/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 Consider a large plane wall of t... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a large plane wall of thickness \(L\) and constant thermal conductivity \(k\). The left side of the wall \((x=0)\) is maintained at a constant temperature \(T_{0}\), while the right surface at \(x=L\) is insulated. Heat is generated in the wall at the rate of \(\dot{e}_{\text {gen }}=a x^{2} \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{3}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wall, \((b)\) by solving the differential equation, obtain a relation for the variation of temperature in the wall \(T(x)\) in terms of \(x, L, k, a\), and \(T_{0}\), and (c) what is the highest temperature \(\left({ }^{\circ} \mathrm{C}\right)\) in the plane wall when: \(L=1 \mathrm{ft}, k=5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ}{ }^{\circ} \mathrm{F}, a=1200 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{5}\), and \(T_{0}=700^{\circ} \mathrm{F}\).

Short Answer

Expert verified
Answer: The highest temperature in the plane wall is \(700°F\).

Step by step solution

01

Set up the differential equation and boundary conditions

We start by applying Fourier's law of heat conduction: \(q_x = -k\frac{dT}{dx}\). Considering the heat generated within the wall, the heat balance equation in a differential control volume is given by \(\frac{d}{dx}(q_x) = \dot{e}_{gen}\). Substituting the Fourier's law, we get: \(\frac{d}{dx} \left(-k\frac{dT}{dx}\right) = ax^2\) Boundary conditions: 1. At \(x=0\), \(T=T_0\) 2. At \(x=L\), the wall is insulated, so \(\frac{dT}{dx}=0\)
02

Solving the differential equation for temperature distribution

To solve the differential equation, we will first integrate once to get: \(-k\frac{dT}{dx} = \frac{1}{3}ax^3 + C_1\) Now we will integrate again, keeping in mind that we need to satisfy the boundary conditions: \(T(x) =-\frac{a}{12}x^4 - C_1x + C_2\) Satisfying the boundary conditions: 1. At \(x=0\), \(T=T_0\): \(T_0 = -\frac{a}{12}(0)^4 - C_1(0) + C_2 \Rightarrow C_2 = T_0\) 2. At \(x=L\), \(\frac{dT}{dx}=0\): \(0 = -k\frac{dT}{dx}= -k\left(-\frac{a}{12}(4L^3) + C_1\right) \Rightarrow C_1 = \frac{aL^3}{3}\) The temperature distribution is given by: \(T(x) = -\frac{a}{12}x^4 + \frac{aL^3}{3}x + T_0\)
03

Find the highest temperature in the wall

Given the values \(L=1\,ft\), \(k=5\,Btu/h\cdot ft\cdot°F\), \(a=1200\,Btu/h\cdot ft^5\), and \(T_0=700°F\), we have the temperature distribution as follows: \(T(x) = -\frac{1200}{12}x^4 + \frac{1200(1)^3}{3}x + 700\) Since the temperature distribution is in terms of \(x^4\), the highest temperature will occur at \(x=0\). Using the given values, let's calculate the highest temperature in the wall: \(T(0) = -\frac{1200}{12}(0)^4 + \frac{1200(1)^3}{3}(0) + 700 = 700°F\) The highest temperature in the plane wall is \(700°F\).

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Key Concepts

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

Differential Equation
A differential equation is a mathematical equation that involves derivatives of a function and is used to describe various physical phenomena. In the context of heat conduction through a wall, it helps us determine how temperature changes throughout the wall. Here, we're looking at how heat moves from one side of the wall to the other.

Think of it this way: a differential equation tells us the rate of change of temperature, which is how quickly or slowly the temperature increases or decreases. The differential equation for heat conduction in this problem is:\[\frac{d}{dx} \left(-k\frac{dT}{dx}\right) = ax^2\]

This equation accounts for the heat generated in the wall, represented by \(ax^2\), and the rate at which heat is conducted across the wall, represented by \(-k\frac{dT}{dx}\). With this equation, we can solve for the temperature distribution across the wall.✨
Boundary Conditions
Boundary conditions are constraints necessary for solving differential equations. They specify the values the solution needs to satisfy at specific points. These conditions help us find a unique solution to the equation based on how the system behaves at its boundaries.

In this exercise, we have two boundary conditions:
  • At \(x=0\): The temperature is maintained at \(T_0\).
  • At \(x=L\): The wall is insulated, meaning no heat passes through, so \(\frac{dT}{dx}=0\).
This setup tells us how the temperature behaves at the edges of the wall and thereby allows us to solve the differential equation for the entire wall.

The first condition ensures that our solution stays true to the given temperature at one side of the wall. The second condition ensures that there is no heat loss (or gain) at the insulated boundary. Based on these, we can identify the constants in the temperature distribution equation.
Fourier's Law
Fourier's Law of heat conduction is a fundamental principle that describes the flow of heat through a material. It states that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which the heat flows. This can be expressed as:\[q_x = -k \frac{dT}{dx}\]

Here, \(q_x\) is the heat transfer rate per unit area, \(\frac{dT}{dx}\) is the temperature gradient across the material, and \(k\) is the thermal conductivity. A higher thermal conductivity means heat passes through the material more easily.

In our problem, Fourier's Law helps establish the relationship between the thermal conductivity of the wall and the change in temperature. By substituting Fourier's Law into the heat balance equation, we derived our differential equation. This principle illuminates how heat conduction varies with different materials and temperature gradients.
Temperature Distribution
Temperature distribution refers to how temperature varies at different points in a material. In the case of heat conduction in a wall, we're interested in knowing what the temperature is at any point along the wall from one side to another. This understanding helps in predicting thermal behavior under different conditions.

The solution to our differential equation gives us the temperature distribution:\[T(x) = -\frac{a}{12}x^4 + \frac{aL^3}{3}x + T_0\]

This equation tells us how the temperature changes inside the wall as a function of \(x\), the distance through the thickness of the wall. The terms in this equation are influenced by the wall’s material properties and the heat generation rate.

By solving this, we can determine where the highest and lowest temperatures occur. For example, in our exercise, the highest temperature can be found at \(x=0\), which aligns with the boundary condition specified at that point. Calculating actual values gives insight into thermal management in engineering applications.

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Most popular questions from this chapter

How do differential equations with constant coefficients differ from those with variable coefficients? Give an example for each type.

A long homogeneous resistance wire of radius \(r_{o}=\) \(0.6 \mathrm{~cm}\) and thermal conductivity \(k=15.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of \(16.4 \mathrm{~W} / \mathrm{cm}^{3}\). The heat generated is transferred to water at \(100^{\circ} \mathrm{C}\) by convection with an average heat transfer coefficient of \(h=3200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wire, \((b)\) obtain a relation for the variation of temperature in the wire by solving the differential equation, and \((c)\) determine the temperature at the centerline of the wire.

Consider a water pipe of length \(L=17 \mathrm{~m}\), inner radius \(r_{1}=15 \mathrm{~cm}\), outer radius \(r_{2}=20 \mathrm{~cm}\), and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the pipe material uniformly by a \(25-\mathrm{kW}\) electric resistance heater. The inner and outer surfaces of the pipe are at \(T_{1}=60^{\circ} \mathrm{C}\) and \(T_{2}=80^{\circ} \mathrm{C}\), respectively. Obtain a general relation for temperature distribution inside the pipe under steady conditions and determine the temperature at the center plane of the pipe.

A cylindrical nuclear fuel rod of \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube of \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(30 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(T_{1}=90^{\circ} \mathrm{C}\) while the right side loses heat by convection to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the rate of heat transfer through the wall. Answer: (c) \(7389 \mathrm{~W}\)

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