Pipe outlet thermal losses - QuickField simulation example
Insulated water pipe with non-insulated outlet.
How to find thermal losses at non-insulated pipe outlets?
Answer Typical applications Geometry
Given
Task
Solution
Results
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).
Engineering question
Solve one quarter of the 3D structure by symmetry, calculate total heat loss with the outlet, and subtract the loss of an equivalent fully insulated pipe to obtain the additional outlet loss.
pipe outlet insulation gaps, pipeline thermal discontinuities, insulation termination regions
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Simulation problem
Problem Type
3D problem of heat transfer.
Thermal conductivity of steel λsteel=40 W/K-m;
Thermal conductivity of insulation λinsulation=0.1 W/K-m;
Water temperature Twater = 90 °C;
Air temperature Tair = -15 °C, convection coefficient: hair = 20 W/K-m²;
Temperature distribution and additional heat loss at an uninsulated outlet of an otherwise insulated hot-water pipe carrying a heated fluid..
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] = (Twater - Tair) / R,
where thermal resistance is R = ln(rsteel/rinner) / 2πλsteel + ln(rinsulation/rsteel) / 2πλinsulation + 1/2πrinsulationhair
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.
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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