#### Question

Seven rods A, B, C, D, E, F and G are joined as shown in the figure. All the rods have equal cross-sectional area A and length *l*. The thermal conductivities of the rods are K_{A} = K_{C} = K_{0}, K_{B} = K_{D} = 2K_{0}, K_{E} = 3K_{0}, K_{F} = 4K_{0} and K_{G} = 5K_{0}. The rod E is kept at a constant temperature T_{1} and the rod G is kept at a constant temperature T_{2} (T_{2} > T_{1}). (a) Show that the rod F has a uniform temperature T = (T_{1} + 2T_{2})/3. (b) Find the rate of heat flowing from the source which maintains the temperature T_{2}.

#### Solution

Given:

K_{A} = K_{C} = K_{0}

K_{B} = K_{D} = 2K_{0}

K_{E} = 3K_{0}, K_{F} = 4K_{0}

K_{9}= 5K_{0}

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

`(K_A·A·(T-T_1))/ ( l ) + (K_c·A(T-T_1))/(l) + (K_B·A( T_2 - T ))/l`

2k_{0} ( T - T_{1 }) = 2 × 2 K_{0} ( T_{2} - T)

Temp of rod F = T = `(T_1 + 2T_2)/3`

(b) To find the rate of flow of heat from the source (rod G), which maintains a temperature T_{2} is given by

Rate of flow of heat, q = `(DeltaT)/("Thermal 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.

`R_A = l/(K_0A)`

`R_B = l/(2K_0A)`

`R_C = l /( K_0A)`

`R_D = l/(2K_0A)`

`R_{AB} = (3l)/(2k_0A)`

`R_{CD} = (3l)/(2K_0A)`

`R_"eff" = ((3l)/(2K_0A)xx(3l)/(2K_0A))/ (((3l)) /( 2K_0A) + (3l)/(2K_0A))`

`=(3l)/(4K_0A)`

`⇒ q = (DeltaT)/("R_eff")`

`= (T_1 - T_2)/((3l)/ (4K_0A))`

`⇒ (4K_0A(T_1 - T_2))/(3L)`