Find the suitable tapping on an auto-transformer starter for an induction motor required to start the motor with 36% of full load torque. The short circuit current of the motor is 5 times the full load current and the full load slip is 4%. Determine also the starting current in the supply mains as

Question

Find the suitable tapping on an auto-transformer starter for an induction motor required to start the motor with 36% of full load torque. The short circuit current of the motor is 5 times the full load current and the full load slip is 4%. Determine also the starting current in the supply mains as a percentage of full load current.

Accepted Answer

Ratio of starting torque to full load torque,

[latex]\frac{T_{st}}{T_{fl}} = 0.36 [/latex]

Ratio of short-circuit current to full load current,

[latex]\frac{I_{st}}{I_{fl}} = 5 [/latex]

Full load slip, [latex]S = 0.04 [/latex]

Now [latex]\frac{T_{st}}{T_{fl}} = K^{2} \left\lgroup\frac{I_{sc}}{I_{fl}} ight group^{2} imes S [/latex]

[latex]0.36 = K^{2} imes \left(5 ight)^{2} imes 0.04 [/latex]

∴ Transformation ratio or tapping of auto-transformer,

[latex]K = \sqrt{\frac{0.036}{5 imes 5 imes 0.04} } = \mathbf{0.6} or \mathbf{60\%} [/latex]

In an auto-transformer starter, ratio of starting current to short-circuit current,

[latex] \frac{I_{s}}{I_{sc}} = K^{2} where I_{sc} = 5 I_{f}[/latex]

[latex] herefore \frac{I_{s}}{I_{f}} = 5 K^{2} = 5 imes \left(0.6 ight)^{2} = \mathbf{1.8} or \mathbf{180\%} [/latex]

Question 10.6: Find the suitable tapping on an auto-transformer starter for...

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