voltage - Calculation of Steady state power limit

I am confused with the above problem. I was thinking shouldn't we maximise \$\mid E\mid sin(\delta)\$ instead to get steady state power limit?

\$\mid E\mid sin(\delta) = 1.8 sin(\theta)\$

maximum will occur when \$\theta=90^\circ \$

EDIT: Is there any software through which one can plot power developed by leftmost voltage source i.e. generator vs \$\delta\$ ?

Answer

Steady state power limit = the maximum that the generator (= a synchronous machine) can push without running out of sync.

In the example V is assumed to be rigid (=no matter, what the single generator does) => If one increases the mechanical power that rotates the generator, the generator will still stay in sync, but the phase angle of its internal induced voltage gets some lead over the phase angle of the rigid grid. Finally, if the mechanical power is increased enough, the lead reaches 90 degrees. That point is the top of the hill. Bigger lead will make the rotation easier and that pulls the generator out of the sync, the system gets unstable.

Why it would be easier? Because del=90 degrees is the phase angle for max. power transmission between 2 voltage sources which are connected together through a reactance and run at same frequency. The rotation resistance (=torque) is the power divided by the angular velocity of the rotation.

So, the internal induced voltage of the generator has 90 degrees phase lead at the theoretical max power. Of course, in practice some safety margin is left, but the theoretical limit is asked.

There's another restriction: Vt should be 1,2pu (=1,2 x the nominal grid voltage). Now you know the phase angle of E (=90 degrees) and the relative rms voltage of Vt. You must find the phase of Vt and the relative value of E. The problem is only a little tricky phasor calculus AC circuit problem.

Vt=1,2pu obviously is the maximum allowed local output voltage of the generator. It's somewhere between the internal induced voltage and the grid voltage.