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Chapter 13: Problem 14

In the rectangular plate problem, we have so far had the temperature specified all around the boundary. We could, instead, have some edges insulated. The heat flow across an edge. is proportional to \(\partial T / \partial n\), where \(n\) is a variable in the direction normal to the edge (see normal derivatives, Chapter 6 , Section 6). For example, the heat flow across an edge lying along the \(x\) axis is proportional to \(\partial T / c y .\) Since the heat flow across an insulated cdge is zero, we must have not \(T\), but a partial derivative of \(T\), equal to zero on an insulated boundary. Use this fact to find the steady-state temperature distribution in a semi-infinite plate of width \(10 \mathrm{~cm}\) if the two long sides are insulated, the far end (at \(\infty\) as in Section 2) is at \(0^{\circ}\), and the bottom cdge is at \(T=f(x)=x-5\) Note that you used \(T \rightarrow 0\) as \(y \rightarrow \infty\) only to discard the solutions \(e^{+k y} ;\) it would be just as satisfactory to say that \(T\) does not become infinite as \(y \rightarrow \infty\). Actually, the temperature (assumed finite) as \(y \rightarrow \infty\) in this problem is determined by the given temperature at \(y=0\). Let \(T=f(x)=x\) at \(y=0\), repeat your calculations above to find the temperature distribution, and find the value of \(T\) for large \(y\). Don't forget the \(k=0\) term in the series!

Short Answer

Expert verified
The steady-state temperature distribution is given by the sum of Fourier series terms: \[ T(x, y) = \sum_{n=0}^{\infty} \left( A_n \cos \frac{n\pi x}{10} + B_n \sin \frac{n\pi x}{10} \right) e^{-\frac{n\pi}{10} y}. \]

Step by step solution

01

Understand the Problem

The task is to find the steady-state temperature distribution in a semi-infinite plate of width 10 cm. The two long sides are insulated, the far end is at 0°C, and the bottom edge has a temperature distribution given by a function.
02

Identify Boundary Conditions

The boundary conditions are: 1. \(\frac{\partial T}{\partial x} = 0\) for the insulated sides. 2. \(T \rightarrow 0\) as \(y \rightarrow \infty\). 3. \(T = x - 5\) at \(y = 0\).
03

Formulate the Heat Equation

The heat equation for steady-state in two dimensions is \[\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0.\]
04

Separation of Variables

Assume the solution has the form \(T(x,y) = X(x)Y(y)\). Substitute into the heat equation to get \[X''(x)Y(y) + X(x)Y''(y) = 0\] which simplifies to \[\frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)} = -k.\]
05

Solve for X(x)

This gives two ordinary differential equations: \[X''(x) + kX(x) = 0,\] the solution for which is \[X(x) = A\cos(kx) + B\sin(kx).\]
06

Insulated Boundary Condition for X

Using the insulated boundary conditions \(\frac{\partial X}{\partial x} = 0\) at the edges, we get \[B\cos(k \cdot 0) + B\cos(k \cdot 10\text{cm}) = 0.\] This simplifies to \[k=\frac{n\pi}{10}.\]
07

Solve for Y(y)

The differential equation for Y becomes: \[Y''(y) - k^2Y(y) = 0,\] with the solution \[Y(y) = Ce^{ky} + De^{-ky}.\]
08

Apply Condition at Infinity

Since \(T \rightarrow 0 \) as \(y \rightarrow \infty\), \(Ce^{ky}\) must be discarded, leaving: \[Y(y) = De^{-ky}.\]
09

Combine the Solutions

Thus, the general solution is: \[T(x, y) = \sum_{n=0}^{\infty} \left[ A_n \cos\left( \frac{n\pi x}{10} \right) + B_n \sin\left( \frac{n\pi x}{10} \right) \right] e^{-k_n y} \]
10

Determine Constants Using Bottom Edge Condition

Given \(T(x, 0) = f(x) = x - 5\), decompose \(f(x)\) using Fourier series. The series expansion is: \[x-5 = \sum_{n=0}^{\infty} \left(A_n \cos \frac{n\pi x}{10} + B_n \sin \frac{n\pi x}{10} \right).\]
11

Calculate Fourier Coefficients

Compute the coefficients \(A_n\) and \(B_n\) using the given boundary function. Set up the integral expressions: \[A_n = \frac{2}{10} \int_0^{10} (x-5) \cos \frac{n\pi x}{10} dx,\] and \[B_n = \frac{2}{10} \int_0^{10} (x-5) \sin \frac{n\pi x}{10} dx.\]
12

Evaluate Kernel Integrals

Solve the integrals to find \(A_n\) and \(B_n\).
13

Construct the Full Solution

Use the computed \(A_n\) and \(B_n\) to form the final temperature distribution: \[ T(x, y) = \sum_{n=0}^{\infty} \left( A_n \cos \frac{n\pi x}{10} + B_n \sin \frac{n\pi x}{10} \right) e^{-\frac{n\pi}{10} y}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boundary Conditions
Boundary conditions are essential for finding unique solutions to differential equations like the heat equation. They describe the behavior of temperature at the plate's edges and are critical in defining the problem.
There are three main types:
* Dirichlet boundary conditions: where the temperature is specified on the boundary.
* Neumann boundary conditions: where the heat flux (related to the temperature gradient) is specified.
* Mixed boundary conditions: a combination of Dirichlet and Neumann conditions.
In this exercise, the bottom edge's temperature is given by a function, and the long sides are insulated. The insulated condition (Neumann) means no heat flow across the sides, thus the temperature gradient perpendicular to those sides is zero. Lastly, the far end being at 0°C is another Dirichlet condition.
Heat Equation
The heat equation describes the distribution of heat (or temperature) in a given region over time. For a steady-state, it simplifies to the Laplace equation, that is The equation used to describe steady-state temperature distribution is

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Most popular questions from this chapter

A sphere initially at \(0^{\circ}\) has its surface kept at \(100^{\circ}\) from \(t=0\) on (for example, a frozen potato in boiling water !). Find the time- dependent temperature distribution. Hint : Subtract \(100^{\circ}\) from all temperatures and solve the problem; then add the \(100^{\circ}\) to the answer. Can you justify this procedure ? Show that the Legendre function required for this problem is \(P_{0}\) and the \(r\) solution is \((1 / \sqrt{r}) J_{1 / 2}\) or \(j_{0}\) [see (17.4) in Chapter 12]. Since spherical Bessel functions can be expressed in terms of elementary functions, the series in this problem can be thought of as cither a Bcssel series or a Fouricr series. Show that the results arc identical.

The following two \(R(r)\) equations arise in various separation of variables problems in polar, cylindrical, or spherical coordinates: $$ \begin{aligned} &r \frac{d}{d r}\left(r \frac{d R}{d r}\right)=n^{2} R \\ &\frac{d}{d r}\left(r^{2} \frac{d R}{d r}\right)=N(l+1) R \end{aligned} $$ There are various ways of solving them: They are a standard kind of equation (often called Euler or Cauchy equations-see Chapter 8, Section 7d); you could use power series methods; given the fact that the solutions are just powers of \(r\), it is easy to find the powers. Choose any method you like, and solve the two equations for future reference. Consider the case \(n=0\) separately. Is this necessary for \(l=0 ?\)

Find the steady-state temperature distribution in a circular annulus (shaded area) of inner radius \(I\) and outer radius 2 if the inner circle is held at \(0^{\circ}\), and the outer circle has half its circumference at \(0^{\circ}\) and half at \(100^{\circ}\). Hint: Don't forget the \(r\) solutions corresponding to \(k=0\).

Water at \(100^{\circ}\) is flowing through a long pipe of radius 1 rapidly enough so that we may assume that the temperature is \(100^{\circ}\) at all points. At \(t=0\), the water is turned off and the. surface of the pipe is maintained at \(40^{\circ}\) from then on (neglect the wall thickness of the pipe). Find the temperature distribution in the water as a function of \(r\) and \(t\). Note that you need only consider a cross section of the pipe.

Find the steady-state temperature distribution in a metal plate \(10 \mathrm{~cm}\) square if one side is held at \(100^{\circ}\) and the other three sides at \(0^{\circ}\). Find the temperature at the center of the plate.
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