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 = 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	~759.111kJ
	672kW + 7kW = 679kW	3s / 2.3125 = 48/37s	679kW * 48/37s = 32,592/37kJ	~880.865kJ
	693kW + 7kW = 700kW	3s / 1.5 = 2s	700kW * 2s = 1,400kJ	1,400.000kJ	1.2	(25MJ/2) * 1.2 - 1,400kJ = 13,600kJ	13,600.000kJ
	441kW + 7kW = 448kW	3s / 0.875 = 24/7s	448kW * 24/7s = 1,536kJ	1,536kJ
	714kW + 7kW = 721kW	3s / 0.6875 = 48/11s	721kW * 48/11s = 34,608/11kJ	~3,146.181kJ	1.3	(25MJ/2) * 1.3 - 34,608/11kJ = 144,142/11kJ	~13,103.818kJ

In a closed power loop, it is most efficient to convert light oil and petroleum gas into solid fuel with 2 productivity 3 modules and 1 speed 3 module.

Using beacons may further increase produced energy, even using 1 beacon for 2 chemical plants gains a little more. It is advisable to use beacons, since this process usually involves more facilities than other steps in this production chain. However, beaconed designs require more planning while placing and operating.

Main rule on using beacons is "full productivity on industry and full speed on beacons". There is some math, that proves it, see threads 1, 2, 3, 4

This tables shows the results of various beacon-per-industry ratios on some reasonable layouts (12 or less chemical plants)

Beacons per industry	Total beacons and industries count	Total modules effect	Energy cost	Time per cycle	Energy cost per cycle	Energy gained per cycle	Energy gained per cycle per 1 industry
1	1 2		2 * (861kW + 7kW) + 480kW = 2,216kW	3s / 1.3125 = 16/7s	2,216kW * 16/7s = 35,456/7kJ	2 * (25MJ/2) * 1.3 - 35,456/7kJ = 192,044/7kJ	192,044/7kJ / 2 = 96,022/7kJ = ~13,717.429kJ
	1 12		12*(861kW + 7kW) + 480kW = 10,896kW		10,896kW * 16/7s = 174,336/7kJ	12 * (25MJ/2) * 1.3 - 174,336/7kJ = 1,190,664/7kJ	1,190,664/7kJ / 12 = 99,222/7kJ = ~14,174.429kJ
2	2 6		6 *(1,008kW + 7kW) + 2 * 480kW = 7,050kW	3s / 1.9375 = 48/31s	7,050kW * 48/31s = 338,400/31kJ	6 * (25MJ/2) * 1.3 - 338,400/31kJ = 2,684,100/31kJ	2,684,100/31kJ / 6 = 447,350/31kJ = ~14,430.645kJ
	3 12		12 * (1,008kW + 7kW) + 3 * 480kW = 13,620kW		13,620kW * 48/31s = 653,760/31kJ	12 * (25MJ/2) * 1.3 - 653,760/31kJ = 5,391,240/31kJ	5,391,240/31kJ / 12 = 449,270/31kJ = ~14,492.581kJ
3	6 12	3 3	12 * (1155kW + 7kW) + 6 * 480kW = 16,824kW	3s / 2.5625 = 48/41s	16,824kW * 48/41s = 807,552/41kJ	12 * (25MJ/2) * 1.3 - 807,552/41kJ = 7,187,448/41kJ	7,187,448/41kJ / 12 = 598,954/41kJ = ~14,608.634kJ
4	9 12	3 4	12 * (1,302kW + 7kW) + 9 * 480kW = 20,028kW	3s / 3.1875 = 16/17s	20,028kW * 16/17s = 320,448/17kJ	12 * (25MJ/2) * 1.3 - 320,448/17kJ = 2,994,552/17kJ	2,944,552/17kJ / 12 = 249,546/17kJ = ~14,679.176kJ
some theoretical upper (non-reachable) values
4	infinite row of 1 2	3 4	2 * (1,302kW + 7kW) + 480kW = 3,098kW	3s / 3.1875 = 16/17s	3,098kW * 16/17s = 49,568/17kJ	2 * (25MJ/2) * 1.3 - 49,568/17kJ = 502,932/17kJ	502,932/17kJ / 2 = 251,466/17kJ = ~14,792.118kJ
8	infinite grid of 1 1	3 8	1,890kW + 7kW + 480kW = 2,377kW	3s / 5.6875 = 48/91s	2,377kW * 48/91s = 114,096/91kJ	(25MJ/2) * 1.3 - 114,096/91kJ = 1,364,654/91kJ	1,364,654/91kJ = ~14,996.198kJ

Heavy oil into light oil

Based on the above tables, 1 light oil will be given an energy worth of 680kJ, since this is the optimal amount of power that can be made when converting into solid fuel in non-beaconed setup.

Combinations without productivity modules are omitted, since the first combination produces more net energy per cycle than 30 units of light oil (20,400kJ) 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	680 * 33 - 8,400/17kJ = 373,080/17kJ	~21,945.882kJ
	693kW + 7kW = 700kW	3s / 1.5 = 2s	700kW * 2s = 1,400kJ	36	680kJ * 36 - 1,400kJ = 23,080kJ	23,080kJ
	714kW + 7kW = 721kW	3s / 0.6875 = 48/11s	721kW * 48/11s = 34,608/11kJ	39	680kJ * 39 - 34,608/11kJ = 257,112/11kJ	~23,373.818kJ

In a closed power loop, it is most efficient to convert heavy oil into light oil with 3 productivity 3 modules. Since optimal result includes maximum productivity modules, this conclusion still stand if you use greater value for light oil gained energy, e.g. from your favorite beaconed setup.

As for previous conversion, using beacons with speed 3 modules will further increase produced energy.

Beacons per industry	Layout and modules	Energy cost per cycle	Energy gained per cycle	Energy gained per cycle per 1 industry
1	1 x 2 + 2 x 3	35,456/7kJ	2 * 680 * 39 - 35,456/7kJ = 335,824/7kJ	167,912/7kJ = ~23,987.429kJ
	1 x 2 + 12 x 3	174,336/7kJ	12 * 680 * 39 - 174,336/7kJ = 2,053,344/7kJ	171,112/7kJ = ~24,444.571kJ
2	2 x 2 + 6 x 3	338,400/31kJ	6 * 680 * 39 - 338,400/31kJ = 4,594,320/31kJ	765,720/31kJ = ~24,700.645kJ
	3 x 2 + 12 x 3	653,760/31kJ	12 * 680 * 39 - 653,760/31kJ = 9,211,680/31kJ	767,640/31kJ = ~24,762.581kJ
3	6 x 2 + 12 x 3	807,552/41kJ	12 * 680 * 39 - 807,552/41kJ = 12,240,288/41kJ	1,020,024/41kJ = ~24,878.634kJ
4	9 x 2 + 12 x 3	320,448/17kJ	12 * 680 * 39 - 320,448/17kJ = 5,089,632/17kJ	424,136/17kJ = ~24,949.176kJ
	1 x 2 + 2 x 3 (infinite row)	49,568/17kJ	2 * 680 * 39 - 49,568/17kJ = 852,112/17kJ	426,056/17kJ = ~25,062.118kJ
8	1 x 2 + 1 x 3 (infinite grid)	114,096/91kJ	680 * 39 - 114,096/91kJ = 2,299,224/91kJ	2,299,224/91kJ = ~25,266.198kJ

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 = 32,139/55kJ (based on optimal non-beaconed conversion to light oil)
• Light oil = 680kJ
• Petroleum gas = 340kJ (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 = 192,834/11kJ
• 30 Light oil = 20,400kJ
• 40 Petroleum gas = 13,600kJ
• Total = 566,834/11kJ = ~51,530.364kJ

Advanced oil processing:

• 10 Heavy oil = 64,278/11kJ
• 45 Light oil = 30,600kJ
• 55 Petroleum gas = 18,700kJ
• Total = 606,578/11kJ = ~55,143.455kJ

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%	606,578/11kJ * 1.1 - 35,000/17kJ = 9,961,826/170kJ	~58,598.976kJ
	1,386kW + 14kW = 1,400kW	5s / 1.2 = 25/6s	1,400kW * 25/6s = 35,000/6kJ	20%	606,578/11kJ * 1.2 - 35,000/6kJ = 9,955,904/165kJ	~60,338.812kJ
	1,428kW + 14kW = 1,442kW	5s / 0.55 = 100/11s	1,442kW * 100/11s = 144,200/11kJ	30%	606,578/11kJ * 1.3 - 144,200/11kJ = 58,577.4kJ	58,577.4kJ

In a closed power loop, it is most efficient to convert crude oil into its products using 2 productivity 3 modules and 1 speed 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.

As for previous processes, beacons may be used and will further increase gained energy.

Beacons per industry	Layout and modules	Energy cost	Time per cycle	Energy cost per cycle	Energy gained per cycle	Energy gained per cycle per 1 industry
1	1 x 2 + 1 x 3	1,722kW + 14kW + 480kW = 2,216kW	5s / 1.05 = 100/21s	2,216kW * 100/21s = 221,600/21kJ	606,578/11kJ * 1.3 - 221,600/21kJ = 70,609,897/1,155kJ	70,609,897/1,155kJ = ~61,134.110kJ
	1 x 2 + 8 x 3	8 * (1,722kW + 14kW) + 480kW = 14,368kW		14,368kW * 100/21s = 1,436,800/21kJ	8 * 606,578/11kJ * 1.3 - 1,436,800/21kJ = 583,359,176/1,155kJ	72,919,897/1,155kJ = ~63,134.110kJ
2	2 x 2 + 4 x 3	4 * (2,016kW + 14kW) + 2 * 480kW = 9,080kW	5s / 1.55 = 100/31s	9,080kW * 100/31s = 908,000/31kJ	4 * 606,578/11kJ * 1.3 - 908,000/31kJ = 438,961,868/1,705kJ	109,740,467/1,705kJ = ~64,363.910kJ
3	3 x 2 + 4 x 3	4 * (2,310kW + 14kW) + 3 * 480kW = 10,736kW	5s / 2.05 = 100/41s	10,736kW * 100/41s = 1,073,600/41kJ	4 * 606,578/11kJ * 1.3 - 1,073,600/41kJ = 587,564,148/2,255kJ	146,891,037/2,255kJ = ~65,140.149kJ
4	5 x 2 + 4 x 3	4 * (2,604kW + 14kW) + 480kW = 12,872kW	5s / 2.55 = 100/51s	12,872kW * 100/51s = 1,287,200/51kJ	4 * 606,578/11kJ * 1.3 - 1,287,200/51kJ = 733,526,428/2,805kJ	183,381,607/2,805kJ = ~65,376.687kJ
5	9 x 2 + 6 x 3	6 * (2,898kW + 14kW) + 9 * 480kW = 21,792kW	5s / 3.05 = 100/61s	21,792kW * 100/61s = 2,179,200/61kJ	6 * 606,578/11kJ * 1.3 - 2,179,200/61kJ = 1,323,193,062/3,355kJ	220,532,177/3,355kJ = ~65,732.393kJ
	2 x 2 + 2 x 3 (infinite row)	2 * (2,898kW + 14kW) + 2 * 480kW = 6,784kW		6,784kW * 100/61s = 678,400/61kJ	2 * 606,578/11kJ * 1.3 - 678,400/61kJ = 443,704,354/3,355kJ	221,852,177/3,355kJ = ~66,125.835kJ
10	2 x 2 + 1 x 3 (infinite grid)	4,368kW + 14kW + 2 * 480kW = 5,342kW	5s / 5.55 = 100/111s	5,342kW * 100/111s = 534,200/111kJ	606,578/11kJ * 1.3 - 534,200/111kJ = 408,265,027/6,105kJ	408,265,027/6,105kJ = ~66,873.879kJ

Pumpjacks

Based on the above table, 100 crude oil will be given an energy worth of 9,955,904/165kJ, since this is the optimal amount of power that can be made when converting into solid fuel in non-beaconed setup.

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
	18kW	1s / 1 = 1s	18kW * 1s = 18kJ	0%	(9,955,904/165kJ * 1 - 18kJ) / 1s = 9,952,934/165kW	~60,320.812kW
	108kW	1s / 1.5 = 2/3s	108kW * 2/3s = 72kJ	0%	(9,955,904/165kJ * 1 - 72kJ) / 2/3s = 14,916,036/165kW	~90,400.218kW
	216kW	1s / 2 = 0.5s	216kW * 0.5s = 108kJ	0%	(9,955,904/165kJ * 1 - 108kJ) / 0.5s = 19,876,168/165kW	~120,461.624kW
	116kW	1s / 0.85 = 20/17s	116kW * 20/17s = 2,320/17kJ	10%	(9,955,904/165kJ * 1.1 - 2,320/17kJ) / 20/17s = 358,917,532/6,375kW	~56,300.789kW
	225kW	1s / 1.35 = 20/27s	225kW * 20/27s = 500/3kJ	10%	(9,955,904/165kJ * 1.1 - 500/3kJ) / 20/27s = 11,172,267/125kW	89,378.136kW
	234kW	1s / 0.7 = 10/7s	234kW * 10/7s = 2,340/7kJ	20%	(9,955,904/165kJ * 1.2 - 2,340/7kJ) / 10/7s = 69,369,578/1,375kW	~50,450.602kW

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
	336kW + 7kW = 343kW	30s / 0.875 = 240/7s	343kW * 240/7s = 11,760kJ	11,760.000kJ	1.2	(225MJ*1.2-250MJ)/2 - 11,760kJ = -1,760kJ	-1,760.000kJ
	588kW + 7kW = 595kW	30s / 1.5 = 20s	595kW * 20s = 11,900kJ	11,900.000kJ
	840kW + 7kW = 847kW	30s / 2.125 = 240/17s	847kW * 240/17s = 203,280/17kJ	~11,957.647kJ
	609kW + 7kW = 616kW	30s / 0.6875 = 480/11s	616kW * 480/11s = 26,880kJ	26,880.000kJ	1.3	(225MJ*1.3-250MJ)/2 - 19,840kJ = 1,410kJ	~1,410kJ
	861kW + 7kW = 868kW	30s / 1.3125 = 160/7s	868kW * 160/7s = 19,840kJ	19,840kJ
	882kW + 7kW = 889kW	30s / 0.5 = 60s	889kW * 60s = 53,340kJ	53,340.000kJ	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 speed 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.

As always, using beacons will further improve efficiency. This table shows optimal combinations for reasonable beacon setups (no more than 12 assemblers), and additionally, upper theoretical limits.

Beacons per industry	Beacons and assemblers	Modules effect	Energy cost	Time per cycle	Energy cost per cycle	Energy gained per cycle	Energy gained per cycle per 1 industry
1	1 + 2	2 3	2 * (1,008kW + 7kW) + 480kW = 2,510kW	30s / 1.9375 = 480/31s	2,510kW * 480/31s = 1,204,800/31kJ	2 * (225MJ*1.3-250MJ)/2 - 1,204,800/31kJ = 112,700/31kJ	56,350/31kJ = ~1,817.742kJ
	1 + 12		12 * (1,008kW + 7kW) + 480kW = 12,660kW		12,660kW * 480/31s = 6,076,800/31kJ	12 * (225MJ*1.3-250MJ)/2 - 6,076,800/31kJ = 1,828,200/31kJ	152,350/31kJ = ~4,914.516kJ
		1 4	12 * (1,029kW + 7kW) + 480kW = 12,912kW	30s / 1.125 = 80/3s	12,912kW * 80/3s = 344,320kJ	12 * (225MJ*1.4-250MJ)/2 - 344,320kJ = 45,680kJ	11,420/3kJ = ~3,806.667kJ
2	2 + 6	2 4	6 * (1,176kW + 7kW) + 2 * 480kW = 8,058kW	30s / 1.75 = 120/7s	8,058kW * 120/7s = 966,960/7kJ	6 * (225MJ*1.4-250MJ)/2 - 966,960/7kJ = 398,040/7kJ	66,340/7kJ = ~9,477.143kJ
	3 + 12		12 * (1,176kW + 7kW) + 3 * 480kW = 15,636kW		15,636kW * 120/7s = 1,876,320/7kJ	12 * (225MJ*1.4-250MJ)/2 - 1,876,320/7kJ = 853,680/7kJ	71,140/7kJ = ~10,162.857kJ
3	6 + 12	3 4	12 * (1,323kW + 7kW) + 6 * 480kW = 18,840kW	30s / 2.375 = 240/19s	18,840kW * 240/19s = 4,521,600/19kJ	12 * (225MJ*1.4-250MJ)/2 - 4,521,600/19kJ = 2,888,400/19kJ	240,700/19kJ = ~12,668.421kJ
4	9 + 12	4 4	12 * (1,470kW + 7kW) + 9 * 480kW = 22,044kW	30s / 3 = 10s	22,044kW * 10s = 220,440kJ	12 * (225MJ*1.4-250MJ)/2 - 220,440kJ = 169,560kJ	14,130kJ
	1 + 2 (infinite row)		2 * (1,470kW + 7kW) + 480kW = 3,434kW		3,434kW * 10s = 34,340kJ	2 * (225MJ*1.4-250MJ)/2 - 34,340kJ = 30,660kJ	15,330kJ
8	1 + 1 (infinite grid)	8 4	2,058kW + 7kW + 480kW = 2,545kW	30s / 5.5 = 60/11s	2,545kW * 60/11s = 152,700/11kJ	(225MJ*1.4-250MJ)/2 - 152,700/11kJ = 204,800/11kJ	204,800/11kJ = ~18,618.182kJ

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

See also

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