Chapter 8: Problem 23
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
Step by step solution
Understand the Problem
Express Heat Flow
Use Fourier's Law of Heat Conduction
Set up the Differential Equation
Solve the Differential Equation
Write the Final Temperature Distribution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fourier's Law of Heat Conduction
\[ \frac{dQ}{dt} = -k A \frac{dT}{dr} \]
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.
Boundary Conditions
- The inside wall, at \( r = 1 \), has a temperature of \( T = 100 \).
- The outside wall, at \( r = 2 \), has a temperature of \( T = 0 \).
Temperature Distribution
\[ \frac{dQ}{dt} = -k (2 \pi r L) \frac{dT}{dr} \]
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
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:
\[ T(r) = C_1 \ln(r) + C_2 \].