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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\$ ?

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  • \$\begingroup\$ Maximum steady-state power transfer occurs when delta is 0 but to calculate the steady state power limit, it only makes sense that delta has to be 90. Steady state power limit is the maximum power allowed before the steady state breaks (not quoting from anything but just my understanding!). Look for the steady state power limit definition, maybe then things will be more clear. \$\endgroup\$ – charansai Mar 12 '18 at 13:14
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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.

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  • \$\begingroup\$ But power drawn from generator is still increasing till del=105 or theta=90. So, why that is not the limit ?? \$\endgroup\$ – Nikhil Kashyap Mar 17 '18 at 19:27
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    \$\begingroup\$ @NikhilKashyap Generator's internal induced voltage E is a voltage source. Rigid grid V is another voltage source. There's no more sources here. Between those sources is j1.5 inductive reactance. Max power transmission, when only del is variable, occurs at del=90 degrees. That's a mathematical fact, it's proved before year 1900, the proof is elementary and it will be valid to eternity just like all pure mathematical facts. Theta is not the phase angle of some voltage source. Theta is a cosequence. \$\endgroup\$ – user287001 Mar 17 '18 at 20:45
  • \$\begingroup\$ @NikhilKashyap plotting software. No need to plot. The transferred real (non reactive) power is a sine pulse Asin(del) where A=(EV)/X. The maximum of sine is 1 at del=90 degrees. At del=0 degrees (=the generator runs freely and is considered to be ideally lossless) the power is zero, only reactive power is generated to grid or sucked from grid depending on voltage E. \$\endgroup\$ – user287001 Mar 17 '18 at 21:23
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According to my simulation, maximum power transfer is limited by voltage difference between Generator and Grid by transmission line impedance 1.5 Xp.u.

Optimum generator Voltage is 1.875Vpu at 90 deg lead relative to grid that results in 1.2Vpu at access point.

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