CN103632031B - A kind of rural area based on load curve decomposition load type load modeling method - Google Patents

Abstract

The present invention provides a kind of rural area based on load curve decomposition load type load modeling method, said method comprising the steps of: decomposes rural area load curve, and calculates the ratio of total amount shared by every kind of load in the load of rural area；Calculating static load equivalent parameters, dynamic load equivalent parameters and distribution network system impedance, and combine the power supply area network topology data of rural area load bus, obtain rural area load type load model.The present invention provides a kind of rural area based on load curve decomposition load type load modeling method, the shortcoming the method overcoming conventional statistics synthesis, can be resident load Site Modeling quickly, easily and accurately, the load model generated by the present invention can improve the degree of accuracy that grid simulation calculates, and ensures power grid security, reliably, run economically.

Description

A kind of rural area based on load curve decomposition load type load modeling method

Technical field

The present invention relates to a kind of modeling method, a kind of rural area based on load curve decomposition load type load modeling Method.

Background technology

Power system digital simulation oneself become the main tool of planning and design of power system, management and running and analysis and research, power train The Mathematical Modeling of each element of uniting and the total system Mathematical Modeling being made up of it are the bases of power system digital simulation, the standard of model Whether true directly affect simulation result and decision scheme based on this.Model and parameter that emulation is used are accuracy of simulation Important decisive factor, generation current machine, excitation system, governing system, transformer, the detail mathematic model of transmission line of electricity and Modeling technique has been obtained for well developing, and electric load model is the simplest comparatively speaking, often from basic physical concept The practical model set out and use and parameter.For many years, each bulk power grid of China when Electrical power system analysis and computing, generally according to Experience selectes certain common load model (such as motor+constant-impedance model or firm power+constant impedance model) qualitatively Determine model parameter.

Power system load modeling, due to factors such as its complexity, distributivity, time variation and randomnesss, determines its mathematical modulo The difficulty that type is set up.Current load modeling method can be basically divided into 3 classes, i.e. Component Based, totally distinguish survey method and fault Fitting process.These three method is respectively arranged with its pluses and minuses.The load model obtained with Component Based has clear physics conception, is prone to The advantage understood by engineering staff, but its core is built upon on the basis of " statistics is complete, and part throttle characteristics is accurate ", This point is often difficult to, and can not often add up, thus cannot consider the time dependent characteristic of load.Always Body examination distinguishes that method avoids substantial amounts of statistical work, it is possible to obtain time dependent online Real-time Load characteristic, asking of its maximum Topic is excessively to rely on disturbance accident, and the model parameter physical significance of identification is indefinite, and another problem is to be difficult in systems All transformer stations are fitted with closing device.The advantage of fault fitting method is parameter determination process and Selection parameter during the calculating of present program Process consistent, and reproduction can be obtained under some fault.But actually it is a kind of method that examination is gathered, under some fault Load parameter whether be applicable to other faults and be difficult to ensure that, and think that total system load parameter is identical, constant and obviously do not meet The essence of load.In three of the above method, Component Based is owing to its physical model is clear, definite conception, it is simple to qualitative understanding is born Lotus characteristic, is widely used.

But there is certain shortcoming in traditional Component Based, including:

(1) to obtain load capacity not consistent with actual load power for inquiry agency, because there is the problem of simultaneity factor, and not all sets Standby is all to come into operation for 24 hours, accordingly, it would be desirable to carry out investigation statistics at times；

(2) As time goes on, actual load power, load structure and network structure all it may happen that change, if right Load is just thought once and for all after carrying out an investigation statistics modeling work, it is difficult to reach accuracy requirement；

(3) investigation work need to add up load composition and the parameter of thousands of users, and workload is huge, and is difficult to obtain standard True statistics.

The random time-dependent of synthetic load constituent is the essential reason that its part throttle characteristics has stochastic time-dependent.This time variation must The daily load curve so causing power consumer and transformer station has time variation.Therefore, the daily load curve of user and transformer station is inevitable Abundant information containing reflection load structure characteristic.

According to the convention of industry load classification, load can be divided into industrial load, Commercial Load and city dweller's load, rural area Load and this 4 class of other loads.Wherein rural area load type is one of more complicated load type, is also important bearing Lotus type, occupies larger specific gravity in each load type website.Rural area load curve has a characteristic that

Basic characteristics containing the rural area load irrigating load are as follows:

(1) rural area load changed also with season and the change of time, contained substantial amounts of irrigation load on spring and daytime in autumn, Contain certain heating load winter, and contain certain cooling load summer, but heating load and cooling load ratio are the most not Greatly.

(2) contain bigger irrigation load in spring and autumn, be the main feature of rural area load.Irrigate load and be generally present in white It the morning and afternoon, 3,4, May and 9,10, November irrigate the large percentage that load occupies；

(3) rural area load is smaller, and the regularity of load curve is not strong, and the proportion that special load curve occupies is higher.

(4) ratio that different rural area charge circuit irrigation loads occupies has certain difference.

(5) for the load curve of every day, peak load is generally present in evening, and the irrigation load on part circuit daytime is more, It is likely to occur by day.For load in evening, the load in general spring and autumn is relatively low, and the load in winter and summer is higher. For the load on daytime, the duty ratio in February is relatively low, other in month load the highest.

By to user and the signature analysis of transformer station's daily load curve, just can determine that each typical case's electricity consumption industry corresponding is used Electricity Constitution ratio of plant example and the electricity consumption industry composition of transformer station.Due to data acquisition and monitoring (SCADA) system and load Control management system (being called for short negative Ore-controlling Role) can provide the daily load curve of real-time transformer station's synthetic load and electric power to use respectively The daily load curve at family, therefore, the synthetic load composition thereby determined that will have online, the character of Real-time modeling set, thus Can fundamentally overcome the inherent shortcoming of conventional statistics synthesis load modeling.

Summary of the invention

In order to overcome above-mentioned the deficiencies in the prior art, the present invention provides a kind of rural area based on load curve decomposition load type load Modeling method, the shortcoming the method overcoming conventional statistics synthesis, can be resident load website quickly, easily and accurately Modeling, the load model generated by the present invention can improve the degree of accuracy that grid simulation calculates, and ensures power grid security, reliable, economic Ground runs.

In order to realize foregoing invention purpose, the present invention adopts the following technical scheme that:

The present invention provides a kind of rural area based on load curve decomposition load type load modeling method, and described method includes following step Rapid:

Step 1: decompose rural area load curve, and calculate the ratio of total amount shared by every kind of load in the load of rural area；

Step 2: calculate static load equivalent parameters, dynamic load equivalent parameters and distribution network system impedance, and it is negative to combine rural area The power supply area network topology data of lotus node, obtain rural area load type load model.

Described step 1 comprises the following steps:

Step 1-1: decompose rural area load curve；

Step 1-1-1: to the transformer station that supplied load type is rural area load, merges this transformer station 10kV or 6kV load outlet Daily load curve；

Step 1-1-2: calculate this transformer station's load Payload number of days monthly, if certain day load has null value, it is believed that this day nothing Effect, if certain day load has opposite sign phenomenon, be also considered as this day invalid；

Step 1-1-3: calculate this transformer station's load average daily load curve monthly and average monthly load curve；

Step 1-1-4: calculating every daily load curve period by day monthly, 6～the maximum of 17, and from February Average daily load curve selects the maximum of daytime period, and determines the load curve scope meeting reference load curve requirement, In the range from the maximum 0.8 of the daytime period calculated again to 1.15 times；

Step 1-1-5: start the load curve in all months is decomposed from February, if certain the daily load song of certain month The maximum of line period by day meets the daily load curve average value ranges that reference load curve requires, then it is assumed that this daily load is bent Line is the curve meeting reference load curve requirement, selects to decompose the curve meeting reference load curve requirement in month, calculates song The averaged curve of line, as reference load curve；

Step 1-1-6: determine the load curve mean value lower limit of satisfied big load curve requirement, this lower limit be daytime period 1.1 times of big value, if the maximum of certain daily load curve period by day is more than this lower limit, then it is assumed that this load song Line is the curve meeting big load curve requirement；

Step 1-1-7: determine the big load curve of this month, and calculate the mean value of datum curve load；

Step 1-2: calculate the ratio of total amount shared by every kind of load in the load of rural area；

Step 1-2-1: calculate each transformer station monthly average load value monthly and specify month to specify the average load value of day；

Step 1-2-2: calculate and specify month, specify day, the load value in appointment moment；

Step 1-2-3: calculate and include irrigating load, cooling load and the separate section load proportion k of heating load_p0And base Quasi-load proportion k_f。

In described step 1-1-3, calculate this transformer station's load average daily load curve monthly, all effective day monthly is born Lotus curve is added, the more effective number of days divided by this month；

Calculate this transformer station's load average monthly load curve monthly, by the load phase in 24 moment in monthly daily load curve every day Add again divided by the curve obtained by 24；In average monthly load curve, each effective day corresponding value is also referred to as daily load curve mean value, Daily load curve mean value is that the load in 24 moment of every day is added, then divided by the mean value obtained by 24；

In described step 1-1-5, if the number meeting the load curve of datum curve this month is less than 3, and above do not have There is the datum curve calculating any one month, then first do not calculate the curve of this month, carry out the calculating of next month curve；As Really this month meets the curve number of datum curve requirement more than or equal to 3, calculates datum curve, and all by this month meet benchmark The daily load curve of curve requirement is added, then meets the curve number of datum curve requirement, calculated datum curve divided by this month Smaller load curve as this month；Otherwise, the datum curve of month is used；

In described step 1-1-7, determine the big load curve of this month, if the load meeting big load curve requirement this month is bent Number of lines is less than 3, then ignore these load curves, do not calculate big load curve this month, does not carry out load curve and separates meter Calculate；If the load curve bar number meeting big load curve requirement this month is more than or equal to 3, calculate big load curve；

Calculate the mean value of datum curve load, big load curve, reference load curve are carried out standardization, average by three Value becomes 1.

In described step 1-2-1, calculate each transformer station monthly average load value monthly, by each in average monthly load curve monthly The value that effectively it is corresponding is added, the more effective number of days divided by this month；

Calculate the average load value specifying month to specify day, the monthly average load value in this month calculated is multiplied by the day specifying day The perunit value of load curve mean value；

In described step 1-2-2, calculate and specify month, specify day, the load value in appointment moment；

1) if only Smaller load curve, then Smaller load curve is used to calculate；

2) if there is big load curve and Smaller load curve simultaneously, the daily load curve mean value size of day is specified according to this month Determine that the big load curve of employing or Smaller load curve calculate；

Setting value is 1.1 times of the reference load mean value in certain month, then have:

If 2-1) the daily load curve mean value of this month is less than this definite value, Smaller load curve is used to calculate；

If 2-2) daily load curve mean value is more than this definite value, using big load curve to calculate, big carry calculation result is The reference load mean value in this month calculated is multiplied by specifies little/big load curve middle finger timing of day to carve corresponding value in this month；

Also need to calculate and specify month, appointment day, the segregational load value in appointment moment, calculate big load curve and reference load curve Difference, as separate after load curve, this value equal to calculate specify month, specify day, specify the moment load value Deduct the reference load mean value in this month to be multiplied by and specify the Smaller load curve middle finger timing of day to carve corresponding value in this month, then divided by Calculate specifies month, appointment day, the load value in appointment moment；

In described step 1-2-3, calculate separate section load proportion k_p0, by segregational load value divided by the appointment moon calculated Part, appointment day, the load value in appointment moment；Then reference load proportion k_fFor 1-k_p0。

In described step 2, it is as follows that static load equivalent parameters calculates process:

By the multinomial static load model that the relationship description between load power and voltage is polynomial equation form, have

P=P₀[a×(V/V₀)²+b×(V/V₀)+c] (1)

Q=Q₀[α×(V/V₀)²+β×(V/V₀)+γ] (2)

Wherein, P and Q is respectively active power and the reactive power of static load, and V is the real-time voltage of static load, V₀For The rated voltage of static load, P₀And Q₀It is respectively and represents at V₀The specified active power of lower static load and reactive power, a, B and c is the active power coefficient of multinomial static load model, and α, β and γ are the idle of multinomial static load model Power coefficient；

The equivalence of static load is mainly a, b, c, P₀With α, β, γ, Q₀Equivalence, to multinomial static load The equivalence of model is based on the sensitivity to static load terminal voltage of the static load power, has

$\frac{\∂P}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}\frac{{\∂P}_i}{\∂V}{|}_{V=V_0}---\left(3\right)$

$\frac{\∂Q}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}\frac{{\∂Q}_i}{\∂V}{|}_{V=V_0}---\left(4\right)$

Wherein, n is static load number, P_iAnd Q_iIt is respectively the meritorious of i-th static load and reactive power,With It is respectively i-th static load active power relative to the partial differential of voltage and i-th static load reactive power relative to voltage Partial differential；

Work as V=V₀Time, have:

$P_0=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}---\left(5\right)$

$Q_0=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}---\left(6\right)$

Wherein, P_0iAnd Q_0iIt is respectively initial active power and the reactive power of i-th static load；

$a=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{a}_i}{P_0}---\left(7\right)$

$b=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{b}_i}{P_0}---\left(8\right)$

$c=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{c}_i}{P_0}---\left(9\right)$

Wherein, a_i、b_iAnd c_iIt is the multinomial static load model active power coefficient of i-th static load；

$\mathrm{\α}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\α}}_i}{Q_0}---\left(10\right)$

$\mathrm{\β}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\β}}_i}{Q_0}---\left(11\right)$

$\mathrm{\γ}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\γ}}_i}{Q_0}---\left(12\right)$

Wherein, α_i、β_iAnd γ_iIt is the multinomial static load model reactive power coefficient of i-th static load.

Described step 2 specifically includes following steps:

Step A: calculate all motor always absorbs active-power P_ΣWith reactive power Q_Σ, total electromagnetic power P_Σem, total rotor Winding copper loss P_Σcu2With total maximum electromagnetic power P_{Σem_max}, specifically have:

$P_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_j$

$Q_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{Q}_i$

$P_{\mathrm{\Σem}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{emj}}---\left(13\right)$

$P_{\mathrm{\Σcu}2}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{cu}2j}$

$P_{\mathrm{\Σem}\_\max }=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{em}\_\max j}$

Wherein P_j、Q_j、P_emj、P_cu2jAnd P_{em_maxj}Respectively represent the active power of jth platform equivalence motor, reactive power, Electromagnetic power, rotor windings copper loss and maximum electromagnetic power；M is equivalent motor number of units；

Step B: calculate the stator winding copper loss P of equivalent motor_Σcu1And the slippage S of equivalence motor, it is expressed as:

P_Σcu1=P_Σ-P_Σem(14)

S=P_Σcu2/P_Σem(15)

Initialize the maximum electromagnetic power P of equivalent motor_{emt_max}, make P_{emt_max}=P_{Σem_max}；

Step C: calculate the stator winding resistance R of equivalent motor_s；

First calculate the stator winding phase current of equivalent motorHave

${\stackrel{\·}{I}}_{\mathrm{\Σ}}=-{\left(\frac{P_{\mathrm{\Σ}}+{\mathrm{jQ}}_{\mathrm{\Σ}}}{\sqrt{3}{\stackrel{\·}{U}}_1}\right)}^{*}---\left(16\right)$

Wherein,Set end voltage for motor；

The then stator winding resistance R of equivalent motor_sIt is expressed as:

$R_s=\frac{P_{\mathrm{\Σcu}1}}{3{I_{\mathrm{\Σ}}}^2}---\left(17\right)$

Step D: calculate the equivalent impedance Z of equivalent machine model_deq:

$Z_{\mathrm{deq}}=\frac{{U_1}^2}{P_{\mathrm{\Σ}}-j{Q}_{\mathrm{\Σ}}}---\left(18\right)$

And have

R_deq=real(Z_deq) (19)

X_deq=imag(Z_deq) (20)

Wherein, R_deqAnd X_deqFor corresponding substitutional resistance and equivalent reactance；

Step E: calculate stator winding leakage reactance X of equivalent motor_sWith rotor windings leakage reactance X_r, have

$X_s=X_r=\frac{X}{2}=-\frac{1}{2}\sqrt{{(\frac{{U_1}^2}{2P_{\mathrm{emt}\_\max }}-R_s)}^2-{R_s}^2}---\left(21\right)$

Step F: through iteration stator winding leakage reactance X to equivalent motor_sWith rotor windings leakage reactance X_rIt is modified；

Step G: calculate rotor windings resistance and the excitatory reactance of equivalence of equivalent motor；

According to calculated R_s、X_s、X_rAnd Z_deqIf, K_r=R_deq-R_s, K_x=X_deq-X_s, the most equivalent electronic The rotor windings resistance R of machine_rReactance X excitatory with equivalence_mIt is expressed as:

$\begin{array}{c}{R}_r=\frac{(K_r+{K_x}^2/K_r-\sqrt{{(K_r+{K_x}^2/K_r)}^2-{{4X}_s}^2})S}{2}\\ X_m=\frac{K_r{X}_s+K_x\frac{R_r}{S}}{\frac{R_r}{S}-K_r}\end{array}---\left(23\right)$

Step H: use iterative method to calculate the maximum electromagnetic power of equivalent motor, and it is modified；

1) the maximum electromagnetic power P of the equivalent motor in calculating kth time iteration_{emt_maxk}, it is expressed as

$P_{\mathrm{emt}\_\max k}=\frac{{U_1}^2}{2(R_s+\sqrt{{R_s}^2+X^2})}---\left(24\right)$

Based on Thevenin's theorem, the Thevenin's equivalence impedance Z of equivalent motor_dpIt is expressed as

$Z_{\mathrm{dp}}={\mathrm{jX}}_r+\frac{{\mathrm{jX}}_m(R_s+{\mathrm{jX}}_s)}{R_s+j(X_s+X_m)}---\left(25\right)$

And have, R_dp=real(Z_dp) and X_dp=imag(Z_dp), R_dpAnd X_dpIt is respectively Thevenin's equivalence resistance and equivalence electricity Anti-；

Equivalent motor produces the condition of maximum electromagnetic power:

$R_{\mathrm{pm}}=\frac{R_r}{S_m}=\sqrt{{R_{\mathrm{dp}}}^2+{X_{\mathrm{dp}}}^2}---\left(26\right)$

Wherein, R_pmThe Thevenin's equivalence impedance magnitude of maximum electromagnetic power, S is produced for correspondence_mFor critical slippage；

Thevenin's equivalence open-circuit voltageFor

${\stackrel{\·}{V}}_0=\frac{{\stackrel{\·}{U}}_1}{\sqrt{3}}\×\frac{{\mathrm{jX}}_m}{R_s+j(X_s+X_m)}---\left(27\right)$

Then, the maximum electromagnetic power of the Thevenin's equivalence motor in kth time iteration is represented by

$P_{\mathrm{em}\_\max k}=\frac{3{U_0}^2{R}_{\mathrm{pm}}}{{(R_{\mathrm{dp}}+R_{\mathrm{pm}})}^2+{X_{\mathrm{dp}}}^2}---\left(28\right)$

2) the maximum electromagnetic power of equivalent motor is revised；

Calculate the correction factor τ of kth time iteration medium value motor maximum electromagnetic power_maxk, it is expressed as:

${\mathrm{\τ}}_{\max k}=\frac{P_{\mathrm{emt}\_\max k}}{P_{\mathrm{em}\_\max k}}---\left(29\right)$

The most revised equivalent motor maximum electromagnetic power P_{emt_max}It is expressed as:

P_{emt_max}=τ_maxk×P_{Σem_max}(30)

With P_{emt_max}With P_{emt_maxk}The iteration error that absolute error is equivalent motor maximum electromagnetic powerHave

${\mathrm{Err}}_{P_{\mathrm{em}\_\max }}=|P_{\mathrm{emt}\_\max }-P_{\mathrm{emt}\_\max k}|---\left(31\right)$

WithFor iteration convergence standard, ifThen recalculate X_s、R_r、X_rAnd X_m；

Step I: calculate equivalent inertia time constant；

Equivalent inertia time constant H is expressed as:

$H=\frac{\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{nj}}{H}_j}{\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{nj}}}---\left(32\right)$

Wherein P_njAnd H_jRated power and inertia time constant for jth platform equivalence motor.

In described step 2, distribution network system impedance meter is shown as:

$Z_{\mathrm{eq}}=\left[\underset{f=1}{\overset{M}{\mathrm{\Σ}}}{u}_f^2{(1/Z_f)}^{*}\right]/{\left(\underset{l=1}{\overset{N}{\mathrm{\Σ}}}{I}_l\right)}^2---\left(33\right)$

Wherein, Z_eqFor distribution network system impedance；u_fRepresent busbar voltage, Z_fIndication transformer and distribution line impedance；I_lTable Show load current, wherein, node total number that M and N is respectively in distribution network system and branch road total number.

Power supply area network topology data in described step 2 include distribution line data, transformer data and reactive-load compensation data.

Compared with prior art, the beneficial effects of the present invention is:

1. this method electrically-based system sales department Load Control Management System gathers the daily load curve of power consumer and number According to the daily load curve of the real-time transformer station synthetic load gathered with monitoring (SCADA) system offer, pass through load curve decomposition Obtain any time rural area Load Substation each electrical equipment type of confession and ratio, and the load utilizing traffic department to provide saves The power supply area network topology data of point, use Component Based, it is achieved the online load modeling of rural area load website, to reach It is the purpose of rural area load Site Modeling exactly, improves the degree of accuracy that grid simulation calculates, ensure power grid security, reliable, warp Ji ground runs.

2. this method daily load curve based on actual electric network data, have carried out online load modeling, have changed conventional statistics comprehensive Method relies on the pattern of off-line manual research, can carry out load modeling quick and easy, exactly, be greatly saved human and material resources, And avoiding the inaccurate problem of the load modeling caused owing to investigation result is inaccurate, this method is that tradition modeling pattern is entered one Step promotes, and provides the effect of important guidance for load modeling work.

3. (day such as the power consumer of marketing Load Control Management System collection is negative by diversified online data resource for this method The daily load curve of the real-time transformer station synthetic load that lotus curve and data acquisition and monitoring (SCADA) system provide and scheduling net Network topology information) realize onlineization of Component Based, load bus can be carried out real-time SLM modeling, model physical significance Clearly, strong adaptability, overcome the problem that traditional all load modeling methods are difficult in adapt to load time variation.

Accompanying drawing explanation

Fig. 1 is the load chart that Luohe becomes 35kV rural area load January；

Fig. 2 is the load chart that Luohe becomes 35kV rural area load May；

Fig. 3 is the load chart that Luohe becomes 35kV rural area load July；

Fig. 4 is the load chart that Luohe becomes 35kV rural area load November；

Fig. 5 is the load chart after Luohe becomes the decomposition in January of 35kV rural area load；

Fig. 6 is the load chart after Luohe becomes the decomposition in May of 35kV rural area load；

Fig. 7 is the load chart after Luohe becomes the decomposition in July of 35kV rural area load；

Fig. 8 is to become the load chart after the decomposition in November of 35kV rural area load for Luohe；

Fig. 9 is rural area based on load curve decomposition load type load modeling method flow diagram；

Figure 10 is the comparison diagram of 220kV load busbar voltage curve；

Figure 11 is the comparison diagram of 220kV load bus load active power curves；

Figure 12 is the comparison diagram of 220kV load bus load reactive capability curve.

Detailed description of the invention

Below in conjunction with the accompanying drawings the present invention is described in further detail.

The present invention provides a kind of rural area based on load curve decomposition load type load modeling method, and described method includes following step Rapid:

Step 1: decompose rural area load curve, and calculate the ratio of total amount shared by every kind of load in the load of rural area；

Step 2: calculate static load equivalent parameters, dynamic load equivalent parameters and distribution network system impedance, and it is negative to combine rural area The power supply area network topology data of lotus node, obtain rural area load type load model.

For the load curve basic characteristics of rural area load, the basic handling method of employing is as follows:

(1) smaller due to rural area load, and regular poor, therefore for the rural area load of same transformer station, Merge.

If do not merged, owing to the regularity of rural area load is poor, hardly result in preferable decomposition result.In Fig. 1-Fig. 4 Load curve be exactly that Luohe becomes the load curve after all 35kV rural areas charge circuit merges, load total after merging is relatively big, The Changing Pattern of load is the strongest.

(2) the rural area load curve decomposition containing irrigation load is base load, cooling load, heating load and irrigation load.

Cooling load mainly appears on the 6 of summer, 7, August, heating load mainly appears on the 12 of winter, 1, February, Irrigate load mainly appear on the 9 of the 3 of spring, 4, May and autumn, 10, November.Therefore, when decomposing, isolate Come load, 6,7, August be considered cooling load, 12,1, February be considered heating load, 3,4,5,9, 10, November be considered irrigate load.But the load separated in the part month of season alternation may be simultaneously present two kinds Load, owing to cannot be further continued for decomposing, is therefore approximately considered only exist that a kind of load, such as August separate negative Lotus may be simultaneously present cooling load and irrigates load, and being approximately considered the load separated August is all cooling load.

(3) basic skills of curve is separated: according to daytime period maximum, curve is decomposed, by the load curve of every month Being divided into two load curves, load curve on the basis of, another is big load curve, and the difference of two curves is as separation Load curve.

For the load curve of every month, it is classified as two suite lines according to the size of load, calculates the average bent of this two suites line Line, as typical load curve, load curve during different load level in approximate simulation every month.One therein Curve is base load curve, and this curve represents the change procedure of base load, should be less according to average load in all months Curve calculate.Urban residents' load contains irrigation load, separates the basic skills of basic load curve when being to use daytime Section maximum is decomposed, and selects the curve that daytime period maximum is less as reference load curve.

Decomposing two periods of main consideration, one is daytime period (6-17 point), and another is the period in the evening, reference load curve Should meet daytime load less, evening, load was the least.But knowable to the basic characteristics of rural area load curve, the 1. usual spring The load in season and evening in autumn is less, and the load in summer and evening in winter is relatively big, but that both differ and not very big；The whitest It load changes greatly in Various Seasonal, particularly season in spring and autumn；3. the load curve that load on daytime is less, evening load not Smaller.If selecting load on daytime and the smaller curve of load in evening as reference load curve, then meet benchmark The curve number of curve is less, can compare accurate decomposition base load curve only with daytime period maximum, therefore only with in vain It load maximum is decomposed.

Determine reference load curve daytime period maximum meet scope use two kinds of methods:

Calculate the maximum of all daily load curve daytime period, therefrom select a range of value, calculate the average of them Value；

Owing to daytime period load in February is minimum, and dispersiveness is less, and therefore calculating February, all daily load curves were white The maximum of it period, then calculates its mean value.

On the basis of above-mentioned mean value, the scope that selection reference load curve daytime period peak load meets, for every month Load, as long as daytime period peak load is within the scope of this, it is believed that be basic load curve；More than the maximum of this scope, It is considered big load curve.The average load curve calculating basic load curve is born greatly as Smaller load typical load curve, calculating The averaged curve of lotus curve is as the typical curve of big load curve, and both differences are as freezing, heating or irrigate load.

In two kinds of above-mentioned methods, first method is it needs to be determined that calculate the scope of datum curve mean value, and this scope is relatively difficult to Determine；Second method is relatively easy, and has certain adaptability.

Decompose rural area load curve process specific as follows:

Step 1-1-1: to the transformer station that supplied load type is rural area load, merges this transformer station 10kV or 6kV load outlet Daily load curve；

Step 1-1-2: calculate this transformer station's load Payload number of days monthly, if certain day load has null value, it is believed that this day nothing Effect, if certain day load has opposite sign phenomenon, be also considered as this day invalid；

Step 1-1-3: calculate this transformer station's load average daily load curve monthly and average monthly load curve；

Calculate this transformer station's load average daily load curve monthly, will be added by all effective daily load curve monthly, then remove Effective number of days with this month；

Calculate this transformer station's load average monthly load curve monthly, will the monthly load in 24 moment in daily load curve every day It is added again divided by the curve obtained by 24；In average monthly load curve, to be also referred to as daily load curve average for each effective day corresponding value Value, daily load curve mean value is the load in 24 moment of every day and is added, then divided by the mean value obtained by 24；

Step 1-1-4: calculating every daily load curve period by day monthly, i.e. 6～the maximum of 17, and from February Average daily load curve in select the maximum of daytime period, and determine the load curve scope meeting reference load curve requirement, In the range from the maximum 0.8 of the daytime period calculated again to 1.15 times；

Step 1-1-5: start the load curve in all months is decomposed from February, if certain the daily load song of certain month The maximum of line period by day meets the daily load curve average value ranges that reference load curve requires, then it is assumed that this daily load is bent Line is the curve meeting reference load curve requirement, selects to decompose the curve meeting reference load curve requirement in month, calculates song The averaged curve of line, as reference load curve；

If the number meeting the load curve of datum curve this month is less than 3, and does not the most calculate any one moon The datum curve of part, then first do not calculate the curve of this month, carry out the calculating of next month curve；If this moon meets datum curve The curve number required, more than or equal to 3, calculates datum curve, all of this month will meet the daily load that datum curve requires Curve is added, then meets the curve number of datum curve requirement divided by this month, and calculated datum curve is as the Smaller load of this month Curve；Otherwise, the datum curve of month is used；

Step 1-1-6: determine the load curve mean value lower limit of satisfied big load curve requirement, this lower limit be daytime period 1.1 times of big value, if the maximum of certain daily load curve period by day is more than this lower limit, then it is assumed that this load song Line is the curve meeting big load curve requirement；

Step 1-1-7: determine the big load curve of this month, and calculate the mean value of datum curve load；

Determine the big load curve of this month, if the load curve bar number meeting big load curve requirement this month is less than 3, then Ignore these load curves, do not calculate big load curve this month, do not carry out load curve and separate calculating；If this month meets The load curve bar number that big load curve requires, more than or equal to 3, calculates big load curve；

Calculate the mean value of datum curve load, big load curve, reference load curve are carried out standardization, will three flat Average becomes 1.

Accompanying drawing 5-8 be Luohe become 35kV rural area load 1,5,7, November decompose after load curve.

As shown in accompanying drawing 1-4 and accompanying drawing 5-8, January, load was relatively big, and dispersiveness is less, does not meet reference load curve The load curve required, uses the basic load curve in December, defines a big load curve, and the load separated is Cooling load.Irrigate the ratio that load occupies in May and November more, irrigate load and mainly appear on the morning and afternoon.7 The load separated in month is cooling load.

Calculate the ratio process of total amount shared by every kind of load in the load of rural area as follows:

Step 1-2-1: calculate each transformer station monthly average load value monthly and specify month to specify the average load value of day；

Calculate each transformer station monthly average load value monthly, will each effective day corresponding value phase in average monthly load curve monthly Add, the more effective number of days divided by this month；

Calculating the average load value specifying month to specify day, the monthly average load value in this month that will calculate is multiplied by specifies day The perunit value of daily load curve mean value；

Step 1-2-2: calculate and specify month, specify day, the load value in appointment moment；

1) if only Smaller load curve, then Smaller load curve is used to calculate；

2) if there is big load curve and Smaller load curve simultaneously, the daily load curve mean value size of day is specified according to this month Determine that the big load curve of employing or Smaller load curve calculate；

Setting value is 1.1 times of the reference load mean value in certain month, then have:

If 2-1) the daily load curve mean value of this month is less than this definite value, Smaller load curve is used to calculate；

If 2-2) daily load curve mean value is more than this definite value, using big load curve to calculate, big carry calculation result is The reference load mean value in this month calculated is multiplied by specifies little/big load curve middle finger timing of day to carve corresponding value in this month；

Also need to calculate and specify month, appointment day, the segregational load value in appointment moment, calculate big load curve and reference load curve Difference, as separate after load curve, this value equal to calculate specify month, specify day, specify the moment load value Deduct the reference load mean value in this month to be multiplied by and specify the Smaller load curve middle finger timing of day to carve corresponding value in this month, then divided by Calculate specifies month, appointment day, the load value in appointment moment；

Step 1-2-3: calculate and include irrigating load, cooling load and the separate section load proportion k of heating load_p0And base Quasi-load proportion k_f。

Calculate separate section load proportion k_p0, will segregational load value divided by the appointment month calculated, specify day, appointment The load value in moment；Then reference load proportion k_fFor 1-k_p0。

In described step 2, it is as follows that static load equivalent parameters calculates process:

By the multinomial static load model that the relationship description between load power and voltage is polynomial equation form, have

P=P₀[a×(V/V₀)²+b×(V/V₀)+c] (1)

Q=Q₀[α×(V/V₀)²+β×(V/V₀)+γ] (2)

Wherein, P and Q is respectively active power and the reactive power of static load, and V is the real-time voltage of static load, V₀For The rated voltage of static load, P₀And Q₀It is respectively and represents at V₀The specified active power of lower static load and reactive power, a, B and c is the active power coefficient of multinomial static load model, and α, β and γ are the idle of multinomial static load model Power coefficient；

The equivalence of static load is mainly a, b, c, P₀With α, β, γ, Q₀Equivalence, to multinomial static load The equivalence of model is based on the sensitivity to static load terminal voltage of the static load power, i.e.

$\frac{\∂P}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}\frac{{\∂P}_i}{\∂V}{|}_{V=V_0}---\left(3\right)$

$\frac{\∂Q}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}\frac{{\∂Q}_i}{\∂V}{|}_{V=V_0}---\left(4\right)$

Wherein, n is static load number, P_iAnd Q_iIt is respectively the meritorious of i-th static load and reactive power,With It is respectively i-th static load active power relative to the partial differential of voltage and i-th static load reactive power relative to voltage Partial differential；

Work as V=V₀Time, have:

$P_0=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}---\left(5\right)$

$Q_0=\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}---\left(6\right)$

Wherein, P_0iAnd Q_0iIt is respectively initial active power and the reactive power of i-th static load；

$a=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{a}_i}{P_0}---\left(7\right)$

$b=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{b}_i}{P_0}---\left(8\right)$

$c=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{P}_{0i}{c}_i}{P_0}---\left(9\right)$

Wherein, a_i、b_iAnd c_iIt is the multinomial static load model active power coefficient of i-th static load；

$\mathrm{\α}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\α}}_i}{Q_0}---\left(10\right)$

$\mathrm{\β}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\β}}_i}{Q_0}---\left(11\right)$

$\mathrm{\γ}=\frac{\underset{i=1}{\overset{n}{\mathrm{\Σ}}}{Q}_{0i}{\mathrm{\γ}}_i}{Q_0}---\left(12\right)$

Wherein, α_i、β_iAnd γ_iIt is the multinomial static load model reactive power coefficient of i-th static load.

Described step 2 specifically includes following steps:

Step A: calculate all motor always absorbs active-power P_ΣWith reactive power Q_Σ, total electromagnetic power P_Σem, total rotor Winding copper loss P_Σcu2With total maximum electromagnetic power P_{Σem_max}, specifically have:

$P_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_j$

$Q_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{Q}_i$

$P_{\mathrm{\Σem}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{emj}}---\left(13\right)$

$P_{\mathrm{\Σcu}2}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{cu}2j}$

$P_{\mathrm{\Σem}\_\max }=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{em}\_\max j}$

Wherein P_j、Q_j、P_emj、P_cu2jAnd P_{em_maxj}Respectively represent the active power of jth platform equivalence motor, reactive power, Electromagnetic power, rotor windings copper loss and maximum electromagnetic power；M is equivalent motor number of units；

Step B: calculate the stator winding copper loss P of equivalent motor_Σcu1And the slippage S of equivalence motor, it is expressed as:

P_Σcu1=P_Σ-P_Σem(14)

S=P_Σcu2/P_Σem(15)

Initialize the maximum electromagnetic power P of equivalent motor_{emt_max}, make P_{emt_max}=P_{Σem_max}；

Step C: calculate the stator winding resistance R of equivalent motor_s；

First calculate the stator winding phase current of equivalent motorHave

${\stackrel{\·}{I}}_{\mathrm{\Σ}}=-{\left(\frac{P_{\mathrm{\Σ}}+{\mathrm{jQ}}_{\mathrm{\Σ}}}{\sqrt{3}{\stackrel{\·}{U}}_1}\right)}^{*}---\left(16\right)$

Wherein,Set end voltage for motor；

The then stator winding resistance R of equivalent motor_sIt is expressed as:

$R_s=\frac{P_{\mathrm{\Σcu}1}}{3{I_{\mathrm{\Σ}}}^2}---\left(17\right)$

Step D: calculate the equivalent impedance Z of equivalent machine model_deq:

$Z_{\mathrm{deq}}=\frac{{U_1}^2}{P_{\mathrm{\Σ}}-{\mathrm{jQ}}_{\mathrm{\Σ}}}---\left(18\right)$

And have

R_deq=real(Z_deq) (19)

X_deq=imag(Z_deq) (20)

Wherein, R_deqAnd X_deqFor corresponding substitutional resistance and equivalent reactance；

Step E: calculate stator winding leakage reactance X of equivalent motor_sWith rotor windings leakage reactance X_r, have

$X_s=X_r=\frac{X}{2}=-\frac{1}{2}\sqrt{{(\frac{{U_1}^2}{2P_{\mathrm{emt}\_\max }}-R_s)}^2-{R_s}^2}---\left(21\right)$

Step F: through iteration stator winding leakage reactance X to equivalent motor_sWith rotor windings leakage reactance X_rIt is modified；

Step G: calculate rotor windings resistance and the excitatory reactance of equivalence of equivalent motor；

According to calculated R_s、X_s、X_rAnd Z_deqIf, K_r=R_deq-R_s, K_x=X_deq-X_s, the most equivalent electronic The rotor windings resistance R of machine_rReactance X excitatory with equivalence_mIt is expressed as:

$\begin{array}{c}{R}_r=\frac{(K_r+{K_x}^2/K_r-\sqrt{{(K_r+{K_x}^2/K_r)}^2-{{4X}_s}^2})S}{2}\\ X_m=\frac{K_r{X}_s+K_x\frac{R_r}{S}}{\frac{R_r}{S}-K_r}\end{array}---\left(23\right)$

Step H: use iterative method to calculate the maximum electromagnetic power of equivalent motor, and it is modified；

1) the maximum electromagnetic power P of the equivalent motor in calculating kth time iteration_{emt_maxk}, it is expressed as

$P_{\mathrm{emt}\_\max k}=\frac{{U_1}^2}{2(R_s+\sqrt{{R_s}^2+X^2})}---\left(24\right)$

Based on Thevenin's theorem, the Thevenin's equivalence impedance Z of equivalent motor_dpIt is expressed as

$Z_{\mathrm{dp}}={\mathrm{jX}}_r+\frac{{\mathrm{jX}}_m(R_s+{\mathrm{jX}}_s)}{R_s+j(X_s+X_m)}---\left(25\right)$

And have, R_dp=real(Z_dp) and X_dp=imag(Z_dp), R_dpAnd X_dpIt is respectively Thevenin's equivalence resistance and equivalence electricity Anti-；

Equivalent motor produces the condition of maximum electromagnetic power:

$R_{\mathrm{pm}}=\frac{R_r}{S_m}=\sqrt{{R_{\mathrm{dp}}}^2+{X_{\mathrm{dp}}}^2}---\left(26\right)$

Wherein, R_pmThe Thevenin's equivalence impedance magnitude of maximum electromagnetic power, S is produced for correspondence_mFor critical slippage；

Thevenin's equivalence open-circuit voltageFor

${\stackrel{\·}{V}}_0=\frac{{\stackrel{\·}{U}}_1}{\sqrt{3}}\×\frac{{\mathrm{jX}}_m}{R_s+j(X_s+X_m)}---\left(27\right)$

Then, the maximum electromagnetic power of the Thevenin's equivalence motor in kth time iteration is represented by

$P_{\mathrm{em}\_\max k}=\frac{3{U_0}^2{R}_{\mathrm{pm}}}{{(R_{\mathrm{dp}}+R_{\mathrm{pm}})}^2+{X_{\mathrm{dp}}}^2}---\left(28\right)$

2) the maximum electromagnetic power of equivalent motor is revised；

Calculate the correction factor τ of kth time iteration medium value motor maximum electromagnetic power_maxk, it is expressed as:

${\mathrm{\τ}}_{\max k}=\frac{P_{\mathrm{emt}\_\max k}}{P_{\mathrm{em}\_\max k}}---\left(29\right)$

The most revised equivalent motor maximum electromagnetic power P_{emt_max}It is expressed as:

P_{emt_max}=τ_maxk×P_{Σem_max}(30)

With P_{emt_max}With P_{emt_maxk}The iteration error that absolute error is equivalent motor maximum electromagnetic powerHave

${\mathrm{Err}}_{P_{\mathrm{em}\_\max }}=|P_{\mathrm{emt}\_\max }-P_{\mathrm{emt}\_\max k}|---\left(31\right)$

WithFor iteration convergence standard, ifThen recalculate X_s、R_r、X_rAnd X_m；

Step I: calculate equivalent inertia time constant；

Equivalent inertia time constant H is expressed as:

$H=\frac{\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{nj}}{H}_j}{\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{\mathrm{nj}}}---\left(32\right)$

Wherein P_njAnd H_jRated power and inertia time constant for jth platform equivalence motor.

In described step 2, distribution network system impedance meter is shown as:

$Z_{\mathrm{eq}}=\left[\underset{f=1}{\overset{M}{\mathrm{\Σ}}}{u}_f^2{(1/Z_f)}^{*}\right]/{\left(\underset{l=1}{\overset{N}{\mathrm{\Σ}}}{I}_l\right)}^2---\left(33\right)$

Wherein, Z_eqFor distribution network system impedance；u_fRepresent busbar voltage, Z_fIndication transformer and distribution line impedance；I_lTable Show load current, wherein, node total number that M and N is respectively in distribution network system and branch road total number.

Power supply area network topology data in described step 2 include distribution line data, transformer data and reactive-load compensation data.

Accompanying drawing 10-12 is the simulation curve comparison diagram with measured curve of the SLM Equivalent Model parameter using the present invention to draw

Accompanying drawing 10 is the comparison diagram of 220kV load busbar voltage curve；Wherein solid line is measured curve, and dotted line is for using the present invention Simulation result during the equivalent SLM model parameter drawn；

Accompanying drawing 11 is the comparison diagram of 220kV load bus load active power curves；Wherein solid line is measured curve, and dotted line is for adopting Simulation result during the equivalent SLM model parameter drawn by the present invention；

Accompanying drawing 12 is the comparison diagram of 220kV load bus load reactive capability curve；Wherein solid line is measured curve, and dotted line is for adopting Simulation result during the equivalent SLM model parameter drawn by the present invention.

Technical scheme is applied in the Henan Electric Power System stable model of Henan Electric Power Company is analyzed and researched.

Finally should be noted that: above example is only in order to illustrate that technical scheme is not intended to limit, although reference The present invention has been described in detail by above-described embodiment, those of ordinary skill in the field it is understood that still can to this Invention detailed description of the invention modify or equivalent, and without departing from spirit and scope of the invention any amendment or etc. With replacing, it all should be contained in the middle of scope of the presently claimed invention.

Claims (7)

1. rural area based on a load curve decomposition load type load modeling method, it is characterised in that: described method include with Lower step:

Step 1: decompose rural area load curve, and calculate the ratio of total amount shared by every kind of load in the load of rural area；

Step 2: calculate static load equivalent parameters, dynamic load equivalent parameters and distribution network system impedance, and it is negative to combine rural area The power supply area network topology data of lotus node, obtain rural area load type load model；

Described step 1 comprises the following steps:

Step 1-1: decompose rural area load curve；

Step 1-1-1: to the transformer station that supplied load type is rural area load, merges this transformer station 10kV or 6kV load outlet Daily load curve；

Step 1-1-2: calculate this transformer station's load Payload number of days monthly, if certain day load has null value, it is believed that this day nothing Effect, if certain day load has opposite sign phenomenon, be also considered as this day invalid；

Step 1-1-3: calculate this transformer station's load average daily load curve monthly and average monthly load curve；

Step 1-1-4: calculating every daily load curve period by day monthly, 6～the maximum of 17, and from February Average daily load curve selects the maximum of daytime period, and determines the load curve scope meeting reference load curve requirement, In the range from the maximum 0.8 of the daytime period calculated again to 1.15 times；

Step 1-1-5: start the load curve in all months is decomposed from February, if certain the daily load song of certain month The maximum of line period by day meets the daily load curve average value ranges that reference load curve requires, then it is assumed that this daily load is bent Line is the curve meeting reference load curve requirement, selects to decompose the curve meeting reference load curve requirement in month, calculates song The averaged curve of line, as reference load curve；

Step 1-1-6: determine the load curve mean value lower limit of satisfied big load curve requirement, this lower limit be daytime period 1.1 times of big value, if the maximum of certain daily load curve period by day is more than this lower limit, then it is assumed that this load song Line is the curve meeting big load curve requirement；

Step 1-1-7: determine the big load curve of this month, and calculate the mean value of datum curve load；

Step 1-2: calculate the ratio of total amount shared by every kind of load in the load of rural area；

Step 1-2-1: calculate each transformer station monthly average load value monthly and specify month to specify the average load value of day；

Step 1-2-2: calculate and specify month, specify day, the load value in appointment moment；

Step 1-2-3: calculate and include irrigating load, cooling load and the separate section load proportion k of heating load_p0And base Quasi-load proportion k_f。

Rural area based on load curve decomposition the most according to claim 1 load type load modeling method, its feature exists In: in described step 1-1-3, calculate this transformer station's load average daily load curve monthly, all effective day monthly is born Lotus curve is added, the more effective number of days divided by this month；

Calculate this transformer station's load average monthly load curve monthly, by the load phase in 24 moment in monthly daily load curve every day Add again divided by the curve obtained by 24；In average monthly load curve, each effective day corresponding value is also referred to as daily load curve mean value, Daily load curve mean value is that the load in 24 moment of every day is added, then divided by the mean value obtained by 24；

In described step 1-1-5, if the number meeting the load curve of datum curve this month is less than 3, and above do not have There is the datum curve calculating any one month, then first do not calculate the curve of this month, carry out the calculating of next month curve；As Really this month meets the curve number of datum curve requirement more than or equal to 3, calculates datum curve, and all by this month meet benchmark The daily load curve of curve requirement is added, then meets the curve number of datum curve requirement, calculated datum curve divided by this month Smaller load curve as this month；Otherwise, the datum curve of month is used；

In described step 1-1-7, determine the big load curve of this month, if the load meeting big load curve requirement this month is bent Number of lines is less than 3, then ignore these load curves, do not calculate big load curve this month, does not carry out load curve and separates meter Calculate；If the load curve bar number meeting big load curve requirement this month is more than or equal to 3, calculate big load curve；

Calculate the mean value of datum curve load, big load curve, reference load curve are carried out standardization, average by three Value becomes 1.

Rural area based on load curve decomposition the most according to claim 1 load type load modeling method, its feature exists In: in described step 1-2-1, calculate each transformer station monthly average load value monthly, by each in average monthly load curve monthly The value that effectively it is corresponding is added, the more effective number of days divided by this month；

Calculate the average load value specifying month to specify day, the monthly average load value in this month calculated is multiplied by the day specifying day The perunit value of load curve mean value；

In described step 1-2-2, calculate and specify month, specify day, the load value in appointment moment；

1) if only Smaller load curve, then Smaller load curve is used to calculate；

2) if there is big load curve and Smaller load curve simultaneously, the daily load curve mean value size of day is specified according to this month Determine that the big load curve of employing or Smaller load curve calculate；

Setting value is 1.1 times of the reference load mean value in certain month, then have:

If 2-1) the daily load curve mean value of this month is less than this definite value, Smaller load curve is used to calculate；

If 2-2) daily load curve mean value is more than this definite value, using big load curve to calculate, big carry calculation result is The reference load mean value in this month calculated is multiplied by specifies little/big load curve middle finger timing of day to carve corresponding value in this month；

Also need to calculate and specify month, appointment day, the segregational load value in appointment moment, calculate big load curve and reference load curve Difference, as separate after load curve, this value equal to calculate specify month, specify day, specify the moment load value Deduct the reference load mean value in this month to be multiplied by and specify the Smaller load curve middle finger timing of day to carve corresponding value in this month, then divided by Calculate specifies month, appointment day, the load value in appointment moment；

In described step 1-2-3, calculate separate section load proportion k_p0, by segregational load value divided by the appointment moon calculated Part, appointment day, the load value in appointment moment；Then reference load proportion k_fFor 1-k_p0。

Rural area based on load curve decomposition the most according to claim 1 load type load modeling method, its feature exists In: in described step 2, it is as follows that static load equivalent parameters calculates process:

By the multinomial static load model that the relationship description between load power and voltage is polynomial equation form, have

P=P₀[a×(V/V₀)²+b×(V/V₀)+c] (1)

Q=Q₀[α×(V/V₀)²+β×(V/V₀)+γ] (2)

Wherein, P and Q is respectively active power and the reactive power of static load, and V is the real-time voltage of static load, V₀For The rated voltage of static load, P₀And Q₀It is respectively and represents at V₀The specified active power of lower static load and reactive power, a, B and c is the active power coefficient of multinomial static load model, and α, β and γ are the idle of multinomial static load model Power coefficient；

The equivalence of static load is mainly a, b, c, P₀With α, β, γ, Q₀Equivalence, to multinomial static load The equivalence of model is based on the sensitivity to static load terminal voltage of the static load power, has

$\frac{\∂P}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\Σ}}\frac{\∂P_i}{\∂V}{|}_{V=V_0}---\left(3\right)$

$\frac{\∂Q}{\∂V}{|}_{V=V_0}=\underset{i=1}{\overset{n}{\Σ}}\frac{\∂Q_i}{\∂V}{|}_{V=V_0}---\left(4\right)$

Wherein, n is static load number, P_iAnd Q_iIt is respectively the meritorious of i-th static load and reactive power,With It is respectively i-th static load active power relative to the partial differential of voltage and i-th static load reactive power relative to voltage Partial differential；

Work as V=V₀Time, have:

$P_0=\underset{i=1}{\overset{n}{\Σ}}{P}_{0i}---\left(5\right)$

$Q_0=\underset{i=1}{\overset{n}{\Σ}}{Q}_{0i}---\left(6\right)$

Wherein, P_0iAnd Q_0iIt is respectively initial active power and the reactive power of i-th static load；

$a=\frac{\underset{i=1}{\overset{n}{\Σ}}{P}_{0i}{a}_i}{P_0}---\left(7\right)$

$b=\frac{\underset{i=1}{\overset{n}{\Σ}}{P}_{0i}{b}_i}{P_0}---\left(8\right)$

$c=\frac{\underset{i=1}{\overset{n}{\Σ}}{P}_{0i}{c}_i}{P_0}---\left(9\right)$

Wherein, a_i、b_iAnd c_iIt is the multinomial static load model active power coefficient of i-th static load；

$\mathrm{\α}=\frac{\underset{i=1}{\overset{n}{\Σ}}{Q}_{0i}{\mathrm{\α}}_i}{Q_0}---\left(10\right)$

$\mathrm{\β}=\frac{\underset{i=1}{\overset{n}{\Σ}}{Q}_{0i}{\mathrm{\β}}_i}{Q_0}---\left(11\right)$

$\mathrm{\γ}=\frac{\underset{i=1}{\overset{n}{\Σ}}{Q}_{0i}{\mathrm{\γ}}_i}{Q_0}---\left(12\right)$

Wherein, α_i、β_iAnd γ_iIt is the multinomial static load model reactive power coefficient of i-th static load.

Rural area based on load curve decomposition the most according to claim 1 load type load modeling method, its feature exists In: described step 2 specifically includes following steps:

Step A: calculate all motor always absorbs active-power P_∑With reactive power Q_∑, total electromagnetic power P_∑em, total rotor Winding copper loss P_∑cu2With total maximum electromagnetic power P_{∑em_max}, specifically have:

$\begin{array}{c}{P}_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_j\\ Q_{\mathrm{\Σ}}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{Q}_i\\ P_{\mathrm{\Σ}em}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{emj}\\ P_{\mathrm{\Σ}cu2}=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{cu2j}\\ P_{\mathrm{\Σ}em\_\max }=\underset{j=1}{\overset{m}{\mathrm{\Σ}}}{P}_{em\_\max j}\end{array}---\left(13\right)$

Wherein P_j、Q_j、P_emj、P_cu2jAnd P_em__maxjRespectively represent the active power of jth platform equivalence motor, reactive power, Electromagnetic power, rotor windings copper loss and maximum electromagnetic power；M is equivalent motor number of units；

Step B: calculate the stator winding copper loss P of equivalent motor_∑cu1And the slippage S of equivalence motor, it is expressed as:

P_∑cu1=P_∑-P_∑em (14)

S=P_∑cu2/P_∑em (15)

Initialize the maximum electromagnetic power P of equivalent motor_{emt_max}, make P_{emt_max}=P_{∑em_max}；

Step C: calculate the stator winding resistance R of equivalent motor_s；

First calculate the stator winding phase current of equivalent motorHave

${\stackrel{\·}{I}}_{\mathrm{\Σ}}=-{\left(\frac{P_{\mathrm{\Σ}}+{\mathrm{jQ}}_{\mathrm{\Σ}}}{\sqrt{3}{\stackrel{\·}{U}}_1}\right)}^{*}---\left(16\right)$

Wherein,Set end voltage for motor；

The then stator winding resistance R of equivalent motor_sIt is expressed as:

$R_s=\frac{P_{\mathrm{\Σ}cu1}}{3{I_{\mathrm{\Σ}}}^2}---\left(17\right)$

Step D: calculate the equivalent impedance Z of equivalent machine model_deq:

$Z_{deq}=\frac{{U_1}^2}{P_{\mathrm{\Σ}}-{\mathrm{jQ}}_{\mathrm{\Σ}}}---\left(18\right)$

And have

R_deq=real (Z_deq) (19)

X_deq=imag (Z_deq) (20)

Wherein, R_deqAnd X_deqFor corresponding substitutional resistance and equivalent reactance；

Step E: calculate stator winding leakage reactance X of equivalent motor_sWith rotor windings leakage reactance X_r, have

$X_s=X_r=\frac{X}{2}=-\frac{1}{2}\sqrt{{(\frac{{U_1}^2}{2P_{emt\_max}}-R_s)}^2-{R_s}^2}---\left(21\right)$

Step F: through iteration stator winding leakage reactance X to equivalent motor_sWith rotor windings leakage reactance X_rIt is modified；

Step G: calculate rotor windings resistance and the excitatory reactance of equivalence of equivalent motor；

According to calculated R_s、X_s、X_rAnd Z_deqIf, K_r=R_deq-R_s, K_x=X_deq-X_s, the most equivalent electronic The rotor windings resistance R of machine_rReactance X excitatory with equivalence_mIt is expressed as:

$\begin{array}{c}{R}_r=\frac{(K_r+{K_x}^2/K_r-\sqrt{{(K_r+{K_x}^2/K_r)}^2-4{X_s}^2})S}{2}\\ X_m=\frac{K_r{X}_s+K_x\frac{R_r}{S}}{\frac{R_r}{S}-K_r}\end{array}---\left(23\right)$

Step H: use iterative method to calculate the maximum electromagnetic power of equivalent motor, and it is modified；

1) the maximum electromagnetic power P of the equivalent motor in calculating kth time iteration_{emt_maxk}, it is expressed as

$P_{emt\_\max k}=\frac{{U_1}^2}{2(R_s+\sqrt{{R_s}^2+X^2})}---\left(24\right)$

Based on Thevenin's theorem, the Thevenin's equivalence impedance Z of equivalent motor_dpIt is expressed as

$Z_{dp}={\mathrm{jX}}_r+\frac{{\mathrm{jX}}_m(R_s+{\mathrm{jX}}_s)}{R_s+j(X_s+X_m)}---\left(25\right)$

And have, R_dp=real (Z_dp) and X_dp=imag (Z_dp), R_dpAnd X_dpIt is respectively Thevenin's equivalence resistance and equivalence electricity Anti-；

Equivalent motor produces the condition of maximum electromagnetic power:

$R_{pm}=\frac{R_r}{S_m}=\sqrt{{R_{dp}}^2+{X_{dp}}^2}---\left(26\right)$

Wherein, R_pmThe Thevenin's equivalence impedance magnitude of maximum electromagnetic power, S is produced for correspondence_mFor critical slippage；

Thevenin's equivalence open-circuit voltageFor

${\stackrel{\·}{V}}_0=\frac{{\stackrel{\·}{U}}_1}{\sqrt{3}}\×\frac{{\mathrm{jX}}_m}{R_s+j\left(X_s+X_m\right)}---\left(27\right)$

Then, the maximum electromagnetic power of the Thevenin's equivalence motor in kth time iteration is represented by

$P_{em\_\max k}=\frac{3{V_0}^2{R}_{pm}}{{(R_{dp}+R_{pm})}^2+{X_{dp}}^2}---\left(28\right)$

2) the maximum electromagnetic power of equivalent motor is revised；

Calculate the correction factor τ of kth time iteration medium value motor maximum electromagnetic power_maxk, it is expressed as:

${\mathrm{\τ}}_{\max k}=\frac{P_{emt\_\max k}}{P_{em\_\max k}}---\left(29\right)$

The most revised equivalent motor maximum electromagnetic power P_{emt_max}It is expressed as:

P_{emt_max}=τ_maxk×P_{∑em_max} (30)

With P_{emt_max}With P_{emt_maxk}The iteration error that absolute error is equivalent motor maximum electromagnetic powerHave

${\mathrm{Err}}_{P_{em\_max}}=|P_{emt\_max}-P_{emt\_\max k}|---\left(31\right)$

WithFor iteration convergence standard, ifThen recalculate X_s、R_r、X_rAnd X_m；

Step I: calculate equivalent inertia time constant；

Equivalent inertia time constant H is expressed as:

$H=\frac{\underset{j=1}{\overset{m}{\Σ}}{P}_{nj}{H}_j}{\underset{j=1}{\overset{m}{\Σ}}{P}_{nj}}---\left(32\right)$

Wherein P_njAnd H_jRated power and inertia time constant for jth platform equivalence motor.

Rural area based on load curve decomposition the most according to claim 1 load type load modeling method, its feature exists In: in described step 2, distribution network system impedance meter is shown as:

$Z_{eq}=\[\underset{f=1}{\overset{M}{\Σ}}{u}_f^2{(1/Z_f)}^{*}\]/{\left(\underset{l=1}{\overset{N}{\Σ}}{I}_l\right)}^2---\left(33\right)$

Wherein, Z_eqFor distribution network system impedance；u_fRepresent busbar voltage, Z_fIndication transformer and distribution line impedance；I_lTable Show load current, wherein, node total number that M and N is respectively in distribution network system and branch road total number.

Rural area based on load curve decomposition the most according to claim 1 load type load modeling method, its feature exists In: the power supply area network topology data in described step 2 include distribution line data, transformer data and reactive-load compensation data.