A rod of length L with sides fully insulated | vedclass

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Question

(A) The heat current $H$ through a rod of cross-sectional area $A$ is given by $H = -KA \frac{dT}{dx}$.Since the rod is insulated,$H$ is constant alon

Vedclass · Accepted Answer

$T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}}$

Vedclass · Answer

(A) The heat current $H$ through a rod of cross-sectional area $A$ is given by $H = -KA \frac{dT}{dx}$.Since the rod is insulated,$H$ is constant along the length.Substituting $K = \frac{\alpha}{T}$,we get $H = -\frac{\alpha A}{T} \frac{dT}{dx}$.Rearranging the terms: $\frac{H}{\alpha A} dx = -\frac{dT}{T}$.Integrating from $x=0$ to $x$ and $T=T_1$ to $T$:$\int_0^x \frac{H}{\alpha A} dx = -\int_{T_1}^T \frac{dT}{T}$.$\frac{H}{\alpha A} x = -(\ln T - \ln T_1) = \ln \frac{T_1}{T}$.At $x=L$,$T=T_2$,so $\frac{H}{\alpha A} L = \ln \frac{T_1}{T_2}$.Thus,$\frac{H}{\alpha A} = \frac{1}{L} \ln \frac{T_1}{T_2}$.Substituting this back: $\frac{x}{L} \ln \frac{T_1}{T_2} = \ln \frac{T_1}{T}$.$\ln \left( \frac{T_1}{T_2} \right)^{\frac{x}{L}} = \ln \frac{T_1}{T}$.Taking the exponential of both sides: $\left( \frac{T_1}{T_2} \right)^{\frac{x}{L}} = \frac{T_1}{T}$.Therefore,$T = T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}}$.

Similar Questions

Two rods of the same length and same cross-sectional area are joined in series. The temperatures at the two ends are as shown in the figure. As we move along the rod,the temperature variation is as shown in the following graph.

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