# Electrical Power

Discussion in 'Physics' started by lendo1, Jun 10, 2011.

1. ### lendo1 Thread Starter Active Member

Apr 24, 2010
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So let's say a power plant is powering a city. At one point in the day the city is using some amount of power, lets say 100 GWatts. At another point in the day the city might be using 10 Gwatts of power. How does the power plant know the difference? I mean, how is the difference in the magnitude of the loads and the power being consumed perceived by the generator? It has to spin at the same speed no matter what, is there a difference in energy required to spin the turbines due to less magnetic opposition when there is a lighter load or what?

I've found a lot on power transmission, transformers, and phasing on the internet, but none of the sources I've found really touch on the exact conversions of energy that happen during the power cycle. If hardly any power is being required by the city, doesn't the power plant still have to convert the same amount of kinetic energy to electrical energy to spin the turbines? I'm confused.

Also, this might be an unrelated question, but how do power plants receive power? I've heard about people decking out their houses with solar equipment and actually getting a paycheck from the electric company.

2. ### studiot AAC Fanatic!

Nov 9, 2007
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515
OK so let's first do a bit of maths.

10KW is a reasonable demand for an average home at any one time.

At this rate 100GW is enough for 10 million homes.

Such a large city would have several power stations.

Each power station would have several generators.

So at any power station generators can be taken on or off line to balance demand.

You are correct in thinking that the magnetic circuit of generators is such that a greater electric load requires a greater mechanical driving input and governor devices are fitted to control this.

go well

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The solar array equipment will generate energy as DC. The electronic hardware associated with the solar equipment will invert this DC to an AC "equivalent" primarily for use in the house. If the household is not using all of the available solar generation capacity, it's possible to synchronize the locally generated AC power and return this unused energy generation capacity to the AC power grid. The energy doesn't have to be returned to the power station - it can be just consumed locally, say by another household or households connected to the same AC power grid. Where the specific energy is finally consumed is completely transparent to both the power generation company and the household with the solar panels. The metering of energy returned to the AC grid from the household with solar generation is done in the household metering system - which has to be somewhat more sophisticated than the conventional one-way metering typical of a normal household consumer electrical installation. The company billing consumers for the electrical generating company will be able to read the amount of energy returned to the AC grid by the household with the solar installation. The household can then be credited with that returned energy at some agreed rate set by the power company or perhaps a national distribution regulatory authority.

Last edited: Jun 10, 2011
4. ### lendo1 Thread Starter Active Member

Apr 24, 2010
34
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I'm trying to understand this from a theoretical, rather than a practical perspective. The most important part to me is how is the difference in the magnitude of the loads and the power being consumed is perceived by the generator. It has to spin at the same speed no matter what, is there a difference in energy required to spin the turbines due to less magnetic opposition when there is a lighter load or what? How so?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
You need to keep in mind that an unloaded synchronous machine [generator or motor] synchronized to the AC infinite bus will spin at synchronous speed under the influence of a synchronizing torque derived from the AC bus. A synchronous generator could notionally run in motoring condition. What determines the power flow between the machine and the AC bus is the DC excitation applied to the machine field winding. As the DC excitation field current is increased, a machine operating in generator mode begins to inject power into the AC infinite bus and simultaneously experiences a commensurate retarding torque (due to electromagnetic field interaction across the machine air gap). In the absence of any assisting mechanical torque (say from a coupled steam turbine) the retarding torque would eventually pull the machine out of synchronous speed - with somewhat undesirable consequences. The mechanical drive torque is therefore regulated to provide sufficient mechanical input energy to account for the total energy being supplied to the infinite AC bus.

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6. ### lendo1 Thread Starter Active Member

Apr 24, 2010
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That was incredibly helpful t_n_k, thanks!

7. ### nsaspook AAC Fanatic!

Aug 27, 2009
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The electrical generation loading problem is causing grief for the wind power companies in the NW now. To keep the water behind the dams at a lower level for spring melt and reduce flooding the BPA is dumping water into the turbines creating a huge surplus in available power. To keep the generators stable requires a base load that now has almost every other generation method offline at times. Almost all coal, thermal, wind and nuclear are down or throttled so hydro can dump power when demand is low.
http://info.bpa.gov/afterhours.aspx

http://ecotrope.opb.org/2011/05/bpa-shuts-off-wind-power-to-make-way-for-hydro/