Question

Heat is escaping at a constant rate \([d Q / d t \text { in }(1.1) \text { is constant }]\) 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 \(r=1\) and temperature \(T=100\) and the outside wall has \(r=2\) and \(T=0.\)

Short answer

The temperature distribution is \[ T(r) = - \frac{100}{\ln(2)} \ln(r) + 100 \].

Step-by-step solution

Understand the Problem

The problem involves heat transfer through the walls of a cylindrical pipe. The goal is to find the temperature distribution as a function of the radial distance from the cylinder's axis, given boundary temperatures at the inside and outside walls.

Express Heat Flow

Since the heat is escaping at a constant rate, denote the heat flow per unit time and per unit length of the cylinder as \(\frac{dQ}{dt}\). This is constant.

Use Fourier's Law of Heat Conduction

Fourier's law states that the heat flow rate \(\frac{dQ}{dt}\) through a cylindrical surface is given by \[ \frac{dQ}{dt} = -k A \frac{dT}{dr} \] where \(k\) is the thermal conductivity, \(A\) is the surface area and \(\frac{dT}{dr}\) is the temperature gradient. For a cylinder, the surface area \(A = 2 \pi r L\), where \(L\) is the length of the cylinder.

Set up the Differential Equation

Substitute the surface area \(A = 2 \pi r L\) into Fourier's law: \[ \frac{dQ}{dt} = -k (2 \pi r L) \frac{dT}{dr} \] Since \(\frac{dQ}{dt}\) is constant, divide both sides by \(2 \pi k L\): \[ \frac{1}{r} \frac{dT}{dr} = C \] where \(C\) is a constant.

Solve the Differential Equation

Rearrange the equation to separate variables: \[ \frac{dT}{dr} = Cr \] Integrate both sides: \[ T(r) = C_1 \ln(r) + C_2 \] where \(C_1\) and \(C_2\) are integration constants.

Write the Final Temperature Distribution

Substitute \(C_1\) and \(C_2\) back into the general solution: \[ T(r) = - \frac{100}{\ln(2)} \ln(r) + 100 \]

Key concepts

Fourier's Law of Heat Conduction

To understand heat conduction in a cylindrical coordinate system, we must first explore Fourier's Law of Heat Conduction. This fundamental law describes how heat flows through materials. According to Fourier's Law, the heat transfer rate per unit time through a material is proportional to the negative of the temperature gradient and the area through which the heat flows. Specifically, for a cylindrical surface, the heat flow rate \( \frac{dQ}{dt} \) is expressed as:
\[ \frac{dQ}{dt} = -k A \frac{dT}{dr} \]
Here:

• \( k \) is the thermal conductivity of the material.
• \( A \) is the surface area, which in the case of a cylinder is \( 2 \pi r L \).
• \( \frac{dT}{dr} \) is the temperature gradient, indicating how temperature changes with radial distance.

By understanding this, we can begin to set up the problem involving heat conduction in cylindrical coordinates.

Boundary Conditions

Boundary conditions are crucial in determining the temperature distribution within a system. In our problem, the cylindrical pipe has specific boundary conditions defined at the inside and outside walls.

• The inside wall, at \( r = 1 \), has a temperature of \( T = 100 \).
• The outside wall, at \( r = 2 \), has a temperature of \( T = 0 \).

These conditions allow us to solve for unknown constants in the general solution of the differential equation governing heat conduction. Without proper boundary conditions, we cannot determine the unique temperature distribution within the pipe.

Temperature Distribution

The temperature distribution describes how temperature varies within the material. To find this, we use the differential equation derived from Fourier's Law of Heat Conduction. After substituting the cylindrical surface area into the law, we get:
\[ \frac{dQ}{dt} = -k (2 \pi r L) \frac{dT}{dr} \]
Assuming the heat flow rate \( \frac{dQ}{dt} \) is constant and simplifying, we arrive at:
\[ \frac{1}{r} \frac{dT}{dr} = C \]
where \( C \) is a constant. By solving this differential equation through integration, we obtain the general temperature distribution equation:
\[ T(r) = C_1 \ln(r) + C_2 \]
Applying the boundary conditions, we can solve for the constants \( C_1 \) and \( C_2 \) to find the specific temperature distribution:
\[ T(r) = - \frac{100}{\ln(2)} \ln(r) + 100 \]

Differential Equations

Differential equations are mathematical equations involving derivatives that describe how a quantity changes over time or space. In this context, the differential equation derived from Fourier's Law helps us understand how temperature varies within the cylindrical pipe.
Starting from:
\[ \frac{1}{r} \frac{dT}{dr} = C \]
we solve for \( T(r) \). This involves:

• Separating the variables to get: \( \frac{dT}{dr} = Cr \).
• Integrating to find the general solution:
  \[ T(r) = C_1 \ln(r) + C_2 \].

These steps show how we convert the physical principles of heat conduction into a solvable mathematical form. Using the boundary conditions, we determine the specific constants. This process illustrates the crucial role of differential equations in modeling and solving real-world engineering problems.