Question

An AISI 316 stainless steel spherical container is mic reaction that provides a uniform heat flux of \(60 \mathrm{~kW} / \mathrm{m}^{2}\) to the container's inner surface. The container has an inner diameter of \(1 \mathrm{~m}\) and a wall thickness of \(5 \mathrm{~cm}\). To prevent thermal burns on individuals working around the container, it is necessary to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). If the ambient temperature is \(23^{\circ} \mathrm{C}\), determine the necessary convection heat transfer coefficient to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). Is the necessary convection heat transfer coefficient feasible with free convection of air? If not, discuss other options to prevent the container's outer surface temperature from causing thermal burns.

Short answer

Answer: The calculated convection heat transfer coefficient needed is 649.87 W/(m^2 K). However, this value is too high for free convection of air, which typically ranges from 5 to 25 W/(m^2 K). As such, passive cooling through free convection of air is not sufficient to maintain the outer surface temperature below 50°C.

Step-by-step solution

Calculate the surface area of the inner and outer surfaces of the container.

Given the inner diameter of the container to be 1 m and the wall thickness to be 5 cm. First, the inner radius (r1) is calculated as half of the inner diameter and the outer radius (r2) is calculated by adding the wall thickness to r1. \(r1 = \frac{1 \, \mathrm{m}}{2} = 0.5 \, \mathrm{m}\) \(r2 = r1 + 0.05 \, \mathrm{m} = 0.55 \, \mathrm{m}\) To calculate the surface areas of the inner and outer surfaces, the following formula is used: \(A = 4\pi r^2\) \(A_{inner} = 4\pi (0.5 \, \mathrm{m})^2 = 3.1416 \, \mathrm{m^2}\) \(A_{outer} = 4\pi (0.55 \, \mathrm{m})^2 = 3.8013 \, \mathrm{m^2}\)

Determine the heat transfer through the container's wall.

The heat is transferred through the container's wall by conduction. The heat transfer rate can be determined by the following formula: \(q = \frac{k \cdot A \cdot (T_{inner} - T_{outer})}{t}\) First, let's substitute the given heat flux value to determine the temperature difference across the container's wall. \(60 \, \mathrm{kW/m^2} = \frac{q}{A_{inner}}\) \(q = 60 \, \mathrm{kW/m^2} \times 3.1416 \, \mathrm{m^2} = 188.496\, \mathrm{kW}\), convert this to Watts: \(q = 188,496 \, \mathrm{W}\) We are given that the outer surface temperature should be below 50°C and the ambient temperature is 23°C. Therefore, \(T_{inner} - T_{outer} = T_{inner} - 50 ^{\circ}\mathrm{C}\) Now that we have all the necessary values, we can determine the temperature of the inner surface and the temperature difference across the wall: \(T_{inner} = \frac{q \cdot t + k \cdot A_{outer} \cdot T_{outer}}{k \cdot A_{outer}}\) Calculate the temperature difference as follows: \(T_{inner} - T_{out} = \frac{188,496 \, \mathrm{W} \cdot 0.05\, \mathrm{m}}{\mathrm{AISI \, 316 \, steel \, conductivity \, (k)}\) We'll need to look up the specific thermal conductivity (k) value for AISI 316 stainless steel. The accepted value is around 16.3 W/(mK). Calculating the temperature difference: \(T_{inner} - T_{out} = 363^{\circ}\mathrm{C}\)

Determine the necessary convection heat transfer coefficient to keep the container's outer surface temperature below 50°C.

Now that the heat transfer rate q is known, we can use the heat transfer due to convection formula to find the required convection heat transfer coefficient (h): \(q = h \cdot A_{outer} \cdot (T_{outer} - T_{ambient})\) Rearranging the formula to find the convection heat transfer coefficient: \(h = \frac{q}{A_{outer} \cdot (T_{outer} - T_{ambient})}\) Substituting the values: \(h = \frac{188,496 \, \mathrm{W}}{3.8013 \, \mathrm{m^2} \cdot (50^{\circ} \mathrm{C} - 23^{\circ} \mathrm{C})} = 649.87 \, \mathrm{W/(m^2 K)}\)

Determine if the calculated convection heat transfer coefficient is feasible with free convection of air. If not, suggest other prevention methods.

The calculated convection heat transfer coefficient of 649.87 W/(m^2 K) is too high for free convection of air, which usually has a heat transfer coefficient ranging from 5 to 25 W/(m^2 K). It indicates that passive cooling through free convection of air is not enough to keep the outer surface temperature below 50°C. Some alternative methods can be considered to prevent the container's outer surface temperature from causing thermal burns: 1. Forced convection: Use fans or blowers to increase the airflow over the container's outer surface, which will enhance the convection and improve heat dissipation. 2. Insulation: Apply insulation material to the outer surface of the container. This will reduce the heat flow through the container's wall, resulting in a lower outer surface temperature. 3. Heat exchangers or cooling systems: Implementing a heat exchanger or a dedicated cooling system, which will help remove heat from the container more effectively.