Question 7.7.3: Radial Conductive Resistance Consider a cylindrical tube who...
Radial Conductive Resistance
Consider a cylindrical tube whose inner and outer radii are r_{i} and r_{o}. Heat flow in the tube wall can occur in the axial direction along the length of the tube and in the radial direction. If the tube surface is insulated, there will be no radial heat flow, and the heat flow in the axial direction is given by
q_{h} = \frac{k A}{L} \Delta T
where L is the length of the tube, ΔT is the temperature difference between the ends a distance L apart, and A is area of the solid cross section (see Figure 7.7.7a).
If only the ends of the tube are insulated, then the heat flow will be entirely radial. Derive an expression for the conductive resistance in the radial direction.
As shown in Figure 7.7.7b the inner and outer temperatures are T_{i} and T_{o}, and are assumed to be constant along the length L of the tube. As shown in part (c) of the figure, from Fourier’s law, the heat flow rate per unit area through an element of thickness dr is proportional to the negative of the temperature gradient dT/dr. Thus, assuming that the temperature inside the tube wall does not change with time, the heat flow rate q_{h} out of the section of thickness dr is the same as the heat flow into the section. Therefore,
\frac{q_{h}}{2 \pi r L} = −k \frac{d T}{d r}
Thus,
q_{h} = −k \frac{d T}{d r} 2 \pi r L = −2 \pi Lk \frac{d T}{d r/r}
or
\int^{r_{o}}_{r_{i}} {q_{h} \frac{d r}{r} = −2 \pi Lk} \int^{T_{o}}_{T_{i}} {d T}
Because q_{h} is constant, the integration yields
q_{h} \ln \frac{r_{o}}{r_{i}} = −2 \pi Lk(T_{o} − T_{i})
or
q_{h} = \frac{2 \pi Lk}{\ln (r_{o}/r_{i})} (T_{i} − T_{o})
The radial resistance is thus given by
R = \frac{\ln (r_{o}/r_{i})}{2 \pi L k} (1)