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.
On the difference between 'load-flow' and 'phasor' studies
A loadflow (power-flow) simulation is a phasor simulation. It is a phasor simulation of a power system at nominal frequency (50Hz or 60Hz.) It assumes that the system is at sinusoidal steady state and that nothing is changing.
The distinction between a 'load flow' study and a 'phasor study' is that a phasor study can be performed at any arbitrary frequency, say 50Hz, 100Hz, 150Hz, whereas a load-flow study is nearly always performed at the power system nominal frequency (50 or 60Hz.)
The generalised 'phasor study' is useful in the study of power system harmonics, which requires simulation of the power system at 50Hz and its harmonic frequencies 100Hz, 150Hz, 200Hz, 250Hz, ... and so on. This is done by running one separate 'phasor study' for each harmonic frequency of interest.
On the difference between load-flow/phasor and dynamic/transient studies
A load-flow study evaluates steady state operation of a power system. We do load-flow studies to check that elements like transformers, overhead lines, and cables won't be overloaded, and that system voltage regulation is within acceptable limits (-6%, +10% for Australian domestic power supply.)
The time scale of interest is hours to days.
The loadflow study is just an exercise in solving a lot of simultaneous linear equations. There is no time dependent element, no differential equations, or anything exciting. You multiply some big matrices together and that's it.
A dynamic/transient study evaluates the behaviour of the power system when a change occurs. The change could be an increase or decrease in load, a line fault, a change in generator output, or a big motor starting.
The objective is to determine if there will be any detrimental effects on the scale of milliseconds to minutes. Detrimental effects might include - voltage spikes/dips, generator frequency slip, protection relay operation.
A dynamic/transient study must take account of the time-dependent response of the electrical and mechanical parts of the power system.
- Generators and motors have a mechanical inertia
- Capacitors and inductors have energy storage
- Iron-cored transformers have remanence/hysteresis
- Protection relays are digital signal processors which decide whether the power system is healthy or not, based on the history of the signals they see.
- Generators have control systems with sophisticated transfer functions for calculating output voltage set point and governor (throttle) set point
Therefore a transient study involves simulating a system of differential equations evolving over time, with a typical time step of 1 millisecond.
The electrical quantities are still voltages and currents, but there are also a lot of variables in things like 'generator inertial energy' and 'motor rotational speed'.
PS: I do power system studies for a living.
Best Answer
What we know:
Note: This is what would be done practically with a generator. They are never run at 100% load, so typically would be used at some under rating. For example if the generator is running at 80% load, the speed would be set to provide 50 Hz at this load point. I've assumed here that this is what is meant by 12 MW @ 50 Hz ...not that 50 Hz is the no load speed.
12/15 = 0.8 (12 MW is 80% of full load)
Therefore Droop at 12 MW is: 0.8 * 5 = 4%
If the speed at 12 MW (50 Hz) has drooped 4 %, then the speed is 100 - 4 = 96% of full speed.
No-Load frequency: (50/96) * 100 = 52.0833 Hz
Note: 50/96 = 0.520833r Hz ...this represents the frequency drop for a 1% change in load. Since it's 6 decimals I chose to use 50/96 elsewhere in the calculations.
We now need to decide if the under frequency trip will operate at 15.5 MW (The question was corrected for this to be 48 Hz)
15.5/15 = 1.03 (15.5 MW is 103% of full load)
Therefore Droop at 15.5 MW is: 1.03 * 5 = 5.17%
Max frequency drop: (50/96) * 5.17 = 2.693 Hz
Frequency at 15.5 MW: 52.0833 - 2.693 = 49.39 Hz .....(this seems to indicate the under frequency trip of 48 Hz would not trip)
**** Speculation now. Let's assume that you have the number wrong in the question and that the frequency trip does trip.
The load on the generator is now as you specified: 2 * 4 + 3.5 = 11.5 MW.
11.5/15 = 0.767 (11.5 MW is 76.7% of full load)
Therefore Droop at 11.5 MW is 0.767 * 5 = 3.833%
Frequency drop at 11.5 MW: (50/96) * 3.833 = 1.997 Hz (we can round this up to 2)
Frequency at 11.5 MW: 52.0833 - 2 = 50.0833 Hz
****This seems to give what you said was the correct answer.....however by my calculations the frequency trip would not actuate.
What is Droop?
Droop is a governor characteristic that reduces fuel/throttle as the load increases. It is fixed at only 1 point, it's defined point. It is a slope defined for the genset. For example a 1500 rpm genset may droop 5% at full load. So the speed will drop 75 rpm to 1425 at full load. Since the frequency of the generator is related to the number of poles and speed....the frequency drops by 5% at full load. At 50 percent load the frequency drop (droop) would be 2.5%.
With the values calculated we can plot the droop for the question: