Heat Transfer Mechanisms — Question 10

Question 10

Adding cylindrical insulation can initially increase heat loss

A long wire of radius $r_i=.00200$ m and length 1.00 m is held at 350 K in air at 300 K. Add insulation with conductivity $k=.0500$ W/(m K), outer radius $r_o$, and outer convection coefficient $h=10.0$ W/(m$^2$ K). Ignore radiation and end effects. Use $R=\ln(r_o/r_i)/(2\pi kL)+1/(h2\pi r_oL)$.

1. Compare heat loss with no insulation to that with $r_o=.00400$ m.

2. Differentiate $R(r_o)$ to find the radius at which heat loss is largest. Explain why the added conduction resistance does not always reduce total heat loss.

Question 10 - Solution

Adding cylindrical insulation can initially increase heat loss

(a) Method and calculation

$$R_{\mathrm{bare}}=\frac1{h2\pi r_iL}=7.95775\,\mathrm{K/W},\quad P_{\mathrm{bare}}=50/R_{\mathrm{bare}}=6.28319\,\mathrm W.$$$$R(.00400)=\frac{\ln2}{2\pi(.0500)}+\frac1{10(2\pi)(.00400)}=6.18523\,\mathrm{K/W}.$$$$P(.00400)=50/6.18523=8.08377\,\mathrm W.$$

(b) Interpretation and further calculation

$$\frac{dR}{dr_o}=\frac1{2\pi L}\left(\frac1{kr_o}-\frac1{hr_o^2}\right)=0\ \Rightarrow\ r_{\mathrm{crit}}=k/h=.00500\,\mathrm m.$$$$r_o<r_{\mathrm{crit}}:\ dR/dr_o<0;\qquad r_o>r_{\mathrm{crit}}:\ dR/dr_o>0.$$Resistance is smallest, and heat loss largest, at 5.00 mm. Initially the extra convecting area lowers convection resistance faster than insulation raises conduction resistance.

Physical check

The critical-radius result assumes fixed $h$ and fixed wire temperature; changing those boundary conditions changes the problem.