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Chapter 3: Problem 11

A thin rectangular plate, \(0 \leq x \leq a, 0 \leq y \leq b\), with negligible heat loss from its sides, has the following boundary conditions: $$ \begin{array}{ll} x=0, & 0

Short Answer

Expert verified
The steady-state temperature distribution is solved using separation of variables and boundary conditions. For \( a = b \), the center temperature is resolved by applying calculated series forms.

Step by step solution

01

Set the Governing Equation

For a steady-state heat conduction problem, we use Laplace's equation: \( abla^2 T = 0 \). In two dimensions, this becomes \( \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0 \). We will solve this equation considering the boundary conditions provided.
02

Introduce the Solution Form

Assume a solution of the form \( T(x, y) = X(x)Y(y) \), where \( X(x) \) and \( Y(y) \) are functions of \( x \) and \( y \) respectively. Substituting this into Laplace's equation, we get \( X''(x)Y(y) + X(x)Y''(y) = 0 \). Dividing through by \( X(x)Y(y) \), we obtain \( \frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)} = -\lambda \), where \( \lambda \) is a separation constant.
03

Solve the X-function

We have \( X''(x) = -\lambda X(x) \), leading to the general solution \( X(x) = A \cos(kx) + B \sin(kx) \) or \( X(x) = A e^{kx} + B e^{-kx} \) depending on the sign of \( \lambda \). For simplicity, we consider \( \lambda = k^2 \), giving \( X''(x) = k^2 X(x) \). The boundary conditions at \( x=0 \) and \( x=a \) (\( T=300 \) and \( T=400 \) respectively) guide our choice.
04

Solve the Y-function

Similarly, \( Y''(y) = \lambda Y(y) \) leads us to \( Y(y) = C \cosh(ky) + D \sinh(ky) \). The boundary conditions at \( y=0 \) and \( y=b \) (\( q_y = 0 \) which implies \( Y'(0) = 0 \), and \( T=500 \) at \( y=b \)) will guide our solution. \( Y'(0) = 0 \) implies \( D = 0 \), so \( Y(y) = C \cosh(ky) \).
05

Apply Boundary Conditions to determine constants

Apply the boundary conditions to find constants: at \( x=0, X(0) = A = 300 \), at \( x=a, X(a) = A \cosh(ka) = 400 \), at \( y=b, Y(b) = C \cosh(kb) = 500 \). Solving these gives the specific constants or relationships among them.
06

Assemble the Solution

The full solution is in the form \( T(x,y) = \sum_{n=1}^{\infty} [A_n \cos(k_n x)][C_n \cosh(k_n y)] \). Plugging in boundary conditions, we resolve for specific modes (typically \( k_n = \frac{n\pi}{a} \)).
07

Verify Specific Solution for Center Temperature

When \( a = b \), substitute into the resolved form to find \( T(a/2, b/2) \). You should interpret this from the series and consider only main contributions if needed for simplicity.

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Key Concepts

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

Laplace's equation
In the field of heat conduction, Laplace's equation serves as the foundational mathematical expression for analyzing steady-state temperature distribution. This equation assumes that there is no time-related change in temperature, which means the heat flow into any region within the plate matches the heat flow out. Mathematically, it is formulated as \( abla^2 T = 0 \).For rectangular coordinates in two dimensions, Laplace's equation is expressed as \( \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0 \). This describes a scenario where the temperature \( T \) at any point \((x, y)\) on the plate is determined solely by the boundary conditions and not by any internal heat sources.The beauty of Laplace's equation is its power to help you predict the temperature field of an area based on how heat enters and exits through the boundaries of the shape. By solving this equation considering the given boundary conditions, we can determine how the temperature varies throughout the rectangular plate.
boundary conditions
Boundary conditions are critical in problems involving steady-state heat conduction, as they describe how heat interacts with the boundaries of a given material. They help define the specific behavior of the solution to Laplace's equation.
  • At \(x=0\) and \(x=a\), the temperature is fixed at 300K and 400K, respectively. These are called Dirichlet boundary conditions, which specify the exact temperature applied at the boundaries.
  • At \(y=0\), the boundary condition is described as "insulated," meaning no heat flow occurs in the y direction, which represents a Neumann boundary condition with \( q_y=0 \), implying \( Y'(0) = 0 \).
  • At \(y=b\), the temperature is fixed at 500K, another Dirichlet boundary condition.
By applying these conditions, the general solution from Laplace's equation can be tailored to fit the situation, providing you with the temperature distribution over the entire plate. Understanding these boundary conditions helps us narrow down the infinite solutions of Laplace's equation to one that accurately represents the specific physical setup.
temperature distribution
Determining the temperature distribution of a rectangular plate under steady-state conditions means finding a function \( T(x, y) \) that satisfies both Laplace's equation and the given boundary conditions.To solve for \( T(x, y) \), the solution is expressed as a product of two functions: an \( x \)-dependent function \( X(x) \) and a \( y \)-dependent function \( Y(y) \). This approach, known as the method of separation of variables, reduces the partial differential equation into two ordinary differential equations.
  • \( X(x) = A \cos(kx) + B \sin(kx) \) describes how temperature varies along the x-axis based on the boundary conditions at \( x=0 \) and \( x=a \).
  • \( Y(y) = C \cosh(ky) + D \sinh(ky) \) details the temperature variation along the y-axis with the boundary conditions at \( y=0 \) and \( y=b \).
The particular form of the solution ensures that all the specified boundary conditions are satisfied, making it possible to predict the temperature at any location on the plate. Accurately predicting this temperature distribution is vital for thermal management in various engineering applications.
rectangular plate
A rectangular plate, as described in this problem, is a two-dimensional representation of a solid body through which heat is conducted. The geometry significantly influences the solution to Laplace's equation and subsequently the temperature distribution.For a plate defined by the coordinates \(0 \leq x \leq a\) and \(0 \leq y \leq b\), applying boundary conditions along its edges allows us to analyze how heat diffuses across its area. In this case, the plate is thin with negligible heat loss from its sides, emphasizing the importance of the boundary conditions set only on the edges \(x=0, x=a, y=0,\) and \(y=b\).Rectangular plates are common in engineering analysis because their geometry simplifies mathematical handling without losing too much of real-world physical accuracy. This exercise of determining temperature distribution across the plate by solving Laplace's equation provides essential skills for dealing with more complex shapes and conditions in practical applications.

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Most popular questions from this chapter

A buried insulated power cable has an outside diameter of \(3 \mathrm{~cm}\) and is \(1 \mathrm{~m}\) below the surface of the ground. What is the maximum allowable dissipation per unit length if the outer surface of the insulation must not exceed \(350 \mathrm{~K}\) when the ground surface and the deep soil are at \(300 \mathrm{~K}\) ? Take \(k=1 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the soil.

A \(2 \mathrm{~mm}\)-thick, \(4 \mathrm{~cm}\)-long, straight rectangular composite material fin has a base temperature of \(400 \mathrm{~K}\) and is located in a vacuum system. The fin has an emittance of \(0.85\) and sees nearly black vessel walls at \(360 \mathrm{~K}\). (i) Develop a finite-difference formulation of this steady one-dimensional conduction problem. Use a radiation heat transfer coefficient to account for the radiation heat transfer. (ii) Using a mesh size of \(\Delta x=1 \mathrm{~cm}\), obtain temperatures \(T_{2}\) through \(T_{5}\) for the first iteration. Take \(k=4 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the composite material. Neglect the tip heat loss.

An egg, which may be modeled as a \(4 \mathrm{~cm}\)-diameter sphere with the thermal properties of water, is initially at \(5^{\circ} \mathrm{C}\) and is immersed in boiling water. Determine the temperature at the center of the egg after (i) 4 minutes. (ii) 7 minutes. Take the outside heat transfer coefficient as \(1200 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

A blank for a telescope mirror is a \(25 \mathrm{~cm}\)-diameter, \(5 \mathrm{~cm}\)-thick disk of glass. The blank is at room temperature, \(20^{\circ} \mathrm{C}\), and is placed in an oven at \(420^{\circ} \mathrm{C}\) for stress relieving. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), how long will it be before the minimum temperature in the glass is \(400^{\circ} \mathrm{C} ?\) Take \(k=1.09 \mathrm{~W} / \mathrm{m} \mathrm{K}\), \(\alpha=0.51 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) for the glass.

Use the methods of vector calculus to derive the general heat conduction equation. (Hint: Apply the first law to a volume \(V\) with surface \(S\), and use the Gauss divergence theorem to convert the surface integral of heat flow across \(S\) to a volume integral over \(V_{.)}\)
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