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 \(T_{0}=700^{\circ} \mathrm{F}\).

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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