A 400 V, 3-phase, star connected induction motor draws 40 A at a line voltage of 80 V with rotor locked. The power drawn by the motor under this condition is 480 W. the DC resistance measured between a pair of stator terminals is 0.2 Ω. if the core loss is 80 W and the ratio of AC to DC resistance

Question

A 400 V, 3-phase, star connected induction motor draws 40 A at a line voltage of 80 V with rotor locked. The power drawn by the motor under this condition is 480 W. the DC resistance measured between a pair of stator terminals is 0.2 Ω. if the core loss is 80 W and the ratio of AC to DC resistance is 1.4, determine the equivalent leakage reactance per phase of the motor as well as stator and rotor resistance per phase.

Accepted Answer

Here, [latex]V_{sc\left(L\right) } = 80 V; I_{sc} = 40 A; P_{sc} = 480 W[/latex]

Phase voltage at short circuit, [latex]V_{sc} = \frac{V_{sc\left(L\right) }}{\sqrt{3} } = \frac{80}{\sqrt{3} } = 46.2 V[/latex]

Equivalent impedance referred to stator side/phase,

[latex]Z_{eq1} = \frac{V_{sc}}{I_{sc}} = \frac{46.2}{40 } = 1.115 \Omega [/latex]

At short-circuit copper loss [latex]P_{c} = P_{sc} - P_{i} = 480 - 80 = 400 W[/latex]

Equivalent resistance referred to stator side/phase,

[latex]Z_{eq1} = \frac{P_{c}}{I_{sc^{2}}} = \frac{400}{\left(40\right) ^{2} } = 0.25 \Omega [/latex]

Equivalent reactance referred to stator side/phase,

[latex]X_{eq1} = \sqrt{Z_{eq1}^{2}- R_{eq1}^{2}} = \sqrt{\left(1.115\right)^{2} - \left(0.25\right)^{2} }\[0.5cm] \qquad = \mathbf{1.087 \Omega } [/latex]

DC resistance of the stator winding/phase,

[latex]R_{dc} = \frac{0.2}{2} = 0.1 \Omega [/latex]

AC resistance of the stator winding/phase,

[latex]R_{ac} = R_{1} = 0.1 \times 1.4 = \mathbf{0.14 \Omega} [/latex]

Rotor resistance referred to stator side/phase,

[latex]R^{\prime}_{2} = 0.25 - 0.14 = \mathbf{0.11 \Omega }[/latex]

Question 9.30: A 400 V, 3-phase, star connected induction motor draws 40 A ...

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