A GLIMPSE OF UTOPIA
I wanted to make a toy model of an electricity grid, to shed some light – at least in my own mind – on some of the issues that arise when a reliable generator is displaced by an intermittent one. Skip straight to the last section if you are allergic to maths.
The baseline is extremely simple. We live in a place called Utopia, where our energy needs are met by a generator called Mr. Reliable. Mr. R supplies a constant power R to the grid, fulfilling all its needs, so R is also the grid capacity. [By way of simplification, there is no hourly variation in demand.] Mr. R is owned by the public, and burns hydrocarbons. His costs, and ours, are
KR + FR
K is Mr. R’s operating costs, Barons per GW per year
R is Mr. R’s capacity (for simplicity he operates at a capacity factor of 1).
F is Mr. R’s fuel cost in Barons per GW year
So far so good. Now into Utopia comes the urgent need to wean our country off its fossil fuel addiction. The new generator is also publicly owned, and his name is Mr. Intermittent. (He isn’t intermittent yet, but still). Mr. I’s costs, and ours, are:
L is Mr. I’s operating costs, Barons per GW per year
I is Mr. I’s capacity
Mr. I has no fuel costs to pay. The fuel saved by adding Mr. I to the grid is:
c is the capacity factor of Mr. I.
So our new cost for the grid, having added Mr. I, is:
KR + FR + LI – cFI
If cFI > LI we have made a saving, or more simply if cF>L.
So if the operational costs of the new generator Mr I are less than his capacity factor times the per-unit cost of fuel he is displacing, we have saved both carbon emissions and money. Now let Mr. R’s actual generation, after Mr. I’s introduction, be A. His costs are now:
KR + FA
Where FA = FR – cFI. To displace all fuel, A must reach 0, which happens when R = cI. This is pretty obvious since for example if c = 0.4 [typical of an offshore windfarm], then R = 0.4I, so you need 2.5 times the capacity of Mr. I to replace Mr. R if Mr. I operates at 40% capacity factor.
MR INTERMITTENT GOT HIS NAME FOR A REASON
The above holds true for the toy model if Mr. I is not intermittent at all, but goes flat out at capacity factor c. When R = cI, what happens in a slightly more realistic world is that half the time Mr. I produces too much power, and half the time he produces too little. Now, c represents not a constant capacity factor, but its average. So that at times when R > cI we have to ask Mr. R to turn back on, and when R < cI we are throwing energy away.
What happens at this stage depends entirely on the distribution of c. I’m taking an extreme example because it’s easy: c is a uniform random variable in the range [0…1] with mean 0.5.
Under this scenario, if Mr. I’s nominal capacity is twice Mr. R’s, I = 2R, and cI > R half the time; half the time Mr I generates more than R, half the time less than R. So half the time Mr. R has to turn on, anywhere from a little bit to full whack. The other half of the time Mr. R is turned off and power is being binned.
For the half of the time Mr. R is turned on, his output averages 0.5R, so that his overall output in the brave new world is 0.25R. For the half of the time power exceeds demand, we are throwing away an average of 0.5R, for an average wastage of 0.25R. Obviously these two things don’t happen at the same time, but all the time, one or other is happening: either Mr. R is turned on, or we’re throwing power away. And we can’t get rid of a watt of Mr. R’s generating capacity, for very occasionally he’s called upon to turn the wick up to full. Thus with twice the intermittent capacity added to the existing reliable capacity, we are still using a quarter of the amount of fuel we did before, and have all of Mr R’s generation still plugged in.
Our costs are now
KR + 0.25FR + 2LR
So we have made a saving if 0.75FR>2LR or if 3F>8L.
THE ANSWER IS STORAGE
Determined as we are to get rid of fossil fuels entirely, running a grid of 3 times the previous capacity and still using a quarter as much fuel is unacceptable. The wisest minds of Utopia propose a new solution: storage.
Mr. R is retired, or as they say in Utopia, blown up by important people to the cheers of hoi polloi. Mr. I still has capacity 2R and produces as for the previous example, his capacity factor an unstructured random variable in the range [0…1], mean 0.5. Remember, half the time we are producing too much, and half the time too little. This implies that all we have to do is to draw on the big battery in times of need, and pay it back in times of plenty. It’s a nice little overdraft facility.
In terms of power, our backup has to be able to cover our needs, i.e. to produce power R. Now I’m going to put some fake numbers in for illustrative purposes. R, the grid power, is 100 GW flat out all the time. Our storage is going to be able to output that for 10 hours, so its capacity is 1000 GWh. For simplicity there is no energy lost in the process.
So the wise minds sit back and relax, knowing they are covered for power and have rid themselves of the dreaded fossil fuel addiction at last. Are they right?
Whoops, no they’re not. Beginning fully brimmed with 1000 GWh in the bank, in the first 500 hours we have hit zero storage and power cuts on ten occasions. But before we ran out of energy altogether we did manage to cap out the storage 6 times. Of course we had nowhere to put the excess at these times so the energy was thrown away. The top figure shows the power Mr I produced each hour (GW). The bottom one shows the energy stored in the big battery (GWh).
This is just one run, but it is as they say in the literature a “representative” example. The average capacity of the series was 0.491. 915 GWh was thrown away when storage was capped out. Actually, of ten runs with this scenario, 8 had power cuts.
Remember if Mr. I was a constant friend and ran at a continuous capacity factor of 0.5 (=100 GW) we would never need Mr. Storage at all.
But perhaps unstructured noise is an unfair test. What about if there is some persistence in Mr. I’s power output? In this example, half of Mr. I’s output depends on the previous value. As before the top figure shows Mr. I’s power output, and the bottom figure shows the energy we have in storage. In this run, storage maxed out 18 times and we had 5 power cuts. Average capacity factor was 0.514. Energy wasted when the storage was brimmed was 1898 GWh.
Power cuts also occur with different scenarios like a bounded random walk with 0.75 of the power dependent on the previous value. Below this situation is illustrated, and here we had 8 power cuts and we threw away 14323 GWh of energy.
So things look bleak for a storage solution, at least in Utopia. Note: The big battery in Hornsdale Australia offers 150 MW/ 193 MWh, and cost A$ 172,000,000. [That’s about a hundred million quid. The cost of the Utopian battery would therefore be about half a trillion quid – that’s 7.1 Barons in Utopian currency.]
Let’s rewind a bit. A minute ago Mr. R was still with us and he was reduced to running at an average of 0.25 of his capacity, thanks to the addition of Mr. I’s generation. Remember, in this scenario Mr. R cannot be got rid of, but has to stand dutifully by for whenever Mr. I is in the doldrums. At the moment this is all fine, although probably quite expensive.
Our costs are
KR + 0.25FR + 2LR
In Utopia where everything is cross-subsidised, we can afford to prop up Mr. R. What about if we inject a little capitalism?
At the beginning, Mr. R’s costs were
KR + FR
And he would be able to cover those costs with sales of R.
But now his costs are
KR+FA, (A being the actual power delivered) which in this case equals KR + 0.25FR
And he can only pay his way with sales of A = 0.25R. The price per unit of power needed to cover costs is (KR + FA)/A, or (KR/A) +F.
You’ll note that if the unit price depends on dividing by the quantity supplied, then as A goes down, the unit price goes up. And as A–> 0, the unit price approaches infinity. This is not surprising as at this stage Mr. R has to cover his costs on an infinitesimal quantity of sales.
I SKIPPED MATHS TO SMOKE A SILK CUT BY THE BIKE SHEDS
What, you may ask, does all this mean? The model is necessarily overly simplistic, but it illustrates the problems of introducing an intermittent generator to a stable grid. In particular what it appears to show to me is that, the more power you get from the intermittent generator, the worse things get. In my first example, with a large amount of random generation and reliable backup, all of the reliable backup had to remain online, and it was still supplying a quarter of the grid’s power.
The storage model showed that even ten hours of flat-out power as stored energy was insufficient to buffer the new intermittent grid once all the reliable generation had been blown up. (That’s an implausible quantity of backup, I think.)
The last bit of maths made the point that, if you have to retain the reliable generator’s capacity, but it is increasingly under-utilised thanks to overcapacity of the intermittent generator, then in a capitalist world the reliable generator has to charge more and more for the power that it is called upon to deliver if it is to break even. In a reverse auction where the last generator to enter sets the price for all, this would be a recipe for very expensive electricity.