Question

A \(1000-W\) iron is left on the ironing board with its base exposed to ambient air at \(26^{\circ} \mathrm{C}\). The base plate of the iron has a thickness of \(L=0.5 \mathrm{~cm}\), base area of \(A=150 \mathrm{~cm}^{2}\), and thermal conductivity of \(k=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate, whose emissivity is \(\varepsilon=0.7\), loses heat by convection to ambient air with an average heat transfer coefficient of $h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {sarr }}=295 \mathrm{~K}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and (c) evaluate the outer surface temperature.

Short answer

Question: Find the temperature of the outer surface of an iron's base plate, given that it is 0.5 cm thick and has an area of 150 cm². The power of the iron is 1000 W. Assume the following properties – thermal conductivity (k) of the iron = 18 W/(m K), the ambient air temperature (T_∞) = 26°C, the convection heat transfer coefficient (h) = 30 W/(m^2 K), surface emissivity (ε) = 0.7, and the surrounding temperature (T_sarr) = 295 K. Use the necessary numerical methods to find the outer surface temperature. Answer: To find the outer surface temperature, first, the temperature expression for the outer surface (T_x=L) should be derived. The expression is: T_x=L = - (q_(x=0) * L) / k + T_∞. Then, solving the boundary condition equation using the given values and substituting the expression for the outer surface temperature, we obtain a nonlinear equation for T_x=L. To solve it, we need to use numerical methods such as the Newton-Raphson method or the fixed-point iteration method. By solving the equation using the appropriate method, we can obtain the numerical value for the outer surface temperature.

Step-by-step solution

(Step 1: Writing the differential equation and boundary conditions)

To write the differential equation for one-dimensional heat conduction, we will start with Fourier's Law of heat conduction: \(q_x = -k\frac{dT}{dx}\) For steady, one-dimensional heat conduction in the x-direction, the temperature distribution can be defined as: \(\frac{d^2T}{dx^2} = 0\) Now, we need to write the boundary conditions for this problem. There are two sides to the base plate: 1. The inner surface (x=0): here, the heat flux generated is given by the power of the iron divided by the base area, i.e., \(q_{x=0} = \frac{1000}{150\times10^{-4}} \mathrm{W/m^2}\). 2. The outer surface (x=L): here, the heat loss is due to both convection and radiation. We can write the boundary condition as: \(q_{x=L} = h(T_{x=L} - T_\infty) + \varepsilon \sigma (T_{x=L}^4 - T_{sarr}^4)\)

(Step 2: Obtaining a relation for the temperature of the outer surface)

Since \(\frac{d^2T}{dx^2} = 0\), the temperature distribution is linear. We can define T in terms of x as: \(T(x) = C_1x + C_2\) Now we can use boundary conditions to find \(C_1\) and \(C_2\). 1. At x=0, \(T(0) = C_2\) \(q_{x=0} = -k\frac{dT}{dx}\Big|_{x=0} = -kC_1\) 2. At x=L, \(T(L) = C_1L + C_2\) \(q_{x=L} = -k\frac{dT}{dx}\Big|_{x=L} = -kC_1\) Now, substituting the value of \(q_{x=0}\), \(C_1 = \frac{-q_{x=0}}{k}\) Substituting the value of \(C_1\) in \(T(L)\): \(T_{x=L} = -\frac{q_{x=0}L}{k} + C_2\) The final temperature expression for the outer surface of the plate is: \(T_{x=L} = -\frac{q_{x=0}L}{k} + T_\infty\)

(Step 3: Evaluating the outer surface temperature)

To obtain the temperature of the outer surface \(T_{x=L}\), we need to solve the boundary condition equation at x=L: \(q_{x=L} = h(T_{x=L} - T_\infty) + \varepsilon \sigma (T_{x=L}^4 - T_{sarr}^4)\) Substituting the temperature expression obtained in Step 2: \(q_{x=L} = h\left(-\frac{q_{x=0}L}{k} + T_\infty - T_\infty\right) + \varepsilon \sigma \left(\left(-\frac{q_{x=0}L}{k} + T_\infty\right)^4 - T_{sarr}^4\right)\) We can plug in the given values to solve for the outer surface temperature: L = 0.5 cm = 0.005 m A = 150 cm^2 = 0.015 m^2 q_{x=0} = \frac{1000}{0.015} = 66666.67 W/m^2 Given values: k = 18 W/(m K), T_∞ = 26°C = 299 K, h = 30 W/(m^2 K), ε = 0.7, and T_sarr = 295 K. After substituting all given values, the equation for the outer surface temperature becomes nonlinear. Therefore, we need to use a numerical method like the Newton-Raphson method or the fixed-point iteration method to obtain the numerical value for \(T_{x=L}\). In this case, you can use any numerical method or software to find the numerical value for the outer surface temperature.