Determine the starting torque of a 3-phase induction motor in terms of full load torque when started by means of:
(i) Star delta starter; and
(ii) An auto-transformer starter with 50% tapings.
The motor draws a starting current of 5 times the full load current when started direct on line. The full load slip is 4 percent.
Ratio of short circuit current to full load current,
\frac{I_{sc}}{I_{fl}} = 5
Full load slip, S = 0.04
(i) For star-delta starter.
Ratio of starting torque to full load torque,
\frac{T_{st}}{T_{fl}} = K^{2} \left\lgroup\frac{I_{sc}}{I_{fl}}\right\rgroup^{2} \times S_{fl} = \frac{1}{3} \left(5\right)^{2} \times 0.04 = 0.3333\therefore \mathbf{Starting torque = 33.33\% of full load torque}
(ii) For Auto-transformer starter:
Transformation ratio, or tapings,
K = 50\% = \frac{1}{2}
Ratio of starting torque to full load torque
\frac{T_{st}}{T_{fl}} = K^{2} \left\lgroup\frac{I_{sc}}{I_{fl}}\right\rgroup^{2} \times S_{fl} = \left(\frac{1}{2}\right)^{1} \times \left(5\right)^{2} \times 0.04 = 0.25\therefore \mathbf{Starting torque = 25\% of full load torque}