Question

A 2-in-diameter spherical ball whose surface is maintained at a temperature of \(170^{\circ} \mathrm{F}\) is suspended in the middle of a room at $70^{\circ} \mathrm{F}\(. If the convection heat transfer coefficient is \)15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{2}{ }^{\circ} \mathrm{F}$ and the emissivity of the surface is \(0.8\), determine the total rate of heat transfer from the ball.

Short answer

The room temperature is 70°F. Answer: The total rate of heat transfer from the ball is approximately 166.77 Btu/h.

Step-by-step solution

Calculate the surface area of the sphere

Using the given diameter of 2 inches, convert it to feet by dividing by 12. With the radius \(r = 1\) inch or \(\frac{1}{12}\) ft, find the surface area \(A_s\) using the formula for the surface area of a sphere, \(A_s = 4\pi r^2\): $$A_s = 4\pi \left(\frac{1}{12}\right)^2 = \frac{\pi}{9} \space \text{ft}^2$$

Calculate the convection heat transfer rate

Using the convection heat transfer coefficient \(h = 15 \frac{\text{Btu}}{\text{h}\cdot \text{ft}^2 \cdot {}^{\circ}\text{F}}\), the surface area \(A_s\), and the temperature difference between the ball and the room \(\Delta T = 170 - 70 = 100^{\circ}\text{F}\), calculate the convection heat transfer rate \(q_{conv}\) using the formula \(q_{conv} = h\cdot A_s \cdot \Delta T\): $$q_{conv} = 15 \cdot \frac{\pi}{9} \cdot 100 = \frac{500\pi}{3} \space \text{Btu/h}$$

Calculate the radiation heat transfer rate

Using the emissivity \(\epsilon = 0.8\), the surface area \(A_s\), the surface temperature \(T_s = 170^{\circ}\text{F} = 529.67\text{R}\) (R stands for Rankine), the room temperature \(T_r = 70^{\circ}\text{F} = 529.67\text{R}\), and the Stefan-Boltzmann constant \(\sigma = 1.714 \times 10^{-9} \frac{\text{Btu}}{\text{h}\cdot \text{ft}^2 \cdot {}^{\circ}\text{R}^4}\), calculate the radiation heat transfer rate \(q_{rad}\) using the formula \(q_{rad} = \epsilon\cdot \sigma\cdot A_s \cdot (T_s^4 - T_r^4)\): $$q_{rad} = 0.8 \cdot 1.714 \times 10^{-9} \cdot \frac{\pi}{9} \cdot (4433900.43 - 76961.17) \approx 3.77 \space \text{Btu/h}$$

Calculate the total rate of heat transfer

Sum up the convection and radiation heat transfer rates to find the total rate of heat transfer \(q_{total}\): $$q_{total} = q_{conv} + q_{rad} = \frac{500\pi}{3} + 3.77 \approx 166.77 \space \text{Btu/h}$$ The total rate of heat transfer from the ball is approximately 166.77 Btu/h.