Oil Cracking Unit: Difference between revisions

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{{Infobox Test
{{cleanup}}
|name=Oil Cracking Unit
|image=OilCrackingUnit.png|alt= A five block long, three block deep, three block tall multiblock made from light grey-blue metallic blocks with rivets. The second and fourth row are copper blocks with diagonal stripes instead. Front and center is an electronic panel with red light, and a light grey hatch marked with a wrench directly below. On the left and right sides are two hatches with four blue arrows pointing in and out. Another hatch with arrows pointing in is centered on the top, between the copper coils.
|mod=Gregtech 5
|type=Tile Entity
|tooltip= Controller Block for the Oil Cracking Unit
|quest=Oil Cracker<div/>Let's Get Crackin!
|tier=HV
|machinesize=5x3x3
|pollutionoutput= None
|blastresistance= 6
|hardness= 1-5
|gteuusage=240+ eu/t
|gtvoltagein= As Energy Hatch
|gtamperage=1-2A per Energy Hatch
}}
== Introduction ==
The '''Oil Cracking Unit''' is a specialized [[multiblock]] that uses Steam or Hydrogen to crack oil-based fuels. Compared to a [[Chemical Reactor]], an Oil Cracking Unit gets an additional 25-50% output bonus for the same recipes but runs them at a higher [[tier]] with similar EU cost/speed. Steam has a 50% bonus output, Hydrogen 25% bonus and 20% less Hydrogen compared to a Chemical Reactor. The Oil Cracking Unit can be further upgraded with improved [[coil]]s and has an advanced variant in the [[Mega Oil Cracker]]. Beyond power production, this multiblock can make Toluene, a component of [[Rocket]] fuel, crack Radox, and break down Naquadah into various noble gasses / radioactive elements.
The Gregtech Oil Cracking Unit is a multi-block that thermally cracks heavy hydrocarbons into lighter fractions. Unlike cracking with a chemical reactor, which cracks with 80% efficiency, an Oil Cracking Unit will quickly crack with 100% efficiency.
 
== How to BuildConstruction ==
An Oil Cracking Unit is a hollow 5x3x3 (WxHxD) structure that requires:
-1* x1 Controller block; front, centered along one of the 5 block wide sides
-1* x1 Maintenance Hatch; anywhereany casing
-1* x1+ Energy hatch; anywhereany casing
-1* x1+ Input Bus/Hatch for the hydrocarbonoil to be crackedproduct; any left/right side (1x3x3) ring casings. This must be on the ringside opposite of the ringproduct withOutput the cracked hydrocarbon output hatchHatch
-1* x1+ Output Hatch for the crackedoil hydrocarbonproduct; any left/right side (1x3x3) ring casings. This must be on the ringside opposite of the ringproduct withInput the hydrocarbon input hatch.Hatch
-1* x1+ Input Hatch for Steam/Hydrogen input; any middle ring (1x3x3) casings
-2* Setsx16 of 8x Cupronickelidentical Coil Blocks,blocks; forming atwo hollow ringvertical onrings eachtouching immediateeither side of the controller
* x18-21 Clean Stainless Steel Machine Casings; everywhere else
 
Note that theThe Oil Cracking Unit, like many other multiblocks, can [[Wallsharing|share walls]] with other Oil Cracking Units, or technically with anything that uses the same parts. A single input/output block of each type is recommended as there is very little allowance for extra non-casing blocks. Coils must all be of the same type or an incomplete structure will result.
 
== Usage ==
The Oil Cracking Unit must have a steady supply of Steam or Hydrogen to work. If there are no [[Steam_Age#Boilers|Steam Boiler]]s available, a [[Fluid Heater]] can make steam from EU and water. IC2 steam is not compatible with this multiblock. Lacking an Input Bus, circuits must be set with a Programmed Circuit item, which is placed in the bottom-right slot of the Controller. Upgraded [[coil]]s reduce power consumption by 10% per tier, up to a maximum discount of 50% with HSS-G Coil Blocks. In practice, this overclocks and will make recipes run faster for about the same power consumption.
 
== Maximizing Output Calculation and Conclusion ==
This section is the challenging part of mathematical theory! Here, we will use the language of linear algebra to calculate the best output method, providing a theoretical basis for petrochemical production lines. If you wish to skip the theoretical part, please moveread tothrough thenon-maths mainpart conclusionsand inthen move to the Petrochemicalcalculator section below.
 
We know that each type of desulfurized fuel can be cracked and distilled into chemicals and two other types of fuels. Therefore, after one cracking and distillation, the fuel is not entirely consumed, and we need to calculate the cracking of the remaining fuel by using linear algebra.
 
The initial approach, which is to maximize the panel output of the target product from each type of oil shown on NEI, might not be the best combination of cracking. We will represent the proportions of the three types of fuel cracking each other using matrices up to the third order. By using matrix multiplication iteratively and summing infinitely, we can calculate the real maximum output. If heavy fuel is distilled into toluene, the matrix becomes of the second order (the target product should not be toluene). If light fuel is stirred to generate diesel for electricity, the matrix degenerates to a scalar.
 
This article is only a theoretical calculation, valuable for reference, butwith therelots isof nodetailed needelements not considered. Please still try to completelydesign your own petrochemical line mimicthat thesuits methodsyour describedown heresave.
A critical point to mention is that heavy cracking under the same pressure slows down the cracking speed. It generally does not affect the use of crude oil; the ultimate bottleneck is still in crude oil distillation. However, for heavy oil, careful planning is needed. If using heavy cracking for heavy fuel, a single cracking unit might not be fast enough. However, heavy distillation of heavy fuel is almost non-existent. If your cracking unit is lagging, please consider increasing the pressure or adding more electrical equipment.
 
The following mathematical theory may be too hard to comprehend. With the general principle understood, it's suggested to directly use the calculator below if you're not into complicated maths.
This article is only a theoretical calculation, valuable for reference, but there is no need to completely mimic the methods described here.
 
=== Variable Definitions and Interpretations ===
First, let's define each variable:
 
<math>\kappa</math>: The ratio of the final product from cracking and distillation per barrel of crude oil (not necessarily petroleum) under a specific combination of cracking methods.
 
<math>\delta</math>: The crude oil distillation ratio.
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<math>\alpha</math>: The yield rate of the target product from each type of fuel under a specific cracking combination.
 
Here, κ is a scalar, while δ and α are column vectors up to three dimensions. Note that we generally use the row vector of δ^T. In this case, we default the order from light to heavy—namely, naphtha, light fuel, and heavy fuel. This order may differ from previous tables, so please pay attention.
 
 
1B of BC crude oil can yield 0.4B of naphtha, 1B of light fuel, and 0.3B of heavy fuel, therefore,
<math>\delta^T = \left( \begin{array}{c} 0.4 \\ 1 \\ 0.3 \end{array} \right)</math>
 
The other parameters are similar, and readers can refer to the previous tables for understanding. In this example, all cracking methods are heavy steam, with the target product being ethylene.
 
If the definitions seem to be not clear enough, just download the calculator.
=== Calculation Formula ===
Now, let's give the specific calculation formula. After inputting a certain amount (a large amount) of a type of crude oil, the first cracking and distillation are performed:
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The conversion ratio of the target product in this instance is
<math>\kappa_0 = \delta^T \alpha</math>
The, and the remaining fuel ratio is
<math>\delta^T A</math>
 
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The conversion ratio of the target product up to this point is
<math>\kappa_1 = \kappa_0 + \delta^T A \alpha = \delta^T (I + A) \alpha</math>
The, and now the remaining fuel ratio is
<math>\delta^T A^2</math>
 
This process can be repeated many times, theoretically infinitely, leading to
<math>\kappa_n = \delta^T \sum_{i=0}^{n} A^i \alpha</math>
 
Taking n to infinity, we get
<math>\kappa = \delta^T \sum_sum^{\infty}^_{n=0} A^n \alpha</math>
 
The infinite sum is similar to a geometric series, and using Taylor's theorem for multivariate functions, we can obtain
<math>\sum^{\infty}_{n=0} A^n = (I - A)^{-1}</math> (Note: The negative exponent represents the matrix inversion operation.)
 
(Note: The negative exponent represents the matrix inversion operation.)
 
 
Thus, the formula simplifies to
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We need to maximize κ under all conditions that meet the constraints (such as not using hydrogen cracking or heavy cracking, etc.). Please note, the type of oil might affect the outcome, but we have not encountered this situation yet.
 
 
When stirring to make diesel but not directly distilling heavy fuel, the fuel cracking matrix A changes. This situation might occur when using heavy oil for electricity generation and cracking heavy fuel to obtain petrochemical by-products in the early to mid-stages. In such cases, the light fuel output in the matrix should be set to zero, and the heavy fuel should be reduced by 0.2 times the light fuel output. Similarly, in δ (crude oil distillation), the light fuel output and 0.2 times the light fuel output in heavy fuel should be removed.
You can also modify the matrix to get results under different constraints.
 
 
WhenFor example, when stirring to make diesel but not directly distilling heavy fuel, the fuel cracking matrix A changes. This situation might occur when using heavy oil for electricity generation and cracking heavy fuel to obtain petrochemical by-products in the early to mid-stages. In such cases, the light fuel output in the matrix should be set to zero, and the heavy fuel should be reduced by 0.2 times the light fuel output. Similarly, in δ (crude oil distillation), the light fuel output and 0.2 times the light fuel output in heavy fuel should be removed.
 
For example, in the matrix of the example, if this operation is performed, then:
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<math>A = \left( \begin{array}{ccc} 0 & 0.1 - 0.1 & 0.125 - 0.1 \times 0.2 \\ 0 & 0 - 0 & 0.1 - 0 \times 0.2 \\ 0.05 & 0 - 0.05 & 0 - 0.05 \times 0.2 \end{array} \right) = \left( \begin{array}{ccc} 0 & 0 & 0.105 \\ 0 & 0 & 0.1 \\ 0.025 & 0 & -0.01 \end{array} \right)</math>
 
(in old version without buffs on OCU)
As light fuel does not produce any chemical products, we can eliminate the second row and second column, making it a second-order matrix and completely ignoring light fuel in subsequent calculations (as it's no longer present).
 
As light fuel does not produce any chemical products, we can eliminate the second row and second column, making it a second-order matrix and completely ignoring light fuel in subsequent calculations (as it's no longer present). But this doesn't really matter.
===== Calculator =====
 
===== Calculator =====
Well, there's a calculator which was mainly contributed by core.exe: [https://github.com/core-exe/GTNHPetroOptim GTNHPetroOptim]
 
It needs Python to run. Feel free to manipulate the parameters in the code and derived the result you want by yourself.
 
[[Category:Multiblocks]][[Category:Gregtech]]
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