A 6.6 kV, Y-connected, 3-phase, synchronous motor operates at constant voltage and excitation. Its synchronous impedance is 2 + j 20 Ω/phase. The motor operates at 0.8 leading power factor while drawing 800 kW from the mains. Find the motor power factor when it is loaded to draw increased power

Question

A 6.6 kV, Y-connected, 3-phase, synchronous motor operates at constant voltage and excitation. Its synchronous impedance is [latex] 2+j 20 \Omega ext { /phase } [/latex]. The motor operates at 0.8 leading power factor while drawing 800 kW from the mains. Find the motor power factor when it is loaded to draw increased power of 1200 kW

Accepted Answer

[latex] \bar{V}_{t}=6.6 / \sqrt{3} \quad \angle 0^{\circ}=3.81 \angle 0^{\circ} kV[/latex]

[latex] \bar{I}_{a}=I_{a} \angle 36.9^{\circ}[/latex]

[latex] I_{a}=\frac{800}{\sqrt{3} imes 6.6 imes 0.8}=87.5 A[/latex]

[latex] \bar{Z}_{s}=2+j 20=20.1 \angle 84.3^{\circ} \Omega, \cos 84.3^{\circ}=0.1[/latex]

[latex] \bar{E}_{f}=3.81-20.1 \angle 84.3^{\circ} imes 0.0875 \angle 36.9^{\circ}=4.724-j 1.504[/latex]

Or [latex] E_{f}=4.96 kV[/latex] (phase)

Power input increases to 1200 kW; no change in excitation

[latex] \bar{I}_{a}=\frac{V_{t} \angle 0^{\circ}-E_{f} \angle-\delta}{Z_{s} \angle heta}=\frac{V_{t}}{Z_{s}} \angle- heta-\frac{E_{f}}{Z_{s}} \angle-(\delta+ heta)[/latex]

[latex] P_{e}( ext { in })=3 \operatorname{Re}\left[V_{t} \angle 0^{\circ} \bar{I}_{a}^{*} ight][/latex]

[latex] =3\left[\frac{V_{t}^{2}}{Z_{s}} \cos heta-\frac{V_{t} E_{f}}{Z_{s}} \cos (\delta+ heta) ight][/latex]

Substituting values

[latex] \frac{1200}{1000 imes 3}=\frac{(3.81)^{2} imes 0.1}{20.1}-\frac{3.81 imes 4.96}{20.1} \cos \left(\delta+84.3^{\circ} ight)[/latex]

Solving we get

[latex] \delta=26.1^{\circ}[/latex]

[latex] \bar{I}_{a}=\frac{3.81-4.96 \angle-26.1^{\circ}}{20.1 \angle 84.3}=113 \angle 22.1^{\circ}[/latex]

[latex] p f=\cos 22.1^{\circ}=0.9265[/latex] leading

Question 8.45: A 6.6 kV, Y-connected, 3-phase, synchronous motor operates a...

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