Question

An opaque horizontal plate is well insulated on the edges and the lower surface. The irradiation on the plate is \(3000 \mathrm{~W} / \mathrm{m}^{2}\), of which \(500 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The plate has a uniform temperature of \(700 \mathrm{~K}\) and has an emissive power of \(5000 \mathrm{~W} / \mathrm{m}^{2}\). Determine the total emissivity and absorptivity of the plate.

Short answer

Answer: The total emissivity of the plate is 0.375 or 37.5%, and the absorptivity of the plate is 0.833 or 83.3%.

Step-by-step solution

Find the amount of absorbed radiation.

The total energy incident on the plate per unit area is the irradiation, which is given as 3000 W/m^2. Also, we are given that 500 W/m^2 is reflected. So, the amount of radiation absorbed by the plate can be found by subtracting the incident radiation from the reflected radiation. Amount of absorbed radiation = Incident radiation - Reflected radiation Amount of absorbed radiation = \(3000 \ \text{W/m}^2 - 500 \ \text{W/m}^2 = 2500 \ \text{W/m}^2\)

Determine the absorptivity of the plate.

Using the absorbed radiation, we can find the absorptivity (α) of the plate. Absorptivity is the ratio between the amount of absorbed radiation and the amount of incident radiation. Absorptivity (α) = (Amount of absorbed radiation) / (Incident radiation) α = 2500 W/m^2 / 3000 W/m^2 α = 0.833 The absorptivity of the plate is 0.833 or 83.3%.

Determine the total emissivity of the plate.

We are given that the emissive power of the plate is 5000 W/m^2, and the plate has a uniform temperature of 700 K. To find the emissivity, we will use the Stefan-Boltzmann law that relates the emitted radiation to temperature: E = εσT^4 Here, ε is the emissivity, σ is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \ \text{W/m}^2 \text{K}^4\)), and T is the temperature. We are given E = 5000 W/m^2 and T = 700 K. We can now solve for ε: ε = E / (σT^4) ε = \((5000 \ \text{W/m}^2) / (5.67 \times 10^{-8} \ \text{W/m}^2 \text{K}^4 \times (700 \ \text{K})^4)\) ε = 0.375 The total emissivity of the plate is 0.375 or 37.5%.

Key concepts

Absorptivity

Absorptivity, denoted by the Greek letter α, describes how well a surface can absorb thermal radiation. It is a dimensionless number that represents the fraction of the incident radiation absorbed by the surface. To calculate absorptivity, we consider the amount of radiation falling onto a surface and the fraction of that radiation the surface absorbs.
According to the formula:

• Amount of absorbed radiation = Incident radiation - Reflected radiation
• Absorptivity (α) = (Amount of absorbed radiation) / (Incident radiation)

In our example, the absorbed radiation is 2500 W/m², while the incident radiation is 3000 W/m², yielding an absorptivity of 0.833 or 83.3%.

This means that 83.3% of the radiation hitting the surface is absorbed, while the rest is either reflected or transmitted. Understanding absorptivity is crucial in designing materials that either maximize or minimize heat absorption, depending on the application.

Emissivity

Emissivity, ε, is a measure of a surface's ability to emit thermal radiation relative to a perfect black body at the same temperature. A black body, which has an emissivity of 1, is an idealized object that perfectly emits and absorbs all radiation at a given temperature.
To calculate the emissivity of a surface, the Stefan-Boltzmann Law is often used, which is given by:

• E = εσT⁴

Where:

• E is the emissive power (in W/m²)
• σ is the Stefan-Boltzmann constant ( 5.67 × 10⁻⁸ W/m²K⁴)
• T is the absolute temperature (in Kelvin)

In the exercise, the emissive power is 5000 W/m² and the temperature is 700 K. Using these figures in the formula yields an emissivity of 0.375 or 37.5%.
This means the surface emits only 37.5% of the radiation that a perfect black body would emit at the same temperature. Emissivity is important for understanding how much heat a material radiates into its environment, which impacts thermal management in processes like heating, cooling, and thermal insulation systems.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in the field of thermal radiation. It states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as emissive power) is directly proportional to the fourth power of the black body's absolute temperature. The law is expressed mathematically as:

• E = σT⁴

Here:

• E is the emissive power (radiated energy per area, in W/m²)
• σ is the Stefan-Boltzmann constant, a universal constant ( 5.67 × 10⁻⁸ W/m²K⁴)
• T is the absolute temperature (in Kelvin)

This law is pivotal in calculating how much heat is radiated from a body at a certain temperature. For real surfaces, the emissivity (ε) factor modifies this relationship to account for surfaces that are not perfect radiators, expressed as: E = εσT⁴.

The Stefan-Boltzmann Law is especially useful for engineers and scientists in fields like thermal energy design, astrophysics, and climate science, where understanding radiant heat transfer is critical. Through this law, we accurately predict and optimize systems based on how they emit energy.