CN103969506B - A kind of three-phase electrical cable harmonic loss computational methods - Google Patents

Abstract

The invention discloses a kind of three-phase electrical cable harmonic loss computational methods, comprise the following steps: A: combine cable practical structures and actual system of laying, consider screen layer electric current and magnetic field coupling condition, three-phase electrical cable harmonic loss formula under derivation particular harmonic number of times electric current and voltage；A2: calculate three-phase electrical cable each layer function of current density under particular harmonic number of times electric current；A3: calculate three-phase electrical cable each layer harmonic wave total losses under particular harmonic number of times electric current and voltage；B: try to achieve three-phase electrical cable each layer harmonic wave total losses under individual harmonic current and voltage by step A；C: the layer each to the three-phase electrical cable obtained in step B harmonic wave total losses under individual harmonic current and voltage carry out summation and obtain three-phase electrical cable harmonic loss.The present invention is beneficial under harmonic wave power cable type selecting under power cable running temperature and the estimation in life-span, beneficially harmonic environment, provides technical support for long distance powedr transmission engineering construction.

Description

A kind of three-phase electrical cable harmonic loss computational methods

Technical field

The present invention relates to electric power quality analysis field, particularly relate to a kind of three-phase electrical cable harmonic loss meter Calculation method.

Background technology

Three-phase electrical cable is as the important medium in electric energy transmitting procedure, and the problem affected by system harmonics is the most prominent Go out.In carrying out the research that three-phase electrical cable is affected by harmonic wave, owing to there is the kelvin effect of wire, approach effect and absolutely The factors such as the distribution capacity that edge medium exists, increase the biggest difficulty all can to research.

Consulting electric relevant criterion, the power cable model that is applicable under harmonic background is many is advised with IEC standard by IEEE Fixed.In IEEE cable model, extract harmonic current size and each harmonic containing ratio when frequency analysis, by each harmonic The correction of lower AC resistance, calculates each harmonic loss respectively, and superposition solves total three-phase electrical cable harmonic loss.Solving During, current algorithm is all with empirical equation correction AC resistance: R (h)=R_dc(0.187+0.532h^0.5), though this algorithm So kelvin effect is considered, but have ignored approach effect, do not accounted for intercoupling between three-phase electrical cable The impact that loss is brought by magnetic field.

IEC proposes with kelvin effect factor y in its standard IEC 60287-1_s, approach effect factor y_pRevise industrial frequency AC Resistance, thus obtain the algorithm of electric parameter under harmonic wave, wherein, the kelvin effect factorApproach effect The factor $y_p=y_s\·{\left(\frac{D_c}{s}\right)}^2\·[0.312{\left(\frac{D_c}{s}\right)}^2+\frac{1.8}{y_s+0.27}].$ Face although the method solves harmonic wave in IEEE cable model The impact that three-phase electrical cable is lost by nearly effect, but not to three-phase electrical cable system of laying and three-phase electrical cable screen Cover layer electric current and reasonably investigate to the impact caused is lost.

Visible, associated international standards there are some computational methods can be to three-phase electrical cable model and three-phase electricity under harmonic wave Power cable loss carries out correlation computations, but is respectively provided with certain limitation, and main cause is:

(1) due to three-phase electrical cable employing overhead transmission line model in existing computational methods, three-phase electrical cable is simplified For long straight conductor, do not set up rational harmonic-model for three-phase electrical cable self structure, thus directly influence three-phase Power cable harmonic loss calculates；

(2) owing to existing computational methods using empirical equation correction to the electric parameter under harmonic wave, to actual three-phase electricity The impact of power cable and system of laying thereof is not paid attention to, and affects computational accuracy.

Three-phase electrical cable is less due to phase spacing when actual motion, it will produces magnetic field and is mutually coupled phenomenon, increases Big running wastage, particularly in the case of harmonic wave, this phenomenon is even more serious.Current all analysis means usually ignore this aspect Problem, system of laying actual to three-phase electrical cable lacks analysis so that long distance powedr transmission cable loss calculate occur serious Defect.Therefore, hazardness three-phase electrical cable run in view of harmonic loss and the shortage of present analysis means, need one badly Can damage in conjunction with the three-phase electrical cable harmonic wave of three-phase electrical cable practical structures, consideration screen layer electric current and alternate laying state Consumption computational methods.

Summary of the invention

It is an object of the invention to provide a kind of three-phase electrical cable harmonic loss computational methods, it is possible to combine three-phase power electricity Cable practical structures and system of laying, consideration screen layer electric current and three-phase electrical cable phased magnetic field coupling condition are comprehensively analyzed Calculate, finally give three-phase electrical cable harmonic loss accurately.The present invention can be power cable running temperature and longevity under harmonic wave Early-stage preparations are made in the estimation of life, provide help for power cable type selecting under harmonic environment, carry for long distance powedr transmission engineering construction For technical support.

The present invention uses following technical proposals:

A kind of three-phase electrical cable harmonic loss computational methods, comprise the following steps:

A: combining three-phase electrical cable practical structures and the actual system of laying of three-phase electrical cable, considering screen layer electric current And on the premise of three-phase electrical cable phased magnetic field coupling condition, three-phase power electricity under derivation particular harmonic number of times electric current and voltage Cable harmonic loss formula, concrete derivation step is as follows:

A1: determine the alternate spacing of each phase of three-phase electrical cable according to the actual system of laying of three-phase electrical cable, calculates Induction electromotive force in each phase screen layer of unit length under particular harmonic number of times electric current, shields then in conjunction with three-phase electrical cable Layer two-terminal-grounding resistance, determines screen layer earth-return according to the screen layer connected mode of three-phase electrical cable and calculates three-phase power Cable shield layer impedance and three-phase electrical cable earth-return impedance, final calculating obtains each phase electric power under particular harmonic number of times electric current Cable shield electric current；

A2: calculate three-phase electrical cable each layer function of current density under particular harmonic number of times electric current；

A3: ask for the cable core loss under particular harmonic number of times of the three-phase electrical cable each layer, insulating barrier loss and screen respectively Cover layer loss, then cable core loss, insulating barrier loss and screen layer loss are sued for peace, finally calculate three-phase electrical cable Each layer harmonic wave total losses under particular harmonic number of times electric current and voltage；

B: by actually detected equipment Inspection, obtain electric current and the voltage of each phase of three-phase electrical cable, then divided by FFT Solve device, analyze and obtain individual harmonic current and the size of harmonic voltage and relative harmonic content；Then by the voltage under each harmonic Current value carries out computing according to step A, is obtained each phase power cable shield layer electricity under individual harmonic current respectively by step A1 Stream, tries to achieve three-phase electrical cable each layer function of current density under individual harmonic current respectively by step A2, passes through step A3 tries to achieve three-phase electrical cable each layer harmonic wave total losses under individual harmonic current and voltage respectively；

C: the layer each to the three-phase electrical cable obtained in step B harmonic wave total losses under individual harmonic current and voltage are entered Row summation processes, and finally obtain and value is three-phase electrical cable harmonic loss.

In described step A1, under particular harmonic number of times electric current, the induction electromotive force in each phase screen layer of unit length includes Each to unit length of induction electromotive force that each phase screen layer of unit length is produced by three-phase cable core electric current and screen layer electric current The induction electromotive force that phase screen layer produces；

The induction electromotive force that three-phase cable core electric current produces in the A phase screen layer P of unit lengthFor:

The induction electromotive force that three-phase cable core electric current produces in the B phase screen layer P of unit lengthFor:

${\stackrel{.}{e}}_{\mathrm{SB}}=2\mathrm{\ωI}\×{10}^{-7}[\frac{\sqrt{3}}{2}\ln \frac{\mathrm{mS}}{R}-j\frac{1}{2}\ln \frac{S}{m\·R}];$

The induction electromotive force that three-phase cable core electric current produces in the C phase screen layer P of unit lengthFor:

${\stackrel{.}{e}}_{\mathrm{SC}}=2\mathrm{\ωI}\×{10}^{-7}[-\frac{\sqrt{3}}{2}\ln \frac{\mathrm{mS}}{R}-j\frac{1}{2}\ln \frac{n^{2}s}{m\·R}];$

Wherein, j is imaginary unit；ω=2 π f, f represents harmonic frequency；P is the conductor being parallel to cable core A, B, C, as Virtual screen layer；The magnetic flux produced in P for three-phase cable core；I is three-phase cable core electric current；Three-phase electrical cable A, C are alternate Away from for S；A, B spaced apart is n times of S, i.e. nS；B, C spaced apart is m times of S, i.e. mS；R is the radius of P；

The induction electromotive force that unit length screen layer is produced by screen layer electric current is:

${\stackrel{.}{e}}_{\mathrm{SA}}^{\′}={\stackrel{.}{I}}_{\mathrm{SB}}\·{\mathrm{jM}}_{\mathrm{AB}}+{\stackrel{\·}{I}}_{\mathrm{SC}}\·{\mathrm{jM}}_{\mathrm{AC}};$

${\stackrel{.}{e}}_{\mathrm{SB}}^{\′}={\stackrel{.}{I}}_{\mathrm{SA}}\·{\mathrm{jM}}_{\mathrm{AB}}+{\stackrel{.}{I}}_{\mathrm{SC}}\·{\mathrm{jM}}_{\mathrm{BC}};$

${\stackrel{.}{e}}_{\mathrm{SC}}^{\′}={\stackrel{.}{I}}_{\mathrm{SA}}\·{\mathrm{jM}}_{\mathrm{AC}}+{\stackrel{.}{I}}_{\mathrm{SB}}\·{\mathrm{jM}}_{\mathrm{BC}};$

Wherein,The sense produced in unit length A phase screen layer for B, C phase screen layer electric current in three-phase electrical cable Answer electromotive force；The induction electric produced in unit length B phase screen layer for three-phase electrical cable A, C phase screen layer electric current Gesture；The induction electromotive force produced in unit length C phase screen layer for three-phase electrical cable A, B phase screen layer electric current； Represent three-phase electrical cable A, B, C phase screen layer electric current respectively；M_AB、M_BC、M_CAThree-phase under representation unit length respectively Under power cable A and B phase screen layer mutual inductance, unit length, under B with C phase screen layer mutual inductance, unit length, C with A phase screen layer is mutual Sense；J is imaginary unit；

Screen layer connected mode according to three-phase electrical cable determines screen layer earth-return, shields in conjunction with three-phase electrical cable Layer two-terminal-grounding resistance, and calculate three-phase electrical cable screen layer impedance, three-phase electrical cable earth-return impedance and induction electric Gesture；Induction electromotive force, three-phase electrical cable screen layer impedance and three-phase electrical cable earth-return impedance are counted by unit length is lower Calculation value is multiplied by line length and obtains；

Final calculating obtains each phase power cable shield layer electric current under particular harmonic number of times electric current.

In described step A2, utilize vector magnetic potential A boundary condition Simultaneous Equations on power cable actual physics layer, Introduce bessel function, use separation of variable solving equation, in conjunction with boundary condition, it is thus achieved that the result of vector magnetic potential A, then lead to Cross relational expression:I.e. can get three-phase electrical cable each layer function of current density under particular harmonic number of times electric current J is:

$J=\underset{n=1}{\overset{\∞}{\mathrm{\Σ}}}{J}_n\left(r\right)\·{\mathrm{\Θ}}_n\left(\mathrm{\θ}\right);$

Wherein, modifying factor ${\mathrm{\Θ}}_n\left(\mathrm{\θ}\right)=\underset{k=1}{\overset{m}{\mathrm{\Σ}}}\frac{I_k}{I}\·\frac{1}{a_k^n}\cos n(\mathrm{\θ}+{\mathrm{\α}}_k),$ M represents the cable of studied Exterior cable The number of phases；b_kRepresent the spacing of kth phase cable and studied cable, define a_k=b_k/b₁；I represents the cable core electric current of research cable；I_k Represent the cable core electric current of kth phase cable beyond studied cable；α_kRepresent the locus of k phase cable；R represents footpath, pole, and θ represents Polar angle.

In described step A3,

By the Joule's law of differential form is integrated, calculate unit length cable under particular harmonic number of times electric current Cable core loss and screen layer loss are

$P_i=\frac{\mathrm{\π}}{g_i}\underset{n=0}{\overset{\∞}{\mathrm{\Σ}}}{\mathrm{\φ}}_M\·{\∫J}_n\left(r\right)\·J_n^{*}\left(r\right)\·\mathrm{rdr};$ Wherein, i=1,3, P₁Represent cable core loss, P₂Represent screen Cover layer loss,

M represents and grinds Study carefully the Exterior cable cable number of phases；a_k=b_k/b₁, b_kRepresent the spacing of kth phase cable and studied cable；Represent the i-th phase cable Cable core current phase；Represent J_nThe conjugate of (r)；

Under particular harmonic number of times voltage, the insulating barrier loss W of unit length cable is:

Wherein, γ is insulant electrical conductivity, and U is harmonic voltage value, r₁For cable core radius, r₂For screen Cover a layer outer radius.

The present invention, according to vector magnetic potential Poisson's equation and Laplace's equation, in conjunction with the actual system of laying of power cable, examines Consider the impact that under harmonic wave, three-phase electrical cable is lost by kelvin effect, approach effect, utilize bessel function and the separation of variable Process vector magnetic potential and set up three-phase electrical cable harmonic loss formula under particular harmonic number of times, finally real according to three-phase electrical cable Border parameter calculates three-phase electrical cable harmonic loss.The present invention is studying the actual system of laying of three-phase electrical cable to loss shadow When ringing, provide introducing modifying factor according to derivation result, can accurately and easily analyze three-phase electrical cable or even heterogeneous electricity Power cable harmonic loss.The present invention can be that under harmonic wave, early-stage preparations are made in the estimation in power cable running temperature and life-span, for humorous Under ripple environment, power cable type selecting provides help, provides technical support for long distance powedr transmission engineering construction.

Accompanying drawing explanation

Fig. 1 is the flow chart of the present invention；

Fig. 2 is screen layer induction electromotive force computation model schematic diagram；

Fig. 3 is three-phase electrical cable screen layer circulation equivalent circuit diagram in the present invention；

Fig. 4 is multi-phase cable spatial distribution schematic diagram in the present invention.

Detailed description of the invention

As it is shown in figure 1, three-phase electrical cable harmonic loss computational methods of the present invention, comprise the following steps:

A: combining three-phase electrical cable practical structures and the actual system of laying of three-phase electrical cable, considering screen layer electric current And on the premise of three-phase electrical cable phased magnetic field coupling condition, three-phase power electricity under derivation particular harmonic number of times electric current and voltage Cable harmonic loss formula, concrete derivation step is as follows:

A1: determine the alternate spacing of each phase of three-phase electrical cable according to the actual system of laying of three-phase electrical cable, calculates Induction electromotive force in each phase screen layer of unit length under particular harmonic number of times electric current, shields then in conjunction with three-phase electrical cable Layer two-terminal-grounding resistance, determines screen layer earth-return according to the screen layer connected mode of three-phase electrical cable and calculates three-phase power Cable shield layer impedance and three-phase electrical cable earth-return impedance, final calculating obtains each phase electric power under particular harmonic number of times electric current Cable shield electric current；

Actual system of laying according to three-phase electrical cable determines the alternate spacing of each phase of three-phase electrical cable, calculates specific Induction electromotive force in each phase screen layer of unit length under overtone order electric current；Unit length under particular harmonic number of times electric current Induction electromotive force in each phase screen layer includes the induction electric that each phase screen layer of unit length is produced by three-phase cable core electric current The induction electromotive force that each phase screen layer of unit length is produced by gesture and screen layer electric current；

Consider the arrangement mode of three-phase electrical cable, in conjunction with Fig. 2, by the magnetic flux asked in screen layer and then try to achieve sensing Electromotive force size.As in figure 2 it is shown, the spaced apart that in three-phase electrical cable, A, C are biphase is S；In three-phase electrical cable, A, B are biphase Spaced apart is n times of S, i.e. nS；The spaced apart that in three-phase electrical cable, B, C are biphase is m times of S, i.e. mS.For convenience of calculating, by P As virtual screen layer；Screen layer P is the conductor being parallel to three-phase electrical cable cable core A, B, C, and screen layer P is in fig. 2 The virtual location of positional representation phase cable shield, R is the radius of screen layer P, A, B, C three-phase and the spacing edge of screen layer P By the definition mode of three spaced apart, respectively D, lD, kD in above-mentioned three-phase electrical cable.In view of soil pcrmeability and Vacuum Magnetic Conductance approximately equal, the magnetic flux that available A, B, C three-phase cable core electric current produces in the screen layer P of unit length is:

Wherein,Represent that A, B, C three-phase cable core electric current produces in the screen layer P of unit length respectively Magnetic flux,Represent A, B, C three-phase cable core electric current respectively.Then three-phase cable core electric current is at the screen layer of unit length The total magnetic flux produced in PFor:

When cable core current induced magnetic field suffered by the A phase screen layer solving unit length affects, can determine that cable core A and shielding The center superposition of layer P.Now, D=R, lD=nS, kD=S, obtain:

Consider three-phase cable core current-symmetrical, orderThen ${\stackrel{.}{I}}_C=(-\frac{1}{2}+j\frac{\sqrt{3}}{2})I,$ Generation Enter formula (3), obtain:

Can obtain what three-phase cable core electric current produced in the A phase screen layer P of unit length according to induction electromotive force definition Induction electromotive forceAs shown in formula (5):

Wherein, j is imaginary unit, and ω=2 π f, f represents harmonic frequency.

When cable core B and screen layer P center superposition, kD=R, D=S, lD=mS, three-phase cable core electric current is in unit length The induction electromotive force produced in B phase screen layer PAs shown in formula (6):

${\stackrel{.}{e}}_{\mathrm{SB}}=2\mathrm{\ωI}\×{10}^{-7}[\frac{\sqrt{3}}{2}\ln \frac{\mathrm{mS}}{R}-j\frac{1}{2}\ln \frac{S}{m\·R}]---\left(6\right);$

In like manner, the induction electromotive force that three-phase cable core electric current produces in the C phase screen layer P of unit lengthSuch as formula (7) Shown in:

${\stackrel{.}{e}}_{\mathrm{SC}}=2\mathrm{\ωI}\×{10}^{-7}[-\frac{\sqrt{3}}{2}\ln \frac{\mathrm{mS}}{R}-j\frac{1}{2}\ln \frac{n^{2}s}{m\·R}]---\left(7\right);$

Owing to a certain phase screen layer electric current can produce induction electromotive force in other phase screen layers, in conjunction with each phase screen layer it Between mutual inductance, the induction electromotive force that can produce in unit length screen layer in the hope of screen layer electric current, screen layer electric current is to list Shown in the induction electromotive force such as formula (8) of bit length screen layer generation, (9), (10).

${\stackrel{.}{e}}_{\mathrm{SA}}^{\′}={\stackrel{.}{I}}_{\mathrm{SB}}\·{\mathrm{jM}}_{\mathrm{AB}}+{\stackrel{\·}{I}}_{\mathrm{SC}}\·{\mathrm{jM}}_{\mathrm{AC}}---\left(8\right);$

${\stackrel{.}{e}}_{\mathrm{SB}}^{\′}={\stackrel{.}{I}}_{\mathrm{SA}}\·{\mathrm{jM}}_{\mathrm{AB}}+{\stackrel{.}{I}}_{\mathrm{SC}}\·{\mathrm{jM}}_{\mathrm{BC}}---\left(9\right);$

${\stackrel{.}{e}}_{\mathrm{SC}}^{\′}={\stackrel{.}{I}}_{\mathrm{SA}}\·{\mathrm{jM}}_{\mathrm{AC}}+{\stackrel{.}{I}}_{\mathrm{SB}}\·{\mathrm{jM}}_{\mathrm{BC}}---\left(10\right);$

Wherein,The sense produced in unit length A phase screen layer for B, C phase screen layer electric current in three-phase electrical cable Answer electromotive force；The induction electric produced in unit length B phase screen layer for three-phase electrical cable A, C phase screen layer electric current Gesture；The induction electromotive force produced in unit length C phase screen layer for three-phase electrical cable A, B phase screen layer electric current； Represent three-phase electrical cable A, B, C phase screen layer electric current respectively；M_AB、M_BC、M_CAThree-phase under representation unit length respectively Under power cable A and B phase screen layer mutual inductance, unit length, under B with C phase screen layer mutual inductance, unit length, C with A phase screen layer is mutual Sense.Introduce Carson formula equivalence degree of depth D_e, any biphase screen layer mutual inductance in three-phase electrical cable under unit length can be obtained M:

$M=2\mathrm{\ω}\×{10}^{-7}\ln \frac{D_e}{S}---\left(11\right);$

$D_e=659.33\sqrt{\frac{\mathrm{\ρ}}{f}}---\left(12\right);$

Wherein, ρ is soil resistivity, and f is harmonic frequency, ρ and f all can draw by using reality to measure device measuring；S For biphase screen layer spacing.

Screen layer connected mode according to three-phase electrical cable determines screen layer earth-return, calculates the shielding of particular harmonic number of times Layer electric current；Three-phase electrical cable screen layer circulation equivalent circuit diagram is as shown in Figure 3.

Owing to, in long distance powedr transmission, screen layer is frequently with the cross interconnected mode of connection of segmentation, and sense in screen layer is asked in segmentation Answer electromotive force, determine the associated loop parameter in screen layer loop, the phase in screen layer loop according to the concrete connected mode of screen layer Close loop parameter and include cable shield layer impedance, cable shield two-terminal-grounding resistance and earth-return impedance.As it is shown on figure 3, R+ JX represents cable shield layer impedance, R₁、R₂Representing screen layer two-terminal-grounding resistance, Re represents earth-return impedance, Represent the induction electromotive force that three-phase cable core electric current produces in A, B, C threephase cable screen layer, E respectively_SA' represent three-phase power electricity The induction electromotive force that in cable, B, C phase screen layer electric current produces in A phase screen layer, E_SB' represent that in three-phase electrical cable, A, C phase is shielded Cover the induction electromotive force that layer electric current produces in B phase screen layer, E_SC' represent that in three-phase electrical cable, A, B phase screen layer electric current is at C The induction electromotive force produced in phase screen layer.In Fig. 3, induction electromotive force, cable shield layer impedance, earth-return impedance are by unit Under length, institute's value of calculation is multiplied by line length acquisition.

Arrange according to Fig. 3 and write equation group:

$\left[\begin{array}{cccccc}{R}_A& R_B& R_B& -X& -X_1& -X_2\\ R_B& R_A& R_B& -X_1& -X& {-X}_1\\ R_B& R_B& R_A& -X_2& -X_1& -X\\ X& X_1& X_2& R_A& R_B& R_B\\ X_1& X& X_1& R_B& R_A& R_B\\ X_2& X_1& X& R_B& R_B& R_A\end{array}\right]\left[\begin{array}{c}{I}_{\mathrm{SAr}}\\ I_{\mathrm{SBr}}\\ I_{\mathrm{SCr}}\\ I_{\mathrm{SAi}}\\ I_{\mathrm{SBi}}\\ I_{\mathrm{SCi}}\end{array}\right]=\left[\begin{array}{c}{E}_{\mathrm{SAr}}\\ E_{\mathrm{SBr}}\\ E_{\mathrm{SCr}}\\ E_{\mathrm{SAi}}\\ E_{\mathrm{SBi}}\\ E_{\mathrm{SCi}}\end{array}\right]---\left(13\right)$

Wherein, R_A=R₁+R₂+R_e+ R, R_B=R₁+R₂+R_e, X, X₁、X₂Represent screen layer induction reactance and mutual inductance, I_SAr、I_SBr、 I_SCr、E_SAr、E_SBr、E_SCrRepresent screen layer three phase circulation, the real part of induction electromotive force, I_SAi、I_SBi、I_SCi、E_SAi、E_SBi、E_SCiTable Show screen layer three phase circulation, the imaginary part of induction electromotive force.

Solve formula (13) and i.e. can get each phase power cable shield layer electric current under particular harmonic number of times electric current.

A2: calculate three-phase electrical cable each layer function of current density under particular harmonic number of times electric current.

Owing to the curl of vector magnetic potential A is magnetic induction density B, magnetic induction density B size direction can dividing with magnetic reaction fields Cloth situation, therefore function of current density and vector magnetic potential also exist relation of equal quantity, can by function of current density ask for be converted into Asking for of vector magnetic potential.Owing to vector magnetic potential meets Laplace's equation formula (14) at insulating barrier and Exterior cable region, Meeting Poisson's equation formula (15) in copper core and screen layer, the cylindrical coordinates form that therefore can get vector magnetic potential is:

$\frac{{\∂}^{2}A}{{\∂r}^2}+\frac{1}{r}\·\frac{\∂A}{\∂r}+\frac{1}{r^2}\·\frac{{\∂}^{2}A}{{\∂\mathrm{\θ}}^2}=0---\left(14\right);$

$\frac{{\∂}^{2}A}{{\∂r}^2}+\frac{1}{r}\·\frac{\∂A}{\∂r}+\frac{1}{r^2}\·\frac{{\∂}^{2}A}{{\∂\mathrm{\θ}}^2}=-m^{2}A=0---\left(15\right);$

Wherein, m=j ω μ g, r represents footpath, pole, and θ represents that polar angle, μ represent that pcrmeability, g represent electrical conductivity.

(i.e. cable core, insulating barrier, metal screen layer, other structures are given at power cable actual physics layer to consider vector magnetic potential A To ignore) on boundary condition Simultaneous Equations, introduce bessel function, use separation of variable solving equation, in conjunction with border Condition, it is possible to obtain the result of vector magnetic potential A, if formula (17) is to shown in formula (20).Then relational expression is passed through:I.e. Available function of current density J.As shown in Figure 4, for polyphase electric power cable, it is considered to actual system of laying is to current phasor function Impact, propose modifying factor Θ_n(θ):

${\mathrm{\Θ}}_n\left(0\right)=\underset{k=1}{\overset{m}{\mathrm{\Σ}}}\frac{I_k}{I}\·\frac{1}{a_k^n}\cos n(\mathrm{\θ}+{\mathrm{\α}}_k)---\left(16\right);$

Complex chart 4 and formula (16), m represents the cable number of phases of studied Exterior cable；b_kRepresent kth phase cable and studied The spacing of cable, defines a_k=b_k/b₁；I represents the cable core electric current of research cable；I_kRepresent kth phase cable beyond studied cable Cable core electric current；α_kRepresent the locus of k phase cable.

Use r₁、r₂、r₃Represent the radius of cable core, insulating barrier, screen layer respectively, only consider I₁To research each physics of cable Layer magnetic field intercouples, and carries out solving of vector magnetic potential, can obtain:

Work as r₁<r≤r₂, the expression formula of vector magnetic potential is

$A=A_0^{\&prime;}\ln r+\underset{n=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}[A_n^{\&prime;}\&CenterDot;r^{-n}+B_n^{\&prime;}\&CenterDot;r^n]\cos \left(\mathrm{n\&theta;}\right)---\left(17\right);$

As r≤r₁, the expression formula of vector magnetic potential is $A(r,\mathrm{\&theta;})=\underset{n=0}{\overset{\&infin;}{\mathrm{\&Sigma;}}}{A}_n{I}_n\left(m_{1}r\right)\cos \left(\mathrm{n\&theta;}\right)---\left(18\right);$

Work as r₂≤r≤r₃, the expression formula of vector magnetic potential is

$A(r,\mathrm{\&theta;})=\underset{n=0}{\overset{\&infin;}{\mathrm{\&Sigma;}}}[B_n{I}_n\left(m_{2}r\right)+C_n{K}_n\left(m_{2}r\right)]\cos \left(\mathrm{n\&theta;}\right)---\left(19\right);$

Work as r₃< r, the expression formula of vector magnetic potential is

$A=-\frac{{\mathrm{\&mu;}}_{0}I}{2\mathrm{\&pi;}}[\ln \left(b\right)-\underset{n=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}\frac{1}{n}{\left(\frac{r}{b}\right)}^n\cos \left(\mathrm{n\&theta;}\right)]+C_0^{\&prime;}\ln \left(r\right)+\underset{n=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}{C}_n^{\&prime;}\&CenterDot;r^{-n}\&CenterDot;\cos \left(\mathrm{n\&theta;}\right)---\left(20\right);$

Wherein,x₁=m₁r₁, x₂=m₂r₂, x₃=m₂r₃,

$B_0=\frac{{\mathrm{\&mu;}}_0{I}_c}{2\mathrm{\&pi;}{m}_2{r}_3}\&CenterDot;\frac{r_3{K}_1\left(x_3\right)-r_2{K}_1\left(x_2\right)}{D_0}-\frac{{\mathrm{\&mu;}}_0{I}_s}{2\mathrm{\&pi;}{m}_2{r}_3}\&CenterDot;\frac{K_1\left(x_2\right)}{D_0}$

$C_0=\frac{{\mathrm{\&mu;}}_0{I}_c}{2\mathrm{\&pi;}{m}_2{r}_3}\&CenterDot;\frac{r_3{I}_1\left(x_2\right)-r_3{I}_1\left(x_3\right)}{D_0}+\frac{{\mathrm{\&mu;}}_0{I}_s}{2\mathrm{\&pi;}{m}_2{r}_3}\&CenterDot;\frac{I_1\left(x_2\right)}{D_0},D_0=K_1\left(x_2\right)I_1\left(x_3\right)-I_1\left(x_2\right)K_1\left(x_3\right),$

$A_n=\frac{{\mathrm{\&mu;}}_{0}I}{\mathrm{\&pi;}{m}_1{r}_1}{\left(\frac{r_1}{b}\right)}^n\&CenterDot;\frac{1}{I_{n-1}\left(x_1\right)}{Z}_n,A_n^{\&prime;}=-\frac{{\mathrm{\&mu;}}_{0}I}{2\mathrm{\&pi;}}\&CenterDot;\frac{r_1^{2n}}{n{b}^n}\&CenterDot;\frac{I_{n+1}\left(x_1\right)}{I_{n-1}\left(x_1\right)}\&CenterDot;Z_n,$

$B_n=\frac{{\mathrm{\&mu;}}_{0}I}{\mathrm{\&pi;}{m}_2{r}_3}{\left(\frac{r_3}{b}\right)}^n\&CenterDot;\frac{K_{n+1}\left(x_2\right)-{\mathrm{\&Delta;}}_n{K}_{n-1}\left(x_2\right)}{D_n^{\&prime;}},$ $B_n^{\&prime;}=\frac{{\mathrm{\&mu;}}_{0}I}{2\mathrm{\&pi;}}\&CenterDot;\frac{1}{{\mathrm{nb}}^n}{Z}_n,$

$C_n=\frac{{\mathrm{\&mu;}}_{0}I}{\mathrm{\&pi;}{m}_2{r}_3}{\left(\frac{r_3}{b}\right)}^n\frac{I_{n+1}\left(x_2\right)-{\mathrm{\&Delta;}}_n{I}_{n-1}\left(x_2\right)}{D_n^{\&prime;}}$

$C_n^{\&prime;}=-\frac{{\mathrm{\&mu;}}_{0}I}{2\mathrm{\&pi;}}\&CenterDot;\frac{r_3^{2n}}{{\mathrm{nb}}^n}\&CenterDot;\frac{1}{D_n^{\&prime;}}\{K_{n+1}\left(x_2\right)I_{n+1}\left(x_3\right)-I_{n+1}\left(x_2\right)K_{n+1}\left(x_3\right)-{\mathrm{\&Delta;}}_n[K_{n-1}\left(x_2\right)I_{n+1}\left(x_3\right)-I_{n-1}\left(x_2\right)K_{n+1}\left(x_3\right)\left]\right\}$

W_n=K_n+1(x₂)I_n-1(x₂)-I_n+1(x₂)K_n-1(x₂), ${\mathrm{\&Delta;}}_n=\frac{I_{n+1}\left(x_1\right)}{I_{n-1}\left(x_1\right)}{\left(\frac{r_1}{r_2}\right)}^{2n},$ E_n=K_n-1(x₂)I_n-1(x₃)-I_n-1 (x₂)K_n-1(x₃), D_n=K_n+1(x₂)I_n-1(x₃)-I_n+1(x₂)K_n-1(x₃),I_c、I_sRepresent respectively and grind Study carefully the cable core electric current of phase cable, screen layer electric current；B is cable core spacing；μ₀Represent permeability of vacuum；m_i=j ω μ₀g_i, g_iRepresent each The electrical conductivity of individual physical layer, i=1,2,3 are respectively cable core, insulating barrier, metal screen layer；I_nX () is that the first kind deforms bessel Function；K_nX () represents n rank Equations of The Second Kind deformation bessel function；D′_nFor D_nFirst derivative.

According to the result of vector magnetic potential A, in conjunction withI.e. can get function of current density, its form isR represents footpath, pole, and θ represents polar angle；Consider to produce modifying factor Θ of impact because of system of laying_n(θ), Available three-phase electrical cable each layer function of current density J under particular harmonic number of times electric current:

$J=\underset{n=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}{J}_n\left(r\right)\&CenterDot;{\mathrm{\&Theta;}}_n\left(\mathrm{\&theta;}\right)---\left(21\right)$

A3: ask for the cable core loss under particular harmonic number of times of the three-phase electrical cable each layer, insulating barrier loss and screen respectively Cover layer loss, then cable core loss, insulating barrier loss and screen layer loss are sued for peace, finally calculate three-phase electrical cable Each layer harmonic wave total losses under particular harmonic number of times electric current and voltage；

Harmonic wave total losses under particular harmonic number of times are calculated by the way of the loss of each layer is superimposed, asks for the most respectively Cable core loss, insulating barrier loss, screen layer loss are sued for peace again.Cable core loss, screen layer loss use differential form The method that Joule's law is integrated calculates.The cable core loss of particular harmonic number of times current unit length cables and screen layer are lost For:

$P_i=\frac{1}{g_i}\underset{0\&RightArrow;2\mathrm{\&pi;}}{\&Integral;\&Integral;}{\left|J(r,\mathrm{\&theta;})\right|}^2\&CenterDot;\mathrm{rdrd\&theta;}---\left(22\right);$

Wherein, i=1,3, P₁Represent cable core loss, P₂Represent screen layer loss.Convolution (21), abbreviation formula (22):

$P_i=\frac{\mathrm{\&pi;}}{g_i}\underset{n=0}{\overset{\&infin;}{\mathrm{\&Sigma;}}}{\mathrm{\&phi;}}_M\&CenterDot;{\&Integral;J}_n\left(r\right)\&CenterDot;J_n^{*}\left(r\right)\&CenterDot;\mathrm{rdr}---\left(23\right)$

Wherein,

M represents research The Exterior cable cable number of phases；a_k=b_k/b₁, b_kRepresent the spacing of kth phase cable and studied cable；Represent the i-th phase cable Core current phase；Represent J_nThe conjugate of (r).

Owing to insulating barrier internal current is by conducting electric current and displacement current two parts form, and the electric field that displacement current produces Power does not become heat to particle work done, but becomes the kinetic energy of particle, so the differential form of Joule's law is the most not Set up, thus cable core, metal screen layer internal loss computational methods can not be used.In view of the insulating barrier between cable core and screen layer Be equivalent to added with harmonic voltage, therefore can get the insulating barrier loss W of unit length cable under particular harmonic number of times voltage:

$W=\frac{2\mathrm{\&pi;\&gamma;}}{\ln \frac{r_2}{r_1}}{U}^2---\left(24\right);$

In formula, γ is insulant electrical conductivity, and U is harmonic voltage value, r₁For cable core radius, r₂For screen layer outer radius.

Being respectively lost mutually under unit length is solved, by cable core loss, screen layer loss, insulating barrier by formula (23), (24) It is lost and superimposed obtains unit length cable harmonic wave total losses, be multiplied by line length, be appreciated that harmonic wave under particular harmonic number of times Total losses.

B: by actually detected equipment Inspection, obtain electric current and the voltage of each phase of three-phase electrical cable, then divided by FFT Solve device, analyze and obtain individual harmonic current and the size of harmonic voltage and relative harmonic content.FFT decomposer can be by the three of input The voltage signal of each phase of phase power cable and current signal are changed into the voltage under each harmonic and electric current, and use amplitude indirect Represent relative harmonic content.Voltage x current value under each harmonic is carried out computing according to step A, is obtained by step A1 each respectively Under subharmonic current, each phase power cable shield layer electric current, tries to achieve each layer of three-phase electrical cable respectively humorous at each time by step A2 Function of current density under ripple electric current, tries to achieve each layer of three-phase electrical cable respectively at individual harmonic current and voltage by step A3 Under harmonic wave total losses.

C: the layer each to the three-phase electrical cable obtained in step B harmonic wave total losses under each harmonic are carried out at summation Reason, finally obtain and value is three-phase electrical cable harmonic loss.

Below in conjunction with specific embodiment, the present invention will be further elaborated:

Embodiment:

Certain power cable model is: 8.7/10kV YJV-50, design parameter is as shown in table 1 below.

Table 1 power cable parameter list

Table1power cable parameter

The cross interconnected two sides earth of 1km power cable shield layer, connects once every 333.3m screen layer commutation.Soil is straight Burying, threephase cable triangular arranged, spacing is 30mm.Soil resistivity is 100 Ω m, and two sides earth resistance is 1 Ω.5 subharmonic current 20A, 5 subharmonic voltage 200V, 7 subharmonic current 15A, 7 subharmonic voltage 100V.

As shown in stepb, for 5 subharmonic, it is first according to step A1, tries to achieve formula (13) coefficient square according to formula (5)-(12) Battle array and induction electromotive force, solve formula (13) and obtain A, B, C phase shielded layer electric current and be 0.493+0.0383j A；

According to step A2, solved the distribution of vector magnetic potential A by formula (17)-(20)；

According to step A3, damage according to 5 subharmonic that formula (23), (24) solve in unit length cable core, screen layer, insulating barrier Consumption, result is as follows: cable core loss 0.1236W/m, screen layer loss 0.0185W/m, insulating barrier loss 5.3286 × 10^-9W/m.Knot Closing line length, obtaining harmonic wave total losses under 5 subharmonic is 1421.11W.In like manner can obtain harmonic wave total losses under 7 subharmonic is 870.25W；

According to step C, suing for peace harmonic wave total losses under each harmonic, in the case of can being somebody's turn to do, harmonic loss is 2291.36W。

Claims (4)

1. three-phase electrical cable harmonic loss computational methods, it is characterised in that comprise the following steps:

A: combining three-phase electrical cable practical structures and the actual system of laying of three-phase electrical cable, considering screen layer electric current and three On the premise of phase power cable phased magnetic field coupling condition, under derivation particular harmonic number of times electric current and voltage, three-phase electrical cable is humorous Ripple loss formula, concrete derivation step is as follows:

A1: determine the alternate spacing of each phase of three-phase electrical cable according to the actual system of laying of three-phase electrical cable, calculate specific Induction electromotive force in each phase screen layer of unit length under overtone order electric current, then in conjunction with three-phase electrical cable screen layer two Termination earth resistance, determines screen layer earth-return according to the screen layer connected mode of three-phase electrical cable and calculates three-phase electrical cable Screen layer impedance and three-phase electrical cable earth-return impedance, final calculating obtains each phase power cable under particular harmonic number of times electric current Screen layer electric current；

A2: calculate three-phase electrical cable each layer function of current density under particular harmonic number of times electric current；

A3: ask for three-phase electrical cable cable core loss under particular harmonic number of times electric current, three-phase electrical cable respectively specific Screen layer loss under overtone order electric current and three-phase electrical cable insulating barrier loss under particular harmonic number of times voltage；So After again by three-phase electrical cable under particular harmonic number of times electric current cable core loss, three-phase electrical cable particular harmonic number of times electricity The insulating barrier loss under particular harmonic number of times voltage of the screen layer loss flowed down and three-phase electrical cable is sued for peace, Obtain three-phase electrical cable each layer harmonic wave total losses under particular harmonic number of times electric current and voltage；

B: by actually detected equipment Inspection, obtain electric current and the voltage of each phase of three-phase electrical cable, then decomposed by FFT Device, analyzes and obtains individual harmonic current and the size of harmonic voltage and relative harmonic content；Then by the voltage electricity under each harmonic Flow valuve carries out computing according to step A, is obtained each phase power cable shield layer electric current under individual harmonic current respectively by step A1, Try to achieve three-phase electrical cable each layer function of current density under individual harmonic current respectively by step A2, divided by step A3 Do not try to achieve three-phase electrical cable each layer harmonic wave total losses under individual harmonic current and voltage；

C: the layer each to the three-phase electrical cable obtained in step B harmonic wave total losses under individual harmonic current and voltage are asked And process, finally obtain and value is three-phase electrical cable harmonic loss.

Three-phase electrical cable harmonic loss computational methods the most according to claim 1, it is characterised in that: described step A1 In, under particular harmonic number of times electric current, the induction electromotive force in each phase screen layer of unit length includes that three-phase cable core electric current is to unit The sensing that each phase screen layer of unit length is produced by the induction electromotive force of each phase screen layer generation of length and screen layer electric current Electromotive force；

The induction electromotive force that three-phase cable core electric current produces in the A phase screen layer P of unit lengthFor:

The induction electromotive force that three-phase cable core electric current produces in the B phase screen layer P of unit lengthFor:

${\stackrel{\&CenterDot;}{e}}_{SB}=2\mathrm{\&omega;}I\&times;{10}^{-7}\&lsqb;\frac{\sqrt{3}}{2}\ln \frac{mS}{R}-j\frac{1}{2}\ln \frac{S}{m\&CenterDot;R}\&rsqb;;$

The induction electromotive force that three-phase cable core electric current produces in the C phase screen layer P of unit lengthFor:

${\stackrel{\&CenterDot;}{e}}_{SC}=2\mathrm{\&omega;}I\&times;{10}^{-7}\&lsqb;-\frac{\sqrt{3}}{2}ln\frac{mS}{R}-j\frac{1}{2}ln\frac{n^{2}S}{m\&CenterDot;R}\&rsqb;;$

Wherein, j is imaginary unit；ω=2 π f, f represents harmonic frequency；P is the conductor being parallel to cable core A, B, C, as virtual Screen layer；The magnetic flux produced in P for three-phase cable core；I is three-phase cable core electric current；Three-phase electrical cable A, C spaced apart is S；A, B spaced apart is n times of S, i.e. nS；B, C spaced apart is m times of S, i.e. mS；R is the radius of P；

The induction electromotive force that unit length screen layer is produced by screen layer electric current is:

${{\stackrel{\&CenterDot;}{e}}_{SA}}^{\&prime;}={\stackrel{\&CenterDot;}{I}}_{SB}\&CenterDot;{\mathrm{jM}}_{AB}+{\stackrel{\&CenterDot;}{I}}_{SC}\&CenterDot;{\mathrm{jM}}_{AC};$

${{\stackrel{\&CenterDot;}{e}}_{SB}}^{\&prime;}={\stackrel{\&CenterDot;}{I}}_{SA}\&CenterDot;{\mathrm{jM}}_{AB}+{\stackrel{\&CenterDot;}{I}}_{SC}\&CenterDot;{\mathrm{jM}}_{BC};$

${{\stackrel{\&CenterDot;}{e}}_{SC}}^{\&prime;}={\stackrel{\&CenterDot;}{I}}_{SA}\&CenterDot;{\mathrm{jM}}_{AC}+{\stackrel{\&CenterDot;}{I}}_{SB}\&CenterDot;{\mathrm{jM}}_{BC};$

Wherein,The induction electric produced in unit length A phase screen layer for B, C phase screen layer electric current in three-phase electrical cable Gesture；The induction electromotive force produced in unit length B phase screen layer for three-phase electrical cable A, C phase screen layer electric current；For The induction electromotive force that three-phase electrical cable A, B phase screen layer electric current produces in unit length C phase screen layer； Represent three-phase electrical cable A, B, C phase screen layer electric current respectively；M_AB、M_BC、M_CAThree-phase electrical cable under representation unit length respectively C Yu A phase screen layer mutual inductance under B Yu C phase screen layer mutual inductance, unit length under A with B phase screen layer mutual inductance, unit length；J is empty Number unit；

Screen layer connected mode according to three-phase electrical cable determines screen layer earth-return, in conjunction with three-phase electrical cable screen layer two Termination earth resistance, and calculate three-phase electrical cable screen layer impedance, three-phase electrical cable earth-return impedance and three-phase electrical cable Total induction electromotive force；The total induction electromotive force of three-phase electrical cable, three-phase electrical cable screen layer impedance and three-phase electrical cable ground Impedance loop is multiplied by line length by institute's value of calculation under unit length and obtains；

Final calculating obtains each phase power cable shield layer electric current under particular harmonic number of times electric current.

Three-phase electrical cable harmonic loss computational methods the most according to claim 2, it is characterised in that: described step A2 In, utilize vector magnetic potential A boundary condition Simultaneous Equations on power cable actual physics layer, introduce bessel function, adopt Use separation of variable solving equation, in conjunction with boundary condition, it is thus achieved that the result of vector magnetic potential A, then pass through relational expression:I.e. can get three-phase electrical cable each layer function of current density J under particular harmonic number of times electric current is:

$J=\underset{n=1}{\overset{\mathrm{\&infin;}}{\&Sigma;}}{J}_n\left(r\right)\&CenterDot;{\mathrm{\&Theta;}}_n\left(\mathrm{\&theta;}\right);$

Wherein, modifying factorM represents the cable phase of studied Exterior cable Number；b_kRepresent the spacing of kth phase cable and studied cable, define a_k=b_k/b₁；I represents the cable core electric current of research cable；I_kTable Show the cable core electric current of kth phase cable beyond studied cable；α_kRepresent the locus of k phase cable；R represents footpath, pole, and θ represents pole Angle.

Three-phase electrical cable harmonic loss computational methods the most according to claim 3, it is characterised in that: described step A3 In,

By the Joule's law of differential form is integrated, calculate the cable core of unit length cable under particular harmonic number of times electric current Loss and screen layer loss are

Wherein, i=1,3, P₁Represent cable core loss, P₂Represent screen layer Loss,

M represents research electricity The cable external cable number of phases；a_k=b_k/b₁, b_kRepresent the spacing of kth phase cable and studied cable；Represent the i-th phase cable conductor Current phase；Represent J_nThe conjugate of (r)；

Under particular harmonic number of times voltage, the insulating barrier loss W of unit length cable is:

Wherein, γ is insulant electrical conductivity, and U is harmonic voltage value, r₁For cable core radius, r₂For screen layer Outer radius.