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Power cable parameters calculation
QuickField, enforced by ActiveField technology may be effectively used for multiphysics analysis of various engineering tasks. This analysis could be highly automated. It even can be implemented as a Microsoft Word document equipped by the set of VBA macros for automatic creation of QuickField problem, solving, postprocessing and report generation. Rather complicated example  analysis of tetracore cable  is available as ActiveField Cable Example. If you have working knowledge of Visual Basic, and understanding of QuickField Object model  you are welcome to analyze source code of these macros.
This document displays the results of cable analysis based on specific modeling parameters. Pictures, tables and graphs below have been automatically calculated by QuickField Professional Edition, controlled by VBA code implemented as MS Word macros. Corresponding QuickField problems can be analyzed by the Student Edition.
 Model description
 Input parameters
 Calculated cable parameters
 Field pictures
1. Model description.
This highvoltage tetracore cable has three triangle sectors with phase conductors and round neutral conductor in the lesser area of the crosssection above. All the conductors are made of aluminum. Each conductor is insulated and the cable as a whole has a threelayered insulation. The cable insulation consists of inner and outer insulators and a protective braiding (steel tape). The sharp corners of the phase conductors are chamfered to reduce the field crown. The corners of the conductors are rounded. Empty space between conductors is filled with some insulator, possibly with an air.
It is often required to design a cable according to parameters of the conductor section areas. Conductor section areas are defined in the Table 1. The tables 2 to 7 describe other input parameters.
2. Input parameters.
Table 1. Conductors' geometric parameters.
Phase conductor area 
120 
mm^{2} 
Neutral conductor area 
35 
mm^{2} 
Thread rounding radius (R) 
2 
mm 
Table 2. Insulator geometric parameters.
Cablecore insulation thickness 
2 
mm 
Inner cable insulation thickness 
1 
mm 
Protective steel braiding thickness 
1 
mm 
Outer cable isolation thickness 
3 
mm 
Table 3. The precision.
Areas calculation reasonable error 
0.001 
mm^{2}. 
Table 4. Conductors' loading.
Current amplitude 
200 
A 
Voltage amplitude (electrostatics) 
6500 
V 
Frequency 
50 
Hz 
Current phase (for static problems) 
0 
deg 
Table 5. Conductors' physical properties.
Relative permeability 
1 

Conductivity 
36000000 
S/m 
Thermal conductivity 
140 
W/K·m 
Young's modulo 
6.9e+10 
N/m^{2} 
Poisson's ratio 
0.33 

Coefficient of thermal expansion 
2.33·10^{5} 
1/K 
Specific density 
2700 
kg/m^{3} 
Table 6. Steel braiding physical properties.
Relative permeability 
1000 

Conductivity 
6000000 
S/m. 
Thermal conductivity 
85 
W/K·m 
Young's modulo 
2·10^{11} 
N/m^{2} 
Poisson's ratio 
0.3 

Coefficient of thermal expansion 
0.000012 
1/K 
Specific density 
7870 
kg/m^{3} 
Table 7. Insulator physical properties.

Core 
Inner 
Outer 

Relative permeability 
1 
1 
1 

Conductivity 
0 
0 
0 
S/m 
Relative electric permittivity 
2.5 
2.5 
2.5 

Thermal conductivity 
0.04 
0.04 
0.04 
W/K·m 
Young's modulo 
10000000 
10000000 
10000000 
N/m^{2} 
Poisson's ratio 
0.3 
0.3 
0.3 

Coefficient of thermal expansion 
0.0001 
0.0001 
0.0001 
1/K 
Specific density 
900 
900 
1050 
kg/m^{3} 
3. Calculated cable parameters.
Cable physical parameters are presented in the next table.
Cable outer diameter is calculated using conductor and insulator geometrical parameters put into Table 1 and Table 2. Cable linear weight per meter is calculated from geometrical parameters and specific densities of the cable components. The whole cable specific density is a total density calculated by taking into account all cable components.
Table 8. Cable physical parameters
Cable outer diameter 
42.8 
mm 
Weight (per meter) 
2.74 
kg 
Cable specific density 
1.90e+03 
kg/m^{2} 
"Conductors' capacitance" table holds self and mutualcapacitances of the cable conductors. These values are calculated in the QuickField electrostatics problem.
q_{1} = c_{11} * (U_{1}  0) + c_{12} * (U_{1}  U_{2}) + ... + c_{1n} * (U_{1}  U_{n})
q_{2} = c_{21} * (U_{2 }  U_{1}) + c_{22} * (U_{2}  0) + ... + c_{2n} * (U_{2}  U_{n})
......
q_{n} = c_{n1} * (U_{n }  U_{1}) + c_{n2} * (U_{n}  U_{2}) + ... + c_{nn} * (U_{n}  0)

Table 9. Conductors' capacitance, pF/m (lumped capacitance)

Conductor1 
Conductor2 
Conductor3 
Nullcord 
Conductor1 
170 
66.1 
9.47 
36.8 
Conductor2 
66.3 
169 
66.3 
0.413 
Conductor3 
9.45 
66.1 
170 
36.8 
Neutral cord 
37.0 
0.534 
37.0 
64.5 
Conductors' inductances are represented in the Table 10. Values in the columns 25 are calculated in the magnetostatic problem at the phase defined in the Table 4. Values in the columns 69 are calculated in AC magnetic problem. All values are calculated using the flux linkage approach by the formula: L_{ij} = F_{j} / I_{i}. The table diagonal elements represent the selfinductance values.
Table 10. Conductors' inductance, uH/m

In magnetostatic problem 
In AC magnetic problem 

C1 
C2 
C3 
0cord 
C1 
C2 
C3 
0cord 
Conductor1 
11.5 
11.2 
11.1 
11.3 
8.73 
8.47 
8.41 
8.51 
Conductor2 
11.2 
11.5 
11.2 
11.1 
8.47 
8.73 
8.47 
8.38 
Conductor3 
11.1 
11.2 
11.5 
11.3 
8.41 
8.47 
8.73 
8.51 
Neutral cord 
11.3 
11.1 
11.3 
117 
8.51 
8.38 
8.51 
8.87 
Table 11 includes the impedance and impedancelike values. In the magnetostatic problem the conductor's impedance (equal to the resistance) per meter is calculated by the formula: R = l / (ρ·S)
Joule heat per meter in magnetostatic problem is calculated by the formula: P = I_{A}^{2} · R, where I_{A} is the rootmeansquare current and R is the conductor impedance.
The conductors' impedances in AC magnetics problem are calculated using the Ohm's law as a complex ratio of the conductor's average potential divided by the conductor total current density. The real part of this ratio represents the resistance, imaginary part  reactance and the modulus  impedance. The Joule heat in the AC magnetic problem is calculated using the corresponding QuickField integral.
Table 11. Conductors' impedance.

In electrostatics problem 
In AC magnetic problem 

Conductors 
Null cord 
Conductor1 
Conductor2 
Conductor3 
Impedance, ω/m 
2.31e04 
7.94e04 
2.40e04 
2.55e04 
2.80e04 
Resistance, ω/m 
2.31e04 
7.94e04 
2.15e04 
2.37e04 
2.59e04 
Reactance, ω/m 
0.00 
0.00 
1.08e04 
9.41e05 
1.06e04 
Joule heat, W/m 
4.63 
0.00 
4.71 
4.74 
4.71 
The generated heat field is exported from the AC magnetics problem into the heat transfer problem. As a result of QuickField simulation you can see the cable exterior surface average temperature, heat flow from the cable surface and the average temperatures of all conductors. Average temperatures are relative numbers presented in Celsius assumed that ambient space temperature is 20 °C.
Table 12. Cable heat parameters
Exterior surface average temperature 
23.5 
°C 
Heat flow 
14.2 
W 
Conductors average temperature, °C 
Conductor1 
Conductor2 
Conductor3 
Nullcord 
45.9 
46.8 
45.9 
39.3 
Stress analysis problem is the utmost one that imports the temperature field from the heat transfer problem and the magnetic forces from the AC magnetic problem. Due to this magnetic and thermal loading the cable components become deformed. The numerical values of these deformations are presented in the next table.
Table 13. Stress analysis problem results.
Maximal displacement 
5.14e02 
Mm 
Maximal Mohr criteria value 
8.16e+07 
N/m^{2} 
The strength value is important for the cable fault analysis.
Table 14. The strength.
Maximal peak strength value 
8.78e+03 
A/m 
The "Strength" field is shown on a figure below as well as the "Total current density", "Energy density", "Momentary flux density", "Temperature" and "Displacement" field pictures.
4. Field pictures.
Current density in the cable
Magnetic field energy distribution in the cable
Magnetic field strength distribution in the cable
Temperature distribution in the cable
Cable mechanical deformation
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