It is very easy to tell what is being asked here. What is needed is specific detail. Two devices are intended to be used in a load sharing arrangement. The 2301A load sharing controller is meant to enforce sharing. Failure to work suggests that the voltage does not droop enough to allow the voltage to fall below that of the second energy source before the load gets too large and trips the overload protection.
The 2301A is a device which allows load sharing between two power generation systems. It creates a sagging load/voltage relationship so that as an alternator etc is loaded it drops below local bus voltage and causes a more lightly loaded and thus higher voltage generator to pick up progressively more load.
Assuming that the equipment is not faulty it seems likely that voltage shaping being applied to one machine is not adequate to drop the voltage enough for the second alternator to load share. The 2301A can have the amount of "droop" under load adjusted with a pot. I'd guess that the pot needs to be adjusted so that the voltage droops more under load than at present.
Recall the old "joke" - What do you call a person who speaks 3 languages / 2 lanhuagrses / one language.
A: Trilingual / Bilingual. guar An Ame....
It's not funny, alas.
This question is worth persevering with for at least 2 excellent reasons.
It is a classic "English as a second language" situation where the OP writes far too little, as expressing the question cogently in English is difficult.
AND some English only speakers who are bright technically are not able to get their brain around the language constructs.
The OP's problem is clear - they have adequately explained what they want to do and the equipment used and what is going wrong.
What is required is clear. The details are not.
A small amount of persuasion will fill in the gaps.
(2) This is a vastly more rewarding question than about 90% that get asked here. This is a real world application of a complex and unusual piece of equipment doing a crucial real world task. Having this topic discussed here will be the first introduction that many people have to this piece of equipment.
Why does reactive power influence the voltage? Suppose you have a (weak) power system with a large reactive load. If you suddenly disconnect the load, you would experience a peak in the voltage.
First, we need to define what exactly is being asked. Now that you have stated this is regarding a utility-scale power system, not the output of a opamp or something, we know what "reactive power" means. This is a shortcut used in the electric power industry. Ideally the load on the system would be resistive, but in reality is is partially inductive. They separate this load into the pure resistive and pure inductive components and refer to what is delivered to the resistance as "real power" and what is delivered to the inductance as "reactive power".
This gives rise to some interesting things, like that a capacitor accross a transmission line is a reative power generator. Yes, that sounds funny, but if you follow the definition of reactive power above, this is all consistant and no physics is violated. In fact, capacitors are sometimes used to "generate" reactive power.
The actual current coming out of a generator is lagging the voltage by a small phase angle. Instead of thinking of this as a magnitude and phase angle, it is thought of as two separate components with separate magnitudes, one at 0 phase and the other lagging at 90° phase. The former is the current that causes real power and the latter reactive power. The two ways of describing the overall current with respect to the voltage are mathematically equivalent (each can be unambiguously converted to the other).
So the question comes down to why does generator current that is lagging the voltage by 90° cause the voltage to go down? I think there are two answers to this.
First, any current, regardless of phase, still causes a voltage drop accross the inevitable resistance in the system. This current crosses 0 at the peak of the voltage, so you might say it shouldn't effect the voltage peak. However, the current is negative right before the voltage peak. This can actually cause a little higher apparent (after the voltage drop on the series resistance) voltage peak immediately before the open-circuit voltage peak. Put another way, due to non-zero source resistance, the apparent output voltage has a different peak in a different place than the open-circuit voltage does.
I think the real answer has to do with unstated assumptions built into the question, which is a control system around the generator. What you are really seeing the reaction to by removing reactive load is not that of the bare generator, but that of the generator with its control system compensating for the change in load. Again, the inevitable resistance in the system times the reactive current causes real losses. Note that some of that "resistance" may not be direct electrical resistance, but mechanical issues projected to the electrical system. Those real losses are going to add to the real load on the generator, so removing the reactive load still relieves some real load.
This mechanism gets more substantial the wider the "system" is that is producing the reactive power. If the system includes a transmission line, then the reactive current is still causing real I2R losses in the transmission line, which cause a real load on the generator.
Best Answer
You are assuming lagging means less voltage. If leading reactive power = lagging reactive power, all power would go to load, pf=1 and unity voltage.
You need 125MW + 150MW + 150MW = 425MW of real power. You produce 182MW + 163MW + 85MW = 430MW of real power.
You need 100MVAR + 80MVAR + 85MVAR = 265MVAR of lagging reactive power. Your capacitive power correction provides 84.2MVAR + 61.2MVAR + 60.4MVAR = 205.8MVAR. You are 59.2MVAR short.
Your over-excited generator on bus 3 provides 9MVAR (an assumption on my part due to the negative), which makes you 50.2VAR short.
So generators provide 425MW and 50.2VAR, which means 427.95MVA and a pf = 0.993. 425MW goes to loads and 50.2VAR goes back and forth between loads and generators. Which is as far as my knowledge goes.
I'd guess if you provide 265MVAR - 9MVAR (bus 3 generator) = 256MVAR of capacitive power, you'd get your unity voltage.
When pf = 1, S = P and \$Q_C = Q_L\$. The voltage of the source drives the real power load. Your unity voltage. Leading reactive power of the capacitors supplies the lagging reactive power of inductors (motors).
When the pf < 1 and \$Q_C < Q_L\$, the load is inductive. Again, the capacitors supply reactive power to the loads, but there is not enough. The source must supply P and \$Q_{NET} = Q_L - Q_C\$. The source voltage, which cannot change, is providing power to drive the MW and some of the MVAR. Less of the voltage is going to the MW real power load.
Same thing would happen if \$Q_C > Q_L\$. The load is capacitive and pf < 1. The capacitors supply all of the reactive power to the loads. Now the source must supply P and \$Q_{NET} = Q_C - Q_L\$. The source voltage is providing power to drive the MW and the extra MVAR for capacitors.
This is not your case, but the same situation applies. Less of the voltage would go to the MW real power load.