I have recently been playing with solar panels and accumulator with varying day/night cycles (SE). I sorted an analytical solution to quickly calculate na, the optimal (minimum) number of accumulator per solar pannel. Thought I'll put it here 'cause I'm proud and it might help someone else.
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Known variables:
- td = 1 day cycle (Default: 416.666 seconds)
- rd, rm, re, rn = daytime ratio for day, morning, evening and night respectively (Default: 0.5, 0.2, 0.2, 0.1)
- Ps = Power of one solar pannel (Default 60kW)
- Pin = Charging power of an accumulator (300kW)
Pout = Output power of an accumulator (300kW)
E = Maximum charge of an accumulator (5MJ)
- r = rd + (rm + re)/2
- Pe = E/td
- a = (1-r)/r
- b = r² - r*rd + rn
- (Pin/a)
- (Pe/b)
- IF (Pe/b) < (Pin/a) AND (Pe/b) < Pout
>>>>>>> nₐ = (r*Ps) / (Pe/b)
- ELSE IF Pout < (Pin/a)
>>>>>>> nₐ = (r*Ps) / Pout
- ELSE
>>>>>>> nₐ = (r*Ps) / (Pin/a)
Notes:
- This solution is a standalone no brain set of equations that will give you the optimal ratio of accumulator to solar panels. Try yourself with the numerical values I gave for the known variables. In less than 3 minutes you'll find the answer nₐ=0.84).
- The average power of one solar panel is r*Ps
- In the solution, the first condition is the case of a charge limited accumulator. The second is an output power limited and third a charging power limited case.
- You can check this topic for more details on the functionment of solar panels and accumulators in factorio: viewtopic.php?f=5&t=5594
- Here is a mode that does all of the above for you: https://mods.factorio.com/mod/solar-calc
Cheers