Oil Cracking Unit: Difference between revisions
Some tedious maths.
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(Some tedious maths.) |
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== 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
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
This article is only a theoretical calculation, valuable for reference,
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
<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
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>
<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>
<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 \
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.)
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.
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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.
Well, there's a calculator which was mainly contributed by core.exe: [https://github.com/core-exe/GTNHPetroOptim GTNHPetroOptim]
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