Anyone use 100% solar power?

[Frost]

(I'm just now researching nuclear power and have never used that yet) but just curious if anyone else uses 100% solar power. Right now I am 100% solar, with 262 solar panels in my main base alone (221 accumulators). I like that it's "free", lower pollution, and unlimited. But it takes up a huge amount of space and takes a lot more maintenance to upgrade so I have power 24 hours a day.

I'm starting to want to go back to steam power since it takes far less space and maintenance to maintain, but does chew up coal.

Just curious.

[Troll Norris]

If you want, I can spoil you the optimal solar/accumulator ratio/MW ratio.

BTW for boilers, it is better to connect all their outputs together and connect the steam generators in the serial way. Using commom parallel way triggers all boilers. This causes high pollution. If you use the serial way, only necessarily ammount of boilers will work.

Most time I use it for self-sufficient Radars. For that I need 8 solar panels and 6 accumulators.

[Fel]

That's kind of wrong, boilers only "work" when they can output steam and generators only consume the steam required to produce the power needed by the base, so you can have your boilers and generators hooked however you want and the amount of pollution will not differ past the initial filling of all of the pipes.

There are reasons to not connect all of the outputs together but it's mostly related to how fluids behave in "loops" and the impact it can have on performance in the late game.

[Troll Norris]

36,75 MW solar power plant.
875 solar panels
735 accumulators

The disadvantages: Too big, lot of resouces for building that.
Advantages: No fuel needed, the immediate power output can be up to 273 MW, just the average daily output must not be higher than 15,3125 GJ (per 416+2/3 seconds)

https://steamcommunity.com/sharedfiles/filedetails/?id=1973036396

[Frost]

Originally posted by ShutEye_DK:

I love solar power. The way I play space is no problem.

My current factory has 45,000 solar panels and 38,000 accus.

Edit: That gives a 194 GJ power bank.

i just got to 1.1k solar panels. 45k?! Wow.

I have a blueprint now that lays down 24 solars 20 accumulators and a station with personal bots, which is very cool combining blueprints with bots.

[Jupiter3927]

You can tile your blueprints into larger blueprints...
It's easy to do if your blueprint is a rectangle.

[impetus_maximus]

Originally posted by Troll Norris:

If you want, I can spoil you the optimal solar/accumulator ratio/MW ratio.

BTW for boilers, it is better to connect all their outputs together and connect the steam generators in the serial way. Using commom parallel way triggers all boilers. This causes high pollution. If you use the serial way, only necessarily ammount of boilers will work.

Most time I use it for self-sufficient Radars. For that I need 8 solar panels and 6 accumulators.

perfect ratio for who? just add panels, or accumulators as needed.

steam engines only work as hard as needed. it doesn't matter if 1 is at 100%, or 5 are at 20%. care to post an example of your method?

you only need 7 solar panels, and 5 accumulators to keep a radar going.
https://steamcommunity.com/sharedfiles/filedetails/?id=1439359468

[Kai]

Would actually be nice if you can use solar to power boilers via electrical heating, then produce Steam which is stored and act as a buffer to Steam Engine.

What actually produces a lot of pollution is the Boilers themselves due to consuming of fuel, but a electrical boiler would play very nicely if you use Solar and Battery to power it.

[Maya-Neko]

Originally posted by Kai:

Would actually be nice if you can use solar to power boilers via electrical heating, then produce Steam which is stored and act as a buffer to Steam Engine.

What actually produces a lot of pollution is the Boilers themselves due to consuming of fuel, but a electrical boiler would play very nicely if you use Solar and Battery to power it.

Heat isn't very good for storing energy, as it's extremely hard to keep it from just radiating away. It's pretty effective in small systems like houses or buildings, but it's definitely not useful in big systems, as it wastes more energy, than it gives back.

Accumulators are way better for this job, as they store the energy without loosing it over time.

[Jan]

I played a lot with 100% solar power. (before nuclear came into the game)
I had each time around 3k solar panels. (lasers need sooo much energy ^^)

Nowadays I switch to nuclear because I don`t need to build the endless fields of solar plants and it`s a lot cheaper because I play "marathon"

[impetus_maximus]

@Kai
yeah when i said 1 at 100% or 5 at 20% i was speaking of the boilers.

[Troll Norris]

Originally posted by impetus maximus:

perfect ratio for who? just add panels, or accumulators as needed.......

Perfect ratio for everyone.

https://steamcommunity.com/app/427520/discussions/0/3799284136642683584/#c3799284136644511693

It all comes from this research which made me and many others:
https://forums.factorio.com/viewtopic.php?f=5&t=5594

One full day has 25000 ticks (416+2/3 seconds)
60 ticks = one second
Now:
Day has 12500 ticks (= 208+1/3 seconds) 100% solar power production
Sunset has 5000 ticks (= 83+1/3 seconds) Solar power production decreasing lineary
Night has 2500 ticks (= 41+2/3 seconds) 0% solar power production
Sunrise has 5000 ticks (= 83+1/3 seconds) Solar power production increasing lineary
https://wiki.factorio.com/Time

Let's say you want to power 6MW by solar power plant.
You need the energy to keep it working for whole day.
(416+2/3)*6= 2500 MJ
This energy you need to generate during day, sunrise and sunset.
That means you have 12500+((5000+5000)/2) = 17500 ticks ( = 291+2/3 seconds) to generate needed amount of energy. Let's calculate how many solar panels do we need:

2500 MJ / (291+2/3 s) = 8+4/7 MW = 8571+3/7 kW. One solar panel produces 60 KW. So (8571+3/7)/60 = 142+6/7 solar panels, rounded 143 solar panels.

Now we have to do some reverse counting and find in which tick the solar panels starts to produce more than 6MW of power to find for how long time the solar panels are not producing enough power. For accurancy we will not use the rounded number. We know the sunrise takes 5000 ticks. That means 0 ticks= 0% ( 0 MW) production and 5000 ticks = 100% (8+4/7 MW)

So: 6/((8+4/7)/5000) = (6*5000)/(8+4/7) = 3500 ticks.
In total there is not enough energy generated for 2500+((3500+3500)/2)= 6000 ticks = 100 seconds.

100 seconds * 6 MW = 600MJ
600 MJ / 5 MJ = 120 Accumulators

The final ratio is (142+6/7) solar panels devided by 120 accumulators results in 25/21..... Here you have the ratio.

120 accumulators devided by 6 MW = 20 accumulators per MW - the second ratio.
Or you can count (142+6/7)/6 = 23+17/21 solar panels per MW

The explanation why for counting sunrise and sunset is used formula (sunrise+sunset)/2 : because in sunrise and sunset there is not produced the full power output, but it is linear. It is easier to show it graphical way. You just "convert" the triange into small rectangle http://i.imgur.com/6qXfqchg.png

Be sure it works. It was tested by many of us. You can try to power the 20 radars with 143 solar panels and 120 accumulators. You will find the accumulators will be in the sunrise almost depleted and after charging they will start almost immediately discharging.

Check:

For example, Actually I have 875 solar panels and 735 accumulators in my factory. It is the 25/21 ratio and is designed to produce 36,75 MW of average daily power output. This is the pro part of solar power plant. It is able to provide up to 273 MW power at instant time. All I have to do is to keep the average output 36,75 MW per 25000 game ticks, Then I can go over value 36,75 up to 273 for a short time.

Math:

36,75*(416+2/3) = 15312,5 MJ - for whole day
15312,5/(291+2/3) = 52,5 MW = 52500 KW - needed power of solar panels
52500/60 = 875 solar panels

36,75/(52,5/5000) = (36,75*5000)/52,5 = 3500 ticks - As you can see, this is exactly the same number which is very important. This number proves that ratio is static and same for every calculation and we always need accumulate power exactly for 100 seconds.

100*36,75 = 3675 MJ for night
3675/5 = 735 accumulators

875/735 = 25/21 - the solar panel/accumulator ratio

735/36,75 = 20 accumulators per MW
875/36,75 = 23+17/21 solar panels per MW

So, I don't have to execute tons of calculations. I just use the ratios to count how much solar panels and accumulators do I need.

About steam boilers. There does matter if they work on 1% or 100% but most probably they always generates the 100% pollution.

[Vyndicu]

On my biggest most recently mega base run I had 3 nuclear power complex and thousands of solar panels.

I also kept dozens of steam turbines, smaller than two steam engines, on regular boilers as an emergency standby mode in case of a brownout.

Nothing like overkill when it came to power/energy production.

[Purpleganja]

Troll Norris: How about having one accumulator for one solar panel to have a buffer for irregular power usage. Setup an alarm to ring if an accumulator goes under 20% to know if you exceed your daily ratio and fuel up some backup boilers or nuclear reactor while you deploy more solar farms.

[Fel]

The ratio is a base that you can adapt to your own situation but 1 to 1 would cost you quite a bit of extra resources.
It is not a bad idea when you are still on a small scale but when you start mass-producing solar pannels and accumulators the difference will consume a significant amount of extra resources.

It would probably be easier to set a similar alarm and backup without such a big increase in accumulators, if your stored power gets under 20% during the night, no matter if it's an irregular night or a normal one it means that you have to start thinking about expanding quite soon especially if you rely on laser turrets for your defenses.

[BoiledInPilk]

I go all solar, got solaris, got chem research production/oil