Chapter 13: Problem 1
Find the steady-state temperature distribution in a rectangular plate covering
the area \(0
Short Answer
Expert verified
The steady-state temperature distribution is given by a series solution involving sine and hyperbolic sine functions; exact coefficients depend on Fourier series expansion of 1 -x.
Step by step solution
01
Understand the problem
Given a rectangular plate with boundaries defined as follows: the plate covers the area where 0<x<1 and 0<y<2. The boundary conditions for temperature are: T=0 for x=0, x=1, and y=2, and T=1-x for y=0. The goal is to find the steady-state temperature distribution within this plate.
02
Set up the heat equation
The steady-state temperature distribution in a region is governed by the two-dimensional Laplace equation for temperature: \[ \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0 \]
03
Apply boundary conditions
Apply the given boundary conditions to this equation: 1. T(0,y) = 0 2. T(1,y) = 0 3. T(x,2) = 0 4. T(x,0) = 1-x
04
Propose a solution form
Propose a general solution for the temperature distribution using separation of variables: let \( T(x, y) = X(x)Y(y) \). Substitute into the Laplace equation, which gives us: \[ X''(x)Y(y) + X(x)Y''(y) = 0 \]
05
Separate the variables
Separate the variables by dividing both sides by \(X(x)Y(y)\): \[ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 \] This can be separated into two ordinary differential equations: \[ \frac{X''(x)}{X(x)} = -\lambda^2 \quad and \quad \frac{Y''(y)}{Y(y)} = \lambda^2 \]
06
Solve for X(x)
Solve for \(X(x)\) with the boundary conditions \(X(0) = 0\) and \(X(1) = 0\): \[ X''(x) + \lambda^2 X(x) = 0 \] The general solution is \[ X(x) = A \sin(\lambda x) \] Applying the boundary conditions: \[ X(0) = 0 \Rightarrow A \sin(0) = 0 \] \[ X(1) = 0 \Rightarrow A \sin(\lambda) = 0 \] Therefore, \(\lambda = n\pi\) for n=1,2,3,...
07
Solve for Y(y)
Solve for \(Y(y)\) with the boundary condition \(Y(2) = 0\): \[ Y''(y) - n^2\pi^2 Y(y) = 0 \] The general solution is: \[ Y(y) = B \sinh(n\pi y) + C \cosh(n\pi y) \] Apply the boundary condition \(Y(2) = 0\): \[ Y(2) = B \sinh(2n\pi) + C \cosh(2n\pi) = 0 \] This implies \(C = 0\) (since \cosh(n\pi y) is never zero). Thus, \[ Y(y) = B \sinh(n\pi y) \]
08
Form overall solution
Combine \(X(x)\) and \(Y(y)\) to form the overall solution: \[ T(x, y) = \sum_{n=1}^{\infty} A_n \sin(n\pi x) \sinh(n\pi y) \]
09
Apply remaining boundary condition
Apply the remaining boundary condition T(x,0) = 1-x to determine coefficients \(A_n\): \( \sum_{n=1}^{\infty} A_n \sin(n\pi x) \sinh(0) = 1-x \Rightarrow A_n \sinh(0) = 1-x \). Since \(\sinh(0)=0\), adjust the boundary condition method accordingly for the non-zero components.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laplace equation
The Laplace equation is a fundamental partial differential equation (PDE) in mathematics, often written as \[ \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0\]. It's crucial in problems involving steady-state heat distribution and other areas such as electrostatics and fluid flow. When we say 'steady-state,' we mean that the temperature at each point in the system doesn’t change over time. Understanding this equation is key to solving problems like the rectangular plate heat problem. Here, you have no dependence on time, which simplifies the problem significantly by focusing only on the spatial variables x and y. Use the Laplace equation to set up the problem before applying the boundary conditions.
boundary conditions
Boundary conditions are the constraints required to solve differential equations such as the Laplace equation. In the rectangular plate problem, we have specific temperature values at the edges:
Each boundary condition translates into mathematical expressions that shape our approach and solution process. For steady-state problems, these conditions are typically set by physical constraints or empirical observations.
- T=0 when x=0, x=1, and y=2
- T=1-x for y=0
Each boundary condition translates into mathematical expressions that shape our approach and solution process. For steady-state problems, these conditions are typically set by physical constraints or empirical observations.
separation of variables
Separation of variables is a method used to solve PDEs like the Laplace equation. It involves assuming that the solution can be written as the product of two functions, each depending on a single coordinate. In this case, let \[ T(x, y) = X(x)Y(y) \]. When we substitute this form into the Laplace equation, we get\[ X''(x)Y(y) + X(x)Y''(y) = 0 \]. Dividing by \[ X(x)Y(y) \] lets us separate the equation into two ordinary differential equations (ODEs): one for \[ X(x) \]and one for \[ Y(y) \]. This technique simplifies solving the PDE since each ODE can be tackled independently before recombining the solutions.
ordinary differential equations
Ordinary Differential Equations (ODEs) are equations involving a function of one variable and its derivatives. In the process of separation of variables, we break the Laplace equation into two ODEs. For the rectangular plate problem, these ODEs are:
- \[ X''(x) + \lambda^2 X(x) = 0 \]
- \[ Y''(y) - \lambda^2 Y(y) = 0 \]
rectangular plate heat problem
The rectangular plate heat problem involves finding the temperature distribution in a rectangular area with given boundary conditions. This problem is framed by the given boundaries and conditions, which in this case, are:
- 0 < x < 1
- 0 < y < 2
- T=0 for x=0, x=1, y=2
- T=1-x for y=0
- Set up the Laplace equation.
- Use separation of variables to propose a solution.
- Solve resulting ODEs for \[ X(x) \]and \[ Y(y) \].
- Combine these findings to form an overall solution.
- Apply boundary conditions to determine specific coefficients.