Tutorial:Producing power from 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.
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 | Solid fuel per cycle | Energy gained per cycle | Result |
---|---|---|---|---|---|---|
168kW + 7kW = 175kW | 3s / 1.0625 = ~2.824s = 48/17s | 175kW * 48/17s = 8,400/17kJ | (25MJ/2) * 1.1 - 8,400/17kJ = 225,350/17kJ | ~13,255.882kJ | ||
420kW + 7kW = 427kW | 3s / 1.687 = 16/9s | 427kW * 16.9s = 6,832/9kJ | ||||
672kW + 7kW = 679kW | 3s / 2.3125 = 48/37s | 672kW * 48/37s = 32,256/37kJ | ||||
692kW + 7kW = 699kW | 3s / 1.5 = 2s | 699kW * 2s = 1,398kJ | (25MJ/2) * 1.2 - 1,398kJ = 12,102kJ | 12,102kJ | ||
440kW + 7kW = 447kW | 3s / 0.875 = 24/7s | 447kW * 24/7s = 10,728/7kJ | ||||
714kW + 7kW = 721kW | 3s / 0.6875 = 48/11s | 721kW * 48/11s = 34,608/11kJ | (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 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 |
---|---|---|---|---|---|---|
168kW + 7kW = 175kW | 3s / 1.0625 = 48/17s | 175kW * 48/17s = 8,400/17kJ | (22,535/34kJ) * 33 - 8,400/17kJ = 726,855/34kJ | ~21,378.088kJ | ||
692kW + 7kW = 699kW | 3s / 1.5 = 2s | 699kW * 2s = 1,398kJ | (22,535/34kJ) * 36 - 1,398kJ = 381,864/17kJ | ~22,462.588kJ | ||
714kW + 7kW = 721kW | 3s / 0.6875 = 48/11s | 721kW * 48/11s = 34,608/11kJ | (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 |
---|---|---|---|---|---|---|
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 | |
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 | |
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.
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 |
---|---|---|---|---|---|---|
18kW | 1s / 1 = 1s | 18kW * 1s = 18kJ | 0% | (784,265,333/13,600kJ * 1 - 18kJ) / 1s = 784,020,533/13,600kW | ~57,648.568kW | |
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 | |
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 | |
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 | |
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 | |
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 |
As shown, it is most efficient to obtain crude oil using 2 speed 3 modules. This only improves with higher levels of productivity research.
Since you will only have a limited number of oil wells, it's 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 | Rocket fuel per cycle | Energy gained per cycle | Result |
---|---|---|---|---|---|---|
336kW + 7kW = 343kW | 30s / 0.875 = 240/7s | 343kW * 240/7s = 11,760kJ | (225MJ*1.2-250MJ)/2 - 11,760kJ = -760kJ | -760kJ | ||
588kW + 7kW = 595kW | 30s / 1.5 = 20s | 595kW * 20s = 11,900kJ | ||||
840kW + 7kW = 847kW | 30s / 2.125 = 240/17s | 847kW * 240/17s = 203,280/17kJ | ||||
609kW + 7kW = 616kW | 30s / 0.6875 = 480/11s | 616kW * 480/11s = 26,880kJ | (225MJ*1.3-250MJ)/2 - 138,720/7kJ = 10,030/7kJ | ~1,432.857kJ | ||
860kW + 7kW = 867kW | 30s / 1.3125 = 160/7s | 867kW * 160/7s = 138,720/7kJ | ||||
882kW + 7kW = 889kW | 30s / 0.5 = 60s | 889kW * 60s = 53,340kJ | (225MJ*1.4-250MJ)/2 - 53,340kJ = -20,840kJ |
As shown, it is most efficient to convert solid fuel rocket fuel with 1 efficiency 3 module and 3 productivity 3 modules.