CN102289545A - Method for calibrating hydrocarbon generation dynamical model parameters by finite first-order parallel reaction - Google Patents

Abstract

The invention discloses a method for calibrating hydrocarbon generation dynamical model parameters by finite first-order parallel reaction and is applied to the technical field of evaluation of hydrocarbon resources. On the basis of a hydrocarbon generation dynamical theory, the method is used for describing a hydrocarbon generation process of an organic matter. By the method, kerogen oiling, kerogen gassing and oil gassing which are greatly different from one another and dynamical processes of different types of organic matters can be separately discussed effectively. Furthermore, by a model calibration math method, an optimization approximate solution is performed; therefore, defects of limited calibration precision improvement and relatively redundant calculation quantity which are caused by infinite first-order parallel reaction can be overcome; and calculation without a precise solution under an experimental condition is realized through approximate solution of dynamical parameters. By the method, the description of a complicated process of organic matter hydrocarbon generation and a quantitative and dynamic description for organic matter hydrocarbon generation characteristics are realized effectively.

Description

Limited parallel first order reaction becomes the scaling method of hydrocarbon kinetic model parameter

Technical field

The invention belongs to the oil and gas resource evaluation technical field, relate to the description that organic matter is become the hydrocarbon kinetic parameter, is to be based upon on the basis of hydrocarbon principle of dynamics in order to describe the organic a kind of technology that becomes the hydrocarbonylation course of reaction.

Background technology

Just beginning to explore both at home and abroad before the eighties of last century the nineties becomes the foundation of hydrocarbon kinetic model and Parameter Optimization to ask for to organic matter.When studying an one-tenth hydrocarbon dynamic characteristic, what at first face is the selection problem of principles of chemical kinetics.The reaction Kinetics Model that has also proposed at present has: (1) overall budget reaction model comprises the non-first order reaction two class hypotypes of overall budget one-level and overall budget; (2) cascade reaction (supposition is along with the process of reaction, and kinetic parameter will change, and kinetic parameter changed when then assumed response proceeded to a certain degree in the practical operation) comprises series connection first order reaction and the non-first order reaction of series connection; (3) parallel first order reaction comprises a unlimited parallel first order reaction and limited parallel first order reaction, and whether identical the pre-exponential factor of answering according to its each reaction pair is, is divided into identical pre-exponential factor and different pre-exponential factor two subclass; (4) serving as theme with cascade reaction, is the model of assisting with parallel reactor.

For parallel first order reaction, unlimited parallel reactor and limited parallel reactor be can be divided into according to the reaction number, wherein parallel reactor with same frequency factor (A) and parallel reactor are divided into again with different frequency factor according to the employing parallel reactor equation medium frequency factor is whether identical.Say that in principle the number of the parallel reactor that sets is many more, just possible more near organic true one-tenth hydrocarbon process.But the parallel reactor number is many more, and the calculated amount that model calibration gets up is big more.Because each parallel reactor process is all had three undetermined parameter (energy of activation E _i, pre-exponential factor A _iWith corresponding reacting dose X _I0).To this, the foreign scholar supposes that then the energy of activation of parallel reactor obeys the distribution of certain functional form, and the pre-exponential factor of all parallel reactors identical (is representative with Burnham) or have the certain funtcional relationship (is representative with IFP researchists such as Behar) or the pre-exponential factor of all parallel reactors to have nothing in common with each other with energy of activation.Can be divided into according to energy of activation distribution function form and to obey the kinetic model that discrete (Discrete) distributes, normal state (Gaussian) distributes, Wei Bu (Weibull) distributes, binomial (Binomial) distributes, gamma (Gamma) distributes, wherein first three plant distributed model research and use in the majority, and in these three kinds of distributed models the Discrete Distribution model research at present and use at most.

Consider the extremely complicacy that the organic source of deposition, composition, structure, of bonding constitute, total packet response all is difficult to approximate reflection with segmented general packet response (cascade reaction) model, and it becomes the essence of hydrocarbon dynamic process.Comparatively speaking, parallel reactor model (the consecutive reaction process that adds in various degree) should be able to comparatively objectively reflect the dynamics essence that it becomes the hydrocarbon process.Although this process is extremely complicated, make people's present (perhaps eternal) it can't be resolved into elementary reaction and investigate its one-tenth hydrocarbon course, become the kinetics equation of each elementary reaction of hydrocarbon mechanism and its essence of foundation reflection, but parallel reactor (adding limited consecutive reaction) model is the most approximate selection that can accomplish at present after all.

When Tissot etc. (1975,1978) proposed this parallel reactor model at first, the energy of activation between the adjacent parallel reaction was at interval up to 10 * 4.187kJ/mol.From dynamic (dynamical) viewpoint, because chemical reaction rate is very responsive to the variation of energy of activation value size, therefore, the of bonding that the interval that 10 * 4.187kJ/mol is big so obviously can not be similar in the reflection kerogen is formed, thereby can not be used to the one-tenth hydrocarbon process of approximate description organic matter.This also be people's such as Tissot model since proposing, one of major reason of failing to be used widely.

In the present invention, according to above-mentioned analysis,, selected the energy of activation interval of less parallel reactor in conjunction with The trial result.The energy of activation distribution range of parallel reactor is also progressively dwindled definite according to The trial result.The kinetic parameter calibrated and calculated of Biao Dinging is comparatively easy like this, is more suitable for describing organic one-tenth hydrocarbon process.

Summary of the invention

The objective of the invention is: provide limited parallel first order reaction to become the scaling method of hydrocarbon kinetic model parameter.Utilize this method can realize organic matter is become quantitative, the dynamic description of hydrocarbon dynamic characteristic.

The technical solution used in the present invention is: limited parallel first order reaction becomes the scaling method of hydrocarbon kinetic model parameter, at first, be created as the chemical dynamic model of hydrocarbon, become the chemical dynamic model of hydrocarbon to comprise that the oily chemical dynamic model of kerogen one-tenth, the chemical dynamic model that kerogen becomes gas become the chemical dynamic model of gas with oil-breaking; Then, observed parameter or experiment parameter are input to the chemical dynamic model of setting up good one-tenth hydrocarbon, the chemical dynamic model that becomes hydrocarbon is demarcated; Secondly, adopt structure objective function and penalty optimization to ask for approximate exact solution, realize organic matter is become the kinetic description of hydrocarbon process; At last, the chemical dynamic model of calibrated one-tenth hydrocarbon, the organic matter that is applied to the exploratory area becomes quantitative, the dynamic evaluation of hydrocarbon history.

Concrete steps comprise: the foundation of model, experiment parameter or observational data obtain the demarcation of kinetic parameter.

One, the foundation of model

A, kerogen become the chemical dynamic model of oil

If kerogen (KEO) becomes oily process to be made up of a series of (NO) individual parallel first order reaction, the energy of activation that each reaction pair is answered is EO _i, unit: kj/mol (kJ/mol), pre-exponential factor AO _i, unit: per minute (/min), and the kerogenic original latent amount of establishing corresponding each reaction is XO _I0, unit, number percent (%), KO _iBe the reaction rate constant that i kerogen becomes the oil reaction, the oil that each reaction generates is O _i, the oil generating quantity of each reaction is XO _i, i=1,2 ..., NO, promptly to time t (minute, min), the time

$\left\{\begin{array}{c}{\mathrm{KEO}}_1\left({\mathrm{XO}}_{10}\right)\stackrel{{\mathrm{KO}}_1}{\→}{O}_1\left({\mathrm{XO}}_1\right)\\ {\mathrm{KEO}}_i\left({\mathrm{XO}}_{10}\right)\stackrel{{\mathrm{KO}}_i}{\→}{O}_i\left({\mathrm{XO}}_i\right)\end{array}\right.$

${\mathrm{KEO}}_{\mathrm{NO}}\left({\mathrm{XO}}_{\mathrm{NO}0}\right)\stackrel{{\mathrm{KO}}_{N0}}{\→}{O}_{\mathrm{NO}}\left({\mathrm{XO}}_{\mathrm{NO}}\right)---\left(1\right)$

$\frac{d{\mathrm{XO}}_i}{\mathrm{dt}}={\mathrm{KO}}_i({\mathrm{XO}}_{i0}-{\mathrm{XO}}_i)---\left(2\right)$

Obtain by Arrhenius (Arrhenius) equation:

${\mathrm{KO}}_i={\mathrm{AO}}_i\exp \left(\frac{-{\mathrm{EO}}_i}{\mathrm{RT}}\right)---\left(3\right)$

i＝1，2，Λ，NC

Wherein R is gas law constant (8.31447kJ/molK); T is a thermodynamic temperature, unit: Kelvin (K); Experiment adopts constant speed to heat up, heating rate D, and unit: degrees celsius/minute (℃/min).

$\frac{\mathrm{dT}}{\mathrm{dt}}=D,$ Promptly $\mathrm{dt}=\frac{\mathrm{dT}}{D}---\left(4\right)$

Get by (2)～(4) formula

$\frac{{\mathrm{dXO}}_i}{{\mathrm{XO}}_{i0}-{\mathrm{XO}}_i}=\frac{{\mathrm{AO}}_i}{D}\·\exp (-\frac{{\mathrm{EO}}_i}{\mathrm{RT}})\mathrm{dT}$

Following formula from T0 → T integration, and is noticed XO _{I (T0)}=0, XO _{I (T)}=XO _i

${\mathrm{XO}}_i={\mathrm{XO}}_{i0}(1-\exp (-{\∫}_{T_0}^T\frac{{\mathrm{AO}}_i}{D}\·\exp (-\frac{{\mathrm{EO}}_i}{\mathrm{RT}})\mathrm{dT}))---\left(5\right)$

Total oil generating quantity XO (relative percentage) of NO parallel reactor then is

$\mathrm{XO}=\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\Σ}}}{\mathrm{XO}}_i=\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\Σ}}}({\mathrm{XO}}_{i0}(1-\exp (-{\∫}_{T_0}^T\frac{{\mathrm{AO}}_i}{D}\·\exp (-\frac{\mathrm{EOi}}{\mathrm{RT}})\mathrm{DT}))---\left(6\right)$

B, kerogen become the chemical dynamic model of gas

In like manner, if establishing kerogen directly becomes the reaction of gas to be made up of (NG) individual parallel reactor, the energy of activation of each parallel reactor is EG _i, unit: kj/mol (kJ/mol), pre-exponential factor AG _i, unit: per minute (/min), the initial amount of diving is XG _I0, unit: dimensionless, D is a heating rate, unit: degrees celsius/minute (℃/min), R is gas law constant (8.31447kJ/molK), get the directly angry computing formula of measuring (number percent) of temperature variant T kerogen in period (KEO) to be

$\mathrm{XG}=\underset{i=1}{\overset{\mathrm{NG}}{\mathrm{\Σ}}}{\mathrm{XG}}_i=\underset{i=1}{\overset{\mathrm{NG}}{\mathrm{\Σ}}}({\mathrm{XG}}_{i0}(1-\exp (-{\∫}_{T_0}^T\frac{{\mathrm{AG}}_i}{D}\·\exp (-\frac{\mathrm{EGi}}{D})\mathrm{dT}))---\left(7\right)$

Compare with (6) formula, (7) formula only is that the pair mark of related variable is different.O represents oil, and G represents gas.

C, oil-breaking become the chemical dynamic model of gas

If oil-breaking becomes the process of gas to be made up of NOG parallel reactor, the energy of activation of each reaction is EOG _i, unit: kj/mol (kJ/mol), pre-exponential factor are AOG _i, unit: per minute (/min), corresponding original latent amount is XOG _I0, unit: dimensionless, it is KOG that i kerogen becomes the reaction rate constant of oil reaction _I0, when reaction proceed to time t (minute, in the time of min), factor of created gase (representing with the percentage that accounts for the overall reaction amount) is XOG _i, unit: dimensionless then has

$\frac{{\mathrm{dXOG}}_i}{\mathrm{dt}}={\mathrm{KOG}}_{i0}\·({\mathrm{XOG}}_{i0}-{\mathrm{XOG}}_i)---\left(8\right)$

Wherein ${\mathrm{KOG}}_i={\mathrm{AOG}}_i\·\exp (-\frac{{\mathrm{EOG}}_i}{\mathrm{RT}})---\left(9\right)$

It is the reaction rate constant that i oil-breaking becomes solid/liquid/gas reactions.

(9) substitution (8) formula is put the back in order to time integral, and notices XOG _{I (t=0)}=0, XOG _{I (t)}=XOG _i, the total angry amount that obtains NOG parallel reactor is

$\mathrm{XOG}=\underset{i=1}{\overset{\mathrm{NOG}}{\mathrm{\Σ}}}{\mathrm{XOG}}_i=\underset{i=1}{\overset{\mathrm{NOG}}{\mathrm{\Σ}}}{\mathrm{XOG}}_{i0}(1-\exp ({\∫}_{T0}^T-({\mathrm{AOG}}_i\·\exp (-\frac{{\mathrm{EOG}}_i}{\mathrm{RT}})\mathrm{dT}))---\left(10\right)$

Equally, the R in the formula is gas law constant (8.31447kJ/molK), and T is a thermodynamic temperature, Kelvin (K).

At last, the demarcation of the chemical dynamic model by becoming hydrocarbon, the organic matter of realizing using the exploratory area become the hydrocarbon history quantitatively, dynamic evaluation.

Two, experiment parameter or observational data obtains

Experiment parameter or observed parameter are mainly obtained the relation between the temperature-time-product hydrocarbon rate of sample, from the pyrolysis experiment parameter under the different condition, the pyrolysis experiment comprises that Rock-Eval experiment, thermogravimetric-mass spectrometry (TG-MS) experiment, gold pipe thermal simulation experiment or MSSV-small size sealing oil gas generate simulated experiment, obtain its heating rate (centigrade per minute respectively, ℃/min) or thermostat temperature (degree centigrade, ℃), experimental period (minute, min), produce hydrocarbon rate (milliliter or gram, ml or g) parameter, step is:

Step 1, representational rock, kerogen or the crude oil sample of selection low-maturity (Ro＜0.6%) high abundance (TOC＞1%);

Step 2 is carried out sample constant temperature or is decided the heating rate thermal simulation experiment;

Step 3, the record experiment parameter, comprise the thermal simulation time (minute, min), thermostat temperature (temperature, ℃) or decide heating rate (centigrade per minute, ℃/min), hydrocarbon production (milliliter or restrain ml or g) parameter.

Step 4, sample is heated to produces the hydrocarbon limit, promptly sample produce hydrocarbon can force failure, the product hydrocarbon amount of each time point added up obtains accumulative total oil offtake and gas production rate, the oil and gas production of each time is obtained produce oil gas rate divided by the accumulative total oil and gas production respectively, comprise kerogen oil productivity XO _Aj(number percent), kerogen factor of created gase XG _Aj(number percent) or crude oil factor of created gase XOG _Aj(number percent), wherein a is a certain heating rate, j is a certain temperature spot.

Three, the demarcation of kinetic parameter

Become the demarcation of the chemical dynamic model of oil to describe with kerogen, the method that kerogen becomes the chemical dynamic model of gas to become the chemical dynamic model of gas to demarcate with oil-breaking is identical; The steps include:

Step 1, the structure objective function

Being located at a certain heating rate 1, is XO1 by the measured oil productivity of experiment when reaching a certain temperature j _1j(number percent) is under identical condition, if there be a certain group of EO _i, unit: kj/mol (kJ/mol), AO _i, unit: per minute (/min), XO _I0The value of (dimensionless) makes all 1, and j has XO1 _1j-XO _1j=0, then should organize EOi, unit: kj/mol (kJ/mol), AO _i, unit: per minute (/min), XO _I0, unit, (number percent) is institute and asks.But because aspects such as experimental errors, this is practically impossible.Therefore, can only ask and make XO1 _1j-XO _1jLittle EO tries one's best _i, AO _i, XO _I0Value.For this reason, the structure objective function is:

$Q({\mathrm{EO}}_i,{\mathrm{AO}}_i,{\mathrm{XO}}_{i0})=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}{\left(\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}}\right)}^2---\left(11\right)$

L0 is the number of different heating rates experiment in the formula, and J0 is the sampling number from the empirical curve.

The number value of parallel reactor is big more, just might comprise more that kerogen becomes all reaction types of oil, thereby should be accurate more, but because the calculated amount that this moment, model calibration reached when model is used subsequently is too big, be difficult to practicability, and, find in the actual process of demarcating, though model to the thermal simulation time under the experiment condition (minute, min), thermostat temperature (degree centigrade, ℃) or decide heating rate (centigrade per minute, ℃/min), hydro carbons productive rate (number percent, %) fitting degree of parameter is generally improved with the segmentation of parallel reactor, but after the number of parallel reactor acquired a certain degree, the improvement of fitting degree was not obvious.Therefore, only need with limited individual parallel reactor with certain intervals.

Like this, because EO _iThe distribution range of the energy of activation by determining parallel reactor and the energy of activation of adjacent parallel reaction are found the solution at interval, and then (11) formula turns to:

$Q({\mathrm{AO}}_i,{\mathrm{XO}}_{i0})=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}{\left(\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}}\right)}^2---\left(12\right)$

In addition, notice that kerogen becomes the AO in oily model (6) formula _i, XO _I0, represent with the percentage that accounts for the overall reaction amount, should satisfy:

Like this, the determining of model (6), i.e. asking for of kinetic parameter, problem just turns to the non-negative minimal point problem of objective function (12) when satisfying constraint condition (13) of asking.

Step 2, the structure penalty

The above-mentioned minimizing problem more complicated of finding the solution that contains constraint condition because except making target function value descends gradually, be also noted that the feasibility of separating, promptly sees to separate whether be within the constraint condition institute restricted portion.Make up by penalty term the band amplification coefficient, reach the purpose that promotes the penalty term constraining force, make the speed of convergence of penalty term accelerate, promote the efficiency and precision of demarcating, here adopt Means of Penalty Function Methods will have constrained extreme-value problem (12), (13) to turn to the unconstrained extrema problem, its step is as follows:

A: to AO _i＞0, this constraint condition, turning to the unconstrained extrema problem has

Be C ₁(AO _i)=[min (0, AO _i)] ²

Amplification coefficient $B_{}$ ${}_1=e^{-{\mathrm{AO}}_i}$

${\mathrm{<figure-callout\; id="1"\; label="G"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">G</figure-callout>}}_1={\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_1\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_1=e^{-{\mathrm{AO}}_i}{\&CenterDot;\left[\min (0,{\mathrm{AO}}_i)\right]}^2$

B: to 1＞XO _I0＞0, this constraint condition, turning to the unconstrained extrema problem has

Amplification coefficient

Promptly

$G_2\left({\mathrm{XO}}_{\mathrm{io}}\right)=C_2\left({\mathrm{XO}}_{\mathrm{io}}\right)\&CenterDot;B_2\left({\mathrm{XO}}_{\mathrm{io}}\right)=e^{-{\mathrm{xo}}_{\mathrm{io}}}\&CenterDot;{\left[\min (0,{\mathrm{XO}}_{\mathrm{io}})\right]}^2+e^{{\mathrm{xo}}_{\mathrm{io}}-1}\&CenterDot;{\left[\min (\mathrm{0,1}-{\mathrm{XO}}_{\mathrm{io}})\right]}^2$

C: right This constraint condition, turning to the unconstrained extrema problem has

Amplification coefficient

Promptly $G_3\left({\mathrm{XO}}_{i0}\right)=C_3\left({\mathrm{XO}}_{\mathrm{io}}\right)\&CenterDot;B_3\left({\mathrm{XO}}_{\mathrm{io}}\right)=e^{\frac{\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_{\mathrm{io}}|}{\mathrm{\&epsiv;}}}\&CenterDot;{\left[\min (0,\frac{\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="NO\; XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">NO</figure-callout>}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_i{0}_{}|}{\mathrm{\&epsiv;}})\right]}^2$

D: the structure of penalty term

$G({\mathrm{<figure-callout\; id="0"\; label="XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_i,{\mathrm{AO}}_i)={\mathrm{<figure-callout\; id="1"\; label="G"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">G</figure-callout>}}_1+G_2+G_3$

$=e^{-{\mathrm{AO}}_i}\&CenterDot;{\left[\min (0,{\mathrm{AO}}_i)\right]}^2+e^{-{\mathrm{xo}}_{\mathrm{io}}}\&CenterDot;{\left[\min (0,{\mathrm{XO}}_{\mathrm{io}})\right]}^2$

$+e^{{\mathrm{xo}}_{\mathrm{io}}-1}\&CenterDot;{\left[\min (\mathrm{0,1}-{\mathrm{XO}}_{\mathrm{io}})\right]}^2+e^{\frac{\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_{\mathrm{io}}|}{\mathrm{\&epsiv;}}}\&CenterDot;{\left[\min (0,\frac{\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="NO\; XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">NO</figure-callout>}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_i{0}_{}|}{\mathrm{\&epsiv;}})\right]}^2---\left(14\right)$

Get an abundant big positive integer R1, construct penalty by (12) and (14) formula

F(AO _i，XO _i0)＝Q(AO _i，XO _i0)+R1·G(XO _i0，AO _i) (15)

If the minimal point of being obtained exceeds constraint condition, then increase R1 gradually, when R1 was fully big, the minimum solution of (14) formula was the minimum solution of objective function (11) formula, the unconstrained extrema problem that so just will have constrained extreme-value problem to turn to relatively easily to find the solution.

Step 3 is asked the single order partial derivative

The necessary condition that minimal value exists is: the single order partial derivative of functional expression (15) is 0.

A, ask local derviation to objective function earlier:

$\frac{\&PartialD;Q}{\&PartialD;{\mathrm{AO}}_m}=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}(-2\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}^{}}\&CenterDot;\frac{\&PartialD;{\mathrm{XO}}_{\mathrm{lj}}}{\&PartialD;{\mathrm{AO}}_m})$

Wherein: $\frac{\&PartialD;{\mathrm{XO}}_{\mathrm{lj}}}{\&PartialD;{\mathrm{AO}}_m}=\frac{\&PartialD;\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}\left({\mathrm{XO}}_{i0}(1-\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AO}}_i}{D_l}\exp (-\frac{{\mathrm{EO}}_i}{\mathrm{RT}})\mathrm{dT}))\right)}{\&PartialD;{\mathrm{AO}}_m}$

${\mathrm{<figure-callout\; id="0"\; label="XO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_m\&CenterDot;\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AO}}_m}{D_l}\exp (-\frac{{\mathrm{EO}}_m}{\mathrm{RT}})\mathrm{dT})\&CenterDot;{\&Integral;}_{T_0}^T\frac{1}{D_l}\exp (-\frac{{\mathrm{EO}}_m}{\mathrm{RT}})\mathrm{dT}$

$\frac{\&PartialD;Q}{\&PartialD;{\mathrm{<figure-callout\; id="0"\; label="AO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">AO</figure-callout>}}_m}=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}(-2\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}^{}}\&CenterDot;\frac{\&PartialD;{\mathrm{XO}}_{\mathrm{lj}}}{\&PartialD;{\mathrm{AO}}_{m0}})$

$=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}(-2\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}^{}}\&CenterDot;(1-\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AO}}_m}{\mathrm{Dl}}\exp (-\frac{{\mathrm{EO}}_m}{\mathrm{RT}})\mathrm{dT})))$

Here m=1,2,3 ..., NO.

B, ask local derviation to penalty term:

$\frac{\&PartialD;Q}{\&PartialD;{\mathrm{AO}}_m}=2\&CenterDot;e^{-{\mathrm{AO}}_m}\&CenterDot;\min (0,{\mathrm{AO}}_m)-{\min (0,{\mathrm{AO}}_m)}^2\&CenterDot;e^{-{\mathrm{AO}}_m}$

$\frac{\&PartialD;Q}{\&PartialD;{\mathrm{XO}}_{\mathrm{mo}}}=2\&CenterDot;\min (0,{\mathrm{XO}}_{\mathrm{mo}})\&CenterDot;e^{-{\mathrm{xo}}_{\mathrm{io}}}-e^{-{\mathrm{xo}}_{\mathrm{io}}}\&CenterDot;{\left[\min (0,{\mathrm{XO}}_{\mathrm{io}})\right]}^2-2\&CenterDot;e^{{\mathrm{xo}}_{\mathrm{io}}-1}\&CenterDot;\min (\mathrm{0,1}-{\mathrm{XO}}_{\mathrm{mo}})$

$+e^{{\mathrm{xo}}_{\mathrm{io}}-1}\&CenterDot;{\left[\min (\mathrm{0,1}-x_{\mathrm{io}})\right]}^2+2\&CenterDot;\min (0,\frac{\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_{\mathrm{io}}|}{\mathrm{\&epsiv;}})\&CenterDot;\mathrm{FN}(\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_{i0}-1)\&CenterDot;e^{\frac{\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="NO\; XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">NO</figure-callout>}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_i{0}_{}|}{\mathrm{\&epsiv;}}}$

$+e^{\frac{\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="NO\; XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">NO</figure-callout>}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_i{0}_{}|}{\mathrm{\&epsiv;}}}\&CenterDot;\min {(0,\frac{\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="NO\; XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">NO</figure-callout>}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_i{0}_{}|}{\mathrm{\&epsiv;}})}^2\&CenterDot;\mathrm{FN}(\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_{i0}-1)$

Here FN is a symbol of getting expression formula in the bracket, promptly

m＝1，2，3，...，NO

Theoretically, the minimal point place should have

$\left\{\begin{array}{c}\frac{\&PartialD;F({\mathrm{AO}}_i,{\mathrm{XO}}_{i0})}{\&PartialD;{\mathrm{AO}}_m}=\frac{\&PartialD;Q}{\&PartialD;{\mathrm{AO}}_m}+\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">R</figure-callout>}1\&CenterDot;\frac{\&PartialD;G}{\&PartialD;{\mathrm{AO}}_m}=0\\ \frac{\&PartialD;F({\mathrm{AO}}_i,{\mathrm{XO}}_{i0})}{\&PartialD;{\mathrm{<figure-callout\; id="0"\; label="XO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_m}=\frac{\&PartialD;Q}{\&PartialD;{\mathrm{<figure-callout\; id="0"\; label="XO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_m}+\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">R</figure-callout>}1\&CenterDot;\frac{\&PartialD;G}{\&PartialD;{\mathrm{<figure-callout\; id="0"\; label="XO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_m}=0\\ m=\mathrm{1,2},\mathrm{\&Lambda;},\mathrm{NO}\end{array}\right.$

Therefore, accurately obtain system of equations (15), promptly solve 2 * NO equation, 2 * NO variable undetermined then draws some possible minimal points, reaches to find the solution 2 * NO kinetic parameter AO undetermined _i, XO _I0, the purpose of peg model (5).

But, can not obtain its exact solution, but can obtain its approximate solution the complicated so non-polynomial function of system of equations (15).

Step 4 is similar to asking for of minimal point

To finding the solution of no binding occurrence problem (15), adopt speed of convergence variable-metric method very fast and that need not to calculate loaded down with trivial details matrix of second derivatives and inverse matrix thereof to be optimized calculating.

The detailed derivation that becomes the dimensional optimization algorithm only is summarized as follows its basic ideas here referring to relevant document [Li Weizheng etc., 1982, operational research]:

Appoint and give an initial point ${\stackrel{\&RightArrow;}{X}}^{\left(0\right)}={({\mathrm{<figure-callout\; id="1"\; label="AO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">AO</figure-callout>}}_1,{\mathrm{AO}}_2,K,{\mathrm{AO}}_{\mathrm{NO}},{\mathrm{XO}}_{10},{\mathrm{XO}}_{20},K,{\mathrm{XO}}_{\mathrm{NO}0})}^T,$ Calculate the first order derivative of function (15) formula by (16) formula at this point Calculate function approximate at the inverse matrix of the matrix of second derivatives of this point by suitable method simultaneously

Prove on the mathematics,

The direction that descends to some extent for the functional value that makes (15) formula.On this direction, carry out linear search, make optimum stepsize λ, promptly obtain an approximate solution of more approaching minimal point Calculate the gradient (mould of first order derivative vector) of this point, if it less than a certain given little positive number ε, thinks that promptly this point is the approximate solution of minimal point.Otherwise, copy said method to find the solution the direction that new some place descends functional value, carry out linear search, till solving the approximate minimal point that satisfies accuracy requirement, demarcate the purpose that kerogen becomes the kinetic model (6) of oil thereby reach.

Beneficial effect of the present invention: limited parallel first order reaction becomes the scaling method of hydrocarbon kinetic model parameter, mainly utilizes the existing description that organic matter is become the hydrocarbon dynamic characteristic of this method.Its principle becomes the hydrocarbon process to be described based on the Arrhenius experimental formula to organic matter, is the chemical reaction of organic cracking on the promptly organic one-tenth hydrocarbon process nature.This principle and scaling method thereof can effectively become oil with kerogen, kerogen becomes gas and oil to become these several difference of gas bigger, and should there be the reaction kinetics process of obvious different characteristic to make a distinction discussion to dissimilar organic matters, in addition, on the mathematical method of the demarcation (asking for of parameter) of model, carried out the optimization approximate solution, the one, abandoned heavy surplus calculated amount that a unlimited parallel first order reaction brings and very limited to the precision improvement of its demarcation, the 2nd, by the approximate solution of kinetic parameter having been realized experiment condition asking for of no exact solution down, the 3rd, in the process that dynamics is demarcated, by making up the penalty term of band amplification coefficient, reach and promote the penalty term constraining force, make the speed of convergence of penalty term accelerate, promote the efficiency and precision of demarcating.Solved the description that organic matter is become the complex process of hydrocarbon effectively, realized quantitatively, dynamically describing the organic matter hydrocarbon generation feature.Demarcate the kinetic parameter that obtains by this method,, can realize that organic matter to this exploratory area becomes quantitative, the dynamic evaluation of hydrocarbon history, the decision-making that this is directly connected to an exploratory area dynamics of investment and explores direction in conjunction with the buried history and the thermal history in exploratory area.

Description of drawings

Fig. 1 is the process flow diagram that limited parallel first order reaction becomes the scaling method of hydrocarbon kinetic model parameter.

Fig. 2 is the oily graph of a relation of testing conversion ratio and theoretical yield and temperature and heating rate of the organic one-tenth of mountain range shale Rock-Eval thermal simulation experiment of getting down from horse.

Fig. 3 is the organic graph of a relation that becomes gas experiment conversion ratio and theoretical yield and temperature and heating rate of mountain range shale Rock-Eval thermal simulation experiment of getting down from horse.

Fig. 4 is the graph of a relation of the southern 57 marine facies crude oil gold of wheel pipe thermal simulation experiment organic one-tenth methane experiment conversion ratio and theoretical yield and temperature and heating rate.

Fig. 5 is the graph of a relation of the total hydrocarbon gas experiment conversion ratio of the organic one-tenth of the southern 57 marine facies crude oil gold of wheel pipe thermal simulation experiment and theoretical yield and temperature and heating rate.

Embodiment

Embodiment 1: with the shut out demarcation of thermal simulation experiment parameter of 13 well mud stone of the distant basin of pine is example, and limited parallel first order reaction become the scaling method of hydrocarbon kinetic model parameter, is described in more detail.

Table 1 Du 13 well mudstone sample information

Concrete steps are, consult Fig. 1:

Step 1: make up the chemical dynamic model that kerogen becomes oil

$\mathrm{XO}=\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_i=\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}({\mathrm{XO}}_{i0}(1-\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AO}}_i}{D}\&CenterDot;\exp (-\frac{\mathrm{EOi}}{\mathrm{RT}})\mathrm{DT}))$

Step 2: make up the chemical dynamic model that kerogen becomes gas

$\mathrm{XG}=\underset{i=1}{\overset{\mathrm{NG}}{\mathrm{\&Sigma;}}}{\mathrm{XG}}_i=\underset{i=1}{\overset{\mathrm{NG}}{\mathrm{\&Sigma;}}}({\mathrm{XG}}_{i0}(1-\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AG}}_i}{D}\&CenterDot;\exp (-\frac{{\mathrm{EG}}_i}{D})\mathrm{dT}))$

Step 3: carry out Rock-Eval (PY-GC) thermal simulation experiment

Adopt Rock-Eval-II type pyrolysis instrument, respectively under the heating rate condition of 10 ℃/min and 50 ℃/min with sample from 200 ℃ of heat temperature raisings to 600 ℃, real time record is produced the hydrocarbon amount, promptly gets into hydrocarbon rate-temperature relation.Then under identical heating-up temperature scope and heating rate condition, collect the pyrolysis product promoting the circulation of qi analysis of hplc (being that PY-GC analyzes) of going forward side by side with 30 ℃ temperature intervals, make each temperature section gas (C from gas chromatogram ₁-C ₅) and liquid (C _{+ 6}) relative content of component, in conjunction with the Rock-Eval experimental result, be about to produce hydrocarbon (oil+gas) rate-time-temperature relation curve and be converted to oil productivity-time-temperature and factor of created gase-time-two curves of temperature relation (consulting Fig. 2 and Fig. 3), for the usefulness of the kinetic parameter of demarcating oil, one-tenth gas respectively.

Step 4: demarcate oil (gas) dynamics initial value to determine

The number of the different heating rates experiments of selecting be 2 (promptly be respectively 10 ℃/min and 50 ℃/min), and the number value of parallel reactor is 19, energy of activation be distributed as 160-340kJ/mol, be spaced apart 10kJ/mol, initial pre-exponential factor value is 1 * 10 ¹⁴Min ^-1, adopt different pre-exponential factors to demarcate precision prescribed ε=0.001.

Step 5: become the demarcation of oil (gas) kinetic parameter

A, structure objective function

$Q({\mathrm{EO}}_i,{\mathrm{AO}}_i,{\mathrm{XO}}_{i0})=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}{\left(\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}}\right)}^2$

Selecting heating rate 1 is 10 ℃/min and 50 ℃/min, is XO1 by the measured oil productivity of experiment when reaching temperature j _1j, L0 is the number of different heating rate experiments, is 2; J0 is the sampling number from the empirical curve, i.e. experimental analysis number of parameters value 50.

B, determine constraint condition

AO in the formula _i, XO _I0(representing with the percentage that accounts for the overall reaction amount) should satisfy:

Get ε=0.001.

C, structure penalty

$G({\mathrm{<figure-callout\; id="0"\; label="XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_i,{\mathrm{AO}}_i)={\mathrm{<figure-callout\; id="1"\; label="G"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">G</figure-callout>}}_1+G_2+G_3$

$={\left[\min (0,{\mathrm{AO}}_i)\right]}^2+{\left[\min (0,{\mathrm{XO}}_{i0})\right]}^2+{\left[\min (\mathrm{0,1}-{\mathrm{Xo}}_{i0})\right]}^2+{\left[\min (0,\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="NO\; XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">NO</figure-callout>}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_i{0}_{}\left|\right)\right]}^2$

D, ask the single order partial derivative

Objective function is asked local derviation:

$\frac{\&PartialD;Q}{\&PartialD;{\mathrm{AO}}_m}=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}(-2\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}^{}}\&CenterDot;\frac{\&PartialD;{\mathrm{XO}}_{\mathrm{lj}}}{\&PartialD;{\mathrm{AO}}_m})$

Wherein: $\frac{\&PartialD;{\mathrm{XO}}_{\mathrm{lj}}}{\&PartialD;{\mathrm{AO}}_m}=\frac{\&PartialD;\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}\left({\mathrm{XO}}_{i0}(1-\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AO}}_i}{D_l}\exp (-\frac{{\mathrm{EO}}_i}{\mathrm{RT}})\mathrm{dT}))\right)}{\&PartialD;{\mathrm{AO}}_m}$

${\mathrm{<figure-callout\; id="0"\; label="XO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_m\&CenterDot;\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AO}}_m}{D_l}\exp (-\frac{{\mathrm{EO}}_m}{\mathrm{RT}})\mathrm{dT})\&CenterDot;{\&Integral;}_{T_0}^T\frac{1}{D_l}\exp (-\frac{{\mathrm{EO}}_m}{\mathrm{RT}})\mathrm{dT}$

$\frac{\&PartialD;Q}{\&PartialD;{\mathrm{<figure-callout\; id="0"\; label="AO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">AO</figure-callout>}}_m}=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}(-2\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}^{}}\&CenterDot;\frac{\&PartialD;{\mathrm{XO}}_{\mathrm{lj}}}{\&PartialD;{\mathrm{AO}}_{m0}})$

$=\underset{l=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="L"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">L</figure-callout>}0}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{J0}{\mathrm{\&Sigma;}}}(-2\frac{{\mathrm{<figure-callout\; id="1"\; label="XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}1}_{\mathrm{lj}}-{\mathrm{<figure-callout\; id="1"\; label="XO\; lj\; XO"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_{\mathrm{lj}}}{{\mathrm{XO}}_{}^{}}\&CenterDot;(1-\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AO}}_m}{\mathrm{Dl}}\exp (-\frac{{\mathrm{EO}}_m}{\mathrm{RT}})\mathrm{dT})))$

Here m=1,2,3 ..., NO.

The local derviation of penalty term is:

$\frac{\&PartialD;Q}{\&PartialD;{\mathrm{AO}}_m}=2\&CenterDot;\min (0,{\mathrm{AO}}_m)$

$\frac{\&PartialD;Q}{\&PartialD;{\mathrm{XO}}_{\mathrm{mo}}}=2\&CenterDot;\min (0,{\mathrm{XO}}_{\mathrm{mo}})-2\&CenterDot;\min (\mathrm{0,1}-{\mathrm{XO}}_{\mathrm{mo}})-2\&CenterDot;\min (0,\mathrm{\&epsiv;}-|1-\underset{i=1}{\overset{\mathrm{<figure-callout\; id="0"\; label="NO\; XO\; i"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">NO</figure-callout>}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_i{0}_{}\left|\right)\&CenterDot;\mathrm{FN}(\underset{i=1}{\overset{\mathrm{NO}}{\mathrm{\&Sigma;}}}{\mathrm{XO}}_{i0}-1)$

Here FN is a symbol of getting expression formula in the bracket, promptly

m＝1，2，3，…，

NO

Theoretically, the minimal point place should have

$\left\{\begin{array}{c}\frac{\&PartialD;F({\mathrm{AO}}_i,{\mathrm{XO}}_{i0})}{\&PartialD;{\mathrm{AO}}_m}=\frac{\&PartialD;Q}{\&PartialD;{\mathrm{AO}}_m}+\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">R</figure-callout>}1\&CenterDot;\frac{\&PartialD;G}{\&PartialD;{\mathrm{AO}}_m}=0\\ \frac{\&PartialD;F({\mathrm{AO}}_i,{\mathrm{XO}}_{i0})}{\&PartialD;{\mathrm{<figure-callout\; id="0"\; label="XO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_m}=\frac{\&PartialD;Q}{\&PartialD;{\mathrm{<figure-callout\; id="0"\; label="XO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_m}+\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">R</figure-callout>}1\&CenterDot;\frac{\&PartialD;G}{\&PartialD;{\mathrm{<figure-callout\; id="0"\; label="XO\; m"\; filenames="HDA0000079711170000021.png,HDA0000079711170000022.png"\; state="\{\{state\}\}">XO</figure-callout>}}_m}=0\\ m=\mathrm{1,2},\mathrm{\&Lambda;},\mathrm{NO}\end{array}\right.$

Asking for of E, approximate minimal point

By initial point

Begin search, calculate the first order derivative of penalty at this point by the local derviation of objective function and penalty term

Calculate function approximate at the inverse matrix of the matrix of second derivatives of this point by suitable method simultaneously

The direction that descends to some extent for the functional value that makes objective function.On this direction, carry out linear search, make optimum stepsize λ, promptly obtain an approximate solution of more approaching minimal point

Calculate the gradient (mould of first order derivative vector) of this point, if it less than a certain given little positive number ε, thinks that promptly this point is the approximate solution of minimal point.Otherwise, copy said method to find the solution the direction that new some place descends functional value, carry out linear search, till solving the approximate minimal point that satisfies accuracy requirement.

Step 5: calibration result

The organic matter that the relevant sample rock-eval that obtains by the calibration principle in the technique scheme tests becomes oil and organic matter directly to become the aerodynamics parameter, sees Table 1-4.Fig. 2 represents is the theoretical yield and the relation of temperature of mountain range shale sample under the corresponding conditions that becomes oily conversion ratio by the organic matter of rock-eval experiment gained under the different heating rate conditions and calculated by institute's peg model of getting down from horse, and Fig. 3 represents to get down from horse that mountain range shale sample is becoming the gas conversion ratio by the organic matter of rock-eval experiment gained under the different heating rate conditions and by the theoretical yield under the corresponding conditions of institute's peg model calculating and the relation of temperature.Experiment parameter and calculating parameter coincide preferably between the two and have tentatively shown the precision and the feasibility of institute's peg model.

The table 1-4 organic kinetic parameter that becomes hydrocarbon reaction of mountain range shale rock-eval experimental calibration of getting down from horse

At last, with the chemical dynamic model of calibrated one-tenth hydrocarbon, the organic matter of using the exploratory area becomes quantitative, the dynamic evaluation of hydrocarbon history.

Embodiment 2: the demarcation with the thermal simulation experiment parameter of taking turns southern 57 well crude oil is an example, and limited parallel first order reaction become the scaling method of hydrocarbon kinetic model parameter, is described in more detail.

Table 2 sample message

The area

The degree of depth

Layer position

Lithology

C(％)

H(％)

N(％)

?S(％)

O(％)

The Tarim Basin

The Ordovician system

Marine facies crude oil

85.16

14.37

0.67

?-

＜0.3

Embodiment 2 is to have utilized the gold pipe thermal simulation experiment under the enclosed system to obtain experiment parameter with embodiment 1 difference, and what mainly demarcate is that crude oil pyrolysis becomes the aerodynamics model, that is:

$\mathrm{XG}=\underset{i=1}{\overset{\mathrm{NG}}{\mathrm{\&Sigma;}}}{\mathrm{XG}}_i=\underset{i=1}{\overset{\mathrm{NG}}{\mathrm{\&Sigma;}}}({\mathrm{XG}}_{i0}(1-\exp (-{\&Integral;}_{T_0}^T\frac{{\mathrm{AG}}_i}{D}\&CenterDot;\exp (-\frac{{\mathrm{EG}}_i}{D})\mathrm{dT}))$

Gold pipe thermal simulation experiment

The qualification system is adopted in experiment, and the wall thickness of golden pipe is 0.2mm, external diameter 4mm, maximum volume 1cm3; The application of sample weight range is 5～100mg oil sample.The golden pipe of dress behind the sample placed the argon gas case, displaces air in tube, carry out sealing with high-frequency induction welder, and the golden pipe that sealing is good to put into water be the autoclave of pressure medium.System inserts 15 autoclaves simultaneously, and each autoclave connects a stop valve and finally is connected in the same pressure system.System pressure is controlled on the spot pressure of setting by control pressurer system, manometric precision is 0.5MPa.Autoclave places constant temperature water tank, and the heating rate constant speed that constant temperature water tank can be certain heats up, and the temperature controller precision is 0.1 ℃.Each gold pipe all is under identical pressure, the temperature conditions like this.Pressure, temperature system all are controlled by the center controlling computer.Under the pressure of 50MPa, the heating rate with 20 ℃/h and 2 ℃/h is warming up to crude oil sample more than 600 ℃ from 200 ℃ respectively.Close the stop valve that some autoclaves connect at target temperature point (is the interval with 20 ℃), this autoclave is taken out from constant temperature oven, take out the gold pipe after cooling.Place special gas collection, quantitative system to carry out accurate measurement golden pipe, and carry out GC with the HP6890 gas chromatograph and analyze.Behind the effective liquid nitrogen frozen of gold, cut off rapidly and put into solvent, ultrasonic concussion 5min, lighter hydrocarbons be not loss fully, and realization is quantitative to oil residues.Get the gas production rate of each experimental point unit mass sample by gas volume and sample size.The highest experimental temperature point when (being higher than 600 ℃) oil-breaking become the gas ability near exhausted, calculate maximum (limit) factor of created gase of the whole cracking into gas of oil sample thus.What each experimental point produced that methane gas rate and the ratio of limit factor of created gase is each point becomes the methane gas conversion ratio, must become methane gas conversion ratio-heating temperature-heating rate relation curve thus, same conversion ratio-heating temperature-heating rate the relation curve that must become total hydrocarbon gas is used when oil becomes the chemical dynamic model of gas for demarcating.

Scaling method is seen embodiment 1, and the crude oil pyrolysis of the relevant sample gold pipe experiment that obtains thus becomes the aerodynamics parameter.The theoretical yield and the relation of temperature that Fig. 4 represents is southern 57 crude oil samples of wheel under the corresponding conditions that becomes the methane gas conversion ratio by the oil of gold pipe experiment gained under the different heating rate conditions and calculated by institute's peg model, Fig. 5 represents, and to be southern 57 crude oil samples of wheel becoming total hydrocarbon gas conversion ratio by the oil of gold pipe experiment gained under the different heating rate conditions and by the theoretical yield under the corresponding conditions of institute's peg model calculating and the relation of temperature.Coincide preferably between the two and tentatively shown the precision and the feasibility of institute's peg model.

Table 2 is taken turns the kinetic parameter that southern 57 well crude oil gold pipe experimental calibration oil becomes methane and total hydrocarbon solid/liquid/gas reactions

At last, the chemical dynamic model of calibrated one-tenth hydrocarbon, the organic matter of using the exploratory area becomes quantitative, the dynamic evaluation of hydrocarbon history.

Claims (4)

1. limited parallel first order reaction becomes the scaling method of hydrocarbon kinetic model parameter, it is characterized in that: at first, be created as the chemical dynamic model of hydrocarbon, become the chemical dynamic model of hydrocarbon to comprise that the oily chemical dynamic model of kerogen one-tenth, the chemical dynamic model that kerogen becomes gas become the chemical dynamic model of gas with oil-breaking; Then, observed parameter or experiment parameter are input to the chemical dynamic model of setting up good one-tenth hydrocarbon, the chemical dynamic model that becomes hydrocarbon is demarcated; Secondly, adopt structure objective function and penalty optimization to ask for approximate exact solution, realize organic matter is become the kinetic description of hydrocarbon process; At last, with the chemical dynamic model of calibrated one-tenth hydrocarbon, the organic matter that is applied to the exploratory area becomes quantitative, the dynamic evaluation of hydrocarbon history.

2. limited parallel first order reaction according to claim 1 becomes the scaling method of hydrocarbon kinetic model parameter, it is characterized in that:

A, kerogen become the chemical dynamic model of oil

If kerogen (KEO) becomes oily process to be made up of a series of (NO) individual parallel first order reaction, the energy of activation that each reaction pair is answered is EO _i, unit: kj/mol (kJ/mol), pre-exponential factor AO _i, unit: per minute, (min), and the kerogenic original latent amount of establishing corresponding each reaction is XO _I0, KO _iBe the reaction rate constant that i kerogen becomes the oil reaction, the oil that each reaction generates is O _i, the oil generating quantity of each reaction is XO _i, i=1,2 ..., NO, promptly to time t (minute, min), the time

Obtain by Arrhenius (Arrhenius) equation:

i＝1，2，Λ，NC

Wherein R is gas law constant (8.31447kJ/molK); T is a thermodynamic temperature, Kelvin (K); Experiment adopts constant speed to heat up, and heating rate is D, unit: degrees celsius/minute (℃/min);

Promptly

Get by (2)～(4) formula

Following formula from T0 → T integration, and is noticed XO _{I (T0)}=0, XO _{I (T)}=XO _i

Total oil generating quantity XO (relative percentage) of NO parallel reactor then is

B, kerogen become the chemical dynamic model of gas

In like manner, if establishing kerogen directly becomes the reaction of gas to be made up of (NG) individual parallel reactor, the energy of activation of each parallel reactor is EG _i, unit: kj/mol (kJ/mol), pre-exponential factor AG _i, unit: per minute (/min), the initial amount of diving is XG _I0, unit: dimensionless, D is a heating rate, unit: degrees celsius/minute (℃/min), R is gas law constant (8.31447kJ/molK), get the directly angry computing formula of measuring (number percent) of temperature variant T kerogen in period (KEO) to be

Compare with (6) formula, (7) formula only is that the pair mark of related variable is different; O represents oil, and G represents gas;

C, oil-breaking become the chemical dynamic model of gas

If oil-breaking becomes the process of gas to be made up of NOG parallel reactor, the energy of activation of each reaction is EOG _i, unit: kj/mol (kJ/mol), pre-exponential factor are AOG _i, unit: per minute (/min), corresponding original latent amount is XOG _I0, unit: dimensionless, it is KOG that i kerogen becomes the reaction rate constant of oil reaction _I0, when reaction proceed to time t (minute, in the time of min), factor of created gase (representing with the percentage that accounts for the overall reaction amount) is XOG _i, unit: dimensionless then has

Wherein

It is the reaction rate constant that i oil-breaking becomes solid/liquid/gas reactions;

(9) substitution (8) formula is put the back in order to time integral, and notices XOG _{I (t=0)}=0, XOG _{I (t)}=XOG _i, the total angry amount that obtains NOG parallel reactor is

Equally, the R in the formula is gas law constant (8.31447kJ/molK), and T is a thermodynamic temperature, Kelvin (K);

At last, the demarcation of the chemical dynamic model by becoming hydrocarbon, the organic matter of realizing using the exploratory area become the hydrocarbon history quantitatively, dynamic evaluation.

3. limited parallel first order reaction according to claim 1 becomes the scaling method of hydrocarbon kinetic model parameter, it is characterized in that: described experiment parameter or observed parameter mainly are meant the relation between temperature-time-product hydrocarbon rate of obtaining sample, from the pyrolysis experiment parameter under the different condition, the pyrolysis experiment comprises the Rock-Eval experiment, thermogravimetric-mass spectrometry (TG-MS) experiment, gold pipe thermal simulation experiment or small size sealing oil gas generate simulation (MSSV) experiment, obtain its heating rate (centigrade per minute respectively, ℃/min) or thermostat temperature (degree centigrade, ℃), experimental period (minute, min), produce hydrocarbon rate (milliliter or gram, ml or g) parameter, step is:

Step 1, representational rock, kerogen or the crude oil sample of selection low-maturity (Ro＜0.6%) high abundance (TOC＞1%);

Step 2 is carried out sample constant temperature or is decided the heating rate thermal simulation experiment;

Step 3, the record experiment parameter, comprise the thermal simulation time (minute, min), thermostat temperature (temperature, ℃) or decide heating rate (centigrade per minute, ℃/min), hydrocarbon production (milliliter or restrain ml or g) parameter;

Step 4, sample is heated to produces the hydrocarbon limit, promptly sample produce hydrocarbon can force failure, product hydrocarbon (oil gas) amount of each time point added up to be obtained accumulative total and produces hydrocarbon (oil gas) amount, the oil and gas production of each time is obtained produce oil gas rate divided by the accumulative total oil and gas production respectively, comprise kerogen oil productivity XO _Aj(number percent), kerogen factor of created gase XG _Aj(number percent) or crude oil factor of created gase XOG _Aj(number percent), wherein a is a certain heating rate, j is a certain temperature spot.

4. limited parallel first order reaction according to claim 1 becomes the scaling method of hydrocarbon kinetic model parameter, it is characterized in that: the step that the chemical dynamic model that becomes hydrocarbon is demarcated is: become the demarcation of the chemical dynamic model of oil to describe with kerogen, the method that kerogen becomes the chemical dynamic model of gas to become the chemical dynamic model of gas to demarcate with oil-breaking is identical;

Step 1, the structure objective function

Being located at a certain heating rate 1, is XO1 by the measured oil productivity of experiment when reaching a certain temperature j _1j(number percent) is under identical condition, if there be a certain group of EO _i, unit: kj/mol (kJ/mol), AO _i, unit: per minute (/min), XO _I0The value of (dimensionless) makes all 1, and j has XO1 _1j-XO _1j=0, then should group EO _i, unit: kj/mol (kJ/mol), AO _i, unit: per minute (/min), XO _I0, unit, (number percent) is institute and asks; But because aspects such as experimental errors, this is practically impossible; Therefore, can only ask and make XO1 _1j-XO _1jLittle EO tries one's best _i, AO _i, XO _I0Value; For this reason, the structure objective function is:

L0 is the number of different heating rates experiment in the formula, and J0 is the sampling number from the empirical curve;

The number value of parallel reactor is big more, just can comprise more that kerogen becomes all reaction types of oil, thereby should be accurate more, but because the calculated amount that this moment, model calibration reached when model is used subsequently is too big, be difficult to practicability, and, find in the actual process of demarcating, though model to the thermal simulation time under the experiment condition (minute, min), thermostat temperature (degree centigrade, ℃) or decide heating rate (centigrade per minute, ℃/min), hydro carbons productive rate (number percent, %) fitting degree of parameter is generally improved with the segmentation of parallel reactor, but after the number of parallel reactor acquired a certain degree, the improvement of fitting degree was not obvious; Therefore, only need with limited individual parallel reactor with certain intervals;

Like this, because EO _iThe distribution range of the energy of activation by determining parallel reactor and the energy of activation of adjacent parallel reaction are found the solution at interval, and then (11) formula turns to:

In addition, notice that kerogen becomes the AO in oily model (6) formula _i, XO _I0, represent with the percentage that accounts for the overall reaction amount, should satisfy:

Like this, the determining of model (6), i.e. asking for of kinetic parameter, problem just turns to the non-negative minimal point problem of objective function (12) when satisfying constraint condition (13) of asking;

Step 2, the structure penalty

The above-mentioned minimizing problem more complicated of finding the solution that contains constraint condition, because except making target function value descends gradually, be also noted that the feasibility of separating, promptly see to separate whether be within the constraint condition institute restricted portion, by making up the penalty term of band amplification coefficient, reach and promote the penalty term constraining force, make the speed of convergence of penalty term accelerate, promote the efficiency and precision of demarcating, here adopt Means of Penalty Function Methods will have constrained extreme-value problem (12), (13) to turn to the unconstrained extrema problem, its step is as follows:

A: to AO _i＞0, this constraint condition, turning to the unconstrained extrema problem has

Be C ₁(AO _i)=[min (0, AO _i)] ²

Amplification coefficient B ₁=e ^-AOi

B: to 1＞XO _I0＞0, this constraint condition, turning to the unconstrained extrema problem has

Amplification coefficient

Promptly

C: right

This constraint condition, turning to the unconstrained extrema problem has

Amplification coefficient

Promptly

D: the structure of penalty term

Get an abundant big positive integer R1, construct penalty by (12) and (14) formula

F(AO _i，XO _i0)＝Q(AO _i，XO _i0)+R1·G(XO _i0，AO _i) (15)

If the minimal point of being obtained exceeds constraint condition, then increase R1 gradually, when R1 was fully big, the minimum solution of (14) formula was the minimum solution of objective function (11) formula, the unconstrained extrema problem that so just will have constrained extreme-value problem to turn to relatively easily to find the solution;

Step 3 is asked the single order partial derivative

The necessary condition that minimal value exists is: the single order partial derivative of functional expression (15) is 0;

A, ask local derviation to objective function earlier:

Wherein:

Here m=1,2,3 ..., NO;

B, ask local derviation to penalty term:

Here FN is a symbol of getting expression formula in the bracket, promptly

m＝1，2，3，...，NO

Theoretically, the minimal point place should have

Therefore, accurately obtain system of equations (15), promptly solve 2 * NO equation, 2 * NO variable undetermined then draws some possible minimal points, reaches to find the solution 2 * NO kinetic parameter AO undetermined _i, XO _I0, the purpose of peg model (5);

But, can not obtain its exact solution, but can obtain its approximate solution the complicated so non-polynomial function of system of equations (15);

Step 4 is similar to asking for of minimal point

To finding the solution of no binding occurrence problem (15), adopt speed of convergence variable-metric method very fast and that need not to calculate loaded down with trivial details matrix of second derivatives and inverse matrix thereof to be optimized calculating.

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