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Tutorial:Producing power from oil

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Oil can be converted into solid fuel (and by extension rocket fuel), which when used to produce power will result in a net profit of power at the cost of oil.

Energy costs and modules

Power cost and power results will be worked out in reverse, with the result that gives the most power being used for each step thereafter.

Light oil and petroleum gas into solid fuel

Petroleum and light oil will be used as-is for producing solid fuel. This table shows the results of various module combinations for a single cycle of the chemical plant for either light oil or petroleum. Since the solid fuel is being used in a closed loop, and therefore is going into boilers, the 25MJ fuel value is halved when used.

Combinations without productivity modules are omitted, since the first combination produces more net energy per cycle than a single piece of solid fuel is worth.

Combinations for each number of productivity modules show their best combination in bold, and only that combination is used to work out energy gained per cycle.

Modules Energy cost Time per cycle Energy cost per cycle Solid fuel per cycle Energy gained per cycle Result
Efficiency module 3.png
Efficiency module 3.png
Productivity module 3.png
168kW + 7kW = 175kW 3s / 1.0625 = ~2.824s = 48/17s 175kW * 48/17s = 8,400/17kJ
Solid fuel.png
1.1
(25MJ/2) * 1.1 - 8,400/17kJ = 225,350/17kJ ~13,255.882kJ
Efficiency module 3.png
Speed module 3.png
Productivity module 3.png
420kW + 7kW = 427kW 3s / 1.687 = 16/9s 427kW * 16.9s = 6,832/9kJ
Speed module 3.png
Speed module 3.png
Productivity module 3.png
672kW + 7kW = 679kW 3s / 2.3125 = 48/37s 672kW * 48/37s = 32,256/37kJ
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
692kW + 7kW = 699kW 3s / 1.5 = 2s 699kW * 2s = 1,398kJ
Solid fuel.png
1.2
(25MJ/2) * 1.2 - 1,398kJ = 12,102kJ 12,102kJ
Efficiency module 3.png
Productivity module 3.png
Productivity module 3.png
440kW + 7kW = 447kW 3s / 0.875 = 24/7s 447kW * 24/7s = 10,728/7kJ
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
714kW + 7kW = 721kW 3s / 0.6875 = 48/11s 721kW * 48/11s = 34,608/11kJ
Solid fuel.png
1.3
(25MJ/2) * 1.3 - 34,608/11kJ = 126,267/11kJ 11,478.8181...kJ

As shown, it is most efficient to convert light oil and petroleum gas into solid fuel with 2 efficiency 3 modules and 1 productivity 3 module.

Heavy oil into light oil

Based on the above table, 1 light oil will be given an energy worth of 22,535/34kJ, since this is the optimal amount of power that can be made when converting into solid fuel.

Combinations without productivity modules are omitted, since the first combination produces more net energy per cycle than a single piece of light oil (~19,883.823kJ) is worth.

Since energy costs per cycle will be the same as above (same machine), only the optimal combination per number of productivity modules will be shown.

Modules Energy cost Time per cycle Energy cost per cycle Light oil per cycle Energy gained per cycle Result
Efficiency module 3.png
Efficiency module 3.png
Productivity module 3.png
168kW + 7kW = 175kW 3s / 1.0625 = 48/17s 175kW * 48/17s = 8,400/17kJ
Light oil.png
33
(22,535/34kJ) * 33 - 8,400/17kJ = 726,855/34kJ ~21,378.088kJ
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
692kW + 7kW = 699kW 3s / 1.5 = 2s 699kW * 2s = 1,398kJ
Light oil.png
36
(22,535/34kJ) * 36 - 1,398kJ = 381,864/17kJ ~22,462.588kJ
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
714kW + 7kW = 721kW 3s / 0.6875 = 48/11s 721kW * 48/11s = 34,608/11kJ
Light oil.png
39
(22,535/34kJ) * 39 - 34,608/11kJ = 8,490,843/374kJ ~22,702.788kJ

As shown, it is most efficient to convert heavy oil into light oil with 3 productivity 3 modules.

Basic vs Advanced oil processing

Crude oil can be processed with either basic or advanced oil processing. Based on the above tables, the following fuel values for each product will be used:

  • Heavy oil = 499,459/880kJ
  • Light oil = 22,535/34kJ
  • Petroleum gas = 22,535/68kJ (half of light oil)

Since all products scale equally based on productivity, each recipe can be expressed solely as the fuel value of the products combined and that value can be scaled based on productivity below.

Basic oil processing:

  • 30 Heavy oil = 1,498,377/80kJ
  • 30 Light oil = 338,025/17kJ
  • 40 Petroleum gas = 225,350/17kJ
  • Total = 70,542,409/1,360kJ = ~51,869.418kJ

Advanced oil processing:

  • 10 Heavy oil = 499,459/80kJ
  • 45 Light oil = 1,014,075/34kJ
  • 55 Petroleum gas = 1,239,425/68kJ
  • Total = 73,842,303/1,360kJ = ~54,295.811kJ

Since advanced oil processing produces more overall, its total fuel value will be used.

Combinations without productivity modules are omitted, since the first combination produces more net energy per cycle than the total fuel value.

Since energy costs per cycle will be the same as above but scaled (same module slot count), only the optimal combination per number of productivity modules will be shown.

Modules Energy cost Time per cycle Energy cost per cycle Productivity level Energy gained per cycle Result
Efficiency module 3.png
Efficiency module 3.png
Productivity module 3.png
336kW + 14kW = 350kW 5s / 0.85 = 100/17s 350kW * 100/17s = 35,000/17kJ 10% 73,842,303/1,360kJ * 1.1 - 35,000/17kJ = 784,265,333/13,600kJ ~57,666.568kJ
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
1,386kW + 14kW = 1,400kW 5s / 1.2 = 25/6s 1,400kW * 25/6s = 35,000/6kJ 20% 73,842,303/1,360kJ * 1.2 - 35,000/6kJ = 2,198,795,999/40,800kJ ~53,892.058kJ
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
1,428kW + 14kW = 1,442kW 5s / 0.55 = 100/11s 1,442kW * 100/11s = 144,200/11kJ 30% 73,842,303/1,360kJ * 1.3 - 144,200/11kJ = 8,598,329,329/149,600kJ ~57,475.463kJ

As shown, it is most efficient to convert crude oil into its products using 2 efficiency 3 modules and 1 productivity 3 module.

This only applies if you use all products for solid fuel production. If you want to use petroleum gas for other means, the optimal combination might change.

Pumpjacks

Based on the above table, 100 crude oil will be given an energy worth of 784,265,333/13,600kJ, since this is the optimal amount of power that can be made when converting into solid fuel.

Results will be given for a depleted oil well, which provides 2 crude oil per second. As the amount of crude oil increases, the importance of optimal modules decreases since the power draw for a given amount of oil output also decreases. Using the minimum amount is important to prove that creating power from crude oil is always possible.

It's also important to note that pumpjacks are affected by mining productivity level. The higher the level, the less effective productivity modules become. Levels 0 (+0%) and 25 (+50%) are shown in separate tables.

Since pumpjacks operate on an infinite resource that has a finite count (oil wells), results will be shown in kW instead of kJ, since the goal here is to produce as much power as possible.

Pumpjacks only have two module slots, so all combinations will be shown. In this instance, results cannot be grouped by number of productivity modules, as the speed is also important.

Modules Energy cost Time per cycle Energy per cycle Productivity level Energy gained per second Result
Efficiency module 3.png
Efficiency module 3.png
18kW 1s / 1 = 1s 18kW * 1s = 18kJ 0% (784,265,333/13,600kJ * 1 - 18kJ) / 1s = 784,020,533/13,600kW ~57,648.568kW
Efficiency module 3.png
Speed module 3.png
108kW 1s / 1.5 = 2/3s 108kW * 2/3s = 72kJ 0% (784,265,333/13,600kJ * 1 - 72kJ) / 2/3s = 2,349,858,399/27,200kW ~86,391.852kW
Speed module 3.png
Speed module 3.png
216kW 1s / 2 = 0.5s 216kW * 0.5s = 108kW 0% (784,265,333/13,600kJ * 1 - 108kJ) / 0.5s = 783,530,933/6,800kW ~115,225.137kW
Efficiency module 3.png
Productivity module 3.png
116kW 1s / 0.85 = 20/17s 116kW * 20/17s = 2,320/17kJ 10% (784,265,333/13,600kJ * 1.1 - 2,320/17kW) / 20/17s =
Speed module 3.png
Productivity module 3.png
225kW 1s / 1.35 = 20/27s 225kW * 20/27s = 500/3kJ 10% (784,265,333/13,600kJ * 1.1 - 500/3kW) / 20/27s =
Productivity module 3.png
Productivity module 3.png
234kW 1s / 0.7 = 10/7s 234kW * 10/7s = 2,340/7kJ 20% (784,265,333/13,600kJ * 1.2 - 2,340/7kW) / 10/7s =

Converting solid fuel into rocket fuel

TODO: Show that only 3x prod3 and 1x speed3 is profitable.

See also