Question

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 \(T_{0^{-}}\)(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} \cdot{ }^{\circ} \mathrm{F}, a=1200 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{5}$, and \(T_{0}=700^{\circ} \mathrm{F} ?\)

Short answer

Answer: The temperature distribution T(x) across the wall is given by the equation T(x) = -\(\frac{a}{12k}x^4 + \frac{aL^3}{3k}x + T_0\). The highest temperature in the wall is 780°F when L = 1ft, k = 5 Btu/h.ft.°F, a = 1200 Btu/h.ft^5, and T_0 = 700°F.

Step-by-step solution

Write the general heat conduction equation

The steady-state, one-dimensional heat conduction equation with heat generation is given by: \(\frac{d}{dx}\left(k\frac{dT}{dx}\right) = -\dot{e}_{gen}\) Since k is constant, we can rewrite the equation as: \(\frac{d^2T}{dx^2} = -\frac{\dot{e}_{gen}}{k}\) We know \(\dot{e}_{gen} = ax^2\). Substitute this into the equation above: \(\frac{d^2T}{dx^2} = -\frac{a}{k}x^2\)

Apply boundary conditions

The boundary conditions are: (1) At x=0: T=T0 (2) At x=L (insulated side): \(\frac{dT}{dx}=0\)

Solve the differential equation to obtain T(x)

To solve the above second-order differential equation, we can use the method of integration. First, integrate \(\frac{d^2T}{dx^2}\) with respect to x: \(\int\frac{d^2T}{dx^2}dx = \int -\frac{a}{k}x^2 dx\) This gives: \(\frac{dT}{dx} = -\frac{a}{k}\int x^2 dx + C_1\) Then, integrate again: \(\int \frac{dT}{dx} dx = -\frac{a}{3k}\int x^3 dx + C_1 \int dx\) This results in: \(T(x) = -\frac{a}{12k}x^4 + C_1x + C_2\) Now we can apply the boundary conditions to find the constants C_1 and C_2: (1) At x=0, T=T0: \(T0 = -\frac{a}{12k}(0)^4 + C_1(0) + C_2\) \(C_2 = T0\) (2) At x=L, \(\frac{dT}{dx}=0\): \(0 = -\frac{a}{3k}(L)^3 + C_1\) \(C_1 = \frac{aL^3}{3k}\) Thus, the temperature distribution T(x) is: \(T(x) = -\frac{a}{12k}x^4 + \frac{aL^3}{3k}x + T_0\)

Find the highest temperature in the wall

Given the values for L, k, a, and T0: \(L = 1ft, k = 5 Btu/h.ft.°F, a = 1200 Btu/h.ft^5, T_0 = 700°F\) Substitute these values into our T(x) equation: \(T(x) = -\frac{1200}{12(5)}x^4 + \frac{1200(1)^3}{3(5)}x + 700\) \(T(x) = -20x^4 + 80x + 700\) To find the highest temperature in the wall, take the derivative of T(x) with respect to x and set it equal to 0: \(\frac{d T(x)}{dx} = -80x^3 + 80 = 0\) Solving for x: \(x = \sqrt[3]{1} = 1\) Now, find the temperature at x=1: \(T(1) = -20(1)^4 + 80(1) + 700\) \(T(1) = 780°\mathrm{F}\) So, the highest temperature in the plane wall is 780°F.