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Question 7.2: The rotational losses for a l000-hp synchronous motor are fo......

The rotational losses for a l000-hp synchronous motor are found to be 18 kW. The motor operates from a 2300-V three-phase supply at 0.8 PF lagging. Find the excitation voltage at full load assuming that the machine’s synchronous reactance is 1.9 Ω and that the armature resistance is negligible.

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The total input power is the sum of the output power and losses. As a result,

P_{\mathrm{in}}=746\times10^{3}+18\times10^{3}=764\times10^{3}\,\mathrm{W}

The current is thus obtained as

I_{a}=\frac{764\times10^{3}}{\sqrt{3}(2300)(0.8)}=239.726\angle\,-36.87\,\,\mathrm{deg\,}\mathrm{A}

We now have

E_{f}=V_{t}-j I_{a}X_{s} \\ ={\frac{2300}{{\sqrt{3}}}}-j(239.726\angle-36.87\ \mathrm{deg})(1.9) \\ =1115.793\angle-19.061\,{\mathrm{deg}}\, V

This is the required excitation voltage per phase.

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