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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 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 = 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 679kW × 48/37s = 32,592/37kJ ~880.865kJ
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
693kW + 7kW = 700kW 3s / 1.5 = 2s 700kW × 2s = 1,400kJ 1,400.000kJ
Solid fuel.png
1.2
(25MJ/2) × 1.2 - 1,400kJ = 13,600kJ 13,600.000kJ
Efficiency module 3.png
Productivity module 3.png
Productivity module 3.png
441kW + 7kW = 448kW 3s / 0.875 = 24/7s 448kW × 24/7s = 1,536kJ 1,536kJ
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 = 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
Beacon.png
1
Chemical plant.png
2
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
Speed module 3.png
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
Beacon.png
1
Chemical plant.png
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
Beacon.png
2
Chemical plant.png
6
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
Speed module 3.png
Speed module 3.png
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
Beacon.png
3
Chemical plant.png
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
Beacon.png
6
Chemical plant.png
12
Productivity module 3.png
3
Speed module 3.png
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
Beacon.png
9
Chemical plant.png
12
Productivity module 3.png
3
Speed module 3.png
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
Beacon.png
1
Chemical plant.png
2
Productivity module 3.png
3
Speed module 3.png
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
Beacon.png
1
Chemical plant.png
1
Productivity module 3.png
3
Speed module 3.png
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
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
680 × 33 - 8,400/17kJ = 373,080/17kJ ~21,945.882kJ
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
693kW + 7kW = 700kW 3s / 1.5 = 2s 700kW × 2s = 1,400kJ
Light oil.png
36
680kJ × 36 - 1,400kJ = 23,080kJ 23,080kJ
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
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
Beacon.png
1
x
Speed module 3.png
2
+
Chemical plant.png
2
x
Productivity module 3.png
3
35,456/7kJ 2 × 680 × 39 - 35,456/7kJ = 335,824/7kJ 167,912/7kJ = ~23,987.429kJ
Beacon.png
1
x
Speed module 3.png
2
+
Chemical plant.png
12
x
Productivity module 3.png
3
174,336/7kJ 12 × 680 × 39 - 174,336/7kJ = 2,053,344/7kJ 171,112/7kJ = ~24,444.571kJ
2
Beacon.png
2
x
Speed module 3.png
2
+
Chemical plant.png
6
x
Productivity module 3.png
3
338,400/31kJ 6 × 680 × 39 - 338,400/31kJ = 4,594,320/31kJ 765,720/31kJ = ~24,700.645kJ
Beacon.png
3
x
Speed module 3.png
2
+
Chemical plant.png
12
x
Productivity module 3.png
3
653,760/31kJ 12 × 680 × 39 - 653,760/31kJ = 9,211,680/31kJ 767,640/31kJ = ~24,762.581kJ
3
Beacon.png
6
x
Speed module 3.png
2
+
Chemical plant.png
12
x
Productivity module 3.png
3
807,552/41kJ 12 × 680 × 39 - 807,552/41kJ = 12,240,288/41kJ 1,020,024/41kJ = ~24,878.634kJ
4
Beacon.png
9
x
Speed module 3.png
2
+
Chemical plant.png
12
x
Productivity module 3.png
3
320,448/17kJ 12 × 680 × 39 - 320,448/17kJ = 5,089,632/17kJ 424,136/17kJ = ~24,949.176kJ
Beacon.png
1
x
Speed module 3.png
2
+
Chemical plant.png
2
x
Productivity module 3.png
3
(infinite row)
49,568/17kJ 2 × 680 × 39 - 49,568/17kJ = 852,112/17kJ 426,056/17kJ = ~25,062.118kJ
8
Beacon.png
1
x
Speed module 3.png
2
+
Chemical plant.png
1
x
Productivity module 3.png
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
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% 606,578/11kJ × 1.1 - 35,000/17kJ = 9,961,826/170kJ ~58,598.976kJ
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% 606,578/11kJ × 1.2 - 35,000/6kJ = 9,955,904/165kJ ~60,338.812kJ
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% 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
Beacon.png
1
x
Speed module 3.png
2
+
Oil refinery.png
1
x
Productivity module 3.png
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
Beacon.png
1
x
Speed module 3.png
2
+
Oil refinery.png
8
x
Productivity module 3.png
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
Beacon.png
2
x
Speed module 3.png
2
+
Oil refinery.png
4
x
Productivity module 3.png
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
Beacon.png
3
x
Speed module 3.png
2
+
Oil refinery.png
4
x
Productivity module 3.png
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
Beacon.png
5
x
Speed module 3.png
2
+
Oil refinery.png
4
x
Productivity module 3.png
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
Beacon.png
9
x
Speed module 3.png
2
+
Oil refinery.png
6
x
Productivity module 3.png
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
Beacon.png
2
x
Speed module 3.png
2
+
Oil refinery.png
2
x
Productivity module 3.png
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
Beacon.png
2
x
Speed module 3.png
2
+
Oil refinery.png
1
x
Productivity module 3.png
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
Efficiency module 3.png
Efficiency module 3.png
18kW 1s / 1 = 1s 18kW × 1s = 18kJ 0% (9,955,904/165kJ × 1 - 18kJ) / 1s = 9,952,934/165kW ~60,320.812kW
Efficiency module 3.png
Speed module 3.png
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
Speed module 3.png
Speed module 3.png
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
Efficiency module 3.png
Productivity module 3.png
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
Speed module 3.png
Productivity module 3.png
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
Productivity module 3.png
Productivity module 3.png
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
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 = -1,760kJ -1,760.000kJ
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 - 19,840kJ = 1,410kJ ~1,410kJ
Speed module 3.png
Productivity module 3.png
Productivity module 3.png
Productivity module 3.png
861kW + 7kW = 868kW 30s / 1.3125 = 160/7s 868kW × 160/7s = 19,840kJ 19,840kJ
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 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
Beacon.png
1
+
Assembling machine 3.png
2
Speed module 3.png
2
Productivity module 3.png
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
Beacon.png
1
+
Assembling machine 3.png
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
Speed module 3.png
1
Productivity module 3.png
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
Beacon.png
2
+
Assembling machine 3.png
6
Speed module 3.png
2
Productivity module 3.png
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
Beacon.png
3
+
Assembling machine 3.png
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
Beacon.png
6
+
Assembling machine 3.png
12
Speed module 3.png
3
Productivity module 3.png
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
Beacon.png
9
+
Assembling machine 3.png
12
Speed module 3.png
4
Productivity module 3.png
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
Beacon.png
1
+
Assembling machine 3.png
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
Beacon.png
1
+
Assembling machine 3.png
1
(infinite grid)
Speed module 3.png
8
Productivity module 3.png
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