Heat is escaping at a constant rate \([d Q / d t\) in \((1.1)\) is constant] through the walls of a long cylindrical pipe. Find the temperature \(T\) at a distance \(r\) from the axis of the cylinder if the inside wall has radius \(r=\mathrm{I}\) and temperature \(T=100\) and the outside wall has \(r=2\) and \(T=0\)

The heat equation is a vital part of understanding heat conduction, especially in solid objects like a cylindrical pipe. It describes how heat diffuses through a material. For a steady-state condition, where the temperature doesn't change over time, the heat equation in cylindrical coordinates is:

\[ \frac{d}{dr} \bigg(k r \frac{dT}{dr} \bigg) = 0 \]
The variable \( T \) represents temperature, \( r \) is the radial distance from the axis of the cylinder, and \( k \) is the thermal conductivity, which remains constant in this problem. When dealing with cylindrical coordinates, the equation expands to incorporate the cylindrical geometry, accounting for the curvature and radius. This form of the equation is essential for correctly modeling heat transfer in cylindrical objects.

Cylindrical coordinates are used for problems involving symmetrical geometries around a central axis, like pipes or cylinders. Instead of using \( x \), \( y \), and \( z \), cylindrical coordinates use \( r \) (radius), \( \theta \) (angle), and \( z \) (height). This simplifies the mathematics when dealing with cylindrical shapes.

In the heat equation given prior, \( r \) signifies the radial distance. We focus on how temperature varies with radius because angles (\( \theta \)) and height (\( z \)) are not significant in this problem. This reduction makes the problem one-dimensional, significantly simplifying our calculations.

A crucial concept in heat conduction is the temperature gradient, which describes how temperature changes with position. In our cylindrical problem, we are interested in how temperature \( T \) varies with the radius \( r \).

Using the heat equation, we introduce a substitution \( u = \frac{dT}{dr} \). This converts our differential equation into a simpler form: \[ \frac{d}{dr}(r u) = 0 \]
This transformation shows that the radial component multiplied by the temperature gradient remains constant. Solving this means integrating with respect to \( r \), giving us the general temperature profile. By integrating again and applying boundary conditions, we find the temperature distribution that adheres to the physical constraints of the system.

Boundary conditions are essential for solving differential equations as they provide specific values that the solution must satisfy at certain points. In this problem, the conditions given are:

• The temperature at the inside wall of the cylinder (radius \( r = 1 \)) is 100°C.
• The temperature at the outside wall of the cylinder (radius \( r = 2 \)) is 0°C.

Using these conditions helps determine the constants in the general solution of the differential equation. For \( r = 1 \), along with \( \text{ln}(1) = 0 \), we find that the integration constant \( D \) equals 100. For \( r = 2 \), setting \( T = 0 \) allows us to solve for the constant \( C \).

Applying these boundary conditions ensures our solution is physically meaningful, accurately depicting the temperature distribution within the cylindrical pipe.