A long steam pipe of outside radius r2 is covered with thermal insulation having an outside radius of r3. The temperature of the outer surface of the pipe, T2, and the temperature of the surrounding air, T∞, are fixed. The energy loss per unit area of outside surface of the insulation is described

Question

A long steam pipe of outside radius [latex]r_{2}[/latex] is covered with thermal insulation having an outside radius of [latex]r_{3}[/latex]. The temperature of the outer surface of the pipe, [latex]T_{2}[/latex], and the temperature of the surrounding air, [latex]T_{\infty}[/latex], are fixed. The energy loss per unit area of outside surface of the insulation is described by the Newton rate equation

[latex]\frac{q_{r}}{A}=h\left(T_{3}-T_{\infty}\right)[/latex] (15-11)

Can the energy loss increase with an increase in the thickness of insulation? If possible, under what conditions will this situation arise? Figure 17.4 may be used to illustrate this composite cylinder.

Accepted Answer

In Example 3 of Chapter 15, the thermal resistance of a hollow cylindrical element was shown to be

[latex]R=\frac{\ln \left(r_{o} / r_{i} ight)}{2 \pi k L}[/latex] (17-10)

In the present example, the total difference in temperature is [latex]T_{2}-T_{\infty}[/latex] and the two resistances, due to the insulation and the surrounding air film, are

[latex]R_{2}=\frac{\ln \left(r_{3} / r_{2} ight)}{2 \pi k_{2} L}[/latex]

for the insulation, and

[latex]R_{3}=\frac{1}{h A}=\frac{1}{h 2 \pi r_{3} L}[/latex]

for the air film.

Substituting these terms into the radial heat flow equation and rearranging, we obtain

[latex]q_{r}=\frac{2 \pi L\left(T_{2}-T_{\infty} ight)}{\left[\ln \left(r_{3} / r_{2} ight) ight] / k_{2}+1 / h r_{3}}[/latex] (17-11)

The dual effect of increasing the resistance to energy transfer by conduction and simultaneously increasing the surface area as [latex]r_{3}[/latex] is increased suggests that, for a pipe of given size, a particular outer radius exists for which the heat loss is maximum. As the ratio [latex]r_{3} / r_{2}[/latex] increases logarithmically, and the term [latex]1 / r_{3}[/latex] decreases as [latex]r_{3}[/latex] increases, the relative importance of each resistance term will change as the insulation thickness is varied. In this example, [latex]L, T_{2}, T_{\infty}, k_{2} h[/latex] and [latex]r_{2}[/latex] are considered constant. Differentiating equation (17-11) with respect to [latex]r_{3}[/latex], we obtain

[latex]\frac{d q_{r}}{d r_{3}}=-\frac{2 \pi L\left(T_{2}-T_{\infty} ight)\left(\frac{1}{k_{2} r_{3}}-\frac{1}{h r_{3}^{2}} ight)}{\left[\frac{1}{k_{2}} \ln \left(\frac{r_{3}}{r_{2}} ight)+\frac{1}{h r_{3}} ight]^{2}}[/latex] (17-12)

The radius of insulation associated with the maximum energy transfer, the critical radius, found by setting [latex]d q_{r} / d r_{3}=0[/latex]; equation (17-12) reduces to

[latex]\left(r_{3} ight)_{ ext {critical }}=\frac{k_{2}}{h}[/latex] (17-13)

Question 17.2: A long steam pipe of outside radius r2 is covered with therm...

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