Asynchronous machine is supplied from a constant-voltage source. At no load, the motor armature current is found to be negligible when the excitation is 1.0 p.u. The p.u. motor constants are Xd = 1.0 and xq = 0.6. (a) If the machine loses synchronism when the angle between the quadrature axis and

Question

Asynchronous machine is supplied from a constant-voltage source. At no load, the motor armature current is found to be negligible when the excitation is 1.0 p.u. The p.u. motor constants are [latex]x_{d}=1.0[/latex] and [latex]x_{q}=0.6[/latex].

(a) If the machine loses synchronism when the angle between the quadrature axis and the terminal voltage phasor direction is 60 electrical degrees, what is the p.u, excitation at pull-out?

(b) What is the load on the machine at pull-out? Assume the same excitations as in part (a).

Accepted Answer

(a)

[latex]P=\frac{V E_{f}}{X_{d}}\,\mathrm{sin}\,\delta+\frac{V^{2}}{2X_{d}X_{q}}(X_{d}-X_{q})\,\mathrm{sin}2\delta \ ={\frac{E_{f}(1)}{1}}\sin\delta+{\frac{1}{2 imes0.6}}(1-0.6)\sin2\delta \ =E_{f}\sin\delta+{\frac{1}{3}}\sin2\delta[/latex]

For pull-out power, we have

[latex]{\frac{\partial P}{\partial\delta}}=E_{f}\cos\delta+{\frac{2}{3}}\cos^{2}\delta=0[/latex]

The pull-out angle is

[latex]\delta_{m}=60~\mathrm{deg}[/latex]

Hence we obtain

[latex]E_{f}={\frac{2}{3}}\,\mathrm{p}.\mathrm{u}.[/latex]

(b) Consequently, the pull-out load is found to be

[latex]P={\frac{2}{3}}\sin60~\mathrm{deg}+{\frac{1}{3}}\sin120~\mathrm{deg}=0.866\,\mathrm{p.u.}[/latex]

Question 7.13: Asynchronous machine is supplied from a constant-voltage sou......

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