Rail body heating device, method and thermal efficiency verification method

Abstract

A kind of steel rail body heating device, method and thermal efficiency verification method, belong to steel rail heating technical field, including generator, DC switching power supply, copper wire, heat preservation layer, the generator is connected with DC switching power supply, the DC switching power supply is connected with copper wire, copper wire is externally provided with insulating heat preservation layer, copper wire is placed at the bottom of steel rail covering the length of steel rail and is electrically insulated with steel rail and is bonded in two ends of steel rail using clamp and steel rail electrical connection, the steel rail and the copper wire bonded with the bottom of steel rail are wrapped with heat preservation layer outside, select DC as power supply.The copper wire is bonded with the steel rail by being wrapped in the heat preservation layer in the present application, and the copper wire and the DC switching power supply are reasonably selected, so that the steel rail heating efficiency is improved.

Description

Technical Field

[0001] This invention belongs to the field of rail heating technology, and particularly relates to a rail body heating device, method, and thermal efficiency verification method. Background Technology

[0002] Temperature changes can cause thermal stress inside the rails. Excessive thermal stress can easily cause rail expansion and cracking, which in turn affects the service life of the rails and traffic safety. Therefore, track maintenance personnel need to heat the rails before replacing them, generally keeping the rail temperature at around the annual average temperature.

[0003] A published patent (CN114606815A) discloses a method and system for heating rails. The method employs a low-voltage, high-current body heating method. This method offers advantages such as non-damaging heating of the rail, uniform heating distribution, and safety and convenience. However, because it relies on a low-voltage, high-current approach, the copper wire connecting the rail and the transformer will inevitably carry a large current, resulting in significant heat generation and power loss. Furthermore, the use of alternating current as the power source will also consume power in the inductive reactance section. Therefore, how to utilize this energy to improve heating efficiency is a problem worthy of further research. Summary of the Invention

[0004] In view of the above-mentioned shortcomings and deficiencies of the prior art, the present invention provides a rail body heating device, method and thermal efficiency verification method, which improves the rail heating efficiency by wrapping copper wires in the insulation layer and attaching them to the rail, and by rationally selecting copper wires and DC switching power supply.

[0005] To achieve the above objectives, the main technical solutions adopted by the present invention include:

[0006] A rail heating device includes a generator, a DC switching power supply, copper wires, and an insulation layer. The generator is connected to the DC switching power supply, which is connected to the copper wires. The copper wires are covered with an insulating insulation layer. The copper wires are placed at the bottom of the rail, covering the length of the rail and fitting snugly against it. The rail and the copper wires fitted to the bottom of the rail are completely wrapped with the insulation layer. DC power is used as the power source.

[0007] Furthermore, the bottom of the track has an installation channel within the insulation layer for installing copper wires.

[0008] Furthermore, the copper wire enters the bottom of the rail from the middle of the rail and extends from the middle of the bottom of the rail to both ends.

[0009] Furthermore, the insulation layer includes an upper insulation layer and a lower insulation layer that are attached to the outside of the rail, and the upper and lower insulation layers completely wrap the rail. The insulation layer is an aluminum silicate insulation cotton insulation layer.

[0010] Furthermore, the maximum cross-sectional area S of the copper conductor is determined according to formula (9). 最大 :

[0011]

[0012] In the formula: I1 is the current required to heat the copper wire to the temperature resistance of the insulation material, T is the heating time of the copper, ρ represents the resistivity of the resistor, L represents the length of the copper wire, and C... 铜 ρ is the specific heat capacity of the copper wire, ρ1 is the density of the copper wire, t0 is the initial temperature of the copper wire, and t1 is the target temperature of the copper wire.

[0013] The minimum cross-sectional area S of the copper conductor is determined according to formula (13). 最小 :

[0014]

[0015] In the formula: S 最小 ρ is the minimum cross-sectional area of ​​the copper conductor. 铜 L is the electrical conductivity of copper. 铜 C is the length of the copper conductor. 钢 M represents the specific heat capacity of the rail. 钢 The weight per meter of rail, l 钢 t is the length of the rail. 钢0 The initial temperature of the rail is t. 钢1 The target temperature for the rail is set; the cross-sectional area S of the copper conductor is selected. 实际 In S 最大 and S 最小 Within the range.

[0016] Furthermore, the DC switching power supply is selected according to the following method:

[0017] According to the formula for calculating the resistance of copper wire:

[0018]

[0019] In the formula: R 铜 The resistance of the copper wire is ρ, where ρ represents the resistivity of the resistance, and L is the resistance of the copper wire. 铜 S represents the length of the resistor. 实际 This represents the cross-sectional area of ​​the copper conductor.

[0020] According to the formula for calculating circuit current:

[0021]

[0022] In the formula: I 总 The total current flowing through the heating element; P 总 R is the total power exerted on the system by the current and voltage. 钢 R is the resistance of the rail. 铜 The resistance of the copper wire;

[0023] According to the voltage formula:

[0024]

[0025] In the formula: U 总 Total voltage for heating;

[0026] R is calculated according to formulas (6), (12), and (14). 铜 I 总 U 总 Then, the DC switching power supply is selected.

[0027] The rail heating method using the aforementioned device includes the following steps:

[0028] Step 1: Calculate the cross-sectional area range of the copper wire.

[0029] Calculate the maximum cross-sectional area S of the copper conductor according to formula (9). 最大 ;

[0030]

[0031] In the formula: I1 is the current required to heat the copper conductor to the temperature resistance of the insulation material, T is the heating time of the copper, ρ represents the resistivity of the resistor, L represents the length of the resistor, and C... 铜 ρ is the specific heat capacity of the copper wire, ρ1 is the density of the copper wire, t0 is the initial temperature of the copper wire, and t1 is the target temperature of the copper wire.

[0032] The minimum cross-sectional area S of the copper conductor is determined according to formula (13). 最小 :

[0033]

[0034] In the formula: S 最小 ρ is the minimum cross-sectional area of ​​the copper conductor. 铜 L is the electrical conductivity of copper. 铜 C is the length of the copper conductor. 钢 M represents the specific heat capacity of the rail. 钢 The weight per meter of rail, l 钢 t is the length of the rail. 钢0 The initial temperature of the rail is t. 钢1 The target temperature for the rail;

[0035] Step 2: Select the cross-sectional area S of the copper conductor according to the site requirements. 实际 Should be in S 最大 and S 最小 Within the range;

[0036] Step 3: Calculate the cross-sectional area as S 实际 The resistance R of the copper wire 铜

[0037]

[0038] In the formula: R 铜 The resistance of the copper wire is ρ, where ρ represents the resistivity of the resistance, and L is the resistance of the copper wire. 铜 S represents the length of the copper conductor. 实际 This represents the cross-sectional area of ​​the copper conductor.

[0039] Step 4: Calculate the total power of the circuit

[0040] P 总 =P 发电机 ξ1 (10)

[0041] In the formula: P 发电机 P is the rated power of the generator. 总 ξ1 represents the total power exerted on the system by the current and voltage, and ξ1 is the conversion coefficient of the DC switching power supply.

[0042] Step 5: Select a DC switching power supply

[0043] Calculate the total operating current and voltage output of the heating system.

[0044]

[0045]

[0046] I calculated according to formulas (12) and (14) 总 U 总 Selecting a DC switching power supply;

[0047] Step 6: Heat the rail using the copper wires and DC switching power supply selected above.

[0048] The thermal efficiency verification method using the aforementioned method includes the following steps:

[0049] Step S01: Calculate the heat dissipation power of the insulation material

[0050] The copper wire and steel rail are placed inside the insulation layer. The heat dissipation power is calculated based on the theory of thermal radiation. The calculation formula is as follows:

[0051]

[0052] In the formula: P 单位辐 λ is the radiative heat dissipation power per square meter per second of the insulation material. 保温棉 δ is the thermal conductivity of the insulation layer, δ is the thickness of the insulation layer, t1 is the high-temperature side of the insulation layer, and t0 is the low-temperature side of the insulation layer.

[0053] Calculate the temperature difference between the high-temperature and low-temperature sides.

[0054] t 差 =t1-t0 (16)

[0055] Step S02: Calculate the temperature rise of the copper wire during heating time T1.

[0056]

[0057] In the formula: Q 铜升T1 The temperature rise of copper is t 铜 Energy C at time 铜 M is the specific heat capacity of the copper wire. 铜 L is the weight of each meter of copper wire. 铜 t is the length of the copper conductor. 铜0 t represents the initial temperature of the copper conductor. 铜 This refers to the temperature of the T1 copper conductor.

[0058]

[0059] In the formula: r 铜 S is the radius of the copper conductor. 实际 The cross-sectional area of ​​a copper wire; π is the mathematical constant pi (π).

[0060] Calculate the surface area of ​​copper according to formula (18).

[0061] S 铜表 =2πr 铜 2 +2πr 铜 L 铜 (18)

[0062] In the formula: r 铜 S is the radius of the copper conductor. 铜表 L represents the surface area of ​​the copper conductor; π represents pi (π). 铜 This is the length of the copper wire;

[0063] Calculate the electrical energy consumed during heating time T1 minus the energy Q generated by radiative heat dissipation. 升T1 ;

[0064]

[0065] In the formula: I 总 The total output current of the system loop is T1, the heating time is t, and the temperature difference is t.

[0066] Calculate the temperature rise Δt1 of the copper wire during heating time T1 according to formula (20);

[0067]

[0068] Step S03: Calculate the temperature rise of the rail during heating time T1.

[0069] Q 钢升T1 =C 钢 M 钢 l 钢 (t 钢T1 -t0) (27)

[0070] In the formula: Q 钢升T1 The temperature rise of the rail is t 钢T1 Energy at time, C 钢 M is the specific heat capacity of the rail, M is the weight per meter of rail, and L is the weight per meter of rail. 钢 t is the length of the rail, t0 is the initial temperature of the rail; t 钢T1 This refers to the temperature of the T1 rail.

[0071] Calculate the electrical energy consumed during heating time T1 minus the energy Q generated by radiative heat dissipation. 钢T1 ;

[0072]

[0073] In the formula: I 总 The total output current of the system loop is T1, the heating time is S. 每米表 L is the surface area per meter of the rail. 钢 The length of the heated rail;

[0074] The temperature rise Δt2 of the copper wire during heating time T1 is calculated according to formula (19);

[0075]

[0076] Step S04: Calculate the temperature difference between the copper wire and the steel rail during heating time T1.

[0077] The temperature difference Δt3 between the copper conductor and the rail at time T1 is calculated using formula (21).

[0078] Δt3=Δt1-Δt2 (21);

[0079] Step S05: Calculate the energy absorbed by the rail through heat conduction from the copper conductor at time T1.

[0080] The heat transfer area A of the copper wire is calculated according to formula (22).

[0081] A = 2r 铜 L 钢 (twenty two)

[0082] In the formula: r 铜 L is the radius of the copper conductor. 钢 For rail length

[0083] The energy Q absorbed by the rail from the copper conductor through heat conduction is calculated according to formula (23). 钢热吸

[0084]

[0085] In the formula: K is the thermal conductivity coefficient of the rail, A represents the heat transfer area, ΔL is the distance from the bottom to the top of the rail, and t represents the temperature difference;

[0086] Step S06: Calculate the energy absorbed by the rail from the copper conductor through heat conduction from the start of heating to time T1. Calculate the total energy Q required for the rail to heat up using formula (24). 钢轨

[0087] Q 钢轨 =Q 钢T1 +Q 钢热吸 (twenty four)

[0088] The temperature rise Δt of the rail after heating for time T1 is calculated according to formula (25).

[0089]

[0090] Step S07: Calculate the heating efficiency η at heating time T1.

[0091]

[0092] In the formula: Δt3 is the temperature rise of the original method when heating T1.

[0093] The beneficial effects of this invention are: This invention uses a fully enclosed type of insulation material to reduce heat loss. By wrapping the copper wire inside the insulation layer and attaching it to the rail, the heat of the copper wire is utilized. The heat is converted to DC by a DC switching power supply. By combining the reasonable selection of the copper wire and the DC switching power supply, the heating efficiency of the rail is improved. Attached Figure Description

[0094] Figure 1 This is a schematic diagram of the rail heating device of the present invention;

[0095] Figure 2 This is a schematic diagram of the process for improving the thermal efficiency of rail body heating according to the present invention.

[0096] Components in the diagram: 1 is the steel rail, 2 is the upper insulation layer, 3 is the lower insulation layer, 4 is the DC switching power supply, 5 is the insulating insulation layer tube, 6 is the clamp, 7 is the copper wire, 8 is the generator, and 9 is the wire. Detailed Implementation

[0097] To better explain and facilitate understanding of the present invention, the present invention will be described in detail below with reference to the accompanying drawings and specific embodiments.

[0098] like Figure 1-2 As shown, this invention relates to a rail body heating device, method, and thermal efficiency verification method, including a generator 8, a DC switching power supply 4, copper wires 7, and an insulation layer. The generator 8 is connected to the DC switching power supply 4, which is connected to the copper wires 7. The copper wires 7 are externally covered with an insulating layer. The copper wires 7 enter the bottom of the rail from the middle and extend from the middle of the bottom of the rail towards both ends. The copper wires 7 are placed at the bottom of the rail, covering the length of the rail and fitting snugly against it. The rail and the copper wires 7 fitted against the bottom of the rail are completely covered by the insulation layer. An installation channel for installing the copper wires 7 is provided within the insulation layer at the bottom of the rail. Direct current (DC) is selected as the power source.

[0099] The insulation layer includes an upper insulation layer 2 and a lower insulation layer 3 that are attached to the outside of the rail. The upper insulation layer 2 and the lower insulation layer 3 completely wrap the rail. The insulation layer is an aluminum silicate insulation cotton insulation layer.

[0100] This invention designs different insulation layers based on the dimensions of existing 60-type and 75-type steel rails. To minimize heat loss and wind protection, the insulation layer is designed with the same shape as the rail for four-sided insulation, and a copper wire installation channel is provided at the bottom of the rail. Aluminum silicate insulation cotton is selected as the insulation material. This insulation layer is waterproof, moisture-proof, heat-insulating, chemically resistant, durable, easy to cut, highly reusable, and environmentally friendly. A 1-meter length insulation layer is used for convenient production, transportation, and installation. The insulation layer is divided into upper and lower parts. The upper part is completely sealed to the top and web of the rail, while the lower part is completely sealed to the bottom of the rail. The upper part of the insulation layer is installed at the connection point, and the lower part is then pressed down to achieve a complete seal. Due to the high plasticity of the aluminum silicate insulation cotton material, the entire insulation material can be quickly applied to the rail.

[0101] The existing copper conductor strip is insulated by wrapping an insulating layer over the braided strip. The heating device is then placed in the middle of the heated rail, and the copper conductor is fixed to both ends of the rail using clamps, with the copper conductor placed close to the bottom of the rail. Fixing the copper conductor to the bottom of the rail not only facilitates grinding and reduces contact resistance but also allows the copper conductor to be more closely attached to the rail, improving heating efficiency.

[0102] The rail body heating method of this invention treats the rail as a load and uses a copper conductor strip as a conductor connected in series with a power source. First, the most suitable copper conductor is selected based on calculations. Due to its DC impedance, the copper conductor consumes power from the power source, and this energy is lost as heat. The surface area and volume of the copper conductor are much smaller than the rail, while the impedance values ​​are not significantly different. Therefore, during actual heating, the temperature of the copper conductor is much higher than that of the rail. Based on this characteristic, the insulation material is designed with the same shape as the rail for four-sided insulation, and an installation channel for the copper conductor is left at the bottom of the rail. An insulating insulation layer is then wrapped around the copper conductor strip for insulation. The copper conductor is then fixed to the bottom of the rail, and finally, the copper conductor and the rail are completely covered with insulation material for insulation. Within the insulation layer, due to heat conduction, the copper conductor transfers energy to the rail, accelerating its heating rate. Theoretical formulas and actual measurements show an improvement in heating efficiency.

[0103] The rail body heating method of the present invention includes the following steps:

[0104] Step 1: Calculate the cross-sectional area range of the copper wire.

[0105] Based on the formulas for calculating current and power, the total current I required for heating the rail is... 总 Power P required for heating rails 钢 for:

[0106]

[0107] I 总 R is the total current in the transformer secondary circuit, U is the total voltage, and R is the total voltage. 钢 R is the resistance of the rail. 铜 The resistance of the copper wire.

[0108]

[0109] P 钢 R is the power exerted by the power source on the rail, U is the total voltage, and R is the total voltage. 钢 R is the resistance of the rail. 铜 Given the resistance of the copper wire, P can be determined from formulas (1) and (2). 钢 and R 铜 It is negatively correlated. Based on the temperature resistance of the insulation material, the maximum temperature of the copper conductor is set. Without considering heat loss, the cross-sectional area of ​​the copper conductor cannot be too small. The energy Q required for the copper conductor to reach its maximum temperature is... 升1 for:

[0110] Q 升1 =C 铜 ML(t1-t0) (3)

[0111] In the formula: C 铜t0 is the specific heat capacity of the copper wire, M is the weight of the copper wire per meter, L is the length of the copper wire, t0 is the initial temperature of the copper wire, and t1 is the target temperature.

[0112] The energy Q required for electrical energy to heat a copper conductor to its maximum temperature 升2 for:

[0113] Q 升2 =I1 2 R 铜 T (4)

[0114] In the formula: R 铜 Let Q be the resistance of the copper wire, T be the heating time of the copper, and I1 be the current required to raise the temperature of the copper wire to the temperature resistance of the insulation material. In this case, without considering heat dissipation, Q... 升1 =Q 升2 ;

[0115] By Q 升1 =Q 升2 Formula (5) is obtained through formulas (3) and (4).

[0116]

[0117] Then, using the resistance formula (6), mass formula (7), and volume formula (8), combined with the formula...

[0118]

[0119] In the formula: R 铜 Let ρ represent the resistance of the copper wire, L represent the resistivity of the resistance, and S represent the length of the copper wire. 最大 This represents the cross-sectional area of ​​the resistor.

[0120] m 铜 =ρ1V 钢 (7)

[0121] Where: m 铜 Let ρ be the mass of the copper conductor, ρ1 be the density of the copper conductor, and V be the mass of the copper conductor. 铜 The volume of the copper wire

[0122] V 铜 =SL (8)

[0123] In the formula: V 铜 Let S be the volume of the copper conductor, S be the cross-sectional area of ​​the copper conductor, and L be the length of the copper conductor. These values ​​are derived from the above formula.

[0124]

[0125] The lower limit of the resistance, i.e., the maximum cross-sectional area S of the copper conductor, can be calculated using the above formula. 最大 .

[0126] After calculating the maximum cross-sectional area of ​​the copper conductor, the minimum cross-sectional area S of the copper conductor is calculated based on the minimum current required to raise the temperature of the rail. 最小 That is, the maximum resistance.

[0127] Calculate the energy Q required for the rail to reach the target temperature. 钢升1

[0128]

[0129] In the formula: C 钢 M represents the specific heat capacity of the rail. 钢 The weight per meter of rail, l 钢 t is the length of the rail. 钢0 The initial temperature of the rail is t. 钢1 The target temperature;

[0130] The heat Q generated by the electric current in the rail is calculated. 钢升2 ;

[0131] Q 钢升2 =I2 2 R 钢 T2 (11)

[0132] In the formula: R 钢 The resistance of the heated rail is given by I2, the current flowing through the rail is given by T2, and the heating time of the rail is given by T2. The current I2 is then calculated.

[0133]

[0134] In the formula: P 总 I2 is the total power exerted on the system by the current and voltage, I2 is the current flowing through the rail, and R0 is the total power exerted by the current and voltage on the system. 铜 The resistance of the copper wire;

[0135] Q 钢升1 =Q 钢升2 Substituting resistance formula 6 and current formula 12 into formula 13, we obtain the minimum cross-sectional area S of the copper conductor.

[0136]

[0137] In the formula: S 最小 ρ is the minimum cross-sectional area of ​​the copper conductor. 铜 L is the electrical conductivity of copper. 铜 The length of the copper conductor;

[0138] Step Two: After calculating the range of copper conductor cross-sectional areas in Step One, select the appropriate cross-sectional area S of the copper conductor based on site requirements. 实际 Should be in S 最大 and S 最小 Within the range;

[0139] Step 3: Calculate the cross-sectional area as S 实际 copper wire resistance

[0140]

[0141] Step 4: Calculate the total power of the circuit

[0142] P 总 =P 发电机 ξ1 (10)

[0143] In the formula: P 发电机 P is the rated power of the generator. 总 ξ1 represents the total power exerted on the system by the current and voltage, and ξ1 is the conversion coefficient of the DC switching power supply.

[0144] Step 5: Select a DC switching power supply

[0145] Calculate the total operating current and voltage output of the heating system.

[0146]

[0147]

[0148] I calculated according to formulas (12) and (14) 总 U 总 Select a DC switching power supply.

[0149] Step 6: Heat the guide rail using the copper wires and DC switching power supply selected above.

[0150] The following are the steps to verify the improvement in thermal efficiency of the rail heating method of the present invention:

[0151] Step S01: Calculate the heat dissipation power of the insulation material

[0152] The copper wire and steel rail are placed inside the insulation layer. The heat dissipation power is calculated based on the theory of thermal radiation. The calculation formula is as follows:

[0153]

[0154] In the formula: P 单位辐 λ is the radiative heat dissipation power per square meter per second of the insulation material. 保温棉 δ is the thermal conductivity of the insulation layer, δ is the thickness of the insulation layer, t1 is the high-temperature side of the insulation layer, and t0 is the low-temperature side of the insulation layer.

[0155] Calculate the temperature difference between the high-temperature and low-temperature sides.

[0156] t 差 =t1-t0 (16)

[0157] Step S02: Calculate the temperature rise of the copper wire during heating time T1.

[0158]

[0159] In the formula: Q 铜升T1 The temperature rise of copper is t 铜 Energy C at time 铜 M is the specific heat capacity of the copper wire. 铜 L is the weight of each meter of copper wire. 铜 t is the length of the copper conductor. 铜0 t represents the initial temperature of the copper conductor. 铜 This refers to the temperature of the T1 copper conductor.

[0160] Given the cross-sectional area of ​​the copper conductor, R can be obtained using the radius formula (17). 铜

[0161]

[0162] In the formula: r 铜 S is the radius of the copper conductor. 实际 The cross-sectional area of ​​a copper wire; π is the mathematical constant pi (π).

[0163] Given the radius of the copper wire, calculate the surface area of ​​the copper wire using formula (18).

[0164] S 铜表 =2πr 铜 2 +2πr 铜 L 铜 (18)

[0165] In the formula: r 铜 S is the radius of the copper conductor. 铜表 Surface area of ​​copper wire; π is pi, L 铜 This is the length of the copper wire;

[0166] Calculate the electrical energy consumed during heating time T1 minus the energy Q generated by radiative heat dissipation. 升T1 ;

[0167]

[0168] In the formula: I 总 The total output current of the system loop is T1, the heating time is t, and the temperature difference is t.

[0169] Taking heat dissipation into account, and ignoring other factors, Q 升T1 =Q 铜升T1 According to formula (20), the temperature rise Δt1 of the copper wire during heating time T1 can be calculated.

[0170]

[0171] Step S03: Calculate the temperature rise of the rail during heating time T1.

[0172] Q 钢升T1 =C 钢 M 钢 l 钢 (t 钢T1 -t0) (27)

[0173] In the formula: Q 钢升T1 The temperature rise of the rail is t 钢T1 Energy at time, C 钢 M is the specific heat capacity of the rail, M is the weight per meter of rail, and L is the weight per meter of rail. 钢 t is the length of the rail, t0 is the initial temperature of the rail; t 钢T1 This refers to the temperature of the T1 rail.

[0174] Calculate the electrical energy consumed during heating time T1 minus the energy Q generated by radiative heat dissipation. 钢T1 ;

[0175]

[0176] In the formula: I 总 The total output current of the system loop is T1, the heating time is S. 每米表 L is the surface area per meter of the rail. 钢 The length of the heated rail;

[0177] Taking heat dissipation into account, and ignoring other factors, Q 升T1 =Q 钢T1 According to formula (19), the temperature rise Δt2 of the copper wire during heating time T1 can be calculated.

[0178]

[0179] Step S04: Calculate the temperature difference between the copper wire and the steel rail during heating time T1.

[0180] According to formula (21), the temperature difference Δt3 between the copper conductor and the rail at time T1 is calculated.

[0181] Δt3=Δt1-Δt2 (21)

[0182] Step S05: Calculate the energy absorbed by the rail through heat conduction from the copper conductor at time T1.

[0183] The heat transfer area A of the copper wire is calculated according to formula (22).

[0184] A = 2r 铜 L 钢 (twenty two)

[0185] In the formula: r 铜 L is the radius of the copper conductor. 钢For rail length

[0186] The energy Q absorbed by the rail from the copper conductor through heat conduction is calculated according to formula (23). 钢热吸

[0187]

[0188] In the formula: K is the thermal conductivity coefficient of the rail, A represents the heat transfer area, ΔL is the distance from the bottom to the top of the rail, and t represents the temperature difference.

[0189] Step S06: Calculate the energy absorbed by the rail from the copper conductor through heat conduction from the start of heating to time T1. Calculate the total energy Q required for the rail to heat up using formula (24). 钢轨

[0190] Q 钢轨 =Q 钢T1 +Q 钢热吸 (twenty four)

[0191] Based on formulas (10), (18), (20)-(24), formula (25) is derived, and the temperature rise Δt of the rail after heating time T1 is calculated.

[0192]

[0193] Step S07: Calculate the heating efficiency η at heating time T1.

[0194]

[0195] In the formula: Δt3 is the temperature rise of the original method when heating T1.

[0196] Example 1

[0197] A 10kW generator was used to heat a 12.5m long 60kg / m steel rail for 1 hour. The copper wire was 14m long, the maximum temperature of the insulation material was 1200℃, the initial temperature of the system before heating was 0℃, and the system heating time was 1 hour. The temperature rise of the steel rail was recorded.

[0198] The range of values ​​for copper conductors is calculated based on theoretical formulas;

[0199] 1. Calculate the maximum cross-sectional area of ​​the copper conductor.

[0200]

[0201]

[0202] Substitute formulas (1) and (2) into formula (3).

[0203]

[0204] In the formula: the DC resistance of a 60kg / m rail at 20℃ is approximately (34.36±0.58)μΩ / m, and the resistance of a 12.5m rail is approximately R. 钢 =5×10 -4 Ω, ρ is the resistivity of copper, 1.75 × 10⁻⁶. -8 S / m, L is the length of the copper wire (14m), T is the heating time (3600s), and ρ1 is the density of the copper wire (8960kg / m). 3 The system output voltage is 8V, the maximum temperature of the copper wire at t1 is 1200℃, and the maximum temperature of the heating environment at t0 is 0℃.

[0205] 2. Calculate the minimum cross-sectional area

[0206] P 总 =P 发电机 ξ1=8Kw (4)

[0207] In the formula: P 发电机 The rated power of the generator is 10kW, and ξ1 is the conversion factor of the DC switching power supply, which is 0.8.

[0208]

[0209] In the formula: Lsteel represents the length of the rail (12.5m), Lcopper represents the length of the copper conductor (14m), T2 represents the heating time (3600s), ρcopper represents the resistivity of the copper conductor (1.75×10-8S / m), Msteel represents the weight of the rail per meter (60kg), and t... 钢1 The maximum temperature of the rail is 40℃, and the initial temperature of the t0 rail is 0℃. 钢 The specific heat capacity of steel is C = 0.46 kJ / kg.

[0210] 3. Select the cross-sectional area of ​​the copper wire.

[0211] Based on calculations and actual site requirements, a copper conductor cross-sectional area of ​​500 mm² was selected. 2 .

[0212] 4. Calculate the resistance value of the copper wire.

[0213]

[0214] 5. Calculate the system output voltage and current.

[0215]

[0216]

[0217] 6. Select a DC switching power supply

[0218] Select a 3000A, 3V DC switching power supply.

[0219] Two 7m long copper wires were threaded through the insulating tube, and the DC switching power supply was placed in the middle of the steel rail. Then, according to... Figure 1 The connection method involves connecting the generator, DC switching power supply, copper wire, and steel rail in series. After the circuit connection is complete, the copper wire is fixed to the bottom of the steel rail, starting from both ends. After the copper wire is fixed, it is covered with insulation material to prevent heat loss.

[0220] 7. Calculate the radiative heat dissipation power per square meter and per second of the insulation material.

[0221]

[0222] In the formula: λ 保温棉 The thermal conductivity of the insulation layer is 0.03 W / mk, the thickness of the insulation layer is 0.05 m, t1 is the high-temperature side of the insulation layer, and t0 is the low-temperature side of the insulation layer, with an ambient temperature of 0℃.

[0223] 8. Calculate the weight of each meter of copper wire.

[0224] V 铜 =SL=5×10 -4 m 3 (10)

[0225] m 铜 =ρ1V 铜 ≈4.42kg (11)

[0226] In the formula: ρ1 is the density of the copper conductor, 8960 kg / m 3 L represents the length of the copper conductor (1m), and s represents the cross-sectional area of ​​the copper conductor (500mm²). 2 .

[0227] 9. Calculate the radius and surface area of ​​the copper conductor.

[0228]

[0229] S 铜 =2πr 铜 2 +2πr 铜 L 铜 ≈0.79m 2 (13)

[0230] In the formula: S 实际 The cross-sectional area of ​​the copper conductor is 500 mm². 2 L 铜 The copper wire is 14m long, and Π is 3.14.

[0231] 10. Calculate the temperature rise of the copper wire in 1 second.

[0232] Q 铜升1 =C 铜 M铜 L 铜 (t 铜 -t0)≈24180t 铜 (14)

[0233] Substitute formula 9 into formula (15)

[0234]

[0235] Q 铜升1 =Q 铜升2 Formula (16) can be derived.

[0236]

[0237] In the formula: C 铜 The specific heat capacity of copper is 0.39 × 10⁻⁶. -3 J / (kg·℃), L 铜 The copper wire is 14m long, t0 is the initial system temperature of 0℃, and M 铜 The weight of 1m of copper wire is 4.42Kg, T1 is the heating time of 1s, and t 铜 The temperature of the copper conductor is also the high-temperature side temperature of the system, I 总 The system current is 2843A, R 铜 The resistance of the copper wire is 4.9 × 10⁻⁶. -4 Ω.

[0238] 11. Calculate the temperature rise of the rail in 1 second.

[0239] Q 钢升1 =C 钢 M 钢 l 钢 (t 钢 -t0)=345000t 钢 (17)

[0240] Substitute formula 9 into formula (18)

[0241]

[0242] Q 钢升1 =Q 钢升2 Formula (19) can be derived.

[0243]

[0244] In the formula: C 钢 The specific heat capacity of copper is 0.39 × 10⁻⁶. -3 J / (kg·℃), L 钢 The rail is 12.5m long, t0 is the initial system temperature of 0℃, and M... 钢 The weight of 1m of rail is 60kg, T1 is the heating time of 1s, and t钢 The temperature of the copper conductor is also the high-temperature side temperature of the system, I 总 With a system current of 2843A, the DC resistance of a 60kg / m rail at 20℃ is approximately (34.36±0.58)μΩ / m, and the resistance of a 12.5m rail is approximately R. 钢 =5×10 -4 Ω, by looking up the parameters, we can find the surface area S per meter of rail. 每米表 Approximately 0.3984m 2 .

[0245] 12. Calculate the temperature difference between the steel rail and the copper wire during the 1-second heating cycle.

[0246] Δt3=Δt1-Δt2≈0.15℃ (20)

[0247] 13. Calculate the energy absorbed by the steel rail through heat conduction from the copper conductor after heating for 1 hour.

[0248] The heat transfer area A of the copper wire is calculated according to formula (21).

[0249] A = 2r 铜 L 铜 ≈0.32m 2 (twenty one)

[0250] In the formula: r 铜 The radius of the copper conductor is 12.61 mm, L 钢 The rail is 12.5m long.

[0251] Calculate the temperature difference between the copper wire and the steel rail after heating for 1 hour.

[0252] Δt4=Δt3×T≈540℃ (22)

[0253] The energy Q absorbed by the rail from the copper conductor through heat conduction is calculated according to formula (21). 钢热吸

[0254]

[0255] In the formula: K is the thermal conductivity coefficient of the rail (45W / ℃), ΔL is the distance from the bottom to the top of the rail (0.176m), and T is the heating time (3600s).

[0256] 14. Calculate the temperature rise of the rail after 1 hour of heating.

[0257] Δt5=Δt2×T≈43.2℃ (23)

[0258]

[0259] Q 钢轨 =Q 钢T1 +Q 钢热吸≈22513414J (25)

[0260]

[0261] In the formula: C 钢 The specific heat capacity of steel is 0.46 × 10⁻⁶. -3 J / (kg·℃), L 钢 The rail is 12.5m long, M 钢 The weight of 1m of rail is 60kg, T is the heating time of 3600s, and I is... 总 With a system current of 2843A, the DC resistance of a 60kg / m rail at 20℃ is approximately (34.36±0.58)μΩ / m, and the resistance of a 12.5m rail is approximately R. 钢 =5×10 -4 Ω.

[0262] 15. Calculate the heating efficiency η

[0263]

[0264] In the formula: Δt3 is the temperature rise of 40℃ in the original method when heating T1.

[0265] Taking heat dissipation into account, this method is 1.63 times more efficient than the original method without considering heat dissipation.

[0266] Although embodiments of the present invention have been shown and described above, it is understood that the above embodiments are exemplary and should not be construed as limiting the present invention. Those skilled in the art can make modifications, alterations, substitutions and variations to the above embodiments within the scope of the present invention.

Claims

1. A rail body heating apparatus characterised in that: The utility model relates to a power generation device, including generator (8), DC switching power supply (4), copper wire (7), heat preservation layer, the generator (8) links to DC switching power supply (4), DC switching power supply (4) links to copper wire (7), copper wire (7) outside is equipped with insulating heat preservation layer, copper wire (7) is placed in the steel rail bottom cover steel rail length and is pasted together with steel rail, the steel rail and copper wire (7) pasted together with steel rail bottom outside are all wrapped with heat preservation layer, select direct current as power supply, copper wire (7) determines copper wire maximum cross -sectional area S according to formula (9) 最大 : （9）； wherein: Ii is the current required to heat the copper conductor to the temperature resistance of the insulation material, T is the heating time of the copper, p represents the resistivity of the resistance, L represents the length of the copper conductor, C 铜 is the specific heat capacity of the copper conductor, is the density of the copper conductor, t0 is the initial temperature of the copper conductor; ti is the target temperature of the copper conductor; The minimum cross-sectional area S of the copper conductor is determined according to equation (13) 最小 : （13）； where: S 最小 is the minimum cross-sectional area of the copper wire, p 铜 is the electrical conductivity of copper, L 铜 is the length of the copper wire, C 钢 is the specific heat capacity of the rail, M 钢 is the weight of the rail per meter, is the length of the rail, t 钢0 is the initial temperature of the rail, t 钢1 is the target temperature of the rail; the cross-sectional area of the copper wire S 实际 is selected to be within the interval S 最大 and S 最小 ; P 总 is the total power supplied to the system by the current voltage, R 钢 is the resistance of the rail, and T2is the length of time the rail is heated.

2. A rail body heating apparatus according to claim 1, characterised in that: The bottom of the track has installation channels in the heat preservation layer for installing copper wires (7).

3. A rail body heating apparatus as claimed in claim 2, wherein: The copper wires (7) enter the bottom of the track from the middle of the track and extend from the middle of the bottom of the track to both ends.

4. A rail body heating apparatus as defined in claim 1, wherein: The heat preservation layer includes an upper heat preservation layer (2) and a lower heat preservation layer (3) that are attached to the outside of the track and completely wrap the track, and the heat preservation layer is an aluminum silicate heat preservation cotton heat preservation layer.

5. A rail body heating apparatus as defined in claim 1, wherein: The DC switching power supply (4) is selected as follows: According to the copper wire resistance calculation formula: （6）； wherein: R 铜 is the resistance of the copper conductor, p represents the resistivity of the resistance, L 铜 represents the length of the resistance, S 实际 represents the cross-sectional area of the copper conductor, According to the circuit current calculation formula: （12）； where: I 总 is the total current flowing through the heating; P 总 is the total power acting on the system by the current voltage, R 钢 is the resistance of the steel rail, R 铜 is the resistance of the copper conductor; According to the voltage formula: （14）； In the formula: U 总 Total voltage of heating; R calculated according to formula (6), (12), (14) 铜 , I 总 , U 总 Further, the DC switching power supply is selected.

6. A method of heating a rail using the apparatus of any one of claims 1 to 5, characterised in that: The steps include: Step one: Calculate the cross-sectional area range of the copper wire; The maximum cross-sectional area S of the copper conductor is calculated according to formula (9) 最大 ; （9）； wherein: Ii is the current required to heat the copper conductor to the temperature resistance of the insulation material, T is the heating time of the copper, p represents the resistivity of the resistance, L represents the length of the resistance, C 铜 is the specific heat capacity of the copper conductor, is the density of the copper conductor, t0 is the initial temperature of the copper conductor; ti is the target temperature of the copper conductor; The minimum cross-sectional area S of the copper conductor is determined according to equation (13) 最小 : （13）； where S 最小 is the minimum cross-sectional area of the copper wire, p 铜 is the electrical conductivity of copper, L 铜 is the length of the copper wire, C 钢 is the specific heat capacity of the steel rail, M 钢 is the weight of the steel rail per meter, is the length of the steel rail, t 钢0 is the initial temperature of the steel rail, t 钢1 is the target temperature of the steel rail; Step two: Select the cross-sectional area S of the copper wire according to the field requirements 实际 S 最大 and S 最小 interval; Step three: Calculate the resistance R of a copper wire with a cross-sectional area of S 实际 铜； ​ （6） ； wherein: R 铜 is the resistance of the copper wire, p represents the resistivity of the resistance, L 铜 represents the length of the copper wire, S 实际 represents the cross-sectional area of the copper wire, Step four: Calculate total power of circuit (10); where: P 发电机 P is the rated power of the generator 总 is the total power acting on the system, and ξ1 is the conversion factor of the DC switching power supply Step five: Select a DC switching power supply; Calculate the total working current and voltage output in the heating system （12）； （14）； I calculated according to formula (12), (14) 总 , U 总 Selecting a DC switching power supply; Step six: Heat the track according to the selected copper wire and DC switching power supply.

7. A heat-efficiency-verification method employing the method of claim 6, characterized in that, The steps include: Step S01: Calculate the heat dissipation power of the heat preservation material; Put the copper wire and the track into the heat preservation layer, calculate the heat dissipation power according to the heat radiation theory, and the calculation formula is as follows: （15）； wherein: P 单位辐 is the radiant heat dissipation power per square meter per second of the thermal insulation material, λ 保温棉 is the thermal conductivity of the thermal insulation layer, δ is the thickness of the thermal insulation layer, t1 is the high temperature side of the thermal insulation layer, and t0 is the low temperature side of the thermal insulation layer. Calculate the temperature difference between high and low temperature sides （16）； Step S02: Calculate the temperature rise of the copper wire during heating T1 time (28)； In the formula: Q 铜升T1 The temperature rise of copper is t 铜 Energy C at time 铜 M is the specific heat capacity of the copper wire. 铜 L is the weight of each meter of copper wire. 铜 t is the length of the copper conductor. 铜0 t represents the initial temperature of the copper conductor. 铜 This refers to the temperature of the T1 copper conductor. （17）； where: r 铜 is the radius of the copper conductor, S 实际 is the cross-sectional area of the copper conductor; and Π is the constant Pi. Calculate the surface area of copper according to formula (18) (18)； where: r 铜 is the radius of the copper wire, S 铜表 is the surface area of the copper wire; Π is the constant Pi, L 铜 is the length of the copper wire; Calculate the heating T1 time electrical energy minus the energy Q generated by the radiation heat dissipation 升T1 ; (19)； In the formula, I 总 T1 is the heating time, and t is the temperature difference. Calculate the temperature rise of the copper wire during heating T1 time according to formula (20); (20)； Step S03: Calculate the temperature rise of the track during heating T1 time (27) In the formula: Q 钢升T1 The temperature rise of the rail is t 钢T1 Energy at time, C 钢 M is the specific heat capacity of the rail, M is the weight per meter of rail, and L is the weight per meter of rail. 钢 t is the length of the rail, t0 is the initial temperature of the rail; t 钢T1 This refers to the temperature of the T1 rail. Calculate the heating T1 time electrical energy minus the energy Q generated by the radiation heat dissipation 钢T1 ; (19)； wherein: I 总 is the total current output of the system circuit, T1 is the heating time length, S 每米表 is the surface area of the steel rail per meter, L 钢 is the steel rail heating length; Calculate the temperature rise of the copper wire during heating T1 time according to formula (19); (20)； Step S04: Calculate the temperature difference between the copper wire and the track during heating T1 time; Calculate the temperature difference between the copper wire and the track during T1 time according to formula (21) (21)； Step S05: Calculate the energy absorbed by the track from the copper wire during T1 time; Calculate the heat transfer area A of the copper wire according to formula (22) (22)； where: r 铜 is the radius of the copper wire, L 钢 is the length of the rail; The energy Q absorbed by the rail from the thermal conduction of the copper wire is calculated according to equation (23) 钢热吸 (23)； In the formula: K is the heat transfer coefficient of the track, A represents the heat transfer area, and ΔL is the distance from the bottom to the top of the track, and t represents the temperature difference; Step S06: Calculate the energy absorbed by the track from the copper wire during T1 time; The total energy Q for the steel rail temperature rise is calculated according to equation (24) 钢轨 (24)； Calculate the temperature rise of the track after heating T1 time according to formula (25) ；； Step S07: Calculate the heating efficiency η during T1 time (26)； In the formula: Δt3 is the temperature rise of the original method during T1 time.

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