Overcoming generator's counter torque?

[Miyz]

Hi all,

When counter torque is produced in a generator, does that mean we have to apply MORE mechanical energy to sustain the output?


Miyz,

 

[Averagesupernova]

Yes.

 

[Miyz]

Yes.

But doesn't that depend on the amount of current drawn from the generator?

Simple examples:

300W of input mechanical energy is applied to a generator and only 250W of electrical energy is draw what is it necessary to add more input in this condition?

Could you give some examples for the sake of my understanding please.

 

[Miyz]

Yes.

+ What if we stayed with the same amount of mechanical input over and over again, as a fact we know that the generator's counter torque will always be less than of the mechanical input torque!

So another example has to be added(Reason for using 100W just to illustrate my understanding and its easier):

200W of MEC.Input = 190W Electrical output drawn from the generator,

If we stayed with 200W of MEC.Input over and over and over again how is it not possible to sustain a 190W output?

 

[Windadct]

The (real) system has to be regulated - if you apply torque with no or less counter torque it will accelerate. For a basic generator with a gas engine there is a speed based throttle. As the electrical generator apples more "counter" torque the gas engine slows down and the control system provides more throttle ( fuel) -- It must always achieve a balance in relatively short time frame.
The fact that there is an electrical generator on there has nothing to do with it. Keep your car in first gear and floor it - you will accelerate until you hit the rev-limiter ( a throttle limit based on engine RPM) -

 

[Miyz]

The (real) system has to be regulated - if you apply torque with no or less counter torque it will accelerate. For a basic generator with a gas engine there is a speed based throttle. As the electrical generator apples more "counter" torque the gas engine slows down and the control system provides more throttle ( fuel) -- It must always achieve a balance in relatively short time frame.
The fact that there is an electrical generator on there has nothing to do with it. Keep your car in first gear and floor it - you will accelerate until you hit the rev-limiter ( a throttle limit based on engine RPM) -

Ok! Based on what you've said as soon as the generator applies its counter torque the engine slows down, OK you apply more throttle can that throttle be the same amount as the previous one that made the generate produce electricity or does it have to be MORE than the original amount of input? Look at the 2 examples I've posted earlier it might help describe my problem.

 

[Miyz]

Anyone?

 

[Averagesupernova]

I'll break it down into chunks. You have a electric generator driven by a gasoline engine. The engine starts and runs up to operating speed of for instance 1800 RPM. At this point the generator is not powering any load whatsoever. NO power is being taken from it. So the torque required is to overcome friction, wind resistance, etc. Now you attach a light bulb to the generator which draws 100 watts for instance from the generator. This puts a back EMF on the generator which will slow the engine and generator down unless more fuel/air is fed into the engine. Anything more you want to know is too specific to answer.

 

[Miyz]

I'll break it down into chunks. You have a electric generator driven by a gasoline engine. The engine starts and runs up to operating speed of for instance 1800 RPM. At this point the generator is not powering any load whatsoever. NO power is being taken from it. So the torque required is to overcome friction, wind resistance, etc. Now you attach a light bulb to the generator which draws 100 watts for instance from the generator. This puts a back EMF on the generator which will slow the engine and generator down unless more fuel/air is fed into the engine. Anything more you want to know is too specific to answer.

In this case... If the engine can NOT apply more effort to the system and the load requires more juice! What can be some alternatives?

Build another engine? Let it supply additional energy? Imagine a situation that the engine has that generator's load and can NOT supply any more energy... What do you think are the solutions for this?

 

[Miyz]

Can there be a way to limit a generator's load?

 

[Averagesupernova]

So is this a hypothetical case? If you have a generator loaded which then laods the engine to its maximum you will have a short lived engine.

 

[russ_watters]

In a steady state, the input must always be greater than the output. This is required by the laws of thermodynamics.

 

[Miyz]

In a steady state, the input must always be greater than the output. This is required by the laws of thermodynamics.

Technically the input would always = output based on the laws of conservation.However, in reality where there are multiple resistive forces we know the fact that the output would be less than the input for sure.


Now, what I'm trying to achieve is the proper understanding of how a generator works under load, and all the factors surrounding it. I know as a fact that there will be a counter torque, the counter torque will always be less than the inputted torque based on the laws of thermodynamics and conservation... It makes no sense to me that I have to apply MORE torque specifically to overcome the counter torque that is produced based on lenz's law.

Now I ask you guys the "experts" to explain me every aspect so I can be enlightened.

 

[Miyz]

So is this a hypothetical case? If you have a generator loaded which then laods the engine to its maximum you will have a short lived engine.

How is it overloading the engine?

Maybe an example is needed? A 100KW is applied mechanically to a generator that's efficiency rate is about 95%(Not sure about the calculations so bear with me).

That's 95KW of electrical power converted to supply that LOAD. Now, due to the counter torque is it necessary to increase the input mechanical power over 100KW? Now here where it gets tricky to, even if you've applied 20KW more = 120KW mechanical input that will still be converted to 114KW of electrical power and the load will STILL apply it's counter torque. You keep adding and adding and adding...

There is something wrong here that I need you guys to help me clear out... Please use my example to explain the system. Thank you!

 

[Windadct]

Every part of the system has a limit. I really do not know what you mean by " due to the counter torque is it necessary to increase the input mechanical power over 100KW" - why are you saying "it is necessary" - do do what? - do you need more electric power to be output?

A 100KW rated engineer can only put out 100KW ( plus a little margin) but that is the limit - if you keep giving more fuel etc - it will fail - blown engine.
An electric motor has a limit - usually a current limit for the winding's - more electric power output at a fixed voltage - requires more current - the current heats up the windings and it overheats / fails ( typical overload issue with an electric motor - but there are others like over speed bearing failures).
If you try to draw more current (power) from the electric motor than the engine can provide - you will stall the engine. The same as in a car.

 

[Averagesupernova]

The power that the load is dissipating IS what causes counter torque. Remove the load and the counter torque goes away. I don't understand how it can be any more clear than this.

 

[SirAskalot]

Newton second law gives the answer:
$$J \frac{d\omega}{dt}=\Delta \tau \\<br /> \text{Output power from salient synchronous generator:} \\<br /> P=\frac{V~E}{X}\sin{\theta} \\<br /> \text{And the relation:}\\<br /> \tau \omega=P$$

Adding these relations together, you can solve for what happens if so and so.

You can also linearize the system around a operating point so you get the relation:
$$2H \frac{d\omega}{dt}=P_{in} - P_{out} ~[per~unit]$$

If i remember correctly. So basically the speed of the generator changes if input power and output power is not equal. Also most generators have speed governor where the input power/torque is a linear function of speed. Hence the steps are:

- Output power drops
- Speed increases (excess energy is transferred into rotating energy)
- Speed governor reacts
- decreases input power
- acceleration stops
- steady state is achieved at a higher rotating speed
( and frequency is proportional to speed)

or vice versa.

Regarding the torque and counter torque refer to eq. 1. The torque acting on the shaft have to be equal and opposite to achieve steady state operation. Mechanical torque is input torque on the turbine. "counter torque" is the electromagnetic torque produced by the generator, this is a function of many factors (current, voltage etc.), in essence the loading of the generator as Averagesupernova pointed out.

 

[Miyz]

Every part of the system has a limit. I really do not know what you mean by " due to the counter torque is it necessary to increase the input mechanical power over 100KW" - why are you saying "it is necessary" - do do what? - do you need more electric power to be output?

A 100KW rated engineer can only put out 100KW ( plus a little margin) but that is the limit - if you keep giving more fuel etc - it will fail - blown engine.
An electric motor has a limit - usually a current limit for the winding's - more electric power output at a fixed voltage - requires more current - the current heats up the windings and it overheats / fails ( typical overload issue with an electric motor - but there are others like over speed bearing failures).
If you try to draw more current (power) from the electric motor than the engine can provide - you will stall the engine. The same as in a car.

Ahh... Due to my confusion when I read most of you said "increase" the mechanical input to overcome the counter torque the indicated to me that if I applied 100KW, I need to add more to the system more than 100KW to overcome the counter torque generated by the load. I'm wrong, it simply means that when load is applied I have to continuously keep the 100KW of Mec.power going on and on and on to resist the counter torque and to supply for the load, I think you all meant that I had to continuously keep the Mec.power going to supply for the generator so that it can continuously keep supplying for the load and resist the counter torque. I think I've understood (HOPEFULLY!)

The power that the load is dissipating IS what causes counter torque. Remove the load and the counter torque goes away. I don't understand how it can be any more clear than this.

You did not understand my point, that's why it may not seem clear to you... Sorry for not explaining more.

Newton second law gives the answer:
$$J \frac{d\omega}{dt}=\Delta \tau \\<br /> \text{Output power from salient synchronous generator:} \\<br /> P=\frac{V~E}{X}\sin{\theta} \\<br /> \text{And the relation:}\\<br /> \tau \omega=P$$

Adding these relations together, you can solve for what happens if so and so.

You can also linearize the system around a operating point so you get the relation:
$$2H \frac{d\omega}{dt}=P_{in} - P_{out} ~[per~unit]$$

If i remember correctly. So basically the speed of the generator changes if input power and output power is not equal. Also most generators have speed governor where the input power/torque is a linear function of speed. Hence the steps are:

- Output power drops
- Speed increases (excess energy is transferred into rotating energy)
- Speed governor reacts
- decreases input power
- acceleration stops
- steady state is achieved at a higher rotating speed
( and frequency is proportional to speed)

or vice versa.

Regarding the torque and counter torque refer to eq. 1. The torque acting on the shaft have to be equal and opposite to achieve steady state operation. Mechanical torque is input torque on the turbine. "counter torque" is the electromagnetic torque produced by the generator, this is a function of many factors (current, voltage etc.), in essence the loading of the generator as Averagesupernova pointed out.

Thank you for that effort. I'll review this and try to relate it! "The torque acting on the shaft have to be equal and opposite to achieve steady state operation" Makes good sense! This also support the notion Ein = Eout!Thank you all,I appreciate all your efforts!
Miyz,

 

[Miyz]

Do correct me if I've understood something wrong.

Thank you.