In a cylindrical pipe where area isn't constant, Equation 16.5 takes the form $H=-k A(d T / d r),$ where $r$ is the radial coordinate measured from the pipe axis. Use this equation to show that the heat-loss rate from a cylindrical pipe of radius $R_{1}$ and length $L$ is $$H=\frac{2 \pi k L\left(T_{1}-T_{2}\right)}{\ln \left(R_{2} / R_{1}\right)}$$ where the pipe is surrounded by insulation of outer radius $R_{2}$ and thermal conductivity $k$ and where $T_{1}$ and $T_{2}$ are the temperatures at the pipe surface and the outer surface of the insulation, respectively. (Hint: Consider the heat flow through a thin section of pipe, with thickness $d r$, as shown in Fig. 16.16 Then integrate.)
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