Pipe outlet thermal losses - QuickField simulation example

Simulation problem

Geometry

Given
Thermal conductivity of steel λ_steel=40 W/K-m;
Thermal conductivity of insulation λ_insulation=0.1 W/K-m;
Water temperature T_water = 90 °C;
Air temperature T_air = -15 °C, convection coefficient: h_air = 20 W/K-m²;

Task
Calculate temperature distribution and heat losses caused by non-insulated pipe outlet.

Solution
Dew to geometrical symmetry we can simplify the model and simulate 1/4 of the full structure.
Additional losses are equal to the difference between losses in fully insulated pipe and the losses in the pipe with outlet.
Heat losses from fully insulated pipe can be calculated analytically:
q [W/m] = (T_water - T_air) / R,
where thermal resistance is R = ln(r_steel/r_inner) / 2πλ_steel + ln(r_insulation/r_steel) / 2πλ_insulation + 1/2πr_insulationh_air

Results
Analytically calculated losses of fully insulated pipe are: q = (90-(-15)) / (ln(0.105/0.1) / 2π*40 + ln(0.155/0.105) / 2π*0.1 + 1/2π*0.155*20) = 156 W/m.
For a piece of pipe L=0.8m the total losses are q * L = 156 * 0.8 = 125 W.

Temperature distribution in a pipe near the outlet. Heat flux value should be quadrupled (since we have only 1/4 of the pipe in the model).
Total heat loss in the pipe with outlet are 34.7 * 4 = 139 W.
The outlet brings additional 139-125 = 14 W of heat losses.

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