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Fundamental and Derived Units

One end of a ...

Question

One end of a rod of length L and cross-sectional area A is kept in a furnace of temperture $T_{1} .$ The other end of the rod is kept at a temperature $T_{2}$ . The thermal conductivity of the material of the rod is K and emissivity of the rod is e. it is given that $T_{2} = T_{S} + Δ T$ , where $Δ T << T_{S}, T_{S}$ being the temperature of the surroundings. If $Δ T \propto (T_{1} - T_{S})$ , find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is $T_{2}$ .

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Solution

Rate of heat conduction through rod = Rate of the heat lost from right end of the rod

$\frac{K A (T_{1} - T_{2})}{L} = e A σ (T_{2}^{4} - T_{s}^{4})$ ........(i)

Given that $T_{2} = T_{s} + Δ T$

$T_{2}^{4} = (T_{s} + Δ)^{4} = T_{s}^{4} (1 + \frac{Δ T}{T_{s})^{4}}$

Using binomial expansion, we have

$T_{2}^{4} = T_{s}^{4} (1 + 4 \frac{Δ T}{T_{s})}$ (as $Δ T << T_{s})$

$T_{2}^{4} - T_{s}^{4} = 4 (Δ T) (T_{s}^{3})$

Substituting in equation (i), we have

$\frac{K (T_{1} - T_{s} - Δ T)}{L} = 4 e σ T_{s}^{3} Δ T$

$⟹ \frac{K (T_{1} - T_{s})}{L} = (4 e σ T_{s}^{3} + \frac{K}{L}) Δ T$

$⟹ Δ T = \frac{K (T_{1} - T_{s})}{(4 e σ L T_{s}^{3} + K)}$

Comparing with the given relation, proportionality constant $= \frac{K}{4 e σ L T_{s}^{3} + K}$

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