Question

A 9-cm-diameter spherical ball whose surface is maintained at a temperature of \(110^{\circ} \mathrm{C}\) is suspended in the middle of a room at \(20^{\circ} \mathrm{C}\). If the convection heat transfer coefficient is \(15 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{C}\) and the emissivity of the surface is \(0.8,\) determine the total rate of heat transfer from the ball.

Short answer

Question: Determine the total rate of heat transfer from a ball with a diameter of 9 cm, a surface temperature of 110°C, an ambient temperature of 20°C, a convection heat transfer coefficient of 15 W/m²°C, and an emissivity of 0.8. Answer: The total rate of heat transfer from the ball is approximately 41.04 W.

Step-by-step solution

Compute the surface area of the ball

To calculate the surface area of the ball, we can use the formula for the surface area of a sphere, which is given by: \(A = 4\pi r^2\) where \(r\) is the radius of the sphere. Given the ball's diameter is 9 cm, we can find its radius: \(r = \frac{d}{2} = \frac{9\ \mathrm{cm}}{2} = 4.5\ \mathrm{cm} = 0.045\ \mathrm{m}\) Now, we can compute the surface area: \(A = 4\pi (0.045)^2 \approx 0.0255\ \mathrm{m^2}\)

Calculate the convection heat transfer rate

Now that we have the surface area of the ball, we can calculate the convection heat transfer rate using the formula: \(Q_{conv} = hA(T_s - T_\infty)\) \(Q_{conv} = 15\ \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{C}} \times 0.0255\ \mathrm{m^2} \times (110 - 20)\ \mathrm{C} \approx 34.45\ \mathrm{W}\)

Calculate the radiation heat transfer rate

First, we need to convert the temperatures to Kelvin: \(T_s = 110\ \mathrm{C} + 273.15 = 383.15\ \mathrm{K}\) \(T_\infty = 20\ \mathrm{C} + 273.15 = 293.15\ \mathrm{K}\) Now, we can use the radiation heat transfer formula: \(Q_{rad} = \epsilon \sigma A (T_s^4 - T_\infty^4)\) \(Q_{rad} = 0.8 \times 5.67 \times 10^{-8}\ \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}^4} \times 0.0255\ \mathrm{m^2} \times (383.15^4 - 293.15^4)\ \mathrm{K^4} \approx 6.59\ \mathrm{W}\)

Calculate the total heat transfer rate

Finally, we can find the total heat transfer rate by summing up the convection and radiation heat transfer rates: \(Q_{total} = Q_{conv} + Q_{rad} = 34.45\ \mathrm{W} + 6.59\ \mathrm{W} \approx 41.04\ \mathrm{W}\) Therefore, the total rate of heat transfer from the ball is approximately \(41.04\ \mathrm{W}\)