CN106329455B - Prefabricated flexible direct current cable termination stress wimble structure - Google Patents

Prefabricated flexible direct current cable termination stress wimble structure Download PDF

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CN106329455B
CN106329455B CN201610763919.5A CN201610763919A CN106329455B CN 106329455 B CN106329455 B CN 106329455B CN 201610763919 A CN201610763919 A CN 201610763919A CN 106329455 B CN106329455 B CN 106329455B
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mrow
mfrac
msup
gamma
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CN106329455A (en
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朱智恩
杨黎明
陈龙啸
李栋
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State Grid Corp of China SGCC
State Grid Jiangsu Electric Power Co Ltd
NARI Group Corp
Economic and Technological Research Institute of State Grid Jiangsu Electric Power Co Ltd
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Nanjing NARI Group Corp
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    • H—ELECTRICITY
    • H02—GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02G—INSTALLATION OF ELECTRIC CABLES OR LINES, OR OF COMBINED OPTICAL AND ELECTRIC CABLES OR LINES
    • H02G15/00—Cable fittings
    • H02G15/02—Cable terminations
    • H—ELECTRICITY
    • H02—GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02G—INSTALLATION OF ELECTRIC CABLES OR LINES, OR OF COMBINED OPTICAL AND ELECTRIC CABLES OR LINES
    • H02G1/00—Methods or apparatus specially adapted for installing, maintaining, repairing or dismantling electric cables or lines
    • H02G1/14—Methods or apparatus specially adapted for installing, maintaining, repairing or dismantling electric cables or lines for joining or terminating cables
    • H—ELECTRICITY
    • H02—GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER
    • H02G—INSTALLATION OF ELECTRIC CABLES OR LINES, OR OF COMBINED OPTICAL AND ELECTRIC CABLES OR LINES
    • H02G15/00—Cable fittings
    • H02G15/02—Cable terminations
    • H02G15/06—Cable terminating boxes, frames or other structures
    • H02G15/064—Cable terminating boxes, frames or other structures with devices for relieving electrical stress

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  • Battery Electrode And Active Subsutance (AREA)
  • Secondary Cells (AREA)
  • Processing Of Terminals (AREA)

Abstract

The present invention provides a kind of prefabricated flexible direct current cable termination stress wimble structure, including reinforced insulation layer, stress cone semi-conductive layer and stress cone curve, wherein described stress cone semi-conductive layer is placed on direct current cables body insulating layer, the reinforced insulation is placed on the stress cone semi-conductive layer, and the stress cone curve is the lower edge of the stress cone semi-conductive layer；On the basis of considering temperature, electric field factor to resistivity effects, propose the electric field distribution in the soft straight cable terminating insulation of prefabricated, provide fundamental basis for soft straight cable stud connector design；Provide a kind of soft straight cable terminal reinforced insulation layer thickness computational methods of prefabricated, it is ensured that cable insulation reinforced insulation bed boundary electric field is in the reasonable scope；Rational stress cone-shaped is devised, soft straight cable terminal potential line concentration phenomenon is solved, and ensures entire stress cone electric fields uniform；Meet entire soft straight cable system to the requirement on electric performance of terminal, ensure that the long-term safety of soft straight cable system is reliable.

Description

Prefabricated flexible direct current cable termination stress wimble structure
Technical field
The present invention relates to a kind of prefabricated flexible direct current cable termination stress wimble structures.
Background technology
Flexible DC power transmission is increasingly taken seriously as a kind of New type of transmission.Flexible direct current cable system is The important component of soft straight transmission system is made of soft straight cable body and soft straight cable connector (connector and terminal).It is soft Straight cable terminal is the key component of cable system and is easier to the link to break down, therefore restricts soft straight cable system It unites and develops to higher voltage grade.
In cable termination design process, it is necessary to consider the electric field distribution in cable stress cone.In ac cable terminal, Dielectric constant of the electric field distribution depending on insulating materials, it is unrelated with Temperature Distribution.In direct current cables terminating insulation, electric field distribution Depending on resistivity distribution, and resistivity is distributed with temperature and electric field there are relation, therefore its electric field distribution situation is increasingly complex.
At present, the not no design theory and design method on soft straight cable terminal, there are no carried based on DC electric field distribution The structure design gone out.In existing soft straight cable terminal structure design, largely apply mechanically AC terminal structure design and carry out slightly Micro-adjustment does not propose specific design theory and design method.
The content of the invention
The present invention provides a kind of prefabricated flexible direct current to solve drawbacks described above and deficiency in the prior art Cable termination stress wimble structure.
In order to solve the above technical problems, the present invention provides a kind of prefabricated flexible direct current cable termination stress wimble structure, bag Reinforced insulation layer, stress cone semi-conductive layer and stress cone curve are included, wherein the stress cone semi-conductive layer is placed in direct current cables sheet On body insulating layer, the reinforced insulation is placed on the stress cone semi-conductive layer, and the stress cone curve is the stress cone The lower edge of semi-conductive layer, the computational methods of wherein stress cone curve comprise the following steps：
Step 1 calculates radial electric field intensity E at stress cone curve y2(y)；
Wherein, ρ2(y) it is the resistivity at reinforced insulation layer y, U bears voltage for reinforced insulation layer, and R (y) is exhausted for enhancing Unit resistance of the edge layer inner surface at stress cone curve y；
Step 2 determines the thickness of reinforced insulation layer；
Reinforced insulation layer radius RsThe radial electric field intensity E at place2(Rs) it is cable insulation maximum functional electric field strength E0's Half, expression formula are as follows：
E2(RS)=0.5E0
With reference to the expression formula and above formula of radial electric field intensity at stress cone curve y, R is calculated to obtains, so as to which enhancing be calculated Thickness of insulating layer Δ n=Rs- R, R are cable insulation radius；
Step 3, identified sign cone curvilinear equation；
The axial electric field strength E of any point on stress cone curvetWith the radial electric field E of the point2There are following relations：
Above formula is integrated to obtainMake EtFor constant, by radial electric field E2Expression formula is brought intoAnd several groups of coordinates (x, y) are obtained using numerical method, obtain stress cone curvilinear equation.
Wherein, in step 1, the electricalresistivityρ at reinforced insulation layer y2(y) calculating process is as follows：
At reinforced insulation layer r' with the temperature difference of cable conductor：
I.e.：
In formula, θ2It is the temperature at r' for reinforced insulation layer outer diameter, θRFor the temperature of cable insulation outer surface, θcFor electricity Cable conductor temperature, R be cable insulation radius, ρT2For the thermal resistivity of reinforced insulation layer, ρT1For the thermal resistance system of cable insulation Number, WcIt is lost for cable conductor；rcFor cable conductor radius；
According to resistivity formula, electricalresistivityρ at reinforced insulation layer r'2(r') expression formula is：
Wherein,ρ2,0For the resistivity of reinforced insulation layer Coefficient；a2For the temperature coefficient of resistivity of reinforced insulation layer；γ2For the resistivity electric field coefficient of reinforced insulation layer；E2(r') it is increasing Radial electric field intensity at strong insulating layer r', according to Ohm's law：Therefore,I should Electric current at power cone curve y；ByWith electricalresistivityρ at reinforced insulation layer r'2(r') expression formula obtains：Bring this formula at reinforced insulation layer r' electricalresistivityρ2(r') expression formula obtains：
Wherein,
In step 1, the calculating process of unit resistance R (y) of the reinforced insulation layer inner surface at stress cone curve y is as follows：
Wherein, resistivity at reinforced insulation layer r'It is electric at cable insulation r Resistance rate
From hot road equation：
I.e.：
According to Ohm's law：Therefore,Electric current at I stress cone curves y；ByWith electricalresistivityρ at cable insulation r1(r) expression formula obtains：This formula is brought into Electricalresistivityρ at cable insulation r1(r) expression formula obtains：
Therefore it is as follows to obtain unit resistance R (y) expression formula of the reinforced insulation layer inner surface at stress cone curve y：
With reference to unit resistance R (y) expression formula and reinforced insulation layer y of the reinforced insulation layer inner surface at stress cone curve y The expression formula of place's resistivity obtains radial electric field intensity E at stress cone curve y2(y) it is：
It is as follows that letter involved in foregoing calculating formula represents meaning：
θcFor cable conductor temperature, θRFor the temperature of cable insulation outer surface, θ1For the temperature at cable insulation outer diameter r Degree, θ2For the temperature at reinforced insulation layer outer diameter r', R is cable insulation radius, WcIt is lost for cable conductor；rcIt is led for cable Body radius；ρT2For the thermal resistivity of reinforced insulation layer, ρT1For the thermal resistivity of cable insulation, ρ1,0For the electricity of cable insulation Hinder rate coefficient, ρ2,0For the resistivity coefficient of reinforced insulation layer, a1For cable insulation temperature coefficient of resistivity, a2For reinforced insulation The temperature coefficient of resistivity of layer, γ1For the resistivity electric field coefficient of cable insulation, γ2For the resistivity electric field of reinforced insulation layer Coefficient, E1(r) it is the radial electric field intensity at cable insulation radius r, E2(r') it is the radial direction electricity at reinforced insulation layer radius r' Field intensity；
The advantageous effects that the present invention is reached：1. on the basis of considering temperature, electric field factor to resistivity effects, It proposes the electric field distribution in the soft straight cable terminating insulation of prefabricated, provides fundamental basis for soft straight cable stud connector design；2. it carries A kind of soft straight cable terminal reinforced insulation layer thickness computational methods of prefabricated are supplied, it is ensured that cable insulation-reinforced insulation stratum boundary Face electric field is in the reasonable scope；3. devising rational stress cone-shaped, solve soft straight cable terminal potential line concentration phenomenon, And ensure entire stress cone electric fields uniform；4. meeting requirement on electric performance of the entire soft straight cable system to terminal, ensure soft straight electricity The long-term safety of cable system is reliable.
Specific embodiment
With reference to specific embodiment, the invention will be further described, and following embodiment is only used for clearly illustrating Technical scheme, and be not intended to limit the protection scope of the present invention and limit the scope of the invention.
The present invention provides a kind of prefabricated flexible direct current cable termination stress wimble structure, including reinforced insulation layer, stress cone Semi-conductive layer and stress cone curve, wherein the stress cone semi-conductive layer is placed on direct current cables insulating layer, the reinforced insulation It is placed on the stress cone semi-conductive layer, the stress cone curve is the lower edge of the stress cone semi-conductive layer, wherein should The computational methods of power cone curve comprise the following steps：
Step 1 calculates radial electric field intensity E at stress cone curve y2(y)；
Wherein, ρ2(y) it is the resistivity at reinforced insulation layer y, U bears voltage for reinforced insulation layer, and R (y) is exhausted for enhancing Unit resistance of the edge layer inner surface at stress cone curve y；
Electricalresistivityρ at one, reinforced insulation layer y2(y) calculating process is as follows：
At reinforced insulation layer r' with the temperature difference of cable conductor：
I.e.：
According to resistivity formula, electricalresistivityρ at reinforced insulation layer r'2(r') expression formula is：
According to Ohm's law：Therefore,Electric current at I stress cone curves y；ByWith electricalresistivityρ at reinforced insulation layer r'2(r') expression formula obtains：It will This formula brings electricalresistivityρ at reinforced insulation layer r' into2(r') expression formula obtains：
Two, the calculating process of unit resistance R (y) of the reinforced insulation layer inner surface at stress cone curve y is as follows：
Wherein, resistivity at reinforced insulation layer r'It is electric at cable insulation r Resistance rate
From hot road equation：
I.e.：
According to Ohm's law：Therefore,Electric current at I stress cone curves y；ByWith electricalresistivityρ at cable insulation r1(r) expression formula obtains：By this formula band Enter electricalresistivityρ at cable insulation r1(r) expression formula obtains：
Therefore it is as follows to obtain unit resistance R (y) expression formula of the reinforced insulation layer inner surface at stress cone curve y：
With reference to unit resistance R (y) expression formula and reinforced insulation layer y of the reinforced insulation layer inner surface at stress cone curve y The expression formula of place's resistivity obtains radial electric field intensity E at stress cone curve y2(y) it is：
It is as follows that letter involved in foregoing calculating formula represents meaning：
θcFor cable conductor temperature, θRFor the temperature of cable insulation outer surface, θ1For the temperature at cable insulation outer diameter r Degree, θ2For the temperature at reinforced insulation layer outer diameter r', R is cable insulation radius, WcIt is lost for cable conductor；rcIt is led for cable Body radius；ρT2For the thermal resistivity of reinforced insulation layer, ρT1For the thermal resistivity of cable insulation, ρ1,0For the electricity of cable insulation Hinder rate coefficient, ρ2,0For the resistivity coefficient of reinforced insulation layer, a1For cable insulation temperature coefficient of resistivity, a2For reinforced insulation The temperature coefficient of resistivity of layer, γ1For the resistivity electric field coefficient of cable insulation, γ2For the resistivity electric field of reinforced insulation layer Coefficient, E1(r) it is the radial electric field intensity at cable insulation radius r, E2(r') it is the radial direction electricity at reinforced insulation layer radius r' Field intensity；
Step 2 determines the thickness of reinforced insulation layer；
Reinforced insulation layer radius RsThe radial electric field intensity E at place2(Rs) it is cable insulation maximum functional electric field strength E0's Half, expression formula are as follows：
E2(RS)=0.5E0
With reference to the expression formula and above formula of radial electric field intensity at stress cone curve y, R is calculated to obtains, so as to which enhancing be calculated Thickness of insulating layer Δ n=Rs- R, R are cable insulation radius；
Step 3, identified sign cone curvilinear equation；
The axial electric field strength E of any point on stress cone curvetWith the radial electric field E of the point2There are following relations：
Above formula is integrated to obtainMake EtFor constant, by radial electric field E2Expression formula is brought intoAnd several groups of coordinates (x, y) are obtained using numerical method, obtain stress cone curvilinear equation.
Embodiment one
Using the soft straight cable terminal of rated voltage ± 320kV prefabricateds.
Radial electric field intensity E at one, stress cone curve y2(y) calculate
For the soft straight cable terminal of ± 320kV prefabricateds, U=320kV；ρ1,0=ρ2,0=1016Ω.m；a1=0.06 DEG C-1, a2=0.05 DEG C-1；θc=90 DEG C, wc=68W；ρT1=3.5K.m/W, ρT2=3.3K.m/W；rc=26.5mm, R=50.5mm； γ1=2.2, γ2=1.69, bring above-mentioned data at stress cone curve y radial electric field intensity E2(y) calculation formula：
Two, reinforced insulation layer thickness
Soft straight cable body insulation thickness R=24mm, maximum field E0=18kV/mm, then E (Rs)=9kV/mm.
E (R are taken to leave enough safety marginss)=7kV/mm, which substitutes into formula above formula, can obtain RS=71.5mm can be enhanced Insulation thickness Δ n=RS- R=21mm.
Three, identified sign cone curvilinear equation
It is computed that several groups of coordinates (x, y) on the soft straight cable terminal stress cone curve of ± 320kV prefabricateds can be obtained, respectively It is：(0,50.5), (11.9,51.5), (23.4,52.5), (34.5,53.5), (45.2,54.5), (55.6,55.5), (65.6,56.5), (75.4,57.5), (84.9,58.5), (94.1,59.5), (103.1,60.5), (111.8,61.5), (120.3,62.5), (128.6,63.5), (136.7,64.5), (144.6,65.5), (152.3,66.5), (159.9, 67.5), (167.3,68.5), (174.6,69.5), (181.7,70.5), (188.7,71.5) finally obtain stress cone shaped form Shape.
Embodiment 2
Using the soft straight cable terminal of rated voltage ± 200kV prefabricateds.
Radial electric field intensity E at one, stress cone curve y2(y) calculate
For the soft straight cable terminal of ± 200kV prefabricateds, U=200kV；ρ1,0=ρ2,0=1016Ω.m；a1=0.06 DEG C-1, a2=0.05 DEG C-1；θc=90 DEG C, wc=66W；ρT1=3.5K.m/W, ρT2=3.3K.m/W；rc=20.4mm, R=36.4mm； γ1=2.2, γ2=1.69, bring above-mentioned data at stress cone curve y radial electric field intensity E2(y) calculation formula：
Two, reinforced insulation layer thickness
Soft straight cable body insulation thickness R=15mm, maximum field E0=20kV/mm, then E (Rs)=10kV/mm.
E (R are taken to leave enough safety marginss)=7kV/mm, which substitutes into formula (22), can obtain RS=46.4mm can be enhanced Insulation thickness Δ n=RS- R=10mm.
Three, identified sign cone curvilinear equation
It is computed that several groups of coordinates (x, y) on the soft straight cable terminal stress cone curve of ± 200kV prefabricateds can be obtained, respectively It is：(0,36.4), (12.2,37.4), (23.7,38.4), (34.6,39.4), (44.9,40.4), (54.7,41.4), (64, 42.4), (72.9,43.4), (81.4,44.4), (89.6,45.4), (97.6,46.4) finally obtain stress cone curve shape.
The above is only the preferred embodiment of the present invention, it is noted that for the ordinary skill people of the art For member, without departing from the technical principles of the invention, several improvement and deformation can also be made, these are improved and deformation Also it should be regarded as protection scope of the present invention.

Claims (1)

1. prefabricated flexible direct current cable termination stress wimble structure, it is characterised in that：Including reinforced insulation layer, stress cone semiconductive Layer and stress cone curve, wherein the stress cone semi-conductive layer is placed on direct current cables body insulating layer, the reinforced insulation layer It is placed on the stress cone semi-conductive layer, the stress cone curve is the lower edge of the stress cone semi-conductive layer, wherein stress The computational methods of cone curve comprise the following steps：
Step 1 calculates radial electric field intensity E at stress cone curve y2(y)；
<mrow> <msub> <mi>E</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>&amp;rho;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> <mi>y</mi> </mrow> </mfrac> <mo>&amp;CenterDot;</mo> <mfrac> <mi>U</mi> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow>
Wherein, ρ2(y) it is the resistivity at reinforced insulation layer y, U bears voltage for reinforced insulation layer, and R (y) is in reinforced insulation layer Unit resistance of the surface at stress cone curve y；
Electricalresistivityρ at reinforced insulation layer y2(y) calculating process is as follows：
At reinforced insulation layer r' with the temperature difference of cable conductor：
<mrow> <msub> <mi>&amp;theta;</mi> <mi>c</mi> </msub> <mo>-</mo> <msub> <mi>&amp;theta;</mi> <mn>2</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>&amp;theta;</mi> <mi>c</mi> </msub> <mo>-</mo> <msub> <mi>&amp;theta;</mi> <mi>R</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mrow> <mo>(</mo> <msub> <mi>&amp;theta;</mi> <mi>R</mi> </msub> <mo>-</mo> <msub> <mi>&amp;theta;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <msub> <mi>W</mi> <mi>c</mi> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>&amp;rho;</mi> <mrow> <mi>T</mi> <mn>1</mn> </mrow> </msub> <mi>l</mi> <mi>n</mi> <mfrac> <mi>R</mi> <msub> <mi>r</mi> <mi>c</mi> </msub> </mfrac> <mo>+</mo> <mfrac> <msub> <mi>W</mi> <mi>c</mi> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>&amp;rho;</mi> <mrow> <mi>T</mi> <mn>2</mn> </mrow> </msub> <mi>l</mi> <mi>n</mi> <mfrac> <msup> <mi>r</mi> <mo>&amp;prime;</mo> </msup> <mi>R</mi> </mfrac> </mrow>
I.e.：
In formula, θ2It is the temperature at r' for reinforced insulation layer outer diameter, θRFor the temperature of cable insulation outer surface, θcIt is led for cable Temperature, R be cable insulation radius, ρT2For the thermal resistivity of reinforced insulation layer, ρT1For the thermal resistivity of cable insulation, Wc It is lost for cable conductor；rcFor cable conductor radius；
According to resistivity formula, electricalresistivityρ at reinforced insulation layer r'2(r') expression formula is：
<mrow> <msub> <mi>&amp;rho;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <msup> <mi>r</mi> <mo>&amp;prime;</mo> </msup> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>&amp;rho;</mi> <mrow> <mn>2</mn> <mo>,</mo> <mn>0</mn> </mrow> </msub> <msup> <mi>e</mi> <mrow> <mo>-</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <msub> <mi>&amp;theta;</mi> <mn>2</mn> </msub> </mrow> </msup> </mrow> <mrow> <msub> <mi>E</mi> <mn>2</mn> </msub> <msup> <mrow> <mo>(</mo> <msup> <mi>r</mi> <mo>&amp;prime;</mo> </msup> <mo>)</mo> </mrow> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </msup> </mrow> </mfrac> <mo>=</mo> <msub> <mi>c</mi> <mn>3</mn> </msub> <mfrac> <msup> <mi>r</mi> <mrow> <mo>&amp;prime;</mo> <msub> <mi>c</mi> <mn>4</mn> </msub> </mrow> </msup> <mrow> <msub> <mi>E</mi> <mn>2</mn> </msub> <msup> <mrow> <mo>(</mo> <msup> <mi>r</mi> <mo>&amp;prime;</mo> </msup> <mo>)</mo> </mrow> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </msup> </mrow> </mfrac> </mrow>
Wherein,ρ2,0For the resistivity coefficient of reinforced insulation layer； a2For the temperature coefficient of resistivity of reinforced insulation layer；γ2For the resistivity electric field coefficient of reinforced insulation layer；E2(r') it is exhausted for enhancing Radial electric field intensity at edge layer r', according to Ohm's law：Therefore,I stress cones Electric current at curve y；ByWith electricalresistivityρ at reinforced insulation layer r'2(r') expression formula obtains：Bring this formula at reinforced insulation layer r' electricalresistivityρ2(r') expression formula obtains：
<mrow> <msub> <mi>&amp;rho;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <msup> <mi>r</mi> <mo>&amp;prime;</mo> </msup> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>3</mn> </msub> <msup> <msub> <mi>c</mi> <mn>7</mn> </msub> <mrow> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </msup> <msup> <mi>r</mi> <mrow> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <msub> <mi>c</mi> <mn>8</mn> </msub> </mrow> </msup> <msup> <mi>I</mi> <mrow> <mo>-</mo> <mfrac> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mrow> </msup> </mrow>
Wherein,
The calculating process of unit resistance R (y) of the reinforced insulation layer inner surface at stress cone curve y is as follows：
<mrow> <mi>R</mi> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <msub> <mi>r</mi> <mi>c</mi> </msub> <mi>R</mi> </munderover> <mfrac> <mrow> <msub> <mi>&amp;rho;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>r</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> <mi>r</mi> </mrow> </mfrac> <mi>d</mi> <mi>r</mi> <mo>+</mo> <munderover> <mo>&amp;Integral;</mo> <mi>R</mi> <mi>y</mi> </munderover> <mfrac> <mrow> <msub> <mi>&amp;rho;</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <msup> <mi>r</mi> <mo>&amp;prime;</mo> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> <mi>r</mi> </mrow> </mfrac> <msup> <mi>dr</mi> <mo>&amp;prime;</mo> </msup> </mrow>
Wherein, resistivity at reinforced insulation layer r'Resistivity at cable insulation r
From hot road equation：
<mrow> <msub> <mi>&amp;theta;</mi> <mi>c</mi> </msub> <mo>-</mo> <msub> <mi>&amp;theta;</mi> <mn>1</mn> </msub> <mo>=</mo> <mfrac> <msub> <mi>W</mi> <mi>c</mi> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>&amp;rho;</mi> <mrow> <mi>T</mi> <mn>1</mn> </mrow> </msub> <mi>l</mi> <mi>n</mi> <mfrac> <mi>r</mi> <msub> <mi>r</mi> <mi>c</mi> </msub> </mfrac> </mrow>
I.e.：
According to Ohm's law：Therefore,Electric current at I stress cone curves y；ByWith electricalresistivityρ at cable insulation r1(r) expression formula obtains：This formula is brought into Electricalresistivityρ at cable insulation r1(r) expression formula obtains：
<mrow> <msub> <mi>&amp;rho;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>r</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>1</mn> </msub> <msup> <msub> <mi>c</mi> <mn>5</mn> </msub> <mrow> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> </mrow> </msup> <msup> <mi>r</mi> <mrow> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> <msub> <mi>c</mi> <mn>6</mn> </msub> </mrow> </msup> <msup> <mi>I</mi> <mrow> <mo>-</mo> <mfrac> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mrow> </msup> </mrow>
Therefore it is as follows to obtain unit resistance R (y) expression formula of the reinforced insulation layer inner surface at stress cone curve y：
<mrow> <mi>R</mi> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>9</mn> </msub> <msup> <mi>I</mi> <mrow> <mo>-</mo> <mfrac> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> </mrow> </mfrac> </mrow> </msup> <mo>+</mo> <msub> <mi>c</mi> <mn>10</mn> </msub> <msup> <mi>y</mi> <mrow> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <msub> <mi>c</mi> <mn>8</mn> </msub> </mrow> </msup> <msup> <mi>I</mi> <mrow> <mo>-</mo> <mfrac> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mrow> </msup> <mo>-</mo> <msub> <mi>c</mi> <mn>11</mn> </msub> <msup> <mi>I</mi> <mrow> <mo>-</mo> <mfrac> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </mfrac> </mrow> </msup> </mrow>
With reference to electricity at unit resistance R (y) expression formulas of the reinforced insulation layer inner surface at stress cone curve y and reinforced insulation layer y The expression formula of resistance rate obtains radial electric field intensity E at stress cone curve y2(y) it is：
<mrow> <msub> <mi>E</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>c</mi> <mn>12</mn> </msub> <msup> <mi>y</mi> <msub> <mi>c</mi> <mn>13</mn> </msub> </msup> <mfrac> <mn>1</mn> <mrow> <msub> <mi>c</mi> <mn>9</mn> </msub> <msup> <mrow> <mo>&amp;lsqb;</mo> <mfrac> <mrow> <msub> <mi>E</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>y</mi> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mi>C</mi> <mn>7</mn> </msub> <msup> <mi>y</mi> <msub> <mi>c</mi> <mn>8</mn> </msub> </msup> </mrow> </mfrac> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mo>&amp;lsqb;</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <mo>-</mo> <mfrac> <mrow> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>&amp;rsqb;</mo> </mrow> </msup> <mo>+</mo> <msub> <mi>c</mi> <mn>10</mn> </msub> <msup> <mi>y</mi> <mrow> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <msub> <mi>c</mi> <mn>8</mn> </msub> </mrow> </msup> <mo>-</mo> <msub> <mi>c</mi> <mn>11</mn> </msub> </mrow> </mfrac> </mrow>
It is as follows that letter involved in foregoing calculating formula represents meaning：
θcFor cable conductor temperature, θRFor the temperature of cable insulation outer surface, θ1For the temperature at cable insulation outer diameter r, θ2 For the temperature at reinforced insulation layer outer diameter r', R is cable insulation radius, WcIt is lost for cable conductor；rcFor cable conductor half Footpath；ρT2For the thermal resistivity of reinforced insulation layer, ρT1For the thermal resistivity of cable insulation, ρ1,0For the resistivity of cable insulation Coefficient, ρ2,0For the resistivity coefficient of reinforced insulation layer, a1For cable insulation temperature coefficient of resistivity, a2For reinforced insulation layer Temperature coefficient of resistivity, γ1For the resistivity electric field coefficient of cable insulation, γ2For the resistivity electric field system of reinforced insulation layer Number, E1(r) it is the radial electric field intensity at cable insulation radius r, E2(r') it is the radial electric field at reinforced insulation layer radius r' Intensity；
<mrow> <msub> <mi>c</mi> <mn>1</mn> </msub> <mo>=</mo> <msub> <mi>&amp;rho;</mi> <mrow> <mn>1</mn> <mo>,</mo> <mn>0</mn> </mrow> </msub> <msup> <mi>e</mi> <mrow> <mo>-</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> <msub> <mi>&amp;theta;</mi> <mi>c</mi> </msub> </mrow> </msup> <msup> <msub> <mi>r</mi> <mi>c</mi> </msub> <mrow> <mo>-</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> <msub> <mi>&amp;rho;</mi> <mrow> <mi>T</mi> <mn>1</mn> </mrow> </msub> <mfrac> <msub> <mi>w</mi> <mi>c</mi> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> </mrow> </msup> <mo>,</mo> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>=</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> <msub> <mi>&amp;rho;</mi> <mrow> <mi>T</mi> <mn>1</mn> </mrow> </msub> <mfrac> <msub> <mi>w</mi> <mi>c</mi> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>c</mi> <mn>3</mn> </msub> <mo>=</mo> <msub> <mi>&amp;rho;</mi> <mrow> <mn>2</mn> <mo>,</mo> <mn>0</mn> </mrow> </msub> <msup> <mi>e</mi> <mrow> <mo>-</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <msub> <mi>&amp;theta;</mi> <mi>c</mi> </msub> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mfrac> <mi>R</mi> <msub> <mi>r</mi> <mi>c</mi> </msub> </mfrac> <mo>)</mo> </mrow> <mrow> <mo>-</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <msub> <mi>&amp;rho;</mi> <mrow> <mi>T</mi> <mn>1</mn> </mrow> </msub> <mfrac> <msub> <mi>w</mi> <mi>c</mi> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> </mrow> </msup> <msup> <mi>R</mi> <mrow> <mo>-</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <msub> <mi>&amp;rho;</mi> <mrow> <mi>T</mi> <mn>2</mn> </mrow> </msub> <mfrac> <msub> <mi>w</mi> <mi>c</mi> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> </mrow> </msup> <mo>,</mo> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>=</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <msub> <mi>&amp;rho;</mi> <mrow> <mi>T</mi> <mn>2</mn> </mrow> </msub> <mfrac> <msub> <mi>w</mi> <mi>c</mi> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <mo>,</mo> <msub> <mi>c</mi> <mn>5</mn> </msub> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mn>1</mn> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <mo>)</mo> </mrow> <mfrac> <mn>1</mn> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> </mrow> </mfrac> </msup> <mo>,</mo> </mrow>
<mrow> <msub> <mi>c</mi> <mn>6</mn> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>-</mo> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>,</mo> <msub> <mi>c</mi> <mn>7</mn> </msub> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mfrac> <msub> <mi>c</mi> <mn>3</mn> </msub> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <mo>)</mo> </mrow> <mfrac> <mn>1</mn> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </mfrac> </msup> <mo>,</mo> <msub> <mi>c</mi> <mn>8</mn> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>-</mo> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </mfrac> <mo>,</mo> <msub> <mi>c</mi> <mn>9</mn> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <msub> <mi>c</mi> <mn>1</mn> </msub> <msup> <msub> <mi>c</mi> <mn>5</mn> </msub> <mrow> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> </mrow> </msup> <mo>&amp;lsqb;</mo> <mfrac> <msup> <mi>R</mi> <mrow> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> <msub> <mi>c</mi> <mn>6</mn> </msub> </mrow> </msup> <mrow> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> <msub> <mi>c</mi> <mn>6</mn> </msub> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <msup> <msub> <mi>r</mi> <mi>c</mi> </msub> <mrow> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> <msub> <mi>c</mi> <mn>6</mn> </msub> </mrow> </msup> </mrow> <mrow> <msub> <mi>c</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>1</mn> </msub> <msub> <mi>c</mi> <mn>6</mn> </msub> </mrow> </mfrac> <mo>&amp;rsqb;</mo> <mo>,</mo> </mrow>
<mrow> <msub> <mi>c</mi> <mn>10</mn> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <mfrac> <mrow> <msub> <mi>c</mi> <mn>3</mn> </msub> <msup> <msub> <mi>c</mi> <mn>7</mn> </msub> <mrow> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </msup> </mrow> <mrow> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <msub> <mi>c</mi> <mn>8</mn> </msub> </mrow> </mfrac> <mo>,</mo> <msub> <mi>c</mi> <mn>11</mn> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <mfrac> <mrow> <msub> <mi>c</mi> <mn>3</mn> </msub> <msup> <msub> <mi>c</mi> <mn>7</mn> </msub> <mrow> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </msup> <msup> <mi>R</mi> <mrow> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <msub> <mi>c</mi> <mn>8</mn> </msub> </mrow> </msup> </mrow> <mrow> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <msub> <mi>c</mi> <mn>8</mn> </msub> </mrow> </mfrac> <mo>,</mo> <msub> <mi>c</mi> <mn>12</mn> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>c</mi> <mn>3</mn> </msub> <msup> <msub> <mi>c</mi> <mn>7</mn> </msub> <mrow> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> </mrow> </msup> <mi>U</mi> </mrow> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> </mrow> </mfrac> <mo>,</mo> <msub> <mi>c</mi> <mn>13</mn> </msub> <mo>=</mo> <msub> <mi>c</mi> <mn>4</mn> </msub> <mo>-</mo> <msub> <mi>&amp;gamma;</mi> <mn>2</mn> </msub> <msub> <mi>c</mi> <mn>8</mn> </msub> <mo>-</mo> <mn>1</mn> <mo>;</mo> </mrow>
Step 2 determines the thickness of reinforced insulation layer；
Reinforced insulation layer radius RsThe radial electric field intensity E at place2(Rs) it is cable insulation maximum functional electric field strength E0Two points One of, expression formula is as follows：
E2(RS)=0.5E0
With reference to the expression formula and above formula of radial electric field intensity at stress cone curve y, R is calculated to obtains, so as to which reinforced insulation be calculated Layer thickness Δ n=Rs- R, R are cable insulation radius；
Step 3, identified sign cone curvilinear equation；
The axial electric field strength E of any point on stress cone curvetWith the radial electric field E of the point2There are following relations：
<mrow> <msub> <mi>E</mi> <mi>t</mi> </msub> <mo>=</mo> <msub> <mi>E</mi> <mn>2</mn> </msub> <mo>&amp;CenterDot;</mo> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>&amp;alpha;</mi> <mo>=</mo> <msub> <mi>E</mi> <mn>2</mn> </msub> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <mi>d</mi> <mi>y</mi> </mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mfrac> </mrow>
Above formula is integrated to obtainMake EtFor constant, by radial electric field E2Expression formula is brought intoAnd several groups of coordinates (x, y) are obtained using numerical method, obtain stress cone curvilinear equation.
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