Question

Seven rods A, B, C, D, E, F and G are joined as shown in figure (28-E9). All the rods have equal cross-sectional area A and length l. The thermal conductivities of the rods are KA = KC = K0, KB = KD = 2K0, KE = 3K0, KF = 4K0 and KG = 5K0. The rod E is kept at a constant temperature T1 and the rod G is kept at a constant temperature T2 (T2 > T1). (a) Show that the rod F has a uniform temperature T = (T1 + 2T2)/3. (b) Find the rate of heat flowing from the source which maintains the temperature T2.
Figure

Answer 1

Explanation:

(a) At steady state both ends of F will have same temp.

$\mathrm{q}_{\mathrm{A}}+\mathrm{q}_{\mathrm{C}}=\mathrm{q}_{\mathrm{B}}+\mathrm{q}_{\mathrm{D}}$

$\frac{K_{A} A\left(T-T_{1}\right)}{l}+\frac{K_{C} A\left(T-T_{1}\right)}{l}=\frac{K_{B} A\left(T_{2}-T\right)}{l}+\frac{K_{D} A\left(T_{2}-T\right)}{l}$

$\left(T-T_{1}\right)\left(K_{A}+K_{C}\right)=\left(T_{2}-T\right)\left(K_{B}+K_{D}\right)$

Put the values of thermal conductivities

$\left(T-T_{1}\right)=2\left(T_{2}-T\right)$

$\mathrm{T}=\left(2 T_{2}+T_{1}\right) / 3$

(b) As F have the same temp. at both the ends, it acts like wheatstone bridge(no heat flows through F)

RA and RB are in series ,RC and RD are in series

Equivalent thermal resistance is$\mathrm{R}=\left(\mathrm{R}_{\mathrm{A}}+\mathrm{R}_{\mathrm{B}}\right)\left(\mathrm{R}_{\mathrm{c}}+\mathrm{R}_{\mathrm{D}}\right) /\left(\mathrm{R}_{\mathrm{A}}+\mathrm{R}_{\mathrm{B}}+\mathrm{R}_{\mathrm{C}}+\mathrm{R}_{\mathrm{D}}\right)$

(thermal resistance=l/KA)

By putting the values of  $\mathrm{R}_{\mathrm{A}}, \mathrm{R}_{\mathrm{B}}, \mathrm{R}_{\mathrm{C}}, \mathrm{R}_{\mathrm{D}}$

$R=\frac{3 l}{4 K_{0} A}$

$\mathrm{q}=\frac{\Delta T}{R}$

$\mathrm{q}=\frac{\left(T_{1}-T_{2}\right) 4 K_{0} A}{3 l}$