# Analysis of the Temperature Characteristics of High-speed Train Bearings Based on a Dynamics Model and Thermal Network Method – Chinese Journal of Mechanical Engineering

Aug 11, 2022

### Isothermal Nodes Division

The heat input in the temperature field needs to be solved by the dynamic model. The detailed modeling process of the dynamic model of the high-speed train bearing rotor system and the parameters of the system have been described in Refs. [25, 26], so this paper will not repeat the description in detail. The bearing system was assembled using different components [27]. It was assumed that in the heat transfer process, the temperatures at different positions in the same components were equal. Based on this, the bearing was divided into several isothermal nodes, as shown in Figure 1(a) and Table 1.

The contact point between the two contact objects of the bearing was also regarded as an isothermal node. According to the Hertz contact theory, the friction loss between the roller and raceway is generated in a small area, with a size much smaller than the diameter of the roller. This phenomenon limits the flow of heat from the contact area to the center of the contact body. That is, there is a ‘bottleneck’ in heat conduction between the surface of the bearing element and the main body [28]. According to Refs. [9, 10], the high-temperature zone of the bearing components is mainly concentrated on a very thin layer on the surface, whereas the temperature of the main body is relatively low. Furthermore, the bogie was constructed like a container with its opening facing down, through which the air around the spindle was enclosed and isolated from the air around the axle box. Therefore, the air around the spindle and the air around the axle box were regarded as two different isothermal nodes.

In the system, heat was transferred between adjacent nodes that were connected by thermal resistance. In bearing operations, part of the heat generated inside the bearing is absorbed by the grease, and the other part is transmitted through various components of the bearing and finally dissipated into the air. The direction of heat flow is shown in Figure 1(b). The thermal resistance is defined by the generalized Ohm’s law as:

$$R = frac{Delta T}{Q},$$

(1)

where R is the thermal resistance, ΔT is the temperature difference, and Q is the heat flow between the two adjacent nodes. A generalized schematic of thermal resistance is shown in Figure 2.

### Contact Force Analysis under Fault Conditions

The calculation of the contact force is the basis of model establishment and temperature field analysis. Existing studies [5, 29, 30] show that when a rolling-element passes through a fault area, the change in the deformation will cause sudden changes in the contact force, as shown in Figure 3. Simultaneously, an impact force was generated and modelled [31, 32]. When a fault exists in the bearing, the contact force will undergo a sudden change and an additional impact force will be generated to further increase the contact force. Based on the model established in this study, the resultant force of the contact force (referred to as the contact force hereafter) between the rolling-elements and the raceway was obtained when the bearing had outer-ring, inner-ring, and rolling-element faults, as shown in Figure 4. In this simulation, the fault size L0 was 1 mm for an actual bearing wear width of approximately 1–3 mm, and the bearing inner-ring speed wc was 1600 r/min, which was approximately 265 km/h, which is slightly lower than the current running speed of the high-speed train.

The characteristic fault frequencies of the bearing are denoted as fBPFO, fBPFI, and fBSF for faults in the outer-ring, inner-ring, and rolling-element, respectively. Each time a rolling-element passed through the fault area, the contact force changed. The interval time of the contact force change tcit is the same as the interval time for the rolling-element passing through the fault area. Furthermore, the rolling-elements contact the fault edge when entering and leaving the fault area, as shown in Figure 3, which causes two sudden changes in the contact force, and the time of the two sudden changes tsct is proportional to the width of the fault. However, when a bearing has a rolling-element fault, the edge of the fault area is always in contact with the raceway. In this case, the deformation and contact force did not change. For the bearings used in the simulation, the following results were obtained.

When the bearing had an outer-ring fault:

left{ begin{aligned} t_{cit} & = frac{1}{{f_{{{text{BPFO}}}} }} = 0.0053{text{ s,}} \ t_{sct} & = frac{{N_{0} arcsin (frac{{L_{0} }}{{2r_{0} }})t_{cit} }}{uppi }{ = }4.78 times 10^{ – 4} {text{ s}}{.} \ end{aligned} right.

(2)

When the bearing had an inner-ring fault:

left{ begin{aligned} t_{cit} & = frac{1}{{f_{{{text{BPFI}}}} }} = 0.0037{text{ s,}} \ t_{sct} & = frac{{N_{0} arcsin (frac{{L_{0} }}{{2r_{0} }})t_{cit} }}{uppi }{ = }1.7 times 10^{ – 4} {text{ s}}{.} \ end{aligned} right.

(3)

When the bearing had a rolling-element fault:

left{ begin{aligned} t_{cit} & = frac{1}{{f_{{{text{BSF}}}} }} = 0.013{text{ s,}} \ t_{sct} & = frac{{N_{0} arcsin (frac{{L_{0} }}{{2r_{0} }})t_{cit} }}{uppi }{ = }3 times 10^{ – 4} {text{ s}}{.} \ end{aligned} right.

(4)

A comparison of Eqs. (2)–(4) and Figure 4 show that the simulation and theoretical calculation results are consistent.

When the bearing speed was constant, the contact force P changed periodically within a certain period of time. Under the law of conservation of energy, it can be replaced by the average value of the contact force over a given period. Figure 8 shows that the equivalent contact force changed when the fault size changed. As shown in Figure 5, when the bearing had no fault, the contact force was usually relatively small. In the presence of a fault, the contact force changed suddenly owing to the impact force. As the size of the fault increased, the contact force also increased. When the bearing had a rolling-element fault, the change in the contact force was more evident, and its characteristics were different from those of other fault scenarios. As the size of the rolling-element fault increased, the contact force dropped rapidly after a sudden change, and then gradually decreased. This is because the fault area became smoother as the width of the fault increased, making the impact force smaller, and thus the contact force smaller.

The internal power loss of rolling bearings was mainly caused by the mutual friction of the internal components of the bearing. This friction, including the friction between the rolling-elements and raceways of the inner- and outer-rings and the friction between the rolling-elements and the cage, is also the main reason for the increase in the bearing temperature. As the dynamics model in this study ignored abnormal bearing behaviors, such as roller slip and cage shaking, only the former friction was considered when calculating the energy loss, and the latter form of friction is equivalent to the contact force. In most cases, the friction torque is used to calculate the energy loss inside the bearing and obtain the total heat generation of the bearing. However, the positions of each roller of the bearing and the force are different. Local methods are more reasonable and easier to use [11]. In this method, the heat generation of each roller was calculated separately and then summed to obtain the overall heat generation of the bearing. The heat generation of each roller was calculated as follows:

$$q_{i} = mu P_{i} nu_{i} ,$$

(5)

where qi is the total frictional heat generation at the i-th roller, µ is the rolling friction coefficient between the roller and raceway, and νi is the relative speed of the roller and raceway. The overall heat generation is given as:

$$Q = sumlimits_{i = 1}^{{N_{0} }} {q_{i} } .$$

(6)

### Thermal Resistance

Heat transfer in the bearing rotor system includes conduction heat transfer, convection heat transfer, and heat radiation. As the heat transfer by thermal radiation is much smaller than that of the other two modes, it can be ignored.

#### Conduction Heat Transfer

Heat conduction is a process in which a large number of molecules, atoms, or electrons collide with each other, and energy is transferred from a relatively high-temperature component or object to a low-temperature component or object.

1. (a)

The bearing spindle, inner-ring, outer-ring, and axle box can be simplified as cylinders with inner- and outer-ring radii rin and rex, respectively. This type of heat conduction relationship can be simplified as a one-dimensional heat conduction problem along the radius [16], and the thermal resistance can be calculated as:

$$R = frac{{ln r_{ex} – ln r_{in} }}{{2uppi kL}},$$

(7)

where k is the thermal conductivity and L is the characteristic length. Thermal resistance between ‘T2&T3’, ‘T3&T4’, ‘T8&T9’, and ‘T9&T10’ in Figure 1 and Table 1, can be determined by Eq. (7).

2. (b)

The contact between the roller and raceway is the Hertz contact, and the size of the contact area is much smaller than the size of the bearing. The thermal resistance of this component can be calculated as [18]:

$$R = frac{1}{uppi }left( frac{a}{b} right)frac{1}{{kasqrt {Pe} }},$$

(8)

where a is the semi-major axis of the ellipse along the rolling direction; b is the semi-minor axis of the ellipse perpendicular to the rolling direction; and Pe is the Peclet number. Eq. (8) is used to determine the thermal resistance between ‘T4&T5’, ‘T5&T6’, ‘T6&T7’, and ‘T7&T8’, as shown in Figure 1 and Table 1.

#### Convection Heat Transfer

Thermal convection is the heat exchange between a fluid and solid when the fluid flows over a solid surface. Heat convection is the most difficult form of heat transfer to quantify. In the bearing system, heat is transferred from the bearing to the lubricant, and then from the lubricant to the bearing seat and other components. Thermal convection also occurs between the outer surface of the bearing and the surrounding fluid, which is usually air. In most cases, the convection thermal resistance can be expressed as:

$$R = frac{1}{{Ah_{c} }}{ = }frac{1}{A}left( frac{L}{kNu} right),$$

(9)

where A is the surface area of the contact surface where heat exchange occurs, k is the thermal conductivity of the fluid, L is the characteristic length, and Nu is the dimensionless Nusselt number, which is determined by different convection conditions, as follows:

1. (a)

The part of the spindle extending beyond the bearing exhibits a convective heat dissipation relationship with the air, and Nu is calculated as follows:

$$Nu = left{ {begin{array}{*{20}c} {0.00308Re + 4.432} & {Re < 7300,} \ {Re^{0.37} } & {7300 le Re < 9600,} \ {30.5Re^{ – 0.0042} } & {Re ge 9600,} \ end{array} } right.$$

(10)

where Re = VDs/νair, νair is the kinematic viscosity of air, V is the spindle line speed, and Ds is the spindle diameter. Eq. (10) is used to determine the thermal resistance between ‘T1&T2’ in Figure 1 and Table 1.

2. (b)

When the bearing ring rotates, it exchanges heat with grease. The heat transfer relationship can be simplified as a concentric rotating cylinder model [33]. In this case, Nu was calculated as follows:

$$Nu = left{ {begin{array}{*{20}c} {begin{array}{*{20}c} 2 \ {0.167Ta^{0.69} Pr^{0.4} } \ {0.401Ta^{0.5} Pr^{0.4} } \ end{array} } & {begin{array}{*{20}c} {Ta < 41,} \ {41 le Ta < 100,} \ {Ta ge 100,} \ end{array} } \ end{array} } right.$$

(11)

where (T_{a} = {{rho varepsilon_{R} sqrt {varepsilon_{R} r_{{{text{ring}}}} } } mathord{left/ {vphantom {{rho varepsilon_{R} sqrt {varepsilon_{R} r_{{{text{ring}}}} } } {nu_{{text{g}}} }}} right. kern-nulldelimiterspace} {nu_{{text{g}}} }}) in which εR is the radial clearance between the inner- and outer-rings, rring is the internal radius of the ring, and νg is the kinematic viscosity of grease. Eq. (11) is used to determine the thermal resistance between ‘T4&T11’, ‘T5&T11’, ′T6&T11’, ‘T7&T11’, and ‘T8&T11’, in Figure 1 and Table 1.

3. (c)

When the train is moving, air blows over the surface of the axle box at the speed of the train, which causes forced convection to dissipate heat. The surface heat transfer coefficient hc can be approximated as

$$h_{c} = 0.03frac{k}{{D_{h} }}({{u_{s} D_{h} } mathord{left/ {vphantom {{u_{s} D_{h} } {nu_{{{text{air}}}} }}} right. kern-nulldelimiterspace} {nu_{{{text{air}}}} }})^{0.57} ,$$

(12)

where k is the air thermal conductivity, and Dh is the outer diameter of the axle box.

As air is a fluid, there is natural convection heat dissipation between the air and axle box. In this case, Nu can be determined using Eq. (10). Assuming that the equivalent thermal resistance of the air and axle box is R, the resistance of forced convection heat dissipation is R1, and the resistance of natural convection heat dissipation is R2. Then, we have

$$frac{1}{R} = frac{1}{{R_{1} }} + frac{1}{{R_{2} }}.$$

(13)

Eq. (13) can be used to determine the thermal resistance between ‘T10&T12’ in Figure 1 and Table 1.

#### Heat Transfer in the Enclosed Air Layer

Under normal circumstances, high-speed trains travel very quickly, which reduces the efficiency of the heat exchange between the air near the spindle of the bogie and the outside atmosphere. The air near the spindle is confined in a closed space and isolated from the outside atmosphere by an air layer. The air near the spindle dissipates the heat absorbed from the spindle to the outside atmosphere through the air layer. The thermal resistance between ‘T1&T12’ in Figure 1 and Table 1 can be calculated by:

$$R = frac{{d_{{{text{air}}}} }}{{lambda_{a} + lambda_{c} + lambda_{r} }},$$

(14)

where dair is the thickness of the enclosed air layer, λa is the thermal conductivity of the air layer, λc is the convective equivalent thermal conductivity of the air layer, and λr is the radiation equivalent thermal conductivity of the air layer.

### Node Temperature Solution

In electricity, the nodal current law states that the sum of the currents flowing into a node is equal to the sum of the currents flowing out of the node. Similarly, in a thermal grid, the sum of the heat flowing into a node is equal to the sum of the heat flowing out of the node. An equivalent schematic is shown in Figure 6. In particular, additional heat is generated near the rolling-elements owing to friction, and the heat relationship at the relevant nodes is:

$$sum {Q_{{{text{in}}}} + sum {Q_{f} – sum {Q_{{{text{out}}}} = 0} } } ,$$

(15)

where Qin is the input heat, Qout is the output heat, and Qf is the heat generated by friction.

According to Ohm’s law, Eq. (15) can be used to determine the heat flow relationship at each node in the thermal network and finally obtain a set of equations representing the direction of heat flow in the bearing rotor system. The thermal resistance between the two nodes is distinguished by the node number, where c is the heat transfer by conduction, and v is the heat transfer by convection. For example, R3c2 represents the conduction thermal resistance between T3 and T2 and the heat flows from T3 to T2. Similarly, R6v11 represents the convection thermal resistance between T6 and T11 and the heat flows from T6 to T11. The thermal resistance of the enclosed air layer is Rair. Based on the analysis above, the heat transfer in the bearing system can be described as follows:

left{ begin{aligned} & frac{{T_{1} – T_{12} }}{{R_{{{text{air}}}} }} – frac{{T_{2} – T_{1} }}{{R_{{2{text{v}}1}} }} = 0, \ & frac{{T_{2} – T_{1} }}{{R_{{{text{2v1}}}} }} – frac{{T_{3} – T_{1} }}{{R_{{3{text{c}}2}} }} = 0, \ & frac{{T_{3} – T_{1} }}{{R_{{{text{3c2}}}} }} – frac{{T_{4} – T_{3} }}{{R_{{{text{4c3}}}} }} = 0, \ & frac{{T_{4} – T_{3} }}{{R_{{{text{4c3}}}} }} – frac{{T_{11} – T_{4} }}{{R_{{11{text{v}}4}} }} – frac{{T_{5} – T_{4} }}{{R_{{5{text{c}}4}} }} = 0, \ & frac{{T_{5} – T_{4} }}{{R_{{{text{5c4}}}} }} + frac{{T_{5} – T_{11} }}{{R_{{{text{5v11}}}} }} + frac{{T_{5} – T_{6} }}{{R_{{5{text{c6}}}} }} = {1 mathord{left/ {vphantom {1 {left( {2Q_{{text{i}}} } right),}}} right. kern-nulldelimiterspace} {left( {2Q_{{text{i}}} } right),}} \ & frac{{T_{6} – T_{11} }}{{R_{{{text{6v11}}}} }} – frac{{T_{5} – T_{6} }}{{R_{{{text{5c6}}}} }} – frac{{T_{7} – T_{6} }}{{R_{{{text{7c6}}}} }} = {1 mathord{left/ {vphantom {1 {left( {2Q_{{text{i}}} } right) + {1 mathord{left/ {vphantom {1 {left( {2Q_{{text{o}}} } right),}}} right. kern-nulldelimiterspace} {left( {2Q_{{text{o}}} } right),}}}}} right. kern-nulldelimiterspace} {left( {2Q_{{text{i}}} } right) + {1 mathord{left/ {vphantom {1 {left( {2Q_{{text{o}}} } right),}}} right. kern-nulldelimiterspace} {left( {2Q_{{text{o}}} } right),}}}} \ & frac{{T_{7} – T_{8} }}{{R_{{{text{7c8}}}} }} + frac{{T_{7} – T_{11} }}{{R_{{{text{7v11}}}} }} + frac{{T_{7} – T_{6} }}{{R_{{{text{7c6}}}} }} = {1 mathord{left/ {vphantom {1 {left( {2Q_{{text{o}}} } right),}}} right. kern-nulldelimiterspace} {left( {2Q_{{text{o}}} } right),}} \ & frac{{T_{8} – T_{9} }}{{R_{{{text{8c9}}}} }} – frac{{T_{11} – T_{8} }}{{R_{{11{text{v8}}}} }} – frac{{T_{7} – T_{8} }}{{R_{{{text{7c8}}}} }} = 0, \ & frac{{T_{9} – T_{10} }}{{R_{{{text{9c10}}}} }} – frac{{T_{8} – T_{9} }}{{R_{{{text{8c9}}}} }} = 0, \ & frac{{T_{10} – T_{12} }}{{R_{{{text{10v12}}}} }} – frac{{T_{9} – T_{10} }}{{R_{{{text{9c10}}}} }} = 0, \ & frac{{T_{11} – T_{4} }}{{R_{{{text{11v4}}}} }} + frac{{T_{11} – T_{8} }}{{R_{{{text{11v8}}}} }} – frac{{T_{7} – T_{11} }}{{R_{{{text{7c11}}}} }} – frac{{T_{6} – T_{11} }}{{R_{{{text{6v11}}}} }} – frac{{T_{5} – T_{11} }}{{R_{{{text{5v11}}}} }}{ = }0, \ end{aligned} right.

(16)

where Qi and Qo are the heat generated by the friction between the rolling-elements and raceway of the inner- and outer-rings, respectively. The temperature of the outside atmosphere, T12, was constant in this study. In Eq. (16), there were 11 equations and 11 unknowns. The temperature of each node can be obtained using the Gauss-Seidel iteration method [34].

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