A rod of length l with thermally insulated lateral surface consists of material whose heat conductivity

Coefficient varies with temperature as k=α/T, where a is a constant. The ends of the rod are kept at

Temperatures T₁ and T₂. Find the function T(x), where x is the distance from the end whose temperature is T₁, and the heat flow density.

To solve the problem of finding the temperature distribution T(x) along a rod with varying thermal conductivity, we need to apply the principles of heat conduction and differential equations. The given thermal conductivity coefficient is k = α/T, where α is a constant. Let's break down the steps to derive the temperature function and the heat flow density.

Understanding the Problem

We have a rod of length l, with one end at temperature T1 and the other at temperature T2. The thermal conductivity of the rod changes with temperature, which complicates the analysis. The goal is to find T(x), the temperature at a distance x from the end at T1, and the heat flow density, which is the rate of heat transfer per unit area.

Setting Up the Heat Equation

The heat conduction in a one-dimensional rod can be described by Fourier's law, which states that the heat flow (q) is proportional to the negative gradient of temperature:

• q = -k * (dT/dx)

Substituting the expression for k, we have:

• q = - (α/T) * (dT/dx)

Applying Steady-State Conditions

In a steady-state condition, the heat flow q is constant along the length of the rod. Therefore, we can set q = - (α/T) * (dT/dx) equal to a constant value, which we can denote as -Q:

• -Q = - (α/T) * (dT/dx)

Rearranging gives us:

• dT/dx = (α/T) * Q

Integrating the Equation

To solve for T, we can separate variables and integrate:

• ∫ T dT = ∫ αQ dx

This leads to:

• (1/2) T^2 = αQx + C

Where C is the integration constant. To find C, we can use the boundary conditions:

• At x = 0, T = T1
• At x = l, T = T2

Finding the Integration Constant

Using the boundary condition at x = 0:

• (1/2) T1^2 = C

Now substituting back into our equation:

• (1/2) T^2 = αQx + (1/2) T1^2

Rearranging gives us:

• T^2 = 2αQx + T1^2

Final Temperature Distribution

Now, we can express T as:

• T(x) = √(2αQx + T1^2)

Calculating Heat Flow Density

To find the heat flow density, we can substitute T(x) back into our expression for q:

• q = - (α/T) * (dT/dx)

Calculating dT/dx from our expression for T(x) will give us the heat flow density. The derivative can be computed using the chain rule:

• dT/dx = (1/2)(2αQ)(2αQx + T1^2)^(-1/2)

Substituting this back into the equation for q will yield the final expression for heat flow density.

Summary

In summary, we derived the temperature distribution along the rod as T(x) = √(2αQx + T1^2) and established a method to calculate the heat flow density. This approach illustrates the interplay between temperature, thermal conductivity, and heat flow in a system where conductivity varies with temperature.