CN108197769A - Single library-multiple station systems water resource optimal allocation the method in library is directly mended under a kind of abundant irrigation conditions - Google Patents
Single library-multiple station systems water resource optimal allocation the method in library is directly mended under a kind of abundant irrigation conditions Download PDFInfo
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Abstract
The invention discloses a kind of single library-multiple station systems water resource optimal allocation methods that library is directly mended under abundant irrigation conditions, it establishes under abundant irrigation conditions and directly mends the single library-multiple station systems water resource optimal allocation mathematical model in library, using the one-dimensional dynamic programming method of " single library-multistation " big system based on the planning polymerization of " mending library Group of Pumping Station " subsystem decomposing level, minimum water deficit in intake area in certain delivery period, corresponding reservoir day part optimal water supply can be obtained, abandon water and respectively mends library pumping plant rate of water make-up process.The present invention distributes with most important theories meaning and actual application value Wall in Plain Reservoir Water Resources Irrigation rationally.
Description
Technical field
The present invention relates to single reservoirs under abundant irrigation conditions and the method for multiple benefit library pumping plant cooperation scheduling, belong to
Water Resources Irrigation distributes technical field rationally.
Background technology
Currently due to water resource spatial and temporal distributions unevenness, the socio-economic development in many areas is restricted, for fully irrigating
For the irrigated area of condition, under limited water resources total amount and water source project scale, to be up to target with water comprehensive benefit, strengthen
The United Dispatching of regional water resources and management, the hydraulic engineering using irrigation system are used and are adjusted as unified entirety, use
Built engineering (such as reservoir and the operation of Group of Pumping Station combined dispatching) makes it play the effect of bigger, is the master for solving irrigated area water shortage problem
Want approach.How single reservoir-more pumping station systems traffic control reasonably uses system water as one content of water resources management
Scheduling of resource, which reaches some target, makes system benefit best, the problem of being relatively conventional in water project management.
For directly mending single reservoir-more pumping station systems water resource optimal allocation in library, although water condition is abundant, by
Then moisturizing is carried out to reservoir by multiple lift pumping stations, in running, how is ensureing that it is sufficient that intake area is supplied water
Under the conditions of, it reduces and mends library Group of Pumping Station overall operation energy consumption, save engineering operation cost, be very important major issue.
Invention content
The present invention considers different water frequencies for single reservoir-more pumping station systems that library is directly mended under abundant irrigation conditions
Under intake area water shortage situation, consider to supplement reservoir water shortage by multiple lift pumping stations for the first time, establish under abundant irrigation conditions directly
Mend the single library-multiple station systems joint optimal operation mathematical model in library.For the single library-multistation cooperation tune of adjusting of specific year
Degree system, when known reservoir supplies water and divides, hop count, initial storage, minimum capacity of a reservoir, utilizable capacity, flood control are corresponding
Storage capacity, year come process water, evaporation and leakage process for water inventory, day part, mend library pumping plant quantity, each benefit library pumping plant year
Allow under water lift total amount and day part intake area water requirement process condition, using decomposed based on " mending library Group of Pumping Station " subsystem-
The one-dimensional dynamic programming method of " single library-multistation " big system of Dynamic Programming polymerization, it is minimum can to obtain intake area in certain delivery period
Water deficit, day part optimal water supply in corresponding reservoir delivery period abandon water and each mend library pumping plant day part rate of water make-up mistake
Journey.
The present invention program is as follows：
Single library-multiple station systems water resource optimal allocation the method in library is directly mended under a kind of abundant irrigation conditions, by multiple benefits
Pump works supplies water to reservoir, from single library-multiple station systems to intake area co-supplying, includes the following steps：
First, model construction includes the following steps 1~step 2：
1. directly to mend the difference of the reservoir yield of day part and intake area water requirement of single library in library-in multiple station systems year
The minimum target of quadratic sum, establish following object function：
In formula：F be research object year in day part the difference for water requirement least square and；Z is in research object year
The quadratic sum of the difference for water requirement of day part；N is the when hop count divided in year；Segment number when i is (i=1,2 ... N)；Gi
Water supply (ten thousand m for the i-th period of reservoir3)；YSiWater requirement (ten thousand m for the i-th period of intake area3)；Object function use square
And expression is to accelerate to reduce the deviation between reservoir yield and intake area water requirement.
2. constraints is set
Including single library-multiple station systems year for water inventory constraints, reservoir operation criterion constraints, reservoir capacity
Constraints and benefit library Group of Pumping Station operation energy consumption least commitment condition.
2nd, model solution
1. data preparation specifically includes：N number of period was divided into, and determine day part length by 1 year；It is initial to measure reservoir
Storage capacity V0；Determine year for water inventory SK, minimum capacity of a reservoir Vmin, the corresponding storage capacity V of flood controlPAnd utilizable capacity Vmin+
△1；It measures and calculates reservoir day part and carry out water LSi, evaporation with leakage EFi；Determine that each benefit library pumping plant year allows water lift total amount
BZk, k=1,2 ... M, M is mend library pumping plant sum；Measure different periods lift HkiThe water lift flow Q of lower operationkiAnd corresponding water
Efficiency of pump ηz,ki, electric efficiency ηmot,k, transmission efficiency ηint,k；Determine the water demand of crop YS of day part intake areai, i=1,
2,…,N。
2. being solved using one-dimensional dynamic programming method to " single library-multistation " large-scale system model, it is optimal to obtain day part reservoir
Water supply process Gi, it is optimal to abandon process water PSiAnd mend the optimal moisturizing total amount process Y of library Group of Pumping Stationi。
3. being solved using decomposition-Dynamic Programming polymerization to " mending library Group of Pumping Station " subsystem model, each benefit library pumping plant is obtained
The optimal rate of water make-up process YB of day partki *。
Further, the constraints includes：
(1) single library-multiple station systems year is for water inventory constraints：In different level year difference fraction, examine
Worry needs water requirement, the water that water supply project may provide；
In formula：SK is the year of reservoir for water inventory (ten thousand m3)；M is total (seat) to mend library pumping plant；K is mends library pumping plant number
(k=1,2 ... M)；BZkBeing mended for kth seat allows water lift total amount (ten thousand m in library pumping plant year3)。
(2) reservoir operation criterion constraints：According to single library-multiple station systems water equation：
Vi=Vi-1+LSi+Yi-PSi-EFi-Gi, (i=1,2, N) (3)
1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir VminWhen, then the i-th period should be by Group of Pumping Station to reservoir
Moisturizing, moisturizing to utilizable capacity (Δ more than reservoir minimum capacity of a reservoir1), i.e.,：
Vi<VminWhen：Yi=Vmin-Vi+Δ1； (4)
This period reservoir abandons water PSi=0.
2. when meeting with flood, the i-th period end pondage is more than the pondage V corresponding to flood controlP
When, then the i-th period reservoir should be carried out to abandon water, abandon water to flood control by reservoir regulation limiting water level, i.e.,：
Vi>VPWhen：PSi=Vi-VP； (5)
This period Group of Pumping Station moisturizing total amount Yi=0.
3. when the i-th period end pondage is between minimum capacity of a reservoir VminWith the pondage corresponding to flood control
VPBetween, then the i-th period reservoir does not need to abandon water, and Group of Pumping Station is also without moisturizing, i.e.,：
Vmin≤Vi≤VPWhen：Yi=PSi=0； (6)
In formula：Vi、Vi-1Respectively reservoir i-th, the reservoir storage of i-1 period Mos (ten thousand m3)；YiGroup of Pumping Station for the i-th period is mended
Library water inventory (ten thousand m3)；LSi、PSi、EFiRespectively the i-th period of reservoir carrys out water (ten thousand m3), abandon water (ten thousand m3), evaporation with
Leakage (ten thousand m3)；Vmin、VPStorage capacity (ten thousand m respectively corresponding to the minimum capacity of a reservoir of reservoir and flood control3)。
(3) reservoir capacity constraints：The pondage of day part should be between reservoir minimum capacity of a reservoir and flood control
Between corresponding storage capacity, i.e.,：
Vmin≤Vi≤VP, (i=1,2, N) (7)
(4) library Group of Pumping Station cooperation energy consumption least commitment condition is mended：On the basis of the constraint of reservoir operation criterion is met, it is
Ensure the abundant irrigation conditions in intake area, to mending library Group of Pumping Station, the cooperation within each water supply period is considered as energy consumption most
It is small, i.e.,：
In formula, LiLibrary Group of Pumping Station combined operation system energy consumption (kWh) is mended for the i-th period；lkiDuring for kth seat pumping plant the i-th
Section operation energy consumption (kWh)；M is mends library pumping plant quantity (seat)；K is numbered for pumping plant；ρ is water density (kg/m3), g adds for gravity
Speed (m/s2)；Qki、Hki、△Tki、ηz,kiRespectively flow (the m of the i-th period of kth seat pumping plant3/ s), when equal lift (m), the period
Length (h) and pump efficiency；ηmot,k、ηint,kThe respectively motor efficiency and transmission efficiency of kth seat pumping plant.
Further, the solution of " single library-multistation " large-scale system model is specifically included using one-dimensional dynamic programming method：Root
According to one-dimensional dynamic programming evaluation principle, obtaining corresponding recurrence equation is：
(1) stage i=1：
f1(λ1)=min (G1-YS1)2 (9)
In formula, λ1For state variable, 1 water supply period reservoir yield (ten thousand m before expression3), it can be in corresponding feasible zone
It is discrete by a fixed step size：To each discrete λ1, decision variable G1It can be feasible in correspondence
It is discrete by a fixed step size in domain, such as 00,000 m3, 200,000 m3, 400,000 m3, 600,000 m3、…G1,maxDeng (G1,maxFor the 1st stage reservoir maximum
Water supply capacity), then G will be met1≥λ1It is required that G1Formula (9) is substituted into respectively, determines to correspond to each discrete λ respectively1During value, most
Excellent G1And its corresponding period minimum water deficit quadratic sum f1(λ1)。
Then, according to formula (3), the 1st stage end reservoir capacity V1=V0+LS1-EF1-G1, not yet consider that Group of Pumping Station is mended at this time
Water or reservoir abandon water, should be tested and be corrected using formula (4)~(6)：
1. work as V1<Vmin, then consider to carry out moisturizing, Group of Pumping Station moisturizing total amount Y to reservoir1=Vmin-V1+Δ1, correct at this time
Storage capacity V1 *=Vmin+△1。
2. work as V1>VP, then need to abandon water to ensure the dispatching requirement of reservoir capacity, PS1=V1-VP, storage capacity V is corrected at this time1 *
=VP。
3. work as Vmin≤V1≤VP, then Y1=PS1=0, storage capacity V is corrected at this time1 *=V1。
By step 1.~3., correct simultaneously determine the 1st stage end reservoir capacity V1 *, while corresponding reservoir can be obtained and abandon water
Measure PS1Or Group of Pumping Station moisturizing total amount Y1。
(2) stage i=2,3 ... N-1：
fi(λi)=min [(Gi-YSi)2+fi-1(λi-1)] (10)
In formula, state variable λiReservoir for the preceding i period supplies water inventory (ten thousand m3), it is equally carried out respectively discrete：To each discrete λi, decision variable GiIt is discrete to be same as above, and should meet：
State transition equation：λi-1=λi-Gi(11) in formula：I=2,3 ..., N-1
To each discrete λi, by each discrete GiValue substitutes into the (G in formula (10) respectivelyi-YSi)2, by state transfer side
Formula (11), to each discrete Gi, before lookup the i-1 stages meetIt is required that preceding i period system minimum lack
Water quadratic sum fi-1(λi-1) value, it thus can obtain (a Gi-YSi)2+minfi-1(λi-1), complete all of above discrete Gi
After optimizing, it can finally obtain and meet min [(Gi-YSi)2+fi-1(λi-1)] requirement fi(λi) value and its corresponding each stage water
Library optimal water supply Gi(i=1 ... i).
Equally, it also needs to determine the i-th stage end storage capacity V using formula (3)i, then tested, corrected using formula (4)~(6)
Determine the i-th period end storage capacity Vi *, while corresponding reservoir can be obtained and abandon process water PSiAnd Group of Pumping Station moisturizing total amount process Yi
(i=1,2 ... ... i), process with step 1.~3..
(3) stage N：
fN(λN)=min [(GN-YSN)2+fN-1(λN-1)] (12)
State variableDecision variable GNIt is equally discrete in corresponding feasible zone, it should meet：λN-1=λN-GN。
Using step (2) the method, final obtain meets the λNIt is required that the optimal water supply process G of reservoiri(i=1 ...
N), corresponding reservoir abandons process water PSi, Group of Pumping Station moisturizing total amount Yi(i=1,2 ... ... N) and master mould object function is most
Figure of merit F=fN(λN)。
Further, " mending library Group of Pumping Station " subsystem model is solved, specifically included using decomposition-Dynamic Programming polymerization
Following steps：
(1) consider to mend the constraint of library Group of Pumping Station cooperation energy consumption minimum criteria, established by formula (8) and mend library Group of Pumping Station economy fortune
Row mathematical model：
Object function：
Period water supply constrains：
Power constraint：
In formula, Nk0The electric drilling match power (kW) of library pumping plant is mended for kth seat；Remaining variables meaning is same as above.
(2) " benefit library Group of Pumping Station " subsystem decomposition-Dynamic Programming polymerization solves：
1. subsystem two level is decomposed：
Above-mentioned subsystem model (13)~(15) are further decomposed, can obtain M single station economical operation two level submodel：
Object function：
Power constraint：
In formula, liFor single i-th period minimum operation energy consumption of standing, unit：kW·h.
2. two level submodel energy consumption determines：
For model above (16)~(17), segment length △ T during the i-th periodiIt is known that it can simultaneously be determined by pumping plant water levels of upstream and downstream
Water lift lift HiAnd its corresponding pump capacity Qi, pump efficiency ηz,i, electric efficiency ηmotWith transmission efficiency ηint, and the period
Maximum rate of water make-up is YBi,max=3600Qi△Ti/ 10000, by the discrete period maximum water lift amount YB of a fixed step sizei,max(i.e. pair
Period duration △ TiCarry out discrete), each water lift amount YB can be obtainedi,mPlace an order station operation energy consumption li,m(m=1,2 ... max).
To other pumping plants, equally using above method, thus to obtain each pumping plant difference water lift amount YBki,mUnder, single station operation
Energy consumption lki,m(k=1,2 ... M, m=1,2 ... max).
3. atomic system Dynamic Programming polymerize：
It is solved by more than two level subsystem, to each benefit library pumping plant, can obtain a series of single station water lift energy consumption lki,m
~mono- stand mends library water YBki,mRelationship (k=1,2 ... M, m=1,2 ... max), thus builds following polymerization model substitution atoms
Model (13)~(15)：
Object function：
Period water supply constrains：
Polymerization model (18)~(19) are similarly one-dimensional dynamic programming model, stage variable for pumping plant number k (k=1,
2,…M)；Decision variable is each i-th period of pumping plant water lift amount YBki, target water when discrete range is single station optimization from
Dissipate range YBki,m(k=1,2 ... M, m=1,2 ... max)；The centrifugal pump that each pumping plant water lift total amount is understood by formula (19) is shape
State variable (λ).The model is solved with reference to above one-dimensional dynamic programming, obtains and meets the i-th period pumping plant target water lift total amount Yi
LiValue and the optimal rate of water make-up combination YB of corresponding each pumping plantki *(k=1,2 ... M).
The invention can be achieved directly to mend the water resource optimization tune of single reservoir-more pumping station systems in library under abundant irrigation conditions
Degree, method precision is reliable, accuracy up to more than 98%, meanwhile, reduce mend more than 15% library pumping station operation energy consumption, reach irrigated area
The purpose of water resource optimal allocation improves irrigated area, society and ecological benefits.
Description of the drawings
Fig. 1 generally changes system schematic directly to mend the single library-multistation water resource in library under abundant irrigation conditions.
Specific embodiment
As shown in Figure 1, it is needed with directly mending the reservoir yield of day part of single library in library-in multiple station systems year and intake area
The minimum target of quadratic sum of the difference of water, the day part reservoir yield divided in year are decision variable, with system year for
Water inventory (i.e. reservoir year for water inventory and draw the sum of water lift total amount in each pumping plant year), reservoir operation criterion, reservoir water balance,
Library Group of Pumping Station cooperation energy consumption minimum etc. is mended as constraints, establishes under abundant irrigation conditions and directly mends single library-multistation in library
System water resources optimal operation model.
First, model construction
1. object function
In formula：F be research object year in day part the difference for water requirement least square and；Z is in research object year
The quadratic sum of the difference for water requirement of day part；N is the when hop count divided in year；Segment number when i is (i=1,2 ... N)；Gi
Water supply (ten thousand m for the i-th period of reservoir3)；YSiWater requirement (ten thousand m for the i-th period of intake area3)；Object function use square
And expression is to accelerate to reduce the deviation between reservoir yield and intake area water requirement.
2. constraints
Including single library-multiple station systems year for water inventory constraints, reservoir operation criterion constraints, reservoir capacity
Constraints and benefit library Group of Pumping Station operation energy consumption least commitment condition.Specifically it is described below：
(1) single library-multiple station systems year constrains for water inventory：In different level year difference fraction, consider to need
Water requirement, the water that water supply project may provide.
In formula：SK is the year of reservoir for water inventory (ten thousand m3)；M is total (seat) to mend library pumping plant；K is mends library pumping plant number
(k=1,2 ... M)；BZkBeing mended for kth seat allows water lift total amount (ten thousand m in library pumping plant year3)；Remaining variables meaning is same as above.
(2) reservoir operation criterion constrains：According to single library-multiple station systems water equation：
Vi=Vi-1+LSi+Yi-PSi-EFi-Gi, (i=1,2, N) (3)
1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir VminWhen, then the i-th period should be by Group of Pumping Station to reservoir
Moisturizing, moisturizing to utilizable capacity (Δ more than reservoir minimum capacity of a reservoir1), i.e.,：
Vi<VminWhen：Yi=Vmin-Vi+Δ1； (4)
This period reservoir abandons water PSi=0.
2. when meeting with flood, the i-th period end pondage is more than the pondage V corresponding to flood controlP
When, then the i-th period reservoir should be carried out to abandon water, abandon water to flood control by reservoir regulation limiting water level, i.e.,：
Vi>VPWhen：PSi=Vi-VP； (5)
This period Group of Pumping Station moisturizing total amount Yi=0.
3. when the i-th period end pondage is between minimum capacity of a reservoir VminWith the pondage corresponding to flood control
VPBetween, then the i-th period reservoir does not need to abandon water, and Group of Pumping Station is also without moisturizing, i.e.,：
Vmin≤Vi≤VPWhen：Yi=PSi=0； (6)
In formula：Vi、Vi-1Respectively reservoir i-th, the reservoir storage of i-1 period Mos (ten thousand m3)；YiGroup of Pumping Station for the i-th period is mended
Library water inventory (ten thousand m3)；LSi、PSi、EFiRespectively the i-th period of reservoir carrys out water (ten thousand m3), abandon water (ten thousand m3), evaporation with
Leakage (ten thousand m3)；Vmin、VPStorage capacity (ten thousand m respectively corresponding to the minimum capacity of a reservoir of reservoir and flood control3)。
(3) reservoir capacity constrains：The pondage of day part should be corresponded between reservoir minimum capacity of a reservoir and flood control
Storage capacity between, i.e.,：
Vmin≤Vi≤VP, (i=1,2, N) (7)
(4) constraint of library Group of Pumping Station cooperation energy consumption minimum criteria is mended：Further, meeting the constraint of reservoir operation criterion
On the basis of, to ensure the abundant irrigation conditions in intake area, to mending library Group of Pumping Station, the cooperation within each water supply period is taken an examination
It is minimum to consider energy consumption, i.e.,：
In formula, LiLibrary Group of Pumping Station combined operation system energy consumption (kWh) is mended for the i-th period；lkiDuring for kth seat pumping plant the i-th
Section operation energy consumption (kWh)；M is mends library pumping plant quantity (seat)；K is numbered for pumping plant；ρ is water density (kg/m3), g adds for gravity
Speed (m/s2)；Qki、Hki、△Tki、ηz,kiRespectively flow (the m of the i-th period of kth seat pumping plant3/ s), when equal lift (m), the period
Length (h) and pump efficiency；ηmot,k、ηint,kThe respectively motor efficiency and transmission efficiency of kth seat pumping plant.
2nd, model feature
(1) in view of should strictly carry out water total amount control in water resources development and utilization, therefore in model constraints
Add in reservoir year library pumping plant year allows water lift total amount to constrain for water inventory and each mend, i.e. formula (2).
(2) constraint of reservoir operation criterion, i.e. formula (4)~(6) are considered in constraints, can realize lift pumping station " idle
Single library-multiple station systems water resources optimal operation mode of benefit library, busy water supply ".If idle pondage is less than the dead library of reservoir
Hold, then consider to carry out moisturizing, the benefit library total Water Y of lift pumping station group to reservoiriIt is stored for reservoir minimum capacity of a reservoir and the i-th period end reservoir
The difference of water, along with minimum capacity of a reservoir more than Δ1, i.e. Yi=Vmin-Vi+Δ1；If reservoir capacity is more than flood control by reservoir regulation limiting water level institute
During corresponding reservoir storage, then need to abandon water to ensure the dispatching requirement of reservoir capacity, reservoir abandons water PSiFor the i-th period end reservoir
The difference of reservoir storage, i.e. PS corresponding to reservoir storage and flood control by reservoir regulation limiting water leveli=Vi-VP.Busy according to pondage situation,
Then consider to be combined to intake area from reservoir, benefit library Group of Pumping Station and supply water, to meet the water demand of water user as far as possible.
(3) for mending library pumping plant co-supplying by Building M when, it is contemplated that unit performance characteristic between standing under each pumping plant difference lift
Difference adds the minimum constraints of Group of Pumping Station cooperation energy consumption, i.e. formula (8).It is excellent by establishing Group of Pumping Station subsystem joint
Change operation mathematical model and solved using decomposition-Dynamic Programming polymerization, by the determining day part pumping plant of large-scale system model optimization
Group's moisturizing total amount YiDistribution is advanced optimized to each benefit library pumping plant YBki(k=1,2 ... M) make the operation of Group of Pumping Station subsystem to the greatest extent may be used
Energy consumption can be reduced.
3rd, model solution
It is the nonlinear mathematical model that can divide in a stage for model above Chinese style (1)~(8), each stage in object function
Intake area water requirement YSiFor it is known that therefore can be with the period i (i=1,2 ..., N) that reservoir delivery period divides for stage variable, respectively
Period reservoir available water GiFor decision variable, preceding i period reservoir is state variable λ for water inventoryi, advised using one-dimensional dynamic
The method of drawing solves.
It is assumed that the initial storage V of annual-storage reservoir0It is known that model Chinese style (2) is Dynamic Programming coupling constraint, formula (3) is
Reservoir operation day part water balance criterion；By to each stage reservoir yield Gi(i=1,2 ..., N) carries out discrete, use
One-dimensional dynamic programming solves, while is tested using formula (4)~(6) to storage capacity, and then each stage initial storage is carried out
It corrects, thereby determines that day part reservoir optimal water supply process Gi, abandon process water PSiAnd mend library Group of Pumping Station moisturizing total amount mistake
Journey Yi(i=1,2 ..., N)；Then consider to mend library Group of Pumping Station cooperation energy consumption minimum criteria constraint (8), establish Group of Pumping Station
System combined optimization operation mathematical model, is solved using decomposition-Dynamic Programming polymerization, by determining day part Group of Pumping Station moisturizing
Total amount YiDistribution is advanced optimized to each benefit library pumping plant, obtains each benefit library pumping plant day part rate of water make-up YBki *, it is thus final to obtain
Study the least square of the difference for water requirement of day part and F, corresponding reservoir day part optimal water supply G in area yeari, abandon water
Process PSiAnd the optimal rate of water make-up YB of each benefit library pumping plant day partki *(i=1,2 ... ... N, k=1,2 ... M).Achievement can be
Researching on Water Resources Optimal Management where directly mending the single library-multiple station systems in library under abundant irrigation conditions provides foundation.
Model (1)~(8) are single decision variable Optimized model, and one-dimensional dynamic programming method can be used and solve, can not only obtain
Obtain optimizing decision variate-value --- each stage reservoir yield G in object functioni；And by using formula (4)~(6) to library
Appearance is tested amendment, can also obtain that day part reservoir is optimal to abandon water PSiWith the optimal moisturizing total amount process Y of Group of Pumping Stationi；It is sharp again
With formula (8), by day part Group of Pumping Station moisturizing total amount YiDistribution is advanced optimized to each benefit library pumping plant, and each benefit library pumping plant can be obtained
The optimal rate of water make-up YB of day partki *(i=1,2 ... ... N, k=1,2 ... M)；It actually solves and obtains M+2 decision variable
Value, enriches such complex nonlinear model solution method.
1st, data preparation is carried out
It specifically includes：N number of period was divided into, and determine day part length by 1 year；According to reservoir initial water level, water is searched
Position-capacity curve measures reservoir initial storage V0；With reference to reservoir operational management provide, determine year for water inventory SK,
Minimum capacity of a reservoir Vmin, the corresponding storage capacity V of flood controlPAnd utilizable capacity Vmin+△1；It is provided according to reservoir locality meteorological model
Material measures and calculates reservoir day part and carrys out water LSi, evaporation with leakage EFi；According to each pumping plant working system, each benefit library is determined
Pumping plant year allows water lift total amount BZk(k=1,2 ... M)；By each water pump in pump station device performance characteristic curve, measure different periods and raise
Journey HkiThe water lift flow Q of lower operationkiAnd corresponding pump efficiency ηz,ki, electric efficiency ηmot,k, transmission efficiency ηint,k；According to by
The data such as pool variety of crops, planting scale, multiple crop index calculate the water demand of crop YS for determining day part intake areai(i
=1,2 ..., N).
2nd, the one-dimensional dynamic programming method of " single library-multistation " large-scale system model solves
With reference to one-dimensional dynamic programming evaluation principle, obtaining corresponding recurrence equation is：
(1) stage i=1：
f1(λ1)=min (G1-YS1)2 (9)
In formula, λ1For state variable, 1 water supply period reservoir yield (ten thousand m before expression3), it can be in corresponding feasible zone
It is discrete by a fixed step size：To each discrete λ1, decision variable G1It can be feasible in correspondence
It is discrete by a fixed step size in domain, such as 00,000 m3, 200,000 m3, 400,000 m3, 600,000 m3、…G1,maxDeng (G1,maxFor the 1st stage reservoir maximum
Water supply capacity), then G will be met1≥λ1It is required that G1Formula (9) is substituted into respectively, can determine to correspond to each discrete λ respectively1During value,
Optimal G1And its corresponding period minimum water deficit quadratic sum f1(λ1)。
Then, according to formula (3), the 1st stage end reservoir capacity V1=V0+LS1-EF1-G1, not yet consider that Group of Pumping Station is mended at this time
Water or reservoir abandon water, should be tested and be corrected using formula (4)~(6)：
1. work as V1<Vmin, then consider to carry out moisturizing, Group of Pumping Station moisturizing total amount Y to reservoir1=Vmin-V1+Δ1, correct at this time
Storage capacity V1 *=Vmin+△1。
2. work as V1>VP, then need to abandon water to ensure the dispatching requirement of reservoir capacity, PS1=V1-VP, storage capacity is corrected at this time
V1 *=VP。
3. work as Vmin≤V1≤VP, then Y1=PS1=0, storage capacity V is corrected at this time1 *=V1。
By step 1.~3., correct simultaneously determine the 1st stage end reservoir capacity V1 *, while corresponding reservoir can be obtained and abandon water
Measure PS1Or Group of Pumping Station moisturizing total amount Y1。
(2) stage i=2,3 ... N-1：
fi(λi)=min [(Gi-YSi)2+fi-1(λi-1)] (10)
In formula, state variable λiReservoir for the preceding i period supplies water inventory (ten thousand m3), it is equally carried out respectively discrete：To each discrete λi, decision variable GiIt is discrete to be same as above, and should meet：
State transition equation：λi-1=λi-Gi (11)
In formula：I=2,3 ..., N-1
To each discrete λi, by each discrete GiValue substitutes into the (G in formula (10) respectivelyi-YSi)2, by state transfer side
Formula (11), to each discrete Gi, before lookup the i-1 stages meetIt is required that preceding i period system minimum lack
Water quadratic sum fi-1(λi-1) value, it thus can obtain (a Gi-YSi)2+minfi-1(λi-1), complete all of above discrete Gi
After optimizing, it can finally obtain and meet min [(Gi-YSi)2+fi-1(λi-1)] requirement fi(λi) value and its corresponding each stage water
Library optimal water supply Gi(i=1 ... i).
Equally, it also needs to determine the i-th stage end storage capacity V using formula (3)i, then tested, corrected using formula (4)~(6)
Determine the i-th period end storage capacity Vi *, while corresponding reservoir can be obtained and abandon process water PSiAnd Group of Pumping Station moisturizing total amount process Yi
(i=1,2 ... ... i), process with step 1.~3..
(3) stage N：
fN(λN)=min [(GN-YSN)2+fN-1(λN-1)] (12)
State variableDecision variable GNIt is equally discrete in corresponding feasible zone, it should meet：λN-1=λN-GN。
Using step (2) the method, final obtain meets the λNIt is required that the optimal water supply process G of reservoiri(i=1 ... N), it is corresponding
Reservoir abandons process water PSi, Group of Pumping Station moisturizing total amount process Yi(i=1,2 ... ... N) and master mould object function optimal value F
=fN(λN)。
3rd, " benefit library Group of Pumping Station " subsystem decomposition-Dynamic Programming polymerization solves
Above after the one-dimensional dynamic plan optimization of " single library-multistation " big system, except the confession for having determined that day part in research area year
The least square and F of the difference of water requirement, reservoir day part optimal water supply GiWith abandon water process PSiOutside, to mending library Group of Pumping Station, only
The optimal moisturizing total amount Y of day part is determinediProcess not yet considers the rate of water make-up YB of specific each benefit library pumping plantki(k=1,
2 ... M), since each benefit library pumping plant unit installation performance has differences, along with day part water lift lift difference, pumping plant is caused to be transported
Row energy consumption is different, and library Group of Pumping Station operation energy consumption is mended to reduce, need to be by YiIt advanced optimizes distribution and mends library pumping plant to each, clearly
Day part respectively mends library pumping plant optimization rate of water make-up YBki, so as to be really achieved single library-multiple station systems water resource optimal allocation purpose.
(1) library Group of Pumping Station subsystem economical operation mathematical model structure is mended
Consider to mend the constraint of library Group of Pumping Station cooperation energy consumption minimum criteria, established by formula (8) and mend library Group of Pumping Station economical operation
Mathematical model：
Object function：
Period water supply constrains：
Power constraint：
In formula, Nk0The electric drilling match power (kW) of library pumping plant is mended for kth seat；Remaining variables meaning is same as above.
(2) " benefit library Group of Pumping Station " subsystem decomposition-Dynamic Programming polymerization solves
1. subsystem two level is decomposed：
Above-mentioned subsystem model (13)~(15) are further decomposed, can obtain M single station economical operation two level submodel：
Object function：
Power constraint：
In formula, liFor single i-th period minimum operation energy consumption of standing, unit：kW·h.
2. two level submodel energy consumption determines：
For model above (16)~(17), segment length △ T during the i-th periodiIt is known that it can simultaneously be determined by pumping plant water levels of upstream and downstream
Water lift lift HiAnd its corresponding pump capacity Qi, pump efficiency ηz,i, electric efficiency ηmotWith transmission efficiency ηint, and the period
Maximum rate of water make-up is YBi,max=3600Qi△Ti/ 10000, by the discrete period maximum water lift amount YB of a fixed step sizei,max(i.e. pair
Period duration △ TiCarry out discrete), each water lift amount YB can be obtainedi,mPlace an order station operation energy consumption li,m(m=1,2 ... max).It is right
Other pumping plants, equally using above method, thus to obtain each pumping plant difference water lift amount YBki,mUnder, single operation energy consumption l that standski,m (k
=1,2 ... M, m=1,2 ... max).
3. atomic system Dynamic Programming polymerize：
It is solved by more than two level subsystem, to each benefit library pumping plant, can obtain a series of single station water lift energy consumption lki,m
~mono- stand mends library water YBki,mRelationship (k=1,2 ... M, m=1,2 ... max) thus can build following polymerization model and substitute original
Submodel (13)~(15)：
Object function：
Period water supply constrains：
Polymerization model (18)~(19) are similarly one-dimensional dynamic programming model, stage variable for pumping plant number k (k=1,
2,…M)；Decision variable is each i-th period of pumping plant water lift amount YBki, target water when discrete range is single station optimization from
Dissipate range YBki,m(k=1,2 ... M, m=1,2 ... max)；The centrifugal pump of each pumping plant water lift total amount is knowable to formula (19)
State variable (λ).The model is solved with reference to above one-dimensional dynamic programming, obtains and meets the i-th period pumping plant target water lift total amount
YiLiValue and the optimal rate of water make-up combination YB of corresponding each pumping plantki *(k=1,2 ... M).
4th, master mould optimal solution is determined
Group of Pumping Station optimal moisturizing total amount Y in library is mended by the day part that step 2 obtainsiAfter (i=1,2 ... N) process；It is right
The Y that each period determinesi, it is both needed to via a step 3 (n times step 3 calculates altogether), you can further by day part YiOptimization point
M benefit library pumping plant is assigned to, obtains the optimal rate of water make-up YB of each benefit library pumping plant day partki *；Thus when each in final acquisition research area year
The least square of the difference for water requirement of section and F, corresponding reservoir day part optimal water supply Gi, abandon water process PSiIt is and each
Mend the optimal rate of water make-up YB of library pumping plant day partki *(i=1,2 ... ... N, k=1,2 ... M).
Claims (4)
1. the single library-multiple station systems water resource optimal allocation method in library is directly mended under a kind of abundant irrigation conditions, by multiple moisturizings
Pumping plant supplies water to reservoir, from single library-multiple station systems to intake area co-supplying, which is characterized in that include the following steps：
First, model construction includes the following steps 1~step 2：
1. directly to mend square of the difference of the reservoir yield of day part and intake area water requirement of single library in library-in multiple station systems year
With minimum target, following object function is established：
In formula：F be research object year in day part the difference for water requirement least square and；When Z is each in research object year
The quadratic sum of the difference for water requirement of section；N is the when hop count divided in year；Segment number when i is, i=1,2 ... N；GiFor
The water supply of i-th period of reservoir, unit are：Ten thousand m3；YSiFor the water requirement of the i-th period of intake area, unit is：Ten thousand m3；Target letter
Number is to accelerate to reduce the deviation between reservoir yield and intake area water requirement using quadratic sum expression.
2. constraints is set
Including single library-multiple station systems year for water inventory constraints, reservoir operation criterion constraints, reservoir capacity constraint item
Part and benefit library Group of Pumping Station operation energy consumption least commitment condition.
2nd, model solution
1. data preparation specifically includes：N number of period was divided into, and determine day part length by 1 year；Measure reservoir initial storage
V0；Determine year for water inventory SK, minimum capacity of a reservoir Vmin, the corresponding storage capacity V of flood controlPAnd utilizable capacity Vmin+Δ1；
It measures and calculates reservoir day part and carry out water LSi, evaporation with leakage EFi；Determine that each benefit library pumping plant year allows water lift total amount BZk,
K=1,2 ... M, M is mend library pumping plant sum；Measure different periods lift HkiThe water lift flow Q of lower operationkiAnd corresponding water pump
Efficiency etaZ, ki, electric efficiency ηMot, k, transmission efficiency ηInt, k；Determine the water demand of crop YS of day part intake areai, i=1,2 ...,
N。
2. being solved using one-dimensional dynamic programming method to " single library-multistation " large-scale system model, the optimal water supply of day part reservoir is obtained
Amount process Gi, it is optimal to abandon process water PSiAnd mend the optimal moisturizing total amount process Y of library Group of Pumping Stationi。
3. " mending library Group of Pumping Station " subsystem model is solved using decomposition-Dynamic Programming polymerization, when each benefit library pumping plant of acquisition is each
The optimal rate of water make-up process YB of sectionki *。
2. according to the method described in claim 1, it is characterized in that, the constraints includes：
(1) single library-multiple station systems year is for water inventory constraints：In different level year difference fraction, consider to need
Water requirement, the water that water supply project may provide；
In formula：SK is the year of reservoir for water inventory, unit：Ten thousand m3；M is mends library pumping plant sum；K is mends library pumping plant number, k=
1,2 ... M；BZkBeing mended for kth seat allows water lift total amount, unit in library pumping plant year：Ten thousand m3。
(2) reservoir operation criterion constraints：According to single library-multiple station systems water equation：
Vi=Vi-1+LSi+Yi-PSi-EFi-Gi, i=1,2 ..., N (3)
1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir VminWhen, then the i-th period should give reservoir moisturizing by Group of Pumping Station,
Δ more than moisturizing to utilizable capacity, that is, reservoir minimum capacity of a reservoir1, i.e.,：
Vi＜ VminWhen：Yi=Vmin-Vi+Δ1； (4)
This period reservoir abandons water PSi=0.
2. when meeting with flood, the i-th period end pondage is more than the pondage V corresponding to flood controlPWhen, then
I-th period should carry out reservoir to abandon water, abandon water to flood control by reservoir regulation limiting water level, i.e.,：
Vi＞ VPWhen：PSi=Vi-VP； (5)
This period Group of Pumping Station moisturizing total amount Yi=0.
3. when the i-th period end pondage is between minimum capacity of a reservoir VminWith the pondage V corresponding to flood controlPIt
Between, then the i-th period reservoir does not need to abandon water, and Group of Pumping Station is also without moisturizing, i.e.,：
Vmin≤Vi≤VPWhen：Yi=PSi=0； (6)
In formula：Vi、Vi-1Respectively reservoir i-th, the reservoir storage of i-1 period Mos, unit：Ten thousand m3；YiGroup of Pumping Station for the i-th period is mended
Library water inventory, unit：Ten thousand m3；LSi、PSi、EFiRespectively the i-th period of reservoir carrys out water, abandons water, evaporation and leakage, single
Position：Ten thousand m3；Vmin、VPStorage capacity respectively corresponding to the minimum capacity of a reservoir of reservoir and flood control, unit：Ten thousand m3。
(3) reservoir capacity constraints：The pondage of day part should be corresponded between reservoir minimum capacity of a reservoir and flood control
Storage capacity between, i.e.,：
Vmin≤Vi≤VP, i=1,2 ..., N (7)
(4) library Group of Pumping Station cooperation energy consumption least commitment condition is mended：On the basis of the constraint of reservoir operation criterion is met, to ensure
The abundant irrigation conditions in intake area, to mending library Group of Pumping Station, the cooperation within each water supply period is considered as energy consumption minimum,
I.e.：
In formula, LiLibrary Group of Pumping Station combined operation system energy consumption, unit are mended for the i-th period：kW·h；lkiDuring for kth seat pumping plant the i-th
Section operation energy consumption, unit：kW·h；M is mends library pumping plant quantity；K is numbered for pumping plant；ρ is water density, unit：kg/m3, g attaches most importance to
Power acceleration, unit：m/s2；Qki、Hki、ΔTki、ηZ, kiRespectively flow (the m of the i-th period of kth seat pumping plant3/ s), when equal lift
(m), Period Length (h) and pump efficiency；ηMot, k、ηInt, kThe respectively motor efficiency and transmission efficiency of kth seat pumping plant.
It is 3. according to the method described in claim 2, it is characterized in that, big to " single library-multistation " using one-dimensional dynamic programming method
System model solution specifically includes：According to one-dimensional dynamic programming evaluation principle, obtaining corresponding recurrence equation is：
(1) stage i=1：
f1(λ1)=min (G1-YS1)2 (9)
In formula, λ1For state variable, 1 water supply period reservoir yield, unit before expression：Ten thousand m3, can be in corresponding feasible zone
It is discrete inside to press a fixed step size：To each discrete λ1, decision variable G1It can be in correspondence
It is discrete by a fixed step size in feasible zone, then G will be met1≥λ1It is required that G1Formula (9) is substituted into respectively, determines to correspond to respectively each
Discrete λ1During value, optimal G1And its corresponding period minimum water deficit quadratic sum f1(λ1)。
Then, according to formula (3), the 1st stage end reservoir capacity V1=V0+LS1-EF1-G1, at this time not yet consider Group of Pumping Station moisturizing or
Reservoir abandons water, should be tested and be corrected using formula (4)~(6)：
1. work as V1＜ Vmin, then consider to carry out moisturizing, Group of Pumping Station moisturizing total amount Y to reservoir1=Vmin-V1+Δ1, storage capacity is corrected at this time
V1 *=Vmin+Δ1。
2. work as V1＞ VP, then need to abandon water to ensure the dispatching requirement of reservoir capacity, PS1=V1-VP, storage capacity V is corrected at this time1 *=
VP。
3. work as Vmin≤V1≤VP, then Y1=PS1=0, storage capacity V is corrected at this time1 *=V1。
By step 1.~3., correct simultaneously determine the 1st stage end reservoir capacity V1 *, that is, correct storage capacity V1 *, while correspondence can be obtained
Reservoir abandon water PS1Or Group of Pumping Station moisturizing total amount Y1。
(2) stage i=2,3 ... N-1：
fi(λi)=min [(Gi-YSi)2+fi-1(λi-1)] (10)
In formula, state variable λiReservoir for the preceding i period is for water inventory, unit：Ten thousand m3, it is equally carried out respectively discrete：To each discrete λi, decision variable GiIt is discrete to be same as above, and should meet：
State transition equation：λi-1=λi-G (11)
In formula：I=2,3 ..., N-1
To each discrete λi, by each discrete GiValue substitutes into the (G in formula (10) respectivelyi-YSi)2, by state transition equation formula
(11), to each discrete Gi, before lookup the i-1 stages meetIt is required that minimum fi-1(λi-1) value, thus it can obtain
Obtain (a Gi-YSi)2+min fi-1(λi-1), complete all of above discrete GiAfter optimizing, it can finally obtain and meet min [(Gi-
YSi)2+fi-1(λi-1)] requirement preceding i period system minimum water deficit quadratic sum fi(λi) value and its corresponding each stage
Reservoir optimal water supply Gi, i=1 ... i.
Equally, it also needs to determine the i-th stage end storage capacity V using formula (3)i, then tested using formula (4)~(6), correct and determine
I-th period end storage capacity Vi *, while corresponding reservoir can be obtained and abandon process water PSiAnd Group of Pumping Station moisturizing total amount process Yi, i=
1,2 ... i, process with step 1.~3..
(3) stage N：
fN(λN)=min [(GN-YSN)2+fN-1(λN-1)] (12)
State variableDecision variable GNIt is equally discrete in corresponding feasible zone, it should meet：λN-1=λN-GN。
Using step (2) the method, final obtain meets the λNIt is required that the optimal water supply process G of reservoiri, i=1 ... N, correspondence
Reservoir abandon process water PSi, Group of Pumping Station moisturizing total amount Yi, i=1,2 ... N and master mould object function optimal value F
=fN(λN)。
4. according to the method described in claim 1, it is characterized in that, using decomposition-Dynamic Programming polymerization to " mending library pumping plant
Group " subsystem model solves, and specifically comprises the following steps：
(1) consider to mend the constraint of library Group of Pumping Station cooperation energy consumption minimum criteria, established by formula (8) and mend library Group of Pumping Station economical operation number
Learn model：
Object function：
Period water supply constrains：
Power constraint：
In formula, Nk0The electric drilling match power of library pumping plant, unit are mended for kth seat：kW.
(2) " benefit library Group of Pumping Station " subsystem decomposition-Dynamic Programming polymerization solves：
1. subsystem two level is decomposed：
Above-mentioned subsystem model (13)~(15) are further decomposed, can obtain M single station economical operation two level submodel：
Object function：
Power constraint：
In formula, liFor single i-th period minimum operation energy consumption of standing, unit：kW·h.
2. two level submodel energy consumption determines：
For model above (16)~(17), segment length's Δ T during the i-th periodiIt is known that simultaneously can water lift be determined by pumping plant water levels of upstream and downstream
Lift HiAnd its corresponding pump capacity Qi, pump efficiency ηZ, i, electric efficiency ηmotWith transmission efficiency ηint, and the period is maximum
Rate of water make-up is YBI, max=3600QiΔTi/ 10000, by the discrete period maximum water lift amount YB of a fixed step sizeI, max, can obtain each
Water lift amount YBI, mPlace an order station operation energy consumption lI, m, m=1,2 ... max.
To other pumping plants, equally using above method, thus to obtain each pumping plant difference water lift amount YBKi, mUnder, single operation energy consumption of standing
lKi, m, k=1,2 ... M, m=1,2 ... max.
3. atomic system Dynamic Programming polymerize：
It is solved by more than two level subsystem, to each benefit library pumping plant, can obtain a series of single station water lift energy consumption lKi, m~mono- station
Mend library water YBKi, mThus relationship, k=1,2 ... M, m=1,2 ... max build following polymerization model substitution atoms model
(13)~(15)：
Object function：
Period water supply constrains：
Polymerization model (18)~(19) are similarly one-dimensional dynamic programming model, and stage variable is pumping plant number k, k=1,2 ...
M；Decision variable is each i-th period of pumping plant water lift amount YBki, target water discrete range when discrete range is single station optimization
YBKi, m, k=1,2 ... M, m=1,2 ... max；The centrifugal pump that each pumping plant water lift total amount is understood by formula (19) is state variable
(λ).The model is solved with reference to above one-dimensional dynamic programming, obtains and meets the i-th period pumping plant target water lift total amount YiLiValue,
And the corresponding optimal rate of water make-up combination YB of each pumping plantki *, k=1,2 ... M.
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