Question

Heat dissipated from a machine in operation hot spots that can cause thermal burns on human skposed hot spots that can cause thermal burns on human skin are considered to be hazards in the workplace. Consider a $1.5-\mathrm{m} \times 1.5-\mathrm{m}\( flat machine surface that is made of \)5-\mathrm{mm}-$ thick aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). During operation, the machine's inner aluminum surface temperature can be as high as \(90^{\circ} \mathrm{C}\), while the outer surface is cooled with $30^{\circ} \mathrm{C}\( air flowing in parallel over it at \)10 \mathrm{~m} / \mathrm{s}$. To protect machine operators from thermal burns, the machine surface can be covered with insulation. The aiuminum/insulation interface has a thermal contact conductance of \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the thickness of insulation (with a thermal conductivity of $0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ ) needed to keep the local outer surface temperature at \(45^{\circ} \mathrm{C}\) or lower? Using appropriate software, plot the required coefficient along the outer surface in parallel with the airflow. coefficient along the outer surface in parallel with the airflow.

Short answer

In this problem, we were asked to find the thickness of insulation needed to keep the local outer surface temperature of a machine at \(45^{\circ} \mathrm{C}\) or lower. Using Fourier's Law of Conduction and the given thermal properties, we calculated the heat transfer rate through the aluminum surface and determined the required heat transfer rate through the insulation to maintain the desired temperature. The required insulation thickness was found to be 0.019 m, or 19 mm. To plot the required coefficient along the outer surface in parallel with the airflow, dedicated software such as MATLAB, Python, or other available tools can be used.

Step-by-step solution

Determine the heat transfer rate through conduction

To find the necessary insulation thickness, we first need to calculate the heat transfer rate through the aluminum surface at the given operating conditions. According to Fourier's Law of Conduction, the heat transfer rate can be calculated using the formula: $$q = k \frac{T_{hot} - T_{cold}}{L}$$ where \(q\) is the heat transfer rate, \(k\) is the thermal conductivity, \(T_{hot}\) and \(T_{cold}\) are the hot and cold temperatures, and \(L\) is the thickness of the material. We are given \(k = 237 \mathrm{~W/m \cdot K}\) for aluminum, \(T_{hot} = 90^{\circ} \mathrm{C}\), \(T_{cold} = 30^{\circ} \mathrm{C}\), and \(L = 5 \mathrm{~mm} = 0.005 \mathrm{~m}\). Calculating the heat transfer rate through the aluminum surface: $$q = 237 \frac{90 - 30}{0.005} = 474 \times 10^3 \mathrm{~W/m^2}$$

Determine the required heat transfer rate through the insulation

Now we need to determine the required heat transfer rate through the insulation to maintain the local outer surface temperature at \(45^\circ\mathrm{C}\). Using the given thermal contact conductance of the aluminum/insulation interface, we can calculate the total heat transfer rate using the following equation: $$q_{required} = h_c \cdot (T_{interface} - T_{cold})$$ where \(h_c\) is the thermal contact conductance, and \(T_{interface}\) and \(T_{cold}\) are the interface and cold temperatures respectively. We are given \(h_c = 3000 \mathrm{~W/m^2 \cdot K}\), \(T_{interface} = 45^{\circ} \mathrm{C}\), and \(T_{cold} = 30^{\circ} \mathrm{C}\). Calculating the required heat transfer rate: $$q_{required} = 3000 \cdot (45 - 30) = 45 \times 10^3 \mathrm{~W/m^2}$$

Determine the insulation thickness

Now we can determine the insulation thickness using Fourier's Law of Conduction and the required heat transfer rate. We are given the thermal conductivity of insulation material as \(k_{ins} = 0.06 \mathrm{~W/m \cdot K}\) and we found the required heat transfer rate in step 2. Let's denote the insulation thickness as \(L_{ins}\). Rearranging Fourier's Law of Conduction: $$ L_{ins} = k_{ins} \frac{T_{interface} - T_{cold}}{q_{required}} $$ Substitute the given values: $$ L_{ins} = 0.06 \frac{45 - 30}{45 \times 10^3} $$ Calculate the insulation thickness: $$ L_{ins} = 0.019 \mathrm{~m} $$ So, the required insulation thickness is \(0.019 \mathrm{~m}\) or \(19 \mathrm{~mm}\). For plotting the required coefficient along the outer surface in parallel with the airflow, you may use dedicated software like MATLAB, Python, or other available tools to input the data and generate the plot accordingly.