CN104179521B - A kind of based on the heat dissipating method having groove box cable tunnel thermal field model - Google Patents

Abstract

The present invention relates to a kind of based on there being the heat dissipating method of groove box cable tunnel thermal field model, comprise the following steps: 1) cable tunnel prospecting, obtain the parameter of each ingredient in cable tunnel；2) mathematical model of cable tunnel thermal field is set up；3) caloric value of this cable tunnel is calculated according to mathematical model；4) draft type and the ventilating system of this electric power tunnel are designed.Compared with prior art, the present invention has advantage the most accurately.

Description

A kind of based on the heat dissipating method having groove box cable tunnel thermal field model

Technical field

The present invention relates to field of power transmission, especially relate to a kind of based on there being groove box cable tunnel thermal field model Heat dissipating method

Background technology

In order to solve increasingly severe powerup issue, plan utilization lays 220kV power cable at the big passage built. For this section of river-crossing tunnel, current tentative programme is as high tension cable using the cable passage of segmentation in every tunnel Tunnel, lays 220kV power cable.

The original scheme in tunnel does not consider lay high voltage power cable, so the design of tunnel ventilation heat radiation is also Do not consider the caloric value of power cable.Power cable generates heat in longtime running, causes cable passage temperature to raise, The safe operation of the miscellaneous equipment that may affect in tunnel in the same space.It addition, the radiating condition of this cable passage Different with conventional tunnel, conventional tunnel four faces up and down the most directly contact soil, can dispel the heat.Shanghai is worshipped In Soviet Union tunnel, cable passage only one side and bottom are soil, and top is highway, and opposite side is that track is handed over Circulation passage.

Summary of the invention

Defect that the purpose of the present invention is contemplated to overcome above-mentioned prior art to exist and provide effectively, a kind of accurately Based on the heat dissipating method having groove box cable tunnel thermal field model

The purpose of the present invention can be achieved through the following technical solutions:

A kind of based on there being the heat dissipating method of groove box cable tunnel thermal field model, comprise the following steps:

1) cable tunnel prospecting, obtains the parameter of each ingredient in cable tunnel；

2) mathematical model of cable tunnel thermal field is set up；

3) caloric value of this cable tunnel is calculated according to mathematical model；

4) draft type and the ventilating system of this cable tunnel are designed.

Described thermal resistance is drawn by R=d/KA, and in formula, R is thermal resistance, and d is the thickness of heat conductor, and A is heat conductor Area of section, K is heat conductivity.

Described step 2) in the Mathematical Models of cable tunnel thermal field include following sub-step:

21) electric cable heating model is set up:

N telegram in reply cable heat radiation power is:

$P=n\frac{I^2\mathrm{\σ}}{A}LC$

Wherein, P is cable heat radiation power, and I is current-carrying capacity, and A is cable cross-sectional area, and σ is cable resistance rate, L be cable length, C be cable radiation loss coefficient value 0.9, n is cable backhaul number；

22) cable thermal losses model is set up:

I²[RT₁+R(1+λ₁)T₂+R(1+λ₁+λ₂)(T₃+T₄)]+W_d[0.5T₁+n(T₂+T₃+T₄)]=θ₁-θ_C

Wherein:

I: current-carrying capacity, θ_C: ambient temperature, θ₁: cable conductor temperature, the conductor exchange under R: operating temperature Resistance, W_d: insulation dielectric loss, λ₁: metal shading loss factor, λ₂: metal armouring loss factor, T₁: Insulating barrier thermal resistance between conductor and protective metal shell, T₂: inner liner thermal resistance between protective metal shell and armor, T₃: cable Outer jacket thermal resistance, T₄: Exterior cable thermal resistance, ambient temperature θ_CRefer to air themperature, it is assumed that θ_CIt is constant, leads Temperature θ₁Refer to the maximum permissible temperature of long-term work, be set to 90 DEG C；θ₁And θ_CIt it is all boundary condition.

Described step 22) in, the calculating formula of described conductor AC resistance R is:

R=R ＇ (1+Y_s+Y_p)

R ＇=R₀[1+α₂₀(θ-20)]

$Y_s=\frac{{X_S}^4}{192+0.8{X_S}^4}$

${X_S}^2=8\mathrm{\π}f\frac{K_s}{R^{\′}}\×{10}^{-7}$

$Y_p=\frac{X_P^4}{192+0.8X_P^4}{\left(\frac{d_c}{s}\right)}^2\[0.312{\left(\frac{d_c}{s}\right)}^2+\frac{1.18}{\frac{X_P^4}{192+0.8X_P^4}+0.27}\]$

${X_P}^2=8\mathrm{\π}f\frac{K_P}{R^{\′}}\×{10}^{-7}$

Wherein:

R ＇: maximum operating temperature lower conductor D.C. resistance, Y_s；Kelvin effect factor, Y_p: kindred effect factor, R₀: conductor DC resistance when 20 DEG C, θ: running temperature, α₂₀: the temperature coefficient of copper conductor when 20 DEG C, for Circular compact conductor K_s=1, d_c: conductor diameter, s: the spacing in each conductor axle center.

Described step 22) in, described dielectric loss W_dCalculating formula be:

$W_d={\mathrm{\ωCU}}_0^{2}tan\mathrm{\δ}$

$C=\frac{\mathrm{\ϵ}\×{10}^{-9}}{18ln\frac{D_i}{d_c}}$

Wherein:

ω=2 π f, U₀: voltage-to-ground, ε=2.3, D_i: for outer insulation diameter, d_c: for inner shield external diameter.

Described step 22) in metal shading loss λ₁Calculating formula be:

λ₁=λ₁′+λ₁″

${\mathrm{\λ}}_1^{\′}=\frac{R_s}{R}\[g_s{\mathrm{\λ}}_0\left(1+{\mathrm{\Δ}}_1+{\mathrm{\Δ}}_2\right)+\frac{\left({\mathrm{\β}}_{1}t\right)}{12\×{10}^{12}}\]$

${\mathrm{\β}}_1=\sqrt{\frac{4\mathrm{\π}\mathrm{\ω}}{24.824\×{10}^{-8}\×{10}^7}}$

$g_s=1+{\left(\frac{t_s}{D_s}\right)}^{1.74}({\mathrm{\β}}_1{D}_s{10}^{-3}-1.6)$

Then have when for rounded projections arranged:

${\mathrm{\λ}}_0=3\left(\frac{m^2}{1+m^2}\right){\left(\frac{D}{2s}\right)}^2$

${\mathrm{\Δ}}_1=\left(1.14m^{2.45}+0.33\right){\left(\frac{D}{2s}\right)}^{(0.92m+1.66)}$

Δ₂=0

$D=\frac{D_{oc}+D_{it}}{2}+t_s$

Wherein:

λ₁'=0, λ₁" for eddy-current loss, ρ: protective metal shell resistivity, R: protective metal shell resistance, t_s: metal protects Set thickness, D: protective metal shell external diameter, D_oc: corrugated aluminium sheath maximum outside diameter, D_it: corrugated aluminium sheath minimum diameter.

Described step 22) in cable internal thermal resistance T₁、T₂、T₃Calculating formula be:

$T_1=\frac{{\mathrm{\ρ}}_{T_1}}{2\mathrm{\π}}Ln(1+\frac{2t_1}{d_c})$

Wherein:

: insulant thermal resistivity, d_c: conductor diameter t1: the insulation thickness between conductor and sheath；

This cable does not has a steel armour, therefore T₂=0；

$T_3=\frac{{\mathrm{\ρ}}_{T_s}}{2\mathrm{\π}}Ln\[\frac{D_{oc}+2t_3}{\frac{(D_{oc}+D_{it})}{2}+t_s}\]$

Wherein:

t₃: oversheath thickness,: nonmetal oversheath thermal resistivity, D_oc: corrugated aluminium sheath maximum outside diameter.

Described step 22) in Exterior cable thermal resistance T₄Groove box inwall thermal resistance T is arrived including cable₅, groove box body heat Resistance T₆, thermal resistance T of groove outer box wall to air₇, thermal resistance T of air to cable passage inwall₈, tunnel is at the bottom of river Soil thermal resistance T₉, described T₅、T₆、T₇、T₈And T₉Calculating formula be:

$T_5=\frac{1}{A_5\mathrm{\π}X(d_2-d_1)/ln(d_2/d_1)}$

$T_6=\frac{{\mathrm{\ρ}}_{T_6}}{A_6}$

$T_7=\frac{1}{\frac{1}{X_{7\_1}{A}_{7\_1}}+\frac{1}{X_{7\_2}{A}_{7\_2}}+\frac{1}{X_{7\_3}{A}_{7\_3}}}$

$T_8=\frac{1}{1/T_{8\_1}+1/T_{8\_2}}$

$T_9=\frac{1}{2{\mathrm{\π\λ}}_t}ln\[\frac{2H}{d_z}+\sqrt{{\left(\frac{2H}{d_z}\right)}^2-1}\]$

$d_z=\sqrt{4\×L\×W/\mathrm{\π}}$

T_{8_1}=1/X₁A₁

T_{8_1}=1/X₂A₂

Wherein:

A₅: the average area of dissipation of interlayer inside and outside wall, d₁: threephase cable equivalent diameter, d₂: the equivalence of groove box inwall Diameter, the convection transfer rate of X: interlayer,: groove box thermal resistivity, A₆: groove box area of dissipation, X_{7_1}、 X_{7_2}、X_{7_3}: groove box side, end face, the convection transfer rate of bottom surface, A_{7_1}、A_{7_2}、A_{7_3}: groove box side, End face, the area of bottom surface, T_{8_1}: thermal resistance T of air to sidewall_{8_1}: the thermal resistance of air to ground, X₁、X₂: The convection transfer rate A of sidewall or bottom₁、A₂: sidewall or the heat exchange area of bottom, λ_t: soil thermal conductivity, H: the tunnel degree of depth, d_z: the diameter of circular tunnel outer surface, L and W is the height and width of Rectangular Tunnel respectively.

Compared with prior art, the invention have the characteristics that:

1, set up model accurate, analyze cable heating heat dispersal situations in tunnel from actual scene, examine It is thorough to consider, and modeling is accurately.

2, good cooling effect, by setting up hair-dryer and other auxiliary equipments at tunnel internal, reaches the mark of cooling Accurate.

Accompanying drawing explanation

Fig. 1 is the method flow diagram of the present invention.

Detailed description of the invention

The present invention is described in detail with specific embodiment below in conjunction with the accompanying drawings.

Embodiment:

A kind of based on there being the heat dissipating method of groove box cable tunnel thermal field model, comprise the following steps:

1) cable tunnel prospecting, obtains the parameter of each ingredient in cable tunnel；

2) mathematical model of cable tunnel thermal field is set up；

3) caloric value of this cable tunnel is calculated according to mathematical model；

4) draft type and the ventilating system of this cable tunnel are designed.

Described thermal resistance is drawn by R=d/KA, and in formula, R is thermal resistance, and d is the thickness of heat conductor, and A is heat conductor Area of section, K is heat conductivity.

Described step 2) in the Mathematical Models of cable tunnel thermal field include following sub-step:

21) electric cable heating model is set up:

N telegram in reply cable heat radiation power is:

$P=n\frac{I^2\mathrm{\σ}}{A}LC$

Wherein, P is cable heat radiation power, and I is current-carrying capacity, and A is cable cross-sectional area, and σ is cable resistance rate, L be cable length, C be cable radiation loss coefficient value 0.9, n is cable backhaul number；

22) cable thermal losses model is set up:

I²[RT₁+R(1+λ₁)T₂+R(1+λ₁+λ₂)(T₃+T₄)]+W_d[0.5T₁+n(T₂+T₃+T₄)]=θ₁-θ_C

Wherein:

I: current-carrying capacity, θ_C: ambient temperature, θ₁: cable conductor temperature, the conductor exchange under R: operating temperature Resistance, W_d: insulation dielectric loss, λ₁: metal shading loss factor, λ₂: metal armouring loss factor, T₁: Insulating barrier thermal resistance between conductor and protective metal shell, T₂: inner liner thermal resistance between protective metal shell and armor, T₃: cable Outer jacket thermal resistance, T₄: Exterior cable thermal resistance, ambient temperature θ_CRefer to air themperature, it is assumed that θ_CIt is constant, leads Temperature θ₁Refer to the maximum permissible temperature of long-term work, be set to 90 DEG C；θ₁And θ_CIt it is all boundary condition.

Described step 22) in, the calculating formula of described conductor AC resistance R is:

R=R ＇ (1+Y_s+Y_p)

R ＇=R₀[1+α₂₀(θ-20)]

$Y_s=\frac{{X_S}^4}{192+0.8{X_S}^4}$

${X_S}^2=8\mathrm{\π}f\frac{K_s}{R^{\′}}\×{10}^{-7}$

$Y_p=\frac{X_P^4}{192+0.8X_P^4}{\left(\frac{d_c}{s}\right)}^2\[0.312{\left(\frac{d_c}{s}\right)}^2+\frac{1.18}{\frac{X_P^4}{192+0.8X_P^4}+0.27}\]$

${X_P}^2=8\mathrm{\π}f\frac{K_P}{R^{\′}}\×{10}^{-7}$

Wherein:

R ＇: maximum operating temperature lower conductor D.C. resistance, Y_s；Kelvin effect factor, Y_p: kindred effect factor, R₀: conductor DC resistance when 20 DEG C, θ: running temperature, α₂₀: the temperature coefficient of copper conductor when 20 DEG C, for Circular compact conductor K_s=1, d_c: conductor diameter, s: the spacing in each conductor axle center.

Described step 22) in, described dielectric loss W_dCalculating formula be:

$W_d={\mathrm{\ωCU}}_0^{2}tan\mathrm{\δ}$

$C=\frac{\mathrm{\ϵ}\×{10}^{-9}}{18ln\frac{D_i}{d_c}}$

Wherein:

ω=2 π f, U₀: voltage-to-ground, ε=2.3, D_i: for outer insulation diameter, d_c: for inner shield external diameter.

Described step 22) in metal shading loss λ₁Calculating formula be:

λ₁=λ₁′+λ₁″

${\mathrm{\λ}}_1^{\′}=\frac{R_s}{R}\[g_s{\mathrm{\λ}}_0\left(1+{\mathrm{\Δ}}_1+{\mathrm{\Δ}}_2\right)+\frac{\left({\mathrm{\β}}_{1}t\right)}{12\×{10}^{12}}\]$

${\mathrm{\β}}_1=\sqrt{\frac{4\mathrm{\π}\mathrm{\ω}}{24.824\×{10}^{-8}\×{10}^7}}$

$g_s=1+{\left(\frac{t_s}{D_s}\right)}^{1.74}({\mathrm{\β}}_1{D}_s{10}^{-3}-1.6)$

Then have when for rounded projections arranged:

${\mathrm{\λ}}_0=3\left(\frac{m^2}{1+m^2}\right){\left(\frac{D}{2s}\right)}^2$

${\mathrm{\Δ}}_1=\left(1.14m^{2.45}+0.33\right){\left(\frac{D}{2s}\right)}^{(0.92m+1.66)}$

Δ₂=0

$D=\frac{D_{oc}+D_{it}}{2}+t_s$

Wherein:

λ₁'=0, λ₁" for eddy-current loss, ρ: protective metal shell resistivity, R: protective metal shell resistance, t_s: metal protects Set thickness, D: protective metal shell external diameter, D_oc: corrugated aluminium sheath maximum outside diameter, D_it: corrugated aluminium sheath minimum diameter.

Described step 22) in cable internal thermal resistance T₁、T₂、T₃Calculating formula be:

$T_1=\frac{{\mathrm{\ρ}}_{T_1}}{2\mathrm{\π}}Ln(1+\frac{2t_1}{d_c})$

Wherein:

: insulant thermal resistivity, d_c: conductor diameter t1: the insulation thickness between conductor and sheath；

This cable does not has a steel armour, therefore T₂=0；

$T_3=\frac{{\mathrm{\ρ}}_{T_s}}{2\mathrm{\π}}Ln\[\frac{D_{oc}+2t_3}{\frac{(D_{oc}+D_{it})}{2}+t_s}\]$

Wherein:

t₃: oversheath thickness,: nonmetal oversheath thermal resistivity, D_oc: corrugated aluminium sheath maximum outside diameter.

The heat exchange that air and the solid such as tunnel wall, cable are carried out belongs to fluid heat transfer free convection process, and fluid is certainly So heat convection is divided into large space heat transfer free convection and confined space heat transfer free convection two class, and so-called large space is certainly So heat convection, refers to the heat transfer free convection that at nearly wall, the development of boundary region is not interfered because space limits, And when space is little to making fluid rising and descending motion of being heated interfere with each other when affecting, the referred to as confined space is natural Heat convection, such as, for height h, interval d two pieces of parallel perpendicular planomurals between heat convection, when < when 0.33, the motion of fluid raising and lowering interferes with each other d/h, it is necessary to calculate by confined space heat transfer free convection.

When fluid flows through the wall differed with its temperature, one layer of temperature will be formed at wall adnexa jumpy Thin layer of fluid, referred to as thermal boundary layer, the flow regime of thermal boundary layer is divided into turbulent flow and laminar flow two kinds, and both flows Convection transfer rate computational methods under state also differ.Thermal boundary layer is turbulent flow or laminar flow, depends on this temperature The temperature difference of air physical parameter, fluid and wall under Du, wall size shape, wall degree of roughness etc. are multiple Factor.

Have the situation of groove box as it is shown in figure 1, described step 22) in Exterior cable thermal resistance T₄Groove is arrived including cable Box inwall thermal resistance T₅, groove box body thermal resistance T₆, thermal resistance T of groove outer box wall to air₇, in air to cable passage Thermal resistance T of wall₈, tunnel is to soil thermal resistance T at the bottom of river₉, described T₅、T₆、T₇、T₈And T₉Calculating formula be:

$T_5=\frac{1}{A_5\mathrm{\&pi;}X(d_2-d_1)/ln(d_2/d_1)}$

$T_6=\frac{{\mathrm{\&rho;}}_{T_6}}{A_6}$

$T_7=\frac{1}{\frac{1}{X_{7\_1}{A}_{7\_1}}+\frac{1}{X_{7\_2}{A}_{7\_2}}+\frac{1}{X_{7\_3}{A}_{7\_3}}}$

$T_8=\frac{1}{1/T_{8\_1}+1/T_{8\_2}}$

$T_9=\frac{1}{2{\mathrm{\&pi;\&lambda;}}_t}ln\&lsqb;\frac{2H}{d_z}+\sqrt{{\left(\frac{2H}{d_z}\right)}^2-1}\&rsqb;$

$d_z=\sqrt{4\&times;L\&times;W/\mathrm{\&pi;}}$

T_{8_1}=1/X₁A₁

T_{8_1}=1/X₂A₂

Wherein:

A₅: the average area of dissipation of interlayer inside and outside wall, d₁: threephase cable equivalent diameter, d₂: the equivalence of groove box inwall Diameter, the convection transfer rate of X: interlayer,: groove box thermal resistivity, A₆: groove box area of dissipation, X_{7_1}、 X_{7_2}、X_{7_3}: groove box side, end face, the convection transfer rate of bottom surface, A_{7_1}、A_{7_2}、A_{7_3}: groove box side, End face, the area of bottom surface, T_{8_1}: thermal resistance T of air to sidewall_{8_1}: the thermal resistance of air to ground, X₁、X₂: The convection transfer rate A of sidewall or bottom₁、A₂: sidewall or the heat exchange area of bottom, λ_t: soil thermal conductivity, H: the tunnel degree of depth, d_z: the diameter of circular tunnel outer surface, L and W is the height and width of Rectangular Tunnel respectively.

Claims (6)

1. one kind based on the heat dissipating method having groove box cable tunnel thermal field model, it is characterised in that include following step Rapid:

1) cable tunnel prospecting, obtains the parameter of each ingredient in cable tunnel；

2) mathematical model of cable tunnel thermal field is set up, including following sub-step:

21) electric cable heating model is set up:

N telegram in reply cable heat radiation power is:

$P=n\frac{I^2\mathrm{\&sigma;}}{A}LC$

Wherein, P is cable heat radiation power, and I is current-carrying capacity, and A is cable cross-sectional area, and σ is cable resistance rate, L be cable length, C be cable radiation loss coefficient value 0.9, n is cable backhaul number；

22) cable thermal losses model is set up:

I²[RT₁+R(1+λ₁)T₂+R(1+λ₁+λ₂)(T₃+T₄)]+W_d[0.5T₁+n(T₂+T₃+T₄)]=θ₁-θ_C

Wherein:

I: current-carrying capacity, θ_C: ambient temperature, θ₁: cable conductor temperature, the conductor exchange under R: operating temperature Resistance, W_d: insulation dielectric loss, λ₁: metal shading loss factor, λ₂: metal armouring loss factor, T₁: Insulating barrier thermal resistance between conductor and protective metal shell, T₂: inner liner thermal resistance between protective metal shell and armor, T₃: cable Outer jacket thermal resistance, T₄: Exterior cable thermal resistance, ambient temperature θ_CRefer to air themperature, it is assumed that θ_CIt is constant, leads Temperature θ₁Refer to the maximum permissible temperature of long-term work, be set to 90 DEG C, θ₁And θ_CIt it is all boundary condition；

3) caloric value of this cable tunnel is calculated according to mathematical model；

4) draft type and the ventilating system of this cable tunnel are designed.

The most according to claim 1 a kind of based on there being the heat dissipating method of groove box cable tunnel thermal field model, its It is characterised by, described step 22) in, the calculating formula of described conductor AC resistance R is:

R=R ＇ (1+Y_s+Y_p)

R ＇=R₀[1+α₂₀(θ-20)]

$Y_s=\frac{{X_S}^4}{192+0.8{X_S}^4}$

${X_S}^2=8\mathrm{\&pi;}f\frac{K_s}{R^{\&prime;}}\&times;{10}^{-7}$

$Y_p=\frac{X_P^4}{192+0.8X_P^4}{\left(\frac{d_c}{s}\right)}^2\&lsqb;0.312{\left(\frac{d_c}{s}\right)}^2+\frac{1.18}{\frac{X_P^4}{192+0.8X_P^4}+0.27}\&rsqb;$

${X_P}^2=8\mathrm{\&pi;}f\frac{K_P}{R^{\&prime;}}\&times;{10}^{-7}$

Wherein:

R ＇: maximum operating temperature lower conductor D.C. resistance, Y_s；Kelvin effect factor, Y_p: kindred effect factor, R₀: conductor DC resistance when 20 DEG C, θ: running temperature, α₂₀: the temperature coefficient of copper conductor when 20 DEG C, for Circular compact conductor K_s=1, d_c: conductor diameter, s: the spacing in each conductor axle center.

The most according to claim 1 a kind of based on there being the heat dissipating method of groove box cable tunnel thermal field model, its It is characterised by, described step 22) in, described dielectric loss W_dCalculating formula be:

$W_d={\mathrm{\&omega;CU}}_0^{2}tan\mathrm{\&delta;}$

$C=\frac{\mathrm{\&epsiv;}\&times;{10}^{-9}}{18ln\frac{D_i}{d_c}}$

Wherein:

ω=2 π f, U₀: voltage-to-ground, ε=2.3, D_i: for outer insulation diameter, d_c: for inner shield external diameter.

The most according to claim 1 a kind of based on there being the heat dissipating method of groove box cable tunnel thermal field model, its Be characterised by, described step 22) in metal shading loss λ₁Calculating formula be:

λ₁=λ₁′+λ″₁

${\mathrm{\&lambda;}}_1^{\&prime;}=\frac{R_s}{R}\&lsqb;g_s{\mathrm{\&lambda;}}_0(1+{\mathrm{\&Delta;}}_1+{\mathrm{\&Delta;}}_2)+\frac{\left({\mathrm{\&beta;}}_{1}t\right)}{12\&times;{10}^{12}}\&rsqb;$

${\mathrm{\&beta;}}_1=\sqrt{\frac{4\mathrm{\&pi;}\mathrm{\&omega;}}{24.824\&times;{10}^{-8}\&times;{10}^7}}$

$g_s=1+{\left(\frac{t_s}{D_s}\right)}^{1.74}({\mathrm{\&beta;}}_1{D}_s{10}^{-3}-1.6)$

Then have when for rounded projections arranged:

${\mathrm{\&lambda;}}_0=3\left(\frac{m^2}{1+m^2}\right){\left(\frac{D}{2s}\right)}^2$

${\mathrm{\&Delta;}}_1=(1.14m^{2.45}+0.33){\left(\frac{D}{2s}\right)}^{(0.92m+1.66)}$

Δ₂=0

$D=\frac{D_{oc}+D_{it}}{2}+t_s$

Wherein:

λ₁'=0, λ "₁For eddy-current loss, ρ: protective metal shell resistivity, R: protective metal shell resistance, t_s: metal protects Set thickness, D: protective metal shell external diameter, D_oc: corrugated aluminium sheath maximum outside diameter, D_it: corrugated aluminium sheath minimum diameter.

The most according to claim 1 a kind of based on there being the heat dissipating method of groove box cable tunnel thermal field model, its Be characterised by, described step 22) in cable internal thermal resistance T₁、T₂、T₃Calculating formula be:

$T_1=\frac{{\mathrm{\&rho;}}_{T_1}}{2\mathrm{\&pi;}}Ln(1+\frac{2t_1}{d_c})$

Wherein:

ρT₁: insulant thermal resistivity, d_c: conductor diameter t₁: the insulation thickness between conductor and sheath；

This cable does not has a steel armour, therefore T₂=0；

$T_3=\frac{{\mathrm{\&rho;}}_{T_s}}{2\mathrm{\&pi;}}Ln\&lsqb;\frac{D_{oc}+2t_3}{\frac{(D_{oc}+D_{it})}{2}+t_s}\&rsqb;$

Wherein:

t₃: oversheath thickness,Nonmetal oversheath thermal resistivity, D_oc: corrugated aluminium sheath maximum outside diameter.

The most according to claim 1 a kind of based on there being the heat dissipating method of groove box cable tunnel thermal field model, its Be characterised by, described step 22) in Exterior cable thermal resistance T₄Groove box inwall thermal resistance T is arrived including cable₅, groove box Body thermal resistance T₆, thermal resistance T of groove outer box wall to air₇, thermal resistance T of air to cable passage inwall₈, tunnel arrives Soil thermal resistance T at the bottom of river₉, described T₅、T₆、T₇、T₈And T₉Calculating formula be:

$T_5=\frac{1}{A_5\mathrm{\&pi;}X(d_2-d_1)/ln(d_2/d_1)}$

$T_6=\frac{{\mathrm{\&rho;}}_{T_6}}{A_6}$

$T_7=\frac{1}{\frac{1}{X_{7\_1}{A}_{7\_1}}+\frac{1}{X_{7\_2}{A}_{7\_2}}+\frac{1}{X_{7\_3}{A}_{7\_3}}}$

$T_8=\frac{1}{1/T_{8\_1}+1/T_{8\_2}}$

$T_9=\frac{1}{2{\mathrm{\&pi;\&lambda;}}_t}ln\&lsqb;\frac{2H}{d_z}+\sqrt{{\left(\frac{2H}{d_z}\right)}^2-1}\&rsqb;$

$d_z=\sqrt{4\&times;L\&times;W/\mathrm{\&pi;}}$

T_{8_1}=1/X₁A₁

T_{8_1}=1/X₂A₂

Wherein:

A₅: the average area of dissipation of interlayer inside and outside wall, d₁: threephase cable equivalent diameter, d₂: the equivalence of groove box inwall Diameter, the convection transfer rate of X: interlayer,Groove box thermal resistivity, A₆: groove box area of dissipation, X_{7_1}、 X_{7_2}、X_{7_3}: groove box side, end face, the convection transfer rate of bottom surface, A_{7_1}、A_{7_2}、A_{7_3}: groove box side, End face, the area of bottom surface, T_{8_1}: thermal resistance T of air to sidewall_{8_1}: the thermal resistance of air to ground, X₁、X₂: The convection transfer rate A of sidewall or bottom₁、A₂: sidewall or the heat exchange area of bottom, λ_t: soil thermal conductivity, H: the tunnel degree of depth, d_z: the diameter of circular tunnel outer surface, L and W is the height and width of Rectangular Tunnel respectively.