Heat Transfer Mechanisms — Question 4

Question 4

Cylindrical conduction depends on a logarithmic radius ratio

A cylindrical insulating shell of length 2.00 m has inner radius .0200 m, outer radius .0400 m, and conductivity .200 W/(m K). Its inner and outer surfaces are held at 360 K and 300 K. Neglect end effects. At radius $r$, use steady $P=-k(2\pi rL)dT/dr$.

1. Integrate to obtain the thermal resistance and heat rate.

2. Find the radius where temperature is 330 K. Compare it with the arithmetic midpoint radius.

Question 4 - Solution

Cylindrical conduction depends on a logarithmic radius ratio

(a) Method and calculation

$$\int_{T_i}^{T_o}dT=-\frac P{2\pi kL}\int_{r_i}^{r_o}\frac{dr}{r}.$$$$R=\frac{\ln(r_o/r_i)}{2\pi kL}=\frac{\ln2}{.8\pi}\approx.275795\,\mathrm{K/W}.$$$$P=\frac{60}{R}\approx217.553\,\mathrm W.$$

(b) Interpretation and further calculation

$$T(r)=T_i-\frac{P}{2\pi kL}\ln(r/r_i).$$$$\frac{T_i-T}{T_i-T_o}=\frac{\ln(r/r_i)}{\ln(r_o/r_i)}=\frac12\ \Rightarrow\ r=\sqrt{r_ir_o}=.0282843\,\mathrm m.$$The arithmetic radius midpoint is .0300 m; equal temperature drops correspond to equal logarithmic radius intervals.

Physical check

The transfer area increases outward, so treating the shell as a plane slab with one arbitrary surface area is not exact.