Question

A 5-in-diameter spherical ball is known to emit radiation at a rate of \(550 \mathrm{Btu} / \mathrm{h}\) when its surface temperature is \(950 \mathrm{R}\). Determine the average emissivity of the ball at this temperature.

Short answer

Based on the given diameter of a spherical ball, the radiation it emits, and its surface temperature, determine the average emissivity of the ball at this temperature. The average emissivity of the ball at this temperature is approximately 0.544.

Step-by-step solution

Find the surface area of the ball

Firstly, we need to find the surface area of the ball. The formula for the surface area of a sphere is: \(A = 4 * \pi * r^2\) The diameter of the ball is 5 in, so the radius is half of the diameter: \(r = \frac{d}{2} = \frac{5 \ \text{in}}{2} = 2.5 \ \text{in}\) Now convert the radius to feet: \(r = 2.5 \ \text{in} * \frac{1 \ \text{ft}}{12 \ \text{in}} = \frac{5}{24} \ \text{ft}\) Now we can find the surface area of the ball: \(A = 4 * \pi * (\frac{5}{24})^2 \approx 1.653 \ \text{ft}^2\)

Apply the Stefan-Boltzmann law and solve for emissivity

Now we have the surface area of the ball (\(A\)). We can apply the Stefan-Boltzmann law and solve for emissivity (\(\epsilon\)): \(Q = \epsilon * A * \sigma * T^4\) Rearrange the equation to solve for emissivity: \(\epsilon = \frac{Q}{A * \sigma * T^4}\) Now plug in the given values: \(\epsilon = \frac{550 \ \frac{\mathrm{Btu}}{\mathrm{h}}}{1.653 \ \mathrm{ft}^2 * 1.714 * 10^{-9} \frac{\mathrm{Btu}}{\mathrm{h} \cdot \mathrm{ft}^2 \cdot \mathrm{R}^4} * (950 \ \mathrm R)^4}\) Solve for emissivity: \(\epsilon \approx 0.544\) The average emissivity of the ball at this temperature is approximately 0.544.

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann law is fundamental in understanding radiation heat transfer. It describes how the power radiated by an object relates to its temperature. For a perfect black body, which is an idealized physical body that absorbs all incident electromagnetic radiation, the law is mathematically expressed as \[Q = A * \, \sigma * T^4\] - **Q** is the total energy radiated per unit time (often measured in Btu/h or watts), - **A** is the surface area of the radiating body, - **\(\sigma\)** is the Stefan-Boltzmann constant, and - **T** is the absolute temperature of the body in Kelvin or Rankine.
The Stefan-Boltzmann constant \(\sigma\) is different depending on the units used. In our example, the constant used is \(1.714 \times 10^{-9} \frac{\mathrm{Btu}}{\mathrm{hr} \, \mathrm{ft}^2 \, \mathrm{R}^4}\). This law helps us predict how much energy an object emits purely due to its temperature, assuming it behaves like a black body.

Emissivity

Emissivity is a measure of an object's ability to emit thermal radiation compared to that of a perfect black body. A black body has the highest possible emissivity value of 1, meaning it emits all thermal radiation possible at its temperature. Real-life materials have emissivity values less than 1, indicating they emit less radiation than a perfect black body.
Emissivity is critical in the modified Stefan-Boltzmann equation: \[Q = \varepsilon * A * \, \sigma * T^4\] where \(\varepsilon\) is the emissivity factor.
A higher emissivity means the material is more efficient at radiating energy. In our example, calculating the emissivity of the spherical ball helps define how well it actually emits thermal energy when compared to a perfect black body. Understanding emissivity is essential in various fields like engineering, architecture, and atmospheric science, where heat transfer efficiency matters.

Surface Area Calculation

Calculating the surface area is foundational in determining the thermal radiation emitted by any object. For a sphere, the formula to find the surface area is \[A = 4 * \pi * r^2\] - **r** is the radius of the sphere.
In our exercise, the radius was found by halving the diameter of the ball: 5 inches converted to feet amounts to \(\frac{5}{24}\) feet, resulting in a surface area of approximately 1.653 square feet for the ball.
Surface area plays a critical role because the larger the area, the more potential there is to emit thermal radiation. Accurate surface area calculation ensures that when using thermal radiation equations, the predictions of radiant energy output are as precise as possible.

Thermal Radiation Equations

Thermal radiation equations form the backbone of understanding how objects emit and exchange heat as radiation. Using the Stefan-Boltzmann law in combination with emissivity, one can determine not just the total energy radiated by a surface but also how efficiently different materials radiate energy under the same conditions.
These equations metaphorically paint a picture of an object's thermal footprint, allowing us to determine:

• How much energy an object radiates
• How temperature affects radiation
• How surface area influences radiative properties

The emphasis on temperature to the fourth power in the formula \(Q = \varepsilon * A * \, \sigma * T^4\) significantly shows temperature's influence on thermal radiation output. This exponential factor means even small changes in temperature can dramatically impact the amount of radiated energy. Hence, applying these equations effectively assesses and optimizes heating and cooling solutions across industries.