# Numerical Algorithms for Calculating Temperature, Layered Stress, and Critical Current of Overhead Conductors

Authors: Yongdou Liu; Zhiwei Chen; Quan Gu
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

Accurate calculation of temperature, stress, sag, and critical current (corresponding to critical temperature) of operational overhead conductors is important for ensuring the strength and sag safety of overhead lines. Based on 2D steady-state heat transfer equations, this article studies the temperature fields of the cross section of typical electrified conductors and establishes numerical simulation methods for calculating the layered stress, sag, and critical temperature. Using the algorithm, the relationship between the critical temperature and characteristics of conductors (e.g., the sag and tensile force) is studied. The results are verified by a comparison with the test results for heat-resistant aluminum alloy conductors JNRLH1/G1A-400/65 and JNRLH1/G1A-630/55. Finally, the paper studies the relationship between the critical temperature of the conductor and its most sensitive factors.


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## Body

## 1. Introduction

During peak times, it is often necessary to dynamically increase the current of overhead lines to meet the power supply demands. In these cases, the stress and sag of conductors must be accurately calculated to avoid breakage due to excessive stress, multicircuit wire mixture [1], or nonenough safety distance to the ground caused by large sag. In addition, the stress at different layers of strands (called layered stress) and the sag are generally affected by the temperature fields on the conductor cross sections. Only when the temperature fields are simulated accurately can the layered stress and sag be calculated actually. Moreover, due to the uneven thermal expansion in different layers during operation, the mechanical properties of the conductor's cross section will greatly change at definite temperature which is defined as the critical temperature (CT) [2, 3]. However, to the author’s knowledge, there is limited research on the safety assessment of overhead conductors at the CT. Therefore, calculation of the temperature fields, layered stress, sag and tensile force, and the analysis of the safety of overhead lines at CT are of great significance.Many experimental studies [4–12] have shown that the radial temperature of conductors is not evenly distributed. IEEE specifications [13] proposed that there is radial temperature gradient among each layer. In recent years, some methods [14–16] based on the finite element method were proposed to simulate the conductor radial temperature field. With these methods, models better conforming to the actual sectional structure of conductors can be established to avoid calculation errors due to improper values for effective thermal conductivity. However, these studies failed to fully consider the impact of cross-section gap distribution, inter-strand contact, and convection conditions on the radial temperature field. Furthermore, there is little research that systematically studied the temperature field, layered stress, sag, and critical temperature of conductors. This paper proposes a complete set of numerical methods to calculate the temperature field, layered stress, and critical current (or temperature) of overhead conductors. The temperature field with voids is calculated by using the finite element method, and the layered stress and sag are calculated by using the temperature results. The CT and the corresponding critical current (CC) are calculated through the temperature and stress calculative method iteratively. By comparing the results of calculation with those of the experiments for a typical ACSR (aluminum-conductor steel reinforced), the algorithms are verified. In order to facilitate the dynamic assessment of safety of the conductors during the operation, this paper studies the relationship between the critical temperature of the conductor and its sensitive factors with the heat-resistant aluminum alloy conductor JNRLH1/G1A-400/65 as an example.


## 2. Calculation of Conductor Temperature Field

The thermal balance equation of the entire cross sectionΩ of the conductor can be established [14]:(1)kT,xx+T,yy+s=0,x,y∈Ω.And the boundary condition on the surface of the conductorΓ can also be given [14]:(2)qn=−kT,xnx+T,yny=αT−Ta,lx,y∈Γ,where T,xx, T,yy, T,x, and T,y indicate the second-order derivatives and first-order derivatives of the 2D temperature field T in the conductor on the x and y coordinates, respectively; qn indicates the heat loss rate along the normal direction n on the outer surface of the conductor; nx and ny are the components of n in the direction of x and y; α indicates the composite heat loss coefficient; Ta indicates the ambient temperature; k is the thermal conductivity, represented by ka in the area of the aluminum strand, ks in the area of the steel core, and kair in the area of the air; and s indicates the rate of heat per unit volume, represented by sa in the area of the aluminum strand, ss in the area of the steel core, and 0 in the area of the air. Herein, ka and ks are irrelevant to temperature field, while kair is relevant to it, and the relationship is as follows:(3)kair=2.42×10−2+7.2×10−5Tav,where Tav indicates the average temperature of the cross section. The voltage of steel core per unit length is the same as that of aluminum strand. The current is inversely proportional to the resistance which is inversely proportional to the conductor cross section and proportional to the resistivity [14]. The ratio between the current Is of steel core and that Ia of aluminum strand is(4)λ=IsIa=AsAaρaρs,where As and ρs are the cross-sectional area and electrical resistivity of steel core, respectively, and Aa and ρa are the area and electrical resistivity of aluminum strand respectively. The rate of heat of steel core per unit volume ss is(5)ss=Pλ/λ+1As.External aluminum strands should consider solar energy [17], so the rate of heat per unit volume sa is(6)sa=P/λ+1Aa+γsSDAa.In equation (6), the first term on the right end of the equation is caused by the current, and the second term is caused by the solar radiation, P indicates its total Joule heat gain [17] which is the function of the current, average temperature, and integrated resistance rate of the conductor, S indicates the solar intensity, γs indicates the solar absorption of the conductor with a value of 0.23–0.9 and is usually valued at 0.5, and D indicates the conductor diameter.An aluminum-conductor steel-reinforced cable (ACSR) mainly loses heat by means of convective coolingPc and radiative cooling Pr. The calculation of heat loss is provided in the literature [17]. According to the principle of constant heat loss per unit volume, the composite heat transfer coefficient of conductor surface in equation (2) can be determined:(7)α=Pc+PrTsur−TaA,where A indicates the lateral surface area of conductor per unit length (that is, the interface of the conductor and the external environment).


## 3. Calculation of Conductor Stress Field

This article assumes that conductors can only be resistant to tension, ignoring their resistance to shear, bend and torsion [18], and the tensile force on each strand of the same layer is equal [19]. Three basic conditions are required when performing calculations for overhead conductors: (1) deformation compatibility; (2) material constitutive relationship; and (3) balance of internal and external forces.


### 3.1. Deformation Compatibility

Using the condition of deformation compatibility, the strain of each strand can be represented by the longitudinal strain. For a strand with a length ofSi, if unfolding it, as shown in Figure 1, the deformation compatibility equation can be obtained as follows [20]:(8)ε0Ls=ΔSisinαi+ΔRiφitanαi,where ε0 indicates the axial strain of the overhead conductor; Ls indicates the length of the overhead conductor corresponding to a strand with a length of Si; αi indicates the angle between the tangent of layer i of the strand and the conductor cross section; ΔRiφi indicates the lateral deformation of the unfolded strand; and φi indicates the ratio between projected length of the unfolded strand on the cross section and the outer radius Ri of layer i of the strand as follows:(9)φi=Sicosαi/Ri.Figure 1
Unfolding view of the strand.From the geometrical relationship, equation (10) can be obtained as follows:(10)sinαi=LsSi.Dividing the left and right sides of equation (8) by Si and simplifying it with equations (9) and (10), the relationship between the strain of layer i of the strand and the conductor ε0 is(11)εi=ε0sin2αi−ΔRiRicos2αi.


### 3.2. Lateral Deformation

From the inside to outside, the lateral deformationΔRi of conductors of layer i includes three parts: (1) the deformation Δri1′ arising from cross-section shrinkage due to the Poisson effect when the strands of layer i are stretched along their own axis [20]; (2) the deformation Δri2′ due to squeezing between layers; and (3) the deformation ΔriT′ due to thermal expansion (or contraction). Studies have shown that the lateral deformation ΔRi has a nonlinear functional relation with axial strain ε0. The calculation of ε0 with ΔRi requires iteration, which leads to complicated procedures. Furthermore, study on the lateral deformation of ACSR (JL/LB1A-300/50) found that the lateral deformation of the first two parts is very small (the lateral deformation of the first two parts is less than 1%) [21] and can be neglected compared with that of the third part. The equation of lateral deformation can be simplified as follows:(12)ΔRi=−ΔriT′.The termΔriT′ in equation (12) indicates the total deformation of layer i due to changes in temperature.(13)ΔriT′=Δr1T,i=1,Δr1T+2∑j=2i−1ΔrjT+ΔriT,i=2,n.The equation ofΔrjT (strand deformation due to thermal expansion or contraction) in equation (13) is as follows:(14)ΔrjT=αjTrjTj−20,j=1,2,…n,where αjT and Tj indicate the coefficient of thermal expansion and the average temperature of jth layer, respectively. The strain εiT due to changes in temperature is as follows:(15)εiT=αiTTi−20.


### 3.3. Balance Equation of Internal and External Force

Linear elastic material is supposed in this article. The resultant axial internal force of the conductor section can be obtained by summing up all axial components of internal forces of strands on the cross section. The internal force of each strand of layeri is as follows:(16)pi=Eiεi−εiTAi,where Ei indicates the modulus of layer i of the conductor; Ai indicates the cross-sectional area of one strand in layer i. What needs to be mentioned is that the temperature strain εiT (see equation (15)) of each layer of strands may differ due to the difference of the thermal expansion coefficient. As the temperature arises, aluminum strands with a higher thermal expansion coefficient may get zero stress, and even some of them get negative stress. As a result, all the external force is resisted by the steel strands. This may cause changes to the composite property of the overhead conductors and will be studied in detail in Section 4.2 later.The equation of the axial internal forces on the cross section can be obtained as follows:(17)Pε0=∑i=1npinisinαi,where ni indicates the number of layer i strands. Figure 2 is the diagrammatic sketch of single span overhead conductor, in which the letters “A”, “B,” and “O” represent the suspension points on the left and right and the lowest point, respectively. The span of the overhead conductor is l and the height difference is h (when suspension point B on the right is higher than A on the left, h is positive, otherwise it is negative). The ratio between h and l is denoted as β. The axial tensile external force Nx at the distance x from the left suspension point A (as shown in Figure 2) can be calculated according to the axial tensile stress at the lowest point O σ0 [18]. In equation (18), γ denotes the gravity of conductor per unit length:(18)Nx=σ0A01+tanβ−γ1−2x2σ0cosβ2.Figure 2
Diagrammatic sketch of the overhead conductors.Thus, the force balance equation of the conductor section at distance ofx from the origin can be obtained as follows:(19)fε0=Nx−Pε0=0.


## 3.1. Deformation Compatibility

Using the condition of deformation compatibility, the strain of each strand can be represented by the longitudinal strain. For a strand with a length ofSi, if unfolding it, as shown in Figure 1, the deformation compatibility equation can be obtained as follows [20]:(8)ε0Ls=ΔSisinαi+ΔRiφitanαi,where ε0 indicates the axial strain of the overhead conductor; Ls indicates the length of the overhead conductor corresponding to a strand with a length of Si; αi indicates the angle between the tangent of layer i of the strand and the conductor cross section; ΔRiφi indicates the lateral deformation of the unfolded strand; and φi indicates the ratio between projected length of the unfolded strand on the cross section and the outer radius Ri of layer i of the strand as follows:(9)φi=Sicosαi/Ri.Figure 1
Unfolding view of the strand.From the geometrical relationship, equation (10) can be obtained as follows:(10)sinαi=LsSi.Dividing the left and right sides of equation (8) by Si and simplifying it with equations (9) and (10), the relationship between the strain of layer i of the strand and the conductor ε0 is(11)εi=ε0sin2αi−ΔRiRicos2αi.


## 3.2. Lateral Deformation

From the inside to outside, the lateral deformationΔRi of conductors of layer i includes three parts: (1) the deformation Δri1′ arising from cross-section shrinkage due to the Poisson effect when the strands of layer i are stretched along their own axis [20]; (2) the deformation Δri2′ due to squeezing between layers; and (3) the deformation ΔriT′ due to thermal expansion (or contraction). Studies have shown that the lateral deformation ΔRi has a nonlinear functional relation with axial strain ε0. The calculation of ε0 with ΔRi requires iteration, which leads to complicated procedures. Furthermore, study on the lateral deformation of ACSR (JL/LB1A-300/50) found that the lateral deformation of the first two parts is very small (the lateral deformation of the first two parts is less than 1%) [21] and can be neglected compared with that of the third part. The equation of lateral deformation can be simplified as follows:(12)ΔRi=−ΔriT′.The termΔriT′ in equation (12) indicates the total deformation of layer i due to changes in temperature.(13)ΔriT′=Δr1T,i=1,Δr1T+2∑j=2i−1ΔrjT+ΔriT,i=2,n.The equation ofΔrjT (strand deformation due to thermal expansion or contraction) in equation (13) is as follows:(14)ΔrjT=αjTrjTj−20,j=1,2,…n,where αjT and Tj indicate the coefficient of thermal expansion and the average temperature of jth layer, respectively. The strain εiT due to changes in temperature is as follows:(15)εiT=αiTTi−20.


## 3.3. Balance Equation of Internal and External Force

Linear elastic material is supposed in this article. The resultant axial internal force of the conductor section can be obtained by summing up all axial components of internal forces of strands on the cross section. The internal force of each strand of layeri is as follows:(16)pi=Eiεi−εiTAi,where Ei indicates the modulus of layer i of the conductor; Ai indicates the cross-sectional area of one strand in layer i. What needs to be mentioned is that the temperature strain εiT (see equation (15)) of each layer of strands may differ due to the difference of the thermal expansion coefficient. As the temperature arises, aluminum strands with a higher thermal expansion coefficient may get zero stress, and even some of them get negative stress. As a result, all the external force is resisted by the steel strands. This may cause changes to the composite property of the overhead conductors and will be studied in detail in Section 4.2 later.The equation of the axial internal forces on the cross section can be obtained as follows:(17)Pε0=∑i=1npinisinαi,where ni indicates the number of layer i strands. Figure 2 is the diagrammatic sketch of single span overhead conductor, in which the letters “A”, “B,” and “O” represent the suspension points on the left and right and the lowest point, respectively. The span of the overhead conductor is l and the height difference is h (when suspension point B on the right is higher than A on the left, h is positive, otherwise it is negative). The ratio between h and l is denoted as β. The axial tensile external force Nx at the distance x from the left suspension point A (as shown in Figure 2) can be calculated according to the axial tensile stress at the lowest point O σ0 [18]. In equation (18), γ denotes the gravity of conductor per unit length:(18)Nx=σ0A01+tanβ−γ1−2x2σ0cosβ2.Figure 2
Diagrammatic sketch of the overhead conductors.Thus, the force balance equation of the conductor section at distance ofx from the origin can be obtained as follows:(19)fε0=Nx−Pε0=0.


## 4. Calculation of Sag and Critical Current (or Temperature)

### 4.1. Sag

Sag refers to the vertical distance between the location of actual overhead lines and the line connecting the left and right suspension points, such as the sagfM at the midpoint of the conductor (shown in Figure 2). Studies showed that the sag of overhead conductors can be approximately represented by a parabola [18]. If suspension point A on the left is regarded as the origin, the sag fx at the location x is as follows:(20)fx=fM⋅4xl−x/l2.The relation between the sagfM in the midpoint of the conductor and its arc length L is as follows [18]:(21)fM=L−lcosβ⋅3l8cos3β.If the deformation (due to the gravity and other external loads) is neglected (that is, the axial rigidity of the conductor is assumed to be infinite), its initial arc lengthL0 is as follows:(22)L0=2σoγshγl2σo2+h2.If the sagfM at the midpoint is known, the horizontal tensile stress can be calculated with the following equation:(23)σ0=γl28fMcosβ.If the sagfM at the midpoint is unknown but the height difference fAO between the left suspension point and the lowest point is known, the horizontal tensile stress can be obtained by solving the following equation:(24)σ0γ1−chγl2σ01+hLh=02+h2+fAO=0.In the operation, the arc length of the overhead line will change with the effect of current and load. If the average strain of the whole overhead conductor isε¯, the arc length L is(25)L=L01+ε¯.


### 4.2. Critical Temperature and Critical Current

Overhead conductors are generally twisted by multilayers of strands. The deformation of each strand is caused by stress or due to changes in temperature. The deformation due to changes in temperature will not cause internal force in the strands. As the coefficients of thermal expansion and temperature of each layer of strands are different, their temperature deformations are also different. With the increase of temperature, the distribution of stresses in the strands of each layer will vary. If the resultant external force on the conductor is constant, when the temperature rises, the tensile stresses of strands with greater thermal deformation (aluminum strands with higher thermal expansion coefficients) will decrease while the stress of other strands (generally steel core with lower coefficients) will increase. When the temperature reaches a certain critical value, the resultant axial forces on all aluminum strands are zero. At this point, the average temperature of the cross section and the corresponding current are defined as critical temperature (CT) and critical current (CC), respectively. Note that the CT is the same concept as the knee-point temperature [22–24]. The CC Ic and CT Tc of the conductor meet the following conditions:(26)PaII=Ic=0,(27)PaI=∑i=1naσiImiAisinαi,(28)Tc=TIc,where PaI indicates the resultant force on the aluminum strands when the current is I; mi indicates the number of aluminum strands of layer I; αi indicates the angle between the layer i aluminum strands and the cross section; Ai indicates the cross-sectional area of a single strand of the layer i aluminum strand; σi indicates the stress of layer i aluminum strands; and na indicates the total layers of aluminum strands. CT and CC are related to span, height difference, external load (including its own weight), ambient temperature, wind speed, and solar heating.


### 4.3. Calculation Process

The CT, temperature field, sag, and layered stress proposed above can be calculated according to the flowcharts shown in Figure3. The main calculative module indicates the process for calculating CT and CC. The submodules involve the calculation process of temperature field (submodule 1 in Figure 3) and the calculation process of sag and stress (submodule 2 in Figure 3). The two submodules need to be repeatedly called in the calculation of critical temperature. The calculation steps of the main module are as follows:Step 1: set the initial values of current step sizedI, number of iterations i=0, and allowable error tol. Calculate the upper and lower limits of current. Perform steps 1.1–1.3:Step 1.1: call the submodule 1 to calculate the temperature field of the conductor cross section with a given currentI=i⋅dI.Step 1.2: call the submodule 2 to calculate the layered stress of the conductor and the resultant forcePa on aluminum strands (see equation (27)).Step 1.3: ifPa≥0, i=i+1, go to Step 1.1. Otherwise, the upper and lower limits of CCI2=i⋅dI and I1=i−1⋅dI and the corresponding resultant force Pa2 and Pa1 on the aluminum strands are obtained.Step 2: calculate the CT and CC with the false position method. Perform steps 2.1∼2.3:Step 2.1: update the CCI=I2−Pa2I2−I1/Pa2−Pa1 according to its upper and lower limits.Step 2.2: ifI−I2≥tol, perform step 2.2.1∼step 2.2.3:Step 2.2.1: call the submodules 1 and 2 to calculate the resultant forcePa on the aluminum strands when the current is I.Step 2.2.2: ifPa⋅Pa2<0, update the lower limit, i.e., I1=I2,Pa1=Pa2.Step 2.2.3: update the upper limit, i.e.,I2=I,Pa2=Pa, and go to step 2.1.Step 2.3: otherwise, output the critical currentIc, critical temperature Tc, and corresponding layered stress.Figure 3
The calculation methods for critical temperature, temperature field, sag, and layered stress.The calculation of temperature field is performed with fixed point iteration, and the calculative process is as follows:Step 1: set the initial values of allowed maximum number of iterationsItmax, allowed error tol, coefficient of thermal conductivity ks for steel core and ka for aluminum strands, current I, ambient temperature Tc, and wind speed vw, respectively. Import the cross-sectional finite element mesh.Step 2: update the number of iterationsi=i+1. If i≤Itmax, continue; otherwise, the calculation fails; exit and prompt an error.Step 3: calculate the average temperatureTav of the conductor cross section and the average temperature Tsur of the conductor surface. Calculate kair, ss, sa, and α with equations (3)–(7).Step 4: solve the temperature field of the conductor cross section by using the finite element program derived from equations (1) and (2).Step 5: if the temperature fieldTi of the iteration is very different from the result Ti−1 of the previous iteration (i.e., Ti−Ti−1≥tol), perform step 2–5.Step 6: otherwise, converge the calculation and save the results.The iterative process for calculating the sag and stress is as follows:Step 1: import the geometric information of weight per unit lengthγ, span l, height difference h, and initial sag fAO. Set the initial value of the iteration number i=0 and the maximum number of iterations Itmax.Step 2: solve equation (24) to obtain the initial value of horizontal tensile stress σ0. Calculate the initial iterative values fm0 of sag at the midpoint according to equations (22) and (21), respectively.Step 3: equally divide the whole span of conductors inton−1 sections (totally n nodes). For each node j∈1,n, calculate the strain and layered stress and the resultant force Paj of aluminum strands. Perform the following steps:Step 3.1: calculate the resultant external forceNx at the node j of x with equation (18).Step 3.2: perform step 3.2.1∼step 3.2.4 (false position method) to solve equation (19).Step 3.2.1: given the upper and lower limits of the axial strainε1 and ε2, read in the cross-sectional parameters of ith layer including Ei (see equation (16)), Ai (in equation (16)), αiT (in equation (15)), αi (in equation (8)), ni (in equation (17)), and Ti (in equation (14)).Step 3.2.2: calculate the lateral deformationΔRi of each layer with equations (12)–(14). Calculate fε1 and fε2 combined with equations (11) and (15)–(19).Step 3.2.3: update the axial strainε=ε2−fε2ε2−ε1/fε2−fε1 according to the upper and lower limits of axial strain.Step 3.2.4: ifε−ε2≥tol, perform steps 3.2.4.1∼ 3.2.4.3:Step 3.2.4.1: calculatefε combined with equations (11) and (15)–(19).Step 3.2.4.2: iffεfε2<0, update the lower limit of strain: ε1=ε2,fε1=fε2.Step 3.2.4.3: update the upper limit of strain:ε2=ε,fε2=fε, go to step 3.2.2.Step 3.2.5: otherwise (i.e.,ε−ε2<tol), output the axial strain εj=ε, resultant force Paj on aluminum strands, and layered stress and strain.Step 4: calculate the averageε¯=∑j=1nεj/n of strain at n nodes. Calculate the arc length Li and sag fmi at iteration i with equations (25) and (21)Step 5: if the sagfmi at the midpoint in the iteration is very different from the result fmi−1 of the previous iteration (i.e., fmi−fmi−1≥tol), execute the steps 5.1–5.3:Step 5.1: update the iteration numberi=i+1.Step 5.2: if the iteration numberi≤Itmax go to step 3.Step 5.3: else the calculation fails, exit and prompt an error.Step 6: otherwise, (i.e.,fmi−fmi−1<tol), output the sag fmi at the midpoint, the resultant force minPaj,j=1∼n on the aluminum strands at the lowest point in the span, etc.


## 4.1. Sag

Sag refers to the vertical distance between the location of actual overhead lines and the line connecting the left and right suspension points, such as the sagfM at the midpoint of the conductor (shown in Figure 2). Studies showed that the sag of overhead conductors can be approximately represented by a parabola [18]. If suspension point A on the left is regarded as the origin, the sag fx at the location x is as follows:(20)fx=fM⋅4xl−x/l2.The relation between the sagfM in the midpoint of the conductor and its arc length L is as follows [18]:(21)fM=L−lcosβ⋅3l8cos3β.If the deformation (due to the gravity and other external loads) is neglected (that is, the axial rigidity of the conductor is assumed to be infinite), its initial arc lengthL0 is as follows:(22)L0=2σoγshγl2σo2+h2.If the sagfM at the midpoint is known, the horizontal tensile stress can be calculated with the following equation:(23)σ0=γl28fMcosβ.If the sagfM at the midpoint is unknown but the height difference fAO between the left suspension point and the lowest point is known, the horizontal tensile stress can be obtained by solving the following equation:(24)σ0γ1−chγl2σ01+hLh=02+h2+fAO=0.In the operation, the arc length of the overhead line will change with the effect of current and load. If the average strain of the whole overhead conductor isε¯, the arc length L is(25)L=L01+ε¯.


## 4.2. Critical Temperature and Critical Current

Overhead conductors are generally twisted by multilayers of strands. The deformation of each strand is caused by stress or due to changes in temperature. The deformation due to changes in temperature will not cause internal force in the strands. As the coefficients of thermal expansion and temperature of each layer of strands are different, their temperature deformations are also different. With the increase of temperature, the distribution of stresses in the strands of each layer will vary. If the resultant external force on the conductor is constant, when the temperature rises, the tensile stresses of strands with greater thermal deformation (aluminum strands with higher thermal expansion coefficients) will decrease while the stress of other strands (generally steel core with lower coefficients) will increase. When the temperature reaches a certain critical value, the resultant axial forces on all aluminum strands are zero. At this point, the average temperature of the cross section and the corresponding current are defined as critical temperature (CT) and critical current (CC), respectively. Note that the CT is the same concept as the knee-point temperature [22–24]. The CC Ic and CT Tc of the conductor meet the following conditions:(26)PaII=Ic=0,(27)PaI=∑i=1naσiImiAisinαi,(28)Tc=TIc,where PaI indicates the resultant force on the aluminum strands when the current is I; mi indicates the number of aluminum strands of layer I; αi indicates the angle between the layer i aluminum strands and the cross section; Ai indicates the cross-sectional area of a single strand of the layer i aluminum strand; σi indicates the stress of layer i aluminum strands; and na indicates the total layers of aluminum strands. CT and CC are related to span, height difference, external load (including its own weight), ambient temperature, wind speed, and solar heating.


## 4.3. Calculation Process

The CT, temperature field, sag, and layered stress proposed above can be calculated according to the flowcharts shown in Figure3. The main calculative module indicates the process for calculating CT and CC. The submodules involve the calculation process of temperature field (submodule 1 in Figure 3) and the calculation process of sag and stress (submodule 2 in Figure 3). The two submodules need to be repeatedly called in the calculation of critical temperature. The calculation steps of the main module are as follows:Step 1: set the initial values of current step sizedI, number of iterations i=0, and allowable error tol. Calculate the upper and lower limits of current. Perform steps 1.1–1.3:Step 1.1: call the submodule 1 to calculate the temperature field of the conductor cross section with a given currentI=i⋅dI.Step 1.2: call the submodule 2 to calculate the layered stress of the conductor and the resultant forcePa on aluminum strands (see equation (27)).Step 1.3: ifPa≥0, i=i+1, go to Step 1.1. Otherwise, the upper and lower limits of CCI2=i⋅dI and I1=i−1⋅dI and the corresponding resultant force Pa2 and Pa1 on the aluminum strands are obtained.Step 2: calculate the CT and CC with the false position method. Perform steps 2.1∼2.3:Step 2.1: update the CCI=I2−Pa2I2−I1/Pa2−Pa1 according to its upper and lower limits.Step 2.2: ifI−I2≥tol, perform step 2.2.1∼step 2.2.3:Step 2.2.1: call the submodules 1 and 2 to calculate the resultant forcePa on the aluminum strands when the current is I.Step 2.2.2: ifPa⋅Pa2<0, update the lower limit, i.e., I1=I2,Pa1=Pa2.Step 2.2.3: update the upper limit, i.e.,I2=I,Pa2=Pa, and go to step 2.1.Step 2.3: otherwise, output the critical currentIc, critical temperature Tc, and corresponding layered stress.Figure 3
The calculation methods for critical temperature, temperature field, sag, and layered stress.The calculation of temperature field is performed with fixed point iteration, and the calculative process is as follows:Step 1: set the initial values of allowed maximum number of iterationsItmax, allowed error tol, coefficient of thermal conductivity ks for steel core and ka for aluminum strands, current I, ambient temperature Tc, and wind speed vw, respectively. Import the cross-sectional finite element mesh.Step 2: update the number of iterationsi=i+1. If i≤Itmax, continue; otherwise, the calculation fails; exit and prompt an error.Step 3: calculate the average temperatureTav of the conductor cross section and the average temperature Tsur of the conductor surface. Calculate kair, ss, sa, and α with equations (3)–(7).Step 4: solve the temperature field of the conductor cross section by using the finite element program derived from equations (1) and (2).Step 5: if the temperature fieldTi of the iteration is very different from the result Ti−1 of the previous iteration (i.e., Ti−Ti−1≥tol), perform step 2–5.Step 6: otherwise, converge the calculation and save the results.The iterative process for calculating the sag and stress is as follows:Step 1: import the geometric information of weight per unit lengthγ, span l, height difference h, and initial sag fAO. Set the initial value of the iteration number i=0 and the maximum number of iterations Itmax.Step 2: solve equation (24) to obtain the initial value of horizontal tensile stress σ0. Calculate the initial iterative values fm0 of sag at the midpoint according to equations (22) and (21), respectively.Step 3: equally divide the whole span of conductors inton−1 sections (totally n nodes). For each node j∈1,n, calculate the strain and layered stress and the resultant force Paj of aluminum strands. Perform the following steps:Step 3.1: calculate the resultant external forceNx at the node j of x with equation (18).Step 3.2: perform step 3.2.1∼step 3.2.4 (false position method) to solve equation (19).Step 3.2.1: given the upper and lower limits of the axial strainε1 and ε2, read in the cross-sectional parameters of ith layer including Ei (see equation (16)), Ai (in equation (16)), αiT (in equation (15)), αi (in equation (8)), ni (in equation (17)), and Ti (in equation (14)).Step 3.2.2: calculate the lateral deformationΔRi of each layer with equations (12)–(14). Calculate fε1 and fε2 combined with equations (11) and (15)–(19).Step 3.2.3: update the axial strainε=ε2−fε2ε2−ε1/fε2−fε1 according to the upper and lower limits of axial strain.Step 3.2.4: ifε−ε2≥tol, perform steps 3.2.4.1∼ 3.2.4.3:Step 3.2.4.1: calculatefε combined with equations (11) and (15)–(19).Step 3.2.4.2: iffεfε2<0, update the lower limit of strain: ε1=ε2,fε1=fε2.Step 3.2.4.3: update the upper limit of strain:ε2=ε,fε2=fε, go to step 3.2.2.Step 3.2.5: otherwise (i.e.,ε−ε2<tol), output the axial strain εj=ε, resultant force Paj on aluminum strands, and layered stress and strain.Step 4: calculate the averageε¯=∑j=1nεj/n of strain at n nodes. Calculate the arc length Li and sag fmi at iteration i with equations (25) and (21)Step 5: if the sagfmi at the midpoint in the iteration is very different from the result fmi−1 of the previous iteration (i.e., fmi−fmi−1≥tol), execute the steps 5.1–5.3:Step 5.1: update the iteration numberi=i+1.Step 5.2: if the iteration numberi≤Itmax go to step 3.Step 5.3: else the calculation fails, exit and prompt an error.Step 6: otherwise, (i.e.,fmi−fmi−1<tol), output the sag fmi at the midpoint, the resultant force minPaj,j=1∼n on the aluminum strands at the lowest point in the span, etc.


## 5. Comparison of Experimental and Numerical Results

This section provides comparative analyses of the results from experiments and numerical simulation, including those for (1) cross-section temperature of ACSR JL/G1A-630/55 [25]; (2) layered stress of ACSR JL/G1A-400/35; and (3) temperature-sag and temperature-tensile force characteristics of heat-resistant aluminum alloy strands JNRLH1/G1A-400/65 and JNRLH1/G1A-630/55 with a span of 60m.


### 5.1. Layered Temperature

With reference to the experimental study [26] of conductor JL/G1A-630/55 in the different current and wind speed conditions, this article conducts numerical simulation of the conductor and sets up a finite element model, as shown in Figure 4. The model is performed with meshing by using triangular elements with a maximum size of 2.4 mm and with 8080 elements and 4184 finite element nodes. The results of numerical simulation and experiments are compared in Table 1. In the measurement of the conductor’s temperature, the surface temperature of the conductors is recorded. However, in the numerical simulation, the temperature field on the whole cross section of the conductor can be calculated and then the average temperature of strands of each layer can be obtained.Figure 4
The finite element mesh of ACSR JL/G1A-630/55/mm.Table 1
The experimental and numerical results of ACSR JL/G1A-630/55.
Wind speed (m/s)
Current (A)
Ambient temperature (°C)
Experimental value (°C)
Numerical result (°C)
Surface temperature
Steel core
Aluminum strand of inner layer
Aluminum strand of outer layer
0
922
31
70
68.34
68.34
68.01
1106
84
82.87
82.86
82.38
1290
98
100.16
100.15
99.48
1
922
33.5
42
46.90
46.90
46.62
1106
54
53.97
53.96
53.56
1290
61
62.73
62.72
62.15
2
922
33.5
41
41.50
41.50
41.25
1106
45
45.97
45.96
45.60
1290
51
51.48
51.47
50.97
4
922
33
38
37.56
37.55
37.33
1106
40
40.11
40.11
39.80
1290
45
43.22
43.21
42.78Table1 and Figure 5 show that (1) the surface temperature simulation results of the conductor at each wind speed fit well with the experimental values apart from individual data when the wind speed is 1 m/s and the current is 922 A. The simulation results are different from the experimental ones by 9.91% and the experimental data at this point are obviously unreasonable. Under the currents of 1106 A and 1290 A, when the wind speed increases from 1 m/s to 2 m/s, the surface temperature decreases by 9°C and 10°C, respectively, compared to a 1°C reduction when the current is 922 A. Therefore, the experimental value when the wind speed is 1 m/s and the current is 922 A is lower than expected. (2) The difference of the cross-sectional temperature is small (about 1°C) and the temperature of the internal strands is slightly higher than that of external strands. Therefore, the temperature gradients can be ignored in the calculation of layered stress. (3) With the increase of current, the cross-section temperature of the conductor rises, while with the increase of wind speed it drops.Figure 5
The experimental and numerical results of ACSR JL/G1A-630/55.


### 5.2. Layered Stress

The layered stress experiment on JL/G1A-400/35 ACSRs is performed through stripping, that is, stretching the conductor to a certain strain and successively stripping the conductors by layer to calculate the difference of axial force, so the layered stress of the conductor can be calculated under the strain. Only the layered stresses of the aluminum strands are studied, as the axial force is mainly resisted by the aluminum strands (more than 3 times than steel core). The experiment is performed at temperature 20°C without applying current.The calculation and test results are compared in Figure6, where the legend layer 1, 2, and 3 represent the external, middle, and internal layers of aluminum strands respectively. As shown in the figure: (1) the numerical simulation results in the elastic phase (with strain of less than 0.2%) and the experimental and numerical results agree each other with an acceptable accuracy; (2) the tensile stresses of the internal aluminum strands are slightly greater than external ones. There are still some differences between the two results (with maximum error of 17.6%) which may be because the displacement deformation of strands under tension is random as strands of each layer cannot be closely arranged without any gap in the ACSR construction. It should be noted that although layered stress does not perform well, the resultant tensile stresses fit well with the experimental results, which will be discussed later.Figure 6
The comparison of layered stress-strain relationship of ACSR JL/G1A-400/35.


### 5.3. Temperature-Sag Characteristics

The sag-temperature curves of heat-resistant aluminum alloy conductor JNRLH1/G1A-400/65 (with initial tension of 38.8 kN) and JNRLH1/G1A-630/55 (with initial tension of 48.5 kN) are compared in Figures7 and 8. The curves reflect the change law of sag (at the midpoint) with the cross-sectional temperature. The experiment is performed as follows: fix the overhead lines (with a height difference of 0) on the equipment with a span of 60 m. Then, gradually increase the temperature of the conductors by charging with electricity and measure the sag at the midpoint.Figure 7
The temperature and sag characteristics of conductor JNRLH1/G1A-400/65.Figure 8
The temperature and sag characteristics of conductor JNRLH1/G1A-630/55.Figures7 and 8 show that the numerical and experimental results of the two examples fit well (the maximum difference is 7.37% and 4.49%, respectively). With the increase of temperature, the sag increases with a rate that initially increase and then decrease. When the temperature increases to a certain value (i.e., Point A in Figures 7 and 8), the resultant tensile force on the aluminum strands reduces to zero and the sag gradually increases. According to equation (26), at point A the temperature reaches CT. The CT and CC of JNRLH1/G1A-400/65 are 87.5°C and 816.2 A, respectively, and those of JNRLH1/G1A-630/55 are 105.4°C and 142.06 A, respectively. Therefore, the slope of the sag-temperature curves of the conductors near the critical temperature is in the transition phase, i.e., the slope of the sag-temperature curves increases below and decreases above the critical temperature, but does not change greatly near the critical temperature. When the conductor temperature is close to the critical temperature, the steel core in the conductor starts to bear all the loads. However, around CT, the sag reaches a relatively large value, in which the axial force along the conductor becomes relatively lower and the rate of sag increase becomes smaller.


### 5.4. Temperature-Tension Force Characteristics

The experimental and numerical temperature-tensile force characteristics of heat-resistant aluminum alloy conductor JNRLH1/G1A-400/65 (with initial tension of 34 kN) and JNRLH1/G1A-630/55 (with initial tension of 42 kN) are compared in Figures9 and 10. The experiment is performed using the same process in Section 5.3.Figure 9
The temperature and tensile force characteristics of conductor JNRLH1/G1A-400/65.Figure 10
The temperature and tensile force characteristics of conductor JNRLH1/G1A-630/55.Figures9 and 10 show that the numerical and experimental results fit well (the maximum difference is 5.19% and 5.63%, respectively). The tensile force of the overhead conductors decreases gradually with the increase of temperature. The CT and CC are, respectively, (83.56°C, 827.64 A) and (100.96°C, 1391.86 A) for the two examples. Below the CT, the tensile force rapidly decreases as the temperature increases and gradually decreases until the critical temperature. It should be noted that the critical point cannot be directly observed according to the curve of temperature vs. tensile force but needs be calculated.


## 5.1. Layered Temperature

With reference to the experimental study [26] of conductor JL/G1A-630/55 in the different current and wind speed conditions, this article conducts numerical simulation of the conductor and sets up a finite element model, as shown in Figure 4. The model is performed with meshing by using triangular elements with a maximum size of 2.4 mm and with 8080 elements and 4184 finite element nodes. The results of numerical simulation and experiments are compared in Table 1. In the measurement of the conductor’s temperature, the surface temperature of the conductors is recorded. However, in the numerical simulation, the temperature field on the whole cross section of the conductor can be calculated and then the average temperature of strands of each layer can be obtained.Figure 4
The finite element mesh of ACSR JL/G1A-630/55/mm.Table 1
The experimental and numerical results of ACSR JL/G1A-630/55.
Wind speed (m/s)
Current (A)
Ambient temperature (°C)
Experimental value (°C)
Numerical result (°C)
Surface temperature
Steel core
Aluminum strand of inner layer
Aluminum strand of outer layer
0
922
31
70
68.34
68.34
68.01
1106
84
82.87
82.86
82.38
1290
98
100.16
100.15
99.48
1
922
33.5
42
46.90
46.90
46.62
1106
54
53.97
53.96
53.56
1290
61
62.73
62.72
62.15
2
922
33.5
41
41.50
41.50
41.25
1106
45
45.97
45.96
45.60
1290
51
51.48
51.47
50.97
4
922
33
38
37.56
37.55
37.33
1106
40
40.11
40.11
39.80
1290
45
43.22
43.21
42.78Table1 and Figure 5 show that (1) the surface temperature simulation results of the conductor at each wind speed fit well with the experimental values apart from individual data when the wind speed is 1 m/s and the current is 922 A. The simulation results are different from the experimental ones by 9.91% and the experimental data at this point are obviously unreasonable. Under the currents of 1106 A and 1290 A, when the wind speed increases from 1 m/s to 2 m/s, the surface temperature decreases by 9°C and 10°C, respectively, compared to a 1°C reduction when the current is 922 A. Therefore, the experimental value when the wind speed is 1 m/s and the current is 922 A is lower than expected. (2) The difference of the cross-sectional temperature is small (about 1°C) and the temperature of the internal strands is slightly higher than that of external strands. Therefore, the temperature gradients can be ignored in the calculation of layered stress. (3) With the increase of current, the cross-section temperature of the conductor rises, while with the increase of wind speed it drops.Figure 5
The experimental and numerical results of ACSR JL/G1A-630/55.


## 5.2. Layered Stress

The layered stress experiment on JL/G1A-400/35 ACSRs is performed through stripping, that is, stretching the conductor to a certain strain and successively stripping the conductors by layer to calculate the difference of axial force, so the layered stress of the conductor can be calculated under the strain. Only the layered stresses of the aluminum strands are studied, as the axial force is mainly resisted by the aluminum strands (more than 3 times than steel core). The experiment is performed at temperature 20°C without applying current.The calculation and test results are compared in Figure6, where the legend layer 1, 2, and 3 represent the external, middle, and internal layers of aluminum strands respectively. As shown in the figure: (1) the numerical simulation results in the elastic phase (with strain of less than 0.2%) and the experimental and numerical results agree each other with an acceptable accuracy; (2) the tensile stresses of the internal aluminum strands are slightly greater than external ones. There are still some differences between the two results (with maximum error of 17.6%) which may be because the displacement deformation of strands under tension is random as strands of each layer cannot be closely arranged without any gap in the ACSR construction. It should be noted that although layered stress does not perform well, the resultant tensile stresses fit well with the experimental results, which will be discussed later.Figure 6
The comparison of layered stress-strain relationship of ACSR JL/G1A-400/35.


## 5.3. Temperature-Sag Characteristics

The sag-temperature curves of heat-resistant aluminum alloy conductor JNRLH1/G1A-400/65 (with initial tension of 38.8 kN) and JNRLH1/G1A-630/55 (with initial tension of 48.5 kN) are compared in Figures7 and 8. The curves reflect the change law of sag (at the midpoint) with the cross-sectional temperature. The experiment is performed as follows: fix the overhead lines (with a height difference of 0) on the equipment with a span of 60 m. Then, gradually increase the temperature of the conductors by charging with electricity and measure the sag at the midpoint.Figure 7
The temperature and sag characteristics of conductor JNRLH1/G1A-400/65.Figure 8
The temperature and sag characteristics of conductor JNRLH1/G1A-630/55.Figures7 and 8 show that the numerical and experimental results of the two examples fit well (the maximum difference is 7.37% and 4.49%, respectively). With the increase of temperature, the sag increases with a rate that initially increase and then decrease. When the temperature increases to a certain value (i.e., Point A in Figures 7 and 8), the resultant tensile force on the aluminum strands reduces to zero and the sag gradually increases. According to equation (26), at point A the temperature reaches CT. The CT and CC of JNRLH1/G1A-400/65 are 87.5°C and 816.2 A, respectively, and those of JNRLH1/G1A-630/55 are 105.4°C and 142.06 A, respectively. Therefore, the slope of the sag-temperature curves of the conductors near the critical temperature is in the transition phase, i.e., the slope of the sag-temperature curves increases below and decreases above the critical temperature, but does not change greatly near the critical temperature. When the conductor temperature is close to the critical temperature, the steel core in the conductor starts to bear all the loads. However, around CT, the sag reaches a relatively large value, in which the axial force along the conductor becomes relatively lower and the rate of sag increase becomes smaller.


## 5.4. Temperature-Tension Force Characteristics

The experimental and numerical temperature-tensile force characteristics of heat-resistant aluminum alloy conductor JNRLH1/G1A-400/65 (with initial tension of 34 kN) and JNRLH1/G1A-630/55 (with initial tension of 42 kN) are compared in Figures9 and 10. The experiment is performed using the same process in Section 5.3.Figure 9
The temperature and tensile force characteristics of conductor JNRLH1/G1A-400/65.Figure 10
The temperature and tensile force characteristics of conductor JNRLH1/G1A-630/55.Figures9 and 10 show that the numerical and experimental results fit well (the maximum difference is 5.19% and 5.63%, respectively). The tensile force of the overhead conductors decreases gradually with the increase of temperature. The CT and CC are, respectively, (83.56°C, 827.64 A) and (100.96°C, 1391.86 A) for the two examples. Below the CT, the tensile force rapidly decreases as the temperature increases and gradually decreases until the critical temperature. It should be noted that the critical point cannot be directly observed according to the curve of temperature vs. tensile force but needs be calculated.


## 6. Study of Critical Temperature and Critical Current

### 6.1. Layered Stress and Sag at the Critical Point

In this section, the experiment in Section5.4 is simulated, and the relationship between the current and the layered stress (and sag) at the midpoint of overhead conductors is studied. At CT (or CC), the characteristics of the layered stress and sag are discussed. Figures 11 and 12 show the law of the current with the layered stress on the cross section at the midpoint of JNRLH1/G1A-400/65 and JNRLH1/G1A-630/55, respectively. The legend layer 1∼2 (in Figures 11 and 12) represent the internal and external layer of steel core, respectively, the legend layer 3∼4 (in Figure 11) and layer 3∼5 (in Figure 12) indicate the internal and external layers of aluminum strands, respectively. Figures 13 and 14 show relationships between their sags at the midpoint and the current, respectively. The “A” point in Figures 11–14 indicates the critical current.Figure 11
Relationship between layered stress of conductor JNRLH1/G1A-400/65 and the current: (a) stress of steel core at each layer and (b) stress of aluminum strand at each layer.
(a)
(b)Figure 12
Relationship between layered stress of conductor JNRLH1/G1A-630/55 and current: (a) stress of steel core at each layer and (b) stress of aluminum strand at each layer.
(a)
(b)Figure 13
Relationship between the sag at the midpoint of conductor JNRLH1/G1A-400/65 and the current.Figure 14
Relationship between the sag at the midpoint of conductor JNRLH1/G1A-630/55 and the current.Figures11 and 12 show that the stresses of the steel core of the internal layers are basically the same and the stresses of the aluminum strands of the external layers are basically the same too. With the increase of current, the stresses of the external aluminum strands gradually decrease and those of the external steel core decrease first and then increase. When the current is greater than the critical current, the stresses of the steel core increase rapidly. In Figure 11, the stress of the internal steel core at critical current is less than the initial stress (no current). However, its stress is far greater than the initial stress in Figure 12. It is shown (in Figures 13 and 14) that the sag (at the midpoint) gradually increases with the increase of current and that, under the critical current, they are 3.8 and 4.7 times more than the initial sag, respectively. This is because when the current is far below the CC, with the increase of current (the temperature rises), the length and sag gradually increase and the tensile force on it decreases leading to gradual reduction in the stress of the internal steel core. When the current (or temperature) is higher than the CC (or CT), the strands (such as aluminum strands) with a larger thermal expansion coefficient will be under pressure rather than tension. The strands (steel core) with a smaller thermal expansion coefficient will resist the tensile force caused by the aluminum strands under pressure but not that of external load, which leads to a rapid increase in the stress of the steel core and linear increase in the sag. It is obvious that this stress state is not conducive to the synergy of each layer of strands. In actual operation, the current (or temperature) of the conductor should be controlled to be lower than the critical current (temperature) as much as possible.


### 6.2. The Sensitive Factors of Critical Temperature and Current

As discussed in Section5.1, the operational current and temperature should be less than the critical values. As the critical temperature and current are related to layered stress and temperature field of conductors, the following factors are assumed to be sensitive: (1) solar intensity S with a unit of W/m2 and range of 0∼1000; (2) solar absorption αs, dimensionless, with a range of 0.23∼0.9, reflecting the age of the conductors; new bright conductors have an absorption rate of 0.23 and old gray conductors have an absorption rate of 0.9, and conductors in service period have an absorption rate between them; (3) ambient temperature Ta with a unit of °C and range of −20∼50; (4) wind speed vw with a unit of m/s and range of 0∼7; (5) initial tensile force F with a unit of kN and range of 0∼0.3Fmax, where Fmax indicates the ultimate bearing capacity. In other words, critical temperature and critical current have a functional relation with these variables Tcαs,Ta,vw,F,S and Icαs,Ta,vw,F,S.This section takes the heat-resistant aluminum alloy conductor JNRLH1/G1A-400/65 as an example. First, the fitting of the functional relationships (e.g., linear/quadratic/exponential functions/trigonometric functions) between critical current and the abovementioned five variables is conducted. The relationships between critical current and each of the five factors are, respectively, shown in Figures15(a)–15(e). Then, the relationship between critical current and its five factors can be postulated as equation (29) according to the fittings of the individual factor. The fitting coefficients ai,i=1∼11 with a confidence level of 95% can be obtained with the multifactor method (see Table 2 for specific data). The relationship between critical current and its influencing factors is shown in Figure 16. The two dotted lines in the figure represent the confidence interval of predicted values (with a corresponding confidence level of 95%):(29)I=a1Fa2+a3sina4vw+a5+a6αs+a7sina8Ta+a9+a10S+a11.Figure 15
Relationship between critical current and its factors. (a) Relationship between critical current and horizontal tensile force. (b) Relationship between critical current and wind speed. (c) Relationship between critical current and solar absorption. (d) Relationship between critical current and ambient temperature. (e) Relationship between critical current and solar intensity.
(a)
(b)
(c)
(d)
(e)Table 2
The values of the fitting coefficientai,i=1∼11.
Coefficients
Value
Confidence bounds
a
1
203.4001
−671.671
1078.472
a
2
0.300959
−0.38589
0.98781
a
3
7833.395
−86675.7
102342.5
a
4
0.091287
−0.50256
0.685134
a
5
0.998746
−2.53519
4.532683
a
6
−39.569
−110.065
30.92676
a
7
2386.236
−44062.9
48835.41
a
8
−0.00761
−0.09472
0.079501
a
9
1.132626
−2.66625
4.931498
a
10
−0.10697
−0.14894
–0.065
a
11
−8308.29
−107465
90848.25Figure 16
Fitting of critical current and its sensitive factors.In the same way, the change law of critical temperature with each influencing factor can be obtained (as shown in Figure17). The results show that the critical temperature has a great correlation with the initial tensile force and little correlation with other factors (i.e., they can be ignored). For function relationship fitting, the following equation can be obtained:(30)Tc=aFb+c,where a=13.54, b=0.4206, and c=24.72.Figure 17
Relationship between critical temperature and factors.Figures15 and 16 show that (1) the sensitive factors of critical current are successively wind speed, ambient temperature, horizontal tensile force, solar intensity, and solar absorption according to their significance level. (2) Horizontal tensile force and wind speed are positively correlated with critical current. Solar absorption, solar intensity, and ambient temperature are negatively correlated with it. Figure 17 shows that the critical temperature is positively correlated with horizontal tensile force and has little correlation with other factors.


## 6.1. Layered Stress and Sag at the Critical Point

In this section, the experiment in Section5.4 is simulated, and the relationship between the current and the layered stress (and sag) at the midpoint of overhead conductors is studied. At CT (or CC), the characteristics of the layered stress and sag are discussed. Figures 11 and 12 show the law of the current with the layered stress on the cross section at the midpoint of JNRLH1/G1A-400/65 and JNRLH1/G1A-630/55, respectively. The legend layer 1∼2 (in Figures 11 and 12) represent the internal and external layer of steel core, respectively, the legend layer 3∼4 (in Figure 11) and layer 3∼5 (in Figure 12) indicate the internal and external layers of aluminum strands, respectively. Figures 13 and 14 show relationships between their sags at the midpoint and the current, respectively. The “A” point in Figures 11–14 indicates the critical current.Figure 11
Relationship between layered stress of conductor JNRLH1/G1A-400/65 and the current: (a) stress of steel core at each layer and (b) stress of aluminum strand at each layer.
(a)
(b)Figure 12
Relationship between layered stress of conductor JNRLH1/G1A-630/55 and current: (a) stress of steel core at each layer and (b) stress of aluminum strand at each layer.
(a)
(b)Figure 13
Relationship between the sag at the midpoint of conductor JNRLH1/G1A-400/65 and the current.Figure 14
Relationship between the sag at the midpoint of conductor JNRLH1/G1A-630/55 and the current.Figures11 and 12 show that the stresses of the steel core of the internal layers are basically the same and the stresses of the aluminum strands of the external layers are basically the same too. With the increase of current, the stresses of the external aluminum strands gradually decrease and those of the external steel core decrease first and then increase. When the current is greater than the critical current, the stresses of the steel core increase rapidly. In Figure 11, the stress of the internal steel core at critical current is less than the initial stress (no current). However, its stress is far greater than the initial stress in Figure 12. It is shown (in Figures 13 and 14) that the sag (at the midpoint) gradually increases with the increase of current and that, under the critical current, they are 3.8 and 4.7 times more than the initial sag, respectively. This is because when the current is far below the CC, with the increase of current (the temperature rises), the length and sag gradually increase and the tensile force on it decreases leading to gradual reduction in the stress of the internal steel core. When the current (or temperature) is higher than the CC (or CT), the strands (such as aluminum strands) with a larger thermal expansion coefficient will be under pressure rather than tension. The strands (steel core) with a smaller thermal expansion coefficient will resist the tensile force caused by the aluminum strands under pressure but not that of external load, which leads to a rapid increase in the stress of the steel core and linear increase in the sag. It is obvious that this stress state is not conducive to the synergy of each layer of strands. In actual operation, the current (or temperature) of the conductor should be controlled to be lower than the critical current (temperature) as much as possible.


## 6.2. The Sensitive Factors of Critical Temperature and Current

As discussed in Section5.1, the operational current and temperature should be less than the critical values. As the critical temperature and current are related to layered stress and temperature field of conductors, the following factors are assumed to be sensitive: (1) solar intensity S with a unit of W/m2 and range of 0∼1000; (2) solar absorption αs, dimensionless, with a range of 0.23∼0.9, reflecting the age of the conductors; new bright conductors have an absorption rate of 0.23 and old gray conductors have an absorption rate of 0.9, and conductors in service period have an absorption rate between them; (3) ambient temperature Ta with a unit of °C and range of −20∼50; (4) wind speed vw with a unit of m/s and range of 0∼7; (5) initial tensile force F with a unit of kN and range of 0∼0.3Fmax, where Fmax indicates the ultimate bearing capacity. In other words, critical temperature and critical current have a functional relation with these variables Tcαs,Ta,vw,F,S and Icαs,Ta,vw,F,S.This section takes the heat-resistant aluminum alloy conductor JNRLH1/G1A-400/65 as an example. First, the fitting of the functional relationships (e.g., linear/quadratic/exponential functions/trigonometric functions) between critical current and the abovementioned five variables is conducted. The relationships between critical current and each of the five factors are, respectively, shown in Figures15(a)–15(e). Then, the relationship between critical current and its five factors can be postulated as equation (29) according to the fittings of the individual factor. The fitting coefficients ai,i=1∼11 with a confidence level of 95% can be obtained with the multifactor method (see Table 2 for specific data). The relationship between critical current and its influencing factors is shown in Figure 16. The two dotted lines in the figure represent the confidence interval of predicted values (with a corresponding confidence level of 95%):(29)I=a1Fa2+a3sina4vw+a5+a6αs+a7sina8Ta+a9+a10S+a11.Figure 15
Relationship between critical current and its factors. (a) Relationship between critical current and horizontal tensile force. (b) Relationship between critical current and wind speed. (c) Relationship between critical current and solar absorption. (d) Relationship between critical current and ambient temperature. (e) Relationship between critical current and solar intensity.
(a)
(b)
(c)
(d)
(e)Table 2
The values of the fitting coefficientai,i=1∼11.
Coefficients
Value
Confidence bounds
a
1
203.4001
−671.671
1078.472
a
2
0.300959
−0.38589
0.98781
a
3
7833.395
−86675.7
102342.5
a
4
0.091287
−0.50256
0.685134
a
5
0.998746
−2.53519
4.532683
a
6
−39.569
−110.065
30.92676
a
7
2386.236
−44062.9
48835.41
a
8
−0.00761
−0.09472
0.079501
a
9
1.132626
−2.66625
4.931498
a
10
−0.10697
−0.14894
–0.065
a
11
−8308.29
−107465
90848.25Figure 16
Fitting of critical current and its sensitive factors.In the same way, the change law of critical temperature with each influencing factor can be obtained (as shown in Figure17). The results show that the critical temperature has a great correlation with the initial tensile force and little correlation with other factors (i.e., they can be ignored). For function relationship fitting, the following equation can be obtained:(30)Tc=aFb+c,where a=13.54, b=0.4206, and c=24.72.Figure 17
Relationship between critical temperature and factors.Figures15 and 16 show that (1) the sensitive factors of critical current are successively wind speed, ambient temperature, horizontal tensile force, solar intensity, and solar absorption according to their significance level. (2) Horizontal tensile force and wind speed are positively correlated with critical current. Solar absorption, solar intensity, and ambient temperature are negatively correlated with it. Figure 17 shows that the critical temperature is positively correlated with horizontal tensile force and has little correlation with other factors.


## 7. Conclusion

This article proposes a set of numerical methods for calculating the temperature field, layered stress, and sag of overhead conductors, based on which the critical temperature and critical current are studied. The calculation methods and procedures are verified through experimental results. The authors also studied the relationship between the five most sensitive factors and the critical temperature (or current). The following conclusions are made:(1)
The results of numerical calculation and experiments for conductor temperature agree well. The temperature gradient on the cross section is relatively small; therefore, its impact on the layered stress can be ignored in the calculation. Meanwhile, the average temperature of cross sections rises with the increase of current and drops with the increase of wind speed.(2)
The numerical simulation and experimental results for layered stress in the elastic phase (with strain of less than 0.2%) agree each other with acceptable accuracy. The tensile stresses of the internal aluminum strands are slightly greater than those of the external ones.(3)
The calculated sages fit well with the experimental results. With the increase of temperature, the sag increases with a rate that first increases and then decreases, that is, when the conductor temperature increases close to the critical temperature, the sag of the conductors gradually increases. The calculated tensile forces agree well with experimental ones. Below the critical temperature, the tensile force decreases rapidly with the increase of temperature, while the changing rate decreases close to the critical temperature. The critical temperature cannot be directly observed according to the curve of temperature vs. tensile force before calculation.(4)
With the increase of current, the stresses of the steel core first decrease and then increase, while those of the aluminum monotonically decrease. When the current (or temperature) of overhead conductors is higher than the critical current (or temperature), the stresses of the steel core rapidly increase and the sag linearly increases with increasing temperature. At the critical current, the sag increases to 3–4 times of its original value. In some cases, the stresses of steel cores in the conductor are far greater than their initial value. In actual operation, the current (or temperature) of conductors should be controlled so that it remains lower than the critical current (or temperature) as much as possible.(5)
The critical temperature is positively correlated with horizontal tensile force and has little correlation with other factors. According to their relative sensitivities, the importance ranking of the influencing factors to critical current are (from high to low): wind speed, ambient temperature, horizontal tensile force, solar intensity, and solar absorption. Horizontal tensile force and wind speed are positively correlated with critical current. Solar absorption, solar intensity, and ambient temperature are negatively correlated.



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*Source: 6019493-2020-04-30.xml*