CN104196555B - A kind of heat dissipating method based on slotless box cable tunnel thermal field model - Google Patents

Abstract

The present invention relates to a kind of heat dissipating method based on slotless box cable tunnel thermal field model, comprise the following steps: 1) cable tunnel is reconnoitred, obtain the parameter of the each part in cable tunnel; 2) set up the Mathematical Modeling of cable tunnel thermal field; 3) go out the caloric value of this cable tunnel according to calculated with mathematical model; 4) design draft type and the ventilating system of this electric power tunnel. Compared with prior art, the present invention has advantage effectively accurately.

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 based on slotless box cable tunnel thermal field modelHeat dissipating method

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 is that the cable passage cut apart in every tunnel is as high-tension cableTunnel, lays 220kV power cable.

In the original scheme in tunnel, do not consider to lay high voltage power cable, so the design of tunnel ventilation heat radiation alsoDo not consider the caloric value of power cable. Power cable, in long-term heating in service, causes cable passage temperature to raise,May affect the safe operation of the miscellaneous equipment in the same space in tunnel. In addition, the radiating condition of this cable passageDifferent with conventional tunnel, conventional tunnel up and down four faces all directly contacts soil, can dispel the heat. Shanghai is highIn Soviet Union tunnel, cable passage only side and bottom is soil, and top is highway, and opposite side is that track is handed overCirculation passage.

Summary of the invention

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

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

Based on a heat dissipating method for slotless box cable tunnel thermal field model, comprise the following steps:

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

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

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

4) design draft type and the ventilating system of this cable 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 conductorArea of section, K is thermal conductivity factor.

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:

$P=n\frac{I^2\mathrm{\σ}}{A}LC$

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 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:

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: environment temperature, θ₁: cable conductor temperature, R: the conductor under operating temperature exchangesResistance, W_d: insulation dielectric loss, λ₁: metallic shield 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: cableOuter jacket thermal resistance, T₄: cable external thermal resistance, environment temperature θ_CRefer to air themperature, suppose θ_CBe constant, leadTemperature θ₁The maximum permissible temperature that refers to long-term work, is made as 90 DEG C; θ₁And θ_CIt 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 20 DEG C time, θ: running temperature, α₂₀: the temperature coefficient of copper conductor 20 DEG C time, forCircular 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:

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

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

Wherein:

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

Described step 22) middle metallic shield loss λ₁Calculating formula be:

λ₁＝λ′₁+λ″₁

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

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

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

When for rounded projections arranged, have:

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

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

Δ₂＝0

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

Wherein:

λ′₁＝0，λ″₁For eddy-current loss, ρ: protective metal shell resistivity, R: protective metal shell resistance, t_s: metal protectsCover 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) middle 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:

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

This cable does not have 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) middle cable external thermal resistance T₄Comprise the thermal resistance T of cable to air₅, air is logical to cableThe thermal resistance T of road inwall₆, tunnel is to the soil thermal resistance T at the bottom of river₇, described T₄、T₅、T₆And T₇Calculating formula be:

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

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

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

$T_7=\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_{6_1}＝1/X₁A₁；

T_{6_2}＝1/X₂A₂

Wherein:

The outside diameter of cable, h: coefficient of heat transfer, Δ θ: exceed 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, X₁、X₂: the heat convection of sidewall or bottomCoefficient, A₁、A₂: 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 outer 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, examineIt is thorough to consider, and modeling is accurate.

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

Brief description of the drawings

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

Detailed description of the invention

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

Embodiment:

Based on a heat dissipating method for slotless box cable tunnel thermal field model, comprise the following steps:

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

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

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

4) design draft type and the ventilating system of this cable 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 conductorArea of section, K is thermal conductivity factor.

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:

$P=n\frac{I^2\mathrm{\σ}}{A}LC$

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 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:

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: environment temperature, θ₁: cable conductor temperature, R: the conductor under operating temperature exchangesResistance, W_d: insulation dielectric loss, λ₁: metallic shield 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: cableOuter jacket thermal resistance, T₄: cable external thermal resistance, environment temperature θ_CRefer to air themperature, suppose θ_CBe constant, leadTemperature θ₁The maximum permissible temperature that refers to long-term work, is made as 90 DEG C; θ₁And θ_CIt 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 20 DEG C time, θ: running temperature, α₂₀: the temperature coefficient of copper conductor 20 DEG C time, forCircular 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:

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

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

Wherein:

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

Described step 22) middle metallic shield loss λ₁Calculating formula be:

λ₁＝λ′₁+λ″₁

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

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

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

When for rounded projections arranged, have:

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

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

Δ₂＝0

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

Wherein:

λ′₁＝0，λ″₁For eddy-current loss, ρ: protective metal shell resistivity, R: protective metal shell resistance, t_s: metal protectsCover 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) middle 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:

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

This cable does not have 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 the solid such as air and tunnel wall, cable carries out belongs to fluid heat transfer free convection process, and fluid certainlySo heat convection is divided into large space heat transfer free convection and confined space heat transfer free convection two classes, and so-called large space certainlySo heat convection, refers to the heat transfer free convection that the development in boundary layer, nearly wall place is not interfered because of spatial limitation,And little of making fluid increase and the descending motion of being heated interferes with each other while affecting when space, be called the confined space naturallyHeat convection, for example, for the heat convection between two of height h, interval d parallel perpendicular planomurals, whenD/h < 0.33 o'clock, fluid rises and descending motion interferes with each other, and must calculate by confined space heat transfer free convection.

In the time that fluid flows through the wall not identical with its temperature, will form one deck temperature at wall annex jumpyThin layer of fluid, is called thermal boundary layer, and the flow regime of thermal boundary layer is divided into turbulent flow and two kinds of laminar flows, and these two kinds mobileConvection transfer rate computational methods under state are not identical yet. Thermal boundary layer is turbulent flow or laminar flow, depends on this temperatureThe temperature differences, wall size shape, wall degree of roughness of air physical parameter, fluid and wall under degree etc. are multipleFactor.

The situation of slotless box as shown in Figure 1, described step 22) in cable external thermal resistance T₄Comprise that cable is to skyThe thermal resistance T of gas₅, air is to the thermal resistance T of cable passage inwall₆, tunnel is to the soil thermal resistance T at the bottom of river₇, describedT₄、T₅、T₆And T₇Calculating formula be:

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

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

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

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

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

T_{6_1}＝1/X₁A₁；

T_{6_2}＝1/X₂A₂

Wherein:

The outside diameter of cable, h: coefficient of heat transfer, Δ θ: exceed 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, X₁、X₂: the heat convection of sidewall or bottomCoefficient, A₁、A₂: 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 outer surface, L and W are respectively the height and width of Rectangular Tunnel.

Claims (6)

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

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

2) set up the Mathematical Modeling 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:

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

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 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:

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: environment temperature, θ₁: cable conductor temperature, R: the conductor under operating temperature exchangesResistance, W_d: insulation dielectric loss, λ₁: metallic shield 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: cableOuter jacket thermal resistance, T₄: cable external thermal resistance, environment temperature θ_CRefer to air themperature, suppose θ_CBe constant, leadTemperature θ₁The maximum permissible temperature that refers to long-term work, is made as 90 DEG C; θ₁And θ_CIt is all boundary condition;

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

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

2. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 1, itsBe characterised in that 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 20 DEG C time, θ: running temperature, α₂₀: the temperature coefficient of copper conductor 20 DEG C time, forCircular compact conductor K_s＝1，d_c: conductor diameter, s: distance between each conductor axle center.

3. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 1, itsBe characterised in that 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＝2πf，U₀: voltage-to-ground, ε=2.3, D_i: be outer insulation diameter, d_c: be inner shield external diameter.

4. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 1, itsBe characterised in that described step 22) middle metallic shield loss λ₁Calculating formula be:

λ₁＝λ′₁+λ″₁

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

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

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

When for rounded projections arranged, have:

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

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

Δ₂＝0

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

Wherein:

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

5. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 1, itsBe characterised in that described step 22) middle 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:

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

This cable does not have 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.

6. a kind of heat dissipating method based on slotless box cable tunnel thermal field model according to claim 1, itsBe characterised in that described step 22) middle cable external thermal resistance T₄Comprise the thermal resistance T of cable to air₅, air arrivesThe thermal resistance T of cable passage inwall₆, tunnel is to the soil thermal resistance T at the bottom of river₇, described T₄、T₅、T₆And T₇MeterFormula is:

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

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

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

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

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

T_{6_1}＝1/X₁A₁；

T_{6_2}＝1/X₂A₂

Wherein:The outside diameter of cable, h: coefficient of heat transfer, Δ θ: exceed cable surface temperature more than environment temperatureRise T_{6_1}: air is to the thermal resistance of sidewall, T_{6_2}: air is to the thermal resistance on ground, X₁、X₂: sidewall or bottomConvection transfer rate, A₁、A₂: the heat exchange area of sidewall and bottom, λ_t: soil thermal conductivity, H: tunnel is darkDegree, d_z: the diameter of circular tunnel outer surface, L and W are respectively the height and width of Rectangular Tunnel, Z, g, E areConstant in IEC60287.