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I'm working with a factory which has a backup generator which is used during grid instabilities or whenever the factory must disconnect from the grid for maintenance. In these cases, the generator is used instead to maintain operations.

However, I'm investigating whether there's another opportunity in using the generator's excess capacity to sell power on the grid.

I'll advance now that this question is not concerned with energy-market regulatory or financial viability issues for this idea (I've simplified the context here). The question is whether this is technically feasible.

The generator would then need to power two separate networks (the factory and the regional network), as seen below.

enter image description here

I had to create this with Powerpoint due to do the particular setup for the generator, which needs to feed into the two circuits without connecting them with each other (otherwise that'd just be the grid powering the factory, when this is precisely in the case when that's not possible or desirable).

So, this is the first part of the question: is this even possible? Can a source power two circuits while keeping them isolated from each other?

Assuming it is possible, we then get to the second part of the question. The generator's primary objective is powering the factory. Whatever is done on the grid is secondary. So the power must satisfy the factory and -- if there's excess capacity on the generator -- then feed into the grid.

Now, my thinking is that the factory has certain power needs (let's assume 1 MW). The generator (let's assume it can generate 1.5 MW) can run at 2/3 capacity and satisfy those needs. However, if the generator is also connected to the grid and is ramped up to generate it's maximum of 1.5 MW, where will those new electrons go? The resistance of the grid will be effectively infinite (when compared to the factory or the generator), so the natural route would seem to be factory. Which would mean the generator fries the factory's circuit and feeds almost nothing to the grid.

So, is it possible to design a circuit or device (including PIDs, microcontrollers, whatever) such that as much power as needed goes to one circuit, and the excess goes to another?

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    \$\begingroup\$ What kind of fuel does your generator run on that it produces energy cheaper than the grid supplies it? Why do you even have a grid connection if this is the case? \$\endgroup\$ – brhans Aug 6 '18 at 18:51
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    \$\begingroup\$ Not at all the case. The grid is very low impedance, and your factory can't just draw excess power from the generator. Any power over and above what your factory requires will be absorbed by the grid, similar to how a home solar installation can run the meter backwards if the house is using less energy than the solar inverter is producing. \$\endgroup\$ – John D Aug 6 '18 at 18:51
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    \$\begingroup\$ @brhans the OP states that the question excludes financial viability, but you're spot on that it will be more expensive than the grid. Even more so when you consider repair and maintenance on the generator. \$\endgroup\$ – John D Aug 6 '18 at 18:54
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    \$\begingroup\$ I don't understand why you think you have two unconnected circuits at your generator – it would really just one circuit. There's no "new electrons" here – that's not how electricity works; current always runs in a circle. Your generator's job is to supply energy to the charges that make up that current (energy per charge is voltage). There's no "excess" power – there's just the power you let your generator produce. The grid doesn't have infinite resistance, it has damn near to zero resistance. \$\endgroup\$ – Marcus Müller Aug 6 '18 at 19:08
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    \$\begingroup\$ @Wasabi What makes you think you can do that? The motor / generator is a mechanical way of transporting power. Whether or not you have electrical connections doesn't actually matter. When the generator produces a higher voltage on its grid connection than the grid has, then power flows out of the mechanical momentum that moves that generator. If the generator produces a lower voltage than the grid has, it draws power from the grid, and actually becomes a motor. There's no separation possible here! It's just one source of energy: Movement of some mass. \$\endgroup\$ – Marcus Müller Aug 6 '18 at 19:19
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Can a source power two circuits while keeping them isolated from each other?

Not when running from the same generator but that's not how you would do it anyway.

What you are proposing is quite common in countries with reliable energy supply but where peak demand is on the edge of system capacity. Industrial users with generation capacity can sign up to receive payment for being available for remote start-up and power generation. Going online requires the following operations:

  • Start of generator engine. This can be manual or remote or automatic.
  • Adjustment of output voltage to match grid voltage.
  • Adjustment of frequency to closely match the grid frequency.
  • Wait until the voltage between the generator phase and the grid phase drops to zero then close the connecting breaker.
  • Now ramp up the generator power.

You are now pushing energy into the grid. Since your load is the closest it will draw the power it requires and the rest feeds into the grid.

where will those new electrons go?

Forget electrons. They just oscillate back and forth in AC and don't "go" anywhere. Think of current.

The resistance of the grid will be effectively infinite (when compared to the factory or the generator), ...

Nope. The resistance of the grid is close to zero. That's why the lights don't dim much when you switch on your big loads.

... so the natural route would seem to be factory. Which would mean the generator fries the factory's circuit and feeds almost nothing to the grid.

Nope. You'll only need a couple of volts higher than the no-load mains voltage to start exporting power.

enter image description here

Figure 1. A synchronisation setup. Source: CircuitGlobe.

I did this in college many years ago where we had a machine (the incoming machine of Figure 1) driven by a DC motor. We ramped it up and all three lamps start blinking slowing down as the difference in frequencies gets smaller. When the lamps go out there is no voltage across the contacts so they can be closed without arcing. We then applied power to the DC machine to generate power and feed back to the grid. A following exercise was to reduce the DC machine drive and then switch it into reverse and ramp up the power to act as a brake on the AC machine which then imported power from the grid to act as a motor. It was a most educational experience.


See also G59/G10 relays.

Commonly known in the UK as G59, and as G10 injection testing in the Republic of Ireland, these regulations apply to power plants and generators used for standby and peak lopping application, or grid parallel use.

G59/G10 testing assesses the performance of the mains protection relay that sits between the generator and the electricity grid. It’s not a test of the generator itself. The relay’s purpose is to disconnect or “decouple” the generator from the electricity grid if it detects instability on either side of the connection.

Source: G59/G10 injection testing by Edina.

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  • \$\begingroup\$ I'd edit out the "electrons" in my answer if they hadn't been quoted here and in the comments. For the record, I know the electrons barely move around, I've just seen that terminology used elsewhere and thought it was interpreted as "cute", not "stupid". Learned two things today! Great answer, btw. \$\endgroup\$ – Wasabi Aug 6 '18 at 19:27
  • \$\begingroup\$ I recommend you unaccept until the earth has spun around a couple of times to give the whole of humanity a chance to read and answer your question. You'll get some different insights that way and possibly a better answer. Thank you anyway. \$\endgroup\$ – Transistor Aug 6 '18 at 19:38
  • \$\begingroup\$ Something just occurred to me. As I mentioned, the generator is mostly used when the factory is momentarily off-grid. So I couldn't "connect" the factory and the regional grid in series, since the factory's meter is offline. Instead, they'd have to be connected in parallel. In this case, would the problem actually be the inverse of what I suggested above? Would the grid's near-zero resistance actually mean most or all of the load would go to the grid instead of the factory? And yeah, I just unaccepted thinking precisely that. \$\endgroup\$ – Wasabi Aug 6 '18 at 19:38
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    \$\begingroup\$ Read the last paragraph of the G59/G10 relay. It must disconnect if mains fails because you would be trying to power the nation with your little generator and you could electrocute a lineman. You have a few problems with terminology: Connecting the generator and grid in series doesn't make any sense. The energy meter doesn't affect power - it monitors energy (power by time). The grid's near zero impedance would present itself as pretty much a short-circuit to your generator. It's circuit breaker should trip with a massive arc. \$\endgroup\$ – Transistor Aug 6 '18 at 19:44
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Yes, this is certainly possible. It would function just like a grid tied solar power system. When power draw of the factory is less than the gen's capacity, excess power can flow into the grid, and when it exceeds the gen's capacity power will flow from the grid. Whether this is economically viable is another question entirely.

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