3-phase cable thermal analysis, CIGRE TB 880 Case #0-1
Temperature field of a three-phase cable computed for the rated current and laying conditions given in CIGRE TB 880 Case #0-1.
How can the temperature distribution be computed for the rated current given in CIGRE TB 880 Case #0-1?
Build a plane-parallel steady-state heat transfer problem, convert the conductor, sheath and dielectric losses from CIGRE TB 880 Table 2 into volumetric heat sources using the cross-sectional areas measured in the model, and solve for the temperature field.
- Three-phase underground cables
- Trefoil-arranged cables
- Buried high-voltage cable systems
Simulation Problem
- Problem Type
- Plane-parallel steady-state heat transfer
- Geometry
- 132 kV XLPE-insulated three-core cable, copper stranded conductor 630 mm², buried at 1 m depth, layers from inside out: copper stranded conductor, inner semiconducting layer, XLPE insulation, outer semiconducting layer, aluminum sheath, HDPE outer jacket.
- Given
- Soil temperature far from the cable 20°C
- Thermal conductivity: XLPE insulation 0.285714, semiconducting layer 0.4, aluminum 237, copper 401, soil 1 W/(m·K)
- Conductor loss Wc = 31.0365 W/m (current I = 886.17 A)
- Sheath loss Ws = 2.4117 W/m
- Dielectric loss Wd = 0.3851 W/m (at 132 kV)
- Task
- Compute the temperature field.
- Solution
The real conductor is made of many stranded wires; the model uses a solid conductor in place of the stranded conductor.
The volumetric heat generation rate is obtained by dividing each component's loss by its cross-sectional area as measured in the QuickField model:
Conductor Qc = Wc / Sc = 31.0365 W/m ÷ 721×10⁻⁶ m² = 43046 W/m³
Sheath Qs = Ws / Ss = 2.4117 W/m ÷ 170.15×10⁻⁶ m² = 14200 W/m³
Dielectric (including semiconducting layers) Qd = Wd / Sd = 0.3851 W/m ÷ 2794.1×10⁻⁶ m² = 137.83 W/m³
- Results
The computed maximum conductor temperature is close to the CIGRE reference value of 90°C.
Reference: CIGRE Working Group B1.56, Power Cable Rating Examples for Calculation Tool Verification (TB 880).
