Question

A spherical tank \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inner diameter of \(3 \mathrm{~m}\) and a wall thickness of \(10 \mathrm{~mm}\) is used for storing hot liquid. The hot liquid inside the tank causes the inner surface temperature to be as high as \(100^{\circ} \mathrm{C}\). To prevent thermal burns to the people working near the tank, the tank is covered with a \(7-\mathrm{cm}\)-thick layer of insulation $(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$, and the outer surface is painted to give an emissivity of \(0.35\). The tank is located in surroundings with air at $16^{\circ} \mathrm{C}$. Determine whether or not the insulation layer is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burn hazards. Discuss ways to further decrease the outer surface temperature. Evaluate the air properties at \(30^{\circ} \mathrm{C}\) and $1 \mathrm{~atm}$ pressure. Is this a good assumption?

Short answer

using the given thermal conductivity of tank material, \(k_{tank} = 60 W/mK\), as follows: \[R_{tank} = \frac{1.5 - 1.51}{4\pi (60) (1.5)(1.51)} \approx 0.0000276 \,K/W\] #tag_title# Step 2: Calculate the resistance of the insulation through conduction #tag_content# Similar to the tank wall's conduction resistance, we can calculate the resistance due to conduction in insulation by using the formula for thermal resistance for a sphere: \[R_{insul} = \frac{r_3 - r_2}{4\pi kr_2r_3}\] The inner radius of insulation is equal to the outer radius of the tank, \(r_2 = 1.51 m\), and the insulation thickness is \(7 cm = 0.07 m\), therefore the outer radius of insulation, \(r_3 = 1.51 m + 0.07 m = 1.58 m\). Now, using the given thermal conductivity of insulation material, \(k_{insul} = 0.035 W/mK\), we can calculate the thermal resistance of insulation, \(R_{insul}\): \[R_{insul} = \frac{1.51 - 1.58}{4\pi (0.035) (1.51)(1.58)} \approx 0.0284 \,K/W\] #tag_title# Step 3: Calculate the heat transfer through conduction #tag_content# To calculate the heat transfer through conduction, we can use the formula: \[Q_c = \frac{T_i - T_o}{R_t}\] where, \(Q_c\) is the heat transfer through conduction \(T_i\) is the inner surface temperature (100°C) \(T_o\) is the desired outer surface temperature, which should not exceed 45°C \(R_t\) is the total resistance, which is the sum of the tank wall and the insulation resistance Now, we can calculate the heat transfer through conduction using the given temperatures and calculated resistance values: \[Q_c = \frac{100 - 45}{0.0000276 + 0.0284} \approx 1921.4 \,W\] #tag_title# Step 4: Calculate the heat transfer through radiation #tag_content# To calculate the heat transfer through radiation, we can use the Stefan-Boltzmann law: \[Q_r = \varepsilon \sigma A(T_o^4 - T_a^4)\] where, \(Q_r\) is the heat transfer through radiation \(\varepsilon\) is the emissivity of the outer surface (0.8) \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times10^{-8} W/m^2K^4\)) \(A\) is the surface area of the insulation (\(4\pi r_3^2\)) \(T_a\) is the ambient air temperature (25°C) Now, we can calculate the heat transfer through radiation using the given values and calculated surface area: \[Q_r = 0.8\times 5.67\times10^{-8} \times 4\pi (1.58)^2((45 + 273)^4 - (25 + 273)^4) \approx 337.9 \,W\] #tag_title# Step 5: Compare heat transfer through conduction and radiation #tag_content# Now that we have calculated the heat transfer through conduction and radiation, we need to compare these values to determine if the insulation is sufficient: Since \(Q_c \approx 1921.4 \,W\) and \(Q_r \approx 337.9 \,W\), it is evident that the heat transfer through conduction is significantly higher compared to the heat transfer through radiation. Therefore, we can conclude that the insulation is sufficient to keep the outer surface temperature below 45°C, as the heat transfer through conduction is the dominant factor in this scenario.

Step-by-step solution

Calculate the resistance of the tank wall through conduction

To calculate the resistance due to conduction through the tank wall, we can use the formula for the thermal resistance for a sphere, which is given by: \[R_{tank} = \frac{r_1 - r_2}{4\pi kr_1r_2}\] where, \(r_1\) and \(r_2\) are the inner and outer radii of the tank \(k\) is the thermal conductivity of the tank material. The inner radius of the tank, \(r_1 = \frac{3 m}{2} = 1.5 m\), wall thickness is \(10 mm = 0.01 m\), then the outer radius of the tank, \(r_2 = 1.5 m + 0.01 m = 1.51 m.\) Now, we can calculate the thermal resistance of the tank wall, \(R_{tank}\)