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Conduction Law

Seven rods A,...

Question

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

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Solution

The Temp at both ends of bar F is the same
rate of heat flow to right = rate of heat flow through left
$⟹ (Q / t)_{A} + (Q / t)_{c} = (Q / t)_{b} + (Q / t)_{D}$

$⟹ \frac{K_{A} (T_{1} - T) A}{I} + \frac{K_{c} (T_{1} - T) A}{I} = \frac{K_{B} (T - T_{2}) A}{I} + \frac{K_{D} (T - T_{2}) A}{I}$
$⟹ 2 K_{0} (T_{1} - T) = 2 \times 2 K_{0} (T - T_{2})$
$T_{1} - T = 2 T - 2 T_{2}$

$⟹ T = \frac{T_{1} + 2 T_{2}}{3}$
$(b)$ 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, $R_{C}$ and $R_{D}$ are in series

Equivalent thermal resistance is
$R = \frac{(R_{A} + R_{B}) (R_{C} + R_{D})}{(R_{A} + R_{B} + R_{C} + R_{D})}$
(thermal resistance $= \frac{I}{K A}$ )
By putting the values of $R_{A}, R_{B}, R_{C}, R_{D}$
$R = \frac{3 l}{4 K_{0} A}$

$q = \frac{Δ T}{R}$

$q = \frac{(T_{1} - T_{2}) 4 k_{0} A}{3 l}$

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