Power cable parameters calculation
QuickField simulation example
QuickField, enforced by ActiveField technology may be effectively used for multi-physics analysis of various engineering tasks. This analysis could be highly automated.
This document displays the results of cable analysis based on specific modeling parameters. Pictures, tables and graphs below have been automatically calculated by QuickField, controlled by VBA code implemented as Microsoft Word macros.
Engineering question
How to find electrical and thermal parameters of power cables?
Engineering answer
Set up a plane-parallel QuickField Multiphysics problem for a power cable system and evaluate electrical and thermal parameters from computed field results.
Typical applications
power transmission cables, underground cable systems, utility power cables
- Download simulation files (files may be viewed using any QuickField Edition).
Problem Type
Plane-parallel problem of AC magnetics.
Geometry
This high-voltage tetra-core cable has three triangle sectors with phase conductors and round neutral conductor in the lesser area of the cross-section above. All the conductors are made of aluminum. Each conductor is insulated and the cable as a whole has a three-layered 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.
| Phase conductor area | 120 mm² |
| Neutral conductor area | 35 mm² |
| Thread rounding radius (R) | 2 mm |
| Cable-core insulation thickness | 2 mm |
| Inner cable insulation thickness | 1 mm |
| Protective steel braiding thickness | 1 mm |
| Outer cable isolation thickness | 3 mm |
Given
| Current amplitude | 200 A |
| Voltage amplitude (electrostatics) | 6500 V |
| Frequency | 50 Hz |
| Conductivity | 36 MS/m |
| Thermal conductivity | 140 W/K-m |
| Young's modulo | 69 GPa |
| Poisson's ratio | 0.33 |
| Coefficient of thermal expansion | 2.33·10-5 1/K |
| Specific density | 2700 kg/m³ |
| Relative permeability | 1000 |
| Conductivity | 6 MS/m. |
| Thermal conductivity | 85 W/K-m |
| Young's modulo | 200 GPa |
| Poisson's ratio | 0.3 |
| Coefficient of thermal expansion | 0.000012 1/K |
| Specific density | 7870 kg/m³ |
| Core | Inner | Outer | |
| Relative electric permittivity | 2.5 | 2.5 | 2.5 |
| Thermal conductivity, W/K-m | 0.04 | 0.04 | 0.04 |
| Young's modulo, MPa | 10 | 10 | 10 |
| Poisson's ratio | 0.3 | 0.3 | 0.3 |
| Coefficient of thermal expansion, 1/K | 0.0001 | 0.0001 | 0.0001 |
| Specific density, kg/m³ | 900 | 900 | 1050 |
Solution
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.
| Cable outer diameter | 42.8 mm |
| Weight (per meter) | 2.74 kg |
| Cable specific density | 1900 kg/m² |
Results
Conductors' capacitance table holds self- and mutual- capacitances of the cable conductors.
| Conductor1 | Conductor2 | Conductor3 | Null-cord | |
| Conductor1 | 170 | 66.1 | 9.47 | 36.8 |
| Conductor2 | 169 | 66.3 | 0.413 | |
| Conductor3 | 170 | 36.8 | ||
| Neutral cord | 64.5 |
Conductors' inductances are calculated using the flux linkage approach by the formula: Lij = Φj / Ii. The table diagonal elements represent the self-inductance values.
| Magnetostatics | AC magnetics | |||||||
|---|---|---|---|---|---|---|---|---|
| C-1 | C-2 | C-3 | 0-cord | C-1 | C-2 | C-3 | 0-cord | |
| Conductor1 | 11.5 | 11.2 | 11.1 | 11.3 | 8.73 | 8.47 | 8.41 | 8.51 |
| Conductor2 | 11.5 | 11.2 | 11.1 | 8.73 | 8.47 | 8.38 | ||
| Conductor3 | 11.5 | 11.3 | 8.73 | 8.51 | ||||
| Neutral cord | 117 | 8.87 | ||||||
In the magnetostatics 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 magnetostatics problem is calculated by the formula: P = I² · R, where I is the root-mean-square 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. The real part of this ratio represents the resistance, imaginary part represents the reactance and the modulus represents the impedance. The Joule heat in the AC magnetic problem is calculated using the corresponding QuickField integral.
| Magnetostatics | AC magnetics | ||||
|---|---|---|---|---|---|
| Conductor | Null cord | Conductor1 | Conductor2 | Conductor3 | |
| Impedance, Ω/m | 2.31e-04 | 7.94e-04 | 2.40e-04 | 2.55e-04 | 2.80e-04 |
| Resistance, Ω/m | 2.31e-04 | 7.94e-04 | 2.15e-04 | 2.37e-04 | 2.59e-04 |
| Reactance, Ω/m | - | - | 1.08e-04 | 9.41e-05 | 1.06e-04 |
| Joule heat, W/m | 4.63 | 0 | 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.
| Exterior surface average temperature | 23.5°C |
| Heat flow | 14.2 W |
| Conductor1 | Conductor2 | Conductor3 | Null-cord |
| 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.
| Maximal displacement | 5.14e-02 mm |
| Maximal Mohr criteria value | 8.16e+07 N/m² |
- Video: Power cable parameters calculation. Watch on YouTube
- View simulation report in PDF