Question

Suppose the bent part of the frame of the previous problem has a thermal conductivity of 780 J s−1 m−1 °C−1 whereas it is 390 J s−1 m−1°C−1 for the straight part. Calculate the ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part.

Answer 1

The ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part is $\frac{q_{2}}{q_{1}}=\frac{12}{7}$

Explanation:

Step 1:

Given,  

Thermal conductivity for bent part $=780 \mathrm{Js}^{-1} \mathrm{m}^{-1}$ $^{\circ} \mathrm{C}^{-1}$

Thermal conductivity for straight part $=390 \mathrm{Js}^{-1} \mathrm{m}^{-1}$$^{\circ} \mathrm{C}^{-1}$

Bent part is consist of three arms so total thermal resistance of bent part is given as sum of all the parts .

Step 2:

Thermal Resistance of bent part $\mathrm{R}_{1}=(5+60+5) \times \frac{10^{-2}}{780 A}=70 \times 10^{-2} / 780 A$Thermal resistance of straight part $\mathrm{R}_{2}=60 \times 10^{-2} / 780 \mathrm{A}$

Let,

q1= rate of heat flow of bent part

q2= rate of heat flow in straight part

Step 3:

straight and bent part are parallel to each other

$\mathrm{q}_{2} \mathrm{R}_{1}=\mathrm{q}_{1} \mathrm{R}_{2}$

by  putting the value of R1 and R2 in above equation

$\frac{q_{2}}{q_{1}}=\frac{12}{7}$