CN104196555A - Heat dissipation method based on slotless box cable tunnel thermal field model - Google Patents

Abstract

The invention relates to a heat dissipation method based on a slotless box cable tunnel thermal field model. The method includes the following steps of (1) exploring a cable tunnel to obtain parameters of all components of the cable tunnel, (2) building a mathematical model of a cable tunnel thermal field, (3) working out the calorific value of the cable tunnel according to the mathematical model and (4) designing the ventilation mode and the ventilation system of the electric power tunnel. Compared with the prior art, the heat dissipation method has the advantages of being effective and accurate.

Description

A kind of heat dissipating method based on slotless box cable tunnel thermal field model

Technical field

The present invention relates to electric power transfer field, especially relate to a kind of heat dissipating method based on slotless box cable tunnel thermal field model

Background technology

In order to solve increasingly severe powerup issue, plan utilization lays 220kV power cable at the major path of building.For this section of river-crossing tunnel, at present tentative programme be the cable passage cut apart in every tunnel as high voltage cable tunnel, lay 220kV power cable.

In the original scheme in tunnel, do not consider to lay high voltage power cable, so the calorific value of power cable is not considered in the design of tunnel ventilation heat radiation yet.Power cable, in long-term heating in service, causes cable passage temperature to raise, and may affect the safe operation of the miscellaneous equipment in the same space in tunnel.In addition, the radiating condition of this cable passage and conventional tunnel are different, and conventional tunnel up and down four faces all directly contacts soil, can dispel the heat.In Chong Su tunnel, Shanghai, cable passage only side and bottom is soil, and top is speedway, and opposite side is track traffic passage.

Summary of the invention

Object of the present invention be exactly in order to overcome the defect that above-mentioned prior art exists, provide effectively, a kind of heat dissipating method based on slotless box cable tunnel thermal field model accurately

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

A heat dissipating method based on slotless box cable tunnel thermal field model, comprises the following steps:

1) cable tunnel is reconnoitred, and obtains the parameter of each ingredient of cable tunnel;

2) set up the Mathematical Modeling of cable tunnel thermal field;

3) according to calculated with mathematical model, go out the calorific value of this cable tunnel;

4) design draft type and the ventilation system of this electric power tunnel.

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

The Mathematical Models of the cable tunnel thermal field described step 2) comprises following sub-step:

21) set up electric cable heating model:

N telegram in reply cable heat radiation power is:

$P=n\frac{I^2\mathrm{\σ}}{A}\mathrm{LC}$

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, and L is that cable length, C are cable radiation loss coefficient value 0.9, and n is cable backhaul number;

22) set up cable thermal losses model:

I ²[RT ₁+R(1+λ ₁)T ₂+R(1+λ ₁+λ ₂)(T ₃+T ₄)]+W _d[0.5T ₁+n(T ₂+T ₃+T ₄)]＝θ ₁-θ _C

Wherein:

I: current-carrying capacity, θ _c: environment temperature, θ ₁: cable conductor temperature, R: the conductor AC resistance under operating temperature, W _d: insulation dielectric loss, λ ₁: metallic shield loss factor, λ ₂: metal armouring loss factor, T ₁: insulating layer thermal resistance between conductor and protective metal shell, T ₂: inner liner thermal resistance between protective metal shell and armor, T ₃: protective coverings of cable thermal resistance, T ₄: cable external thermal resistance, environment temperature θ _crefer to air themperature, suppose θ _cconstant, conductor temperature θ ₁the maximum permissible temperature that refers to long-term work, is made as 90 ℃; θ ₁and θ _cit is all fringe conditions.

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

R＝R′(1+Y _s+Y _p)

R′＝R ₀[1+α ₂₀(θ-20)]

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

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

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

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

Wherein:

R ': maximum operating temperature lower conductor D.C. resistance, Y _s; Kelvin effect factor, Y _p: kindred effect factor, R ₀: conductor DC resistance in the time of 20 ℃, θ: running temperature, α ₂₀: the temperature coefficient of copper conductor in the time of 20 ℃, for circular compact conductor K _s=1, d _c: conductor diameter, s: distance between each conductor axle center.

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

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

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

Wherein:

ω=2 π f, U ₀: voltage-to-ground, ε=2.3, D _i: be outer insulation diameter, d _c: be inner shield external diameter.

Described step 22) metallic shield loss λ in ₁calculating formula be:

λ ₁＝λ′ ₁+λ″ ₁

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

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

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

When for rounded projections arranged, have:

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

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

Δ ₂＝0

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

Wherein:

λ ' ₁=0, λ " ₁for eddy-current loss, ρ: protective metal shell resistivity, R: protective metal shell resistance, t _s: protective metal shell 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) cable internal thermal resistance T in ₁, T ₂, T ₃calculating formula be:

$T_1=\frac{{\mathrm{\ρ}}_{T_1}}{2\mathrm{\π}}\mathrm{Ln}(1+\frac{{2t}_1}{d_c})$

Wherein:

insulation materials thermal resistivity, d _c: conductor diameter t ₁: the insulation thickness between conductor and sheath;

This cable does not have steel armour, therefore T ₂=0;

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

Wherein:

T ₃: oversheath thickness, oversheath (nonmetal) thermal resistivity, D _oc: corrugated aluminium sheath maximum outside diameter.

Described step 22) cable external thermal resistance T in ₄comprise that cable is to the thermal resistance T of air ₅, air is to the thermal resistance T of cable passage inwall ₆, tunnel is to the soil thermal resistance T at the bottom of river ₇, described T ₄, T ₅, T ₆and T ₇calculating formula be:

$T_5=\frac{1}{{\mathrm{\πD}}_{e}h{\left(\mathrm{\Δ\θ}\right)}^{0.25}}$

$h=\frac{Z}{{\left(D_e^{*}\right)}^8}+E$

$T_6=\frac{1}{1/T_{6\_1}+1/T_{6\_2}}$

$T_7=\frac{1}{2\mathrm{\π}{\mathrm{\λ}}_t}\ln [\frac{2H}{d_z}+\sqrt{{\left(\frac{2H}{d_z}\right)}^2-1}]$

$d_z=\sqrt{4\×L\×W/\mathrm{\π}}$

T _{6_1}＝1/X ₁A ₁；

T _{6_2}＝1/X ₂A ₂

Wherein:

the outside diameter of cable, h: coefficient of heat transfer, Δ θ: surpass cable surface temperature rise more than environment temperature, T _{6_1}: air is to the thermal resistance of sidewall, T _{6_2}: air is to the thermal resistance on ground, X ₁, X ₂: the convection transfer rate of sidewall or bottom, A ₁, A ₂: the heat exchange area of sidewall and bottom, λ _t: soil thermal conductivity, H: the tunnel degree of depth, d _z: the diameter of circular tunnel external surface, L and W are respectively the height and width of Rectangular Tunnel.

Compared with prior art, the present invention has following characteristics:

1, set up model accurate, analyze the heating heat radiation situation of cable in tunnel from actual scene, it is thorough to consider, modeling is accurate.

2, good cooling effect, by set up hair-dryer and other auxiliary equipment at tunnel internal, reaches the standard of cooling.

Accompanying drawing explanation

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

The specific embodiment

Below in conjunction with the drawings and specific embodiments, the present invention is described in detail.

Embodiment:

A heat dissipating method based on slotless box cable tunnel thermal field model, comprises the following steps:

1) cable tunnel is reconnoitred, and obtains the parameter of each ingredient of cable tunnel;

2) set up the Mathematical Modeling of cable tunnel thermal field;

3) according to calculated with mathematical model, go out the calorific value of this cable tunnel;

4) design draft type and the ventilation system of this electric power tunnel.

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

The Mathematical Models of the cable tunnel thermal field described step 2) comprises following sub-step:

21) set up electric cable heating model:

N telegram in reply cable heat radiation power is:

$P=n\frac{I^2\mathrm{\σ}}{A}\mathrm{LC}$

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, and L is that cable length, C are cable radiation loss coefficient value 0.9, and n is cable backhaul number;

22) set up cable thermal losses model:

I ²[RT ₁+R(1+λ ₁)T ₂+R(1+λ ₁+λ ₂)(T ₃+T ₄)]+W _d[0.5T ₁+n(T ₂+T ₃+T ₄)]＝θ ₁-θ _C

Wherein:

I: current-carrying capacity, θ _c: environment temperature, θ ₁: cable conductor temperature, R: the conductor AC resistance under operating temperature, W _d: insulation dielectric loss, λ ₁: metallic shield loss factor, λ ₂: metal armouring loss factor, T ₁: insulating layer thermal resistance between conductor and protective metal shell, T ₂: inner liner thermal resistance between protective metal shell and armor, T ₃: protective coverings of cable thermal resistance, T ₄: cable external thermal resistance, environment temperature θ _crefer to air themperature, suppose θ _cconstant, conductor temperature θ ₁the maximum permissible temperature that refers to long-term work, is made as 90 ℃; θ ₁and θ _cit is all fringe conditions.

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

R＝R′(1+Y _s+Y _p)

R′＝R ₀[1+α ₂₀(θ-20)]

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

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

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

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

Wherein:

R ': maximum operating temperature lower conductor D.C. resistance, Y _s; Kelvin effect factor, Y _p: kindred effect factor, R ₀: conductor DC resistance in the time of 20 ℃, θ: running temperature, α ₂₀: the temperature coefficient of copper conductor in the time of 20 ℃, for circular compact conductor K _s=1, d _c: conductor diameter, s: distance between each conductor axle center.

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

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

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

Wherein:

ω=2 π f, U ₀: voltage-to-ground, ε=2.3, D _i: be outer insulation diameter, d _c: be inner shield external diameter.

Described step 22) metallic shield loss λ in ₁calculating formula be:

λ ₁＝λ′ ₁+λ″ ₁

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

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

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

When for rounded projections arranged, have:

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

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

Δ ₂＝0

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

Wherein:

λ ' ₁=0, λ " ₁for eddy-current loss, ρ: protective metal shell resistivity, R: protective metal shell resistance, t _s: protective metal shell 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) cable internal thermal resistance T in ₁, T ₂, T ₃calculating formula be:

$T_1=\frac{{\mathrm{\ρ}}_{T_1}}{2\mathrm{\π}}\mathrm{Ln}(1+\frac{{2t}_1}{d_c})$

Wherein:

insulation materials thermal resistivity, d _c: conductor diameter t ₁: the insulation thickness between conductor and sheath;

This cable does not have steel armour, therefore T ₂=0;

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

Wherein:

T ₃: oversheath thickness, oversheath (nonmetal) thermal resistivity, D _oc: corrugated aluminium sheath maximum outside diameter.

Air and tunnel wall, the interchange of heat that the solids such as cable carry out belongs to fluid heat transfer free convection process, fluid heat transfer free convection is divided into large space heat transfer free convection and confined space heat transfer free convection two classes, so-called large space heat transfer free convection, refer to the heat transfer free convection that the development in boundary layer, nearly wall place is not interfered because of spatial limitation, and when space little of making fluid increase and the descending motion of being heated interferes with each other while affecting, be called confined space heat transfer free convection, for example, for height h, heat convection between two parallel perpendicular planomurals of interval d, when d/h < 0.33, fluid rises and descending motion interferes with each other, must calculate by confined space heat transfer free convection.

When fluid flows through the wall not identical with its temperature, at wall annex, will form one deck temperature thin layer of fluid jumpy, be called thermal boundary layer, the flow regime of thermal boundary layer is divided into turbulent flow and two kinds of laminar flows, and the convection transfer rate computational methods under these two kinds of flow regimes are not identical yet.Thermal boundary layer is turbulent flow or laminar flow, depends on the many factors such as temperature difference, wall size shape, wall degree of roughness of air physical parameter, fluid and wall at this temperature.

The situation of slotless box as shown in Figure 1, described step 22) in cable external thermal resistance T ₄comprise that cable is to the thermal resistance T of air ₅, air is to the thermal resistance T of cable passage inwall ₆, tunnel is to the soil thermal resistance T at the bottom of river ₇, described T ₄, T ₅, T ₆and T ₇calculating formula be:

$T_5=\frac{1}{{\mathrm{\&pi;D}}_{e}h{\left(\mathrm{\&Delta;\&theta;}\right)}^{0.25}}$

$h=\frac{Z}{{\left(D_e^{*}\right)}^8}+E$

$T_6=\frac{1}{1/T_{6\_1}+1/T_{6\_2}}$

$T_7=\frac{1}{2\mathrm{\&pi;}{\mathrm{\&lambda;}}_t}\ln [\frac{2H}{d_z}+\sqrt{{\left(\frac{2H}{d_z}\right)}^2-1}]$

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

T _{6_1}＝1/X ₁A ₁；

T _{6_2}＝1/X ₂A ₂

Wherein:

the outside diameter of cable, h: coefficient of heat transfer, Δ θ: surpass cable surface temperature rise more than environment temperature, T _{6_1}: air is to the thermal resistance of sidewall, T _{6_2}: air is to the thermal resistance on ground, X ₁, X ₂: the convection transfer rate of sidewall or bottom, A ₁, A ₂: the heat exchange area of sidewall and bottom, λ _t: soil thermal conductivity, H: the tunnel degree of depth, d _z: the diameter of circular tunnel external surface, L and W are respectively the height and width of Rectangular Tunnel.

Claims (8)

1. the heat dissipating method based on slotless box cable tunnel thermal field model, is characterized in that, comprises the following steps:

1) cable tunnel is reconnoitred, and obtains the parameter of each ingredient of cable tunnel;

2) set up the Mathematical Modeling of cable tunnel thermal field;

3) according to calculated with mathematical model, go out the calorific value of this cable tunnel;

4) design draft type and the ventilation system of this electric power tunnel.

2. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 1, is characterized in that, described thermal resistance draws by R=d/KA, and in formula, R is thermal resistance, the thickness that d is heat conductor, and A is heat conductor section area, K is coefficient of thermal conductivity.

3. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 1, is characterized in that described step 2) in the Mathematical Models of cable tunnel thermal field comprise following sub-step:

21) set up electric cable heating model:

N telegram in reply cable heat radiation power is:

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

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, and L is that cable length, C are cable radiation loss coefficient value 0.9, and n is cable backhaul number;

22) set up cable thermal losses model:

I ²[RT ₁+R(1+λ ₁)T ₂+R(1+λ ₁+λ ₂)(T ₃+T ₄)]+W _d[0.5T ₁+n(T ₂+T ₃+T ₄)]＝θ ₁-θ _C

Wherein:

I: current-carrying capacity, θ _c: environment temperature, θ ₁: cable conductor temperature, R: the conductor AC resistance under operating temperature, W _d: insulation dielectric loss, λ ₁: metallic shield loss factor, λ ₂: metal armouring loss factor, T ₁: insulating layer thermal resistance between conductor and protective metal shell, T ₂: inner liner thermal resistance between protective metal shell and armor, T ₃: protective coverings of cable thermal resistance, T ₄: cable external thermal resistance, environment temperature θ _crefer to air themperature, suppose θ _cconstant, conductor temperature θ ₁the maximum permissible temperature that refers to long-term work, is made as 90 ℃; θ ₁and θ _cit is all fringe conditions.

4. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 3, is characterized in that described step 22) in, the calculating formula of described conductor AC resistance R is:

R＝R′(1+Y _s+Y _p)

R′＝R ₀[1+α ₂₀(θ-20)]

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

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

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

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

Wherein:

R ': maximum operating temperature lower conductor D.C. resistance, Y _s; Kelvin effect factor, Y _p: kindred effect factor, R ₀: conductor DC resistance in the time of 20 ℃, θ: running temperature, α ₂₀: the temperature coefficient of copper conductor in the time of 20 ℃, for circular compact conductor K _s=1, d _c: conductor diameter, s: distance between each conductor axle center.

5. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 3, is characterized in that described step 22) in, described dielectric loss W _dcalculating formula be:

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

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

Wherein:

ω=2 π f, U ₀: voltage-to-ground, ε=2.3, D _i: be outer insulation diameter, d _c: be inner shield external diameter.

6. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 3, is characterized in that described step 22) middle metallic shield loss λ ₁calculating formula be:

λ ₁＝λ′ ₁+λ″ ₁

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

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

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

When for rounded projections arranged, have:

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

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

Δ ₂＝0

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

Wherein:

λ ' ₁=0, λ " ₁for eddy-current loss, ρ: protective metal shell resistivity, R: protective metal shell resistance, t _s: protective metal shell thickness, D: protective metal shell external diameter, D _oc: corrugated aluminium sheath maximum outside diameter, D _it: corrugated aluminium sheath minimum diameter.

7. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 3, is characterized in that described step 22) middle cable internal thermal resistance T ₁, T ₂, T ₃calculating formula be:

$T_1=\frac{{\mathrm{\&rho;}}_{T_1}}{2\mathrm{\&pi;}}\mathrm{Ln}(1+\frac{{2t}_1}{d_c})$

Wherein:

insulation materials thermal resistivity, d _c: conductor diameter t ₁: the insulation thickness between conductor and sheath;

This cable does not have steel armour, therefore T ₂=0;

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

Wherein:

T ₃: oversheath thickness, oversheath (nonmetal) thermal resistivity, D _oc: corrugated aluminium sheath maximum outside diameter.

8. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 3, is characterized in that described step 22) middle cable external thermal resistance T ₄comprise that cable is to the thermal resistance T of air ₅, air is to the thermal resistance T of cable passage inwall ₆, tunnel is to the soil thermal resistance T at the bottom of river ₇, described T ₄, T ₅, T ₆and T ₇calculating formula be:

$T_5=\frac{1}{{\mathrm{\&pi;D}}_e^{*}h{\left(\mathrm{\&Delta;\&theta;}\right)}^{0.25}}$

$h=\frac{Z}{{\left(D_e^{*}\right)}^8}+E$

$T_6=\frac{1}{1/T_{6\_1}+1/T_{6\_2}}$

$T_7=\frac{1}{2\mathrm{\&pi;}{\mathrm{\&lambda;}}_t}\ln [\frac{2H}{d_z}+\sqrt{{\left(\frac{2H}{d_z}\right)}^2-1}]$

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

T _{6_1}＝1/X ₁A ₁；

T _{6_2}＝1/X ₂A ₂

Wherein: the outside diameter of cable, h: coefficient of heat transfer, Δ θ: surpass cable surface temperature rise more than environment temperature, T _{6_1}: air is to the thermal resistance of sidewall, T _{6_2}: air is to the thermal resistance on ground, X ₁, X ₂: the convection transfer rate of sidewall or bottom, A ₁, A ₂: the heat exchange area of sidewall and bottom, λ _t: soil thermal conductivity, H: the tunnel degree of depth, d _z: the diameter of circular tunnel external surface, L and W are respectively the height and width of Rectangular Tunnel, Z, g, E are the constant in IEC60287.