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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.
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.



Revision as of 13:37, 20 November 2017

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 gas and light oil will be used as-is for producing solid fuel. Light oil is not cracked since it takes twice as much petroleum gas to make one 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 Cost 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 ~494.117kJ
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 ~759.111kJ
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 ~871.783kJ
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
692kW + 7kW = 699kW 3s / 1.5 = 2s 699kW * 2s = 1,398kJ 1,398.000kJ
Solid fuel.png
1.2
(25MJ/2) * 1.2 - 1,398kJ = 12,102kJ 12,102.000kJ
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 ~1,532.571kJ
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 ~3,146.181kJ
Solid fuel.png
1.3
(25MJ/2) * 1.3 - 34,608/11kJ = 126,267/11kJ ~11,478.818.kJ

In a closed power loop, 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 30 units 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

In a closed power loop, 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

In a closed power loop, 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 all products are used for solid fuel production. If petroleum gas is being used for anything other than solid fuel, 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 is also important to note that pumpjacks are affected by mining productivity level. The higher the level, the less effective productivity modules become.

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 = 8,311,793,063/160,000kW ~51,948.706kW
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 = 232,314,803,901/2,720,000kW ~85,409.854kW
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 = 16,390,011,993/340,000kW ~48,205.917kW

In a closed power loop, it is most efficient to obtain crude oil using 2 speed 3 modules. This also improves with higher levels of productivity research.

Since there are a limited number of oil wells, it is advisable to use beacons in order to increase the amount of crude oil being collected. However, due to the nature of oil wells in the world and beacons affecting multiple pumpjacks at once, there will not be a table showing this.

Converting solid fuel into rocket fuel

Solid fuel can be converted into rocket fuel in order to increase the fuel value. Normally this would result in a loss since 10 solid fuel (250MJ) is worth more than 1 rocket fuel (225MJ), but productivity modules can be used to increase yield.

At least 2 productivity 3 modules must be used in order to increase yield, so combinations with fewer are omitted.

Modules Energy cost Time per cycle Energy cost per cycle Cost Rocket fuel per cycle Energy gained per cycle Result
Efficiency module 3.png
Efficiency module 3.png
Productivity module 3.png
Productivity module 3.png
336kW + 7kW = 343kW 30s / 0.875 = 240/7s 343kW * 240/7s = 11,760kJ 11,760.000kJ
Rocket fuel.png
1.2
(225MJ*1.2-250MJ)/2 - 11,760kJ = -760kJ -760.000kJ
Efficiency module 3.png
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
588kW + 7kW = 595kW 30s / 1.5 = 20s 595kW * 20s = 11,900kJ 11,900.000kJ
Speed module 3.png
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
840kW + 7kW = 847kW 30s / 2.125 = 240/17s 847kW * 240/17s = 203,280/17kJ ~11,957.647kJ
Efficiency module 3.png
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
609kW + 7kW = 616kW 30s / 0.6875 = 480/11s 616kW * 480/11s = 26,880kJ 26,880.000kJ
Rocket fuel.png
1.3
(225MJ*1.3-250MJ)/2 - 138,720/7kJ = 10,030/7kJ ~1,432.857kJ
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
860kW + 7kW = 867kW 30s / 1.3125 = 160/7s 867kW * 160/7s = 138,720/7kJ ~19,817.142kJ
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
882kW + 7kW = 889kW 30s / 0.5 = 60s 889kW * 60s = 53,340kJ 53,340.000kJ
Rocket fuel.png
1.4
(225MJ*1.4-250MJ)/2 - 53,340kJ = -20,840kJ -20,840.000kJ

In a closed power loop, it is most efficient to convert solid fuel rocket fuel with 1 efficiency 3 module and 3 productivity 3 modules. In fact, this is the only combination of modules that produces a net positive when accounting for boiler inefficiency.

This is also applicable for rocket fuel production for trains, however the results are different since locomotives are 100% fuel efficient.

See also