CN107742166A

CN107742166A - More storehouse multiple station systems water resource optimal allocation methods in storehouse are directly mended under a kind of fully irrigation conditions

Info

Publication number: CN107742166A
Application number: CN201710978806.1A
Authority: CN
Inventors: 龚懿; 程吉林; 陈兴; 蒋晓红; 张礼华; 袁承斌; 程浩淼; 周建康
Original assignee: Yangzhou University
Current assignee: Yangzhou University
Priority date: 2017-10-19
Filing date: 2017-10-19
Publication date: 2018-02-27
Anticipated expiration: 2037-10-19
Also published as: CN107742166B

Abstract

The invention discloses a kind of more storehouses-multiple station systems water resource optimal allocation method that storehouse is directly mended under fully irrigation conditions, using the method for solving of " more storehouse-multistations " large system decomposition-directly one-dimensional dynamic plan optimization of " single storehouse-multistation " subsystem backward-" mending storehouse Group of Pumping Station " two level subsystem decomposing level planning polymerization in benefit storehouse, all intake area minimum water deficits in certain delivery period, corresponding each reservoir day part optimal water supply can be obtained, abandon water, outer water diversion volume, and each benefit storehouse pumping plant rate of water make-up process.The present invention is distributed rationally with most important theories meaning and actual application value to Wall in Plain Reservoir Water Resources Irrigation.

Description

More storehouses-multiple station systems the water resource optimization in storehouse is directly mended under a kind of fully irrigation conditions Collocation method

Technical field

The present invention relates to abundant irrigation conditions lower step multi-reservoir with it is multiple benefit storehouse Group of Pumping Station cooperations scheduling methods, Belong to Water Resources Irrigation and distribute technical field rationally.

Background technology

Currently because water resource spatial and temporal distributions are uneven, 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 adjusted as unified entirety, use Built engineering (such as multiple multi-reservoirs and the operation of multiple Group of Pumping Station combined dispatchings) makes it play bigger effect, is to solve irrigated area to lack The main path of water problems.How more reservoirs-more pumping station systems traffic control reasonably transports as one content of water resources management Reached some target with the scheduling of system water resource and made system benefit optimal, the problem of being relatively conventional in water project management.

More storehouses-multiple station systems water resource optimal allocation for directly mending storehouse, supplied water from multiple benefit storehouse pumping plants to single reservoir, Form one and directly mend single storehouse-multiple station systems in storehouse, then be made up of single storehouse-multiple station systems of multiple series connection and directly mend the more of storehouse Storehouse-multiple station systems, combine and supplied water to multiple intake areas.Although water condition is abundant, due to being related to multiple multi-reservoirs, in system During actual motion, in addition to final stage list storehouse-multiple station systems only supply water to intake area, remaining single storehouse-multiple station systems at different levels is also additional outer Assign the output of next stage list storehouse-multiple station systems；And for each single storehouse-multiple station systems, and include multiple benefit storehouse pumps Stand group, how under the conditions of ensureing that intake area is supplied water sufficiently, reduces and mends storehouse Group of Pumping Station system operation energy consumption, save engineering operation Cost, it is very important major issue.

The content of the invention

The present invention supplies water in known reservoir and drawn for adjusting more reservoirs-more pumping plants cooperation scheduling system specific year Point when hop count, reservoir quantity, the initial storage of each reservoir, minimum capacity of a reservoir, utilizable capacity, storage capacity corresponding to flood control, It is available for water inventory, day part to come process water, evaporation and leakage process in year, each reservoir mends storehouse pumping plant quantity, each benefit storehouse pumping plant Year allows under water lift total amount, and each intake area water requirement process condition of day part, using " more storehouse-multistations " large system decomposition one The one-dimensional dynamic plan optimization of " single storehouse-multistation " subsystem backward in storehouse-" mending storehouse Group of Pumping Station " two level subsystem decomposes-is directly mended to move The method for solving of state planning polymerization, can obtain minimum water deficit in intake area in certain delivery period, in corresponding each reservoir delivery period Day part optimal water supply, abandon water, outer water diversion volume, and each benefit storehouse pumping plant day part rate of water make-up process.

The present invention program is as follows：

More storehouses-multiple station systems water resource optimal allocation the method in storehouse is directly mended under a kind of fully irrigation conditions, by multiple benefits Storehouse pumping plant supplies water to single reservoir, forms the single storehouse-multiple station systems for directly mending storehouse, then single storehouse-multistation system by multiple series connection System form directly mend storehouse more storehouse-multiple station systems, combine to multiple intake areas supply water, its water resource optimal allocation method including with Lower step：

First, model construction, 1~step 2 is comprised the following steps：

1. directly to mend the difference of each reservoir yield of day part and intake area water requirement of more storehouses in storehouse-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 supply and demand water difference least square and；Z is in research object year The quadratic sum of the difference of the supply and demand water of day part；R is reservoir quantity；H numbers for reservoir, h=1, and 2 ... ..R；N is to be drawn in year The when hop count divided；Segment number when i is, i=1,2 ... N；G_{H, i}、YS_{H, i}The respectively output of the i-th period of h seats reservoir With the water requirement of the corresponding period of intake area i-th, unit：Ten thousand m³；Object function is to accelerate to reduce using quadratic sum expression Deviation between reservoir yield and intake area water requirement.

2. constraints is set

Be available for water inventory constraints single storehouse-multiple station systems year including directly mend storehouse, consider higher level's reservoir call in from Single reservoir operation criterion constraints of the outer water diversion volume requirement of body, single reservoir capacity constraints, and mend the operation of storehouse Group of Pumping Station Energy consumption least commitment condition.

2nd, model solution

1. data prepare, specifically include：N number of period was divided into by 1 year, and determines day part length；At the beginning of determining each reservoir Beginning storage capacity V_{H, 0}；Determine that each reservoir year is available for water inventory SK_h, minimum capacity of a reservoir V_{H, min}, storage capacity V corresponding to flood control_{H, P}And Utilizable capacity V_{H, min}+Δ_{H, 1}；Measure and calculate each benefit storehouse reservoir day part and carry out water LS_{H, i}, evaporation with leakage EF_{H, i}, h=1, 2 ... R；Determine that each benefit storehouse pumping plant year allows water lift total amount BZ_{H, k}, k=1,2 ... M；Determine different periods lift H_{H, k, i}Lower operation Water lift flow Q_{H, k, i}And corresponding pump efficiency η_{Z, h, k, i}, electric efficiency η_{Mot, h, k}, transmission efficiency η_{Int, h, k}；Determine day part The water demand of crop YS of each intake area_{H, i}, h=1,2 ... R；I=1,2 ..., N.

2. direct " more storehouse-multistations " large-scale system model for mending storehouse is decomposed into the R single storehouse-multistation water resources for directly mending storehouse Distribute mathematical modeling rationally, object function is：

3. the single storehouse-multistation water resource optimal allocation mathematical modeling in pair directly benefit storehouse optimizes, from final stage without outer water transfer Single storehouse-multiple station systems that amount requires set out, and carry out single storehouse-multiple station systems water resource optimal allocation successively, specifically include：

(1) single seat reservoir-multiple station systems water resource optimal allocation submodel is solved using one-dimensional dynamic programming, obtained Day part reservoir optimal water supply process G_{H, i}, it is optimal to abandon process water PS_{H, i}, and mend the optimal moisturizing total amount process of storehouse Group of Pumping Station Y_{H, i}, h=1,2 ... R, i=1,2 ... N；

(2) solved using decomposition-Dynamic Programming polymerization to mending storehouse Group of Pumping Station two level subsystem model, obtain each benefit storehouse The optimal rate of water make-up YB of pumping plant day part_{H, k, i} ^*；

4. obtain each reservoir day part output G in more reservoir-more pumping station systems_{H, i}, outer water diversion volume WD_{H, i}, abandon water PS_{H, i}And corresponding each benefit storehouse pumping plant rate of water make-up YB_{H, k, i}Process, h=1,2 ..., R.

Further, the constraints in step (1) includes：

(1) that directly mends storehouse is available for water inventory constraints in single storehouse-multiple station systems year：In varying level year difference fraction In the case of, consideration needs water requirement, the water that water supply project provides；Wherein, in addition to final stage reservoir only needs to supply water to place intake area, Remaining reservoir at different levels also needs to undertake outer water transfer task, supplies next stage reservoir, i.e.,：

To the 1st~the R-1 seat reservoirs：

To Building R reservoir：

In formula：WD_{H, i}To assign the output of next stage reservoir, unit outside h seats the i-th period of reservoir：Ten thousand m³；SK_hFor h Seat reservoir is available for water inventory, unit year：Ten thousand m³；M be h seat reservoirs benefit storehouse pumping plant quantity, unit：Seat；K is benefit storehouse pumping plant Numbering, k=1,2 ... M；BZ_{H, k}Storehouse pumping plant year permission water lift total amount, unit are mended for the kth seat of h seat reservoirs：Ten thousand m³；Its Remaining variable implication is same as above.

(2) consider that higher level's reservoir calls in single reservoir operation criterion constraints with itself outer water diversion volume requirement：For appointing Meaning pumping plant more than h-th directly mends single water reservoir system in storehouse, it also is contemplated that the outer water diversion volume requirement of reservoir, according to water balance equation：

To the 1st~the R-1 seat reservoirs：V_{H, i}=V_{H, i-1}+LS_{H, i}+Y_{H, i}-PS_{H, i}-EF_{H, i}-G_{H, i}-WD_{H, i} (4)

To Building R reservoir：V_{R, i}=V_{R, i-1}+LS_{R, i}+Y_{R, i}-PS_{R, i}-EF_{R, i}-G_{R, i} (5)

On this basis, reservoir operation criterion is as follows：

1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir V_{H, min}When, then the i-th period should be by Group of Pumping Station to reservoir Moisturizing, Δ more than moisturizing to utilizable capacity, that is, reservoir minimum capacity of a reservoir_{H, 1}, i.e.,：

V_{H, i}＜ V_{H, min}When：Y_{H, i}=V_{H, min}-V_{H, i}+Δ_{H, 1} (6)

This period reservoir abandons water PS_{H, i}=0.

2. when meeting with flood, the i-th period end pondage is more than pondage V corresponding to flood control_{H, P} 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.,：

V_{H, i}＞ V_{H, P}When：PS_{H, i}=V_{H, i}-V_{H, P} (7)

This period Group of Pumping Station rate of water make-up Y_{H, i}=0.

3. when the i-th period end pondage is between minimum capacity of a reservoir V_{H, min}With flood control corresponding to pondage V_{H, P}Between, then the i-th period reservoir need not abandon water, and Group of Pumping Station is also without moisturizing, i.e.,：

V_{H, min}≤V_{H, i}≤V_{H, P}When：Y_{H, i}=PS_{H, i}=0 (8)

In formula：V_{H, i}、V_{H, i-1}Respectively h seats reservoir i-th, the reservoir storage of i-1 period Mos, unit：Ten thousand m³；Y_{H, i}For h The Group of Pumping Station of the seat period of reservoir i-th mends storehouse water, unit：Ten thousand m³, to the 2nd~the Building R reservoir, consider that it includes higher level's reservoir Its benefit storehouse water is fed by mending storehouse pumping plant；LS_{H, i}、PS_{H, i}、EF_{H, i}Respectively the i-th period of h seats reservoir carrys out water, abandoned Water, evaporation and leakage, unit：Ten thousand m³；V_{H, min}、V_{H, P}The respectively minimum capacity of a reservoir of h seats reservoir and flood control institute is right The storage capacity answered, unit：Ten thousand m³。

(3) single reservoir capacity constraints：The pondage of day part should be between reservoir minimum capacity of a reservoir and flood control limitation water Between the corresponding storage capacity in position, i.e.,：

V_{H, min}≤V_{H, i}≤V_{H, P}, i=1,2 ..., N (9)

(4) storehouse 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 the benefit storehouse Group of Pumping Station of any h seat reservoirs, its joint fortune within each water supply period Row is considered as energy consumption minimum, i.e.,：

In formula, L_{H, i}Storehouse Group of Pumping Station combined operation system energy consumption, unit are mended for h seats the i-th period of reservoir：kW·h；l_{H, k, i} For to kth seat the i-th period of pumping plant operation energy consumption of h seat reservoir moisturizings, unit：kW·h；M is the benefit storehouse pumping plant of h seat reservoirs Quantity, unit：Seat；ρ is water density, unit：kg/m³, g is acceleration of gravity, unit：m/s²；Q_{H, k, i}、H_{H, k, i}、ΔT_{H, k, i}、 η_{Z, h, k, i}Respectively to h seat reservoir moisturizings kth seat the i-th period of pumping plant flow (m³/ s), when equal lift (m), Period Length And pump efficiency (h)；η_{Mot, h, k}、η_{Int, h, k}Respectively to h seat reservoir moisturizings kth seat pumping plant motor efficiency and transmission Efficiency.

Further, in step (2), the pact of the single single storehouse-multistation water resource optimal allocation mathematical modeling for directly mending storehouse Beam condition is as follows：

(1) that directly mends storehouse is available for water inventory constraints in single storehouse-multiple station systems year：

1. for the 1st~the R-1 seat reservoirs, in addition to being supplied water to intake area, there is the outer output for assigning subordinate's reservoir will Ask, therefore：

In formula, h=1,2 ... R-1.

2. for Building R reservoir, only to outside the water supply of intake area, nothing assigns the output requirement of subordinate's reservoir outside, therefore：

(2) consider that higher level's reservoir calls in single reservoir operation criterion constraints with itself outer water diversion volume requirement：

1. similarly for the 1st~the R-1 seat reservoirs, it is according to water balance equation：

V_{H, i}=V_{H, i-1}+LS_{H, i}+Y_{H, i}-PS_{H, i}-EF_{H, i}-G_{H, i}-WD_{H, i} (14)

2. for Building R reservoir, it is according to water balance equation：

V_{R, i}=V_{R, i-1}+LS_{R, i}+Y_{R, i}-PS_{R, i}-EF_{R, i}-G_{R, i} (15)

The same formula of reservoir operation criterion (6)~(8) on this basis.

(3) single reservoir capacity constraints：Same formula (9).

(4) storehouse Group of Pumping Station cooperation energy consumption least commitment condition is mended：Same formula (10).

Further, the 3rd step of step (2) comprises the following steps that：

(1) Building R Dan Ku-multiple station systems water resource optimal allocation submodel solves

1) the one-dimensional dynamic programming evaluation of subsystem

With reference to one-dimensional dynamic programming evaluation principle, must correspond to recurrence equation is：

1. stage i=1：

f₁(λ₁)=min (G_{R, 1}-YS_{R, 1})² (16)

In formula, λ₁For state variable, 1 period reservoir yield, unit before expression：Ten thousand m³, it is pressed in corresponding feasible zone One fixed step size is discrete：λ₁=0, W₁, W₂...,To each discrete λ₁, decision variable G_{R, 1}In corresponding feasible zone It is discrete inside to press a fixed step size, then G will be met₁≥λ₁It is required that G₁Formula (16) is substituted into respectively, determines each discrete λ respectively₁During value, Optimal G_{R, 1}And its corresponding period minimum water deficit quadratic sum f₁(λ₁)。

Then, according to formula (15), the 1st stage end reservoir capacity V_{R, 1}=V_{R, 0}+LS_{R, 1}-EF_{R, 1}-G_{R, 1}, now not yet consider Group of Pumping Station moisturizing or reservoir abandon water, should be tested and be corrected using formula (6)~(8)：

I) V is worked as_{R, 1}＜ V_{R, min}, then consider to carry out moisturizing, moisturizing total amount Y to reservoir_{R, 1}=V_{R, min}-V_{R, 1}+Δ_{R, 1}, now repair Positive storage capacity V_{R, 1} ^*=V_{R, min}+Δ_{R, 1}。

Ii V) is worked as_{R, 1}＞ V_{R, P}, then need to abandon water to ensure the dispatching requirement of reservoir capacity, PS_{R, 1}=V_{R, 1}-V_{R, P}, now repair Positive storage capacity V_{R, 1} ^*=V_{R, P}。

Iii V) is worked as_{R, min}≤V_{R, 1}≤V_{R, P}, then Y_{R, 1}=PS_{R, 1}=0, now correct storage capacity V_{R, 1} ^*=V_{R, 1}。

By step i)~iii), correct and determine the 1st stage end reservoir capacity V_{R, 1} ^*, while corresponding reservoir can be obtained Abandon water PS_{R, 1}Or Group of Pumping Station moisturizing total amount Y_{R, 1}。

2. stage i=2,3 ... N-1：

f_i(λ_i)=min [(G_{R, i}-YS_{R, i})²+f_i-1(λ_i-1)] (17)

In formula, state variable λ_iFor the reservoir water supply total amount of preceding i period, unit：Ten thousand m³, it is equally carried out respectively from Dissipate：λ₁=0, W₁, W₂...,To each discrete λ_i, decision variable G_{R, i}It is discrete to be same as above, and should meet：

State transition equation：λ_i-1=λ_i-G_{R, i} (18)

In formula：I=2,3 ..., N-1

To each discrete λ_i, by each discrete G_{R, i}(the G that value is substituted into formula (17) respectively_{R, i}-YS_{R, i})², turned by state Equation (18) is moved, to each discrete G_{R, i}, before lookup the i-1 stages meetIt is required that minimum f_i-1(λ_i-1) Value, thus can obtain (a G_{R, i}-YS_{R, i})²+minf_i-1(λ_i-1), complete all of above discrete G_{R, i}After optimizing, it can finally obtain Min [(G must be met_{R, i}-YS_{R, i})²+f_i-1(λ_i-1)] require preceding i period subsystem minimum water deficit quadratic sum f_i(λ_i) value, And its corresponding each stage reservoir optimal water supply G_{R, i}, i=1 ... i.

Equally, also need to determine the i-th stage end storage capacity V using formula (15)_{R, i}, then tested, repaiied using formula (6)~(8) It is positive to determine the i-th period end storage capacity V_{R, i} ^*, while corresponding reservoir can be obtained and abandon process water PS_{R, i}And Group of Pumping Station moisturizing total amount mistake Journey Y_{R, i}, i=1,2 ... i, process is with step i)~iii).

3. stage N：

f_N(λ_N)=min [(G_{R, N}-YS_{R, N})²+f_N-1(λ_N-1)] (19)

State variableDecision variable G_{R, N}It is equally discrete in corresponding feasible zone, it should meet：λ_N-1=λ_N- G_{R, N}。

Using step, 2. methods described, final obtain meet the λ_NIt is required that the optimal water supply process G of reservoir_{R, i}, i=1 ... N, corresponding reservoir abandon process water PS_{R, i}, Group of Pumping Station moisturizing total amount process Y_{R, i}, i=1,2 ... N, and the submodel Object function optimal value F_R=f_N(λ_N)。

2) storehouse Group of Pumping Station two level subsystem model decomposition-Dynamic Programming polymerization is mended to solve

1. mend storehouse Group of Pumping Station two level subsystem economical operation mathematical modeling structure

Consider to mend storehouse Group of Pumping Station cooperation energy consumption least commitment condition, established by formula (10) and mend storehouse Group of Pumping Station economical operation Mathematical modeling：

Object function：

Period output constrains：

Power constraint：

In formula, N_{R, k, 0}The electric drilling match power of storehouse pumping plant, unit are mended for the kth seat of Building R reservoir：kW；Remaining variables contain Justice is same as above.

2. mend storehouse Group of Pumping Station subsystem decomposition-Dynamic Programming polymerization to solve

I) two level subsystem decomposes：

Above-mentioned subsystem model (20)~(22) are further decomposed, can obtain M single station economical operation three-level submodel：

Object function：

Power constraint：

In formula, l_{R, i}For the i-th period minimum operation energy consumption of single station of Building R reservoir, unit：kW·h.

Ii) three-level submodel energy consumption determines：

For model above (23)~(24), segment length's Δ T during the i-th period_{R, i}, it is known that and determined by pumping plant water levels of upstream and downstream Water lift lift H_{R, i}, and its corresponding Q_{R, i}、η_{Z, R, i}、η_{Mot, R}And η_{Int, R}, and the period maximum rate of water make-up is YB_{R, i, max}= 3600Q_{R, i}ΔT_{R, i}/ 10000, by the discrete period maximum water lift amount YB of a fixed step size_{R, i, max}, obtain each water lift amount YB_{R, i, m}Under Single station operation energy consumption l_{R, i, m}, m=1,2 ... max.

To other pumping plants of the reservoir, equally using above method, each pumping plant difference water lift amount YB is derived from_{R, k, i, m} Under, single operation energy consumption l that stands_{R, k, i, m}, k=1,2 ... M, m=1,2 ... max.

Iii) former two level subsystem Dynamic Programming polymerization：

Solved by above three-level subsystem, to each benefit storehouse pumping plant, can obtain a series of single station water lift energy consumptions l_{R, k, i, m}~mono- stand mends storehouse water YB_{R, k, i, m}Relation, k=1,2 ... M, m=1,2 ... max, it thus can build following polymerization model Substitute former two level submodel (20)~(22)：

Object function：

Period water quantity restraint：

Polymerization model (25)~(26) are similarly one-dimensional dynamic programming model, and stage variable is the benefit storehouse of Building R reservoir Pumping plant numbering k, k=1,2 ... M；Decision variable is each i-th period of pumping plant water lift amount YB_{R, k, i}, its discrete range is that single station is excellent Target water discrete range YB during change_{R, k, i, m}, k=1,2 ... M, m=1,2 ... max；Each pumping plant water lift is understood by formula (26) The centrifugal pump of total amount is state variable λ, solves the model with reference to one-dimensional dynamic programming, obtains and meet the i-th period pumping plant mesh Mark water lift total amount Y_{R, i}L_{R, i}Value, and the optimal rate of water make-up combination YB of corresponding each pumping plant_{R, k, i} ^*, k=1,2 ... M.

3) Building R Dan Ku-multiple station systems submodel optimal solution is determined

The day part obtained by step 1) mends the optimal moisturizing total amount Y of storehouse Group of Pumping Station_{R, i}After process, each period is determined Y_{R, i}, altogether need to be via n times step 2), further by day part Y_{R, i}Optimization distribution obtains each benefit storehouse pumping plant to M benefit storehouse pumping plant The optimal rate of water make-up YB of day part_{R, k, i} ^*；Thus the difference of the supply and demand water of day part of R levels list storehouse-in multiple station systems year is finally obtained Least square and F_R, corresponding reservoir day part optimal water supply G_{R, i}, abandon water process PS_{R, i}, and it is each mend storehouse pumping plant it is each when The optimal rate of water make-up process YB of section_{R, k, i} ^*, i=1,2 ... N, k=1,2 ... M.

(2) R-1, R-2 ..., 1 single storehouse-multiple station systems water resource optimal allocation submodel solve

For R-1, R-2 ..., 1 directly mend storehouse single storehouse-multiple station systems, equally using submodel (11)~(15), (6)~(10), it is with reservoir day part output G_{H, i}One-dimensional nonlinear model can be divided for the stage of decision variable, with reference to step (1), solved using one-dimensional dynamic programming；Differ only in：

1) that directly mends storehouse is available for water inventory to constrain in single storehouse-multiple station systems year：

Reservoir, which supplies water, should supply the output G of intake area_{H, i}, it is also contemplated that the outer output WD for assigning next stage reservoir_{H, i} Factor, then it is available for water inventory constraint to use formula (12) in single storehouse-multiple station systems year, i.e.,： Due to its outer water diversion volume WD_{H, i}It is to mend storehouse pumping plant by a certain seat in the reservoir of Building R to provide：

WD_{H, i}=YB_{H+1,1, i} (27)

Then formula (12) can be converted into such as following formula (28), so as to obtain reservoir yield restriction range.

2) consider that higher level's reservoir is called in constrain with single reservoir operation criterion of itself outer water diversion volume requirement：

Outer water diversion volume WD is considered as to each stage water balance equation_{H, i}Factor, formula (14) should be used, i.e.,：V_{H, i}=V_{H, i-1}+ LS_{H, i}+Y_{H, i}-PS_{H, i}-EF_{H, i}-G_{H, i}-WD_{H, i}, with reason formula (27), formula (14) is converted into following form：

V_{H, i}=V_{H, i-1}+LS_{H, i}+Y_{H, i}-PS_{H, i}-EF_{H, i}-G_{H, i}-YB_{H+1,1, i} (29)

On this basis, still each stage storage capacity is modified using formula (6)~(8).

The water resource optimization that the invention can be achieved directly to mend more reservoirs-more pumping station systems in storehouse under abundant irrigation conditions is adjusted Degree, method precision is reliable, rate of accuracy reached more than 95%, reduces and mends storehouse pumping station operation energy consumption more than 20%, reaches irrigated area water money The purpose of source optimization configuration, improves irrigated area, society and ecological benefits.

Brief description of the drawings

Fig. 1 is that more storehouses-multistation water resource that storehouse is directly mended under abundant irrigation conditions generally changes system schematic.

Embodiment

As shown in figure 1, needed with directly mending each reservoir yield of day part of more storehouses in storehouse-in multiple station systems year and intake area The minimum target of quadratic sum of the difference of water, each reservoir yield of day part of division is decision variable in year, with each Dan Ku-more It is available for water inventory (be available for water inventory i.e. single reservoir year and draw water lift total amount sum in each pumping plant year) system of standing year, considers higher level's reservoir Call in and single reservoir operation criterion of itself outer water diversion volume requirement, single reservoir water balance, benefit storehouse Group of Pumping Station cooperation energy consumption Minimum etc. is constraints, establishes under abundant irrigation conditions and directly mends the more storehouses-multiple station systems water resources optimal operation model in storehouse.

First, model construction

1. object function

In formula：F be research object year in day part supply and demand water difference least square and；Z is in research object year The quadratic sum of the difference of the supply and demand water of day part；R is reservoir quantity (seat)；H be reservoir numbering (h=1,2 ... R)；N is The when hop count of division in year；Segment number when i is (i=1,2 ... N)；G_{H, i}、YS_{H, i}Respectively the i-th period of h seats reservoir Output (ten thousand m³) and the period of corresponding intake area i-th water requirement (ten thousand m³).Object function is to add using quadratic sum expression Speed reduces the deviation between reservoir yield and intake area water requirement.

2. constraints

Be available for water inventory constraints single storehouse-multiple station systems year including directly mend storehouse, consider higher level's reservoir call in from Single reservoir operation criterion constraints of the outer water diversion volume requirement of body, single reservoir capacity constraints, and mend the operation of storehouse Group of Pumping Station Energy consumption least commitment condition.Specifically it is described below：

(1) that directly mends storehouse is available for water inventory to constrain in single storehouse-multiple station systems year：In varying level year difference fraction situation Under, consideration needs water requirement, the water that water supply project may provide.Wherein, in addition to final stage reservoir only needs to supply water to place intake area, Remaining reservoir at different levels also needs to undertake outer water transfer task, supplies next stage reservoir, i.e.,：

To the 1st~the R-1 seat reservoirs：

To Building R reservoir：

In formula：WD_{H, i}To assign output (ten thousand m of next stage reservoir outside h seats the i-th period of reservoir³)；SK_hFor h seat reservoirs Be available for water inventory (ten thousand m years³)；M is the benefit storehouse pumping plant quantity (seat) of h seat reservoirs；K for mend storehouse pumping plant numbering (k=1, 2 ... M)；BZ_{H, k}Storehouse pumping plant year permission water lift total amount (ten thousand m is mended for the kth seat of h seat reservoirs³)；Remaining variables implication is same On.

(2) consider that higher level's reservoir is called in constrain with single reservoir operation criterion of itself outer water diversion volume requirement：For any h Individual more pumping plants directly mend single water reservoir system in storehouse, it also is contemplated that the outer water diversion volume requirement of reservoir, according to water balance equation：

On this basis, reservoir operation criterion is as follows：

1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir V_{H, min}When, 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 reservoir_{H, 1}), i.e.,：

V_{H, i}＜ V_{H, min}When：Y_{H, i}=V_{H, min}-V_{H, i}+Δ_{H, 1} (6)

This period reservoir abandons water PS_{H, i}=0.

V_{H, i}＞ V_{H, P}When：PS_{H, i}=V_{H, i}-V_{H, P} (7)

This period Group of Pumping Station rate of water make-up Y_{H, i}=0.

V_{H, min}≤V_{H, i}≤V_{H, P}When：Y_{H, i}=PS_{H, i}=0 (8)

In formula：V_{H, i}、V_{H, i-1}Respectively h seats reservoir i-th, reservoir storage (ten thousand m of i-1 period Mos³)；Y_{H, i}For h seat water The Group of Pumping Station of the period of storehouse i-th mends storehouse water (ten thousand m³), to the 2nd~the Building R reservoir, consider it comprising higher level's reservoir by mending storehouse Pumping plant feeds its benefit storehouse water；LS_{H, i}、PS_{H, i}、EF_{H, i}Respectively the i-th period of h seats reservoir carrys out water (ten thousand m³), abandon water Measure (ten thousand m³), evaporation with leakage (ten thousand m³)；V_{H, min}、V_{H, P}The respectively minimum capacity of a reservoir of h seats reservoir and flood control institute is right Storage capacity (ten thousand m answered³)。

(3) single reservoir capacity constraint：The pondage of day part should be between reservoir minimum capacity of a reservoir and flood control pair Between answering storage capacity, i.e.,：

V_{H, min}≤V_{H, i}≤V_{H, P}, (i=1,2 ..., N) (9)

(4) constraint of storehouse Group of Pumping Station cooperation energy consumption minimum criteria is mended：Further, the constraint of reservoir operation criterion is being met On the basis of, to ensure the abundant irrigation conditions in intake area, to the benefit storehouse Group of Pumping Station of any h seat reservoirs, it is in each water supply period Interior cooperation is considered as energy consumption minimum, i.e.,：

In formula, L_{H, i}Storehouse Group of Pumping Station combined operation system energy consumption (kWh) is mended for h seats the i-th period of reservoir；l_{H, k, i}For to Kth seat the i-th period of pumping plant operation energy consumption (kWh) of h seat reservoir moisturizings；M is the benefit storehouse pumping plant quantity of h seat reservoirs (seat)；ρ is water density (kg/m³), g is acceleration of gravity (m/s²)；Q_{H, k, i}、H_{H, k, i}、ΔT_{H, k, i}、η_{Z, h, k, i}Respectively to h Flow (the m of kth seat the i-th period of pumping plant of seat reservoir moisturizing³/ s), when equal lift (m), Period Length (h) and pump efficiency； η_{Mot, h, k}、η_{Int, h, k}Respectively to h seat reservoir moisturizings kth seat pumping plant motor efficiency and transmission efficiency.

2nd, model feature

(1) consider strictly carry out water total amount control in water resources development and utilization, therefore in model constraints Add direct benefit storehouse is available for water inventory to constrain in single storehouse-multiple station systems year, i.e. formula (2)~(3).

(2) the single reservoir operation for considering that higher level's reservoir is called in itself outer water diversion volume requirement is added in constraints first Criterion is constrained, i.e. formula (4)~(8), and water diversion volume outside reservoirs at different levels is mutually connected with mending storehouse pumping plant rate of water make-up, for using backward dynamic Plan optimization provides feasibility；At the same time it can also realize more storehouse-multistations of lift pumping station group " idle mends storehouse, busy supplies water " System water resources optimal operation mode.If idle pondage is less than reservoir minimum capacity of a reservoir, considers to carry out moisturizing to reservoir, carry The benefit storehouse total Water Y of pump works group_{H, i}For reservoir minimum capacity of a reservoir and the difference of the i-th period end pondage, more than minimum capacity of a reservoir Δ_{H, 1}, i.e. Y_{H, i}=V_{H, min}-V_{H, i}+Δ_{H, 1}；If reservoir capacity exceedes corresponding to flood control by reservoir regulation limiting water level during reservoir storage, need Water is abandoned to ensure the dispatching requirement of reservoir capacity, reservoir abandons water PS_{H, i}Limited for the i-th period end pondage and flood control by reservoir regulation The difference of reservoir storage, i.e. PS corresponding to water level processed_{H, i}=V_{H, i}-V_{H, P}.Busy then considers by reservoir, mended according to pondage situation Storehouse pumping plant is combined to supply water to intake area, to meet the water demand of water user as far as possible.

(3) for from Building M mend storehouse pumping plant combine to certain one-level list reservoir supply water when, it is contemplated that stood under each pumping plant difference lift Between unit performance property difference, add the minimum constraints of Group of Pumping Station cooperation energy consumption, i.e. formula (10).By establishing pumping plant Group's two level subsystem combined optimization operation mathematical modeling is simultaneously solved using decomposition-Dynamic Programming polymerization, by one-level subsystem mould The day part Group of Pumping Station moisturizing total amount Y that type optimization determines_{H, i}Further optimization distribution is carried out to each benefit storehouse pumping plant YB_{H, k, i}(k=1, 2 ... M), the operation of Group of Pumping Station subsystem is reduced energy consumption as far as possible.

(4) model (1)~(10) use " more storehouse-multistations " big though being in form one-dimensional complex nonlinear mathematical modeling System decomposition-directly mends the one-dimensional dynamic plan optimization of " single storehouse-multistation " subsystem backward-" mending storehouse Group of Pumping Station " the two level subsystem in storehouse The method for solving of system decomposition-Dynamic Programming polymerization, can not only obtain the optimizing decision variate-value in object function --- each stage Each reservoir yield G_{H, i}；And consider to meet that higher level's reservoir calls in single reservoir operation criterion with itself outer water diversion volume requirement Constraint, storage capacity is tested amendment by using formula (4)~(8), can also obtain that day part reservoir is optimal to abandon water PS_{H, i}, it is outer Water diversion volume WD_{H, i}With the optimal moisturizing total amount process Y of Group of Pumping Station_{H, i}；Recycling formula (10), by day part Group of Pumping Station moisturizing total amount Y_{H, i} Further optimization distribution can obtain the optimal rate of water make-up YB of each benefit storehouse pumping plant day part again to each benefit storehouse pumping plant_{H, k, i} ^*；Actually ask Solution obtains (M+3) R-1 decision variable value, enriches such complex nonlinear model solution method.

3rd, model solution

1st, data preparation is carried out

Specifically include：N number of period was divided into by 1 year, and determines day part length；Measure determines each reservoir initial storage V_{H, 0}；Determine that each reservoir year is available for water inventory SK_h, minimum capacity of a reservoir V_{H, min}, storage capacity V corresponding to flood control_{H, P}And Xing Liku Hold V_{H, min}+Δ_{H, 1}；Measurement and calculating determine that each benefit storehouse reservoir day part carrys out water LS_{H, i}, evaporation with leakage EF_{H, i}(h=1, 2 ... R)；Determine that each benefit storehouse pumping plant year allows water lift total amount BZ_{H, k}(k=i, 2 ... M)；Measure determines different periods lift H_{H, k, i} The water lift flow Q of lower operation_{H, k, i}And corresponding pump efficiency η_{Z, h, k, i}, electric efficiency η_{Mot, h, k}, transmission efficiency η_{Int, h, k}；It is determined that The water demand of crop YS of each intake area of day part_{H, i}(h=1,2 ... R；I=1,2 ..., N).

2nd, " more storehouse-multistations " large-scale system model for directly mending storehouse decomposes

Model above (1)~(10) are with each reservoir day part output G_{H, i}(h=1,2 ... R) it is decision variable R dimension complex nonlinear mathematical modelings, consider that each reservoir supplies water to respective intake area respectively, can be R by model decomposition Directly mend the single storehouse-multistation water resource optimal allocation mathematical modeling in storehouse.

(1) object function：

(2) that directly mends storehouse is available for water inventory to constrain in single storehouse-multiple station systems year：

In formula, h=1,2 ... R-1.

(3) consider that higher level's reservoir is called in constrain with single reservoir operation criterion of itself outer water diversion volume requirement：

2. for Building R reservoir, it is according to water balance equation：

V_{R, i}=V_{R, i-1}+LS_{R, i}+Y_{R, i}-PS_{R, i}-EF_{R, i}-G_{R, i} (15)

The same formula of reservoir operation criterion (6)~(8) on this basis.

(4) single reservoir capacity constraint：Same formula (9).

(5) constraint of storehouse Group of Pumping Station cooperation energy consumption minimum criteria is mended：Same formula (10).

3rd, the single storehouse-multistation subsystem optimization in storehouse is directly mended

Because the 1st~the R-1 seats reservoir also undertakes the outer output WD for assigning next stage reservoir_{H, i}, actual is subordinate's reservoir Mend storehouse Group of Pumping Station in a certain seat pumping plant rate of water make-up YB_{H+1, k, i}(k ∈ M), i.e. WD_{H, i}=YB_{H+1, k, i}(YB_{H+1, k, i}∈Y_{H+1, i})。 And YB_{H+1, k, i}Criterion need to be run by reservoir to be judged, belong to coupling constraint.Therefore hysterology is considered as, from final stage without investigation mission outside the city or town Single storehouse-multiple station systems of water requirement set out, and carry out single storehouse-multiple station systems water resource optimal allocation successively.

It is with reservoir day part output G for submodel (11)~(15), (6)~(10)_{R, i}For decision variable Stage can divide one-dimensional nonlinear model, one-dimensional dynamic programming can be used to solve.

1) the one-dimensional dynamic programming evaluation of subsystem

1. stage i=1：

f₁(λ₁)=min (G_{R, 1}-YS_{R, 1})² (16)

In formula, λ₁For state variable, 1 period reservoir yield (ten thousand m before expression³), it can press one in corresponding feasible zone Fixed step size is discrete：λ₁=0, W₁, W₂...,To each discrete λ₁, decision variable G_{R, 1}Can be corresponding feasible It is discrete by a fixed step size in domain, such as 00,000 m³, 200,000 m³, 400,000 m³, 600,000 m³、…G_{R, 1max}Deng (G_{R, 1, max}For the 1st stage reservoir most Big water supply capacity), then G will be met₁≥λ₁It is required that G₁Formula (16) is substituted into respectively, can determine each discrete λ respectively₁It is optimal during value G_{R, 1}And its corresponding period minimum water deficit quadratic sum f₁(λ₁)。

2. stage i=2,3 ... N-1：

f_i(λ_i)=min [(G_{R, i}-YS_{R, i})²+f_i-1(λ_i-1)] (17)

In formula, state variable λ_iFor reservoir water supply total amount (ten thousand m of preceding i period³), it is equally carried out respectively discrete：λ_i =0, W₁, W₂...,To each discrete λ_i, decision variable G_{R, i}It is discrete to be same as above, and should meet：

State transition equation：λ_i-1=λ_i-G_{R, i} (18)

In formula：I=2,3 ..., N-1

To each discrete λ_i, by each discrete G_{R, i}(the G that value is substituted into formula (17) respectively_{R, i}-YS_{R, i})², turned by state Equation (18) is moved, to each discrete G_{R, i}, before lookup the i-1 stages meetIt is required that minimum f_i-1(λ_i-1) Value, thus can obtain (a G_{R, i}-Ys_{R, i})²+minf_i-1(λ_i-1), complete all of above discrete G_{R, i}After optimizing, it can finally obtain Min [(G must be met_{R, i}-YS_{R, i})²+f_i-1(λ_i-1)] require preceding i period subsystem minimum water deficit quadratic sum f_i(λ_i) value, And its corresponding each stage reservoir optimal water supply G_{R, i}(i=1 ... i).

Equally, also need to determine the i-th stage end storage capacity V using formula (15)_{R, i}, then tested, repaiied using formula (6)~(8) It is positive to determine the i-th period end storage capacity V_{R, i} ^*, while corresponding reservoir can be obtained and abandon process water PS_{R, i}And Group of Pumping Station moisturizing total amount mistake Journey Y_{R, i}(i=1,2 ... i), process is with step i)~iii).

3. stage N：

f_N(λ_N)=min [(G_{R, N}-YS_{R, N})²+f_N-1(λ_N-1)] (19)

Using step, 2. methods described, final obtain meet the λ_NIt is required that the optimal water supply process G of reservoir_{R, i}(i=1 ... N), corresponding reservoir abandons process water PS_{R, i}, Group of Pumping Station moisturizing total amount process Y_{R, i}(i=1,2 ... N), and the submodule Type object function optimal value F_R=f_N(λ_N)。

After the one-dimensional dynamic plan optimization of above list reservoir-big system of more pumping plants, day part in area's year is studied except having determined that The least square and F of the difference of supply and demand water_R, reservoir day part optimal water supply G_{R, i}With abandon water process PS_{R, i}Outside, to mending storehouse pumping plant Group, only determines the optimal moisturizing total amount Y of day part_{R, i}Process, not yet consider the rate of water make-up YB of specific each benefit storehouse pumping plant_{R, k, i} (k=1,2 ... M), because each benefit storehouse pumping plant set performance has differences, along with day part water lift lift is different, cause Pumping station operation energy consumption is different, and storehouse Group of Pumping Station operation energy consumption is mended to reduce, need to be by Y_{R, i}Further optimization distribution mends storehouse pump to each Stand, specify day part and respectively mend storehouse pumping plant optimization rate of water make-up YB_{R, k, i}, match somebody with somebody so as to be really achieved single storehouse-multiple station systems water resource optimization Put purpose.

Consider to mend the constraint of storehouse Group of Pumping Station cooperation energy consumption minimum criteria, established by formula (10) and mend storehouse Group of Pumping Station economical operation Mathematical modeling：

Object function：

Period output constrains：

Power constraint：

In formula, N_{R, k, 0}The electric drilling match power (kW) of storehouse pumping plant is mended for the kth seat of Building R reservoir；Remaining variables implication is same On.

I) two level subsystem decomposes：

Object function：

Power constraint：

Ii) three-level submodel energy consumption determines：

For model above (23)~(24), segment length's Δ T during the i-th period_{R, i}, it is known that simultaneously can be true by pumping plant water levels of upstream and downstream Determine water lift lift H_{R, i}, and its corresponding Q_{R, i}、η_{Z, R, i}、η_{Mot, R}And η_{Int, R}, and the period maximum rate of water make-up is YB_{R, i, max}= 3600Q_{R, i}Δ_{R, i}/ 10000, by the discrete period maximum water lift amount YB of a fixed step size_{R, i, max}(i.e. to period duration Δ R_{T, i}Enter Row is discrete), each water lift amount YB can be obtained_{R, i, m}Place an order station operation energy consumption l_{R, i, m}(m=1,2 ... max).

To other pumping plants of the reservoir, equally using above method, each pumping plant difference water lift amount YB is derived from_{R, k, i, m} Under, single operation energy consumption l that stands_{R, k, i, m}(k=1,2 ... M, m=1,2 ... max).

Iii) former two level subsystem Dynamic Programming polymerization：

Solved by above three-level subsystem, to each benefit storehouse pumping plant, can obtain a series of single station water lift energy consumptions l_{R, k, i, m}~mono- stand mends storehouse water YB_{R, k, i, m}Relation (k=1,2 ... M, m=1,2 ... max), it thus can build following polymerization mould Type substitutes former two level submodel (20)~(22)：

Object function：

Period water quantity restraint：

Polymerization model (25)~(26) are similarly one-dimensional dynamic programming model, and stage variable is the benefit storehouse of Building R reservoir Pumping plant numbering k (k=1,2 ... M)；Decision variable is each i-th period of pumping plant water lift amount YB_{R, k, i}, its discrete range is that single station is excellent Target water discrete range YB during change_{R, k, i, m}(k=1,2 ... M, m=1,2 ... max)；Each pumping plant water lift is understood by formula (26) The centrifugal pump of total amount is state variable (λ).One-dimensional dynamic programming solves the model with reference to more than, obtains and met for the i-th period Pumping plant target water lift total amount Y_{R, i}L_{R, i}Value, and the optimal rate of water make-up combination YB of corresponding each pumping plant_{R, k, i} ^*(k=1,2 ... M).

3) the multiple station systems submodel optimal solutions of Building R Dan Ku mono- are determined

The day part obtained by step 1) mends the optimal moisturizing total amount Y of storehouse Group of Pumping Station_{R, i}(i=1,2 ... N) process Afterwards；The Y determined to each period_{R, i}, altogether need to be via n times step 2), you can further by day part Y_{R, i}Optimization is distributed to M benefit Storehouse pumping plant, obtain the optimal rate of water make-up YB of each benefit storehouse pumping plant day part_{R, k, i} ^*；Thus R levels list storehouse-multiple station systems year is finally obtained The least square and F of the difference of the supply and demand water of interior day part_R, corresponding reservoir day part optimal water supply G_{R, i}, abandon water process PS_{R, i}, and the optimal rate of water make-up process YB of each benefit storehouse pumping plant day part_{R, k, i} ^*(i=1,2 ... N, k=1,2 ... M).

For R-1, R-2 ..., 1 directly mend storehouse single storehouse-multiple station systems, equally using submodel (11)~(15), (6)~(10), it is with reservoir day part output G_{H, i}One-dimensional nonlinear model can be divided for the stage of decision variable, with reference to step (1), can still one-dimensional dynamic programming be used to solve.Differ only in：

Reservoir, which supplies water, should supply the output G of intake area_{H, i}, it is also contemplated that the outer output WD for assigning next stage reservoir_{H, i} Factor, then it is available for water inventory constraint to use formula (12) in single storehouse-multiple station systems year, i.e.,： Due to its outer water diversion volume WD_{H, i}It is to be mended storehouse pumping plant by a certain seat in the reservoir of Building R and provided (herein to unify with the of Building R reservoir Mend storehouse pumping plant for 1 and mend storehouse pumping plant as the outer water diversion volume of higher level's reservoir), it can obtain：

WD_{H, i}=YB_{H+1,1, i} (27)

Outer water diversion volume WD is considered as to each stage water balance equation_{H, i}Factor, formula (14) should be used, i.e.,：V_{H, i}=V_{H, i-1}+ LS_{H, i}+Y_{H, i}-PS_{H, i}-EF_{H, i}-G_{H, i}-WD_{H, i}, with reason formula (27), formula (14) can be converted following form：

Due to YB_{H+1,1, i}Determine, therefore use during preceding once single storehouse-multistation subsystem water resource optimal allocation " single storehouse-multistation " subsystem backward-dimension Dynamic Programming optimization is feasible.

4th, master mould optimal solution is determined

Thus, to each single storehouse-multistation subsystem model for directly mending storehouse, single storehouse-multiple station systems submodel one is respectively adopted Dimension Dynamic Programming, which solves, mends storehouse Group of Pumping Station two level subsystem model decomposition-Dynamic Programming polymerization solves, determine each single storehouse- Multiple station systems submodel optimal solution, finally give the optimal reservoir day part output G of each subsystem_{H, i}, outer water diversion volume WD,_{H, i}、 Abandon water PS_{H, i}And corresponding each benefit storehouse pumping plant rate of water make-up YB_{H, k, i}Process (h=1,2 ..., R).

Claims

1. the more storehouses-multiple station systems water resource optimal allocation method in storehouse is directly mended under a kind of fully irrigation conditions, by multiple benefit storehouses Pumping plant supplies water to single reservoir, forms the single storehouse-multiple station systems for directly mending storehouse, then single storehouse-multiple station systems by multiple series connection More storehouse-the multiple station systems for directly mending storehouse are formed, combines and is supplied water to multiple intake areas, it is characterised in that water resource optimal allocation side Method comprises the following steps：

First, model construction, 1~step 2 is comprised the following steps：

1. directly to mend the flat of the difference of each reservoir yield of day part and intake area water requirement of more storehouses in storehouse-in multiple station systems year Square and minimum target, establishes following object function：

<mrow> <mi>F</mi> <mo>=</mo> <mi>min</mi> <mi> </mi> <mi>Z</mi> <mo>=</mo> <mi>m</mi> <mi>i</mi> <mi>n</mi> <munderover> <mo>&Sigma;</mo> <mrow> <mi>h</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>R</mi> </munderover> <munderover> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <msup> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>YS</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow>

In formula：F be research object year in day part supply and demand water difference least square and；When Z is each in research object year The quadratic sum of the difference of the supply and demand water of section；R is reservoir quantity；H numbers for reservoir, h=1, and 2 ... R；N is division in year When hop count；Segment number when i is, i=1,2 ... N；G_{H, i}、YS_{H, i}The respectively output of the i-th period of h seats reservoir and right The water requirement for the period of intake area i-th answered, unit：Ten thousand m³；Object function is to accelerate to reduce reservoir using quadratic sum expression Deviation between output and intake area water requirement.

2. constraints is set

Be available for water inventory constraints single storehouse-multiple station systems year including directly mend storehouse, consider higher level's reservoir call in outside itself Single reservoir operation criterion constraints of water diversion volume requirement, single reservoir capacity constraints, and mend storehouse Group of Pumping Station operation energy consumption Least commitment condition.

2nd, model solution

1. data prepare, specifically include：N number of period was divided into by 1 year, and determines day part length；Determine the initial storehouse of each reservoir Hold V_{H, 0}；Determine that each reservoir year is available for water inventory SK_h, minimum capacity of a reservoir V_{H, min}, storage capacity V corresponding to flood control_{H, P}And Xing Li Storage capacity V_{H, min}+Δ_{H, 1}；Measure and calculate each benefit storehouse reservoir day part and carry out water LS_{H, i}, evaporation with leakage EF_{H, i}, h=1,2 ... R；Determine that each benefit storehouse pumping plant year allows water lift total amount BZ_{H, k}, k=1,2 ... M；Determine different periods lift H_{H, k, i}Lower operation carries Water-carrying capacity Q_{H, k, i}And corresponding pump efficiency η_{Z, h, k, i}, electric efficiency η_{Mot, h, k}, transmission efficiency η_{Int, h, k}；Determine day part respectively by The water demand of crop YS in pool_{H, i}, h=1,2 ... R；I=1,2 ..., N.

2. direct " more storehouse-multistations " large-scale system model for mending storehouse is decomposed into the R single storehouse-multistation water resource optimizations for directly mending storehouse Mathematical modeling is configured, object function is：

<mrow> <msub> <mi>F</mi> <mi>h</mi> </msub> <mo>=</mo> <mi>m</mi> <mi>i</mi> <mi>n</mi> <mi> </mi> <msub> <mi>Z</mi> <mi>h</mi> </msub> <mo>=</mo> <mi>m</mi> <mi>i</mi> <mi>n</mi> <munderover> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <msup> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>YS</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow>

3. the single storehouse-multistation water resource optimal allocation mathematical modeling in pair directly benefit storehouse optimizes, will without outer water diversion volume from final stage Single storehouse-the multiple station systems asked set out, and carry out single storehouse-multiple station systems water resource optimal allocation successively, specifically include：

(1) single seat reservoir-multiple station systems water resource optimal allocation submodel is solved using one-dimensional dynamic programming, when obtaining each Duan Shuiku optimal water supply processes G_{H, i}, it is optimal to abandon process water PS_{H, i}, and mend the optimal moisturizing total amount process Y of storehouse Group of Pumping Station_{H, i}, H=1,2 ... R, i=1,2 ... N；

(2) solved using decomposition-Dynamic Programming polymerization to mending storehouse Group of Pumping Station two level subsystem model, obtain each benefit storehouse pumping plant The optimal rate of water make-up YB of day part_{H, k, i} ^*；

4. obtain each reservoir day part output G in more reservoir-more pumping station systems_{H, i}, outer water diversion volume WD_{H, i}, abandon water PS_{H, i}、 And corresponding each benefit storehouse pumping plant rate of water make-up YB_{H, k, i}Process, h=1,2 ..., R.

2. according to the method for claim 1, it is characterised in that the constraints in step (1) includes：

(1) that directly mends storehouse is available for water inventory constraints in single storehouse-multiple station systems year：In varying level year difference fraction situation Under, consideration needs water requirement, the water that water supply project provides；Wherein, in addition to final stage reservoir only needs to supply water to place intake area, remaining Reservoirs at different levels also need to undertake outer water transfer task, supply next stage reservoir, i.e.,：

To the 1st~the R-1 seat reservoirs：

To Building R reservoir：

In formula：WD_{H, i}To assign the output of next stage reservoir, unit outside h seats the i-th period of reservoir：Ten thousand m³；SK_hFor h seat water The year in storehouse is available for water inventory, unit：Ten thousand m³；M be h seat reservoirs benefit storehouse pumping plant quantity, unit：Seat；K compiles to mend storehouse pumping plant Number, k=1,2 ... M；BZ_{H, k}Storehouse pumping plant year permission water lift total amount, unit are mended for the kth seat of h seat reservoirs：Ten thousand m³；Remaining Variable implication is same as above.

(2) consider that higher level's reservoir calls in single reservoir operation criterion constraints with itself outer water diversion volume requirement：For any h Individual more pumping plants directly mend single water reservoir system in storehouse, it also is contemplated that the outer water diversion volume requirement of reservoir, according to water balance equation：

On this basis, reservoir operation criterion is as follows：

1. when the i-th period end pondage is less than reservoir minimum capacity of a reservoir V_{H, min}When, then the i-th period should by Group of Pumping Station give reservoir mend Water, Δ more than moisturizing to utilizable capacity, that is, reservoir minimum capacity of a reservoir_{H, 1}, i.e.,：

V_{H, i}＜ V_{H, min}When：Y_{H, i}=V_{H, min}-V_{H, i}+Δ_{H, 1} (6)

This period reservoir abandons water PS_{H, i}=0.

2. when meeting with flood, the i-th period end pondage is more than pondage V corresponding to flood control_{H, P}When, then I-th period should carry out abandoning water to reservoir, abandon water to flood control by reservoir regulation limiting water level, i.e.,：

V_{H, i}＞ V_{H, P}When：PS_{H, i}=V_{H, i}-V_{H, P} (7)

This period Group of Pumping Station rate of water make-up Y_{H, i}=0.

3. when the i-th period end pondage is between minimum capacity of a reservoir V_{H, min}With flood control corresponding to pondage V_{H, P}It Between, then the i-th period reservoir need not abandon water, and Group of Pumping Station is also without moisturizing, i.e.,：

V_{H, min}≤V_{H, i}≤V_{H, P}When：Y_{H, i}=PS_{H, i}=0 (8)

In formula：V_{H, i}、V_{H, i-1}Respectively h seats reservoir i-th, the reservoir storage of i-1 period Mos, unit：Ten thousand m³；Y_{H, i}For h seat water The Group of Pumping Station of the period of storehouse i-th mends storehouse water, unit：Ten thousand m³, to the 2nd~the Building R reservoir, consider that it passes through comprising higher level's reservoir Mend storehouse pumping plant and feed its benefit storehouse water；LS_{H, i}、PS_{H, i}、EF_{H, i}Respectively the i-th period of h seats reservoir come water, abandon water, Evaporation and leakage, unit：Ten thousand m³；V_{H, min}、V_{H, P}Respectively corresponding to the minimum capacity of a reservoir of h seats reservoir and flood control Storage capacity, unit：Ten thousand m³。

(3) single reservoir capacity constraints：The pondage of day part should be between reservoir minimum capacity of a reservoir and flood control pair Between answering storage capacity, i.e.,：

V_{H, min}≤V_{H, i}≤V_{H, P}, i=1,2 ..., N (9)

(4) storehouse 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 the benefit storehouse Group of Pumping Station of any h seat reservoirs, its cooperation within each water supply period should Consider that energy consumption is minimum, i.e.,：

<mrow> <msub> <mi>L</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>=</mo> <munderover> <mo>&Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>M</mi> </munderover> <msub> <mi>l</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>k</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>=</mo> <munderover> <mo>&Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>M</mi> </munderover> <mfrac> <mrow> <msub> <mi>&rho;gQ</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>k</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <msub> <mi>H</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>k</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> </mrow> <mrow> <mn>1000</mn> <msub> <mi>&eta;</mi> <mrow> <mi>z</mi> <mo>,</mo> <mi>h</mi> <mo>,</mo> <mi>k</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <msub> <mi>&eta;</mi> <mrow> <mi>m</mi> <mi>o</mi> <mi>t</mi> <mo>,</mo> <mi>h</mi> <mo>,</mo> <mi>k</mi> </mrow> </msub> <msub> <mi>&eta;</mi> <mrow> <mi>int</mi> <mo>,</mo> <mi>h</mi> <mo>,</mo> <mi>k</mi> </mrow> </msub> </mrow> </mfrac> <msub> <mi>&Delta;T</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>k</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>=</mo> <mi>m</mi> <mi>i</mi> <mi>n</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow>

In formula, L_{H, i}Storehouse Group of Pumping Station combined operation system energy consumption, unit are mended for h seats the i-th period of reservoir：kW·h；l_{H, k, i}For to Kth seat the i-th period of pumping plant operation energy consumption of h seat reservoir moisturizings, unit：kW·h；M is the benefit storehouse pumping plant number of h seat reservoirs Amount, unit：Seat；ρ is water density, unit：kg/m³, g is acceleration of gravity, unit：m/s²；Q_{H, k, i}、H_{H, k, i}、ΔT_{H, k, i}、 η_{Z, h, k, i}Respectively to h seat reservoir moisturizings kth seat the i-th period of pumping plant flow (m³/ s), when equal lift (m), Period Length And pump efficiency (h)；η_{Mot, h, k}、η_{Int, h, k}Respectively to h seat reservoir moisturizings kth seat pumping plant motor efficiency and transmission Efficiency.

3. according to the method for claim 1, it is characterised in that in step (2), the single single storehouse-multistation water for directly mending storehouse The constraints of most optimum distribution of resources mathematical modeling is as follows：

1. for the 1st~the R-1 seat reservoirs, in addition to being supplied water to intake area, there is the outer output requirement for assigning subordinate's reservoir, because This：

<mrow> <munderover> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>WD</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&le;</mo> <msub> <mi>SK</mi> <mi>h</mi> </msub> <mo>+</mo> <munderover> <mo>&Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>M</mi> </munderover> <msub> <mi>BZ</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>k</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow>

In formula, h=1,2 ... R-1.

<mrow> <munderover> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <msub> <mi>G</mi> <mrow> <mi>R</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>&le;</mo> <msub> <mi>SK</mi> <mi>R</mi> </msub> <mo>+</mo> <munderover> <mo>&Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>M</mi> </munderover> <msub> <mi>BZ</mi> <mrow> <mi>R</mi> <mo>,</mo> <mi>k</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow>

2. for Building R reservoir, it is according to water balance equation：

V_{R, i}=V_{R, i-1}+LS_{R, i}+Y_{R, i}-PS_{R, i}-EF_{R, i}-G_{R, i} (15)

The same formula of reservoir operation criterion (6)~(8) on this basis.

(3) single reservoir capacity constraints：Same formula (9).

4. according to the method for claim 1, it is characterised in that the 3rd step of step (2) comprises the following steps that：

1) the one-dimensional dynamic programming evaluation of subsystem

1. stage i=1：

f₁(λ₁)=min (G_{R, 1}-YS_{R, 1})² (16)

In formula, λ₁For state variable, 1 period reservoir yield, unit before expression：Ten thousand m³, it is in corresponding feasible zone by certain Step-length is discrete：To each discrete λ₁, decision variable G_{R, 1}Pressed in corresponding feasible zone One fixed step size is discrete, then will meet G₁≥λ₁It is required that G₁Formula (16) is substituted into respectively, determines each discrete λ respectively₁It is optimal during value G_{R, 1}And its corresponding period minimum water deficit quadratic sum f₁(λ₁)。

Then, according to formula (15), the 1st stage end reservoir capacity V_{R, 1}=V_{R, 0}+LS_{R, 1}-EF_{R, 1}-G_{R, 1}, now not yet consider pumping plant Group's moisturizing or reservoir abandon water, should be tested and be corrected using formula (6)~(8)：

I) V is worked as_{R, 1}＜ V_{R, min}, then consider to carry out moisturizing, moisturizing total amount Y to reservoir_{R, 1}=V_{R, min}-V_{R, 1}+Δ_{R, 1}, now correct storehouse Hold V_{R, 1} ^*=V_{R, min}+Δ_{R, 1}。

Ii V) is worked as_{R, 1}＞ V_{R, P}, then need to abandon water to ensure the dispatching requirement of reservoir capacity, PS_{R, 1}=V_{R, 1}-V_{R, P}, now correct storehouse Hold V_{R, 1} ^*=V_{R, P}。

Iii V) is worked as_{R, min}≤V_{R, 1}≤V_{R, P}, then V_{R, 1}=PS_{R, 1}=0, the now amendment storage capacity V of the period Mo of Building R reservoir the 1st_{R, 1} ^* =V_{R, 1}。

By step i)~iii), correct and determine the 1st stage end reservoir amendment storage capacity V_{R, 1} ^*, while corresponding reservoir can be obtained Abandon water PS_{R, 1}Or Group of Pumping Station moisturizing total amount Y_{R, 1}。

2. stage i=2,3 ... N-1：

<mrow> <msub> <mi>f</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>&lambda;</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mi>m</mi> <mi>i</mi> <mi>n</mi> <mo>&lsqb;</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>R</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>YS</mi> <mrow> <mi>R</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>f</mi> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mrow> <mo>(</mo> <msub> <mi>&lambda;</mi> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>17</mn> <mo>)</mo> </mrow> </mrow>

In formula, state variable λ_iFor the reservoir water supply total amount of preceding i period, unit：Ten thousand m³, it is equally carried out respectively discrete：To each discrete λ_i, decision variable G_{R, i}It is discrete to be same as above, and should meet：

State transition equation：

In formula：I=2,3 ..., N-1

To each discrete λ_i, by each discrete G_{R, i}(the G that value is substituted into formula (17) respectively_{R, i}-YS_{R, i})², by state transfer side Formula (18), to each discrete G_{R, i}, before lookup the i-1 stages meetIt is required that minimum f_i-1(λ_i-1) value, by This can obtain (a G_{R, i}-YS_{R, i})²+minf_i-1(λ_i-1), complete all of above discrete G_{R, i}After optimizing, it can finally be expired Sufficient min [(G_{R, i}-YS_{R, i})²+f_i-1(λ_i-1)] require preceding i period subsystem minimum water deficit quadratic sum f_i(λ_i) value, and its Corresponding each stage reservoir optimal water supply G_{R, i}, i=1 ... i.

Equally, also need to determine the i-th stage end storage capacity V using formula (15)_{R, i}, then tested using formula (6)~(8), amendment is true Fixed i-th period end storage capacity V_{R, i} ^*, while corresponding reservoir can be obtained and abandon process water PS_{R, i}And Group of Pumping Station moisturizing total amount process T_{R, i}, i=1,2 ... i, process is with step i)~iii).

3. stage N：

<mrow> <msub> <mi>f</mi> <mi>N</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>&lambda;</mi> <mi>N</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mi>m</mi> <mi>i</mi> <mi>n</mi> <mo>&lsqb;</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>R</mi> <mo>,</mo> <mi>N</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>YS</mi> <mrow> <mi>R</mi> <mo>,</mo> <mi>N</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>f</mi> <mrow> <mi>N</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mrow> <mo>(</mo> <msub> <mi>&lambda;</mi> <mrow> <mi>N</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>19</mn> <mo>)</mo> </mrow> </mrow>

State variableDecision variable G_{R, N}It is equally discrete in corresponding feasible zone, it should meet：λ_N-1=λ_N-G_{R, N}。

Using step, 2. methods described, final obtain meet the λ_NIt is required that the optimal water supply process G of reservoir_{R, i}, i=1 ... N, correspond to Reservoir abandon process water PS_{R, i}, Group of Pumping Station moisturizing total amount process Y_{R, i}, i=1,2 ... N, and the submodel target letter Number optimal value F_R=f_N(λ_N)。

Consider to mend storehouse Group of Pumping Station cooperation energy consumption least commitment condition, established by formula (10) and mend storehouse Group of Pumping Station economical operation mathematics Model：

Object function：

Period output constrains：

Power constraint：

In formula, N_{R, k, 0}The electric drilling match power of storehouse pumping plant, unit are mended for the kth seat of Building R reservoir：kW；Remaining variables implication is same On.

I) two level subsystem decomposes：

Object function：

Power constraint：

Ii) three-level submodel energy consumption determines：

For model above (23)~(24), segment length's Δ T during the i-th period_{R, i}, it is known that and water lift is determined by pumping plant water levels of upstream and downstream Lift H_{R, i}, and its corresponding Q_{R, i}、η_{Z, R, i}、η_{Mot, R}And η_{Int, R}, and the period maximum rate of water make-up is YB_{R, i, max}=3600Q_{R, i}Δ T_{R, i}/ 10000, by the discrete period maximum water lift amount YB of a fixed step size_{R, i, max}, obtain each water lift amount YB_{R, i, m}Place an order station operation energy Consume l_{R, i, m}, m=1,2 ... max.

To other pumping plants of the reservoir, equally using above method, each pumping plant difference water lift amount YB is derived from_{R, k, i, m}Under, single station Operation energy consumption l_{R, k, i, m}, k=1,2 ... M, m=1,2 ... max.

Iii) former two level subsystem Dynamic Programming polymerization：

Solved by above three-level subsystem, to each benefit storehouse pumping plant, can obtain a series of single station water lift energy consumption l_{R, k, i, m}~mono- Stand and mend storehouse water YB_{R, k, i, m}Relation, k=1,2 ... M, m=1,2 ... max, it thus can build following polymerization model and substitute original two Level submodel (20)~(22)：

Object function：

Period water quantity restraint：

Polymerization model (25)~(26) are similarly one-dimensional dynamic programming model, and stage variable is the benefit storehouse pumping plant of Building R reservoir Numbering k, k=1,2 ... M；Decision variable is each i-th period of pumping plant water lift amount YB_{R, k, i}, when its discrete range is single station optimization Target water discrete range YB_{R, k, i, m}, k=1,2 ... M, m=1,2 ... max；Each pumping plant water lift total amount is understood by formula (26) Centrifugal pump be state variable λ, solve the model with reference to one-dimensional dynamic programming, obtain and meet that the i-th period pumping plant target carries Water inventory Y_{R, i}L_{R, i}Value, and the optimal rate of water make-up combination YB of corresponding each pumping plant_{R, k, i} ^*, k=1,2 ... M.

The day part obtained by step 1) mends the optimal moisturizing total amount Y of storehouse Group of Pumping Station_{R, i}After process, each period is determined Y_{R, i}, altogether need to be via n times step 2), further by day part Y_{R, i}It is each to obtain each benefit storehouse pumping plant to M benefit storehouse pumping plant for optimization distribution Period optimal rate of water make-up YB_{R, k, i} ^*；Thus the difference of the supply and demand water of day part of R levels list storehouse-in multiple station systems year is finally obtained Least square and F_R, corresponding reservoir day part optimal water supply G_{R, i}, abandon water process PS_{R, i}, and each benefit storehouse pumping plant day part Optimal rate of water make-up process YB_{R, k, i} ^*, i=1,2 ... N, k=1,2 ... M.

Single storehouse-multiple station systems in storehouse are directly mended for R-1, R-2 ..., 1, equally using submodel (11)~(15), (6) ~(10), it is with reservoir day part output G_{H, i}It can divide one-dimensional nonlinear model for the stage of decision variable, with reference to step (1), Solved using one-dimensional dynamic programming；Differ only in：

Reservoir, which supplies water, should supply the output G of intake area_{H, i}, it is also contemplated that the outer output WD for assigning next stage reservoir_{H, i}Cause Element, then it is available for water inventory constraint to use formula (12) in single storehouse-multiple station systems year, i.e.,： Due to its outer water diversion volume WD_{H, i}It is to mend storehouse pumping plant by a certain seat in the reservoir of Building R to provide：

WD_{H, i}=YB_{H+1,1, i} (27)

<mrow> <munderover> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>N</mi> </munderover> <mrow> <mo>(</mo> <msub> <mi>G</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>YB</mi> <mrow> <mi>h</mi> <mo>+</mo> <mn>1</mn> <mo>,</mo> <mn>1</mn> <mo>,</mo> <mi>i</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&le;</mo> <msub> <mi>SK</mi> <mi>h</mi> </msub> <mo>+</mo> <munderover> <mo>&Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>M</mi> </munderover> <msub> <mi>BZ</mi> <mrow> <mi>h</mi> <mo>,</mo> <mi>k</mi> </mrow> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>28</mn> <mo>)</mo> </mrow> </mrow>

Outer water diversion volume WD is considered as to each stage water balance equation_{H, i}Factor, formula (14) should be used, i.e.,：V_{H, i}=V_{H, i-1}+LS_{H, i} +Y_{H, i}-PS_{H, i}-EF_{H, i}-G_{H, i}-WD_{H, i}, with reason formula (27), formula (14) is converted into following form：