Discussion
That machine delivering 0.7 per unit power has a load angle of
Answer: D
Correct answer: 30 degrees
\[ 0.7 = 1.4\sin\delta \quad\Rightarrow\quad \sin\delta = 0.5 \quad\Rightarrow\quad \delta = 30^\circ. \] Half the pull-out power is delivered at only thirty degrees, not forty-five, because the relation is sinusoidal - and the machine still retains \( \cos 30^\circ \), or 87 per cent, of its maximum synchronising stiffness.
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