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 K_A = K_C = K₀, K_B = K_D = 2K₀, K_E = 3K₀, K_F = 4K₀ and K_G = 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.

Answer 1

Given:
K_A = K_C = K₀
K_B = K_D = 2K₀
K_E = 3K₀, K_F = 4K₀
K₉= 5K₀
Here, K denotes the thermal conductivity of the respective rods.

In steady state, temperature at the ends of rod F will be same.

Rate of flow of heat through rod A + rod C = Rate of flow of heat through rod B + rod D

$$\begin{gathered} \frac{K_{\mathrm{A}}·A·\left(T-T_1\right)}{l}+\frac{K_{\mathrm{C}}·A\left(T-T_1\right)}{l}=\frac{K_{\mathrm{B}}·A·\left(T_2-T\right)}{l}+\frac{K_{\mathrm{D}}·A\left(T_2-T\right)}{l} \\ 2K_0\left(T-T_1\right)\mathit{}=2\times 2K_0\left(T_2-T\right) \\ \mathrm{Temp}\mathrm{of}\mathrm{rod}\mathrm{F}=T\mathit{}=\frac{T_1+2T_2}{3} \end{gathered}$$
(b) To find the rate of flow of heat from the source (rod G), which maintains a temperature T₂ is given by
Rate of flow of heat, q = $\frac{∆T}{\mathrm{Thermal}\mathrm{resistance}}$
First, we will find the effective thermal resistance of the circuit.
From the diagram, we can see that it forms a balanced Wheatstone bridge.
Also, as the ends of rod F are maintained at the same temperature, no heat current flows through rod F.
Hence, for simplification, we can remove this branch.

From the diagram, we find that R_A and R_B are connected in series.
∴ R_AB = R_A + R_B
R_C and R_D are also connected in series.
∴ R_CD = R_C + RD
Then, R_AB and R_CD are in parallel connection.
$$\begin{gathered} R_{\mathrm{A}}=\frac{l}{K_{0}A} \\ {R}_{\mathrm{B}}=\frac{l}{2K_{0}A} \\ {R}_{\mathrm{C}}=\frac{l}{K_{0}A} \\ {R}_{\mathrm{D}}=\frac{l}{2K_{0}A} \\ {R}_{\mathrm{AB}}=\frac{3l}{2K_{0}A} \\ {R}_{\mathrm{CD}}=\frac{3l}{2K_{0}A} \\ {R}_{\mathrm{eff}}=\frac{\frac{3l}{2K_{0}A}\times \frac{3l}{2K_{0}A}}{\frac{3l}{2K_{0}A}+\frac{3l}{2K_{0}A}} \\ =\frac{3l}{4K_{0}A} \end{gathered}$$
$$\begin{gathered} \Rightarrow q=\frac{∆T}{R_{eff}} \\ =\frac{T_1-T_2}{{\displaystyle \frac{3l}{4K_{0}A}}} \\ =\frac{4K_{0}A(T_1-T_2)}{3l} \end{gathered}$$