CN106311761B - A method of reducing mill load fluctuation - Google Patents
A method of reducing mill load fluctuation Download PDFInfo
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- CN106311761B CN106311761B CN201510371386.1A CN201510371386A CN106311761B CN 106311761 B CN106311761 B CN 106311761B CN 201510371386 A CN201510371386 A CN 201510371386A CN 106311761 B CN106311761 B CN 106311761B
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Abstract
The present invention relates to a kind of methods of reduction mill load fluctuation, the described method comprises the following steps: 1) calculate 1-5 rack initial depression load, 2) according to deformation resistance, tension, coefficient of friction, roll-force adaptation coefficient calculate 1-5 racks roll-force P1-P5；3) roll-force coefficient of balance is calculated；4) roll-force coefficient of balance Bal is checked；5) load for checking 1,2 racks, compensates roll-force coefficient of balance ComPa；If 1, the load of 2 racks meets several situations above, after compensation, go to the first step and recalculate, otherwise calculating terminates.This method improves acid and rolls load setting precision, reduces load fluctuation, improves acid and rolls product lumber recovery.
Description
Technical field
The present invention relates to a kind of methods of reduction mill load fluctuation, belong to cold continuous rolling automation field.
Background technique
The balance of control parameter can play the maximum capacity of equipment under the premise of guaranteeing product quality, according to equipment
For ability, as long as taking the balance of power, but the unit for producing Thin Strip Steel, since deflection is comparable
Greatly, especially 1 rack causes contact arc angle of the roll with strip to become larger, if only motor power is taken to balance, after 1 rack
Roll marks can be generated on the surface of strip, in order to avoid this kind of situation, it is main right in Process Control System to roll in plum steel acid
The roll-force of the power of motor and 1 rack is balanced calculating, gives the roll-force reasonable distribution of 1 rack to 2 racks, while by 2
The method that each rack of rack to 5 racks takes power-balance, can well solve this problem, but this mode due to
It is influenced by tube rolling simulation result, load fluctuation is larger, and especially before and after the independent roll change of 1# or 2# rack, 1,2 racks are born
Lotus is more than 50% or lower than 35%.
According to tube rolling simulation formula: p=f (r, sk, ten, sr, myu, zp)
Wherein: r is roller diameter
Sk is material deformation resistance
Ten is tension
Sr is relative deformation (load)
Myu is coefficient of friction
Zp is roll-force adaptation coefficient
Roll-force and coefficient of friction myu and adaptation coefficient zp are proportional
There are two the principal elements for influencing mill load fluctuation:
1) force adaptive is rolled
2) fluctuation of roll change front and back coefficient of friction
It is illustrated by taking the independent roll change of 2# rack as an example below:
Coefficient of friction is related with rolling quantity, and after roll change, the compensation of 2# rack coefficient of friction becomes 1.3,2# rack meter from 1.0
It calculates roll-force to increase, rolls dynamic balance (0.98 < P1/P2/A < 1.02) according to 1,2 racks, A is a constant, at this time P1/P2/Bal
< 0.98, it can only be realized by increasing the load of 1# rack, so the load of 1# rack can reach 50 or more.It is automatic by model
The result of calculating at all can not steel rolling, can only intervene manually, manual intervention rate is 95% or more, operative employee's large labor intensity, plate
The bad enclosed of shape is high.
By retrieval, the patent of load distribution method related with this case is as follows: number of patent application is
CN200910057514.X, the invention discloses a kind of stainless steel Load Distribution for Tandem Cold Rolling methods of adjustment, first by each rack phase
Setting value is initialized as to reduction ratio, the roll-force equilibrium valve of the 1st rack is then calculated, calculates separately the 2nd rack to M rack
Motor power equilibrium valve, according to the motor function of the roll-force equilibrium valve for the 1st rack being calculated, the 2nd rack to M rack
Rate equilibrium valve is judged, is adjusted to each rack with respect to reduction ratio.This method, can be according to item practical in the practical operation of rolling
Part adjusts the opposite reduction ratio of each rack, reaches each rack sharing of load balance.
CN200910235653.7, the present invention relates to a kind of pressing load distribution method for double-stander medium plate production line, belong to
In Medium and Heavy Plate Rolling Production line production process optimization control field.The present invention is on the basis of meeting to warm condition, according to penalizing letter
Number solve belt restraining problems basic thoughts, construct be up to based on rolling line production efficiency optimization aim majorized function, adopt
It is iterated optimizing with Fibonacci method, the sharing of load for carrying out overall reduction between Stand Mill is realized, effectively raises
The production efficiency of rolling line.
Above two patent is all based on the basis of performance milling train maximum production capacity the sharing of load for calculating each rack of milling train,
But load setting precision is rolled for how to improve acid, and load fluctuation is reduced, acid is improved and rolls product lumber recovery, in the prior art one
There is no very good solution method directly.
Summary of the invention
The present invention exactly for the technical problems in the prior art, provides a kind of side of reduction mill load fluctuation
Method, this method improve acid and roll load setting precision, reduce load fluctuation, improve acid and roll product lumber recovery.
To achieve the goals above, the technical solution adopted by the present invention is as follows, a method of mill load fluctuation is reduced,
It is characterized in that, the described method comprises the following steps:
1) the initial depression load sr of 1-5 rack is calculatedi
According to formula above, the initial outlet thickness of each rack is calculated, wherein H1For supplied materials thickness, HLFor product outlet thickness
Degree, srtriTo read identical material and specification from pressure loading data sheet (table 5) according to the material rate and specification of strip
Depress data
Further according to the exit thickness of a rack, the initial depression load of each rack is calculated, calculation formula is as follows:
2) according to deformation resistance, tension, coefficient of friction, roll-force adaptation coefficient calculate 1,2 racks roll-force P1,
P2；
Wherein: kp is average resistance of deformation in formula, and Dp is influence of the frictional force to roll-force, and k is tension to rolling
The influence of powerB is the width of strip, and zp is that roll-force is adaptive
Coefficient is answered, R ' is to flatten radius, and H is rack inlet thickness, and h is rack outlet thickness；
3) roll-force coefficient of balance is calculated；
Bal=100*P1/P2/ (A+ComPa) roll-force that wherein P1 calculates for 1# rack, P2 are rolling for 2# rack calculating
Power processed, A are a fixed values in 50-150, are obtained from data form, and ComPa is roll-force coefficient of balance, are initially 0；
4) roll-force coefficient of balance Bal is checked；1-α≤Bal≤1+α
Wherein, α takes 2%, if coefficient of balance Bal meets above-mentioned condition, calculating goes to the 4th step, otherwise corrects 1,2
Load sr1, sr2 of rack, correction amount are ± sr/ (1+3*loop), and loop is cycle-index, is counted again since the first step
It calculates；
5) load for checking 1,2 racks, compensates roll-force coefficient of balance ComPa
Compensation principle: following sr1 is the load of 1# rack, and sr2 is the load of 2# rack:
Exit thickness is less than 0.35mm under smooth roll rolling mode,
Sr1-sr2 > 1, it is automatic to reduce roll-force coefficient of balance ComPa=ComPa-1；
Sr2-sr1 > 4, it is automatic to increase roll-force coefficient of balance ComPa=ComPa+1；
Exit thickness is greater than 0.35mm under hair roller rolling mode, and 5# rack is under permanent roll-force constant pressure,
Sr1-sr2 > 3, it is automatic to reduce roll-force coefficient of balance ComPa=ComPa-1
Sr1-sr2 < 1, it is automatic to increase roll-force coefficient of balance ComPa=ComPa+1
If 1, the load of 2 racks meets several situations above, after compensation, goes to second step and recalculate, otherwise calculate
Terminate.
Compared with the existing technology, advantages of the present invention is as follows, and this method increase roll-force coefficients of balance to compensate skill automatically
Art, it is ensured that within 2%, manual intervention rate is reduced to current 1% from original 95% for the load fluctuation of each rack of same specification
Within, the ratio that exit thickness deviation is no more than exit thickness 0.5% improves 2.1%, and exit thickness deviation is no more than outlet
The ratio of thickness 1% improves 1.5%, realizes the stable rolling that acid rolls thin gauge strip, provides premise for the improvement of plate shape
Condition improves sour rolling mill lumber recovery.
Detailed description of the invention
Fig. 1 is the T4 material of volume 100, the sharing of load result figure of every volume.
Fig. 2 is the T5 material of volume 100, the sharing of load result figure of every volume.
Specific embodiment
In order to deepen the understanding of the present invention and recognize, the invention will be further described below in conjunction with the accompanying drawings and introduces.
The present invention is applied in the automatic control of 1,420 5 rack of plum steel, six roller UCM acid milling train, can also be applied to
Other kinds of milling train.
1: Mei Gang 1420 acid of table rolls milling equipment parameter
| Serial number | Device name | Parameter |
| 1 | Working roll size mm | φ435/φ385X1420 |
| 2 | Intermediate calender rolls size mm | φ490/φ440X1400 |
| 3 | Support roller size mm | φ1300/φ1150×1420 |
| 4 | Power of motor kw | 4600 |
| 5 | Motor speed rpm | 400/1200 |
| 6 | Intermediate roll shifting amount mm | 0/380 |
| 7 | Work roll bending kn | -180/400 |
| 8 | Intermediate roll bending kn | 10/500 |
Acid rolls in production, and there are many product specification, different according to product width and thickness for the ease of carry calculation, in conjunction with
Milling train product mix is divided into different brackets, division methods: of different size according to strip is divided into three width grades, often
11 level of thickness are divided into according to exit thickness in a width grade.
Embodiment 1: Fig. 1 is that the T4 of volume 100 expect, entrance 2.0mm, width 800-900mm, and exit thickness 0.201mm specification is every
The sharing of load of volume is as a result, sr1-sr5 is the load of 1-5 rack.
2 width grade of table corresponds to table；
| Grade | Width range (mm) |
| 1 | 700≤width < 800 |
| 2 | 800≤width < 900 |
| 3 | 900≤width < 1000 |
3 level of thickness of table corresponds to table；
| Grade | Thickness range (mm) |
| 1 | 0≤thickness < 0.17 |
| 2 | 0.17≤thickness < 0.18 |
| 3 | 0.18≤thickness < 0.19 |
| 4 | 0.19≤thickness < 0.20 |
| 5 | 0.20≤thickness < 0.22 |
| 6 | 0.22≤thickness < 0.25 |
| 7 | 0.25≤thickness < 0.30 |
| 8 | 0.30≤thickness < 0.35 |
| 9 | 0.35≤thickness < 0.40 |
| 10 | 0.40≤thickness < 0.45 |
| 11 | 0.45≤thickness < 0.55 |
4 T4 of table expects 800-900mm width 1-5 coefficient of balance, wherein the 1st is classified as 1,2 rack roll-force coefficients of balance, 2-5
It is classified as 2-5 rack power-balance coefficient
5 initial depression coefficient table of table
T4 material, specification: inlet thickness 2.0mm, exit thickness 0.201mm, width 856mm, 2# rack are individually more
It changes jobs and is illustrated for roller, calculate that steps are as follows:
1) the initial depression load of 1-5 rack is calculated
The specification is checked according to the material rate, width, thickness of T4 material
Initial depression coefficient is 0.2241,0.4591,0.6492,0.8294,1, according to the following formula, going out for 1-5 rack is calculated
Mouth thickness are as follows:
H1=2*e(0.2241*ln(0.201/2))=1.195
H2=2*e(0.4591*ln(0.201/2))=0.679
H3=2*e(0.6492*ln(0.201/2))=0.450
H4=2*e(0.8294*ln(0.201/2))=0.297
H5=2*e(1*ln(0.201/2))=0.201
The exit thickness for calculating 1-5 rack, further according to formula
The load of 1-5 rack is calculated are as follows:
Sr1=(2-1.195)/2=40.2%
Sr2=(1.195-0.679)/1.195=41.7%
Sr3=(0.679-0.450)/0.679=35.4%
Sr4=(0.450-0.297)/0.450=33.9%
Sr5=(0.297-0.201)/0.297=32.4%
2) after the independent roll change of 2# rack, since 2# rack coefficient of friction changes, coefficient of friction penalty coefficient is improved
The roll-force of 30%, 2# rack greatly improves, and according to tube rolling simulation formula, the roll-force that 1,2 racks are calculated is respectively
847 tons, 889 tons；
3) roll-force coefficient of balance is calculated
According to width, thickness, checking in the coefficient of balance between 1,2 racks in table 4 is 110, then
Bal=100*847/889/ (110+0)
=0.86
4) roll-force coefficient of balance Bal is checked；
α takes 2%, and due to bal < 0.98, if coefficient of balance Bal meets above-mentioned condition, calculating goes to the 4th step, otherwise
Load sr1, sr2 of 1,2 racks are corrected, correction amount is ± sr/ (10+3*loop), and loop is cycle-index, since second step
It recalculates, by 11 cycle calculations, as a result the load of 1-5 rack is 50.2,43.9,30.1,28.9,27.6；
It checks the load of 1,2 racks, compensates roll-force coefficient of balance ComPa；
Steel grade exit thickness under smooth roll rolling mode is less than 0.35mm,
Sr1-sr2 > 1, it is automatic to reduce roll-force coefficient of balance ComPa=ComPa-1；
If 1, the load of 2 racks meets several situations above, after compensation, goes to the first step and recalculate, otherwise calculate
Terminate, the load for finally obtaining 1-5 rack is 43.4,45.8,36.1,30.4,26.4.
As can be seen from Figure 1 the 1-5 rack sharing of load situation of volume 100, the load for realizing same steel grade same specification are basic
It remains unchanged, without intervention, setting accuracy is improved at scene.
Embodiment 2: Fig. 2 is that the T5 of volume 100 expect, entrance 2.0mm, width 800-900mm, and exit thickness 0.183mm specification is every
The sharing of load of volume is as a result, sr1-sr5 is the load of 1-5 rack.
6 T5 of table expects 800-900mm width 1-5 coefficient of balance, wherein the 1st is classified as 1,2 rack roll-force coefficients of balance, 2-5
It is classified as 2-5 rack power-balance coefficient
Expected with T5, specification: inlet thickness 2.0mm, exit thickness 0.183mm, width are illustrated for 836mm, 1#
Rack is illustrated for being individually replaced working roll, and steps are as follows for calculating:
1) the initial depression load of 1-5 rack is calculated
The specification is checked according to the material rate, width, thickness of T5 material
Initial depression coefficient be 0.2336,0.4751,0.6665,0.8356,1, according to the following formula, 1-5 rack is calculated
Exit thickness are as follows:
H1=2*e(0.2336*ln(0.201/2))=1.144
H2=2*e(0.4751*ln(0.201/2))=0.642
H3=2*e(0.6665*ln(0.201/2))=0.406
H4=2*e(0.8356*ln(0.201/2))=0.271
H5=2*e(1*ln(0.201/2))=0.183
The exit thickness for calculating 1-5 rack, further according to formula
The load of 1-5 rack is calculated are as follows:
Sr1=(2-1.144)/2=42.8%
Sr2=(1.144-0.642)/1.144=43.9%
Sr3=(0.642-0.406)/0.642=36.7%
Sr4=(0.406-0.271)/0.406=33.3%
Sr5=(0.271-0.183)/0.271=32.5%
2) after the independent roll change of 1# rack, since 1# rack coefficient of friction changes, coefficient of friction penalty coefficient is improved
The roll-force of 30%, 2# rack greatly improves, and is calculated according to deformation resistance, tension, coefficient of friction, roll-force adaptation coefficient
Roll-force to 1,2 racks is respectively 1027 tons, 825 tons；
3) roll-force coefficient of balance is calculated；
According to width, thickness, it is 105 that the coefficient of balance between 1,2 racks is checked in table 6, then
Bal=100*1027/825/ (105+0)
=1.18
4) roll-force coefficient of balance Bal is checked；
α takes 2%, and due to bal > 1.02, if coefficient of balance Bal meets above-mentioned condition, calculating goes to the 4th step, otherwise
Load sr1, sr2 of 1,2 racks are corrected, correction amount is ± sr/ (10+3*loop), and loop is cycle-index, since the first step
It recalculates, by 9 cycle calculations, as a result the load of 1-5 rack is 40.6,50.2,34.8,31.5,30.8
5) load for checking 1,2 racks, compensates roll-force coefficient of balance ComPa
Steel grade exit thickness under smooth roll rolling mode is less than 0.35mm；
Sr2-sr1 > 4, it is automatic to increase roll-force coefficient of balance ComPa=ComPa+1
After compensation, go to second step and recalculate, otherwise calculate terminate, finally obtain 1-5 rack load be 46.4,
48.0、35.0、31.2、26.2。
As can be seen from Figure 2 the 1-5 rack sharing of load situation of volume 100, the load for realizing same steel grade same specification are basic
It remains unchanged, without intervention, setting accuracy is improved at scene.
It should be noted that above-described embodiment, is not intended to limit the scope of protection of the present invention, in above-mentioned technical proposal
On the basis of made equivalents or substitution each fall within the range that the claims in the present invention are protected.
Claims (1)
1. a kind of method for reducing mill load fluctuation, which is characterized in that the described method comprises the following steps: 1) calculating 1-5 machine
The initial depression load of frame uses roll-force balance mode between 1-2 rack, uses power balance system between 2-5 rack；
2) roll-force P1-P5 of 1-5 racks is calculated according to deformation resistance, tension, coefficient of friction, roll-force adaptation coefficient；
Wherein: kp is average resistance of deformation in formula, and Dp is influence of the frictional force to roll-force, and k is tension to roll-force
It influences, b is the width of strip, and zp is roll-force adaptation coefficient, and R ' is to flatten radius, and H is rack inlet thickness, and h is rack
Exit thickness；
3) roll-force coefficient of balance is calculated；
Bal=100*P1/P2/(A+ComPa)
Wherein P1 is the roll-force that 1# rack calculates, and P2 is the roll-force that 2# rack calculates, and A is a fixation in 50-150
Value is obtained from data form, and being specifically shown in Table 4, ComPa is roll-force coefficient of balance, is initially 0；
1-α≤Bal≤1+α
4) roll-force coefficient of balance Bal is checked；
Wherein, α takes 2%, if coefficient of balance Bal meets above-mentioned condition, calculating goes to step 5), otherwise corrects 1,2 racks
Load sr1, sr2, correction amount be ± sr/ (1+3*loop), loop is cycle-index, is recalculated since step 1)；
5) load for checking 1,2 racks, compensates roll-force coefficient of balance ComPa
Compensation principle: following sr1 is the load of 1# rack, and sr2 is the load of 2# rack:
Exit thickness is less than 0.35mm under smooth roll rolling mode,
Sr1-sr2 > 1, it is automatic to reduce roll-force coefficient of balance ComPa=ComPa-1；
Sr2-sr1 > 4, it is automatic to increase roll-force coefficient of balance ComPa=ComPa+1；
Exit thickness is greater than 0.35mm under hair roller rolling mode, and 5# rack is under permanent roll-force constant pressure,
Sr1-sr2 > 3, it is automatic to reduce roll-force coefficient of balance ComPa=ComPa-1
Sr1-sr2 < 1, it is automatic to increase roll-force coefficient of balance ComPa=ComPa+1
If 1, the load of 2 racks meets several situations above, after compensation, go to the first step and recalculate, otherwise calculating terminates；
Table 4
Wherein, T4 expects 800-900mm width 1-5 coefficient of balance, wherein the 1st is classified as 1,2 rack roll-force coefficients of balance, 2-5 column
For 2-5 rack power-balance coefficient.
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