CN104635018A

CN104635018A - Method for calculating current carrying capacity of submarine cable based on equivalent circuit model

Info

Publication number: CN104635018A
Application number: CN201510085366.8A
Authority: CN
Inventors: 李健; 赵远涛; 黄欲成; 郑伟; 吴高波; 吴庆华; 赵全江; 马凌
Original assignee: China Power Engineering Consultant Group Central Southern China Electric Power Design Institute Corp
Current assignee: China Power Engineering Consultant Group Central Southern China Electric Power Design Institute Corp
Priority date: 2015-02-15
Filing date: 2015-02-15
Publication date: 2015-05-20

Abstract

The invention discloses a method for calculating current carrying capacity of a submarine cable based on an equivalent circuit model. The method comprises the following steps of (1) according to the structure of a circuit, establishing a corresponding equivalent circuit; (2) according to the equivalent circuit in the step (1), listing an equation set; (3) solving the equation set in the step (2), so as to obtain a ratio lambda1 of metal sheath loss and conductor loss and a ratio lambda2 of armor loss and conductor loss; (4) substituting the obtained ratio lambda1 of metal sheath loss and conductor loss and the ratio lambda2 of armor loss and conductor loss in the step (3) into a formula, so as to obtain the current carrying capacity of a submarine backflow conductor cable. The method has the advantages that the structure is simple, convenient and reasonable, the defect that the existing cable current carrying capacity formula is only suitable for the cable with given structure, arrangement method and grounding type is overcome, the key problem of design of the existing mainstream submarine cable product is effectively solved, and the positive function is realized for promoting the progress and innovation of the submarine cable industry.

Description

Based on the subsea cable carrying current calculation method of equivalent-circuit model

Technical field

The present invention relates to subsea cable carrying current calculation method, specifically refer to a kind of subsea cable carrying current calculation method based on equivalent-circuit model.

Background technology

Current-carrying capacity is one of topmost parameter of cable line, no matter its accurate Calculation is to the design of power cable or runs all important in inhibitings.Electric Design department carries out type selecting and cable line design according to current-carrying capacity to cable, and operation department controls according to the load of current-carrying capacity to cable line.

At present, interchange subsea cable carrying current calculation current in the world carries out according to IEC60287-1-1 " the specified continuous current-carrying capacity of cable " clause 1.4, and computing formula is as follows:

I = {[\frac{θ_{c} θ_{a} - W_{d} \cdot [\frac{T_{1}}{2} + n \cdot (T_{2} + T_{3} + T_{4})]}{R \cdot T_{1} + n \cdot R \cdot (1 + λ_{1}) \cdot T_{2} + n \cdot R (1 + λ_{1} + λ_{2}) \cdot (T_{3} + T_{4})}]}^{\frac{1}{2}}

(formula 1)

Wherein:

I is conductor current;

θ _cfor conductor temperature;

θ _afor cable place environment temperature;

W _dfor the dielectric loss of conductor insulation unit head, according to IEC 60287-1-1, clause 2.2 calculates;

T ₁for the thermal resistance between conductor and sheath, according to IEC 60287-2-1, clause 2.1.1 calculates;

T ₂for the thermal resistance of lining between sheath and armouring, according to IEC 60287-2-1, clause 2.1.2 calculates;

T ₃for the thermal resistance of protective coverings of cable, according to IEC 60287-2-1, clause 2.1.3 calculates;

T ₄for the thermal resistivity between cable surface and surrounding medium, according to IEC 60287-2-1, clause 2.2 calculates;

R is running temperature lower conductor AC resistance, according to IEC 60287-1-1, and the resistance of conductor when clause 2.1 calculates or directly can obtain 20 DEG C according to IEC 60228;

λ ₁for the loss factor of protective metal shell, i.e. the ratio of protective metal shell loss and conductor losses, according to IEC 60287-1-1, clause 2.3 calculates;

λ ₂for the loss factor of armouring, i.e. the ratio of loss induced by armour and conductor losses, according to IEC60287-1-1, clause 2.4 calculates.

Temperature, dielectric loss, thermal resistance, resistance etc. in above formula (1) are all by the given formulae discovery of IEC, IEC is for several typical construction of cable, arrangement mode and earthing mode, directly give the loss factor computing formula of each Rotating fields, and do not explain formula origin.In addition, when the construction of cable changes, when the situation different from the given formula of IEC appears in arrangement mode, earthing mode, IEC formulae discovery current-carrying capacity of cable cannot directly be utilized.

Summary of the invention

Object of the present invention will provide a kind of armor to adopt Steel material exactly, set up backflow conductor based on the subsea cable carrying current calculation method of equivalent-circuit model, this method solve the defect that IEC60287-1-1 " the specified continuous current-carrying capacity of cable " clause 1.4 cannot be applicable to take back current conductor subsea cable.

For achieving the above object, the subsea cable carrying current calculation method based on equivalent-circuit model designed by the present invention, comprises the steps:

(1) armor adopts Steel material, sets up backflow conductor in circuit, and backflow conductor is in parallel with armor, sets up corresponding equivalent electrical circuit according to the structure of circuit;

(2) according to the equivalent electrical circuit in step (1), because the electric current in armor, backflow conductor, protective metal shell is all responded to by conductor current, following system of equations is listed:

\begin{matrix} jXs \cdot \overset{\cdot}{I} c + \overset{\cdot}{Z} s \cdot \overset{\cdot}{I} s + {jX}_{R} \cdot {\overset{\cdot}{I}}_{R} + {jX}_{A} \cdot {\overset{\cdot}{I}}_{A} = 0 \\ {jX}_{R} \cdot \overset{\cdot}{I} c + {jX}_{R} \cdot \overset{\cdot}{I} s + {\overset{\cdot}{Z}}_{R} + {jX}_{A} \cdot {\overset{\cdot}{I}}_{A} = 0 \\ {jX}_{A} \cdot \overset{\cdot}{I} c + {jX}_{A} \cdot \overset{\cdot}{I} s + {jX}_{A} \cdot {\overset{\cdot}{I}}_{R} + {\overset{\cdot}{Z}}_{A} \cdot {\overset{\cdot}{I}}_{A} = 0 \end{matrix}

(formula 2)

Wherein, X _sfor the reactance of protective metal shell; X _rfor the reactance of the conductor that refluxes; X _afor the reactance of armor; Z _sfor the impedance of protective metal shell; Z _rfor impedance, the Z of the conductor that refluxes _afor the impedance of armor,

(3) separate the system of equations in step (2), obtain protective metal shell electric current I _s, backflow conductor current I _rwith armor electric current I _awith conductor current I _cratio, thus obtain the ratio P of protective metal shell loss and conductor losses _s, backflow conductor losses and the ratio P of conductor losses _rwith the ratio P of armor loss and conductor losses _a, and then obtain the ratio λ of protective metal shell loss and conductor losses ₁with the ratio λ of loss induced by armour and conductor losses ₂;

(4) by the ratio λ of the protective metal shell loss that obtains in step (3) and conductor losses ₁with the ratio λ of loss induced by armour and conductor losses ₂substitute in IEC60287-1-1 " the specified continuous current-carrying capacity of cable " clause 1.4, seabed backflow conductor cable current-carrying capacity can be tried to achieve.

In technique scheme, in described step (3), the ratio P of protective metal shell loss and conductor losses _s, backflow conductor losses and the ratio P of conductor losses _awith the ratio P of armor loss and conductor losses _rtried to achieve by following formula respectively,

\begin{matrix} P_{S} = {| \frac{{\overset{\cdot}{I}}_{S}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{S}}{R}) \\ P_{R} = {| \frac{{\overset{\cdot}{I}}_{R}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{R}}{R}) \\ P_{A} = {| \frac{{\overset{\cdot}{I}}_{A}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{A}}{R}) \end{matrix}

(formula 4)

Wherein, R _sfor the resistance of protective metal shell; R _rfor the resistance of the conductor that refluxes; R _afor the resistance of armor; R is the resistance of conductor,

The then ratio λ of protective metal shell loss and conductor losses ₁with the ratio λ of loss induced by armour and conductor losses ₂obtained by following formula:

λ ₁＝P _S，λ ₂＝P _R+P _A

Advantage of the present invention is mainly reflected in following several respects:

One, compensate for IEC 60287-1-1 " the specified continuous current-carrying capacity of cable " clause 1.4 formula is only applicable to the deficiency of the cable of fixed structure, arrangement mode and Contact patch, efficiently solve the key issue of current main flow subsea cable product design, for promotion subsea cable industry progress and innovation, there is positive role.

Its two, computing method are rationally easy.

Accompanying drawing explanation

Fig. 1 is the schematic equivalent circuit of the subsea cable that have employed backflow conductor.

In figure: conductor 1; Plumbous cover 2; Backflow conductor 3; Armor 4.

Embodiment

Below in conjunction with the drawings and specific embodiments, the present invention is described in further detail, but this embodiment should not be construed limitation of the present invention.

The schematic equivalent circuit that have employed the subsea cable of backflow conductor shown in figure, comprises conductor 1, plumbous cover 2, backflow conductor 3, armor 4, and the electric current in plumbous cover 2, backflow conductor 3 and armor 4 is all responded to by the electric current of conductor 1.

Based on the subsea cable carrying current calculation method of equivalent-circuit model, comprise the steps:

(1) armor 4 adopts Steel material, in order to increase the intensity of subsea cable, has just set up backflow conductor 3, and backflow conductor 3 is in parallel with armor 4, according to circuit structure, sets up the equivalent electrical circuit of backflow conductor cable;

(2) according to the equivalent electrical circuit described in step (1), following system of equations is listed:

\begin{matrix} jXs \cdot \overset{\cdot}{I} c + \overset{\cdot}{Z} s \cdot \overset{\cdot}{I} s + {jX}_{R} \cdot {\overset{\cdot}{I}}_{R} + {jX}_{A} \cdot {\overset{\cdot}{I}}_{A} = 0 \\ {jX}_{R} \cdot \overset{\cdot}{I} c + {jX}_{R} \cdot \overset{\cdot}{I} s + {\overset{\cdot}{Z}}_{R} + {jX}_{A} \cdot {\overset{\cdot}{I}}_{A} = 0 \\ {jX}_{A} \cdot \overset{\cdot}{I} c + {jX}_{A} \cdot \overset{\cdot}{I} s + {jX}_{A} \cdot {\overset{\cdot}{I}}_{R} + {\overset{\cdot}{Z}}_{A} \cdot {\overset{\cdot}{I}}_{A} = 0 \end{matrix}

(formula 2)

Wherein, X _sfor the reactance of lead cover; X _rfor the reactance of the conductor that refluxes; X _afor the reactance of armor; Z _sfor the impedance of lead cover; Z _rfor impedance, the Z of the conductor that refluxes _afor the impedance of armor,

Be out of shape to obtain following system of equations:

\begin{matrix} {\overset{\cdot}{Z}}_{s} (\frac{{\overset{\cdot}{I}}_{s}}{{\overset{\cdot}{I}}_{c}}) + {jX}_{R} (\frac{{\overset{\cdot}{I}}_{R}}{{\overset{\cdot}{I}}_{C}}) + {jX}_{A} (\frac{{\overset{\cdot}{I}}_{A}}{{\overset{\cdot}{I}}_{C}}) = {- jX}_{S} \\ {jX}_{R} (\frac{{\overset{\cdot}{I}}_{S}}{{\overset{\cdot}{I}}_{C}}) + {\overset{\cdot}{Z}}_{R} (\frac{{\overset{\cdot}{I}}_{R}}{{\overset{\cdot}{I}}_{C}}) + {jX}_{A} = {- jX}_{R} \\ {jX}_{A} (\frac{{\overset{\cdot}{I}}_{S}}{{\overset{\cdot}{I}}_{C}}) + {jX}_{A} (\frac{{\overset{\cdot}{I}}_{R}}{{\overset{\cdot}{I}}_{C}}) + {\overset{\cdot}{Z}}_{A} (\frac{{\overset{\cdot}{I}}_{A}}{{\overset{\cdot}{I}}_{C}}) = {- jX}_{A} \end{matrix}

(formula 3)

(3) separate the system of equations described in step (2), obtain plumbous cover electric current I _s, backflow conductor current I _rwith armor electric current I _awith conductor current I _cratio, thus obtain the ratio P of the loss of plumbous cover and conductor losses _s, backflow conductor losses and the ratio P of conductor losses _rwith the ratio P of armor loss and conductor losses _a, and then obtain the ratio λ of the loss of plumbous cover and conductor losses ₁with the ratio λ of loss induced by armour and conductor losses ₂;

The ratio P of the loss of plumbous cover and conductor losses _s, backflow conductor losses and the ratio P of conductor losses _awith the ratio P of armor loss and conductor losses _rtried to achieve by following formula respectively,

\begin{matrix} P_{S} = {| \frac{{\overset{\cdot}{I}}_{S}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{S}}{R}) \\ P_{R} = {| \frac{{\overset{\cdot}{I}}_{R}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{R}}{R}) \\ P_{A} = {| \frac{{\overset{\cdot}{I}}_{A}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{A}}{R}) \end{matrix}

(formula 4)

Wherein, R _sfor the resistance of lead cover; R _rfor the resistance of the conductor that refluxes; R _afor the resistance of armor; R is the resistance of conductor,

The then ratio λ of the loss of plumbous cover and conductor losses ₁with the ratio λ of loss induced by armour and conductor losses ₂obtained by following formula:

λ ₁＝P _S，λ ₂＝P _R+P _A

(4) lead obtained in step (3) is overlapped the ratio λ of loss and conductor losses ₁with the ratio λ of loss induced by armour and conductor losses ₂substitute in IEC60287-1-1 " the specified continuous current-carrying capacity of cable " clause 1.4, seabed backflow conductor cable current-carrying capacity can be tried to achieve;

The content be not described in detail in this instructions, belongs to the known prior art of those skilled in the art.

Claims

1., based on the subsea cable carrying current calculation method of equivalent-circuit model, it is characterized in that: comprise the steps:

(2) according to the equivalent electrical circuit in step (1), following system of equations is listed:

jXs \cdot \overset{\cdot}{I} c + {\overset{\cdot}{Z}}_{S} \cdot \overset{\cdot}{I} s + j X_{R} \cdot {\overset{\cdot}{I}}_{R} + j X_{A} \cdot {\overset{\cdot}{I}}_{A} = 0

j X_{R} \cdot \overset{\cdot}{I} c + j X_{R} \cdot \overset{\cdot}{I} s + {\overset{\cdot}{Z}}_{R} \cdot {\overset{\cdot}{I}}_{R} + j X_{A} \cdot {\overset{\cdot}{I}}_{A} = 0

(formula 2)

j X_{A} \cdot \overset{\cdot}{I} c + j X_{A} \cdot \overset{\cdot}{I} s + j X_{A} \cdot {\overset{\cdot}{I}}_{R} + {\overset{\cdot}{Z}}_{A} \cdot {\overset{\cdot}{I}}_{A} = 0

(3) separate the system of equations in step (2), obtain protective metal shell electric current I _swith conductor current I _cratio, backflow conductor current I _rwith conductor current I _cratio and armor electric current I _awith conductor current I _cratio, thus obtain the ratio P of protective metal shell loss and conductor losses _s, backflow conductor losses and the ratio P of conductor losses _rwith the ratio P of armor loss and conductor losses _a, and then obtain the ratio λ of protective metal shell loss and conductor losses ₁with the ratio λ of loss induced by armour and conductor losses ₂;

(4) by the ratio λ of the protective metal shell loss that obtains in step (3) and conductor losses ₁with the ratio λ of loss induced by armour and conductor losses ₂substitute in the specified continuous current-carrying capacity formula of cable, seabed backflow conductor cable current-carrying capacity can be tried to achieve.

2. according to the subsea cable carrying current calculation method based on equivalent-circuit model described in claim 1, it is characterized in that: in described step (3), the ratio P of protective metal shell loss and conductor losses _s, backflow conductor losses and the ratio P of conductor losses _awith the ratio P of armor loss and conductor losses _rtried to achieve by following formula respectively,

P_{S} = {| \frac{{\overset{\cdot}{I}}_{S}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{S}}{R})

P_{R} = {| \frac{{\overset{\cdot}{I}}_{R}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{R}}{R})

(formula 4)

P_{A} = {| \frac{{\overset{\cdot}{I}}_{A}}{{\overset{\cdot}{I}}_{C}} |}^{2} \cdot (\frac{R_{A}}{R})

λ ₁＝P _S，λ ₂＝P _R+P _A。