ENERGY SERVICE SYSTEM OF MULTI-MACHINE PRODUCTION LINE AND DESIGN METHOD OF SHARED DRIVE SYSTEM

Abstract

Disclosed are an energy service system of a multi-machine production line and a control method thereof, the method includes reorganizing respective machines in the production line into three types of controllable entities: a drive system, an energy supply bus and an execution device, equipping them with a control center, and selecting a sub-drive system that is idle and is capable of completing the work stage with high energy efficiency to supply energy service for the corresponding execution device through the energy supply bus. Further disclosed is a design method of a multi-machine shared drive system of a production line, which increases the number of basic flow units of each drive unit to a maximum value one by one, and coordinates action time to form a variety of scheduling schemes, and selects a configuration scheme whose total time and total energy consumption are less as the shared drive system.

Claims

1. An energy service system of a multi-machine production line, comprising: a drive system, an execution device, an energy supply bus and a control center; the drive system is composed of m drive units, AG={AG.sub.1, AG.sub.2, . . ., AG.sub.i, . . ., AG.sub.m}, and is configured to supply energy for the execution device, wherein AG, represents the i-th drive unit, i ∈{1,2, . . .,m}; a rated power of the i-th drive unit AG.sub.i is recorded as PR.sub.i; the execution device is composed of n execution units, CL={CL.sub.1, CL.sub.2, . . ., CL.sub.j, . . ., CL.sub.n}, and is configured to complete a processing task in the production line, wherein CL.sub.j represents the j-th execution unit, j ∈{1,2, . . .,n}; the control center is configured to monitor and control on-off state of each drive unit and the energy supply bus; the energy supply bus is composed of m independent units, and is configured to convert an output energy form of the i-th drive unit AG.sub.i into an energy form required by the j-th execution unit; and each independent unit is made to consist of an inlet A end and an outlet B end, wherein the inlet A end connected to the i-th drive unit AG.sub.i is recorded as IN.sub.i; and, the outlet B end corresponding to the i-th inlet A end IN.sub.i has n outlets, which are sequentially connected to the first execution unit CL.sub.1 to the n-th execution unit CL.sub.n, wherein an outlet connected to the j-th execution unit CL.sub.1 is recorded as OU.sub.j; and, a switch SW.sub.ij is provided on the outlet OU.sub.ij, and when the switch SW.sub.ij is off, the i-th inlet A end IN.sub.i is connected to the outlet OU.sub.ij, and the i-th drive unit AG.sub.i is capable of supplying energy for the j-th execution unit CL.sub.j; the processing task is made to composed of n subtasks, TA={TA.sub.1, TA.sub.2, . . ., TA.sub.j, . . ., TA.sub.n}, wherein TA.sub.1 represents the j-th subtask; the j-th subtask TA.sub.j completes processing on the j-th execution unit CL.sub.j; the j-th subtask TA.sub.j is composed of processing stages with similar power requirements in k.sub.j different processes, wherein a power similarity Sm.sub.j.sup.α of any α-th processing stage is obtained by formula (1), and satisfies formula (2): $\begin{array}{cc}{\mathrm{Sm}}_j^a=\left(\frac{{\int }_{{\mathrm{st}}_j^a}^{{\mathrm{ct}}_j^a}{\mathrm{Te}}_j^a{\mathrm{Tf}}_j^a\mathrm{dt}}{{\mathrm{ct}}_j^a-{\mathrm{st}}_j^a}\right)/\left(\frac{{.Math.}_{a=1}^{k_j}{\int }_{{\mathrm{st}}_j^a}^{{\mathrm{ct}}_j^a}{\mathrm{Te}}_j^a{\mathrm{Tf}}_j^a\mathrm{dt}}{{.Math.}_{a=1}^{k_j}{\mathrm{ct}}_j^a-{\mathrm{st}}_j^a}\right)& \left(1\right)\\ {\mathrm{Sm}}_j^a\in {\mathrm{Sy}}_j& \left(2\right)\end{array}$ in formula (1), ct.sub.j.sup.α and st.sub.j.sup.α are an end time and a start time of the α-th processing stage, Te.sub.j.sup.α is an effort required by the α-th processing stage, Tƒ.sub.j.sup.α is a flow required by the α-th processing stage, α∈{1, 2,. . .,k.sub.j}; in formula (2), Sy.sub.j is a range of a j-th given power similarity, and between [0, 1]; a process is composed of D process steps {PRO.sub.3, PRO.sub.2, . . ., PRO.sub.d, . . ., PRO.sub.D}, wherein PRO.sub.d represents the d-th process step, the d-th process step PRO.sub.d is made to comprise k.sub.d processing stages, and to satisfy formula (3): $\begin{array}{cc}\underset{d=1}{\overset{D}{.Math.}}{k}_d=\underset{j=1}{\overset{n}{.Math.}}{k}_j.& \left(3\right)\end{array}$ the processing stages of the d-th process step PRO.sub.d are completed sequentially and successively, and satisfies formula (4):
ct(.sub.d.sup.α=st.sub.d.sup.α+1  (4) TAPR.sub.jd.sup.αr is made to represent that the r-th processing stage in the d-th process step PRO.sub.d is the α-th processing stage in the j-th subtask TA.sub.j; an output power Pn.sub.j of the j-th execution unit CL.sub.j and a required power of a corresponding processing task satisfy formula (5): $\begin{array}{cc}{\mathrm{Pn}}_j\in \left[\underset{a}{\min }\left\{{\mathrm{Te}}_j^{a}T{f}_j^a\right\},\underset{a}{\max }\left\{{\mathrm{Te}}_j^{a}T{f}_j^a\right\}\right].& \left(5\right)\end{array}$

2. A control method of the energy service system according to claim 1, being implemented as the following steps: step 1: starting m drive units AG.sub.1, AG.sub.2, . . ., AG.sub.i, . . ., AG.sub.m in the drive system; and turning on all switches {SW.sub.ij|i=1,2,. . .,m; j=1,2,. . .,n} of the energy supply bus, so that each drive unit is in an idle state, and sending current idle states to the control center; step 2: defining and initializing a loop variable x=2; step 3: initializing d=1; step 4: initializing r=1; step 5: determining whether r+d is equal to x, if yes, performing step 6-step 9, otherwise, performing step 10; step 6: calculating, by utilizing formula (6), an average power PA.sub.d.sup.r of the r-th processing stage TAPR.sub.jd.sup.αr in the d-th process step PRO.sub.d:
PA.sub.d.sup.r=∫.sub.st.sub.d.sub.r.sup.ct.sup.d.sup.rTe.sub.d.sup.rTƒ.sub.d.sup.rdt/(ct.sub.d.sup.r−st.sub.d.sup.r)  (6) step 7: sending, by the j-th execution unit CL.sub.j, an energy service request to the control center, according to the average power PA.sub.d.sup.r; selecting, by the control center, a drive unit AG.sub.ζwhich is in the idle state and with the highest drive efficiency from a current drive system by utilizing formula (7) according to the average power PA.sub.d.sup.r; , to supply energy for the j-th execution unit CL.sub.j, wherein AG.sub.ζ∈AG:
min|PR.sub.t−∫.sub.st.sub.d.sub.r.sup.ct.sup.d.sup.rTe.sub.d.sup.rTƒ.sub.d.sup.rdt/(ct.sub.d.sup.r−st.sub.d.sup.r)|  (7) step 8: responding, by the drive system, to the energy service request, and turning off a switch SW.sub.ζj through which a smallest drive unit AG.sub.ζ drives the j-th execution device CL.sub.j in the energy supply bus, so as to start to perform the processing stage TAPR.sub.jd.sup.αr, and marking a state of the smallest drive unit AG.sub.ζ as busy in the control center; step 9: turning on the corresponding switch SW.sub.ζj after the processing stage TAPR.sub.jd.sup.αr is completed, and marking the state of the smallest drive unit AG.sub.ζ as idle; step 10: determining whether r>k.sub.d holds after assigning r+1 to r, if yes, performing step 11, otherwise, performing step 5; step 11: determining whether d>D holds after assigning d+1 to d, if yes, performing step 12, otherwise, performing step 4; step 12: determining whether x>(max {k.sub.d}+D) holds after assigning x+1 to x, if yes, performing step 13, otherwise, performing step 3; and step 13: turning off m drive units AG.sub.1, AG.sub.2, . . ., AG.sub.i, AG.sub.m in the drive system, and turning off all the switches of the energy supply bus.

3. A design method of a multi-machine shared drive system of a production line, the production line being composed of K machines, which collectively complete a production task, wherein the design method is implemented as the following steps: step 1, designing a process: designing, according to a production task of the production line, K processes for completing the production task, and sequencing them according to a sequential order of the processes; and, calculating, according to a material to be processed, a load curve of each process, recorded as l.sub.1, l.sub.2, . . ., l.sub.k, . . ., l.sub.K, l.sub.k represents a load curve of the k-th process, k=1, 2, . . ., K; letting the k-th machine process the k-th process; step 2, designing a task: step 2.1: dividing, according to a load difference of the load curve of each process, each process into a plurality of processing stages, and sequencing the processing stages according to a sequential order; step 2.2: obtaining, by utilizing formula (8), a power similarity index Ps.sub.α of the α-th processing stage: $\begin{array}{cc}{\mathrm{Pn}}_j\in \left[\underset{a}{\min }\left\{{\mathrm{Te}}_j^a{\mathrm{Tf}}_j^a\right\},\underset{a}{\max }\left\{{\mathrm{Te}}_j^a{\mathrm{Tf}}_j^a\right\}\right].& \left(5\right)\end{array}$ in formula (8), ct.sub.α and st.sub.α are an end time and a start time of the α-th processing stage, respectively, p.sub.α(t) is a required pressure of the α-th processing stage, q.sub.α(t) is a required flow of the α-th processing stage, and A is a total number of processing stages; step 2.3: calculating the power similarity index of each processing stage, and constituting a task by processing stages corresponding to a power similarity index within a set range, so as to obtain m tasks, recording any i-th task as TA.sub.i; step 3, designing a schedule: step 3.1: scheduling, by utilizing formula (9), m drive units, so that the m drive units is capable of continuously completing processing stages of a process without conflict:
ct(α.sub.l.sub.k.sup.b(TA.sub.i))=st(α.sub.l.sub.k.sup.b+1(TA.sub.z))  (9) in formula (9), l.sub.k is the k-th process curve, k=1, 2, . . ., K; b represents a serial number of a processing stage of the k-th process curve, ct(α.sub.l.sub.k.sup.b(TA.sub.i)) is an end time of the b-th processing stage belonging to the k-th process curve l.sub.k in the i-th task TA.sub.i, and st(α.sub.l.sub.k.sup.b+1(TA.sub.z)) is a start time of the b+1-th processing stage belonging to the k-th process curve l.sub.k in the z-th task TA.sub.z, z=1, 2, . . ., m; step 3.2: scheduling, by utilizing formula (10), m drive units and all processing stages in the m tasks, so that the m drive units complete an entire production task in sequence according to a set process order without conflict:
ct(α.sub.l.sub.k.sup.last(TA.sub.i))≤st(α.sub.l.sub.k+1.sup.l(TA.sub.z))  (10) in formula (10), ct(α.sub.l.sub.k.sup.last(TA.sub.i)) is an end time of the last processing stage belonging to the k-th process curve l.sub.k in the i-th task TA.sub.i, and st(α.sub.l.sub.k+1.sup.1(TA.sub.z)) is a start time of the first processing stage belonging to the k+1-th process curve in the z-th task TA.sub.z; step 4, designing a drive unit: step 4.1: designing, according to the required pressure and the required flow of each processing stage in the i-th task TA.sub.i, the i-th drive unit DU.sub.i, which completes all processing stages in the i-th task TA.sub.i, and is composed of the i-th group of motors and the i-th group of pumps, i=1, 2, . . ., m; m represents a total number of drive units; step 4.2: selecting, from a set collection of pumps and for the i-th drive unit DU.sub.i, a hydraulic pump with the smallest rated displacement and using as the i-th basic flow unit Q.sub.i; when a corresponding task is completed, an output pressure of the i-th basic flow unit Q.sub.i is recorded as p.sub.i, an output flow is recorded as q.sub.i, an output power is recorded as P.sub.Qi, and an energy efficiency is recorded as η.sub.i; step 4.3: selecting, from the set collection of motors, a motor that matches the i-th basic flow unit Q.sub.i, so that a load rate of the selected motor when driving a corresponding task is within a set interval, so as to obtain a basic motor unit D.sub.i in the i-th drive unit DU.sub.i; and obtaining, by utilizing formula (11), a load rate β.sub.i of the basic motor unit D.sub.i when driving a corresponding action:
β.sub.i=P.sub.Qi/(P.sub.Di×η.sub.i)  (11)
in formula (11), P.sub.Di is a rated power of the i-th basic motor unit D.sub.i; step 5, designing a shared drive system: step 5.1: obtaining, by utilizing formula (12), a maximum value MAX.sub.i of the i-th basic flow unit Q.sub.i, and increasing the number of the i-th basic flow unit Q.sub.i in the i-th drive unit DU.sub.i to a set maximum value MAX.sub.i one by one:
MAX.sub.i=[Q.sub.max/Q.sub.i]+1  (12) in formula (12), Q.sub.max is a maximum flow value output by a single machine; step 5.2: increasing sequentially the number of the i-th basic motor unit D.sub.i one by one, and matching it with the i-th basic flow unit Q.sub.i; step 5.3: calculating a time t.sub.i.sup.n.sup.i(k) and an energy consumption E.sub.i.sup.n.sup.i(k) for completing a corresponding task of the k-th machine, wherein n is the number of the i-th basic flow unit Q.sub.i added in the i-th drive unit DU.sub.i, n.sub.i=1, 2, . . ., MAX.sub.i; step 5.4: obtaining, by utilizing formula (13), the total configuration schemes number M of the m drive units:
M=Π.sub.i=1.sup.mMAX.sub.i  (13) step 5.5: making, by scheduling run time and idle time of drive units in each configuration scheme, the m drive units complete processing stages in all tasks successively and sequentially, and forming M scheduling schemes, so as to utilize formula (14) to obtain a total time T.sub.j of the j-th scheduling scheme:
T.sub.j=Σ.sub.i=1.sup.mt.sub.i.sup.n.sup.i(k).sub.j+t(idle).sub.j  (14) in formula (14), t.sub.i.sup.n.sup.i(k).sub.j is a time for the i-th drive unit DU.sub.i in the j-th scheduling scheme to complete a corresponding task of the k-th machine, t(idle).sub.j is a total unloading time of all drive units in the j-th scheduling scheme, j=1, 2, . . ., M; step 5.6: obtaining, by utilizing formula (15), a total energy consumption E.sub.j of the j-th scheduling scheme:
E.sub.j=Σ.sub.k=l.sup.KΣ.sub.i=l.sup.mE.sub.i.sup.n.sup.i(k).sub.j+E(idle).sub.j  (15) in formula (15), K is a total number of machines in the production line, E.sub.i.sup.n.sup.i(k).sub.j is an energy consumption of the i-th drive unit DU.sub.i in the j-th scheduling scheme to complete the corresponding task of the k-th machine, and E(idle).sub.j is a total unloading energy consumption of all drive units in the j-th scheduling scheme; step 5.7: selecting, from the total time and the total energy consumption of M scheduling schemes, m drive units corresponding to a scheduling scheme with less total time and total energy consumption and using as a drive system shared by K machines, so as to achieve efficient production.

4. The design method of a multi-machine shared drive system of a production line according to claim 3, further comprising: calculating a sum of theoretical displacement of all basic flow units in the i-th drive unit DU.sub.i in the shared drive system; and selecting, from the set collection of pumps, a single pump having large displacement, whose theoretical displacement is equal to the sum of the theoretical displacement of all basic flow units in the i-th drive unit DU.sub.i, to replace all basic flow units in the i-th drive unit DU.sub.i, and to serve as a pump of the i-th drive unit; selecting, from the set collection of motors, a motor that matches the pump of the i-th drive unit DU.sub.i and using as a motor in the i-th drive unit, a load rate of the motor in the i-th drive unit DU.sub.i when driving a corresponding task is within a set interval, so that an energy efficiency of the selected motor when driving the corresponding task is not lower than a set efficiency value, so as to achieve a high efficiency operation.

Description

BRIEF DESCRIPTION OF THE DRAWINGS

[0075] FIG. 1 is a schematic diagram which shows a drive system, an execution device and an energy supply bus according to the present disclosure.

[0076] FIG. 2 is a flow chart which shows a control method of the energy service system according to the present disclosure.

[0077] FIG. 3 is a flow chart which shows a design method of a multi-machine shared drive system of a production line according to the present disclosure.

DESCRIPTION OF EMBODIMENTS

[0078] Embodiment 1: an energy service system of a multi-machine production line includes: a drive system, an execution device, an energy supply bus and a control center;

[0079] the drive system is composed of m drive units, AG={AG.sub.1, AG.sub.2, . . ., AG.sub.1, . . ., AG.sub.m}, and is configured to supply energy for the execution device, where, AG.sub.i represents the i-th drive unit, i ∈{1,2, . . .,m}; a rated power of the i-th drive unit AG.sub.i is recorded as PR.sub.i; the composition of a drive unit is configured according to an energy form required by the execution device, the output energy form can be hydraulic energy or mechanical energy, and implementation forms are a motor-pump and an electric motor, respectively.

[0080] The execution device is composed of n execution units, CL={CL.sub.1, CL.sub.2, . . ., CL.sub.j, . . ., CL}, and is configured to complete a processing task in the production line, where, CL.sub.j represents the j-th execution unit, j ∈{1,2, . . .,n}; the execution unit is a mechanical structure part of a machine tool.

[0081] The control center is configured to monitor and control on-off state of each drive unit and the energy supply bus, which is implemented by real-time monitoring and controlling each component of the system.

[0082] The energy supply bus is composed of m independent units, and is configured to convert an output energy form of the i-th drive unit AG.sub.i into an energy form required by the j-th execution unit; and each independent unit is made to consist of an inlet A end and an outlet B end, wherein, the inlet A end connected to the i-th drive unit AG.sub.i is recorded as IN.sub.i; and, the outlet B end corresponding to the i-th inlet A end IN.sub.i has n outlets, which are sequentially connected to the first execution unit CL.sub.1 to the n-th execution unit CL.sub.n, where, an outlet connected to the j-th execution unit CL.sub.j is recorded as OU.sub.ij; and, a switch SW.sub.ij is provided on the outlet OU.sub.ij, and when the switch SW.sub.ij is off, the i-th inlet A end IN.sub.i is connected to the outlet OU.sub.ij, and the i-th drive unit AG.sub.i is capable of supplying energy for the j-th execution unit CL.sub.j; as shown in FIG. 1.

[0083] The processing task is made to composed of n subtasks, TA={TA.sub.1, TA.sub.2, . . ., TA.sub.j, . . ., TA.sub.n}, where TA.sub.j represents the j-th subtask; the j-th subtask TA.sub.j completes processing on the j-th execution unit CL.sub.j; the j-th subtask TA.sub.j is composed of processing stages with similar power requirements in k.sub.j different processes, where a power similarity Sm.sub.j.sup. of any α-th processing stage is obtained by formula (1), and satisfies formula (2):

[00005]$\begin{array}{cc}{\mathrm{Sm}}_j^a=\left(\frac{{\int }_{{\mathrm{st}}_j^a}^{{\mathrm{ct}}_j^a}{\mathrm{Te}}_j^a{\mathrm{Tf}}_j^a\mathrm{dt}}{{\mathrm{ct}}_j^a-{\mathrm{st}}_j^a}\right)/\left(\frac{{.Math.}_{a=1}^{k_j}{\int }_{{\mathrm{st}}_j^a}^{{\mathrm{ct}}_j^a}{\mathrm{Te}}_j^a{\mathrm{Tf}}_j^a\mathrm{dt}}{{.Math.}_{a=1}^{k_j}{\mathrm{ct}}_j^a-{\mathrm{st}}_j^a}\right)& \left(1\right)\\ {\mathrm{Sm}}_j^a\in {\mathrm{Sy}}_j& \left(2\right)\end{array}$

[0084] in formula (1), ct.sub.j.sup.α and st.sub.j.sup.α are an end time and a start time of the α-th processing stage, Te.sup.α; is an effort required by the α-th processing stage, Tƒ.sub.j.sup.α is a flow required by the α-th processing stage, α∈{1,2, . . ., k.sub.j};

[0085] in formula (2), S.sub.j is a range of a j-th given power similarity, and between [0, 1].

[0086] the process is composed of D process steps {PRO.sub.1, PRO.sub.2, . . ., PRO.sub.d, . . ., PRO.sub.D}, where PRO.sub.d represents the d-th process step, the d-th process step PRO.sub.d is made to include k.sub.d processing stages, and to satisfy formula (3):

[00006]$\begin{array}{cc}\underset{d=1}{\overset{D}{.Math.}}{k}_d=\underset{j=1}{\overset{n}{.Math.}}{k}_j.& \left(3\right)\end{array}$

[0087] The processing stages of the d-th process step PRO.sub.d are completed sequentially and successively, and satisfies formula (4):

ct.sub.d.sup.α=st.sub.d.sup.α+1  (4)

[0088] TAPR.sub.jd.sup.αr is made to represent that the r-th processing stage in the d-th process step PRO.sub.d is the α-th processing stage in the j-th subtask TA.sub.j.

[0089] An output power Pn.sub.j of the j-th execution unit CL.sub.j and a required power of a corresponding processing task satisfy formula (5):

[00007]$\begin{array}{cc}{\mathrm{Pn}}_j\in \left[\underset{a}{\min }\left\{{\mathrm{Te}}_j^a{\mathrm{Tf}}_j^a\right\},\underset{a}{\max }\left\{{\mathrm{Te}}_j^a{\mathrm{Tf}}_j^a\right\}\right].& \left(5\right)\end{array}$

[0090] In the present embodiment, as shown in FIG. 2, a control method of the energy service system is implemented as the following steps: step 1: starting m drive units AG.sub.1, AG.sub.2, . . ., AG.sub.i, . . ., AG.sub.m in the drive system; and turning on all switches {SW.sub.ij|i=1,2,. . ., m; j=1,2, . . ., n} of the energy supply bus, so that each drive unit is in an idle state, and sending current idle states to the control center;

[0091] Step 2: defining and initializing a loop variable x=2.

[0092] Step 3: initializing d=1.

[0093] Step 4: initializing r=1.

[0094] Step 5: determining whether r+d is equal to x, if yes, performing step 6-step 9, otherwise, performing step 10.

[0095] Step 6: calculating, by utilizing formula (6), an average power PA.sub.d.sup.r of the r-th processing stage TAPR.sub.jd.sup.αr in the d-th process step PRO.sub.d:

PA.sub.d.sup.r=∫.sub.st.sub.d.sub.r.sup.ct.sup.d.sup.rTe.sub.d.sup.rTƒ.sub.d.sup.rdt/(ct.sub.d.sup.r−st.sub.d.sup.r)  6).

[0096] Step 7: sending, by the j-th execution unit CL.sub.1, an energy service request to the control center, according to the average power PA.sub.d.sup.r; selecting by the control center, a drive unit AG.sub.ζ which is in the idle state and with the highest drive efficiency from a current drive system by utilizing formula (7) according to the average power PA.sub.d.sup.r to supply energy for the j-th execution unit CL.sub.j, where AG.sub.ζ∈AG:

min|PR.sub.i−∫.sub.st.sub.d.sub.r.sup.ct.sup.d.sup.rTe.sub.d.sup.rTƒ.sub.d.sup.rdt/(ct.sub.d.sup.r−st.sub.d.sup.r)  (7).

[0097] Step 8: responding, by the drive system, to the energy service request, and turning off a switch SW.sub.ζj through which a smallest drive unit AG.sub.ζ drives the j-th execution device CL.sub.j in the energy supply bus, so as to start to perform the processing stage TAPR.sub.jd.sup.αr, and marking a state of the smallest drive unit AG.sub.ζ as busy in the control center.

[0098] Step 9: turning on the corresponding switch SW i after the processing stage TAPR.sub.jd.sup.αr is completed, and marking the state of the smallest drive unit AG.sub.ζ as idle.

[0099] Step 10: determining whether r>k.sub.d holds after assigning r+1 to d, if yes, performing step 11, otherwise, performing step 5.

[0100] Step 11: determining whether d>D holds after assigning d+1 to d, if yes, performing step 12, otherwise, performing step 4.

[0101] Step 12: determining whether x>(max {k.sub.d}+D) holds after assigning x+1 to x, if yes, performing step 13, otherwise, performing step 3; and step 13: turning off m drive units AG.sub.1, AG.sub.2, . . ., AG.sub.i, . . ., AG.sub.m in the drive system, and turning off all the switches of the energy supply bus. At the end of the task, the state of each switch can be set to be on or off according to the requirement of a drive unit.

[0102] Embodiment 2: a production line is composed of K machines, which collectively complete a production task, a plurality of machines share one drive system, the one drive system supplies the energy required for all machines to complete a production task, the shared drive system is composed of a plurality of drive units, each of the drive units has different installed power; a flow chart of a design method of a multi-machine shared drive system of a production line is shown in FIG. 3, and the design method is implemented as the following steps:

[0103] step 1, designing a process: designing, according to a production task of the production line, K processes for completing the production task, and sequencing them according to a sequential order of the processes; and, calculating, according to a material to be processed, a load curve of each process, recorded as l.sub.1,l.sub.2, . . ., l.sub.k, . . ., l.sub.K, l.sub.k represents a load curve of the k-th process, k=1, 2, . . ., K; letting the k-th machine process the k-th process.

[0104] Step 2, designing a task:

[0105] step 2.1: dividing, according to a load difference of the load curve of each process, each process into a plurality of processing stages, and sequencing the processing stages according to a sequential order; in a production process, each processing stage is completed with high energy efficiency by a drive unit with a higher matching degree the same;

[0106] step 2.2: obtaining, by utilizing formula (8), a power similarity index Ps.sub.α of the α-th processing stage:

[00008]$\begin{array}{cc}{\mathrm{Ps}}_a=\left(\frac{{\int }_{{\mathrm{st}}_a}^{{\mathrm{ct}}_a}{p}_a\left(t\right)q_a\left(t\right)\mathrm{dt}}{{\mathrm{ct}}_a-{\mathrm{st}}_a}\right)/\left(\underset{1\le a<A}{\max }\frac{{\int }_{{\mathrm{st}}_a}^{{\mathrm{ct}}_a}{p}_a\left(t\right)q_a\left(t\right)\mathrm{dt}}{{\mathrm{ct}}_a-{\mathrm{st}}_a}\right),a\in \left\{1,2,\mathrm{.Math.},A\right\}& \left(8\right)\end{array}$

[0107] in formula (8), ct.sub.α and st.sub.α are an end time and a start time of the α-th processing stage, respectively, p.sub.α(t) is a required pressure of the α-th processing stage, q.sub.α(t) is a required flow of the α-th processing stage, and A is a total number of processing stages;

[0108] step 2.3: calculating the power similarity index of each processing stage, and constituting a task by processing stages corresponding to a power similarity index within a set range, so as to obtain m tasks, recording any i-th task as TA.sub.i.

[0109] Step 3, designing a schedule:

[0110] step 3.1: scheduling, by utilizing formula (9), m drive units, so that them drive units is capable of continuously completing processing stages of a process without conflict:

ct (α.sub.j.sub.k.sup.b(TA.sub.1))=st(α.sub.l.sub.k.sup.b+1(TA.sub.z))  (9)

[0111] in formula (9), l.sub.k is the k-th process curve, k=1,2, . . ., K; b represents a serial number of a processing stage of the k-th process curve, ct(a.sub.l.sub.i.sup.b(TA.sub.i)) is an end time of the b-th processing stage in the k-th process curve l.sub.k in the i-th task TA.sub.i, and st(α.sub.l.sub.k.sup.b+1(TA.sub.z)) is a start time of the b+1-th processing stage belonging to the k-th process curve l.sub.k in the z-th task TA.sub.z, z=1,2, . . .,m;

[0112] step 3.2: for all processes, since some processes can only be performed after a specific process is completed, scheduling, by utilizing formula (10), m drive units and all processing stages in the m tasks, so that the m drive units complete an entire production task in sequence according to a set process order without conflict:

ct(α.sub.l.sub.z.sup.fast(TA.sub.i))≤st (α.sub.l.sub.k+1.sup.1(TA.sub.z))  (10)

[0113] in formula (10), ct(α.sub.l.sub.k.sup.last(TA.sub.i)) is an end time of the last processing stage belonging to the k-th process curve l.sub.k in the i-th task TA.sub.i, and st(α.sub.l.sub.k.sup.1(TA.sub.z)) is a start time of the first processing stage belonging to the k+1-th process curve in the z-th task TA.sub.z;

[0114] Step 4, designing a drive unit:

[0115] step 4.1: designing, according to the required pressure and the required flow of each processing stage in the i-th task TA.sub.i, the i-th drive unit DU.sub.i, which completes all processing stages in the i-th task TA.sub.i and is composed of the i-th group of motors and the i-th group of pumps, i=1,2, . . .,m; m represents a total number of drive units;

[0116] step 4.2: selecting, from a set collection of pumps and for the i-th drive unit DU.sub.i, a hydraulic pump with the smallest rated displacement to be used as the i-th basic flow unit Q.sub.i; when a corresponding task is completed, an output pressure of the i-th basic flow unit Q.sub.i is recorded as p.sub.i, an output flow is recorded as q.sub.i, an output power is recorded as P.sub.Qi and an energy efficiency is recorded as η.sub.i;

[0117] step 4.3: selecting, from the set collection of motors, a motor that matches the i-th basic flow unit Q.sub.iso that a load rate of the selected motor when driving a corresponding task is within a set interval, so as to obtain a basic motor unit D.sub.i in the i-th drive unit DU.sub.i; and obtaining, by utilizing formula (11), a load rate β.sub.i of the basic motor unit D.sub.i when driving a corresponding action:

β.sub.i=P.sub.Qi/(P.sub.Di×η.sub.i)  (11)

[0118] in formula (11), P.sub.Di is a rated power of the i-th basic motor unit D.sub.i.

[0119] Step 5, designing a shared drive system:

[0120] step 5.1: obtaining, by utilizing Formula (12), a maximum value MAX.sub.i of the i-th basic flow unit Q.sub.iand increasing the number of the i-th basic flow unit Q.sub.i in the i-th drive unit DU.sub.i to a set maximum value MAX.sub.i one by one:

MAX.sub.i=[Q.sub.max/Q.sub.i]+1  (12)

[0121] in formula (12), Q.sub.max is a maximum flow value output by a single machine;

[0122] step 5.2: increasing sequentially the number of the i-th basic motor unit D.sub.i one by one, and matching it with the i-th basic flow unit Q.sub.i;

[0123] step 5.3: calculating a time t.sub.i.sup.n.sup.i(k) and an energy consumption E.sub.i.sup.n.sup.i(k) for completing a corresponding task of the k-th machine, wherein, n.sub.i is the number of the i-th basic flow unit Q.sub.i added in the i-th drive unit DU.sub.i, n.sub.i=1,2, . . ., MAX.sub.i;

[0124] step 5.4: obtaining, by utilizing formula (13), the total configuration schemes number M of the m drive units:

M=Π.sub.i=1.sup.mMAX.sub.i  (13)

[0125] step 5.5: making, by scheduling run time and idle time of drive units in each configuration scheme, the m drive units complete processing stages in all tasks successively and sequentially, and forming M scheduling schemes, so as to utilize formula (14) to obtain a total time T.sub.j of the j-th scheduling scheme:

T.sub.j=Σ.sub.i=1.sup.mt.sub.i.sup.n.sup.l(k).sub.j+t(idle).sub.j  (14)

[0126] in formula (14), t.sub.i.sup.n.sup.i(k).sub.j is a time for the i-th drive unit DU.sub.i in the j-th scheduling scheme to complete a corresponding task of the k-th machine, t(idle), is a total unloading time of all drive units in the j-th scheduling scheme, j=1,2, . . ., M;

[0127] step 5.6: obtaining, by utilizing formula (15), a total energy consumption E.sub.j of the j-th scheduling scheme:

E.sub.j=Σ.sub.k−1.sup.KΣ.sub.i=1.sup.mE.sub.i.sup.n.sup.i(k).sub.j+E(idle).sub.j  (15)

[0128] in formula (15), K is a total number of machines in the production line, E.sub.i.sup.n.sup.i(k).sub.j is an energy consumption of the i-th drive unit DU.sub.i in the j-th scheduling scheme to complete the corresponding task of the k-th machine, and E(idle).sub.j is a total unloading energy consumption of all drive units in the j-th scheduling scheme;

[0129] step 5.7: selecting, from the total time and the total energy consumption of M scheduling schemes, m drive units corresponding to scheduling schemes with less total time and total energy consumption and using it as the drive system shared by K machines, so as to achieve efficient production.

[0130] With respect to the shared drive system of a production line designed by the design method according to the present disclosure, calculating a sum of theoretical displacement of all basic flow units in the i-th drive unit DU.sub.i in the shared drive system; and selecting, from the set collection of pumps, a single pump having large displacement, whose theoretical displacement is equal to the sum of the theoretical displacement of all basic flow units in the i-th drive unit DU.sub.i, to replace all basic flow units in the i-th drive unit DU.sub.i and to serve as a pump of the i-th drive unit;

[0131] selecting, from the set collection of motors, a motor that matches the pump of the i-th drive unit DU.sub.i and using it as a motor in the i-th drive unit, a load rate of the motor in the i-th drive unit DU.sub.i when driving a corresponding task is within a set interval, so that an energy efficiency of the selected motor when driving the corresponding task is not lower than a set efficiency value, so as to achieve a high efficiency operation.

[0132] Taking a production line of hydraulic presses as an example, the production line has four hydraulic presses, each hydraulic actuator has the same parameters, configured to collectively complete a production task of a housing of a certain clutch. A multi-machine shared drive system of the production line is designed according to the following steps:

[0133] step 1, designing a process:

[0134] designing, according to the production task, 4 processes, and letting them sequenced according to a sequential order of processing; each hydraulic press completes a process correspondingly.

[0135] Step 2, designing a task:

[0136] dividing, according to a load difference of a load curve of each process, each process into 5 processing stages, the time duration, the required pressure and the required flow of each processing stage of each process are shown in Table 1, in which, the required flow and the required pressure of an unloading stage and a standby stage of each process are zero.

TABLE-US-00001 TABLE 1 Processing stage Falling Forming Returning Flow Flow Unloading Flow Standby Time Pressure (L/ Time Pressure (L/ Time Time Pressure (L/ Time Process (s) (MPa) min) (s) (MPa) min) (s) (s) (MPa) min) (s) l.sub.1 2.5 2.66 114 0.50 16 46.44 1.88 2.21 2.87 65.46 2.91 l.sub.2 2.5 2.66 114 1.50 8.46 45.10 0.64 2.46 2.87 65.46 2.90 l.sub.3 2.5 2.66 114 0.50 9.94 45.76 1.83 2.22 2.87 65.46 2.95 l.sub.4 2.5 2.66 114 0.50 7.67 46.94 1.88 2.14 2.87 65.46 2.98

[0137] The processing stages of each process are sequenced according to a sequential order, and formula (8) is utilized to calculate a power similarity index of each processing stage, as shown in Table 2.

TABLE-US-00002 TABLE 2 Power similarity index Process Falling Forming Unloading Returning Standby l.sub.1 Processing 0.41 Processing 1 Processing 0 Processing 0.25 Processing 0 stage 1 stage 2 stage 3 stage 4 stage 5 l.sub.2 Processing 0.41 Processing 0.51 Processing 0 Processing 0.25 Processing 0 stage 6 stage 7 stage 8 stage 9 stage 10 l.sub.3 Processing 0.41 Processing 0.61 Processing 0 Processing 0.25 Processing 0 stage 11 stage 12 stage 13 stage 14 stage 15 l.sub.4 Processing 0.41 Processing 0.48 Processing 0 Processing 0.25 Processing 0 stage 16 stage 17 stage 18 stage 19 stage 20

[0138] Constituting a task by processing stages with the power similarity index in an interval [0.3, 0.45], in other words, processing stages 1, 6, 11 and 16 constitute task 1, recorded as TA.sub.1; constituting a task by processing stages with the power similarity index in an interval [0.46, 1], in other words, processing stages 2, 7, 12 and 17 constitute task 2, recorded as TA.sub.2; constituting a task by processing stages with the power similarity index in a range [0, 0.3], in other words, processing stages 4, 9, 14 and 19 constitute task 3, recorded as TA.sub.3; constituting a task by processing stages with the power similarity index equal to 0, in other words, processing stages 3, 5, 8, 10, 13, 15, 18 and 20 constitute task 4, recorded as TA.sub.4. As a result, four tasks are obtained, and each task is completed by one drive unit, and the shared drive system includes four drive units.

[0139] Step 3, designing a schedule:

[0140] completing, by scheduling the four drive units and all the processing stages in the four tasks, each process according to the order of the processing stages in the process, and at the same time, completing the entire production task according to the order of processes 1, 2, 3 and 4.

[0141] Step 4, designing a drive unit:

[0142] designing, according to the required pressure and the required flow of each processing stage in the i-th task TA.sub.i, the i-th drive unit DU.sub.i which completes all the processing stages in the i-th task TA.sub.i. Selecting, from the set collection of pumps (a certain series of plunger pumps with constant power), the basic flow unit of each drive unit. Afterwards, selecting, according to the selected basic flow unit, a basic motor unit that matches it; setting the load rate of the selected basic motor unit when driving a corresponding task to be within an interval [0.4, 1.0], and the selection results are shown in Table 3. Since the loads of the fourth task are all zero and the fourth drive unit does not need to provide energy, the configuration within the fourth drive unit is empty. With respect to PV023 plunger pump, the minimum flow is 40 L/min when its pressure is 2.66 MPa, the minimum flow is 35 L/min when the pressure is 2.87 MPa; with respect to PV032 plunger pump, the minimum flow is 24 L/min when the pressure is 16 MPa.

TABLE-US-00003 TABLE 3 1st Drive 2nd Drive 3rd Drive 4th Drive Drive unit unit unit unit unit Basic flow PV023 PV032 PV023 — unit Basic motor Y112M Y160M Y112M — unit

[0143] Step 5, designing a shared drive system:

[0144] with respect to the four hydraulic presses of the production line, the maximum flow output by a single machine is 120 L/min, and, based on formulas (12)-(13), a total of 60 configuration schemes for 4 drive units are calculated. By scheduling run time and idle time of drive units in each configuration scheme, the 4 drive units is made to complete processing stages in all tasks successively and sequentially, so as to form 60 scheduling schemes.

[0145] Formulas (14)-(15) are utilized to calculate a total time and a total energy consumption of each scheduling scheme, 4 drive units corresponding to scheduling schemes with less total time and total energy consumption are selected as the drive system shared by 4 machines, the configuration of the shared drive system is shown in Table 4.

TABLE-US-00004 TABLE 4 1st Drive 2nd Drive 3rd Drive 4th Drive Shared drive system unit unit unit unit Number of basic flow units 3 2 2 — Number of basic motor units 3 2 2 — Total time (s) 40 Total energy consumption 170.59 (kJ)

TABLE-US-00005 TABLE 5 1st Drive 2nd Drive 3rd Drive 4th Drive Shared drive system unit unit unit unit Pump PV063 PV046 PV032 — Motor Y160M Y160L Y132S —

[0146] Calculating, for the designed shared drive system of a multi-machine production line, a sum of theoretical displacement of all basic flow units in each drive unit; selecting, from the set collection of pumps, a single pump having large displacement whose theoretical displacement is equal to the sum of the theoretical displacement of basic flow units in the i-th drive unit to replace all basic flow units in the i-th drive unit and to serve as a pump of the i-th drive unit; and, selecting, from the set collection of motors, a motor that matches the pump of the i-th drive unit, so that a load rate of the motor when driving the corresponding task is within an interval [0.4, 1.0], in order to make a selected motor complete the corresponding task with high efficiency, the selection results are shown in Table 5.