CN102553943B - Method for controlling helper rolls of loop for carbon steel continuous annealing unit - Google Patents

Abstract

The invention provides a method for controlling helper rolls of a loop for a carbon steel continuous annealing unit, which includes steps of 1), receiving data of the loop; 2), computing tension loss caused by inherent inertia of each helper roll according to the data received in the step 1); 3), zoning the helper rolls of the loop so that each zone comprises one driving roll and a non-driving roll close to the same; and 4), computing compensation torques which are used for compensating affection to tension caused by inertia of all helper rolls, steels trip inertia and inherent inertia of a motor of the driving roll and are required on a motor shaft of the driving roll within each zone, and then applying the three types of compensating torques for execution of the motor of the driving roll within the zone. The helper rolls are zoned as units, and various tension losses within different zones are computed and can be compensated by the driving rolls within the zones, so that affection to tension, caused by the length of the steel strip and the mechanical inertia of the loop, can be reduced.

Description

A kind of control method that connects the kink help roll that moves back unit for carbon steel

Technical field

The invention belongs to cold rolling continuous processing line field, be specifically related to a kind of control method that connects the kink help roll that moves back unit for carbon steel.

Background technology

In cold rolling continuous processing line unit process function be generally the first uncoiling of stock roll, weld with the band steel that moves ahead after, enter process section and process, after PROCESS FOR TREATMENT completes, cut by coiling machine and be rolled into finished product volume at outlet section.The processing of process section generally has the technologies such as pickling, annealing, zinc-plated, zinc-plated, degreasing, color coating.In the production process of cold rolling continuous processing line, in order to ensure the Quality and yield of cold-rolled products, the speed of service of technique section strip steel should keep stablizing constant as far as possible, to meet the production technology of cold-rolling treatment product.And there is the conversion operation process such as uncoiling, crop, back-end crop and welding due to stock roll in entrance, cannot remain consistent speed running with process section, therefore an entry loop is set between entrance and process section, be used for storing a certain amount of band steel, to ensure that the conversion operation of entrance can not cause process section to stop or reduction of speed.Equally, outlet section, because existence such as cuts, samples, batches at the conversion operation process, also arranges an outlet looping that plays same purpose between outlet section and process section.

One of target of cold rolling processing line unit is to ensure that production process is continuously continual all the time.Need the equipment that adopts kink relevant for this reason.With kink as inlet region, the buffering between process island and outlet area.These bufferings can make to keep in regular hour amount with steel the continuous operation of production.For example: for inlet region kink, be filled the band steel of certain cover amount while starting upper volume, if a new coil of strip is put into entrance (entrance velocity=0), this kink can make process section continue operation; For outlet area kink, when beginning, by emptying, carrying out between coil of strip conversion operational period, this kink can absorb the band steel from process section.It is the assembly pulley with steel that kink can be understood as weight abstractively.

Kink is divided into vertical loop and horizontal loop according to installation.As shown in Figure 1, its entirety moves up and down by looping car (kink movable frame) is mobile and realizes vertical loop.Looping car is by rope traction, and the convergent-divergent of steel wire rope is by the winding drum realization of reeling, thereby kink main driving motor is realized moving up and down of looping car by the coiling action of gear-box control winding drum, and then control kink cover amount.Bottom or top at kink also have a series of fixing roller to be, are help roll, are divided into live-roller and non-drive roller.

The Position Control of kink and tension force control mainly realize by the main driving motor of control kink.But between kink moving period, or in the kink accelerating period, for the machinery inertial of band steel length (weight) or kink is reduced to a minimum on an impact for tension force performance, must accurately control live-roller.

Summary of the invention

The technical problem to be solved in the present invention is: a kind of control method that connects the kink help roll that moves back unit for carbon steel is provided, can compensates the moment of live-roller motor, to reduce the impact on tension force performance with steel length and kink machinery inertial.

The present invention for solving the problems of the technologies described above taked technical scheme is: a kind of control method that connects the kink help roll that moves back unit for carbon steel, is characterized in that: it comprises the following steps:

1) receive kink data, comprising: kink number of plies Nr_Loops, band steel density Spec_Weight, belt steel thickness Strip_Thickness, strip width Strip_Width, kink entrance velocity V_In, the entry accelerated degree of kink A_In, kink muzzle velocity V_Out, kink outlet acceleration A _ Out and kink physical location S_act;

2) according to step 1) data that receive calculate the loss of tension that the intrinsic inertia of each help roll causes;

3) kink help roll is carried out to subregion, each subregion comprises a live-roller and the non-drive roller near this live-roller;

4), in each subregion, calculate the impact on tension force performance causing by the inertia of all help rolls, with steel inertia and the intrinsic inertia of live-roller motor in order to compensate, the compensating torque needing respectively on live-roller motor shaft; Above-mentioned three kinds of compensating torques are added to the live-roller motor in this subregion and carried out.

Press such scheme, described step 2) concrete computational process be: the acceleration that first calculates single help roll:

$A\_\mathrm{Roll}=A\_\mathrm{In}+(A\_\mathrm{Out}-A\_\mathrm{In})\×\frac{\mathrm{Position}\_\mathrm{Roll}}{\mathrm{Nr}\_\mathrm{Loops}}[m/s^2]---\left(2\right)$

In formula, the acceleration that A_Roll is help roll; A_In is the entry accelerated degree of kink; A_Out is kink outlet acceleration; Nr_Loops is the kink number of plies; Position_Roll is the positional value of help roll, can be roller 0,1,2,3

The loss of tension that the intrinsic inertia of single help roll causes is:

$T\_\mathrm{Inertia}=\frac{J\×4\×A\_\mathrm{Roll}}{D^2}\left[N\right]---\left(3\right)$

In formula, T_Inertia is the loss of tension being caused by the intrinsic inertia of help roll; J is the inertia of roller; A_Roll is the acceleration of help roll; D is the diameter of help roll.

Press such scheme, described step 4) the loss of tension that caused by the inertia of all help rolls in subregion on the impact of tension force performance be:

$T\_\mathrm{In}\_\mathrm{zone}=\frac{T\_\mathrm{In}\_R1}{2}+T\_\mathrm{In}\_R2+T\_\mathrm{In}\_R3+...+\frac{T\_\mathrm{In}\_\mathrm{Rlast}}{2}\left[N\right]---\left(4\right)$

In formula, T_In_zone is the loss of tension being caused by the inertia of all help rolls in subregion; T_In_Rn is the loss of tension being caused by the inertia of n help roll in subregion, and T_In_Rn is obtained by formula 3.

Press such scheme, described step 4) affected by the loss of tension causing with steel inertia in subregion and be:

T_lose＝Nr_Loops_Zn×S_act×Spec_Weight×S_Width×S_Thick×A_average_Zn[N]?(5)

In formula, T_lose is by the loss of tension causing with steel inertia in n subregion; Nr_Loops_Zn is the number of plies of kink in n subregion; S_act is the physical location of kink; Spec_Weight is the density with steel; S_Width is the width with steel; S_Thick is the thickness with steel; A_average_Zn is the average acceleration with steel in n subregion, is calculated by following formula:

$\frac{A\_\mathrm{First}\_\mathrm{Roll}\_\mathrm{Zn}+A\_\mathrm{Last}\_\mathrm{Roll}\_\mathrm{Zn}}{2}---\left(6\right),$

In formula, A_average_Zn is the average acceleration with steel in n subregion; A_First_Roll_Zn is the acceleration of the 1st help roll in n subregion, and use formula 2 is calculated; A_Last_Roll_Zn is the acceleration of last 1 help roll in n subregion, and use formula 2 is calculated.

Press such scheme, described step 4) the loss of tension that caused by the inertia of all help rolls in subregion affect the compensating torque needing on live-roller motor shaft causing and be:

$\mathrm{TQ}\_\mathrm{Inertia}\_\mathrm{Roll}\_\mathrm{Zn}=\frac{T\_\mathrm{Inertia}\_\mathrm{Roll}\_\mathrm{Zn}\×D}{2\×i}[N\·m]---\left(7\right),$

In formula, TQ_Inertia_Roll_Zn is in n subregion, for compensation 4 losses of tension that cause, and needed compensating torque on live-roller motor shaft; T_Inertia_Roll_Zn is the loss of tension being caused by the inertia of all help rolls in n subregion, and use formula 4 is calculated; D is the diameter with transmission help roll; I is speed reducing ratio.

Press such scheme, described step 4) affect by the loss of tension causing with steel inertia in subregion the compensating torque needing causing on live-roller motor shaft and be:

$\mathrm{TQ}\_\mathrm{Inertia}\_\mathrm{Srtip}\_\mathrm{Zn}=\frac{T\_\mathrm{Inertia}\_\mathrm{Strip}\_\mathrm{Zn}\×D}{2\×i}[N\·m]---\left(8\right),$

In formula, TQ_Inertia_Strip_Zn is in n subregion, for compensation 5 losses of tension that cause, and needed compensating torque on live-roller motor shaft; T_Inertia_Strip_Zn is by the loss of tension causing with steel inertia in n subregion, and use formula 5 is calculated; D is the diameter with transmission help roll; I is speed reducing ratio.

Press such scheme, described step 4) the loss of tension that caused by the intrinsic inertia of live-roller motor in subregion affect the compensating torque needing on live-roller motor shaft causing and be:

$\mathrm{TQ}\_\mathrm{Inertia}\_\mathrm{Mot}\_\mathrm{Zn}=J\_\mathrm{Mot}\×\frac{2\×A\_\mathrm{Roll}\×i}{\mathrm{Diam}}[N\·m]---\left(9\right),$

In formula, TQ_Inertia_Mot_Zn is the compensating torque being caused by the intrinsic inertia of live-roller motor in n subregion; J_Mot is the intrinsic inertia of live-roller motor; A_Roll is the acceleration of live-roller, and use formula 2 is calculated; Diam is the diameter of motor shaft; I is speed reducing ratio.

Beneficial effect of the present invention is:

1, because the help roll in kink is divided into live-roller and non-drive roller, inertia loss on non-drive roller cannot supplement by self, only have by nearest live-roller and compensate, design subregion for this reason and processed this problem, the division of subregion is divided according to live-roller, and each subregion is made up of live-roller and near non-drive roller.Taking subregion as unit, calculate the various losses of tension in each subregion, then make up these loss of tension by the live-roller torque compensation in subregion, to reduce the impact on tension force performance with steel length and kink machinery inertial.

Brief description of the drawings

Fig. 1 is the schematic diagram of vertical loop.

Fig. 2 is the division schematic diagram of vertical loop subregion.

Fig. 3 is the control block diagram of live-roller motor in the arbitrary subregion of vertical loop.

Detailed description of the invention

In the arbitrary subregion of vertical loop, as shown in Figure 3, the control core of live-roller motor is speed regulator and torque adjusting device to the control block diagram of live-roller motor, and the effect of speed regulator is the speed that regulates motor, makes the speed of motor reach motor speed setting value.The effect of torque adjusting device is the output of accepting speed regulator, forms moment given after other torque compensations that simultaneously superpose, and makes motor export corresponding moment by adjuster.

The control method that connects the kink help roll that moves back unit for carbon steel comprises the following steps:

1) receive kink data, comprising: kink number of plies Nr_Loops, band steel density Spec_Weight, belt steel thickness Strip_Thickness, strip width Strip_Width, kink entrance velocity V_In, the entry accelerated degree of kink A_In, kink muzzle velocity V_Out, kink outlet acceleration A _ Out and kink physical location S_act.Use these data, all rollers in kink are done to some calculating.

2) according to step 1) data that receive calculate the loss of tension that the intrinsic inertia of each help roll causes.

For each help roll (transmission or non-drive), all by the calculating of carrying out below.The result of these calculating is ephemeral data, and these ephemeral datas calculate the help roll for after a while.

Help roll has the loss of tension.In addition, the speed of help roll or acceleration are to being influential with the calculating of steel inertia.

Speed is calculated:

$V\_\mathrm{Roll}=V\_\mathrm{In}+(V\_\mathrm{Out}-V\_\mathrm{In})\×\frac{\mathrm{Position}\_\mathrm{Roll}}{\mathrm{Nr}\_\mathrm{Loops}}[m/s]---\left(1\right)$

In formula, the speed that V_Roll is help roll; V_In is kink entrance velocity; V_Out is kink muzzle velocity; Nr_Loops is the number of plies of kink; Position_Roll is the positional value of help roll, can be roller 0,1,2,3 ...

The acceleration of help roll:

$A\_\mathrm{Roll}=A\_\mathrm{In}+(A\_\mathrm{Out}-A\_\mathrm{In})\×\frac{\mathrm{Position}\_\mathrm{Roll}}{\mathrm{Nr}\_\mathrm{Loops}}[m/s^2]---\left(2\right)$

In formula, the acceleration that A_Roll is help roll; A_In is the entry accelerated degree of kink; A_Out is kink outlet acceleration; Nr_Loops is the number of plies of kink; Position_Roll is the positional value of help roll, can be roller 0,1,2,3

The loss of tension that the intrinsic inertia of help roll causes is:

$T\_\mathrm{Inertia}=\frac{J\×4\×A\_\mathrm{Roll}}{D^2}\left[N\right]---\left(3\right)$

In formula, T_Inertia is the loss of tension being caused by the intrinsic inertia of help roll; J is the inertia of help roll; A_Roll is the acceleration of help roll; D is the diameter of help roll.

3) kink help roll is carried out to subregion, each subregion comprises a live-roller and the non-drive roller near this live-roller.Fig. 2 is the division schematic diagram of the vertical loop subregion of belt drive roller, and the present embodiment comprises 7 help rolls, is wherein numbered 2,4 for live-roller, is numbered 0,1,3,5,6 for non-drive roller.This kink is divided into two subregions: the 1st subregion and the 2nd subregion.In the 1st subregion, comprise 0,1,2, No. 3 help roll, the loss of local area internal tension realizes by the transmission device of No. 2 help rolls of compensation.In the 2nd subregion, comprise 3,4,5, No. 6 help roll, the loss of local area internal tension realizes by the transmission device of No. 4 help rolls of compensation.No. 3 help roll had both belonged to the 1st subregion, also belonged to the 2nd subregion, and the loss of tension causing on No. 3 help rolls is divided the 1st subregion and the 2nd subregion equally.

4), in each subregion, calculate the impact on tension force performance causing by the inertia of all help rolls, with steel inertia and the intrinsic inertia of live-roller motor in order to compensate, the compensating torque needing respectively on live-roller motor shaft; Above-mentioned three kinds of compensating torques are added to the live-roller motor in this subregion and carried out.

The loss of tension formula 3 being caused by the inertia (band grade of steel is other) of single help roll is calculated.

The loss of tension formula 4 being caused by the inertia (band grade of steel is other) of all help rolls in definition is calculated:

$T\_\mathrm{In}\_\mathrm{zone}=\frac{T\_\mathrm{In}\_R1}{2}+T\_\mathrm{In}\_R2+T\_\mathrm{In}\_R3+...+\frac{T\_\mathrm{In}\_\mathrm{Rlast}}{2}\left[N\right]---\left(4\right)$

In formula, T_In_zone is the loss of tension (band grade of steel is other) being caused by the inertia of all help rolls in subregion; T_In_Rn is the loss of tension (band grade of steel is other) being caused by the inertia of n help roll in subregion, and the computing formula of T_In_Rn is shown in formula 3.

By the loss of tension causing with steel inertia in definition be:

T_lose＝Nr_Loops_Zn×S_act×Spec_Weight×S_Width×S_Thick×A_average_Zn[N]?(5)

In formula, T_lose is the loss of tension causing with steel inertia by definition; Nr_Loops_Zn is the number of plies of kink in n subregion; S_act is the physical location of kink; Spec_Weight is the density with steel; S_Width is the width with steel; S_Thick is the thickness with steel; A_average_Zn is the average acceleration with steel in n subregion, is calculated by following formula:

$\frac{A\_\mathrm{First}\_\mathrm{Roll}\_\mathrm{Zn}+A\_\mathrm{Last}\_\mathrm{Roll}\_\mathrm{Zn}}{2}---\left(6\right),$

In formula, A_average_Zn is the average acceleration with steel in n subregion; A_First_Roll_Zn is the acceleration of the 1st help roll in n subregion, and use formula 2 is calculated; A_Last_Roll_Zn is the acceleration of last 1 help roll in n subregion, and use formula 2 is calculated.

In n subregion, for compensation 4 and formula 5 described in the impacts on tension force performance of two kinds of loss of tension, on live-roller motor shaft, needed compensating torque is:

$\mathrm{TQ}\_\mathrm{Inertia}\_\mathrm{Roll}\_\mathrm{Zn}=\frac{T\_\mathrm{Inertia}\_\mathrm{Roll}\_\mathrm{Zn}\×D}{2\×i}[N\·m]---\left(7\right),$

$\mathrm{TQ}\_\mathrm{Inertia}\_\mathrm{Srtip}\_\mathrm{Zn}=\frac{T\_\mathrm{Inertia}\_\mathrm{Strip}\_\mathrm{Zn}\×D}{2\×i}[N\·m]---\left(8\right),$

In formula, TQ_Inertia_Roll_Zn is in n subregion, for compensation 4 losses of tension that cause, and needed compensating torque on live-roller motor shaft; TQ_Inertia_Strip_Zn is in n subregion, for compensation 5 losses of tension that cause, and needed compensating torque on live-roller motor shaft; T_Inertia_Roll_Zn is the loss of tension (band grade of steel is other) being caused by the inertia of all help rolls in n subregion, and use formula 4 is calculated; T_Inertia_Strip_Zn is by the loss of tension causing with steel inertia in n subregion, and use formula 5 is calculated; D is the diameter with transmission help roll; I is speed reducing ratio.

In n subregion, in order to compensate the impact of the intrinsic inertia of live-roller motor on tension force performance, on live-roller motor shaft, needed compensating torque is:

$\mathrm{TQ}\_\mathrm{Inertia}\_\mathrm{Mot}\_\mathrm{Zn}=J\_\mathrm{Mot}\×\frac{2\×A\_\mathrm{Roll}\×i}{\mathrm{Diam}}[N\·m]---\left(9\right),$

In formula, TQ_Inertia_Mot_Zn is the compensating torque being caused by the intrinsic inertia of live-roller motor in n subregion; J_Mot is the intrinsic inertia of live-roller motor; A_Roll is the acceleration of live-roller, and use formula 2 is calculated; Diam is the diameter of motor shaft; I is speed reducing ratio.

Claims (4)

1. a control method that connects the kink help roll that moves back unit for carbon steel, is characterized in that: it comprises the following steps:

1) receive kink data, comprising: kink number of plies Nr_Loops, band steel density Spec_Weight, belt steel thickness S_Thick, strip width Strip_Width, single kink help roll entrance velocity V_In, the entry accelerated degree of single kink help roll A_In, single kink help roll muzzle velocity V_Out, single kink help roll outlet acceleration A _ Out and single kink help roll physical location S_act;

2) according to step 1) data that receive calculate the loss of tension that the intrinsic inertia of single kink help roll causes;

3) kink help roll is carried out to subregion, each subregion comprises a live-roller and the non-drive roller near this live-roller;

4) in each subregion, the loss of tension of calculating the inertia of all kink help rolls, causing with steel inertia and the intrinsic inertia of live-roller motor; In order to compensate above-mentioned three losses of tension, calculate respectively above-mentioned three losses of tension needed compensating torque on live-roller motor shaft; Above-mentioned three compensating torques are added to the live-roller motor in this subregion and carried out;

Described step 2) concrete computational process be:

First calculate the acceleration of single kink help roll in n subregion:

$A\_\mathrm{Roll}=A\_\mathrm{In}+(A\_\mathrm{Out}-A\_\mathrm{In})\×\frac{\mathrm{Position}\_\mathrm{Roll}}{\mathrm{Nr}\_\mathrm{Loops}},m/s^2,$ Formula 2),

In formula, Position_Roll is the positional value of single kink help roll;

The loss of tension that the intrinsic inertia of single kink help roll causes is:

$T\_\mathrm{Inertia}=\frac{J\×4\×A\_\mathrm{Roll}}{D^2},N,$ Formula 3),

In formula, J is the intrinsic inertia of single kink help roll; D is the diameter of single kink help roll.

2. the control method that connects the kink help roll that moves back unit for carbon steel according to claim 1, is characterized in that: described step 4) in, the loss of tension causing with steel inertia in n subregion is:

T_lose=Nr_Loops_Zn × S_act × Spec_Weight × S_Width × S_Thick × A_average_Zn, N, formula 5), in formula, Nr_Loops_Zn is the kink number of plies in n subregion; A_average_Zn is the average acceleration with steel in n subregion, is calculated by following formula:

$A\_\mathrm{average}\_\mathrm{Zn}=\frac{A\_\mathrm{First}\_\mathrm{Roll}\_\mathrm{Zn}+A\_\mathrm{Last}\_\mathrm{Roll}\_\mathrm{Zn}}{2},m/s^2,$ Formula 6),

In formula, A_First_Roll_Zn is the acceleration of first kink help roll in n subregion, use formula 2) calculate; A_Last_Roll_Zn is the acceleration of last kink help roll in n subregion, use formula 2) calculate.

3. the control method that connects the kink help roll that moves back unit for carbon steel according to claim 2, it is characterized in that: described step 4) in, the compensating torque needing on live-roller motor shaft that the loss of tension causing with steel inertia in n subregion causes is:

$\mathrm{TQ}\_\mathrm{Inertia}\_\mathrm{Strip}\_\mathrm{Zn}=\frac{T\_\mathrm{lose}\×D}{2\×i},N\·m,$ Formula 8),

In formula, i is speed reducing ratio.

4. the control method that connects the kink help roll that moves back unit for carbon steel according to claim 1, it is characterized in that: described step 4) in, the compensating torque needing on live-roller motor shaft that the loss of tension that the intrinsic inertia of live-roller motor in n subregion causes causes is:

$\mathrm{TQ}\_\mathrm{Inertia}\_\mathrm{Mot}\_\mathrm{Zn}=J\_\mathrm{Mot}\×\frac{2\×A\_\mathrm{Roll}\×i}{\mathrm{Diam}},N\·m,$ Formula 9),

In formula, J_Mot is the intrinsic inertia of live-roller motor; Diam is the diameter of live-roller motor shaft; I is speed reducing ratio.