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	Cost	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	~494.117kJ	1.1	(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	759.111...kJ
	672kW + 7kW = 679kW	3s / 2.3125 = 48/37s	672kW * 48/37s = 32,256/37kJ	871.783783...kJ
	692kW + 7kW = 699kW	3s / 1.5 = 2s	699kW * 2s = 1,398kJ	1,398kJ	1.2	(25MJ/2) * 1.2 - 1,398kJ = 12,102kJ	12,102kJ
	440kW + 7kW = 447kW	3s / 0.875 = 24/7s	447kW * 24/7s = 10,728/7kJ	~1,532.571kJ
	714kW + 7kW = 721kW	3s / 0.6875 = 48/11s	721kW * 48/11s = 34,608/11kJ	3,146.1818...kJ	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 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	33	(22,535/34kJ) * 33 - 8,400/17kJ = 726,855/34kJ	~21,378.088kJ
	692kW + 7kW = 699kW	3s / 1.5 = 2s	699kW * 2s = 1,398kJ	36	(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	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
	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	1.2	(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	1.3	(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	1.4	(225MJ*1.4-250MJ)/2 - 53,340kJ = -20,840kJ	-20,840kJ

As shown, 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.

See also

• Crude oil
• Solid fuel
• Rocket fuel
• Electric system