Calculation method of temperature field of naturally cooled symmetrical UI inductor based on thermal resistance network

Abstract

The present invention relates to a method for quickly calculating the temperature field of a symmetrical UI inductor based on a thermal resistance network under natural heat dissipation conditions. The method comprises the following steps: firstly, determining the size of each component of the symmetrical UI inductor, dividing the symmetrical UI inductor into thermal node modules, drawing a thermal resistance network diagram for each thermal node module, determining each thermal resistance calculation formula in the thermal resistance network diagram, substituting each parameter into the corresponding thermal resistance calculation formula and calculating each thermal resistance value in the thermal resistance network, and then substituting the thermal resistance value into the thermal resistance network for iterative solution to obtain the final temperature result. The method of the present invention takes into account the influence of the dimensional parameters of the inductor such as the magnetic core and coil on the heat transfer, and the established model is consistent with the actual symmetrical UI inductor model. The thermal resistance network method is combined to quickly and accurately solve the inductor temperature field, and the calculated inductor temperature has an error within 10% of the finite element result. The present invention solves the problem of fast calculation of the inductor temperature field distribution.

Description

Technical Field

[0001] The invention belongs to the technical field of power electronic inductor design, and relates to a temperature field calculation method of a naturally cooled symmetrical UI inductor based on a thermal resistance network. Background Art

[0002] During the use of inductors, their closed structure is prone to excessive hot spot temperatures in the windings, posing a threat to the service life of the inductors. Therefore, the prediction and calculation of the hot spot temperature of the inductors is particularly important. The empirical formula calculation method is simple and convenient, but the accuracy is relatively low. With the development of numerical analysis methods, the use of finite element method to simulate the temperature field of the inductor can significantly improve the calculation accuracy. However, the calculation results of the finite element method are overly dependent on the quality of the mesh. Considering the calculation cost and time cost required by the finite element method, it is generally not suitable for actual engineering situations. The equivalent thermal circuit model based on thermoelectric analogy takes into account the actual structure of the inductor to calculate the thermal resistance, and the amount of hot spot temperature calculation is greatly reduced. Therefore, it is suitable for accurately calculating the hot spot temperature of the transformer winding in engineering. Summary of the invention

[0003] In view of this, the purpose of the present invention is to provide a fast calculation method for the temperature field of a symmetrical UI inductor based on a thermal resistance network under natural heat dissipation conditions, so as to significantly reduce the large amount of trial work of designers, improve work efficiency, and reduce manual calculation errors.

[0004] In order to achieve the above object, the present invention provides the following technical solutions:

[0005] A method for calculating the temperature field of a naturally cooled symmetrical UI inductor based on a thermal resistance network comprises the following steps:

[0006] S1: Material preparation: prepare the physical materials required for calculation, including geometric dimensions and material physical property parameters;

[0007] S2: Divide thermal resistance nodes: Divide the thermal resistance nodes of the symmetrical UI inductor according to the inductor geometry and the coil winding connection form;

[0008] S3: Draw a thermal resistance network diagram: Draw a symmetrical UI inductor thermal resistance network diagram based on the structural characteristics and the distribution of thermal resistance nodes;

[0009] S4: Calculate thermal resistance value: Calculate the thermal resistance value of each thermal resistor in the thermal resistance network according to geometric parameters and material properties;

[0010] S5: Solve the thermal resistance network: Substitute the thermal resistance value into the drawn thermal resistance network. After iterative calculation, the final temperature value of each node is obtained, thereby obtaining the symmetrical UI inductor temperature field distribution.

[0011] Furthermore, in step S1, the geometric dimensions of each component inside the symmetrical UI inductor are determined, and the connection method and physical parameters of each component are determined; the physical parameters include the core material, coil material and winding method, ambient temperature, air thermal parameters, inductor operating parameters, and the heating value of the core and coil.

[0012] Furthermore, in step S2, the magnetic core, coil and air gap of the symmetrical UI inductor are equivalent to different nodes according to the structural characteristic environment.

[0013] Further, step S3 includes: setting a temperature point between two contacting nodes of the same material; setting a contact thermal resistance between two contacting nodes of different materials; setting a convection thermal resistance on the surface in contact with the air and then grounding it; connecting the position where there is a slit with a slit thermal resistance; and indicating that there is no heat flow on the symmetric surface with a 0 connection;

[0014] Nodes without temperature difference are directly connected by lines, nodes with temperature difference are connected by contact thermal resistance, locations with slits are connected by slit thermal resistance, blocks in contact with air are connected by convection thermal resistance, and each node, line, and thermal resistance are interconnected to form a thermal resistance network model.

[0015] Further, in step S4, the symmetrical UI inductor model is simplified to a 1 / 4 model to construct a thermal resistance network, and the Grashof number Gr, Rayleigh number Ra, and Prandtl number Pr of each thermal resistance node module are calculated:

[0016]

[0017]

[0018] Where: g is the acceleration of gravity, v is the kinematic viscosity, 1 is the characteristic length, ΔT is the temperature difference between the object temperature and the ambient temperature, β is the volume change coefficient, T m To calculate the qualitative temperature;

[0019]

[0020] Among them, Gr * is a new type of characteristic number, q is the heat flux, Nu is the Nusselt number, λ c is the thermal conductivity of air;

[0021] R a =G r ·P r

[0022] Where Pr is Prandtl's constant;

[0023] For Nusselt number Nu and characteristic length l, there are five types of calculations: vertical surface, horizontal top surface, horizontal bottom surface, horizontal slit, vertical slit;

[0024] (1) Vertical surface

[0025]

[0026] Where Ra = 10 -6 ~∞;

[0027] l=h

[0028] (2) Horizontal top surface

[0029]

[0030] Among them, 6.37×10 5 <Gr * <1.12×10;

[0031]

[0032] When the surface is a circle,

[0033]

[0034] (3) Horizontal bottom surface

[0035] Nu=0.747(Gr * ·Pr) 1 / 6

[0036] Among them, 6.37×10 5 <Gr * <1.12×10 8

[0037]

[0038] When the surface is a circle:

[0039]

[0040] (4) Air interlayer

[0041] 1) When the vertical air interlayer Gr≤2860 and the horizontal interlayer (the bottom surface is the hot surface) Gr≤2430, the equivalent thermal conductivity of the air gap hc=λc, λc is the thermal conductivity of air;

[0042] 2) When the vertical air interlayer Gr>2860 and the horizontal interlayer Gr>2430:

[0043] ① When the air gap ratio hour,

[0044] Vertical slits:

[0045]

[0046]

[0047] Horizontal slits:

[0048] Nu=0.059(Gr·Pr) 0.4 2.4×10 3 ≤Gr≤1.0×10 4

[0049] Nu=0.212(Gr·Pr) 1 / 4 1.0×10 4 ≤Gr≤4.6×10 5

[0050] Nu=0.061(Gr·Pr) 1 / 3 Gr>4.6×10 5

[0051] ② When the air gap ratio hour,

[0052] Vertical slit

[0053] Nu=0.59(Gr·Pr) 1 / 4 1.0×10 4 ≤Gr.Pr≤1.0×10 9

[0054] Nu=0.1(Gr·Pr) 1 / 3 1.0×10 9 ≤Gr.Pr≤1.0×1012

[0055] Horizontal flat plate with hot side up or cold side down

[0056] Nu=0.54(Gr·Pr) 1 / 4 1.0×10 5 ≤Gr.Pr≤2.0×10 7

[0057] Nu=0.14(Gr·Pr) 1 / 3 2.0×10 7 ≤Gr.Pr≤3.0×10 10

[0058] Horizontal flat plate with cold side up or hot side down

[0059] Nu=0.27(Gr·Pr) 1 / 4 3.0×10 5≤Gr.Pr≤2.0×10 10

[0060] 3) When Gr exceeds the above range, a separate air gap model is established, the temperature field of the air gap is calculated using the finite element method, the air gap equivalent heat transfer coefficient is extracted, and substituted into the thermal resistance network for calculation;

[0061] l=s

[0062] Calculate the convection heat transfer coefficient h c :

[0063]

[0064] When the thermal resistance network is established, there is no heat flux flowing through the symmetric surface, and the nodes are unified to 0;

[0065] The heat flux at the coil corner is transferred along the path Z, X is the normal direction, Sor is in contact with the air, and the corner is treated as a rectangle by the area equivalent. In the thermal resistance network, both boundaries in the X direction are in contact with the air.

[0066] Further, the step S5 specifically includes the following steps:

[0067] Calculate the heat source size P in the core and coil of each thermal node module n , assuming that the initial thermal power P 0 and thermal conductivity are G 0 ;

[0068] like Figure 3 As shown, referring to the thermal resistance network diagram, the thermal power and thermal conductivity of each node are calculated in turn; mi is the positive node of branch i, ni is the negative node of branch i, Pi is the heat source on the node, and Gi is the thermal conductivity;

[0069] When the number of positive nodes m i >0 o'clock,

[0070] G mi，mi =G mi，mi +G i

[0071] P mi =P mi +p i

[0072] When the number of negative nodes n i >0 o'clock,

[0073] G ni，ni =G ni，ni +G i

[0074] P mi =P mi- pi

[0075] When the number of positive nodes and the number of negative nodes are both greater than 0

[0076] G pi，ni =G pi，ni -G i

[0077] G ni，pi =G ni，pi -G i

[0078] Run the symmetrical UI inductor natural cooling method to obtain the temperature values ​​on all hot nodes.

[0079] The beneficial effect of the present invention is that the method proposed by the present invention can quickly calculate the temperature field of the symmetrical UI inductor, thereby greatly reducing the large amount of trial work of designers, improving work efficiency, and reducing manual calculation errors.

[0080] Other advantages, objectives and features of the present invention will be described in the following description to some extent, and to some extent, will be obvious to those skilled in the art based on the following examination and study, or can be taught from the practice of the present invention. The objectives and other advantages of the present invention can be realized and obtained through the following description. BRIEF DESCRIPTION OF THE DRAWINGS

[0081] In order to make the purpose, technical solutions and advantages of the present invention more clear, the present invention will be described in detail below in conjunction with the accompanying drawings, wherein:

[0082] Figure 1 This is the overall natural cooling process of the symmetrical UI inductor of the present invention;

[0083] Figure 2 The symmetrical UI inductor thermal resistance calculation sub-process of the present invention, (a) is the vertical plane calculation process, (b) is the horizontal plane calculation process, (c) is the horizontal air gap calculation process, and (d) is the vertical air gap calculation process;

[0084] Figure 3 The calculation process of the symmetrical UI inductive thermal node module of the present invention;

[0085] Figure 4 This is a schematic diagram of the division of symmetrical UI inductor nodes of the present invention;

[0086] Figure 5 It is the thermal resistance network diagram of the symmetrical UI inductor 1 / 4 model of the present invention;

[0087] Figure 6 It is the temperature cloud diagram of the symmetrical inductor 1 / 4 model;

[0088] Figure 7 is the residual curve of coil average temperature;

[0089] Figure 8 is the residual curve of the average temperature of the core;

[0090] Fig. 9 Schematic diagram of the shape surface, (a) is the vertical surface, (b) is the horizontal top surface, (c) is the horizontal bottom surface, (d) is the horizontal slit, and (e) is the vertical slit;

[0091] Fig.10 Schematic diagram of coil corner processing method. DETAILED DESCRIPTION

[0092] The following describes the embodiments of the present invention by specific examples, and those skilled in the art can easily understand other advantages and effects of the present invention from the contents disclosed in this specification. The present invention can also be implemented or applied through other different specific embodiments, and the details in this specification can also be modified or changed in various ways based on different viewpoints and applications without departing from the spirit of the present invention. It should be noted that the illustrations provided in the following embodiments only illustrate the basic concept of the present invention in a schematic manner, and the following embodiments and features in the embodiments can be combined with each other without conflict.

[0093] Among them, the drawings are only used for illustrative explanations, and they only represent schematic diagrams rather than actual pictures, and should not be understood as limitations on the present invention. In order to better illustrate the embodiments of the present invention, some parts of the drawings may be omitted, enlarged or reduced, and do not represent the size of actual products. For those skilled in the art, it is understandable that some well-known structures and their descriptions in the drawings may be omitted.

[0094] The same or similar numbers in the drawings of the embodiments of the present invention correspond to the same or similar parts; in the description of the present invention, it should be understood that if the terms "upper", "lower", "left", "right", "front", "rear", etc. indicate the orientation or position relationship, they are based on the orientation or position relationship shown in the drawings, which is only for the convenience of describing the present invention and simplifying the description, rather than indicating or implying that the device or element referred to must have a specific orientation, be constructed and operate in a specific orientation. Therefore, the terms describing the position relationship in the drawings are only used for illustrative purposes and cannot be understood as limiting the present invention. For ordinary technicians in this field, the specific meanings of the above terms can be understood according to specific circumstances.

[0095] 1. Material preparation: Before entering the main process, first prepare the physical materials required for calculation. Air density, specific heat capacity, heat transfer coefficient, coil material, core material. Coil heat flux density 2000W / m 3 , core heat flux density 1000W / m 3 .

[0096] The material parameters of the symmetrical UI inductor are shown in Table 1.

[0097] Table 1

[0098]

[0099] 2. Dividing Thermal Resistance Nodes for Symmetrical UI Inductors

[0100] 1. According to Figure 4 The node division diagram of the symmetrical UI inductor 1 / 4 model is shown in the figure, which divides the thermal resistance node module.

[0101] 2. Divide the thermal resistance nodes of the magnetic core, coil, air, etc. according to the structural characteristics. Generally, the nodes are differentiated at the interface of different structures, such as slits, magnetic core interfaces, etc.

[0102] 3. In the example of this application, the magnetic core is divided into several modules: A, B, F, L, and M; the coil is divided into several modules: E, G, O, P, and Q; and the slit between the coils is divided into module S.

[0103] 3. Draw a thermal resistance network diagram

[0104] 1. Draw a symmetrical UI inductor thermal resistance network diagram based on the structural characteristics and the distribution of thermal resistance nodes

[0105] 2. Set the temperature point between two contacting nodes of the same material. Figure 5 In the embodiment, the core module A and the core module B are connected by a temperature point 46 .

[0106] 3. Set the contact thermal resistance between two nodes where different materials are in contact. Figure 5 As shown in the figure, between the coil module E and the core module F, the contact thermal resistance R EF The temperature point 55 is the boundary temperature of module E, and the temperature point 56 is the boundary temperature of module F.

[0107] 4. On the surface in contact with the air, set a convection thermal resistor and then ground it. Figure 5 As shown in the figure, the negative X reverse boundaries of the new modules A, L and coil module E are in direct contact with the air, so R Aoxa ,R Eoxa ,R Loxa Thermal resistance, then make a ground connection.

[0108] 5. Where there is a slit, use a slit thermal resistor to connect. Figure 5 In the figure, there is an air gap between the core module B and the coil module G, and the gap thermal resistance R is used between the temperature point 52 and the temperature point 53. BG connect.

[0109] 6. There is no heat flow on the symmetry surface, which is indicated by a 0 connection. Figure 5 As shown in , the positive X direction of the core module B and the negative Y direction of the coil module E are both symmetrical interfaces, represented by 0 connection.

[0110] 7. The results of the thermal resistance network diagram of the symmetrical UI inductor 1 / 4 model are as follows Figure 5 shown.

[0111] 4. Calculate the Grashof number (Gr), Rayleigh number (Ra), and Prandtl number (Pr) of each thermal resistance node module:

[0112] 1. Calculate Gr

[0113]

[0114]

[0115] Where: g: 9.8 m / s 2 ,v kinematic viscosity m 2 / s, 1 characteristic length m, ΔT is the temperature difference between the object temperature and the ambient temperature, β is the volume change coefficient.

[0116] 2. Calculate Gr *

[0117]

[0118] Where q is the heat flux w / m 2 , Nu is the Nusselt number, λ c is the thermal conductivity of air.

[0119] 3. Calculate Ra

[0120] R a =G r ·P r Formula 1-3

[0121] Where: Pr: Prandtl constant, obtained from Table 2.

[0122] Table 2 Thermophysical properties of dry air (P = 1.01325 × 10 5 Pa=760mmHg)

[0123]

[0124]

[0125] 4. Calculate Nu and l, which are divided into five different types, such as Fig. 9 (a)-(e) show the vertical plane, horizontal top plane, horizontal bottom plane, horizontal slit, and vertical slit, respectively.

[0126] (1) Vertical plane calculation formula

[0127] (A) Nu value

[0128]

[0129] Where Ra = 10 -6 ~∞

[0130] (B) Feature length 1

[0131] l=h Formula 1.1-2

[0132] (2) Calculation formula for horizontal top surface:

[0133] (A) Nu value

[0134]

[0135] Among them, 6.37×10 5 <Gr * <1.12×10 8

[0136] (B) Feature length 1

[0137]

[0138] When the surface is a circle,

[0139]

[0140] (3) Calculation formula for horizontal bottom surface:

[0141] (A) Nu value

[0142] Nu=0.747(Gr * ·Pr) 1 / 6 Formula 1.3-1

[0143] Among them, 6.37×10 5 <Gr * <1.12×10 8

[0144] (B) Feature length 1

[0145]

[0146] When the surface is a circle:

[0147]

[0148] (4) Air interlayer calculation formula

[0149] a) When Gr≤2860 for vertical air interlayer and Gr≤2430 for horizontal interlayer (bottom is hot surface), the equivalent thermal conductivity of air layer hc=λ c ,λ c Thermal conductivity of air

[0150] b) When the vertical air interlayer Gr>2860 and for the horizontal interlayer Gr>2430,

[0151] ◆When the air gap ratio hour,

[0152] Vertical slits:

[0153]

[0154]

[0155] Horizontal slits:

[0156] Nu=0.059(Gr·Pr) 0.4 2.4×10 3 ≤Gr≤1.0×10 4 Formula 1.4-3

[0157] Nu=0.212(Gr·Pr) 1 / 4 1.0×10 4 ≤Gr≤4.6×10 5 Formula 1.4-4

[0158] Nu=0.061(Gr·Pr) 1 / 3 Gr>4.6×10 5 Formula 1.4-5

[0159] ◆When the air gap ratio hour,

[0160] Vertical slit

[0161] Nu=0.59(Gr·Pr) 1 / 4 1.0×10 4 ≤Gr.Pr≤1.0×10 9 Formula 1.4-6

[0162] Nu=0.1(Gr·Pr) 1 / 3 1.0×10 9 ≤Gr.Pr≤1.0×10 12 Formula 1.4-7

[0163] Horizontal flat plate with hot side up or cold side down

[0164] Nu=0.54(Gr·Pr) 1 / 41.0×10 5 ≤Gr.Pr≤2.0×10 7 Formula 1.4-8

[0165] Nu=0.14(Gr·Pr) 1 / 3 2.0×10 7 ≤Gr.Pr≤3.0×10 10 Formula 1.4-9

[0166] Horizontal flat plate with cold side up or hot side down

[0167] Nu=0.27(Gr·Pr) 1 / 4 3.0×10 5 ≤Gr.Pr≤2.0×10 10 Formula 1.4-10

[0168] c) When Gr exceeds the above range, a separate air gap model is established, the temperature field of the air gap is calculated using the finite element method, the air gap equivalent heat transfer coefficient is extracted, and substituted into the thermal resistance network for calculation.

[0169] (B) Feature length 1:

[0170] l=s Formula 1.4-11

[0171] 5. Calculate the convection heat transfer coefficient h c

[0172]

[0173] Where: c is the thermal conductivity of air.

[0174] Coil corner processing method Fig.10 As shown, the heat flux at the coil corner is transferred along the path Z, and X is the normal direction. Sor is in contact with the air. The corner is treated as a rectangle by the area equivalent. In the thermal resistance network, both boundaries in the X direction are in contact with the air.

[0175] 5. Solving the Thermal Resistance Network

[0176] 1. Run Figure 1 Overall flow chart of natural cooling of symmetrical UI inductor.

[0177] 2. Assume that the initial temperature T0 = 50°C and the ambient temperature Tw = 20°C. Set the maximum number of cycles Kmax = 3000 and the maximum allowable error emax = 1.

[0178] 3. According to the ambient temperature Tw and the initial temperature T0, calculate the average temperature Tm = (Tw + T0) / 2 = 35 ° C. According to the Tm value, look up the air properties table in Table 2 to obtain the thermal conductivity λ c=0.0276w / (mk), kinematic viscosity v=1.696e-5m2 / s, Planck number Pr=0.699.

[0179] 4. According to the different structural forms of the divided thermal resistance node modules, run them separately Figure 2 The heat transfer coefficient calculation process sub-processes a, b, c, and d calculate the thermal resistance of all thermal node modules in the thermal resistance network. Taking the core module A as an example, LAx = 0.023m, LAy = 0.075m, and LAz = 0.028m. Solve the thermal resistance values ​​of the core module A in the six directions of X, Y, and Z respectively. Get h_Alya = 3.8053, h_Alza = 3.5278, and h_Aoxa = 3.8053. The positive direction of X, the negative direction of Y, and the negative direction of Z are directly connected to other modules, and the thermal resistance value is not calculated.

[0180] 5. Run the thermal node module calculation process 3 to calculate the temperature values ​​of all thermal resistance node modules. When the average temperature is T0-Tm>emax, k<Kmax, k=k+1, T0=Tm. Repeat step 1.

[0181] 6. When T0-Tm<emax, the process ends and the result is output.

[0182] 7. After calculation, the calculation results of the thermal resistance network are:

[0183] a) The temperatures of the core modules are 32.1648℃, 32.1658℃, 32.3778℃, 32.3444℃, and 32.3632℃ respectively. The average temperature is 32.28℃.

[0184] The temperatures of the coil modules are 32.5791℃, 32.5998℃, 32.5812℃, 32.5841℃, and 32.5885℃ respectively. The average temperature is 32.59℃.

[0185] b) Finite element calculation results and residual curves are as follows Figure 6 , Figure 7 , Figure 8 As shown in the figure, the average temperature of the core is 29.68℃, and the average temperature of the coil is 29.77℃.

[0186] c) The relative error of the core temperature is 8.06%, and the relative error of the coil temperature is 8.65%. The errors are both within 10%.

[0187] Finally, it should be noted that the above embodiments are only used to illustrate the technical solution of the present invention rather than to limit it. Although the present invention has been described in detail with reference to the preferred embodiments, those skilled in the art should understand that the technical solution of the present invention can be modified or replaced by equivalents without departing from the purpose and scope of the technical solution, which should be included in the scope of the claims of the present invention.

Claims

1. A method for calculating the temperature field of a naturally cooled symmetrical UI inductor based on a thermal resistance network, characterized in that: The following steps are involved: S1: Material preparation: prepare the physical materials required for calculation, including geometric dimensions and material physical property parameters; S2: Divide thermal resistance nodes: Divide the thermal resistance nodes of the symmetrical UI inductor according to the inductor geometry and the coil winding connection form; S3: Draw a thermal resistance network diagram: Draw a symmetrical UI inductor thermal resistance network diagram based on the structural characteristics and the distribution of thermal resistance nodes; S4: Calculate thermal resistance: Calculate the thermal resistance of each thermal resistor in the thermal resistance network based on geometric parameters and material properties; In step S4, the symmetrical UI inductor model is simplified to a 1 / 4 model to construct the thermal resistance network, and calculate the Grashof number Gr, Rayleigh number Ra, and Prandtl number Pr of each thermal resistance node module: Where: g is the acceleration of gravity, v is the kinematic viscosity, l is the characteristic length, ΔT is the temperature difference between the object temperature and the ambient temperature, β is the volume change coefficient, T m To calculate the qualitative temperature; Among them, Gr * is a new type of characteristic number, q is the heat flux, Nu is the Nusselt number, λ c is the thermal conductivity of air; R a =G r ·P r Where Pr is Prandtl's constant; For Nusselt number Nu and characteristic length l, there are five types of calculations: vertical surface, horizontal top surface, horizontal bottom surface, horizontal slit, vertical slit; (1) Vertical surface Where Ra = 10 -6 ~∞; l＝h (2) Horizontal top surface Among them, 6.37×10 5 <Gr * <1.12×10; When the surface is a circle, (3) Horizontal bottom surface No=0.t4t(Gr * ·Pr) 1 / 6 Among them, 6.37×10 5 <Gr * <1.12×10 8 When the surface is a circle: (4) Air interlayer 1) When the vertical air interlayer Gr≤2860, and the horizontal interlayer, the bottom surface is the hot surface, Gr≤2430, the equivalent thermal conductivity of the air gap hc=λc, λc is the thermal conductivity of air; 2) When the vertical air interlayer Gr>2860 and for the horizontal interlayer Gr>2430: ① When the air gap ratio hour, Vertical slits: Horizontal slits: No=0.059(Gr·Pr) 0.4 2.4×10 3 ≤Gr≤1.0×10 4 No=0.212(Gr·Pr) 1 / 4 1.0×10 4 ≤Gr≤4.6×10 5 No=0.061(Gr·Pr) 1 / 3 Gr>4.6×10 5 ② When the air gap ratio hour, Vertical slit No=0.59(Gr·Pr) 1 / 4 1.0×10 4 ≤Gr.Pr≤1.0×10 9 No=0.1)Gr·Pr( 1 / 3 1.0×10 9 ≤Gr.Pr≤1.0×10 12 Horizontal flat plate with hot side up or cold side down No=0.54(Gr·Pr) 1 / 4 1.0×10 5 ≤Gr.Pr≤2.0×10 t No=0.14(Gr·Pr) 1 / 3 2.0×10 t ≤Gr.Pr≤3.0×10 10 Horizontal flat plate with cold side up or hot side down No=0.2t(Gr·Pr) 1 / 4 3.0×10 5 ≤Gr.Pr≤2.0×10 10 3) When Gr exceeds the above range, a separate air gap model is established, the temperature field of the air gap is calculated using the finite element method, the air gap equivalent heat transfer coefficient is extracted, and substituted into the thermal resistance network for calculation; l＝s Calculate the convection heat transfer coefficient h c : When the thermal resistance network is established, there is no heat flux flowing through the symmetric surface, and the nodes are unified to 0; The heat flux at the coil corner is transferred along the path Z, X is the normal direction, Sor is in contact with the air, and the corner is treated as a rectangle by the area equivalent. In the thermal resistance network, both boundaries in the X direction are in contact with the air; S5: Solve the thermal resistance network: Substitute the thermal resistance value into the drawn thermal resistance network. After iterative calculation, the final temperature value of each node is obtained, thereby obtaining the symmetrical UI inductor temperature field distribution.

2. The method for calculating the temperature field of a naturally cooled symmetrical UI inductor based on a thermal resistance network according to claim 1 is characterized in that: In the step S1, the geometric dimensions of each component inside the symmetrical UI inductor are determined, and the connection method and physical parameters of each component are determined; the physical parameters include the core material, coil material and winding method, ambient temperature, air thermal parameters, inductor operating parameters, and the heating value of the core and coil.

3. The method for calculating the temperature field of a naturally cooled symmetrical UI inductor based on a thermal resistance network according to claim 1 is characterized in that: In the step S2, the magnetic core, coil and air gap of the symmetrical UI inductor are equivalent to different nodes according to the structural characteristic environment.

4. The method for calculating the temperature field of a naturally cooled symmetrical UI inductor based on a thermal resistance network according to claim 1 is characterized in that: Step S3 includes: setting a temperature point between two contacting nodes of the same material; setting a contact thermal resistance between two contacting nodes of different materials; setting a convection thermal resistance and then grounding on the surface in contact with the air; connecting the position where there is a slit with a slit thermal resistance; and indicating that there is no heat flow on the symmetric surface with a 0 connection; Nodes without temperature difference are directly connected by lines, nodes with temperature difference are connected by contact thermal resistance, locations with slits are connected by slit thermal resistance, blocks in contact with air are connected by convection thermal resistance, and each node, line, and thermal resistance are interconnected to form a thermal resistance network model.

5. The method for calculating the temperature field of a naturally cooled symmetrical UI inductor based on a thermal resistance network according to claim 1 is characterized in that: The step S5 specifically comprises the following steps: Calculate the heat source size P in the core and coil of each thermal node module n , assuming that the initial thermal power P0 and thermal conductivity are both G0; Referring to the thermal resistance network diagram, calculate the thermal power and thermal conductivity of each node in turn; mi is the positive node of branch i, ni is the negative node of branch i, Pi is the heat source on the node, and Gi is the thermal conductivity; When the number of positive nodes m i >0, G mi,mi =G mi,mi +G i P mi =P mi +p i When the number of negative nodes n i >0, G ni,ni =G ni,ni +G i P mi =P mi -p i When the number of positive nodes and the number of negative nodes are both > 0 G pi,ni =G pi,ni -G i G ni,pi =G ni,pi -G i Run the symmetrical UI inductor natural cooling method to obtain the temperature values ​​on all hot nodes.

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