Question

Consider the situation shown in figure (28-E5). The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross section of the bent part if the total heat taken out per second from the end at 100°C is 130 J.
Figure

Answer 1

Explanation:

Step 1:

Thermal resistance of bent part is

$\mathrm{R}_{1}=(5+60+5) / \mathrm{KA}=70 / \mathrm{KA}$

Thermal resistance of straight part is

$\mathrm{R}_{2}=60 / \mathrm{KA}$

Step 2:

Let q1 and q2 be the rate of heat flow through R1 and R2 respectively,

Then total amount of rate of heat flow is $\mathrm{q}=\mathrm{q}_{1}+\mathrm{q}_{2}$

As R1 and R2 are parallel to each other $\mathrm{q}_{2} \mathrm{R}_{2}=\mathrm{q}_{2} \mathrm{R}_{1}$

$\mathrm{q}_{2}(70 / \mathrm{KA})=\mathrm{q}_{1}(60 / \mathrm{KA})$

$q_{1}=7 q_{2} / 6$

Step 3:

by using $\mathrm{q}=\mathrm{q}_{1}+\mathrm{q}_{2}$

$q=q_{2}+\frac{7 q_{2}}{6}$

$q=\frac{13 q_{2}}{6}$

given  $\mathrm{q}=130 \mathrm{J} / \mathrm{s}$    

$\mathrm{q}_{2}=60 \mathrm{J} / \mathrm{s}$