Speed of Synchronous Generator

jannatul,

you one basic question needs many corrections. The size of an engine is equal to the volumetric capacity--not the amount of fuel to be supplied. You just need to supply sufficient fuel to achieve the proper stoichiometric mixture for combustion (presuming you're referring to an internal combustion engine and not a steam engine).

As for your second question, that, too, really requires a lot of clarification I'm not sure you're prepared to make. Suffice it to say that if the engine is driving a synchronous generator and synchronized to a grid with other synchronous generators and their prime movers the speed of the engine and generator will be controlled by the frequency of the grid. To increase power output you would need to increase the fuel flow-rate to the engine.

If the engine-generator were operating independently of any other synchronous generators the only way to increase the power output while maintaining a constant speed and frequency is to turn on more motors or lights or televisions or computers and computer monitors. Then you--or the engine governor (control system)--would see an immediate decrease in speed and an increase in load. To return to the same speed while supplying the increased load it would be necessary to increase the fuel flow-rate to the engine.

There are numerous threads on control.com about droop and isochronous speed control--which is essentially what you are enquiring about. Use the cleverly-hidden 'Search' feature at the far right of every control.com webpage (after looking at the Search 'Help' function) and you will find lots of information about speed control and frequency and load changes. Look for threads with the bicycle analogy, that may help you to get a handle on the topic a little faster.

Hope this helps!
 
> also if my rpm is fixed, how may i get more power by
> increasing load and fuel? what would be the explanation.

The explanation is: Torque. Even though speed is fixed, when the energy flow-rate is increased the torque will increase. It's like riding a bicycle at a steady speed on a flat road. When it's just on the bike, the pressure required on the pedals to create the torque required to maintain a constant speed is less the pressure required to create the torque to maintain the same speed when your brother or your friend is on the bike (or you're carrying some heavy packages). The principle is exactly the same as for an engine supplying torque to a synchronous generator at a constant speed to maintain a constant frequency with different loads. If you just increase the pressure you are applying to the pedals of the bike the speed will increase. If you are riding the same road and wanting to travel at the same speed but carrying a heavier load if you don't increase the pressure you are applying to pedals to create more torque then you won't travel at the same speed.

If you want to increase the power output of a synchronous generator that is not synchronized with any other generators, if you just increase the torque from the engine without starting any motors or turning on any lights the speed of the engine will increase and so will the frequency. The only way to increase the generator output while maintaining frequency is to increase load.

In the case where the generator is synchronized with other generators the speed of the engine and generator is fixed by the frequency of the grid and if you want to increase the power output of the generator you have to increase the fuel flow-rate to the engine. (If all the other generators and their prime movers maintain the same output and the load on the grid is constant (the number of running motors and lights and televisions and computers and computer monitors remains unchanged) the grid frequency will increase slightly--so the grid regulators will reduce the load on another unit by the same amount you increased the output of your engine-generator set to return the frequency to normal.)

But, when speed is fixed increasing the fuel flow-rate to the engine--which would normally increase the speed--increases the torque being applied to the generator which increase the output of the generator. When a single prime mover and generator is driving loads, increasing the energy flow-rate to the prime mover without any change in the load will result in an increase in speed and frequency. The only way to increase the output of a single generator and its prime mover powering loads is to increase the load--which will require an increase in the energy flow-rate to the prime mover in order to maintain speed and frequency.

A single prime mover governor controlling a single synchronous generator powering a "load" must operate in Isochronous mode to maintain frequency as load changes. A single prime mover governor a single synchronous generator synchronized to a grid with other prime movers driving generators must operate in Droop mode to produce power at a stable output. In Droop mode, changes in load have little effect on individual generator-set load but will cause changes in frequency if not monitored and responded to by grid regulators. (This presumes a large grid with lots of load and many generators and their prime movers.)

Hope this helps!
 
Sir,

I have query regarding output of the generator.

let's have a grid of 50,000 MW capacity working at 50 HZ grid supply if all the generators are working at full capacity and 10 MW generator trip due to fault. Now due to load and demand difference frequency will fall and new steady state will be achieve. Because grid is already working at full capacity frequency has to fall to get steady state.

In my opinion, at this condition all the generators will provide same mechanical output as previous but electrical output will be increase due to increase in current. Please rectify me if i am wrong here.
 
Sir,

I will be happy to supply the required rectification.

Every synchronous generator connected to the grid (of ANY size) requires a certain amount of energy (torque) just to get to synchronous speed. Any torque over that amount is converted to amperes by the generator.

If every prime mover providing torque to its generator is already producing maximum torque and one of the prime movers and the generator it is driving is separated from the grid, then there is no way any of the remaining prime movers can provide any more torque to the generators. In this case, some of the torque that is being used just to maintain synchronous speed is converted to amperes--because the load hasn't changed (that is, the number of motors and lights and computers and computer monitors and televisions didn't change when the generator and its prime mover was separated from the grid). So, some of the energy (torque) required to maintain synchronous speed (rated system frequency) must be used to supply the load, which causes the system frequency to decrease by an amount directly proportional to the amount of generation that was lost (in your example 10 MW out of 50,000 MW). You can do the maths to determine how much frequency will decrease.

The electrical output will not increase--the number of lights and motors and computers and computer monitors and televisions hasn't changed--because the electrical load has not changed. The electrical output will remain the same--just at a lower frequency. And, again, a certain amount of the torque produced by the prime mover driving the generator is required to maintain the prime mover and generator at synchronous speed/frequency (0.0 MW). If the load requires that some of that energy (torque) that is required to maintain rated frequency is required to maintain rated load, then load wins--and frequency suffers (loses).

Is that sufficient rectification?

Here's the part that most people seem to have a problem with. Let's use your 50,000 MW grid where ever generator is producing maximum output (because the prime movers driving the generators are at maximum torque production). A 20 MW generator synchronizes to the grid and starts increasing it's power output to 20 MW--but none of the other generator prime movers decreases its power output, and the number of motors and lights and computers and computer monitors and televisions remains unchanged. So, the load is constant but now there is an excess of generation but the electrical load remains constant (unchanged)--the grid load remains at 50,000 MW. What happens in this case is that there is more torque than is required to maintain synchronous speed/frequency and so the grid frequency increases. The grid operator responds to this by decreasing the amount of one or more of the other generators by 20 MW--and this returns the grid frequency back to rated.

The concept of an AC grid is that many generators are all acting as one "generator"--and many motors and lights and computers and computer monitors and televisions are all acting as a one "load". If the amount of torque being supplied to the "generator" is equal to the amount of torque required by the "load" then the grid frequency is at rated. If the amount of torque being supplied to the "generator" decreases but the "load" remains the same, then the grid frequency decreases because some of the torque that's required just to maintain grid frequency is used to supply the "load."

If the amount of torque that's being supplied to the "generator" increases but the "load" remains the same, then there is more torque than required to supply the "load" and that results in an increase in speed/frequency.

When people synchronize their generator to the grid and load it up, they don't really look at the frequency meter--but the grid operators do. They have frequency meters with two or three or four decimal places, and they see an increase in frequency. So to maintain rated frequency they have to reduce the load of one or more generators to return the grid frequency to normal.

When people unload their generator and open the generator breaker (or the breaker trips and separates the generator from the grid), they don't look at the frequency meter. But the grid operators see the loss of generation as a decrease in frequency. And, to return the grid frequency to normal they have to increase the output of one or more generators by an equal amount to the generation that was lost.

Hope this helps!
 
Sir,

Thank you for your response.

I think mechanical output coming from 50 HZ supply will be more than from output coming from 49.9 Hz supply for the same voltage, current. This way you will get two different mechanical output for the same electrical output as frequency is not counted for measurement of electrical power consumption on households.
 
At 57Hz grid frequency, since the mechanical "power in" will remain the same (no further governor action as being at the end of the speed-droop curve already). While the grid frequency declined, the mechanical "torque in" will increase accordingly. This additional torque will accelerate the rotor, and thus the load angle will advance. Given sufficient headroom is available (not breaching angular stability limit), the load angle will stabilise at a higher operating value. At this point, generator output will stabilise at 275MW and will remain in synchronism at 57Hz.
 
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