Answer to Question #231323 in Molecular Physics | Thermodynamics for Unknown346307

Question

A small electric heating application

uses wire of 2 mm diameter with 0.8 mm thick insulation (k

= 0.12 W/m°C). The heat transfer coefficient (ho) on the in￾sulated surface is 35 W/m2°C. Determine the critical thick￾ness of insulation in this case and the percentage change in

the heat transfer rate if the critical thickness is used, assum￾ing the temperature difference between the surface of the

wire and surrounding air remains unchanged.

Answer 1

Gives

$k=0.12W/m°c$

$h_0=35W/m^2°c$

$r_1=2mm\\r_2=1+0.8=2.8mm$

$\dot{Q_1}=\frac{∆t}{\frac{1}{2\pi kl}ln(\frac{r_2}{r_1})+\frac{1}{2\pi r_2l h_0}}$

$\dot{Q_1}=\frac{2\pi l∆t}{\frac{1}{k}ln(\frac{r_2}{r_1})+\frac{1}{r_2h_0}}$

Put value

$\dot{Q_1}=\frac{2\pi l∆t}{\frac{1}{0.12}ln(\frac{2.8}{2})+\frac{1}{0.0028\times35}}$

$\dot{Q_1}=\frac{2\pi l∆t}{0.02803+10.2040}$

$\dot{Q_1}=\frac{2\pi l∆t}{10.2321}$

$\dot{Q_1}=2\pi l∆t \times0.09773$

Critical radius

$$r_c=\frac{k}{h_0}=\frac{0.12}{35}=0.0034285m=3.4285mm$$

$\dot{Q_2}=\frac{2\pi l∆t}{\frac{1}{k}ln(\frac{r_c}{r_1})+\frac{1}{r_ch_0}}$

Put value

$\dot{Q_2}=\frac{2\pi l∆t}{\frac{1}{0.12}ln(\frac{3.4285}{2})+\frac{1}{0.0034285\times35}}$

$\dot{Q_2}=\frac{2\pi l∆t}{4.491+8.333}$

$\dot{Q_2}=\frac{2\pi l∆t}{12.8245}$

$\dot{Q_2}={2\pi l∆t}\times0.07797$

Percentage increase and heat loss

$$=\frac{\dot{Q_2}-\dot{Q_1}}{\dot{Q_1}}\times100=\frac{0.07797-0.09773}{0.09773}\times100=20.21$$

Heat loss=20.21%