heat flow in a annular cylinder of insulating material

Question

Maybe this belongs in DIY, because ... I'm trying to analyze the amount of heat required to prevent water supply pipes from freezing. The pipes would be protected by electrical heating tape and insulated by tubular pipe insulation.

I believe it's an engineering question, since the standard heat flow calculation using R-values will not work; that's because it's unclear what is the area over which the heat exchange is taking place, since the outside of the cylinder has larger area than the inside.

So, how do I determine the required wattage of pipe-heating tape per unit length of pipe, from the required temperature differential and the insulating capability of the material composing the insulation tubes ?

Answer 1

Model this as an annular cylinder of insulating material with a heat source in the center. The required heat is given by the thermal resistance of the insulation, and the temperature differential which must be maintained (the difference between ambient temperature and zero-degrees Celsius). We'll ignore 2^nd order effects such as the temperature of any water flow that's occurring, and the heat of crystallization which must be removed from the water, once it reaches freezing, before any damage is done.

Thermal resistance R has units of Cm² / W (where C is the temperature differential, m² is area, and W is watts), or more commonly (in the US) of F*ft² / btu/hr. The heat energy E required to maintain a given temperature differential is given by Eqn. 1:

(1) E = C * A / R, where A is the area over which the heat exchange takes place.

This is the calculation used for evaluating heat loss through elements of a building's envelope, for example. But in the case of the annular cylinder, it's not clear what A would be, since the areas of the inside and outside of the cylinder (with ratios r₁ and r₂, respectively) are different.

More useful is to note that the R value depends on the thickness of the insulation (e.g. fiberglass batts) so a measure of the intrinsic insulating ability of a material is thermal resistivity S, with units Cm² / Wm, where the m in the denominator represents the thickness of the material. So Eqn. 1 becomes:

(2) E = (C * A) / (S * D), where D is the thickness of the insulation.

If we note that the area A is equal to the length of the pipe L multiplied by the ill-defined effective width W of the insulation, then we can write the equation as:

(3) E = (C * L) / (S * D/W)

Again, it's not obvious what W should be, for the annular cylinder. But we CAN compute D/W, the effective ratio T of thickness to width (the aspect ratio, in some sense) of a cross-sectional slice of the cylinder. By thinking of the annular cylinder as a infinitude of infinitely-thin walled concentric cylinders, we can set up an integral:

(4) T = integral-from-r₁-to-r₂ of (dr / 2πr)

which evaluates to

(5) T = ln (r₂/r₁) / 2π

where ln is natural logarithm. Note that T is dimensionless.

So, from Eqn. 3,

(6) E = (C * L) / (S * T)

and

(7) E/L = C / (S * T)

and writing this using units commonly used in US plumbing, HVAC, etc,

(8) watts-per-foot = (32 - ambient_temp_degrees_F) / (S * T)

... where thermal resistivity S is given in awkward mixed units of degrees-F * feet / watts. This value could be computed from the R-value of a specified thickness of the material used in the pipe insulation tubes, or may be directly available for the material in question.