Theory

When the thickness is above 0, the insulation becomes the second planar layer after the shell (thicknessLayers, kappaLayers). With a conductivity of 0.06 W/(m K), 10 mm of wrap is a resistance of about 0.17 m² K/W, comparable to or larger than the outer film. The insulation also moves the outer surface outwards in the cylinder correlations, equation (2) of Shell thickness.

Because the layers are planar, the curvature of a wrap around a small pipe is not represented: a cylindrical layer has the resistance of equation (2), lower than $t/k$ when the wrap is thick relative to the radius.

Equations

R_{ins} = \frac{t_{ins}}{k_{ins}} \quad \text{(planar, as solved)}

(1)

R_{ins,cyl} = \frac{r_i}{k_{ins}}\ln\frac{r_i + t_{ins}}{r_i} \quad \text{(cylindrical, per unit inner area; for comparison)}

(2)

where	meaning	unit
$t_{ins}$	Insulation thickness (this setting)	m
$k_{ins}$	Insulation conductivity	W/(m K)
$r_i$	Radius at the inside of the insulation	m

Default and rule

Default: 0 mm (no insulation)
Rule: 0 mm.
Source: Validation: for a 0.3 m pipe with a 10 mm wrap, the planar and cylindrical treatments differ by 3.1 % in the temperature drop.

Valid range ≥ 0; 0 = no insulation layer

Thick wraps on small pipes

The planar layer overestimates the resistance of a thick wrap on a small radius.

Theory

Equations

Default and rule

Valid range ≥ 0; 0 = no insulation layer

Related 4