Question 15.2: Consider a hollow cylindrical heat-transfer medium having in......
Consider a hollow cylindrical heat-transfer medium having inside and outside radii of r_{i} and r_{o} with the corresponding surface temperatures T_{i} and T_{o}. If the thermal-conductivity variation may be described as a linear function of temperature according to
k=k_{o}(1+\beta T)
then calculate the steady-state heat-transfer rate in the radial direction, using the above relation for the thermal conductivity, and compare the result with that using a k value calculated at the arithmetic mean temperature.
Figure 15.3 applies. The equation to be solved is now
q_{r}=-[k_{o}(1+\beta T)](2\pi r L)\frac{d T}{d r}
which, upon separation and integration, becomes
q_{r}\int_{r_{i}}^{r_{o}}{\frac{d r}{r}}=-2\pi k_{o}L\int_{T_{i}}^{T_{o}}(1+\beta T)d T
=2\pi k_{o}L\int_{T_{o}}^{T_{i}}(1+\beta T)d T
q_{r}=\ \frac{2\pi k_{o}L}{\ln r_{o}/ r_{i}}\left[T+\frac{\beta T^{2}}{2}\right]_{T_{o}}^{T_{i}}
q_{r}={\frac{2\pi k_{o}L}{\ln{r_{o}}/{r_{i}}}}\Big[1+{\frac{\beta}{2}}(T_{i}+T_{o})\Big](T_{i}-T_{o}) (15-10)
Noting that the arithmetic average value of k would be
k_{\mathrm{avg}}=k_{o}\left[1+\frac{\beta}{2}(T i+T_{o})\right]
we see that equation (15-10) could also be written as
q_{r}={\frac{2\pi k_{\mathrm{avg}}L}{\ln r_{o}/r_{i}}}(T_{i}-T_{o})
Thus, the two methods give identical results.
