Pipe outlet thermal losses

QuickField simulation example

Insulated water pipe with non-insulated outlet.

Engineering question
How to find thermal losses at non-insulated pipe outlets?

Engineering answer
Set up a 3D QuickField Steady-state Heat Transfer problem for a pipe outlet and evaluate thermal losses from computed field results.

Typical applications
pipe outlet insulation gaps, pipeline thermal discontinuities, insulation termination regions

• Download simulation files (files may be viewed using any QuickField Edition).

Problem Type
3D problem of heat transfer.

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.