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Chapter 5: Problem 5

Find the temperature distribution in a solid whose boundaries are two parallel planes, \(d\) units apart, kept at temperatures \(T_{1}, T_{2}\), respectively. (Hint: Take the boundaries to be the planes \(x=0, x=d\) and note that, by symmetry, \(T\) must be independent of \(y\) and \(z\), .)

Short Answer

Expert verified
Question: Derive the temperature distribution within a solid bounded by two parallel planes, \(x=0\) and \(x=d\), which are kept at temperatures \(T_1\) and \(T_2\). Answer: The temperature distribution within the solid is given by the equation \(T(x) = \frac{T_2-T_1}{d}x + T_1\).

Step by step solution

01

Assumption

Due to symmetry, the temperature distribution \(T\) depends only on the position along the \(x\) axis. Hence, the temperature \(T\) is a function of \(x\) only: \(T= T(x)\).
02

Derive the equation for the temperature

As the temperature distribution in the solid is due to heat conduction, we apply the heat equation in one dimension. In the steady-state, the heat equation is given by: \(\frac{d^2T}{dx^2}=0\)
03

Solve the heat equation

We solve the second-order ordinary differential equation (ODE): \(\frac{dT}{dx}=C_1\) Integrating once again: \(T(x)=C_1x+C_2\)
04

Apply boundary conditions

The problem states that the solid's boundaries are at temperatures \(T_1\) at \(x=0\) and \(T_2\) at \(x=d\). We apply these as boundary conditions: For \(x=0\), \(T(0)=T_1\): \(T_1 = C_1(0)+C_2\), so \(C_2=T_1\) For \(x=d\), \(T(d)=T_2\): \(T_2 = C_1d+T_1\)
05

Solve for the constants

We find the values of \(C_1\) and \(C_2\): We already found \(C_2 = T_1\) in Step 4. Next, we find the value of \(C_1\): \(T_2 = C_1d + T_1\) \(C_1 = \frac{T_2-T_1}{d}\)
06

Write the final temperature distribution equation

Inserting the values of \(C_1\) and \(C_2\) into the general solution for \(T(x)\), we get: \(T(x) = \frac{T_2-T_1}{d}x + T_1\) This equation describes the temperature distribution in the solid between the two parallel planes kept at temperatures \(T_1\) and \(T_2\).

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Key Concepts

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

Heat Equation
The heat equation is a fundamental piece of mathematics applied in physics and engineering to predict how heat diffuses through a given medium. It's based on Fourier's law of heat conduction, which states that heat flow is proportional to the negative gradient of the temperature and the material's thermal conductivity. The one-dimensional steady-state heat equation can be written as \[ \frac{d^2T}{dx^2}=0 \. \] In the context of our exercise, the heat equation helps us understand how temperature varies within a solid when its boundaries are held at constant temperatures. Unlike transient scenarios, the steady-state condition simplifies the equation significantly, reflecting the fact that the temperature does not change with time; it solely depends on the position within the solid.
Boundary Conditions
Boundary conditions are essential for the solution of differential equations, as they specify the values at the domain's edges that a solution must satisfy. In the context of the heat equation, they define the temperatures at which the material's boundaries are maintained. For our temperature distribution problem, we have two types of boundary conditions known as Dirichlet conditions:
  • At \(x = 0\), the temperature is \(T_1\), indicating the temperature at one surface of the solid.
  • At \(x = d\), the temperature is \(T_2\), pertaining to the opposing surface's temperature.
By applying these boundary conditions to the general solution of the heat equation, we can determine the constants of integration which ultimately leads us to a specific solution that maps the temperature distribution across the solid.
Ordinary Differential Equation
An Ordinary Differential Equation (ODE) is an equation involving derivatives of a function with respect to one independent variable. In the exercise at hand, the temperature distribution is governed by a second-order linear ODE,
\[ \frac{d^2T}{dx^2} = 0 \],
indicating that the second derivative of the temperature with respect to \(x\) is zero. This simplification stems from the assumption of a steady-state process and one-dimensional heat flow. To solve the ODE, we integrate it with respect to \(x\), first finding a general solution and then applying boundary conditions to derive the constants involved. The final linear expression for temperature \(T(x)\) describes how the temperature varies linearly between the two boundaries, which is consistent with our physical intuition for heat distribution in a homogenous solid under steady-state conditions.

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Most popular questions from this chapter

Evaluate the following line integrals: a) \(\int_{(0,-1)}^{(0,1)} y^{2} d x+x^{2} d y\), where \(C\) is the semicircle \(x=\sqrt{1-y^{2}}\); b) \(\int_{(0,0)}^{(2,4)} y d x+x d y\), where \(C\) is the parabola \(y=x^{2}\); c) \(\int_{(1,0)}^{(0,1)} \frac{y d x-x d y}{x^{2}+y^{2}}\), where \(C\) is the curve \(x=\cos ^{3} t, y=\sin ^{3} t, 0 \leq t \leq \frac{\pi}{2}\). \(C\)

Determine all values of the integral $$ \int_{(1,0)}^{(2,2)} \frac{-y d x+x d y}{x^{2}+y^{2}} $$ B on a path not passing through the origin.

Prove the rule \(\alpha \beta=(-1)^{m_{1} m_{2}} \beta \alpha\) of (5.207). [Hint: By the distributive laws (5.208) it suffices to prove it for $$ \alpha=A d x^{i_{1}} \ldots d x^{i_{n_{1}}}, \quad \beta=B d x^{j_{1}} \ldots d x^{j_{N_{2}}} . $$ If \(i_{k}=j \ell\) for some \(k, \ell\), both sides are 0. Otherwise, as in Example 2 of Section 5.19, one has to verify that one can go from \(\left(i_{1}, \ldots, i_{m_{1}}, j_{1}, \ldots, j_{m_{2}}\right)\) to \(\left(j_{1}, \ldots, j_{m_{2}}, i_{1}, \ldots, i_{m_{1}}\right)\) by \(m_{1} m_{2}\) interchanges.]

Evaluate by the divergence theorem: a) \(\iint_{S} x d y d z+y d z d x+z d x d y\), where \(S\) is the sphere \(x^{2}+y^{2}+z^{2}=1\) and \(\mathbf{n}\) is the outer normal; b) \(\iint_{S} v_{n} d \sigma\), where \(\mathbf{v}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}, \mathbf{n}\) is the outer normal and \(S\) is the surface of the cube \(0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\); c) \(\iint_{S} e^{y} \cos z d y d z+e^{x} \sin z d z d x+e^{x} \cos y d x d y\), with \(S\) and \(\mathbf{n}\) as in (a); d) \(\iint_{S} \nabla F \cdot \mathbf{n} d \sigma\) if \(F=x^{2}+y^{2}+z^{2}, \mathbf{n}\) is the exterior normal, and \(S\) bounds a solid region \(R\); e) \(\iint_{S} \nabla F \cdot \mathbf{n} d \sigma\) if \(F=2 x^{2}-y^{2}-z^{2}\), with \(\mathbf{n}\) and \(S\) as in (d); f) \(\iint_{S} \nabla F \cdot \mathbf{n} d \sigma\) if \(F=\left[(x-2)^{2}+y^{2}+z^{2}\right]^{-1 / 2}\) and \(S\) and \(\mathbf{n}\) are as in (a).

Let \(S\) be an oriented surface in space that is planar; that is, \(S\) lies in a plane. With \(S\) one can associate the vector \(\mathbf{S}\), which has the direction of the normal chosen on \(S\) and has a length equal to the area of \(S\). a) Show that if \(S_{1}, S_{2}, S_{3}, S_{4}\) are the faces of a tetrahedron, oriented so that the normal is the exterior normal, then $$ \mathbf{S}_{1}+\mathbf{S}_{2}+\mathbf{S}_{3}+\mathbf{S}_{4}=\mathbf{0} \text {. } $$ [Hint: Let \(\mathbf{S}_{i}=A_{i} \mathbf{n}_{i}\left(A_{i}>0\right)\) for \(i=1, \ldots, 4\) and let \(\mathbf{S}_{1}+\cdots+\mathbf{S}_{4}=\mathbf{b}\). Let \(p_{1}\) be the foot of the altitude on face \(S_{1}\) and join \(p_{1}\) to the vertices of \(S_{1}\) to form three triangles of areas \(A_{12}, \ldots, A_{14}\). Show that, for proper numbering, \(A_{1 j}=\pm A_{j} \mathbf{n}_{j} \cdot \mathbf{n}_{1}\), with \(+\) or - according as \(\mathbf{n}_{j} \cdot \mathbf{n}_{1}>0\) or \(<0\), and \(A_{1 j}=0\) if \(\mathbf{n}_{j} \cdot \mathbf{n}_{1}=0(j=2,3,4)\). Hence deduce that \(\mathbf{b} \cdot \mathbf{n}_{j}=0\) for \(j=2,3,4\) and thus \(\mathbf{b} \cdot \mathbf{b}=0\).] b) Show that the result of (a) extends to an arbitrary convex polyhedron with faces \(S_{1}, \ldots, S_{n}\), that is, that $$ \mathbf{S}_{1}+\mathbf{S}_{2}+\cdots+\mathbf{S}_{n}=\mathbf{0}, $$ when the orientation is that of the exterior normal. c) Using the result of (b), indicate a reasoning to justify the relation $$ \iint_{S} \mathbf{v} \cdot d \boldsymbol{\sigma}=0 $$ for any convex closed surface \(S\) (such as the surface of a sphere or ellipsoid), provided that \(\mathbf{v}\) is a constant vector. d) Apply the result of (b) to a triangular prism whose edges represent the vectors \(\mathbf{a}, \mathbf{b}\), \(\mathbf{a}+\mathbf{b}\), c to prove the distributive law (Equation (1.19) $$ \mathbf{c} \times(\mathbf{a}+\mathbf{b})=\mathbf{c} \times \mathbf{a}+\mathbf{c} \times \mathbf{b} $$ for the vector product. This is the method used by Gibbs (cf. the book by Gibbs listed at the end of this chapter).
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