Method for calculating current-carrying capacity of multi-loop cable in ventilation tunnel based on thermal circuit model

Abstract

The invention provides a method for calculating the current-carrying capacity of a multi-loop cable in a ventilation tunnel based on a thermal circuit model. The method comprises the following steps of segmenting the cable of the ventilation tunnel, calculating the temperature of the cable of the mth segment, constructing equivalent external thermal resistance, obtaining expected current and obtaining the current-carrying capacity. Compared with the prior art, the method for calculating the current-carrying capacity of the multi-loop cable in the ventilation tunnel based on the thermal circuit model can be applied to actual engineering, and compared with finite element modeling calculation, the method is simpler and more suitable for engineering.

Description

Technical Field

[0001] The present invention relates to circuit calculation, and more particularly to a method for calculating the current-carrying capacity of multi-circuit cables in a ventilated tunnel with a thermal circuit model. Background Art

[0002] The current-carrying capacity of high-voltage power cables is an important parameter that needs to be determined in engineering design, which determines the transmission capacity of the entire power cable for safe operation. Therefore, it is necessary to calculate the current-carrying capacity of the cable.

[0003] Currently, the internationally common method for calculating the current-carrying capacity is the IEC 60287 CALCULATION OF THE CURRENT RATING series regulations. International projects usually require a current-carrying capacity calculation report that complies with this regulation. This regulation uses the method of an equivalent thermal circuit model to calculate the current-carrying capacity. However, this regulation is not comprehensive and only gives the calculation methods for some parameters, resulting in difficulties in implementation in engineering; moreover, it is only applicable to single-circuit cable ventilated tunnels and is not applicable to the calculation of the current-carrying capacity of multi-circuit cables in ventilated tunnels.

[0004] The idea of calculating the current-carrying capacity of cables using the thermal circuit method is as Figure 1 shown. It uses concepts and methods similar to those of an electric circuit to describe a system of heat conduction and heat transfer processes. The thermal circuit model is based on the principles of heat conduction and heat transfer, regarding heat as current, temperature as electric potential, and thermal resistance as resistance, and analyzing the heat flow and temperature distribution by analogy with the relationship between current and voltage in an electric circuit.

[0005] The existing standard IEC 60287-2-3 2016 CALCULATION OF THE CURRENT RATING–Part 2-3:Thermal resistance-Cables installed in ventilated tunnels gives the calculation methods for some parameters, but it is only applicable to single-circuit cable ventilated tunnels and is not applicable to the calculation of the current-carrying capacity of multi-circuit cables in ventilated tunnels.

[0006] The current-carrying capacity of multi-circuit ventilated tunnel cables can currently be calculated using the finite element method, but it requires finite element calculation software and has a high operation difficulty, so it is difficult to apply in engineering. Summary of the Invention

[0007] The purpose of the present invention is to provide a solution method to solve the problem of solving the non-linear system of the current-carrying capacity of multi-circuit cables. The present invention proposes the following solution method:

[0008] The present application discloses a method for calculating the current-carrying capacity of multi-circuit cables in a ventilated tunnel based on a thermal circuit model, characterized in that the calculation method includes the following steps:

[0009] (1) Segment the cables in the ventilation tunnel.

[0010] Divide it into m segments according to the total length of the cables in the ventilation tunnel.

[0011] (2) Calculate the temperature of the cables in the m-th segment.

[0012] The heat flux of the cables in the m-th segment consists of the heat flux due to thermal radiation and the heat flux due to convective air. For m cables, m equations of Equation (1) are constructed based on the common tunnel wall temperature nodes, the average tunnel air temperature nodes, the thermal resistance from air to tunnel wall, and Ohm's law.

[0013]

[0014] Among them, W irad (n) is the heat flux due to thermal radiation, in W ictoa (n) is the heat flux due to air, in T st is the thermal radiation resistance from the cable to the tunnel wall, in T as is the heat conduction resistance from the cable to the air, in T at is the heat conduction resistance from the air to the tunnel wall.

[0015] At the same time, calculate the heat carried away by the air, and obtain the air temperature of the cables in the m-th segment expressed by Equation (2):

[0016]

[0017] Among them, θ it (n) are the common tunnel wall temperature nodes, θ at (n) are the average tunnel air temperature nodes, θ a is the ambient temperature, and Te is the external thermal resistance outside the tunnel.

[0018] Thus, the heat flux due to thermal radiation and the heat flux due to convective air of each cable, the temperature at the outlet of the m-th segment, and the heat carried away by ventilation are obtained.

[0019] (3) Calculate the overall temperature distribution of the entire ventilation tunnel.

[0020] Assume the initial current I 0 , and calculate from the first small segment to the m-th segment of the ventilation tunnel by the method in step (2), so as to obtain the temperature distribution of the entire cable tunnel.

[0021] (4) Construct an equivalent external thermal resistance;

[0022] Construct an equivalent external thermal resistance, equivalent resistance, metal sheath loss coefficient, and metal armor loss coefficient. Among them, the equivalent resistance is calculated based on the heat generation values of the cable conductors in all small sections of the entire tunnel. The equivalent metal sheath loss coefficient and metal armor loss coefficient are calculated based on the sum of the heat generations of the cable sheaths and the sum of the heat generations of the cable armors in all small sections of the entire tunnel. Finally, the equivalent external thermal resistance is calculated based on the equivalent resistance, metal sheath loss coefficient, and metal armor loss coefficient.

[0023] (5) Obtain the expected current

[0024] Subtract the temperature of the highest cable conductor temperature in all loops of the ventilation tunnel from the allowable temperature of the corresponding loop, and take the absolute value to obtain a temperature difference array. For the loop where the phase with the largest temperature difference is located, calculate the current value corresponding to the three phases of this loop reaching the limit temperature under the condition that the current influence of all other phases on this phase, the thermal resistance, and the sheath armor loss coefficient remain unchanged. If at least one of the three-phase currents is negative, it means that the current values of the multi-loop cables are currently too high. Multiply the cable current of this loop by k and return to step (2) for calculation. If all three-phase currents are positive, take the minimum value of the three-phase circuits as the current value of this loop, and check whether the change amount of the cable current meets the accuracy. The current is obtained through the following formula

[0025]

[0026] Among them, θ is the maximum allowable temperature of the cable, θ a is the ambient temperature, Δθ o is the temperature rise caused by other cables to this calculated cable, W d is the dielectric loss, R is the conductor resistance, T 1 is the thermal resistance between the conductor and the metal sheath, T 2 is the thermal resistance between the sheath and the armor, T 3 is the thermal resistance of the cable body outside the armor, T 4 is the ambient thermal resistance, λ 1 is the loss coefficient of the sheath, λ 2 is the loss coefficient of the armor;

[0027] (6) Obtain the ampacity

[0028] Calculate the ampacity through the obtained current

[0029] In a preferred example, the (2) includes the following sub-steps: During the calculation, assume an initial temperature to obtain the thermal radiation thermal resistance of the cable to the tunnel wall, the heat conduction thermal resistance of the cable to the air, and the heat conduction thermal resistance of the air to the tunnel wall, and use the finally calculated temperature as the temperature of A and D for multiple iterations.

[0030] In a preferred example, the equivalent resistance in step (4) is calculated according to the following formula:

[0031]

[0032] where R i is the equivalent resistance, W iR is the sum of the heat generated by the conductors of all small sections of the entire tunnel cable, and I 0 is the assumed initial current value.

[0033] In a preferred example, the equivalent metal sheath loss coefficient and the metal armor loss coefficient in step (4) are calculated according to the following formula:

[0034]

[0035] where λ i1 is the equivalent metal sheath loss coefficient, λ i2 is the metal armor loss coefficient, W iS is the sum of the heat generated by the sheaths of all small sections of the tunnel cable, and W iA is the sum of the heat generated by the cable armor.

[0036] In a preferred example, the equivalent external resistance in step (4) is calculated by the following formula:

[0037]

[0038] In a preferred example, the value of k is not less than 0.9.

[0039] In a preferred example, the heat carried away by the air in step (2) is calculated by the following sub-steps:

[0040] The heat carried away by the air and the temperature rise of the air on each small section of the tunnel can be expressed as:

[0041]

[0042] where W a (z) is the heat carried away by the air, and Δθ at (n) is the temperature rise of the air on each small section ΔL of the tunnel; C vair is the volumetric specific heat capacity, that is, the heat absorbed by unit volume of air per one-degree temperature rise. V is the air flow velocity in the tunnel, and A is the cross-sectional area of the tunnel;

[0043] Also, since the change in air temperature at the outlet of each section can be expressed as the air temperature at the outlet of this section minus the air temperature at the outlet of the previous section, that is:

[0044] Δθ at (n) = θ at (n) - θat (n - 1)

[0045] Therefore, through combination, we can obtain:

[0046]

[0047] In a preferred example, the assumed value of the initial temperature is from 0 to 100 °C.

[0048] The advantages of the present invention are as follows:

[0049] 1) It gives a specific solution method for solving the current - carrying capacity of multi - circuit cables.

[0050] 2) Based on solving the current - carrying capacity of multi - circuit cables, a calculation method for solving the current - carrying capacity of multi - circuit ventilation tunnels is developed. Compared with the finite - element modeling calculation, it is simpler and more applicable to engineering. BRIEF DESCRIPTION OF THE DRAWINGS

[0051] Figure 1 It is a schematic diagram of the idea of calculating the current - carrying capacity of cables by the thermal - circuit method in the prior art through IEC60287;

[0052] Figure 2 It is a calculation flow chart of the current - carrying capacity of multi - circuits of the present invention;

[0053] Figure 3 It is a partial thermal - circuit diagram of cables laid in a tunnel; DETAILED DESCRIPTION OF THE EMBODIMENTS

[0054] The inventor of the present invention has invented a calculation method for the current - carrying capacity of multi - circuit cables in a ventilation tunnel based on a thermal - circuit model. Compared with the prior art, it is simpler and more applicable to engineering.

[0055] Terms.

[0056] It should be noted that in the invention document of this patent, relational terms such as first and second are only used to distinguish one entity or operation from another entity or operation, and do not necessarily require or imply any actual relationship or order between these entities or operations. Moreover, the term "comprising", "including" or any other variant thereof is intended to cover non-exclusive inclusion, so that a process, method, article or device comprising a series of elements not only includes those elements, but also includes other elements not expressly listed, or also includes elements inherent in such process, method, article or device. Without further limitation, an element defined by the statement "comprising one" does not exclude the presence of additional identical elements in the process, method, article or device comprising such element. In the invention document of this patent, if it is mentioned that an act is performed according to a certain element, it means that the act is performed at least according to that element, including two cases: the act is performed only according to that element, and the act is performed according to that element and other elements. Expressions such as multiple, many times, various, etc. include 2, 2 times, 2 kinds, as well as more than 2, more than 2 times, more than 2 kinds.

[0057] Multi-circuit cable current-carrying capacity

[0058] Under normal circumstances, the limiting condition of the current-carrying capacity is the conductor temperature. As the conductor current-carrying capacity increases, the heat generated by the cable increases, the conductor temperature rises, and the insulating material tightly wraps around the conductor. In order to ensure that the insulating material is not damaged during long-term use, there is usually a limiting temperature, and the limiting temperatures of different insulating materials are different.

[0059] When the current-carrying capacity of a certain cable in each circuit of the cable reaches the limiting temperature, and the other cables in the same circuit are less than or equal to the limiting temperature, the calculation can be regarded as completed. Usually, there are more than one cable in one circuit: usually three for the AC system and two for the DC system. Since the positions of each cable are different, their environmental thermal resistances are different, and the heat generation is also different. Therefore, the temperatures of each cable in the same circuit are not the same. As long as one cable in each circuit reaches the temperature limit, the current in this circuit cannot continue to increase.

[0060] Since the cable current determines the conductor resistance, which in turn determines the conductor temperature and the environmental thermal resistance, and the conductor temperature and the environmental thermal resistance in turn limit the cable current. Therefore, solving the current-carrying capacity of multi-circuit cables is actually a problem of solving a non-linear black-box system, which is difficult to solve by analytical methods. The present invention proposes the following solution method.

[0061] Steps for calculating the current-carrying capacity:

[0062] 1) Given the initial temperature and current. Calculate through the initial temperature, current, and voltage: A. The resistance of the conductor, and then calculate the heat generated by the conductor; B. The resistance of the metal sheath and the heat generated; C. The resistance of the metal reinforcement layer of the sheath and the heat generated; D. The resistance of the armor and the heat generated; E. The environmental thermal resistance; F. The eddy current circulation loss; Temperature and current independent parameters: A. The thermal resistance from the conductor to the metal sheath; B. The thermal resistance from the metal sheath to the armor; C. The heat generated by the armor to the medium. The method proposed by the present invention has low requirements for the initial temperature and current. Usually, the initial temperature can converge between 0 and 100°C, and the current can converge between 0 and 5000A.

[0063] 2) Substitute the temperature calculated in 1) back into 1) again. After several iterations, when the calculation result no longer changes, the actual temperatures of each layer of the cable under this current are obtained.

[0064] 3) Subtract the temperature corresponding to the maximum temperature of the cable conductor in all loops from the allowable temperature of the corresponding loop, and take the absolute value to obtain a temperature difference array. For the loop where the phase with the largest temperature difference is located, calculate the current value corresponding to the three phases of this loop reaching the limit temperature under the condition that the influence of all other phases (including the phases of other loops) on this phase remains unchanged, the thermal resistance remains unchanged, and the sheath armor loss coefficient remains unchanged. If all three-phase currents are positive, take the minimum value of the three-phase circuit as the current value of this loop and substitute it into the next round of loop (step 1); if at least one of the three-phase currents is negative, it means that the current values of the current multi-circuit cables are too high. Multiply the current of this cable by k and enter the next round of calculation (step 1). The smaller the k value, the faster the calculation speed but it may not converge. The larger the k value, the slower the calculation speed but the better the convergence. Through calculation experiments, it is found that when the k value is greater than or equal to 0.9, there is better convergence and calculation speed.

[0065]

[0066] In the formula, θ is the maximum allowable temperature of the cable, θ a is the ambient temperature, Δθ o is the temperature rise caused by other cables to the cable being calculated, W d is the dielectric loss, R is the conductor resistance, T 1 is the thermal resistance between the conductor and the metal sheath, T 2 is the thermal resistance between the sheath and the armor, T 3 is the thermal resistance of the cable body outside the armor, T 4 is the environmental thermal resistance, λ 1 is the loss coefficient of the sheath, λ 2 is the loss coefficient of the armor.

[0067] To more conveniently explain the meaning of the above formula, the above formula can be transformed as follows:

[0068] I 2 R[T 1 +(1 + λ1 )T 2 +(1 + λ 1 + λ 2 )(T 3 + T 4 )] + W d (0.5T 1 + T 2 + T 3 + T 4 ) = θ - θ a - Δθ o

[0069] The left side of the equation is the temperature rise of the conductor relative to the external environment caused by electrical and thermal losses, and the right side is the allowable temperature rise of the cable. Since there is a corresponding initial current for each calculation, the influence of other phases (including the phases of other circuits) on this phase remains unchanged, the thermal resistance remains unchanged, and the sheath armor loss coefficient remains unchanged, that is, Δθ o 、W d 、R、T 1 、T 2 、T 3 、T 4 、λ 1 、λ 2 Adopt the calculated values in steps 1) and 2), and θ a belongs to the external known conditions, and the expected current I of this iteration of this cable can be obtained.

[0070] The features of this method are: 1) Iterate by setting the current, and each time correct the current of each phase cable in the loop with the largest absolute value of the temperature difference. 2) When the current is too large, that is, when the corrected current is less than 0, use the k-value correction coefficient, recommended to be above 0.9. 3) When the corrected current is greater than 0, use the corrected current for iteration.

[0071] Equivalent Parameter Algorithm for Multi-Circuit Cables in Ventilation Tunnels

[0072] The multi-circuit cable ampacity calculation method in the ventilation tunnel based on the thermal circuit model described in the present invention is the equivalent parameter algorithm for multi-circuit cables in the ventilation tunnel combined with the multi-circuit cable ampacity algorithm. The purpose of calculating the environmental thermal resistance is to obtain the environmental thermal resistance T4.

[0073] One of the characteristics of the cables laid in the ventilation tunnel is that as the heat accumulates in the tunnel, the temperatures of the air, the cables at the tunnel exit are different from those at the tunnel entrance. Therefore, the tunnel is calculated in segments. Let the total length of the tunnel be L, the calculation length of each segment be ΔL, and the current calculation position of the cable be z = n×ΔL (the current calculation is the nth segment).

[0074] 1) For each small segment ΔL, the following calculations need to be performed:

[0075] There are two ways for the heat of a cable to dissipate outward: heat conduction from the cable to the surrounding air and heat radiation from the cable to the surrounding environment. Let the heat flow of the i-th cable be W ik (n) (adopting I 0 initial value; assumed), the heat flow assigned to heat radiation is W irad (n), and the heat flow assigned to convective air is W ictoa (n). Then the heat circuit diagram is as shown in the figure, and T st is the heat radiation thermal resistance from the cable to the tunnel wall, and T as is the heat conduction thermal resistance from the cable to the air, and T at is the heat conduction thermal resistance from the air to the tunnel wall. The formulas for these three can be obtained from IEC 60287-2-3. The heat flow from the air to the tunnel wall in the figure is W iatow (n). So for each cable:

[0076] W ik (n) = W irad (n) + W ictoa (n) Equation 3

[0077] For m cables, the common tunnel wall temperature node is θ it (n), the average tunnel air temperature node is θ at (n), and the heat resistance from the air to the tunnel wall is T at . According to Ohm's law, m equations can be constructed:

[0078]

[0079] At the same time, the heat carried away by the air W a (z) and the temperature rise Δθ at (n) of the air on each small section ΔL of the tunnel can be expressed as:

[0080]

[0081] The temperature change of the air at the outlet of each section is equal to the air temperature at the outlet of this section minus the air temperature at the outlet of the previous section. Therefore,

[0082] Δθ at (n) = θ at (n) - θ at (n - 1) Equation 6

[0083] Combining Equation 5 and Equation 6 gives:

[0084]

[0085] C vair is the volumetric specific heat capacity, that is, the heat absorbed per unit volume of air per degree of temperature rise. V is the air flow velocity in the tunnel, and A is the cross-sectional area of the tunnel.

[0086] And since all the cables share the tunnel wall temperature node θ it (n), the average tunnel air temperature node θ at (n), the thermal resistance T from air to tunnel wall at and the external tunnel thermal resistance T e . The air temperature at the outlet of this small section can be expressed by the following formula:

[0087]

[0088] where Te is the external tunnel thermal resistance, which can be obtained from IEC60287, and θ a is the ambient temperature, which is an input known quantity.

[0089] For m cables, m equations can be obtained from Equation 3, m equations can be obtained from Equation 4, 1 equation for Equation 7, and 1 equation for Equation 8, for a total of 2m + 2 equations. Each cable has two unknowns, W irad (n) and W ictoa (n), for a total of 2m unknowns, θ at (n) and W a (n) are two unknowns, for a total of 2m + 2 unknowns. And the system of equations is a system of linear equations of the first order, so it is very easy to obtain W irad (n) and W ictoa (n) for each cable, the temperature θ at (n) at the outlet of this small section and the heat carried away by ventilation W a (n).

[0090] Note that when calculating the nth section for the above equations, θ at (n - 1) is regarded as a known quantity, and the reason is as follows. θ at (0) is the air temperature at the inlet, which is a known quantity. Therefore, for n = 1, θ at (n - 1) is known, and then θ at (n) can be solved through the system of equations composed of Equations 3, 4, 7, and 8. Calculating sequentially from n = 1, then θ at (n - 1) for each section is the result calculated in the previous section, so it can be regarded as a known quantity.

[0091] Because in the formula, T st , T as , T at and the cable resistance are temperature-related quantities, and temperature is a function of current. Therefore, it is also necessary to first assume an initial temperature. Then calculate a single small section to obtain the temperature distribution. Then iterate the calculated temperature for recalculation. Usually, three iterations can achieve a very high accuracy.

[0092] 2) Overall calculation of the entire tunnel. As can be seen from 1), if the initial current I is assumed0 , starting from the first small section, it can be calculated up to the last section of the tunnel, thereby obtaining the temperature distribution of the entire cable tunnel. However, at this time, it is still impossible to substitute into Figure 1 's calculation method to solve the multi-loop current. The reasons are as follows:

[0093] A. The resistance R of the i-th cable i (n), the metal sheath loss coefficient λ i1 (n) and the metal armor loss coefficient λ i2 (n) are calculated separately for each small section, and their values may be different for each ΔL along the entire tunnel path.

[0094] B. T 4 (n) has not been explicitly calculated. It can be seen from Figure 2 that the external thermal resistance is jointly composed of the thermal resistance network formed by T st , T as , T at and T e , and each section of W a (n) is different, resulting in different T 4 (n) for each small section.

[0095] To solve the above problems and thus solve the multi-loop current-carrying capacity, the present invention proposes the following method: construct an equivalent external thermal resistance T i4 , an equivalent resistance R i , a metal sheath loss coefficient λ i1 and a metal armor loss coefficient λ i2 .

[0096] Calculation method of the equivalent resistance R i : First, find the sum W of the heat generated by the cable conductors in all small sections of the entire tunnel iR , and then substitute it into Equation 9 to obtain.

[0097]

[0098] Calculation method of the equivalent metal sheath loss coefficient λ i1 and the metal armor loss coefficient λ i2 : First, find the sum W of the heat generated by the cable sheaths in all small sections of the entire tunnel iS , the sum W of the heat generated by the cable armors iA , and then substitute them into Equations 10 and 11 to obtain.

[0099]

[0100]

[0101] Then the equivalent external thermal resistance T i4 can be obtained:

[0102]

[0103] All the parameters in the formula have the same meaning as those in formula 1, and θ takes the temperature of the cable conductor at the tunnel exit.

[0104] 3) With the equivalent external thermal resistance T i4 , the equivalent resistance R i , the metal sheath loss factor λ i1 and the metal armor loss factor λ i2 , they can be substituted into Figure 2 in the flow chart to solve the current-carrying capacity of each loop.

[0105] The present invention will be described in detail below with reference to the embodiments shown in the accompanying drawings. It should be noted, however, that these embodiments do not limit the present invention, and any equivalent transformation or substitution in function, method, or structure made by those of ordinary skill in the art based on these embodiments shall fall within the protection scope of the present invention.

[0106] The specific implementation process of the present invention will be described below through embodiments.

[0107] Example:

[0108] An embodiment of the present invention is as Figures 1 - 3 shown. This embodiment calculates the current-carrying capacity of multi-loop cables in a ventilation tunnel through the following method:

[0109] (1) Segment the cables in the ventilation tunnel

[0110] According to the total length of the cables in the ventilation tunnel, divide it into m segments;

[0111] (2) Calculate the temperature of the cables in the m-th segment

[0112] The heat flow of the cables in the m-th segment consists of the heat flow to thermal radiation and the heat flow to convective air. For m cables, m formulas 1 are constructed based on the temperature nodes of the shared tunnel wall, the average temperature node of the tunnel air, the thermal resistance from air to tunnel wall, and Ohm's law;

[0113]

[0114] Among them, W irad (n) is the heat flow to thermal radiation, W ictoa (n) is the heat flow to air, T st is the thermal radiation thermal resistance from the cable to the tunnel wall, T as is the heat conduction thermal resistance from the cable to air, T at is the heat conduction thermal resistance from air to the tunnel wall;

[0115] Calculate the heat carried away by the air at the same time, and obtain the air temperature of the m-th section of the cable as expressed by Equation 2:

[0116]

[0117] where, θ it (n) is the temperature node of the shared tunnel wall, θ at (n) is the average temperature node of the tunnel air, θ a is the ambient temperature, Te is the external thermal resistance outside the tunnel

[0118] Thus, the heat flux to thermal radiation and the heat flux to convective air of each cable, the temperature at the outlet of the m-th section, and the heat carried away by ventilation are obtained.

[0119] Optionally, in an embodiment, by assuming an initial temperature, the thermal radiation thermal resistance of the cable to the tunnel wall, the heat conduction thermal resistance of the cable to the air, and the heat conduction thermal resistance of the air to the tunnel wall are obtained, and the finally calculated temperature is used as the assumed temperature again for multiple iterations. Optionally, in an embodiment, the assumed value of the initial temperature is 0 to 100 °C.

[0120] (3) Perform a general calculation on the entire ventilation tunnel to obtain the temperature distribution

[0121] Assume an initial current I 0 , and calculate from the first small section of the ventilation tunnel to the m-th section by the method of step (2), so as to obtain the temperature distribution of the entire cable tunnel. Optionally, in an embodiment, the obtained temperature is substituted back into

[0122] (4) Construct an equivalent external thermal resistance;

[0123] Construct an equivalent external thermal resistance, an equivalent resistance, a metal sheath loss coefficient, and a metal armor loss coefficient, where the equivalent resistance is calculated according to the heat generation value of the cable conductors of all small sections of the entire tunnel, and the equivalent metal sheath loss coefficient and the metal armor loss coefficient are calculated according to the sum of the heat generation of the cable sheaths and the sum of the heat generation of the cable armors of all small sections of the entire tunnel, and finally the equivalent external thermal resistance is calculated according to the equivalent resistance, the metal sheath loss coefficient, and the metal armor loss coefficient;

[0124] (5) Obtain the expected current

[0125] Subtract the temperature of the highest cable conductor temperature in all circuits of the ventilation tunnel from the allowable temperature of the corresponding circuit, take the absolute value, and obtain a temperature difference array. For the circuit where the phase with the largest temperature difference is located, calculate the current value corresponding to the three phases of this circuit reaching the limit temperature under the condition that the influence of all other phases on this phase, the thermal resistance, and the sheath armor loss coefficient remain unchanged at present; if at least one of the three-phase currents is negative, it means that the current values of the current multi-circuit cables are all too high. Multiply the cable current of this circuit by k and return to step (2) for calculation. If all three-phase currents are positive, take the minimum value of the three-phase circuits as the current value of this circuit, and check whether the change in the cable current meets the accuracy; the current is obtained through the following formula

[0126]

[0127] Among them, θ is the maximum allowable temperature of the cable, θ a is the ambient temperature, Δθ o is the temperature rise caused by other cables to this calculated cable, W d is the dielectric loss, R is the conductor resistance, T 1 is the thermal resistance between the conductor and the metal sheath, T 2 is the thermal resistance between the sheath and the armor, T 3 is the thermal resistance of the cable body outside the armor, T 4 is the ambient thermal resistance, λ 1 is the loss coefficient of the sheath, λ 2 is the loss coefficient of the armor;

[0128] (6) Obtain the ampacity

[0129] Calculate the ampacity through the obtained current

[0130] During the calculation, assume an initial temperature to obtain the thermal radiation thermal resistance of the cable to the tunnel wall, the heat conduction thermal resistance of the cable to the air, and the heat conduction thermal resistance of the air to the tunnel wall, and use the finally calculated temperature as the assumed temperature for multiple iterations.

[0131] Optionally, in an embodiment, the equivalent resistance in step (4) is calculated according to the following formula:

[0132]

[0133] Among them, R i is the equivalent resistance, W iR is the sum of the heat generated by the cable conductors of all small sections of the entire tunnel, I 0 is the assumed initial current value.

[0134] Optionally, in an embodiment, the equivalent metal sheath loss coefficient and the metal armor loss coefficient in step (4) are calculated according to the following formula:

[0135]

[0136] Among them, λ i1 is the equivalent metal sheath loss coefficient, and λ i2 is the metal armor loss coefficient, and W iS is the heat generated by the cable sheaths of all small sections of a tunnel, and W iA is the sum of the heat generated by the cable armor.

[0137] Optionally, in one embodiment, it is characterized in that the equivalent external resistance in the step (4) is calculated by the following formula:

[0138]

[0139] Optionally, in one embodiment, the value of k is not less than 0.9.

[0140] Optionally, in one embodiment, the heat taken away by the air in the step (2) is calculated by the following sub-steps:

[0141] The heat taken away by the air and the temperature rise of the air on each small section of the tunnel can be expressed as:

[0142]

[0143] Among them, W a (z) is the heat taken away by the air, and Δθ at (n) is the temperature rise of the air on each small section ΔL of the tunnel; C vair is the volume specific heat capacity, that is, the heat absorbed by unit volume of air per degree of temperature rise. V is the air flow velocity in the tunnel, and A is the cross-sectional area of the tunnel;

[0144] Also, because the temperature change amount of the air at the outlet of each section can be expressed as the air temperature at the outlet of this section minus the air temperature at the outlet of the previous section, that is:

[0145] Δθ at (n) = θ at (n) - θ at (n - 1)

[0146] Therefore, through combination, we can obtain:

[0147]

[0148] The series of detailed descriptions listed above are only specific descriptions of the feasible implementation manners of the present invention, and they are not intended to limit the protection scope of the present invention. Any equivalent implementation manners or modifications made without departing from the technical spirit of the present invention should be included within the protection scope of the present invention.

[0149] It will be apparent to those skilled in the art that the present invention is not limited to the details of the above-described exemplary embodiments, and that the present invention can be implemented in other specific forms without departing from the spirit or essential characteristics thereof. Therefore, in any respect, the embodiments should be regarded as exemplary and non-limiting. The scope of the present invention is defined by the appended claims rather than the above description. Accordingly, all changes that fall within the meaning and scope of the equivalent elements of the claims are intended to be embraced within the present invention. Any reference signs in the claims should not be construed as limiting the claims concerned.

[0150] In addition, it should be understood that although this specification is described in terms of embodiments, not every embodiment contains only one independent technical solution. This narrative manner of the specification is merely for clarity. Those skilled in the art should regard the specification as a whole, and the technical solutions in each embodiment can also be appropriately combined to form other embodiments that can be understood by those skilled in the art.

Claims

1. A calculation method for the current-carrying capacity of multi-circuit cables in a ventilation tunnel based on a thermal circuit model, characterized in that, the calculation method includes the following steps: (1) Segment the cables in the ventilation tunnel According to the total length of the cables in the ventilation tunnel, divide them into m segments; (2) Calculate the temperature of the cables in the m-th segment The heat flow of the cables in the m-th segment consists of the heat flow to thermal radiation and the heat flow to convective air. For m cables, m equations (1) are constructed based on the common tunnel wall temperature nodes, the average tunnel air temperature nodes, the thermal resistance from air to tunnel wall, and Ohm's law; Among them, W irad (n) is the heat flux to thermal radiation, W ictoa (n) is the heat flux to air, T st is the thermal radiation thermal resistance from the cable to the tunnel wall, T as is the heat conduction thermal resistance from the cable to air, T at is the heat conduction thermal resistance from air to the tunnel wall; At the same time, calculate the heat carried away by the air, and obtain the air temperature of the m-th segment of cables expressed by Equation (2): where θ it (n) is the temperature node of the shared tunnel wall, and θ at (n) is the average temperature node of the tunnel air. θ a is the ambient temperature, and Te is the external thermal resistance of the tunnel Thus, obtain the heat flow to thermal radiation and the heat flow to convective air of each cable, the temperature at the outlet of the m-th segment, and the heat carried away by ventilation; (3) Perform a general calculation for the entire ventilation tunnel to obtain the temperature distribution Assume the initial current I 0 , calculate the temperature distribution of the entire cable tunnel from the first small section to the m-th section of the ventilation tunnel by the method of step (2), so as to obtain the temperature distribution of the entire cable tunnel; (4) Construct an equivalent external thermal resistance; Construct an equivalent external thermal resistance, an equivalent resistance, a metal sheath loss coefficient, and a metal armor loss coefficient. Among them, the equivalent resistance is calculated based on the heat generation values of the conductors of all small segments of cables in the entire tunnel, and the equivalent metal sheath loss coefficient and metal armor loss coefficient are calculated based on the sum of the heat generation of the sheaths of all small segments of cables and the sum of the heat generation of the cable armors in the entire tunnel. Finally, calculate and obtain the equivalent external thermal resistance based on the equivalent resistance, the metal sheath loss coefficient, and the metal armor loss coefficient; (5) Obtain the expected current Subtract the temperature of the highest temperature of the cable conductors in all circuits of the ventilation tunnel from the allowable temperature of the corresponding circuit, and take the absolute value to obtain a temperature difference array; for the circuit where the phase with the largest temperature difference is located, calculate the current value corresponding to the three phases reaching the limit temperature under the condition that the current influence of all other phases on this phase remains unchanged, the thermal resistance remains unchanged, and the sheath armor loss coefficient remains unchanged; if at least one of the three-phase currents is negative, it means that the current values of the current multi-circuit cables are all too high. Multiply the cable current of this circuit by k and return to step (2) for calculation. If all three-phase currents are positive, take the minimum value of the three-phase circuits as the current value of this circuit, and check whether the change amount of the cable current meets the accuracy; the current is obtained through the following formula Among them, θ is the maximum allowable temperature of the cable, θ a is the ambient temperature, Δθ o is the temperature rise caused by other cables to this calculated cable, W d is the dielectric loss, R is the conductor resistance, T 1 is the thermal resistance between the conductor and the metal sheath, T 2 is the thermal resistance between the sheath and the armor, T 3 is the thermal resistance of the cable body outside the armor, T 4 is the ambient thermal resistance, λ 1 is the loss coefficient of the sheath, λ 2 is the loss coefficient of the armor; (6) Obtain the current-carrying capacity Calculate the current-carrying capacity through the obtained current.

2. The calculation method for the current-carrying capacity of multi-circuit cables in a ventilation tunnel based on a thermal circuit model according to claim 1, characterized in that, the (2) includes the following sub-steps: during the calculation, assume an initial temperature to obtain the thermal radiation thermal resistance from the cables to the tunnel wall, the heat conduction thermal resistance from the cables to the air, and the heat conduction thermal resistance from the air to the tunnel wall, and use the finally calculated temperature as the initial temperature for multiple iterations.

3. The calculation method for the current-carrying capacity of multi-circuit cables in a ventilation tunnel based on a thermal circuit model according to claim 1, the equivalent resistance in step (4) is calculated according to the following formula: wherein, R i is the equivalent resistance, W iR is the sum of the heat generated by the conductors of all small sections of the entire tunnel, I 0 is the assumed initial current value.

4. The calculation method for the current-carrying capacity of multi-circuit cables in a ventilation tunnel based on a thermal circuit model according to claim 1, characterized in that, the equivalent metal sheath loss coefficient and metal armor loss coefficient in step (4) are calculated according to the following formula: Among them, λ i1 is the equivalent metal sheath loss coefficient, λ i2 is the metal armor loss coefficient, W iS is the heat generated by the cable sheaths of all small sections of a tunnel, W iA is the sum of the heat generated by the cable armors.

5. The calculation method of the current-carrying capacity of multi-circuit cables in a ventilation tunnel based on a thermal circuit model according to claim 1, characterized in that, the equivalent external resistance in step (4) is calculated by the following formula:

6. The calculation method of the current-carrying capacity of multi-circuit cables in a ventilation tunnel based on a thermal circuit model according to claim 1, characterized in that, the value of k is not less than 0.

9.

7. The calculation method of the current-carrying capacity of multi-circuit cables in a ventilation tunnel based on a thermal circuit model according to claim 1, characterized in that, the heat carried away by the air in step (2) is calculated by the following sub-steps: The heat carried away by the air and the temperature rise of the air on each small section of the tunnel can be expressed as: Among which W a (z) is the heat carried away by air, Δθ at (n) is the temperature rise of air on each small section ΔL of the tunnel; C vair is the volumetric specific heat capacity, that is, the heat absorbed by unit volume of air for each degree of temperature rise; V is the air flow velocity in the tunnel, and A is the cross-sectional area of the tunnel; Also, since the change in air temperature at the outlet of each section can be expressed as the air temperature at the outlet of this section minus the air temperature at the outlet of the previous section, that is: Δθ at θ(n) = θ at θ(n) - θ at (n - 1) Therefore, through combination, we can obtain:

8. The calculation method of the current-carrying capacity of multi-circuit cables in a ventilation tunnel based on a thermal circuit model according to claim 2, characterized in that, the assumed value of the initial temperature is from 0 to 100 °C.

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