Question 17.2: A long steam pipe of outside radius r2 is covered with therm...
A long steam pipe of outside radius r_{2} is covered with thermal insulation having an outside radius of r_{3}. The temperature of the outer surface of the pipe, T_{2}, and the temperature of the surrounding air, T_{\infty}, are fixed. The energy loss per unit area of outside surface of the insulation is described by the Newton rate equation
\frac{q_{r}}{A}=h\left(T_{3}-T_{\infty}\right) (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.
In Example 3 of Chapter 15, the thermal resistance of a hollow cylindrical element was shown to be
R=\frac{\ln \left(r_{o} / r_{i}\right)}{2 \pi k L} (17-10)
In the present example, the total difference in temperature is T_{2}-T_{\infty} and the two resistances, due to the insulation and the surrounding air film, are
R_{2}=\frac{\ln \left(r_{3} / r_{2}\right)}{2 \pi k_{2} L}
for the insulation, and
R_{3}=\frac{1}{h A}=\frac{1}{h 2 \pi r_{3} L}
for the air film.
Substituting these terms into the radial heat flow equation and rearranging, we obtain
q_{r}=\frac{2 \pi L\left(T_{2}-T_{\infty}\right)}{\left[\ln \left(r_{3} / r_{2}\right)\right] / k_{2}+1 / h r_{3}} (17-11)
The dual effect of increasing the resistance to energy transfer by conduction and simultaneously increasing the surface area as r_{3} 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 r_{3} / r_{2} increases logarithmically, and the term 1 / r_{3} decreases as r_{3} increases, the relative importance of each resistance term will change as the insulation thickness is varied. In this example, L, T_{2}, T_{\infty}, k_{2} h and r_{2} are considered constant. Differentiating equation (17-11) with respect to r_{3}, we obtain
\frac{d q_{r}}{d r_{3}}=-\frac{2 \pi L\left(T_{2}-T_{\infty}\right)\left(\frac{1}{k_{2} r_{3}}-\frac{1}{h r_{3}^{2}}\right)}{\left[\frac{1}{k_{2}} \ln \left(\frac{r_{3}}{r_{2}}\right)+\frac{1}{h r_{3}}\right]^{2}} (17-12)
The radius of insulation associated with the maximum energy transfer, the critical radius, found by setting d q_{r} / d r_{3}=0; equation (17-12) reduces to
\left(r_{3}\right)_{\text {critical }}=\frac{k_{2}}{h} (17-13)