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Magazine Chapter 15: Problem 36
A uniform solid cylinder has radius 5 and height \(10 .\) and its temperature \(u\) is proportional to the square of the dislance from its vertical axis, with \(u=100\) at the outer curved boundary of the cylinder. If the heat conductivity of the cylinder is \(K=2 .\) find the ratc of flow of heat across a concentric cylinder of radius 3 and height 10 .
Short Answer
Expert verified
The rate of heat flow across the cylinder of radius 3 is \( 960\pi \).
Step by step solution
01
Understand the Problem
We need to find the rate of flow of heat across a cylinder with a smaller radius of 3 inside the larger cylinder with a radius of 5. The temperature distribution is given by a relation to the square of the distance from the vertical axis, and the heat conductivity is provided as 2.
02
Formulate Temperature Distribution
Given that the temperature is proportional to the square of the distance from the vertical axis, we express it as \( u = kr^2 \), where \( k \) is a proportionality constant. Since \( u = 100 \) at the boundary \( r = 5 \), we have \( 100 = k \times 5^2 \). From this, \( k = 4 \).
03
Calculate the Temperature Function
Substitute the value of \( k \) back into the temperature function: \( u = 4r^2 \). This is our temperature distribution across the cylinder in terms of the radius \( r \).
04
Determine the Temperature Gradient
The temperature gradient \( abla u \) is the derivative of \( u \) with respect to \( r \), which gives us \( \frac{du}{dr} = 8r \). This gradient will help us in calculating the heat flow.
05
Calculate Rate of Heat Flow
Using Fourier's law for cylindrical coordinates, the rate of heat flow \( Q \) is given by \( Q = -K \int (abla u) \, dA \), where \( dA \) is the area element of the lateral surface: \( 2\pi r dz \). Substitute \( abla u = 8r \), \( K = 2 \), and \( r = 3 \). Integrate along the height of the cylinder (0 to 10): \[ Q = -2 \cdot \int_{0}^{10} 8 \times 3 \times 2\pi \times 3 \, dz = -2 \cdot 48\pi \int_{0}^{10} dz = -2 \cdot 48\pi \times 10 = -960\pi. \]
06
Interpret the Result
The negative sign indicates the direction of heat flow is outward from the surface at radius 3. The magnitude \( 960\pi \) represents the rate of heat flow in absolute value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Temperature Distribution
In analyzing heat flow in cylinders, understanding temperature distribution is crucial. This concept generally refers to how temperature varies across different parts of a cylinder. In our exercise, temperature distribution was characterized by the formula \( u = kr^2 \). This means that the temperature \( u \) changes outwards from the vertical axis, proportionally to the square of the distance \( r \).
Here are the key steps on how this relationship was formed:
Here are the key steps on how this relationship was formed:
- We were given that the temperature \( u \) is proportional to the square of the distance from the axis, forming \( u = kr^2 \).
- To find \( k \), utilize the boundary condition where \( u = 100 \) at \( r = 5 \). Substitute to get \( 100 = 25k \), leading to \( k = 4 \).
- Substitute \( k \) into the relationship to complete the temperature distribution: \( u = 4r^2 \).
Heat Conductivity
Heat conductivity is a measure of a material's ability to conduct heat. In our cylinder problem, the heat conductivity was given as \( K = 2 \). This number plays a key role in determining how efficiently heat is transferred within the cylinder.
- When materials have higher heat conductivity, they transfer heat more efficiently.
- The value of heat conductivity directly impacts the rate of heat flow through the cylinder, as observed in our calculations.
- In the context of Fourier's Law, a higher conductivity \( K \) results in a larger rate of heat flow for a given temperature gradient.
Fourier's Law
Fourier's Law is a fundamental principle in heat transfer. It relates the heat flow to the temperature gradient and the material's heat conductivity. The law states that the heat flow rate \( Q \) is proportional to the negative temperature gradient \( \abla u \) times the conductivity \( K \). In our problem, Fourier's Law in cylindrical coordinates was expressed as:
- \( Q = -K \int (abla u) \, dA \).
- The law describes how heat will flow from regions of high temperature to low temperature until equilibrium is met.
- The negative sign indicates that heat flows in the direction opposite to the increase of temperature.
Cylindrical Coordinates
Cylindrical coordinates are an extension of the polar coordinate system to three dimensions and are ideal for problems involving cylindrical symmetry. They emphasize radial and angular components, which is often the form naturally aligning with cylindrical problems, like ours. In the current exercise, cylindrical coordinates allowed us to express the area element for integration conveniently as \( dA = 2\pi r \, dz \).
- The radial component \( r \) measures the distance from the axis.
- The angular and vertical components are crucial for three-dimensional heat flow calculations.
- This system simplifies the expression and integration when dealing with symmetry concerning a central axis, reflecting real-world structures like pipes and cables.
