Question

Three rods of material x and three of material y are connected as shown in figure. All the rods are identical in length and cross-sectional area. If the end A is maintained at 60${}^{\circ }$C and the junction E at 10 ${}^{\circ }$ C, calculate the temperature of the junction B. The thermal conductivity of x is 800 W/M- ${}^{\circ }$ C and

that of y is 400 W/M- ${}^{\circ }$ C

• A. ${50}^{\circ }C$
• B. ${20}^{\circ }C$
• C. ${60}^{\circ }C$
• D. ${40}^{\circ }C$

Answer 1

The correct option is D

${40}^{\circ }C$

It is clear from the symmetry of the figure that the points C and D are equivalent in all respect and hence, they are at the same temperature, say$\theta$. No heat will flow through the rod CD. We can, therefore, neglect this rod in further analysis. Let l and A be the length and the area of cross-section of each rod. The thermal resistances of AB, BC and BD are equal. Each has a value

$R_1=\frac{1}{K_x}\frac{l}{A}$ ........(i)

Similarly, thermal resistance of CE and DE are equal, each having a value $R_2=\frac{1}{K_y}\frac{l}{A}$ ...(ii)

As the rod CD has no effect, we can say that the rods BC and CE are joined in series. Their equivalent thermal resistance is

$R_3=R_{BC}+R_{CE}=R_1+R_2$.

Also the rods BD and DE together have an equivalent thermal resistance $R_4=R_{BD}+R_{DE}=R_1+R_2$.

The resistances $R_{3}and{R}_4$ are joined in parallel and hence their equivalent thermal resistance is given by

$\frac{1}{R_5}=\frac{1}{R_3}+\frac{1}{R_4}=\frac{2}{R_3}$

or $R_5=\frac{R_3}{2}=\frac{R_1+R_2}{2}$

This resistance $R_5$ is connected in series with AB. Thus, the total arrangement is equivalent to a thermal resistance

$\frac{R_1+R_2}{2}=\frac{3R_1+R_2}{2}$

$R=R_{AB}+R_5=R_1+$

Figures below shows the successive steps in this reduction.

The heat current through A is.

$i=\frac{{\theta }_A-{\theta }_E}{R}=\frac{2({\theta }_A-{\theta }_E)}{3R_1+R_2}$

The current passes through the rod AB. We have

$i=\frac{{\theta }_A-{\theta }_B}{R_{AB}}$

or, ${\theta }_A-{\theta }_B=\left(R_{AB}\right)i$

$=R_1\frac{2({\theta }_A-{\theta }_E)}{3R_1+R_2}$

Putting from (i) and (ii),

${\theta }_A-{\theta }_B=\frac{2K_y({\theta }_A-{\theta }_E)}{K_x+3K_y}$

$=\frac{2\times 400}{800+3\times 400}\times {50}^{\circ }C={20}^{\circ }C$

or, ${\theta }_B={\theta }_A-{20}^{\circ }C={40}^{\circ }C$.