Worked example: an insulated hot water main

Insulation on a hot pipe is easy to specify and easy to under-value, until you see the temperature drop on a bare section next to a lagged one. Heat leaves a pipe through three resistances in series: the film inside the pipe, the pipe wall, and whatever lags the outside. A bare steel pipe has almost nothing on the outside to slow the heat, so it loses heat quickly. Wrap it in even a modest layer of insulation and the dominant resistance moves to the lagging, cutting the loss sharply. This worked example puts the two side by side.

The setup

An 80 degree Celsius supply main runs in three 400 m segments of DN100 schedule 40 steel pipe (102.26 mm internal bore, 0.15 mm roughness), with heat transfer switched on:

• Segment one, insulated: 30 mm of mineral wool at a conductivity of 0.04 W/m.K, with an outside film coefficient of 12 W/m squared.K and an ambient of 10 degrees Celsius.
• Segment two, bare: no insulation, the same outside film of 12 W/m squared.K to the same 10 degree ambient.
• Segment three, insulated: identical to segment one.

The reservoir supplies water at 80 degrees Celsius at 60 m head, and the line ends in a small draw-off of 0.004 cubic metres per second (4 L/s). The fluid is water at 80 degrees, with its conductivity and specific heat taken at that temperature. The flow is deliberately modest, because a low flow gives the heat more time to leak out per metre and sharpens the comparison.

The physics and the method

This is the Advanced heat-transfer mode. The solver first solves the hydraulics, then builds each segment's overall heat-loss coefficient from the resistances in series: the inside convective film from the Gnielinski correlation in turbulent flow (with the laminar Nusselt-number floor applied automatically below turbulent), the conduction through the steel wall as a cylindrical shell, and, on the insulated segments, the mineral wool plus the outside film. With that coefficient, the temperature falls along each segment by the standard exponential decay of a fluid cooling to a fixed ambient, and the flow and temperature fields are solved together with an energy balance reported as a residual on every solve.

The contrast is structural. On the insulated segments the 30 mm of mineral wool is by far the largest resistance, so the heat loss is small and the temperature barely moves. On the bare segment the only resistances are the inside film and the thin steel wall, both small, so heat pours out and the temperature falls steeply, by more over that one bare segment than over both insulated segments combined.

What you learn

Solving the example gives you the temperature at each junction and the heat lost in each segment in watts, alongside the flows and pressures. Colour the network by temperature and the bare middle segment stands out as the weak point. The direct experiment is to add insulation to the bare segment, or thicken the lagging on the others, and re-solve to see how much temperature you keep. This is the calculation behind sizing lagging on a hot water or process line, set out in full on the pipe heat loss application page.

Heat transfer is part of the Advanced plan, A$50/month or A$500/year. The glossary explains the film coefficient, the overall heat-transfer coefficient and the resistances in series.

Open this example in FNS and add insulation to the bare segment to see the temperature you recover.