CN102728624A - Method for setting load distribution of finish rolling band steel - Google Patents

Abstract

The invention relates to a method for setting the load distribution of finish rolling band steel, belonging to the field of production control of finish rolling band steel, and solving the problems that the rolling force proportion of stands is large in fluctuation when the rolling condition is changed under the pressed-down distribution mode, and the thickness distribution of the stands is large in fluctuation when the rolling force model is large in deviation under the rolling force distribution mode. According to the invention, the pressed-down distribution mode and the rolling force distribution mode are combined; the rolling force distribution result is restricted by the distribution range determined by pressed-down distribution; the requirement that the rolling force of stands is decreased according to proportion is met while the thickness distribution fluctuation of the stands is reduced; the results of two distribution modes are gradually consistent, and the iterations are gradually reduced by learning a pressed-down rate learning table when the deviation of the rolling force model is small; and the fluctuation range of the pressed-down rate of the stands is effectively controlled through the pressed-down distribution amplitude limit when the deviation of the rolling force model is large. By adopting the method, the advantages of the two modes are combined and the shortages are overcome, so that the rolling schedule setting precision and the rolling stability of the finish rolling band steel are effectively improved.

Description

A kind of fine-rolling strip steel sharing of load establishing method

Technical field

The present invention relates to fine-rolling strip steel production control field, relate in particular to a kind of fine-rolling strip steel sharing of load establishing method.

Background technology

In the hot strip rolling unit, the formulation of rolling procedure plays an important role for improving the quality of products, and rationally whether it all have direct influence to the stability of difficulty or ease, unit maintainability and the production process of the height of product quality, rolling equipment adjustment.Rolling procedure generally comprises reduction system, speed system, tension force system, temperature schedule etc.Wherein reduction system refers to the distribution system of each frame (or each passage) drafts, is referred to as sharing of load or thickness in the industry usually and distributes.

Sharing of load is the core of rolling procedure, and it directly has influence on product qualities such as plate shape, thickness of slab precision, and sharing of load also has material impact to item indexs such as the stability of required power, roller consumption, production process and operating rates.The load distribution method of hot strip rolling unit from before the experience schedule method begin, also experienced several stages such as energy consumption method, dynamic load apportion design, sharing of load Y-factor method Y, artificial intelligence approach.

At present, the method that modernized hot continuous rolling unit sharing of load more generally adopts is the sharing of load Y-factor method Y, and it is by the drafts and the rolled piece thickness of certain given each frame of sharing of load coefficient.The sharing of load coefficient can be confirmed with reference to kind and specification, also can adopt the method for offline optimization to confirm.The sharing of load Y-factor method Y mainly comprises types such as drafts, reduction ratio, roll-force and power.Wherein roll-force, power type need iterative computation, for convenient the analysis all is classified as the roll-force allocation model; Drafts, reduction ratio type then can directly obtain the drafts and the thickness of each frame, all are classified as and depress allocation model.The pluses and minuses of these two kinds of patterns:

The advantage of depressing distribution is that each frame thickness distributes fluctuation little, and computational process is simple; Each frame roll-force ratio fluctuation can not be eliminated the roll-force reversal of the natural order of things greatly when shortcoming was rolling situation change (like roll change), can not guarantee production board shape.

The advantage that roll-force is distributed is each frame roll-force decline ratio substantially constant, helps production board shape and rolling stability; Each frame thickness distributed fluctuation greatly, the computational process complicacy when shortcoming was that big or process conditions change greatly when the rolling force model deviation.

At present; Some disclosed patent documents also appear in the establishing method of relevant fine-rolling strip steel sharing of load; Example, name is called the method for putting down in writing in " a kind of method of dynamically setting tandem rolling schedule of hot rolling strip " file (CN101733289A): will depress distribution coefficient as the optimizing starting point, and calculate corresponding load this moment; And compare with the target load distribution coefficient, according to each reduction in pass of difference adjustment; Through repeatedly comparing and adjustment, reach assumed load and distribute the purpose of approaching the target load distribution.Its limitation is: 1. only adjust the distribution of two frames, harmony is bad between frame at every turn, and adjustment efficient is lower; 2. only will depress distribution coefficient as the optimizing starting point, still a kind of in essence roll-force is distributed, and can not overcome the shortcoming that roll-force is distributed.

Summary of the invention

The objective of the invention is to propose a kind of fine-rolling strip steel sharing of load establishing method, thereby improve fine-rolling strip steel rolling procedure setting accuracy and rolling stability in order to solve the problem that prior art exists.

The objective of the invention is to realize through following technical scheme:

A kind of fine-rolling strip steel sharing of load establishing method may further comprise the steps:

S1, according to reduction ratio distribution coefficient table and reduction ratio learning table; Calculate the allocation result of depressing of each frame of finish rolling by depressing allocation model; Said depress under the allocation result finger pressure reduction ratio

that distributes and depress exit thickness

S2 of distribution, the exit thickness

of above-mentioned each frame that obtains being depressed distribution is as initial value; Absolute draft amount by each frame of roll-force allocation model iterative computation; Satisfy target call up to the roll-force ratio; Obtain the roll-force allocation result of each frame; Said roll-force allocation result comprises reduction ratio

S3 that exit thickness

that roll-force is distributed and roll-force distribute, the roll-force allocation result of said each frame is carried out amplitude limiting processing; It is departed from saidly depress allocation result within the specific limits; Thereby the final reduction ratio of each frame is being distributed in the certain limit of confirming by depressing; Just do not get the roll-force allocation result as long as do not go beyond the scope, going beyond the scope then is limited in boundary with it; After S4, band steel toe portion wear the band completion and collect actual roll-force, judge the deviation of each frame rolling force setup value and actual roll-force; S5 judges each frame deviate whether in θ, and θ is the Deviation Control parameter; Get 5%-10%; If do not satisfy each frame roll-force deviation all at θ with interior condition, then do not carry out reduction ratio study, at this moment; Other data of equivalent layer are not upgraded in the said reduction ratio learning table, and flow process gets into end step; S6, if each frame deviate is then learnt the reduction ratio of this setting in θ, it is other that learning value is updated to reduction ratio study form equivalent layer, through the study process, the other roll-force of this layer can progressively reach the requirement of target roll-force ratio; S7, sharing of load set and finish.

Said step S1 may further comprise the steps:

S11; get table; get the reduction ratio and the reduction ratio learning value of each frame from reduction ratio distribution coefficient table and reduction ratio learning table; both carry out addition, obtain the reduction ratio allocation table scale value of each frame

S12 carries out relativization and calculates, and obtains the allocation result of depressing of each frame: must equal the overall reduction of finish rolling according to the drafts sum of each frame of finish rolling, the scale factor xrk that obtains reduction ratio is:

$\mathrm{Xrk}=\frac{-\mathrm{Kb}\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="Xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">Xs</figure-callout>}}_1+\sqrt{{(\mathrm{Kb}\&CenterDot;{\mathrm{Xs}}_1)}^2+4\&CenterDot;\mathrm{Ka}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="Xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">Xs</figure-callout>}}_2\&CenterDot;\mathrm{<figure-callout\; id="2"\; label="Ln\; x"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">Ln</figure-callout>}x}}{2\&CenterDot;\mathrm{Ka}\&CenterDot;{\mathrm{Xs}}_2}$ Formula 1

Wherein,

h _nBe finish to gauge thickness, h ₀Be intermediate blank thickness;

${\mathrm{<figure-callout\; id="1"\; label="Xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">Xs</figure-callout>}}_1=\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}_0^i,$ ${\mathrm{<figure-callout\; id="2"\; label="Xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">Xs</figure-callout>}}_2=\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{\left({\mathrm{\&epsiv;}}_0^i\right)}^2,$ N is the finishing pass number, and i is a shelf number;

kb＝0.906501，ka＝0.959597。

Reduction ratio after the reduction ratio allocation table scale value relativization of each frame is a reduction ratio

of depressing distribution

${\mathrm{\&epsiv;}}_{\mathrm{Red}}^i=\mathrm{Xrk}\&CenterDot;{\mathrm{\&epsiv;}}_0^i$ Formula 2

According to depressed the assigned rolling reduction rate

to obtain the front of n-1 rack respective depression of assigned to exit thickness

$h_{\mathrm{Red}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i),i=\mathrm{1,2},\mathrm{\&Lambda;},n-1.$ Formula 3

The reduction ratio allocation table scale value of each frame among the said step S11

The process of obtaining is: according to big type of steel grade, level of thickness and the width class information of current band steel, do not get the reduction ratio of each frame of finish rolling from depressing distribution coefficient table equivalent layer

Wherein i is a shelf number; According to tapping mark, level of thickness and the width class information of current band steel, do not get the reduction ratio learning value of each frame of finish rolling from reduction ratio learning table equivalent layer

Wherein i is a shelf number, and said reduction ratio learning table is the tables of data that is obtained by program automatic learning, and data are brought in constant renewal in through study in the table, and initial value gets 0; The reduction ratio allocation table scale value of said each frame

For: ${\mathrm{\&epsiv;}}_0^i={\mathrm{\&epsiv;}}_{\mathrm{Ori}}^i+{\mathrm{\&epsiv;}}_{\mathrm{Lrn}}^i.$

Said step S2 may further comprise the steps:

S21, big type of steel grade and level of thickness information according to current band steel do not get access to the initial roll-force distribution coefficient of each frame α from roll-force distribution coefficient table equivalent layer _0i

S22 operates correcting process, obtains the revised roll-force distribution coefficient of each frame α _1i: the operation modifying factor ζ that collects each frame from operation screen HMI _i, the revised roll-force distribution coefficient of each frame α _1iFor:

α _1i=α _0i(1+ ζ _i) formula 4

Wherein, ζ _iSpan is-20%～20%;

Carry out normalization, the roll-force distribution coefficient α after the normalization _iFor: ${\mathrm{\&alpha;}}_i=\frac{{\mathrm{\&alpha;}}_{1i}}{\mathrm{Max}\left({\mathrm{\&alpha;}}_{1i}\right)};$

S23; Calculating distributes roll-force and the roll-force of each frame down derivative to the absolute draft amount at current thickness: calculating roll-force

is established: i is a shelf number; J is an iterations;

is the roll-force of the j time iteration of i frame; The exit thickness of

the j time iteration of i frame, the exit thickness of

the j time iteration of i-1 frame; The exit thickness of depressing distribution

that the iterative computation preceding i frame exit thickness initial value first time

obtains for above-mentioned formula (3) is under the situation that other technological parameter is fixed; Roll-force is the function of inlet thickness and exit thickness, exists:

Calculate the derivative of roll-force to the absolute draft amount

$\frac{\&PartialD;p_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)}=\frac{p_{i1}^j-p_{i2}^j}{\mathrm{\&Delta;}{h}_i^j\&CenterDot;\mathrm{Dt}}$

Where,

is the absolute amount of reduction is perturbed rolling force;

is the absolute reduction of negative perturbed rolling force,

is the i j-th iteration rack absolute reduction; has the following function:

where

After a positive absolute reduction disturbance,

is negative absolute reduction after disturbance,

$\mathrm{\&Delta;}{h}_{i1}^j=(1+\frac{\mathrm{Dt}}{2})\&CenterDot;\mathrm{\&Delta;}{h}_i^j,$ $\mathrm{\&Delta;}{h}_{i2}^j=(1-\frac{\mathrm{Dt}}{2})\&CenterDot;\mathrm{\&Delta;}{h}_i^j,$ Wherein dt is disturbance, gets 1%;

S24, according to the roll-force of above-mentioned each frame and the roll-force result of calculation to absolute draft amount derivative, calculate the correction value of each frame absolute draft amount:

Rolling distribution iterative computation formula:

$\mathrm{\&delta;}\left(\mathrm{\&Delta;}{h}_i^j\right)=(\mathrm{\&Sigma;}{p}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{\&PartialD;p_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)}-\mathrm{\&Sigma;}((\mathrm{\&Sigma;}{p}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{\&PartialD;p_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)})\&CenterDot;\frac{\mathrm{\&Delta;}{h}_i^j}{\mathrm{\&Sigma;\; \&Delta;}{h}_i^j}$ Formula 5

Wherein:

is the correction value of the absolute draft amount of the j time iteration of i frame;

S25; After the correction value of each frame absolute draft amount

obtains; Calculate the updating value of each frame absolute draft amount, this updating value is that absolute draft amount computing formula of the j+1 time iteration is:

$\mathrm{\&Delta;}{h}_i^{j+1}=\mathrm{\&Delta;}{h}_i^j+{\mathrm{Damp}}_j\&CenterDot;\mathrm{\&delta;}\left(\mathrm{\&Delta;}{h}_i^j\right)$ Formula 6

Damp wherein _jBe damped coefficient, damp _j=damp_mpy β+(1-β) (1-e ^-j),

Damp_mpy gets 1.0, and β gets 0.6;

S26; After the updating value of each frame absolute draft amount

obtains; Calculate the updating value of each frame exit thickness, this updating value is the exit thickness

of the j+1 time iteration

And intermediate blank thickness h ₀Be known quantity, according to

First frame only

All the other all are the inputs that previous frame is output as a back frame, calculate last frame from first frame, obtain the updating value of each frame exit thickness, and this value will be used for next iteration and calculate roll-force usefulness, wherein

The exit thickness that refers to the j+1 time iteration of i-1 frame;

S27, judge whether the roll-force ratio satisfies the condition of convergence:

Roll-force distributes the condition of convergence of iterative computation to be:

$|\frac{p_i^j}{\mathrm{\&Sigma;}{p}_i^j}-\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}|\&le;\frac{\mathrm{\&tau;}}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}$ Formula 7

Wherein, τ is a positive number, gets 0.01; Set up when formula (7), finishing iteration was calculated when perhaps iterations j surpassed set point number, otherwise continued to carry out said S23 step, and iterations j adds up 1 simultaneously;

S28, if the roll-force ratio satisfies the condition of convergence among the above-mentioned steps S27, judge then whether the roll-force of certain frame surpasses the maximum rolling force that equipment allows:

If exist roll-force to transfinite:

Roll-force p when the i frame _iSurpass the maximum rolling force that equipment allows

The time, get into the S281 step, adjust the roll-force distribution coefficient of this frame, the distribution coefficient of all the other frames does not adjust, and exists:

$p_i>p_i^{\mathrm{Max}}$ The time, then carry out ${\mathrm{\&alpha;}}_i^{\&prime;}={\mathrm{\&alpha;}}_i\&CenterDot;\frac{1}{(p_i/p_i^{\mathrm{Max}}-1)\&CenterDot;0.8+1}$ Formula 8

Wherein, α _iBe the roll-force distribution coefficient α after the said normalization _i, α _i' be the adjusted roll-force distribution coefficient that transfinites;

α _i' get back to step S23 after obtaining, with adjusted roll-force distribution coefficient α _i' carry out absolute draft amount iterative computation again, iterations j assignment 0 simultaneously, gets the absolute draft amount of the last iteration of step S27

And exit thickness

Initial value as absolute draft amount and exit thickness;

If do not exist roll-force to transfinite, then directly carry out next step;

S29; Absolute draft amount iterative computation finishes with the inspection of transfiniting; Obtain the roll-force allocation result; The exit thickness

that get the last iteration of step S27 exit thickness

distributes as each frame roll-force obtains the reduction ratio

that each frame roll-force distributes simultaneously: is the exit thickness that i frame roll-force is distributed in

formula, and

is the exit thickness that i-1 frame roll-force is @ REPEAT[D

Said step 3 may further comprise the steps:

S31, carry out amplitude limiting processing to the roll-force allocation result:

${\mathrm{\&epsiv;}}_{\mathrm{Xf}}^i=\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1+\mathrm{\&eta;}){\mathrm{\&epsiv;}}_{\mathrm{Rfd}}^i>{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{Rfd}}^i{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1-\mathrm{\&eta;})\&le;{\mathrm{\&epsiv;}}_{\mathrm{Rfd}}^i\&le;{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1-\mathrm{\&eta;}){\mathrm{\&epsiv;}}_{\mathrm{Rfd}}^i<{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1-\mathrm{\&eta;})\end{array}\right.$ Formula 9

Wherein

is reduction ratio after the amplitude limiting processing; η is an amplitude limit control parameter, gets 5%～10%;

S32, carry out relativization to reduction ratio after the amplitude limiting processing

and calculate:

If reduction ratio is modified among the above-mentioned steps S31, calculating finish to gauge thickness h ' _nWith target finish to gauge thickness h _nBetween can have certain deviation, also to carry out taking turns relativization and calculate:

Calculating finish to gauge thickness h ' _n:

${h_n}^{\&prime;}=h_0\&CenterDot;\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i)$

Calculate the finish to gauge relativization factor (1-xc), wherein

$\mathrm{xc}=\frac{\ln \frac{h_n}{{h_n}^{\&prime;}}}{\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}\frac{{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}{1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}}$

Reduction ratio after the finish to gauge relativization

is called the final assignment reduction ratio:

${\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i=(1-\mathrm{xc})\&CenterDot;{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i$

Get

As the reduction ratio of the final output of each frame, obtain the final outlet thickness of each frame simultaneously $h_{\mathrm{Xdh}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{Xdh}}^i),i=\mathrm{1,2},\mathrm{\&Lambda;},n-1,$

According to the final allocation of reduction ratio

and the final exit thickness for finishing strip rolling schedule setting calculation.

Among the said step S6, the process that reduction ratio is learnt is:

Ask each frame final assignment and poor

that depress the reduction ratio of distribution

${\mathrm{\&epsiv;}}_{\mathrm{Dif}}^i=\frac{{\mathrm{\&epsiv;}}_{\mathrm{Xdh}}^i-{\mathrm{\&epsiv;}}_{\mathrm{Red}}^{\mathrm{<figure-callout\; id="10"\; label="i\; Xrk\; Formula"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">i</figure-callout>}}}{\mathrm{Xrk}}$ Formula 10

Wherein,

is the reduction ratio of final assignment;

for to depress the reduction ratio of distribution, xrk is the scale factor of the reduction ratio in the formula (1);

Obtain rolling procedure and set the learning value

that back reduction ratio learning table equivalent layer does not upgrade

${\mathrm{\&epsiv;}}_{\mathrm{Lrn}}^{i,\mathrm{New}}={\mathrm{\&epsiv;}}_{\mathrm{Lrn}}^i\&CenterDot;\mathrm{\&mu;}+{\mathrm{\&epsiv;}}_{\mathrm{Dif}}^i\&CenterDot;(1-\mathrm{\&mu;})$ Formula 11

Wherein, μ is a smoothing factor, gets 0.7;

sets the preceding not original learning value of reduction ratio learning table equivalent layer for rolling procedure.

Beneficial effect of the present invention:

Fine-rolling strip steel sharing of load establishing method of the present invention; Realize the aggregative model of " depressing distribution+roll-force distributes "; Distribute the distribution of confirming to retrain the roll-force allocation result with depressing, combine the advantage of two kinds of patterns, overcome both shortcomings separately simultaneously; Reduce to satisfy the requirement that each frame roll-force descends in proportion when each frame thickness distributes fluctuation as far as possible, reach the purpose that improves fine-rolling strip steel rolling procedure setting accuracy and rolling stability.

For further specifying above-mentioned purpose of the present invention, design feature and effect, below will combine accompanying drawing that the present invention is elaborated.

Description of drawings

Fig. 1 is a fine-rolling strip steel sharing of load establishing method flow chart of the present invention;

Fig. 2 is the particular flow sheet that calculates each frame absolute draft amount in Fig. 1 flow process;

The reduction ratio fluctuation that Fig. 3 distributes for roll-force and final assignment is caused by the rolling force model deviation.

The specific embodiment

Accompanying drawing specific embodiments of the invention below in conjunction with embodiment is elaborated.

Referring to Fig. 1, Fig. 1 is a fine-rolling strip steel sharing of load establishing method flow chart of the present invention, and fine-rolling strip steel sharing of load establishing method of the present invention may further comprise the steps:

S1, according to reduction ratio distribution coefficient table and reduction ratio learning table; Calculate the allocation result of depressing of each frame of finish rolling by depressing allocation model, said depress the reduction ratio

of the distribution allocation result finger pressure under and depress distribution exit thickness

its may further comprise the steps:

S11 gets table, gets the reduction ratio and the reduction ratio learning value of each frame from reduction ratio distribution coefficient table and reduction ratio learning table, and both carry out addition, obtain the reduction ratio allocation table scale value of each frame, and detailed process is:

Big type of steel grade, level of thickness and width class information according to current band steel; Do not get each frame of finish rolling (reduction ratio (i is a shelf number) of example: F1～F7) from depressing distribution coefficient table equivalent layer; Said reduction ratio distribution coefficient table is a table commonly used in the industry; Data are confirmed before production in the table; Be generally the empirical data of considering capacity of equipment and product performance, can collect the excellent case in the actual production;

Tapping mark, level of thickness and width class information according to current band steel; Do not get each frame of finish rolling (reduction ratio learning value (i is a shelf number) of F1～F7) from reduction ratio learning table equivalent layer; Said reduction ratio learning table is the tables of data that is obtained by program automatic learning; Data are brought in constant renewal in through study in the table, and initial value gets 0;

Obtained by adding the two finishing each rack reduction rate allocation table values

is

(i is the rack number).

The benefit of two-layer form design is can reflect through the reduction ratio learning table difference (the big class of steel grade comprises some tapping marks) of big type of down different tapping marks of same steel grade.

S12; Carrying out relativization calculates; Obtain the allocation result of depressing of each frame, exit thickness

detailed process of promptly depressing the reduction ratio

of distribution and depressing distribution is:

In order to obtain the concrete reduction ratio data under various intermediate blank thickness and the finish to gauge thickness; Need carry out relativization to above-mentioned reduction ratio allocation table scale value calculates; Must equal the overall reduction of finish rolling according to the drafts sum of each frame of finish rolling, the scale factor xrk that obtains reduction ratio is:

$\mathrm{xrk}=\frac{-\mathrm{kb}\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">xs</figure-callout>}}_1+\sqrt{{(\mathrm{kb}\&CenterDot;{\mathrm{xs}}_1)}^2+4\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">xs</figure-callout>}}_2\&CenterDot;\ln x}}{2\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{xs}}_2}---\left(1\right)$

Wherein,

h _nBe finish to gauge thickness, h ₀Be intermediate blank thickness (h _nWith h ₀Be known value before the finish rolling rolling procedure set-up and calculated);

${\mathrm{<figure-callout\; id="1"\; label="Xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">Xs</figure-callout>}}_1=\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}_0^i,$ ${\mathrm{<figure-callout\; id="2"\; label="Xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">Xs</figure-callout>}}_2=\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{\left({\mathrm{\&epsiv;}}_0^i\right)}^2,$ N is the finishing pass number, and i is a shelf number;

kb＝0.906501，ka＝0.959597。

Reduction ratio (promptly depressing the reduction ratio of distribution)

after the reduction ratio allocation table scale value relativization of each frame is:

${\mathrm{\&epsiv;}}_{\mathrm{red}}^i=\mathrm{xrk}\&CenterDot;{\mathrm{\&epsiv;}}_0^i---\left(2\right)$

According to the assigned pressure reduction rate n-1 is obtained in front of the rolling rack assigned to the respective exit thickness

$h_{\mathrm{red}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{red}}^i),i=\mathrm{1,2},\mathrm{\&Lambda;},n-1---\left(3\right)$

Formula (3) is to obtain according to the relation between each frame inlet thickness, exit thickness and the reduction ratio, is the fundamental formular of using in the industry.

S2, the exit thickness of depressing distribution with each frame that obtains among the above-mentioned S12 are initial value; Absolute draft amount by each frame of roll-force allocation model iterative computation; Satisfy target call up to the roll-force ratio; Obtain the roll-force allocation result of each frame; Said roll-force allocation result comprises reduction ratio

idiographic flow that exit thickness

that roll-force is distributed and roll-force distribute referring to Fig. 2, comprising absolute draft amount iterative computation process (S21-S27) and the checking process that transfinites (S28):

S21, big type of steel grade and level of thickness information according to current band steel do not get access to the initial roll-force distribution coefficient of each frame α from roll-force distribution coefficient table equivalent layer _0i

Data are confirmed before production in the above-mentioned distribution coefficient table, mainly consider factors such as belt steel rolling stability and plate shape index, can on the empirical data basis, constantly collect outstanding rolling case and be optimized.Reduction ratio distribution coefficient table among the above-mentioned steps S1 is necessary for depressing allocation model institute; Roll-force distribution coefficient table in this step is that roll-force allocation model institute is necessary; It also is known in the industry tables of data; Both data are independently, all are the technology empirical datas, and the present invention has realized the comprehensive of these two kinds of distribution.

S22 operates correcting process, obtains the revised roll-force distribution coefficient of each frame α _1i

Collect the operation modifying factor ζ of each frame from operation screen HMI _i, the revised roll-force distribution coefficient of each frame α _1iFor:

α _1i＝α _0i·(1+ζ _i) (4)

Wherein, ζ _iSpan is-20%～20%.

Carry out normalization, the roll-force distribution coefficient α after the normalization _iFor: ${\mathrm{\&alpha;}}_i=\frac{{\mathrm{\&alpha;}}_{1i}}{\mathrm{Max}\left({\mathrm{\&alpha;}}_{1i}\right)}.$

The normalization here refers to α maximum in institute's organic frame (routine F1～F7 is totally 7 frames) _1iValue is classified as 1, and all the other projects are carried out handled.

S23, calculate roll-force and the roll-force of each frame under current thickness the distributes derivative to the absolute draft amount:

(1) calculates roll-force

The variable declaration that is provided with in the following formula: i is a shelf number; J is an iterations;

is the roll-force of the j time iteration of i frame; The exit thickness of

the j time iteration of i frame, the exit thickness of

the j time iteration of i-1 frame.

The first iteration of the i-th frame before calculating the thickness of the initial value of exports to the above formula (3) obtained in the outlet pressure distribution of the thickness of

is

Under the situation that other technological parameter is fixed; Roll-force is the function of inlet thickness and exit thickness; Promptly exist: (this functional relation is known in the industry; Do not explain at this), the size of roll-force receives the influence of rolling force model precision; Be noted that; The variation that each frame thickness distributes can cause each frame variation of temperature, and before calculating each frame roll-force, (temperature and roll-force interact should to recomputate once the temperature of each frame; Its interactive relation is not explained at this by knowing in the industry);

(2) calculate the derivative

of roll-force to the absolute draft amount

Computing formula is:

the formula principle of this computational process be that the roll-force variable quantity is divided by absolute draft amount variable quantity; It is the definition of derivative; Be known in the industry computational process, briefly explain at this.

In the formula,

is the roll-force after the positive disturbance of absolute draft amount;

roll-force after for the negative disturbance of absolute draft amount,

is the absolute draft amount of the j time iteration of i frame;

There is following functional relation: $p_{i1}^j=p(h_{i-1}^j,h_{i-1}^j-\mathrm{\&Delta;}{h}_{i1}^j),$ $p_{i2}^j=p(h_{i-1}^j,h_{i-1}^j-\mathrm{\&Delta;}{h}_{i2}^j)$

Wherein,

is the absolute draft amount after the positive disturbance;

is the absolute draft amount after the negative disturbance

$\mathrm{\&Delta;}{h}_{i1}^j=(1+\frac{\mathrm{dt}}{2})\&CenterDot;\mathrm{\&Delta;}{h}_i^j,$ $\mathrm{\&Delta;}{h}_{i2}^j=(1-\frac{\mathrm{dt}}{2})\&CenterDot;\mathrm{\&Delta;}{h}_i^j$

Wherein dt is very little disturbance, desirable 1%.

S24 according to the roll-force of above-mentioned each frame and the roll-force result of calculation to absolute draft amount derivative, calculates the correction value of each frame absolute draft amount.

The core formula of rolling distribution iterative computation:

$\mathrm{\&delta;}\left(\mathrm{\&Delta;}{h}_i^j\right)=(\mathrm{\&Sigma;}{p}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{\&PartialD;p_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)}-\mathrm{\&Sigma;}((\mathrm{\&Sigma;}{p}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{\&PartialD;p_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)})\&CenterDot;\frac{\mathrm{\&Delta;}{h}_i^j}{\mathrm{\&Sigma;\&Delta;}{h}_i^j}---\left(5\right)$

Wherein:

is the correction value of the absolute draft amount of the j time iteration of i frame.

Above-mentioned formula (5) is the result who derives according to a kind of Newton of improvement method, and the absolute draft amount of selecting each passage for use is as variable, and the nonlinear equation of n dimension is become n nonlinear equation of one dimension, and each nonlinear equation is found the solution with the Newton iterative method.It has used for reference the characteristics that the Newton-Raphson method has good convergence property, can avoid again asking Jacobian matrix and inverse matrix thereof simultaneously, and algorithm is effectively simplified, and computational speed is very fast, and precision is higher.

S25; After the correction value of each frame absolute draft amount

obtains; Calculate the updating value of each frame absolute draft amount, just the absolute draft amount of the j+1 time iteration

Computing formula is:

$\mathrm{\&Delta;}{h}_i^{j+1}=\mathrm{\&Delta;}{h}_i^j+{\mathrm{damp}}_j\&CenterDot;\mathrm{\&delta;}\left(\mathrm{\&Delta;}{h}_i^j\right)---\left(6\right)$

Damp wherein _jBe damped coefficient, damp _j=damp_mpy β+(1-β) (1-e ^-j),

Damp_mpy is desirable 1.0, and β desirable 0.6.

Only absorb the certain percentage of absolute draft amount correction value during each iteration, along with iterations increases, percent absorption also increases.

S26; After the updating value of each frame absolute draft amount

obtains; Calculate the updating value of each frame exit thickness, just the exit thickness of the j+1 time iteration

And intermediate blank thickness h ₀Be known quantity, according to

First frame only

All the other all are the inputs that previous frame is output as a back frame, calculate last frame from first frame, obtain the updating value of each frame exit thickness, and this value will be used for next iteration and calculate roll-force usefulness, wherein

The exit thickness that refers to the j+1 time iteration of i-1 frame;

Example is to the F1 frame; I=1;

be which time iteration no matter, has

constant;

To the F2 frame; I=2,

F1 output is imported as F2;

To the F3 frame, i=3, recursion successively .....

So reduce only when F1, all the other all are the input that previous frame is output as a back frame for

.

S27 judges whether the roll-force ratio satisfies the condition of convergence.

Roll-force distributes the condition of convergence of iterative computation to be:

$|\frac{p_i^j}{\mathrm{\&Sigma;}{p}_i^j}-\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}|\&le;\frac{\mathrm{\&tau;}}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}---\left(7\right)$

Wherein, τ is very little positive number, desirable 0.01.

Set up when formula (7), finishing iteration was calculated when perhaps iterations j surpassed set point number, otherwise continued to carry out the S23 step, and iterations j adds up 1 simultaneously.For example, the set point number of iterations j is 6 times, and promptly the span of j is that 0≤j≤5, the 1 time iterative computation j gets 0, when formula (7) set up or iterations j when surpassing 6 times iterative computation finish; If iterative computation does not also reach the condition of convergence for the first time, continue to carry out the S23 step, iterations j adds up 1 simultaneously.

After iterative computation finishes, can obtain the absolute draft amount

and the exit thickness of each frame

S28, if the roll-force ratio satisfies the condition of convergence among the above-mentioned steps S27, judge then whether the roll-force of certain frame surpasses the maximum rolling force that equipment allows:

If exist roll-force to transfinite:

Roll-force p when the i frame _iSurpass the maximum rolling force that equipment allows

The time, get into the S281 step, adjust the roll-force distribution coefficient of this frame, the distribution coefficient of all the other frames does not adjust, and exists:

$p_i>p_i^{\mathrm{Max}}$ The time, then carry out ${\mathrm{\&alpha;}}_i^{\&prime;}={\mathrm{\&alpha;}}_i\&CenterDot;\frac{1}{(p_i/p_i^{\mathrm{Max}}-1)\&CenterDot;0.8+1}---\left(8\right)$

Wherein, α _iBe the roll-force distribution coefficient α after the said normalization _i, α _i' be the adjusted roll-force distribution coefficient that transfinites;

α _i' get back to step S23 after obtaining, with adjusted roll-force distribution coefficient α _i' carry out absolute draft amount iterative computation again.Iterations j assignment 0 is simultaneously got absolute draft amount

and the exit thickness

of the last iteration of the S27 initial value as absolute draft amount and exit thickness;

If do not exist roll-force to transfinite, then directly carry out next step S29.

S29; Absolute draft amount iterative computation finishes with the inspection of transfiniting; Obtain the roll-force allocation result; The exit thickness

that get the last iteration of step S27 exit thickness distributes as each frame roll-force obtains the reduction ratio that each frame roll-force distributes simultaneously: is the exit thickness that i frame roll-force is distributed in

formula, and

is the exit thickness that i-1 frame roll-force is @ REPEAT[D

In the S2 step, be that the initial value that roll-force is distributed can reduce iterations to depress allocation result, roll-force is distributed can control the ratio that each frame roll-force descends, and then improves the rolling stability and the plate shape index of fine-rolling strip steel.

S3, the roll-force allocation result of each frame among the above-mentioned steps S2 is carried out amplitude limiting processing; Make it depart from above-mentioned steps S1 and depress allocation result within the specific limits; Thereby the final reduction ratio of each frame is being distributed in the certain limit of confirming by depressing; Just do not get the roll-force allocation result as long as do not go beyond the scope, going beyond the scope then is limited in boundary with it.

S31, carry out amplitude limiting processing to the roll-force allocation result:

${\mathrm{\&epsiv;}}_{\mathrm{xf}}^i=\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;}){\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i>{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})\&le;{\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i\&le;{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;}){\mathrm{\&epsiv;}}_{\mathrm{rfd}}^i<{\mathrm{\&epsiv;}}_{\mathrm{red}}^i\&CenterDot;(1-\mathrm{\&eta;})\end{array}\right.---\left(9\right)$

Wherein

is reduction ratio after the amplitude limiting processing; η is an amplitude limit control parameter, gets 5%～10%;

S32, carry out relativization to reduction ratio after the amplitude limiting processing

and calculate:

If reduction ratio is modified among the above-mentioned steps S31, calculating finish to gauge thickness h ' _nWith target finish to gauge thickness h _nBetween can have certain deviation, also to carry out taking turns relativization and calculate:

Calculating finish to gauge thickness h ' _n:

${h_n}^{\&prime;}=h_0\&CenterDot;\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i)$

Calculate the finish to gauge relativization factor (1-xc), wherein

$\mathrm{xc}=\frac{\ln \frac{h_n}{{h_n}^{\&prime;}}}{\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}\frac{{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}{1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}}$

Reduction ratio after the finish to gauge relativization

is called the final assignment reduction ratio:

${\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i=(1-\mathrm{xc})\&CenterDot;{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i$

Get

As the reduction ratio of the final output of each frame, obtain the final outlet thickness of each frame simultaneously

$h_{\mathrm{Xdh}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{Xdh}}^i),i=\mathrm{1,2},\mathrm{\&Lambda;},n-1,$

According to the final allocation of reduction ratio

and the final exit thickness

for finishing strip rolling schedule setting calculation.

S4; After band steel toe portion wears the band completion and collects actual roll-force; Judge the deviation of rolling force setup value (rolling force model when this rolling force setup value is calculated by rolling procedure calculates, and is publicly-owned knowledge, and is the same with calculating roll-force among the above-mentioned S23) and actual roll-force.

S5 judges each frame deviate whether in θ, and θ is the Deviation Control parameter; Desirable 5%-10%; If do not satisfy each frame roll-force deviation all at θ with interior condition, then do not carry out reduction ratio study, at this moment; Other data of equivalent layer are not upgraded in the reduction ratio learning table (following table 2), and flow process gets into end step.

S6, if each frame deviate is then learnt the reduction ratio of this setting in θ, it is other that learning value is updated to reduction ratio study form equivalent layer, for:

${\mathrm{\&epsiv;}}_{\mathrm{dif}}^i=\frac{{\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i-{\mathrm{\&epsiv;}}_{\mathrm{red}}^i}{\mathrm{xrk}}---\left(10\right)$

Wherein

is the poor of i frame final assignment and the reduction ratio of depressing distribution, and xrk is the scale factor of the reduction ratio in the formula (1).

Obtaining rolling procedure sets the learning value that back reduction ratio learning table equivalent layer do not upgrade and is:

${\mathrm{\&epsiv;}}_{\mathrm{lrn}}^{i,\mathrm{new}}={\mathrm{\&epsiv;}}_{\mathrm{lrn}}^i\&CenterDot;\mathrm{\&mu;}+{\mathrm{\&epsiv;}}_{\mathrm{dif}}^i\&CenterDot;(1-\mathrm{\&mu;})---\left(11\right)$

Wherein, μ is a smoothing factor, desirable 0.7;

is the not original learning value of reduction ratio learning table (following table 2) equivalent layer (rolling procedure is set preceding).

S7; Calculate and finish;

is updated to reduction ratio learning table (form 2) equivalent layer not after; Because function of learning, the other roll-force of this layer can progressively reach the requirement of target roll-force ratio.

Lift the embodiment explanation below.

Embodiment 1:

Stress the The whole calculations flow process.Final result is that the roll-force after excessive rolling distributes amplitude limit is distributed; Overcome in (this kind situation generation when big and process conditions change greatly in the rolling force model error of single drawback that the fluctuation of frame reduction ratio is big when distributing with roll-force; Roll-force is distributed will be through regulating the requirement that reduction ratio reaches each frame roll-force ratio, so the reduction ratio fluctuation is big; And use the reduction ratio of depressing each frame of branch timing almost to remain unchanged), help improving fine-rolling strip steel rolling procedure setting accuracy and rolling stability.

Certain band steel information is following: intermediate blank thickness 40.69mm, and finish to gauge thickness 2.0mm, strip width 1233.6mm, finish rolling inlet temperature FET is 1000 ℃, finish to gauge target temperature FDT is 880 ℃.The finish rolling rolling pass is counted n=7.The other information of layer: steel grade is 11002 for big type, and level of thickness is 6, and the width grade is 3, and the tapping mark is GV4924E1.It is following to get other data of equivalent layer from 3 forms:

Table 1: reduction ratio distribution coefficient table

Table 2: reduction ratio learning table:

Table 3: roll-force distribution coefficient table

Above-mentioned steps S1: according to depressing the distribution coefficient table, calculate the data addition that the thickness distribution of each frame will be depressed distribution coefficient table and reduction ratio learning table, obtain following table 4: depress the allocation table scale value by depressing allocation model

In order to obtain at intermediate blank thickness is that 40.69mm and finish to gauge thickness are the concrete reduction ratio data under the 2.0mm, need carry out relativization to

and calculate.

According to formula (1), calculate the scale factor xrk of reduction ratio

$x=\frac{h_0}{{\mathrm{<figure-callout\; id="7"\; label="h"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">h</figure-callout>}}_7}=20.346,$ ${\mathrm{<figure-callout\; id="1"\; label="xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">xs</figure-callout>}}_1=\underset{i=1}{\overset{7}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}_0^i=2.36,$ ${\mathrm{<figure-callout\; id="2"\; label="xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">xs</figure-callout>}}_2=\underset{i=1}{\overset{7}{\mathrm{\&Sigma;}}}{\left({\mathrm{\&epsiv;}}_0^i\right)}^2=0.9768$

$\mathrm{xrk}=\frac{-\mathrm{kb}\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">xs</figure-callout>}}_1+\sqrt{{(\mathrm{kb}\&CenterDot;{\mathrm{xs}}_1)}^2+4\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">xs</figure-callout>}}_2\&CenterDot;\ln x}}{2\&CenterDot;\mathrm{ka}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="xs"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">xs</figure-callout>}}_2}=0.984$

Obtain following table 5: reduction ratio after each frame relativization and exit thickness data

Above-mentioned steps S2: to depress allocation result is initial value, obtains the roll-force allocation result of each frame.

(S23～S27), obtain following table 6: the roll-force ratio of each time iteration changes to carry out absolute draft amount iterative computation process earlier

Iterations	F1	F2	F3	F4	F5	F6	F7
									1	1	0.90617	0.85469	0.75396	0.55891	0.45492	0.31708
2	1	0.93258	0.86329	0.75989	0.5853	0.45067	0.33803
								3	1	0.95718	0.87182	0.76879	0.60745	0.44919	0.35346
4	1	0.97063	0.87697	0.7759	0.61707	0.44984	0.35862
								Target	1	0.98	0.88	0.78	0.62	0.45	0.36

Carry out the checking process (S28) that transfinites again, each frame roll-force does not all exceed the maximum rolling force of its permission, so need not come back to S23.

Following table 7: reduction ratio and exit thickness that roll-force is distributed

Above-mentioned steps S3: the roll-force allocation result is done amplitude limiting processing, it is departed from depress allocation result within the specific limits.

1, the roll-force allocation result is done amplitude limiting processing.Get η=6%, find institute's organic frame roll-force allocation result all depress allocation result ± 6% in, so the thickness distribution is constant.

2, because data do not change in above-mentioned 1, calculate so need not relativization.

Above-mentioned steps S4-S6: after band steel toe portion wears band and accomplishes and collect actual roll-force, judge the deviation of rolling force setup value and measured value, if each frame deviation all in θ, is then learnt the reduction ratio of this setting.

After band steel toe portion wears band and accomplishes and collect actual roll-force, judge the deviation of rolling force setup value and measured value, find that each frame deviation all 5% with interior (getting θ=5%), obtains following table 8: carry out reduction ratio and learn

The equivalent layer that

is updated in the reduction ratio learning table is other.Because function of learning, this layer are not depressed distribution technology table and roll-force distribution technology table data can be more and more identical, promptly depress the corresponding roll-force of allocation result can be distributed target proportion more near roll-force requirement.

This situation is ideal, and not only the roll-force distribution reaches the target proportion requirement, and departs from and depress allocation result within the specific limits, need not amplitude limiting processing, and final assignment is got the roll-force distribution result.

Hypothesis is to same embodiment below, and the input data are constant, and only the increase of rolling force model deviation is following table 9:

Identical computational process is carried out one time.

Step S1 result is constant.Because the deviation of rolling force model is different, step S2 result of calculation is different.Iteration just satisfies end condition 5 times, and the roll-force ratio of each time iteration changes like following table 10:

Iterations	F1	F2	F3	F4	F5	F6	F7
									1	1	0.86302	0.7326	0.64625	0.5323	0.47658	0.27178
2	1	0.89134	0.77522	0.69524	0.57158	0.46032	0.3122
								3	1	0.92722	0.82495	0.74563	0.60658	0.45129	0.34476
4	1	0.95498	0.85953	0.77391	0.62107	0.45078	0.35705
								5	1	0.97054	0.87596	0.78306	0.62332	0.45095	0.3599
Target	1	0.98	0.88	0.78	0.62	0.45	0.36

Reduction ratio and exit thickness such as following table 11 that roll-force is distributed:

Above-mentioned steps S3 carries out amplitude limiting processing to the roll-force allocation result, find F1, F3, F4, F6, F7 frame all exceed depress allocation result ± 6% (η=6%), amplitude limiting processing result such as following table 12:

Will do relativization after the amplitude limiting processing calculates.Carry out and satisfy termination condition 1 time, xc=0.0079, reduction ratio and the exit thickness such as the following table 13 of final output:

At last, the roll-force of the two kinds of situation in front and back distributed with the reduction ratio fluctuation of final assignment do contrast like following table 14:

The reduction ratio fluctuation that Fig. 3 distributes for roll-force and final assignment is caused by the rolling force model deviation.It is thus clear that; If directly roll-force is distributed as final assignment, the latter is very big with respect to the former fluctuation so, and F6, F7 nearly 15%; Promptly all constant only because of the rolling force model deviation just can cause so big fluctuation in other condition, all unfavorable to rolling procedure set-up and calculated and rolling stability; But the final assignment after the amplitude limiting processing, the latter is also little with respect to the former fluctuation, and basic controlling is in 5%.This patent proposition that Here it is will be with depressing the reason of distributing the distribution of confirming to retrain the roll-force allocation result; In the past the technology otherwise be simple " depressing distribution "; Be simple " roll-force distribution ", this patent has been realized " depressing distribution+roll-force distributes ", and its advantage is: when the rolling force model deviation is little; Study through form 2 makes two kinds of allocation model results more and more consistent, and iterations is fewer and feweri; When big or process conditions change greatly in the rolling force model deviation, effectively control the fluctuation range of each frame reduction ratio, help improving finish rolling rules set-up and calculated precision and rolling stability through the amplitude limit interaction energy of depressing distribution.

Embodiment 2:

The chassis equipment limit check.

Band steel information: intermediate blank thickness 42.308mm, finish to gauge thickness 4.597mm, road number of times n=7.Roll-force distribution coefficient that requires among the step S2 such as following table 15:

SGF	THICK	α1	α2	α3	α4	α5	α6	α7
									11002	16	1	0.98	0.88	0.78	0.62	0.45	0.36

The roll-force inspection of transfiniting

For ease of relatively, calculate two kinds of sharing of loads under the situation respectively: a kind of is not restriction of maximum rolling force to F2; A kind of is that maximum rolling force to F2 limits, and supposes that

result of calculation is relatively like following table 16.

	F1	F2	F3	F4	F5	F6	F7
								α
i	1	0.98	0.88	0.78	0.62	0.45	0.36
								α iAfter the modification	1	0.94	0.88	0.78	0.62	0.45	0.36

It is thus clear that when the maximum rolling force restriction was set, F2 frame roll-force satisfied $p_{}$ ${}_2\&ap;{\mathrm{<figure-callout\; id="2"\; label="p"\; filenames="HDA0000055008610000031.png"\; state="\{\{state\}\}">p</figure-callout>}}_2^{\mathrm{Max}}.$

In sum; The present invention is directed to the deficiency of depressing distribution and roll-force distribution in the prior art; Propose a kind of new fine-rolling strip steel load distribution method, realize " depressing distribution+roll-force distributes ", distribute the distribution of confirming to retrain the roll-force allocation result with depressing; The advantage of comprehensive two kinds of patterns; Overcome both shortcomings separately simultaneously,, reach the purpose that improves fine-rolling strip steel rolling procedure setting accuracy and rolling stability reduce to satisfy the requirement that each frame roll-force descends in proportion when each frame thickness distributes fluctuation as far as possible.Main innovate point explanation: see from single step, be mainly step S2,3; In general, realized the combination of two kinds of allocation models, advantage: when the rolling force model deviation was little, the study through form 2 made two kinds of allocation model results more and more consistent, and iterations is fewer and feweri; When big or process conditions change greatly in the rolling force model deviation, effectively control the fluctuation range of each frame reduction ratio, help improving finish rolling rules set-up and calculated precision and rolling stability through the amplitude limit interaction energy of depressing distribution.

Those of ordinary skill in the art will be appreciated that; Above embodiment is used for explaining the object of the invention; And be not with opposing qualification of the present invention; As long as in essential scope of the present invention, all will drop in the scope of claim of the present invention variation, the modification of the above embodiment.

Claims (6)

1. fine-rolling strip steel sharing of load establishing method is characterized in that may further comprise the steps:

S1, according to reduction ratio distribution coefficient table and reduction ratio learning table; Calculate the allocation result of depressing of each frame of finish rolling by depressing allocation model, saidly depress the reduction ratio of the distribution allocation result finger pressure under and depress the exit thickness

of distribution

S2, above-mentioned each frame that obtains is depressed distribution exit thickness

as initial value; Absolute draft amount by each frame of roll-force allocation model iterative computation; Satisfy target call up to the roll-force ratio; Obtain the roll-force allocation result of each frame, said roll-force allocation result comprises the exit thickness

of roll-force distribution and the reduction ratio that roll-force is distributed

S3, the roll-force allocation result of said each frame is carried out amplitude limiting processing; It is departed from saidly depress allocation result within the specific limits; Thereby the final reduction ratio of each frame is being distributed in the certain limit of confirming by depressing; Just do not get the roll-force allocation result as long as do not go beyond the scope, going beyond the scope then is limited in boundary with it;

After S4, band steel toe portion wear the band completion and collect actual roll-force, judge the deviation of each frame rolling force setup value and actual roll-force;

S5 judges each frame deviate whether in θ, and θ is the Deviation Control parameter; Get 5%-10%; If do not satisfy each frame roll-force deviation all at θ with interior condition, then do not carry out reduction ratio study, at this moment; Other data of equivalent layer are not upgraded in the said reduction ratio learning table, and flow process gets into end step;

S6, if each frame deviate is then learnt the reduction ratio of this setting in θ, it is other that learning value is updated to reduction ratio study form equivalent layer, through the study process, the other roll-force of this layer can progressively reach the requirement of target roll-force ratio;

S7, sharing of load set and finish.

2. fine-rolling strip steel sharing of load establishing method as claimed in claim 1 is characterized in that:

Said step S1 may further comprise the steps:

S11; get table; get the reduction ratio and the reduction ratio learning value of each frame from reduction ratio distribution coefficient table and reduction ratio learning table; both carry out addition, obtain the reduction ratio allocation table scale value

of each frame

S12 carries out relativization and calculates, and obtains the allocation result of depressing of each frame:

Must equal the overall reduction of finish rolling according to the drafts sum of each frame of finish rolling, the scale factor xrk that obtains reduction ratio is:

$\mathrm{Xrk}=\frac{-\mathrm{Kb}\&CenterDot;{\mathrm{Xs}}_1+\sqrt{{(\mathrm{Kb}\&CenterDot;{\mathrm{Xs}}_1)}^2+4\&CenterDot;\mathrm{Ka}\&CenterDot;{\mathrm{Xs}}_2\&CenterDot;\mathrm{Ln}x}}{2\&CenterDot;\mathrm{Ka}\&CenterDot;{\mathrm{Xs}}_2}$ Formula 1

Wherein,

h _nBe finish to gauge thickness, h ₀Be intermediate blank thickness;

${\mathrm{Xs}}_1=\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}_0^i,$ ${\mathrm{Xs}}_2=\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{\left({\mathrm{\&epsiv;}}_0^i\right)}^2,$ N is the finishing pass number, and i is a shelf number;

kb＝0.906501，ka＝0.959597。

Reduction ratio after the reduction ratio allocation table scale value relativization of each frame is a reduction ratio of depressing distribution

${\mathrm{\&epsiv;}}_{\mathrm{Red}}^i=\mathrm{Xrk}\&CenterDot;{\mathrm{\&epsiv;}}_0^i$ Formula 2

According to the assigned pressure reduction rate to obtain n-1 preceding the rolling rack assigned to each exit thickness

$h_{\mathrm{Red}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i),i=\mathrm{1,2},\mathrm{\&Lambda;},n-1.$ Formula 3

3. fine-rolling strip steel sharing of load establishing method as claimed in claim 2 is characterized in that:

The reduction ratio allocation table scale value of each frame process of obtaining is among the said step S11:

According to big type of the steel grade of current band steel, level of thickness and width class information, from depress distribution coefficient table equivalent layer do not get the reduction ratio of each frame of finish rolling

wherein i be shelf number;

Tapping mark, level of thickness and width class information according to current band steel; The reduction ratio learning value of not getting each frame of finish rolling from reduction ratio learning table equivalent layer wherein i is a shelf number; Said reduction ratio learning table is the tables of data that is obtained by program automatic learning; Data are brought in constant renewal in through study in the table, and initial value gets 0;

The reduction ratio allocation table scale value of said each frame

For: ${\mathrm{\&epsiv;}}_0^i={\mathrm{\&epsiv;}}_{\mathrm{Ori}}^i+{\mathrm{\&epsiv;}}_{\mathrm{Lrn}}^i.$

4. fine-rolling strip steel sharing of load establishing method as claimed in claim 1 is characterized in that:

Said step S2 may further comprise the steps:

S21, big type of steel grade and level of thickness information according to current band steel do not get access to the initial roll-force distribution coefficient of each frame α from roll-force distribution coefficient table equivalent layer _0i

S22 operates correcting process, obtains the revised roll-force distribution coefficient of each frame α _1i:

Collect the operation modifying factor ζ of each frame from operation screen HMI _i, the revised roll-force distribution coefficient of each frame α _1iFor:

α _1i=α _0i(1+ ζ _i) formula 4

Wherein, ζ _iSpan is-20%～20%;

Carry out normalization, the roll-force distribution coefficient α after the normalization _iFor: ${\mathrm{\&alpha;}}_i=\frac{{\mathrm{\&alpha;}}_{1i}}{\mathrm{Max}\left({\mathrm{\&alpha;}}_{1i}\right)};$

S23, calculate roll-force and the roll-force of each frame under current thickness the distributes derivative to the absolute draft amount:

Calculate roll-force

If: i is a shelf number; J is an iterations;

is the roll-force of the j time iteration of i frame; The exit thickness of

the j time iteration of i frame, the exit thickness of the j time iteration of i-1 frame;

The first iteration of the i-th frame before calculating the thickness of the initial value of exports

is obtained by pressing the above formula 3 exit thickness distribution

Under the situation that other technological parameter is fixed; Roll-force is the function of inlet thickness and exit thickness, exists:

Calculate the derivative of roll-force to the absolute draft amount

$\frac{\&PartialD;p_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)}=\frac{p_{i1}^j-p_{i2}^j}{\mathrm{\&Delta;}{h}_i^j\&CenterDot;\mathrm{Dt}}$

Wherein, is the roll-force after the positive disturbance of absolute draft amount;

roll-force after for the negative disturbance of absolute draft amount,

is the absolute draft amount of the j time iteration of i frame;

There is following functional relation: $p_{i1}^j=p(h_{i-1}^j,h_{i-1}^j-\mathrm{\&Delta;}{h}_{i1}^j),$ $p_{i2}^j=p(h_{i-1}^j,h_{i-1}^j-\mathrm{\&Delta;}{h}_{i2}^j)$

Wherein,

is the absolute draft amount after the positive disturbance;

is the absolute draft amount after the negative disturbance

$\mathrm{\&Delta;}{h}_{i1}^j=(1+\frac{\mathrm{dt}}{2})\&CenterDot;\mathrm{\&Delta;}{h}_i^j,$ $\mathrm{\&Delta;}{h}_{i2}^j=(1-\frac{\mathrm{dt}}{2})\&CenterDot;\mathrm{\&Delta;}{h}_i^j$

Wherein dt is disturbance, gets 1%;

S24, according to the roll-force of above-mentioned each frame and the roll-force result of calculation to absolute draft amount derivative, calculate the correction value of each frame absolute draft amount:

Rolling distribution iterative computation formula:

$\mathrm{\&delta;}\left(\mathrm{\&Delta;}{h}_i^j\right)=(\mathrm{\&Sigma;}{p}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{\&PartialD;p_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)}-\mathrm{\&Sigma;}((\mathrm{\&Sigma;}{p}_i^j\&CenterDot;\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}-p_i^j)/\frac{\&PartialD;p_i^j}{\&PartialD;\left(\mathrm{\&Delta;}{h}_i^j\right)})\&CenterDot;\frac{\mathrm{\&Delta;}{h}_i^j}{\mathrm{\&Sigma;\; \&Delta;}{h}_i^j}$ Formula 5

Wherein:

is the correction value of the absolute draft amount of the j time iteration of i frame;

S25; After the correction value of each frame absolute draft amount

obtains; Calculate the updating value of each frame absolute draft amount, this updating value is the absolute draft amount

of the j+1 time iteration

Computing formula is:

$\mathrm{\&Delta;}{h}_i^{j+1}=\mathrm{\&Delta;}{h}_i^j+{\mathrm{Damp}}_j\&CenterDot;\mathrm{\&delta;}\left(\mathrm{\&Delta;}{h}_i^j\right)$ Formula 6

Damp wherein _jBe damped coefficient, damp _j=damp_mpy β+(1-β) (1-e ^-j),

Damp_mpy gets 1.0, and β gets 0.6;

S26; After the updating value of each frame absolute draft amount

obtains; Calculate the updating value of each frame exit thickness, this updating value is the exit thickness

of the j+1 time iteration

And intermediate blank thickness h ₀Be known quantity, according to

First frame only

All the other all are the inputs that previous frame is output as a back frame, calculate last frame from first frame, obtain the updating value of each frame exit thickness, and this value will be used for next iteration and calculate roll-force usefulness, wherein The exit thickness that refers to the j+1 time iteration of i-1 frame;

S27, judge whether the roll-force ratio satisfies the condition of convergence:

Roll-force distributes the condition of convergence of iterative computation to be:

$|\frac{p_i^j}{\mathrm{\&Sigma;}{p}_i^j}-\frac{{\mathrm{\&alpha;}}_i}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}|\&le;\frac{\mathrm{\&tau;}}{\mathrm{\&Sigma;}{\mathrm{\&alpha;}}_i}$ Formula 7

Wherein, τ is a positive number, gets 0.01;

Set up when formula (7), finishing iteration was calculated when perhaps iterations j surpassed set point number, otherwise continued to carry out said S23 step, and iterations j adds up 1 simultaneously;

S28, if the roll-force ratio satisfies the condition of convergence among the above-mentioned steps S27, judge then whether the roll-force of certain frame surpasses the maximum rolling force that equipment allows:

If exist roll-force to transfinite:

Roll-force p when the i frame _iSurpass the maximum rolling force that equipment allows

The time, get into the S281 step, adjust the roll-force distribution coefficient of this frame, the distribution coefficient of all the other frames does not adjust, and exists:

$p_i>p_i^{\mathrm{Max}}$ The time, then carry out ${\mathrm{\&alpha;}}_i^{\&prime;}={\mathrm{\&alpha;}}_i\&CenterDot;\frac{1}{(p_i/p_i^{\mathrm{Max}}-1)\&CenterDot;0.8+1}$ Formula 8

Wherein, α _iBe the roll-force distribution coefficient α after the said normalization _i, α _i' be the adjusted roll-force distribution coefficient that transfinites;

α _i' get back to step S23 after obtaining, with adjusted roll-force distribution coefficient α _i' carry out absolute draft amount iterative computation again, iterations j assignment 0 simultaneously, gets the absolute draft amount of the last iteration of step S27

And exit thickness

Initial value as absolute draft amount and exit thickness;

If do not exist roll-force to transfinite, then directly carry out next step;

S29; Absolute draft amount iterative computation finishes with the inspection of transfiniting; Obtain the roll-force allocation result; The exit thickness

that the exit thickness of getting the last iteration of step S27 distributes as each frame roll-force obtains the reduction ratio

that each frame roll-force distributes simultaneously:

is the exit thickness that i frame roll-force is distributed in

formula, and

is the exit thickness that i-1 frame roll-force is chosen.

5. fine-rolling strip steel sharing of load establishing method as claimed in claim 1 is characterized in that:

Said step 3 may further comprise the steps:

S31, carry out amplitude limiting processing to the roll-force allocation result:

${\mathrm{\&epsiv;}}_{\mathrm{Xf}}^i=\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1+\mathrm{\&eta;}){\mathrm{\&epsiv;}}_{\mathrm{Rfd}}^i>{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{Rfd}}^i{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1-\mathrm{\&eta;})\&le;{\mathrm{\&epsiv;}}_{\mathrm{Rfd}}^i\&le;{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1+\mathrm{\&eta;})\\ {\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1-\mathrm{\&eta;}){\mathrm{\&epsiv;}}_{\mathrm{Rfd}}^i<{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i\&CenterDot;(1-\mathrm{\&eta;})\end{array}\right.$ Formula 9

Wherein

is reduction ratio after the amplitude limiting processing; η is an amplitude limit control parameter, gets 5%～10%;

S32, carry out relativization to reduction ratio after the amplitude limiting processing

and calculate:

If reduction ratio is modified among the above-mentioned steps S31, calculating finish to gauge thickness h ' _nWith target finish to gauge thickness h _nBetween can have certain deviation, also to carry out taking turns relativization and calculate:

Calculating finish to gauge thickness h ' _n:

${h_n}^{\&prime;}=h_0\&CenterDot;\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i)$

Calculate the finish to gauge relativization factor (1-xc), wherein

$\mathrm{xc}=\frac{\ln \frac{h_n}{{h_n}^{\&prime;}}}{\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}\frac{{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}{1-{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i}}$

Reduction ratio after the finish to gauge relativization

is called the final assignment reduction ratio:

${\mathrm{\&epsiv;}}_{\mathrm{xdh}}^i=(1-\mathrm{xc})\&CenterDot;{\mathrm{\&epsiv;}}_{\mathrm{xf}}^i$

Get

As the reduction ratio of the final output of each frame, obtain the final outlet thickness of each frame simultaneously

$h_{\mathrm{Xdh}}^i=h_0\&CenterDot;\underset{i=1}{\overset{i}{\mathrm{\&Pi;}}}(1-{\mathrm{\&epsiv;}}_{\mathrm{Xdh}}^i),i=\mathrm{1,2},\mathrm{\&Lambda;},n-1,$

According to the final allocation of reduction ratio

and the final exit thickness

for finishing strip rolling schedule setting calculation.

6. fine-rolling strip steel sharing of load establishing method as claimed in claim 1 is characterized in that:

Among the said step S6, the process that reduction ratio is learnt is:

Ask each frame final assignment and poor

that depress the reduction ratio of distribution

${\mathrm{\&epsiv;}}_{\mathrm{Dif}}^i=\frac{{\mathrm{\&epsiv;}}_{\mathrm{Xdh}}^i-{\mathrm{\&epsiv;}}_{\mathrm{Red}}^i}{\mathrm{Xrk}}$ Formula 10

Wherein,

is the reduction ratio of final assignment;

for to depress the reduction ratio of distribution, xrk is the scale factor of the reduction ratio in the formula (1);

Obtain rolling procedure and set the learning value

that back reduction ratio learning table equivalent layer does not upgrade

${\mathrm{\&epsiv;}}_{\mathrm{Lrn}}^{i,\mathrm{New}}={\mathrm{\&epsiv;}}_{\mathrm{Lrn}}^i\&CenterDot;\mathrm{\&mu;}+{\mathrm{\&epsiv;}}_{\mathrm{Dif}}^i\&CenterDot;(1-\mathrm{\&mu;})$ Formula 11

Wherein, μ is a smoothing factor, gets 0.7;

sets the preceding not original learning value of reduction ratio learning table equivalent layer for rolling procedure.

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