CN104297272B

CN104297272B - A kind of CT characterizing method of coal directly-liquefied residue sample

Info

Publication number: CN104297272B
Application number: CN201410633492.8A
Authority: CN
Inventors: 王海鹏; 杨玉双; 蒋兴家; 杨建丽; 聂行; 聂一行
Original assignee: Shanxi University
Current assignee: Shanxi University
Priority date: 2014-11-11
Filing date: 2014-11-11
Publication date: 2016-09-28
Anticipated expiration: 2034-11-11
Also published as: CN104297272A

Abstract

The invention relates to a visual characterization technology for the distribution of organic components of a direct coal liquefaction residue sample, in particular to a CT characterization method for a direct coal liquefaction residue sample. A CT characterization method of a coal direct liquefaction residue sample, comprising the following steps: the first step, sampling and pre-testing of the coal direct liquefaction residue sample; the second step, the second step, the X-ray absorption characteristics of each ash component of the coal direct liquefaction residue sample Analysis; the third step, the second coal direct liquefaction residue sample component grouping; the fourth step, the CT experiment of the second coal direct liquefaction residue sample; the fifth step, the establishment of a physical model to distinguish and identify different components in the CT slice ; The sixth step is to remove the inorganic components, further distinguish the organic components, and use different colors to display the distribution of different organic components in the model, so as to realize the visual representation of the distribution of organic components in the direct coal liquefaction residue sample. The physical model adopted in the present invention can greatly improve the calculation efficiency, and can be used for the characterization of a large number of samples.

Description

A method for CT characterization of coal direct liquefaction residue samples

技术领域technical field

本发明涉及煤直接液化残渣样品有机组分分布的可视化表征技术，具体是一种煤直接液化残渣样品的CT表征方法。The invention relates to a visual characterization technology for the distribution of organic components of a direct coal liquefaction residue sample, in particular to a CT characterization method for a direct coal liquefaction residue sample.

背景技术Background technique

煤直接液化残渣是煤炭液化后的产物，其成分可分为有机组分和无机组分两大类。其中有机组分包含沥青类物质、重油及未转化的煤基质；无机组分包含原煤中的矿物质和液化过程中加入的催化剂。煤直接液化残渣样品中的有机组分具有较高的发热量和利用价值，对煤直接液化残渣样品有机组分的分布形式和结合方式进行可视化表征有助于发展新的残渣中有用物质的回收方法。Coal direct liquefaction residue is the product of coal liquefaction, and its components can be divided into two categories: organic components and inorganic components. The organic components include bituminous substances, heavy oil and unconverted coal matrix; the inorganic components include minerals in raw coal and catalysts added in the liquefaction process. The organic components in coal direct liquefaction residue samples have high calorific value and utilization value, and the visual characterization of the distribution and combination of organic components in coal direct liquefaction residue samples is helpful for the development of new recovery of useful substances in residues method.

目前煤直接液化残渣样品的可视化表征技术主要包括：扫描电镜法、光学显微镜法等。扫描电镜结合能谱可以对感兴趣的样品区域中的元素分布进行可视化鉴别，光学显微镜可以较为便捷地观察样品不同组分的形貌。但是无论是扫描电镜还是光学显微镜，都很难观察煤直接液化残渣样品中不同有机组分的分布形式和相互结合方式。At present, the visual characterization techniques of coal direct liquefaction residue samples mainly include: scanning electron microscopy, optical microscopy, etc. The scanning electron microscope combined with the energy spectrum can visually identify the element distribution in the sample area of interest, and the optical microscope can conveniently observe the morphology of different components of the sample. However, whether it is scanning electron microscope or optical microscope, it is difficult to observe the distribution and mutual combination of different organic components in the direct coal liquefaction residue sample.

X射线CT成像是一种重要的材料形貌、结构可视化表征技术。尤其是微焦点CT和同步辐射CT技术的出现可以实现多种材料的定量CT表征。但是在煤直接液化残渣样品的CT表征中存在诸多困难。残渣中的不同有机组分具有相近的X射线吸收系数，现有的图像阈值分割法是直接依赖于样品不同组分的X射线吸收系数值，由于实验误差及部分体积效应的存在，很难对残渣中不同的有机组分进行区分。另外，残渣中矿物组分的具体分子式很难确定，而且有大量小于CT分辨率的矿物组分存在，这些因素会导致CT切片上像元所表现出来的X射线吸收系数与残渣组分难以建立直接的关联，增加了残渣中有机组分区分的难度。X-ray CT imaging is an important technique for visualization and characterization of material morphology and structure. In particular, the emergence of microfocus CT and synchrotron radiation CT technologies can realize quantitative CT characterization of various materials. However, there are many difficulties in the CT characterization of coal direct liquefaction residue samples. Different organic components in the residue have similar X-ray absorption coefficients. The existing image threshold segmentation method directly depends on the X-ray absorption coefficient values of different components of the sample. Due to the existence of experimental errors and partial volume effects, it is difficult to The different organic components in the residue are distinguished. In addition, it is difficult to determine the specific molecular formula of the mineral components in the residue, and there are a large number of mineral components that are smaller than the resolution of CT. The direct correlation increases the difficulty of distinguishing the organic components in the residue.

发明内容Contents of the invention

本发明为解决现有煤直接液化残渣样品的可视化表征技术存在的有机组分难以区分的技术问题，提供了一种煤直接液化残渣样品的CT表征方法。The invention provides a CT characterization method for the direct coal liquefaction residue sample in order to solve the technical problem that the organic components are difficult to distinguish in the existing visual characterization technology of the direct coal liquefaction residue sample.

本发明是通过以下技术方案实现的：一种煤直接液化残渣样品的CT表征方法，包括如下步骤：第一步、煤直接液化残渣样品的取样及预测试：于同一批煤直接液化残渣样品中选取用于有机组分分子式、矿物灰组成测试的第一煤直接液化残渣样品，以及用于CT实验的第二煤直接液化残渣样品；通过实验手段获得第一煤直接液化残渣样品基本性质测试数据，所述煤直接液化残渣样品基本性质测试数据包括煤直接液化残渣样品中灰分含量、灰分成分及灰分成分含量、煤直接液化残渣样品中未转化煤基质、重油与沥青类物质分子式；根据煤直接液化残渣样品灰分成分含量推断各灰分成分在总的灰分中的体积分数；第二步、第二煤直接液化残渣样品各灰分成分的X射线吸收特性分析：根据公式(1)计算不同X射线能量下第二煤直接液化残渣样品中各灰分成分的X射线吸收量的比值，The present invention is achieved through the following technical solutions: a CT characterization method of a direct coal liquefaction residue sample, comprising the following steps: the first step, sampling and pre-testing of a coal direct liquefaction residue sample: in the same batch of coal direct liquefaction residue samples Select the first coal direct liquefaction residue sample for organic component molecular formula and mineral ash composition test, and the second coal direct liquefaction residue sample for CT experiment; obtain the basic property test data of the first coal direct liquefaction residue sample by experimental means , the basic property test data of the direct coal liquefaction residue sample includes ash content, ash content and ash content in the direct coal liquefaction residue sample, unconverted coal matrix, heavy oil and bituminous substance molecular formula in the direct coal liquefaction residue sample; according to the direct coal liquefaction residue sample The volume fraction of each ash component in the total ash is inferred from the ash content of the liquefaction residue sample; the second step, the X-ray absorption characteristic analysis of each ash component of the second coal direct liquefaction residue sample: calculate different X-ray energies according to formula (1) The ratio of the X-ray absorption of each ash component in the second coal direct liquefaction residue sample,

${y the y}_{11} = = \frac{{μ μ}_{i i} ((x x)) \times \times {V V}_{i i}}{{μ μ}_{m m} ((x x)) \times \times {V V}_{m m}} - - - - - - ((11))$

公式(1)中i代表不同的灰分成分；m代表参照灰分成分，所述参照灰分成分是灰分中对X射线吸收最多的成分；x代表X射线能量；μ_i(x)和μ_m(x)分别代表灰分成分i和参照灰分成分在能量为x keV的X射线线性吸收系数；V_i和V_m分别代表灰分成分i和参照灰分成分的体积分数；忽略掉比值小于或等于CT实验噪声水平的灰分成分，剩余部分灰分成分；所述的剩余部分灰分成分是由一个参照灰分成分和其余灰分成分构成的；第三步、第二煤直接液化残渣样品组分分组：根据公式(2)计算不同X射线能量下剩余部分灰分成分与未转化煤基质的X射线线性吸收系数的比值，In the formula (1), i represents different ash components; m represents the reference ash component, and the reference ash component is the component that absorbs the most X-rays in the ash; x represents the X-ray energy; μ _i (x) and μ _m (x ) represent the X-ray linear absorption coefficients of the ash component i and the reference ash component at energy x keV; V _i and V _m represent the volume fractions of the ash component i and the reference ash component respectively; ignore the ratio less than or equal to the noise level of the CT experiment The ash composition of the remaining part, the remaining part of the ash composition; the remaining part of the ash composition is composed of a reference ash composition and the rest of the ash composition; the third step, the second coal direct liquefaction residue sample component grouping: according to formula (2) calculation The ratio of the X-ray linear absorption coefficient of the remaining part of the ash component to the unconverted coal matrix under different X-ray energies,

${y the y}_{22} = = \frac{{μ μ}_{α α}^{r r} ((x x))}{{μ μ}_{00}^{r r} ((x x))} - - - - - - ((22))$

公式(2)中的α代表不同的剩余部分灰分成分，代表剩余部分灰分成分α在能量为x keV的X射线线性吸收系数；代表残渣中未转化煤基质在能量为x keV的X射线线性吸收系数；做出不同X射线能量下各剩余部分灰分成分与未转化煤基质的X射线线性吸收系数比值曲线，将剩余部分灰分成分进行分组，X射线线性吸收系数曲线相互平行的灰分成分设为一个组；第四步、第二煤直接液化残渣样品的CT实验：找出不同组别之间的X射线线性吸收系数曲线彼此最不平行的能量段，于上述能量段内选取三个X射线实验能量分别进行CT实验，获得多个投影像，CT实验投影像的成像分辨率为a；将上述投影像进行CT切片重构，重构过程中扣除投影像中的亮背景和暗背景，CT切片的最小可分辨单元的尺寸为a×a，在三个能量的CT切片中选取对应样品相同位置的三张CT切片，在这三张CT切片上选取图像质量较好并且孔隙及矿物较少的区域进行下一步计算，切割出来的区域像素尺寸为c×d；α in formula (2) represents different residual ash components, Represents the X-ray linear absorption coefficient of the remaining ash component α at energy x keV; Represents the X-ray linear absorption coefficient of the unconverted coal matrix in the residue at an energy of x keV; draw the ratio curve of the X-ray linear absorption coefficient of the remaining part of the ash component to the unconverted coal matrix under different X-ray energies, and divide the remaining part of the ash component Carry out grouping, and the ash components whose X-ray linear absorption coefficient curves are parallel to each other are set as a group; the fourth step, the CT experiment of the second coal direct liquefaction residue sample: find out the X-ray linear absorption coefficient curves between different groups are most mutually In the non-parallel energy segment, three X-ray experimental energies are selected in the above energy segment to conduct CT experiments respectively, and multiple projection images are obtained. The imaging resolution of the CT experiment projection images is a; the above projection images are reconstructed by CT slices, In the reconstruction process, the bright background and dark background in the projection image are deducted, and the size of the smallest resolvable unit of the CT slice is a×a. Three CT slices corresponding to the same position of the sample are selected in the three energy CT slices. Select the area with better image quality and fewer pores and minerals on the three CT slices for the next step of calculation, and the pixel size of the cut out area is c×d;

第五步、建立物理模型，对CT切片中不同组分进行区分鉴别：模型由N(N＝e×f)个简立方格子构成，每个简立方格子与CT切片上所选区域的像素点一一对应，每个简立方格子的尺寸为a×a×a，对所有简立方格子进行如(3)式所示的线性最优规划计算：The fifth step is to establish a physical model to distinguish and identify different components in the CT slice: the model is composed of N (N=e×f) simple cubic grids, and each simple cubic grid is connected to the pixel points of the selected area on the CT slice One-to-one correspondence, the size of each simple cubic grid is a×a×a, and the linear optimal programming calculation shown in (3) is performed on all simple cubic grids:

$\{\begin{matrix} {Σ Σ}_{c c = = 11}^{C C} {μ μ}^{((11,, c c))} {V V}_{n no}^{((c c))} \approx \approx {\overset{^^}{μ μ}}_{n no}^{((11))} \\ {Σ Σ}_{c c = = 11}^{C C} {μ μ}^{((22,, c c))} {V V}_{n no}^{((c c))} \approx \approx {\overset{^^}{μ μ}}_{n no}^{((22))} \\ {Σ Σ}_{c c = = 11}^{C C} {μ μ}^{((33,, c c))} {V V}_{n no}^{((c c))} \approx \approx {\overset{^^}{μ μ}}_{n no}^{((33))} \\ {Σ Σ}_{c c = = 00}^{C C} {V V}_{n no}^{((c c))} = = 11 \\ 00 \leq \leq {V V}_{n no}^{((c c))} \leq \leq 11 \end{matrix} - - - - - - ((33))$

其中c(＝0，1，2，....C)的不同值分别对应不同的分组，c＝0对应孔隙组；c＝1对应有机物组，c＝2,3,...C对应剩余部分灰分分成的灰分组分组；表示第n个简立方格子中分组c的体积分数；μ^(1,c)、μ^(2,c)、μ^(3,c)分别表示分组c在三个CT实验能量下的X射线线性吸收系数，c＝1时，μ^(1,c)、μ^(2,c)、μ^(3,c)等于有机物中的煤基质、重油、沥青类物质在相应X射线能量下吸收系数的平均值，c＝2,3,...C时，μ^(1,c)、μ^(2,c)、μ^(3,c)等于分组c中各组分在相应能量下的X射线线性吸收系数乘以该组分在分组中的体积分数，加总求和得到总值，再除以分组c总的体积分数；分别表示实验得到的第n个简立方格子在三个实验能量下的X射线线性吸收系数；利用计算机编程，将三个能量下CT切片所选区域(图4中方框)中各点的X射线线性吸收系数输入(3)式对应的模型并利用数学中常规的单纯形方法对模型进行线性最优化求解，可得到模型中各点的不同组分组体积分数将模型中灰分组分组对应的区域找出来，该部分区域对应残渣中的矿物组分；第六步、剔除无机组分，进一步区分有机组分：根据第五步的计算结果，将模型中计算得到的灰分组分组对应区域设定为只读，不参与进一步的计算；将有机组分中的重油与沥青类物质设定为一个组分组；将未转化的煤基质设定为一个组分组；孔隙单独设定为一个组分组；重复第五步的计算，即进行如(4)式所示线性最优规划计算：Among them, different values of c (=0, 1, 2, ... C) correspond to different groups, c = 0 corresponds to the pore group; c = 1 corresponds to the organic group, c = 2, 3, ... C corresponds to The remaining ash is divided into ash groups; Indicates the volume fraction of group c in the nth simple cubic lattice; μ ^(1,c) , μ ^(2,c) , μ ^(3,c) respectively represent the X-ray linear absorption of group c under three CT experimental energies Coefficient, when c=1, μ ^(1,c) , μ ^(2,c) , μ ^(3,c) are equal to the average value of the absorption coefficients of coal matrix, heavy oil, and bituminous substances in organic matter under the corresponding X-ray energy , c=2,3,...C, μ ^(1,c) , μ ^(2,c) , μ ^(3,c) are equal to the X-ray linear absorption coefficients of each component in group c at the corresponding energy Multiply by the volume fraction of the component in the grouping, add up and sum to get the total value, and then divide by the total volume fraction of grouping c; Respectively represent the X-ray linear absorption coefficients of the nth simple cubic lattice obtained in the experiment under three experimental energies; using computer programming, the X-rays of each point in the selected area of the CT slice (the box in Figure 4) under the three energies linear absorption coefficient Input the model corresponding to formula (3) and use the conventional simplex method in mathematics to solve the model linearly, and the volume fraction of different components of each point in the model can be obtained Find out the area corresponding to the ash group in the model, which corresponds to the mineral component in the residue; the sixth step is to remove the inorganic component and further distinguish the organic component: according to the calculation result of the fifth step, the calculated The corresponding area of the obtained ash group is set as read-only and does not participate in further calculations; the heavy oil and bituminous substances in the organic components are set as a component group; the unconverted coal matrix is set as a component group; Pores are individually set as a component group; repeat the calculation of the fifth step, that is, carry out the linear optimal programming calculation as shown in (4):

$\{\begin{matrix} {Σ Σ}_{k k = = 11}^{22} {μ μ}^{((11,, k k))} {v v}_{n no}^{((k k))} \approx \approx {\overset{^^}{μ μ}}_{n no}^{((11))} \\ {Σ Σ}_{k k = = 11}^{22} {μ μ}^{((22,, k k))} {v v}_{n no}^{((k k))} \approx \approx {\overset{^^}{μ μ}}_{n no}^{((22))} \\ {Σ Σ}_{k k = = 11}^{22} {μ μ}^{((33,, k k))} {v v}_{n no}^{((k k))} \approx \approx {\overset{^^}{μ μ}}_{n no}^{((33))} \\ {Σ Σ}_{k k = = 00}^{22} {v v}_{n no}^{((k k))} = = 11 \\ 00 \leq \leq {v v}_{n no}^{((k k))} \leq \leq 11 \end{matrix} - - - - - - ((44))$

其中k(＝0，1，2)的不同值分别对应不同的分组，k＝0对应孔隙组，k＝1对应未转化的煤基质组，k＝2对应重油与沥青类物质组；表示第n个简立方格子中分组k的体积分数；μ^(1,k)、μ^(2,k)、μ^(3,k)分别表示分组k在三个CT实验能量下的X射线线性吸收系数，重油与沥青类物质组在相应能量下的X射线线性吸收系数等于重油与沥青类物质在相应能量下吸收系数的平均值；分别表示实验得到的第n个简立方格子在三个实验能量下的X射线线性吸收系数；通过单纯形方法可对最优规划问题(4)式进行求解，求得CT切片上各简立方格子中各分组的体积分数；利用不同颜色在模型中显示不同有机组分的分布，实现煤直接液化残渣样品有机组分分布形式的可视化表征。Among them, different values of k (=0, 1, 2) correspond to different groups, k=0 corresponds to the pore group, k=1 corresponds to the unconverted coal matrix group, and k=2 corresponds to the heavy oil and asphalt group; Indicates the volume fraction of group k in the nth simple cubic lattice; μ ^(1,k) , μ ^(2,k) , μ ^(3,k) respectively represent the X-ray linear absorption of group k under three CT experimental energies coefficient, the X-ray linear absorption coefficient of heavy oil and asphalt-like substances at corresponding energies is equal to the average value of the absorption coefficients of heavy oil and asphalt-like substances at corresponding energies; respectively represent the X-ray linear absorption coefficients of the nth simple cubic lattice obtained in the experiment under three experimental energies; the optimal programming problem (4) can be solved by the simplex method, and each simple cubic lattice on the CT slice can be obtained The volume fraction of each group in the model; using different colors to display the distribution of different organic components in the model, to realize the visual representation of the distribution of organic components in the coal direct liquefaction residue sample.

第一步中煤直接液化残渣样品中灰分含量、灰分成分及灰分成分含量、煤直接液化残渣样品中重油与沥青类物质分子式是通过工业分析、元素分析、灰分分析等实验手段获得的。煤直接液化残渣样品中有机组分(未转化煤基质、重油、沥青类物质)分子式是依据有机元素比例写出。上述实验手段为本领域常规使用手段。In the first step, the ash content, ash composition and ash content in the direct coal liquefaction residue sample, and the molecular formula of heavy oil and asphalt in the direct coal liquefaction residue sample were obtained through industrial analysis, elemental analysis, ash analysis and other experimental means. The molecular formula of the organic components (unconverted coal matrix, heavy oil, bituminous substances) in the residue samples of direct coal liquefaction is written according to the ratio of organic elements. The above experimental means are routinely used in this field.

第二步中第二煤直接液化残渣样品各组分的X射线吸收特性分析，是根据第一煤直接液化残渣样品的组分分析结果进行的。第一煤直接液化残渣样品与第二煤直接液化残渣样品从同一批的煤直接液化残渣中选取，是为了减少不同批次残渣样品基本性质差异所带来的误差。The X-ray absorption characteristic analysis of each component of the second coal direct liquefaction residue sample in the second step is carried out based on the component analysis results of the first coal direct liquefaction residue sample. The first coal direct liquefaction residue sample and the second coal direct liquefaction residue sample are selected from the same batch of coal direct liquefaction residue, in order to reduce the error caused by the difference in the basic properties of different batches of residue samples.

CT实验噪声是由CT实验装置及实验参数引起的，CT实验装置及实验参数不同，CT实验噪声水平就不同，本领域技术人员可根据实际使用的装置以及相关参数确定CT实验噪声的数值。CT experimental noise is caused by CT experimental equipment and experimental parameters. Different CT experimental equipment and experimental parameters have different CT experimental noise levels. Those skilled in the art can determine the value of CT experimental noise according to the actual equipment and related parameters.

第四步中，在CT切片上选取矿物组分和孔隙较少的区域进行计算是为了尽可能减少由于残渣中矿物成分的不确定性以及大面积孔隙对残渣中有机组分CT鉴别的影响。In the fourth step, the area with fewer mineral components and pores is selected for calculation on the CT slice in order to minimize the impact of the uncertainty of the mineral composition in the residue and the large area of pores on the CT identification of organic components in the residue.

第五步中，在(3)式中的前三个方程中使用约等于号，是由于用残渣中灰分成分代替样品中的矿物成分存在近似，利用本发明所述方法计算得到样品灰分组分组的X射线线性吸收系数与残渣中矿物的真实X射线线性吸收系数并不严格相等，所以方程两边并不是严格相等。这使得(3)式中的求解并不是简单的线性方程组求解，而是属于数学上的线性最优规划问题。这使得残渣CT切片中不同组分的鉴别不是单纯依赖于各组分的线性吸收系数理论值与实验值的对应关系，而是以三个不同能量的CT实验数据以及(3)式中的后两式作为约束条件，通过线性最优规划对残渣中不同组分进行鉴别。这样可以解决残渣中矿物组分分子式难以确定的难题。In the fifth step, the approximately equal sign is used in the first three equations in (3), because there is an approximation in replacing the mineral components in the sample with the ash components in the residue, and the method of the present invention is used to calculate the sample ash group group The X-ray linear absorption coefficient of is not strictly equal to the real X-ray linear absorption coefficient of the minerals in the residue, so both sides of the equation are not strictly equal. This makes (3) where The solution of is not a simple solution of linear equations, but belongs to the linear optimal programming problem in mathematics. This makes the identification of different components in residue CT slices not simply rely on the corresponding relationship between the theoretical value of the linear absorption coefficient of each component and the experimental value, but on the basis of the CT experimental data of three different energies and the latter in (3) The two equations are used as constraints, and the different components in the residue are identified by linear optimal programming. This can solve the problem that the molecular formula of the mineral components in the residue is difficult to determine.

第六步中，将模型中计算得到的灰分组分组对应区域设置为只读，对有机组分进行进一步的组分鉴别，是因为无机组分与有机组分X射线线性吸收系数差异远大于有机组分中不同组分之间的差异，如果在第五步中一次性对所有组分(不同的有机组分和不同的矿物组分)同时进行最优化计算，则很难对有机组分进行相互区分。同时，对残渣组分分两次进行分组和线性最优规划计算，可以减少每一次最优化计算中的组分个数，在最优化计算过程中有利于获得可靠的计算结果。In the sixth step, the area corresponding to the gray group calculated in the model is set as read-only, and the organic component is further identified because the difference between the X-ray linear absorption coefficient of the inorganic component and the organic component is much larger than that of the organic component. The difference between different components in the composition, if the optimization calculation is performed simultaneously for all components (different organic components and different mineral components) at the same time in the fifth step, it is difficult to calculate the distinguish each other. At the same time, the grouping and linear optimal programming calculation of the residue components can be divided into two groups, which can reduce the number of components in each optimization calculation, and is conducive to obtaining reliable calculation results during the optimization calculation process.

本发明所述的CT表征方法采用多个单色X射线实验能量下获得的CT数据联合分析，分两次对煤直接液化残渣样品的组分进行分组与CT鉴别，利用残渣的灰分成分代替残渣中的矿物组分成分，将残渣样品CT图片上不同组分的鉴别形成一个线性最优规划问题，通过单纯形的方法对该问题进行求解，可实现对煤直接液化残渣样品CT切片中有机组分的区分与鉴别。通过在单个CT体元上建立物理模型，可以有效检出有机组分中的微小无机组分颗粒。对模型中的无机组分设置为只读后再次对有机组分进行分组计算，通过多个CT实验能量的数据约束，利用线性最优规划的方法可以区分出有机组分中未转化的煤基质与重油及沥青类物质，减弱了残渣样品中无机组分对有机组分鉴别过程的影响。本发明所采用的以线性最优规划问题为核心的物理模型，可以极大提高计算效率，可用于大量样品的表征。本发明所用方法对认识煤直接液化残渣中有机组分的相互结合方式，发展新的残渣中有用物质回收方法具有重大的科技和经济意义。The CT characterization method of the present invention adopts the combined analysis of CT data obtained under multiple monochromatic X-ray experimental energies, and performs grouping and CT identification of the components of the coal direct liquefaction residue sample twice, and uses the ash component of the residue to replace the residue The identification of different components on the CT image of the residue sample forms a linear optimal programming problem, and the simplex method is used to solve the problem, which can realize the organic group in the CT slice of the direct coal liquefaction residue sample. Separation and identification. By establishing a physical model on a single CT volume, tiny inorganic component particles in organic components can be effectively detected. After setting the inorganic components in the model as read-only, the organic components are grouped and calculated again. Through the data constraints of multiple CT experiment energies, the unconverted coal matrix in the organic components can be distinguished by using the linear optimal programming method. Compared with heavy oil and bituminous substances, the influence of inorganic components in residue samples on the identification process of organic components is weakened. The physical model with the linear optimal programming problem as the core adopted by the present invention can greatly improve the calculation efficiency and can be used for the characterization of a large number of samples. The method used in the invention has great scientific and economic significance for understanding the mutual combination mode of organic components in the coal direct liquefaction residue and developing a new method for recovering useful substances in the residue.

附图说明Description of drawings

图1残渣中Al₂O₃、SiO₂及CaO相对Fe₂O₃的X射线吸收比图像。Fig. 1 X-ray absorption ratio image of Al ₂ O ₃ , SiO ₂ and CaO relative to Fe ₂ O ₃ in the residue.

图2残渣中K₂O、Na₂O、P₂O₅、MgO及TiO₂相对Fe₂O₃的X射线吸收比图像。Fig. 2 X-ray absorption ratio images of K ₂ O, Na ₂ O, P ₂ O ₅ , MgO and TiO ₂ relative to Fe ₂ O ₃ in the residue.

图3残渣组分相对未转化煤基质的X射线吸收特性图像。Figure 3 X-ray absorption characteristic images of residue components relative to unconverted coal matrix.

图4 16keV的X射线能量下重构得到的一张CT切片。Fig. 4 A CT slice reconstructed under the X-ray energy of 16keV.

图5残渣中矿物、有机组分组、孔隙的分布图。Fig. 5 Distribution map of minerals, organic groups and pores in the residue.

图6不同有机组分在残渣中的分布图。Fig. 6 Distribution diagram of different organic components in the residue.

具体实施方式detailed description

本发明所述的表征方法可表征不同液化技术方案获得的煤直接液化残渣样品。下面以某一液化技术方案获得的煤直接液化残渣为实施例对本发明进行详细的说明。该实施例只是为了进一步阐述本发明，但并不限制本发明所保护的范围。The characterization method described in the present invention can characterize the coal direct liquefaction residue samples obtained by different liquefaction technical schemes. In the following, the present invention will be described in detail by taking the coal direct liquefaction residue obtained by a certain liquefaction technical scheme as an example. This embodiment is only to further illustrate the present invention, but does not limit the protection scope of the present invention.

一种煤直接液化残渣样品的CT表征方法，包括如下步骤：第一步、煤直接液化残渣样品的取样及预测试：于同一批煤直接液化残渣样品中选取用于有机组分分子式、矿物灰组成测试的第一煤直接液化残渣样品，以及用于CT实验的第二煤直接液化残渣样品；通过实验手段获得第一煤直接液化残渣样品基本性质测试数据，所述煤直接液化残渣样品基本性质测试数据包括煤直接液化残渣样品中灰分含量、灰分成分及灰分成分含量、煤直接液化残渣样品中未转化煤基质、重油与沥青类物质分子式。根据煤直接液化残渣样品灰分成分含量推断各灰分成分在总的灰分中的体积分数；进行煤直接液化残渣样品工业分析、元素分析、灰分分析等行业常规测试手段，其中第一煤直接液化残渣样品的基本性质测试数据见表1。残渣有机物分子式中，重油与沥青类物质的分子式是根据文献(谷小会，周铭，史士东.煤炭学报神华煤直接液化残渣中重质油组分的分子结构[J].煤炭学报，2006,31(1):76-80.谷小会，史士东，周铭.神华煤直接液化残渣中沥青烯组分的分子结构研究[J].煤炭学报,2006，31(6):785-789.)报道方式测试得到，未转化的煤基质是根据用于液化的原煤中有机元素比直接写出。A CT characterization method of a coal direct liquefaction residue sample, comprising the following steps: the first step, sampling and pre-testing of the coal direct liquefaction residue sample: selecting organic component molecular formula, mineral ash The first coal direct liquefaction residue sample for composition testing, and the second coal direct liquefaction residue sample for CT experiments; the basic property test data of the first coal direct liquefaction residue sample was obtained through experimental means, and the basic properties of the coal direct liquefaction residue sample The test data include ash content, ash composition and ash content in the direct coal liquefaction residue sample, unconverted coal matrix, heavy oil and bituminous substance molecular formula in the direct coal liquefaction residue sample. Infer the volume fraction of each ash component in the total ash according to the ash component content of the direct coal liquefaction residue sample; carry out industrial analysis, element analysis, ash analysis and other industry routine testing methods for the direct coal liquefaction residue sample, among which the first coal direct liquefaction residue sample The basic properties of the test data are shown in Table 1. In the molecular formula of residue organic matter, the molecular formula of heavy oil and bituminous substances is based on the literature (Gu Xiaohui, Zhou Ming, Shi Shidong. Molecular structure of heavy oil components in Shenhua Coal Direct Liquefaction Residue, Journal of Coal Science Journal of Coal Science, 2006, 31(1):76-80. Gu Xiaohui, Shi Shidong, Zhou Ming. Molecular structure of asphaltenes in Shenhua coal direct liquefaction residue[J]. Coal Journal, 2006, 31(6):785-789. ) is reported by testing, and the unconverted coal matrix is directly written according to the ratio of organic elements in the raw coal used for liquefaction.

表1第一煤直接液化残渣样品基本性质Table 1 Basic properties of the first coal direct liquefaction residue samples

表2是根据表1中煤直接液化残渣样品灰分成分含量推断得到各灰分成分在总的灰分中的体积分数。推断方法为根据各灰分成分的质量含量与密度相比得到数值即为各灰分成分的体积比，根据体积比可算出灰分中各灰分成分的体积含量。Table 2 is the volume fraction of each ash component in the total ash obtained by inferring the ash component content of the coal direct liquefaction residue sample in Table 1. The inference method is to obtain a value based on the mass content of each ash component compared with the density, which is the volume ratio of each ash component, and the volume content of each ash component in the ash can be calculated according to the volume ratio.

表2计算得到残渣中各灰分成分占总灰分的体积分数Table 2 calculates the volume fraction of each ash component in the total ash in the residue

根据CT装置样品台要求，用于CT实验的第二煤直接液化残渣样品需磨制成合适的形状，保证样品表面没有尖锐的棱角，且在实验所用X射线能量下样品的X射线透射率在30％-70％。本实施例中将用于CT实验的第二煤直接液化残渣样品用金相砂纸单方向手工磨制成直径4mm，高0.8mm的圆柱。除圆柱状，第二煤直接液化残渣样品还可磨制成其他符合CT装置样品台要求的形状。According to the requirements of the sample table of the CT device, the second coal direct liquefaction residue sample used in the CT experiment needs to be ground into a suitable shape to ensure that the surface of the sample has no sharp edges and corners, and the X-ray transmittance of the sample under the X-ray energy used in the experiment is 30%-70%. In this example, the second coal direct liquefaction residue sample used in the CT experiment was manually ground in one direction with metallographic sandpaper to form a cylinder with a diameter of 4 mm and a height of 0.8 mm. In addition to the cylindrical shape, the second coal direct liquefaction residue sample can also be ground into other shapes that meet the requirements of the CT device sample stage.

第二步、第二煤直接液化残渣样品各灰分成分的X射线吸收特性分析：根据公式(1)计算不同X射线能量下第二煤直接液化残渣样品中各灰分成分的X射线吸收量的比值，The second step, X-ray absorption characteristic analysis of each ash component of the second coal direct liquefaction residue sample: calculate the ratio of the X-ray absorption of each ash component in the second coal direct liquefaction residue sample under different X-ray energies according to formula (1) ,

公式(1)中i代表不同的灰分成分；m代表参照灰分成分，所述参照灰分成分是灰分中对X射线吸收最多的成分；x代表X射线能量；μ_i(x)和μ_m(x)分别代表灰分成分i和参照灰分成分在能量为x keV的X射线线性吸收系数；V_i和V_m分别代表灰分成分i和参照灰分成分的体积分数；忽略掉比值小于或等于CT实验噪声水平的灰分成分，剩余部分灰分成分；所述的剩余部分灰分成分是由一个参照灰分成分和其余灰分成分构成的。In the formula (1), i represents different ash components; m represents the reference ash component, and the reference ash component is the component that absorbs the most X-rays in the ash; x represents the X-ray energy; μ _i (x) and μ _m (x ) represent the X-ray linear absorption coefficients of the ash component i and the reference ash component at energy x keV; V _i and V _m represent the volume fractions of the ash component i and the reference ash component respectively; ignore the ratio less than or equal to the noise level of the CT experiment The ash composition of the remaining part, the remaining part of the ash composition; the remaining part of the ash composition is composed of a reference ash composition and the rest of the ash composition.

图1和图2是根据公式(1)得到的图像。从图2可以看到，与Fe₂O₃相比，K₂O、Na₂O、P₂O₅、MgO及TiO₂引起的X射线吸收量很少。考虑到本实施中采用的CT实验装置的CT实验噪声水平为2％，这些组分(K₂O、Na₂O、P₂O₅、MgO、TiO₂)很难被探测到。故而，在后期的CT分析中这些组分将被忽略。所以，剩余部分灰分成分为Fe₂O₃、Al₂O₃、SiO₂及CaO。Figure 1 and Figure 2 are images obtained according to formula (1). It can be seen from Fig. 2 that, compared with Fe ₂ O ₃ , the amount of X-ray absorption caused by K ₂ O, Na ₂ O, P ₂ O ₅ , MgO, and TiO ₂ is small. Considering that the CT experimental noise level of the CT experimental setup used in this implementation is 2%, these components (K ₂ O, Na ₂ O, P ₂ O ₅ , MgO, TiO ₂ ) are difficult to detect. Therefore, these components will be ignored in the later CT analysis. Therefore, the remaining ash is divided into Fe ₂ O ₃ , Al ₂ O ₃ , SiO ₂ and CaO.

第三步、第二煤直接液化残渣样品组分分组：根据公式(2)计算不同X射线能量下剩余部分灰分成分与未转化煤基质的X射线线性吸收系数的比值，The third step, the second coal direct liquefaction residue sample component grouping: according to the formula (2), calculate the ratio of the remaining ash component and the X-ray linear absorption coefficient of the unconverted coal matrix under different X-ray energies,

公式(2)中的α代表不同的剩余部分灰分成分，代表剩余部分灰分成分α在能量为x keV的X射线线性吸收系数；代表残渣中未转化煤基质在能量为x keV的X射线线性吸收系数；做出不同X射线能量下各剩余部分灰分成分与未转化煤基质的X射线线性吸收系数比值曲线，将剩余部分灰分成分进行分组，X射线线性吸收系数曲线相互平行的灰分成分设为一个组。α in formula (2) represents different residual ash components, Represents the X-ray linear absorption coefficient of the remaining ash component α at energy x keV; Represents the X-ray linear absorption coefficient of the unconverted coal matrix in the residue at an energy of x keV; draw the ratio curve of the X-ray linear absorption coefficient of the remaining part of the ash component to the unconverted coal matrix under different X-ray energies, and divide the remaining part of the ash component Grouping is carried out, and the ash components whose X-ray linear absorption coefficient curves are parallel to each other are set as a group.

图3为残渣中剩余部分灰分成分(Fe₂O₃、Al₂O₃、SiO₂、CaO)相对未转化煤基质的X射线线性吸收系数曲线。为了便于比较不同组分吸收系数曲线的斜率，图中将各组分的吸收系数曲线进行了整体的放大或缩小，使得不同组分的吸收系数曲线彼此相互靠近。Fig. 3 is the X-ray linear absorption coefficient curve of the remaining ash components (Fe ₂ O ₃ , Al ₂ O ₃ , SiO ₂ , CaO) in the residue relative to the unconverted coal matrix. In order to facilitate the comparison of the slopes of the absorption coefficient curves of different components, the absorption coefficient curves of each component are enlarged or reduced as a whole in the figure, so that the absorption coefficient curves of different components are close to each other.

由图3可看到，Fe₂O₃与CaO的吸收系数曲线相互平行，Al₂O₃与SiO₂的X射线吸收系数曲线相互平行。根据不同组分的X射线吸收特性可将剩余部分灰分成分分为：a)Fe₂O₃与CaO、b)Al₂O₃与SiO₂两个组分组。It can be seen from Figure 3 that the absorption coefficient curves of Fe ₂ O ₃ and CaO are parallel to each other, and the X-ray absorption coefficient curves of Al ₂ O ₃ and SiO ₂ are parallel to each other. According to the X-ray absorption characteristics of different components, the remaining ash components can be divided into two groups: a) Fe ₂ O ₃ and CaO, b) Al ₂ O ₃ and SiO ₂ .

第四步、煤直接液化残渣样品的CT实验：找出不同组别之间的X射线线性吸收系数曲线彼此最不平行的能量段，于上述能量段内选取三个X射线实验能量分别进行CT实验，获得多个投影像，CT实验投影像的成像分辨率为a；将上述投影像进行CT切片重构，重构过程中扣除投影像中的亮背景和暗背景，CT切片的最小可分辨单元的尺寸为a×a，在三个能量的CT切片中选取对应样品相同位置的三张CT切片，在这三张CT切片上选取图像质量较好并且孔隙及矿物较少的区域进行下一步计算，切割出来的区域像素尺寸为e×f；The fourth step, CT experiment of coal direct liquefaction residue samples: find out the energy segment where the X-ray linear absorption coefficient curves of different groups are most non-parallel to each other, and select three X-ray experimental energies in the above energy segment for CT In the experiment, multiple projection images were obtained, and the imaging resolution of the CT experimental projection images was a; the above projection images were reconstructed by CT slices, and the bright and dark backgrounds in the projection images were deducted during the reconstruction process, the minimum resolvable CT slice The size of the unit is a×a, select three CT slices corresponding to the same position of the sample in the three energy CT slices, and select the area with better image quality and less pores and minerals on these three CT slices for the next step Calculate, the pixel size of the cut out area is e×f;

具体实施时，第四步中进行的CT实验是在同步辐射或能提供良好单色X射线的装置上进行的。本发明的CT实验是在上海光源同步辐射BL13W线站获得，其成像分辨率为a＝3.7μm，在成像前后均采集暗背景和亮背景。选取14keV、16keV、20keV三个(与样品中非孔隙分组个数相同)X射线实验能量，于上述三个X射线实验能量下，分别采集1080张投影像。During specific implementation, the CT experiment carried out in the fourth step is carried out on synchrotron radiation or a device which can provide good monochromatic X-rays. The CT experiment of the present invention is obtained at the Shanghai Light Source Synchrotron Radiation BL13W line station, and its imaging resolution is a=3.7 μm, and dark background and bright background are collected before and after imaging. Select three X-ray experimental energies of 14keV, 16keV, and 20keV (the same as the number of non-pore groups in the sample), and collect 1080 projection images under the above-mentioned three X-ray experimental energies.

将所有投影像进行CT切片重构，重构过程中扣除了投影像中的亮背景及暗背景。图4为16keV的X射线能量下重构得到的一张CT切片，图像上最小可分辨的单元尺寸为a×a＝3.7μm×3.7μm。图4(实质审查参考资料彩图4)中不同灰度值对应不同的X射线吸收系数，白色区域表示较高的X射线吸收系数值对应矿物，灰色区域表示较低的吸收系数值对应有机物。在三个能量重构得到的CT切片中选取图4方框内区域进行下一步的计算，所选区域的像素尺寸为：e×f＝224×224。在所选区域内矿物分布较少，且无明显大尺寸孔隙。All projection images were reconstructed from CT slices, and the bright and dark backgrounds in the projection images were deducted during the reconstruction process. Fig. 4 is a CT slice reconstructed under the X-ray energy of 16keV, and the smallest resolvable unit size on the image is a×a=3.7μm×3.7μm. In Figure 4 (Color Figure 4 of substantive review reference materials), different gray values correspond to different X-ray absorption coefficients. The white area indicates that the higher X-ray absorption coefficient value corresponds to minerals, and the gray area indicates that the lower absorption coefficient value corresponds to organic matter. In the three energy reconstructed CT slices, select the area inside the box in Fig. 4 for the next calculation, and the pixel size of the selected area is: e×f=224×224. Mineral distribution is less in the selected area, and there are no obvious large-size pores.

其中c(＝0，1，2，....C)的不同值分别对应不同的分组，c＝0对应孔隙组；c＝1对应有机物组，c＝2,3,...C对应剩余部分灰分分成的灰分组分组；表示第n个简立方格子中分组c的体积分数；μ^(1,c)、μ^(2,c)、μ^(3,c)分别表示分组c在三个CT实验能量下的X射线线性吸收系数，c＝1时，μ^(1,c)、μ^(2,c)、μ^(3,c)等于有机物中的煤基质、重油、沥青类物质在相应X射线能量下吸收系数的平均值，c＝2,3,...C时，μ^(1,c)、μ^(2,c)、μ^(3,c)等于分组c中各组分在相应能量下的X射线线性吸收系数乘以该组分在分组中的体积分数，加总求和得到总值，再除以分组c总的体积分数；分别表示实验得到的第n个简立方格子在三个实验能量下的X射线线性吸收系数；利用计算机编程，将三个能量下CT切片所选区域(实质审查参考资料图4中方框)中各点的X射线线性吸收系数输入(3)式对应的模型并利用数学中常规的单纯形方法对模型进行线性最优化求解，可得到模型中各点的不同组分组体积分数将模型中灰分组分组对应的区域找出来，该部分区域表示残渣中的矿物组分。Among them, different values of c (=0, 1, 2, ... C) correspond to different groups, c = 0 corresponds to the pore group; c = 1 corresponds to the organic group, c = 2, 3, ... C corresponds to The remaining ash is divided into ash groups; Indicates the volume fraction of group c in the nth simple cubic lattice; μ ^(1,c) , μ ^(2,c) , μ ^(3,c) respectively represent the X-ray linear absorption of group c under three CT experimental energies Coefficient, when c=1, μ ^(1,c) , μ ^(2,c) , μ ^(3,c) are equal to the average value of the absorption coefficients of coal matrix, heavy oil, and bituminous substances in organic matter under the corresponding X-ray energy , c=2,3,...C, μ ^(1,c) , μ ^(2,c) , μ ^(3,c) are equal to the X-ray linear absorption coefficients of each component in group c at the corresponding energy Multiply by the volume fraction of the component in the grouping, add up and sum to get the total value, and then divide by the total volume fraction of grouping c; Respectively represent the X-ray linear absorption coefficients of the nth simple cubic lattice obtained in the experiment under the three experimental energies; using computer programming, each of the selected areas of the CT slice under the three energies (the box in Figure 4 of the substantive review reference material) X-ray linear absorption coefficient of point Input the model corresponding to formula (3) and use the conventional simplex method in mathematics to solve the model linearly, and the volume fraction of different components of each point in the model can be obtained Find the area corresponding to the ash group in the model, which represents the mineral components in the residue.

其中N＝e×f为N＝224×224，每个简立方格子的尺寸为3.7×3.7×3.7μm³。(3)式中，C＝3；c(＝0,1,2,3)分别对应孔隙组、有机物组、Al₂O₃与SiO₂组及Fe₂O₃与CaO组。Where N=e×f is N=224×224, and the size of each simple cubic lattice is 3.7×3.7×3.7 μm ³ . (3) In formula, C=3; c(=0,1,2,3) correspond to pore group, organic group, Al ₂ O ₃ and SiO ₂ group and Fe ₂ O ₃ and CaO group respectively.

非孔隙组c(＝1、2、3)在三个实验能量下的X射线线性吸收系数取值如表3所示：The X-ray linear absorption coefficient values of the non-porous group c (=1, 2, 3) under the three experimental energies are shown in Table 3:

表3第一次线性最优化计算中各组分组线性吸收系数取值Table 3 Values of linear absorption coefficients of each component group in the first linear optimization calculation

其中，c＝1时，对应的μ^(1,c)、μ^(2,c)、μ^(3,c)取值为残渣中未转化的煤基质、重油、沥青类物质X射线线性吸收系数的平均值，c＝2,3时，对应的μ^(1,c)、μ^(2,c)、μ^(3,c)取值为组分组内各组分X射线线性吸收系数的体积加权平均值。的值可在三个CT实验能量下重构得到的CT切片中如图4所示的方框区域内各点读取到，利用计算机编程，将μ^(1,c)、μ^(2,c)、μ^(3,c)、的数值输入(3)式对应的模型并利用单纯形的方法对模型进行线性最优化求解，可得到模型中各点的不同组分组体积分数。Among them, when c=1, the corresponding values of μ ^(1,c) , μ ^(2,c) and μ ^(3,c) are the X-ray linear absorption coefficients of unconverted coal matrix, heavy oil, and bituminous substances in the residue When c=2,3, the corresponding values of μ ^(1,c) , μ ^(2,c) and μ ^(3,c) are the volume-weighted values of the X-ray linear absorption coefficients of each component in the component group average value. The value of can be read at each point in the framed area shown in Figure 4 in the CT slice reconstructed under the three CT experimental energies. Using computer programming, μ ^(1,c) , μ ^{(2,c )} , μ ^(3,c) , Input the numerical value of (3) into the model corresponding to formula (3) and use the simplex method to solve the model by linear optimization, and the volume fraction of different components at each point in the model can be obtained.

图5(实质审查参考资料彩图5)是对(3)式进行计算后得到的残渣中矿物、有机组分组、孔隙的分布情况。图中矿物组分组都用蓝色显示，有机物组和孔隙组用红色显示，图中各点颜色的显示强度正比于各组分在该点位置处计算得到的体积分数。Figure 5 (Color Figure 5 of Substantive Review Reference Materials) shows the distribution of minerals, organic groupings, and pores in the residue obtained after calculating formula (3). In the figure, the mineral component groups are shown in blue, and the organic matter group and pore group are shown in red. The display intensity of the color of each point in the figure is proportional to the volume fraction of each component calculated at the point position.

第六步、剔除无机组分，进一步区分有机组分：根据第五步的计算结果，将模型中计算得到的灰分组分组对应区域设定为只读，不参与进一步的计算。将有机组分中的重油与沥青类物质设定为一个组分组；将未转化的煤基质设定为一个组分组；孔隙单独设定为一个组分组；重复第五步的计算，即进行如(4)式所示线性最优规划计算：The sixth step is to eliminate the inorganic components and further distinguish the organic components: according to the calculation results of the fifth step, set the corresponding area of the gray group calculated in the model as read-only, and do not participate in further calculations. Set the heavy oil and bituminous substances in the organic components as a component group; set the unconverted coal matrix as a component group; set the pores alone as a component group; repeat the calculation of the fifth step, that is, the following The linear optimal programming calculation shown in (4) formula:

具体实施中，μ^(1,k)、μ^(2,k)、μ^(3,k)的取值如表4所示。In specific implementation, the values of μ ^(1,k) , μ ^(2,k) and μ ^(3,k) are shown in Table 4.

表4第二次线性最优化计算中各组分组线性吸收系数取值Table 4 Values of linear absorption coefficients of each component group in the second linear optimization calculation

利用计算机编程，将表4中数值以及三个CT实验能量重构切片中图5(实质审查参考资料彩图5)红色区域对应位置的X射线线性吸收系数代入(4)式所示模型中，对图5所示图片中红色区域进行最优化计算，得到如图6所示的重油与沥青类物质组及未转化煤基质组的分布图。图6(实质审查参考资料彩图6)中红色表示未转化的煤基质，绿色表示重油与沥青类物质，各点颜色显示强度正比于各个简立方格子中各组分的体积分数，图中未显示孔隙组与图5中矿物组分组的分布。Using computer programming, the values in Table 4 and the X-ray linear absorption coefficients corresponding to the positions in the red area in Figure 5 (Color Figure 5 of the substantive review reference material) in the three CT experiment energy reconstruction slices Substituting into the model shown in formula (4), the red area in the picture shown in Figure 5 is optimized and calculated, and the distribution map of heavy oil and bituminous substances group and unconverted coal matrix group is obtained as shown in Figure 6. In Figure 6 (Color Figure 6 of Substantive Review Reference Materials), red represents unconverted coal matrix, green represents heavy oil and bituminous substances, and the intensity of each point color is proportional to the volume fraction of each component in each simple cubic grid. The distribution of pore groups and mineral component groups in Fig. 5 is shown.

Claims

1. the CT characterizing method of a coal directly-liquefied residue sample, it is characterised in that include as Lower step: the first step, the sampling of coal directly-liquefied residue sample and pretest: in a collection of coal Direct liquefaction residue sample is chosen for organic component molecular formula, the of mineral grey composition test One coal directly-liquefied residue sample, and the second coal directly-liquefied residue sample for CT experiment Product；The first coal directly-liquefied residue sample fundamental property test data are obtained by laboratory facilities, Described coal directly-liquefied residue sample fundamental property test data include coal directly-liquefied residue sample In product in content of ashes, ash composition and ash component content, coal directly-liquefied residue sample not Convert matrix of coal, heavy oil and bitumen molecular formula；According to coal directly-liquefied residue sample ash Component content is divided to infer each ash composition volume fraction in total ash；Second step, second The X-ray absorption specificity analysis of coal directly-liquefied residue sample each ash composition: according to formula (1) each ash composition in the second coal directly-liquefied residue sample is calculated under different x-ray energy The ratio of X-ray absorption amount,

y_{1} = \frac{μ_{i} (x) \times V_{i}}{μ_{m} (x) \times V_{m}} - - - (1)

In formula (1), i represents different ash compositions；M represents with reference to ash composition, described reference Ash composition is composition most to X-ray absorption in ash；X represents X-ray energy； μ_i(x) and μ_mX () represents ash component i respectively and is x keV with reference to ash composition at energy X-ray linear absorption coefficient；V_iAnd V_mRepresent ash component i respectively and with reference to ash composition Volume fraction；Neglect the ratio ash composition less than or equal to CT experimental noise level, Remainder ash composition；Described remainder ash composition is with reference to ash composition by one With remaining ash composition composition；3rd step, the second coal directly-liquefied residue sample component packet: Remainder ash composition and unconverted coal under different x-ray energy is calculated according to formula (2) The ratio of the X-ray linear absorption coefficient of substrate,

y_{2} = \frac{μ_{α}^{r} (x)}{μ_{0}^{r} (x)} - - - (2)

α in formula (2) represents different remainder ash compositions,Represent residue Part of ash composition α is at the X-ray linear absorption coefficient that energy is x keV；Represent In residue, unconverted matrix of coal is at the X-ray linear absorption coefficient that energy is x keV；Make Each remainder ash composition and the X-ray line of unconverted matrix of coal under different x-ray energy Property absorptance ratio curve, is grouped remainder ash composition, and X-ray is linearly inhaled The ash that receipts coefficient curve is parallel to each other becomes to be divided into a group；4th step, the second direct liquid of coal Change the CT experiment of residue sample: find out the X-ray linear absorption coefficient between different group bent Line the most uneven energy section, chooses three X-ray experiment energy in above-mentioned energy section Amount carries out CT experiment respectively, it is thus achieved that multiple projection images, and the imaging of CT experimental projection picture is differentiated Rate is a；Above-mentioned projection image is carried out CT section reconstruct, restructuring procedure is deducted in projection image Bright background and dark background, the size of the distinguishable unit of minimum of CT section is a × a, The CT section of three energy is chosen three CT sections of counter sample same position, at this Picture quality is chosen preferably and under hole and the less region of mineral carry out in three CT section One step calculates, the area pixel cut out a size of c × d；

5th step, setting up physical model, in cutting into slices CT, different component makes a distinction discriminating: model Being made up of N (N=e × f) individual simple cubic lattice, each simple cubic lattice is selected with in CT section The pixel one_to_one corresponding in region, the size of each simple cubic lattice is a × a × a, to all The linear optimal planning that simple cubic lattice is carried out as shown in (3) formula calculates:

\{\begin{matrix} Σ_{c = 1}^{C} μ^{(1, c)} V_{n}^{(c)} \approx {\hat{μ}}_{n}^{(1)} \\ Σ_{c = 1}^{C} μ^{(2, c)} V_{n}^{(c)} \approx {\hat{μ}}_{n}^{(2)} \\ Σ_{c = 1}^{C} μ^{(3, c)} V_{n}^{(c)} \approx {\hat{μ}}_{n}^{(3)} \\ Σ_{c = 0}^{C} V_{n}^{(c)} = 1 \\ 0 \leq V_{n}^{(c)} \leq 1 \end{matrix} - - - (3)

Wherein c (=0,1,2 ... .C) different value respectively corresponding different packet, c=0 pair Answer hole group；C=1 correspondence Organic substance group, c=2,3 ... C correspondence remainder ash is divided into Ash component group；Represent the volume fraction being grouped c in the n-th simple cubic lattice；μ^(1,c)、 μ^(2,c)、μ^(3,c)Represent packet c X-ray linear absorption under three CT test energy respectively Coefficient, during c=1, μ^(1,c)、μ^(2,c)、μ^(3,c)Equal to the matrix of coal in Organic substance, heavy oil, drip Blue or green class material meansigma methods of absorptance under corresponding X-ray energy, c=2,3 ... during C, μ^(1,c)、 μ^(2,c)、μ^(3,c)Equal to each component X-ray linear absorption coefficient under corresponding energy in packet c It is multiplied by this component volume fraction in a packet, adds up summation and obtain total value, more total divided by packet c Volume fraction；Represent that the n-th simple cubic lattice that experiment obtains exists respectively X-ray linear absorption coefficient under three experiment energy；Utilize computer programming, by three The X-ray linear absorption coefficient of each point in CT section selected areas under energy Input model corresponding to (3) formula also utilizes simplex method conventional in mathematics to model Carry out linear optimization to solve, the different component group volume fraction of each point in available model Region corresponding for model ash component group is found out, the ore deposit in this subregion correspondence residue Thing component；6th step, rejecting inorganic component, further discriminate between organic component: according to the 5th step Result of calculation, ash component group corresponding region calculated in model is set as read-only, It is not involved in further calculating；Heavy oil in organic component and bitumen are set as one Component group；Unconverted matrix of coal is set as a component group；Hole is individually set as one Component group；Repeat the calculating of the 5th step, i.e. carry out linear optimal planning meter as shown in (4) formula Calculate:

\{\begin{matrix} Σ_{k = 1}^{2} μ^{(1, k)} v_{n}^{(k)} \approx {\hat{μ}}_{n}^{(1)} \\ Σ_{k = 1}^{2} μ^{(2, k)} v_{n}^{(k)} \approx {\hat{μ}}_{n}^{(2)} \\ Σ_{k = 1}^{2} μ^{(3, k)} v_{n}^{(k)} \approx {\hat{μ}}_{n}^{(3)} \\ Σ_{k = 0}^{2} v_{n}^{(k)} = 1 \\ 0 \leq v_{n}^{(k)} \leq 1 \end{matrix} - - - (4)

The wherein the most corresponding different packet of the different value of k (=0,1,2), k=0 correspondence hole group, The corresponding unconverted matrix of coal group of k=1, k=2 correspondence heavy oil and bitumen group；Table Show the volume fraction being grouped k in the n-th simple cubic lattice；μ^(1,k)、μ^(2,k)、μ^(3,k)Represent respectively Packet k X-ray linear absorption coefficient under three CT test energy, heavy oil and Colophonium class Material group X-ray linear absorption coefficient under corresponding energy is equal to heavy oil and bitumen The meansigma methods of absorptance under corresponding energy；Represent that experiment obtains respectively The n-th simple cubic lattice three experiment energy under X-ray linear absorption coefficients；Pass through Optimum programming problem (4) formula can be solved by simplex method, tries to achieve in CT section each The volume fraction of each packet in simple cubic lattice；Difference has to utilize different colours to show in a model The distribution of machine component, it is achieved the visualization of coal directly-liquefied residue sample organic component distribution form Characterize.