One-dimensional steady state heat conduction takes place through a solid whose crosssectional area varies linearly in the direction of heat transfer. Assume there is no heat generation in the solid and the thermal conductivity of the material is constant and independent of temperature. The temperature distribution in the solid is

The correct answer is Logarithmic.

To find the temperature distribution in the solid, we can analyze the governing equation for one-dimensional steady-state heat conduction. Let the direction of heat transfer be along the x-axis.

According to Fourier's law of heat conduction, the rate of heat transfer $Q$ is given by:
$Q=-kA\left(x\right)\frac{dT}{dx}$
where:
• $k$ is the thermal conductivity of the material (constant),
• $A\left(x\right)$ is the cross-sectional area as a function of position $x$, and
• $\frac{dT}{dx}$ is the temperature gradient.

Since there is no heat generation and the process is at steady-state, the rate of heat transfer $Q$ must remain constant along the direction of heat transfer.

We are given that the cross-sectional area varies linearly in the direction of heat transfer, which can be mathematically represented as:
$A\left(x\right)=c_{1}x+c_2$
where $c_1$ and $c_2$ are constants.

Substituting the linear area equation into Fourier's law, we obtain:
$Q=-k(c_{1}x+c_2)\frac{dT}{dx}$

Rearranging the equation to separate the variables $T$ and $x$ gives:
$dT=-\frac{Q}{k(c_{1}x+c_2)}dx$

Integrating both sides with respect to their respective variables:
$\int dT=-\frac{Q}{k}\int \frac{1}{c_{1}x+c_2}dx$

Performing the integration yields:
$T\left(x\right)=-\frac{Q}{k{c}_1}ln(c_{1}x+c_2)+c_3$
where $c_3$ is the constant of integration.

This result shows that the temperature distribution $T\left(x\right)$ in the solid is logarithmic in nature.