What is equivalent thermal conductivity of a compound rod? (and its derivation/formula)

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Solution

Dear Student,

Heat Transfer( heat current ) $\dot{Q}$ is directly proportional to the thermal conductivity of a material ( it is specific to a material ), drawing analogy from current flowing through different conductors with different current condivity we can deduce that the flow of heat through a compound system ( made with more than one material ) will be different and will depend on the effective conductivity co-efficient of the system ( same as that in current electricity )

using this knowledge and the fact

$\dot{Q} = - \frac{K A (T' - T)}{L}$
we define our thermal resistance,

$\dot{Q} = \frac{(T - T')}{\frac{L}{K A}} d r a w i n g a n a \log y f r o m I = \frac{V}{R} w e d e f i n e t h e r m a l r e s i s \tan c e R_{t h} = \frac{L}{K A} w h e r e L = l e n g t h o f c o n d u c t o r, A = a r e a o f c r o s s s e c t i o n a n d K i t s t h e r m a l c o n d u c t i v i t y$
Now there can be two situations, conductors placed end-end(series ) or one on top of another ( parallel )

(a) Series

$R_{T h_{s e r i e s}} = R_{t h_{1}} + R_{t h_{2}} (s a m e a s c u r r e n t e l e c t r i c i t y r e s i s \tan c e i n s e r i e s) \frac{L'}{K_{e q} A'} = \frac{L_{1}}{K_{1} A_{1}} + \frac{L_{2}}{K_{2} A_{2}} l e t s a s s u m e L_{1} = L_{2} = L t h u s L' = 2 L a n d A_{1} = A_{2} = A t h u s A' = A \frac{2 L}{K_{e q} A} = \frac{L}{K_{1} A} + \frac{L}{K_{2} A} \Rightarrow \frac{2}{K_{e q}} = \frac{1}{K_{1}} + \frac{1}{K_{2}} o r, K_{e q} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

(b) parallel

$\frac{1}{R_{t h_{p a r a l l e l}}} = \frac{1}{R_{t h_{1}}} + \frac{1}{R_{t h_{2}}} \frac{K_{e q} A'}{L'} = \frac{K_{1} A_{1}}{L_{1}} + \frac{K_{2} A_{2}}{L_{2}} f o r L_{1} = L_{2} = L a n d A_{1} = A_{2} = A w e h a v e \frac{K_{e q} 2 A}{L} = \frac{K_{1} A}{L} + \frac{K_{2} A}{L} o r, K_{e q} = \frac{K_{1} + K_{2}}{2}$

Regards.

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