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I've heard many times that the grid electricity production and consumption must be balanced and imbalances lead to overvoltage or undervoltage. I've never seen an explanation of that phenomena from energy conservation law standpoint.

Suppose I have a grid with exactly one gigawatt generation and exactly one gigawatt consumption. Let's pretend there're no losses - all the conductors are superconductors. So whatever is generated is fully consumed.

Now someone switches his one thousand watts space heater off. Generation now exceeds consumption. Although it's a relatively small increases there's still some imbalance.

What happens? Where does this excess power go?

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  • \$\begingroup\$ I think in your scenario the frequency goes up (less load to generators, they tend to spin up) and the guy begin the (country's) control panel quickly switches of some generators. Saw a documentary on tv showing how it's done. \$\endgroup\$ – Dan Feb 24 '15 at 8:27
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    \$\begingroup\$ The "superconductor" part of the question interferes with the normal mechanism for load adjustment. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 24 '15 at 12:42
  • \$\begingroup\$ @IgnacioVazquez-Abrams I'm not sure I see the difference. If a heater is connected with resistive wires we just have to add their resistance to the resistance of the heater. So when the heater goes off both it and the wires are excluded. \$\endgroup\$ – sharptooth Feb 24 '15 at 12:48
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    \$\begingroup\$ The transmission wires are part of a resistor divider. With no resistance, things go strange. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 24 '15 at 12:50
  • \$\begingroup\$ @IgnacioVazquez-Abrams: Even a superconducting generator has a source impedance, which allows the terminal voltage to vary with load. But it's true that the resistance of a real transmission system plays an important role in damping out disturbances caused by load variations. A purely inductive (and capacitive) distribution system would be no fun! \$\endgroup\$ – Dave Tweed Feb 24 '15 at 12:58
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For tiny variations of load, the grid voltage changes slightly, which causes the power consumption of all of the other loads to change accordingly.

For example, if your 1 GW load suddenly becomes 0.999999 GW (-1 ppm), it only takes a voltage rise of +0.5 ppm to bring the consumption back up to 1 GW. This is enough of a rise to cause the power that was being consumed by the switched-off load to be distributed across all of the other loads.

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The generated Power comes from electric generators. I think the round thing inside is called rotor in english. Anyways, that rotors are rotating very fast (most commom are 3000 or 1500 rpm) and they have a very big weight. Therefore in all this rotors is a LOT of kinetic energy. If now less energy is produced as consumed, the energy comes from this kinetic energy. As the shafts loose kinetic energy, they rotate slower. The frequenzy of the grid is proprotional to the roational frequenzy. This drop in frequenzy is measuerd an all powerplants together are powering up to make up for the difference. (Some power plants do much and some very little)

If you want to know more, you should read a bit about Primary reserve / primary Control which is this uppowering of the power plants (in the first seconds). When bigger/longer differences happens, also the secondary reserve is needed.

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  • \$\begingroup\$ No, large utility generators have many poles, and normally rotate at only a few hundred RPM at most. But it's still true that there's a significant amount of mechanical inertia in the system. \$\endgroup\$ – Dave Tweed Feb 24 '15 at 12:42
  • \$\begingroup\$ Yeah i know, thats why i used "most commen". I am only aware of big power plants using water-power to have many poles and rotating at low speed. All other power plants i know go as i said. Also this slow ones are much bigger and have more weight and therfore also a lot of energy, even as the go slower. \$\endgroup\$ – Kai Feb 24 '15 at 12:45

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