AC Mains Generator -- RPM changes based on load changes

[Ocata]

TL;DR Summary

For an electrical system powered by a generator, how does adding electrical load to the system affect the generator's rotational speed? Thanks

Hi,

Scenario:

Given a powerplant is being provided a constant power source - like steam is being produced at a steady rate, so that the generator rotates at a constant speed, and in turn, produces a constant electrical frequency.

And, suppose, hypothetically, all the power being used within the grid (like all the home electricity and city electricity) remains constant - like all electrical appliances and equipment remain in whatever state they are in (whether on or off). So that a steady amount of power is being used.

So now suppose someone comes along and plugs in or turns on some piece of electrical equipment. This is known as "increasing the load," I believe..

Question:

Does turning on an appliance, or increasing the load on the electrical system, impact the generator's rotational speed? If so, how does "increasing the load" on the generator affect the rotational speed of the generator?

Thanks

 

[Tom.G]

There is a governor on the steam feed to the generator. When the generator slows down. its frequency tends to drop. The governor opens the steam valve a little bit to match the generator frequency to the frequency of the power grid.

There are a lot more details that can probably be filled in by @anorlunda here.

Cheers,
Tom

 

[Ocata]

Hi Tom,

You mention "When the generator slows down."

So, my question is, When does the generator slow down? What makes the generator slow down ..or speed up, initially? Does it have to do with loads (electrical devices) being added and removed from the system (circuit)?

Thanks

 

[anorlunda]

@Tom.G is right, this is my favorite topic.

You can look at the question from the electrical side, or from the mechanical side. The whole truth marries both views.

Mechanically: The spinning turbine and generator make a huge flywheel storing kinetic energy via rotation. Energy is conserved every instant, so when you increase load but it takes time for the power plant's boiler to make more power, where does the missing energy come from? (I think that is a restatement of your question in the OP.) It comes from kinetic energy in that flywheel. The rotor slows down releasing some of that kinetic energy to serve the load. The governor senses the drop in speed and signals the boiler to make more power. Later the boiler catches up and speeds up the flywheel again.

Electrically: The power transmitted from point A to point B in the grid is proportional to the difference in phase angle of the voltage at A compared to the voltage at B. To change the relative phase, one end must change speed temporarily until the new phase is reached then reduce speed again to keep the phase difference constant.

Visualize yourself riding on a Ferris wheel. A second Ferris wheel is alto turning right beside you. Your friend is riding on the other wheel. You and your friend each hold one end of an elastic band. The band stretches the least, when you and your friend are at the same point in the at the same time, say at the very bottom. If you and your friend are not at the same point, how do we get them together? We temporarily speed up (or slow down) one wheel until the two friends are at the same point. After that, the two wheels must turn at exactly the same speeds to keep the two side-by-side.

I have a couple of Insights articles if you want to learn more.

The first one uses no math. It traces all the changes triggered when you add load such as turning on a light. Power produced must equal power consumed every microsecond, every hour, every year, every decade. So it makes a long story. I think your question in the OP asks only about seconds.
https://www.physicsforums.com/threads/what-happens-when-you-flip-the-light-switch-comments.829441/These two use some math.
https://www.physicsforums.com/threads/ac-power-analysis-part-1-basics-comments.863224/

https://www.physicsforums.com/threads/ac-power-analysis-part-2-network-analysis-comments.864530/

 

[essenmein]

Hi Tom,

You mention "When the generator slows down."

So, my question is, When does the generator slow down? What makes the generator slow down ..or speed up, initially? Does it have to do with loads (electrical devices) being added and removed from the system (circuit)?

Thanks

Maybe the grid is a hard way to see what is going on as managing it is quite complicated keeping a bunch of different generators all synchronized etc.

If you consider a basic gas or diesel backup generator, one that is not connected to the grid. Just consider your generator and your load.

The small cheap ones generally have a simple governor, ie a device that tries to maintain constant engine speed regardless of load by changing the throttle (or fuel pump). These are simple proportional controllers, that is they typically don't integrate an error. So these will have a steady state speed error as load changes.

So in this case, as you add load, the generator will slow down a little, and conversely, speed up again as load is removed. My older diesel will run at ~63Hz no load (~1900rpm), and as you load it down, this drops to about 57Hz at full load (~1700rpm), this is maximum fuel delivery for the injector pump. Add more load and the generator stalls because the engine can no longer support the load. Another way of looking at it, this governor needs a speed error of ~200rpm for 0-100% range on injector pump fuel delivery.

If you make this governor, or speed controller, more fancy, and add an integrator (ie going from a proportional controller to proportional-integral controller), then at the instant the load is added, the generator will slow down, the integrator will keep adding more power to the engine/turbine etc to bring the speed back up to the exact demanded speed (to keep the frequency fixed). This is because the characteristic of an integrator is a zero steady state error. This response time is determined by control stability criteria for the whole system (generally quite slow due to all the mechanical inertia).

This creates another problem, when load is removed, the speed will overshoot and it takes time to bring the speed back down.

 

[Merlin3189]

If so, how does "increasing the load" on the generator affect the rotational speed of the generator?

I'm hesitant to comment in the presence of people who clearly know much more about this than I, but it seems to me that a very basic point has been omitted. So FWIW

How does the turbine know there's any load at all? Why does it have to provide torque, beyond that to accelerate from rest and then overcome friction? It must have to provide this extra torque and do a lot of work, else the electric power users are getting free energy.

The cause is the current they draw, which flows through the windings in the generator. A current through a wire in a magnetic field produces a force. That force is what the the driving torque is opposed by.

When someone plugs in more load, more current flows and this opposing force increases, decelerating the rotor. At least until some control system persuades the turbine to push harder. *
-----------------------------------------
If I can also speculate, anticipating correction by others,
it seems to me that your adding a conductance in parallel with other uses in your part of the grid must reduce the voltage for them (albeit by a tiny amount) and hence reduce the current provided to their load. So initially you get your new load current by stealing a tiny bit from thousands of other users.
Of course any lowering of voltage propagates back to the generator, thus allowing it to supply extra current and thereby put greater mechanical load on the turbine.

* (or the lower speed reduces the output voltage and thence the current that users draw, or the increasing phase lag allows other generators to supply more power to this grid?)

Apologies if this is so beneath people as to be irrelevant (or evn wrong), but It's what came to my mind when I read the question above (apparently reiterated in OP's last post 3.)