
The starting (s = 1) torque is: TST = Pg,ST/ws = (|Ir,ST|2Rr/sST)/ws » (|Is,ST|2Rr/1)/ws (II.)
(II.) / (I.) yields: TST / TFL » sFL(|Is,ST|/|Is,FL|)2 (III.)
Which for |Is,ST|/|Is,FL| = 6 and |Is,ST|/|Is,FL| = 3 yields: TST / TFL = 1.8 and 0.45 respectively (Note: 1.8 = 4´0.45)
An induction motor is designed to run at 5% slip on full load. If the motor draws 6 times its full-load current on starting (at rated voltage), estimate the ratio of the starting torque to the full-load torque. Repeat for a starting current of 3 times rated full-load current.