Civil Engineering Example: Temperature In A Concrete Slab (18.9) - Separation of Variables, Use of Fourier Series
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Civil Engineering Example: Temperature in a Concrete Slab

Civil Engineering Example: Temperature in a Concrete Slab

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Interactive Audio Lesson

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Understanding Temperature Distribution

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Teacher
Teacher Instructor

Today, we're discussing how temperature distributions can be modeled in materials like concrete. We will focus on a one-dimensional slab.

Student 1
Student 1

What factors influence the temperature distribution in a concrete slab?

Teacher
Teacher Instructor

Great question! Factors include the initial temperature, external environment, and the material's thermal properties. For our example, we're assuming the slab is initially described by a function.

Student 2
Student 2

Can you explain how we set up this initial temperature function?

Teacher
Teacher Instructor

Certainly! The initial condition is defined as $$u(x,0) = f(x) = 100 imes ext{sin} igg( rac{3 ext{π}x}{10}igg)$$. This function defines how the temperature varies along the length of the slab.

Student 3
Student 3

What do those parameters mean?

Teacher
Teacher Instructor

The coefficient 100 represents the amplitude of the temperature distribution, and the argument of the sine function accounts for the spatial variation, with 3 being the frequency.

Student 4
Student 4

So, is the temperature uniform across the slab initially?

Teacher
Teacher Instructor

Not at all! The sine function suggests a variation, peaking and dipping across the slab.

Teacher
Teacher Instructor

To summarize, ensure you understand both how we define initial conditions and their significance in modeling.

Boundary Conditions and Separation of Variables

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Teacher
Teacher Instructor

Now let's consider the boundary conditions applied to our slab, which are crucial for solving the problem.

Student 1
Student 1

What are these boundary conditions exactly?

Teacher
Teacher Instructor

In this case, we have insulated boundaries: $$u(0,t)=0$$ and $$u(10,t)=0$$, meaning no heat escapes at either end.

Student 2
Student 2

How does that affect our equation?

Teacher
Teacher Instructor

The separation of variables allows us to write our solution as a product $$u(x,t) = X(x)T(t)$$, where $$X(x)$$ and $$T(t)$$ are functions of space and time, respectively. This greatly simplifies our partial differential equation.

Student 3
Student 3

How do we actually separate these variables?

Teacher
Teacher Instructor

After substituting into the PDE and rearranging, we arrive at equations that depend on single variables $$(T(t))$$ and $$(X(x))$$, allowing us to solve them independently.

Student 4
Student 4

And how do we select values for $$λ$$?

Teacher
Teacher Instructor

By applying boundary conditions to the spatial equation, we can find the eigenvalues and corresponding eigenfunctions.

Interpreting the Solution

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Teacher
Teacher Instructor

Let's take a closer look at our final solution: $$u(x,t)=100 ext{sin}igg( rac{3πx}{10}igg)e^{-α^2( rac{3π}{10})^2t}$$.

Student 1
Student 1

What does this solution tell us about temperature over time?

Teacher
Teacher Instructor

As time increases, the term $$e^{-α^2( rac{3π}{10})^2t}$$ shows that the amplitude of our temperature wave decays exponentially.

Student 2
Student 2

Why does it approach zero?

Teacher
Teacher Instructor

Because eventually, as heat dissipates, the slab will reach a steady state where temperature is uniform across the slab—essentially zero, considering our boundary conditions.

Student 3
Student 3

So the mode shape stays the same even if temperature drops?

Teacher
Teacher Instructor

Exactly! The shape of our sine wave remains constant, reflecting the spatial distribution of temperature. It's a consistent pattern, just diminishing in intensity.

Teacher
Teacher Instructor

To wrap it all up, our analysis illustrated how Fourier expansions assist in solving PDEs for practical applications like heat transfer in concrete.

Introduction & Overview

Read summaries of the section's main ideas at different levels of detail.

Quick Overview

This section discusses the temperature distribution in a concrete slab using the separation of variables method in partial differential equations.

Standard

In this section, we analyze the temperature distribution in a concrete slab of length 10 meters with insulated boundaries. We use the Fourier sine series through separation of variables to derive the solution, demonstrating how the temperature wave decays over time as it approaches a steady state.

Detailed

Civil Engineering Example: Temperature in a Concrete Slab

In this section, we examine a practical application of the separation of variables method within the context of a one-dimensional heat conduction problem. The problem involves a concrete slab of length L = 10 m with an initial temperature distribution described by the function:

$$u(x,0) = f(x) = 100 imes ext{sin}igg( rac{3 ext{π}x}{10}igg)$$

This represents the temperature across the slab at time t = 0, while it is indicated that the boundaries of the slab are insulated, which leads to the boundary conditions:

  • $$u(0,t) = 0$$
  • $$u(10,t) = 0$$

Application of Separation of Variables

By employing the separation of variables, we derive that only the n=3 term is significant in the Fourier sine expansion. Therefore, our solution can be formatted as:

$$u(x,t) = 100 imes ext{sin}igg( rac{3 ext{π}x}{10}igg) e^{-α^2( rac{3 ext{π}}{10})^2t}$$

Interpretation of Results

This solution illustrates a decaying temperature wave within the slab. As time approaches infinity (t → ∞), the temperature within the slab converges to zero, indicating that a steady state has been reached. Despite the decay in amplitude, the spatial mode shape remains constant, demonstrating how temperature distribution evolves over time. The layered breakdown of how this mathematical modeling is applied in civil engineering highlights the utility of separation of variables and Fourier series in solving partial differential equations effectively.

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Audio Book

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Problem Statement

Chapter 1 of 3

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Chapter Content

A concrete slab of length L=10m initially has a temperature distribution given by:

\( u(x,0) = f(x) = 100 \sin\left(\frac{3\pi x}{10}\right) \text{ with insulated boundaries } (u(0,t)=u(10,t)=0).

Detailed Explanation

The problem addresses the temperature distribution in a concrete slab, specifically one that is 10 meters long. Initially, the temperature at any point x along the length is determined by a sine function, which shows that the temperature varies along the length of the slab. The boundaries of the slab are insulated, meaning that no heat can enter or leave the slab at its ends (at x=0 and x=10 meters). Therefore, at these points, the temperature is fixed at zero.

Examples & Analogies

Imagine a long, heated metal rod that you leave outside during winter. Initially, the middle part of the rod is warm, while the ends are cold because they are in contact with the freezing air. If you insulate the ends by wrapping them, no heat can escape that way. The temperature on the rod, like on our slab, would create a gradient where the center is warmest, similar to our sine function representing temperature distribution.

Using Separation of Variables

Chapter 2 of 3

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Chapter Content

Using Separation of Variables:

We identify that:
- Only the n=3 term is non-zero in Fourier sine expansion.
- Thus, the solution is:

\( u(x,t) = 100 \sin\left(\frac{3\pi x}{10}\right) e^{-\alpha^2\left(\frac{3\pi}{10}\right)^2 t} \)

Detailed Explanation

In this chunk, we use the method of separation of variables to solve our heat equation. By translating the temperature distribution into a Fourier sine series, we discover that only the third harmonic (n=3) contributes to the solution because it's the only term that remains significant in this context. The resulting solution shows how temperature changes over time; the sine function indicates the shape of the temperature distribution along the slab, and the exponential part signifies that this distribution gradually decays over time as heat dissipates.

Examples & Analogies

Think of it like a guitar string – when you pluck it, the string vibrates in various modes. However, depending on how you pluck it, some vibrations become dominant while others fade. Here, just like how a specific mode stands out in the sound, the third harmonic describes how the temperature profile behaves over time.

Interpretation of the Solution

Chapter 3 of 3

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Chapter Content

Interpretation:
- This represents a decaying temperature wave.
- As t→∞, \( u(x,t)→0 \) (steady state reached).
- The mode shape (spatial distribution) does not change; only amplitude decays.

Detailed Explanation

The interpretation of the solution reveals key behaviors of the temperature profile over time. The decaying wave indicates that as time progresses, the amplitude of the temperature distribution decreases, eventually reaching a steady state where the temperature across the slab becomes uniform (essentially zero due to insulation). The shape of the temperature function remains consistent throughout this process, suggesting that the ‘shape’ of the thermal response (the sine wave) does not change, even though its intensity diminishes.

Examples & Analogies

This can be compared to watching a candle burn. The flame's flicker (or pattern) remains the same, but as it burns, the height of the flame decreases until it goes out. Similarly, the temperature gradient in our slab stays consistent while the overall temperature becomes less intense over time.

Key Concepts

  • Temperature Distribution: The variation of temperature within a slab illustrated through a sinusoidal function.

  • Fourier Sine Expansion: Only specific terms contribute to the solution based on the imposed initial and boundary conditions.

  • Decay Over Time: The temperature wave diminishes in intensity but maintains its shape as it approaches steady state.

Examples & Applications

The temperature initially peaks at specific points and gradually decreases towards zero as time progresses, indicating heat dissipation in the concrete slab.

An insulated concrete slab subjected to an initial temperature distribution that follows a sine function illustrates how temperature varies along its length.

Memory Aids

Interactive tools to help you remember key concepts

🎵

Rhymes

In the slab, heat breathes and flows, sine wave's curve in beauty grows, watch it fade, as time bestows, to zero's calm, the warmth it throws.

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Stories

Once there was a concrete slab standing tall, with initial heat distribution just so and so. As time went on, its warmth did fade, settling quietly to zero, a peaceful state made.

🧠

Memory Tools

SINE: Sine terms, Insulated edges, N waves, Energy decay — remember these for the Fourier way!

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Acronyms

F.E.A.D

Fourier

Eigenfunctions

Amplitude decay — key concepts that hold sway.

Flash Cards

Glossary

Temperature Distribution

The variation of temperature within a material or at different points in a given space.

Fourier Sine Series

A mathematical series that represents a function as the sum of sine terms, often used in solving boundary value problems.

Separation of Variables

A method to solve partial differential equations by separating the variables into independent functions.

Eigenvalue

A value for which there exists a nontrivial solution to a system of equations, particularly in relation to linear transformations.

Boundary Conditions

Constraints that define the behavior of a differential equation at the boundaries of the domain.

Reference links

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